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S such that y?‘(0) = a and (,;)} (n > 0). Let g : N —> A be the function defined by g(0) = a\ ffO) = *#(0) = x\a\; • • • ; g(n + 1) = Xvwl------------Use the order properties of N and the following facts to verify that g is injective: 0) g(f0 e v?(«) for all n > 0; (ii) g(n) £ if(n — 1) for all n > 1; (iii) g(n) £ Kh S(H) and f~l(J) is a subgroup of G for every subgroup J of H. Since J < H implies Ker / < f~\J) and f(f~'U)) = J, F' such that the diagram: H G i such iel F0 given by 11—> 1F and Xi*1- • -xnSn (-+ |xi6l| - • • |-x-n5n| is clearly a surjection such that y?(H>,w2) = . Consequently, the normal subgroup N generated by Y in F is contained in Ker 0). If we denote XX & suC'h that -kup = S such that ip(0) = a and n/1, then is clearly a monomorphism that embeds Z in Q. Furthermore, for every nonzero n, S such that ^ | /?[xi, . . . , xk] — Tp and p(xt) = Si for / = k + 1,. .., n. By construction lp \ R = S is a homo morphism such that yp I R = *p a°d ^(x,) = Si for / = 1,2Then the same ar , xA] gument used in the proof of uniqueness in Theorem 5.5 shows that \p ! = ip. Therefore, the uniqueness statement of Theorem 5.5 (applied to /?[xi,..., xA]) implies that yp = p. Consequently, /?[xi,. . ., x*-][x*+i, . . ., x„] has the desired uni versal mapping property, whence /?[xi,. . ., xt][x*+i,. . . , x„] = /?[xi,. . . , x„] by Theorem 5.5. The other isomorphism is proved similarly. ■ 1: (a) (Mat„/?)[x] = Matn/?[x]. (b) (Mat„/?)[[x]] =* Mat„/?[[x]]. 3. Let R be a ring and G an infinite multiplicative cyclic group with generator de noted x. Is the group ring R(G) (see page 117) isomorphic to the polynomial ring in one indeterminate over /?? 4. (a) Let 5 be a nonempty set and let Ns be the set of all functions N such that ( I r) implies f(c) = (h(c) — q(c))(c — c) = 0. Commutativity is not required for the converse; use Corollary 6.3. ■ S a ring homomorphism. Then every S-moduIe A can be made into an .R-module by defining rx (x e A) to be A; such that irup = UAi which has the desired property, given by tp(c) = { A be given by *p = ixiri i27r2' + - - • + i„7rn' and \p : A —-» Ax ©• • - © An by = iiVi + t2'7r2 -+-•••-+- in^n. Then /n A and $ : B —* D K-module homomorphisms. Then the map 6 : //cw7r(A,B) —* //owr(C,D) given by f ipfif is a homomorphism of abelian groups. SKETCH OF PROOF. 6 is well defined since composition of /^-module homo morphisms is an /^-module homomorphism. 6 is a homomorphism since such com position of homomorphisms is distributive with respect to addition. ■ The map 6 of Theorem 4.1 is usually denoted Hom(v?,^) and called the homomor phism induced by and ip. Observe that for homomorphisms ^/and is denoted yp. We now examine the behavior of Homfi with respect to exact sequences. \pg = 0 => Im g d Ker yp = Im A, then h e Hom/j(Z),v4) and g = A. Therefore, for every x a D = Ker yp, x — j(x) = tpf(x) s Im C is exact. ■ D is an /^-module homomorphism such that J(f A^>B-^C—>0 is split exact by Proposition 4.3 and Theorem 1.18. The other implications are proved similarly. ■ { fu], is a homomorphism such that A' is a homomorphism of left R-modules, then the induced map l,B) is actually a right S-module is tedious but straight-forward; similarly for (iii). (ii) ip is an abelian group homomorphism by Theorem 4.1. If / e Homs(/l/,B), a e A and s e S, then Homn(A,R) = A* is a homomorphism of right R-modules. (ii) There is an R-module isomorphism (A (J) C)* A* (J) C*. (iii) 7/R is a division ring and 0->A->B->C->0«a short exact sequence of A' is an homomorphism of right R-modules, then the induced map <jp : HornR(A',B) —> HornK(A,B) is an homomorphism at left 5-modules. (c) Hom«(C,D) is a right 5-module, with the action of 5 given by (gs)(c) = g(sc). (d) If \J/ : D —> D' is an homomorphism of right Jf?-modules, then \f/ : Horn/* (C,D) —► HomR(C,D') is an homomorphism of right 5-modules. 5. Let R be a ring with identity; then there is a ring isomorphism HomR(R,R) ^ R°v where Hom« denotes left R-module homomorphisms (see Exercises III.1.17 and 1.7). In particular, if R is commutative, then there is a ring isomorphism HomR(/?,/?) = R. 6. Let 5 be a nonempty subset of a vector space V over a division ring. The annihilator of 5 is the subset 5° of V* given by 5° = { /e V* \ (s,f) = 0 for all s e 5}. (a) 0° = V*\ V° = 0; 5 {0} =*5° V*. (b) If W is a subspace of V, then IV0 is a subspace of V*. (c) If W is a subspace of V and dim V is finite, then dim W° = dim V — dim W. (d) Let W,V be as in (c). There is an isomorphism W*_= V*/W°. (e) Let Wy be as in (c) and identify V with V** under the isomorphism 6 of Theorem 4.12. Then (W70)0 = W CZ V**. 7. If V is a vector space over a division ring and /e V*, let W = {a e V \ (a,f) = 0}, then W is a subspace of V. If dim V is finite, what is dim W? 8. If R has an identity and we denote the left /^-module R by rR and the right R-module R by Rr, then ( r R)* = R r and (R r)* ^ r R. 9. For any homomorphism / : A —> B of left R-modules the diagram F given by f/g\-^ f(u)/g(u) — f(u)g(u)1 is a well-defined monomorphism of fields which is the identity on K. But Tm K[u] given by gh s(u) is a nonzero ring epimorphism by Theorems III.5.5. and 1.3. Since K[jc] is a principal ideal domain (Corollary III.6.4), Ker ) e K(z)[y\. By Lemma III.6.13 f/g induces a homomorphism a : K(x) —» K(x) such that ip(x)/}p(x) |—> P is a monomorphism and by Corollary III.4.6 there is a monomorphism of fields (p : Q —»P. As before, we must have Q = Im ip = P. ■ upr is a Z.p-monomorphism of fields. If ¥ is finite, then F given by u |—> uv is a Zp-automorphism by Lemma 5.5. Clearly tpn is the identity and no lower power k of 2. (b) Find all n > 0 such that and translate the result into statements about xp, using the facts just developed. ■ In order to use the preceding results to obtain a set of canonical forms for the relation of similarity on Mat^X we need : E—-> E a linear transformation with invariant factors qx = x4 — 4x3 + 5x2 — 4x + 4 = ( x ~ 2)2(x2 + 1) e R[x] and q2 = x1 + 6xe + \4x& — 20x4 + 25a3 — 22a2 + 12a — 8 = (a — 2)3(a2 + 1 )2 e R[ vj. By Theorem 4.6(i) dim*E = 11 and the minimal poly nomial of 0 is q2. The remarks after Theorem 4.2 show that the elementary divisors Warning: rational and Jordan canonical forms are defined somewhat differently by some authors. 3 5 such that oid R such that V>(0) = a and
set of N consisting of all those n e N for which there exists a unique xneS such that (n,xn) e R. We shall prove M = N by induction. If 0 \M, then there exists (0,b) e R with b 3^ a and the set R — {(0,6)} CZ N X S is in 9. Consequently R = fl Y Y*S
a R — {(0,6)}, which is a contradiction. Therefore, 0 e M. Suppose inductively that n e M (that is, (n,x„) e R for a unique xn e S). Then (n + I ,/„(-*„)) e R also. If (n + I,c) e R with c # f„(x„) then / ? - { ( « + I,c)}e9 (verify!), which leads to a contradiction as above. Therefore, jc„+i = fn(xn) is the unique element of S such that (n + 1,jc„+i) e R. Therefore by induction (Theorem 6.1) N = M, whence the
6. THE INTEGERS
11
assignment n |—> xn defines a function tp : N —*S with graph R. Since (0,o) e R we must have <^(0) = a. For each n e N, (n,xn) = («,
Theorem 6.3. (Division Algorithm) If a,b, e Z and a ^ 0, then there exists unique integers q and r such that b = aq + r, and 0 < r < |a|.
SKETCH OF PROOF. Show that the set5 = {b — ax \ x e Z, b — ax > 0} is a nonempty subset of N and therefore contains a least element r = b — aq (for some q e Z). Thus b = aq + r. Use the fact that r is the least element in S to show 0 < r < \a\ and the uniqueness of q,r. ■ We say that an integer a^O divides an integer b (written a\b)'\i there is an integer k such that ak = b. If a does not divide b we write aJ(b.
Definition 6.4. The positive integer c is said to be the greatest common divisor of the integers ai,a2,. . . , a„ if:
(1) c | a; for 1 < i < n; (2) d e Z and d | a4 for 1 < i < n => d | c. c is denoted (ai,a2, . .., £„).
Theorem 6.5. If a j , a 2 , . . . , a n are integers, not all 0, then ( a i , a 2 , . . . , an) exists.
Furthermore there are integers kj,k2,. . . , kn such that ( a i , a 2 , . . . , a„) = kiai + k2a2 H-------- h kDanSKETCH OF PROOF. Use the Division Algorithm to show that the least posi tive element of the nonempty set S = {xtai + x<2a2 + ■ ■ • + xnan | .xv e Z, x.o, > 0} i
is the greatest common divisor of au . .., an. For details see Shockley [51,p.l0]. ■ The integers aua2, are said to be relatively prime if (aua2,. . . , aT) = 1. A positive integer p > 1 is said to be prime if its only divisors are ±1 and ±p. Thus if p is prime and a e Z, either (a,p) = p (if p | a) or (a,p) = 1 (if pJf a).
Theorem 6.6. //a and b are relatively prime integers and a | be, then a | c. If p is
prime and p | aia2- • - a n , then p | a i for some i .
12
PREREQUISITES AND PRELIMINARIES
SKETCH OF PROOF. By Theorem 6.5 1 — ra + sb, whence c = rac + sbc. Therefore a ] c. The second statement now follows by induction on n. ■
Theorem 6.7. (Fundamental Theorem of Arithmetic) Any positive integer n > 1 may be written uniquely in the form n = pitlp2tj- • Pktk, where pi < p2 < • • • < Pk arc primes and ti > 0 for all i.
The proof, which proceeds by induction, may be found in Shockley [51, p.17]. Let m > 0 be a fixed integer. If a,b s Z and m \ (a — b) then a is said to be con gruent to b modulo m. This is denoted by a = b (mod m).
Theorem 6.8. Let m > 0 be an integer and a,b,c,d e Z.
(i) Congruence modulo m is an equivalence relation on the set of integers Z, which has precisely m equivalence classes. (ii) If a = b (mod m) and c = d (mod m), then a + c = b + d (mod m) and ac -- bd (mod m). (iii) 7/ab = ac (mod m) and a and m are relatively prime, then b s c (mod m). PROOF, (i) The fact that congruence modulo m is an equivalence relation is an easy consequence of the appropriate definitions. Denote the equivalence class of an integer a by a and recall property (20), which can be stated in this context as: a = b <=> a = b (mod m).
(20')
Given any a e Z, there are integers q and r, with 0 < r < m, such that a = mq + r. Hence a — r = mq and a = r (mod m); therefore, a = r by (20'). Since a was ar bitrary and 0 < r < m, it follows that every equivalence class must be one of 0,1,2,3,... ,(m — 1). However, these m equivalence classes are distinct: for if 0 < i < j < m, then 0 < (j — i) < m and m Jf (j — i). Thus / ^ j (mod m) and hence 1 ^ J by (20')- Therefore, there are exactly m equivalence classes. (ii) We are given m \ a — b and m | c — d. Hence m divides (a — b) + (c — d) = (a + c) — (b + d) and therefore a + c = b + d (mod m). Likewise, m divides (a — b)c + (c — d)b and therefore divides ac — be + cb — db — ac — bd; thus ac - bd (mod m). (iii) Since ab = ac (mod m), m \ a(b — c). Since (m,a) = 1, m \ b — c by Theo rem 6.6, and thus b = c (mod m). ■
7. THE AXIOM OF CHOICE, ORDER, AND ZORN’S LEMMA Note. In this section we deal only with sets. No proper classes are involved. If /^0 and {At \ i s /} is a family of sets such that Ai ^ 0 for all i e I, then we would like to know that Ai 0. It has been proved that this apparently inieZ
nocuous conclusion cannot be deduced from the usual axioms of set theory (al though it is not inconsistent with them — see P. J. Cohen [59]). Consequently we shall assume
7. THE AXIOM OF CHOICE, ORDER, AND ZORN'S LEMMA
13
The Axiom of Choice. The product of a family of nonempty sets indexed by a non
empty set is nonempty. See Exercise 4 for another version of the Axiom of Choice. There are two propo sitions equivalent to the Axiom of Choice that are essential in the proofs of a number of important theorems. In order to state these equivalent propositions we must introduce some additional concepts. A partially ordered set is a nonempty set A together with a relation R on A X A (called a partial ordering of A) which is reflexive and transitive (see (15), (17) in section 4) and antisymmetric: (a,b) e R and (b,a) e R => a = b.
(31)
If R is a partial ordering of A, then we usually write a < b in place of (ajti) e R. In this notation the conditions (15), (17), and (31) become (for all a,b,c e A): a < a; a < b and b < c => a < c; a < b and b < a => a = b. We write a < b if a < b and a tA b. Elements a,b e A are said to be comparable, provided a < b or b < a. However, two given elements of a partially ordered set need not be comparable. A partial ordering of a set A such that any two elements are comparable is called a linear (or total or simple) ordering. EXAMPLE. Let A be the power set (set of all subsets) of {1,2,3,4,5}. Define C < D if and only if C CL D. Then A is partially ordered, but not linearly ordered (for example, {1,2} and {3,4} are not comparable). Let (A ,< ) be a partially ordered set. An element a e A is maximal in A if for every c e A which is comparable to a, c < a; in other words, for all c e A, a < c => a = c. Note that if a is maximal, it need not be the case that c < a for all c e A (there may exist c £ A that are not comparable to a). Furthermore, a given set may have many maximal elements (Exercise 5) or none at all (for example, Z with its usual ordering). An upper bound of a nonempty subset B of A is an element d e A such that b < d for every b e B. A nonempty subset B of A that is linearly ordered by < is called a chain in A.
Zorn’s Lemma. If A is a nonempty partially ordered set such that every chain in A
has an upper bound in A, then A contains a maximal element. Assuming that all the other usual axioms of set theory hold, it can be proved that Zorn’s Lemma is true if and only if the Axiom of Choice holds; that is, the two are equivalent — see E. Hewitt and K. Stromberg [57: p. 14]. Zorn’s Lemma is a power ful tool and will be used frequently in the sequel. Let B be a nonempty subset of a partially ordered set (A,<). An element c e B is a least (or minimum) element of B provided c < b for every beB. If every nonempty subset of A has a least element, then A is said to be well ordered. Every well-ordered set is linearly ordered (but not vice versa) since for all a,b e A the subset \a,b\ must
PREREQUISITES AND PRELIMINARIES
14
have a least element; that is, a < b or b < a. Here is another statement that can be proved to be equivalent to the Axiom of Choice (see E. Hewitt and K. Stromberg [57; p.14]).
The Well Ordering Principle. If A is a nonempty set, then there exists a linear
ordering < of A such that ( A , < ) is well ordered. EXAMPLES. We have already assumed (Section 6) that the set N of natural numbers is well ordered. The set Z of all integers with the usual ordering by magni tude is linearly ordered but not well ordered (for example, the subset of negative integers has no least element). However, each of the following is a well ordering of Z (where by definition a < b <=> a is to the left of by. (i) 0,1,—1,2,—2,3, — 3 , . . . , n,—n,. . . ; (ii) 0,1,3,5,7,... , 2,4,6,8,.. . , - 1 , - 2 , - 3 , - 4 , . . . ; (iii) 0,3,4,5,6,. . ., —1,-2, —3, — 4 , . . . , 1,2. These orderings are quite different from one another. Every nonzero element a in ordering (i) has an immediate predecessor (that is an element c such that a is the least element in the subset {x | c < x ) ) . But the elements — 1 and 2 in ordering (ii) and — 1 and 1 in ordering (iii) have no immediate predecessors. There are no maximal ele ments in orderings (i) and (ii), but 2 is a maximal element in ordering (iii). The element 0 is the least element in all three orderings. The chief advantage of the well-ordering principle is that it enables us to extend the principle of mathematical induction for positive integers (Theorem 6.1) to any well ordered set.
Theorem 7.1. (Principle of Transfinite Induction) If B is a subset of a well-ordered
set (A, <) such that for every a £ A, { c £ A | c < a } C B => a e B, then B = A. PROOF. If A — B 0, then there is a least element a £ A — B. By the defini tions of least element and A — B we must have {c e A \ c < a] CZ B. By hypothesis then, a £ B so that a £ B D (A — B) = 0, which is a contradiction. Therefore, A — B = 0 and A = B. ■
EXERCISES
1. Let (A, < ) be a partially ordered set and B a nonempty subset. A lower bound of B is an element d £ A such that d < b for every beB. A greatest lower bound (g.l.b.) of B is a lower bound do of B such that d < d0 for every other lower bound d of B. A least upper bound (l.u.b.) of B is an upper bound t0 of B such that f0 < t for every other upper bound / of B. (A,<) is a lattice if for all a,b e A the set {a,b\ has both a greatest lower bound and a least upper bound.
8. CARDINAL NUMBERS
15
(a) If S ^ 0, then the power set P(S) ordered by set-theoretic inclusion is a lattice, which has a unique maximal element. (b) Give an example of a partially ordered set which is not a lattice. (c) Give an example of a lattice w.ith no maximal element and an example of a partially ordered set with two maximal elements. 2. A lattice (A, <) (see Exercise 1) is said to be complete if every nonempty subset of A has both a least upper bound and a greatest lower bound. A map of partially ordered sets f: A B is said to preserve order if a < a' in A implies f(a) < /(o') in B. Prove that an order-preserving map /of a complete lattice A into itself has at least one fixed element (that is, an a s A such that f(a) = a). 3. Exhibit a well ordering of the set Q of rational numbers. 4. Let S be a set. A choice function for S is a function / from the set of all nonempty subsets of 5 to S such that f(A) e A for all A ^ 0, A C S. Show that the Axiom of Choice is equivalent to the statement that every set 5 has a choice function. 5. Let S be the set of all points (jr,>) in the plane with >’ < 0. Define an ordering by (*i J>i) < (^2,j'v) <£=> xi = x-2 and >’i < y-2. Show that this is a partial ordering of S, and that 5 has infinitely many maximal elements. 6. Prove that if all the sets in the family {At \ i e / ^ 0} are nonempty, then each
of the projections ivk :
n Ai —■* A is surjective. iel
k
7. Let (A, < ) be a linearly ordered set. The immediate successor of at A (if it exists) is the least element in the set {x e A | a < *}. Prove that if A is well ordered by <, then at most one element of A has no immediate successor. Give an example of a linearly ordered set in which precisely two elements have no immediate successor.
8. CARDINAL NUMBERS The definition and elementary properties of cardinal numbers will be needed fre quently in the sequel. The remainder of this section (beginning with Theorem 8.5), however, will be used only occasionally (Theorems II.1.2 and IV.2.6; Lemma V.3.5; Theorems V.3.6 and VI.1.9). It may be omitted for the present, if desired. Two sets, A and B, are said to be equipollent, if there exists a bijective map A-+B; in this case we write A ^ B. Theorem 8.1. Equipollence is an equivalence relation on the class S of all sets.
PROOF. Exercise; note that 0 ~ 0 since 0 (Z 0 X 0 is a relation that is (vacuously) a bijective function.3 ■ Let 70 = 0 and for each n e N* let /n = { 1 , 2 , 3 , . . . , « } . It is not difficult to prove that Im and In are equipollent if and only if m = n (Exercise 1). To say that a set A 3See
page 6.
16
PREREQUISITES AND PRELIMINARIES
has precisely n elements means that A and In are equipollent, that is, that A and /„ are in the same equivalence class under the relation of equipollence. Such a set A (with A ~ /„ for some unique n > 0) is said to be finite; a set that is not finite is infinite. Thus, for a finite set A, the equivalence class of A under equipollence provides an answer to the question: how many elements are contained in A? These considerations motivate
Definition 8.2. The cardinal number (or cardinality) of a set A, denoted |A|, is the
equivalence class of A under the equivalence relation ofequipollence. |A| is an infinite or finite cardinal according as A is an infinite or finite set. Cardinal numbers will also be denoted by lower case Greek letters: a,/3,y, etc. For the reasons indicated in the preceding paragraph we shall identify the integer n > 0 with the cardinal number |/„| and write |/„| = n, so that the cardinal number of a finite set is precisely the number of elements in the set. Cardinal numbers are frequently defined somewhat differently than we have done so that a cardinal number is in fact a set (instead of a proper class as in Definition 8.2). We have chosen this definition both to save time and because it better reflects the intuitive notion of “the number of elements in a set.” No matter what definition of cardinality is used, cardinal numbers possess the following properties (the first two of which are, in our case, immediate consequences of Theorem 8.1 and Defini tion 8.2). (i) Every set has a unique cardinal number; (ii) two sets have the same cardinal number if and only if they are equipollent (Ml = \B\
Definition 8.3. Let a and (3 be cardinal numbers. The sum a + /? is defined to be the
cardinal number |A U B|, where A and B are disjoint sets such that |A| = a and |B| = fi. The product a(i is defined to be the cardinal number |A X B|. It is not actually necessary for A and B to be disjoint in the definition of the product «/3 (Exercise 4). By the definition of a cardinal number a there always exists a set A such that Ml = «- It is easy to verify that disjoint sets, as required for the definition of a + /?, always exist and that the sum a + /3 and product afi are in dependent of the choice of the sets A,B (Exercise 4). Addition and multiplication of cardinals are associative and commutative, and the distributive laws hold (Exercise 5). Furthermore, addition and multiplication of finite cardinals agree with addition
8. CARDINAL NUMBERS
i7#
and multiplication of the nonnegative integers with which they are identified; for if A has m elements, B has n elements and A fl B = 0, then A U B has m + n ele ments and A X B has mn elements (for more precision, see Exercise 6).
Definition 8.4. Let a%f3 be cardinal numbers and A, B sets such that |A| = «, |B| = 0. a is-h?ss than or equal to /?, denoted a. < (3 or 13 > a, if A is equipollent with a subset of B (that is, there is an injective map A —* B). a is strictly less than [3, denoted a < jS or (3 > a, if a < (3 and a 9* (3.
It is easy to verify that the definition of < does not depend on the choice of A and B (Exercise 7). It is shown in Theorem 8.7 that the class of all cardinal numbers is linearly ordered by <. For finite cardinals < agrees with the usual ordering of the nonnegative integers (Exercise 1). The fact that there is no largest cardinal number is an immediate consequence of
Theorem 8.5. If A is a set and P(A) its power set, then |A| < |P(A)|.
SKETCH OF PROOF. The assignment a\-> [a] defines an injective map A —> P(A) so that \A\ < \P(A)\. If there were a bijective map /: A —> P(A), then for some a0 e A, f(ao) = B, where B = [a e A \ a \ f(a)} CZ A. But this yields a con tradiction: «0 e B and a0 $ B. Therefore\A\ \P(A)\ and hence \A\ < \P(A)\. ■ REMARK. By Theorem 8.5, = |N| < |P(N)|. It can be shown that |P(N)[ = |R[, where R is the set of real numbers. The conjecture that there is no cardinal number (3 such that < 0 < |P(N)I = |R| is called the Continuum Hy pothesis. It has been proved to be independent of the Axiom of Choice and of the other basic axioms of set theory; see P. J. Cohen [59]. The remainder of this section is devoted to developing certain facts that will be needed at several points in the sequel (see the first paragraph of this section).
Theorem 8.6. (Schroeder-Bernstein) If A and B are sets such that |A| < |B| and |BJ < |A|, then |A| = |B|.
SKETCH OF PROOF. By hypothesis there are injective maps f : A ^ B and g :B—> A. We shall use / and g to construct a bijection h : A-+B. This will imply that A ~B and hence \A\ = |B|. If a e A, then since g is injective the set g~\a) is either empty (in which case we say that a is parentless) or consists of exactly one ele ment b e B (in which case we write g-1(o) = b and say that b is the parent of a). Similarly for beB, we have either f~\b) = 0 (b is parentless) or f~\b) = a' e A (a' is the parent of b). If we continue to trace back the “ancestry” of an element as A in this manner, one of three things must happen. Either we reach a parentless ele ment in A (an ancestor of a e A), or we reach a parentless element in B (an ancestor
PREREQUISITES AND PRELIMINARIES
18
of a), or the ancestry of a e A can be traced back forever (infinite ancestry). Now de fine three subsets of A [resp. B] as follows: A\ = {a e A [ a has a parentless ancestor in A}; A2 = {a e A | a has a parentless ancestor in B}; A3~ {a e A | a has infinite ancestry}; Bi = [ b e B \ b has a parentless ancestor i n A } ; B 2 = [ b e B \ b has a parentless ancestor in B ); B 3 — { b e B | b has infinite ancestry}. Verify that the Ai [resp. Bt] are pairwise disjoint, that their union is A [resp. B] \ that /1 At is a bijection Ai —> Bi for / = 1 , 3 ; and that g \ B2 is a bijection B2 —> A2. Con sequently the map h : A —> B given as follows is a well-defined bijection: h(n\ = \f^ if a e A l U As> K ) \g-Ka) if a e A . ' 2 U
Theorem 8.7. The class of all cardinal numbers is linearly ordered by <. If a and (3
are cardinal numbers, then exactly one of the following is true: a < f3\ a = (3] (3 < a (TrichotomyLaw). SKETCH OF PROOF. It is easy to verify that < is a partial ordering. Let a,(3 be cardinals and A,B be sets such that \A\ = a, \B\ = ft. We shall show that < is a linear ordering (that is, either a < (3 or /? < a) by applying Zorn’s Lemma to the set fF of all pairs (fX), where X CZ A and f : X — > B is an injective map. Verify that fF t6 0 and that the ordering of fF given by (/i.A'i) < (f,X2) if and only if Xx (Z X2 and f21 X\ = f is a partial ordering of fF. If [(/,A'l) | / e /} is a chain in fF, let X — (J Xi and define f : X — + B by f(x) — f(x) for x eA',-. Show that /is a well-deiel
fined injective map, and that (fX) is an upper bound in fF of the given chain. There fore by Zorn’s Lemma there is a maximal element (g,X) of fF. We claim that either X = A or Im g = B. For if both of these statements were false we could find aeA — X and b e B — Im g and define an injective map h : X U \a} —» B by h(x) = g(x) for x eX and h(a) = b. Then (h,X U { a } ) e fF a n d (g,X) < (h,X U ( a ( ) , which contra dicts the maximality of (g,X). Therefore either^ = A so that |y4| < \B\ or Im g = B 0-1 in which case the injective mapfl —*X CZ A shows that |Z?| < \A\. Use these facts, the Schroeder-Bernstein Theorem 8.6 and Definition 8.4 to prove the Trichotomy Law. ■ REMARKS. A family of functions partially ordered as in the proof of Theorem 8.7 is said to be ordered by extension. The proof of the theorem is a typical example of the use of Zorn’s Lemma. The details of similar arguments in the sequel will fre quently be abbreviated.
Theorem 8.8. Every infinite set has a denumerable subset. In particular, Ho < « for
every infinite cardinal number a.
8. CARDINAL NUMBERS
19
SKETCH OF PROOF. If B is a finite subset of the infinite set A, then A — B is nonempty. For each finite subset B of A, choose an element xB e A — B (Axiom of Choice). Let F be the set of all finite subsets of A and define a map /: F —> F by f(B) = B U {A'ii}. Choose as A. By the Recursion Theorem 6.2 (with fn — f for all n) there exists a function
Therefore Im g is a subset of A such that |Im g| = |N| = S0. ■
Lemma 8.9. IfA is an infinite set and F a finite set then |A U F| = \A\. In particular,
a + n = a for every infinite cardinal number a and every natural number (finite cardinal) n. SKETCH OF PROOF. It suffices to assume A fl F = 0 (replace F by F — A if necessary). If F = j bubi, . . . , bn} and D = J Xi | / e N* | is a denumerable subset of A (Theorem 8.8), verify that f : A — > A U Fis a bijection, where /is given by f(x)
bi for x = Xi, 1 < i < n\ Xi_n for x = Xi, i > n; x for x e A — D.
Theorem 8.10. If a and /3 are cardinal numbers such that < a and a is infinite,
then a -f- (3 = a. SKETCH OF PROOF. It suffices to prove a + a = a (simply verify that a < a + / 3 < a + a = a and apply the Schroeder-Bernstein Theorem to conclude ct + /3 = a). Let A be a set with \A\ = a and let ‘S be the set of all pairs ( f X ) , where A" C A and f : X X {0,1} — > A " i s a bijection. Partially order JF by extension (as in the proof of Theorem 8.7) and verify that the hypotheses of Zorn’s Lemma are satisfied. The only difficulty is showing that fF . To do this note that the map NX {0,1} —> N given by («,0) |—> 2 n and (n, 1) |—> 2n + 1 is a bijection. Use this fact to construct a bijection /: D X {0,1} —» D, where D is a denumerable subset of A (that is, |D| = |N|; see Theorem 8.8). Therefore by Zorn’s Lemma there is a maximal element (g,C) £ fF.
0
Clearly C0 = {(c,0) | c e C\ and Ci = {(c,l) | c e Cj are disjoint sets such that |C0| = |C| = |Ci| and C X | 0 , 1 | = C 0 U Cx. The map g : C X {0,1} -> C is a bi jection. Therefore by Definition 8.3,
PREREQUISITES AND PRELIMINARIES
20
To complete the proof we shall show that |C| = a. If A — C were infinite, it would contain a denumerable subset B by Theorem 8.8, and as above, there would be a bi jection f : B X {0,1} —* B. By combining f with g, we could then construct a bijec tion h : (C U B) X J 0 , l J - > C U B s o that (g,C) < (h,C U B) e JF, which would contradict the maximality of (g,C). Therefore A — C must be finite. Since A is in finite and A = C U (A — C), C must also be infinite. Thus by Lemma 8.9, |C| = |C U (A - C)| = \A\ = a. ■ Theorem 8.11. If a and yS are cardinal numbers such that 0
and a is
infinite, then afi = a; in particular, aX0 = a and if yS is finite Kq/? = K0. SKETCH OF PROOF. Since a < a/3 < aa it suffices (as in the proof of Theo rem 8.10) to prove aa = a. Let A be an infinite set with \A\ = a and let 'S be the set of all bijections /: X X X - + X , where X is an infinite subset of A. To show that J 5^ 0, use the facts that A has a denumerable subset D (so that \D\ = |N| = |N*|) and that the map N* X N * ^ N * given by (/«,«)(—> 2m_1(2/j — 1) is a bijection. Partially order 3 by extension and use Zorn’s Lemma to obtain a maximal element g : B X B —'«B. By the definition of g, |fl| |fl| = \B X B\ = \B\. To complete the proof we shall show that |5| = \A\ = a. Suppose \A — B\ > |2?|. Then by Definition 8.4 there is a subset C of A — B such that |C| = |£|. Verify that |C| = l«l = \B X B\ = \B X C\ = |C X B\ = \C X C| and that these sets are mutually disjoint. Consequently by Definition 8.3 and Theo rem 8.10 |(5 U C) X (B U C)\ = |(5 X B) U (B X C) U (C X B) U (C X C)l = \B X B\ + \B X C| + |C X B\ + \C X C| = (|fi| + |fl|) + (|C| + |C|) = |5| + |C| = \B U C| and there is a bijection (B U C) X (B U C) —> (B U C), which con tradicts the maximality of g in 5. Therefore, by Theorems 8.7 and 8.101A — B \ | B \ and \B\ = \ A - B \ + |5| = |(A - B) U B\ = \A\ = a. ■ Theorem 8.12. Let Abe a set and for each integer n > 1 let An = A X A X - ■ ■ X A (n factors). (i) If A is finite, then |An| = |A|n, and if A is infinite, then |An| = |A|. (ii) | A"| = «0|A|.
u
neN*
SKETCH OF PROOF, (i) is trivial if \A\ is finite and may be proved by induc tion on n if \A\ is infinite (the case n = 2 is given by Theorem 8.11). (ii) The sets An(n > 1) are mutually disjoint. If A is infinite, then by (i) there is for each n a bijec tion/ : An —> A. The map An —> N* X A, which sends u e An onto («,/(«)), is a neN*
bijection. Therefore | U
An\
= |N* X A\ = |N*||.4| = ^oM|. (ii) is obviously true
neN*
if A = 0. Suppose, therefore, that A is nonempty and finite. Then each An is non empty and it is easy to show that = |N*| < | U An\- Furthermore each An is 7J6N*
finite and there is for each n an injective map gn : An —> N*. The map U An —>
neN*
N* X N*, which sends u e An onto (n,g„(u)) is injective so that | U An\ < |N* X N*| ne. N*
= |N*| = «o by Theorem 8.11. Therefore by the Schroeder-Bernstein Theorem | |J An| = «0. But «0 = since A is finite (Theorem 8.11). ■ neN*
8. CARDINAL NUMBERS
21
Corollary 8.13. If A is an infinite set and F(A) the set of all finite subsets of A, then |F(A)| = |A|. PROOF. The map A —> F(A) given by a\—> [a] is injective so that \A\ < \F(A)\. For each ^-element subset 5 of A, choose (au . . . , an) e An such that S — j «i,..., a„}. This defines an injective map F(A) —> U An so that \F(A)\ < I U A*\ = = \A\ weN*
neN*
by Theorems 8.11 and 8.12. Therefore, \A\ = \F(A)\ by the Schroeder-Bernstein Theorem 8.6. ■ EXERCISES
1. Let h — 0 and for each n e N* let /„ = {1,2,3, (a) In is not equipollent to any of its proper subsets [Hint: induction]. (b) Im and In are equipollent if and only if m = n. (c) Im is equipollent to a subset of /„ but ln is not equipollent to any subset of In if and only if m < n. 2. (a) Every infinite set is equipollent to one of its proper subsets. (b) A set is finite if and only if it is not equipollent to one of its proper subsets [see Exercise 1]. 3. (a) Z is a denumerable set. (b) The set Q of rational numbers is denumerable. [Hint: show that |Z| < |Q| < |Z X Z| = |Z|.] 4. If A,A',B,B' are sets such that \A\ = \A'\ and |B| = \B'\, then \A X B\ = |A' X B'\. If in addition A fl B = 0 = A' fl B', then \A U B\ = |A' U B'\. Therefore multiplication and addition of cardinals is well defined. 5. For all cardinal numbers a,(3,y: (a) a -f /3 = /S a and a/3 = (3a (commutative laws). (b) (a + (3) -f- 7 = a + OS + 7) and (a(3)7 = a((3y) (associative laws). (c) a(i3 + 7) = a(3 + «7 and (a + (3)y ~ ay + (3y (distributive laws). (d) a + 0 = a and al = a. (e) If a ^ 0, then there is no (3 such that a + (3 = 0 and if a ^ 1, then there is no (3 such that a/3 = 1. Therefore subtraction and division of cardinal num bers cannot be defined.
0, then (A U B) ~ Irn+n and A X B ~ Imn. Thus if we identify \A\ with m and |5| with n, then \A\ -\r \B\ = m + n and \A\\B\ = mn.
6. Let h be as in Exercise 1. If A ~ Im and B ~ /„ and A (1 B =
7. If A ~ A', B ~ B' and f : A — * B is injective, then there is an injective map A' —> B'_ Therefore the relation < on cardinal numbers is well defined. 8. An infinite subset of a denumerable set is denumerable.
9. The infinite set of real numbers R is not denumerable (that is, No < |R|). [Hint: it suffices to show that the open interval (0,1) is not denumerable by Exercise 8. You may assume each real number can be written as an infinite decimal. If (0,1) is denumerable there is a bijection /: N * — > ( 0,1). Construct an infinite decimal (rea 1 number) .aiar • • in (0,1) such that an is not the «th digit in the decimal expansion of /(«). This number cannot be in Im /.]
22
PREREQUISITES AND PRELIMINARIES
10. If a,(3 are cardinals, define a? to be the cardinal number of the set of all functions B—> A, where A,B are sets such that \A\ = a, \B\ = /3. (a) is independent of the choice of A,B. e+y (b) a = (a^)(ay); (aP)y = (a7)(/37); a&y = (a^)y. (c) If a < |8, then a7 < jS7. (d) If a,fi are finite with a > 1, /? > 1 and y is infinite, then ay = (e) For every finite cardinal n, an = aa ■ ■ a (n factors). Hence a" = a if a is infinite. (f) If P(A) is the power set of a set A, then \P(A)\ — 2,x|. 11. If /is an infinite set, and for each iel Ai is a finite set, then HJ Ai\ < |/|. iel
12. Let a be a fixed cardinal number and suppose that for every / e I, A{ is a set with \Ai\ - a. Then |U A{\ < |/|a. iel
CHAPTER
GROUPS
The concept of a group is of fundamental importance in the study of algebra. Groups which are, from the point of view of algebraic structure, essentially the same are said to be isomorphic. Ideally the goal in studying groups is to classify all groups up to isomorphism, which in practice means finding necessary and sufficient conditions for two groups to be isomorphic. At present there is little hope of classifying arbitrary groups. But it is possible to obtain complete structure theorems for various restricted classes of groups, such as cyclic groups (Section 3), finitely generated abelian groups (Section II.2), groups satisfying chain conditions (Section II.3) and finite groups of small order (Section II.6). In order to prove even these limited structure theorems, it is necessary to develop a large amount of miscellaneous information about the structure of (more or less) arbitrary groups (Sections 1, 2, 4, 5, and 8 of Chapter I and Sections 4 and 5 of Chapter II). In addition we shall study some classes of groups whose structure is known in large part and which have useful applications in other areas of mathematics, such as symmetric groups (Section 6), free [abelian] groups (Sections 9 and II. 1), nilpotent and solvable groups (Sections 11.7 and II.8). There is a basic t r u t h that applies not only to groups but also to many other algebraic objects (for example, rings, modules, vector spaces, fields): in order to study effectively an object with a given algebraic structure, it is necessary to study as well the functions that preserve the given algebraic structure (such functions are called homomorphisms). Indeed a number of concepts that are common to the theory of groups, rings, modules, etc. may be described completely in terms of ob jects and homomorphisms. In order to provide a convenient language and a useful conceptual framework in which to view these common concepts, the notion of a category is introduced in Section 7 and used frequently thereafter. Of course it is quite possible to study groups, rings, etc. without ever mentioning categories. How ever, the small amount of effort needed to comprehend this notion now will pay large dividends later in terms of increased understanding of the fundamental relationships among the various algebraic structures to be encountered. With occasional exceptions such as Section 7, each section in this chapter de pends on the sections preceding it. 23
CHAPTER I GROUPS
24
1. SEMIGROUPS, MONOIDS AND GROUPS If G is a nonempty set, a binary operation on G is a function C X G —> G. There are several commonly used notations for the image of (a,b) under a binary operation: ab (multiplicative notation), a + b (additive notation), ab, a* b, etc. For con venience we shall generally use the multiplicative notation throughout this chapter and refer to ab as the product of a and b. A set may have several binary operations defined on it (for example, ordinary addition and multiplication on Z given by (a,b) [—> a + b and (a,b) |—> ab respectively).
Definition 1.1. A semigroup is a nonempty set G together with a binary operation
on G which is (i) associative: a(bc) = (ab)c for all a, b, c e G; a monoid is a semigroup G which contains a (ii) (two-sided) identity element e e G such that ae = ea = a for all a e G. A group is a monoid G such that (iii) for every a e G there exists a (two-sided) inverse element a-1 e G such that a-1a = aa-1 = e. A semigroup G is said to be abelian or commutative if its binary operation is (iv) commutative: ab = ba for all a,b e G. Our principal interest is in groups. However, semigroups and monoids are con venient for stating certain theorems in the greatest generality. Examples are given below. The order of a group G is the cardinal number |G|. G is said to be finite [resp. infinite] if |G| is finite [resp. infinite].
Theorem 1.2. 7/G is a monoid, then the identity element e is unique. If G is a group,
then (i) c e G and cc = c =* c = e; (ii) for all a, b, c e G ab = ac => b = c and ba = ca =* b = c (left and rightcancellation)', (iii) for each a e G, the inverse element a-1 is unique; (iv) for each a e G, (a-1)-1 = a; (v) for a, b e G, (ab)-1 = b_1a_1; (vi) for a, b £ G the equations ax = b and ya = b have unique solutions in G : x = a-1b and y = ba-1. SKETCH OF PROOF. If e' is also a two-sided identity, then e = ee’ = e’. (i) cc = c => c~\cc) = c-1c —> (c~lc)c = e 'c => ec = e=* c = e\ (ii), (iii) and (vi) are proved similarly, (v) (ab)(trlarx) = a{bb~l)arl = (ae)a~l = aa~1 = e => (ab)'1 = by (iii); (iv) is proved similarly. ■
1. SEMIGROUPS, MONOIDS AND GROUPS
25
If G is a monoid and the binary operation is written multiplicatively, then the identity element of G w i l l always be denoted e. If the binary operation is written additively, then a + b (a, b s G) is called the sum of a and b, and the identity element is denoted 0; if G is a g r o u p the inverse of a s G is denoted by — a. We write a — b for a + ( — />)■ Abelian groups are frequently written additively. The axioms used in Definition 1.1 to define a group can actually be weakened considerably.
Proposition 1.3. Let G be a semigroup. Then G is a group if and only if the following
conditions hold: (i) there exists an element e £ G such that ea = a for all a £ G (left identity element)', (ii) for each a s G, there exists an element a-1 £ G such that a-1a = e (left inverse). REMARK. An analogous result holds for “right inverses” and a “right identity.” SKETCH OF PROOF OF 1.3. (=*) Trivial. (<=) Note that Theorem 1.2(i) is t r u e under these hypotheses. G ^ 0 since e £ G. If a e G, then by (ii) (aa l)(aa *) = a ( a ~ l a ) a — a(ea~l) = aar1 and hence aa~1 = e by Theorem 1.2(i). Thus a~l is a two-sided inverse of a. Since ae = a(a~la) = (aa^)a = ea = a for every a £ G, e is a two-sided identity. Therefore G is a group by Definition 1.1. ■
Proposition 1.4. Let G be a semigroup. Then G is a group if and only if for all a, b £ G the equations ax = b and ya = b have solutions in G.
PROOF. Exercise; use Proposition 1.3. ■ EXAMPLES. The integers Z, the rational numbers Q, and the real numbers R are each infinite abelian groups under ordinary addition. Each is a monoid under ordinary multiplication, but not a group (0 has no inverse). However, the nonzero elements of Q and R respectively form infinite abelian groups under multiplication. The even integers under multiplication form a semigroup that is not a monoid. EXAMPLE. Consider the square with vertices consecutively numbered 1,2,3,4,
center at the origin of the x-y plane, and sides parallel to the axes. )’ 1
2
4
3
Let Z)4* be the following set of “transformations” of the square. Dt* = {R,R2,R3J,Tx,Tv,Ti.3,7^.4}, where R is a counterclockwise rotation about the center of 90°, R2 a counterclockwise rotation of 180°, Rs a counterclockwise rotation of 270°
26
CHAPTER I GROUPS
and / a rotation of 360° ( = 0°); Tx is a reflection about the a- axis, 7i,3 a reflection about the diagonal through vertices 1 and 3; similarly for Tv and'T^. Note that each Us D* is a bijection of the square onto itself. Define the binary operation in £>4* to be composition of functions: for U, V e D4*, U ° V is the transformation V fol lowed by the transformation U. Z)4* is a nonabelian group of order 8 called the group of symmetries of the square. Notice that each symmetry (element of A*) is com pletely determined by its action on the vertices. EXAMPLE. Let 5 be a nonempty set and A(S) the set of all bijections S — > S . Under the operation of composition of functions, f ° g, A(S) is a group, since com position is associative, composition of bijections is a bijection, Is is a bijection, and every bijection has an inverse (see (13) of Introduction, Section 3). The elements of A(S) are called permutations and A(S) is called the group of permutations on the set S. If 5 = { 1 , 2 , 3 , . . . , nj, then A(S) is called the symmetric group on n letters and denoted SH. Verify that |S„| = n\ (Exercise 5). The groups S7, play an important role in the theory of finite groups. Since an element a of Sn is a function on the finite set 5 = {1,2, it can be described by listing the elements of S on a line and the image of each element under a 1
(
2
3
w
). The product ar of two elements of 5* is the il k /3 in/ composition function r followed by a; that is, the function on S given by A-1—> a(r(k)).1 For instance, let a = Q ^ ^ 4^ anc* r ~ Q \ \ 3) ^ e^ements *^4- T^en under ar, 1 h ^ W O ) = «K4) = 4, etc.; thus ar = Q J ^ 4^(4 123) (\ 2 3 4\. . .. . A 2 3 4 \ / l 2 3 4\ _ (\ 2 3 4 \ Simi y Ta \4 3 1 2/’ ’ 1 2 3 / V 3 1 2 4j \ 2 4 1 3 / This example also shows that S„ need not be abelian. Another source of examples is the following method of constructing new groups from old. Let G and H be groups with identities eG, en respectively, and define the direct product of G and H to be the group whose underlying set is G X H and whose binary operation is given by: 0a,b)(a',b') = (aa',bb'), where a,a' e G; b,b' e H.
Observe that there are three different operations in G, H and G X H involved in this statement. It is easy to verify that G X H is, in fact, a group that is abelian if both G and H are; (eG,eH) is the identity and (arx,b~l) the inverse of (a,b). Clearly |G X H\ = |G||//| (Introduction, Definition 8.3). If G and H are written additively, then we write G @ H in place of G X H. Theorem 1.5. Let R (~) be an equivalence relation on a monoid G such that ai ~ a2
and bi ~ b2 imply a,bt ~ a2b2 for all a;,b; e G. Then the set G/R of all equivalence classes of G under R is a monoid under the binary operation defined by (a)(b) = ab, where x denotes the equivalence class of x e G. //G is an [abelian] group, then so is G / R . xIn
many books, however, the product <jt is defined to be “o- followed by t."
1. SEMIGROUPS, MONOIDS AND GROUPS
An equivalence relation on a monoid G that satisfies the hypothesis of the theo rem is called a congruence relation on G. PROOF OF 1.5. If ax = a2 and b\ = b2 (ot, bi e G), then ax ~ a2 and b\ ~ b2 by (20) of Introduction, Section 4. Then by hypothesis axbx ~ a2b2 so that a.\bx = a2b2 by (20) again. Therefore the binary operation in G/R is well defined (that is, inde pendent of_the choice of equivalence class representatives). It is associative since a(b c) = a(bc) = a(bc) = (ab)c = (ab)c = (a b)c. e is the identity element since (a)(e) = ae = a = ea = (e)(a). Therefore G/R is a monoid. If G is a group, then a e G/R clearly has inverse o_1 so that G/R is also a group. Similarly, G abelian im plies G/R abelian. ■ EXAMPLE. Let m be a fixed integer. Congruence modulo m is a congruence re lation on the additive group Z by Introduction, Theorem 6.8. Let Zm denote the set of equivalence classes of Z under congruence modulo m. By Theorem 1.5 (with addi tive not at ion) Zm is an abelian group, with addition given by a + b = a + b(a,b e Z). The proof of Introduction, Theorem 6.8 shows that Zm = { 0 , 1 , . . . , m — 1} so that Zm is a finite group of order m under addition. Zm is called the (additive) group of integers modulo m. Similarly since Z is a commutative monoid under multiplication, and congruence modulo m is also a congruence relation with respect to multiplica tion (Introduction, Theorem 6.8), Zm is a commutative monoid, with multiplication given by (a)(5) = aE (a,b e Z). Verify that for all a, b, caZm\ a(b + c) = ab + ac and (a + b)c = ac + be (distributivity). Furthermore if p is prime, then the nonzero elements of Zp form a multiplicative group of order p — 1 (Exercise 7). It is customary to denote the elements of Zm as 0 , 1 , . . . , m — 1 rather than 0 , 1 , . . . , m — 1. I n context this ambiguous notation will cause no difficulty and will be used whenever convenient. EXAMPLE. The following relation on the additive group Q of rational numbers is a congruence relation (Exercise 8): a ~ b *=> a — be Z. By Theorem 1.5 the set of equivalence classes (denoted Q/Z) is an (infinite) abelian group, with addition given by a + b = a + b. Q/Z is called the group of rationals modulo one. Given a\,. . . , a* e G (n > 3) it is intuitively plausible that there are many ways of inserting parentheses in the expression axar ■ -an so as to yield a “meaningful” product in G of these n elements in this order. Furthermore it is plausible that any two such products can be proved equal by repeated use of the associative law. A necessary prerequisite for further study of groups and rings is a precise statement and proof of these conjectures and related ones. Given any sequence of elements of a semigroup G, {aua2 . . . | define inductively a meaningful product of au • . . , an (in this order) as follows. If n = 1, the only mean ingful product is a\. If n > 1, then a meaningful product is defined to be any product of the form (a\ - • • am)(am+\ ■ ■ ■ a„) where m < n and (oi • • • am) and (om+i • • • an) are meaningful products of m and n — m elements respectively.2 Note that for each 2To
show that this definition is in fact well defined requires a stronger version of the Recursion Theorem 6.2 of the Introduction; see C. W. Burrill [56; p. 57].
28
CHAPTER I GROUPS
n > 3 there may be many meaningful products of au . . . , an. For each n e N * we single out a particular meaningful product by defining inductively the standard n n product ai of au ■ ■ ■, an as follows: i=i 7i
1
H=
au
/n—l\
and for n > 1, JI= ( II W
i=li=l\i=l /
The fact that this definition defines for each n e N* a unique element of G (which is clearly a meaningful product) is a consequence of the Recursion Theorem 6.2 of the Introduction (Exercise 16).
Theorem 1.6. (Generalized Associative Law) If G is a semigroup and ai,. . . , a„ e G,
then any two meaningful products o / a i , . . . , an in this order are equal. PROOF. We use induction to show that for every n any meaningful product 71
a\ • ■ • a„. is equal to the standard n product H This is certainly true for n = 1 , 2 . »=i If n > 2, then by definition (oi- • •£„) = (ai - ■ -a^Xo^+i- • -a„) for some m < n. Therefore, by induction and associativity: /m \ /n—m \ (a, ■ ■ ■ a„) = (a, • • • am){am,, • • -o„) = ( ^ II am+i)
O
m \ / /n—m— 1
\ \ / / m \ /n—m — 1
\\
[T )( ( II a^+ijanj — II II Jan /n — 1 \
( II
ai )an
\t=l / 1=1
71
= II a{.
In view of Theorem 1.6 we can and do write any meaningful product of f l i , . . . , an e G (G a semigroup) as «ifl2- ■ an without parentheses or ambiguity.
Corollary 1.7. (Generalized Commutative Law) 7/G is a commutative semigroup and ai......... an e G, then for any permutation ii, .. . , in of 1, 2, . . . n, aia2- • -a„ = ai,ai2 • • -ain.
PROOF. Exercise. ■
Definition 1.8. Let G be a semigroup, a e G andn e N*. The element an e G is defined n
to be the standard n product n ai with ai = a for 1 < i < n. If G is a monoid, a0 is i = 1
defined to be the identity element e. If G is a group, then for each n e N*, a n is defined to be (a-1)n e G. The remarks preceding Theorem 1.6 and Exercise 16 show that exponentiation is well defined. By definition, then, a1 = a, a2 — aa, a3 = (aa)a = aaa, . .., an = an~Ya
1. SEMIGROUPS, MONOIDS AND GROUPS
29
= aa- ■ a (n factors). Note that we may have am = an with m j* n (for example, in
C, -1 = i2 = i8). ADDITIVE NOTATION. If the binary operation in G is written additively, then we write na in place of an. Thus Oa = 0, la = a, na = (n — I)a + a, etc.
Theorem 1.9. If G is a group [resp. semigroup, monoid] and a e G, then for all
m, n e Z [resp. N*, N]: (i) aman = am+n (additive notation: ma + na = (m + n)a); (ii) (am)n = amn (additive notation: n(ma) = mna). SKETCH OF PROOF. Verify that (an)~1 = (a-1)" for all neN and that a~n = (o_1)n for all n e Z. (i) is true for m > 0 and n > 0 since the product of a standard n product and a standard m product is a meaningful product equal to the standard (m + n) product by Theorem 1.6. For m < 0, and n < 0 replace a, m,n by a~l, —m, —n and use the preceding argument. The case m = 0 or n = 0 is trivial and the cases m > 0, n < 0 and m < 0, n > 0 are handled by induction on m and n re spectively. (ii) is trivial if m = 0. The case when m > 0 and n e Z is proved by induc tion on m. Use this result to prove the case m < 0 and n e Z . ■
EXERCISES
1. Give examples other than those in the text of semigroups and monoids that are not groups. 2. Let G be a group (written additively), S a nonempty set, and M(S,G) the set of all functions /: S —> G. Define addition in M(S,G) as follows: (/+ g) : S —> G is given by s|—»f ( s ) + g(s) e G. Prove that M(S,G) is a group, which is abelian if G is. 3. Is it true that a semigroup which has a left identity element and in which every element has a right inverse (see Proposition 1.3) is a group? 4. Write out a multiplication table for the group Z)4*. 5. Prove that the symmetric group on n letters, Sn, has order n\. 6. Write out an addition table for Z2 @Z2.
is called the Klein four group.
7. If p is prime, then the nonzero elements of ZT form a group of order p — 1 under multiplication. [Hint: a 9^ 0 => (a,p) = 1; use Introduction, Theorem 6.5.] Show that this statement is false if p is not prime. 8. (a) The relation given by a ~ b<^> a — f t e Z i s a congruence relation on the
additive group Q [see Theorem 1.5]. (b) The set Q/Z of equivalence classes is an infinite abelian group. 9. Let p be a fixed prime. Let Rp be the set of all those rational numbers whose de nominator is relatively prime to p. Let R* be the set of rationals whose de nominator is a power of p (p\ i > 0). Prove that both Rp and Rp are abelian groups under ordinary addition of rationals.
30
CHAPTER I GROUPS
10. Let p be a prime and let Z(pm) be the following subset of the group Q/Z (see Pg. 27): Z(pCD) = {a/b e Q/Z | a,b e Z and b = pi for some / > 0 | . Show that Z(jT) is an infinite group under the addition operation of Q/Z. 11. The following conditions on a group G are equivalent: (i) G is abelian; (ii) (ab)2 = a2b2 for all a,b e G; (iii) (ab)~l = arxb~x for all a,b e G; (iv) (ab)n = anbn for all n e Z and all a,b e G; (v) (ab)n = anbn for three consecutive integers n and all a,b e G. Show that (v) => (i) is false if “three” is replaced by “two.” 12. If G is a group, a,b e G and b a b 1 = ar for some r e N, then bjab~i = arl for all yeN. 13. If a2 = e for all elements a of a group G, then G is abelian. 14. If G is a finite group of even order, then G contains an element a 5* e such that a2 = e.
15. Let G be a nonempty finite set with an associative binary operation such that for all a,b,c e . G a b = ac=*b = c and ba = ca=> b = c. Then G is a group. Show that this conclusion may be false if G is infinite. 16. Let 01,02, . . . be a sequence of elements in a semigroup G. Then there exists a unique function ^ : N* —» G such that ^(1) = au xp(2) = oio2, ^(3) = (0102)03 and for n > 1, \p(n + 1) = (ip(n))an+1. Note that \p(n) is precisely the standard 71 n product a,-. [Hint: Applying the Recursion Theorem 6.2 of the Introduci=i tion with a = au S = G and fn : G —> G given by jc|—> xo„+2 yields a function ip : N —> G. Let = ip6, where 6 : N* —» N is given by A: |—> A: — 1.]
n
2. HOMOMORPHISMS AND SUBGROUPS Essential to the study of any class of algebraic objects are the functions that pre serve the given algebraic structure in the following sense.
Definition 2.1. Let G and H be semigroups. A function f: G —> H is a homomorphism provided f(ab) = f(a)f(b) for all a,b e G. If f is injective as a map of sets, f is said to be a monomorphism. Iff is surjective, f is called an epimorphism. Iff is bijective, f is called an isomorphism. In this case G and H are said to be isomorphic (written G = H). A homomorphism f : G —» G is called an endomorphism of G and an isomorphism f : G —> G is called an automorphism of G. If f : G ^ H and g : H —> K are homomorphisms of semigroups, it is easy to see that g f : G —» K is also a homomorphism. Likewise the composition of monomorphisms is a monomorphism; similarly for epimorphisms, isomorphisms and auto morphisms. If G and H are groups with identities e G and e H respectively and
2. HOMOMORPHISMS AND SUBGROUPS
31
/: G —> H i s a homomorphism, then f(eG) = e#; however, this is not true for mon oids (Exercise 1). Furthermore f(a~l) = /(a)-1 for all a e G (Exercise 1). EXAMPLE. The map /: Z —> Zm given by x |—> x (that is, each integer is mapped onto its equivalence class in Zm) is an epimorphism of additive groups, /is called the canonical epimorphism of Z onto Zm. Similarly, the map g : Q —» Q/Z given by |—> r is also an epimorphism of additive groups. EXAMPLE. If A is an abelian group, then the map given by a |—> o-1 is an auto morphism of A. The map given by a }—► a2 is an endomorphism of A. EXAMPLE. Let 1 < m, k e N*. The map g : Zm —►Z*,* given by X |—»kx is a monomorphism. EXAMPLE. Given groups G and H, there are four homomorphisms: G t = ± G X H ± = * H , given by ii(g) = (g,e); t2(h) = (e,h); n(g,h) = g\ ^(gji) = h. 7T1
7T2
ii is a monomorphism and 7r;- is an epimorphism (ij = 1,2).
Definition 2.2. Let f : G —» H be a homomorphism of groups. The kernel of f (de noted Ker f) is {a e G | f(a) = e e H}. If A is a subset of G, then f(A) = {b e H | b = f(a) for some a e A} is the image of A. f(G) is called the image of f and denoted Im f. If B is a subset o/H, f-1(B) = {a e G | f(a) e B} is the inverse image of B.
Theorem 2.3. Let f : G —» H be a homomorphism of groups. Then
(i) f is a monomorphism if and only i f Ker f = {e}; (ii) f is an isomorphism if and only if there is a homomorphism f-1 : H —> G such that ff-1 = Ih and f_1f = 1gPROOF, (i) If / is a monomorphism and a e Ker /, then f ( a ) — en = f(e), whence a = e and Ker / = {e\. If Ker f = [e] and f ( a ) = f(b), then efI = f ( a ) f(b)~l = f ( a ) f(b~l) = f(ab~l) so that ab'1 e Ker /. Therefore, ab~x = e (that is, a = b) and /is a monomorphism. (ii) If /is an isomorphism, then by (13) of Introduction, Section 3 there is a map of sets /-1 : H —> G such that f 'f = \G and /f~x = 1 h- /-1 is easily seen to be a homomorphism. The converse is an immediate consequence of (13) of Introduction, Section 3 and Definition 2.1. ■ Let G be a semigroup and H a nonempty subset of G. If for every a,b e H we have ab e H, we say that H is closed under the product in G. This amounts to saying that the binary operation on G, when restricted to H, is in fact a binary operation on H.
Definition 2.4. Let G be a group and H a nonempty subset that is closed under the product in G. // H is itself a group under the product in G, then H is said to be a sub group o f G . This is denoted by H < G.
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CHAPTER I GROUPS
Two examples of subgroups of a group G are G itself and the trivial subgroup (e) consisting only of the identity element. A subgroup H such that H G, H (e) is called a proper subgroup. EXAMPLE. The set of all multiples of some fixed integer n is a subgroup of Z, which is isomorphic to Z (Exercise 7). EXAMPLE. In Sn, the group of all permutations of {1,2, . . . , n \ , the set of all permutations that leave n fixed forms a subgroup isomorphic to Sn_i (Exercise 8). EXAMPLE. InZ6 = {0,1,2,3,4,5), both {0,3} and {0,2,4} are subgroups under addition. If p is prime, (Zp,+) has no proper subgroups. EXAMPLE. If /: G — * H is a homomorphism of groups, then Ker / is a sub group of G . If A is a subgroup of G , f ( A ) is a subgroup of H; in particular Im /is a subgroup of H. If B is a subgroup of H, f~\B) is a subgroup of G (Exercise 9). EXAMPLE. If G is a group, then the set Aut G of all automorphisms of G is a group, with composition of functions as binary operation (Exercise 15). By Theorem 1.2 the identity element of any subgroup H is the identity element of G and the inverse of a e H is the inverse arl of a in G.
Theorem 2.5. Let H be a nonempty subset of a group G. Then H is a subgroup of G
if and only //ab-1 e H for all a,b e H. PROOF. (<=) There exists a e //and hence e = aa 1 e H. Thus for any b e H, Zr1 = eb~l e H. If a,b e H, then b~l e H and hence ab = a(6-1)-1 e H. The product in H is associative since G is a group. Therefore H is a (sub)group. The converse is trivial. ■
Corollary 2.6. // G is a group and {Hi [ i e l } is a nonempty family of subgroups, then
fl Hi is a subgroup of G. izi
PROOF. Exercise. ■
Definition 2.7. Let G be a group and~X a subset of G. Let {Hi | i e l } be the family of
all subgroups of G which contain X. Then Hi is called the subgroup of G generated iel
by the set X and denoted (X). The elements of X are the generators of the subgroup ( X ) , which may also be generated by other subsets (that is, we may have ( X ) = (F) with X ^ Y ) . If X = {a u . . . , a n ) , we write ( a u . . . , a n ) in place of (A"). If G = (oi, . . . , a n), (a. e G ) , G is said to be finitely generated. If a e G , the subgroup ( a ) is called the cyclic (sub group generated by a .
2. HOMOMORPHISMS AND SUBGROUPS
33
Theorem 2.8. I f G is a group and X is a nonempty subset of G, then the subgroup (X)
generated by X consists ofallfinite products aiT11a2I1J ■ • • atnt (ai e X; rii e Z). In particular for every a e G, (a) = {an | n e Z\. SKETCH OF PROOF. Show that the set H of all such products is a subgroup of G that contains X and is contained in every subgroup containing X. Therefore H < (X) < H. u EXAMPLES. The additive group Z is an infinite cyclic group with generator 1, since by Definition 1.8 (additive notation), ml = m for all me Z. Of course the “powers” of the generating element need not all be distinct as they are in Z. The trivial subgroup (e) of any group is cyclic; the multiplicative subgroup (i) in C is cyclic of order 4 and for each m the additive group Zm is cyclic of order m with generator 1 e Zm. In Section 3 we shall prove that every cyclic subgroup is isomorphic either to Z or Zm for some m. Also, see Exercise 12. If {Hi | / e /} is a family of subgroups of a group G, then (J Hi is not a subgroup iel of G in general. The subgroup ((J //,) generated by the set (J Hi is called the subiel
iel
group generated by the groups {Hi [ i e I}. If H and K are subgroups, the subgroup (H U K) generated by H and K is called the join of H and K and is denoted H V K (additive notation: H + K).
EXERCISES
1. I f / : G —» //is a homomorphism of groups, then f(eG) = en and /(a-1) = /(o)-1 for all ae G. Show by example that the first conclusion may be false if G, H are monoids that are not groups. 2. A group G is abelian if and only if the map G —> G given by x |—► x~l is an auto morphism. 3. Let C?8 be the group (under ordinary matrix multiplication) generated by the com plex matrices A = ^ ® ^ and B —
where /2 = —I. Show that Q»
is a nonabelian group of order 8. Q8 is called the quaternion group. [Hint: Observe that BA = ASB, whence every element of Qh is of the form A'Bk Note also that A4 = B4 = /, where / = ^ ^ is the identity element of Q%.\ 4. Let H be the group (under matrix multiplication) of real matrices generated by and D =
Cl)
. Show that //is a nonabelian group of order 8
which is not isomorphic to the quaternion group of Exercise 3, but is isomorphic to the group Z)4*. 5. Let 5 be a nonempty subset of a group G and define a relation on G by a ~ b if and only if ab~l e S. Show that ~ is an equivalence relation if and only if 5 is a subgroup of G.
CHAPTER I GROUPS
34
6.
A nonempty finite subset of a group is a subgroup if and only if it is closed under the product in G.
1. If n is a fixed integer, then {kn | k e Zj CL Z is an additive subgroup of Z, which
is isomorphic to Z. 8.
The set [a zSn |
9. Let /: G —► // be a homomorphism of groups, A a subgroup of G, and B a sub group of H. (a) Ker /and f~\B) are subgroups of G. (b) f(A) is a subgroup of H. 10. List all subgroups of Z2 ©Z2. IsZ2 ©Z2 isomorphic to Z4? 11. If G is a group, then C = \a e G | ax = xa for all x e G) is an abelian subgroup of G. C is called the center of G. 12. The group D* is not cyclic, but can be generated by two elements. The same is true of Sn (nontrivial). What is the minimal number of generators of the additive group Z ® Z? 13. If G = (a) is a cyclic group and H is any group, then every homomorphism /: G —► // is completely determined by the element f(a) e //. 14. The following cyclic subgroups are all isomorphic: the multiplicative group (/') in C, the additive group Z4 and the subgroup
of St.
15. Let G be a group and Aut G the set of all automorphisms of G. (a) Aut G is a group with composition of functions as binary operation. [Hint: 1 g e Aut G is an identity; inverses exist by Theorem 2.3.] (b) Aut Z =lZi and Aut Z6 = Z2; Aut Z8 = Z2 (J) Z2; Aut Zv ~ Zp_i (/? prime). (c) What is Aut Z„ for arbitrary n e N*? 16. For each prime/? the additive subgroup ZO?”) qf Q/Z (Exercise 1.10) is genera ted by the set {\/pn \ n e N*}. 17. Let G be an abelian group and let H,K be subgroups of G. Show that the join H V K is the set [ab \ a e H, b e. K}. Extend this result to any finite number of subgroups of G. 18. (a) Let G be a group and {Hi | / e /} a family of subgroups. State and prove a condition that will imply that U H is a subgroup, that is, that [J H = (IJ Hi). iel
iel
iel
(b) Give an example of a group G and a family of subgroups {Hi \ iel | such that U Hi 9*
iel
19. (a) The set of all subgroups of a group G, partially ordered by set theoretic in clusion, forms a complete lattice (Introduction, Exercises 7.1 and 7.2) in which the g.l.b. of [Hi | / e /} is Q Hi and the l.u.b. is (U Hi). iel
iel
(b) Exhibit the lattice of subgroups of the groups S3, Z)4*, Z6, Z27, and Z36.
3. CYCLIC GROUPS
35
3. CYCLIC GROUPS The structure of cyclic groups is relatively simple. We shall completely char acterize all cyclic groups (up to isomorphism).
Theorem 3.1. Every subgroup H of the additive group Z is cyclic. Either H = (0) or H = (m), where m is the least positive integer in H. //H 5^ (0), then H is infinite.
PROOF. Either H = (0) or H contains a least positive integer m. Clearly (m) = I km | k e Zj d H. Conversely if he H, then h = qm + r with q, re Z and 0 < r < m (division algorithm). Since r = h — qm e H the minimality of m implies r = 0 and h = qm. Hence H CZ (m). If H 5* (0), it is clear that H = (m) is in finite. ■
Theorem 3.2. Every infinite cyclic group is isomorphic to the additive group Z and every finite cyclic group of order rr: is isomorphic to the additive group Zm.
PROOF. If G = (a) is a cyclic group then the map a : Z —> G given by k > ak is an epimorphism by Theorems 1.9 and 2.8. If Ker a = 0, then Z = G by Theorem 2.3 (i). Otherwise Ker a is a nontrivial subgroup of Z (Exercise 2.9) and hence Ker a = (m), where m is the least positive integer such that am = e (Theorem 3.1). For all r, s e Z, ar = as <=> ar~s = e <=> r — s e Ker a = (m)
<=> m | (r — s) <=> r = s in Zm, (where k is the congruence class of k £ Z). Therefore the map (3 :Zm—>G given by k (—* ak is a well-defined epimorphism. Since (3(k) = e <=?> ak = e = a° <=?> k = 0 in Zm, (3 is a monomorphism (Theorem 2.3(i)), and hence an i s o m o r p h i s m = G. ■
Definition 3.3. Let G be a group and a e G. The order of a is the order of the cyclic subgroup (a) and is denoted |a|.
Theorem 3.4. Let G be a group and a e G. If a has infinite order, then
(i) ak = e if and only ifk = 0; (ii) the elements ak (k e Z) are all distinct. If a has finite order m > 0, then
(iii) (iv) (v) (vi) (vii)
m is the least positive integer such that am = e; ak = e if and only if m \ k; ar = as if and only if r = s (mod m); (a) consists of the distinct elements a,a2,. . . , am_1,ain = e; for each k such that k | m, |ak| = m/k.
36
CHAPTER I GROUPS
SKETCH OF PROOF, (i)-(vi) are immediate consequences of the proof of Theorem 3.2. (vii) (ak)mlk = am = e afid (ak)T 9^ e for all 0 < r < m/k since other wise akr = e with kr < k(m/k) = m contradicting (iii). Therefore, \ak\ = m/k by (iii). ■
Theorem 3.5. Every homomorphic image and every subgroup of a cyclic group G is cyclic. In particular, if H is a nontrivial subgroup of G = (a) and m is the least positive integer such that am e H, then H = (am).
SKETCH OF PROOF. If f:G—*K is a homomorphism of groups, then Im / = (/(a)). To prove the second statement simply translate the proof of Theorem 3.1 into multiplicative notation (that is, replace every t e Z by a1 throughout). This proof works even if G is finite. ■ Recall that two distinct elements in a group may generate the same cyclic sub group.
Theorem 3.6. Let G = (a) be a cyclic group. If G is infinite, then a and a-1 are the only generators of G. // G is finite of order m, then ak is a generator of G if and only if (k,m) = 1.
SKETCH OF PROOF. It suffices to assume either that G = Z, in which case the conclusion is easy to prove, or that G = Z,„. If (k,m) = 1, there are c,d e Z such that ck + dm = 1; use this fact to show that k generates Zm. If (k,m) = r > 1, show that for n = m/r < m,nk = nk = 0 and hence k cannot generate Z,„. ■ A naive hope might be that the techniques used above could be extended to groups with two generators and eventually to all finitely generated groups, and thus provide a description of the structure of such groups. Unfortunately, however, even groups with only two generators may have a very complex structure. (They need not be abelian for one thing; see Exercises 2.3 and 2.4.) Eventually we shall be able to characterize all finitely generated abelian groups, but even this will require a great deal more machinery.
EXERCISES 1. Let a,b be elements of group G. Show that \a\ = |a-1|; \ab\ = \ba\, and \a\ = \cac~l\ for all c e G. 2. Let G be an abelian group containing elements a and b of orders m and n re spectively. Show that G contains an element whose order is the least common multiple of m and n. [Hint: first try the case when (m,n) = 1 . ]
3. Let G be an abelian group of order pq, with (p,q) = 1. Assume there exist a,b e G such that \a\ — p, \b\ = q and show that G is cyclic. 4. I f / : G —* H is a homomorphism, a e G, and /(o) has finite order in H, then 'a\ is infinite or | /(o)| divides \a\.
4. COSETS AND COUNTING
37
5. Let G be the multiplicative group of all nonsingular 2X2 matrices with rational entries. Show that a =
has order 4 and b = (-, -0 has order 3,
but ah has infinite order. Conversely, show that the additive group Z2 0) Z con tains nonzero elements a,b of infinite order such that a + b has finite order. 6.
If G is a cyclic group of order « and k \ n, then G has exactly one subgroup of order k.
7. Let p be prime and H a subgroup of Z(p°°) (Exercise 1.10). (a) Every element of Z(pco) has finite order pn for some n > 0. (b) If at least one element of H has order pk and no element of H has order greater than pk, then H is the cyclic subgroup generated by 1 /pk, whence //= Zpk. (c) If there is no upper bound on the orders of elements of H, then H = Z(pco); [see Exercise 2.16]. (d) The only proper subgroups of Z(pco) are the finite cyclic groups Cn = (1 /pn) (n = 1,2, . . .). Furthermore, (0) = C0 < Ci < C2 < C3 < • ■ •. (e) Let xi,x2 , . . . be elements of an abelian group G such that |jci| = p, px2 = xu px3 = x2,..., pxn+i = xn,... ■ The subgroup generated by the Xi (i > 1) is isomorphic to Z(pm). [Hint: Verify that the map induced by |—> \/pi is a well-defined isomorphism.] 8.
A group that has only a finite number of subgroups must be finite.
9. If G is an abelian group, then the set T of all elements of G with finite order is a subgroup of G. [Compare Exercise 5.] 10. An infinite group is cyclic if and only if it is isomorphic to each of its proper subgroups.
4. COSETS AND COUNTING In this section we obtain the first significant theorems relating the structure of a finite group G with the number theoretic properties of its order | G|. We begin by ex tending the concept of congruence modulo m in the group Z. By definition a = b (mod m) if and only if m \ a — b, that is, if and only if a — b is an element of the subgroup (m) = {mk \ k eX\. More generally (and in multiplicative notation) we have
Definition 4.1. Let W be a subgroup of a group G and a,b e G. a is right congruent to b modulo H, denoted a = r b (mod H) if ab-1 e H. a is left congruent to b modulo H, denoted a =i b (mod H), if a_1b e H.
If G is abelian, then right and left congruence modulo H coincide (since ab~x e H <=> (ab~l)~l e H and (ab~x)~x = ba~x = a~lb). There also exist nonabelian groups G and subgroups H such that right and left congruence coincide (Section 5), but this is not true in general.
38
CHAPTER I GROUPS
Theorem 4.2. Let W be a subgroup of a group G.
(i) Right [resp. left] congruence modulo H is an equivalence relation on G. (ii) The equivalence class of a e G under right [resp. left] congruence modulo H is the set Ha = {ha | h e H} [resp. aH = {ah | h e H}]. (iii) |Ha| = |H| = |aH| for all a e G. The set Ha is called a right coset of i/in G and aH is called a left coset of Hin G. In general it is not the case that a right coset is also a left coset (Exercise 2). PROOF OF 4.2. We write a = b for a b (mod H) and prove the theorem for right congruence and right cose ts. Analogous arguments apply to left congruence. (i) Let a,b,c e G. Then a = a since aa~l = e e H; hence .= is reflexive. = is clearly symmetric (a = b => ab1 e H => (ab1)1 e H => bar1 e H => b = a). Finally a = b and b = c imply oZr1 e H and bc~l e H. Thus oc-1 = (ab~1){fxrr) e H and a = c; hence = is transitive. Therefore, right congruence modulo H is an equivalence relation. (ii) The equivalence class of ae G under right congruence is { a e G | a = o } = {a s G | a a~l e H] = {* e G | a a~l = h e H| = {a e G | a = ha; h e Hj = [ha | heH} = Ha. (iii) The map Ha H given by ha > h is easily seen to be a bijection. ■
Corollary 4.3. Let H be a subgroup of a group G.
(i) (ii) (iii) (iv)
G is the union of the right [resp. left] cosets of H in G. Two right [resp. left] cosets of H in G are either disjoint or equal. For all a,b e G, Ha = Hb ab-1 e H and aH = bH <=> a-1b e H. If (R is the set of distinct right cosets of H in G and £ is the set of distinct left cosets of H in G, then |(R| = |£|. PROOF, (i)-(iii) are immediate consequences of the theorem and statements (19)—(21) of Introduction, Section 4. (iv) The map (R —> £ given by Ha ► or1// is a bijection since Ha = Hb abs //<=> (a'^b1 e H <=> a^H = b~lH. m ADDITIVE NOTATION. If H is a subgroup of an additive group, then right congruence modulo His defined by: a =r b (mod H) <=> a — b e H. The equivalence class of a e G is the right coset H + a = [h a \ h e. H) \ similarly for left congru ence and left cosets.
Definition 4.4. Let H be a subgroup of a group G. The index of H in G, denoted [G : H], is the cardinal number of the set of distinct right [resp. left] cosets of H in G.
In view of Corollary 4.3 (iv), [G : H] does not depend on whether right or left cosets are used in the definition. Our principal interest is in the case when [G : H] is finite, which can occur even when G and H are infinite groups (for example, [Z : {m)] = m by Introduction, Theorem 6.8(i)). Note that if H = (e), then Ha = {<:/} for every ae G and [G : H] = |G|.
4. COSETS AND COUNTING
39
A complete set of right coset representatives of a subgroup H in a group G is a set {ai] consisting of precisely one element from each right coset of H in G. Clearly the set {ai} has cardinality [G : H], Note that such a set contains exactly one element of H since H = He is itself a right coset. Analogous statements apply to left cosets.
Theorem 4.5. //K,H,G are groups with K < H < G, then [G : K] = [G : H][H : K], If any two of these indices are finite, then so is the third.
PROOF. By Corollary 4.3 G = U Hai with ax e G, |/| = [G : H] and the cosets iel
Hai mutually disjoint (that is, Hai = Ha, <=> / = j). Similarly H = [J Kbi with bj e H, jeJ
|J| ^ [H : K] and the cosets Kb, are mutually disjoint. Therefore G = (J Hai =
U (U
Kbj)ai
iel jeJ
= (J
iel
KbjOi. It suffices to show that the cosets KbjOi are mutually
( i,j)eI>J
disjoint. For then by Corollary 4.3. we must have [G : K] = \I X J|, whence [G : K| = \I X J\ = |/||J| = [G : H][H: K}. If Kbfii = Kbrat, then bjtu = kbrat (k e K). Since bj,bT,k e H we have Hai = HbjOi = Hkbrat = Hat; hence i = t and bj = kbr. Thus Kbj = Kkbr = Kbr and j = r. Therefore, the cosets KbjOi are mutually disjoint. The last statement of the theorem is obvious. ■
Corollary 4.6. (Lagrange). If H is a subgroup of a group G, then |G| = [G : H]|H|. In particular if G is finite, the order |a| of a e G divides |G|.
PROOF. Apply the theorem with K = (e) for the first statement. The second is a special case of the first with H = (a). ■ A number of proofs in the theory of (finite) groups rely on various “counting” techniques, some of which we now introduce. If G is a group and H,K are subsets of G, we denote by HK the set {ab \ a e H, b e K\; a right or left coset of a subgroup is a special case. If H,K are subgroups, HK may not be a subgroup (Exercise 7).
Theorem 4.7. Let H and K be finite subgroups of a group G. Then |HK| =
|H||K|/|H fl K|. SKETCH OF PROOF. C = H fl K is a subgroup of K of index n = \K\/\H fl K| and K is the disjoint union of right cosets Cki U Ck2 U • • • U Ckn for some ki e K. Since HC = H, this implies that HK is the disjoint union Hk\ U Hk2 U • • • U Hkn. Therefore, \HK\ = \H\-n = \H\\K\/\H (1 tf|. ■
Proposition 4.8. If H and K are subgroups of a group G, then [H : H fl K] < [G : K], If [G : K] is finite, then [H : H fl K] = [G : K] if and only if G = KH.
SKETCH OF PROOF. Let A be the set of all right cosets of H fl K'mH and B the set of all right cosets of K in G. The map
CHAPTER I GROUPS
40
(h e IT) is well defined since (H PI K)h' = (H CI K)h implies h'h~l e H fl K d K and hence Kh' — Kh. Show that
Proposition 4.9. Let H and K be subgroups of finite index of a group G. Then [G : H D K] is finite and [G : H fl K] < [G : H][G : K], Furthermore, [G : H D K] = [G : H][G : K] if and only ifG — HK.
PROOF. Exercise; use Theorem 4.5 and Proposition 4.8. ■
EXERCISES 1. Let G be a group and {Hi \ is. I) a family of subgroups. Then for any a£ G,
(fl
Hi)a
= fl
Ha.
ii*
(
1 2 3\ 2 1 3/
Then no left coset of H (except H itself) is also a right coset. There exists such that aH fl Ha = {a}. 1 2 3\ 2 3 1 /’ ^en every left coset of K is also a right coset of K.
(
3. The following conditions on a finite group G are equivalent. (i) | G| is prime. (ii) G 9^ (e) and G has no proper subgroups. (iii) G ^ Zp for some prime p. 4. (Euler-Fermat) Let a be an integer and p a prime such that pJfa. Then av~l = 1 (mod p). [Hint: Consider az.Zv and the multiplicative group of nonzero elements of Zp; see Exercise 1.7.] It follows that av = a (mod p) for any integer a. 5. Prove that there are only two distinct groups of order 4 (up to isomorphism), namely Z4 and Z2@Z2. [Hint: By Lagrange’s Theorem 4.6 a group of order 4 that is not cyclic must consist of an identity and three elements of order 2.] 6.
Let H,K be subgroups of a group G. Then HK is a subgroup of G if and only if HK = KH.
7. Let G be a group of order pkm, with p prime and (p/ri) = 1. Let H be a subgroup of order pk and K a subgroup of order pd, with 0 < d < k and K H. Show that HK is not a subgroup of G. 8.
If H and K are subgroups of finite index of a group G such that [G : H] and [G : K\ are relatively prime, then G = HK.
9. If H,K and N are subgroups of a group G such that H < N, then HK fl N = H{K n N).
5. NORMALITY, QUOTIENT GROUPS, AND HOMOMORPHISMS
41
10. Let H,K,N be subgroups of a group G such that H < K, H V\ N — K C\ N, and HN = KN. Show that H = K. 11. Let G be a group of order 2n; then G contains an element of order 2. If « is odd and G abelian, there is only one element of order 2. 12. If H and K are subgroups of a group G, then [H V K: H] > [K : H fl K\. 13. If p > q are primes, a group of order pq has at most one subgroup of order p. [Hint: Suppose H,K are distinct subgroups of order p. Show H fl K = (e); use Exercise 12 to get a contradiction.] 14. Let G be a group and a,b e G such that (i) \a\ = 4 = \b\; (ii) a2 = b2; (iii) ba = a3b = a~xb\ (iv) a 9^ b; (v) G = (a,b). Show that |G| = 8 and G ^ (See Exercise 2.3; observe that the generators A,B of Qs also satisfy (i)-(v).)
5. NORMALITY, QUOTIENT GROUPS, AND HOMOMORPHISMS We shall study those subgroups N of a group G such that left and right con gruence modulo N coincide. Such subgroups play an important role in determining both the structure of a group G and the nature of homomorphisms with domain G.
Theorem 5.1. If N is a subgroup of a group G, then the following conditions are equivalent.
(i) Left and right congruence modulo N coincide (that is, define the same equiva lence relation on G); (ii) every left coset of N in G is a right coset of hi in G; (iii) aN = Na for all a e G; (iv) for all a e G, aNa-1 C N, where aNa-1 = {ana-11 n e N}; (v) for all a e G, aNa~1 = N. PROOF, (i) <=> (iii) Two equivalence relations R and S' are identical if and only if the equivalence class of each element under R is equal to its equivalence class under S. In this case the equivalence classes are the left and right cosets respectively of N. (ii) => (iii) If aN = Nb for some b e G, then a e Nb fl Na, which implies Nb = Na since two right cosets are either disjoint or equal, (iii) => (iv) is trivial. (iv)=>(v) We have aNa~l
Definition 5.2. A subgroup N of a group G which satisfies the equivalent conditions of Theorem 5.1 is said to be normal in G (or a normal subgroup ofG); we write N <3 G if N is normal in G.
In view of Theorem 5.1 we may omit the subscripts ‘V” and when denoting congruence modulo a normal subgroup.
CHAPTER I GROUPS
42
EXAMPLES. Every subgroup of an abelian group is trivially normal. The sub 2 3\ ) in Sz is normal (Exercise 4.2). More generally any
)
subgroup N of index 2 in a group G is normal (Exercise 1). The intersection of any family of normal subgroups is a normal subgroup (Exercise 2). If G is a group with subgroups N and M such that N M and M <0 G, it does not follow that N <\ G (Exercise 10). However, it is easy to see that if Aris normal in G, then N is normal in every subgroup of G containing N. Recall that the join H V K of two subgroups is the subgroup (H U K ) generated by H and K.
Theorem 5.3. Let K and N be subgroups of a group G with N normal in G. Then
(i) (ii) (iii) (iv)
N (1 K is a normal subgroup of K; N is a normal subgroup of N V K; NK = N V K = KN; if K is normal in G and K fl N = (e), then nk = kn for all k e K and n e N .
PROOF, (i) If n e N fl K and a e K, then ana~l e N since N <\ G and ana~1 e K since K < G. Thus a(N fl K)a~l CZ N C\ K and N C\ K <] K. (ii) is trivial since N < N V K. (iii) Clearly NK CZ N V K. An element x of N V K is a product of the form mkimki- ■ nrkT, with m e N, ki£ K (Theorem 2.8). Since N <] G, n^k, = k,n/, m' e N and therefore x can be written in the form n(kY ■ - • kr), n e N. Thus N V KCZ NK. Similarly KN = N V K. (iv) Let kzK and n e N. Then nkn~l e K since K <\ G and kn ~xk~x e N since N <\ G. Hence (nkn l)k~l = n(krrlk l) e N fl K = (e), which implies kn = nk. ■
Theorem 5.4. // N is a normal subgroup of a group G and G/N is the set of all (left) cosets of N in G, then G/N is a group of order [G : N] under the binary operation given by (aN)(bN) = abN.
PROOF. Since the coset aN [resp. bN, abN} is simply the equivalence class of ae G [resp. b e G, ab e G] under the equivalence relation of congruence modulo N, it suffices by Theorem 1.5 to show that congruence modulo N is a congruence relation, that is, that ai = a (mod N) and bx = b (mod N) imply a\b\ = ab (mod N). By assumption a^r1 = th e N and h\b~x = m e N. Hence (aib\)(abyx — a\bjrxarl = (oi«2)o_1. But since N is normal, OiA^ = M?, which implies that ai«2 = «30i for some ns e N. Consequently (oifei)(oZ?)_1 = (ain^)a~1 = nza\0~l = e N, whence aibi = ab (mod AO. ■ If TV is a normal subgroup of a group G, then the group G/ N, as in Theorem 5.4, is called the quotient group or factor group of G by N. If G is written additively, then the group operation in G/N is given by (a + N) + (b + AO = (a + b) + A^. REMARK. If m > 1 is a (fixed) integer and k e Z, then the remarks preceding Definition 4.1 show that the equivalence class of k under congruence modulo m is
5. NORMALITY, QUOTIENT GROUPS, AND HOMOMORPHISMS
43
precisely the coset of (m) in Z which contains k; that is, as sets,ZTO = Z/(m). Theo rems 1.5 and 5.4 show that the group operations coincide, whence Zm = Z/(m) as groups. We now explore the relationships between normal subgroups, quotient groups, and homomorphisms.
Theorem 5.5. If /: G —*• H is a homomorphism of groups, then the kernel off is a normal subgroup of G. Conversely, if N is a normal subgroup of G, then the map 7r : G —» G/N given by 7r(a) = aN is an epimorphism with kernel N.
PROOF. If a- e Ker / and ae G, then f(axa-') = f(a) f(x) f(av) = f(a)ef(a)~l = e
and axa~l e Ker /. Therefore o(Ker /)o_1 C Ker / and Ker /
The map ir: G —> G/N is called the canonical epimorphism or projection. Here after unless stated otherwise G —> G/N (N <1 G) always denotes the canonical epimorphism.
Theorem 5.6. If f : G —*• H is a homomorphism of groups and N is a normal subgroup of G contained in the kernel off, then there is a unique homomorphism f : G/N —» H such that f(aN) = f(a) for all a e G. Im f = Im f and Ker f = (Ker f)/N. f is an iso morphism if and only iff is an epimorphism and N = Ker f.
The essential part of the conclusion may be rephrased: there exists a unique homomorphism /: G/N —> H such that the diagram:
G/N
is commutative. Corollary 5.8 below may also be stated in terms of commutative diagrams. PROOF OF 5.6. If b e aN, then b = an, n e N, and f(b) = f(an) = f(a) /(«) = f(a)e = fid), since N < Ker /. Therefore, / has the same effect on every element of the coset oNand the map /: G/N-+ //given by J(aN) = f(a) is a well-defined function. Since JiaNbN) = f(abN) = f(ab) = f(a) f(b) = f(aN) J(bN), f is a homomorphism. Clearly Im / = Im /and
CHAPTER I GROUPS
44
aN e Ker / <=> f(a) = e <=> azKerf
whence Ker /= {aA^I a e Ker /) = (Ker f)/N. f is unique since it is completely determined by /. Finally it is clear that /is an epimorphism if and only if /is. By Theorem 2.3 / is a monomorphism if and only if Ker / = (Ker f)/Nis the trivial sub group of G/N. which occurs if and only if Ker f = N. ■
Corollary 5.7. (First Isomorphism Theorem) If f : G —> H is a homomorphism of groups, then f induces an isomorphism G/Ker f ~ Im f.
PROOF. f:G-*+ Im /is an epimorphism. Apply Theorem 5.6 with N = Ker f ■
Corollary 5.8. Iff : G —* H is a homomorphism of groups, N <3 G, M H/M, A>’ aN (—> f(a)M. f is an isomorphism if and only iflm f V M = H and f-1(M) CZ N. In particular if f is an epimorphism such that f(N) = M and Ker f CZ N, then f is an isomorphism.
SKETCH OF PROOF. Consider the composition G -^> H A H/M and verify that N CZ /-1(M) = Ker tt/. By Theorem 5.6 (applied to irf) the map G/N —> H/M given by aN\-+ (-rrf)(a) = f(a)M is a homomorphism that is an isomorphism if and only if 7r/is an epimorphism and N = Ker tt/. But the latter conditions hold if and only if Im /V M = H and f~\M) CZ N. If /is an epimorphism, then H = Im / = Im / V M. If /(AQ = M and Ker fCZN, then /-1(M) CZ N, whence / is an isomorphism. ■
Corollary 5.9. (Second Isomorphism Theorem) If K and N are subgroups of a group
G, with N normal in G, then K/(N D K) = NK/N. PROOF. N <3 NK = N V K by Theorem 5.3. The composition K NK NK/N is a homomorphism /with kernel K f) N, whence /: K/K fl Af^Tm /by Corollary 5.7. Every element in NK/N is of the form nkN (n e N.k e K). The normal ity of N implies that nk = kn\ (m e N), whence nkN = knxN = kN = f(k). There fore /is an epimorphism and hence Im /= NK/N. ■
Corollary 5.10. (Third Isomorphism Theorem). If H and K are normal subgroups of a group G such that K < H, then H/K is a normal subgroup of G/K and
(G/K)/(H/K) ^ G/H. PROOF. The identity map \G : G —»• G has lc(^) < H and therefore induces an epimorphism / : G/K—* G/H, with l(aK) = aH. Since H = I(aK) if and only if aeH, Ker I = | aeH] = H/K. Hence H/K <3 G/K by Theorem 5.5 and G/H = Im / ^ (G/K)/Ker / = (G/K)/(H/K) by Corollary 5.7. ■
5. NORMALITY, QUOTIENT GROUPS, AND HOMOMORPHISMS
45
Theorem 5.11. If f : G —> H is an epimorphism of groups, then the assignment K |—> f(K) defines a one-to-one correspondence between the set Sf(G) of all subgroups Ko/G which contain Ker f and the set S(H) of all subgroups of H. Under this corre spondence normal subgroups correspond to normal subgroups.
SKETCH OF PROOF. By Exercise 2.9 the assignment /Tt—> f(K) defines a function
Corollary 5.12. If N is a normal subgroup of a group G, then every subgroup of G/N is of the form K/N, where K is a subgroup of G that contains N. Furthermore, K/N is normal in G/N if and only if K is normal in G.
PROOF. Apply Theorem 5.11 to the canonical epimorphism N
tt
: G —* G/N. If
EXERCISES 1. If N is a subgroup of index 2 in a group G, then N is normal in G. 2. If j Ni | / s /} is a family of normal subgroups of a group G, then H N{ is a
iti
normal subgroup of G. 3. Let N be a subgroup of a group G. N is normal in G if and only if (right) con gruence modulo N is a congruence relation on G. 4. Let ~ be an equivalence relation on a group G and l e t N = { o e G \ a ~ e). Then ~ is a congruence relation on G if and only if /V is a normal subgroup of G and ~ is congruence modulo N. 5. Let N < Si consist of all those permutations a such that tr(4) = 4. Is N normal in SP. 6.
Let H < G; then the set aHa~1 is a subgroup for each ae G, and H = aHa~l.
7. Let G be a finite group and H a subgroup of G of order n. If H is the only sub group of G of order n, then H is normal in G. 8.
All subgroups of the quaternion group are normal (see Exercises 2.3 and 4.14).
9. (a) If G is a group, then the center of G is a normal subgroup of G (see Ex ercise 2.11); (b) the center of Sn is the identity subgroup for all n > 2. 10. Find subgroups H and K of D4* such that H <] K and K < D4*, but H is not normal in D4*. 11. If His a cyclic subgroup of a group G and His normal in G, then every subgroup of H is normal in G. [Compare Exercise 10.]
46
CHAPTER I GROUPS
12. If H is a normal subgroup of a group G such that H and G/H are finitely gen erated, then so is G. 13. (a) Let H <] G, K <1G. Show that H VK is normal in G. (b) Prove that the set of all normal subgroups of G forms a complete lattice under inclusion (Introduction, Exercise 7.2). 14. If M <1 Gu < G2 then (M X W2) < (G, X G2) and (Gt X G2)/(M X M) ^ (G./M) X (G2/AT2). 15. Let A^<1 G and K <] G. If N f] K = (e> and N V K = G, then G/N ^ K 16. I f / : G —► // is a homomorphism, H is abelian and N is a subgroup of G con taining Ker /, then A'' is normal in G. 17. (a) Consider the subgroups (6) and (30) of Z and show that (6)/(30) =Z5. (b) For any k,m > 0, {k)/{km) =Zm; in particular, Z/(m) = (1 )/{m) =Zn. 18. If /: G —> H is a homomorphism with kernel N and K < G, then prove that f'KfUO) = KN. Hence f~'(f(K)) = K if and only if N < K. 19. If N <1 G, [G : N] finite, H < G, \H\ finite, and [G : N] and \H\ are relatively prime, then H < N. 20. If N <1G, |A^| finite, H < G, [G : H] finite, and [G : H] and |jV| are relatively prime, then N < H. 21. If H is a subgroup of Z(pm) and H ^ Z(jf°), then Z(pm)/ H = Z(pm). [Hint: if H = (1 /pn), let Xi = 1 /pn+i + H and apply Exercise 3.7(e).]
6. SYMMETRIC, ALTERNATING, AND DIHEDRAL GROUPS In this section we shall study in some detail the symmetric group Sn and certain of its subgroups. By definition Sn is the group of all bijections /„ —> /„, where In = { 1 , 2 , . . . , « } . The elements of Sn are called permutations. In addition to the notation given on page 26 for permutations inS„ there is another standard notation:
Definition 6.1. Let ii,i2,. . . , ir, (r < n) be distinct elements of In = (1,2, . . . n}. Then (iii2i3- - -ir) denotes the permutation that maps ii |—> i2, i2J — > i 3, to |—*■ U, - - -. ir_i •—> ir, and ir'—> ii, and maps every other element of In onto itself. (iii2- • • ir) is called a cycle of length r or an r-cycle; a 2-cycle is called a transposition.
The cycle notation is not unique (see below); indeed, strictly speaking, the cycle notation is ambiguous since (/'i • • • ir) may be an element of any Sn, n > r. In context, however, this will cause no confusion. A 1-cycle (A;) is the identity permutation. Clearly, an r-cycle is an element of order r in Sn. Also observe that if t is a cycle and t(x ) 5^ x for some x e /„, then r = (xr(x)r2(x) ■ - • rd(x)) for some d > 1. The inverse of the cycle (zW • • ir) is the cycle (irir-iir-i ■ • hid = (/ifWr-ifV-2- ■ - ^(verify!). EXAMPLES. The permutation r = ^ ^ ^
3^
’s
a
4
"cycle:
T
=
(1432)
= (4321) = (3214) = (2143). If cr is the 3-cycle (125), then ar = (125)(1432) = (1435)
6. SYMMETRIC, ALTERNATING, AND DIHEDRAL GROUPS
47
(remember: permutations are functions and ot means r followed by o)\ similarly to = (1432)(125) = (2543) so that or to. There is one case, however, when two permutations do commute.
Definition 6.2. The permutations 01,02,. . . , oT of Sn are said to be disjoint provided that for each 1 < i < r, and every k e In, <7i(k) 9^ k implies <jj(k) = k for all j 9^ i.
In other words ouoi,. . . ,or are disjoint if and only if no element of /„ is moved by more than one of
Theorem 6.3. Every nonidentity permutation in Sn is uniquely (up to the order of the factors) a product of disjoint cycles, each of which has length at least 2.
SKETCH OF PROOF. Let oeSn, o 9^ (1). Verify that the following is an equivalence relation on /„: for x,y e /„, x ~ y if and only if y = om(x) for some me Z. The equivalence classes {B:\ 1 < / < 5} of this equivalence relation are called the orbits of cr and form a partition of /„ (Introduction, Theorem 4.1). Note that if x e Bt, then Bi = \u \ x ~ u\ = {om(x) | m e ZJ. Let B\,Bi,. .., Br (1 < r < s) be those orbits that contain more than one element each (r > 1 since o 9^ (1)). For each i < r define cr; e Sn by:
Each d is a well-defined nonidentity permutation of /„ since o \ Bi is a bijection Bi —» Bi. cri,cr2, . . . , oT are disjoint permutations since the sets Bu . .., Br are mu tually disjoint. Finally verify that o = o\ 02- ■ • or; (note that x e Bi implies o(x) = ot(x) if i < r and o(x) = x if i > r; use disjointness). We must show that each o,- is a cycle. If x e Bi (i < r), then since Bi is finite there is a least positive integer d such that = oj(x) for some j (0 < j < d). Since od~j(x) = x and 0 < d — j < d, we must have 7 = 0 and od(x) = x. Hence (xo(x)o%x)- ■ ■ od_1(x)) is a well-defined cycle of length at least 2. If om(x) e Bi, then m = ad + b for some a,b e Z such that 0 < b < d. Hence om(x) = ob+ad(x) = oboad(x) = ob(x) e {x,o(x),o%x) , . . . , ad_1(x)} .Therefore Bi = [ x,o(x),o%x) , . . . , <7d_1(jr)| and it follows that cr, is the cycle od(x)
. (x<7(x)ct2(a:) • • •od *(a)). Suppose ti, ... ,Tt are disjoint cycles such that o = tit2- • Tt. Let x e /„ be such that a(x) 9± x. By disjointness there exists a unique y(l < j < r)with cr(x) = r,(x). Since ctt, = t o, we have ok(x) = T,k(x) for all k e Z. Therefore, the orbit of x under r, is precisely the orbit of x under o, say B>. Consequently, rt(y) = o(y) for every y e Bi (since y = on(x) = Tjn(x) for some n e Z). Since r3 is a cycle it has only one nontrivial orbit (verify!), which must be Bt since x 9^ o(x) = Tj(x). Therefore Tj(y) = y for all y $ Bi, whence r,- =
CHAPTER I GROUPS
48
Corollary 6.4. The order of a permutation a e S„ is the least common multiple of the orders of its disjoint cycles.
PROOF. Let cr = ov • .(Tr, with {o-,-j disjoint cycles. Since disjoint cycles com mute, am = o\m* • - Cr™ for all m e Z and am = (J) if and only if am = (1) for all i. Therefore am = (1) if and only if [o-,| divides m for all / (Theorem 3.4). Since |cr| is the least such m, the conclusion follows. ■
Corollary 6.5. Every permutation in Sn can be written as a product of (not necessarily disjoint) transpositions.
PROOF. It suffices by Theorem 6.3 to show that every cycle is a product of transpositions. This is easy: (xi) = (x^Xxix;*) and for r > 1, (xix2x3- • -xr) = (xixr)(xlxr_i)- • • (XiX3)(XiX2). ■ Definition 6.6. A permutation r e S„ is said to be even [resp. odd] if t can be written as a product of an even [resp. odd] number of transpositions.
The sign of a permutation r, denoted sgn r, is 1 or — 1 according as r is even or odd. The fact that sgn t is well defined is an immediate consequence of
Theorem 6.7. A permutation in S„ (n > 2) cannot be both even and odd.
PROOF. Let /i,/2, . . . , / „ be the integers 1,2,. .., n in some order and define A ( / i , . . . , /„) to be the integer JJ (/, — /*), where the product is taken over all pairs O’,A:) such that 1 < j < k < n. Note that A(/i, . . . , / „ ) ^ 0. We first compute A(crO'i), . . . , cr(in)) when a e Sn is a transposition, say a = (idd) with c < d. We have A(/i, . . . , / „ ) = (ic — id)ABCDEFG, where A = n (ii — 4); B = XI (ii — ic); C = JI (/,- — id); j
j
j
D = II (ii - id)', E = H (ic — /*); F = H (ic - ik);
c <j
c
d
G = H (id — ii).
We write a(A) for
n
d
(cr(ij) — cr(ikj) and similarly for a(B), o-(C), etc. Verify that
j
a(A) = A; a(B) = C and a(C) = B, u(D) = (-1 y-*-'E and u(E) = (-l)^"1/); <j(F) = G, and a(G) = F. Finally, c(ic — id) = a(ic) — a(id) = id — ic = ~(ic — id).
Consequently, A(cr(zi), . . . , a(in)) = a(ic - id)a(A)a(B) ■ ■ u(G) = (-1 Y+^-^(ic - id)ABCDEFG A(/l, . . . , in). Suppose for some reSn, t = n- • • rr and t — ui- ■ -crs with r», cr, transposi tions, r even and s odd. Then for (/\, . . . , / „ ) = (1,2,... , n) the previous paragraph implies A(r(l), . . . , r(n)) = A(tv • rr( 1), . . ., t - ■ Tr(n)) = -A(t - ■ -rr( 1), . . . , x
2
6. SYMMETRIC, ALTERNATING, AND DIHEDRAL GROUPS
49
Ti--- rr(«» = • ■ ■ = (— l)rA(l,2,
= A ( 1 , 2 , . . . , n). Similarly A(t(1), .. ., t(«)) = . . . , « ) = — A ( l , 2 , . . . , « ) , whence A(l,2, . . . , « ) = — A(l,2,. .., n). This is a contradiction since A ( l , 2 , . . . , « ) ^ 0 . ■ (—1)SA(1,2,
Theorem 6.8. For each n > 2, let An be the set of all even permutations of Sn. Then is a normal subgroup of S„ of index 2 and order |Sn|/2 = n!/2. Furthermore An is the only subgroup ofSnof index 2.
The group An is called the alternating group on n letters or the alternating group of degree n. SKETCH OF PROOF OF 6.8. Let C be the multiplicative subgroup {1,-1} of the integers. Define a map /: Sn —» C by a (—» sgn a and verify that /is an epimor phism of groups. Since the kernel of /is clearly An, An is normal in Sn. By the First Isomorphism Theorem Sn/An = C, which implies [S„ : An\ = 2 and \An\ = \Sn\/2. An is the unique subgroup of Sn of index 2 by Exercise 6. ■
Definition 6.9. A group G is said to be simple if G has no proper normal subgroups.
The only simple abelian groups are theZv with p prime (Exercise 4.3). There are a number of nonabelian simple groups; in particular, we have
Theorem 6.10. The alternating group An is simple if and only if n ^ 4 .
The proof we shall give is quite elementary. It will be preceded by two lemmas. Recall that if r is a 2-cycle, r2 = (1) and hence t = r_1.
Lemma 6.11. Let r,s be distinct elements of {1,2, . . . , n}. Then A„(n > 3) is gen erated by the 3 -cycles {(rsk) | 1 < k < n, k ^ r,s|.
PROOF. Assume n > 3 (the case n = 3 is trivial). Every element of An is a product of terms of the form (abXcd) or (ab)(ac), where a,b,c,d are distinct elements of { 1 , 2 , . Since (ab)(cd) = (acb)(acd) and (ab)(ac) = (acb), An is generated by the set of all 3-cycles. Any 3-cycle is of the form (rsa), (ras), (rab), (sab), or (abc), where a,b,c are distinct and a,b,c ^ r,s. Since (ras) = (rsa)2, (rab) = (rsb)(rsay, (sab) = (rsb)2(rsa), and (abc) = (rsa)2(rsc)(rsb)2(rsa), An is generated by {(rsk) | 1 < k < n, k r,^}. ■
Lemma 6.12. //N is a normal subgroup of An (n > 3) and N contains a 3-cycle, then
N = An. PROOF. If (rsc) e N, then for any k ^ r,s,c, (rsk) = (rs)(ck)(rsc)2(ck)(rs) = [(r.v)(c/:)](r.vc)2[(r.v)(cA)]^1 e N. Hence N = An by Lemma 6.11. ■
CHAPTER I GROUPS
50
PROOF OF THEOREM 6.10. A2 = (1) and A3 is the simple cyclic group of order 3. It is easy to verify that {(1),(12)(34),(13)(24),(14)(23)} is a normal subgroup of Ai (Exercise 7). If n > 5 and Ar is a nontrivial normal subgroup of An we shall show N = An by considering the possible cases. CASE 1. N contains a 3-cycle; hence N = An by Lemma 6.12. CASE 2. N contains an element cr, the product of disjoint cycles, at least one of which has length r > 4. Thus a = (ciia-i■ ■ •ar)r (disjoint). Let S = (aiaaa^ e An. Then cr_1(5cr6_J) e N by normality. But o--1(5cr5-1) = T~l(aiarcir-i- ■ ■ a2)(aia2a3)(aia2- ■ • flr)r(flifl3fl2) = (aia3ar) e N. Hence N = An by Lemma 6.12. CASE 3. N contains an element a, the product of disjoint cycles, at least two of which have length 3, so that a = (aia2<33)(fl4fl5fl6)T-(disjoint). Let 8 = (aia2a4) e An. Then as above, N contains o--1(5crS_1) = r'^aeasXaias^aia^Xai^tasX^asae)?(aia^) = (ciiaia‘2af,as). Hence N = An by case 2. CASE 4. N contains an element cr that is the product of one 3-cycle and some 2-cycies, say cr = (aia2a3)T (disjoint), with r a product of disjoint 2-cycles. Then cr2 e A^and cr2 = (aia2a3)T(aia2a3)T = (aia2a3)2T2 = (aia^Qzf = (01^2), whence N = A„ by Lemma 6.12. CASE 5. Every element of N is the product of (an even number of) disjoint 2-cycles. Let a e N, with cr = (aia2)(a3ai)T (disjoint). Let S = (aia2a3) e An; then cr_1(5cr5—1) e N as above. Now cr_1(5o-5_1) = T^iaza^iaia^iaia^a^iaia^iaza^Tiaiaza^ = (flifl3)(«204). Since n > 5, there is an element b e {1,2,. .., n) distinct from aua2,as,a4. Since £ = (cha3b) e An and f = (aia3)(a2ai) e N, f(£f£_1) e Ar. But f^fiT1) = (aia^(a2a^(aia3b)(aiai){a^ii)(ja\bai) = (axa3b) e N. Hence N = An by Lemma 6.12. Since the cases listed cover all the possibilities, An has no proper normal sub groups and hence is simple. ■ Another important subgroup of Sn (w > 3) is the subgroup Dn generated by a = (123- • •«) and
^ _ /l 2 3 4 5 - - - / • - • 71 — 1 /A \ 1 n Ti — l ti — 2 72 — 3 -
n (/'
71
n2—i
3 2/
+ 2 — i). Dn is called the dihedral group of degree n. The group
2
Dn is isomorphic to and usually identified with the group of all symmetries of a regular polygon with n sides (Exercise 13). In particular D4 is (isomorphic to) the group D*
of symmetries of the square (see pages 25-26).
Theorem 6.13. For each n > 3 the dihedral group D„ is a group of order 2n whose generators a and b satisfy:
(i) an = (1); b2 = (1); ak ^ (1) if 0 < k < n; (ii) ba = a~*b.
6. SYMMETRIC, ALTERNATING, AND DIHEDRAL GROUPS
51
Any group G which is generated by elements a,b e G satisfying (i) and (ii) for some n > 3 (with e e G in place of (I)) is isomorphic to Dn.
SKETCH OF PROOF. Verify that a,b e Dn as defined above satisfy (i) and (ii), whence Dn = (a,b) = (aibi | 0 < /' < n; j = 0,1} (see Theorem 2.8). Then verify that the 2n elements aibi (0 < / < « ; / = 0,1) are all distinct (just check their action on 1 and 2), whence \Dn\ = 2n. Suppose G is a group generated by a,b e G and a,b satisfy (i) and (ii) for some n> 3. By Theorem 2.8 every element of G is a finite product am'bm*amsbmt • ■ ■ bmk (mi e Z). By repeated use of (i) and (ii) any such product may be written in the form a'b’ with 0 < i < n and j = 0,1 (in particular note that b2 = e and (ii) imply b = b~l and ab = ba~l). Denote the generators of Dn by ax,bx to avoid confusion and verify that the map f: Dn—> G given by ajbj —* a'bj is an epimorphism of groups. To complete the proof we show that /is a monomorphism. Suppose /(afbi’) = a{b’ = e e G with 0 < i < n and j = 0,1. If j = 1, then ai = 6 and by (ii) ai+1 = a*a = ba = a~*b = a~la{ = a'-1, which implies a2 = e. This contradicts (i) since n > 3. Therefore j = 0 and e = a'b° = a1 with 0 < / < n, which implies / = 0 by (i). Thus f(a,'bi>) = e implies a^bi1 = a^b^ = (1). Therefore / is a monomorphism by Theorem 2.3. n This theorem is an example of a characterization of a group in terms of “genera tors and relations.” A detailed discussion of this idea will be given in Section 9.
EXERCISES 1. Find four different subgroups of S4 that are isomorphic to S3 and nine iso morphic to 2. (a) Sn is generated by the n—l transpositions (12), (13), (14),.. ., (1«). [Hint: (l/Xl/Xl/) = (tf).] (b) Sn is generated by the n — l transpositions (12), (23), ( 3 4 ) , . . . , ( « — 1 n). [Hint: (1j) = (I j - l)(j - 1 j)(l j - 1); use (a).] 3.' If cr = (/*i/2* • • ir) £ Sn and t eSn, then tot~ is the r-cycle (r0i)r(/2)- • r(/V)). 1
4. (a) Sn is generated by
is the only subgroup of Sn of index 2. [Hint: Show that a subgroup of index 2 must contain all 3-cycles of Sn and apply Lemma 6.11.]
7. Show that N = {(1),(12)(34),(13)(24),( 14)(23)} is a normal subgroup ofS4 con tained in A4 such that S4/ N = Ss and AJN ^ Z3. 8.
The group A4 has no subgroup of order 6.
9. For n > 3 let Gn be the multiplicative group of complex matrices generated by /0 1\ , /e2rri/n where i2 = —1. Show that Gn = Dn. andy o)
~\o (Hint: recall that e27ri = 1 and ek2vi ^ 1, where k is real, unless k e Z.)
CHAPTER I GROUPS
52
10. Let a be the generator of order n of Dn. Show that (a) < D„ and Dn/{a) 11. Find all normal subgroups of Dn. 12. The center (Exercise 2.11) of the group Dn is (e) if n is odd and isomorphic toZ2 if n is even. 13. For each n > 3 let Pn be a regular polygon of n sides (for n = 3, Pn is an equi lateral triangle; for n = 4, a square). A symmetry of Pn is a bijection Pn —» Pn that preserves distances and maps adjacent vertices onto adjacent vertices. (a) The set Dn* of all symmetries of Pn is a group under the binary operation of composition of functions. (b) Every ft Dn* is completely determined by its action on the vertices of PnNumber the vertices consecutively 1 , 2 , t h e n each ft Dn* determines a unique permutation oy of { 1 , 2 , . . . , « } . The assignment /(—► 07 defines a mono morphism of groups (p : Dn* —> Sn. (c) Dn* is generated by /and g, where /is a rotation of 2-n/n degrees about the center of Pn and g is a reflection about the “diameter” through the center and vertex 1. (d)
= (123- • •«) and ae = (\ 2 3 ■■■ « - 1 «\ \1 n « — 1 ••• 3 2/ Im ip = Dn and Dr* = Dn. 07
7. CATEGORIES: PRODUCTS, COPRODUCTS, AND FREE OBJECTS Since we now have several examples at hand, this is an appropriate time to intro duce the concept of a category. Categories will serve as a useful language and provide a general context for dealing with a number of different mathematical situations. They are studied in more detail in Chapter X. The intuitive idea underlying the definition of a category is that several of the mathematical objects already introduced (sets, groups, monoids) or to be introduced (rings, modules) together with the appropriate maps of these objects (functions for sets; homomorphisms for groups, etc.) have a number of formal properties in com mon. For example, in each case composition of maps (when defined) is associative; each object A has an identity map 1^ : A —*■ A with certain properties. These notions are formalized in
Definition 7.1. A category is a class C of objects (denoted A,B,C,...) together with
(i) a class of disjoint sets, denoted hom( A,B), one for each pair of objects in G ;(an element’f of hom(A,B) is called a morphism from A to B and is denoted f : A —> B); (ii) for each triple (A,B,C) of objects of Q a function hom(B,C) X hom(A,B) —> hom(A,C); (for morphisms f : A —> B, g : B —> C, this function is written (g,f) 1—> g g 0 f: A —> C is called the composite of f and g); all subject to the two axioms:
0
f and
(I) Associativity. Iff : A—> B g : B — > C , h : C — > D are morphisms of Q, then h o (g o f) = (h o g) o f.
7. CATEGORIES: PRODUCTS, COPRODUCTS, AND FREE OBJECTS 53
(II)
Identity. For each object B of Q there exists a morphism 1b : B —> B such that for any f : A —» B, g : B —» C, 1B
° f = f and g o 1B = g.
In a category Q a morphism / : A —> B is called an equivalence if there is in C a morphism g :B —> A such that g ° /= \A and f ° g = \b- The composite of two equivalences, when defined, is an equivalence. If / : A—* B is an equivalence, A and B are said to be equivalent. EXAMPLE. Let S be the class of all sets; for A,B e S, hom(v4,Z?) is the set of all functions / : A ^ B. Then S is easily seen to be a category. By (13) of Introduc tion, Section 3, a morphism /of S is an equivalence if and only if /is a bijection. EXAMPLE. Let 9 be the category whose objects are all groups; hom(/l^B) is the set of all group homomorphisms / : A —» B. By Theorem 2.3, a morphism / is an equivalence if and only if / is an isomorphism. The category 6L of all abelian groups is defined similarly. EXAMPLE. A (multiplicative) group G can be considered as a category with one object, G. Let hom(G,G) be the set of elements of G; composition of morphisms a,b is simply the composition ab given by the binary operation in G. Every morphism is an equivalence (since every element of G has an inverse). 1g is the identity element e of G. EXAMPLE. Let the objects be all partially ordered sets (S,<). A morphism (T,<) is a function/ :S —> Tsuch that for x,y eS, x < y => f(x) < f(y). EXAMPLE. Let Q be any category and define the category 3D whose objects are all morphisms of C. If / : A —» B and g : C —> D are morphisms of G, then hom(/,g) consists of all pairs (c*,/3), where a : A —» C, /3 : B —> D are morphisms of C such that the following diagram is commutative: /
A-------- 1-------- *-B «1
C-------- -------- *-D Definition 7.2. Let © be a category and {Ai | is I) a family of objects of Q. A product for the family {Ai | i e 1} is an object P of C together with a family of mor phisms {7i"i : P —> Ai | i e l } such that for any object B and family of morphisms [
A product P of {Ai | i e /} is usually denoted JI A- It is sometimes helpful to deie/
scribe a product in terms of commutative diagrams, especially in the case / = 11,2}. A product for \AUA2\ is a diagram (of objects and morphisms) Ax V~P A2 such that: for any other diagram of the form Ai^-B^ A2, there is a unique morphism ip : B^> P such that the following diagram is commutative:
CHAPTER I GROUPS
54
A family of objects in a category need not have a product. In several familiar categories, however, products always exist. For example, in the category of sets the Cartesian product U Ai is a product of the family {Ai \ i e /} by Introduction, is/
Theorem 5.2. In the next section we shall show that products exist in the category of groups.
Theorem 7.3. 7/(P,{7Ti}) and (Q,{\M) are both products of the family {A; | i e l ) of objects of a category C, then P and Q are equivalent.
PROOF. Since P and Q are both products, there exist morphisms / : P —> Q and g : Q —> P such that the following diagrams are commutative for each i e I:
Composing these gives for each iel a commutative diagram:
Thus go f ; P —> p is a morphism such that tti° (g ° f) = tti for all i e /. But by the definition of product there is a unique morphism with this property. Since the map lp : P —> P is also such that 7r, o lP = ^ for all i e I, we must have g°f= lp by uniqueness. Similarly, using the fact that Q is a product, one shows that f° g = 1«. Hence / : P —> Q is an equivalence. ■ Since abstract categories involve only objects and morphisms (no elements), every statement about them has a dual statement, obtained by reversing all the arrows (morphisms) in the original statement. For example, the dual of Definition 7.2 is
Definition 7.4. A coproduct (or sum) for the family { A 4 1 i e I } of objects in a cate gory C is an object S of G, together with a family of morphisms { i i : A ; — ► S | i e 1 } such that for any object B and family of morphisms { ^ i : A t — ► B | i e I J , there is a unique morphism $ : S — > B such that ^ ° i i = for all i e I .
7. CATEGORIES: PRODUCTS, COPRODUCTS, AND FREE OBJECTS 55
There is no uniform notation for coproducts, although JJA is sometimes used. iel
In the next two sections we shall discuss coproducts in the category g of groups and the category <1 of abelian groups. The following theorem may be proved by using the “dual argument” to the one used to prove Theorem 7.3 (do it!).
Theorem 7.5. 7/(S,{ ti}) and (S', {\i))are both coproducts for the family {Ai | i e 1} of objects of a category G, then S and S' are equivalent.
In several of the categories mentioned above (for example, groups), every object in the category is in fact a set (usually with some additional structure) and every morphism / : A —> B in the category is a function on the “underlying sets” (usually with some other properties as well). We formalize this idea in
Definition 7.6. A concrete category is a category C together with a function a that assigns to each object A of G a set cr(A) (called the underlying set of A) in such a way that:
(i) every morphism A —» B of G is a function on the underlying sets o(A) —> cr(B); (ii) the identity morphism of each object A of Q is the identity function on the underlying set u(A); (iii) composition of morphisms in G agrees with composition of functions on the underlying sets.
EXAMPLES. The category of groups, equipped with the function that assigns to each group its underlying set in the usual sense, is a concrete category. Similarly the categories of abelian groups and partially ordered sets, with the obvious underlying sets, are concrete categories. However, in the third example after Definition 7.1, if the function a assigns to the group G the usual underlying set G, then the categoi y in question is not a concrete category (since the morphisms are not functions on the set G). Concrete categories are frequently useful since one has available not only the properties of a category, but also certain properties of sets, subsets, etc. Since in virtually every concrete category we are interested in, the function a assigns to an object its underlying set in the usual sense (as in the examples above), we shall denote both the object and its underlying set by the same symbol and omit any explicit refer ence to a. There is little chance of confusion since we shall be careful in a concrete category G to distinguish morphisms of G (which are by definition also functions on the underlying sets) and maps (functions on the underlying sets, which may not be morphisms of Q).
Definition 7.7. Let F be an object in a concrete category G, X a nonempty set, and i : X —> F a map (of sets). F is free on the set X provided that for any object A of G and map (of sets) f : X —» A, there exists a unique morphism of G, f :F —* A,such that fi = f (as a map of sets X —*■ A).
CHAPTER I GROUPS
56
The essential fact about a free object F is that in order to define a morphism with domain F, it suffices to specify (he image of the subset i(X) as is seen in the following examples.
EXAMPLES. Let G be any group and g e G. Then the map / : Z —> G defined by /(«) = gn is easily seen to be the unique homomorphism Z —► <7 such that 11—> g. Consequently, if X = {1} and i: X —> Z is the inclusion map, then Z is free on X in the category of groups; (given f :X-+ G, let g = /(l) and define f as above). In other words, to determine a unique homomorphism from Z to G we need only specify the image ofl e Z (that is, the image of i(X)). The (additive) group Q of ra tional numbers does not have this property. It is not difficult to show that there is no nontrivial homomorphism Q —* S3. Thus for any set X, function /: X —> Q and func tion / :X—> S3 with f(xi) ^ (1) for some xi e X, there is no homomorphism f :Q—>S3 with fi = f.
Theorem 7.8. If <3 is a concrete category, F and F' are objects of C such that F is free on the set X and F' is free on the set X' and |X| = |X'|, then F is equivalent to F'.
Note that the hypotheses are satisfied when F and F' are both free on the same set X. PROOF OF 7.8. Since F, F' are free and \X\ = \X'\, there is a bijection / : X —»X' and maps /: A" —> F and j : X' —> F'. Consider the map jf :X—+F'. Since Fis free, there is a morphism
F---------
f hi
X---------►A"
/ is commutative. Similarly, since the bijection /has an inverse f l :X' —>X and F' is free, there is a morphism \f/: F' —> F such that: * %
X’-------- +-X n
is commutative. Combining these gives a commutative diagram:
7. CATEGORIES: PRODUCTS, COPRODUCTS, AND FREE OBJECTS 57
Hence (\p °
Definition 7.9. An object I in a category Q is said to be universal {or initial) if for each object C o / C there exists one and only one morphism I —> C. An object T of Q is said to be couniversal (or terminal) if for each object C of Q there exists one and only one morphism C —* T.
We shall show below that products, coproducts, and free objects may be con sidered as (co)universal objects in suitably chosen categories. However, this char acterization is not needed in the sequel. Since universal objects will not be mentioned again (except in occasional exercises) until Sections III.4, III.5, and IV.5, the reader may wish to omit the following material for the present.
Theorem 7.10. Any two universal [resp. couniversal] objects in a category Q are equivalent.
PROOF. Let / and J be universal objects in G. Since / is universal, there is a unique morphism / : /—»/. Similarly, since J is universal, there is a unique morphism g :J —> /. The composition go f : I —> I is a morphism of Q. But 1 / : / —* / is also a morphism of C. The universality of /implies that there is a unique morphism / —+ /, whence g ° / = 1 /. Similarly the universality of J implies that f°g = 1 j. Therefore / : / —* J is an equivalence. The proof for couniversal objects is analogous. ■ EXAMPLE. The trivial group (e) is both universal and couniversal in the cate gory of groups. EXAMPLE. Let F be a free object on the set X (with /: X —* F) in a concrete category C. Define a new category 3D as follows. The objects of 3D are all maps of sets / : X—* A, where A is (the underlying set of) an object of C. A morphism in 3D from / :X —* A \og ‘.X —>B is defined to be a morphism h : A —>B of C such that the diagram:
58
CHAPTER I GROUPS
is commutative (that is, hf = g). Verify that 1A ' A —> A is the identity morphism from /to / in 3D and that h is an equivalence in 3D if and only if h is an equivalence in C. Since F is free on the set X, there is for each map / : X —* A a unique mor phism / : F —> A such that fi = / This is precisely the statement that / : X —* F is a universal object in the category 3D. EXAMPLE. Let [Ai\izl\ be a family of objects in a category 6. Define a category £ whose objects are all pairs (£,{ / | / e / j), where B is an object of G and for each i, f:B —> Ai is a morphism of C. A morphism in £ from (B,\ f\ i el}) to (D,{gi | / e /}) is defined to be a morphism h : B —+ D of G such that gi ° fi = /■ for every i e 1. Verify that 1 B is the identity morphism from (B, {/}) to (B, {/•}) in £ and that h is an equivalence in £ if and only if h is an equivalence in G. If a product exists in G for the family f [ / e /} (with maps irk : —> Ak for each k el), then for every (i?,{ / )) in £ there exists a unique morphism f : B —> such that 77\ ° / = fi for every i e I. But this says that (JJA-,{ | / e / }) is a couniversal object in the category £. Similarly the coproduct of a family of objects in G may be considered as a universal object in an appropriately constructed category. Since a product YJ.Ai of a family {A% | / e /[ in a category may be considered as a couniversal object in a suitable category, it follows immediately from Theorem 7.10 that Y\Ai is uniquely determined up to equivalence. Analogous results hold for co products and free objects.
EXERCISES 1. A pointed set is a pair (£,jc) with S a set and x & S. A morphism of pointed sets (S,x) —> (£',*') is a triple (f,x,x'), where /: S —*S' is a function such that f(x) = x\ Show that pointed sets form a category. 2. If / : A —* B is an equivalence in a category G and g : B —> A is the morphism such that g o /= fo g = lBj show that g is unique. 3. In the category g of groups, show that the group Gi X G2 together with the homomorphisms 7ri : Gi X G2 —* Gi and 7r2: Gi X G2 —* G2 (as in the Example preceding Definition 2.2) is a product for { Gi,G2\. 4. In the category Cl of abelian groups, show that the group A, X A2, together with the homomorphisms ii : Ax —» A\ X A2 and i2: A2—+ Ax X A2 (as in the Example preceding Definition 2.2) is a coproduct for {AUA2}.
5. Every family {A< | / s /} in the category of sets has a coproduct. [Hint: consider U Ai = {(a,i) e ( U Ai) X / | a e A,} with Ai —> CJ Ai given by a\—> (a,i). U A is called the disjoint union of the sets Ai.] 6.
(a) Show that in the category S* of pointed sets (see Exercise 1) products always exist; describe them. (b) Show that in S* every family of objects has a coproduct (often called a “wedge product”); describe this coproduct.
7. Let F be a free object on a set X (i: X —> F) in a concrete category G. If C con tains an object whose underlying set has at least two elements in it, then / is an in jective map of sets.
8. DIRECT PRODUCTS AND DIRECT SUMS 8.
59
Suppose X is a set and F is a free object on X (with i: X —* F) in the category of groups (the existence of F is proved in Section 9). Prove that i(X) is a set of generators for the group F. [Hint: If G is the subgroup of F generated by i(X), then there is a homomorphism
8. DIRECT PRODUCTS AND DIRECT SUMS In this section we study products in the category of groups and coproducts in the category of abelian groups. These products and coproducts are important not only as a means of constructing new groups from old, but also for describing the structure of certain groups in terms of particular subgroups (whose structure, for instance, may already be known). We begin by extending the definition of the direct product G X H of groups G and H (see page 26) to an arbitrary (possibly infinite) family of groups {Gi \ i e /}. Define a binary operation on the Cartesian product (of sets) Gi as follows. If
n
tel
fg e XI Gi (that is, f,g : I —> U Gi and f(i),g(i) e G, for each /'), then/g : / —> |J Gi is iel iel iel the function given by /' —>/(/)#(/). Since each Gi is a group, /(/)#(/) e Gi for every /, whence fg e Gi by Introduction, Definition 5.1. If we identify /eJJ Gi with its iel iel
n
image {£;}(£»• = /(/) for each / e /) as is usually done in the case when / is finite, then the binary operation in H Gi is the familiar component-wise multiplication: {a,} {6*} iel — {aibi]. H G*> together with this binary operation, is called the direct product iel
(or complete direct sum) of the family of groups {Gt- | / e /}. If I = { 1 , 2 , . . . , n\, H Gi is usually denoted Gi X G2 X • • • X G„ (or in additive notation, Gi @ G2 ©•••©G„).
Theorem 8.1. //{Gi | ie 1} is a family of groups, then
(i) the direct product
n Gi iel
(ii) for each \lz\, the map irk :
is a group;
n Gi —> Gk iel
given by f H f(k) [or {ai} H ak] is an
epimorphism of groups.
PROOF. Exercise. ■ The maps irk in Theorem 8.1 are called the canonical projections of the direct product. Theorem 8.2. Let {Gi | i e l } be a family of groups and {
that 7T i ip = (fi for all i e l and this property determines X I G i uniquely up to isomorphism. In other words,
n Gi iel
iel
is a product in the category of groups.
CHAPTER I GROUPS
60
PROOF. By Introduction, Theorem 5.2, the map of sets ip : H
—> n
Gt given by
iel
J Gi is the unique function such that iutp =
iel
easy to verify that
sense) and therefore determined up to isomorphism (equivalence) by Theorem 7.3. ■ Since the direct product of abelian groups is clearly abelian, it follows that the direct product of abelian groups is a product in the category of abelian groups also.
Definition 8.3. The (external) weak direct product of a family of groups {Gi | i e l } , denoted IT Gi, is the set of all f e Gi such that f(i) = ei, the identity in Gi, for all iel
itl
(
)
hut a finite number ofi e I. If all the groups Gi are additive abelian,
IT G; is usually iel
called the (external) direct sum and is denoted Giis/
If I is finite, the weak direct product coincides with the direct product. In any case, we have
Theorem 8.4. //{Gi | i e 1} is a family of groups, then 0)
II” Gi is a normal subgroup °/II Gi;
iel
iel
(ii) for each k e I, the map ik : Gk —*■ Gi given by ik(a) = {a^ iei, where ai = e iel
for i 5^ k and ak = a, is a monomorphism of groups; (iii) for each i e l , ii(G0 is a normal subgroup of JJ Gi. iel
PROOF. Exercise. ■ The maps ik in Theorem 8.4 are called the canonical injections.
Theorem 8.5. Let {A; | i s I] be a family of abelian groups (written additively). If B is an abelian group and {\pi : Ai —» B \ i e Ij a family of homomorphisms, then there is a unique homomorphism ^ ^ A; —* B such that \fsii = for oil i s I and this property iel
determines ^ A; uniquely up to isomorphism. In other words, ^ Ai is a coproduct in iel
iel
the category of abelian groups.
REMARK. The theorem is false if the word abelian is omitted. The external weak direct product is not a coproduct in the category of all groups (Exercise 4). PROOF OF 8.5. Throughout this proof all groups will be written additively. If 0 ^ [ai} e then only finitely many of the a, are nonzero, say a^,a, 2 , . . . , a,r. Define ^ A{ —> B by \p\0! = 0 and \p({ <7,}) = ^(a,,) + &2(a,2) H------------------h ^ir(air) = ^ \pi(aj), where /0 is the set [i\,h,. . . , ir\ = {/ e / | 0 j. Since B is abelian, ie/o
8. DIRECT PRODUCTS AND DIRECT SUMS
61
it is readily verified that \p is a homomorphism and that \pLi = for all i e I. For each {«, } s {ai\ u(ai), /o finite as above. If £ : ^ A, —> B is a homomorie/o phism such that = \pi for all / then £({fl,|) = g(^ = X) &("*) /0
=
*>(«>)) = /o
/o
/o
hence £ = \p and \p is unique. Therefore yi A
/o
is a coproduct in the category of abelian groups and hence is determined up to iso morphism (equivalence) by Theorem 7.5. ■ Next we investigate conditions under which a group G is isomorphic to the weak direct product of a family of its subgroups.
Theorem 8.6. Let {Ns [ i e I} be a family of normal subgroups of a group G such that
(i) G = ( U Ni); ie/ (ii) for each k e I, Nk D ( U Ni) = (e). i
Then G ^ II" Ni. ie/
Before proving the theorem we note a special case that is frequently used.' Ob serve that for normal subgroups Ni,N2, . . ., Nr of a group G, (M U N-2 U • • • U Nr) = NiN2- - -Nr = {«i«2- ■ -nr | «, e Ni\ by an easily proved generalization of Theorem 5.3. In additive notation NiN2- • • Nr is written Ni + N2-\— - + Nr. It may be help ful for the reader to keep the following corollary in mind since the proof of the general case is essentially the same.
Corollary 8.7. If Ni,N2, . . . , Nr are normal subgroups of a group G such that G = NiN2- • -Nr and for each 1 < k < r, Nk fl (Ni - ■ -Nk-iNk+i- • Nr) = (e), then G = N, X N2 X ■ ■ ■ X Nr. ■ then Ui = e for a11 but a finite PROOF OF THEOREM 8.6. If {a^ e number of i e I. Let /0 be the finite set {ie 11 a, ^ e\. Then JX a, is a well-defined eleiel o
ment of G, since for ae Ni and b e N„ (i ^ j), ab = ba by Theorem 5.3(iv). Conse quently the map tp : H“M- —» G, given by (a,} H( XI e ^ I el H e), is a homoiel o
morphism such that ipifai) = ai for a, e Nt. Since G is generated by the subgroups N, every element a of G is a finite product of elements from various N{. Since elements of Nt and Nj commute (for i ^ j), a can be written as a product a,, where a,- e N and /0 is some finite subset of I. Hence
n
izi o
II ti(fl,-) e H^TV, and <^(H <.,(«,)) = II = II a% = a. Therefore,
0
0
E 0
e G. Clearly we may assume for convenience of noUI°
t
T tation that I0 = {1,2, . . . , « } . Then a, = a^a-r • -a„ = e, with a, e N,. Hence
iel o
ar1 = fl-j - - ■ fl„ e Ni fl ( U M) = (e) and therefore Ai = e. Repetition of this argu-
15*1 ment shows that a* = e for all / e I. Hence ^ is a monomorphism. ■
CHAPTER I GROUPS
62
Theorem 8.6 motivates
Definition 8.8. Let {N; | i e 1} be a family of normal subgroups of a group G such that G = (U Ni) and for each k e I, Nk fl ( U Ni) = (e). Then G is said to be the internal iel
weak direct product of the family {Ni | i e l } (or the internal direct sum ifG is (additive) abelian).
As an easy corollary of Theorem internal weak direct products.
8.6
we have the following characterization of
Theorem 8.9. Let {Ni | i e IJ be a family of normal subgroups O f a group G. G is the internal weak direct product of the family {N; | i e 1} if and only if every nonidentity element of G is a unique product a^a^- • -ain with i i , . . . , in distinct elements of I and e t* aik e Nik for each k = 1,2, . . . , n.
PROOF. Exercise. | There is a distinction between internal and external weak direct products. If a group G is the internal weak direct product of groups Ni, then by definition each Ni is actually a subgroup of G and G is isomorphic to the external weak direct product wN . However, the external weak direct product Ni does not actually contain t
n
IT
itI
iel
the groups Ni, but only isomorphic copies of them (namely the i,(N,) — see Theorem 8.4 and Exercise 10). Practically speaking, this distinction is not very important and the adjectives “internal” and “external” will be omitted whenever no confusion is possible. In fact we shall use the following notation. NOTATION. We write G = IJ’W, to indicate that the group G is the internal ui
weak direct product of the family of its subgroups {Nt | / e /}.
Theorem 8.10. Let {fi : Gi —> Hi | i e 1} be a family of homomorphisms of groups and let f = JJfi be the map JJ Gi —» Hi, given by { a i l f fi(aO}. Then f is a homoiel
iel
morphism of groups such that f(J Jw Gi) d JI™ Hi, Ker f = Ker fi and Im f
-nze/ Im fi.
iel
iel
iel
Consequently f is a monomorphism [resp. epimorphism] if and only if each
fi is. PROOF. Exercise. | Corollary 8.11. Let {Gi | i e I) and {Ni | i s I j be families of groups such that Ni is a normal subgroup of Gi for each i e l .
(i)
Ni is a normal subgroup o f n Gi and IlGi/n N i ^ J I C i / N i . is T
iel
iel
ieT
iel
(ii) j][w Ni is a normal subgroup of Gi and JJW Gi/JJw Ni = JJW Gi/Ni. iel
iel
iel
iel
iel
8. DIRECT PRODUCTS AND DIRECT SUMS
63
PROOF, (i) For each /', let 7Ti : G, —> Gi/Ni be the canonical epimorphism. By Theorem 8.10, the map : Gi/Ni is an epimorphism with kernel
n►n
itl
iel
n Ni. Therefore XIg,/XJm = Y\dl/Nl by the First Isomorphism Theorem, (ii) iel
is similar. ■
EXERCISES 1. S3 is noi the direct product of any family of its proper subgroups. The same is true of Zvn (p prime, n > 1) and Z. 2. Give an example of groups Hi, K} such that Hi X H2 = Kx x K2 and no H is isomorphic to any Kjm 3. Let G be an (additive) abelian group with subgroups H and K. Show that ffl
/-s
7r*
G= H(+) K if and only if there are homomorphisms H fc? G <=± K such that
LI ft* TTiti = 1 h, 7T2i2 = 1 k, ^1*2 = 0 and 7T2ti = 0, where 0 is the map sending every element onto the zero (identity) element, and ii7Ti(jc) + i27r2(jc) = x for all x e G. 4. Give an example to show that the weak direct product is not a coproduct in the category of all groups. (Hint: it suffices to consider the case of two factors G X H.) 5. Let G, H be finite cyclic groups. Then G X His cyclic if and only if (|G|,|//|) = 1. 6.
Every finitely generated abelian group G ^ (e) in which every element (except e) has order p (p prime) is isomorphic to Zv @ZV ©• ■ - @ZV (n summands) for some n > 1. [Hint: Let A = {ai,..., a„] be a set of generators such that no proper subset of A generates G. Show that (o») = Zp and G = (ai) X (^2) X • • • X (fln).]
7. Let H,K,N be nontrivial normal subgroups of a group G and suppose G = H X K. Prove that N is in the center of G or N intersects one of H,K nontrivially. Give examples to show that both possibilities can actually occur when G is nonabelian. 8.
Corollary 8.7 is false if one of the TV, is not normal.
9. If a group G is the (internal) direct product of its subgroups H,K, then H ^ G/K and G/H= K. 10. If j Gi | i e /} is a family of groups, then its subgroups {ia(G,) | / e /}.
is the internal weak direct product
11. Let {Ni | i e /} be a family of subgroups of a group G. Then G is the internal weak direct product of {M | / e /} if and only if: (i) a,a, = a,a, for all /' ^ j and e Ni, a, e 7V2 ; (ii) every nonidentity element of G is uniquely a product atl ■ ■ ■ ain, where iu are distinct elements of I and e ^ e Nik for each k. [Compare Theorem 8.9.] 12. A normal subgroup Hof a group G is said to be a direct factor (direct summand if G is additive abelian) if there exists a (normal) subgroup K of G such that G = HX K.
64
CHAPTER I GROUPS
(a) If H is a direct factor of K and AT is a direct factor of G, then H is normal in G. [Compare Exercise 5.10.] (b) If H is a direct factor of G, then every homomorphism H —> G may be ex tended to an endomorphism G —» G. However, a monomorphism H —> G need not be extendible to an automorphism G —* G. 13. Let | Gi | / e /} be a family of groups and J CZ /. The map « : n c , - n c , jeJ
iel
given by {a,} |—> (6,}, where bj = cij for ye/ and bi = e2 (identity of G,) for / ^ /, is a monomorphism of groups and JJ Gi/adi °>) = n iel
jeJ
tel — J
14. For / = 1,2 let Hi <1 Gi and give examples to show that each of the following statements may be false: (a) G\ = C2 and Hi = H2 => Gi/Hx = G2/H-2. (b) Gi = Gi and Gi/Hx G2/H2 =» //, ^ H2. (c) //, ^ H2 and C,///, ^ G2/H2 ^ G] = G2.
9. FREE GROUPS, FREE PRODUCTS, AND GENERATORS AND RELATIONS We shall show that free objects (free groups) exist in the (concrete) category of groups, and we shall use these to develop a method of describing groups in terms of “generators and relations.” In addition, we indicate how to construct coproducts (free products) in the category of groups. Given a set X we shall construct a group F that is free on the set X in the sense of Definition 7.7. If X = 0, F is the trivial group (e). \IX ^ 0, let X~* be a set disjoint from X such that \X\ — \X~l[. Choose a bijection X —* X~l and denote the image of xe Jbyx - 1 . Finally choose a set that is disjoint from X U X~l and has exactly one element; denote this element by 1. A word on X is a sequence (alfa2, • •.) with a, e X U X~1 U {1} such that for some n e N*, ak— 1 for all k> n. The constant sequence (1,1,. . .) is called the empty word and is denoted 1. (This ambiguous notation will cause no confusion.) A word (aua2,. ..) on X is said to be reduced provided that (i) for all x bX, x and x~l are not adjacent (that is, a, = jc => a,+1 ^ x~l and Oi = x_1 => ai+1 9^ x for all / s N*, x e l ) and (ii) ak = 1 implies a, = 1 for all / > A. In particular, the empty word 1 is reduced. Every nonempty reduced word is of the form (xix,,x2X2,. . ., x n X n , l , l , . . . ) , where n e N*, Xi eX and X, = ±1 (and by convention xJ denotes x for all x eX). Hereafter we shall denote this word by x^lx2^ ■ ■ - xnx\ This new notation is both more tractable' and more suggestive. Observe that the definition of equality of sequences shows that two reduced words *iXl ■ • • xmXm and >’iSl • • ■ yj71 (*;,>', e X; X„<5, = ± 1) are equal if and only if both are 1 or m = n and x, = >v, X, = Si for each / = 1 , 2 , . , n. Consequently the map from A" into the set F(X) of all reduced words onA" given by x I—> x1 = x is in jective. We shall identify X with its image and consider A" to be a subset of F(X). Next we define a binary operation on the set F = F(X) of all reduced words on X. The empty word 1 is to act as an identity element (wl = l»v = w for all w e F). In formally, we would like to have the product of nonempty reduced words to be given by juxtaposition, that is, (xiXl- • •xmXm)(>'ifil- • ■yn6n) = *iXl- • ^Xm>’ifil- • -ynSn.
9. FREE GROUPS, FREE PRODUCTS, GENERATORS AND RELATIONS 65
Unfortunately the word on the right side of the equation may not be reduced (for example, if xmXm = yrSl). Therefore, we define the product to be given by juxtaposi tion and (if necessary) cancellation of adjacent terms of the form xx-1 or x~lx; for example (jcVav 1-X31)(*3-1X21-*'41) ~ More precisely, if *iXl- • ■xmXm and are nonempty reduced words on X with m < n, let k be the largest integer (0 < k < m) such that x*™z] = y~~s£\ for j = 0,1,. . . , k — 1. Then define Xi Xl'
(aiX>-
••
XmXm)(>’16‘
• ■ xm-km~kyk+\ k+1- • -ynn if k < m; • ■ •>'„*") ymJ=m+x- ■ ■ynn if k = m
If m > n, the product is defined analogously. The definition insures that the product of reduced words is a reduced word.
Theorem 9.1. If X is a nonempty set and F = F(X) is the set of all reduced words on X, then F is a group under the binary operation defined above and F = (X).
The group F = F(X) is called the free group on the set X. (The terminology “free” is explained by Theorem 9.2 below.) SKETCH OF PROOF OF 9.1. Since 1 is an identity element a n d x i S l - - ■x„5n has inverse x~Sn • • • xx~Sl, we need only verify associativity. This may be done by in duction and a tedious examination of cases or by the following more elegant device. For each x bX and 8 = ±1 let |jc*| be the map F —> F given by 11—> xs and rj,...vs„K
I • • • *«*- if Xs + xf6'; j x2^-■ • ■ xnSn if xs = xt~Sl (= I if n = 1).
Since I*!!*-1! = I7- = every |.v6| is a permutation (bijection) of F (with in -i verse |x |) by (13) of Introduction, Section 3. Let A(JF) be the group of all permuta tions of F (see page 26) and F0 the subgroup generated by {\jc| | x e A"}. The map
Certain properties of free groups are easily derived. For instance if \X\ > 2, then the free group on X is nonabelian (x,y e X and x ^ y => x^y^xy is reduced => x~xy~xxy 9^ 1 => xy 5* yx). Similarly every element (except 1) in a free group has infinite order (Exercise 1). If X = {a\, then the free group on X is the infinite cyclic group {a) (Exercise 2). A decidedly nontrivial fact is that every subgroup of a free group is itself a free group on some set (see J. Rotman [19]).
Theorem 9.2. Let F be the free group on a set X and t : X —► F the inclusion map. If G is a group and f: X —*■ G a map of sets, then there exists a unique homomorphism of groups f : F —> G such that ft = f. In other words, F is a free object on the set X in the category of groups.
66
CHAPTER I GROUPS
REMARK. If F' is another free object on the set A' in the category of groups (with X : X —» F'), then Theorems 7.8 and 9.2 imply that there is an isomorphism
PROOF. Let X be a set of generators of G and let F be the free group on the set X. By Theorem 9.2 the inclusion map X —» G induces a homomorphism / : F —*■ G such that x\->xeG. Since G = (X), the proof of Theorem 9.2 shows that / is an epimorphism. ■ An immediate consequence of Corollary 9.3 and the First Isomorphism Theorem is that any group G is isomorphic to a quotient group F/N, where G = (X), F is the free group on X and N is the kernel of the epimorphism F —* G of Corollary 9.3. Therefore, in order to describe G up to isomorphism we need only specify X, F, and N. But F is determined up to isomorphism by X (Theorem 7.8) and N is determined by any subset that generates it as a subgroup of F. Now if w = xi®1 ■ ■ • xnbn s F is a generator of N, then under the epimorphism F—> G, w)-h- • • • xn6r‘ = esG. The equation jciSi- • -xnbn = e in G is called a relation on the generators Xi. Clearly a given group G may be completely described by specifying a set X of generators of G and a suitable set R of relations on these generators. This description is not unique since there are many possible choices of both X and R for a given group G (see Exercises 6 and 9). Conversely, suppose we are given a set X and a set Y of (reduced) words on the elements ofX. Question: does there exist a group G such that G is generated byXand all the relations w = e (w s Y) are valid (where w = jciSi • • -xjn now denotes a product in G)? We shall see that the answer is yes, providing one allows for the possibility that in the group G the elements ofX may not all be distinct. For instance, if a,b eX and alb x is a (reduced) word in Y, then any group containing a,b and satisfying c£b~Y = e must have a = b. Given a set of “generators” X and a set Y of (reduced) words on the elements of X. we construct such a group as follows. Let F be the free group on X and TV the normal subgroup of F generated by Y.3 Let G be the quotient group F/N and identify X with its image in F/N under the map X d F—> F/N\ as noted above, this may involve identifying some elements of X with one another. Then G is a group generated by X (subject to identifications) and by construction all the relations w = e (w s Y) are satisfied (w = jfi®1- • -A-„®n s Y =» a:!®1- • ■xnbn s N => xbxN- ■ ■xbnN = N\ that is, -Xi®1- • •xnbn = e in G = F/N). 3The normal subgroup generated by a set S cz F is the intersection of all normal subgroups of F that contain S\ see Exercise 5.2.
9. FREE GROUPS, FREE PRODUCTS, GENERATORS AND RELATIONS 67
Definition 9.4. Let X be a set and Y a set of (reduced) words on X. A group G is said to be the group defined by the generators x s X and relations w = e (w e Y) provided G = F/N, where F is the free group on X and N the normal subgroup of F generated by Y. One says that (X | Y) is a presentation of G.
The preceding discussion shows that the group defined by given generators and relations always exists. Furthermore it is the largest possible such group in the following sense.
Theorem 9.5. (Van Dyck) Let X be a set, Y a set of (reduced) words on X and G the group defined by the generators x s X and relations w = e ( w £ Y). If H is any group such that H = (X) and H satisfies all the relations w = e (w s Y), then there is an epimorphism G —> H.
REMARK. The elements of Y are being interpreted as words on X, products in G, and products in H as the context indicates. PROOF OF 9.5. If F is the free group on X then the inclusion map X —* H in duces an epimorphism
bm
EXAMPLE. The group defined by one generator b and the single relation = e (m e N*) is ZrrL (Exercise 9).
EXAMPLE. The free group F on a set X is the group defined by the generators x eX and no relations (recall that (0) = (e) by Definition 2.7). The terminology “free” arises from the fact that F is relation-free.
68
CHAPTER I GROUPS
We close this section with a brief discussion of coproducts (free products) in the category of groups. Most of the details are left to the reader since the process is quite similar to the construction of free groups. Given a family of groups {Gi \ i e /} we may assume (by relabeling if necessary) that the Gi are mutually disjoint sets. Let A- = U Gl and let {1} be a one-element set iel
disjoint from A". A word on A" is any sequence (ai,a2,. . .) such that a{ e X U {1} and for some n s N*, at = 1 for all i > n. A word (ai,a2,. . .) is reduced provided: (i) no ai s A is the identity element in its group G3; (ii) for all ij > 1, a, and ai+1 are not in the same group G,; (iii) ak = 1 implies a, = 1 for all i > k. In particular 1 = (1,1,...) is reduced. Every reduced word (?^1) may be written uniquely as a\a2- • an — (01,02, - - - , - • •)> where a, eX. Let IT Gi (or Gi * Gi * • • ■ * G« if / is finite) be the set of all reduced words on A. iel
JJ*G, forms a group, called the free product of the family {Gt | / e /}, under the iel
binary operation defined as follows. 1 is the identity element and the product of two reduced words (^ 1) essentially is to be given by juxtaposition. Since the juxtaposed product of two reduced words may not be reduced, one must make the necessary cancellations and contractions. For example, if ai,bi e G, for / = 1,2,3, then (a&asXa^bzbibs) = aic2bibs = (auc2,bub3,i ,...), where c-2 = a2b2z G2. Finally, for each k s / the map i* : G* —» ri*G, given by e (—> 1 and a |—> a — (a,l ,1,. . .) is a iel
monomorphism of groups. Consequently, we sometimes identify Gk with its iso morphic image in JI*G, (for example Exercise 15). isl
Theorem 9.6. Let {Gi | i e 1} be a family of groups and J^*Gi their free product. If iel
{^i : Gi —» H | i s 1} is a family of group homomorphisms, then there exists a unique homomorphism \p : XI *G; —> H such that i/'t; = \p\ for all i e I and this property deteriel
mines IT Gi uniquely up to isomorphism. In other words, IT Gi is a coproduct in the iel
iel
category of groups.
SKETCH OF PROOF. If aia2- ■ ■ aTI is a reduced word in XI*G» with ak s Gik, iel
define \p(ax ■ • • a,) to be \pil(ai)\pii(d2) ■ ■ -&„(°") a H. ■
EXERCISES 1. Every nonidentity element in a free group F has infinite order. 2. Show that the free group on the set {a) is an infinite cyclic group, and hence isomorphic to Z. 3. Let F be a free group and let N be the subgroup generated by the set {xn \ x s F, n a fixed integer}. Show that N <] F. 4. Let F be the free group on the set X, and let Y (Z X. If H is the smallest normal subgroup of F containing Y, then F/H is a free group.
9. FREE GROUPS, FREE PRODUCTS, GENERATORS AND RELATIONS 69
5. The group defined by generators a,b and relations a8 = b2a4 = ab~xab = e has order at most 16. 6.
The cyclic group of order 6 is the group defined by generators a,b and relations a2 = b3 = crxb~xab = e.
7. Show that the group defined by generators a,b and relations a2 = e, b3 — e is in finite and nonabelian. 8.
The group defined by generators a,b and relations an = e (3 < n e N*), b2 = e and abab = e is the dihedral group D„. [See Theorem 6.13.]
9. The group defined by the generator b and the relation bm = e (me. N*) is the cyclic group Zm. 10. The operation of free product is commutative and associative: for any groups A,B,C, A * B ^ B * A and A * (B * C) ^ (A * B) * C. 11. If N is the normal subgroup of A *B generated by A, then (A * B)/N = B. 12. If G and H each have more than one element, then G * H is an infinite group with center (e). 13. A free group is a free product of infinite cyclic groups. 14. If G is the group defined by generators a,b and relations a2 = e, b3 = e, then G = Zi *Z3. [See Exercise 12 and compare Exercise 6.] 15. If / : Gi —> Gi and g : M —> H2 are homomorphisms of groups, then there is a unique homomorphism h : Gi * Hi —> G2 * Hi such that h \ Gi = f and h | Hi = g.
