Galois Theory

2.5 Splitting Fields Definition 2.5.1. (1) Let f(X) e F[X] be a nonconstant polynomial. If E 2 F is such that f{X) factors into (possibly repeated) linear factors in E[X], then / ( Z ) splits in E[X]. (2) If f(X) splits in E[X] but not in B[X] for any proper subfield B of E, then E is a splitting field of / (X). o Note that X — a is SL factor of f(X) if and only if f(a) = 0, i.e., if and only if Of is a root of f(X). Thus, we regard f(X) as splitting in E if "all" the roots of f(X), i.e., in E, and we regard E as a splitting field if "all" the roots of f(X) lie in E but not in any proper subfield of E. Lemma 2.5.2. L^^ f{X) has a splitting field.

e F[X] be a nonconstant polynomial. Then f{X)

Proof. By induction on deg / ( X ) , and all fields simultaneously. Without loss of generality we may assume that f{X) is monic. If d e g / ( X ) = 1, then f{X) = X-ae F[Z] and F is a splitting field for / ( X ) . Assume the lemma holds for all polynomials of degree less than d, and all fields, and let fiX) e F[X] have degree d. By Theorem 2.2.6, there is an extension field B of F in which f(X) has a root ai, and moreover B = F(ai). Then f{X) = (X - ai)giX) e B[X]. By induction, g(X) has a splitting field E over B. Hence g(X) = (X—^2) • • • (X — ad) e E[X] and / ( X ) = (X - aO • • • (X - a^) e E[X] splits. Clearly E 2 B(Qf2, • • •, oin), and then E = B ( a 2 , . . . , ««), since certainly g{X) does not split over any field not containing each of a 2 , . . . , a„. Then E = F(Qfi)(Qf2,..., otn) = F ( a i , . . . , an), and / ( X ) does not split in any proper subfield of E, by the same logic, so E is a splitting field for f(X). n Note that in general it may well be the case that an extension B of F contains a root of an irreducible polynomial / ( X ) e F[X] but that polynomial

2.5 Splitting Fields

23

does not split in B[X], i.e., that "some" but not "all" of the roots of f{X) are in B. As an example of this, we may take B = Q ( ^ ) . Then the polynomial Z"^ - 2 is irreducible in Q[X] with roots v ^ , / ^ , - v ^ , and - / v ^ , and B contains just two of these four roots {\/2 and —\fT). We shall soon see that any two splitting fields of f{X) G F [ X ] are isomorphic. Granting that, for the moment, we have the following result. Lemma 2.5.3. Let f{X) e F[X] be a nonconstantpolynomial, with deg f(X) = d. Let Ebe a splitting field of f(X). If f(X) is irreducible, (E/F) is divisible by d. In any case, (E/F)
Example 2.5.4. (1) Let F = Q and let E be the sphtting field of the irreducible polynomial X^ -2e F[X]. Then E = F(w, v ^ ) where co^ + co + 1 =0,md (E/F) = 6 = 3!. (2) Let F = Q(a>) and let E be the splitting field of the irreducible polynomial X^ -2e F[X]. Then E = F ( ^ ) , and (E/F) = 3 < 3!. o Corollary 2.5.5. Let Ebe a splitting field of a nonconstant polynomial f{X) € F[Z]. Then E/F is algebraic. Proof E / F is

finite.

n

Remark 2,5.6. Suppose that / ( X ) e F[Z] splits in some extension A of F, and let a i , . . . , Qfjt be the roots of f(X) in A (so that fiX) = (X - aO^^i • • • (Z a^t)^^ e A[X]). Then E = F(Qfi,..., a^t) is a splitting field for / ( X ) , as f(X) visibly splits over E, and any field over which f(X) splits must contain Qfi,..., Qfj^, and hence must contain E. o

24

2 Field Theory and Galois Theory

2.6 Extending Isomorphisms We now prove two results about extensions of field isomorphisms that will be key technical tools. Lemma 2.6.1. Let CTQ : Fi -> F2 be an isomorphism of fields. Let f\ (X) € Fi [X] be irreducible and let Ei = Fi (^1) where f\ (fii) = 0. Let E2 = F2(^2) where f2{^2) = 0. Then GQ extends to a unique isomorphism a : Ei -> E2 witha(fii) = ^2Proof. Clearly ao gives an isomorphism ao : Fi[X] -^ F2[X] and f2(X) is irreducible if and only if fi(X) is. Now, by Lemma 2.4.13, we have isomorphisms ^i : FdX]/{f(X))

-> F,(A) = E,-, / = 1, 2,

and we let cr =(P2cro(Pi^'

•

Corollary 2.6.2. Lef f(X) e F[X] be irreducible. Suppose ^\ and ^2 cire both roots of fiX) and let E^ = F(^), / = 1,2. Then there is a unique isomorphism a : Ei ^^ E2 with a | F = id and a(fii) = ^2- In particular, if E = F(^i) = F()62), there is a unique automorphism a ofK with a | F = id anda(fii) = ^2D This corollary says that, algebraically, there is no distinction between roots of the same irreducible polynomial — choosing any two roots gives us isomorphic extensions. Lemma 2.6.3. Let ao : Fi ^- F2 be an isomorphism of fields. Let f\ {X) e FiIX] andletf2(X) = ao(fi(X)) e F2[X]. LetEi be a splitting field of Fi(X) and let E2 be a splitting field of f2(X). Then ao extends to an isomorphism o '. El —> E2. Proof. Factor f\{X) into irreducibles in F[X], and let there be k factors. Set dY(fi) = deg(/i) — k. We prove the theorem by induction on J F ( / I ) , and all fields. If d^ifi) = 0, then fi(X) is a product of linear factors, and Ei = Fi, E 2 = F 2 , s o or = CTQ.

Suppose J F C / I ) > 0. Then fi(X) has an irreducible factor gi(X) of degree greater than 1. Let ai be a root of gi(X) in Ei and let 0^2 be a root of aoigiiX)) in E2. Then Ei D F(ai) and E2 2 F(a2). By Lemma 2.6.1, there is an isomorphism ai : F(ai) -^ F(cr2) with ai | F = CTO and cri(ai) = 012.

2.7 Normal, Separable, and Galois Extensions

25

Set B = F(ai), and consider / i ( Z ) e B[X]. Now gi(X) e B[X] has the factor X — Qfi, so fi(X) has at least k + I irreducible factors in B[X] and hence d^{fi) < d^ifi). Since Ei was a splitting field of fi(X), regarded as polynomial in Fi[X], and B c Ei, then, by Remark 2.5.6, Ei is still a splitting field of fi(X) regarded as a polynomial in B[X], and similarly for E2, so by induction ai extends to a, as claimed. n Corollary 2.6.4. Let f(X) isomorphic.

e F[X], Then any two splitting fields of f(X) are D

Note that we are certainly not claiming that the isomorphism a of Lemma 2.6.3 is unique! In fact, studying automorphisms E that fix F is what Galois theory is all about. To give a simple example of this, let F = Q and E = Q ( \ / D ) . Then E has an automorphism a given hy a{a + b's/D) = a — h\fD.

1.1 Normal, Separable, and Galois Extensions We begin by defining two (independent) properties of field extensions. Definition 2.7.1. An algebraic extension E of F is a normal extension if every irreducible polynomial f(X) € F[X] with / ( a ) = 0 for some a € E splits in E[Z]. o Remark 2 J,2. Observe that Definition 2.7.1 is equivalent to: An algebraic extension E of F is normal if m^ {X) splits in E[X] for every a G E. o Definition 2.7.3. (1) If the irreducible factors of f{X) e F[X] split into a product of distinct linear factors in some (hence in any) splitting field for / ( Z ) , then f(X) is a separable polynomial. Otherwise, it is an inseparable polynomial. (2) Let E be an algebraic extension of F. Then a G E is a separable element if ma (X) is a separable polynomial. Otherwise, a is an inseparable element. (3) Let E be an algebraic extension of F. Then E is a separable extension of F if every a G E is a separable element. Otherwise, it is an inseparable extension. o Remark 2,7,4. We are most interested in extensions that are both normal and separable, as these have the closest connection with Galois theory, so we will defer our consideration of extensions that satisfy only one of these two properties until later. However, some discussion of both of these properties is in order now.

26

2 Field Theory and Galois Theory

First, we might observe that the definitions of normal and separable are rather discouraging on the face of it, as they each require us to verify a condition on niotiX) for each a € E. But, as we will soon see (in Theorem 2.7.14), we have an easy criterion for extensions to be both normal and separable. Even if we are interested in extensions that are normal or separable, the situation is not so bad. We note some results that we shall prove later on. Theorem 5.2.1. E is a finite normal extension ofF if and only ifE is the splitting field of a polynomial f(X) e F[X]. Theorem 5.1.9. E is a finite separable extension ofF if and only if E is obtained from F by adjoining root(s) of a separable polynomial f(X)eF[Xl Actually, sometimes separability is automatic. There are some fields F all of whose algebraic extensions are separable. Such a field F is called perfect, and we will show below that the following is true. Theorem 3.2.6. F is a perfect field if and only ifF is afield of characteristic Q orF is a finite field, o We now define one of the most important objects in mathematics. Definition 2.7.5. Let E be an algebraic extension of F. The Galois group Gal(E/F) is the group of all automorphisms of E that restrict to the identity onF, Gal(E/F) = {a: E -^ E automorphism | a | F = id}.

o

(Strictly speaking, we should have defined Gal(E/F) to be the set of these automorphisms, but it is easy to verify that this set is indeed a group under composition.) Lemma 2.7.6. Let E be an algebraic extension ofF and let B be afield with F C B c E. Then Gal(E/B) is a subgroup o/Gal(E/F). Proof Any automorphism a : E ^- E with a | B = id certainly satisfies a I F = id. D Definition 2.7.7. Let G be a group of automorphisms of a field E. Then the fixed field Fix(G) = {a G E | a ( a ) = or for all a e G). o Lemma 2.7.8. Let G be a group of automorphisms of afield E. Then Fix(G) is a subfieldofF, D

2.7 Normal, Separable, and Galois Extensions

27

Note that, by definition, F c Fix(Gal(E/F)). Definition 2.7.9. Let E be an algebraic extension of F. Then E is a Galois extension of F if Fix(Gal(E/F)) = F. o Our first main goal is the Fundamental Theorem of Galois Theory (FTGT), which, for a finite Galois extension E of F, establishes a one-to-one correspondence between intermediate fields B (i.e., fields B with F c B c E) and subgroups H of G = Gal(E/F). On the way there, we need to obtain a criterion for E to be a Galois extension of F, an important result in its own right. Definition 2.7.10. Let E be an algebraic extension of F and let a G E. Then {or (a) I a e Gal(E/F)} is the set of (Galois) conjugates of a in E. o Lemma 2.7.11. Let H = {a e Gal(E/F) \ a(a) = a}. Then H is a subgroup ofG = Gal(E/F) and the number of conjugates of a in E is equal to the index [G : HI U The following lemma is particularly useful. Lemma 2.7.12. Let ^be a Galois extension ofF and let a G E. Then a has finitely many conjugates in E. /f {a/}r=i,...,r is the set of conjugates of a = ai, then

ia(X) =

Yl{X-ai),

Proof Since m^iX) e F[Z], for each a e G = Gal(E/F), a (a) is a root of a(ma{X)) = ma(X). Since ma(X) has only finitely many roots, {cr(a)} must be a finite set. Now let na(X) = YY^^i(X — at). Since a e G permutes {a/}, a{na{X)) = na(X), In other words, every a e G fixes each coefficient of na{X). Since E is a Galois extension of F, this implies that each of these coefficients is in F, i.e., that na(X) e ¥[X]. As we have observed, maiat) = 0 for each /, so X—at divides m^ (X) for each /, and hence n^ (X) divides m^ (X) in E[X], and hence in F[X], by Lenmia 2.2.7. But ma{X) is irreducible in F[Xlsoma(X)=na(X). D Corollary 2.1.H. Let F be a Galois extension of F and let a G E. Then (F(a)/F) is equal to the number of conjugates of a in E. Proof (F(Qf)/F) = degmc(Z) by Proposition 2.4.6 (2) and degm«(X) is equal to the number of conjugates of a in E by Lemma 2.7.12. D The following theorem is a key result.

28

2 Field Theory and Galois Theory

Theorem 2.7.14. Let E be a finite extension ofF. The following are equivalent: (1)E is a Galois extension ofF. (2) E is a normal and separable extension ofF. (3) E is the splitting field of a separable polynomial f(X) e F[X], Proof (1) =^ (2): Let E be a Galois extension of F and let of G E. Let {of/} be the set of conjugates of a in E. By Lemma 2.7.12, ma(X) = YYi=i(^ ~ ^t)^ and this is a separable polynomial (as its roots are distinct) that splits in E. (2) => (3): Let {6,} be a basis for E over F, and let f{X) be the product of the distinct elements of {m^. {X)]. Then f{X) is separable, as each m^. (X) has distinct roots, and f{X) certainly splits in E. Moreover, any field in which f{X) splits must contain each 6^, so E is the splitting field of / ( X ) . (3) =^ (1): We prove this by induction on (E/F) and all fields simultaneously. Let E be the splitting field of the separable polynomial f{X). If (E/F) = 1, then E = F, Gal(E/F) = {id}, and we are done. Suppose now that the theorem is true for all extensions of degree less than n and let (E/F) = n > 1. Factor f{X) into irreducibles / ( X ) = /i(X) • • • fk{X) in F[Z]. Some fi{X) has degree greater than one, as otherwise f{X) would split in F[X] and so E = F. Renumbering if necessary, we may suppose that this factor is fi(X). Let it have degree r. Let a = ai be a root of fi(X), a G E. Then fi{X) = ma(X), since fi(X) is irreducible. Of course, of ^ F. By assumption, ma(X) is separable, so has distinct roots a i , . . . , ofr- Now at is a root of the irreducible polynomial m^ (X) by assumption, and of/ is a root of the irreducible polynomial mc^. (X) by definition, so, by Lemma 2.4.3, mo(i(X) = ma(X), for each /. Furthermore, by Corollary 2.6.2, there are isomorphisms (Ti: F(ai) -> F(ai) with at \ F = id and or^(ai) = at, i = l,...,r. Consider E 3 F(a,) D F. Then E is the splitting field of ma(X) G F(ai)[X], so, by Lemma 2.6.3, there is an automorphism of E extending the isomorphism at. Denote this extension by at as well. Then a^ | F = id, so Gi G Gal(E/F). Note that (j/(F(ay)) = F(ayt) for some k, as the automorphism Gi of E must permute the roots of the irreducible polynomial m^ (X), Now consider E as an extension of F(a). Then (E/F(a)) < (E/F), and E is the splitting field of the separable polynomial ma(X) e F(a)[X], so by induction E is a Galois extension of F(a), i.e., F(a) is the fixed field of Gal(E/F(a)). By Lemma 2.7.6, Gal(E/F(a)) is a subgroup of Gal(E/F), so this implies that B = Fix(Gal(E/F)) c F(Qf). We wish to show that B = F.

2.8 The Fundamental Theorem of Galois Theory

29

Let rhaiX) e B[X] be the minimum polynomial of a regarded as an element of B. Since B c F(a), we see that B(a) = F(a), and then degmJX) = (B(a)/B) = (F(cy)/B) < (F(a)/F) = degm«(X) with equality if and only if B = F. Thus it remains to show that deg rha (X) = dcg ma(X), In fact we show that rhaiX) = ma(X). Now rhaiX) divides ma(X) and ma(X) = YYi=i(^ ~ ^i) t>y Lemma 2.7.12. Consider a/. We have constructed an element at e Gal(E/F) with at (a) = at, i.e., there is an automorphism a/ of E with at {a) = at and at | F = id. But B = Fix(Gal(E/F)) so a, | B = id, and hence a, e Gal(E/B). Then atirhaiX)) = rhaiX), as ai(ma(X)) e B[X]; thus rhaiat) = rhaicxiiot)) = atirhaia)) = a,(0) = 0, so at is a root of rhaiX) for each /, i.e., X — at is a factor of rhaiX) for each /, and so rhaiX) = niaiX), as required. D

2.8 The Fundamental Theorem of Galois Theory In this section we reach our first main goal. Before we get there, we have some work to do. We let F* denote the set of nonzero elements of F, and recall that F* is a group under multiplication. Definition 2.8.1. A character of a group G in a field F is a homomorphism aiG^F*. o Example 2.8.2. (1) Let G = F*. Then id: G ^ F* is a character. (2) Let G = F*. For any automorphism a of F, a : G -^ F* is a character. (3) Let G = F*. For any map of fields a : F ^ character.

E, a : G ^

E* is a o

Definition 2.8.3. A set of characters {a^ }/=!,...,« is dependent if there is a set of elements {a/}i=i,...,„ of F, not all zero, with aicri -\ hanCfn = 0 (i.e., with (aiOTi H \-cincrn)(x) = 0 for every x e G). Otherwise it is independent, o Theorem 2.8.4 (Dirichlet). Let {cri}i=i,,..,n be a set of mutually distinct characters ofG in F. Then {a/} is independent.

30

2 Field Theory and Galois Theory

Proof (Artin). By induction on n. Suppose n = 1. Then a\a\ = 0 implies a\ =0 (as a ( l ) = 1). Suppose now that the theorem is true for any set of fewer than n characters and consider {cr/}/=i,...,«. Suppose a\G\ -\ h GnCfn = 0. Some at is nonzero. Let that be a\. Since ai and a„ are distinct, there is an jc G G with G\{X) ^ Gnix). Now by hypothesis, for every y e G, diGiiy) H

h anOniy) = 0,

so, multiplying by (Jn(x), (*) aian(x)ai(y) H

h«nOr„(x)or„(y) = 0.

On the other hand, for every y e G, aiaiixy) -\

h a„a„(xy) = 0;

using the fact that at (xy) = at (x)ai (y), (**) ai(riix)ai(y)

+ • • • + ancrn(x)an(y) = 0,

and then, subtracting (*) from (**), aiiaiix)

- anix))ai(y) -\

\-an-i(crn-i(x) - a„(x))a„_i(>;) = 0

for every y e G.By induction, each coefficient is zero, so in particular cri(x) = a„ (jc), a contradiction. n Theorem 2.8.5. Let E = {a^, / = 1,.,, ,n} be a group of mutually distinct automorphisms of afield E, and let F be the fixed field ofTi, Then (E/F) = n. Proof (Artin) First we show n < (E/F) (which only uses the fact that S is a set) and then we show n > (E/F) (which uses the fact that S is a group). We prove both inequalities by contradiction. Let r = (E/F). For the first step, suppose r < n. Let { 6 i , . . . , 6^} be a basis for E over F. Consider the system of equations in E: "^l(^l) ••• Crn(€i)

ot\

= 0. CTli^r) ••• Cr^i^r)

Otn

This system has more unknowns than equations, so it has a nontrivial solution which we still denote by a i , . . . , a„. Then

2.8 The Fundamental Theorem of Galois Theory

31

n

y ^ Qf/CT/{€k) = 0 for each k = \, , . . , r. /=i

Let 6 € E be arbitrary. Since [ek] is a basis, we have that 6 = Xl^=i /A:^it with fk € F, and since F is the fixed field of E, at (fk) = fk for each /, k. Then

i=0

i=0 r

k=l n

i,k

i,k r

= E hC£^^i^i^^k)) = Y. A(0) = 0, A:=l

/=1

k=\

SO Xl/^=i ^/^^ "= 0' contradicting Theorem 2.8.4. For the second step, suppose r > n. Since S is a group, some at is the identity. Let it be ai. Let {^i,..., €„+i} be a set of F-linearly independent elements of E. Consider the system of equations in E: 'cfiiei) ••• o-i(€„+i)'

Pi = 0.

CTni^l) '"

Cfn(€n+l)

Pn+1

This system has more unknowns than equations, so it has a nontrivial solution which we still denote by ^ i , . . . , ^n+i- Among all such solutions choose one with the smallest number s of nonzero values. By permuting the subscripts, we may assume this solution is )Si,..., ^^, 0 , . . . , 0. Note that 5 > 1, since if 5 = 1, 0 = Picfiiei) = )Si6i implies ^i = 0 as €i 7^ 0. Then we may further assume (multiplying by P~^ if necessary) that ^8^ = L Finally, not all Pi are in F, since if they were, we would have 0 = Piai (^i) -h h Ps^^i i^s) = iSi^i H h^s^s^ an F-linear dependence of {^i,..., 6„}. Thus we may finally assume that jSi ^ F. Then we have (*r) iSior/(6i) + '" + Ps-icri{es-i) + ati^s) = 0 , / = 1 , . . . , n. Since jSi ^ F = Fix(E), there is a OTJ^ G S with crk(Pi) i=- P\. Since S is a group, for any oi € S there is a or^ G S with oi = 0^0j. Then applying ak to (*j), we have (**0 (rk{Pi)ai{ei) + "' + ak(Ps-i)(yi(^s-i) + or.fe) = 0, for / = 1 , . . . , n. Subtracting (**/) from (*j) gives (* * */) (A - crk(Pi))ai(ei) -h • • • + (Ps-i - or,fe-i))a,fe_i) = 0, for / = 1 , . . . , n. Now ^1 — cfkiPi) # 0 and this equation has fewer than s nonzero coefficients, a contradiction. D

32

2 Field Theory and Galois Theory

Corollary 2.8.6. (1) Let G be a finite group of automorphisms ofE and let F = Fix(G). Then any automorphism of Infixing F must belong to G. (2) Let Gi and G2 be two distinct finite groups of automorphisms ofE, and let Fi = Fix(Gi), F2 = Fix(G2). Then Fi and F2 are distinct. Proof (1) (E/F) = \G\ = n, say, by Theorem 2.8.5, so if there were some automorphism cr of E fixing F that was not in G, then F would be fixed by at least n + 1 > (E/F) automorphisms of E, a contradiction. (2) We prove the contrapositive: If Fi = F2, then Fi is fixed by G2, so G2 ^ Gi by part (1), and vice versa, so Gi = G2. • Lemma 2.8.7. Let G be a group of automorphisms of F fixing F and let H be a subgroup of G. LetB = Fix(H), Thenforanya e G, cr(B) = Fix(aHa~^). Proof Let ^ G B and r e H. Then ara-\a(fi)) = a r ( ^ ) = a()8), so or(B) CFixCa/fa-i). On the other hand, if orror-HiSO = ^'for some ^S' G E and all r € //, then ra-Hb') = o'-^^O; so a-^iSO = P e Fix(H) = B and fi' = a(fi), so Fix(aHa-^) c a(B). D Here is the most important theorem in this book, and one of the fundamental theorems in all of mathematics. Theorem 2.8.8 (Fundamental Theorem of Galois Theory = FTGT). Let E be a finite Galois extension ofF and let G = Gal (E/F). (1) There is a one-to-one correspondence between intermediate fields E 2 B 2 F and subgroups {1} ^ GB ^ G given by B = Fix(GB). (2) B is a normal extension ofF if and only ifG^ is a normal subgroup of G. This is the case if and only ifB is a Galois extension ofF. In this case Gal(B/F) = G / G B . (3) For each E 2 B 3 F, (B/F) = [G : GB] and (E/B) = |GB|. We say that the intermediate field B and the subgroup GB "belong" to each other. Before proving this theorem, we make some observations. Remark 2.8.9. (1) If Bi c B2, then GBJ 2 GBJ, SO the correspondence is "inclusion-reversing". (2) E is always a Galois extension of B and GB = Gal (E/B), both claims being true by the definition of a Galois extension (Definition 2.7.9).

2.8 The Fundamental Theorem of Galois Theory

33

(3) Since E / F is separable, B/F is separable for any B intermediate between F and E (recall Definition 2.8.3 (3)), so, by Theorem 2.7.14, B/F is normal if and only if B/F is Galois. This justifies the second sentence of Theorem 2.8.8 (2). (4) As we shall see from the proof, in case B is a Galois extension of F, the quotient map Gal(E/F) -> Gal(B/F) is simply given by restriction,

a \-^ a \B.

o

Proof. (1) For each subgroup H and G, let

BH=Fix(H). This gives a map r : {subgroups of G} -^ {fields intermediate between F and E}. We show r is a one-to-one correspondence. r is one-to-one: If Hi 7^ H2, then Fix(//i) 7^ Fix(H2) by Corollary 2.8.6 (2). r is onto: Let F c B c E and let H = {a eG\

a | B = id} = Gal(E/B) c Gal(E/F).

Since E / F is Galois, it is the splitting field of a separable polynomial f(X) e ¥[X] by Theorem 2.7.14. But F c B so f(X) e B[X]. Then E is the splitting field of the separable polynomial f(X) e B[X], so E/B is Galois, again by Theorem 2.7.14, and hence B = Fix (Gal (E/B)) = Fix(H). Thus r is one-to-one and onto, and if r~^(B) = GB, then B = Fix(GB). (3) Since E/B is Galois, we know that (E/B) = | Gal(E/B) | = | GR I, by Theorem 2.8.5, and (E/F) = (E/B)(B/F), while \G\ = | G B | [ G : G B ] , SO (B/F) = [G : G B ] . (2) Suppose that GB is a normal subgroup of G. For any a e Gal (E/F), or(B) = Fix(orGBcr"^) = Fix(GB) = B, by Lenrnia 2.8.7. Hence we have the restriction map R : Gal(E/F) -> Gal(B/F) given by a i-> a | B, and Ker(R) = {or G G | a | B = id} = GB by definition. Thus Im(R) = Gal(E/F)/Ker(/?) = G / G B - But let CXQ e Gal(B/F). Then E/B is Galois, so E is the splitting field of a (separable) polynomial f(X). Then OTQ: B ^- B extends to a : E -^ E by Lemma 2.6.3, and a | F = id, so a e Gal(E/F). Then R(a) = ao, so R: Gal(E/F) -> Gal(B/F) is onto and hence Gal(B/F) = G/GB.

34

2 Field Theory and Galois Theory

Conversely, suppose that B/F is Galois. Then B is the splitting field of a separable polynomial f{X), Let / ( X ) have roots ^ i , . . . , ^^ in B, and note, by Remark 2.5.6, that B = F()Si,..., )8,). Let a G G = Gal(E/F). Now f{X) e F[X], so or(f{X)) = f(X) and hence a permutes the roots of f(X), i.e., for every /, cr(Pi) = fij for some j . In particular this implies that a(B) = B. But now, by Lemma 2.8.7, Fix(GB) = B = or(B) = Fix(aGBor"^), so, by part (1), GB = (TGBO^"^. Since a is arbitrary, GB is a normal subgroup of G. n In the next section, we shall give a number of examples of the FTGT, but first we continue here with our theoretical development. Definition 2.8.10. Two fields Bi and B2 intermediate between E and B are conjugate if there is a a e Gal(E/F) with a(Bi) = B2. o Corollary 2.8.11. (1) Two intermediate fields Bi and B2 are conjugate if and only I/GBI ^^^ ^^2 ^^^ conjugate subgroups of Gai(K/F). (2) B is its only conjugate if and only if GB is a normal subgroup of Gal(E/F). Proof Immediate from Lemma 2.8.7 and Theorem 2.8.8.

n

Returning to the situation of the Fundamental Theorem of Galois Theory, let E be a finite Galois extension of F, and let B be an intermediate field between E and F. Definition 2.8.12. Let QB = {o-o: B -> B' an isomorphism | F c B' c E, ao | F = id}. Let A: 2 B ^^ {left cosets of GB in G] be defined as follows: Let o-Q € 2 B - Since E is a splitting field, GQ extends to a G Gal(E/F) by Lemma 2.6.3. Then set A(ao) = [a^i e G / G B .

O

Proposition 2.8.13. (1) A is well defined, (2) A is one-to-one and onto, (3) Gal (B/F) c Q^ with equality if and only ifG^ is normal, or equivalent if and only if B/F is Galois, In this case, A is an isomorphism of groups, A:QB^G/GB^

Proof (1) We need to check that A(oro) is independent of the choice of extension a. Let a' be another extension of OTQ. Then a' | B = ao = a | B, i.e., a'(l3) = a(fi) for all yS G B, so (a-^a')(P) = ^ for all )S G B, and hence a~ V G G B , SO a^ e GGB and [a^] = [a].

2.8 The Fundamental Theorem of Galois Theory

35

(2) Suppose A(oro) = [cr] and A(ro) = [r] with [a] = [r]. Then x = ap for some p € GB, so T | B = a p | B . But p G GB so yo | B = id; hence r I B = (J I B, i.e., to = (TQ, and A is one-to-one. Also, let a G s be a left coset of GB in G, and let ao = a | B, ao: B ^ B' = ao(B) c E. Then A(or) = [ao], so A is onto. (3) Clearly Gal(B/F) c Qg as Gal(B/F) = {ao G GB I B' = B}. The first sentence then follows directly from Corollary 2.8.11. As for the second, if A(a) = [ao] and A(r) = to, then a | B = ao and r | B = to, so a r I B = aoTo and hence A ( a r ) = [aoro]. But if GB is normal, (o^oGB)(roGB) = OTOTOGB, i.e., [aoTo] = [oro][ao], so A ( a r ) = A(a)A(r). Thus A is a group homomorphism that is one-to-one and onto as a map of sets, so A is an isomorphism. n Proposition 2.8.14. Let E be a finite Galois extension ofF and let Bi and B2 be intermediate fields with Bi = Fix(//i) and'Ri = Fix(//2)Then: (1) B1B2 = Fix(Hi n H2) (2) Bi n B2 = Fix{H^) where H^ is the subgroup generated by H\ and H2. Proof. (1) If a G Jf/i n H2, then a G / / i , so a fixes Bi, and a G H2, so a fixes B2; hence a fixes B1B2. On the other hand, if a fixes B1B2, then a fixes Bi, so a G Hi, and a fixes B2, so a G H2\ hence a G if1 Pi H2. (2) If a ( ^ ) = ^ for every a G i/3, then a ( ^ ) = ^ for every a G Hi, so P G Bi, and a ( ^ ) = ^ for every a e H2,so fi e B2; hence ^ G Bi fi B2. On the other hand, if )S G Bi fl B2, then & G Bi, so a ( ^ ) = p for every a e Hi, and fc G B2, so a ( ^ ) = )S for every a e H2; hence a(P) = ^ for every a G H3.

D

In the original development of group theory, and in particular in the work of Galois, groups were not regarded as abstract groups, but rather as groups of synmietries, and in particular as groups of permutations of the roots of a polynomial. This viewpoint, still a most useful one, is encapsulated in the following result. In order to state it, we first recall that an action of a group G on a set 5 = {si} is transitive if for any two elements st and Sj of 5, there is an element a of G with a(^0 = Sj, Proposition 2.8.15. Let E be a finite Galois extension ofV, so E is the splitting field of a separable polynomial f(X) e F[X], Then G = Gal(E/F) is isomorphic to a permutation group on the roots of f{X). If f{X) is irreducible, then G is isomorphic to a transitive permutation group on the roots of f(X),

36

2 Field Theory and Galois Theory

Proof. Let QTi,..., Qfj be the roots of f{X) in E. For any a e Gal(E/F), a{(Xi) is a root of or (/(X)) = / ( X ) , so Gal(E/F) permutes {ai]. Furthermore, E = F ( a i , . . . , ad) (Remark 2.5.6), so if a(cy/) = a/ for each /, then a = id. Now if f{X) is irreducible, and of/ and Uj are any two roots of / ( X ) , there is isomorphism GQ: F(a/) -> ¥{otj) with a | F = id and cfQipti) = ay (Lemma 2.6.1), and (JQ extends to a : E -> E (Lemma 2.6.3). n Remark 2.8.16, Note Proposition 2.8.15 implies Lemma 2.5.3 in case f(X) is separable. o Remark 2.8.17. Note that Proposition 2.8.15 is about how the Galois group is realized, not about its structure as an abstract group. For example, consider E = Q(V2, V3) and F = Q. Then G = Gal(E/F) = (Z/2Z) 0 (Z/2Z), and E is the splitting field of the polynomial (X^ — 2)(X^ — 3), which is not irreducible, so G does not act transitively on {V2, — V2, \ / 3 , — v ^ } . On the other hand E = Q(\/2 + V3) and E is the splitting field of the polynomial

(X - (V2 + V3))(X - (V2 - V3))(X - (-V2 + V3))(Z - (-V2 - V3)) = X"^ — lOX^ + 1, which is irreducible, and G does indeed act transitively on

{V2 + V3, V2 - V3, -V2 + V3, -V2 - V3}.

o

Corollary 2.8.18. Let f{X) G F [ X ] be a separable polynomial and write f(X) = fi(Xy^ '' • fkiXy^y a factorization into irreducibles in F[X]. Let fi(X) have degree dt. Let E be the splitting field of f(X) e F[X]. Then Gal(E/F) is isomorphic to a subgroup of Sd^ x - - - x Sdj, (where Sd denotes the symmetric group on d elements). In particular, if f(X) e ¥[X] is a separable polynomial of degree d, and E is the splitting field of f(X) e F[X], then Gal(E/F) is isomorphic to a subgroup ofSd^ • Theorem 2.8.19. Let E be a finite Galois extension of F of degree d. Then G = Gal(E/F) is isomorphic to a subgroup of Sd, the symmetric group on d elements. Also, G has a subgroup H of index d. Proof. Since E is a Galois extension of F, it is certainly a separable extension of F (Proposition 2.8.14).We now apply a result we shall prove later (Corollary 3.5.3) to conclude that E = F(a) for some a G E. Then d = (E/F) = (F(Qf)/F) = dtgma(X) (Proposition 2.4.6 (2)), so the theorem follows from Proposition 2.8.15 and the following general fact: If G acts transitively on S = [si], a set of cardinality d, then H = [a e G | a(5'i) = ^i} is a subgroup of index (i. D

2.9 Examples

37

2.9 Examples In this section we give a number of examples of the Fundamental Theorem of Galois Theory, and compute a number of Galois groups. Example 2.9.1. See Example 1.1.1.

o

Example 2.9.2. See Example 1.1.2.

o

Example 2.9.3. See Example 1.1.3.

o

Example 2.9.4. See Example 1.1.4.

o

Example 2.9.5. Let E^ be the splitting field of f(X) — X^ —a over Q, where for simplicity we take a to be an integer. There are several cases. a = 0: Then f{X) = {X){X){X) is a product of linear factors, so E^ = Q. (E^ is obtained from Q by adjoining 0 three times, but adjoining 0 does nothing.) Note (E«/Q) = 1. a 7^ 0 a perfect cube: Let a = Z?^Then/(X) = X^-b^ = b\(X/b)^-l) so E^ = El. Thus we need only find the splitting field of f(X) = X^-l, This we have done already. Ei = Q((i>) where o) = (—1 + i^/3)/2 is a primitive cube root of 1. Note (E«/Q) = 2 and Gal(E«/Q) = Z/2Z. a 7^ 0 not a perfect cube: This situation is entirely analogous to Example 1.1.3. Ea = Q(co, ^/a) and (E«/Q) = 6. Also, G = Gal(E^/Q) = De. the dihedral group of order 6. Note that we can determine G without doing any computation. We know that \G\ = (E^/Q) = 6, and there are only two groups of order 6, Z/6Z, which is abelian, and D^, which is not. Now E« D Q(v^) and Q(^) is not a normal extension of Q, so G has a subgroup that is not normal and hence G is nonabelian. o Example 2.9.6. E is the splitting field of f(X) = X^ -3 over Q. Then E = Q(f, v^) where f is a primitive sixth root of 1 and (E/Q) = 12. We may choose f = (1 + /V3)/2. Then co = ^^ and - 1 = f ^. But then notice also that ^ = 1 + a; so in particular Q(^) = Q(o)). An element a of Gal (E/Q) is determined by its effect on ^ and on ^/3. One can check that G = Gal(E/Q) has elements a and r with:

Clearly a^ = 1 and r^ = 1. AlsoorrCf) = f^ = T - V ( f ) andarCv^) = f - i ^ = r~^a(\/3). Since any element of G must take ^ to ^ or f ~ \ and

38

2 Field Theory and Galois Theory

must take ^ to f ^ ^ for some /, we see that a and r generate G. Thus we conclude that G = (a, r I or^ = 1, r^ = 1, / ax = r~^a) is isomorphic to the dihedral group D^. It is then routine but lengthy to check that G has 15 subgroups, which we list below, where subgroups denoted by the some letter are mutually conjugate (so in particular Ai, Di, Ei, Gi, Hi and /i are normal): Ai ={1} Bi = [I, a), B2 = {1, r^cT}, B^ = [I, r V } Ci = {1, ra}, Ci = {1, r V } , C3 = {1, T V } El = {1, r\ r^} Fi = 1, a, T^, T^cr}, F2 = {1, Tcr, T^, T V } , F3 = {1, T^c, T^, T^a} Gi = {1, T^, r"*, cr, r^or, T V } HI

/i

= [1, T, r2, T^ r4, T^} =G

These subgroups have the following fixed fields: Fix(Ai) = E Fix(Bi) = Q ( ^ ) , Fix(B2) = Q(?v^), Fix(B3) = Q(ft^v^) Fix(Ci) = Q((l + O^), Fix(C2) = Q ( i ^ ) , Fix(C3) = Q((l + r')^) Fix(Di) = Q ( ^ ^ ) Fix(£i) = Q ( ^ V3) Fix(Fi) = Q ( ^ ) , Fix(F2) = Q(f v ^ ) , FixCF?) = Qico^) Fix(Gi) = Q(V3) Fix(//i) = Q(C) Fix(/i) = Q The 6 subfields fixed by normal subgroups are Galois extensions of Q, so are splitting fields of separable polynomials: Fix(Ai) = splitting field of X^ - 3 Fix(Di) = sphtting field of X ' - 3 FixCFi) = splitting field of (X^-X + 1)(X^- 3) Fix(Gi) = sphtting field of X^ - 3 Fix(ifi) = sphtting field of Z^ - X + 1 Fix(/i) = splitting field of 1 Note here there are also interesting relationships between the fields intermediate between E and Q. For example, let F = Fix(Fi) = Q ( ^ ) . Then Bu

2.10 Exercises

39

C2, and Di ait all normal subgroups of Fi (although Bi and C2 are not normal subgroups of G). Hence Fix(5i)/Fi, Fix(C2)/Fi, and Fix(Di)/Fi are all Galois extensions, all of degree two, hence v/ith Galois groups isomorphic to Z/2Z, and they are the splitting fields of the polynomials X^ - v ^ , X^ + 4 ^ , mdX^ + X + l e Q(v^)[X], respectively. o Example 2.9.7. E is the splitting field of Z^ + 3 over Q. Again E = Q(f, \/^) where f is a primitive sixth root of 1. However, ( v ^ ^ ) ^ = / \ / 3 , so f G Q(^^^) and E = Q(V-3). Thus in this case (E/Q) = 6, and here Gal(E/Q) = De by an argument as in Example 2.9.4. o Example 2.9.8. If (E/Q) = 3, then, by Theorem 2.8.19, Gal (E/Q) is a subgroup of 53 whose order is divisible by 3, so it is either 5*3 or Z/3Z. The construction in part (4) of this example always gave 53. Here is a case where the Galois group is Z/3Z. Let f be a primitive ninth root of 1 and consider the polynomial f(X) = (X - (? + r ' ) ) ( ^ - ( ? ' + r ' ) ) ( ^ - ( ? ' + r ' ) ) = X^ — 3X + 1 € Q[X]. This polynomial is irreducible over Q (as otherwise it would have a linear factor, i.e., a root in Q, and it does not). Let AQ, AI, and X2 be the roots of this polynomial, in the order written, and let E = F(Ao, Ai, A2) be its splitting field. Then (E/Q) = deg f(X) = 3, and since | Gal(E/Q)| = (E/Q), the Galois group must be isomorphic to Z/3Z. We can see this concretely as follows: Note that Xl = ki+2 and A^ = A2 + 2 (and also kl = Xo + 2), so in fact E = F(Ao), and a e Gal(E/Q) is determined by a(Xo). Since /(X) is irreducible, there is an element at € Gal (E/Q) with at (XQ) = Xi,i =0,1, 2. Clearly ao = id.Ifori(Ao) = A.i,thenori(A.i) = ai(A§-2) = ai(Xo)^-cri(2) = A ^ - 2 = A.2 and similarly 0-1(^2) = XQ, Then crfiXo) = cri(ai(Xo)) = 0^1 (A-i) = X2, so af = a2\ similarly aliX^) = XQ, SO af = (TQ, and we see explicitly that Gal(E/Q) = {l, au af}, o

2.10 Exercises Exercise 2.10.1. In each case, find a polynomial in Q[X] that has a as a root. Then find all complex roots of this polynomial. (a) a = 1 + V2. (b) a = V2 + V3. (c) a = y i + V2. (d) a = ^ 3 - 2V2.

(e) a = /r+7^.

40

2 Field Theory and Galois Theory

(f) a = ^l + V2. (g) a = V T + ^ . (h) a = y 6 + x/3 + ^ 6 - V s . (i) a = 7 4 + V7 + y 4 - V 7 . Exercise 2.10.2. Let / ( X ) = X^ + 7X + 1 G Q [ X ] . This polynomial is irreducible. Let E = Q[a] for some root a of f(X), Consider the following elements of E: ^i = of + 3, ^2 = c^^ + 1, ^3=oi^ -2a + 3. (a) Express each of the following in the form gia) for some polynomial

g(X)eQ[X]:^^\

p-\

p^\ ^,^2. Pi^3. ^2^3^

(b)Findm^^(Z), m^,(Z),

m^,{X)eQ[Xl

Exercise 2.10.3. Let f{X) = X^ +X^ + 2 e FsLX]. This polynomial is irreducible. Let E = F5[a] for some root a of f(X). Consider the following elements of E: )8i = 2a + 1, ^2 = a^ - 1, ^63 = a^ - a -f 2. (a) Express each of the following in the form g(a) for some polynomial

g(X)eFs[X]:^-\

p^\ p^\ ^1^2, ^3.

^2^3-

(b)Findm^^(Z), m^,(Z), m^,{X) € FsLX]. Exercise 2.10.4. As we shall see in Section 3.3, for every prime power q = p^, there is a unique (up to isomorphism) field F^ with q elements. (a) Write down an explicit addition and multiplication table for F2, F3, and F5. (b) Write down an explicit addition and multiplication table for F4 and F9. (c) For each element a G F4, find its minimum polynomial niaiX) e F2[X], and for each element a G F9, find its minimum polynomial ma(X) e F3[Z]. (d) Explicitly compute the Frobenius endomorphism <[> on F4 and F9. (e) Write down an explicit addition and multiplication table for Fg. (f) For each element a G Fg, find its minimum polynomial ma(X) e F2[X]. (g) Explicitly compute the Frobenius endomorphism O on Fg. Exercise 2.10.5. (a) Find all irreducible quadratic, cubic, and quartic polynomials f(X) G F2[X]. (b) Find all irreducible quadratic, and cubic polynomials f(X) e F3[Z]. (c) Find all irreducible quadratic polynomials f(X) e F5[Z]. (d) Find all irreducible quadratic polynomials /(X) e F4[Z]. Exercise 2.10.6. Let / ( X ) = a^X'^ + • • • -f ao G F[X] split in E. Show that (1) implies (2): (1) For every a G E, if a is a root of / ( Z ) , then —a G E is also a root of

f(X). (2) Gn-k = 0 for all odd k.

2.10 Exercises

41

Exercise 2.10.7. Let f{X) = anX"" -\ h ^o ^ F[^] split in E. Show that (1) implies (2): (1) For every a 7^ 0 G E, if a is a root of / ( X ) , then l/of € E is also a root of (2) For ^ = ± 1 , an-k = eau for all k. Exercise 2.10.8. (a) Let f{X) e Z[X] be a monic polynomial. Show that if its mod p reduction f(X) e Fp[X] is irreducible for some /?, then f(X) is irreducible. (b) Use this to show that f(X) = Z^ - 5X + 36 is irreducible. Exercise 2.10.9. A priori, there is a distinction between polynomial expressions and polynomisil functions, i.e., between f(X) regarded as an element of F[X] and f(X) as defining a function on F. Clearly, if f(X) = g(X) as elements of F[X], then f(X) = g(X) as functions on F[X] (i.e., f(a) = g(a) for every a eF), (a) Show that, if F is infinite, the converse is true: If f(a) = g(a) for every a € F, then f(X) = g(X) e F[X]. (b) Find a counterexample to the converse if F is finite. That is, if F is finite, find polynomials f(X) ^ g(X) e F[X] but with f(a) = g{a) for every aeF, (c) Use the result of (b) to show that no finite field is algebraically closed, i.e., that if F is any finite field, there is a nonconstant polynomial f{X) € F[X] that does not have a root in F. Exercise 2.10.10. (a) Let f{X) G F [ Z ] have roots ofi,..., a„ in some splitting field E. Show that f{X) is irreducible in F[X] if and only if

Y\{X-ai)iF[X] ateT

for any nonempty proper subset T of 5 = { a i , . . . , a„}. (b) Observe that E = Q[V3, V5] is the splitting field of/i (X) = X'^-46X^ + 289, with roots Vs + 2^/5, VS - 2V5, - V S + 2x/5, and - Vs - 2V5, and also of f2(X) = X4 - 6X^ + 5X^ + lOX + 2, with roots 1 + VS, 1 - VS, 2 + VS, and 2 — VS. Use part (a) to show that fi(X) is irreducible in Q[X] but/2(X)isnot. Exercise 2.10.11. Let / ( X ) , g{X) G F[X] be irreducible polynomials. Show that if f(X) and g(X) have a common root in some extension E of F, then

f(X) = g(X), Exercise 2.10.12. Let E 2 F and let a 7^^ 0 G E. Show that a is algebraic over F if and only if 1/a = f(a) for some polynomial f(X) e F[X],

42

2 Field Theory and Galois Theory

Exercise 2.10.13. Let /? be a subring of F and let E be an extension of F. Let Of G E be algebraic over F. If niaiX) e R[X], then a is called integral over R. In case R = Z and F = Q, show that a G E is integral over R if and only if f(a) = 0 for some monic polynomial f(X) e Z[X]. (Such a complex number is called an algebraic integer.) Exercise 2.10.14. Suppose that /? = Z and F = Q. Show that, for any of G E, there is an integer n ^0 eZ such that na is an algebraic integer. Exercise 2.10.15. Let E be a field and let /? be a subring of E. If every element of E is integral over R, show that /? is a field. Exercise 2.10.16. Let E be an extension of F and let f(X) e F[X] be an irreducible polynomial. Let f(X) = fi(X)f2(X) • • • /,(X) G E[X] be factored as a product of irreducible polynomials. (a) If E is a Galois extension of F, show that {/i(X), / 2 ( X ) , . . . , ft(X)} are mutually conjugate in E[X]. (Two polynomials f\{X), fiiX) e E[X] are mutually conjugate if there exists an element a e Gal(E/F) with a ( / i ( X ) ) = fiiX),) (b) Give a counterexample to this if E is not a Galois extension of F. Exercise 2.10.17. Let E be a splitting field of the separable irreducible polynomial f(X) G F[Z]. (a) For a root a G E of / ( X ) , let r(a) be the number of roots of /(X) in F(a). Show that r(a) is independent of the choice of a. Call this common value r. (b) Let d be the number of distinct fields {F(a) | a G E a root of f(X)}. Show thatrfr = d e g / ( X ) . (c) Give examples where r = 1, 1 < r < deg f(X), and r = deg f(X). Exercise 2.10.18. Let E be the splitting field of the separable irreducible polynomial f(X) G F[X] and let B be a Galois extension of F with F c B c E. Let f(X) = fiiX) • • • fk(X) G B[Z] be a factorization into irreducible polynomials. Show that each fi (X) has the same degree d. Furthermore, if a G E is any root of / ( X ) , show that d = (B(a)/B) and that k = (Bn F(a)/F). Exercise 2.10.19. Let f(X) e Q[X] be an irreducible cubic with roots a, ^,y in some splitting field E. Show that one of the following two alternatives holds: (1) Gal(E/Q) = Z/3Z, and there is no field B with Q C B c E. (2) Gal(E/Q) = 53, and there are exactly four fields B with Q C B C E, these fields being Q(a), Q(^), Q ( K ) , and Q ( \ / D ) for some D eQ that is not a perfect square. (In fact, it is possible to decide between these two alternatives, and to find D, without finding the roots of / ( X ) . See Section 4.8.)

