Maximal Orders (London Mathematical Society Monographs New Series)

(1.2)

char. pol.K

IT hi (X). i= 1

On the other hand, let f(X) E K[ X]; then f(
min. pol.K

From (1.2) and (1.3) we have at once 0.4) THEOREM. Let

f(X) = min. pol.K
g(X) = char. pol.K <po

Then f(X) divides g(X) in K [X]; furthermore, f(X) and g(X) have the same irreducible factors in K[X], apart from multiplicities. Finally, if h(X)EK[X], then h(
Now let A be a finite dimensional K-algebra. Each ct E A determines a K-linear transformation ctL on A, where ct L is the left multiplication x ....... LXX, X EA. Clearly (1.5)

for all r, s E K, ct, {3 EA. Further, ct L = 0 if and only if ct = 0, since A has a unity element. Therefore the map ct ....... ctL , ct E A, is a K-algebra isomorphism of A into the ring HomK(A, A) of K-linear transformations on A. Now define (1.6) min. pol.K ct = min. pol.K ctL '

char. pol.A/K ct = char. pol.K ct L .

For h(X) E K[ X] we have h(ct L) = h(ct)L' and thus h(ct) = 0 if and only if h(ct L) = O. This shows that min. pol.K ct is the unique monic polynomial f(X) E K[ X], of least degree, such that f(ct) = O. For later use, we record explicitly the following consequence of (1.4): (1.7) THEOREM. Let A be a finite dimensional K -algebra. For ct E A, let

f(X) = min. pol.K ct,

g(X) = char. POl.A/K ct.

Then f(X) is the unique monic polynomial in K[X], of least degree, such that f(ct) = 0; it has the property that f(X) Ih(X) for each h(X) E K[ X] such that

MINIMUM AND CHAR.POL.; NORM, TRACE

(1.8)

3

h(rJ.) = O. Further, f(X) 1 g(X), and the polynomials f(X), g(X) have the same irreducible factors in K[ X], apart from multiplicities.

A remark about notation seems called for. If B is a K-algebra containing A, then each rJ. E A also lies in B. Viewing rJ. as element of B does not affect the calculation of min. pol.K rJ., and so we have suppressed the subscript A in this notation. On the other hand, char. pol.AIK rJ. certainly depends on the choice of A, since it must be computed by letting rJ. act on a K-basis for A; indeed, char. POl.AIK rJ. #- char. pol.BIK rJ. if B #- A, since char. pol.AIK rJ. has degree (A: K). Suppose now that A =

f: Ku;, and let

rJ. E

A. We may write

i= 1 m

rJ.'

1

u.J = "L.,'J a ..u., i= 1

I

~j ~

m.

Then char. pol.K rJ. = det (bijX - a i)

(1.8)

=

xm -

(TAlK rJ.)X m- 1

+ ... + (-It NAIK rJ..

We call TAlK the trace map, and N AIK the norm map. Clearly

TAlK

rJ.

= trace of rJ.L ,

so from (1.5) we deduce

T(rrJ. + sf3) = rT(rJ.) + sT(f3) { N(rJ.f3) = N(rJ.)· N(f3), N(rrJ.) = rmN(rJ.),

(l.9) for r, s E K,

rJ.,

f3 E A. Here, m = (A:K).

lb Integral elements Now let R be an integral domain with quotient field K, and let A be a ·finite dimensional K-algebra. An element rJ. E A is integral over R if f(rJ.) = 0 for some monic polynomial f(X) E R[ Xl (1.10) THEOREM. For an element

rJ.EA, the following three conditions are equivalent: (i) rJ. is integral over R. (ii) R[ rJ.] is a finitely generated R-module. (iii) rJ. is contained in some subring B of A, such that B is a finitely generated R-module.

4

INTEGRAL CLOSURE

(1.11 )

Proof If (i) holds, then there is a relation aiER. It is then obvious that

R[a] = R

+ Ra+

...

+ Ran-I,

a finit~ly generated R-module. Thus (i) implies (ii). Clearly (ii) implies (iii), by choosmgB = R[a]. To prove that (iii) implies (i), let us write n

B=

I 1 Rf3"

i=

Since B is a ring, and a E B, it follows that each af3i E B. Therefore we have n

I

af3 i =

aijf3j,

j=l

We may rewrite these equations as n

I

(aD ij - aij )f3 j

=

0,

1

~

i

~

n.

j=l

°

Let d = det (e«\ - au) E B. By Cramer's Formula, it follows that d . f3 j = for 1 ~ j ~ n. Therefore d . B = 0, and hence d = 0, since B is a ring with unity element. This shows that a is a zero of the monic polynomial det (X Dij - ai) E R[ X], and thus a is integral over R. We have now shown that (iii) implies (i), and the theorem is proved. (1.11) COROLLARY. Let IX, f3 E A be such that af3 = f3a. If both a and f3 are integral over R, then so are a ± f3 and af3.

Proof Since a and f3 are integral over R, by (1.10) we may write k

R[a] =

I Ru" ;=1

m

R[f3] =

I 1 Rv ..

j=

J

Since af3 = f3a, every element of R[a] commutes with every element of R[f3], and therefore R[a, f3] = I. . Ru;vJ.. ',J

Thus R[a, f3] is a finitely generated R-module, and is a subring of A. It follows from (1.10) that every element of R[a, f3] is integral over R, and hence the corollary is established.

Remarks. (i) A slightly simpler proof can be given when R is a noetherian

(1.12)

5

INTEGRAL CLOSURE

domain (see section 2a). Keep the above notation, and let x E R[ a, 13]. Then R[x] is an R-submodule of the finitely generated R-module R[a, P]. Hence R[x] is also finitely generated as R-module, and thus x is integral over R by (1.10). (ii) If a and p are integral elements which do not commute, it may well occur that a + Pis not integral. For example, let R = Z, K = Q, A = M 2(Q) (the algebra of all 2 x 2 matrices with entries in Q). Let

a= (

°0 1.)~,

Both a and f3 satisfy the equation X 2 = 0, and so are integral over Z. However, min. pol.Q(a + (J) = X 2 - ¢ Z[X].

±

Therefore a

+ Pis not integral over Z, by (1.14) below. Ie Integral closure

The integral closure of R in A is the set of all elements of A which are integral over R. If A is a commutative K-algebra, then by (1.11) the integral closure of R in A is a subring of A. If A is not commutative, the integral closure need not be a ring. (1.12) DEFINITION. The integral domain R is integrally closed if the integral closure of R in its quotient field K coincides with R. Thus, R is integrally closed if for a E K, f(a)

=

0, f(X)

E

R[X], f(X) monic

==>-

a E R.

(1.13) GAUSS' LEMMA. Let R be an integrally closed domain with quotient field K. Let f(X) E R[X] be a monic polynomial, and suppose that f(X)

=

g(X)' h(X),

where g(X) and h(X) are monic polynomials in K[Xj. Then both g(X) and h(X) lie in R[X]. Proof. Let K' be a finite extension of K which is a splitting field for f(X), and write f(X) = (X - Xi ),

n

Let R' denote the integral closure of RinK'. Since f(X) f(x i ) = for each i, it follows that each Xi E R'. In K'[ X] we have

°

g(X) . h(X)

=

n (X -

Xi)'

E

R[X] is monic, and

6

INTEGRAL CLOSURE

(1.14)

whence both g(X) and h(X) are expressible as partial products IT(X - xJ. But R' is a ring by (1.11), so both g(X) and h(X) have coefficients in R'. The coefficients of g(X) and h(X) are thus in R' n K. However, R' n K = R, because R is integrally closed. Thus both g(X) and h(X) lie in R[X], and the lemma is proved. In order to determine whether an element rx E A is integral over R, we must in theory consider all monic polynomials f(X) E K[X] such that f(rx) = 0, and then we must see whether any such f(X) lies in R[X]. In practice, however, this task is considerably simplified by use of the following result. (1.14) THEOREM. Let R be an integrally closed domain with quotient field K,

and let A be a finite dimensional K -algebra. An element rx E A is integral over R if and only if min. pol.K rx E R[X]. Proof. If min. pol.K rx E R[ X], it is clear that rx is integral over R. Conversely, let g(X) = min. pol.K ex, and assume that rx is integral over R. Then f(rx) = 0 for some monic polynomial f(X) E R[ X]. Therefore f(X) = g(X) . h(X) for some monic polynomial h(X) E K[X]. It then follows from Gauss' Lemma that g(X) ER[X], as desired. We shall conclude this section by listing some examples of integrally closed domains. Proofs are available in standard references such as Bourbaki [4, Ch. 5, § 1] and Zariski-Samuel [1, Ch. V, § 3]. (1.15) Examples of integrally closed domains. Every principal ideal domain is integrally closed. Every unique factorization domain is integrally closed. Every Dedekind domain is integrally closed (see §4a). Let R be a domain with quotient field K, and let K' be any extension field of K. Then the integral closure of R in K' is always integrally closed. (e) Let S be any multiplicative subset of R (see § 3a). If R is integrally closed, so is the ring of quotients S-l R. (f) Let K[ Xl' ... , Xn] be the polynomial domain in n indeterminates over a field K (or more generally, over a unique factorization domain K). Then K[x 1 , .•. , xn] is also a unique factorization domain, and thus is integrally closed by (b). (a) (b) (c) (d)

7

EXERCISES

EXERCISES 1. Keep the notation and hypotheses of(I.14). Show that an element 11. E A is integral over R if and only if char. pol.AIK 11. E R[ X]. [Hint: Use (1. 7) and (1.14).]

2. Let A be a finite dimensional K-algebra, and let L be any extension field of K. Each element a E A gives rise to an element 1 @ a of the L-algebra L @K A. Prove that char. pol. L0 KA,L 1 @ a In other words, char. to L.

POI.A/K

char. pol.

=

AIK

a.

a is not affected by the change of ground field from K

3. Let E ::J K be fields, and let (E: K) = n. Each 11. E E corresponds to a matrix Mn(K), obtained by letting 11. act as left multiplication on a K-basis of E. Show by induction on r that for each matrix (lJ. i ) E Mr FJ. E

det(FJ.ij)

=

{det(lJ.i ) } ·

NE/K

4. Let L be a finite galois extension field of K, with galois group G. Show that for aEL, char. pol.L/K a

f1

=

(X - a"),

"eG

and hence

"

",,' £..., a,

TLIK a =

NL/Ka =

f1

a".

ueG

O"EG

[Hint: Let n

f(X)= min. pol.K a

f1

=

(X - a;l,

i=l

Then

f1

(X - a") E K[ X], hence is a power of f(X). Then compare degrees of

ueG

char. POI.L/K a and

f1

(X - aU).]


5. Keep the notation of Exercise 3, and let A be a finite dimensional E-algebra. Prove that for a E A,

[Hint: Let A = L'Ex i ,

aX j =

L

lJ. ij X i ,

lJ.ij E E.

Relative to a suitable K-basis of A, the K-linear transformation a L on A corresponds to the matrix (FJ. ij ). Hence TA/Ka = trace of (iiij) =

L trace of fl.ii

A similar result holds for norms, by Exercise 3.J

8

(2.1)

HOMOLOGICAL ALGEBRA

2. HOMOLOGICAL ALGEBRA This section contains a brief review of several topics from homological algebra. In most cases, proofs will be omitted. For details, we refer the reader to standard references such as Cartan-Eilenberg [1], Hilton-Stammbach [1], Jaos [1], Northcott [1], Rotman [1 J. We shall use the same terminology and notation as the last of these references. 2a Modules: bomomorpbisms, pullbacks, pusbouts, tensor products, chain conditions

Throughout this section let A be a ring, not necessarily commutative, having a unity element 1. All A-modules are left A-modules, unless otherwise specified, and each A-module M is assumed unital (that is, 1· m = m, mE M). Given a pair of A-modules Land M, denote by HomA(L, M) the additive group consisting of all A-homomorphisms from L into M. For each qJ E HomA(L, M), let ker qJ denote the kernel of qJ, im qJ the image of qJ, and cok qJ the cokernel of qJ, that is, cok qJ = M/qJ(L). When there is no danger of confusion, we shall write Hom instead of HomA. A sequence of A-modules and A-homomorphisms (2.1)

X1

1,

~

X2

j~

----fo'

X3~

f"_1 • .. -----+

X PI

is exact at Xi if ker 1; = im 1; _ 1 . The sequence is exact if it is exact at each place, that is, ker 1; = im1;_l'

2~i~n-1.

An exact sequence of A-modules and A-homomorphisms will sometimes be called a A-exact sequence. A short exact sequence is an exact sequence of the form (2.2)

0 - L-LM ~ N----+ O.

Exactness of (2.2) is equivalent to the following three conditions: f is monic, ker 9 = imf, 9 is epic. In this case, 9 induces an isomorphism M /f(L) ~ N. The short exact sequence (2.2) is split if f(L) is a direct summand of M. This is equivalent to either of the following: (i) There exists an hE HomA (M, L) such that h . f = 1L' where 1L denotes the identity map on L. (ii) There exists an hi E HomA (N, M) such that g' hi = IN' A diagram of A-modules and A-homomorphisms

9

MODULES

(2.3)

L

---.L. M

~ N1 9

is commutative if 9 . f = h. The same terminology applies to more complicated diagrams. Given a A-homomorphism f:L ...... M, and an arbitrary A-module X, there are two maps induced by f, namely f*: Hom(X, L) ...... Hom(X, M), { f*:Hom(M,X) ...... Hom(L,X).

(2.3)

These are defined by the commutative diagrams X ---'L.. L

r~

i f

M,

L

II

.

~ .p

M~ X.

Note that Hom A (X, L), Hom A (X, M), etc., are additive groups (but not usually A-modules), and the maps f*, f* are additive homomorphisms. The map f* goes "in the same direction" as f, namely from L to M. Since f* arises from a change in the second variable occurring in Hom, we say that Hom is covariant in the second variable. Similarly, f* goes "in the opposite direction" from f. Since f* arises from a change in the first variable in Hom, we call Hom contravariant in the first variable. If f: L ...... M and g: M ...... N are A-homomorphisms, then (2.4)

(gf)* = f* . g* .

(Contravariance reverses order of composition of mappings!) The concepts of covariance and contravariance are extremely useful in working with modules. Thus, for example, let ~ be another ring, and let M = AMtJ. be a (A, ~)-bimodule, that is, M is a left A-module and right ~-module, with the actions of A and ~ on M commuting: (},m)b

If L = AL is a left A-module, then HomA (M, L) can be given the structure of a ~-module, because of the bimodule structure on M. Since M is a right ~-module, and M appears in a contravariant position, it is automatically true that HomA(M,L) must be a left ~-module. To be explicit, for each f E Hom A (M, L) and each b E~, we define (bf)m = f(mb).

This is the only "natural" definition, and it must use the fact that M is a right

10

(2.5)

HOMOLOGICAL ALGEBRA

.1-module. The hypothesis that M be a (A, .1)-bimodule is needed to verify that of is again a A-homomorphism. Analogously, HomA (L, M) can be made into a right .1-module, by setting (go)x = g(x) . 0,

XEL.

We state without proof a theorem describing the effect of applying HomA(X,') or HomA (', Y) to A-exact sequences.

(2.5)

THEOREM.

(i) For each A-module X and each A-exact sequence O - L ~M~N,

the sequence of additive groups

is exact. (ii) For each A-module Yand each A-exact sequence L ~ M -----i.... N

---+

0,

the sequence of additive groups

is exact. It is sometimes convenient to use the concepts of pushouts and pullbacks of pairs of maps. Given a pair of A-homomorphisms J; : A -- Mi' i = 1,2, we define a A-module X, called the pushout of the pair (f1 ' f2 ), by the formula X =(M t -i-M 2 )/{(fla,

-f2 a):aEA}.

Thus X is obtained from the external direct sum M 1 -i- M 2 by identifying the image of A in M 1 with the image of A in M 2 . There is a commutative diagram (called a pushout diagram or fibre sum)

A~M

./21

I;,

M2 ----g;-+ X

where gi is defined by composition: Mi -- M1 -i- M2 -- X, i = 1,2. The pushout X is characterized up to isomorphism by a "universal mapping property", as follows: given any commutative diagram of A-modules and A-homomorphisms

11

MODULES

(2.5)

there is a unique A-homomorphism t{1:X ....... X' such that t{1gi = g;, i = 1,2. Dually, given A-homomorphisms f, : Mi ....... B, i = 1,2, we can form their pullback (or fibre product) Y

=

{(m 1 , m 2

) E Ml

-l-

M2

:f1ml = f 2 m2

}·

There is a commutative pullback diagram

where gi (m 1 ' m 2 ) = m i , i = 1,2. The pullback Y is also characterized by a universal mapping property: given any commutative uiagram

there is a unique ({J: Y' ....... Y such that g; = gi({J, i = 1,2. Next, we recall some facts about tensor products (see Curtis-Reiner [1, §12], for example). Given a right A-module L and a left A-module M (notation: LA' AM), we may form their tensor product L ®A M. This is an additive group, but not usually a A-module. The tensor product L ®A M is covariant in each variable: given A-homomorphisms f:L 1 ....... L 2, g: M 1 ....... M 2 ' there is an additive homomorphism

f®g:L 1 ®A M l

.......

L 2®A M 2·

Consequently, if A is another ring with unity element, and if I1LA is a bimodule, then for each AM we can make L ®A M into a left A-module in a natural way. The action of an element bE A on L ® A M is given by

In particular, if A is commutative and L, M are viewed as two-sided A-modules, then L ®A M is also a A-module. The effect of applying X ® A . or . ® A Y to exact sequences is described in the following result, stated without proof:

12

(2.6)

HOMOLOGICAL ALGEBRA

(2.6) THEOREM. (i) Given a right A-module X and an exact sequence of left A-modules L ----> M ----> N ----> 0, there is an exact sequence of additive groups

X ® AL

X ® AM

---->

X ® AN

---->

---->

0.

(ii) Given A Y and an exact sequence of right A-modules A there is an exact sequence of additive groups A ®AY

---->

B ® AY

C ®AY

---->

---->

---->

B

---->

C

---->

0,

0.

Next we list a number of natural isomorphisms involving Hom and ®. In most cases there is no difficulty in writing down explicitly the maps which yield these isomorphisms, and we shall leave this task to the reader. It will also be clear from the context whether the modules which occur are left A-modules or right A-modules, and we shall suppress such information in some of the statements below. (2.7)

THEOREM.

(i) Given AM, there is a left A-isomorphism Hom A (A, M)

~

M,

obtained by mapping each f E HomA (A, M) onto f(1). The left A-structure of Hom A (A, M) arises from the A-A-bimodule structure of A. (ii) For each index set I, HomA

(Io M;,

N) ~ IT

HomA (M;, N),

ieI

iEI

Hom A (M,

IT N;) ~ IT Hom

A

(M, N;).

ieI

iEI

Choosing I to be the finite set {I, ... , k}, it follows as special cases of these formulas that Hom commutes with finite direct sums in either variable: HomA

ct:

HomA ( M, (2.8)

THEOREM.

J: it: ~ it Mi , N)

N)

~

Hom A (M;, N), Hom A

(M, N;).

(i) Given AM, there is a left A-isomorphism A®AM~M,

obtained by mapping I A; ® m; E A ® A M onto I A;m; EM. The left A-structure of A ® A M arises from the A-A-bimodule structure of A. (ii) Tensor product commutes with direct sums:

(2.9)

MODULES

2::" M;) ( leI

®A N

M ®A(I"N;) ;eI

13

~ 2::" (M; ®A N), iEI

~ I"

eM ®A N )'

;eI

for any index set I. Finally we give some results on chain conditions. A A-module M is noetherian if the submodules of M satisfy the ascending chain condition (ACC.), that is, if every ascending chain of sub modules terminates The ring A is left noetherian if the left ideals of A satisfy the A.e.C The following result is used repeatedly (see Curtis-Reiner [1, § 11 ]): (2.9) THEOREM. For a ring A, the following are equivalent: (i) A is left noetherian.

(ii) Every finitely generated left A-module is noetherian. (iii) Every submodule of a finitely generated left A-module is finitely generated. (iv) Every nonempty collection of submodules of a finitely generated left A-module contains a maximal element. (2.1 0) COROLLARY. Let R be a commutative noetherian ring, and let A be an

R-algebra which is finitely generated as R-module. Then A is left and right noetherian as a ring. Proof. Every left ideal of A is an R-submodule of the finitely generated R-module A. Since R is noetherian, it follows from (2.9 (ii)) that the R-submodules of A satisfy the Ae.e. Hence the left ideals of A satisfy the Ae.e., so A is left noetherian. Likewise A is right noetherian. Next, a A-module M is artinian if its sub modules satisfy the descending chain condition (ne.e.), that is, if every descending chain of submodules terminates. The ring A is left artinian if its left ideals satisfy the D.e.C We have (see Curtis-Reiner [1, § 11]) (2.11) THEOREM. The following statements are equivalent: (i) A is left artinian.

(ii) Every finitely generated left A-module is artinian. (iii) Every nonempty collection of submodules of a finitely generated A-module contains a minimal element. Just as in the preceding discussion, we obtain (2.12) COROLLARY. Let R be a commutative art in ian ring, and let A be an

14

HOMOLOGICAL ALGEBRA

(2.13)

R-algebra which is finitely generated as R-module. Then A is left and right artinian as a ring.

2b Functors and categories In order to simplify the remaining discussion of homological algebra, and also for later use in the chapter on Morita equivalence, it is convenient to introduce some of the concepts of category theory. For simplicity, we shall restrict our attention to categories of modules. (For a more detailed account of category theory, see MacLane [1 J.) If A is a ring with unity element, let vitA denote the category of right A-modules. The objects of vitA are right A-modules, and for each L, M E vitA, there is a set of morphisms (or maps) from L to M, namely HomA (L, M). Let d b{a category with objects A, A', ... , and :JI a category with objects B, B ', ... A (covariant) functor F: d ....co:JI carries each A Ed onto an object FA E!!J, and carries each IX E Hom,j (A, A') onto a map FIX E Hom£l6' (FA, FA'), in such a way as to preserve identity maps and compositions. We assume always that our functors are additive, that is, F(IXl + IX 2) = FIXl + FIX 2 . A contravariant functor F: d....co fJI carries A Ed onto FA E f!J, and carries each IX E HomAA, A') onto Fa E Homgj'(F A', FA); such an F reverses composition of maps. A covariant functor F is right exact if it carries each exact sequence Al -;. A 2 ....co A3 -> 0 in d onto an exact sequence FAt ....co FA2 -;. FA3 -;. 0 in f!J. For example, if d6- denotes the category of abelian groups, and X A is some fixed right A~module, there is a functor F:AJit-> $6, given by M ....co X ®A M, and IX -;. 1 ® IX. We often write F = X ®A'. Theorem 2.6 is just the assertion that F is a right exact functor. On the other hand, let AX be a fixed left A-module, and define F:AJII -;. $6by M -;. HomA (X, M), ME A.H. We must also describe the action of F on maps; if IX E HomA(M, M '), we set FIX = IX* (see (2.3». Then F is covariant, and (2.5) asserts that F is a left exact functor. We often write F = HomA(X, '). One can also define right- or left-exactness of contravariant functors. We shall not give the definition here, but remark only that the functor F:AJII -;. $6-, given by F = HomA (', X), is a contravariant functor which is left exact (by (2.5)). A functor F: d -;. f!J is exact if F is both left and right exact. We state without proof: (2.13) THEOREM. Let F: d -;. f!J be a covariant functor. The following conditions are equivalent: (i) F is an exact Junctor. (ii) F preserves exactness of short exact sequences, that is, if

(2.14)

PROJECTIVE AND FLAT MODULES

15

o

-? Al -? A2 -? A3 ---> 0 is an d'-exact sequence, then 0---> F Al ---> F A2 ---> F A3--->0 is an exact sequence in f!4. (iii) F preserves exactness of sequences, that is, if A' -? A -+ A" IS d -exact, then F A' -? FA ---> FA" is !!.I-exact.

2c Projective and flat modules

A (left) A-module F is said to have a free A-basis .~ xJ, where rx ranges over some index set, ifthere exist elements x~ E F such that each x E F is expressible as a finite sum x = LC~Xa with uniquely determined coefficients ca E A. Such modules F are called free A-modules. Equivalently, F is free if and only if F is isomorphic to a direct sum of copies of the left A-module AA. Every A-module M is a homomorphic image of some free module F. Indeed, choose any set of elements ma E M such that M = LAma, and let F be free with basis {xo:}; then define an epimorphism ({): F --'> M by letting ({)(x~) = mo:. Note that if M is finitely generated as A-module, then F may be chosen free with a finite basis. A projective A-module is a direct summand of a free module. For example, if A = J 1 EB J 2 , where each J i is a left ideal of A, then J 1 and J 2 are projective left A-modules. Obviously, every free module is projective. Let {MJ be any family of A-modules. It is clear from the definition of projectivity that L' Mi is projective if and only if each M, is projective. As we shall see, projective modules play a basic role in homological algebra, and in other areas of algebra as well. We state without proof (see references listed at the beginning of §2): 2.14) THEOREM. Let P be any A-module. The following statements are equivalent: (i) P is projective. (ii) Every short exact sequence 0 ---> X ---> Y --'> P -? 0 must split. (iii) Given any diagram with exact bottom row P

;/ jl X It Y

--'>

0,

there exists a A-homomorphism h such that f = gh. (iv) Given any epimorphism g:X ---> Y, the induced map g*:HomA(P,X)

is also an epimorphism.

--->

HomA(P, Y)

16

HOMOLOGICAL ALGEBRA

(2.15)

(v) Hom"..(P,·) is an exact functor.

Remark. There is a concept of injective modules, dual to that of projective modules: a A-module L is injective if every short exact sequence o -> L -> X -> Y -> 0 is necessarily split. The analogue of (2.14) holds true. We shall seldom need to use injective modules in this book.

A right A-module X is flat if X ®A' preserves exactness, that is, if for every exact sequence of left A-modules f 9 N L->M-> ,

the sequence of additive groups X®ALl!PJ. X®AM

1

is also exact. We may rephrase this definition as follows: the module X A determines a covariant functor F: A..;/{ -> dlJ as in §2(b), by setting F(M) = X ®AM, M EA..;/{' Then X is flat if and only if F is an exact functor. We have already remarked in (2.6) that F is always right exact, and the only question to be decided is whether F is also left exact. Hence we have (2.15) THEOREM. A module X A is flat if and only if X (8) A' preserves monomorphisms, that is, for each monomorphi!>m f:L -> M of left A-modules, the additive homomorphism

is also monic.

(2.16)

COROLLARY.

Every projective module is flat.

Proof The right A-module AA is flat, because of the isomorphism given in (2.8(i)). Since tensor product commutes with direct sum (see (2.8)), it follows that every free right A-module is flat. Now let X Abe projective; then X is a direct summand of some free right A-moduleF, and we may writeF = X EB X' for some module X'. Let f:L -> M be a monomorphism of left A-modules. Then there is a commutative diagram

X'®A L

11x- 0 f X' ®A L.

Since F is flat, IF ®f is monic. Hence also Ix ®f is monic, and therefore X is flat, as claimed. (2.17) Remark. The techni que of the preceding proof can be appli ed to many

(2.18)

PROJECTIVE AND FLAT MODULES

17

similar situations. In order to prove certain types of theorems about projective modules, one first treats the case of a free module on one generator, then the case of an arbitrary free module, and finally the case of projective modules. Of course, one needs to know that the functors which occur commute with the operation of forming direct sums. In many arguments, it suffices to restrict attention to finite direct sums. (2.18) Remark. We shall show in §3c that every ring of quotients R' of a commutative ring R is a flat R-module. From (2.15) it then follows that for each inclusion of R-modules L c M, there is an inclusion of R'-modules R' ®R L c R' ®R M. Further, there is an R' -isomorphism R' ®R(M/L) ~ (R' ®RM)/(R' ®R L ),

a result to be used repeatedly. Next we note (2.19) THEOREM. A right A-module X is flat ~f every finitely generated submodule of X isflat. Proof. See Rotman [1, Corollary 3.31]' The converse of this theorem is false.

Using this, we prove (see also Exercise 5.3): (2.20) THEOREM. Let R be a principal ideal domain. Then every torsionfreet R-module is flat. Proof Submodules of torsionfree modules are also torsionfree. Hence by (2.19), it suffices to prove that every finitely generated torsionfree R-module X is flat. But by the Structure Theorem for modules over principal ideal domains (see Curtis-Reiner [1, §16]), each such X is R-free. Hence X is R-flat, by (2.16), and the theorem is proved.

As a matter offact, the converse of (2.20) also holds: if X is a flat R-module, then X must be torsionfree (see Rotman [1, Exercise 3.11 ]). A flat right A -module X is faithfully flat if for each left A -module L, the equality X ® /\ L = 0 implies that L = O. To test for faithful flatness, we may use the result: (2.21) THEOREM. Let X/\ be flat. The following statements are equivalent: (i) X is faithfully flat. (ii) A sequence of left A-modules is exact if and only ~f it becomes exact after applying X ®/\. to it. tAn R-module X is torsionfree iffor each nonzero x

E

X and each nonzero r E R, also rx '" O.

18

HOMOLOGICAL ALGEBRAS

(2.22)

(iii) For each left A-homomorphism f: L --+ M, the map 1 ® f: X ®A L --+ X ®A M is the zero map if and only iff = O. (iv) For each maximal left ideal m of A, we have X =1= Xm. Proof See Bourbaki [2, §3, no. 1J.

(2.22) COROLLARYt. Let R be a principal ideal domain, P a prime ideal of R, and R* the P-adic completion of R. Then R* is a flat R-module. If R is a discrete valuation ring, then R* is fait~fully flat as R-module. Proof (See §4e for the relevant definitions.) There is an embedding R --+ R* which makes R* into an R-module. Then R* is R-torsionfree, since R* is also an integral domain. Therefore R* is R-flat, by (2.20). Now suppose that R is a discrete valuation ring, with unique maximal ideal P. Then PR* is the unique maximal ideal of R*, so R* # PR*. It follows from (2.21 (iv)) that R* is faithfully flat, as claimed. COROLLARY. Let R* be the completion of a discrete valuation ring R. For each left R-module M let M* = R* ®RM, a left R*-module. Then R* is faithfully flat as R-module. This means (i) For every exact sequence of R-modules

(2.23)

o --+ L

--+

M

--+

N

--+

0,

the sequence ofR*-modules 0--+ L*

is also exact, and (ii) L* = 0 if and only

--+

M*

--+

N*

--+

0

if L = O.

As a matter of fact, the results of (2.23) hold in a more general situation: given any commutative noetherian ring R having a unique maximal ideal P, we may form the P-adic completion R* of R. It is then true that R* is a faithfully flat R-module (see Bourbaki [3]).

2d Extensions of modules

We shall first give an informal discussion of Ext~ (N, L) in terms of equivalence classes of extensions of a left A-module N by a left A-module L. Then we shall define Ext~ in terms of projective resolutions, and shall list standard properties of Ext. In what follows, all modules are left A -modules, and we shall write Hom instead of Hom" for convenience.

t See also Exercises 5.3, 5.4.

(2.24)

EXTENSIONS OF MODULES

19

A A-exact sequence (2.24) is called an extension of N by L; sometimes we refer to the module X itself as such an extension. Another extension 0 --+ L --+ X' --+ N --+ 0 is equivalent to that in (2.24) if there exists a A-isomorphism e making the following diagram commutative:

where IL' IN are identity maps. This defines an equivalence relation among extensions of N by L. Caution: it may happen that X ~ X' even though the extensions are inequivalent. We shall introduce an additive group Extl(N, L) whose elements are in one-to-one correspondence with the equivalence classes of extensions of N by L. To begin with, choose an epimorphism cp: P --+ N, where P is Aprojective (we could even choose P A-free, if desired), and let K = ker cpo This gives a A-exact sequence

O---->K

i/J

---->

rp

P---->N ---->0,

with IjI the inclusion map. Keeping the A-module L fixed, let (J' range over the elements of Hom(K, L). For each such J, let X" denote the pushout of the pair of maps IjI:K ----> P, (J':K ----> L (see §2a). It follows readily from Exercise 2.2 that there is a commutative diagram with exact rows:

O---->K~P~N---->O o---->l---->l---->tN---->o. " Thus each (J' E Hom(K, L) yields an extension X" of N by L. The mapljl induces a map 1jI* : Hom(P, L) ----> Hom(K, L). For each T E imljl*, it is easily verified that the extension XO'+< is equivalent to X". Hence, each element of the factor group Hom(K, L)/im 1jI* determines an equivalence class of extensions of N by L. Conversely, we show that every class can be obtained in this manner. Given any extension (2.24), it follows from the fact that P is projective that there exists a homomorphism p making the diagram below commutative:

20

(2.25)

HOMOLOGICAL ALGEBRA

Clearly p induces a rna p (J: K ---+ L, and it can be verified that the extension X is equivalent to X". One also checks that the class of X uniquely determines (J modulo the image of t/J*. Hence there is a bijection between the set of equivalence classes of extensions of N by L, and the additive group given by (2.25)

Ext~(N,

L) = Hom(K, L)/t/J*{Hom(P, L)}.

The split extension corresponds to 0 in this group. (One can define an additive structure on the set of equivalence classes of extensions, by using "Baer sums", as in Rotman [1 J. In that case, the above bijection is in fact an additive homomorphism, and the expression (2.25) defines Ext up to isomorphism, regardless of the choice of P and cpo We shall not go into further details of this matter, however, but pass at OIice to the definition of Ext" by means of projective resolutions.) Given a left A -module N, chose an epimorphism CPo: F 0 ---+ N, with F 0 free. Then choose an epimorphism CP1: F 1 ---+ ker CPo' with F 1 free, and so on. This yields a free resolution of N, that is, a A -exact sequence

.. ·---+F 2 ~F (~F ~N---+O 10 in which each Fi is A-free. It is often more convenient to use a more general concept: a projective resolution of N is an exact sequence (2.26) in which each Pi is A-projective. (If it happens that ker CPm is projective for some m, then the process of forming such a resolution can be terminated, giving a finite resolution 0---+ ker CPm ---+ Pm ---+ P m-1 ---+ ... ---+ Po ---+ N ---+ O. In this case, we say that N has finite homological dimension.) Given a projective resolution (2.26), the subsequence '1',

'I'

P: ",-+P 2 --->-P 1 -4Po-+O is called a deleted projective resolution of N; it is an exact sequence except possibly at Po . Now let L be a A-module, and form the sequence of additive groups Hom(P, L): 0 ---+ Hom(P 0' L) '€.4 Hom(P l ' L) ~ Hom(P 2' L) ---+ ... , where (as usual) Hom is an abbreviation for Hom A . By (2.4) we have

(CPi+ 1)* . (cp)* = (CPi' CPi+ 1)* = 0,

i

~

1,

and therefore im CP{ c ker CP{ + l' Now define (2.27)

Ext~ (N, L)

= ker cp! + l/im cp!,

n ~ l.

This yields a sequence of additive groups Ext~(N, L), Ext~(N, L), ... , the first of which coincides with that given in (2.25).

(2.28)

21

EXTENSIONS OF MODULES

(2.28) Remarks. (i) Up to isomorphism, the group Ext~(N,L) depends only upon the modules Nand L, and not upon the choice of the projective resolution (2.26). (See references.) (ii) We could have defined Ext~(N, L) to be ker
PI '1'-4 P [) '1'~ N -> 0

implies the exactness of the sequence 0-> Hom(N,L)

cp*

~

(,0*

Hom(Po,L)4 Hom(P1,L),

by (2.5(ii)). Therefore ker
~

im
~

Hom (N, L).

It is therefore customary to identify ExtO(N, L) with Hom(N, L).

(iii) The groups Extn(N, L) may also be computed by keeping N fixed, and using an injective resolution of L (see references). (iv) The groups Extn(N, L), n ~ 1, have many of the same properties as Hom(N, L). For instance, Extn is additive in each variable:

+ N 2, L) ~ Extn(Nl'L) -1- Extn(N2' L), Extn(N, Ll + L 2) ~ Ext"(N, L 1) -1- Ext"(N, L2)' Extn(Nl

Further, Ext" is contravariant in the first variable, and covariant in the second variable. Thus, each fE Hom(N, N') induces additive homomorphisms

fn*: Ext"(N', L) -> Ext"(N, L).

n

~

l.

Likewise, each g E Hom(L, 1.:) induces additive homomorphisms n

~

1.

(v) If N is a projective module, then 0 -> N -> N -> 0 gives a projective resolution of N. It follows at once from Definition (2.27) that Ext"(N, L) = 0 for all L and all n ~ l. Conversely, it may be shown that if Ext1(N, L) = 0 for all L, then N is projective. One of the most important properties of the groups Ext" is as follows (see references) : (2.29) THEOREM. (Long exact sequence for Ext). Let

O->Al.Bi.C->O be a short exact sequence of A-modules, and let L be any A-module. Then there is a long exact sequence of extension groups

22

(2.30)

HOMOLOGICAL ALGEBRA

0--+ Hom(C, L) ~ Hom(B, L) L; Hom(A, L) ~ Extl(C, L) ~ Ext1(B, L) C Ext1(A, L)

a-4 Ext 2 (C, L)

g* --+

Ext 2 (B, L)

j*

"-+

Ext 2 (A, L) a-4 ....