THE STRUCTURE OF GROUPS
We continue our study of groups according to the plan outlined in the introduction of Chapter I. The chief emphasis will be on obtaining structure theorems of some depth for certain classes of abelian groups and for various classes of (possibly non abelian) groups that share some desirable properties with abelian groups. The chapter has three main divisions which are essentially independent of one another, except that results from one may be used as examples or motivation in the others. The interdependence of the sections is as follows. 1
3
4
I
\
2
5 6^\7_8
Most of Section 8 is independent of the rest of the chapter.
1. FREE ABELIAN GROUPS We shall investigate free objects in the category of abelian groups. As is the usual custom when dealing with abelian groups additive notation is used throughout this section. The following dictionary may be helpful. ab............................................................. a + b a^1........................................................... — a e............................................................... 0 an ............................................................ na ab~l.......................................................... a — b HK........................................................... H+K aH............................................................ a + H 70
1. FREE ABELIAN GROUPS
71
GXH.........................................................G@H H\J K....................................................... H + K iel
.................................................. g<
iel
weak direct product.................................direct sum For any group G in additive notation, (m + n)a = ma + na(ae G; m,n s Z). If the group is abelian, then m(a + b) = ma + mb. If A- is a nonempty subset of G, then by Theorem 1.2.8 the subgroup (X) generated by A- in additive notation consists of all linear combinations niXi + «2 x2 H--------1- nkxk (nt s Z, Xi e A'). In particular, the cyclic group (x) is {nx \ n e Z}. A basis of an abelian group F is a subset X of F such that (i) F — (X); and (ii) for distinct xi,x2,. . . , xk e X and tti s Z, niXi + n2x2 H— • + nkxk = 0 => /i, = 0 for
every i.
The reader should not be misled by the tempting analogy with bases of vector spaces (Exercise 2).
Theorem 1.1. The following conditions on an abelian group F are equivalent. (i) (ii) (iii) (iv)
F has a nonempty basis. F is the (internal) direct sum of a family of infinite cyclic subgroups. F is (isomorphic to) a direct sum of copies of the additive group Z of integers. There exists a nonempty set X and a function i : X —> F with the following property: given an abelian group G and function f : X —> G, there exists a unique homo morphism of groups f: F —► G such that fi = f. In other words, F is a free object in the category of abelian groups.
An abelian group F that satisfies the conditions of Theorem 1.1 is called a free abelian group (on the set A-). By definition the trivial group 0 is the free abelian group on the null set 0. SKETCH OF PROOF OF 1.1. (i) => (ii) If A is a basis of F, then for each x s X, nx = 0 if and only if n = 0. Hence each subgroup (x) (x s A) is infinite cyclic (and normal since F is abelian). Since F = (A), we also have F = (|J (x)). If for xtX
some z s A, (z) ft (U (x)) ^ 0, then for some nonzero neZ,nz = «i*i H— • + nkxk xeX x^z
with z,xi , . . . , xk distinct elements of X, which contradicts the fact that A- is a basis. Therefore (z) fl (|J (x)) = 0 and hence F = ^ (x) by Definition 1.8.8. xeX
xtX
(ii) => (iii) Theorems 1.3.2, 1.8.6, and 1.8.10. (iii) => (i) Suppose F ^ ^ Z and the copies of Z are indexed by a set A. For each x e X, let 6X be the element {«, } of ^ Z, where w, = 0 for i ^ *, and ux = 1. Verify that [dI \ x eX] is a basis of ^ Z and use the isomorphism F = ^ Z to obtain a basis of F. (i) => (iv) Let A- be a basis of F and i : A' —> F the inclusion map. Suppose we are given a map f:X-+G. If u s F, then u = n^ H----------------- b nkXk (m s Z; Xi eX) since A k
generates F. If u = miXi + - —f mkxk, (mk e Z), then («< — w,)jf, = 0, whence
CHAPTER II THE STRUCTURE OF GROUPS
72
m = tm for every / since A is a basis. Consequently the map / : F —> G, given by m=/
= n, f(xi) + ■■■ + «*; f{xk), is a well-defined function such that
Jl = /. Since G is abelian / is easily seen to be a homomorphism. Since A' generates F, any homomorphism F —■* G is completely detemined by its action on A. Thus if g : F —■+ G is a homomorphism such that gi = f then for any x eX g(x) = g(i(x)) = f(x) = f(x), whence g = j and / is unique. Therefore, by Definition 1.7.7 F is a free object on the set X in the category of abelian groups. (iv) => (iii). Given i: X —* F, construct the direct sum ^ Z with the copies of Z indexed by X. Let Y = {6X | a- eA} be a basis of£ Z as in the proof of (iii) => (i). The proof of (iii) => (i) => (iv) shows that 23 Z is a free object on the set Y. Since we clearly have \X\ = \Y\, F = ^ Z by Theorem 1.7.8. ■
Given any set A, the proof of Theorem 1.1 indicates how to construct a free abelian group F with basis X. Simply let F be the direct sum^Z, with the copies of Z indexed by A. As in the proof of (iii) => (i), {6X \ x s A| is a basis of F = ^Z, and F is free on the set \6X \ x zX\. Since the map i : A —» F given by a (-> 6X is injective it follows easily that Fis free on A in the sense of condition (iv) of Theorem 1.1. In this situation we shall identify A with its image under i so that A C Fand the cyclic sub group (6X) = {n6x | n e Z) = ZBX is written (x) = Zv. In this notation F = ^ (6X) is xs.X
written F = ^ Zx, and a typical element of F has the form mx\ -j— • + nkxk xeX
(nt s Z, Xi e A). In particular, X = t(A) is a basis of F.
Theorem 1.2. Any two bases of a free abelian group F have the same cardinality.
The cardinal number of any basis ant of F\ \X\ is called the rank of F.
A
of the free abelian group F is thus an invari
SKETCH OF PROOF OF 1.2. First suppose F has a basis A of finite cardinal ity n so that F = Z ©• • - © Z (« summands). For any subgroup G of F verify that 2G = \2u | u e G J is a subgroup of G. Verify that the restriction of the isomorphism F = Z ©- ■ •© Z to 2F is an isomorphism 2F ^ 2Z 0 - • •© 2Z, whence F/2F = Z/2Z ©- ■ ■ © Z/2Z = Z2 ©• • -©Z2 (n summands) by Corollary 1.8.11. Therefore |F/2F[ = 2". If Y is another basis of F a n d r any integer such that |F| > r. then a similar argument shows that |F/2F| > 2r, whence 2r < 2n and r < n. It follows that |KJ = m < n and |F/2F| — 2m. Therefore 2m = 2n and |A| = n = m = \Y\. If one basis of F is infinite, then all bases are infinite by the previous paragraph. Consequently, in order to complete the proof it suffices to show that |A| = |F|, if A is any infinite basis of F. Clearly |A| < |F|. Let 5 = U A", where A" = X X • • • X A rceN*
(n factors). For each s = (xu .. . , xn) e S let Gs be the subgroup (xu . . ., xn). Then Gs = Zyi ©• • •© Zyt where y\, ... ,yt (t < n) are the distinct elements of { j f i , . . .,xn\. Therefore, |GS| = \7J\ = |Z| =Ko by Introduction, Theorem 8.12. Since F = U Gs, we have |F| = |(J Gs[ < by Introduction, Exercise 8.12. seiS fieS But by Introduction, Theorems 8.11 and 8.12, |£| = |AT|, whence |f| < |a| ^ 0 = |A|. Therefore |F| = \X\ by the Schroeder-Bernstein Theorem. ■
1. FREE ABELIAN GROUPS
73
Proposition 1.3. Let Fi be the free abelian group on the set Xi and F2 the free abelian group on the set X2. Then Fi ~ F2 if and only if Fi and F2 have the same rank (that is,
IX,| = |X2|). REMARK. Proposition 1.3 is also true for arbitrary nonabelian free groups (as in Section 1.9); see Exercise 12. SKETCH OF PROOF OF 1.3. If a :Fi~ F2, then <*(*,) is a basis of F2, whence J A", j = 1ck(A^)| = \X2\ by Theorem 1.2. The converse is Theorem 1.7.8. ■
Theorem 1.4. Every abelian group G is the homomorphic image of a free abelian group of rank |X|, where X is a set of generators of G.
PROOF. Let F be the free abelian group on the set X. Then F = ^ Zx and rank F = \X\. By Theorem 1.1 the inclusion map X—> G induces a homomorphism / : F —> G such that \x (—> x e G, whence X C Im f Since X generates G we must have Im f = G. m
We now prove a theorem that will be extremely useful in analyzing the structure of finitely generated abelian groups (Section 2). We shall need
Lemma 1.5. //{x1?. . . , xn} is a basis of a free abelian group F and a s Z, then for all i 9^ j { x , , . . . , Xj_i,Xj + ax;,Xj+i,. . ., xnJ is also a basis of F.
PROOF. Since x, = —axi -f- (xj + axi), it follows that F = (*,,. . ., jc,_,,x3 + axi,xj+u . . . , xn). If /c,x, H------------- f- kjixj + ax^ H----------f- knxn = 0 (ki s Z), then kiXi — ■ + (h + kja)xi + • —|- kjXj H------------ f- knxn = 0 , which implies that kt = 0 for all t. m
Theorem 1.6. // F is a free abelian group of finite rank n andG is a nonzero subgroup of F, then there exists a basis { x , , . . . , xn} of F, an integer r (1 < r < n) and positive integers d,, . . . , dr such that d, | d2 | • • • | dr and G is free abelian with basis
{d,x,, . . ., drxr}. REMARKS. Every subgroup of a free abelian group of (possibly infinite) rank a is free of rank at most a; see Theorem IV.6.1. The notation “d\ | d21 . . . | d,” means “di divides d2, d2 divides d3, etc.” PROOF OF 1.6. If n = 1, then F = (x,) ^ Z a n d G = (dlX,) ~ZUI, e N*) by Theorems 1.3.5, 1.3.1, and 1.3.2. Proceeding inductively, assume the theorem is true for all free abelian groups of rank less than n. Let 5 be the set of all those integers s such that there exists a basis {>,, . . . ,yn\ of F and an element in G of the form ■s^i + k2y2 H------ (- knyn (ki e Z). Note that in this case , yn] is also a basis of F, whence k2eS; similarly kj eS for j = 3 , 4 , S i n c e G ^ 0, we have S 0. Hence 5 contains a least positive integer d\ and for some basis { j > i , . . . , yn]
CHAPTER II THE STRUCTURE OF GROUPS
74
of F there exists v s G such that v = dxyi -f- k2y2 + ■ ■ • + k„yn. By the division algorithm for each /' = 2 , /c, = diql -f r, with 0 < r,- < d\, whence v = di(yi + q2y2 4----------h qny„) + r2y2 H--------h r„yn. Let x: = yi -f- q^y-i -\--------h qny*; then by Lemma 1.5 W = {xx,y2, is a basis of F. Since v s G, r, < di and W in any order is a basis of F, the minimality of d\ in S implies that 0 = r2 = r3 = - - ■ = r„ so that diXi = v s G. Let H — (y2,y3, . . ., yn). Then H is a free abelian group of rank n—l such that F = (jci) © H. Furthermore we claim that G = (v) ©(G Pi H) = (diXi) ©(G (1 H). Since f xuy2 , . . . , ! is a basis of F, (v) fl (G D H) = 0. If u = txXi + t2y2 + • ■ - + tnyn e G (ti s Z), then by the division algorithm tY — dxqi + n with 0 < n < d\. Thus G contains u — qxv = r^x\ + t2y2 + ■ • ■ + tnyn. The minimality of d\ in 5 im plies that = 0, whence t2y2 + ■ ■ ■ + tnyn s G D H and u = qxv + (t2y2 4-------------1 - tnyn). Hence G = (v) + (G D FT), which proves our assertion (Definition 1.8.8). Either G fl H = 0, in which case G = (diXj) and the theorem is true or G Cl H 9^ 0. Then by the inductive assumption there is a basis j x2,x2,... ,xn} of H and positive integers r,d2,d3,... ,dr such that d21 d31 - ■ ■ | dr and G D H is free abelian with basis {d2x2, . . . , drxr). Since F = {xx) © Hand G = (d\X\) © (G fl H), it follows easily that {xi,x2,. . . , xn} is a basis of F and {dxxu . . . , drxr} is a basis of G. To complete the inductive step of the proof we need only show that di | d2. By the division algorithm d2 = qdi + r0 with 0 < r0 < dx. Since \x2,xi + qx2,x3, . . . , xn\ is a basis of F by Lemma 1.5 and r0x2 + d\(xx + qx2) = diXi + d2x-2 s G, the mini mality of di in 5 implies that r0 = 0 , whence di \ d2. ■
Corollary 1.7. 7/G is a finitely generated abelian group generated by n elements, then every subgroup H of G may be generated by m elements with m < n. The corollary is false if the word abelian is omitted (Exercise 8 ). PROOF OF 1.7. By Theorem 1.4 there is a free abelian group F of rank n and an epimorphism tv : F —> G. ir~l(H) is a subgroup of F, and therefore, free of rank m < « by Theorem 1.6 . The image under 7 r of any basis of tt~1(H) is a set of at most m elements that generates 7t(tt~\H)) = H. ■
EXERCISES 1. (a) If G is an abelian group and m s Z, then mG = {mu | k e G ) is a sub group of G. (b) 1 f G ~ ^ Gi, then mG = mGi and G/mG ~ ^ Gl/mG{. iel
iel
iel
2. A subset A'of an abelian group F is said to be linearly independent if ntx{ + - - - + nkxh — 0 always implies nt = 0 for all /' (where s Z and x\,... ,xk are distinct elements of X). (a) X is linearly independent if and only if every nonzero element of the sub group (X) may be written uniquely in the form n\Xi [ • • • - ( - nkxk («, e Z,«,• ^ 0, x\,. .. ,xk distinct elements of A"). (b) IfF is free abelian of finite rank n, it is not true that every linearly independent subset of n elements is a basis [Hint: consider F = Z].
1. FREE ABELIAN GROUPS
75
(c) If F is free abelian, it is not true that every linearly independent subset of F may be extended to a basis of F. (d) If F is free abelian, it is not true that every generating set of F contains a basis of F. However, if F is also finitely generated by n elements, F has rank m < n. 3. Let A' = | ai | i e /) be a set. Then the free abelian group on A- is (isomorphic to) the group defined by the generators X and the relations (in multiplicative no tation) {aiajai~1af'1 = e\ ij el}. 4. A free abelian group is a free group (Section 1.9) if and only if it is cyclic. 5. The direct sum of a family of free abelian groups is a free abelian group. (A direct product of free abelian groups need not be free abelian; see L. Fuchs [13, p. 168].) 6.
If F = £ Zjc is a free abelian group, and G is the subgroup with basis
xzX X' = X — j a'o} for some x0eX, then F/G = Zx0. Generalize this result to ar bitrary subsets X' of X.
7. A nonzero free abelian group has a subgroup of index n for every positive integer n. 8.
Let G be the multiplicative group generated by the real matrices a = d ‘ )
and b = G ! ) • , f H is the set of all matrices in G whose (main) diagonal entries are 1, then H is a subgroup that is not finitely generated. 9. Let G be a finitely generated abelian group in which no element (except 0) has finite order. Then G is a free abelian group. [Hint: Theorem 1.6.] 10. (a) Show that the additive group of rationals Q is not finitely generated. (b) Show that Q is not free. (c) Conclude that Exercise 9 is false if the hypothesis “finitely generated” is omitted. 11. (a) Let G be the additive group of all polynomials in x with integer coefficients. Show that G is isomorphic to the group Q* of all positive rationals (under multiplication). [Hint: Use the Fundamental Theorem of Arithmetic to con struct an isomorphism.] (b) The group Q* is free abelian with basis {p\p is prime in Z}. 12. Let F be the free (not necessarily abelian) group on a set A" (as in Section 1.9) and G the free group on a set Y. Let F' be the subgroup of F generated by \aba~lb~l\a,beF} and similarly for G'. (a) F' <1 F, G' O G and F/F', G/G' are abelian [see Theorem 7.8 below], (b) F/F' [resp. G/G'\ is a free abelian group of rank \X\ [resp. |F|], [Hint: {xF' | x eX\ is a basis of F/F'.] (c) F ^ G if and only if \X\ = |F|. [Hint: if
76
CHAPTER II THE STRUCTURE OF GROUPS
2. FINITELY GENERATED ABELIAN GROUPS We begin by proving two different structure theorems for finitely generated abelian groups. A uniqueness theorem (2.6) then shows that each structure theorem provides a set of numerical invariants for a given group (that is, two groups have the same invariants if and only if they are isomorphic). Thus each structure theorem leads to a complete classification (up to isomorphism) of all finitely generated abelian groups. As in Section 1, all groups are written additively. Many of the results (though not the proofs) in this section may be extended to certain abelian groups that are not finitely generated; see L. Fuchs [13] or I. Kaplansky [17]. All of the structure theorems to be proved here are special cases of corresponding theorems for finitely generated modules over a principal ideal domain (Section IV.6 ). Some readers may prefer the method of proof used in Section IV. 6 to the one used here, which depends heavily on Theorem 1.6.
Theorem 2.1. Every finitely generated abelian group G is (isomorphic to) a finite direct sum of cyclic groups in which the finite cyclic summands (if any) are of orders m i , . . . , mt, where mx > 1 and mi | m2 1 • - • |mt.
PROOF. If G 9^ 0 and G is generated by n elements, then there is a free abelian group F of rank n and an epimorphism 7 r : F —> G by Theorem 1.4. If tt is an iso morphism, then G = / r = Z © - - - © Z ( « summands). If not, then by Theorem 1.6 there is a basis {xi,.. ., xn i of F and positive integers d\,... ,dr such that 1 < r < n, n
d\ | d«! • • • | dr and \d\Xi, . . . , drx,\ is a basis of K = Ker it. Now F = ^ (xt) and r
i = l
K = ^ (diXi), where (xt) = Z and under the same isomorphism (dtXi) = diZ 1 = 1 = {diU | u e Zj. For i = r + 1, r + 2 , . . . , n let di = 0 so that K = ^ (,c/,a\).
n t=i
Then by Corollaries 1.5.7,1.5.8, and 1.8.11 n
In
<*.■> i =1 I i —1
G9=F/K = X
n
1=1
n
= Z ZMZ. i=l
If di = 1, then Z/t/,Z = Z/Z = 0; if di> 1, then Z/^Z = Zdl; if di = 0, then Z/JtZ = Z/0 = Z. Let m i , . . . , m, be those di (in order) such that di ^ 0, 1 ancHet s be the number of d% such that di ~ 0. Then G^Zmi©---@Zmt©( Z©---©Z), b
where mi > 1, m\ \ m21 • • • [ mt and (Z © • • • © Z) has rank s. ■
Theorem 2.2. Every finitely generated abelian group G is (isomorphic to) a finite direct sum of cyclic groups, each of which is either infinite or of order a power of a prime.
2.1
SKETCH OF PROOF. The theorem is an immediate consequence of Theorem and the following lemma. Another proof is sketched in Exercise 4. ■
2. FINITELY GENERATED ABELIAN GROUPS
77
Lemma 2.3. If m is a positive integer and m = p1 n'p2n!- • -ptnt (pi, . . . , pt distinct primes and each n; > 0), then Zm = Zpi°i © ZP2ej ©• • • © ZPtnt.
SKETCH OF PROOF. Use induction on the number t of primes in the prime decomposition of m and the fact that ZTn ^ Zr © ZK whenever (r,n) = 1,
which we now prove. The element n — n\ eZTn has order r (Theorem 1.3.4 (vii», whence Zr ^ («1) < Zrn and the map Zr —>Zrn given by k H nk is a monomor phism. Similarly the map ^2 :Zn —>Zr„ given by k |—> rk is a monomorphism. By the proof of Theorem 1.8.5 the map \p : Zr © Z„ —»Zrn given by (*,>') H 'PvU) + 'Pz(y) = nx + ry is a well-defined homomorphism. Since (r,n) = 1 , ra + nb = 1 for some a,b 6 Z (Introduction, Theorem 6.5). Hence k = rak + nbk = \p(bk,ak) for all k £Zrn and \p is an epimorphism. Since |Zr ©Z„| = rn = \ZTri\, \p must also be a monomoiphism. ■
Corollary 2.4. IfG is a finite abelian group oforder n, then G has a subgroup of order m for every positive integer m that divides n. k
SKETCH OF PROOF. Use Theorem 2.2 and observe that G = XI implies i=i that | G\ = | Gi11 G2\ ■ ■ ■ | <7*| and for / < r,pr~lZpT = Zpi by Lemma 2.5 (v) below. ■ REMARK. Corollary 2.4 may be false if G is not abelian (Exercise 1.6.8). In Theorem 2.6 below we shall show that the orders of the cyclic summands in the decompositions of Theorems 2.1 and 2.2 are in fact uniquely determined by the group G. First we collect a number of miscellaneous facts about abelian groups that will be used in the proof.
Lemma 2.5. Let G be an abelian group, m an integer and p a prime integer. Then each of the following is a subgroup of G: (i) inG = {mu | u e G}; (ii) G[m] = { u e G | mu = 0}; (iii) G(p) = {u e G | |u| = pn for some n > 0}; (iv) Gt = {u e G | |u| is finite}. In particular there are isomorphisms (v) ZpE[p] ^ Zp (n > 1) and pmZpD ^ ZpD-m (m < n). Let H and Gj (i e I) be abelian groups. (vi) If % : G —* Gi is an isomorphism, then the restrictions of g to mG and G[m] iel
respectively are isomorphisms mG = mGi and G[m] ^ ^ Gi[m]. i’l iti
(vii) Iff : G —* H is an isomorphism, then the restrictions off to Gt and <J(p) re spectively are isomorphisms Gt = Ht and G(p) == H(p).
78
CHAPTER II THE STRUCTURE OF GROUPS
SKETCH OF PROOF, (i)-(iv) are exercises; the hypothesis that G is abelian is essential (S3 provides counterexamples for (i)—(iii) and Exercise 1.3.5 for (iv)). (v) pn~lzZvn has order p by Theorem 1.3.4 (vii), whence (/?"_1 ) = Z P and (pn~x) < ZpTt[p]. If u e Zp„[p], then pu = 0 in Zp„ so that pu = 0 (mod pn) in Z. But pn j pu impliespn~x \ u. Therefore, inZp„, u e (pn~l) andZpn[p\ < (pn~l). For the second state ment note that pTneZpn has order pn~m by Theorem 1.3.4 (vii). Therefore pmZpn = (pm) = Zp1l-m. (vi) is an exercise, (vii) If f:G —* His a homomorphism and x e G(p) has order pn, then pnf(x) = f(pnx) = /(0) = 0. Therefore f(x) e H(p). Hence / : G(p) —» H(p). If / is an isomorphism then the same argument shows that f~l : H(p) —> G(p). Since ff~x = l/,(p) and/-1/ = \GM, G(p) ^ H(p). The other con clusion of (vii) is proved similarly. ■ If G is an abelian group, then the subgroup Gt defined in Lemma 2.5 is called the torsion subgroup of G. If G = Gt, then G is said to be a torsion group. If Gt = 0, then G is said to be torsion-free. For a complete classification of all denumerable torsion groups, see I. Kaplansky [17].
Theorem 2.6. Let G be a finitely generated abelian group.
(i) There is a unique nonnegative integer s such that the number of infinite cyclic summands in any decomposition of G as a direct sum of cyclic groups is precisely s; (ii) either G is free abelian or there is a unique list of (not necessarily distinct) positive integers m i , . . . , i w t such that m i > 1 , m i | m 2 1 • • • [ m t and G == Zmi ©• • - ©Zmt©F with F free abelian; (iii) either G is free abelian or there is a list of positive integers pi81,. . . , Pk8*, which is unique except for the order of its members, such that pi,. . ., pk are (not necessarily distinct) primes, Si,. . . , Sk are (not necessarily distinct) positive integers and
G = Zpls. 0 . . . © Z p k * © F with F free abelian.
PROOF, (i) Any decomposition of G as a direct sum of cyclic groups (and there is at least one by Theorem 2.1) yields an isomorphism G = H © F, where His a direct sum of finite cyclic groups (possibly 0) and F is a free abelian group whose rank is precisely the number s of infinite cyclic summands in the decomposition. If i : H —> H © F is the canonical injection (h (h,0)), then clearly l(H) is the torsion subgroup of H © F. By Lemma 2.5, Gt = i(H) under the isomorphism G = H © F. Consequently by Corollary 1.5.8, G/Gt = (F © H)/l(H) = F. Therefore, any decomposition of G leads to the conclusion that G/Gt is a free abelian group whose rank is the number s of infinite cyclic summands in the decomposition. Since G/Gt does not depend on the particular decomposition and the rank of G/Gt is an invariant by Theorem 1.2, s is uniquely determined. (iii) Suppose G has two decompositions, say r
d
G^Y,Zni@F and G = XX © F', i=i j—i
2. FINITELY GENERATED ABELIAN GROUPS
79
with each m,kj a power of a prime (different primes may occur) and F,F' free abelian; (there is at least one such decomposition by Theorem 2.2). We must show that r = d and (after reordering) n, = A:, for every /. It is easy to see that the torsion subgroup of Zni F is (isomorphic to) and similarly for the other decomposition.
©
r
Hence
X
d
X
X
(X
= Zkj by Lemma 2.5. For each prime p, Z^X/?) is obvii=i i=i ously (isomorphic to) the direct sum of those Zfli such t h a t i s a power of p and sim ilarly for the other decomposition. Since Zni)(p) = Zki)(j)) for each prime p by Lemma 2.5, it suffices to assume that G = Gt and each m,kj is a power of a fixed prime p (so that G = G(j?)). Hence we have
(X!
(2
d
r
X! Z °i
1 = 1 j= 1
p
=
G^
X
< 0 2 < • • • < f l r i 1 < C l < C 2 < • • < Cd).
We first show that in any two such decompositions of a group we must have r = d. Lemma 2.5 and the first decomposition of G show that r
G[p\ =
X
t=i
z
P°i[p]
= @ - • @ZP (r summands),
whence \G[p]\ = A similar argument with the second decomposition shows that \G\p}\ = pd. Therefore, pr = pd and r = d. Let v (1 < v < r) be the first integer such that = a for all / < v and av ^ cv. We may assume that av < cv. Since pavZpai = 0 for < av, the first decomposi tion and Lemma 2.5 imply that r pOvG
^
XP
r
avZva<
—
i=l i = i>+l
X
Zp°i~av,
with av + 1 — av < ar+i — av <■■■< ar — av. Clearly, there are at most r — (v + 1) + 1 = r — v nonzero summands. Similarly since ai = c, for i < v and av < cv the second decomposition implies that r PUvG
^ XI i—v
with 1 < cv — av < rr+i — av <■■■ < cr — a, - Obviously there are at least r — v + 1 nonzero summands. Therefore, we have two decompositions of the group pavG as a direct sum of cyclic groups of prime power order and the number of summands in the first decomposition is less than the number of summands in the second. This contradicts the part of the Theorem proved in the previous paragraph (and applied here to pavG). Hence we must have a, = c, for all i. (ii) Suppose G has two decompositions, say G =:Zmi ©• - -® Zmt © F and G = Zkl ©• • - ®Zkd ®) F'
with mi > 1, mi | W2 j • • • | mi, ki > 1, kx | k21 • • -1 ka and F, F' free abelian; (one such decomposition exists by Theorem 2.1). Each m^kj has a prime decomposition and by inserting factors of the form p° we may assume that the same (distinct) primes Pi, .. ., pr occur in all the factorizations, say
80
CHAPTER II THE STRUCTURE OF GROUPS m
__ nflunaii— .Pi PnairPr
rv,
___
kl
2
...
.
-.ail an rflir M2 ~ P\ P%
_____
-Mtl an natr
k2
**P
mt
=
r
PTpT' pTpT-
I, _____ — P\ Pi Pr Kd — Pi Pi Pr ■
•
-Pr
■ P?
-.Cdl-Cdi. . . ncdr
Since nti | /n21 • - -1 mt, we must have for each j, 0 < aij < o2, < • • • < a t j. Similarly 0 < cij < Cij < ■ ■ ■ < c d j for each j . By Lemmas 2.3 and 2.5 ( d
= 5Z
J2 Zv?'> i,j » = 1 1 = 1 i.j
z-i
zpfa,
where some summands may be zero. It follows that for each./ = 1,2,... ,r td
X! z Pi *i = G(pi) ^ X! z v, c '>- ' i= 1 i=l
t
Since m\ > 1, there is some pj such that 1 < a\j <• - • < a tJ -, whence ^ Z Vj a n has t d» = i
nonzero summands. By (iii) Z„/.y has exactly r nonzero summands, whence t=i t < d. Similarly > 1 implies that d < / and hence d = t. By (iii) we now must have fly = cy for all zj, which implies that mt = kt for / = 1 ,2 , . . . , /. g If G is a finitely generated abelian group, then the uniquely determined integers m i , . . . , mt as in Theorem 2.6 (ii) are called the invariant factors of G. The uniquely determined prime powers as in Theorem 2.6 (iii) are called the elementary divisors of G.
Corollary 2.7. Two finitely generated abelian groups G and H are isomorphic if and only if G/Gt and H/Ht have the same rank and G and H have the same invariant factors [resp. elementary divisors].
PROOF. Exercise. ■ EXAMPLE. All finite abelian groups of order 1500 may be determined up to isomorphism as follows. Since the product of the elementary divisors of a finite group C must be | G| and 1500 = 2 2 -3-53, the only possible families of elementary di visors are {2,2,3,53|, {2,2,3,5,52}, {2,2,3,5,5,5}, {22 ,3,53|, |22 ,3,5,52} and \2\3,5,5,5}. Each of these six families determines an abelian group of order 1500 (for example, {2,2,3,53} determinesZ2 ©Z2 ®Z3 ®Zi25). By Theorem 2.2 every abelian group of order 1500 is isomorphic to one of these six groups and no two of the six are iso morphic by Corollary 2.7. If the invariant factors m\,. . . , mt of a finitely generated abelian group G known, then the proof of Theorem 2.6 shows that the elementary divisors of G the prime powers pn (n > 0 ) which appear in the prime factorizations of mu . . . , Conversely if the elementary divisors of G are known, they may be arranged in following way (after the insertion of some terms of the form p° if necessary):
are are mt.
the
2. FINITELY GENERATED ABELIAN GROUPS
81
P?\P?\ • • - , P r U
P™\P™, • • • ,Prtr
P?‘\P?\ ■ ■ • , P r ‘ r -
where pi,... ,pr are distinct primes; for each j = 1 ,2 , . . . , r, 0 < ni, < ni} <■■■ < nti with some na ^ 0; and finally «i3 0 for some j. By the definition of elementary tr
divisors (Theorem 2.6 (iii)), G =
,nij- (J) F where F is free abelian (and some i=li=1
finite summands are 0, namely those with p*** = p? = 1 ) . For each i — 1,2 let Wi = p™lp™- ■ Pr" (that is, nii is the product of the /th row in the array above). Since some j* 4 0, /wi > 1 and by construction /m | m2 1 • • • | mt. By Lemma 2.3 G = X! I Y. Zpjnij ) ffi F = yi Zmi © F. Therefore, wi,..., m t are the invariant i=l \y=i /
i=l
factors of G by Theorem 2.6 (ii). EXAMPLE. If G is the group Z5 © Z1 5 ® Z25 © Z36 ® ZM, then by Lemma 2.3 G ^ Z5 © (Z5 © Z3) © Z25 © (Z9 ® Z4) © (Z27 © Z2). Hence the elementary divi sors of G are 2,22 ,3,32 ,33 ,5,5,52 which may be arranged as explained above: 2°, 3, 5 2, 32, 5 22, 33, 52. Consequently the invariant factors of G are 1 - 3 - 5 = 1 5 , 2 - 3 2 - 5 = 90, and 22 -33- 52 = 2700 so that G ^Z15 ©Z90 ©Z^ooA topic that would fit naturally into this section is the determination of the struc ture of a finitely generated abelian group which is described by generators and rela tions. However, since certain matrix techniques are probably the best way to handle this question, it will be treated in the Appendix to Section VII.2. The interested reader should have little or no difficulty in reading that material at the present time.
EXERCISES 1. Show that a finite abelian group that is not cyclic contains a subgroup which is isomorphic to Zv ©Zp for some prime p. 2. Let G be a finite abelian group and x an element of maximal order. Show that (x)
is a direct summand of G. Use this to obtain another proof of Theorem 2.1. 3. Suppose G is a finite abelian p-group (Exercise 7) and x e G has maximal order. If y e G/(x) has order p\ then there is a representative y e G of the coset p such that |.v| = pr. [Note that if |x| = p‘, then p‘G = 0.] 4. Use Exercises 3 and 7 to obtain a proof of Theorem 2.2 which is independent of Theorem 2.1. [Hint: If G is a p-group, let x t G have maximal order; G/(x) is a direct sum of cyclics by induction, G/(x) — (x,) ( + ) - - - © (x„), with | x, | = p"
82
CHAPTER II THE STRUCTURE OF GROUPS
and 1 < r\ < r2 < • • • < rn. Choose representatives Xi of Xi such that |xi| = |x,-|. Show that G = (xi) ©■ ■ • © (xn) © (x) is the desired decomposition.] 5. If G is a finitely generated abelian group such that G/Gt has rank n, and H is a subgroup of G such that H/Ht has rank m, then m < n and (G/H)/(G/H)t has rank n — m. 6.
Let k,m e N*. If (k,m) = l , t h e n kZm = Zm andZm[k] = 0. If k \ m, say m = kd, then kZm ~ Zd and Z„\k\ Zk.
7. A (sub)group in which every element has order a power of a fixed prime p is called a p-(sub)group (note: |0| = 1 = p°). Let G be an abelian torsion group. (a) G(p) is the unique maximum /?-subgroup of G (that is, every p-subgroup of G is contained in G(p)). (b) G = X G(p), where the sum is over all primes p such that G(p) 0. [Hint: If |w| = pinl ■ ■ Ptnt, let m, = \u\/pini. There exist c, e Z such that c\tn\ H— • + ctnh = 1, whence u = c\tn\U + ■ ■ • + Ctfntu; but dmxu e G(pi).] (c) If H is another abelian torsion group, then G = H if and only if G(p) = H(p) for all primes p. 8.
A finite abelian p-group (Exercise 7) is generated by its elements of maximal order.
9. How many subgroups of order p2 does the abelian group Zp3 ©Zp2 have? 10. (a) Let G be a finite abelian p-group (Exercise 7). Show that for each n > 0, pn+1G fl G[p] is a subgroup of pnG D G[p\. (b) Show that (pnG D G[p])/(pn+1G fl G[p\) is a direct sum of copies ofZp; let k be the number of copies. (c) Write G as a direct sum of cyclics; show that the number k of part (b) is the number of summands of order pn+1. 11. Let G, H, and K be finitely generated abelian groups. (a) If G © G ^ //© H, then G ^ H . (b) If G © H ^ G © K, then //= K . (c) If Gi is a free abelian group of rank N0, then G, © Z © Z ~ G, © Z, but Z © Z ^ Z. Note: there exists an infinitely generated denumerable torsion-free abelian group G such that G ^ G © G © G, but G ^ G © G, whence (a) fails to hold with H = G © G. See A.L.S. Corner [60]. Also see Exercises 3.11, 3.12, andIV.3.12. 12. (a) What are the elementary divisors of the group Z2 ©Z9 ©Zi5; what are its invariant factors? Do the same for Z,(i©z4, © z « © Z200 © Z1000. (b) Determine up to isomorphism all abelian groups of order 64; do the same for order 96. (c) Determine all abelian groups of order n for n < 20. 13. Show that the invariant factors of Zm©Zn are (m,ri) and [m,n] (the greatest common divisor and the least common multiple) if (m,n) > 1 and mn if (m,n) = 1. 14. If H is a subgroup of a finite abelian group G, then G has a subgroup that is isomorphic to G / H . 15. Every finite subgroup of Q/Z is cyclic [see Exercises 1.3.7 and 7].
3. THE KRULL-SCHMIDT THEOREM
83
3. THE KRULL-SCHMIDT THEOREM The groups Z and Zp„ ( p prime) are indecomposable, in the sense that neither is a direct sum of two of its proper subgroups (Exercise 1.8.1). Consequently, Theorems 2.2 and 2.6(iii) may be rephrased as: every finitely generated abelian group is the direct sum of a finite number of indecomposable groups and these indecomposable summands are uniquely determined up to isomorphism. We shall now extend this result to a large class of (not necessarily abelian) groups.1 For the remainder of this chapter we return to the use of multiplicative notation for an arbitrary group.
Definition 3.1. A group G /^indecomposable i/G^ (e) and G is not the (internal) direct product of two of its proper subgroups.
Thus G is indecomposable if and only if G ^ {e) and G = H H = (e) or K = {e) (Exercise 1).
X K
implies
EXAMPLES. Every simple group (for example, An, n ^ 4) is indecomposable. However indecomposable groups need not be simple: Z, Zpn (p prime) and Sn are in decomposable but not simple (Exercises 2 and 1.8.1).
Definition 3.2. A group G is said to satisfy the ascending chain condition (ACC) on [normal] subgroups if for every chain Gi < G2 < • • • of [normal] subgroups of G there is an integer n such that G; = Girfor all i > n. G is said to satisfy the descending chain condition (DCC) on [normal] subgroups if for every chain Gj > G2 > • • • of [normal] subgroups ofG there is an integer n such that G, = Gn for all i > n.
EXAMPLES. Every finite group satisfies both chain conditions. Z satisfies the ascending but not the descending chain condition (Exercise 5) and Z(pm) satisfies the descending but not the ascending chain condition (Exercise 13).
Theorem 3.3. Ifagroup G satisfies either the ascending or descending chain condition on normal subgroups, then G is the direct product of a finite number of indecomposable subgroups.
SKETCH OF PROOF. Suppose G is not a finite direct product of indecom posable subgroups. Let S be the set of all normal subgroups H of G such that H is a direct factor of G (that is, G = H X TH for some subgroup TH of G) and H is not a finite direct product of indecomposable subgroups. Clearly G eS. If He S, then His not indecomposable, whence there must exist proper subgroups KH and Jh of H such that H — Kh X Jh (= Jh X KH). Furthermore, one of these groups, say KH, must lie in S (in particular, KH is normal in G by Exercise 1.8.12). Let f : S — > S be the map 'The results of this section are not needed in the sequel.
84
CHAPTER II THE STRUCTURE OF GROUPS
defined by f ( H ) = KH. By the Recursion Theorem 6.2 of the Introduction (with fn = / for all n) there exists a function
If G satisfies the ascending chain condition on normal subgroups, this is a con tradiction. ■ In order to determine conditions under which the decomposition of Theorem 3.3 is unique, several definitions and lemmas are needed. An endomorphism /of a group G is called a normal endomorphism if o/(Z>)o_1 = /(aba-1) for all a,b e G.
Lemma 3.4. Let G be a group that satisfies the ascending [resp. descending] chain condition on normal subgroups and f a [normal] endomorphism ofG. Then f is an auto morphism if and only iff is an epimorphism [resp. monomorphism].
PROOF. Suppose G satisfies the ACC and /is an epimorphism. The ascending chain of normal subgroups (e) < Ker / < Ker f2 < ■ ■ ■ (where fk= jf - f) must become constant, say Ker fn = Ker fn+1. Since /is an epimorphism, so is /". If a e G and f(a) = e, then a = fn(b) for some be G and e = f(a) = fn+\b). Consequently b e Ker /" f 1 = Ker /", which implies that a = fn(b) = e. Therefore, /is a monomor phism and hence an automorphism. Suppose G satisfies the DCC and /is a monomorphism. For each k > 1, Im fk is normal in G since /is a normal endomorphism. Consequently, the descending chain G > I m / > Im/2 > ■ ■ • must become constant, say Im fn = Im fn+1. Thus for any a e G, fn(a) = fn+'(b) for some be G. Since /is a monomorphism, so is fn and hence fn(d) = fn+\b) = /'(f(b)) implies a = f(b). Therefore / is an epimorphism, and hence an automorphism. ■
Lemma 3.5. (Fitting) //G is a group that satisfies both the ascending and descending chain conditions on normal subgroups and f is a normal endomorphism of G, then for some n > 1, G = Ker fn X Im fn.
3. THE KRULL-SCHMIDT THEOREM
85
PROOF. Since /is a normal endomorphism each Im fk (k > 1) is normal in G. Hence we have two chains of normal subgroups: G > Im /> Imp > ■ • • and (e) < Ker /< Ker p < ■ ■ -.
By hypothesis there is an n such that Im fk = Im /” and Ker fk = Ker /” for all k > n. Suppose a s Ker /” fl Im fn. Then a = fn(b) for some be G and Pn(b) = fn( fn(b)) = fn(a) = e. Consequently, b e Ker/2” = Ker/” so that a = fn(b) = e. Therefore, Ker /” fl Im /" = (e). For any c e G, /”(c) s Im /" = Im /*”, whence /"(c) = Pn(d) for some de G. Thus fn(cfn(d~1)) = fn(c) Pn(,d~l) = /"(c) Pn(d)~l = /"(c)/"(c)"1 = e and hence cfn(d~l) e Ker /”. Since c = (cfn(d~1)) fn(d), we conclude that G = (Ker /”)(Im /"). Therefore G = Ker /” X Im /" by Definition 1. 8. 8.
■
An endomorphism /of a group G is said to be nilpotent if there exists a positive integer n such that /"(g) = e for all g e G.
Corollary 3.6. 7/G is an indecomposable group that satisfies both the ascending and descending chain conditions on normal subgroups and f is a normal endomorphism of G, then either f is nilpotent or f is an automorphism.
PROOF. For some n > 1, G = Ker fn X Im /” by Fitting’s Lemma. Since G is indecomposable either Ker /" = (e) or Im /" = (e). The latter implies that / is nilpotent. If Ker fn = (e), then Ker / = (e) and /is a monomorphism. Therefore,/is an automorphism by Lemma 3.4. ■ If G is a group and f, g are functions from G to G, then f + g denotes the function G —> G given by a H> f(a)g(a). Verify that the set of all functions from G to G is a group under + (with identity the map 0o:G —> G given by a H> e for all a e G). When f and g are endomorphisms of G, f + g need not be an endomorphism (Exercise 7). So the subset of endomorphisms is not in general a subgroup.
Corollary 3.7. Let G (9^ (e)) be an indecomposable group that satisfies both the as cending and descending chain conditions on normal subgroups. If f i , . . . , fn are normal nilpotent endomorphisms ofG such that every f;i + • • • + fir (1 < ii < i2 < • ■ • < ir < n) is an endomorphism, then fi + f2 H--------- f- fn is nilpotent. SKETCH OF PROOF. Since each /, H— • + fr is an endomorphism that is normal (Exercise 8(c)), the proof will follow by induction once the case n = 2 is established. I f / i + /2 is not nilpotent, it is an automorphism by Corollary 3.6. Verify that the inverse g of f + /2 is a normal automorphism. If #1 = fg and g2 = fig, then 1g = gi + g2 and for all x e G, x~l = (gY + ^X*'1) = gi(A-1)g2(x-1). Hence x = teiC^Mx-1)]-1 = g2(x)g-i(x) = (g 2 + £i)(x) and 1G = g2 + gi. Therefore, £1 + g2 = g2 + gi and gi(gi + g2) = gilG = lGgi = (g 1 + g2)gi, which implies that S1S2 = g2gi. A separate inductive argument now shows that for each m > 1, m
(gi +
gi)m
= X) Cig^g”-* (c, e Z), i=o
86
CHAPTER II THE STRUCTURE OF GROUPS
where the c, are the binomial coefficients (see Theorem III. 1.6) and Cih means h + h H------- 1- h fa summands). Since each fi is nilpotent, gi = fig has a nontrivial kernel, whence gi is nilpotent by Corollary 3.6. Therefore for large enough m and all m
ae G, (gi +
g2)m{a)
=X
cig?g
l 2~ (fl)
i-0 i = 0
= XT
m
e°i = e-
®ut this contradicts the facts
that gi + 82 = 1g and G j* (e). ■ The next theorem will make use of the following facts. If a group G is the internal direct product of its subgroups G i , . . . , G s then by the proof of Theorem 1.8.6 there is an isomorphism < p : G i X -X G , ^ G given by ( g i , .. ., g») H> gig2- ■ g*. Con sequently, every element of G may be written uniquely as a product gig2 ■ ■ ■ gs (gi e Gi). For each i the map xt : G —> Gi given by gig2- • -g,|—> is a well-defined epimor phism; (it is the composition of ip~l with the canonical projection Gi X • • • X G, —» Gi.) We shall refer to the maps x, as the canonical epimorphisms associated with the internal direct product G = Gi X ■ ■ ■ X G,.
Theorem 3.8. (Krull-Schmidt) Let G be a group that satisfies both the ascending and descending chain conditions on normal subgroups. If G = Gi X G2 X ■ • • X G„ and G = Hi X H2 X • • ■ X Ht with each Gi,Hj indecomposable, then s = t and after reindexing Gj = H; for every i and for each r < t. G = Gi X • • • X Gr X Hr+1 X - X Ht.
REMARKS. G has at least one such decomposition by Theorem 3.3. The unique ness statement here is stronger than simply saying that the indecomposable factors are determined up to isomorphism. SKETCH OF PROOF OF 3.8. Let P(0) be the statement G = Hi X-X Ht. For 1 < r < min (s,t) let P(r) be the statement: there is a reindexing of Hu . . . , / / < such that Gi — Hi for / = 1,2, ... ,r and G = Gi X - ■ X Gr X Hr+1 X ■ ■ • X Ht (or G = Gi X • • ■ X Gt if r = t). We shall show inductively that P(r) is true for all r such that 0 < r < min (s,t). P(0) is true by hypothesis, and so we assume that P(r — 1) is true: after some reindexing Gi= Hi for i = 1, . . . , r — 1 and G = Gi X - - X Gr-1 X Hr X ■ ■ ■ X Ht. Let x i , . . . , x, [resp. x / , . . . , x/] be the canonical epimorphisms associated with the internal direct product G = Gi X ■ • • X Gs [resp. G = Gi X ■ ■ ■ X Gr-1 X Hr X - X Ht\
as in the paragraph preceding the statement of the Theorem. Let A» [resp. A/] be the inclusion maps sending the /th factor into G. For each / let
=
*Pi‘j
ViVi = (i 7* j)2',
H-------h = 1g; Mr = Tpu Mi = 0 ^ i)\ Im
It follows that <prTpi = Of? for all i < r (since \pi(x) e Gi so that <pr\f/i(x) = = iPrMi(x) = e). 2See
the paragraph preceding Corollary 3.7.
3. THE KRULL-SCHMIDT THEOREM
87
The preceding identities show that <pr = ipAc =
EXERCISES 1. A group G is indecomposable if and only if G ^ (e) and G = H X K implies H = (e) or K = (e). 2. Sn is indecomposable for all n > 2. [Hint: If« > 5 Theorems 1.6.8 and 1.6.10 and
Exercise 1.8.7 may be helpful.] 3. The additive group Q is indecomposable. 4. A nontrivial homomorphic image of an indecomposable group need not be in decomposable. 5. (a) Z satisfies the ACC but not the DCC on subgroups. (b) Every finitely generated abelian group satisfies the ACC on subgroups. 6.
Let H,K be normal subgroups of a group G such that G = H X K. (a) If N is a normal subgroup of H, then N is normal in G (compare Exercise 1.5.10). (b) If G satisfies the ACC or DCC on normal subgroups, then so do H and K.
CHAPTER II THE STRUCTURE OF GROUPS
88
7. If /and g are endomorphisms of a group G, then /+ g need not be an endo morphism. [Hint: Let a = (123), b = (132) eS3 and define f ( x ) = axcr1, g(x) = bxb^1.] 8.
Let /and g be normal endomorphisms of a group G. (a) fg is a normal endomorphism. (b) H <3 G implies f ( H ) <3 G . (c) I f / + g is an endomorphism, then it is normal.
9. Let G = G.i X • • ■ X G n . For each / let X;: Gi —> G be the inclusion map and 7T,-: G —> Gi the canonical projection (see page 59). Let
4. THE ACTION OF A GROUP ON A SET The techniques developed in this section will be used in the following sections to develop structure theorems for (nonabelian finite) groups.
Definition 4.1. An action of a group G on a set S is a function G x S —> S (iusually denoted by (g,x) |—> gx) such that for all x e S and gi,g2 e G:
ex = x and (gig2)x = gi(g2x). When such an action is given, we say that G acts on the set S. Since there may be many different actions of a group G on a given set S, the nota tion gx is ambiguous. In context, however, this will not cause any difficulty. EXAMPLE. An action of the symmetric group Sn on the set /„ = {1,2_______ ,n) is given by (ct,x) —* ct(x). EXAMPLES. Let G be a group and H a subgroup. An action of the group H on the set G is given by (h,x) |—> hx, where hx is the product in G. The action oi he. H on G is called a (left) translation. If K is another subgroup of G and S is the set of all left cosets of K in G, then H acts on 5 by translation: (h , x K ) |—» hxK. EXAMPLES. Let H be a subgroup of a group G. An action of H on the set G is given by (h,x) |—> hxh-1; to avoid confusion with the product in G, this action of h e H is always denoted hxhrx and not hx. This action of h e H on G is called conjugation by
4. THE ACTION OF A GROUP ON A SET
89
h and the element hxhr1 is said to be a conjugate of x. If K is any subgroup of G and h e H, then h K f r 1 is a subgroup of G isomorphic to K (Exercise 1.5.6). Hence H acts on the set S' of all subgroups of G by conjugation: ( h , K ) |—» hKfr1. The group hKh"1 is said to be conjugate to K.
Theorem 4.2. Let G be a group that acts on a set S.
(i) The relation on S defined by x ~ x' <=> gx = x' for some g e G is an equivalence relation. (ii) For each x e S , Gx = { g e G | gx = x ) is a subgroup of G.
PROOF. Exercise. ■ The equivalence classes of the equivalence relation of Theorem 4.2(i) are called the orbits3 of G on 5; the orbit of x s S is denoted x. The subgroup Gx is called vari ously the subgroup fixing x, the isotropy group of x or the stabilizer of x. EXAMPLES. If a group G acts on itself by conjugation, then the orbit { g x g ~11 g e G} of x e G is called the conjugacy class of x . If a subgroup H acts on G by conjugation the isotropy group H x = { h e H \ hxhr1 — x} = {h e H \ hx = xh} is called the centralizer of x in H and is denoted Cn(x). If H = G, Cg(x) is simply called the centralizer of x. If H acts by conjugation on the set S’ of all subgroups of G, then the subgroup of H fixing K e 5, namely {h e H | hKhr1 = J^}, is called the normalizer of K in H and denoted Nu(K). The group Ng(K) is simply called the normalizer of K. Clearly every subgroup K is normal in Nc(K) ; K is normal in G if and only if Ng(K) = G.
Theorem 4.3. If a group G acts on a set S, then the cardinal number of the orbit of xeS is the index [G : Gx].
PROOF. Let g,h e G. Since gx = hx g~xhx = x <=> g~lh e G j « hGx = gGx,
it follows that the map given by gGx ( — » g x is a well-defined bijection of the set of co sets of G x in G onto the orbit x = [ g x \ g £ G ] . Hence [G : G x] = |x|. ■
Corollary 4.4. Let G be a finite group and K a subgroup of G.
(i) The number of elements in the conjugacy class o / x e G is [G : Cg(x)], which divides |G|; (ii) ifxi , . . . , xn (x; e G) are the distinct conjugacy classes of G, then 3This
agrees with our previous use of the term orbit in the proof of Theorem 1.6.3, where the special case of a cyclic subgroup (a-) of Sn acting on the set /„ was considered.
90
CHAPTER II THE STRUCTURE OF GROUPS
|G| = £[G:CG(Xi)]; i=i (iii) the number of subgroups of G conjugate to K is [G : Ng(K)], which divides |G|. PROOF, (i) and (iii) follow immediately from the preceding Theorem and Lagrange’s Theorem 1.4.6. Since conjugacy is an equivalence relation on G (Theorem 4.2), G is the disjoint union of the conjugacy classes xu . . . , xn, whence (ii) follows from (i). ■ n The equation |G[ = X ' Q?(*01 as in Corollary 4.4 (ii) is called the class i= 1 equation of the finite group G.
Theorem 4.5. If a group G acts on a set S, then this action induces a homomorphism G —» A(S), where A(S) is the group of all permutations of S.
PROOF. If g e G, define re : S —> S by x \—> gx. Since x — g(.g~xx) for all x s S, t is surjective. Similarly gx = gy ( x , y e S ) implies x = g~Kgx) — g K g y ) = y , whence re is injective and therefore a bijection (permutation of 5). Since t > = t t > : 5 —> 5 for all g , g ' s G , the map G —»^(5) given by g (—> t „ is a homomorphism. ■ b
0(/
v b
Corollary 4.6. (Cayley) If G is a group, then there is a monomorphism G —> A(G). Hence every group is isomorphic to a group of permutations. In particular every finite group is isomorphic to a subgroup of Sn with n = |G|.
PROOF. Let G act on itself by left translation and apply Theorem 4.5 to obtain a homomorphism r : G —> A ( G ) . If r(g) = t = 1<7, then gx = re(x) = x for all .v e G . In particular ge = e , whence g — e and r is a monomorphism. To prove the last statement note if | (?[ = n, then A(G) ^ Sn. ■ b
Recall that if G is a group, then the set Aut G of all automorphisms of G is a group with composition of functions as binary operation (Exercise 1.2.15).
Corollary 4.7. Let G be a group.
(i) For each g e G, conjugation by g induces an automorphism of G. (ii) There is a homomorphism G —* Aut G whose kernel is C(G) = {g e G | gx = xg for all x e G |. PROOF. (1) If G acts on itself by conjugation, then for each g s G, the map : G —> G given by t (x) = gxg~x is a bijection by the proof of Theorem 4.5. It is easy to see that re is also a homomorphism and hence an automorphism, (ii) Let G act on itself by conjugation. By (i) the image of the homomorphism r : G —» A(G) of Theorem 4.5 is contained in Aut G. Clearly tb
b
g e Ker r <=> rp = 1g <=> gxg~1 - r^ix) = x for all x s G.
But gxg~l = x if and only if gx = xg, whence Ker r = C(G). ■
4. THE ACTION OF A GROUP ON A SET
91
The automorphism re of Corollary 4.7(i) is called the inner automorphism in duced by g. The normal subgroup C(G) = Ker r is called the center of G. An element ge G is in C(G) if and only if the conjugacy class of g consists of g alone. Thus if G is finite and x e C(G), then [G : Cg(x)] = 1 (Corollary 4.4). Consequently, the class equation of G (Corollary 4.4(H)) may be written m
\G\ = |C(G)| + Z [G : Coixi)], 1
where Xi,. . . , Xm (x, [G : CG(*,)] > 1.
e
G — C(G)) are distinct conjugacy classes of G and each
Proposition 4.8. Let H be a subgroup of a group G and let G act on the set S of all left cosets of H in G by left translation. Then the kernel of the induced homomorphism G —> A(S) is contained in H.
PROOF. The induced homomorphism G —> /4(S) is given by g |—» re, where : 5 —»5 and t (xH) = g x H . If g is in the kernel, then t b = ls and g x H = x H for all x e G ; in particular for x = e , g e H = e H = H , which implies g e H . ■ v
Corollary 4.9. I f H is a subgroup of index n in a group G and no nontrivial normal subgroup of G is contained in H, then G is isomorphic to a subgroup of Sn.
PROOF. Apply Proposition 4.8 to H; the kernel of G —> /4(S) is a normal sub group of G contained in H and must therefore be (e) by hypothesis. Hence, G —> A(S) is a monomorphism. Therefore G is isomorphic to a subgroup of the group of all permutations of the n left cosets of H, and this latter group is clearly isomorphic to Sn. U Corollary 4.10. If H is a subgroup of a finite group G of index p, where p is the small est prime dividing the order of G, then H is normal in G.
PROOF. Let S be the set of all left cosets of H in G. Then /4(S) = Sp since [G : H] = p. If K is the kernel of the homomorphism G —> >1(5) of Proposition 4.8, then K is normal in G and contained in H. Furthermore G/K is isomorphic to a sub group of Sp. Hence \ G / K \ divides |5P| = p \ But every divisor of \ G / K \ = [ G : Jt] must divide |G| = |J£| [G : K], Since no number smaller than p (except 1) can divide |G|, we must have \ G / K \ = p or 1. However \ G / K \ = [ G : K \ = [G : H ] [ H : K ] = p [ H : J^J > p. Therefore \ G / K \ = p and [ H : K ] — 1, whence H = K . But K is normal in G. ■
EXERCISES 1. Let G be a group and A a normal abelian subgroup. Show that G / A operates on A by conjugation and obtain a homomorphism G / A —> Aut A . 2 . If H , K are subgroups of G such that H <] K , show that K <
Ng(H).
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CHAPTER II THE STRUCTURE OF GROUPS
3. If a group G contains an element a having exactly two conjugates, then G has a proper normal subgroup N ^ (
Let G be a group acting on a set S containing at least two elements. Assjume that G is transitive; that is, given any xj'e.S, there exists gzG such that gx =y. Prove (a) for x sS, the orbit x of x is S; (b) all the stabilizers Gz (for x eS) are conjugate; (c) if G has the property: [g e G | gx = x for all x eS) = (e) (which is the case if G < Sn for some n and 5 = { 1 , 2 , . . . , « } ) a n d i f N < ] G a n d N < Gx for some x e S, then N = (^); (d) for x eS, [S| = [G : G*]; hence |5J divides |G|.
7. Let G be a group and let In G be the set of all inner automorphisms of G. Show that In G is a normal subgroup of Aut G. 8.
Exhibit an automorphism of Zc that is not an inner automorphism.
9. If G/C(G) is cyclic, then G is abelian. 10. Show that the center of S4 is ( e ) ; conclude that S4 is isomorphic to the group of all inner automorphisms of 54. 11. Let G be a group containing an element a not of order 1 or 2. Show that G has a nonidentity automorphism. [Hint: Exercise 1.2.2 and Corollary 4.7.] 12. Any finite group is isomorphic to a subgroup of An for some n. 13. If a group G contains a subgroup (^ G) of finite index, it contains a normal sub group (^ G) of finite index. 14. If |G| = pn, with p > n, p prime, and H is a subgroup of order p, then H is normal in G. 15. If a normal subgroup N of order p (p prime) is contained in a group G of order pn, then N is in the center of G.
5. THE SYLOW THEOREMS Nonabelian finite groups are vastly more complicated than finite abelian groups, which were completely classified (up to isomorphism) in Section 2. The Sylow Theo rems are a basic first step in understanding the structure of an arbitrary finite group. Our motivation is the question: if a positive integer m divides the order of a group G, does G have a subgroup of order ml This is the converse of Lagrange’s Theorem 1.4.6. It is true for abelian groups (Corollary 2.4) but may be false for arbitrary groups (Exercise 1.6.8). We first consider the special case when m is prime (Theorem 5.2), and then proceed to the first Sylow Theorem which states that the answer to our question is affirmative whenever m is a power of a prime. This leads naturally to a
5. THE SYLOW THEOREMS
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discussion of subgroups of maximal prime power order (second and third Sylow Theorems).
Lemma 5.1. If a group H of order pn (p prime) acts on a finite set S and if S0 = {x e S | hx = x for all h e H), then [S| = |S0| (mod p).
REMARK. This lemma (and the notation 50) will be used frequently in the sequel.4 PROOF OF 5.1. An orbit x contains exactly one element if and only if x e 50. Hence 5 can be written as a disjoint union 5 = S0 U xx U x2 U • • • U xn, with \xi\ > 1 for all i. Hence |S| = |50| + |*i| + \x2\ H— ■ + 1**1. Now p | |x;| for each i since |Jc»| > 1 and |x,| = [H : HxJ divides \H\ = pn. Therefore |5| = |50| (mod p). ■
Theorem 5.2. (Cauchy) If G is a finite group whose order is divisible by a prime p, then G contains an element of order p.
PROOF. (J. H. McKay) Let 5 be the set of p-tuples of group elements {(oi,a2,... ,ap)\ ai£ G and oio2- - -civ = e\. Since av is uniquely determined as (fli02• • • op_i)-1, it follows that |£| = n^1, where |G| = n. Sincep \ n, |S| =0 (modp). Let the group Zp act on 5 by cyclic permutation; that is, for k eZp, k(ai,a2 ,..., ap) = (ak+i,ak+2,. .. , ap,au . . ., ak). Verify that (ak+uak+2,.. . ,ak)sS (use the fact that in a grouped = e implies ba = (ala)(ba) = ar\ab)a — e). Verify that for 0,k,k' eZp and x e S, Ox = x and (k + k')x = k(k'x) (additive notation for a group action on a set!). Therefore the action of Zv on S is well defined. Now (ai,. . . , op) e So if and only if ay = o2 = ■ • ■ = ap\ clearly (e,e,.. . , e) e S0 and hence |S0| ^ 0. By Lemma 5.1, 0 = |S| = |50| (mod p). Since |50| ^ 0 there must be at least p elements in 50; that is, there is a e such that (a,a,... ,a)eS0 and hence av = e. Since p is prime, \a\ = p. ■ A group in which every element has order a power (> 0) of some fixed prime/? is called a p-group. If H is a subgroup of a group G and H is a p-group, H is said to be a p-subgroup of G. In particular (e) is a /7-subgroup of G for every prime p since |(
Corollary 5.3. A finite group G is a p-group if and only //|G| is a power of p.
PROOF. If G is a /7-group and q a prime which divides |G|, then G contains an element of order q by Cauchy’s Theorem. Since every element of G has order a power °f A <7 P- Hence |G| is a power of p. The converse is an immediate consequence of Lagrange’s Theorem 1.4.6. ■ —
4I
am indebted to R. J. Nunke for suggesting this line of proof.
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CHAPTER II THE STRUCTURE OF GROUPS
Corollary 5.4. The center C(G) of a nontrivial finite p-group G contains more than one element. PROOF. Consider the class equation of G (see page 91):
|G| = |C(G)|+£[G:CgU)]Since each [G : CG(xi)] > 1 and divides |G| = pn (n > 1), p divides each [G : CgC*,-)] and |G| and therefore divides |C(G)|. Since |C(G)| > 1, C(G) has at least p ele ments. ■
Lemma 5.5. If H is a p-subgroup of a finite group G, then [Ng(H) : H] = [G : H] Cmod p). PROOF. Let 5 be the set of left cosets of H in G and let H act on 5 by (left) translation. Then |S| = [G : H], Also,
xH e 50 hxH = xH for all h e H x-'hxH = H for all hzH<=> x^hx e H for all h e H
<=> jt'Hx = H<^ xHx-1 = Htt x e NG(H). Therefore |50| is the number of cosets a//with a- e NG(H); that is, |50| = [NG(H): H]. By Lemma 5.1 [A^^O : H] = |50| s [5| = [G : H] (mod p). ■
Corollary 5.6. If H is p-subgroup of a finite group G such that p divides [G : H], then Ng(H) ^ H. PROOF. 0 = [G : H] = [NG(H) : H] (mod p). Since [NG(H) :H]>\ in any case, we must have {NG(H) : H] > 1. Therefore NG(H) j* H. ■
Theorem 5.7. (First Sylow Theorem) Let G be a group of order p'‘m, with n > 1, p prime, and (p,m) = 1. Then G contains a subgroup of order p1 for each 1 < i < n and every subgroup of G of order p‘ (i < n) is normal in some subgroup of order p1+1. PROOF. Sincep\ |G|, G contains an element a, and therefore, a subgroup (a) of order p by Cauchy’s Theorem. Proceeding by induction assume H is a subgroup of G of order pi (1 < i < n). Then p \ [G : H] and by Lemma 5.5 and Corollary 5.6 H is normal in NG(H), H ^ NG(H) and 1 < \Na(H)/H\ = [NG(H) : H\ = [G : H] = 0 (mod p). Hence p | \NG(H)/H\ and NG(H)/H contains a subgroup of order p as above. By Corollary 1.5.12 this group is of the form H\/H where Hx is a subgroup of NG(H) containing H. Since H is normal in NG(H), H is necessarily normal in Hi. Finally \H\ = \H\\Hi/H\ = p'p = pi+l. ■ A subgroup P of a group G is said to be a Sylow p-subgroup (p prime) if P is a maximal /p-subgroup of G (that is, P < H < G wilh H a p-group implies P = H). Sylow /^-subgroups always exist, though they may be trivial, and every /7-subgroup is contained in a Sylow p-subgroup (Zorn’s Lemma is needed to show this for infinite
5. THE SYLOW THEOREMS
95
groups). Theorem 5.7 shows that a finite group G has a nontrivial Sylow p-subgroup for every primep that divides |G|. Furthermore, we have
Corollary 5.8. Let G be a group oforder pnmwith p prime, n > 1 and (m,p) = 1 .Let H be a p-subgroup of G.
(i) H is a Sylow p-subgroup ofG if and only //|H( = pn. (ii) Every conjugate of a Sylow p-subgroup is a Sylow p-subgroup. (iii) If there is only one Sylow p-subgroup P, then P is normal in G. SKETCH OF PROOF, (i) Corollaries 1.4.6 and 5.3 and Theorem 5.7. (ii) Exer cise 1.5.6 and (i). (iii) follows from (ii). ■ As a converse to Corollary 5.8 (ii) we have
Theorem 5.9. (Second Sylow Theorem) If H is a p-subgroup of a finite group G, and P is any Sylow p-subgroup of G, then there exists x e G such that H < xPx-1. In par ticular, any two Sylow p-subgroups of G are conjugate.
PROOF. Let 5 be the set of left cosets of P in G and let H act on 5 by (left) trans lation. |So| = |£| = [<7 : jP] (mod p) by Lemma 5.1. But p)f[Gm.P]\ therefore |S0| 0 and there exists xP e S0. xP eSo<^> hxP = xP for all h z H
<=> x~'hxP = P for all h e H <=> x~lHx < P <=> H < xPx~l. If //is a Sylow p-subgroup \H\ = |P| = |xPx-1| and hence H = xPx~l. ■
Theorem 5.10. (Third Sylow Theorem) If G is a finite group and p a prime, then the number of Sylow p-subgroups of G divides |G| and is of the form kp -\ 1 for some k > 0.
PROOF. By the second Sylow Theorem the number of Sylow p-subgroups is the number of conjugates of any one of them, say P. But this number is [G : Na(P)], a divisor of | G|, by Corollary 4.4. Let 5 be the set of all Sylow /7-subgroups of G and let P act on S by conjugation. Then Q e SG if and only if xQx~} = Q for all x e P. The latter condition holds if and only if P < N g (Q). Both P and Q are Sylow /^-subgroups of G and hence of N g (Q) and are therefore conjugate in N g (Q). But since Q is normal in NdQ), this can only occur if Q = P. Therefore, 50 = |Pj and by Lemma 5.1, |£| = |S0| = 1 (mod p). Hence |S| = kp + 1. ■ Theorem 5.11. If P is a Sylow p-subgroup of a finite group G, then Ng(Ng(P)) = Ng(P).
PROOF. Every conjugate of P is a Sylow p-subgroup of G and of any subgroup of G that contains it. Since P is normal in N = Nq(P), P is the only Sylow p-subgroup of N by Theorem 5.9. Therefore,
CHAPTER II THE STRUCTURE OF GROUPS
96
x e NG(N) => xNx-1 = W=> xPx~l < W=> xPx~' = P => * e N. Hence NG(NG(P)) < N; the other inclusion is obvious. ■
EXERCISES 1. If N <3 G and N, G/N are both p-groups, then G is a p-group. 2. If G is a finite p-group, H <3G and H ^ {e), then H D C(G) ^ (e). 3. Let |G[ = p". For each k, 0 < k < n, G has a normal subgroup of order pk. 4. If G is an infinite p-group (p prime), then either G has a subgroup of order p” for each n > 1 or there exists m e N* such that every finite subgroup of G has order
m.
5. If P is a normal Sylow p-subgroup of a finite group G and / : G —» G is an endo morphism, then /(P) < P. 6. If H is a normal subgroup of order pk of a finite group G, then J¥ is contained in every Sylow p-subgroup of G. 7. Find the Sylow 2-subgroups and Sylow 3-subgroups of S3, Si, S$. 8. If every Sylow p-subgroup of a finite group G is normal for every prime p, then G is the direct product of its Sylow subgroups. 9. If |G| = pnq, with p > q primes, then G contains a unique normal subgroup of index q. 10. Every group of order 12, 28, 56, and 200 must contain a normal Sylow subgroup, and hence is not simple. 11. How many elements of order 7 are there in a simple group of order 168? 12. Show that every automorphism of St is an inner automorphism, and hence S4 = Aut S4. [Hint: see Exercise 4.10. Every automorphism of Si induces a per mutation of the set \Pi,P2,P3,Pi} of Sylow 3-subgroups of S4. If /e Aut S4 has f(Pi) = Pi for all i, then / = ls4.] 13. Every group G of order p2 (p prime) is abelian [Hint: Exercise 4.9 and Corollary 5.4],
6. CLASSIFICATION OF FINITE GROUPS We shall classify up to isomorphism all groups of order pq (p,q primes) and all groups of small order (n < 15). Admittedly, these are not very far reaching results; but even the effort involved in doing this much will indicate the difficulty in deter mining the structure of an arbitrary (finite) group. The results of this section are not needed in the sequel.
Proposition 6.1. Let p and q be primes such that p > q. //q-f'P — 1 > then every group of order pq is isomorphic to the cyclic group ZPQ. //q | p — 1, then there are (up
6. CLASSIFICATION OF FINITE GROUPS
97
to isomorphism) exactly two distinct groups of order pq: the cyclic group ZPQ and a nonabelian group K generated by elements c and d such that
I c | = p; |d| = q; dc = cBd, where s ^ 1 (mod p) and sq = 1 (mod p).
SKETCH OF PROOF. A nonabelian group K of order pq as described in the proposition does exist (Exercise 2). Given G of order pq, G contains elements a,b with \a\ = p,\b\ = q by Cauchy’s Theorem 5.2. Furthermore, 5 = (a) is normal in G (by Corollary 4.10 or by counting Sylow p-subgroups, as below). The coset bS has order q in the group G/S. Since |G/S| = q, G/S is cyclic with generator bS, G/S = (bS). Therefore every element of G can be written in the form biai and G = (a,b). The number of Sylow ^-subgroups is A:^ -f- 1 and divides pq. Hence it is 1 or p. If it is 1 (as it must be if q\p — 1), then (b) is also normal in G. Lagrange’s Theorem 1.4.6 shows that (a) fl (b) = (e). Thus by Theorems 1.3.2,1.8.6,1.8.10 and Exercise 1.8.5, G = (a) X {b) = Zp © Zq = Zpq. If the number is p, (which can only occur if P I Q — 1). then bab~1 = aT (since (a) <3 G) and r ^ 1 (mod p) (otherwise G would be abelian by Theorem I.3.4(v) and hence have a unique Sylow ^-subgroup). Since bab~l = ar, it follows by induction that biab~i = arl. In particular for j = q, a = ar'\ which implies r5 s 1 (mod p) by Theorem 1.3.4 (v). In order to complete the proof we must show that if q | p — 1 and G is the non abelian group described in the preceding paragraph, then G is isomorphic to K. We shall need some results from number theory. The congruence xQ = 1 (mod p) has exactly q distinct solutions modulo p (see J. E. Shockley [51; Corollary 6.1, p. 67]). If r is a solution and k is the least positive integer such that rk = 1 (mod p), then k \ q (see J.E. Shockley [51; Theorem 8, p. 70]). In our case r ^ 1 (mod p), whence k = q. It follows that 1 ,r,r2,. .. , are all the distinct solutions modulo p of xq = 1 (mod p). Consequently, s rl (mod p) for some t (1 < t < q — 1). If b\ = bl e G, then |fci| = q. Our work above (with b\ in place of b) shows that G = {a,bi); that every element of G can be written bi'a’; that \a\ = p\ and that biabr1 *= blab~‘ = ari = as (Theorem I.3.4(v)). Therefore, bxa = asb\. Verify that the map G —»K given by a |—> c and is an isomorphism. ■
Corollary 6.2. // p is an odd prime, then every group of order 2p is isomorphic either to the cyclic group Z2p or the dihedral group Dp.
PROOF. Apply Proposition 6.1 with q = 2. If G is not cyclic, the conditions on s imply s = — 1 (mod p). Hence G = {c,d), \d\ = 2, |c| = p, and dc = c~xd by Theorem I.3.4(v). Therefore, G = Dp by Theorem 1.6.13. ■
Proposition 6.3. There are (up to isomorphism) exactly two distinct nonabelian groups of order 8: the quaternion group Q8 and the dihedral group D4.
REMARK. The quaternion group Qn is described in Exercise 1.2.3.
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CHAPTER II THE STRUCTURE OF GROUPS
SKETCH OF PROOF OF 6.3. Verify that Z)4 ^ Q8 (Exercise 10). If a group G of order 8 is nonabelian, then it cannot contain an element of order 8 or have every nonidentity element of order 2 (Exercise 1.1.13). Hence G contains an element a of order 4. The group (a) of index 2 is normal. Choose b § (a). Then b2 e (a) since | G/(a)\ = 2. Show that the only possibilities are b2 = a2 or b2 = e. Since (a) is nor mal in G, bab~x e (a); the only possibility is bab_1 = a3 = a-1. It follows that every element of G can be written b‘aj. Hence G = (a,b). In one case we have \a\ = 4, b2 = a2, ba = a~lb, and G = Qs by Exercise 1.4.14.; in the other case, \a\ = 4, |6| = 2, ba = a~xb and G = £>4 by Theorem 1.6.13. ■
Proposition 6-4. There are (up to isomorphism) exactly three distinct nonabelian groups of order 12: the dihedral group D6, the alternating group A4, and a group T generated by elements a,b such that |a| = 6, b2 = a3, and ba = a-1b.
SKETCH OF PROOF. Verify that there is a group T of order 12 as stated (Exercise 5) and that no two of De,AiyT are isomorphic (Exercise 6). If G is a non abelian group of order 12, let P be a Sylow 3-subgroup of G. Then |P| = 3 and [G : P] = 4. By Proposition 4.8 there is a homomorphism / : G —»S4 whose kernel K is contained in P, whence K = P or (e). If K = (e), /is a monomorphism and G is isomorphic to a subgroup of order 12 of S4, which must be Ai by Theorem 1.6.8. Otherwise K = P and P is normal in G. In this case P is the unique Sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then [G : Cg(c)] = 1 or 2 since [G : Cc(c)] is the number of conjugates of c and every con jugate of c has order 3. Hence Cg(c) is a group of order 12 or 6. In either case there is d e Cg(c) of order 2 by Cauchy’s Theorem. Verify that \cd\ = 6. Let a = cd; then (a) is normal in G and \G/(a)\ = 2. Hence there is an element be G such that b $ (a), b ^ e, b2 e (a), and bab~l e (a). Since G is nonabelian and \a\ = 6, bab~x = ab = a~l is the only possibility; that is, ba = a~lb. There are six possibilities for b2 e (a), b2 = a2 or b2 = o4 lead to contradictions; b2 = a or b2 = o5 imply |6| = 12 and G abelian. Therefore, the only possibilities are (i) \a\ = 6; b2 = e; ba — a~xb, whence G = D6 by Theorem 1.6.13; (ii) \a\ = 6; b2 = a3; ba = a~lb, whence G T by Exercise 5(b). ■ The table below lists (up to isomorphism) all distinct groups of small order. There are 14 distinct groups of order 16 and 51 of order 32; see M. Hall and J.K. Senior [16], There is no known formula giving the number of distinct groups of order «, for every n. Order
1 2 3 4 5 6 7
Distinct Groups
<«>
z2 z3
Z2©Z2, Zj
z6
Z^, D3
z7
Reference
Exercise 1.4.3 Exercise 1.4.3 Exercise 1.4.5 Exercise 1.4.3 Corollary 6.2 Exercise 1.4.3
6. CLASSIFICATION OF FINITE GROUPS
Order
Distinct Groups Z2©Z2©Z2, Z2®Z4, Z8, Qs, Di
8 9
Z3©Z3, Z9
10 11 12
Zio, Dt Zn z2 © Z6, Z12, At, D$, T
13 14 15
z13
Zh, Di
z16
99
Reference Theorem 2.1 and Proposition 6.3 Exercise 5.13 and Theorem 2.1 Corollary 6.2 Exercise 1.4.3 Theorem 2.1 and Proposition 6.4 Exercise 1.4.3 Corollary 6.2 Proposition 6.1
EXERCISES 1. Let G and H be groups and 6 : H —» Aut G a homomorphism. Let G Xe H be the set G x H with the following binary operation: (g,h)(g',h') = (g[6(/i)(g')],hh'). Show that G Xe H is a group with identity element (e,e) and (g,h)~l = /?“')• G Xe H is called the semidirect product of G and H.
2. Let Cp = (a) and CQ = (b) be (multiplicative) cyclic groups of prime orders p and q respectively such that p > q and q | p — 1. Let s be an integer such that s ^ 1 (mod p) and sQ = 1 (mod p), which implies s ^ 0 (mod p). Elementary number theory shows that such an s exists (see J.E. Shockley [51; Corollary 6.1, p. 67]). (a) The map a : Cp —» Cp given by a1|—> asi is an automorphism. (b) The map 0: C„ —» Aut Cp given by O(b') = a* (a as in part (a)) is a homo morphism (a0 = \Cp). (c) If we write a for (a,e) and b for (e,b), then the group Cp Xe Cq (see Exer cise 1) is a group of order pq, generated by a and b subject to the relations: \a\ = p, \b\ = q,ba = asb, where s ^ 1 (mod p), and sq = 1 (mod p). The group Cp Xe Cq is called the metacyclic group. 3. Consider the set G = {±l,±i',±/,±A:} with multiplication given by /'2 = f = k2 — — 1; ij = k; jk = i, ki = j; ji = —k, kj = —i, ik = —j, and the usual rules for multiplying by ±1. Show that G is a group isomorphic to the quaternion group Q8. 4. What is the center of the quaternion group Q81 Show that Q&/CCQ&) is abelian. 5. (a) Show that there is a nonabelian subgroup T of S» X Z4 of order 12 generated by elements a,b such that \a\ = 6, az = b2, ba = a~lb. (b) Any group of order 12 with generators a,b such that \a\ = 6, a3 = b1, ba = a~'b is isomorphic to T. 6. No two of De, and T are isomorphic, where T is the group of order 12 de scribed in Proposition 6.4 and Exercise 5. 7. If G is a nonabelian group of order p3 {p prime), then the center of G is the sub group generated by all elements of the form aba~1b~1 (a,b e G). 8. Let p be an odd prime. Prove that there are, at most, two nonabelian groups of order p3. [One has generators a,b satisfying |a| = p2', |6[ = p; b~lab = a1+p;
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CHAPTER II THE STRUCTURE OF GROUPS
the other has generators a,b,c satisfying \a\ = \b\ — \c\ = p; c = a xb xab; ca = ac\ cb -- be.] 9. Classify up to isomorphism all groups of order 18. Do the same for orders 20 and 30. 10. Show that Dt is not isomorphic to Q*. [Hint: Count elements of order 2.]
7. NILPOTENT AND SOLVABLE GROUPS Consider the following conditions on a finite group G. (i) G is the direct product of its Sylow subgroups. (ii) If m divides |G|, then G has a subgroup of order m. (iii) //|G| = mn with (m,n) = 1, then G has a subgroup of order m. Conditions (ii) and (iii) may be considered as modifications of the First Sylow Theo rem. It is not difficult to show that (i) => (ii) and obviously (ii) => (iii). The fact that every finite abelian group satisfies (i) is an easy corollary of Theorem 2.2. Every pgroup satisfies (i) trivially. On the other hand, A4 satisfies (iii) but not (ii), and Ss satisfies (ii) but not (i) (Exercise 1). Given the rather striking results achieved thus far with finite abelian and p-groups, the classes of groups satisfying (i), (ii), and (iii) respectively would appear to be excellent candidates for investigation. We shall re strict our attention to those groups that satisfy (i) or (iii). We shall first define nilpotent and solvable groups in terms of certain “normal series” of subgroups. In the case of finite groups, nilpotent groups are characterized by condition (i) (Proposition 7.5) and solvable ones by condition (iii) (Proposition 7.14). This approach will also demonstrate that there is a connection between nilpotent and solvable groups and commutativity. Other characterizations of nilpotent and solvable groups are given in Section 8. Our treatment of solvable groups is purely group theoretical. Historically, how ever, solvable groups first occurred in connection with the problem of determining the roots of a polynomial with coefficients in a field (see Section V.9). Let G be a group. The center C(G) of G is a normal subgroup (Corollary 4.7). Let C2(G) be the inverse image of C(G/C(G)) under the canonical projection G —> G/C(G). Then by (the proof of) Theorem 1.5.11 C2(G) is normal in G and con tains C(G). Continue this process by defining inductively: Ci(G) = C(G') and C,(G) is the inverse image of C(G/C;_i(G)) under the canonical projection G —» G/C,_i(G). Thus we obtain a sequence of normal subgroups of G, called the ascending central series of G: (e) < Ci(G) < C2(G) < • - ■
Definition 7.1. A group G is nilpotent //Cr,(G) = G for some n.
Every abelian group G is nilpotent since G = C(G) = Ci(G).
Theorem 7.2. Every finite p-group is nilpotent.
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PROOF. G and all its nontrivial quotients arep-groups, and therefore, have non trivial centers by Corollary 5.4. This implies that if G ^ C,(G), then Ci(G') is strictly contained in Cl+j(G). Since G is finite, C„(G) must be G for some n. ■
Theorem 7.3. The direct product of a finite number of nilpotent groups is nilpotent.
PROOF. Suppose for convenience that G = H X K, the proof for more than two factors being similar. Assume inductively that C,(G) = C,(//) X Ci(K) (the case i = 1 is obvious). Let im be the canonical epimorphism H—* H/Ci(H) and similarly for irK. Verify that the canonical epimorphism v : G —> G/Ci(G) is the composition G = H X H/Ci(H) X K/Q(K) ^------------- —X----- = HxK = G/Q(G), Q(H) X Q(K) Ci(H X K) where tv = irI{ X ttk (Theorem 1.8.10), and \p is the isomorphism of Corollary 1.8.11. Consequently, C1+1(G) = tpl[C(G/ Ci(G))} = tt-^-1[C(G/C(G))1 = ir^[C(H/Ci(H) X K/Q(K))] = Tr~x[C(H/Ci(H)) X C{K/CAK))\ = 7T//1[C(///CI(//))] X TrK-'[C(K/Ci(K))\ = Ci+1(H) X Ci+l(K). Thus the inductive step is proved and C,(G) = Ci(H) X Ct(fQ for all i. Since H,K are nilpotent, there exists n e N* such that C„(//) = H and CV(K) = K, whence Cr,(G) = H X K = G. Therefore, G is nilpotent. ■
Lemma 7.4. If H is a proper subgroup of a nilpotent group G, then H is a proper sub group of its normalizer Ng(H).
PROOF. Let C(,(G) = (e) and let n be the largest index such that C„(G) < //; (there is such an n since G is nilpotent and H a proper subgroup). Choose a s Cr,+ )(G) with a $ H. Then for every h e H, Cnah = (Cna)(CJi) — (Crih)(Cna) = CJiti in G/C„(G) since Cna is in the center by the definition of C^+^G). Thus ah = h'ha, where h' e Cn(G) < H. Hence aha'1 e H and a e Ng(H). Since a £ H, H is a proper subgroup of Ng(H). ■
Proposition 7.5. A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups.
PROOF. If G is the direct product of its Sylow /;-subgroups, then G is nilpotent by Theorems 7.2 and 7.3. If G is nilpotent and P is a Sylow p-subgroup of G for some prime p, then either P = G (and we are done) or P is a proper subgroup of G. In the latter case P is a proper subgroup of NC(P) by Lemma 7.4. Since NG(P) is its own normalizer by Theorem 5.11, we must have NC(P) = G by Lemma 7.4. Thus P is normal in G, and hence the unique Sylow p-subgroup of G by Theorem 5.9. Let
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\G\ = pini- ■ ■pknk (pi distinct primes, «, > 0) and let Pi,P2,..., Pk be the corre sponding (proper normal) Sylow subgroups of G. Since |P,-| = pini for each /', Pi D Pj = {e) for i ^ j. By Theorem 1.5.3 xy = yx for every x e P„ y e P, (/' ^ j). It follows that for each i, PiP2- ■ -Pi_iP,+i ■ Pk is a subgroup in which every element has order dividing p,"1- • ■ p^p^1 ■ ■ Pknk■ Consequently, Pi fl (Pi - ■ - Pr-\Pi+\ ■ Pk) = (e) and PiP2- • P* = Pi X • • • X Pk. Since |G| = pi"1- • -pknk = |Pi X - • • X Pk\ — |Pi • • -Pk| we must have G = PiP2- • -P* = Pi X • - • X P*. ■
Corollary 7.6. If G is a finite nilpotent group and m divides |G|, then G has a sub group of order m.
PROOF. Exercise. ■
Definition 7.7. Let G be a group. The subgroup of G generated by the set j aba_1b_1 | a,b e G| is called the commutator subgroup of G and denoted G'.
The elements abarlb~x (a,b e G) are called commutators. The commutators only generate G', so that G' may well contain elements that are not commutators. G is abelian if and only if G' = {e). In a sense, G' provides a measure of how much G differs from an abelian group.
Theorem 7.8. //G is a group, then G' is a normal subgroup ofG and G/G' is abelian. 7/N is a normal subgroup of G, then G/N is abelian if and only if N contains G'.
PROOF. Let / : G —* G be any automorphism. Then f(aba lb ') = f(a)f(b)f(a) 'f(b) ' e G'.
It follows that f(G') < G'. In particular, if /is the automorphism given by conjuga tion by a e G, then aG'ar1 = f(G') < G', whence G’ is normal in G by Theorem 1.5.1. Since (ab)(ba)~1 — aba~lb~l e G‘, abG' = baG' and hence G/G' is abelian. If G/N is abelian, then abN = baN for all a,b e G, whence ab(ba)_1 = aba1b '1 e N. There fore, N contains all commutators and G’ < N. The converse is easy. ■ Let G be a group and let G(1) be G'. Then for / > 1, define G(l) by G(i) = (G(i~l))'. G(i) is called /th derived subgroup of G. This gives a sequence of subgroups of G, each normal in the preceding one: G > G(l) > G(2) > • - •. Actually each Gw is a normal subgroup of G (Exercise 13).
Definition 7.9. A group G is said to be solvable //G(n) = (e) for some n.
Every abelian group is trivially solvable. More generally, we have
Proposition 7.10. Every nilpotent group is solvable.
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PROOF. Since by the definition of Ct(G) C£«7)/C,-_i(G) = C(G/Ci_1(G)) is abelian, Ci(G)' < Ci_i(G) for all / > 1 and Ci(G)' = C(G)' = (e). For some n, G = C„(G). Therefore, C(G/Cn^(G)) = Cn(G)/Cn_,(G) = G/C„_i(G) is abelian and hence G(,) = G‘ < Cn_,(G). Therefore, G(2> = G(1)' < C„_i(G)' < Cn_2(G); similarly G(3) < C„_2(G)' < C„_3(G);. .. , G(""« < C2(G)' < Ci(G); G‘»> < G(G)' = (e>. Hence G is solvable. ■
Theorem 7.11. (i) Every subgroup and every homomorphic image of a solvable group is solvable. (ii) //N is a normal subgroup of a group G such that N and G/N are solvable, then G is solvable.
SKETCH OF PROOF, (i) If / : G —» H is a homomorphism [epimorphism], verify that /(G(,)) < //(i)[/(G(<)) = Hw] for all i. Suppose /is an epimorphism, and G is solvable. Then for some n, (e) — f(e) — f(G(n)) = H(n), whence H is solvable. The proof for a subgroup is similar. (ii) Let / : G —» G/N be the canonical epimorphism. Since G/N is solvable, for some n /(G('°) = (G/N)(n) = (e). Hence G(n) < Ker f = N. Since G(n) is solvable by (i), there exists k e N* such that G(nAh) — (G(n))(W = (e). Therefore, G is solvable. ■
Corollary 7.12. Ifn > 5, then the symmetric group S„ is not solvable.
PROOF. If Sn were solvable, then An would be solvable. Since An is nonabelian, A„r ^ (1). Since An' is normal in An (Theorem 7.8) and An is simple (Theorem 1.6.10), we must have An' = An. Therefore An(i) = An ^ (1) for all / > 1, whence An is not solvable. ■ NOTE. The remainder of this section is not needed in the sequel. In order to prove a generalization of the Sylow theorems for finite solvable groups (as mentioned in the first paragraph of this section) we need some definitions and a lemma. A subgroup H of a group G is said to be characteristic [resp. fully in variant] if /(//) < H for every automorphism [resp. endomorphism] / : G —* G. Clearly every fully invariant subgroup is characteristic and every characteristic sub group is normal (since conjugation is an automorphism). A minimal normal subgroup of a group G is a nontrivial normal subgroup that contains no proper subgroup which is normal in G.
Lemma 7.13. Let N be a normal subgroup of a finite group G and H any sub group of G.
(i) 7/H is a characteristic subgroup o/N, then H is normal in G.
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(ii) Every normal Sylow p-subgroup of G is fully invariant. (iii) //G is solvable and N is a minimal normal subgroup, then N is an abelian pgroup for some prime p. PROOF, (i) Since aNa~~l = N for all a t G, conjugation by a is an automor phism of N. Since H is characteristic in N, aHa~l < H for all a e G. Hence H is normal in G by Theorem 1.5.1. (ii) is an exercise, (iii) It is easy to see that N' is fully invariant in A', whence N' is normal in G by (i). Since N is a minimal normal subgroup, either N' = (e) or N' = N. Since N is solvable (Theorem 7.11), N' N. Hence N' = (e) and TV is a nontrivial abelian group. Let P be a nontrivial Sylow p-subgroup of N for some prime p. Since N is abelian, P is normal in N and hence fully invariant in N by (ii). Consequently P is normal in G by (i). Since N is minimal and P nontrivial we must have P - N. ■
Proposition 7.14. (P. Hall) Let G be a finite solvable group of order mn, with (m,n) = 1. Then
(i) G contains a subgroup of order m; (ii) any two subgroups of G of order m are conjugate; (iii) any subgroup of G of order k, where k | m, is contained in a subgroup of order m. REMARKS. If m is a prime power, this theorem merely restates several results contained in the Sylow theorems. P. Hall has also proved the converse of (i): if G is a finite group such that whenever | G| = mn with (m,n) ~ 1, G has a subgroup of order m, then G is solvable. The proof is beyond the scope of this book (see M. Hall [15; p. 143]). PROOF OF 7.14. The proof proceeds by induction on [G|, the orders < 5 being trivial. There are two cases. CASE 1. There is a proper normal subgroup H of G whose order is not divisible by n. (i) | H\ = wi«i, where/?7i | m,nx | n, and«, < n. G/H is a solvable group of order (m/ni\)(n/n\) < mn, with (jn/m\,n/n\) = l. Therefore by induction G/H contains a subgroup A/H of order (m/m\) (where A is a subgroup of G — see Corollary 1.5.12). Then |/1| = \H\[A : H] = (min\)(/n/mi) = mni < mn. A is solvable (Theorem 7.11) and by induction contains a subgroup of order m. (ii) Suppose B,C are subgroups of G of order m. Since H is normal in G, HB is a subgroup (Theorem 1.5.3), whose order k necessarily divides |G| = mn. Since k = |HB\ = \H\\B\/\H 0 B\ = nnn,m/\H D B\, we have k\H D B\ = /mmm, whence k | mxmm. Since (mi,n) = 1, there are integers x,y such that m\x + ny = 1, and hence mn\m\x + mmny = mrii. Consequently k | mn,. By Lagrange’s Theorem 1.4.6 m = \B\ and nnn\ = \H\ divide k. Thus (m,n) = 1 implies nun I k. Therefore k = mn,-, similarly \HC\ = mn\. Thus HB/H and HC/H are subgroups of G/H of
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order m/nu. By induction they are conjugate: for some x e G/H(where x is the coset of x e G), x(HB/H)x l = HC/H. It follows that xHBxx — HC. Consequently xBx x and C are subgroups of HC of order m and are therefore conjugate in HC by induction. Hence B and C are conjugate in G. (iii) If a subgroup K of G has order k dividing m, then HK/H = K/H D K has order dividing k. Since HK/H is a subgroup of G/H, its order also divides \G/H\ = (m/mi)(n/n\). (k,n) = 1 implies that the order of HK/H divides m/mu By induc tion there is a subgroup A/H of G/H of order m/m\ which contains HK/H (where A < Gas above). Clearly K is a subgroup of A. Since \A\ = \H\\A/H\ = min/tm/im) = mrii < mn, K is contained in a subgroup of A (and hence of G) of order m by in duction. CASE 2. Every proper normal subgroup of G has order divisible by n. If H is a minimal normal subgroup (such groups exist since G is finite), then \H\ = pr for some prime p by Lemma 7.13 (iii). Since (m,n) = 1 and n\\H\, it follows that n = pT and hence that H is a Sylow /^-subgroup of G. Since H is normal in G, H is the unique Sylow p-subgroup of G. This argument shows that H is the only minimal normal subgroup of G (otherwise n = pT and n = qs for distinct primes p,cj). In par ticular, every nontrivial normal subgroup of G contains H. (i) Let AT be a normal subgroup of G such that K/H is a minimal normal sub group of G/H (Corollary 1.5.12). By Lemma 7.13 (iii) |K/H\ = q* (q prime, q ^ p), so that \K\ = prqs. Let 5 be a Sylow ^-subgroup of K and let M be the normalizer of S in G. We shall show that \M\ = m. Since H is normal in K, HS is a subgroup of K. Clearly H flS = (e) so that |//S| = \H\\S\/\H D S\ = prq* = |ff|, whence K = HS. Since K is normal in G and S < K, every conjugate of S in G lies in K. Since 5 is a Sylow subgroup of K, all these subgroups are already conjugate in K. Let N = Nk(S); then the number c of conjugates of 5 in G is [G : M] = : TV] by Corollary 4.4. Since S < N < K, K > HN > HS — K, so that K = HN and c = [G:M] = [K:N] = [HN : N\ = [H: H D N] (Corollary 1.5.9). We shall show that H fl N = (e), which implies c = |//| — pT and hence \M\ = |G|/[G : M] = mpr/pr = m. We do this by showing first that H f) N < C(K) and second that C(K) = (e). Let xeH f) N and k e K. Since K = HS, k = hs (h e H, s e 5). Since H is abelian (Lemma 7.13 (iii)) and x e H, we need only show xs = sx in order to have xk = kx and x e C{K). Now (xsx-1)5"1 £ S since x e N — Nh(S). But x(xx_1j_1) e H since x s H and H is normal in G. Thus xsx-1.^1 e H fl 5 = (e), which implies xs = sx. It is easy to see that C(K) is a characteristic subgroup of K. Since K is normal in G, C(K) is normal in G by Lemma 7.13 (i). If C(K) ^ (e), then C(K) necessarily concontains H. This together with K = HS implies that S is normal in K. By Lemma 7.13 (ii) and (i) S is fully invariant in K and hence normal in G (since K <3 G). This implies H < S which is a contradiction. Hence C(K) = (e). (ii) Let M be as in (i) and suppose B is a subgroup of G of order m. Now |BK\ is divisible by \B\ = m and |AT| = prqs. Since im,p) = 1, is divisible by pTm = nm = |G[. Hence G = BK. Consequently G/K = BK/K ^ B/B D K (Corollary 1.5.9), which implies that \B D AT| = \B\/\G/K\ = qs. By the Second Sylow Theorem B fl K is conjugate to S in K. Furthermore B fl K is normal in B (since K <3 G) and hence B is contained in No(B D K). Verify that conjugate subgroups have conjugate
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normalizers. Hence Ng(B 0 K) and Ng(S) = M are conjugate in G. Thus \Ng(B fl A')| = |M\ = m. But |Z?j = m\ therefore B < Ng(B fl K) implies B = Ng(B 0 K). Hence B and M are conjugate. (iii) Let D < G, where |.D| = k and k [ m. Let M (of order m) and H (of order r p , with (p,m) = 1) be as in (i). Then D fl H = (e) and \DH\ = \D\\H\/\D fl H\ — kpT. We also have |G| = mpr, M 0 H — (e) and MH = G (since |MH\ = \M\\H\/\M ft H\ = mpr = \G\). Hence M(DH) = G and therefore |M fl DH\ = \M\\DH\/\MDH\ = m(kpr)/mpr = k. Let M* = M fl DH\thenM* and D are conjugate (by (ii) applied to the group DH). For some a e G, aM*a~l = D. Since M* < M, D is contained in aMarl, a conjugate of M, and thus a subgroup of order m. ■ We close this section by mentioning a longstanding conjecture of Burnside: every finite group of odd order is solvable. This remarkable result was first proved by W. Feit and J. Thompson [61] in 1963.
EXERCISES 1. (a) Ai is not the direct product of its Sylow subgroups, but A4 does have the property: mn =12 and (m,n) = 1 imply there is a subgroup of order m. (b) S3 has subgroups of orders 1, 2, 3, and 6 but is not the direct product of its Sylow subgroups. 2. Let G be a group and a,b e G. Denote the commutator aba e G by [a,b\Show that for any a,b,c, e G, [ab,c] = a[b,c]a~ \a,c\. 3. If H and K are subgroups of a group G, let (H,K) be the subgroup of G generated by the elements {hkh~lk~l | h e H, k e K\. Show that (a) (H,K) is normal in H V K. (b) If (H,G') = (e), then (H',G) = (e). (c) H <\ G if and only if (//,G) < H. (d) Let K <\ G and K < H; then H/K < C(G/K) if and only if (H,G) < K. 4. Define a chain of subgroups yi(G) of a group G as follows: 71(G) = G, 72(G) = (G,G), 7,(G) = (7,_i(G),G) (see Exercise 3). Show that G is nilpotent if and only if 7m(G) = (e) for some m.
5. Every subgroup and every quotient group of a nilpotent group is nilpotent. [Hint: Theorem 7.5 or Exercise 4.].
6. (Wielandt) Prove that a finite group G is nilpotent if and only if every maximal proper subgroup of G is normal. Conclude that every maximal proper subgroup has prime index. [Hint: if P is a Sylow p-subgroup of G, show that any subgroup containing Ng(P) is its own normalizer; see Theorem 5.11.] 7. If ./Vis a nontrivial normal subgroup of a nilpotent group G, then N fl C(G) ^ (e).
8. If Dn is the dihedral group with generators a of order n and b of order 2, then (a) a* e Dn’. (b) If n is odd, Dn' ^ Zn. (c) If n is even, Dn' ^ Zm, where 2m = n. (d) Dn is nilpotent if and only if n is a power of 2.
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9. Show that the commutator subgroup of is A4. What is the commutator group of AP. 10. Sn is solvable for n < 4, but Ss and S4 are not nilpotent. 11. A nontrivial finite solvable group G contains a normal abelian subgroup H j* (e). If G is not solvable then G contains a normal subgroup H such that //' = H. 12. There is no group G such that G' = S4- [Hint: Exercises 9 and 5.12 may be helpful.] 13. If G is a group, then the ith derived subgroup G(i) is a fully invariant subgroup, whence G{i) is normal. 14. If TV < G and TV PI G' = (e), then N < C(G). 15. If H is a maximal proper subgroup of a finite solvable group G, then [G : H] is a prime power. 16. For any group G, C(G) is characteristic, but not necessarily fully invariant. 17. If G is an abelian p-group, then the subgroup G[p} (see Lemma 2.5) is fully in variant in G. 18. If G is a finite nilpotent group, then every minimal normal subgroup of G is con tained in C(G) and has prime order.
8. NORMAL AND SUBNORMAL SERIES The usefulness of the ascending central series and the series of derived subgroups of a group suggests that other such series of subgroups should be investigated. We do this next and obtain still other characterizations of nilpotent and solvable groups, as well as the famous theorem of Jordan-Holder. Definition 8.1. A subnormal series of a group G is a chain of subgroups G = G0 > Gi > • • • > Gn such that Gi+i is normal in Gi for 0 < i < n. The factors of the series are the quotient groups Gi/Gi+i. The length of the series is the number of strict inclu sions (or alternatively, the number of nonidentity factors). A subnormal series such that G, is normal in G for all i is said to be normal.5 A subnormal series need not be normal (Exercise 1.5.10). EXAMPLES. The derived series G > Gw > • • • > G
Definition 8.2. Let G = G0 > Gi >••• > G n be a subnormal series. A one-step re finement of this series is any series of the form G = G 0 > - > G j > N > Gi+i > - • • 6Some
authors use the terms “normal” where we use “subnormal.”
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> G„ or G = G0 > ■ • • > Gn > N, where N is a normal subgroup ofGx and (if i < n) Gi+i is normal in N. A refinement of a subnormal series S is any subnormal series ob tained from S by a finite sequence of one-step refinements. A refinement of S is said to be proper if its length is larger than the length of S.
Definition 8.3. A subnormal series G = G0 > Gi > - - > Gn = (e) is a composi tion series if each factor G,/G,+i is simple. A subnormal series G = G0 > Gi >••■ > Gn = (e) is a solvable series if each factor is abelian.
The following fact is used frequently when dealing with composition series: if N'is a normal subgroup of a group G, then every normal subgroup of G/ N is of the form H/N where His a normal subgroup of G which contains N (Corollary 1.5.12). There fore, when G N, G/N is simple if and only if N is a maximal in the set of all normal subgroups M of G with M ^ G (such a subgroup N is called a maximal normal subgroup of G).
Theorem 8.4. (i) Every finite group G has a composition series. (ii) Every refinement of a solvable series is a solvable series. (iii) A subnormal series is a composition series if and only if it has no proper re finements.
PROOF, (i) Let Gi be a maximal normal subgroup of G; then G/Gi is simple by Corollary 1.5.12. Let G2 be a maximal normal subgroup of Gu and so on. Since G is finite, this process must end with G„ = (e). Thus G > Gi > ■ ■ ■ > Gn = (e) is a composition series. (ii) If Gi/Gi+1 is abelian and Gi+i <]//<] then H/Gi+J is abelian since it is a subgroup of Gi/Gi+i and Gi/H is abelian since it is isomorphic to the quotient (Gi/Gi+\)/(H/Gi+1) by the Third Isomorphism Theorem 1.5.10. The conclusion now follows immediately. (iii) If Gi+i <1H Gi are groups, then H/Gi+\ is a proper normal subgroup of G,/Gi+1 and every proper normal subgroup of Gi/Gi+1 has this form by Corollary 1.5.12. The conclusion now follows from the observation that a subnormal series G = G0 > Gi >■■■> Gr, = (e)'has a proper refinement if and only if there is a subgroup H such that for some /, G i+1 <]//<] Gi. ■
Theorem 8.5. A group G is solvable if and only if it has a solvable series.
PROOF. If G is solvable, then the derived series G > Ga) > Gm > • • • > G(n) = (e) is a solvable series by Theorem 7.8. If G = G0 > Gi > ■ ■ ■ > Gn = {e) is a solvable series for G, then G/G\ abelian implies that G\ > Ga) by Theorem 7.8; G1/G2 abelian implies G2 > Gi > G(2>. Continue by induction and conclude that Gt > G(i) for all /; in particular (e) = Gn > G{n) and G is solvable. ■ EXAMPLES- The dihedral group Dn is solvable since D„ > (a) > {e) is a solv able series, where a is the generator of order n (so that Dn/{a) =Z2). Similarly if
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|G[ = pq (p > q primes), then G contains an element a of order p and (a) is normal in G (Corollary 4.10). Thus G > (a) > (e) is a solvable series and G is solvable. More generally we have
Proposition 8.6. A finite group G is solvable if and only ifG has a composition series whose factors are cyclic of prime order.
PROOF. A (composition) series with cyclic factors is a solvable series. Con versely, assume G = G0 > Gx > ■ ■ • > Gn = {e) is a solvable series for G. If G0 ^ Gu let Hi be a maximal normal subgroup of G = G0 which contains Gu If Hx ^ Gu let Hi be a maximal normal subgroup of H which contains Gi, and so on. Since G is finite, this gives a series G > Hx > H2 > ■ ■ ■ > Hk > Gi with each subgroup a maxi mal normal subgroup of the preceding, whence each factor is simple. Doing this for each pair (Gi,G1+i) gives a solvable refinement G = N0 > N, > • ■ ■ > Nr = (e) of the original series by Theorem 8.4 (ii). Each factor of this series is abelian and simple and hence cyclic of prime order (Exercise 1.4.3). Therefore, G > ArJ > • ■ • > Arr = (e) is a composition series. ■ A given group may have many subnormal or solvable series. Likewise it may have several different composition series (Exercise 1). However we shall now show that any two composition series of a group are equivalent in the following sense.
Definition 8.7. Two subnormal series S and T of a group G are equivalent /f there is a one-to-one correspondence between the nontrivial factors ofS and the nontrivial factors ofT such that corresponding factors are isomorphic groups.
Two subnormal series need not have the same number of terms in order to be equivalent, but they must have the same length (that is, the same number of non trivial factors). Clearly, equivalence of subnormal series is an equivalence relation.
Lemma 8.8. If S is a composition series of a group G, then any refinement ofS is equivalent to S.
PROOF. Let S be denoted G = G0 > Gi > • • • > Gn = (e). By Theorem 8.4 (iii) S has no proper refinements. This implies that the only possible refinements of S are obtained by inserting additional copies of each Gi. Consequently any re finement of S has exactly the same nontrivial factors as S and is therefore equivalent to S. ■ The next lemma is quite technical. Its value will be immediately apparent in the proof of Theorem 8.10.
Lemma 8.9. (Zassenhaus) Let A*, A, B*, B be subgroups of a group G such that A* is normal in A and B* is normal in B.
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(i) A*(A fl B*) is a normal subgroup of A*(A fl B); (ii) B*(A* fl B) is a normal subgroup o/B*(A fl B); (iii) A*(A D B)/A*(A D B*) ^ B*(A D B)/B*(A* D B). PROOF. Since B* is normal in B, A D B* = (A fl B) D B* is a normal sub group of A fl B (Theorem 1.5.3 (i)); similarly A* fl B is normal in A fl B. Con sequently D = (A* fl B)(A fl B*) is a normal subgroup of A fl B (Theorem 1.5.3 iii) and Exercise 1.5.13). Theorem 1.5.3 (iii) also implies that A *(A fl B) and B\A fl B) are subgroups of A and B respectively. We shall define an epimorphism / : A*(A fl B) —» (A fl B)/D with kernel A*(A fl B*). This will imply that A*(A fl B*) is normal in A*(A fl B) (Theorem 1.5.5) and that A*(A D B)/A\A n B*)9^(A D B)/D (Corollary 1.5.7). Define / : A*(A fl B) —> (A fl B)/D as follows. If a e A*, ceA fl B, let f(ac) = Dc. Then /is well defined since ac = aiCi (a,ax e A*; c,ci e A fl B) implies cic-1 = ay-'a e (A fl B) fl A* = A* fl B < D, whence Dei = Dc. /is clearly surjective. / is an epimorphism since f[(aiCi)(aiC?)] = fiaiazcvc^) = Dcic2 = DciDci = f(aiC\)f(a
Theorem 8.10. (Schreier) Any two subnormal [resp. normal] series of a group G have subnormal [resp. normal] refinements that are equivalent.
PROOF. Let G = Gu > Gi >••• > Gn and G — H0 > Hi > ■ ■ ■> Hm be sub normal [resp. normal] series. Let Gn+1 = (e) = Hm+l and for each 0 < i < n con sider the groups Gi = Gi+i(Gi fl H0) > Gi+i(Gi fl //,)>•••> Gi+i(Gi D H,) > G,+1(G, D Hi+l) > ■ - ■ > Gi+i(Gi D Hm) > Gi+\(Gi D Hm+1) = Gi+i.
For each 0 < j < m, the Zassenhaus Lemma (applied to Gi+uGi,Hj+\, and Hj) shows that G1+i(G, fl H}+1) is normal in Gi+1(Gi fl Hj). [If the original series were both normal, then each Gj+i(G, fl Hj) is normal in G by Theorem 1.5.3 (iii) and Exercises 1.5.2 and 1.5.13.] Inserting these groups between each G, and Gi+J, and denoting Gi+i(G, fl //,) by G(i,j) thus gives a subnormal [resp. normal] refinement of the series G0 > Gi > - • - > G„: G = G(0,0) > G(0,1) > > G(0,m) > G(1,0) > G(l,l) > G(1,2) > > G(l,m) > G(2,0) > - > G(n - 1 ,m) > G(«,0) > > G(n,m), where G(/,0) = Gi. Note that this refinement has {n + 1 )(m + 1) (not necessarily distinct) terms. A symmetric argument shows that there is a refinement of G = H0 > H >■■■> Hm (where H(iJ) = //J+1(G, fl Hj) and H(0,j) = Hj): G = //(0,0) > H(l,0) > ■ ■ - > H(n,0) > //(0,1) > //(1,1) > //(2,1) > • • • > Hiti, 1) > H(0,2) > > H(n,m - 1) > H(0,m) > ■ ■ > H(n,m).
8. NORMAL AND SUBNORMAL SERIES
111
This refinement also has (« + l)(m + 1) terms. For each pair (i,j) (0
This provides the desired one-to-one correspondence of the factors and shows that the refinements are equivalent. ■
Theorem 8.11. (Jordan-Holder) Any two composition series of a group G are equivalent. Therefore every group having a composition series determines a unique list of simple groups.
REMARK. The theorem does not state the existence of a composition series for a given group. PROOF OF 8.11. Since composition series are subnormal series, any two com position series have equivalent refinements by the Theorem 8.10. But every refine ment of a composition series S is equivalent to S by Lemma 8.8. It follows that any two composition series are equivalent. ■ The Jordan-Holder Theorem indicates that some knowledge of simple groups might be useful. A major achievement in recent years has been the complete classifi cation of all finite simple groups. This remarkable result is based on the work of a large number of group theorists. For an introduction to the problem and an outline of the method of proof, see Finite Simple Groups by Daniel Gorenstein (Plenum Publishing Corp., 1982). Nonabelian simple groups of small order are quite rare. It can be proved that there are (up to isomorphism) only two nonabelian simple groups of order less than 200, namely A5 and a subgroup of S7 of order 168 (see Exercises 13 -20).
EXERCISES 1. (a) Find a normal series of D4 consisting of 4 subgroups. (b) Find all composition series of the group D4(c) Do part (b) for the group A4. (d) Do part (b) for the group s3 x z*. (e) Find all composition factors of S4 and De. 2. If G = Go > Gi > • • ■ > Gn is a subnormal series of a finite group G, then \G\ = (ni
3. If N is a simple normal subgroup of a group G and G/N has a composition series, then G has a composition series. 4. A composition series of a group is a subnormal series of maximal (finite) length.
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CHAPTER II THE STRUCTURE OF GROUPS
5. An abelian group has a composition series if and only if it is finite.
6. If H <] G, where G has a composition series, then G has a composition series one of whose terms is H. 7. A solvable group with a composition series is finite.
8. If //and K are solvable subgroups of G with H < G, then HK is a solvable sub group of G. 9. Any group of order p2q (p,q primes) is solvable. 10. A group G is nilpotent if and only if there is a normal series G — G0 > Gi > ■ ■ ■ > Gn = (e) such that GJGl+i < C(G/Gl+1) for every /. 11. (a) Show that the analogue of Theorem 7.11 is false for nilpotent groups [Consider S8]. (b) If H < C(G) and G/H is nilpotent, then G is nilpptent. 12. Prove the Fundamental Theorem of Arithmetic, Introduction, Theorem 6.7, by applying the Jordan-Holder Theorem to the group Zn. 13. Any simple group G of order 60 is isomorphic to As. [Hint: use Corollary 4.9; if H < G, then [G : H] > 5 (since |S„| < 60 for n < 4); if [G : H] = 5 then G = As by Theorem 1.6.8. The assumption that there is no subgroup of index 5 leads to a contradiction.] 14. There are no nonabelian simple groups of order < 60. 15. Let G be the subgroup of S7 generated by (1234567) and (26)(34). Show that |G| = 168. Exercises 16-20 outline a proof of the fact that the group G of Exercise 15 is simple. We consider G as acting on the set 5 = {1,2,3,4,5,6,7} as in the first example after Definition 4.1 and make use of Exercise 4.6. 16. The group G is transitive (see Exercise 4.6). 17. For each x eS, Gx is a maximal (proper) subgroup of G. The proof of this fact proceeds in several steps: (a) A block of G is a subset Tof 5 such that for each g e G either gT fl T = 0 or gT — T, where gT = {gx \ x e T). Show that if T is a block, then [7r'| divides 7. [Hint: let H= (g e G\gT= T) and show that for x e T, GX
8. NORMAL AND SUBNORMAL SERIES
113
19. The group G contains a subgroup P of order 7 such that the smallest normal sub group of G containing P is G itself. 20. If (1) 5^ <3 G, then N = G; hence G is simple. [Use Exercise 1.5.19 and Exercise 18 to show P < N; apply Exercise 19.]
CHAPTER
III
RINGS
Another fundamental concept in the study of algebra is that of a ring. The problem of classifying all rings (in a given class) up to isomorphism is far more complicated than the corresponding problem for groups. It will be partially dealt with in Chapter IX. The present chapter is concerned, for the most part, with presenting those facts in the theory of rings that are most frequently used in several areas of algebra. The first two sections deal with rings, homomorphisms and ideals. Much (but not all) of this material is simply a straightforward generalization to rings of concepts which have proven useful in group theory. Sections 3 and 4 are concerned with commuta tive rings that resemble the ring of integers in various ways. Divisibility, factoriza tion, Euclidean rings, principal ideal domains, and unique factorization are studied in Section 3. In Section 4 the familiar construction of the field of rational numbers from the ring of integers is generalized and rings of quotients of an arbitrary com mutative ring are considered in some detail. In the last two sections the ring of poly nomials in n indeterminates over a ring R is studied. In particular, the concepts of Section 3 are studied in the context of polynomial rings (Section 6). The approximate interdependence of the sections of this chapter is as follows: 1
I 2
6
Section 6 requires only certain parts of Sections 4 and 5. 114
1. RINGS AND HOMOMORPHISMS
115
1. RINGS AND HOMOMORPHISMS The basic concepts in the theory of rings are defined and numerous examples given. Several frequently used calculational facts are presented. The only difficulty with this material is the large quantity of terminology that must be absorbed in a short period of time.
Definition 1.1. A ring is a nonempty set R together with two binary operations (usually denoted as addition (+) and multiplication) such that:
(i) (R,+) is an abelian group; (ii) (ab)c = a(bc) for all a,b,c e R (associative multiplication); (iii) a(b + c) = ab + ac and (a + b)c = ac + be (left and right distributive laws). If in addition:
(iv) ab = ba for all a,b e R, then R is said to be a commutative ring. 7/R contains an element I such that (v) lRa = alR = a for all a e R, r
then R is said to be a ring with identity.
REMARK. The symbol \r is also used to denote the identity map R —> R. In context this usage will not be ambiguous. The additive identity element of a ring is called the zero element and denoted 0. If R is a ring, a e R and n e Z, then na has its usual meaning for additive groups (Definition 1.1.8); for example, na = a + a +• —\- a (n summands) when n > 0. Before giving examples of rings we record
Theorem 1.2. Let R be a ring. Then
(i) (ii) (iii) (iv)
0a = aO = 0 for all a e R; (—a)b = a(—b) = — (ab) for all a,beR; (—a)(—b) = ab for all a,b e R; (na)b = a(nb) = n(ab) for all n e Z and all a,b e R;
(
n m
n \ / 77i \
X a i)(H b i)
=
aibj
for all ai,bj e R.
i = l / \j = l / » = 1 j = l
SKETCH OF PROOF, (i) 0a = (0 + 0)a = 0a + 0a, whence Oo = 0. (ii) ab + ( — a)b = (a + (—a))b = 0b = 0, whence ( — a)b = —(ab) by Theorem I.1.2(iii). (ii) implies (iii). (v) is proved by induction and includes (iv) as a special case. ■ The next two definitions introduce some more terminology; after which some examples will be given.
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CHAPTER III RINGS
Definition 1.3. A nonzero element a in a ring R is said to be a left [resp. right] zero divisor if there exists a nonzero b e R such that ab = 0 [resp. ba = 0], A zero divisor is an element of R which is both a left and a right zero divisor.
It is easy to verify that a ring R has no zero divisors if and only if the right and left cancellation laws hold in R; that is, for all a,b,c e R with a^O, ab — ac ox ba — ca => b = c.
Definition 1.4. An element a in a ring R with identity is said to be left [resp. right] in vertible if there exists c e R [resp. b e R] such that ca = 1r [resp. ab = 1r]. The ele ment c [resp. b] is called a left [resp. right] inverse of a. An element a e R that is both left and right invertible is said to be invertible or to be a unit.
REMARKS, (i) The left and right inverses of a unit a in a ring R with identity necessarily coincide (since ab = \r = ca implies b = 1 Rb = (ca)b = c(ab) — c\R = c). (ii) The set of units in a ring R with identity forms a group under multiplication.
Definition 1.5. A commutative ring R with identity Ik ^ 0 and no zero divisors is called an integral domain. A ring D with identity Id 5^ 0 in which every nonzero ele ment is a unit is called a division ring. A field is a commutative division ring.
REMARKS, (i) Every integral domain and every division ring has at least two elements (namely 0 and 1«). (ii) A ring R with identity is a division ring if and only if the nonzero elements of R form a group under multiplication (see Remark (ii) after Definition 1.4). (iii) Every field F is an integral domain since ab = 0 and a ^ 0 imply that b — 1 Fb = (a~~ld)b - arl(ab) — a"10 = 0. EXAMPLES. The ring Z of integers is an integral domain. The set E of even integers is a commutative ring without identity. Each of Q (rationals), R (real numbers), and C (complex numbers) is a field under the usual operations of addition and multiplication. The n X n matrices over Q (or R or C) form a noncommutative ring with identity. The units in this ring are precisely the nonsingular matrices. EXAMPLE. For each positive integer n the setZ,, of integers modulo n is a ring. See the example after Theorem 1.1.5 for details. If // is not prime, say n = kr with k > 1, r > 1, then k ^ 0, r ^ 0 and kr = kr = n = 0 in Z„, whence A and r are zero divisors. If p is prime, then Zr is a field by Exercise 1.1.7. EXAMPLE. Let A be an abelian group and let End A be the set of endomorphisms / : A A. Define addition in End A by (/+ g)(a) = f(a) + g(a). Verify that/+ g z End A. Since A is abelian, this makes End A an abelian group. Let multi plication in End A be given by composition of functions. Then End A is a (possibly noncommutative) ring with identity l.i : A —-> A. -
EXAMPLE. Let G be a (multiplicative) group and R a ring. Let R(G) be the additive abelian group 23 R (one copy of R for each g z G). It will be convenient to VtG
1. RINGS AND HOMOMORPHISMS
117
adopt a new notation for the elements of R(G). An element x = j rg} gtG of R(G) has only finitely many nonzero coordinates, say rB l ,..., r„n (gi e G). Denote x by the n
formal sum reigi + re2g2 + • ■ - + rBngn or ^ We also allow the possibility that i=i some of the rgi are zero or that some gt are repeated, so that an element of R(G) may be written in formally different ways (for example, r{g{ + 0g2 — r\g\ or = (r, + 5,) g,). In this notation, addition in the group R(G) is given by: n
n
'H rBi8i +
5Z
i = l i= 1 i = l
n
s0igi
= H (r + s )gi; Bi
ei
(by inserting zero coefficients if necessary we can always assume that two formal sums involve exactly the same indices gi,. .., g„). Define multiplication in R(G) by
(X
\/m
n
\n m
(X / jj-1 j) X X i-1 ‘/‘)\j~ j-1 r8
s h
=
(V/Xftfy);
this makes sense since there is a product defined in both R (r,Sj) and G(gihJ) and thus the expression on the right is a formal sum as desired. With these operations R(G) is a ring, called the group ring of G over R. R(G) is commutative if and only if both R and G are commutative. If R has an identity 1«, and e is the identity element of G, then \Re is the identity element of R(G). EXAMPLE. Let R be the field of real numbers and S the set of symbols 1 ,i,j,k. Let K be the additive abelian group R @ R © R © R and write the elements of K as formal sums (a0,aua2,a3) = a01 + ad + a2j + a%k. Then aQ 1 + ad + a2j + azk = Z?nl + bii + b2j + bzk if and only if a, = £>, for every i. We adopt the conventions that a<,l e K is identified with a(l £ R and that terms with zero coefficients may be omitted (for example, 4 + 2y = 4 • 1 + 0/ + 2y + Ok and / = 0 + 1/ + Oy + Ok). Then addition in K is given by (flo + ad
a2j + Ozk) + (bo -f- b\i -f- b2j + b-Ji)
~ (#o + bo) + (a\ + b\)i -f- (a2 + bi)j -f- (as + bs)k.
Define multiplication in K by (flo + ad + a-ij + a?,k)(bu + bd + b2j + bsk) = ((i(\bd aibi — a>b2 — a%bz) -j- (a^b\ -f- aibo -f- a?.bz — asb2)i + (aob-i + a2bo + a^bi — a\hs)j + (a§bz + a^bo + a\b2 — a2b\)k. This product formula is obtained by multiplying the formal sums term by term sub ject to the following relations: (i) associativity; (ii) ri — ir; rj = jr, rk = kr (for all r s R); (iii) i2 = j2 = k‘l = ijk = — 1; ij = —ji = k; jk = —kj = /'; ki = —ik = j. Under this product A' is a noncommutative division ring in which the multiplicative inverse of a« + ad + a2j + ask is (a0/d) — (a\/d)i — (a2/d)j — (az/d)k, where d = tv* + + a-i2 + ai1. K is called the division ring of real quaternions. The quaternions may also be interpreted as a certain subring of the ring of all 2 X 2 matrices over the field C of complex numbers (Exercise 8). Definition 1.1 shows that under multiplication the elements of a ring R form a semigroup (a monoid if R has an identity). Consequently Definition 1.1.8 is appli cable and exponentiation is defined in R. We have for each a e R and n e N*, an — a- ■ -a (n factors) and a0 = 1* if R has an identity. By Theorem 1.1.9 aman = am+n and (am)n = amn
CHAPTER III RINGS
118
Subtraction in a ring R is defined in the usual way: a — b = a + (~b). Clearly a(Jb — c) = ab — ac and (a — b)c = ac — be for all a,b,c e R. The next theorem is frequently useful in computations. Recall that if k and n are integers with 0 < k < n, then the binomial coefficient
is the number
n\f{n — k)\kl, where 0! = 1 and n\ = n(n — 1 )(n — 2)- ■ -2-1 for n > 1. is
actually an integer (Exercise 10).
Theorem 1.6. (Binomial Theorem). Let R be a ring with identity, n a positive integer, and a,b,ai,a2, . . . , aB e R.
(i) 7/ab = ba, then (a + b)n = Y. ^j^akbn_k; (ii) 7/aiaj = aja; for alii and}, then (ai + a2 -I--- 1- a.,)n = ^ n‘ . a^W*- • - a.,*8, OiO-• ObO where the sum is over all s-tuples (ii,i2,. . . , is) such that ii + i2 + • • • + i8 = n.
SKETCH OF PROOF, (i) Use induction on n and the fact that
j^
I ?”*”! ) f°r k < n (Exercise 10(c)); the distributive law and the commutativity of U+l / a and b are essential, (ii) Use induction on s. The case s = 2 is just part (i) since =
n f yi\
n\
(ai + a2)n = T] I i )a\kaTk = Y. -rr~. afatf. If the theorem is true for s, note (p0 \k/
ktjtn k\j!
that n
{ai + • • • + as + fls+O" = ({(ai + • ■ ■ + as) +
tfs+i)n
= 53 k=
0
_^ ^ I = k
23 777 (fli + ‘' • + as)kaLi by part (i). Apply the induction hypothesis and
=nk
.J .
compute. ■
Definition 1.7. Let R and S be rings. A function f : R —» S is a homomorphism of rings provided that for all a,b e R:
f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b). REMARK. It is easy to see that the class of all rings together with all ring homo morphisms forms a (concrete) category. When the context is clear then we shall frequently write “homomorphism” in place of “homomorphism of rings.” A homomorphism of rings is, in particular, a homomorphism of the underlying additive groups. Consequently the same termi nology is used: a monomorphism [resp. epimorphism, isomorphism] of rings is a homo
1. RINGS AND HOMOMORPHISMS
119
morphism of rings which is an injective [resp. surjective, bijective] map. A mono morphism of rings R —> S is sometimes called an embedding of R in S. An isomor phism R —> R is called an automorphism of R. The kernel of a homomorphism of rings / : R —> S is its kernel as a map of addi tive groups; that is, Ker / = {r e R | f(r) = 0}. Similarly the image of /, denoted Im f is {s e S \ s = f(r) for some r e R}. If R and S both have identities 1B and ls, we do not require that a homomorphism of rings map to Is (see Exercises 15, 16). EXAMPLES. The canonical rnap^Z —>Zm given by k I—* k is an epimorphism of rings. The map Z3 —» Z6 given by k \-+ 4k is a well-defined monomorphism of rings. EXAMPLE. Let G and H be multiplicative groups and / : G —> H a homomor phism of groups. Let R be a ring and define a map on the group rings/: R(G)—*R(H) by: /( 5Zngt) = 5Z rtf(gi). \i=i / ;=i Then / is a homomorphism of rings.
Definition 1.8. Let R be a ring. If there is a least positive integer n such that na = 0 for all a e R, then R is saicl to have characteristic n. If no such n exists R is said to have characteristic zero. (Notation: char R = n).
Theorem 1.9. Let R be a ring with identity 1r and characteristic n > 0.
(i) If (p : Z —> R is the map given by m (—> mlR, then
SKETCH OF PROOF, (ii) If A: is the least positive integer such that klR = 0, then for all a e R: ka — k(\Ra) = (klR)a = 0-a = 0 by Theorem 1.2. (iii) If n — kr with \ < k < n, 1 < r < n, then 0 = n\R = (kr)\R\n = (A: 1 /s)(rl/e) implies that k 1 n = 0 or rl« = 0, which contradicts (ii). ■
Theorem 1.10. Every ring R may be embedded in a ring S with identity. The ring S (which is not unique) may be chosen to be either of characteristic zero or of the same characteristic as R.
SKETCH OF PROOF. Let S.be the additive abelian group R 0) Z and define multiplication in 5 by (ruki)(r2,k2) = (nr-i + k2ri + kir2,kik2), (r, s R; ki e Z).
CHAPTER III RINGS
120
Verify that S is a ring with identity (0,1) and characteristic zero and that the map R —»S given by r H (r,0) is a ring monomorphism (embedding). If char R = n > 0, use a similar proof with S = R ©Z„ and multiplication defined by (ri,ki)(r2,k2) = (ri/-2 + k2ri + k^kih),
where /•; e R and £, e Zn is the image of fcsZ under the canonical map. Then charS = n. ■
EXERCISES 1. (a) Let G be an (additive) abelian group. Define an operation of multiplication in G by ab = 0 (for all a,b e G). Then G is a ring. (b) Let S be the set of all subsets of some fixed set U. For A,B e S, define A + B = (A ~ B) U (B — A) and AB = A fl B. Then S is a ring. Is 5 com mutative? Does it have an identity? 2. Let {Ri | / e /} be a family of rings with identity. Make the direct sum of abelian groups ^ into a ring by defining multiplication coordinatewise. Does ^ R, iel
iel
have an identity? 3. A ring R such that a2 — a for all a e R is called a Boolean ring. Prove that every Boolean ring R is commutative and a + a = 0 for all a e R. [For an example of a Boolean ring, see Exercise 1(b).] 4. Let R be a ring and S a nonempty set. Then the group M(S,R) (Exercise 1.1.2) is a ring with multiplication defined as follows: the product of fge M(S,R) is the function S —> R given by s (-> f(s)g(s). 5. If A is the abelian group Z @ Z, then End A is a noncommutative ring (see page 116). 6. A finite ring with more than one element and no zero divisors is a division ring. (Special case: a finite integral domain is a field.) 7. Let R be a ring with more than one element such that for each nonzero a e R there is a unique be R such that aba = a. Prove: (a) R has no zero divisors. (b) bab = b. (c) R has an identity. (d) R is a division ring. 8. Let R be the set of all 2 x 2 matrices over the complex field C of the form
(-:
'■>
where z,w are the complex conjugates of z and w respectively (that is, c = a + byf^l <=> c = a — byj— 1). Then R is a division ring that is isomorphic to the division ring K of real quaternions. [Hint: Define an isomorphism K —► R by letting the images of 1 ,i,j,k e K be respectively the matrices / 1 0W V^T °\ / 0 1\/ 0 V^T\
Vo lM 0
V -1 oM
0)■
1. RINGS AND HOMOMORPHISMS
121
9. (a) The subset G = {1,— — j,k,— k} of the division ring K of real quaternions forms a group under multiplication. (b) G is isomorphic to the quaternion group (Exercises 1.4.14 and 1.2.3). (c) What is the difference between the ring K and the group ring R(G) (R the field of real numbers)? 10. Let k,n be integers such that 0 < k < n and the binomial coefficient »!/(« — k)\k\, where 0! = 1 and for n > 0, n\ = n(n — 1)(« — 2)- • -2-1. (a)
(;) (:)
(b>W
*
+
(d)
*<«•
is an integer.
(e) if p is prime and 1 < k < pv — 1, then ^ ^ is divisible by p. ["""S;(b) observe , hat
= (”)^;0d) note tha, (“) = (™) = 1
and use induction on n in part (c).] 11. (The Freshman’s Dream1). Let R be a commutative ring with identity of prime characteristic p. If a,b e R, then (a ± by” = a1'" ± bp,‘ for all integers n > 0 [see Theorem 1.6 and Exercise 10; note that b = —b if p = 2], 12. An element of a ring is nilpotent if a’• = 0 for some /?. Prove that in a commuta tive ring a + b is nilpotent if a and b are. Show that this result may be false if R is not commutative. 13. In a ring R the following conditions are equivalent. (a) R has no nonzero nilpotent elements (see Exercise 12). (b) If a s R and a2 = 0, then u = 0. 14. Let R be a commutative ring with identity and prime characteristic p. The map R R given by r (—> rp is a homomorphism of rings called the Frobenius homo morphism [see Exercise 11]. 15. (a) Give an example of a nonzero homomorphism f : R —* S of rings with identity such that /(Ir) ^ Is. (b) 11/ \ R^S is an epimorphism of rings with identity, then /(I.r) = Is(c) If /:/?—> 5 is a homomorphism of rings with identity and u is a unit in R such that f(u) is a unit in 5, then /(I R) = 1 s and f(url) = f(u)~K [Note: there are easy examples which show that f(u) need not be a unit in 5 even though u is a unit in /?.] 16. Let / : R —be a homomorphism of rings such that f(r) ^ 0 for some non zero r c R. If R has an identity and S has no zero divisors, then 5 is a ring with identity /(!«). ’Terminology due to V. O. McBrien.
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CHAPTER III RINGS
17. (a) If R is a ring, then so is R1'. where R r is defined as follows. The underlying set of R"p is precisely R and addition in R“>' coincides with addition in R. Multi plication in R"1’, denoted is defined hv a ° b = ba, where ba is the product in R. is called the opposite ring of R. (b) R has an identity if and only if R ‘‘ does. (c) R is a division ring if and only if R">' is. (d) (R"liy» = R. (e) If 5 is a ring, then R = S if and only if R'r 2= S"r. 18. Let Q be the field of rational numbers and R any ring. If fg : Q —> R are homo morphisms of rings such that /j Z = g j Z, then f = g. [Hint: show that for n s Z (n ^ 0), f(l/nMfi) = g(l), whence /(I, n) = g(\/ri).]
2. IDEALS Just as normal subgroups played a crucial role in the theory of groups, so ideals play an analogous role in the study of rings. The basic properties of ideals are de veloped, including a characterization of principal ideals (Theorem 2.5) and the vari ous isomorphism theorems (2.9-2.13; these correspond to the isomorphism theorems for groups). Prime and maximal ideals are characterized in several ways. Direct products in the category of rings are discussed and the Chinese Remainder Theorem is proved. Definition 2.1. Let R be a ring and S a nonempty subset of R that is closed under the operations of addition and multiplication in R. If S is itself a ring under these operations then S is called a subring ofR. A subring I of a ring R is a left ideal provided
r s R and x e I => rx e I; I is a right ideal provided r £ R and x £ I => xr £ I; I is an ideal if it is both a left and right ideal. Whenever a statement is made about left ideals it is to be understood that the analogous statement holds for right ideals. EXAMPLE. If R is any ring, then the center of R is the set C = {c s R | cr = rc for all r £ /?}. C is easily seen to be a subring of /?, but may not be an ideal (Exer cise 6). EXAMPLE. If / : R—>S is a homomorphism of rings, then Ker /is an ideal in R (Theorem 2.8 below) and Im / is a subring of S. Im / need not be an ideal in S. EXAMPLE. For each integer n the cyclic subgroup (n) = \kn\keZ} is an ideal in Z. EXAMPLE. In the ring R of n X n matrices over a division ring D, let I k be the set of all matrices that have nonzero entries only in column A . Then I k is a left ideal,
2. IDEALS
123
hut not a right ideal. If Jk consist of those matrices with nonzero entries only in row k, then is a right ideal but not a left ideal. EXAMPLE. Two ideals of a ring R are R itself and the trivial ideal (denoted 0), which consists only of the zero element. REMARKS. A [left] ideal / of R such that / ^ 0 and / ^ R is called a proper [left] ideal. Observe that if R has an identity 1R and / is a [left] ideal of R, then / = R if and only if 1 h s /. Consequently, a nonzero [left] ideal / of R is proper if and only if / con tains no units of R; (for if u £ R is a unit and u £ /, then l/f = u~lu £ /). In particular, a division ring D has no proper left (or right) ideals since every nonzero element of D is a unit. For the converse, see Exercise 7. The ring of n X n matrices over a division ring has proper left and right ideals (see above), but no proper (two-sided) ideals (Exercise 9).
Theorem 2.2. A nonempty subset I of a ring R is a left [resp. right] ideal if and only if for all a,b s I and rsR:
(i) a,b d => a — b s I; and (ii) a s I, r s R => ra£l [resp. ar e I]. PROOF. Exercise; see Theorem 1.2.5. ■
Corollary 2.3. Let 1 A, 1 i £ 11 be a family of [left] ideals in a ring R. Then P| Ai is iti
also a [left] ideal.
PROOF. Exercise. ■ Definition 2.4. Let X be a subset of a ring R. Let {A; | i e 1} be the family of all [left] ideals in R which contain X. Then P| A, is called the [left] ideal generated by X. 1*7
This ideal is denoted (X).
The elements of X are called generators of the ideal (A'). If X = {jfi,. . ., xn\, then the ideal (.X) is denoted by (.Vi,.\o,. . . , .v„) and said to be finitely generated. An ideal (.v) generated by a single element is called a principal ideal. A principal ideal ring is a ring in which every ideal is principal. A principal ideal ring which is an integral domain is called a principal ideal domain.2
Theorem 2.5. Let R be a ring a £ R and X C R.
(i) The principal ideal (a) consists of all elements of the form ra -I- as + na + in 2 rjas; (r,s,ri,s; e R; m e N*; and n e Z). i—1 2The
term “principal ideal ring” is sometimes used in the literature to denote what we have called a principal ideal domain.
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CHAPTER III RINGS
(ii) If R has an identity, then (a) =
53 nasi | r;,Si e R; n e N* \. . i=i J (iii) If a is in the center of R, then (a) = {ra + na | r e R, n e Z|. (iv) Ra = {ra | r e R) [resp. aR = {ar | r e R} ] is a left [resp. right) ideal in R (which may not contain a). If R has an identity, then a e Ra and a e aR. (v) If R has an identity and a is in the center of R, then Ra == (a) = aR. (vi) If R has an identity cmdX is in the center of R, then the ideal (X) consists of all finite sums riaL +-••-(- r„an (n s N*; n e R; a; e X). REMARK. The hypothesis of (iii) is always satisfied in a commutative ring. SKETCH OF PROOF OF 2.5. (i) Show that the set m | r as r s s /= ra + as -j- na -j- 53 < i I > Si, i £ R',n e Z; m s N*, i=i J is an ideal containing a and contained in every ideal containing a. Then / = (a), (ii) follows from the facts that ra = ralR, as = 1 pas, and na = n(\Ra) = (n\R)a, with n\n e R. ■ Let Ai,A-h ..., An be nonempty subsets of a ring R. Denote by Ai + A2 -\— • + An the set {ai + a2 + • —j- an | a, e A, for i = 1 , 2 , I f A and B are nonempty subsets of R let AB denote the set of all finite sums {oA + • • ■ + onbn | n e N*; a,- e A; bi e B\. If A consists of a single element a, we write aB for AB. Similarly if B = {b\, we write Ab for AB. Observe that if B [resp. A] is closed under addition, then aB = {ab\bzB\ [resp. Ab = \ab\as. A\). More generally let A^Ao-■ • A n denote the set of all finite sums of elements of the form aia2■ ■ an (oi e A{ for i = 1,2,. . . , n). In the special case when all Ai (1 < i < n) are the same set A we denote AtAn - • ■ A n — A A ■ ■ ■ A by A n . Theorem 2.6. Let A,Ai,A2, . . . , A„, B and C be [left] ideals in a ring R.
(i) Ai -j- A2 + • • • + An and AiA2- ■ ■ A„ are [left] ideals; (ii) (A + B) + C = A + (B + C); (iii) (AB)C = ABC = A(BC); (iv) B(Ai Ac +■■■-(- An) = BAi -f- BA2 BA„; and (Ai -f- A2 + ■ ■ ■ + A„)C = AiC -f- A2C -(-••• + AnC. SKETCH OF PROOF. Use Theorem 2.2 for (i). (iii) is a bit complicated but straightforward argument using the definitions. Use induction to prove (iv) by first showing that A(B + C) = AB + AC and (A + B)C = AC + BC. ■ Ideals play approximately the same role in the theory of rings as normal sub groups do in the theory of groups. For instance, let R be a ring and I an ideal of R. Since the additive group of R is abelian, / is a normal subgroup. Consequently, by Theorem 1.5.4 there is a well-defined quotient group R/I in which addition is given by: (a + /) + (b + /) = (a + b) + /. R/I can in fact be made into a ring.
2. IDEALS
125
Theorem 2.7. Let R be a ring and I an ideal of R. Then the additive quotient group R/I is a ring with multiplication given by
(a + I)(b + I) = ab + I. If R is commutative or has an identity, then the same is true of R/I.
SKETCH OF PROOF OF 2.7. Once we have shown that multiplication in R/I is well defined, the proof that R/I is a ring is routine. (For example, if R has identity lfl, then + / is the identity in R/I.) Suppose a + I = a' + I and b + / = b‘ + /. We must show that ab + / = a'b' + I. Since a's a' + / = a + /, a' = a + i for some i s I. Similarly b' = b + j with j e I. Consequently a'b' = (a + 0(b + j) = ab + ib + aj -j- ij. Since I is an ideal, a'b' — ab — ib + aj + ij e I.
Therefore a'b' + / = ab + / by Corollary 1.4.3, whence multiplication in R/I is well defined. ■ As one might suspect from the analogy with groups, ideals and homomorphisms of rings are closely related.
Theorem 2.8. Iff : R —> S is a homomorphism of rings, then the kernel off is an ideal in R. Conversely if I is an ideal in R, then the map ir : R —» R/I given £>>’ r 1—> r -j— I is an epimorphism of rings with kernel I.
The map ir is called the canonical epimorphism (or projection). PROOF OF 2.8. Ker /is an additive subgroup of R. If * £ Ker /and r e R, then f(rx) = f(r) fix) = /(r)0 = 0, whence rx s Ker / Similarly, xr e Ker / Therefore, Ker /is an ideal. By Theorem 1.5.5 the map ir is an epimorphism of groups with kernel I. Since ir(ab) = ab + 1 = (a + I)(b + /) = Tv(a)-jv(b) for all a,b e R, ir is also an epimorphism of rings. ■ In view of the preceding results it is not surprising that the various isomorphism theorems for groups (Theorems 1.5.6-1.5.12) carry over to rings with normal sub groups and groups replaced by ideals and rings respectively. In each case the desired isomorphism is known to exist for additive abelian groups. If the groups involved are, in fact, rings and the normal subgroups ideals, then one need only verify that the known isomorphism of groups is also a homomorphism and hence an isomor phism of rings. Caution: in the proofs of the isomorphism theorems for groups all groups and cosets are written multiplicatively, whereas the additive group of a ring and the cosets of an ideal are written additively.
Theorem 2.9. Iff: R —> S is a homomorphism of rings and\ is an idea/ of R which is contained in the kernel off, then there is a unique homomorphism of rings f : R/I —> S such that f(a + I) = f(a) for all asR./mf = Im f and Ker f = {Ker f)/I. f is an iso morphism if and only iff is an epimorphism and I = Ker f.
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PROOF. Exercise; see Theorem 1.5.6. ■
Corollary 2.10. (First Isomorphism Theorem) Iff : R —> S is a homomorphism of rings, then f induces an isomorphism of rings K/Ker f = Im f.
PROOF. Exercise; see Corollary 1.5.7. ■
Corollary2.11. Iff:R-> S is a homomorphism of rings, I is an ideal in R and J is an ideal in S such that f(I) CZ J, then f induces a homomorphism of rings f : R/I —> S/J, given by a + I f(a) + J. f is an isomorphism if and only i//m f+J = S and f-1(J) CZ I. In particular, iffis an epimorphism such that f(I) = J and Ker f CZ I, then f is an isomorphism.
PROOF. Exercise; see Corollary 1.5.8. ■
Theorem 2.12. Let I and J be ideals in a ring R.
(i) (Second Isomorphism Theorem) There is an isomorphisms of rings 1/(1 fl J) — (i + J)/J; (ii) (Third Isomorphism Theorem) if I CZ J, then J/I is an ideal in R/I and there is an isomorphism of rings (R/I)/(J/I) ~ R/J. PROOF. Exercise; see Corollaries 1.5.9 and 1.5.10. ■
Theorem 2.13. If I is an ideal in a ring R, then there is a one-to-one correspondence between the set of all ideals of R which contain I and the set of all ideals of R/I, given by J (—> J/I. Hence every ideal in R/I is of the form J/I, where J is an ideal of R which contains I.
PROOF. Exercise; see Theorem 1.5.11, Corollary 1.5.12 and Exercise 13. ■ Next we shall characterize in several ways two kinds of ideals (prime and maxi mal), which are frequently of interest.
Definition 2.14. An ideal P in a ring R is said to be prime ifP ^ R and for any ideals A,B in R
A B C P = > A C P or B C P . The definition of prime ideal excludes the ideal R for both historical and technical reasons. Here is a very useful characterization of prime ideals; other characteriza tions are given in Exercise 14.
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2. IDEALS
Theorem 2.15. If P is an ideal in a ring R such that P^R and for all a,b e R
abeP ^ aeP or beP,
(1)
then P is prime. Conversely if P is prime and R is commutative, then P satisfies con dition (1).
REMARK. Commutativity is necessary for the converse (Exercise 9 (b)). PROOF OF 2.15. If A and B are ideals such that AB CZ P and A 0. P, then there exists an element ae A — P. For every beB, ab e AB CZ P, whence aeP or be P. Since a\P, we must have b e P for all beB; that is, B CZ P. Therefore, P is prime. Conversely, if P is any ideal and ab e P, then the principal ideal (ab) is con tained in P by Definition 2.4. If R is commutative, then Theorem 2.5 implies that (ia)(b) CZ (ab), whence (a)(b) CZ P. If P is prime, then either (a) CZ P or (b) CZ P, whence a e P or b e P. ■ EXAMPLES. The zero ideal in any integral domain is prime since ab = 0 if and only if a = 0 or b = 0. If p is a prime integer, then the principal ideal (p) in Z is prime since ab e (p) => p\ab => p\a or p \ b => ae (p) or be (p).
Theorem 2.16. In a commutative ring R with identity Ir j* 0 an ideal P is prime if and only if the quotient ring R/P is an integral domain.
PROOF. R/P is a commutative ring with identity I r + P and zero element 0 + P = P by Theorem 2.7. If P is prime, then 1R + P P since P R. Further more, R/P has no zero divisors since (a + P)(b + P) = P => ab + P = P => ab e P => aeP or beP => a -\- P = P or b + P — P.
Therefore, R/P is an integral domain. Conversely, if R/P is an integral domain, then \r + P 0 + P, whence Ir i P. Therefore, P ^ R. Since R/P has no zero divisors, ab e P => ab + P = P => (a + P)(b + P) = P => a -\- P = P or b + P = P => aeP or b e P.
Therefore, P is prime by Theorem 2.15. ■
Definition 2.17. An ideal [resp. left idealJ M in a ring R is said to be maximal if R and for every ideal [resp. left ideal] N such that M CZ N CZ R, either N = M or N = R.
EXAMPLE. The ideal (3) is maximal in Z; but the ideal (4) is not since (4) CZ (2)
CZ Z.
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REMARK. If R is a ring and S is the set of all ideals I of R such that / 5* R, then Sis partially ordered by set-theoretic inclusion. M is a maximal ideal (Definition 2.17) if and only if M is a maximal element in the partially ordered set S in the sense of Introduction, Section 7. More generally one sometimes speaks of an ideal / that is maximal with respect to a given property, meaning that under the partial ordering of set theoretic inclusion, I is maximal in the set of all ideals of R which have the given property. In this case / need not be maximal in the sense of Definition 2.17.
Theorem 2.18. In a nonzero ring R with identity maximal [left] ideals always exist. In fact every [left] ideal in R (except R itself) is contained in a maximal [left] ideal.
PROOF. Since 0 is an ideal and 0 R, it suffices to prove the second statement. The proof is a straightforward application of Zorn’s Lemma. If A is a [left] ideal in R such that A ^ R, let S be the set of all [left] ideals B in R such that A CZ B R. S is nonempty since A s S. Partially order S by set theoretic inclusion (that is, Bi < B2 <=> Bx CZ B2). In order to apply Zorn’s Lemma we must show that every chain (2 = {C, | / e /} of [left] ideals in S has an upper bound in S. Let C (J C,. iel
We claim that C is a [left] ideal. If a,b e C, then for some i,j e /, a e Ct and h s C,. Since G is a chain, either C* CZ C3 or C, (Z C,; say the latter. Hence a,b e C,. Since C, is a left ideal, a — b e C* and ra s C,- for all r e R (if Ci is an ideal ar s C, as well). Therefore, a,b e C imply a — b and ra are in C< CI C. Consequently, C is a [left] ideal by Theorem 2.2. Since A d Q for every i, A CZ (J C, = C. Since each C, is in S, Cij^R for all /e I. Consequently, 1 r ^ Ci for every i (otherwise C, = R), whence 1r $ (J Ci = C. Therefore, C ^ R and hence, C s S. Clearly C is an upper bound of the chain G. Thus the hypotheses of Zorn’s Lemma are satisfied and hence S contains a maximal element. But a maximal element of S is obviously a maximal [left] ideal in R that contains A. ■
Theorem 2.19. //R is a commutative ring such that R2 = R (in particular ifR has an • identity), then every maximal ideal M in R is prime.
REMARK. The converse of Theorem 2.19 is false. For example, 0 is a prime ideal in Z, but not a maximal ideal. See also Exercise 9. PROOF OF 2.19. Suppose ab e M but a \ M and b If. M. Then each of the ideals M + (a) and M -f (b) properly contains M. By maximality M + (a) = R = M + (b). Since R is commutative and ab e M, Theorem 2.5 implies that (a)(b) (Z (ab) (Z M. Therefore, R = R2 = (M + (a))(M f (b)) (Z M'1 (a)M + M(b) + (a)(b) CZ M. This contradicts the fact that M ^ R (since M is maximal). Therefore, a e M or b s M, whence M is prime by Theorem 2.15. ■ Maximal ideals, like prime ideals, may be characterized in terms of their quotient rings.
Theorem 2.20. Let M be an ideal in a ring R with identity 1r ?£ 0.
2. IDEALS
129
(i) If M is maximal and R is commutative, then the quotient ring R/M is a field. (ii) If the quotient ring R/M is a division ring, then M is maximal. REMARKS, (i) is false if R does not have an identity (Exercise 19). If Mis maxi mal and R is not commutative, then R/M need not be a division ring (Exercise 9). PROOF OF 2.20. (i) If M is maximal, then M is prime (Theorem 2.19), whence R/M is an integral domain by Theorem 2.16. Thus we need only show that if a + M M, then a + M has a multiplicative inverse in R/M. Now a + M ^ M implies that a\M, whence M is properly contained in the ideal M + (a). Since M is maximal, we must have M + (a) = R. Therefore, since R is commutative, 1« = m + ra for some me M and r e R, by Theorem 2.5(v). Thus 1R — ra = m e M, whence 1R + M = ra + M = (r + M)(a + M). Thus r + M is a multiplicative inverse of a + M in R/M, whence R/M is a field, (ii) If R/M is a division ring, then lfl + M 0 + M, whence 1 R § M and R. If N is an ideal such that M (Z N, let a e N — M. Then a + M has a multi plicative inverse in R/M, say (a + M)(b + M) = 1R + M. Consequently, ab + M = 1R + M and ab — \R = c s M. But an N and M CL N imply that 1« e N. Thus N = R. Therefore, M is maximal. ■
Corollary 2.21. The following conditions on a commutative ring R with identity 1r 5^ 0 are equivalent.
(i) (ii) (iii) (iv)
R is a field; R has no proper ideals; 0 is a maximal ideal in R; every nonzero homomorphism of rings R —> S is a monomorphism.
REMARK. The analogue of Corollary 2.21 for division rings is false (Exercise 9). PROOF OF 2.21. This result may be proved directly (Exercise 7) or as follows. R = R/0 is a field if and only if 0 is maximal by Theorem 2.20. But clearly 0 is maxi mal if and only if R has no proper ideals. Finally, for every ideal I(^R) the canonical map 7r: /? —► R/I is a nonzero homomorphism with kernel I (Theorem 2.8). Since tt is a monomorphism if and only if / = 0, (iv) holds if and only if R has no proper ideals. ■ We now consider (direct) products in the category of rings. Their existence and basic properties are easily proved, using the corresponding facts for groups. Co products of rings, however, are decidedly more complicated. Furthermore co products in the category of rings are of less use than, for example, coproducts (direct sums) in the category of abelian groups.
Theorem 2.22. Let
{ I Ri
i e I) be a nonempty family of rings and
product of the additive abelian groups Rij
n itl
Ri the direct
CHAPTER III RINGS
130
o) n Ri is a ring with multiplication defined by {aijia {bi}»i = {aibi} iei;
iel
(ii) //Ri has an identity [resp. is commutative] for every iel, then H Rj has an iel
identity [resp. is commutative]', (iii) for each k e I the canonical projection 7Tki XT Ri —» Rk given by {a; \ |-> ak, is ie.1
cm epimorphism of rings; (iv) for each k e I the canonical injection tk : Rk —» XT Ri, given by ak H> {a;} iel
(where a, = 0 for i 9^ k), is a monomorphism of rings. PROOF. Exercise. | T Ri is called the (external) direct product of the family of rings {Ri | /e 7}. If the iel
index set is finite, say I = {1, , then we sometimes write Ri X Rz X • • ■ X R„ instead of XX If {Ri | iel] is a family of rings and for each i e I, Ai is an ideal in Ri, then it is easy to see that n Ai is an ideal in XX If A = 0 for all i ^ k, then the ideal
n
iel
iel
Ai is precisely ik(Ak). If the index set 7 is finite and each R{ has an identity, then
iel
every ideal in XX Ri is of the form XX ^ with an ideal R* (Exercise 22). iel
iel
Theorem 2.23. Let {R; | i e l ) be a nonempty family of rings, S a ring and {v?i : S — > Rj | i e I} a family of homomorphisms of rings. Then there is a unique homo morphism of rings
n
n
iel
iel
uniquely determined up to isomorphism by this property. In other words XX Ri
a
iti
product in the category of rings.