2.10 Exercises

43

Exercise 2.10.20. Let f{X) = X"" - p e Q[X]. By Proposition 4.1.7 below, f{X) is irreducible. Let E be a splitting field of f(X). Show that for every n > 3, the Galois group G = Gal(E/Q) is not abelian. Exercise 2.10.21. Let E be a splitting field of the separable irreducible polynomial f(X) e F[Z]. Let a be any root of f(X) in E. Let /? be a prime dividing (E/F). Show there is a field F c B c E with E = B(a) and with (E/B) = p. Exercise 2.10.22. Let F be an arbitrary field. Let m and n be positive integers and let d = gcd(m, n). Let f(X) = X'" - 1, g(X) = X"" - I e F[X]. Show that gcd(/(X), g(X)) = h(X) where h(X) = X^ - 1 € F[X]. Exercise 2.10.23. Let / ( X ) e Q[Xl If f(X) e Z[X], then certainly f(n)eZ for every n e Z. The converse of this is false, as the example f(X) = X(X + l)/2 shows. Let co(X) = 1 and let Ck(X) = X(X - 1) • • • (X-(kl))/k\ for A: > 0. Show that any / ( X ) e Q[X] with the property that f(n)eZ for every n eZ can be written a linear combination of {ck(X)} with integer coefficients. (This remains true if Q is replaced by any field of characteristic 0.) Note this implies that if f(n) is a polynomial of degree at most k with f(n) an integer divisible by k\ for every integer n, then f(X) e Z[X],

Development and Applications of Galois Theory

3.1 Symmetric Functions and the Symmetric Group We now apply our general theory to the case of symmetric functions. We let D be an arbitrary field and set E = D ( X i , . . . , Xd), the field of rational functions in the variables X i , . . . , Xj. Then the symmetric group 5j acts on E by permuting { X i , . . . , Zj}. Definition 3.1.1. The subfield F of E fixed under the action of Sd is the field of symmetric functions in X i , . . . , Xj. (F = { / ( X i , . . . , X^) e E | / ( X ^ ( i ) , . . . , X^(j)) = / ( X i , . . . , Xj) for every a e Sd}.) A polynomial / ( X i , . . . , Xd) G F is a symmetric polynomial. o Since F = Fix(5j), by definition E is a Galois extension of F with Galois group Sd. We may use this observation to easily show that any finite group is the Galois group of some field extension. Lemma 3.1.2. (1) Let G be a finite group. Then there is a Galois extension E/B with Gal(E/B) = G. (2)There is a Galois extension E/B with Gal(E/B) = G and with E (and hence BJ a subfield of the complex numbers C. Proof (1) Any finite group G is isomorphic to a subgroup of some symmetric group Sd. Identify G with its image under this isomorphism. Let E = D ( X i , . . . , Xd), as above, and set B = Fix(G). Then E/B is Galois with Galois group isomorphic to G. (2) Let D = Q. Then we need only show there is subfield of C isomorphic to Q ( X i , . . . , Xj). Let ti,,,, ,td be algebraically independent elements of C, i.e., elements that do not satisfy any polynomial equation.

46

3 Development and Applications of Galois Theory

Let (p: Q ( X i , . . . , X j ) -^ Q ( ^ i , . . . , ^j) be defined by (^(X,) = f,. If r ( Z i , . . . , X ^ ) = / ( X i , . . . , Z ^ ) / g ( X i , . . . , Z ^ ) G Q ( Z i , . . . , X j ) , then ^ ( r ( X i , . . . , Xd)) = fih,..., ^ j ) / g a i , . . . , ^^) = r ( r i , . . . , fj). Then cp is well defined as the denominator g ( ^ i , . . . , ^j) is never zero. It is obviously onto and it is 1-1 as the numerator fih,... ,td) is never zero, so cp is an isomorphism. Then set E = Q ( ^ i , . . . , ^^) and let B be the fixed field of G acting by permuting ^ i , . . . , r^. It remains only to shov^ that we can indeed find algebraically independent elements t\,.., ,td of Q. Let Eo = Q and consider the extension field Fi = {z G C I z is algebraic over EQ}. EQ is countable; so there are only countably many polynomials in Eo[Xi], each of which has only finitely many roots. Hence Fi is countable. But C is uncountable. Thus we may choose any t\ ^ Fi and let Ei = Eo(^i). Then X\ \-^ t\ induces an isomorphism Eo(Xi) -^ Ei. But again Eo(Xi) has only countably many elements, so Ei is countable and we may iterate this process. n We now return to the study of symmetric functions. Definition 3.1.3. The elementary symmetric polynomials 5 i , . . . , ^^ are Sk=

^

Xi^Xi^-'Xi^,

fc=l,...,J.

o

\
For example, ifd = A: s\ = Xi + X2 + X2, + X4, 5*2 = X\X2 + XiX'^ + X\X^ + X2X3 + X2X4 + X3X4, •^3 = X1X2X3 + X1X2X4 + X1X3X4 + X2X3X4, S4 = X 1 X 2 X 3 X 4 .

Lemma 3.1.4. The field of symmetric fiinctions F = \y{si,...,

Sd)-

Proof, Let B = D ( 5 i , . . . , 5^). Clearly F 2 B, as each st is fixed by SdNow (E/F) = \Sd\ = dl so (E/B) > dl We will show that F is the splitting field of a separable polynomial f{X) G B[X] of degree d. Then, by Lemma 2.8.18, (E/B) = | Gal(E/B)| < d\. Hence (E/B) = d\ and F = B. Now f{X) is simply the polynomial f{X) = {X-

Xi)(Z -X2)"'{X-

Xd).

which obviously has splitting field E, as its roots are X i , . . . , X^.

n

3.1 Symmetric Functions and the Symmetric Group

47

Remark 3.1.5, (1) Note in particular that we have constructed an explicit polynomial f{X) of degree d whose Galois group is 5^. Explicit computation shows that d

(2) Let us write f(X) = X"^ + ad-\X^~^ H hflo- Then we see from the above expression that the coefficients of f{X) are, up to sign, the elementary symmetric functions of its roots, ad-i = (-lysi,

i =

l,...,d,

and, in fact, this is true for any polynomial f(X).

o

Lemma 3.1.6. Let Y be the monomial Y =

X\xl^-X'f\=X^,X\-^-X'f').

Then E = F(F). Proof. Obviously, the only element of Sd that fixes Y is the identity, so Y has d\ conjugates. Then, by Corollary 2.7.13, (F(F)/F) = d\ = (E/F), so E = F(F). D Corollary 3.1.7. Any element f(Xi,...,

Xd) ofE can be written uniquely as d\-l

f(Xi,...,Xd)

= J2si(^i^'"^'d)Y' i=0

where gt (^i, . . . , Sd) is a rational function insi,...,

Sd.

Proof By Lemma 3.1.6 and Proposition 2.4.6 (1), { 1 , T , . . . , y^'-i} is a vector space basis for E over F, and any element of F is a rational function of 5*1,..., Sd, by Lemma 3.1.4. n Remark 3.1.8. In the language of Section 3.5, Lemma 3.1.6 shows that E is a simple extension of F and that F is a primitive element of E. o Lemma 3.1.9. Let Y be the monomial Y = Xl' " X^~^ (as in Lemma 3.1.6) and for a e Sd, let Y^ be the monomial y yl yj—1 ^or — A^(2) • • ' ^a(d)'

Then {Y^ \ cr e Sd} is a (vector space) basis for E over F.

48

3 Development and Applications of Galois Theory

Proof. We prove the claim by induction on d. For d = I this is simply the claim that F = 1 is a basis for E over E, which is trivial. Now assume the result is true fovd — 1, and suppose that / , C(j{X\, . . . , Xd)Xfj{2) ' ' ' ^a(^) — ^' aeSd

with Ca(Xi,..., Xj) G F. We wish to show that c ^ ( X i , . . . , Xd) = 0 for each a. Assume not. By "clearing denominators" we may assume that {Ca (Xi,..., Xd)} are polynomials in F with no common polynomial factor. For convenience, write the above sum as J^- ^^^ ^ = 1 , . . . , ^, let i/^ = [a eSd\ cf{i) = 1}, / = 1 , . . . , rf. Then E = E i + * *' + E j ' where X / "^ X / ^ ^ ( ^ 1 ' • • • ' ^d)Xa{2) ' ' ' X^^ld)' i

ere Hi

We may further write E i = E^ + E ^ ' ^ — 1 , . . . , rf, where E / i^ taken over those a e Hi with c^iXi,..., Xj) not divisible by Z/, and E ^ is taken over those a e Hi with c ^ ( X i , . . . , Xj) divisible by X/. Call these two subsets Hj and ///^ respectively. Let us consider the equation E == 0- Note that the exponent of Xi is at least one for each term in E^» ^ 7^ 1 (as if a (0 ^ 1, then a(j) = 1 for some j with 2 < j < d), and also for each term in E i • Thus, considering the terms in E that are not divisible by Xi, we thus see that we must have E i = 0' i-^-' that /

, Ca{Xi,

. . . , X^)Xcr(2) • • • ^ a ( j ) = ^•

aeH[

Observe that Hi is isomorphic to 5j_i, permuting the subset { 2 , . . . , J} of { 1 , . . . , (i} (and each Hi is a left coset of ^ i ) . Let us set Xi = 0 in E ' l ' thereby obtaining 0 = 2_^ Ca(0, X 2 , . . . , Xd)X(j(2)... X^^^ = ( 2^ <^a(0, X 2 , . . . , Xd)Xcj{y)''' x^^^ JX2 • • • Xj. Now we apply the inductive hypothesis. 1X^(3) • • • X^^^} is a basis for D(X2,..., Xd) over the subfield of synunetric functions in X 2 , . . . , X^, so in particular they are linearly independent. Hence we have that each coefficient

3.1 Symmetric Functions and the Symmetric Group

49

CaiO, X2,..., Xd) = 0. In other words, each Ca(Xi,..., Xd) is a polynomial that vanishes when we set Xi = 0, so it must have Zi as a factor, contradicting the definition of H[, Hence we see that H[ = 0, and so H[^ = Hi. In other words, Ca(Xi,..., Xd) is divisible by Xi for every a e Hi. But Ca (Xi,..., Xd) € F is a symmetric polynomial in Z i , . . . , X^, so this implies that Ca(Xi,..., Xd) is also divisible by X 2 , . . . , X^, and hence by the product Xi' " Xd, for each a e Hi. Similarly, we see, for each / = 1 , . . . , J, that Hj = 0, Hl^ = Hi, so Ca(Xi,..., Xd) is divisible by Xi, and thus by Xi • • • Xd, for each a e Hi. Thus {C(y(Xi,..., Xd) \ cr € Sd} have the common polynomial factor Xi'' • Xd, contradicting our choice of {CaiXi, . . . , Xd)}. Thus, {^a(2)''' ^t^d)^ is linearly independent, and thus is a vector space basis for E over F (as it has d\ = (E/F) elements), completing the inductive step. n Corollary 3.1.10. Any element f(Xi,...,

Xd) ofE can be written uniquely

as

f(Xi, . . . , Xd) = 22 ^^('^l' • • • ' Sd)Xa(2) ' ' • X^'^J^ creSd

where ha(si, ..., Sd) is a rational function ofsi, ,.. ,Sd. Proof. Lenrnia 3.1.4 and Lemma 3.1.9.

n

Remark 3.1.11. In the language of Section 3.6, Lemma 3.1.9 shows that {^a(2) • • • ^t'{d) I 0^ € 5^} is a normal basis of E over F. o Our discussion here is not yet complete. Corollaries 3.1.7 and 3.1.10 show how to write rational functions uniquely in terms of synmietric functions. What we have not yet shown is that symmetric functions can be written uniquely in terms of the elementary symmetric functions. We do that now. Lemma 3.1.12. L^^ f{Xi,...,Xd) E F be a symmetric function. Then f(Xi,...,Xd) may be written uniquely as a rational function e(si,... ,Sd) of the elementary symmetric functions si, ..., SdProof. We prove this by induction on d. In case d = 1, ^-i = Zi and ^(^i) = f(Xi) is the unique expression. Now suppose the result is true for J — 1. We first need to make an observation about the elementary synmietric functions. We really should have written Sk as Sk,d^ as the expressions in Definition 3.1.4 depended on d, but for simplicity we did not. We do that now. We then observe that if we let Sk,d be the expression obtained from Sk,d by setting Xd = 0, then

50

3 Development and Applications of Galois Theory Sk^ = Sk,d-\ fork
By equating different expressions for f{Xi,..., Xd) and "clearing denominators", it is easy to see that to prove the lemma it suffices to show that the only polynomial eisi^,..., Sd,d) that is identically zero is the zero polynomial. Thus suppose eisi^,..., Sd,d) = 0 but eisi^,..., Sd,d) is not the zero polynomial. Since F is a field and Sd,d is a nonzero element of F, e{si4,..., Sd,d) = 0 if and only if fe,j)"^e(5i,^,..., Sd,d) = 0 for any k. Hence, factoring out by a power of Sd,d if necessary, we may assume, expanding e(si4,..., Sd,d) in powers of Sd,d^ that the constant term is nonzero. In other words, we have that n

0 = e(si4, . . . , Sd,d) = ^ei(si4,

. . . , Sd-\4)s'^^

i=0

with ei(si4,... ,Sd-i,d) polynomials in si^,... ,Sd-i,d^ and furthermore ^o(^i,j, • • •. Sd-i,d) not the zero polynomial. Now set Xd = 0. Then the above expressions become 0 = e(si4,

,.,,Sd-i,d,0)

= ^o('^l,J. • • • . Sd-l,d) = ^o('yi,rf-l. • • • »

Sd-l,d-l),

contradicting thtd — I case.

D

Remark3.1.13. We have just shown that the only polynomial in ^-i,..., 5^ that vanishes is the zero polynomial. This is usually phrased as saying that ^ 1 , . . . , 5"^ are "algebraically independent" over D. o We can also strengthen Lerrmia 3.1.12 as follows. Lemma 3.1.14. Let / ( X i , . . . , Xd) e F be a symmetric polynomial. Then / ( X i , . . . , Xd) may be written uniquely as a polynomial e{si,... ,Sd) in the elementary symmetric functions s\, ... .Sd^ Proof. Uniqueness follows from Lemma 3.1.12 so we must show existence. We do so by induction on d. We include in our inductive hypothesis that every term in e{s\,..., Sd), when regarded as a polynomial in X i , . . . , Xj, has degree at most the degree of / ( X i , . . . , Xj). For d = \,s\ = Xi, and the result is trivial. Assume the result is true for d — 1. We proceed by induction on the degree k of the symmetric polynomial. \fk = Q the result is trivial. Assume it is true

3.2 Separable Extensions

51

for all symmetric polynomials in Z i , . . . , X^ of degree less than k, and let f(Xi,..., Xd) have degree k. Now f{X\,..., Xd-\, 0) is a synmietric polynomial in Z i , . . . , Xd-\ so by induction / ( Z i , . . . , Xd-\) = g{s\^d-\^ • • •, Sd-\,d-\) for some polynomial g (where we use the notation of the proof of Lemma 3.1.12). Then / ( Z i , . . . , X j ) - g(^i,...,5^_i) = / i ( X i , . . . , Xj) is a symmetric polynomial in X i , . . . , X^ with / i ( X i , . . . , Xj_i, 0) = 0. Hence / i ( Z i , . . . , Xd) is divisible by Xd in D [ X i , . . . , Xj], and hence, since this polynomial is symmetric, it is divisible by the product X\.. .Xd as well. Hence / i ( X i , . . . , Xd) = (Xi • • • Xd) fiiXu • . . , X^) with / 2 ( X i , . . . , X^) a polynomial of lower degree. By induction / 2 ( X i , . . . , Xd) = h(si,..., Sd) for some symmetric polynomial h(si,... ,Sd) satisfying our inductive hypothesis, and of course Xi • • • Xj = Sd is a symmetric polynomial satisfying our hypothesis as well. Then / ( X i , . . . , Xj) = g(su . . . , Sd-i) + Sdh(su . . . , ^j) = e(su

'",Sd)

is a polynomial in ^-i,..., 5*^, as required.

n

We have been considering polynomials with coefficients in a field. But, in fact, we also have a result valid for polynomials with integer coefficients. Lemma 3.1.15. Let / ( X i , . . . , Xd) be a symmetric polynomial with integer coefficients. Then / ( X i , . . . , Xd) may be written uniquely as e{si, ..., Sd) where e(si, ..., Sd) is a polynomial in the elementary symmetric functions si,..., Sd with integer coefficients. Proof. The proof of Lenmia 3.1.14 goes through unchanged to prove this result. n

3.2 Separable Extensions In this section we develop criteria for polynomials and extensions to be separable. Recall that an irreducible polynomial / ( X ) e F[X] is separable if / ( X ) splits into a product of distinct linear factors over a splitting field E. Recall that a root a of a polynomial / ( X ) is called simple if X — a divides / ( X ) but (X — a)^ does not, (otherwise a is called multiple) so an irreducible polynomial / ( X ) is separable if and only if all of its roots in a splitting field E are simple.

52

3 Development and Applications of Galois Theory For a polynomial f(X)

= Yl^=oCiX\ we let f\X)

be its derivative

Lemma 3.2.1. Let E be an extension of¥, let a be an algebraic element ofK, and let f(X) e F[X] a nonzero polynomial with f(a) = 0. Then a is a simple root of f(X) if and only if f'{ot) ^ 0. Proof Write f{X) = {X - oi)g{X), a factorization in E[X]. Then, by the usual rules for derivatives, f'{X) = (X - ot)g\X) + g{X) so f\a) = g(a). By the division algorithm, we may write g(X) = (X — a)h(X) + g(a), so here g(X) = (X - a)h(X) + fipt) and then f{X) = (X - a)g{X) = (X - afh(X) + (X - a)f(a). Thus (X - af divides f(X) if and only if

f(a) = 0.

n

Proposition 3.2.2. Let f(X) e F[X] be irreducible. If f\X) # 0, then f(X) is separable. In particular: (1) If ch2ir(F) = 0, then f{X) is separable, (2) If char (F) = p, then f(X) is separable or f(X) is of the form k

/(X) =

^QX^>.

Proof Let a be any root of f{X) in an extension field E of F. Since f{X) is irreducible, f{X) is the unique monic polynomial in F[X] of lowest degree with / ( a ) = 0 (and indeed, f(X) = m^(X)). Thus, if f(X) ^ 0, then f(X) is a polynomial of lower degree, so f\oi) ^^ 0 and a is a simple root of f(X) by Lemma 3.2.L Hence every root of f(X) in a splitting field is simple and f(X) is separable. Now if fiX) is a nonconstant polynomial and char F = 0, then / ' ( Z ) ^ 0. If charF = p, then f\X) 7^ 0 unless the exponent of every power of X appearing in f{X) is a multiple of p, proving the proposition. n Recall that an algebraic extension E of F is separable if ma(X) e F[X] is a separable polynomial for every a G E. Thus we see: Corollary 3.2.3. TfcharF = 0, every algebraic extension ofF is separable, D Example 3.2.4. Here is an example of an inseparable polynomial and, correspondingly, an inseparable extension. Let F = Fp(t), the field of rational functions in the variable t with coefficients in Fp. Let E be the field obtained by adjoining a root of the polynomial X^ — t,E = F(s) with s^ = t. Then, in F[Xl XP-t = XP-sP = (X-s)P,sos is not a simple root of X^ -1, Note that E is in fact a splitting field for this polynomial. To complete this example we need to show that X^ — t is an irreducible polynomial in F[Z], and this follows immediately from Lemma 3.2.8 below. o

3.2 Separable Extensions

53

Definition 3.2.5. A field F is a perfect field if every algebraic extension of F is separable. o We have a simple criterion for a field to be perfect. In order to state it, recall that for a field F of characteristic p, we have the Frobenius map : F -> F defined by ^(a) = a^. We let F^ = Im($) and note that, since is an endomorphism, F^ is a subfield of F. (This is easy to verify directly: 1 = 1^, if c = aP and d = b^, then c + d = (a + by and cd = (ab)P, and if Theorem 3.2.6. F is a perfect field if and only if: (1) char(F) = 0, or f2Jchar(F) = pandF = FP. Proof By Corollary 3.2.3, if char(F) = 0, then F is perfect. Let char(F) = p. First assume F = F^. Let a € E be inseparable and consider m^iX). By Proposition 3.2.2 (2), ma{X) must be of the form X^Lo Q^'^Since F = F^, for each Q there is a J^ G F with df = ct. Then k

k

k

contradicting the irreducibility of ma(X). Hence a is separable and F is perfect. Next assume F is perfect. (The argument here parallels Example 3.2.4.) Let a € F, a ^ F^ and consider f(X) = XP - a e F[X], By Lemma 3.2.8 below, f(X) is irreducible. Let E = F(a) where a is a root of / ( Z ) . Then, in E, aP = a so XP — a = XP — aP = (X — a)P and so a G E is an inseparable element, contradicting the fact that F is perfect. Hence F = F^. n Corollary 3.2.7. Every finite field is perfect. Proof Immediate from Corollary 2.L11 and Theorem 3.2.6.

n

Lemma 3.2.8. L^f char(F) = p, and let a e F, a ^ FP. Then f{X) XP' — a e F[X] is irreducible for every t > I.

=

Proof Factor f(X) into monic irreducible polynomials in F[X], f{X) = gi(X). "gk(X), Let E = F(a) with gi(a) = 0. Then gi(X) = m,(X) e F[X] as gi(X) is irreducible. Now 0 = giia) implies 0 = f(a) = aP' — a so a = aP' and f(X)

= XP' -a = XP' - otP' = {X-

a)P' e E[Z].

54

3 Development and Applications of Galois Theory

For any /, gi(X) divides /(X) in F[X], and hence gi(X) divides f(X) in E[X] by Lemma 2.2.7, i.e., gi(X) divides (X - a^ in E[Z], so gi{X) is a power of X — a and hence giipt) = 0. Since gi(a) = 0, ^/(X) is divisible by rriaiX) = gi(X), for every i. Since the choice of gi{X) was arbitrary, we see that giiX) = g2(X) = '" = gk(X) and further that f(X) = g{X)^ where g(X) = (X — ay for some j . Thus XP' -a = ((X - a)^y e E[X] so y = p' and A: = p^"^ for some s
3.3 Finite Fields Let us use the Fundamental Theorem of Galois Theory to completely determine the structure of finite fields. Recall we defined the Frobenius map O in Definition 2.L10: Let F be a field of characteristic p. Then 0 : F -^ F is the map defined by * ( a ) = aP. Theorem 3.3.1. Let p be a prime and r a positive integer. (1) There is afieldFpr of p^ elements, unique up to isomorphism. (2) Let s divide r. Then Fpr contains a unique subfield of p^ elements. Ifs does not divide r, then Fpr does not contain such afield. (3)IfE = Fpr and F = Fps for s dividing r, then Gal(E/F) is cyclic of order r/s, generated by ^ = 4>*, O the Frobenius map. (Then * ( a ) = aP\) In particular, ifF = Fpr and F = Fp, then Gal(E/F) is cyclic of order r, generated by the Frobenius map O. Proof (1) Set^ = p\ Let / ( X ) = X^ - X e Fp[X], andletFbethe splitting field of this polynomial. We know that a splitting field exists and is unique up to isomorphism (Lemma 2.5.2 and Corollary 2.6.4). Let g(X) = X^~^ — 1, so / ( X ) = X^(X). Now fiX) = - 1 so, by Lemma 3.2.1, the roots of / ( X ) are all simple. Thus / ( X ) has deg / ( X ) = q distinct roots in F, so in particular F must have at least q elements. Now let D c F be the subset of F consisting of the roots of / ( X ) and note that |D| = deg / ( X ) = q. Now we have that D = {a eF \ a^ = a} and we observe that D is a field: l^ = l,ifa^ = a and M = b, then (aby = ab and (a + by = a^ + Z?^ = a + b, and for a ^^ 0, (a~^y = a'K Thus / ( X ) splits in D and since F is a splitting field, F c D, and so F = D. Hence F = D = F^ is a field of q elements.

3.3 Finite Fields

55

On the other hand, if B is any field of q elements, then the multiplicative group of B has order ^ — 1, so every nonzero element of B is a root of g(X) = X^-^ - 1; hence every element of B is a root of f(X), so B is a splitting field of fiX) (again as f(X) cannot split in any proper subfield of D) and hence B is isomorphic to F. We have observed that the roots of f(X) are all simple, and hence f(X) is separable. (Alternatively, this follows from from Proposition 3.2.2 (2).) Now F^ is the splitting field of the separable polynomial f(X) e Fp[X], so F^ is a Galois extension of F^ (Theorem 2.7.14) and | Gal(F^/Fp)| = (F^/F^) = dimp^ F^ = r. Now $ : F^ -> F^ and <^ \Fp = idasaP = a for every a eFp (by Fermat's Little Theorem, see Corollary 2.1.12). Hence 4> e Gal(F^/Fp). Also, <^^(a) = aP' = a for every of G F, as every a G F^ is a root of / ( X ) , i.e., CD" = id : F^ ^ F^. On the other hand, ^' # id for 5 < r as ^ (or) = a^' = Of if and only if a is a root of the polynomial X^' — X, and this polynomial cannot have p^ roots by Lemma 2.2.1. Hence {1, ,..., O'*"^} is a subgroup of Gal(F^/Fp). Since this subgroup has r = | Gal(F^/Fp)| elements, it must be the entire Galois group. Now we may prove (2) and (3) by a direct application of the Fundamental Theorem of Galois Theory. The subfields of F^ are in 1 — 1 correspondence with the subgroups of the cyclic group {1, * , . . . , ^~^} and each subgroup is a cyclic group, of order r/s, generated by ^ = O^ for some divisor s of r. Thus these subfields are the fixed fields of the cyclic subgroups generated by * , i.e., the fixed fields of * . But * ( a ) = a is the equation a^' = a, whose solutions are the roots of X^' — X, and these are the fields F^. as in (1). n Remark 3.3,2. Theorem 3.3.1 (2) has an alternate, more elementary proof. Consider the multiplicative group G of F^^, a group of order p^ — I. Suppose Fpr contains a subfield isomorphic to F^., and let H be the multiplicative group of that field, of order p^ — I. Now, by elementary number theory, p^ — I divides p^ — I if and only if s divides r. Hence if s does not divide r, \H\ does not divide |G|, so no subgroup H, and hence no such subfield Fps can exist. On the other hand, we also know, by Corollary 2.2.2, that G is cyclic, so if s divides r, and hence p^ — I divides p^ — I, there is a unique subgroup H of order p^ — 1, and then H U {0} is the unique subfield of Fpr with p^ elements. o Corollary 3.3.3. Let F = Fps and let f(X) e F[X] be an irreducible polynomial of degree n. Let r = ns. Then E = Fpr is a splitting field for f(X). Furthermore, Gal(E/F) = Z/nZ generated by "if = ^ Proof. Let EQ = F(a) be the field obtained by adjoining a single root a of f{X) to F. Then, since deg f{X) = n, ( E Q / F ) = n, and EQ = Fpr.

56

3 Development and Applications of Galois Theory

Let E 2 Eo be a splitting field for f{X). We claim that E = EQ. On the one hand, this claim follows from Theorem 3.3.1. By Theorem 3.3.1, Eo is a Galois extension of F, so by Theorem 2.7.14, it is in particular a normal extension of F. But f(X) has a root in EQ; by the definition of a normal extension this implies that f(X) splits in EQ. On the other hand, we can verify this claim directly here. ^ is an automorphism of Eo = ¥(a) of order n, and that implies that a, ^ ( a ) , . . . , ^^~^(a) are all distinct. But for any k, 0 = ^^(0) = ^^(f(a)) = / ( * ^ ( a ) ) , so f(X) has the n distinct roots a,^ia),.,,, vi/«-i(a) in Eo; therefore f(X) = YYkZoi^ — ^^(ot)) e Eo[X] is a product of linear factors, i.e., f(X) splits in EQ.

The last sentence in the statement of the corollary is then simply a restatement of Theorem 3.3.1 (3). n Corollary 3.3.4, (1) For any p, s, and n, there is an irreducible monic polynomial f{X) G Fp5[X] of degree n. (2) Let k(p^, n) be the cardinality of the set K = {\
(p""' - 1)/ gcd(A:, p""' - 1) does not divide p^^ — I for any proper divisor m ofn}.

Then the number of distinct irreducible monic polynomials in ¥ps [X] of degree n is k{p\ n)/n. Proof (1) By Corollary 2.2.2, the multiplicative group G of Fpns is cyclic. Let ao be a generator of G, so F^n. = F^. (ao). Then f(X) = m^^ (X) e F^. [X] is an irreducible monic polynomial of degree n. (2) First we claim that k{p^, n) is the cardinality of the set K' = [a e ¥pns \ a ^ any proper subfield of F^n. containing Fp4To see this, consider a nonzero element a G ¥pns. Then, as an element of G, a has order dividing p^^ — 1. By Theorem 3.3.1, the proper subfields of ¥pns containing F^. are the fields F^m. with m a proper divisor of n, and a is in such a subfield if and only if its order, as an element of G, divides p^^ — 1. Now a = c^Q for some k, and then a, has order {p^^ — 1)/ gcd(A:, p^^ — 1). Thus we see that K' = {al\ke K], Now observe that Fp.(a) = ¥pns if and only if a e /C^ and also that Fp.(a) = ¥pns if and only if degmQ,(X) = n. In this case f{X) = maiX) is an irreducible monic polynomial of degree n, and every irreducible monic polynomial of degree n arises in this way. Also, every such polynomial has

3.4 Disjoint Extensions

57

distinct roots (as it divides X^"' — X, which has distinct roots), and no two distinct such polynomials have a root in common (since if fi(a) = f2(oi) = 0, then fi(X) = ma(X) = f2{X)). Since each such polynomial has n distinct roots in IC\ and /C' has k(p\ n) elements, there are k(p\ n)/n such polynomials. D Now let us investigate k(p^, n) a bit further. Lemma 3.3.5. (1) k{p\ n) > (p(p^^ — 1), where (p denotes the Euler totient function. (2) Ifn isprimey k{p\ n) = p^^ — p\ (3) For any positive integer n,

k(p\n) = J2f^(^/d)p''. d\n

where /JL is the Mobius /x function, defined by /ji(j) = 0 if j is divisible by the square of a prime, and otherwise /ji{j) = (—1)^ where i is the number of distinct prime factors of j . Proof (1) Observe that it G ^ for any 1 < it < p'^' - 1 with gcd(it, p""'-\) = 1, and by definition (p{p^^ — 1) is the number of such integers. (2) If n is a prime, the only fields intermediate between F^. and ¥pns are these two fields themselves, so in this case K^ = [a e ¥pns, a ^ F^^} and there are p^^ — p^ such elements. (3) Let f{p\ n) = p^^ denote the cardinality of the field Fpns, Then

fip^n) = Y^k(p\d) d\n

as any element a of ¥pns generates F^^^ for exactly one value of d dividing n. (This is the value of d for which a G F^^^ but a ^ F^^ for any proper divisor c of d,) But then the Mobius Inversion Formula from elementary number theory shows that

k{p\n) =

Y,l^(ri/d)f{p\d), d\n

as claimed.

3.4 Disjoint Extensions In this section, we assume that all fields are subfields of some field A.

n

58

3 Development and Applications of Galois Theory

Definition 3.4.1. Let Bi and B2 be extensions of F. Then Bi and B2 are disjoint extensions of F if Bi n B2 = F. o Of course, Bi and B2 are always disjoint extensions of Bi fl B2. We first note the following case, which is very special, but easy and useful. Note also that in this case we can strengthen Lemma 2.3.4. Lemma 3.4.2. Let Bi and ^2 be finite extensions ofF, with di = (B//F), / = 1,2. Ifdi and J2 ^^^ relatively prime, then Bi and B2 are disjoint extensions ofF, and, furthermore, (B1B2/F) = did2,

(B1B2/B2) = du and (B1B2/B1) = ^2.

Proof. Let BQ = Bi H B2 and JQ = (Bo/F). Then (B,/F) = (B,/Bo)(Bo/F), for / = 1, 2, so do divides both d\ and t/2, so JQ = 1 and BQ = F. Now let d = (B1B2/F). We observe that d = (BiB2/B,)(B//F) for / = 1,2. From this we immediately see that d is divisible by both d\ and d2, and from Lemma 2.3.4 we see that d < did2, so d = d\d2 from which the remaining two equalities inmiediately follow. n As we shall now see, we can say quite a bit about the composite of disjoint extensions when at least one of the extensions is finite Galois, and even more when both are. Theorem 3.4.3. Suppose that B2 is a finite Galois extension o/Bi fl B2. Then B1B2 is a finite Galois extension o/Bi, and Gal(BiB2/Bi) = Gal(B2/Bi nB2), with the isomorphism given by a \-^ a \ B2. Proof Let BQ = Bi nB2. Since B2 is a Galois extension of BQ, it is the splitting field of a separable polynomial f{X) e Bo[X]. If f{X) has roots ofi,..., a^^, then B2 = B o ( a i , . . . , otk) (Remark 2.5.6), and so B1B2 = B i ( a i , . . . , a^). Then B1B2 is the splitting field of the separable polynomial f{X) G B I [X], so B1B2 is a finite Galois extension of Bi. Let a e Gal(BiB2/Bi). Then a | Bi = id, so certainly a | BQ = id, and hence a(f(X)) = f(X). In particular, a permutes the roots a i , . . . , a;^ of / ( X ) , so cr(B2) = Bo(or(ai) . . . , a(ak)) = BoCa^,..., a^) = B2, where {ofp . . . , Qf^} is a permutation of { a i , . . . , ak}. Thus a | B2 : B2 ^^ B2, and or I Bo = id, so or I B2 G Gal(B2/Bo). We claim that the map a i-> a | B2 is an isomorphism.

3.4 Disjoint Extensions

59

First, suppose that a | B2 = id. Then a (at) = at for each /. Also, since a e Gal(BiB2/Bi), a | Bi = id by definition. But we have observed that B1B2 = B i ( a i , . . . , ak), soa = id. Next, let r e Gal(B2/Bo). Then B1B2 = Bi(Qfi,..., ak) is the splitting field of f(X) e Bi[X], so, by Lemma 2.6.3, r extends to a : B1B2 -^ B1B2, and then (J | B2 = r. n Under our hypothesis we may strengthen Lemma 2.3.4. Corollary 3.4.4. Suppose that B2 is a finite Galois extension of^\ fl B2. Then (BiB2/Bi) = ( B 2 / B i n B 2 ) ,

(BiB2/B2) = ( B i / B i n B 2 ) , and (B1B2/B1 n B 2 ) = ( B i / B i nB2)(B2/Bi n B 2 ) . Proof. The first equality is immediate from Theorem 3.4.3 and then the second and third follow as (B1B2/B1 n B 2 ) = ( B i B 2 / B i ) ( B i / B i n B 2 ) = (BiB2/B2)(B2/BinB2).

D

We also make the following observations. Corollary 3.4.5. Let Bi and B2 be finite extensions ofF and suppose that B2 is a Galois extension ofF, Then (B1B2/F) = (Bi/F)(B2/F) if and only if Bi and B2 are disjoint extensions ofF. Proof

By Corollary 3.4.4,

(Bi/F)(B2/F) = (Bi/Bi n B 2 ) ( B i n B 2 / F ) ( B 2 / B i n B 2 ) ( B i n B 2 / F ) = (B1B2/B1 n B 2 ) ( B i n B 2 / F ) 2 = (BiB2/F)(BinB2/F) so we have the equality in the statement of the corollary if and only if (Bi n B2/F) = 1, i.e., if and only if Bi n B2 = F. n

60

3 Development and Applications of Galois Theory

Corollary 3.4.6. (Theorem on Natural Irrationalities) Let f(X) e F[X] be a separable polynomial and let B 3 F. Let l^be a splitting field for f(X) over F. Then BE is a splitting field for f(X) over B and Gal(BE/B) is isomorphic to Gal(E/E n B), a subgroup of Gal(E/F), with the isomorphism given by a i-> or I E. Proof Let f(X) have roots ofi,..., a/:, so E = F(Qfi,..., ak) (Remark 2.5.6); then BE = B ( a i , . . . , a^^) is a splitting field for f(X) e B[Z]. Nov^ BE is a Galois extension of B as it is the splitting field of a separable polynomial, and E is a Galois extension of F for the same reason. Then, setting D = B fl E, F c D c E, E = D ( a i , . . . , a^), and so E is a Galois extension of D, also for the same reason. Then, by Theorem 3.4.3, Gal(BE/B) is isomorphic, via the restriction map, to Gal(E/D), which is a subgroup of Gal(E/F). D Theorem 3.4.7. (1) Let Bi and B2 be disjoint Galois extensions of¥. Then E = B1B2 is a Galois extension of F with Gal(E/F) = Gal(Bi/F) x Gal(B2/F). (2) Let E be a Galois extension ofF and suppose that Gal(E/F) = Gi x G2. LetBi = Fix(Gi) andB2 = Fix(G2). Then Bi andB2 are disjoint Galois extensions ofF and E = B1B2. Proof (1) By Theorem 3.4.3, we have isomorphisms Gal(E/B,-) -^ Gal(B^-/F) given hy a \-^ a \Bj, for / = 1 and j = 2, and vice versa. In other words, any aj e Gal(By/F) has a unique extension to aj e Gal(E/B,) c Gal(E/F), j = 1, 2. Now any e eF may be written ase = J2 ^\^1 with b] G Bi and &? G B2. Thenaia2(^) = E^i^2(&/^') = Y.^ia2{b])a,a2{b^) = T.^i{b])a2ib^) = J2or2dri(bl)a2d^i(bj) = Y1^20^i{blbj) = a2Cf\{e), so a\ and 02 commute and we have a map Gal(Bi/F) X Gal(B2/F) ^

Gal(E/F)

defined by (ori,Or2) l-> ^ 1 ^ 2 .

Next, this map is an injection: Suppose ori6r2 = id • E -> E. Then in particular a 102 \ B2 = id, but or2 | B2 = id by definition, so ai = ai | Bi = id, and similarly 0-2 = ^2 | B2 = id. Finally, on the one hand, as Gal(E/F) contains Gal(Bi/F) x Gal(B2/F),

3.4 Disjoint Extensions |Gal(E/F)| > = = = =

61

|Gal(Bi/F) x Gal(B2/F)| |Gal(Bi/F)||Gal(B2/F)| (Bi/F)(B2/F) (Bi/F)(E/Bi) (E/F),

and, on the other hand, | Gal(E/F)| = (E/D) < (E/F), where D 2 F is the fixed field of Gal(E/F), so | Gal(E/F) | = (E/F). Thus this map is an injection from one group to another group of the same order, and hence an isomorphism; furthermore E is a Galois extension of F. (2) Since G = G1G2 and {1} = Gi n G2, this follows directly from Proposition 2.8.14. D Corollary 3.4.8. Let Bi and B2 be finite Galois extensions ofF and let E = B1B2, Bo = Bi n B2. Then E is a Galois extension ofF and Gal(E/F) = {(ai, a2) e Gal(Bi/F) x Gal(B2/F) | ai | BQ = or2 | BQ}. Proof. Since Bi is a Galois extension of F, it is the splitting field of a separable polynomial fi{X) e F[X], by Theorem 2.7.14. Similarly, B2 is the splitting field of a separable polynomial f2(X) e F[X], Then E is the splitting field of the separable polynomial f(X) = fi(X)f2(X) € F[X], so, by Theorem 2.7.14, it is a Galois extension of F. As in the proof of Theorem 3.4.3, a(B,) = B, for each a e Gal(E/F), / = 1, 2 (as B/ is the splitting field of a separable polynomial whose roots are permuted by a). Thus we have a map a h^ (cri, (T2) = (cr I Bi, a \ B2). Again this map is an injection, as E = B1B2, and clearly any (ai, a2) in image of this map satisfies ai | BQ = or2 | BQ. Then (E/F) = (E/Bo)(Bo/F) = (E/B2)(B2/Bo)(Bo/F) = (Bi/Bo)(B2/Bo)(Bo/F) = (Bi/F)(B2/F)/(Bo/F), and these are the orders of the groups in the statement of the corollary.

n

Corollary 3.4.9. Let B and E be disjoint extensions ofF, with E a finite Galois extension ofF. Let r : Gal(BE/B) -> Gal (E/F) be the restriction isomorphism, r{a) = a \F. Let H C Gal(BE/B) be a subgroup. Then, for the fixed fields, we have Fix(H) = B F i x ( r ( H ) ) .

62

3 Development and Applications of Galois Theory

Proof, First observe that r is an isomorphism by Theorem 3.4.3. Now^ let K = r(H) and D = Fix(i^). Then certainly Fix(H) 3 BD. Nov^ (BE/B) = (BE/BD)(BD/B) and (E/F) = (E/D)(D/F). Now (BE/B) = (E/F) as the Galois groups of these two extensions are isomorphic, again by Theorem 3.4.3. On the other hand, by Lemma 2.3.4, (BE/BD) < (E/D) and (BD/B) < (D/F), so both inequalities must be equalities. But, since r is an isomorphism between the groups H and K, (BE/Fix(7f)) = \H\ = \K\ = (E/Fix(K)) so (BE/ Fix(H)) = (BE/BD) and hence Fix(H) = BD.

= (E/D) n

Theorem 3.4.10. Let p and E be disjoint extensions ofF, (1) Assume that both B and E are finite Galois extensions ofF. Let Ni = Gal (BE/B) and N2 = Gal(BE/E). Then BE is a Galois extension ofF, and N\ and N2 are normal subgroups <9/Gal(BE/F). Furthermore, Gal(BE/F) is the direct product Ni x N2 of the subgroups Ni and A^2(2) Assume that E is a finite Galois extension ofF. Let H = Gal (BE/B) and N = Gal(BE/E). Then BE is a Galois extension ofF and N is a normal subgroup of Gal(BE/F). Furthermore, Gal(BE/F) is the semidirect product HN of the subgroups H and N. Proof (1) This follows immediately from Theorem 3.4.3 and Corollary 3.4.6. (2) E is a Galois extension of F, by hypothesis, so N = Gal(BE/E) is a normal subgroup of Gal(BE/F), with quotient HQ = Gal(E/F). By Theorem 3.4.3 the subgroup H = Gal(BE/B) maps isomorphically onto HQ under the projection G -> G/N. But then G = HN. a Lemma 3.4.11. Let a G A with separable minimum polynomial maiX) e F[X]. Let E be a splitting field ofm^iX). IfB be an extension ofF such that B andE are disjoint, then ma{X) is irreducible in B[X]. Proof. Let rha (X) be the minimum polynomial of a over B. Of course, m^ (X) is irreducible in B[X] (as is ma(X) in F[X]). Let {ofi,..., a^} be the roots of ma(X) in E. E is a Galois extension of F by Theorem 2.7.14. Now Gal (E/F) acts transitively on { a j , by Proposition 2.8.15, and every ao G Gal (E/F) extends to a G Gal(EB/B), by Theorem 3.4.3, so Gal(EB/B) acts transitively on {of/}. Then, by Lemma 2.7.12, rhaiX) = YlLiiX - ott) = ma(X). u

3.5 Simple Extensions

63

Remark 3,4.12. The converse of Lemma 3.4.11 is false in general. Here is an example. Let F = Q and A = Q{o), \/2) (co a primitive cube root of 1). Let a = ^ , so ma(X) = X^ - 2 e F[X] with splitting field E = A. Let B = Q(co). Then m^CX) is irreducible in B[X], but B fl E = B D F. o However, we do have the following partial converse. Lemma 3.4.13. Let a G A with separable minimum polynomial m^ (X) e F[X] and suppose that E = F(Qf) is a splitting field ofm^iX). IfB is an extension ofF such that m^ (X) is irreducible in B[Z], then B and E are disjoint extensions ofF. Proof. Since m^iX) is irreducible in B[Z], rhaiX) = ma(X). By Corollary 3.4.4, (B(a)/B) = (BF(a)/B) = (F(a)/B D F(a)) = (E/B H E). But (B(a)/B) = degm,(X) = degm,(X) = (F(a)/F) = (E/F), so B n E = F.