(2.30) Remarks. (i) We have written Hom, Ext instead of HomA , ExtA , for convenience. The subscripts on the various maps g*,f* have also been omitted; in accordance with the notation of (2.28(iv)), we should have denoted by g: the map Ext"(C, L) --+ Ext"(B, L) induced by g. (ii) It is instructive to compare Theorem 2.29 with (2.5(ii)); we now see that (2.5(ii)) gives the start of the long exact sequence of groups occurring in (2.29). Furthennore, we can describe the homomorphism 80 explicitly, provided we view Extl(C, L) as the set of equivalence classes of extensions of C by L. To be precise, given any h E Hom(A, L), we define 80 (h) to be the equivalence class of the extension occurring in the bottom row of the commutative diagram

Here, X represents the pushout of the pair of maps (f, h), and the dotted arrows denote maps defined as in the discussion preceding (2.25). (iii) It is somewhat more difficult to describe the additive homomorphisms 81' 8 2 "" explicitly. One procedure is to define them recursively, using the dimension-shifting techniques given in (iv) below. Another method uses the fact (2.28(iii)) that we can compute all of the groups Ext"(·, L) by starting with an injective resolution of L; in this context, the long exact sequence of (2.29) arises naturally from an exact triangle associated with an exact sequence of graded complexes. (iv) Formula 2.25 now appears in a more natural setting. From the short exact sequence o --+ K '4 P ~ N --+ 0, P projective, we obtain a long exact sequence of additive groups Hom(P, L) 't:. Hom(K, L) --+ Ext1(N, L) --+ Ext1(P, L) --+ ....

But Ext1(P, L) = 0 by (2.28v), and so we obtain an isomorphism between the expressions given in (2.25). Furthermore, since Ext"(P, L) = 0, n ~ 1, we obtain an isomorphism Ext"+ l(N, L) ~ Ext"(K, L),

n

~

1,

for each module L. By means of this formula, the calculation ofExt"+ 1 may

(2.31)

EXTENSIONS OF MODULES

23

be reduced to that of Ext". This procedure is called dimension shifting. A more general version is as follows: Given a A-exact sequence (2.31) where Pi is projective, 0 ~ i ~ n, then for each A-module L and for each m ~ 1, there is an isomorphism of additive groups Extm+n+ l(N, L) ~ Extm(K", L).

(2.32)

(v) Let N be a A-module, and let o~ALB!4.C~O

be a A-exact sequence. An analogue of Theorem 2.29 states that there is a long exact sequence of extension groups

o ~ Hom(N, A) ~ Hom(N, B) ~ Hom(N, C) ..!4 Ext 1 (N, A) ~ Ext 1 (N, B)~ Ext 1 (N, C)..!4 Ext2(N, A) ~ .... We shall not attempt to describe the maps a occurring here. Another basic property of extension groups is as follows (see references): (2.33) THEOREM (Ladder theorem). Given a commutative diagram of A-modules and A-homomorphisms with exact rows: O~A~B~C~O

! 1 1

o ~ A' ~ B' ~ C' ~ 0, and given any A-module L, there is a commutative diagram of additive groups and homomorphisms with exact rows: O~Hom(C,

t

L) ~ Hom(B, L) ~ Hom(A, L) ~ Ext 1 (C, L) ~ Ext 1 (B, L) ~ ... t t t t

O~Hom(C', L)~Hom(B', L)~Hom(A', L)~Extl(C', L)~Extl(B', L)~ ....

We conclude with some results on finite generation of extension groups. Let R be a commutative ring, and let A be an R-algebra. Then A itself, and all A-modules, can be viewed as R-modules. If L, M, ... denote A-modules, then the additive groups HomA(L, M), L ® AM, Ext~(L, M) (n ~ 1) are also R-modules, and all of the various maps so far mentioned in this section are R -homomorphisms. In particular, suppose that R is a noetherian commutative ring, and that A is finitely generated as R-module. For any A-module N which is finitely generated over R, all of the A-modules occurring in a projective resolution of N may be chosen finitely generated over R. This readily implies

24

(2.34)

HOMOLOGICAL ALGEBRA

(2.34) THEOREM. Let A be an R-algebra, finitely generated as module over the commutative noetherian ring R. Let L, M, .. , denote A-modules which are finitely generated over R (or equivalently, over A). Then the R-modules Ext~(L,

HomA(L, M),

M),

n

~

1,

are also finitely generated over R.

2e Change of rings in Hom and Ext

Throughout this section let R be a commutative ring with unity element. Suppose for the moment that R' is a ring of quotients of R with respect to some multiplicative subset of R (see§ 3a). For each R-module M, we may form the R'-module M'

=

R'@RM.

It is often necessary to know how Hom and Ext behave under passage from R to R'. The aim of this section is to establish formulas such as R @R HomR(M, N)

~

HomR.(M', N')

and likewise for Ext, assuming that the R-modules M and N satisfy some mild hypotheses. A significant role is played by the fact that R' is R-flat (see §2c). We shall prove formulas of the above type in somewhat more general circumstances, which will in fact be met later on. Let Nr) denote the left A-module which is the external direct sum of r copies of A If M is any finitely generated left A-module, there is an epimorphism CfJo: A(r) -+ M for some r. If A is left noetherian, then by (2.9) ker CfJo is also finitely genera ted, and so there is an epimorphism CfJ 1: A(s) -+ ker CfJo for some s. Thus we obtain a A-exact sequence (2.35) Even if A is not necessarily noetherian, it may happen that the module M occurs in some sequence of this type, with r, s finite. In such case, we say that M is a finitely presented A-module. We emphasize that when A is left noetherian, every finitely generated left A-module is necessarily finitely presented. Now let A and r be R-algebras and let M, N, ... be left A-modules. Set

so M' is a left module over the R-algebra N. There exists an R-homomorphism (2.36)

Ct.:

r

@RHomA(M,N)

-+

HomA' (M', N'),

given by 'yEr,

f

E

Hom A(M, N),

(2.37)

25

CHANGE OF RINGS IN HOM AND EXT

where 'rr denotes the right multiplication by y acting on r. By virtue of the bimodule structures rr Rand N(M')r ' both sides of (2.36) have the structure of left r -modules. It is easily checked that IX is a left r -homomorphism. Analogously, the bimodule structures rr.R and N(N')r make the expressions in (2.36) into_right r-modules, and IX is also a right r-homomorphism.

r be R-algebras, M and N left A-modules. Suppose that r is R-flat, and that M is finitely generated as A-module. Then the map IX in (2.36) is a two-sided r -monomorphism. (2.37) THEOREM. Let A and

Proof. There is a A-exact sequence

o -+ K

-+ A (r) -+

At

-+ 0

for some r. It follows from (2.5) that the sequence of R-modules

o -+ HomA(M, N) -+ Hom A(A

(r),

N)

is also exact. Since r is R-flat, exactness is preserved when the functor r @R' is applied to the above two sequences. We obtain two new exact sequences: o -+ K' -+ A,(r) -+ M' -+ 0, and 0-+ r@R HomA(M, N) -+ r @R HomA(Nr), N). By virtue of (2.5), the first of these sequences gives rise to an exact sequence

o -+ HomA' (M', N') -+ HomA' (A'(r), N'). Therefore we obtain a commutative diagram with exact rows: 0-+

r@RHomA(M,N)

0-+

HomA' (M', N') -

-+

1

r@RHomA(Nr),N)

1~1

a

HomA' (A'(r), N'),

in which the vertical maps IX, IX! are precisely those given by (2.36). But an isomorphism, since (2.7) gives r @R HomA(A~), N) It follows at once that

IX

~

r @R N(r)

~

(N')(r)

~

IX!

is

HomN(A'(r), N').

is monic, and the theorem is proved.

A variant of the preceding result is as follows:

r be R-algebras, M and N left A-modules. Suppose that r is R-flat, and that M is a finitely presented A-module. Then the map IX in (2.36) is a two-sided r-isomorphism.

(2.38) Theorem. Let A and

26

(2.39)

HOMOLOGICAL ALGEBRA

Proof There is a A-exact sequence A(s) ----> A(r) ----> M ----> 0 for some rand s. By (2.5), the sequence 0----> HomA(M, N) ----> HomA(A(r), N) ----> HomA(A(S), N) is R-exact. As in the proof of (2.37), we obtain a commutative diagram with exact rows: 0----> n8)RHomA(M, N) ----> r ®RHomA(A(r), N) ----> r® RHomA(A(S), N)

1

1

la,

a,

a

0----> Hom".(M', N') -

HomN(A'(r), N') --HomA,(A'(S), N').

As in (2.37), both maps a l and a 2 are isomorphisms. This implies at onct<,that a is an isomorphism, and the theorem is proved. (2.39) Theorem (Change of rings). Let R be a commutative ring, and let l~ and A be R-algebras such that A is left noetherian and r is Rllat. Let M be a finitely generated left A-module, and let N be arbitrary. Then the two-sided r-isomorphism given by (2.36) is the first of a family of two-sided r-isomorphisms r ®RExt~(M, N) ~ Ext~.(M', N'),

n

~

0,

where

= r®RA,

A'

M' = r®RM,

N' = r ®RN,

Proof. The hypotheses on M and A imply that M is finitely presented, so the map a in (2.36) is an isomorphism by (2.38). Now let - - P 2 ~P~P-M-O 1 0 be a free A-resolution of M, chosen so that each Pi is A-free on a finite number of generators. (Such a resolution exists since A is left noetherian.) Then forming the sequence

(2.40) we have (by (2.27)) Ext~

(M, N) ~ ker CP: + 1 lim CP: ,

n~l.

Since r is R-flat, we have r ® R Ext~ (M, N) ~ ker 1® CP:+ l/im 1® CP:,

(2.41) where for n

~

1,

(2.42)

27

HEREDITARY RINGS

1 ® cP;: [®R Hom A (Pn-l' N)

---->

[®R Hom A (Pn,N).

By (2.38), the vertical arrows in the following commutative diagram are two-sided [-isomorphisms:

(2.42)

[ ®R Hom A (P n - 1 , N) t Hom A , (P~ _ 1 ' N')

~

HomA' (P~ , N ').

The map t/J~ occurring above is obtained as follows: applying the exact functor [ ® R . to the original free resolution of M, we obtain a free A' -resolution of M ' , namely, ---->

P~ ~ p 1 ~ P~ I

------+

M' - - O.

This in turn yields a sequence

O - - Hom A , (P i0' NT!)

I/Ji ~

H omA' (Pil ' N') ~

"',

in which the maps t/J; occur. By definition, Ext~, (M', N')

= ker t/J;+l/im t/J;.

Comparing this formula with that given in (2.41), and using the fact that the vertical arrows in the commutative diagram (2.42) are isomorphisms, we conclude at once that

[ ® R Ext~ (M, N)

~ Ext~, (M ' , N '),

n

~

1.

This is a two-sided [-isomorphism, since all of the maps occurring in the proof are two-sided [-maps. The result also holds when n = 0, if we set Ext O = Hom. This completes the proof. (2.43) COROLLARY. Let R be a commutative ring, and let R' be the ring of quotientst of R with respect to some multiplicative subset of R. Let A be any left noetherian R-algebra. Then for each finitely generated left A-module M, and any A-module N, there is an R'-isomorphism R' ® R Ex...t~ (M, N) ~ Ext~, (M ' , N '),

n

~

0,

where A' = R' ®RA, M' = R' ®RM, and so on. This isomorphism extends that given by (2.36) for the case n = O.

2f Hereditary rings

A ring A with unity element is left hereditary if every left ideal of A is a projective A-module. As we shall see eventually, maximal orders are heredit See §3a.

28

HOMOLOGICAL ALGEBRA

(2.44)

tary rings, and some of the important facts about maximal orders are actually special cases of assertions about hereditary rings. THEOREM. If A is a left hereditary ring, then every submodule of a free left A-module M is isomorphic to an external direct sum of left ideals of A, and is therefore projective.

(2.44)

Proof.t Let {ml ' ... , m k } be a free A-basis of M, and let N be any submodule of M. We shall use induction on k to show that N is isomorphic to an external direct sum ofleft ideals of A The result is obvious for k = 1, so now let k > 1, and assume that the theorem holds for submodules of M ' , where M' = Am2 + ... + Am k · Each n EN is expressible in the form

with uniquely determined coefficients {rJ. Let J be the set of all first coefficients r 1 which occur as n ranges over all elements of N. Then J is a left ideal of A, and there is an epimorphism cp: N ----> J given by cp(n) = r l ' n EN. Clearly ker cp = N !l M ', so there is a A-exact sequence

o

------4

N!l M'

-------+

N ~ J - - - O.

Since A is left hereditary, the left ideal J is A-projective, and so the above sequence splits by (2.14). Therefore N

~

J

+

N!l M'.

But N !l M' is a submodule of M ', and thus N !l M' is isomorphic to an external direct sum of left ideals of A, by the induction hypothesis. Hence N is also isomorphic to such a sum. Finally, each summand is projective, whence so is N. (2.45) Rpmarks. (i) If the ring A has the property that submodules of free left modules are projective, then in particular all left ideals of A are projective. By (2.44), we may conclude that A is left hereditary if and only if submodules of free modules are projective. (ii) The importance of (2.44) is that it gives us a structure theorem for sub modules of free A-modules when A is hereditary. This structure theorem generalizes the one valid for modules over a principal ideal domain, since clearly every such domain is hereditary. The above proof shows further that t We give the proof only for the case where M is finitely generated: it is this special case which arises most frequently in practice. For the general case, see Cartan-Eilenberg [I J.

29

EXERCISES

a sub module of a free module on k generators is isomorphic to an external direct sum of at most k left ideals of A (iii) There are examples of rings which are left hereditary but not right hereditary. However, if A is left and right noetherian, then A is left hereditary if and only if A is right hereditary (see Rotman [1, Corollary 9.20]; the result is due to M. Auslander). (iv) The proof of (2.44), for the case where M is not finitely generated, proceeds in much the same manner as the above proof. The main difference is that k must be allowed to be transfinite, and the proof uses transfinite induction. EXERCISES 1. Let Bl

y

g,l

M2

M1

V,

-

M

j~

be a pullback diagram of A-modules. Prove (i) 92 induces an isomorphism ker 91 ~ ker fz· (ii) If J2 is epic, then so is 9 1 . 2. Let

be a pushout diagram of A-modules. Prove (i) 9 1 induces an isomorphism cok f 1 ~ cok 92 . (ii) If J;, is monic, then so is 92 . 3. Let rand ,i be rings, and suppose we have modules M" LA' and a bimodule AN, . Show that there is a well-defined additive homomorphism

a: L ® A Hom,(M, N) defined by I ® f

-->

-->

Hom,(M, L ® AN),

(I, fl, where

(I,flm = I ® fm,

IE L,

fE Hom,(M, N),

mE M.

Prove that a is an isomorphism whenever L is ,i-flat and M is a finitely presented r -module. [Hint: Imitate the proof of (2.38), using ® A in place of ® R .J 4. Let ,i be a ring, and let LA be flat, and M prove that

A

finitely presented. Use Exercise 3 to

30

LOCALIZA TION

5. Let !J. be a ring, Lt!. any module, and let Mt!. be a finitely generated projective module. Prove that L®AHomA(M,!J.)

~

Hom{}(M,L).

[Hint: Use (2.17). ] 6. Let q>: R ---> S be a homomorphism of commutative rings, and let J be an ideal in R. Prove that as S-modules. 7. Prove Schanuel's Lemma: Given two exact sequences ofleft A-modules

o ---> M ---> p !. X

0,

--->

o ---> M' ---> pi t

X ---> 0,

in which P and P' are projective, there is a A-isomorphism M'

-i-

p ~M

-i-

P'.

[Hint: Let A be the pullback of the pair of maps q>, 1/1. Then there exist A-exact sequences 0---> K

with K P

-i-

~

ker 1/1

K :::;: pi

-i-

~

--->

M', K'

A

--->

~

P

--->

ker q>

0, ~

o -> K'

--->

A

--->

pi

--->

0,

M. Since P and pi are projective, A

~

K'.]

8. Prove the Snake Lemma: Given a commutative diagram of modules, with exact rows:

there is an exact sequence Ip ker a-----!,.ker {3

--->

" kef )i~cok a

--->

tP cok (J-..!+cok)i.

If q> is monic, so is q> * . If 1/1 is epic, so is 1/1 * . The lemma remains true for a diagram of groups and group homomorphisms, if each of the groups A', B' and C' is commutative. [Hint: Define c5 by c5(c) = a' + im a, where

a'

--->

b',

b' = {3(b),

b

--->

c.]

3. LOCALIZAnON 3a Rings of quotients

Throughout this section, R denotes a commutative ring. A multiplicative subset of R is a subset S closed under multiplication, and such that 0 rj S, 1 E S. We introduce an equivalence relation on the Cartesian product

(3.1)

31

RINGS OF QUOTIENTS

R x S by setting (a, s) ~ (a', s') if and only if t(sa' - s' a) = 0 for some t E S. Let a/s (or s -I a) denote the equivalence class of the pair (a, s). Now define

b -as += - t

ta

+ sb -

a b s t

st

ab st

verifying that these operations are indeed well-defined. The set of symbols {a/s} then forms a commutative ring, denoted by S--IR, with unity element 1/1 and zero element 0/1. For s E S, we have (s/l) . (l/s) = 1/1, so s/l is a unit in S-I R, with inverse l/s. Denoting l/s by s- \ we obtain

a/s = (l/s) (a/l) =

S-I

(a/l).

We shall call S - 1 R a ring of quotients of R. An element a E R is an S-torsion element if ta = 0 for some t E S. The above discussion shows at once that (a, s) lies in the zero class of S- 1 R if and only if a is an S-torsion element of R. There is a ring homomorphism i:R -+ S-IR,

given by i(x) = x/I, x E R. Then ker i is precisely the set of S-torsion elements of R. The map i permits us to view S-I R as R-module. If R is an integral domain, the map i may be regarded as an embedding of R into S- 1 R; in particular, if S = R - {O}, then S-I R is precisely the quotient field of R. For an arbitrary ring R, the map i is monic if and only if R is S-torsionfree (that is, R contains no nonzero S-torsion elements, or equivalently, S contains no divisors of zero). (3.1) THEOREM. Let S be a multiplicative subset of R. Then any ring homomorphism cp: R -+ R' which maps each element of S onto a unit of R', extends uniquely to a ring homomorphism ljJ: S- 1 R -+ R' such that ljJ i = cp. 0

Proof. Define ljJ(a/s)

=

cp(a)/ cp(s),

aEA,

SE

S.

It is easily checked that ljJ is the desired homomorphism. If also ljJ' then ljJ'(a/s) = ljJ'(a/l) . ljJ'(l/s) = cp(a)/cp(s) = ljJ(a/s),

so ljJ

=

0

i = cp,

ljJ'.

(3.2) COROLLARY. Let cp: R -+ R' be a ring homomorphism mapping a multiplicative subset S of R into a multiplicative subset S' of R'. Then there is a

32

(3.3)

LOCALIZA nON

unique ring homomorphism Ij;:S-lR ___ S'-lR' which extends cp, that is, Ij;(a/l) = cp(a)/l, a E R.

3b Modules of quotients

Let S be a multiplicative subset of the commutative ring R, and Many R-module. We view S-lR as R-module by means of the canonical map i:R --- S-lR, and we form the S-lR-module

= S-lR ®R M .

S-lM

We shall call S- 1 M a module of quotients. Each XES -1 M is then expressible as a finite sum x =

Set s =

TI

Si

I

ais i- 1 ® mi

'

E S; then

x =" S-1· at .s t = S-l 'r:;:.m s:-1 ®m.t = S-l ®"a.ss:-1m. ~ ~l-t C', 1

say. Denote S-l ® m by m/s; we have thus shown that every element of S-l M is of the form m/s for some mE M, s E S. It is easily checked that m/s

±

m'/s' = (s'm

±

sm')/ss'.

We shall call an element m of an R-module M an S-torsion element if sm = 0 for some s E S. The collection of all S-torsion elements of M is called the S-torsion sub module of M, and is an R-submodule of M. (3.3) THEOREM. Let M' be the S-torsion submodule of M. Then m/s = 0 in S- 1 M if and only if m EM'. Proof. If mE M', then tm = 0 for some t E S, whence m/s

= S-l ® m = (St)-l ® tm = O.

Conversely, suppose that S-l ® mo = 0 in S-l M. Then 1 ® mo = = 0, and therefore 1 ® mo is expressible as a finite sum of elements of S- 1 M of the following three types:

S(S-l ® mo) (u

+ u') ® m -

u ® m - u' ® m,

u ® (m

+ m')

- u ® m - u ® m',

ua ® m - u ® am,

with u, u' E S-l R, m, m' EM, a E R. Choose t E S to be a common denomina tor for all of the u's which occur in these terms, so tha t tu E R for each such u (or more precisely, tu E i(R)). Let R1

= t- 1 . i(R) = {a/t:aER},

(3.4)

33

MODULES OF QUOTIENTS

an R-submodule of S-1 R. We shall now define a homomorphism f:Rl ®R M

-+

MIM'

by setting f(x®m)

=

txm

+ M ',

This map f is well-defined, since if x = alt = bit in Rl ' then a - b is an S-torsion element of R, and hence (a - b)m EM'. Consider now the element I ® mo . From the manner in which Rl was chosen, it is clear that I ® mo = 0 in Rl ® R M. Therefore f(l ® mo) = 0 in MIM', that is, tmo EM'. This implies that mo also lies in M ', which completes the proof of the theorem. (3.4)

COROLLARY. The module of quotients S-1 M coincides with the moduli? whose elements are equivalence classes of ordered pairs (m, s), with (m, s) ~ (m', S') if and only if s'm - sm' is an S-torsion element of M. The class of (m, s) may be identified with the element mls = S-1 ® m in S- 1 M.

The S-1 R-module S-IM may be viewed as an R-module by use of the homomorphism i: R -+ S - 1R; in other words, an element a E R acts on S-1 M as left multiplication by the element all of S-1 R. We shall say that the element a E R acts invertibly on an R-module X if the mapping x -+ ax, x E X, gives a one-to-one map of X onto itself. It is then clear that every element of S acts invertibly on the module of quotients S-1 M. On the other hand, given any R-module X on which every element of S acts invertibly, we can make X into an S-l R-module by defining (3.5)

(als)'x=s-I'ax,

aER,

SES,

XEX.

This is the unique way in which X can be made into an S-1 R-module so as to preserve the action of R on X. (3.6) THEOREM. Let X be an R-module on which every element of S actsinvertibly, and make X into an S-1 R-module by means of (3.5). Then there is an S-1 Risomor phism S-IR ®RX ~ X. Further, for each R-module Y, every R-homomorphism f: Y -+ X extends canonically to an S-1 R-homomorphism f' :S-1 Y -+ X, given by the formula f'(u ® y) = u . f(y), Proof. By (3.4), every element of S-IR ®RX is expressible as S-1 ®x, with s E S, X E X. Denote this element by xis. By (3.4) we have xis = x'ls ' if and

34

(3.7)

LOCALIZA nON

only if s'x - sx' is an S-torsion element of X. But S acts invertibly on X, so X is S-torsion free. Thus xis = x'/s' if and only if s'x = sx'. Now define a mapping e:S-lR ®RX --> X by setting e(x/s) = S-lX, x EX, S E S. The preceding remarks imply that e is well-defined; for if xis = x'/s', then S-lX = S,-lX'. Clearly e is epic, and is an S-lR-homomorphism. Finally, e is monic, since if 8(x/s) = 0, then s- lX = 0 and hence also x = O. This proves the first part of the theorem. We omit the straightforward proof of the second part. 3c Flatness Recall that an R-module is flat if tensoring with it preserves exactness (see §2c). We shall now prove that every ring of quotients S-lR is a flat R-module. Given any R-homomorphism f:L --> M of R-modules, there is an S-lR-homomorphism j~: S-- 1 L --> S-l M, defined by setting j~ = 1 ® f, where Thus (3.7)

f*(I/s)

= f(l)/s,

tEL,

SES.

U sing this we prove

(3.8) THEOREM. Let L 1. M ~ N be an R-exact sequence and let f 9 be , ' *' * defined as in (3.7). Then S-lL ~S-lM ~S-lN is an exact sequence of S-l R-modules. Proof. Since gf = 0, we have 9 * f* = (gf)* = write x = mis, mE M, s E S. Then

o=

g*(m/s)

o.

Now let x E ker 9 * ' and

= s-lg(m),

so 1 ® g(m) = 0 in S-l N. Hence there exists an element t E S such that t . g(m) = 0, that is, g(tm) = O. But ker 9 = im f, and so we may write tm = f(l) for some 1E L. Then x = m/s = tm/ts = f*(l/ts) E im f*. This shows that ker g* = im f* . Finally, it is clear that f* and g* are S-l Rhomomorphisms, which completes the proof. For later use, we state a special case of (3.8):

(3.9)

LOCALIZA nON AT PRIME IDEALS

35

(3.9) COROLLARY. An inclusion L eM of R-modules induces an inclusion S-lL c S-lM of S-lR-modules, and there is an S-lR-isomorphism

S-l(M/L)

~

S-lM/S- 1 L.

Recall that an R-module M is noetherian if its submodules satisfy the ascending chain condition. A ring A is (left) noetherian if A is noetherian as left A-module. Analogously, M is artinian if its submodules satisfy the descending chain condition. We wish to prove that the properties of being noetherian or artinian are preserved under formation of rings or modules of quotients. This is best accomplished by using the concept of "saturation". Let L eM be R-modules; call L an S-saluratedt submodule of M if MIL is S-torsionfree, that is, MIL has no nonzero S-torsion elements. Equivalently, L is S-saturated in M if for mE M, s E S, the inclusion sm E L implies that mEL. (3.10) THEOREM. Let i:M ---> S-lM be dejined by i(m) = I ® m, mEM. There is a one-to-one inclusion-preserving correspondence between the set of S-saturated submodules X of the R-module M, and the set of S-lR-submodules X' of S-l M, given by Proof. Straightforward exercise for the reader.

As an immediate consequence of (3.10), we obtain (3.11) THEOREM. If M is a noetherian R-module, then S-lM is a noetherian S-l R-module. In particular, if R is a noetherian ring, so is S-l R. The analogous results hold true with "noetherian" replaced by "artinian".

3d Localization at prime ideals

Let R be a commutative ring, and let J, J' denote ideals of R. We write J < J' to indicate that J is properly contained in J'. Let us set R - J

= {x:xER,x¢:J}.

(3.12) Dejinitions. (i) J is a proper ideal if J < R. (ii) J is a maximal ideal if J < R, and if there is no ideal J' such that J < J' < R. (iii) J is a prime ideal if J < R, and R/ J has no zero divisors.

Clearly, a proper ideal J is prime if and only if its complement R - J is a mUltiplicative set. Starting with a prime ideal P of R, we may fonn the multiplicative set S = R - P, and then define a ring of quotients S- 1R. t See Bourbaki, Algebre Commuti ve, page 69 for this terminology.

36

LOCALIZA TION

(3.13)

This ring, hereafter denoted by R p , is called the localization of Rat P. Since every element of R - P is invertible in R p , it is easily verified that Rp has a unique maximal ideal, namely p. Rp. We should remark that the ring homomorphism i: R -> Rp , defined in § 3a, enables us to view Rp (and all Rp-modules) as R-modules. Thus P . Rp is the same as i(P)' Rp . If R happens to be an integral domain, then the mapping i:R -> Rp is an embedding. In particular, when P = {O} then Rp is precisely the quotient field of the domain R. Returning to the general case, let M be any R-module. We define M p = Rp ®R M, an Rp-module called the localization of Mat P. (3.13) THEOREM. Localization at maximal ideals does not affect residue class modules. SpecUical/y, let P be a maximal ideal of R, and let M be an R-module. Then for each r :> 0, we may view M / pro M as an Rp-module, and there is an Rp-isomorphism M/pr'M ~ Mp/P'·M p . Proof Let S = R - P, and put M = M/ P'M. We show first that M can be made into an Rp-module. For each s E S we have Rs + P = R, since P is maximal. Hence there exist elements a E R, n E P, such that as + n = 1. Taking rth powers, we obtain

f3s + nr =

1

for some f3 E R. Therefore s is a unit in the ring R/P, and hence acts invertibly on the (R/P)-module M. From (3.6) we deduce that M can be viewed as Rp-module, and that there is an Rp-isomorphism M

~

Rp®RM

= (M)p.

But by (3.9) we have

= (M/pr'M)p

~

Mp/P'·M p .

Hence there is an Rp-isomorphism M proved.

~

Mp/P'Mp, and the theorem is

(M)p

The preceding theorem is used most frequently for the special case where r = 1, and yields the (R/P)-isomorphism (3.14 )

for each maximal ideal P of R, and each R-module M. We often refer to problems concerning R-modules and R-homomorphisms as global problems, whereas those involving Rp-modules are called local problems. A fundamental technique in algebraic number theory, and in its generalizations considered in this book, is the method of solving global

37

LOCALIZATION AT PRIME IDEALS

(3.l5)

questions by first settling the local case, and then applying this local information to the global case. The next few theorems provide some connections between local and global information. For each R-module M, and each prime ideal P of R, there is an R-homomorphism

ip:M

~

Mp,

given by ip(m) = 1 ® mE Rp ®R M, mE M. Consequently there is an R-homomorphism M ~ IT M P' where P ranges over all maximal ideals of R. p

(3.15) THEOREM. The map M ~

only

if M p

IT M

p

is monic. In particular, M

= 0

if and

p

0 for each maximal ideal P of R.

=

Proof. Denote the above map by f, and suppose that mE ker f. Set J = {aER:a.m = O}, an ideal in R. If J < R, then t J is contained in some maximal ideal P of R. But ip(m) = 0, so by (3.3) there exists an s E R - P such that s . m = O. Hence s E J, s ~ P, which contradicts the fact that J c P. This shows that J = R, and thus m = 0 as claimed. The second assertion in the theorem is then obvious. (3.16) COROLLARY. Let f E Hom R (L, M) induce fp: Lp (i) There are Rp-isomorphisms

~

M p' Then

(ker f)p ~ ker f p , (im f)p ~ im f p , (cok f)p ~ cok f p,

for each maximal ideal P of R. (ii) f is monic if and only if each fp is monic. (iii) f is epic if and only if each fp is epic. (iv) A sequence of R-modules

O~L~M~N~O is exact

if and only if the sequence

is exact for each P. Proof. Given any f

E

HomR (L, M), there is an R-exact sequence

t This statement is a consequence of Zorn's Lemma (see Curtis-Reiner [1 J); it does not require the hypothesis that R be noetherian,

38

(3.17)

LOCALIZA TION

o --. ker f --. L .4 M. Since Rp is R-flat, it follows that the sequence

o --. (ker f)p --. Lp &. M p is Rp-exact. This gives an Rp-isomorphism (ker f)p ~ ker fp. Similar arguments yield the other assertions in (i). If f is monic, then ker f = 0, whence ker fp = 0 for each P, by (i). Conversely, if each fp is monic, then by (i) and (3.15), also f is monic. This proves (ii), and the remaining assertions in the theorem follow in a similar manner.

(3.17) COROLLARY. Let R be an integral domain with quotient field K, and regard R and all of its localizations Rp as embedded in K. Then

R = riRp, p

where P ranges over all maximal ideals of R. Proof. Obviously, R is contained in the intersection. Conversely, let x E K be such that x E Rp for all P. Writing x = alb, a, bE R, it follows that a E bR p for all P. Let f:R --.R be defined by fer) = rb, rER. Then the image of a in cok fp is zero for all P. But by (3.15) and (3.16), the map

cok f --.

TI cok fp p

is monic. Hence a has zero image in cokf, that is, a E bR. This shows that x E R, so R contains nR p , and the proof is complete.

The reader should be cautioned that (3.17) does not hold for arbitrary commutative rings R, nor does it generalize to the case of R-modules unless additional hypotheses are imposed. Let us next record the following special cases of (2.38) and the "Change of rings" Theorem 2.43. (3.18) THEOREM. Let A be an R-algebra, where R is a commutative ring, and let M be any finitely presented left A-module. Then for every A-module N, and for each prime ideal P of R, there is an Rp-isomorphism

Rp

®R

HomA (M, N)

~

HomA p (Mp, N p ).

(3.19) THEOREM. Let A be a left noetherian R-algebra, where R is a commutative ring, and let P be any prime ideal of R. Then there is an Rp-isomorphism

Rp ®R

Ext~ (M, N) ~ Ext~ p(Mp, N p ),

n ~ 0,

for each pair of left A-modules M, N such that M is finitely generated over A.

(3.20)

LOCALIZA TION AT PRIME IDEALS

39

We may emphasize that when R is noetherian, and A is finitely generated as R-module, then A is left and right noetherian, and every finitely generated A-module is automatically finitely presented. Now let A be any R-algebra; we shall show that the question, as to whether a given A-exact sequence is A-split, is a "local" question, that is, the answer is determined by the answers for the corresponding Ap-sequences, where P ranges over the maximal ideals of R. (3.20) THEOREM. Let A be an algebra over the commutative ring R, and let (3.21)

o ~ LJ.. M .4 N

~0

be a A-exact sequence, where N is a finitely presented A-module. Then the sequence is A-split if and only if for each maximal ideal P of R, the Ap-exact sequence (3.22)

Proof. The epimorphism g induces a map g*:HomA (N, M)

~

Hom A(N, N).

We claim that the sequence (3.21) is split if and only if g* is epic. Indeed, if h: N ~ M is such that gh = 1N' the identity map on N, then for each cp E HomA(N, N) we have hcp E HomA (N, M), and

g*(hcp) = gh' cp = cp. Thus if (3.21) splits, then g* is epic. Conversely, if g* is epic, there exists an hE Hom A (N, M) such that g * h = 1N' that is, gh = 1N' This completes the proof of the claim. Now let us define (gp)*:HomA)Np , Mp)

~

Hom Ap (Np, N p),

where g induces gp:Mp ~ N p . By virtue of (3.18), we may identify (gp)* with (g*)p . We have seen above that (3.21) is A-split if and only if g* is epic, while analogously (3.22) is Ap-split if and only if (g *)p is epic. The theorem then follow immediately from (3.16). (3.23) COROLLARY. Let A be an R-algebra, N afinitely presented left A-module. Then N is A-projective if and only if Np is Ap-projective for each P.

Proof. Choose a free A-module F mapping onto N, so there is a A-exact

40

(3.24)

LOCALIZATION

sequence 0 -4 L -4 F -4 N -4 O. If N is projective, then NIF, whence NplF p and Np is Ap-projective. Conversely, if each Np is projective, then each localization of the above sequence is split. Hence the original sequence splits, by (3.22), and so N is projective. Recall that A is a left hereditary ring if every left ideal of A is projective. We have at once COROLLARY. Let A be a left noetherian R-algebra, where R is a commutative ring. Then A is left hereditary if and only if the ring Ap is left hereditary for every maximal ideal P of R.

(3.24)

Proof. Suppose that Ap is left hereditary for each P, and let L be any left ideal of A Then L is finitely presented as A-module, since A is left noetherian. Further, Lp is a left ideal of Ap , and hence Lp is projective. Therefore L is projective, by (3.23). Conversely, let A be left hereditary, and let P be any maximal ideal of R. By (3.1 0), every left ideal of Ap is of the form L p , for some left ideal L of A Since L is projective as A-module, it follows as in (3.23) that Lp is Ap-projective, as desired. This complete the proof of the corollary.

In order to generalize (3.24), we introduce the concept of homological dimension, which is sometimes referred to as "projective dimension". The homological dimension hdAM of a left A-module M is the least positive integer n for which there exists a A-exact sequence (3.25)

0 -4 Xn

-4

Xn -

1

-4 . . . -4

Xo

-4

M

-4

0, Xi projective.

We set hdAM = 00 if no such n exists. Once (3.25) is given, it follows directly from definition (2.27) that Ext~ (M, .) = 0, r ~ n + 1. (The notation Ext~ (M,') = 0 means that Ext~(M, L) = 0 for each A-module L.) Now consider a A-exact sequence (3.26)

0

-4

Kn

-4

Xn -

1

-4 . . . -4

Xo

-4

M

-4

0, Xi projective.

By (2.32) there is an isomorphism Ext~+k (M, . ) ~ Ext~ (Kn ' . ).

But by (2.28 v), Kn is projective if and only if Ext~ (Kn ,.) = O. This gives (3.27) THEOREM. Let M be a left A-module. The following statements are equivalent:

(i) hdAM (ii)

Ext~

~

n.

(M, .)

= 0 for

r ~ n

+

1.

(3.28)

41

LOCALIZA TION AT PRIME IDEALS

(iii) For every exact sequence (3.26) in which X 0' also Kn is projective.