SKETCH OF PROOF. By Theorem 1.8.2 there is a unique homomorphism of groups
homomorphism. Thus XX a product in the category of rings (Definition 1.7.2) iel
and therefore determined up to isomorphism by Theorem 1.7.3. ■
Theorem 2.24. Let Ai,A2, ... ,An be ideals in a ring R such that (i) Ai + A2 + • ■ + An = R and (ii) for each k (1 < k < n), Ak fl (Ai + ■ • • + Ak-i + Ak+i H------1- An) = 0. Then there is a ring isomorphism R = Ai X A2 X • • • X An.
PROOF. By the proof of Theorem 1.8.6 the map : Ai X A2 X ■ ■ ■ X An —> R given by (au - - - , an) Oi + «2 H---- h an is an isomorphism of additive abelian groups. We need only verify that
2. IDEALS
131
If R is a ring and Au . . . , An are ideals in R that satisfy the hypotheses of Theo rem 2.24, then R is said to be the (internal) direct product of the ideals Ai. As in the case of groups, there is a distinction between internal and external direct products. If a ring R is the internal direct product of ideals Au . . ., A„, then each of the Ai is actually an ideal contained in R and R is isomorphic to the external direct product Ax X ■ • • X An. However, the external direct product A, X ■■■ X An does not contain the Ai, but only isomorphic copies of them (namely the — see Theorem 2.22). Since this distinction is unimportant in practice, the adjectives “internal” and “external” will be omitted whenever the context is clear and the following notation will be used. NOTATION. We write R — XT A, or R = AxX A2 X ■ • • X An to indicate that the ring R is the internal direct product of its ideals Au ..., A n . Other characterizations of finite direct products are given in Exercise 24. We close this section with a result that will be needed in Chapters VIII and IX. Let A be an ideal in a ring R and a,b e R. The element a is said to be congruent to b modulo A (denoted a = b (mod A)) if a — be.*4. Thus a = b (mod A) <=> a — be A <=> a + A = b + A.
Since R/A is a ring by Theorem 2.7, a\ - a2 (mod A) and b\ = b2 (mod A) =>
fli + bi = a2 b2 (mod A) and axbx = a2b2 (mod A).
Theorem 2.25. (Chinese Remainder Theorem) Let Ai, . . . , An be ideals in a ring R such that R2 + Ai = R for all i and A; + Aj = R for all i ^ j. //bi,.. ., b„ e R, then there exists b e R such that
b = bi (mod Ai) (i = 1,2,..., n). Furthermore b is uniquely determined up to congruence modulo the ideal
Ai n a2 n - - - n a„. REMARK. If R has an identity, then R2 = R, whence R2 + A = R for every ideal A of R. SKETCH OF PROOF OF 2.25. Since Ai + A2 = R and Ay + A3 - R, R 2 = ( A \ -J- A 2) ( A i -f- A s ) = A i2 -f- A 1A 3 -f- A 2A \ -)- A 2A s
<= A, + A2Az C A + (A2 fl As). Consequently, since R = Ai + R2, R = Ai R2 Ai (Ai + (A2 PI ^3)) = A, + (A2 ft A3) CZ R.
Therefore, R = Ai + (A2 fl A3). Assume inductively that r = A\ -j- (a2 n a3 n • - - n a^i).
Then
CHAPTER III RINGS
132 r
2
= (Ai + (a2 n n A/c-i))(Ai + Atc) a. ai + (/^2 n As n• ■ ■ n a)
and hence R = R2 + A, d.A, + (A 2 D • • ■ fl A k ) d R
Therefore, R — A x -f- (A 2 D • ■ ■ fl A k ) and the induction step is proved. Con sequently, R = Ai + (A 2 fl • • - fl Aj) = A\ + (H A). A similar argument jVI
shows that for each k = 1,2, ...,«,/? = >4* + (H /I,-). Consequently for each A: i^k
there exist elements a* e Ak and r k e O A such that = a* + r*. Furthermore rk = (mod A) and rk = 0 (mod A{) for / ^ A.
Let b — ri + r2 + • • ■ + r„ and use the remarks preceding the theorem to verify that b = bi (mod At) for evary Finally if c e R is such that c = bi (mod Ai) for every /, 71
then b = c (mod Ai) for each i, whence b — c e A, for all i. Therefore, b — c e (~) A{ i—1
and b = c ^mod p|
■
The Chinese Remainder Theorem is so named because it is a generalization of the following fact from elementary number theory, which was known to Chinese mathe maticians in the first century A.D.
Corollary 2.26. Let m1,m2,. . . , mn be positive integers such that (mi,mj) = 1 for i ^ j. If bi,b2,. . . , bn are any integers, then the system of congruences
x = bi (mod mi); x = b2 (mod m2); ; x = b„ (mod m„) has an integral solution that is uniquely determined modulo m = mim 2 --m n . 71
SKETCH OF PROOF. Let Ax = (/«,); then fl = (m). Show that i=i (mi,m,) = 1 implies Ai + A, = Z and apply Theorem 2.25. ■
Corollary 2.27. If A,,..., A n are ideals in a ring R, then there is a monomorphism of rings e : R/(Ai D - - ■
n A„) -» R/A, X R/A2 X • • • X R/A„.
If R2 + A, = R for all i and A; + Aj = R for all i ^ j, then 6 is an isomorphism of rings.
SKETCH OF PROOF. By Theorem 2.23 the canonical epimorphisms tt k : R—> R/A k (k = 1,...,«) induce a homomorphism of rings 6i : R—> R/Ai X • • • X R/A n with 6i(r) = (r + Au ■ - -, r + A„). Clearly ker 0i = A fl • ■ • ft A„. Therefore, 0i induces a monomorphism of rings 6 : R/(A\ fl • • • fl A n ) —> R/A j X - • • X R/A n (Theorem 2.9). The map 0 need not be surjective (Exercise 26). However, if the hypotheses of Theorem 2.25 are satisfied and (b\ + A, . . . , b n + A) e R/Ai
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X • • • X R/A n , then there exists b e R such that b = bi (mod At) for all i. Thus 6(b + Q Ai) = (b + A u .. . , b + An) = (bi + A u . . ., bn + An), whence 6 is an i
epimorphism. ■
EXERCISES 1. The set of all nilpotent elements in a commutative ring forms an ideal [see Exercise 1.12], 2. Let I be an ideal in a commutative ring R and let Rad I = {r e R \ rn s / for some n}. Show that Rad / is an ideal. 3. If R is a ring and a e R, then J = [r K = {r e R \ ar = 0} is a right ideal in R.
e
R | ra
=
0} is a left ideal and
4. If / is a left ideal of R, then A(I) = {r e R \ rx = 0 for every x e /} is an ideal in R. 5. If / is an ideal in a ring R, let [/? : /] = {r s R \ xr e / for every x e R |. Prove that [/? : /] is an ideal of R which contains I.
6. (a) The center of the ring S of all 2 X 2 matrices over a field F consists of all matrices of the form (^ \l) a (b) The center of S is not an ideal in S. (c) What is the center of the ring of all n X n matrices over a division ring? 7. (a) A ring R with identity is a division ring if and only if R has no proper left ideals. [Proposition 1.1.3 may be helpful.] (b) If S is a ring (possibly without identity) with no proper left ideals, then either S2 = 0 or 5 is a division ring. [Hint: show that {a zS | Sa = 0} is an ideal. If cd 0, show that {r e S' | rd — 0] =0. Find e eS such that ed — d and show that e is a (two-sided) identity.] 8. Let R be a ring with identity and S the ring of all n X n matrices over R. J is an ideal of S if and only if J is the ring of all n X n matrices over I for some ideal / in R. [Hint: Given J, let / be the set of all those elements of R that appear as the row 1-column 1 entry of some matrix in J. Use the matrices where 1
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11. If A'' is the ideal of all nilpotent elements in a commutative ring R (see Exercise 1), then R/N is a ring with no nonzero nilpotent elements. 12. Let R be a ring without identity and with no zero divisors. Let S be the ring whose additive group is R X Z as in the proof of Theorem 1.10. Let A = } (r,n) e S | rx + nx = 0 for every x e R ]. (a) A is an ideal in 5. (b) S/A has an identity and contains a subring isomorphic to R. (c) S/A has no zero divisors. 13. Let / : R —* S be a homomorphism of rings, I an ideal in R, and J an ideal in S. (a) f~\J) is an ideal in R that contains Ker /. (b) If /is an epimorphism, then /(/) is an ideal in S. If /is not surjective, /(/) need not be an ideal in 5. 14. If P is an ideal in a not necessarily commutative ring R, then the following con ditions are equivalent. (a) P is a prime ideal. (b) If r,s e R are such that rRs CZ P, then r e P or s e P. [Hint: If (a) holds and rRs CZ P, then (RrR)(RsR) CZ P, whence RrR CZ P or RsR CZ P, say RrR CZ P. If A = (r), then A3 CZ RrR CZ P, whence re A Cl P.] (c) If (r) and (s) are principal ideals of R such that (/•)($) CZ P, then reP or s e P. (d) If U and V are right ideals in R such that UV CZ P, then U CZ P or V CZ P. (e) If U and V are left ideals in R such that UV CZ P, then U CZ P or V CZ P. 15. The set consisting of zero and all zero divisors in a commutative ring with identity contains at least one prime ideal. 16. Let R be a commutative ring with identity and suppose that the ideal A of R is contained in a finite union of prime ideals P\ U ■ • • U Pn. Show that A CZ Pi for some i. [Hint: otherwise one may assume that A D Pj jzf |J Pi for all j. Let i^j cij e (A fl Pj) — (IJ Pi). Then ai + g2«3- ■ -an is in A but not in Pj U - - - U Pn.] 17. Let / : R —>S be an epimorphism of rings with kernel K. (a) If P is a prime ideal in R that contains K, then f(P) is a prime ideal in S [see Exercise 13]. (b) If Q is a prime ideal in 5, then f~l(Q) is a prime ideal in R that contains K. (c) There is a one-to-one correspondence between the set of all prime ideals in R that contain K and the set of all prime ideals in S, given by P |—> f(P). (d) If I is an ideal in a ring R, then every prime ideal in R/I is of the form P/1, where P is a prime ideal in R that contains /. 18. An ideal M R in a commutative ring R with identity is maximal if and only if for every r e R — M, there exists x e R such that 1 r — rx e M. 19. The ring E of even integers contains a maximal ideal M such that E/M is not a field. 20. In the ring Z the following conditions on a nonzero ideal I are equivalent: (i) I is prime; (ii) I is maximal; (iii) I = (p) with p prime. 21. Determine all prime and maximal ideals in the ringZ™.
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22. (a) If Ru . . . , Rn are rings with identity and I is an ideal in Ri X ■ ■ • X R„, then I - Ai X • • ■ X Am, where each A{ is an ideal in Ri. [Hint: Given / let Ak = irk(I), where irk'.R\ X - • - X R n —> R k is the canonical epimorphism.] (b) Show that the conclusion of (a) need not hold if the rings Ri do not have identities. 23. An element e in a ring R is said to be idempotent if c2 = e. An element of the center of the ring R is said to be central. If e is a central idempotent in a ring R with identity, then (a) lR — e is a central idempotent; (b) eR and (lj? — e)R are ideals in R such that R = eR X (Ib — e)R. 24. Idempotent elements eu . . . , en in a ring R [see Exercise 23] are said to be orthogonal if eiei = 0 for i j. If R, Ru . .., Rn are rings with identity, then the following conditions are equivalent: (a) R s Ri X • • • X Rn. (b) R contains a set of orthogonal central idempotents [Exercise 23] | eu . . . , en } such that ex + e2 + • — en = 1 * and etR = Ri for each i. (c) R is the internal direct product R = A\ X ■ ■ • X An where each Ai is an ideal of R such that A{ ^ Ri. [Hint: (a) => (b) The elements ex = (1^,0, . . . , 0), e2 = (0,1k2,0, . .., 0),. .., en = (0,..., O.lflJ are orthogonal central idempotents in S = Ri X • • • X Rn such that ex + • • • + en — Is and &S =. Ri. (b) => (c) Note that Ak = ekR is the principal ideal (ek) in R and that ekR is itself a ring with identity ek.] 25. IfmeZ has a prime decomposition m = pihl- - -ptkt (kt > 0; pi distinct primes), then there is an isomorphism of rings Z„, = Zp,fci X • • • X ZPlkt. [Hint: Corollary 2.27.] 26. If R — Z, Ai — (6) and A2 = (4), then the map 6 :R/A X fl A 2 —+R/A X X R/A 2 of Corollary 2.27 is not surjective.
3. FACTORIZATION IN COMMUTATIVE RINGS In this section we extend the concepts of divisibility, greatest common divisor and prime in the ring of integers to arbitrary commutative rings and study those integral domains in which an analogue of the Fundamental Theorem of Arithmetic (Intro duction, Theorem 6.7) holds. The chief result is that every principal ideal domain is such a unique factorization domain. In addition we study those commutative rings in which an analogue of the division algorithm is valid (Euclidean rings).
Definition 3.1. A nonzero element a of a commutative ring R is said to divide an element b e R (notation: a | b) if there exists x e R such that ax = b. Elements a,b of R are said to be associates if a | b and b | a.
Virtually all statements about divisibility may be phrased in terms of principal ideals as we now see.
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Theorem 3.2. Let a,b and u be elements of a commutative ring R with identity.
(i) a | b if and only //(b) CZ (a). (ii) a and b are associates if and only if (a) = (b). (iii) u is a unit if and only if u | r for all r e R. (iv) u is a unit if and only if( u) = R. (v) The relation “a is an associate ofb" is an equivalence relation on R. (vi) If a — br with r s R a unit, then a and b are associates. If R is an integral domain, the converse is true. PROOF. Exercise; Theorem 2.5(v) may be helpful for (i) and (ii). ■
Definition 3.3. Let R be a commutative ring with identity. An element c o/R is irreducible provided that:
(i) c is a nonzero nonunit; (ii) c = ab => a or b is a unit. An element p o/R is prime provided that:
(i) p is a nonzero nonunit; (ii) p | ab => p | a or p | b. EXAMPLES. If p is an ordinary prime integer, then both p and —p are irre ducible and prime in Z in the sense of Definition 3.3. In the ringZ6, 2 is easily seen to be a prime. However 2 eZ6 is not irreducible since 2 = 2-4 and neither 2 nor 4 are units inZ6 (indeed they are zero divisors). For an example of an irreducible element which is not prime, see Exercise 3. There is a close connection between prime [resp. irreducible] elements in a ring R and prime [resp. maximal] principal ideals in R.
Theorem 3.4. Let p and c be nonzero elements in an integral domain R.
(i) p is prime if and only if( p) is nonzero prime ideal; (ii) c is irreducible if and only //(c) is maximal in the set S of all proper principal ideals ofR. (iii) Every prime element of R is irreducible. (iv) If R is a principal ideal domain, then p is prime if and only if p is irreducible. (v) Every associate of an irreducible [resp. prime] element of R is irreducible [resp. prime]. (vi) The only divisors of an irreducible element of R are its associates and the units of R. REMARK. Several parts of Theorem 3.4 are true for any commutative ring with identity, as is seen in the following proof. SKETCH OF PROOF OF 3.4. (i) Use Definition 3.3 and Theorem 2.15. (ii) If c is irreducible then (c) is a proper ideal of R by Theorem 3.2. If (c) C (d), then
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c = dx. Since c is irreducible either d is a unit (whence (d) — R) or x is a unit (whence (c) = (d) by Theorem 3.2). Hence (c) is maximal in S. Conversely if (c) is maximal in S, then c is a (nonzero) nonunit in R by Theorem 3.2. If c — ab, then (c) CZ (a), whence (c) = (a) or (a) = R. If (a) = R, then a is a unit (Theorem 3.2). If (c) = (a), then a = cy and hence c = ab = cyb. Since R is an integral domain 1 = yb, whence b is a unit. Therefore, c is irreducible, (iii) If p = ab, then p | a or p | b; say p | a. Then px = a and p = ab = pxb, which implies that 1 = xb. Therefore, b is a unit, (iv) If p is irreducible, use (ii), Theorem 2.19 and (i) to show that p is prime, (v) If c is irreducible and d is an associate of c, then c = du with we/? a unit (Theorem 3.2). If d — ab, then c — abu, whence a is a unit or bu is a unit. But if bu is a unit, so is b. Hence d is irreducible, (vi) If c is irreducible and a \ c, then (c) CZ (a), whence (c) = (a) or (a) = R by (ii). Therefore, a is either an associate of c or a unit by Theorem 3.2. ■
We have now developed the analogues in an arbitrary integral domain of the concepts of divisibility and prime integers in the ring Z. Recall that every element in Z is a product of a finite number of irreducible elements (prime integers or their negatives) according to the Fundamental Theorem of Arithmetic (Introduction, Theorem 6.7). Furthermore this factorization is essentially unique (except for the order of the irreducible factors). Consequently, Z is an example of:
Definition 3.5. An integral domain R is a unique factorization domain provided that:
(i) every nonzero nonunit element a of R can be written a = CiC2- ■ cn, with Ci, . . . , cn irreducible. (ii) If a = CiC-2- • -c„ and a = did2- • dm (Ci,dj irreducible), then n = m and for some permutation o of{ 1,2,. . ., nj, Ci and dP(i) are associates for every i. REMARK. Every irreducible element in a unique factorization domain is neces sarily prime by (ii). Consequently, irreducible and prime elements coincide by Theorem 3.4 (iii). Definition 3.5 is nontrivial in the sense that there are integral domains in which every element is a finite product of irreducible elements, but this factorization is not unique (that is, Definition 3.5 (ii) fails to hold); see Exercise 4. Indeed one of the historical reasons for introducing the concept of ideal was to obtain some sort of unique factorization theorems (for ideals) in rings of algebraic integers in which factorization of elements was not necessarily unique; see Chapter VIII. In view of the relationship between irreducible elements and principal ideals (Theorem 3.4) and the example of the integers, it seems plausible that every principal ideal domain is a unique factorization domain. In order to prove that this is indeed the case we need:
Lemma 3.6. IfR is a principal ideal ring and (ai) C (a2) CZ - • • is a chain of ideals in R, then for some positive integer n, (aj) = (an) for all j > n.
CHAPTER III RINGS
138
PROOF. Let A = U (a,-). We claim that A is an ideal. If b,c e A, then b e (a,) i>\ and c e (a,). Either / < j or / > j; say / > j. Consequently (a,)
Theorem 3.7. Every principal ideal domain R is a unique factorization domain.
REMARK. The converse of Theorem 3.7 is false. For example the polynomial ring Z[x\ can be shown to be a unique factorization domain (Theorem 6.14 below), but Z[x] is not a principal ideal domain (Exercise 6.1). SKETCH OF PROOF OF 3.7. Let S be the set of all nonzero nonunit ele ments of R which cannot be factored as a finite product of irreducible elements. We shall first show that S is empty, whence every nonzero nonunit element of R has at least one factorization as a finite product of irreducibles. Suppose S is not empty and a e 5. Then (a) is a proper ideal by Theorem 3.2(iv) and is contained in a maximal ideal (c) by Theorem 2.18. The element c e R is irreducible by Theorem 3.4(ii). Since (a) CZ (c), c divides a. Therefore, it is possible to choose for each a e S an irreducible divisor ca of a (Axiom of Choice). Since R is an integral domain, ca uniquely deter mines a nonzero xa e R such that caxa = a. We claim that xa e S. For if xa were a unit, then a = caxa would be irreducible by Theorems 3.2(vi) and 3.4(v). If xa is a non unit and not in S, then x„ has a factorization as a product of irreducibles, whence a also does. Since aeS this is a contradiction. Hence xa eS. Furthermore, we claim that the ideal (a) is properly contained in the ideal (xa). Since xa \ a, (a) CZ (xa) by Theorem 3.2(i). But (a) = (xa) implies that xa = ay for some y s R, whence a = xaca — ayca and 1 = yca. This contradicts the fact that ca is irreducible (and hence a nonunit). Therefore (a) Cl (xa). The preceding remarks show that the function f :S —>S given by f(a) — xa is well defined. By the Recursion Theorem 6.2 of the Introduction (with /= fn for all n) there exists a function
If we denote
@2
> ^n+1
***■
Consequently, the preceding paragraph shows that there is an ascending chain of ideals (a) C= (ai) (Z (a2) C (a,) 9 • • •,
contradicting Lemma 3.6. Therefore, the set 5 must be empty, whence every nonzero nonunit element in R has a factorization as a finite product of irreducibles.
3. FACTORIZATION IN COMMUTATIVE RINGS
139
Finally if cxc2- ■ -cn = a = dxd2- ■ dm (c„ N such that:
(i) if a,b e R and ab 0, then (p{a) < $?(ab); (ii) ifa,b e R and b ^ 0, then there exist q,r e R such that a = qb + r with r = 0, or r 0 and
Theorem 3.9. Every Euclidean ring R is a principal ideal ring with identity. Con sequently every Euclidean domain is a unique factorization domain.
REMARK. The converse of Theorem 3.9 is false since there are principal ideal domains that are not Euclidean domains (Exercise 8). PROOF OF 3.9. If / is a nonzero ideal in R, choose as. I such that
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Definition 3.10. Let X be a nonempty subset of a commutative ring R. An element deRtffl greatest common divisor ofX provided:
(i) d | a for all a e X; (ii) c | a for all a e X => c | d. Greatest common divisors do not always exist. For example, in the ring E of even integers 2 has no divisors at all, whence 2 and 4 have no (greatest) common divisor. Even when a greatest common divisor of a i , , a n exists, it need not be unique. However, any two greatest common divisors of X are clearly associates by (ii). Furthermore any associate of a greatest common divisor of X is easily seen to be a greatest common divisor ofX. If R has an identity and a t ,a 2 ,. • •, a„ have 1* as a greatest common divisor, then ax,a2 ,... a„ are said to be relatively prime. Theorem 3.11. Let ai,.. ., a„ be elements of a commutative ring R with identity.
(i) d e R is a greatest common divisor of {ai, . . . , an| such that d = riai H--- 1- rnan for some r; e R if and only if (d) = (at) + (a2) H---- (- (an); (ii) ifK is a principal ideal ring, then a greatest common divisor of ai,. . . , an exists and every one is of the form nai + • • • + rnan (rj e R); (iii) if R is a unique factorization domain, then there exists a greatest common divisor of ai,..., a„. REMARK. Theorem 3.1 l(i) does not state that every greatest common divisor of ai,... ,a n is expressible as a linear combination of a\,. . . , a„. In general this is not the case (Exercise 6.15). See also Exercise 12. SKETCH OF PROOF OF 3.11. (i) Use Definition 3.10 and Theorem 2.5. (ii) follows from (i). (iii) Each a, has a factorization: a,.r= • - • c™" withe,,... ,c, distinct irreducible elements and each ma > 0. Show that d = cik,c2kl- ■ ctkl is a greatest common divisor of a\,. . ., an, where kj = min {mi„m2„m33, . . ., m n i }. ■
EXERCISES 1. A nonzero ideal in a principal ideal domain is maximal if and only if it is prime. 2. An integral domain R is a unique factorization domain if and only if every non zero prime ideal in R contains a nonzero principal ideal that is prime. 3. Let R be the subring {a 4- b^l 0 | a,b e Z} of the field of real numbers. (a) The map N \ R — * Z . given by a + b^lO l—> (a + b'\[l0)(a — b^lO) = a2 — 10b2 is such that N(uv) = N(u)N(v) for all u,v e R and N(u) = 0 if and only if u = 0. (b) u is a unit in R if and onlyjf N(u) = ± 1. (c) 2, 3, 4 + VTO and 4 — ^10 are irreducible elements of R. (d) 2, 3, 4 + yj 10 and 4 — yjlO are not prime elements of R. [Hint: 3-2 = 6 = ^4 + Vl0)(4 - \10)-] 4. Show that in the integral domain of Exercise 3 every element can be factored into a product of irreducibles, but this factorization need not be unique (in the sense of Definition 3.5 (ii)).
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141
5. Let R be a principal ideal domain. (a) Every proper ideal is a product PiP2• • P„ of maximal ideals, which are uniquely determined up to order. (b) An ideal P in R is said to be primary if ab e P and a 4 P imply bn e P for some n. Show that P is primary if and only if for some n,P = where p e R is prime (= irreducible) or p = 0. (c) If Pi,P2, . . . , P„ are primary ideals such that Pi = (p?‘)and the p, are distinct primes, then PXP2- ■ Pn = Px fi P2 fi •• • fi Pn. (d) Every proper ideal in R can be expressed (uniquely up to order) as the intersection of a finite number of primary ideals. 6. (a) If a and n are integers, n > 0, then there exist integers q and r such that a = qn + r, where |r| < n/2. (b) The Gaussian integers Z[/] form a Euclidean domain with tp(a + bi) = a2 + b2. [Hint: to show that Definition 3.8(ii) holds, first let y — a + bi and assume x is a positive integer. By part (a) there are integers such that a = qix + rx and b = q2x + r2, with Ini < x/2, |r2| < x/2. Let# = q\ + qdand r = rx + then y = qx + r, with r = 0 or
rk
— qk+\rk+i + fk+2, with r*+2 = 0 or v>(r*+2) <
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CHAPTER III RINGS
Let r0 = b and let n be the least integer such that rn+x = 0 (such an n exists since the
4. RINGS OF QUOTIENTS AND LOCALIZATION In the first part of this section the familiar construction of the field of rational numbers from the ring of integers is considerably generalized. The rings of quotients so constructed from any commutative ring are characterized by a universal mapping property (Theorem 4.5). The last part of this section, which is referred to only oc casionally in the sequel, deals with the (prime) ideal structure of rings of quotients and introduces localization at a prime ideal.
Definition 4.1. A nonempty subset S of a ring R is multiplicative provided that
a,b e S =* ab s S. EXAMPLES. The set S of all elements in a nonzero ring with identity that are not zero divisors is multiplicative. In particular, the set of all nonzero elements in an integral domain is multiplicative. The set of units in any ring with identity is a multiplicative set. If P is a prime ideal in a commutative ring R, then both P and S = R ~ P are multiplicative sets by Theorem 2.15. The motivation for what follows may be seen most easily in the ring Z of integers and the field Q of rational numbers. The set S of all nonzero integers is clearly a multiplicative subset of Z. Intuitively the field Q is thought of as consisting of all fractions a/b with a e Z and b e S, subject to the requirement a/b = c/d <=> ad = be (or ad — be = 0).
More precisely, Q may be constructed as follows (details of the proof will be supplied later). The relation on the set Z X S defined by (a,b) ~ (c,d) <=> ad — be = 0 is easily seen to be an equivalence relation. Q is defined to be the set of equivalence classes of Z X S under this equivalence relation. The equivalence class of (a,b) is denoted a/b and addition and multiplication are defined in the usual way. One verifies that these operations are well defined and that Q is a field. The map Z —■» Q given by al—> a/1 is easily seen to be a monomorphism (embedding). We shall now extend the construction just outlined to an arbitrary multiplicative subset of any commutative ring R (possibly without identity). We shall construct a commutative ring S~'R with identity and a homomorphism
Theorem 4.2. Let S be a multiplicative subset of a commutative ring R. The relation defined on the set R X S by
(r,s) ~ (r',s') <=> si(rs' — r's) = 0 for some Si e S
4. RINGS OF QUOTIENTS AND LOCALIZATION
143
is an equivalence relation. Furthermore if R has no zero divisors and 0 | S, then
(r,s) ~ (ris') <=> rs' — r's = 0. PROOF. Exercise. ■ Let 5 be a multiplicative subset of a commutative ring R and ~ the equivalence relation of Theorem 4.2. The equivalence class of (r,s) e R X S will be denoted r/s. The set of all equivalence classes of R X S under ~ will be denoted by S~*R. Verify that (i) r/s = r'/s' <=> s\{rs' — r's) = 0 for some Si e S', (ii) tr/ts = r/s for all r e R and s,t eS; (iii) If 0 e S, then S~*R consists of a single equivalence class.
Theorem 4.3. Let S be a multiplicative subset of a commutative ring R and let S *R be the set of equivalence classes of R X S under the equivalence relation of Theorem 4.2.
(i) S-1R is a commutative ring with identity, where addition and multiplication are defined by r/s + r'/s' = (rs' + r's)/ss' and (r/s)(r'/s') = rr'/ss'. (ii) 7/R is a nonzero ring with no zero divisors and 0 ^ S, then S_1R is an integral domain. (iii) 7/R is a nonzero ring with no zero divisors and S is the set of all nonzero ele ments of R, then S-1R is a field. SKETCH OF PROOF, (i) Once we know that addition and multiplication in are well-defined binary operations (independent of the choice of r,s,r',s'), the rest of the proof of (i) is routine. In particular, for all s,s' e S, 0/s = 0/s' and 0/s is the additive identity. The additive inverse of r/s is —r/s. For any s,s' zS,s/s = s'/s' and s/s is the multiplicative identity in S~*R. To show that addition is well defined, observe first that since S is multiplicative (rs' + r's)/ss' is an element of S~lR. If r/s = rjsy and r'/s' = r\/s\, we must show that (rs' + r's)/ss' = (ns/ + ri'sO/sis/. By hypothesis there exist s2,s3 e S such that
S_1R
s2(rsi — ns) = 0, s3(r's,' — r/s') — 0. Multiply the first equation by s3s's/ and the second by s2ssL. Add the resulting equa tions to obtain s2s3[(rs' + r's)sisi' — (r,s/ + ri'si)ss'] = 0. Therefore, (rs' + r's)/ss' = (ns/ + r/s^/sis/ (since s-zs3 s 5). The proof that multiplication is independent of the choice of r,s,r',s' is similar. (ii) If R has no zero divisors and 0 $ S, then r/s = 0/s if and only if r = 0 in R. Consequently, (/-/sXr'/s') = Oin S~lR if and only if rr' = Oin R. Since rr* — 0 if and only if r = 0 or r' = 0, it follows that S~lR is an integral domain, (iii) If r 0, then the multiplicative inverse of r/s e S~XR is s/re S~XR. ■
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CHAPTER III RINGS
The ring S_1R in Theorem 4.3 is called the ring of quotients or ring of fractions or quotient ring of R by S. An important special case occurs when 5 is the set of all non zero elements in an integral domain R. Then S~*R is a field (Theorem 4.3(iii)) which is called the quotient field of the integral domain R. Thus if R = Z, the quotient field is precisely the field Q of rational numbers. More generally suppose R is any non zero commutative ring and 5 is the set of all nonzero elements of R that are not zero divisors. If 5 is nonempty (as is always the case if R has an identity), then S~*R is called the complete (or full) ring of quotients (or fractions) of the ring R* Theorem 4.3 (iii) may be rephrased: if a nonzero ring R has no zero divisors, then the complete ring of quotients of R is a field. Clearly the complete ring of quotients of an integral domain is just its quotient field. If
Theorem 4.4. Let S be a multiplicative subset of a commutative ring R.
(i) The map
Theorem 4.5. Let S be a multiplicative subset of a commutative ring R and let T be any commutative ring with identity. Iff : R - > T is a homomorphism of rings such that f(s) is a unit in T for all s e S, then there exists a unique homomorphism of rings f : S ]R - > T such that fips = f- The ring S 'R is completely determined (up to iso morphism) by this property.
SKETCH OF PROOF. Verify that the map / : S~'R -> T given_ by f(r/s) = /(/■) f(s) 1 is a well-defined homomorphism of rings such that ftps — /• If 3For
the noncommutative analogue, see Definition 1X.4.7.
4. RINGS OF QUOTIENTS AND LOCALIZATION
145
g : S~*R —> T is another homomorphism such that gtps = fi then for every s e S, g(
= g(
= /W/Cj)"1 = /(r/s). Therefore, f = g. To prove the last statement of the theorem let C be the category whose objects are all ( fT), where T is a commutative ring with identity and f : R — > T a homomor phism of rings such that f(s) is a unit in T for every s e S. Define a morphism in Q from (/i,7'i) to (fi,T2) to be a homomorphism of rings g :Ti—>Ti such that gfi= fi. Verify that C is a category and that a morphism g in Q (fi,T\) —> (fStTi) is an equiv alence if and only if g : T\ —> 7^ is an isomorphism of rings. The preceding paragraph shows that (
Corollary 4.6. Let R be an integral domain considered as a subring of its quotient field F. 7/E is a field and f : R —> E a monomorphism of rings, then there is a unique monomorphism of fields f : F —> E such that f | R = f. In particular any field Ei con taining R contains an isomorphic copy Fi of F with R CZ Fi CL E^
SKETCH OF PROOF. Let S be the set of all nonzero elements of R and apply Theorem 4.5 to/: R —> E. Then there is a homomorphism / :S~*R — F —*E such that fcps = /. Verify that /is a monomorphism. Since R is identified with (ps(R), this means that /1 R = /. The last statement of the theorem is the special case when / : R —> Ei is the inclusion map. ■ Theorems 4.7-4.11 deal with the ideal structure of rings of quotients. This material will be used only in Section VIII.6. Theorem 4.13, which does not depend on Theorems 4.7-4.11, will be referred to in the sequel. Theorem 4.7. Let S be a multiplicative subset of a commutative ring R.
(i) If I is an ideal in R, then S-1I = {a/s|a s I ; s s S j is an ideal in S-1R. (ii) If 3 is another ideal in R, then S-i(I + J) = ST + S^J; S-KU) = (S_1I)(S_1J);
s-xi n j) = s n s i3. REMARKS. S~'I is called the extension of I in S 'R. Note that r/s e S1/ need not imply that re I since it is possible to have a/s — r/s with as I, r if. I. SKETCH OF PROOF OF 4.7. Use the facts that in S 'R, £ (a/s) n \m
(
m
23 )A’ X! (Ojbj/s) = 53 (a,/s)(bjs/s); and i=1 / y-i y-i c'i
146
CHAPTER III RINGS
CkSlS2• • ■ Sk—lSk+1 ■ • 'St
1^/5152
■St. ■
Theorem 4.8. Let S be a multiplicative subset of a commutative ring R with identity and let I be an ideal of R. Then S 'I = S _ 1 R if and only ifS D I 5^ 0.
PROOF. If seS fl /, then ls-]ie = s/seS-1! and hence S~lI = S-1/?. Con versely, if S1! = S~XR, then
Lemma 4.9. Let S be a multiplicative subset of a commutative ring R with identity and let I be an ideal in R.
(i) I d ps-KS-1!). (ii) If I = <£s ‘(J) for some ideal J in S XR, then S ‘I = J. In other words every ideal in S_1R is of the form S_1I for some ideal I in R. (iii) 7/P is a prime ideal in R and S fl P = 0, then S-1P is a prime ideal in S XR and vs-KS-'V) = P. PROOF, (i) If ae I, then as e I for every seS. Consequently,
Theorem 4.10. Let S be a multiplicative subset of a commutative ring R with identity. Then there is a one-to-one correspondence between the set H of prime ideals of R which are disjoint from S and the set V of prime ideals of S-1R, given by P|—* S-IP.
PROOF. By Lemma 4.9(iii) the assignment P\- + SXP defines an injective map H —> V. We need only show that it is surjective as well. Let J be a prime ideal of S~lR and let P = Since 5_1P = J by Lemma 4.9(ii), it suffices to show that P is prime. If ab e P, then (ps(a)
4. RINGS OF QUOTIENTS AND LOCALIZATION
147
in S lR, either
Theorem 4.11. Let P be a prime ideal in a commutative ring R with identity.
(i) There is a one-to-one correspondence between the set of prime ideals of R which are contained in P and the set of prime ideals of Rp, given by Q |—» Qp; (ii) the ideal Pp in Rp is the unique maximal ideal of Rp. PROOF. Since the prime ideals of J? contained in P are precisely those which are disjoint fromS = R — P, (i) is an immediate consequence of Theorem 4.10. If Mis a maximal ideal of RP, then M is prime by Theorem 2.19, whence M = Qp for some prime ideal Q of R with Q CZ P. But Q d P implies Qp d PP. Since Pp ^ RP by Theorem 4.8, we must have QP = Pp. Therefore, Pp is the unique maximal ideal in RP. ■ Rings with a unique maximal ideal, such as RP in Theorem 4.11, are of some interest in their own right.
Definition 4.12. A local ring is a commutative ring with identity which has a unique maximal ideal.
REMARK. Since every ideal in a ring with identity is contained in some maximal ideal (Theorem 2.18), the unique maximal ideal of a local ring R must contain every ideal of R (except of course R itself). EXAMPLE. If p is prime and n > 1, then Zvn is a local ring with unique maxi mal ideal (p).
Theorem 4.13. If R is a commutative ring with identity then the following conditions are equivalent.
(i) R is a local ring; (ii) all nonunits of R are contained in some ideal M?^R; (iii) the nonunits of R form an ideal. SKETCH OF PROOF. If I is an ideal of R and a e I, then (a) Cl / by Theorem 2.5. Consequently, I 9* Rif and only if I consists only of nonunits (Theorem 3.2(iv)). (ii) ==> (iii) and (iii) => (i) follow from this fact, (i) => (ii) If a e R is a nonunit, then (a) R. Therefore, (a) (and hence a) is contained in the unique maximal ideal of R by the remark after Definition 4.12. ■
148
CHAPTER III RINGS
EXERCISES 1. Determine the complete ring of quotients of the ring Z„ for each n > 2. 2. LetS be a multiplicative subset of a commutative ring R with identity and let Tbe a multiplicative subset of the ring S~lR. Let S* = {r e R | r/s e T for some s e £]. Then S* is a multiplicative subset of R and there is a ring isomorphism S* \R £* T^S'R). 3. (a) The set E of positive even integers is a multiplicative subset of Z such that E~\Z) is the field of rational numbers. (b) State and prove condition(s) on a multiplicative subset S of Z which insure that S~lZ is the field of rationals. 4. If 5 = {2,4 J and R = Z6, thenS-1/? is isomorphic to the field Z3. Consequently, the converse of Theorem 4.3(ii) is false. 5. Let R be an integral domain with quotient field F. If T.is an integral domain such that R CZ T CZ F, then F is (isomorphic to) the quotient field of T. 6. Let S be a multiplicative subset of an integral domain R such that 0 ^ S. If R is a principal ideal domain [resp. unique factorization domain], then so is S~lR. 7. Let Ri and R2 be integral domains with quotient fields Fi and F2 respectively. If / : /?i —> /?2 is an isomorphism, then / extends to an isomorphism Fx = F2. [Hint: Corollary 4.6.] 8. Let R be a commutative ring with identity, / an ideal of R and ir: R R/I the canonical projection. (a) If S is a multiplicative subset of R, then ttS = 7r(S) is a multiplicative subset of R/1. (b) The mapping 6 :S~'R —»(irS)~l(R/1) given by r/ih ir(r)/ir(s) is a welldefined function. (c) 8 is a ring epimorphism with kernel S~'I and hence induces a ring iso morphism S-'R/S-'I (ttS)-'(R/I). 9. Let 5 be a multiplicative subset of a commutative ring R with identity. If I is an ideal in R, then S-1(Rad /) = Rad (S-1/). [See Exercise 2.2.] 10. Let R be an integral domain and for each maximal ideal M (which is also prime, of course), consider RM as a subring of the quotient field of R. Show that f"l Rm — R, where the intersection is taken over all maximal ideals M of R. 11. Let p be a prime in Z; then (p) is a prime ideal. What can be said about the rela tionship of Zp and the localization Z(p)? 12. A commutative ring with identity is local if and only if for all r,s e R, r + s = 1« implies r or s is a unit. 13. The ring R consisting of all rational numbers with denominators not divisible by some (fixed) prime p is a local ring. 14. If M is a maximal ideal in a commutative ring R with identity and n is a positive integer, then the ring R/M n has a unique prime ideal and therefore is local. 15. In a commutative ring R with identity the following conditions are equivalent: (i) R has a unique prime ideal; (ii) every nonunit is nilpotent (see Exercise 1.12);
5. RINGS OF POLYNOMIALS AND FORMAL POWER SERIES
149
(iii) R has a minimal prime ideal which contains all zero divisors, and all non units of R are zero divisors. 16. Every nonzero homomorphic image of a local ring is local.
5. RINGS OF POLYNOMIALS AND FORMAL POWER SERIES We begin by defining and developing notation for polynomials in one indeter minate over a ring R. Next the ring of polynomials in n indeterminates over R is defined and its basic properties are developed. The last part of the section, which is not needed in the sequel, is a brief introduction to the ring of formal power series in one indeterminate over R.
Theorem 5.1. Let R be a ring and let /?[x] denote the set of all sequences of elements of R (a0,ai,. . .) such that a; = 0 for all but a finite number of indices i.
(i) R[x] is a ring with addition and multiplication defined by: (a0,ai,. ..) + (b0,bi,...) = (ao + bo,ai bi,...) and
(a0,ai,.. .)(b0,bi,...) = (co,Oi,.. .), where n
Cn = 53 &n—ibi = anb0 + an_ibi + • • - +aibn-i + aobn = 53 akbj. i=0
k 4-j = n
(ii) If R is commutative [resp. a ring with identity or a ring with no zero divisors or an integral domain], then so is R[x]. (iii) The map R —* R[x] given by r (—■» (r,0,0,. . .) is a monomorphism of rings. PROOF. Exercise. If R has an identity 1K, then (1 a,0,0,. . .) is an identity in /?[*]. Observe that if (a0,fli,.. •)> (b0,bu . . .) e i?[x] and k [resp../] is the smallest index such that ak 5* 0 [resp. b, 5* 0], then (ao,ai, . . .)(bo,bu . . .) = (0,. . . ,0,akb,,ak+ibj + «*£>/+1, . . .). ■
The ring /?[*] of Theorem 5.1 is called the ring of polynomials over R. Its elements are called polynomials. The notation /?[*] is explained below. In view of Theorem 5.1 (iii) we shall identify R with its isomorphic image in /?[x] and write (r,0,0,. . .) simply as r. Note that r(a0,a\,. . .) = (ra0,rau . . .). We now develop a more familiar notation for polynomials.
Theorem 5.2. Let R be a ring with identity and denote by x the element (0,1r,0,0, ...) o/R[x].
(i) xn = (0,0, . . . ,0,1r,0, . . .), where 1r is the (n -(- l)st coordinate. (ii) If r e R, then for each n > 0, rxn = xnr = (0, . . . ,0,r,0, . . .), where r is the (n + l)st coordinate.
CHAPTER III RINGS
150
(iii) For every nonzero polynomial f in R[x] there exists an integer n e N and ele ments a 0 ) .. ., a„ e R such that f = a0x° + aiX1 H------\- anxn. The integer n and elements ai are unique in the sense that f = b0x° + biX1 -\— • + bmxm (bi s R) implies m > n; ai = bi for i = 1,2,..., n; and bi = 0 for n < i < m. SKETCH OF PROOF. Use induction for (i) and straightforward computation for (ii). (iii) If/= (fl0,fli, - - -) e there must be a largest index n such that an ^ 0. Then a 0 ,ai,... ,a„sR are the desired elements. ■ If R has an identity, then x° = I r (as in any ring with identity) and we write the polynomial / = aox° + aix1 H---- 1- anxn as / = a0 + + • • • + anxn. It will be convenient to extend the notation of Theorem 5.2 to rings without identity as follows. If R is a ring without identity, then R may be embedded in a ring S with identity by Theorem 1.10. Identify R with its image under the embedding map so that R is a sub ring of S. Then /?[*] is clearly a subring of S[x]. Consequently, every polynomial / = (a0,ai ,...)£ i?[x] may be written uniquely as/= a0 + a{xl + • • • + a„xn, where Oi s R C S, On 0, and x = (0,ls,0,0,...) eS[x]. The only important difference between this and the case when R has an identity is that in this case the element x is not in /?[*]. Hereafter a polynomial /over a ring R (with or without identity) will always be written in the form / = a0 + aix + a^x2 + • —(- anxn (a, e R). In this notation addi tion and multiplication in /?[*] are given by the familiar rules: n
n
n
E a'x' + 53 biX1 = 53 (fl. + bi)xi 1=0i=0i=0
(
\/m
n
\m
53 J ( 53 bjxi) = 53 ckXk, where ck = 53 i = 0 / \i>=0 / k = 0 i+j = k a<xi
71
If / =
53 ' '
ax e
^W> then the elements a, e R are called the coefficients of /. The
i=0
element a0 is called the'constant term. Elements of R, which all have the form n
r = (r, 0,0,.. .) = rx° are called constant polynomials. If / =
53 ' ' = «oi += 0 ax
Aix + ■ —h anxn = anxn ----- 1- a,x + flo has an ^ 0, then an is called the leading coefficient of /. If R has an identity and leading coefficient 1R, then /is said to be a monic polynomial. Let R be a ring (with identity). For historical reasons the element x = (0,1 R,0,...) of /?[*] is called an indeterminate. One speaks of polynomials in the indeterminate If S' is another ring (with identity), then the indeterminate x s S[jc] is not the same ele ment as x e /?[*]. In context this ambiguous notation will cause no confusion. If R is any ring, it is sometimes convenient to distinguish one copy of the poly nomial ring over R from another. In this situation the indeterminate in one copy is denoted by one symbol, say x, and in the other copy by a different symbol, say >\ In the latter case the polynomial ring is denoted /?[>>] and its elements have the form flo + axy H--- f- anyn. We shall now define polynomials in more than one indeterminate. For con venience the discussion here is restricted to the case of a finite number of indeterminates. For the general case see Exercise 4. The definition is motivated by the fact that a polynomial in one indeterminate is by definition a particular kind of sequence,
5. RINGS OF POLYNOMIALS AND FORMAL POWER SERIES
151
that is, a function N —* R. For each positive integer n let Nn = N X ■ • • X N (« factors). The elements of Nn are ordered n tuples of elements of N. Nn is clearly an additive abelian monoid under coordinate-wise addition.
Theorem 5.3. Let R be a ring and denote by R[xi,. . ., xn] the set of all functions f : Nn —■* R such that f(u) ^ 0 for at most a finite number of elements u of Nn.
(i) R[xj,. . ., xn] is a ring with addition and multiplication defined by (f + g)(u) = f(u) + g(u) and (fg)(u) = X f(v)g(w), V +w*=tt
v,weNn
where f,g e R[xi,. . ., xn] and u e Nn. (ii) //R is commutative [resp. a ring with identity or a ring without zero divisors or an integral domain], then so is R[xi, . . . , xn]. (iii) The map R —> R[xi,.. . , xn] given by r H- fr, where fr(0,. . ., 0) = r and f(u) = 0 for all other u e Nn, is a monomorphism of rings.
PROOF. Exercise. ■ The ring /?[*i, . .., xn] of Theorem 5.3 is called the ring of polynomials in n indeterminates over R. R is identified with its isomorphic image under the map of Theorem 5.3(iii) and considered as a subring of . . . , *„]. If n = 1, then /?[jci] is precisely the ring of polynomials as in Theorem 5.1. As in the case of polynomials in one indeterminate, there is a more convenient notation for elements of /?[*i,... ,>„’]. Let n be a positive integer and for each i = 1,2,..., n, let Bi = (0,..., 0,1,0,. . ., 0) e Nn,
where 1 is the z'th coordinate of e*. If k e N, let fce, = (0,.. ., 0,k,0, . . . 0). Then every element of N" may be written in the form kiBi + fc2e2 + - ■ ■ + knen.
Theorem 5.4. Let R be a ring with identity and n a positive integer. For each i = 1,2,. . . , n let x; e R[xi,. . . , x„] be defined by x^eO = 1r and x;(u) = 0 for u ^ ei.
(i) For each integer k e N, Xik(ks;) = 1r and Xik(u) = 0 for u ke,; (ii) for each (ki, . . . , kn) e Nn, xik,x2k2- • -xnk°(kiei -\----- (- knen) = 1r and kl xi x2kj- ■ -Xn^u) = 0 for u ^ kiei -|--- b knen; (iii) x^xj1 = xfxf for all s,t e N and all i,j = 1,2, ... , n; (iv) x^r = rxi1 for all r e R and all t e N; (v) for every polynomial f in R[x,,... ,xn] there exist unique elements a k l ,.. ,,kn e R, indexed by all (k,,... ,kn) e N" and nonzero for at most a finite number of (k,,... ,k„) e Nn, such that f = 2 • • • • v**'- • -x"n’ where the sum is over all (k,,... ,kj e N n .
SKETCH OF PROOF, (v) Let akJ,
f(k,,..., kj. U
CHAPTER III RINGS
152
If R is a ring with identity, then the elements xi,x2,. . . , xn e /?[xi,. . . , a„] as in Theorem 5.4 are called indeterminates. As in the case of one indeterminate symbols different than xu .. . , xn may be used to denote indeterminates whenever convenient. The elements ao,au . . . , am in Theorem 5.4(v) are called the coefficients of the poly nomial /. A polynomial of the form axj'xf1- ■ ■xnkn (a e R) is called a monomial in xi,x2, . . . , xn. Theorem 5.4(v) shows that every polynomial is a sum of monomials. It is customary to omit those Xi that appear with exponent zero in a monomial. For ex ample, ao^i°-V20^30 + aiXi‘2X2°Xi + a^x^xs is written a0 + aix,2xz + a2x\xix3. The notation and terminology of Theorem 5.4 is extended to polynomial ring /?[ai, . . . , xn], where R has no identity, just as in the case of one indeterminate. The ring R is embedded in a ringS with identity and /?[xi,. . . , x7l] is considered as a sub ring of S[ai, . . . , A-*]. If R has no identity then the indeterminates Ai,A2, . . ., a„ and the monomials xxklx2kl • • • xnK (k, e N) are not elements of R [ x r , , x„]. m If R is any ring, then the map /?[ai] —* -R[*i,. .., a„] defined by H m m i—0 T. ciixix-P■ ■ -Xn° = X) e Mai, ■ • • , xn] is easily seen to be a monomorphism t=0i=0 of rings. Similarly, for any subset [h,. .., ik] of {1,2,. . ., n\ there is a monomor phism /?[*,!,. . . , a,-J —» . . . , xn\. . . ., A,*] is usually identified with its isomorphic image and considered to be a subring of -R[xi,. .. , x„]. Let
By Theorem 5.4 f = Y alxkl" • • • a*1" with a, e R and ki} s N. Omit all a, that appear m
i=0
with exponent zero. Then
i=0
that is,
Theorem 5.5. Let R and S be commutative rings with identity and ip : R —» S a homo morphism of rings such that <^(1r) = Is. If Si,s2,. . . , sn e S, then there is a unique homomorphism of rings Tp : R[xi, . . . , x„] —> S such that Tp \ R = ip and ^(x;) = Si for i = 1,2, . . . , n. This property completely determines the polynomial ring R[xi, . . . , xn] up to isomorphism. SKETCH OF PROOF. If/e /?[a,, . . . , a„], then
m /= 2 atxi" • * • xkn‘" (a, e R foj e N) /—0 by Theorem 5.4. The map ip given by ip(f) =
5. RINGS OF POLYNOMIALS AND FORMAL POWER SERIES
153
phism of rings. Suppose that \p ■ - - ■ > *«] —»S is a homomorphism such that \f/ | R =
m
= 52
i=0 m = 52
whence \p = *P and ^ is unique. Finally in order to show that R[xi,. . . , xn\ is com pletely determined by this mapping property define a category © whose objects are all (n + 2)-tuples (ip,K,si,. . . , sn) where K is a commutative ring with identity, Si e K and \p : R —> K is a homomorphism with \[/(lH) = 1*. A morphism in C from (ip,K,Si, to (6,T,tu ...,/„) is a homomorphism of rings f : K—+T such that f(l a) = 1 t,W = 0 and = tx for i = 1,2,. . . , n. Verify that f is an equivalence in C if and only if f is an isomorphism of rings. If i : /? —* R[xu . . . , jf„] is the in clusion map, then the first part of the proof shows that ..., *„],jci, ... , xn) is a universal object in C. Therefore, . . . , xv\ is completely determined up to isomorphism by Theorem 1.7.10. ■
Corollary 5.6. If ip : R —* S is a homomorphism of commutative rings and si,s2, . . . , s„ e S, then the map R[xi,. . . , x„] —> S given by f (—» ^>f(Si,. . . , sn) is a homomorphism of rings. SKETCH OF PROOF OF 5.6. The proof of Theorem 5.5 showing that the assignment f\->
Corollary 5.7. Let R be a commutative ring with identity and n a positive integer. For each k (1 < k < n) there are isomorphisms of rings R[xi,. . . , xk][xk+i, . . ., x„] = R[xi, . . . , x„] ^ R[xk+1--- , x„][xi, . . . , xkJ.
CHAPTER Hi RINGS
154
PROOF. The corollary may be proved by directly constructing the isomor phisms or by using the universal mapping property of Theorem 5.5 as follows. Given a homomorphism : R—+S of commutative rings with identity and elements slt. . . ,snsS, there exists a homomorphism ip : /?[xi, . . . , x*] —> S such that ip | R =
Since R[xi,. .., xj is usually considered as a subring of /?[x1;..., xK] (see page 152) it is customary to identify the various polynomial rings in Corollary 5.6 under the isomorphisms stated there and write, for example, /?[xi,. .., x*][x*+i,.. ., x„] = R[Xl,. . ., x„]. We close this section with a brief introduction to rings of formal power series, which is not needed in the sequel.
Proposition 5.8. Let R be a ring and denote by R[[x]] the set of all sequences of ele ments of R (a0,ai,. ..). (i) R[[x]] is a ring with addition and multiplication defined by: (a0,ai, . . .) + (b0,bi,. . .) = (a0 + b0,ax + bi, . . .) and (a0,ai,. . .)(b0,bi,. . .) = (c0,ci, . . .), where n
Cn =
53
i=0
n
3ibn—i
=
53 akt>j.
k -\-j — n
(ii) The polynomial ring R[x] is a subring o/R[[x]]. (iii) If R is commutative [resp. a ring with identity or a ring with no zero divisors or an integral domain], then so is R[[x]].
PROOF. Exercise; see Theorem 5.1. ■ The ring /?[[x]] of Proposition 5.8 is called the ring of formal power series over the ring R. Its elements are called power series. If R has an identity then the polynomial x = (0,1 /j,0,. . .) e /?[[x]] is called an indeterminate. It is easy to verify that xV = rx1 for all r e R and i e N. If (a0,fli,. - .) e /?[[x]], then for each n, (a0,ai, • • • , fln,0,0, . . .) is a polynomial, whence (a0,. . . , an,0,0,. . .) = a0 + a,x + «2X2 -f - - ■ + onxn by Theorem 5.2. Consequently, we shall adopt the following notation. The power series co
(a0,«i, • • .) e R[[x]] is denoted by the formal sum
53 a‘xt- The elements a, are called
i=0
coefficients and a<> is called the constant term. Just as in the case of polynomials this notation is used even when R does not have an identity (in which case x ^ /?[[x]]).
5. RINGS OF POLYNOMIALS AND FORMAL POWER SERIES
155
Proposition 5.9. Let R be a ring with identity and f = 22 aix* 5 R[[x]]i=0
(i) f is a unit in R[[x]] if and only if its constant term a0 is a unit in R. (ii) If a0 is irreducible in R, then f is irreducible in R[[x]]. REMARK. If fe /?[[*]] is actually a polynomial with irreducible [resp. unit] con stant term then /need not be irreducible [resp. a unit] in the polynomial ring /?[x] (Exercise 8). PROOF OF 5.9. (i) If there exists g = 22^iA;i e MM] such that fg
= g f = Ik e J?[[x]],
it follows immediately that a0b0 = boao = 1R, whence a0 is a unit in R. Now suppose a0 is a unit in R. If there were an element g = Y.bjXi e /?[[x]] such that fg = I r, then the following equations would hold:
a0b0 = Ir aobi -j- aib0 = 0
Conversely if a solution (bn,bi,b-2,. ..) for this system of equations in R exists, then oo g = biXi e /?[[*]] clearly has the property that fg = lft. Since a0 is a unit (with »=o
multiplicative inverse o0_1), the first equation can be solved: b0 = cuTl\ similarly, bi = a0~l(~aib0) = ao_1(_oioo-1). Proceeding inductively, if b0 ,..., £>„_i are determined in terms of the o„ then a0bn = — i--------------- anb0 implies that bn = a0~l( — aibn _i — ■ • • — anb0). Thus, if a0 is a unit this system of equations can be solved and there is a g such that fg = 1« e /?[[jt]]. A similar argument shows that there exists h e /?[[*]] such that hf = 1R. But h = hiJt = h(fg) = (hf)g = \Rg = g, whence g is a two-sided inverse of /. Therefore /is a unit in /?[[*]]. (ii) is ar. immediate consequence of (i). ■
Corollary 5.10. If R is a division ring, then the units in R[[x]] are precisely mose power series with nonzero constant term. The principal ideal (x) consists precisely of the nonunits in R[[x]] and is the unique maximal ideal of R[[x]]. Thus if R is a field, R[[x]] is a local ring. PROOF. The first statement follows from Proposition 5.9 (i) and the fact that every nonzero element of R is a unit. Since x is in the center of /?[[*]], M = {*/I/£*[[*]]} by Theorem 2.5. Consequently, every element xf of (x) has zero constant term,
156
CHAPTER III RINGS
whence xf is a nonunit. Conversely every nonunit /e /?[[*]] is necessarily of the form co
co
53 °o 0. Let g = 53 hx' where bi = a,+i for all i. Then xg = /, i=0 i=0 whence /e(x). Therefore, (x) is -the set of nonunits. Finally, since \r | (x), (x) 7^ /?[[x]]. Furthermore, every ideal I of /?[[*]] with I j* /?[[jc]] necessarily consists of nonunits (Remarks, p. 123). Thus every ideal of /?[[x]] except /?[[*]] is contained in (x). Therefore, (x) is the unique maximal ideal of /?[[*]]. ■ / =
aix'
=
EXERCISES 1. (a) If
e Ns),
where the sum is over all pairs (0,f) such that =
6. FACTORIZATION IN POLYNOMIAL RINGS
157
of monoids Ns = N" given by
6. FACTORIZATION IN POLYNOMIAL RINGS We now consider the topics introduced in Section 3 (divisibility, irreducibility, and unique factorization) in the context of polynomial rings over a commutative ring. We begin with two basic tools: the concept of the degree of a polynomial and the division algorithm. Factors of degree one of a polynomial are then studied; finding such factors is equivalent to finding roots of the polynomial. Finally we con sider irreducible factors of higher degree: Eisenstein’s irreducibility criterion is proved and it is shown that the polynomial domain D[xi,. .., xn] is a unique factor ization domain if D is. Let R be a ring. The degree of a nonzero monomial ax}klx2kq- ■ ■ xnkn e R[xu . . ., xn] is the nonnegative integer kx + + • • ■ + kn. If / is a nonzero polynomial in m . . . , xn], then/ = ^ by Theorem 5.4. The (total) degree of the i=0
polynomial / is the maximum of the degrees of the monomials ajt". . . x*‘" such
CHAPTER III RINGS
158
that ai ^ 0 (/ = 1 , 2 , , m). The (total) degree of /is denoted deg / Clearly a nonzero polynomial / has degree zero if and only if / is a constant polynomial / = ao = aoXi0- ■ -xn0. A polynomial which is a sum of monomials, each of which has degree k, is said to be homogeneous of degree k. Recall that for each k (1 < k < ri), /?[*!,. .., xk-Uxk+lt. . . , xn] is a subring of /?[xi,. .., xn] (see page 152). The degree of f in Xk is the degree of /considered as a polynomial in one indeterminate xk over the ring i?[xi,..., Xk-uXb+u • • • > xn]. EXAMPLE. The polynomial 3xi2x22x32 + 3xix34 — 6x^x3 e Z[x] has degree 2 in Xi, degree 3 in x2, degree 4 in x3 and total degree 6. For technical reasons it is convenient to define the degree of the zero polynomial to be — co and to adopt the following conventions about the symbol deg 0 = — « : (—00) < n and (—<») + «= —o° = « + (— <=o) for every integer n; (—°°) + (— 00 ) == — CO '
Theorem 6.1. Let R be a ring and f,g e R[xi,. . . , xn]. (i) deg(i + g) < max (deg f, deg g). (ii) deg(fg) < deg f + deg g. (iii) If R has no zero divisors, deg(fg) = deg f + deg g. (iv) If n = 1 and the leading coefficient ofi or g is not a zero divisor in R (in par ticular, if it is a unit), then deg(fg) = deg f + deg g. REMARK. The theorem is also true if deg /is taken to mean “degree of /in xk." SKETCH OF PROOF OF 6.1. Since we shall apply this theorem primarily when n = 1 we shall prove only that case, (i) is easy (ii) is trivial if / = 0 or g = 0. If n
m
0 ^ / = 23 aixi has degree n and 0 ^ g = 23 hx* has degree m, then fg = a0b0 i—0 i= 0 + • • • + (an-ibm + anbm-i)xn+m~1 + anbmxm+n has degree at most m + n. Since an 0 7^ bm, fg has degree m + n if one of an,bm is not a zero divisor. ■
Theorem 6.2. (The Division Algorithm) Let R be a ring with identity and f,g e R[x] nonzero polynomials such that the leading coefficient of % is a unit in R. Then there exist unique polynomials q,re R[x] such that f = Qg + r and deg r < deg g. n
PROOF. If deg g > deg/, let q = Oandr = /.If deg g < deg/, then/ =
23
m
i=0
g = 53 blxi, with a„ 0, bm ^ 0, m < n, and bm a unit in R. Proceed by induction i=0
on n = deg /. If n = 0, then m = 0, / = a0, g = b0 and b0 is a unit. Let q = a0b0_1 and r = 0; then deg r < deg g and qg + r = (a06o_1)^o = a0 = f. Assume that the existence part of the theorem is true for polynomials of degree less than n = deg /. A straightforward calculation shows that the polynomial (anbr,r1xn~m)K has degree n and leading coefficient an. Hence
6. FACTORIZATION IN POLYNOMIAL RINGS
159
f — (anbm 1xn m)g - (anxn H------- h a0) — (anxn -{---- b anbm 1b0xn m) is a polynomial of degree less than n. By the induction hypothesis there are poly nomials q' and r such that /— (anbm~lxn-m)g = q'g -f r and deg r < deg g. Therefore, ifq = anbm~lxn~m + q', then / = (,anbm~lxnm)g + q'g + r = qg + r. (Uniqueness) Suppose / = qig + ru and / = q^g + r2 with deg n < deg g and deg r
(qi — qz)g = r2 — n. Since the leading coefficient bm of g is a unit, Theorem 6.1 implies deg (
Corollary 6.3. (Remainder Theorem) Let R be a ring with identity and n
f(x) = 53 aixi s R[x]. j
=0
For any c e R there exists a unique q(x) e R[x] such that f(x) = q(x)(x — c) + f(c). PROOF. If / = 0 let q = 0. Suppose then that f ^ 0. Theorem 6.2 implies that there exist unique polynomials q(x), r(x) in /?[.*] such that f(x) = q(x)(x — c) + K*) and deg r(x) < deg(x — c) = 1. Thus r(x) = r is a constant polynomial (possibly 0). n—l
n—l
If q(x) =
53
bjX1,
then f(x) = q(x)(x - c) + r = — boC +
j—0k=1
53 (~bkC -f- bk_i)xk +
bn^xn + r, whence n—l
/(<") = — - f - 5 3 ( — “ l k=1 n —1
=
—53 A- = 0
bk~\)ck
-f-
bn iCn
-}-
r
n
+
53
+ r = 0 + r = ,. ■
A- = 1
Corollary 6,4. //F is a field, then the polynomial ring F[x] /.v a Euclidean domain, whence F[x] /a a principal ideal domain and a unique factorization domain. The units in F|x] are precisely the nonzero constant polynomials. SKETCH OF PROOF. F[jr] is an integral domain by Theorem 5.1. Define : /■’[*] — {0 j —> N by <£(/) = deg /. Since every nonzero element of F is a unit, Theorems 6.1(iv) and 6.2 imply that F[jt] is a Euclidean domain. Therefore, F[a] is a principal ideal domain and a unique factorization domain (Theorem 3.9). Finally Theorem 6.1 (iv) implies that every unit /in /^x] has degree zero, whence /is a non zero constant. The converse is obvious. ■
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160
If F is a field, then F[xi,. . ., xn] is not a principal ideal domain (Exercise 1), but it is a unique factorization domain (Theorem 6.14 below). Before proving this latter fact we shall discuss factors of degree one in polynomial rings.
Definition 6.5. Let R be a subring of a commutative ring S, Ci,C2, . . . , cn e S and 771
f = 22 °i‘ •**'"
e
R[*1,. . . , xn] a polynomial such that f(ci,c2,. .. c„) = 0.
i=0
Then (ci,c2,. . . , cn) is said to be a root or zero off (or a solution of the polynomial equation f(xi, . . . , xn) = 0).4
Theorem 6.6. Let R be a commutative ring with identity and f e R[x]. Then c e R is a root off if and only if x — c divides f. SKETCH OF PROOF. We have f(x) = q(x)(x — c) + f(c) by Corollary 6.3. If x — c | f(x), then h(x)(x — c) = f(x) = cj(x)(x — c) + f(c) with h e /?[x], whence (h(x) — q(x))(x — c) = f(c). Since R is commutative, Corollary 5.6 (with
Theorem 6.7. If D is an integral domain contained in an integral domain E and f e D[x] has degree n, then f has at most n distinct roots in E. SKETCH OF PROOF. Let ci,c 2 ,... be the distinct roots of /in E. By Theorem 6.6 f(x) = qi(x)(x — ci), whence 0 = f(c2) = qi(c2)(c2 — ci) by Corollary 5.6. Since ci ^ C2 and E is an integral domain, qi(c2) = 0. Therefore, x — c2 divides q2 and fix') = q-i(x)(x — c2)(x — ci). An inductive argument now shows that whenever ci, ... ,cm are distinct roots of / in E, then gm = (x — ct)(x — c2)- • (x — c„,) divides / But deg gm = m by Theorem 6.1. Therefore m < n by Theorem 6.1 again. ■ REMARK. Theorem 6.7 may be false without the hypothesis of commutativity. For example, x2 + 1 has an infinite number of distinct roots in the division ring of real quaternions (including ±i, ±j and ±k). If D is a unique factorization domain with quotient field Fand /e D[x], then the roots of /in F may be found via
Proposition 6.8. Let D be a unique factorization domain with quotient field F and let n f = X ajX1 e D[x], If u = c/d e F with c and d relatively prime, and u is a root off, i=o then c divides a0 and d divides a„. 4Commutativity
is not essential in the definition provided one distinguishes “left roots” and “right roots” (the latter occur when / is written / = 22 ' ' ' x%nQi)•
6. FACTORIZATION IN POLYNOMIAL RINGS
SKETCH ( DF PROOF, /(w) = 0 implies that a^11 =
-ancn =12.,
/V
\i = 0
161
c^52
(—a»)ci_Wn and
c{dn i~^jd. Consequently, if (c,d) = \R then c | a0 and d\an by
Exercise 3.10. ■ EXAMPLE. If / = x4 — 2x8 — Ix2 — (ll/3)x — 4/3 e Q[x], then/has the same roots in Q as does 3/= 3x4 — 6x3 — 21x2 — llx — 4e Z[x], By Proposition 6.8 the only possible rational roots of 3/are dhl, ±2, ±4, ±1/3, ±2/3 and ±4/3. Sub stitution shows that 4 is the only rational root. Let D be an integral domain and /e D[x). If c e D and c is a root of f then re peated application of Theorem 6.6 together with Theorem 6.7 shows that there is a greatest integer m (0 < m < deg /) such that /(x) = (x - c)ra#(*), where g(x) e /?[x] and x — cJfg(x) (that is, g(c) ^ 0). The integer m is called the multiplicity of the root c of /. If c has multiplicity 1, c is said to be a simple root. If c has multiplicity m > 1, c is called a multiple root. In order to determine when a poly nomial has multiple roots we need:
Lemma 6.9. Let D be an integral domain and f = 52 aix‘ e DM- Let f' e D[x] be the n
1=0
polynomial ? = 52 ka^x1"-1 = ai + 2a2x + 3a3x2 + • - • + nanxn_1. Then for all k=l f,g e D[x] and c e D: (i) (ii) (iii) (iv)
(cf)' = cf'; (f + gY = V + g'; (fg)' = f'g + fg'; (gn)' = ngn_1g'.
PROOF. Exercise. ■ The polynomial /' is called the formal derivative of / The word “formal” em phasizes the fact that the definition of /' does not involve the concept of limits. According to Definition 3.3 a nonzero polynomial /e/?[x] is irreducible pro vided /is not a unit and in every factorization / = gh, either g or h is a unit in /?[x].
Theorem 6.10. Let D be an integral domain which is a subring of an integral domain E. Let f e D[x] and c e E. (i) c is a multiple root ofi if and only if f(c) = 0 and f '(c) = 0. (ii) //D is afield and f is relatively prime to f', then f has no multiple roots in E. (iii) If D is afield, f is irreducible in D[x] and E contains a root of f, then f has no multiple roots in E if and only iff' ^ 0. PROOF, (i) f(x) = (x — c)mg(x) where m is the multiplicity of f(m > 0) and g(c) 9^ 0. By Lemma 6.9 /'(x) = m(x — c)m^1g(x) + (x — c)mg'(x). If c is a multiple
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CHAPTER III RINGS
root of/ then m > 1, whence /'(c) = 0. Conversely, if/(c) = 0, then m > 1 (Theo rem 6.6). If m = 1, then /'(*) = #(jc) + (x — c)g'(x). Consequently, if f'(c) = 0, then 0 = /'(c) = g(c) by Corollary 5.6, which is a contradiction. Therefore, m > 1. (ii) By Corollary 6.4 and Theorem 3.11 kf-\- hf = 1 d for some k,h e D[x\ d E[x). If c is a multiple root of/ then by Corollary 5.6 and (i) 1 d = k(c) /(c) + h(c) /'(c) = 0, which is a contradiction. Hence c is simple root. (iii) If /is irreducible and /' ^ 0, then /and /' are relatively prime since deg f < deg / Therefore, /has no multiple roots in E by (ii). Conversely, suppose /has no multiple roots in E and b is a root of /in E. If /' = 0, then b is a multiple root by (i), which is a contradiction. Hence /' ^ 0. ■ This completes the discussion of linear factors of polynomials. We now consider the more general question of determining the units and irreducible elements in the polynomial ring D[x), where D is an integral domain. In general this is quite difficult, but certain facts are easily established: (i) The units in D[x\ are precisely the constant polynomials that are units in D [see the proof of Corollary 6.4]. (ii) If c e D and c is irreducible in D, then the constant polynomial c is irreducible in D[x] [use Theorem 6.1 and (i)]. (iii) Every first degree polynomial whose leading coefficient is a unit in D is irre ducible in D[x). In particular, every first degree polynomial over a field is irreducible. (iv) Suppose D is a subring of an integral domain E and /e D[x] CZ E[x\. Then / may be irreducible in £]x] but not in D[x] and vice versa, as is seen in the following examples. EXAMPLES. 2x + 2 is irreducible in Q[x] by (iii) above. However, 2x + 2 = 2(x + 1) and neither 2 nor x + 1 is a unit in Z[x] by (i), whence 2x -f 2 is re ducible in Z[x], x2 + 1 is irreducible over the real field, but factors over the complex field as (x + i)(x — i). Since x + / and x — i are not units in C[x] by (i), x2 + 1 is reducible in C[x]. In order to obtain what few general results there are in this area the rest of the discussion will be restricted to polynomials ovei" a unique factorization domain D. We shall eventually prove that D[xu . . . , xn] is also a unique factorization domain. The proof requires some preliminaries, which will also provide a criterion for irreducibility in D[x). 71 Let D be a unique factorization domain and / = 53 Qixi a nonzero polynomial in i=0 D[x]. A greatest common divisor of the coefficients a0,ai,. . ., a7l is called a content of /and is denoted C(/). Strictly speaking, the notation C(f) is ambiguous since great est common divisors are not unique. But any two contents of /are necessarily associ ates and any associate of a content of /is also a content of /. We shall write b — c whenever b and c are associates in D. Now = is an equivalence relation on D and since D is an integral domain, b ~ c if and only if b = cu for some unit u e D by Theorem 3.2 (vi). If a e D and /e D\x), then C(af) = aC(f) (Exercise 4). If/e D[x] and C(/) is a unit in D, then /is said to be primitive. Clearly for any polynomial g e D[jr], g = C(g)gi with g} primitive.
Lemma 6.11. (Gauss) If D is a unique factorization domain and f,g e D[x], then C(fg) = C(f)C(g). In particular, the product of primitive polynomials is primitive.
6. FACTORIZATION IN POLYNOMIAL RINGS
163
PROOF./ = C(/)/ and g = C(g)gi with f,gi primitive. Consequently, C(fg) = C(C(f )flC(g)gl) ~ C(f)C(g)C(/gi). Hence it suffices to prove that fgi is n
m
primitive (that is, C(fgi) is a unit). If f = and gi = 22 then m-\-n i=0 j«=0 c xk fgi = 22 k with ck = 22 ^/i-?i is not primitive, then there exists an irrek = 0 i +j = k ducible element p in R such that p | ck for all k. Since C(/) is a unit p\C(/), whence there is a least integer s such that
aixi
b,XK
p | a, for i < s and pJfas. Similarly there is a least integer t such that
p [ bj for j < t and pJfbt. Since p divides cs+« = aobs+t + • - ■ + as-\bt+i + asbt + as+\bt—i + ■ ■ • + as+tbo, p must divide asbt. Since every irreducible element in D is prime, p \ as or p \ bt. This is a contradiction. Therefore fig\ is primitive. ■
Lemma 6.12. Let D be a unique factorization domain with quotient field F and let f and g be primitive polynomials in D[x]. Then f and g are associates in D[x] if and only if they are associates in F[x], PROOF. If /and g are associates in the integral domain F[x], then / = gu for some unit u e F[;c] (Theorem 3.2 (vi)). By Corollary 6.4 u e F, whence u — b/c with b,c e D and c ^ O . Therefore, cf = bg. Since C(/) and C(g) are units in D,
c = cC(f) = C(cf) = C(bg) - bC(g) - b. Therefore, b = cv for some unit v s D and cf = bg = vcg. Consequently, f = vg (since c ^ 0), whence / and g are associates in D[x], The converse is trivial. ■
Lemma 6.13. Let D be a unique factorization domain with quotient field F and f a primitive polynomial of positive degree in D[x], Then f is irreducible in D[x] if and only iff is irreducible in F[x], SKETCH OF PROOF. Suppose / is irreducible in D[x] and f — gh with
g,h e F[x] and deg g > 1, deg h > 1. Then g = 22
n
(^/bl)xi
m
and h =
22 (Pi/d^x*
j=o with ai,bi,Cj,dj e D and bi ^ 0, dj ^ 0. Let b = b0bi ■ ■ ■ bn and for each i let n bi* = bobi ■ ■ ■bt-ibi+i - ■ bn. If gi = 22 aj\*xi e D[x\, then g\ = ag2 with a = C(gi), i=0
i=0
g2 e D[x] and g2 primitive. Verify that g = (lD/b)gi = (a/b)g2 and deg g = deg g2. Similarly h — (c/d)h2 with c,d £ £>, /z2 s D[x], h2 primitive and deg h = deg h2. Con sequently,/ = gh = (a/b){c/d)g2h2, whence bdf = acg2h2. Since /is primitive by hy pothesis and g2h2 is primitive by Lemma 6.11, bd = bdC(f) = C(bdf) = C(acg2h2) = acC(g2h-2) = ac. As in the proof of Lemma 6.12, bd and ac associates in D imply that /and g2h2 are
164
CHAPTER III RINGS
associates in D[x\. Consequently, / is reducible in D[x\, which is a contradiction. Therefore, /is irreducible in F[x], Conversely if /is irreducible in F[jr] and / = gh with g,h e D[x\, then one of g,h (say g) is a constant by Corollary 6.4. Thus C(f) = gC(h). Since /is primitive, g must be a unit in D and hence in D[x\. Therefore, /is irreducible in D[x\. ■
Theorem 6.14. If D is a unique factorization domain, then so is the polynomial ring D[xi,. . . , xn], REMARK. Since a field F is trivially a unique factorization domain, F[xu . . . , xn] is a unique factorization domain. SKETCH OF PROOF OF 6.14. We shall prove only that D[x) is a unique factorization domain. Since D[xi,.. . , xn] = D[xi , . . . , xn_i][xn\ by Corollary 5.7, a routine inductive argument then completes the proof. If/e D[x] has positive degree, then /= C(/) f with f a primitive polynomial in D[x] of positive degree. Since D is a unique factorization domain, either C(/) is a unit or C(/) = CiC2- • ■ cm with each c, irreducible in D and hence in Z>[jc]. Let F be the quotient field of D. Since F[x] is a unique factorization domain (Corollary 6.4) which contains D[x],f = pi*p2*- ■ -p„* with each p,* an irreducible polynomial in F[x]. The proof of Lemma 6.13 shows that for each i, p{* = (ai/hl)pi with a^hi s D, bi ^ 0, ajb, e F, p, e D[x] and pl primitive. Clearly each pi is irreducible in FM, whence each p, is irreducible in D\x] by Lemma 6.13. If a = Oifl2- • •an and b = bib2• • -b„, then f = (a/b)pxp2• • •pn. Consequently, bf = apip2• • •pn. Since f and pipr ■ pn are primitive (Lemma 6.11), it follows (as in the proof of Lemma 6.12) that a and b are associates in D. Thus a/b = u with u a unit in D. Therefore, if C(/) is a nonunit, /= C(f)f = cic2• - -Cm{upi)p2 - - 'Pn with each Ci,pi, and upx irreducible in D[x\. Similarly, if C(/) is a unit, /is a product of irreducible elements in D[x). (Uniqueness) Suppose /is a nonprimitive polynomial in D[x] of positive degree. Verify that any factorization of /as a product of irreducible elements may be written / = cic2- • ■cmp\ ■ • -pr, with each d irreducible in D, C(/) = cx- • -cm and each p, irre ducible (and hence primitive) in D[x] of positive degree. Suppose f — dx ■ ■ •dTq\ • ■ qs with each d} irreducible in D, C(/) — d\ - • dr and each q} irreducible primitive in D[x] of positive degree. Then c\c2- ■ c„ and d\d2- ■ dr are associates in D. Unique factorization in D implies that n = r, and (after reindexing) each c, is an associate of di. Consequently, pi/v • p« and qiq2■ ■ qs are associates in D[x] and hence in F[jr]. Since eachpx [resp. q,] is irreducible in F[x\ by Lemma 6.13, unique factorization in F\x\ (Corollary 6.4) implies that n = s and (after reindexing) each p, is an associate of ql in T7!*]. By Lemma 6.12 each pi is an associate of qi in D[x\. ■
Theorem 6.15. (Eisenstein's Criterion). Let D be a unique factorization domain with n quotient field F. Iff = 52 aixi E deg f > 1 and p is an irreducible element of D i=0 such that p^an; p|a £ for i = 0,1,--------- n — l; p2^a0,
6. FACTORIZATION IN POLYNOMIAL RINGS
165
then f is irreducible in F[x], Ifiis primitive, then f is irreducible in D[x]. PROOF. / = C(f) / with fi primitive in D[x] and C(/) s D\ (in particular fi = f if /is primitive). Since C(/) is a unit in F (Corollary 6.4), it suffices to show that / is irreducible in F[x]. By Lemma 6.13 we need only prove that / is irreducible in D[x\. Suppose on the contrary that f = gh with
g = brxT H---- H bo e £>[*], deg g = r > 1; and h = csxs + • • • + Co e D[x], deg h = s > 1. n
Now p does not divide C(/) (since pJfan), whence the coefficients of / = 5Z a**xi »=o satisfy the same divisibility conditions with respect to p as do the coefficients of / Since p divides aQ* = b0c0 and every irreducible in D is prime, either p \ bo or p | c0, say p | b0. Since p2Jfa0*, c0 is not divisible by p. Now some coefficient bk of g is not divisible by p (otherwise p would divide every coefficient of gh = /, which would be a contradiction). Let k be the least integer such that
p | bi for i < k and pJfbk. Then 1 < k < r < n. Since ak* = bock + + • - ■ + bk-ici + bkCo and p | ak*, p must divide bkc0, whence p divides bk or c0. Since this is a contradiction, f must be irreducible in D\x]. ■ EXAMPLE. If /= 2xh — 6x3 + 9x2 — 15 e Z[x], then the Eisenstein Criterion with p = 3 shows that / is irreducible in both Q[x] and Z[*]. EXAMPLE. Let / = y3 + x2y2 + x3y + x e /?[.*,}•] with R a unique factorization domain. Then x is irreducible in /?[x] and /considered as an element of (/?[x])[> ] is primitive. Therefore, / is irreducible in by Theorem 6.14 and Eisenstein’s Criterion (with p = x and D = /?[*]). For another application of Eisenstein’s Criterion see Exercise 10. There is a lengthy method, due to Kronecker, for finding all the irreducible factors of a poly nomial over a unique factorization domain, which has only a finite number of units, such as Z (Exercise 13). Other examples and techniques appear in Exercises 6-9.
EXERCISES 1. (a) If D is an integral domain and c is an irreducible element in D, then D[x] is not a principal ideal domain. [Hint: consider the ideal (x,c) generated by x and 2, then ^[xi, is not a principal ideal domain. [Hint: show that Xi is irreducible in F[xu . . . , x„_i].] 2. If F is a field and fg s F[x] with deg g > 1, then there exist unique polynomials f0,f,. . . , / e F[x] such that deg f < deg g for all /' and f=fo + fig + fg2+ • • • + frgT •
3. Let/be a polynomial of positive degree over an integral domain D. (a) If char D = 0, then f ^ 0.
CHAPTER III RINGS
166
(b) If char D = p 0, then /' = 0 if and only if /is a polynomial in xp (that is, / = «o + aPxv + a2vx2p H----- \-ajpX’p). 4. If D is a unique factorization domain, a e D and /e D[x], then C(af) and aC(f) are associates in D. n
5. Let R be a commutative ring with identity and / =
22
a'xi e
Then /is a
i=0
unit in /?[jt] if and only if is a unit in R and ax, . . . , an are nilpotent elements of R (Exercise 1.12). 6. [Probably impossible with the tools at hand.] Let p e Z be a prime; let F be a field and let c e F. Then xp — c is irreducible in F[jt] if and only if xp — c has no root in F. [Hint: consider two cases: char F — p and char F p.] 7. Iff = 22 a<x4 e Z[jc] and/? is prime, let / = 22 5iX' where a is the image of a under the canonical epimorphism Z —* Zp. (a) If /is monic and /is irreducible in Zp[x] for sdme prime p, then /is irre ducible in Z[x], (b) Give an example to show that (a) may be false if /is not monic. (c) Extend (a) to polynomials over a unique factorization domain. 8. [Probably impossible with the tools at hand.] (a) Let c e F, where F is a field of characteristic p (p prime). Then xp — x — c is irreducible in /•’[*] if and only if xp — x — c has no root in F. (b) If char F = 0, part (a) is false. 9. Let / =
22 aix< e Z-M have degree n. Suppose that for some k(Q < k < n) and i=0
some prime p : p\an\ p\ak\p \ ai for all 0 < i < k — 1; and p^Xa^. Show that / has a factor g of degree at least k that is irreducible in Z[x]. n 10. (a) Let D be an integral domain and c e D. Let /(x) = 22 a^i e and
i = 0
n
f(x — c) =
22 a*(x ~ e ^[Jc]- Then f(x) is irreducible in D[x] if and only if i=0
f(x — c) is irreducible. (b) For each prime p, the cyclotomic polynomial / = xp_1 + xp~2 -|----h x + 1 is irreducible in Z[x]. [Hint: observe that / = (xp — l)/(x — 1), whence f(x -)- 1) = ((x + l)p — l)/x. Use the Binomial Theorem 1.6 and Eisenstein’s Criterion to show that f(x + 1) is irreducible in Z[x].] 11. If c0, ci, . . . , cn are distinct elements of an integral domain D and do, . . . , d„ are any elements of D, then there is at most one polynomial / of degree < n in /)[.x] such that f(ci) = dt for i = 0,1,..., n. [For the existence of /, see Exercise 12]. 12. Lagrange's Interpolation Formula. If F is a field, a0,au . . . , ar are distinct ele ments of F and c0,ci,. . ., cn are any elements of F, then /( ) _ y' (* — «(>)•••(* — fl.-i)(-x ~ Qi+i)- • •(* — an) ^ 1 = 0 (fli Oo) ■ • ■ (a, fl»_l)(fli fli+i)- • (ai On)
is the unique polynomial of degree < n in F[x] such that f(at) = c, for all i [see Exercise 11]. 13. Let D be a unique factorization domain with a finite number of units and quotient field F. If /e D[x\ has degree n and c0,c\,. . ., cn are n + 1 distinct ele
6. FACTORIZATION IN POLYNOMIAL RINGS
167
ments of D, then /is completely determined by /(c0),/(ci), . . . ,/(c„) according to Exercise 11. Here is Kronecker’s Method for finding all the irreducible factors of/in D[x\. (a) It suffices to find only those factors g of degree at most «/2. (b) If g is a factor of /, then g(c) is a factor of /(c) for all c e D. (c) Let m be the largest integer
CHAPTER
IV
MODULES
Modules over a ring are a generalization of abelian groups (which are modules over Z). They are basic in the further study of algebra. Section 1 is mostly devoted to carrying over to modules various concepts and results of group theory. Although the classification (up to isomorphism) of modules over an arbitrary ring is quite difficult, we do have substantially complete results for free modules over a ring (Section 2) and finitely generated modules over a principal ideal domain (Section 6). Free modules, of which vector spaces over a division ring are a special case, have widespread applica tions and are studied thoroughly in Section 2. Projective modules (a generalization of free modules) are considered in Section 3; this material is needed only in Section VIII.6 and Chapter IX. With the exception of Sections 2 and 6, we shall concentrate on external struc tures involving modules rather than on the internal structure of modules. Of particu lar interest are certain categorical aspects of the theory of modules: exact sequences (Section 1) and module homomorphisms (Section 4). In addition we shall study vari ous constructions involving modules such as the tensor product (Section 5). Algebras over a commutative ring K with identity are introduced in Section 7. The approximate interdependence of the sections of this chapter is as follows: 1
7 A broken arrow A--- > B indicates that an occasional result from Section A is used in Section B, but that Section B is essentially independent of Section A. 168
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169
1. MODULES, HOMOMORPHISMS AND EXACT SEQUENCES Modules over a ring are a generalization of abelian groups (which are modules over Z). Consequently, the first part of this section is primarily concerned with carrying over to modules various concepts and results of group theory. The re mainder of the section presents the basic facts about exact sequences.
Definition 1.1. Let R be a ring. A (left) R-module is an additive abelian group A to gether with a function R X A —» A (the image of (r,,a) being denoted by ra) such that for all r,s e R and a,b e A: (i) r(a + b) = ra + rb. (ii) (r + s)a = ra + sa. (iii) r(sa) = (rs)a. If R has an identity element 1r and (iv) lrta = a for all a e A, then A is said to be a unitary R-module. If R is a division ring, then a unitary R-module is called a (left) vector space. A (unitary) right /^-module is defined similarly via a function A X R —>► A de noted (a,r) |—> ar and satisfying the obvious analogues of (i)-(iv). From now on, un less specified otherwise, “R-module” means “left R-module” and it is understood that all theorems about left R-moduIes also hold, mutatis mutandis, for right Rmodules. A given group A may have many different R-module structures (both left and right). If R is commutative, it is easy to verify that every left R-module A can be given the structure of a right R-module by defining ar = ra for re R,a e A (commutativity is needed for (iii); for a generalization of this idea to arbitrary rings, see Exercise 16). Unless specified otherwise, every module A over a commutative ring R is assumed to be both a left and a right module with ar = ra for all r s R, a e A. If A is a module with additive identity element Oa over a ring R with additive identity 0*, then it is easy to show that for all r e R, a e A:
rOA = Oa and 0 Ra = 0A. In the sequel 0A,0R,0 e Z and the trivial module {0) will all be denoted 0. It also is easy to verify that for all r e R, n e Z and a e A:
( — r)a = —(ra) = r(—a) and n(ra) = r(na), where na has its usual meaning for groups (Definition 1.1.8, additive notation). EXAMPLE. Every additive abelian group G is a unitary Z-module, with na (n e Z,a e G) given by Definition 1.1.8.
EXAMPLE. If S is a ring and R is a subring, then 5 is an R-module (but not vice versa!) with ra (r e R,a eS) being multiplication in 5. In particular, the rings R[jti,. . . , xm] and /?[[*]] are R-modules.
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EXAMPLES. If / is a left ideal of a ring R, then / is a left R-module with ra (r e R,a e I) being the ordinary product in R. In particular, 0 and R are .R-modules. Furthermore, since I is an additive subgroup of R, R/1 is an (abelian) group. R/I is an R-module with r(ry + /) = rn + I. R/I need not be a ring, however, unless / is a two-sided ideal. EXAMPLE. Let R and S be rings and
Definition 1.2. Let A and B be modules over a ring R. A function f : A —» B is an R-module homomorphism provided that for all a,c e A and r e R: f(a + c) = f(a) + f(c) and f(ra) = rf(a). //R is a division ring, then an R-module homomorphism is called a linear trans formation. When the context is clear .R-module homomorphisms are called simply homo morphisms. Observe that an R-module homomorphism /: /4 —> Z? is necessarily a homomorphism of additive abelian groups. Consequently the same terminology is used: /is an R-module monomorphism [resp. epimorphism, isomorphism] if it is in jective [resp. surjective, bijective] as a map of sets. The kernel of /is its kernel as a homomorphism of abelian groups, namely Ker /= {aeA | f(a) = 0[. Similarly the image of /is the set Im / = {b e B J b = f{a) for some aeA}. Finally, Theorem 1.2.3 implies: (i) /is an .R-module monomorphism if and only if Ker /= 0; (ii) / : A —* B is an .R-module isomorphism if and only if there is an .R-module homomorphism g : B —» A such that gf = \A and fg = 1«. EXAMPLES. For any modules the zero map 0 : A - > B given by a 1—* 0 (a e A) is a module homomorphism. Every homomorphism of abelian groups is a Z-module homomorphism. If R is a ring, the map /?[*] —■* R[a] given by / xf(for example, (x2 + 1)1—» a(x2 + 1)) is an .R-module homomorphism, but not a ring homo morphism. REMARK. For a given ring R the class of all R-modules [resp. unitary R-modules] and R-module homomorphisms clearly forms a (concrete) category. In fact, one can define epimorphisms and monomorphisms strictly in categorical terms (objects and morphisms only — no elements); see Exercise 2.
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171
Definition 1.3. Let R be a ring, A an R-module andB a nonempty subset of A. B is a submodule of A provided that B is an additive subgroup of A and rb e B for all r e R, b e B. A submodule of a vector space over a division ring is called a subspace. Note that a submodule is itself a module. Also a submodule of a unitary module over a ring with identity is necessarily unitary. EXAMPLES. If R is a ring and /: A —»B is an i?-module homomorphism, then Ker /is a submodule of A and Im /is a submodule of B. If C is any submodule of B, then f~l(C) = \a e A | f(d) eC) is a submodule of A. EXAMPLE. Let / be a left ideal of the ring R, A an /?-module and S
a nonempty
r« subset of A. Then IS — j 53 riGi I I;oitS; n e. N* is a submodule of A (ExerU=i cise 3). Similarly if a e A, then la = [ra \ r e /) is a submodule of A. EXAMPLE. If {Bi | / e /} is a family of submodules of a module A, then p) Bt is iel
easily seen to be a submodule of A.
Definition 1.4. IfX is a subset of a module A over a ring R, then the intersection of all submodules of A containing X is called the submodule generated by X (or spanned by X). If X is finite, and X generates the module B, B is said to be finitely generated. If X = 0, then X clearly generates the zero module. If X consists of a single element, X = [a], then the submodule generated by X is called the cyclic (sub)module gen erated by a. Finally, if {Bi \ i e /} is a family of submodules of A, then the submodule generated by X = (J Bi is called the sum of the modules Bi. If the index set / is finite, iel
the sum of Bu . . . , Bn is denoted Bi + B2 + • • • + B„.
Theorem 1.5. Let R be a ring, A an K-moduIe, X a subset of A, {B; | i e IJ a family of submodules of A and a e A. Let Ra = {ra | r s R). (i) Ra is a submodule of A and the map R —> Ra given by r |—> ra is an R-module epimorphism. (ii) The cyclic submodule C generated by a is {ra + na | r e R; n e Z}. If R has an identity and C is unitary, then C = Ra. (iii) The submodule D generated by X is
St
53
i=1
1 ria
;
+
53 3=1
nibi
I s,t e N*; a^bj s X ;r; s R; nj e Zf •
// R has an identity and A is unitary, then S
D = RX =
53 riai | s s N*; a; e X; T; e R f.
i
= 1
|
CHAPTER IV MODULES
172
(iv) bit e Bik.
The sum of the family {Bi | i e I} consists of all finite sums b;i +■■ • + bi„ with
PROOF. Exercise; note that if R has an identity 1 for all n e Z and na = («1 r)o for all a e A. ■
R
and A is unitary, then n\K e R
Theorem 1.6. Let B be a submodule of a module A over a ring R. Then the quotient group A/B is an K-module with the action of R on A/B given by: r(a + B) = ra + B for all r e R,a e A.
The map tt : A —-> A/B given by a |—> a + B is an K-module epimorphism with kernel B. The map ir is called the canonical epimorphism (or projection). SKETCH OF PROOF OF 1.6. Since A is an additive abelian group, B is a normal subgroup, and A/B is a well-defined abelian group. If a + B = a' + B, then a — a' zB. Since B is a submodule ra — ra' = r(a — a') eB for all r e R. Thus ra + B = ra' + B by Corollary 1.4.3 and the action of R on A/B is well defined. The remainder of the proof is now easy. ■ In view of the preceding results it is not surprising that the various isomorphism theorems for groups (Theorems 1.5.6-1.5.12) are valid, mutatis mutandis, for modules. One need only check at each stage of the proof to see that every subgroup or homo morphism is in fact a submodule or module homomorphism. For convenience we list these results here.
Theorem 1.7. If R is a ring and f : A —> B is an K-module homomorphism andC is a submodule of Ker f, then there is a unique K-module homomorphism f : A/C —» B such that f (a -f- C) = f(a) for all aeA;/mf = Im f and Ker f = Ker f/C. f is an K-module isomorphism if and only iff is an K-module epimorphism and C = Ker f. In particular, A/Ker fs Im f. PROOF. See Theorem 1.5.6 and Corollary 1.5.7. ■
Corollary 1.8. If R is a ring and A' is a submodule of the K-module A and B' a submodule of the K-module B and f : A —> B is an K-module homomorphism such that f(A') CZ B', then f induces an K-module homomorphism f : A/A' —* B/B' given by a + A' (—> f(a) + B'. f is an K-module isomorphism if and only iflm f -j- B' = B and f_1(B') CZ A'. In particular if f is an epimorphism such that f(A') = B' and Ker f CZ A', then f is an K-module isomorphism. PROOF. See Corollary 1.5.8. ■
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173
Theorem 1.9. Let B and C be submodules of a module A over a ring R. (i) There is an R-module isomorphism B/(B fl C) = (B + C)/C; (ii) ifC d B, then B/C is a submodule of A/C, and there is an R-module isomor phism (A/C)/(B/C) ^ A/B. PROOF. See Corollaries 1.5.9 and 1.5.10. ■
Theorem 1.10. If R is a ring and B is a submodule of an R-module A, then there is a one-to-one correspondence between the set of all submodules of A containing B and the set of all submodules of A/B, given by C f-> C/B. Hence every submodule of A/B is of the form C/B, where C is a submodule of A which contains B. PROOF. See Theorem 1.5.11 and Corollary 1.5.12. ■ Next we show that products and coproducts always exist in the category of R-modules.
n
52
Theorem 1.11. Let R be a ring and j A, | i £ I) a nonempty family of R-modules, Ai the direct product of the abelian groups A;, and Ai the direct sum of the iel
iel
abelian groups Ai. o) n Ai is an R-module with the action of R given by r{ai} = {ra;}. iel
(ii) 52 A; is a submodule of n a,. iel
iel
(iii) For each k s I, the canonical projection 7rk : riA,-A k (Theorem 1.8.1) is an R-module epimorphism. (iv) For each k e I, the canonical injection '• Ak —> 52 Ai (Theorem 1.8.4) is an R-module monomorphism. PROOF. Exercise. ■ n Ai is called the (external) direct product of the family of R-modules {Af | / s /} iel
and 52 ^ is its (external) direct sum. If the index set is finite, say / = {1,2,...,«}, iel
then the direct product and direct sum coincide and will be written A\ - ©^nThe maps irk [resp. i*] are called the canonical projections [resp. injections].
Tneorem 1.12. If R is a ring, {Ai I i e I] a family ofR-moduIes, C an R-module, and {<£i : C —> A; | i e 1} a family of R-module homomorphisms, then there is a unique R-module homomorphism
n iel
iel
uniquely determined up to isomorphism by this property. In other words, n A; is a iel
product in the category of R-modules. PROOF. By Theorem 1.8.2 there is a unique group homomorphism
CHAPTER IV MODULES
174
module homomorphism,
Theorem 1.13. If R is a ring, {A; | i e I) a family of R-modules, D an R-module, and {^i : Ai —> D | i e 1} a family of R-module homomorphisms, then there is a unique R-module homomorphism )p : 53 —> D such that y]/ii = i/', for all i e I. 53 ^ is
_
iel
iel
uniquely determined up to isomorphism by this property. In other words, 53 's a co~ iel
product in the category ofR-modules. PROOF. By Theorem 1.8.5 there is a unique abelian group homomorphism : 53 D with the desired property, given by \J/( {ai () = 53 &(c»)> where the sum i
is taken over the finite set of indices / such that a, ^ 0. It is easy to see that \p is an R-module map. Hence 53^» *s a coproduct in the category of R-modules (Definition 1.7.4), and therefore, determined up to isomorphism by Theorem 1.7.5. ■ Finite direct sums occur so frequently that a further description of them will be useful. We first observe that if /and g are R-module homomorphisms from an Rmodule A to an R-module B, then the map / + g : A —► B given byah> f(a) + g{a) is also an R-module homomorphism. It is easy to verify that the set Homn(^,/?) of all R-module homomorphisms A —► B is an abelian group under this addition (Exer cise 7). Furthermore addition of module homomorphisms is distributive with respect to composition of functions; that is,
h(f+ g) = hf+ hg and (/+ g)k = fk + gk, where fig : A —> B, h : B —» C, k : D —► A.
Theorem 1.14. Let R be a ring and A,Ai,A2, . . . , An R-modules. Then A == Ai © A2 © — ■ © A„ // and only if for each i = 1,2, . . . , n there are R-module homomor phisms 7Ti : A —» A; and i; : A; —» A such that (i) TTjii = lAifor i = 1,2, . . . , n; (ii) TTjij = 0 for i ^ j; (iii) li7Ti -f- l27T2 + • • + LnTTn = 1aPROOF. (=*) If A is the module A\ © A2 ©• • •© An, then the canonical in jections ^ and projections ir» satisfy (i)—(iii) as the reader may easily verify. Likewise if A ^ Ax ©- - •© An, under an isomorphism /: A —> Ax ©• • •© An, then the homomorphisms 7r,/: A —> Ai and f~ln : Ai—* A satisfy (i)-(iii). (<=) Let 7n : A —y Ai and u : A{ —» A (/ = 1,2,..., n) satisfy (i)-(iii). Let 7r,' : Ax ©• • - @ An —* Ai and u : At —► Ay ©• • • © An be the canonical projections and injections. Let
\/n
\
n n
n
‘f*P — (53 ) (53 t/7ri) = 53 53 v*r> = 53 v*«
V =1 / \j = 1 / i=l } = l
1=1
1. MODULES, HOMOMORPHISMS AND EXACT SEQUENCES
—^ i=l 1=1
175
^ ' t-t'TT* 1 A
-
nn
n
22 521=1 i/tt.IjTt/ = 22 t/7r»' = 1^, © • - • © a„. Therefore, <£> is an i=li=l
Similarly =
isomorphism by Theorem 1.2.3. ■
Theorem 1.15. Let R be a ring and {A; | i e I} a family of submodules of an K-module A such that (i) A is the sum of the family {As | i e l } ; (ii) for each k e I, Ak H Ak* = 0, where Ak* is the sum of the family {A; | i ^ k}. Then there is an isomorphism A = 22 A;. iel
PROOF. Exercise; see Theorem 1.8.6. ■ A module A is said to be the (internal) direct sum of a family of submodules {At | iel] provided that A and {A{] satisfy the hypotheses of Theorem 1.15. As in the case of groups, there is a distinction between internal and external direct sums. If a module A is the internal direct sum of modules Air then by definition each of the Ai is actually a submodule of A and A is isomorphic to the external direct sum 22 A_ iti However the external direct sum 22 A does not contain the modules Ai, but only iel
isomorphic copies of them (namely the u(Ai) — see Theorem 1.11 and Exercise T.8.10). Since this distinction is unimportant in practice, the adjectives “internal” and “external” will be omitted whenever the context is clear and the following nota tion will be used. NOTATION. We write A = 22 Ai to indicate that the module A is the internal iel
direct sum of the family of submodules {A{ | / s /}.
Definition 1.16. A pair of module homomorphisms, A —> B -^* C, is said to be exact at B provided Im f = Ker g. A finite sequence of module homomorphisms, A0^> Ax—> A2 —»• • • An_, -A An, is exact provided Im f; = Ker fi+i for i = 1,2,..., n — 1 .An
infinite sequence of module homomorphisms, ■ • A;_! —> A; Ai+1 - • is exact provided Im fi = Ker fi+1 for all i e Z. When convenient we shall abuse the language slightly and refer to an exact se quence of modules rather than an exact sequence of module homomorphisms. EXAMPLES. Note first that for any module A, there are unique module homo morphisms 0 —* A and A — 0. If A and B are any modules then the sequences ►
and 0 — » 0 are exact, where the t’s and 7 r’s are the canonical injections and projections respectively. Similarly, if C is a submodule of D, then the sequence 0 —» C —> D —> D/C —» 0 is exact, where / is the
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inclusion map and p the canonical epimorphism. If /: A —► B is a module homo morphism, then A/Ker /[resp. B/Im /] is called the coimage of /[resp. cokernel of /] and denoted Coim / [resp. Coker /]. Each of the following sequences is exact: 0 —> Ker f—>A—* Coim /—>0, 0 —> Im /—>5—>Coker /—> 0 and 0 —> Ker /—*
A^> B —> Coker /—* 0, where the unlabeled maps are the obvious inclusions and projections. REMARKS. 0 —> /I i? is an exact sequence of module homomorphisms if and only if /is a module monomorphism. Similarly, B C —> 0 is exact if and only if g is a module epimorphism. If A B -^> C is exact, then gf = 0. Finally if /i B -^» C —* 0 is exact, then Coker / = 5/Im / = B/Ker g = Coim g = C. An exact se quence of the form 0—> A B C —> 0 is called a short exact sequence; note that / is a monomorphism and g an epimorphism. The preceding remarks show that a short exact sequence is just another way of presenting a submodule (A = Im /) and its quotient module (B/Im / - B[Ker g ~ C).
Lemma 1.17. (The Short Five Lemma) Let R he a ring and
a
Jf3 r g' 0—~A BC'-+- 0 r
a commutative diagram of R-modules and R-module homomorphisms such that each row is a short exact sequence. Then (i) a,7 monomorphisms => /3 is a monomorphism; (ii) a,7 epimorphisms => ft is an epimorphism; (iii) a, 7 isomorphisms => (3 is an isomorphism. PROOF, (i) Let beB and suppose fi(b) = 0; we must show that b = 0. By com mutativity we have 7g(b) = g'13(b) = g'(0) - 0. This implies g(b) = 0, since 7 is a monomorphism. By exactness of the top row at B, we have b e Ker g = Im f say b = f(a), aeA. By commutativity,
f'a(a) = 0f(a) = (3(b) = 0. By exactness of the bottom row at A', f is a monomorphism (Theorem 1.2.3(i)); hence a(a) = 0. But a is a monomorphism; therefore a = 0 and hence b = f(a) - /(0) = 0. Thus (3 is a monomorphism. (ii) Let b' eB’. Then g'(b') e C' ; since 7 is an epimorphism g'(b') = 7 (c) for some c e C. By exactness of the top row at C, g is an epimorphism; hence c = g(b) for some beB. By commutativity,
g'(3(b) = 7
g(b) = y(c) = g'(b').
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Thus g'[(3(b) — b'] = 0 and (3(b) — b' e Ker g' = Im /' by exactness, say /'(«') = /3(A) — b', o' s /!'. Since a is an epimorphism, a' = a(a) for some ae/1. Consider 6 — /(a) e /3[A - /(a)] = m - (if (a). By commutativity, (if (a) = f'a(a) = f'(a') = /8(6) — 6'; hence
0[b - f(a)] = /3(A) - |8/(fl) = (3(b) - ((3(b) - b') = b' and /3 is an epimorphism. (iii) is an immediate consequence of (i) and (ii). ■ Two short exact sequences are said to be isomorphic if there is a commutative diagram of module homomorphisms 0—-A —~B—-C—-0
f 0 —~A
g B'—-C—-0
such that fg, and h are isomorphisms. In this case, it is easy to verify that the diagram
0—*~ A —-B—*~C—►O f1 0—
8
-l
h-i a—-0
(with the same horizontal maps) is also commutative. In fact, isomorphism of short exact sequences is an equivalence relation (Exercise 14).
Theorem 1.18. Let R be a ring and 0—» Ai B A2 —> 0a short exact sequence of K-module homomorphisms. Then the following conditions are equivalent. (i) There is an K-module homomorphism h : A2 —> B with gh = l A z ; (ii) There is an K-module homomorphism k : B —> A! with kf = 1a,; (iii) the given sequence is isomorphic (with identity maps on Al and A2) to the
direct sum short exact sequence 0 —> A, -A Ai © A2 —* A2 —> 0; in particular B =* A! @ A2. A short exact sequence that satisfies the equivalent conditions of Theorem 1.18 is said to be split or a split exact sequence. SKETCH OF PROOF OF 1.18. (i) => (iii) By Theorem 1.13 the homomor phisms / and h induce a module homomorphism ip : A\ (+) A2 —> B, given by (ai,a.) > f(ai) + h(a2). Verify that the diagram
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178
0
— Ay -H-A!
I
0-^Ai
,
0 At U2
1
0
—?-+A2-+~ 0
is commutative (use the fact that gf — 0 and gh = 1^.,). By the Short Five Lemma ip is an isomorphism. (ii) => (iii) The diagram
f
0-^A I Ay 0 -+A
is commutative, where \p is the module homomorphism given by \p(b) = (k(b),g(b)) (see Theorem 1.12). Hence the short Five Lemma implies \p is an isomorphism. (iii) ■=> (i), (ii) Given a commutative diagram with exact rows and
0
__ TT
2
+-A i A j /12 -* - A‘i—►O 7Ti
Uy
0 — Ay f
B
‘’I
Ai -0,
define h : A2—> B to be ^>i2 and k : B —> At to be irytp'1. Use the commutativity to show that kf = \ and of the diagram and the facts w,i,- = 1^,, = Al gh = \a . U 2
EXERCISES Note: R is a ring. 1. If A is an abelian group and n > 0 an integer such that na — 0 for all aeA, then A is a unitary Zn-module, with the action of Z„ on A given by ka = ka, where k eZ and k |—> k eZ„ under the canonical projection Z —>Z„. 2. Let /: A —> B be an R-module homomorphism. (a) /is a monomorphism if and only if for every pair of i?-module homomor phisms g,h : D —*■ A such that fg = fh, we have g = h. [Hint: to prove (<=), let D = Ker f with g the inclusion map and h the zero map.] (b) /is an epimorphism if and only if for every pair of /^-module homomor phisms k,t : B —> C such that kf = tf, we have k = t. [Hint: to prove (<=), let k be the canonical epimorphism B —» B/Im /and t the zero map.] 3. Let / be a left ideal of a ring R and A an /?-module. (a) If 5 is a nonempty subset of A, then IS — 53 r,o,-1 n e N*; r, e T, a* e S f ,» = i J is a submodule of A. Note that if 5 = {a}, then IS = la = [ra\r el\.
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179
(b) If / is a two-sided ideal, then A/1A is an /?//-module with the action of R/I given by (r + I)(a + I A) = ra + I A. 4. If R has an identity, then every unitary cyclic /^-module is isomorphic to an /?-module of the form R/J, where 7 is a left ideal of R. 5. If/? has an identity, then a nonzero unitary /^-module A is simple if its only submodules are 0 and A. (a) Every simple /^-module is cyclic. (b) If A is simple every /?-module endomorphism is either the zero map or an isomorphism. 6. A finitely generated i?-module need not be finitely generated as an abelian group. [Hint: Exercise II.1.10.] 7. (a) If A and B are /^-modules, then the set Homft(/l,6) of all /^-module homo morphisms A —* B is an abelian group with / + g given on a e A by (/+ g)(a) = f(a) + g(a) e B. The identity element is the zero map. (b) HomR(A,A) is a ring with identity, where multiplication is composition of functions. HomR(A,A) is called the endomorphism ring of A. (c) A is a left Homfi(A,A)-module with fa defined to be
f(a) (a e A,ft HomR(>l,/l)). 8. Prove that the obvious analogues of Theorem 1.8.10 and Corollary 1.8.11 are valid for /^-modules. 9. Iff:A->A is an i?-module homomorphism such that Jf = f then
A = Ker /© Im /. 10. Let A,AU . . . , An be i?-modules. Then A = At ©■ • ■© A„ if and only if for each / = 1,2, ... ,n there is an /^-module homomorphism <£. : A —> A such that Im tpi ^ Ai\ (pupj = 0 for i ^ j\ and
Q'r
0 -» T(A) -U T(B) -A T(C). (f) If g : B —> C is an /?-module epimorphism, then gT : T(B) —> T(C) need not be an epimorphism. [Hint: consider abelian groups.]
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12. (The Five Lemma). Let
Aj
|
— ► £*1 |a2 .61
A‘2,
10
— — *~A$ : 3
|«4OCb
Bi —*~B$ ~~*~Bi —►Bi
be a commutative diagram of /^-modules and /^-module homomorphisms, with exact rows. Prove that: (a) c*i an epimorphism and <*2 ,0 : 4 monomorphisms ==> a3 is a monomorphism; (b) as a monomorphism and 012,0(4 epimorphisms => as is an epimorphism. 13. (a) If 0 —> A —* B —> C —> 0 and 0 —> C Z) —»£ —> 0 are short exact sequences of modules, then the sequence 0->A-+B^D-^E—> 0is exact. (b) Show that every exact sequence may be obtained by splicing together suit able short exact sequences as in (a). 14. Show that isomorphism of short exact sequences is an equivalence relation. 15. If / : A —► B and g : B —» A are /^-module homomorphisms such that gf = \ A , then B = Im /© Ker g. 16. Let R be a ring and Rop its opposite ring (Exercise III.1.17). If A is a left [resp. right] /^-module, then A is a right [resp. left] /^-module such that ra — ar for all a e A, r s R, r e Rop. 17. (a) If R has an identity and A is an R-module, then there are submodules B and C of A such that B is unitary, RC = 0 and A = B © C. [Hint: let B = {1 Ra | a e A] and C = [a e A | 1 Ka = 0| and observe that for all a e A, a — 1 Ra e C.] (b) Let Ai be another /?-module, with Ai = Bx © Ci (B\ unitary, RCi = 0). If / : A —* Ai is an i?-module homomorphism then f(B)
2. FREE MODULES AND VECTOR SPACES In this section we study free objects in the category of modules over a ring. Such free modules, the most important examples of which are vector spaces over a division ring (Theorem 2.4), have widespread applications in many areas of mathematics. The special case of free abelian groups (Z-modules) will serve as a model for the first part of this section. The remainder of the section consists of a discussion of the di mension (or rank) of a free module (Theorems 2.6-2.12) and an investigation of
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the special properties of the dimension of a vector space (Theorems and Corollaries 2.13-2.16). A subset X of an .R-module A is said to be linearly independent provided that for distinct Xi,.. ., xn eX and rt- e R.
nxi + rtx2
H--- f- rnxn = 0 => n = 0 for every i.
A set that is not linearly independent is said to be linearly dependent. If A is generated as an R-module by a set Y, then we say that Y spans A. If R has an identity and A is unitary, Y spans A if and only if every element of A may be written as a linear com bination: rly1 + r2y2 H----- 1- rnyn (r, e R,yt e F); see Theorem 1.5. A linearly inde pendent subset of A that spans A is called a basis of A. Observe that the empty set is (vacuously) linearly independent and is a basis of the zero module (see Defini tion 1.4).
Theorem 2.1. Let R be a ring with identity. The following conditions on a unitary R-module F are equivalent: (i) F has a nonempty basis; (ii) F is the internal direct sum of a family of cyclic R-modules, each of which is isomorphic as a left R-module to R; (iii) F is R-module isomorphic to a direct sum of copies of the left R-module R; (iv) there exists a nonempty set X and a function i : X —> F with the following property: given any unitary R-module A and function f : X —» A, there exists a unique R-module homomorphism f: F —> A such that f i = f. In other words, F is a free object in the category of unitary R-modules. The theorem is proved below. A unitary module F over a ring R with identity, which satisfies the equivalent conditions of Theorem 2.1, is called a free R-module on the set X. By Theorem 2.1 (iv), F is a free object in the category of all unitary left R-modules. But such an F is not a free object in the category of all left R-modules (Exercise 15). By definition the zero module is the free module on the empty set. It is possible to define free modules in the category of all left R-modules over an arbitrary ring R (possibly without identity); see Exercise 2. Such a free module is not isomorphic to a direct sum of copies of R, even when R does have an identity (Exer cise 2). In a few carefully noted instances below, certain results are also valid for these free modules in the category of all left R-modules. However, unless stated otherwise, the term “free module” will always mean a unitary free module in the sense of Theorem 2.1. SKETCH OF PROOF OF 2.1. (i) => (ii) Let X be a basis of F and * el. The map R —> Rx, given by r\-*rx, is an R-module epimorphism by Theorem 1.5. If rx = 0, then r = 0 by linear independence, whence the map is a monomorphism and R = Rx as left R-modules. Verify that F is the internal direct sum of the cyclic modules Rx (x eX). (ii) => (iii) Theorem 1.15 and Exercise 1.8. (iii) => (i) Suppose F == Y.R and the copies of R are indexed by a set X. For each X eXlet 6X be the element {/v} of^^R, where n = Ofori ^ x and rx = 1K. Verify that \6X | x eX] is a basis of 52^ anc* use isomorphism F = Y.R to obtain a basis of F. (i) => (iv) Let A" be a basis of F and i : X —» F the inclusion map. Suppose we are
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given a map / : X—*■ A. If u e F, then u = 22 r^* (r»
n
u=
i=l
e e^0
since X spans F. If
22 (* e ^), then 22 (r* ~ s»)** = 0, whence r, = for every / by linear ini = l
i
dependence. Consequently, the map / : F —* A given by n
f(u) = 7
n
\
=
22 ) 52 = 1 / t‘=l r'xi
rif(Xi)
i
is a well-defined function such that fi = /. Verify that /is an /{-module homomor phism. Since A' generates F, any /{-module homomorphism F —> A is uniquely deter mined by its action on X. Thus, if g : F—> A is an /{-module homomorphism such that gi = /, then for every x e X, g(x) = g(i(x)) = f(x) = /(x), whence g = f and /is unique. Therefore, by Definition 1.7.7 F is a free object on the set X in the cate gory of unitary /{-modules. (iv) => (iii) Given i : X —* F construct the direct sum 52^> w'th one copy of R for each x eX. Let Y = {8X \ x eX} be the basis of the (unitary) /{-module 22^ as *n the proof of (iii) => (i). The proof of (iii) => (i) => (iv) shows that 22^ is a free object on the set Y in the category of /{-modules (with Y —>22R the inclusion map). Since \X\ = \Y\, the proof of Theorem 1.7.8 implies that there is an /^-module isomorphism / : F = 2> such that /(i(X)) = Y. ■ REMARKS, (a) If F is a free /?-module on a set X(l : X —* F), then the proof of (iv) => (iii) of Theorem 2.1 implies that i(X) is actually a basis of F. (b) Conversely, the proof of (i) => (iv) of Theorem 2.1 shows that if X is a basis of a unitary module F over a ring R with identity, then Fis free onX, with i : X —> Fthe inclusion map. (c) If X is any nonempty set and R is a ring with identity, then the proof of Theorem 2.1 shows how to construct a free /{-module on the set X. Simply let F be the direct sum 22^> w^h the copies of R indexed by the set A". In the notation of the proof, {8X | x e X} is a basis of F so that F = 22 Since the map i : A —> F, given xzX
by x (—» 6X, is injective it follows easily that F is free on X in the sense of condition (iv) of Theorem 2.1. In this situation we shall usually identify X with its image under t, writing x in place of 8X, so that X C F. In this notation F = 22 written as xeX
22 Rx ar*d a typical element of F has the form nxi + • ■ • + rnxr (r, e R;x{ eX). In xeX
particular, X — l(X) is a basis of F. (d) The existence of free modules on a given set in the category of all modules over an arbitrary ring (possibly without identity) is proved in Exercise 2.
Corollary 2.2. Every (unitary) module A over a ring R (with identity) is the homomor phic image of a free K-module F. If A is finitely generated, then F may be chosen to be finitely generated. REMARK. Corollary 2.2 and its proof are valid if the words in parentheses are deleted and “free module” is taken to mean a free module in the category of all left modules over an arbitrary ring (as defined in Exercise 2).
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183
SKETCH OF PROOF OF 2.2. Let X be a set of generators of A and F the free ^-module on the set X. Then the inclusion map X —» A induces an /^-module homo morphism / : F —> A such that X CZ Im /(Theorem 2.1 (iv)). Since X generates A, we must have Im f= A. ■ REMARK. Unlike the situation with free abelian groups, a submodule of a free module over an arbitrary ring need not be free. For instance {0,2,4} is a submodule ofZ6, but is clearly not a freeZ6-module; compare Theorem II. 1.6 and Theorem 6.1 below. Vector spaces over a division ring D (Definition 1.1) are important, among other reasons, because every vector space over D is in fact a free Z)-module. To prove this we need
Lemma 2.3. A maximal linearly independent subset X of a vector space V over a division ring D is a basis of V. PROOF. Let W be the subspace of V spanned by the set A'. Since A" is linearly in dependent and spans W, X is a basis of W. If W = V, we are done. If not, then there exists a nonzero a eV with a\W. Consider the set X U {a}. If ra + rYXi -\---- (rnxn = 0 (r,n e D,Xi eX) and r 9* 0, then a = r l(ra) = —r~lr 1 * 1 ------ — r_1/-nA:n e IV, which contradicts the choice of a. Hence r = 0, which implies n = 0 for all i since X is linearly independent. Consequently X \J (a) is a linearly independent subset of V, contradicting the maximality of X. Therefore W = V and X is a basis. ■
Theorem 2.4. Every vector space V over a division ring D has a basis and is therefore a free D-module. More generally every linearly independent subset of V is contained in a basis ofV. The converse of Theorem 2.4 is also true, namely, if every unitary module over a ring D with identity is free, then D is a division ring (Exercise 3.14). SKETCH OF PROOF OF 2.4. The first statement is an immediate con sequence of the second since the null set is a linearly independent subset of every vector space. Consequently, we assume that X is any linearly independent subset of V and let S be the set of all linearly independent subsets of V that contain X. Since X e S, S ^ 0. Partially order S by set theoretic inclusion. If {C, | / e I] is a chain in S verify that the set C = (J C, is linearly independent and hence an element of S. iel
Clearly C is an upper bound for the chain { C | / e /}. By Zorn’s Lemma S contains a maximal element B that contains X and is necessarily a maximal linearly inde pendent subset of V. By Lemma 2.3 B is a basis of V. ■
Theorem 2.5. //V is a vector space over a division ring D and X is a subset that spans V, then X contains a basis of\.
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SKETCH OF PROOF. Partially order the set S of all linearly independent subsets of A by inclusion. Zorn’s Lemma implies the existence of a maximal linearly independent subset Y of X. Every element of A1 is a linear combination of elements of Y (otherwise, as in Lemma 2.3, we could construct a linearly independent subset of X that properly contained Y, contradicting maximality). Since X spans V, so does Y. Hence Y is a basis of V. ■ In the case of free abelian groups (Z-modules) we know that any two bases of a free Z-module have the same cardinality (Theorem II.1.2). Unfortunately, this is not true for free modules over arbitrary rings with identity (Exercise 13). We shall now show that vector spaces over a division ring and free modules over a commutative ring with identity have this property.
Theorem 2.6. Let R be a ring with identity and F a free R-module with an infinite basis X. Then every basis of F has the same cardinality as X. PROOF. If Y is another basis of F, then we claim that Y is infinite. Suppose on the contrary that Y were finite. Since Y generates F and every element of Y is a linear combination of a finite number of elements of X, it follows that there is a finite subset {jfi,. .., xm} o(X, which generates F. Since X is infinite, there exists
x e X — {xi,. . . , xm}. Then for some neR, x = nx\ + • •■ + rmxm, which contradicts the linear inde pendence of X. Therefore, Y is infinite. Let K(Y) be the set of all finite subsets of Y. Define a map / : X —> K(Y) by x f—> {, .Vr.}, where x = nyi H— - + and ^ 0 for all Since Y is a basis, the yi are uniquely determined and /is a well-defined function, (which need not be injective). If Im /were finite, then U S would be a finite subset of Y that would Seim j generated and hence F. This leads to a contradiction of the linear independence of Y as in the preceding paragraph. Hence Im /is infinite. Next we show that f \T) is a finite subset of X for every T e Im f d K(Y). If x e f~KT), then x is contained in the submodule Ft of F generated by T; that is, f~KT) CZ Ft (see Theorem 1.5). Since T is finite and each y e T is a linear combina tion of a finite number of elements of A', there is a finite subset S of A" such that FT is contained in the submodule Fs of F generated by S. Thus x e f~\T) implies x eFs and x is a linear combination of elements of 5 (Theorem 1.5). Since x e X and 5 CI A, this contradicts the linear independence of A unless x e 5. Therefore, f \T) CZ S, whence f~l(T) is finite. For each T e Im / order the elements of f~\T), say xj,. . ., xn, and define an in jective map gT : f~\T) —> Im /XN by xk H (T,k). Verify that the sets f~\T) (Te Im /) form a partition of A. It follows that the map A—► Im /X N defined by x gr(x), where x e f~\T), is a well-defined injective function, whence |A| < jlm/X N|. Therefore by Definition 8.3, Theorem 8.11, and Corollary 8.13 of the Introduction: IAJ < |Im/X N| = |Im/| N0 = |Im/| < \K(Y)\ = |y|.
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Interchanging A and Y in the preceding argument shows that |F| < \X\. Therefore |F| = \X\ by the Schroeder-Bernstein Theorem. ■
Theorem 2.7. If V is a vector space over a division ring D, then any two bases of V have the same cardinality. PROOF. Let X and Y be bases of V. If either A" or Y is infinite, then \X\ — |3^| by Theorem 2.6. Hence we assume X and Y are finite, say X = {Xi, . . ., xn}, and Y = \yu • • • ,ym\- Since X and Y are bases, 0 ym = rYx\ + ■ —|- rnxn for some t\ e D. If rk is the first nonzero n, then xk = rk~'ym — rk-lrk+\xk+i----- - — rk~lrnxri. Therefore, the set X' = [ym,xu . . . , **.+!,. . . , xn\ spans V (since X does). In particular ym—1 = Smym -f- t\X\ + ' • • + /k—\Xfc—1 -|- tk+iXk+i + ■ ■ • + tnXji (Sm,ti 6 D).
Not all of the U are zero (otherwise ym_i — smym = 0, which contradicts the linear in dependence of Y). If tj is the first nonzero u, then x, is a linear combination of ym-i,ym and those x,- with / ^ j,k. Consequently, the set [ym-i,ym\ U [xi \ i ^ j,k\ spans V (since A' does). In particular, ym-z is a linear combination of ym-i,ym and the Xi with / ^ j,k. The above process of adding a y and eliminating an x may therefore be repeated. At the end of the A:th step we have a set consisting of ym,ym-1, . . . , ym-k+1 and n — k of the Xi, which spans V. If n < m, then at the end of n steps we would conclude that {ym, . . . , ym-„+1} spans V. Since m — n + 1 >2,y Y would be a linear combination of y m , , ym-n+1, which would contradict the linear independence of Y. Therefore, we must have m < n. A similar argument with the roles of X and Y re versed shows that n < m and hence m = n. ■
Definition 2.8. Let R be a ring with identity such that for every free R-module F, any two bases of F have the same cardinality. Then R is said to have the invariant dimension property and the cardinal number of any basis of F is called the dimension (or rank) of F over R. Theorem 2.7 states that every division ring has the invariant dimension property. We shall follow the widespread (but not universal) practice of using “dimension” when referring to vector spaces over a division ring and “rank” when referring to free modules over other rings. The dimension of a vector space V over a division ring D will be denoted here by dimDV. The properties of dimDV will be investigated after Corollary 2.12. Results 2.9-2.12 are not needed in the sequel, except in Sections IV.6 and VII.5.
Proposition 2.9. Let E and F be free modules over a ring R that has the invariant dimension property. Then E == F if and only if E and F have the same rank. PROOF. Exercise; see Proposition II. 1.3. ■
Lemma 2.10. Let R be a ring with identity, I (^ R) an ideal o/R, F a free R-module with basis X and it : F —» F/IF the canonical epimorphism. Then F/IF is a free R/Imodule with basis 7r(X) and [tt(X)| = |X|.
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Recall that IF = | ^3 nm | rt e /, ai e F, n e N* | and that the action of R/I on
F/IF is given by (r + I)(a + IF) = ra + IF (Exercise 1.3).
PROOF OF 2.10. If u + IF e F/IF, then u — 23
rixi
71 with ri e xi e* since 3= 1
u e F and A is a basis of F. Consequently, u + IF = (23 nxd + IF = 23 + //r> y i = 23 O'i + + IF) = 23 (ri + I)7r(xi)-> whence tt(X) generates F/IF as an
j
ym
R//-module. On the other hand, if 23 I)^(xk) = 0 with r* e R and Xi,. . . , xm k=l
23 + I)^(kxk) = 23 (r* + I)(Xk + IF) _______ k = 23 rkxk + IF, whence 23 r^Xk e IF. Thus 23 rkxk = 23 sjui witb s, e I, Uj e F. kk kj Since each «3 is a linear combination of elements of X and I is an ideal, 23 siui is a ym linear combination of elements of X with coefficients in I. Consequently, 23 d fc = 1 = 23 siui = 53 r«>’( with ct e I, }U eX. The linear independence of X implies that distinct elements of X, then 0 =
j t= 1
(after reindexing and inserting terms Ox*, 0yt if necessary) m = d, xk = yk and rk = cke I for every k. Hence rk + I = 0 in R/I for every k and tt(X) is linearly in dependent over R/I. Thus F/IF is a free /^//-module with basis 7r(X) (Theorem 2.1). Finally if x, a:' eA'and7r(jf) = tt(x') in F/IF, then(lfl + I)tt(x) — (1* + I)^(x') = 0If x x', the preceding argument implies that 1 re I, which contradicts the fact that I 9^ R. Therefore, x = x' and the map tt : X —» ir(X) is a bijection, whence 1*1 = wm\. u
Proposition 2.11. Let f : R —» S be a nonzero epimorphism of rings with identity. If S has the invariant dimension property, then so does R. PROOF. Let I = Ker /;thenS == R/I (Corollary III.2.10). LetAand Fbe bases of the free /^-module F and 7r : F —» F/IF the canonical epimorphism. By Lemma 2.10 F/IF is a free R//-module (and hence a free S-moduIe) with bases tt(X) and 7t(Y) such that \X\ = |-7r(A")|, | Y\ = |-7r(F)|. Since S has the invariant dimension property, k(*)| = | tt(Y)\. Therefore, \X\ = \Y\ and R has the invariant dimension property. ■
Corollary 2.12. If R is a ring with identity that has a homomorphic image which is a division ring, then R has the invariant dimension property. In particular, every com mutative ring with identity has the invariant dimension property. PROOF. The first statement follows from Theorem 2.7 and Proposition 2.11. If R is commutative with identity, then R contains a maximal ideal M (Theorem III.2.18) and R/M is a field (Theorem III.2.20). Thus the second statement is a special case of the first. ■ We return now to vector spaces over a division ring and investigate the properties of dimension. A vector space V over a division ring D is said to be finite dimensional if dimdV is finite.
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Theorem 2.13. Let W be a subspace of a vector space V over a division ring D. (i) dinioW/ < dinioV', (ii) if dinioW = dimjyW and dimoV is finite, then W = V; (iii) dimjyW = dininW + ^/wd(V/W).
SKETCH OF PROOF, (i) Let Y be a basis of W. By Theorem 2.4 there is a basisXof ^containing Y. Therefore, = |Y| < [A| = d\mDV. (ii) If |F| = 1^1 and | A| is finite, then since Y C X we must have Y = X, whence W = V. (iii) We shall show that U = {* + tV \ x eX — Y\ is a basis of V/ W. This will imply (by Defini tion 8.3 of the Introduction) that d\mDV = |A^| = |K| + \X — Y\ = |F| + |t/| e Y; = dimDW + dimfi(F/(V). If v s V, then v = Y ny>i + 53 sixi
Xj s X — Y) so that v + W =
53 ■*;(-*■ j + W7).
Therefore, U spans V/ W. If j 53 ry(AV + **0 = 0 (r, e D, Xj eX — Y), then53 e whence 53 r i x i — 53 jjjk (sk £ D;yk e Y). This contradicts the linear independence of X = Y U (X — Y) unless r, = 0, sk = 0 for all j,k. Therefore, U is linearly independent and | U\ = \X — Y\. ■
Corollary 2.14. Iff : V —»V' is a linear transformation of vector spaces over a divi sion ring D, then there exists a basis X of V such that X fl Ker f is a basis of Ker f and {f(x) | f(x) 9^ 0, x £ X j is a basis of Im f. In particular, dimr>\ = dimxAKer f) + dimr>(Im f).
SKETCH OF PROOF. To prove the first statement let W = Ker /and let Y,X be as in the proof of Theorem 2.13. The second statement follows from Theorem 2.13 (iii) since V/W 2= Im/by Theorem 1.7. ■
Corol lary 2.15. // V and W are finite dimensional subspaces of a vector space over a division ring D, then
dimr>V + dmriW = dimr>(V fl W) -f- dimn(V + W).
SKETCH OF PROOF. Let A" be a basis of V fl W,Y a (finite) basis of V that contains X, and Z a (finite) basis of W that contains X (Theorem 2.4). Show that X U (F — X) U (Z — X) is a basis of V + W, whence dimn(V + W) = |A| + \ Y — X\ + |Z - X\ = dimn(V fl W) + (dim;>^ — dimD(V fl IV)) + (dimjjW — dim^K fl W)). ■ Recall that if a division ring R is contained in a division ring S, then S is a vector space over R with rs (s £ S,r e R) the ordinary product in S. The following theorem will be needed for the study of field extensions in Chapter V.
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Theorem 2.16. Let R,S,T be division rings such that R Cl S CZ T. Then dirnnT = (dimffT)(dimRS). Furthermore, dim^Y is finite if and only if dim sT and dim rS are finite. PROOF. Let U be a basis of T over S, and let V a basis of 5 over R. It suffices to show that [vu | v e V,u e £/} is a basis of Tover R. For the elements vu are all distinct by the linear independence of U over S. Consequently, we may conclude that dim^r = \U\\V\= (dim,s7’Xdim«.S'). The last statement of the theorem then follows immediately since the product of two finite cardinal numbers is finite and the product of an infinite with a finite cardinal number is infinite (Introduction, Theorem 8 .11 ). n
If u e T, then u — 53 (& e e £0 since U spans T as a vector space over 5. 1=1 m, Since 5 is a vector space over R each st may be written as s< = 53 r'iv> (r*ye R>vie ^)i=i r v u Thus u = 5) SM = 5) (2 rijVjfiii = 5) 51 u J i- Therefore, {um|i; £ £ t/} I 'i i> spans r as a vector space over i?. n 77i Suppose that EI rtj(v,u,) = 0 (r„ e e V,Ui e C/). For each i, let m i=lj=l r v Si = 53 *i i e S. Then 0 = 53 53 **,(»,«,) = 53 (53 r'ivdu< = 53 The linear i=l*i»i* independence of f/ over 5 implies that for each 0 = st =/, n-;Vj. The linear indei pendence of V over R implies that r„- = 0 for all ij. Therefore, j vu | v e V,u s (/) is linearly independent over R and hence a basis. ■
EXERCISES 1. (a) A set of vectors {xx , . . . , *„) in a vector space V over a division ring R is linearly dependent if and only if some xk is a linear combination of the pre ceding Xi. (b) If {xi,x2,x3} is a linearly independent subset of V, then the set {*i + x2, x2 + *3, *3 + *1} is linearly independent if and only if Char R ^ 2. [See Defini tion III. 1.8]. 2. Let R be any ring (possibly without identity) and X a nonempty set. In this exer cise an /?-module F is called a free module on X if F is a free object on X in the category of all left i?-modules. Thus by Definition 1.7.7, F is the free module on X if there is a function i : X —> F such that for any left i?-module A and function / :X—> A there is a unique .ft-module homomorphism / : F —»A with fi = f. (a) Let {Xi | /' e /) be a collection of mutually disjoint sets and for each i e I, suppose Fi is a free module on Xi, with i, : Xi —» Let X = [J Xt and F = 53 it/
iel
Fi, with fa : Fi —► Fthe canonical injection. Define i : X —» F by i(x) = faiXx) for x e Xi", (l is well defined since the Xi are disjoint). Prove that F is a free module on X. [Hint: Theorem 1.13 may be useful.] (b) Assume R has an identity. Let the abelian group Z be given the trivial /^-module structure (rm = 0 for all r e R, m e Z), so that R © Z is an /?-module with r(r',m) = (rr0) for all r,r' e R, m e Z. If X is any one element set,X = {/},
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let t : X —» R (0 Z be given by i(r) = (l/j,l). Prove that R® Z is a free module on X. [Hint: given f:X-^A, let A = B® C as in Exercise 1.17, so that fit) = b + c (b e B,c e C). Define fir,m) = rb + me.] (c) If R is an arbitrary ring and X is any set, then there exists a free module on X. [Hint. Since X is the disjoint union of the sets {/} with t e X, it suffices by (a) to assume X has only one element. If R has an identity, use (b). If R has no identity, embed R in a ring S with identity and characteristic 0 as in the proof of Theorem III.1.10. Use Exercise 1.18 to show that S is a free R-module on X.] 3. Let R be any ring (possibly without identity) and F a free /^-module on the set X, with i : X —* F, as in Exercise 2. Show that i(X) is a set of generators of the R-module F. [Hint: let G be the submodule of F generated by i(X) and use the definition of “free module” to show that there is a module homomorphism ip such that
is commutative. Conclude that
R/ip) is a field (Theorems III.2.20 and III.3.4). pA and A[p] are submodules of A. A/pA is a vector space over R/(p), with (r + (p))ia -f- pA) = ra + pA. A[p] is a vector space over R/(p), with (r + (p))a = ra.
5. Let V be a vector space over a division ring D and S the set of all subspaces of V, partially ordered by set theoretic inclusion. (a) S is a complete lattice (see Introduction, Exercise 7.2; the l.u.b. of Vi,V2 is V, + V2 and the g.l.b. F, fl V2). (b) S is a complemented lattice; that is, for each Vx s S there exists V 2 eS such that V - Vi + V2 and V\ fl V2 = 0, so that V = Vi 0 V2. (c) S is a modular lattice; that is, if Vxy2yz eS and V3 CI Vu then
vx n iv% + v3) = (Fi n v2) + v3. 6. Let R and C be the fields of real and complex numbers respectively. ( a ) dimRC = 2 and dimnR = 1. (b) There is no field K such that R C K C C. 7. If G is a nontrivial group that is not cyclic of order 2, then G has a nonidentity automorphism. [Hint: Exercise II.4.11 and Exercise 4(d) above.] 8. If V is a finite dimensional vector space and Vm is the vector space F © F 0- ■ • 0 V (m summands), then for each m > 1, Fm is finite dimensional and dim Vm = «(dim V).
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9. If Fi and F2 are free modules over a ring with the invariant dimension property, then rank (Fi © F2) = rank Fi + rank F2. 10. Let R be a ring with no zero divisors such that for all r,s e R there exist a,b e R, not both zero, with ar + bs = 0. (a) If R = K 0 L (module direct sum), then K = 0 or L = 0. (b) If R has an identity, then R has the invariant dimension property. 11. Let F be a free module of infinite rank a over a ring R that has the invariant di mension property. For each cardinal /3 such that 0 < /3 < a, F has infinitely many proper free submodules of rank /3. 12. If F is a free module over a ring with identity such that F has a basis of finite cardinality n > 1 and another basis of cardinality n + 1, then F has a basis of cardinality m for every m > n (m e N*). 13. Let K be a ring with identity and F a free X-module with an infinite denumerable basis {eue2,. .. |. Then R = Hornk(F,F) is a ring by Exercise 1.7(b). If n is any positive integer, then the free left R-module R has a basis of n elements; that is, as an .R-module, R = .R0---0R for any finite number of summands. [Hint: {1/e) is a basis of one element; { /i,/2j is a basis of two elements, where f(e2n) = er, /i(e2rl_i) = 0, /2(e2n) = 0 and /2(e2n_ i) = er. Note that for any g e R, g = gifi + £2/2, where gi(en) = g(e2n) and g^en) = g(e2n-i).] 14. Let / : V —» V be a linear transformation of finite dimensional vector spaces V and V’ such that dim V = dim V'. Then the following conditions are equivalent: (i) /is an isomorphism; (ii) /is an epimorphism; (iii) /is a monomorphism. [Hint: Corollary 2.14.] 15. Let R be a ring with identity. Show that R is not a free module on any set in the category of all R-modules (as defined in Exercise 2). [Hint. Consider a nonzero abelian group A with the trivial R-module structure (ra = 0 for all r e R, a e A). Observe that the only module homomorphism R —> A is the zero map.]
3. PROJECTIVE AND INJECTIVE MODULES Every free module is projective and arbitrary projective modules (which need not be free) have some of the same properties as free modules. Projective modules are especially useful in a categorical setting since they are defined solely in terms of modules and homomorphisms. Injectivity, which is also studied here, is the dual notion to projectivity.
Definition 3.1. A module P over a ring R is said to be projective if given any diagram of R-module homomorphisms P
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with bottom row exact (that is, g an epimorphism), there exists an K-module homo morphism h : P —> A such that the diagram P
is commutative (that is, gh = f). The theorems below will provide several examples of projective modules. We note first that if R has an identity and P is unitary, then P is projective if and only if for every pair of unitary modules A, B and diagram of /?-module homomorphisms
P f " A ^B 0 g
with g an epimorphism, there exists a homomorphism h : PA with gh = /. For by Exercise 1.17, A = Ax © A2 and B = BX(£)B2 with AUBX unitary and RA2 = 0 = RB2. Exercise 1.17 shows further that f(P) d Bx and g \ Ax is an epimorphism Ax —> Bu so that we have a diagram of unitary modules:
P
f \ 4^ ^
a1
b1 o
Thus the existence of h :P—> A with gh = f is equivalent to the existence of h : P —> A\ with gh = f.
Theorem 3.2. Every free module F over a ring R with identity is projective. REMARK. The Theorem is true if the words “with identity” are deleted and F is a free module in the category of all left /?-modules (as defined in Exercise 2.2). The proof below carries over verbatim, provided Exercise 2.2 is used in place of Theo rem 2.1 and the word “unitary” deleted. PROOF OF 3.2. In view of the remarks preceding the theorem we may assume that we are given a diagram of homomorphisms of unitary /^-modules:
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with g an epimorphism and F a free 7?-module on the set X (i :X^> F). For each x eX, f(i(x)) e B. Since g is an epimorphism, there exists ax e A with g(ax) = /(t(x)). Since F is free, the map X —> A given by x ax induces an /?-module homomor phism h : F —> A such that h{i{x)) = ax for all x eX. Consequently, ghi(x) = g(ax) = fi(x) for all x e X so that gfu = fi : X —» B. By the uniqueness part of Theorem 2.1 (iv) we have#/? = f Therefore F is projective. ■
Corollary 3.3. Every module A over a ring R is the homomorphic image of a projec tive R-module. PROOF. Immediate from Theorem 3.2 and Corollary 2.2. ■
Theorem 3.4. Let R be a ring. The following conditions on an R-module P are equivalent. (i) P is projective;
fg
(ii) every short exact sequence 0 —> A —> B —> P —>0 is split exact (hence B = A © P); (iii) there is a free module F and an R-module K such that F ~ K © P.
REMARK. The words “free module” in condition (iii) may be interpreted in the sense of Theorem 2.1 if R has an identity and P is unitary, and in the sense of Exercise 2.2 otherwise. The proof is the same in either case. PROOF OF 3.4. (i) => (ii) Consider the diagram
with bottom row exact by hypothesis. Since P is projective there is an /^-module homomorphism h : P —> B such that gh = 1 r. Therefore, the short exact sequence 0
—
>
h
0
i
s
split exact by Theorem 1.18 and B = A © P.
(ii) => (iii) By Corollary 2.2 there is a free /?-module F and an epimorphism
g : F —P. If K = Ker g, then 0 —1> K p—>> 0 is exact. By hypothesis the sequence splits so that F = AT© P by Theorem 1.18. (iii) => (i) Let tt be the composition F ~ K © P —► P where the second map is the canonical projection. Similarly let i be the composition P—* K@P = F with the first map the canonical injection. Given a diagram of /?-module homomorphisms P f aJLb-+
o
3. PROJECTIVE AND INJECTIVE MODULES
193
with exact bottom row, consider the diagram
F Trjfi P
Since F is projective by Theorem 3.2, there is an /^-module homomorphism hi m .F-^A such that gh\ = fir. Let h = h\i: P—> A. Then gh = ghxi = (fir)i = /(7ri) = flp = f. Therefore, P is projective. ■ EXAMPLE. If R = Z6, then Z3 and Z2 are Z6-modules (see Exercise 1.1) and there is aZ6-module isomorphism Z6 =Z2 (+)Z3. Hence both Zi andZ3 are projec tive Z6-moduIes that are not free Z6-modules.
Proposition 3.5. Let R be a ring. A direct sum ofR-modules 23 's projective if ill and only if each P; is projective. SKETCH OF PROOF. Suppose 23p* »s projective. Since the proof of (iii) => (i) in Theorem 3.4 uses only the fact that F is projective, it remains valid with P„
T. P, and Pj in place of F,K, and P respectively. The converse is proved by similar »Vi techniques using the diagram Pi lMtTj
T,p< v
gt a£b-*~ o If each Pj is projective, then for each j there exists h3 :P,—> A such that gh, — /t2 By Theorem 1.13 there is a unique homomorphism h :^Pi—+A with htj = A, for every j. Verify that gh — f. ■ Recall that the dual of a concept defined in a category (that is, a concept defined in terms of objects and morphisms), is obtained by “reversing all the arrows.” Pushing this idea a bit further one might say that a monomorphism is the dual of an epimorphism, since A —»B is a monomorphism if and only if 0 —» A —> B is exact and B —v A is an epimorphism if and only if ^ —> ^4 —> 0 (arrows reversed!) is exact. This leads us to define the dual notion of projectivity as follows.
Definition 3.6. A module J over a ring R is said to be injective if given any diagram of K-module homomorphisms
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B f J with top row exact (that is, g a monomorphism), there exists an R-module homomor phism h : B —> J such that the diagram B
J is commutative (that is, hg = f). Remarks analogous to those in the paragraph following Definition 3.1 apply here to unitary injective modules over a ring with identity. It is not surprising that the duals of many (but not all) of the preceding propositions may be readily proved. For example since in a category products are the dual concept of coproducts (direct sums), the dual of Proposition 3.5 is
Proposition 3.7. A direct product of R-modules
JJ
Ji is injective if and only ifi, is
injective for every i e l . PROOF. Exercise; see Proposition 3.5. ■ Since the concept of a free module cannot be dualized (Exercise 13), there are no analogues of Theorems 3.2 or 3.4 (iii) for injective modules. However, Corollary 3.3 can be dualized. It states, in effect, that for every module A there is a projective module P and an exact sequence P —> A —> 0. The dual of this statement is that for every module A there is an injective module J and an exact sequence 0 —» A —> J; in other words, every module may be embedded in an injective module. The remainder of this section, which is not needed in the sequel, is devoted to proving this fact for unitary modules over a ring with identity. Once this has been done the dual of Theo rem 3.4 (i), (ii), is easily proved (Proposition 3.13). We begin by characterizing in jective /?-modules in terms of left ideals (submodules) of the ring R.
Lemma 3.8. Let R be a ring with identity. A unitary R-module J is injective if and only if for every left ideal L of R, any R-module homomorphism L —* J may be ex tended to an R-module homomorphism R —> J. SKETCH OF PROOF. To say that f :L-^J may be extended to R means that there is a homomorphism h : R—>J such that the diagram
3. PROJECTIVE AND INJECTIVE MODULES
195
0-^L 5 -R
J is commutative. Clearly, such an h always exists if J is injective. Conversely, suppose J has the stated extension property and suppose we are given a diagram of module homomorphisms
0 +A *Lb
f 't J with top row exact. To show that J is injective we must find a homomorphism h :B -^>J with hg = f. Let S be the set of all R-module homomorphisms h C —»J, where Im g CI C C. B. S is nonempty since fg~* : Im g —* J is an element of S (g is a monomorphism). Partially order S by extension: hi < h2 if and only if Dom hi (Z Dom hi and hi | Dom hi = hi. Verify that the hypotheses of Zorn’s Lemma are satis fied and conclude that S contains a maximal element h : H—+J with hg = /. We shall complete the proof by showing H = B. If H 9^ B and beB — H, then L = (r e R \ rb s H\ is a left ideal of R. The map L —> J given by r |—► h(rb) is a well-defined R-module homomorphism. By hypothesis there is an R-module homomorphism k : R —> / such that k(r) = h(rb) for all r s L. Let c = k{\ r) and define a map h : H + Rb —*► J by a + rb (—> h(a) + rc. We claim that h is well defined. For if ai + nb = o2 + r2b e H -f- Rb, then ai — a2 = (r2 — n) b e H D Rb. Hence r2 — n e L and h(ax) — h(a2) = h(ax — a2) = h((r2 — r{)b) = k (r2 — ri) = (r2 — ri)k(lit) = (r2 — n)c. Therefore, h(ai + nb) = h(ai) + nc = h(a-2) + r2c = h(a2 + r2b) and h is well defined. Verify that h\ H Rb-^J is an R-module homomorphism that is an element of the set S. This contradicts the maximality of h since b § H and hence H (Z H + Rb. Therefore, H = B and J is injective. ■
An abelian group D is said to be divisible if given any y e D and 0 ^ n s Z, there exists x z D such that nx = y. For example, the additive group Q is divisible, but Z is not (Exercise 4). It is easy to prove that a direct sum of abelian groups is divisible if and only if each summand is divisible and that the homomorphic image of a divisible group is divisible (Exercise 7).
Lemma 3.9. An abelian group D is divisible if and only if D is an injective (unitary) Z.-module. PROOF. If D is injective, y e D and 0 3^ n e Z, let /: (n) —► D be the unique homomorphism determined by « H1- K ((«> is a free Z-module by Theorems 1.3.2 and II.1.1). Since D is injective, there is a homomorphism h : Z —> D such that the diagram
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o-H-z
A? D is commutative. If x = /?(1), then nx — nh( 1) = h(n) = fin) = y. Therefore, D is divisible. To prove the converse note that the only left ideals of Z are the cyclic groups («), n e Z. If D is divisible and / : (n) —» D is a homomorphism, then there exists x e D with nx = /(«). Define //: Z —> £> by 1|—> x and verify that h is a homomorphism that extends /. Therefore, D is injective by Lemma 3.8. ■ REMARK. A complete characterization of divisible abelian groups (injective unitary Z-modules) is given in Exercise 11.
Lemma 3.10. Every abelian group A may be embedded in a divisible abelian group. PROOF. By Theorem II. 1.4 there is a free Z-module F and an epimorphism F —> A with kernel K so that F/K= A. Since F is a direct sum of copies of Z (Theorem II. 1.1) and Z C Q , Fmay be embedded in a direct sum D of copies of the rationals Q (Theorem 1.8.10). But D is a divisible group by Proposition 3.7, Lemma 3.9, and the remarks preceding it. If / : F —> D is the embedding monomor phism, then /induces an isomorphism F/K ™ f(F)/f(K) by Corollary 1.5.8. Thus the composition A ~ F/K = /(F)/.f(K) C D/f(K) is a monomorphism. But D/f(K) is divisible since it is the homomorphic image of a divisible group. ■ If R is a ring with identity and J is an abelian group, then Hom7(/?,7), the set of all Z-module homomorphisms R —»J, is an abelian group (Exercise 1.7). Verify that Horn7//?,./) is a unitary left /^-module with the action of R defined by (r/)(x) = f(xr), (r,x e R ; fe Homz(R,J)).
Lemma 3.11. If 3 is a divisible abelian group and R is a ring with identity, then //omz(R,J) is an injective left R-module. SKETCH OF PROOF. By Lemma 3.8 it suffices to show that for each left ideal L of R, every R-module homomorphism / : L —» Homz(/?,7) may be extended to an /^-module homomorphism h : /? —► Homz(/?,7). The map g : L —>7 given by g(a) = [ /(«)](!n) is a group homomorphism. Since 7 is an injective Z-module by Lemma'3.9 and we have the diagram 0+z,S*
g J there is a group homomorphism g : R—+J such that g \ L = g. Define h. R—> Homy//?,7) by rh> h(r), where h(r) : /? —> 7 is the map given by [//(r)](x) = g(xr)
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197
(.x b R). Verify that his a well-defined function (that is, each h(r) is a group homo morphism /?—>./) and that h is a group homomorphism R —* Homz(R,J). If s,r,x b R, then h(.sr)(x) = g(x(sr)) = g({xs)r) = h(r)(xs). By the definition of the /^-module structure of Homz(/?,J), h(r)(xs) = [s/»(r)](.x), whence h(sr) = sh(r) and h is an /?-module homomorphism. Finally suppose r b L and x b R. Then xr bL and
h(r)(x) = g(xr) = g(xr) = [/(^r)](lfi). Since /is an R-module homomorphism and Homy(R,J) an /^-module, [/(*r)]U*) = [x/(r)](l«) = f(r)(lRx) = f(r)(x). Therefore, h(r) = f(r) for rzL and h is an extension of/. ■ We are now able to prove the duals of Corollary 3.3 and Theorem 3.4.