D

3.5 Simple Extensions Let E be a finite extension of F. Then, choosing a basis {ofi,..., ak] for E as an F-vector space, E = F ( a i , . . . , ak). We may think of E as obtained from F in (at most) k steps, successively adjoining ofi, then 0^2, • • •, and finally ak. It is natural to ask when we can obtain E in one step. Definition 3.5.1. Let E be an algebraic extension of F such that E = F(a) for some a in E. Then E is a simple extension of F and a is a primitive element ofE. o We derive a criterion for E to be a simple extension of F. Theorem 3.5.2. Let E be a finite extension ofF. Then E is a simple extension ofF if and only if there are a finite number of intermediate fields F C B C E. Proof. First, we consider the case when F is finite. Then E is finite, so on the one hand there are (trivially) only finitely many intermediate fields, and on the other hand, the multiplicative group of E is cyclic (Corollary 2.2.2), so it consists of the powers of some a G E, and then E = F(Qf). Now we consider the case when F is infinite. First, assume that E is simple, E = F(a). Then a has minimum polynomial ma(X) e F[X]. Let F c B c E and consider mc^ (X) e B[X], the minimum polynomial of a overB. ThenE =

64

3 Development and Applications of Galois Theory

B(a), so (E/B) = deg rhaiX), Let D be the field obtained from F by adjoining the coefficients of rhaiX). Then F c D c B. As rhaiX) is irreducible in B[Z], it is certainly irreducible in D[Z], so rhaiX) is the minimum polynomial of a over D, and E = D(Qf), so (E/D) = degm«(Z) = (E/B) and hence B = D. Thus the intermediate field B is determined by the polynomial rhaiX). But rhaiX) divides niaiX) in B[X] (as nicdct) = 0) and hence rhaiX) divides rriaiX) in E[X]. But ma(X) has only finitely many factors in E[X], so there are only finitely many possibilities for rhaiX) and hence for B. Conversely, suppose there are only finitely many fields between F and E. We show that given any a and ^ in E there is a yi in E with F(a, fi) = ¥(yi). Since E is a finite extension of F, it is obtained by adjoining a finite number k of elements, and then the theorem follows by induction on k. Consider the fields F(a + a^) for a e F. Since there are infinitely many elements of F and only finitely many intermediate fields, there must exist distinct elements ai and ^2 of F with F(a+ai^) = F(a+a2P). Set yi = a+aifi, y2 = a+a2^. ThenF(]/i) = Fiyi)^ so y2 G F{yi), and certainly y\ G F(yi), so y2 — yi = (a2-ai)fi e F()/i), and hence ^S e F(yi) andyi—ai^ = a e F(yi), Thus F(yi) c F(a, ^) c F(yi) and they are equal. n While Theorem 3.5.2 does not look easy to apply, in fact it readily gives a strong general result. Corollary 3.5.3. If F is a finite separable extension of F, then F is a simple extension ofF. Proof Let E = F ( a i , . . . , a^). By the definition of a separable extension (Definition 2.7.3) each moj. (X) is a separable polynomial and hence (also by Definition 2.7.3) the product f{X) = ma^(X) • • • ma,,(X) is also a separable polynomial. Let E' 3 E be an extension of F that is a splitting field of the polynomial f{X). Then E^ is a Galois extension of F (Theorem 2.7.14), so by the Fundamental Theorem of Galois Theory there are only finitely many fields intermediate between F and E' (corresponding to the subgroups of Gal(E7F)), and hence there are certainly only finitely many fields between F and E. Then, by Theorem 3.5.2, E is a simple extension of F. n In case E is a Galois extension of F, we have the following criterion for an element of E to be primitive. Lemma 3.5.4. Let E be a finite Galois extension ofF and let a G E. Then a is primitive, i.e., F = F(a), if and only ifcr(a) ^ a for every a G Gal(E/F), a 7^ id.

3.5 Simple Extensions

65

First proof. Let d = (F(a)/F) and n = (E/F), so E = F(a) if and only if d = n. Now n = |Gal(E/F)| (Theorem 2.8.5) and d = dcgma(X) (Lemma 2.4.6 (2)). But, by Lemma 2.7.12, ma(X) = n f = i ( ^ - ^t) where ai = Of, 0^2,..., a j are the Galois conjugates of a in E. Hence E = F(a) if and only if a has n conjugates, and this is true if and only if a (a) — or, a e Gal (E/F), implies a = id. Second Proof Let B = F(a) so B is intermediate between F and E. By the FTGT, F = Fix(H) for H a subgroup of Gal(E/F), and furthermore H = {a e Gal(E/F) | a \B = id} = {a e Gal(E/F) \a(a) = a}. Then, also by the FTGT, B = E if and only if H = {id}. n The following proposition is very useful in constructing primitive elements. Proposition 3.5.5. Let F be afield of characteristic 0 and let Bi and B2 be disjoint finite Galois extensions ofF. Set B = B1B2. Let a\ be a primitive element o/Bi and let ot^ be a primitive element ofB^- Then a = ofi + a2 is a primitive element ofB. Proof We first observe that B is a Galois extension of F (Theorem 3.4.7), so, by Lemma 3.5.4, it suffices to show that the Galois conjugates of a are all distinct. In order to show this, we need only show that if a (a) = a for some a € G = Gal(B/F), then a = id. Thus, let a € G with a (a) = a. By Theorem 3.4.7, a = (ai, a2) with at € Gal(B,/F), / = 1, 2. Then ai+a2

= oi = a(a) = cri(ai) + ^2(^2),

so we see that cri(ai) — ai = cr2(a2) — ^2- But or/(a/) — at e B/, / = 1, 2, and Bi and B2 are disjoint extensions of F, so ai (ai) — ai = a2(a2) — 0^2 = ^ ^ F. Thus ai(a) = a + a, from which we see that a^(a) = a + ka for every k. But ai has finite order m, so a = a^(a) = a + ma, so ma = 0 and hence a = 0. Thus cTi (ai) = a i , and then cr2(a2) = 0L2 as well. But a\ is a primitive element of Bi, so ai(ai) —oi\ implies o\ = id on Bi; similarly 02 = id on B2 as well, so a = id on B as required. n Remark 3.5,6. In order to effectively apply Proposition 3.5.5 we need to develop some more background, which we do in the next chapter. But to show its applicability, as well as to provide some interesting examples of primitive elements, we shall quote an example from the next chapter here. Example 4.6.21. (1) Let E = Q(V2, 4 ^ , v ^ , >/2). Then E is an extension of Q of degree 210 with primitive element V 2 + V ^ + \ / 2 + A / 2 .

66

3 Development and Applications of Galois Theory (2) Let E = Q(V2, V3, VS, V7). Then E is an extension of Q of degree 16 with primitive element \/2 + \/3 + \/5 + V?. Similarly, if E = Q ( ^ , v ^ , A/S, >y7), then E is an extension of Q of degree 81 with primitive element yfl + \/3 + \ll> + ^ , (3) Let E = Q(V2, ^ , V3, ^ ) . Then E is an extension of Q of degree 36 with primitive element \fl + \fl + V^ + \fi. o

Example 3.5.7. Here is an example that shows that the conclusion of Proposition 3.5.5 may be false if Bi and B2 are not Galois extensions of F. Let F = Q, Bi = Q(v^), and B2 = Q(co^). Then B = Q(a), ^ ) . Let ai = - ^ , a primitive element of Bi, and let a2 = —co\/2, a primitive element of B2. Then Of = (Yi + Qf2 = o)^\/2 is not a primitive element of B. o Example 3.5.8. Here is an example of an extension that is not simple. Let F = Fp(s,t), the field of rational functions in the variables s and t over F^. Let E be the splitting field of the polynomial (X^ — S){XP — f). More simply, E = F(^, Y) where ^P = s and y^ = t. Then [E/F] = p^. On the other hand, it is easy to check that for any of € E, a^ € F, so [F(a) : F] = 1 or /?, and in particular ¥{a) c E. o

3.6 The Normal Basis Theorem Let E be a Galois extension of F of degree n. Then, as an F-vector space, E has a basis of n elements. On the other hand, Gal(E/F) has n elements, and so it natural to ask if there is some element a of E whose conjugates {a {a) I G G Gal (E/F)} form a basis for E over F. In fact, this is always the case, as we shall show in this section. Definition3.6.1. Let E be a Galois extension of F. A basis {oti]i=i,...,n of E as a vector space over F is a normal basis for E over F if {aJ = [at {a)} for some a G E, where {orJ,=i,.. ,„ = Gal(E/F). o There is one case of this theorem that follows directly from linear algebra. Theorem 3.6.2. Let E be a finite Galois extension ofF. Suppose Gal (E/F) is cyclic. Then E has a normal basis. Proof. Let a be a generator of Gal(E/F) and let T : E -> E be given by T(a) = a(a). Then T is a linear transformation that satisfies T^ — I = 0, We claim the minimum polynomial mjiX) = X^ — \. Certainly mjiX) divides X"" - 1. Suppose mriX) # X'^ - 1. Then mriX) = J^Lo^t^' with d < n. In other words.

3.6 The Normal Basis Theorem ao id H

67

h ad-icr^~^ = 0

as a map from E to E, and hence as a map from E* to E. But {id, a , . . . , a^~^} are a set of distinct characters of E* in E, so by Theorem 2.8.4 are linearly independent, so ai = 0 for each /, a contradiction. Now order the elements of Gal(E/F) as (id, a , . . . , a"^'^}. Then (by [AW, Theorem 3.7.1]) there is an element a of E with Ann(Qf) = {rriTiX)) (i.e., where the annihilator ideal of a in F[Z] is the ideal generated by mriX)), Then, since degmr(Z) = n, {a, r ( a ) , . . . , 7 " - ^ ^ ) } = {of, ( 7 ( a ) , . . . , or"~^ (a)} are linearly independent, hence form a basis, and this is a normal basis. (Alternatively phrased, mriX) = rricdX), by Proposition 2.4.6 (3). But E has a basis over F in which the matrix of T is the companion matrix of mriX). But then if ai is the first element of the basis, cr^(Qfi) = T^{a\) = a\^j for 7 = 0 , . . . , n — 1 ([AW, Corollary IV.4.15]), so [at] is a normal basis.) D The argument for the general case also uses linear algebra, but is more involved. Lemma 3.6.3. Let E be a Galois extension ofF of degree n, and let at, i = I,.,. ,n, be the elements o/Gal(E/F). Let ofi,..., a„ fee elements ofE. Then {ai,,,, ,an} is a basis for E over F if and only if the matrix A = (of/y), aij = (yi{otj)y isnonsingular. Proof Let { a i , . . . , a„} be a basis for E over F, and suppose some linear combination of the rows of A is zero, say YTi=i ^t^tj = 0 ^ ^ ^^^^ J^ i-^-» Jll=i^i^i(^) = 0, Of = ai, .,.,aj. But each at is an F-linear map, so Xl"=i <^i^i{^) = 0 for every a € E. But the at are distinct characters of E* in E, so by Theorem 2.8.4 are linearly independent; so ci = • • = c„ = 0. Thus the rows of A are linearly independent and A is nonsingular. On the other, suppose { a i , . . . , a„} is not a basis. Then it is not linearly independent, so there is a nontrivial linear combination X!y=i ^j^j — 0- Then Ey=i djaij = X;^^! djOiiaj) = o',(E"=i djaj) = or,(0) = 0 for each i, so the columns of A are linearly dependent, and A is singular. D Lemma 3.6.4. Let F be an infinite field, let E be an extension of F, and let / ( Z i , . . . , X^) e E[Zi, ,,,,Xk] be a polynomial. If f(au . . . , ^it) =Ofor every ( a i , . . . , aj^) € F^, then f(Xi,.,,,Xk)is the zero polynomial Proof By induction on k. For k = 1, let f(Xi) e E[Zi] be a nonzero polynomial with f{ai) = 0 for every ai e F. Then f(Xi) has more than deg / ( X i ) roots, contradicting Lemma 2.2.1. Now assume the lemma is true for A: — 1 and

68

3 Development and Applications of Galois Theory

consider / ( X i , . . . , Xjt). Regard this as a polynoniial in Xk with coefficients inXu.,,,Xk-uSO n

For any fixed elements ai,.,,, a^-x of F, f(Xk) = f{a\,..., ak-\, Xk) has more than d roots, so is identically zero by the k = I case, so each coefficient gi(Xi,..., Xk-i) is zero for every ( ^ i , . . . , ak-i), and hence is identically zero by the A: — 1 case. n Theorem 2.8.4 gives us linear independence of characters. In fact, when F is infinite and E is a finite Galois extension of F, we may derive a stronger kind of independence, algebraic independence. Theorem 3.6.5. Let F be an infinite field and let Gal(E/F) = {or/}, / = 1, . . . , n. Then {aj are algebraically independent, i.e., if f{X\,..., X^) G E[Xi, . . . , Xk\ is a polynomial with f(ai(a), . . . , crk(a)) = 0 for every a e E, then f{X\,,.,,Xk)isthe zero polynomial Proof Fix a basis {ofi,..., a„} for E over F. By Lenmia 3.6.3, the matrix A = A ( a i , . . . , an) is nonsingular, where A = (ptij) = (ai(aj)). In this proof we shall identify an n-tuple ( ^ i , . . . , fin) of elements of E with the column vector [^i,..., ^6^]^ (Purely for typographical convenience, we write this column vector [fix,..., ^nV as the transpose of the row vector [)6i,...,i8n].) For an arbitrary n-tuple ( c i , . . . , c„) of elements of F, let a = 5Zy=i ^j^jThen a/(a) = Yl^i ^j^ii^j) = I^j=i ^^7^/^^^ ^^^^ i = I,... ,n, i.e., [ori(a),..., cTnia)]^ = A [ c i , . . . , CnV. Now suppose that / ( a i ( a ) , . . . , a„(a)) = 0 for every a eE. Then, in our notation, 0 = / ( [ o r i ( a ) , . . . , an(a)r) = f(A[cu

. . . , Cnf)

for every ( c i , . . . , c„) G F". Thus the polynomial ^(Xi,...,X,) = /(A[Xi,...,X,r) is zero for every ( c i , . . . , c„) € F", so, by Lemma 3.6.4, it must be the zero polynomial in E [ X i , . . . , X„], and g ( y i , . . . , y„) = 0 for every (yu ..., Yn) G E". We claim also that f{X\,..., X„) is the zero polynomial, and, again

3.6 The Normal Basis Theorem

69

by Lemma 3.6.4, that will follow if we show /(&!,...,&«) = 0 for every (fci,..., bn) e F". Thus, let (&i,..., &„) e F". Since A is invertible, we may define ( y i , . . . , K„) by [ y i , . . . , ynV = A'^lbu . • •, &n]^ and then f(bu ...,bn) = f([bu . . . , bnV) = f(A[yu . . . , K.r) = giVu . . . , y.) = 0, as claimed. D Remark 3.6,6, Theorem 3.6.5 is false for finite fields. Here is a counterexample. Let F = F2 = {0, 1} and E = F4 = {0, 1, ao, otQ + 1} with al = aQ + L Let Gal(E/F) = {ai, 02} with a\ = id and 02 given by 0^2(ofo) = or -f- L Let / ( Z i , Z2) be the polynomial / ( X i , X2) = (X2 - Xi)(X2 - Xi - 1). Then p{ai{(x)a2{ot)) = {a2{ot) — cfi{a)){(j2{pt) — ai{a) — 1) = 0 for every a G E. o Theorem 3.6.7. Let E be a finite Galois extension ofF. Then E has a normal basis. Proof. If F is finite, then E is a cyclic extension of F (Theorem 3.3.1) so the result is a special case of Theorem 3.6.2. Assume henceforth that F is infinite. Let Gal(E/F) = { a i , . . . , a„}. Let B = B{Xi,..., Xfi) be the matrix whose entries are indeterminates Xk, with bij = Xk ifcfiGj = Gk. Then det(5) is a polynomial mX\,,.. ,Xk, det(5) = d{Xi,..., Xk). Note each Xt appears exactly once in every row and colunm. Thus, setting Zi = 1 and Xt = 0 for / > 1, we see det(fi) = ± 1 , i.e., t/(l, 0 , . . . , 0) = ± L Thus d is not the zero polynomial. Hence by Theorem 3.6.5 there is an of G E with d(ai(a),..., crn(a)) ^ 0. Let aj = cfj{ot), 7 = 1 , . . . , n. Now let A = B{a\{qt),... , an{oi)). If A = {atj), then atj — ak where GiOjia) = (Xkipt), or, in other words, aij = aiGj{a) = cxtiaj). Thus A is a matrix as in Lemma 3.6.3, and det(A) = d{ai{a),..., or„(a)) 7^ 0, so A is nonsingular, and then, by Lemma 3.6.3, {ofi,..., a„} = {(y\{a),..., or„(Qf)} is a basis for E over F. n Remark 3.6.8. Let E be a Galois extension of F and let a G E. Then a is a primitive element if the conjugates of a are all distinct (Lemma 3.5.4), but for a to be part of a normal basis, the conjugates of a must be linearly independent. This latter is definitely a stronger condition, as the following simple example shows. Let F = Q and E = Q(V2). Let a = \ / 2 . Then the conjugates of Of are a and —a, so the conjugates are distinct (and E = ¥(a)) but they are certainly not linearly independent (so {a, —a} is not a normal basis for E over F). o

70

3 Development and Applications of Galois Theory

3.7 Abelian Extensions and Kummer Fields Let E be a Galois extension of F. Then E is called an abelian extension of F if the Galois group Gal(E/F) is abelian. Since abelian groups are easier to understand than arbitrary groups, we may hope that abelian extensions are easier to understand than arbitrary Galois extensions. As we shall see in this section, under the appropriate conditions on F, we can indeed describe abelian extensions very concretely. Recall that for a field F, FQ denotes the prime field: FQ = Q if char(F) = 0 and Fo = F^ if char(F) = p. Definition 3.7.1. An element f G F is a primitive n^^ root of 1 in F if f" = 1, but f'" 7^ 1 for any I < m < n, o Lemma 3.7.2. (1) IfF has a primitive n^^ root of 1, then char(F) = 0 or is prime to n. (2) The following are equivalent: (a) F has a primitive n^^ root of I. (b) F has n distinct n^^ roots of L Proof (1) Suppose char(F) = p and p divides n. Let ^'^ = L Then f is a root o f X « - L B u t X « - l = ( X ^ / ^ ^ - l = (X'^/^-l)^so^isarootofX'^/^-l, i.e., f "/^ = 1 and f is not primitive. (2) If f is a primitive n^^ root of 1, then 1 , ^ , . . . , ^ " " ^ are distinct and are all n^^ roots of L On the other hand, the group of n^^ roots of 1 in F is a subgroup of the multiplicative group of F, so is cyclic (Corollary 2.2.2). Thus if F has n distinct n^^ roots of 1, a generator of this group is a primitive n^^ root of L D Henceforth we assume that char(F) = 0 or that p = char(F) is relatively prime to n and we let f„ denote an arbitrary but fixed primitive n^^ root of L Observe that ^„ = exp(2;r//n) is a primitive n^^ root of 1 in C. Lemma 3.7.3. Fo(fn) is a Galois extension of FQ, and Gal(Fo(f„)/Fo) is isomorphic to a subgroup ofi^jn'Ef, In particular this group is abelian. Proof F(f„) is the splitting field of the separable polynomial X" — 1, so it is a Galois extension of FQ. If (Ji € Gal(Fo(f„)/Fo), then o\ (f„) is a primitive n^^ root of 1, so o\ (f„) = ^^1 for some k\ with (fci, n) = L Similarly, 02 e Gal(F(fo)/F) satisfies a2(U) = ^n^ where (k2,n) = 1. Then aia2(^n) = ^n'^' = cr2cri(^n). and a e Gal(Fo(f„)/Fo) is determined by a(f„); so we see that we have an injection a^ h-> kt from Gal(Fo(f„)/Fo) into (Z/nZ)*. a

3.7 Abelian Extensions and Kummer Fields

71

Remark 3 J.4, We will see in Corollary 4.2.7 below that if FQ = Q, then Gal(Fo(^n)/Fo) = {Z/nljY, This is false in general. In the most extreme case, if p — 1 is a multiple of n, then ^„ € FQ, SO in this case Gal(Fo(^«)/Fo) = {id}. o We now introduce some nonstandard but very useful language. Definition 3.7.5. Let a € F. Then a is n-powerless in F if or is not an m^^ power in F for any m dividing n, m > 1. o Proposition 3.7.6. Let F 2 Fo(?«). Let E be the splitting field of X"^ - a € F[Z]. Then Gal(E/F) is isomorphic to a subgroup ofL/nTj, In particular, this group is cyclic. Furthermore, the following are equivalent: (1) a is n-powerless in F. (2) Gal(E/F) = Z/nZ. (3) X^ — a is irreducible in ¥[X\ Proof Let Of be aroot of Z'^ - a in E. Then X''-a = n"=o (^ - ^L^) so E = F(a), and m«(X) divides Z" - a in F[Z] and hence in E[Z] (Lemma 2.2.7) so m«(Z) = Y^kl^iX - l;l^a) for some subset [ik] of { 0 , . . . , n - 1}. Since f^^a and a are both roots of the irreducible polynomial ma(X), by Lemma 2.6.1 there is an element ak € Gal(E/F) with crk(a) = f^*a, and this equation determines ak uniquely. Furthermore, if cri(a) = f^^a, then
72

3 Development and Applications of Galois Theory This proposition has a partial converse.

Proposition 3.7.7. Let F 5 Fo(fn)- Let E be an extension ofF. The following are equivalent: (1) E is the splitting field ofX^—a e F[X], for some a that is n-powerless in¥. (2) E is a Galois extension ofF with Gal(E/F) = Z/nZ. Proof (1) implies (2) is (1) implies (2) of Proposition 3.7.6. Now suppose (2) is true. Let a be a generator of Gal(E/F). By Theorem 3.6.2, E has a normal basis {a^ (a) | / = 0 , . . . , n — 1} for some a € E. Let

Since {a^ (a)} are independent, ^8 7^ 0; then direct calculation show^s a(P) = a(a) + ^'V^Ca) + • • • + C^""'^« = ^nP, and so a'(fi) = i;^^ for each /. Thus {a'{P)] are all distinct, so E = F(^) by Lemma 3.5.4. Furthermore, n-l

^ M ^ ) = n ^ ^ - ^«^) = X" - iS'^ G F m . Thus, if a = ^", a G F[Z] and E is the splitting field of X" - a. Finally, a is n-powerless in F as in the proof of Proposition 3.7.6: If a = b^,n = mj, then m^ (X) would have X^ — & as a factor, and m^ (X) is irreducible so j = n andm = L D Remark 3,7,8, Consider the case FQ = Q. In order to apply Propositions 3.7.6 and 3.7.7, we must do arithmetic in the field Q(^«), which may not be so easy. It would be a lot simpler if we only had to do arithmetic in Q, which is easy. Fortunately, this is often, and indeed usually, the case. We have the following results, which we shall prove later on. Lemma 4.6.5. (1) Let n be an odd integer, and let a be n-powerless in Q. Then a is a n-powerless in Q(^N)for every N. (2) Let n be an even integer, and let a be n-powerless in Q, If a is negative, assume also that —a is not a square in Q. Then one of the following two alternatives holds: (a) a is n-powerless in Q(^N)for every N, (b) a = b^ for some b e Q(^N)> for some N, and b is n/2-powerless in Q(^N')for every multiple N^ ofN, Corollary 4.5.5. Let a be a square-free integer If a is positive and odd, set a' — a. Otherwise set a' — 4|a|. Then a — b^ for some b € Qi^n) if and only ifa^ divides n, o

3.7 Abelian Extensions and Kummer Fields

73

Definition 3.7.9. Let F 3 Fo(f„). A Kummer field E is the splitting field of a polynomial of the form f(X) = (Z'^ - ai)(Z« - a2) • • • (X'^ - a,), at eF, i = I,...

,t.

o

Recall that the exponent ^ of a finite group G is the smallest positive integer with the property that g^ = I for every g e G, and that the exponent of a group divides its order. Theorem 3.7.10. Let F 2 Fo(f„), and let E be an extension ofF. Then E is a Kummer field if and only if: (1) E is a Galois extension ofF. (2) Gal(E/F) is abelian, (3) exponent (Gal(E/F)) divides n. Proof First suppose that E is a Kummer field, the splitting field of f(X) = (X"" - ai) • • • (X"" - at). Let a, e E with < = a, for each /. Then X"" - at = YTj=\ (^ ~ ^n^i)^ SO X^ — ai splits into a product of distinct linear factors and is separable; then f(X), the product of separable polynomials, is separable as well. Thus E is the splitting field of a separable polynomial in F[Z], so E is a Galois extension ofF, proving (1). We prove (2) and (3) by induction on t. The t = 1 case follows immediately from Proposition 3.7.6. Assume the theorem is true for f - 1. Let Bi c E be the splitting field of fi(X) = (X""-ai)--• (X" at-i) over F and let B2 c E be the splitting field of /2(X) = X" - a^ over F. Then E = B1B2. Then, by Corollary 3.4.7, Gal(E/F) is a subgroup of Gal(Bi/F) X Gal(B2/F), so by the inductive hypothesis (for /i(X)) and the t = I case (for fiiX)) Gal(E/F) is a subgroup of an abelian group of exponent dividing n, so Gal(E/F) is itself an abelian group of exponent dividing n, and by induction the result follows. Conversely, suppose G = Gal(E/F) is as claimed. Write G as a direct sum of cyclic groups G = //i © • • • © //^. Since the exponent of G divides n, for each / the order m/ of Hi divides n. We prove the theorem by induction on 5. If 5* = 1, then G = H\ is cyclic of order m, so, by Proposition 3.7.7, E is the splitting field of X^ — b\ for some b\ € F or equivalently (since F 2 Fo(C«)) of X^ — ai where ai = b^ e F. Now suppose the theorem is true for 5 — 1 and let G be written as above. Set Bi = Fix(Hi © • • • © Hs-i) and B2 = Fix(Hs). Then, by the inductive hypothesis, Bi is the splitting field of a polynomial gi(X) of the form gi(X) = (X"" - ai)(Z« - a2) • • • (X« - ar) and, by the 5 = 1 case, B2 is the splitting field of a polynomial g2(X) of the form gziX) = X" — a^+i. Then, by Proposition 2.8.14 (1), B1B2 = Fix((/fi © • • © Hs-i) D Hs) = Fix({l}) = E and E = B1B2 is the splitting field of gi{X)g2{X) = (X" — ai)(X^ — a2) " ' (X^ — ar+i)» and by induction the result follows. D

74

3 Development and Applications of Galois Theory

Let us now explicitly determine the Galois group of a Kummer extension. For a field F, recall that F* denotes the group (under multiplication) of nonzero elements of n. We let (F*)" denote those elements that are n^^ powers, and observe that (F*)" is a subgroup of F*. Theorem 3.7.11. LetF ^ Fo(f«) and let E be the splitting field of /(X) = (X"-ai)«^-(Z«-a,) with each at ^ 0. Then Gal(E/F) is isomorphic to ( a i , . . . , at) c F*/(F*)" (Le.y to the subgroup of¥*/(¥*)^ generated by a\,,,,, at). Proof, Let at € E* with a" = at, i = 1 , . . . , n. Let G =-. ( a i , . . . , a^) C E 7 F * and let G^ = ( a i , . . . , « , ) c F7(F*)". Let ^ : G ^ G' by (p{a) = a\ We claim cp is an isomorphism. Clearly, (p is an epimorphism, so we need only show it is a monomorphism. Suppose (p(a) = L Then a^ e (F*)", so a" = / " for some / € F*; then a = f^/ € F* for some k (as ^„ € F by hypothesis). Thus instead of considering G' we may consider G. Now write G as a direct sum of cyclic subgroups G = //i © • • • © //^ and let Pi € E be a generator of Hi. Let Hi have order m^ dividing n. Set bi = p^', so bi e F and bi is mrpowerless in F. (Then also G' = / / ( © • • • © H^ with Hj the subgroup generated by bi.) We prove the corollary by induction on s. We include in the inductive hypothesis that Gal(E/F) is generated by elements r]i with r]i (Pi) = ^rmPi and rji(Pj) = Pj for J 7^ i.If s = 1, this hypothesis is just the conclusion of Proposition 3.7.7. Now assume the hypothesis is true for 5 — L Let B = F()Si,..., Ps-i) and D = F(^,). Claim: B and D are disjoint extensions of F with composite BD = E. Assuming this claim, the inductive step, and hence the corollary then follows immediately from Theorem 3.4.7. Proof of claim: E = F ( a i , . . . , a^) and ( a i , . . . , of^) = {^i,..., Ps) C E*/F*, so certainly BD = E. Let Do = B n D. By the FTGT, BQ = Fix(K) for some subgroup K of Gal(B/F), which is a cyclic group of order mg. From this we see that Do = ^((Ps)"^) = F ( \ ^ ) » where mm^ = m^, for some m dividing m^. We want to showm = I, Let us set p = ^/S^ = (Ps)"^ - Then ^ e B, so by the inductive hypothesis we may write uniquely

ii,...,is-i

3.7 Abelian Extensions and Kummer Fields

75

with ij = 0 , . . . , my — 1, for each j = 1 , . . . , 5* — 1, and with each Ci^,,„j^_^ €F. Now let a € Gal(B/F). Then cr(P) = 0 for some root of unity f, and, similarly, a(p[' • • • plz\) = ^'Pi • • • PsZ\ for some root of unity f'. Thus we see that we must have ^' = f for every term with Q,,. .,4_j ^^ 0. But, by the inductive hypothesis, we may choose a with cr(fij) = ^jfij where each ^y is an arbitrary m^^ root of 1. Hence, if this summation has more than one nonzero term, we may find a a which has the effect of multiplying one term by f and another term by f' with t,' ^ t,, which is impossible. Hence we can conclude there is only one term in the summation, and that

;s = (ys,r' = c,„...,,-_.;s;'...^irl; so we see that (yS^)""' e i/i © . . . © Hs-\ C E*/F*. But by assumption G = H\® • • • ® Hs\'va. particular {H\ © . . . ® Hs~\) HHs = {1}, and hence we must have m' = nts and hence m = 1, as claimed. D Corollary 3.7,12. In the situation of, and in notation of the proof of Theorem 3.7.11, the isomorphism ^^^^iX/miX) -> Gal(E/F) is given by 0\,...,

js) H> a where a{fii) = Kt^"" Pi ( = f^^A), i =

Proof. This follows directly from the proof of Theorem 3.7.11.

l,...,s. n

Example 3.7.13. (1) Let E be the sphtting field of (X^ - 2){X^ - ?,){X^ - 5) over Q. Then Gal(E/Q) = (Z/2Z)^ as (2, 3, 5) c Q7(Q*)2 is isomorphic to (Z/2Z)^ (2) Let E be the splitting field of (X^ - 2)(Z2 - 3)(X2 - 6) over Q. Then Gal(E/Q) = (Z/2Z)2 as (2, 3, 6) c Q7(Q*)2 is isomorphic to (Z/2Z)2. (3) Let E be the sphtting field of Z^^ _ 5 ^^^^ Q(f 12). Then Gal(E/Q(Ci2)) = Z/12Z as (5) c Q(fi2)7(Q(fi2)*)^^ is isomorphic to Z/12Z. The polynomial X*2 — 5 is irreducible over Q and also over Q(Ci2)(4) Let E be the splitting field of X^^ - 3 over Q(fi2). Then Gal(E/Q(f 12)) = Z/6Z as (3) c Q(cr2)/(Q(?i2)*)^^ is isomorphic to Z/6Z. Note 4l € Q(f 12). The polynomial X^^ — 3 is irreducible over Q. Over Q(fi2), it factors as the product of two irreducible factors (Z^ - N / 3 ) ( Z ^ + v ^ ) . (5) Let E be the splitting field of Z"* + 1 over Q(^4). Then E = Q^-g), Gal(E/Q(f4)) = Z/2Z, and (-1) c Q.{UTli^iUYf = Z/2Z as - 1 is a square but not a 4* power in Q(^4). The polynomial Z"* + 1 is irreducible over Q. Over Q(f4), it factors as the product of two irreducible polynomials

(X^ +

i)(X^-i).

(6) Let E be the sphtting field of Z'* + 4 over Q(f4). In fact E = Q(f4) as Z^ + 4 = (Z - (1 + i))(X - (1 - i))(X - ( - 1 + i))(X - ( - 1 - 0 ) ; thus

76

3 Development and Applications of Galois Theory

Gal(E/Q) = {id}, and (-4) c Q(?4)7(Q(?4)*)^ = {id} as - 4 = (1 + i)\ an equation in Q(f4). In this example we have made a number of claims about arithmetic in cyclotomic fields (i.e., in fields Q(f„)). They will be verified in Section 4.2. o

3.8 The Norm and Trace Let E be a finite extension of F. We shall define two important functions from E to F, the norm and the trace. We shall begin by defining them in case E is a Galois extension of F, which is the most important case, and the case to which we denote most of our attention. At the end, we will show how to generalize the definition. Definition 3.8.1. Let E be a finite Galois extension of F with Galois group G = Gal(E/F). For a G E, the norm (from E to F) of a is

NE/FC^)=n^*^^^ aeG

and the trace (from E to F) of a is TrE/F(c^) = ^ o r ( a ) .

o

aeG

Lemma 3.8.2. For any a € E, NE/F(C^) (^^d TrE/F(c^) cire elements ofF, Proof. N E / F ( « ) and TrE/F(Q^) are obviously invariant under G.

n

Lemma 3.8.3. L^f a e E with dtgniaiX) J2'^Q a X'. Let (E/F) = n. Then

=

NE/F(«)

= r, and write niaiX)

= ( - i r c ^ ^ ' and TrE/F(a) = - ( n / r ) Q _ i .

Proof, If a has distinct conjugates a i , . . . , or^ (with ai = of), then niodX) = n ^ ^ i ( ^ - oci) by Lenmia 2.7.12, so Co = (-^Ycci

"Oir and Cr-i = - ( a i H

h a^).

Lemma 3.8.4. For a e F, let Ta : E ^ - E Z?e the linear transformation of F-vector spaces given by T^i^) = a^. Then NE/F(a) = Det(rc,) and TrE/F(c^) = Trace(ra,).

•

3.8 The Norm and Trace

77

Proof. Let B = F(a). Then B has F-basis B = {1, a , . . . , a'^"^} where r = degm«(X). If (E/F) = n, then (E/B) = n / r . Choose a B-basis {1 = 6 i , . . . , €„/r} for E over B. Then C = {1, Of, . . . , a ' ' " \ . . . , €n/r, €n/rOi, . . . , ^^/^Of'""^}

is a basis for E over F. Let Sa = Ta \ F(a). Now S^ : F(Qf) ^ F(Qf) and the matrix of 5^ in the basis B is just C(mQ,(Z)), the companion matrix of the polynomial ma(X) (see the proof of Proposition 2.4.6(3)). Then the matrix of Ta in the basis C is a block diagonal matrix consisting of n/r diagonal blocks, each equal to C(ma (X)), and the result then follows from Lemma 3.8.3. n Theorem 3.8.5. Let E be a finite Galois extension ofF of degree n. Let B be a field intermediate between F and E. (Then E is a Galois extension ofB,) Assume further that B is a Galois extension of F, Then for any element aofF and any elements a, P ofE: (1) N E / F ( ^ ) = (^^ andTvE/Fi^) = ^ci(2) NE/F(c^iS) = NE/F(c^)NE/F(i8) andTtE/Ficc + fi) = TrE/F(c^) + TrE/F(^)-

(3) NE/F(C^) = NB/F(NE/B(C^)) and TrE/F(c^) = TrB/F(TrE/B(c^)). Proof (1) and (2) are clear. We prove (3). Let H = Gal(E/B) = {r,} have left coset representatives {pj}. Then we may identify G/H with {pj}. Then NB/F(NE/B(a)) = Uj Pj(Uiri(a)) = Uij PM^C) = UaeG^(^) = NE/F(c^). and similarly for TYE/F (oi). • Lemma 3.8.6. There is an element a G E with TrE/F(c»^) # 0. Proof If char(F) = 0 orn = (E/F) is relatively prime to p = char(F), this is completely trivial, as we may simply choose a = 1. In any case, for each a e Gal (E/F), a : E* -^ E* is a character of the group E* in E in the sense of Definition 2.8.1, so the result follows immediately from Dirichlet's theorem on the independence of characters (Theorem 2.8.4). D Theorem 3.8.7 (Hilbert's Theorem 90). Let E be a cyclic extension ofF of degree n and let a be a generator ofG = Gal (E/F). (1) If ^ is any element ofF with ^E/F{P) = 1» then P = a/a(a)for some a GE. (2) If P is any element ofF with TTE/F{P) = 0> then fi = a — a(a) for some a € E.

78

3 Development and Applications of Galois Theory

Proof, (1) By Theorem 2.8.4, {1, a , . . . , a""^} are independent characters of E* in E, so for any elements ^ o , . . . , 6„_i of E, there is an element (5 of E with « = E"=o ^/^'(^) 7^ 0- Now let

e, = a%^)a\p)...

a'-\p),

/ = 1,..., n - 1.

Note that Pa{€i) = 6/+i for / = 1 , . . . , n — 2, and 6„_i = NE/F()8) = 1 so also Pa{en-\) = ^ = cr^(P) = ei. Let a be as above and note then that Pa(a) = a so ^ = a/a (a), (2) By Lemma 3.8.6, there is an element of y of E with TrE/rCy) / 0. Set S = ^ a ( y ) + (^ + a(^))aHy) + . . . + (^ + . . . + or'^-2()S))or'^-ny). Then, by direct calculation, S - or(y) = ^(or(y) + . . . + or'^-^C)/)) - (or(^) + • • • + ^^^-^^8))/

= iSTrE/F(y) - Y TrE/F(iS) = iSTrE/rCK), so if Of = 6/ TrE/pCy), we have that fi = a — a (a).

D

Now we extend the definitions of norm and trace. First we consider separable extensions. Here we assume familiarity with Section 5.2 below. Definition 3.8.8. Let E be a finite separable extension of F. Then NE/F

= ND/F and TTE/F = Tro/F

where D is any normal closure of E.

o

Remark 3,8.9, (1) It is easy to see that NE/F and TT^/F are well defined, i.e., independent of the choice of D. (2) The results of this section through Theorem 3.8.5 go through unchanged in this more general situation. o Now we consider arbitrary finite extensions. Here we further assume familiarity with Section 5.1 below. Definition 3.8.10. Let E be a finite extension of F. Then NE/F

= (ND/F)^ and TrE/F = d Tro/F,

where D is any normal closure of E and d = (Fix(Gal(D/F))/F).

o

Remark 3,8,11, (1) Let F, = Fix(Gal(D/F)). Then D is a Galois extension of F/ and F/ is a purely inseparable extension of F. If E is a separable extension of F, then D is a Galois extension of F, F/ = F, and d = l,In particular this is always the case for a perfect field F. (2) If char(F) = p and E is not a separable extension of F, then J is a power of p and so in this case TYE/F = 0(3) Again the results of this section through Theorem 3.8.5 continue to hold. o

3.9 Exercises

79

3.9 Exercises Exercise 3.9.1. Give an example of an extension E of F of degree n and a divisor dofn for which there does not exist a field B intermediate between F and E with (B/F) = d. Exercise 3.9.2. Let F be a field with char(F) ^ 2. Call a rational function f(Xu . . . , X,) G F ( Z i , . . . , X„) alternating if f(a(Xi),..., a(Xn)) = sign{a)f(Xi,..., Xn) for every a € 5„. Let ^ i , . . . , ^-^ denote (as usual) the elementary symmetric functions in Z i , . . . , X„. Let S be as in Definition 4.8.L (a) Show that every alternating rational function f(Xi,..., Z„) can be written as a rational function in ^ i , . . . , 5„ and 5 in a unique way. (b) Show that in fact every alternating rational function f(Xi,..., X„) can be written as f(Xi,..., Xn) = g(si,..., Sn)S for some rational function g(Si, . . . , 5 „ ) of 5-1, . . . , 5 „ .

(c) Show that every alternating polynomial f(Xi,..., X„) can be written as a polynomial in 5 i , . . . , 5^ and 5 in a unique way. (d) Show that in fact every alternating polynomial f(Xi,..., Xn) can be written as f(Xi,..., Xn) = g(si,..., Sn)S for some polynomial function g(si,.,,,Sn)msu ...,^«. (e) Show that every rational function f(Xi,..., X„) invariant under the action of A„ (i.e., every rational function / ( Z i , . . . , Xn) with f{a{X\),..., a{Xn)) = f{X\,..., Xn) for every a e An) can be written in a unique way as f(Xu . . . , Z„) = MXu . . . , X,) + fa(Xu . . . , X,) where / , ( X i , . . . , Z J is symmetric and fa{X\,..., X„) is alternating. Furthermore, if / ( X i , . . . , Z„) is a polynomial, then so are fs{X\,..., X„) and fa(Xi,..., X„). Exercise 3.9.3. Show that (2) implies (1) in Exercise 2.10.6. Exercise 3.9.4. Show that (2) implies (1) in Exercise 2.10.7. Exercise 3.9.5. Interpret a in Exercise 2.10.1 as being in an extension of F5, find an irreducible polynomial in F5[X] that has a as a root, and find all roots of that polynomial in a splitting field E. Also, find G = Gal(E/F). (In the definition of a, there may be an ambiguity because of the roots. Handle all cases.) Exercise 3.9.6. Interpret a in Exercise 2.10.1 as being in an extension of F7, find an irreducible polynomial in F7[X] that has a as a root, and find all roots of that polynomial in a splitting field E. Also, find G = Gal(E/F). (In the definition of a, there may be an ambiguity because of the roots. Handle all cases.)

80

3 Development and Applications of Galois Theory

Exercise 3.9.7. In this exercise, let E be a splitting field of the polynomial f(X) e F[X] and let G = Gal(E/F). Find (E/F) and find G. (a) f(X) = X'-le F3[X]. ih)f{X) = X'-leF5[Xl (c) f(X) = X'-le F,[X], (d)f(X) = X'-3eFs[Xl (e) f(X) = X^+4X + 2e F ^ m , (f) f(X) = X^ + 2X^ + 3Z2 + 3Z + 4 € FslX], (g) f(X) = X^+ 3X^ + 4 eFslXl (h) f(X) = X^ + 3X^ + 3e F5[X]. Exercise 3.9.8. Let F = F^ and let f(X) = XP -X-a for some a^OeF, Let E be the splitting field of F and let a be a root of f(X) in E. Show that a, Of + 1 , . . . , Of + (p — 1) are the roots of f{X) in E. Conclude that f(X) e F[X] is irreducible and that Gal (E/F) is cyclic of order p with generator a the automorphism of E given by a (a) = a + L Then the Frobenius automorphism 4> of F must be given by (a) = a + k for some k ^0 eF. Show that a can be chosen so that 4>(a) = of + L (E is called an Artin-Schreier extension ofF.) Exercise 3.9.9. Let f(X) e Z[X] be a monic polynomial and let f{X) be its image in F^[X] (identifying F^ with Z/pZ). Let E be a splitting field for f(X) and let E be a splitting field for f(X). Show that if, for some p, all the roots of f(X) in E are simple, then all the roots of f(X) in E are simple. Exercise 3.9.10. Let Bi and B2 be Galois extensions of F, both contained in some field A. Let E be the composite E = B1B2. (a) Show that every a e Gal(Bi/F) extends to a (not necessarily unique) automorphism a e Gal (E/F). (b) Show that Bi and B2 are disjoint extensions of F if and only if every a G Gal(Bi/F) extends to an automorphism a e Gal (E/F) with a | B2 = id. Exercise 3.9.11. Find an example of a field F and pairwise disjoint Galois extensions BQ, B I , B2, of F with BQ and B1B2 not disjoint extensions of F. Exercise 3.9.12. Let D be a Galois extension of F and let B be an extension of F. Suppose that D and B are disjoint extensions of F. (a) Show that D and B' are disjoint extensions of F for any Galois conjugate BofB. (b) Let B be the Galois closure of B. Give an example to show that D and B need not be disjoint.

3.9 Exercises

81

Exercise 3.9.13. Let Bi and B2 be finite Galois extensions of F. Show that Bi n B2 is a Galois extension of F. Exercise 3.9.14. (a) Let a eF and suppose that a is not a p^^ power in F. Let B be an extension of F of degree m, and suppose that m and p are relatively prime. Show that a is not a p^^ power in B. (b) More generally, let a € F and suppose that a is n-powerless in F. Let B be an extension of F of degree m, and suppose that m and n are relatively prime. Show that a is n-powerless in B. Exercise 3.9.15. Suppose that X"^ — a and X^ — a ait both irreducible polynomials in F[X]. (a) Suppose that m and n are relatively prime. Show that X^'^ — a is an irreducible polynomial in F[X]. (b) More generally, let i = lcm(m, n). Show that X^ — ais an irreducible polynomial in ¥[X]. Exercise 3.9.16. (a) Let E = Q(V2, VS, v ^ , V 7 , . . . ) . Show that E / Q is infinite. (b) Let E = Q(V2, v ^ , .^2, ^ ^ 2 , . . . ) . Show that E / Q is infinite. Exercise 3.9.17. (a) Let F 2 Fo(f2) (and consequently char(F) ^ 2) and let f(X) = X^ - a e F[X] be irreducible. Let E = F(a) where a^ = a. Let fi = bo + bia with bo, b\ e F. Show that the conjugates of ^ form a normal basis for E if and only iffeo&i# 0. (b) Let F 2 Fo(?3) (and consequently char(F) ^ 3) and let f(X) = X^ a e F[X] be irreducible. Let E = F(Qf) where a^ = a. Let fi = bo+bia+b20t^ with bo, b\,b2 € F. Show that the conjugates of ^ form a normal basis for E if and only if bobibi # 0. Exercise 3.9.18. Let F c B c E with E / F Galois. Show that there exist irreducible polynomials f{X) e F[X] and g{X) G B [ X ] such that E is a splitting field of / ( X ) , and E is also a splitting field of g(Z), and g{X) divides /(X)inB[Z]. Exercise 3.9.19. (a) Let E be a separable extension of F and suppose there is an integer n such that deg m^ (X) < n for every a € E. Show that E is a finite extension of F. Furthermore, show that (E/F) < n. (b) Give an explicit counterexample to (a) in case E is not a separable extension of F. Exercise 3.9.20. Let E be an extension of F of degree 2, where F = F^ is a finite field, with q odd. Let E = F(a), where a^ e F. Show that, for any a,b eF,

82

3 Development and Applications of Galois Theory (a + ba)^ = a — hot

or, equivalently,

NE/FC^)

= (« + ba)^^^.