... ,

Xn -1 are projective,

We note that hdAM = 0 if and only if M is projective. The following result is thus a generalization of (3.23): (3.28) THEOREM. Let A be a left noetherian R-algebra, where R is a commutative ring, and let M be any finitely generated left A-module. Then

hd AM = sup {hdA Mp}, p p where P ranges over all maximal ideals of R. Proof. Denote the sup by s, and let n = hdAM. We show first that n ~ s. The result is clear if n = 00, so assume n finite. Then there is an exact sequence (3.25) with each Xi projective. Localizing at P, we obtain a projective resolution of M p, and therefore hdA M p ~ n. This holds for each P, whence s ~ n, as claimed. P Conversely, we prove that n ~ s, and we may assume that s is finite. From (3.19) we obtain Rp ®R Ext~+ 1 (M, N) = 0

for all P and for all A-modules N. It now follows from (3.15) that Ext~+ 1 (M, .) = 0, and so hdAM ~ s. This completes the proof of the theorem. The left global dimension of A is defined as dim A = sup {hdAM: M = left A-module}. (We should really use the notation 1. gl. dim. A, but will employ the simpler notation dim A when there is no danger of confusion.) Clearly, dim A = 0 if and only if every left A-module is projective, that is, if and only if A is a semisimple artinian ring (see § 7a). We claim that dim A ~ 1 if and only if A is left hereditary. Suppose that dim A ~ 1, and let J be any left ideal of A There is a A-exact sequence

o -+ J -+ A -+ A/ J -+ 0, in which A is projective. Since hd AAI J ~ dim A ~ 1, it follows from (3.27) that J is projective, as desired. Conversely, let A be left hereditary, and let M be any left A-module. There is a A-exact sequence

o -+ K

-+

F

-+

M

-+

0,

in which F is A-free. Since A is hereditary, the submodule K of the free module F must be projective, by (2.44). Therefore hdAM ~ 1 for each M, whence also dim A ~ 1.

42

(3.29)

LOCALIZATION

We are now ready to give a generalization of (3.24). (3.29) THEOREM. Let A be a left noetherian R-algebra, where R is a commutative ring. Then dim A = sup {dim Ap}, p

where P ranges over all maximal ideals of R. Proof. We shall use a formula due to M. Auslander (see Rotman [1, Th. 9.14]):

dim A = sup {hdAAI J: J = left ideal of A}. Bu t A/l is a finitely generated A-module, so by (3.28) hdAA/J = sup {hdA Ap/J p }. p

P

Further, by (3.1 0), as J ranges over all left ideals of A, J p ranges over all left ideals of Ap . Therefore we have dim A = sup {hd A Ap/Jp} = sup {dim Ap}, l,P

P

P

as claimed. To conclude this section, let us show that homological and global dimensions are unchanged by passage to P-adic completions. The preceding theorems reduce the calculations of hdAM and dim A to the case where the ground ring is R p , a local ring. If we make the additional hypothesis that R is noetherian, then each Rp is also noetherian. The P-adic completion Rp of the ring Rp is then a faithfully flat Rp-module (see remark following (2.23), or Exercise 5.3). Using this fact, we prove (3.30) THEOREM. Let Rp be the P-adic completion of the local noetherian ring R p , and let Ap be a left noetherian Rp-algebra. Set

Ap = Rp ®R Ap, P

Mp = Rp ®R M p , p

where Mp is any finitely generated left Ap-module. Then

hd AP Mp = hd Ap Mp for each M p . Furthermore

dim Ap = dim Ap . Proof As in the proof of (3.28), we find at once that hdM p ::;;; hdM p' On the other hand, if s = hdM P' then for each Ap-module N p we have

Rp®R Ext~"'l(Mp, N p ) P P

O.

43

EXERCISES

Since Rp is faithfully flat as Rp-module, it follows that Ext s + 1 (M p' .) = 0, and therefore hd M p ~ s. The remaining assertion in (3.30) follows from the formula of Auslander used in the proof of (3.29). EXERCISES 1. Let R be a domain with quotient field K, and let M be a direct summand of a free R-module F of rank n. Show that M and all of its localizations M P' where P ranges over the maximal ideals of R, may be embedded in the n-dimensional vector space K ® RF over K. Then use (3.17) to prove the formula M =(\M p

,

the intersection being taken over all maximal ideals P of R. 2. Let R be a domain with quotient field K, and let X, Y be submodules of an R-module M. Show that if X p = Yp in M p for each P, then X = Y. [Hint: For each P, {(X

+

Y)/Y}p =

0.]

3. Let R be a commutative ring, and let S, T be multiplicative subsets of R such that SeT. Prove that T -1 R is isomorphic to the ring of quotients of S -1 R relative to the multiplicative set which is the image of Tin S -1 R. 4. Let P 1 ' ... , p" be a set of prime ideals of the commutative ring R, no one of which contains another, and let S = R - (P 1 U .. ,

u P n)·

(i) Show that S is a multiplicative subset of R, and that the distinct maximal ideals of S-1R are Pi' S-1R, I ~ i ~ n. (ii) Show that R p , is canonically isomorphic to the localization of S-1 R at the maximal ideal Pi . S - 1 R. (iii) Show that

" Rp i S-lR=n i==l

if R is an integral domain, the intersection being formed within the quotient field of R. (Reference: Bourbaki [2, Ch. II, § 3, no. 5].)

5. Let S be a multiplicative subset of the commutative ring R, and let R' = S- 1 R. Let A be an R-algebra, and set N = R' ®R A, an R'-algebra. Show that every finitely generated left N-module X contains a finitely generated left A-module M such that N®;\M [Hint: The ring homomorphism A X

=

L" i=1

N

Xi'

--+

~

N M

=

X.

A' makes X into a left A-module. Let

and set M

=

L"

A Xi' so N M

=

X.

i=1

The inclusion Me X yields an inclusion N ®;\ MeN ®;\ X ~ X, since R' is R-flat. Hence the map N ®AM --+ NM is monic, and thus is an isomorphism. For another approach, see Exercise 26.5.]

44

DE DE KIND DOMAINS

4. DEDEKIND DOMAINS Throughout this section R denotes an integral domain with quotient field K. To avoid trivialities, we assume always that R #- K. We shall define Dede-

kind domains and shall state, without proof or detailed references, some of their most important properties. Proofs and complete expositions are readily available in texts such as Curtis-Reiner [1], Janusz [1], O'Meara [1], Ribenboim [1], Samuel [1], Weiss [1], Zariski-Samuel [ll From the standpoint of ideal theory, Dedekind domains are the simplest type of domain beyond principal ideal domains, and share many of their arithmetical properties. Dedekind domains arise naturally, as follows: let R be a principal ideal domain with quotient field K, let L be a finite extension of K, and let S be the integral closure of R in L. Then S is a Dedekind domain with quotient field L. As a matter offact (see (4.4)), this same conclusion holds under the weaker hypothesis that R itself is a Dedekind domain. We may also remark that Dedekind domains are a special kind of maximal order. One of the main topics of this book is the study of arithmetical properties of maximal orders analogous to those of Dedekind domains. Many of the results stated in this section are special cases of the theorems to be developed in Chapter VI. Let us fix some notation, to be used throughout this section. The annihilator of an R-module M is defined by annRM

= {aER:a'M = a}.

annRm

= {aER:a·m = O},

Likewise we set for mE M. Call mE M an R-torsion element if ann R m #- O. The set of all R-torsion elements of M is the R-torsion submodule of M, and is the kernel of the R-homomorphism M -+ K ® R M given by m -+ 1 ® m, mE M (see §3b). The R-rank of M is defined as rankRM = dim K (K ® R M). We call M R-torsionfree if M contains no torsion element except O. In this case, we may identify M with its image 1 ® M in K ® R M; we may then write K ® R M in the simpler form K . M, where K . M denotes the set of all finite sums {~aimi: ai E K, mi EM}. In the same way, for each ring of quotients R' of R, the R-torsionfree R-module M is embedded in the R' -module R' ® R M; after making the obvious identification, we may denote R' ®R M by R'· M. An R-lattice is a finitely generated R-torsionfree R-module. Each Rlattice M is an R-submodule of a finite dimensional vector space V over K, namely V = K· M. We call M a full R-lattice in V, the adjective indicating that M contains a K-basis of V Let M and N be a pair of full R-lattices in a

(4.0)

IDEAL THEORY, MODULES, ORDER IDEALS

45

K-space V. Since N contains a K-basis for V, it is clear that for each x E M there is a nonzero r E R such that r· x EN. But M is finitely generated as R-module. Therefore we can choose r E R, r # 0, such that r· MeN. Let M be an R-Iattice, and let N be a sublattice of M. We shall call N an R-pure sublattice of M if MIN is R-torsionfree. Equivalently, N is R-pure in M if for each nonzero a E R, N naM=aN.

Let S be the multiplicative set R - {o}. Then MIN is R-torsionfree if and only if N is an S-saturatedt R-submodule of M. As a special case of (3.10), we thus obtain (4.0). THEOREM. Let M be an R-Iattice. There is a one-to-one inclusion-preservmg correspondence N ~ W between the set of R-pure sub lattices N of M, and the set of K-subspaces W of KM, given by

W=KN,

N= W nM.

4a Ideal theory, modules, order ideals

Let R be an integral domain with quotient field K. We call R a Dedekind domain if it satisfies anyone of the following three equivalent conditions: (4.1) R is a hereditary ring, that is, every ideal of R is a projective R-module (see §2f). (4.2) R is a noetherian integrally closed domain such that every nonzero prime ideal of R is a maximal ideal. (4.3) Every proper nonzero ideal of R is uniquely expressible as a product of nonzero prime ideals of R, apart from order of occurrence of the factors. We recall that R is noetherian if its ideals satisfy the A.c.e. (see (2.9)). The ring R is integrally closed if each element of K integral over R necessarily lies in R. A prime ideal of R is an ideal J properly contained in R (notation: J < R) such that RIJ has no zero divisors (see (3.12)). For the proof that these three conditions are equivalent, and also for the proofs of theorems stated below, the reader may consult the references listed at the beginning of §4. We shall seldom specify the exact location in these references where the proofs may be found. (4.4) TH:OREM. Le~ R be a Dedekind domain with quotient field K, and let L ~e any flel~ extensIOn of K of finite degree. Then the integral closure of R in L IS a Dedekmd domain S with quotient field L.

t This concept has been defined immediately preceding (3.10).

46

(4.5)

DEDEKIND DOMAINS

Remarks. (i) Recall from §1 that S consists of all elements of L which are zeros

of monic polynomials in R[ Xl By (1.11) we know that S is a ring. The assertion that S is a Dedekind domain contains the additional information that S is integrally closed. We may also point out that each x E L is expressible as a quotient x = y/r with YES, r E R; namely, we need only choose r to be a multiple of the denominators of the coefficients which occur in min.pol.K x. (ii) The integral closure of R in K is R itself, since R is integrally closed. (iii) Clearly S is a torsionfree R-module, since R is a subring of the integral domainS. (iv) Let K be an algebraic number jield, that is, a finite extension of the rational field Q. By (4.4), the integral closure ofZ in K is a Dedekind domain, hereafter denoted by alg. int. {K}. Its elements are called algebraic integers. (v) In many texts, Theorem 4.4 is proved only when L is separable over K. For the general version, see Zariski-Samuel [1, Chapter V, §8, Theorem 19]. Let L be a finite field extension of K, and let TL/K be the trace map from L to K, defined as in §la. From T we can construct a bilinear trace form ,:L x L --> K by setting (4.5)

x, Y E L.

Then, is a symmetric K-bilinear form. The form, is called nondegenerate if for nonzero XE L, the map ,(x, '):L --> K is not the zero map. We may represent, by a symmetric matrix

T

as follows: let L =

f' Kx

i,

where

i= 1

n = (L:K),andput T

= (,(Xi' X))l"i.i"" = (TL/K(X i X))l "i,j"" .

Then, is nondegenerate if and only if the matrix result in field theory is as follows:

"t

is nonsingular. A standard

THEOREM. Let L be a jinite jield extension of K. The following statements are equivalent: (i) L is separable over K. (ii) The bilinear trace form from L x L to K is non degenerate. (iii) There exists an x E L such that TL/K(X) = 1. (iv) TLIK:L --> K isan epimorphism.

(4.6)

(We remind the reader that if char K = 0, or more generally if char K does not divide the degree (L: K), then L is necessarily separable over K. The same is true whenever K is a finite field.) We give an addendum to (4.4) for the case of separable extensions. (4.7) THEOREM. Let R be a Dedekind domain with quotient jield K, and let S be

(4.8)

IDEAL THEORY, MODULES, ORDER IDEALS

47

the integral closure of R in a jinite separable jield extension L of K. Then S is a jinitely generated torsionfree R-module of R-rank (L:K).

We tum now to the ideal theory of a Dedekind domain R. A fractional R-ideal is a full R-Iattice in K, that is, a finitely generated R-submodule J contained in K, such that K . J = K. Since R is noetherian, every nonzero ideal of R is necessarily a fractional ideal, and will be called an integral ideal. Given fractional R-ideals J and J', we may define J

+ J'

= {x

+ y : x E J,

y E J'},

J. J' = { l:XiYi: Xi E J, Yi

E

J'},

finite

r1={XEK:xJcR}.

Then J + J', J n J', J. J' and J- 1 are fractional R-ideals as well. Relative to the multiplication J . J' defined above, the set of fractional R-ideals forms a multiplicative group with identity element R, and with J- 1 the inverse of J. Property (4.3) of Dedekind domains extends readily to the factorization of fractional ideals, as follows: (4.8) THEOREM. The set of fractional R-ideals in K is afree abelian group, with free generators the nonzero prime ideals of R. In other words, every fractional ideal is uniquely expressible in the form ei E Z, J =~ e, .. , Pt e" where the {PJ are distinct nonzero prime ideals of R, and where no ei is O. (We agree to write R itself as an "empty product", with t = 0).

TI

TI

Consider next the factorization J = Pi ai, J' = p/;, of a pair of fractional ideals. If a i :( bi for each i, we say that J divides J' (notation: J I1'), and in this case it is clear that J :::> J', Conversely, if J :::> J' then J- 1J' c R, whence by (4.8) a i :( bi for each i. Given any pair of fractional ideals J, J', we call J + J' their greatest common divisor, and J n J' their least common multiple. If the factorizations of J and J' are as above, it follows at once from the preceding paragraph that (4.9) J + J' = Pimin (ai, bi!, J n J' = p;max(a i ' b<J,

TI

TI

Two ideals J,J' of R are relatively prime if J + J' = R. By (4,9), if J and J' are nonzero, they are relatively prime if and only if they have no prime ideal factor in common. A set of nonzero integral ideals {J l' . , . , J n} is pairwise relatively prime if J i + J k = R for 1 :( i < k:( n. (4.10) THEOREM. (Chinese Remainder Theorem), Let {J1' ... , J.} be a set of pairwise relatively prime integral ideals of R. Then there is a ring isomorphism

48

DEDEKIND DOMAINS

RIJ1

•.•

I n ~ (RIJ 1 )

(4.11)

-i- ... + (RIJJ

An easy consequence of (4.10) is the extremely important

(4.11) THEOREM. (Strong Approximation Theorem). Let P 1"" P n be distinct nonzero prime ideals of the Dedekind domainR, and let the elements al ' ... , an EK be given, as well as positive integers r 1" .. , rn' Then there exists an element b E K such that b - a. E (PJ' . R p , 1 ~ i ~ n, {

b E R;

for all p' =1= PI" .. ,Pn'

where P denotes a variable nonzero prime ideal of R. We shall also record (4.12) THEOREM. Let J be an integral ideal of the Dedekind domain R, and A any fractional R-ideal. Then there is an R-isomorphism

RIJ

~

AIJA.

We may partition the set offractional ideals of R into ideal classes, putting J and J' into the same class whenever J ~ J' as R-modules. Note that J ~ J' if and only if J' = Jx for some x E K. The ideal classes form a multiplicative group, the ideal class group of R, hereafter denoted by CI R. Multiplication of ideal classes is given by [J][J'] = [J. J'], where [J] is the class containing J. A basic result tells us that Cl R is a finite group whenever R = alg. int. {K}, K an algebraic number field. (This is a special case ofthe Jord&n-Zassenhaus Theorem 26.4). Next we recall some properties of R-Iattices. Let M be an R-Iattice of R-rank n, and set V = K· M, an n-dimensional vector space over K. We may write n

V=

.I"1 Kv"

and set N =

I

RVi' a free full R-Iattice in V. Since M and N are

1=

full lattices in the same space V, there exists a nonzero r E R such that r' MeN. But M ~ r' M, and thus we conclude that M may be embedded in a free R-Iattice on n generators. This argument holds for any domain R. If however R is a Dedekind domain, then by (4.1) R is hereditary. Hence M is R-projective by (2.44), and there is an R-isomorphism M ~ J1

-i- ... -+-

In,

where the {J;} are fractional R-ideals. There must be n summands because rankR M = n. To complete the discussion of the structure of lattices over a Dedekind domain, it is necessary to know under what conditions two such external direct sums of fractional ideals are isomorphic. The complete answer is given by a theorem of Steinitz:

(4.13)

49

IDEAL THEORY, MODULES, ORDER IDEALS

(4.13) THEOREM. Let R be a Dedekind domain. Each R-lattice M is R -projective, and is isomorphic to an external direct sum M ~ J1 -1- J.,

+...

where the {J,} are fractional R-ideals, and n = rankRM. Further, two such •

m

i= 1

i= 1

sums Jj

L' J i and L' J; are R-isomorphic if and only if m =

•••

n, and the products

J n , J; ... J~ are in the same ideal class.

There is a generalization of the invariant factor theorem for modules over principal ideal rings, as follows: (4.14) THEOREM (Invariant Factor Theorem). Let R be a Dedekind domain, and let M, N be R-lattices such that N c K· M. Then there exist elements {m,} in M, and fractional R-ideals {J,} and {E,}, such that

where r = rank R M, s = rank R N, and where El then R ::> El .

::>

E2

::> ••• ::>

Es' If N eM,

The ideals El ' . " , Es occurring above are called the invariant factors of the pair M, N; theyareuniquelydetermined by the inclusion map N c K· M. If N c M and K· N = K· M, then MIN is an R-torsion module, and there is an R-isomorphism (4.15)

MIN

t

t

~ i= 1 (JJE;J) ~ i= 1 (RIEJ,

the latter isomorphism being a consequence of (4.12). Then (see below) the order ideal of MIN is given by

ordMIN=Ej"·E r · Let us define the order ideal ord X of a finitely generated R-module X as follows: (i) If X = 0, ord X = R. (ii) If X is not an R-torsion module, ord X = O. (iii) If X is a nonzero R-torsion module, then X has an R-composition series (see Exercise 4.1); whose composition factors are {RIP,}, with Pi ranging over some set of maximal ideals of R. We set ord X = Pi' where the number of factors equals the number of composition factors of X. The ideal P., can be recovered from the composition factor RIP;, namely (4.16)

TI

Pi = annRRIP;. Since the set of composition factors is uniquely determined by X, according to the Jordan-Holder Theorem, it follows that ord X is well-defined.

50

(4.17)

DEDEKIND DOMAINS

We observe that if X is an R-torsion module, then ord X = R if and only if X = o. Further, we prove (4.17) THEOREM. (i) For each exact sequence of finitely generated R-modules

o -+ X' -+ X

-+

X"

-+

0,

we have ord X = (ord X') (ord X"). (ii) For each nonzero ideal J of R, ord (RIJ) = J.

Proof. (i) is clear from the fact that the composition factors of X are those of X' together with those of X". For (ii), we can use the factorization J = P7 i in (4.8) to write an explicit composition series for RjJ, from which it is evident that the composition factors of RjJ are the {RIP;} with multiplicities {eJ

TI

From (4.14) we obtain a structure theorem for finitely generated Rmodules: (4.18) THEOREM. Every finitely generated R-module is isomorphic to a finite external direct sum of ideals of Rand cyc/ict modules RjJ, with J an integral ideal of R.

Proof Given a finitely generated R-module X, we can find an R-exact sequence

o -+ N

-+

M

-+

X

-+

0,

with M R-free on r generators, r finite. We then have (in the notation of (4.14))

X

~

MIN

~

f

i=r+l

Ji

+ I:CRlE

i) ,

i=l

as claimed. Note that the expression 'i.·Ii does not occur if Xis an R-torsion module 4b Localizations, valuations

Throughout let R be an integral domain with quotient field K, R # K. We shall describe some properties of the ring R, and of R-modules, with respect to localization at prime ideals of R. In cases where proofs are omitted, the reader may consult the references listed at the beginning of §4. For the first part of this subsection, we shall restrict our attention to Dedekind

t An

R-module is cyclic if it can be generated by one element.

(4.19)

51

LOCALIZATIONS, VALUATIONS

domains. In the second part, we shall consider the more general situation in which R is a noetherian integrally closed domain. This more general material may be skipped in a first reading, since it will occur only peripherally in later chapters. Let us introduce some concepts from valuation theory (we shall consider only rank one valuations !). Let R be the real field, R + the set of nonnegative real numbers. (4.19) Definition. A valuation of K is a mapping
If the valuation also satisfies the stronger condition (iv)
The trivial valuation is defined by the formulas
in §§4-5, so hereafter the term "valuation" always means "non-trivial valuation". The value group of a valuation

if and only if l/I(a)

~

1.

Each valuation

E

K:
B},

where B ranges over all positive real numbers. Equivalent valuations give the same topology on K. Given any non-archimedean valuation

R = {a E K:
~

I}.

Then R is a ring, and is called the valuation ring of <po The set p = {a E K:
is the unique maximal ideal of R. If

52

DEDEKIND DOMAINS

(4.19)

principal ideal, namely P = Rn where n is any element of P such that cp(n) < 1 and cp(n) generates the value group of cp. In this case R is a discrete valuation ring, by which we shall mean a principal ideal domain having a unique maximal ideal P, and such that P "# O. One way of obtaining archimedean valuations is as follows: the ordinary absolute value I I on the complex field C is an archimedian valuation, whose restriction to any subfield of C is an archimedean valuation on that subfield. Now let K be a field which can be embedded in C, and let tt:K ~ C be an embedding. Define cp:K ~ R+ by setting cp(a) = 1tt(a) I,

aEK.

Then cp is an archimedean valuation on K. In particular, every algebraic number field K can be embedded in C. For we may write K = Q(a), and let f(X) = min. pol.lia). Then there exist elements {IX;} in C such that f(X) = TI(X - IXi ), and we may define embeddings tti: K ~ C by tti{g(a)}

= g(rx i),

f or each

g(X) E Q[ Xl

If f(X) has r 1 real zeros IXl" .. , IXr! and 2r2 nonreal zeros in C, arranged in pairs of complex conjugates IX" + 1 , iXr! + 1" .• , IXr! + r2 ' iX r, + '2' then there are exactly r1 + r2 inequivalent archimedean valuations of K. These are giyen by the above construction, using the r1 + r2 embeddings K ~ C defined by letting a ~ IX;, 1 :( i :( r 1 + r2' (The embedding in which a ~ iXi is equivalent to that where a ~ lXi' for r1 + 1 :( i :( r1 + r2 .) The non-archimedean valuations of algebraic number fields can be described in a similar way, by means of embeddings into P-adic fields, which are fields complete with respect to certain non-archimedean valuations. We shall say more about this point of view in §5. In the present subsection, we shall instead give the connection between prime ideals of Dedekind domains and non-archimedean valuations. Assume now tha t R is a Dedekind domain, and let P be a nonzero prime ideal of R, or equivalently, a maximal ideal of R. For each nonzero a E K, we may factor the principal ideal Ra into a product of powers of prime ideals, as in (4.8). Let vp(a) denote the exponent to which P occurs in this factorization. If P does not occur, set vp(a) = O. Also, we put vp(O) = + 00. Now fix some K E R"', K > 1, and define

aEK,

a"# 0,

and cpp(O) = O. Then CPP is a discrete non-archimedean valuation on K, whose value group is the cyclic group generated by K. (If instead of K we used another real number K', with K' > 1, the valuation cp~ thus obtained would be equivalent to the above-defined valuation cpp.)

(4.19a)

53

LOCALIZA TIONS, VAL U ATIONS

We refer to vp as the exponential valuation associated with P. The properties of qJ p listed in (4.19) are consequences of the following properties of vp: (4.19a) For any elements a, b E K, we have (i) vp(a) = 00 if and only if a = O. (ii) vp(ab) = vp(a) + vp(b). (iii) vp(a + b) ~ min (vp(a), vp(b)), with equality whenever vp(a) -# vp(b). As in §3d, we shall denote by Rp the localization of Rat P, defined by Rp = {xis: xER,SER- P}. This ring is in fact the valuation ring of the P-adic valuation qJp on K, and its unique maximal ideal is precisely P . Rpo Thus Rp is a discrete valuation ring, and is automatically a principal ideal domain. We may choose a prime element n of the ring R p , that is, an element n E Rp such that nRp = p. Rp. Indeed, n may be chosen to lie in R. The fractional Rp-ideals of K are {nnRp:nEZ}. It follows from (3.13) that localization does not affect residue class fields, that is, RIP ~ Rpl p. Rp. This isomorphism is not only an R-isomorphism as asserted in (3.13), but is in fact an isomorphism of fields. More generally, there are ring isomorphisms n;:;'l.

Keeping the above notation, let J be any fractional R-ideal in K. Let vp(J) denote the exponent to which P occurs in the factorization of J into a product of powers of prime ideals. (4.20) THEOREM. Let P be a maximal ideal of the Dedekind domain R.

(i) For each fractional R-ideal J of K, the localization J p is given by

J p = (p. Rp)Vp(J) (ii) For any finitely generated R-module X, (ord X)p = ord (X p). P

(iii) For any finitely generated R-module M, + annRM = R.

we

have M p = 0

if and only if

Proof Statement (i) is well known (see references), and can easily be deduced from the formula vp(J) = min{vp(a):aEJ}. As a consequence of (i), we have R p = J p for each integral ideal J of R relatively prime to P.

54

(4.21)

DEDEKIND DOMAINS

Statement (ii) follows from the preceding remarks. First of all, unless X is an R-torsion module, both ord X and ord X pare O. Next, if X is a nonzero R-torsion module, it has a composition series X

=:>

X'

=:>

X"

=:> ••• =:>

X(k)

=:>

whose composition factors are (say) R/Pl'R/P 2 maximal ideals of R. Then Xp

=:>

(X')p

=:> ••• =:>

(X(k»)p

=:>

0,

, ...

,R/Pk , with the {PJ

0

is a descending chain of R-modules, possibly with repetitions, and with factor modules {(R/PJp}. But (R/PJp

~ Rp/(PJp ~ {~/P'

Therefore

as claimed. (See also Exercise 4.4). Finally we prove (iii), remarking in advance that the proof will be valid for the more general case where R is any commutative ring, and P a maximal ideal of R. Let S = R - P, a multiplicative subject of R. By definition, M p = S-l M; thus M p = 0 if and only if M is an S-torsion module. Since M is finitely generated as R-module, we see that M is an S-torsion module if and only if there exists an s E S such that s . M = O. But this occurs if and only if ann RM cf P, that is, if and only if P + annRM = R. This completes the proof. So far we have considered only localizations at a single maximal ideal P of R. In order to study relationships between "local" and "global" questions, one often needs results of the following type, in which the entire set of localizations are considered simultaneously. (4.21) THEOREM. Let R be a Dedekind ring with quotient field K, and let M be any R-lattice. Vzew M and its localizations {M p} as embedded in the vector space KM over K. Then M =

OM p, p

the intersection being formed within KM, and where P ranges over all maximal ideals of R. Proof By (4.13), the R-Iattice Mis R-projective. The desired assertion is now a consequence of Exercise 3.1.

In the same vein, we prove

(4.22)

55

LOCALIZA nONS, VALVA nONS

(4.22) Theorem. Let R be a Dedekind domain, Man R-lattice, and let V = KM. F or each maximal ideal P of R, let there be given a full Rp-lattice X(P) in V, such that X(P) = Mp a.e.t Dejine N = (jX(P), p

the intersection being formed within V. Then N is a full R-lattice in V, and N p = X(P) Proof Let V =

for all P.

f' KVi' and set L = t

i= 1

i= 1

Rvi . Then L is a full R-lattice in V,

and so Lp = Mp a.e.t (See Exercise 4.6). Therefore Lp = X(P) a.e., and so we may choose nonzero a, bE R such that a' X(P) c L pcb' X(P) for all P. Then N = nX(p) c a-I nLp = a-1L, so N is an R-lattice in V. Likewise L c bN, so N is a full R-lattice in V. It remains for us to show that N p = X(P) for each P. Since N c X(P), it follows that N pC Rp' X(P) = X(P). To prove the reverse inclusion, let us first write N = Rni + ... + Rn., s

where of course s ;:;:: r. Then each x E X(P) is expressible as a sum x =

I

ain i ,

i= 1

ai E K. By the Strong Approximation Theorem (4.11), we may find elements

bl

,.·.,

b s EK such that

bi - ai ER p , bi E Rp' (P'=f P), 1 ~ i ~ s, where P' ranges over all maximal ideals of R distinct from P. Let us put y = I bini' Then for pi =f P, we have y E Rp" N C X(P on the other hand, I

);

y - x = I(b i - aJniERp'N c N p ' so also YEX(P). ThereforeYEN,and so x = Y - (y - x) E N p' This shows that X(P) c N p' and completes the proof. Let M be an R-module, where R is any domain, and define the R-modules (4.23)

M* = HomR(M, R),

The evaluation map rp: M

--->

M** = HomR(M*, R).

M** is given by

{rp(m)} f = f(m},

fE M*.

Clearly rp = Oifandonly if M* = O. We call M rejlexiveifrp is an isomorphism. t "a.e." means "almost everywhere", that is, for all but a finite number of P's.

56

DEDEKIND DOMAINS

(4.24)

(4.24) THEOREM. Let R be a Dedekind domain. Then every R-lattice is reflexive, that is, M

~

M**.

Proof. By (4.13) each R-lattice M is isomorphic to an external direct sum of fractional R-ideals J of K. Since the functors involved in (4.23) commute with finite direct sums, it suffices to prove the theorem for the case where M = J. Now .1* = HomR(J, R), and each f E.1* extends to a map f' E HomK(K, K). Each such f' is given by a right multiplication by an element a E K, and so there is an identification J*

= {aEK:J·a

c R}.

Therefore J* = J-l, and so J** = (r 1)-1 = J. Thus the result holds for M = J, and hence also for every R-lattice M. (See also Exercise 4.17.) Combining the results of (4.24) and (4.21), we find that M**

= OMp p

for any lattice M over a Dedekind domain R. An analogous result holds under somewhat more general circumstances, described below. Let R be any domain; a minimal prime of R is a nonzero prime ideal of R which is minimal among the set of all nonzero prime ideals of R. Let M be an R-lattice, where R is a noetherian domain, and let V = KM. Then define dual spaces V** = HomK(V*, K).

The evaluation map gives a K-isomorphism V ~ V**. Since R is noetherian, M* and M** are also R-lattices, and there are embeddings M* c V*, M** c V**. Explicitly we have M**

= {vEV:f(v)ER forallfEM*}.

(4.25) THEOREM. Let R be a noetherian integrally closed domain, M an R-lattice, and let P range over the minimal primes of R. Then (i) For each P, Rp is a discrete valuation ring (that is, a principal ideal domain with unique maximal ideal p. Rp). (ii) R = r) Rp. (iii) For each nonzero x E R, x is a unit in Rp a.e. (iv) M** = (J M p. p

Remarks. The proof of (i) may be found in Serre [1, p. 19, Prop. 3]' As general

reference, see Bourbaki [5], Fossum [4]. Conditions (iHiii) are the postulates for a Krull ring. Thus the theorem asserts that every noetherian integrally closed domain is a Krull ring. The

(4.26)

RAMIFICATION INDEX; RESIDUE CLASS DEGREE

57

converse is false: Krull rings are always integrally closed, but need not be noetherian. For example Z[Xl' X 2 , •.• ] is a non-noetherian Krull ring. Every unique factorization domain is automatically a Krull ring. As examples of noetherian integrally closed domains, we list Z [X 1 ' ... , X n] and K[Xl' ... 'X n

J.

The following generalization of (4.22) is proved in the above-named references (see, for instance, Bourbaki [5, Ch. 7, §4, No.3]): (4.26) THEOREM. Let R be a noetherian integrally closed domain, and let M and L be full R-lattices in the K-space V. Let P range over the minimal primes of R. Then M p = Lp a.e. Further, let there be given for each P a full Rp-lattice X(P) in V, such that X(P) = M p a.e. Set N

= (J X(P). p

Then N is a full reflexive R-lattice in V, and N p = X(P) for all P. 4c Ramification index; residue class degree

Throughout this section, R is a Dedekind domain with quotient field K, and L is a finite separable field extension of K. We shall denote by S the integral closure of R in L, so by (4.4) and (4.7), S is a Dedekind domain with quotient field L, finitely generated as R-module, of R-rank (L: K). To avoid trivialities, we assume always that R =f K and L =f S. Given a maximal ideal P of R, the integral ideal P . S can be written as 9

(4.27)

p·S

=

n Pt,

i= 1

where the {PJ are distinct maximal ideals of S, and the {eJ are positive integers. Call ei the ramification index of Pi for the extension LIK, and write (4.28)

ei

=

e(Pi , LjK),

I

~

i

~ g.

We say that Pi is unramified in LIK if ei = 1 and SIP i is a separable extension of RIP. Likewise, we say that Pis unramified in the extension LIK if each Pi is unramified in L I K. Turning to residue class fields, we observe at once that for 1 ~ i ~ g, the field SIP; is an extension field of RIP. We set (4.29)

J; = f(P i , LIK) = (SIP i : RIP),

and call J; the residue class degree of Pi relative to the extension LI K. The finiteness of the {J;} is a consequence of the following basic result:

58 (4.30)

(4.30)

DEDEKIND DOMAINS THEOREM.

Keeping the above notation, we have 9

L eJ = (L:K).

i;;;; 1

Proof. Let n = (L: K), so rankRS = n. Let Sp = Rp . S; then Sp is a torsionfree Rp-module of Rp-rank n. But Rp is a principal ideal domain, and so Sp ~ Rp(n). Therefore SIPS

~

SpIP'S p

~

(Rpl p. Rp)(n)

~

(RIP)(n)

by virtue of (3.14). This shows that the dimension of SIPS as vector space over the field RIP is precisely n. We may also compute this dimension as follows: by (4.10) we have !J

SIPS = S/IlP/'

L'(SIP/').

;=1

There is a descending chain of RIP-modules SIP; e,

::>

PjP i e,

::> .•• ::>

Pi e,-lI P ;e,

::>

0,

each of whose factor modules is isomorphic to SIP; by (4.12). Hence dimR/p(SIP i e,) = ei ' dimR/p(SIPJ = eJ;, ' for each i, 1 :( i :( g. This gives fJ

I eJ;,

i=l

and the theorem is proved. (4.31) Comments. (i) It is easily verified that Sp is the integral closure of Rp in L, and that { Pi' Sp : 1 :( i :( g} ~re the distinct maximal ideals of Spo Furthermore,

where S p is the localization of S at Pi' and where S p = Rp' S. Note that Rp is a discr'ete valuation ring, but S p is not (unless 9 = 1). In connection with this remark, see Exercise 3.4. (ii) If (4.27) holds, then we claim that Pi 11 R = P, 1 :( i :( g. Indeed, Pi 11 R is a prime ideal of R containing P, or else possibly Pi 11 R = R. The latter alternative cannot occur, since 1 ¢ Pi' On the other hand, starting with any maximal ideal Pi of S, we may set P = Pi 11 R, a prime ideal of R. Surely P -# 0, since for each nonzero a E Pi' the constant term in min.pol.K (a) is a nonzero element in P. Thus Pi ::> P,

(4.32)

RAMIFICATION INDEX; RESIDUE CLASS DEGREE

59

and so Pi :::;) p. S, whence Pi occurs in the factorization of P . S into maximal ideals ofS. (iii) We may define a relative norm map N L/K which associates to each fractional S-ideal J of L, a fractional R-ideal NL/K(J) of K. For each maximal ideal Pi of S, we set (4.32)

N L/K(P) = pi" where P = Pi

1\

R, j,

=

f(P" LI K).

Since these {P,) form a free basis for the abelian group of fractional S-ideals in L, we may then define N L/K on this group by requiring it to be multiplicative: N L/K(JJ1 = N L/K(J) . N L/K(l').

We could also have defined the norm map NL/K as follows: for each nonzero ideal J of S, put N L/K(J) = ordR(Sp),

where ord Rdenotes R-order ideal (see §4a). If J' is another nonzero ideal of S, there is aU R-exact sequence 0-> JIJJ'

->

SIJJ'

S/l-> 0,

->

and by (4.12) the term JIJJ' is isomorphic to SIJ'. From (4.17) we then conclude ordR(S/ll') = ordR(S/l)'ordR(S/l'),

that is,

N(ll') = N(l)·N(l').