Proposition 3.12. Every unitary module A over a ring R with identity may be em bedded in an injective K-module. SKETCH OF PROOF. Since A is an abelian group, there is a divisible group J and a group monomorphism / : A —> J by Lemma 3.10. The map / : Homz(/?,.4) —» Hornz(RJ) given on g e Homz(/?,/4) by /(g) = /ge Homz(/?,./) is easily seen to be an R-module monomorphism. Since every /^-module homomorphism is a Z-module homomorphism, we have HomK(/?,/4) C Homz(/?,/4). In fact, it is easy to see that HomR(/?,^) is an K-submodule of Homz(/?,v4). Finally, verify that the map A —> Homfi(/?,/l) given by flh/„, where fjjr) = ra, is an /^-module monomorphism (in fact it is an isomorphism). Composing these maps yields an /^-module monomorphism
A —> HomECR,/4) -^> Hom^/^,/4) Homz(/?,/). Since Homz(/?,J) is an injective /^-module by Lemma 3.11, we have embedded A in an injective. ■
Proposition 3.13. Let K be a ring with identity. The following conditions on a unitary K-module J are equivalent. (i) J is injective; (ii) every short exact sequence 0 — » J — > B - ^ » C — > 0 is split exact (hence B s J © C); (iii) J is a direct summand of any module B of which it is a submodule. SKETCH OF PROOF, (i) => (ii) Dualize the proof of (i) => (ii) of Theorem 3.4. I—
(ii) =* (iii) since the sequence 0 —»./—»2? —>B/J—> 0 is split exact, there is a homo morphism g:B/J—> B such that ng= \B/J. By Theorem 1.18 ((i) =* (iii)) there is an isomorphism J © B/J ^ B given by (x,y) \—► x + g(y). It follows easily that B is the
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internal direct sum of J and g(B/J). (iii) (i) It follows from Proposition 3.12 that J is a submodule of an injective module Q. Proposition 3.7 and (iii) imply that J is injective. ■
EXERCISES Note: R is a ring. If R has an identity, all /^-modules are assumed to be unitary. 1. The following conditions on a ring R [with identity] are equivalent: (a) Every [unitary] /?-module is projective. (b) Every short exact sequence of [unitary] /^-modules is split exact. (c) Every [unitary] /?-module is injective. 2. Let R be a ring with identity. An /^-module A is injective if and only if for every left ideal L of R and /^-module homomorphism g : L —> A, there exists aeA such that g(r) = ra for every r e L. 3. Every vector space over a division ring D is both a projective and an injective Z)-module. [See Exercise 1.] 4. (a) For each prime p,Z{jF) (see Exercise 1.1.10) is a divisible group. (b) No nonzero finite abelian group is divisible. (c) No nonzero free abelian group is divisible. (d) Q is a divisible abelian group. 5. Q is not a projective Z-module. 6. If G is an abelian group, then G = D @ N, with D divisible and N reduced (meaning that N has no nontrivial divisible subgroups). [Hint: Let D be the sub group generated by the set theoretic union of all divisible subgroups of G'.] 7. Without using Lemma 3.9 prove that: (a) Every homomorphic image of a divisible abelian group is divisible. (b) Every direct summand (Exercise 1.8.12) of a divisible abelian group is divisible. (c) A direct sum of divisible abelian groups is divisible. 8. Every torsion-free divisible abelian group D is a direct sum of copies of the ra tionals Q. [Hint: if 0 ^ n e Z and a e D, then there exists a unique b e D such that nb = a. Denote b by (1 /n)a. For in, n s Z (n ^ 0), define (m/ri)a = m(\/n)a. Then D is a vector space over Q. Use Theorem 2.4.] 9. (a) If D is an abelian group with torsion subgroup D t , then D/D t is torsion free, (b) If D is divisible, then so is D t , whence D = D t @E, with E torsion free. 10. Let p be a prime and D a divisible abelian p-group. Then D is a direct sum of copies of Zip03). [Hint: let X be a basis of the vector space D[p) over Zv (see Exercise 2.4). If xeX, then there exists xi,xz,x3,. . . e D such that xt = .v, |xi| = p, px-2 = xi, px3 = x 2 ,... ,px n+1 = x„, . . . . If Hx is the subgroup generated by the jc,, then Hx = Z(p° ) by Exercise 1.3.7. Show that D = 53 Hx.] xeX
11. Every divisible abelian group is a direct sum of copies of the rationals Q and copies of Z(p<*>) for various primes p. [Hint: apply Exercise 9 to D and Exercises
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7 and 8 to the torsion-free summand so obtained. The other summand Dt is a direct sum of copies of various Zf/?”) by Exercises 7, 10 and H.2.7.] 12. Let G,H,K be divisible abelian groups. (a) If G @ G = H @ H, then G = H [see Exercise 11]. (b) If G © H^ G ® K, then Hg^K [see Exercises 11 and IL2.11.]. 13. If one attempted to dualize the notion of free module over a ring R (and called the object so defined “co-free”) the definition would read: An /^-module F is cofree on a set X if there exists a function i : F—>X such that for any /^-module A and function / : A-+X, there exists a unique module homomorphism / : A—* F such that if = /(see Theorem 2.1(iv)). Show that for any set X with 1^1 > 2 no such /^-module F exists. If \X\ = 1, then 0 is the only co-free module. [Hint: If F exists and 1^1 > 2, arrive at a contradiction by considering possible images of 0 and constructing f :R-^X such that if ^ / for every homomorphism f'.R^F.] 14. If D is a ring with identity such that every unitary /5-module is free, then D is a division ring. [Hint: it suffices by Exercise III.2.7 and Theorem III.2.18 to show that D has no nonzero maximal left ideals. Note that every left ideal of D is a free /5-module and hence a (module) direct summand of D by Theorem 3.2, Exercise 1, and Proposition 3.13.]
4. HOM AND DUALITY We first discuss the behavior of Homfi(/4,Z?) with respect to induced maps, exact sequences, direct sums, and direct products. The last part of the section, which is essentially independent of the first part, deals with duality. Recall that if A and B are modules over a ring R, then HomB(/4,Z?) is the set of all /^-module homomorphisms / : A —» B. If R = Z we shall usually write Hom(/4,fl) in place of Homz(^,B). Homft(/4,B) is an abelian group under addition and this addi tion is distributive with respect to composition of functions (see p. 174).
Theorem 4.1. Let A,B,C,D be modules over a ring R and
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by and is denoted ip. Similarly if A = C and
Theorem 4.2. Let R be a ring. 0 —» A B C is an exact sequence of R-ntodules if and only if for every R-module D 0 —> Homn(D,A) Z/o/wr(D,B) —> HomR(D,C)
is an exact sequence of abelian groups. PROOF. If 0 A ^ B U C is exact we must prove: (i) Ker
= 0 => ipf(x) — 0 for all x e D. Since 0 —» A B is exact, ip is a monomorphism, whence /(x) = 0 for all x e D and / = 0. Therefore,- Ker
8 i* Proposition 4.3. Let R be a ring. A —> B —> C —> 0 is an exact sequence of R-mod ules if and only if for every R-module D 0 —* Hom\i(C,Xy) Homn(B,D) //omR(A,D)
is an exact sequence of abelian groups. SKETCH OF PROOF. If A B C —> 0 is exact, we shall show that Ker 6 d Im f. If /e Ker 6, then 0 = 6(f) = f6, whence 0 = /(Im 6) = /(Ker f)- By Theorem 1.7 /induces a -homomorphism / : B/Ker f —» D such that f(b -f- Ker f) = f(b). By Theorem 1.7 again there is an isomorphism
7t — 0, whence C = Im f and B C —» 0 is exact. Similarly, show that Ker f C Im0
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201
by considering D = BjIm 6 and the canonical epimorphism B—*D. Finally, if
D — C, then 0 = 0f(lc) = whence Im 6 C Ker f. Therefore, A B C —> 0 is exact. ■ One sometimes summarizes the two preceding results by saying that WomR(A,B) is left exact. It is not true in general that a short exact sequence 0 —>A—>B—>C—>0 induces a short exact sequence 0 —» Hom/e(D,A) —> Homfl(Z),5) —» HomB(Z),C) —> 0 (and similarly in the first variable; see Exercise 3). However, the next three theorems show that this result does hold in several cases.
Proposition 4.4. The following conditions on modules over a ring R are equivalent. (i) 0—>A-^+B-^+C—>0 is a split exact sequence of R-modules; (ii) 0 —> //cwr(D,A) —► Homr(D,B) Hornr(D,C) —> 0 is a split exact se quence of abelian groups for every R-module D; (iii) 0 —> Homn(C,D) Hom r(B,D) ^ //owr(A,D) —> 0 is a split exact se quence of abelian groups for every R-module D. SKETCH OF PROOF, (i) => (iii) By Theorem 1.18 there is a homomorphism a : Z? —► A such that cap = \A. Verify that the induced-homomorphism
a : Hom/e(/l,Z)) —> Hom^(5,D) is such that Tpa. = lnom^A.o. Consequently, Tp is an epimorphism (Introduction, Theorem 3.1) and the Homfl sequence is split exact by Proposition 4.3 and Theorem 1.18. (iii) => (i) If D = A and f : B —> A is such that 1a =
Theorem 4.5. The following conditions on a module P over a ring R are equivalent (i) P is projective; (ii) iffy : B —»C is any R-module epimorphism then $ : //owr(P,B) —» Hom r(P,C) is an epimorphism of abelian groups; (iii) if 0 —> A —> B —» C —» 0 is any short exact sequence of R-modules, then 0 —> Homn(P,A) //owR(P,B) —► //cwr(P,C) —> 0 is an exact sequence of abelian groups. SKETCH OF PROOF, (i) <=> (ii) The map £ : Homfl(P,£) -» Hom/e(P,C) (given by g \pg) is an epimorphism if and only if for every /^-module homomor phism / :P—► C, there is an /?-module homomorphism g : P—> B such that the diagram
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is commutative (that is, / = ypg = \p(g)). (ii) => (iii) Theorem 4.2. (iii) => (ii) Given an epimorphism \p : B —> C let A = Ker yp and apply (iii) to the short exact sequence
0^A^>B-tC-^0. m
Proposition 4.6. The following conditions on a module J over a ring R are equivalent. (i) J is injective; (ii) if 6 :A —> B is any K-module monomorphism, then 6 :7/o/wr(B,J) —» //o/?2r(A,J) is an epimorphism of abelian groups; (iii) if 0 —» A —* B —> C —> 0 is any short exact sequence of R-modules, then 0 —> Homn(C,]) —> ffomR(B,J) —> HomniA,J) —> 0 is an exact sequence of abelian groups. PROOF. The proof is dual to that of Theorem 4.5 and is left as an exercise. ■
Theorem 4.7. Let A,B, {A; | i elj and {Bj | j e J} be modules over a ring R. Then there are isomorphisms of abelian groups:
(i) //owr(23iel Ai,B) = II Homn( A ,B); iel ;
(ii) //OWr(A,
//owR(A,Bj). jeJ
jeJ
REMARKS. If I and J are finite, then T" At = JI Ai and 23 iel
iel
=
II ^ ^ jeJ
jeJ
and J are infinite, however, the theorem may be false if the direct product H is re placed by the direct sum 23 (see Exercise 10). SKETCH OF PROOF OF 4.7. (i) For each i e
/ let i, : A, -> 23 Ai
b
e
the
iel
canonical injection (Theorem 1.11). Given {g, } e H Homfl(/l„fi), there is a unique iel
/^-module homomorphism g :
23 Ai —* B such that gu = g{ for every iel (Theorem iel
_
1.11). Verify that the map \p : H Homfi(A,£) —» Homff(y^/ltvB) given by |?,jh? is a homomorphism. Show that the map
r(as) = (ra)s for all a e A, r e R, s e S. We sometimes write rAs to indicate the fact that A is an R-S indicates a left /^-module B and Cs a right 5-module C.
rB
bimodule. Similarly
4. HOM AND DUALITY
203
EXAMPLES. Every ring R has associative multiplication and hence is an R-R bimodule. Every left module A over a commutative ring R is an R-R bimodule with ra = ar (a s A, r e R).
Theorem 4.8. Let R and S be rings and let rA, rBs, rCs, rD be (bi)modules as in dicated. (i) HomniA,B) is a right S-module, with the action of S given by (fs)(a) = (f(a))s (s e S; a e A; f e //cwjr(A,B)). (ii) If
irf){a) = fiar) = /(ra) = rf(a) = (f(a))r = (fr)(a). It follows that HomR(/l,B) is an R-R bimodule with rf = fr for all r e R, /e Homft(v4,Z?).
Theorem 4.9. If A is a unitary left module over a ring R with identity then there is an isomorphism of left R-modules A = Homu(R,A). SKETCH OF PROOF. Since R is an R-R bimodule, the left module structure of HomR(/?,/4) is given by Theorem 4.8(iii). Verify that the map : HomR(/?,/l) —* A given by f\-> /(li?) is an /^-module homomorphism. Define a map \p : A —> HomR(/?,/4) by a (—► /, where f{r) = ra. Verify that is a well-defined .ft-module homomorphism such that
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Theorem 4.10. Let A,B and C be left modules over a ring R. (i) If
left vector spaces, then 0 —»C* —> B* —> A* —> 0 is a short exact sequence of right vector spaces. PROOF. Exercise; see Theorems 2.4, 3.2, 4.1, 4.5, and 4.7. The map ip of (i) is called the dual map of
(nai + r2a2,f) = n(aij) + r2(a2,f) (n eR,fe A*, a, e A).
(1)
Similarly since A* is a right R-module with (fr)(a) = f(a)r, we have
(a, fn + f2r2) = (a,f)n + (aj2)r2 (n e R, fie A*, a e A).
(2)
In the proofs below we shall use the brackets notation for linear functionals as well as the Kronecker delta notation: for any index set I and ring R with identity the symbol 8,} (i,j e I) denotes 0 e R if / j and 1« e R if / = j.
Theorem 4.11. Let ¥ be a free left module over a ring R with identity. Let X be a basis of F and for each x s X let fx : F > R be given by fx(y) = 6XV (y e X). Then (i) {fx | x e X] is a linearly independent subset of F* of cardinality |X|; (ii) i/X is finite, then F* is a free right R-module with basis {fx | x e X}. REMARKS. The homomorphisms fx are well defined since F is free with basis X (Theorem 2.1). In part (ii), { fx \ x e A| is called the dual basis to X. This theorem is clearly true for any vector space V over a division ring by Theorem 2.4. In particular, if V is finite dimensional, then Proposition 2.9 and Theorem 4.11 imply that dim V = dim V* and V ^ V*. However, if V is infinite dimensional then dim V* > dim V (Exercise 12). More generally, if F is a free module over an arbitrary ring (for ex ample, Z), F* need not be free (see Exercise 10). PROOF OF 4.11. (i) If fxjx + fX2r2+- ■ ■ + fnrn = 0 (n e R; .v, e X), then for each j = 0,1,2,. . . , n, 0 = (xh0) = \ X3, X] f*Si / = 23 (xi’f*i)ri = 23 = ri\ i=l / i i
Since = 0 for all j, \ fx \ x eX\ is linearly independent. If x ^ y e X, then fx(x) = Ir 0 = fv(x), whence fx ^ fy. Therefore, |A| = |{ fx \ x eX} |. (ii) If X is finite, say X = j x\,. . . , .v„}, and fe F*, let s, = f(xt) = (ae R and denote fXj by fi. If u e F, then u = n.vi + nx2 H-----b rnxn s F for some /v e R and
4. HOM AND DUALITY
^ ' f]Sj ) j
= 1 / \i = l > =
205
fiXi, ^ 1 fjSj
232] ri(xi,f)sj =23 23 'I'&A=23 r*v* * jiii
= 23 ri<^»/) = (53 ^«,/> = («,/)■ Therefore, / = /uv, + /2s2 H— • + /.s,, and { f} = {fx \ xzX] generates F*. Hence { fx | x e X\ is a basis and F* is free. ■ The process of forming duals may be repeated. If A is a left /^-module, then A* is a right i?-module and A** = (A*)* = Hom^Hom^A,R),R) (where the left hand HomB indicates all right /^-module homomorphisms) is a left jR-module (see Exercise 4(a)). A** is called the double dual of A.
Theorem 4.12. Let A be a left module over a ring R. (i) There is an K-module homomorphism 6 : A —> A**. (ii) If R has an identity and A is free, then 6 is a monomorphism. (iii) If R has an identity and A is free with a finite basis, then 6 is an isomorphism. A module A such that 6 : A —> A** is an isomorphism is said to be reflexive. PROOF OF 4.12. (i) For each a e A let 6(a) : A* —> R be the map defined by [#(«)](/) = («,/) £ R- Statement (2) after Theorem 4.10 shows that 6(a) is a homo morphism of right R-modules (that is, 6(a) e A**). The map 6 : A —> A** given by a f—► 6(a) is a left /^-module homomorphism by (1) after Theorem 4.10. (ii) Let Abe a basis of A. If a e A, then a = riXi + r2x 2 H— • + rnxn (ri e R; x:»eA). If 6(a) = 0, then for all /e A*, 0 = (a,f) =
= 23 r<{xi,f).
In particular, for / = fX} (j = 1,2,..., n), 0 = 23 ri(xi,fXj) = 23 fiSij = r,. i i Therefore, a = 23 rix< = 23 Ojci = 0 and 6 is a monomorphism. i i (iii) If A is a finite basis of A, then A* is free on the (finite) dual basis { fx | x e A} by Theorem 4.11. Similarly /4** is free on the (finite) dual basis \gx \ x eX], where for each x e X, gx : A * —> R is the homomorphism that is uniquely determined by the condition: gx(fy) = Sxy (y eX). But 0(a) e A** is a homomorphism A* —> R such that for every y e A
e(x)(fy) = (x,fv) = Sxy = gx(fv). Hence gx = 6(x) and {6(x) | x e Aj is a basis of A**. This implies that Im 6 = A**, whence 6 is an epimorphism. ■
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206
EXERCISES Note: R is a ring. 1. (a) For any abelian group A and positive integer m, Horn(Zm,A) = A[m\ = {a e A | ma = 0}. (b) Hom(Z„„Z„) (c) The Z-module Zm has Zm* = 0. (d) For each A; > 1, Zm is aZm*-module (Exercise 1.1); as aZmt-module,Zm* =Zm. 2. If A,B are abelian groups and m,n integers such that mA = 0 = nB, then every element of Horn(A,B) has order dividing (m,n). 3. Let 7r: Z —>Z2 be the canonical epimorphism. The induced map tt : Hom(Z2,Z) —> Hom(Z2,Z2) is the zero map. Since Hom(Z2,Z2) ^ 0 (Exercise 1(b)), tt is not an epimorphism. 4. Let R,5 be rings and AR, SBR, sCR, DR (bi)modules as indicated. Let HomK de note all right R-module homomorphisms. (a) Hornr(.A,B) is a left 5-module, with the action of 5 given by (sf)(a) = s(f(a)). (b) If
5. TENSOR PRODUCTS
207
is commutative, where dA,6/t are as in Theorem 4.12 and f* is the map induced on A** — Hom«(Hom«(/4,R),/?) by the map / : Homft(B,/?) —> Hom^/l,/?). 10. Let F — 52 Z* be a free Z-module with an infinite basis X. Then \fx | x e X\ xeX (Theorem 4.11) does not form a basis of F*. [Hint: by Theorems 4.7 and 4.9, F* = JT Z-v; but under this isomorphism f„\—> 18xyxj e JX Zr.] xzX
xeX
Note: F* = IIz, is not a free Z-module; see L. Fuchs [13; p. 168]. 11. If R has an identity and P is a finitely generated projective unitary left R-module, then (a) P* is a finitely generated projective right .R-module. (b) P is reflexive. This proposition may be false if the words “finitely generated” are omitted; see Exercise 10. 12. Let F be a field, X an infinite set, and V the free left /-"-module (vector space) on the set X. Let Fx be the set of all functions. / : X —> F. (a) Fx is a (right) vector space over F (with (/+ g)(-v) = f(x) + g(x) and (fr){x) = rf(x)). (b) There is a vector-space isomorphism V* ~ Fx. (c) dim/. Fx = |F[|A| (see Introduction, Exercise 8.10). (d) dim,. V* > dimy. V [Hint: by Introduction, Exercise 8.10 and Introduc tion, Theorem 8.5 dim/,- V* = dim/.- Fx = |F|IX| > 2|A | = |jP(A')| > \X\ = dimF V.]
5. TENSOR PRODUCTS The tensor product A (X)/e B of modules AR and rB over a ring R is a certain abelian group, which plays an important role in the study of multilinear algebra. It is frequently useful to view the tensor product A (X)/f B as a universal object in a certain category (Theorem 5.2). On the other hand, it is also convenient to think of A (x)R B as a sort of dual notion to Homr(A,B). We shall do this and consider such topics as induced maps and module structures for A (x)r B as well as the behavior of tensor products with respect to exact sequences and direct sums. If Ak and rB are modules over a ring R, and C is an (additive) abelian group, then a middle linear map from A X B to C is a function / : A X B —> C such that for all a,a, e A, b,bi s B, and r e R\ /(fli + ai,b) = f(aub) + f(a-,,b);
(3)
/(«,/>! -f bi) = f(ajh) + f(aJH)\
(4)
f(ar, b) = f(a,rb).
(5)
For fixed AH,,tB consider the category ?HT(/4,B) whose objects are all middle linear maps on A X B. By definition a morphism in 31l(/l,£) from the middle linear map / : A X B * C to the middle linear map g : A X B —> D is a group homomorphism h : C —> D such that the diagram
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CHAPTER IV MODULES
is commutative. Verify that 9TC(/i,B) is a category, that lc is the identity morphism from / to /, and that h is an equivalence in 9TC(/1,B) if and only if h is an isomorphism of groups. In Theorem 5.2 we shall construct a universal object in the category 9TC(/i,B) (see Definition 1.7.9). First, however, we need
Definition 5.1. Let A be a right module and B a lefi module oier a ring R. Let F be the free abelian group on the set A X B. Let K be the subgroup of F generated by all elements of the following forms (for all a,a' e A; b,b' e B; r e R): (i) (a + a',b) - (a,b) - (a',b); (ii) (a,b + b') - (a,b) - (a,b'); (iii) (ar,b) - (a,rb).
The quotient group F/K is called the tensor product of A and B; it is denoted A (X)r B (or simply A (X) B ifR = Z). The coset (a,b) + K. of the element (a,b) in F is denoted a (X) b; the coset of(0,0) is denoted 0. Since F is generated by the set A X B, the quotient group F/K = A (x)^ B is generated by all elements (cosets) of the form «(X) b (a e A, b e B). But it is not true that every element of A (x)/j B is of the form a (x) b (Exercise 4). For the typical eler
ment of F is a sum 23 thUuM) (rti e Z, a, e A, bi e B) and hence its coset in A (XV B i=1
= F/K is of the form
r
23 ® ^«)- Furthermore, since it is possible to choose i=1
different representatives for a coset, one may have a(£) b = a' (x) b' in A (x)^ B, but a ^ a' and b ^ b' (Exercise 4). It is also possible to have A (X^ B = 0 even though A 0 and B ^ 0 (Exercise 3). Definition 5.1 implies that the generators «(x)6 of A (x)rB satisfy the follow ing relations (for all a,ai s A, b,b, e B, and r s R): («i + a2) 0 b = ai (X) b + a2 (X) b\
(6)
a (X) (bi + 62) = a (X) b\ + a (X) bi\
(7)
ar (X) b — a(£)rb.
(8)
The proof of these facts is straightforward; for example, since (ai + a2,b) — (aub) — (a2,b) e K, the “zero coset,” we have [(«i + a2,b) + K) - [(aub) + K] - [(a2,b) + K] = K~, or in the notation (a,b) + K = « (x) b,
(ai + (X) b — ax (X) b — a2 (X) b = 0. Indeed an alternate definition of A (X)/? B is that it is the abelian group with genera tors all symbols a (X) b (a e A, b e B), subject to the relations (6)-(8) above. Further more, since 0 is the only element of a group satisfying x + x = x, it is easy to see that for all a s A, b s B: «®0 = 0(X)6 = 0®0 = 0.
5. TENSOR PRODUCTS
209
Given modules AR and RB over a ring R, it is easy to verify that the map i: A X B —* A (X)K B given by (a,b) a (x) b is a middle linear map. The map / is called the canonical middle linear map. Its importance is seen in
Theorem 5.2. Let Ar and rB be modules over a ring R, and let C be an abelian group. Ifg : A X B —► C is a middle linear map, then there exists a unique group homomor phism g : A (x)r B —► C such that gi = g, where i : A X B —■* A (X)r B is the canonical middle linear map. A (x)r B is uniquely determined up to isomorphism by this property. In other words i : A X B —» A (X)R B is universal in the category 3TC(A,B) of all middle linear maps on A X B. SKETCH OF PROOF. Let F be the free abelian group on the set A X B, and let K be the subgroup described in Definition 5.1. Since F is free, the assignment (a,b) I—> g(a,b) e C determines a unique homomorphism g, : F —* C by Theorem 2.1 (iv). Use the fact that g is middle linear to show that g\ maps every generator of K to 0. Hence K d Ker gy. By Theorem 1.7 gi induces a homomorphism g : F/K—» C such that g[(a,b) + K] = gi[(a,b)\ — g(a,b). But F/K — A (x)R B and (a,b) -J- K = a (X) b. Therefore, g : A (X)K B —► C is a homomorphism such that gi(a,b) = g(a (x) b) = g(a,b) for all (a,b) e A X B; that is, gi = g. If h : A (x)R B —+ C is any homomorphism with hi = g, then for any generator a (X) b of A (x)K B,
h(a (X) b) = hi(a,b) = g(a,b) = gi(a,b) = g(a ® b). Since h and g are homomorphisms that agree on the generators of A (x)/? B, we must have h = g, whence g is unique. This proves that i: A X B —► A (X)R B is a universal object in the category of all middle linear maps on A X B, whence A (x)K B is uniquely determined up to isomorphism (equivalence) by Theorem 1.7.10. ■
Corollary 5.3. If Ar, Ar', rB and rB' are modules over a ring R and f : A —» A', g : B —* B' are R-module homomorphisms, then there is a unique group homomorphism A (X)r B —> A' (X)r B' such that a (X) b H> f(a) (X) g(b) for all a e A, b e B. SKETCH OF PROOF. Verify that the assignment (a,b) |—> f(a) (X) g(b) defines a middle linear map h : A X B —> C = A' (X)R B'. By Theorem 5.2 there is a unique homomorphism h : A (X)K B —> A' (X)R B' such that h(a (X) b) = hi(a,b) = h(a,b) — f(°) ® 8(b) for all at A, be B. ■ The unique homomorphism of Corollary 5.3 is denoted /(X) g : A (x)^ B —► A' (y)it B'. If /' : AR' —> Ah" and g' : RB' —> RB" are also R-module homomorphisms, then it is easy to verify that (/' ® g'X f® g) = (/7(g) g'g) : A (g)R B ^ A" (X)R B". It follows readily that if /and g are R-module isomorphisms, then /(X) g is a group isomorphism with inverse / 1 (X) g_1.
Proposition 5.4. If A —^ B --> C —> 0 is an exact sequence of left modules over a ring R and D is a right R-module, then
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210
D(X) r A
D ®R B
D ®R C -» 0
is an exact sequence of abelian groups. An analogous statement holds for an exact se quence in the first variable. PROOF. We must prove: (i) Tm (I/;®#) = D ®w C; (ii) Im (1 C (X)/)G Ker (1 D ®g)\ and (iii) Ker (1 D ® g) CI Im (lz> (X)/). (i) Since g is an epimorphism by hypothesis every generator d ® c of D ®« C is of the form d ® g(b) = (lc (X) g)(d(x) b) for some beB. Thus Im (lz> (X) g) contains all generators of D ®K C, whence Im (1D ® g) = D ®fi C. (ii) Since Ker g = Im / we have gf = 0 and (1D ® g)(lD ®/) = 1 D ® gf= 1 D ® 0 = 0, whence Im (lz> (X)/) C Ker (l/j®g). (iii) Let tt : D (x)n B(D (X)R B)/Im ( I d ® / ) be the canonical epimorphism. By (ii) and Theorem 1.7 there is a homomorphism a : (D ®« B)/Im (1 ^ ® /) -^ Z) ®« C such that a(?r(^ ® 6)) = (1« ® g)(d ® Z>) = c/® g(b). We shall show that a is an isomorphism. This fact and Theorem 1.7 will imply Ker (1D ® g) = Im (1 D ® /) and thus complete the proof. We show first that the map (3 : D X C —» (D ®K B)/Im (1 c ® /) given by (tf,c) H> 7r(d(x)b), where g(b) = c, is independent of the choice of b. Note that there is at least one such b since g is an epimorphism. If g(b') = c, then g(b — b') = 0 and b — b' e Ker g = Im /, whence b — b' = f(a) for some a e A . Since
Theorem 5.5. Let R and S be rings and s A r , (i) A
®r
rB, Cr,
rDs (bi)modules as indicated.
B is a left S-module such that s(a ® b) = sa ® b for all s e S, a e A,
beB.
(ii) // f : A —> A' is a homomorphism of S-R bimodules and g : B - > B' is an R-module homomorphism, then the induced map f ® g : A ®r B —> A' ®r B' is a homomorphism of left S-modules. (iii) C ®r D is a right S-module such that (c ® d)s = c ® ds for all c e C, d e D, s e S. (iv) If h : C —» C' is an R-module homomorphism and k : D —> D' a homomor phism of R-S bimodules, then the induced map h ® k : C ®ft D —* C' ®r D' is a homomorphism of right S-modules.
211
5. TENSOR PRODUCTS
SKETCH OF PROOF, (i) For each 5 e 5 the map A X B -> A (g)fi B given by (1a,b) H- sa (x) b is K-middle linear, and therefore induces a unique group homomor phism as : A (x)K B —* A (x)/j B such that as(a (X) b) = sa (X) b. For.each element n
n
u = a* ® ^ s A (X)r £ define su to be the element a„(u) — 23 «*(«*' ® A.) 1—1 1=1
n 1
= 23 ® Since as is a homomorphism, this action of 5 is well defined (that is,
=1
independent of how u is written as a sum of generators). It is now easy to verify that A (x)ij B is a left 5-module. ■ REMARK. An important special case of Theorem 5.5 occurs when R is a com mutative ring and hence every K-module A is an R-R bimodule with ra = ar (r e R,a e A). In this case A (Xk B is also an R~R bimodule with
r(a (X) b) = ra^)b = ar(^)b = a(^)rb = a(^)br = (a (X) b)r for all r e R, a e A, b e B. If R is a commutative ring, then the tensor product of /^-modules may be char acterized by a useful variation of Theorem 5.2. Let A,B,C be modules over a com mutative ring R. A bilinear map from A X B to C is a function / : A X B —» C such that for all a,ai e A, b,bi s B, and r e R:
f(ui + a>,b) = f(aub) + f(a2,b); f(a, by + />,) = f(aA) + RciM)\ f(ra,b) = rf(a,b) = f(a,rb).
(9) (10) (11)
Conditions (9) and (10) are simply a restatement of (3) and (4) above. For modules over a commutative ring (11) clearly implies condition (5) above, whence every bi linear map is middle linear. EXAMPLE. If A* is the dual of a module A over a commutative ring R, then the map A X A* —> R given by (a,f) H> f(a) = («,/) is bilinear (see p. 204). EXAMPLE. If A and B are modules over a commutative ring R, then so is A (x)b B and the canonical middle linear map i: A X B —> A (x)^ B is easily seen to be bilinear. In this context i is called the canonical bilinear map.
Theorem 5.6. I f A,B,C are modules over a commutative ring R andg : A X B-^C is a bilinear map, then there is a unique K-module homomorphism g : A (X)r B —> C such that gi = g, where i : A X B —* A (x)r B is the canonical bilinear map. The module A (x)r B is uniquely determined up to isomorphism by this property. SKETCH OF PROOF. Verify that the unique homomorphism of abelian groups g : A (X)R B —> C given by Theorem 5.2 is actually an K-module homomor phism. To prove the last statement let <Sb(A,B) be the category of all bilinear maps on A X B (defined by replacing the groups C,D and group homomorphism h : C —» D by modules and module homomorphisms in the definition of 9fR(A,B) on p. 207).
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Then first part of the Theorem shows that i: A X B —► A (x)K B is a universal object in (B(A,B), whence A (x)K B is uniquely determined up to isomorphism by Theo rem 1.7.10. ■ Theorem 5.6 may also be used to provide an alternate definition of A B when R is a commutative ring with identity. Let Fi be the free R-module on the set A X B and Ki the submodule generated by all elements of the forms: (,a + a', b) — (a,b) — (o', b);
(
Theorem 5.7. 7/R is a ring with identity and An, there are R-module isomorphisms A(x) r R^A and R
rB
are unitary R-modules, then
®r B = B.
SKETCH OF PROOF. Since R is an R-R bimodule R (x)« B is a left Rmodule by Theorem 5.5. The assignment (r,b) \—> rb defines a middle linear map R X B—> B. By Theorem 5.2 there is a group homomorphism a : R (x)R B^B such that a(r (x) b) = rb. Verify that a is in fact a homomorphism of left R-modules. Then verify that the map (3 :B —► R (x)fi B given by b (—> \R (X) b is an R-module homomorphism such that a/3 = \u and /3a = 1/2®*./?. Hence a : R (X)R B = B. The isomorphism A (x)R R ^ A is constructed similarly. ■ If R and s' are rings and Ar, rBs, sC are (bi)modules, then A (x)rB is a right S-module and B (x)s C is a left R-module by Theorem 5.5. Consequently, both (A B) 0 SC and A (x)R (B (x) ,SC) are well-defined abelian groups.
Theorem 5.8. 7/R and S are rings and isomorphism
Ar, rBs,
sC are (bi)modules, then there is an
(A ®r B) (X)s C ^ A ®R (B 0s C). PROOF. By definition every element v of (A 0R B) ®s C is a finite sum n mi T. (u( e A K B, d e C). Since each «, e A R B is a finite sum 52 ai> ® ki i=l }=1 (a,, e A, bi, e B), we have
«,0c,
v
0
0
= 52 i ® < = 52 (52 n 0 ba) 0 = 52 52 [fe, 0 />„) 0 cj. u
c
a
5. TENSOR PRODUCTS
213
Therefore, (A (y)K B) (x)s C is generated by all elements of the form (a (x) b) (x) c (a e A, b e B, c e C). Similarly, ^4 (X)K (f? (x)s C) is generated by all a (x) (b (X) c) with
a £ A, b e B, c e C. Verify that the assignment (22 cu (X) £>„c) H 22 ® ® <")] Vi= 1
/ i=l
defines an S-middle linear map (A (x)K B) X C —> A (x)K (B ®5 C). Therefore, by Theorem 5.2 there is a homomorphism a :(A (x)R B) (x)s CA (g)R (B (g)s C) with a[(a (x) b) (X) c] = a (x) (b (X) c) for all a e ,4, 6 e B, c e C. Similarly there is an /^-middle linear map A X (B (x)s C) —> (/I(x)K B) (x)s C that induces a homo morphism
®rB)®sC such that (3[a (X) (/> (X) <-)] = (a (X) £>) (X) c for all o e ,4, £» e B,c e C. For every genera tor (a (X) A) (X) c of (/I 5) (X)s C, (3a[(a (X) b) (X) c] = (a (x) b) (x) c, whence /3a; is the identity map on (A (x)fi B) (X)s C. A similar argument shows that /3a is the identity on A (x)fi (B ®s Q- Therefore, a and /3 are isomorphisms. ■ In the future we shall identify (A (X)K B) ®s C and A (x)fi (B ®s C) under the isomorphism of Theorem 5.8 and simply write A (X)/j B ®s C. It is now possible to define recursively the «-fold tensor product:
A2 (x)r2 • • • (x)ftn An+\ where Ru ..., Rn are rings and ARll,RlAR2,..., RnAn+1 are (bi)modules. Such iter ated tensor products may also be characterized in terms of universal n-linear maps (Exercise 10).
Theorem 5.9. Let R be a ring, A and {A; | i s 1} right R-modules, B and {Bj | j e J} left R-modules. Then there are group isomorphisms.
(22 Ai) (X) B == 22 ® B); A ®R (22 By) ^22 (A ® B ). r
iel
r
iel
r }
jeJ
jtJ
PROOF. Let Lk, 7rfc be the canonical injections and projections of
22 A{. By i(I
Theorem 1.8.5 the family of homomorphisms ik (X) ls : Ak (y)r B —> induce a homomorphism a : J2 (Ai (x)R B) —*■ _
=
22o
(ti("i)
iel
® b) =
(22
__
ti(fl«)) iel
o
ie/
(22
iel
B such that a[{ a{ (x) b\]
iel
® b, where l0 = {/ e / | ai (X) b 0 j. The assign-
ment (u,b) ► ) *■»(«) (X) b\lzi defines a middle linear map -
22 (^*
(22 A)(x)*B
(22 Aj) x B —>► ie/
an^
thus induces a homomorphism /3 : (22 ^0 B —> 22 (^
iel
such that /3(w (x) £>) = { tt,(w) (x) b) . We shall show that a/3 and /3a are the respec tive identity maps, whence a is an isomorphism. itI
CHAPTER IV MODULES
214
Recall that if u e52 A- and 70 = {/e /! 7r,(w) ^ 0}, then « = 52 l*7r*(M). Thus ie/o
for every generator «(x) b of (52 ^«) ^
we
have
a/3(w (x) Z>) .= o:[{ 7T,(tt) (X) 6} ] = (52 t£7T»<M)) ® b = u ® />.
ie/o
Consequently a/3 is the identity map. For each ye/ let 1,* : /!,- (x) R/? —» 52 (^* the canonical injection and _i verify that 52 (A< *s generated by all elements of the form i*{a (x) b) =
i
{7TiL,{a) (X) b\id (j e I, ae A„ b e B). For each such generator we have (tt,t,(o)) (X) b — 0 if / 9^ j and (7r,-i,-(o)) (X) b = a (X) 6, whence /Sa[i3*(« 0 6)] = /3a[{7r,i,{fl) ® 6}] = j8[i,7r/i,{a) (X) 6] (X) 6] = {7Tit,(o) (X) 6}le; = i,*(a ® b).
=
Consequently the map (3a must be the identity. The second isomorphism is proved similarly. ■
Theorem 5.10. (Adjoint Associativity) Let R and S be rings and Ar, rBs, Cs (bi)modules. Then there is an isomorphism of abelian groups
a : Homs(A (X)r B,C) = Homn(A,Homs(B,C)), defined for each f : A
(X)r
B > C by [(afXa)J(b) = f(a(x)b).
Note that Homfi(____,__ ) and Hom,s(__ ,__ ) consist of homomorphisms of right modules. Recall that the R-module structure of Hom,s(B,C) is given by: (gr)(b) = g{rb) (for r e R, b £ B, g e Homs(B,C); see Exercise 4.4(c)). SKETCH OF PROOF OF 5.10. The proof is a straightforward exercise in the use of the appropriate definitions. The following items must be checked. (i) For each aeA, and fe Homs(/4 (X)R B,C), (af)(a) : B —* C is an S-module homomorphism. (ii) (af) : A —> Homs(R,C) is an R-module homomorphism. Thus a is a welldefined function. (iii) a is a group homomorphism (that is, a(J\ + /2) = a(fi) + a(f)). To show that a is an isomorphism, construct an inverse map /3 : Homft(/4,Homs(R,C)) —» Hom,s(y4 (x)K B,C) by defining
(fig)(a (X) b) = \g{a)\{b), where a e A, b e B, and g e Hom/c(A,Homs(B,C)). Verify that (iv) /3g as defined above on the generators determines a unique 5-module homo morphism A (x)/j B —» C. (v) (3 is a homomorphism. (vi) (3a and a(3 are the respective identities. Thus a is an isomorphism. ■ We close this section with an investigation of the tensor product of free modules. Except for an occasional exercise this material will be used only in Section IX.6.
5. TENSOR PRODUCTS
215
Theorem 5.11. Let R be a ring with identity. If A is a unitary right R-module and F is a free left R-module with basis Y, then every element u of A (X)r F may be written n uniquely'in the form u = 53 ai ® Yi> where as e A and the y, are distinct elementsofY. i= 1
t
m
REMARK. Given u = 53 ak®.yk and v = 53 (X) z,- (a^M e A, yk,z, e F), k = 1 7=1 we may, if necessary, insert terms of the form 0 (X) y (y e Y) and assume that n
n
u = 53 ® >’*' ar*d v = 53 ® The word “uniquely” in Theorem 5.11 means i=i i=i n n n a that if 53 *' = 53 ^ ® >'*, then u, = /?, for every /. In particular, if 53 ® >’> » = 1 f=l «=1 K = 0 = 53 0 (8) >v, then a, = 0 for every /. i= 1
PROOF OF 5.11. For each y e Y, let Ay be a copy of A and consider the direct sum 53 Av. We first construct an isomorphism 6 : A (X)« F = 53 A as follows. J/eF , yeF Since Y is a basis, {>>} is a linearly independent set for each y zY. Consequently, the R-module epimorphism ip : R —» Ry given by r |—> ry (Theorem 1.5) is actually an isomorphism. Therefore, by Theorem 5.7 there is for each y zY an isomorphism ___
1 A^
_
A (g)B Ry----- > A R ^ A = Au.
Thus by Theorems 5.9 and 1.8.10 there is an isomorphism 6:
A (X)« F = A (X)fl (53 Ry) = 53 yeY
yzY
A
Ry = 53 Av yeY
Verify that for every a s A, z e Y, 6(a (x) z) = {uy 1 e 53 where uz = a and uv = 0 for y z; in other words, 0(« (X) z) = i£(a), with iz : Az —> 53 Ay the canonical in jection. Now every nonzero re53 A *s a finite sum v = 1^(01) -1— •+ iy„(an) = 0(ai (X) >1) + • —1- 6(a„ ® >'n) with yi, . . . , yn distinct elements of Y and ai uniquely determined nonzero elements of A. It follows that every element of A (X)R F n (which is necessarily 6~\v) for some v) may be written uniquely as 53 Qi ® )’<■ ■ 1
=1
Corollary 5.12. If R is a ring with identity and Ar andrB are free R-modules with bases X and Y respectively, then A (x)R B is a free (right) R-module with basis W = {x (X) y | x e X,y e Y j of cardinality |X||Y|. ~ REMARKS. Since R is an R-R bimodule, so is every direct sum of copies of R. In particular, every free left R-module is also a free right R-module and vice versa. However, it is not true in general that a free (left) R-module is a free object in the cat egory of R-R bimodules (Exercise 12). SKETCH OF PROOF OF 5.12. By the proof of Theorem 5.11 and by Theo rem 2.1 (for right R-modules) there is a group isomorphism
CHAPTER IV MODULES
216
Since fi is an R-R bimodule by the remark preceding the proof, A (x)/e fi is a right /?-module by Theorem 5.5. Verify that 6 is an isomorphism of right /?-modules such that 6(W) is a basis of the free right /^-module xR). Therefore, A (x)/e fi is a YX
free right R-module with basis W. Since the elements of W are all distinct by Theo rem 5.11, \IV\ = \X\\Y\. m
Corollary 5.13. Let S be a ring with identity andR a subring of S that contains Is- If F is a free left R-module with basis X, then S (x)r F is a free left S-module with basis {Is (x) x | x e X} of cardinality |X|. SKETCH OF PROOF. Since S is clearly an S-R bimodule, S (x)/e F is a left S-module by Theorem 5.5. The proof of Theorem 5.11 shows that there is a group isomorphism 6 : 5 (x)R F = 52 w'th each Sx = S. Furthermore, if for z e X, xeX
c-.s-s.^Z Sx is the canonical injection, then 0(ls (x) z) = iz(ls) for each zsX. xeX
Verify that 6 is in fact an isomorphism of left S-modules. Clearly, {i^(ls) | xsX} is a basis of cardinality \X\ of the free left 5-module 52 whence S (x)R F is xeX
a free 5-module with basis {lsCx)*!*^} of cardinality \X\. ■
EXERCISES Note: R is a ring and ® = (x)z1. If R = Z, then condition (iii) of Definition 5.1 is superfluous (that is, (i) and(ii) imply (iii)). 2. Let A and fi be abelian groups. (a) For each m > 0, A (x)Zm ■= A/mA. (b) Zm (x) Zn =ZC, where c - (m,n). (c) Describe A (x) fi, when A and fi are finitely generated. 3. If A is a torsion abelian group and Q the (additive) group of rationals, then (a) A®Q = 0. (b) Q (x) Q = Q. 4. Give examples to show that each of the following may actually occur for suitable rings R and modules AR, RB. (a) A ®RB 5* A ®z fi. (b) u e A (x)fl fi, but u o (X) b for any a e A, b e fi. (c) a (x) b = Ai (x) bi but a 5^ fli and b bi. 5. If A' is a submodule of the right /?-module A and B' is a submodule of the left /?-module fi, then A/A' (x)fl fi/fi' = (A (x)R B)/C, where C is the subgroup of A (x)fl fi generated by all elements a' (x) b and fl (x) b' with a e A, a' e A', b e fi, b' e B'. 6. Let / : AR-^> AR and g : RB —> RB' be /?-module homomorphisms. What is the difference between the homomorphism/(x) g (as given by Corollary 5.3) and the element f®)g of the tensor product of abelian groups HornR(A,A') (X) HornR{B,B’)1
5. TENSOR PRODUCTS
217
7. The usual injection a : Z2 —> Z4 is a monomorphism of abelian groups. Show that 1 (x) a : Z2 ® Z2 —*• Z2 (x) Z4 is the zero map (but Z2 ® Z2 ^ 0 andZ2 (x) Z4 5^ 0; see Exercise 2). 8. Let 0 —> /I —> K —* C —► 0 be a short exact sequence of left K-modules and D a right K-module. Then 0 —> D 0)R /I I) 0)H B D (x)K C —> 0 is a short exact sequence of abelian groups under any one of the following hypotheses: (a) 0 —* A -4 B -^> C —> 0 is split exact. (b) K has an identity and Z) is a free right K-module. (c) K has an identity and D is a projective unitary right K-module. 9. (a) If / is a right ideal of a ring K with identity and B a left K-module, then there is a group isomorphism R/I0)r B ^ B/IB, where IB is the subgroup of B generated by all elements rb with r e /, bsB. (b) If K is commutative and I,J are ideals of K, then there is an K-module iso morphism R/I (x)/{ R/J = K/(/ + J). 10. If R,S are rings, AR, RBS, sC are (bi)modules and D an abelian group, define a middle linear map to be a function f\ AXBXC-^D such that (i) (ii) (iii) (iv) (v)
f(a + a',b,c) = f(a,b,c) + f{a’,b,c); f(a,b + b’,c) = f(a,b,c) + f(a,b',c); f(a,b,c + c') = f(a,b,c) + f(a,b,c'); f(ar,b,c) = f(a,rb,c) for r e K; f(a,bs,c) = f(a,b,sc) for s e 5.
(a) The map i: A X B X C—>(A (x)R B) (x),s C given by (a,b,c))—1• (a (x) A) (x) c is middle linear. (b) The middle linear map i is universal; that is, given a middle linear map g : A X B X C —> D, there exists a unique group homomorphism g : (A (x)fl B) ®s C D such that gi = g. (c) The map j : A X B X C —> A (x)K (B (x)s O given by ia,b,c) |—> a (X) (b (x) c) is also a universal middle linear map. (d) (A (X)R B) 0s C QZ A (x)R (B 0s O by (b), (c), and Theorem 1.7.10. (e) Define a middle linear function on n (bi)modules (n > 4) in the obvious way and sketch a proof of the extension of the above results to the case of n (bi)modules (over n — 1 rings). (f) If K = S, R is commutative and A,B,C,D are K-modules, define a trilinear map A X B X C —> D and extend the results of (a),(b),(c) to such maps. 11. Let A,B,C be modules over a commutative ring K. (a) The' set £(A,B)C) of all K-bilinear maps A X B —> C is an K-module with (/+ X)(a,b) = f(a,b) + g(a,b) and (rf)(a,b) = rf(a,b). (b) Each one of the following K-modules is isomorphic to £(A,B;C): (i) Homfl(/l 0R B,C); (ii) Hom/j(^,Homfi(K,C)); (iii) Hom/e(K,Homfi(^,C))12. Assume K has an identity. Let G be the category of all unitary K-K bimodules and bimodule homomorphisms (that is, group homomorphisms / : A —► B such that f(ras) = rf(a)s for all r,s e K). Let A7 = j lfl} and let 1 : X —* R be the in clusion map.
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CHAPTER IV MODULES
(a) If R is noncommutative, then R (equipped with l :X^>R) is not a free object on the set X in the category G. (b) R®ZR is an R-R bimodule (Theorem 5.5). If i : X —* R (x)z R is given by 1R I—» \R (x) 1*, then R (x)z R is a free object on the set X in the cate gory e.
6. MODULES OVER A PRINCIPAL IDEAL DOMAIN The chief purpose of this section, which will be used again only in Sections VII.2 and VII.4, is to determine the structure of all finitely generated modules over a principal ideal domain. Virtually all of the structure theorems for finitely generated abelian groups (Sections 11.1,11.2) carry over to such modules. In fact, most of the proofs in Sections II. 1 and II.2 extend immediately to modules over Euclidean domains. However, several of them must be extensively modified in order to be valid for modules over an arbitrary principal ideal domain. Consequently, we shall use a different approach in proving the structure theorems here. We shall show that just as in the case of abelian groups every finitely generated module may be decomposed in two ways as a direct sum of cyclic submodules (Theorem 6.12). Each decomposition provides a set of invariants for the given module (that is, two modules have the same invariants if and only if they are isomorphic (Corollary 6.13)). Thus each method of decomposition leads to a complete classification (up to isomorphism) of all finitely generated modules over a principal ideal domain. Here and throughout this section
'‘'’module" means “unitary module We begin with free modules over a principal ideal domain R. Since R has the in variant dimension property by Corollary 2.12, the rank of a free /?-module (Defini tion 2.8) is well defined. In particular, two free /^-modules are isomorphic if and only if they have the same rank (Proposition 2.9). Furthermore we have the follow ing generalization of Theorem II.1.6.
Theorem 6.1. Let F be a free module over a principal ideal domain R and G a sub module of F. Then G is a free R-module and rank G < rank F. SKETCH OF PROOF. Let {*, | ie /} be a basis of F. Then F = £ Rx{ with iel
each Rxi isomorphic to R (as a left /?-module). Choose a well ordering < of the set I (Introduction, Section 7). For each is I denote the immediate successor of i by /' + 1 (Introduction, Exercise 7.7). Let J = I U {a}, where a \ I and by definition i < a for all i e I. Then J is well ordered and every element of I has an immediate successor in J.1 For each j e J define F;to be the submodule of F generated by the set \xx \ i < j\. Verify that the submodules F, have the following properties: (i) j < k<=$ F, G F*; (ii) U Fi = F, jeJ
set J is a technical device needed to cope with the possibility that some (necessarily unique) element of / has no immediate successor in /. This occurs, for example, when / is finite. JThe
6. MODULES OVER A PRINCIPAL IDEAL DOMAIN
219
(iii) for each i e /, F,+i/F, = Rxi = R. [Apply Theorem 1.7 to the canonical pro jection Fi+i = 52 Rxk —> /?*».] k
<» + l
For each j e J let G, = C fI F, and verify that: (i\)j
(v) U Gi = G; jeJ
(vi) for each / e I, Gi = Gi+i fl Fi. Property (vi) and Theorem 1.9(i) imply that G,-+i/G,- = Gi+1/(Gi+i fl Fi) = (Gi+i + Fi)/Fi. But (Gi+1 + Fi)/Fi is a submodule of F,+i/F,-. Therefore, Gi+i/Gi is isomorphic to a submodule of R by (iii). But every submodule of R is necessarily an ideal of R and hence of the form (c) = Rc for some c e R. If c ^ 0, then the R-module epimorphism R —> Rc of Theorem 1.5(i) is actually an isomor phism. Thus every submodule of R (and hence each Gi+i/G,) is free of rank 0 or 1. By Theorems 3.2 and 3.4 the sequence 0 —> Gi Gt+i —> Gt+i/G,- —*• 0 is split exact for every z e I. Theorem 1.18 and Exercise 1.15 imply that each Gi+1 is an internal direct sum G, 11 = Gi © Rbi, where bi e Gi+1 — Gi and Rb{ ^ R if G,+i ^ Gi, and bi = 0 if Gi+1 = Gi (that is, Gi+X/Gi = 0). Thus bi e G is defined for each i e I. Let B = [bi | bi ^ 0|. Then \B\ < |/| = rank F. To complete the proof we need only show that B is a basis of G. Suppose u = 52 = 0 (j e I; r} e R; finite sum). Let k be the largest index (if i
one exists) such that rk ^ 0. Then u = / , + rkbk eG 4 0 Rbk = Gk+X. But j
u — 0 implies that rk = 0, which is a contradiction. Hence r, = 0 for all j. Therefore, B is linearly independent. Finally we must prove that B spans G. It suffices by (v) to prove that for each k e J the subset Bk = {bj e B \ j < k} of B spans Gk. We shall use transfinite induc tion (Introduction, Theorem 7.1). Suppose, therefore, that Bj spans Gj for all j
Corollary 6.2. Let R be a principal ideal domain. If A is a finitely generated R-module generated by n elements, then every submodule of A may be generated by m elements with m < n. PROOF. Exercise; see Corollary II. 1.7 and Corollary 2.2. ■
Corollary 6.3. A unitary module A over a principal ideal domain is free if and only if
A is projective.
220
CHAPTER IV MODULES
PROOF. (=>) Theorem 3.2. (<=) There is a short exact sequence 0 —>■ F A —» 0 with F free, /an epimorphism and X = ker /by Corollary 2.2. If A is projec tive, then F S AT © A by Theorem 3.4. Therefore, A is isomorphic to a submodule of F, whence A is free by Theorem 6.1. ■ We now develop the analogues of the order of an element in a group and of the torsion subgroup of an abelian group.
Theorem 6.4. Let A be a left module over an integral domain R and for each aeA let 0a = I r e R | ra = 0}. (i) ©a is an ideal of R for each aeA. (ii) At = {a e A | 0a ^ 0} is a submodule of A. (iii) For each aeA there is an isomorphism of left modules R/0„ = Ra = {ra | r e Rj.
Let R be a principal ideal domain and p e R a prime. (iv) /f p*a = 0 (equivalently (p1) CI 0a), then 0a = (pO with 0 < j < i. (v) If (3a = (p1), then pja ^ 0 for all j such that 0 < j < i. REMARK. Prime and irreducible elements coincide in a principal ideal domain by Theorem III.3.4. SKETCH OF PROOF OF 6.4. (iii) Use Theorems 1.5(i) and 1.7. (iv) By hy pothesis Qa = (r) for some re/?. Since pi s 0a, r divides p\ Unique factorization in R (Theorem III.3.7) implies that r = p’u with 0 < j < i and u a unit. Hence 6a = (r) = (p;u) = (p’~) by Theorem III.3.2. (v) If p’a = 0 with j < i, then pi e (3„ = (pl), whence pi | p\ This contradicts unique factorization in R. ■ Let A be a module over an integral domain. The ideal Qa in Theorem 6.4 is called the order ideal of a e A . The submodule At in Theorem 6.4 is called the torsion submodule of A. A is said to be a torsion module if A = A t and to be torsionfree if A t =0. Every free module is torsion-free, but not vice versa (Exercise 2). Let A be a module over a principal ideal domain R. The order ideal of a e A is a principal ideal of R, say 6a = (r), and a is said to have order r. The element r is unique only up to multiplication by a unit (Theorem III.3.2). The cyclic submodule Ra generated by a (Theorem 1.5) is said to be cyclic of order r. Theorem 6.4(iii) shows that a e A has order 0 (that is, Ra is a cyclic module of order 0) if and only if Ra = R (that is, Ra is free of rank one). Also a e A has order r, with r a unit, if and only if o = 0; (for a = \Ra = r~\ra) = r-10 = 0). EXAMPLE. If R is a principal ideal domain and r e R, then the quotient ring R/(r) is a cyclic /?-module with generator a = 1« + (r). Clearly ©a = (r), whence a has order r and R/(r) is cyclic of order r. Theorem 6.4(iii) shows that every cyclic module C over a principal ideal domain R is isomorphic to R/(r), where (r) = 0a and o is a generator of C.
221
6. MODULES OVER A PRINCIPAL IDEAL DOMAIN
EXAMPLE. Let R = Z and let A be an (additive) abelian group. Suppose the group theoretic order of a e A (Definition 1.3.3) is finite. Then 6a = («), where \n\ is the group theoretic order of a. If a e A has infinite order, then 0„ = (0). In either case Za is the cyclic subgroup (a) generated by a (Theorem 1.2.8). Furthermore, Zxi = Z/(«) == Z„ if Ga = («), n ^ 0; and Tm ^ Z/(0) ^ Z if = (0).
Theorem 6.5. A finitely generated torsion-free module A over a principal ideal do main R is free. REMARK. The hypothesis that A is finitely generated is essential (Exercise II. 1.10). PROOF OF 6.5. We may assume A ^ 0. Let X be a finite set of nonzero generators of A. If x e X, then rx = 0 (re/?) if and only if r = 0 since A is torsion-free. Consequently, there is a nonempty subset S = {jci, ..., xk) of X that is maximal with respect to the property:
nxi -f- • • ■ + rkxk = 0 (ri e R) => = 0 for all /. The submodule F generated by S is clearly a free /?-module with basis S. If y eX — S, then by maximality there exist ry,ri,..., rk e R, not all zero, such that rvy + rxxi
k + • ■ ■ + rkxk = 0. Then rvy = —23 r<-*» e F. Furthermore, ry 0 since otherwise i=i ^ = 0 for every /. Since X is finite, there exists a nonzero reR (namely r = rv) yeX-S
such that rX = {rx | x e X] is contained in F. Therefore, rA = {ra | a e A} CZ F. The map / : A —> A given by a (-^ ra is easily seen to be an /?-module homomorphism with image rA. Since A is torsion-free Ker / = 0, whence A ~ Im / = rA CZ F. Therefore, A is free by Theorem 6.1. ■ Determining the structure of a finitely generated module A over a principal ideal domain now proceeds in three steps. We show first that A is a direct sum of a torsion module and a free module (Theorem 6.6). Every torsion module is a direct sum of “/^-primary modules” (Theorem 6.7). Finally every /^-primary module is a direct sum of cyclic modules (Theorem 6.9).
Theorem 6.6. If A is a finitely generated module over a principal ideal domain R, then A = At © F, where F is a free K-module of finite rank and F = A/A t . SKETCH OF PROOF. The quotient module A/At is torsion-free since for each r 0,
r(a + At) = At =» ra e At => n(ra) = 0 for some n 0 => a e At Furthermore, A/At is finitely generated since A is. Therefore, A/A, is free of finite rank by Theorem 6.5. Consequently, the exact sequence 0 ~^> At A A/At —> 0 is split exact and A ^ At © A/At (Theorems 3.2 and 3.4). Under the isomorphism At © A/At = A of Theorem 3.4 the image of A, is A, and the image of A/A, is a submodule Fof A, which is necessarily free of finite rank. It follows that A is the internal direct sum A = At © F (see Theorem 1.15). ■
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Theorem 6.7. Let A be a torsion module over a principal ideal domain R and for each prime p e R let A(p) = {a e A | a has order a power of p). (i) A(p) is a submodule of A for each prime p e R ; (ii) A = 52 A(p), where the sum is over all primes p e R. If A is finitely gener ated, only finitely many of the A(p) are nonzero. PROOF, (i) Let a,b s A(p). If Ga = (pr) and 6b = (ps) let k = max (r,s). Then p\a + b) = 0, whence 0a+b = (p‘) with 0 < / < k by Theorem 6.4(iv). Therefore, a,b e A(p) imply a + b e A(p). A similar argument shows that a e A(p) and reR imply ra e A(p). Therefore, A(p) is a submodule. (ii) Let 0 9^ aeA with 0a = (r). By Theorem III.3.7 r = pxni- ■ ■ pknk with pi dis tinct primes in R and each «,■ > 0. For each z, let n = p™1- ■ ' 'Pknk- Then r x , . . . , r k are relatively prime and there exist s x , . . . , s k z R such that sxri -f • ■ • -fskrk = 1 it (Theorem III.3.11). Consequently, a = 1 Ra — SirYa + ■—f- skrka. But pl,"siria = s^a = 0, whence Sina s A(pi). We have proved that the submodules A(p) (p prime) generate the module A. Let p £ R be prime and let Ax be the submodule of A generated by all A(q) with q 5^ p. Suppose a £ A(p) D Ax. Thenpma = 0 for some m > 0 and a = ax + • ■ - -f- at with ai e A(qi) for some primes qx,... ,qt all distinct from p. Since a, e A(qt), there are integers w,- such that qimiai = 0, whence (qimi- ■ ■qtmt)a = 0. If d = q™x■ ■ q™1, then pm and d are relatively prime and rpm -f sd = Ir for some r,s e R. Consequently, a = 1 Ra — rpma + sda — 0. Therefore, A(p) fl = 0 and A = 52 ^ (p) by Theo rem 1.15. The last statement of the Theorem is a consequence of the easily verified fact that a direct sum of modules with infinitely many nonzero summands cannot be finitely generated. For each generator has only finitely many nonzero coordi nates. ■ In order to determine the structure of finitely generated modules in which every element has order a power of a prime p (such as A(p) in Theorem 6.7), we shall need a lemma. If A is an /?-module and r e R, then rA is the set {ra \ a e A J.
Lemma 6.8. Let A be a module over a principal ideal domain R such that pnA = 0 and pn-1A 0 for some prime p e R and positive integer n. Let a be an element of A of order pn. (i) If A 9^ Ra, then there exists a nonzero b e A such that Ra D Rb = 0. (ii) There is a submodule C of A such that A = Ra © C. REMARK. The following proof is quite elementary. A more elegant proof of (ii), which uses the concept of injectivity, is given in Exercise 7. PROOF OF 6.8. (G. S. Monk) (i) If A ^ Ra, then there exists c s A — Ra. Since pnc e pnA = 0, there is a least positive integer j such that pjc e Ra,- whence pi lc £ Ra and p>c = na (n e R). Since R is a unique factorization domain n = rp* for some k > 0 and r e R such that p\ r. Consequently, 0 = pnc = p^Kp’c) = pn~’rpka. Since p)fr and pn~la ^ 0 (Theorem 6.4(v)), we must have n — j + k > n, whence k >j > 1. Therefore, b = pj~lc — rpk~la is a well-defined element of A.
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223
Furthermore, b ^ 0 (since pJ_1c ^ Ra) and pb = p'c — rpka — p’c — na = 0. If Ra D Rb ^ 0, then there exists s e R such that sb e Ra and sb 0. Since sb 9^ 0 and pb = 0, p does not divide s. Therefore, s and pn are relatively prime and sx -f- pny = ljj for some x,y e R (Theorem III.3.11). Thus since pnA = 0, b = 1 Rb = sxb + pnyb = x(sb) e Ra. Consequently, pj~lc = b -f- rpk~*a e Ra. If j — 1 9* 0, this contradicts the minimality of j, and if 7 — 1 =0, this contradicts the fact that c ^ Ra. Therefore, Ra fl Rb = 0. (ii) If A = Ra, let C = 0. If A Ra, then let S be the set of all submodules B of A such that Ra fl B = 0. S is nonempty since by (i) there is a nonzero b e A such that Ra fl Rb = 0. Partially order S by set-theoretic inclusion and verify that every chain in S has an upper bound in S. By Zorn’s Lemma there exists a submodule C of A that is maximal in S. Consider the quotient module A/C. Clearly pn(A/C) = 0 and pn(a + C) = 0. Since Ra fl C = 0 and pn~Ya 5^ 0, we have pn~\a + C)^C, whence a + C has order pn in A/C and pn~\A/C) 9^ 0. Now if A/C is not the cyclic R-module generated by a + C (that is, A/C R(a -f- C)), then by (i) there exists d + C e A/C such that d + C 9* C and R(a + C) fl R(d + C) = C. Since Ra fl C = 0, it follows that Ra ft (Rd + C) = 0. Since d $ C, Rd + C is in S and properly contains C, which contradicts the maximality of C. Therefore, A/C is the cyclic R-module generated by a + C (that is, A/C = R(a + C)). Consequently, A = Ra + C, whence A = Ra © C by Theorem 1.15. ■
Theorem 6.9. Let A be a finitely generated module over a principal ideal domain R such that every element of A has order a power of some prime p e R. Then A is a direct sum of cyclic R-modules of orders p n l , . . . , pnk respectively, where ni > n2 > ■ • • > nk > 1PROOF. The proof proceeds by induction on the number r of generators of A, with the case r = 1 being trivial. If r > 1, then A is generated by elements au ... ,ar whose orders are respectively pnl,pm2,pm\ . . . , pmr. We may assume that m = max J ni,m2,. .., mT I. Then pn'A = 0 and pni lA 9^ 0. By Lemma 6.8 there is a submodule C of A such that A = Rdi © C. Let 7r be the canonical epimorphism it : A —* C. Since A is generated by ai,a2,. . . , aT, C must be generated by -rr(ai),7r(a2), - - - , Har). But 7r(«i) = 0, whence C may be generated by r — 1 or fewer elements. Consequently, the induction hypothesis implies that C is a direct sum of cyclic R-modules of orders pn\pn\ ... ,pnk respectively with «2 > ns > ■ • • > nk > 1. Thus C contains an element of order n2. Since pniA = 0, we have pnlC — 0, whence n\ > n2. Since Roi is a cyclic R-module of order pni, A is a direct sum of cyclic R-modules of orders pn',pn2,. . . , p"k respectively with «i > «2 > • • ■ > nk > 1. ■ Theorems 6.6, 6.7, and 6.9 immediately yield a structure theorem for finitely generated modules over a principal ideal domain (see Theorem 6.12(ii) below). Just as in the case of abelian groups (Section II.2), there is a second way of decomposing a finitely generated module as a direct sum of cyclic submodules. In order to obtain this second decomposition and to prove a uniqueness theorem about each of the de compositions, we need two lemmas.
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Lemma 6.10. Let A,B, and A; (i el) be modules over a principal ideal domain R. Let r e R and let p e R be prime. (i) rA = {ra | a e A j and A[r] = {a e A | ra = 0} are submodules of A. (ii) R/(p) is a field and A[p] is a vector space over R/(p). (iii) For each positive integer n there are K-module isomorphisms (R/(pn))[p] = R/(p) and pm(R/(pn)) = R/(pn_m) (0 < m < n). (iv) If A = 23 Ai, then r A = 23 rA; and A[r] ^ 23 A;[r]. teJ
izl
(v) Iff : A —» B is an K-module isomorphism, then f: At = Bt and f: A(p) = B(p). SKETCH OF PROOF, (ii) Exercise 2.4. (v) See Lemma II.2.5 (vii). (iii) The first example preceding Theorem 6.5 may be helpful.. Verify that (R/(pn))[p] is generated as an K-module (and hence as a vector space over R/(p)) by the single nonzero element pn~l + (pn)- Therefore, (R/(pn))\p\ = R/(p) by Theorems 2.5 and 2.1. The submodule of R/(pn) generated by pm + (pn) is precisely pm(R/(pT'))- Since Pm + (Pn) has order pn~m, we have pm(R/(pn)) = R/(pn~m) by Theorem 6.4(iii). ■
Lemma 6.11. Let R be a principal ideal domain. If r e R factors as r = pim- • Pknk with Pi, . . . , Pk e R distinct primes and each n, > 0, then there is an K-module iso morphism R/(r) S R/(Pini) ©■ ■ ■© R/(Pknk).
Consequently every cyclic K-module of order r is a direct sum of k cyclic K-modules of orders pini,. . ., pknk respectively. SKETCH OF PROOF. We shall prove that if s,t e R are relatively prime, then R/(st) = K/(,s) © R/(t). The first part of the lemma then follows by induction on the number of distinct primes in the prime decomposition of r. The last statement of the lemma is an immediate consequence of the fact that R/(c) is a cyclic K-module of order c for each c e R by Theorem 6.4. The map 6 : R —> R given by xH ?jf is an /?-module monomorphism that takes the ideal (s) onto the ideal (s/). By Corollary 1.8 6 induces an K-module homomorphism R/(s) —> R/(st) given by x -f- (5) (—> tx + (st). Similarly there is a homomorphism R/(t)—> R/(st) given by x + (r)[—>5* + Csf). By the proof of Theorem 1.13 the map a : R/(s) © R/0) —* R/(st) given by (x + COj7 + (0)h~* \tx + ■'>’] + (st) is a well-defined K-module homomorphism. Since (s,t) = lfi, there exist u,v e K such that su -f- tv = 1« (Theorem III.3.11). If c e K, then c = sue + tve, whence a(vc -f- (^), uc + (0) = c + (^0- Therefore, a is an epimorphism. In order to show that a is a monomorphism we must show that <x(x + (s), y + (0) = 0 => x e (5) and y e (t). If <*(a- + (s), y + (r)) = 0, then tx + sy = stb e (st) for some b e K. Hence utx -J- usy = ustb. But>' = \Ry — (su + to)y, whence utx + (y — tvy) = ustb and y = ustb — utx + tvy e (r). A similar argument shows that x e (5). ■
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225
Theorem 6.12. Let A be a finitely generated module over a principal ideal domain R. (i) A is the direct sum of a free submodule F of finite rank and a finite number of cyclic torsion modules. The cyclic torsion summands (if any) are of orders ri, . . . , rt, where ri, . . . , rt are (not necessarily distinct) nonzero nonunit elements of R such that Ti | r21 ■ ■ | rt. The rank ofF and the list of ideals (r,), • • - , (rt) are uniquely determined by A. (ii) A is the direct sum of a free submodule E of finite rank and a finite number of cyclic torsion modules. The cyclic torsion summands (ifany) are of orders pi B l ,..., Pk8k, where pi, . . . , pk are (not necessarily distinct) primes in R and Si, . . . , St are (not necessarily distinct) positive integers. The rank of E and the list of ideals (p^1),... ,(pkBk) are uniquely determined by A (except for the order of the p;). The notation ri|r21 • -|r, means n divides r2, r2 divides rs, etc. The elements ru . . . , rt in Theorem 6.12 are called the invariant factors of the module A just as in the special case of abelian groups. Similarly pis\ . .., pksk are called the elementary divisors of A.
SKETCH OF PROOF OF 6.12. The existence of a direct sum decomposition of the type described in (ii) is an immediate consequence of Theorems 6.6, 6.7, and 6.9. Thus A is the direct sum of a free module and a finite family of cyclic R-modules, each of which has order a power of a prime. In the case of abelian groups these prime powers are precisely the elementary divisors of A. The method of calculating the in variant factors of an abelian group from its elementary divisors (see pp. 80-81) may be used here, mutatis mutandis, to prove the existence of a direct sum decompo sition of A of the type described in (i). One need only make the following modifica tions. The role of Zpn = Z/(pn) (p e Z prime) is played by a cyclic torsion submodule of A of order pn (p e R prime). Such a cyclic torsion module is isomorphic to R/(p") by Theorem 6.4(iii). Lemma II.2.3 is replaced by Lemma 6.11.
The proof of the uniqueness of the direct sum decompositions in (i) and(ii) is essentially the same as the proof of the corresponding facts for abelian groups (Theorem II.2.6). The following modifications of the argument are necessary. First of all prime factorization in R is unique only up to multiplication by a unit (Definition III.3.5 and Theorem III.3.7). This causes no difficulty in Z since the only units are ± 1 and primes are defined to be positive. In an arbitrary principal ideal domain R, however, an element a e R may have order p and order q with p,q distinct primes. However, since (p) = Qa = (ci), p and q are associates by Theorem III.3.2; that is, q = pu with u e R a unit. Hence the uniqueness statements in (i) and (ii) deal with ideals rather than elements. Note that a ^ 0 implies that Oa ^ R and that a cyclic module Ra is free if and only if 0„ = (0). Thus the elements r, in (i) are non zero nonunits. Other modifications: as above replace each finite cyclic summand Z7l = Z/(«) with n > 1 by a cyclic torsion module R/(r) (r e R a nonzero nonunit). Replace the subgroup generated by the infinite cyclic summands Z by a free R-module of finite rank. Use Lemmas 6.10 and 6.11 in place of Lemmas 11.2.3 and II.2.5. Instead of the counting argument on p. 79 (showing that r = d) use the fact that A[p\ is a vector space over R/(p). Hence the number of summands R/(p) is pre cisely dimrkp)A\p], which is invariant by Theorem 2.7. ■
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Corollary 6.13. Two finitely generated modules over a principal ideal domain, A and B, are isomorphic if and only ifA/A t andB/Bt have the same rank and A and B have the same invariant factors [resp. elementary divisors]. PROOF. Exercise. ■
EXERCISES Note: Unless stated otherwise, K is a principal ideal domain and all modules are unitary. 1. If R is a nonzero commutative ring with identity and every submodule of every free K-module is free, then K is a principal ideal domain. [Hint: Every ideal / of R is a free K-module. If u,v e / (w ^ 0,v # 0), then uv + (—v)u = 0, which implies that I has a basis of one element; that is, / is principal.] 2. Every free module over an arbitrary integral domain with identity is torsion-free. The converse is false (Exercise II.1.10). 3. Let A be a cyclic K-module of order r e K. (a) If j e K is relatively prime to r, then sA = A and ^l[s] = 0. (b) If s divides r, say sk = r, then sA = R/(k) and A[s] = R/(s). 4. If A is a cyclic K-module of order r, then (i) every submodule of A is cyclic, with order dividing r; (ii) for every ideal (.s) containing (r), A has exactly one submodule, which is cyclic of order s. 5. If A is a finitely generated torsion module, then {re K | rA = 0} is a nonzero ideal in K, say (n). n is called the minimal annihilator of A. Let A be a finite abelian group with minimal annihilator me Z. Show that a cyclic subgroup of A of order properly dividing m need not be a direct summand of A. 6. If A and B are cyclic modules over K of nonzero orders r and s respectively, and r is not relatively prime to s, then the invariant factors of A © B are the greatest common divisor of r,s and the least common multiple of r,s. 7. Let A and a e A satisfy the hypotheses of Lemma 6.8. (a) Every K-submodule of A is an K/(/?n)-module with (r (pn))a = ra. Con versely, every K/(/?”)-submodule of A is an K-submodule by pullback along K — K/(p"). (b) The submodule Ra is isomorphic to R/(pn). (c) The only proper ideals of the ring R/(pn) are the ideals generated by P* + (/>") (/' = 1 , 2 , — 1). (d) R/(pn) (and hence Ra) is an injective K/(/>n)-module. [Hint: use (c) and Lemma 3.8.] (e) There exists an K-submodule C of A such that A = Ra® C. [Hint: Propo sition 3.13.]
7. ALGEBRAS Algebras are introduced and their basic properties developed. Tensor products are used extensively in this discussion. Algebras will be studied further in Chapter IX.
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111
Definition 7.1. Let K be a commutative ring with identity. A K-algebra (or algebra over K) A is a ring A such that: (i) (A,+) is a unitary (left) K-module; (ii) k(ab) = (ka)b = a(kb) for all k e K and a,b e A.
A K-algebra A which, as a ring, is a division ring, is called a division algebra. The classical theory of algebras deals with algebras over a field K. Such an algebra is a vector space over K and hence various results of linear algebra are ap plicable. An algebra over a field K that is finite dimensional as a vector space over K is called a finite dimensional algebra over K. EXAMPLE. Every ring R is an additive abelian group and hence a Z-module. It is easy to see that R is actually a Z-algebra. EXAMPLES. If K is a commutative ring with identity, then the polynomial ring K[xu ... ,x„] and the power series ring K[[jc]] are K-algebras, with the respective K-module structures given in the usual way. EXAMPLE. If V is a vector space over a field F, then the endomorphism ring WomF(V,V) (Exercise 1.7) is an F-algebra. The F-module structure of HomF(K,K) is discussed in the Remark after Theorem 4.8. EXAMPLES. Let A be a ring with identity and K a subring of the center of A such that 1 a b K. Then A is a K-algebra, with the .K-module structure being given by multiplication in A. In particular, every commutative ring K with identity is a K-algebra. EXAMPLE. Both the field of complex numbers C and the division ring of real quaternions (p. 117) are division algebras over the field R of real numbers. EXAMPLE. Let G be a multiplicative group and K a commutative ring with identity. Then the group ring K(G) (p. 117) is actually a K-algebra with K-module structure given by
k&ngi) = ^2(kr,)g, (k,r{ e K; gx e G). K(G) is called the group algebra of G over K. EXAMPLE. If K is a commutative ring with identity, then the ring MatnK of all n X n matrices over K is a K-algebra with the K-module action of K given in the usual way. More generally, if A is a K-algebra, then so is Mat„/4. REMARK. Since K is commutative, every left K-module (and hence every K-algebra) A is also a right K module with ka = ak for all a e A, k e K. This fact is implicitly assumed in Theorems 7.2 and 7.4 below, where tensor products are used. The motivation for the next theorem, which provides another means of defining K algebras, is the fact that for any ring R the unique map R (x)z R —> R, defined on a generator r(x)s by r (x) .s rs, is a homomorphism of additive abelian groups. Since rings are simply Z-algebras, this fact is a special case of
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Theorem 7.2. Let K be a commutative ring with identity and A a unitary left K-module. Then A is a K-algebra ifand only if there exists a K-module homomorphism 7r : A (x)k A —> A such that the diagram
A ®)k A (x)/c A 7r
is commutative. In this case the K-algebra A has an identity if and only if there is a K-module homomorphism I : K —* A such that the diagram
is commutative, where f,6 are the isomorphisms of Theorem 5.7. SKETCH OF PROOF. If A is a A'-algebra, then the map A X A —> A given by (a,b) h-* ab is AT-bilinear, whence there is a AT-module homomorphism 7r : A (XV A —> A by Theorem 5.6. Verify that ir has the required properties. If A has an identity 1^, then the map I: K—> A given by k [—> k\A is easily seen to be a /if-module homo morphism with the required properties. Conversely, given A and the map 7r : A (X)/c A —> A, define ab = ir(a(X) b) and verify that A is a /("-algebra. If / : K—> A is also given, then 1(1 k) is an identity for A. ■ The homomorphism 7r of Theorem 7.2 is called the product map of the /C-algebra A. The homomorphism / is called the unit map.
Definition 7.3. Let K be a commutative ring with identity and A, B K-algebras. (i) A subalgebra of A is a subring of A that is also a K-submodu/e of A. (ii) A (left, right, two-sided) algebra ideal of A is a (left, right, two-sided) ideal of
the ring A that is also a K-submodule of A. (iii) A homomorphism [resp. isomorphism] of K-algebras f : A —» B is a ring ho momorphism [isomorphism] that is also a Y^-module homomorphism [isomorphism]. REMARKS. If A is a /^-algebra, an ideal of the ring A need not be an algebra ideal of A (Exercise 4). If, however, A has an identity, then for all k e K and a e A
ka = k(\Aa) = (k\A)a and ka = (ka)\A = a(k\A), with k\A e A. Consequently, for a left [resp. right] ideal J in the ring A,
kJ = (A: 1.4)J
7. ALGEBRAS
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Therefore, if A has an identity, every (left, right, two-sided) ideal is also a (left, right,
two-sided) algebra ideal. The quotient algebra of a K-algebra A by an algebra ideal / is now defined in the obvious way, as are the direct product and direct sum of a family of K-algebras. Tensor products furnish another way to manufacture new algebras. We first observe that if A and B are K-modules, then there is a K-module isomorphism a : A (gk B —> B (gk A such that a(a (g) b) = b(x) a (as A,b e B); see Exercise 2.
Theorem 7.4. Let A and B be algebras [with identity ] over a commutative ring K with identity. Let tt be the composition (A
(gk B) (gk (A (gk B)------------>
(A
(gk A) (gk (B (g)K B)--------
where irA, 7Tr are the product maps of A and B respectively. Then A algebra [with identity] with product map ir. PROOF. Exercise; note that for generators a (x) b and a\ product is defined to be
(g) b\
>A (gk
(gk: B,
B is a K-
of A
(gk B
the
(a (g) b)(ax (g) bi) = ir(a (g) b (g) a, (g) b\) = aax (g) bbx. Thus if A and B have identities 1^, I# respectively, then 1^ in A (gk B. ■
(g)
1b is the identity
The K-algebra A (g)A- B of Theorem 7.4 is called the tensor product of the Kalgebras A and B. Tensor products of algebras are useful in studying the structure of division algebras over a field K (Section IX.6).
EXERCISES Note: K is always a commutative ring with identity. 1. Let C be the category whose objects are all commutative K-algebras with identity and whose morphisms are all K-algebra homomorphisms f:A—+B such that f( 1 a) = 1 b- Then any two K-algebras A, B of C have a coproduct. [Hint: consider A—> A (gk B B, where a I—► a (g) \B and b h-> 1A (g) b.] 2. If A and B are unitary K-modules [resp. K-algebras], then there is an isomorphism of K-modules [resp. K-algebras] a : A (gk B —> B (gk A such that a(a (g) b) = b (g) a for all a e A,b e B. 3. Let A be a ring with identity. Then A is a K-algebra with identity if and only if there is a ring homomorphism of K into the center of A such that 1* (—> \A.
4. Let A be a one-dimensional vector space over the rational field Q. If we define ab = 0 for all a,b e A, then A is a Q-algebra. Every proper additive subgroup of A is an ideal of the ring A, but not an algebra ideal. 5. Let C be the category of Exercise 1. If X is the set then the poly nomial algebra K[*i, ...,*„] is a free object on the set X in the category C. [Hint: Given an algebra A in C and a map g : {*1,..., xn} —» A, apply Theorem III.5.5 to the unit map / : K —> A and the elements #Ui),. . . , g(xn) s A.]
I
CHAPTER V
FIELDS AND GALOIS THEORY
The first principal theme of this chapter is the structure theory of fields. We shall study a field F in terms of a specified subfield K (F is said to be an extension field of K). The basic facts about field extensions are developed in Section 1, in particular, the distinction between algebraic and transcendental extensions. For the most part we deal only with algebraic extensions in this chapter. Arbitrary field extensions are considered in Chapter VI. The structure of certain fields and field extensions is thoroughly analyzed: simple extensions (Section 1); splitting fields (normal exten sions) and algebraic closures (Section 3); finite fields (Section 5); and separable algebraic extensions (Sections 3 and 6). The Galois theory of field extensions (the other main theme of this chapter) had its historical origin in a classical problem in the theory of equations, which is dis cussed in detail in Sections 4 and 9. Various results of Galois theory have important applications, especially in the study of algebraic numbers (see E. Artin [48]) and algebraic geometry (see S. Lang [54]). The key idea of Galois theory is to relate a field extension K CL F to the group of all automorphisms of Fthat fix K elementwise (the Galois group of the extension). A Galois field extension may be defined in terms of its Galois group (Section 2) or in terms of the internal structure of the extension (Section 3). The Fundamental Theo rem of Galois theory (Section 2) states that there is a one-to-one correspondence between the intermediate fields of a (finite dimensional) Galois field extension and the subgroups of the Galois group of the extension. This theorem allows us to trans late properties and problems involving fields, polynomials, and field extensions into group theoretic terms. Frequently, the corresponding problem in groups has a solu tion, whence the original problem in field theory can be solved. This is the case, for instance, with the classical problem in the theory of equations mentioned in the pre vious paragraph. We shall characterize those Galois field extensions whose Galois groups are finite cyclic (Section 7) or solvable (Section 9). The approximate interdependence of the sections of this chapter is as follows: 230
1. FIELD EXTENSIONS
231
1 1 2
A broken arrow A-- > B indicates that an occasional result from section A is used in section B, but that section B is essentially independent of section A. See page xviii for a description of a short basic course in fields and Galois theory.
1. FIELD EXTENSIONS The basic facts needed for the study of field extensions are presented first, followed by a discussion of simple extensions. Finally a number of essential proper ties of algebraic extensions are proved. In the appendix, which is not used in the sequel, several famous geometric problems of antiquity are settled, such as the tri section of an angle by ruler and compass constructions.
Definition 1.1. A field F is said to be an extension field of K (or simply an extension of K) provided that K is a subfield of F.
If F is an extension field of K, then it is easy to see that 1a = 1 f - Furthermore, F is a vector space over K (Definition IV.1.1). Throughout this chapter the dimension of the K-vector space F will be denoted by [F : K] rather than dim^Fas previously. F is said to be a finite dimensional extension or infinite dimensional extension of K according as [F : KJ is finite or infinite.
Theorem 1.2. Let F be an extension field ofE and E an extension field of K. Then [F : K] = [F : E][E : K]. Furthermore [F : K] is finite if and only if[ F : E] and [E : K]
are finite. PROOF. This is a restatement of Theorem IV.2.16. ■ In the situation K CZ E CZ F of Theorem 1.2, E is said to be an intermediate field of K and F. If F is a field and X CZ F, then the subfield [resp. subring] generated by X is the intersection of all subfields [resp. subrings] of F that contain X. If F is an extension
CHAPTER V FIELDS AND GALOIS THEORY
232
field of K and X CZ F, then the subfield [resp. subring] generated by K U X is called the subfield [resp. subring] generated by X over K and is denoted K(X) [resp. K[X]\. Note that K[X] is necessarily an integral domain. If X = then the subfield K(X) [resp. subring K[X}\ of F is denoted K(uu . . . ,un) [resp. K[u\, .. ., «„]]. The field K(ux,. . . , un) is said to be a finitely generated extension of K (but it need not be finite dimensional over K\ see Exercise 2). If X = {«}, then K(u) is said to be a simple extension of K. A routine verification shows that neither K(uu . . . , un) nor K[u\,. . . , m„] depends on the order of the Ui and that K{uu . .., un^i)(un) = K(uu. . . , u„) and K[m,..., M„_i][Mn] = K[uu . . . , un] (Exercise 4). These facts will be used frequently in the sequel without explicit mention. NOTATION. If F is a field u,v s F, and v ^ 0, then uv~r e F will sometimes be denoted by u/v.
Theorem 1.3. If F is an extension field of a field K, u, Ui e F, and X CZ F, then (i) the subring K[u] consists of all elements of the form f(u), where f is a poly nomial with coefficients in K (that is, f e K[x]); (ii) the subring K[ui, . . . , um] consists of all elements of the form g(ui,u 2 ,..., um), where g is a polynomial in m indeterminates with coefficients in K (that is, geK[x!, . . . , xj); (iii) the subring K[X] consists of all elements of the form h(ui, . . . , un), where each Uj e X, n is a positive integer, andh is a polynomial in n indeterminates with coefficients in K (that is, n e N*, h e K[xi,. . . , xj); (iv) the subfield K(u) consists of all elements of the form f(u)/g(u) = f(u)g(u)_1, where f,g e K[x] and g(u) ^ 0; (v) the subfield K(ui, . . . , u,„) consists of all elements of the form h(ui,. . ., um)/k(ui,. . ., um) = h(ui, . . . , u„,)k(ui,. . . , um)_1,
where h,k e K[x,,. . . , xm] and k(u l 5 . . . , u,n) ^ 0; (vi) the subfield K(X) consists of all elements of the form f(ui, . . . , un)/g(u!, ...,u n ) = f(u,,. . . , Un)g(u,,. . . , u,,)-1
where n e N*, f,g e K[xi, . . . , x„], Uj, . . . , u» e X and g(ui5 . . . , u„) ^ 0. (vii) For each v s K(X) (resp. K[X]) there is a finite subset X' of X such that v e K(X') (resp. K[X']). SKETCH OF PROOF, (vi) Every field that contains K and X must contain the set E = {/(hi, . . . , un)/g(uu . . . ,un) | ne N*; fg e K[xu . . . , xrJ; u,■ bX; g(uj,. . ., «B) ^ 0], whence K(X) ZD E. Conversely, if fg e ^[xi,. . . , xm] and f,gi e K[xu. .., xn], then define h,k e K[xu. . . , xm+„] by
h(xi, . . . , Xm_j_n) f(x 1, - • - , Xm)gi(AVj_j-i5 . . . 5 -^rn+n) ..•)
=
• • • j ^rn)SlC^wt+l* • • • 9
Then for any Wi, ..u»t9 , vn eX such that g(uu ... >Km) ^ 0, g\(vu ..., i?n) 5* 0, /(«1, . - . ,
Um)
...,
Vn) h(u 1, . . . , 01, - . . , V )
n ---------- — ---------- = —-----------------£ E.
g(tfl, • • • ? Mm) £l(^l> • * - 9 Vn) k{u.\y • * • 5
• • • 9 ^n)
1. FIELD EXTENSIONS
233
Therefore, E is a group under addition (Theorem 1.2.5). Similarly the nonzero ele ments of E form a group under multiplication, whence E is a field. Since X d Fand K a E, we have K(X) C E. Therefore, K(X) = E. (vii) If u s K(X), then by (vi) u = f(uu . . . , w„) g(u i, . . . , //„) s K(X'), where X' = j uu . . . , u„ | CI ■ If L and M are subfields of a field F, the composite of L and M in F, denoted LM is the subfield generated by the set L U M. An immediate consequence of this defini tion is that LM = Z.(M) = M(L). It is easy to show that if AT is a subfield of L D M such that M = K(S) where S CI M, then LM = L(S) (Exercise 5). The relationships of the dimensions [L : A"], [M : K], [LM : A"], etc. are considered in Exercises 20-21. The composite of any finite number of subfields EuE->, . . . ,En is defined to be the subfield generated by the set £\ U E-, U • • ■ U En and is denoted EXE2- ■ Er (see Exercise 5). The next step in the study of field extensions is to distinguish two fundamentally different situations that occur.
Definition 1.4. Let F be an extension field of K. An element u of F is said to be algebraic over K provided that u is a root of some nonzero polynomial f e K[xl. If u is not a root of any nonzero f e K[x], u is said to be transcendental over K. F is called an algebraic extension of K if every element of F is algebraic over K. F is called a trans cendental extension if at least one element of F is transcendental over K. REMARKS. If u e K, then u is a root of .v — u e A'[jc] and therefore algebraic over K. If u e F is algebraic over some subfield K' of K, then u is algebraic over K since A"'[jc] C A'[x]. If u e F is a root of /s K[x] with leading coefficient c ^ 0, then u is also a root of r-1/, which is a monic polynomial in A'fjc], A transcendental extension may contain elements that are algebraic over K (in addition to the elements of K itself). EXAMPLES. Let Q,R and C’ be the fields of rational, real, and complex numbers respectively. Then / s C is algebraic over Q and hence over R; in fact, C = R(/). It is a nontrivial fact that it, e s R are transcendental over Q; see, for instance, I. Herstein [4]. EXAMPLE. If A" is a field, then the polynomial ring A'[xi, is an integral domain (Theorem III.5.3). The quotient field of K[x\, . . . , xn] is denoted A(xi,. . . , x„). It consists of all fractions f/g, with fig e K\xu . . . , xn] and g ^ 0, and the usual addition and multiplication (see Theorem III.4.3). K(xi, . . . , x„) is called the field of rational functions in X\, . . ., xn over K. In the field extension
K C K(xu . . . , xn) each x, is easily seen to be transcendental over K. In fact, every element of K(xu . . . , Xr,) not in K itself is transcendental over K (Exercise 6). In the next two theorems we shall characterize all simple field extensions up to isomorphism.
Theorem 1.5. If F is an extension field of K and u e F is transcendental over K, then there is an isomorphism of fields K(u) = K(x) which is the identity on K.
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SKETCH OF PROOF. Since u is transcendental f(u) 0, g(u) 0 for all nonzero f,g e Afc]. Consequently, the map
Theorem 1.6. If F is an extension field of K and ueF is algebraic over K, then (i) K(u) = K[u]; (ii) K(u) = K[x]/(f), where f e K[x] is an irreducible monic polynomial of degree n > 1 uniquely determined by the conditions that f(u) = 0 and g(u) = 0 (g e K[x]) if and only if f divides g; (iii) [K(u) : K] = n; (iv) { 1k,u,u2, . . ., u""1} is a basis of the vector space K(u) over K; (v) every element of K(u) can be written uniquely in the form an + aiU + • • ■ + an-iu""1 (a; e K). PROOF, (i) and(ii) The map
K[x]/{f) = K[x]/Ker
= K[u],
Since K[u] is an integral domain, the ideal (/) is prime in by Theorem III.2.16. Theorem III.3.4 implies that /is irreducible and hence that the ideal (/) is maximal. Consequently, K[x]/(J) is a field (Theorem III.2.20). Since K(u) is the smallest subfield of F containing K and u and since K(u) ZD K[u] = K[x]/(f), we must have K(u) = K[u\. The uniqueness of /follows from the facts that /is monic and g(w) = 0 <=> g£ Ker /divides g. (iv) Every element of K(u) = K[u\ is of the form g(u) for some g e K[jc] by Theo rem 1.3. The division algorithm shows that g — q/+ h with q,h e K[.y] and deg h < deg/. Therefore, g{u) — q(u) f(u) + h{u) = 0 + h{u) = h(u) — b0 4- b,u + ■ • • + bmum with m < n = deg /. Thus { 1a,m, . . . , m”~M spans the K-vector space K(u). To see that | \ k , u , . . . , un~1} is linearly independent over K and hence a basis, suppose
a0 + a\u + • • ■ + an^unl = 0 fee K). Then g = a0 + axx + ■ • ■ + on_iJtn-1 e K[a-] has u as a root and has degree < n — 1. Since / | g by (ii) and deg f = n, we must have g = 0; that is, a* = 0 for all /, whence { 1a-,«, . . - , w“—1} is linearly independent. Therefore, { 1a,«, . . . , un~l\ is a basis of
K(u). (iii) is an immediate consequence of (iv). The equivalence of (iv) and (v) is a routine exercise. ■ Definition 1.7. Let F be an extension field of K and u e F algebraic over K. The monic irreducible polynomial f of Theorem 1.6 is called the irreducible (or minimal or minimum) polynomial of u. The degree of u over K is deg f = [K(u) : K],
1. FIELD EXTENSIONS
235
The following example illustrates how Theorem 1.6 and the techniques of its proof may be used for specific computations. EXAMPLE. The polynomial a3 — 3a — 1 is irreducible over Q (Theorem 111.6.6 and Proposition III.6.8) and has real root u (Exercise 111.6.16(d)). By Theorem 1.6 u has degree 3 over Q and \\,u,u2\ is a basis of Q(u) over Q. The element m4 + 2m3 + 3 s Q(m) = Q[m] may be expressed as a linear combination (over Q) of the basis elements as follows. The division algorithm (that is, ordinary long division) in the ring Q[a] shows that a4 + 2a3 + 3 = (a + 2)(x3 — 3a — 1) + (3a2 + Ix + 5), whence m4
+2
m3
+3=
(m
+ 2)(m3 — 3m — 1) + (3 u2 + lu + 5)
=
(m
+ 2)0 + (3m2 + lu + 5)
=3
m2
+ lu + 5.
The multiplicative inverse of 3u2 + lu -f- 5 in Q(m) may be calculated as follows. Since a3 — 3a — 1 is irreducible in Q[a], the polynomials a3 — 3a — 1 and 3x2 + 7a + 5 are relatively prime in Q[x], Consequently, by Theorem III.3.11 there exist g(x), h(x) e Q[a] such that (a3
Therefore, since
m3
- 3a -
l)g(A)
+ (3a2 + 7x + 5)/z(a) = 1.
— 3m — 1 = 0 we have (3m2 + lu + 5 )h(u) = 1
so that /i(m) e Q[m] is the inverse of 3m2 + lu + 5. The polynomials g and h may be explicitly computed via the Euclidean algorithm (Exercise III.3.13): g(x) — —1/31 x -|- 29/111, and h(x) = 7/111 a2 - 26/111 a + 28/111. Hence //(m) = 7/111 m2 26/111
m
+ 28/111.
Suppose E is an extension field of K, F is an extension field of L, and a : K —> L is an isomorphism of fields. A recurrent question in the study of field extensions is: under what conditions can a be extended to an isomorphism of E onto F. In other words, is there an isomorphism r : E —> Fsuch that r | K =
this map extends a. We shall denote the extended map K[a] —■> S[a] by a also and the image of /e /?[a] by trf.
Theorem 1.8. Let a : K —> L be an isomorphism of fields, u an element of some ex tension field of K and v an element of some extension field of L. Assume either (i) u is transcendental over K and v is transcendental over L; or (ii) u is a root of an irreducible polynomial f e K[x] and v is a root of af e L[x], Then a extends to an isomorphism of fields K(u) ^ L(v) which maps u onto v.
CHAPTER V FIELDS AND GALOIS THEORY
236
SKETCH OF PROOF, (i) By the remarks preceding the theorem o extends to an isomorphism K[x] = Z,[jc], Verify that this map in turn extends to an isomorphism K(x) —> L(x) given by h/g }—> ah/ag. Therefore, by Theorem 1.5 we have K(u) = K(x) = L(x) = L(v). The composite map extends a and maps u onto v. (ii) It suffices to assume that / is monic. Since a : K[x] = L[x] this implies that erf e L[x] is monic irreducible. By the proof of Theorem 1.6 the maps : K[x]/( /) -> K[u) = K{u) and * : L[x]/(af) - L[v] = L(v), given respectively by tf[g + (/)] = giu) and \p[h -f- (cr/)] = h(v), are isomorphisms. The map 6 : K[x]/(f) —> L[x]/(af) given by 6[g + (/)] = ag + (a/) is an isomor phism by Corollary III.2.11. Therefore the composite
K(u) C K[x]/(f) ^ L[x)/(af) ^ L(v) is an isomorphism of fields such that g(w)[—» (crg)(i;). In particular, agrees with a on K and maps u onto v (since <7(1*) = \L by Exercise III. 1.15). ■
Corollary 1.9. Let E and F each be extension fields of K and let u e E and v e F be algebraic over K. Then u and v are roots of the same irreducible polynomial f £ K[x] if and only if there L an isomorphism offields K(u) = K(v) which sends u onto v and is the identity on K. PROOF. (=>) Apply Theorem 1.8 with a = Ia (so that er/= /for all /e K[a]). (<=) Suppose a : K(u) =. K(v) with
Up to this point we have always dealt with a root of a polynomial /s K[jc] in some given extension field F of K. The next theorem shows that it really is not necessary to have F given in advance. Theorem 1.10. If K is a field and f e K[x] polynomial of degree n, then there exists a simple extension field F = K(u) of K such that: (i) u e F is a root o/f; (ii) [K(u) : K] < n, with equality holding if and only iff is irreducible in K[x]; (iii) if f is irreducible in K[x], then K(u) is unique up to an isomorphism which is the identity on K. REMARK. In view of (iii) it is customary to speak of the field F obtained by ad joining a root of the irreducible polynomial /e K[x) to the field K. SKETCH OF PROOF OF 1.10. We may assume that /is irreducible (if not, replace / by one of its irreducible factors). Then the ideal (/) is maximal in (Theorem III.3.4 and Corollary III.6.4) and the quotient ring F = K[x)/(f) is a
1. FIELD EXTENSIONS
237
field (Theorem III.2.20). Furthermore, the canonical projection ir : K[x] -* K[x\/(f) — F, when restricted to K, is a monomorphism (since 0 is the only constant in a maximal ideal of K[x]). Thus F contains ir(K) = K, and therefore may be considered as an extension field of K (providing that K is identified with ir(K) under the iso morphism). For x e AT[jc], let u = ir(x) s F. Verify that F = K(u) and that f(u) = 0 in F. Theorem 1.6 implies statement (ii) and Corollary 1.9 gives (iii). ■ In the remainder of this section we shall develop the essential basic facts about algebraic field extensions.
Theorem 1.11. 7/F is a finite dimensional extension field of K, then F is finitely generated and algebraic over K. PROOF. If [Z7: AT] = n and ue F, then the set of n + 1 elements {1 k,u,u2, ..., un) must be linearly dependent. Hence there are a, e K, not all zero, such that a0 + a\u -fa2u2 + ■ - • + anun = 0, which implies that u is algebraic over K. Since u was arbi trary, F is algebraic over K. If {vu ..., vn I is a basis of F over K, then it is easy to see that F = K(vi,. . . , vv). ■
Theorem 1.12. 7/F is an extension field of K and X is a subset of F such that F = K(X) and every element ofX is algebraic over K, then F is an algebraic extension of K. IfX is a finite set, then F is finite dimensional over K. PROOF. If v e F, then v e K(uu . . . , wn) for some m, e X (Theorem 1.3) and there is a tower of subfields:
K a K(u0 CZ K(uuu2) CZ ■ ■■ CZ K(uu wn_i) CZ K(uu w„). Since «, is algebraic over K, it is necessarily algebraic over K(u\,... , «,•_ i) for each / > 2, say of degree rt. Since K(u\, . . . , «,_i)(Mi) = K(ui, - - ■ , «.•) we have [K(uu K(uu . . . , Mt_i)] = n by Theorem 1.6. Let ri be the degree of «i over K; then repeated application of Theorem 1.2 shows that [K{u\, ...,«„): K\ — nr2- - ■/■„. By Theorem 1.11 K(ui,. . . ,un) (and hence v) is algebraic over K. Since ve F was arbitrary, F is algebraic over K. If X = {ui,..., «„} is finite, the same proof (with F = K(ut,. . ., un)) shows that [F \ K] = r\r2- • -rn is finite. ■
Theorem 1.13. If F is an algebraic extension field of E sion field of K, then F is an algebraic extension of K.
and
E is an algebraic exten
PROOF. Let u e F; since u is algebraic over E, bnun + ■ • • + b\u + 6o = 0 for some bi eE(bn ^ 0). Therefore, u is algebraic over the subfield K(b0,. . . , bn). Con sequently, there is a tower of fields
K CZK(b0,. . . ,bn) CZK(b0, bn)(u), with [K(b0 ,..., b„)(u):K(b0 ,..., i>„)] finite by Theorem 1.6 (since u is algebraic over K(bc,. . ., bn)) and [AX£>o,. . ., bn) : K] finite by Theorem 1.12 (since each bi e E is algebraic over K). Therefore, [K(b0 ,.. ., bn)(u) : K] is finite (Theorem 1.2). Hence
CHAPTER V FIELDS AND GALOIS THEORY
238
u e K(bo,..., bn)(u) is algebraic over K (Theorem 1.11). Since u was arbitrary, F is algebraic over K. ■
Theorem 1.14. Let F be an extension field of K and E the set of all elements of F which are algebraic over K. Then E is a subfield of F (which is, of course, algebraic over K). Clearly the subfield E is the unique maximal algebraic extension of K contained in F. PROOF OF 1.14. If u,v e E, then K(u,v) is an algebraic extension field of K by Theorem 1.12. Therefore, since u — v and uv~1 (c ^ 0) are in K(u,v), u — v and uv~* e E. This implies that E is a field (see Theorem 1.2.5). ■
APPENDIX: RULER AND COMPASS CONSTRUCTIONS The word “ruler” is to be considered as a synonym for straightedge (as is cus tomary in geometric discussions). We shall use field extensions to settle two famous problems of antiquity: (A) Is it possible to trisect an arbitrary angle by ruler and compass constructions? (B) Is it possible via ruler and compass constructions to duplicate an arbitrary cube (that is, to construct the side of a cube having twice the volume of the given cube)? We shall assume as known all the standard ruler and compass constructions as presented in almost any plane geometry text. Example: given a straight line L and a point P not on L, the unique straight line through P and parallel L [resp. perpen dicular to L] is constructible. Here and below “constructible” means “constructible by ruler and compass constructions.” Furthermore we shall adopt the viewpoint of analytic geometry as follows. Clearly we may construct with ruler and compass two perpendicular straight lines (axes). Choose a unit length. Then we can construct all points of the plane with integer coordinates (that is, locate them precisely as the intersection of suitable con structible straight lines parallel to the axes). As will be seen presently, the solution to the stated problems will result from a knowledge of what other points in the plane can be constructed via ruler and compass constructions. If F is a subfield of the field R of real numbers, the plane of F is the subset of the plane consisting of all points (c,d) with c e F, de F. If P,Q are distinct points in the plane of F, the unique line through P and Q is called a line in F and the circle with center P and radius the line segment PQ is called a circle in F. It is readily verified that every straight line in F has an equation of the form ax + by + c = 0 ia,b,c e F) and every circle in F an equation of the form x2 + y2 + ax + by + c = 0 (a,b,c e F) (Exercise 24).
Lemma 1.15. Let ¥ be a subfield of the field R of real numbers and let LX,L2 be nonparallel lines in F and Ci,C2 distinct circles in F. Then
APPENDIX: RULER AND COMPASS CONSTRUCTIONS
239
(i) Li fl L2 is a point in the plane of F; (ii) Li fl Ci = 0 or consists of one or two points in the plane of¥(-\ju) for some u e F (u > 0); (iii) Ci D C2 = 0 or consists of one or two points in the plane o/F(^u) for some u e F (u > 0). SKETCH OF PROOF, (i) Exercise, (iii) If the circles are Ci : x2 + y2 + axx + b\y + Ci = 0 and C2: x2 + y2 + a^x + b2y + c2 = 0 (a„/>„c,- e F by the remarks pre ceding the lemma), show that G fl C2 is the same as the intersection of G or C2 with the straight line L : (ax — a2)x + (b\ — b2)y + (ci — c2) = 0. Verify that L is a line in F; then case (iii) reduces to case (ii). (ii) Suppose Lx has the equation dx + ey + / = 0 (d,e,f s F). The case d = 0 is left as an exercise; if d ^ 0, we can assume d = 1 (why?), so that x = ( — ey — f). If (x,y)zLx fl Ci, then substitution gives the equation of G as 0 = ( — ey — f)2 + y2 + Ci (—ey — f) + bxy + Ci = Ay2 + By + C = 0, with A,B,C s F. If ,4 = 0, then y e f ; hence x e F and x,y e F(^l) = F. If A ^ 0, we may assume A — 1. Then y2 + By + C = 0 and completing the square yields (y + B/2)2 + (C — B2/4) = 0. This implies that either Lx fl Ci = 0 or e F(*\]u) with u = — C + B2/4 >0. ■ A real number c will be said to be constructible if the point (c,0) can be con structed (precisely located) by a finite sequence of ruler and compass constructions that begin with points with integer coordinates. The constructibility of c (or (c,0)) is clearly equivalent to the constructibility (via ruler and compass) of a line segment of length | c |. Furthermore the point (c,d) in the plane may be constructed via ruler and compass if and only if both c and d are constructible real numbers. The integers are obviously constructible, and it is not difficult to prove the following facts (see Exercise 25): (i) every rational number is constructible; (ii) if c > 0 is constructible, so is -yjc] (iii) if c,d are constructible, then c ± d, cd, and c/d (d ^ 0) are constructible, so that the constructible numbers form a subfield of the real numbers that contains the rationals.
Proposition 1.16. If a real number c is constructible, then c is algebraic of degree a power of 2 over the field Q of rationals. PROOF. The preceding remarks show that we may as well take the plane of Q as given. To say that c is constructible then means that (c,0) may be located (con structed) by a finite sequence of allowable ruler and compass constructions be ginning with the plane of Q. In the course of these constructions various points of the plane will be determined as the intersections of lines and/or circles used in the con struction process. For this is the only way to arrive at new points using only a ruler and compass. The first step in the process is the construction of a line or circle, either of which is completely determined by two points (center P and radius PT for the circle). Either these points are given as being in the plane of Q or else they may be chosen arbitrarily, in which case they may be taken to be in the plane of Q also. Similarly at each stage of the construction the two points that determine the line or circle used may be taken to be either points in the plane of Q or points constructed
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CHAPTER V FIELDS AND GALOIS THEORY
in previous steps. In view of Lemma 1.15 the first new point so constructed lies in the plane of an extension field Q(V«) °f Q» with w e Q, or equivalently in the plane of an extension Q(i) with v- e Q. Such an extension has degree 1 = 2° or 2 over Q (de pending on whether or not v e Q). Similarly the next new point constructed lies in the plane of Q(i;,w) = Q(r)(w) with w2 e Q(r). It follows that a finite sequence of ruler and compass constructions gives rise to a finite tower of fields: Q d Q(i>i) d
Q(vuv2)
d-d Q(vu . . . , un)
with e Q(vu . . . , v,-i) and [Q(^i,. . . , vt) : Q(vi,. . ., = 1 or 2 (2 < / < n). The point (c,0) constructed by this process then lies in the plane of F = Q(vi,..., vn). By Theorem 1.2, [F : Q] is a power of two. Therefore, c is algebraic over Q (Theo rem 1.11). Now Q d Q(c) d F implies that [Q(c) : Q] divides [F : Q] (Theorem 1.2), whence the degree [Q(c) : Q] of c over Q is a power of 2. ■
Corollary 1.17. An angle of 60° cannot be trisected by ruler and compass con structions. PROOF. If it were possible to trisect a 60° angle, we would then be able to construct a right triangle with one acute angle of 20°. It would then be possible to construct the real number (ratio) cos 20° (Exercise 25). However for any angle a, elementary trigonometry shows that cos 3a = 4 cos3 a — 3 cos a. Thus if a = 20°, then cos 3a = cos 60° = § and cos 20° is a root of the equation x = 4x3 — 3x and hence of the polynomial 8x3 — 6x — 1. But this polynomial is irreducible in Q[x] (see Theorem III. 6.6 and Proposition III.6.8). Therefore cos 20° has degree 3 over Q and cannot be constructible by Proposition 1.16. ■
Corollary 1.18. It is impossible by ruler and compass constructions to duplicate a cube of side length 1 (that is, to construct the side of a cube of volume 2). PROOF. If s is the side length of a cube of volume 2, then s is a root of x3 — 2, which is irreducible in Q[x] by Eisenstein’s Criterion (Theorem III.6.15). Therefore s is not constructible by Proposition 1.16. ■
EXERCISES Note: Unless specified otherwise F is always an extension field of the field K and Q,R,C denote the fields of rational, real, and complex numbers respectively. 1. (a) [F : K] = 1 if and only if F = K. (b) If [F : K] is prime, then there are no intermediate fields between F and K. (c) If u e F has degree n over K, then n divides \F : K].
2. Give an example of a finitely generated field extension, which is not finite di mensional. [Hint: think transcendental.]
APPENDIX: RULER AND COMPASS CONSTRUCTIONS
241
3. If Mi, , un e Fthen the field K(uu ... ,un) is (isomorphic to) the quotient field of the ring K[uu . . . , «„]. 4. (a) For any uu . . . , un e F and any permutation a e Sn, K(uu . . . , « „ ) — K(u„w,. . ., ua^)(b) K(UU . . . , W„_l)(«r,) = K(uu . .., un). (c) State and prove the analogues of (a) and (b) for K[uu . . . , «„]. (d) If each m, is algebraic over K, then K(uu ...,«„) = K[uu . . . , «„]. 5. Let L and M be subfields of F and LM their composite. (a) If K CZL fl M and M = K(S) for some S CZM, then LM - L(S). (b) When is it true that LM is the set theoretic union L U M? (c) If £i,..., En are subfields of F, show that
LxE2- • -En = E&E4E&. • (£n_i(£n)))- ■ •). 6. Every element of K(xi,. . ., xn) which is not in K is transcendental over K. 7. If v is algebraic over K(u) for some u s F and v is transcendental over K, then u is algebraic over K(v). 8. If u e F is algebraic of odd degree over K, then so is u2 and K(u) = K(u2).
9. If xn — ae ^[x] is irreducible and u e F is a root of xn — a and m divides n, then prove that the degree of um over K is n/m. What is the irreducible polynomial for um over K? 10. If F is algebraic over K and D is an integral domain such that K CZD CZF, then D is a field. 11. (a) Give an example of a field extension K CLF such that u,v e F are transcen dental over K, but K(u,v) K(xi,x2). [Hint: consider v over the field ^(w).] (b) State and prove a generalization of Theorem 1.5 to the case of n transcen dental elements Ui,... ,uv. 12. If d > 0 is an integer that is not a square describe the field Q(-\Jd) and find a set of elements that generate the whole field. 13. (a) Consider the extension Q(«) of Q generated by a real root u of x3 — 6x2-\9x + 3. (Why is this irreducible?) Express each of the following elements in terms of the basis {1 ,u,u2} : m4;m5;3m5 — m4 + 2; (m + I)-1; («2 — 6m + 8)-1. (b) Do the same with respect to the basis {l,u,u2,u3,u*\ of Q(m) where u is a real root of x5 + 2x + 2 and the elements in question are: (m2 + 2)(m3 + 3m);m_1; m*(m4 + 3m2 + lu + 5 );(m + 2)(m2 + 3)"1. 14. (a) If F = Q(-\j2,yj3), find [Z7 : Q] and a basis of F over Q. (b) Do the same for F = Q(/,\f3,to), where / e C, i2 = —1, and co is a com plex (nonreal) cube root of 1. 15. In the field K(x), let u = x3/(x + 1). Show that K(x) is a simple extension of the field K(u). What is [^(x) : K(u)]l 16. In the field C, Q(/) and Q(\^2) are isomorphic as vector spaces, but not as fields. 17. Find an irreducible polynomial /of degree 2 over the field Z2. Adjoin a root u of / toZ2 to obtain a field Z2(«) of order 4. Use the same method to construct a field of order 8.
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CHAPTER V FIELDS AND GALOIS THEORY
18. A complex number is said to be an algebraic number if it is algebraic over Q and an algebraic integer if it is the root of a monic polynomial in Z[x]. (a) If u is an algebraic number, there exists an integer n such that nu is an algebraic integer. (b) If r e Q is an algebraic integer, then re Z. (c) If u is an algebraic integer and neZ, then u + n and nu are algebraic integers. (d) The sum and product of two algebraic integers are algebraic integers. 19. If u,v e F are algebraic over K of degrees m and n respectively, then [K(u,v) : K] < mn. If (m,n) = 1, then [K(u,v) : K] = mn. 20. Let L and M be intermediate fields in the extension K d F. (a) [LM : K] is finite if and only if [L : K] and [M : K] are finite. (b) If [LM : K] is finite, then [L : K] and [M : K] divide [LM : K] and
[LM : K] < [L : K][M : K]. (c) If [L : K] and [M : K] are finite and relatively prime, then
[LM : K] = [L : K][M : K]. (d) If L and M are algebraic over K, then so is LM.
21. (a) Let L and M be intermediate fields of the extension K C F, of finite dimen sion over K. Assume that [LM : K] = [L : K][M : K] and prove that L D M = K. (b) The converse of (a) holds if [L : K] or [M : K) is 2. (c) Using a real and a nonreal cube root of 2 give an example where L D M = K, [L:K] = [M:K] = 3, but [LM : K] <9. 22. F is an algebraic extension of K if and only if for every intermediate field E every monomorphism a : E-* E which is the identity on Kis in fact an automorphism of E. 23. If u e F is algebraic over K(X) for some X CZ F then there exists a finite subset X' dX such that u is algebraic over K(X').
24. Let F be a subfield of R and P,Q points in the Euclidean plane whose coordinates lie in F. (a) The straight line through P and Q has an equation of the form ax + by + c = 0, with a,b,c e F. (b) The circle with center P and radius the line segment PQ has an equation of the form x2 + y2 + ax + by + c = 0 with a,b,c e F. 25. Let c,d be constructible real numbers. (a) c + d and c — d are constructible. (b) If d 0, then c/d is constructible. [Hint: If (x,0) is the intersection of the x axis and the straight line through (0,1) that is parallel the line through (0,d) and (c,0), then x = c/d.] (c) cd is constructible [Hint: use (b)J. (d) The constructible real numbers form a subfield containing Q. (e) Ifc > 0, then \fc is constructible. [Hint: If y is the length of the straight line segment perpendicular to the x axis that joins (1,0) with the (upper half of the) circle with center ((c + l)/2,0) and radius (c + l)/2 then y = yjc.]
2. THE FUNDAMENTAL THEOREM
243
26. Let Ex and F2 be subfields of F and X a subset of F. If every element of Ex is algebraic over E2, then every element of Ex(X) is algebraic over E2(X). [Hint: Ei(X) d (E2(X))(Ei); use Theorem 1.12.]
2. THE FUNDAMENTAL THEOREM The Galois group of an arbitrary field extension is defined and the concept of a Galois extension is defined in terms of the Galois group. The remainder of the section is devoted to proving the Fundamental Theorem of Galois Theory (Theorem 2.5), which enables us to translate problems involving fields, polynomials, and extensions into group theoretical terms. An appendix at the end of the section deals with sym metric rational functions and provides examples of extensions having any given finite group as Galois group. Let F be a field. The set Aut F of all (field) automorphisms F —> F forms a group under the operation of composition of functions (Exercise 1). In general, it is not abelian. It was Galois’ remarkable discovery that many questions about fields (especially about the roots of polynomials over a field) are in fact equivalent to cer tain group-theoretical questions in the automorphism group of the field. When these questions arise, they usually involve not only F, but also a (suitably chosen) subfield of F; in other words we deal with field extensions. If F is an extension field of K, we have seen in Section 1 that the K-module (vector space) structure of F is of much significance. Consequently, it seems natural to con sider those automorphisms of F that are also .K-module maps. Clearly the set of all such automorphisms is a subgroup of Aut F. More generally let E and F be extension fields of a field K. If a : E —* F is a non zero homomorphism of fields, then a{\E) = 1/- by Exercise III.1.15. If cr is also a K-module homomorphism, then for every k e K <j(k) = u(klE) = k<j{\E) = k\y = k.
Conversely, if a homomorphism of fields a : E—> F fixes K elementwise (that is, cr(k) = k for all k s K), then a is nonzero and for any u e E, a(ku) =
whence a is a K-module homomorphism.
Definition 2.1. Let E and F be extension fields of a field K. A nonzero map a : E —* F which is both a field and a Y^-module homomorphism is called a K-homomorphism. Similarly if a field automorphism a e Aut F is a K-homomorphism, then a is called a K-automorphism of F. The group of all K-automorphisms of F is called the Galois group of F over K and is denoted AuIkF. REMARKS. AT-monomorphisms and A'-isomorphisms are defined in the obvious way. Here and below the identity element of AutAF and its identity subgroup will both be denoted by 1. EXAMPLE. Let F = K(x), with K any field. For each a s. K with a ^ 0 the map cra : F —* F given by f(x)/g(x) f(ax)/g(ax) is a AT-automorphism of F; (this may
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be verified directly or via Corollaries III.2.21(iv), III.4.6, and III.5.6, and Theorem III.4.4(ii)). If K is infinite, then there are infinitely many distinct automorphisms ca, whence Aut/
Theorem 2.2. Let F be an extension field of K and f e K[x], //us F is a root of i and o e AutRp, then o-(u) e F is also a root off. n
PROOF. If / =
52 kiX\ then f(u) = 0 implies 0 = a( /(«)) = ai^kjU') i=i
= 52
2. THE FUNDAMENTAL THEOREM
245
Theorem 2.3. Let F be an extension field of K, E an intermediate field and H a sub group of Aut^F. Then (i) H' = {v e F | er(v) = v for all a s H | is an intermediate field of the extension; (ii) E' = {as AutK F |
F' = AutpF = 1 and K' = Aut kF = G; and on the other, 1' = F (that is, F is the fixed field of the identity subgroup). It is not necessarily true, however, that G' = K (as can be seen in the first examples after Theorem 2.2, where G = 1 and hence G' = F ^ K; also see Exercise 2).
Definition 2.4. Let F be an extension field of K such that the fixed field of the Galois group Aut\<JF is K itself. Then F is said to be a Galois extension (field) of K or to be Galois over K.1 REMARKS. F is Galois over K if and only if for any «eF - K, there exists a /^-automorphism a s Aut^F such that o(u) ^ u. If F is an arbitrary extension field of K and K0 is the fixed field of Aut^F (possibly K0 j6- K), then it is easy to see that F is Galois over K0, that K C K0, and that Aut*F = Aut^F. EXAMPLES. C is Galois over R and Q(^3) is Galois over Q (Exercise 5). If K is an infinite field, then K(x) is Galois over K (Exercise 9). Although a proof is still some distance away, it is now possible to state the Fundamental Theorem of Galois Theory, so that the reader will be able to see just where the subsequent discussion is headed. If L,M are intermediate fields of an ex tension with L C M, the dimension [M : L] is called the relative dimension of L and M. Similarly, if H,J are subgroups of the Galois group with H < J, the index [J : H] is called the relative index of H and J.
Theorem 2.5. (Fundamental Theorem of Galois Theory) If F is a finite dimensional Galois extension of K, then there is a one-to-one correspondence between the set of all JA
Galois extension is frequently required to be finite dimensional or at least algebraic and is defined in terms of normality and separability, which will be discussed in Section 3. In the finite dimensional case our definition is equivalent to the usual o.ie. Our definition is essentially due to Artin, except that he calls such an extension “normal.” Since this use of “normal” conflicts (in case char F 5^ 0) with the definition of “normal” used by many other authors, we have chosen to follow Artin's basic approach, but to retain the (more or less) conventional terminology.
CHAPTER V FIELDS AND GALOIS THEORY
246
intermediate fields of the extension and the set of all subgroups of the Galois group AutkF (given by E |—» E' = Aut^F) such that: (i) the relative dimension of two intermediate fields is equal to the relative index of the corresponding subgroups; in particular, Aut kF has order [F : KJ; (ii) F is Galois over every intermediate field E, but E is Galois over K if and only if the corresponding subgroup E' = AutEF is normal in G = AutKF; in this case G/E' is (
A
M (----- ► M' U
A
F-*------- J 1 U
A
//'-«-------J II U
A
L |-----J'-*-------------------------------------------- 1 J A
U
A
K |---- ► G,
K
G.
U
Formally, the basic facts about the priming operations are given by
Lemma 2.6. Let F be an extension field of K with intermediate fields L and M. Let H and J be subgroups of G = AutKF. Then: (i) F' = 1 andYJ = G; O') 1' = F; (ii) L C M => M' < L'; (ii') H < J => J' C H'; (iii) L C= L" and H < H" (where L" = (L')' and H" = (H')'); (iv) L' = L'" and H' = H"\ SKETCH OF PROOF, (i)—(iii) follow directly from the appropriate definitions. To prove the first part of (iv) observe that (iii) and (ii) imply U" < L' and that (iii) applied with L' in place of H implies L‘ < L"’. The other part is proved similarly. ■
2. THE FUNDAMENTAL THEOREM
247
REMARKS. It is quite possible for L" to contain L properly (similarly for H" and H). F is Galois over K (by definition) if G' = K. Thus since K' = G in any case, F is Galois over K if and only if K = K". Similarly F is Galois over an intermediate field E if and only if E = E". Let X be an intermediate field or subgroup of the Galois group. X will be called closed provided X = X". Note that F is Galois over K if and only if K is closed.
Theorem 2.7. If F is an extension field of K, then there is a one-to-one correspondence between the closed intermediate fields of the extension and the closed subgroups of the Galois group, given by E (—*• E' = Aut^F. PROOF. Exercise; the inverse of the correspondence is given by assigning to each closed subgroup H its fixed field H'. Note that by Lemma 2.6(iv) all primed objects are closed. ■ This theorem is not very helpful until we have some more specific information as to which intermediate fields and which subgroups are closed. Eventually we shall show that in an algebraic Galois extension all intermediate fields are closed and that in the finite dimensional case all subgroups of the Galois group are closed as well. We begin with some technical lemmas that give us estimates of various relative di mensions.
Lemma 2.8. Let F be an extension field of K and L,M intermediate fields with L C M. If [M : L] is finite, then [L' : M'] < [M : L]. In particular, if [F : K] is finite, then |^h/kF| < [F : K]. PROOF. We proceed by induction on n = [M : L], with the case n = 1 being trivial. If n > 1 and the theorem is true for all i < n, choose uzM with u $ L. Since [M : L] is finite, u is algebraic over L (Theorem 1.11) with irreducible polynomial fs L[x] of degree k > 1. By Theorems 1.6 and 1.1, [L(u) : L] = k and [M : £(«)] = n/k. Schematically we have:
n/k
rm i— u
—► M' A
-L{u) |——► L(u)' k
A u L|---- —► L'.
There are now two cases. If k < n, then 1 < n/k < n and by induction [L1 : L(u)'] < k and \L(u)' : M'J < n/k. Hence \L' : M'] = [L' : L(m)'][L(«)' : M'\ < k(n/k) = n = [M : L] and the theorem is proved. On the other hand if k = n, then [M : L(u)\ = 1 and M = L(u). In order to complete the proof in this case, we shall construct an in jective map from the set of 5 of all left cosets of M' in L' to the set T of all distinct
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CHAPTER V FIELDS AND GALOIS THEORY
roots (in F) of the polynomial feL[x], whence \S\ < \T\. Since [7’[ < n by Theorem III.6.7 and |5| = [L’ : M '] by definition, this will show that [L' : M'] < \ T \ < n = [M : L\. The final statement of the theorem then follows immediately since |Aut*F| = [AutxF : 1] = [K' : F'] < [F : K]. Let tM' be a left coset of M' in L'. If a e M' = Aut a//7, then since u s M, t<j(u) = t(u). Thus every element of the coset tM' has the same effect on u and maps u (—> r(w). Since t e V = AuU/% and u is a root of /s L[x], t(u) is also a root of /by Theorem 2.2. This implies that the map 5 —> T given by tM' [—► t(u) is well defined. If t(h) = t0(«) (r,r0 s V), then r0_1r(t/) = u and hence t0~1t fixes u. Therefore, TiTlr fixes L(u) — M elementwise (see Theorem 1.6(iv)) and r0_1r e A/'. Consequently by Corollary 1.4.3 r0M' = tM' and the map S —* T is injective. ■ Several important applications of Lemma 2.8 are treated in tne appendix. We now prove an analogue of Lemma 2.8 for subgroups of the Galois group.
Lemma 2.9. Let F be an extension field of K and let H,J be subgroups of the Galois group AutKF with H < J. If [3 : H] is finite, then [H' : J'] < [J : H]. PROOF. Let [J : H] = n and suppose that [//' : J'] > n. Then there exist «i,«2, - - - , un+1 s H' that are linearly independent over J'. Let {ri,r2,. . ., rn} be a complete set of representatives of the left cosets of H in J (that is, J = TjH U t2H U • • • U rnH and 77-'r, e H if and only if i = j) and consider the system of n homo geneous linear equations in n + 1 unknowns with coefficients t,(w3) in the field F: T\{u\)x\ 4“ Tl(w2)-^2 4“ T\(Uz)Xi + ■ • • + T\(un+\)xn+i = 0 Ti(ui)x 1 -j- r2(t/2Xv2 -f- T2(u-j)Xz + • • • + T-£un+l)Xn+1 =
0 (i)
rn(ui)xi -f- Tn(u2)x2 -f- Tn(,u-i)Xs -f- ■ • ■ -f- T„(un+i).xr„+i = 0
Such a system always has a nontrivial solution (that is, one. different from the zero solution xi = x2 = ■ ■ ■ = x„+1 = 0; see Exercise VII.2.4(d)). Among all such non trivial solutions choose one, say xx = a\,. . . , xn+i = an+1 with a minimal number of nonzero a,. By reindexing if necessary we may assume that xi = at, . . ., xr = ar, 0 (a, 9^ 0). Since every multiple of a solution is also a solution AY-fl * -^n+l we may also assume ai = (if not multiply through by ai-1)We shall show below that the hypothesis that uu . . ., un+l e //' are linearly inde pendent over J' (that is, that [H' :/'] > n) implies that there exists a sJ such that xi = uai, x-i = o-a2, . . . , xr =
2. THE FUNDAMENTAL THEOREM
249
To complete the proof we must find a e J with the desired properties. Now exactly one of the t„ say n, is in H by definition; therefore ti(«,) = Ui e H' for all i. Since the a, form a solution of (1), the first equation of the system yields:
U\di -(- W2Q2 “I- • • • “t- uraT = 0. The linear independence of the Ui over J' and the fact that the a, are nonzero imply that some say a2, is not in J'. Therefore there exists a eJ such that <7«2 ^ a2. Next consider the system of equations
(TTi(«l)jCl + dTi(u2)X2 H---------1-
(2)
Lemma 2.10. Let F be an extension field of K, L and M intermediate fields with L C M, and H,J subgroups of the Galois group Aut kF with H < J. (i) If L is closed and [M : L] finite, then M is closed and [U : M'] = [M : L]; (ii) // H is closed and [J : H] finite, then J is closed and [H' : J'] = [J : H]; (iii) if F is a finite dimensional Galois extension of K, then all intermediate fields and all subgroups of the Galois group are closed and Aut^F has order [F : K]. Note that (ii) (with H = 1) implies that every finite subgroup of Aut^Fis closed.
SKETCH OF PROOF OF 2.10. (ii) Applying successively the facts that J CZ J" and H = H" and Lemmas 2.8 and 2.9 yields
[J : H] < [.I" : H) = [J" : H") < [H' :/'] < [J : H\; this implies that J = J” and [H' : J'] = [J : H], (i) is proved similarly. (iii) If E is an intermediate field then [E : K] is finite (since [F : X] is). Since F is Galois over K, K is closed and (i) implies that E is closed and [K' : E'] = [E : K\. In particular, if E = F, then |AutA'F| = [Aut*F; l] = [K' : F'] = [F: KJ is finite.
CHAPTER V FIELDS AND GALOIS THEORY
250
Therefore, every subgroup J of Aut^F is finite. Since 1 is closed (ii) implies that J is closed. ■ The first part of the Fundamental Theorem 2.5 can easily be derived from Theo rem 2.7 and Lemma 2.10. In order to prove part (ii) of Theorem 2.5 we must deter mine which intermediate fields correspond to normal subgroups of the Galois group under the Galois correspondence. This will be done in the next lemma. If E is an intermediate field of the extension K d F, E is said to be stable (relative to K and F) if every /('-automorphism a s Aut«F maps E into itself. If E is stable and o--1 s Aut/vF is the inverse automorphism, then also maps E into itself. This im plies that a | E is in fact a /('-automorphism of E (that is, a | E s Aut*F) with inverse o--11 E. It will turn out that in the finite dimensional case E is stable if and only if E is Galois over K.
Lemma 2.11. Let F be an extension field of K. (i) If E is a stable intermediate field of the extension, then E' = Aut^F is a normal subgroup of the Galois group /Im/rF; (ii) if H is a normal subgroup of /4 m/rF, then the fixed field H' of H is a stable
intermediate field of the extension. PROOF, (i) If u e E and a e AutAF, then tr(u) e E by stability and hence = ct(u) for any t eE' = AuteF. Therefore, for any a e AuU-F, t e E' and u s E, = o- V(w) = u. Consequently, er Vo- e E' and hence E' is normal in AutA'F(ii) If cr s AutsF and re//, then a^ra e H by normality. Therefore, for any u e //', o--1rcr(t/) = u, which implies that ra(u) — cr(u) for all r e / / . Thus a(u) e H' for any u e //', which means that H’ is stable. ■ tct(u)
In the next three lemmas we explore in some detail the relationships between stable intermediate fields and Galois extensions and the relationship of both to the Galois group.
Lemma 2.12. If F is a Galois extension field of K and E is a stable intermediate field of the extension, then E is Galois over K. PROOF. If u s E — K, then there exists a e AutxF such that tr(u) ^ u since F is Galois over K. But a \ E s Aut^F by stability. Therefore, E is Galois over K by the Remarks after Definition 2.4. ■
Lemma 2.13. If F is an extension field of K and E in an intermediate field of the ex tension such that E is algebraic and Galois over K, then E is stable (relative to F and¥L). REMARK. The hypothesis that E is algebraic is essential; see Exercise 13. PROOF OF 2.13. If u s E, let /s AT[jc:] be the irreducible polynomial of u and let « = «i, «2,. . ■ , «r be the distinct roots of / that lie in E. Then r < n = deg/by Theo
2. THE FUNDAMENTAL THEOREM
251
rem III.6.7. If r e AutA'F, then it follows from Theorem 2.2 that t simply permutes the Ui. This implies that the coefficients of the monic polynomial #(*) = (a: — u\) (x — u/)- ■ (x — ur) e E[x] are fixed by every t s AuIkE. Since E is Galois over K, we must have g e K[x], Now u = ux is a root of g and hence / | g (Theorem 1.6(ii)). Since g is monic and deg g < deg f we must have f = g. Consequently, all the roots of/are distinct and lie in E. Now if o-e Aut*F, then a(u) is a root of/by Theorem 2.2, whence a(u) e E. Therefore, E is stable relative to F and K. ■ Let E be an intermediate field of the extension K C F. A K-automorphism t e AutxE is said to be extendible to F if there exists a e AuU-F such that a \ E = r. It is easy to see that the extendible K-automorphisms form a subgroup of AuuF. Recall that if E is stable, E' = Aut^F is a normal subgroup of <7 = Aut/cF (Lemma 2.11). Consequently, the quotient group G/E' is defined.
Lemma 2.14. Let F be an extension field of K and E a stable intermediate field of the extension. Then the quotient group A u t / Aut^F is isomorphic to the group of all K-automorphisms of E that are extendible to F. SKETCH OF PROOF. Since E is stable, the assignment a (—>
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CHAPTER V FIELDS AND GALOIS THEORY
Theorem 2.15. (Art in) Let ¥ be a field, G a group of automorphisms of F and K the fixed field ofG in F. Then F is Galois over K. If G is finite, then F is a finite dimen sional Galois extension of K with Galois group G. PROOF. In any case G is a subgroup of Aut*/7. If u s F — K, then there must be a a s G such that
APPENDIX: SYMMETRIC RATIONAL FUNCTIONS Let K be a field, K[xi, . . . , *„] the polynomial domain and K(xu . . . , x„) the field of rational functions (see the example preceding Theorem 1.5). Since K(xi, . . . , x„) is by definition the quotient field of K[xi, . . . , x„], we have K[xu . . . , xr] Cl K( xi,. . . , xn) (under the usual identification of /with f/l a). LetS„ be the symmetric group on n letters. A rational function
/ = *1 + *2 + • ■ • + *« = Xi\ i=
fi =
1
52 1
fa = 52 *.*;** J 1
fn = XiX2- ■ The verification that the / are indeed symmetric follows from the fact that they are simply the coefficients of y in the polynomial g(>0 s K[xi, . . . , *„][>'], where g(>) = (>’ - *iX>' - x2)(y - *3)• • •(>'- *„)
= yn - fiy+ /v71”2----------h ( -1 Y^fi-iy + (-1)71/If tr s Sn, then the assignments x, f—> xa^(i = 1,2,...,«) and
fixi , . - - , Xn)/g(Xi, . . . , Xn)| ./(Xff(1), . . . , *ff(n))/#(*ff(i), . . . , *(7(n))
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253
define a AT-automorphism of the field K(xi, . . . , xn) which will also be denoted a (Exercise 16). The map SK —» AutjcKOa, . .. , xn) given by a |—> a is clearly a mono morphism of groups, whencemay be considered to be a subgroup of the Galois group AutxK(x,,..., xn). Clearly, the fixed field E of Sn in K(xi,. . ., x„) consists precisely of the symmetric functions; that is, the set of all symmetric functions is a subfield of K(xu . . . , xn) containing K. Therefore, by Artin’s Theorem 2.15 K(xi,. .., xn) is a Galois extension of E with Galois group Sn and dimension \Sn\ = «!.
Proposition 2.16. If G is a finite group, then there exists a Galois field extension with Galois group isomorphic to G. PROOF. Cayley’s Theorem II.4.6 states that for n = |G|, G is isomorphic to a subgroup of Sn (also denoted G). Let K be any field and E the subfield of symmetric rational functions in K(xu . . . , x„). The discussion preceding the theorem shows that K(xi,. . . , x\) is a Galois extension of E with Galois group S„. The proof of the Fundamental Theorem 2.5 shows that K(xi,.. ., xn) is a Galois extension of the fixed field Ex of G such that Aut£li^(xi, . . . , xn) = G. ■ The remainder of this appendix (which will be used only in the appendix to Sec tion 9) is devoted to proving two classical theorems about symmetric functions. Throughout this discussion n is a positive integer, K an arbitrary field, E the subfield of symmetric rational functions in K(xu . . . , x„) and f,...,fneE the elementary symmetric functions in xi,. . . , xn over K. We have a tower of fields:
Kd ...,/„) C £ <= K(xu . . . , xn). In Theorem 2.18 we shall show that E = K(fu . .. ,fn). If Mi,. . ., ur e K(xi, . . . , x„), then every element of K(ui, . . . , ur) is of the form g(uu . . ., uT)/h(ui,. .., ur) with g, he K[xi, • - - , XA by Theorem 1.3.Consequently, an element of K(u\,. . . , ur) [resp. K[uu .. . , ur]] is usually called a rational function [resp. polynomial] in Mi,..., ur over K. Thus the statement E = K(f,. . . , fn) may be rephrased as: every rational symmetric function is in fact a rational function of the elementary symmetric functions f,. . . , fn over K. In order to prove that E = K(f, . . . , fn) we need
Lemma 2.17. Let K be a field, fi,. . . , f„ the elementary symmetric functions in Xi,. . . , xn over K andkan integer with 1 < k < n — 1. //hi, . . . , hk e K[xi, . . . , xn] are the elementary symmetric functions in x,, ... , xk, then each hj can be written as a polynomial over K in fi,f2, . . . , f„ and xk+i,xk+2, . . . , xn. SKETCH OF PROOF. The theorem is true when k = n — 1 since in that case hi = fi — xn and hj = fi — h,-ixn (2 < j < ri). Complete the proof by induction on k in reverse order: assume that the theorem is true when k = r + 1 and r + 1 < n — 1. Let gu , gr+i be the elementary symmetric functions in Xi,. .., xT+i and hi, . . . , hT the elementary symmetric functions in x\,. .., xr. Since hi = gi — *r+i and hj = gj — hj-ixT+i (2 < j < r), it follows that the theorem is also true for k = r. ■
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Theorem 2.18. If K is a field, E the subfield of all symmetric rational functions in K(xi, . . . , x„) and fi, . . . , f„ the elementary symmetric functions, then E = K(f,,. . . , fn). SKETCH OF PROOF. Since . . . , xn) : E\ = «! and K(f,...,fn) d E d K(xi,. . . , xn), it suffices by Theorem 1.2 to show that [K(a'i, . . . , x„) : K(f,..., fn)\ < n\. Let F — K(fu...,fn) and consider the tower of fields:
F d F(xn) d F(xn-i,xn) d - d F(x2, . . . , x„) d F(xi, ...,*„) = K(xu . . . , xn). Since F(xk,xk+i, . . ., xn) = F(xk+l,. . . , x„)(xk), it suffices by Theorems 1.2 and 1.6 to show that xn is algebraic over F of degree < n and for each A < n, xk is algebraic of degree
gn(y) = (y - *1 )(y - x2) • • ■ (y - xn) = yn - fiy”-1 + ■ ■ ■ + (-1 Yfn. Since gn s F[y] has degree n and xn is a root of gr,, xn is algebraic of degree at most n over F = K( f , . . . ,f„) by Theorem 1.6. Now for each k (1 < k < ri) define a monic polynomial: ^(> ) = gn(y)/(y - xk+,)- ■ •(>- — xn) = (>■ — Ai)(y — x2)- • -(> — xk). Clearly each gk(y) has degree k, xk is a root of gk(y) and the coefficients of gk{y) are precisely the elementary symmetric functions in xi,.. . , xk. By Lemma 2.17 each gk(y) lies in F(xk+1, . . . , jr„)[y], whence xk is algebraic of degree at most k over F(xk+1, . . . , jt„). ■ We shall now prove an analogue of Theorem 2.18 for symmetric polynomial func tions, namely: every symmetric polynomial in x\,. . ., xn over K is in fact a poly nomial in the elementary symmetric functions fi,. . . ,fn over K. In other words, every symmetric polynomial in K[xu . . . , *„] lies in K[f,. . . First we need
Lemma 2.19. Let K be a field and E the subfield of all symmetric rational functions in K(xi, . . . , x„). Then the set X = {xii*x2i2- • ■ xn*n | 0 < ik < k for each kj is a basis of K(xi, . . . , xn) over E. SKETCH OF PROOF. Since [K(Xl,..., x„) :£] = «! and \X\ = «!, it suffices to show that X spans K(xi,. . . , x„) (see Theorem IV.2.5). Consider the tower of fields E d E{xn) d E(xn^,xn) d - d E(xy,. .., xn) = K(xu . . . , xn). Since is al gebraic of degree < n over E (by the proof of Theorem 2.18), the set {x„j \ 0 < j < n} spans E(x„) over E (Theorem 1.6). Since E(xn__uxn) = E(xr )(x„ i), and xn^\ is algebra ic of degree < n — 1 over £■(*„), the set {x*^ | 0 < i < n — 1} spans E(xn^i,xn) over E(xn). The argument in the second paragraph of the proof of Theorem IV.2.16 shows that the set [x^xj \ 0 < / < n — 1;0 <j < n \ spans E{xn^x,xn) over E. This is the first step in an inductive proof, which is completed by similar arguments. ■
Proposition 2.20. Let K be a field and let f(, . . . , f., be the elementary symmetric functions in K(xi, . . . , x„).
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(i) Every polynomial in K[xi, . . . , x„] can be written uniquely as a linear combina tion of the n! elements x^x-/2- • xn‘n(0 < ik < k for each k) with coefficients in K[fi, . . . , f,J; (ii) every symmetric polynomial in K[xi, . . . , x„] lies in K[fi, . . . , f„], PROOF. Letg* (>)(& = 1,...,«) be as in the proof of Theorem 2.18. As noted there the coefficients of gk(y) are polynomials (over K) in fu . .. , fn and xk+u . .. ,xn. Since gk is monic of degree k and gk(xk) = 0, xkk can be expressed as a polynomial over K in f,...,f„, xk+\,. . . , xn and xk{ (/ < k — 1). If we proceed step by step beginning with k = 1 and substitute this expression for xkk in a polynomial h e K[x\,. .. , x„], the result is a polynomial in f , x \ , . .., xn in which the highest exponent of any xk is k — 1. In other words h is a linear combination of the n \ elements x\hxh- ■ -x„in (ik < k for each k) with coefficients in K[f,. . . F u r thermore these coefficient polynomials are uniquely determined since jjfi1'1- • -xnin | 0 < ik < k for each k) is linearly independent over E = K(f,... ,fn) by Lemma 2.19. This proves (i) and also implies that if a polynomial h e K[xi , . . . , x„] is a linear combination of the xi’1- - -xnil1 (4 < k) with coefficients in K(f,. . . ,fn), then the coefficients are in fact polynomials in K[f,. . . ,f„]. In particular, if h is a symmetric polynomial (that is, h sE = K(f,. . . , f,)), then h = hxi°x2°- • Xr.0 necessarily lies in K[f, T h i s proves (ii). ■
EXERCISES Note: Unless stated otherwise F is always an extension field of the field K and E is an intermediate field of the extension. 1. (a) If F is a field and a : F F a (ring) homomorphism, then a = 0 or a is a monomorphism. If a ^ 0, then
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Since f/g is transcendental over K (why?) we may for convenience replace K(f/g) by K(z) (z an indeterminate) and consider
7. Let G be the subset of AuU A^x) consisting of the three automorphisms induced (as in 6 (c)) by x f—> x, x (—> I k / O k — x), x [—> (x — 1 k ) / x . Then G is a subgroup of A\x\kK(x). Determine the fixed field of G. 8. Assume char K = 0 and let G be the subgroup of AutAAr(x) that is generated by the automorphism induced by x|—> * + 1A-. Then G is an infinite cyclic group. Determine the fixed field E of G. What is [Aj(x) : £]? 9. (a) If K is an infinite field, then K(x) is Galois over K. [Hint: If K(x) is not Galois over K, then K(x) is finite dimensional over the fixed field E of AuU A^x) by Exercise 6(b). But Aut£^(x) = AutA-A^x) is infinite by Exercise 6(d), which con tradicts Lemma 2.8.] (b) If K is finite, then K(x) is not Galois over K. [Hint: If K(x) were Galois over K, then AutA'X(x) would be infinite by Lemma 2.9. But AuU /f(x) is finite by Exercise 6(d).]
10. If AT is an infinite field, then the only closed subgroups of AutxAT(x) are itself and its finite subgroups. [Hint: see Exercises 6(b) and 9.] 11. In the extension of Q by Q(x), the intermediate field Q(x2) is closed, but Q(x3) is not.
12. If £ is an intermediate field of the extension such that E is Galois over K, Fis Galois over E, and every cr e AutAf is extendible to F, then F is Galois over K. 13. In the extension of an infinite field K by K(x,y), the intermediate field K(x) is Galois over K, but not stable (relative to K(x,y) and AT). [See Exercise 9; compare this result with Lemma 2.13.] 14. Let F be a finite dimensional Galois extension of K and let L and M be two inter mediate fields. (a) AuUa/F = AuUF D AuWF; (b) Auttn^/F = Autz.F V AuWF; (c) What conclusion can be drawn if AutzF D AutwF =1? 15. If F is a finite dimensional Galois extension of K and F is an intermediate field, then there is a unique smallest field L such that E d L d F and L is Galois over AT; furthermore
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AutiF = Pi cr(Aut£F)o-_1,
C
where a runs over Aut/fF. 16. If cr e Sn, then the map K(xi,. . . , —» K(xu . . ., xn) given by ,/X^l 1 ■ • • ?
/(^"ffCl)* • • • j -^"cCrO)
---------- )—* —-------------
gOi,. . ., x„) g(xca),. .., xff(K))
is a K-automorphism of K(xi,. . ., xn).
3. SPLITTING FIELDS, ALGEBRAIC CLOSURE AND NORMALITY We turn now to the problem of identifying and/or constructing Galois exten sions. Splitting fields, which constitute the principal theme of this section, will enable us to do this. We first develop the basic properties of splitting fields and algebraic closures (a special case of splitting fields). Then algebraic Galois extensions are char acterized in terms that do not explicitly mention the Galois group (Theorem 3.11), and the Fundamental Theorem is extended to the infinite dimensional algebraic case (Theorem 3.12). Finally normality and other characterizations of splitting fields are discussed. The so-called fundamental theorem of algebra (every polynomial equation over the complex numbers has a solution) is proved in the appendix. Let F be a field and / s F[x\ a polynomial of positive degree, /is said to split over F (or to split in F[x]) if/can be written as a product of linear factors in F[x]; that is, / = Uo(x — Wi)(x — w2) • • • (x — un) with m, e F.
Definition 3.1. Let K be a field and f s K[x] a polynomial of positive degree. An ex tension field F of K is said to be a splitting field over K of the polynomial f iff splits in F[x] and F = K(ui, . . . , u„) where u,, . . . , u„ are the roots off in F. Let S be a set of polynomials of positive degree in K[x]. An extension field F ofK is said to be a splitting field over K of the set S of polynomials if every polynomial in S splits in F[x] and F is generated over K by the roots of all the polynomials in S. EXAMPLES. The only roots of x2 — 2 over Q are \/2 and — \^2 and x2 — 2 = (x — -^2)(x + -\J2). Therefore Q(\^2) = Q('\j2, — ^2) is a splitting field of x2 — 2 over Q. Similarly C is a splitting field of x2 + 1 over R. However, if u is a root of an irreducible /e K[x], K(u) need not be a splitting field of /. For instance if u is the real cube root of 2 (the others being complex), then Q(«) CI R, whence Q(w) is not a splitting field of x3 — 2 over Q. REMARKS. If F is a splitting field of 5 over K, then F = K(X), where X is the set of all roots of polynomials in the subset 5 of K[x]. Theorem 1.12 immediately implies that F is algebraic over K (and finite dimensional if S, and henceX, is a finite set). Note that if 5 is finite, say S = \ f,fi, ...,/, 1, then a splitting field of 5 coin cides with a splitting field of the single polynomial / = /,/- • -/„ (Exercise 1). This fact will be used frequently in the sequel without explicit mention. Thus the splitting field of a set 5 of polynomials will be chiefly of interest when S either consists of a single polynomial or is infinite. It will turn out that every [finite dimensional]
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algebraic Galois extension is in fact a particular kind of splitting field of a [finite] set of polynomials. The obvious question to be answered next is whether every set of polynomials has a splitting field. In the case of a single polynomial (or equivalently a finite set of polynomials), the answer is relatively easy.
Theorem 3.2. 7/K is a field and f e K[x] has degree n > 1, then there exists a splitting field F off with [F : K] < n! SKETCH OF PROOF. Use induction on n = deg /. If n = 1 or if /splits over K, then F = K is a splitting field. If n > 1 and /does not split over K, let g e K[x] be an irreducible factor of f of degree greater than one. By Theorem 1.10 there is a simple extension field K(u) of K such that u is a root of g and [K(u) : K] = deg g > 1. Then by Theorem III. 6.6, / = (x — u)h with h e K(u)[x] of degree n — 1. By induction there exists a splitting field F of h over K(u) of dimension at most (« — 1)! Show that Fisa splitting field of / over K (Exercise 3) of dimension [F: K] = [F : K(u)][K(u) : K] < ( n — 1)! (deg g) < n! ■ Proving the existence of a splitting field of an infinite set of polynomials is con siderably more difficult. We approach the proof obliquely by introducing a special case of such a splitting field (Theorem 3.4) which is of great importance in its own right. Note: The reader who is interested only in splitting fields of a single polynomial (i.e. finite dimensional splitting fields) should skip to Theorem 3.8. Theorem 3.12 should be omitted and Theorems 3.8-3.16 read in the finite dimensional case. The proof of each of these results is either divided in two cases (finite and infinite dimen sional) or is directly applicable to both cases. The only exception is the proof of (ii) => (i) in Theorem 3.14; an alternate proof is suggested in Exercise 25.
Theorem 3.3. The following conditions on a field F are equivalent.
Every nonconstant polynomial f e F[x] has a root in F; every nonconstant polynomial f e F[x] splits over F; every irreducible polynomial in F[x] has degree one; there is no algebraic extension field of F (except F itself)', there exists a subfield Ko/F such that F is algebraic over K and every poly nomial in K[x] splits in F[x]. (i) (ii) (iii) (iv) (v)
PROOF. Exercise; see Section III. 6 and Theorems 1.6, 1.10,1.12 and 1.13. ■ A field that satisfies the equivalent conditions of Theorem 3.3 is said to be algebraically closed. For example, we shall show that the field C of complex num bers is algebraically closed (Theorem 3.19).
Theorem 3.4. If F is an extension field of K, then the following conditions are equivalent.
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(i) F is algebraic over K and F is algebraically closed; (ii) F is a splitting field over K of the set of all [irreducible] polynomials in K[x], PROOF. Exercise; also see Exercises 9, 10. ■ An extension field F of a field K that satisfies the equivalent conditions of Theo rem 3.4 is called an algebraic closure of K. For example, C = R(z) is an algebraic closure of R. Clearly, if F is an algebraic closure of K and S is any set of polynomials in K[x], then the subfield E of F generated by K and all roots of polynomials in 5 is a splitting field of S over K by Theorems 3.3 and 3.4. Thus the existence of arbitrary splitting fields over a field K is equivalent to the existence of an algebraic closure of K. The chief difficulty in proving that every field K has an algebraic closure is settheoretic rather than algebraic. The basic idea is to apply Zorn’s Lemma to a suitably chosen set of algebraic extension fields of K.2' To do this we need
Lemma 3.5. IfF is an algebraic extension field of K, then |F| < K0|K|. SKETCH OF PROOF. Let T be the set of monic polynomials of positive de gree in AT[x]. We first show that \T\ = K0|AT|. For each n e N* let Tn be the set of all polynomials in T of degree n. Then \Tn\ = IA-71!, where Kn = K X K X ■■■ K (n factors), since every polynomial / = xn + ixn~l + • • • + o0 e Tn is completely determined by its n coefficients a0,ai,. . ., an-i e K. For each n e N* let fn:Tn~-> Kn be a bijection. Since the sets Tn [resp. Kn] are mutually disjoint, the map / : T = 1J Tn —> U Kn, given by f(u) = fn(u) for u e T„, is a well-defined bijection. 7ieN*
ij«N*
Therefore \T\ = | U Kn\ = K0|K] by Introduction, Theorem 8.12(ii). neN* Next we show that |F| < |7’|, which will complete the proof. For each irreducible /s T, choose an ordering of the distinct roots of /in F. Define a map F —» T X N* as follows. If a e F, then a is algebraic over K by hypothesis, and there exists a unique irreducible monic polynomial /e T with f(a) = 0 (Theorem 1.6). Assign to ae F the pair (fi) e T X N* where a is the /th root of /in the previously chosen ordering of the roots of /in F. Verify that this map F->TX N* is well defined and injective. Since T is infinite, |F| < |TX N*| = |Z'HN*! = |7]N 0= | T\ by Theorem 8.11 of the Introduction. ■
Theorem 3.6. Every field K has an algebraic closure. Any two algebraic closures of K are K-isomorphic. SKETCH OF PROOF. Choose a set S such that N0|K] < |lS] (this can always be done by Theorem 8.5 of the Introduction). Since |X| < K0|AT| (Introduction, Theorem 8.11) there is by Definition 8.4 of the Introduction an injective map 6 : K —»5. Consequently we may assume K CZ S (if not, replace S by the union of S — Im 6 and K). 2As
anyone familiar with the paradoxes of set theory (Introduction, Section 2) might suspect, the class of all algebraic extension fields of K need not be a set, and therefore, cannot be used in such an argument.