Exercise 3.9.21. Let E be an extension of F of degree 2, where char(F) / 2. Let E = F(3), where 5^ = rf G F. Let 7 : E -> E be an F-Iinear transformation. Let a = T(l) and ^ = T(S)/S, (Note a, ^ e E.) (a) If a = di)8, show that there exists an / G F with NE/F(r(6)) = / NE/F(e) for all e G E and in fact / = N E / F ( « ) = NE/F()S). (b) If a 7^ d=^, show there does not exist such an / . Exercise 3.9.22. Let E be an extension of F of degree 2, where char(F) ^ 2. Let E = F(5), where 5^ = rf G F, and let a be the nontrivial element of Gal(E/F). Prove Hilbert's Theorem 90 (Theorem 3.8.7 (1)) directly in this case. That is, if y = a +fo(5G E with NE/F(K) = a^ — b^d = 1, then there is an element 6 = x + yS eE with _ e _x + yS ^ ~ ~a(e) ~ x-y8' Exercise 3.9.23. Let F = F^. and let E = F^r for some multiple r of 5^. Set m = (p"" - l)/ip' - 1). Recall that Gal(E/F) is cyclic of order r/s with generator ^ given by ^ ( a ) = a^\ (a) Show that NE/F(a) = Qf"" for every a G E. (b) Show that NE/F : E -> F is an epimorphism. (c) Prove Hilbert's Theorem 90 (Theorem 3.8.7 (1)) directly in this case: If )6 G E with NE/F(iS) = 1, then )S = (a)(a-P') for some a G E. Exercise 3.9.24. (a) Fix a polynomial k^Y) e ¥[¥], Let f(X) G F [ X ] be any separable polynomial. Let f{X) split in E with roots a i , . . . , a;„. Let QkiY)(f(X)) be the polynomial QkiY)(f(X)) =

Y\(X-k(ai)), i

Evidently Qk(Y){f{X)) e E[X]. Show that in fact Qk(Y){f{X)) e F[X]. (b) Fix a polynomial k{Y, Z) 6 F[F, Z]. Let / ( Z ) , g(,X) € F[X] be any separable polynomials. Let f(X) and g(X) split in E with roots ai,... ,am and )8i,...,fi„,respectively. Let Qk(Y,z)(fiX), giX)) be the polynomial QHY,z)if(X),g(X)) = Y[(X - kiUi, Pj)).

3.9 Exercises

83

Evidently Qk{Y,z)(f(X), g(X)) e E[X]. Show that in fact QkiY,z)(f(X),g(X))

eF[Xl

(This generalizes to arbitrarily many polynomials.) Exercise 3.9.25. (a) Let {5/} be the elementary symmetric polynomials in {at}. Show that for any k(Y) e F[F], QkiY)(fiX)) is a polynomial whose coefficients are polynomials in {si}, and hence polynomials in the coefficients of

f(X). (b) Let {Si} be the elementary symmetric polynomials in {at} and let {sj} be the elementary symmetric polynomials in {^j}. Show that for any k(Y, Z) G F[y, Z], QkiY,z)(f(X), g(X)) is a polynomial whose coefficients are polynomials in {st} and {5^}, and hence polynomials in the coefficients of f(X) and g(X). (This generalizes to arbitrarily many polynomials.) Exercise 3.9.26. (a) Show that if k(Y) e Z[Y] and f(X) e Z[X], then QkiY)(f(X)) €Z[Xl (b) Show that if k(Y, Z) G Z[F, Z] and / ( Z ) , g(X) e Z[Z], then QkiY,z)(f(X), g(X)) G Z[X]. (This generalizes to arbitrarily many polynomials.) Exercise 3.9.27. Let f(X) = X^ + aX + b, g(X) = X^ + cX + d, and h(X) = X^ + eX^ + fX + g be polynomials in F[X]. Without finding the roots of these polynomials: (a) Find Gy+i(/(X)), Q2Y(f(X)). QY^ifiX)), and QY^(f(X)). (b) Find QY+iiHX)), Q2Y(HX)), and QY2ih(X)). (c) Find e r + z ( / ( X ) , g(X)) and QYz(f(X), g(X)), (d) Find Qy+zifiX), h{X)) and QYz(f(X), h{X)). Exercise 3.9.28. Let /? be a subring of F and let E be an extension of F. Let 5 = {a G E I a is integral over /?}. Show that 5 is a subring of E. Exercise 3.9.29. Let E be a field and let /? c E. Suppose that E = /?[«] for some a G E. Show that there is a nonzero element s e R such that /?[l/5'] is a field. Exercise 3.9.30. (a) Let {^/}/=i, ..,j be the elementary symmetric polynomials in X i , . . . , Z j . Let tt =X\-\ h Xj for / > 0. Prove Newton's identities:

84

3 Development and Applications of Galois Theory ti-

si =0

h — s\t\ + 2s2 = 0 h -sit2

+ S2h - 3 ^ 3 = 0

td - sitd-i + '" + {-lY-^Sd-xh + i-l^dsd td-^i-sitd + '-' + i-lfsdh

=0 =0

td+2 - Sitd+l + • • • + {-lYsdti = 0

(b) Conclude that tk is a polynomial in ^-i,..., 5y with integer coefficients, where j = inin(A:, d), for every A: > 0. (c) Let R be the ring of symmetric polynomials in X i , . . . , Xj with coefficients in F. Let f{X) be a polynomial in R of degree at most k, and let j be as above. If char(F) = 0 or char(F) > j , show that f(X) can be written uniquely as a polynomial in ti,,,. ,tk. In particular, if char(F) = 0 or char(F) > d, show that every polynomial in R can be written uniquely as a polynomial in ti,... ,td.

Extensions of the Field of Rational Numbers

In this chapter we assume that all fields we consider are of characteristic 0.

4.1 Polynomials in Q[Z] In this section we deal with a number of questions about polynomials in Q[X] related to factorization and irreducibility. We begin with a couple of elementary results. Lemma 4.1.1. L^^ F c R and let f(X) e F[X]. Ify eCisa root of f{X\ then its complex conjugate y is also a root off(X). Consequently, iff(X) has odd degree, then it has an odd number of real roots, counted with multiplicity, and if f{X) has even degree, then it has an even number of real roots, counted with multiplicity. Proof f(y)

= f{y) = f{y) = 0 = 0.

D

Lemma 4.1.2. Let f(X) e Z[X], f(X) = Yll=o ^t^'- Suppose that r eQis a root of f(X) (or equivalently, that X — r is a factor of f(X)), Ifr= s/t, with s and t relatively prime, then s divides ao and t divides an- In particular, if f(X) is monic, then r is an integer dividing ao. Proof 0 = f{s/t) = f'fis/t) = ^"^1 aisU''-\ Now 0 is divisible by s, and every term of this sunmiation except possibly a^t^ is divisible by 5, so that term must be divisible by s as well, and since t is relatively prime to s, this implies that a^ is divisible by 5. By the same logic applied to the term a„5", a„ is divisible by f. n

86

4 Extensions of the Field of Rational Numbers

Note that this lemma, used recursively, provides an effective method for finding all linear factors of f(X) in Q[X], Corollary 4.1.6 below is a standard result, but one we shall use repeatedly. First, some preliminaries. Definition 4.1.3. A polynomial f(X) = anX^ -\ ifgcd(a„,...,ao) = 1.

\-ao € Z[X] is primitive o

Lemma 4.1.4. (Gauss's Lemma) Ifg{X) andh(X) are primitive polynomials in Z[Z], and f(X) = g(X)h(X), then f(X) is a primitive polynomial in

nx\ Proof, Write g{X) = bmX"^ + • • • + &o and h{X) = CkX^ + • • • + CQ. Let f(X) = g(X)h(X) = a„X" + • • • + ao. Suppose f(X) is not primitive and let d = gcd(a„,..., ao). Pick a prime p dividing d. By hypothesis, p does not divide all of the coefficients of g(X), so let / be the smallest index with bt not divisible by p. Similarly, p does not divide all of the coefficients of h(X), so let j be the smallest index with Cj not divisible by p. Consider a^+y, the coefficient of X^"*"^ in f(X). Then at^j = biCj + (bi^iCj-i

+ bi+2Cj-i H

) + (bi-iCj^i

+ bi-2Cj^2 + • • • ) .

By hypothesis, p divides a^+y, and by construction p divides each of the terms in the two parenthesized expressions, so p divides btCj as well, and hence p divides bt or p divides Cj, a contradiction. a Proposition 4.1.5. Let f{X) be a primitive polynomial in Z[Z] and suppose f(X) = g(X)h(X) with g(X),h(X) € Q[X]. Then f(X) = giiX)hi(X) with gi(X), hi(X) e Z[Z], where gi(X) is a constant multiple of g(X), and h\{X) is a constant multiple ofh(X), Proof We may write f(X) = efo(X) where fo{X) e Z[X] is primitive and e e Z, We may write g{X) = cgo(X) where go{X) € Z[X] is primitive and c G Q. Similarly we may write h(X) = dhoiX) where ho(X) e Z[X] is primitive and d e Q. Then efo(X) = f(X) = g(X)h(X) = cdgo(X)ho(X). By Lemma 4.L4, go(X)ho(X) is primitive so we must have e = ±cd, so fo(X) = ±go(X)ho(X) and f(X) = gi(X)hi(X) where gi(X) = ±ego(X) = (±e/c)g(X) and hi(X) = ho(X) = {\/d)h{X), U Corollary 4.1.6. Let f(X) be a monic polynomial in Z[X] and suppose f(X) = g{X)h(X) with g(X),h(X) e Q[X] monic polynomials. Then giX), h{X) € Z[X]. Consequently, iff(X) is irreducible in Z[X], then f{X) is irreducible in Q[X].

4.1 Polynomials in Q[X]

87

Proof. Apply Proposition 4.1.5, and note that we must have g\{X) = g{X) and h\{X) = h{X) (up to a common sign) as f{X), g(X), and h(X) are monic. The second claim follows by contraposition. D We have the following useful criterion for a polynomial to be irreducible over Q. Proposition 4.1.7 (Eisenstein's Criterion). Let f(X) be a polynomial with integer coefficients, f(X) = anX^ + an-iX"~^ -\ h ^Q. Suppose there is a prime p with p not dividing a„, p dividing ao,,.,, cin-h (^^d p^ not dividing UQ. Then f(X) is irreducible in Q[X]. Proof, By Corollary 4.1.6, it suffices to show that fiX) is irreducible in Z[Z]. Suppose f{X) = g(X)h(X) with g(X) = bmX"^ + bm-iX"^'^ + • • • + Z^o and h(X) = CkX^+Ck-iX^~^-] hco. Then bmCk = «„, so neither is divisible by p. Since p divides ao = boco but p^ does not divide ao, by hypothesis, p must divide exactly one of &o and CQ. Let that one be &o- Now ai = bico+boci. Since pdividesai andfcobut not co,/? must divide Z?i. Then ^2 = b2Co + biCi+boC2, so by similar logic p must divide Z72. Proceeding inductively, we find that p divides bm, contradicting fo^ = 1. D Example 4.1.8. We will use Eisenstein's Criterion to show that for a prime p the polynomial ^p(X) = XP'^ + • • + X + 1 = {XP - 1)/(Z - 1) is irreducible. Of course, Eisenstein's Criterion does not apply directly. But observe that ^p{X) is irreducible if and only if ^ ^ ( X + 1) is irreducible, ^piX + 1) = {{X + \)P - 1)/Z. Expanding by the Binomial Theorem, we see that Op(X + 1) = X]f=i (f)^^~^ ^^ ^ monic polynomial of degree p — I with the coefficient of every power of X other than XP~^ divisible by p , and with constant term p divisible by p but not p^, so is irreducible. {^p{X) is the p^^ cyclotomic polynomial. For n composite, ^n{X) is not defined by the above formula. We deal with cyclotomic polynomials in Section 4.2.) o Let a be an integer of the form a = pq, where ;? is a prime and p and q are relatively prime. Then, by Eisenstein's Criterion (Proposition 4.1.7), the polynomial X^—ais irreducible over Q for any n. Thus, for example, X^ — 2 is irreducible over Q. However, Eisenstein's Criterion does not apply to X^ — 4, which is also irreducible. Let us develop a criterion that does. Lemma 4.1.9. Let a be an integer of the form a = p^q, where p is a prime and p and q are relatively prime. Let n be any integer that is relatively prime to m. Then the polynomial X^ — a is irreducible over Q.

88

4 Extensions of the Field of Rational Numbers

Proof. Let a = ^ and E = Q(a). Then (E/Q) = deg rriaiX) by Proposition 2.4.6 (2). Thus (E/Q) = n if and only degm«(X) = n. Now m«(X) divides X^ — a (as Of is a root of this polynomial), so deg m^ (X) = n if and only if Z" — a is irreducible. Thus, putting these together, we see that (E/Q) = nif and only ifX^—a is irreducible. Let j be an integer with jm = l(modn), so jm = nk + \ for some k. Clearly Q{ot) 2 Qiot^ (and in fact they are equal). Now {ot^Y = a^^ = aJ = p^^qj = p^^~^^qj so ^ = a-^/p^ satisfies ^" = p^^. In other words, )S is a root of the polynomial X^ — pqK and this polynomial is irreducible by Eisenstein's Criterion. Thus (Q(af^)/Q) = (Q(iS)/Q) = n so (E/Q) = n as required. D Corollary 4.1.10. (1) Let p be an odd prime. If a E Z is not a p^^ power, then X^ — a is irreducible for every t >Q. (2) If a e Z is positive and not a square, then X^ — a is irreducible for every ^ > 0. Proof In these cases a satisfies the hypotheses of Lenmia 4.1.9 (changing notation so that n = pin case (1) and n = 2 in case (2).) D Remark 4.1.11. In the excluded case, X^' — a may not be irreducible. For example, X4 + 4 = (X^ + 2X + 2)(X^ - 2X + 2). o Lemma 4.1.9 and Corollary 4.1.10 do not apply to the polynomial X^ — 72. This polynomial is also irreducible, but it will require considerably more work to show this. We do this in Section 4.6, where we show that X^ —a is irreducible in considerable generality. Remark 4.1.12. There is an algorithm, due to Kronecker, for factoring polynomials f(X) e Q[X]. It clearly suffices to consider the case that f(X) e Z[X] is a primitive polynomial. Then, by Corollary 4.1.6, it suffices to factor f(X) in Z[X]. Let f(X) have degree n. We begin by recalling the Lagrange Interpolation Formula. If ai,... ,a,n are distinct, and & i , . . . , &^ are arbitrary, there is a unique polynomial g(X) of degree at most m — \ with g{ai) = bi for each /, given by {X - aj)_

t-f

4 ,f (at — ai) '

Using this we proceed as follows, looking for a factor g(X) of f(X). If there is no such factor, then f(X) is irreducible. If there is such a factor, write

4.2 Cyclotomic Fields

89

f(X) = g{X)fi(X) and proceed recursively (next factoring fi(X),... )• To find such a factor, let m = 2 , . . . , [m/2] + 1. Pick distinct integers a i , . . . , a^. If f(ai) = 0 for some /, then we have found a linear factor X — at. Assume not and let Bt = [bi] be the finite set of divisors of /(«/), for each /. For each choice of elements bt e 5/, let g{X) be the polynomial given by the Lagrange Interpolation Formula with g{ai) = bt. If g{X) has degree less than m — 1, or does not have integer coefficients, there is nothing to check. If g{X) is a polynomial of degree m — \ with integer coefficients, test whether f{X) is divisible by g (X). o Remark 4,1.13. For future reference, we record an extension of this algorithm. Let R be the polynomial ring R = Z [ Z i , . . . , X„] and let F be its quotient field, the field of rational functions in Z i , . . . , Z„ with coefficients in Q. Let f(Xn-^i) e F[Xn-\-i] be a polynomial. Then there is an algorithm for factoring fiXn^il We proceed by induction. For n = 0 this is just the algorithm of Remark 4.1.n. Assume we have an algorithm for n = k and consider n = k + I. Now /? is a UFD (a Unique Factorization Domain), so the analogs of Gauss's Lemma (Lemma 4.1.4), Proposition 4.1.5, and Corollary 4.1.6 continue to hold for R. Thus we may, as in Remark 4.1.11, assume that /(X„+i) is a primitive polynomial in /?[X„+i]. Then proceed as in Remark 4.1.11. The only thing we need to see is that there is an algorithm for finding each of the sets Bt. But that follows by induction, since in factoring / ( a / ) , we are in the case n = k. o

4.2 Cyclotomic Fields In this section we study the splitting fields of the polynomials X" — 1 over Q. Recall that f„ is a primitive n^^ root of 1 in a field E if ^^ = 1 but f^ 7^ 1 for any 1 < m < n. If E is the splitting field of X" - 1 over Q, then E = Q(fn)In this case, we may simply take f„ = cxp(2ni/n), and we do this henceforth. Definition 4.2.1. 4>„(X) = ]~[(^ ~ ?) where the product is taken over all primitive n^^ roots of 1. ^n(X) is the n^^ cyclotomic polynomial and E = Q(f„) is the n^^ cyclotomic field. o Remark 4.2.2. Observe that a>„(X) and X" - 1 = n!'=o(^ " O ^^^ both products of distinct linear factors in E[X], and that <^n(X) divides X'^ — 1 in E[X]. o We let (p(n) (as usual) denote the Euler totient function, (p(n) being the number of integers between 1 and n that are relatively prime to n.

90

4 Extensions of the Field of Rational Numbers

Lemma 4.2.3. deg 4>„(X) = (p(n). Proof We observe that f is a primitive n^^ root of 1 if and only if f = i;;f for some k relatively prime to n, so {primitive n^^ roots of 1} = {f ^ | 1 < k < n, (k,n) = 1} and the cardinality of this set, which by Definition 4.2.1 is the degree of m^ (X), is (p(n), D Lemma 4.2.4. (1) ^n(X) e Q[X]for every n. (2) If Hi 7^ ^2, then „j (X) and ^miX) are relatively prime in Q[X]. Proof (1) Let E = Q(f„). By Remark 4.2.2, E is the splitting field of the separable polynomial X^ — \, Thus, by Theorem 2.7.14, E is a Galois extension of Q. If f is a primitive n^^ root of 1 in E, then a (i:) is also a primitive n^^ root of 1 for any a G Gal(E/Q), so ^n{X) is invariant under Gal(E/Q), and hence

^n{X)eQ[Xl (2) Let E = Q(?„i„2)- ^^^^ Definition 4.2.1 we see that 4>„^(Z) and „2 iX) are relatively prime in E[Z], and hence also in Q[X] by Lemma 2.2.7. (3) Every n^^ root of 1 is a primitive cfi^ root of 1 for a unique d dividing n. n Corollary 4.2.5. ci>„(X) € Z[X]. Proof From Lemma 4.1.4 we see that ^n{X) divides Z'* — 1 in Q[X], and hence, since ^n{X) is monic, „(Z) G Z[X] by Corollary 4.1.6. D Theorem 4.2.6. <E>„(X) is irreducible for every n. First Proof Suppose 4>„(Z) is not irreducible, and consider a monic irreducible factor f(X) of 0„(X) that has f„ as a root. Then f{X) does not have every primitive n^^ root of unity as a root, so there is some k > I with fi^n) ^ 0. Choose the smallest such k, and let p be any prime factor of k. Since k is relatively prime to n, p is relatively prime to n as well. Set f = ^n^^' Then, by the minimality of Jt, / ( f ) = 0 but / ( f ^) 7^ 0. Now f(X) is irreducible and monic, and / ( X ) divides X" — 1 in Q[X], so in fact X"" -I = f(X)g(X) where both f(X) and ^(X) are in Z[Z], by Corollary 4.1.6. Then 0 = i^Py - 1 = fingi^n

and / ( f 0 # 0, so g(f 0 = 0.

Writing g(X) = J2^iX\ Ci e Z, we see that 0 = Ylct^P' = E ^ K ? ^ ) ' , so h(^) = 0 where h(X) = g(XP). Since / ( O = 0 and / ( X ) is irreducible

4.2 Cyclotomic Fields

91

and monic, f(X) divides h(X) in Q[X] and hence in Z[X], again by Corollary 4.1.6. Now let 7T: Z[X] -> (Z/pZ)[X] be the map induced by reducing the coefficients mod p. Then n(f(X)) divides 7t(h(X)) — 7t{g{XP)) = 7t(g(X)y in (Z/pZ)[Xl so 7T(f(X)) and 7r(g(X)) have an irreducible factor q{X) in common. But X'^ - 1 = 7r(X" - 1) =

7t(f{X))7t(g(X)),

This would imply that X" - 1 is divisible by q(X)^ in (Z/pZ)[X], and hence that X" — 1 has a repeated linear factor in E^, its splitting field over Z/pZ = Fp. However, X^ — I has distinct roots in E^,, a contradiction. (This follows inmiediately from Lemma 3.2.1, but is easy to see directly.) Second Proof (Landau). Let m^^ (X) have degree d. Since m^^ (X) is monic and divides X" - 1 in Q[X], m^„ (X) G Z[X], by Corollary 4.1.6. We claim that, for any k, ^^ = niO for a unique polynomial rk(X) e Z[X] of degree less than d. To see this, note that, by the division algorithm, X^ = m^„ {X)qk(X)+rk(X) with qk(X), rk(X) e Z[X] unique polynomials with degryt(X) < d, and then set X = f„. It follows immediately that for any polynomial h(X) e Z[X], h{l;^) can be written as a unique polynomial in f„ of degree less than d with integer coefficients. In particular, we may write m^„(f ^) = gkiO with gk{X) a polynomial of this form. Since f "+^ = ^^, we see that gn+k{X) = gk(X), Now it follows immediately from the Multinomial Theorem and Fermat's Little Theorem (a^ = a(modp) for every a e Z, for p a. prime) that for any prime p and for any polynomial / ( X ) e Z[X], / ( X ^ ) - f(Xy has all of its coefficients divisible by /?. In particular this is true for / ( X ) = m^„(X), so m^^iXP) - m^^iXy = /7^(X) for some polynomial e(X) e Z[X]. Setting X = f, we see that gp(0 = m^^(^P) = pe{^P), Now e(^^) can be written as a polynomial in ^ of degree less than d with integer coefficients in a unique way, and gp{0 is also such a polynomial, so gp(f) = pe{^P), Thus we see that the coefficients of gp(X) are all divisible by p. There are only finitely many distinct gk{X), so there is an upper bound A for the absolute values of all of their coefficients. Let p be any prime with p > A. Then we see that gp{X) is the zero polynomial, so 0 = gp{^) = m^^(^n), and hence f„ and ^n have the same minimum polynomial m^^ (X) (as m^^ (X) is irreducible). We may iterate this argument to conclude that m^ (f ^) = 0 for any s not divisible by any prime p < A. Let r be any integer relatively prime to n. Let p i , . . . , /?^ be the primes less than or equal to A that do not divide r, and set 5 = r + npi " - pt^ Then s is such an integer. (If p < A is a prime, then either p is equal to pt for some /, in which case p divides the second term but not the first, or it is not, in which case p divides the first term but not the second.) Then s = r (modn), so

92

4 Extensions of the Field of Rational Numbers

m^^ (^^) = m^^ (f ^) = 0. Thus every primitive n^^ root of 1 is a root of m^„ (X) and hence *„(X) = m^^(X) is irreducible. D Corollary 4.2.7. (1) m^„(X) = ^n(X) and (Q(?„)/Q) = (pin). (2) Gal(Q(f„)/Q) ^ (Z/nZ)*. In particular, Gal(Q(f„)/Q) is abelian. Proof, (1) Since 0« (f„) = 0, m^„ {X) divides 4>„ (Z). But O^ (X) is irreducible by Theorem 4.2.6 so they are equal. Then (Q(frt)/Q) = degm^^(Z) = (p{n) where the first equality is Proposition 2.4.6 (2) and the second is Lenmia 4.2.3. (2) For any k with {k, n) = 1, f^ is also a primitive n^^ root of 1, i.e, a root of m^^ (X), so by Lemma 2.6.1 there is a map a^ : Q(Cn) -> Q(^n) with orjt(^„) = f^ and (J I Q = id, and a^ is determined by its value on f„. n Corollary 4.2.8. Let m and n be positive integers and set d = gcd(m, n) and I = lcm(m, n). Then Q(f^)Q(?J = Q{^d andQ(U) H Q & ) = Q(fj). In particular ifm andn are relatively prime, Q(fm)Q(fn) = Q(?m«) <^w<^ Q(fm) ^ Q(Cn) = Q» ^-^-^ Q(?m) «^^ Q(fn) ^^^ disjoint extensions ofQ. Proof. As f^ = f//" and f„ = ^'/\ clearly Q(f,) 3 Q(?^)Q(f«). Now l/m and £/n are relatively prime, so let s and t be integers with s(i/m) + t(i/n) = 1, so 1/i = s/m + t/n. Then f^ = ^^C„' so f^ e Q(?^)Q(?„) and

Qto) = Q(Cm)Qfe). Observe that, for any integers r and s with r dividing 5*, Q(^r) ^ QC?^)* so (Q(?.)/Q(?r))(Q(?r)/Q) = (Q(?.)/Q), and hence, by Corollary 4.2.7, (Q(?.)/Q(?r)) = (^(^)/(r). As frf = C^^ and ^d = Kn\ clearly Q(?^) c Q(f^) n Q(f,). Now, by Corollary 3.4.3 and the above observation,

(Q(?n)/(QaM) nQa„))) = (Q(?^)Q(f„)/Q(?^)) = (Q(?£)/Q(?m)) = (p{l)/(p{m). Furthermore, again using the above observation, (Q(f«)/(Q(?m) n Q(?«)))((Q(?^) n Q(?„))/Q(?^)) = (Q(?„)/Q(?^)) = (p(n)/(p{d), so ((Q(?m) n Q & ) ) / Q ( ? ^ ) ) = (^(n)/(p(rf))/(^(£))/(p(m)) = ((^(m)(^(n))/((p(rf))^(£)).

4.3 Solvable Extensions and Solvable Groups

93

But it is a fact from elementary number theory that for any two positive integers m and n, (p{m)(p{n) = (p{d)(p{i), so Q(fm) H Q(f„) = Q(^j), as claimed. n We conclude this section by giving a formula for „(Z). Proposition 4.2.9. For any positive integer n, the n-th cyclotomic polynomial ^n(X) is given by d\n

where JJL is the Mobius jji function, defined by jjiij) = 0 if j is divisible by the square ofaprimey and otherwise fi(j) = (—1)^ where i is the number of distinct prime factors of j . Proof Since, by Lemma 4.2.4 (3), X"" - \ = H^in * j ( ^ ) , the result follows immediately from the multiplicative version of the Mobius Inversion Formula. n

4.3 Solvable Extensions and Solvable Groups In this section, we show that an equation is solvable by radicals if and only if its group is a solvable group. Of course, we must begin by defining these terms. We are most interested in the case F = Q, but the proofs apply more generally. We develop the necessary field theory and the connections between field theory and group theory here. We refer the reader to Section A.l where we develop the necessary group theory. Definition 4.3.1. (1) A polynomial f{X) e F[X] is a radical polynomial if f{X) = X^ — a for some positive integer n and element a of F. An extension E of F is a radical extension (or an extension by radicals) if E = F(a) for a a root of a radical polynomial. (2) An equation f{X) = 0, f{X) e F[X], is solvable by radicals if there is a sequence of extensions EQ = F c Ei c . . . c Ej^ with E/ the splitting field of a radical polynomial fi(X) e E/_i(Z) and with E^: 2 E, a splitting field o f / ( X ) . o (Radical polynomial is not a standard term, but radical extension is.) Before proving our main theorem, Theorem 4.3.4, let us observe a special case.

94

4 Extensions of the Field of Rational Numbers

Lemma 4.3.2. Let f{X) = X^'-a e F[Z], letE be the splitting field of and let G = Gal(E/F). Then G is a solvable group.

f(Xl

Proof, E = ¥{^n.
4.3 Solvable Extensions and Solvable Groups

95

Conversely, assume that f{X) = 0 is solvable by radicals. Then there is a sequence of fields F = BQ c Bi c . . . c B)t with B/ the splitting field of X^' —at, at e B/_i, and with Bk 2 E. Then Gal(B//B/_i) is a solvable group, by Lemma 4.3.2. Suppose for the moment that k = 2. Then we have that Gal(B2/Bi) and Gal(Bi/Bo) are solvable groups, and that B2/B1 and B I / B Q are Galois extensions. If we knew that B2/B0 was a Galois extension, we could apply Lemma A.L4 (3) to conclude that G = Gal(B2/Bo) was a solvable group. However, a sequence of Galois extensions may not be Galois (and indeed, one can construct such an example precisely in this case). Thus in order to proceed we must enlarge the fields we are dealing with. We proceed inductively. We begin by setting BQ = BQ = F and B\ = Bi, and observe that B\ is a Galois extension of F. Suppose BJ_j is defined and is a Galois extension of F. We let B^ be the splitting field of riaGGaicB^ /B')(^'^' ~ ^(^i)) ^ B/-i[^]- Observc that this polynomial is invariant under Gal(B;_i/B[)) so it is in fact in B'Q[X] = F[X]. Thus B ; is the splitting field of a separable polynomial in F[X] so it is a Galois extension of F. Thus we obtain a sequence of fields F = BQ c Bj c . . . c BjJ, with each B^ a Galois extension of F and with BjJ, 2 B)t 2 E. Let G = Gal(B^/F) and let G, = Gal(B^_./B[)) for / = 0 , . . . , k. Then G = Go^"DGk = {!}. Also, G,_i/G, = Gal(B;^_.+i/B;_.). By Lemma 4.3.3, Gi^i/Gi is solvable for each / = 1 , . . . , A:. Then, by Lemma A.L3, G = Gal(B^/F) is solvable. Now F c E c B^ and E is a Galois extension of F, so Gal(E/F) = Gal(B^/F)/ Gal(B;^/E) is a quotient of a solvable group and hence is solvable, by Lemma A. 1.4 (2). D Corollary 4.3.5. (Abel) The general equation of degree n is not solvable by radicals for n > 5. Proof Let fniX) e F[X] be a polynomial with splitting field E, where Gal(E/F) = Sn, the synmietric group on n elements. Such a polynomial exists by Lenmia 3.L2, and in fact we have explicitly exhibited one in Remark 3.L5. Now, by Corollary A.3.6, 5„ is not solvable for n > 5. Thus, for n > 5, fn(X) is not solvable by radicals, by Theorem 4.3.4. D While we credit Theorem 4.3.4 to Galois, in fact he did not state it that way. We shall now derive Galois's original result. One the one hand, this is of historical interest, but on the other hand, it provides an easy way of exhibiting a polynomial in Q[X] whose Galois group is not solvable. Theorem 4.3.6. (Galois) Let f(X) e F[X] be an irreducible polynomial of prime degree p. Then f(X) = 0 is solvable by radicals if and only if all of its roots are rational functions of any two of them.

96

4 Extensions of the Field of Rational Numbers

Proof, By Theorem 4.3.4 we know that f{X) = 0 is solvable by radicals if and only if G = Gal(E/F) is a solvable group, where E is a splitting field of f{X). Since f{X) is irreducible, G acts transitively on the roots of f{X), First, suppose that all of the roots of f{X) are rational functions of any two of them. Regard G as a subgroup of the synmietric group Sp by its action permuting the roots of / ( X ) , and let a e GAfai ^ ai are roots, then for any € € E, a{e) is determined by a{a\) and or(a2). There are at most p{p — 1) choices for <j(ai) and cf{a2), so G has order at most p{p — 1), and so by Corollary A.1.9 G is solvable. Conversely, suppose that G is solvable. Let a\ and 0^2 be any two distinct roots of / ( X ) , and let B = F(ai, ^2) ^ E. We need to show that B = E. We apply the Fundamental Theorem of Galois Theory (FTGT), Theorem 2.8.8. B is a field intermediate between F and E, so G' = Gal(E/B) c G = Gal(E/F). By definition, Gal(E/B) = {a e G \ a{P) = ^ for every p e B}. But the FTGT establishes a 1 — 1 correspondence between subgroups of G and fields intermediate between F and E, so B = E if and only if G' = {id}. We proceed to show this. Let a e G. Note that a G G' if and only if or(ai) = ofi and a(a2) = af2. By Proposition A.L7 (and using the notation there), we may identify G with GH for some subgroup H of F*, and ai and a2 with [^J and [^^] for some ii 7^ 12. Let <^ = [0 1 ] ^ ^H with (j(ai) = a\ and a{oi2) = Qf2- Then

p i ] _ \h ri\ U{\ _ r/i/i + n\

[ij-[oij[ij-[

n2l _ \h n\ r/2l _ \hi2 + n\

1 J' LiJ"LoiJLiJ-L

1 J'

and these two equations readily imply /i = 1, n = 0, i.e., a = id. Thus we see that G' = {id}, completing the proof. D Corollary 4.3.7. Let F c R and let f(X) e F[X] be an irreducible polynomial of prime degree p with k real roots, I < k < p. Then f(X) = 0 is not solvable by radicals. Proof Suppose f(X) has real roots or 1 7^ 0^2 and a nonreal root ^8. Then ^ is not a rational function of ai and ^2, so, by Theorem 4.3.6, / ( X ) = 0 is not solvable by radicals. n Remark 4.3.8, In Example 4.7.5 we will see that, for each odd prime p , there is an irreducible polynomial fp{X) e Q[X] with p — 2 real roots. Thus, by Corollary 4.3.7, fp{X) is not solvable by radicals for p > 5. But in fact we will show the stronger result that the Galois group of fp(X) over Q is isomorphic to Sp, thus providing an example of Corollary 4.3.5 over Q. We will

4.4 Geometric Constructions

97

further provide an example of such a polynomial /„ (X) over Q for every w > 5 in Proposition 4.7.10, but the construction there is considerably more complicated, o

4.4 Geometric Constructions In this section we will investigate the question of constructibility by straightedge and compass. We will see that the three classical questions of Greek antiquity (trisecting the angle, duplicating the cube, and squaring the circle) are unsolvable by these methods. We will also see when it is possible to construct a regular n-gon by these methods, a result originally due to Gauss. In particular, we will show how to construct a regular 17-gon. We begin by drawing an arbitrary line L, and choose two distinct points O and A on it. We normalize coordinates by letting O be the origin and A be the point (1, 0), so in particular the line segment OA has length 1. Thus the line L is the ;c-axis. We recall that, using straightedge and compass, it is possible to construct a line through a given point parallel to, or perpendicular to, a given line. Thus, in particular, we may construct the y-axis and hence we have the usual Cartesian coordinates in the plane. There are two ways to regard points in the plane. We may consider that a point P has coordinates (x,y), or that P is represented by the complex number z = x + iy.Ifwt have constructed the point P represented by z = x + iy, then we may drop perpendiculars to obtain the real and imaginary parts x and y of P ; conversely, given real numbers x and y we may certainly obtain the point (x, y) corresponding to z = x + iy. Also, all arithmetic operations on complex numbers (addition, subtraction, multiplication, division) may be performed by doing arithmetic operations on their real and imaginary parts. Thus we shall pass freely between these two different descriptions. Our first main result is the following: Theorem 4.4.1. The complex number z can be constructed by straightedge and compass if and only if z is algebraic over Q and there is a sequence of fields Fo = Q C Fi C • • • C F^ with Q(z) c Yk and with (F^/F/_i) = 2 for each i = I,..

.,k.

Proof First, we observe that we may perform all arithmetic operations with straightedge and compass. Clearly we can add and subtract (positive) real numbers X and y,Wc may multiply them as follows: Let the line segment AB have

98

4 Extensions of the Field of Rational Numbers

length X and extend AB to a line L. Let M be another line through A and let C and D be points on the same side of A with the line segments AC and AD having lengths 1 and y, respectively. Construct the line A^ through D parallel to the line segment CB and let this line intersect the line L in the point E. Then, by similar triangles, the line segment AE has length xy. To divide we perform a similar construction, this time letting N be the line through C parallel to the line segment DB, and then AE has length x/y. We also observe that we may take the square root of a (positive) real number X using straightedge and compass. To do so, let line segment AB have length X and extend A 5 to a line L. Let C be a point on the opposite side of A from B with the length of the line segment AC equal to L Construct a semicircle with diameter CB, and a line M perpendicular to CB at A. Let D be the point at which the line M intersects this semicircle. Then, by similar triangles, AD has length ^/x. With this in hand, we turn to the proof of the theorem. First, suppose there is a sequence of fields as indicated. By our first observation, every x € Q = FQ can be constructed. Since ( F / F Q ) = 2, F I = Fo(ai) where ai is a root of an irreducible quadratic equation with coefficients in FQ. Then, by the quadratic formula, Fo(ai) = Foi^/oi) for some ai e FQ, and by our second observation, y/oi can be constructed, and hence every x G Fi can be constructed. Then proceed by induction. Conversely, suppose a point can be constructed with a straightedge and compass. Then it is obtained by a succession of the following operations: (1) Finding the intersection of two lines. (2) Finding the intersection of a circle and a line. (3) Finding the intersection of two circles. In case of operation (1), let the two lines be given by the equations aix + biy = c\ and a2X + b2y = C2 with ai, ^2, &i, b2, ci, C2 G F for some field F. Then the solution (x, y) has x e F and y € F. In case of operation (2), let the circle be given by (x — ai)^ + (y — bi)^ = cj and the line be given by a2X + 62J = C2 with ai,a2,bi,b2,ci,C2 e F. Solving for y in terms of X (or vice versa) in the linear equation, substituting in the quadratic equation, and solving, shows x e F(d) and y e V{d) where (F(J)/F) = 1 or 2. In case of operation (3), let the two circles have equations (x — ai)^ + (y — bi)^ = c\ and (x — ^2)^ + {y — ^2)^ = c^ with ai, ^2,fci,b2, c\,C2 G F. Subtracting the second equation from the first gives a linear equation a^x + b^y = C3 for some as, 63, C3 € F, so combining the equation of the first circle (say) with this linear equation, we are reduced to operation (2). This yields the theorem. n

4.4 Geometric Constructions

99

Corollary 4.4.2. It is impossible to perform the following constructions using straightedge and compass: (1) (Trisecting the angle) Trisecting an arbitrary angle. (2) (Duplicating the cube) Constructing a cube with volume twice that of a given cube. (3) (Squaring the circle) Constructing a square with area that of a given circle. Proof. (1) Given an angle 0, let one side of the angle be the x-axis and let the other side of the angle intersect the unit circle at P. Then P has coordinates (cos^, sin^). Thus we may construct the angle ^/3 if and only if we may construct the real number cos^/3. Recall the formula cos 3^ = Acos^ (p — 3 cos (p. Letting cp = 6/3 and x = cos cp, we see that x satisfies the cubic equation 4x^ — 3x = cos 6. We may certainly construct an angle 0 = 7t/3 (= 60°), and cos(7r/3) = | . Thus jc is a root of the polynomial f(X) = 8X^ — 6Z — 1. Using Lemma 4.1.2, it is easy to check that f(X) has no linear factor in Q(X), and hence is irreducible, so (Q(x)/Q) = 3. But then we cannot have Q(x) c F/: with Fk as in Theorem 4.4.1, as (Fk/Q) is a power of 2 and hence not divisible by 3. (2) Trivially, we may construct a segment of length the side of a cube of volume 1. But then to construct a segment of length the side of a cube of volume 2 would be to construct x = \fl. But (Q(v^)/Q) = 3 does not divide any power of 2, so this is impossible as in part (1). (3) Trivially, we may construct a segment of length the radius of a circle of area n. But then to construct a segment of length the side of a square of area n would be to construct x = ^y/n. But it is a famous theorem of Lindemann that n is transcendental (i.e., not algebraic, or equivalently that jt and hence ^/n does not satisfy any algebraic equation) so (Q(V^)/Q) is infinite and hence y/n is not an element of any finite extension of Q (regardless of degree). D We have an equivalent formulation of the condition in Theorem 4.4.1. Lemma 4,4.3. For a complex number z, the following are equivalent: (1) There is a sequence of fields FQ = Q C Fi C • • • C F^; with Q(z) c Fk and with (F^/F^_i) = 2 for each i = I,... ,k. (2) IfFis a splitting field ofm^iX), then (E/Q) is a power of 2. Proof (2) implies (1): Let G = Gal(E/Q) and let G have order 2^. By Corollary A.2.3 there is a sequence of subgroups G = Go D Gi D -- - D Gk = {1} with [G : Gi] = 2\ Let F, = Fix(G/). Then {F,} is as required. (1) implies (2): We prove this by induction on k. For A: = 0 it is trivial and for A: = 1 it is immediate, as every quadratic extension is a splitting field. Let (F^/Fj^-i) = 2. Then F], = Fk-iiak) for some ak € Fk with

100

4 Extensions of the Field of Rational Numbers

deg rria,^ (X) = 2. In the course of the proof we shall have occasion to replace the field F^t-i by another field Ejt_i. Observe that for any field E^_i 2 Fk-i, (Ek-i(ak)/Ek-i) = 1 or 2 by Lemma 2.4.7. Let Eo = Fo and Ei = Fi. If a^t ^ E^_i, then E^ = Ek-i(ak) = E^_i and there is nothing to do. Otherwise, let G = Gal(E^_i/Eo) and let

f(X) = Yla(ma,iX)). aeG

Then f{X) is invariant under G, so f{X) e Eo[X]. Let E^: be a splitting field of f{X) containing Fj^. Then f{X) has roots ^\ = ak,..., ^m- Now nioikiX) has degree 2, so aimaiX)) has degree 2 for each aeG, and hence (Ek-i(Pi)/Ek-i) = 2 for each /. Now consider E/:_i c Ek-i(^i)

c Ek-i(^i, ^2) ^ • • • ^ Ek-iiPi,...,

Pm) = ^k-

Set E^_i = Ek-i(fiu ^.., A - i ) . By Remark 2.3.6, Ek-i(^u . . . , A) = E ^ _ i ( ^ i , . . . , A-i)(A) = E^-i(A) is the composite E'j^_^Ek-i{Pi), so, by Lemma 2.3.8, ( E , _ i ( ^ i , . . . , A ) / E , _ i ( ^ i , . . . , A_i)) = (E^_i(A)/E^-i) < ( E , _ i ( A ) / E , _ i ) = 2, so each successive extension is of degree 1 (i.e., is trivial) or of degree 2. Thus (Ek/Q) is of degree a power of 2, and is a splitting field, hence a Galois extension of Q. Now z G Ej^, so mz(X) splits in Ej^ and hence m^(Z) has a splitting field E with Q c E c E^. Then (E/Q) divides (Ek/Q), so (E/Q) is a power of 2. D Remark 4,4.4, For an arbitrary prime p, the implication (2) implies (1) of Lemma 4.4.3 holds, with the identical proof, but the implication (1) implies (2) does not. The proof breaks down as Lemma 2.3.8 merely guarantees a degree between 1 and p, but it may be strictly between 1 and p, A counterexample to the implication (1) implies (2) of Lenmia 4.4.3 is given by z = \/2. Then (Q(z)/Q) = 3 but m^(X) = X^ -2, whose splitting field E over Q has (E/Q) = 6, which is not a power of 3. (Compare Example 2.3.7.) We now do a bit of elementary number theory. Let n = 2^ + 1 for some positive integer t. If t has an odd factor r > 1, then, writing t = rs, n is divisible by 2^ + 1 (as we see by substituting X = 2* in the algebraic identity X' + l = (X + l)(X'-^ - X'-^ + • • • + 1), valid for any odd r). Thus the only possible primes of this form are when t = 2^ for some k. Set Fk = 2^ + 1. If Fk is prime, it is called a Fermat prime. This terminology is due to the

4.4 Geometric Constructions

101

fact that Fermat claimed F^ is prime for every k. Now FQ = 3, F I = 5, F2 = 17, F3 = 257, and F4 = 65537 are indeed prime, but Euler discovered that F5 = 4294967297 is divisible by 641 and hence is composite. In fact, there is no known value of A: > 4 for which Fk is prime. o Here is our second main result: Theorem 4.4.5. (Gauss) A regular n-gon is constructible by straightedge and compass if and only ifn is of the form n = 2V1

"Pj

with {pi} distinct Fermat primes. Proof Clearly a regular n-gon is constructible if and only if a regular n-gon inscribed in a circle of radius 1 is constructible, and this is true if and only if its vertices are constructible, and this is true if and only if f„ = txp(2ni/n) is constructible. By Theorem 4.4.1 and Lemma 4.4.3, this is true if and only if the splitting field of m^^ (X) has degree a power of 2. But this splitting field is simply Q(exp(27r/(n)), the n^^ cyclotomic field, which has degree (p(n) by Corollary 4.2.7. Ifn = 2^p\pl.. •, then (p{n) = 2''-\pi - l)Pi~\p2 - ^)PT^ • • •, SO we see that (p{n) can be a power of 2 if and only if ft = c = • • • = 1 and each of Pi, P2,' " is more than a power of 2, and hence a Fermat prime. D Example 4.4.6. (1) Everyone knows how to construct an equilateral triangle. But in any case, ^3 satisfies the equation X^ + X + 1 = 0. In fact, constructing ^3 = (—1 + iV3)/2 gives a construction of an equilateral triangle that is different from the usual one. It is the following (using the notation at the beginning of this section): Extend the line segment A O so that it intersects the unit circle at the point D. Construct the midpoint E of the line segment OD, and then construct the line M perpendicular to the segment OD at E. Let this line intersect the unit circle at the points B and C. Then AfiC is an equilateral triangle. (2) We essentially showed how to construct a regular pentagon in Example 1.1.4. To review that, let 0 = ^5 + ^^ = ^5 + ^^K Then 6 satisfies the equation Z^ - X - 1 = 0 and f5 satisfies the equation X^ -OX + 1=0. (3) We show how to construct a regular 17-gon. Let ^ = ^17 = exp(27r//17). Let G = Go = Gal(Q(f )/Q). Then we know that G is cyclic of order 16. Write G = GQ D Gi D G2 D G3 D G4 = {1} where [G : G/] = 2\ and note that G/ is cyclic of order 2^''. Let F, = Fix(G/)

102

4 Extensions of the Field of Rational Numbers

so Fo = Q and F4 = Q(f). G is isomorphic to F*^, and direct calculation shows that 3 is a generator of this group. (In general, a generator of F* is called a primitive root mod p.) Let crd^) = ^^\i = 0 , . . . , 15. Then Go = {OTQ, OTi, . . . , oris}, Gi = {ao, a 2 , . . . , a u } , G2 = {<^o. ^^4, erg, ai2), G3 = {ao, org}, and G4 = {CTQ}. Also, (F//F,_i) = 2 for / = 1 , . . . , 4. Write E = F4 = Q ( 0 . In order to simplify our computations, we recall the following elementary fact: If r and s are the roots of the quadratic equation X^ — aX + b = 0, then r + s = -amdrs = b, (Proof: (Z - r){X - s) = X^ - {r + s)X + rs.) We work from the bottom up. ( F I / F Q ) = 2 SO F I = Fo(a) for any a G E fixed under G\ but not under Go, i.e., any a with a2(a) = a but G\{pt) ^ a. Let «O = E L O ? ' "

=^ + ^' + ^'' + ^'' + i'' + ^' + ^' + ^\

If a = Qfo or ai, then a2(a) = ot but a\{a) 7^ a. Direct calculation shows ao + Qfi = — 1 and ofoc^^i = —4, so ao and a\ are the roots of X^ + X — 4 = 0. Now (F2/F1) = 2 so F2 = Fi(^) for any ^ G E fixed under G2 but not Gi, i.e., any ^ with a^i^) = ^ but a2{^) ^ ^. Let ^ 1 = ^ 3 + ^ 5 + ^14 + ^12^

^2 = f^ + ?l^ + ?^ + C^

^ 3 = ^ 1 0 + ^ 1 1 + ^ 7 + ^6^

If )6 = )So, ^ 1 , )62, or ^3, then a^{P) = p but a2(fi) # )S. Then ^0 + ^2 = c^o. )6o)S2 = —1, and ^1^3 = ai, ^1^3 = —1, so ^0 and ^2 are the roots of X^ -otoX- 1 = O a n d ^ i and^3 are theroots of X^ - aiX - 1 = 0 . Next (F3/F2) = 2 and F3 = F2(y). Let 74 = ?^^ + ?^

If K = yo or }/4, then a^{y) = y but or4(]/) 7^ y. Then yo + 74 = /^o, YQY^ = ^1, SO yo and 3/4 are the roots of X^ — PQX + ^1 = 0. Finally, let So = ? and 5g = f ^^. If 5 = So or 5g, then ag(5) # 5. Set F4 = F3(5o) = F3(f). Note that f satisfies <• + ^^^ = yo, f ?^^ = 1, so f is a rootofX^-yoX + 1 = 0 . Summarizing, we construct ^ = ^17 as follows: (1) Solve the quadratic X^ + Z — 4 = 0; call the roots ao and a i .