If we now put -

-1

= N- L/K(J 1 )' N- L/K(J2 ) -1 for 1 1 ,12 nonzero ideals of S, it follows that NL/K is well defined on the group N L/K(J 1 · J 2

)

of fractional S-ideals in L. We now show that N L/K coincides with NI~/K'

It suffices to verify that

(4.33) using the notation of (4.32). -But SIP i is a field extension of RIP of degree J;, so as R-modules we have SIP i ~ (RIP)(fi) , and then (4.33) is obvious. (iv) In the special case where K = Q, R = Z, one defines a counting norm .JV2!Q ' by setting

1;

= f(P i , LIQ),

where p is the positive rational prime such that pZ = Pi 1\ Z. Thus ./VL/Q(P,) is the number of elements in the finite field SIP i . As usual, ./VL / Q is defined on all fractional S-ideals in L, so as to be multiplicative. Then for each nonzero ideal 1 in S,

60

(4.33)

DEDEKIND DOMAINS ,Af~/Q(J)

= card (S/J),

where card (SjJ) denotes the number of elements in the ring SjJ. (v) If L is a galois extension of K, with galois group G, then for each (J E GwehaveS" = S, whereS" = {X":XE S}. It is well known (see references) tha t G acts transitively on the set of prime ideals {PJ occurring in (4.27), that is, there exist elements {(Ji} E G such that Pi = (P 1)"', 1 ~ i ~ g. Furthermore, in this case we have e 1 = .. ·=eg, !l="'=!g' 4d Different, discriminant

As in §4c, let L be a finite separable field extension of K, let R be a Dedekind domain with quotient field K, and let S be the integral closure of R in L. Let T:L x L -+ K be the nondegenerate bilinear trace form defined in (4.5), that is,

x, YEL. For any set J c L, put T(X,1) = TL/K(xJ) = {T(X, y): y E J}.

We prove at once that TLlK(S) c: R. Indeed, each a E S is integral over R, so char. pol.L/K a E R[X] by Exercise 1.1. Therefore TL/K a E R, as desired. Given any full R-Iattice M in L, we can define a complementary lattice

M = {x EL:T(X,M)

c

R}.

The map M -+ M is inclusion-reversing. Let us verify that M is a full R-Iattice in L. First observe that M is contained 10 some full free R-Iattice n

N =

L· Rxi'

where n = (L: K). There exist elements {yJ

E

L such that

i= 1

T(Xi' Yj)

= bij' 1

~ i,j ~ n.

Then n

M~N= LORy .. j= 1

J

Similarly, 1\1 is contained in some full R-Iattice in L. Therefore M is itself a full R-Iattice in L, as claimed In particular, if J is a fractional S-ideal in L, then J is an S-submodule of L which is an R-Iattice, and thus J is also a fractional S-ideal of L; we call J the complementary ideal of J. For the case where J is S itself, we note that '8 ~ s, so that '8- 1 is an integral ideal of S. Let us define the d!fferent of S with respect to R as and then

'8 =

'J:J(S/R) = '8- \ 'J:J(S/R)-l may be referred to as the inverse d!fferent.

61

DIFFERENT, DISCRIMINANT

(4.34)

(4.34) THEOREM. Let J be a fractional S-ideal in L. Then

Proof Let us abbreviate TL/K as T, and !)(S/R) as !). We have T(J· rl!)-l) = T(S' !)-1) C R,

so J

::::J

r

1 !) - 1.

On the other hand, R

::::J

T(JJ) = T(JJ· S),

whence JJ c;: !)- \ so J c r 1!)- 1. This proves that formula for j is then obvious (see also Exercise 4.l2).

J=

r

1. !) -1.

The

The discriminant of S with respect to R is defined as d(S/R) = N L/K(!) (S/R)),

the norm of the different !)(S/R). Thus d(S/R) is a nonzero ideal of R, often called the discriminant ideal of SIR. (4.35) THEOREM. (i) For each maximal ideal P of R,

{d(S/R)}P = d(SP/Rp).

(ii) If n

S=

L:' Rx" i= 1

n

= (L:K),

then

Proof For P a maximal ideal of R, and J any S-ideal in L, we put J p = Rp' J, an S p-ideal of L As remarked earlier, Sp is the integral closure of Rp in L, so in the passage from the pair R, S to the pair R p, Sp, we have introduced a new pair of Dedekind domains, with quotient fields K, L, respectively. The advantage of this procedure is that the new domain Rp is a principal ideal domain, and thus Sp is Rp-free on n generators, where n = (L: K). Since the trace map TLIK is K-linear, it follows easily that

(4.36) Furthermore, for each S-ideal J in L, we have NL/K(J p) = {NL/K(J)}P'

that is, forming norms commutes with localization. Applying N L/K to (4.36) yields assertion (i) of the theorem.

62

(4.37)

DEDEKIND DOMAINS

f: Rx

Now suppose that S happens to be R-free, say S =

i= 1

elements {Y j } EL with TL/K(XiY) =

(5ij'

i•

There exist

1 ~ i,j ~ n. Then

!:J(S/R) = S- 1, where S =

t

j=l

RyJ •

We have d(S/R) = NL/K(!:J(S/R)) = ordR(S/!:J(S/R))

= ordR(S/S-l) = ordR(S/S) (by (4.12)). Write n

Xi =

I

j=l

aijYj

'

Then (Exercise 4.2) ordR(S/S) = R· det(a i ) . But au = TL/K(X i X j)' which completes the proof of (ii). We may remark that this theorem enables us to compute the discriminant d(S/R) by calculating each "local" component d(S/R)p, and that by (ii), each local component can be obtained once an Rp-basis of Sp is known. The different !:J(S/R) and the discriminant d(S/R) are of great importance because they give information about ramification of prime ideals in the extension L/K. Recall that a maximal ideal Pi of Sis unramifled in L/K if e(P" L/K) = 1, and S/P i is a separable extension of RIP. Likewise, a maximal ideal P of R is un ramified in LIK if p. S = product of distinct unramified prime ideals of S. (4.37) THEOREM. A maximal ideal Pi of S is

unram~tled

if and only if Pi t !:J(S/R). A maximal ideal P of R and only if Ptd(S/R).

TI Pi' a

in the extension L/ K

is unramifled in L/K

if

Proof Let R = R/P, S = SIPS, n = (L: K); then S is an algebra over the field R, of dimension n. For f(X) E R[X], let f(X) denote its image in R[X]. We claim that for a E S, (4.38)

char. pol.L/K a = char. POl.SIR a.

To prove this, first replace R by R p' and S by Sp. Such a change does not affect R or S. Then we may write n

Sp =

I" Rpx"

i= 1

n

S=

I" Rx

i=1

i·

(4.38)

63

GLOBAL FIELDS

For a E S, we have aX j

=

L

(XijX i ,

aX j

=

L ,xili'

(Xu E

R.

Then char. pol. (a) is the characteristic polynomial of the matrix ((Xij)' while char. poL (a) is that of (,xij)' which establishes (4.38) at once. Now let Pi be a maxImal ideal of S, and let p. S = Pt J, where e ~ 1 and where J is an integral ideal of S relatively prime to Pl' For each a E Pi J we have ae = 0 in S, and therefore char. pol.sil~ a =X". Hence from (4.38) we conclude that TL1K(P J) c P. Writing::D instead of:D(SIR) for convenience, we then obtain Pi J c P::D - 1, and consequently ::D c Pi e -1. Thus Pi e -ll::D, and therefore P 11::D whenever e > 1. To complete the proof, we now assume that e = 1, and we must show that Pi I::D if and only if SIP 1 is inseparable over R. Since e = 1, we have

S = SIPS ~ SIP 1 -i- SIJ, and each a E J maps onto an ordered pair (a', 0), where a' is the image of a in SIP 1 . Set SIP 1 = S' for convenience of notation, and let k = (S/J:R). Then for a E J, char. pol. siR

a = X k • char.

pol.s' (Ii a'.

This implies at once that

TS(ii a = TS'iii ai, a E J. If S'IR is inseparable, then TS'IR is the- zero map, and thus by (4.38) we have TL1K(J) c P. This yields J c P::D-l, whence P 11::D. The reverse argument is equally simple: suppose that S'IR is separable. Then there exists an a E J such that TS'ik a' ;/; 0, and therefore TL1K(J) ¢ P. Hence J ¢. P::D- 1 , that is, P 11::D. This completes the proof of the first assertion of the theorem. The second assertion is an obvious consequence of the first.

4e Global fields A global field K is either an algebraic number field (that is, a finite extension of the rational field Q), or else a function jield (that is, a finite extension of a field k(X) of rational functions in an indeterminate X over a finite field k). A prime of K is an equivalence class of valuations of K. We exclude once and for all the "trivial" valuation
64

(4.39)

DEDEKIND DOMAINS

If K is a function field, there are no archimedean primes, and the nonarchimedean or finite primes arise from discrete valuation rings in K. In this case, as well as in the number field case, the residue class field associated with a finite prime P is a finite field; let AlP) denote the number of elements in this residue class field (see (4.31 (iv)). For each finite prime P of a global field K, we normalize the P-adic valuation q>p by setting

q>p(a)

a E K,

= AlPtvp(al,

a =I 0,

q>p(O) = 0 (see §4b). If P is a real prime of K, that is, an infinite prime arising from an embedding /1: K --;. R, let q>p(a) = 1/1(a)l, a E K. If P is a complex prime of K, that is, an infinite prime arising from an embedding /1: K --;. C such that /1(K) rj:. R, let q>p(a) = 1/1(aW,

a E K.

Thus we have a normalized valuation q>p for each prime P of K. (4.39) THEOREM (Product Formula). Let K be a global field. Then for each nonzero a EK, q>p(a) = 1 a.e., and

TIp q>p(a) =

1,

where P ranges over all primes of K. For global fields, there is a stronger version of(4.11): (4.40) THEOREM (Very Strong Approximation Theorem). Let PI"'" P n be distinct primes of K (finite or il'!flnite), and let Po be a fixed prime distinct from these. Given any elements a l , ... E K and any c > 0, there exists an element a E K such that

,an

1

~

i ~ n,

P =I Po' PI" .. ,Pn '

(4.41) COROLLARY. Let {P I " ' " Pn} be distinct primes of the algebraic number jield K, and suppose that there exists an infinite prime Po of K distinct from PI' ... , Pn· Let aI' ... , an E K be such that ai E R p, if Pi is a finite prime, where R = alg. int.{K}. Let 0 < c < 1. Then there exists an aER such that

q>p,(a - at) < c, q>ia) < 1,

P infinite,

1 ~ i ~ n, P =I PO' PI"'" P n '

65

EXERCISES

EXERCISES Throughout, R is an integral domain with quotient field K. Unless otherwise stated, R is assumed to be a Dedekind domain.

1. Let J be a nonzero ideal of R. Using (4.9), find all ideals of R which contain J. Prove that RjJ is an artinian and noetherian ring. Deduce from this that every finitely generated R-torsion R-module possesses an R-composition series.

2. Let R be a principal ideal domain, and let N Suppose that

I'Rm;,

M be R-lattices with KN

=

KM.

r

r

M =

c

N = I'Rnj' j= 1

i=l

Show that ord R MIN

= R'

det(lXij)'

[Hint: After change of R-bases of M and N, we may assume that the matrix (lX i ) is diagonal. Such basis changes have the effect ofmultiplyingdet(lXi ) by a unit factor from R, and thus do not change R . det(IX;). Finally, if (IX) is diagonal, it follows from (4.17) that ord R MIN is the principal ideal generated by the product of the diagonal entries.J 3. Let X be a finitely generated R-torsion module. Show that ordRX and annRX have the same prime ideal factors, apart from multiplicities. Explain the connection between this result and Theorem 1.7. 4. Let ReS be an inclusion of Dedekind domains, and let X be any finitely generated R-module. Prove that S0

R

ordRX

=

ord s (S 0

R

X).

[Hint: By (4.1S) it suffices to handle the cases X = J and X = RjJ, with J a nonzero ideal of R. The result is obvious for X = J, while for X = RjJ it follows from Exercise 2.6.J 5. Let I E HomR(M, M), where M is an R-lattice. Prove that ord R Mlf(M)

= R'

detj,

where detl is computed by extending Ito a K-linear transformation on KM. [Hint: Use (4.20(ii)) to reduce the problem to the case where R is a principal ideal domain, and then use Exercise 2.J 6. Let M, N be full R-lattices in a K-space V. Prove that Mp exist nonzero IX, 13 E R such that IXM c N c f3M.J

=

N p a.e. [Hint: There

7. Let R be any domain, M a finitely generated R-module, and let cp:M --+ M** be the evaluation map (see (4.23)). Show that if Mis R-torsionfree, then cp is monic. Show that M* = 0 if and only if M is an R-torsion module. S. Let R be a noetherian integrally closed domain, and let M, N be full R-lattices in a K-space V. Suppose that M peN p for each minimal prime P of R, and let N be reflexive. Prove that MeN.

66

(4.42)

DEDEKIND DOMAINS

9. Keeping the hypotheses and notation of §4c, prove that N L/K(Sa) = R· N L/K(a),

a E L.

[Hint: Use (4.31(iii» and Exercise 5.] 10. Prove that the symbols e(P" Lj K) and f(P" Lj K) defined in §4c are mUltiplicative for towers of separable extensions. ll. Prove that the different !)(SjR) defined in §4d is mUltiplicative for towers of separable extensions. [Hint: Let L2 => Ll => K be such a tower, and let S, be the integral closure of R in L" i = 1,2. Set

T,

= T LdK ,

!), =

!)(SJR),

To prove that !)2 = !)1 !)', show the equivalence of the following statements, for x E L2 (use Exercise l.5): T 1 (SI' T'(xS 2 »

c

R;

X!)I c !)'-1.]

12. Let r: V x V ~ K be a nondegenerate symmetric K-bilinear form on a finite dimensional K-space V, and let M be a full R-lattice in V. Define

At

=

{x E V:r(x, M) c R}.

Show that At is a full R-lattice in V, and that M = M. [Hint: If N ~ L'Rx, is a full R-lattice in V, then iii = I'RYi' where r(x"y) = (jii' and then iii = I'Rx, = N. Hence if N 1 c MeN 2 with N I ' N 2 R-free, we have iii 1 => At => iii 2' whence At is a full R-lattice in V. To prove that M= M in general, show that the process of forming At from M commutes with localization, and thus the problem can be reduced to the case of Rp-lattices.] 13. Keeping the notation of Exercise 12, let n = rank R M. Define the discriminant ideal of M with respect to r to be the R-ideal d(M) in K generated by {det (r(xi ,X)I"i.i".:

XI""

,X. EM}. n

Prove that d(M p)

=

d(M)p for each maximal ideal P of R. Show that if M = L'Rxi' i= 1

then d(M) = R . det (r(x" xi»! "'. i".' If N

c

M, where N is another full R-lattice in V, prove that

(4.42)

d(N) = (ord R MjN)2. d(M),

and that d(N) c d(M). Deduce from this that d(M) = d(N) if and only if M = N. [Hint: To prove (4.42), first localize, and then write R in place of Rp. Use the notation of Exercise 2. Then

14. Let L

= K(IJ.) be

a finite separable extension of K, and let

67

COMPLETIONS

IT" (X

f(X) = min. pol.K IX =

- IX;),

i= 1

where IX 1 ( = IX), IX 2 , criminant of IX as

••• ,

IXn are the algebraic conjugates of IX over K. Define the dis-

Prove that 1 disc IX

=

det [

TI

1

~i<j~n

1]2 .I

IX]··· IX~-

~ ~2.:: 1

an·· . a~- 1

(IX; - IX

Y

=

± N L/K(f'(IX)),

where f'(X) is the formal derivative of f(X). 15. Keeping the above notation, let f(X) the integral closure of R in L. Show that

E

R[ X], so IX is integral over R. Let S be

R EBRIX EB··· EBRIX n-

1

c

S,

and hence prove that the discriminant d(SjR) divides R· disc IX.

t

16. Let OJ be a primitive mth root of lover K, where char K m. Show that the only prime ideals of R dividing disc OJ are those which divide m. Hence if P t Rm, then P is unramified in the extension K(OJ) of K. l7. Give another proof of (4.24) by showing that every finitely generated free Rmodule is reflexive, and then proving that every finitely generated projective R-module is reflexive.

5.

COMPLETIONS AND VALVA TIONS 5a Completions

We shall review, without proof for the most part, some elementary facts about completions. The reader is referred to the standard references, listed at the start of § 4, for details and proofs. Let K be a field with a valuation cp, topologized as in §4b. Let K denote the completion of K relative to this topology; then K is a field whose elements are equivalence classes of Cauchy sequences of elements of K, two sequences being equivalent if their difference is a null sequence. The field K is embedded in K, and the valuation cp extends to a valuation cp on K. The field K is complete relative to the topology induced by cp, that is, every Cauchy sequence from K has a limit in K. If cp is an archimedean valuation, then so is cp, and K is a complete field with respect to an archimedean valuation. The only possibilities for K are

68

COMPLETIONS AND VALUATIONS

(5.1)

R (the real field) or C (the complex field), and in each case ip is equivalent to the usual absolute value. If ({J is non-archimedean, so also is ip; the two valuations have the same value group, and the same residue class field (up to isomorphism). In particular, let R be a Dedekind domain with quotient field K, where K "# R, and let P be a maximal ideal of R. The completion of K with respect to the P-adic valuation ({Jp on K (see §4b) will be denoted by Kp (or just K, ifthere is no danger of confusion). Call K p a P-adic field, and its elements P-adic numbers. The discrete valuation ({Jp extends to a discrete valuation (pp on Kp. We have already remarked that the valuation ring of ({Jp is the localization Rp; let Rp be the valuation ring of ipp. Every element of Rp can be represented by a Cauchy sequence from Rp (or from R, for that matter). If n is a prime element of R p , then n is also a prime element of R p . Let :7' temporarily denote a full set of residue class representatives in R of the residue class field R = RIP, with 0 E:7'. Each x E Rp is uniquely expressible as

x = IXo

+ IXl n + IX2 n 2 + ... ,

IX; E :7',

and each y E Kp is uniquely of the form y = n . x, k E Z, where uniqueness is guaranteed by the condition that IXo "# O. (If y = 0, take k = - ro). We note that K is dense in K p , that is, given any IXEKp and any 6 > 0, there exists an element a E K such that k

ipp(a - IX) <

6.

Likewise R is dense in R P' and so also is R p .

(5.1) THEOREM. (i) Rp is a faithfully flat Rp-module. (ii) For each k ~ 1, there are R-isomorphisms

RIPk

~

Rplnk Rp

~

~/k~ Rp n R p,

where n is a prime element of Rp. Proof. The first assertion is contained in (2.22) (see also Exercise 5.3). The first part of (ii) is proved in (3.13). Finally, the composition of maps R p -+ R p -+ R pi n k R p yields a map A.: R p -+ R pi nk R P' and A. is an epimorphism since Rp is dense in Rp. Clearly ker A. = Rp (\ nk Rp = nk R p, which completes the proof. For any R-module M, let Mp

= Rp®RM,

Thus the passage from M to 1\1 p can be accomplished in two steps: localization (from M to M p), and completion (from M p to 1\1 p). In the second step, we

(5.2)

69

COMPLETIONS

start with a discrete valuation ring R p, and pass to its completion Rp. We shall analyze this step in some detail below. (5.2) THEOREM. Let R be a discrete valuation ring with prime element n, quotient field K, and let R, K denote completions. (i) For each finitely generated R-module M, the map M -+ R ®R M is an inclusion. Denote R ®R M by RM hereafter. (ii) Let V be a finite dimensional K-space, and set V = K ®K V. Theformulas T=RM,

M

= Tn V,

give a. one-to-.one inclusion-preserving~correspondence between the set of full R-lattlces M In V, and the set of full R-lattice Tin V. Proof. For (i), we note that M is isomorphic to a direct sum of cyclic modules of the form R/nkR, k ~ 0, so it suffices to prove the result for M = R/nkR. But then M -+ R ®R M = R/nkR is surely monic (obvious when k = 0, and true for k > by (5.1)). To prove (ii), use bases! Suppose first that M is a full R-lattice in V. We may write

°

n

M

j;

n

n

= I"Rx

V

j ,

1

=

I" Kxp j;

M=

1

I" Rxp j;

1

where n = (V: K). Then

Mn

=

V

I" (R n

K)Xj

as desired. Conversely, let T be a full R-lattice in

=

V,

I" RXi = M,

and write n

n

T

=

I"

V

RXi'

=

I" KYj

i; 1

(note that (v: K) = (V: K)). Then y!.. = I CTijXj' CT ij E K, 1 ~ i ~ n, and S = (CT ij ) is an invertible matrix over K. Choose T = ('l:jj) with entries in K close to those of S-l (in the topology induced by the valuation); then TS is a unit in the matrix ring Mn(R). Set

y; = Then we have

and so

I'l:ijYj

=

I

'l:ijCTjkX k •

70

(5.3)

COMPLETIONS AND VALUATIONS

Thus Tn V is a full R-Iattice in V such that R(T n V) = T. This completes the proof of the theorem. We may now prove analogues of(4.21) and (4.22): (5.3) Theorem. Let R be a Dedekind domain with quotient field K, let M be an R-lattice, and let V = KM. Let P range over all maximal ideals of R. (i) We have M = KMn{nMp}' p

(ii) For each P, let there be given a full R p-lattice Yep) in = M p a.e. D~fine

Yep)

N

=

V

n{n

Kp V, such that

Yep)}·

p

Then N is a full R-lattice in V, and Np = Yep) for all P. Proof. By (5.2), M p = KM n M p for each P. Assertion (i) now follows at once from (4.21). To prove (ii), set X(P) = V n Yep) for each P. By (5.2), X(P) is a full R p-Iattice in V. Clearly X(P) = V n M p = M p a.e., and N = X(P). By (4.22) N is a full R-Iattice in V, and N p = X(P) for all P.

n p

Therefore proved.

Np =

Rp . X(P) = Yep) for all P, by (5.2), and the theorem is

We indicate the generalization of (5.3) to the case where R is a noetherian integrally closed domain. Recall that for each minimal prime P of such a domain R, the localization Rp is a discrete valuation ring, and so we may fonn its completion Rp just as above. (5.4) THEOREM. Let M be a full R-lattice in a finite dimensional K-space V, where R is a noetherian integrally closed domain with quotient field K. Suppose that for each minimal prime P of R, we are given an R p-iattice Yep) in Vp ,

such that Yep)

= Mp

a.e. Set N

=

V

n{n

Y(P)}.

p

Then N is a full R-lattice in V, N is reflexive, and Np = Yep) for all P. Proof. The theorem follows readily from (4.26) and (5.2). 5b Extensions of complete fields

Throughout this section, let K be a field complete with respect to a valuation cp, either archimedean or not, and let K be an algebraic ,5losure of K. Then (see references) we may extend cp to a valuation rp on K, as follows:

(5.5)

EXTENSIONS OF COMPLETE FIELDS

71

each a E K lies in some field L with K c L c K, (L :K) finite (for example, L = K(a) will do). Set (5.5) q>(a) = {cp(NL1Ka)pI(L:K). Then we find that the value q>(a) is independent of the choice of L, and that every finite extension of K contained in K is complete with respect to the valuation q>. When cp is archimedean, there are only two possibilities: (i) K = C = K, cp = q> (ii) K = R, K = C, q> extends cp, where cp and q> are the usual absolute values on R or C. If cp is non-archimedean, so is q>. However, ijJ need not be a discrete valuation, even if cp is discrete. We shall consider discrete valuations in more detail below, and for this purpose let us adopt a more systematic notation. If cp is a discrete valuation on K, denote by OK its valuation ring, and by PK the maximal ideal of OK . Let OK = 0K/PK be the residue class field, and let PK = n K' OK' so n K is a prime element of OK . Let vK be the exponential valuation on K, defined by setting aR = p~K(a), a E K, a i= 0, and vK(O) = + 00. Any finite extension L of K can be embedded in K, and the restriction of q> to L gives a discrete valuation r/J which extends cp. We define the ramification index e = e(L/K) and residue class degree f = f(L/K) by the formulas

(5.6) THEOREM. Let L be a finite extension of the complete field K, and keep the

above notation. Then (i) 0L is the integral closure of OK in L. (ii) Both e(L/K) and f(L/K) are finite, and e(L/K)' f(L/K) = (L:K). (iii) For each a E L, (iv) Let a 1 , •.. , af E 0L be such that 0L = 2:' 0Kai' and let no, n 1 , · · · , n e - 1 E 0L be such that vL(n) = j, 0 ~ j ~ e - 1, where e = e(L/K), f = f(L/K). Then the e felements rain) form a free 0K-basisfCK 0c For future use, we state (5.7) THEOREM. (Hensel's Lemma). Let f(X)

0K[ X] there is a factorization

E

0K[X] and suppose that in

72

COMPLETIONS AND VALUATIONS

](X) = u(X) v(X),

(S.8)

u(X) monic,

where u(X) and v(X) are relatively prime. Then there is a factorization in OK

[X]:

f(X) = g(X)h(X), such that 9 = u, Ii = v.

g(X) monic,

Recall that L is an unramified extension of K if e(L/K) = 1 and 0L is a separable extension of OK' The study of unramified extensions reduces at once to the study of separable extensions of the residue class field OK by virtue of the following basic result (see references): (S.8) THEOREM. (i) There is a one-to-one inclusion-preserving correspondence between the set fields L such that

K c L c K, L = finite unramified extension of K, and the set of fields k finite separable over OK' This correspondence assigns to each such field L the field k given by k = L . (ii) If L = K(a), where a is a zero of a monic f(X) E 0K[ X] such that a is a simple zero of J(X), then L is unramified over K, and

°

0L = 0K[a], 0L = 0K(a),

(L:K) = (OL:OK)'

Conversely, every unramlfted extension L of K is of this form. (iii) Given k = 0KW, let f(X) E 0K[ X] be any monic polynomial such that J(X) = min. POl.PK(' Then there exists a field L = K(a) c K such that L is unramlfted over K, and 0L = k, and further f(X) = min. pol.K a,

a= (

in 0L'

(iv) L/K is a galois extension if and only if adaK is a galois extension. If G is the galois group of L/ K, then each (J E G induces an element (j in the yalois group G of 0L/OK . The map (J -+ ij gives an isomorphism G ~ G. (S.9) COROLLARY. Let (E:K) < with respect to the properties

00.

There is a unique field W which is maximal

K eWe E, W is unramified over K. This field W is the inertia field of the extension E/K; it is characterized by the facts that W /K is unramified, and Ow is the separable closure of OK in 0L' that is, Ow = {x E 0L : x is separable over OK}' (S.lO) THEOREM. Suppose that the complete field K has finite residue class field OK' and let q = card OK' Then for each positive integer f, there is a

(S.11)

73

EXTENSIONS OF COMPLETE FIELDS

unique unramified extension W of K such that (W:K) = (ow: OK) = f, namely, W = K(w) for w a primitive (qf - l)-th root ofl over K. Furthermore,

Ow

= 0K[ w],

Ow

= 0K(W),

Proof. There is a unique separable extension of OK of degree f, namely is a primitive (qf - l)-th root of lover OK' Hence by (S.8) there is a unique unramified extension W of K such that (W: K) = (ow: K) = f. We need only show that K(w) is an unramified extension of K such that (K(w): K) = f. However, w is a zero of the polynomial 0K(~)' where

e

°

f(X)

= X Qf - 1

-

1 EOK[X].

Since f(X) has no repeated zeros, it follows from (S.8ii) that K(w) is unramified over K. Further

where 1 ~ i ~ qf I in each product. Hence W is a primitive (qf - l)-th root of 1, so by (S.8 ii) (K(w):K)

= (OK[W]:0K) = f·

This completes the proof of the theorem. (S.11) COROLLARY. Keep the notation of (S.10). Then W/K and OW/OK are galois extensions, with galois groups G, G cyclic of order f. The group G has generator 0': w ~ w q , while G is generated by a: W ~ wq• Call 0' the Frobenius automorphism of the extension W /K. Proof. It is well-known that G is cyclic of order f with generator a, since G is the galois group of 0K(W)/OK' and 0K(W) is the splitting field of f(X) over OK' If g(X) = min. poLoK (w), then f-1

g(X)

=

TI

(X -

w

Qi ).

;=0

Therefore f-1

min. poLK w =

TI

(X - w

qi

),

i=O

and hence O':w

-+

w Q is a generator of the group G.

The extreme opposites of unramified extensions are the completely ramified extensions. Recall that L is a completely ramified extension of K if L = OK' that is, f(L/ K) = 1. Such extensions can be classified, as follows: an Eisenstein polynomial is one of the form

°

74

(S.12)

COMPLETIONS AND VALUATIONS

f(X) = xn

+ a 1Xn-1 + ... + an EOK[X],

where each a; E PK' and an ¢ pi· By Exercise S.1, f(X) is irreducible in K[ X]. (S.12) THEOREM. (i) Let L be a completely ramified extension of K of degree n. Then min. pol.KnL is an Eisenstein polynomial over OK' and L = K(n L ). (ii) Let f(X) be an n-th degree Eisenstein polynomial over OK' and let L = K(rx), where min. pol.K rx = f(X). Then L is completely ramified over K, (L:K) = n, and rx is a prime element OfOL' Furthermore; 0L = 0K[rxJ. The proof is straightforward (see references), and we omit it. We call L/K an Eisenstein extension if it is of the type described in (S.12). Now let E be any finite extension of K, and suppose that the residue class field OK is perfect, that is, OK has no inseparable extensions. This certainly is the case when OK is a finite field. If W is the inertia field of the extension E/ K, then we have K eWe E,

oft =

0E'

f(W/K) = f(E/K),

e(W/K) = 1,

f(E/W) = 1,

e(E/W) = e(E/K).

Thus the step from K to E is divided into an unramified step from K to W, followed by a completely ramified step from W to E. We have seen that E is an Eisenstein extension of W, and if OK is finite, then W is a cyclotomic extension of E. 5c Extensions of valuations

Throughout this section, K denotes a field with a valuation cp, archimedean or not, and (jJ the extension of cp to the algebraic closure n of the completion K, as in § Sb. Given a finite separable extension Lover K, we wish to determine all extensions ofthe valuation cp from K to L. Each such extension determines an embedding of L in n which preserves the embedding of K in K. Two embeddings jl, jl' of Linn are called equivalent if there exists a K -isomorphism Cf: p(L) ~ p' (L) such that Cftl = p'. Let P1 , ... , p, be a full set of inequivalent isomorphisms of L into n which preserve the embedding of K in K. Let Li = K . p,(L), the composite of K and Pi(L) in n (that is, the smallest subfield of n containing both), and set n i = (L;:K). Then (see references) there are precisely r inequivalent valuations rjJ l ' . . . ,If, of L which extend cp, and these are given by the formula (S.13)

1 ~ i ~ r.

In order to obtain a full set of inequivalent embeddings of L in n, write L = K(fJ), f(X) = min. pol.K fJ, so f(X) is a separable polynomial. Factor

75

EXTENSIONS OF VALUATIONS

(5.14)

f(X) into irreducible factors in K[X]: f(X) =

TI J;(X),

!;(X) irreducible, J;(X) E K[X].

i= 1

The {J;} are necessarily distinct, and we may choose elements {tlJ En such thatf;(tli) = O. Any K-isomorphism WL -+ n must carry (J into a K-conjugate of some tl, , and since we are interested in embeddings only up to equivalence, we may indeed choose 11, so that Il,(tl) = tl,. In this way we obtain r inequivalent embeddings III , ... , I1r of L in n, and for each i, 1 :( i :( r, we have

Ii = K(OJ, Now

L = K(tl) ~ K[X]/(f(X)), and thus

K @KL ~ K[X]/(f(X)) ~

r

r

,= 1

i= 1

L' K[X]/U;(X)) ~ L . I,.

Hence for each a E L, we have r

char.

(5.14)

POl.LIK

K

a

=

TI char. POl.f.;/K JI,(a), ,= 1

L

since a acts on @K L just as 11,(a) acts on from (5.14) the important formulas TLIK a

=

L, Tfi!F:' p,(a),

LO I,. In particular, we obtain

N LIK a

=

TI, N Lil it pla).

Briefly, global trace is the sum of local traces, and global norm the product of local norms. Suppose now that R is a Dedekind domain with quotient field K, and S the integral closure of R in L. For each maximal ideal P of R, let r

p·S =

TI P/i

,=1

be the factorization of p. S into a product of powers of distinct maximal ideals {PJ of S. Then there are precisely r inequivalent val~ations t/1 l , .. . ,.t/1r on L which extend the P-adic valuation q>p on K, obtamed by choosmg .1,. to be the P ,.-adic valuation on L. The fields I, defined above are precisely '1', the P,-adic completions of L, and we have

n,

=

(I,:Kp)

f;

=

eJ;,

e, = e(P" L/K) = e(I/Kp),

= f(P" L/K) = f(IJK p),

1 :( i :( r.

76

(5.14)

COMPLETIONS AND VALUATIONS

If K is a global field, so that the residue class fields RIP, SIP; are finite, we may normalize the P-adic valuation ({Jp of K, and the P;-adic valuation ({J Pi of L, by setting ({J p(a)

=

.Af{p)-vp(a), f/Jp;(b)

=

.Af{P)-vp,
a E K,

bEL,

where .;v{P)

= card RIP,

In this case, ({JPi = ({J~i on K, so ({Jp.l/n; is the valuation on Li which extends ({Jp on Kp. Thus ' Hence for a E L, f/Jp(NL/Ka) =

IT f/Jp(Ni;/KfJ-i a) = IT f/Jp;(a), i= 1

i~

a EL.

1

(This also holds if P is archimedean, with the standard normalizations of ({J p, f/J Pi as in § 4e).

EXERCISES 1. Let P be a maximal ideal of the Dedekind domain R, and let

f(X) = xn

+ a1 xn- 1 + ... + an E R[ X],

where each ai E P, and an ¢ p2. (Call f an Eisenstein polynomial). Show that f(X) is irreducible in K[ Xl [Hint: By Gauss' Lemma, it suffices to show that f(X) is irreducible in R[X]. Let bars denote passage to RIP = R. If f = gh in R[X], then xn = I = 9 . fi in R[X], so g(O), h(O) E P, whence an E P2.] 2. Let R be a complete discrete valuation ring, R its residue class field, and suppose that R is a finite field with q elements. Using Hensel's Lemma (5.7), show that the polynomial xq -1 - 1 splits into q - 1 distinct linear factors in R[ XJ. 3. Let R be the P-adic completion of the discrete valuation ring R. Prove directlyt that R is faithfully flat as R-module. [Hint: We show that R is R-flat, that is, for each inclusion i: L c M of R-modules, the map 1 ® i: R ® L ...... R ® M is also an inclusion, where ® is over R. If (1 ® i)u = 0, this equation in R ® M holds as a consequence of finitely many defining relations of the tensor product R ® M. Hence it suffices to prove that i': R' ® L ...... R' ® M is monic, where R' is some finitely generated R-submodule of R. But then R' is R-free, and so i' is clearly monic. To show that R is faithfully flat, we show that R ® M -+ 0 if M -+ O. Choose any nonzero mE M. Since R is'R-flat, R ® Rm is a submodule of R ® M. But Rm ~ RIJ, J = annRm < R, and then

R ® (RfJ) ~ RfJ R =I 0.] 4. Let Rp be the P-adic completion of a Dedekind domain R. Prove directlyt that

t

That is, without using (2.23).

(6.1 )

77

RADICALS OF RINGS

Rp is R-flat. [Hint: Flatness is transitive, since Rp ®R'

=

Rp ®RjRp ®R')'

But Rp is R-flat, and Rp is Rp-t1at, by Exercise 3.] 5. Does (5.2 i) remain true without the hypothesis that M be finitely generated? 6. Keeping the notation of (5.2), show that if T is an R-lattice of V with rank R T < (V:K), then it may happen that Tn V = 0.

7. Let R be a discrete valuation ring with maximal ideal P = nR, and let R be its P-adic completion. For M a finitely generated R-module, we set M = R ®R M. Prove (i) MlnkM ~ M/nkM, k ~ 1. (ii) M n nk M = n;k M, k ~ 0, where we view M as a submodule of M as in (5.2). [Hint: We can express M as a direct sum of cyclic modules of the form RIP", n ~ 0, and it suffices to prove the results when M = RIP". For M = R, (i) follows from (5.1), while (ii) is clear. For M = RIP", n ;;. 1, we have M ~ RIP"R, so M ~ M by (5.1). Identifying M with M, it is then obvious that (i) and (ii) hold true.]

6. RADICALS OF RINGS Throughout this section, A is a ring (with unity element), not necessarily commutative. All modules are unital. For an A-module M, define its annihilator as annAM = {aEA: aM = O}, and omit the subscript A when there is no danger of confusion. We shall also write Hom instead of Hom A • By an epimorphism A ~ B of the ring A onto the ring B, we shall mean always a ring homomorphism which is surjective. (This represents a change from the terminology used in category theory.) As general reference for the material in this section, we cite Jacobson

[1,2]' 6a Jacobson radical

"*

An A-module M is irreducible (or simple) if M 0 and M contains no submodules except 0 and M. If M is simple, then for each nonzero mE M we have M = Am. The module epimorphism A ~ M defined by a ~ am, a E A, has kernel a left ideal L in A, and thus M ~ AIL. Since M is simple, L must be a maximal left ideal of A. Conversely, for each such L, the A-module AI L is simple. The Jacobson radical of A, denoted by fad A, is defined as (6.1) rad A = nann M = n ann AIL, M

L

where M ranges over all simple left A-modules, and L over all maximal left ideals of A. Since each ann M is a two-sided ideal of A, so is rad A.