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Let S be the class of all fields E such that J? is a subset of S and E is an algebraic extension field of K. Such a field E is completely determined by the subset E of S and the binary operations of addition and multiplication in E. Now addition [resp. multiplication] is a function
Corollary 3.7. 7/K is a field and S a set of polynomials (ofpositive degree) in K[x], then there exists a splitting field of S over K. PROOF. Exercise. ■ We turn now to the question of the uniqueness of splitting fields and algebraic closures. The answer will be an immediate consequence of the following result on the extendibility of isomorphisms (see Theorem 1.8 and the remarks preceding it).
Theorem 3.8. Let c : K —» L be an isomorphism of fields, S = {f;} a set of poly nomials (of positive degree) in K[x], and S' = {erf;} the corresponding set of poly nomials in L[x], 7/F is a splitting field ofS over K and M is a splitting field of S' over L, then a is extendible to an isomorphism F = M. SKETCH OF PROOF. Suppose first that 5 consists of a single polynomial /e K[x] and proceed by induction on n = [F : K]. If n = 1, then F = K and /splits over K. This implies that
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reducible in L\x]. If v is a root of eg in M, then by Theorem 1.8 a extends to an iso morphism r : K(u) 9= L(v) with t(h) = v. Since [K(u) : K] = deg g > 1 (Theorem 1.6), we must have [Z7 : K(w)] < n (Theorem 1.2). Since F is a splitting field of /over K(u) and M is a splitting field of cf over L{v) (Exercise 2), the induction hypothesis implies that r extends to an isomorphism F = M. If 5 is arbitrary, let S consist of all triples (E,N,t), where E is an intermediate field of Fand K, N is an intermediate field of Mand L, and t :E—* N is an isomorphism that extends a. Define (£i,M,ti) < {E2,N2,ti) if Et CZ E2, Ni CZ N2 and r21 Ex = n. Verify that S is a nonempty partially ordered set in which every chain has an upper bound in 8. By Zorn’s Lemma there is a maximal element (F0,M0,t0) of S. We claim that F0 = F and M0 = M, so that t0 : F = M is the desired extension of a. If F0 ^ F, then some feS does not split over Fn. Since all the roots of/lie in F, F contains a splitting field Fi of /over F0. Similarly, M contains a splitting field Mx of r0/ = cf over M0. The first part of the proof shows that r0 can be extended to an isomorphism ri : Fi = Mi. But this means that (FuMun) e S and (F0,M0,t0) < (Fi,Mi,ti) which contradicts the maximality of (F0,M0,t0). A similar argument using r0“J works if 5^ M. ■
Corollary 3.9. Let K be a field and S a set ofpolynomials {ofpositive degree) in K[x], Then any two splitting fields of S over K are K-isomorphic. In particular, any two algebraic closures of K are K-isomorphic. SKETCH OF PROOF. Apply Theorem 3.8 with a = Ik. The last statement is then an immediate consequence of Theorem 3.4(ii). ■ In order to characterize Galois extensions in terms of splitting fields, we must first consider a phenomenon that occurs only in the case of fields of nonzero char acteristic. Recall that if K is any field,/is a nonzero polynomial in K[x], and c is a root of f then /= (x — c)mg(x) where g(c) ^ 0 and m is a uniquely determined positive integer. The element c is a simple or multiple root of /according as m = 1 or m > 1 (see p. 161).
Definition 3.10. Let K be a field and f s K[x] an irreducible polynomial. The poly nomial f is said to be separable if in some splitting field of f over K every root of f is a simple root. If F is an extension field of K and u e F is algebraic over K, then u is said to be separable over K provided its irreducible polynomial is separable. If every element of F is separable over K, then F is said to be a separable extension of K. REMARKS, (i) In view of Corollary 3.9 it is clear that a separable polynomial /e K[x] has no multiple roots in any splitting field of /over K. (ii) Theorem III.6.10 shows that an irreducible polynomial in is separable if and only if its derivative is nonzero, whence every irreducible polynomial is separable if char K = 0 (Exercise III.6.3). Hence every algebraic extension field of a field of characteristic 0 is separable. (iii) Separability is defined here only for irreducible polynomials, (iv) According to Definition 3.10 a separable extension field of K is necessarily algebraic over K. There
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is a definition of separability for possibly nonalgebraic extension fields that agrees with this one in the algebraic case (Section VI.2). Throughout this chapter, however, we shall use only Definition 3.10. EXAMPLES, x2 + 1 e Q[x] is separable since x2 + 1 = (x + i)(x — /) in C[x]. On the other hand, the polynomial x2 + 1 overZ2 has no simple roots; in fact it is not even irreducible since x2 + 1 = (x + l)2 in Z2[x].
Theorem 3.11. // F is an extension field o/K, then the following statements are equivalent. (i) F is algebraic and Galois over K; (ii) F is separable over K and F is a splitting field over K of a set S ofpolynomials in K[x]; (iii) F is a splitting field over K of a set T of separable polynomials in K[x], REMARKS. If F is finite dimensional over K, then statements (ii) and (iii) can be slightly sharpened. In particular (iii) may be replaced by: F is a splitting field over K of a polynomial /e K\x] whose irreducible factors are separable (Exercise 13). PROOF OF 3.11. (i) => (ii) and (iii). If u e F has irreducible polynomial f then the first part of the proof of Lemma 2.13 (withF = F) carries over verbatim and shows that /splits in F[x] into a product of distinct linear factors. Hence u is separ able over K. Let {i\ | / e /} be a basis of F over K and for each i e / let f e K[x] be the irreducible polynomial of Vi. The preceding remarks show that each f is separable and splits in F[x]. Therefore F is a splitting field over K of S = { f \ i e /}. (ii) => (iii) Let fieS and let ge A"[x] be a monic irreducible factor of /. Since / splits in Ffx], g must be the irreducible polynomial of some u e F. Since F is separable over K, g is necessarily separable. It follows that F is a splitting field over K of the set T of separable polynomials consisting of all monic irreducible factors (in AT[x]) of polynomials in S' (see Exercise 4). (iii) => (i) F is algebraic over K since any splitting field over K is an algebraic ex tension. If u e F — K, then u e K(vi, . . . , v„) with each Vi a root of some f e T by the definition of a splitting field and Theorem 1.3(vii). Thus ueE = K(uu . . . , ur) where the w, are all the roots of f,. . ., fn in F. Hence [E : K] is finite by Theorem 1.12. Since each fi splits in F,E is a splitting field over K of the finite set { fi,, or equivalently, of / = fifi - ■ ■ fi. Assume for now that the theorem is true in the finite dimensional case. Then E is Galois over K and hence there exists t e AutA-F such that t(w) 9^ u. Since F is a splitting field of T over F (Exercise 2), r extends to an automorphism cr e AutKF such that a(u) = t(u) ^ u by Theorem 3.8. Therefore, u (which was an arbitrary element of F — K) is not in the fixed field of AuUF; that is, F is Galois over K. The argument in the preceding paragraph shows that we need only prove the theorem when [F : K] is finite. In this case there exist a finite number of polynomials gu . . . , gt e T such that F is a splitting field of {gi, . . . , gt} over K (otherwise F would be infinite dimensional over K). Furthermore AutAF is a finite group by Lemma 2.8. If K0 is the fixed field of Aut^F, then F is a Galois extension of K0 with
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[F: /£0] = lAut/c/7! by Artin’s Theorem 2.15 and the Fundamental Theorem. Thus in order to show that F is Galois over K (that is, K = A'0) it suffices to show that [F :K] = \AutKF\. We proceed by induction on n = [F: K], with the case n — l being trivial. If n > 1, then one of the gi, say gu has degree s > 1 (otherwise all the roots of the gi lie in K and F = K). Let u e F be a root of gi', then [A(w) : K] = deg g\ — s by Theo rem 1.6 and the number of distinct roots of gi is s since gi is separable. The second paragraph of the proof of Lemma 2.8 (with L = K, M = K(u) and / = gi) shows that there is an injective map from the set of all left cosets of H = AutK(u)F in AutjtF to the set of all roots of g\ in F, given by aH (-> a(u). Therefore, [Aut*/7 '• H] < s. Now if v e F is any other root of gu there is an isomorphism r : K(u) K(v) with r(w) = v and r | K = Ik by Corollary 1.9. Since F is a splitting field of jgi,. . . , gt) over K(u) and over K(v) (Exercise 2), r extends to an automor phism a e AutxF with a(u) = v (Theorem 3.8). Therefore, every root of g\ is the im age of some coset of H and [Aut^f : H] = s. Furthermore, F is a splitting field over K(u) of the set of all irreducible factors hj (in AT(w)[x]) of the polynomials gi (Exer cise 4). Each hj is clearly separable since it divides some gi. Since [F : K(u)] = n/s < n, the induction hypothesis implies that [Z7: K(m)] = |Aut/c(U)F| = \H\. Therefore,
[F : K) = [F : A:(h)][K(h) : K] = \H\s = (//([Aut*/7 : H] = |Aut*F| and the proof is complete. ■
Theorem 3.12. (Generalized Fundamental Theorem) If F is an algebraic Galois ex tension field of K, then there is a one-to-one correspondence between the set of all intermediate fields of the extension and the set of all closed subgroups of the Galois group /4«/kF (given by Ef—> E' = Aut^F) such that: (ii') F is Galois over every intermediate field E, but E is Galois over K if and only if the corresponding subgroup E' is normal in G = AutkF; in this case G/E' is (iso morphic to) the Galois group /4wKE of E over K. REMARKS. Compare this Theorem, which is proved below, with Theorem 2.5. The analogue of (i) in the Fundamental Theorem is false in the infinite dimensional case (Exercise 16). If [F : AT] is infinite there are always subgroups of Aut/cF that are not closed. The proof of this fact depends on an observation of Krull[64]: when F is algebraic over K, it is possible to make AutkF into a compact topological group in such a way that a subgroup is topologically closed if and only if it is closed in the sense of Section 2 (that is, H = H"). It is not difficult to show that some infinite compact topological groups contain subgroups that are not topologically closed. A fuller discussion, with examples, is given in P. J. McCarthy [40; pp. 60-63]. Also see Exercise 5.11 below.
PROOF OF 3.12. In view of Theorem 2.7 we need only show that every inter mediate field E is closed in order to establish the one-to-one correspondence. By Theorem 3.11 F is the splitting field over A" of a set T of separable polynomials. Therefore, F is also a splitting field of T over E (Exercise 2). Hence by Theorem 3.11 again, F is Galois over E; that is, E is closed.
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CHAPTER V FIELDS AND GALOIS THEORY
(ii') Since every intermediate field E is algebraic over K, the first paragraph of the proof of Theorem 2.5(ii) carries over to show that E is Galois over K if and only if E' is normal in Aui kF. If E = E" is Galois over K, so that E' is normal in G = Aut*/7, then £ is a stable intermediate field by Lemma 2.11. Therefore, Lemma 2.14 implies that G/E' = AutA'F/AutfcT7 is isomorphic to the subgroup of AutA-£ consisting of those automorphisms that are extendible to F. But F is a splitting field over K (Theorem 3.11) and hence over E also (Exercise 2). Therefore, every K-automorphism in AutA'is extends to F by Theorem 3.8 and G/E' = Aut*E. u We return now to splitting fields and characterize them in terms of a property that has already been used on several occasions.
Definition 3.13. An algebraic extension field F o/K is normal over K (or a normal extension) if every irreducible polynomial in K[x] that has a root in F actually splits in F[x],
Theorem 3.14. If F is an algebraic extension field o/K, then the following statements are equivalent. (i) F is normal over K; (ii) F is a splitting field over K of some set of polynomials in K[x]; (iii) if K is any algebraic closure of K containing F, then for any K-monomorphism offields
APPENDIX: THE FUNDAMENTAL THEOREM OF ALGEBRA
265
Corollary 3.15. Let F be an algebraic extension field of K. Then F is Galois over K if and only if F is normal and separable over K. If char K = 0, then F is Galois over K if and only if F is normal over K. PROOF. Exercise; use Theorems 3.11 and 3.14. ■
Theorem 3.16. IfEis an algebraic extension field of K, then there exists an extension field F of E such that (i) F is normal over K; (ii) no proper subfield of F containing E is normal over K; (iii) //E is separable over K, then F is Galois over K; (iv) [F : K] is finite if and only if[ E : K] is finite. The field F is uniquely determined up to an E-isomorphism. The field F in Theorem 3.16 is sometimes called the normal closure of E over K. PROOF OF 3.16. (i) Let X = {«,-[/ e /1 be a basis of Eover K and let f e K[x] be the irreducible polynomial of If F is a splitting field of S = \ f | / e /| over E, then F is also a splitting field of 5 over K (Exercise 3), whence F is normal over K by Theorem 3.14. (iii) If E is separable over K, then each f is separable. Therefore F is Galois over K by Theorem 3.11. (iv) If [E : K] is finite, then so is Xand hence S. This implies that [F : K] is finite (by the Remarks after Definition 3.1). (ii) A subfield F0 of F that contains E necessarily contains the root iu of fzS for every /. If F0 is normal over K (so that each f splits in F0 by definition), then FCF 0 and hence F = F0. Finally let Fx be another extension field of E with properties (i) and (ii). Since Fi is normal over K and contains each w„ Fi must contain a splitting field F2 of 5 over K with E C Fi. F2 is normal over K (Theorem 3.14), whence F2 = Fi by (ii). Therefore both F and F\ are splitting fields of 5 over K and hence of S over E (Exercise 2). By Theorem 3.8 the identity map on E extends to an E-isomorphism F=Fi. ■
APPENDIX: THE FUNDAMENTAL THEOREM OF ALGEBRA The theorem referred to in the title states that the field C of complex numbers is algebraically closed (that is, every polynomial equation over C can be completely solved.) Every known proof of this fact depends at some point on results from analysis. We shall assume: (A) every positive real number has a real positive square root; (B) every polynomial in R[x] of odd degree has a root in R (that is, every irre ducible polynomial in R[a] of degree greater than one has even degree). Assumption (A) follows from the construction of the real numbers from the rationals and assumption (B) is a corollary of the Intermediate Value Theorem of elementary calculus; see Exercise III.6.16. We begin by proving a special case of a theorem that will be discussed below (Proposition 6.15).
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Lemma 3.17. If F is a finite dimensional separable extension of an infinite field K, then F = K(u) for some u e F. SKETCH OF PROOF. By Theorem 3.16 there is a finite dimensional Galois extension field Ft of K that contains F. The Fundamental Theorem 2.5 implies that AuU-F, is finite and that the extension of A by F\ has only finitely many intermediate fields. Therefore, there can be only a finite number of intermediate fields in the ex tension of K by F. Since [£ : K] is finite, we can choose u e F such that [A(t/) : A"] is maximal. If K(u) j* F, there exists re£ — K(u). Consider all intermediate fields of the form K(u + ac) with a e A. Since A is infinite and there are only finitely many intermediate fields, there exist a,b e A' such that a ^ b and A(w -f- av) = K(u -f- bv). Therefore (a — b)v = (u + av) — (w + be) e A'(« + av). Since a ^ b, we have v (a — b)~] (a — b)v e K(u -f- au), whence u — (u -f- ac) — av e K( u -(- ac). Conse quently A C K(u) CZ K(u -f- ac), whence \K(u + ac) : AJ > [K(u) : /CJ. This contra dicts the choice of u. Hence A'(w) = F. ■
Lemma 3.18. There are no extension fields of dimension 2 ocer the field of complex numbers. SKETCH OF PROOF. It is easy to see that any extension field F of dimension 2 over C would necessarily be of the form F = C(w) for any u z F — C. By Theorem 1.6 u would be the root of an irreducible monic polynomial /£ C[a] of degree 2. To complete the proof we need only show that no such /can exist. For each a -f- bi e C = R(/> the positive real numbers ](a \a2 -f- b2)'2 and j( — a -f- \a2 -h b2), 2| have real positive square roots c and d respectively by as sumption (A). Verify that with a proper choice of signs (±r ± di)2 = a -f bi. Hence every element of C has a square root in C. Consequently, if / = ,v2 4- sx: -f- / £ C[.v], then /has roots ( — 5 db ys2 — At), 2 in C, whence /splits over C. Thus there are no irreducible monic polynomials of degree 2 in C[jr], ■
Theorem 3.19. (The Fundamental Theorem of Algebra) The field of complex numbers is algebraically closed. PROOF. In order to show that every nonconstant /eC[*] splits over C, it suffices by Theorem 1.10 to prove that C has no finite dimensional extensions except itself. Since [C : R] = 2 and char R = 0 every finite dimensional extension field E] of C is a finite dimensional separable extension of R (Theorem 1.2). Consequently, £, is contained in a finite dimensional Galois extension field F of R by Theorem 3.16. We need only show that F = C in order to conclude E} = C. The Fundamental Theorem 2.5 shows that AutR/ 7 is a finite group. By Theorems II.5.7 and 2.5 AutnF has a Sylow 2-subgroup H of order 2" (n > 0) and odd index, whose fixed field E has odd dimension, [E : R] = [Aut«£ : H). E is separable over R (since char R = 0), whence E = R(u) by Lemma 3.17. Thus the irreducible poly nomial of u has odd degree l£ : RJ = [R(w) : R] (Theorem 1.6). This degree must be 1
APPENDIX: THE FUNDAMENTAL THEOREM OF ALGEBRA
267
by assumption (B). Therefore, heR and [Aut RF ://] = [E : R] = 1, whence AutfiF = H and |AutnF| = 2n. Consequently, the subgroup AutcF of Aut R F has order 2m for some m (0 < m < n). Suppose m > 0. Then by the First Sylow Theorem II.5.7 Autc;/7has a subgroup/ of index 2; let E» be the fixed field of J. By the Fundamental Theorem F0 is an ex tension of C with dimension [AutcF:./] = 2, which contradicts Lemma 3.18. Therefore, m = 0 and Aut*;/7 = 1. The Fundamental Theorem 2.5 implies that [F : C] = [Aut,:F : IJ = |Aut,^F| = 1, whence F = C. ■
Corollary 3.20. Every proper algebraic extension field of the field of real numbers is isomorphic to the field of complex numbers. PROOF. If F is an algebraic extension of R and u e F — R has irreducible poly nomial /= of degree greater than one, then /splits over C by Theorem 3.19. If v e C is a root of fi then by Corollary 1.9 the identity map on R extends to an isomor phism R(u) = R(r)
EXERCISES Note: Unless stated otherwise F is always an extension field of the field K and S is a set of polynomials (of positive degree) in K'[a']. 1. Fis a splitting field over K of a finite set \ fu . . . , f i , } of polynomials in K[x] if and only if F is a splitting field over K of the single polynomial / = /,/2- • - f i .
2 . If F is a splitting field of 5 over K and F is an intermediate field, then F is a splitting field of 5 over F. 3. (a) Let F be an intermediate field of the extension K C Z F and assume that E = K(uu ■ ■ ■ , ur) where the ut are (some of the) roots of fe K[x], Then F is a splitting field of /over K if and only if F is a splitting field of /over E. (b) Extend part (a) to splitting fields of arbitrary sets of polynomials. 4. If F is a splitting field over K of S, then F is also a splitting field over K of the set T of all irreducible factors of polynomials in S. 5. If/£ K[x] has degree n and Fisa splitting field of /over K, then [F : X] divides n\.
6. Let K be a field such that for every extension field F the maximal algebraic ex tension of K contained in F(see Theorem 1.14) is K itself. Then Kis algebraically closed. 7. If F is algebraically closed and E consists of all elements in F that are algebraic over K, then E is an algebraic closure of K [see Theorem 1.14], 8. No finite field K is algebraically closed. [Hint: If K = {a0, . . . , an} consider Ci + (* — tfo)(x — Oi)- ■ C* — a n ) e K[x], where a x ^ 0.]
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CHAPTER V FIELDS AND GALOIS THEORY
9. F is an algebraic closure of K if and only if F is algebraic over K and for every algebraic extension E of K there exists a K-monomorphism ZT—» F. 10. F is an algebraic closure of K if and only if F is algebraic over K and for every algebraic field extension E of another field K, and isomorphism of fields a : K, —> K, a extends to a monomorphism E—>F.
11. (a) If ui, . . . , e F are separable over. K, then K(uu . . . , « „ ) is a separable ex tension of K. (b) If F is generated by a (possibly infinite) set of separable elements over K, then F is a separable extension of K. 12. Let E be an intermediate field. (a) If u e F is separable over K, then u is separable over E. (b) If F is separable over K, then F is separable over E and E is separable over K. 13. Suppose [F : K] is finite. Then the following conditions are equivalent: (i) F is Galois over K; (ii) F is separable over K and a splitting field of a polynomial f e AT*]; (iii) F is a splitting field over K of a polynomial / e K[.v] whose irreducible factors are separable. 14. (Lagrange’s Theorem on Natural Irrationalities). If L and M are intermediate fields such that L is a finite dimensional Galois extension of K, then LM is finite dimensional and Galois over M and AutmLM = AutLn mL. 15. Let E be an intermediate field. (a) If F is algebraic Galois over K, then F is algebraic Galois over E. [Exercises 2.9 and 2.11 show that the “algebraic” hypothesis is necessary.] (b) If F is Galois over E, E is Galois over K and F is a splitting field over £ of a family of polynomials in K[x\, then F is Galois over K [see Exercise 2.12]. 16. Let F be an algebraic closure of the field Q of rational numbers and let E Cl F be a splitting field over Q of the set S = {x2 + a \ a e Q} so that E is algebraic and Galois over Q (Theorem 3.11). (a) E = Q(X) whereX — {"\]p\ p = — 1 or p is a prime integer |. (b) If creAutQ£, then a- = lt-. Therefore, the group AutQ£ is actually a vector space overZ2 [see Exercises 1.1.13 and IV. 1.1]. (c) AiUq£ is infinite and not denumerable. [Hint: for each subsetY of Adhere exists a e AutQ^ such that o(^p) = —^p for sjpzY and a(y]p) — yjp for -yjp eX - Y. Therefore, |AutQ£| = |/>(A’)[ > [A'l by Introduction, Theorem 8.5. But |*| = No.] (d) If B is a basis of Aut(j£ over Z2, then B is infinite and not denumerable. (e) AutQ£ has an infinite nondenumerable number of subgroups of index 2. [Hint: If b e B, then B — \b] generates a subgroup of index 2.] (0 The set of extension fields of Q contained in E of dimension 2 over Q is denumerable. (g) The set of closed subgroups of index 2 in Aut<>E is denumerable. (h) [E : Q] < N0, whence [E : Q] < |Aut<j£|.
4. THE GALOIS GROUP OF A POLYNOMIAL
269
17. If an intermediate field E is normal over K, then E is stable (relative to F and K). 18. Let F be normal over K and E an intermediate field. Then E is normal over K if and only if E is stable [see Exercise 17], Furthermore Aui^F/E' = AutKE. 19. Part (ii) or (ii)' of the Fundamental Theorem (2.5 or 3.12) is equivalent to: an intermediate field E is normal over K if and only if the corresponding subgroup E' is normal in G = AuthF in which case G/E' = AutkE. [See Exercise 18.] 20. If F is normal over an intermediate field E and E is normal over K, then F need not be normal over K. [Hint: Let \ 2 be a real fourth root of 2 and consider Q(^2) 3 Q(V2) 3 Q ; use Exercise 23.] Compare Exercise 2. 21. Let Fbe algebraic over K. F is normal over K if and only if for every /C-monomorphism of fields a : F —> N, where N is any normal extension of K containing F, cr(F) = F so that a is a /("-automorphism of F. [Hint: Adapt the proof of Theo rem 3.14, using Theorem 3.16.] 22. If F is algebraic over K and every element of F belongs to an intermediate field that is normal over K, then F is normal over K. 23. If [F : K] = 2, then F is normal over K. 24. An algebraic extension F of K is normal over K if and only if for every irre ducible fz A'[.v], /factors in F[.v] as a product of irreducible factors all of which have the same degree. 25. Let Fbe a splitting field of/e A[.v], Without using Theorem 3.14 show that Fis normal over K. [Hints: if an irreducible g £ K[x\ has a root «eF, but does not split in F, then show that there is a /(-isomorphism
4. THE GALOIS GROUP OF A POLYNOMIAL The primary purpose of this section is to provide some applications and examples of the concepts introduced in the preceding sections. With two exceptions this ma terial is not needed in the sequel. Definition 4.1 and Theorem 4.12, which depends only on Theorem 4.2, are used in Section 9, where we shall consider the solvability by radicals of a polynomial equation.
Definition 4.1. Let K be a field. The Galois group of a polynomial f e K[x] is the group Aut|
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270
Theorem 4.2. Let K be a field and f e K[x] a polynomial with Galois group G. (i) G is isomorphic to a subgroup of some symmetric group S„. (ii) Iff is (irreducible) separable of degree n, then n divides |G| and G is isomor phic to a transitive subgroup of S„; SKETCH OF PROOF, (i) If un are the distinct roots of / in some splitting field F(1 < n < deg /), then Theorem 2.2 implies that every a e Aut^F in duces a unique permutation of \u\,. . ., un] (but not necessarily vice versa!). Consider Sn as the group of all permutations of {u\, . . . , «„) and verify that the assignment of a e AutA-F to the permutation it induces defines a monomorphism AutKF —>Sn. (Note that F = K(uu. . . , «„).) (ii) Fis Galois over K(Theorem 3.11) and [K(uy) : K] = n = deg/(Theorem 1.6). Therefore, G has a subgroup of index n by the Fundamental Theorem 2.5, whence n | | G\. For any i ^ j there is a ^-isomorphism or : K(ui) = K(u3) such that a(ut) = ut (Corollary 1.9). a extends to a K-automorphism of F by'Theorem 3.8, whence G is isomorphic to a transitive subgroup of Sv. ■ Hereafter the Galois group of polynomial /will frequently be identified with the isomorphic subgroup of Sn and considered as a group of permutations of the roots of / Furthermore we shall deal primarily with polynomials fzK[x] all of whose roots are distinct in some splitting field. This implies that the irreducible factors of /are separable. Consequently by Theorem 3.11 (and Exercise 3.13) the splitting field Fof /is Galois over K. If the Galois groups of such polynomials can always be calculated, then it is possible (in principle at least) to calculate the Galois group of an arbitrary polynomial (Exercise 1).
Corollary 4.3. Let K be a field and f e K[x] an irreducible polynomial of degree 2 with Galois group G. If f is separable (as is always the case when char K 5^ 2), then G = Z;‘, otherwise G = 1. SKETCH OF PROOF. Note that S2 = Z2. Use Remark (ii) after Definition 3.10 and Theorem 4.2. ■ Theorem 4.2 (ii) immediately yields the fact that the Galois group of a separable polynomial of degree 3 is either S3 or As (the only transitive subgroups of S3). In order to get a somewhat sharper result, we introduce a more general consideration.
Definition 4.4. Let K be a field with char K ^ 2 and f e K[x] a polynomial of degree n with n distinct roots Ui, . . . , u„ in some splitting field F of f over K. Let A = JJ (Uj — uj) = (ui — LiaXui — U 3) • - ■ (u„_i — u„) 8 F; the discriminant of f is * <3
the element D = A2.
Note that A is an element of a specific splitting field F and therefore, a priori, D = A2 is also in F. However, we have
4. THE GALOIS GROUP OF A POLYNOMIAL
271
Proposition 4.5. Let K, f, F and A be as in Definition 4.4. (l) The discriminant A2 off actually lies in K. (ii) For each a e AutvF < S„, u is an even [resp. odd] permutation if and only if
Corollary 4.6. Let K, f, F, A be as in Definition 4.4 (so that F is Galois over K) and consider G = Autn¥ as a subgroup of S„. In the Galois correspondence (Theorem 2.5) the subfield K(A) corresponds to the subgroup G fl A„. In particular, G consists of even permutations if and only if A e K . PROOF. Exercise. ■
Corollary 4.7. Let K be a field and f e K[x] an (irreducible) separable polynomial of degree 3. The Galois group off is either S3 or A3. If char K ^ 2, it is A3 if and only if the discriminant off is the square of an element of K. PROOF. Exercise; use Theorem 4.2 and Corollary 4.6. ■ If the base held K is a subfield of the field of real numbers, then the discriminant of a cubic polynomial /e K[x] can be used to find out how many real roots /has (Exercise 2). Let / be as in Corollary 4.7. If the Galois group of /is A3 = Z3 there are, of course, no intermediate fields. If it is S3, then there are four proper intermediate fields, K(A), K(u}), K(u2), and K(u3) where uuuo,u3 are the roots of f. K(A) corresponds to A-i and K(ut) corresponds to the subgroup j (1),(_/A)| (/ ^ j,k) of S.t, which has order 2 and index 3 (Exercise 3). Except in the case of characteristic 2, then, computing the Galois group of a separable cubic reduces to computing the discriminant and determining whether or not it is a square in K. The following result is sometimes helpful.
Proposition 4.8. Let K be a field with char K ^ 2,3. If f(x) — x3 -j- bx2 cx + d s K[x] has three distinct roots in some splitting field, then the polynomial g(x) = f(x — b 3) e K[x] has the form x3 px + q and the discriminant of f is — 4p3 - 27q2. SKETCH OF PROOF. Let F be a splitting field of /over K and verify that u e F is a root of /if and only if u 4 b/7> is a root of g = f(x — b/3). This implies that g has the same discriminant as/ Verify that g has the form „v3 + px + q (p,q s K). Let fi,r2,r3 be the roots of g in F. Then (,v — n)(.Y — v,)(x — r3) = g(x) = a3 + px + q which implies
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272
^1 ~l~ Vi -(- t>3 — 0; t\ih + viv3 + v2v3 = p; — V 1V 2V 3 — q.
Since each i>, is a root of g Vi3 = — pVi — q (z = 1,2,3).
The fact that the discriminant A2 of g is —4ps — 21q'z now follows from a gruesome computation involving the definition A2 = (vi — i\)2(v 1 — v3)2(v2 — vs)*, the equa tions above and the fact that (i\ — v,)2 = (v{ + v,)2 — 4ViV,. ■ EXAMPLE. The polynomial x3 — 3x + 1 sQM is irreducible by Theorem III.6.6 and Proposition [II.6.8 and separable since char Q = 0. The discriminant is —4( — 3)3 — 27(1)2 = 108 — 27 = 81 which is a square in Q. Hence the Galois group is Az by Corollary 4.7. EXAMPLE. If f(x) = xs + 3x2 - x - 1 s Q[x], then gix) = f(x - 3/3) = f(x - 1) = x3 - 4x + 2,
which is irreducible by Eisenstein’s Criterion (Theorem III.6.15). By Proposition 4.8 the discriminant of / is —4(—4)3 — 27(2)2 = 256 — 108 = 148, which is not a square in Q. Therefore the Galois group is S3. We turn now to polynomials of degree four (quartics) over a field K. As above, we shall deal only with those /e%] that have distinct roots Mi,m2,«s,«4 in some splitting field F. Consequently, F is Galois over K and the Galois group of /may be considered as a group of permutations of | uuu2,u3,ih\ and a subgroup of.SV The sub set V = {(1),(12)(34),( 13)(24),( 14)(23)} is a normal subgroup of Si (Exercise 1.6.7), which will play an important role in the discussion. Note that V is isomorphic to the four groupZ20Z2 and V fl G is a normal subgroup of G = AutKF < 54.
Lemma 4.9.
Let
K,
f, F, u;, V, and G
=
AntkF
<
S4 be as in the preceding para
graph. If a = UiU> + U3U4, /3 = U1U3 + U2U4, 7 = U]U4 + U2U3 e F, then under the Galois correspondence (Theorem 2.5) the subfield K(a,/3,7) corresponds to the normal subgroup V H G. Hence K(a,fi,y) is Galois ocer K and Aut\iK{a4'i,y) = G/(G fl V). SKETCH
Kia,i3,7).
In
OF
PROOF.
order
to
Clearly
complete
every the
element
proof
it
in
C
fl
suffices,
in
F
fixes
view
of
a,(3,y the
and
hence
Fundamental
Theorem, to show that every element of G not in V moves at least one of ct,(8,7. For instance
if
a
=
(12)
e
G
and
=
0,
then
«2«3
+
=
W1W3
+
and
hence
Uiiu3 — w4) = Wi(m3 — u4). Consequently, u\ = u> or u3 = iu, either of which is a contradiction.
Therefore
a((3)
v*
(3.
The
other
possibilities
are
handled
similarly.
[Hint: Rather than check all 20 possibilities show that it suffices to consider only one representative from each coset of V in S4]. ■ Let K, f F, Ui and a,(3,y be as in Lemma 4.9. The elements a,fi,y play a crucial role
in
determining
the
(x — a)(j: — (3)ix — 7) e
Galois
groups
of
arbitrary
quartics.
The
polynomial
K(a,l3,y)[x] is called the resolvant cubic of /. The resolvant
cubic is actually a polynomial over K:
4. THE GALOIS GROUP OF A POLYNOMIAL
273
Lemma 4.10. If K is afield and f = x4 4- bx3 -j- cx2 + dx + e e K[x], then the resolcant cubic of f is the polynomial x3 — cx2 -f- (bd — 4e)x — b2e + 4ce — d2 e K[x]. SKETCH OF PROOF. Let /have roots uu . . . , «4 in some splitting field F. Then use the fact that / = (x — wi)(x — u2)(x — us)(x — u4) to express b,c,d,e in terms of the wt . Expand the resolvant cubic (x — a)(x — fi)(x — y) and make appro priate substitutions, using the definition of a,fi,y (Lemma 4.9) and the expressions for b,c,d,e obtained above. ■
We are now in a position to compute the Galois group of any (irreducible) separable quartic /s K[x]. Since its Galois group G is a transitive subgroup of S4 whose order is divisible by 4 (Theorem 4.2), G must have order 24,12, 8 or 4. Verify that the only transitive subgroups of orders 24,12, ahd 4 are S4, A4, V (=Z2 ©Z2) and the various cyclic subgroups of order 4 generated by 4-cycles; see Exer cise 1.4.5 and Theorem 1.6.8. One transitive subgroup of S4 of order 8 is the dihedral group Z)4 generated by (1234) and (24) (page 50). Since D4 is not normal in S4, and since every subgroup of order 8 is a Sylow 2-subgroup, it follows from the second and third Sylow Theorems that S4 has precisely three subgroups of order 8, each isomorphic to A
Proposition 4.11. Let K be a field and fsK[x] an (irreducible) separable quartic with Galois group G (considered as a subgroup of S4). Let a,fi,y be the roots of the resolvant cubic of f and let m = [K(q:,/3,7) : K], Then: (i) m = 6 <=> G = S4; (ii) m = 3 <=> G = A4; (iii) m = 1 <=> G = V; (iv) m = 2 <=> G = D4 or G = Z4; in this case G = D4 iff is irreducible over K(a,/3,y) and G = Z4 otherwise. SKETCH OF PROOF. Since K(a,(3,y) is a splitting field over K of a cubic, the only possibilities for rn are 1,2,3, and 6. In view of this and the discussion preceding the theorem, it suffices to prove only the implications <= in each case. We use the fact that m = \K(a,fi,y') : K] — \G/G fl V\ by Lemma 4.9. If G = then G D V = V and m = \G/V\ = \G\/\V\ = 3. Similarly, if G = S4, then m — 6. If G = V, then G fl V = G and m - |G/G) = 1. If G ~ D4, then G fl V = Vsince Vis contained in every Sylow 2-subgroup of S4 and m = \G/V\ ~ \G\/\V\ = 2. If G is cyclic of order 4, then G is generated by a 4-cycle whose square must be in V so that | G fl V\ = 2andw = |G/G D V\ = |G]/|G fl V\ = 2. Since /is either irreducible or reducible and D4 ^Z4, it suffices to prove the con verse of the last statement. Let mi,w2,"3,m4 be the roots of / in some splitting field F and suppose G = D4, so that G D V = V. Since V is a transitive subgroup and G fl V = AutK(„.e.7)F (Lemma 4.9), there exists for each pair i j (1 < U < 4) a ae G fl V which induces an isomorphism K(a,/3,y)(ui) = K(a,(3,y)(uj) such that a(u,) = Uj and a \ K(a,i3,y) is the identity. Consequently for each i ^ j, Ui and w, are roots of the same irreducible polynomial over K(a,(3,y) by Corollary 1.9. It follows that /is irreducible over K(a,(3,y). On the other hand if G =Z4, then G f) V =
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274
AutK(a.p.y)F has order 2 and is not transitive. Hence for some / ^ j there is no a s G fl V such that But since F is a splitting field over K(a,@ ,y)(Ui) and K(a,(3,y)(uj), if there were an isomorphism K(a,fi,y)(ut) ~ K(a,^,y)(iij), which was the identity on K(a,(3,y) and sent to u„ it would be the restriction of some a s AutKia,p.y>F = G fl V by Theorem 3.8. Therefore, no such isomorphism exists, whence and u, cannot be roots of the same irreducible polynomial over K(a,fi,y) by Corollary 1.9. Consequently, /must be reducible over K(a,(3,y). ■
EXAMPLE. The polynomial / = a4 + 4a2 + 2 e Q[a] is irreducible by Eisen stein’s Criterion (Theorem III.6.15); /is separable since char Q = 0. Using Lemma 4.10 the resolvant cubic is found to be x3 — 4x2 — 8a + 32 = (a — 4)(x2 — 8) so that a = 4, (3 = yJS, y = -^8 and Q(a,(3,y) = Q(\/8) = Q(2yj2) = Q(^2) is of dimension 2 over Q. Hence the Galois group is (isomorphic to) Z)4 or Z4. A substitu tion z = v2 reduces/to z2 + 4z -|- 2 whose roots are easily seen to be z = — 2 ± y2; thus the roots of /are a — ±-\Jz =
/= ( x
~
2 ± \/2- Hence
V-2 + V2)(a + + \/2)Cx - V-2 - V^)(^ + V-2 - V^) = (a2 - (-2 + V2) O - (-2 - V2>) e Q(V2)[a].
Therefore, / is reducible over Q(-\j2) and hence the Galois group is cyclic of order 4 by Proposition 4.11 (iv). EXAMPLE. To find the Galois group of / = a4 — 10a2 -f- 4 e Q[a] we first verify that/is irreducible (and hence separable as well). Now/has no roots in Q, and thus no linear or cubic factors, by Theorem IJI.6.6 and Proposition III.6.8. To check for quadratic factors it suffices by Lemma III.6.13 to show that /has no quadratic factors in Z[a]. It is easy to verify that there are no integers a,b,c,d such that / = (a2 -f- ax + b)(a2 + cx -f d). Thus /is irreducible in Q[a], The resolvant cubic of /is a3 + 10a2 — 16a — 160 — (a + 10)(a + 4)(a — 4), all of whose roots are in Q. Therefore, tn = [Q(a,P,y) : Q] = 1 and the Galois group of /is V (==Z2@Z>) by Proposition 4.11. EXAMPLE. The polynomial a4 — 2 e Q[a] is irreducible (and separable) by Eisenstein’s Criterion. The resolvant cubic is a3 + 8a = a(x + 2^j2i)(x — 2^2i) and Q(a,(3,y) = Q(yj2i) has dimension 2 over Q. Verify that a4 — 2 is irre ducible over Q(\/2/) (since yj2, V 2 $ Q(V20)- Therefore the Galois group is isomorphic to the dihedral group D4 by Proposition 4.11. EXAMPLE. Consider / = a4 — 5a2 -f- 6 e Q[a]. Observe that /is reducible over Q, namely / = (a2 — 2)(a2 — 3). Thus Proposition 4.11 is not applicable here. Clearly F = Q(\j2,^]3) is a splitting field of/over Q and since a2 — 3 is irreducible over Q(V2), \F : Q] = [F: Q(yj2)] [Q(^2) : QJ = 2-2 = 4. Therefore Aut0F, the Galois group of f has order 4 by the Fundamental Theorem. It follows from the proof of Theorem 4.2 and Corollary 4.3 that AutoQ(^2) consists of two elements: the identity map 1 and a map a with cr(^2) = — \]2. By Corollary 1.9,1 and a each extend to a Q-automorphism of F in two different ways (depending on whether ^3 f—> yj3 or yj3 f—» — ^3). This gives four distinct elements of Aut<jF (determined by the four possible combinations: -yj2\—> ±-\]2 and -\/3t—> ±-\J3). Since |Aut<jF| = 4 and each of these automorphisms has order 2 the Galois group of / must be isomor phic to the four group Z2 ©Z2 by Exercise 1.4.5.
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275
Determining the intermediate fields and corresponding subgroups of the Galois group of a separable quartic is more complicated than doing the same for a separable cubic. Among other things one may have K(in) = K(u3) even though ^ (see the last example above). There is no easily stated proposition to cover the quartic case and each situation must be attacked on an ad hoc basis. EXAMPLE. Let F C C be a splitting field over Q of / = x4 — 2 e Q[x]. If u is the positive real fourth root of 2, then the roots of /are u, — u, ui, —ui. In order to consider the Galois group G = Autq F of /as a subgroup of Si, we must choose an ordering of the roots, say ux = u, u2 = —u,u3 = ui, w4 = —ui. We know from the third example after Proposition 4.11 that G is one of the three subgroups of order 8 in Si, each of which is isomorphic to the dihedral group A. Observe that complex conjugation is an R-automorphism of C which clearly sends u\—>u, —u\-^—u, ui |—» —ui and —ui |—> ui. Thus it induces a Q-automorphism r of F = Q(u,ui). As an element of S4, t = (34). Now every subgroup of order 8 in S4 is conjugate to Z)4 (Second Sylow Theorem) and an easy calculation shows that the only one containing (34) is the subgroup D generated by a = (1324) and t = (34). It is easy to see that F = Q(u,ui) = Q(",0, so that every Q-automorphism of F is completely determined by its action on u and /. Thus the elements of D may be described either in terms of o and t or by their action on u and This information is summarized in the table: (1)
(34) T
u v—> u i l-> i
U
—i
(1324) (12X34) (1423) a cr2 as ui i
—u i
— ui i
(13X24) OT
ui —i
(12) c 2t
—u —i
(14)(23) cr3r — ui —i
It is left to the reader to verify that the subgroup lattice of D and the lattice of intermediate fields are as given below, with fields and subgroups in the same relative position corresponding to one another in the Galois correspondence. Subgroup lattice (H —» K means H < K):
Intermediate field lattice (M —* N means M d N):
CHAPTER V FIELDS AND GALOIS THEORY
276
F = Q («,/)
Qf«) Q(w) Q(«V) Q((l + /» Q((l - /» X
Q(«2) 0(0 Q(u2i)
Q Specific techniques for computing Galois groups of polynomials of degree greater than 4 over arbitrary fields are rather scarce. We shall be content with a very special case.
Theorem 4.12. If p is prime and f is an irreducible polynomial of degree p over the field of rational numbers which has precisely two nonreal roots in the field of complex numbers, then the Galois group off is (isomorphic to) Sp.
SKETCH OF PROOF. Let G be the Galois group of / considered as a sub group of Sp. Since p | |G| (Theorem 4.2), G contains an element a of order p by Cauchy’s Theorem II.5.2. a is a /7-cycle by Corollary 1.6.4. Now complex conjuga tion (a + bi (—* a — bi) is an R-automorphism of C that moves every nonreal ele ment. Therefore, by Theorem 2.2 it interchanges the two nonreal roots of /and fixes all the others. This implies that G contains a transposition r = (ab). Since a can be written a = (aj2■ - jP), some power of a is of the form ak — (abir ■ -iP) e G. By changing notation, if necessary, we may assume r = (12) and ak = (123- • -p). But these two elements generate Sv by Exercise 1.6.4. Therefore G = Sv. ■ EXAMPLE. An inspection of the graph of / = jr5 — 4x + 2 e Q[x] shows that it has only three real roots. The polynomial / is irreducible by Eisenstein’s Criterion (Theorem I I I . 6 . 15) and its Galois group is S 6 by Theorem 4.12. It is still an open question as to whether or not there exists for every finite group G a Galois extension field of Q with Galois group G. If G — S„, however, the answer is affirmative (Exercise 14).
EXERCISES Note: Unless stated otherwise K is a field, fe K[x] and F is a splitting field of j over K.
1. Suppose /e K[x] splits in F as/= (x — Wi)"1- ■ (x — uk)nk («, distinct; m > 1). Let va, . . . , vk be the coefficients of the polynomial g = (x — u,)(x — w2) . . . (x — uk) and let E = K(v0, . . . vk). Then
4. THE GALOIS GROUP OF A POLYNOMIAL
277
(a) F is a splitting field of g over E. (b) F is Galois over E. (c) Aut/;/7 = AutkF. 2. Suppose K is a subfield of R (so that F may be taken to be a subfield of C) and that /is irreducible of degree 3. Let D be the discriminant of / Then (a) D > 0 if and only if /has three real roots. (b) D < 0 if and only if /has precisely one real root.
3. Let /be a separable cubic with Galois group S3 and roots uuu2,u3 e F. Then the distinct intermediate fields of the extension of K by F are F, K(A), K(ih), K(u2), K(u3), K. The corresponding subgroups of the Galois group are l,/f3, Tu T2, T3 and S3 where Ti = {(1 [ j i ^ k). 4. If char K ^ 2,3 then the discriminant of a3 + bx2 + cx + cl is —4c3 — 27d2 + - 4 bd) + 18 bed.
5. If char K 9^ 2 and /e K[a] is a cubic whose discriminant is a square in K, then / is either irreducible or factors completely in K. 6. Over any base field K,
a3
— 3* + 1 is either irreducible or splits over K.
7. S4 has no transitive subgroup of order 6. 8. Let j be an (irreducible) separable quartic over K and u a root of / There is no field properly between K and K(u) if and only if the Galois group of/is either A or 9. Let .v4 + ax2 + b e A[x] (with char K 2) be irreducible with Galois group G. (a) If b is a square in K, then G = V. (b) If b is not a square in K and b(a2 — 4b) is a square in K, then G =Z4. (c) If neither b nor bid2 — 4b) is a square in K, then G ~ A10. Determine the Galois groups of the following polynomials over the fields indicated: (a) xi — 5 over Q; over Q(\5); over QiyjSi). (b) (a-3 - 2)(x2 - 3)(a2 - 5)(x2 - 7) over Q. (c) a-3 — a — 1 over Q; over Q(\^23/'). (d) a3 — 10 over Q; over Qiyj2). (e) a4 + 3a3 + 3a — 2 over Q. (f) a5 — 6a + 3 over Q. (g) a3 — 2 over Q. (h) (a3 — 2)(a2 — 5) over Q. (i) a4 — 4a2 + 5 over Q. (j) x4 + 2a2 + a + 3 over Q. 11. Determine all the subgroups of the Galois group and all of the intermediate fields of the splitting field (overQ) of the polynomial (a3 — 2)(a2 — 3)eQ[.*]. 12. Let K be a subfield of the real numbers and /e K[x] an irreducible quartic. If / has exactly two real roots, the Galois group of /is or Dt. 13. Assume that fix) s K[x] has distinct roots «i,w2, . . . , u„ in the splitting field F and let G = AutKF < Sn be the Galois group of / Let >1, be indeter minates and define:
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CHAPTER V FIELDS AND GALOIS THEORY
g(X)
=
XI ix ~ (UcWyi +■ «
ozSn
(a) Show that g(x) — XT (X ~ (M1>V(1) GzSn
+■ W2>V(2) + ' ' ' + WtiJVt.)))-
(b) Show that g(x) e K\yx, . . . , yn,x]. (c) Suppose g(x) factors as g\(x)g2(x) • • gr(x) with gt(x) e K(yx, . . . , J monic irreducible. If * — 53 w"<‘>■>'* *s a factor of #i(x), then show that i
gxix)
= IT ( x - x reG
/
Show that this implies that deg gt(x) = |CL (d) If K = Q, /e Z[x] is monic, and p is a prime, let feZp[x] be the poly nomial obtained from / by reducing the coefficients of /(mod p). Assume /has distinct roots in some splitting field F over Zp. Show that g(x)
= IT ( x - Y a , y ? u ) ) £ F [ x y i , reSn
i
■ ■ ■, ynl
If the m, are suitably ordered, then prove that the Galois group G of / is a sub group of the Galois group G of / (e) Show that x6 + 22x& — 9x4 + 12x3 — 37x2 — 29x — 15 e Q[x] has Galois group S6. [Hint: apply (d) with p = 2, 3, 5.] (0 The Galois group of x5 — x — 1 e Q[x] is S5. 14. Here is a method for constructing a polynomial /e Q[x] with Galois group for a given n > 3. It depends on the fact that there exist irreducible polynomials of every degree inZp[x] (p prime; Corollary 5.9 below). First choose/,/,/ e Z[x] such that: (i) deg f\ = n and / eZ2[x] is irreducible (notation as in 13(d)). (ii) deg fi — n and / sZ3[x] factors in Z3[x] as gh with g an irreducible of degree n — 1 and h linear; (iii) deg /3 = n and /3eZ5[x] factors as gh or ghih2 with# an irreducible quadratic in Z5[x] and h,hi,h-, irreducible polynomials of odd degree in Z5[x], (a) Let /= -15/ + 10/ + 6/3. Then f=f (mod 2), f=f2 (mod 3), and f=h (mod 5). (b) The Galois group G of /is transitive (since/is irreducible inZ2[x]). (c) G contains a cycle of the type f = (ixi, ■ ■ ■ in^x) and element
5. FINITE FIELDS In this section finite fields (sometimes called Galois fields) are characterized in terms of splitting fields and their structure completely determined. The Galois group of an extension of a finite field by a finite field is shown to be cyclic and its generator is given explicitly.
5. FINITE FIELDS
279
We begin with two theorems and a lemma that apply to fields which need not be finite. In each case, of course, we are interested primarily in the implications for finite fields.
Theorem 5.1. Let ¥ be a field and let P be the intersection of all subfields of F. Then P is a field with no proper subfields. If char F = p (prime), then P = Zp. If char F = 0, then P = Q, the field of rational numbers. The field P is called the prime subfield of F. SKETCH OF PROOF OF 5.1. Note that every subfield of F must contain 0 and 1F. It follows readily that P is a field that has no proper subfields. Clearly P con tains all elements of the form m\y (m e Z). To complete the proof one may either show directly that P = {m\f | m e Zj if char F = p and P = {(ml F)(n I*-)-1 | m,n e Z, n 9^ 0} if char F = 0 or one may argue as follows. By Theorem III. 1.9 the map (p : Z — > P given by m h m l f is a ring homomorphism with kernel («), where n = char F a n d n = 0 or n is prime. If n = p (prime), thenZ„ = ZKp) = Z/Ker
Corollary 5.2. If¥is a finite field, then char F = p 7^ 0 for some prime p and | Ff = pn for some integer n > 1. PROOF. Theorem III.1.9 and Theorem 5.1 imply that F has prime character istic p 9^ 0. Since F is a finite dimensional vector space over its prime subfield Z,,, F = ZP(+)- ■ - ©Zp (n summands) by Theorem IV.2.4 and hence |F| = pn. ■ In the sequel the prime subfield of a field F of characteristic p will always be identified with Zv under the isomorphism of Theorem 5.1. For example, we shall write Zp CI F; in particular, I*- coincides with 1 eZp.
Theorem 5.3. If ¥ is a field and G is a finite subgroup of the multiplicative group of nonzero elements of¥, then G is a cyclic group. In particular, the multiplicative group of all nonzero elements of a finite field is cyclic.
PROOF. If G (^ 1) is a finite abelian group, G = Z,„, ©Zm2 ©• ■ -@Zmk where W] > 1 and mx I m21 • • • | mk by Theorem II.2.1. Since mk(^ Zmi) — 0, it follows that every h e G is a root of the polynomial xmk — 17 e F[x] (G is a multiplicative group). Since this polynomial has at most mk distinct roots in F (Theorem III.6.7), we must have k = 1 and G = Zmk. ■
Corollary 5.4. lf¥ is a finite field, then F is a simple extension of its prime subfield Zp; that is, F = Zp(u) for some u e F.
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CHAPTER V FIELDS AND GALOIS THEORY
SKETCH OF PROOF. Let u be a generator of the multiplicative group of nonzero elements of F. ■
Lemma 5.5. 7/F is a field of characteristic p and r > 1 is an integer, then the map
Zv-automorphism of F.
SKETCH OF PROOF. The key fact is that for characteristic p, (u ± v)vT = ± vpT for all u,v e F (Exercise III.l .11). Since 1P I—» 1F,
We can now give a useful characterization of finite fields.
Proposition 5.6. Let p be a prime and n > 1 an integer. Then F is a finite field with pn elements if and only if F is a splitting field of xp" — x over Zp. PROOF. If |F| = p n , then the multiplicative group of nonzero elements of F h a s order pn — 1 and hence every nonzero u = F satisfies wp,,_1 = 1/.-. Thus every nonzero u e F is a root of xp,'_1 — 1 /. and therefore a root of jc(x*,"_1 — 1/.) = xv" — x sZ,,[xj as well. Since 0 e Fis also a root of xJ,n — x, xpn — x haspn distinct roots in F(that is, it splits over F by Theorem I1I.6.7) and these roots are precisely the elements of F. Therefore, F is a splitting field of x1>n — x over Zp. If F is a splitting field of / = x'l'n — x over Zv, then since char F = char Zv = p, f = — 1 and /is relatively prime to f. Therefore /has p" distinct roots in F by Theo rem III.6.10(ii). If ip is the monomorphism of Lemma 5.5 (with r = n), it is easy to see that u e F is a root of /if and only if
Corollary 5.7. If p is a prime and n > 1 an integer, then there exists a field with pn elements. Any two finite fields with the same number of elements are isomorphic.
PROOF. Given p and n, a splitting field F of x*’7‘ — x over Zv exists by Theorem 3.2 and has order pn by Proposition 5.6. Since every finite field of order p" is a splitting field of xp" - x over Zp by Proposition 5.6, any two such are isomorphic by Corollary 3.9. ■
Corollary 5.8. 7/K is a finite field and n > 1 is an integer, then there exists a simple extension field ¥ — K(u) of K such that F is finite and [F : K] = n. Any two n -dimen sional extension fields of K are K-isomorphic.
5. FINITE FIELDS
281
SKETCH OF PROOF. Given K of order pr let F be a splitting field of / = xpr" — x over K. By Proposition 5.6 every u e K satisfies upr = u and it follows in ductively that ufn = u for all u e K. Therefore, F is actually a splitting field of / over Zp (Exercise 3.3). The proof of Proposition 5.6 shows that F consists of precisely the pnT distinct roots of /. Thus pnT = I/7! = |K|(F:K1 = (pr)XF'K], whence [F : K] = n. Corollary 5.4 implies that F is a simple extension of K. If F\ is another extension field of K with [F : K] = n, then [Fi :ZP] = n[K :ZP] — nr, whence |Fi| = pnr. By Proposition 5.6 Fi is a splitting field of xp7ir — x over Zv and hence over K. Conse quently, F and Fi are A-isomorphic by Corollary 3.9. ■
Corollary 5.9. If K is a finite field and n > 1 an integer, then there exists an irre ducible polynomial of degree n in K[x], PROOF. Exercise; use Corollary 5.8 and Theorem 1.6. ■
Proposition 5.10. If F is a finite dimensional extension field of a finite fieldYL, then F is finite and is Galois over K. The Galois group /4«/kF is cyclic. SKETCH OF PROOF. Let Zp be the prime subfield of K. Then F is finite di mensional over Zp (Theorem 1.2), say of dimension n, which implies that |F| = pn. By the proof of Proposition 5.6 and Exercise 3.2 F is a splitting field over Zp and hence over K, of xpn — x, all of whose roots are distinct. Theorem 3.11 implies that F is Galois over K. The map
EXERCISES Note: F always denotes an extension field of a field K.
1. If K is a finite field of characteristic p, describe the structure of the additive group of K. 2. (Fermat) If p e Z is prime, then ap = a for all azZp or equivalently, (mod p) for all c e Z.
cv
=c
3. If [AT| = pn, then every element of K has a unique pth root in K. 4. If the roots of a monic polynomial /e (in some splitting field of /over K) are distinct and form a field, then char K = p and / = xv* — x for some n > 1. 5. (a) Construct a field with 9 elements and give its addition and multiplication tables. (b) Do the same for a field of 25 elements. 6. If [ ATJ = cj and {n,q) = 1 and F is a splitting field of xn — 1* over K, then [F : K] is the least positive integer k such that n | (c/fc — 1).
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CHAPTER V FIELDS AND GALOIS THEORY
7. If |K| = 3, then x2" + x + 1 is reducible over Z2. 10. Every element in a finite field may be written as the sum of two squares. 11. Let F be an algebraic closure of Zv (p prime). (a) F is algebraic Galois over Zp. (b) The map : F F given by u \~* up is a nonidentity Z„-automorphism of F. (c) The subgroup H = {(p) is a proper subgroup of AutzpF whose fixed field is Zp, which is also the fixed field of AutzpF by (a). 12. If K is finite and F is an algebraic closure of K, then AutKF is abelian. Every ele ment of AuU-F (except 1 /.-) has infinite order.
6. SEPARABILITY Our study of separability will be greatly facilitated by the simultaneous con sideration of a concept that is, in a sense, the complete opposite of separability. Consequently the section begins with purely inseparable extensions, which are char acterized in several different ways in Theorem 6.4. These ideas are then used to prove all the important facts about separability of algebraic extensions (principally Theo rem 6.7). The degree of (in)separability of an algebraic extension is discussed in detail (most of this material, however, is not needed in the sequel). Finally the Primitive Element Theorem is proved (Proposition 6.15). This result is independent of the rest of the section and may be read at any time.
Definition 6.1. Let F be an extension field of K. An algebraic element u e F is purely inseparable over K if its irreducible polynomial f in K[x] factors in F[x] as f = (x — u)m. F is a purely inseparable extension ofKif every element of F is purely inseparable over K. Thus u is separable over K if its irreducible polynomial /of degree n has n distinct roots (in some splitting field) and purely inseparable over K i f / h a s precisely one root. It is possible to have an element that is neither separable nor purely inseparable over K.
Theorem 6.2. Let F be an extension field of K. Then u e F is both separable and purely inseparable over K if and only if u e K. PROOF. The element u e F is separable and purely inseparable over K if and only if its irreducible polynomial is of the form (x — u)m and has m distinct roots in some splitting field. Clearly this occurs only when m — 1 so that x — uz K[x] and u e K. ■
6. SEPARABILITY
283
If char K = 0, then every algebraic element over K is separable over K. There fore, Theorem 6.2 implies that the only elements that are purely inseparable over K are the elements of K itself. Thus purely inseparable extensions of K are trivial if char K = 0. Consequently, we usually restrict our attention to the case of nonzero (prime) characteristic. We shall frequently use the following fact about characteristic p without explicit mention: if char K = p ^ 0 and u,v e K, then (« ± v)pTl = upn =h vpn for all // > 0 (Exercise III.1.11). In order to characterize purely inseparable exten sions we need:
Lemma 6.3. Let F be an extension field of K with char K = p / 0. // u £ F is algebraic over K, then upn is separable over K for some n > 0. SKETCH OF PROOF. Use induction on the degree of it over K: If deg u = 1 or u is separable, the lemma is true. If /is the irreducible polynomial of a nonseparable u of degree greater than one, then /' = 0 (Theorem III.6.10), whence /is a poly nomial in xp (Exercise III.6.3). Therefore, up is algebraic of degree less than deg u over K, whence by induction (up)pm is separable over K for some m > 0. ■
Theorem 6.4. If F is an algebraic extension field of a fieldK. of characteristic p ^ 0,
then the following statements are equivalent :
(i) (ii) (iii) (iv) (v)
F is purely inseparable over K; the irreducible polynomial of any u e F is of the form xp" — a e K[x]; if u e F, then ur" e K for some n > 0; the only elements of F which are separable over K are the elements of K itself; F is generated over K by a set of purely inseparable elements.
SKETCH OF PROOF OF 6.4. (l) => (ii) Let (x - u)m be the irreducible poly nomial of u s F and let m — npT with (n,p) = 1. Then (a- — u)m = ( x — ii)prn = ( x p r — u p T ) n by Exercise I I I . 1.11. Since ( x — u ) m e X|a], the coefficient of x pr(7,-1>, namely rfcnupT (Theorem III.1.6), must lie in K. Now (p,n) = 1 implies that upT e K (Exercise 1). Since (at — u ) m = (xpT — u p r ) n is irreducible in K[x], we must have n = 1 and (a- — u)m — x p T — a, where a = upT e K. The implications (ii) => (iii) and (i) => (v) are trivial, (iii) => (i) by Exercise III. 1.11; (i) => (iv) by Theorem 6.2; and (iv) => (iii) by Lemma 6.3. (v) => (iii) If u is purely inseparable over K, then the proof of (i) => (ii) shows that upTl e K for some n > 0. If u b F is arbitrary use Theorem 1.3 and Exercise III.l.ll. ■
Corollary 6.5. // F is a finite dimensional purely inseparable extension field of K and char K = p ^ 0, then [F : K] = pn for some n > 0. PROOF. By Theorem 1.11 F = K(uy,. . ., «„,). By hypothesis each Ui is purely inseparable over Kand hence over K(uu . . . , as well (Exercise 2). Theorems 1.6 and 6.4 (ii) imply that every step in the tower K d K(ut) CZ K(uuu2) d - CZ K(ux,. . . , um) = F has dimension a power of p. Therefore [F : K] = pn by Theo rem 1.2. ■
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284
One more preliminary is needed for the principal theorem on separability.
Lemma 6.6 7/F is an extension field of K, X is a subset of F such that F = K(X), and every element ofX is separable over K, then F is a separable extension of K. PROOF. If ve F, then there exist uu .. . , un eX such that v s K(uu . - - , Hr.) by Theorem 1.3. Let fi e /([*] be the irreducible separable polynomial of Ui and E a splitting field of { fi, ..., f,\ over K(uu . . . , u„). Then E is also a splitting field of | fi, . . . ,/„! over K (Exercise 3.3). By Theorem 3.11 E is separable (in fact Galois) over K, which implies that v e K(u\,. . . , u„) Cl E is separable over K. ■
Theorem 6.7. Let F be an algebraic extension field of K, S the set of all elements oj F which are separable over K, and P the set of all elements of F which are purely in separable over K. (i) S is a separable extension field of K. (ii) F is purely inseparable over S. (iii) P is a purely inseparable extension field of K. (iv) P S = K. (v) F is separable over P if and only if F = SP. (vi) If F is normal over K, then S is Galois over K, F is Galois over P and Aut\<$ ^ Aut pF = /4«/kF.
n
REMARKS. It is clear that S' is the unique largest subfield of F separable over K and that S' contains every intermediate field that is separable over K; similarly for P and purely inseparable intermediate fields. If char K — 0, then S = F and P = K (Theorem 6.2). SKETCH OF PROOF OF 6.7. (i) If u, v eS and v ^ 0, then K(u,v) is separable over K by Lemma 6.6, which implies that u — c, uv 1 e S. Therefore, S is a subfield. Lemma 6.3 and Theorem 6.4 imply (ii). (iii) is a routine exercise using Exercise III.l.l 1 if char K = p and the fact that P = K if char K = 0. Theorem 6.2 implies (iv). (v) If F is separable over P, then F is separable over the composite field SP (Exer cise 3.12) and purely inseparable overSP((ii) and Exercise 2). Therefore, F = SP by Theorem 6.2. Conversely, if F = SP = P(S), then F is separable over P by Exercise 3.12 and Lemma 6.6. (vi) We show first that the fixed field K0 of Aut^Fis in fact P, which immediately implies that Fis Galois over P and AutpF = Aut/cF. Let u s F have irreducible poly nomial / over K and let a e AutA-F; a{u) is a root of / (Theorem 2.2). If u e P, then / = (x — uY1 and hence u{u) = u. Therefore, P CI K0. If u z K0 and v s F is any other root of fi then there is a /(-isomorphism r : K(u) —» K(v) such that r(«) = c (Corollary 1.9). By Theorems 3.8 and 3.14 and Exercise 3.2 r extends to a /(-automorphism of F. Since u e K0, we have u = t(u) = v. Since / splits in F[a] by normality, this argument shows that f = (x — n)™ for some m. Therefore, u e P and P ID K0. Hence P = Ku. Every a e Aut pF = AuU-F must send separable elements to separable elements (Theorem 2.2). Therefore, the assignment cr (—> <x | defines a homomorphism 6 : Aut pF—» AuUS. Since F is normal over S, 8 is an epimorphism (Theorems 3.8
6. SEPARABILITY
285
and 3.14 and Exercise 3.2). Since F is Galois over P, F = SP by (v), which implies that 6 is a monomorphism. Hence Aut/F^ AuU-S. Finally suppose u zS is fixed by all a e Aut/vS. Since 6 is an epimorphism, u is in the fixed field P of AutpF, whence u e P fl 5 = K. Therefore, S is Galois over K. ■
Corollary 6.8. If F is a separable extension field of E and E is a separable extension field of K, then F is separable over K. PROOF. If S is as in Theorem 6.7, then E CZ S and F is purely inseparable over S. But F is separable over E and hence over S (Exercise 3.12). Therefore, F = S by Theorem 6.2. ■ Let F be a field of characteristic p ^ 0. Lemma 5.5 shows that for each n > 1, the set Fv>' = {u,,n | u £ F] is a subfield of F. By Theorem 6.4 (iii), F is purely inseparable over F,,n and hence over any intermediate field as well (Exercise 2).
Corollary 6.9. Let F be an algebraic extension field of K, with char K = p ^ 0. If F is separable over K, then F = KFpn/«r each n > 1. If [F : K] is finite and F = KFP, then F is separable over K. In particular, usF is separable over K if and only if K(up) = K(u). SKETCH OF PROOF. Let S be as in Theorem 6.7. If [F : K] is finite, then F = K{uu • • • , u„,) = S(ui, . . . , u,„) by Theorem 1.11. Since each in is purely in separable over S (Theorem 6.7), there is an n > 1 such that w,7'" e 5 for every /. Since F = S(u1, . . . , wm),ExerciseIII.l.l 1 and Theorem 1.3 imply thatF”" C S. Clearly every element of S is purely inseparable over Fpn, and hence over KFpn. S is separ able over K, and hence over KFpn. Therefore S = KFpn by Theorem 6.2. Use the fact that char K — p and Theorem 1.3 to show that for any t > 1, Fp< = [K(wj, . . . , wm)]p< = Kpt(uipl, . . . , umpt). Consequently for any i> 1 we have KFpt = K(Kpt(ulpt, . . . , umpl) = K(iiipt, . . . , umpt). Note that this argument works for any generators ux, . . . , um of F over K. Now if F = KFP, then K(ut, . . . , u,„) = F = KFP = K(uiP, . . . , ump). An iterated argument with the generators u?1 in place of ut [/ = 1, 2, . . . , «] shows that F = AX«i, • - • , «m) = K(u1p’', . . . , umpn) — KFpn = S, whence F is separable over K. Conversely, if F is separable over K, then F is both separable and purely inseparable over KFptl (for any n > 1). Therefore F = KFptl by Theorem 6.2. ■ Next we consider separability and inseparability from a somewhat different point of view. Although Proposition 6.12 is used at one point in Section 7, all that is really essential for understanding the sequel is Definition 6.10 and the subsequent remarks.
Definition 6.10. Let F be an algebraic extension field of K and S the largest subfield of F separable over K (as in Theorem 6.7). The dimension |S : K] is called the separable degree of F over K and is denoted IF : K]a. The dimension [F : S] is called the in separable degree (or degree of inseparability) of F over K and is denoted [F : K|;. REMARKS. [F : K]„ = [F : A] and |F : A], = 1 if and only if Fis separable over K. [F : K], = 1 and [F : K], = [F : K] if and only if Fis purely inseparable over K. In
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CHAPTER V FIELDS AND GALOIS THEORY
any case, [F: K] = [F : K],[F : K]< by Theorem 1.2. If [F : X] is finite and char K = p y* 0, then [F: A"], is a power of p by Corollary 6.5 and Theorem 6.7(ii). The following lemma will enable us to give an alternate description of [F: AT], and to show that for any intermediate field E, [F : E]S[E : AT], = [F : K]„.
Lemma 6.11. Let F be an extension field of E, E an extension field of K and N a normal extension field of K containing F. If r is the cardinal number ofdistinct E-monomorphisms F —> N and t is the cardinal number of distinct K-monomorphisms E —> N, then rt is the cardinal number of distinct K-monomorphisms F —» N. PROOF. For convenience we assume that r, t are finite. The same proof will work in the general case with only slight modifications of notation. Let n , . . . , rr be all the distinct E-monomorphisms F —> TV and
Proposition 6.12. Let F be a finite dimensional extension field of K and N a normal extension field of K containing F. The number of distinct K-monomorphisms F —* N is precisely [F : K]„, the separable degree of F over K. SKETCH OF PROOF. Let 5 be the maximal subfield of F separable over K (Theorem 6.7(i)). Every K-monomorphism S —+ N extends to a K-automorphism of ^(Theorems 3.8 and 3.14 and Exercise 3.2) and hence (by restriction) to a K-mono morphism F N. We claim that the number of distinct K-monomorphisms F —» N is the same as the number of distinct K-monomorphisms 5 —* N. This is trivially true if char K = 0 since F = S in that case. So let char K = p ^ 0 and suppose c, r are K-monomorphisms F —» N such that a 15 = r | S. If u e F, then upn e S for some n > 0 by Theorems 6.4 and 6.7(ii). Therefore,
6. SEPARABILITY
287
Since v is a root of / (as in Theorem 2.2) there are at most [F : K] = deg /such Jf-monomorphisms. Since /splits in N b y normality and is separable, Corollary 1.9 shows that there are exactly [F : K] distinct AT-monomorphisms F —» N. ■
Corollary 6.13. If F is an extension field ofE and E is an extension field of K, then [F : E].[E : K], = [F : KJ. and [F : E].[E : Kh = [F : KJ,. PROOF. Exercise; use Lemma 6.11 and Proposition 6.12. ■
Corollary 6.14. Let f e K[x] be an irreducible monic polynomial over a field K, F a splitting field off over K and Ui a root off in F. Then (i) every root off has multiplicity (K(ui) : K]i so that in F[x], f = [(x — ui)- (x — un)]lK(u,)'Kli, where u i f . . . , un are all the distinct roots off and n = [K(ui) : Kjs;
(ii) mIK(U|) K1> is separable over K. SKETCH OF PROOF. Assume char K — p 0 since the case char K = 0 is trivial, (i) For any / > 1 there is a ^-isomorphism a : K(ut) = K(u,) with o-(ui) = «, that extends to a ^-isomorphism a of F (Corollary 1.9, Theorem 3.8, and Exercise 3.2). Since /e K\x] we have by Theorem 2.2 (jl - U i p ■ ■ ( x - wn)r' = / = erf = ( x - aiuOp ■ ■ ■ (x - crfa))'".
Since u\,. . ., un are distinct and cr is injective, unique factorization in AT[jc] implies that (x — u,)r' = (x — o(«i))r\ whence rx = r.. This shows that every root of /has multiplicity r = rx so that /= (x — ui)r■ ■ (x — u„)r and [K(ui) : K] — deg /= nr. Now Corollary 1.9 and Theorem 2.2 imply that there are n distinct AT-monomorphisms K(ux) —► F, whence [K(wi) : K], = n by Proposition 6.12 and Theorem 3.14. Therefore, [A(wi) : K], = [AT(w!) ; K\/[K(u{) : K], = nr/n = r. (ii) Since r is a power of p = char K, we have / = (x — u,)r- ■ (x — u„)r = (xr — ur)- - (xT — unr). Thus / is a polynomial in .vr with coefficients in K, say n
n
/—
Consequently, uir is a root of g(x) = a'x* = (x ~ uir)' '(x ~ u"r) »=»0 i—0 e AT[jc]. Since u\, . . . , u„ are distinct, g(x) e A[.v] is separable. Therefore ur— \k ut) ki. js separable over K. ■ u
The following result is independent of the preceding material and is not needed in the sequel.
Proposition 6.15. (Primitive Element Theorem) Let F be a finite dimensional ex tension field of K.
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CHAPTER V FIELDS AND GALOIS THEORY
(i) If F is separable over K, then F is a simple extension of K. (ii) (Art in) More generally, F is a simple extension o/K if and only if there are only finitely many intermediate fields. REMARK. An element u such that F = K(u) is said to be primitive. SKETCH OF PROOF OF 6.15. The first paragraph of the proof of Lemma 3.17, which is valid even if the field K is finite, shows that a separable extension has only finitely many intermediate fields. Thus it suffices to prove (ii). Since (ii) clearly holds if K is finite (Corollary 5.8), we assume that K is infinite. One implication of (ii) is proved in the second paragraph of the proof of Lemma 3.17. Conversely assume F = K(u) with u algebraic over K (since [F : K] is finite). Let E be an intermediate field and g e E[x] the irreducible monic polynomial of u over E. If g = xn + i*"-1 -\----- \-aix + a0, then [F : F] = n. Show that F = K(a0,ai, - . ., 1) by verifying that [F : K(a0 , . . . , an-1)] = n. Thus every intermediate field E is uniquely deter mined by the irreducible monic polynomial g of u over E. If /is the monic irreducible polynomial of u over K, then g \ f by Theorem 1.6. Since / factors uniquely in any splitting field (Corollary III.6.4), /can have only a finite number of distinct monic divisors. Consequently, there are only a finite number of intermediate fields. ■
EXERCISES Note: Unless stated otherwise F is always an extension field of a field K.
1. Let char K = p ^ 0 and let n > 1 be an integer such that (p,ri) = 1. If v e F and nv e K, then v e K. 2. If u e F is purely inseparable over K, then u is purely inseparable over any inter mediate field E. Hence if F is purely inseparable over K, then F is purely in separable over E. 3. If F is purely inseparable over an intermediate field E and E is purely inseparable over K, then F is purely inseparable over K. 4.
If u £ F is separable over K and v z F is purely inseparable over K, then K(u,v) = K(u + v). If u 9^ 0, v 9^ 0, then K(u,v) — K(uv).
5.
If char K = p 0 and a z K but a every n > 1.
|
K‘\ then x>‘" — a z
K[.v]
is irreducible for
6. If/e K\x] is monic irreducible, deg / > 2, and /has all its roots equal (in a splitting field), then char K = p ^ 0 and / = xpn — a for some n > 1 and a z K. 7. Let F, K, 5, P be as in Theorem 6.7 and suppose E is an intermediate field. Then (a) F is purely inseparable over E if and only if 5 C E. (b) If F is separable over F, then P Cl E. (c) If E D 5 = K, then Ed P. 8. If char K = p ^ 0 and [F : K] is finite and not divisible by p, then F is separable over K. 9. Let char K = p 0. Then an algebraic element tt z F is separable over K if and only if K(u) = K(up") for all n > 1.
7. CYCLIC EXTENSIONS
289
10. Let char K = p 0 and let /e K[x] be irreducible of degree n. Let m be the largest nonnegative integer such that / is a polynomial in xpm but is not a poly nomial in xpm+l. Then n = n0pm. If u is a root of /, then [K(«) : K]s = n0 and [K(u) : KU = Pm-
11. If /e K[x] is irreducible of degree m > 0, and char K does not divide m, then /is separable. 12. F is purely inseparable over K if and only if F is algebraic over K and for any ex tension field E of F, the only A'-monomorphism F —» F is the inclusion map. 13. (a) The following conditions on a field K are equivalent: (i) (ii) (iii) (iv)
every irreducible polynomial in K[x] is separable; every algebraic closure K of K is Galois over K; every algebraic extension field of K is separable over K; either char K = 0 or char K = p and K = Kp.
A field K that satisfies (i)—(iv) is said to be perfect. (b) Every finite field is perfect. 14. If F = K(u,v) with u,v algebraic over K and u separable over K, then F is a simple extension of K. 15. Let char K — p ^ 0 and assume F = K(u,c) where u v e K, vv s K and [F : K] = p2. Then Fis not a simple extension of K. Exhibit an infinite number of intermediate fields. 16. Let F be an algebraic extension of K such that every polynomial in A[x] has a root in F. Then F is an algebraic closure of K. [Hint: Theorems 3.14 and 6.7 and Proposition 6.15 may be helpful.]
7. CYCLIC EXTENSIONS The basic idea in Sections 7-9 is to analyze Galois field extensions whose Galois groups have a prescribed structure (for example, cyclic or solvable). In this section we shall characterize most finite dimensional Galois extensions with cyclic Galois groups (Propositions 7.7 and 7.8; Theorem 7.11). In order to do this it is first necessary to develop some information about the trace and norm.
Definition 7.1. Let F be a finite dimensional extension field o/K and K an algebraic closure of K containing F. Let
Tkf(u) = [F : K]i(
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CHAPTER V FIELDS AND GALOIS THEORY
places K by any normal extension of K containing F (Exercise 1). K is normal over K (Theorems 3.4 and 3.14), whence r = [F: A]5 is finite by Proposition 6.12. If the con text is clear Nkf and TKF will sometimes be written simply as N and T. Note that the trace is essentially the additive analogue of the norm. In many in stances this means that a proof involving the one will translate directly into a proof of the analogous fact for the other. There are some exceptions, however. For instance if Fis not separable over K, then char K = p ^ 0 and [F : A], = p1 (t > 1). Consequently, Tkf(ju) = 0 for every u s F, but NKF(u) may not be zero. EXAMPLE. Let F = C a r d A ' = R and take K = C. It is easy to see that the only R-monomorphisms C —» C are the identity and complex conjugation. Conse quently N(a + bi) — [(a + bi)(a — hi)]1 = a2 -f- b2. The principal applications to be given here of the norm and trace occur when F is Galois over K. In this case the Galois group is finite and there is a more convenient description of the norm and trace, which is sometimes taken as a definition.
Theorem 7.2. //F is a finite dimensional Galois extension field of K and AuIkF
= {CTi, . . . , <Jn },
then for any u s F, N k f( u)
=
Tkf(u) = cn(u) + er2(u) + • ■ • +
PROOF. Let K be an algebraic closure of K which contains F. Since F is normal over K (Corollary 3.15), the ZT-monomorphisms F —> K are precisely the elements of AutA-F by Theorem 3.14. Since F is also separable over K (Corollary 3.15), [F: K]i = 1. The conclusion of the theorem now follows directly from Defini tion 7.1. ■ Suppose F is Galois over K and Aut^-F = { < r i , . . . ,
Theorem 7.3. Let F be a finite dimensional extension field o/K. Then for all u,v e F: (i)
Nkf(u)Nkf(v) = NKF(uv) and TKF(u) + TKF(v) = TKF(u + v);
if u s K, then Nkf(u) = u!F:K) o«rfTKF(u) = [F : KJu; (iii) Nkf(u) and TKF(u) are elements of K. More precisely, (ii)
Nkf(u) = (( — l)na0)tF:K(u)1 e K a^Tkf(u) = -[F : K(u)]a„_i s K,
where f = x" + a„_i xri ~‘ + ■ —b a0 e K[x] is the irreducible polynomial of u; (iv) if E is an intermediate field, then
NKE(NEF(u)) = NKF(u)
o/7c/Tke(Tef(u))
= TKF(u).
7. CYCLIC EXTENSIONS
291
SKETCH OF PROOF, (i) and (ii) follow directly from Definition 7.1 and the facts that r = [F : K ] , and [F : K],[F : K ] { = [F: KJ. (iii) Let E = K(u). An algebraic closure K of K which contains F is also an algebraic closure of E. The proof of Lemma 6.11 shows that the distinct AT-monomorphisms F—> X are precisely the maps er^r, (1 < k < /;1 < j < r), where the
= (n
Use (ii) and Corollary 6.13 to show that NKF(u) = {II o*(u) J
1
an(j
T k f ( u ) = [F : FJ[F : X],^23 ffk(u)J. Since
plies that
fft(«))p=*>.
If [E: K ] i = 1, then n = t and the conclusion is immediate. If [E: K ] i > 1, then [E: AT], is a positive power of p = char X. It is easy to calculate a0 and to see that cin-i = 0 = Tkf(u); use Exercise III.l.ll. (iv) Use the notation in the first paragraph of the proof of (iii), with E any inter mediate field. Apply the appropriate definitions and use Corollary 6.13. ■ In addition to the trace and norm we shall need
Definition 7.4. Let S be a nonempty set of automorphisms of a field F. S is linearly independent provided that for any au . . . , a „ e F and ai, . . . , cr„ e S (n > 1): aicri(u) -f ■ • ■ -f a„crn(u) = 0 for all u e F => a-, = 0 for every i.
Lemma 7.5. If S is a set of distinct automorphisms of a field F, then S is linearly independent.
PROOF. If S is not linearly independent then there exist nonzero ai e F and distinct
(1)
Among all such “dependence relations” choose one with n minimal; clearly n > 1. Since ax and er2 are distinct, there exists v e F with
(2)
and multiplying (1) by
The difference of (2) and (3) is a relation: a2[o2(v) - cv(v)\c2(u) + a3[o3(v) - oy(v)](T3(u) + • • • + an[cn(v) - cl(v)]on(u) = 0
(3)
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CHAPTER V FIELDS AND GALOIS THEORY
for all uzF. Since a-z 9^ 0 and er2(t>) ^ a,(v) not all the coefficients are zero and this contradicts the minimality of n. ■ An extension field F of a field K is said to be cyclic [resp. abelian] if F is algebraic and Galois over K and Aut a/7 is a cyclic [resp. abelian] group. If in this situation Aut/cF is a finite cyclic group of order n, then F is said to be a cyclic extension of degree n (and [F : K] = n by the Fundamental Theorem 2.5). For example, Theorem 5.10 states that every finite dimensional extension of a finite field is a cyclic extension. The next theorem is the crucial link between cyclic extensions and the norm and trace.
Theorem 7.6. Let F be a cyclic extension field of K of degree n, a a generator of /4«/kF and u e F. Then (i) Tkf(u) = 0 if and only if u = v — cr(v) for some v e F; (ii) (Hilbert's Theorem 90) Nrf(u) = Ik if and only if u = vcr(v)-1 for some nonzero V e F . SKETCH OF PROOF. For convenience write c(x) = ax. Since a generates AutkF, it has order n and a,a2,az,. . ., an~l,an = \F =
to show that T(u) = 0. Conversely suppose T(u) = 0. Choose w s F such that T(w) = Ik as follows. By Lemma 7.5 (since Ik 7^ 0) there exists zzF such that 0 ^ 1 pz -f- az - f - a2z + • • • -f- an~1z = T(z). Since T(z) e K by the remarks after Theorem 7.2, we have
Now let v = uw + («-f au)(aw) (u -\- au a7u)(ahv)
+ (« + au +
+ u(an~lw) = uT(w) =
u
Ik
u(an~2w)
= «•
(ii) If u — va(v)~l, then since a is an automorphism of order «, an(v~l) = u_1, l a(v~ ) = a(v)1 and for each 1 < / < n — 1, a'(va(v)~l) = a{(v)ai+l(v)1. Hence: N(u) = (va(«)“'')(aua2( u)~l)(a‘lua3( v)1) • • ■ (o-n_1i?
7. CYCLIC EXTENSIONS
293
Conversely suppose N(u) = 1*, which implies 0. By Lemma 7.5 there exists yef such that the element v given by v = uy -f- (uau)ay -f- (uau
+ (uau - - - an~lii)an~ly is nonzero. Since the last summand of v is Ar(w)o-n~1y = 1 K<*n~ly = o'n_1>’, it is easy to verify that u~lv = av, whence u = io{v) 1 (a(v) ^ 0 since v ^ 0 and a is injec tive). ■ We now have at hand all the necessary equipment for an analysis of cyclic ex tensions. We begin by reducing the problem to simpler form.
Proposition 7.7. Let ¥ be a cyclic extension field of K of degree n and suppose n = mp1 where 0 ^ p = char K and (m,p) = 1. Then there is a chain of intermediate fields F D E0 D Ei D • • O Et_i Z) Et = K such that F is a cyclic extension of E0 of degree m and for each 0 < i < t, Ej_, is a cyclic extension ofE{ of degree p. SKETCH OF PROOF. By hypothesis F is Galois over K and Aut*F is cyclic (abelian) so that every subgroup is normal. Recall that every subgroup and quotient group of a cyclic group is cyclic (Theorem 1.3.5). Consequently, the Fundamental Theorem 2.5(ii) implies that for any intermediate field E, F is cyclic over E and E is cyclic over K. It follows that for any pair L,M of intermediate fields with L
Proposition 7.8. Let K be a field of characteristic p ^ 0. F is a cyclic extension field of K of degree p if and only if F is a splitting field over K of an irreducible polynomial o f the form xp — x — a £ K[x]. In this case F = K(u) where u is any root o/xp — x — a. PROOF. (=>) If a is a generator of the cyclic group Aut*F, then Tx^(lx) = [F: = plK = 0
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CHAPTER V FIELDS AND GALOIS THEORY
by Theorem 7.3(ii), whence 1 x = v — < j ( v ) for some v e F by Theorem 7.6(i). If u — — v, then cr(u) = u U ^ «, whence u £ K. Since [F: K] = p there are no intermediate fields, and we must have F = K(u). Note that c(up) = (u + lx)p = + 1xp = uv + lx which implies that cr(uv — u) = (uv + lx) — (w + Ix) = uv — u. Since F is Galois over K and AutxF = (cr), a = uv — u must be in K. Therefore, u is a root of xv — x — a e A[x], which is necessarily the irreducible polynomial of u over K since the degree of u over K is [A(u) : K] = [F : K] = p. Recall that the prime subfieldZp of K consists of the p distinct elements 0,1 = lx, 2 = lx + lx, - • ■ ,P — 1 = lx + ■ ■ • + lx (Theorem 5.1). The first paragraph of the proof of Theorem 5.6 shows that ip — i for all / e Zv. Since u is a root of xv — x — a, we have for each / zZv\ (u + i)v — (u + /') — a = uv + iv — u — i a = (up — u — a) + (ip — i) = 0 + 0 = 0. Thus u + i e K(u) = F is a root of xv — x — a for each i eZP, whence F contains p distinct roots of xv — x — a. Therefore, F = K(u) is a splitting field over K of xv — x — a. Finally if u + i (/' e Zp CZ K) is any root of xv — x — a, then clearly K(u + i) = K(u) = F. (<=) Suppose F is a splitting field over K of xv — x — a e A[x], We shall not as sume that xv — x — a is irreducible and shall prove somewhat more than is stated in the theorem. If u is a root of xp — x — a, then the preceding paragraph shows that K(u) contains p distinct roots of xv — x — a: u, u + 1, . . . , u (p — 1) s K(u). But xp — x — a has at most p roots in F and these roots generate F over K. There fore, F = K(u), the irreducible factors of xv — x — a are separable and F is Galois over K (Theorem 3.11 and Exercise 3.13). Every r s AutxF = AutkK(u) is completely determined by r(w). Theorem 2.2 implies that r(«) = u + i for some izZp d K. Verify that the assignment rhi defines a monomorphism of groups 6 : AutxF —»Z,,. Consequently, AutxF = Im 6 is either 1 or ZP. If AutxF = 1, then [F : A'] = 1 by the Fundamental Theorem 2.5, whence u e K and xp — x — a splits in Afc]. Thus if xp — x — a is irreducible over K, we must have AutxF = Zv. In this case, therefore, F is cyclic over K of degree p. ■
Corollary 7.9. // K is a field of characteristic p ^ 0 and xp — x — a £ K[x], then xp — x — a is either irreducible or splits in K[x], PROOF. We use the notation of Proposition 7.8. In view of the last paragraph of that proof it suffices to prove that if AutxF = Im 6 = Zp, then xp — x — a is irre ducible. If u and v = u + i (/' eZp d K) are roots of x7' — x — a, then there exists r £ AutxF such that t(u) = v and hence r : K(u) = K(i) (choose r with 6(r) = i). Therefore, u and v are roots of the same irreducible polynomial in K[x] (Corollary 1.9). Since v was arbitrary this implies that xv — x — a is irreducible. ■ Proposition 7.8 completely describes the structure of a cyclic extension of the first type mentioned on p. 293. In order to determine the structure of a cyclic exten sion of degree n of the second type it will be necessary to introduce an additional assumption on the ground field K. Let A' be a field and n a positive integer. An element f e A' is said to be an nth root of unity provided f" = 1A (that is, f is a root of xn — lx e A^.y]). It is easy to see that the set of all nth roots of unity in K forms a multiplicative subgroup of the multiplica tive group of nonzero elements of K. This subgroup is cyclic by Theorem 5.3 and has
7. CYCLIC EXTENSIONS
295
order at most n by Theorem III.6.7. f e K is said to be a primitive nth root of unity provided f is an nth root of unity and f has order n in the multiplicative group of nth roots of unity. In particular, a primitive nth root of unity generates the cyclic group of all nth roots of unity. REMARKS. If char K = p and p \ n, then n = pkm with (p,m) = 1 and m < n. Thus a" — 1a- = (xm — 1 a)"* (Exercise III. 1.11). Consequently the nth roots of unity in K coincide with the mth roots of unity in K. Since m < n, there can be no primitive nth root of unity in K. Conversely, if char K\n (in particular, if char K = 0), then nx"-1 ^ 0, whence a" — 1K is relatively prime to its derivative. Therefore - 1a has n distinct roots in any splitting field F of xn — 1a over K (Theorem III.6.10). Thus the cyclic group of nth roots of unity in Fhas order n and F(but not necessarily K) contains a primitive nth root of unity. Note that if K does contain a primitive nth root of unity, then K contains n distinct roots of xn — 1K, whence F = K. EXAMPLES. 1a is an nth root of unity in the field K for all n > 1. If char K = p 0 and n = pk, then 1a is the only nth root of unity in K. The subfield Q(/') of C contains both primitive fourth roots of unity (±i) but no cube roots of unity except 1, (the others being —1/2 =fc ^3 i/2). For each n > 0, e2riln e C is a primitive nth root of unity. In order to finish our characterization of cyclic extensions we need
Lemma 7.10. Let n be a positive integer and K a field which contains a primitive nth root o f unity f. (i) //'d|n, then Cn/d= t] is a primitive dth root of unity in K. (ii) If d | n and u is a nonzero root of xd — a e K[x], then xd — a has d distinct roots, namely u,j?u,tj2u, . . . , i7d-1u, where t] e K is a primitive dth root of unity. Fur thermore K(u) is a splitting field ofxd — a over K and is Galois over K. PROOF, (i) f generates a multiplicative cyclic group of order n by definition. If d | n, then 77 = £nld has order d by Theorem 1.3.4, whence rj is a primitive dih root of unity, (ii) If u is a root of xd — a, then so is rj'u. The elements 17°= 1a, 17, ... , are distinct (Theorem 1.3.4). Consequently since 77 e K, the roots w, rju,. . ., 7f~lu of xd — a are distinct elements of K{u). Thus K(u) is a splitting field of’ xd — a over K. The irreducible factors of ad — a are separable since all the roots are distinct, whence K(u) is Galois over K by Theorem 3.11 and Exercise 3.13. ■
Theorem 7.11. Let n be a positive integer and K a field which contains a primitive
nth root of unity f. Then the following conditions on an extension field F of K are equiv alent.
(i) F is cyclic of degree d, where d | n; (ii) F is a splitting field over K of a polynomial of the form xn — a e K[x] (in which case F = K(u), for any root u of xn — a); (iii) F is a splitting field over K of an irreducible polynomial of the form xd — b 3 K[x], where d | n (in which case F = K(v), for any root v of xd — b).
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CHAPTER V FIELDS AND GALOIS THEORY
PROOF, (ii) => (i) Lemma 7.10 shows that F = K(u) and F is Galois over K for any root u of xn — a. If a e AutkF = A\AkK(u), then a is completely determined by oiu), which is a root of x71 — a by Theorem 2.2. Therefore, a(u) = for some / (0 < i < n — 1) by Lemma 7.10. Verify that the assignment crH> defines a mono morphism from Aut^F to the multiplicative cyclic group (of order n) of «th roots of unity in K. Consequently, AutAF is a cyclic group whose order d divides n (Theorem 1.3.5 and Corollary 1.4.6). Hence F is cyclic of degree d over K. (i) => (iii) By hypothesis AutKF is cyclic of order d = [F : K] with generator a. Let 17 = z K be a primitive dth root of unity. Since Nk*(v) = V11':K] = Vd = I a-, Theorem 7.6(ii) implies that 77 = woiw)-1 for some w e F. If v — w~l, then a(v) = -qv and o(vd) = (-qvY = ifc'1 = vd. Since F is Galois over K, 1/ = b must lie in K so that v is a root of xd — b e X[x]. By Lemma 7.10 K(v) CZ F and K(v) is a splitting field over K of xd — b (whose distinct roots are , r}d~lv). Furthermore for each i (0 < / < d — 1), aKc) = rfv so that a' : K(c) ^ K^v). By Corollary 1.9 v and ifv are roots of the same irreducible polynomial over K. Consequently, xd — b is irre ducible in X[x]. Therefore, [X(u) : K] = d = [F : X], whence/7 = K(v). (iii) => (ii) If £ F is a root of xd — b e K[x\, then F = K(v) by Lemma 7.10. Now (ft;)" = = \Kcd(nld) = bnld s K so that ft; is a root of x” — a s. K[x], where a = bnld. By Lemma 7.10 again K(£v) is a splitting field of x" — a over K. But f e K implies that F = K(v) = K(£v). ■
It is clear that the primitive «th roots of unity play an important role in the proof of the preceding results. Characterization of the splitting fields of polynomials of the form x" — a e K[x] is considerably more difficult when K does not contain a primitive «th root of unity. The case when a = l/v is considered in Section 8.
EXERCISES 1. If K is replaced by any normal extension N of K containing F in Definition 7.1, then this new definition of norm and trace is equivalent to the original one. In particular, the new definition does not depend on the choice of N. See Exercise 3.21. 2. Let F be a finite dimensional extension of a finite field K. The norm Nkf and the trace TKF (considered as maps F —> K) are surjective. 3. Let Q be a (fixed) algebraic closure of Q and 1 s Q, c | Q. Let E be a subfield of Q maximal with respect to the condition v ^ E. Prove that every finite dimen sional extension of E is cyclic. 4.
Let K be a field, K an algebraic closure of K and a £ Aut/vA'. Let F = ! u £ K | cr(u) = u!.
Then Fis a field and every finite dimensional extension of F is cyclic. 5.
If F is a cyclic extension of K of degree p" (p prime) and L is an intermediate field such that F = L ( u ) and L is cyclic over K of degree pn l, then F = K(u).
6. If char K = p 0, let Kp = j u” — u | u £ K\. (a) A cyclic extension field F of K of degree p exists if and only if K #
8. CYCLOTOMIC EXTENSIONS
297
(b) If there exists a cyclic extension of degree p of K, then there exists a cyclic extension of degree pn for every n > 1. [Hint: Use induction; if E is cyclic over K of degree pn~l with Auu-£ generated by a, show that there exist u,v e E such that Tke(v) = 1* and o(u) — u = vp — v. Then xp — x —he E[x] is irreducible and if w is a root, then K(w) is cyclic of degree pn over A.] 7. If n is an odd integer such that K contains a primitive «th root of unity and char K 7^ 2, then K also contains a primitive 2«th root of unity. 8. If F is a finite dimensional extension of Q, then F contains only a finite number
of roots of unity. 9. Which roots of unity are contained in the following fields: Q(/), Q(*\j2), Q(V5X Q(\/Z~2), Q(V~3)? 10. (a) Let p be a prime and assume either (i) char K = p or (ii) char A ^ p and K contains a primitive />th root of unity. Then xv — a e K[x] is either irreducible or splits in AT[a]. (b) If char A = p ^ 0, then for any root u of X* — a e A[a], K(u) ^ K(uv) if and only if [K(u) : K] = p .
8. CYCLOTOMIC EXTENSIONS Except for Theorem 8.1 this section is not needed in the sequel. We shall examine splitting fields of the polynomial xn — 1 a , with special attention to the case K = Q. These splitting fields turn out to be abelian extensions whose Galois groups are well known. A splitting field F over a field A of xn — 1* s K[a] (where n > 1) is called a cyclotomic extension of order n. If char K = p 9* 0 and n = mp' with (p,m) = 1, thenA" — Ik = (a"* — I)71'(Exercise III. 1.11) so that a cyclotomic extension of order n coincides with one of order nu Thus we shall usually assume that char A" does not divide// (that is, char K = 0 or is relatively prime to//). The dimension of a cyclotomic extension field of order n is related to the Euler function of elementary number theory, which assigns to each positive integer n the number
Theorem 8.1. Lei n be a positive integer, K a field such that char K does not divide n and F a cyclotomic extension of K of order n. (i) F = K(f). where £ e F is a primitive nth root of unity. (ii) F is an abelian extension of dimension d, where d j <^(n) (^ the Euler function); if n is prime F is actually a cyclic extension. (iii) Autu F is isomorphic to a subgroup of order d of the multiplicative group of units of Z„.
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CHAPTER V FIELDS AND GALOIS THEORY
REMARKS. Recall that an abelian extension is an algebraic Galois extension whose Galois group is abelian. The dimension of F over K may be strictly less than For example, if f is a primitive 5th root of unity in C, then R d R(f) CI C, whence, [R(f) : R] = 2 < 4 = v?(5). If K = Q, then the structure of the group Auto/7 completely determined in Exercise 7. SKETCH OF PROOF OF 8.1. (i) The remarks preceding Lemma 7.10 show that F contains a primitive nth root of unity f. By definition 1 a , . . . , e K(£) are the n distinct roots of xn — 1a, whence F = K(£). (ii) and (iii). Since the irreducible factors of x" — \k are clearly separable, Theorem 3.11 and Exercise 3.13 imply that Fis Galois over K. If a e Aut*F, then a is completely determined by cr(f). For some / (1 < i < n — 1), oin = by Theorem 2.2. Similarly <7-1(f) = f3 so that f =
Proposition 8.2. Let n be a positive integer, K a field such that char divide n and gn(x) the nth cyclotomic polynomial over K.
K
does not
(i) xn - 1K = II gd(x). d'n (ii) The coefficients of gn(x) lie in the prime subfield P of K. If char K = 0 and P is identified with the field Q of rationals, then the coefficients are actually integers. (iii) Deg g„(x) = yr(n), where
n
n 1a= tjeG
* -
II (x - rj) = II ( II (* - v)) = II gd(x). d rjeG d\n
d d\n
8. CYCLOTOMIC EXTENSIONS
299
(ii) We prove the first statement by induction on n. Clearly gi(x) = x — 1K e P[x], Assume that (ii) is true for all k < n and let /(x) = gd(x). Then /s P[x] by the d d\n d< n
induction hypothesis and in F[x], xn — 1* = f(x)gn(x) by (i). On the other hand xn — Ik eP[x] and /is monic. Consequently, the division algorithm in P[x] implies that xn — Ik = fh-\-r for some h, r e P[x] CI F[x], Therefore by the uniqueness of quotient and remainder (of the division algorithm applied in F[x]) we must have r = 0 and g„(x) = h e P[x\. This completes the induction. If char K = 0 and P = Q, then a similar inductive argument using the division algorithm in Z[x] and Q[x] (instead of P[x], F[x]) shows that gn(x) e Z[x], (iii) deg gn is clearly the number of primitive «th roots of unity. Let f be such a primitive root so that every other (primitive) root is a power of f. Then (1 < / < ri) is a primitive /zth root of unity (that is, a generator of G) if and only if (/',«) = 1 by Theorem 1.3.6. But the number of such i is by definition precisely
a
REMARKS. Part (i) of the theorem gives a recursive method for determining g„(x) since f\g'()
~
1k
n
d d\n d
g&y
For example if p is prime, then g p ( x ) = (x” — \ K ) / g \ ( x ) = ( x p — 1 k ) / ( x — Ik) = xp_1 + xp~2 + • • • 4- x2 -f x + Ik. Using the example preceding Theorem 8.2 we have for K = Q: ge(x) = (xe - l)/gi(x)g2(x)g3(x) = ( X 6 - l ) / ( x - l ) ( x + l)(x2 + X + 1 ) = x 2 — a: + 1;
similarly gA*) = (x12 ~ l)/(* — 0(^ + 0(^2 + x + l)(x2 + 1)(*2 — x + 1) = x4 — X2 + 1.
When the base field is the field Q, we can strengthen the previous results somewhat.
Proposition 8.3. Let F be a eyclotomic extension of order n of the field Q of rational numbers and g„(x) the nth cyclotomic polynomial over Q. Then (i) gn(x) is irreducible in Q[x], (ii) [F : Q] = <^(n), where
CHAPTER V FIELDS AND GALOIS THEORY
300
We shall show first that is also a root of h. Since f is a root of gn(*), f is a primitive /ith root of unity. The proof of Proposition 8.2(iii) implies that is also a primitive nth root of unity and therefore a root of either /or h. Suppose is not a r
root of h. Then
fp
r
is a root of f(x) = 52
°ix< and
hence f is a root of
f(xp)
= 52 <*iXxp.
i=0 i=0
Since h is irreducible in Q[x] (Lemma III.6.13) and has f as a root, h must divide f(xp) by Theorem 1.6, say f(xp) = h(x)k(x) with k e Q[x]. By the division algorithm in Z[x], f(xp) = h(x)k\(x) + rx(jt) with kx,rx e Z[x], The uniqueness statement of the division algorithm in Q[x] shows that k(x) = ki(x) e Z[xJ. Recall that the canonical projection Z —> Zv (denoted on elements by b (—» b) induces a ring epimorphism tt Z[x]—*Zp[x] defined by g = 52 hx* (—► g = 52 fox* (Exercise III.5.1). Consei=0 i=0 quently, in Zp[x], f(xp) = h(x)k(x). But in Z,,[x],f(xp) — f(x)p (since char Zp = p). Therefore, f(x)p = h(x)k(x) £ Zp[*].
Consequently, some irreducible factor of h(x) of positive degree must divide f(x)p and hence f(x) inZp[jc], On the other hand, since gn(x) is a factor of xn — 1, we have xn — 1 = gn(x)r(x) = f(x)h(x)r(x) for some r(x) e Z[x]. Thus in Zv\x\ xn — I = xn — 1 = f(x)h(x)r(x).
Since /and h have a common factor, xn — T eZp[x] must have a multiple root. This contradicts the fact that the roots of x” — I are all distinct since (p,n) = 1 (see the Remarks preceding Lemma 7.10). Therefore is a root of h(x). If r s Z is such that 1 < r < n and (r,n) = 1, then r = pvkl- ■ -p„ki where ki > 0 and each pi is a prime such that (pi,n) = 1. Repeated application of the fact that is a root of h whenever f is, shows that is a root of h(x). But the (1 < r < n and (r,n) = 1) are precisely all of the primitive nth roots of unity by the proof of Proposi r = g (x), whence gX*) = h(x). tion 8.2(iii). Thus h(x) is divisible by (x n i
n
f)
(r,n) = 1
Therefore, gn(x) is irreducible. (ii) Lemma 7.10 shows that F = Q(f), whence [ F: Q} = [Q(D : Q] = deg gn = M
by Proposition 8.2 and (i). (iii) is a consequence of (ii), Theorem 8.1, and Exer cise 1. ■ REMARK. A nontrivial theorem of Kronecker states that every abelian exten sion of Q is contained in a cyclotomic extension.
EXERCISES 1. If / e Z, let 7 denote the image of i in Zn under the canonical projection Z —»Z„. Prove that 7 is a unit in the ringZn if and only if (/,«) = 1. Therefore the multipli cative group of units in Z„ has order <£(«). 2. Establish the following properties of the Euler function
8. CYCLOTOMIC EXTENSIONS (c)
If
n
pikl---prkr
=
(p i
distinct
primes;
301 fc,
>
0),
then
n( 1 - l/p,)(l - l/p2)- • (1 - 1//V).
(d) 52
d\n
1 if m(«)
h
=1
( — 1)' if n is a product of t distinct primes
=
if p 2 divides n for some prime p .
0
3. Let
(c) If n is odd, then g2n(x) = gn( ~x). (d) If p is a prime and pJf n, then ^^(a) = gn(xp)/gn(x). (e) gn(x) = (xnld — 1)mW), where n is the Moebius function of Exercise 2 (e).
n
d\n
(f) 8n( 1)
=
P if
n
— ih > 0), 0 if n = 1, and 1 otherwise.
6. Calculate the «th cyclotomic polynomials over Q for all positive n with n < 20. 7. Let Fn be a cyclotomic extension of Q of order n. Determine the structure of AutQF* for every n. [Hint: if Un* denotes the multiplicative group of units inZ„, r
then show that Un* = VPini* where n has prime decomposition n = pini- ■ ■prnr. i—i
Apply Exercise 4.] 8. Let Fn be a cyclotomic extension of Q of order n. (a) Determine AutQF5 and all intermediate fields. (b) Do the same for Fs. (c) Do the same for F7; if f is a primitive 7th root of unity what is the irre ducible polynomial over Q of f -f- f-1? 9. If n > 2 and f is a primitive «th root of unity over Q, then [Q(f + r^) : QJ =
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CHAPTER V FIELDS AND GALOIS THEORY
10. (Wedderburn) A finite division ring D is a field. Here is an outline of the proof (in which E* denotes the multiplicative group of nonzero elements of a division ring E). (a) The center K of D is a field and D is a vector space over K, whence |£)| = qn where q = \K\ > 2. (b) If 0 5* a e D, then N(a) = [de D \ da = ad] is a subdivision ring of D containing K. Furthermore, |N(a)| = qT where r \ n. (c) If 0 ^ a e D — K, then N(a)* is the centralizer of a in the group D* and [D* : N(a)*] = (qn — l)/(<7r — 1) for some r such that 1 < r < n and r | n. (d) qn — 1 = q — 1 + 22 (qn — l)/0yr — 1), where the last sum taken over a T
finite number of integers r such that 1 < r < n and r \ n. [Hint: use the class equation of D*; see pp. 90-91.] (e) For each primitive «th root of unity f e C, \q — f | > q — 1, where |a + bi| = yja2 + b2 for a + bi e C. Consequently, |g*(t/)| > q — 1, where gn is the Hth cyclotomic polynomial over Q. (f) The equation in (d) is impossible unless n = 1, whence K = D. [Hint: Use Proposition 8.2 to show that for each positive divisor r of n with r ^ n, ft x) = (,vn — 1 )/(xr — 1) is in Z[a] and fT(x) = gn(x)hr(x) for some h,(x) e Z[x]. Consequently, for each such r g„(q) divides f(q) in Z, whence gn(q) I (q — 1) by (d). This contradicts (e).]
9. RADICAL EXTENSIONS Galois theory had its historical origin in a classical problem in the theory of equations, which may be intuitively but reasonably accurately stated as follows. Given a field K, does there exist an explicit “formula” (involving only field opera tions and the extraction of «th roots) which gives all the solutions of an arbitrary polynomial equation f(x) = 0 (/e /f[x])? If the degree of /is at most four, the answer is affirmative (for example, the familiar “quadratic formula” when deg / = 2 and char K 2; see also Exercise 5). We shall show, however, that the answer is negative in general (Proposition 9.8). In doing so we shall characterize certain field extensions whose Galois groups are solvable (Theorem 9.4 and Proposition 9.6). The first task is to formulate a precise statement of the classical problem in fieldtheoretic terms. Throughout the discussion we shall work in a fixed algebraic closure of the given base field K. Intuitively the existence of a “formula” for solving a specific polynomial equation f(x) = 0 means that there is a finite sequence of steps, each step being a field operation (addition, multiplication, inverses) or the extraction of an Hth root, which yields all solutions of the given equation. Performing a field operation leaves the base field unchanged, but the extraction of an «th root of an element c in a field E amounts to constructing an extension field £(«) with un sE (that is, u = \/r). Thus the existence of a “formula” for solving/(x) = 0 would in effect imply the existence of a finite tower of fields K = Eu a Ei a ■ ■ ■ ci En
such that E„ contains a splitting field of /over K and for each i > 1, Ei = £i_i(w,) with some positive power of w, lying in £,_i.
9. RADICAL EXTENSIONS
303
Conversely suppose that there exists such a tower of fields and that En contains a splitting field of /(that is, En contains all solutions of fix) = 0). Then En K{ui, . . » , wn)
and each solution is of the form f(uu . . . , u„)/giuu
ifg
e
K[xi,. . . , x„])
by Theorem 1.3. Thus each solution is expressible in terms of a finite number of ele ments of K, a finite number of field operations and uu . . . , un (which are obtained by extracting roots). But this amounts to saying that there is a “formula” for the solu tions of the particular given equations. These considerations motivate the next two definitions.
Definition 9.1. An extension field F of a field K is a radical extension oj K if F = K(ui, . . . , u„), some power of Ui lies in K and for each i > 2, some power of u; lies in K(ui, . . . , U;_i). REMARKS. If ur e Kinu .. . , 0 then is a root of ,vm — Uim e Kim, . . . , «i_i)[jr]. Hence Kiut,. . . , «,) is finite dimensional algebraic over K(uu .. . , 0 by Theorem 1.12. Therefore every radical extension Fof Kis finite dimensional algebraic over K by Theorems 1.2 and 1.11.
Definition 9.2. Let K be a field and f e K[x], The equation f(x) = 0 is solvable by radicals if there exists a radical extension ¥ of K and a splitting field E off over K
such that F D E D K.
Definition 9.2 is the first step in the formulation of the classical problem of find ing a “formula” for the solutions of fix) = 0 that is valid for every polynomial fe K[x\ of a given degree r (such as the quadratic formula for r = 2). For whatever the precise definition of such a “formula” might be, it is clear from the discussion preceding Definition 9.1 that the existence of such a “formula” should imply that every polynomial equation of degree r is solvable by radicals. Thus in order to demonstrate the nonexistence of such a formula, it suffices to prove that a specific polynomial equation is not solvable by radicals. We shall now develop the necessary information in order to do this (Corollary 9.5) and shall leave the precise formulation of the classical problem for the appendix.
Lemma 9.3. //F is a radical extension field o/K and N is a normal closure of F over K {Theorem 3.16), then N is a radical extension o/K. SKETCH OF PROOF. The proof consists of combining two facts, (i) If F is any finite dimensional extension of K (not necessarily radical) and N is the normal closure of F over K, then N is the composite field EXE2- ■ ■Er, where each Ei is a sub field of N which is /T-isomorphic to F. (ii) IfEi, . . . , Er are each radical extensions of
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CHAPTER V FIELDS AND GALOIS THEORY
X (as is the case here since F is radical), then the composite field EjE2 ■ ■ -E r is a radical extension of X. These statements are jusiified as follows. (i) Let {Wi,. . ., wn\ be a basis of F over X and let fi be the irreducible poly nomial of Wi over X. The proof of Theorem 3.16 shows that A is a splitting field of [f i,... ,f n \ over X. Let v be any root of f in A. Then there is a X-isomorphism a : K(wj) = K(v) such that a{wj) = v by Theorem 1.8. By Theorem 3.8 er extends to a X-automorphism r of N. Clearly r(F) is a subfield of A which is X-isomorphic to F and contains t(w,) = cr(Wj) = v. In this way we can find for every root v of every fi a subfield E of N such that v e E and E is X-isomorphic to F. If E x ,..., E r are the subfields so obtained, then Ei£2- • ■Er is a subfield of A which contains all the roots of f,h, ■ ■ ■ ,fn, whence E X E 2 ■ E r — N. (ii) Suppose r = 2, E, = K(uu . . ., uk) and E2 = K(vu . .., vm) as in Definition 9.1. Then EiE2 = K(ui,. . . , uk,vi, . . . , vm) is clearly a radical extension of K. The general case is similar. ■
Theorem 9.4. 7/F is a radical extension field of K and E is an intermediate field, then AuIkE is a solvable group.
PROOF. If K0 is the fixed field of E relative to the group Aut^E, then E is Galois over K0, Autk0E = AutKE and F is a radical extension of K0 (Exercise 1). Thus we may assume to begin with that E is algebraic Galois over K. Let N be a normal closure of Fover K(Theorem 3.16). Then A^is a radical extension of Xby Lemma 9.3 and E is a stable intermediate field by Lemma 2.13. Consequently, restriction (cr |—>• <j | E) induces a homomorphism 6 : AutKN—> AutA-E. Since is a splitting field over K (and hence over E) every a s AutA'E extends to a X-automorphism of N by Theorem 3.8. Therefore 6 is an epimorphism. Since the homomorphic image of a solvable group is solvable (Theorem II.7.11), it suffices to prove that Aut*:A is solvable. If Xj is the fixed field of N relative to Aut^A, then A is a radical Galois extension of Xi (Exercise 1) and AutA, A = AutA-A. Therefore, we may return to our original notation and with no loss of generality assume that F = E and F is a Galois radical extension of X. If F = K(ui.......... un) with «im* e X and Uimi s K(uu . . . , i) for i > 2, then we may assume that char X does not divide m,-. This is obvious if char X = 0. If char X = p 0 and nii = rp1 with (r,p) = 1, then e K(uu . . . , t/j_i) so that uxr is purely inseparable over K(uu . . . , «»_]). But F is Galois and thus separable over X (Theo rem 3.11), whence F is separable over X(«u . . . , «i_i) (Exercise 3.12). Therefore n't e K ( u { , . . . , j) by Theorem 6.2, and we may assume mt = r. If m = m\tn2- ■ ntn, then by the previous paragraph char X (= char F) does not divide m. Consider the cyclotomic extension F(f) of F, where f is a primitive wth root of unity (Theorem 8.1). The situation is this:
9. RADICAL EXTENSIONS
305
where F(f) is Galois over F (Theorem 8.1) and hence over K as well (Exercise 3.15(b)). The Fundamental Theorem 2.5 shows that AutaF = AutA'F(f)/Aut/F(f). Consequently, it suffices by Theorem II.7.11 to prove that Aut*F(f) is solvable. Ob serve that K(£) is an abelian Galois extension of K (Theorem 8.1), whence AutKK(£) = AutA-^n/AutAxn^f) by the Fundamental Theorem 2.5. If we knew that AutA(f)F(0 were solvable, then Theorem II.7.11 would imply that AutA'F(f) is solvable (since AuiKK(£) is abelian, hence solvable). Thus we need only prove that AutA(f)F(D is solvable. By assumption, F(f) is Galois over K and hence over any intermediate field. Let F0 = K(£) and Ei = K($,u u ■ ■ ■ , Ui) (i = 1,2, . . . , n) so that En = K(£,uu . . . , « „ ) = F(f). Let Hi = Aut£(F(f), the corresponding sub group of Autjc(nF(f) under the Galois correspondence. Schematically we have: F«) = En I---------- ► Hn = 1
Ei 1------- ► Hi = Aut£iF(f)
u
Ei~11----- ► = AuU-^Ftf)
m) = E01----------- ► Ho = AutA(f)F(f)
By Lemma 7.10(i) K(f) contains a primitive m,th root of unity for each i (i = 1,2, ... , n). Since Uimi e E,-_i and Ei — F;_i(h,), each Fa is a cyclic extension of Ei_i by Lemma 7.10 (ii) (with d = m.) and Theorem 7.11 (ii) (with n = mj. In par ticular, £, is Galois over F,_i. The Fundamental Theorem 2.5 implies that for each i = l,2,Hi <1 //,_! and H-i/H ~ Aut^F,, whence H^\/H l is cyclic abelian. Consequently, 1 = < Hr ,—i < • • ■ < Ha = AuU(f,F(r) is a solvable series (Definition II.8.3). Therefore, AutA(n^(f) *s solvable by Theo rem II.8.5. ■
Corollary 9.5. Let K be a field and f z Kfx], If the equation f(x) = 0 is solvable by radicals, then the Galois group off is a solvable group. PROOF. Immediate from Theorem 9.4 and Definition 9.2. ■ EXAMPLE. The polynomial /= xb — 4x -j- 2 e Q[jc] has Galois group Sb (see the example following Theorem 4.12), which is not a solvable group (Corollary II.7.12). Therefore, xs — 4x + 2 = 0 is not solvable by radicals and there can be no “formula” (involving only field operations and extraction of roots) for its solutions.
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CHAPTER V FIELDS AND GALOIS THEORY
Observe that the base field plays an important role here. The polynomial — 4x + 2 = 0 is not solvable by radicals over Q, but it is solvable by radicals over the field R of real numbers. In fact, every polynomial equation over R is solvable by radicals since all the solutions lie in the algebraic closure C = R(z') which is a radical extension of R.
x5
We close this section by proving a partial converse to Theorem 9.4. There is no difficulty if K has characteristic zero. But if char K is positive, it will be necessary to place some restrictions on it (or alternatively to redefine “radical extension” — see Exercise 2).
Proposition 9.6. Let E be a finite dimensional Galois extension field of K with solvable Galois group /)w/kE. Assume that char K does not divide [E : KJ. Then there exists a radical extension F o/K such that F ID E ID K. REMARK. The requirement that E be Galois over K is essential (Exercise 3). SKETCH OF PROOF OF 9.6. Since AutA-E is a finite solvable group, it has a normal subgroup H of prime index p by Proposition II.8.6. Since E is Galois over K, | Aut^l = [E : K] (Theorem 2.5), so that char K Jf p. Let N = E(f) be a cyclotomic extension of E, where f is a primitive /?th root of unity (Theorem 8.1). Let M = AT(f); then we have n = sen
K N is finite dimensional Galois over E (Theorem 8.1) and hence over K as well (Exer cise 3.15(b)). Now M is clearly a radical extension of K . Consequently, it will suffice (by Exercise 4) to show that there is a radical extension of M that contains N. First observe that Eis a stable intermediate field of A and K (Lemma 2.13). Thus restriction (
(i) Autm N is isomorphic under 6 to a proper subgroup of AutA'E; (ii) 6 : Autm N = AutAE. In either case Aut.^N is solvable (Theorem II.7.11) and TV is a finite dimensional Galois extension of K and hence of M. In case (i) [// : M] = |AutA///| < |Aut«E| = [E : K] = n, whence the inductive hypothesis implies that there is a radical exten sion of M that contains N. As remarked in the first paragraph, this proves the theo rem in case (i). In case (ii), lety = Since H is normal of indexpin AiUaE,./is normal of index p in AutA/N. Furthermore J is solvable by Theorem 11.7.11. If P is the fixed field of J (relative to AutA/N), then we have:
APPENDIX: THE GENERAL EQUATION OF DEGREE n
307
AU------------ 1 UA P ----------J = AutpA^
UA M -------- ► AutAiN
The Fundamental Theorem 2.5 implies that P is Galois over M and that AutMP = Aut M N/J. But [AutA/N \J] = p by construction, whence AutM P = Z p . Therefore, P is a cyclic extension of M and P = A/(m), where u is a root of some (irre ducible) x v — a e M[x\ (Theorem 7.11). Thus P is a radical extension of M and [Ar: P] < [N : M] = [F \ K] = n. Since A\x\PN = J is solvable and N is Galois over P (Theorem 2.5), the induction hypothesis implies that there is a radical extension F of P that contains N. F is a clearly radical extension of M (Exercise 4). This completes the proof of case (ii). ■
Corollary 9.7. Let K be a field and f e K[x] a polynomial of degree n > 0, where char K does not divide n! (which is always true when char K = 0). Then the equation f(x) = 0 is solvable by radicals if and only if the Galois group off is solvable. SKETCH OF PROOF. (<=) Let £ be a splitting field of / over K. In view of Proposition 9.6 we need only show that E is Galois over K and char KJ([E : X]. Since char K Jf n\ the irreducible factors of /are separable by Theorem III.6.10 and Exercise III.6.3, whence E is Galois over K (Theorem 3.11 and Exercise 3.13). Since every prime that divides [E: K] necessarily divides nl (Theorem 3.2), we conclude that char K )f[E : X], ■
APPENDIX: THE GENERAL EQUATION OF DEGREE n The motivation for our discussion can best be seen by examining polynomial equations of degree 2 over a field K with char K j* 2. Here and below there will be no loss of generality in restricting consideration to monic polynomials. If h and t2 are indeterminates, then the equation x2
—
tlX
+ ^2 = 0
over the field X(/i,/2) of rational functions in ti,t2 is called the general quadratic equa tion over K. Any (monic) quadratic equation over K may be obtained from the general quadratic equation by substituting appropriate elements of K for ti and /2. It is easy to verify that the solutions of the general quadratic equation (in some algebraic closure of K(ti,t2)) are given by: _ 1 1 ± -yjti 2 — 4/2
CHAPTER V FIELDS AND GALOIS THEORY
308
where n = h\k for be Z. This is the well known quadratic formula. It shows that the solutions of the general quadratic equation lie in the radical extension field K(ti,t2)(u) with u2 = 112 — 4t2. In order to find the solutions of x2 — bx + c — 0 (b,c e K) one need only substitute b,c for tut2. Clearly the solutions lie in the radical extension K(u) with u2 = b2 — 4c £ K. We now generalize these ideas to polynomial equations of arbitrary degree. Let K be a field and n a positive integer. Consider the field K(tu of ra tional functions over K in the indeterminates ti,..., tn. The polynomial P n ( x ) = X n — /I*"-1 + t2Xn~2 H-------------------1- (— I)"-1/„_!.* + (— l)ntn
s K(tu . . . ,
/„)[*]
is called the general polynomial of degree n over K and the equation p„(x) = 0 is called the general equation of degree n over K} Note that any (monic) polynomial of degree n in K[ x\, say f(x) = xn + axxn~x + • • • + an~ix + a„ may be obtained from the general polynomial p„(x) by substituting (—1)^, for The preceding discussion makes the following definition quite natural. We say that there is a formula for the solutions of the general equation of degree n provided that this equation is solvable by radicals over the field K(tu . . ., tn). If pn(x) = 0 is solvable by radicals, then the solutions of any (monic) polynomial equation of degree n over K may be found by appropriate substitutions in the solutions of pn(x) = 0. Having precisely formulated it, we can now settle the classical problem with which this section was introduced.
Proposition 9.8. (Abel) Let K be a field and n a positive integer. The general equa tion of degree n is solvable by radicals only if n < 4. REMARKS. The words “only if” in Proposition 9.8 may be replaced by “if and only if’ when char K = 0. If radical extensions are defined as in Exercise 2, then “only if” may be replaced by “if and only if” for every characteristic. The fact that the general equation of degree n is not solvable by radicals for n > 5 does not exclude the possibility that a particular polynomial equation over K of degree n > 5 is solvable by radicals. SKETCH OF PROOF OF 9.8. Let the notation be as above and let « i , . . . , un be the roots of pn(x) in some splitting field F = K(tu . .., t n )(ui,..., un). Since P n ( x ) = (x — ui) (x — u 2 )- ( x — un) in F, a direct calculation shows that 71 fi =
i —1
t2 =
22
1
, tn = uiu2- -
that is, ti = fX ui,. . ., un) w h e r e / , a r e the elementary symmetric functions in n indeterminates (see the appendix to Section 2). It follows that F = K(uu . . . , un). Now consider a new set of indeterminates {xu ■ ■ -, xn} and the field K(xu . . . , x„). Let E be the subfield of all symmetric rational functions in K(xi, . . . , x„). The basic idea of the proof is to construct an isomorphism of fields F K(xu . . . , xn) such that K(ti,. . ., O is mapped onto E. Then the Galois group AutAUl........ t n )F, of p„(x) will be isomorphic to AuU’AT(.vi,.. ., x„). But AutKX(xi, . . . , x„) is isomorphic to 3The
signs ( — 1Y are inserted for convenience in order to simplify certain calculations
APPENDIX: THE GENERAL EQUATION OF DEGREE n
309
Sn (see p. 253). Sn is solvable if and only if n < 4 (Corollary II.7.12 and Exercise II.7.10). Therefore, if pn(x) = 0 is solvable by radicals then n < 4 by Corollary 9.5. [Conversely if n < 4 and char K = 0, then p„ix) = 0 is solvable by radicals by
Corollary 9.7.] In order to construct the isomorphism F S K(xi,. . . , jt„) we first observe that the subfield E of K(xu . .., x„) is precisely K(f u . . . , / , ) by Theorem 2.18, where are the elementary symmetric functions. Next we establish a ring isomor phism K[t u . . . , t n ] = K[f , . . . , / , ] as follows. By Theorem III.5.5 the assignment g(fi,. . ., r n ) t -* gifi, ...,/„) (in particular u \-> f) defines an epimorphism of rings 8 : K[t\,. .., t n ] -* K [ f , . . . ,/„]. Suppose g(tu . . . , ?„)h0, so that£(/,...,/,) = () in K [ f , ...,/„] CZ K(xu . . . , xn). By definition ft: fkiXl, • ■ • , Xn)
^1
X{ l Xi2'
' ‘ Xifc
1
and hence 0 = g(/i, . . . , / , ) = g(f(x u • ■ • , -vn), - - - , fix i, . . . , jv„)). Since gi f i, . . . , / „ ) is a polynomial in the indeterminates x u . . . , x n over K and F = Ki u\,. . . , un) is a field containing K, substitution of «, for jv, yields: 0 = g(/l(«l, • - • , Un), - - - , fin 1, ■ • • , Un)) = git 1, • - ■ , tn)',
thus 8 is a monomorphism and hence an isomorphism. Furthermore 8 extends to an isomorphism of quotient fields 6 : K{tu . . . , t„) ~ Kif u . . ., fn) = E (Exercise III.4.7). Now F = Kiu,,. . . ,u n ) is a splitting field over K{tu . . . , / „ ) of pn{x) and under the obvious map on polynomials induced by 8,pn{x) [-> Pnix) = xn — /i*"'1 + f2xn~2 — - • •+ ( — 1)”/ = ix — joX* — x2) ■ ix — xn) (see p. 252). Clearly K{xi,. . . , xn) is a splitting field of p n {x) over Kif u ...,f n ) = E. Therefore by Theo rem 3.8 the isomorphism 6 extends to an isomorphism F ~ Kixu . . . , x„) which by construction maps K{tx , . . . , r„) onto E as desired. ■
EXERCISES 1. If F is a radical extension field of K and E is an intermediate field, then F is a radical extension of E. 2. Suppose that “radical extension” is defined as follows: F is a radical extension of K if there is a finite tower of fields K = E() d E, (Z • ■ CZ En = F such that for each 1 < / ' < « , Ei = Ej^iui) and one of the following is true: (i) Uimi e E{_, for some mi > 0; (ii) char K = p and uv — we i. State and prove the analogues of
Theorem 9.4. Proposition 9.6, Corollary 9.7, and Proposition 9.8. 3. Let K be a field, /= K[x] an irreducible polynomial of degree n > 5 and F a.split ting field of/over K. Assume that Aut*/7 = S n . (See the example following Theo rem 4.12). Let u be a root of /in F. Then (a) K{u) is not Galois over K\ [Kiu) : K] = n and AutA^(«) = 1 (and hence is solvable). (b) Every normal closure over K that contains u also contains an isomorphic copy of F. (c) There is no radical extension field E of K such that E 3 K{u) ZD K. 4. If F is a radical extension field of E and E is a radical extension field of K, then F is a radical extension of K.
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CHAPTER V FIELDS AND GALOIS THEORY
5. (Cardan) Let K be a field with char K 2,3 and consider the cubic equation n , r °12 J 2fll3 «lfl2 , x3 + a\x 2 + o 2 x + a% = 0 (a* e K). Let p = a 2 --------and q = —-------------- f- a 3 . Let P = V—q /2 + V /?3/27 + g2/4 and Q = V—q /2 — \ f p 3 /27 + q 2 /A (with cube roots chosen properly). Then the solutions of the given equation are P + Q — ai/3; 00P + u?Q — «i/3; and a>2P + <*>Q — Ci/3 where a> is a primitive cube root of unity.
CHAPTER
VI
THE STRUCTURE OF FIELDS
In this chapter we shall analyze arbitrary extension fields of a given field. Since algebraic extensions were studied in some detail in Chapter V, the emphasis here will be on transcendental extensions. As the first step in this analysis, we shall show that every field extension K CZ F is in fact a two-step extension K d E CZ F, with F algebraic over E and £ purely transcendental over K (Section 1). The basic concept used here is that of a transcendence base, whose cardinality (called the transcendence degree) turns out to be an invariant of the extension of K by F (Section 1). The notion of separability is extended to (possibly) nonalgebraic extensions in Section 2 and separable extensions are characterized in several ways.
1. TRANSCENDENCE BASES The first part of this section is concerned with the concept of algebraic inde pendence, which generalizes the idea of linear independence. A transcendence base of a field F over a subfield K is the analogue (with respect to algebraic independence) of a vector space basis of F over K (with respect to linear independence). The cardi nality of a transcendence base of F over K (the transcendence degree) is shown to be an invariant and its properties are studied. In this section we shall frequently use the notation u/v for uv 1 , where u,v are elements of a field and v ^ 0. Throughout this section K denotes a field.
Definition 1.1. Let F be an extension field ofK and S a subset of F. S is algebraically dependent over K if for some positive integer n there exists a nonzero polynomial f e K[xi, . . . , xn] such that f(si, — , S„) = 0 for some distinct Si, . . . , s„ e S. S is algebraically independent over K ifS is not algebraically dependent over K. 311
312
CHAPTER VI THE STRUCTURE OF FIELDS
REMARKS. The phrase “over K" is frequently omitted when the context is clear. A subset S' of F is algebraically independent over K if for all n > 0, / e A T [ x i , a n d distinct si, — , s „ e S, f(si,. . . ,
Sn)
= o => /=
0.
Every subset of an algebraically independent set is algebraically independent. In particular, the null set is algebraically independent. Every subset of K is clearly algebraically dependent. The set {«} is algebraically dependent over K if and only if u is algebraic over K. Clearly every element of an algebraically independent set is necessarily transcendental over K. Hence if Fis algebraic over K, the null set is the only algebraically independent subset of F. Algebraic (in)dependence may be viewed as an extension of the concept of linear (in)dependence. For a set S is linearly dependent over K provided that for some positive integer n there is a nonzero polynomial f of degree one in K[x\ , . . . , xn] such that f(si, . . . , * , ) = 0 for some distinct Si e S. Consequently, every algebraically /^dependent set is also linearly independent, but not vice versa; (see the Example after Definition 1.4 below). EXAMPLE. Let AT be a field. In the field of rational functions K(xi , . . . , xn) the set of indeterminates {xi,_____ _ xn] is algebraically independent over K. More generally, we have:
Theorem 1.2. Let F be an extension field o/'K and { S i , . . . , s„} a subset ofF which is algebraically independent over K. Then there is K-isomorphism K(si,. . . , sn) = K(xi, . . . , xn). SKETCH OF PROOF. The assignment 1—> Si defines a AT-epimorphism of rings 6 : K[x x , X[s,, by Theorems III.5.5 and V.1.3. The algebra ic independence of {st, . . . , s„) implies that 8 is a monomorphism. By Corol lary III.4.6 6 extends to a Af-monomorphism of fields (also denoted 8) K(xu . . . , xn) —> K(su . . . , sn) such that 8( f/g) = f(s u . . . , s n )/g(s u .. ,,s n ) = f( s u . . . , s„)g(si, . . . , s„)~l. 6 is an epimorphism by Theorem V.1.3(v). ■
Corollary 1.3. For i = 1,2 let F, be an extension field of Ki and Si CZ Fi with Si algebraically independent over Ki. If (p : Si —♦ S2 is an injective map of sets and a : Ki —> K2 a monomorphism of fields, then a extends to a monomorphism of fields d : Ki(Si) —» K2(S2) such that ct(s) =
REMARK. In particular, the corollary implies that every permutation of an algebraically independent set S over a field K extends to a X-automorphism of K(S); (just let Ki = K = AT2 and a = 1A). SKETCH OF PROOF OF 1.3. For each n > 1 cr induces a monomorphism of rings Ki [xi, . . ., jc„] —> Ay*i, - - - , x 7 L ] (also denoted a; see p. 235). Every element of
1. TRANSCENDENCE BASES
313
Ki(Si) is of the form f(su . . . , sn)/g(su .. . ,s n ) (sf e Si) by Theorem V.1.3. For con venience we write
1
j • * • j Sn)fg( s 11 • • ■ i Sn) | * Crf(i£ S\, . * . , tpSji)/(Tg(tpS\, • • • , ^n) £ K(S'/)•
For any finite subset {Sj,. .., sr} of S', the restriction of a to Kt(su . . ., s,) is the composition Kl(si, . . . , sr) —* Ai(Xi, . . . , Xr) —*• Kz(xi, . . . , Xr) —* K2(ifS\, . . . ,
where the 0, are the Ari-isomorphims of Theorem 1.2 and a is the unique monomor phism of quotient fields induced by a\K x [x u .. ., x r ] —1- K 2 [x u . . ., x rJ and given by &( f/g) = (of)/(erg) (Corollary III.4.6). It follows that a is a well-defined monomor phism of fields. By construction a extends a and agrees with
Definition 1.4. Let F be an extension field ofK. A transcendence base (or basis) of F over K is a subset S ofF which is algebraically independent over K and is maximal (with respect to set-theoretic inclusion) in the set of all algebraically independent sub sets of F.
The fact that transcendence bases always exist follows immediately from a Zorn’s Lemma argument (Exercise 2). If we recall the analogy between algebraic and linear independence, then a transcendence base is the analogue of a vector-space basis (since such a basis is precisely a maximal linearly independent subset by Lemma IV.2.3). Note, however, that a transcendence base is not a vector-space basis, al though as a linearly independent set it is contained in a basis (Theorem IV.2.4). EXAMPLE. If f/g = f( x)/ g(x) e K(x) with fig?* 0, then the nonzero poly nomial h(yuy2) = g(y\)y2 - /(>'i) e K[yuy2] is such that h(x,f/g) = g(x)[f(x)/g(x)] — f(x ) = 0. Thus {x, f/g} is algebraically dependent in K(x). This argument shows that {x} is a transcendence base of K(x) over K. The set {x| is not a basis since {l*,x,x2,x3,. ..} is linearly independent in K(x). In order to obtain a useful characterization of transcendence bases we need
Theorem 1.5. Let F be an extension field of K, S a subset of F algebraically inde pendent over K, and u e F — K(S). Then S U juj is algebraically independent over K if and only if u is transcendental over K(S). PROOF. (<=) If there exist distinct s t , , sn-i e S and an /e K[xi , . . . , xn] such that f(su . .., sn-Uu) = 0, then u is a root of f(su . .., 5„_1,xri) e K(S)[x n \. Now fe K[x u - - - , x„] = K[x u . . . , x„_i][x„], whence / = hTxnr -f hr~ixr~l H--------------- [hixn -f h0 with each hi e K[xi,. .., xn_i]. Since u is transcendental over K(S), we have f(si,..., vi,x„) = 0. Consequently, hi(si,..., sn_{) = 0 for every i. The algebraic independence of S implies that hi = 0 for every /, whence/ = 0. Therefore S U j«) is algebraically independent.
314
CHAPTER VI THE STRUCTURE OF FIELDS n
(=>) Suppose f(u) = 0 where / =
52 s
By Theorem V.1.3 there is a
i=0
finite subset { j i , . . . , sr} of 5 such that a, e K(su . . . , sr) for every i, whence at = fi(su ..s r )/gi(su . .. ,s r ) for some f,gi e A:[x,,.. ., x r ]. Let g = gig2- ■ gn z_K[xi, . . . , x,] and for each i let / = figi - ■ -gi ig,>i- • gn e K[xi ,. .., xr\. Then a, = f(s, , . . . , Sr)/g (si,..., s r ) and f(x) =
52 = 52
» Sr)/g(su . . . , Sr)xl
gi.S\ , > » • j Sr) ^ l j - • - > Sr)x ).
(All we have done is to factor out a “common denominator” for the coefficients of /.) Let h(xi , . . . , xr,x) = 53 Mxu - - - , xr)x{ e AT[jci, .. . , xr,x]. Since f(u) = 0 and g ( 5 l 5 . . . , .sy)-1 5** 0, we must have h(s\ , . . . , sT,u) = 0. The algebraic independence of 5 U («) implies that h = 0, whence / = 0 for every /. Thus each a, = 0 and / = 0. Therefore u is transcendental over K(S). ■
Corollary 1.6. Ler F an extension field o/K anJS « subset o/F that is algebraically independent over K. Then S is a transcendence base of F over K if and only if F is algebraic over K(S). PROOF. Exercise. ■ REMARKS. A field F is called a purely transcendental extension of a field K if F — K(S), where S CZ F and S is algebraically independent over K. In this case S is necessarily a transcendence base of F over K by Corollary 1.6. If F is an arbitrary ex tension field of K, let S be a transcendence base of F over K and let E = K{S). Corollary 1.6 shows that F is algebraic over E and E is purely transcendental over K. Finally Corollary 1.6 and the remarks after Definition 1.1 show that F is an algebraic extension of K if and only if the null set is a transcendence base of F over K. In this case the null set is clearly the unique transcendence base of F over K.
Corollary 1.7. If F is an extension field o/K and F is algebraic over K(X) for some subset X of F (in particular, if F = K(X)), then X contains a transcendence base ofF over K. PROOF. LetS be a maximal algebraically independent subset of X (S exists by a routine Zorn’s Lemma argument). Then every u eX — S is algebraic over K(S) by Theorem 1.5, whence K(X) is algebraic over K(S) by Theorem V.1.12. Consequently, F is algebraic over K(S) by Theorem V.1.13. Therefore, S is a transcendence base of F over K by Corollary 1.6. ■ As one might suspect from the analogy with linear independence and bases, any two transcendence bases have the same cardinality. As in the case of vector spaces, we break the proof into two parts.
Theorem 1.8. Let F be an extension field of K. // S is a finite transcendence base of F over K, then every transcendence base of F over K has the same number of elements as S.
1. TRANSCENDENCE BASES
315
SKETCH OF PROOF. Let 5 = {si,________ , s„} and let T be any transcendence base. We claim that some h e T is transcendental over K(s2 , . . . , s„). Otherwise every element of T is algebraic over K(s2 , . . . , .v„), whence K(s2 , . . . , sn)(T) is algebraic over K(s- > , . . . , sn) by Theorem V.1.12. Since F is algebraic over K(T) by Corollary 1.6, F is necessarily algebraic over K(T)(s2,. . . , s„) = K(s2 , . . . , sn)(T). Therefore, F is algebraic over K(s2 , . . . , s„) by Theorem V.1.13. In particular, sx is algebraic over K(s2 , . . . , sn), which is a contradiction (Theorem 1.5). Hence some t\ e T is transcen dental over K(s2, . . . , sn). Consequently, {h,s2 , . . . , 5n I is algebraically independent by Theorem 1.5. Now if si were transcendental over K(tus2 , . . . , j„), then j ti,sus2, . . . , j„) would be algebraically independent by Theorem 1.5. This is obviously impossible since S is a transcendence base. Therefore, 5) is algebraic over K(tus2 , . . . , sn). Consequently, K(S)(ti) = K(tus2 , . . . , sn)(si) is algebraic over K(ti,s 2 ,... ,s n ) (Theorem V.1.12), whence F is algebraic over K(ti,s2 , . . . , s„) (Theorem V.1.13 and Corollary 1.6). Therefore, . . . , sn) is a transcendence base of F over K by Corollary 1.6. A similar argument shows that some t2 e T is transcendental over K(tus3 , . . . , sn), whence {t2,ti,s3 , . . . , sn } is a transcendence base. Proceeding inductively (inserting a U and omitting an Si at each stage) we eventually obtain t\,t2, e T such that {fi,.. ., tn] is a transcendence base of F over K. Clearly, we must have T= and hence |S| = |7’|. ■
Theorem 1.9. Let F be an extension field ofK. 7/S is an infinite transcendence base of F over K, then every transcendence base of F over K has the same cardinality as S.
PROOF. If T is another transcendence base, then T is infinite by Theorem 1.8. If s s S, then s is algebraic over K(T) by Corollary 1.6. The coefficients of the irreducible polynomial /of s over K(T) all lie in K(TS) for some finite subset Ts of T (Theorem V.1.3). Consequently,/e A'(7's)[a] and s is algebraic over K(TS). Choose such a finite subset Ts of T for each s e S. We shall show that (J T„ is a transcendence base of F over K. Since U T, a T, stS
s
this will imply that IJ Ts = T. As a subset of T the set U T. is algebraically inS
S
dependent. Furthermore every element of 5 is algebraic over X(U Ts). ConseS
quently, X((J Ts)(S) is algebraic over X((J 7^) by Theorem V.1.12. Since K(S) CZ S 8 HU Ts)(S), every element of K(S) is algebraic over X((J Ts). Since F is algebraic S
s
over K(S) by Corollary 1.6, F is also algebraic over X(|J Ts) (see Theorem V.1.13). £ Therefore, by Corollary 1.6 again U T„ is a transcendence base, whence \J T, = T. S
s
Finally we shall show that \T\ < |S|. The sets Ts need not be mutually disjoint and we remedy this as follows. Well order the set S (Introduction, Section 7) and de note its first element by 1. Let 7V = Tx and for each 1 < s eS, define T„‘ = T, — U Ti - Clearly each T/ is finite. Verify that U T» — U Ts' and that the T,' are « <S
S
s
mutually disjoint. For each s tS, choose a fixed ordering of the elements of 77 : tut2,
316
CHAPTER VI THE STRUCTURE OF FIELDS
. . . , tk„. The assignment u H► CM) defines an injective map U ^'->5 X N*. ThereS
fore by Definitions 8.3 and 8.4 and Theorem 8.11 of the Introduction we have: \T\ = IU n = IU T11 < IS X N*| = |S||N*|= |S|K0= |S|. fi fi
Reversing the roles of S and T in the preceding argument shows that |S| < \T\ , whence |S|= |7'| by the Schroeder-Bemstein Theorem 8.6 of the Introduction. ■
Definition 1.10. Let F be an extension field of K. The transcendence degree o/F over K (denoted tr.d.F/¥L) is the cardinal number |S), where S is any transcendence base o/F over K. The two preceding theorems show that tr.d .F/K is independent of the choice of S. In the analogy between algebraic and linear independence tr.d .F/K is the analogue of the vector space dimension [F.K ], The remarks and examples after Definition 1.4 show that tr.d .F/K < [F : K] and that tr.d .F/K = 0 if and only if F is algebraic over K.
Theorem 1.11. If F is an extension field of E and E an extension field of K, then tr.d. F/K = (tr.d. F/E) + (tr.d.E/K).
PROOF. Let S be a transcendence base of E over K and T a transcendence base of F over E. Since S d E, S is algebraically dependent over E, whence S fl T = 0. It suffices to show that S U T is a transcendence base of F over K, since in that case Definition 1.10 and Definition 8.3 of the Introduction imply tr.d .F/K = |S U T\ = |7*| + \S\ = (tr.d .F/E) + (tr.d .E/K). First of all every element of E is algebraic over K(S) (Corollary 1.6) and hence over K(S U T). Thus K(S (J T)(E) is algebraic over K(S (J T) by Theorem V.1.12. Since K(S (J D = K(S)(T) CZ E(T) d K(S U T)(E), E(T) is algebraic over K(S U T). But F is algebraic over E(T) (Corollary 1.6) and therefore algebraic over K(S U T) by Theorem V.1.13. Consequently, it suffices by Corollary 1.6 to show that S U Tis algebraically independent over K. Let / be a polynomial over K in n + m variables (denoted for convenience xi, . . . , xn,yu . . ., ym) such that f(su . . . , s n ,ti ,... ,t m ) — 0 for some distinct Si,. . ., s n eS, h , . . . , tm e.T. Let g = g(yi,. . ,,y m ) = f(su . . ., Sn,yi,..., >w) s tf(S)[y,, . . . , y n ] d E[y u y m ). Since g(tu . . . , tn) = 0, the algebraic inde pendence of T over E implies that g = 0. Now / = f(xi , . . . , xn,yi, . . . , ym) r
= 52 hi(xi , . . . , xn)ki(yu . .. , ym) with hx e AT[x,,. .. , xn], /c, e K[y\,. . ., >’mi. Hence » = i
o = g(yu .. • ,y m ) = f(su . . . , Sn,yu • ■ ■ ,y m ) implies that hi(su ...,s n ) = 0 for every i. The algebraic independence of S over K implies that hi — 0 for all whence f( x i, . . . , xnvyi,. .. , y m ) = 0. Therefore S U T is algebraically independent over K. ■
1. TRANSCENDENCE BASES
317
If Kv and K2 are fields with algebraic closures, Fj,F2 respectively, then Theorem V.3.8 implies that every isomorphism Ki ^ K2 extends to an isomorphism Fi = F2. Under suitable hypotheses this result can now be extended to the case where the fields Fi are algebraically closed, but not necessarily algebraic over Ki.
Theorem 1.12. Let Fi [resp. F2] be an algebraically closed field extension of a field Ki [resp. K2], If tr.d.Fi/Ki = /r.^.F2/K2, then every isomorphism of fields Ki = K2 extends to an isomorphism Fi = F2. PROOF. Let Si be a transcendence base of F,- over Kt. Since |Si| = |S2|, a : Kx ^ K2 extends to an isomorphism a : Ki(Si) = K^S2) by Corollary 1.3. F, is algebraically closed and algebraic over A,(St) (Corollary 1.6) and hence an algebraic closure of Kt{Si). Therefore a extends to an isomorphism Fi = F2 by Theorems V.3.4 and V.3.8. ■
EXERCISES Note: F is always an extension field of a field K.
1. (Exchange property) Let S' be a subset of F. If u e F is algebraic over K(S) and u is not algebraic over K(S — { u } ) , where v e S , then v is algebraic over K((S- M)U { « } ) . 2. (a) Use Zorn’s Lemma to show that every field extension possesses a trans cendence base. (b) Every algebraically independent subset of F is contained in a transcendence base. 3. {xi , . . . , x„} is a transcendence base of K(x i y . . . , xn). 4. If EiJEz are intermediate fields, then (i) ir.d.EiEi/K > tr.d.Ei/K for i = 1,2; (ii) tr.d.E1E2/K < (tr.d.EJK) + (tr.d.E2/K). 5. If F = K(uu . . . , «n) is a finitely generated extension of K and E is an inter mediate field, then E is a finitely generated extension of K. [Note: the algebraic case is trivial by Theorems V.1.11 and V.1.12.] 6. (a) If S is a transcendence base of the field C of complex numbers over the field Q of rationals, then S is infinite. [Hint: Show that if S is finite, then 10(5)1 = IQ(*1, • • •, *»)l = IQ[*>, - - -, *»]| = IQI < |C| (see Exercises 8.3 and 8.9 of the Introduction and Theorem 1.2). But Lemma V.3.5 implies |Q(S)| = |C|.] (b) There are infinitely many distinct automorphisms of the field C. (c) tr.d.C/Q = |C|. 7. If F is algebraically closed and E an intermediate field such that tr.d.E/K is finite, then any A'-monomorphism E- ^F extends to a ^-automorphism of F. 8. If F is algebraically closed and tr.d.F/K is finite, then every A'-monomorphism F—» F is in fact an automorphism.
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CHAPTER VI THE STRUCTURE OF FIELDS
2. LINEAR DISJOINTNESS AND SEPARABILITY The chief purpose of this section is to extend the concept of separability to (possibly) nonalgebraic field extensions. This more general concept of separability will agree with our previous definition in the case of algebraic extensions (Theorem 2.8). We first introduce the idea of linear disjointness and develop its basic properties (Theorems 2.2-2.7). Separability is defined in terms of linear disjointness and char acterized in several ways (Theorem 2.10). Other properties of separability are de veloped in the corollaries of Theorem 2.10. In the following discussion all fields are assumed to be subfields of some (fixed) algebraically closed field C.
Definition 2.1. Let C be an algebraically closed field with subfields K,E,F such that K C E fl F. E and F are linearly disjoint over K if every subset ofE which is linearly independent over K is also linearly independent over F. REMARKS. An alternate definition in terms of tensor products is given in Ex ercise 1. Note that a subsets of E is linearly independent over a subfield of C if and only if every finite subset of A" is. Consequently, when proving linear disjointness, we need only deal with finite linearly independent sets. EXAMPLE. If K d E then E and K are trivially linearly disjoint over K. This fact will be used in several proofs. Other less trivial examples appear in the theorems and exercises below. The wording of Definition 2.1 suggests that the definition of linear disjointness is in fact symmetric in E and F. We now prove this fact.
Theorem 2.2. Let C be an algebraically closed field with subfields K,E,F such that K C E fl F. Then E and F are linearly disjoint over K if and only if F and E are linearly disjoint over K. PROOF. It suffices to assume Eand F linearly disjoint and show that F and E are linearly disjoint. Suppose X d F is linearly independent over K, but not over E so that n«i + • • • + rr,un = 0 for some u t eX and r, e E not all zero. Choose a subset of {n, } which is maximal with respect to linear independence over K ; reindex if t necessary so that this set is \r u r?,. . . , r t \ (t > 1). Then for each j > t, rj = 23 with aij e K (Exercise IV.2.1). After a harmless change of index we have: n
t
o = 23
r>ui
j=1
- 23 (+
=X
k = 1\j
71 ri“t
j=1
/t
+ 23 ( 23
23 a kjUj =(+i /
\
Oi,n
j = t+1 V = i /
V
Jr/:•
Since E and Fare linearly disjoint, { r i , . . . , r t ] is linearly independent over F which
2. LINEAR DISJOINTNESS AND SEPARABILITY
implies that u k + 22
319
= 0 for every k < t. This contradicts the linear inde-
j=t+1
pendence of X over K. Therefore X is linearly independent over E. ■ The following lemma and theorem provide some useful criteria for two fields to be linearly disjoint.
Lemma 2.3. Let C be an algebraically closed field with subfields K,E,F such that K C E fl F. Let R be a subring ofE such that K(R) = E and K CZ R (which implies that R is a vector space over K). Then the following conditions are equivalent: (i) E and F are linearly disjoint over K; (ii) every subset of R that is linearly independent over K is also linearly inde pendent over F; (iii) there exists a basis of R over K which is linearly independent over F. REMARK. The lemma is true with somewhat weaker hypotheses (Exercise 2) but this is all that we shall need. PROOF OF 2.3. (i) => (ii) and (i) => (iii) are trivial, (ii) => (i) Let X = { w i , } be a finite subset of E which is linearly independent over K. We must show that A' is linearly independent over F. Since w< e E = K(R) each iu is of the form iu = c\df1 = d/d,, where c, = f(r u - • - , r t i ), 0 ^ dx = g,(n,. . . , rti) with r3 e R and f,gi e K[ xu (Theorem V.1.3). Let d=did 2 -- d n and for each /' let i\ = Cidi- ■ -di^id,+i- ■ -d„ e R. Then ux = Vid~l and the subset X' = {t>i, . . . , i\,} of R is linearly independent over a subfield of C if and only if X is. By hypothesis X and hence X' is linearly independent over K. Consequently, (ii) implies that X' is linearly independent over F, whence X is linearly independent over F. (iii) => (ii) Let U b e a basis of R over K which is linearly independent over F. We must show that every finite subset X of R that is linearly independent over K is also linearly independent over F. Since A' is finite, there is a finite subset U\ of U such that X is contained in the A"-subspace V of R spanned by tA; (note that tA is a basis of V over K). Let V\ be the vector space spanned by £A over F. U and hence U\ is linearly independent over F by (iii). Therefore U x is a basis of V\ over F and d.\mKV = dinvFi. Now X is contained in some finite basis W of V over K (Theorem IV.2.4). Since W certainly spans Vt as a vector space over F, W contains a basis W\ of V\ over F. Thus \Wi\ < \IV\ = dim/,-!/ = = \ Wi\, whence W = W,. Therefore, the subset A" of W is necessarily linearly independent over F. ■
Theorem 2.4. Let C be an algebraically closed field with subfields K,E,L,F such that K CZ E and K (Z L (Z F. Then E and F are linearly disjoint over K if and only ifiS) E and L are linearly disjoint over K and (ii) EL and F are linearly disjoint over L. PROOF. The situation looks like this:
320
CHAPTER VI THE STRUCTURE OF FIELDS
EF
EL
F
K
(<=) If a subset * of E is linearly independent over K, then* is linearly inde pendent over L by (i). Therefore (since* CZ E CZ EL),X is linearly independent over F by (ii). (=>) If E and F are linearly disjoint over K, then E and L are automatically linearly disjoint over K. To prove (ii) observe that EL = L(R), where R is the subring L[E\ of C generated by L and E. By Theorem V.l .3 every element of R is of the form f( ei, . . . , e„) ( et eE;fz L[x u . . . , x„]). Therefore, any basis U of E over K spans R considered as a vector space over L . Since E and L are linearly disjoint over K , U is linearly independent over L. Hence U is a basis of R over L. But U is linearly inde pendent over F by the linear disjointness of E and F. Therefore, EL and F are linearly disjoint over L by Lemma 2.3. ■ Next we explore linear disjointness with respect to certain extension fields of K that will play an important part in the definition of separability.