4.5 Quadratic Extensions of Q (2a) (2b) (3) (4)

103

Solve the quadratic X^ — aoX — 1 = 0 ; call the roots fio and ^2. Solve the quadratic X^ — ofiX — 1 = 0; call the roots fix and ^3. Solve the quadratic X^ — P^X + ^1 = 0; call the roots yo and y^. Solve the quadratic X^ — yo^ + 74 = 0; its roots are f = 5o and ^^^ = 58.

With enough patience (and careful attention to sign) the reader may obtain an explicit formula for f which involves square roots nested four deep. (To start off, CYQ = ( - 1 + vT7)/2 and ofi = ( - 1 - vT7)/2, showing that Q(VT7) c Q(f 17). Compare Corollary 4.5.2.) o

4.5 Quadratic Extensions of Q We recall a bit of elementary number theory. Let /? be a prime. Identifying F^ with Z//>Z, we know that G = F* = { l , . . . , / 7 — l } i s a cyclic group, and a generator r of this group is called a primitive root mod p. For example, 3 is a primitive root mod? as 3^ = 3 (mod?), 3^ = 2 (mod?), 3^ = 6 (mod?), 34 = 4 (mod?), 3^ = 5 (mod?), and 3^ = 1 (mod?). On the other hand, 2 is not a primitive root mod? as 2^ = 1 (mod?). (We must be careful not to confuse a primitive root mod p with a primitive p^^ root of 1. Observe that ^p is a primitive p^^ root of 1 for any k not divisible by p,) Observe also that (Z/4Z)* = {1, 3}, so 3 is a primitive root mod 4. If p is an odd prime, then G is cyclic of even order p — 1, so it has a unique subgroup H of index 2. Let TT: G -> G/H = Z/2Z = {0, 1} be the quotient map. For an integer k relatively prime to p, we regard k as an element of Z/pZ, Then Xp^ the quadratic residue character mod p, is defined by XpC^) = (—1)^^^^. (For fixed p, we shall write x for Xp-) There is a more traditional definition of the quadratic residue character. Recalling that G is a multiplicative group, the subgroup H consists of the (quadratic) residues, i.e., of those elements of G that are squares, and its complement consists of the (quadratic) nonresidues, i.e., of those elements of G that are nonsquares. Then x(k) = 1 if ^ is a residue and x (^) = — 1 if ^ is a nonresidue. Furthermore, if r is a generator of G, i.e., a primitive root modp, then X(k) = lifk is an even power of r and x (^) = — 1 if ^ is an odd power of r (and this is independent of the choice of r). For example, if p = ?, 1 = (±1)^ 4 = (±2)^, and 2 = (it3)^ are the residues and 3, 5, and 6 are the nonresidues. Choosing the primitive root 3, we see that 1 = 3 ^ 2 = 3 ^ and 4 = 3^, while 3 = 3 ^ 5 = 3 ^ and 6 = 3 ^ It is then easy to see that the quadratic residue character has the following properties:

104

4 Extensions of the Field of Rational Numbers

(1) X(7^) = xU)x(k). (x is a homomorphism to the multiplicative group {±1}.) (2) J2^Zl x(k) = 0. (There are (p - l)/2 residues and (p - l)/2 nonresidues.) (3) x ( - l ) = (-l)^^-i>/^ = 1 if /7 = 1 (mod 4) and = -lif p = 3 (mod4). (If r is a primitive root mod/?, let a = r^P-^^^, Then a ^ 1 but a^ = 1. On the other hand, the equation X^ — 1 = 0 has only two solutions in F„, X = ± 1 . Hence a = —I. Thus if p = I (mod4), (p — l)/2 is even, a is an even power of r, and x (^) = 1, while if p = 3 (mod4), (p — l)/2 is odd, a is an odd power of r, and x (^) = — 1 •) Theorem 4.5.1. Let E be a quadratic extension ofQ, Then E is intermediate between Q and some cyclotomic field. In particular: ( 7 J Q ( V ^ ) = Q(f4). (2) Q(V2) C Q(?8) and Q ( V ^ ) C Q(?8). (3) If p is a prime congruent to 1 modulo 4, then Q(.y/p) ^ Q(?/7)(4) If p is a prime congruent to 3 modulo 4, then Q(V~P) — Q(?p)Pmo/ Observe that if Q ( v ^ ) c Q(^^) andQ(V&) Q Q(^«),thenQ(Va^) ^ Q(V^, V&) c Q(^^, f„) c Q(f^„), so in order to prove the general statement it suffices to prove the particular statements (1), (2), (3), and (4). (1) is obvious. (2) is direct calculation: ^g = (1 + / ) / A / 2 . We prove (3) and (4) simultaneously. Actually, we present two proofs. Let p be an odd prime and let p-i

Sp = J2x(k)^P k=\

Then

k=l

7=1

k=l

j=l

Set j = km (mod p) and notice that as j runs over the nonzero congruence classes modp, so does m. Also, x(kj) = x(k^^) = Xi^)- Thus

k=l

m=\

p-\

p-l

m=l

k=l

4.5 Quadratic Extensions of Q

105

Now 1 + ^p H h ^p~ = 0, so as long as m 7^ p — 1, the inner sum is — 1. If m = p — 1, then the inner sum is of course p — I. Thus

^

m=l

/

^^t 5Zm=i X (^) ~ ^' ^^ ^'^^ ^^^^ ^^^"^ ^^ +X (~ 1) ^iid we see

5j = px(-i), as required. Our second proof uses more theory and a minimum of calculation. Let G = Gal(Q(fp)/Q) and recall that G is generated by CTQ, where o^oCfp) = Cp with r a primitive root mod p . Since Q(^p) is a Galois extension of Q, any a e Q(^p) with cro(a) = a (and hence with cr(a) = a for every a e G)is in fact an element of Q. Now (Q(^p)/Q) = p — I and Q(fp) is certainly spanned by {1, f^,..., ^p~ }' We know one relation among these p elements: 1 + f^ + • • • + f/~ = 0. Hence {f^,..., ^/~ } form a basis for Q(^p), so in particular this set is Q-linearly independent. Thus if of e Q(^p) with cxoia) = a, then, writing a = XlfJi ^i^p with a/ G Q, we must have ai = - - - = ap-i = a, for some a. (Of course, in this case a = —a e Q.) With these observations in mind, let us set (/7-l)/2

(p-l)/2

Observe that the exponents in ao are all quadratic residues mod p, and that the exponents in ofi are all quadratic nonresidues mod p. Then oio + cxi = ^p-{

f^~^ = - 1 and a(ao) = au cr(ai) = (XQ,

Lotb = aooti. Then ao(b) = oro(aoQfi) = ao(ao)(ro(ai) = aiao = b, sob e Q. We now compute b.

106

4 Extensions of the Field of Rational Numbers

First, suppose p = I (mod 4). There are (p—1)/2 terms in ao and (p—l)/2 terms in ofi, so there are (p — 1)^/4 terms in the product. Each term is of the form f/^^ with j a quadratic residue and k a quadratic nonresidue. Since p = 1 (mod4), x(-j) = X ( - I ) x ( y ) = X(y). so j and - y are always either both residues or both nonresidues, thus f^ 7^ (f/)~^ and hence no term f/f^ is equal to 1. Then, writing b = J2f=i ^/fp> ^ ^ see that Yl^t = (P ~ 1)^/4 and that ai = - - = Up-i = ^, for some a, so J]) a^ = (p — \)a = (p — 1)^/4 and hence a = (p — l)/4. Then p-i

p-i

b = aJ2^' = ((p- l)/4) ^C^ = dp - l)/4)(-l) = ~(/7 - l)/4.

Now suppose p = 3 (mod4). Then x(—j) = —X(7)» so one of 7 and —j is always a quadratic residue and the other is a quadratic nonresidue. Hence each term in ao "pairs up" with one of the terms in ai to give a product of 1. This takes care of (p — l)/2 of the (p — 1)^/4 terms. Thus we may write b = (p — l)/2 + b^ where b^ is the sum of the remaining (p ~ 1)^/4 — (p — l)/2 = (p — l)(p — 3)/4 terms. We now apply the same logic to b\ Writing b' = Efj/^/^j^' we see that J2a'. = (p - l)(p - 3)/4 and that a[ = " ' = a^_j = a\ for some a^ so ^ a- = (p — l)a^ z= (p — l)(p -^ 3)/4, and hence a^ = (p — 3)/4. Then

&' = a'J2^' = ((^ - 3)/4) 5^?^" = ((P - 3)/4)(-l) = -(p ~ 3)/4,

and then b=^(p-

l ) / 2 + Z>^ = (/7 - l ) / 2 - (/7 ^ 3)/4 = (p + l)/4.

Now (X - ao)(X - ai) = X^ - (ofo + ctri)X + aoai

X^ + Z + &, so (Z - ao)(X - ai) = X^ + Z - (j9 ~- l)/4

if p = 1 (mod4),

(X - ao)(Z - QTi) = X^ + Z + (77 + l ) / 4

if /? = 3 (mod4).

4.5 Quadratic Extensions of Q

107

In other words, ao and ofi are the roots of this quadratic equation. But we may find these roots by the quadratic formula, and we obtain {ao, oci} = {(-1 ± V P ) / 2 }

if P = 1 (mod4),

{oio,oci} = {(-1 ± V ^ ) / 2 }

ifp = 3 (mod4).

Corollary 4.5.2. For p an odd prime, Qiy/Xp(~^)p) field intermediate between Q and Q(fp).

D

i^ ^he unique quadratic

Proof. Gal(Q/^p)/Q) = Z / ( p — 1)Z contains a unique subgroup of index two, so there is a unique quadratic field intermediate between Q and Q(fp), and we have just identified that field. D Remark 4.5.3. Let us cite some results from number theory that will enable us to extend Corollary 4.5.2. They are: (1) If mi and m2 are relatively prime, then (Z/(mim2)Z)* = (Z/miZ)* 0 (Z/m2Z)*. (This follows easily from the Chinese Remainder Theorem.) (2) For p an odd prime and k a positive integer, (Z/p^Z)*, a group of order (p(p^) = (p — l ) p ^ ~ \ is cyclic. (This is usually phrased as saying there is a primitive root mod p^.) (3) (Z/4Z)* = Z/2Z and for k > 2, (Z/2^Z)*, a group of order (p(2^) = 2^~^ is isomorphic to the direct sum of a cyclic group of order 2 and a cyclic group of order 2^~^. o Corollary 4.5.4. Le^ n > 2 be an integer Define a set A = {a/}^=i ...,^ as follows: If p is an odd prime factor ofn, then Xp(~^)P ^ ^- If^ i^ divisible by 4, then —1 e A. If n is divisible by 8, then 2 e A. Then Q(^n) contains 2^ — 1 quadratic extensions ofQ, and they are Q(^/ni) for m any nontrivial product of distinct elements of A. Proof. By the Fundamental Theorem of Galois Theory, the quadratic extensions of Q contained in Q(fn) are in 1 — 1 correspondence with the subgroups of index 2 of Gal(Q(f„)/Q = (Z/nZ)*. Now if n = 2"23"^5^5 _^^ (Z/nZ)* = (Z/2^2^)* X (Z/3^^Z)* X (Z/S^^Z)* X . . . . For /7 odd and k>h (Z/p^Z)* is a cyclic group of even order. Also, (Z/2)* is trivial, (Z/4)* = Z/2Z, and (Z/2^Z)* = (Z/2Z) 0 (Z/2^-^Z). Thus (Z/nZ)* contains 2' - 1 subgroups of index 2, where t is as in the statement of the corollary. (A subgroup of index 2 is the kernel of an epimorphism x/r: Gal(Q(f„)/Q) -^ Z/2Z and since Gal(Q(f„)/Q) is isomorphic to the direct sum of t cyclic groups of even order, there are 2^ homomorphisms from Gal(Q(fw)/Q) to Z/2Z, one of which is the trivial one.) Since Q ( V ^ ) ^ Q(?«) for each of the 2^ — 1 values of m in the statement of the corollary, by Theorem 4.5.3, we see that these are all the quadratic subfields of Q(f«). •

108

4 Extensions of the Field of Rational Numbers

Corollary 4.5.5. Let m be a square-free integer Ifm is odd and positive, set m' = m. Otherwise set m' = 4|m|. Then Q(>/m) ^ Q(?n) if and only ifn is a multiple ofm'. Proof Corollary 4.5.4 gives all the quadratic subfields of Q(^„).

n

Remark 4,5,6. We have shown in the proof of Theorem 4.5.1 that for an odd prime /?, Sp = ±y/xp(—l)P' It is a famous theorem of Gauss that in every case the sign is positive. It is easy to check that Sp = 2ao(p) + 1, so we see that aoip) = (—1 + y/xp(—'i~)p)/2 in every case as well. o

4.6 Radical Polynomials and Related Topics In this section we investigate radical polynomials f(X) e Q[X], i.e., polynomials of the form X" — a, their associated radical extensions Q(^/a), and their splitting fields Q ( v ^ , ^n)- The results we obtain are interesting in their own right, and, as we shall see, also have applications to cyclotomic fields and Kummer fields. First we determine the Galois groups of their splitting fields. We remind the reader of our discussion of primitive roots at the beginning of Section 4.5. Lemma 4.6.1. Let p be an odd prime. Let a e Q with a not a p^^ power in Q, and let E be the splitting field ofX^ — a. Let G = Gal(E/Q). Then G=

(or, T I or^ = 1, T^'^ = 1, t a r " ^ = a')

for some primitive root r mod p. Proof Certainly E = Q(CY, ^p) where a G E, of^ = a. We claim that Q(a) n Q(^p) = Q, To see this, let B = Q(a) fl Q(fp). Then Q c B c Q(fp), so (B/Q) divides (Q(^p)/Q) = p - 1, and Q c B c Q(a), so (B/Q) divides (Q(Qf)/Q) = p. Thus (B/Q) = 1, i.e., B = Q. In particular, a ^ Q(Cp) and (E/Q(fp)) = p. Thus, if rhaiX) denotes the minimum polynomial of a over Q(fp), and m^ (X) denotes the minimum polynomial of a over Q, we see that p-i

maiX) = m,(X) = Yl(X - ^ a ) e E[X]. i=0

Then, by either Lemma 2.6.1 or Lemma 2.6.3, there is an automorphism a of E with cr(a) = fpQf,

a(^p) = ^p.

4.6 Radical Polynomials and Related Topics

109

On the other hand, consider a generator TQ of Gal(Q(fp)/Q). This is given by to(fp) = ^p for r a primitive root mod/?. Again, by either Lemma 2.6.1 or Lemma 2.6.3, there is an automorphism TQ of E extending TQ. Then ro(a) = f^a for some k. Set r = cr"^ to. Then r is an automorphism of E with r(Qf) = a ,

r(fp) = ^'p.

Then direct calculation shows a'r{a)

= or''(r(a)) = or''(a) = ^;a

while ra(a)

= r(a(a))

= T(^pa) = f^a

rcr(^p) = T(a(^p)) = r(^p)

= ?;,

so a^T = TO e Gal(E/Q), i.e., xax~^ = a^, as claimed.

n

Remark 4.6,2, (1) We see that, in the situation of Lemma 4.6.1, G = HN is the semidirect product of the normal subgroup N = {a \ a^ = I), isomorphic to Z/pZ, and the subgroup H = {r \ r^~^ = 1), isomorphic to Z/(p — \)Z, (2) The structure of G is independent of the primitive root r. (There is a choice involved in the generators of G. Varying this choice varies r among all the primitive roots.) (3) Note that G is nonabelian. Note also that Lemma 4.6.1 also holds for p = 2, when it recovers the much easier fact that in this case G = Z/2Z. o Lemma 4.6.3. Let a e Q with a 7^ ±b^ for any b e Q^ and let E be the splitting field of X^ - a. Let G = Gal(E/Q). Then G=

(o-, r I o-^ = 1, r^ = 1, r a r " ^ = a^).

Proof, Let a € E with a"^ = a. Then E = Q(a, ^4). First we must show that Q(a) H Q(f4) = Q. Assume Q(a) 0 Q(^4) D Q Since (Q(f4)/Q) = 2, this implies that Q(Qf) 2 Q(?4) and hence that E = Q(Qf). Then (E/Q) = 4 and (Q(f4)/Q) = 2, and so Q(f4) is the fixed field of a subgroup {id, a} of Gal (E/Q). Since a is a root of the polynomial X'^—a, a (a) must also be a root of this polynomial, so a (a) = ^^a for some k. But cr(^4) = ^4 and a 7^ id but a^ = id, so ^^a 7^ a but f|^a = a. Hence A: = 2 and a (a) = —a. Now E = {co + cia + C2a^ + csa^ \ ci e Q} and Fix(or) = {CQ + C20c^ \ ct e Q}. We claim that / = f4 ^ Fix(a). Suppose i = Co + C20t^. Then — 1 = /^ = (CQ + C2a^)^ = (CQ + c^a) + 2coC2a^. Since a is not a square, of^ ^ Q, so C0C2 = 0 and either CQ = 0,

110

4 Extensions of the Field of Rational Numbers

c\ = —\/a, which is impossible since —a is not a square, or C2 = 0, CQ = — 1, which is certainly impossible. Thus Q(Qf) fi Q(f4) = Q. The rest of the proof entirely parallels the proof of Lemma 4.6.1. n Lemma 4.6.1 and 4.6.3 have a considerable generalization in Theorem 4.6.6. We have seen in Section 4.5 that every quadratic field Q(Va) is contained in some cyclotomic field. We now use the theory we have developed to show that no cubic, quartic, . . . field is contained in a cyclotomic field. Corollary 4.6.4. Let a e Q and letn>2 be an integer. Suppose that a is not a p^^ power for some odd prime p dividing n or that dia is not a square ifn is divisible by 4. Then Q(v^) is not contained in any cyclotomic field. Proof. Let k = p or k = A according as n is divisible by the odd prime p or by 4. Suppose that Q ( ^ ) c Q(f^). Since ^ = ( ^ ) ' ^ / ^ Q ( ^ ) c Q ( ^ ) . Then E = Q ( ^ , fy^) c Q(f^, f^) c Q{^rnk)- Now E is the splitting field of Z^ —fl,so is a Galois extension of Q. Then, by the Fundamental Theorem of Galois Theory, Gal(E/Q) = Gal(Q(f^,)/Q)/Gal(Q(?^,)/E). In particular, Gal(E/Q) is a quotient of Gal(Q(^^jt)/Q). But we have that Gal(Q(f^jt)/Q) is abelian, by Corollary 4.2.7, while under our hypothesis Gal(E/Q) is nonabelian, by Lemma 4.6.1 or Lenmia 4.6.3, which is impossible. D We now prove a result which allows us to weaken the hypothesis of Lenmia 3.7.7 in our discussion of Kunmier fields, as well as to further investigate the Galois groups of splitting fields of radical polynomials. Lemma 4.6.5. (1) Let n be an odd integer, and let a be n-powerless in Q. Then a is n-powerless in Q(^N)for every N. (2) Let n be an even integer, and let a be n-powerless in Q. If a is negative, assume also that —a is not a square in Q. Then one of the following two alternatives holds: (a) a is n-powerless in Q(^N)for every N. (b)a = b^for some b G Q(fA^), for some N, and b is n/2-powerless in Q(^N')for every multiple N^ ofN, Proof If a is w-powerless in Q(fA^), there is nothing to prove. Otherwise, a is an m^^ power in QC^A^) for some m dividing n,m > I. We shall show that the only possibility is m = 2, from which the lemma follows.

4.6 Radical Polynomials and Related Topics

111

Suppose m > 2. Then m is divisible by k, where k is either an odd prime or 4. Write a = b^ where b e Q(^N)' Then we see that Q ( ^ ) is contained in some cyclotomic field. But, under our hypothesis on a, this is impossible, by Corollary 4.6.4. n Theorem 4.6.6. Let n be an integer and let a be n-powerless in Q. Let E be the splitting field ofX^ —a over Q, and let G = Gal(E/Q). (1) If a is n-powerless in Q(fn)> then G = HN, the semidirect product of a normal subgroup N isomorphic toTj/nX and a subgroup H isomorphic to (Z/nZ)*. (2) If a = b^ with b n/2-powerless in Q(Cn), then G = HNy the semidirect product of a normal subgroup N isomorphic to Z/(n/2)Z and a subgroup H isomorphic to {Ij/nXy. Proof (1) We have that E = Q(f„, a), where a"" = a. Let B = Q(f„). Then Z'^ — a is irreducible over B, so by Proposition 3.7.6 the Galois group N = Gal(E/B) is isomorphic to Z/nZ. Now N is a subgroup of G and is normal as B is a Galois extension of Q. Also, H = Gal(E/Q(a)) is isomorphic to Gal(B/Q) which is isomorphic to (Z/AIZ)*. Also, (E/Q) = (E/B)(B/Q) = (Q(a)/Q)(B/Q) so Q(a) and B are disjoint extensions of Q by Corollary 3.4.5, so G = ^A^ by Theorem 3.4.10. (2) Let F = Q(b) c Q(^„). Then (F/Q) = 2, and so G' = Gal(E/F) is a subgroup of index 2 of G. The proof of part (1) goes through with Q replaced by F and n replaced by m to give that G' = H'N with H' = Gal(Q(f„)/Q(fo)) a subgroup ofH = Gal(Q(f^)/Q) of index 2. Now [G : G'] = [H : H'] = 2. To show G = HN it suffices to show that a ^ N where a is any element of H not in H\ But for any a e H, except a = id, a is non-trivial on B = Q(?n), while every element of N acts trivially onB SLS N = Gal(E/B), so a ^ N, SLS required. D Remark 4.6.7. As we have seen in our discussion of quadratic fields, both cases in Lemma 4.6.5 (2) may indeed arise. For example, 5 is 6-powerless in Q and VS ^ Q(f6)» so we are in case (a) for n = 6, while 5 is 10-powerless in Q but Vs G Q(fio), so we are in case (b) for n = 10. o We now wish to determine when the polynomial X'^ — a is irreducible. Clearly if a is not n-powerless it is not, so we need only consider the case when a is n-powerless. Recall that in Corollary 4.L10 we showed that X^ —a is irreducible if a is /^-powerless for p an odd prime, or if a is positive and a is 2-powerless. We shall generalize this result in several stages.

112

4 Extensions of the Field of Rational Numbers

Corollary 4.6.8. Let a be n-powerless in Q.Ifn X^ — a is irreducible in Q[X].

is odd, then the polynomial

Proof. Since a is n-powerless in Q, and n is odd, we have, by Lemma 4.6.5, that a is n-powerless in Q(^„). Then, by Proposition 3.7.6, X" —a is irreducible in Q(f„)[Z], and hence X^ — ais certainly irreducible in Q[X]. n Lemma 4.6.9. Let n = 2^ and assume that a is not a square in Q.Ift>2, assuw.e also that —4a is not a 4^^ power in Q. Then the polynomial X^ — a is irreducible in Q[X], Proof. Clearly the lemma holds when ^ = 1; so suppose t >2. Let Of be a root of X^' — a in some extension E of Q. Consider Q = Q(a) = Q(a'^) c Q(a2^-^) c . . . c Q(a). Claim: (Q(a^''')/Q(a^')) = 2 for it = 1 , . . . , ^ Assuming this claim, we then immediately have that (Q(a)/Q) = 2^ and hence that m^iX) is a polynomial of degree 2^ But a is a root of X^' — a, so ma(X) = X^' — a and hence X^' — a is irreducible. Proof of claim: We prove this by downward induction on k. \ik = t, then a^ is a root of X^—a, and a is not a square in Q = Q{ct^ ), Now suppose the claim is true for all integers between k and t and consider k—l. We wish to show (Q(Q^^^^)/Q(Qf^^^)) = 2. We prove this by contradiction. Assume it is not the case. Then Q(a^ ) = Q(a^ ). For simplicity, set ^ = oP'^~\ y = a^'~\ and 8 = a^\ Now ( Q ( K ) / Q ( 5 ) ) = 2 by the inductive hypothesis, so Q(y) has basis {1, y} as a Q(5)-vector space. Hence we may write P = c + dy uniquely with c,d e Q(5). Then y = ^^ z= (c + dy)^ = (c^ + d^S) + 2cdy, so (c^ + d^S) = 0 and 2cd = I. Solving for 8 yields 8 = —4c^ with c e Q(8). We claim this is impossible. If ^ - 1 = t -I, then k = t and Q(5) = Q. But then 8 = - 4 c ^ implies —45 = I6c^ = (2c)'^, contradicting the hypotheses of the lemma. If ^ - 1
4.6 Radical Polynomials and Related Topics

113

Proof, Let n = n\n2 where ni is odd and n2 is a power of 2. Let a be a root of X^ — a in some extension field E of Q. Then a"^ jg ^ root of X^^ — a, which is irreducible by Lemma 4.6.9, so (Q(Qf"^)/Q) = n\ divides (Q(Qf)/Q). Similarly, a"^ is a root of X"^ — a, which is irreducible by Lemma 4.6.10, so (Q(a'^0/Q) = ^2 divides (Q(Qf)/Q). But n\ and ^2 are relatively prime, so n = ^1^2 divides (Q(a)/Q). But certainly (Q(Qf)/Q) < n, so (Q(Qf)/Q) = n, and X^ — a = Moi {X) is irreducible. D .Remark4.6.11, Theorem 4.6.10 is sharp. If n = 4m and —4a = b^, then b = 2c and X'^ - a = (X^"^ + 20^^^ + 2c^)(X^'^ - IcX"^ + 2c^), so X'^ — a is not irreducible in this case.

o

Remark 4.6.12, (1) Theorem 4.6.10 and its proof show that the hypothesis in Lemma 4.6.5 (2) may be weakened to: If a is a negative integer, assume also that —4a is not a 4^^ power in Z. However, it may not be weakened further. For example, let n = 4 and a = —4. Then a = (2/)^ in Q(^4) and 2/ = (1 + /)^ in Qa4). (2) A similar observation applies to Theorem 4.6.6. If E is the splitting field of X"^ + 5, then | Gal(E/Q) | = 8 and we are in case (1) of that theorem. If E is the splitting field of X"^ + 9, then | Gal(E/Q)| = 4 and we are in case (2) of that theorem. On the other hand, the roots of X"^ + 4 are ±(1 ± /), so if E is the splitting field of X^ + 4, then | Gal(E/Q) \=2, o We conclude this section by considering multiple radical extensions. First we show they have the degrees we expect, and then we find primitive elements. Again we proceed in stages. Theorem 4.6.13. Let {n^}, / = 1, • •., t, be pairwise relatively prime integers and let {at}, i = I, ,., ,t be integers with at nt-powerless for each i. If some nt is divisible by 4, assume that —Aai is not a 4^^ power in Z. Then Q( V ^ ' • • •' y/^t) is an extension ofQ of degree ni - - - nt. Proof, Immediate from Theorem 4.6.12 and Lenmia 3.4.2.

n

Lemma 4.6.14. Let A = {a\,..., at) be a set of integers with the property that no product of the elements of any nonempty subset of A is a square, (In particular, this holds if a\,,,. ,at are pairwise relatively prime, none of which is a square, and at most one of which is the negative of a square.) Then G a l ( Q ( V ^ , . . . , V5;)/Q) = (Z/2Z)^ In particular, ( Q ( v ^ , . . . , V ^ ) / Q ) = 2^ First proof. This is a special case of Theorem 3.7.11.

114

4 Extensions of the Field of Rational Numbers

Second proof. We proceed by induction on t. For / = 0 there is nothing to prove, and for ^ = 1 the result is obvious. By induction, it suffices to prove that Q{^/a\,..., y/c^~\) and Qi^/ai) are disjoint extensions of Q. Since (Q(V^)/Q) = 2, to show this it suffices to show that Q ( v ^ ) ^ Q ( y ^ , . . . , v^o^IT). Suppose ^ ^ X e Q ( v ^ , • • •, V ^ ^ ) - Write x = y + z^y/a^Ii with y,z e Q(^,..., .Ja^Ii). Squaring, at = {y^ + z^fl^-i) + lyz^yja^. Since every term but the last is contained in QC^y/oT,..., ^at-i), by induction the last term must be zero, so either z = 0, in which case X G Q ( v ^ » • • •» ^cit-\). contradicting the inductive hypothesis, or y = 0, in which case x = z^at-\ and a^ = x^ = z^a^-i with z G Q, which is certainly impossible. D Theorem 4.6.15. Letn\,,.. ,nt be integers and letai,,,. ,at bepairwise relatively prime integers with at nt-powerless, i = 1 , . . . , f. If some nt is even, assume that no at is the negative of a square. Then (Q( " V ^ , . . . , V ^ ) / Q ) with basis {01=1 ^/

=ni'"nt

' \ 0 < kt < nt).

Proof Let n = lcm(ni, ... ,nt). By hypothesis, each at is nrpowerless. We will first prove the theorem under the stronger hypothesis that each at is npowerless, and then show how to remove this stronger hypothesis. Thus, assume that each ai is n-powerless. Observe that in this case it suffices to prove the theorem for ni = - --= nt = n SLS clearly (Q( "V^u . . . , ^ ) / Q )


and ( Q ( ^ , . . . , ^ ) / Q ( V ^ , . . . , V ^ ) ) < (n/m) •.. (n/nt), so ( Q ( ^ , . . . , ^ ) / Q ) = n' forces both inequalities to be equalities. First, suppose n is odd. We prove the theorem by induction on n. It is trivially true for n = 1, so assume it is true for n' < n and consider the case n' = n. Let F = Q(?iv) where A^ is any multiple of n and let E = ¥(.^/a[,..., .y^^). (We need to consider not only N = n but also N a multiple of n for the inductive step below.) Then, by Theorem 3.7.11, Gal(E/F) is isomorphic

4.6 Radical Polynomials and Related Topics

115

to ( a i , . . . , at) C F7(F*)". Suppose a = a^^ --- a'i e (F*)'^, i.e., a = a"" for some Of G F. In other words, a is not n-powerless in F, so, by the contrapositive of Lemma 4.6.5, a is not n-powerless in Q, and under our hypothesis on ai,.,. ,at this is true if and only if each of / i , . . . , it is a multiple of n^^ for some n^' dividing n, n'' > 1. We claim that each / i , . . . , /^ is a multiple of n. If n^^ = n we are done, so suppose not. Set n' = n/n'\ Let a' = a'/^"' • • • al'^''\ Then (aY' = a = a"" = (a'^/"'^", so a' = t;l„a^l''" for some it, i.e., a' = {C fot'"' for a^ = C > ,so ^/ ^ (F*)"'.Nown^ < n, so by induction each 4/n^^ is a multiple of n^ = nin", i.e., each /^ is a multiple of n, as claimed. Since, by Lemma 2.3.4, n' = I Gal(E/F)| = (E/F) < ( Q ^ , . . . , ^ ) / Q ) < n^ we must have equality. Next, suppose that n is even and let m = n/2. Choose A^ such that Q(fiv) 5 Q(^n, V ^ , ' " , \^), which is possible by Theorem 4.5.1, and set F = QCCAT), E = F ( ^ , . . . , ^ ) . Again, by Theorem 3.7.11, Gal(E/F) is isomorphic to ( a i , . . . , at) C F*/(F*)'^, and now by the same argument as in the case n odd,

m' = I Gal(E/F)| = (E/F) < ( Q ( ^ . . . , ^ ) / Q ( V ^ , . . . , V^)) < m^ and we have equality. But (Q(.y/a7,..., v ^ ) / Q ) = 2^ by Lemma 4.6.14, so ( Q ( v ^ . . . , ^ / ^ ) / Q ) = 2^^^ = n^ completing the proof in this case. Now let us see that we may in fact assume that the at are all n-powerless. To do so we must use an argument from elementary number theory. First let / = 1, and let ai have prime factorization ai = pf ' ' * K ^ ^^d let di = gcd(^i,,.. ,ek). Since a\ is ^i-powerless, gcd(Ji, ni) = 1. Let fii be the product of the primes p that divide n but that do not divide ni. By the Chinese Remainder Theorem, for each / there is an integer e^- satisfying the congruences e- = ei(modni),

e\ = l(modni).

Let d = gcd(^j,..., ^^). We claim that gcd{d, n) = 1. To see this, consider a prime p dividing n,lf p divides ni, then, since gcd(Ji, nO = 1, p does not divide ^/ for some /. But e[ = ei(modni) implies e^ = edmodp), so p does not divide e^-. If p does not divide ni, then p divides ni. But ^- = l(modni) for every i, and e^ = l(modni) implies e- = l(modp), so p does not divide e^.

116

4 Extensions of the Field of Rational Numbers

Now let a[ = p^* • • • p ^ \ Then a[ is n-powerless. Also, writing e^- = et + x/ni, note that -\-xini

et+Xtn\

' ' ' Pt

= (pT'"Pt')i(pT-"Pl')

= rrV^i

whereri = (pf . . . p f O e Q . Proceeding in exactly the same fashion, we may obtain a[,... ^a^ with each a- w-powerless, and with "l/al = rt ^oj" with n € Q for each /. But then Q( ^ o ^ , . . . , ^fa^t) = Q(\/^^ • • •» v ^ ) . so we may replace at by a- and argue as above. n Remark 4.6,16. (1) To illustrate a point in the proof of Theorem 4.6.15, let ai = 8 = 2^, wi = 5,a2 = 3,n2 = 3. Then n = 15 and ai is not 15-powerless, but we may rechoose ai = 256 = 2^ = 2^~^^ which is 15-powerless. (2) Clearly Theorem 4.6.15 is true somewhat more generally, but the best result is clumsy to state. Our hypothesis were chosen so that we could directly appeal to Lemma 4.6.5. o Corollary 4.6.17. Let n be an integer and let ai, ..., at bepairwise relatively prime n-powerless integers. Ifn is even, suppose that no at is the negative of a square and that n is relatively prime to each ai. Let E be the splitting field of f(X) = Y\i=i(^^ —cii) overQ. Then Gal(E/Q) is the semidirect product of a group isomorphic to {^jn'Ef and a normal subgroup isomorphic to {^TLjn'Ef. Proof Let B = Q(.
4.6 Radical Polynomials and Related Topics

117

First, suppose each n/ is odd. Then B/ and F are disjoint extensions of Q by Corollary 4.6.4, and F is a Galois extension of Q, so, by Corollary 3.4.4, (D//F) = Hi, Since the {ni} are pairwise relatively prime, we have, applying Lemma 3.4.2 successively, that D2 is disjoint from Di (as extensions of F), and then that D3 is disjoint from D1D2, Consequently, (D/F) = N. Clearly at = 'ii/ai is a primitive element of D/ as an extension of F. Then, applying Proposition 3.5.5 successively, we have that ofi + 0^2 is a primitive element of D1D2,..., and finally that of is a primitive element of D as an extension of F. It remains to show that or is a primitive element of B as an extension of Q. Let B' = Q(a). Then, by Lemma 2.3.4, A^ = (D/F) = (FB7F) < (B7Q) < (B/Q) = N, so we must have equality and B' = B. Now suppose some n/, say ni, is even. The same argument as above applies, except that (Di/F) = ni or ni/2, enabling us to conclude that (D/F) = N or N/2 and hence that B = B' or that (B/BO = 2. But in the latter case we must have B^ = Q( "1^(^07, '^5/02,..., V ^ ) , which is impossible as a is not an element of this field. n Lemma 4.6.19. Let ni,.,. ,nt be integers and let ai,.,. ,at be pairwise relatively prime integers with at ni-powerless, i = 1, . . . , f. If some nt is even, assume that no at is the negative of a square. Then a = "^/oi + • • • + "i^ is a primitive element ofQ( ^^j/aT,..., ^^o^). Proof The proof proceeds along the same lines as the proof of Lemma 4.6.18. We need to know, in the notation of that proof, that (D/F) = N,ln the easy case when ni = ••• = nt = 2 , F = Q and that follows immediately from Lemma 4.6.14. In the general case, we have shown that in the proof of Theorem 4.6.15. We then apply that fact to conclude that D2 is disjoint from Di (as extensions of F), and then that D3 is disjoint from D1D2 . . . , (as otherwise (D/F) < N), The remainder of the argument is the same. n We assemble these last two results into one theorem. Theorem 4.6.20. Let n\,.,, ,nt be integers and let ai,.,, ,at be positive integers with at nt-powerless, / = 1, . . . , f. Ifnt and nj are not relatively prime, assume that at and aj are relatively prime. Then a = ' y ^ + • • • + '^i^/a^ is a primitive elementofQ( ""^fox,..., ''4/a't)' Proof We proceed by induction on t. The / = 1 case is trivial. Now assume the theorem is true for all positive integers at most t. Consider the case ^ + 1. Let A = {n^ | / = 1 , . . . , r},let Ai = {n^ G A | gcd(n/, nt) — 1},

118

4 Extensions of the Field of Rational Numbers

and let A2 = {rit e A \ gcd(n/, rit) > 1}. If Ai = A, then this is just Lemma 4.6.18, and if A2 = A, this is just Lemma 4.6.19. Thus we must consider the "mixed" case. Reordering if necessary, we may assume Ai = { n i , . . . , n^} and A2 = {n^+i,..., n j for some s with I < s < t. Then, by Lemma 4.6.18, Qfi = n^i/oi + • • • + V ^ ^^ ^ primitive element of Bi = Q( ^^i/oT,..., ^5j/^), and, by Lemma 4.6.19, 0^2 = "^i^a^+i,..., "?iya^+i is a primitive element of B2 = Q ( "^i^as-^i,...,

"^i^a^+i).

Furthermore, if F = Q(fiv) for A/^ = ^i • • • n^+i, BiF and B2F are disjoint Galois extensions of F, so we may apply Proposition 3.5.5 as above to conclude that a = ofi + 0^2 is a primitive element of B1B2F; then we continue to argue as above to conclude that or is a primitive element of B1B2 = Q( ".^/oi,..., "^i^a^+i), completing the inductive step. n We conclude this section with an example that illustrates these last several results. Example 4.6.21. (1) Let E = Q(V2, v ^ , ^ , ^). Then E is an extension of Q of degree 210 with primitive element V2 + ^^2 + v ^ + ^2. (2) Let E = Q(V2, V3, V5, V?). Then E is an extension of Q of degree 16 with primitive element V2 + VS + VS + V?. Similarly, if E = Q(v^, \/3, y/5, v ^ ) , then E is an extension of Q of degree 81 with primitive element ^ + ^ + ^ + ^ , (3) Let E = Q(V2, v ^ , V3, ^). Then E is an extension of Q of degree 36 with primitive element V2 + v ^ + V3 + v ^ . o

4.7 Galois Groups of Extensions of Q We showed in Lemma 3.1.2 that every finite group is the Galois group of some Galois extension. It is natural to ask whether every finite group is the Galois group of some Galois extension of Q. The answer to this question is unknown. In this section we shall investigate two special cases: abelian extensions, i.e., extensions with abelian Galois group, and symmetric extensions, i.e., the splitting fields of polynomials of degree n with Galois group the symmetric group 5„. (Abelian extension is standard terminology but synmietric extension is not.) Theorem 4.7.1. Every finite abelian group is the Galois group of a Galois extension EofQ. Proof. Let H be an abelian group and write H as the direct sum of cyclic groups of prime power order, H = //i © • • • © //y. Let Hi have order m^

4.7 Galois Groups of Extensions of Q

119

Choose distinct primes {/?/} with pt = l(modm/), / = 1,... ,s. Such primes Pi exist by Lenmia C. 1.1. Let no = I and rit = pi-- - pi for / > 0. Then Q(^„._i) and Q(fp,) are disjoint extensions of Q for each i > 0, by Corollary 4.2.8, so by Theorem 3.4.7, applied inductively, we see that G = Gal(Q(^«J/Q) = Gi 0 • • • 0 G^ where G/ = '^/(pt — 1)Z. Since the order of Hi divides the order of G/ for each /, we see that / / is a quotient of G, so by the Fundamental Theorem of Galois Theory there is a field E intermediate between Q and Q(f«,) with Gal(E/Q) = H. D Remark 4.7.2. Instead of using Lenmia C.Ll, we may use a deep and famous theorem of Dirichlet: Every arithmetic progression ax + b with a and b relatively prime contains infinitely many primes. However, we have chosen to include Lemma C.Ll in order to give a self-contained and elementary proof of Theorem 4.7.1, rather than to simply quote this deep result. o Remark 4.7.3. We can ask: When can a Galois extension E of Q be a subfield of a cyclotomic field Q(^„) for some n? Since Gal(Q(f„)/Q) is abelian and E c Q(f„) implies Gal(E/Q) is a quotient of Gal(Q(f„)/Q), an obvious necessary condition is that E be an abelian extension of Q, i.e., that Gal(E/Q) be abelian. This necessary condition turns out to be sufficient. This is the famous Kronecker-Weber Theorem: Every abelian extension of Q is a subfield of a cyclotomic field. The proof of this deep theorem is far beyond the confines of this book. (Note that we proved a very special case by direct construction. We showed in Theorem 4.5.1 that every quadratic extension of Q is a subfield of a cyclotomic field. Of course, our construction in Theorem 4.7.1 produced a subfield of a cyclotomic field as well.) o Now we turn to the case of synmietric extensions. First, we treat the case of a polynomial of prime degree, which is relatively simple. Then we handle the case of a polynomial of arbitrary degree, using a construction of van der Waerden. Theorem 4.7.4. Let f(X) e Q[X] be an irreducible polynomial of degree p, p a prime, with exactly p — 2 real roots. Let E be the splitting field of f{X). Then Gal(E/Q) = Spy the symmetric group on p elements. Proof. Recall that since f{X) is irreducible and char(Q) = 0, f{X) must have p distinct roots in C. Since f{X) e R[X], the two nonreal roots of f{X) must be conjugates of each other. Let the roots of f{X) in C be a i , . . . , of^ with a\ and Qf2 nonreal. Let r be complex conjugation. Then x{a\) = a^, Tioti) = a\, ziott) = oct for / > 2.

120

4 Extensions of the Field of Rational Numbers

We identify G = Gal(E/Q) with a subgroup of Sp by its action permuting {ai,,., ,ap}. Under this identification, r is the transposition (12). Now, since f(X) is irreducible, G acts transitively on { a i , . . . , ap}. Thus we see that G is a transitive subgroup of Sp, p a. prime, that contains a transposition. Then, by Lenmia A.3.1, G = Sp. n Example 4.7.5. We now show that there exists a polynomial satisfying the hypotheses of Theorem 4.7.4, for each prime p. For /? = 2 the result is trivial. Suppose now that p is an odd prime. Fix p and let f(X) = X(X - 2)(X - 2a)(X - 4a) • • • (Z - 2(p - 2)a) - {b + 2). We will show that with the proper choice of a and b, f{X) is irreducible and has exactly p — 2 real roots. Let g{X) = X{X - 2){X - 2a){X - 4a) • • • (Z - 2{p - 2)a), so g{X) is a polynomial of degree p with roots 0, 2, 2a, 4a, . . . , 2{p — 2)a. Then, by elementary calculus, g\X) = f\X), a polynomial of degree /? — 1, has roots i*!,..., Sp-\ with 0 < ^i < 2 < ^2 < 2a < • • • < Sp-\ < 2(p — 2)a. Then we see that f{X) is increasing on the interval (—oo, ^i), decreasing on the interval (^i, 52), increasing on the interval {s2, ^3), . . . , and increasing on the interval (^^-i, oc). We also see that the maximum value of f{X) on the interval [0, 2] is at most (2)(2)(2a)(4a) • • • {2{p - 2)a) = 2PaP-^{p - 2)!, and that gOa) > a^, ^(7a) > a^, g ( l l a ) > a^, and so forth. Thus we choose a = 2P{p - 2)! and b = 2^a^-^(p - 2)!. First, with this choice of a and b, f(X) is irreducible by Eisenstein's Criterion (Proposition 4.1.7). Next, with this choice of a and &, f(X) < 0 for x < 2, and / ( 3 a ) > 0, / ( 7 a ) > 0, /(11a) > 0, and so forth. Now si < 2 < S2 < 2a < s^. Thus f(X) is decreasing on the interval [2, ^2], increasing on the interval [52, 2a], with / ( 2 ) < 0, / ( 2 a ) < 0, so f{X) has no roots on the interval [2, 2a]. Then 52 < 2a < ^3 < 4a < ^4. Thus f(X) is increasing on the interval [2a, 53], decreasing on the interval [^3, 4a], with / ( 2 a ) < 0, / ( 4 a ) < 0, and f(s3) > fOo) > 0, so f(X) has exactly two roots ri and r2 on the interval [2a, 4a]. By the same argument, f(X) has no roots on the interval [4a, 6a] and exactly two roots r^ and r4 on the interval [6a, 8a]. Proceeding in this fashion we find roots ri, r 2 , . . . , rp_3 of f(X) on the interval (—00, 2(p — 2)]. But Sp-i < 2(p — 2)a, so f(X) is increasing on the interval [2(p — 2)a, 00), and f(2{p — 2)a) < 0, so there is exactly one more root r^_2 of f{X) in that interval, and lastly f(X) has a total of p — 2 real roots, as claimed. o

4.7 Galois Groups of Extensions of Q

121

Now we turn to the case of arbitrary degree, which is considerably harder. Let f{X) e ¥[X] be a separable polynomial and let E be a splitting field of f(X), Let G = Gal(E/F). If the irreducible factors of f(X) are all distinct, then f(X) has distinct roots, and eliminating repeated irreducible factors will not change E, so we assume f(X) has distinct roots { a i , . . . , of^}. Then G permutes {«!,...,«„} (and this determines the action of G on E). Let { F i , . . . , y„} be a set of indeterminates and observe that the symmetric group Sn acts in a natural way as a group of permutations of this set. Then there is an isomorphism G ^^ G^ c. Sn given by a i-> cr' where a^(Yi) = Yj if

a(ai)=aj. For brevity, let F denote the field of rational functions F = F ( F i , . . . , ? « ) and let E denote the field of rational functions E = E ( F i , . . . , F„). Definition 4.7.6. In the above situation, let e = aiYi + ' " + anYn G E and let

F ( z ) = n (^ - ^'(^>) ^ ^f^]-

^

The polynomial F(Z) is a symmetric function of its roots, so by Lemma 3.L12 its coefficients are functions of the elementary symmetric functions of its roots and hence of F i , . . . , Y„ and the coefficients of f{X), Thus, F(Z) e F[Z]. Now we may factor F(Z) into a product of irreducibles in F[Z], F(Z) = F i ( Z ) . . . F , ( Z ) . One of these is divisible by Z — ^ in E(Z). Renumbering if necessary we may assume it is Fi (Z). Since F{Z) is invariant under 5"^, the action of Sn permutes these factors. Lemma 4.7.7. In the above situatioriy let H' C Sn be the subgroup leaving F\{Z) invarianty i.e., H' = W eSn\T\Fi(Z))

= Fi(Z)}.