78

RADICALS OF RINGS

(6.2)

The ring A is semisimple if rad A = O. (Some authors call such flngs "semisimple in the sense of Jacobson".) (6.2) THEOREM. A/rad A is a semisimple ring. Proof. Let A = A/rad A; since (rad A) M = 0 for each simple left A -module M, we may view M as a simple A-module. Now let x E A be such that x + rad A E rad A. Then (x + rad A) M = 0, whence xM = O. Hence x E rad A, which shows that rad A = 0, and so A is semisimple.

(6.3) THEOREM. rad A is the intersection of all maximal left ideals of A. Proof. For each maximal left ideal L of A we have ann(A/L) c L, since A has a unity element. Therefore

radA = n ann (AIL) c n L L

L

For the reverse inclusion, let M be any simple A-module. Then ann M = nann m. mill

Bu t if mE M, m =F 0, then ann m is the kernel of the epimorphism A ---+ Am = M, and so ann m is a maximal left ideal of A. Hence annM::::J nL, L

where L ranges over all maximal left ideals of A. Therefore rad A = nann M ::::J n L, and the theorem is proved.

M

L

An element u E A is a unit if there exists an element v E A such that = vu = 1. We shall denote by u(A) the group of units of A. In most of the cases arising in this' book, each element of A having a one-sided inverse is necessarily a unit of A. To prove this, we begin with a result of independent interest: uv

(6.3a) THEOREM. Let M be a noetherian left A-module, and let f E HomA(M, M) be such that f(M)

=

M. Then f is an isomorphism.

Proof. We need only show that f is monic. For n ;;;:, 1, set

Kn = ker 1" = {x EM: f"(x) = O}. Then 0 c Kl

C

K2

C

'"

is an ascending chain of submodules of M, and

(6.4)

JACOBSON RADICAL

79

hence Kn = Kn+ 1 for some n, since M is noetherian. Now let x EM be such that f(x) = O. From the hypothesis thatf(M) = M, it follows that M = f"(M), and so we may write x = fn(y) for some y E M. Then 0 = f(x) = fn+ l(y), whence y E Kn + 1 • But then y E K n , and thus x = fn(y) = o. This completes the proof that f is monic, and establishes the theorem. As an immediate consequence of the above, we have (6.4) THEOREM. Let A be a left noetherian ring, and let ab = 1 in A. Then also = 1.

ba

Proof Let M = AA, a noetherian left A-module; by (6.3a), every A-epimorphism f : M --> M must be an isomorphism. Since A

=

Aab c Ab c A,

we have A = Ab. Let f: M ---+ M be the epimorphism defined by setting = xb, x EM. Then f is an isomorphism. However, f(1 - ba) = (1 - ba)b = 0, and therefore I - ba = o. This completes the proof.

f(x)

Returning to the general case, we prove next (6.5) THEOREM. For any ring A,

radA = {xEA: 1 - axbEu(A)foralla,bEA}. Proof. (i) Let x E rad A, and let a, bE A; since rad A is a two-sided ideal of A, we have y = axb E rad A. We now show that 1 - y E u(A) for each y E rad A. If A(l - y) < A, there is a maximal left ideal L of A such that A(l - y) c L, and then 1 - y E L. Since rad A c L, this gives 1 E L, a contradiction. Therefore A(1 - y) = A for all y E rad A, and so t(1 - y) = I for some tEA. Therefore 1 - t = - ty E rad A, whence (as above) A(l - (1 - t» = A, that is At = A. Hence there exists u E A such that ut = 1, and thus u = ut(1 - y) = 1 - y.

This proves that t is a two-sided inverse of 1 - y, so I - Y E u(A), as desired. (ii) Let x E A be such that 1 - axb E u(A) for all a, bE A. To show that x E rad A, we need prove that xM = 0 for every simple left A-module M. Let mEM, m i= 0; if xm i= 0, then M = A ·xm, so m = axm for some a E A, and thus (1 - ax)m = O. But 1 - ax E u(A), and hence m = 0, a contradiction. This proves that xm = 0 for each nonzero mE M, and therefore xM = 0, completing the proof of the theorem. (6.6) COROLLARY. If we compute a radical of A by using right A-modules rather than left A-modules, we obtain the same rad A.

80

(6.7)

RADICALS OF RINGS

Proof. The preceding theorem gives a left-right symmetric characterization of the radical.

A left ideal N of A is nilpotent ifthere exists a positive integer k such that

,-N . IVy ... L'V, = 0,

°

k

or equivalently, Xl ... x k = for all {xJ EN. An element X E A is nilpotent 2 if ~ = for some k. An element e E A is idempotent if e¥-O and e = e. Obviously, a nilpotent ideal cannot contain any idempotent elements.

°

(6.7) THEOREM. Let N be a nilpotent left ideal of A, and let X E A be a nonnilpotent element such that x 2 - x E N. Then the left ideal Ax contains an idempotent y such that y - x E N.

Proof. Let N k = 0, and set n l = x 2 we are done. If n1 ¥- 0, let Xl

=

X

+ nl

X E

-

-

N. If nl

2xnl

E

= 0,

choose y

= x,

and

Ax.

Then x, Xl and nl commute with each other, and hence if Xl is nilpotent, so also is x = Xl - nl + 2xnl ' a contradiction. Thus Xl is a non-nilpotent element of Ax, and direct calculation gives 3 Xl2 - Xl4nl -

=

3n 2l

·

The element n 2 = xf - Xl is nilpotent, contains nf as factor, and commutes with Xl' Continuing in this manner, we may construct a sequence {xJ of non-nilpotent elements of Ax, such that nf occurs as a factor in x; - Xi' If we choose i so that 2i ~ k, then we have x; - Xi = O. Further, \ ¥- 0 since Xi is non-nilpotent. Thus we may take y = Xi' the desired idempotent in Ax such 'that y - x E N. (6.8) COROLLARY. Let L be a non-nilpotent left ideal in a left artinian ring A. Then L contains an idempotent element.

Proof. Let LI be a minimal member of the (non-empty) set of non-nilpotent = L I . Now let I be muumal III the set of left ideals contained in L I such that L I . I ..,..., -'- 0 and choose a E I so that Ll . a ¥- O. Then L j • L j a¥-O and L j a c I, whence LI a = I. Hence a = xa for some x E L 1 , and thus a = xka for all k ~ O. Therefore I contains the non-nilpotent element x. Ifwe now set

lef~ ~deals. contained in L; then L/ eLl' whence L/

(6.9)

81

JACOBSON RADICAL

then x 2 - x E N. Further, Li a = I¥-O soN < L i . Then N is nilpotent, by the way in which Li was chosen. Thus Ax contains an idempotent element by (6.7), whence so does L, since L ;:) Li ;:) Ax. (6.9) THEOREM. If A is left artinian, then radA is the largest nilpotent left ideal of A. Proof. If L is a nilpotent left ideal, and x E L, then for all a, nilpotent; namely, (axb)k

= ax' (baxt- 1 . b

E

bE

A also axb is

V· b = O.

But then I - axb E u(A), since 1

+ axb + (axb)2 + ...

is a two-sided inverse of 1 - axb. This proves that L c rad A, by (6.5). On the other hand, rad A must itself be nilpotent; for, if not, then by (6.8) rad A contains an idempotent e. Hence 1 - e E u(A) by (6.5), which is impossible since (1 - e)e = 0 but e ¥- O. We may remark that if A is left artinian, then rad A is also the largest nilpotent right ideaL For if J is any nilpotent right ideal of A, then AJ is a nilpotent two-sided ideal, whence J c AJ c rad A. Now we prove some easy properties of radicals. ---> B is an epimorphism of rings, thenf(rad A) c radE. Hence f induces an epimorphism A/rad A ---> B/rad B.

(6.10) THEOREM. If f:A

Proof. Let x E rad A, y = f(x), and let b i ' b2 E B. Then there exist a i ' a 2 E A such that bi = f(a;), i = 1,2. But 1 - a i xa 2 E u(A), whence 1 - b i yb 2 E u(B). Hence y E rad B, by (6.5).

(6.11) THEOREM (Nakayama's Lemma). Let M be a finitely generated left = M. Then M = O.

A-module such that (rad A)M

Proof. If M ¥- 0, let m i , ... , mk be a minimal set of generators of the left A-module M. Since m i EM = (rad A)M, we may write mi

= r i m i + .,. + rkmk,riEradA.

But then 1 - r i E u(A), so we can solve for m i in terms ofm 2 , ... , m k , thereby reducing the number of generators of M. (6.12) COROLLARY. Let M be a finitely generated left A-module, and N a submodule such that

82

RADICALS OF RINGS N

ThenN

+ (radA)M =

(6.13)

M.

= M.

Proof. The hypothesis implies that M/N is finitely generated, and (rad A) (M/N) = M/N. The result then follows from (6.11).

(6.13) COROLLARY. Every maximal two-sided ideal J of A contains rad A. Proof. If J :p rad A, then J + rad A is a two-sided ideal of A properly containing J. Hence J + rad A = A, whence J = A by applying (6.12) with N = J,M = A.

6b Local rings The ring A is local (or completely primary) if A has a unique maximal left ideal. (6.14) THEOREM. The following are equivalent: (i) A is local. (ii) The set S of non-units of A is a left ideal of A. (iii) A/rad A is a skewjield (that is, a ring whose nonzero elements form a multiplicative group). Proof. Suppose first that A is local, with J its unique maximal left ideal, so of course J c S. Since rad A is the intersection of all maximal left ideals of A, we have rad A = J, and so J is a two-sided ideal. We claim that S = J. For let XES; if Ax oF A, then Ax lies in a maximal left ideal of A, whence Ax c J, so X E J. On the other hand, if Ax = A then for some YEA, yx = 1. Clearly y ¢ J, otherwise, 1 = yx E rad A. Hence Ay = A, so there exists an element z E A with zy = 1, and hence z = x. Thus x E u(A), a contradiction. This proves that S = J, and shows also that each element of A - J is a unit in A, whence A/rad A is a skew field. Next, suppose that S is a left ideal of A. Since every proper one-sided ideal of A lies in S, it is clear that S is the unique maximal left ideal (and also right ideal), so A is local. Finally, suppose that A/rad A is a skewfield. Clearly rad A c S, and we need only prove the reverse inclusion. If x E A - rad A, there exists YEA such that yx = 1 + n, for some n E rad A. Hence yx E u(A), and x has a left inverse in A. Analogously, x has a right inverse, whence x E u(A). Thus A - rad A c u(A), so rad A c S, as desired. This completes the proof.

(6.15) THEOREM. Let R be a commutative local ring, with maximal ideal

(6.15)

83

LOCAL RINGS

P = radR, and residue class field R = R/P. Let A be an R-algebra, finitely generated as R -module, and let rp:A -+ A = A/ P A be the natural e pimor phism of A onto the finite dimensional R-algebra A. Then we have PA c rp-l (radA)

= radA,

and (rad A)f cPA for some t. Further, rp induces a ring isomorphism

A/rad A ~ A/rad A. Proof. We show first that rad A :::J P A. Let M be any simple left A-module; then M = Am for each nonzero mE M, whence M is finitely generated as R-module. Now PM is an A-submodule of M, hence is 0 or M. But PM -# M, otherwise by Nakayama's Lemma we find that M = O. Thus PM = 0, and so P A annihilates every such M. Therefore PAc rad A. By (6.10), rp induces an epimorphism

A/rad A

-+

A/rad A.

On the other hand, there is an epimorphism t/J: A -+ A/rad A, since PAc rad A. By (6.2) we have t/J(rad A) c rad(A/rad A) = 0, and thus t/J induces an epimorphism A/rad A -+ A/rad A. Both of these quotients are finite dimensional R-algebras, and thus are isomorphic. This shows that rad A = rp-l(radA), as well. Next, rad A is a nilpotent ideal of the left (and right) artinian ring A, by (6.9), and hence (rad A)I = 0 for some n. Thus rp{ rad An = 0, whence (rad A)I cPA, as claimed. We shall need some standard results on completions of a commutative k

local ring R, and of modules over such a ring. Suppose first that X =

I" Rx; i~

1

is a free R-module, and let P denote the maximal ideal of R. We define a P-adic topology on X, by taking as basis for the neighborhoods of a point x E X the sets {x + pmx: m = 0, 1,2, ... }. Given a sequence {Yn} from X, we may write + ... + rlk)X n 1 n k' Yn-- r(1)x Then {Yn} is a Cauchy sequence in X if and only if, for each i, {r~i)} is a Cauchy sequence in R. Further, {yJ converges if and only if each {r~)} converges, and we have n-.oo

n~CXJ

Thus if R is complete in the P-adic topology, then so is X. When R is not complete, we may form its P-adic completion R.. Then R. ®R X is a free

84

RADICALS OF RINGS

(6.16)

R-module, and can be topologized as above. If X denotes the P-adic completion of X, the above discussion shows that there is an R-isomorphism

X ~ R®RX, and that this is a topological isomorphism. The following generalization of the above holds true, and we refer the reader to Bourbaki [2, Ch. 3] or Matsumura [1 ] for proofs: (6.16) THEOREM. Let R be a commutative noetherian local ring, with maximal ideal P, and let R be the P-adic completion of R. Let X be a finitely generated R-module, and let X be the completion of X relative to the P-adic topology on X, defined by taking {x + pmx} as basis for the neighborhoods of x EX. Let R ®R X be topologized by taking g + pm(R ® X)} as basis for the neighborhoods of ~ E R ® X. Then there is a topological R-isomorphism

X ~ R®RX, and

X is a complete Hausdorff space

relative to its P-adic topology.

(6.17) COROLLARY. Let R be a complete commutative noetherian local ring, and let A be an R-algebra which is finitely generated as R-module. Then A is a complete Hausdorff space relative to the P-adic topology on A.

For the most part, we shall be dealing with the case where R is a complete discrete valuation ring, and A is an R-order. In this case, A is of course R-free, and we need not use (6.16) and (6.17) in their full generality. 6c Lifting idem po tents

We shall discuss here the relationship between idempotents in a ring A, and idempotents in a factor ring A = A/N, where N is a two-sided ideal of A contained in rad A. To begin with, suppose that there is a direct sum decomposition A = Ll EB'" EB Lk '

where each Li is an indecomposable left ideal of A. We may write

For each x E L;, we obtain x =

L xe;,

whence xe i = x and xe j = 0 for

j # i. This gives

Thus each e; is idempotent, and we call {e 1

, ., .,

ek } a full set of orthogonal

(6.18)

85

LIFTING IDEMPOTENTS

idempotents in A. Further, each e i is primitive, that is, e i cannot be expressed as a sum e' + e" of orthogonal idempotents; for if e i = e' + e", then Li = Ae' EB Ae" gives a decomposition of L, . We have thus shown that there is a one-to-one correspondence between decompositions of A into direct sums of indecomposable left ideals, and decompositions of 1 A into a sum of primitive orthogonal idempotents. Now let A = A/N, where N is a two-sided ideal of A such that N c rad A, and let a E A have image aE A. If e E A is idempotent, then e is an idempotent in A; indeed, e Z = e implies that e Z = e, while if e = 0 then e E rad A, an impossibility (see proof of (6.9)). Thus each decomposition 1 = Lei into orthogonal idempotents in A gives a corresponding such decomposition 1 = Lei in A. In general, however, it may happen that ei is primitive but ei is not. Our aim here is to show that under suitable hypotheses, ei is primitive whenever e i is primitive, and indeed, that every decomposition I = Lei into a sum of primitive orthogonal idempotents "lifts" to a decomposition 1 = Lei into primitive orthogonal idem po tents, such that e i = ei for each i. Given any two-sided ideal N of A, we may give A the N -adic topology by taking as basis for the open neighbourhoods of an element a E A the sets {a + Nr: r= 0,1, ... }. We call A complete in the N-adic topology if each Cauchy sequence (relative to this topology) converges to an element of A. If N c rad A, there are two obvious cases where A is necessarily complete in the N -adic topology: (i) if A is a left artinian ring, then N is a nilpotent ideal by (6.9), and hence any Cauchy sequence from A, relative to the N-adic topology, is constant from some point on; (ii) if A is an R-algebra, finitely generated as R-module, where R is a complete noetherian local ring with maximal ideal P, then by (6.15) we have Nt c (radAY c PA

for some t. Hence any Cauchy sequence (from A) relative to the N -adic topology is also a Cauchy sequence in the P-adic topology and thus converges by (6.17). For the remainder of this section, we assume that A is complete in the N -adic topology. (6.18) THEOREM. Each idem potent e E A can be lffted to an idem potent e E A, that is, e = c. If el' e z are idempotents of A, then Ae l if and only if Ael ~ Ae z as left A-modules.

~

Ae z as left A-modules

Proof. We examine the proof of (6.7). Given an idempotent e E A, choose any E A with Xl = c. Then nl = Xl EN. Once Xi and n i are chosen, let

Xl

xi -

86

RADICALS OF RINGS

(6.19)

2i

Then n i E N , and so {xJ is a Cauchy sequence in the N-adic topology of A. Let e = .lim x·, E A. Then ,--+ ro e2

-

e = lim n i = 0,

and furthermore e = Xl = ", so e i= 0, and e is the desired idempotent in A. Now let e 1 , e2 E A be idempotent. Any isomorphism Ae 1 ~ Ae 2 carries Ne 1 onto Ne 2 , and induces an isomorphism Ae 1 ~ Ae 2 . Conversely, let j:Ae l ~ Ae 2 , and let 9 = J- 1 • Then in the diagram

_____ 1 ______>

j"'-----:------

r

the vertical arrows are epic. Since A = Aei EB A(l - eJ, each Aei is Aprojective. Hence we can find A-homomorphisms j; g liftingJ, g, respectively. We claim that f is an isomorphism, and need only show that 8 = g' f is an automorphism of Ae 1 (for -then, by symmetry, f· g is an automorphism of Ae 2 ). Clearly 8 lifts 9 'J, and hence (8 - 1)Ae 1 c Ne 1 . Let ~ = 1 - 8; then ~k. Ae1 c N k e 1 ; thus 1 + ~ + + ... is a well defined endomorphism of Ae 1 ' and is a two-sided inverse of 1 - ~. Since 1 - ~ = 8, this shows that 8 is an A-automorphism of Ael' as claimed. This completes the proof.!

e

(6.19) THEOREM. Let idempotents in such that

1=

"I + ... + "n be a decomposition of I into orthogonal

A. Then there exist orthogonal idempotents

e1 , •.. , en

E

A

Proof. We proceed by induction on n, the result being trivial when n = l. Assume _that n > 1 and that the result holds at n - 1. Set b = "n-1 + "II' so now 1 = "I + ... + 2 + b is an orthogonal decomposition. By the induction hypothesis there exists an orthogonal decomposition

"n-

1

= e1 + ... + en -

2

+ e,

Choose a E A such that a = ".-1' and set b = eae. Then b = b· ".-1 . b = and eib = bei = 0, 1 ~ i ~ n - 2. In the proof of (6.18), we take Xl = b. Then the sequence {xJ converges to an idempotent e'- l E A such that e._ 1 = and eie._ 1 = e._ 1 ei = 0, 1 ~ i ~ n - 2. Finally, set

".-1'

".-1'

t

For more general theorems of this type, see Azumaya [1, Th. 24 J.

(6.20)

87

LIFTING IDEMPOTENTS

= e - en - 1 ; then 1 = e 1 + ... + en is the desired orthogonal decomposition in A such that ei = 6 i for 1 ~ i ~ n.

en

(6.20) COROLLARY. An idempotent e E A is primitive if and only

eE A is primitive.

if

its image

Prool If e is imprimitive, there is an orthogonal decomposition e = e1 + e2 into idempotents e l' e 2 EA. Then e = e1 + e2 is an orthogonal decomposition of e. Conversely, let e = 6 1 + 6 2 be an orthogonal decomposition into idempotents of A. The proof of (6.19) shows that there exist orthogonal idempotents e1 ,e2 E A such that e1 = 6 1 , e2 = 6 2 , and e = e1 + e 2 • This completes the proof.

(6.21) THEOREM. Let A be a left artinian ring, and let A = A/rad A. Let 1 = e1 + ... + en be a decomposition of 1 E A into orthogonal primitive idempotents in A. Then A = Ae 1 EB ... EBAen is a decomposition of A into indecomposable left ideals of A, and

A

=

A~\

EB ... EB Aen

is a_ decomposition of A into minimal left ideals. Further, Aei ~ Ae.J if and only _ if Aei ~ Ae j • n

Proof. Since A is left art in ian, it can be expressed as a finite direct sum

L

L' Li 1

of indecomposable left ideals of A. Writing I = ep ei E Lp we obtain a decomposition of I into orthogonal primitive idem po tents. Then I = is such a decomposition in A, and so each Aei is an indecomposable left ideal of A. But A is a semisimple artinian ring, so (see (7.1)) every indecomposable left ideal of A is a minimal left ideal. The last statement in the theorem has already been established in (6.18), so the proof is completed.

Lei

(6.22) COROLLARY. Let A be an R-algebra, finitely generated as R-module, where R is a complete noetherian commutative local ring. Then the assertions in the preceding theorem remain valid for this case. Proof. In the proof of (6.21), we used the hypothesis there that A be left artinian, in order to guarantee that (i) A is expressible as a finite direct sum of indecomposable left ideals, and (ii) A/rad A is a semisimple artinian ring. However, both of these remain valid under the hypotheses of the present

88

RADICALS OF RINGS

corollary. The first is true, since in the present case A is a left noetherian ring. The second holds true, by virtue of (6.15). Thus the proof of (6.21) applies equally well in this case, and the corollary is established. EXERCISES L Let J be a two-sided ideal of a ring A, such that A/l is semisimple. Prove that J => rad A. [Hint: By (6.10), (J + rad A)/ J c: rad (A/ J) = O.J

2. Let x E A map onto X E A/rad A. Show that x E urAl if and only if x E u(A/rad A). 3. Keep the notation and hypotheses of (6.15), and let J be any two-sided ideal of A such that 1'" c: P A. Show that J c: rad A. [Hint: cp(J) is a nilpotent ideal in the artinian ring A, whence cp(J) c: rad A, so J c: cp - 1 (rad A) = rad A.J In Exercises 4--8 below, let R be a complete noetherian commutative local ring, and let A be an R-algebra which is finitely generated as R-module. We include here the special case where R is a field, and {O} is its maximal ideal. 4. Prove that A is a local ring if and only if 1 is the only idempotent in A. [Hint: Let A be local, and let e E A be idempotent, e ¥ L Since e( I - e) = 0, both e and 1 - e

are non-units, whence so is their sum; this is a contradiction. Conversely, if A is not local, then A/rad A is not a skewfield. If P = rad R, then by (6.15) we know that A/rad A is a finite dimensional semisimple (R/P)-algebra. Hence by § 7, A/rad A contains an idempotent other than 1. Therefore A contains an idempotent different from 1, by (6.18).J 5. An indecomposable left A-module is a nonzero left A-module which is not expressible as a direct sum of non-trivial submodules. Let M be a nonzero finitely generated left A-module, and set E = Hom A (M, M). Show that E is an R-algebra which is finitely generated as R-module. Us.ing Exercise 6.4, deduce that M is indecomposable as A-module if and only if E is a local ring. [Hint: Given a non-trivial decomposition M = M 1 ffi M 2 into A-submodules, let 7t 1 : M -+ M 1 be the corresponding projection of M onto MI' Then 7tl E E is an idempotent, and 7tl =I- 1. Conversely, if e E E is an idempotent different from 1, then M = eM ffi(1 - elM gives a decomposition of M.J 6. Prove the Krull-Schmidt Theorem: Every finitely generated left A-module M is expressible as a finite direct sum of indecomposable modules, and these direct summands are uniquely determined by M, up to A-isomorphism and order of occurrence. [Hint: Since M is finitely generated over the commutative noetherian ring R, M is itself a noetherian module. The process of decomposing M into direct summands must therefore terminate. Now let M = Ml ffi··· ffiM, = N} ffi··· ffi N u

'

where the {MJ and {N) are indecomposable. Let Il i : M -+ Mp vj:M -+ N j , be the projections associated with these decompositions. Restrict the operators in the equation III = L III Vj to M}, getting the relation 1 = L III Vj in the local ring Hom A (M I'M 1)' Hence some f)1 Vj must be a unit III this ring, that is, must be an automorphism of MI' If (say) 111 VI is an automorphism cp of M l ' then in the diagram

89

EXERCISES

"

Ml ~NI' 'P 'J1, VI is monic and rp-l ltl · VI is the identity on MI' Hence there is a splitting N1 ~ M1 EBker rp-1 111 , and therefore N1 ~ MI' Thus both VI and I"l are isomorphisms. Next verify that the sum M' =N 1 EBM 2 EB"'EBM,

is direct. Then show that M'

M, since for each x EN 1 '

=

Il l X

=

X -Illx - ... -Il,XEM',

and thus M1 = 11 I(N 1) c M'. Prove next that the map

defines an A-isomorphism of M onto M', carrying M 1 onto N l ' Therefore p induces an A-isomorphism Ml EB ... EBM, ~ MINI ~ Nl EB .. ·EB N u '

An induction argument completes the proof.] 7. Let L, M, N be finitely generated left A-modules such that L+M~L+N.

Prove that M ~ N. [Hint: Express L, M, N as direct sums of indecomposable modules, and then use the Krull-Schmidt Theorem proved above.] 8. Let {Mb ... , Mtl be finitely generated indecomposable left A-modules, and let L be a left A-module which is isomorphic to adirect summand ofM! M t. Show that

+... +

L ~. M. 1 ~

where {i 1

,

r

i= 1

, ... , i r }

Mi ~ L

+". +M,. , .-

is some subset of {l, ... , t}. [Hint: The hypothesis implies that

+r;, where both Land r:, are finitely generated left A-modules. Now

express Land r; as direct sums of indecomposable modules, and then use the KrullSchmidt Theorem.] 9. Let A be an R-algebra as in (6.22), and let P be the unique maximal ideal of R. Assume that P does not annihilate any nonzero R-submodule of A. Let J = rad A, and set A = AIJ. Show that every simple left A-module occurs as a direct summand of the left A-module JIJ 2 . [Hint: Assume the result false. By § 7, there is a central idempotent e in the semisimple artinian ring A such that e' (JIJl ) = O. By (6.18) there exists an idempotent e E A mapping onto e. Then eJ c ]2, whence multiplying by e we obtain eJ = eJ2 = eJ· J. Therefore eJ = 0 by Nakayama's Lemma. But J ::::J P A by (6.15), and so p. eA = O. This gives eA = 0, so e = 0 and e = 0, a contradiction.]

90

SEMI SIMPLE RINGS AND SIMPLE ALGEBRAS

(7.1)

7. SEMISIMPLE RINGS AND SIMPLE ALGEBRAS 7a Wedderburn's Theorem

Recall that a ring A is semisimple if rad A = O. Of particular interest, because they occur so frequently in practice, are the left artinian semisimple rings. The algebraic structure of such rings can be described explicitly, as we shall see below. We shall omit proofs of some of the relatively well known theorems stated below , referring the reader to Curtis-Reiner [I] or Bourbaki [1] for details. A minimal left ideal of A is any minimal member of the set of nonzero left ideals of A. Equivalently, a minimal left ideal is a simple submodule of AA. (7.1) THEOREM. Let A be a left artinian ring. Thefollowing are equivalent: (i) A is semisimple. (ii) A is a finite direct sum of minimal left ideals. (iii) Every exact sequence of left A-modules is split. (iv) Every left ideal of A is a direct summand of A. (v) A contains no nonzero nilpotent ideals.

(vi) Every left A-module is a direct sum of simple A-modules. Given a left artinian semisimple ring A, we may write A = Ll EB··· EB L" L; = minimal left ideal in A. Set

l=e 1 + .. ·+e" From the equation ei =

I

e;EL;.

etej we deduce at once that

j

e? = ei , e/ j = 0

i"# j,

for

L; = Ae;.

Call {eJ a full system of primitive idempotents of A. Let Bl denote the sum of those {L j } which are isomorphic to L 1 , B2 the sum of those {L j } isomorphic to a remaining summand (if any), and so on. Then we have A = Bl EB ... EB B t

for some t :( r. It is easily shown (see references) that each B; is a subring of A, and that B;' B j = 0 for i "# j, so A is the ring direct sumt of the {BJ Each B; is a simple ring, that is, a ring whose only two-sided ideals are 0 and B;; t The ring

direct sum

I' Bi of the rings {B.} is defined to be the Cartesian product Bl x ... x B" i= 1

with component wise addition and multiplication. This ring is in fact the direct product (in the

,

sense of category theory), and is often denoted by "ring direct sum" throughout this book.

nB i= 1

i•

We shall use the older terminology

(7.2)

WEDDERBURN'S THEOREM

91

of course, Bi is left artinian (since A is). Call these {BJ the simple components of A; they are uniquely determined inside A, although the {LJ are not. Further, every simple A-module is isomorphic to a minimal left ideal of A, hence to some minimal left ideal of some B; . Conversely, if Bl ' ... , Bt are simple left artinian rings, then their ring direct B; is a semisimple left artinian ring whose simple components are sum the {BJ. Now let D be a skewfield (or division ring), that is, a ring whose nonzero elements form a multiplicative group. Denote by M.(D) the ring of all n x n matrices with entries in D. We call Mn(D) a full matrix algebra over D. Let DO (or DOPP) denote the opposite ring of D; the elements of DO are the symbols {xo:xED}, with (X O ± yO) = (x ± y)O, x,YED.

I"

Then DO is also a skew field. (7.2) THEOREM. (i) If V =

t

Dv; is a left vector space over the skewjield D,

1

then Hom D(V, V)

~

Mn(DO).

n

(ii) If V

=

I"

vp is a right vector space over the skewjield D, then

1

HomD(V, V) ~ M n(D).

Proof. In case (i), for a E HomD(V, V), let n

aV j

=

L

rt;jV;, rt;j

ED.

i= 1

Then the desired isomorphism is given by a In case (ii), let

--+

(rtf).

n

avo J

=

L v;rt;j' rtij ED, ;= 1

(7.3) THEOREM. For each n, the matrix ring A = Mn(D) is a simple ring, both left and right artinian. Every minimal left ideal of A is A-isomorphic to the left A-module L of n-component column vectors with entries in D. Further, Mn(D) ~ Dnl as left A-modules and DO ~ HomA(L, L). If we view L as right D-space, then A ~ Hom D (L, L). (7.4) THEOREM (Wedderburn Structure Theorem). Every left artinian simple ring A is isomorphic to an algebra of n x n matrices over a skewfield. The

92

SEMISIMPLE RINGS AND SIMPLE ALGEBRAS

( ;.5)

ring A determines n uniquely, and determines the skewjield up to isomorphism. If L is any minimal left ideal of A, then D = Rom A (L, L) is a skewjield, and A = Romv(L, L)

~

Mn(DO),

where n is the dimension of the left D-space L.

(7.5) COROLLARY. Every left artinian simple ring is also right artinian and is left and right noetherian. The same holds for left artinian semisimple rings. By virtue of (7.5), we may speak unambiguously of artinian simple or semisimple rings, without having to distinguish between "left" and "right". A quick proof of (7.4) is sketched in Exercise 7.4. See also Exercises 16.9 and 16.10. 7b Splitting fields

Let K, L, E denote fields, and D, D' skewfields. The center of an algebra A will be denoted by Ae. A central simple K-algebra is a simple K-algebra A for which Ae = K and (A: K) is finite. Call D a skewjield over K if De ~ K and (D: K) is finite. By Wedderburn's Theorem, every central simple K -algebra A is of the form Mn(D) for some skew field Dover K. Further, A C = {ctI n: ct E DC}

~

De,

so D has center K, and (A:K) = n2(D:K). (7.6) THEOREM. Let A be a central simple K-algebra, B an artinian simple Kalgebra, not necessarily jinite dimensional over K. Then B Q9 KA is an artinian simple algebra with center Be,

Proof. We shall first compute (B Q9 A)C, where we write Q9 instead of Q9 K for brevity. Let B = L' Ke i , so B Q9 A = L' ei Q9 A. If u = Lei Q9 ai E (B Q9 A)e, then u commutes with 1 Q9 a for all a E A, whence Lei Q9 (aa i - aia) = O. Therefore aa i = ala for all a E A, whence each ai E Ae = K, and so u = bo Q9 1 for some bo E B. But now u commutes with b Q9 1 for all bE B, whence bo E Be, Identifying B with B Q9 1 in B Q9 A, we have thus shown that (B Q9 A)< c Be. The reverse inclusion is obvious, and therefore (B Q9 A)< = Be. Since B is artinian and (A: K) is finite, it is clear that B Q9 A is also artinian. It remains to show that B Q9 A is simple. We may write B = Mr(D), A = M.(D'), where (D')' = K. Now for any skewfields D, D' whose centers contain K, we have (7.7)

Mr(D) Q9 KMs(D')

~

Mr(K) Q9 Ms(K) Q9 D Q9 D'

~

Mrs(K) Q9 (D Q9 D')

~

Mrs(D Q9KD').

(7.8)

93

SPLITTING FIELDS

Hence to prove that B 0 A is simple, it suffices to prove that D 0 D' is simple. Let D = L" Kd; , so D 0 D' = L" d; 0 D'. Let X # 0 be a two-sided ideal m

of D 0 D', and choose a shortest x = x(l 0

b~

L d; 0 c\

in X,

c\ ED'.

Then

1

1) E X, so after changing notation, we have m

X

= d 1 0 1 + L d; 0 b; EX. 2

Now let bED', b # 0; theny = (1 0b)'x'(1 0b- 1 ) EX,and m

y

= d 1 01

+ Ld;

0bb;b- 1 •

2

Since x - y E X and x - y is shorter than x, this gives y = x. Therefore each bi commutes with each nonzero bED', whence each b; E K. Thus xED, and so x is a unit in D 0 D', whence X = D 0 D'. This completes the proof. (7.8) CoROLLARY. Let A be a central simple K-algebra, L is a central simple L-algebra.

~

K. Then L 0KA

Let A be any ring, M a left A-module. Each a E A determines a left multiplication'aL:m --+ am, mE M. The map a --+ aL is a ring homomorphism of A onto a subring AL of Homz(M, M). Call M a fait~ful A-module if A --+ AL is monic, that is, if a . M = 0 implies that a = O. We shall say that the pair (A, M) has the double centralizer property if AL = Homv(M, M), where D = Hom jM, M). It is trivially verified that (A, AA) has the double centralizer property, since in this case D = A R , the set of right multiplications by elements of A. THEOREM. (i) For each A-module N, the pair (A, AA -+ N) has the double centralizer property. (ii) Let M be an A-module, k a positive integer. If (A, M(k») has the double centralizer property, thea so does (A, M).

(7.9)

Proof. (i) Let D = HomA(M, M), where M = AA -+ N. Each dE D is represented by a 2 x 2 matrix (d u) of A-homomorphisms, where dll:A--->A,

Now let

qJ

d 12 :A--->N,

E Homv(M, M), and write

homomorphisms. Since

qJ

d21 :N--+A, qJ

=

(qJ;j)'

d 22 :N--+N.

a 2 x 2 matrix of additive

commutes with the element (~

~) of D, it follows

94

SEMISIMPLE RINGS AND SIMPLE ALGEBRAS

(7.10)

that ({J12 = 0, ({J21 = O. Next, fix no EN, and define dna ED by dna (a, n) = (0, ano)' Then while

Thus ({Jl! a)no = ({J22(an O) for all (a, no) E M, ({Jl1(1))n o

=

({J22(n O)' But ({J commutes with

whence

(taking a = 1)

(~R ~)ED

for each aEA,

whence ({Jll commutes with each aR , and so ({Jll = (aO)L for some a o EA. Thence ({J22(n O) = aon o' and so ({J is (aO)L on M. This completes the proof of (i). To prove (ii), let D = HomA(M, M), and let V = M(k) be the set of I x k row vectors with entries in M. If we think of D as a domain of right operators on M, then M may be viewed as a bimodule AM D' In that case, E = Hom A(V, V) ~ M k(D), and we may view V as a bimodule AVE' Now let f E HomD(M, M); we must prove that f = aL for some a EA. Define f*:V-+ Vby It is easily checked that f*

E HomE(V, V). But (A, V) has the double centralizer property, whence f* = aL for some a E A. Then also f = aL (on M), as desired.

(7.10) COROLLARY. Let M be any faithful jinitely generated left module over a semisimple artinian ring A. Then (A, M) has the double centralizer property.