Definition 2.5. Let K be a field of characteristic p ^ 0 and let C be an algebraically closed field containing K. For each integer n > 0 K1/pn = {u eC | upn e K}. K1/pCO = U K*'pn = {u e C | upn £ K for some n > 0|. n >0
REMARKS. Since (u ± v)vn = up*‘ ± vpn in a field of characteristic p (Exercise III.l.ll) each KllpTl is actually a field. Since K = KvC K l / p n CZ K11?™ CZ Ku** for all n,m such that 0 < n < m, it follows readily that KllpCO is also a field. The fact that C is algebraically closed implies that K11^ is a splitting field over K of the set of poly nomials — k | k e K} (Exercise 5). In particular, every k e K is of the form vvTl for some v £ Kllpn. Since K11^ is a splitting field over K, it is essentially independent of C (that is, another choice C would yield an isomorphic copy of K 1 / p T I by Theorem V.3.8).
Lemma 2.6. If F is an extension field of K of characteristic p ^ 0 and C is an alge braically closed field containing F, then for any n > 0 a subset X o/F is linearly inde pendent over K1/pn if and only //Xpn = |up" | u e X} is linearly independent over K. Furthermore X is linearly independent over K1/pCO if and only ifX is linearly independent over K1/pn for all n > 0.
2. LINEAR DISJOINTNESS AND SEPARABILITY
321
SKETCH OF PROOF. Every a s K is of the form a = vpTl for some v e Kl>pn (Exercise 5). For the first statement note that 23 cnUipn = 0 (o, e K; m eA")o T. VipnUiPn = 0 (ii £ K1,pn and Vipn = ai) <=> (23 i\Ui)pn = 0 <=> 2Z i ii
=
F°r the
t second statement observe that if 23 WiUi = 0 (w, e KllpCO; Ui £ X), then for a large i=i enough n, w\, . . . , wt £ KllpTl. ■
Theorem 2.7. Let ¥ be a field contained in an algebraically closed field C. IfF is a purely transcendental extension of a field K of characteristic p ^ 0, then F and K1/pn are linearly disjoint over K for all n > 0 and F and¥JlrF° are linearly disjoint over K. PROOF. Let F = K(S) with S a transcendence base of F over K. If S = 0, then F = K and every linearly independent subset of F over K consists of exactly one nonzero element of K. Such a nonzero singleton is clearly linearly independent over any subfield of C whence the theorem is true if 5 = 0. If 5 is not empty let M be the set of monomials over 5 (that is, the set of all finite products of elements of S). Then M is linearly independent over K since 5 is algebraically independent over K. By Theorem V.1.3 M spans the subring X[S] (considered as a vector space over K). Therefore, M is a basis of A|S] over K. The algebraic independence of 5 implies that for every n > 0, Mpn = | mpn | m e Mj is linearly independent over K. By Lemma 2.6 M is linearly independent over K V p n for every n and hence over KllpC°. Therefore, for each 0 < /7 < co j Fand KUpTI are linearly disjoint over K by Lemma 2.3 (with X[S], F, KllpT> in place of R, E, F respectively). ■ The next theorem shows the connection between linear disjointness and separable algebraic extensions and will motivate a definition of separability in the case of ar bitrary (possibly nonalgebraic) extensions.
Theorem 2.8. Let F be an algebraic extension field of a field K of characteristic p^O and C an algebraically closed field containing F. Then F is separable over K if and only if F and K1/p are linearly disjoint over K. PROOF. We shall prove here only that separability implies that F and Kllp are linearly disjoint. The other half of the proof will be an easy consequence of a result below (see the Remarks after Theorem 2.10). LetX = \uu . .., un\ be a finite subset of F which is linearly independent over K. We must show that X is linearly inde pendent over K U p . The subfield E = K (u u - - - , « * ) is finite dimensional over K (Theorem V.1.12) and has a basis \ uu . . . , u„,u„+u . . . , ur\ which contains^(Theor
rem, IV.2.4). If i a E and k is a positive integer, then vk = 23 (a* £ and hence i=i vkl’ = (23 Oi^y = 23° ivu,v. Since v is separable over K, K (c) is both separable i
algebraic and purely inseparable over K(v v ) (Theorem V.6.4 and Lemma V.6.6). whence K(v ) = K(v”) = K[v p ] (Theorems V.1.6 and V.6.2). Thus v is a linear com bination of the vk>’ and hence of the Therefore E is spanned by j«ip, . . . , u r v \ . Since [E: K ] = r, [u\ p , . . . , u T v \ must be a basis by Theorems IV.2.5 and IV.2.7.
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CHAPTER VI THE STRUCTURE OF FIELDS
Thus {, u r p 1 and henceX p is linearly independent over K. By Lemma 2.6* is linearly independent over *1/7', whence F and K l l p are linearly disjoint over K. u
Definition 2.9. Let F be an extension field o/K. A transcendence base S o/F over K is called a separating transcendence base of F over K if¥ is separable algebraic over K(S). If F has a separating transcendence base over K, then F is said to be separably generated over K. REMARKS. Recall that Fis algebraic over K(S) (Corollary 1.6). If Fis separably generated over K, it is not true that every transcendence base of F over K is neces sarily a separating transcendence base (Exercise 8). EXAMPLES. If F is separable algebraic over K, then the null set is a separating transcendence base. Every purely transcendental extension is trivially separably generated since F = K(S). In order to make the principal theorem meaningful in the case of characteristic zero we define (for any field K of characteristic 0) A-1'0 = K110" = Kll0CD = K.
Theorem 2.10. If F is an extension field of a field K of characteristic p > 0 and C is an algebraically closed field containing F, then the following conditions are equivalent. (i) F and K1/p are linearly disjoint over K; (ii) F and K1/ptl are linearly disjoint over K for some n > 1; (iii) F and KJ/pC° are linearly disjoint over K; (iv) every finitely generated intermediate field E is separably generated over K; (v) K0 and F are linearly disjoint over K, where K0 is the fixed field (relative to C and K) of AutvfZREMARKS. The theorem is proved below. The implication (i) => (iv) provides a proof of the second half of Theorem 2.8 as follows. For every u eF, K(u) is a finitely generated intermediate field and thus separably generated over K. But F (and hence K( u)) is assumed algebraic over K and the only transcendence base of an algebraic extension is the null set. Therefore K(u) is separable algebraic over K(0) = K. Hence every u eF is separable algebraic over K. SKETCH OF PROOF OF 2.10. Except in proving (iii) <=> (v) we shall assume that char K = p ^ 0 since the case when char K = 0 is trivial otherwise, (iii) => (ii) => (i) is immediate since K l l p d K1,pn d KllpC° for every n > 1. (i) => (iv) Let E = K(s\, . . . , s n ) and tr.d.E/K = r. By Corollary 1.7 r < n and some subset of | ^i,. . ., sn} is a transcendence basis of E over K, say {.vi, . . . , s r ] . If r = n , then | su . .., sn | is trivially a separating transcendence base, whence (iv) holds. If r < n, then sr+1 is algebraic over K(su . . ., sT) (Corollary 1.6) and therefore m the root of an irreducible monic polynomial f(x) = £ K(su ■ ■ ■ , Sr)M. A »=i “least common denominator argument” such as that used in the proof of Theorem
2. LINEAR DISJOINTNESS AND SEPARABILITY
323
/m
1.5 shows that /(x) = d^l 22
iiX x j with 0 ^ d e A'fii, . . . , ir]) Vi = hi(s\, . . . , sr) m
and hi e K[xi , . . . , x r ]. Thus /i = 22 hi(*u • • • > x^xl+i e K\xu • • ■ xr+1] and »=o f\(su - - - , ^r,-Sr+i) = 0. It follows that there exists a polynomial g e A[xi, . . . , xr+i] of least positive degree such that g(si, . . . , ir+i) = 0. Clearly g is irreducible in A'lxi, . . . , xr+i]. Recall that X, is said to occur in g(xi,. . . , xj if some nonzero term of g contains a factor x,m with m > 1. We claim that some Xi occurs in g with an exponent that is not divisible by p. Otherwise g = c0 + r,wi(xi, . . . , x r + \) p -4-------h ekmk(xu . . . , x r+i) p , where c, e K, the C j are not all zero, and each m/xi, . . . , xr+i) is a monomial in x l 5 . . . , x r + i . Let »7o(xi, . . ., xr+1) = 1A- and for each j > 0 choose d) e K 1 / p such that df = c,. Then g = ( 22 djfftjixi, \j—0
>) k ^ - djfrijisi , ■ • • j | i) 0, i =0
whence the subset {m,(,si,. . . , sr+i) \j > 0) of F is linearly dependent over Kl,p. But , Sr+i) 17 > 0} is necessarily linearly independent over K (otherwise there would exist a g, e K[xi, . . . , av+i] with deg gi < deg g and gifo,. . . , sr+i) = 0). This fact contradicts the linear disjointness of F and Kllp. Therefore some x», say xi, occurs in g with an exponent that is not divisible by p. The polynomial g(x,s2, . . . , sr+i) e K(s z ,.. ., sr+i)[x] is necessarily nonzero. Otherwise, since x x occurs in g(xu . . . , xr+i) by the previous paragraph, we could obtain a polynomial g2 e K[x u . . . , xr+1] such that 0 < deg g2 < deg g and g‘i(5i,52, ■ ■ ■ , sr+i) = 0. Such a gi would contradict the choice of g. Therefore, g(x,s2,. . . , jv^i) 0. Since g(si,s2,. . . , sv+i) = 0, si is algebraic over K(s2, . . . , jv+i)But s2, . . . , Sr+i are obviously algebraic over K(s2,. . . , sr+i) and E is algebraic over K( si,. . . , 5,+i). By Theorems V.l .12 and V.1.13 E is algebraic over K(si, . . . , sr+i). Since tr.d.E /K = r, {s 2 ,..., sr+1} is a transcendence base of E over K (Corollary 1.7). The proof of Theorem 1.2 shows that the assignment x, ( -> s t determines a ^-iso morphism
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CHAPTER VI THE STRUCTURE OF FIELDS
and }j<+i, .. ., 5*) is a transcendence base of E over K. Therefore E is separably generated over K. (iv) => (iii) Let W be a finite subset of F that is linearly independent over K. We must show that W is linearly independent over K11^. Let E = K(IV). We need only show that E and KllpC° are linearly disjoint over AT, since this fact immediately implies that W is linearly independent over AT17*’00. Since W is finite, E has a separating tran scendence base S over K by (iv). We shall prove the linear disjointness of E and KllpC° by applying Theorem 2.4 to the extensions K d KUpCO and AT d K(S) d E as follows. K(S) and K1/pa> are linearly disjoint over K by Theorem 2.7. Let A'be a subset off that is linearly independent over K(S). Since E is separable algebraic over K(S), X is linearly independent over AT(S)1/p by the half of Theorem 2.8 already proved. There fore Xp is linearly independent over K(S) by Lemma 2.6. The last three sentences form the heart of an inductive argument which shows that Xpm is linearly indepen dent over K(S) for all m > 0; (note that (X p T ) p = Xpr+1). Hence X is linearly inde pendent over K(S)llpm for all m > 0 by Lemma 2.6 again. Therefore X is linearly independent over K(S)1,pC° and hence over its subfield K'lprrjK(S). We have proved that E and KllpmK(S) are linearly disjoint over K(S). Consequently E and K l / p r j ‘ are linearly disjoint over AT by Theorem 2.4. (iii) <=> (v). It suffices to prove that AT0 = AT1/p“. Let u e K0. If u is transcendental over K, then there exists v e C with v ^ u and v transcendental over K (for example, take v = it2). The composition AT(«) = K(x) = K(v) (where the isomorphisms are given by Theorem V.1.5) is a AT-isomorphism a such that a (it) = v. We thus have 1 = tr.d.K(x)/Ar = t r.d. K(u) /K = tf. d.K(v)/K. Theorem 1.11 (and Introduction, Lemma 8.9 if tr.d.C/AT(«) is infinite) implies that tr.d.C/A'(«) = tr.d,C/K(v). There fore cr extends to a AT-automorphism of C by Theorem 1.12. But c(u) = v j* u, which contradicts the fact that u e K 0 . Therefore, u must be algebraic over K with irre ducible polynomial fe AT[a:]. If v e C is another root of /, then there is a Ar-isomorphism t : K(u) ^ K( v) such thatV(«) = v (Corollary V.1.9). An argument similar to the one in the transcendental case shows that r extends to a AT-automorphism of C. Since u e K 0 we must have u = t (u) — v, whence /has only one root in C. Thus u is purely inseparable over K. If char K = 0, then / (which is necessarily separable) must have degree 1. Hence uzK = KIICP>. If char K = p 0, then upTi e AT for some n > 0 by Theorem V.6.4. Thus u e K1,pn d AT1'?00. We have proved that K0 d Kllp'x'. Conversely suppose that char K = p 0, u e K U p n d KVp'x‘ and a e AutAC- Then a(u) p n = a( u p n ) = upn, whence 0 = a(it)p,‘ — uv" = (a(u) — it)pn and cr(it) = u. Therefore, KllpCO d K0. ■
Definition 2.11. An extension field F of a field K is said to be separable over K (or a separable extension of K) if F satisfies the equivalent conditions of Theorem 2.10. REMARKS. Theorem 2.8 shows that this definition is compatible with our previous use of the term “separable” in the case of algebraic extensions (Definition V.3.10). Since the first condition of Theorem 2.10 is trivially satisfied when char K = 0, every extension field of characteristic 0 is separable. The basic properties of separability are developed in the following corollaries of Theorem 2.10.
2. LINEAR DISJOINTNESS AND SEPARABILITY
325
Corollary 2.12. (Mac Lane's Criterion) If F is an extension field of a fieldK. an d¥ is separably generated over K, then F is separable over K. Conversely, if F is separable and finitely generated over K, say F = K(ui, . . . , un), then F is separably generated over K. In fact some subset of {ui ,. . . , un) is a separating transcendence base of F over K. SKETCH OF PROOF. The proof of (iv) => (iii) => (i) in Theorem 2.10 is valid here with F = E since it uses only the fact that E is separably generated. The last two statements are consequences of the proof of (i) => (iv) in Theorem 2.10. ■
Corollary 2.13. Let F be an extension field of K and E an intermediate field. (i) If F is separable over K, then E is separable over K; (ii) if F is separable over E andE is separable over K, then F is separable over K; (iii) if F is separable over K and E is algebraic over K, then F is separable over E. REMARK, (iii) may be false if E is not algebraic over K (see Exercise 8). SKETCH OF PROOF OF 2.13. (ii) Use Theorems 2.4 and 2.10. (iii) If char K = p 0, let X be a subset of F which is linearly independent over E. Extend X to a basis U of F over E and let V be a basis of E over K. The proof of Theorem IV.2.16 shows that UV = | uv \ u e U,v e V\ is a basis of F over K, whence UV is linearly inde pendent over Kllp by separability. Lemma 2.6 implies that (UV) P = {u p v p | u e U,v e V) is linearly independent over K. We claim furthermore that V p is a basis of E over K. For E is separable over K by (i). Consequently, the linear disjointness of E and K l l p shows that V is linearly independent over Kl,v, whence Vp is linearly independent over K by Lemma 2.6. Since E = KEp by Corollary V.6.9, Vp necessarily spans E over K. Therefore, Vp is a basis of E over A". To complete the proof we must show that X is linearly independent over Ellp. If 52 aiUi = 0 (a, e E 1 / p ]Ui e X CZ U), then i
T. atputp = 0. Since each atp e E is of the form 52 (<"«> e K; v3 eV) we have i j 0 = 53 (52 C i j V j p ) u i p = 52 c l j u 1 p v j p . The linear independence of ( U V ) P implies * 1 i,j that = 0 for all ij and hence that a = 0 for all i. Therefore, X is linearly inde pendent over El,p. ■
EXERCISES Note. E and F are always extension fields of a field K, and C is an algebraically closed field containing E and F.
1. The subring E[F] generated by E and F is a vector space over K in the obvious way. The tensor product E (xk F is also a A'-vector space (see Theorem IV.5.5 and Corollary IV.5.12). E and F are linearly disjoint over K if and only if the A'-linear transformation E (x)A- F -> E[F] (given on generators of E (x)K F by ab) is an isomorphism. 2. Assume Eand Fare the quotient fields of integral domains R and 5 respectively. Then C is an /^-module and an 5-module in the obvious way.
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CHAPTER VI THE STRUCTURE OF FIELDS
(a) E and F are linearly disjoint over K if and only if every subset of R that is linearly independent over K is also linearly independent over S. (b) Assume further that R is a vector space over K. Then E and F are linearly disjoint over AT if and only if every basis of R over K is linearly independent over F. (c) Assume that both R and S' are vector spaces over K. Then E and F are linearly disjoint over K if and only if for every basis X of R over K and basis Y of 5 over K, the set {uv | u eX; c e y ) is linearly independent over K. 3. Use Exercise 1 to prove Theorem 2.2. 4. Use Exercise 1 and the associativity of the tensor product to prove Theorem 2.4. 5. If char K = p ^ 0, then (a) K 1 / p n is a field for every n > 0. See Exercise III. 1.11. (b) AT*** is a field. (c) K l l f n is a splitting field over K of {x*,n — k j k e AT}.
6. If is algebraically independent over F, then Fand K(uu . . . , un) are linearly disjoint over AT. 7. If E is a purely transcendental extension of K and F is algebraic over K, then F and F are linearly disjoint over K. 8. Let AT = Zp, F = Z p (x), and E = Z v (x v ). (a) F is separably generated and separable over K. (b) E F. (c) F is algebraic and purely inseparable over E. (d) { jcp) is a transcendence base of F over K which is not a separating tran scendence base. 9. Let char K = p ^ 0 and let u be transcendental over K. Suppose F is generated over K by j u,vi,v2,. . .j, where v, is a root of xpl — u e X(f/)[A] for / = 1 , 2 , . . . . Then F is separable over K, but F is not separably generated over K. 10. (a) AT is a perfect field if and only if every field extension of K is separable (see Exercise V.6.13). (b) (Mac Lane) Assume AT is a perfect field, F is not perfect and tr.d.F/K — 1. Then F is separably generated over K. 11. F is purely inseparable over K if and only if the only A'-monomorphism F —» C is the inclusion map. ✓
12. E and F are free over K if every subset X of E that is algebraically independent over AT is also algebraically independent over F. (a) The definition is symmetric (that is, E and F are free over K if and only if F and E are free over AT). (b) If E and F are linearly disjoint over K, then E and F are free over K. Show by example that the converse is false. (c) If E is separable over K and E and F are free over K , then EF is separable over F. (d) If Fand Fare free over AT and both separable over AT, then EF is separable over AT.
CHAPTER
VII
LINEAR ALGEBRA
Linear algebra is an essential tool in many branches of mathematics and has wide applications. A large part of the subject consists of the study of homomorphisms of (finitely generated) free modules (in particular, linear transformations of finite di mensional vector spaces). There is a crucial relationship between such homomor phisms and matrices (Section 1). The investigation of the connection between two matrices that represent the same homomorphism (relative to different bases) leads to the concepts of equivalence and similarity of matrices (Sections 2 and 4). Certain important invariants of matrices under similarity are considered in Section 5. Deter minants of matrices (Section 3) are quite useful at several points in the discussion. Since there is much interest in the applications of linear algebra, a great deal of material of a calculational nature is included in this chapter. For many readers the inclusion of such material will be well worth the burden of additional length. How ever, the chapter is so arranged that the reader who wishes only to cover the im portant basic facts of the theory may do so in a relatively short time. He need only omit those results labeled as propositions and observe the comments in the text as to which material is needed in the sequel. The approximate interdependence of the sections of this chapter is as follows: 1
3--►4-*--2
As usual a broken arrow A ----> B indicates that an occasional result of Section A is used in Section B, but that Section B is essentially independent of Section A. 327
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CHAPTER VII LINEAR ALGEBRA
1. MATRICES AND MAPS The basic properties of matrices are briefly reviewed. Then the all important rela tionship between matrices and homomorphisms of free modules is explored. Except in Theorem 1.1 all rings are assumed to have identity, but no other restrictions are imposed. Except for the discussion of duality at the end of the section all of this material is needed in the remainder of the chapter. Let R be a ring. An array of elements of the form
with e R, n rows (horizontal), and m columns (vertical), is called an n X m matrix over R. An n X n matrix is called a square matrix. For brevity of notation an arbi trary matrix is usually denoted by a capital letter, A,B,C or by (an ), which indicates that the i-jth entry (row /, column j) is the element alt e R. Two n X m matrices («,,) and (bn) are equal if and only if at, = bi, in R for all i,j. The elements 011,022,033, - - are said to form the main diagonal of the matrix (an). An n X n matrix with an = 0 for all / 9^ j is called a diagonal matrix. If R has an identity element, the identity matrix /„ is the n X n diagonal matrix with \R in each entry on the main diagonal; that is, In = (8ij) where 8 is the Kronecker delta. The n X m matrices with all entries 0 are called zero matrices. The set of all n X n matrices over R is denoted MatnR. The transpose of an n X m matrix A = (ai}) is the m X n matrix A' = (bi,) (note size!) such that bn = an for all If A = (an) and B = (bn) are n X m matrices, then the sum A + B is defined to be the n X m matrix (cn), where c,, = + bt,. If A = (an) is an m X n matrix and B — ibid is an n X p matrix then the product AB is defined to be the m X p matrix n (Cij) where ci} = ^ aikbkJ. Multiplication is not commutative in general. If A = (atJ) is an n X m matrix and r z R, rA is the n X m matrix (ra,,) and Ar is the n X m matrix (a,-,r); rlr is called a scalar matrix. If the matrix product AB is defined, then so is the product of transpose matrices l B‘A . If R is commutative, then (AB) 1 = B‘A'. This conclusion may be false if R is noncommutative (Exercise 1).
Theorem 1.1. If R is a ring, then the set of all n X m matrices over R forms an R-R bimodule under addition, with the n X m zero matrix as the additive identity. Multiplication of matrices, when defined, is associative and distributive over addition. For each n > 0, MatnR is a ring. If R has an identity, so does MatnR (namely the identity matrix I„). PROOF. Exercise. |
One of the important uses of matrices is in describing homomorphisms of free modules.
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1. MATRICES AND MAPS
Theorem 1.2. Let R be a ring with identity. Let Fbe a free left R-module with a finite basis of n elements and F a free left R-module with a finite basis of m elements. Let M be the left R-module of all n X m matrices over R. Then there is an isomorphism of abelian groups: //owr(E,F) ^ M. 7/R is commutative this is an isomorphism of left R-modules.
PROOF. Let \uu . . . , un\ be a basis of E, {ui, . . . , vm\ a basis of F and /e Homr(E,F). There are elements ri3 of R such that
f(u i) = /"lit!! -f- r\
2
f(un) — rniV\ -f- rn2V2 + ' • • + trnmvm. The ri3 are uniquely determined since {ui, . . . , vm\ is a basis of F. Define a map : Homfi(£,F) —» M by /(-> A, where A is the n X m matrix It is easy to verify that is an additive homomorphism. If /3(/) = 0, then f(ui) = 0 for every basis element whence / = 0. Thus /3 is a monomorphism. Given a matrix (r,,-) e M, de fine f :E—+F by /(«,-) = r{1v 1 + ri2v2 + • • ■ + rimvm (/ = 1,2, ... , n). Since E is free, this uniquely determines /as an element of Homfl(£,F) by Theorem IV.2.1. By con struction /3(/) = (/■{]). Therefore /3 is surjective and hence an isomorphism. If R is commutative, then Homs(£,F) is a left 7?-module with (rf)(x) = r(f(x)) by the Remark after Theorem IV.4.8. It is easy to verify that is an 7?-module isomor phism. ■
Let R,E,F and (3 be as in Theorem 1.2. The matrix of a homomorphism /= Homfl(£,F) relative to the ordered bases U = {uy, 1 of E and V = {ui,. .. , vm] of F is the n X m matrix (/•„) = (3(f) as in the proof of Theorem 1.2. Thus the z'th row of the matrix of /consists of the coefficients of /(w,-) e F relative to the ordered basis {ui, . . . , vm}. In the special case when E — F and V = V we refer to the matrix of the endomorphism /relative to the ordered basis U.
REMARK. Let E,F, fU,V be as in the previous paragraph. The image under /of an arbitrary element of E may be conveniently calculated from the matrix A = (rtj) of /as follows. If u = xiui + + • ■ ■ + xnun e E (x, e R), then
( 22 n
\
i=i / i= 1
n
22 (22 Kira )vj = yM,
m/n
=
) = 22 *./(«.) = 22 h
x*ui
\
t=
22 u i /m
xi[
r v
1 \j = 1
m
j = l \i = 1 / j-1
n
where yt =
22 Thus if X is the 1 X n matrix (xi x2 - - • xn) and Y is the 1 X m
t=i matrix (>1 >2 ■ ■ • ym), then Yis precisely the matrix product XA. Xand Y are some times called row vectors.
CHAPTER VII LINEAR ALGEBRA
330
Theorem 1.3. Let R be a ring with identity and let E,F,G, be free left K-modules with finite ordered bases U = (Ui, . . . , u„}, V = {Vi, . . . , vm}, W = {Wi, . . . , wp| re spectively. Iff e //o/?7r(E,F) has n X m matrix A (relative to bases U and V) and g e Hornr(F,G) has m X p matrix B (relative to bases V and W), then gf e //ootr(E,G) has n X p matrix AB (relative to bases U and W). PROOF. If A = (/-,,) and B = (sk]), then for each /' = 1,2, ... ,n (rn
in.
\
m /v
\
23 Wk) = 23 rikg(Dk) = 23 23 Wi J A=1
/ fc = l
k = 1 \>= 1
/
V' (V1 ^ j = 1 \t=l /
m Therefore the matrix of gf relative to V and W has /-yth entry 23 r^s a-j- But this is a-
=1
precisely the i-jth entry of the matrix AB. ■ Let R be a ring with identity and E a free left /^-module with a finite basis U of n elements. Then Homfi(£^) is a ring with identity, where the product of maps /and g is simply the composite function fg\E^>E (Exercise IV. 1.7). We wish to note for future reference the connection between the ring Horn;,.(£,£) and the matrix ring Mat„/?. If S and T are any rings, then a function 6 : S —► T is said to be an antiisomorphism if 6 is an isomorphism of additive groups such that 8(sis2) = 6(s j8(si) for all SieS. The map Hom/((£,£) -> MatnZ? which assigns to each /e Hom;,(£,£) its matrix (relative to U) is an anti-isomorphism of rings by Theorems 1.2 and 1.3. It would be convenient if Hom/((£,E) were actually isomorphic to some matrix ring. In order to show that this is indeed the case, we need a new concept. If R is a ring, then the opposite ring of R, denoted R np , is the ring that has the same set of elements as R, the same addition as R, and multiplication ° given by
a ° b = ba, where ba is the product in R; (see Exercise III. 1.17). The map given by r(—> r is clearly an anti-isomorphism R —> R°p. If A = (ai}) and B = (bt,) are n X n matrices over R, then A and B may also be considered to be matrices over R VT . Note that in n Mat,,/?, AB (ca) where etl =
=
23 tiikbku but in MatnR"v, AB (dlJ), where =
k=i
dij 23 @ik ° bkj 23 bkjaikk=1 k=1
Theorem 1.4. Let R be a ring with identity and E a free left K-module with a finite basis of n elements. Then there is an isomorphism of rings: HomR(E,E) ^ A/afn(Rop). In particular, this isomorphism exists for every n-dimensional vector space E over a division ring R, in which case Rop is also a division ring. REMARK. The conclusion of Theorem 1.4 takes a somewhat nicer form when R is commutative, since in that case R = Rop.
1. MATRICES AND MAPS
331
SKETCH OF PROOF OF 1.4. Let 0 : Homfi(£,£) Mat,/? be the anti isomorphism that assigns to each map /its matrix relative to the given basis. Verify that the map ip : Mat„/? —» Matr,/?,,p given by \]/(A) = A‘ is an anti-isomorphism of rings. Then the composite map : Hornu(E,E) —> Mat„/?',p is an isomorphism of rings. The last statement of the theorem is a consequence of Theorem IV.2.4 and Exercise III.1.17. ■ Let R be a ring with identity and A e MatnR. A is said to be invertible or non singular if there exists B e Matn/? such that AB = In = BA. The inverse matrix B, if it exists, is easily seen to be unique; it is usually denoted A~x. Clearly B = A~1 is in vertible and (A~l)~' = A. The product AC of two invertible matrices is invertible with (/4C)_1 = C~1A~1. If A is an invertible matrix over a commutative ring, then so is its transpose and (A1^1 = (A_1)' (Exercise 1). The matrix of a homomorphism of free ^-modules clearly depends on the choice of (ordered) bases in both the domain and range. Consequently, it will be helpful to know the relationship between matrices that represent the same map relative to different pairs of ordered bases.
Lemma 1.5. Let R be a ring with identity and E,F free left R-modules with ordered bases U,V respectively such that |U| = n = |V|. Let A e MatnR. Then A is invertible if and only if A is the matrix of an isomorphism f: E —»F relative to U and V. In this case A-1 is the matrix of f_1 relative to V and U. SKETCH OF PROOF. An /^-module homomorphism /: E —» F is an isomor phism if and only if there exists an /?-module homomorphism /_1 : F —> E such that f~lf = 1 n and ff~l = 1/.' (see Theorem 1.2.3). Suppose /is an isomorphism with matrix A relative to U and V. Let B be the matrix of /_1 relative to V and U. Sche matically we have
f f-1
map:
module: E-------- * F----- * E basis: U matrix:
V A
U B
By Theorem 1.3 AB is the matrix of f~lf = 1F. relative to U. But /„ is clearly the matrix of lE relative to U. Hence AB = In by the proof of Theorem 1.2. Similarly BA — In, whence A is invertible and B = A~ x . The converse implication is left as an exercise. ■
Theorem 1.6. Let R be a ring with identity. Let E and F be free left R-modules with finite ordered bases U and V respectively such that |U| = n, |V| = m. Let f e HomR(E,F) have n X m matrix A relative to U and V. Then f has n X m matrix B relative to another pair of ordered bases of E and F if and only if B = PAQ for some invertible matrices P and Q. PROOF. (=») If B is the nX m matrix of/relative to the bases U of E and V' of F, then \U'\ = n and \ V'\ = m. Let P be the n X n matrix of the identity map 1 e rela
CHAPTER VII LINEAR ALGEBRA
332
tive to the ordered bases U' and U. P is invertible by Lemma 1.5. Similarly let Q be the m X m invertible matrix of 1^ relative to V and V (note order). Schematically we have:
\E
map:
f
\F
module: E ---------> E ---- *F ----- * F basis: V
U P
matrix:
V A
V' Q
By Theorem 1.3 the matrix of /= l^/l^ relative to U’ and V' is precisely PAQ. Therefore B = PAQ by the proof of Theorem 1.2. (<=) We are given U,V,fA as above and B = PAQ with P,Q invertible. Let g : E E be the isomorphism with matrix P relative to U and h : F —* F the iso morphism with matrix Qrx relative to V (Lemma 1.5).- If U = {uu I, then g(U) = (#(wi),. . . ,g(u n ) j is also an ordered basis of E and P is the matrix of 1£ relative to the ordered bases g(U) and U. Similarly Q~* is the matrix of 1F relative to the ordered bases h(V) and V , whence Q = (C?-1)-1 is the matrix of 1/- relative to V and h(V) (Lemma 1.5). Schematically we have map:
module:
\E f lF E----- >E----- *F----- >F
basis: g(U) matrix:
U
V
h(V)
PAQ
By Theorem 1.3 the matrix of /= 1f/1e relative to the ordered bases g(U) and h(V) is PA Q = B. ■
Corollary 1.7 Let R be a ring with identity and E a free left R-module with an ordered basis U of finite cardinality n. Let A be the n X n matrix offe HomR(E,E) relative to U. Then f has n X n matrix B relative to another ordered basis ofE if and only if B = PAP-1 for some invertible matrix P. SKETCH OF PROOF. If E = F,U = F,and(7' = V' in the proof of Theorem 1.6, then Q = P~l by Lemma 1.5. ■ The preceding results motivate:
Definition 1.8. Let R be a ring with identity. Two matrices A,B e MatnR are said to be similar if there exists an invertible matrix P such that B = PAP-1. Two n X m matrices C,D are said to be equivalent if there exists invertible matrices P and Q such that D = PCQ. Theorem 1.6 and Corollary 1.7 may now be reworded in terms of equivalence and similarity. Equivalence and similarity are each equivalence relations (Exercise 7) and will be studied in more detail in Sections 2 and 4.
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1. MATRICES AND MAPS
We close this section with a discussion of right modules and duality. If R is commutative, then the preceding results are equally valid for right Rmodules. There are important cases, however, in which R is not commutative (for example, vector spaces over a division ring). In order to prove the analogue of Theorem 1.3 for right modules in the noncommutative case it is necessary to define the matrix of a homomorphism somewhat differently. Let R be a ring with identity and let E and F be free right 7?-modules with finite ordered bases U = [u x ,..., u n j and V = {ei, . . . , c m ] respectively. The matrix of the homomorphism /e Hom^E/) relative to U and V is defined to be the m X n matrix (note size): 'ill ■S'12
Sn Sn
^2 ’ ' ’ Sjrm / 5 where the sit e R are uniquely determined by the equations: /(«l) — UlSll + £>2521 + tsSsi +• • ■ + Vmsm
1
/(«n) — ViS]n + ViSin + VsSin + ' ' ' + VmSmn.
Thus the coefficients of f(uj) with respect to the ordered basis V form they'th column of the m X n matrix ($,-,•) of /(compare the proof of Theorem 1.2). The action of /may be described in terms of matrices as follows. Let u — u\x + u2x2 + 1 • • + unxn (x, e R) be any element of E and let X be the n X 1 matrix (or
column vector)
. Let A be the matrix of /relative to the bases U and V. Then
/(«) = fi.vi + v2y2 + • ■ ■ + v„,y„„ where >v s R and
is the m X 1 matrix
ov (column vector) AX. The analogues of results 1.2-1.5 above are now easily proved, in particular,
Theorem 1.9. Let R be a ring with identity and E,F free right K-modules with finite bases U andV of cardinality n and m respectively. Let N be the right K-module of all m X n matrices over R. (i) There is an isomorphism of abelian groups Homn(E,F) = N, which is an iso morphism of right K-modules if R is commutative;
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334
(ii) let G be a free right R-module with a finite basis W of cardinality p. If f e Homn(E,F) has m X n matrix A (relative to U and V) and g e //<wir(F,G) has p X rn matrix B (relative to V and W), then gf e //o/«r(E,G) has p X n matrix BA (relative to U and W); (iii) there is an isomorphism of rings //owr(E,E) =: Matn R. PROOF. Exercise; see Theorems 1.2-1.4. Note that for right modules (iii) is actually an isomorphism rather than an anti-isomorphism. ■
Proposition 1.10. Let R be a ring with identity and f : E —> F a homomorphism of finitely generated free left R-modules. If A is the matrix off relative to (ordered) bases U and\, then A is also the matrix of the dual homomorphism f : F* —» E* offree right R-modules relative to the dual bases V* and U*. REMARK. Dual maps and dual bases are defined in Theorems IV.4.10 and I V.4.11. If R is commutative (for example, a field) it is customary to consider the dual M* of a left Z?-module M as a left Z?-module (with rm* = m*r for r e R, m* e M* as usual). In this case the matrix of the dual map /is the transpose A‘ (Exercise 8). PROOF OF 1.10. Recall that the dual basis V* = {i?i F* = Hom^F,/?) is determined by:
iv,*J of
Vi*(vi) = 8n (Kronecker delta; 1 < ij < m), and similarly for the dual basis U* = jw]*,. . ., un*\ of E* (TheoremIV.4.11). Ac cording to the definition of the matrix of a map of right F-modules we must show Tl that for each j = 1,2,... t m, J(v*) = 53 */»j> where A = (/•,,) is the n X m matrix i=i of /: E —» F relative to U and V. Since both sides of the preceding equation are maps E-* R, it suffices to check their action on each uk e U. By Theorem IV.4.10 we have: /m \ m r f(vi*)(uk) = v,*( f(uk)) = t3*( 53 r kt Vt) = 53 ktVj*(vt) = rki. V=i / < = i On the other hand,
(
n
\
n
53 ^**^3 k^fc) 53 ^ rkj. i j=i / t=i
EXERCISES Note: All matrices are assumed to have entries in a ring R with identity. 1. Let R be commutative. (a) If the matrix product AB is defined, then so is the product B‘Al and (AB)‘ = BlAl. (b) If A is invertible, then so is A1 and (A')-1 = (A-1)'. (c) If R is not commutative, then (a) and (b) may be false.
2. RANK AND EQUIVALENCE
335
2. A matrix (o,-,) e Mat„i? is said to be
(upper) triangular <=> a,, = 0 for j < i; strictly triangular <=> a„ = 0 for j < i. Prove that the set of all diagonal matrices is a subring of Mat„R which is (ring) isomorphic to R x ■ • • x R(n factors). Show that the set T of all triangular matrices is a subring of Mat„/? and the set I of all strictly triangular matrices is an ideal in T. Identify the quotient ring T/I. 3. (a) The center of the ring MatnR consists of all matrices of the form rln, where r is in the center of R. [Hint: every matrix in the center of Mat*/? must commute with each of the matrices Br s, where Br s has 1 r in position (r,s) and 0 elsewhere.] (b) The center of Mat„/? is isomorphic to the center of R. 4. The set of all m X n matrices over R is a free .R-module with a basis of mn ele ments. 5. A matrix A e Mat„i? is symmetric if A = A‘ and skew-symmetric if A = —A1. (a) If A and B are [skew] symmetric, then A + B is [skew] symmetric. (b) Let R be commutative. If A,B are symmetric, then AB is symmetric if and only if AB = BA. Also show that for any matrix B e Mat„i?, BB‘ and B + B' are symmetric and B — B‘ is skew-symmetric. 6. If R is a division ring and A,B e Mat„7? are such that BA = /„, then AB = /„ and B = A~l. [Hint: use linear transformations.] 7. Similarity of matrices is an equivalence relation on Mat„R. Equivalence of ma trices is an equivalence relation on the set of all m X n matrices over R. 8. Let E,F be finite dimensional (left) vector spaces over a field and consider the dual spaces to be left vector spaces in the usual way. If A is the matrix of a linear trans formation / : £ —► F, then A1 is the matrix of the dual map / : F* —► E*.
2. RANK AND EQUIVALENCE The main purpose of this section is to find necessary and sufficient conditions for matrices over a division ring or a principal ideal domain to be equivalent. One such condition involves the concept of rank. In addition, useful sets of canonical forms for such matrices are presented (Theorem 2.6 and Proposition 2.11). Finally, practical techniques are developed for finding these canonical forms and for calculating the inverse of an invertible matrix over a division ring. Applications to finitely generated abelian groups are considered in an appendix, which is not needed in the sequel.
Definition 2.1. Let f : E —> F be a linear transformation of (left) vector spaces over a division ring D. The rank of f is the dimension oflmi and the nullity off is the dimen sion of Ker f. REMARK. If /:£—>F is as in Definition 2.1, then by Corollary IV.2.14, (rank f) + (nullity f) = dimcE.
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If R is a ring with identity and n a positive integer, then Rn will denote the free /?-module /?©--- © R (n summands). The standard (ordered) basis of Rn consists of the elements Ei = (1*,0, . . ., 0), e2 = (0,1*,0,. . . , 0), . . . , e„ = (0,... , 0,1*).
Definition 2.2. The row space [resp. column space] of an n X m matrix A over a ring R with identity is the submodule of the free left [resp. right] module Rm [resp. Rn] generated by the rows [resp. columns] of A considered as elements of Rm [resp. Rn]. If R is a division ring, then the row rank [resp. column rank] of A is the dimension of the row [resp. column] space of A.
Theorem 2.3. Let f : E —+ F be a linear transformation of finite dimensional left [resp. right] vector spaces over a division ring D. If A is the matrix off relative to some pair of ordered bases, then the rank off is equal to the row [resp. column] rank of AREMARK. “Row rank” is replaced by “column rank” in the case of right vector spaces because of the definition of the matrix of a map of right vector spaces (p. 333). PROOF OF 2.3. Let A be the n X m (resp. m X n] matrix of /relative to or dered bases U = {Wi,. . . , un | of E and V = { V\,.. . , vm } of F. Then under the usual isomorphism Dm given by W H (ru ... ,r m ) the elements /(«i),. . . ,f(u„) i are mapped onto the rows [resp. columns] of A (considered as vectors in Dm). Since Im /is spanned by /(«i),. .., f(un), Im / is isomorphic to the row [resp. column] space of A, whence the rank of /is equal to the row [resp. column] rank of A. ■ We now digress briefly to prove that the row and column rank of a matrix over a division ring are in fact equal. This fact, which is proved in Corollary 2.5, is not essential for understanding the sequel since “row rank” is all that is actually used hereafter.
Proposition 2.4. Any linear transformation f : E —» F of finite dimensional left vector spaces over a division ring D has the same rank as its dual map F : F* —► E*. The dual map is defined in Theorem IV.4.10. PROOF OF 2.4. Let rank / — r. By Corollary IV.2.14 there is a basis X = \uu ... ,u n \ such that \u r+ 1, is a basis of Ker / and Yi = I /(«i), - - • >/("r) ( is a basis of Im /. Extend Y\ to a basis Y = {/i= f{u i), ..., t r = f(u r ),t r+l ,t m ) of F. Consider the dual bases X* of E* and Y* of F* (Theorem IV.4.11). Verify that for each / = 1,2,. . . ,m,
Ku*){u,) = /,*(/(«,)) =
ti*0i) = Sij if 7=1,2, ...,r; tz*(0) = 0 if j = r+ l,r + 2, ...,«.
where is the Kronecker delta. Consequently for each j — 1, 2,. .., n, = ",*(«,) if i = \,2, ... ,r
2. RANK AND EQUIVALENCE
337
Therefore, f(h*) = Ui* for i = 1,2,,r and /(/;*) = 0 for i = r + 1..................................m. Im / is spanned by f(Y*) and hence by {«i*,. .., u r *\. Since {«i*,..., ur*\ is a subset of**, it is linearly independent in E*. Therefore {wi*,..., u r *\ is a basis of Im /, whence rank f=r = rank /. ■
Corollary 2.5. If A is an n X rn matrix over a division ring D, then row rank A — column rank A. PROOF. Let / : Dn —+ Dm be a linear transformation of left vector spaces with matrix A relative to the standard bases. Then the dual map /of right vector spaces also has matrix A (Proposition 1.10). By Theorem 2.3 and Proposition 2.4 row rank A = rank / = rank / = column rank A. ■ REMARK. Corollary 2.5 immediately implies that row rank A = row rank A1 for any matrix A over a field. In view of Corollary 2.5 we shall hereafter omit the adjectives “row” and “column” and refer simply to the rank of a matrix over a division ring. In Theorem 2.6 below equivalent matrices over a division ring D will be char acterized in terms of rank and in terms of the following matrices. If m,n are positive integers, then El,m is defined to be the n X m zero matrix. For each r (1 < r < min (n,m)), E*'m is defined to be the n X m matrix whose first r rows are the standard basis vectors ... ,er of Dm and whose remaining rows are zero: 'I* 0 0 ..................................................... 0 1* 0 ....................................................
£"- m =© 0 ........................... 0 1* 0 0 ...................... 0 0 0
0i Clearly rank/s"’"1 = r. Furthermore if EnT'm is the matrix of an /?-module homo of E and morphism / :E—*F of free /^-modules, relative to bases {v u . . . , v m} of F, then /(«0
Vi if / = 1,2, . . . , r; 0 if / = r + 1 ,r + 2,. . . , n.
An immediate consequence of Theorem 1.6 and Theorem 2.6 below is that every linear transformation of finite dimensional vector spaces has this convenient form for some pair of bases (Exercise 6). A set of canonical forms for an equivalence relation R on a set* is a subset C of* that consists of exactly one element from each equivalence class of R. In other words, for every x e* there is a unique c e C such that x is equivalent to c under R. We now show that the matrices form a set of canonical forms for the relation of equiva lence on the set of all n X m matrices over a division ring.
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Theorem 2.6. Let M be the set of all n X m matrices over a division ring D and let A,B e M. (i) A is equivalent to E"-'11 if and only if rank A = r. (ii) A is equivalent to B if and only if rank A = rank B. (iii) The matrices E'1"' (r = 1,2, . . . , min (n,m)) constitute a set of canonical forms for the r elution of equivalence on M. SKETCH OF PROOF, (i) A is the matrix of some linear transformation / : Dn —> Dm relative to some pair of bases by Theorem 1.2. If rank A — r, then Corollary IV.2.14 implies that there exist bases U = {wi, . . . , w„j of Dn and V - {fi,. . ., vm } of Dm such that f(ui) = ii for /' = 1,2,... ,r and /(«,•) = 0 for / = r + 1, . . . , n. Clearly the matrix of /relative to U and V is £"•’". Therefore A is equivalent to £”’m by Theorem 1.6. Conversely suppose A is equivalent to £"•”'. By Theorem 1.6 there is a linear transformation g : Dn —> Dm such that A is the matrix of g relative to one pair of bases and £",m is the matrix of g relative to another pair of bases. Consequently, rank A = rank g = rank £"•’" = r by Theorem 2.3. (ii) and (iii) are consequences of (i). ■ The following definition, theorem, and corollaries have a number of useful con sequences, including practical methods for constructing: (i) canonical forms under equivalence for matrices over a principal ideal do main (Proposition 2.11); (ii) the canonical forms E”-7” under equivalence for matrices over a division ring; (iii) the inverse of an invertible matrix over a division ring (Proposition 2.12). Proposition 2.11 is used only in the proof of Proposition 4.9 below. The remainder of the material is independent of Proposition 2.11 and is not needed in the sequel. We shall frequently consider the rows [resp. columns] of a given n X m matrix over a ring R as being elements of Rm [resp. Rn]. We shall speak of adding a scalar multiple of one row [resp. column] to another; for example,
r(aua2, . . . ,a m ) -f (bu . . . , bm) = (rax + bu ... , ram + bm).
Definition 2.7. Let A be a matrix over a ring R with identity. Each of the following is called an elementary row operation on A: (i) interchange two rows of A; (ii) left multiply a row of A by a unit c e R; (iii) for r s R and i ^ j, add r times row j to row i.
Elementary column operations on A are defined analogously (with left multiplication in (ii), (iii) replaced by right multiplication). An n X n elementary (transformation) matrix is a matrix that is obtained by performing exactly one elementary row (or column) operation on the identity matrix In. Theorem 2.8. Let A be an n X m matrix over a ring R with identity and let En [resp. E,„] be the elementary matrix obtained by performing an elementary row [resp. column] operation T on In [resp. Im], Then EnA [resp. AEni] is the matrix obtained by performing the operation T on A. PROOF. Exercise. ■
2. RANK AND EQUIVALENCE
339
Corollary 2.9. Every n X n elementary matrix E over a ring R with identity is in vertible and its inverse is an elementary matrix. SKETCH OF PROOF. Verify that /,, may be obtained from E by performing a single elementary row operation T. If F is the elementary matrix obtained by per forming T on /„, then FE = In by Theorem 2.8. Verify directly that EF = /„. ■
Corollary 2.10. If B is the matrix obtained from an n X m matrix A over a ring R with identity by performing a finite sequence of elementary row and column operations, then B is equivalent to A. PROOF. Since each row [column] operation used to obtain B from A is given by left [right] multiplication by an appropriate elementary matrix (Theorem 2.8), we have B = (Ep■ - -E\)AiFx- ■ ■FQ) = PAQ with each E{ Fj an elementary matrix and P = EP- ■ Ei, Q = Fi - ■ -Fq.P and Q are products of invertible matrices (Corollary 2.9) and hence invertible. ■ We now consider canonical forms under equivalence of matrices over a principal ideal domain R. The rank of a free module over R is a well-defined invariant by Corollary IV.2.12. Since every submodule of a free /?-module is free (Theorem IV.6.1), we may define the rank of a homomorphism / : E —* F of free F-modules to be the rank of Im /. Similarly the row rank of a matrix A over R is defined to be the rank of the row space A (see Definition 2.2). The proof of Theorem 2.3 is easily seen to be valid here, whence the rank of a map /of finitely generated free J?-modules is the row rank of any matrix of /relative to some pair of bases. Consequently, if A is equivalent to a matrix B, then row rank A = row rank B. For A and B are matrices of the same homomorphism / :/?”—> Rm relative to different pairs of bases by Theo rem 1.6, whence row rank A = rank / = row rank B. Here is the analogue of Theo rem 2.6 for matrices over a principal ideal domain.
Proposition 2.11. If A is an n X m matrix of rank r > 0 over a principal ideal domain R, then A is equivalent to a matrix
of the form ( q
where L is an r X r
diagonal matrix with nonzero diagonal entries d1} . . . , dr such that di | d21 • - -1 dr. The ideals (d t ),. . ., (dr) in R are uniquely determined by the equivalence class of A. REMARKS. The proposition provides sets of canonical forms for the relation of equivalence on the set ofn X m matrices over a principal ideal domain (Exercise 5). If R is actually a Euclidean domain, then the following proof together with Exercise 7 and Theorem 2.8 shows that the matrix
may be obtained from A by a
finite sequence of elementary row and column operations. SKETCH OF PROOF OF 2.11. (i) Recall that a,b e R are associates if a \ b and b \ a. By Theorem III.3.2 a and b are associates if and only if a = bu with u a unit. We say that c e R is a proper divisor of a e R if c \ a and c is not an associate of a (that is, a^c). By a slight abuse of language we say that two proper divisors ci and c2
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CHAPTER VII LINEAR ALGEBRA
of an element a are distinct if c\ and c2 are not associates. Now R is a unique fac torization domain by Theorem III.3.7. If a = p™lp%2. . . p(A distinct irreducibles and each «,• > 0), then every divisor of a is an associate of an element of the form p^pf ...p\ l with 0 < ki < tii for each i. Consequently a nonzero element of R has only finitely many distinct proper divisors. (ii) If a and b are nonzero elements of R, let c be their greatest common divisor. By Definition III.3.10 and Theorem III.3.11 there exist r,s s R such that ar + bs = c, cai — a and cbi = 6, whence a\r + b x s = 1« and ha\ — ab\ = 0. Consequently the m X m matrix
is invertible with inverse
If the first row of A is (a,b,ai3,. . . , then A is equivalent to AT = InAT, whose first row is (c,0,fli3,. . . , a\m). If the first column1 of A is (a,d,a31,. .., o„i)', then an analogous procedure yields an invertible matrix S such that A is equivalent to SA and SA has first column (e,0,a3l, . .., a„i)‘, where e is the greatest common divisor of a and d. A matrix such as S or T is called a secondary matrix. (iii) Since A ^ 0 a suitable sequence of row and column interchanges and multi plications on the right by secondary matrices changes A into a matrix A, which has first row (oi,0,0, . . . , 0) with ai 7^ 0. A is equivalent to Ax by (ii) and Corollary 2.10. (iv) If Oi divides all entries in the first column of Ah then a finite sequence of elementary row operations produces a matrix B of the form
which is equivalent to Au and hence to A, by Corollary 2.10. (v) If a 1 does not divide some first column entry b of Au then a sequence of row and column interchanges and multiplications on the left by secondary matrices changes A\ into a matrix A2 which has first column feAO, . . . , 0)' with at a common divisor of a\ and b (see (ii)). Note that A2 may well have many nonzero entries in the 1For
typographical convenience we shall frequently write an n X 1 column vector as the transpose of a 1 X n row vector; for example, ( 1 J = (aia2)'. \a2/
2. RANK AND EQUIVALENCE
341
first row. However since a21 au a2 \ b and ax)(b, a2 is a proper divisor of ax by (i). A2 is equivalent to Au and hence to A, by (ii) and Corollary 2.10. (vi) If a2 divides every entry in the first row of A2 then a sequence of elementary column operations produces a matrix equivalent to A and of the same general form as B above. (vii) If a2 fails to divide some entry k in the first row of A2, then repeat (iii) and obtain a matrix A3 which is equivalent to A and has first row (a3,0,0, .. . , 0) with a3 a common divisor of a2 and k. A3 may have nonzero entries in its first column. But since a3 \ a2, a31 k and a:)(k, a3 is a proper divisor of a2 by (i). Furthermore, a2 and a3 are distinct proper divisors of ax by (v). A3 is equivalent to A2, and hence to A, by (ii) and Corollary 2.10. (viii) Since ai has only finitely many distinct proper divisors, a finite number of repetitions of (iii)-(vii) must yield a matrix C which is equivalent to A and has the form 5l 0 0 c22
0 cn2 with Si 9* 0. (ix) If si does not divide some c,„ add row / to row 1 and repeat (iii)-(viii). The result is a matrix D that is equivalent to A, has the same general form as the matrix C above, and has for its (1,1) entry an element s2 which is a common divisor of Si and Cij and a proper divisor of st. (x) If s2 does not divide every entry in D, then a repetition of (ix) yields a matrix that is equivalent to A, has the same general form as Cand has (1,1) entry s3 such that s3 is a proper divisor of s2, whence s2 and s3 are distinct proper divisors of si. Since .Si has only finitely many distinct proper divisors, a finite number of repetitions of this process produces a matrix that is equivalent to A, has the same general form as C, and has a (1,1) entry which divides all other entries of the matrix. (xi) Use induction and (x) to show that A is equivalent to a diagonal matrix '-(!•
ti
as in the statement of the theorem. Since the rank of F is obviously
r, the rank of A is r by Theorem 2.6. (xii) (uniqueness) Let A and F be as in (xi), with du . . . , dr, the diagonal ele ments of Lr. Suppose M is a matrix equivalent to A (so that rank M = r) and A^is a matrix equivalent to M of the form (:■' s) where Z./ is an r X r diagonal matrix with nonzero diagonal entries ki such that ki | k21 • ■ • | kT. By Theorem 1.2 F is the matrix of a homomorphism /:/?“-+ Rn relative to bases of Rn and m {vu .. . , vm | of R . Consequently, /(«.) = d{Vi for z = 1,2,... ,r and /(«,-) = 0 for = r + \,... ,n, whence Im /= Rdxvx ©• • •© Rdrvr. By the analogue for modules of Corollary 1.8.11, i?m/Im/ RvJRd^ @ • ■ ■ @ Rvr/Rdrvr @ Rvr+l © • • • 0 Rv„ = R/{dx) © • • • © R/(dr) ® R © - - - © / ? (m summands; di\d 21 • • • I dr). Since F is equivalent to N by hypothesis, Theorem 1.6 implies that N is the matrix of f relative to a different pair of bases. A repetition of the preceding argument then shows that i?m/Im / = R/(k J © • • • © R/(kr) © R © • • • © R
i
CHAPTER VII LINEAR ALGEBRA
342
(ra summands; k i \k 2 \- - ’\kr). The structure Theorem IV.6.12 for modules over a principal ideal domain implies that (dt) = (/c,) for i = 1,2,..., r. ■ A simplified version of the techniques used in the proof of Proposition 2.11 may be used to obtain the canonical form E?'m of an n X m matrix A over a division ring D. If A = 0 = there is nothing to prove. If a^ is a nonzero entry in A, then interchanging rows / and 1 and columns j and 1 moves to position (1,1). Multi plying row 1 by a,-,-1 yields a matrix with first row of the form (1 *,c 2 ,..., cm). Sub tract suitable multiples of row 1 [resp. column 1] from each subsequent row [resp. column] and obtain a matrix of the form:
If every = 0, we are done. If some ^ 0, then we may repeat the above proce dure on the (/? — 1) X (m — 1) submatrix (c,-,). Since row [column] operations on rows 2,... , n [columns 2,..., m\ do not affect the first row or column, we obtain a matrix '1ft 0 0 Ik 0 0
0 0 d$3
0
dfi3
0
Continuing this process eventually yields the matrix E^'m for some r. By Corollary 2.10 A is equivalent to E?’m, whence r = rank A and m is the canonical form of A under equivalence by Theorem 2.6. A modified version of the preceding technique gives a constructive method for finding the inverse of an invertible matrix, as is seen in the proof of;
Proposition 2.12. The following conditions on an n X n matrix A over a division ring D are equivalent: (i) (ii) (iii) (iv)
rank A = n; A is equivalent to the identity matrix In; A is invertible; A is the product of elementary transformation matrices.
SKETCH OF PROOF, (i) <=> (ii) by Theorem 2.6 since ££’” = /„. (i) => (iii) The rows of any matrix of rank n are necessarily linearly independent (see Theorem IV.2.5 and Definition 2.2.) Consequently, the first row of A — (a„) is not the zero vector and a,, ^ 0 for some j. Interchange columns j and 1 and multiply the new column 1 by a x ~ l . Subtracting suitable multiples of column 1 from each succeeding column yields a matrix
APPENDIX: ABELIAN GROUPS DEFINED BY GENERATORS AND RELATIONS 343
'I d 0 bn 622 B=
. . . Un /,
Ai] Un 2
B is equivalent to A by Corollary 2.10. Assume inductively that there is a sequence of elementary column operations that changes A to a (necessarily equivalent) matrix Ik -1 0 Ck 1 * OfcA: C=
' ' ‘ Cnk For some j > A, ^ 0 since otherwise row k would be a linear combination of rows 1,2,... ,k — 1. This would contradict the fact that rank C = rank A — n by Theorem 2.6. Interchange columns j and k, multiply the new column k by c k ~ l and subtract a suitable multiple of column k from each of columns 1,2,... ,k — 1, k + 1,..., n. The result is a matrix D that is equivalent to A (Corollary 2.10): 0
fc+l1 •••Uk 1 fc+i•■-Ulc+l n +
D=
*
UnR
This completes the induction and shows that when k — n, A is changed to /n by a finite sequence of elementary column operations. Therefore by Theorem 2.8 A(JF\F2• • • Ft) = In with each Fi an elementary matrix. The matrix Fi/v • • Ft is a twosided inverse of A by Exercise 1.7, whence A is invertible. Corollary 2.9 and the fact that A = Fr1- ■ Ff'Fr1 show that (i) => (iv). (iii) => (i) by Lemma 1.5 and Theo rem 2.3. (iv) => (iii) by Corollary 2.9. ■ REMARK. The proof of (i) (iii) shows that A~l = FiF2- - Ft is the matrix ob tained by performing on /„ the same sequence of elementary column operations used to change A to 7n. As a rule this is a more convenient way of computing inverses than the use of determinants (Section 3).
APPENDIX: ABELIAN GROUPS DEFINED BY GENERATORS AND RELATIONS An abelian group G is said to be the abelian group defined by the generators ai,... ,a m e G) and the relations
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344
fncii + rncii 4~ • • • 4” fimam — 0,
4~ r-nfii + ■ • ■ + r2mam = 0,
fnldl 4~ f-ntdl 4” ' ' ’ 4" fnmdm — 0, (r,, s Z) provided that G = F/K, where F is the free abelian group on the set |«i, . . ., «„,} and K is the subgroup of F generated by bi = rnai 4-- ■ - 4- rimam, £>2 — f2iCii + ■ ■ ■ + r2mam i - - - » bn fniai + ■■■■!■ fnmam - Note that the same symbol at denotes both an element of the group G and a basis element of the free abelian group F(see Theorem II. 1.1). This definition is consistent with the concept of generators and relations discussed in Section 1.9 (see Exercise 10). The basic problem is to determine the structure of the abelian group G defined by a given finite set of generators and relations. Since G is finitely generated, G is necessarily a direct sum of cyclic groups (Theorem II.2.1). We shall now determine the orders of these cyclic summands. Let G be the group defined by generators au ... ,am and relations 22 ruai = 0 3
as above. We shall denote this situation by the n X m matrix A = (ri,). The rows of A represent the generators bi,... ,b n of the subgroup K relative to the ordered basis {ai,. . . , am} of F. We claim that elementary row and column operations performed on A have the following effect. (i) If B = (Si,) is obtained from A by an elementary row operation, then the elements ci = 5nai + • • • + Simam, . . . , cn = snXa\ 4- - - • 4- snmam of F (that is, the rows of B) generate the subgroup K. (Exercise 11 (a)). (ii) If B = (sn) is obtained from A by an elementary column operation, then there is an easily determined basis {a i , . . . ,am') of F such that bi = snai + Sna2 4■ ■ • 4- simam' for every z (Exercise 11 (b), (c)). If K 9^ 0, then by Proposition 2.11 and Exercise 7, A may be changed via a finite sequence of elementary row and column operations, to a diagonal matrix
0
dr 0 00
such that di ^ 0 for all / and dx | \ di | • • • | dr. In other words a finite sequence of elementary operations yields a basis \ uu . . ., um \ of Fsuch that {diui,d2u-i,. . . , dTuT\ generates K. Consequently by Corollary 1.8.11
G = F/K = (Z«! © • • • © Zum)/(Zdltii ® ■ ■ • ® Zdriir © 0 ©•••©' 0)
^ Z/diZ©- • -® ZMZ© z/o ©■ • •© z/o =
^©---©^©Z©---©Z,
where the rank of (Z ©• ■ •© Z) is m — r and di \ d2 \ • • • | dr (see Theorem II.2.6).
APPENDIX: ABELIAN GROUPS DEFINED BY GENERATORS AND RELATIONS 345
EXAMPLE. Determine the structure of the abelian group G defined by genera tors a,b,c and relations 3a -f 9b + 9c = 0 and 9a — 3b + 9c = 0. Let F be the free abelian group 2a -f Z6 Zc and K the subgroup generated by bi = 3a -\- 9b + 9c and 62 = 9a — 3b + 9c. Then G is isomorphic to F/K and we have the matrix
We indicate below the various stages in the diagonalization of the matrix A by ele mentary operations; (sometimes several operations are performed in a single step). At each stage we indicate the basis of F and the generators of K represented by the given matrix; (this can be tricky; see Exercise 11).
Matrix
Generators of K, expressed as Ordered basis of F linear combinations of this basis
(3 9 9\ \9 -3 9/
a; b; c
bi = 3a + 9b + 9c bi = 9a — 3b + 9c
/3 0 9\ \9 -30 9J
a + 3b; b; c
bi = 3(a + 3b) + 9c b2 = 9(a + 3b) - 30b + 9c
(3 0 0\ \9 -30 —18/
a + 3b + 3c; 6; c
6i = 3(a + 36 + 3c) 62 = 9(o + 3b + 3c) - 30b - 18c
(3 0 0\ \0 -30 —18/
a + 3b + 3c; b\c bi = 3(a + 3b + 3c) b2 - 3bx = -30b- 18c
(3 0 0\ \0 18 30/
a + 36 + 3c; c; b b\ = 3(a -J- 3b + 3c) ~(b2 - 360 = 18c + 306
(3 \0
0 0\ 18 12/
a + 36 + 3c; c + 6; 6 6i = 3(a + 36 + 3c) — 62 "h 36i = 18(c -f- 6) —|— 126
(3 \0
0 0\ 6 12/
a -\- 36 -\- 3c; c -f- 6; 61 = 3(a -\- 36 -J- 3c) 6 + (c + 6) -62 + 3 61 = 6(c + 6) + 12(26 + c)
(3 \0
0 0\ 6 0/
a + 36 4- 3c; 56 + 3c; 61 = 3(a + 36 + 3c) 26 -I- c -62 + 36i = 6(56 + 3c)
Therefore G ^ F/K ^ Z/3Z ® Z/6Z ® Z/0Z ^ Z3 ® Z6 ® Z. If veG is the image of v + K e F/K under the isomorphism F/K =G, then G is the internal direct sum of a cyclic subgroup of order three with generator a + 36 + 3c, a cyclic sub group of order six with generator 56 + 3c, and an infinite cyclic subgroup with generator 26 + c.
EXERCISES 1. Let / g : E E, h : E F, k : F G be linear transformations of left vector spaces over a division ring D with dim^E = n, dim^F = m, dimr,G = p. (a) Rank (/+ g) < rank / + rank g. (b) Rank (kh) < min {rank A, rank k}. (c) Nullity kh < nullity h + nullity k. (d) Rank / + rank g — n < rankZs < min {rank/, rank #}.
346
CHAPTER VII LINEAR ALGEBRA
(e) Max {nullity#, nullity h\ < nullity hg. (f) If m rt, then (e) is false for h and k. 2. An n X m matrix A over a division ring D has an m X n left inverse B (that is, BA = Im) if and only if rank A = m. A has an rn X n right inverse C (with AC = In) if and only if rank A = n. 3. If (c»i,c»2- • ■Ci.n) is a nonzero row of a matrix (c„), then its leading entry is cu where t is the first integer such that cit ^ 0. A matrix C = (c,-,-) over a division ring D is said to be in reduced row echelon form provided: (i) for some r > 0 the first r rows of C are nonzero (row vectors) and all other rows are zero; (ii) the leading entry of each nonzero row is \D\ (iii) if c,j = \D is the leading entry of row /', then ckj = 0 for all k ^ /; (iv) if , cr3r are the leading entries of rows 1,2, , r , then j\ < U <■ • • < jr. (a) If C is in reduced row echelon form, then rank C is the number of nonzero rows. (b) If A is any matrix over D, then A may be changed to a matrix in reduced row echelon form by a finite sequence of elementary row operations. 4. (a) The system of n linear equations in m unknowns x, over a field K
ci\\X\ + cinXo + ■ ■ ■ -f a\wXm = b\
CtnlXy Cin‘lX'2 I"'I(111);Xjh: bn
has a (simultaneous) solution if and only if the matrix equation AX = B has a solution X, where A is the n X m matrix (oiy), X is the m X 1 column vector (xix 2 - ■ xmY and B is the «X1 column vector (bib2- ■ - bn)1. (b) If Ax,By are matrices obtained from A,B respectively by performing the same sequence of elementary row operations on both Ax and Bx then X is a solu tion of AX = B if and only if X is a solution of AxX = Bx. (c) Let C be the n X (m + 1) matrix ^11 * * * Cl\m
1*
&nm
Then AX = B has solution if and only if rank A = rank C. In this case the solu tion is unique if and only if rank A = m. [Hint: use (b) and Exercise 3.] (d) The system AX = B is homogeneous if B is the zero column vector. A homogeneous system AX — B has a nontrivial solution (that is, not all x, = 0) if and only if rank A < m (in particular, if n < m). 5. Let R be a principal ideal domain. For each positive integer r and sequence of nonzero ideals I\ ZD I, Z) • • • ID Ir choose a sequence du . . . , dr e R such that (dj) = Ij and dx | d21 • • • | dT. For a given pair of positive integers (n,m), let S be the set of all n X m matrices of the form &:) , where r = 1,2,..., min (n,m) and Lr is an r X r diagonal matrix with main diagonal one of the chosen se-
APPENDIX: ABELIAN GROUPS DEFINED BY GENERATORS AND RELATIONS 347
quences du ■ . ■, dr. Show that S is a set of canonical forms under equivalence for the set of all n X m matrices over R. 6. (a) If / : E —> F is a linear transformation of finite dimensional vector spaces over a division ring, then there exist bases {uu . .., un} of E and {i>i,..., vm } of F and an integer r (r < min (m,n)) such that /(«,-) = Vi for / = 1,2, . . ., r and /(«,) = 0 for /' = r + 1, . . . , n. (b) State and prove a similar result for free modules of finite rank over a principal ideal domain [see Proposition 2.11], 7. Let R be a Euclidean domain with “degree function” 4> '• R — {0} —»N (Definition III.3.8). (For example, let R = Z). (a) UA = (:i) is a 2 X 2 matrix over R then A can be changed to a diagonal matrix D by a finite sequence of elementary row and column operations. [Hint: If a 7* 0, h 9^ 0, then b = aq + r with r = 0, or r 9^ 0 and 0(r) < 0(c). Performing suitable elementary column operations yields:
(
a b\ fab — aq\ _ fa r\ fr c\ c d)~*\c d — cq) ~ \c
Since (/-) < («), repetitions of this argument change A to B = (: •) with (*) < 0(a) if s 0. If w ^ 0, a similar argument with rows changes B to t hA q + ) with (r) < <£Cs) < 4>(a) if t 9^ 0; (and possibly w 9^ 0). Since
(
the degrees of the (1,1) entries are strictly decreasing, a repetition of these argu ments must yield a diagonal matrix D =
(^l
^ ^ after a finite number of steps.]
(b) If A is invertible, then A is a product of elementary matrices. [Hint: By (a) and the proof of Corollary 2.10 D — PA Q with P,Q invertible, whence D is in vertible and dud2 are units in R. Thus A =
^ ^0^
use
Corollary 2.9.] (c) Every n X m secondary matrix (see the proof of Proposition 2.11) over a Euclidean domain is a product of elementary matrices. 8. (a) An invertible matrix over a principal ideal domain is a product of elementary and secondary matrices. (b) An invertible matrix over a Euclidean domain is a product of elementary matrices [see Exercise 7], 9. Let «i, n2, . . . , nt, n be positive integers such that m + n2 +•••+«* = n and for each i let Mi be an n, X /i, matrix. Let M be the n X n matrix
' My
m2 0
0
Mt
CHAPTER VII LINEAR ALGEBRA
348
where the main diagonal of each Mx lies on the main diagonal of M. For each permutation a of {1,2,.. ., t), M is similar to the matrix
/o In\ [Hint: If / = 3, a = (13), and P = I /„2 J, then P~l = I /„2 j and
/0 /„A
V„, 0 / \ln3 0/ PMP~l = oM. In the general case adapt the proof of results 2.8-2.10.] 10. Given the set [a x ,. . ., a n \ and the words w u w 2 ,. . . ,w r (on the at), let F* be the free (nonabelian multiplicative) group on the set \a x , ..., a n \ and let M be the normal subgroup generated by the words w u w 2 ,..., wT (see Section 1.9). Let N be the normal subgroup generated by all words of the form aia}arlafl. (a) F*/M is the group defined by generators \a x , ...,a n \ and relations {wi = w2 = ■ ■ ■ — wr = e\ (Definition 1.9.4). (b) F*/N is the free abelian group on [a x , . . . , a n \ (see Exercise II. 1.12). (c) F*/(M V AO is (in multiplicative notation) the abelian group defined by generators \a x ,... ,a n \ and relations {w x = w 2 = ■ ■ ■ = w r = e\ (see p. 343). (d) There are group epimorphisms F* —> F*/N —> F*/(M V N). 11. Let F be a free abelian group with basis [a x ,... ,a m \. Let K be the subgroup of F generated by bx = rnax H----- (- rimom,. . ., bn = rnlaj, -\---- h rnmam (r„ e Z). (a) For each both \ b i , , bt-X, —bi,bi+x,... ,bn} and [bu..., + rbh bi+u ■ . - , bn] (r e Z; iV j) generate K. [See Lemma II. 1.5.] (b) For each i[a x ,..., o,_i,—ai,ai+u_______, fln} is a basis of F relative to which bj = r jiai + • • • + ry,i_iOi_i — rJt( —a,-) + /■/,»+iO»+i + • • • + f (c) For each / and j ^ i \ a x , . . . , a^x,a, — roi,ai+u . .. , am) (r e Z) is a basis of F relative to which bk = rkxa\ H--- b + (rki + rrk])cn + rk^ai+x + • ■ • ■+■
4”
— rai) + r k 'j +xcij +x + - - • -f- r kma m.
12. Determine the structure of the abelian group G defined by generators [ a,b | and relations 2a + Ab = 0 and 3b = 0. Do the same for the group with generators \a,b,c,d) and relations 2a 4 3b = Aa = 5c + lid = 0 and for the group with generators \a,b,c,d,e\ and relations {a - lb 4- lAd - 21c = 0; 5c - lb - 2c + 10d ~ I5e = 0; 3a - 3b - 2c + 6d — 9e = 0; a — b 4- 2d — 3e = 0).
3. DETERMINANTS The determinant function Mat„7? —» R is defined as a particular kind of /?-multilinear function and its elementary properties are developed (Theorem 3.5). The re mainder of the section is devoted to techniques for calculating determinants and the connection between determinants and invertibility. With minor exceptions this ma
3. DETERMINANTS
349
terial is not needed in the sequel. Throughout this section all rings are commutative
with identity and all modules are unitary. If B is an 7?-module and n > 1 an integer, Bn will denote the 7?-module B © B ©''' © B (« summands). Of course, the underlying set of the module Bn is just the cartesian product B X - ■ ■ X B.
Definition 3.1. Let Bi,. . . , Bn and C be modules over a commutative ring R with identity. A function f : Bj X • ■ • X Bn —> C is said to be R-multilinear if for each i = 1,2,. .. , n and all r,s e R, bj e Bj and b,b' e B;: f(bi,. .., bi_i,rb + sb',bi+i,.. ., bn) = rf (bi, .. ., bi_i,b,bi+i,..., bn) + s f(bi,... , bi_i,b',bi+i,..., b„).
IfC = R, then f is called an n-linear or R-multilinear form. 7/C = R and Bj = B2 = • • • = Bn = B, then f is called an R-multilinear form on B. The 2-linear functions are usually called bilinear (see Theorem IV.5.6). Let B and C be 7?-modules and / : Bn —> C an 7?-multilinear function. Then /is said to be sym metric if
f(ba 1,. . ., ban) = f(bu . .., bn) for every permutation <7 eSn, and skew-symmetric if /(^
f is said to be alternating if f(bi,. . ., bn) = 0 whenever bt =■ bj for some / j. EXAMPLE. Let B be the free 7?-module 7? © 7? and let d : B X B —» 7? be de fined by ((Gn,fli2),(021,022)) |—> O11O22 — O12 021. Then d is a skew-symmetric alternating bilinear form on 7?. If one thinks of the elements of B as rows of 2 X 2 matrices over R, then d is simply the ordinary determinant function.
Theorem 3.2. 7/B and C are modules over a commutative ring R with identity, then every alternating R-multilinear function f : Bn —» C is skew-symmetric. SKETCH OF PROOF. In the special case when n = 2 and a = (1 2), we have: 0 = f(bi + bi,bi + bi) = f(bi,bi) + f(bi,b-i) + f(bi,bi) + f(b-2,b-2) =
0 + fibiybi) + f(b2,bi) + 0,
whence f(b2,bi) = —f(bi,bi) = (sgn a) f(bi,b2). In the general case, show that it suffices to assume a is a transposition. Then the proof is an easy generalization of the case n = 2. u Our chief interest is in alternating /7-linear forms on the free 7?-module /?". Such a form is a function from (Rn)n = /?"©•••© /?"(« summands) to R.
CHAPTER VII LINEAR ALGEBRA
350
Theorem 3.3. 7/R is a commutative ring with identity and r e R, then there exists a unique alternating K-multilinear form f : (Rn)n —* R such that f(£i,£2,. . ., e„) = r, where {ei, ..., en} is the standard basis of Rn. REMARK. The standard basis is defined after Definition 2.1. The following facts may be helpful in understanding the proof. Since the elements of Rn may be identified with 1 X n row vectors, it is clear that there is an /?-module isomorphism (RnY = Mat„/? given by (*,*2,. - -, Xn) |—*■ A, where A is the matrix with rows A",,*,..., Xn. If {ei,..., s„} is the standard basis of Rn, then (ei,e 2 ,..., e„) {-> /„ under this isomorphism. Thus the multilinear form /of Theorem 3.3 may be thought of as a function whose n arguments are the rows of n X n matrices.
PROOF OF 3.3. (Uniqueness) If such an alternating n-linear form /exists and if (A'i,... ,Xn) e (/?")", then for each / there exist ai} e R such that*,- = (0ii,0i 2 ,..., ain)
n
—
52 Oifij. (In other words, under the isomorphism (/?“)" ^ Mat„/?, {Xu . . . ,*„)(—>
1 (uii).) Therefore by multilinearity, j=
fiXij ■ • • > Xn) =
/(52 ii
n
- Z Z - Z i\ h
alha2n'
in
in
' ' ^nJn/(en>eJ2» • • • * eJn)-
Since /is alternating the only possible nonzero terms in the final sum are those where jy,h, - - - ,jn are all distinct; that is, {y'i,. . . ,jn} is simply the set {1,2,...,«) in some order, so that for some permutation a eSn, ij\,. . . ,/„) = (oT,. . . , crn). Conse quently by Theorem 3.2,
fiX 1, . . . ,X n ) = 52 a U\ a 1c2' • • Qnan /(e
= 52 (sen
oeSn
‘
Gnon
/(e 1 ,£2, . . . , e„).
Since /(e 1,. .., e„) = r, we have
fiXu ■ • • > Xn) =
52 (s8n o')ra\aianc2
’ ^ntrn*
(1)
aeSn
Equation (1) shows that f(Xu ..., Xn) is uniquely determined by *1, and r. (Existence) It suffices to define a function / : (/?")" —» R by formula (1) (where Xi = (a,-i, . . . , ain)) and verify that / is an alternating /j-linear form with fie 1,..., e„) = r. Since for each fixed k every summand of 5Z (sgn a)ra\ai ' &nan <7E&n contains exactly one factor ai} with i = k, it follows easily that / is R-multilinear. n
Since e, =
52 •Mj (Kronecker delta), /(e 1,...,
e„) = r. Finally we must show that
i= l
fiXu ... ,X n ) = 0 if Xi = X 3 and i ^ j. Assume for convenience of notation that / = l,y = 2. If p = (12), then the map A„ —> S n given by a |—► ap is an injective func tion whose image is the set of all odd permutations (since a even implies op odd and \A ri \ = |S„|/2). ThusS,, is a union of mutually disjoint pairs {er.crp J with cr e A n . If a is even, then the summand of fiXuXr,X3,. . . , Xn) corresponding to a is + raiaiOi
351
3. DETERMINANTS
Since Xi — X2, a XaX — .2ai, and a2ai — aU2, whence the summand corresponding to the odd permutation crp is: ““ ^*^l<rpl^2irp2^3cTp3 * * '&nopn
/"^Itr2^2
Thus the summands of f(XuXi,X3,... ,Xn) cancel pairwise and
f(XuXuX3, ...,X n ) = 0. Therefore /is alternating. ■ We can now use Theorem 3.3 and the Remark following it to define determinants. In particular, we shall frequently identify Mat„/? and (R n ) n under the isomorphism (given in the Remark), which maps (ei, . . ., e«)|—> /«. Consequently, a multilinear form on Mat„R is an /^-multilinear form on (R n ) n whose arguments are the rows of n X n matrices considered as elements of Rn.
Definition 3.4. Let R be a commutative ring with identity. The unique alternating R-multilinear form d : MatnR —> R such that d(In) = 1r is called the determinant function on Matn R. The determinant of a matrix A e Matn R is the element d(A) e R and is denoted |A|.
Theorem 3.5. Let R be a commutative ring with identity and A,B e MatnR. (i) Every alternating R-multilinear form f on MatnR is a unique scalar multiple of the determinant function d. (ii) If A = (a;j), then |A| = (sSn o')aiffia2„2- ■ -an
r
352
CHAPTER VII LINEAR ALGEBRA
(iv) AA~ X = /„ implies MIM_1I = \AA~ l \ = |/„| = 1 by (iii). Hence\A\ isaunit in R with \A\~ l — \A~l |. (v) Similarly, B = PAP~l implies JB\ = |P||/4||F|_1 = \A\ since R is commutative. (vi) Let A = (ai,). If iu are the integers 1,2,..., n in some order, then since R is commutative any product fltli«t22 • • ■ cunn may be written as atna-2n ■ ■ ■ On,n. If a is the permutation such that
jA^j ~ ^ 1 (sgn o)biffim * ■ bnan ^ j (sgn ffcSn ITzSn • ■ -ana~K = \A\.
= 2Z (sgn C~ltSn
(vii) By hypothesis either at, = 0 for all j < i or at] — 0 for all j > /. In either case show that if a s.Sn and
= —d(Xi,. . . ,Xi,.. Similarly if B has rowsXu ... , X r X and alternating
.,Xj,. . . ,X n) x
+X X
n
= — Mlthen since dis multilinear
|fi| = d(Xu ... ,Xi,..., rXi + Xj, ...,X n ) = rd(Xu . . . ,Xi, . . . ,Xi, . . . ,X n ) + d(Xu . ■ • i Xi, . . . , Xj, . . . , Xn) = r0+ Ml = MlThe other statement is proved similarly; use (v) for the corresponding statements about columns. ■ If R is a field, then the last part of Theorem 3.5 provides a method of calculating
Ml- Use elementary row and column operations to change A into a diagonal matrix B = (bij), keeping track at each stage (via (viii)) of what happens to Ml- By (viii), |Z?| = r\A\ for some 0 ^ r e R. Hence rMI = b^b^- ■ bnn by (vii) and
Ml = r~lbn - • •b„n. More generally the determinant of an n X n matrix A over any commutative ring with identity may be calculated as follows. For each pair (i,j) let A„ be the (« — 1)X(« — 1) matrix obtained by deleting row / and column j from A. Then Mill e R is called the minor of A = (all) at position (i,j) and ( — 1),+,M»jI e R is called the cofactor of a,-,-.
Proposition 3.6. If A is an n X n matrix over a commutative ring R with identity, then for each i = 1,2, . . . , n,
n IAI = E (-lY+iaij\Aij\ j=
i
3. DETERMINANTS
353
and for each j = 1,2, . . . , n,
71
Ml = i= 521( — l)‘+ya,3M,yfThe first [second] formula for |y<| is called the expansion of |A| along row i [column j], PROOF OF 3.6. We let j be fixed and prove the second statement. By Theorem 3.3 and Definition 3.4 it suffices to show that the map <£ : Mat„/? —> R given by
71
A = (an) |-»
52
( — l)*+,Oi,-|^t-,-| is an alternating /?-multilinear form such that t=l <£(/„) = lfl- Let X,, ... ,X„ be the rows of A. If Xk = Xt with 1 < k < t < n, then \Aij\ = 0 for / ^ k,t since it is the determinant of a matrix with two identical rows. Since Ak, may be obtained from At] by interchanging row t successively with rows t — 1,. .. ,A + 1, \A k y| = (—1) ‘-b-'lAt,} by Theorem 3.5. Thus <}>(A) = ( — \)k+i\Akj\ +- (—iy +i \A t j\ = (—lfc—ij^ t y j ( — = o. Hence <£ is alternating. If for some k, Xk = rYk + sWk, let B = (bi,) and C = (c„) be the matrices with rows Xu. .. ,Xn_uYk,Xk+1,... ,X n , and Xu ... ,Xk_ulVk,Xk+u ...,*„ respectively. To prove that $ is /?-multilinear we need only show that 4>(A) = r(j>(B) + s4>(C). If /' = k, then \A kj \ = \B kj \ = |CfcJ-|, whence a kj \A kj \ = (rbkj -+■ sck,)\Ak,-\ = rbkj\Bk]-\ + sckj\Ck}\. If /' ^ k, then since each \Aij\ is a multilinear function of the rows of Ai} and an = hi = ca for i k, we have a \A,j\ = aij(r\Bi,\ + j|C«- -|) = rb j\Bi,\ + 5Ci,|C,-,|. 13 t 3 It follows that
Proposition 3.7. If A = (aij) is an n X n matrix over a commutative ring R with identity and Aa = (bij) is the n X n matrix with bij = (— l)1+i|Aji|, then AAa = |A|In = AaA. Furthermore A is invertible in MatnR if and only if |A| is a unit in R, in which case A-1 = |A|^1Aa. The matrix Aa is called the classical adjoint of A. Note that if R is a field, then \A\ is a unit if and only if \A\ ^ 0. PROOF OF 3.7. The (/,_/) entry of
AAa
is =
52 ( —
71 l)1+kaik\Ajk\.
If i = j,
k=l
then ca = by Proposition 3.6. If i ^ j (say i < j) and A has rows*,. . . , l e t B = (bl}) be the matrix with rows *, ,*_i, Xi,X j+ i,... ,X„. Then brk = alk = b, k and\A ik \ = \B,k\ for all k; in particular, \B\ = 0 since the determinant is an alternating form. Hence n
di = E
n
(-l)i+kOik\A]k\
k=
1
k=
1
= Z (-lV+kb,k\Bik\ = \B\ = 0.
Therefore, c1? = 8i,\A\ (Kronecker delta) and A A' 1 = \A\I r . In particular, the last statemeni holds with A1 in place of A : A'(A')n = \A'\I n . Since (A°)1 = (A')“, we have \A\I n = \A'\I n = A‘(A‘)a = Al(Aa)‘ = (AaA)‘, whence AaA = (\A\I n ) 1 = \A\I n . Thus if \A\ is a unit in R, \A\~^A a e Mat„/? and clearly (|Aj^A^A = In = A(jA\-1Aa). Hence A is invertible with (necessarily unique) inverse A~x = \Al~1Aa. Conversely if A is invertible, then \A\ is a unit by Theorem 3.5. ■
354
CHAPTER VII. LINEAR ALGEBRA
Corollary 3.8. (Cramer's Rule) Let A = (a;j) be the matrix of coefficients of the system of n linear equations in n unknowns anXi + 812X2 • • ■ + ainXn — b:
a„ix, -)- an2x2
a„nxn — b.
over a field K. lf\ A| ^ 0, then the system has a unique solution which is given by:
PROOF. Clearly the given system has a solution if and only if the matrix equa tion AX = B has a solution, where X and B are the column vectors A' = (av • •xn)‘, B = (bi - ■ bny. Since \A\ ^ 0, A is invertible by Proposition 3.7, whenceX — A~lB is a solution. It is the unique solution since AY — /? implies Y = A~lB. To obtain the formula for a, simply compute, using the equation
X = AlB = ([AY-'A^B = \A\~\A a B). u
EXERCISES Note: Unless stated otherwise all matrices have entries in a commutative ring R with identity. 1. If r + r ^ 0 for all nonzero r e R, then prove that an n-linear form Bn —* R is alternating if and only if it is skew-symmetric. What if char R = 2? 2. (a) If m > n, then every alternating /^-multilinear form on (Rn)n is zero. (b) If m < n, then there is a nonzero alternating /?-multilinear form on (Rn)m. 3. Use Exercise 2 to prove directly that if there is an /^-module isomorphism Rm ~ Rn, then m = n. 4. If As. Mat„/?, then \A a \ = M|"_1 and (Aa)a — \A\ n ~ 2 A. 5. If R is a field and A,B e Mat„/? are invertible then the matrix A rB is invertible for all but a finite number of re R. 6. Let A be an n X n matrix over a field. Without using Proposition 3.7 prove that A
is invertible if and only if | A\ ^ 0. [Hint: Theorems 2.6 and 3.5 (viii) and Proposi tion 2.12.J 7. Let F be a free /?-module with basis V — {uu . . ., un}. If
equations
4. DECOMPOSITION OF A SINGLE LINEAR TRANSFORMATION Gll-Xl + ■ • ■ + ci\n Xn =
Un lX|
4
ClnnXn
355
0
0
and that A = (a,,) is the n X n matrix of coefficients. Then \A\b x = 0 for every [Hint: If Bi is the n X n diagonal matrix with diagonal entries 1 R ,. . . , 1*A, lfl,. . . , Iff, then \ ABi\ = \A\bi. To show that \AB t \ = 0 add b, times column j of ABi to column i for every j j* i. The resulting matrix has determinant \ABi\ and (k,i) entry auibx + a*262 -----h aknbn = 0 for k = 1,2
4. DECOMPOSITION OF A SINGLE LINEAR TRANSFORMATION AND SIMILARITY The structure of a finite dimensional vector space E over a field K relative to a linear transformation E—*E is investigated. The linear transformation induces a de composition of E as a direct sum of certain subspaces and associates with each such decomposition of £ a set of polynomial invariants in A^x] (Theorem 4.2). These sets of polynomial invariants enable one to choose various bases of E relative to each of which the matrix of the given linear transformation is of a certain type (Theorem 4.6). This leads to several different sets of canonical forms for the relation of similar ity in Matn-K (Corollary 4.7). Note. The results of this section depend heavily on the structure theorems for finitely generated modules over a principal ideal domain (Section IV.6). Let K be a field and <}>:E-+E a linear transformation of an ^-dimensional /^-vector space E. We first recall some facts about the structure of Homx(£,£) and MatnA'. HomK(£,£) is not only a ring with identity (Exercise IV. 1.7), but also a vector space over K with (k\p)(u) = k\p(u) (k e K,u e E,\p e HomK(£,£)); see the Re mark after Theorem IV.4.8). Therefore if / = 22 is a polynomial in AT[x], then /WO = 22 kift is a well-defined element of HomA(£,£) (where 0° = 1E as usual). Similarly the ring MatnA" is also a vector space over K. If A e MatnA, then f(A) = 22 kiA{ is a well-defined n X n matrix over K (with A0 = /„).
Theorem 4.1. Let E be an n-dimensional vector space over a field K, 0 : E —» E a linear transformation and A an n X n matrix over K. (i) There exists a unique monic polynomial of positive degree, q# e K[x], such that = 0 and | f for all f e K[x] such that f(<£) = 0. (ii) There exists a unique monic polynomial of positive degree, qA e K[x], such that qA(A) = 0 and qA | f for all f e K[x] such that f(A) = 0. (iii) If A is the matrix of <j> relative to some basis of E, then qA = q
356
CHAPTER VII. LINEAR ALGEBRA
sional over K. Since X[jc] is infinite dimensional over K, we must have Ker f ^ 0 by Corollary IV.2.14. Since ATM is a principal ideal domain whose units are precisely the nonzero elements of K (Corollary III.6.4), Ker f = (q) for some monic q e ATf*]. Since f is not the zero map, (q) K[x], whence deg q > 1. If Ker f = (qi) with q, e K[x\ monic. then q | qv and q x \qby Theorem III.3.2, whence q = qx since both are monic. Therefore q$ = q has the stated properties. (ii) The proof is the same as (i) with A in place of and Mat„X in place of HorriA-(£,£). qA e K[x] is the unique monic polynomial such that (qA) = Ker £A, where f a : K[x] —■» Mat„AT is the unique ring homomorphism given by /H f(A). (iii) Let A be the matrix of relative to a basis U of E and let 6 : Hom^E) = Mat„/? be the isomorphism of Theorem 1.2, so that 8(
is commutative by Theorem III.5.5 since 6^(x) = 0(4>) — A — £a(x) and 6^(k) = 0tklE) = kln = £A(k) for all k e AT. Since G is an isomorphism, ( are as in Theorem 4.1, then the polynomial q$ [resp. qA] is called the minimal polynomial of the linear transformation <£ [matrix A], In general, q^ is not irreducible. Corollary 1.7 and Theorem 4.1(iii) immediately imply that similar matrices have the same minimal polynomial. Let AT, E, and 4> be as above. Then induces a (left) Af[jr]-module structure on E as follows. If ft K[x] and u e E, then /() e Homk(£,£) and fu is defined by fu = /(<£)(«). A AT-subspace F of E is said to be invariant under 4> (or <j>-invariant) if 4>{F) (I F. Clearly F is a ^-invariant AT-subspace if and only if F is a Af[x]-submodule of E. In particular, for any v e E the subspace E(
Theorem 4.2. Let 4> : E —> E be a linear transformation of an n-dimensional vector space E over a field K. (i) There exist monic polynomials of positive degree qi,q2, . . . , qt e K[x] and <j>-cyclic subspaces Ei, . . . , Et of E such that E = Ei © E2 ©■ • •© Et and Qi [ q2 [ • • • | qt- Furthermore q, is the minimal polynomial of and qt is the minimal polynomial of 4>. (ii) There exist monic irreducible polynomials pi, . . . , pe e K[x] and ^-cyclic subs ki spaces En, . • • , Eik1,E2i,. . . , E2k.,E3i, . . . , EBke of E such that E = Eij and »= l j—l for each i there is a nonincreasing sequence of integers mu > mi2 > ■ - • > m;^ > 0 such that p'nij is the minimal polynomial of
22 22
4. DECOMPOSITION OF A SINGLE LINEAR TRANSFORMATION
357
nomials {p™" | 1 < i < s; 1 < j < k; J is uniquely determined by E and 4> and pmupmj,. . . p%i is the minimal polynomial of 4>. The polynomials qu . . . ,qt in part (i) of the theorem are called the invariant factors of the linear transformation
358
CHAPTER VII LINEAR ALGEBRA
Theorem 4.3. Let : E —* E be a linear transformation of a finite dimensional vec tor space E over a field K. Then E is a 0-cyclic space and <J>has minimal polynomial q = xr + ar_iXr_1 + • - - + a0 e K[x] if and only if dim kE = r and E has an ordered basis V relative to which the matrix of
1 ^
0
Ik
0
0
0
0
0
0
0
Ik
0
0
0
0
0
0
0
Ik
0
0
0
0 0 0 0 0 ••• 0 1k / —3o —3i —32 — a3 —84 •
■ • — ar—2 —3r —1
J■
In this case V = {v,0(v),02(v), . . . , <£r_1(v)) for some v e E. The matrix A is called the companion matrix of the monic polynomial q e A'| >r].2 Note that if q = x + an, then A = (—r/0). PROOF OF 4.3. (=>)If Eis 0-cyclic, then the remarks preceding Theorem 4.2 show that for some v 5 E, E is the cyclic AT[jr]-moduIe K[x]v, with the A'fxJ-module structure induced by <£. If k0v -f ki<j>(v) + ■ ■ ■ -f- kr-i
[ v,
E is considered as a right /^-vector space and matrices of maps are constructed accord ingly (as on p. 333) then the companion matrix of q must be defined to be A‘ in order to make the theorem true. 2If
4. DECOMPOSITION OF A SINGLE LINEAR TRANSFORMATION
359
Corollary 4.4. Let ip : E —> E be a linear transformation of a finite dimensional vector space E over a field K. Then E is a ip-eyelie space and xp has minimal polynomial q = (x — b)r (b e K) if and only if c//wkE = r andE has an ordered basis relative to which the matrix of xp is b
Ik
0
0
•
■ ■ 0
0
•■ 0 ■0
0
0
b
Ik
0 •
0
0
b
Ik
0
0 0 0 b 1K 0 000• • • O b
0
B=
The r X r matrix B is called the elementary Jordan matrix associated with (x — b)r e K[x). Note that for r = 1,B = (b). SKETCH OF PROOF OF 4.4. Let = xp - b\E e HornK(E,E). Then q = (x — b)r is the minimal polynomial of xp if and only if xr is the minimal poly nomial of 0 (for example, <$>r = (xp — b\ E ) r = q(\p) = 0). E has two XfjcJ-module structures induced by
Lemma 4.5. Let : E —» E be a linear transformation of an n-dimensional vector space E over a field K. For each i = 1, . . . , t let M; be an n; X nj matrix over K, with nt + n2 + • • • + nt = n. Then E = Ei © E2 ©• • ■© Et, where each Es is a <£-in variant subspace o/E and for each i, Mi is the matrix of 4> \ E; relative to some ordered basis of E,, if and only if the matrix of relative to some ordered basis of E is
where the main diagonal of each M; lies on the main diagonal of M.
360
CHAPTER VII LINEAR ALGEBRA
A matrix of the form M as in Lemma 4.5 is said to be the direct sum of the ma trices Mi,..., Mt (in this order). SKETCH OF PROOF OF 4.5. (=>) For each i let Vl be an ordered basis of Ei such that the matrix of 0 j £, relative to Vt is M,-. Since E = Ei ©• • •© Et, it t follows easily that V = U V, is a basis of E. Verify that Mis the matrix of 0 relai=1 tive to V (where V is ordered in the obvious way). (<=0 Conversely suppose V = {Ui,. . . , u n} is a basis of E and M the matrix of 0 relative to V. Let E x be the sub space of E with basis | u u ... ,u ni \ and for i > 1 let £, be the subspace of E with basis | Mr+i, • • •, Ur+m] where r = m + w2 + • • • + «»-i. Then E = Ei ©£ 2 © ■ ■ ■ @£*, each Ei is ^-invariant and Mi is the matrix of 0 | Ei relative to { Wr+l, nr+nt- } . ■
Theorem 4.6. Let 0 : E —> E be a linear transformation of an n-dimensional vector space E over a field K. (i) E has a basis relative to which the matrix of 0 is the direct sum of the com panion matrices of the invariant factors qi, . . . , qt e K[x] of 0. (ii) E has a basis relative to which the matrix of 0 is the direct sum of the com panion matrices of the elementary divisors p"1", . . . , p™8ka s K[x] o/0. (iii) Ifthe minimal polynomial q of0 factors as q = (x— b,)ri(x — b2)ri- ■ - (x — b<0rd (bi s K), which is always the case if K is algebraically closed, then every elementary divisor of 0 is of the form (x — b;)j (j < rj) and E has a basis relative to which the matrix of 0 is the direct sum of the elementary Jordan matrices associated with the ele mentary divisors of 0. The proof, which is an immediate consequence of results 4.2-4.5 (and unique factorization in X[x] for (iii)), is left to the reader. The next corollary immediately yields two (or three if K is algebraically closed) sets of canonical forms for the rela tion of similarity on MatnK.
Corollary 4.7. Let A be an n X n matrix over a field K. (i) A is similar to a matrix D such that D is the direct sum of the companion matrices of a unique family of polynomials qi, . . ., qt. £ K[x] such that qi | q2 [ • • • | qtThe matrix D is uniquely determined. (ii) A is similar to a matrix M such that M is the direct sum of the companion matrices of a unique family ofprime power polynomials p™11, . . . , p™BkB e K[x], where each Pi is prime (irreducible) in K[x], M is uniquely determined except for the order of the companion matrices of the p"‘ij along its main diagonal. (iii) 7/K is algebraically closed, then A is similar to a matrix J such that J is a direct sum of the elementary Jordan matrices associated with a unique family of polynomials of the form (x — b)m (b e K). J is uniquely determined except for the order of the ele mentary Jordan matrices along its main diagonal. The proof is given below. The matrix D in part (i), is said to be in rational canoni cal form or to be the rational canonical form of the matrix A. Similarly, the matrix M
4. DECOMPOSITION OF A SINGLE LINEAR TRANSFORMATION
361
in part (ii) is said to be in primary rational canonical form and the matrix yin (iii) is said to be in Jordan canonical form.3 The word “rational” refers to the fact that the similarity of matrices occurs in the given field A and not in an extension field of K (see Exercise 7). The uniquely determined polynomials qu . .. ,qt in part (i) are called the invariant factors of the matrix A. Similarly, the unique prime power poly nomials p™’ in part (ii) are called the elementary divisors of the matrix A. SKETCH OF PROOF OF 4.7. (ii) Let : K" -* Kn be the linear transforma tion with matrix A relative to the standard basis (Theorem 1.2). Corollary 1.7 and Theorem 4.6 show that A is similar to the matrix D that is the direct sum in some order of the companion matrices of the elementary divisors p™'* of <j>. If A is also similar to Du where A is the direct sum of the companion matrices of a family of prime power polynomials /,...,/&£ K[x\, then A is the matrix of 4> relative to some basis of Kn (Corollary 1.7). By Theorem 4.3 and Lemma 4.5 Kn = Ex ©£2 ©• • - ©Et, where each £, is a 0-cyclic subspace and/is the minimal polynomial of 0 | The uniqueness statement of Theorem 4.2 implies that the polynomials / are precisely the elementary divisors p"1" of <£, whence D differs from A only in the order of the companion matrices of thep™'1' along the main diagonal. The proof of(i) and (iii) is similar, except that in (i) a stronger uniqueness statement is possible since the invariant factors (unlike the elementary divisors) may be uniquely ordered by divisibility. ■
Corollary 4.8. Let 4> ■ E —> E be a linear transformation of an n-dimensional vector space E over a field K. (i) If
CHAPTER VII LINEAR ALGEBRA
362
are (x — 2)3 = x3 — 6x2 + 12a — 8, (x — 2)2 = a2 — 4a -(- 4, + = + 2a2 + 1, and a2 + 1. By Theorem 4.6 £ has two bases relative to which the respective matrices of 0 are of 0 in
(a2
l)2
R[a]
a4
0
1
0
0
0
0
1
0
0
0
0
4
-4
-5
0
1
4
D ■=
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
8
-12
22
-25
20
-14
1 0 010 -12 6
01 -4 4 0 0 0 -1
M=1
1
\\°
1 0 0 0
0 1 0 -2
0 0 1 0 0 -1 (
1
The matrix D is in rational canonical form and M is in primary rational canonical form. If Eis actually a complex vector space and ip : E —* £is a linear transformation with the same invariant factors= (a — 2)2(a2+ l)eC[x]and<72 = (a — 2)3(x2 + l)2 e C[a], then since a2 + 1 = (a + /)(x — i) in C[a], the elementary divisors of in C[x] are (a — 2)3, (a — 2)2, (a f i)2, (a + i), (a — /)2, and (a — /')- Therefore, relative to some basis of E, \p has the following matrix in Jordan canonical form
0 21 02
J=1
-it 0 -/
—i i1 0/
REMARK. The invariant factors in A"[a] of a matrix A e Mat„K are the same as the invariant factors of A in F[x], where F is an extension field of K (Exercise 6). As the previous example illustrates, however, the elementary divisors of A over K may differ from the elementary divisors of A over F.
4. DECOMPOSITION OF A SINGLE LINEAR TRANSFORMATION
363
We close this section by presenting a method of calculating the invariant factors of a given matrix A, and hence by Corollary 4.8 of any linear transformation that has matrix A relative to some basis. This method is a consequence of
Proposition 4.9. Let A be an n X n matrix aver a fields. Then the matrix of poly nomials xln — A e MatnK[x] is equivalent (over K[x]) to a diagonal matrix D with nonzero diagonal entries fi, . . . , fn e K[x] such that each fi is monic and fi | f21 • • • | f„. Those polynomials fi which are not constants are the invariant factors of A. REMARK. If K is a field, then K[a] is a Euclidean domain (Corollary III.6.4). Consequently, the following proof together with the Remarks after Proposition 2.11 show that the matrix D may be obtained from xln — A by a finite sequence of ele mentary row and column operations. Thus Proposition 4.9 actually provides a con structive method for finding invariant factors. An example is given after the proof. SKETCH OF PROOF OF 4.9. Let <£ : Kn —> Kn be the A-linear transforma tion with matrix A = (ai}) relative to the standard basis {e, } of Kn. As usual Kn is a ^[A]-module with structure induced by . Let F be a free /C[A]-module with basis V = {Mi, I and let 7r : F —* Kn be the unique #[A]-module homomorphism such that ir[ui) = e, for i = 1 , 2 ( T h e o r e m IV.2.1). Let ^ : F—> F be the n a ^ui- Then the unique /C[A]-module homomorphism such that ip(u,) = am, — y=i matrix of ^ relative to the basis U is a/„ — A.
22
We claim that the sequence of /C[A]-modules F F A Kn —> 0 is exact. Clearly ir is a AT[x]-module epimorphism. Since A is the matrix of 4> and the Jf[x]-module structure of Kn is induced by
7t(a«,) = A7r(Ui) = XEi =
0(sa) = 22 Oifii;=l
Consequently, for each i = 7T^AUi — 22 OijUj^ = ir(xUi) — ^2
22 Clij-j 2 Qifii ~ o, i
i
whence Im ^ CZ Ker ir. To show that Ker ir CZ Im ^ it suffices to prove that every n element w of F is of the form w = \J/(v) + kiui (v e T, kj e K). For in this case if >=i w s Ker ir, then
22
0 = tt(m’) = mp(v) + 7r(22 k,-Uj) =0 + ii
22 ki-i-
Since {e, } is a basis of Kn, kj = 0 for all j. Consequently, w = \p(v) and hence Ker 7r CZ Im \p. Since every element of F is a sum of terms of the form /«, with /e Af[x], we need only show that for each i and t, there exist vit e F and kj e K such n
that xlUi = \J/(vit) +
22 kjuj. For each i and t = 1, we have au{ = ^(w,) + 22 aaui j=l j
364
CHAPTER VII LINEAR ALGEBRA
(fli} e K). Proceeding inductively assume that for each j there exist v, t ~\ e Fand k jr e K
n
such that
-f
22 kjru r . Then for each i
r=1
XlUi = X'^(xU%) = X1 '{ip(Ui) +
= =
\p(xl-lUi)
22
22
+
y
+
Thus xlUi = 0(f*c) +
r
22 i
OiiVj.t-l) +
i
= tix^Ui) +
+ 22r ^/r«r)
22r j (22 l l
i
Qi}kji)ur.
22 °r r with vit = x ~ Ui + 22 j u
22 ai]Xl~lUj
aiivi.t~i
e Fand cr =
22
e
K
i
and the induction is complete. Therefore F —> F —► ATn —> 0 is exact and hence X" £* F/Ker tt = F/Im Since K[x] is a principal ideal domain, Proposition 2.11 shows that xln — A is equivalent to a diagonal matrix D ~
ft 9
, where r is the rank of xln — ^ and Lr
is an r X rdiagonal matrix with nonzero diagonal entries fu ..., fr e A"[x] such that f I A I • • ■ I fr- We may assume each f is monic (if necessary, perform suitable ele mentary row operations on D). Clearly the determinant \xln — A j in K[x] is a monic polynomial of degree n. In particular, \xln — A | ^ 0. By Definition 1.8 and Theorem 3.5(iii), (iv), |D| is a unit multiple of |xl„ — A\, whence |D| ^ 0. Consequently, all the diagonal entries of D are nonzero. Thus Lr = D and r = n. Since D is equivalent to xl„ — A,D is the matrix of \p relative to some pair of ordered bases V = {vu..., vn} and W = {wi,. . . , wn } of F (Theorem 1.6). This means that ifivt) ~ f*w' f°r each /and Im = K[x] fiwi © - • •© K[x]fnwn. Consequently, X[x]h,i (J)■ • ■ (J) K\x\wn K" ^ F/Ker tt = F/Im ^ = — 1 ------ ----- br 'r77~ K[x] /iM’i © • • • © K[x] fnwn •
-(S)K[x]Wn/K[x]fnWn
= K[xV(f )©---©XM/(/«), where each f is monic and f | f21 ■ • ■ | fn. For some / (0 < t < n), f = f2 = ■ ■ ■ — f = 1a- and f+1,... ,fi. are nonconstant. Thus for / < t, K[x]/(fy = K[x]/(\k) — 0 and for > t, K[x\/(f) is a cyclic A'f.xj-module of order fi. Therefore, Kn is the in ternal direct sum of nonzero torsion cyclic Ar[x]-submodules (0-cyclic subspaces) Et+1,..., En of orders fi+u respectively such that ft+\ | fi+■> | | fn. Since the n A'[x]-module structure of K is induced by 0, 0 = fEi — f(4>)Et. It follows readily that fi is the minimal polynomial of 0 | Therefore, fi+1, a r e the invariant factors of 0 (and hence of A) by Theorem 4.2. ■
i
EXAMPLE. If 0 : Q3 —> Q3 is a linear transformation and relative to some basis / 0 4 2\ /x -4 -2\ the matrix of 0 is A = I — 1 —4 — 1 I, then xh — >4 = 11 x + 4 \ 0 0 -2/ \0 Performing suitable elementary row and column operations yields:
lj. 0 x + 2/
4. DECOMPOSITION OF A SINGLE LINEAR TRANSFORMATION
365
1 a+ 4 0 -4 - a(a + 4) ,0 0
( 0
0
1 0 0 \ /l 0 0 \ -(x + 2)2 0 ) —> I 0 a + 2 0 I
0
a
+ 2/ \0 0 (x + 2)2/.
Therefore by Corollary 4.8 and Proposition 4.9 the invariant factors of A and are + 2 and (a + 2)2 and their minimal polynomial is (a + 2)2.
a
EXERCISES Note: Unless stated otherwise, E is an ^-dimensional vector space over a field K. 1. If A and B are n X n matrices over K with minimum polynomials q\ and q2 re spectively, then the minimal polynomial of the direct sum of A and B (a 2n X 2n matrix) is the least common multiple of q\ and q2. 2. The 0 linear transformation E—+E has invariant factors [resp. elementary divisors] q\ = a, q2 — x, . . ., qn — a. 3. (a) Let a,b,c be distinct elements of K and let D e Mat6A" be the diagonal matrix with main diagonal a,a,a,b,b,c. Then the invariant factors of D are qi = a — a, qi = (a — a)(a — b) and q3 = (a — a)(x — b){x — c). (b) Describe the invariant factors of any diagonal matrix in Mat„A'. 4. If q is the minimal polynomial of a linear transformation
366
CHAPTER VII. LINEAR ALGEBRA
11. If A is the companion matrix of a monic polynomial /e K[x], with deg f — n, show explicitly that A — xln is similar to a diagonal matrix with main diagonal i/c»ix) * * * 5 ix,y. 12. A e MatnK is idempotent provided A2 = A. Show that two idempotent matrices in Mat„jRT are similar if and only if they are equivalent. 13. An n X n matrix A is similar to its transpose A1.
5. THE CHARACTERISTIC POLYNOMIAL, EIGENVECTORS AND EIGENVALUES In this section we investigate some more invariants of a linear transformation of a finite dimensional vector space over a field. Since several of these results are valid more generally we shall deal whenever possible with free modules of finite rank over a commutative ring with identity. If A is an n X n matrix over a commutative ring K with identity, then xln — A is an n X n matrix over K[x], whence the determinant |.v/„ — AI is an element of K[x}. The characteristic polynomial of the matrix A is the polynomial pA = \xln — A | e K[x\. Clearly, pA is a monic polynomial of degree n. If B e Matr; A is similar to A, say B = PAP 1 , then since xln is in the center of the ring Mat^fjc],
p B = |xh -B\ = |xln - PAP '\ = |P(xln - A)P~'\ = |Pl|,v/n - A\\P\~' = \xh - A\ = p A \ that is, similar matrices have the same characteristic polynomial. Let
Lemma 5.1. (i) If A],A2,. . . , Ar are square matrices (of various sizes) over a com mutative ring K with identity and Pj e K[x] is the characteristic polynomial of Ai, then pip 2 -•-p r e Kfx] is the characteristic polynomial of the matrix direct sum of Ai,A2, . . . , Ar. (ii) The companion matrix C of a monic polynomial f e K[x] has characteristic polynomial f. SKETCH OF PROOF, (i) If A e Mat„AT and B e MatmK, then (o ft) ' (o IX0 b)-
whence
A 0 0B
A 0 0 /„
In 0 = \A\\B\. 0B
An inductive argument now shows that the determinant of a direct sum of matrices B u ... ,B k is \B X \\B 2 \ ■ ■ \B k \. (ii) To show that/is the characteristic polynomial of C, expand \xj n — C | along the last row. ■
5. THE CHARACTERISTIC POLYNOMIAL, EIGENVECTORS AND EIGENVALUES 367
Theorem 5.2. Let
=
QiQs- ■ ■ Qt
=
qiq2- * -qt—lq^-
(ii) (Cayley-Hamilton)
Definition 5.3. Let : E —► E be a linear transformation of a vector space E over a field K. A nonzero vector u e E is an eigenvector (or characteristic vector or proper vector) of
Theorem 5.4. Let
368
CHAPTER VII LINEAR ALGEBRA
If k e K is an eigenvalue of an endomorphism 0 of a K-vector space E, then it is easy to see that C(4>,k) — {c e E |
Theorem 5.5. Let
that v =
52 an eigenvector of
pendent and
n
n
n
52 = kv =
for all /. Thus for each / such that r, ^ 0, k = ki\ (since v ^ 0, at least one r, j6 0). Therefore, ki,... ,k n are the only eigenvalues of 0. Furthermore, if k is an eigen value of 4> that appears t times on the diagonal of D and u h ,..., uit are those ele ments of V with eigenvalue k, then this argument shows that {m,„ spans C(
Proposition 5.6. Let K be a commutative ring with identity. Let <{> be an endomor phism of a free K-module of rank n and let A = (a,j) s Mat„K be the matrix of 4> relative to some ordered basis. If the characteristic polynomial of (j> and A is P
5. THE CHARACTERISTIC POLYNOMIAL, EIGENVECTORS AND EIGENVALUES 369
Let K be a commutative ring with identity. The trace of an n X n matrix A = (a„) over K is an + a
EXERCISES Note: Unless stated otherwise X is a commutative ring with identity. 1. Prove directly that a matrix over K and its transpose have the same characteristic polynomial. 2. (Cayley-Hamilton) If 0 is an endomorphism of a free K-module E of finite rank, thenp^(0) = 0. [Hint:if A is the matrix of 0and B = xln — A,thenB a B = l#| h = Pi>ln in MatnX[jr]. If £ is a K[jc]-module with structure induced by 0 and 0 is the X[x]-module endomorphism E-^E with matrix B, then \p(u) = xu —
(m + n) X (m 4- n) matrices over
B
and observe that \CD\ = \DC\.\ 4. (a) Exhibit three 3X3 matrices over Q no two of which are similar such that —2 is the only eigenvalue of each of the matrices. (b) Exhibit a 4 x 4 matrix whose eigenvalues over R are ± 1 and whose eigen values over C are ±1 and dti. 5. Let K be a field and A e MatnA-. (a) 0 is an eigenvalue of A if and only if A is not invertible. (b) If ki, - -. - kr s K are the (not necessarily distinct) eigenvalues of A and /e AT[x], then f(A) e Mat*# has eigenvalues f(k\),. . . ,f(k r ). 6. If 0 and \p are endomorphisms of a finite dimensional vector space over an algebraically closed field K such that 0^ = i//0, then 0 and have a common eigenvector. 7. (a) Let 0 and 0 be endomorphisms of a finite dimensional vector space E such that 00 = 00. If E has a basis of eigenvectors of 0 and a basis of eigen vectors of 0, then E has a basis consisting of vectors that are eigenvectors for both 0 and 0. (b) Interpret (a) as a statement about matrices that are similar to a diagonal matrix.
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CHAPTER VI! LINEAR ALGEBRA
8. Let 0 : E —»Zsbe a linear transformation of a vector space E over a field K. If U is a set of eigenvectors of 0 whose corresponding eigenvalues are all distinct, then U is linearly independent. [Hint: If U were linearly dependent, there would be a relation r x u x -\-b rtut = 0 (m, e U; 0 ^ r, e K) with t minimal. Apply the transformation Aril*’ — 0, where = k x u u and reach a contradiction.] 9. Let F be an extension field of a field K and u e F. Let 0 : F —> F be the endo morphism of the vector space F given by cl—> uv. (a) Then Tr0 is the trace of u, TKF(u), as in Definition V.7.1. [Hint: first try the case when F = K(u)\. (b) The determinant of 0 is the norm of u, NKr(u). 10. Let K be a field and A e MatnA". (a) If A is nilpotent (that is, Am = 0 for some m), then T t A t = 0 for all r > 1. [Hint: the minimal polynomial of Ar has the form xl and Ar is similar to a matrix in rational or Jordan canonical form.] (b) If char K = 0 and Tr Ar = 0 for all r > 1, then A is nilpotent.
CHAPTER
VIII
COMMUTATIVE RINGS AND MODULES
For the most part this chapter is a brief introduction to what is frequently called commutative algebra. We begin with chain conditions (Section 1) and prime ideals (Section 2), both of which play a central role in the study of commutative rings. Actually no commutativity restrictions are made in Section 1 since this material is also essential in the study of arbitrary rings (Chapter IX). The theory of commutative rings follows a familiar pattern: we attempt to obtain a structure theory for those rings that possess, at least in some generalized form, properties that have proven useful in various well-known rings. Thus primary de composition of ideals (the analogue of factorization of elements in an integral do main) is considered in Sections 2 and 3. We then study rings that share certain de sirable properties with the ring of integers, such as Dedekind domains (Section 6) and Noetherian rings (Section 4). The analysis of Dedekind domains requires some knowledge about ring extensions (Section 5). This information is also used in proving the Hilbert Nullstellensatz (Section 7), a famous classical result dealing with ideals of the polynomial ring K[xu .. ., *„]. Except in Section 1, all rings are commutative. The approximate interdepen dence of the sections of this chapter (subject to the remarks below) is as follows:
A broken arrow A---> B indicates that an occasional result from Section A is used in Section B, but that Section B is essentially independent of Section A. Section 1 is not needed for Section 5 but is needed for Section 4. Only one important result in Section 4 depends on Sections 2 and 3. This dependence can be eliminated by using an al ternate proof, which is indicated in the exercises. 371
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
1. CHAIN CONDITIONS In this section we summarize the basic facts about the ascending and descending chain conditions for modules and rings that will be needed in the remainder of this chapter and in Chapter IX. Rings are not assumed to be commutative, nor to have identity elements.
Definition 1.1. A module A is said to satisfy the ascending chain condition (ACC) on submodules (or to be Noetherian) if for every chain A, CZ A2 (Z A3 CZ- • of submod ules of A, there is an integer n such that A, = A„ for all i > n. A module B is said to satisfy; the descending chain condition (DCC) on submodules (ior to be Artinian) iffor every chain Bi Z) B2 Z) B3 Z) • - • of submodules of B, there is an integer m such that B; = Bm for all i > m. EXAMPLE. The Z-module (abelian group) Z satisfies the ascending but not the descending chain condition on submodules (Exercise II.3.5). The Z-module Zip") satisfies the descending but not the ascending chain condition (Exercise II.3.13). If a ring R is considered as a left [resp. right] module over itself, then it is easy to see that the submodules of R are precisely the left [resp. right] ideals of R. Con sequently, in this case it is customary to speak of chain conditions on left or right ideals rather than submodules.
Definition 1.2. A ring R is left [resp. right] Noetherian z/R satisfies the ascending chain condition on left [resp. right] ideals. R is said to be Noetherian if R is both left and right Noetherian. A ring R is left [resp. right] Artinian //R satisfies the descending chain condition on left [resp. right] ideals. R is said to be Artinian //R is both left and right Artinian. In other words, a ring R is (left or right) Noetherian if it is a (left or right) Noe therian /?-module, and similarly for Artinian. Consequently, all subsequent defini tions and results about modules that satisfy the ascending or descending chain condition on submodules apply, mutatis mutandis, to (left or right) Noetherian or Artinian rings. EXAMPLES. A division ring D is both Noetherian and Artinian since the only left or right ideals are D and 0, (Exercise III.2.7). Every commutative principal ideal ring is Noetherian (Lemma III.3.6); special cases include Z, Zn, and Fix] with F a field. EXAMPLE. The ring Mat„£> of all n X n matrices over a division ring is both Noetherian and Artinian (Corollary 1.12 below). REMARKS. A right Noetherian [Artinian] ring need not be left Noetherian [Artinian] (Exercise 1). Exercise II.3.5 shows that a Noetherian ring need not be Artinian. However every left [right] Artinian ring with identity is left [right] Noether ian (Exercise IX.3.13 below).
1. CHAIN CONDITIONS
373
A maximal element in a partially ordered set (C, < ) was defined in Section 7 of the Introduction. A minimal element is defined similarly: be C is minimal if for every c e C which is comparable to b, b < c. Note that it is not necessarily true that b < c for all c eC. Furthermore, C may contain many minimal elements or none at all.
Definition 1.3. A module A is said to satisfy the maximum condition [resp. minimum condition] on submodules if every nonempty set of submodules of A contains a maximal [resp. minimal] element (with respect to set theoretic inclusion).
Theorem 1.4. A module A satisfies the ascending [resp. descending] chain condition on submodules if and only if A satisfies the maximal [resp. minimal] condition on submodules. PROOF. Suppose A satisfies the minimal condition on submodules and A\ Z) Ai Z) • • ■ is a chain of submodules. Then the set {Al | i > 1 j has a minimal element, say A„. Consequently, for i > n we have An Z) Ai by hypothesis and An CZ Ar by minimality, whence A, = An for each / > n. Therefore, A satisfies the descending chain condition. Conversely suppose A satisfies the descending chain condition, and S is a non empty set of submodules of A. Then there exists B0 e S. If 5 has no minimal element, then for each submodule B in S there exists at least one submodule B' in S such that B Z) B'. For each B in S, choose one such B' (Axiom of Choice). This choice then de fines a function / : S —>* S by B |—> B'. By the Recursion Theorem 6.2 of the Introduc tion (with / = fn for all n) there is a function : N —*■ S such that V?(0) = Bu and
Bi Z) • • - - This contradicts the descending chain condition. Therefore, 5 must have a minimal element, whence A satisfies the minimum condition. The proof for the ascending chain and maximum conditions is analogous. ■
Theorem 1.5. Let 0—>A^B-^+C —* 0 be a short exact sequence of modules. Then B satisfies the ascending [resp. descending] chain condition on submodules if and only if A and C satisfy it. SKETCH OF PROOF. If B satisfies the ascending chain condition, then so does its submodule f(A). By exactness A is isomorphic to f(A), whence A satisfies the ascending chain condition. If Ci CZ C2 C • • • is a chain of submodules of C, then g_1(Ci) CZ £-1(C2) CZ • • • is a chain of submodules of B. Therefore, there is an n such that g-1(G) = g-1(C„) for all / > n. Since g is an epimorphism by exactness, it follows that Ci = Cn for all i > n. Therefore, C satisfies the ascending chain condition. Suppose A and C satisfy the ascending chain condition and Bi C fis C • ■ • is a chain of submodules of B. For each i let
A t = f Kf(A) fl Bi) and C = g(Bi).
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
Let fi = f | Ai and gi — g | 7?t. Verify that for each / the following sequence is exact: 0 -+Ai -*-Bi ^Ci—~0. Verify that Ax C A2 C ■ • • and G C C2 C • • •. By hypothesis there exists an integer n such that A{ = An and C, = Cn for all i > n. For each i > n there is a commutative diagram with exact rows:
0— An + Bn + Cr,-+ 0 fr fi gi 0 —Ai —Bi — Q — 0,
where a and y are the respective identity maps and /?, is the inclusion map. The Short Five Lemma IV.l. 17 implies that /3, is the identity map, whence B satisfies the ascend ing chain condition. The proof for descending chain condition is analogous. ■
Corollary 1.6. If A is a submodule of a module B, then B satisfies the ascending [resp. descending] chain condition if and only if A and B/A satisfiy it. PROOF. Apply Theorem 1.5 to the sequence 0 A —> B —> B/A —» 0. ■
Corollary 1.7. If Ai, . . . , An are modules, then the direct sum Ax © A2 ©• • •© An satisfies the ascending [resp. descending] chain condition on submodules if and only if each Ai satisfies it. SKETCH OF PROOF. Use induction on n. If n = 2, apply Theorem 1.5 to the sequence 0 —» Ax A Ax © A? —^ A2 —» 0. ■
Theorem 1.8. 7/R is a left Noetherian [resp. Artinian] ring with identity, then every finitely generated unitary left K-module A satisfies the ascending [resp. descending] chain condition on submodules. An analogous statement is true with “left” replaced by “right.” PROOF OF 1.8. If A is finitely generated, then by Corollary IV.2.2 there is a free ^-module F with a finite basis and an epimorphism tt : F A. Since Fis a direct sum of a finite number of copies of R by Theorem IV.2.1, F is left Noetherian [resp. Artinian] by Corollary 1.7. Therefore A = F/Ker tt is Noetherian [resp. Artinian] by Corollary 1.6. ■ Here is a characterization of the ascending chain condition that has no analogue for the descending chain condition.
1. CHAIN CONDITIONS
375
Theorem 1.9. A module A satisfies the ascending chain condition on submodules if and only if every submodule of A is finitely generated. In particular, a commutative ring R is Noetherian if and only if every ideal of R is finitely generated. PROOF. (=>) If B is a submodule of A, let S be the set of all finitely generated submodules of B. Since S is nonempty (0 e S), S contains a maximal element C by Theorem 1.4. C is finitely generated by ci,c2,. . ., c„. For each b e B let Db be the submodule of B generated by b,cx,c2 ,..., cn. Then D b eS and C CL D h . Since C is maxi mal, D b = C for every beB, whence b e D b = C for every beB and B CL C. Since C CL B by construction, B = C and thus B is finitely generated. (<=) Given a chain of submodules Ax CL A2 CL A3 CL ■ • ■ , then it is easy to verify that (J Ai is also a submodule of A and therefore finitely generated, say by »>i au , a k . Since each a % is an element of some A,, there is an index n such that a t e A„ for /' = 1,2,... ,k. Consequently, IJ At (Z An, whence Ai = An for i > n. ■ We close this section by carrying over to modules the principal results of Section II.8 on subnormal series for groups. This material is introduced in order to prove Corollary 1.12, which will be useful in Chapter IX. We begin with a host of defini tions, most of which are identical to those given for groups in Section II.8. A normal series for a module A is a chain of submodules: A = A0 3 Ax 3 Ai 3 ■ ■ ■ 3 A„. The factors of the series are the quotient modules
Ai/Ai+1 (/' = 0,1,..., n — 1). The length of the series is the number of proper inclusions (= number of nontrivial factors). A refinement of the normal series A 3 /I, 3 • • - 3 /ln is a normal series obtained by inserting a finite number of additional submodules between the given ones. A proper refinement is one which has length larger than the original series. Two normal series are equivalent if there is a one-to-one correspondence between the non trivial factors such that corresponding factors are isomorphic modules. Thus equivalent series necessarily have the same length. A composition series for A is a normal series A = Ac 3 Ax Z) A2 3 ■ • • D An = 0 such that each factor Ak/Ak+X (k = 0 , 1 , — 1) is a nonzero module with no proper submodules.1 The various results in Section II.8 carry over readily to modules. For example, a composition series has no proper refinements and therefore is equivalent to any of its refinements (see Theorems IV.1.10 and II.8.4 and Lemma II.8.8). Theorems of Schreier, Zassenhaus, and Jordan-Holder are valid for modules:
Theorem 1.10. Any two normal series of a module A have refinements that are equivalent. Any two composition series of A are equivalent. PROOF. See the corresponding results for groups (Lemma II.8.9 and Theorems II.8.10 and II.8.11). ■ JIf
R has an identity, then a nonzero unitary module with no proper submodules is said to be simple. In this case a composition series is a normal series A = A0 3 • • • zd An = 0 with simple factors. If R has no identity simplicity is defined somewhat differently; see Defini tion IX. 1.1 and the subsequent Remarks.
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
Theorem 1.11. A nonzero module A has a composition series if and only if A satisfies both the ascending and descending chain conditions on submodules.
PROOF. (=>) Suppose A has a composition series 5 of length n. If either chain condition fails to hold, one, can find submodules f A — Aq Z) A\ ZD Ao ZD ■ • ■ ZD An ATl+1, ^ ^ ^ ^ ^ which form a normal series T of length n + 1. By Theorem 1.105 and T have refine ments that are equivalent. This is a contradiction since equivalent series have equal length. For every refinement of the composition series 5 has the same length n as 5, but every refinement of T necessarily has length at least n + 1. Therefore, A satisfies both chain conditions. (<=) If fi is a nonzero submodule of A, let S(B) be the set of all submodules C of B such that C ^ B. Thus if B has no proper submodules, S(B) = {0}. Also define 5(0) = {0}. For each B there is a maximal element B' of S(B) by Theorem 1.4. Let 5 be the set of all submodules of A and define a map / : 5 —>5 by f(B) = B'; (the Axiom of Choice is needed for the simultaneous selection of the B'). By the Recur sion Theorem 6.2 of the Introduction (with / = fn for all n) there is a function
Corollary 1.12. If D is a division ring, then the ring Mat,, D of all n X n matrices over D is both Artinian and Noetherian.
SKETCH OF PROOF. In view of Definition 1.2 and Theorem 1.11 it suffices to show that R = Mat„D has a composition series of left /^-modules and a composi tion of right .K-modules. For each /' let ct e R be the matrix with In in position (/',/') and 0elsewhere. Verify that Re, = {Aei | A e 7?} is a left ideal (submodule) of R con sisting of all matrices in R with column j zero for all j ^ /'. Show that Rei is a minimal nonzero left ideal (that is, has no proper submodules). One way to do this is via ele mentary transformation matrices (Definition VII.2.7 and Theorem VII.2.8). Let Mo = 0 and for / > 1 let M, = R(e, + e-2 -|------b ei). Verify that each Mi is a left ideal of R and that = Rei, whence R — M„ ZD ZD ■ ■ • 3 M\ Z) M0 = 0 is a composition series of left ft-modules. A similar argument with the right ideals eiR ~ {e,A | A e R | shows that R has a composition series of right /^-modules. ■
2. PRIME AND PRIMARY IDEALS
377
EXERCISES 1. (a) The ring of all 2 X 2 matrices
ft 9
such that a is an integer and b,c are
rational is right Noetherian but not left Noetherian. (b) The ring of all 2 X 2 matrices ft;) such that
d
is rational and
r,s
are real
is right Artinian but not left Artinian. 2. If / is a nonzero ideal in a principal ideal domain R, then the ring R/I is both Noetherian and Artinian. 3. Let 5 be a multiplicative subset of a commutative Noetherian ring R with identity. Then the ring S~lR is Noetherian. 4. Let R be a commutative ring with identity. If an ideal / of R is not finitely gener ated, then there is an infinite properly ascending chain of ideals J\ CZ J2 CZ - ■ ■ such that Jk CZ I for all k. The union of the Jk need not be I. 5. Every homomorphic image of a left Noetherian [resp. Artinian] ring is left Noetherian [resp. Artinian]. 6. A ring R is left Noetherian [resp. Artinian] if and only if Mat*/? is left Noetherian [resp. Artinian] for every n ^ 1 [nontrivial]. 7. An Artinian integral domain is a field. sider (a) ZD (a2) 3 (a3) Z) • • •.]
[Hint: to find an inverse for a ^ 0, con
2. PRIME AND PRIMARY IDEALS Our main purpose is to study the ideal structure of certain commutative rings. The basic properties of prime ideals are developed. The radical of an ideal is intro duced and primary ideals are defined. Finally primary decomposition of ideals is discussed. Except for Theorem 2.2, all rings are commutative. We begin with some background material that will serve both as a motivation and as a source of familiar examples of the concepts to be introduced. The motiva tion for much of this section arises from the study of principal ideal domains. In particular such a domain D is a unique factorization domain (Theorem III.3.7). The unique factorization property of D can be stated in terms of ideals: every proper ideal of D is a product of maximal (hence prime) ideals, which are deter mined uniquely up to order (Exercise III.3.5). Every nonzero prime ideal of D is of the form (p) with p prime ( = irreducible) by Theorem III.3.4 and (p) n — (p n ). Consequently, every proper ideal (a) of D can be written uniquely (up to order) in the form ^ =
. .^)
=
f| (p7™) D • • • D (pr"*),
where each > 0 and the pi are distinct primes (Exercise III.3.5). Now an ideal Q = (pn) (p prime) has the property: abzQ and a\Q imply b k e Q for some k (Exercise III.3.5). Such an ideal is called primary. The preceding discussion shows that every ideal in a principal ideal domain is the intersection of a finite number of
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
primary ideals in a unique way. Furthermore there is an obvious connection between these primary ideals and the prime ideals of D; in fact every primary ideal (pn) = (p)n is a power of a prime ideal. In the approach just outlined the viewpoint has switched from consideration of unique factorization of elements as products of primes in D to a consideration of the “primary decomposition” of ideals in the principal ideal domain D. We shall now investigate the “primary decomposition” of ideals in more general commutative rings (where, for instance, ideals need not be principal and primary ideals may not be powers of prime ideals). We begin with some facts about prime ideals.
Theorem 2.1. An ideal P (s^R) in a commutative ring R is prime ifand only ifR — P is a multiplicative set. PROOF. This is simply a restatement of Theorem III.2.15; see Definition III.4.1. ■ REMARK. The set of all prime ideals in a ring R is called the spectrum of R.
Theorem 2.2. If S is a multiplicative subset of a ring R which is disjoint from an ideal I of R, then there exists an ideal P which is maximal in the set of all ideals of R disjoint from S and containing I. Furthermore any such ideal P is prime. The theorem is frequently used in the case 7 = 0. SKETCH OF PROOF OF 2.2. The set S of all ideals of R that are disjoint from S and contain 7 is nonempty since 7 e S. Since S 9* 0 (Definition III.4.1) every ideal in S is properly contained in R. S is partially ordered by inclusion. By Zorn’s Lemma there is an ideal P which is maximal in S. Let A,B be ideals of R such that AB CZ P. If A P and B jzf P, then each of the ideals P A and P + B properly contains P and hence must meet S. Consequently, for some PitP, aeA, beB
Pi + a = Si e S and p2 + b = s2 e S. Thus SiS2 = P\Pi + pib + ap-i + ab e P + AB C P. This is a contradiction since SiSi e S and 5 0 / * = 0. Therefore A C P or B C P, whence P is prime. ■
Theorem 2.3. Let K be a subring of a commutative ring R. If Pi,. . . , Pn are prime ideals of R such that K C Pj U P2 U • • ■ U Pn, then K CZ Pi for some i. REMARK. In the case n < 2, the following proof does not use the hypothesis that each Pi is prime; the hypothesis is needed for n > 2. PROOF OF 2.3. Assume K (Zf P, for every i. It then suffices to assume that n > 1 and n is minimal; that is, for each /, K $ZL U Pjm For each i there exists j a l eK — U Pj. Since K CZ U 7\, each o, e Pi. The element ai + o2a3- • -a n lies in K
2. PRIME AND PRIMARY IDEALS
379
and hence in (J P- Therefore a\ +a-2a3- -an = bj with bj e Pj. If j > 1, then ax e Pj, i
which is a contradiction. If / =1, then a2a3- ■ an e Px, whence a, e Px for some i >1 by Theorem III.2.15. This also is a contradiction. ■
Proposition 2.4. If R is a commutative ring with identity and P is an ideal which is maximal in the set of all ideals of R which are not finitely generated, then P is prime. PROOF. Suppose ab e P but a\P and b £ P. Then P + (a) and P + (b) are ideals properly containing P and therefore finitely generated (by maximality). Consequently P + (a) = (pi + na,. .. ,p n + rna) and P + (b) = (/?/ + n'b, Pm' + rm'b) for some pi,p/ e P and /■„/■»' e R (see Theorems III.2.5 and III.2.6). If J = \r e R \ ra e P\, then J is an ideal. Since ab e P, (pi + rt'b)a = p/a + r/ab e P for all i, whence P C P + (b) CZ J. By maximality, J is finitely generated, say J — ^ 71 (7*1,.. . ,jk). If x eP, then x e P + (a) and hence for some s, e R, x = 22 •'.(ft + na) n
»=1
n
= 22 + 22 sSia. Consequently, (22 s,rt)a = x — 22 i =1i —1i i i SiP*
n
Thus for some /, e R, 22
k
=
n
22 /
and
x=
e
whence 22
e
J-
k
22 + 22 Therefore, P is
1 i=l i=l i=l
generated by />i,..., pn,ha, ... ,jka, which is a contradiction. Thus a eP or beP and P is prime by Theorem III.2.15. ■
Definition 2.5. Let I be an ideal in a commutative ring R. The radical (or nilradical) of I, denoted Rad I, is the ideal fl P, where the intersection is taken over all prime ideals P which contain I. If the set of prime ideals containing I is empty, then Rad I is defined to be R. REMARKS. If R has an identity, every ideal I (^R) is contained in a maximal ideal M by Theorem III.2.18. Since M ^ R and M is necessarily prime by Theorem III.2.19, Rad / R. Despite the inconsistency of terminology, the radical of the zero ideal is sometimes called the nilradical or prime radical of the ring R. EXAMPLES. In any integral domain the zero ideal is prime; hence Rad 0 = 0. In the ring Z, Rad (12) = (2) fl (3) = (6) and Rad (4) = (2) = Rad (32).
Theorem 2.6. If I is an ideal in a commutative ring R, then Rad I = {reR|r n eI for some n > 0}. PROOF. If Rad I = R, then {r e R [ r" e /} CZ Rad I. Assume Rad I ^ R. If rn e / and P is any prime ideal containing I, then rn eP whence reP by Theorem III.2.15. Thus \r e R \ rn e I] CZ Rad /. Conversely, if t e R and tn \ I for all n > 0, then S = {/" + x | n e N*; jc e /) is a multiplicative set such that S fl / = 0. By Theorem 2.2 there is a prime ideal P dis joint from S that contains I. By construction, t \ P and hence t ^ Rad I. Thus t £ {r e R | rn e /} implies t £ Rad I, whence Rad / C Z j r s ^ l r ^ e / } . ■
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
Theorem 2.7. If I, Il5 I2,. . . , I„ are ideals in a commutative ring R, then: (i) Rad (Rad X) = Rad I; (ii) Rad( I,I2 - -I n ) = Rad ( f) Ij V=i (iii) Rad (Im) = Rad I.
)
= f) Pad I,; }=i
SKETCH OF PROOF. In each case we prove one of the two required contain ments. (i) If r e Rad (Rad I), then rn e Rad I and hence rnm = (rn)m e I for some n, m > 0. Therefore, r e Rad I and Rad(Rad /) CZ Rad I. (ii) If r e Q Rad /„ then i
there are mi,m 2 ,... ,m n > 0 such that r mi s /, for each j. If m = my + m 2 + • ■ • + m„, then r m = r ml r mi - ■ -r mn e hh- ■ ■/«, whence f~) Rad /, CZ Rad (I, ■ ■ ■/„)• Finally since i /]•••/„ CZ f') /„ we have Rad(/r--/„) CZ Rad(Q /,). (iii) is a special case of ij
(ii)- ■ Definition 2.8. An idea! Q (^ R) in a commutative ring R is primary if for any a,b e R: ab e Q and a ^ Q bn e Q for some n > 0. EXAMPLE. Every prime ideal is clearly primary. If p is a prime integer and n > 2 a positive integer, then (p)n — (p n ) is a primary ideal in Z which is not prime (Exercise 17). In general, a power P n of a prime ideal P need not be primary. EXAMPLE. If F is a field, the ideal (atj) is maximal in F[x,y) (Exercise 12) and therefore prime (Theorem III.2.19). Furthermore^,^)2 = (x2,^,^2) CZ (x2,.v) CZ (x,y). The ideal (x2,y) is primary and (x,y) is the only (proper) prime ideal containing (x2,^) (Exercise 12). Hence the primary ideal (x-,>0 is not a power of any prime ideal in /r[x,>’].
In the rest of this section all rings have identity. Theorem 2.9. If Q is a primary ideal in a commutative ring R, then Rad Q is a prime ideal. PROOF. Suppose ab e Rad Q and a £ Rad Q. Then anbn = (ab)n e Q for some n. Since a $ Rad Q, an £ Q. Since Q is a primary, there is an integer m > 0 such that (An)m e Q, whence b e Rad Q. Therefore, Rad Q is prime by Theorem III.2.15. ■ In view of Theorem 2.9 we shall adopt the following terminology. If Q is a primary ideal in a commutative ring R, then the radical P of Q is called the associated prime ideal of Q. One says that Q is a primary ideal belonging to the prime P or that Q is primary for P or that Q is P-primary. For a given primary ideal Q, the associated prime ideal Rad Q is clearly unique. However, a given prime ideal P may be the associated prime of several different primary ideals. EXAMPLE. If p is a prime in Z, then each of the primary ideals (p'% (p3 ),... belongs to the prime ideal (p). In the ring Z[x,y] the ideals (x2,^), (x2,y2), (x2,y3). etc. are all primary ideals belonging to the prime ideal (x,j ) (Exercise 13).
2. PRIME AND PRIMARY IDEALS
381
Theorem 2.10. Let Q and P be ideals in a commutative ring R. Then Q is primary for P if and only if: (i) Q CZ P CZ Rad Q; and (ii) if ah e Q and a | Q, then be P. SKETCH OF PROOF. Suppose (i) and (ii) hold. If ab e Q with a\ Q, then beP CZ Rad Q, whence br e Q for some n > 0. Therefore Q is primary. To show that Q is primary for P we need only show P = Rad Q. By (i), P CZ Rad Q. If b e Rad Q, let n be the least integer such that bn e Q. If n = 1, b e Q (Z P. If n > 1, then b^b = bn e Q, with bn~l If. Q by the minimality of n. By (ii), be P. Thus b e Rad Q implies b e P, whence Rad Q CZ P. The converse implication is easy. ■
Theorem 2.11. 7/Qi,Q2,. .. , Q„ are primary ideals in a commutative ring R, all of n
which are primary for the prime ideal P, then P) Qs is also a primary ideal belonging i=i to P. n n PROOF. Let Q = f) Qi- Then by Theorem 2.7(ii), Rad Q = 0 Qi n i=1 i=l = n P = P; in particular, Q CZ P CZ Rad Q. If ab e Q and a 4 Q, then ab e Qi and i=i a | Qi for some i. Since Q, is P-primary, b s P by Theorem 2.10(ii). Consequently, Q itself is P-primary by Theorem 2.10. ■
Definition 2.12. An ideal I in a commutative ring R has a primary decomposition if I = Qi fl Q2 fl - - - fl Q„ with each Qi primary. If no Qi contains Qi fl - - - fl Q;_i fl Qi+1 fl • ■ - D Qn and the radicals of the Q; are alt distinct, then the primary decomposi tion is said to be reduced (or irredundant).
Theorem 2.13. Let I be an ideal in a commutative ring R. If I has a primary decom position, then I has a reduced primary decomposition. PROOF. If / = Qi D ■ ■ • fl Qn (Qi primary) and some Qi contains Q\ fl • • ■ fl Qi-1 fl Qi+l D ■ • ■ ft Q,„ then I — Qx H • • ■ H (3,_i H Qi+l D - • • fl Q„ is also a primary decomposition. By thus eliminating the superfluous Qi (and reindexing) we have / = Qi fl • • • fl Qk with no Qt containing the intersection of the other Let P\, ... ,Pr be the distinct prime ideals in the set {Rad Qu . .., Rad Q k \. Let Qi (1 < / < r) be the intersection of all the Q 's that belong to the prime Pt . By Theo rem 2.11 each Q is primary for P,. Clearly no Qi contains the intersection of all the k
r
other Q/. Therefore, I = O Q; = O Qi', whence / has a reduced primary de2—1 2=1 composition. ■ At this point there are two obvious questions to ask. Which ideals have a reduced primary decomposition? Is a reduced primary decomposition unique in any way? Both questions will be answered in a more general setting in the next section (Theo rems 3.5 and 3.6).
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EXERCISES Note: R is always a commutative ring. 1. Let R be a commutative Artinian ring with identity. (a) Every prime ideal of R is maximal [Hint: Theorems III.2.16 and III.2.20 and Exercises 1.5 and 1.7]. (b) R has only a finite number of distinct prime ideals. 2. If R has an identity and {P, | / e /] is a nonempty family of prime ideals of R which is linearly ordered by inclusion, then (J P, and Q P. are prime ideals.
iel
iel
3. If P i,P2) ... ,Pn are prime ideals in R and I is any ideal such that / P, for all /, then there exists re/ such that r £ Pi for all /. 4. If R has an identity and A/i,..., Mr are distinct maximal ideals in R, then show that Mi D M2 fl • • • fl Mr = A/iM2- • - Mr. Is this true if “maximal” is replaced by “prime”? 5. If R has an identity, then the set of all zero divisors of R is a union of prime ideals. 6. Let R have an identity. A prime ideal P in R is called a minimal prime ideal of the ideal I if I CZ P and there is no prime ideal P' such that / CZ P' CZ P. (a) If an ideal / of R is contained in a prime ideal P of R, then P contains a minimal prime ideal of I. [Hint: Zornify the set of all prime ideals P' such that I CZ P' CZ P.] (b) Every proper ideal possesses at least one minimal prime ideal. 7. The radical of an ideal / in a ring R with identity is the intersection of all its min imal prime ideals [see Exercise 6]. 8. If R has an identity, I is an ideal and J is a finitely generated ideal such that J C Rad /, then there exists a positive integer n such that Jn CZ I. 9. What is the radical of the zero ideal in Z„? 10. If S is a multiplicative subset of a commutative ring R and I is an ideal of R, then £-1(Rad /) = Rad (S'-1/) in the ring S~lR. 11. Let Q R) be an ideal in R. Then Q is primary if and only if every zero divisor in R/Q is nilpotent (see Exercise III. 1.12). 12. If Fis a field, then: (a) the ideal (x,y) is maximal in F[x,y]\ (b) (x,y)2 = (x2,xy,y2) CZ (x2,y) CZ {x,y)\ (c) the ideal (a2jO is primary and the only proper prime ideal containing it is (x,y). 13. In the ring Z[x,y] the ideals (A-2,>’),(*V2),(jr2,>3),. . . , CvSyOi • • ■ are all primary ideals belonging to the prime ideal (a,j). 14. The conclusion of Theorem 2.11 is false if infinite intersections are allowed. [Hint: consider Z.]
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383
15. Let f\R —> S be an epimorphism of commutative rings with identity. If ./is an ideal of S, let I — (a) Then / is primary in R if and only if J is primary in 5. (b) If J is primary for P, then I is primary for the prime ideal 16. Find a reduced primary decomposition for the ideal I — (x2,xy,2) in Z[x,y] and determine the associated primes of the primary ideals appearing in this decom position. 17. (a) If p is prime and n > 1, then (pn) is a primary, but not a prime ideal of Z. (b) Obtain a reduced primary decomposition of the ideal (12600) in Z. 18. If F is a field and I is the ideal (x2,xy) in F[x,y], then there are at least three dis tinct reduced primary decompositions of /; three such are: (i) / = (x) 0 (x\y); (ii) / = (x) 0 (x\x + >-); (iii) / = (x) fl (x\xy,y% 19. (a) In the ring Z[x), the following are primary decompositions: (4,2*,a2) = (4,a) fl (2,*2); (9,3a- + 3) = (3) fl (9,a + 1).
(b) Are the primary decompositions of part (a) reduced?
3. PRIMARY DECOMPOSITION We shall extend the results of Section 2 in a natural way to modules. A unique ness statement for reduced primary decompositions (of submodules or ideals) is proved as well as the fact that every submodule [ideal] of a Noetherian module [ring] has a primary decomposition. Throughout this section all rings are commutative with identity and all modules are unitary.
Definition 3.1. Let R be a commutative ring with identity and B an R-module. A submodule A (^ B) is primary provided that r e R, b $ A and rb e A => rnB C A for some positive integer n. EXAMPLE. Consider the ring R as an 7?-module and let Q be a primary ideal (and hence a submodule) of R. If rb e Q with r e R and b\ Q, then rn e Q for some n. Since Q is an ideal, this implies rnR C Q. Hence Q is a primary submodule of the module R. Conversely every primary submodule of R is a primary ideal (Exercise 1). Therefore, all results about primary submodules apply to primary ideals as well.
Theorem 3.2. Let R be a commutative ring with identity and A a primary submodule of an K-module B. Then Qa = {r e R | rB
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Let R,A,B, and Qa be as in Theorem 3.2. By Theorem 2.9 Rad QA = Pi is a prime ideal. It is easy to see that Pi = {r e R \ rnB C A for some n > 0}. A primary submodule A of a module B is said to belong to a prime ideal P or to be a P-primary submodule of B if P = Rad Qa = {re R I rnB CZ A for some n > 0}. This termi nology is consistent with that used for ideals. In particular, if J is a primary ideal, then Qj = J.
Definition 3.3. Let R be a commutative ring with identity and B an R-module. A submodule C ofB has a primary decomposition ifC = Aj D A2 D - - * fl An, with each A; a P-primary submodule of B for some prime ideal Pi of R. If no A, contains Ai n ... n Aj_i D Ai+1 n • • • n A„ and if the ideals Pi, . . . , Pn are all distinct, then the primary decomposition is said to be reduced. Again the terminology here is consistent with that used for ideals. If C,At and Pi are as in the definition and Pj {Zf P, for all j i, then P, is said to be an isolated prime ideal of C. In other words, P, is isolated if it is minimal in the set {Pi,.. . , P„}. If Pi is not isolated it is said to be embedded.
Theorem 3.4. Let R be a commutative ring with identity and B an R-module. If a submodule C of B has a primary decomposition, then C has a reduced primary decom position. SKETCH OF PROOF. The proof is similar to that of Theorem 2.13. Note that 71
if QA = \r e R | rB CZ A}, then f) Qa,• = Qha,- Thus if A x , . . . , A T are all »=1
r
P-primary submodules for the same prime ideal P, then D A, is also P-primary by i=i Theorem 2.11. ■
Theorem 3.5. Let R be a commutative ring with identity and B an R-module. Let C (?^B) be a submodule of B with two reduced primary decompositions, a,
n a2 n - - n Ak = c = a/ n a./ n • • • n a/,
where A\ is Prprimary and A' is P,'-primary. Then k = s and (after reordering if necessary) P; = P; 'fori = 1,2, . . . , k. Furthermore if A, and A/ both are P-primary and P, is an isolated prime, then A, = Ai'. PROOF. By changing notation if necessary we may assume that Pi is maximal in the set {Pi, . . . , P^,P/,. . ., P/J. We shall first show that Pi = P/ for some j. Suppose, on the contrary, that Pi ^ P/ for j = 1,2,..., s. Since Pi is maximal we have P, $Z^ P/ for j — 1,2, . . . , s. Since the first decomposition is reduced, Pi,P2......... P k are distinct, whence P P> for /' = 2,3, . . . , k. By the contrapositive of Theorem 2.3, Pi JZ' P2 U • • • U P* U P/ U - - - U P/. Consequently, there exists r £ Pi such that r $ P, (/ > 2) and r \ P(j > 1). Since A i is Pi-primary r n B (Z A x for some positive integer n. Let C* be the submodule {x e B | r n x e Cj. If A: = 1, then C = A, and hence C* = B. We claim that for k > 1, C* ~ C and for k > 1,
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C* = A 2 fl • • • fl A k . Now it is easy to see that A 2 D • • • D A k CZ C* and Ay D A 2 ' f) ■ ■ ■ f) A/ = C (Z C* for k > 1. Conversely, if jc \ A t (/ > 2), then r n x | Ai (otherwise r n s P { since A { is Pi-primary, whence r e Pi since Pi is prime). Con sequently, r n x | C, whence x ^ C*. Therefore, C* CZ A 2 fl • • • fl A k for k > 1. A similar argument shows that C* C Ay D A 2 ' D - - - fl A/ = C, so that C* = C (k > l)andC* = A 2 fl • ■ • fl A k (k > 1). If A: = 1, then as observed above C* = B. Thus C = C* = B, which contradicts the fact that C ^ B. If k > 1, then As
n•••n
Ak
= c* = c =
Ax
n
a2
n•••n
Ak,
whence A 2 fl • ■ • D A k CZ Ay. This conclusion contradicts the fact that the first de composition is reduced. Thus the assumption that Py ^ P/ for every j always leads to a contradiction. Therefore Py = P/ for some j, say 7 = 1 . The proof now proceeds by induction on k. If k = 1, then we claim 5=1 also. For if s > 1, then the argument above with Py = Py' and the roles of A t ,A/ reversed) shows that B = C* = A fl • • • fl A/, whence A/ = B for some j > 2. Thus the second decomposition of C is not reduced, a contradiction. Therefore, s = 1 = k and Ay = C = Ay. Now assume that k > 1 and the theorem is true for all submodules that have a reduced primary decomposition of less than k terms. The argu ment of the preceding paragraph (with Py = Py) shows that for k > 1 the submodule C* has two reduced primary decompositions: 2
At n a3 n ■ • • n Ak = c* = as n • • - n a/. By induction k = s, and (after reindexing) P, = P/ for all z. This completes the in duction and the proof of the first part of the theorem. Suppose Ai and Ai are both /^-primary and Pi is an isolated prime. For con venience of notation assume i = 1. Since Py is isolated, there exists for each j > 2, r, e Pj — Py. Then t = r2r3 ■ ■ -r k z P, for j > 1, but t § Py. Since A 3 is P,-primary, there exists for each j > 2 an integer n, such that tniB CZ Ajm Similarly, for each j > 2 there is an m3 such that tmiB CZ A/. Let n = max {n 2 ,... , n k ,m 2 ,. . . , m k); then t n B C Aj and i n B C A/ for all j > 2. Let D be the submodule {x e B \ t n x e C\. To complete the uniqueness proof we shall show Ay = D = A/. If x e Ay, then t n xeAy fl A 2 D • • • fl A k. = C, whence x eD and/ii CZ D. If x e D, then t n x e C C Ay. Since Ay is Pi-primary and t \ Py, we have t m B (Zf Ay, for all m > 0. Since Ay is primary, we must have x e Ay, (otherwise t n x e Ay and x \ A x imply tnqB CZ Ay for some positive q by Definition 2.1). Hence D = Ay. An identical argument shows that Ay — D. Therefore, Ay = Ay'. ■
Thus far we have worked with a module that was assumed to have a primary de composition. Now we give a partial answer to the question: which modules [ideals] have primary decompositions?