Then G^ = H\ and hence G is isomorphic to H\ Proof. We begin by making several observations. Note that, for any r' e Sn, x'{Z - 6) divides r'(Fi(Z)), so H' = {T' e Sn I T\Z - 0) divides Fi(Z)}.

122

4 Extensions of the Field of Rational Numbers

Note also that GxG^ ^ GxSn acts on E, and for any a e G, aa'{0) = 0 (as aa^ simply permutes the terms of the sum aiYi + • • • + anYn), and so

cr(e) = {aTHe), Consider

Gi(Z) =

Y\(Z-a(e)). creG

This is obviously invariant under G. Since the fix set of G operating on E is F, the fix set of G operating on E is F, by Corollary 3.4.9, so Gi(Z) e F[Z]. Since { a i , . . . , a„} is distinct, {cr(0) | a G G} is distinct. Hence, by Lemma 2.7.12, Gi(Z) is irreducible in F[Z]. Since G\{Z) has Z — ^ as a factor in F[Z], we must have that Gi(Z) = Fi(Z). Also, by Lemma 3.5.4, E = F((9). Hence the action of Sn on E is determined by its action on 0, Suppose r' e H\ Then T\Z - 6) is also a factor of Fi[Z] = Gi[Z], so it mustbeZ-or(6>)forsomea G G . But r'(Z-6>) = Z-x\e) = Z-a(e) = Z - (aTHO), so r' = (aT^. But a' G G' so r' G G\ On the other hand, if r' ^ H\ then r'(Z - 6) = Z - x\e) is not a factor of Fi(Z) = Gi(Z), so it is not equal to Z - o-((9) = Z - {aY^{e) for any a' G G'. Thus T^ ^ (aO"^ for any a^ G G' and r^ ^ G', u Remark 4,7.8. From the proof of Lemma 4.7.7, we see that

Fi(Z)=

W{Z-a\e)). cr'eG'

and in fact

F,(z)= Yl (^--^'c^)) T'GT/G'

where the product is taken over a left coset of G\ In particular, since {ofi,..., a„} are distinct, {r'(^) | r' G 5„} are also all distinct. We see from this that each Fi{Z) has distinct roots and no two Fi{Z) and Fj(Z) have a root in conmion, for i ^ j . Furthermore, we also see from this that the action of Sn permutes {Ft (Z)} transitively. o Lemma 4.7.9. Let p be a prime. Let f(X) G Z [ X ] be a monic polynomial and let f(X) be its image in ¥p[X] (identifying Fp with Z/pZ). Let E be a splitting field of f(X) over Q and let F be a splitting field of f{X) over ¥p. Suppose that all roots of f{X) in E are simple. Then K = Gal(E/Fp) is isomorphic to a subgroup ofG = Gal(E/Q).

4.7 Galois Groups of Extensions of Q

123

Proof. Begin with the polynomial f{X) and form F{Z) as in Definition 4.7.6. Since F(Z) is a polynomial in F i , . . . , F^, Z, its factorization in the rational function field Q ( F i , . . . , F„) is actually a factorization in the polynomial ring Z [ F i , . . . , F„]. This follows as the coefficients of F(Z) are in fact polynomials in the coefficients of f{X) by Lemma 3.1.14, and by the fact that Corollary 4.3.8 generalizes from Z to Z [ F i , . . . , F„] as Z [ F i , . . . , F„] is a unique factorization domain. Thus we have F(Z) = F i ( Z ) . . . F , ( Z ) with each F/ (Z) irreducible. Then reducing mod p gives a factorization (where the factors may not be irreducible) F(Z) = F i ( Z ) . . . F , ( Z ) . Let {Fij{X)] be the irreducible factors of Fi(X), fori = I,... ,t. Let K^ be the subgroup of 5„ consisting of those permutations in 5„ that preserve the irreducible factor Fii(X) of Fi(X). Now the action of 5„ permutes {Fi(X)} and hence {Fij(X)}. By Remark 4.7.8, applied to E, the Fi(X) have no common root and hence no common irreducible factor, so if Fn(X) is preserved, Fi(X) must be preserved. In other words, K^ = {r^ e S„ | T\Fn{X)) = Fn(X)} c {T^ e Sn I r(Fi(X)) = Fi(X)} = H\ But by the same logic as before, K' is isomorphic io K = Gal(E/Fp). n Proposition 4.7.10. For every positive integer n there exists a polynomial fniX) e Z[X] of degree n whose splitting field E has Gal(E/Q) isomorphic toS„ •'nProof Choose a monic polynomial fiiX) e Z[X] of degree n whose mod2 reduction f2(X) e F2[X] is irreducible, a monic polynomial fsiX) G Z[X] of degree n whose mod 3 reduction fsiX) e F3[X] is the product of an irreducible polynomial and a linear factor, and a monic polynomial fs (X) e Z[X] whose mods reduction fsiX) e Fs[X] is the product of an irreducible quadratic and other factors that are distinct irreducible polynomials of odd degree. Let f(X) = -l5f2(X)

+ 10/3(X) + 6/5(X).

Then f(X) is a monic polynomial that reduces modp to fp(X) for p = 2, 3, 5. Since f2(X) is irreducible, f{X) is irreducible, and hence its roots are distinct. Let Gp be the Galois group of fp(X) over F^, p = 2,3, 5. By Lemma 4.7.9, each Gp is isomorphic to a subgroup of G = Gal(E/Q).

124

4 Extensions of the Field of Rational Numbers

Since /2(X) is irreducible, G2 operates transitively on its roots, so G is a transitive subgroup of 5^. Since /s (X) has an irreducible factor of degree n — l, and the Galois group of a finite extension of a finite field is cyclic, G3 and hence G contains an (n — l)-cycle. By the same logic, G5 and hence G contains an element that is the product of a transposition and one or two cycles of odd order, so some (odd) power of it is a transposition in G. Hence, by Lemma A32, G = Sn• Remark4.7.11. Our work actually gives an algorithm for computing Galois groups of polynomials f(X) e Q[X]: Form the polynomial F(Z) as in Definition 4.7.6. Note that we do not need to know the roots of f(X) to do so, since, as we have observed, its coefficients are functions of F i , . . . , F„ and the coefficients of f(X). Then factor F(Z) into a product of irreducibles F(Z) = Fi(Z)'' • Ft(Z) inF[Z]. Remark 4.1.13 gives an algorithm for doing so. As in Lemma 4.7.7, let Hj be the stabilizer of Fi(Z) in 5„. Note that, by Remark 4.7.8, the {///} are mutually conjugate subgroups of Sn so they are all isomorphic as abstract groups. (This is a general fact. If a group H acts transitively on a set 5 = {st} with Hi the stabilizer of st, then, if h e H is any element with h(si) = Sj, we have that Hj = hHih~^.) Thus we see from Lemma 4.7.7 that, for any i, the Galois group of f{X) (i.e., the Galois group Gal(E/Q), where E is a splitting field of f{X)) is isomorphic to H^. Thus we may simply pick any factor Fi{X), and go through all of the (finitely many) elements of Sn, seeing which of them leave Fi(X) invariant, to obtain Hj and hence the Galois group. It is evident that this algorithm is wholly impractical. We include it to show that the problem of computing Galois groups is algorithmically solvable (not to give a practical method for their computation). o

4.8 The Discrimmant Let f{X) e F[X] be a polynomial of degree n, with splitting field E. As we have seen, the Galois group Gal(E/F) is a subgroup of the symmetric group Sn. In this section we develop a criterion for deciding whether G is a subgroup of An, the alternating group. Definition 4.8.1. Let f(X) e F[X] be a polynomial of degree n with roots Qfi,..., a„ in some splitting field E. Let

4.8 The Discriminant

125

5 = f](af,- - aj) and let A = A(/(X)) = S\ A is the discriminant of f(X).

o

Remark 4.8.2. (1) Note that 8 depends on the order of the roots ofi,..., «„ but A(/(X)) does not, so A(/(X)) is indeed an invariant of f(X). (2) If f(X) is irreducible, then it has distinct roots, so in this case A ( / ( Z ) ) # 0. (3) A(/(X)) is a symmetric polynomial in the roots of f(X), so, by Remark 3.1.5 and Lemma 3.1.14, it is a polynomial in the coefficients of f(X). Thus A(/(X)) G F (and it is independent of the choice of splitting field E). (4) A(f(X)) is visibly a square in E, but may or may not be a square inF. o Lemma 4.8.3. Let f(X) e F[Z] be an irreducible polynomial with splitting field E and with Galois group G = Gal(E/F). Let A = A(f(X)) be the discriminant of f(X). (1) If A is a square in F, i.e., if 8 € F, then G C A„. (2) If A is not a square in F, i.e., if 8 ^ F, then G ^ An^ Proof. We observe that E is a Galois extension of F so Fix(G) = F. Let a e Sn. Recall that a G A„ if and only if sign(a) = 1. Also, sign(a) = (—1)^ where / is the number of inversions in a, i.e., / is the number of elements in the set X = {(/, j) \ i < j but ^(7) < a(i)}. From this it follows immediately that a(8) = sign(a)5. Thus, if 8 G F, then cr(8) = 8 for every a G G, so G c A„. On the other hand, if 5 ^ F, then ao(8) 7^ 8 for some ao G G, and ao ^ A„, so G ^ A„. n Corollary 4.8.4. In the situation of Lemma 4.8.3, suppose 8 ^ F. Let B = F(5). Also, let H = G n An. Then E is a Galois extension ofB with Galois group Gal(E/B) = H, and B is a Galois extension ofF with Galois group Gal(B/F) = G/H = Z/2Z. n Corollary 4.8.5. Let f(X) G F[X] be an irreducible cubic with splitting field E, and let G = Gal(E/F). IfA(f(X)) ^ F, then G = S3, while ifA(f(X)) e F, then G = A3 = Z/3Z. Proof. By Theorem 2.8.19, G must be either 53 or A3, and then Lenmia 4.8.3 decides between the two. n

126

4 Extensions of the Field of Rational Numbers

In order to effectively use Lemma 4.8.3 we must be able to compute the discriminant. We do that now, in low degrees. Proposition 4.8.6. Let f(X) e F[X], and let A = A ( / ( Z ) ) . (1) Iff(X) = X^ + aX + b, then A = a^ - Ab. (2a) IffiX) = X^+bX + c, then A = -4b^ - 27cl (2b) IffiX) = X^+ aX^ + bX + c, then A = -4a^c + a^b^ + ISabc 4b^ - 27c^. Proof By Remark 4.8.2 (3), A is a polynomial in the elementary symmetric functions si of the roots of f(X), or, equivalently, in the coefficients of f{X), (1) From Remark 3.1.5 we see that a = —si = —(ai + 0C2) and that b = S2 = ot\oi2^ Directly from Definition 4.8.1 we see that A = (ai — 0^2)^ = Qfj — 2Qfia2 + a\ = al + 2aia2 + a | — 4aia2 = a^ — 4b. (2) From Remark 3.1.5 we see that a = —si = —(ai + 0^2 + ^3), that b = S2 = oi\a2 + 0i2Oiz + oi\a^, and that c = —S3 = —(Qfia2Qf3). In principle, we can then simply find A by direct computation, but the computations get very messy, so we look for a better way. (2a) We observe that A is a homogenous polynomial of degree 6 in ^-i, 52, and 53, where Si has degree /, or, equivalently, that A is a homogenous polynomial of degree 6 in a, &, and c, where a has degree 1, b has degree 2, and a has degree 3. Since a = 0 in this case, A must be of the form A = sb^ + tc^ for some s and t. First, consider the polynomial f(X) = X^ — X with roots 1, 0, and —1. Direct computation shows 4 = A = s(—l)^ + t(0)^, so 5 = —4. Next consider / ( Z ) = X^ — I with roots 1, co, and co^. Direct computation shows - 2 7 = A = s(0)^ + r ( - l ) ^ , so t = - 2 7 . (2b) Note that adding the same value to each root leaves A unchanged, as it only depends on the difference between roots. If f(X) = X^ +aX^+ bX + c, let X = Y-a/3, Then f(X) = giY) = Y^+b'Y + c' and so A = -4(&0^ 27(cO^. But, by elementary algebra, b^ and c' can be expressed in terms of a, b and c, and doing so and substituting yields the result. D Example 4.8.7. (1) Let / ( X ) = X^ + X + le Q[X] and let f(X) have splitting field E. Then f(X) is irreducible and A(/(X)) = - 3 1 so Gal(E/Q) = 53. On the other hand, if B = Q[\/^3T], then Gal(E/B) = A3. (2) Let f(X) = X^-3X + 1 e Q[X] and let f(X) have splitting field E. Then f(X) is irreducible and A(/(X)) = 81, so Gal(E/Q) = A3 = Z/3Z. Note we have already encountered this polynomial in Example 2.9.8. o

4.9 Practical Computation of Galois Groups

127

4.9 Practical Computation of Galois Groups Our focus in this book has been the theoretical development of Galois theory, rather than its computational aspects. However, in the course of this development we have proved a number of results that are useful in (hand) computation of Galois groups of polynomials f(X) e F[X], By this (common) language we mean the Galois groups Gal(E/F), where E is a splitting field of f(X). For the convenience of the reader, we shall collect most of these results in this section. We shall keep the original numbering to make it easy for the reader to refer back to the place they originally appeared. It is not only the individual results we have derived that are useful, but also their interplay, and this better illustrated than summarized. Thus we conclude this section by working out a specific example that illustrates the application of many of these results together. We begin by considering Galois groups of specific polynomials. Since we are in characteristic 0, the prime field FQ = Q and we may choose as primitive n^^ root of unity f„ = txp(2ni/n). Recall we have the following definition: Definition 3.7.5. Let a € F. Then a is w-powerless in F if a is not an m^^ power in F for any m dividing n,m > 1. In determining whether an element of F is n-powerless, the following results are useful: Lemma 4.6.5. (1) Let n be an odd integer^ and let a be n-powerless in Q. Then a is a n-powerless in Q(^N)for every N, (2) Let n be an even integer, and let a be n-powerless in Q. If a is negative, assume also that —a is not a square in Q. Then one of the following two alternatives holds: (a) a is n-powerless in Q(^N)for every N. (b) a = b^ for some b e Q(^N)* for some N, and b is n/2-powerless in Q(^N')for every multiple N^ ofN. Corollary 4.5.5. Let a be a square-free integer If a is positive and oddy set a' = a. Otherwise set a' — 4|a|. Then a = b^ for some b e Q(fn) if and only if a' divides n. In terms of this definition, we have the following result: Proposition 3.7.7. Let F 2 ^oi^n)- Let E be an extension ofF. The following are equivalent: (1) E is the splitting field ofX^—a G F[X], for some a that is npowerless in F.

128

4 Extensions of the Field of Rational Numbers

(2) E is a Galois extension ofF with Gal(E/F) = Z/nZ. This result has a generalization: Theorem 3.7.11. Let F 2 Fo(fn) and let E be the splitting field of /(X) = (Z'^-ai)...(X'^-a,) with each at ^ 0. Then Gal(E/F) is isomorphic to {a\,.,. ,at) c F*/(F*)'' (i.e., to the subgroup o/F*/(F*)" generated byai,..., at). Now we turn to Galois groups of polynomials over Q itself. We observe that the n^^ cyclotomic polynomial 0„ e Q[X], and the polynomial X"" -I e Q[Z], both have splitting field Q(f„). Recall that (Z/nZ)* denotes the multiplicative group of units in (Z/nZ), of order (p(n). Then we have: Corollary 4.2.7. Gal(Q(f„)/Q) = (Z/nZ)*. Now we have some results on extensions whose Galois groups are nonabelian. Lemma 4.6.1. Let p be an odd prime. Let a G Q with a not a p^^ power in Q, and let E be the splitting field of X^ — a. Let G = Gal(E/Q). Then G=

(a, r I or^ = 1, xP-^ = 1, rcrr-^ = a')

for some primitive root r mod p. Lemma 4.6.3. Let a e Q with a ^ ±b^ for any b e Q, and let E be the splitting field ofX"^ - a. Let G =^ Gal(E/Q). Then G=

(a, r I cT^ = 1, T^ = 1, rar-^

= a^).

These two results have a common generalization: Corollary 4.6.17. Let n be an integer and let ai,... ,at be pairwise relatively prime n-powerless integers, Ifn is even, suppose that no at is the negative of a square and that n is relatively prime to each at. Let E be the splitting field off(X) = UUii^"" " ^^) over Q. Then Gal(E/Q) is the semidirect product of a group isomorphic to (Z/nZY and a normal subgroup isomorphic to (Z//iZ)*. Now we have two separate results that are useful in showing that Galois groups are large:

4.9 Practical Computation of Galois Groups

129

Theorem 4.7.4. Let f(X) e Q[X] be an irreducible polynomial of degree py p a prime, with exactly p — 2 real roots. Let E be the splitting field of f(X). Then Gal(E/Q) = Sp, the symmetric group on p elements. Lemma 4.7.9. Let p be a prime. Let f(X) e Z[X] be a monic polynomial and let f(X) be its image in Fp[X] (identifying Fp with Z/pZ). Suppose that all roots of both f(X) and f(X) are simple. Let E be the splitting field of f(X) over Q and let E be the splitting field of f(X) over Fp. Then K = Gal(E/F^) is isomorphic to a subgroup of G = Gal(E/Q). We now return to general fields F. For a polynomial f(X) e F[X] we defined the discriminant A ( / ( Z ) ) in Definition 4.8.1. We then have: Lemma 4.8.3. Let f(X) e F[X] be an irreducible polynomial with splitting field E and with Galois group G = Gal(E/F). Let A = A(f(X)) be the discriminant of f (X). (1) If A is a square in F, i.e., if 5 e F, then G c A„. (2) If A is not a square in F, i.e., ifS ^ F, then G ^ A„. In low degrees we have the following computation: Proposition 4.8.6. Let f(X) e F[X], and let A = A(/(X)). (I) Iff(X) = X^ + aX + b, then A = a^ - 4b. (2a) Iff(X) = X^ + bX + c, then A = -4P - 21c^. (2b) If f(X) = X^ + aX^ + bX + c, then A = -4a^c + a^b^ + lSabc-4b^ -27c^. For irreducible cubics (and for quadratics, too, but there the result is trivial), the discriminant gives us the complete answer. Corollary 4.8.5. Let f(X) e F[X] be an irreducible cubic with splitting field E, and let G = Gal(E/F). / / A ( / ( X ) ) ^ F then G = S3 while ifA(f(X)) e F then G = A3 = Z/3Z. Now we turn to more general results which are useful in considering polynomials that are not irreducible. Let f(X) G F[X] with f(X) = fi(X)f2(X). Let B, be a splitting field of fi{X), i = 1,2, with Bi and B2 both subfields of some field A. Let G/ = Gal(B,/F), / = 1,2, and let G = Gal(E/F), where E = B1B2 is a splitting field of f(X). If Bi and B2 are disjoint extensions of F, we can apply the following result.

130

4 Extensions of the Field of Rational Numbers

Theorem 3.4.7. Let Bi and B2 be disjoint Galois extensions of F. Then E = B1B2 /^ a Galois extension of F with Gal(E/F) = Gal(Bi/F) X Gal(B2/F). In Section 3.4 we gave some criteria for determining when two extensions are disjoint. But, in any case, we have the following generalization of Theorem 3.4.7. Corollary 3.4.8. Let Bi and B2 be finite Galois extensions of F and let E = B1B2, Bo = Bi n B2. Then E is a Galois extension of F ant/Gal(E/F) = {(o'i,a2) G Gal(Bi/F) x Gal(B2/F) | ai | Bo = ^2 I Bo}. Finally, we have a result that enables us to compare Galois groups over different fields. Corollary 3.4.6. (Theorem on Natural Irrationalities) Let f(X) e F[X] be a separable polynomial and let B ^ F. Let ^ be a splitting field for f(X) over F. Then EB is a splittingfieldfor f{X) over B and Gal(BE/B) is isomorphic to Gal(E/E fl B), a subgroup o/Gal(E/F), with the isomorphism given by a \-^ a \^, We now present a single, rather elaborate, example, which shows how, with cleverness, hard work, and the use of a number of the results we have proved so far, we can determine the structure of a reasonably large Galois group. (We advise the reader to write out the diagram of intermediate fields and their inclusions in order to follow the argument here.) Example 4.9.1. Let a = VlHh\/3 and let E be the splitting field of m,, (X) e Q[X]. We determine G = Gal(E/Q). We observe that a^ = 1 + \ / 3 , so a^ — 1 = ^/3 and (a^ — 1)^ = 3 and hence a^ - 2a^ - 2 = 0. Thus a is a root of the polynomial X^ - 2X^ - 2 = 0. This polynomial is irreducible by Eisenstein's criterion, so we see m^iX) = X^ - 2X^ - 2. We can readily find all the roots of this polynomial. Certainly (coa)^ = 1 + A/3 and (co^a)^ = 1 + \ / 3 , so coa and o/a are also roots. But we can also see that if ^^ — 1 = —\/3 then ()8^ — 1)^ = 3 and hence )8^ — 2)8^ — 2 = 0, so ^ is also a root of m^iX) then co^ and oP'P are roots as well. Thus E = Q(a, coot, o?a, ^, a>^, 0?^) — Q(ct), a, )S). Clearly E contains the fields Q(V3), Q(a>) = Q ( V ^ ) , and Q(/) as well, as / = ( \ / 3 ) ( \ / ^ ) / 3 . These are all extensions of Q of degree 2, and so we see that G contains at least 3 subgroups of index 2. Since ma(X) is an irreducible

4.9 Practical Computation of Galois Groups

131

polynomial of degree 6, (Q(a)/Q) = 6, so (Q(Qf)/Q(\/3)) = 3, and similarly (Q{P)/QiV^)) = 3. Since Q(ft>) and Q(V3) are disjoint Galois extensions of Q, both of degree 2, Q(co, \/3) is a Galois extension of Q of degree 4, and Q(a), \/3) is a Galois extension of Q(V3) of degree 2. Since 2 and 3 are relatively prime, Q(co, V3) and Q(Qf) are disjoint extensions of Q(V3), and so Q(ct>, a) = Q(co, a, V3) is an extension of Q(V^) of degree 6. In fact, it is a Galois extension as it is the splitting field of the polynomial Z^ — (1 + V3). Similarly Q(co, fi) is a Galois extension of Q(\/3) of degree 6. We now make an important observation. We have just noted that Q((o, \/3) and Q(a) are disjoint extensions of Q(V3), so in particular CD ^ Q(a). Thus Q(Qf) contains the root a of rria (X) but does not contain the root coa, so Q(af) is not a normal extension of Q ( A / 3 ) . Hence we see that Gal(Q(w, Q ' ) / ( Q ( V 3 ) ) is a nonabelian group of order 6, as is Gal(Q(6f;, )S)/(Q(V3)). Furthermore, since these two Galois groups are quotients of subgroups of G, we conclude that G is not abelian. We claim ^ ^ Q((o, a). For if )S 6 Q(a>, a), we would have E = Q(a>, a) and then (E/Q) = (Q(^, a)/Q(V3))(Q(V3)/Q) = (6)(2) = 12, so G would be a nonabelian group of order 12 with at least 3 subgroups of order 6, and no such group exists. Similarly, a ^ Q(a>, ^). Now6 = (Q(o;, a)/Q(V3)) = (Q(a;, a)/Q(a;, V3))(Q(c^, V3)/Q(V3)), so we see that (Q(CL>, a)/Q{co, A/3)) = (Qico, a)/Q(a))) = 3, and similarly we see that (Q(a), P)/Qio))) = 3. We claim that Qico, a) and Q(co, P) are disjoint extensions of Q((o, V3). For we have the inclusions Q(co, A/3) C Q((O, a) D Qico, y6) c Q(co, a), so (Qico, a) n Qico, P)/Qico, A/3)) divides (Q((W, a)/Qico, V3)) = 3, so it is either 1 or 3, i.e., Qico, a) n Qico, j6) = Qico, a) or Qico, A/3). But Qico, a) D Qico, fi) ^ Qico, a) as a ^ Qico, jS). Since Qico, a) and Qico, P) are disjoint Galois extensions of Qico, A/3), we have that iQico, a, fi)/Qico, V3)) = 9 and hence (E/Q) = 36. We also know that 9=

iQico,a,P)/Qico,^))

< iQia, i6)/Q(V3)) < ( Q ( C ^ ) / Q ( V 3 ) ) ( Q ( ) 8 ) / Q ( A / 3 ) ) = (3)(3) = 9, so we see that iQia, fi)/QiV3)) (E/Q(a, ^)) = 2. Since

= 9. Hence (Q(a, iS)/Q) =

18 and

132

4 Extensions of the Field of Rational Numbers

36 = (E/Q) =

{Qico,a,P)/Q)

= ( Q M / Q ) ( Q ( a , )6)/Q) = (2)(18), we see that Q(co) and Q(Qf, ^) are disjoint extensions of Q. We now assemble all of this information to find the structure of G. First, since (E/Q) = 36, G is a group of order 36. As we have observed, Gal(Q(a;, a)/Q(V^)) is a nonabelian group of order 6. By the Theorem on Natural Irrationalities, Gal(E/Q(y6)) is isomorphic to a subgroup of this group. But (E/Q(yS)) = 6 so these two groups are isomorphic. In other words, every automorphism of Q(CJL>, a) fixing Q ( A / 3 ) extends to an automorphism of E fixing Q(jS), and similarly every automorphism of Q(CJL>, P) fixing Q(V3) extends to an automorphism of E fixing Q(a). Similarly, since (E/Q(a, yS)) = (Q(CL>)/Q) = 2, we may again apply the Theorem on Natural Irrationalities to extend the nontrivial element of Gal(Q((w)/Q) to an automorphism of E fixing Q(a, ^S). The elements we have found so far generate a subgroup H of G of order 18. Finally, although Q(a, P) is not a Galois extension of Q, E is a splitting field of Ma (X), an irreducible polynomial two of whose roots are a and fi, so there is an automorphism of E taking a to ^6. This automorphism is not unique, but if we choose one such, we may multiply it by any element of H, (Note every element of H leaves each of the sets {a, coa, co^a} and {^, co^, ^^P) invariant while this new automorphism interchanges these two sets.) Thus, assembling all of this information, we find that G is a group of order 36, generated by an element p of order 2, two elements ai and a2 of order 3, and an element r of order 2, whose actions on E are given by: p(co) = (jo^, cri((o) = 0),

p(a)=a,

p(p) = fi

ori(a) = coa, cfiiP) = ^

02(0)) = CO, or2(a) = a, r(6t>) = CO, r(a) = )6,

cTii^) = o)^ x{^) = a.

(Actually, H is generated by p, ai, and (J2, and G is generated by p, ai, and r, as (J2 = yOCTi yo" \ but the more synmietric set of generators better reveals the structure of G.) Finally, if we number the roots a,o)a, O?OL, ^,CO^, O?^ of m^ (X) as 1 , . . . , 6, the generators of G are given by the permutations p = (23)(56), cTi = (123), 02 = (456), and r = (14)(25)(36).

o

4.10 Exercises

133

4.10 Exercises Exercise 4.10.1. Show, without using Dirichlet's theorem about the existence of infinitely many primes in arithmetic progressions, and without using Lemma C.l.l,that: (a) For any integer n, there exists a Galois extension E of Q of degree n, (b) For any integer n, and any algebraic number field F, there exists a Galois extension E of F of degree n. Exercise 4.10.2. For a rational number x = r/s in lowest terms, say that p divides x if p divides r. Show that Eisenstein's Criterion (Proposition 4.1.7) remains valid for polynomials in Q[X] with this definition of divisibility. Exercise 4.10.3. Let f(X) = UnX"" + • • • + ^o G Z[X] and let r e Q with / ( r ) = 0. Show that each of the n rational numbers a„r,

anV^ + an-\r,

anP + a„-ir^ + a„_2r, ...,

anr"" + an-ir""'^ H

\-air

is in fact an integer. (This exercise is taken from the 2004 Putnam competition.) Exercise 4.10.4. Use Exercise 2.10.10 to show that the polynomial X^ — 72 is irreducible in Q[X]. Exercise 4.10.5. We might call an ordinary compass a 2-compass since, along with a straightedge, it allows us to geometrically find square roots of complex numbers. Suppose we also had a 3-compass, an instrument that allows us to geometrically find cube roots of complex numbers. For what values of n < 1000 could we construct a regular n-gon with these instruments? Exercise 4.10.6. (a) Let n be an integer and let k be the product of the distinct prime factors of n. Show that 0„(X) = ^kiX""^^), (b) Let n be an integer not divisible by p. Show that ^np(X)

=

^n(Xn/<^niX).

Exercise 4.10.7. (a) Find the n-the cyclotomic polynomial ^n(X) for n = p^ a prime power. (b) Find *«(X) for 2 < n < 16. (c) Examining n (X) for small values of n might lead one to conjecture that the coefficients of ^n(X) are always 0, 1, or —1. Show, by hand computation, that n = 105 is a counterexample to this conjecture. (This is known to be the

134

4 Extensions of the Field of Rational Numbers

smallest counterexample.) Note that 4>io5(X) is a polynomial of degree 48. With enough patience, this entire polynomial can be computed by hand. But only a partial, and much more manageable, computation is necessary to find a coefficient that is not 0, 1, or —1. Exercise 4.10.8. Fix n > 2 and let E = Q(^n) be the n-th cyclotomic field. Let B = E n R. (a) Show that (E/B) = 2 and that B = Q(f^ + ^~^), (b) Show that B is the unique subfield of E with (E/B) = 2 if and only if n = 4, p^, or 2p^ for p an odd prime. (c) Find 3 subfields B of E with (E/B) = 2 in case n = 8. (d) Show that there are exactly 3 fields B with (E/B) = 2 if n = 2^ for any k>3. Exercise 4.10.9. Let E = Q(fp) be the p-ih cyclotomic field. Let a, Z? G Q be arbitrary. Find NE/Q(a + b^p) and TrE/Q(a + bl;p). Exercise 4.10.10. (a) Of course, ^i = 1, ^2 = - 1 , fs = ^ = ( - 1 + /V3)/2, and ^4 = «. Find ^5, ^5, ?8, ?io, ?i2, ?i5, and ^i6. (b) Find ^17. (This is an intricate computation. See Example 4.4.6 (3).) (c) Find ^7, ^9, ^13, and ^14. (This uses the solution of the cubic in Exercise 4.10.25.) Exercise 4.10.11. Throughout this exercise, we assume n > 2. (a) Show that 0„(x) > 0 for all real numbers x. (b) \fn = p^ for some k, show that „(1) = p, (c) If n is not a prime power, show that <E>„(1) = 1. (d) If n = 2^ for some it, show that <&„(-l) = 2. (e) If n = 2p^ for some odd prime p, show that
^n{-l) = 1. (g) Let E = Q(?«). Show that the computations in (b) and (c) give the value of NE/Q(1 — ^n) and that the computations in (d), (e), and (f) give the value of NE/Q(-1-Cn). Exercise 4.10.12. (a) Let F = F^ be the field with p elements and let let m be an integer relatively prime to p. Let 0^(X) e ¥[X] be the mod p reduction of the cyclotomic polynomial ^rn(X) and let g(X) = X"^ - I e F[X]. Show that ^m(X) and g(X) have the same splitting field E, and furthermore that E = Fpr where r is the smallest positive integer such that p"^ = l(modm). (b) Let ^m(X) = fi(X)"' fk(X) e F[X] be a factorization of 0^(X) into a product of irreducible polynomials. Show that each ft (X) has degree r, and hence that k = (p{m)/r.

4.10 Exercises

135

Exercise 4.10.13. Referring to Exercise 3.9.24 (b), let k(Y, Z) = YZ. Let m and n be arbitrary positive integers and let d — gcd(m,n) and i = lcm(m, n). Show that Qyzi^miX), d(X)). In particular, if m and n are relatively prime, show that QYzi^miX), „(X)) = QYz{<^n,niX),^l{X))

=

^„,r,{X).

Exercise 4.10.14. Factor each of the following polynomials into a product of irreducible polynomials in Q[X]. (This includes the possibility that the given polynomial is irreducible.) (a) fiX) = Z3 - IX^ + 36. (b) fiX) = X^ + 5X^ - lOZ - 8. (c) f{X) = X^ - 4X^ - 5X + 6. (d) f(X) = X^ + 2X^ + 9. (e) f(X) = Z^ + 7X^ + 17Z2 + 18Z + 6. (f) f(X) = Z'^ - 8X2 ^ 15 (g)f(X)^X'-4X^ + 16. (h) f(X) = X'^ + X^ + 7X^ + 15Z + 15. (i) f(X) = X^ + X'^ + 4X^ + 4X^ + 4X + 4. (j)f(X) = X'-2X'-l. (k) fiX) = X^ - IX^ + 6X + 30. Exercise 4.10.15. For each a in Exercise 2.10.1, find m„(X) € Q[X]. (Consider the polynomial you found in solving that problem. If it is irreducible, it is nia {X). Thus, you must decide whether it is irreducible. If it is, you are done. If not, you have more work to do.) Exercise 4.10.16. For each a in Exercise 4.10.15, let E be a splitting field of MociX). Find (E/Q) and find G ^ Gal(E/Q). Exercise 4.10.17. In this exercise, let E be a splitting field of the polynomial f{X) e Q[X] and let G = Gal(E/Q). (1) Find (E/Q) and find G. (2) Find all subgroups H of G, and the fields B to which they belong. (3) For each field B in (b), find a basis for E as a vector space over B and a basis for B as a vector space over Q. (4) Determine which subfields B in (b) are Galois extensions of Q. For each of these fields B, find a polynomial g(X) e Q[X] with B a splitting field for g(X), and find Gal(B/Q). (a) f(X) = X^-2. (b) f(X) = X'-2. (c) f(X) = (X3 - 2)(Z3 - 3). (d)/(X) = ( Z 2 - 3 ) ( X 3 - l ) .

136

4 Extensions of the Field of Rational Numbers

(c)f(X) = (X^ + 3)(X'-l). (f) f(X) = X' + 3, (g)f(X) = X^ + 5. (h) f(X) = Z^ - 2. (i) f{X) = Z^ - 3. Exercise 4.10.18. (a) Let f(X) =^ X^ + bX + c e Q[X] be irreducible. Suppose b > 0, Show that the Galois group of f(X) is isomorphic to 53. (b) More generally, let f(X) = X^ + aX^ + bX + c e Q[X] be irreducible. Suppose b > a^/3. Show that the Galois group of f(X) is isomorphic to 53. Exercise 4.10.19. (a) Let p be an odd prime and let f(X) = XHX - 2)(X _ 4 ) . . . (X - 2(p - 2)) - 2. Show that, for k sufficiently large, f(X) p — 2 real roots. (b) Let p be an odd prime and let

is an irreducible polynomial with

f(X) = (X^ + 2)(XXX - 2 ) . . . (X - 2(p - 3)) + 2/(2^ + 1). Show that, for k sufficiently large, /(X) is an irreducible polynomial with p — 2 real roots. (Then, choosing 2A: + 1 to be a /?* power, i.e., 2k + I = s^ for some integer s, we may make the substitution F = ^-X to obtain a monic irreducible polynomial with integer coefficients g(Y) = (2k + l)f(X) =

sPf(X).) (c) Let p be an odd prime and let f(X)

= (X^ + 2k)(X)(X - 2 ) . . . (X - 2(p - 3)) + 2.

Show that, for k sufficiently large, / ( X ) is an irreducible polynomial with p — 2 real roots. Hence, in any of these three cases, / ( X ) has Galois group Sp. Compare Theorem 4.7.4 and Example 4.7.5. Exercise 4.10.20. Let / ( X ) e Q[X] be a reducible cubic. Show that / ( X ) has splitting field Q(5), where 8 is as in Definition 4.8.1. Exercise 4.10.21. Let a = ^1 + 5 V2. Then certainly (a^ - 7)^ = 50, so we see that a is a root of / ( X ) = 0, where / ( X ) = X^ - 14X^ - 1 e Q[X]. Observe that ot = p^, where ^ = 1 + V2. Use this observation to find all the roots of / ( X ) , the factorization of / ( X ) as a product of irreducible polynomials in Q[X], the splitting field of / ( X ) , and the Galois group of / ( X ) .

4.10 Exercises

137

Exercise 4.10.22. Let f{X) = X^ - 15X^ + 15X^ - 9 Z ^ - 16X - 5 6 € Q[X]. Show that the Galois group of f{X) is 55. Exercise 4.10.23. Let f{X) = 4X^ - lOSX^ + 840X^ - 2160^^ + 6000 e Q[Z]. Show that the Galois group of f{X) is 55. Exercise 4.10.24. Let G be a finite group of order n and suppose G can be written as G = //i x • • x / ^ with Hi a group of prime power order, for each i. Such a group is called nilpotent. (This is not the definition of a nilpotent group but rather a characterization of finite nilpotent groups.) (a) Show that G is solvable. (b) Let E be an extension of F with Gal(E/F) = G. Show that, for every divisor d of n, there is a field B with F c B c E and with (B/F) = d. (c) Find a counterexample to (b) for G a solvable group. Exercise 4.10.25. In this exercise we will show how to find the roots of a general cubic polynomial f(X). Clearly it suffices to consider the case that f(X) is monic, so let f(X) = X^ + aX^ + bX + c. By making the substitution Y = X - a / 3 , we may transform / ( Z ) to a cubic giY) = Y^ + b'Y + c\ i.e., to one in which the quadratic term is absent. Thus it suffices to handle this case, so we assume f{X) = X^ -[-bX + c. Recall from Section 4.8 that f{X) has discriminant A = —Ab^ — 27c^. Let E be the splitting field of f{X) and let D = E(6o). Then ( E / Q C V A ) ) = 1 or 3 so we see from Section 3.7 that if B = Q((^, V A ) , then either D = B or that D = B(6) with 6^ = e G B, i.e., that D is the splitting field of the polynomial X^ — e e B[Z]. We shall not try to find 6 or ^ directly, but these considerations guide the form of our answer. Indeed, with these in mind, and the observation that e must be of the form e = u -\- v^fK for some u,v e Q(ct)), we look for roots of f{X) of the form a = yj u + v^TK + yu — f v A , ^ = (D^^u + i;VA + a>^Y w — i»vA, Y = (x?\^u + i;vA + co\^u — f v A . (Observe that a, ^8, and y are all fixed by Gal(B/Q), as we expect.) Now recall that the coefficients of f(X) are, up to sign, the elementary symmetric functions in the roots o f / ( X ) . Since l+co+o/ = 0, wehavethata+yS+y =0, which is consistent with the quadratic term of f(X) being absent. Examining the linear and constant terms we obtain the equations (x^ + fiy +ay = b,

a^y = —c.

138

4 Extensions of the Field of Rational Numbers

Show that these equations yield u = -c/2,

V= v^/18.

and verify by direct substitution that, with these values of u and v,a, ^, and y are roots of f(X) = X^+aX + b. Exercise 4.10.26. Use the formula in Exercise 4.10.25 to find the roots of the following cubics: (a)f(X) = X'-3X-L (b) f(X) = X^ - 2\X - 37. (c) f(X) =:X^ - \1X + 18. (d) / ( X ) = Z^ - 11X + 30. (e) f(X) = X^ + 3X - 4. (f) f(X) = X^ - 13X - 12. Exercise 4.10.27. Observe that the polynomials in Exercise 4.10.26 parts (e) and (f) are not irreducible. Find their roots directly. Compare your answers here with your answers in Exercise 4.10.26. You may be surprised! Exercise 4.10.28. (a) Let F c R and let E = F(a) where a"" = a e F and Of G R. Let B be any Galois extension of F with B c E. Show that (B/F) = 1 or 2. (b) Use part (a) to show that the solution of a cubic must involve nonreal radicals even when there are only real roots. (Examples of this are provided by parts (a) and (d) of Exercise 4.10.26.)

Further Topics in Field Theory

5.1 Separable and Inseparable Extensions We now wish to further investigate questions related to separabiHty and inseparabiUty of algebraic extensions. Recall from Corollary 3.2.3 that every algebraic extension in characteristic 0 is separable, so in this case there is nothing more to be said. Thus in this section we shall assume that char(F) = /? > 0. Recall we have the Frobenius endomorphism <J>: F -^ F given by 0(x) = xP, and F^ = Im(4>) = {y e¥ \ y = x^ fox some x G F} is a subfield of F. (See Definition 2.1.10 and Corollary 2.1.11.) If F is finite, then F^ = F, but in this case, too, every algebraic extension of F is separable (Theorem 3.2.6), so we will be interested in the case that F is infinite. First, we shall deal with extensions that are separable, and then with extensions that are not. We begin with a technical lemma. Lemma 5.1.1. Let E be an extension ofF and leta eE,a

^ F. Then a^ ^ F^.

Proof, Suppose aP e F^, i.e., a^ = ^P for some ^ e F. Then 0 = aP - fiP = (a — fi)P so Of = )S. But a ^ F and ^ G F, so this is impossible. n We also remind the reader of the following result. Lemma 5.1.2. Leta e¥,a for every ^ > 1. Proof Lemma 3.2.8.

i F^. Then f(X) = XP' -a e F[X] is irreducible D

Lemma 5.1.3. Let E be a finite extension ofF. IfFFP = E, then E is a separable extension ofF.

140

5 Further Topics in Field Theory

Proof, We claim that if [ai} is a set of elements of E that is linearly independent over F, then {af} is also linearly independent over F. To see this, note that, extending [at] if necessary, we may assume that {Qf/}/=:i,...,j, d = (E/F), is a basis for E over F. We will show that {af }/=i,. .,j spans E over F. Hence, since (E/F) = d is finite, {off }/=i,. .,j is a basis for E over F and hence this set is linearly independent. Let y eE. Since E = FE^, we may write y = ba^

for some b GF, a G E .

Since {at} is a basis, a = c\a\ H

+ CdOid

with c i , . . . , Q G F ,

and then y = to^ = biciax • • • + caOLdY = {hc\)oL{ + • • • + (&c^)a^, proving the claim. Now suppose a G E is not separable. Then, by Proposition 3.2.2, its minimum polynomial mQ^ (X) is of the form tc({X) = y ^ CiX^^ for some k and {c/}, not all zero. i=0

Thus k

0 = J2cia'P, i=0

so 1 = a^, a ^ , . . . , a^P are linearly dependent and hence, by the contrapositive of our claim, 1 = a^, a , . . . , a^ are linearly dependent as well. Thus k

0 = y ^ diat

with not all dt equal to zero.

/=o

so if

/(X)-^rf,X\ then /(of) = 0. But deg f(X) < dtgma(X), which is impossible.

D

5.1 Separable and Inseparable Extensions

141

Lemma 5.1.4. Let or e E. Then a is separable over F if and only if F(a) = F(aPy Proof Suppose that a is separable over F. Clearly F(a^) c F(a), so we need only show the reverse inclusion. To show that, it suffices to show that a e F(aP). Let niaiX) e F[Z] be the minimum polynomial of a over F, and let rhaiX) e F(Qf^)[Z] be the minimum polynomial of ot over F(Qf^). Then rhaiX) divides m^^CX) in F(a^)[X]. Clearly a is a root of the polynomial f{X) = XP - aP € F(aP)[Xl Suppose a ^ F(a^). Then aP ^ (F(aP))P by Lemma 5.1.1, so / ( Z ) = Z^ aP is irreducible in F(aP)[X] by Lemma 5.1.2. Since f(X) is irreducible, rhaiX) = XP - aP. Since m«(Z) divides ma(X) in F(aP)[Xl rfiaiX) divides ma(X) in F(Qf)[X]. Hence m^iX) is divisible by XP - aP = (X - a)P in F(a)[X], contradicting the separability of of. Conversely, if F(Qf) = F(a^), then F(a) = F(a^) = F(F(a))^ so, by Lemma 5.1.3, F(a) is a separable extension of F, i.e., a is a separable element of E. D Corollary 5.1.5. IfE is obtained from F by adjoining a root of a separable polynomial, then E is a separable extension ofF. Proof Let E = F(a) where a is a root of a separable polynomial. Then F(a) = FiptP) by Lemma 5.1.4, and F(a^) = F(F(a))^, so F(cy) = F(F(a))^ and Fipt) is a separable extension of F by Lenrnia 5.1.3. D Lemma 5.1.6. Let B be afield intermediate between F and F.IfF is a separable extension ofF, then B is a separable extension ofF and E is a separable extension o/B. Proof As B c E, B is trivially a separable extension of F. Let Of € E. Since E is a separable extension of F, of is a simple root of a polynomial f(X) e F[X]. But F c B so f(X) e B[X] and a is separable over B. n Lemma 5.1.3 has a converse (under weaker hypotheses, in fact). Lemma 5.1.7. Let E be an extension ofF.IfF then FEP = E.

is a separable extension ofF,

142

5 Further Topics in Field Theory

Proof. Let B = FE^ so that B is intermediate between F and E. By Lemma 5.L6, E is a separable extension of B. Suppose B c E, and let a G E, a ^ B. Then a = of^ G B so a is a root of XP — a e B[X]. But, by Lemma 5.1.2, this polynomial is irreducible, so it is equal to m^ (X), the minimum polynomial of a over B. Then this polynomial rhaiX) = XP — a = XP — a^ = (X — aY has repeated roots, contradicting the separability of E over B. D Theorem 5.1.8. Let B be afield intermediate between F and E. TjfE is separable extension ofB and B is a separable extension ofF, then E is a separable extension of¥. Proof. First, assume both (E/B) and (B/F) are finite. Since B is a separable extension of F, we have B = FB^, and since E is a separable extension of B, we also have E = BE^, both by Lenmia 5.1.7. Then E = BE^ = FB^(BE^)^ c FE^, so E = FE^ and (E/F) = (E/B) (B/F) is finite, and so E is a separable extension of F by Lemma 5.1.3. Now for the general case. Let a G E, and let rhaiX) e B[X] be the minimum polynomial of a over B. Let BQ be the subfield of B obtained by adjoining the coefficients of rhaiX) to F. Then BQ is a finite extension of F and F ^ Bo c B with B a separable extension of F, by hypothesis, so BQ is a separable extension of F by Lemma 5.1.6. Let EQ = Bo(Qf). Then EQ is a finite extension of BQ. Since E is a separable extension of B, by hypothesis, rfiaiX) is a separable polynomial, so, by Corollary 5.1.5, EQ is a separable extension of BQ. Then, by the finite case, EQ is a separable extension of F. Since a is arbitrary, E is separable. D Theorem 5.1.9. A finite extension E o / F is separable if and only ifK is obtained from F by adjoining root(s) of a separable polynomial f{X) G F[X]. Proof. E is obtained from F by successively adjoining finitely many elements. If one of them is not separable, then certainly E is not separable. If each of them is a root of a separable polynomial, then the theorem follows immediately from Corollary 5.1.5 and Theorem 5.1.8. D Lemma 5.1.10. Let E be an extension ofF. Then Fs = {a eF \ a is separable over F} is afield intermediate between F and E.