Proof Let Bl , ... , Br be the simple components of A, and let M; be a simple

left B(module. Then by (7.1) we may write M = L. ~. M.(r;). Since B.' M. = 0 J 1

1

1

for j -# i, and since M is a faithful A-module, it follows that each rl > O. Further, we know that B; ~ M?') as left B(modules, for some sr It follows that if k = max {Sf}, then AA is a direct summand of the module M(k). Hence by (7.9(i)) the pair (A, M(k» has the double centralizer property, whence so does (A, M) by (7.9(ii». We are now ready to apply this corollary to the study of simple subalgebras of a central simple algebra A, and eventually to the study of maximal subfields of A. We begin with a fundamental result: (7.11) Theorem. Let K c B c A, where B is a simple subring of the central simple K-algebra A. Let

(7.12)

SPLITTING FIELDS

B' = {x E A: xb = bx

95

for all b E B},

the centralizer of Bin A. Then B' is a simple artinian ring, and B is its centralizer inA. Proof Let V be a simple left A-module, and let D = HomiV, V), viewed as left operator domain on V. Then A = HomD(V, V), and D is a skew field with center K. For a E A, dE D, let aL and dL denote left multiplications on V. Then bL ' dL = dL . bL for all bE B, dE D, and therefore we may make V into a left D (8) K B-module by setting (d(8)b)v=d L bL v,

vEV.

By (7.6), D (8) KB is a simple artinian algebra of finite K-dimension. Both DL and BL are K-subalgebras of HomK(V, V), and the elements of DL commute with those of B L. Hence the map D (8) KB ~ DL . BL is a K-algebra isomorphism (the kernel must be a two-sided ideal of D (8) K B), and so letting S denote DL . B L , it follows that S is a simple algebra. Further, V is a finitely generated faithful S-module, so the pair (S, V) has the double centralizer property, by (7.10). Let q> E Homs(V, V); then q> E Hom D (V, V) = A L , so q> = aL for some a E A which centralizes B L' Since both A Land B act faithfully on V, this shows tha t a E B', and therefore (B')L = HomsCV, V). Since B' ~ (B')L' it follows from Exercise 7.2a that B' is a simple artinian ring. Further, since (S, V) has the double centralizer property, we have (7.12) S = DL . BL = Hom(B')r,(V' V). Now let xEA centralize B'; then XL centralizes (B')L' so XLES by (7.12). On the other hand, since XL E A L , it follows that XL centralizes DL in S. We now make use of the isomorphism S ~ D (8)KB. The proof of(7.6) shows that the elements in D (8) KB which centralize D (8) I are contained in DC (8) B, that is, in B itself (since DC = K). Therefore XL E B L, whence X E B, as desired. For later use, we give two important consequences of(7.12): COROLLARY. Keeping the notation of (7.11), let V be a simple left A-module, and set D = HomA(V, V). Then

(7.13)

(B:K)(B':K) = (A:K). Proof The first assertion is a restatement of (7.12). To prove the second, let W be a simple left B'-module. Since V is a left B'-module, we have V~ W(k)

96

(7.14)

SEMISIMPLE RINGS AND SIMPLE ALGEBRAS

where k = dim V/dim W, and dim denotes dim K . Let do = dim Do, where Do = Homn·UV, W) is the skewfield part of B'. Let us write dimD = d,

dim A = dn 2 ,

dim V= dn,

dimB' = d m2

dimDo=do'

o

'

for some m, n. Then k = dim VJdim W = dn/d o m, while from the first part of the corollary, d· dim B = dim Homn.(W(k), W(k)) = dim Mk(Do) = k 2do ' Consequently (dim B) (dim B') = (k 2do/d) . dom 2 = k2d~m2 /d

= d 2 n 2 /d = dim A, which completes the proof. (7.14) COROLLARY. In the notation of (7. I 1), we have A0 Further, B 0 K B'

~

K

A

F

~

Mr(B'),

where

r

= (B: K).

if B has center K.

Proof Keeping the terminology of the proof of(7.13), we have A

~

Mn(D

O ),

so

On the other hand D 0 B ~ Mk(D o)' whence A 0 B ~ Mkn(D~PP). Since B' ~ Mm(D~PP), we need only show that dim B = kn/m. But this is clear, since O

dim B = k 2do/d = k . (kdo/d) = k . (n/m). For the second part of the corollary, we observe that B is a central simple K-algebra, and thus B 0 B' is simple. Therefore the algebra homomorphism B0 K B'--+B'B'cA must be monic. Comparison of dimensions then shows that B 0 B' and the corollary is established.

~

A,

Now let A = Mn(D) be a central simple K-algebra, where D is a skew field with center K. We say that an extension field E of K splits A, or is a splitting field for A, if

(7.15)

97

SPLITTING FIELDS

E 0 KA

~

AJr(E)

for some r.

Since

E 0KMn(D) ~ Mn(E 0KD), it follows from the uniqueness part of Wedderburn's Theorem that E splits A if and only if E splits D. We may further remark that if E splits A, then so also does every field Q =:J E, since Q0 KA

~ Q

0

E

(E 0 K A )

~ Q

0

E

M ,CE)

~

Mr(Q)·

We are going to show that splitting fields always exist. Indeed, if K is any algebraic closure of K, then necessarily K splits A. To see this, observe that K 0 KA is a central simple K-algebra, hence is of the form Mr(D') for some skewfield D' over 1(. But then D' = K, since each element of D' is algebraic over K, and hence lies in K (see Exercise 31.6). The important problem for us is to show the existence of splitting fields which are finite separable extensions of K. This fact is contained in the following basic result: (7.15) Theorem. Let D be a skewjleld with center K, and let (D: K) be finite. (i) Every maximal subjleld E of D contains K, and is a splitting jleld for D. Further, if m = (E: K), then E 0 KD ~ Mm(E). (D:K) = m 2, (ii) There exists a maximal subjleld L of D which is separable over K.

Proof (i) Since (D:K) is finite, D contains maximal subfields. Let E be a maximal subfield of D. Clearly E =:J K, otherwise E(K) is a larger subfield of D. Now choose A = D, V = DD, B = E in (7.11). Obviously B' =:J E, and equality must hold, since for each x E B', E(x) is a sub field of D containing E. Thus B' = B = E, and (7.13) becomes D0

K

E ~ HomE(V, V) ~ A1)E),

(E :K)2 = (D:K),

where r = (V: E). But

r2 = (D 0 KE: E) = (D:K) = (E:K)2, so r = (E: K), and (i) is proved. (ii) We use induction on (D: K), and let (D: K) = n > 1, the result being trivial when n = 1. It clearly suffices to handle the case where char K = P -# O. Let us show first that D contains a separable extension E of K, with E =I K. For each xED - K we have K < K(x); call x purely inseparable over K if min. pol.K x is of the form XP' - a for some a E K. If x is not purely inseparable, then

98

SEMISIMPLE RINGS AND SIMPLE ALGEBRAS

min. POl.K x

(7.16)

= f(Xpe)

for some separable polynomial f(X) E K[X]. In this case, we need only choose E = K(x P "). So now suppose that each XED - K is purely inseparable over K. Then for each xED, min. pol. KX is of the form Xpe - a, a E K, e ~ O. We note that pi n, since n is a multiple of (K(x): K) for each XED - K. Choose F to be any maximal subfield of D, so by (i) there is an isomorphism of F -alge bras fl: F @K D ~ M/F)

for some r.

pe

Then [fl(1 @ x)]pe = fl(l @ x ) = a . In' whence all of the characteristic roots of fl(1 @ x) are equal, and thus (since pin) Tr fl(l @ x) = O. This holds for each x ED. But the matrices {fl(l @ x):x ED} form an F-basis for Mr(F). Hence every matrix in M/F) has zero trace, which is impossible. This completes the proof that there exists a field E such that K < E c D, E separable over K. Continuing with the proof, we let D' = {aED: ax

= xaforallaEE},

the centralizer of E in D. Then D' is a sub-skewfield of D, and we claim that E is the center of D'. By (7.11), E is the centralizer of D' in D. Any x E (D')" centralizes D', hence lies in E. This proves that (D'Y c E, and the reverse inclusion is obvious, because D' centralizes E. We have thus obtained a central simple E-algebra D', and (D' :E) < (D :K). By the induction hypothesis, there is a maximal subfield L of D' such that L is separable over E. Since E is separable over K, it follows (by transitivity of separability) that L is separable over K. Finally, we show that L is a maximal subfield of D. If not, then there exists an XED - L centralizing L. Then x centralizes E, so xED', and then L < L(x) cD', so L(x) is a subfield of D' properly containing L. This is impossible, and therefore L must be a maximal subfield of D, as desired. This completes the proof of the theorem. To conclude this subsection, we present a basic result concerning the behavior of K-algebras under separable extensions of the ground field: (7.16) THEOREM. Let L be a finite separable field extension of K, and let A be a

finite dimensional semisimple E-algebra, where E is any field containing K (possibl y (E: K) is in:finite). Then L@ KA is also a finite dimensional semisimple E-algebra. Proof It is clear that L@KA is an E-algebra of dimension (L:K)(A:E). It

(7.17)

99

SEP ARABLE ALG EBRAS

suffices to prove the result when A is a central simple E-algebra. We may write L ~ K[ X]/(f(Xn where f(X) is an irreducible separable polynomial in K[X]. Let f(X) = I1J;(X) be the factorization of f(X) into irreducible polynomials in E[X]. Since f(X) is separable, the {fJX)} are distinct, and therefore

L@K E ~ E[ X]/(f(X)) ~ L'E[ X]/Cf;(X)). Setting Fi = E[X]/(J;(X)), we see that each Fi is a field containing E. Therefore L @ K A ~ (L @K E) @E A ~ L' Fi @E A.

Butnow each Fi@EAisa central simple F(algebra by (7.8), whence L@KA is a semisimple E-algebra, as claimed. (7.17) COROLLARY. Let A be a finite dimensional E-algebra, where E and let L be a finite separable extension of K. Then

:::J

K,

Proof Let N = rad A. Then L @ N is a nilpotent ideal of L @ A, hence lies in rad (L @ A). On the other hand, L @ (AI N)

~

(L

@

A)/(L

@

N),

and L@ (AIN) is semisimple by (7.16). Hence L @ N => rad (L@ A), by Exercise 6.4. Thus L @ N = rad (L @ A), as claimed.

7c Separable algebras Throughout this section, let K, E, L denote fields, D and D' skewfields, and A, B finite dimensional K-algebras. If L is a finite extension of K, the assumption that L be separable over K has many important consequences (see (4.7), (7.16) and (7.17), for example). There is a more general concept of separability for K-algebras, and in this section we shall present this definition and some of its consequences. Indeed, one can even define separability of R-algebras, wllere R is a commutative ring (see for instance DeMeyerIngraham [IJ or Exercise 7.9). The most direct definition of separability is as follows: a separable Kalgebra is a finite dimensional semisimple K-algebra A, such that the center of each simple component of A is a separable field extension of K. Note that separability is not an intrinsic property of A, but depends on the choice of ground field. If(K: E) is finite and A is K-separable, then A will be E-separable precisely when K is separable over E.

100

(7.18)

SEMISIMPLE RINGS AND SIMPLE ALGEBRAS

(7.18) THEOREM. Let A be ajinite dimensional K-algebra. Thefollowing statements are equivalent: (i) A is a separable K-algebra. (ii) There exists a jinite separable jield extension E of K such that E Q9 KA is a direct sum of full matrix algebras over E. (iii) For every jield F ::J K, F Q9 KA is semisimple. Proof. Let A be a separable K-algebra with simple components Bl , ... ,Bt' and let K; be the center of B;, so K; is separable over K, 1 ~ i ~ t. For each i, by (7.15) we may choose a separable finite extension field E; of K; which splits B; , say E; Q9K, B; ~ Mr,{EJ Let us set K; ~ K[ X]/U;(X)), where J;(X) is a separable polynomial over K, and let E; be the splitting field of J;(X) over K. Then E; is a finite separable extension of K for each i. By Exercise 7.6, we can find a finite separable extension E of K which contains the fields E 1 ' ..• , Et , E~ , ... ,E; (or, more precisely, E contains K-isomorphic copies of the {EJ and {E;}). Therefore E Q9 K K; ~ E[X]/U;(X)) ~ I'E, where the number of summands on the right equals the degree of J;,(X). WethenhaveEQ9KA = I'EQ9 KB p and E Q9 K Bj ~ (E Q9 K KJ Q9 K,B; ~ I'EQ9 K,B; ~ '\"E Q9 Ei (E.t Q9K' B.) ~ '\" ~ )' Mr (E). ~ i l L... E Q9E' M r ,(E.) l i.-J I

1

I

This shows that (i) implies (ii). To prove that (ii) implies (iii), suppose that E Q9 K A ~

and let F

::J

I' M

r;

(E),

K. By Exercise 7.6, we may find a field L containing both E and

F, and then

L Q9K A ~ L Q9E(E Q9K A) ~ I'Mr,(L).

On the other hand, L Q9KA ~ L Q9F(F Q9KA).

If F Q9 K A were not semisimple, it would contain a nonzero nilpotent ideal N, and then L Q9 F N would be a nonzero nilpotent ideal in the semisimple ring L Q9 K A. This is impossible, and so F Q9 K A is semisimple, as claimed. Finally, we show that (iii) implies (i), by proving that if (i) is false, then so is (iii). If (i) is false, then either

101

SEPARABLE ALGEBRAS

(7.19)

(a) rad A ¥ 0, or (b) A is semisimple, but some Ki is inseparable over K, (using the notation of the first part of the proof). In case (a), (iii) is false Uust pick F = K 1). In case (b), Ki contains an element y such that min. pol.K y = (Xpr

+ exl(Xpr 1 + ... + ex" E K[ X],

where char K = p ¥ 0. Choose F to be the field K(ex/ IP , ... , ex" liP), and let z

= 1 ® y" + ex l l/p ® y"-l + .,. + ex" lip ® 1 E F ® K Ki .

Then z ¥ 0, since {I, y, ... ,y"} are K -linearly independent elements of K i • Further, zP = 1 ® yP" + ex ® yp(" - 1) + . . . + ex ® 1 = 0. l

"

Thus F ® K Ki contains a nonzero nilpotent element z. Since

F ®KBi ~ (F ®KKi) ®K,B i , it follows that z ® 1 is a nonzero nilpotent element in the center of F ® K B i , and thus generates a nonzero nilpotent ideal in F ® K Bi . Therefore neither F ® K Bi nor F ® K A can be semisimple. This completes the proof of the theorem. (7.19) COROLLARY. If A and B are separable K-algebras, so is A ®KB.

Proal Choose fields F, F' containing K such that F®KA ~ Mr(F),

F' ®KB ~ Ms(F'),

and then choose a field E containing both F and F' (Exercise 7.6). Then

E®KA ~ M.(E), and so by (7.7)

E ®K (A ®KB) ~ (E ®K A) ®E(E ®K B ) ~ Mr,(E). Thus A ® K B is separable over K, as claimed. Let AO denote the opposite ring of A, that is, AO

=

{xo:x EA}, with x,YEA.

Then A and AO have the same center. We set

Ae = A ®KAo, a finite dimensional K-algebra called the enveloping algebra of A. We may

102

SEMISIMPLE RINGS AND SIMPLE ALGEBRAS

(7.20)

view A as a left Ae-module, by means of the formula x,aEA,

(Indeed, every two-sided A-A-bimodule can be viewed as a left Ae-module.) There is a left Ae-epimorphism fl:A e

-4

A,

defined by setting fl(x Q9 yO) = xy. Clearly A is a projective left Ae-module if and only if there exists a map v E HomA.(A, A e) such that fl· v = 1. It is obvious that such a map v exists if and only if for each K-basis {xl' ... , xn} of A, there exist elements {y;} of A such that (i) L x;Y; = 1, and (ii) For each a E A, if ax; = L rI.;jXj' 1 :( i :( n, with the {rI.;j} E K, then y;a = Lrl.j;Y j , 1 :(

i:( n. n

Indeed, once v is given, we set v(1) =

L

x; Q9 y~; condition (i) asserts that

i= 1

fl· V = 1, while (ii) asserts that v is an Ae-homomorphism. Conversely, given a K-basis {xJ of A, and given elements {yJ E A satisfying (i) and (ii), we need only set v(a) = a· LX; Q9 y~, a EA. (7.20) THEOREM. Let A be a finite dimensional K-algebra. Then A is a separable

K-algebra

if and only if A is Ae-projective.

Proof. If A is K-separable, so is AO, and thus A e is K-separable by (7.19). In particular, then, A e is semisimple, and so every Ae-module is projective. Conversely, suppose that A is Ae-projective. We shall prove that A is Kseparable by showing that B = F Q9 K A is semisimple for each field F ::::> K. We have Be = B Q9 F SO = (F Q9 KA) Q9 F (F Q9 AO) ~

F Q9K(A Q9KAO) = F Q9KAe.

Let fl'=lQ9fl:FQ9KAe-4FQ9KA, and likewise let v'=lQ9v. Then v' is a Be-homomorphism such that fl' . v' = 1; let us set Xi,YiEB,

so LX;Yi = 1. In order to prove that B is semisimple, we show that if N c M are left B-modules, thenN is a direct summand of M. We may view HomK(M, M) as a B-B-bimodule, hence as a left Be-module, with x, YEB,

(7.21 ) Let n: M

103

SKOLEM-NOETHER THEOREM -4

N be a K-projection of M onto N, so n 2 = n. We set n' = v'(l) . n =

L XjnYj .

Since n = 1 on N, for each n E N we have n'(n) =

L xjn(Yjn) = L xjYjn =

n,

whence (n')2 = n'. Finally, for bE B we have bn' = (b (8) 1) . v'( l)n = v'((b (8) 1) . l)n

= v'((1

(8) be) .

l)n = (1 (8) bO)n' = n'b.

Thus n' is a B-homomorphism, whence N is a B-direct summand of M as claimed, and the theorem is proved. 7d Skolem-Noether Theorem

Throughout this subsection, A denotes a central simple K-algebra. Each invertible element a E A determines an inner automorphism of A, given by x -4 axa- 1 , x E A, and this automorphism fixes each element of K. The Sko1em-Noether Theorem, to be proved below, asserts the converse: every algebra automorphism of A fixing each element of K must be an inner automorphism. We shall prove a slightly more general version of this result, namely: (7.21) Theorem. (Sko1em-Noether). Let K c Be A, where B is a simple subring of the central simple K-algebra A. Then every K-isomorphism cp of B onto a subalgebra B of A extends to an inner automorphism of A, that is, there exists an invertible a E A such that

Proof. Let (8) stand for (8)K throughout this proof, and let us use the notation introduced in the first paragraph of the proof of (7.11). We have seen there that a simple left A-module V is a left D (8) B-modu1e, under the action (d(8)b)v = d·b·v,

VEV.

In the same way, we may view Vas left D (8) B module. We now make use of the structure of V as left D (8) B-modu1e, together with the K-a1gebra isomorphism 1 (8)cp:D(8)B~D(8)B,

to define an action (denoted by *) of D (8) B on V. We then obtain a left ii, having the same elements as V, but with the action of

D (8) B-modu1e

104 D (8) B on

SEMI SIMPLE RINGS AND SIMPLE ALGEBRAS

(7.22)

V given by (d (8) b)* v = d· cp(b) . v,

v E V.

Then V and V are a pair of left modules over the simple artinian ring D (8) B, and have the same K-dimension. It follows at once that there is a left D (8) B-isomorphism e: V ~ V. In other words, there is an isomorphism eE HomK(V, V) such that (7.22)

e(d' b· v) = (d (8) b) * e(v) = d· cp(b)' e(v),

VE

V.

Taking b = 1, it follows that eE HomD (V, V) = A, so e is given by a left multiplication by an element a E A. Since e is an isomorphism, a is invertible. Formula (7.22) becomes a . (d . b 'v) = d' cp(b) . av,

v E V.

If we put d = 1, and remember that A acts faithfully on V, we may conclude that a . b = cp(b) . a,

bE B,

which proves the theorem. (7.23) COROLLARY. (i) Every K-automorphism of A is inner. (ii) Any two K -isomorphic subfields Land L of A are conjugate, that is, L = aLa- l for some invertible a E A. (7.24) Theorem. (Wedderburn). Every finite skewfield is a field. Proof. Let D be a finite skewfield with center K, and let D* = D - {O}, K* = K - {O}. If L is a maximal subfield of D, then (D:K) = n 2 , (L:K) = n. Any other maximal subfield L is a finite field, and (L: K) = n, so card L = card L. Thus L is K-isomorphic to L, whence by (7.23) L is conjugate to L. Since every XED lies in a subfield K(x) of D, and hence in some maximal subfield, we conclude that (7.25)

D*

= U aL*a- \

where a ranges over some set of invertible elements of D. But for x E U, (ax)L*(ax)-l = aUa-l,

so in (7.25) it suffices to let a range over the left coset representatives of L* in D*. Hence there are (D*: L*) sets {aL*a- l } occurring in (7.25), each of cardinality card L*; these sets are not disjoint, since each contains 1: Thus D* cannot be their union, unless there is only one such set, that is, D* = L*. Thus D = L = K, as desired.

105

EXERCISES

EXERCISES 1. Prove Schur's Lemma: if M is a simple left A-module, then HomA(M, M) is a skewfield. [Hint: For each nonzero IE Hom)M, M), both ker I and im I are submodules of M, and so ker I = 0, im I = M.] 2. Let V(k) be the direct sum of k copies of the left A-module V, and let E = HomA(V, V). Prove that HomA(V(k), V(k)) ;::;< Mk(E).

2a. Let M be a finitely generated left A-module, where A is a simple artinian ring. Show that HomA(M, M) is also a simple artinian ring. [Hint: Write M ;::;< V(k), where V is a simple left A-module. Then use the preceding exercises and (7.3).] 2b. Prove that the result in Exercise 2a remains valid when "simple" is replaced by "semisimple". t

I" Bi ' where each Bi is a ring having no two-sided ideals except 0 and B .. Show that every two-sided ideal J of A is a subsum I" B .. [Hint: J. Bi = 0 or B: for each i, and J = I" JBi.J 3. Let A be a ring direct sum

i= 1

Iv

4. Let L be a minimal left ideal of the simple left artinian ring A, and let D = Hom A (L, L). Show that A = HomD(L, L). Using Exercises 1 and 2, prove that D is a skewfield, and that A ;::;< MJDO), where n is the di;nension of L as left D-space. [Hint: This is the harder part of the Wedderburn Structure Theorem (7.4). The following quick proof is due to Milnor [1]. Since L· A is a two-sided ideal of A, we have

L·A =A.Let

with m minimal. There is a left A -epimorphism m

I

xia i . The map is monic, since if I xia i

=

=

Ax!a l

C

m

)

0 with (say)

that

La l

n

La 2

contradicting the minimality of m. Thus A ;::;<

->

Xl

A, given by (Xl"'" Xm)

->

oF 0, then Ax! = L implies

+ ... + Lam'

nm) as left A-modules, whence

A°;::;< HomA(A,A);::;< HomAWm),n m));::;< Mm(D), so A ;::;< Mm(DO). Note that m

=

n.]

5. Let A be a semisimple finite dimensional K-algebra, and let E be a separable algebraic extension field of K, that is, for each field L with K c LeE, (L:K) finite, the extension L/K is separable. Prove that E ®KA is semisimple. [Hint: Let N = rad(E ®KA). For each L, N n (L ®KA) is a nilpotent ideal of L ®KA, hence is zero. But each X E N lies in some L ®K A.J 6. Let L 1 ' ... , Ln be a set of field extensions of K. Show that there exists a field 0 containing K, for which there exist K-isomorphisms l1 i :Li -> 0, 1 .:;; i .:;; n Further, if each (Li: K) is finite, the field 0 may be chosen so that (0: K) is finite. If in addition

106

SEMI SIMPLE RINGS AND SIMPLE ALGEBRAS

each L; is separable over K, then 0 may be chosen finite separable over K. [Hint: Let A = LI ®K'" ®K L., a commutative K-algebra. By Zorn's Lemma, A has a maximal ideal M. Set 0 = AIM, a field containing K, and define J-l; by composition of maps: Li -+ A -+ O. If each (L;: K) is finite, so is (0: K). If each L; is finite separable over K, then by (7.19) A is a direct sum of fields OJ, each of which is a finite separable extension of K, and we may pick 0 to be any 0rJ 7. Let A be a separable K-algebra, B any semisimple finite dimensional K-algebra. Prove that A ®KB is semisimple. [Hint: We may assume that both A and B are simple, and set L = A', E = B'. As in the proof of (7.16), we have L ®K E ~ L' F;, since L is separable over K, where each F; is a field containing K-isomorphic copies of Land E. Hence A ®KB ~ A ®L(L ®KE) ®EB ~

L' (A ®LF;) ®EB.

But A ®L F; is a central simple F;-algebra, and B is a central simple E-algebra, so each summand (A ®LF;) ®EB is simple by (7.6).J

8. Let G = {g 1 ' ... , g.} be a finite group of order n, and let K be a field. Show that the group algebra KG isa separableK-algebraifand only if char Kt n. [Hint: If char Kin, then L" go is a nilpotent element in the center of KG, so KG is not semisimple. If 1

char Ktn, set A

=

KG, and define v:A

v(l)

=

-+

A ®KAo = A e by

n- I Lg; ®g;-I,

v(a) = a' v(I), a E A. Show that v is an Ae-homomorphism such that J-lV = I, where WAe -+ A is as in § 7c.]

9. Let A be an R-algebra, where R is a commutative ring. Call A R-separable if A is N-projective, where N = A ®R A0. Show that if A is R-separable, then every R-projective A-module is A-projective. [Hint: Imitate the proof of (7.20)]' 10. Let R be a commutative ring, G a group of order n. Show that RG is R-separable if n is a unit in R. II. Let e, e' be idempotents in the simple artinian ring A, such that Ae ~ Ae' as left A-modules. Show that there exists a unit u E A such that e' = ueu- I . [Hint: Let A = Hom D (V, V), where V is a right vector space over the skew field D. Since V = e V EB (I - e) V is a D-decomposition of V, then relative to a suitable D-basis of V, e is represented by a diagonal matrix diag (1, ... , 1,0, ... , 0). The same holds for e', with the same number of I's, since Ae ~ Ae'. If u E A gives a suitable change of D-bases of V, then e' = ueu- I .] 12. If e is a primitive idempotent of the simple artinian ring A, show that Ae is a minimal left ideal of A, and A(I - e) a maximal left ideal. Further, let Ax be a maximal left ideal of A. Show that xA is a maximal right ideal. Prove that if Ax and Ay are any pair of maximal left ideals in A, then U· Ax . u- I = Ay for some unit u E A. [Hint: Let Ax be maximal, and write A = Ae EB Ax with e a primitive idempotent. Set I = e + ax, a E A. Then A = eA EB axA, so axA is a maximal left ideal. But axA is a homomorphic image of xA, and xA # A, whence xA is also a maximal right ideal. Further, if e and e' are primitive idempotents of A, then e' = ueu- I for some unit

EXERCISES

107

U E A, by Exercise II. But every maximal left ideal of A is of the form A(1 - e) for some primitive idempotent e.] 13. Let A be a separable K-algebra, L any extension field of K. Show that L <8>K A is a separable L-algebra. [Hint: For any field F => L, we have

F <8>L (L <8>K A) ~ F <8>K A,

and the latter is semisimple by (7.18).] 14. Let the separable K-algebra A have simple components {AJ, and let Ki be the center of AI. Let F be a field containing K, where possibly (F: K) is infinite. Prove that there is an F-algebra isomorphism F <8>KA ~

L:" L:" Fij <8>K,A;, ;

j

where each F .. is a field containing K-isomorphic copies of K j and F, and where (F;j: F) is finit~. Show that each summand F;) <8> K, A; is a central simple F;{algebra. [Hint: We have F <8>K A ~

L:" (F <8>K K;) <8>K, A;. i

Since K./K is finite separable, the F -algebra F <8> K K; is semisimple by (7.16), hence is expre~sible as a direct sum of fields Fij. Each Fi] contains K-isomorphic copies of F <8> 1 and 1 <8> K j

.J

2. Orders Throughout this chapter let R denote a noetherian integral domain with quotient field K, and let A be a finite dimensional K -algebra. Our aim here is to define orders, and to develop some of their basic properties. In later chapters we shall concentrate on maximal orders. Usually R will be a Dedekind domain, or at the very least, an integrally closed domain (see (1.12)), but occasionally theorems will be stated for the general situation in which R is only assumed to be noetherian.

8. DEFINITIONS AND EXAMPLES For any finite dimensional K-space V, a full R-Iattice in V is a finitely generated R-submodule M in V such that K· M = V, where K'M

=

CL:

(Xi

m! (finite sum): (Xi E K, mi EM}.

An R-order in the K-algebra A is a subring A of A, having the same unity element as A, and such that A is a full R-Iattice in A. Note that A is both left and right noetherian, since A is finitely generated over the noetherian domain R. Let us give some examples of orders: . (i) If A = M n(K), ~he algebra of all n x n matrices over K, then A = M n(R) IS an R-order in A. (ii) Let R be a Dedekind domain, and let L be a finite separable extension of K. Denote by S the integral closure of R in L. Then S is an R-order in L ~see (4.7)). In particular, taking R = Z, we see that alg. into {L} is a Z-order mL. (iii) Let a E A be integral over R, that is, a is a zero of a monic polynomial over R. Then the ring R[a] is an R-order in the K-algebra K[aJ. (iv) Let G be a finite group, and let A = KG be its group algebra over K that is, KG consists of all formal sums (Xx' x, C1." E K, with '

L:

xeG

(L: (Xx' x) + (L: Px ' x) = L: ((Xx + PJ' x, (L: (Xx' x) (L: P Y) = x.L (XxP (xy). y'

y

108

y'

(8.1)

109

DEFINITIONS AND EXAMPLES

Then RG =

{L

rtxx: rtx

E R} is an R-order in A.

xEG

From example (ii) it is evident that the study of orders includes as a special case the classical subject of algebraic number theory. On the other hand, in dealing with representations of a finite group G by means of matrices with entries in a domain R, one of the first steps is to pass from matrix theory to the study of RG-modules (see Curtis-Reiner [1J), so as to be able to take advantage of the ring structure of RG. Thus the study of orders also includes the theory of integral representations of finite groups. Let us show at once that every K-algebra A contains R-orders. Let M be anx full R-Iattice in A; such M's exist, and indeed if A = L· KXi' then RXi is a full R-Iattice in A. We define the left order of Mas

L

(8.1 )

O,(M) = {xEA:xM eM}.

Clearly 0lM) is a sub ring of A, and is an R-module. Let us verify that O,(M) is a full R-lattice in A. For each YEA, yM is an R-lattice in A, and so by §4 there exists a nonzero r E R such that r . yM c M. Thus ry E O,(M), which proves that K· O,(M) = A. Next, there exists a nonzero S E R such that s· 1REM. Therefore O,(M)' (S . 1R) c M, whence ,(M) c S -1 M. Since R is noetherian and s - 1 M is an R-Iattice, the above implies that O,(M) is also an R-lattice. This completes the proof that O,(M) is an R-order in A. Likewise, the right order

°

(8.2)

OJM) = {x E A:Mx

c

M}

is an R-order in A. For a full R-lattice M in A, and an R-order r in A such that r· M c M, let us compute Homr(M, M). Each cP E Homr(M, M) extends uniquely to an element of HomA(KM, KM~ hence is given by a right multiplication by an element of Or(M). Thus we have an identification (8.3)

where M is a left r-module. Note that Homr(M, M) does not depend on the choice of the R-order r in A. Now let R' be an integral domain containing R, with quotient field K'. Then R' ® RM is a full R' -lattice in the K' -algebra K' ® K A. Since M is finitely generated over the noetherian ring r, we have by (2.39) R'®RHomr(M,M) ~ HomR®R' (R'®RM,R'®R M ), provided that R' is R-flat. It then follows from (8.3) that there is a ring isomorphism (8.4)

110

ORDERS

(8.5)

This holds in particular if R' is any ring of quotients of R. Hence (8.5)

for each prime ideal P of R. Here, the subscript P indicates localization at P (see § 3d). The following result is fundamental: (8.6) THEOREM. Every element of an R-order A is integral over R. Furthermore, is integrally closed, then for each a E A we have

if R

min. pol.K a E R[ X], char. poL AtKa E R[ Xl Proof. For each a E A, it follows from the inclusion R[ a] c A that R[ a] is a finitely generated R-module, so a is integral over R by (1.10). If we assume in addition that R is integrally closed, the remaining assertions are consequences of (1.14) and Exercise 1.1.

A maximal R-order in A is an R-order which is not properly contained in any other R-order in A. If the integral closure Rcl of R in A happens to be an R-order in A, then by (8.6) Wi is automatically the unique maximal R-order in A. However, most of the time Rei is not a ring; furthermore, even in the case where A is commutative, so that Rei is a ring by (1.11), it may happen that Rei is not an R-Iattice. Thus, in example (ii) of maximal orders given above, if we drop the hypothesis of separability of Lover K, then S need not be finitely generated as R-module (see Artin [1]); in this case, there are no maximal R-orders in L. By way of reassurance, as well as for later use, we now prove (8.7) THEOREM. If A is a maximal R-order in A, then for each n, M.(A) is a maximal R-order in M.(A). If R is integrally closed, then M.(R) is a maximal R-order in M.(K). Proof. Let B = M.(A), r = M.(A), where A is a maximal R-order in A. Clearly r is an R-order in B. Let r c r', where r' is an R-order in B, and let N be the set of all those elements of A which occur as some entry of some matrix in r'. Obviously A c N c A, and we shall show that N is an R-order in A. Let AE N occur in the (r, s)-position of a matrix X E r'. We can find permutation matrices P 1 ' P z E r such that A occurs in the (l,1)-position of P1XP z Er'. Let E;. = diag(A,O, ... ,O)EMn(A). Then E;.

Since

= E1P1XPzE 1 Er'

for allAE N.

(8.8)

DEFINITIONS AND EXAMPLES

111

it follows that A' is a ring containing R. Further, the map A -+ E)., AE N, embeds Nin the R-lattice r ', and hence A' is also an R-lattice. The above shows that A' is an R-order in A. Therefore A' = A, since A is maximaL Hence every entry of every matrix in r ' lies in A, whence r ' c r. This shows that r = r ', and proves that r is amaximalorder. If R is integrally closed, then R is a maximal R-order in K, so the second assertion in the theorem is the special case of the first assertion, in which A = K, A = R. We shall see in §§ 17-18 that under suitable hypotheses on R and A, the converse of (8.7) is true: every maximal R-order in M.(A) is of the form M.(A) for some maximal R-order A in A. To conclude this section, let us consider the situation where an integral domain R is contained in a larger integral domain R' having the same quotient field as R. For convenience, we treat only the case where R is a Dedekind domain. We shall show that every R'-order A' comes from an R-order A by extension of the ground ring from R to R', and that likewise every A'-lattice comes from some A-lattice. (8.8) THEOREM. Let R be a Dedekind ring with quotient field K, and let R' be an integral domain such that R c R' c K. Then (i) Every R'-lattice M' contains an R-lattice M such that M' = R'M. (ii) Every R'-order N (in a finite dimensional K-algebra A) contains an R-order A in A, such that A' = R'A. (iii) Let A' = R'A, as in (ii). Then every left A' -lattice M' contains a left A-lattice M such that M' = NM = R'M.

Proof. It is clear that K is the quotient field of R'. The domain R' is also a Dedekind domain (see references listed in §4), though we shall not require this fact here. To prove (i), we note that any R'-lattice M' can be written as a sum S

M' =

L

R'mi'

i= 1

Set M = L Rmi ; then M is an R-lattice in M' such that R'M = M ', as desired. As usual, we identify an R-lattice M with its image 1 (8) Min K (8)R M, and denote the latter by KM. We may then compute the R'-module R'M in KM. Let us show that (8.9)

R' (8)R M

~

R'M c KM.

Indeed, the above R'-isomorphism holds when M = R, and hence by (2.17) it also holds true for every projective R-module. Since R is a Dedekind domain, by (4.13) every R-lattice is R-projective. This proves (8.9), and we may hereafter identify R' (8)R M with R'M, by means of the map IX (8) m -+ IXm, IXER' ,

mEM.