Theorem 3.6. Let R be a commutative ring with identity and B an R-module satisfy ing the ascending chain condition on submodules. Then every submodule A (9^ B) has a reduced primary decomposition. In particular, every submodule A (^B) of a finitely generated module B over a commutative Noetherian ring R and every ideal (r^R) of R has a reduced primary decomposition.
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PROOF OF 3.6. Let S be the set of all submodules of B that do not have a primary decomposition. Clearly no primary submodule is in S. We must show that S is actually empty. If S is nonempty, then S contains a maximal element C by Theorem L4. Since C is not primary, there exist r e R and b eB — C such that rb e C but rnB fZf C for all n > 0. Let B„ = \x tB\ rnx eC j. Then each Bn is a submodule of B and By a B2 d Bs CZ - ■ ■. By hypothesis there exists k > 0 such that Bi = Bk for > k. Let D be the submodule {x e B | x = rky + c for some y e B,c e C). Clearly C CZ Bk fl D. Conversely, if x e Bk fl D, then x = rhy + c and rkx e C, whence r2kv = rk(rky) = rk(x — c) = rkx — rkc e C. Therefore, y s B2k = Bk. Consequently, rky z C and hence x = rky + c e C. Therefore Bk D D CZ C, whence Bk fl D = C. Now C ^ Bk 9^ B and C 9^ D ^ B since b e Bk — C and rkB CZ.C. By the maximal ity of C in S, Bk and D must have primary decompositions. Thus C has a primary decomposition, which is a contradiction. Therefore S is empty and every submodule has a primary decomposition. Consequently, every submodule has a reduced primary decomposition by Theorem 3.4. The last statement of the theorem is now an immediate consequence of Theorems 1.8 and 1.9. ■
EXERCISES Note: Unless otherwise stated R is always a commutative ring with identity. 1. Consider the ring R as an /^-module. If Q is a primary submodule of R, then Q is a primary ideal. 2. (a) Let / : B —» D be an i?-module epimorphism and C (9^ D) a submodule of D. Then C is a primary submodule of D if and only if /_1(C) is a primary submodule of B. (b) If C and /_1(C) are primary, then they both belong to the same prime ideal P. 3. If A (9±B) is a submodule of the /^-module B and P is an ideal of R such that (i) rx e A and x | A (r s R,x e B) => r z P\ and (ii) r e P => rnB (Z A for some positive integer n, then P is a prime ideal and A is a P-primary submodule of B. 4. If A is a P-primary submodule of an /^-module B and rx e A (r either r e P or x s A.
e R,x e
B), then
5. If A is a P-primary submodule of an P-module B and C is any submodule of B such that C A then jr e R j rC CZ A\ is a P-primary ideal. [Hint: Exercise 3 may be helpful.] 6. Let A be a P-primary submodule of the /^-module B and let C be any submodule of B such that C %L A. Then A fl C is a P-primary submodule of C. [Hint: Exer cise 3 may be helpful.] 7. If B is an P-module and x e B, the annihilator of x, denoted ann x, is | r e R | rx = 01. Show that ann x is an ideal. 8. If B 9^ 0 is an /?-module and P is maximal in the set of ideals j ann x | 0 ^ x s B |
(see Exercise 7), then P is prime.
4. NOETHERIAN RINGS AND MODULES
387
9. Let R be Noetherian and let B be an P-module. If P is a prime ideal such that P — ann x for some nonzero x e B (see Exercise 7), then P is called an associated prime of B. (a) If B 7^ 0, then there exists an associated prime of B. [Hint: use Exercise 8.] (b) If B 0 and B satisfies the ascending chain condition on submodules, then there exist prime ideals Pi, . . . , Pr-X and a sequence of submodules B = Bi ZD B2 ZD ■ ■ • ZD Br = 0 such that Bi/BiS R/Pi for each / < r. 10. Let R and B be as in Exercise 9(b). Then the following conditions on r e R are equivalent: (i) for each xzB there exists a positive integer n(x) such that rn(x)x = 0; (ii) r lies in every associated prime of B (see Exercises 9 and 15). 11. Let R be Noetherian, r a R, and B an R-module. Then rx = 0 (x e B) implies x = 0 if and only if r does not lie in any associated prime of B (see Exercises 8 and 9). 12. Let R be Noetherian and let B be an R-module satisfying the ascending chain condition on submodules. Then the following are equivalent: (i) There exists exactly one associated prime of B (see Exercise 9); (ii) B 0 and for each r s R one of the following is true: either rx = 0 im plies x = 0 for all x e B or for each x e B there exists a positive integer n(x) such that r n{x) x = 0. [See Exercises 10 and 11.] 13. Let R and B be as in Exercise 12. Then a submodule A of Bis primary if and only if B/A has exactly one associated prime P and in that case A is P-primary; (see Exercises 9 and 12). 14. Let R and B be as in Exercise 12. If A B ) is a submodule of B, then every associated prime of A is an associated prime of P. Every associated prime of B is an associated prime of either A or B/A; (see Exercise 9). 15. Let R and B be as in Exercise 12. Then the associated primes of B are precisely the primes P u ... ,P„ t where 0 = A } D-D A n is a reduced primary de composition of 0 with each A{ P,-primary. In particular, the set of associated primes of B is finite. [Hint: see Exercises 9, 13, 14.] 16. Let S be a multiplicative subset of R and let A be a P-primary submodule of an R-module B. If P fl S’ = 0, then S-M is an S_1P-primary submodule of the S-1R-module S-1P.
4. NOETHERIAN RINGS AND MODULES This section consists of two independent parts. The first part deals primarily with Noetherian modules (that is, modules satisfying the ascending chain condition). A rather strong form of the Krull Intersection Theorem is proved. Nakayama’s Lemma and several of its interesting consequences are presented. In the second part of this section, which does not depend on the first part, we prove that if R is a commutative Noetherian ring with identity, then so are the polynomial ring R[xi,. . . , x„] and the power series ring R[[x]]. With few exceptions all rings are commutative with identity. We begin by recalling that a commutative ring R is Noetherian if and only if R
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satisfies the maximum condition on (two-sided) ideals (Definition 1.2 and Theorem 1.4), or equivalently if and only if every ideal of R is finitely generated (Theorem 1.9). As a matter of fact, one need only consider prime ideals of R:
Proposition 4.1. (/. S. Cohen). A commutative ring R with identity is Noetherian if and only if every prime ideal of R is finitely generated. SKETCH OF PROOF. (<=) Let S be the set of all ideals of R which are not finitely generated. If S is nonempty, then use Zorn’s Lemma to find a maximal ele ment P of S. Pis prime by Proposition 2.4 and hence finitely generated by hypothesis. This is a contradiction unless S = 0. Therefore, R is Noetherian by Theorem 1.9. ■ We now develop the preliminaries needed to prove the Krull Intersection Theo rem. If B is a module over a commutative ring R, then it is easy to see that I=[reR\rb = 0 for all b e B] is an ideal of R. The ideal / is called the annihilator of B in R.
Lemma 4.2. Let B be a finitely generated module over a commutative ring R with identity and let 1 be the annihilator of B in R. Then B satisfies the ascending [resp. descending] chain condition on submodules if and only if R/I is a Noetherian [resp. Artinian] ring. SKETCH OF PROOF. Let B be generated by bi,. . ., br and assume B satisfies the ascending chain condition. Then B = Rbi H----- \- Rbn by Theorem IV.l .5. Con sequently, / = /1 fl h fl • - - D /„, where /, is the annihilator of the submodule Rbj. By Corollary III.2.27 there is a monomorphism of rings 6 : R/I —> R/Ix X • ■ • X R/h. It is easy to see that 6 is also an F-module monomorphism. Verify that for each j the map R/lj —> Rbj given by r + /, I—► rb, is an isomorphism of /^-modules. Since the submodule Rb, of B necessarily satisfies the ascending chain condition, so does R/I,. Therefore, R/Ix ® • • ■ © R/In satisfies the ascending chain condition on /?-submodules by Corollary 1.7. Consequently its submodule Im 6 = R/I satisfies the ascending chain condition on F-submodules. But every ideal of the ring R/I is an F-submodule of R/I. Therefore, R/I is Noetherian. Conversely suppose R/I is Noetherian. Verify that B is an ^//-module with (r + I)b = rb and that the R/I submodules of B are precisely the /?-submodules. Consequently, B satisfies the ascending chain condition by Theorem 1.8. ■ Recall that if I is any ideal in a ring R with identity and B is an F-module, then n 1 IB = 52 r*bi I ri £ !'■> bi e B; n e N* [ is a submodule of B (Exercise IV.l.3). »'=i
Lemma 4.3. Let P be a prime ideal in a commutative ring R with identity. IfC is a P-primary submodule of the Noetherian R-module A, then there exists a positive integer m such that PmA Cl C.
4. NOETHERIAN RINGS AND MODULES
389
PROOF. Let I be the annihilator of A in R and consider the ring 7? = R/I. De note the coset r + I eRbyr. Clearly Id [r e R \ rA d C\ d P, whence P = P/I is an ideal of R. A and C are each R-modules with ra = ra (r e R,a s A). We claim that C is a primary P-submodule of A. If fa e C with r e R and a e A — C, then ra e C. Since C is a primary P-submodule, rnA d C for some n, whence rnA d C and C is ^-primary. Since \r e 7? | rkA CI C for some k > 0} = \r £ R | rkA d Cj = {r e R | r e P) = P, P is a prime ideal of R and C is a P-primary 7?-submodule of A (see Theorems 2.9 and 3.2). Since R is Noetherian by Lemma 4.2, P is finitely generated by Theorem 1.9. Let pi,. . ., p, (pi £ P) be the generators of P. For each i there exists in such that piniA d C. If m = «i +• ■ ■ + «„ then it follows from Theorems III.1.2(v) and III.2.5(vi) that PmA d C. The facts-that P = P/I and IA — 0 now imply that PmA d C. ■
Theorem 4.4. (Krull Intersection J'heorem). Let R be a commutative ring with CD
identity, I an ideal of R and A a Noetherian K-module. 7/B = InA, then IB = B. 77 = 1
Theorem 4.4 was first proved in the case where R is a Noetherian local ring with maximal ideal I. The proof we shall give depends on primary decomposition (as did the original proof). However, if one assumes that R is Noetherian, there are a num ber of proofs that do not use primary decomposition (Exercise 2). PROOF OF 4.4. If IB = A, then A = IB d B, whence B = A = IB. If IB A, then by Theorem 3.6 IB has a primary decomposition: ib = At
n a2 n • ■ • n as,
where each A, is a P,-primary submodule of A for some prime ideal P, of R. Since IB d B in any case, we need only show that B d Ai for every i in order to conclude that B d IB and hence that B = IB. Let i (1 < i < s) be fixed. Suppose first that / d Pi. By Lemma 4.3 there is an integer m such that PimA d Ai, whence B = (~) lnA CZ lmA CZ P™A d Ai. Now n
suppose / (£ Pi. Then there exists r e / — P,. If B cf: Au then there exists b eB — Au Since rb e IB d A,, b ^ A, and A, is primary, rnA d Ai for some n > 0. Conse quently, r £ Pi since Ai is a Pi-primary submodule. This contradicts the choice of r e I — Pi. Therefore B d Ai. ■
Lemma 4.5. (Nakayama) If J is an ideal in a commutative ring R with identity, then the following conditions are equivalent. (i) (ii) (iii) (iv) then A
J is contained in every maximal ideal of R; 1 r — j is a unit for every j e J; If A is a finitely generated K-module such that JA = A, then A = 0; //B is a submodule of a finitely generated K-module A such that A = JA + B, = B.
REMARK. The Lemma is true even when R is noncommutative, provided that (i) is replaced by the condition that J is contained in the Jacobson radical of R (Exercise IX.2.17).
CHAPTER VIII COMMUTATIVE RINGS AND MODULES
390
PROOF OF 4.5. (i) => (ii) if j e J and 1 r — j is not a unit, then the ideal (Ik — j) is not R itself (Theorem III.3.2) and therefore is contained in a maximal ideal M ^ R (Theorem III.2.18). But \R — j e M and j e J CL M imply that 1 r e M, which is a contradiction. Therefore, 1^* — j is a unit. (ii) => (iii) Since A is finitely generated, there must be a minimal generating set X = {ai, . . . , an] of A (that is, no proper subset of X generates A). If A ^ 0, then a! ^ 0 by minimality. Since JA = A, ax — j\U\ + y'2o2 + • • • + jnan (7, e J ) , whence 1 Rd! = so that (1K - jjdi - 0 if n — 1 and (Iff —
=
J2CI2 + ■ ’ ’ + jnOn i f n >
1.
Since \K — ji is a unit in R, ai = ( I r — ji)~l(lu — y’Ooi. Thus if n = 1, then ai = 0 which is a contradiction. If n > 1, then ai is a linear combination of a2,. . . , ci„. Consequently, [a2, j generates A, which contradicts the choice of X. (iii) => (iv) Verify that the quotient module A/B is such that J(A/B) = A/B, whence A B = 0 and A = B by (iii). (iv) => (i) If M is any maximal ideal, then the ideal JR + M contains M. But JR -f M R (otherwise R = M by (iv)). Consequently, JR + M = M by maxi mality. Therefore J = JR CL M. ■ We now give several applications of Nakayama’s Lemma, beginning with a result that is the starting point of the theory of completions.
Proposition 4.6. Let J be an ideal in a commutative ring R with identity. Then J is contained in every maximal ideal ofR if and only if for every R-module A satisfying co
the ascending chain condition on submodules, P| JnA = 0. 11= 1
PROOF. (=>) If B = Pi JnA, then JB = B by Theorem 4.4. Since B is finitely n generated by Theorem 1.9, B = 0 by Nakayama’s Lemma 4.5. 0=) We may assume R ^ 0. If M is any maximal ideal of R, then M ^ R and A = R Mis a nonzero /?-module that has no proper submodules (Theorem IV. 1.10). Thus A trivially satisfies the ascending chain condition, whence Q JnA = 0 by hyn
pothesis. Since JA is a submodule of A, either JA = A or JA = 0. If JA = A, then JnA = A for all n. Consequently, P) J"A = A ^ 0, which is a contradiction. Hence n
JA — 0. But 0 = JA = J(R/M) implies that J CL JR CL M. ■
Corollary 4.7. //R is a Noetherian local ring with maximal idealM, then 0 Mn = 0. n=1
PROOF. If J — M and A = R, then JnA = Mn; apply Proposition 4.6. ■
Proposition 4.8. IfR is a local ring, then every finitely generated projective R-mod ule is free.
4. NOETHERIAN RINGS AND MODULES
391
Actually a much stronger result due to I. Kaplansky [63] is true, namely: every projective module over a (not necessarily commutative) local ring is free. PROOF OF 4.8. If P is a finitely generated projective /?-module, then by
Corollary IV.2.2 there exists a free F-module F with a finite basis and an epimor phism 7r : F —> P. Among all the free P-modules F with this property choose one with a basis | xt,x2, . . . , xn | that has a minimal number of elements. Since tt is an epimor phism |7r(xi), . . ., 7t(a„)) necessarily generate P. We shall first show that K = Ker7r is contained in MF, where M is the unique maximal ideal of R. If K jzf MF, then there exists k e K with k ^ MF. Now k = r\X\ + r2x2 + ■ ■ • + rnXn with r, e R uniquely determined. Since k | MF, some say n, is not an element of M. By Theo rem III.4.13, n is a unit, whence — rr'k = — r\~lr2X2 — • — ri_1/-„x„. Consen
(
\n
y, —rC^iXi ) =
52 —r,rlTr(xi).
i=2 / i=2
Therefore, {tt(x2), . . . , 7r(x„) J generates P. Thus if F' is the free submodule of F with basis x„} and tt' : F’ —» P the restriction of ir to F', then it' is an epimor phism. This contradicts the choice of F as having a basis of minimal cardinality. Hence K CZ MF. Since 0 — » F A p — > O i s exact and P is projective K © P F by Theorem IV.3.4. Under this isomorphism (A:,0)H> k for all k e K (see the proof of Theorem IV.l.18), whence F is the internal direct sum F — K (£) P' with P' = P. Thus F = K -\- P'd MF + P'. If ueF, then « = 52 v* + Pi with m, e M, Vi e F, p, e P'. i
Consequently, in the P-module F/P', U + P' =
52 w* + P' = 52 + P') S M(F/P'),
ii
whence M(F/P') = F/P'. Since F is finitely generated, so is F/P'. Therefore K = F/P' = 0 by Nakayama’s Lemma 4.5. Thus P =. P' = F and P is free. ■ We close this section with two well known theorems. The proofs are independent of the preceding part of this chapter.
Theorem 4.9. (Hilbert Basis Theorem) 7/R is a commutative Noetherian ring with identity, then so is R[xu . . . , x„], PROOF. Clearly it suffices to show that 7?[jc] is Noetherian. By Theorem 1.9 we need only show that every ideal J in F[x] is finitely generated. For each n > 0, let I„ be the set of all r s R such that r = 0 or r is the leading co efficient of a polynomial ftJ of degree n. Verify that each /„ is an ideal of R. If r is a nonzero element of /„ and fzJ is a polynomial of degree n with leading coefficient r, then r is also the leading coefficient of xf, which is a polynomial in J of degree n + 1. Hence 70 Cl lx CZ /2 Cl ■ ■ ■. Since R is Noetherian, there exists an integer t such that In = /j for all« > /; furthermore, by Theorem 1.9 each /„ (n > 0) is finitely generated say In = (/■„!,rn2, . . . , rIlin). For each rnJ with 0 < n < t and 1 < j < /„, let £ J be a polynomial of degree n with leading coefficient r7ll. Observe that f0j = rQ] eRCZ P[x], We shall show that the ideal J of F[jr] is generated by the finite set of polynomials X = [ fn] \ 0 < n < f, \ <j< i„}.
CHAPTER VIII COMMUTATIVE RINGS AND MODULES
392
Clearly (X) CZ J. Conversely, the polynomials of degree 0 in J are precisely the elements of 70 and hence are contained in (X). Proceeding by induction assume that (A) contains all polynomials of J of degree less than k and let g e J have degree k and leading coefficient r ^ 0. If k < t, then r e Ik and hence r = strki -f s2rk2 + ■ - ■ + sikrkiI. for some s, e R. ik Therefore the polynomial 22 sifk-i E C^) ^as leading coefficient r and degree k. Con3=1
sequently, g —
has degree at most k — 1. By the induction hypothesis 3
g - 22 Sjf k j s (A), whence g s (X). j it it
If k > t, then r s Ik = It and r = 22 (-Sy £ R)- Furthermore 22 sixk~lfti £ (A) 3=1 j=1 has leading coefficient r and degree k. Thus g — 22 sixk~‘fi has degree at most 3
k — 1 and lies in (X) by the induction assumption. Consequently, g e (A-) and the in duction is complete. Therefore, J = (A). ■
Proposition 4.10. If R is a commutative Noetherian ring with identity, then so is R[[x]]. REMARK. Our proof makes use of Proposition 4.1. Although we shall not do so, the technique used to prove Theorem 4.9 may also be used here, with nonzero coefficients of lowest degree replacing those of highest degree in the argument. How ever, great care must be used to insure that certain power series constructed in ductively in the course of the proof are in fact validly defined. The Axiom of Choice and some version of the Recursion Theorem are necessary (this part is frequently obscured in many published proofs of Proposition 4.10). PROOF OF 4.10. It suffices by Proposition 4.1 to prove that every prime ideal P in is finitely generated. Define an epimorphism of rings /?[[*]] —» R by co
mapping each power series/ =
22
onto its
constant term a0. Let P* be the image
i=0
of P under this map. Then P* is a finitely generated ideal in R (Exercise III.2.13 and Theorem 1.9), say P* = (/-,,..., /-„). For each /-, choose feP with constant term /-,. If x e P, we claim that P is generated by r1? . . . , rn,x. First note that if CO
f = rk +
22 GiX', then r
/go
i=i
k
= f: — x\
22
\
Vj =o /
CO
) £ P- If g = ^2 blxi s P, then »=o n
b0 = ii/-, + - - - + s„rn for some s, £ R. Consequently, g —
term; that is, g —
22
22 has 0 constant 22 + 8' *
t == 1 ■*«ft = xgi (gi £ 7?[[x]]). Therefore g = i i
generated by n, . . . , rv,x.
s
oo
22 s R. Consequently, h — 22 i= i
If x £ P, we claim that P is generated by f,...,f n s P. If h = n
Co = t,n + ■ ■ ■ + tnrn for some tt
and p
x
CiX%
e
then
i=0
lif = xb*
f°r some
h* e /?[[*]]. Since x £ P and xh* = h -22 uf e P and P is prime, we have h* e P. i
m
For each h e P, choose u e R and h* e P such that h = i—1
22 hf + xh* (Axiom of
4. NOETHERIAN RINGS AND MODULES
393
Choice). Let X : P —> P be the map defined by h |—> h*. Let g be any element of P. Then by the Recursion Theorem 6.2 of the Introduction (with X = fi, for all ri) there is a function 0 : N —» P such that 0(0) = g and 4>(k + 1) = X(0(*)) = 4>(k)* Let 0(A) = hk e i?[[x]] and denote by tki the previously chosen elements of R such that n
n
hk ^ 1 fki f xhk ^ 1 tkifi “f- xhk+1.
t=i i=i
CO
For each /' (1 < /' < ri) let g, =
22 Then k=o 6
n / co
\
co / n
\
gi/i h—h = 22 (22 f = 22 ( 22 )•**
i=l \fc=0 / fc = 0 \i = 1 / CO
=
22 “ xh k=0
k+l)x
k.
Consequently, for each m > 0 the coefficient of xrn in gif + • ■ • + gnfi is the same m m as the coefficient of x in (hk — xhk+i)xk. Since lc = 0 m (hk — a//a+i)a* = An — xm+1hm+1 = g — a1”4 xhmtn i,
22
22
fc = 0
the coefficient of xm in fgY + ■ ■ ■ + fig,, is precisely the coefficient of xm in g. There fore, g = gif + g2f H---------- 1-gnfi and f,...,f „ generate P. ■
EXERCISES 1. Let R be a commutative ring with identity and / a finitely generated ideal of R. Let C be a submodule of an 7?-module A. A ssume that for each r el there exists a positive integer m (depending on r) such that r”'A d C. Show that for some integer n, InA d C. [Hint: see Theorems III.1.2(v) and III.2.5(vi)]. 2. Without using primary decomposition, prove this version of the Krull Inter section Theorem. If R is a commutative Noetherian ring with identity, / an ideal co
of R, A a finitely generated i?-moduIe, and B = P) InA, then IB = B. [Hints: n=
1
Let C be maximal in the set S of all submodules S of A such that B D S = IB.lt suffices to show I™A d C for some tn. By Exercise 1 it suffices to show that for each re I,r n A ClCfor some n (depending on r). For each k, let Dk = joe A | rkaz C\. D0 d Di d D2 d ■ ■ ■ is an ascending chain of /?-submodules; hence for some n, Dk — Dn for all k > n. Show that (rnA + C) fl B = IB. The maximality of C implies rnA + C = C, that is, rnA d C.[ 3. Let R be a Noetherian local ring with maximal ideal M. If the ideal M/M2 in R/M 2 is generated by {oj + M 2 , + M'2}, then the ideal M is generated in R by {ax, . . . , «„}.
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
4. (Nakayama’s Lemma, second version) Let R be a commutative ring with identity, J an ideal that is contained in every maximal ideal of R, and A a finitely generated fi-module. If R/J A = 0, then A = 0. [Hint: use the exact sequence 0 —»./—» R —> R/J —> 0 and the natural isomorphism R (X)* A ^ A to show JA = A.] 5. Let R and J be as in Exercise 4; let A be a finitely generated fi-module and / : C. —» A an /?-module homomorphism. Then / induces a homomorphism / : C//C —> A/JA in the usual way (Corollary IV. 1.8). Show that if /is an epimor phism, then /is an epimorphism. 6. (a) Let R be a commutative ring with identity. If every ideal of R can be generated by a finite or denumerable subset, then the same is true of /?[*]. (b) State and prove an analogue of part (a) for /?[[*]]; (the answer is not quite the same here). 7. Let R be a commutative ring with identity and let fg e /?[[*]]. Denote by In f the co
initial degree of /(that is, the smallest n such that a„ ^ 0, where / = 52 «»*0i=0 Show that (a) In (/+ *) > min (In f In g). (b) In (f g ) > In /+ In*. (c) If R is an integral domain, In (fg ) = In /+ In g . 8. Let R be a commutative Noetherian ring with identity and let Q\ fl - - - fl Qn = 0 be a reduced primary decomposition of the ideal 0 of R with Qi belonging to the prime ideal Pt. Then P, U P2 U ■ ■ • U Pn is the set of zero divisors in R. 9. Let R be a commutative ring with identity. If every maximal ideal of R is of the form (c), where c1 = r, for some c e R, then R is Noetherian. [Hint: show that every primary ideal is maximal; use Proposition 4.1.]
5. RING EXTENSIONS In the first part of this section ring extensions are defined and the essential properties of integral extensions are developed. The last part is devoted to the study of the relations between prime ideals in rings R andS, where S is an extension ring of R. Throughout this section all rings are commutative with identity.
Definition 5.1. Let S be a commutative ring with identity and R a subring of S con taining Is- Then S is said to be an extension ring of R. EXAMPLES. Every extension field F of a field K is obviously an extension ring of K. If R is a commutative ring with identity, then /?[[x]] and /?[xi, . . . , a„] are ex tension rings of R. The ring Z is not an extension of the subring E of even integers since E does not contain 1.
5. RING EXTENSIONS
395
Definition 5.2. Let S be an extension ring of R and s e S. If there exists a monic polynomial f(x) e R[x] such that s is a root off (that is, f(s) = 0), then s is said to be integral over R. If every element of S is integral over R, S is said to be an integral ex tension of R. The key feature of Definition 5.2 is the requirement that /be monic.
EXAMPLES. Every algebraic extension field F of a field K is an integral exten sion ring (see the Remarks after Definition V.1.4). The ring R is integral over itself since r £ R is a root of x — r e In the extension of Z by the real field R, i/\ ! 3 is algebraic over Z since it is a root of 3x 2 — 1 but l/\^3 is not integral over Z. How ever, l/\/3 is integral over the rational field Q since it is a root of x 2 — 1/3. Let 5 be an extension ring of R and A" a subset of S. Then the subring generated by X over R is the intersection of all subrings of S that contain X U R; it is denoted R[ X]. The first half of Theorem V.1.3 is valid for rings and shows that /?[A']consists of all elements f( s\, .. . , s„) with n e N*, fe /?[j£j,. . . , x„] and Si e X. In par ticular, for any s u . . . , s t eS the subring generated by |si, . . . , i'd over R, which is denoted /?[si,. . . , <sj, consists of all elements f(su . . . , st) with /e /?[xi,.. . , x<]. An element of /?[si, . . . , s«] is sometimes called a polynomial in 5, , . . . , ,st. Despite this terminology R [s u . . . , sf] need not be isomorphic to the polynomial ring R[x u . . . , xt] (for example, f(su . . . , s,) may be zero even though /is a nonzero polynomial). It is easy to see that for each i (1 < i < /), /?[si, . .., s^ x][.svJ = . . . , s,-]. Since /?[si,. . . , sf] is a ring containing R, .. ., is an /?-module in the obvious way. Likewise every module over /?[si, . . . , 5<] is obviously an i?-module.
Theorem 5.3. Let S be an extension ring of R and s e S. Then the following conditions are equivalent.
(i) s is integral over R; (ii) R[s] is a finitely generated R-module', (iii) there is a subring T ofS containing 1 s andR\$\ which is finitely generated as an K-module', (iv) there is an R[s]-submodule B of S which is finitely generated as an K-module and whose annihilator in R[s] is zero.
SKETCH OF PROOF, (i) => (ii) Suppose 5 is a root of the monic polynomial /s /?[*] of degree n. We claim that 1« = s°,s,s2, . . . , s""1 generate i?[s] as an /^-module. As observed above, every element of ./?[$] is of the form #(.s) for some g e R[x]. By the Division Algorithm I1I.6.2 g(x) = f(x)q(x) + r(x) with deg r < deg/. Therefore in 5, #(5) = /(.v)c/(.v) + r(s) = 0 + r(s) = r(s). Hence g(s) is an /^-linear combination of 1 n,s,s2,. . ., s”‘ with m = deg r < deg f=n. (ii) => (iii) Let T = /?[s], (iii) =?- (iv) Let B be the subring T. Since R d 7?[^] CZ T, B is an /?[.y]-module that is finitely generated as an 7?-module by (iii). Since ls e B, uB — 0 for any uzS implies u = uls = 0; that is, the annihilator of B in /?[s] is 0. (iv) => (i) Let B be generated over R by b u .. ., b„. Since B is an /?[s|-module sbi e B for each i. Therefore there exist r, 3 e R such that
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CHAPTER VUI COMMUTATIVE RINGS AND MODULES
sbi — rnbt + fnbi + ■ ■ ■ + T\ „b„ sbi = f'i\b\ + rnb‘i + • • • + r%nbn
sbn = rn\b\ + rn-ibi + • - • + rnnbn.
Consequently, (/" 11 — s)b\ + rn,b-i + • • ■ + r\vbn = 0
f2lb\ + (/*22 — s)b-i + • • ■ + K2nbn = 0
rn\b\ + rn-i,bi + • • • + — s)bn = 0.
Let M be the n X n matrix (ru) and let d e /?[.s] be the determinant of the matrix M — sln. Then db, = 0 for all / by Exercise VI1.3.8. Since B is generated by the bi, dB = 0. Since the annihilator of B in /?[s] is zero by (iv) we must have d = 0. If / is the polynomial |M — xln\ in P[jt], then one of f —/ is monic and ± f{s) - ±| M — .s/„| - ±<7 = 0. Therefore s is integral over R. ■
Corollary 5.4. IfSisaring^extensionofR and S is finitely generated as an R-module, then S is an integral extension of R. PROOF. For any s e S let S = T in part (iii) of Theorem 5.3. Then j is integral over R by Theorem 5.3(i). ■ The proofs of the next propositions depend on the following fact. If R Cl S Cl T are rings (with \T e R) such that T is a finitely generated 5-module and 5 is a finitely generated .R-module, then Tis a finitely generated /?-module. The second paragraph of the proof of Theorem IV.2.I6 contains a proof of this fact, mutatis mutandis.
Theorem 5.5. 7/S is an extension ring of R and si, . . . , st e S are integral over R, then R[ii, . . . , St] is a finitely generated R-module and an integral extension ring ofR. PROOF. We have a tower of extension rings: R C fl[si] C /?[i,,i2] C - • - C /?[*,, . . . , s t ].
For each i, s , is integral over R and hence integral over R[s l t . . . ,.$V-i]. Since R[s u ■ ■ ., s.] = /?[$,,. . •, 5,-_i][jt], /?[5i,. . • , s;] is a finitely generated module over R[s u - - - , -sv-i] by Theorem 5.3 (i), (ii). Repeated application of the remarks preced ing the theorem shows that /f[$i,.. . , j„] is a finitely generated /f-module. Therefore, 7?[si, . . . , s n ] is an integral extension ring of R by Corollary 5.4. ■
5. RING EXTENSIONS
397
Theorem 5.6. If T is an integral extension ring of S and S is an integral extension ring of R, then T is an integral extension ring of R. PROOF. T is obviously an extension ring of R. If t s T, then t is integral overS 71
and therefore the root of some monic polynomial /eS[x], say / = 22
SiX'-
Since /is
i=0
also a polynomial over the ring R[s0,.si, ■ . - , sn-i], t is integral over R[s 0 ,. . . , s„_i]. By Theorem 5.3 R[s 0,. . . , Sn-i][/] is a finitely generated R U o , . . . , s„_i]-module. But since 5 is integral over R, R[-y0, - . . , .s«-i] is a finitely generated .R-module by Theorem 5.5. The remarks preceding Theorem 5.5 show that R[s0, . . . , JK_i][/] = R[s0, • ■ . , Sn-l,/]
is a finitely generated R-module. Since R[/] CZ R[s0 , . . . , s„_i,/], t is integral over R by Theorem 5.3(iii). ■
Theorem 5.7. Let S be an extension ring ofR and let R be the set of all elements ofS that are integral over R. Then R is an integral extension ring o/R which contains every subring of S that is integral over R. PROOF. If s,t e R, then s,t e R[s,r], whence t — s e .R[5,r] and ts e R[s,t]. Since s and t are integral over R, so is the ring R[.s,?] (Theorem 5.5). Therefore t — s eR and ts e R. Consequently, R is a subring of S’ (see Theorem 1.2.5). R contains R since every element of R is trivially integral over R. The definition of R insures that R is integral over R and contains all subrings of >S that are integral over R. ■ IfS is an extension ring of R, then the ring R of Theorem 5.7 is called the integral closure of R in S. If R = R, then R is said to be integrally closed in S.
REMARKS, (i) Since 1« e R CZ R, S is an extension ring of A. Theorems 5.6 and 5.7 imply that R is itself integrally closed in S. (ii) The concepts of integral closure and integrally closed rings are relative notions and refer to a given ring R and a par ticular extension ringS. Thus the phrase “R is integrally closed” is ambiguous unless an extension ring S is specified. There is one case, however, in which the ring S is understood without specific mention. An integral domain R is said to be integrally closed provided R is integrally closed in its quotient field (see p. 144). EXAMPLE. The integral domain Z is integrally closed (in the rational field Q; Exercise 8). However, Z is not integrally closed in the field C of complex numbers since / e C is integral over Z. EXAMPLE. More generally, every unique factorization domain is integrally closed (Exercise 8). In particular, the polynomial ring F[x i,. . . , x„] (F a field) is integrally closed in its quotient field £(*1,. . . , a„). The following theorem is used only in the proof of Theorem 6.10.
Theorem 5.8. Let T be a multiplicative subset of an integral domain R such that 0 ^ T. If R is integrally closed, then T-1R is an integrally closed integral domain.
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
SKETCH OF PROOF. T lR is an integral domain (Theorem III.4.3(ii)) and R may be identified with a subring of T lR (Theorem III.4.4(ii)). Extending this identi fication, the quotient field Q(R) of R may be considered as a subfield of the quotient field QiT 'R) of T l R. Verify that Q(R) = Q(T~lR). Let u e Q(T^R) be integral over T 'R; then for some r% e R and s, e T, un -f- (/v i/s,, i)wn_1 + • • • + (r 1/S\)u -f- (r0/s0) — 0.
Multiply through this equation by sn, where s = sosi • • ■ j„.-i e T, and conclude that su is integral over R. Since su e Q(T~lR) = Q(R) and R is integrally closed, su e R. Therefore, u = su/s e T~lR, whence T lR is integrally closed. ■
The remainder of this section is devoted to exploring the relationships between (prime) ideals in rings R and S, where S is an extension ring of R. The only point in the sequel where this material is used is in the proof of Lemma 7.3. If S is an extension ring of R and / (^S) is an ideal of S, it is easy to see that / fl R ^ R and 1 fl R is an ideal of R (Exercise 10). The ideal J = I f1 R is called the contraction of / to R and / is said to lie over J. If Q is a prime ideal in an extension ring 51 of a ring R, then the contraction Q fl R of Q to R is a prime ideal of R (Exercise 10). The converse problem is: given a prime ideal P in R does there exist a prime ideal Q in 51 that lies over P (that is, Q fl R = P)? There are many examples where the answer is negative (for example, the extension of Z by the field Q of rationals). A partial solution to the problem is given by the next theorem, which is due to Cohen-Seidenberg.
Theorem 5.9. (Lying-over Theorem) Let S be an integral extension ring of R and P a prime ideal of R. Then there exists a prime ideal Q in S which lies over P (that is, Q (1 R = P).
PROOF. Since P is prime, R — P is a multiplicative subset of R (Theorem 2.1) and hence a multiplicative subset of S. Clearly 0 4 R — P. By Theorem 2.2 there is an ideal Q of S' that is maximal in the set of all ideals / of S such that / fl (R — P) = 0; furthermore any such ideal Q is prime in S. Clearly Q D R (Z P. If Q fl R ^ P, choose u e P such that u £ Q. Then the ideal Q + («) in S properly contains Q. By maximality there exists c e (Q + («)) D (R — P), say c = q + su(q e Q\ s e S). Since .s is integral over R, there exist n e R such that sn + rr.^isn~1 + • • • + ns + rb — 0.
Multiplying this equation by un yields (su)n + \u(su)n~l + • ■ ■ + r,un^l(su) -(- r0un = 0.
Since su = c — q the Binomial Theorem III.1.6 implies that v = cn + r,L- 1 ucn 1 + - • ■ + rxun~lc + r0u" e Q.
But c e R and hence c e R fl Q CZ. P. But u £ P and c e P imply cn prime, c must lie in P, which is a contradiction. ■
s
P. Since P is
5. RING EXTENSIONS
399
Corollary 5.10. (Going-up Theorem) Let S be an integral extension ring of R and Pi, P prime ideals in R such that Pi CZ P. //Qi is a prime ideal ofS lying over Pi, then there exists a prime ideal Q of S such that Qi (Z Q and Q lies over P. SKETCH OF PROOF. As in the proof of Theorem 5.9, R — P is a multiplica tive set in S'. Since Qi fl R = Pi d P, we have Qx fl (R — P) = 0. By Theorem 2.2 there is a prime ideal Q of S that contains Qi and is maximal in the set of all ideals I of 5 such that Qx d I and / D (R — P) = 0. The proof of Theorem 5.9 now carries over verbatim to show that Q 0 R = P. ■
Theorem 5.11. Let S be an integral extension ring o/R and P a prime ideal in R. //Q and Q' are prime ideals in S such that Q CZ Q' and both Q and Q' lie over P, then Q = Q'PROOF. It suffices to prove the following statement: if Q is a prime ideal in S such that Q fl R — P, then Q is maximal in the set S of all ideals / in S with the property / fl (R — P) = 0. If Q is not maximal in S, then there is an ideal / in S with Q d I and / fl (R - P) = 0.
Consequently, / D R d P. Choose u e / — Q. Since u is integral over R, the set of all monic polynomials /e /?[.*] such that deg/> 1 and /(«) e Q is nonempty. Choose n
such an /of least degree, say / = 52 riXi• Then i=0 un +
+ • ■ ■ + riM + r0 e Q d /,
whence r o s / f l R d P = Q f ) R d Q . Therefore u(un~x + rn_iun~2 --------[- riu + r) e Q.
By the minimality of deg f («"_1 + iu"~2 + ■ ■ ■ + n) \ Q, and u \ Q by choice. This is a contradiction since Q is prime (Theorem III.2.15). Therefore Q is maximal in S. m
Theorem 5.12. Let S be an integral extension ring o/R and let Q be a prime ideal in S which lies over a prime ideal P m R . Then Q is maximal in S if and only if P is maximal in R. PROOF. Suppose Q is maximal in S. By Theorem III.2.18 there is a maximal ideal M of R that contains P. M is prime by Theorem III.2.19. By Corollary 5.10 there is a prime ideal Q’ in S' such that Q d Q’ and Q' lies over M. Since Q' is prime, Q'S (Definition III.2.14). The maximality of Q implies that Q = Q', whence P —- Q fl R = Q’ fl R = M. Therefore, P is maximal in R. Conversely suppose P is maximal in R. Since Q is prime in S,Q ^ S and there is a maximal ideal N of S containing Q (Theorem III.2.18). N is prime by Theorem III.2.19, whence in = Is 4 N. Since P = R D Q d R fl N d R, we must have P = R fl N by maximality. Thus Q and N both lie over P and Q d N. Therefore, Q = N by Theorem 5.11. ■
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
EXERCISES Note: Unless otherwise specified, S' is always an extension ring of R.
1. Let 5 be an integral extension ring of R and suppose R and S are integral do mains. Then £ is a field if and only if R is a field. [Hint: Corollary III.2.21.] 2. Let R be an integral domain. If the quotient field F of R is integral over R, then R is a field. 3. Let R be an integral domain with quotient field F. If 0 integral over R, then a is a unit in R.
^
a
e R and 1
r/
a e F is
4. (a) Let R be an integral domain with quotient field F. If 0 ^ a e R, then the following are equivalent: (i) every nonzero prime ideal of R contains a; (ii) every nonzero ideal of R contains some power of a: (iii) F = R[ l R / a\ (ring extension). An integral domain R that contains an element satisfying (i)—(iii) is called a Goldmann ring. (b) A principal ideal domain is a Goldmann ring if and only if it has only finitely many distinct primes. (c) Is the homomorphic image of a Goldmann ring also a Goldmann ring? 5. If S is an integral extension ring of R and / :S—*S is a ring homomorphism, such that /(Is) = Is, then f(S ) is an integral extension ring of f(R). 6. If S is an integral extension ring of R , then 5[jri, . . . , x v ] is an integral extension ring of R[x u . .. , *„]. 7. If S' is an integral extension ring of R and F i s a multiplicative subset of R (0 ^ T), then T~lS is an integral extension of T lR. [Hint: If s/t £ T~XS, then s/t =
i
10. (a) If I (?^S) is an ideal of S, then / fl R ^ R and / fl R is an ideal of R. (b) If Q is a prime ideal of S, then Q D R is a prime ideal of R.
6. DEDEKIND DOMAINS In this section we examine the class of Dedekind domains. It lies properly be tween the class of principal ideal domains and the class of Noetherian integral domains. Dedekind domains are important in algebraic number theory and the algebraic theory of curves. The chief result is Theorem 6.10 which characterizes Dedekind domains in several different ways.
6. DEDEKIND DOMAINS
401
The definition of a Dedekind domain to be given below is motivated by the following facts. Every principal ideal domain D is Noetherian (Lemma III.3.6). Con sequently, every ideal (^ D) has a primary decomposition (Theorem 3.6). The intro duction to Section 2 shows that a particularly strong form of primary decomposition holds in a principal ideal domain, namely: every proper ideal is (uniquely) a prod uct of prime ideals.
Definition 6.1 .A Dedekind domain is an integral domain R in which every ideal( R) is the product of a finite number of prime ideals. EXAMPLE. The preceding discussion shows that every principal ideal domain is Dedekind. The converse, however, is false. There is an example after Theorem 6.10 below of a Dedekind domain that is not a principal ideal domain. It is not immediately evident from the definition that every Dedekind domain is in fact Noetherian. In order to prove this fact and to develop other properties of Dedekind domains we must introduce the concept of a fractional ideal.
Definition 6.2. Let R be an integral domain with quotient fieldK. A fractional ideal of R is a nonzero R-submodule I o/K such that ai C R for some nonzero a eR. EXAMPLE. Every ordinary nonzero ideal I in an integral domain R is an /?-submodule of R and hence a fractional ideal of R. Conversely, every fractional ideal of R that is contained in R is an ordinary ideal of R. EXAMPLE. Every nonzero finitely generated -submodule / of K is a fractional ideal of R. For if /is generated by bi, . . . , bn e K, then I = Rbi + • • ■ + Rbn and for each bi = Ci/ai with 0 ^ r, £ R. Let a = u\ar ■ ■an. Then a ^ 0 and al — Rut ■ anC\ + • —1- Rax ■ ■ -an~icn C R. REMARK. If / is a fractional ideal of a domain R and al C R (0 ^ a e R), then al is an ordinary ideal in R and the map I al given by jc |—» ax is an .R-module isomorphism.
Theorem 6.3. If R is an integral domain with quotient field K, then the set of all fractional ideals of R forms a commutative monoid, with identity R and multiplication given by IJ = aibi | a; e I; bj e J; n.e N* j.
PROOF. Exercise; note that if I and J are ideals in R, then IJ is the usual product of ideals. ■ A fractional ideal I of an integral domain R is said to be invertible if IJ = R for some fractional ideal J of R. Thus the invertible fractional ideals2 are precisely those that have inverses in the monoid of all fractional ideals. 2In
the literature invertible fractional ideals are sometimes called simply invertible ideals.
402
CHAPTER VIII COMMUTATIVE RINGS AND MODULES
REMARKS, (i) The inverse of an invertible fractional ideal / is unique and is 7_1 = {a e K \ ai CL R}. Indeed for any fractional ideal 7 the set 7_1 = \azK\al CL R\ is easily seen to be a fractional ideal such that 7_17 = 77_1 CL R. If 7 is invertible and IJ = JI = R, then clearly J C I~\ Conversely, since 7_1 and J are /^-submodules of /C, 71 = R I 1 = (JI)I 1 = Jill'1) CL JR = RJ C J, whence J = I1. (ii) If I,A,B are fractional ideals of R such that IA = IB and 7 is invertible, then A = RA = (l~lI)A = I~\IB) = RB = B. (iii) If 7 is an ordinary ideal in R, then R CZ I_1. EXAMPLE. Every nonzero principal ideal in an integral domain R is invertible. If K is the quotient field of R and 7 = (b) with b ^ 0, let J = Rc C. K where c = 1R /b. Then J is a fractional ideal of R such that IJ = R. Invertible fractional ideals play a key role in characterizing Dedekind domains. The next five results develop some facts about them.
Lemma 6.4. Let I, Ii, I2, . . . , I„ be ideals in an integral domain R. (i) The ideal Iil2- ■ - In is invertible if and only if each Ij is invertible. (ii) 7/Pi - • Pm = I = Qr • Q,,, where the Pi and Qj are prime ideals in R and every Y*iis invertible, then m = n and (after reindexing) Pi = Q, for each i = l , . . . , m . PROOF, (i) If J is a fractional ideal such that J(h ■ ■ -7,,) = R, then for each j = 1,2, . . . , « , I } (J1 1 - ■ f-if+i■ ■ L) = R, whence Ij is invertible. Conversely, if each Ij is invertible, then (7i ■ ■ ■ 7„) (7i_1 • • ■ 7„_1) = R, whence 7i ■ • • 7„ is invertible. (ii) The proof is by induction on m with the case m = 1 being left to the reader. If m > 1, choose one of the P t , say P u such that Pi does not properly contain P t for i — 2,.. ., m. Since Qi • - ■ Qn = Pi ■ ■ ■ Pm CL Pi and Pi is prime some Q„ say Qu is contained in Pi (Definition III.2.14). Similarly since Px - ■ Pm = Qr ■ ■ Qn C Q,, Pi CL Qi for some Hence Pi C. Qi C. Pt. By the minimality of Pi we must have Pi = Qi = Pi. Since Pi = Qx is invertible, Remark (ii) after Theorem 6.3 implies
P2PS- • * Pm = Q'iQv, • ■ • QnTherefore by the induction hypothesis m = n and (after reindexing) P, = Qi for i = 1,2, ... ,m. ■ The example preceding Lemma 6.4 and Theorem III.3.4 show that every nonzero prime ideal in a principal ideal domain is both invertible and maximal. More generally we have
Theorem 6.5. 7/R is a Dedekind domain, then every nonzero prime ideal of R is in vertible and maximal. PROOF. We show first that every invertible prime ideal P is maximal. If a e R — P, we must show that the ideal P Ra generated by P and a is R. If P + Ra ^ R, then since R is Dedekind, there exist prime ideals P, and Qj such that p + Ru = p , p 2 . - -P „ and P + Rd1 = Q X Q 2 ■ ■ ■ Qn. Let tt : R —> R/P be the canoni cal epimorphism and consider the principal ideals in R/P generated respectively by ir(a) and ir(a2). Clearly (7r(a))
= tt(P 1) • • • ir(Pm) and (7r(a 2 )) = MQi)- - HQ*).
6. DEDEKIND DOMAINS
403
Since ker 7r = P C P, and P d Qi for each z, the ideals iv(Pl) and MQi) are prime in R/P (Exercise 111.2.17(a)). Since R /P is an integral domain (Theorem III.2.16), every principal ideal in R/P is invertible (see the example preceding Lemma 6.4). Con sequently, 7r(Pt) and tt (Q,) are invertible by Lemma 6.4(i). Since tKGi)- • -7KQn) = (7r(a2)) = (tt(a))2 = tt(P1)2- • -tt(Pm)2, Lemma 6.4(ii) implies n = 2m and (after reindexing) 7r(Pt) = 7r(£)2l) = 7r(<92l_,) for i = 1,2, , m . Since Ker 7t = P d Pi and P Cl <2/ for all /,/, Pi = 7r_1(7r(P,)) = 7r_1(7r(02i)) = (?2t and similarly P, = £>2l_i for 1 = 1,2,. . ., m. Consequently, P + Ra2 = (P + Pa)2 and PCP + Pa2 CZ (P + Pa)2 Cl P2 + Ra. If b = c + ra £ P (c e P2,r e P), then ra e P. Thus r e P since P is prime and a $ P. Therefore, P d P2 + Pa d P, which implies P — P2 Pa = P(P -j- Pa). Since P is invertible, P = P_1P = P~1P(P -(- Pa) = P(P + Pa) = P + Pa. This is a contradiction. Therefore every invertible prime ideal P is maximal. Now suppose P is any nonzero prime ideal in P and c is a nonzero element of P. Then (c) = PiP2- ■ • Pn for some prime ideals Pi. Since P\Py ■ ■ Pn = (c) d P, we have for some k, Pkd P (Definition III.2.14). The principal ideal (c) is invertible and hence so is Pk (Lemma 6.4(i)). By the first part of the proof Pk is maximal, whence Pic = P. Therefore, P is maximal and invertible. ■ EXAMPLE. If F is a field, then the principal ideals C*i) and (jc2) in the poly nomial domain F[xi,x2] are prime but not maximal (since (jt5) d (jci,x2) d P[xi,x 2]). 9* Ti Consequently, F[xux^ is not Dedekind (Theorem 6.5). Since F[x u x 2 } is Noetherian by Theorem 4.9, the class of Dedekind domains is properly contained in the class of Noetherian domains.
Lemma 6.6. //I is a fractional ideal of an integral domain R with quotient field K and f e HoniR(l,R), then for all a,b £ I: af(b) = bf(a). PROOF. Now a = r/s and b = v/t (r,s,v,t e P; s,t 9^ 0) so sa = r and tb = v. Hence sab = rb £ / and tab = ca e I. Thus sf(tab) = f(stab) = tf(sab) in R. Therefore, af (b) — saf(b)/s — f(sab)/s = f (tab)/t = tbf (a)/t — bf (a). ■
Lemma 6.7. Every invertible fractional ideal of an integral domain R with quotient field K is a finitely generated R-module. n
PROOF. Since 7-1/ = P, there exist a, £ I~l,bi e / such that 1 n
r
= 53 ^
i=l
c £ /, then c = 52 (.cai)bi. Furthermore each ca, £ P since a* s I~l — {azK \a I d R\.
1—1 Therefore / is generated as an P-module by b u ... , b n (Theorem IV.1.5(iii)). ■
We have seen that every nonzero ideal / in a principal ideal domain D is in vertible. Furthermore I is isomorphic to D as a Z)-module (see Theorem IV.1.5(i)). Thus / is a free and hence projective Z)-module. This result also holds in arbitrary integral domains.
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
Theorem 6.8. Let R be an integral domain andl a fractional ideal of R. Then I is in vertible if and only if I is a projective R-module. PROOF. (=>) By Lemma 6.7 and Theorem IV. 1.5, 7 = Rbi H---------- f Rbn with 71
bi e I and lR =
22
a&i («. e I 1 ). Let F be a free R-module with a basis of n elements i=l eu ... , en. Then the map ir : F —> I defined by a |—► bi is an R-module epimorphism
(see Theorem IV.2.1), and there is a short exact sequence: 0 —* Ker tt —* F^ I—* 0. Define f : 7 —* F by f (c) = ca^x + - ■ • + canen (c e 7) and verify that f is an /?-module homomorphism such that 7rt = 1/i (note that cai e R for each i since ai e 7_1). Con sequently the exact sequence splits and 7 is a direct summand of a free .R-module (Theorem IV.1.18). Therefore, 7 is projective by Theorem IV.3.4. (<=) Let X = \bj \jeJ\ be a (possibly infinite) set of nonzero generators of the projective R-module 7. Let b0 be a fixed element of X. Let F be a free .R-module with basis {e1 j e J| and let 4> : F —* 1 be the R-module epimorphism defined by c, H> bi (Theorem IV.2.1). Since I is projective there is an R-module homomorphism \p : 1 —> F such that = 1 /. For each j e J let nr,- : T7 —► Re, £= R be the canonical r*e* £ F onto r - e R (see Theorem IV.2.1). Then for each j the projection that maps 3
22i
map 6j = 7r: / —> R is an R-module homomorphism. Let c, = d,(bo). For any ce l, cc, = c8j(b0) = boOj(c) by Lemma 6.6, whence in the quotient field K of R, c( c j/bo) = c cj /ba — bo8j(c)/bo = B3(c) e R. Therefore Cj/ boe I
1
= {ae^|fl/CIjRJ.
Consequently, for any cel \Hc) = 22 0X0*7 = 22 c(ci/bo)ej, jeJ l
jeJ i
where J\ is the finite subset \j e J | 6j(c) ^ 0|. Therefore, for any nonzero cel, c = 4>i(c) = 0(22 c(c]/b0)e3) = 22 cCcj/b 0 )bj = c(22 (Ci/bo)bj), jeJ i
whence 1« =
22 ( i/bv)bj with Cj/b c
ieJi
jeJ i
0
jeJi
e 7-1. It follows that R CZ 7-17. Since l~ll d R
is always true, R = 7_17. Therefore I is invertible. ■ The characterization of Dedekind domains to be given below requires us to intro duce another concept. A discrete valuation ring is a principal ideal domain that has exactly one nonzero prime ideal; (the zero ideal is prime in any integral domain).
Lemma 6.9. 7/R is a Noetherian, integrally closed integral domain and R has a unique nonzero prime ideal P, then R is a discrete valuation ring. PROOF. We need only show that every proper ideal in R is principal. This re quires the following facts, which are proved below: (i) Let K be the quotient field of R. For every fractional ideal 7 of 7? the set 7 = {a s K | al C 7) is precisely R; (ii) R a P-1; (iii) P is invertible;
6. DEDEKIND DOMAINS
(iv)
405
0^ = 0;
neN*
(v) P is principal. Assuming (i)-(v) for now, let I be any proper ideal of R. Then / is contained in a non zero maximal ideal M of R (Theorem [II.2.18), which is necessarily prime (Theorem III.2.19). By uniqueness M = P, whence 1 d P. Since H Pn = 0 by (iv), there is a 71EN* largest integer m such that I d Pm and I jzf Pm+1. Choose b e / — Pm+1. Since P = (a) for some a e R by (v), Pm = (a) m = (a m). Since b e Pm, b - uam. Further more, u^P = (a) (otherwise b e Pm+1 = (am+1)). Consequently, u is a unit in R; (otherwise (u) would be a proper ideal by Theorem III.3.2 and hence contained in P by the argument used above). Therefore by Theorem III.3.2 Pm = (am) = (uam) = (b) CZ I, whence I is the principal ideal Pm = (am). Statements (i)-(v) are justified as follows. (i) Clearly R d I. It is easy to see that 7 is a subring of K and a fractional ideal of R, whence 7 is isomorphic (as an 7?-module) to an ideal of R (Remark preceding Theorem 6.3). Thus since R is Noetherian, 7 is finitely generated (Theorem 1.9). Theorem 5.3 (with T — 7 ) implies that every element of 7 is integral over R. There fore, 7 C R since R is integrally closed. Hence 7 = R. (ii) Recall that R d J1 for every ideal J in R. Let 5 be the set of all ideals J in R such that R d J~K Since P is a proper ideal (Definition III.2.14), every nonzero ele ment of P is a nonunit by Theorem III.3.2. If J = (a), (0 a e P), then 1R /a e/_1, but 1 r /a | R, whence R d J~l. Therefore, J is nonempty. Since R is Noetherian, JF contains a maximal element M (Theorem 1.4). We claim M is a prime ideal of R. If ab e M with a,b e R and a $ M, choose c e M~l — R. Then c(ab) e R, whence bc(aR + M) d R and be e (aR + M)_1. Therefore, be £ R (otherwise, aR + A/eJ, contradicting the maximality of M). Consequently, c(bR + M) CZ R, and thus c £ (bR + M)~l. Since c $ R the maximality of M implies that bR + M = M, whence b e M. Therefore M is prime by Theorem III.2.15. Since M y* 0, we must have P = M by uniqueness. Thus R d A/-1 = P~l. (iii) Clearly P d PP_1 CZ R. The argument in the first paragraph of the proof shows that P is the unique maximal ideal in R, whence P = PP~l or PP~l — R. But if P = PP_1, then P~l d P and by (i) and (ii), R d P_1 d P = R, which is a contradiction. Therefore PP~l = R and P is invertible. (iv) If fl Pn 5* 0, then p| Pn is a fractional ideal of R. Verify that neN*
P-1
d p|
Pn.
mN*
Then by (i) and (ii) R d P^1 d p| Pn = R, which is a contra-
mN*
*
neN*
diction. (v) There exists a £ P such that a ^ P2; (otherwise P = P2, whence p| Pn = neN*
P 0 contradicting (iv)). Then aP~x is a nonzero ideal in R such that aP~x JZ. P (otherwise, a e aR = aP~'P d P2). The first paragraph of the proof shows that every proper ideal in R is contained in P, whence aP_1 = R. Therefore by (iii), (a) = (a)R = (a)P~lP = (aP- l)P — RP = P. m
Theorem .10. The following conditions on an integral domain R are equivalent. (i) R is a Dedekind domain;
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
(ii) every proper ideal in R is uniquely a product of a finite number of prime ideals; (iii) every nonzero ideal in R is invertible; (iv) every fractional ideal of R is invertible; (v) the set of all fractional ideals of R is a group under multiplication; (vi) every ideal in R is projective; (vii) every fractional ideal of R is projective; (viii) R is Noetherian, integrally closed and every nonzero prime ideal is maximal; (ix) R is Noetherian and for every nonzero prime ideal Po/R, the localization Rp of R at P is a discrete valuation ring.
PROOF. The equivalence (iv) <=> (v) is trivial (see Theorem 6.3). (i) => (ii) and (ii) => (iii) follow from Lemma 6.4 and Theorem 6.5. (iii) <=> (vi) and (vii) <=> (iv) are immediate consequences of Theorem 6.8. (vi) => (vii) follows from the Remark preceding Theorem 6.3. In order to complete the proof we need only prove the implications (iv) => (viii), (viii) =* (ix) and (ix) => (i). (iv) => (viii) Every ideal of R is invertible by (iv) and hence finitely generated by Lemma 6.7. Therefore R is Noetherian by Theorem 1.9. Let K be the quotient field of R. If u e K is integral over R, then /?[«] is a finitely generated P-submodule of K by Theorem 5.3. Consequently, the second example after Definition 6.2 shows that P[w] is a fractional ideal of R. Therefore, P[w] is invertible by (iv). Thus since P[w]P[w] = P[m],P[m] = RR[u] = (P[m]_1P[m])P[m] = P[m]_1P[m] = R, whence u e R. Therefore R is integrally closed. Finally if P is a nonzero prime ideal in R, then there is a maximal ideal M of R that contains P (Theorem III.2.18). M is invertible by (iv). Consequently M~XP is a fractional ideal of R with MlP C M~'M = R, whence M~lP is an ideal in R. Since M(M_1P) = RP = P and P is prime; either M CL P or M~lP C P. But if M~'P C P, then R C M~l = M~lR = M-xPP~l C PP1 C R, whence M-1 = R. Thus R = MM~X = MR = M, which contradicts the fact that M is maximal. Therefore M CL P and hence M = P. Therefore, P is maximal. (viii) => (ix) Rp is an integrally closed integral domain by Theorem 5.8. By Lemma III.4.9 every ideal in Rp is of the form Ip = {i/s | i e I;s £ P], where / is an ideal of R. Since every ideal of R is finitely generated by (viii) and Theorem 1.9, it follows that every ideal of Rp is finitely generated. Therefore, Rp is Noetherian by Theorem 1.9. By Theorem III.4.11 every nonzero prime ideal of Rp is of the form Ip, where / is a nonzero prime ideal of R that is contained in P. Since every nonzero prime ideal of R is maximal by (viii), Pp must be the unique nonzero prime ideal in Rp. Therefore, Rp is a discrete valuation ring by Lemma 6.9. (ix) => (i) We first show that every ideal I (^0) is invertible. Ii1 is a fractional ideal of R contained in R (Remark (i) after Theorem 6.3), whence //-1 is an ideal in R. If //_1 ^ R, then there is a maximal ideal M containing //-1 (Theorem III.2.18). Since M is prime (Theorem III.2.19), the ideal IM in Rm is principal by (ix); say hi = (a/s) with a e I and s e R — M. Since R is Noetherian, I is finitely generated, say I = (bi, . . . , b n ), by Theorem 1.9. For each i , b-/l J { e I m, whence in R M , bi/ in = (ri/s i)(a/s) for some n s R, s t £R — M. Therefore s1sbi = r l a £l . Let t = ssis2- ■ -s„ . Since R — M is multiplicative, t e R — M. In the quotient field of R we have for every /, (t/a)bi = tbi/a = ■ si^1si+1 ■ ■ ■ s„rt e R, whence t/a e I1. Consequently t = (t/a)a e I~lI C M, which contradicts the fact that t £ R — M. Therefore Il~l = R and I is invertible. For each ideal / (j^P) of R choose a maximal ideal Mi of R such that
6. DEDEKIND DOMAINS
407
I d M, d R (Theorem III.2.18; Axiom of Choice). If / = R, let Mr = R. Then Mr1 is a fractional ideal of R with IMr1 d MiMr1 d R. Therefore, IMr1 is an ideal of R that clearly contains I. Also, if / is proper, then I d IMr1 (otherwise I
since I and Mi are invertible, R = RR = (I~lI)(MrlMi) = I~x(IMrl)Mi = TlIM, = RMi = Mj, which contradicts the choice of Mi). Let S be the set of all ideals of R and define a function / : 5 —> S by /H> IMr1. Given a proper ideal J, there exists by the Recursion Theorem 6.2 of the Introduction (with fn = / for all n) a function
^^^^
Thus Jk ^k • "The remarks above show that this can occur only if Jk = R. Consequently, R = Jk = f(Jk-1) = Jk-\M^_V whence Jk-1 = Jk—\R — Jk-\Mk\ Mk-i = RMk-1 = Mk-1.
Since Mk~i = Jk-i d Jk = R, Mk~i is a maximal ideal. The minimality of k insures that each of M„, . . . , M,t_2 is also maximal (otherwise M, = R, whence 1 = JjR = J )- It is easy to verify that }
J3+i = J3Mr1 = JjR
Mk_i = Jk_! = Jk-iMt* = Jk-SM-k'_3Mk\z = ■ ■ • - JMo 'Mr1- • ■ M;_V
Consequently, since each M, is invertible, Mk_x(Mo- ■ -Mk-2) = JMr1 ■ ■ -MtiMo- - - Mk-o) = J.
Thus J is the product maximal (hence prime) ideals. Therefore R is Dedekind. ■ We close with an example showing that the class of principal ideal domains is properly contained in the class of Dedekind domains. EXAMPLE. The integral domain Z[-yfT0] = [a + 6^10 I a& £ ^as Quotient field Q(\^10) = {r + s^lQ | r,s e Q}. A tedious calculation and elementary number theory show_that Z[\^10] is integrally closed (Exercise 14). Since the evaluation map Z[x\ —> Z[-y/l0] given by /(.v) |—> /(VlO) is an epimorphism and Z[x] is Noetherian (Theorem 4.9), Z[\/l0] is also Noetherian (Exercise 1.5). Finally it is not difficult to prove that every nonzero prime ideal of Z[\/T0] is maximal (Exercise 15). Therefore Z[-y/l0] is a Dedekind domain by Theorem 6.10(viii). However Z l y l O ] is not a principal ideal domain (Theorem IIL3.7 and Exercise III.3.4).
EXERCISES 1. The ideal generated by 3 and 1 + \5i in the subdomain Z[\f5/'] of C is in vertible. 2. An invertible ideal in an integral domain that is a local ring is principal. 3. If I is an invertible ideal in an integral domain R and 5 is a multiplicative set in R with 0 ^ S, then S~*I is invertible in S~'R.
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CHAPTER VIII COMMUTATIVE RINGS AND MODULES
4. Let R be any ring with identity and P an P-module. Then P is projective if and only if there exist sets {a,-1 / e /} CZ P and { f \ i e 1 } CZ Hornr(P,R) such that for all a e P, a = 22 f(o)ai. [See the proof of Theorem 6.8.] iii
5. (Converse of Lemma 6.9) A discrete valuation ring R is Noetherian and in tegrally closed. [Hint: Exercise 5.8.] 6. (a) If every prime ideal in an integral domain R is invertible, then R is Dedekind, (b) If R is a Noetherian integral domain in which every maximal ideal is in vertible, then R is Dedekind. 7. If S is a multiplicative subset of a Dedekind domain R (with \R eS,0 $S), then S lR is a Dedekind domain. 8. If R is an integral domain and P a prime ideal in P[x] such that.P fl R = 0, then P[x]p is a discrete valuation ring. 9. If a Dedekind domain R has only a finite number of nonzero prime ideals Pi, ... , P„, then R is a principal ideal domain. [Hint: There exists a{ e P, — P,2 and by the Chinese Remainder Theorem III.2.25 there exists bi e P, such that bi = ai (mod Pi) and bt = \R (mod P3) for j /'. Show that Pl = (bi), which im plies that every ideal is principal.] 10. If / is a nonzero ideal in a Dedekind domain R, then R/I is an Artinian ring. 11. Every proper ideal in a Dedekind domain may be generated by at most two elements. 12. An P-module A is divisible if rA = A for all nonzero r e R. If R is a Dedekind domain, every divisible P-module is injective. [N.B. the converse is also true, but harder.] 13. (Nontrivial) If R is a Dedekind domain with quotient field K, F is a finite di mensional extension field of K andS is the integral closure of R in F (that is, the ring of all elements of F that are integral over R), then S is a Dedekind domain. 14. (a) Prove that the integral domain Z[\/l0] is an integral extension ring of Z with quotient field Q(^10). (b) Let u s Q(\'l0) be integral over Z[^/i0]. Then u is integral over Z (Theorem 5.6). Furthermore if u e Q, then u e Z (Exercise 5.8). Prove that if u e Q(^10) and u $ Q, then u is the root of an irreducible monic polynomial of degree 2 in Z[x\. [Hint: Corollary III.6.13^and Theorem V.I.6.] (c) Prove that if u = r + s^J 10 £ Q(^10) and u is a root of x2 + ax + b £ Z[x], then a = —2r and b = r 2 — 10s2. [Hint: note that u2 — 2ru -f- (r2 — 1 Os2) = 0; if u | Q use Theorem V.I.6.] (d) Prove that Z^lO] is integrally closed. [Hi nt: if u = r + s-yj 10 £ Q(\ft0) is a root of x 2 -\- ax -\- b z Z[x] and a is even, then re Z by (c); it follows that s e Z. The assumption that a is odd leads to a contradiction.] 15. (a) If P is a nonzero prime ideal of the ring Z[\/l0], then P (1 Z is a nonzero prime ideal of Z. [Hint: if 0 ^ u e P , then u is a root of x 2 + ax + b £ Z[x ] by Exercise 14. Show that one of a ,b is nonzero and lies in P.] (b) Every nonzero prime ideal of Z[^To] is maximal. [Use (a), Theorem III.3.4 and either an easy direct argument or Theorem 5.12.]
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16. A valuation domain is an integral domain R such that for all a ,b e R either a | b or b | a. (Clearly a discrete valuation ring is a valuation domain.) A Priifer domain is an integral domain in which every finitely generated ideal is invertible. (a) The following are equivalent: (i) R is a Priifer domain; (ii) for every prime ideal P in R, Rp is a valuation domain; (iii) for every maximal ideal Min R, R M is a valuation domain. (b) A Priifer domain is Dedekind if and only if it is Noetherian. (c) If R is a Priifer domain with quotient field K , then any domains such that R CZ S CZ K is Priifer.
7. THE HILBERT NULLSTELLENSATZ The results of Section VI. 1 and Section 5 are used to prove a famous result of classical algebraic geometry, the Nullstellensatz (Zeros Theorem) of Hilbert. Along the way we also prove the Noether Normalization Lemma. We begin with a very brief sketch of the geometric background (this discussion is continued at the end of the section). Classical algebraic geometry is the study of simultaneous solutions of systems of polynomial equations: f(x u x 2 , . . . , * „ ) = 0 ( f e S)
where K is a field and S CZ A f x , , . . . , *„]. A solution of this system is an /i-tuple, ( f l i , . . . ,a n )e F n = F X F X • • • X F (« factors), where F is an algebraically closed ex tension field of K and f(a x , . . . , « „ ) = 0 for all feS. Such a solution is called a zero of S in F". The set of all zeros of S is called the affine K-variety (or algebraic set) in Fn defined by S and is denoted V(S). Thus V(S) = {(a,, . . . , « „ ) £ F " | f(a u . . . , * „ ) = 0 for a l l feS } .
Note that if / is the ideal of A[xi, . . . , x n] generated by S, then V(I ) = F(S). The assignment S V(S) defines a function from the set of all subsets of A[xi, . . ., x„] to the set of all subsets of Fn. Conversely, define a function from the set of subsets of Fr‘ to the set of subsets of A[xt,. . . , x„] by Y |—► J(F), where Y CZ F n and J{Y) = { f e K[x u . . . , x„] I f(a u . . . , « „ ) = 0 for a l l (a u ... ,a n )eY}.
Note that J (Y) is actually an ideal of AT[xi,. . . , x„]. The correspondence given by V and J has the same formal properties as does the Galois correspondence (priming operations) between intermediate fields of an extension and subgroups of the Galois group. In other words we have the following analogue of Lemma V.2.6.
Lemma 7.1. Lei F be an algebraically closed extension field of K and let S, T be sub sets o/K[xi, . . . , x„] and X,Y subsets of Fn. Then (i) (ii) (iii) (iv)
V(K[x,,. . . , x,,]) = 0; J(Fn) = 0; J(0) = K[xi----------------- xj; S C T => V(T) C= V(S) and X C Y => J(Y) C J(X); S C= J(V(S)) andY CZ V(J(Y)); V(S) = V(J(V(S))) and J(Y) = J(V(J(Y))).
PROOF. Exercise. ■
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It is natural to ask which objects are closed under this correspondence, that is, which S and Y satisfy J(V(S)) = S and V(J (Y)) = Y. Closed subsets of F n are easily described (Exercise 1), but the characterization of closed subsets of K [ x x , . . . , x n ] re quires the Nullstellensatz, which states that J(V(I )) = Rad I for every proper ideal 1 of K[*i,. . . , x n ]. In order to prove the Nullstellensatz we need two preliminary results, the first of which is of interest in its own right.
Theorem 7.2. (Noether Normalization Lemma) Let R be an integral domain which is a finitely generated extension ring of a field K and let r be the transcendence degree over K of the quotient field F of R. Then there exists an algebraically independent subset {ti,t2, . . . , tr} of R such that R is integral over K[ti, . . . , tr]. PROOF. Let R = K[ui, . . . , «„]; then F = K(uu . . . , un). If { « t , . . . , un\ is algebraically independent over K, , un} is a transcendence base of F over K by Corollary VI.1.6, whence r = n and the theorem is trivially true. If {uu ..., u n j is algebraically dependent over K , then r < n — 1 (Corollary VI.1.7) and iWs-
23
(ti, . . .
• -u,r = o,
where 1 is a finite set of distinct ^-tuples of nonnegative integers and A,, i n is a non zero element of K for every (z'i, . . . , / ' „ ) e I. Let c be a positive integer that is greater than every component i s of every element ( / , , . . . , / „ ) of /. If (/],...,/'„), ( j u ... , jn) e I are such that /’i + ci-> -J- c2/’s 4~ • • ■ 4- cn 1 i„ = ji -f- c/2 4~ c2js 4~ ■ ■ 4~ cn lj„, then c | ii — y'i which is impossible unless i\ = ji (since c > A > 0 and c > jt > 0 imply c > 11*! — y,|). Consequently, i2 + ci3 4---- 1- cn~Hn = h + cjs 4------- 1 cn~2jn. As before c \ i2 — h, whence h = ji. Repetition of this argument shows that (/i, . . . , / „ ) = (Ju ■ - - yjn). Therefore, the set {i\ 4 eii 4 c2i3 4-------cn~xin \ (i’i, . . . , /„) 6 /} "
-
consists of |/| distinct nonnegative integers; in particular, it has a unique maximum element jt 4- cj2 4- • ■ • 4- cn ljn for some (y'i,.. . ,j n ) e I . Let Vz = «2 —
U\c,
v3 =
U3
— u /, . . . ,
Vn
=
Un
—
Kf
If we expand the algebraic dependence relation above, after making the substitutions Ui = v\ 4- ti? 1 (2 < /' < «), we obtain kh...jnu1Mc”+‘*”+ -
+cn lin
+ f(u u v 2,V3, .. ., c„) = 0,
where the degree of fe K[x u . . . , jr„] in .vi is strictly less than ji + cj? 4- • • • 4- cn~lj„. Therefore, ux is a root of the monic polynomial xh+ch+
. ■ .+c*-h-n _f_ kjl...jnf(x,v2, . . . , c „ ) c K[v-2,-u„][jfJ.
Consequently, ux is integral over K[v2, . . . , v„]. By Theorem 5.5 K [u u V 2,. . . , c„] = K[vz, . . . , Dn][«i] is integral over K[t72,. . - , y»i]- Since each w, (2 < i < n ) is ob viously integral over K[u u vi , . . . , v n ], Theorems 5.5 and 5.6 imply that P A [, . . . ,
Un]
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is integral over K[v 2 , . . . , v n ] (whence F is algebraic over K(vi, . . . , v n )). If {i?2,. . ., un} is algebraically independent, then r = n — 1 by Corollary VI.1.6 and the theorem is proved. If not, the preceding argument with K [v 2, . . . , v n ] in place of R shows that for some w 3 , . . . , w „ s R, K [ u 2 , . . . , v n \ is integral over K [w 3 , . . . , wn\. By Theorem 5.6 R is integral over K[w?J, . . . , (whence F is algebraic over K(w3, . .. ,w„) and r < n — 2). If { w 3 , . . . , w„) is algebraically independent, we are finished. If not, the preceding process may be repeated and an inductive argument will yield an algebraically independent subset { z „ _ r + i , . . . , z „ } of r elements of R such that R is integral over K[zn-r+u . . . , znJ. ■ Now let K be a field and Fan algebraically closed extension field of K. If a proper ideal / of K[*i,. . ., is finitely generated, say I = ( g i , . . . , gk), then the affine variety V(I) clearly consists of every (au . . . , an) e F” that is a common root of gu - ■ ■ ,g k (see Exercise 4). If n = 1, Kfci] is a principal ideal domain and it is ob vious that V(T) is nonempty. More generally (and somewhat surprisingly) we have:
Lemma 7.3. If F is an algebraically closed extension field of a field K and I is a proper ideal o/K[xu . . . , xn], then the affine variety V(I) defined by I in Fn is nonempty. PROOF. By Theorems III.2.18 and III.2.19 / is contained in a proper prime ideal P, whence V(P) CZ V(l). Consequently, it suffices to prove that V(P) is non empty for every proper prime ideal P of ^[xi,.. ., x „]. Observe that P fl K = 0; (otherwise 0 ^ a e P D K, whence 1a- = a l a z P, contradicting the fact that P is proper). Let R be the integral domain K[x u . . . , x n ]/P (see Theorem III.2.16) and let t t : K[x i,. .., x„] —» R be the canonical epimorphism. If we denote t t (a,) e R by then R = 7r(A)[H!, . . . , «„]. Furthermore since K fl P — 0, i t maps K isomorphically onto in particular, 7r(AT) is a field. By the Noether Normalization Lemma there exists a subset { / i , . . . , t T ) of R such that [t u • • •, t T ) is algebraically independent over t v (K) and R is integral over S = i r(K)[t u . . . , tr\. If M is the ideal of 5 generated by tu . . . , tr, then the map ir(K) —> S/M given by 7r(a) j—> -w(a) + M is an isomorphism (see Theorem VI. 1.2). Consequently Mis a maximal ideal ofS by Theorem III.2.20. Therefore, there is a maximal ideal N of R such that N C\ S = M (Theorems 5.9 and 5.12). Let t : R —» R/N be the canonical epimorphism. Then t ( R) = R /N is a field by Theorem III.2.20. The Second Isomorphism Theorem III.2.12 together with the maps defined above now yields an isomorphism K ^ tr(A) =£ S/M = S/(N fl S)^(S + N)/N = t(S),
which is given by a |—> ir(a) \—> tt(o) + M |—> ir(a) + N = t(7t(o)). Let t(R) b e a n algebraic closure of t(R). Since R is integral over S, t(R) is an algebraic field exten sion of t(S), whence t(R) is also an algebraic closure of r(5) (Theorem V.3.4). Now F contains an algebraic closure K of K (Exercise V.3.7). By Theorem V.3.8 the isomor phism K = t(S) extends to an isomorphism K = t(R). Restriction of the inverse of this isomorphism yields a monomorphism a : t(R) K CZ F. Let 4> be the compo sition K.[x\,. . ., x„\ A R t(R) A F and verify that 0 | K = \ K and < t> | P = 0. Consequently, for any f(x u . . ., x„) e P
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0(/(jfi,. . ., x„)) = 0, whence (
Proposition 7.4. (Hilbert Nullstellensatz) Let F be an algebraically closed extension field of a field K and I a proper ideal o/K[xi, . . . , xn]. Let V(I) = { ( a i , . . . , an) e Fn | g(ai, . .. , a„) = 0 for all g e l ) . Then Rad I = J(V(I))
= { f s K [ x i , . . . , xn] | f(ai,. . . , a„) = 0 for all (ai,. . . , a„) e V(I)j. In other words, f(ai, . . . , an) = 0 for every zero (ai, . . . , a„) of I in Fn if and only if f1" e T for some m > 1.
REMARK. We shall use Lemma 7.3 to prove the theorem. Since the theorem im plies the lemma (Exercise 6), the two are actually equivalent. PROOF OF 7.4. If /e Rad I, then fm e / for some m > 1 (Theorem 2.6). If (ax , . . . , an) is a zero of I in Fn, then 0 = f m (ai ,. . . , a r ) = (f(a u . . . , ai))m. Con sequently, since F is a field, f(au . . . , a») = 0. Therefore, Rad I CZ J V(I). Conversely, suppose /e JV(I). We may assume /V 0 since 0 e Rad /. Consider K[x i, . . . , a-„] as a subring of the ring K[x u . . . , xn,y] of polynomials in n + 1 in determinates over K. Let L be the nonzero ideal of K[xi, . . . , x v ,y] generated by I and yf— 1 f - Clearly if (ai, . . . , an,b) is a zero of L in Fn+1 then (a u . . . , a„ ) must be a zero of /in F n . But (yf — 1 f)(o\ , . - • , an,b) = bf(au . .., a„) — If = —If for all zeros (au .. . , a„) of / in Fn. Therefore, L has no zeros in Fn+l\ that is, V(L) is empty. Consequently, L = K[x u - - -, x n ,y] by Lemma 7.3, whence 1F eL. Thus «-1 If = 2Z Ri f + 8 t(yf- I f ), t=i where f e I (1 < / < t — 1) and g i e K[x u ■ . ■ , JCr.,3’]- Define an evaluation homomorphism AT[jci, . . . , xn,_y] —>■ K (x u . . . , x n ) by Xi (—» x, and y\ -> f 1 = 1a//(x!, . . . , x n ) (Corollary III.5.6). Then in the field K (x u . . f , xn) t-i
1/*' ^ 1 gi(-^ 11 • * * , X n , f ) fi(xi, . . . , A'ti). i=1 Let m be a positive integer larger than the degree of gi in y for every / (1 < i < t — 1). Then for each /, fm(xu ■ ■., xn)gl(xu . .., x „,/-1) lies in K [x\ ,.. . , xn], whence t-1 m fm = f If = X! f m (xi, . . . , xm)gi(x 1, . . . , A„, f~1)fi(xu . . . , X n ) e I. Therefore i=i /e Rad / and hence JV(I) CZ Rad /. ■ The determination of closed objects as mentioned in the introduction of this section is now straightforward (Exercises 1-3). We close this section with an informal attempt to establish the connection be tween geometry and algebra which characterizes the classical approach to algebraic geometry. Let A" be a field. Every polynomial f e K[x u . . . , x„] determines a function F n —* F by substitution: (a u . .. ,a n ) (—»f (a u - ■ ■ , a n )- l fV= V(I ) is an affine variety contained in F n , the restriction of this function to Fis called a regular function on V. The regular functions V —* F form a ring Y(V) which is isomorphic to
7. THE HILBERT NULLSTELLENSATZ
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K[.xu . . . , x n ]/J(V(I))
(Exercise 10). This ring is called the coordinate ring of V. Since I Cl J(V(I)) — Rad / the ring r(K) has no nonzero nilpotent elements. Furthermore r(K) is a finitely generated algebra over K (since K[x i,. . ., x„] and the ideal J( V(I)) are; see Section IV.7). Conversely it can be proved that every finitely generated AT-algebra with no nonzero nilpotent elements is the coordinate ring of some affine variety. Therefore, there is a one-to-one correspondence between affine varieties and a rather special class of commutative rings. With a suitable definition of morphisms the affine varieties form a category as do the commutative rings in question and this corre spondence is actually an “equivalence” of categories. Thus statements about affine varieties are equivalent to certain statements of commutative algebra. For further information see W. Fulton [53] and I. G. MacDonald [55].
EXERCISES Note: F is always an algebraically closed extension field of a field K;J,V, and F n
are as above. 1. A subset Y of F n is closed (that is, V(J(Y)) — Y) if and only if Y is an affine /^-variety determined by some subset S' of . K [ x i , . . . , x„]. 2. A subset S of A [ x , , . . . , jcn] is closed (that is, y(F(S)) = S) if and only if S is a radical ideal (that is, S is an ideal and S = Rad S). 3. There is a one-to-one inclusion reversing correspondence between the set of affine A-varieties in F n and the set of radical ideals of K[x u ..., x„]. [See Exer cises 1, 2.) 4. Every affine A-variety in F n is of the form V(S) where S is a finite subset of A[xi,. . . , a-J. [Hint: Theorems 1.9 and 4.9 and Exercise 3.] 5. If Vi ID Vi ID - • ■ is a descending chain of A-varieties in Fn, then Vm = Vn+1 = ■ • ■ for some m. [Hint: Theorem 4.9 and Exercise 3.] 6. Show that the Nullstellensatz implies Lemma 7.3. 7. If 7 i , . . . , lk are ideals of K[xi, . . . , x„], then V(h fl h D ■ • • D I k ) = V(h) U V(h) U ■ ■ • U V(lk) and V( hh • ■/*) = V(I X ) fl V(I 2 ) D — fl V(I k ). 8. A AT-variety V in F n is irreducible provided that whenever V = W\ U W 2 with each Wi a /f-variety in F n , either V — W\ or V = W 2 . (a) Prove that V is irreducible if and only if J(V) is a prime ideal in /f[xi, . . . , x„], (b) Let F — C and S = { x i 2 — 2x 2 2 }. Then F(S) is irreducible as a Q-variety but not as an R-variety. 9. Every nonempty A'-variety in F n may be written uniquely as a finite union Vi U Vi U • ■ • U V k of affine A-varieties in F n such that V, $L V, for i ^ j and each Vi is irreducible (Exercise 8). 10. The coordinate ring of an affine/^-variety V(I) is isomorphic toAfxi,... ,x n ]/J(V(I))-
CHAPTER
IX
THE STRUCTURE OF RINGS
In the first part of this chapter a general structure theory for rings is presented. Al though the concepts and techniques introduced have widespread application, com plete structure theorems are available only for certain classes of rings. The basic method for determining such a class of rings might be described intuitively as follows. One singles out an “undesirable” property P that satisfies certain conditions, in particular, that every ring has an ideal which is maximal with respect to having property P. This ideal is called the P-radical of the ring. One then attempts to find structure theorems for the class of rings with zero P-radical. Frequently one must in clude additional hypotheses (such as appropriate chain conditions) in order to obtain really strong structure theorems. These ideas are discussed in full detail in the intro ductions to Sections 1 and 2 below. The reader would do well to read both these dis cussions before beginning serious study of the chapter. We shall investigate two different radicals, the Jacobson radical (Section 2) and the prime radical (Section 4). Very deep and useful structure theorems are obtained for left Artinian semisimple rings (that is, left Artinian rings with zero Jacobson radical) in Section 3. Goldie’s Theorem is discussed in Section 4. It includes a char acterization of left Noetherian semiprime rings (that is, left Noetherian rings with zero prime radical). The basic building blocks for all of these structure theorems are the endomorphism rings of vector spaces over division rings and certain “dense” sub rings of such rings (Section 1). The last two sections of the chapter deal with algebras over a commutative ring with identity. The Jacobson radical and related concepts and results are carried over to algebras (Section 5). Division algebras are studied in Section 6. A theme that occurs continually in this chapter is the close interconnection be tween the structure of a ring and the structure of modules over the ring. The use of modules in the study of rings has resulted in a host of new insights and deep theo rems. 414
1. SIMPLE AND PRIMITIVE RINGS
415
The interdependence of the sections of this chapter is as follows: 1
* 2
i /\ Much of the discussion here depends on the results of Section VIII.l (Chain con ditions).
1. SIMPLE AND PRIMITIVE RINGS In this section we study those rings that will be used as the basic building blocks in the structure theory of rings. We begin by recalling several facts that motivate a large part of this chapter. (i) If V is a vector space over a division ring D, then Homd (V,V) is a ring (Exer cise IV. 1.7), called the endomorphism ring of V. (ii) The endomorphism ring of a finite dimensional vector space over a division ring is isomorphic to the ring of all n X n matrices over a (possibly different) division ring (Theorem VII. 1.4). (iii) If D is a division ring, then Matn D is simple (that is, has no proper ideals; Exercise III.2.9) and is both left and right Artinian (Corollary VIII.l.12). Conse quently by (ii) every endomorphism ring of a finite dimensional vector space over a division ring is both simple and Artinian. (iv) The endomorphism ring of an infinite dimensional vector space over a divi sion ring is neither simple nor Artinian (Exercise 3). However, such a ring is primi tive, in a sense to be defined below. Matrix rings and endomorphism rings of vector spaces over division rings arise naturally in many different contexts. They are extremely useful mathematical con cepts. Consequently it seems reasonable to take such rings, or at least rings that closely resemble them, as the basis of a structure theory and to attempt to describe arbitrary rings in terms of these basic rings. With the advantage of hindsight we single out two fundamental properties of the endomorphism ring of a vector space V: simplicity (Definition 1.1) and primitivity (Definition 1.5). As noted above these two concepts roughly correspond to the cases when V is finite or infinite dimensional respectively. In this section we shall analyze simple and primitive rings and show that in several important cases they coincide with endomorphism rings. In other cases they come as close to being endomorphism rings as is reasonably possible. More precisely, an arbitrary primitive ring R is shown to be isomorphic to a par ticular kind of subring (called a dense subring) of the endomorphism ring of a vector space V over a division ring D (Theorem 1.12). R is left Artinian if and only if dimpV
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CHAPTER IX THE STRUCTURE OF RINGS
is finite (Theorem 1.9). In this classical case, simple and primitive rings coincide and R is actually isomorphic to the complete endomorphism ring of V (Theorem 1.14). Furthermore in this situation dimnV is uniquely determined and V is determined up to isomorphism (Proposition 1.17). These results amply justify the designation of simplicity and primitivity as fundamental concepts. As noted in the introduction to this chapter modules play a crucial role in ring theory. Consequently we begin by defining and developing the elementary properties of simplicity for both rings and modules.
Definition 1.1. A (left) module A over a ring R is simple (or irreducible) provided RA / 0 and A has no proper submodules. A ring R is simple if R2 ^ 0 and R has no proper (two-sided) ideals.
REMARKS, (i) Every simple module [ring] is nonzero. (ii) Every simple module over a ring with identity is unitary (Exercise IV.1.17). A unitary module A over a ring R with identity has RA 0, whence A is simple if and only if A has no proper submodules. (iii) Every simple module A is cyclic; in fact, A = Ra for every nonzero a e A. [Proof: both Ra (a e A) and B = {c e A \ Rc = 0} are submodules of A, whence each is either 0 or A by simplicity. But RA 0 implies B ^ A. Consequently B = 0, whence Ra = A for all nonzero a e A.] However a cyclic module need not be simple (for example, the cyclic Z-module Z6). (iv) The definitions of “simple” for groups, modules, and rings can be subsumed into one general definition, which might be roughly stated as: an algebraic object C that is nontrivial in some reasonable sense (for example, RA ^ 0 or R2 ^ 0) is simple, provided that every homomorphism with domain C has kernel 0 or C. The point here is that the absence of nontrivial kernels is equivalent to the absence of proper normal subgroups of a group or proper submodules of a module or proper ideals of a ring as the case may be. EXAMPLE. Every division ring is a simple ring and a simple Z)-module (see the Remarks preceding Theorem III.2.2). EXAMPLE. Let D be a division ring and let R = Mat„D (n > 1). For each k (1 < k < n), Ik = ! (an) e R \ ai, = 0 for j ^ k\ is a simple left /?-module (see the proof of Corollary VIII. 1.12). EXAMPLE. The preceding example shows that Mat„Z) (D a division ring) is not a simple left module over itself if n > 1. However, the ring Mat„£> (n > 1) is simple by Exercise III.2.9. Thus by Theorem VII. 1.4 the endomorphism ring of any finite dimensional vector space over a division ring is a simple ring. EXAMPLE. A left ideal / of a ring R is said to be a minimal left ideal if / ^ 0 and for every left ideal J such that 0 CZ J CZ /, either J = 0 or J = /. A left ideal / of R such that RI ^ 0 is a simple left /?-module if and only if / is a minimal left ideal. EXAMPLE. Let F be a field of characteristic zero and R the additive group of polynomials /'’[*■,>■]. Define multiplication in R by requiring that multiplication be
1. SIMPLE AND PRIMITIVE RINGS
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distributive and that xy = yx + 1 and ax = xa, ay = ya for ae F. Then R is a welldefined simple ring that has no zero divisors and is not a division ring (Exercise 1). Let A = Ra be a cyclic R-module. The map 6 : R —> A defined by rj—> ra is an R-module epimorphism whose kernel / is a left ideal (submodule) of R (Theorem IV.1.5). By the First Isomorphism Theorem IV.1.7 R/I is isomorphic to A. By Theorem IV.1.10 every submodule of R/I is of the form J/I, where / is a left ideal of R that contains I. Consequently R/1 (and hence A) has no proper submodules if and only if / is a maximal left ideal of R. Since every simple R-module is cyclic by Re mark (iii) above, every simple R-module is isomorphic to R/I for some maximal left ideal I. Conversely, if / is a maximal left ideal of R, R/I will be simple provided R( R/I) 9^ 0. A condition that guarantees that R(R//) ^ 0 is given by
Definition 1.2. A left ideal I in a ring R is regular (or modular) if there exists e e R such that r — re e I for every r e R. Similarly, a right ideal J is regular if there exists e e R such that r — er e J for every r e R. REMARK. Every left ideal in a ring R with identity is regular (let e = lfl).
Theorem 1.3. A left module A over a ring R is simple if and only if A is isomorphic to R/I for some regular maximal left ideal I. REMARKS. If R has an identity, the theorem is an immediate consequence of the discussion above. The theorem is true if “left” is replaced by “right” throughout. PROOF OF 1.3. The discussion preceding Definition 1.2 shows that if A is simple, then A = Ra = R/I where the maximal left ideal I is the kernel of 6. Since A = Ra, a = ea for some e e R. Consequently, for any r e R, ra = rea or (r — re)a = 0, whence r — re e Ker 6 = 1. Therefore I is regular. Conversely let / be a regular maximal left ideal of R such that A ^ R//.In view of the discussion preceding Definition 1.2 it suffices to prove that R(R/l ) 9^ 0. If this is not the case, then for all r e R r(e + /) = /, whence re s I. Since r — re e I, we have r e I . Thus R = /, contradicting the maximality of I. ■ Having developed the necessary facts about simplicity we now turn to primitivity. In order to define primitive rings we need:
Theorem 1.4. Let B be a subset of a lefi module A over a ring R. Then Ct(B) = {r e R | rb = 0 for all b e B | is a left ideal of R. If B is a submodule of A, then 0t(B) is an ideal. d(B) is called the (left) annihilator of B. The right annihilator of a right module is defined analogously.
SKETCH OF PROOF OF 1.4. It is easy to verify that d(B) is a left ideal. Let B be a submodule. If r e R and s z <X(B), then for every beB (sr)b = s(rb) = 0 since rb £ B. Consequently, sr e Ci(B) , whence d(B) is also a right ideal. ■
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Definition 1.5. A (left) module A is faithful if its (left) annihilator (5(A) is 0. A ring R is (left) primitive if there exists a simple faithful left K-module. Right primitive rings are defined analogously. There do exist right primitive rings that are not left primitive (see G. Bergman [58]). Hereafter “primitive” will always mean “left primitive.” However, all results proved for left primitive rings are true, mutatis mutandis, for right primitive rings. EXAMPLE. Let V be a (possibly infinite dimensional) vector space over a divi sion ring D and let K be the endomorphism ring Hornd (V,V) of V. Recall that V is a left K-module with 8v = 8(v) for v s V, 8e R (Exercise IV. 1.7). If u is a nonzero vector in V, then there is a basis of V that contains u (Theorem IV.2.4). If v e V, then there exists 8ve R such that 8vu = v (just define 6v(u) — v and 6v(w) = 0 for all other basis elements w; then 6 V e R by Theorems IV.2.1 and IV.2.4). Therefore Ru = V for any nonzero u e V , whence V has no proper K-submodules. Since R has an identity, RV 0. Thus V is a simple K-module. If 6V = 0 (8 e K), then clearly 0 = 0, whence GL(V) = 0 and Kis a faithful K-module. Therefore, R is primitive. If Vis finite dimen sional over D, then R is simple by Exercise III.2.9 and Theorem VII. 1.4. But if V is infinite dimensional over D , then R is not simple: the set of all 8 e R such that Im 8 is finite dimensional subspace of V is a proper ideal of R (Exercise 3). The next two results provide other examples of primitive rings.
Proposition 1.6. A simple ring R with identity is primitive. PROOF. R contains a maximal left ideal I by Theorem III.2.18. Since R has an identity / is regular, whence R/I is a simple K-module by Theorem 1.3. Since d(R/I ) is an ideal of R that does not contain 1/?, d(R/I) = 0 by simplicity. Therefore R/I is faithful. ■
Proposition 1.7. A commutative ring R is primitive if and only if R is a field. PROOF. A field is primitive by Proposition 1.6. Conversely, let A be a faithful simple left K-module. Then A ~ R/I for some regular maximal left ideal I of K. Since K is commutative, I is in fact an ideal and I C d(R/I) = Q(A) = 0. Since / = 0 is regular, there is an e e R such that r = re (= er) for all r e K. Thus K is a commutative ring with identity. Since / = 0 is maximal, K is a field by Corollary III.2.21. ■ In order to characterize noncommutative primitive rings we need the concept of density.
Definition 1.8. Let V be a (left) vector space over a division ring D. A subring R of the endomorphism ring Homj) (V,V) is called a dense ring of endomorphisms ofV (or a dense subring of Homn(V,V)) if for every positive integer n, every linearly independent subset { U i , . . . , u,,} o/V and every arbitrary subset {vu . . . , vn} of V, there exists 8 e R such that 0(uO = vs (i = 1,2, , n).
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EXAMPLE. Won\ D (V,V) is a dense subring of itself. For if { « i , . . . , u n ] is a linearly independent subset of V, then there is.a basis U of V that contains u x ,... , u n by Theorem IV.2.4. If v u ... ,v m e V , then the map 8 : V —> V defined by 8(u t ) = i >, and 8(u) = 0 for ueU — is a well-defined element of Hornd(F,F) by Theorems IV.2.1 and IV.2.4. In the finite dimensional case, Horn.d (V,V) is the only dense subring as we see in
Theorem 1.9. Let R be a dense ring of endomorphisms of a vector space V over a division ring D. Then R is left [resp. right] Artinian if and only if dim^St is finite, in which case R = /tomD(V,V). PROOF. If R is left Artinian and A\mr>V is infinite, then there exists an infinite linearly independent subset \uuu-i, . . .} of V. By Exercise IV. 1.7 V is a left Hom0(f/,f/)-module and hence a left /?-module. For each n let /„ be the left an nihilator in R of the set {uu . . ., u7 l } . By Theorem 1.4,IX 3 /2 3 - • • is a descending chain of left ideals of R. Let w be any nonzero element of V. Since {uu . . ., un+,} is linearly independent for each n and R is dense, there exists 8 e R such that
6ui = 0 for i = 1,2,. . . , n and 6un+i = w ^ 0. Consequently 6 eln but 6 ^ In+l. Therefore A Z) /2 3 - - - is a properly descending chain, which is a contradiction. Hence dim^F is finite. Conversely if &\mDV is finite, then V has a finite basis { i ? i , . . . , vm } . If /is any element of Hornd(F,F), then /is completely determined by its action on a,.. ., vm by Theorems IV.2.1 and IV.2.4. Since R is dense, there exists 8 e R such that
6(Vi) = f(v i) for i = 1,2,... ,m, whence / = 8 e R. Therefore Hornd(F,F) = R. But Horni>(V,V) is Artinian by Theorem VII. 1.4 and Corollary VIII.1.12. ■ In order to prove that an arbitrary primitive ring is isomorphic to a dense ring of endomorphisms of a suitable vector space we need two lemmas.
Lemma 1.10. (Schur) Let A. be a simple module over a ring R and let B be any K-module.
(i) Every nonzero K-module homomorphism f : A —> B is a monomorphism; (ii) every nonzero K-module homomorphism g : B —» A is an epimorphism; (iii) the endomorphism ring D = //owr(A,A) is a division ring. PROOF, (i) Ker /is a submodule of A and Ker /^ A since /^ 0. Therefore Ker / = 0 by simplicity, (ii) Im g is a nonzero submodule of A since g 5* 0, whence Im g = A by simplicity, (iii) If he D and h ^ 0, then h is an isomorphism by (i) and (ii). Thus / has a two-sided inverse /-1 e V\om R (A,A) = D (see the paragraph after Definition IV.1.2). Consequently every nonzero element of D is a unit, whence D is a division ring. ■
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REMARK. If A is a simple R-module, then A is a vector space over the division ring Homr ( A , A ) with fa = f(a) (Exercise IV.1.7 and Lemma 1.10).
Lemma 1.11. Let A be a simple module over a ring R. Consider A as a vector space over the division ring D = /7<7Wr(A,A). IfV is a finite dimensional D-subspace of the T>-vector space A and a e A — V, then there exists r e R such that ra j* 0 and rV = 0. PROOF. The proof is by induction on n = dim^K If n — 0, then V = 0 and a ^ 0. Since A is simple, A = Ra by Remark (iii) after Definition 1.1. Consequently, there exists r e R such that ra = a ^ 0 and rV = r0 = 0. Suppose dimDV = n > 0 and the theorem is true for dimensions less than n. Let {uu . . ., be a D-basis of V and let W be the (n — l)-dimensional £>-subspace spanned by \uu . . . , i| (IV = 0 if n = 1). Then V = IV@ Du (vector space direct sum). Now W may not be an R-submodule of A, but in any case the left annihilator I = <2( W) in R of Wis a left ideal of R by Theorem 1.4. Consequently, lu is an R-submodule of A (Exercise IV.1.3). Since u z A — W, the induction hypothesis implies that there exists r e R such that ru 0 and rW = 0 (that is, r e / = CL(l¥)). Consequently 0 ^ rue lu. whence lu ^ 0. Therefore A = lu by simplicity. [Note: The contrapositive of the inductive argument used above shows that if v e A and rv = 0 for all r e / , then v e W.] We must find r e R such that ra j* 0 and rV = 0. If no such r exists, then we can define a map 6 : A —* A as follows. For ru e lu — A let 6(ru) = ra s A. We claim that 6 is well defined. If ryu = r2u (r, e I = CL(W)), then (rt — r2)« = 0, whence (ri — r2)V = (n — r 2 )(tV 0 Du) = 0. Consequently by hypothesis (r, — r2)a = 0. Therefore, d(riu) = na = r2a = d(r2u). Verify that 6 e Homr (A,A) = D. Then for every r e /, 0 = 6(ru) — ra = r8(u) — ra = r(6(u) — a). Therefore 6(u) — a e W by the parenthetical Note above. Consequently a = 8u — (8u — a) £ Du + W = V,
which contradicts the fact that a\V. Therefore, there exists r £ R such that ra ^ 0 and rV = 0. ■
Theorem 1.12. (Jacobson Density Theorem) Let R be a primitive ring and A a faithful simple R-module. Consider A as a vector space over the division ring Hom^(A,A) = D. Then R is isomorphic to a dense ring of endomorphisms of the T)-vector space A. REMARK. A converse of Theorem 1.12 is also true, in fact in a much stronger form (Exercise 4). PROOF OF 1.12. For each r e R the map ar : A —> A given by aT(a) = ra is easily seen to be a D-endomorphism of A: that is, ar £ HomD (A,A). Furthermore for all r,s e R Q!(r+S) = aT + a* and Q!rs = atras. Consequently the map « : / ? — > HomD (A,A) defined by a(r) = a T is a well-defined homomorphism of rings. Since A is a faithful R-module, ar = 0 if and only if
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r e Cl(y4) = 0. Therefore a is a monomorphism, whence R is isomorphic to the sub ring Im a of Hornj>(A,A). To complete the proof we must show that Im a is a dense subring of HornD (A,A). Given a D-linearly independent subset U — {u u . . ., u n J of A and an arbitrary sub set { f i , . . . , v n } of A we must find a , e l m a such that ar(«,) = v t for i = 1,2 For each i let V, be the D-subspace of A spanned by \ u , . . . , un\. y Since U is D-linearly independent, ux \ Vi. Consequently, by Lemma 1.11 there exists r, e R such that ^ 0 and nVi = 0. We next apply Lemma 1.11 to the zero sub space and the nonzero element nm: there exists s{ e R such that s^rau 5* 0 and .v.O = 0. Since strtUi 5^ 0, the K-submodule Rr2iti of A is nonzero, whence Rr%Ui = A by simplicity. Therefore exists ti e R such that = i;,. Let r = t\T 1 + ttf2 + ■ • ■ + tnrn e R.
Recall that for / ^ j, Ui e Vj, whence e ^(rjVj) = tjO = 0. Consequently for each i = 1,2 ar(Ui) = (hn H------- (- tT,rn)Ui = t,rtm = Vi.
Therefore Im a is a dense ring of endomorphisms of the D-vector space A. ■ REMARK. The only point in the proof of Theorem 1.12 at which the faithfulness of A is used is to show that a is a monomorphism. Consequently the proof shows that any ring that has a simple module A also has a homomorphic image that is a dense ring of endomorphisms of the D-vector space A.
Corollary 1.13. If R is a primitive ring, then for some division ring D either R is isomorphic to the endomorphism ring of a finite dimensioned vector space over D or for every positive integer m there is a subring Rm of R and an epimorphism of rings R„, —> //omr)(Vr,„V,„), where Vm is an m-dimensional vector space over D. REMARK. The Corollary may also be phrased in terms of matrix rings over a division ring via Theorem VII. 1.4. SKETCH OF PROOF OF 1.13. In the notation of Theorem 1.12, a : R —> Hom D (A,A)
is a monomorphism such that R = Im a and Im a is dense in Y\om D (A,A). If dim»A = n is finite, then Im a = Won\ D (A,A) by Theorem 1.9. If dimD A is infinite and \u u u 2,. . .) is an infinite linearly independent set, let V m be the m-dimensional D-subspace of A spanned by J u u . . . . u m \. Verify that R m = {r e R | rV m Cl V m ) is a subring of R. Use the density of R = Im a in Hom D (A,A) to show that the map Rm —> Hom»(V„„Vn) given by r h a r | Vm is a well-defined ring epimorphism. ■
Theorem 1.14. ( Wedderburn-Artin) The following conditions on a left Artinian ring R are equivalent.
(i) R is simple; (ii) R is primitive;
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(iii) R is isomorphic to the endomorphism ring of a nonzero finite dimensional vector space V over a division ring D; (iv) for some positive integer n, R is isomorphic to the ring of all n X n matrices over a division ring.
PROOF, (i) => (ii) We first observe that / = {r e R j Rr = 0} is an ideal of R, whence / = R or / = 0. Since R2 ^ 0, we must have / = 0. Since R is left Artinian the set of all nonzero left ideals of R contains a minimal left ideal J. J has no proper i?-submodules, (an i?-submodule of / is a left ideal of R). We claim that the left annihilator GL(J) of / in R is zero. Otherwise (i(J) = R by simplicity and Ru = 0 for every nonzero u e J. Consequently, each such nonzero u is contained in / = 0, which is a contradiction. Therefore Ct(J) = 0 and RJ ^ 0. Thus J is a faithful simple /^-module, whence R is primitive. (ii) => (iii) By Theorem 1.12 R is isomorphic to a dense ring T of endomorphisms of a vector space V over a division ring D. Since R is left Artinian, R = T~ Hom D (V,V) by Theorem 1.9. (iii) <=> (iv) Theorem VTI. 1.4. (iv) => (i) Exercise III.2.9. ■ We close this section by proving that for a simple left Artinian ring R the integers dimDV and n in Theorem 1.14 are uniquely determined and the division rings in Theorem 1.14 (iii) and (iv) are determined up to isomorphism. We need two lemmas.
Lemma 1.15. Let V be a finite dimensional vector space over a division ring D. 7/A and B are simple faithful modules over the endomorphism ring R = //owd(V,V), then A and B are isomorphic K-modules. PROOF. By Theorems VII.1.4, VIII.1.4 and Corollary VIII.1.12, the ring R contains a (nonzero) minimal left ideal I. Since A is faithful, there exists a e A such that la 0. Thus la is a nonzero submodule of A (Exercise IV. 1.3), whence la = A by simplicity. The map 6 : / —> la = A given by i |—> ia is a nonzero /^-module epi morphism. By Lemma 1.10 6 is an isomorphism. Similarly / = B. ■
Lemma 1.16. Let V be a nonzero vector space over a division ring D and let R be the endomorphism ring Homn(V,V). If g : V —> V is a homomorphism of additive groups such that gr = rg for all r e R, then there exists d e D such that g(v) = dv for all v e V . PROOF. Let u be a nonzero element of V. We claim that u and g (u) are linearly dependent over D. If dimDV = 1, this is trivial. Suppose dim^I7 > 2 and {«,#(«)} is linearly independent. Since R is dense in itself (Example after Definition 1.8), there exists reR such that r(w) = 0 and r(g(u )) ^ 0. But by hypothesis r(g(u)) = rg(u) = gr(u) = g(r(u )) = g( 0) = 0,
which is a contradiction. Therefore for some de D, g (u) = du . If v e V, then there exists s e R such that s(u) = v by density. Consequently, since s e R = Hom D (V,V), g( v) = g (s(u)) = gs(u) = sg(u) = s(du) = ds(u) = dv. ■
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Proposition 1.17. For i = 1,2 let Vi be a vector space of finite dimension rij over the
division ring Di.
(i) If there is an isomorphism of rings //owdi(Vi,Vi) ^ f/cw;D2(V2,V2), then d/mDjVi = J/WD2V2 and Di is isomorphic to D2. (ii) If there is an isomorphism of rings Matni Di = Matn2 D2, then na = n2 andT>i is isomorphic to D2. SKETCH OF PROOF, (i) For i — 1,2 the example after Definition 1.5 shows that Vi is a faithful simple Hom^^F^-module. Let R = HomDl(I/i,l''1) and let a : R—> HomD2(V2,V2)
be an isomorphism. Then V2 is a faithful simple /?-module by pullback along a (that is, rv = cr(r)v for reR,ve V 2 ). By Lemma 1.15 there is an /?-module isomorphism 0 : Vi —* V 2 . For each veVi and f £ R, 0[/(t0] = f
1
= o-( /)
as a homomorphism of additive groups V 2 —► V 2 . For each d e A let a d : V { —> V { be the homomorphism of additive groups defined by x |—» d x. Clearly a d = 0 if and only if d = 0. For every fe R = Hom^Fi.Ki) and every d e D u fa d = ad f Consequently, [4>ad<prl]{af) =
=
Verify that r is a monomorphism of rings. Reversing the roles of A and D2 in the preceding argument (and replacing
Use this fact to show that {uu . . ., uk\ is A-linearly independent in Vx if and only if |0(«i),..., <})(uk) | is A-linearly independent in V2. It follows that dim^F = <\imD2V2. (ii) Use (i), Exercise 111.1.17(e) and Theorem VII.1.4. ■
EXERCISES 1. Let F be a field of characteristic 0 and R = the additive group of poly nomials in two indeterminates. Define multiplication in R by requiring that multiplication be distributive, that ax = xa, ay = ya for all a e F, that the product of x and y (in that order) be the polynomial xy as usual, but that the product of y and x be the polynomial xy + 1.
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(a) R is a ring. (b) yxk = xky + kxk 1 and ykx = xyk + kyk_1. (c) R is simple. (Hint: Let /be a nonzero element in an ideal / of R; then either /has no terms involving y or g = xf — fx is a nonzero element of / that has lower degree in y than does / In the latter case, consider xg — gx. Eventually, find a nonzero h e /, which is free of y. If h is nonconstant, consider hy — yh. In a finite number of steps, obtain a nonzero constant element of /; hence I =R.) (d) R has no zero divisors. (e) R is not a division ring. 2. (a) If A is an .R-module, then A is also a well-defined /?/(2(/i)-module with (r + 0L(A )) a = ra (a e A). (b) If ^4 is a simple left /^-module, then R/d(A ) is a primitive ring. 3. Let V be an infinite dimensional vector space over a division ring D. (a) If Fis the set of all 8 e Horn d(V,V) such that Im 8 is finite dimensional, then F is a proper ideal of Homc(K,K). Therefore Hom D (V,V) is not simple. (b) F is itself a simple ring. (c) F is contained in every nonzero ideal of Hom D (V,V). (d) HornD (V,V) is not (left) Artinian. 4. Let I^be a vector space over a division ring D. A subring R of Hornd(V,V) is said to be n-fold transitive if for every k (1 < k < n) and every linearly independent subset { « i , . . . , uk\ of Vand every arbitrary subset {vx , . . . , vk\ of V, there exists 8 £ R such that 0(w,) = t f o r /' = 1,2,. . . , k. (a) If R is one-fold transitive, then R is primitive. [Hin t: examine the example after Definition 1.5.] (b) If R is two-fold transitive, then R is dense in Hom D (V,V). [Hints: Use (a) to show that R is a dense subring of HomA(f/,f/), where A = \\omn(y,V). Use two fold transitivity to show that A = \/3 d \ d£ D], where f3 d : V —> V is given by x |—> dx . Consequently Horna (V, V) = Homy)(I/,J/)-] 5. If R is a primitive ring such that for all a,b e R, a{a b — ba) = (ab — ba)a, then R is a division ring. [Hint: show that R is isomorphic to a dense ring of endomor phisms of a vector space V over a division ring D with dimD V = 1, whence R = D.] 6. If R is a primitive ring with identity and e e R is such that e2 — e ^ 0, then (a) eRe is a subring of R, with identity e. (b) eRe is primitive. [Hint: if R is isomorphic to a dense ring of endomorphisms of the vector space V over a division ring D, then Ve is a D-vector space and eRe is isomorphic to a dense ring of endomorphisms of Ve.] 7. If R is a dense ring of endomorphisms of a vector space V and A" is a nonzero ideal of R , then K is also a dense ring of endomorphisms of V.
2. THE JACOBSON RADICAL The Jacobson radical is defined (Theorem 2.3) and its basic properties are de veloped (Theorems 2.12-2.16). The interrelationships of simple, primitive, and semi simple rings are examined (Theorem 2.10) and numerous examples are given.
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Before pursuing further our study of the structure of rings, we summarize the general technique that we shall use. There is little hope at present of classifying all rings up to isomorphism. Consequently we shall attempt to discover classes of rings for which some reasonable structure theorems are obtainable. Here is a classic method of determining such a class. Single out some “bad” or “undesirable” property of rings and study only those rings that do not have this property. In order to make this method workable in practice one must make some additional as sumptions. Let P be a property of rings and call an ideal [ring] I a P-ideal [P-ring] if / has property P. Assume that (i) the homomorphic image of a P-ring is a P-ring; (ii) every ring R (or at least every ring in some specified class C) contains a P-ideal P(R) (called the P-radical of R) that contains all other P-ideals of R; (iii) the P-radical of the quotient ring R/P(R) is zero; (iv) the P-radical of the ring P(P) is P(P). A property P that satisfies (i)—(iv) is called a radical property. The P-radical may be thought of as measuring the degree to which a given ring possesses the “undesirable” property P. If we have chosen a radical property P, we then attempt to find structure theorems for those “nice” rings whose P-radical is zero. Such a ring is said to be P-radical free orP-semisimple. In actual practice we are usually more concerned with the P-radical itself rather than the radical property P from which it arises. By condition (iii) every ring that has a P-radical has a P-semisimple quotient ring. Thus the larger P-radical is, the more one discards (or factors out) when studying P-semisimple rings. The basic problem is to find radicals that en able us to discard as little as possible and yet to obtain reasonably deep structure theorems. Wedderburn first introduced a radical in the study of finite dimensional algebras. His results were later extended to (left) Artinian rings. However, the radical of Wedderburn (namely the maximal nilpotent ideal) and the remarkably strong struc ture theorems that resulted applied only to (left) Artinian rings. In subsequent years many other radicals were introduced. Generally speaking each of these coincided with the radical of Wedderburn in the left Artinian case, but were also defined for non-Artinian rings. The chief purpose of this section is to study one such radical, the Jacobson radical. Another radical, the prime radical, is discussed in Section 4; see also Ex ercise 4.11. For an extensive treatment of radicals see N. J. Divinsky [22] or M. Gray [23], The host of striking theorems that have resulted from its use provide ample justification for studying the Jacobson radical in some detail. Indeed Section 1 was developed with the Jacobson radical in mind. Rings that are Jacobson semisimple (that is, have zero Jacobson radical) can be described in terms of simple and primi tive rings (Section 3). Two preliminaries are needed before we define the Jacobson radical.
Definition 2.1. An ideal P of a ring R is said to be left [resp. right] primitive if the quotient ring R/P is a left [resp. right] primitive ring. REMARK. Since the zero ring has no simple modules and hence is not primitive, R itself is not a left (or right) primitive ideal.
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Definition 2.2. An element a in a ring R is said to be left quasi-regular if there exists r e R such that r + a + ra = 0. The element r is called a left quasi-inverse of a. A (right, left or two-sided) ideal \ of R is said to be left quasi-regular if every element of I is left quasi-regular. Similarly, a e R is said to be right quasi-regular if there exists r e R such that a + r + ar = 0. Right quasi-inverses and right quasi-regular ideals are defined analogously.
REMARKS. It is sometimes convenient to write r ° a for r + a + ra. If R has an identity, then a is left [resp. right] quasi-regular if and only if lfl + a is left [resp. right] invertible (Exercise 1). In order to simplify the statement of several results, we shall adopt the following convention (which is actually a theorem of axiomatic set theory). If the class C of those subsets of a ring R that satisfy a given property is empty, then P) I is defined to be R. Tee
Theorem 2.3. If R is a ring, then there is an ideal J(R) of R such that: (i) J(R) is the intersection of all the left annihilators of simple left R-modules; (ii) J(R) is the intersection of all the regular maximal left ideals of R; (iii) J(R) is the intersection of all the left primitive ideals of R; (iv) J(R) is a left quasi-regular left ideal which contains every left quasi-regular left ideal ofR , (v) Statements (i)-(iv) are also true if “left” is replaced by “right Theorem 2.3 is proved below (p. 428). The ideal J(R) is called the Jacobson radical of the ring R. Historically it was first defined in terms of quasi-regularity (Theorem 2.3 (iv)), which turns out to be a radical property as defined in the intro ductory remarks above (see p. 431). As the importance of the role of modules in the study of rings became clearer the other descriptions of J(R) were developed (Theo rem 2.3 (i)—(iii)). REMARKS. According to Theorem 2.3 (i) and the convention adopted above, J( R) = R if R has no simple left /^-modules (and hence no annihilators of same). If R
has an identity, then every ideal is regular and maximal left ideals always exist (Theorem III.2.18), whence J(R) R by Theorem 2.3(ii). Theorem 2.3(iv) does not imply that J (R ) contains every left quasi-regular element of R; see Exercise 4. The proof of Theorem 2.3 (which begins on p. 428) requires five preliminary lemmas. The lemmas are stated and proved for left ideals. However, each of Lemmas 2.4-2.8 is valid with “left" replaced by “right” throughout. Examples are given after the proof of Theorem 2.3.
Lemma 2.4. 7/1 (5^ R) is a regular left ideal of a ring R, then I is contained in a maximal left ideal which is regular.
SKETCH OF PROOF. Since 7is regular, there exists e e R such that r — res I for all r sR. Thus any left ideal J containing / is also regular (with the same element
2. THE JACOBSON RADICAL
427
e s R). If / Cl 7 and e e J, then r — re e / d J implies r e J for every r e R, whence R = J. Use this fact to verify that Zorn’s Lemma is applicable to the set S of all left ideals L such that I d L d R, partially ordered by inclusion. A maximal element of
S is a regular maximal left ideal containing 7. ■
Lemma 2.5. Let R be a ring and let K be the intersection of all regular maximal left ideals of R. Then K is a left quasi-regular left ideal of R. PROOF. K is obviously a left ideal. If a e K let T = {r + ra | r e R}. If T = R, then there exists r e R such that r -\- ra = —a. Consequently r + a + ra — 0 and hence a is left quasi-regular. Thus it suffices to show that T = R. Verify that Tis a regular left ideal of R (with e = —a).\fTj£R, then T is con tained in a regular maximal left ideal 70 by Lemma 2.4. (Thus T R is impossible if R has no regular maximal left ideals.) Since a e K d I 0 , ra e 70 for all r e R. Thus since r + ra e T d 70, we must have r e 70 for all r e R. Consequently, R = I 0 , which contradicts the maximality of I 0 . Therefore T = R. ■
Lemma 2.6. Let R be a ring that has a simple left R-module. If I is a left quasi regular left ideal of R, then I is contained in the intersection of all the left annihilators of simple left R-modules. PROOF. If 7 fl Q(A), where the intersection is taken over all simple left R-modules A, then IB 9^ 0 for some simple left R-module B, whence lb 0 for some nonzero b e B. Since 7 is a left ideal, lb is a nonzero submodule of B. Con sequently B = lb by simplicity and hence ab = —b for some a e 7. Since I is left quasi-regular, there exists r e R such that r -\- a -\- ra = 0. Therefore, 0 = Ob = (r + a + ra) b = rb ab + rab — rb — b — rb = —b. Since this conclusion contradicts the fact that b 7^ 0, we must have 7 CZ fl (2(/l). ■
Lemma 2.7. An ideal P of a ring R is left primitive if and only ifP is the left annihila tor of a simple left R-module.
PROOF. If P is a left primitive ideal, let A be a simple faithful R/P-module. Verify that A is an R-module, with ra (r e R,a e A) defined to be (r + P)a. Then RA = (R/P)A 9^ 0 and every R-submodule of A is an R/P-submodule of A, whence A is a simple R-module. If r e R, then rA = 0 if and only if (r + P)A = 0. But (r + P)A = 0 if and only if r sP since A is a faithful R/P-module. Therefore L* is the left annihilator of the simple R-module A. Conversely suppose that P is the left annihilator of a simple R-module B. Verify that B is a simple R/R-module with (r + P)b = rb for r e R,b zB. Furthermore if (r + P)B = 0, then rB = 0, whence r e d(B) = P and r + P = 0 in R/P. Conse quently, B is a faithful R/P-module. Therefore R/P is a left primitive ring, whence P is a left primitive ideal of R. ■
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CHAPTER IX THE STRUCTURE OF RINGS
Lemma 2.8. Let I be a left ideal of a ring R. 7/1 is left quasi-regular, then I is right quasi-regular.
PROOF. If 7 is left quasi-regular and a e I, then there exists reR such that r°a = r-\-a-\-r a = 0. Since r = —a — ra e 7, there exists s e R such that o r = 5 -j- r + .vr = 0, whence s is right quasi-regular. The operation ° is easily seen to be associative. Consequently a = 0°a = (s °r)°a = s°(r° a) = s°0 = s.
Therefore a, and hence 7, is right quasi-regular. ■
PROOF OF THEOREM 2.3. Let J(R) be the intersection of all the left annilators of simple left /^-modules. If R has no simple left /^-modules, then J(R) = R by the convention adopted above. J(R) is an ideal by Theorem 1.4. We now show that statements (ii)-(iv) are true for all left ideals. We first observe that R itself cannot be the annihilator of a simple left 7?-module A (otherwise RA = 0). This fact together with Theorem 1.3 and Lemma 2.7 implies that the following conditions are equivalent: (a) (b) (c) (d)
J(R) = R ; R has no simple left /^-modules; R has no regular maximal left ideals; R has no left primitive ideals.
Therefore by the convention adopted above, (ii), (iii), and (iv) are true if J(R) — R. (ii) Assume J(R) 9^ R and let K be the intersection of all the regular maximal left ideals of R. Then K C J(R) by Lemmas 2.5 and 2.6. Conversely suppose c e J(R). By Theorem 1.3, J(R) is the intersection of the left annihilators of the quotients R/ 7, where 7 runs over all regular maximal left ideals of R. For each regular maximal ideal 7 there exists e e R such that c — cee. I. Since c e (x(R/I ), cr e 7 for all r e / ? ; in particular, c e e 7. Consequently, c e 7 for every regular maximal ideal 7. Thus J(R) C D 7 = K. Therefore J(R) = K. (iii) is an immediate consequence of Lemma 2.7. (iv) J (R) is a left quasi-regular left ideal by (ii) and Lemma 2.5. 7(7?) contains every left quasi-regular left ideal by Lemma 2.6. To complete the proof we must show that (i)—(iv) are true with “right” in place of “left.” Let J\(R) be the intersection of the right annihilators of all simple right /^-modules. Then the preceding proof is valid with “right” in place of “left,” whence (i)-(iv) hold for the ideal Ji(R). Since J(R ) is right quasi-regular by (iv) and Lemma 2.8, J (R) C Ji(R) by (iv). Similarly A(R ) is left quasi-regular, whence Ji(R) C J(R). Therefore, J(R) = Ji(R). m EXAMPLE. Let 7? be a local ring with unique maximal ideal M (consisting of all nonunits of R; see Theorem III.4.13). We shall show that J(R) = M. Since R has an identity, J(R) ^ R. Since a proper ideal contains only nonunits by Theorem III.3.2, J (R) d M. On the other hand if re M, then \ R + r $ M (otherwise lfi e M). Conse quently, + ris a unit, whence r is left quasi-regular (Exercise 1). Thus M CI J(R) by Theorem 2.3 (iv). Therefore J(R) = M. Here are two special cases:
2. THE JACOBSON RADICAL
429
EXAMPLE. The power series ring F[[a]] over a field F is a local ring with principal maximal ideal (a) by Corollary III.5.10. Therefore 7(F[[a]]) = (a). EXAMPLE. If p is prime, thenZpn in > 2) is a local ring with principal maximal ideal (/;), which is isomorphic as an abelian group to ZF„-i. Therefore J(Zpn) = ip). The radical of Z m im arbitrary) is considered in Exercise 10.
Definition 2.9. A ring R is said to be iJacobson) semisimple if its Jacobson radical J(R) is zero. R is said to be a radical ring // J(R) = R. REMARK. Throughout this book “radical” always means “Jacobson radical” and “semisimple” always means “Jacobson semisimple.” When reading the literature in ring theory, one must determine which notion of radical and semisimplicity is being used in a particular theorem. A number of definitions of radical (and semi simplicity) require that the ring be (left) Artinian. This is not the case with the Jacob son radical, which is defined for every ring. EXAMPLE. Every division ring is semisimple by Theorem 2.3 (ii) since the only regular maximal left ideal is the zero ideal. EXAMPLE. Every maximal ideal in Z is of the form ip) with p prime by Theo rem III.3.4. Consequently, JiZ) = 0 (p) = 0, whence Z is Jacobson semisimple. v For a generalization, see Exercise 9. EXAMPLE. If D is a division ring, then the polynomial ring R = D[a,,a2, . .., xm]
is semisimple. For if fe JiR), then /is both right and left quasi-regular by Theorem 2.3 (iv). Consequently \ R + /= \ D + /is a unit in R by Exercise 1. Since the only units in R are the nonzero elements of D (see Theorem III.6.1), it follows that fe D. Thus JiR) is an ideal of D, whence J iR ) = 0 or JiR) = D by the simplicity of D. Since — \n is not left quasi-regular (verify!), — \ d ^ JiR)- Therefore J(R) — 0 and R is semisimple.
Theorem 2.10. Let R be a ring. (i) If R is primitive, then R is semisimple. (ii) IfR is simple and semisimple, then R is primitive. (iii) If R is simple, then R is either a primitive semisimple or a radical ring. PROOF, (i) R has a faithful simple left /^-module A, whence JiR) Cl CLiA) = 0. (ii) R 9^ 0 by simplicity. There must exist a simple left R-module A; (otherwise by Theorem 2.3 (i) JiR) = R ^ 0, contradicting semisimplicity). The left annihilator &iA) is an ideal of R by Theorem 1.4 and GC4) ^ R (since RA ^ 0). Consequently (±iA) = 0 by simplicity, whence A is a simple faithful R-module. Therefore R is primitive.
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CHAPTER IX THE STRUCTURE OF RINGS
(iii) If R is simple then the ideal J(R) is either R or zero. In the former case R is a radical ring and in the latter R is semisimple and primitive by (ii). ■ EXAMPLES. The endomorphism ring of a (left) vector space over a division ring is semisimple by Theorem 2.10 (i) and the example after Definition 1.5. Conse quently by Theorem VII.1.4 the ring of all n X n matrices over a division ring is semisimple. EXAMPLE. An example of a simple radical ring is given in E. Sasiada and P. M. Cohn [66]. The classical radical of Wedderburn (in a left Artinian ring) is the maximal nilpotent ideal. We now explore the connection between this radical and the Jacobson radical.
Definition 2.11. An element a of a ring R is nilpotent if an — 0 for some positive integer n. A {left, right, two-sided) ideal I ofR is nil if every element ofl is nilpotent; I is nilpotent if In = 0 for some integer n. Every nilpotent ideal is nil since /" = 0 implies an = 0 for all a e /. It is possible, however, to have a nil ideal that is not nilpotent (Exercise 11).
Theorem 2.12. 7/R is a ring, then every nil right or left ideal is contained in the radical J(R). REMARK. The theorem immediately implies that every nil ring is a radical ring. PROOF OF 2.12. If a71 = 0, let r = — a + a2 — a3 + • ■ • + (—l)B-1aB_1. Verify that r+a + ra = 0 = a-\-r + ar, whence a is both left and right quasi regular. Therefore every nil left [right] ideal is left [right] quasi-regular and hence is contained in J (R) by Theorem 2.3 (iv). ■
Proposition 2.13. 7/R is a left [resp. right] Artinian ring, then the radical J(R) is a nilpotent ideal. Consequently every nil left or right ideal of R is nilpotent and J(R) is the unique maximal nilpotent left (or right) ideal of R. REMARK. If R is left [resp. right] Noetherian, then every nil left or right ideal is nilpotent (Exercise 16). PROOF OF 2.13. Let J — J (R ) and consider the chain of (left) ideals J Z) J 2 ZD J 3 Z> - - •. By hypothesis there exists k such that J* = J k for all i > k. We claim that Jk = 0. If J k ^ 0, then the set S of all left ideals I such that JkI 9^ 0 is non empty (since 7V* = J'ik = Jk ^ 0). By Theorem VIII. 1.4 S has a minimal element 70. Since Jkh ^ 0, there is a nonzero a e /0 such t h a t y f c o ^ 0. Clearly Jka is a left ideal of R that is contained in I0. Furthermore Jka eS since Jk(Jka) = J2ka = Jka 0. Con
2. THE JACOBSON RADICAL
431
sequently Jka = I0 by minimality. Thus for some nonzero r e Jk, ra = a. Since — reJ k CI J( R) , — r is left quasi-regular, whence s — r — sr = 0 for some .v e R. Consequently, a = ra = — [—ra] = — [—ra + 0] = — [ — ra + sa — sa]
= — [ — r a + sa — s(ra)] = — [ — r + 5 — .vr]o = — 0a = 0. This contradicts the fact that a ^ 0. Therefore J k = 0. The last statement of the theorem is now an immediate consequence of Theorem 2.12. ■ Finally we wish to show that left quasi-regularity is a radical property as defined in the introduction to this section. By Theorem 2.3 (iv) its associated radical is clearly the Jacobson radical and a left quasi-regular ring is precisely a radical ring (Defini tion 2.9). Since a ring homomorphism necessarily maps left quasi-regular elements onto left quasi-regular elements, the homomorphic image of a radical ring is also a radical ring. To complete the discussion we must show that R/J(R ) is semisimple and that J(R) is a radical ring.
Theorem 2.14. 7/R is a ring, then the quotient ring R/J(R) is semisimple. PROOF. Let 7r : R —* R/J(R ) be the canonical epimorphism and denote ir{r) by r (r e R). Let G be the set of all regular maximal left ideals of R. If / e C, then J (R) C I by Theorem 2.3 (ii) and w(I) = I/J(R ) is a maximal left ideal of R/J {R) by Theorem IV.l .10. If e e R is such that r — reel for all reR, then r — fee t t (I) for all r e R/J(R). Therefore, 7r(7) is regular for every I in Q. Since J(R) = P) /it is easy to verify that if lee
r e Pi 7r(/) = P) I/ J(R), then r e J(R ). Consequently, by Theorem 2.3 (ii) (applied to /eC
IeC
R/J(R)) J(R/J(R)) <= P) tr(/) C ttG/(/?)) - 0, lee
whence R /J (R ) is semisimple. ■
Lemma 2.15. Let R be a ring and a e R. (i) // — a2 is left quasi-regular, then so is a. (ii) a e J(R) if and only if Ra is a left quasi-regular left ideal. PROOF, (i) If r + ( — a2) + r(—a2) = 0, let s = r — a — ra. Verify that s + a + sa = 0, whence a is left quasi-regular. (ii) If a e J (R), then Ra CI J(R). Therefore, Ra is left quasi-regular since J (R) is. Conversely suppose Ra is left quasi-regular. Verify that K = \ ra n a \ r e R,n eZ ) is a left ideal of R that contains a and R a. If s = ra + na , then — s 2 e Ra. By hy pothesis —s 2 is left quasi-regular and hence so is 5 by (i). Thus K i sa left quasi regular left ideal. Therefore a e K CZ J(R ) by Theorem 2.3 (iv). ■
Theorem 2.16. (i) If an ideal I of a ring R is itself considered as a ring, then J(I) = I fl J(R).
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CHAPTER IX THE STRUCTURE OF RINGS
(ii) If R is semisimple, then so is every ideal of R. (iii) J(R) is a radical ring. PROOF, (i) I D 7(7?) is clearly an ideal of /. If a e / fl 7(7?), then ois left quasi regular in R, whence r + a + ra = 0 for some r e R. But r = —a — ra e I. Thus every element of / D J(R) is left quasi-regular in I. Therefore I D J(R) C J{I) by Theorem 2.3 (iv) (applied to I). Suppose a e 7(7). For any r e R, — (ra) 2 = — (rar)a e IJ(I) CZ 7(7), whence —(ra)2 is left quasi-regular in 7 by Theorem 2.3 (iv). Consequently by Lemma 2.15 (i) ra is left quasi-regular in 7 and hence in R. Thus Ra is a left quasi-regular left ideal of R , whence a s J( R) by Lemma 2.15 (ii). Therefore a e 7(7) P J(R) C 7 fl 7(7?). Conse quently 7(7) C 7 D 7(7?), which completes the proof that 7(7) = 7 fl 7(7?). State ments (ii) and (iii) are now immediate consequences of (i). ■
Theorem 2.17. 7/{R; | i s I) is a family of rings, then J(XT 7?j) = n w>. iel
ie/
SKETCH OF PROOF. Verify that an element {at \ e is left quasi-regular in if and only if a, is left quasi-regular in R{ for each i. Consequently Ij[7(7?;) is a left quasi-regular ideal of JTt?,, whence IIW) CZ 7(11*) by Theorem 2.3 (iv). For each k s 7, let trk : —> Rk be the canonical projection. Verify that Ik = 7Tjt(7(Xj7?i)) is a left quasi-regular ideal of Rk. It follows that Ik CI 7(7?;) and therefore that 7(J^7?,) CZ Xj7(7?i). ■
n*.
EXERCISES Note: R is always a ring.
1. For each a,b e 7? let a ° b = a + b + ab. (a) ° is an associative binary operation with identity element 0 e 7?. (b) The set G of all elements of 7? that are both left and right quasi-regular forms a group under (c) If 7? has an identity, then a s 7? is left [resp. right] quasi-regular if and only if 1 r + a is left [resp. right] invertible. [Hint: (\R + r)(lfi + a) = 1 r + r ° a and /-(I*- + a) — \R = (r — lfi) ° a.] 2. (Kaplansky) 7? is a division ring if and only if every element of 7? except one is left quasi-regular. [Note that the only element in a division ring D that is not left quasi-regular is — 11>; also see Exercise 1.] 3. Let 7 be a left ideal of 7? and let ( 7 : 7 ? ) = {r e 7? | r7? C 7 } .
(a) (7 : 7?) is a n ideal of R. If 7 is regular, then (7 : 7?) is the largest ideal of 7? that is contained in 7. (b) If 7 is a regular maximal left ideal of 7? and A = R/I, then GL(A) = (7 : 7?). Therefore 7(7?) = fl (7 : 7?), where I runs over all the regular maximal left ideals of R. 4. The radical 7(7?) contains no nonzero idempotents. However, a nonzero idempotent may be left quasi-regular. [Hint: Exercises I and 2],
2. THE JACOBSON RADICAL
433
5. If R has an identity, then (a) J( R) = { r e / ? | l / e + 5 r i s left invertible for all 5 e R }. (b) J( R) is the largest ideal K such that for all r e K , lfi + r is a unit. 6. (a) The homomorphic image of a semisimple ring need not be semisimple. (b) If / : / ? — > S is a ring epimorphism, then f (J(R)) d J(S). 7. If R is the ring of all rational numbers with odd denominators, then J(R) consists of all rational numbers with odd denominator and even numerator. 8. Let R be the ring of all upper triangular n X n matrices over a division ring D (see Exercise VII. 1.2). Find J(R) and prove that R/J(R) is isomorphic to the direct product D X D X • • • X D in factors). [Hint: show that a strictly tri angular matrix is nilpotent.] 9. A principal ideal domain R is semisimple if and only if R is a field or R contains an infinite number of distinct nonassociate irreducible elements. 10. Let D be a principal ideal domain and d a nonzero nonunit element of D. Let R be the quotient ring D/{d). (a) R is semisimple if and only if d is the product of distinct nonassociate irreducible elements of D. [Hint: Exercise VIII. 1.2.] (b) What is J(R)1 11. If p is a prime, let R be the subring 53 Zvn of XT Zvn- The ideal I = yi n
>i
n>
i
>i
where /„ is the ideal of Zp71 generated by p eZ p n , is a nil ideal of R that is not nilpotent. 12. Let R be a ring without identity. Embed R in a ring 5 with identity which has characteristic zero, as in Theorem III. 1.10. Prove that./(/?) = J(S). Consequently every semisimple ring may be embedded in a semisimple ring with identity. 13. J(MatnR) = Mat^/?). Here is an outline of a proof: (a) If A is a left K-module, consider the elements of An = A ® A ® ■ ■■ (£) A (n summands) as column vectors; then A n is a left (Mat„/?)-module (under ordinary matrix multiplication). (b) If A is a simple K-module, An is a simple (MatnR)-module. (c) 7(Mat„K) C= Mat„y(K). (d) Matn/(R) C J(Mat„K). [Hint: prove that Mat„J(K) is a left quasi-regular ideal of MatnK as follows. For each k = 1,2,. . ., n let Kk consist of all matrices (atj) such that a z j e J(R) and — 0 ify j* k. Show that Kk is a left quasi-regular left ideal of MatnR and observe that K L + K2 + • —\- Kn = Matny(K).] 14. (a) Let / be a nonzero ideal of K[a] and p ( x ) a nonzero polynomial of least degree in / with leading coefficient a . If /(a) e K[x] and a m f ( x ) = 0, then a n ' l p { x ) /(a) = 0.
(b) If a ring K has no nonzero nil ideals (in particular, if R is semisimple), then R[a] is semisimple. [Hint: Let M be the set of nonzero polynomials of least degree in 7(K[a]). Let N be the set consisting of 0 and the leading coefficients of polynomials in M. Use (a) to show that N is a nil ideal of K, whence 7(K[a]) = 0.] (c) There exist rings K such that R[a] is semisimple, but R is not. [Hint, consider R = F[[a]], with F a field.]
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CHAPTER IX THE STRUCTURE OF RINGS
15. Let L be a left ideal and K a right ideal of R. Let M(R) be the ideal generated by all nilpotent ideals of R. (a) L + LR is an ideal such that (L + LR) n C L n + L n R for all n > 1. (b) K + RK is an ideal such that (K + RK) n d K n + RK n for all n > 1. (c) If L [resp. K ] is nilpotent, so is the ideal L + LR [resp. K + RK], whence L C= M( R) [resp. K <= M(R)\ . (d) If N is a maximal nilpotent ideal of R, then R/N has no nonzero nilpotent left or right ideals. [Hint: first show that R/N has no nonzero nilpotent ideals; then apply (c) to the ring R/N.] (e) If K [resp. L] is nil, but not nilpotent and ir : R —► R/N is the canonical epimorphism, then7r(AT) [resp. 7r(L)] is a nil right [resp. left] ideal of R/N which is not nilpotent. 16. (Levitsky) Every nil left or right ideal / in a left Noetherian ring R is nilpotent. [Sketch of Proof. It suffices by Exercise 15 to assume that R has no nonzero nilpotent left or right ideals. Suppose / is a left or a right ideal which is not nilpo tent and Show that aR is a nil right ideal (even though I may be a left ideal), whence the left ideal G(«) is nonzero for all u e aR. There exists a nonzero «o e aR with G(w0) maximal, whence G(wfl) = G(wo*) for all x e R such that UqX ^ 0. Show that (u0y)u0 = 0 for a l l y e R, so that (Ru0)2 = 0. Therefore Ru0 = 0, which implies that {r e R \ Rr = 0} isa nonzero nilpotent right ideal of R: contradiction.] 17. Show that Nakayama’s Lemma VIII.4.5 is valid for any ring R with identity, provided condition (i) is replaced by the condition (i') J is contained in the Jacobson radical of R. [Hint: Use Theorem 2.3(iv) and Exercise 1 (c) to show (i') => (ii).]
3. SEMISIMPLE RINGS In accordance with the theory of radicals outlined in the first part of Section 2 we now restrict our study to rings that are Jacobson semisimple. Arbitrary semisimple rings are characterized as particular kinds of subrings of direct products of primitive rings (Proposition 3.2). Much stronger results are proved for semisimple (left) Artinian rings. Such rings are actually finite direct products of simple rings (Theorem 3.3). They may also be characterized in numerous ways in terms of modules (Theo rem 3.7). Along the way semisimple modules over arbitrary rings are defined and their basic properties developed (Theorem 3.6).
Definition 3.1. A ring R is said to be a subdirect product of the family of rings {Rj | i s 11 ifR is a subring of the direct product JT Ri such that 7Tk(R) = Rk for every iel
k e I, where 7Tk : Ri —> Rk is the canonical epimorphism. iel
REMARK. A ring S is isomorphic to a subdirect product of the family of rings { Ri | i s /} if and only if there is a monomorphism of rings
7i-k
k
e I.
3. SEMISIMPLE RINGS
435
EXAMPLE. Let P be the set of prime integers. For each I c e Z and p e P let kp zZp be the image of k under the canonical epimorphism Z —>ZP. Then the map 0 : Z —■+ Zp given by k {k p } p l p is a monomorphism of rings such that pel'
7tp<£(Z) = Zp for every p e P. Therefore Z is isomorphic to a subdirect product of the family of fields \Z P | p e P) . More generally we have:
Proposition 3.2. A nonzero ring R is semisimple if and only if R is isomorphic to a subdirect product of primitive rings.
REMARK. Propositions 1.7 and 3.2 imply that a nonzero commutative semi simple ring is a subdirect product of fields. SKETCH OF PROOF OF 3.2. Suppose R is nonzero semisimple and let (P be the set of all left primitive ideals of R. Then for each PetP, R/P is a primitive ring (Definition 2.1). By Theorem 2.3 (iii), 0 = J(R) = P| P. For each P let Xp : R —> R/P Pe6>
and 7rp :
n R/Q —> R/P be the respective canonical epimorphisms. The map
(Jeff
II P/P
Pe(?
given by r \-■> {Xp(r)}pE(p = {r + P\p,(? is a monomorphism of
rings such that 7rp<£(7?) = R/P for every P e (P. Conversely suppose there is a family of primitive rings {7?, | i e 7} and a mono morphism of rings (f> : R —* PJ Ri such that 71*0(7?) = Rk for each k e I. Let \pk be the iel
epimorphism irk<}>. Then 7?/Ker \pk is isomorphic to the primitive ring Rk (Corollary III.2.10), whence Ker \pk is a left primitive ideal of R (Definition 2.1). Therefore J( R) d P) Ker by Theorem 2.3 (iii). However, if r e R and \pk(r) = 0, then the /cth kel
component of $(r) in JJt?,- is zero. Thus if r e H Ker ^k, we must have
Since 4> is a monomorphism r = 0. Therefore J(R) d P| Ker \pk = 0, whence R is kel
semisimple. ■ In view of the results on primitive rings in Section 1, we can now characterize semisimple rings as those rings that are isomorphic to subdirect products of families of rings, each of which is a dense ring of endomorphisms of a vector space over a division ring. Unfortunately subdirect products (and dense rings of endomorphisms) are not always the most tractable objects with which to deal. But in the absence of further restrictions this is probably the best one can do. In the case of (left) Artinian rings, however, these results can be considerably sharpened.
Theorem 3.3. (Wedderburn-Artin). The following conditions on a ring R are equivalent.
(i) R is a nonzero semisimple left Artinian ring; (ii) R is a direct product of a finite number of simple ideals each of which is iso morphic to the endomorphism ring of a finite dimensional vector space over a division ring; (iii) there exist division rings Di,. . . , Dt and positive integers ni, . . . , nt such that R is isomorphic to the ring MatniDi X MatniD2 X • • • X MatntDt.
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REMARK. By a simple ideal of R we mean an ideal that is itself a simple ring. PROOF OF 3.3. (ii) <=> (iii) Exercise III.2.9 and Theorem VII. 1.4. t (ii) => (i) By hypothesis R = IX Ri with each R t the endomorphism ring of a i= 1 vector space. The example after Definition 1.5 shows that each/?, is primitive, whence J (R i) = 0 by Theorem 2.10 (i). Consequently by Theorem 2.17 t j(R)= n j(Rt) = o. t=l
Therefore R is semisimple. R is left Artinian by Theorem VII.1.4 and Corollaries VIII.1.7 and VIII.l.12. (i) => (ii) Since R t * 0 and J(R) = 0, R has left primitive ideals by Theorem 2.3 (iii). Suppose that R has only finitely many distinct left primitive ideals: P u P 2 , . . . , Pt. Then each R/Pi is a primitive ring (Definition 2.1) that is left Artinian (Corollary VIII.1.6). Consequently, by Theorem 1.14 each R/Pi is a simple ring isomorphic to an endomorphism ring of a finite dimensional left vector space over a division ring. Since R/Pi is simple, each P t is a maximal ideal of R (Theorem III.2.13). Furthermore R 2 £Z! Pi (otherwise (R/Pif = 0), whence R 2 + Pi = R by maximality. Likewise if / 9* j, then Pi + P, = R by maximality. Consequently by Corollary III.2.27 (of the Chinese Remainder Theorem) and Theorem 2.3 (iii) there is an isomorphism of rings: t R = R/0 = R/J(R) = R/C\ Pi g* R/Pi X ■ ■ • X R/P t .
i=i
t If i k : R/P k —> n R/Pi is the canonical monomorphism (Theorem III.2.22), then i=l t t
n
n
each Lk(R/Pk) is a simple ideal of R/Pi. Under the isomorphism R/Pi =r, 1=1 1=1 the images of the i k (R/Pk ) are simple ideals of R. Clearly R is the (internal) direct product of these ideals. To complete the proof we need only show that R cannot have an infinite number of distinct left primitive ideals. Suppose, on the contrary, that P u P2, P3, • • • is a se quence of distinct left primitive ideals of R. Since Pi
z> Py n p2 z> a n p2 n p3 z> • - •
is a descending chain of (left) ideals there is an integer n such that Pi D • • • fl P n = Pi fl - - • fl P„ D Pn+U whence Pi fl ■ • ■ fl P„ CI Pn+i. The previous paragraph shows that R 2 + Pi = R and Pi + P3 = R (/' 9* j) for i,j = 1,2,. . . , « + 1. The proof of Theorem III.2.25 shows that P n + 1 + (Pi fl • • • fl Pn) = R. Consequently Pn+i = R, which contradicts the fact that Pn+i is left primitive (see the Remark after Definition 2.1). Therefore R has only finitely many distinct primitive ideals and the proof is complete. ■
Corollary 3.4. (i) A semisimple left Artinian ring has an identity. (ii) A semisimple ring is left Artinian if and only if it is right Artinian. (iii) A semisimple left Artinian ring is both left and right Noetherian.
3. SEMISIMPLE RINGS
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REMARK. Somewhat more is actually true: any left Artinian ring with identity is left Noetherian (Exercise 13). SKETCH OF PROOF OF 3.4. (i) Theorem 3.3. (ii) Theorem 3.3 is valid with “left” replaced by “right” throughout. Consequently the equivalence of condi tions (i) and (iii) of Theorem 3.3 implies that R is left Artinian if and only if R is right Artinian. (iii) Corollaries VIII.1.7 and VIII. 1.12 and Theorem 3.3 (iii). ■ The following corollary is not needed in the sequel. Recall that an element e of a ring R is said to be idempotent if el = e.
Corollary 3.5. Ifl is an ideal in a semisimple left Artinian ring R, then I = Re, where e is an idempotent which is in the center of R. SKETCH OF PROOF. By Theorem 3.3 R is a (ring) direct product of simple ideals, R = h X • • • X /„. For each j, I D /, is either 0 or /3 by simplicity. After re indexing if necessary we may assume that / fl /,- = I, for j = 1,2,... ,t and / H /, = 0 for j = t + \,... ,n. Since R has an identity by Corollary 3.4, there exist ej e Ij such that 1/; = ei + e2 H— ■ + en. Since /,-/* = 0 for j ^ k we have ei + e2 H------ en = \ R = (I*)2 = eY2 + e22 H------------ b e 2 ,
whence e,2 = e, for each j. It is easy to verify that each e, lies in the center of R and that e = e\ + e2 + • • ■ + et is an idempotent in / which is in the center of R. Since I is an ideal, Re Cl I. Conversely if u s /, then u = u\ R = uei 4- • • • + uen. But for j > /, uej s / fl Ij = 0. Thus u = uei ■ ■ ■ + uet = ue. Therefore I CL Re. ■ Theorem 3.3 is a characterization of semisimple left Artinian rings in ring theoretic terms. As one might suspect from the close interrelationship of rings and modules, such rings can also be characterized strictly in terms of modules. In order to obtain these characterizations we need a theorem that is valid for modules over an arbitrary ring.
Theorem 3.6. The following conditions on a nonzero module A over a ring R are equivalent.
(i) A is the sum of a family of simple submodules. (ii) A is the (internal) direct sum of a family of simple submodules. (iii) For every nonzero element a of A, Ra ^ 0; and every submodule Bo/A is a direct summand (that i'j, A = B@C for some submodule C). A module that satisfies the equivalent conditions of Theorem 3.6 is said to be semisimple or completely reducible. The terminology semisimple is motivated by Theorem 3.3 (ii) and the fact (to be proved below) that every module over a (left) Artinian semisimple ring is semisimple.
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SKETCH OF PROOF OF 3.6. (i) => (ii) Suppose A is the sum of the family [Bi | i e /} of simple submodules (that is, A is generated by IJ Bi). Use Zorn’s Lemma iel
to show that there is a nonempty subset J of I which is maximal with respect to the property: the submodule generated by [Bj \j e J) is in fact a direct sum Bj. We
23
jeJ
23 Bj. To prove this we need only show that B, d ^3 Bj for every jeJ i e /. Since Bi is simple and Bi fl (23 Bj) is a submodule of Bi, either Bi fl (23 B ) = Bi, which implies Bi d 23 Bj, or B fl (23 B,) = 0. The second case cannot occur. claim that A = jeJ
}
i
For if it did, K = {/} U J would be a set such that the submodule generated by [Bk [ k s K) is a direct sum (Theorem IV.l .15). which contradicts the maximality of J. (ii) => (iii) Suppose A is the direct sum 23 Bi with each Bi a simple submodule. If iel
a is a nonzero element of A, then a = bn +■■■ + b i k with 0 ^ b i k e B i k (/,,. . . , ik e /). Clearly Ra = 0 if and only if Rbik = 0 for each ik. But Remark (iii) after Definition 1.1 shows that Rbik — Bik ^ 0. Therefore Ra ^ 0. Let B be a nonzero submodule of A. By simplicity B fl Bi is either 0 or Br. If B fl Bi = Bi for all /', then A — B and B is trivially a direct summand, A = B (J) 0. Otherwise B fl 6, = 0 for some i. Use Zorn’s Lemma to find a subset J of I which is maximal with respect to the property: B B j ) = 0. We claim that
fl (23
jeJ
(23 Bj). It suffices by Theorem IV. 1.15 to show that Bi d B © (23 Bj) jeJ ___ jeJ for each /'. If / e J, then Bi d ^3 B, and we are done. If /' ^ J and Bi (Zf B © 23 Bj, A = B®
jeJ
jeJ
then Bi H (fi @ 23 Bj) = 0 by the simplicity of B t . It follows that J U {/'} is a set jeJ
that contradicts the maximality of J. Therefore Bi d B © 23 B,. jeJ
(iii) => (i) We first observe that if N is any submodule of A, then every submodule K of N is a direct summand of N. For by hypothesis K is a direct summand of A, say A = K@L. Verify that N = N (1 A (N D K)@{N D L) = K © ( N D L). Next we show that A has simple submodules. Since A ^ 0, there exists a nonzero element a of A. Use Zorn’s Lemma to find a submodule B of A that is maximal with respect to the property that a\B. By hypothesis A — B © C for some nonzero submodule C and RC 0. We claim that C is simple. If it were not, then C would have a proper submodule D, which would be a direct summand of C by the previous para graph. Consequently C = with E ^ 0, whence A = 5©C = £ ? © £ ) © £ , with D^0 and E ^ 0. Now B © D and B © E both contain B properly. Therefore by the maximality of B we must have a £ B © D and a e B (£) E. Thus b -\- d = a = b' + e (b,b' e B; d s D\ e e E). Now 0 = a — a = (b — b')-\-d—eeB(£)D(£)E implies that d = 0, e = 0, and b — b' = 0. Consequently, a = b e B which is a con tradiction. Therefore C is simple. Let A 0 be the submodule of A generated by all the simple submodules of A. Then A - A 0 © N for some submodule N . N satisfies the same hypotheses as A by the paragraph before last. If N ^ 0, then the argument in the immediately preceding paragraph shows that N contains a nonzero simple submodule T. Since T is a simple submodule of A, T d A 0 . Thus T d A 0 D TV = 0, which is a contradiction. There fore N = 0, whence A = A 0 is the sum of a family of sfmple submodules. ■ We are now able to give numerous characterizations of semisimple left Artinian rings in terms of modules. Since the submodules of a ring R (considered as a left
3. SEMISIMPLE RINGS
439
/?-module) are precisely the left ideals of R, some of these characterizations are stated in terms of left ideals. A subset \e u . . ., e m ] of R is a set of orthogonal idempotents if el 1 = e,- for all /' and e,e, = 0 for all i ^ j.