5.1 Separable and Inseparable Extensions

143

Proof, Let a, ^ e F,. Then a is a simple root ofm^{X) e F[X] c F(fi)[Xl so F()S)(Qf) = ¥(a, yS) is separable over F(^). Also, F{P) is separable over F. Hence, by Theorem 5.1.8, F(a, fi) is separable over F. Thus a + fi eFs and Qf)6 € F^. Also, if a 7^ 0 then 1/a € F(Qf) so 1/a G F^. Thus F^ is a field, n Definition 5.1.11. The field F^ of Lemma 5.1.10 is called the separable closure of F in E. o We now turn our attention to inseparable extensions. Definition 5.1.12. Let E be an extension of F. An element a G E is purely inseparable over F ifma(X) = (X — a)^ € E[X] for some m. The integer m is called the degree of inseparability of a. o Remark 5.1.13, If a € E is both separable and purely inseparable over F, then (m = 1 and) a G F. o Lemma 5.1.14. Let a G E be purely inseparable over F with degree of inseparability m. Then m = p^ for some r. Proof Write m = kp' with (k, p) = I, Then, in E[X], MaiX) = (X- afP' = ((X - aY'f = {XP'f-kaP\XP'f-^+

= (X^' - a^'f '" G F [ X ]

by the Binomial Theorem. Thus kaP' G F and so aP' e F, Thus f(X) = XP' - aP' G F[X] and f(X) divides m«(Z) in E[X] and hence in F[Z], by Lenmia 2.2.7. Since m^iX) is irreducible in F[X], ma(X) = f(X) and k= 1. D Corollary 5.1.15. If a is purely inseparable overF, then (F(a)/F) = m = p^ for some r. Proof (F(a)/F) = degma(X),

D

Definition 5.1.16. An extension E of F is purely inseparable if every a G E is inseparable over F. o Lemma 5.1.17. Let E be an extension of F, If a e E is purely inseparable, then F(a) is a purely inseparable extension ofF, Proof, First observe that ma(X) e F[X] factors as ma(X) = (X — a)P' in E[X], by Lenmia 5.1.14, so {X - a)P' = XP'' - aP' e F[X] and a = aP'^ e F.

144

5 Further Topics in Field Theory

Now let ^ G F(a). If ^S G F, there is nothing to prove (as then m^(X) = X - )S) so assume )S ^ F. Then ^ = J2T=o ^t^' with bi e F (and m = p'\ so m—1 /=0

r

m—1

m—1

/=0

/=0

so F = F(pP') and hence F(^^') C F(^). Then, by Lemma 5.1.4, ^ is inseparable over F. (If fi were separable over F, we would have F(^) = F(yS^) = F((pP)P) = . . . = F(pp'), a contradiction.) D Lemma 5.1.18. Let E be a purely inseparable extension of¥. Then every a e E is purely inseparable over F. Proof. Let a € E. If a € F, there is nothing to prove, so let a ^ F. Consider its minimum polynomial nia (X) and let p^ be the highest power of p dividing all the exponents of the nonzero terms in ma(X). Then ntaiX) = f(XP') for some polynomial f(X), and f(X) is irreducible as ma(X) is. Note that f\X) 7^ 0 as some term in f(X) has positive degree not divisible by /?, so, by Proposition 3.2.2, f(X) is separable. Now a^' is a root of f(X), so a^' is a separable element of E. Hence a = a^' e F, and a satisfies the polynomial XP' - a. Let / be the smallest integer with a' = a^^ e F. Then a' ^ F^, by Lenmia 5.1.1, so, by Lemma 5.1.2, g(X) = XP —a' is irreducible, and gioi) = 0. Hence m«(X) = g(X) = XP^ - a' = (X - a)P^ and a is purely inseparable (and f = e). D Proposition 5.1.19. Let E be a purely inseparable extension of F and let B be any field intermediate between F and E. Then B is a purely inseparable extension ofF and E is a purely inseparable extension ofF. Proof Clearly B is a purely inseparable extension of F. Let a € F. Then a has minimum polynomial mc^(X) € F[X] which factors as (X — a)^ e F[X]. Let rhaiX) e B[X] be the minimum polynomial of a over B. Then ma(X) divides m«(X) in E[X], so ffiaiX) = (X - a^' for some m\ and a is purely inseparable over B. Since a is arbitrary, E is purely inseparable over B. D Corollary 5.1.20. Let F be a finite, purely inseparable extension of F. Then (E/F) is a power of p. Proof Let { a i , . . . , ak] be a basis for E over F, so E = F ( a i , . . . , ak). We prove the corollary by induction on k. If A: = 1, then E = F(ai) and the result is inmiediate from Corollary 5.1.15.

5.1 Separable and Inseparable Extensions

145

Assume the result is true for k — I, and let B = F ( a i , . . . , ofj^-i). Since E is a purely inseparable extension of F, E is a purely inseparable extension of B and B is a purely inseparable extension of F, by Proposition 5.1.19. Since (E/F) = (E/B) (B/F), the result follows. n Lemma 5.1.21. Let E be an algebraic extension ofF and let F^ be the separable closure ofF in E. Then E is a purely inseparable extension ofFg. Proof. Suppose a € E is separable over F^. Then F^(a) is a separable extension of Fj and F^ is a separable extension ofF so, by Theorem 5.1.8, Fs{ot) is a separable extension of F, and then a G F^ by the definition of F^. n Corollary 5.1.22. Let E be an algebraic extension ofF. Then there is a unique field B intermediate between F and E with B a separable extension ofF and E a purely inseparable extension o/B. Proof TakeB = F,.

n

Note that Corollary 5.1.22 says that any algebraic extension of F can be obtained by first taking a separable extension, and then taking a purely inseparable extension. Definition 5.1.23. Let E be an algebraic extension of F and let F^ be the separable closure of F in E. Then (F^/F) is the separable degree of E over F and (FJ/FS) is the inseparable degree of E over F. o Remark 5.L24. (1) Of course, (E/F) = (E/F,)(F,/F). (2) E is a separable extension of F if and only if E = F^, or, equivalently, if and only if (F,/F) = (E/F). (3) E is a purely inseparable extension of F if and only if F^ = F, or, equivalently, if and only if (Fs/F) = 1. (4) By Corollary 5.1.20, (E/F,) is a power of p. (5) If (E/F) is relatively prime to p, then E is a separable extension ofF (as in this case (E/F,), which is a power of p, divides (E/F), which is relatively prime to p, so we must have (E/F,) = 1 and hence E = F,). o Remark 5.L25. Observe that every purely inseparable extension of F is normal. (If Of € E is purely inseparable, then mc^(Z) = (X — a)^ G E[X] is a product of linear factors.) o Theorem 5.1.26. Let B be a purely inseparable extension ofF and let F be a purely inseparable extension of B. Then E is a purely inseparable extension OfF.

146

5 Further Topics in Field Theory

Proof. Let a G E and let ma(X) G B[X] be the minimum polynomial of a over B. Let BQ be the extension of F obtained by adjoining the coefficients of rha {X) to F. Then F c BQ c B so BQ is purely inseparable over F by Proposition 5.1.19, and BQ is certainly a finite extension of F. We now use Theorem 5.2.1 below to conclude that BQ is the splitting field of a polynomial f{X) G F[Z]. Let {^1,..., Pk) be the roots of / ( X ) , so BQ = F ( ^ i , . . . , ^^). Note also that rhaiX) = (X — a)^ e E[X], so a is purely inseparable overBo. Let Eo be a splitting field of ma(X) G F[X] with F(a) c EQ. Let a' G EQ be any root of niaiX), Since ma(X) is irreducible, by Lemma 2.6.1 there is a a: F(a) —^ F(aO with or | F = id and a (a) = a\ Since BQ is the splitting field of / ( X ) , by Lemma 2.6.3 a extends to a: F(Qf)()Si,,.., Pk) -> F(a^ ^ 1 , . . . , ^ , ) . Now F ( a ) ( ^ i , . . . , ^,) = F ( ^ i , . . . , ^^)(a) = Bo(a). Similarly, F(a^ Pi,.,., Pk) = Bo(c^O- Now or(^/) = fit as a niust take a root of m^.(X) to a root of m^.(X), and ^/ is the only root of m^.(X), Hence cr I Bo = id. Thus a: Bo(a) -^ Bo(aO with cr | BQ = id, a (a) = a\ Then 0 = a(0) = a(ma{oi)) = ma(cr{a)) = m^CaO. But a is purely inseparable over Bo so the only root of m^ (X) is a. Hence a^ = a(a) = a, and so a is the only root of m^ (X) and hence a is purely inseparable over F. Since a is arbitrary, E is purely inseparable over F. n We now extend the second part of Definition 5.1.12. Definition 5.1.27. Let E be an algebraic extension of F and let of G E. The degree of inseparability / (a) is the multiplicity of a as a root of nia (X). o Lemma 5.1.28. (1) i(a) is independent ofE, (2) If a and a! are two roots of the irreducible polynomial f{X) in E, then i(a) = i(a^).

G F[X]

Proof (1) Let E' D E, / = i(a) in E, i' = iia) in E'. In E[X], ma(X) = {X — aYgiX) with g(X) relatively prime to X — a, and then g(X) is relatively prime to X — a in E'[X] as well (Lemma 2.2.7), so i^ = i, (2) By Lemma 2.6.1, there is an isomorphism a: F(a) -> F(aO with or I F = id and a (a) = a\ Then a(f(X)) = f(X), and also if we write f(X) = (X - ayg(X) G F(a)[X] with g(a) y^ 0, then we see that f(X) = (X - aJgiX) G F ( a O m with g(a') ^0. n Lemma 5.1.29. Let E be an algebraic extension ofF and let a G E. Then i (a) is a power of p. More precisely, i(a) is the highest power of p dividing the exponent of every nonzero term in m^iX).

5.2 Normal Extensions

147

Proof, Let p"^ be the highest power of p such that f{X) = g{XP'), g(X) e F[X]. Since niaiX) is irreducible, so is g{X), and g\X) 7^ 0, so g{X) is separable (Proposition 3.2.2) and hence a^' is a simple root of g{X), Hence g{X) = iX-aP')h(X)withh(aP') 7^ 0, i.e., with X - a ^ ^ not dividing/i(X). Thus, one the one hand, X^' —a^' = (X — a)^' divides f(X) and, on the other hand, this is the highest power of X — a dividing f(X). D Corollary 5.1.30. Let E be an algebraic extension of F and let a G E. 7f ma(X) splits in E[X], then

m,(X) = ( Z - a l r ^ . . ( X - a , ) ^ ^ with at = a for some /, p^ = i(a), andkp^ = dtgma(X).

D

5.2 Normal Extensions In this section we investigate normal extensions. Theorem 5.2.1. E is a finite normal extension ofF if and only ifE is the splitting field of a polynomial f(X) e F[X]. Proof If E is normal and {6i,...,6„} is a basis for E over F, then m^.(X) e F[X] splits into a product of linear factors for each /, so f(X) = m^. (X) • • • m^^ {X) € F[X] splits into a product of linear factors, and then E is the splitting field of f{X), as any field in which f{X) splits must contain Suppose now that E is the splitting field of / ( X ) G F [ X ] and let ofi,..., aj^ be the roots of f{X) in E, so E = F(Qfi,..., ofy^). Let g{X) G F [ X ] have a root )S G E. We need to show g{X) splits into a product of linear factors in E[Z]. Clearly it suffices to consider the case that g{X) is irreducible. Let E 5 E be a splitting field of the irreducible polynomial g{X) and let y3 G E be a root of ^(X). Then, by Lemma 2.6.1, there is a unique isomorphism a: F(/3) -> F(/3) with a ( ^ ) = ^ and or | F = id. Now E is a splitting field for f{X) over F(^) and E(^) = F ( a i , . . . , a^t, )S) is a splitting field for f{X) over F(^) so by Lemma 2.6.3 there is an isomorphism T: E = E()6) -^ E(^) extending cr, and hence with r (^) = ^, Now r extends a, which is the identity on F, so r is itself the identity on F. Thus r permutes { a i , . . . , ak), as it must take any root of f{X) to a root of / ( X ) . Since )6 6 E = F ( a i , . . . , au), P = h(ai,..., au) for some polynomial / i ( Z i , . . . , X ^ ) G F[Xi^...,X^]. T h e n ^ = T ( ^ ) = r ( / i ( a i , . . . , a^)) = /z(r(Qfi),..., T(ajt)) so )6 G E. n

148

5 Further Topics in Field Theory

Corollary 5.2.2. Let E be a normal extension of F and let a: B -> B' be any isomorphism of fields B and B' intermediate between E and F such that or I F = id. Then a extends to an automorphism 5^: E ^ E. Proof First, suppose that E is a finite extension of F. Then E is the splitting field of a polynomial f(X) e F[X]. We may regard E as the splitting field of f(X) e B[X] and note that a(f{X)) = f{X), Then the corollary follows immediately from Lemma 2.6.3. The general case then follows by a Zom's Lemma argument. (See the proof of Theorem 5.4.2 below.) D Definition 5.2.3. Let E be an algebraic extension of F. An extension D of E is a normal closure of E if D is a normal extension of F but no field intermediate between D and E is a normal extension of F. o Lemma 5.2.4. Every algebraic extension E o/F has a normal closure D, and any two normal closures ofE are isomorphic. Proof First, suppose E is a finite extension of F. Let E = F ( a i , . . . , a^) and let f(X) = ma, (X) • - • ma, (X). Let D 2 E be a splitting field of / ( Z ) . Then D is a normal extension of F by Theorem 5.1.1, and f(X) does not split in any proper subfield of D containing E, so D is a normal closure of E. Furthermore, D is unique up to isomorphism as any two splitting fields of f(X) are isomorphic (Lemma 2.6A). The general case then follows by a Zom's Lemma argument. D Corollary 5.2.5. Let Ebe a separable extension ofF. Then the normal closure D o/E is a Galois extension of¥. Proof First, suppose E is a finite extension of F. Then, following the proof of Lemma 5.2.4, D is a splitting field of the separable polynomial f(X) and so, by Theorem 3.7.14, D is a Galois extension of F. The general case then follows from Corollary 5.4.3 below. n This corollary justifies the following definition. Definition 5.2.6. Let E be a separable extension of F. A normal closure D of E is called a Galois closure of E. o Remark 5.2.7. (1) Note that if E is normal over F and B is an intermediate field between F and E, then E is normal over B, but B need not be normal over F. E is normal over B as for any a G E, maiX) splits into linear factors in E, and hence rhaiX), the minimum polynomial of a over B, which divides ma(X),

5.2 Normal Extensions

149

also splits into linear factors in E. On the other hand, if E = Q(o), \/2) is the splitting field of X^ - 2 G Q[X], and B = Q ( ^ ) , F = Q, then E is normal over B but B is not normal over F. (2) If B is a normal extension of F and E is a normal extension of B, then E need not be a normal extension of F. Taking F = Q, B = Q ( A / 2 ) , and E = Q(v^) provides an example of this. o Now we turn to inseparable extensions. We first recall from Remark 5.1.25 that a purely inseparable extension is automatically normal. Lemma 5.2.8. Let E be a normal extension of F. Then E is separable if and only //'Fix(Gal(E/F)) = F, and E is purely inseparable if and only if Fix(Gal(E/F)) = E, or, equivalently, if and only /jTGaKE/F) = {id}. Proof The first claim is inmiediate from the fact that E is a Galois extension of F if and only if E is normal and separable (Theorem 2.7.14 in case (E/F) is finite and Theorem 5.4.2 below in case (E/F) is infinite). For the second claim, first suppose that (E/F) is finite. Suppose next that E is a purely inseparable extension of F. Let a G E. Then niot {X) = (X — a)^ in E[X] for some m. Now for any a e Gal(E/F), a(a) is also a root of ma(X), But the only root of mc^(X) is a, so a (a) = a. Since a is arbitrary, a = id. The general case follows by a Zom's lemma argument. Conversely, let E be normal and finite over F, and suppose that E is not purely inseparable over F. Then there is an a G E such that nia (X) has (at least) two distinct roots a and a\ By Lenmia 2.6.1 there is a a : F(Qf) -^ F(aO with a (a) = a' and, by Corollary 5.2.2, a extends to a : E ^^ E with or (a) = a' and a | F = id, so a G Gal (E/F) and a 7^ id. n We have seen in Section 5.1 that any algebraic extension may be obtained by first taking a separable extension and then taking a purely inseparable extension. In the case of a normal extension, the order may be reversed. Proposition 5.2.9. Let E be a normal extension of F. If Ft = Fix(Gal(E/F)) then Fi is a purely inseparable extension ofF and E is a separable extension ofFi, Furthermore, E is a Galois extension of Ft and Gal (E/F/) = Gal (E/F). Proof By definition, if G is a group of automorphisms of E, then E is a Galois extension of Fix(G). Apply that here with G = Gal(E/F). Then E is a Galois extension of F/ and hence E is a separable extension of F (Theorem 2.7.14 in case (E/F/) is finite and Theorem 5.4.2 below in case (E/F/) is infinite). Now we consider the extension F/ of F. If F/ is not purely inseparable then, as in the proof of Lemma 5.2.8, there is an element a G F/ and an isomorphism

150

5 Further Topics in Field Theory

o: F(Qf) -> F(aO with a | F = id and a ( a ) = a' 7^ a. Then, by Corollary 5.2.2, a extends to a : E -> E with a | F = id, i.e., to a G Gal(E/F) with o{oi) ^ a, contradicting the definition of F/. n In fact, in this situation we may say more. Recall that F^ was defined in Definition 5.1.11 and characterized in Corollary 5.1.22. Theorem 5.2.10. Let Ebea normal extension ofF, Then F^ and F/ are disjoint extensions ofF andF = F^F/. Furthermore, F^ is a Galois extension ofF and Gal(F,/F) = Gal(E/F,). Also, (E/F,) = (F,/F) and (E/F,) = (F,/F). Proof. Let a eFs DFi. Then a is both separable and purely inseparable over F, so, by Remark 5.1.13, a € F. Thus F^ and F/ are disjoint extensions ofF. Let Of G F^. Then a has separable minimum polynomial ma(X). Since E is a normal extension of F, ma (X) splits into a product of linear factors (X — ai) " ' (X — at) in E[X], But each at is a root of a separable polynomial, namely ma (X), so a/ e F^ for each /, and hence ma (X) splits into a product of linear factors (X—ai) • • • (Z—of^) in F^ [X]. Thus F^ is a normal and separable extension of F, so is a Galois extension of F (by Theorem 2.7.14 in case E is a finite extension of F, and by Theorem 5.4.2 below in general). Suppose that E is a finite extension of F. By Theorem 3.4.6, Gal(E/F/) is isomorphic to a subgroup of Gal(F^/F). But let cr e Gal(F^/F). Now E is a purely inseparable extension of F^, so it is a normal extension of F (Remark 5.1.15) and hence it is the splitting field of some polynomial f(X) e F[X] by Theorem 5.2.1. Hence, by Lenuna 2.6.3, a extends to an automorphism of E, i.e., to an element of Gal(E/F) = Gal(E/F/), so these two groups are isomorphic. But then also (E/F) = (E/F,)(F,/F) = (E/F,)(F,/F) giving the claimed equalities between the degrees of the extensions. Furthermore, since F^ is a Galois extension of F, we have from Theorem 2.4.5 that (F,F,/F) = (F,/F)(F,/F) = (E/F), so F,F,- = E. Now for the general case. Let a € E and let EQ C E be the splitting field of ma (X), Then EQ is a finite extension of F, so we may apply the above argument to Eo, F^ D Eo, and F/ Pi EQ. Then we may apply Zom's Lemma to obtain the conclusion of the theorem for E. (The final point to see is that we have equalities among the degrees even if they are not all finite, when we cannot simply use the argument above. But (E/F^) < (F//F) by Lemma

5.3 The Algebraic Cslosure

151

2.3.4. If we did not have equality, then there would be some nontrivial linear combination of basis elements of E over F^ that would be zero, and this linear combination would involve only finitely many basis elements and hence would have coefficients in some finite extension of F, contradicting equality of the degrees in the finite case.) n

5.3 The Algebraic Cslosure In this section we show that every field F has an algebraic closure F that is unique up to isomorphism. Definition 5.3.1. A field E is algebraically closed if the only algebraic extension of E is E itself. o Lemma 5.3.2. The following are equivalent: (1) E is algebraically closed. (2) Every nonconstant polynomial f(X) s E[X] is a product of linear factors in E[X]. (3) Every nonconstant polynomial f{X) € E[X] has a root in E. Proof (1) implies (2): Let E be algebraically closed. Let f{X) e E[X] be irreducible. Then f(X) has a root a in some algebraic extension E(a) of E. But E(a) = E by assumption, so X — a divides / ( X ) in E and hence f{X) = X — a. Now let g{X) e E[X] be arbitrary. Then g(X) splits into a product of irreducible and hence linear factors in E, yielding (2). (2) implies (1): Let a be an element of some extension of E with a algebraic over E. Then ma(X) = E[X] splits into a product of linear factors in E, one of which must be X — a. Hence a G E, yielding (1). (2) implies (3): Let f(X) e E[X]. Then f(X) = {X - ai) ---{X - a^) for some a i , . . . , aj^ G E, so /(ofi) = 0, yielding (3). (3) implies (2): By induction onk = deg f{X). For k = I this is trivial. Assume it is true for ^ — 1 and let f(X) have degree k. Then f(X) has a root ai so f(X) = (X - ai)g{X) with deg^(X) = ^ - 1. By the inductive hypothesis g(X) = (X — Qf2) • • • (X — ak) for some ^2, • • •, ock, and then / ( X ) = (X - ai)(X - a2) • • • (X - ak), yielding (2). n Definition 5.3.3. Let F be a field. An algebraic closure F of F is an algebraic extension of F that is algebraically closed. o Theorem 5.3.4. Let F be an arbitrary field. Then F has an algebraic closure F and F is unique up to isomorphism.

152

5 Further Topics in Field Theory

Proof. Both existence and uniqueness are Zom's Lemma arguments. The idea of the proof is clear. We wish to take the set of all algebraic extensions of F, order this set by inclusion, use Zom's Lenrnia to conclude this set has a maximal element, and show this maximal element is an algebraic closure of F. But, in carefully considering this argument, set-theoretic difficulties appear, as, a priori, it is not clear what inclusion means. (We can say, for example, that Q(V2) C Q(V2, \/3) as we can regard these both as subfields of C. But this is because we already know that C exists. In the abstract, what we mean by this inclusion is the following: Let Fi be the field obtained from Q by adjoining an element a with of^ = 2 and let F2 be the field obtained from Q by adjoining elements ySi and ^2 with ^\ = 2 and jS| = 3. Then there is a map a : Fi -> F2 given by aipt) = fii and using this map we may identify Fi with a subfield of F2. But this is not quite the same thing as saying that Fi is a subset of F2.) The way out of this dilemma is to assume that all fields we need to consider are contained in some universal set. (Note we cannot assume they are all contained in some universal field as that would be presuming the existence of the algebraic closure, which is what we are trying to prove.) Thus we proceed as follows in order to show the existence of F. Given a set U, and a field E, we shall write E c ZY if E is a subset of U. Then the field operations (addition, multiplication, inversion) are defined on this subset of U. For two such fields Ei and E2, we shall write Ei c E2 if El c E2 as subsets of U and the restrictions of the field operations on E2 to El agree with the field operations on Ei. We now set Uo = 2^, the set of subsets of F. If F is infinite we let ZY = Wolf F is finite we let U = 2^\ Regard F c W by identifying a with {a}, a G F, if F is infinite, or with {{a}}, if F is finite. As is well known, for any set A, the cardinality of 2^ is greater than the cardinality of A, so the cardinality of F is less than the cardinality of U. Furthermore, by construction, U is uncountable. Let S = {algebraic extensions E of F with E c W} and order S by inclusion. Then every chain has a maximal element: If Ei c E2 ^ • • •, then E = UE/ is algebraic. (Any a G E is an element of some E/ and hence is algebraic over F.) Hence, by Zom's Lemma, S has a maximal element F. Claim: F is algebraically closed. Proof of claim: Let F c F with F an algebraic extension of F. Then, by Theorem 2.4.12, F is an algebraic extension of F. To complete the existence

5.3 The Algebraic Cslosure

153

part of the proof we must show that it is possible to embed F in U, Assuming that, by the maximality of F we must have that this image is just F, and then F = F, so we see that F is algebraically closed. Claim: Let E be an algebraic extension of F. If F is finite, E is finite or countably infinite. If F is infinite, E has the same cardinality as F. Assuming this claim, F c W is a set of smaller cardinality than that of U and F c F with F also a set of smaller cardinality than ZY, so we may choose an arbitrary embedding of F — F in ZY as a set and then define the field operations on this set to agree with those on F. Proof of claim: We consider the set V of all monic polynomials in F[X]. Then V = U ^ i ^w where Vn is the set of all monic polynomials of degree n. Now there is a bijection between F" and Vn given by ( a o , . . . , a«-i) ^—> ao + a\X H h an-\X^~^ + X^. If F is finite, so is F" for each n, and hence V is countable union of finite sets and so is countable. If F is infinite, then F^ has the same cardinality as F for each n, and then ;P is a countable union of such sets, so also has the same cardinality as F. Now if E is an algebraic extension of F, we have a map p:¥> -> V given by pipt) = moiiX). Then p is a finite-one mapping as if p{a) = f(X), then f(a) = 0, and any polynomial only has finitely many roots. But, in general, if p : A ^- JS is a finite-to-one mapping of the set A onto the infinite set B, then the cardinality of A is equal to the cardinality of B, This completes the proof of the existence of F. Now we must show that F is unique up to isomorphism. Thus let Fi and F2 be two algebraic closures of F. Let 5 = {(E,or) l E c F i , a: E ^ F2} where a is a map of fields. Order S by (Ei, cri) < (E2, 0^2) if Ei c E2 and (J2 | Ei = ai. Again by Zom's Lemma every chain has a maximal element so S has a maximal element (Fi,a). Claim: Fi = F i . Proof of claim: Suppose not, and let a € Fi, ai ^ Fi. Let or (Fi) = F2 ^ F2. Since ofi is algebraic over F, it is algebraic over Fi, so let rha^ (X) e Fi (X) be its minimum polynomial over Fi. Since F2 is algebraically closed, there is a root 0^2 of a{ma^ (X)) in F2. Then, by Lemma 2.6.1, a extends to a map r : Fi(ai) -> ^2(0^2) ^ F2 (with f(ai) = 0^2), contradicting the maximality of Fi. Thus Fi = Fi and we have a map a : Fi ^^ F2. As a is a map of fields, it is injective. To complete the proof we must show it is surjective. Let ^2 € F2- Then ^2 is algebraic over F, so it is algebraic over

154

5 Further Topics in Field Theory

or(Fi). But or(Fi) is algebraically closed so ^2 ^ or(Fi), i-^., Pi = ^(^1) for some P\ € Fi, as required. D We now wish to examine some concrete algebraically closed fields. But first we shall prove a lemma that enables us to weaken the hypotheses in Lemma 5.3.2. Lemma 5.3.5. Let E be an algebraic extension ofF. Then E is algebraically closed if and only if every polynomial f(X) e F[X] is a product of linear factors in E[Z]. IfE is a normal extension o/F, then E is algebraically closed if and only if every polynomial f(X) e F[X] has a root in E. Proof First, let E be an algebraic extension of F. Then the "only i f part is clear. For the " i f part, let a be algebraic over E. Then E(a) is an algebraic extension of E (Corollary 2.4.9) and E is algebraic over F (by hypothesis), so E(a) is algebraic over F (Theorem 2.4.12). Hence a is a root of ma(X) e F[X]. But ma(X) = (X -ai)"'(X - a„) in E[Z], so a = a, G E for some /. Now if E is a normal extension of F, then by definition, any polynomial in F[X] with a root in E splits into a product of linear factors in E[X], so we are back in the previous case. n Theorem 5.3.6. The field of rational numbers Q has an algebraic closure Q that is a subfield of the field of complex numbers C. Proof Let f{X) € Q[X] be an arbitrary irreducible polynomial and let E be a splitting field of f{X). Let G = Gal(E/Q), and let H be the 2-Sylow subgroup of G. Set B = Fix(if). Then E is a Galois extension of B with Galois group H. Also, d = [G : H] = (B/Q) is odd. By Corollary 3.5.3, there is an element a of B with B = Q(a). Then maiX) e Q[X] is a monic irreducible polynomial of odd degree d. By elementary calculus, ma {X) has a real root XQ. (Proof: lim;c->-oo ^a (•^) = —oc, so ma{x-) < 0 for some x- G R, and limjc->oo^a(^) = 00, so moiix^) > 0 for some x+ G R. Then, by the Intermediate Value Theorem, ffia (XQ) = 0 for some XQ G R . ) Let B' = B(xo) c R. Then f(X) has a splitting field E' 2 B'. Let H' = Gal(E7B0. By Corollary 3.4.6 (the Theorem on Natural Irrationalities), H' is isomorphic to a subgroup of H, so H^ is also a 2-group. Thus, by Corollary A.2.3, there is a sequence H' = Hl^ D H[ D - • D H^^ = {1} with [//,_i : Hi] = 2, and, setting E ; = Fix(Hl), B' = E^ C E ; C .. .E^ = E' with (E-/E^_j) = 2. In other words, for each /, E^ is a quadratic extension of E^_j, i.e., E;. = E;_i(7a7) for some a, G E^.^. But if F c C and D = F(Vz) for some z G F, then D is isomorphic to a subfield of C. (If z = re^^, D =

5.3 The Algebraic Cslosure

155

F(v^^^^/^).) Thus, proceeding inductively, we see that f{X) has a splitting field that is a subfield of C. Now let {fi{X)} be the set of all monic irreducible polynomials in Q[X] and letEi c C be the splitting field of ft (X), Let Q be the composite of {E^}. Then Q is algebraic over F and every f(X) e Q[X] splits in Q, so, by Lemma 5.3.5, Q is an algebraic closure of Q. n Actually, a similar argument enables us to prove a stronger result. Theorem 5.3.7. (Fundamental Theorem of Algebra) The field of complex numbers C is algebraically closed. Proof. First, observe that it suffices to prove that every polynomial f{X) e R[Z] has a root in C. On the one hand, this follows directly from Lemma 5.3.5, but on the other hand, we have the following direct argument in this case: Let g{X) e C[X] be an arbitrary polynomial, and set f{X) = g{X)g{X). Then f{X) G R[X]. If f{X) has a root y, then ^(y) = 0, in which case y is a root of g{X), or giy) = 0, in which case g(y) = 0 and y is a root of g(X), Thus let f(X) e R[X] be an arbitrary irreducible polynomial and let E be a splitting field of f{X), Let G = Gal(E/R), and let H be the 2-Sylow subgroup of G. Set B = Fix(H). Then E is a Galois extension of B with Galois group H. Also, d = [G : H] = (B/Q) is odd. By Corollary 3.5.3, there is an element a of B with B = R(a). Then ma(X) e R[X] is an irreducible polynomial of odd degree d. But, by the same elementary calculus argument as in the proof of Theorem 5.3.6, ma(X) has a real root XQ. Thus (X — xo) divides maiX). Since m^iX) is irreducible, we must have m^iX) = {X — XQ), and so (i == 1 and B = R. Then E is a splitting field of f{X) with (E/R) = T" for some n, and, arguing as in the proof of Theorem 5.3.6, we see that E is isomorphic to a subfield of C, so f{X) has a root in C, and we are done. n Remark 5.3,8. (1) Q has an algebraic closure Q c C but Q 7^ C as Q is countable, being an algebraic extension of Q, but C is uncountable. (2) Theorem 5.3.7 implies Theorem 5.3.6, for, once we know C is algebraically closed, it is easy to see that {z G C | z is algebraic over Q} is an algebraic closure of Q. This field is known as the field of algebraic numbers. (3) The Fundamental Theorem of Algebra cannot have a purely algebraic proof, as the definitions of R and C are not purely algebraic. We have given a proof that uses only a minimum of analysis (the argument from elementary calculus given in the proof of Theorem 5.3.6). But this important theorem has many proofs. Here is one well-known proof that uses results from complex analysis:

156

5 Further Topics in Field Theory

L e t / ( X ) € C[Z] be a nonconstant polynomial,/(X) = Yll=o^i^'-^^^^ for any z G C, \f(z)\ > l^nllzT — Yll=o I^^I^T from which it easily follows that there is an A such that |z| > A implies \f(z)\ > 1. Now suppose f(X) does not have a root in C. Consider g(X) = l/f(X), Then \g(z)\ < 1 for \z\ > A. On the other hand, {z \ \z\ < A} is a compact set so there is a C with \g(z)\ < C for all Z with |z| < A. Thus g(z) is a bounded analytic function and hence, by Liouville's theorem, a constant. But that implies f(z) is a constant, which is absurd. o

5.4 Infinite Galois Extensions In this section we will consider infinite algebraic extensions. Our main goals will be to show that Theorem 2.7.14 and Theorem 2.8.8 (the Fundamental Theorem of Galois Theory) have appropriate generalizations to this case. First, we must generalize Lemma 2.6.3. Lemma 5.4.1. Let K be a normal extension ofF and let a : B -> B^ be any isomorphism with a | F = id, where B and B^ are any fields intermediate between F and E. Then a extends to an automorphism ofK, i.e., to an element o/Gal(E/F). Proof. This is an application of Zom's Lemma. Let S = {(D, r) I B c D, r : D -> D, r | B = a } . Order S by (Di, rO < (D2, r2) if Di c D2 and T2 | Di = ti. Then every chain {(D/, r/)} in S has a maximal element (D, r) given by

D = U D,, r(d) = Tiid) ifd e D,. Hence by Zom's Lemma S has a maximal element (EQ, TQ). We claim Eo = E. Suppose not and let a € E, a ^ EQ. Consider its minimum polynomial ma(X) over F. Since ma(X) e F[X], ro(ma(X)) = m^iX). Now consider ma(X) as an element of Eo[X], and let Ei c E be the splitting field ofma(X) over EQ. Note EQ C Ei, as a ^ EQ but a G Ei. Then, by Lemma 2.6.3, there is an automorphism ti of Ei extending to, contradicting the maximality of (EQ, TQ). D Now we generalize Theorem 2.7.14.

5.4 Infinite Galois Extensions

157

Theorem 5.4.2. Let E be an algebraic extension of¥. The following are equivalent: (i) E is a Galois extension of¥. (2) E is a normal and separable extension of¥, (3) Every field B intermediate between F and E with B a finite extension of F is contained in afield D intermediate between E and F that is the splitting field of a separable polynomial in F[Z], or^ equivalently, that is a finite Galois extension ofF. Proof Although this not the logically most economical proof, we shall first show that (1) and (2) are equivalent and then that (3) is equivalent to both of these. First, we show that (1) implies (2). Let a G E. Then a is algebraic over F, so has minimum polynomial ma (X). Let { a i , . . . , a^^} be the roots of ma (X) in E, with Qfi = a. For any a e Gal(E/F), ma(cr(a)) = ainiaict)) = a(0) = 0 so a (a) € {ai, . . . , a ^ } , and thus the action of Gal(E/F) is to permute { a i , . . . , a^}. Consider f{X) = n ? = i ( ^ "" ^i)- This polynomial is invariant under Gal(E/F) so f(X) e F[X], as E is a Galois extension of F. If g(X) e F[Z] is any polynomial with g(a) = 0, then by the same logic g(ai) = 0 for each /, so f{X) divides g(X) in E[X] and hence in F[X], Thus f(X) is irreducible and so f(X) = ma{X). Hence ma(X) is a separable polynomial that splits in E. Next, we show that (2) implies (1). Let a G E, and consider ma(X) e F[X].Thenma(X) = n L i ( ^ - Q f , ) in E[Z] for some {CYI, . . . , a^^} with ai = a. Thus if we let B = F ( a i , . . . , of^), we have that ma(X) = O L i C ^ ~ ^i) in B[X]. Thus we see that B is a finite extension of F and B is the splitting field of a separable polynomial, so B is a Galois extension of F. Hence Fix(Gal(B/F)) = F. Now let a ^ F. Then a ^ Fix(Gal(B/F)), and so there is an automorphism a of B with cr |F = id and with
158

5 Further Topics in Field Theory

Thus D is a Galois extension of F, so there is an automorphism a e Gal(D/F) with a((Y) 7^ a. By Lemma 5.4.1, a extends to an element r G Gal(E/F), and r(a) = a{a) 7^ a, so of ^ Fix(Gal(E/F)). Hence, as or ^ F is arbitrary, we have that Fix(Gal(E/F)) = F, i.e., E is a Galois extension of F. n Corollary 5.4.3. Let E be an algebraic extension of F. The following are equivalent: (1)E is a Galois extension ofF. (2) E is the splitting field of a set of separable polynomials in ¥[X]. Proof Suppose that E is a Galois extension of F. Then, by Theorem 5.4.2, E is normal and separable. Thus for every a G E, ntaiX) G F[X] is a separable polynomial that splits in E[X]. Then E is the splitting field of the set of separable polynomials {moi{X) | a G E}. On the other hand, let E be the splitting field of {/(Z)}/^/, a set of separable polynomials in F[X]. First, observe that any £ G E is contained in the splitting field of a finite subset of {f (X)}/^/. Now let B be any field intermediate between F and E that is a finite extension of F. Then B = F(6i , . . . , € „ ) for some finite set of elements {€1,..., £„} of E, so, by the above observation, B is contained in the splitting field D c E of a finite subset {/i(X),..., fk{X)} of {/;(X)},e/. Then D is the splitting field of the separable polynomial f{X) = f\{X) • • • fuiX) e F[X], so, by Theorem 5.4.2, E is a Galois extension of F. n Corollary 5.4.4. Let Kbe a Galois extension of¥ and let B be any field intermediate between F and E. Then E is a Galois extension ofB. Proof This follows from Theorem 5.4.2. We restate condition (3) of this theorem in order to avoid confusion of notation: Every field A intermediate between F and E with A a finite extension of F is contained in a field D intermediate between E and F that is the splitting field of a separable polynomial. We want to show this remains true when F is replaced by B. Let A be a finite extension of B, so that A = B ( a i , . . . , a«). Let A' = F ( a i , . . . , an). We claim that A' is a finite extension of F. (Of course, if B is a finite extension of F, then A is a finite extension of F and A' c A, so this is clear, but we need to prove this also in case B is not a finite extension of F.) Since A' = F(Qfi) • • • F(a„), in order to prove this it suffices to prove that F(a/) is a finite extension of F, for each i. Thus, fix / and consider F(Qf/). Let rhoa (^) ^ B[Z] be the minimum polynomial of at regarded as an element of B. Let B/ be the field obtained by adjoining the coefficients of mo,. (Z) to F. Then B/ (a) is a finite extension of B/ and B^ is a finite extension of F, so B^ (a/)

5.4 Infinite Galois Extensions

159

is a finite extension of F, and F(a/) c B/ (a/), so F(a^) is a finite extension of F, as claimed. Then, arguing as in the proof of Corollary 5.4.3, A' c D^ c E, where D' is an extension of F that is the splitting field of a separable polynomial f{X) e ¥[X], But A = BA' c BD', and BD' is an extension of B that is the splitting field of the separable polynomial f(X) e B[X], as required. n In order to generalize Theorem 2.8.8, the FTGT, we must introduce a topology on the Galois group. Definition 5.4.5. Let E be an algebraic extension of F, and let G = Gal(E/F). The Krull topology on E is defined as follows: The identity element of G has a neighborhood basis {C/R = Gal(E/B) | B is a finite extension of F}. An element a e G has neighborhood basis {CTUB}. O Remark 5.4.6. We explicitly observe that x eU^if r € (TC/B if and only if r | B = a | B.

and only if r | B = id, and o

Remark 5.4.7. If E is a finite extension of F, then we may let B = E in the above definition and we see that every element cr of G has a neighborhood consisting of a alone. In other words, in this case G has the discrete topology. All of the results in this section remain true in this case, but the topology on G yields no additional information. o Definition 5.4.5 is not complete until we show these sets define a topology on G. In fact we show a bit more. Proposition 5.4.8. The sets {CTC/B} define a topology on G. Under this topology, G is a topological group. Proof. Let [/, = C/R,, / = 1, 2, and let C/3 = C/i n C/2. Let B3 = B1B2. Then B3 is a finite extension of F, and, by Proposition 2.8.14, B3 = Fix(C/i fl U2) = Fix(C/3). Furthermore, it is easy to check that if a ^ C/3, then a | B3 is not the identity. Hence C/3 = Gal(E/B3), i.e., U3 = U^^. This shows that {C/B} form a neighborhood basis of the identity, and hence that {crU^} form a basis for a topology on G. Note that if a € G, then aU^a~^ = Ua{B) and cr(B) is a finite extension of F. Then UBO^ = cr{a~^UBcr) = or[/cr-i(B) so {OTC/B

I O- G G , B / F finite} = {UBCT \ cr e G, B / F , finite}

so, in defining the topology, we may take left cosets, right cosets, or both left and right cosets.

160

5 Further Topics in Field Theory

To show G is a topological group we must show multiplication and inversion are continuous. If m = G x G ^ - G is multiplication, and a r t / is a neighborhoodof o r , thenm~^(crT[/) D (a (r C/r~^)) x (rC/), a neighborhood of (a, r) in G X G, and if /: G —> G is inversion, and a~^U is a neighborhood of or~^ then/~^((j~^t/) = U~^a~^ = C/or~^ a neighborhood of a"^ inG. n Lemma 5.4.9. Let E be a Galois extension ofF and let B be any field intermediate between F and E. Then the Krull topology in Gal(E/B) is the induced topology on Gal(E/B) as a sub space (7/Gal(E/F) with the Krull topology. Proof. Recall that, by Corollary 5.4.4, E is a Galois extension of B. Let U be a neighborhood of the identity in Gal(E/B), so C/ = Gal(E/A) for some finite extension A of B. Then A = B(a) for some a G E, by Corollary 3.5.3. Let D = F(a), so A = BD, and let U = Gal(E/D). Then U = Gal(E/A) = Gal(E/BD) = Gal(E/B) n Gal(E/D) = Gal(E/B) n [/, and, similarly, if f/ is a neighborhood of the identity in Gal(E/F), then U = Gal(E/D) with D a finite extension of F, so U = Gal(E/B) nU = Gal(E/B) n Gal(E/D) = Gal(E/BD) is a neighborhood of the identity in Gal(E/B), as BD is a finite extension ofB. D Now we generalize the Fundamental Theorem of Galois Theory (Theorem 2.8.8) to infinite Galois extensions. Theorem 5.4.10 (Fundamental Theorem of Infinite Galois Theory). Let E be a Galois extension ofF and let G = Gal(E/F) have the Krull topology. (1) There is a 1-1 correspondence between intermediate fields E 2 B 2 F and closed subgroups {1} Q GB ^ G given by B = Fix(GB). (2) B is a normal extension ofF if and only ifG^ is a normal subgroup of G. This is the case if and only ifB is a Galois extension ofF. In this case there is an isomorphism of topological groups Gal(B/F) = G / G B .