112

ORDERS

(8.9)

In order to apply the "Change of Rings" Theorem (2.37), we shall need to know that R' is R-flat. If X is any finitely generated R-submodule of R', then X is an R-lattice. Therefore X is R-projective, and hence X is R-flat by (2.16). This shows that every finitely generated R-submodule of R' is R-flat, whence also R' is R-flat by (2.19). Now let A' be an R'-order in A. By (i), we may choose an R-lattice L in A' such that A' = R'L. We set

A = O/L) = {xEA:Lx c L}, Then A and rare R-orders in A, and there is an identification A Homr(L, L), where L is viewed as left r-module. By (2.37) we obtain

=

R'A = R' ®R A = HomRT(R'L, R'L) = HomRT(A', A') = O/A') = A'. Therefore R' A = N, and (ii) is proved. Finally, any left N-lattice M' may be written as

i= 1

Then M = I Ami is a left A-lattice in M' such that M' = N M. Furthermore, R'M = R'(AM) = NM. This completes the proof of the theorem. EXERCISE 1. Let R be a noetherian integrally closed domain with quotient field K, and let r be a subring of A containing R, and suppose that r is a finitely generated R-module. Show that r is contained in an R-order in A. [Hint: if M is any full R-lattice in A, then so is M' r, and r c Qr(M' r).] A be a finite dimensional K-algebra. Let

9. REDUCED NORMS AND TRACES 9a Central simple algebras

Throughout this subsection, let A denote a central simple K-algebra, that is, a simple K-algebra with center K, such that (A:K) is finite. We have seen in § la how to assign to each a E K a characteristic polynomial, norm and trace, as follows: let aL E HomK (A, A) be left multiplication by a on A, and set char. pol.AIK a = char. pol. of a L

=

xm - (TAlK a)xm-l

+ ... + (-It NA/Ka,

where m = (A:K). Here T = TAlK is the trace map, and N = NA/K the norm map. For any X, Y E Mm(K) we know that trace (XY) = trace (YX). Therefore

(9.1)

REDUCED NORMS AND TRACES

(9.1)

T(ab) = T(ba), T(a + b) = T(a) + T(b), T(ra) = rT(a), { N(ab) = N(a) N(b), N(ra) = rm N(a),

113

fora,bEA,rEK. The above definitions make no use of the assumption that A is a central simple algebra. For central simple algebras, it is possible to introduce more useful concepts, namely those of reduced characteristic polynomial, reduced norm and reduced trace. We proceed to show how this may be done. Since A is a simple algebra with center K, there exists by § 7b an extension field E of K which splits A. This means that there is an isomorphism of E-algebras (9.2)

If 9 is another such isomorphism, then h· 9 -1 is an E-automorphism of Mn(E). Each such automorphism is inner, by (7.23), and so there exists an invertible t E M n(E) such that

h(u) = t·g(u)·t-t,

uEE®KA.

Consequently the matrices g(u) and h(u) have the same characterlstic polynomial. This show that char. pol. h(u) does not depend on the choice of the E-isomorphism h in (9.2). For a E A, define its reduced characteristic polynomial as red. char. pol.AtKa = char. pol. h(l ® a). We shall omit the subscript A/K when there is no danger of confusion. (9.3) THEOREM. For each a E A, red. char. pol.AtK a lies in K[X], and is independent of the choice of the splitting field E of A used to define red. char. pol. Proof. Let us show first that if F ::::::> K is another splitting field for A, then red. char. poL obtained using F is the same as that obtained by using E. By Exercise 7.6, there exists K-embeddings E ----> n, F ----> n, where n is some field containing K. It suffices to prove that red. char. poL via E is the same as red. char. pol. via n. We have an n-algebra isomorphism

After canonical identifications, we may rewrite the above n-isomorphism as (say)

so n is also a splitting field for A. Any element u E E ®K A may also be viewed as element of n ®K A. When this is done, h'(u) is precisely the matrix h(u)

114

(9.4)

ORDERS

viewed as element of Mn(Q). Consequently char. pol. h(u) = char. pol. h'(u), This holds in particular when u = I ® a, a E A, and so red. char. pol.AIK a is the same when computed via E as via n. This completes the proof that red. char. pol. does not depend on the choice of splitting field. In order to prove that red. char. pol. a E K[ X] for a E A, we observe that by (7.15) we may find a finite separable extension E of K which splits A. Since any extension field of E also splits A, we may in fact take E to be a finite galois extension of K. Let G be the galois group of E over K; then each (J EGis a K-automorphism of E, and K is the subfield of E fixed by G. To clarify the following argument, we start with any K-isomorphism f:E ~ E' of fields containing K, and later on we will choose E' = E, and f = (J E G. Consider the diagram E®KA

hi Mn(E)

-

E'~KA ,

-

Mn(E').

fCSl

r

h'i

+

Here, h is the E-isomorphism in (9.2), and f* takes a matrix in M n(E) onto a matrix in Mn(E') by applying f to each of its entries. The map h' is then defined by requiring the diagram to be commutative. Then h' is an isomorphism of E'-algebras, since h is an E-isomorphism. Hence for a E A, f*{h(l ® a)} = {h'(f ® I)} (1 ® a) = h'(l ® a). Therefore red. char. pol. a (via E') = char. pol. h'(l ® a)

= char. pol. f* {h(l ® a)} = f {char. pol. h(l ® a)} = f {red. char. pol. a (via E)}, where f is applied to a polynomial in E[ X] by acting on each of its coefficients. Now let E' = E, and let f range over all elements of G. Then the above shows that for a E A, each coefficient in red. char. pol. a is fixed by f. Therefore each coefficient lies in K, and so red. char. pol. a E K[ X], as claimed. This completes the proof. (9.4) Example. Let A = Q EBQi ® Qj ® Qk be the skewfield of quaternions over Q, where

k = ij

=

-ji.

(9.5)

115

REDUCED NORMS AND TRACES

Clearly A has center Q, and E = Q(i) is a maximal subfield of A, hence is a splitting field for A. Each a E A is uniquely expressible as a

=

IX

+ f3j,

13 E E.

IX,

There is an E-isomorphism

given by

where bars denote complex conjugation. Therefore red. char. poL A1Q a = char. poL ( _

= det

(X, -=f3

= X2

-

(IX

P ~)

IX

-

13 )

X-ii

+ ii)X + (lXii + f3P) EQ[X].

Hence (see (9.6a) below

tr A1Q a

= IX + ii,

nr AIQ a

= rxii + 1313·

Returning to the general case, we have (9.5)

THEOREM.

If(A:K)

=

n 2 , then

char. poL A1K a = {red. char. poL A1K a}n, Proof. Let E

:::J

K split A, so E ® K A

~

a E A.

M n(E). By Exercise 1.2,

char. poL A1K a = char. poL E®AIE 1 ® a,

a E A,

where ® denotes ® K • By (7.3) there is a left E ® A-isomorphism E ® A yen), where V is a minimal left ideal of E ® A. Hence

~

char. poLE®AIE 1 ® a = f(x)n, where f(X) is the characteristic polynomial of the matrix (1 ® a)L which describes the action of 1 ® a on some E-basis of V. On the other hand, the E-isomorphism in (9.2) can be obtained by mapping each Y E E ® A onto the matrix h giving the action of y on some E-basis of V, and thus char. poL h(y) = char. poL y L ' Taking y

=

1 ® a with a E A, we obtain

YEE®A.

116

(9.6)

ORDERS

red. char. pol.A/K a = char. pol. h(l ® a)

(9.6)

= char. pol. (1 ®a)L' Hence char. pol.A/K a = f(X)", as desired. Let us write (9.6a)

red. char. pol.AIKa =

xn -

(tra)X n -

1

+ ... + (-l)"nr(a),

aEA.

We call tr(a) the reduced trace of a, and nr(a) the reduced norm of a. It follows at once from (9.5) that (9.7)

Furthermore, using the notation of (9.6), the map a ---+ (1 ® a)L' a E K, gives a K-a1gebra isomorphism of A into HomE (V, V). Since tr(a) = trace of (1 ® a)L'

nr(a) = determinant of (1 ® a)L'

we conclude that (9.8)

tr(a + b) = tr(a) + tr(b), tr(ab) = tr(ba), { nr(ab) = nr(a)' nr(b), nr(sa) = s" . nr(a),

tr(sa) = s· tr(a),

fora, b E A, s E K.

(9.9) THEOREM. The reduced trace map gives rise to a symmetric nondegenerate K-bilinear form ,:A x A ---+ K, by setting ,(a, b) = tr(ab),

a,bEA.

Further, ' is associative, that is, ,(ab, c) = ,(a, bc) for a, b, c E A. Proof. It is clear that, is symmetric and associative, and we need only prove that, is nondegenerate. Let E ::::> K split A, so E ®KA ~ Mn(E). We may extend, to an E-bilinear map

,': (E ® A) x (E ® A)

---+

E,

and indeed from the definition ofreduced trace, we have ,I(U, v) = trace (uv),

u, VEE ® A

~

Mn(E),

where trace (uv) is the ordinary trace of the matrix uv. Furthermore, r nondegenerate if and only if " is non degenerate. Let B = {u E Mn(E): trace (uv) = 0 for all v E Mn(E)}.

IS

(9.10)

117

REDUCED NORMS AND TRACES

Since " is symmetric, B is a two-sided ideal in the ring M n(E). But B does not contain the identity matrix In' and thus B = O. Hence,' is nondegenerate, whence so is T. This completes the proof. If char K divides n, it follows from (9.7) that the ordinary trace map TAlK is the zero map. Hence in this case it is essential to use the reduced trace, rather than the ordinary trace, in Theorem 9.9 and its applications. 9b Semisimple algebras

In the preceding subsection we have defined the reduced characteristic polynomial, reduced norm and reduced trace, for elements of a central simple K-algebra A. The aim of this subsection is to generalize these concepts to the case where A is an arbitrary finite dimensional semisimple K-algebra. We begin the discussion with some elementary results from linear algebra. LetK c Lbefields,with(L:K) = n. Each a ELmaps onto a matrix fiEMn(K) describing the action ofleft multiplication by a on some K-basis of L. Given f(X) = define l(x) =

I

I

aiX i E L[X],

fiiX i E Mn(K[X]).

We now define a norm map NL/K:L[X] N L/K f(X) = det ](X),

-->

K[X] by setting f(X) E L[ Xl

Let us establish some basic properties of this norm: (9.10) THEOREM. Let K c L bejields, with (L:K) = n. Then (i) N L/K is multiplicative. (ii) For IX} , ... , as E L, we have (9.11)

N L/K(X S

+ a 1 x s - 1 + ... + a,} = xns + (TL/K IX1)X ns - 1 + ... + N L/K as.

(iii) Let V be an s-dimensional L-space, and let qJ be an L-endomorphism of V. We may view qJ as a K-endomorphism of V. Then (9.12)

char. pol.VIK

qJ

= N L/K (char. pol"V/L qJ).

Proof. Write N in place of N L/K ' and let f'.. 9 E L[ Xl The mapa is a ring homomorphism. Therefore Jg = f g, and so N(fg) = det(fg) = (det]) (det f7) = (N f)(N g).

-->

fi, a E L,

118

(9.12)

ORDERS

This establishes (i). Next we have N(X' + CC x s- 1 + 1

... + ccs) =

+ IX 1 x s- 1 + ... + IXs) + (tr IX 1 )X ns - 1 + ... + det IX s'

det(lX S

= xns

with the second equality following from an obvious calculation. This yields assertion (ii), since by § 1 we have tr IXl = T L/K cc 1 ' To prove (iii), we first write V V such that cp(~) c ~. Then

det IXs = N L/K CCs .

L' ~ where each

=

char. POl.V/K cp =

~ is an L-subspace of

IT char. POl.V;/K cP, i

and likewise with K replaced by L. In view of assertion (i), it therefore suffices to establish (9.12) for each ~. Changing notation, we may hereafter assume that V is an indecomposable K[ cp J-module. Thus we may write

V = L[X]/(f(X)) for some f(X)

E

L[X],

with cp acting on Vas left multiplication by X. Then char. pol.v/I. cp = f(X) = xs

+

CCI XS -1

+ ... + CCs

(say).

n

Now let L =

L 'Ku

j ,

and let us calculate the action of cp on the K-basis

j= 1

of V consisting of the elements {u jXi: 1 ~ j ~ n, 0 ~ i ~ s - I}. Relative to a suitable indexing of these basis elements, the action of cp is given by the matrix 0 0 0 -IX s fl

=

I

0

0

- IX s- 1

0

I

0

-IX s- 2

...................... 0 . I -IX 1

0

"

This matrix fl is an s x s array of blocks, each of size n x n. We have

Xl -I

6

. Xl ... .,

6 0

char. pol.V/K cp = det (X I - fl) =

6 6

6 Xl 6 ... -1

iXs IXs-l

(9,1 3)

REDUCED NORMS AND TRACES

119

Now perform elementary row operations on this determinant, from the bottom up: add X times the sth row of blocks to the (s - l)th row, then X times the (s - l)th row of blocks to the (s - 2)th row, and so on. This gives

6 5 -1 5 '" char. pol.V/K q> =

5 J(X) 6 *

....................... .

6 6 ... 5 * 6 6 -1 * = detj(X) =

N L,Kf(x).

This completes the proof of (iii), and establishes the theorem. Any semisimple K-algebra A may be written as a direct sum L' Ai of simple algebras Ai' whose centers are finite extensions of K. In order to define red. char. pol.A/K a for each a E A, we shall first treat the case in which A is a simple K-algebra; the extension to the semisimple case will be straightforward.

(9.13) Dejinition. Let B denote a central simple L-algebra with (B:L) = m2 , and let K be a subfield of L with (L: K) = n. For each bE B, we define its reduced characteristic polynomial relative to K by red. char. POLB/K b = N L/K (red. char. pol.B/L b)

= xmn - (tr B/K b)Xmn-l + ... + (-ltnnrB/Kb. We call tr H/K the relative reduced trace, and nr B/K the relative reduced norm. When L = K, the preceding concepts reduce to those given in § 9a. Before proceeding to the case of semisimple K-algebras, let us establish some basic properties of relative reduced traces, norms, and characteristic polynomials. (9.14) THEOREM. Let B be a central simple L-algebra, K a subjield of L, and let (B :L) = m2 , (L:K) = n. For each b EB, we have (9.15) (9.16) (9. [7)

trB/Kb = TL/K (trB/L b),

nrB/K b = NL/K (nr B/L b),

char. pol.B/K b = {red. char. pol.B/K b}m,

TB/K b = m' tr B/K b,

NB/K b = {nrB/K b}m.

Proof. Given b E B, let us set h(X) = red. char. pol.B/L b = xm - a 1 xm-l

+ ... + (-lta m .

120

ORDERS

(9.18)

Then III = tr BIL b, Il m = nr BIL b, according to the definitions therefore

III

§9a. But

red. char. pol.BIK b = N LIK h(X)

= xmn - (TLIK 1( 1)x mn-1 + '" + (-l)mnN LIKllm' by (9.11). Comparing this expression with that given in (9.13), we conclude at once that

This yields the desired formulas (9.15). Next, we may compute char. POl.BIK b by considering the left multiplication by the element b acting on a K-basis for B; a corresponding statement holds with K replaced by L. Therefore char. pol.BIK b = N LIK (char. pol.BIL b)

by (9.12)

= N LIK {(red. char. POl.BIL br}

by (9.5)

= {NLIK (red. char.pol.BILb)}m

by (9.10)

=

{red. char. pol.BIK b}m.

This proves (9.16); since (9.17) is an obvious consequence of (9.16), the theorem is established. (9.18) COROLLARY. The analogues of formulas (9.8) hold for tr BIK and nr BIK . Proof. The assertion is clear from (9.8) and (9.15). We remark merely that nr B/K Ilb = a

mn

.

nr B/K b,

IlEK,

bEB.

We are now ready to generalize Theorem 9.9, as follows: (9.19) THEOREM. Let B be a simple K-algebra whose center L is separable over K. Then the map tr BIK, gives rise to a symmetric associative nondegenerate

bilinear form from B x B to K. Proof. The form is given by r(a,b) = trBIK(ab), a,bEB, and we need only check that r is non degenerate. If r(a, b) = 0 for all bE B, then TLIK(Il·trBIL(ab)) = TLIK(trBIL(Il'ab)) =OforallIlEL,

bEB.

But TLIK gives a non degenerate trace form, since L is separable over K, and

(9.20)

121

REDUCED NORMS AND TRACES

hence we have trB/Lab = 0 for all bEB. Therefore a = 0 by (9.9), which completes the proof. (We remark that the existence of a symmetric associative bilinear form on B to K tells us that B is a Frobenius algebra over K (see Curtis-Reiner [1, (61.3)]); we know this anyway, since every simple K-algebra is a Frobenius algebra. The significance of (9.19) lies in the fact that the form arises from the reduced trace map tr B/K') Suppose now that A is any semisimple K-algebra, and write A Ki

= Al E8 ... E8 At (simple components), = center of Ai' (Ai:KJ = 1:( i

m;,

:( t.

Each K t is a finite extension field of K. Every element a E A is uniquely expressible as a = a 1 + ... + at' al E Ai' We define t

(9.20)

red. char. pol.A/K a = n red. char. pol. A;/K

Qj

i= 1 t

= n N K;/K (red. char. pol.AdK' a). i= 1

As usual, we define tr A/K and nr A/K by the equation (9.21) red.char.pol.A/Ka = X' - (trA/Ka)X'-l

+ ... + (-I)'nrA/Ka

for each a E A, where r = L mi (Ki: K). For future use, we record the following obvious consequences of these definitions: t

(9.22)

tr A/K a =

t

L

tr AdK a j =

i= 1 t

(9.23)

L j=

TKdK (trAdK, a),

1

t

nrA/Ka = nnrA;/Kai = fINK,'K (nrAdK,a). i= 1

i= 1

From (9.16), we also obtain t

(9.24)

t

char. pol.A/K a = n char. pol.AdK ai = n {red. char. Pol.Ai/Ka;}m" 1=1

i=1

t

t

(9.25)

TA/K a =

L i= 1

mi trAdK ai'

N A/K a =

n {nrA;/K aJm,.

1= 1

Of course, the analogues of formulas (9.8) remain true for tr A / K and nrA / K From (9.19) we have at once

.

(9.26) THEOREM. Let A be a separable K -algebra. Then the map tr AIK gives rise to a symmetric associative nondegenerate bilinear trace form from A x A to K.

122

ORDERS

(9.27)

Finally, we show that reduced characteristic polynomials, norms and traces, are not affected by change of ground field. We prove (9.27) THEOREM. Let A be a separable K-algebra, and let F be any field containing K (not necessarily finite dimensional over K). View A as embedded in the separable F -algebra F (8) A, where (8) means (8) K. Then for all a E A, red. char. pol.F ® A/F a = red. char. pol.A/K a,

(9.28) (9.29)

Proof. By Exercise 7.13, F (8) A is a separable F-algebra. We have

char. pol. F0A/F a = char. pol. A/K a

(9.30)

by Exercise 1.2, a fact which we shall use presently. It suffices to prove the theorem for the case where A is a central simple L-algebra, with L a finite separable extension of K. By Exercise 7.14, we may write d

F(8)A ~

I' Bj' j= 1

where each F j is a field containing Land F. Here, each B j is a central simple F[algebra, and (B;= F j ) = (A:. L) = m 2 , say. For a E A, write a = b 1 + ... + bd , bjEB j . Then (9.24) YIelds d

char. pol.F ®A/F a =

TI {red. char. poLBj!F bj}m j= 1

= {red. char. Pol.F®A/Fa}m.

Since

char. pol.A/K a = {red. char. pol.A/K aJm

by (9.16), we now deduce (9.28) as a direct consequence of(9.30). The formulas in (9.29) follow from (9.28), and the theorem is proved. For each element a of a K-algebra A, we have computed char. pol.A/K a by letting a act from the left on a K-basis of A. We shall conclude this section by showing that when A is a separable K-algebra, we could equally well have calculated the characteristic polynomial of a by letting a act from the right. For each a E A, we defined aL : x --> ax, x E A, aR : x --> xa, x E A, so a L and aR are a pair of K-linear transformations on A. In § 1 we defined char. pol.A/K a = char. pol.K a L We now prove

·

(9.31) (9.31)

123

REDUCED NORMS AND TRACES THEOREM.

Let A be a separable K-algebra. Then

char. pol.KaL = char. po1. K aR for each a EA. n

L . Kxp and

Proof. Let A =

let {y l '

... ,

Yn } be a dual basis of A relative

i= 1

to the reduced trace form, so n

A =

L· Ky

j

1 :( i, j :( n,

,

j= 1

where tr is an abbreviation for tr AIK . We shall compute char. pol.KaL by letting a L act on the K-basis {Xi}' and we shall compute char. pol.KaR by letting a R act on the K-basis {yj }. For 1 :( j :( n, let aL(x j ) = ax] =

L IXijX i ,

aR(y) = YP =

i

L PijY i , i

where the IX ij , Pi] E K. Then char. pol.KaL = det(X(\ - IX i ) , char. pol.KaR = det(Xc5 ij - Pi)' However, for each i,j we have (Xij = tr yiax j'

Pi] = tr YPXi .

Thus the matrices (IXi') and (Pi]) are transposes of one another, and hence have the same chara~teristic polynomiaL This completes the proof of the theorem. (9.32) COROLLARY. Let A =

LO KXi' and let

xp

=

LYijx, '

Yij E K.

Then N AIK a

= det 6' ij ), TAlK a = trace of (y ij)'

It should be pointed out that (9.31) and (9.32) need not be true for arbitrary finite dimensional K-algebras A. The following example is given in Bourbaki [1, §12, no. 3]: let A = KEEl Kx EEl Ky, where

xy

=

y,

Relative to the basis {I, x, Y}, we have XL

0 ° 0) (0 0 1

= 1 1 0,

YX = 0,

124

(9.33)

ORDERS

and XL' x R have different characteristic polynomials. Furthermore, XL and x R have different traces, and if char K "# 2, then (1 + X)L and (1 + X)R have different determinants. EXERCISES 1. Let A be a central simple K-algebra, and let E be a subfield of A such that (A:K) = (E:K)2. Show that for every aEE, red. char. poL A1K a

(min. poLK a)(E: K(a)).

=

[Hint: Letf(X) = min. poLK a; thenf(X) has degree (K(a):K). But red. char. poLA1Ka is a power of f(X), and has degree (E: K).] 2. Let A be a central simple K-algebra, where K is any perfect field (that is, K has no inseparable extensions). Using (9.5), prove directly that red. char. poL A1K a E K[ X] f or each a EA. [Hint: Show that if E is an extension field of K, and f(X) E E[ X] is such that f(X)" E K[X], then also f(X) E K[X].] 3. Let A be a separable K-algebra, and choose a field E

:::>

K such that

d

E®KA=

LOB

j

I

,

~oj ~

d.

j= !

d

(9.33)

red. char. poLA1Ka

=

I1 char. pol. q?/l ® a). j=!

[Hint: Let f(X) denote the expression on the right hand side of(9.33). As in the proof of (9.3), the isomorphism ({Jj is unique up to an inner automorphism of Mnj(E), and hence f(X) does not depend on the choice of the maps {({Jj}' The proof of (9.3) carries over to this case, and shows that f(X) does not depend on the choice of E, and that f(X)EK[X]. To prove (9.33), it suffices to treat the case where A is a central simple L-algebra, with Lj K separable. Choose E :::> L so that E ®IA ~ Mm(E),

where

(A:L) = m 2,

where

(L :K)

and so that n

E ®KL ~

L

0

Ei ,

n,

=

i=l

and where each Ei equals E and contains an isomorphic copy of L. Then n

E ®KA ~ (E ®KL) ®LA ~

L £=1

n

0

Ei ®LA

=

L' M .. (EJ i=l

(10.1 )

EXISTENCE 0 F MAXIMAL ORDERS; DISCRIMINANTS

125

But then by (9.28) red. char. pol.AIK a

=

red. char. POI.1H'iJKAIE 1 ® a

=

n char. pol. (!/t,(1 ® a))

.

=

f(X),

i= 1

where

!/I;: E

®K A

--->

Mm(EJ]

4. Let L be a finite separable extension of K with (L: K) = n, and let t/I l ' ... , t/I. be the distinct embeddings of L into an algebraic closure of K. Prove that for each f(X) EL[X],

N L/d(X)

=

n" t/lJ(X),

i= 1

where t/lJ(X) is obtained from f(X) by applying

5. Let A a = (lX'j) E

t/I i to each of its co\efficients.

M,(D), where D is a skewfield with center K. Let a E A be given by M,(D). Show that =

r

trA/Ka =

[Hint: Choose E

:::>

--->

trD/K(IX,,).

K to be a splitting field for D, and let fl:D

Then p': A

L ;;;1

E ®KA

~Mrs(E)

--->

E ®KD ~ Ms(E).

is given by

/l'(a) = (p(lX ij ))1 and

<; ;.j <;r'

,

trA/Ka = trace of pr(a) =

2:

trace of {1(IX,,)

1=1

, =

L

trD/K(lX jj ) ' ]

i=l

10. EXISTENCE OF MAXIMAL ORDERS; DISCRIMINANTS Throughout this section, A is a separable K-algebra, where K is the quotient field of a noetherian integrally closed domain R. Let A be an R-order in A. In dealing with reduced characteristic polynomials and reduced norms and traces, we shall omit the subscripts AI K when there is no danger of confusion. (10.1) THEOREM. For each a E A we have

red. char. poL a E R[X],

tra

E

R,

nraER.

Proof. By (8.6), char. pol. a E R[ XJ. The same holds for red. char. pol. a by

126

ORDERS

(10.2)

(9.24) and Gauss' Lemma. The assertions about tr, nr are then clear from (9.21 ). In brief, (10.1) asserts that integral elements have integral reduced norms and reduced traces. We shall use this fact repeatedly. Let us define the discriminant of A to be the ideal d(A) of R generated by the set of elements where m = (A:K). Each x;xjEA, so the m x m matrix (trx.x.) has entries in R, and thus d(A) c R. Furthermore, if Xl ' ... , xrn are cho~~n so as to be linearly independent over K, then since the bilinear reduced trace form A x A ~ K is nondegenerate, it follows at once that det(tr x.x.) '# O. Thus d(A) is a nonzero ideal of R. I J (10.2) THEOREM. Let A have a free R-basis Ul d(A) is the principal ideal R' det(tr uiu j ).

' ... , Urn'

Then the discriminant

Proof. This principal ideal surely lies in d(A). On the other hand, let Xl' ... , Xm E A. We may write Xi = riku k , rik E R, and then

L

tr xix j = Therefore

L rikrjl' tr ukul'

1 ~ i, j ~ m.

k,/

det(tr xix j ) = det(r ik )' det(truku/)' det (r j /)', where ' denotes transpose. But det(r ik ) E R, and thus we deduce that d(A) cR' det (tr uiu j ), which completes the proof. If P is any prime ideal of R, and Rp is the localization of R at P, then Ap = Rp ®R A is an Rp-order in A. It follows at once from Exercise 4.13 that d(Ap) = {d(A)}p. If R is a Dedekind domain, then each Rp is a principal ideal domain, and Ap has a free Rp-basis. Thus we may use (10.2) to compute d(Ap), and then we have (by Exercise 3.1 or (4.21)) d(A)

=

n

d(Ap).

p

Returning to the case where R is any noetherian integrally closed domain, we prove (10.3) THEOREM. Let r be a subring of A containing R, such that K· r = A, and suppose that each a E r is integral over R. Then r is an R-order in A. Conversely, every R-order in A has these properties. Proof. We already know that every R-order has the indicated properties. Now

let

r

be a subring satisfying the given conditions; we need only check that

EXISTENCE OF MAXIMAL ORDERS; DISCRIMINANTS

(10.4)

r

127

is an R-lattice. Let m

A

=

L' Ku

ex = det(tru,u')l_" ER. 1. J ~!,J~m

i ,

i= 1

Then a i= 0 since A is separable over K. We claim that m

rca - 1 .

L

Ru;.

i= 1

Let x E r, and write x =

I

riu i , r i E K. Then m

tr xU j

=

L r i ' tr uiu

1 ~j

j '

~

m.

i= 1

Since each xU j E r, we have tr xU j E R. If we use Cramer's Rule to solve the above system of equations for the {rJ, we thus obtain ri

= (element of R)/a,

This proves that rca-i.

L Ru

i

,

1

~ i ~

m.

and hence that r is an R-lattice.

(10.4) Corollary. Every R-order in A is contained in a maximal R-order in A. There exists at least one maximal R-order in A. Proof. Let A be an R-order in A, and let C be the collection of R-orders in A containing A. Then C is non-empty. If {Aa} is a chain of orders containing

A, let

N =

LA" = IjA". a

a

Then A' is a subring of A containing R, and K· N = A. Each x E A' lies in some Aa , hence is integral over R. It follows from (10.3) that A' is an R-order in A. We have now shown that any increasing chain of elements of C has an upper bound in C. By Zorn's Lemma, C has a maximal element A o ' which is then obviously a maximal R-order in A. It remains to prove that A contains at least one maximal R-order. The discussion preceding (8.1) shows that there exists a full R-lattice M in A. Setting A = DcCM), it follows from §8 that A is an R-order in A. Hence A is contained in some maximal R-order Ao in A, by the first part of this proof, and the corollary is established. We remark that the crucial step in the proof of (10.3) is the fact that (tr uiu ) is a nonsingular matrix, or equivalently that trAIK gives rise to a nondegenerate bilinear form A x A --+ K. If char K = 0, it is clear from (9.25) that the ordinary trace TAlK already gives a nondegenerate form, and so the proof of (10.3) could have been given using ordinary traces instead of reduced

128

(I 0.5)

ORDERS

traces. Nevertheless, even when char K = 0, the reduced trace and reduced norm will playa basic role in our later discussions. (Without the hypothesis that A be a separable K-algebra, it may happen that there are no maximal R-orders in A, as already pointed out in §8. The following example, due to Faddeev [1. ~25J (see Roggenkamp and HuberDyson [I, pp. 201-202J) shows how to construct an infinite strictly increasing chain of orders, in case A is not semisimple. Let A be any R-order in the K-algebra A, and assume that rad A i= O. Set Lk = A n (rad A)\ k ~ 1. Then each Lk is an R-pure A-submodule of A (see (4.0)), and is a full R-lattice in (rad At Since rad A is nilpotent, there exists an integer n ~ 1 such that (rad A)" i= 0, (rad A)n+l = O. Let r be a non-unit in R, and set Ak = A

+ r-kL 1 +

r-zkL z

+ ... + r-nkLn ,

k ~ l.

Then each Ak is an R-order in A, and A = Ao C Al C Az C . . . . We show that the chain is strictly increasing. For suppose that Ak = Ak+ 1 for some k. Multiplying by r"(k+ I), it follows that Ln C rk A. Therefore Ln C ~A n(radA)n = ~Ln' the equality following from the fact that Ln is R-pure in A. This gives Ln = ~L. ' a contradiction since L. is an R-lattice and r is not a unit of R. In particular, the above shows that A is always strictly contained in the larger R-order Al ' and thus A has no maximal R-orders whatsoever.)

Let us now return to the case where A is a separable K-algebra. We shall now prove that a decomposition of A into simple components A = Al E9 ... E9 At yields a corresponding decomposition of maximal orders and hereditary orders in A. We may write 1 = e 1 + ... + ep e i E Ai; the {eJ are then central idempotents of A such that eie j ~ 0, i =1= j, and Ai = AeJoreach i. (10.5) THEOREM. Let A be a separable K-algebra with simple components {AJ, and let R; be the integral closure of R in the center Ki of Ai' Then (i) For each maximal R-order J\. in A, we have A = L' Ae p where the {eJ are the central idempotents oj A such that Ai = Ae i . Further, each Ae i is a maximal R -order in Ai . (ii) If Ai is a maximal R-order in Ai' 1 ~ i ~ t, then L' Ai is a maximal R-order in A. (iii) An R-order Ai in Ai is a maximal R-order if and only if Ai is a maximal R(order. (iv) When R is a complete discrete valuation ring, all the preceding assertions remain true under the weaker hypothesis that A is a semisimple K-algebra, not necessarily K-separable.

(10.6)

EXISTENCE OF MAXIMAL ORDERS; DISCRIMINANTS

129

Proof. Suppose that A is a separable K-algebra until further notice, and let A be a maximal R-order in A. Then A = A(e 1 + ... + e,) c Ae 1 EB ". EB Ae, ' and each Ae. is an R-order in A .. If Ae. c r. c A., where r. is a maximal R-order, the~ L' ~ is an R-orde; in A ~ontaining 'A. Theref~re A = L'~' and so Ae i = ~ for each i. Conversely,letA i c Ai beamaximaIR-order, I ~ i ~ t,andletA = L' Ai' If A e r e A with r maximal, then Ai c rei' whence Ai = rei for each i. Therefore r = L' rei = A, proving that A is a maximal order. We claim next that Ri is an R-Iattice, and is also a noetherian integrally closed domain. Indeed, Ri is an R-Iattice by (10.3), since Ki is a separable K-algebra. Thus Ri is noetherian. Further, if a E Ki is integral over Rp then RJa] is a finitely generated Ri-module, hence is also finitely generated over R. But then it follows by (1.10) that every element of Ri[a] is integral over R. Thus a is integral over R, and so a E R i. This completes the proof that Ri is integrally closed. Now let Ai be a maximal R-order in Ai' Since the elements of R j commute with those of Ai' and both Ri and Ai are R-Iattices, also R j • Ai is an R-Iattice. Thus R i ' Ai is an R-order in Ai' whence Ai = Ri . Ai' which shows that Ai is necessarily an Rj-order in Ai' But any Rj-order in Ai is also an R-order in Ai' since Ri is finitely generated over R. Hence Ai must be a maximal R(order in Ai' Conversely, the same remarks show that every maximal R,-order in Ai is also a maximal R-order in Ai' This completes the proof of (i)-(iii ). To prove (iv), let us now assume that R is a complete discrete valuation ring, and that A is a semisimple K-algebra. The hypothesis of separability was used to establish that each Ri is an R-Iattice, and is a noetherian integrally closed domain. But these facts hold even without this hypothesis, by § 5, since we are assuming that R is complete. Indeed, each Ri is also a complete discrete valuation ring, and is finitely generated over R by Theorem 5.6. The preceding proofs of (i)-(iii) thus hold equally well in the present case, and the proof of the theorem is complete.

The preceding theorem shows that the study of maximal orders always reduces to the case of simple algebras, and even to the central simple case if the ground ring is a Dedekind domain. As we shall see eventually, when R is a Dedekind domain, every maximal R-order is hereditary. We shall not use this result in the discussion below, however. Let A be any R-order in A; by a left A-lattice we mean a left A-module M which is an R-Iattice. In analogy with the results in §4a, we prove (10.6)

THEOREM.

Let A be an R-order in a semisimple K-algebra A. Then

130

(10.7)

ORDERS

every left A-lattice M is A-isomorphic to a submodule of a free module Nk) for some k. Proof. We may embed Min KM ( = K ®R M), and KM is a finitely generated left A-module. Hence there is an A-exact sequence A(k) -> KM -> 0

for some k, and this sequence splits since A is semisimple. Thus there is an A-isomorphism cp of KM into A(k); this isomorphism carries M into a A-lattice cp(M) in A(k). Since A (k) is a full R-Iattice in A (k), there exists a nonzero rER such that r' cp(M) c A(k). The map defined by m -> r' cp(m), mEM, then gives the desired embedding of M into A(k). The ring A is left hereditary if every left ideal of A is projective, or equivalently (see §2f) if sub modules of free A-modules are projective. The above result gives at once COROLLARY. The R-order A is left hereditary A-lattice is projective.

(10.7)

if and

only

if every

left

(10.8) THEOREM. If A is a left hereditary R-order in the semisimple K-algebra A,

I'

with central idempotents {e i } as in (10.5), then A = Ae i , and each Ae i is a left hereditary R-order in Ai' Conversely, the direct sum of left hereditary orders is left hereditary. Proof. Let A be a left hereditary order. There is an exact sequence of left A-modules o -> A' -> A -> Aei -> 0, and A' = A n A(l - e i ) is an R-order in the algebra A' = A EB··· EB At' It is easily seen that A' is also left hereditary. Now Ae i is a left A-lattice, hence is A-projective by (10.7), and so the above sequence splits, that is, A' is a direct summand of A:

A = A' EB J,

J

=

left ideal of A.

Multiplying. by ep we find that J = Ae,. An induction argument then proves that A = I Ae i . Suppose conversely that Ai is a left hereditary R-order in Ai' 1 ~ i ~ t, and let A = I' Ai' Each left A-lattice M then decomposes as M = I' eiM, and eiM is a left A(lattice. Thus each eiM is ArProjective, and so M is A -projective. It should be pointed out that an R-order A is left and right noetherian, so

131

LOCALIZA nON OF .ORDERS

(10.9)

by the discussion in §2f, the ring A is left hereditary if and only if A is right hereditary. We may remark that hereditary orders enjoy yet another property of maximal orders; in (40.7) we shall prove the following result of Harada [2J: (10.9) THEOREM. Every hereditary R-order Ai in Ai is also an R(order, where R; is the integral closure of R in the center of Ai' as in (10.5).