Theorem 3.7. The following conditions on a nonzero ring R with identity are equiv alent.
(i) R is semisimple left Artinian; (ii) every unitary left K-module is projective; (iii) every unitary left K-module is injective; (iv) every short exact sequence of unitary left K-modules is split exact; (v) every nonzero unitary left K-module is semisimple; (vi) R is itself a unitary semisimple left K-module; (vii) every left ideal of R is of the form Re with e idempotent; (viii) R is the (internal) direct sum (as a left K-module) of minimal left ideals K i , . . . , Km such that K, = Re* (e* e R) for i = 1,2,. . ., m and { e i , . . . , e m } is a set of orthogonal idempotent s with ei + e2 -j- • ■ • + em = 1r. REMARKS. Since a semisimple ring is left Artinian if and only if it is right Artinian (Corollary 3.4), each condition in Theorem 3.7 is equivalent to its obvious analogue for right modules or right ideals. There is no loss of generality in assuming R has an identity, since every semisimple left Artinian ring necessarily has one by Corollary 3.4. The theorem is false if the word “unitary” is omitted (Exercise 10). SKETCH OF PROOF OF 3.7. (ii) <=> (iii) <=> (iv) is Exercise IV.3.1. To com plete the proof we shall prove the implications (iv) <=> (v) and (v) => (vii) - ( v i ) => (i) => (viii) => (v). (iv) => (v) If B is a submodule of a nonzero unitary R-module A, then 0 -* B ^ A -> A/B -> 0 is a short exact sequence, which splits by hypothesis. The proof of Theorem IV.1.18 shows that A = B © C with C = A/B . Since A is unitary, Ra ^ 0 for every non zero a e A. Therefore A is semisimple by Theorem 3.6. (v) => (iv) Let 0 —> A B —C —* 0 be a short exact sequence of unitary K-mod ules. Then / : A—* f(A) is an isomorphism. Since B is semisimple by (v), f(A) is a direct summand of B by Theorem 3.6. If t t : B —> f(A) is the canonical epimorphism, then7r/= /and/_17r \B—*A is an K-module homomorphism such that(/_17r)/= \ A . Therefore the sequence splits by Theorem IV.l.18. (v) => (vii) The left ideals of K are precisely its submodules. If L is a left ideal, then K = L © / for some left ideal / by (v) and Theorem 3.6. Consequently, there are e, e L and e2 e / such that 1/e = ei + e2. Since e x e L, Rei CL L. If r e L, then r = re, + re2, whence re2 = r — re, s L D / = 0. Thus r = rei for every r s L; in particular, exei = e, and L C Re x . Therefore, L = Re, with e, idempotent. (vii) => (vi) A submodule L of K is a left ideal, whence L = Re with e idempotent. Verify that K(1k — e) is a left ideal of K such that K = Re © R(lfl — e). Therefore, K is semisimple by Theorem 3.6. (vi) => (i) By hypothesis K is a direct sum 52 Bi, with each /?, a simple submodule iel
(left ideal) of K. Consequently there is a finite subset I 0 of / (whose elements will be
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CHAPTER IX THE STRUCTURE OF RINGS
labeled 1,2,.. . , k for convenience) such that 1# = e\ + e2 H— • + ek (e, e Bi). Thus kk for every r e R, r = rex + re2 + • ■ • + re* whence R = B%. If r eJ(R), »=1 i=1 then rBi = 0 for all i by Theorem 2.3 (i). Consequently,
e22
22
r = rlR = rei + re 2 H---------h re* = 0. Therefore, J(R) = 0 and R is semisimple. Since Bi is simple and (B, 0 ■ - • © Bi)/(B, 0 • • • 0 £._,) S Bu
the series R = B \ 0 • • • 0 B k ZD Bi 0 • ■ • 0 Bk -1 ZD • • • ZD B\ 0 B 2 ZD B\ D 0
is a composition series for R. Therefore, R is left Artinian by Theorem VIII.1.11. t (i) => (viii) In view of Theorem 3.3 it suffices to assume that R = MatniA
n
t=i
with each > 0 and each £>, a division ring. For each fixed / and each j = 1,2 let eij be the matrix in Mat„j£>,- with 1/), in position (j,j) and 0 elsewhere. Then {e { 1 ,..., eini) is a set of orthogonal idempotents in Mat„4D, = Ri whose sum is the identity matrix. The proof of Corollary VIII.1.12 shows that each is a minimal left ideal of Ri and RL = Rten 0 • • -0 Rie,^. Since R is the ring direct product Ri X • • ■ X Rt, it follows that R { Rj = 0 for i ^ j; that Rev, = Riethat is a minimal left ideal of R; and that | 1 < / ' < / ; 1 < j < ( is a set of orthogonal t
idempotents in R whose sum is t
=
t
52 (22 l
j i
=
ea) i
»
= =
tm
22 lie. = 1«- Clearly R = 22 22 ij
=
Rea-
i
(viii) => (v) Let A be a unitary 7?-module. For each a e A and each /, Kia is a submodule of A (Exercise IV. 1.3) and a = Irci = exa H— • + ^mCi e Kta + • ■ • + Kma. Consequently the submodules Kid (a e A, 1 < / < m) generate A. For each as. A and each i, the map / : K{ —> K,a given by k H ka is an R-module epimorphism. Since Kt is a minimal left ideal of a ring with identity, Kr is a simple /?-module. Con sequently if Kia 7* 0, then / is an isomorphism by Schur’s Lemma 1.10. Thus | Kia | 1 < / < m\ aeA; Kia ^ 0} is a family of simple submodules whose sum is A. Therefore A is semisimple by Theorem 3.6. ■ Theorems 3.3 and 3.7 show that a semisimple left Artinian ring may be decom posed as a direct product [resp. sum] of simple ideals [resp. minimal left ideals]. We turn now to the question of the uniqueness of these decompositions.
Proposition 3.8. Let R be a semisimple left Artinian ring. (i) R = Ii X ■ • ■ X In where each Ij is a simple ideal of R. (ii) If J is any simple ideal of R, then J = Ik for some k. (iii) If R = J, X • ■ • X J,„ with each Jk a simple ideal of R, then n = m and (after reindexing) Ik = Jk for k = 1,2,. . . , n. REMARKS. The conclusion J = /, [resp. Jk = I k ] is considerably stronger than the statement “J [resp. A] is isomorphic to h." The uniquely determined simple ideals A,. . ., /„ in Proposition 3.8 are called the simple components of R.
3. SEMISIMPLE RINGS
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PROOF OF 3.8. (i) is true by Theorem 3.3. (ii) If J is a simple ideal of R, then RJ 0, whence I k J ^ 0 for some k. Since I k J is a nonzero ideal that is contained in both I k and 7, the simplicity of I k and /implies I k = IkJ = J. (iii) The ideals I u ... ,I n [resp. Ji,..., J m ] are nonzero and mutually disjoint by hypothesis. Define a map 8 from the t n element set {Ji,... ,J m \ to the n element set ( A , . . . , /„} by J k |—> I k , where J k = h■ 8 is well defined and injective by (ii), whence m < n . The same argu ment with the roles of J k and I k reversed shows that n < m. Therefore n = m and 6 is a bijection. ■ A semisimple left Artinian ring R is a direct sum of minimal left ideals by Theo rem 3.7 (viii). The uniqueness (up to isomorphism) of this decomposition will be an immediate consequence of the following proposition. For R is a semisimple R-mod ule (Theorem 3.7 (vi)) and the minimal left ideals of R are precisely its simple submodules.
Proposition 3.9. Let Abe a semisimple module over a ring R. If there are direct sum decompositions
A = Bi © ■ ■ • © Bm and A = Ci © ■ ■ ■ © Cn, where each B i , Q is a simple submodule of A, then m = n and ( after reindexing) B ; ^ Q for i = 1,2,. . ., m.
REMARK. The uniqueness statement here is weaker than the one in Proposi tion 3.8. Proposition 3.9 is false if “fi, = C,” is replaced by “fii = C” (Exercise 11). PROOF OF 3.9. The series A =fi1®---@fimZ>fi2©---©fimZ>---Z)fimZ>0
is a composition series for A with simple factors fi,, f i z , . . . , Bm (see p. 375). Similarly A = Ci ©• • -© Cr, ID C2 ©• - -© Cm ID - - - ID Cm ID 0 is a composition series for A with simple factors Cu . . . , Cn. The Jordan-Holder Theorem VIII.1.10 implies that m = n and (after reindexing) fi, ^ Ci for i = \,2 ,... ,m. U The following theorem will be used only in the proof of Theorem 6.7.
Theorem 3.10. Let R be a semisimple left Artinian ring. (i) Every simple left [resp. right] K-module is isomorphic to a minimal left [resp. right] ideal of R. (ii) The number of nonisomorphic simple left [resp. right] K-modules is the same as the number of simple components of R. PROOF. R is right Artinian by Corollary 3.4. Since the preceding results are left-right symmetric, it suffices to prove the theorem for left modules. (i) By Theorem 3.7, R = A", ©• ■ • © K.m with each Kt a nonzero minimal left ideal (simple submodule) of R. R has an identity (Corollary 3.4) and every simple R-module A is unitary by Remark (ii) after Definition 1.1. The proof of (viii) => (v)
CHAPTER IX THE STRUCTURE OF RINGS
442
of Theorem 3.7 shows that for some i (1 < i < m) and at A, A contains a nonzero submodule K{a such that Kta = Ki. The simplicity of A implies that A — Kid = Ki. (ii) The simple components of R are the unique simple ideals I, of R such that R = Ix X • • • X /„ (Proposition 3.8). In view of (i) it suffices to prove: (a) each Ki is contained in some It; (b) each It contains some K\ (c) Ki ^ Kj as /^-modules if and only if K{ and Kj are contained in the same simple component /«. These statements are proved as follows. (a) Since R has an identity, K = RKi = hKi X-X InKi. Since each I jK i is a left ideal of R contained in Ki, we must have l t K = Ki for some t and I jK i = 0 for j 9^ t by minimality. Therefore K = ItKi d It. (b) If It contains no K, then R = y,Kj is contained in 7l X • • - X /,_! X It+1 X • • • X In
by (a). Since It ^ 0 by simplicity and R = m.
o ^ i, = /i n r = /, n a, x • • • x x /(+. x ■ • ■ x /„) = o, which is a contradiction. (c) If Ki d Itl and Kj d I t 2 with u 9^ ti, then by (a), 0 Kt = ItlKi and 0 9^ Kj = I t 2 Kj. Since R = IJ/„ hj t 2 = 0 = Consequently, there can be no /^-module isomorphism K = Kj. Conversely suppose K d h and K j d I t . Then K t and Kj are /(-modules. Since I t is simple and 0 ^ AT, = I f K t by (a), the left anni hilator ideal of Ki in It must be zero. Consequently, K,Kt 9^ 0 since 0 9^ K} d It. Thus for some a s K, Kja ^ 0. Since Kt and Kt are left ideals of R, Kja is a nonzero left ideal of R and K}a d K{. Therefore K3a = Ki by minimality. The proof (viii) => (v) of Theorem 3.7 shows that K}u = Kj, whence K{ ~ Kj. ■
EXERCISES 1. A ring R is isomorphic to a subdirect product of the family of rings {/?; | / s /} if and only if there exists for each / e / an ideal Kt of R such that R/Ki = R, and n ^ = 0. iel
2 . A ring R is subdirectly irreducible if the intersection of all nonzero ideals of R is
nonzero. (a) R is subdirectly irreducible if and only if whenever R is isomorphic to a subdirect product of {R t \i e /}, R = R t for some / e I [see Exercise lj. (b) (BirkhofT) Every ring is isomorphic to a subdirect product of a family of subdirectly irreducible rings. (c) The zero divisors in a commutative subdirectly irreducible ring (together with 0) form an ideal. 3. A commutative semisimple left Artinian ring is a direct product of fields. 4. Determine up to isomorphism all semisimple rings of order 1008. How many of them are commutative? [Hint: Exercise V.8.10.] 5. An element a of a ring R is regular (in the sense of Von Neumann) if there exists x e R such that axa = a. If every element of R is regular, then R is said to be a regular ring.
3. SEMISIMPLE RINGS
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(a) Every division ring is regular. (b) A finite direct product of regular rings is regular. (c) Every regular ring is semisimple. [The converse is false (for example, Z).] (d) The ring of all linear transformations on a vector space (not necessarily finite dimensional) over a division ring is regular. (e) A semisimple left Artinian ring is regular. (f) R is regular if and only if every principal left [resp. right] ideal of R is generated by an idempotent element. (g) A nonzero regular ring R with identity is a division ring if and only if its only idempotents are 0 and ljj. 6. (a) Every nonzero homomorphic image and every nonzero submodule of a semisimple module is semisimple. (b) The intersection of two semisimple submodules is 0 or semisimple. 7. The following conditions on a semisimple module A are equivalent: (a) A is finitely generated. (b) A is a direct sum of a finite number of simple submodules. (c) A has a composition series (see p. 375). (d) A satisfies both the ascending and descending chain conditions on submodules (see Theorem VIII.l.11). 8. Let A be a module over a left Artinian ring R such that Ra ^ 0 for all nonzero aeA and let J = J(R). Then JA = 0 if and only if A is semisimple. [Hints: if JA = 0, then A is an R/7-module, with R/J semisimple left Artinian; see Exercise IV. 1.17.] 9. Let R be a ring that (as a left /?-module) is the sum of its minimal left ideals. Assume that {r s R | Rr = 0} = 0. If A is an R-module such that RA = A, then A is semisimple. [Hint: if /is a minimal left ideal and a e A , show that la is either zero or a simple submodule of A.] 10. Show that a nonzero /?-module A such that RA = 0 is not semisimple, but may be projective. Consequently Theorem 3.7 may be false if the word “unitary” is omitted. [See Exercise IV.2.2, Theorem IV.3.2 and Proposition IV.3.5.] 11. Let R be the ring of 2 x 2 matrices over an infinite field. (a) R has an infinite number of distinct proper left ideals, any two of which are isomorphic as left R-modules. (b) There are infinitely many distinct pairs (B,C) such that B and C are mini mal left ideals of R and R = B © C. 12. A left Artinian ring R has the same number of nonisomorphic simple left R-mod ules as nonisomorphic simple right .R-modules. [Hint: Show that A is a simple R-module if and only if A is a simple P/ZC^-module; use Theorem 2.14 and Theorem 3.10.] 13. (a) (Hopkins) If R is a left Artinian ring with identity, then R is left Noetherian. [Hints: Let n be the least positive integer such that Jn = 0 (Proposition 2.13). Let J° = R. Since J(Jl/J^ ') = 0 and R is left Artinian each Ji/Ji+1 (0 < / ' < « — 1) has a composition series by Exercises 7 and 8. Use these and Theorem IV. 1.10 to construct a composition series for R; apply Theorem VIII.1.11.]
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CHAPTER IX THE STRUCTURE OF RINGS
Remark. Hopkins’ Theorem is valid even if the hypothesis “R has an identity” is replaced by the much weaker hypothesis that {r s R \ rR = 0 and Rr = 0} =
0; see L. Fuchs [13; pp. 283-286]. (b) The converse of Hopkins’ Theorem is false.
4. THE PRIME RADICAL; PRIME AND SEMIPRIME RINGS We now introduce the prime radical of a ring and call a ring semiprime if it has 2ero prime radical (Definition 4.1). We then develop the analogues of the results proved in Sections 2 and 3 for the Jacobson radical and semisimple rings (Proposi tions 4.2-4.4). There is a strong analogy between the prime radical, prime ideals, semiprime rings, prime rings, and the Jacobson radical, left primitive ideals, semi simple rings, and primitive rings respectively. The remainder of the section is devoted to a discussion of Goldie’s Theorem 4.8, which is a structure theorem for semiprime rings satisfying the ascending chain con dition on certain types of left ideals. Goldie’s Theorem plays the same role here as do the Wedderburn-Artin Theorems 1.14 and 3.3 for rings with the descending chain condition on left ideals. In fact Goldie’s Theorem may be considered as an extension of the Wedderburn-Artin Theorems to a wider class of rings. A fuller explanation of these statements is contained in discussions after Proposition 4.4, preceding Theo rem 4.8 and after Corollary 4.9. This section is not needed in the sequel.
Definition 4.1. The prime radical P(R) of a ring R is the intersection of all prime ideals of R. If R has no-prime ideals, then P(R) = R. A ring R such that P(R) = 0 is said to be semiprime. REMARKS. The prime radical (also called the Baer lower radical or the McCoy radical) is the radical with respect to a certain radical property, as defined in the in troduction to Section 2; for details, see Exercises 1 and 2. A semiprime ring is one that is semisimple with respect to the prime radical (see the introduction to Section 2). We use the term “semiprime” to avoid both awkward phrasing and confusion with Jacobson semisimplicity. The relationship of the prime radical with the Jacobson radical is discussed in Exercise 3. Just as in the case of the Jacobson radical, there is a close connection between the prime radical of a ring R and the nilpotent ideals of R. In order to prove one such result, we must recall some terminology. Let £ be a subset of a ring R. By Theorem 1.4 the set {r s R | rS — 0} is a left ideal of R, which is actually an ideal if £ is a left ideal. The set \r s R \ rS = 0} is called the left annihilator of S and is denoted (2(S). Similarly the set (MS) = [ r z R \ S r = 0} is a right ideal of R that is an ideal if S is a right ideal. <2r(S) is called the right annihilator of S . A left [resp. right] ideal / of R is said to be a left [resp. right] annihilator if / = C£(S) [resp. I = (2r(S)] for some subset S of R .
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REMARK. The intersection of two left [resp. right] annihilators is also a left [resp. right] annihilator since d(S ) fl d(T) = (i(S U T). If 5 and T are actually left ideals, then d(S) fl d(T) = G(S U T) = G(S + T).
Proposition 4.2. A ring R is semiprime if and only if R has no nonzero nilpotent ideals.
SKETCH OF PROOF. (=>) If / is a nilpotent ideal and K is any prime ideal, then for some n, In = 0 s K, whence / d K. Therefore I d P(R). Consequently, if R is semiprime, so that P(R) = 0, then the only nilpotent ideal is the zero ideal. (<=) Conversely suppose that R has no nonzero nilpotent ideals. We must show that P(R) = 0. It suffices to prove that for every nonzero element a of R there is a prime ideal K such that a \ K , whence a 4 P(R). We first observe that d (R) fl R is a nilpotent ideal of R since (acR)
n R)(d(R)
n r)cz d(R)R =
o.
Consequently, G(/?) = d(R) fl R = 0. Similarly (Jr(i?) = 0. If b is any nonzero element of R, we claim that RbR ^ 0. Otherwise Rb d d(R) = 0, whence Rb = 0. Thus b e Cir(R) = 0, which is a contradiction. Therefore RbR is a nonzero ideal of R and hence not nilpotent. Consequently bRb ^ 0 (otherwise (RbR) 2 d RbRbR = 0). For each nonzero b s R choose f(b) s bR b such that f(b) ^ 0. Then by the Recursion Theorem 6.2 of the Introduction there is a function
Since am+i ^ K, we must have AB (Zf K. Therefore K is a prime ideal. ■ A ring R is said to be a prime ring if the zero ideal is a prime ideal (that is, if /, J are ideals such that IJ = 0, then / = 0 or J = 0). The relationships among prime ideals, prime rings, and semiprime rings are analogous to the relationships between left primitive ideals, primitive rings, and semisimple rings. In particular, we note the following: (i) The prime [resp. Jacobson] radical is the intersection of all prime [resp. primitive] ideals (see Theorem 2.3(iii)). (ii) Every prime ring is semiprime since 0 is a prime ideal. This corresponds to the fact that every primitive ring is semisimple (Theorem 2.10(i)).
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Proposition 4.3. K is a prime ideal of a ring R if and only if R/K is a prime ring. /
REMARK. This is the analogue of Definition 2.1 (left primitive ideals). SKETCH OF PROOF OF 4.3. If R/K is prime, let tt : R -* R/K be the canonical epimorphism. If / and J are ideals of R such that IJ CZ K, then 7r(/), ir(J) are ideals of R/K (Exercise 111.2.13(b)) such that 7r(/)x(7) = 7r(/7) = 0. Since R/K is prime, either nil) = 0 or ir(J) = 0; that is, / CZ K or J CZ K . Therefore, AT is a prime ideal (Definition III.2.14). The converse is an easy consequence of Theorem III.2.13 and Definition III.2.14. ■ The final part of the semiprime-semisimple analogy is given by
Proposition 4.4. A ring R is semiprime if and only //R is isomorphic to a subdirect product of prime rings. SKETCH OF PROOF. Proposition 4.4 is simply Proposition 3.2 with the words “semisimple” and “primitive” changed to “semiprime” and “prime” re spectively. With this change and the use of Proposition 4.3 in place of Definition 2.1, the proof of Proposition 3.2 carries over verbatim to the present case. ■ We have seen that primitive rings are the basic building blocks for semisimple rings. Proposition 4.4 shows that the basic building blocks for semiprime rings are the prime rings. At this point the analogy between primitive and prime rings fails. Primitive rings may be characterized in terms of familiar matrix rings and endomor phism rings of vector spaces (Section 1). There are no comparable results for prime rings. But the situation is not completely hopeless. We have obtained very striking results for primitive and semisimple left Artinian rings (Sections 1 and 3). Conse quently it seems plausible that one could obtain useful characterizations of prime and semiprime rings that satisfy certain chain conditions. We shall now do precisely that. We first observe that in a left Artinian ring the prime radical coincides with the Jacobson radical (Exercise 3(c)). Consequently, left Artinian semiprime rings are also semisimple, whence their structure is determined by the Wedderburn-Artin Theorem 3.3. Since every semiprime (semisimple) left Artinian ring is also left Noetherian by Corollary 3.4, the next obvious candidate to consider is the class of semiprime left Noetherian rings (that is, semiprime rings that satisfy the ascending chain condition on left ideals). Note that there are semiprime left Noetherian rings that are not left Artinian (for example, Z). Consequently, a characterization of semi prime left Noetherian rings would be a genuine extension of our previous results. We shall actually characterize a wider class of rings that properly includes the class of all semiprime left Noetherian rings. The class in question is the class of all semiprime left Goldie rings, which we now define. A family of left ideals of R [1, \ j z J \ is said to be independent provided that for each k e J , l k fl h * = 0, where h * is the left ideal generated by {/, \ j ^ k \ . In other words, {/, \ j e J \ is independent if and only if the left ideal I generated by { l j \ j e J ] is actually the internal direct sum I = ^ 2 1 , (see Theorem IV. 1.15). jeJ
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Definition 4.5. A ring R is said to be a (left) Goldie ring if (i) R satisfies the ascending chain condition on left annihilators; (ii) every independent set of left ideals of R is finite. REMARKS, (i) Condition (i) of Definition 4.5 means that given any chain of left annihilators Ct(SY)
Definition 4.6. A nonzero element a in a ring R is said to be regular if a is neither a left nor right zero divisor.
Definition 4.7. A ring Q(R) with identity is said to be a left quotient ring of a ring R if (i) R C Q(R); (ii) every regular element in R is a unit in Q(R); (iii) every element c ofQ(R) is ofthe forme = a_1b, where a,b e R and a is regular. REMARKS, (i) A ring R need not have a left quotient ring. If it does, however, it is easy to see that Q(R) is determined up to isomorphism by Definition 4.7. (ii) A right quotient ring of R is defined in the same way, except that “c = a~lb” is replaced by.“c = ba~in condition (iii). A ring may have a right quotient ring, but no left quotient ring (see N. J. Divinsky [22; p. 71]). (iii) If R is a ring that has a left quotient ring Q(R) = T, then R is said to be a left order in T. EXAMPLE. Let R be a commutative ring that has at least one regular element. Let S be the set of all regular elements of R. Then the complete ring of quotients S~*R
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is a ring with identity (Theorem III.4.3) that contains an isomorphic copy
Theorem 4.8. (Goldie) R is a semiprime [resp. prime] left Goldie ring if and only if R has a left quotient ring Q(R) which is semisimple [resp. simple] left Artinian. Theorem 4.8 will not be proved here for reasons of space. One of the best proofs is due to C. Procesi and L. Small [65]; a slightly expanded version appears in I. Herstein [24]. Although long, this proof is no more difficult than many proofs presented earlier in this chapter. It does use Ore’s Theorem, a proof of which is sketched in I. N. Herstein [24; p. 170] and given in detail in N. J. Divinsky [22; p. 66]. Since the structure of semisimple left Artinian rings has been completely deter mined, Theorem 4.8 gives as good a description as we are likely to get of semiprime left Goldie rings (special case: semiprime left Noetherian rings). The “distance” between the rings R and Q(R) is the price that must be paid for replacing the descending chain condition with the ascending chain condition. For as we observed in the discussion after Proposition 4.4 and in Exercise 3.13, the latter is a consider ably weaker condition than the former.
Corollary 4.9. R is a semiprime [resp. prime] left Goldie ring if and only if R has a quotient ring Q(R) such that Q(R) = MatniDi X ■ ■ ■ X MatnkDk, [resp. Q(R) = MafniDi], where nl5. . ., nt are positive integers and Du . . . , Dn are division rings. PROOF. Theorems 1.14, 3.3, and 4.8. ■ Goldie’s Theorem, as rephrased in Corollary 4.9, may be thought of as an exten sion of the Wedderburn-Artin Theorems 1.14 and 3.3 to a wider class of rings. For instance, Theorem 3.3.states that a semisimple left Artinian ring is a direct product of matrix rings over division rings. Goldie’s Theorem states that every semiprime left Goldie ring has a quotient ring that is a direct product of matrix rings over division rings. But every semisimple left Artinian ring is a semiprime left Goldie ring (Corol lary 3.4, Exercise 3(a), and the Example after Definition 4.5). Furthermore every semisimple left Artinian ring is its own quotient ring (Exercise 6). Thus Goldie’s Theorem reduces to the Wedderburn-Artin Theorem in this case. An analogous ar gument holds for simple left Artinian rings and Theorem 1.14.
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EXERCISES Note: R is always a ring. 1. A subset T of R is said to be an m-system (generalized multiplicative system) if
c,d eT => cxd e T for some x e R. (a) P is a prime ideal of R if and only if R — P is an m-system. [Hint: Exercise III.2.14.] (b) Let I be an ideal of R that is disjoint from an m-system T. Show that I is contained in an ideal Q which is maximal respect to the property that Q fl T = 0. Then show that Q is a prime ideal. [Hint: Adapt the proof of Theorem VIII.2.2] (c) An element r of R is said to have the zero property if every m-system that contains r also contains 0. Show that the prime radical P(R) is the set M of all elements of R that have the zero property. [Hint: use (a) to show M d P(R) and (b) to show P(R) d M.] (d) Every element c of P(R) is nilpotent. [Hint: }c* | / > 1} is an m-system.] If R is commutative, P(R) consists of all nilpotent elements of R. 2. (a) If I is an ideal of R, then P(I) = / fl P(R). In particular, P(P(R)) = P(R). [Hint: Exercise 1(c).] (b) P(R) is the smallest ideal K of R such that P(R/K) = 0. In particular, P(R/P(R)) — 0, whence R/P(R) is semiprime. [Hint: Exercise 111.2.17(d).] (c) An ideal / is said to have the zero property if every element of I has the zero property (Exercise 1(c)). Show that the zero property is a radical property (as defined in the introduction to Section 2), whose radical is precisely P(R). 3. (a) Every semisimple ring is semiprime. (b) P(R) d J(R). [Hint: Exercise 1(d); or (a) and Exercise 2(b).] (c) If R is left Artinian, P(R) = J(R). [Hint: Proposition 2.13.] 4. R is semiprime if and only if for all ideals A, B
AB = 0 A (1 B = 0. 5. (a) Let R be a ring with identity. The matrix ring Mat„i? is prime if and only if R is prime. (b) If R is any ring, then P(Mat„/?) = Matr,P(R). [Hint: Use Exercise 2 and part (a) if R has an identity. In the general case, embed R in a ring 5 with identity via Theorem III. 1.10; then P(R) = R fl P(S) by Exercise 2.] 6. If R is semisimple left Artinian, then R is its own quotient ring. [Hint: Since R has an identity by Theorem 3.3, it suffices to show that every regular element ofi? is actually a unit. By Theorem 3.3 and a direct argument it suffices to assume R = Mat„D for some division ring D. Theorem VII.2.6 and Proposition VII.2.12 may be helpful.] 7. The following are equivalent: (a) R is prime; (b) a,b e R and aRb = 0 imply a = 0 or b = 0; (c) the right annihilator of every nonzero right ideal of R is 0; (d) the left annihilator of every nonzero left ideal of R is 0.
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8. Every primitive ring is prime [see Exercise 7],
9. The center of a prime ring with identity is an integral domain. [See Exercise 7; for the converse see Exercise 10.] 10. Let J be an integral domain and let F be the complete field of quotients of J. Let R be the set of all infinite matrices (row, columns indexed byN*) of the form
where An s Matn(F) and d e J C F. (a) R is a ring. (b) The center of R is the set of all matrices of the form
with de J and hence is isomorphic to J. (c) R is primitive (and hence prime by Exercise 8). 11. The nil radical N(R) of R is the ideal generated by the set of all nil ideals of R. (a) N(R) is a nil ideal. (b) N(N(R)) = N(R). (c) N(R/N(R)) = 0. (d) P(R) CZ N(R) CZ J(R). (e) If R is left Artinian, P(R) = N(R) = J(R). (f) If R is commutative P(R) = N(R).
5. ALGEBRAS The concepts and results of Sections 1-3 are carried over to algebras over a com mutative ring K with identity. In particular, the Wedderburn-Artin Theorem is proved for AT-algebras (Theorem 5.4). The latter part of the section deals with algebras over a field, including algebraic algebras and the group algebra of a finite group. Throughout this section K is always a commutatice ring with identity. The first step in carrying over the results of Sections 1-3 to K-algebras is to review the definitions of a K-algebra, a homomorphism of A"-algebras, a subalgebra and an algebra ideal (Section IV.7). We recall that if a K-algebra A has an identity, then (left, right, two-sided) algebra ideals coincide with (left, right, two-sided) ideals of the ring A (see the Remarks after Definition IV.7.3). This fact will be used frequently without explicit mention.
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A left Artinian K-algebra is a AT-algebra that satisfies the descending chain condi tion on left algebra ideals. A left Artinian AT-algebra may not be a left Artinian ring (Exercise 1). EXAMPLE. If D is a division algebra over K, then Mat,,/) is a A'-algebra (p. 227) which is left Artinian by Corollary VIII.l .12.
Definition 5.1. Let A be an algebra over a commutatuve ring K with identity. (i) A left (algebra) A-module is a unitary left K-module M such that M is a left module over the ring A and k(rc) = (kr)c = r(kc) for all k s K, r e A, c e M. (ii) An A-submoduIe of an A-module M is a subset ofM which is itself an algebra A-module (under the operations in M). (iii) An algebra A-module M is simple (or irreducible) if AM ^ 0 and M has no proper A-submodules. (iv) A homomorphism f : M —> N of algebra A-modules is a map that is both a K-module and an A-module homomorphism. REMARKS. If A is a A-algebra the term “/4-module” will always indicate an algebra /4-module. Modules over the ring A will be so labeled. A right /4-module N is defined analogously and satisfies k(cr) — (kc)r = c(kr) for all k e K, r e A, c e N. Simple /C-algebras, primitive /C-algebras, the Jacobson radical of a AT-algebra, semisimple /^-algebras, etc. are now defined in the same way the corresponding con cepts for rings were defined, with algebra ideals, modules, homomorphisms, etc. in place of ring ideals, modules, and homomorphisms. In order to carry over the results of Sections 1-3 to /^-algebras (in particular, the Wedderburn-Artin Theorems) the following two theorems are helpful.
Theorem 5.2. Let A be a K-algebra. (i) A subset I of A is a regular maximal left algebra ideal if and only if I is a regular maximal left ideal of the ring A. (ii) The Jacobson radical of the ring A coincides with the Jacobson radical of the algebra A. In particular A is a semisimple ring if and only if A is a semisimple algebra. REMARK. Theorem 5.2 is trivial if A has an identity since algebra ideals and ring ideals coincide in this case. PROOF OF 5.2. (i) If / is a regular maximal left ideal of the ring A, it suffices to show that ki CZ / for all k e K. Suppose ki $Zl 1 for some k e K. Since r(kl) = k(rl) by Definition 5.1(i), / + ki is a left ideal of A that properly contains /. Therefore, A = I -f ki by maximality. By hypothesis there exists e e A such that r — re el for all r e A. Let e = a + kb (a,b £ I). Then
e2 = e(a + kb) = ea -f e(kb) = ea + (ke)b e I. Since e — e2 e I and e2 £ /, we must have eel. Consequently, the fact that r — re el for all r e A implies A — I. This contradicts the maximality of I. Therefore, ki CZ / for all k £ K.
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Conversely let / be a regular maximal left algebra ideal and hence a regular left ideal of the ring A. By Lemma 2.4 I is contained in a regular maximal left ideal h of the ring A. The previous paragraph shows that h is actually a regular left algebra ideal, whence I = h by maximality. (ii) follows from (i) and Theorem 2.3(ii). ■
Theorem 5.3. Let A be a K-algebra. Every simple algebra A-module is a simple module over the ring A. Every simple module M over the ring A can be given a unique K-module structure in such a way that M is a simple algebra A-module. PROOF. Let TV be a simple algebra /4-module, whence AN ^ 0. If M is a submodule of N, then ANi is an algebra submodule of N, whence ANX = N or ANX — 0. If ANi = N, then Ni = N. If ANY = 0, then Nx CI D = (c e N | /4c = 0J. But D is an algebra submodule of N and D ^ N since AN ^ 0.. Therefore D = 0 by sim plicity, whence M = 0. Consequently, N has no proper submodules and hence is a simple module over the ring A. If M is a simple module over the ring A, then M is cyclic, say M = Ac (c e M), by Remark (iii) after Definition 1.1. Define a K-module structure on M = Ac by
k(rc) = (kr)c, {k e K, r e A). Since kr e A, (kr)c is an element of Ac = M. In order to show that the action of K on M is well defined we must show that
rc = nc => (kr)c = (kr^c, (k e K; r,rt e A). Clearly it will suffice to prove
rc = 0 => (kr)c = 0, (k e K, r s A). Now by the proof of Theorem 1.3, M = A/I where the regular maximal left ideal I is the kernel of the map A —* Ac = M given by x xc. Consequently, rc = 0 implies r e I. But I is an algebra ideal by Theorem 5.4, whence kr s I. Therefore (kr)c = 0 and the action of K on M is well defined. It is now easy to verify that M is a K-module and an algebra ^-module. The K-module structure of M is uniquely determined since any K-module structure on M that makes M = Ac an /4-module necessarily satisfies k(rc) = (kr)c for all k e K, r e A. ■
Theorem 5.4. A is a semisimple left Artinian K-algebra if and only if there is an isomorphism ofti-algebras A ^ MatniDj X MatniD2 X • • • X Mat„tT>t, where each ns is a positive integer and each D, a division algebra over K. REMARK. Theorem 5.4 is valid for any semisimple finite dimensional algebra A over a field K since any such A is left Artinian (Exercise 2). SKETCH OF PROOF OF 5.4. Use Theorems 5.2 and 5.3 and Exercises 3 and 4 to carry over the proof of the Wedderburn-Artin Theorem 3.3 to K-algebras. ■
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The remainder of this section deals with selected topics involving algebras over a field. We first obtain a sharper version of Theorem 5.4 in case K is an algebraically closed field and finally we consider group algebras over a field. If A is a nonzero algebra with identity over a field K, then the map a : K —> A, de defined by k\-^k\A, is easily seen to be a homomorphism of A-algebras. Since £*(1*) = 1 a 9^ 0, ker a j* K. But the field AT has no proper ideals, whence Ker a =0. Thus a is a monomorphism. Furthermore the image of a lies in the center of A since for all k e K, r e A:
a(k)r = (klA)r = k{\Ar)\A = (1 Ar){k\A) = ra(k). Consequently we adopt the following convention:
If A is a nonzero algebra with identity over a field K, then K is to be identified with Im a and considered to be a subalgebra of the center of A. Under this identification the A'-module action of K on A coincides with multiplica tion by elements of the subalgebra Kin A since ka = (k\jda = a(k)a.
Definition 5.5. An element a of un algebra A over a field K is said to be algebraic over K if a is the root of some polynomial in K[x], A is said to be an algebraic algebra over K if every element of A is algebraic over K. EXAMPLE. If >1 is finite dimensional then A is an algebraic algebra. For if dimA-/4 = n and as A, then the n + 1 elements a,a2,az, . . ., an+l must be linearly dependent. Thus kia + k^a2 + ■ • • + kn+ian+i = 0 for some ki e K, not all zero. Thus f{a) = 0 where / is the nonzero polynomial kYx + k^x2 + ■ ■ ■ + kn+ixn+l e K[x]. EXAMPLE. The algebra of countably infinite matrices over a field K with only a finite number of nonzero entries is an infinite dimensional simple algebraic algebra (Exercise 5). REMARK. The radical of an algebraic algebra is nil (Exercise 6).
Lemma 5.6. If D is an algebraic division algebra over an algebraically closedfield K, then D = K. PROOF. K is contained in the center of D by the convention adopted above. If a e D, then f(a) = 0 for some /e K[xJ. Since K is algebraically closed f(x) = k(x — &i)(x — k2)- ■ (x — kn) (k,ki e K; k 9^ 0), whence 0 = f(a) = k(a — ki)(a — k2) ■ ■ ■ (a — kn). Since D is a division ring, a — ki = 0, for some /. Therefore a = ki£ K and thus D C K. ■
Theorem 5.7. Let A be a finite dimensional semisimple algebra over an algebraically closed field K. Then there are positive integers ni, . . . , nt and an isomorphism of K-algebras
A = MatniK X • ■ ■ X MatntK.
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PROOF. By Theorem 5.4 (and the subsequent Remark) A = Mat„,A X Mat„2A X • • ■ X MatntA where each A is a division algebra over K. Each A is necessarily finite dimensional over K; (otherwise MatKiA and hence A would be infinite dimensional). Therefore A = K for every i by Lemma 5.6. ■ A great deal of research over the years has been devoted to group algebras over a field (see p. 227). They are useful, among other reasons, because they make it possible to exploit ring-theoretic techniques in the study of groups.
Proposition 5.8. (Maschke) Let K(G) be the group algebra of a finite group G over a field K. IfK has characteristic 0, then K(G) is semisimple. 7/K has prime characteristic p, then K(G) is semisimple if and only if p does not divide |G|. SKETCH OF PROOF. Suppose char K = 0 or p,-where p)(\G\. If B is any /^-algebra with identity (in particular K(G)), verify that there is a well-defined mono morphism of A'-algebras a :B —> Homk(B,B) given as follows: a(b) is defined to be the map m : B —> B, where m,(x) = bx. If g e G, we denote the element 1 Kg of K(G) simply by g. By definition K(G) is a tf-vector space with basis X = {g | g e G) and finite dimension n = |G|. For each u e K(G) let Mu be the matrix of au relative to the basis X. Let g e G with g ^ e. Then for all gi e G, aff(gi) = ggi gi (since G is a group). Thus aB simply permutes the elements of the basis X and leaves no basis element fixed. Consequently, the matrix Me of a„ relative to the basis X may be obtained from the identity matrix /„ by an appropriate permutation of the rows that leaves no row fixed (see Theorem VII. 1.2). Recall that the trace, Tr Mu, is the sum of the main diagonal entries of Mu (see p. 369). It is easy to see that (i) Tr M0 = 0 for g e G, g ^ 2; (ii) Me = L, whence Tr Me = nlK', (iii) if u = k\g\ H---1- kngn e K(G), then n
n
au = kiO.Bi and Tr Mu = ^ ki Tr Mei. i=I i= 1 If the radical J of K(G) is nonzero, then there is a nonzero element v aj with v = kigi H---- (- kngn. We may assume gi = e and k\ = Ik (if not, replace v by ki~lgi lv, where ki 0, and relabel). Since K(G) is finite dimensional over K, K(G) is left Artinian (Exercise 2). Consequently J is nilpotent by Proposition 2.13 (for algebras). Therefore vbJ is nilpotent, whence av is nilpotent. Thus by Theorem VII.1.3 Mv is a nilpotent matrix. Therefore Tr Mv = 0 (Exercise VII.5.10). On the other hand (i)—(iii) above imply n
n
Tr Mv = ^ Tr Mei = \K Tr Me + J] kt Tr Mui i=l i=2 = Tr Me + 0 = n\K. But h1 a- ^ 0 since char A: = 0 orchard = p and/; does not divide |G| = n. This is a contradiction. Therefore 7=0 and K(G) is semisimple. Conversely suppose char K = p and p | «. Let be the sum in K(G) of all the ele ments of the basis X; that is, w = gi + g2 H---- h gn e K(G). Clearly w ^ 0. Verify
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that wg = gw for all g e G, which implies that w is in the center of K(G). Show that w2 = nw = («1 a)k', whence vv2 = 0 (since p \ n). Thus (K(G)w)(K(G)w) = 0 so that the nonzero left ideal K(G)w is nilpotent. Since K(G)w CZ / by Theorem 2.12, J ^ 0. Therefore K(G) is not semisimple. ■ The following corollary (with K the field of complex numbers) is quite useful in the study of representations and characters of finite groups.
Corollary 5.9. Let K(G) be the group algebra of a finite group G over an algebraically closed field K. If char K = 0 or char K = p and p/f |G|, then there exist positive integers ni,. . . , nt and an isomorphism ofK-algebras K(G) 9* MatnlK X • • • X Mat^K.
PROOF. Since G is finite, K(G) is a finite dimensional AT-algebra and hence left Artinian (Exercise 2). Apply Theorem 5.7 and Proposition 5.8. ■
EXERCISES Note: K is always a commutative ring with identity and A a A'-algebra. 1. The Q-algebra A of Exercise IV.7.4 is a left Artinian Q-algebra that is not a left Artinian ring. 2. A finite dimensional algebra over a field K satisfies both the ascending and de scending chain conditions on left and right algebra ideals. 3. (a) If M is a left algebra /1-module, then G(M) = {r e A \ rc = 0 for all c e M) is an algebra ideal of A. (b) An algebra ideal P of A is said to be primitive if the quotient algebra R/P is primitive (that is, has a faithful simple algebra R/P-module). Show that every primitive algebra ideal is a primitive ideal of the ring A and vice versa. 4. Let M be a simple algebra ,4-module. (a) D = is a division algebra over K, where HomA(M,M) de notes all endomorphisms of the algebra ,4-mOdule M. (b) M is a left algebra D-module. (c) The ring HornD(M,M) of all D-algebra endomorphisms of Mis a ^-algebra. (d) The map A —> Hom/)(A/,M) given by r\-> a.T (where ctT(x) — rx) is a ^-algebra homomorphism. 5. Let A be the set of all denumerably infinite matrices over a field K (that is, ma trices with rows and columns indexed by N*) which have only a finite number of nonzero entries. (a) A is a simple ^-algebra. (b) A is an infinite dimensional algebraic A'-algebra. 6. The radical J of an algebraic algebra A over a field K is nil. [Hint: if re J and knrn + kn^irn l + ■ • • + ktr‘ = 0 (kt ^ 0), then r1 = r‘u with u = —krlknrn~l — ■ ■ • —krlkt+ir, whence —u is right quasi-regular, say — u + v — uv = 0. Show that 0 = /-'( — « + v — uv) = —/■'.]
CHAPTER IX THE STRUCTURE OF RINGS
456
7. Let A be a A'-algebra and C the center of the ring A. (a) C is a A'-subalgebra of A. (b) If K is an algebraically closed field and A is finite dimensional semisimple, then the number t of simple components of A (as in Theorem 5.7) is precisely dim^C.
6. DIVISION ALGEBRAS We first consider certain simple algebras over a field and then turn to the special case of division algebras over a field. We show that the structure of a division algebra is greatly influenced by its maximal subfields. Finally the Noether-Skolem Theorem (6.7) is proved. It has as corollaries two famous theorems due to Frobenius and Wedderburn respectively (Corollaries 6.8 and 6.9). The tensor product of algebras (Section IV.7) is used extensively throughout this section.
Definition 6.1. An algebra A with identity over a field K is said to be central simple if A is a simple K-algebra and the center of A is precisely K. EXAMPLE. Let D be a division ring and let K be the center of D. It is easy to verify that if d is a nonzero element of K, then d l e K. Consequently A" is a field. Clearly D is an algebra over K (with K acting by ordinary multiplication in D). Furthermore since D is a simple ring with identity, it is also simple as an algebra. Thus D is a central simple algebra over K. Recall that if A and B are A'-algebras with identities, then so is their tensor prod uct A (X)A- B (Theorem IV.7.4). The product of a (x) b and aT (x) b\ is aai (X) bb\. Here and below we shall denote the set j \A (x) b \ b z B | by 1 ,t (x)a Z?and j a (x) ly, | a z A ! by A (X)A- 1 b. Note that A (x)A B = (A (x)K 1b)(1a B); see p. 124.
Theorem 6.2. If A is a central simple algebra over a field K and B is a simple K-algebra with identity, then A (x)k B is a simple K-algebra. PROOF. Since B is a vector space over K, it has a basis Y and by Theorem n
IV.5.11 every element u of A (x)K B can be written
53 °i ® w^b y, e Y and the at
i= 1
unique. If U is any nonzero ideal of A (X)x B, choose a nonzero uzU such that 71
u=
52 ai 0
y* with all a, 0 and n minimal. Since A is simple with identity and i=i AaiA is a nonzero ideal, AaiA = A. Consequently there are elements ru . . . , t
,s t s A such that 1^ = 52 Since V is an ideal, the element v = t j= l 53 (fj ® 1 b)u(s3 0 Is) is in U. Now i=l rt,S\,...
ria'si-
6. DIVISION ALGEBRAS
v
457
= 52j i(ni j® 1b)(52 < ® -ViX-sj ® is) = 52 (52 '■jtf.-s,) (x) a
n
=
52
t where a{ = 52
r^isJ
___
n
® + 52 (52 nciiSj) (x) y{ = i A 0 yi + 52 (8) y*> j i =2
ji=2
By the minimality of n, at ^ 0 for all / > 2. If a e A, then
3=1
the element w = (a (X) lB)u — v(a (x) 1B) is in U and
w = (a (X) y + 52 aa' ® y'ij — ® 3>i + ^2 °ia ® yi^ x = 52 (a°i — «.a) (x)y». i=2
By the minimality of«, w = 0 and aa, — a,a = 0 for all / > 2. Thus oo,- = UiCi for all aeA and each ai is in the center of A, which by assumption is precisely K. Therefore n V =
n
lA (x) yt + 52 «*■ ® ^ (X) yi + 52 1^ ® Wi = U (x) 6, i=2 i=2
where b = y\ + a2yz + ■ —h e P. Since each a, 5^ 0 and the y, are linearly in dependent over K,b 5* 0. Thus, since B has an identity, the ideal BbB is precisely B by simplicity. Therefore,
I a (XV B = 1 ^ (X) BbB = (1 a ®k B)(1a ® b)(\A ®K B) = (1^ (g)/c B)v{\
a
®k B) C {/.
Consequently, ^ ®k B = (/I (X)* 1b)(U B) <= (/4 1 B)U d U. Therefore U = A (x)k # and there is only one nonzero ideal of A (x)k B. Since A (x)a- B has an identity 1A (x) 1 B,(A (X)A- Z?)2 ^ 0, whence ^ (X)A- B is simple. ■ We now consider division rings. If D is a division ring and F is a subring of D containing 1D that is a field, F is called a subfield of D. Clearly D is a vector space over any subfield F. A subfield F of D is said to be a maximal subfield if it is not properly contained in any other subfield of D. Maximal subfields always exist (Exer cise 4). Every maximal subfield F of D contains the center K of D (otherwise F and K would generate a subfield of D properly containing F\ Exercise 3). It is easy to see that F is actually a simple A-algebra. The maximal subfields of a division ring strongly influence the structure of the division ring itself, as the following theorems indicate.
Theorem 6.3. Let ID be a division ring With center K and let ¥ be a maximal subfield of D. Then D (X)K F is isomorphic (as a K-algebra) to a dense subalgebra of HomF(D,T>), where D is considered as a vector space over F. PROOF. Homy,(£),D) is an F-algebra (third example after Definition IV.7.1) and hence a A-algebra. For each a e D let aa : D —> D be defined by «„(-*:) = xa. For each c e F let /3C : D —> D be defined by (ic(x) = ex. Verify that aQ,/3c e Hon\F(D,D)
CHAPTER IX THE STRUCTURE OF RINGS
458
and that aa(3c = /3ca0 for all a e D, c e F. Verify that the map D X F —» Honv(D,D) given by (o,c) |—> ao/Sc is /('-bilinear. By Theorem IV.5.6 this map induces a A-module homomorphism 6 : D F —» Homf(Z),D) such that 71
71
G (23 ® Ci) = i=1 i=l ai
23
£ D, cte F).
Verify that 0 is a A'-algebra homomorphism, which is not zero (since 0(1 D (x) 1 D) is the identity map on D). Since D is a central simple and F a simple A'-algebra, D (x)A- F is simple by Theorem 6.2. Since 0^0 and Ker 0 is an algebra ideal, Ker 0 = 0, whence 6 is a monomorphism. Therefore D (x)a- F is isomorphic to the A-subalgebra Im 6 of HomF(D,D). We must show that A = Im 6 is dense in Homj,(D,D). D is clearly a left module over Homj.(Z),Z)) with fd = f(d)(fe Horn F(D,D), d e D). Consequently D is a left module over A = Im 6. If d is a nonzero element of D, then since D is a division ring,
Ad = {6(u)(d) | « e D (X) KF| =
(23 cidat | / e N*; c» e F; e £)} = D. i
Consequently, D has no nontrivial /4-submodules, whence D is a simple /4-module. Furthermore D is a faithful /4-module since the zero map is the only element /of Hom;,(/),/)) such that fD = 0. Therefore by the Density Theorem 1.12 A is isomor phic to a dense subring of HomA(AD), where A is the division ring Honu(£),£)) and D is a lett A-vector space. Under the monomorphism A —> HomA( £>,£>) the image of /e A is /considered as an element of HomA(Z),Z)). We now construct an isomorphism of rings F = A. Let /? : F —* A = Horru(Z),Z)) be given by c |—> /3C (notation as above). Verify that fic e A and that /S is a monomor phism of rings. If /e A and x e D, then ax = 6(x (x) 1 d) £ A and
fix') = filux) = f[oix(\n)\ = «x(/( Id)) = fi}i>)x = (ic(x), where c = /(lz>). In order to show that 13 is an epimorphism it suffices to prove that ceF; for in that case f(x) = cx = f3c(x) for all x e D, whence / = /Sc = (3(c). If y £ F, then /3y = 0(1 D (x) y) e A and ay = 6(y (x) 1 d) £ A and
cy = fOn)y = ay(f(lD)) = f(ay( lD)) = f(\Dy) = f(ylD) = /(AX Id)) = (3yf(li>) = (3y(c) = yc. Therefore c commutes with every element of F. If c | F, then c and F generate a sub field of D that properly contains the maximal subfield F (Exercise 3). Since this would be a contradiction, we must have c e F . Therefore f3:F ~ A. To complete the proof, let vi, ... ,v n zD and let be a subset of D that is linearly independent over F. We claim that {uu is also linearly inden
pendent over A. If
23 = 0, (gi e A), tiien i=i
0 23 EM = 23 &ci(ui) = 23 ciu^ =
where c, e F and gi = (3(ci) = (3ei. The F-linear independence of {uu . .., un} implies that every = 0, whence gi = 13(0) = 0 for all /. Therefore \uu . . ., u„] is linearly independent over A. By the density of A in Hoitia(£>,/)) (Definition 1.7), there exists h £ A such that h(u{) = Vi for every /. Therefore A is dense in Hom/(AZ)). ■ Theorem 6.3 has an interesting corollary that requires two preliminary lemmas.
6. DIVISION ALGEBRAS
459
Lemma 6.4. Let A be an algebra with identity over a field K and F a field containing K; then A (x)k F is an ¥-algebra such that dim^A = dim?(A (X)K F). SKETCH OF PROOF. Since F is commutative and a K-F bimodule, A (x)K F is a vector space over F with b(a(x) bi) = (a (x) b\)b = a (X) bib (a e A; b,b\ e F; see Theorem IV.5.5 and the subsequent Remark). A (X)A- Fis a K-algebra by Theo rem IV.7.4 and is easily seen to be an F-algebra as well. If A" is a basis of A over K, then by (the obvious analogue of) Theorem IV.5.11 every element of A (x)A F can be written
52 0 d = 52 (*»' iii
0
1
/ )c' = 52 (X) 1/-) (jfi e A'; Ci s F),
with the elements xt- and c, uniquely determined. It follows that X(g)K If
= U® 1^ |
xeX]
is a basis of A 0k F over F. Clearly dim^/l = jA'I = lA'(x)/v-1 jr| = dimA(/4 (X)A- F) ■
Lemma 6.5. Let D be a division algebra over a field K and A a finite dimensional K-algebra with identity. Then D (x)K A is a left Artinian K-algebra. SKETCH OF PROOF. D (X)A A is a vector space over D with the action of ds, D on a generator dy (x)aof D (x)A A given by d(dx 0 a) = dd\ (X) a = (
Theorem 6.6. Let Vi be a division ring with center K and maximal subfield F. Then dimrD is finite if and only if dimK¥ is finite, in which case dim^¥i = dim^¥ and dimKD = (dim kF)2. PROOF. If dimAF is infinite, so is dimAD. If dimAF is finite, then D 0* F is a left Artinian K-algebra by Lemma 6.5. Thus D (x)A F is isomorphic to a dense left Artinian subalgebra of Hom, (D,D) by Theorem 6.3. The proof of Theorem 6.3 shows that this isomorphism is actually an F-algebra isomorphism. Consequently, there is an F-algebra isomorphism D 0A F= HomA(D,D) and n = dim^D is finite by Theorem 1.9. Therefore D(x)A F^ HomA(D,D) = Mat„F by Theorem VII.1.4 (and the subsequent Remark). Lemma 6.4 now implies dimAD = dim/(D (X)A- F) = dim/;(MatrF) = n2 = (dimFDY. On the other hand dimAD = (dimipD)(dimAF) by Theorem IV.2.16. Therefore dimAF = dim/D. ■ Recall that if u is a unit in a ring R with identity, then the map R—>R given by r(-> uru~' is an automorphism of the ring R. It is called the inner automorphism in duced by u.
CHAPTER IX THE STRUCTURE OF RINGS
460
Theorem 6.7. (Noether-Skolem) Let R be a simple left Artinian ring and let K be the center of R (so that R is a K-algebra). Let A and B be finite dimensional simple K-subalgebras of R that contain K. If a : A —-> B is a K-algebra isomorphism that leaves K fixed elementwise, then a extends to an inner automorphism of R. PROOF. It suffices by the Wedderburn-Artin Theorem 1.14 to assume R = HomD(V,V), where V is an ^-dimensional vector space over the division ring D. The remarks after Theorem VII. 1.3 show that there is an anti-isomorphism of rings R = HomD(V,V) —> MatnD. Under this map the center K of R is necessarily mapped isomorphically onto the center of MatnD. But the center of MatnD is isomorphic to the center of D by Exercise VII. 1.3. Consequently we shall identify K with the center of D so that D is a central simple K-algebra. Observe that V is a left K-module with rc — r(v) (v e V; r e R = Homu(V,V)). Since V is a left Z)-vector space, it follows that V is a left algebra module over the K-algebra D (x)K R, with the action of a generator d(x) r of D (X)k R on v e V given by
(d (X) r)v = d(rv) — d(r(v)) = r(dv).
(i)
If A is the subalgebra D (x)*- A of D (x)A R, then V is clearly a left /4-module. Simi larly if B = D (x)k B, then V is a left 5-module. Now the map a = lD (£) a : A —> B is an isomorphism of K-algebras. Consequently, V has a second ^-module structure given by pullback along a; (that is, av is defined to be a(a)t- for v s V, a e A; see p. 170). Under this second ^-module structure the action of a generator d(£)r of A — D (x)k A on v e V is given by
(d(X) r)v = a(d(x) r)v = (c/(X) a(r))v = d(a(r)(v)) = a(r)(dv).
(ii)
By Theorem 6.2 and Lemma 6.5 A is a simple left Artinian K-algebra. Conse quently by Theorem 3.10 there is (up to isomorphism) only one simple /4-module. Now V with either the ^-module structure (i) or (ii) is semisimple by Theorem 3.7. Consequently there are ^-module isomorphisms
V = ZU (corresponding to structure (i)) and
(iii)
iel
V
52 (corresponding to structure (ii)),
=
(iv)
jtJ
with each W} a simple /4-module and Ut = W, for all ij. Since dv = (d (x) 1 R)v (d s D,v s V), every /4-submodule of V is a Z)-subspace of V and every /4-module iso morphism is an isomorphism of D-vector spaces. Since &imDV = n is finite, each Ui, Wj has finite dimension t over D and the index sets I, J are finite, say / = {1,2,..., m) and / = {1,2,. .. , 5). Therefore m\m 52 ^ ) = 52 dimDUi = mt, and 1*1 / 1*1
( dim^F = dime
(52 Wj) = 52 dimo^, = st, \j=1
/
j- 1
m
m
whence m = s. Since Ui = Wj for all i,j,
52 ^
—
52 Wj.
This isomorphism com
6. DIVISION ALGEBRAS
461
bined with the isomorphisms (iii) and (iv) above yields an ^-module isomorphism j3 of V (with the ^-module structure (i)) and V (with the /4-module structure (ii)). Thus for all a e A and v e V
13(av) = a(a)(J3(v)). In particular, for d e D and 5 = d (g) 1A e A,
(3(dv) = f}(3v) = a(d)(j3(v)) = (d(x) 1 B)/?(») = d{3(v), whence j8 e Homd(V,V) = R- Since (3 is an isomorphism, /3 is a unit in R. Further more for r e A and r = 1 z> (x) r e /4,
/3r(v) = /3[r(v)] = (3{rv\ = a(r)(3(v) = (In (X) a(r))(3(v) =
a(r)[(3(v)) = [«(r)/3](r),
whence (3r = a(r)(3 in R = \\omD(V,V). In other words,
(3r(3~l = «(/■) for all r s A. Therefore the inner automorphism of R induced by (3 extends the map a'.A-^B. ■ The division algebra of real quaternions, which is mentioned in the following corollary is defined on pages 117 and 227.
Corollary 6.8. (Frobenius) Let D be an algebraic division algebra over the field R of real numbers. Then D is isomorphic to either R or the field C of complex numbers or the division algebra T of real quaternions. SKETCH OF PROOF. Let K be the center of D and F a maximal subfield. We have R C K C F CI D, with F an algebraic field extension of R. Consequently dim a/7 < dim RF < 2 by Corollary V.3.20. By Theorem 6.6 dim^D = dim a/7 and dirndl) = (dim*-/7)2. Thus the only possibilities are dim^D = 1 and dirriA D = 4. If dimx/} = 1, then D = F, and D is isomorphic to R or C by Corollary V.3.20. If dimx-D = 4, then dim*:/7 = 2 = dimFD, whence K = R and F is isomorphic to C by Corollary V.3.20. Furthermore D is noncommutative; otherwise D would be a proper algebraic extension field of the algebraically closed field C. Since F is iso morphic to C, F = R(/) for some i e F such that i2 = — 1. The map F —> F given by a + bi b-> a — bi is a nonidentity automorphism of F that fixes R elementwise. By Theorem 6.7 it extends to an inner automorphism /3 of D, given by (3(x) = dxdr1 for some nonzero dzD. Since — / = (3(i) = did~r, —id = di and hence id2 = dli. Consequently d2 s. D commutes with every element of F = R(i). Therefore d2 e F; otherwise d2 and F would generate a subfield of D that properly contained the maximal subfield F. Since the only elements of F that are fixed by (3 are the elements of R and (3(
462
CHAPTER IX THE STRUCTURE OF RINGS
Corollary 6.9. (Wedderburn) Every finite division ring D is a field. REMARK. An elementary proof of this fact, via cyclotomic polynomials, is given in Exercise V.8.10. PROOF OF 6.9. Let K be the center of D and F any maximal subfield. By Theorem 6.6 dimA£) = «2, where dim*/7 = n. Thus every maximal subfield is a finite field of order qn, where q = |K|. Hence any two maximal subfields F and F' are iso morphic under an isomorphism /3 :F —> F‘ that fixes K elementwise (Corollary V.5.8). By Theorem 6.7, /3 is given by an inner automorphism of D. Thus F' = aFar1 for some nonzero a e D. lfueD, then K(u) is a subfield of D (Exercise 3). K(u) is contained in some maximal subfield that is of the form aFa~l (for some a e D). Thus D = U aFa~1 0
and D* = (J aF*a(where D*,F* are the multiplicative groups of nonzero ele<mD*
mentsof D, F respectively). This is impossible unless F = D according to Lemma 6.10 below. ■
Lemma 6.10. If G is a finite (multiplicative) group and H is a proper subgroup, then U xHx~l d G. XIG
^
PROOF. The number of distinct conjugates of H is [G : N], where N is the normalizer of H in G (Corollary II.4.4). Since H < N < G and H ^ G, [G : N] < [G : H] and \G : H\ > 1. If r is the number of distinct elements in (J xHx-1, then XeG
r< 1 + (|//| - 1)[G : N) < 1 + (|//| - 1)[G : H] = 1 + \H\[G : H] - [G : H] = 1 + |G| - [G : H] < |C|, since [G : H] > 1. ■
EXERCISES 1. If A is a finite dimensional central simple algebra over the field K, then A (xj/v A"r = Matn K, where n = dim/c/1 and A"r is defined in Exercise III. 1.17. 2. If A and B are central simple algebras over a field K, then so is A (x)A B. 3. Let D be a division ring and Fa subfield. If de D commutes with every element of F, then the subdivision ring F(d) generated by F and d (the intersection of all subdivision rings of D containing F and d) is a subfield. [See Theorem V.l .3.] 4. If D is a division ring, then D contains a maximal subfield. 5. If A is a finite dimensional central simple algebra over a field K, then dimA/4 is a perfect square. 6. If A and B are left Artinian algebras over a field K, then A (x)a- B need not be left Artinian. [Hint: let A be a division algebra with center K and maximal subfield B such that dimBA is infinite.]
6. DIVISION ALGEBRAS
463
7. If D is finite dimensional division algebra over its center K and F is a maximal subfield of D, then there is a /('-algebra isomorphism D (xk F = Mat,,/7, where n = dinvD. 8. If A is a simple algebra finite dimensional over its center, then any automorphism of A that leaves the center fixed elementwise is an inner automorphism. 9. (Dickson) Let D be a division ring with center K. If a,b e D are algebraic over the field K and have the same minimal polynomial, then b = dad~l for some ds. D.
CHAPTER X
CATEGORIES
This chapter completes the introduction to the theory of categories, which was begun in Section 1.7. Categories and functors first appeared in the work of Eilenberg-MacLane in algebraic topology in the 1940s. It was soon apparent that these concepts had far wider applications. Many different mathematical topics may be interpreted in terms of categories so that the techniques and theorems of the theory of categories may be applied to these topics. For example, two proofs in disparate areas frequently use “similar” methods. Categorical algebra provides a means of precisely expressing these similarities- Consequently it is frequently possible to provide a proof in a cate gorical setting, which has as special cases the previously known results from two different areas. This unification process provides a means of comprehending wider areas of mathematics as well as new topics whose fundamentals are expressible in categorical terms. In this book category theory is used primarily in the manner just described — as a convenient language of unification. In recent years, however, category theory has begun to emerge as a mathematical discipline in its own right. Frequently the source of inspiration for advances in category theory now comes to a considerable extent from within the theory itself. This wider development of category theory is only hinted at in this chapter. The basic notions of functor and natural transformation are thoroughly dis cussed in Section 1. Two especially important types of functors are representable functors (Section 1) and adjoint pairs of functors (Section 2). Section 3 is devoted to carrying over to arbitrary categories as many concepts as possible from well-known categories, such as the category of modules over a ring. This chapter depends on Section 1.7, but is independent of the rest of this book, except for certain examples. Sections 1 and 3 are essentially independent. Section 1 is a prerequisite for Section 2. 464
1. FUNCTORS AND NATURAL TRANSFORMATIONS
465
1. FUNCTORS AND NATURAL TRANSFORMATIONS As we have observed frequently in previous chapters the study of any mathe matical object necessarily requires consideration of the “maps” of such objects. In the present case the mathematical objects in question are categories (Section 1.7). A functor may be roughly described as a “map” from one category to another which preserves the appropriate structure. A natural transformation, in turn, is a “map” from one functor to another. We begin with the definition of covariant and contra variant functors and numer ous examples. Natural transformations are then introduced and more examples given. The last part of the section is devoted to some important functors in the theory of categories, the representable functors. The reader should review the basic properties of categories (Section 1.7), par ticularly the notion of universal object (which is needed in the study of representable functors). We shall frequently be dealing with several categories simultaneously. Consequently, if A and B are objects of a category C, the set of all morphisms in C from A to B will sometimes be denoted by homeC4,£?) rather than hom(/4,B) as previ ously.
Definition 1.1. Let C and 3D be categories. A covariant functor T from G to 2D (de noted T : C —»SD) is a pair of functions (both denoted by T), an object function that assigns to each object C of G an object T(C) of 3D and a morphism function which as signs to each morphism f :C—+C of & a morphism T(f) : T(C) -» T(C')
of 3), such that (i) T(1 r) = It(o for every identity morphism lc of C; (ii) T(g o f) = T(g) ° T(f) for any two morphisms f, g of G whose composite g °f is defined. EXAMPLE. The (covariant) identity functor /e : C —» C assigns each object and each morphism of the category C to itself. EXAMPLE. Let R be a ring and A a fixed left /?-module. For each R-module C, let 7X0 = Horn 1{(A,C). For each R-module homomorphism / : C —> C', let T(f) be the usual induced map/ : Homj{(A,C)—> HomJe(A,C') (see the remarks after Theo rem IV.4.1). Then r is a covariant functor from the category of left R-modules to the category of abelian groups. EXAMPLE. More generally, let A be a fixed object in a category G. Define a co variant functor hA from G to the category S of sets by assigning to an object C of G the set /;,i(0 = hom(A,C) of all morphisms in G from A to C. If / : C —>• C' is a morphism of G, let h A (f) ’■ hom(/l,0 —»hom (A,C') be the function given by g(—► fog (g e hom(A,C)). The functor hA, which will be discussed in some detail below, is called the covariant hom functor. EXAMPLE. Let Fbe the following covariant functor from the category of sets to the category of left modules over a ring R with identity. For each set X, F(X) is
466
CHAPTER X CATEGORIES
the free /?-module on X (see the Remarks after Theorem IV.2.1). If / :X —*X' is a function, let F( f) : F(X) —> F(X') be the unique module homomorphism / : F(X) -> F(X') such that fi = f where /' is the inclusion map X —> F(X) (Theorem IV.2.1). EXAMPLE. Let C be a concrete category (Definition 1.7.6), such as the category of left /^-modules or groups or rings. The (covariant) forgetful functor from G to the category S of sets assigns to each object A its underlying set (also denoted A) and to each morphism / : A —> A' the function / : A —> A' (see Definition 1.7.6).
Definition 1.2. Let G and 3D be categories. A contravariant functorS from G to 3D (idenoted S : G —> 3D) is a pair o f functions (both denoted by S), an object function which assigns to each object C ofG an object S(C) o/3D and a morphism function which as signs to each morphism f:C—► C'o/Cc/ morphism S(f) : S(C') -» S(C)
of 3D such that (i) S(lc) = 1 scC) for every identity morphism lc of G; (ii) S(g of) = S(f) ° S(g) for any two morphisms f,gof G, whose composite g°f is defined. Thus the morphism function of a contravariant functor S : G —► 3D reverses the direction of morphisms. EXAMPLE. Let R be a ring and B a fixed left /^-module. Define a contravariant functorS from the category of left /?-moduIes to the category of abelian groups by defining S(C) = HomR(C,Z?) for each /^-module C. If / : C —> C' is an /?-module homomorphism, then S(f) is the induced map / : HomJi(C',B) —> HomA*(C,/?) (see the Remarks after Theorem IV.4.1). EXAMPLE. More generally, let B be a fixed object in a category G. Define a con travariant functor hB from G to the category S of sets by assigning to each object C of C the set hB(C) = hom(C,fi) of all morphisms in G from C to B. If / : C —* C' is a morphism of G, let hB( f) : hom(C',Z?) —* hom(C,Z?) be the function given by g |—> g ° / (g s hom(C,B)). The functor hB is called the contravariant hom functor. The following method may be used to reduce the study of contravariant functors to the study of covariant functors. If G is a category, then the opposite (or dual) cate gory of G, denoted ©op, is defined as follows. The objects of Cop are the same as the objects of C. The set homeup(/J,fi) of morphisms in Cop from A to Bis defined to be the set home (B,A) of morphisms in G from B to A. When a morphism /e home(Z?,/1) is considered as a morphism in homc°p(yf,Z?), we denote it by /np. Composition of morphisms in Cop is defined by o y'°p = (f o IfS :C—‘JDisa contravariant functor, let S : Cop —» S) be the unique covariant functor defined by 5(A) == 5(A) and S(/op) = S(f)
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467
for each object A and morphism /of Cop. Conversely, it is easy to verify that every co variant functor on Cop arises in this way from a contravariant functor on G. Recall that every statement involving objects and morphisms in a category has a dual statement obtained by reversing the direction of the morphisms (see p. 54). It follows readily that a statement is true in a category G if and only if the dual state ment is true in Cop. Consequently a statement involving objects, morphisms and a contravariant functorS on G is true provided the dual statement is true for the co variant functor S on eop. For this reason many results in the sequel will be proved only for covariant functors, the contravariant case being easily proved by dualization. In order to define functors of several variables, it is convenient to introduce the concept of a product category. If G and 3D are categories, their product is the category 6X!D whose objects are all pairs (C,D), where C and D are objects of G and 3D re spectively. A morphism (C,D) —» (C',D') of 6 X 3D is a pair (fg), where / : C —* C' is a morphism of G and g : D —> D’ is a morphism of 3D. Composition is given by (f,g') ° (fg) = (f ° f g' ° g). The axioms for a category are readily verified. The product of more than two categories is defined similarly. Functors of several variables are defined on an appropriate product category. Such a functor may be covariant in some variables and contravariant in others. For example, if C,3D,£ are categories, a functor T of two variables (contravariant in the first and covariant in the second variable) from C X 3D to £ consists of an object function, which assigns to each pair of objects (C,D) in C X 3) an object T(C,D) of £, and a morphism function, which assigns to each pair of morphisms / : C —> C', g:D->Z)'ofex£)a morphism of £:
T(fg) : T(C',D) -» T(C,D'), subject to the conditions: (i) T(lc,lv) = 1 Tu:.D) for all (C,D) in G X 3D; (ii) T(f'°fg'og) = T(fg') o T(f',g), whenever the compositions f'°fg'°g are defined in G and £) respectively. The second condition implies that for each fixed object C of G the object function T(C, —) and the morphism function r(lc, —) con stitute a covariant functor 3) —> £. Similarly for each fixed object D of 3D, T( — ,D) and T( — ,l/>) constitute a contravariant functor G —> £. EXAMPLE. Hom^(-,-) is a functor of two variables, contravariant in the first and covariant in the second, from the category 911 of left K-modules1 to the cate gory of abelian groups. EXAMPLE. More generally let 6 be any category. Consider the functor that assigns to each pair (A,B) of objects of G the set home(/4,£) and to each pair of mor phisms / : A —» A', g : B —> B' the function hom(/#j : hom^.^',/?) —> homc(/4,fi') given by h (—»g ° h ° / Then homc( —, — ) is a functor of two variables from G to the category S of sets, contravariant in the first variable and covariant in the second. Note that for a fixed object A, home(/l,—) is just the covariant hom functor hA and hA(g) = hom(lA,g). Similarly for fixed B home( — ,B) is the contravariant hom functor hB and h B (f) = hom(/,l/<). Strictly speaking Horrid—, —) is a functor on 311 X 3TC, but this abuse of language is common and causes no confusion.
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EXAMPLE. Let A" be a commutative ring with identity. Then the functor given by T(AU ..., A„) = Ai (g)K■ ■ -®A- A„ T(f,...,fn)=f®---®fn
is a functor of n covariant variables from the category of /^-modules to itself. If Tx : 6 —» 3D and T2 : 3D —» 8 are functors, then their composite (denoted 7’27'i) is the functor from C to S with object and morphism functions given by
c -»7’2(7’,(0); /-> rmf)).
TiT\ is covariant if 7\ and r2 are both covariant or both contravariant. T/Ti is contravariant if one Ti is covariant and the other is contravariant.
Definition 1.3. Let Q and 3D be categories and S:C—>3D, T :C—>D covariant functors. A natural transformation a : S —> T is a function that assigns to each object C of & a morphism ac : S(C) —»T(C) of fO in such a way that for every morphism f : C —> C of C, the diagram 5(C)—^—^T(C) js(/) | T(f)
S(C')-----t^T(C') ac in S) is commutative. If ac is an equivalence for every C in C, then a is a natural iso morphism (or natural equivalence) of the functors S and T. A natural transformation [isomorphism) /3 : S —» T of contravariant functors S,T : C —> 3D is defined in the same way, except that the required commutative dia gram is:
T(C)
COS(f)
5(0-
T(f) T(C'), fic'
for each morphism / : C —* C' of 6. REMARKS. The composition of two natural transformations is clearly a natural transformation. Natural transformations of functors of several variables are defined analogously. EXAMPLE. If T : C —■» C is any functor, then the assignment C\—> lrco defines a natural isomorphism IT : T —» T, called the identity natural isomorphism.
1. FUNCTORS AND NATURAL TRANSFORMATIONS
469
EXAMPLE. Let ^11 be the category of left modules over a ring R and F: 317 —» 3TC the double dual functor, which assigns to each module A its double dual module //** = Homff(Hom;i(^,R),R)- For each module A let dA '■ A —* A** be the homo morphism of Theorem IV.4.12. Then the assignment A |—> 0A defines a natural trans formation from the identity functor Z$n to the functor F (Exercise IV.4.9). If the cate gory ;W is replaced by the category V of all finite dimensional left vector spaces over a division ring and F considered as a functor V —> V, then the assignment A\^> dA (A e V) defines a natural isomorphism from to F by Theorem IV.4.12 (iii). Also see Exercise 5. Natural transformations frequently appear in disguised form in specific cate gories. For example, in the category of /^-modules (and similarly for groups, rings, etc.), a statement may be made that a certain homomorphism is natural, without any mention of functors. This is usually a shorthand statement that means: there are two (reasonably obvious) functors and a natural transformation between them. EXAMPLE. If 7? is a unitary left module over a ring R with identity, then there is a natural isomorphism of modules an : R (x)R F = 7? (see Theorem IV.5.7). It is easy to verify that for any module homomorphism /: B —> C, the diagram
R® r B
B f
R®rC
c
is commutative. Thus the phrase “natural isomorphism” means that the assignment B |—> an defines a natural isomorphism a : F —> 7^, where 3TC is the category of uni tary left 7?-modules and F : 371 —* 3TC is given by B |—> 7? (x)R B and /(-» 1« (x) /. EXAMPLE. If A,B,C are left modules over a ring R, then the isomorphism of abelian groups : Homft(/l © B,C) = Homft(/l,C) © Homfi(B,C) of Theorem IV.4.7 is natural. One may interpret the word “natural” here by fixing any two variables, say A and C, and observing that for each module homomorphism f\B^>B' the diagram
Homa(>4 © B,C) —^ Hom*04,0 © HomR(B,C) is commutative, where ^ © / ^ © ^ ^ ^ © f i ' i s given by (a,b) H (a,f(b)). Thus 4> defines a natural isomorphism of the contravariant functors 5 and F, where
S(B) = Homri(A © B,C) and T(B) = WomH{A,C) © Hom*(fi,C). One says that the isomorphism is natural in B. A similar argument shows that 4> is natural in A and C as well. Other examples are given in Exercise 4.
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Definition 1.4. Let T be a covariant functor from a category G to the category S of sets. T is said to be a representable functor if there is an object A in G and a natural isomorphism a from the covariant hom functor hA = home(A, — ) to the functor T. The pair (A,a) is called a representation of T and T is said to be represented by the object A. Similarly a contravariant functor S : G —> S is said to be representable if there is an object B of 6 and a natural isomorphism fi : hB —» S, where hB = homc( —,B). The pair (B,/3) is said to be a representation of S. EXAMPLE. Let A and B be unitary modules over a commutative ring K with identity and for each K-module C let T(C) be the set of all K-bilinear maps A XB^C. If / : C —> C' is a K-module homomorphism, let T( f) : T(C) —> T(C') be the function that sends a bilinear map g : A X B —» C to the bilinear map fg : A X B —» C. Then Tis a covariant functor from the category mi of K-modules to the category S of sets. We claim that T is represented by the K-module A (x)K B. To see this, define for each K-module C a function «c : HornK(A 0A- B,C) T(C) by cic(f) = fi, where i: A X B —► A (X)A'Z? is the canonical bilinear map (see p. 211). Now ac(f) : A X -B —> C is obviously bilinear for each fe HomA(/4 (§)A- B,C). By Theorem IV.5.6 every bilinear map g : A X B —C is of the form gi for a unique K-module homomorphism g : A (X)A B —► C. Therefore ac is a bijection of sets (that is, an equivalence in the category S). It is easy to verify that the assignment Ch«c defines a natural isomorphism from h A ® K n to T, whence (A (x)A B,a) is a representa tion of T. It is not just coincidence that A (x)A- B is a universal object in an appropri ate category (Theorem IV.5.6). We shall now show that a similar fact is true for any representable functor. Let (A,a) be a representation of a covariant functor T : C —* S. Let QT be the category with objects all pairs (C,s), where C is an object of G and s e T(C). A mor phism in Qr from (C,s) to (D,t) is defined to be a morphism / : C —> D of Q such that T(f)(s) = t s T(D). Note that/is an equivalence in GT if and only if /is an equiva lence in 6. A universal object in the category Gt (see Definition 1.7.9) is called a universal element of the functor T. EXAMPLE. In the example after Definition 1.4 the statement that (A (x)A B,a) is a representation of the functor T : Dll —> S clearly implies that for each K-module C and bilinear map / : A X B —> C (that is, for each pair (C,/) with /e r(C)), there is a unique K-module homomorphism / : A (x)/c B —> C such that fi = f (that is, such that n/XO = / with / = aM®Aji(1.4
Lemma 1.5. Let T : 6 —> S be a covariant functor from a category G to the category § of sets and let A be an object of G.
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471
(i) If a : hA —» T is a natural transformation from the covariant hom functor hA to T and u = «a(1a) £ T(A), then for any object C of G and g e home(A,C) «c(g) = T(g)(u). (ii) If u s T(A) and for each object Co/C {3c : home( A,C) —> T(C) is the map de fined by g H T(g)(u), then /3 : hA > T is a natural transformation such that /3a(1 a) = u. PROOF, (i) Let C be an object of C and g e home(/4,C). By hypothesis the diagram CtA !ia(A)
= hom^A,A)-*-T(A)
fUg) J Iia(C)
\
T(g)
= hom e(A,C)-*-T(C) ac
is commutative. Consequently,
ac(g) = aC(g °1a) = ac[hA{g)(\A)} = WMh)] ( \ a ) = (7’(g)«A)(i^) = TOMU)] = T(g)(u). (ii) We must show that for every morphism k : B —» C of Q the diagram
fiii hA(B) = homc(/4,Z?) —^T(B) hA{k) \
| T(k)
hA(C) = homc(/4,C) -►7’(C) 13c is commutative. This fact follows immediately since for any f ehom^A,B)
[/3chA(k)](f) = (3 c (kof) = T(k ° /)(«) = [T(k)T(f)](u) = T(k)[T( f)(u)\ = T(k)[(3ij( f)\ = [T(k)pB](f). Therefore /S is a natural transformation. Finally, &i(1.a) = 7’(1^)(«) = \t{A){u) = u. ■
Theorem 1.6. Let T : C —> &be a covariant functor from a category Q to the category S of sets. There is a one-to-one correspondence between the class X of all representa tions o/T and the class Y of all universal elements ofT, given by (A,a) ► (A,g:a(1a)). REMARK. Since aA '■ home(A,A) —> T(A), a^l^) is an element of T(A). PROOF OF 1.6. Let (A,a) be a representation of T and let a^(l^) = u e T(A). Suppose (B,s) is an object of GT- By hypothesis aB : hA(B) = homc(,4,i?) —» T(B) is a bijection, whence .v = aB(f) for a unique morphism f : A—> B. By Lemma 1.5, T(f)(u) = a>n( f) = s. Therefore, /is a morphism in C7 from (A,u) to (B,s). If g is an other morphism in QT from (A,u) to (B,s) then g e home(A,B) and T(g)(u) = s.
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Consequently, by Lemma 1.5 ceB(g) = T(g)(u) = s = aB(f)- Since aB is a bijection, / = g. Therefore, /is the unique morphism in QT from (A,u) to (B,s), whence (A,u) is universal in QT. Thus (A,u) is a universal element of T. Conversely suppose (A,u) is a universal element of T. Let (3: h A —*Tbe the natural transformation of Lemma 1.5 (ii) such that for any object C of G, /0c : home(A,C) —> T(C) is given by (3c(f) = T(f)(u). If s e T(C), then (C,s) is in CT. Since (A,u) is univer sal in Qt, there exists fehome(A,C) such that s = T(f)(u) = /3C(/). Therefore (3C is surjective. If fSc(fi) = (3c(ft), then T(f)(u) = (3c(f) = /8c(/2) = T(fj(u), whence f and /2 are both morphisms in QT from (A,u) to (C,T(/)(«)) = (C,T(/)(«)). Conse quently, / = fit by universality. Therefore each is injective and hence a bijection (equivalence in S). Thus is a natural isomorphism, whence (A,(3) is a representation of T. To complete the proof use Lemma 1.5 to verify that <{)\p = ly and \p<{) = lx, where
Corollary 1.7. Let T : 6 —► S be a covariant functor from a category G to the category S of sets. If (A,a) and (B,/3) are representations o/T, then there is a unique equivalence f: A —*• B such that the following diagram is commutative for all objects C of Q: ht(C) = homc(fi,C) (3C T(C)
hom(/Ic)
hA(C) = homc04,C) ac PROOF. Let u = a A (l A ) and v — (3b0b)- By Theorem 1.6 (A,u) and (B,v) are universal elements of T, whence by Lemma 1.7.10 there is a unique equivalence / : A —* B in Q such that T(f)(u) = v. Lemma 1.5 (i) implies that for any object C of C and g e home(i?,C) [a'chom(/,lc)](g) = a, (g of) = T(g ° /)(«) = inx)T(f)](u) = T(g)[T(f)(u)] = T(g)(v)
= Mg), so that the required diagram is commutative. Furthermore if f : A —> B also makes the diagram commutative, then for C = B and g = 1 b, 7T/i)(w) = oib (J i ) = «&(1# ° fi) = Mhom (/i,lfl)(l/<)]
=
(3b(\b) = £>•
Therefore f = /by uniqueness. ■
Corollary 1.8. (Yoneda) Let T : C —* S be a covariant functor from a category Q to the category S of sets and let A be an object of C. Then there is a one-to-one corre spondence between the set T(A) and the set Nat( 1ia,T) of all natural transformations from the covariant hom functor lu to the functor T. This bijection is natural in A andT. SKETCH OF PROOF. Define a function # = #A : Nat(hA,T) -> T(A) by a t—> a A (l A ) £ T(A)
1. FUNCTORS AND NATURAL TRANSFORMATIONS
473
and a function $ : T(A) —» Nat(/u,7’) by
u{-*fi, where fi is given by Lemma 1.5 (ii). Verify that and \p4> are the respective identity maps. Therefore \pA is a bijection. The naturality statement of the corollary means that the diagrams
Nat {hA,T)*±^T(A)
N*{f) |
| T(f)
Nat (hB,T)—+T(B) Wb
Nat(hA,T) *±*T(A) N*(a) |
|^
Nat(/u,S) ——*~S(A)
Va
are commutative, where / : A —* B is any morphism of Q, a : T —»S is any natural transformation of functors and N*(f), N*(a) are defined as follows. For each object C of C and fi e Nat(/^,r), : hB(C) = homc(fl,C) -» T(C) is given by g f—> fidg ° /)• The map N*(a) : Nat(hA,T) —> Nat(hA£) is given by
fi H ccfi. m A representable functor is a functor of one variable that is naturally isomor phic to the covariant (or contravariant) hom functor. But for a given category SD, homj)( —, —) is a functor of two variables. We now investigate conditions under which a functor T of two variables is naturally isomorphic to homj)(—,—). We shall deal with the following somewhat more general situation. Let Q and SD be categories and T : C X 3D —* S a functor that is contravariant in the first variable and covariant in the second. If S : C —»SD is a covariant functor, then it is easy to verify that the assignments (C,D) (—» homai(5(C),Z)) and (fg) f- > homx>(S( /),#) de fine a functor 6X3)^S that is contravariant in the first variable and covariant in the second.
Theorem 1.9. Let Q and Si be categories and T a functor from the product category C X 3D to the category S of sets, contravariant in the first variable and covariant in the
second, such that for each object C of <3, the covariant functor T(C,—) : 3D —» S has a representation (Ac,ac). Then there is a unique covariant functor S : C —> 33 such that S(C) = Ac and there is a natural isomorphism from hom^S( —), —) to T, given by acD : /70W;d(S(C),D) -» T(C,D). REMARK ON NOTATION. For each object C of G, Ac is an object in 3D and ac is a natural isomorphism from hom^Ac,—) to T{C—). Thus for each D in .2) there is an equivalence acD : homD(Ac,D) —* T(C,D).
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PROOF OF 1.9. The object function of the functor S is defined by 5(C) = Ac for each object C of 6. The morphism function of 5 is defined as follows. For each object C of 6 aCAC '■ hom^c,^) —;► T(C,Ac) and uc = acAc(\Ac) £ T(C,Ac). By Theorem 1.6 (Ac,uc) is a universal element of the functor T(C, — ). If / : C —> C' is a morphism of Q, let v = T(f\Ac>)(uc) e T(C,Ac). By the universality of (Ac,uc) in D there exists a unique morphism f:A c —* Ac, in S such that
T(\c,f)(uc) = v = T(fl Ac 'Xuc)Define S(f) to be the morphism / Clearly 5(lc) = 1 ac = lsco- If CC' C" are morphisms of C, then by definition 5(g) is the unique morphism g : Ac- —> AC" such that
n\c,g)(uc>) = T(g,\AC"){uC”). Similarly S(g ° /) is the unique morphism h : A c —> Ac- such that
Ti\c,h){uc) = T(g ° f,\Ac")(uc>'). Consequently 5(g) °S(f) = g ° f is a morphism A c —> A C " such that
n\ c ,gof)(uc) = n\c,g)T(\c,f)(uc) = T(\c,g)T(f\Ac’)(uC') = n/gX^cO = T( f 1
Ac")T(\C’
,g){uc)
= T(f\Ac")T(g,\Ac")(Mc") = T(g ° f,\Ac")(uc~)
= n\cMuc). Therefore by the uniqueness property of universal objects in SDrcc.-) we must have S(g)°S{f) = gof=h=S{gof).
Thus 5 : C —> £> is a covariant functor. In order to show that a : homa}(S( — ), — ) —» Tis a natural isomorphism we need only show that for morphisms / : C —» C in C and g : D —> D' in SD the diagram
hom ^AC’,D)
CLC'd
nc,D)
hom (5(/),U) hom 3D IAC,D)
T(fAn) olcd
nc,D)
hom(li4c,g)
hom^AcD1)
n\c,g) c
T(C,Dr)
is commutative. The lower square is commutative since for fixed C,
ac : homage, — ) —> T(C, — ) is a natural isomorphism by hypothesis. As for the upper square let k e hom^Ac',/))Then by Lemma 1.5 (i):
1. FUNCTORS AND NATURAL TRANSFORMATIONS
475
nf\ D )cf D (k) = nf\ D )n\cMu c o = nfk)M = T(\c,k)T(f,\Ac.){uc.) = T(\c,k)T{\cJ)(uc) = T(lc,k o f)(uc) = T(\ c ,k o S(/))(«<■) = acD.(k o S(fj) = ac/rhom(5(/),lD)(/:). ■
EXERCISES Note: In these exercises S is the category of sets and functions; (R is the category of rings and ring homomorphisms; R is a ring; 911 is the category of left /?-modules and /^-module homomorphisms; g is the category of groups and group homomorphisms. 1. Construct functors as follows: (a) A covariant functor g —»S that assigns to each group the set of all its subgroups. (b) A covariant functor (R —* (R that assigns to each ring N the polynomial ring Ar[x]. (c) A functor, covariant in both variables 3H X 9TC —3TC such that
(d) A covariant functor g —» g that assigns to each group G its commutator subgroup G' (Definition II.7.7). 2. (a) If T: G —> 3D is a covariant functor, let Im T consist of the objects {7XC) I C s C) and the morphisms {T(f) : T(C) —> 7'(C/) | / : C —► C' a mor phism ine(. Then show that Im T need not be a category. (b) If the object function of T is injective, then show that Im T is a category. 3. (a) If S : C —* 3D is a functor, let cr(S) = 1 if S is covariant and — 1 if S is con travariant. If 7’: 3D —* 8 is another functor, show that TS is a functor from G to £ whose variance is given by cr(TS) = a(T)a(S). (b) Generalize part (a) to any finite number of functors, Si : Ci —» G2, S2 :G2 —>
e3,.. -, s n : e„ e„+i.
4. (a) If A,B,C are sets, then there are natural bijections: A X B B X A and (A X B) X C -» A X (B X C). (b) Prove that the isomorphisms of Theorems IV.4.9, IV.5.8, IV.5.9, and IV.5.10 are all natural. 5. Let V be the category whose objects are all finite dimensional vector spaces over a field F (of characteristic ^2,3) and whose morphisms are all vector-space isomorphisms. Consider the dual space V* of a left vector space V as a left vector space (see the Remark after Proposition VII.1.10). (a) If
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(d) The isomorphism av is not natural; that is, the assignment V f—> av is not a natural isomorphism from the identity functor /(x) = cx with c ^ 0, ±1^.] 6. (a) Let S : G —»SD and T : G —»SD be covariant functors and a :S —> Ta natural isomorphism. Then there is a natural isomorphism (3 :T—*S such that (3a = I s and a(3 = I t , where I s '-S —>S is the identity natural isomorphism and similarly for I T . [Hint: for each C of G, ac : S(C) —* T(C) is an equivalence and hence has an inverse morphism (3C : T(C) —> S(C).] (b) Extend (a) to functors of several variables. 7. Covariant representable functors from S to S preserve surjective maps. 8. (a) The forgetful functor 3TI —» S (see the Example preceding Definition 1.2) is representable. (b) The forgetful functor g —* S is representable. 9. (a) Let P : S —> S be the functor that assigns to each set Aits power set (set of all subsets) P{X) and to each function f:A~*B the map P(f): P(B) —■* P(A) that sends a subset X of B onto f~\X) d A. Then P is a representable contravariant functor. (b) Let the object function of Q : S —»S be defined by Q(A) = P(A). If /: A —* B, let Q(f) : Q(A) —> Q(B) be given by f(X). Then Q is a covariant functor. Is Q representable? 10. Let (A,a) and (B,(3) be representations of the covariant functors S : G —* S and T: G —> S respectively. If r :5 —*• 7" is a natural transformation, then there is a unique morphism / : A A in G such that the following diagram is commuta tive for every object C of G:
hom(/,lc)
|
home(^,C) 5(C)
J
rc
homc(/i,C)——► T(C) pc
2. ADJOINT FUNCTORS Adjoint pairs of functors are defined and discussed. Although they occur in many branches of mathematics formal descriptions of them are relatively recent. Let 5 : G —> SD and T: SD —» G be covariant functors. As observed in the dis cussion preceding Theorem 1.9, the assignments (C,D) (—+ homx>(S(0,Z)) and hom£>(S(/),g) define a functor G X SD —> S which is contravariant in the first variable and covariant in the second. We denote this functor by hom£>(S(—),—). Similarly the functor home(—,T(—)) :G X © —> S is defined by
(C,D) \—> home(C,7'(Z))) and (fg)\^home(f,T(g)).
2. ADJOINT FUNCTORS
477
Definition 2.1. Let S : C —»SD and T : SD —* C be covariant functors. S is said to be a left adjoint ofT (or T a right adjoint ofS, or (S,T) an adjoint pair) if there is a natural isomorphism from the functor hom^f S(—), —) to the functor home(—,T(—)). Thus if 5 is a left adjoint of T, there is for each C of G and D of SD a bijection
ac.D : homx)(S(0,D) —> home(C,7~(D)), which is natural in C and D. The theory of adjoint functors was first suggested by the following example. EXAMPLE. Let R, S be rings and AR, RBS, Cs (bi)modules as indicated. By Theorem IV.5.10 there is an isomorphism of abelian groups Homs(/i (X)fl B,C) = Homfl(/i, Hom5(fi,C)), which is easily shown to be natural in A and C (also in B). Note that A (x)R B is a right S-module by Theorem IV.5.5 (iii) and Homs(6,C) a right /?-module by Exercise IV.4.4 (c). Let B be a fixed R-S bimodule. Let G be the category of right /?-modules and SD the category of right 5-modules so that home(.Y,Y) = Homfl(A',y) and homs (U,V) = Homs(U, V). Then the isomorphism above simply states that the functor —(x)rB from G to SD is a left adjoint of the functor hom,$(/?,—) from SD to G. EXAMPLE. Let R be a ring with identity and SHI the category of unitary left R-modules. Let T : 9TI —» S be the forgetful functor, which assigns to each module its underlying set. Then for each set X and module A, homg(X.T(A)) is just the set of all functions X —> A. Let F : S —» SHI be the functor that assigns to each X the free /?-module F(X) on the set X (see p. 182). Let ix -X —> F(X) be the canonical map. For each set A'and module A, the map
ax.A : Homfl(F(A'),/i) —> homs(^,7’(/4)) defined by g\-+gix is easily seen to be natural in A' and A. Since F(X) is free on A, ax.A is injective (Theorem IV.2.1 (iv)). Furthermore every function / : X —» T(A) is of the form / = fix for a unique homomorphism / : F(X) —> A (Theorem IV.2.1 (iv)). Con sequently ax.A is surjective and hence a bijection. Therefore Fis a left adjoint of T. Other examples are given in the exercises. There is a close connection between adjoint pairs of functors and representable functors.
Proposition 2.2. A covariant functor T : SD —> C has a left adjoint if and only iffor each object C in G the functor hom^(C,T(—)) : SD —» S is representable. PROOF. If 5 : G —»SD is a left adjoint of T, then there is for each object C of G and D of SD a bijection ac.D
: hom£)CS(C),Z>) —> home(C,7~(D)),
which is natural in C and D. Thus for a fixed C, (5(C),<*
CHAPTER X CATEGORIES
478
Conversely suppose that for each C, Ac is an object of 2D that represents homcCC,^—)). By Theorem 1.9 there is a covariant functor S :C—>£> such that S(C) = Ac and there is a natural isomorphism of functors homj)(5(—), —) —► home( —,7X —)). Therefore S is a left adjoint of T. ■
Corollary 2.3. A covariant functor T : 3D —> C has a left adjoint if and only if there exists for each object C of <3 an object S(C) o/S) and a morphism uc : C —> T(S(C)) such that (S(C),uc) is a universal element of the functor home(C,T( —)) : 3D —> 8. PROOF. Exercise; see Theorem 1.6. ■
Corollary 2.4. Any two left adjoints of a covariant functor T : 3D —» C are naturally isomorphic. PROOF. If Si : Q —» 5) and S2: (B —> 5) are left adjoints of T, then there are natural isomorphisms
a : hom^CS^-),-)—►home(-,7'(-)), /3 : hom^CS^-),-)—>home(-,7X-)). For each object C of C the objects £i(C) and S2(C) both represent the functor hom^C,^—)) by the first part of the proof of Proposition 2.2. Consequently for each object C of (B there is by Corollary 1.7 an equivalence fc : Si(C) —* S2(C). We need only show that/,- is natural in C; that is, given a morphism g : C —* C' of C we must prove that
£,(C)_^S2(C)
S£g) Si(C')—7*~ SifC') Jc
is commutative. We claim that it suffices to prove that
hom^CCWC')) ^m(/r,,1) hom^C'X^C'))
1 homjXS.CO^C')) T-77— homdiS^QMC’))
hom(S2(g),l)
hom(S,(g),l)
hom(7c,l)
is commutative (where 1 = I^cco)- For the image of ls2cc') in one direction is SAg) ° fc and in the other direction fc ° Si(g). Consider the following three-dimensional diagram (in which 1 = ls2(c'>, ctx = ax,s2<.c') and the induced map hom(A:,l) is denoted k for simplicity):
2. ADJOINT FUNCTORS
479
homsX^CC'X^CC'))
homa^CCWCO)-^-
home(C',7S2(C'))
Sjg)
g
hom2D(S2(C),S2(C')) h omj/S i (C) ,S2( C'))
hom^C./X^C')).
We must prove that the left rear rectangle is commutative. The top and bottom tri angles are commutative by Corollary 1.7. The front and right rear rectangles are commutative since a and 0 respectively are natural. Consequently
acSi(g) fc = gotc' fc — gfic — PcS'/g) — etc fcS2ig). Since ac = ac,s2(C”) is injective by hypothesis, we must have Si(g)fc = fcS2(g)Therefore the left rear rectangle is commutative. ■
EXERCISES Note: S denotes the category of sets. 1. If T : C —»■ S is a covariant functor that has a left adjoint, then T is representable. 2. Let 6 be a concrete category and T : C —» S the forgetful functor. If T has a left adjoint F : S —» G, then Fis called a free-object functor and F(X) (X e S) is called a free F-object on X. (a) The category of groups has a free-object functor. (b) The category of commutative rings with identity and identity preserving homomorphisms has a free-object functor. [If A'is finite, use Exercise III.5.11 to define F(X).] 3. Let X be a fixed set and define a functor S :§>—> S by F|—► A X Y. Then 5 is a left adjoint of the covariant hom functor hx = homs(A', —). 4. Let g be the category of groups, (i the category of abelian groups, fF the category of fields, SD the category of integral domains, 911 the category of unitary left A-modules, and (B the category of unitary K-K bimodules (K,R rings with identity). In each of the following cases let T be the appropriate forgetful functor (for example, T —> 2D sends each field F to itself, considered as an integral domain). Show that (S,T) is an adjoint pair. (a) T :&—>§, S : g —> ft, where S(G) = G/G' with G' the commutator sub group of G (Definition II.7.7). (b) T: $ —► 2D, 5 :2D—> ff, where S(D) is the field of quotients of D (Section III.4). (c) T :911—»<2, 51:ft—► 911,where SX/l) = K (x)z A (see Theorem IV.5.5). (d) T : ffi -» 911, S : 9R -» ffi, where S(M) = M (x)z R.
CHAPTER X CATEGORIES
480
3. MORPHISMS A significant part of the elementary theory of categories is the attempt to general ize as many concepts as possible from well-known categories (for example, sets or modules) to arbitrary categories. In this section we extend to (more or less) arbitrary categories the concepts of monomorphisms, epimorphisms, kernels and cokernels of morphisms. NOTATION. Hereafter we shall usually denote the composite of two mor phisms of a category by ^/instead of g ° /as previously. We begin by recalling that a morphism / : C —» D in a category is an equivalence if and only if there is a morphism g : D —* C such that gf = lc and fg = ID. This definition is simply a reflection of the fact that a homomorphism in the category of groups (or rings, or modules, etc.) is an isomorphism if and only if it has a two sided inverse (see Theorem 1.2.3). In a similar fashion we may extend the concepts of monomorphisms and epimorphisms to arbitrary categories as follows.
Definition 3.1. A morphism f : C —» D of a category C is monic (or a monomor phism) if fh = fg => h = g
for all objects B and morphisms g,h e hom(B,C). The morphism f is epic (or an epi morphism) if kf = tf => k = t
for all objects E and morphisms k, t e hom(Y),E). EXAMPLE. A morphism in the category of sets is monic [resp. epic] if and only if it is injective [resp. surjective] (Exercise 1). EXAMPLES. Let G be any one of the following categories: groups, rings, left modules over a ring. If / : C —»D and g,h :B—> C are homomorphisms (that is, morphisms of C), then by Exercise IV.l .2(a), fh = fg implies h = g if and only if /is an injective homomorphism (that is, a monomorphism in the usual sense).2 Thus the categorical definition of monomorphism agrees with the previous definition in these familiar categories. EXAMPLES. Exercise IV.l.2(b) shows that a morphism /in the category of left modules over a ring R is epic if and only if /is a surjective homomorphism (that is, an epimorphism in the usual sense). The same fact is true in the category of groups, but the proof is more difficult (Exercise 2). Thus the categorical definition of epimor phism agrees with the previous definition in these two categories. EXAMPLES. In the category of rings every surjective homomorphism is easily seen to be epic. However, if fg : Q —► R are homomorphisms of rings such that 2The
rings.
Exercise deals only with modules, but the same argument is valid for groups and
482
CHAPTER X CATEGORIES
REMARK. Oc.d is called a zero morphism. PROOF OF 3.4. (Uniqueness) If |OciDl and {0c.d} are two families of mor phisms with the stated properties, then for each pair C,D 0C.D — Od.dOc.D — 0
C,D-
(Existence) For each object A of C let iA ‘ 0 —> A and irA A —» 0 be the unique morphisms. For any /e hom(Z>,JE), fiD = i*: 0 —► Z: by universality. For any g e hom(2?,C) ncg = ttb '■ B —► 0 by couniversality. Define 0 .d to be the composition c
C 0 D. Then for /s hom(Z),£), /° 0C the other case. ■
D
= fiDirc = iettc = Oc.e and similarly in
The final step in extending properties of morphisms in familiar categories to mor phisms in arbitrary categories is to develop reasonable definitions of kernels and co kernels of morphisms. We begin in a somewhat more general setting.
Definition 3.5. Let f : C —► D and g : C —> D be morphisms of a category C. A difference kernel (or equalizer) for the pair (f,g) is a morphism i : B —> C such that: (i) fi = gi; (ii) if h : A —> C is a morphism with fh = gh, then there exists a unique morphism K : A —» B such that iE = h. A difference cokernel (or coequalizer) for the pair (f,g) is a morphism j : D —» E such that: (iii) jf = jg; (iv) z'/k : D —> F is a morphism with kf = kg, then there exists a unique morphism Ic : E —* F such that kj = k. EXAMPLES. In the category S of sets a difference kernel of / : C —» D and g : C —* D is the inclusion map B —» C, where B = (c s C | f(c) = #(c)). The same construction shows that every pair of morphisms has a difference kernel in the cate gories of groups, rings, and modules respectively. EXAMPLE. Let / : G —> H and g : G —» H be homomorphisms of groups. Let N be the smallest normal subgroup of H containing | f(a)g(aYl \ ae G\. Then the canonical epimorphism H —> H/N is a difference cokernel of ( fg) by Theorem 1.5.6.
Proposition 3.6. Let f : C —> D and g : C —> D be morphisms of a category &. (i) 7/i : B —> C is a difference kernel of (f,g), then i is a monomorphism. (ii) 7/i : B —*• C and] : A —> C are difference kernels of(f,g), then there is a unique equivalence h : A —» B such that ih = j. PROOF, (i) Let h, k : F —> B be morphisms such that ih = ik. Then f(ih) = (fi)h = (gi)h = g(ih). Since i is a difference kernel of (fg), there is a unique morphism t : F —»B such that it = ih. But both t = h and t = k satisfy this condi tion, whence h = k by uniqueness. Therefore / is monic.
3. MORPHISMS
483
(ii) By hypothesis there exist unique morphisms h: A^B and k :B —» A such that ih = j and jk = i respectively. Consequently ihk = jk = i = i ° \ B and jkh = ih = j = j ° I a - Since i and j are monomorphisms by (i), hk = l B and kh = 1 a . Therefore h is an equivalence. ■
REMARK. Difference cokernels are epimorphisms and the dual of Proposition 3.6 (ii) holds for difference cokernels.
Suppose that C is a category with a zero object 0 and hence zero morphisms (Proposition 3.4). A kernel of a morphism / : C —> D (if one exists) is defined to be any difference kernel of the pair (fOc.n); it is sometimes denoted Ker /. Definition 3.5 and Propositions 3.4 and 3.6 show that k : K —» C is a kernel off : C —» D if and only if (i) A: is a monomorphism with fk = Qk,c; and (ii) if h : B —> C is a morphism such that fh = 0 .d, then there is a unique mor phism h :B —> K such that kh = h. By Proposition 3.6 K is unique up to equivalence. b
A cokernel t: D —»E of a morphism / : C —» D is defined dually as a difference cokernel of the pair (fOc.v); it is sometimes denoted Coker /. As above t is char acterized by the conditions: (iii) t is an epimorphism with tf = 0c.e\ and (iv) if g : D —» Fis a morphism such that gf= 0 .f, then there is a unique mor phism g : F —> F such that gt = g. c
EXAMPLES. In the categories of groups, rings and modules, a kernel of the morphism / : C —» D is the inclusion map K —» C, where K is the usual kernel, K = j c e C | /(c) = 0}. In the category of modules, the canonical epimorphism D —> D/Im /is a cokernel of /.'
EXERCISES 1. A morphism in the category of sets is monic [resp. epic] if and only if it is injective [resp. surjective]. 2. A morphism / : G —* H in the category of groups is epic if and only if /is a sur jective homomorphism (that is, an epimorphism in the usual sense). [Hint : If /is epic, K = Im /, and j : K-+ H is the inclusion map, then j is epic by Proposition 3.2. Show that /is surjective (that is, K = H) as follows. Let S be the set of left co sets of Kin H\ let T — S U {u\ with u $ S. Let A be the group of all permutations of T. Let t: H —> A be given by t(h)(h'K) = hh'K and t(h)(u) = u. Let s : H —» A be given by ot(K)o, where a e A is the transposition interchanging u and K. Show that s and t are homomorphisms such that sj = tj, whence s = t. Show that hK = K for all h e H; therefore K = H.]
484
CHAPTER X CATEGORIES
3. A commutative diagram
of morphisms of a category <3 is called a pullback for f and /2 if for every pair of morphisms hi : B' —> G, h2 : B' —» G such that /i^i = /2/?2 there exists a unique morphism t \B' —> B such that Ai = gif and h2 = g2f. (a) If there is another pullback diagram for /u/2 with B, in the upper left-hand corner, then B and B, are equivalent. (b) In the pullback diagram above, if /2 is a monomorphism, then so is gi. (c) Every pair of functions f : G —> D, f2: G —> D in the category of sets has a pullback. 4. Show that every pair of functions /, g : C —> D has a difference cokernel in the category of sets. 5. Let f g : C —> D be morphisms of a category G. For each X in Q let Eq(A',/,g) = [h e hom(A',C) | fh = gh\. (a) Eq( —,/g) is a contravariant functor from G to the category of sets. (b) A morphism /: K —> C is a difference kernel of ( fg) if and only if Eq( —, fg) is representable with representing object K (that is, there is a natural isomorphism r : homc(—,K) —* Eq(—,fg)). [Him: show that for h :X—» K, TX(h) = ih, where 1
i = ta( a).]
6. If each square in the following diagram is a pullback and B' —> B is a monomor phism, then the outer rectangle is a pullback. [Hint: See Exercise 3.]
P---- ------ *~B'
\I »
A-----/---- *~B. 7. In a category with a zero object, the kernel of a monomorphism is a zero mor phism.
List of Symbols
SYMBOL
MEANING
Q
field of rational numbers
1
R
field of real numbers
1
C
field of complex numbers
1
=>
implies
1
<=>
if and only if
1
e
is an element of
2
4
is not an element of
2
{x | POc)}
the class of all x such that P(x) is true
2
CZ
is a subclass (or subset) of
2
0
empty set
3
P(A) or 2A
power set of A
3
U Ai hi fl Ai ui B—A
union of the sets Ai
3
intersection of the sets Ai
3
relative complement of A in B
3
A'
complement of A
3
f: A-^B
f is a function from A to B
3
a f(a)
the function /maps a to f(a)
3
/1 S
restriction of the function /to S
4
j
| identity function on the set A {identity element of the ring A
4 115
%J
[ composite function of /and g [ composite morphism of /and g
4 52
Im /
image of the function /
f~KT)
inverse image of the set T
A f
PAGE REFERENCE
485
4, 31 4
LIST OF SYMBOLS
486
SYMBOL MEANING
PAGE REFERENCE
f Cartesian product of sets A and B I direct product of groups A and B
AX B\
6 26
[ is equivalent to 1 is equipollent with
6 15
a equivalence class of a
6
(Cartesian) product of the sets Ail
7
yy product of the family of objects {Ai | i e l )
53
direct product of the family of groups
1
[or rings or modules] [ A i \ i e l )
59, 130, 173
Z set of integers
9
N set of nonnegative integers (natural numbers)
9
N* set of positive integers
a\b
a
9 divides
b
11,135
aJf'b a does not divide b 11, 135 (auck, . . . , an) | greatest common divisor of au . . . , a„ [ideal generated by au . . . , an
11 123
a = b (mod ni) a is congruent to b modulo m
12
cardinal number of the set A | A\ order of the group A determinant of the matrix A aleph-naught
16 24 351 16
£)4* group of symmetries of the square
26
Sn symmetric group on n letters
26
G © H direct sum of additive groups G and H
26
Zm integers modulo m
27
Q/Z group of rationals modulo one
Z(pm)
27
Sylow /7-subgroup of Q/Z 30,37
^ is isomorphic to
30
Ker / kernel of the homomorphism / 31, 119, 170
H < G H is a subgroup of G
31
<X> subgroup generated by the set X cyclic (sub)group generated by a
32 32
H V K, H + K the join of subgroups H and K
33
Q& quaternion group
33
\a\ order of the element a
35
a= r b(modH) ab~l c H
37
a=ib(mo<\H)
a~lb
eH
37
Ha, aH right and left cosets of a
38
[G://] index of a subgroup H in a group G
38
HK [ah \a € H, b € K}
39
N <] G N is a normal subgroup of G
41
G/N factor group of G by Af
42
LIST OF SYMBOLS
487
SYMBOL
MEANING
sgn
sign of the permutation r
48
An
alternating group on n letters
49
Dn
dihedral group of degree n
50
U Ai
disjoint union of the sets Ai
58
t
PAGE REFERENCE
iel
J ui T, Gi ui XI* Gi ui G[m]
weak direct product of the groups Gi direct sum of the groups (or modules) Gi free product of the groups G»
{ueG\mu =0\
60 60, 173 68 77,224
G(p)
\ u e G \ u has order a power of p \
77, 222
Gt
torsion subgroup [submodule] of G
78, 220
Gx
stabilizer of x
89
CH(x)
centralizer of x in H
89
Nh(K)
normalizer of K in H
89
C({r)
center of G
91
Cn(G)
«-th term of ascending central series
100
G'
commutator subgroup of G
102
G(n)
M-th derived subgroup of G
102
End A
endomorphism ring of A
116
binomial coefficient
118
char R
characteristic of the ring R
Rop
opposite ring of R
(X)
ideal generated by the set A'
123
(a)
principal ideal generated by a
123
S~lR
ring of quotients of R by 5
143
RP
localization of R at P
147
R[x]
ring of polynomials over R
149
i?[xi, . . . , x„]
ring of polynomials in n indeterminates over R
151
R[[x]]
ring of formal power series over R
154
deg /
degree of the polynomial /
157, 158
C(/)
content of the polynomial /
162
HomR(A,B)
set of all /^-module homomorphisms A — > B
174
dimjo^
dimension of the /}-vector space V
185
rAs
R-S bimodule A
202
rA,[Ar]
left [resp. right] i?-module A
202
A*
dual module of A
203
f(a)
204
8a
Kronecker delta
204
SdX(/l,fi)
category of middle linear maps on A X B
207
119 122,330
LIST OF SYMBOLS
488
SYMBOL
MEANING
PAGE REFERENCE
A (x)fl B
tensor product of modules A and B
208
/(g) g
induced map on the tensor product
209
0a
order ideal of a
220
[F:K]
dimension of field Fas a A'-vector space
231
K[uu . . . , un], [resp. /t'[A']]
subring generated by K and uu ... ,un [resp. X]
232
K(uu . . . , un) subfield generated by K and uu ... ,un [resp. K(X)] [resp. X]
232
K(xi, . . . , xn)
field of rational functions in n indeterminates
233
Aut/cF
Galois group of F over K
243
A
discriminant of a polynomial
270
Fpn
{upn | u e F; char F = p]
285
[F:K]S
separable degree of F over K
285
[F\K]i
inseparable degree of F over K
285
N (u)
norm of«
289
T (u)
trace of«
289
gn(x)
«-th cyclotomic polynomial
298
tr.d. F/K
transcendence degree of F over K
316
K llpn
\u eC \ mp” e K\
320
K1,pCO
\u e C | upn e K for some n > 0[
320
I„
n X n identity matrix
328
Mat„/?
ring of n X n matrices over R
328
A‘
transpose of the matrix A
328
A~1
inverse of the invertible matrix A
331
En'm
a certain matrix
337
Aa
classical adjoint of the matrix A
353
q
minimal polynomial of
356
Tr A
trace of the matrix A
369
Rad I
radical of the ideal I
379
F(5)
affine variety determined by S
409
d(B)
left annihilator of B
417
r° a J(R)
r + « + ra Jacobson radical of R
426 426
P(R)
prime radical of R
444
hom(/4,Z?) or homc(/l,Z?)
set of morphisms A —> B in a category C
hA
covariant hom functor
465
hB
contravariant hom functor
466
Cr
category formed from C and T
470
0C,D
zero morphism from C to D
482
kf
f k
T
52, 465
Bibliography All books and articles actually referred to in the text are listed below. The list also includes a number of other books that may prove to be useful references. No attempt has been made to make the list complete. It contains only a reasonable selection of English language books in algebra and related areas. Almost all of these books are accessible to anyone able to read this text, though in some cases certain parts of this text are prerequisites. For the reader’s convenience the books are classified by topic. However, this classification is not a rigid one. For instance, several books classified as “general” contain a fairly complete treatment of group theory as well as fields and Galois theory. Other books, such as [26] and [39], readily fit into more than one classifica tion.
BOOKS GENERAL 1. Chevalley, C., Fundamental Concepts of Algebra. NewYork: Academic Press, Inc., 1956. 2. Faith, C., Algebra: Rings, Modules and Categories I. Berlin: Springer-Verlag, 1973. 3. Goldhaber, J. and G. Ehrlich, Algebra. New York: The Macmillan Compa ny, 1970. 4. Herstein, I., Topics in Algebra. Waltham, Mass.: Blaisdell Publishing Company, 1964. 5. Lang, S., Algebra. Reading, Mass.: Addison-Wesley, Publishing Company, Inc., 1965. 6. MacLane, S. and G. BirkhofT, Algebra. New York: The Macmillan Company, 1967. 7. Van der Waerden, B. L., Algebra. (7th ed., 2 vols.), New York: Frederick Ungar Publishing Co., 1970. SET THEORY Eisenberg, M., Axiomatic Theory of Sets and Classes. New York: Holt, Rinehart and Winston, Inc., 1971. 9. Halmos, P., Naive Set Theory, Princeton, N. J.: D. Van Nostrand Company Inc., 1960. 10. Suppes, P., Axiomatic Set Theory. Princeton, N. J.: D. Van Nostrand Com pany, Inc., 1960. 8.
489
490
BIBLIOGRAPHY
GROUPS 11. Curtis, C. W. and I. Reiner, Representation Theory of Finite Groups and Asso ciative Algebras. New York: Interscience Publishers, 1962. 12. Dixon, J., Problems in Group Theory. Waltham, Mass.: Blaisdell Publishing Company, 1967. 13. Fuchs, L., Infinite Abelian Groups. New York: Academic Press, Inc., 1970. 14. Gorenstein, D., Finite Groups. New York: Harper and Row, Publishers, 1968. 15. Hall, M., The Theory of Groups. New York: The Macmillan Company, 1959. 16. Hall, M. and J. K. Senior, The Groups of Order 2"(« < 6). New York: The Macmillan Company, 1964. 17. Kaplansky, I., Infinite Abelian Groups (2d ed.), Ann Arbor, Mich.: University of Michigan Press, 1969. 18. Kurosh, A. G., The Theory of Groups (2 vols.), New York: Chelsea Publishing Company, 1960. 19. Rotman, J., The Theory of Groups (2d ed.). Boston: Allyn and Bacon, Inc., 1973. 20. Scott, W. R., Group Theory. Englewood Cliffs, N. J.: Prentice-Hall, Inc., 1964. 21. Zassenhaus, H., The Theory of Groups. New York: Chelsea Publishing Com pany, 1958. RINGS AND MODULES 22. Divinsky, N. J., Rings and Radicals. Toronto: University of Toronto Press, 1965. 23. Gray, M., A Radical Approach to Algebra. Reading, Mass.: Addison-Wesley Publishing Company, Inc., 1970. 24. Herstein, I. N., Noncommutative Rings. Math. Assoc, of America, distributed by J. Wiley, 1968. 25. Jacobson N., Structure of Rings. Amer. Math. Soc. Colloq. Publ., vol. 37,1964. 26. Jans, J., Rings and Homology. New York: Holt, Rinehart and Winston, Inc., 1964. 27. Lambek, J., Lectures on Rings and Modules. Waltham, Mass.: Blaisdell Pub lishing Company, 1966. 28. McCoy, N., Theory of Rings. New York: The Macmillan Company, 1964. 29. Northcott, D. G., Lessons on Rings, Modules and Multiplicity. New York: Cambridge University Press, 1968. COMMUTATIVE ALGEBRA 30. Atiyah, M. F. and I. G. MacDonald, Introduction to Commutative Algebra. Reading, Mass.: Addison-Wesley Publishing Company, Inc., 1969. 31. Kaplansky, I., Commutative Rings. Boston: Allyn and Bacon, Inc., 1970. 32. Larsen, M. and P. J. McCarthy, Multiplicative Theory of Ideals. New York: Academic Press, Inc., 1971. 33. Zariski, O. and P. Samuel, Commutative Algebra (vols. I and II). Princeton N. J., D. Van Nostrand Company, Inc., 1958. 1960. HOMOLOGICAL ALGEBRA 34. Hilton, P. J. and U. Stammbach, A Course in Homological Algebra. Berlin: Springer-Ver lag, 1971. 35. S. MacLane, Homology. Berlin: Springer-Verlag, 1963.
BIBLIOGRAPHY
491
FIELDS 36. Artin, E., Galois Theory. Notre Dame, Ind.: Notre Dame Mathematical Lectures No. 2 (2d ed.), 1944. 37. Gaal, L., Classical Galois Theory with Examples. Chicago: Markham, 1971. 38. Jacobson, N., Lectures in Abstract Algebra (vol. III). Princeton, N. J.: D. Van Nostrand Company, Inc., 1964. 39. Kaplansky, I., Fields and Rings (2d ed.). Chicago: University of Chicago Press, 1972. 40. McCarthy, P. J., Algebraic Extensions of Fields. Waltham, Mass.: Blaisdell Publishing Company, 1966. LINEAR AND MULTILINEAR ALGEBRA 41. 42. 43. 44.
Greub, W., Linear Algebra (3rd ed.). Berlin: Springer-Verlag, 1967. Greub, W., Multilinear Algebra. Berlin: Springer-Verlag, 1967. Halmos, P. R., Finite Dimensional Vector Spaces (2d ed.). Princeton, N. J.: D. Van Nostrand Company, Inc., 1958. Jacobson, N., Lectures in Abstract Algebra (vol. II). Princeton, N. J.: D. Van Nostrand Company, Inc., 1953.
CATEGORIES 45. 46. 47.
MacLane, S., Categories for the Working Mathematician. Berlin: SpringerVerlag, 1972. Mitchell, B., Theory of Categories. New York: Academic Press, Inc., 1965. Pareigis, B., Categories and Functors. New York: Academic Press, Inc., 1970.
NUMBER THEORY Artin E., Algebraic Numbers and Algebraic Functions. New York: Gordon and Breach, 1967. 49. Lang, S., Algebraic Number Theory. Reading Mass.: Addison-Wesley Pub lishing Company, Inc., 1970. 50. O’Meara, O. T., Introduction to Quadratic Forms. Berlin: Springer-Verlag, 1963. 51. Shockley, J. E., Introduction to Number Theory. New York: Holt, Rinehart and Winston, Inc., 1967. 52. Weiss E., Algebraic Number Theory. New York: McGraw-Hill Inc., 1963. 48.
ALGEBRAIC GEOMETRY 53. Fulton, W., An Introduction to Algebraic Geometry. New York: W. A. Benja min Inc., 1969. 54. Lang, S., Introduction to Algebraic Geometry. New York: Interscience Pub lishers, 1959. 55. MacDonald, I. G., Algebraic Geometry: Introduction to Schemes. New York: W. A. Benjamin Inc., 1968. ANALYSIS 56.
Burrill, C. W., Foundations of Real Numbers. New York: McGraw-Hill, Inc., 1967.
492
57.
BIBLIOGRAPHY
Hewitt, E. and K. Stromberg, Real and Abstract Analysis. Berlin: SpringerVerlag, 1969.
ARTICLES AND NOTES 58. 59. 60.
61. 62. 63. 64. 65. 66.
Bergman, G., “A Ring Primitive on the Right but Not on the Left,” Proc. Amer. Math. Soc., 15 (1964), pp. 473-475; correction, pg. 1000. Cohen, P. J., Set Theory and the Continuum Hypothesis, New York: W. A. Benjamin Inc., 1966. Comer, A. L. S., “On a Conjecture of Pierce Concerning Direct Decomposition of Abelian Groups,” Proc. Colloq. on Abelian Groups, Budapest, 1964, pp. 43-48. Feit, W. and J. Thompson, “Solvability of Groups of Odd Order,” Pac. Jour. Math., 13 (1963), pp. 775-1029. Goldie, A. W., “Semiprime rings with maximum condition,” Proc. Lond. Math. Soc., 10(1960), pp. 201-220. Kaplansky, I., “Projective Modules,” Math. Ann., 68 (1958), pp. 372-377. Krull, W., “Galoissche Theorie der unendlichen algebraischen Erweiterungen,” Math. Ann., 100 (1928), pp. 687-698. Procesi, C. and L. Small, “On a theorem of Goldie,” Jour, of Algebra, 2 (1965), pp. 80-84. Sasiada, E. and P. M. Cohn, “An Example of a Simple Radical Ring,” Jour, of Algebra, 5 (1967), pp. 373-377.
Index A. C. C. see ascending chain condition Abel’s Theorem 308 abelian field extension 292 abelian group 24 —defined by generators and relations 343ff divisible—195ff finitely generated— 76ff free— 71 ff action of group on set 88ff additive notation for groups 25, 29, 38, 70 adjoining a root 236 adjoint —associativity 214 classical—matrix 353 —pair of functors 477ff affine variety 409 algebra(s) 226ff, 450ff division— 456ff Fundamental Theorem of— 266 group— 227 homomorphism of— 228 —ideal 228 left Artinian— 451 tensor product of— 229 algebraic —algebra 453 —closure 259ff —element 233, 453 —field extension 233ff —number 242 —set 409 algebraically —closed field 258 —(in)dependent 31 Iff algorithm division— 11, 158 Euclidean— 141 alternating —group 49
—multilinear function 349 annihilator 206, 388, 417 anti-isomorphism 330 antisymmetric relation 13 Artin, E. vi, 251, 252, 288 Artinian left—algebras 451 —module 372ff —rings 372ff, 4l9ff, 435ff ascending —central series 100 —chain condition 83, 372ff associate 135 associated prime ideal 380, 387 associative —binary operation 24 generalized—law 28 Aut a-/7 243 automorphism 30, 119 extendible— 251, 260 inner— 91, 459 /("-automorphism 243 —of a group 30, 90 axiom —of choice 13 —of class formation 2 —of extensionality 2 —of pair formation 6 power— 3 Baer lower radical 444ff base transcendence— 313ff basis 71, 181 dual— 204 Hilbert—Theorem 391 matrix relative to a— 329ff standard— 336 transcendence— 313ff 493
bijective function 5 bilinear map 211, 349 canonical— 211 bimodule 202 binary operation 24 closed under a— 31 binomial —coefficient 118 —theorem 118 Boolean ring 120 bound (greatest) lower— 14 least upper— 14 upper— 13 cancellation 24, 116 canonical —bilinear map 211 —epimorphism 43, 125, 172 —forms 337 —injection 60, 130, 173 Jordan—form 361 —middle linear map 209 primary rational—form 361 —projection 59,130,173 rational—form 360 Cardan’s formulas 310 cardinal number(s) 16ff exponentiation of— 22 —inequalities 17 product of— 16 sum of— 16 trichotomy law for— 18 cardinality 16 Cartesian product 6, 7ff category 52ff, 464ff concrete 55 dual— 466 opposite— 466 product— 467 Cauchy’s Theorem 93 Cayley’s Theorem 90
494
Cayley-Hamilton Theorem 367, 369 center —of a group 34, 91 —of a ring 122 central ascending—series 100 —idempotents 135 —simple division algebra 456 centralizer 89 chain 13 chain conditions 83ff, 372ff char R 119 characterisl' : —of a ring 119 —polynomial 366ff —vectors 367 —values 367 —space 368 —subgroups 103 Chinese Remainder Theorem 131 choice axiom of— 13 —function 15 class 2ff axiom of—formation 2 —equation 90, 91 equivalence— 6 proper— 2 quotient— 6 classical adjoint 353 classification of finite groups 96ff closed —intermediate field 247 —object 410 —subgroups 247 ■—under a binary operation 31 closure algebraic— 259ff integral— 397 normal— 265 codomain 3 coequalizer 482 cofactor 352 Cohen, I. S. 388 coimage 176 cokernel 176 difference— 482 —of a morphism 483
INDEX
column elementary—operation 338 —rank 336 —space 336 —vector 333 commutative —binary operation 24 —diagram 4 generalized—law 28 —group 24 —ring 115 commutator 102 —subgroup 102 companion matrix 358 comparable elements 13 complement 3 complete —direct sum 59 —lattice 15 —ring of quotients 144 —set of coset representatives 39 completely reducible module 437 composite —function 4 —functor 468 —morphism 52 —subfield 233 composition series 108, 375 concrete category 55 congruence —modulo m 12 —modulo a subgroup 37 —relation 27 conjugacy class 89 conjugate elements or subgroups 89 conjugation 88 constant —polynomial 150 —term 150, 154 constructible number 239 content of a polynomial 162 Continuum Hypothesis 17 contraction of an ideal 146, 398 contravariant —functor 466 —hom functor 466 coordinate ring 413 coproduct 54ff
correspondence Galois— 246 coset 38ff complete set of— representatives 39 couniversal object 57 covariant —functor 465 —hom functor 465 Cramer’s Rule 354 cubic resolvant— 272 cycles 46 disjoint— 47 Conway, C. T., xiii cyclic —field extension 289fF —group 32, 35ff —module 171 —subspace 356 cyclotomic —extension 297ff —polynomial 166, 298 D. C. C.—, see descending chain condition Dedekind domain 401 ff degree —of a field element 234 —of inseparability 285ff —of a polynomial 157, 158 transcendence— 316 delta Kronecker— 204 DeMorgan’s Laws 3 dense ring 418 Density Theorem 420 denumerable set 16 dependence algebraic— 311 ff linear— 181 derivative 161 derived subgroup 102 descending chain condition 83, 372ff determinant 35 Iff expansion of a—along a row 353 —of an endomorphism 354 diagonal matrix 328 diagram commutative— 4
INDEX
difference —kernel 482 —cokernel 482 dihedral group 50 dimension 185ff relative— 245 invariant—property 185 direct —factor 63 —product 26, 59ff, 130, 131, 173 —sum 60, 62ff, 173, 175 —sum of matrices 360 —summand 63, 437 discriminant 270 discrete valuation ring 404 disjoint linearly—subfields 318fF —permutations 47 —sets 3 —union 58 divisible group 195ff division —algebras 227, 456ff —algorithm 11, 158 finite—ring 462 —ring 116 divisor(s) elementary— 80, 225, 357, 361 greatest common— 11, 140 zero— 116 domain Euclidean— 139 integral— 116 —of a function 3 principal ideal— 123 Prufer— 409 valuation— 409 unique factorization— 137 double dual 205 dual —basis 204 —category 466 double— 205 —map 204 —module 203ff —statement 54 echelon reduced row—form 346 eigenspace 368 eigenvalues 367ff
eigenvectors 367ff Eisenstein’s Criterion 164 elementary —column operation 338 —divisors 80, 225, 357, 361 —Jordan matrix 359 —row operation 338 —symmetric functions 252 —transformation matrix 338 embedded prime 384 embedding 119 empty —set 3 —word 64 End A 116 endomorphism(s) characteristic polynomial of an— 366ff dense ring of— 418 determinant of an— 354 matrix of an— 329ff, 333 nilpotent— 84 normal— 85 —of a group 30 —ring 116, 179,415 trace of an— 369 epic morphism 480 epimorphism, 30, 118, 170 canonical— 43, 125, 172 —in a category 480 equality 2 equalizer 482 equation(s) general—of degree n 308ff linear— 346 equipollent sets 15 equivalence —class 6 —in a category 53 natural— 468 —relation 6 equivalent —matrices 332, 335ff —series 109, 375 Euclidean —algorithm 141 —ring 139 Euler function 297 evaluation homomorphism 153
495
even permutation 48 exact sequence 175ff short— 176 split— 177 exponentiation —in a group 28 —of cardinal numbers 22 extendible automorphism 251, 260 extension field 23 Iff abelian— 292 algebraic— 233ff cyclic— 289ff, 292 cyclotomic— 297ff finitely generated— 231 Galois— 245 normal— 264ff purely inseparable— 282ff purely transcendental— 314 radical— 303ff separable— 261, 282ff, 324 separably generated— 322 simple— 232 transcendental— 233, 31 Iff extension integral— 395 —of an ideal 145 —ring 394ff extensionality axiom of— 2 external —direct sum 60, 173 —direct product 130 —weak direct product 60 factor group 42ff factorization —in commutative rings 135ff —in polynomial rings 157ff unique ' factorization— 137ff factors invariant—80, 225, 357, 361 —of a series 107, 375 faithful module 418 field 116, 230ff
496
algebraically closed— 258 —extension 23 Iff finite— 278ff, 462 fixed— 245 intermediate— 231 —of rational functions 233 perfect— 289 splitting— 257ff finite dimensional —algebra 227 —field extension 231 —vector space 186 finite field 278ff, 462 finite group 24, 92ff, 96ff finite set 16 finitely generated —abelian groups 76ff, 343ff —extension field 232 —group 32 —ideal 123 —module 171 first isomorphism theorem 44, 126, 172 Fitting’s Lemma 84 five lemma 180 short— 176 fixed field 245 forgetful functor 466 form Jordan canonical—361 multilinear— 349ff primary rational canoni cal— 361 rational canonical— 360 formal —derivative 161 —power series 154ff four group 29 fractional ideal 401 free —abelian group 7Iff —group 65ff —module 181 ff, 188 —object functor 479 —object in a category 55ff —product 68 rank or dimension of a— module 185 freshman’s dream 121 Frobenius Theorem 461
INDEX
fully invariant subgroup 103 function(s) 3ff bijective— 5 bilinear— 211, 349 choice— 15 field of rational— 233 graph of a— 6 injective— 4 left or right inverse of a — 5 Moebius— 301 multilinear— 349ff surjective— 5 symmetric rational— 252ff two sided inverse of a— 5 functor 465ff adjoint— 477ff composite— 468 forgetful— 466 free object— 479 hom— 465, 466 representable— 470ff Fundamental Theorem —of Algebra 265ff —of Arithmetic 12 —of Galois Theory 245ff, 263 g.l.b. 14 Galois —correspondence 246 —extension 245 —fields 278ff Fundamental Theorem of— Theory 245ff, 263 —groups 243 —group of a polynomial 269ff —Theory 230ff Gauss Lemma 162 Gaussian Integers 139 general —equation of degree n 308ff —polynomial of degree n 308 generalized —associative law 28 —commutative law 28 —fundamental theorem of Galois Theory 263
—multiplicative system 449 generators and relations 67ff, 343ff generators —of a group 32 —of an ideal 123 —of a module 171 Going-up Theorem 399 Goldie —ring 447ff —’s Theorem 448 Goldmann ring 400 graph 6 greatest —common divisor 11, 140 —lower bound 14 group(s) 24ff abelian— 24, 70ff, 76ff —algebra 227 alternating— 49 center of a— 34, 91 cyclic— 32, 35ff —defined by generators and relations 67, 343ff dihedral— 50 direct product of— 26, 59ff direct sum of— 60, 62 divisible— 195ff factor— 42 finite— 92ff finitely generated— 32, 76ff free— 65ff free abelian— 7 Iff free product of— 68 Galois— 243 Galois—of a polynomial 269ff homomorphism of— 30ff, 43ff identity element of a— 24 indecomposable— 83 inverses in— 24 isomorphic— 30 isotropy— 89 join of— 33 metacyclic— 99 nilpotent— 100 —of integers modulo m, 27
INDEX
—of rationals modulo one 27 —of small order 98 --- of symmetries of the square 26 p-group 93 quaternion— 33, 97, 99 reduced— 198 —ring 117 simple— 49 solvable— 102ff, 108 symmetric— 26, 46ff torsion— 78 weak direct product of— 60, 62 Hall’s Theorem 104 Halmos, P. ix -Hamilton Cayley—Theorem 367, 369 Hilbert —Basis Theorem 391 —Nullstellensatz 409ff, 412 —’s Theorem 90 292 -Holder Jordan—Theorem 111, 375 Hom,IA,B) 174, 199ff hom functor 465, 466 homogeneous —of degree k 158 —system of equations 346 homorphism induced— 199, 209 /^-homomorphism 243 kernel of a— 31, 119, 170 matrix of a— 329ff, 333 natural— 469 —of algebras 228 —of groups 30ff, 43ff —of modules 170ff, 199ff —of rings 118 substitution— 153 Hopkin’s Theorem 443 ideal(s) 122ff algebra— 228 finitely generated— 123 fractional— 401 —generated by a set 123 invertible— 401
left quasi-regular— 426 maximal— 127ff nil— 430 nilpotent— 430 order— 220 primary— 380ff prime— 126ff, 377ff primitive— 425 principal—ring 123 proper— 123 regular left— 417 idempotent —element 135, 437 orthogonal—s 135, 439 identity 24, 115 —function 4 —functor 465 —matrix 328 image inverse— 4, 31 —of a function 4 —of a homomorphism 31, 119,170 —of a set 4 inclusion map 4 indecomposable group 83 independence algebraic— 31 Iff linear— 74, 181 independent set of ideals 446 indeterminate 150, 152 index —of a subgroup 38 relative— 245 induced homomorphism 199 209 induction mathematical— 10 transfinite— 14 infinite set 16 initial object 57 injections canonical— 60, 130, 173 injective —function 4 —module 193ff inner automorphism 91, 459 inseparable —degree 285ff purely—extension 282ff integers 9ff Gaussian— 139 —modulo m 21, 116
497
integral —closure 397 —domain 116 —element 395 —ring extension 395 integrally closed ring 397 intermediate field 231 closed— 247 stable— 250 intermediate value theorem 167 internal —direct product 131 —direct sum 62, 175 —weak direct product 62 Interpolation Lagrange’s—Formula, 166 intersection 3 invariant —dimension property 185 —factors 80, 225, 357, 361 —subspace 356 inverse 24, 116 —image 4, 31 —of a matrix 331 two sided—of a function 5 invertible —element 116 —fractional ideal 401 —matrix 331 irreducible —element 136 —module 416 —polynomial 161, 234 —variety, 413 irredundant primary de composition 381 isolated prime ideal 384 isomorphism 30, 118, 170 —theorems 44, 125ff,
172ff natural— 468 isotropy group 89 join of groups 33 Jordan —canonical form 361 elementary—matrix 359 —Holder Theorem 111, 375 Jacobson
498
—Density Theorem 420 —radical 426ff —semisimple 429 A'-algebra, see algebra A'-homomorphism 243 Kaplansky, I. vi, 251, 391, 432 kernel, 31, 119, 170 difference— 482 —of a morphism 483 Klein four group 29 Kronecker —delta 204 —"s method 167 Krull —Intersection Theorem, 389 — Schmidt Theorem, 86 l.u.b., 14 Lagrange’s —Interpolation Formula 166 —Theorem 39 lattice 14 complete— 15 Law of Well Ordering 10 leading coefficient I 50 least —element 13 —upper bound 14 left —adjoint 477IT —annihilator 444 —coset 38 —exact 201 —Goldie ring 447AF —ideal 122 —inverse of a function 5 —invertible element 116 —order 447 —quasi-regular 426 —quotient ring 447 length of a series 107, 375 Levitsky’s Theorem 434 linear —algebra 327fT —combination 71 —dependence 181 —equations 346 —functional 203 —independence 74, 181, 291 —ordering 13
INDEX
linear transformation 170 characteristic polynomi al of a— 366ff decomposition of a— 355ff eigenvalues of a— 367 eigenvector of a— 367 elementary divisors of a — 357 invariant factors of a— 357 matrix of a— 329, 333 nullity of a— 335 rank of a— 335ff linearly disjoint subfields 3l8ff linearly independent —automorphisms, 291 —elements 74, 181 local ring 147 localization 147 logic 1 lower bound 14 Lying-over Theorem 398 m-system 449 Mac Lane, S. xiii, 325, 464 main diagonal 328 map, see function Maschke’s Theorem 454 mathematical induction 10 matrices, 328ff characteristic polynomi als of— 366fl companion— 358 determinants of— 35Iff direct sum of— 360 eigenvalues of— 368 eigenvectors of— 368 elementary— 338 elementary divisors of — 361 elementary Jordan— 359 equivalent— 332, 335ff —in row echelon form 346 invariant factors of— 361 invertible— 331 minors of— 352 —of a homomorphism 329ff,333 rank of— 336, 337 row spaces of— 336 secondary— 340
similar— 332, 355ff skew-symmetric— 335 symmetric— 335 trace of— 369 triangular— 335 maximal —element 13 —ideal 127ff —normal subgroup 108 —subfield 457 maximum condition 373 McBrien, V.O. xiii, 121 McCoy radical 444ff McKay, JJH. 93 meaningful product 27 membership 2 metacyclic group 99 middle linear map 207, 217 canonical— 209 minimal —annihilator 226 —element 373 —left ideal 416 —normal subgroup 103 —polynomial 234, 356 —prime ideal 382 minimum —condition 373 —element 13 —polynomial 234 minor of a matrix 352 modular left ideal 417 module(s) J68ff, 371 algebra— 451 Artinian— 372ff completely reducible— 437 direct sum of— 173, 175 direct product of— 173 dual— 203ff exact sequence of— 175ff faithful—418 free— 181 fT, 188 homomorphism of— 17 Off, 199ff injective— 193ff isomorphism theorems for— 172ff Noetherian—372ff —over a principal ideal domain 218ff projective— 190ff rank of— 185 reflexive— 205 semisimple— 437
INDEX
simple— 416 sum of— 171 tensor product of— 208ff torsion— 179, 220 torsion-free— 220 trivial— 170 unitary— 169 Moebius function 301 monic —morphism 480 —polynomial 150 Monk, G.S. ix, 222 monoid 24 monomial 152 degree of a— 157 monomorphism 30, 118, 170 —in a category 480 morphism 52, 480ff cokernel of a— 483 epic— 480 kernel of a— 483 monic— 480 zero— 482 multilinear —form 349ff —function 349ff multiple root 161, 261 multiplicative set 142 multiplicity of a root 161 mutatis mutandis xv «-th root of unity, see unity /i-fold transitive 424 Nakayama’s Lemma 389 natural —homomorphism 469 —isomorphism 468 —numbers 9 —transformation 468ff nil —ideal 430 —radical 379, 450 nilpotent —element 121, 430 —endomorphism 85 —group lOOff —ideal 430 Noether —Normalization Lemma 410 -- Skolem Theorem 460 Noetherian —modules 372ff, 387ff —rings 372, 387ff
nonsingular matrix 331 norm 289 normal —closure 265 —endomorphism 84 —extension field 264ff —series 107ff, 375 —subgroup 41ff normalization lemma 410 normalizer 89 null set 3 nullity 335 Nullstellensatz 412 number cardinal— I6ff natural— 9 Nunke, R. J. x, xiii, 93 object —in a category 52 initial— 57 terminal— 57 odd permutation 48 one-to-one —correspondence 5 —function 4 onto function 5 operation binary— 24 elementary row or column— 338 opposite —category 466 —ring, 122, 330 orbit 89 —of a permutation 47 order —ideal 220 left— 447 —of an element 35, 220 —of a group 24 partial— 13 ordered pair 6 ordering —by extension 18 linear— 13 partial— 13 simple— 13 total— 13 orthogonal idempotents 135, 439
see principal ideal domain p-group 93
499
P-primary 380, 384 P-radical 425 pair —formation 6 ordered— 6 partial ordering 13 partition 7 perfect field 289 permutation(s) 26, 46ff disjoint— 47 even— 48 odd— 48 orbits of— 47 sign of a— 48 Polynomials) characteristic— 366ff coefficients of— 150 content of a— 162 cyclotomic— 166, 298 degree of— 157, 158 discriminant of— 270 Galois group of a— 269ff general—of degree n 308 —in n indeterminates 151 irreducible— 161, 234 minimal— 234, 356 minimum— 234 monic— 150 primitive— 162 ring of— 149ff, 151, 156 roots of— 160 separable— 261 power —axiom 3 —series 154ff —set 3 predecessor immediate— 14 presentation of a group 67 primary —decomposition, 381, 383ff —ideal 380ff P-primary ideal 380 P-primary submodule 384 —rational canonical form 361 —submodule 383ff prime associated—ideal 380, 387 —element 136
500
embedded—ideal 384 —ideal 126ff, 377ff —integer 11 isolated—ideal 384 minimal—ideal 382 —radical 379, 444ff relatively—integers 11 relatively—elements 140 —ring 445ff subfield 279 primitive —element theorem 287, 288 —ideal 425 —polynomial 162 —ring 418ff —root of unity 295 principal —ideal 123 —ideal ring 123 —ideal domain 123 modules over a—ideal domain 218ff principle —of mathematical induction 10 —of transfinite induc tion 14 —of well ordering 14 product —category 467 Cartesian— 6, 7ff direct— 59ff, 130, 131, 173 —in a category 53ff —map 228 semidirect— 99 subdirect— 434 weak direct— 60, 62 projection canonical— 8, 59, 130, 173 projective module 190ff proper —class 2 —ideal 123 —refinement 108, 375 —subgroup 32 —values 367 —vectors 367 Prufer domain 409 pullback 170, 484 purely inseparable —element 282 —extension 282ff
INDEX
purely transcendental ex tension 314 quasi-inverse, 426 quasi-regular 426 quaternion group 33,97,99 quaternions division ring of real— 117, 461 quotient —class 6 —field 144 —group 42ff —ring 125, 144ff, 447 r-cycle 46 module, see module Rad 1379 radical —extension field 303ff Jacobson— 426ff nil— 379, 450 —of an ideal 379 /’-radical 425 prime— 379, 444ff —property 425 —ring 429 solvable by—s 303 range of a function 3 rank column— 336, 339 —of a free abelian group 72 —of a free module 185 —of a homomorphism 335ff, 339 row— 336, 339 —of a matrix 337ff rational —canonical form 360 —function 233 symmetric—function 252ff rationals modulo one 27 Recursion Theorem 10 reduced —abelian group 198 —primary decomposi tion 381, 384 —row echelon form 346 —word 64, 68 refinement of a series 108, 375 reflexive —module 205 —relation 6
regular —element 447 —function 412 —left ideal 417 —left quasi---, 426 Von Neumann—ring 442 relative complement 3 relation 6, 66 antisymmetric— 13 congruence— 27 equivalence— 6 generators and—s, 67ff, 343ff reflexive— 6 symmetric— 6 transitive— 6 relative —dimension 245 —index 245 relatively prime —integers 11 —ring elements 140 Remainder Theorem 159 Chinese— 131 representable functor 470 representation 470 resolvant cubic 272 restriction 4 right —adjoint 477 —annihilator 444 —coset 38 —Goldie ring 447 —ideal 122 —inverse of a function 5 —invertible element 116 —quasi-regular 426 —quotient ring 447 ring(s) 115ff, 371 ff, 414ff Artinian— 372, 42l, 435 Boolean— 120 commutative— 115, 371ff direct product of— 130, 131 discrete valuation— 401 division— 116, 462 endomorphism— 415 Euclidean— 139 —extensions 394ff Goldmann— 400 group— 117 homomorphism of— 118
INDEX
integrally closed— 397 left quotient— 447 local— 147 Noetherian— 372, 387ff —of formal power series 154ff —of polynomials 149ff —of quotients or frac tions 144ff opposite— 122, 330 primitive— 418ff prime— 445ff quotient— 125, 447 radical— 429 regular— 442 semiprime— 444ff semisimple— 429, 434ff simple— 4l6ff subdirectly irreducible— 442 root adjoining a— 236 multiple— 161, 261 multiplicity of a— 161 —of unity 294 simple— 161, 261 row —echelon form 346 elementary—operation 338 —rank 336, 339 —space 336 —vector 329 ruler and compass con structions 238 Russell’s Paradox 2 scalar matrix 328 Schreier’s Theorem 110, 375 Schroeder-Bemstein Theorem 17 Schur's Lemma 419 second isomorphism theorem 44, 126, 173 secondary matrix 340 semidirect product 99 semigroup 24 semiprime ring 444ff semisimple —module 437 —ring 429, 434ff separable —degree 285ff —element 261
—extension 261, 282ff, 324 —polynomial 261 separably generated exten sion 322 separating transcendence base 322 sequence 11 exact— 175ff short exact— 176 series ascending central— 100 composition— 108, 375 equivalent— 109, 375 formal power— 154ff normal— 107ff, 375 refinement of a— 108, 375 solvable— 108 subnormal— 107ff set(s) denumerable— 16 disjoint— 3 empty— 3 equipollent— 15 finite— 16 infinite— 16 linearly ordered— 13 multiplicative— 142 null— 3 partially ordered— 13 power— 3 underlying— 55 well ordered— 13 Short —Five Lemma 176 —exact sequence 176 sign of a permutation 48 similar matrices 332, 355ff simple —components 440 —extension field 232 —group 49 —module 179, 375,416ff —ordering 13 —ring 416ff —root, 161, 261 singleton 6 skew-symmetric —matrix 335 —multilinear form 349 solvable —by radicals 303 —group 102ff, 108 —series 108
501
span 181 spectrum 378 split exact sequence 177 splitting fields 257ff stabilizer 89 stable intermediate field 250 standard —basis of Ra 336 —/i-product 28 subalgebra 228 subclass 2 subdirect product 434 subdirectly irreducible ring 442 subfield(s) composite— 233 —generated by a set 231, 232 linearly disjoint— 318ff maximal— 457 prime— 279 subgroup(s) 3Iff characteristic— 103 closed— 247 commutator— 102 cyclic— 32 derived— 102 fully invariant— 103 —generated by a set 32 —generated by groups 33 join of— 33 maximal normal— 108 minimal normal— 103 normal— 41 ff proper— 32 Sylow— 94 transitive— 92, 269 trivial— 32 submodule(s) 171 chain conditions on— 372ff cyclic— 171 finitely generated—171 —generated (or spanned) by a set 171 primary— 383ff sum of— 171 torsion— 220 subnormal series 107ff subring 122 —generated by a set 231, 232, 395
502
subset 3 subspace 171 -invariant— 356 substitution homomor phism 153 successor immediate— 15 sum ( = coproduct) 54ff sum direct— 60, 62, 173, 175 —of submodules 171 summand direct— 63, 437 surjective function 5 Swords, R.J., xiii Sylow —/^-subgroup 94 —theorems 92ff symmetric —group 26, 46ff —matrix 335 —multilinear function 349 —rational function 252ff —relation 6 symmetries of the square 26 tensor product 208ff —of algebras 229 induced homomorphism on the— 209 terminal object 57 third isomorphism theorem 44,126,173 torsion —group 78 —module 179, 220 —subgroup 78 —submodule 179, 220 torsion-free —group 78 —module 220
INDEX
total —degree 157 —ordering 13 trace 289, 369 transcendence —base 313ff —degree 316 separating—base 322 transcendental —element 233 —extension 233, 31 Iff purely—extension 314 transfinite induction 14 transformation linear— 170, 355ff natural— 468ff transitive —relation 6 —subgroup 92, 269 —subring 424 translation 88 transpose of a matrix 328 transposition 46 triangular matrix 335 trichotomy law 18 trivial ideal 123 U.F.D ., see unique factori zation domain underlying set 55 union 3 disjoint— 58 unique factorization do main 137 unit 116 —map 228 unitary module 169 unity root of— 294 primitive root of— 295 universal —element 470
—mapping property 9, 57 —object 57 upper bound 13 valuation —domain 409 discrete—ring 404 Van Dyck’s Theorem 67 variety 409 vector column— 333 row— 329 vector space 169, 180ff finite dimensional— 186 Von Neumann regular ring 442 weak direct product 60, 62 Wedderburn’s Theorem on finite division rings 462 Wedderburn-Artin Theorems 421, 435 well ordered set 13 well ordering law of— 10 —principle 14 word 64, 68 reduced— 64, 68 empty— 64 Yoneda, N. 472 Zassenhaus Lemma, 109, 375 zero 409 —divisor 116 —element 115 —matrix 328 —morphism 482 —object 481 —of a polynomial 160 Zorn’s Lemma, 13
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