5.4 Infinite Galois Extensions

161

Proof. (1) For each closed subgroup H of G, let B^=Fix(//). This gives a map r : {closed subgroups of G} -> {fields intermediate between F and E}. We show r is a 1-1 correspondence. In this case it is easiest to work with the inverse correspondence. Thus, for each field B intermediate between F and E, we let GB = A ( B ) = Gal(E/B). First, we claim that GB is a closed subgroup of G. Let a e GB, the closure of GB- Let y3 G B be arbitrary. Then F(y6) is a finite extension of F, so a has a neighborhood a^ where N = Gal(E/F()6)). Since every neighborhood of cr has a nonempty intersection with GB, we may choose r € GB n aN. Thus r = av for some v e N, and so, in particular, r(^) = (Jvip), Since r e GB, r(fi) = ^, and, by the definition of A^, v(fi) = )6. Then a(fi) = a(v(y6)) = av(^) = r(^) = ^. Since yS G B is arbitrary, we see that a fixes B, Then, since a G GB is arbitrary, we see that GB fixes B. Now by definition GB consists of all automorphisms of E fixing B. Hence GB = GB and so GB is closed. Thus we have a map A: {fields B intermediate between F and E} -^ {closed subgroups of G} and this map is 1-1 since if GB^ = A(Bi) = A(B2) = GB2, then, by Corollary 5.4.4, E is a Galois extension of Bi and of B2, so Bi = Fix(GBi) = Fix(GB,) = B2. It remains to show A is onto. Thus, let / / be a closed subgroup of G and let B = Fix(H), Then K = Gal(E/B) 2 H and we need only show they are equal. Let a e K he arbitrary and consider an arbitrary basic open neighborhood all of a. We claim aUHH j^ 0. Assuming this claim, we have that a is in the closure H of H. Hence, as a G J^ is arbitrary, K c. H, But H is closed, so H = H and hence K = H. To see the claim, let U = Gal(E/A) with A a finite extension of F. By Corollary 5.4.4, E is a Galois extension of B, and, by Lemma 2.3.4, AB is a finite extension of B, so, by Theorem 5.4.2, AB is contained in a field D that is the splitting field of a separable polynomial f(X) G B [ X ] . Thus every element of Gal(E/B) restricts to an element of Gal(E/D), as it must take any

162

5 Further Topics in Field Theory

root of f{X) to a root of / ( X ) , and conversely any element of Gal(D/B) extends to an element (not, in general, a unique element, but that is irrelevant) of Gal(E/B), by Lemma 5.4.1. In other words, restriction to D give a welldefined epimorphism R: K = Gal(E/B) -^ Gal(D/B). Let ao = R(cr) e Gal(D/B). Now B c AB c D and D is a finite Galois extension of B. Let HQ = R{H) c Gal(D/B). Then Fix(ifo) = Fix(H) =B = Fix(Gal(D/B)), so by the Fundamental Theorem of (finite) Galois Theory, Ho = Gal(D/B). Thus there is an element r e H with R(r) = ao = R(cr), i.e., with r | B = a | B, and we see, by Remark 5.4.6, that r e aUB,so all D H 7^ 0, as claimed. (2) We have proved much of this in proving (1). Just as in the finite case, for any a e Gal(E/F), a(B) = Fix(aGBor~^), so if GB is a normal subgroup of G, then a(B) = B for every a e G, while if GB is not a normal subgroup of G, then cr(B) 7^ B for some a G G, and conversely. If B is a normal extension of F, then for every ^ € B, m^(X) e F[X] splits in B. Now for any a e Gal(E/F), ^ ' = a(fi) is a root ofa(m^(X)) = m^(X), so P' e B, and so or(B) = B. On the other hand, if B is not a normal extension of F, then there is some ^ G B such that m^(X) has a root ^^ € E, ^' ^ B, and then there is an element a of Gal(E/F) with a(P) = ^\ by Lemma 5.4.1, and hence a(B)#B. Now if B is normal, then, just as before, R: Gal(E/F) -^ Gal(B/F), given by R(cx) — o I B, is a well-defined epimorphism, whose kernel is clearly GB = Gal(E/B). It remains only to show that the KruU topology on Gal(B/F) is the quotient topology on G / G B Let JJo be a basic neighborhood of the identity in Gal(B/F). Then C/Q = {a G Gal(B/F) | a | A = id} for some finite extension A of F (where of course A c B) and JJo is the image under the quotient map R of the basic neighborhood of the identity [/A ^ Gal(E/F). On the other hand, if C/A is a basic neighborhood of the identity in Gal(E/F), its image under the quotient map /? is C/o = {or G Gal(B/F) | a | B n A = id}, and B fi A c A is a finite extension of F, so f/o is a basic neighborhood of the identity in Gal(B/F). n Proposition 5.4.11. Let Ebe a Galois extension ofF and let B be an intermediate field between F and E. Let G = Gal(E/F). Then: (1) (E/B) is finite if and only iflG^l is finite. In this case, (E/B) = |GB|. (2) (B/F) is finite if and only if[G : GB] is finite. In this case, (B/F) = [G : GB]. (3) GB is always a closed subgroup ofG. GB is an open subgroup ofG if and only if[G : GB] is finite. Proof. (1) This follows immediately from Theorem 2.8.5.

5.4 Infinite Galois Extensions

163

(2)Assume B/F is finite and let B c D with D/F finite Galois (Theorem 5.4.2). Then, by the FTGT, (B/F) = [Gal(D/F) : Gal(D/B)]. But Gal(E/F) -^ Gal(D/F) is an epimorphism so [G : GB] = [Gal(E/F) : Gal(E/B)] = [Gal(D/F) : Gal(D/B)] = (B/F). Assume B/F is infinite and let {^i} be an infinite set of elements of B with F c F(^i) C F(^i, )S2) C • • • C B. By Lemma 5.4.1, for each / = 1, 2 , . . . there is an element or, G Gal(E/F) with a, {^j) = pj for j < i but at (fit) ^ ^/, so we see that a/ 7^ aj for / 7^ 7. Since two elements of Gal(E/F) are in the same left coset of GB if and only if they agree on B, we see that [G : GB] is infinite. (3) The fact that GB is closed is part of Theorem 5.4.10 (1). Suppose [G : GB] is finite and let {ai,... ,ak} be a set of left coset representatives of GB with ai = I. Since GB is closed, the finite union O^2GB U • • (jy^GB is closed and hence its complement (TIGB = GB is open. Suppose GB is open. Then it contains a basic open set UA = Gal(E/A), with A a finite extension of F. Then B = Fix(GB) c Fix(C/A) = A, so (B/F) is finite and hence [G : GB] is finite. D We now further investigate the KruU topology on G. Lemma 5.4.12. Each basic open setaU^ in G is also closed. Proof. Let r ^ crU^. Then 0 = TB fl aC/B (as C/B is a subgroup of G), i.e., ^t/fi ^ G — (JC/B- In other words, the complement ofaU^ contains a neighborhood of each of its points, so is an open set, and hence cr C/B is a closed set. D We now prove a lenmia we will need in describing the topology of Gal(E/F), but one which is interesting in its own right. Lemma 5.4.13. Let E be an algebraic extension ofF and let a : E -^ E be a map of fields with cr | F = id. Then o is an automorphism ofF (and hence a e Gal(E/F)). Proof. Since a (I) = l,crisan injection, so we need only show a is a surjection. Let a € E be arbitrary. Set S = {roots ofma(X) in E} = {^ € E | m^ifi) = 0}. Then for any ^ e S, ma{cf{^)) = a{{ma{^)) = or(0) = 0 so o'(^) e S. Thus a I 5 is an injective map from S to itself. Since S is finite (as any polynomial has only finitely many roots), a | 5 is surjective as well, so there is a ^0 ^ 5" with or(^o) = a, as required. D

164

5 Further Topics in Field Theory

Remark 5,4,14, This lemma is false for arbitrary extensions. For example, let E = F(X), the field of rational functions over F in the variable X, Then a (X) = X^ induces a : E -^ E, an injective, but not surjective, map of fields. o Theorem 5.4.15. Let Ebea Galois extension ofF and let G = Gal(E/F) have the Krull topology. Then G is Hausdorff, totally disconnected, and compact. Proof. We shall prove this in several steps. (1) G is Hausdorff: Since G is a topological group, to show G is Hausdorff it suffices to show that

f]UB = {id} where the intersection is taken over all finite extensions B of F. Let or € G, a 7^ id. Then for some a € E, a(a) ^ a. Let B = F(a). Then or ^ f/B» as required. (2) G is totally disconnected: Let a, r G G, a 7^ r. Then r ^ orf/s for some UB, and then G = af/fi U (G — CTUB) is a union of two relatively open sets with a (resp. r) in the first (resp. second). (CTUB is open by definition and G — af/fi is open as in the proof of Lenmia 5.4.12.) Thus a and r are in different components of G, so the only components of G are single points. (3) G is compact: This will require considerably more work. For each a G E, let /?c. = {^ G E | maifi) = 0}. Let

G = Y\Ra. aeE

By identifying an element OaeE^^a)" ^^^^ / : E -> E by / ( a ) = ^a for each a G E, we may regard ^ as a set of functions from E to itself. (This is identifying a function with its graph.) Note that /?« = {«} for a G F, so every f E G satisfies f(a)=a for every a e ¥. Note also that Ra is finite for each a. We give E the discrete topology, and each Ra the induced topology as a subset of E, which is also the discrete topology. Then R^ is a finite discrete space, so is compact. We give Q the product topology. Then, by Tychonoff's Theorem, Q is the product of compact sets and hence is compact. (Recall that if Z = n ^i is a product of spaces, the product topology has a basis of open sets n ^i where each Yt is open and Yt = Xt for all but finitely many /. If each Xi is discrete, then every subset of Xt is open, so this condition reduces to Yi = Xi for all but finitely many i.) Since every o G Gal(E/F) satisfies maicfipt)) = 0 for every a G E, we may regard G as a subset of Q.

5.4 Infinite Galois Extensions

165

Claim: The induced topology on G as a subset of Q is the KruU topology onG. Proof of claim: Let af/fi be a basic open set in G. Let {^i,..., ^Sj^} be a basis for B over F. Then for r G G, r G aU^ if and only if r(^/) = cr(Pi), i = 1 , . . . , ^. Thus aUB = G HV where V = HaeE ^cx with 5« = /?« for a ^ Pi and 5^. = {cr(Pi)], and so V is an open set in GConversely, let V be a basic open set in G- Then V = YlaeE '^ct with Sa = Ra for all but finitely many a. Since each Ra is finite, we see that V is a finite union of open sets of the form V = YlaeE '^(^ ^^^^ '^ct = Ra for all but finitely many a, say for or 7^ ofi,..., ckf^, and 5^. = {a^-} is a single point for each / = 1 , . . . , A:. Let B = F(Qfi,..., of^), a finite extension of F. If a G G satisfies a (at) = ot[,i = I,... ,k, then G DV = aU^ is open, while if there is no a G G satisfying or(Qf/) = a-, / = 1 , . . . , A:, then G fl V is the empty set, which is also open. Claim: ^ is a Hausdorff space. Proof of claim: Let fi, fi e G with fii^fi. Then /i(ao) # /2(ofo) for some Qfo G E (regarding ^ as a space of functions). Let A\ — flaGE ^ot with Sa = Ra for Of 7^ Qfo and 5«o = {/i (ofo)}, and let A2 = OCGE ^^^ with 5^ = /?« for a 7^ Qfo and 5^0 = {/2(ckfo)}- Then / i G A I , /2 G A2, and Ai and A2 are disjoint open sets. (Since G is a subspace of a Hausdorff space, it is Hausdorff, but for simplicity we chose to prove that directly in step (1). Observe that the proofs that G and G are Hausdorff are very similar. In the same way, we may adapt the proof that G is totally disconnected to prove that G is totally disconnected. However, it is G and not G that we are interested in.) Claim: G is a closed subset of GProof of claim: Let f e G, f ^ G. We need to show / has an open neighborhood V that is disjoint from G. Note every f e G satisfies / | F = id, so if / ^ G, / is not an automorphism of E. If / : E -> E is an injective map of fields, then, by Lemma 5.4.13, / is an automorphism of E, contradicting our assumption. Thus there are two possibilities: (1) / is not an injection; or (2) / is an injection but not a map of fields. In case (1) letQfi, a2 G E with f{pt\) = fipci)^ ot\ 7^ a2. Let V = FlaeE ^^ with Sa, = {/(ai)}, Sa2 = {/(af2)}, and Sa = Ra for a # aua2. Then / G y , y is open, and G fl V is the empty set (as any ^ G G is an injection, so g(ai) ^ gioci))' In case (2) we must have one of the following: f{ai + Qf2) 7^ /(o^i) + /(Qf2) for some ai,Qf2 G E, /(Qfia2) # fipt\)f{ot2) for some a i , a 2 ^ E,

166

5 Further Topics in Field Theory

or / ( a f ^) # f{oii)~^ for some 0 7^ ofi G E. For simplicity assume it is the first of these. Let V = UaeE^a with Sa, = {/(ai)}, 5,,, = {/(a2)}, Sai+a2 = {/(«i + oi2)}, and Sa = Ra foT a y^ a i , ^2, Oil + ci2- Then f e V, V is open, and G DV is the empty set (as any g G G is a map of fields, so g(ai + Qf2) = g(ai) + g{a2). Claim: G is compact in the KruU topology. Proof of claim: We merely need to assemble our previous work. The KruU topology on G is the topology G inherits as a subspace of G- We have shown that ^ is a compact Hausdorff space and that G is a closed subspace of G- But it is a basic topological fact that a closed subspace of a compact Hausdorff space is compact. n Remark 5.4.16, While we do not need this for our purposes, it is worthwhile to put things in a larger context. An important topology on function spaces is the compact-open topology. If J^ = {f: X -^ Y} is the set of continuous functions from X to Y, the compact-open topology on T has as subbases the sets ZA,B = {f e J^ \ / ( A ) c B} where A c X is compact and B c. Y is open. In our case, E is discrete so every / : E ^- E is continuous, A c E is compact if and only if it is finite, and every S c E is open. Now G ^ ^ and it is easy to see that the topology we have defined on G is the restriction of the compact-open topology on T to G, i.e., that this topology is the compact-open topology on G, and then we further see that the KruU topology on G agrees with the compact-open topology on G. o Let {B^} be a set of fields intermediate between E and F. Generalizing Definition 3.4.1, we say that {B/} are disjoint extensions of F if B/ fi D^ = F for each /, where D^ is the composite of the fields {B^} for j ^ i. Then we have the following generalization of Theorem 3.4.7. Theorem 5.4.17. Let E be an extension of F and let {B^ c E} be disjoint Galois extensions ofF with Gt = Gal(B^/F). Let B be the composite o/{BJ and let G = Gal(B/F). Then B is a Galois extension ofF and G = Yit ^t with the product topology. Proof A Zom's Lemma argument allows us to generalize Theorem 3.4.7 and conclude that B is a Galois extension of F and G = Ylt ^t • We show G has the product topology. Let [/ be a basic open neighborhood of the identity in the product topology. Then, by the definition of the product topology, U = Yitei ^i where Ut c G/ is a basic open set and Ut = Gf except for finitely many /. Then, for each i e I,Ui = Gal(B/A^) where A^ is a finite extension ofF, and A^ = F except for finitely many /. Let A be the composite of the {AJ. Then A is a finite

5.5 Exercises

167

extension of F, and U = Ux is a. basic open neighborhood of the identity in the KruU topology on G. Conversely, let [/ be a basic open neighborhood of the identity in the KruU topology on G. Then, by definition, U = UA for some finite extension A of F. Let A/ = A n Bi and note that, since {BJ are disjoint extensions of F, A/ = F for all but finitely many /. Then, setting Uf = Up^., we have that each Ut is open, Ui = Gi for all but finitely many /, and U = Yltei ^i ^^ ^ basic open neighborhood of the identity in the product topology. n Example 5.4.18. Here are two cases of infinite Galois extensions E of F and two subgroups Hi ^ H2 of G = Gal(E/F) with Fix(fl^i) = Fix(//2). (Of course, H\ = H2.) (1) Let J = {—1, 2, 3, 5 , . . . } consist of —1 and the primes. For each j e J,lctBj = Q(V7) C C, and let B be the composite ofthe {By}. Note {By} are disjoint extensions of Q. Let Gj = Gal(By/Q). Note Gj is isomorphic to Z/2Zforeach7, so, by Theorem 5.4.17, G = Gal(B/Q) = Yljeji^/^'^)^^^^ G is uncountable. Hence G has uncountably many subgroups of index 2. But for any subgroup H of index 2, (Fix(7/)/Q) = 2 by Proposition 5.4.11. However, Q has only countably many extensions of degree 2 (as each is obtained by adjoining a root of a quadratic equation to Q, and there are only countably many quadratic equations with rational coefficients). Hence there are two (and indeed, uncountably many) distinct subgroups Hi and H2 of G of index 2 with Fix(Hi) = Fix(H2). (2) Fix a prime p and an algebraic closure F^ of F^. Then F^ is the composite of the fields B„ = Fpn contained in F^. Let G = Gal(Fp/Fp) and let

(Qf) = a^. Let H be the subgroup of G generated by O. Then Fix(//) = F^ = Fix(G). We claim H j^ G.To see this, let A be the composite of the fields B„ where n is a power of 2, and let D be the composite of the fields B„ for n odd. Then, by Lemma 3.4.2, A and D are disjoint extensions of F^, and AD = F^. Hence G = Gal(A/Fp) x Gal(D/Fp). Let I A = OA and 0 | D = 4>D. Then neither <[>A nor OD is the identity map. Now / / is a cyclic group, so H = {0^} = {A*D)- ^^^ ^ ^^^^ ^^^ ^^^ subgroups {^^} and {^} (and uncountably many others), so H ^ G, o

5.5 Exercises Exercise 5.5.1. Let Bi and B2 be finite extensions of F satisfying property "P". Show that B1B2 and Bi n B2 also satisfy property "P". Here "P" is any one of the following: finite separable, finite purely inseparable, finite normal, separable, purely inseparable, normal, Galois. ("P"" finite Galois is handled in Corollary 3.4.8 and in Exercise 3.9.13.)

168

5 Further Topics in Field Theory

Exercise 5.5.2. (a) Let AQ be the field obtained by adjoining all the complex roots of every irreducible polynomial f(X) e Q[X] of odd degree. For i > 0, let Ai be the field obtained from A/_i by adjoining all complex square roots of all elements of A/_i. Show that AQ C Ai C A2 C • • • and that

Q = UA
(b) Show that the conclusion of part (a) remains true if AQ is the field obtained by adjoining a single real root of every irreducible polynomial f(X) e Q[X] of odd degree. Exercise 5.5.3. Let Q^^^^ = {^ e C | m^(X) is solvable by radicals}. LetBo = Q. For / > 0, let B/ be the field obtained from B/_i by adjoining all complex n^^-roots of all elements of B/_i, for all positive integers n. Show that BQ C Bi C B2 C • • • and that

Q-iv^ljB,. Exercise 5.5.4. Recall that Q denotes the field of algebraic numbers. Show that the only nontrivial element of finite order in Gal(Q/Q) is the automorphism of Q given by complex conjugation (an element of order 2). Exercise 5.5.5. Let F = C and let E = C(X). Note that E is not an algebraic extension of F. Let G = {a: E -> E | a | F = id}. Show that Fix(G) = F. (This shows that in Section 5.4, it was necessary to require that E be an algebraic extension of F. Otherwise we would have nonalgebraic "Galois" extensions and the theory developed there would not hold.) Exercise 5.5.6. In the notation of Example 5.4.18 (1), show that the closed subgroups of G of index 2 are in 1-1 correspondence with the nonempty finite subsets of J, Exercise 5.5.7. Explicitly exhibit uncountably many subgroups of the group GofExample 5.4.18 (2).

Some Results from Group Theory

A.l Solvable Groups Definition A.1.1. A composition series for a group G is a sequence of subgroups G = G o 2 G i 2 . . . G A : = {1} with G/ a normal subgroup of G/_i for each /. A group G is solvable if it has a composition series with each quotient Gi/Gi-i abehan. o Example A.1.2. (1) Every abeUan group G is solvable (as we may choose Go = G a n d G i = {1}). (2) If G is the group of Lemma 4.6.1, G = (a, r | a^ = 1, r^~^ = 1, r a r " ^ = <j^} where r is a primitive root mod/?, then G is solvable: G = G O D G I D G 2 = {1} with Gi the cyclic subgroup generated by a. Then Gi is a normal subgroup of G with G/Gi cyclic of order p — 1 and Gi cyclic of order p . (3) If G is the group of Lemma 4.6.3, G = (a, r | a"^ = 1, r^ = 1, rorr"^ = a^) , then G is solvable: G = Go D Gi D G2 = {1} with Gi the cyclic subgroup generated by a. Then Gi is a normal subgroup of G with G/Gi cyclic of order 2 and Gi cyclic of order 4. (4) If G is a /?-group, then G is solvable. This is Lenmia A.2.4 below. (5) If G = Sn, the synmietric group on { 1 , . . . , n}, then G is solvable for n < 4. ^1 is trivial. S2 is abelian. ^3 has the composition series ^3 D A3 D {1}. 54 has the composition series ^4 D A4 D V4 D {1} where V4 = {1, (12)(34), (13)(24), (14)(23)}. (Here An denotes the alternating group.) (6) If G = Sn or G = An, then G is not solvable for n > 5. This is Corollary A.3.6 below. o Lemma A.1.3. A finite group G is solvable if and only if it has a composition series satisfying any one of the following conditions:

170

A Some Results from Group Theory (1) (2) (3) (4)

Gi-i/Gi Gi-i/Gi Gi-i/Gi Gi-i/Gi

is is is is

solvable for each i. abelianfor each i. cyclic for each i. cyclic of prime order for each i.

Proof If Gi-i/Gi is solvable then we may "refine" the original sequence to G 2 • • 2 Gi-i = Ho 2 ^ 1 2 • • • 2 ^m = G/ 2 • • 2 G^ = {1} with Hj normal in Hj-i and Hj-i/Hj abelian; similarly we may refine a sequence with abelian quotients to one with cyclic quotients and one with cyclic quotients to one with quotients cyclic of prime order. On the other hand, if Gt-i/Gt is cyclic of prime order it is certainly solvable. n Lemma A.1.4. (1) Let G be a solvable group and let H be a subgroup of G. Then H is a solvable group, (2) Let G be a solvable group and let N be a normal subgroup ofG. Then G/N is a solvable group, (3) Let G be a group and let N be a normal subgroup ofG.IfN and G/N are solvable, then G is solvable. Proof Consider G = Go 2 Gi 2 • • • 2 G^ c {1}. Then H = HQ ^ Hi ^ • • • 2 ^A: 2 {1} where Hi = H D G/. It is easy to check that Hi is a normal subgroup of Hi-I, and then by standard theorems of group theory Hi_,/Hi

= (Hn

Gi.i)/H

n Gi = (Hn =

Gi-i)/(H

n G,_i) n G,

(//nG,_i)G,/G,CG,_i/G,

is isomorphic to a subgroup of an abelian group so it is abelian. (2) Let 7t:G-^ G/N = g be the quotient map and consider Q = QQ :D . . . 2 e ^ = {1} where Qi = n{Gi), Then Qi = GiN/N = Gi/Gi n N, so 7r(G,_i)/7r(G,) = =

(Gi.iN/N)/iGiN/N) (Gi-i/Gi-inN)/(Gi/GinN)

is isomorphic to a quotient of the abelian group Gi-i/Gi so it is abelian. (3) Let TT: G -^ G/N = Q, Lei N = No ^ -" ^ Nj = {1} and G = Go 2 • • • 2 2A: = {1} be as in Definition A. 1.1. Then the sequence of subgroups G = 7c-\Qo)

2 7t-\Qi)

shows that G is solvable.

2 • • • 2 yr'^Qk)

= iVo 2 M 2 • • • 2 iV^ = {1} n

A. 1 Solvable Groups

171

So far, we have considered abstract solvable groups. Hov^ever, when we consider solvable subgroups of symmetric groups, we can get very precise descriptions. Lemma A.1.5. Let G be a subgroup of Sp of order divisible by p, p a prime. Then any nontrivial normal subgroup KofG has order divisible by p. Proof Since G has order divisible by p, it must contain an element of order p, which must be a /?-cycle. Regard Sp as operating on T = { 1 , . . . , p]. Write T = Ti U • • • U T^, a decomposition onto orbits of K, In other words, /, y, G Tm for some m if and only if there is some x e K with x{i) = j . Now for any i, j e T there is a a e G with or(/) = j . This implies that the action of G permutes { T i , . . . , T^} so in particular each T^ has the same cardinality. This cardinality then divides p, so must be 1 or /?, and cannot be 1 as ^ is nontrivial. Thus we see that K operates transitively on 7 , so has order divisible by p (as the subgroup of K fixing 1, say, has index p). D Lenvma A.1.6. Let G be a solvable subgroup of Sp of order divisible by p, p a prime. Then G contains a unique subgroup N of order p. Proof Let G = Go D Gi D • • D Gjt = {1} as in Definition A.LL We prove the lenmia by induction on k. \fk = 1, G = Go D Gi = {1} and G is solvable, so G is abelian, and hence has a unique p-Sylow subgroup N, of order p. Suppose the lemma is true for A: — L Then Gi is a normal subgroup of G, so applying Lemma A.L5 with A' = Gi we see that G\ has order divisible by p, and then by the inductive hypothesis that G\ contains a unique subgroup N of order p. Now let A^' be any subgroup of G of order p. Then N' and N are both p-Sylow subgroups, so are conjugate. Let A^' = aNa~^. As N c Gi, A^' = aNcr~^ c aGia~^ = Gi as Gi is normal in G, so N^ = N, as required. D Let a group G operate on a set T. This action is effective if the only element a or G with a(t) = ^ for all ^ G T is a = id. Let a group Gi operate on a set Ti, and a group G2 operate on a set T2. These operations are permutation isomorphic if there is an isomorphism of groups xlr: Gi ^ G2 and an isomorphism of sets (i.e., a bijection) * : Ti -> T2 with * ( a ( 0 ) = ^(or)(vl/(0) for every a eGut e Ti. Proposition A.1.7. Let G be a solvable group operating effectively and transitively on a set T of cardinality p, p a prime. Then there is a subgroup H of F* such that this operation is permutation isomorphic to

172

A Some Results from Group Theory

^^ = l [ o i ] ' ^e^'^eFp} operating on

=![;]'••

€F,)

by left multiplication. Proof. As we have observed, in this situation G has order divisible by p. (Let ^ € r be arbitrary and consider [a e G \ a{t) = t]. Since G acts transitively on T, this subgroup of G has index p.) Since G operates effectively on T, G is isomorphic to a subgroup of the symmetric group 5^,, so, by Lemma A. 1.6, G contains a unique subgroup N of order p, which is generated by a p-cycle. Reordering, if necessary, we have that A^ and T are permutation isomorphic to j L . I

| n e F^ I operating on

I / € Fp I by left multiplication, as a suitable generator v of N takes

MV]

mod p. For simplicity of notation, we simply identify N with

this group and T with this set. Since A^ is the unique subgroup of order p of G, it is certainly a normal subgroup of G. Let a € G be arbitrary. Then ava~^ = v^ for some h ^ 0. Then, for any /,

Let

so

^([;])=[o"][i] and hence

heV;, Thus, if H is the subgroup of F* given by

neFpY

A.2 /7-Groups

173

A.=|*^ni[*„?]^4 then G = GH as claimed.

n

Corollary A.1.8. Let G be a subgroup of Spy p a prime, of order divisible by p. The following are equivalent: (1) G is solvable. (2) The order ofG divides p(p — 1). (3) The order ofG is at most p(p — I). Proof (1) implies (2): If G is solvable, then, by Proposition A. 1.7, G = GH for some H and then \G\ divides p(p — I). (2) implies (3): Trivial. (3) implies (1): G has a /7-Sylow subgroup A^, and N is unique, as if G had another /?-Sylow subgroup N\ then G would have order at least p^. Then the last part of the proof of Proposition A. 1.7 shows that G = GH for some H, But GH ::> N D {1} with A^ normal in G. N is abeUan and GH/N is isomorphic to H which is abelian, so G is solvable. n

A.2 p-Groups Throughout this section, p denotes a prime. Definition A.2.1. A group G is a /? — group if |G| is a power of /?.

o

The center Z(G) of a group G is defined by Z(G) = {a e G \ at = ra for every r e G}. Lemma A.2.2. Let G be a p-group. Then G has a nontrivial center (i.e., Z{G) ^ {1}). Proof Write G = Ci U • U Q , a union of conjugacy classes. If at e Ct, then IQI = [G : Hi] where Hi = {r e G \ r a / t - ^ = a,}. If a, e Z(G), then Q = {a/} and |C/| = 1 (and Hi = G). If a/ ^ Z(G), then /f/ is a proper subgroup of G, so IC/1 = [G^ : Hi ] is a positive power of p. Of course | G | is a positive power of p. But one of the conjugacy classes, say Ci, consists of {id} alone, and |Ci| = 1 is not divisible by p . Now \G\ = \Ci\-\ h | Q | . Hence there must exist other classes Ci with |Q | not divisible by p, and hence with \Ci I = 1, and if Q = {a,} then a, e Z(G). D

174

A Some Results from Group Theory

Corollary A.2.3. Let G be a p-groupy \G\ = p^. Then there is a sequence of subgroups G = Go D Gi D - - D G „ = {1} with Gt a normal subgroup ofG of index p\ for each i = 1, . . . , n. Proof By induction on n. If n = 1, the result is trivial. Now suppose the result is true for all groups of order p^~^, and let G be an arbitrary group of order p^. By Lemma A.2.2, the center Z(G) of G is nontrivial. Since |Z(G)| divides |G|, |Z(G)| is also a power of p, so, in particular, Z(G) contains an element of order p. The cyclic subgroup H of G generated by that element is a subgroup of G of order p, and, since H c Z(G), H is a normal subgroup of G. Let Q = G/H he the quotient group, and let TT: G ^^ 2 be the canonical projection. Now 2 is a group of order p^~^, sohy the inductive hypothesis there is a sequence of subgroups Q = Qo D Qi D - " D Qn-i = {1} with Qt a normal subgroup of Q of index p^ for each / = 1 , . . . , n — 1. Let Gt = 7t~^(Qi) for / = 1 , . . . , n - 1 and G„ = {1}. Then G = Go D Gi D • • D G„ = {1} form a sequence as claimed. (Clearly Gt has index p^. Also, Gt is a normal subgroup of G as if go e Gt and g e G, then ggog~^ e Gt as n(ggog~^) = ^(g)^(go)(^(8))~^ ^ Qi as Qi is a normal subgroup of Q.) n Lemma A.2.4. Let G be a p-group. Then G is solvable. Proof By Corollary A.2.3, there is a sequence of subgroups G = Go D Gi Z) • • • D G„ = {1} with Gi a normal subgroup of G, and hence certainly a normal subgroup of G^_i, for each /. Furthermore, since [G : Gt] = p^ for each /, [Gi-i : Gi] = p and hence |G^_i/G/| = p, and so G/_i/G/ is cyclic of order p and, in particular, is abelian, for each /. Hence, by Definition A. 1.1, G is solvable. n

A.3 Symmetric and Alternating Groups In this section we prove several results about symmetric and alternating groups. We let 5„ be the symmetric group on { 1 , . . . , n} and we let A„ c 5„ be the alternating group. We regard the elements of 5„ as functions on this set and recall that functions are composed by applying the rightmost function first. Thus, for example, (1 2) (1 3) = (1 3 2). We recall that every element of Sn can be written as a product of disjoint cycles, and that disjoint cycles conmiute. The first two results we prove allow us to conclude that certain subgroups of Sn are in fact equal to 5^.

A. 3 Symmetric and Alternating Groups

175

Lemma A.3.1. Let p be a prime and let G be a transitive subgroup of Sp that contains a transposition. Then G = Sp. Proof. By renumbering if necessary, we may assume the transposition is r = (12). Since G acts transitively on { 1 , . . . , p}, it has order divisible by p and hence it has an element OTQ of order p. Since p is a prime, ao is a p-cycle. Thus there is some power a = a^ of (JQ with a (I) = 2. By renumbering if necessary, we may then assume a = (1 2 - - - p). Now direct calculation shows that aJra-^

= ( 1 2 - . . pY{I 2)(1 2 • • • p)"^' = ((j + 1) {j + 2))

for 7 = 0 , . . . , / ? — 2. Direct calculation then shows that (2 3)(12)(2 3) = (13), (3 4)(13)(3 4) = (14),

i(p - 1) p){l (p - imip

-l)p)

= (1 p)

and furthermore that (l/:)(l7)(l/:) = 0/:)forany;7^/:. Thus G contains every transposition. But Sp is generated by transpositions, so G = Sp. n Lemma A.3.2. Let G be a transitive subgroup of Sn that contains an (n — l)cycle and a transposition. Then G = Sn. Proof. Let p be the n — 1-cycle. By renumbering if necessary, we may assume that p = (12 " ' (n — I)). Then the transposition is r = (/ j) for some / and j . Since G operates transitively on {I,... ,n}, there is an element y of G with y(j) = n. Then y{i) = k for some k. Direct calculation shows that yry~^ = (k n) = Xk and then that p^XkP~^ = ((k + i) n) = Xk+t, where A: + / is taken mod n — \. Finally, direct calculation shows that XiXjxr^ = (i j) for any / ^ j . Thus G contains every transposition. But Sn is generated by transpositions, so G = 5„. D

176

A Some Results from Group Theory

Next we show some results on the structure of A„ and 5„. We first need the following technical lemma. Lemma A.3.3. IfG is a normal subgroup of An that contains a S-cycle^ then Proof. We shall first show that if G contains a single 3-cycle, then it contains every 3-cycle. Note that if G contains a 3-cycle (/ j k), it contains {i j k)~^ = {kji). By renumbering if necessary, we may assume that G contains the 3-cycle (1 2 3). Then for each / > 3 (observing that an element of order 2 is its own inverse) G also contains the elements ((12)(3 0)(3 2 1)((12)(3 0 ) = (12 0 , ((13)(2 0 ) ( 1 2 3 ) ( ( 1 3 ) ( 2 / ) ) = : ( 1 3 0 , ((2 3)(10)(3 2 1)((2 3)(10) = (2 3/); then for each distinct i, j > 3, G also contains the elements (12 7 ) ( 2 / l ) = ( l / 7 ) , (2 3 7 ) ( 3 / 2 ) = ( 2 / 7 ) , (3 1 7)(1 / 3) = (3 / 7); finally, for each distinct i, j,k > 3, G also contains the elements (kil)(lij)

=

(ijk),

so G contains all 3-cycles, as claimed. But we now claim that any element of A„ can be written as a product of 3cycles. Since any element of An can be written as a product of an even number of transpositions, to show this it suffices to show that the product of any two transpositions can be written as a product of 3-cycles, and we see this from (/ j)(ik) (ij)(kl) completing the proof.

= =

(ikj), (ikj)(iki), n

Theorem A.3.4. For n > 5, A„ is a simple group (Le,y it has no normal subgroups other than An and {1}).

A. 3 Symmetric and Alternating Groups

177

Proof. Let n > 5 and let G 7^ {1} be a normal subgroup of A„. We will show By Lemma A.3.3, it suffices to show that G contains a 3-cycle, so that is what we shall do. Let a € G, or 7^ L Then a has order m > L Let /? be a prime dividing m and let y = cr^^^^ Then y has order p. Now y can be written as a product of disjoint cycles, y = y^ • • • y^, and since y has order /?, each of y i , . . . , yyt is a p-cycle. Now we must consider several cases: Case \: p > 5: By renumbering if necessary, we may assume that yi = (1 2 3 4 5 6 7 . . . /7). Let n = (1 2)(3 4). Then fix = xmT7^ = ( 1 4 3 5 6 7 • • • /? 2) and yi^f ^ = ( 1 3 5 4 2 ) . Letting ^ = Xiyr~^ and using the fact that disjoint cycles commute, it easily follows that y^~^ = ( 1 3 5 4 2). Let T2 = (1 3)(2 4). Then r2(l 3 5 4 2)x2^ = ( 1 5 2 4 3 ) and (13 5 42)(1 5 2 4 3) = (145). Case II: /? = 3: If A: = 1, i.e., there is only a single 3-cycle, there is nothing to do, so assume ^ > 2. By renumbering if necessary, we may assume that yi = (1 2 3) and y2 = (4 5 6). Let yi2 = yiy2 = (1 2 3)(4 5 6). Let n = (1 2)(3 4). Then Sn = xiynt^^ = (14 2)(3 5 6) and yi25-^ = ( 1 3 4 2 5 ) . Letting 8 = r i y r f \ it then follows, as in Case I, that yS~^ = ( 1 3 4 2 5). Thus G contains a 5-cycle and we are back in Case I. Case III: p = 2: Since we are in A„, A: is even, so A: > 2. By renumbering if necessary, we may assume that yi = (12) and y2 = (3 4). Let yi2 = YiYi = (1 2)(3 4). Let p = (1 2 3). Then ^n = pynP'^ = d 4)(2 3) and yi2^^2^ = (1 3)(2 4). Letting ^ = pyp~^, it then follows, as in Case I, that yf-^ = 6> = (1 3)(2 4). Let rs = (1 3)(2 5). (Note here is where we need n > 5.) Then tser-^ = 6' = (I 3)(4 5). But then 66' = (2 4 5). D Theorem A.3.5. For n > 5, the only normal subgroups of Sn are 5„, A„, and {!}. Proof Let n > 5 and let G 7^ {1} be a normal subgroup of 5^. We will show G ^ An. Since An is a subgroup of Sn of index 2, this shows that G = An or Sn.

The proof of this is almost identical to the proof of Theorem A.3.4. We begin in the same way. Let a e G,a ^ I, Then a has order m > 1. Let p be a prime dividing m and let y = a"^^^. Then y has order p. If we are in Case I, Case II, or Case III with A: > 2 of the proof of Theorem A.3.4, we conclude in exactly the same way that G 2 A„. This leaves only the case where p = 2 and k = I, i.e., where y is a single transposition. But all transpositions in 5„ are conjugate, and Sn is generated by transpositions, so in this case G = S^. D

178

A Some Results from Group Theory

Corollary A.3.6. For n > 5, neither A„ nor Sn is solvable. Proof. Let n > 5. By the definition of a solvable group (Definition A. 1.1), if G = A„ or Sn were solvable, it would have a composition series with abelian quotients. But by Theorem A.3.4, the only possible composition series for A„ is A„ C {1}, and by Theorem A.3.5, the only possible composition series for Sn are Sn C {1} and Sn^ AnZ:> {1}, and A„ is not abelian. n

B A Lemma on Constructing Fields

In Section 2.2 we showed: If fiX) e F[X] is an irreducible polynomial, then ^[X]/(f(X)) is a field. We did so by using arguments particular to polynomials. In this appendix we shall show that this result follow from more general results in ring theory. In particular, in this section we show how to construct fields from integral domains. Definition B.1.1. Let R be an integral domain. (1) An ideal / C /? is a maximal ideal if / c y c /?, 7 an ideal, implies J = IorJ

= R,

(2) An ideal / c /? is a prime ideal if ab e / , with a,b e /?, implies aelorbeL o Lemma B.1.2. (1) Let R be an integral domain. Then every maximal ideal is prime. (2) Let R be a PID (principal ideal domain). Then every prime ideal is maximal. Proof. (1) Let / be a maximal ideal and let ab = i e I. If a e / , there is nothing to prove. Otherwise, let / = {a, I) (the ideal generated by a and / ) . Then I C J and / is maximal, so / = /?. In particular, 1 € / . Then xa + /' = 1 for some x e R and i^ e I. Then b = l-b = ixa + i')b = X{ab) + i'b = iX + i'b e L (2) Suppose / is a prime ideal and / is an ideal with I c. J, Then I = {p) for some p e R and / = {q) for some q e R. Since I c.J,peJ so p = qr for some r e R. Since / is prime, we conclude that: (a) ^ e / in which case J = {q) c. I so J = I,or

180

B A Lemma on Constructing Fields

(b) r € / so r = pr^ for some r' e R'm which case p = qr = qpr' = piqr') so 1 = qr^ and hence 1 € 7, so / = /?. D Lemma B.1.3. Let R be an integral domain and I d R an ideal. Then I is maximal if and only if R/I is afield. Proof Let n : R ^^ R/I be the canonical projection. First suppose / is maximal. Let a e R/I, a 7^ 0. Then a = 7t(a) for some a e R, a ^ I. Since / is maximal, R = {a, I), so 1 = jca + / for some x e R and some / e I. Then, if x = 7t(x), 1 = 7r(l) = n{xa + i) = xa + 0 = xd, so a is invertible. Now suppose / is not maximal, and let / c / C /?. Let j e J, j ^ I, and let j = n(j). We claim j is not invertible. For suppose jk = 1, 7, k e R/I, and let k = 7t(k), Then jk = I + i for some / € / , so 1 = jk — i and hence 1 G / , a contradiction. D Corollary B.1.4. Let R be a PID and let I = (i) be the ideal generated by the element i of R, Then R/I is afield if and only ifi is irreducible. Proof. In a PID, an element / is irreducible if and only if / is prime, and an ideal / = (/} is a prime ideal if and only if the element / is prime, so the corollary follows directly from Lemma B. 1.2 and Lemma B.1.3. n Corollary B.1.5. L^^ f{X) F[X]/{f{X)) is afield.

e F[Z] be an irreducible polynomial. Then

Proof. Since F[X] is a PID, this is a special case of Corollary B.1.4.

n

A Lemma from Elementary Number Theory

In this appendix, we prove a lemma that is used in the proof of Theorem 4.7.1. Lemma C.1.1. Let p be a prime and let t be a positive integer. Then there are infinitely many primes congruent to \(moA.p^). Proof. We begin by recalling the following definition and facts: Let q be a prime and let a be an integer relatively prime to q. Then ord^(fl), the order of a(mod^), is defined to be the smallest positive integer x such that a^ = l(modqr). Then, for an arbitrary integer y, a^ = l(mod^) if and only if ord^(a) divides y. Also, since, by Fermat's Little Theorem, a^~^ = l(mod^), we have that ord^(a) divides q — I. We now proceed with the proof, which we divide into two cases: p = 2 and p odd. Case I: p = 2. For a nonnegative integer k, let Fk = 2^^ + 1 . Note that, if j < it, Fj = 2^' + 1 divides 2^' - 1 = (2^' )^'"' - 1 = F)t - 2, so gcd(F^, Fk) = gcd(Fj, 2) = 1, i.e., {Fi, F 2 , . . . } are pairwise relatively prime. Thus, if qk is a prime factor of Fk, then {^1, ^2, • •} are all distinct. Now let ^ = qs for s > ^ — 1. By definition, q divides Fy, i.e., 2^' + 1 = O(mod^), 2^' = - l ( m o d ^ ) , and hence 2^""' = l(mod^). Thus ord^(2) divides 2^"*"^ but does not divide 2^, so ord^(2) = 2^"'"^ Then, as we have observed above, 2*+^ divides ^ — 1, so ^ = l(mod2^+^) and hence q = l(mod20. Thus we see that {qt-.i,qt,...} are distinct primes congruent to l(mod20. Case II: p odd. In this case we proceed by contradiction. Assume there are only finitely many primes = l(modp^), and let these bcqi,,.. ,qr. Let a = 2qx'"qr,

c = a^"',

N = a^' - 1 = c^ - I,

182

C A Lemma from Elementary Number Theory

and write N = (c — 1)M where M = CP~^ -\ hi. Claim: c — \ and M are relatively prime. Proof of claim: M = CP-\ + • • • + 1 = (c^-^ - 1) + (c^-^ - ! ) + ••• + (c — 1) +p, and each term except possibly the last one is divisible by c — 1, so gcd(c — 1, M) = gcd(c — 1, p) = I or p. But a = 2(mod/?) so c = 2P'~\modp), By Fermat's Little Theorem, 2^"^ = l(modp), and p^~^ = l(modp — 1), so c = 2^ = 2(modp) and so c — 1 = l(mod/7). Thus p does not divide c — 1 and so gcd(c — I, M) = gcd(c — I, p) = I. Now let q be any prime dividing M. Then q divides A^ = c^ — 1, so cP = l(mod^), i.e., a^' = l(mod^). So ord^(a) divides p\ i.e., ord^(a) = 1, / 7 , . . . , p^~^ or /7^ But if ord^(fl) = 1, / ? , . . . , or /?^~\ then a^' = l(mod^), i.e., c = l(mod^), so q divides c — 1, contradicting the fact that c — 1 and M are relatively prime. Hence ord^(a) = pK Then, as we have observed above, p^ divides q — I, i.e., q = l(modp^). But certainly q does not divides a, while each of qi,... ,qr does divide a, so q ^ qi,... ,qr, contradicting the hypothesis that qi,... ,qr are all the primes congruent to l(mod/?0- Hence there must be infinitely many such primes. D

Index

17-gon

101

Abel 95 adjoining 17 algebraic integer 42 algebraic number 18, 20 algebraic number field 20 algebraically closed 155 algorithm for computing Galois groups 124 algorithm for factoring polynomials 88 Artin 30 Artin-Schreier extension 80 basis normal, 49 character 29 characteristic 9 closure algebraic, 151, 154 Galois, 148 normal, 148 separable, 143, 145 conjugate Galois, 27 degree 9, 16, 17, 19, 23, 27, 58, 59, 113, 114,144 inseparable, 145

separable, 145 Dirichlet 29,119 discriminant 125 division algorithm 11 domain Euclidean, 11 integral, 10, 179 principal ideal, 11, 179 duplicating the cube 99 Eisenstein's Criterion 87 element algebraic, 18, 21 inseparable, 25 irreducible, 180 primitive, 47, 63, 64 purely inseparable, 143, 144 separable, 25, 141 extension 9 abelian,70, 118, 119 algebraic, 18, 20, 21 composite, 166 finite, 9 Galois, 27, 28, 32, 35, 58, 59, 61, 157, 158, 166, 167 inseparable, 25, 52 nonsimple, 66 normal, 25, 26, 28, 32, 145, 147, 148, 157

184

Index

purely inseparable, 143-145 quadratic, 104 radical, 108 separable, 25, 26, 28, 51, 52, 139, 141, 142, 157 simple, 47, 63, 64 symmetric, 118, 119 extension by radicals see radical extension extensions conjugate, 34 disjoint, 58-60, 92, 166 Fermat prime 100,101 Fermat's Little Theorem 10 field 7, 13, 180 algebraically closed, 151, 154 composite, 16 cyclotomic, 89, 104, 108, 110, 119 extension, 14, 15 finite, 53, 54 fixed, 61 Kummer, 73, 108 perfect, 26, 53 quadratic, 107, 108 splitting, 22, 24-26, 28, 147, 157, 158 Frobenius map 10, 53, 54, 139 FTGT see Fundamental Theorem of Galois Theory function synmietric, 45 Fundamental Theorem of Algebra 155 Fundamental Theorem of Galois Theory 2, 27, 32, 37, 55, 160 Fundamental Theorem of Infinite Galois Theory 160 Galois 95 Gauss 101, 108 Gauss's Lenmia 86 gcd see greatest common divisor greatest common divisor 11,15 group abelian, 118 alternating, 174,176

cyclic, 12 Galois, 1, 26, 32, 36, 37,45, 55, 58, 61,62,70-72,74,92,108,109, 111,113,116,118,119,122,123, 164, 167 nonsolvable, 178 /7-group, 173, 174 permutation, 35 simple, 176 solvable, 94, 169-171, 173, 174 symmetric, 119, 123, 171, 173-175, 177 topological, 159 Hubert's Theorem 90

77

ideal maximal, 179, 180 prime, 179 inseparability degree of, 146 inseparability, degree of 143 Invariance of gcd under Field Extensions 15 Kronecker 14,88 Kronecker-Weber Theorem

119

Lagrange Interpolation Formula Landau 91 Lindemann 99 Mobius Inversion Formula

88

57, 93

n-powerless 71, 72, 110 Newton's identities 83 norm 76 normal basis 66, 69 polynomial cyclotomic, 87, 89, 90 elementary symmetric, 46 inseparable, 25, 52 irreducible, 56, 86-88, 90, 112,180 minimum, 18, 19, 27, 62, 63, 76, 147

Index primitive, 86 radical, 93, 108 separable, 25, 26, 28, 51, 52, 141, 142, 157,158 symmetric, 45 polynomial ring 11 radical extension

93

solvable by radicals 93-95 not, 95, 96 splits 22 squaring the circle 99

straightedge and compass

97, 99, 101

Theorem on Natural Irrationalities 154 topology compact-open, 166 Krull, 159, 163, 164, 166 product, 166 trace 76 transformation, Hnear 19 trisecting the angle 99 van der Waerden

119

185

60,