EXERCISES Throughout, R is a Dedekind domain with quotient field K, and A is a separable K-algebra. 1. Let tr: A -> K be the reduced trace map. If a E A is integral over R, show that tr(a m ) E R for each positive integer m. Is the converse true? 2. Find deAl for A = ZffiZiffiZjffiZk (see (9.4)). Set AD = A +Za, where a = (1 + i + j + k)/2, and find d(Ao)' Prove that AD is a maximal Z-order in A, where A = QA. 3. Let A c A' be a pair of R-orders in A. Show that deN) IdeAl, and that deN) = deAl if and only if N = A. [Hint: First localize so as to reduce the problem to the case where R is a principal ideal domain.J 4. Let A be any R-order in A. Deduce from Exercise 3 that A is contained in a maximal R-order in A. 5. Let r be a subring of A containing R, such that r is finitely generated as Rmodule. Prove that r lies in a maximal R-order in A. [Hint: Use Exercise 8.1.] 6. Let A be an R-order in a separable K-algebra A, and let S be a domain containing ®R A is an S-order in L ® K A. Prove the discriminant formula:

R, with quotient field L. Then S

where the last equality is an identification. 7. Let A be an R-order in A, and let L be a left A-lattice in A, M a right A-lattice in A, such that K . L = K· M = A. Prove that for each a E A,

ordRL/aL

= R' NA/Ka =

ordRM/Ma.

[Hint: By (4.20) or Exercise 4.4, it suffices to prove the result when R is a discrete valuation ring. In this case, Land M have free R-bases, which are also K-bases for A If L = RXi and we write ax j = L a.ijX i , a.ij E R, then ord R L/aL = R . det(a. i ) by Exercise 4.2. But det(cxij) = N A/K a by § 1. For the corresponding statement about M, use (9.32).J .

r

11. LOCALIZATION OF ORDERS Here we investigate the relation between an R-order A

III

A and its

132

ORDERS

(11.1)

localizations Ap and completions Ap , where P ranges over the prime ideals of R. We keep the hypotheses of section 10. (11.1) THEOREM. Let S be a multiplicatively closed subset of R. For each maximal R-order A in A, S-I A is a maximal S-I R-order in A. Proof. By (USe) and (3.11), we know that S-I R is noetherian and integrally closed, since R has these properties by hypothesis. Let i: A --+ S-I A be the homomorphism x --+ 1 ® x, X E A, defined as in (3.10). Now S-I A is an S-I R-orderin A; let r be any larger order inA, and choose a nonzero a E R such that a' r c: S-I A. Then Ll = i-I (ar) is an S-saturated submodule of A such that ar = S- 1 Ll. Since i is a two-sided A-homomorphism, Ll is a two-sided A-module. Then the right order O.(Ll) coincides with A, since O,(Ll) :::::> A and A is maximal. Hence by (8.4) we have

S-IA = O,(S-ILl) = O/ar) =r, which completes the proof. (11.2) COROLLARY. An R-order A in A is maximal if and only if for each maximal ideal P of R, Ap is a maximal Rp-order in A. Proof. If A is maximal, so is each Ap by (11.1). Conversely, assume that each Ap is maximal, and let A c: r, where r is any R-order in A. Then Ap c: r p ,

so that Ap = rp for all P. But then (r/A)p = 0 for all P, whence r = A by (3.15). The following material, up to (11.5), can be skipped by the reader interested only in the case where R is a Dedekind domain. The above result shows that the question of whether an R-order A is maximal is a "local" one, and can be decided by knowing whether the localizations Ap are maximal orders, where P ranges over the maximal ideals of R. Now if P' c: P are a pair of prime ideals of R, we have an inclusion Ap c: Ap" and indeed Ap'

~

(Rp)p" Rp ®RpAp,

that is, Ap' is a localization of Ap at the prime ideal P' . Rp of Rp. Hence if Ap is a maximal Rp-order, then by (1Ll), Ar is necessarily a maximal Rp,-order. Thus the hypothesis that Ap be a maximal order for each maximal ideal P of R is a rather strong one, since it implies that Ap' is a maximal Rp,-order for every prime ideal P' of R. The weakest hypothesis would be the assumption that Ap' is a maximal order for every minimal prime P' of R. (Recall that a minimal prime of R is any minimal member in the set of nonzero prime ideals of R.)

(1l.3)

133

LOCALIZA nON OF ORDERS

The following result shows that in order to decide whether A is maximal, it suffices to consider the ring structure of the localizations of A at the minimal primes of R, together with the structure of A as an- R-module. We recall (see page 55) that A is a reflexive R-module if A = A **, where A*

= HomR(A,R),

A**

= HomR(A*,R).

We saw in (4.25) that for any noetherian integrally closed domain R, we have A** = nAp,

(1l.3)

p

where P ranges over the minimal left ideals of R. Using this, we prove (11.4) THEOREM. (Auslander-Goldman [1]). An R-order A is maximal if and only if A is a reflexive R-module such that for each minimal prime P of R, Ap is a maximal Rp-order. Proof. Obviously A c: A ** since A c: Ap for each P. By (4.26), A ** is a full R-Iattice in A. Furthermore, A ** is a ring by (11.3), since it is an intersection of rings. Thus A ** is an R-order in A containing A. If A is a maximal R-order, then we have A ** = A, and hence A is reflexive. The fact that each Ap is maximal has already been shown in (11.1). Conversely, let A be reflexive, and let each Ap be maximal. Then A

= A** = nAp,

P minimal.

p

If A c: r, with r an R-order in A, then Ap c: rp for each P, whence Ap = rp for each P. Thence r c: n rp

= nAp

=

A,

p

so A is maximal. This completes the proof. In applying (11.2) or (11.4), we are faced with the problem of deciding whether the Rp-order Ap in A is maximal. The underlying ring Rp is an integrally closed local noetherian domain, with unique maximal ideal p. Rp. For such a domain, we now show that we may pass to the completion, and then consider the problem in the complete case.

(11.5) THEOREM. Let R be an integrally closed local noetherian domain, R its completion, and K the quotient field of R. Let A be an R-order in A, and set

A = R ® RA, so A is an R-order in A. Then A is a maximal R-order in a maximal R-order in A.

A

if and only if A is

134

ORDERS

Proof. Suppose A maximal, and let A c L1, where L1 is an R-order in A. Then r = A n L1 is an R-order in A and t = L1, by (5.4). Since A c r, we deduce that A = r, and thus A = L1. This proves that A is maximal. Conversely, let A be maximal, and let A c r for some R-order r in A. Then Act, so A = t, and thus by (5.4) A = A n A = Ant = r. This completes the proof.

(11.6) COROLLARY. Let R be a Dedekind domain, and A an R-order in A. Then A is a maximal order if and only if for each prime ideal P of R, the P-adic completion Ap is a maximal Rp-order in Ap. Proof. This follows from 01.2) and 01.5).

EXERCISE 1. Let R' be a ring of quotients of the domain R, and let M be an R-lattice, X a full R'-lattice in KM. Show that M () X is an R-lattice in M of the same R-rank as M.

3. Maximal Orders in Skewfields (Local Case) Throughout this chapter, let R be a complete discrete valuation ring, that is, R is a principal ideal domain with a unique maximal ideal P = nR #- 0, and R is complete relative to the P-adic valuation. Let K be the quotient field of R, and R = RIP its residue class field. We call K a local field. Let D be a skewfield whose center contains K, such that (D: K) is finite. The aim of this chapter is to study the structure of maximal R-orders in D. As we shall see, the results in this case are surprisingly explicit, and generalize several of the theorems listed in § 5. The presentation below is based on a beautifully written article by Hasse [I].

12. UNIQUENESS OF MAXIMAL ORDERS We are going to show that D contains a unique maximal R-order L1. The argument will depend strongly on the use of Hensel's Lemma (5.7). Let v be the exponential P-adic valuation defined on K (see § 5b, where v was denoted by vK ). Then it is easily seen that for a, b EK we have (i) (ii) . { (iii) (iv)

(12 I)

v(a) = 00 if and only if a = 0 v(ab) = v(a) + v(b) = v(ba) v(a + b) ~ min (v(a), v(b)) The value group of v is infinite cyclic,

where the value group of v is defined to be {v(a):a #- o}. (12.2) THEOREM. Let f(X)

= ctoxn +

ct l

xn-l

+ ... + ctn E K[ X] be irreducible.

Then Proof. Let t = min {v(aJ 0 ~ i ~ n}. If the theorem is false, then t < min (v(a o ), v(an )). Let r be the largest subscript for which v(a,) = t. Then r i= 0, r i= n, and ct; 1 . f(X) = poxn

+ ... +

p,xn- r +

... + Po,

Pi E R,

where {J, = I and (J, + 1 ' ..• , Pn E P. However, in R[ X] the right hand expression is a product xn-'(I + ~r-l X + ... + ~oX') of two relatively prime 135

136

(12.3)

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

polynomials, with the first factor monic. It follows by Hensel's Lemma that rx; 1 . f(X) is reducible, whence so is f(X), a contradiction. This completes the proof. Now let N DIK and TDIK be the norm and trace maps, defined by char.pol.DIKa = xm - (TDIKa)xm-l

+ ... + (-lfNDIKa,

aEA,

where for the rest of this section we put (D: K) = m. We now set w(a) = m- 1 'v(ND1Ka),

(12.3)

aED.

(12.4) THEOREM. For a E D, let f(X) = min. pol.Ka. Then w(a) = (K(a):K)-l'v(f(O)) = (K(a):K)-l ·v(NK(a)IKa). Proof. Since D is a skewfield,f(X) is irreducible, and therefore char. pol.DIK a = f(Xt ,n , where n = degree of f(X) = (K(a): K). Therefore N DIKa = (-It· {j(0)}m1n, whence w(a) = m- 1 . v({j(0)}m1n) = n- 1 . v(f(O)),

which completes the proof. Using (12.2) and (12.4), we may prove the fundamental (12.5) THEOREM. An element a E D is integral over R or equivalently, if and only if w(a) ~ O.

if and only if N DIK a E R,

Proof. It is obvious that w(a) ~ 0 if and only if Na E R, where we write N for N DIK' If a is integral over R, then char. pol.DIK a E R[ X] by Exercise 1.1, whence Na ER. Conversely, let Na ER, and set f(X) = xn + rx1X n- 1 + ... + rxn = min.pol.Ka.

Then f(O) = rxn , and since ± N a is a power of f(O), it follows that rxn E R. But f(X) is irreducible in K[X], so by (12.2) we deduce that min (v(I), v(rxn)) = 0, 0 :( i:( n. Hence f(X) E R[ X], so a is integral over R. This completes the proof. v(rxJ

~

(12.6) THEOREM. The map w is a discrete valuation on D which extends v, that is, w satisfies conditions (l2.l), and w( rx) = v( rx), rx E K. Proof. Consider the conditions (12.1) on w; (i) clearly holds, since a = 0 if and only if N a = O. Condition (ii) is satisfied since N is multiplicative, and

(12.7)

UNIQUENESS OF MAXIMAL ORDERS

137

v satisfies (ii). For IX E K, we have W(IX) = V(IX) by (12.4). Next, the value group of w, {w(a): a E D, a #- O}, is a nonzero additive subgroup of m -lZ, and is therefore isomorphic to Z. It remains for us to show that w satisfies (iii), and for this it suffices to consider the case where a = 1 and w(b) ~ 0, with bED. But then b is integral over R by (12.5), so by (1.11) also 1 + b is integral over R. Hence w(1

+ b)

~

0 = min (w(1), w(b)),

as desired. This completes the proof. Let us set (12.7)

~

= {aED:w(a)

~

O} = {aED:ND/KaER}.

Then ~ is a ring containing R, by (12.6). We call ~ the valuation ring of w. We shall show in (13.3) that ~ is finitely generated as R-module. Taking this for granted, we have (12.8) Theorem. ~ is the unique maximal R-order in D, and is the integral closure of R in D. Proof Clearly K . ~ = D, and so ~ is an R-order in D. But by (8.6), every element of an R-order in D is integral over R, hence lies in ~ by (12.5). Therefore ~ is the unique maximal R-order in D, and the result is proved.

We are now going to prove that w is the unique extension ofv to D satisfying (12.1), and begin with (l2.9)LEMMAt. Let f(X) Then

= xn +

1X1xn-1

+ ... + IXnEK[X]

be irreducible.

1 :>:; k:>:; n.

Proof. Let E be a splitting field for f(X) over K, and write f(X) = (X - P1) ... (X - Pn)' Pi E E. Let v' be the extension of v to E defined as in (12.3), namely, v'(P) = (E: K)- 1. v(N E/K P), pEE.

But min. pol.K Pj = f(X), so by (12.4) we have v'(P) = (K(Pj):K)-l. v(f(O)) = n- 1 . v(lXn)

for each j. However, t For a generalization of section 3.1 J.

± IXk

is the kth elementary symmetric polynomial in

this lemma, see the discussion of Newton's polygon in Weiss [1,

l38

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

(12.10)

P1 ' ... , Pn ' so by (12.1) ii) and iii) we obtain V(lX k ) = V'(lX k )

k v(a ) n n

~ - •

for each k, as desired. (12.10) THEOREM. The valuation w is the unique extension of v to D having properties (12.1).

Proof Let u: D ~ R u {CfJ} be another such extension of v. If u ;/; w, there exists a nonzero a E D such that u(a) ;/; w(a). Replacing a by a- 1 if need be, we may assume that u(a) > w(a). Let f(X) = min. pol.K a =

xn + 1X1X·- 1 + ... + IX. E K[X].

Then w(a) = v(lXn)/n by (12.4), while V(IX) irreducible. Therefore n u(IX.·a J

for 0

~

.

J)

~

(jfn) V(IX) by (12.9), since f(X) is n - j

j

= V(IX.) + (n - j) u(a) > _. + --·V(IX) = V(IX), J n V(IX) n n n n

j ~ n - 1, where il

n

-

1X0

= 1. But now

il,,_l·

a-

11..-

2

• a'·

- ... - an,

whence

:0

~j ~

n - I} > v(an).

This gives a contradiction, so the theorem is proved. The preceding discussion is of course valid for the special case where D is an extension field of the complete P-adic field K. It shows that the P-adic valuation on K can be extended to a discrete valuation on D, by means of formula (12.3), and that this extension is unique (up to equivalence).

13. RAMIFICATION INDEX, INERTIAL DEGREE Keeping the notation of §12, let D be a skew field of finite dimension over a complete P-adic field K contained in the center of D, and let 11 be the unique maximal R-order in D. We shall define the ramification index e = e(D/K) and inertial degree f = f(D/ K), and shall show that the analogue of Theorem 5.6 holds true in this case. In fact, the discussion below does not make use of any of the earlier results from §5, and so provides a proof of Theorem 5.6, by choosing D to be a field. Let v be the exponential valuation on K, and let w be the extension of v to the skewfield D. We have seen that the value group ofv is Z, while the value

(13.1)

RAMIFICATION INDEX, INERTIAL DEGREE

139

group of w is a subgroup of m - lZ, where m = (D: K). Hence the value group ofw equals e- lZ, for some positive integer e dividing m. We write e = e(DIK), and call e the ramUication index of Dover K. Thus, the least positive number in the set {w(a):aEL1} is precisely lie. We shall now normalize the valuation w, by replacing w by the equivalent valuation vD' where vD = e' w. Then vD has value group Z, and for a E D,

vD(a) = (elm) . v(N D/K a )

(13.1)

{

= e'(K(a):K)-l v(f(O)), where f(X) = min.pol.Ka.

Let n E R be a prime element, so v(n) = 1. We may choose n DE L1 so that vD(n D) = 1; call n D a prime element of L1. We have at once

vD(n) = e. The group of units of L1 is given by

u(L1) = {aEL1:v D(a) = O} = {aEL1:N D/K aEu(R)}. Each nonzero a ED is uniquely expressible as

a = n; . a' = a" . n; ,

n = vD(a),

where

a', a" E u(L1).

Note that a' and a" need not be equal.

(13.2) THEOREM. Let n D be a prime element of L1, and set p = n DL1. Then every nonzero one-sided ideal of L1 is a two-sided ideal, and is a power of p. The residue class ring L1/p is a skewfield over the field R, and p n R = P. Proof For L any nonzero left ideal in L1, let

I = min {vD(a): a E L}, and choose an element x E L such that vD(x) = l. Then for each y E L we have yx- 1 E L1, so Y E L1x c L. This proves that L = L1x. Furthermore, L1x = xL1 SInce

vD(x-1ax) = vD(a),

a E D.

Thus L is a two-sided ideal of L1, and clearly

L = n~L1 = L1n~ = pi,

since

X'

n;;l E L1.

The above shows that p is a maximal left ideal of L1. The ring K = L1/p therefore has no left ideals except 0 and K, and thus K is a skewfield. Finally, p n R is a prime ideal of R, and is nonzero since p is a full R-Iattice in D. Thus p n R = P, whence P . X = 0, so X is a skewfield over R.

140

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

(13.3)

We now define the inertial degree of Dover K as f

= f(D/K) = (iLR).

Note that 3; = 6./p is a vector space over R, since Pcp. The fact that f is finite follows from the theorem below, which generalizes Theorem 5.6, and indeed provides an independent proof thereof. (13.3) THEOREM. Let D be a skewfield of finite dimension m over a complete P-adic field K contained in the center of D. Let e = e(D/K) be the ramf{ication index of Dover K, and f = f(D/K) the inertial degree of Dover K. Both e and f are finite, and e' f = (D:K)= m. Let a l' ... , af E 6. be such that 3; = L' Ra; , and let bo , ... , be- 1 E 6. be such that vD(b j ) = j, 0 ~ j ~ e - 1. Then the e'f products {a;b j } form a free R -basis for 6.. ~roof. F~rst, let {a;: 1 ~ i ~s} be elements in 6. whose images {aJ in is are 1mearly l~dependent over R, and let {b/ 1 ~ j ~ t} be elements of K such that the mtegers {vD(b j )} are incongruent mod e. We claim that the s' t products {a;bJ are linearly independent over K. If not, there is a relation

(13.4)

(IlXliaJb 1

+ ... + (IlXt;aJb, = 0,

lXj;EK,

where we may assume that for each bj , at least one of the coefficients lXjl ' ... , lXjs is nonzero. Write

Then each f3 j; E R, and for each j there is an index k for which f3 jk ¢ P. We may rewrite (13.4) as (If3 li a)nn'b 1

+ .,. + (If3tia)nn'bt =

O.

Since vD(n) = e, replacing each bj by nnJb j does not affect the hypotheses or the conclusion. After making this replacement, and renumbering the a's and b's if need be, we obtain a relation

(I l'Jia)b

1

+ ... + (l: ytia)bt =

0,

where each l')1.. E R, and and l'11

rt P. Therefore

2: (~l'j;a,)bj/bl )

1

= 0, and b)b1 E P for 2 ~ j ~ t.

(13.5)

141

RAMIFICATION INDEX, INERTIAL DEGREE

L

Passing to the residue class ring /1, we obtain the relation )Ilia; = 0, with )111 # 0, which contradicts the linear independence of the {aJ over R. We have therefore established that the s . t products {a;b;} are elements of D linearly independent over K, so st ~ m. This proves that f is finite, and shows also that ef ~ m. Now let a1 ' ... , af' bo ' ... , be- I E /1 be elements satisfying the hypotheses of the theorem. The e' f elements {a;b j } are linearly independent over K, by the first part of the proof, and we need only show tha t /1 = L Ra;b j ' Let X E

/1, x # 0, and let vD(x) = ke x

+ j,

where

= (nkb)u,

°

i, j

~ j ~

e - 1. Then

u E u(/1).

Since the {ai } form an R-basis for L\, we may find elements {rJ in R such that u = r l a l + ... + rfa f + z, with Z E p. Hence we have

x

= (Lr;nka)b j + xl'

where

i

If x 1 # 0, we may repeat this procedure with Xl ' and continue if need be with x 2' x 3 ' .• , • Note that k is the greatest integer in vD(x)je. Hence after n' e steps, we obtain an equality e-l

(13.5) x = L

f

L (rWnk

+ riJl nk+ I + '" +

r~~? nk+n-l)aib j

+ Yn ,

j=O i= I

where vD(Yn ) ~ vD(x) + ne. For fixed i, j, the sequence k ·Ji)nk r(1)n tj ,Ttl

k+ 1 + r(2)n ij , ...

is a Cauchy sequence from R, relative to the P-adic valuation of R. Since R is assumed complete, this sequence has a limit sij E R. If we set x' = ') s 'J..a.b. E /1, then for each n we have from (13.5) , J ~

i, j

Thus vD(x - x') = proved.

00,

so x = x'. Therefore x E L Ra;bj , and the theorem is

We have now established that L\ is finitely generated as R-module, and indeed that /1 is a free R-module on m generators. This shows that Theorem 12.8 is valid, that is, L\ is the unique maximal R-order in D. (13.6) COROLLARY. The skewfield D is complete relative to the valuation vD.

142

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

Proof Keeping the notation of(13.3), we have D = be a Cauchy sequence from D, whence vD(x n r(n) 'J

r" Kaibj' Let )

Xn -1) -+ 00

-

EK,

as n

(13.7) {Xl' X , ... } 2 -+ 00.

Write

n;;::: 1.

Set VD(X n -

so also n_ lim kn = if

00.

=

X n _ 1)

e'

kn

+ In'

We have then Xn -

n kn . y n'

Xn - 1 =

YEA n 0.

Write Yn as an R-linear combination of the {ajb j from the above equation that r\n) _

r(n-1)

')

};

then by (13.3), it follows

EnknR

I)

.

Hence {r~~), r~f), ... } is a Cauchy sequence from K, with limit p .. , say. Then 'J obviously lim xn =

n-+oo

.

I

i~j

PiPibj'

Let K be the center of the skewjield D, and set (D:K) = n2 , e = e(DIK), f = f(DIK). Then

(13.7)

THEOREM.

eln,

nlf

Proof Let w be the extension of v to D defined in (12.3). By definition of e, we have wenD) = lie. Then K(n D) is a subfield of D, and has ramification index e. Hence by (5.6) or (13.3), we have el (K(nD):K). But K(nD) lies in some maximal subfield E of D, whence (K(nD):K)I(E:K). Since (E:K) = n by (7.15), the result follows. EXERCISES 1. Let Y be any full set of representatives in t1 of the residue classes in K, where we assume that 0 E Y. Prove that every nonzero XED is uniquely expressible as

x

= n~D(X)·(so

+ SInD + s2n~ + ... ),

SiEY,

2. Prove that the order ideal ordR ii eq uals pI, where f [Hint: K ~ RV) as R-modules.]

=

# O. feD! K). So

3. Prove that v(N DfK n D) = f, and that v(m DfK n D) = .J]Te. [Hint: e- 1 = m- 1 v(N DfK n D), m = (D:K), so v(NDfKn D ) = f. Next, N DfK n D = (m nD),,(iJ:K),

so f

=

.J(D: K) v(m n D) =

jeJ. v(nr n D).]

(14.1)

FINITE RESIDUE CLASS FIELD CASE

143

4. A fractional L\-ideal is a full left L\-lattice in D. Show that the set of fractional L\-ideals, with the obvious definition of multiplication, is an infinite cyclic group generated by the ideal p. Prove also that p = rad L\.

14. FINITE RESIDUE CLASS FIELD CASE Throughout this section we assume that R is a finite field with q elements. Let D be a skewfield with center K, and let (D : K) = n 2 . We call n the index of D. Denote by v the valuation vK on K, and by vD that on D, as in (13.1). Let ~ be the unique maximal R-order in D. We shall see here that the structures of D and ~ can be described explicitly in this case, and depend only on the index n and some integer r such that 1 ::s; r ::s; n, (I', n) = l. The discussion to follow depends heavily on the results listed in §5b, and we shall use the notation introduced there. Since R is finite, we know from (5.10) and (5.11) that for each positive integer f, there is a unique unramified extension W of K with (W: K) = f, given by W = K(w), where w is a primitive (qf - 1)th root of 1. If Ow denotes the valuation ring of W, and Ow the residue class field of ow' then Ow

= R[w],

Further, lf7K is a galois extension with cyclic galois group of order j; generated by the Frobenius automorphism a defined by w -+ w q. Likewise, owlR is a galois extension, with galois group cyclic of order f, generated by the automorphism (j which maps (jj onto (jjq. Before turning to the study of skew fields, we need one more preliminary result. (14.1) THEOREM. Let W be an unramified extension of K of degreef, and let v be the valuation on K. GiOen any element a E K, the equation NW'KX

= a,

is solvable for x if and only iff divides v(a). Proof The prime element n of R is also a prime element for ow' Each nonzero x E W is thus expressible as x = nkx o , Xo E u(ow). Then vw(x o) = 0, whence N Xo E u(R), where N denotes N W/K' But

Nx

=

nnf N(x o)'

and thus the condition that f divides v(cr) is a necessary one. To prove sufficiency, it is enough to deal with the case where a E u(R). Since the galois groups of WIK and of 0wlR are isomorphic, it follows at

144

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

04.2)

once from Exercise 1.4 that

N(a)

T(et) = T(~),

NCa),

=

where N, T are the norm and trace maps from Ow to R, while Nand Tare those from W to K. We shall show that both Nand T are epic. We know this for T, since Ow is separable over R (see (4.6)). Next, the multiplicative group of Ow is cyclic of order qI - 1, with generator 05. Since Gal (ow/R) is cyclic with generator if: 05 --> O5q, the distinct algebraic conjugates of 05 over Rare { O5q' : 0 ~ i ~

f - I}.

Each nonzero element of Ow is of the form O5t, 1 Exercise l.4) we have N(wt)

= O5t(l+q+'" +qf-')

~ t ~ qI -

1, and (by

= W t(qf- 1 )/(q-l).

Thus N(05~ = 1 if and only if (q - 1)1 t. Hence exactly (qI - l)/(q - 1) elements of Ow have norm 1, so there are q - 1 distinct nonzero norms. This proves that N is epic. Continuing with the proof, let et E u(R). Since N is epic, we may choose b o E Ow such that et := Nb o mod nR. For i ;:: 1, suppose that we have found an element such that (mod n'R).

et := N ai

Of course ai E u(ow) since Na i E u(R). We proceed to construct the next approximation, and set a i+1 = ai + bni, with b E Ow to be determined so that (14.2)

et := Na i + 1 (mod n i + I R).

Since i ;:: 1, by Exercise 1.4 we have I-I Na i+1 =

IT

:= Na i

where for each s, Cs

(c;S(aJ

+

+ c;S(b)ni)

I-I ni

I

s=o

Cs

(mod n i +1 R),

= ai·c;(ai)···c;S-I(ai)·c;S+I(ai)···O'I-l(ai)·O'S(b).

If we set y = O'(a)' O'2(a,) . .. 0'1- 1 (aJ, then Cs = O'S(y)' O'S(b) = O'S(yb), whence I-I I Cs = TW/K(yb). s=o

(14.3)

FINITE RESIDUE CLASS FIELD CASE

145

Condition (14.2) thus becomes rx == Na,

+ ni TW/K(yb)

(modn i + 1 R).

But y is a unit in ow' and T is epic, so we can always choose b E Ow satisfying the above condition. We have therefore obtained our next approximation a.+ l ' If we set x = lim a., then x E Ow and rx = N W/K x, which completes l i-X) ~ the proof of the theorem. Now we are ready to consider skewfields. The first important consequence of the hypothesis that R be finite is as follows: (14.3) Theorem. If(D:K) = n 2 ,then e(DIK) = f(DIK) = n. Proof. Li is a skewfield of finite dimension f over R, hence is a finite skewfield. By Wedderburn's Theorem (7.24), Li must be a field. Hence Li = R(a) for some a E 11, so by (13.3) or (5.6) we have (K(a):K) ~ (Li:R) = f.

But K(a) is a sub field of D, whence (K(a): K) ~ n by (7.15). Therefore f ~ n, so f = n by (13.7). Finally, ef = n2 implies that e = n, and the proof is complete. As a consequence of the above theorem, we now know that nl1 = pO,

Li =

R(~),

where p is the unique maximal ideal of 11, n is a prime element of R, and ~ is a primitive (q" - l)th root of lover R. We wish to prove, in analogy with the results of § 5b for the case of fields, that the skewfield D comes from an unramified extension, followed by a completely ramified extension. To begin with, choose any element WE 11 such that w = ~, and set W = K(w). Then W is a subfield of D, and by (5.8) W is an umramified extension of K such that (W: K) = (Li: R) = n, and hence W is a maximal subfie1d of D. Further, by (5.10) W is a cyclotomic extension of K containing all of the (q" - l)th roots of 1. Hence we could have chosen w to be a primitive (q" - l)th root of lover K. We shall call Wan inertia field of D. We now know that every skewfield D of index n contains an inertia field K(w) as maximal sub field, where w is a primitive (q" - l)th root of lover K. To what extent is the subfield W of D unique? It is unique up to K-isomorph ism by (5.10), but there may be many mutually K-isomorphic subfie1ds of D. Obviously W = K(w i ) for any j relatively prime to q" - 1, so replacing

146

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

(14.4)

w by such a power wi does not affect W However, (see Exercise 14.1) there are usually more than q" - 1 distinct (q" - l)th roots of 1 in D. Indeed, if ex E D is nonzero, then exwex- 1 is also a (q" - 1)th root of 1, and need not be a power of w. We call the elements w, exwex- 1 conjugate in D, and likewise the fields Wand ex Wex -1 are conjugate subfields of D.

(14.4) THEOREM. The inertia field of D is unique up to conjugacy.

Proof. By (5.10), W is unique up to K-isomorphism. Hence W is unique up to conjugacy in D by (7.23). We remark that if K is an infinite field, then by Exercise 14.1, there are infinitely many primitive (q" - l)th roots of 1 in D, and thus there are infinitely many distinct inertia fields in D. Let nD be a prime element of ~. Since n~ = n~~, the field K(n D) is a completely ramified extension of K of degree n, and is a maximal subfield of D. By (13.3) we have "-1

~

=

L'

Rwin£ = R[w, n D],

i.j= 0

Thus D is obtained by adjoining the element n D to any of its inertia fields K(w), or equivalently, by adjoining w to the field K(n D). Note that wand nD do not commute, unless n = 1. The inertia field K(w) is uniquely determined up to K-isomorphism by the index n, and our next task is to improve our choice of the prime element nD • (14.5) THEOREM. Let WED be a primitive (q" - l)th root of 1, and let n be any prime element of R. Then there exists a prime element n DE ~ such that

n; = n, where r is a positive integer such that 1 ~ r ~ n, (r, n) = 1. The integer r is uniquely determined by D, and does not depend upon the choice of w or n. Proof. By Exercise 14.2, there exists a nonzero ex ED such that awex- 1 = w q • Replacing ex by nla if need be, we may assume that ex E~. Now let vD(a) = k, and put h = n/(k, n). Then bEZ, B E u(~), and h is the least positive integer for which such an expression for exh is possible. Then

(14.5)

FINITE RESIDUE CLASS FIELD CASE

147

since 3: = /)./p is a field. But w is a primitive (qn - l)th root of 1 in 3:. whence n = h, and so (k, n) = 1. Now choose r, t EZ with kr - nt = 1, 1 :( r :( n, so surely (r, n) = 1. Setting we have vD(n D) = rvD(ct) - tvD(n) = rk - tn = 1,

so n D is indeed a prime element of /).. Furthermore,

n;

n

Therefore commutes with w, and certainly also commutes with D • Thus lies in the center of D, and is then obviously a prime element of R. We do not know at this point that equals a preassigned prime element n of R, and we show next that we can modify the above n D so as to accomplish this. For any AE U(Ow), AnD is also a prime element of /)., and

n;

n;

(AnD)' W· (An D)-1

= A' w qr . ).-1 =

w qr ,

since A lies in the inertia field K(w). It remains to prove that A may be chosen so that ().n D )" n. Now the galois group Gal(W/K) is cyclic of order n, generated by the automorphism a:w -. w q • Since (r, n) = 1, the automorphism p:w -. w qr also generates this galois group. But p(x) = nDxn~t, x E W. Therefore by Exercise 1.4,

This shows that for each A E U(Ow), (AnD)"

=

(N W/K A) .

n; .

Now any preassigned prime element n of R is expressible as n = f3. n; for some f3Eu(R). By (14.1) we may choose AEU(Ow) such that NW/KA = f3. Therefore (AnD)" = n, as desired. Finally, we prove the uniqueness of r. To begin with, if Wi is another primitive (qn - l)th root of 1 in D, then there exists a nonzero f3 ED such that Wi = f3w t f3- 1 for some t. Let n~ = f3n Df3-l, another prime element of /).. Then we have (n~)"

= f3nf3-1 = n,

n~wln'D-l

= f3. w tqr . {3-1 = (w'r,

so the choice of r does not depend on which root of unity is used. We must still show that r does not depend on the choice of the prime element n D • Let n ' be any prime element of /)., and suppose that

148

(14.6)

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

for some s between 1 and n. We may write n' = yn Dwith y E u(d), and then w qS = ynD'w'(ynD)-l = Y·Wqr'y-l. Passing to A we obtain the equality O5qS = O5qr. Therefore s = r, since 05 is a primitive (qn - l)th root of unity. This completes the proof of the theorem. The above shows that once the complete field K is given, the skewfield D is completely determined by its index n, and by the integer r. Indeed, we first form the field K(w), with w any primitive (qn - l)th root of 1. Then we pick any prime n E R, and adjoin to the field W an element nD satisfying the conditions listed in (14.5). This determines the skewfield D = K(w, n D ) up to K-isomorphism. We call the fraction r/n the Hasse invariantt of D. We are still left with the problem of deciding whether each fraction r/n arises from some skewfield. (14.6) THEOREM. Let 1 ~ r ~ n, (r, n) = 1. Given the complete field K, there exists a skewfield D with center K, index n, and Hasse invariant r/n. Proof. Let W = K(w), where w is a primitive (qn - l)th root of 1. If there is such a skewfield D with the desired properties, then W must be a maximal subfield of D, and hence W splits D. Therefore W ®KD ~ M.(W), and hence each element d ED is representable by a matrix d* E Mn(W), We will therefore try to find a set of matrices in M n(W) which constitute a skew field having the desired properties. Let () be the automorphism of W for which ()(w) = w qr , and let n be a prime element of R. For r:t. E W, define

r:t. 0 r:t.* = 0 0

0 ()(r:t.)

0 0 ()2(r:t.)

0

o o

0 0 0 n 0 0

o

0 0 , n*D-

0 ()n-l(r:t.)

0

0

0 1

0

0 0

The map a. ---> a.*, a. E W, gives a K-isomorphism of W onto the field W* = K( w*) c M n(K), where we identify each }, E K with the scalar matrix AIn E Mn(K). It is easily checked that (n~)n

= nln'

n~'

w*- (n~)-l = (w*)qr.

We now set

n-1

D

= K[w*, n~] =

L

K(w*Y(n~Y,

i,i= 0

t rr'

Later on, we shall call the fraction r'/n the Hasse invariant of D, where r' E Z is chosen so that 1 (mod n), 1 <:; r' <:; n.

=

149

FINITE RESIDUE CLASS FIELD CASE

(14.7)

a K-subalgebra of Mn(W), We shall verify that D is the desired skewfield. 2 Each element a E D is expressible as a K-linear combination of the n i elements {(W*)i. (n~Y}. Since (W*)i = (w )*, it follows that a is expressible in the form "-1

a

L

=

Ltr(n~)j,

j=O

Since

we obtain

a=

Lt _ n 1

)

8(Lt1) 8 2( Lto )

82(Ltn _ 3)

8(Lt o)

n8(Lt n -1)

(14.7)

Lt2

Lt1

Lto 2

2

8(Ltn- 2)

n8 (Lt n _ 2)

n8 (Lt n _

n8 n - 2(Lt2)

n8 n - 2( Lt

)

n8n - 2(Lt4 )

nf)n-1(Lt

n8 n - 1 (Lt 2 )

n8n -1(Lt 3)

1

)

1

3

8n - 2( Lt 1) 8n - 1 ( Lto)

Hence if a = 0, then each Lti = 0. This shows that D is a vector space over W 2 with basis {(n~Y:O:!( j:!( n - I}, and therefore (D:K) = n . Next suppose that a ED is a zero divisor, but that a =I- 0. Let a be given as above; after replacing a by n1a with suitable I, we may assume that each Lt E ow' and that at least one Ltj is a unit in ow' (This depends on the observaj tion that n is also a prime element for ow.) Now (by Exercise 1.4), But det a = 0, since a is a zero divisor in D. Therefore N W/KLtO E now ( l R, whence Lto is not a unit in ow' Hence we may write Lto = nf30 for some f3 0 E ow' But then det a == n . N W/K Lt1

2

mod n ow'

so also ~1 = nf3 1 for some f3 1 E ow' Continuing in this way, it follows that each Lt j IS a multiple of n, which is a contradiction. This proves that D has no zer? divisors except 0. Hence for each nonzero a ED, the K-homomorphIsm x ---> ax, xED, is monic. Since (D: K) is finite, this shows that D = aD = Da for each nonzero a E D, and hence D is a skew field. Observe n~xt that W* is a sub field of D, and that (W*: K) = n. If a ED commutes WIth w*, then since the diagonal entries w, 8(w), ... , 8n - 1 (w) of w* are pairwise distinct, it follows that a is a diagonal matrix. Hence by

150

MAXIMAL ORDERS TN SKEWFIELDS (LOCAL CASE)

(14.8)

(14.7) we have a E W*, and so we have shown that W* is a maximal subfie1d of D. On the other hand, if a also commutes with n~. then writing a = et~ with et o E W, it follows at once that e(et o ) = et o ' Hence et o E K, and a = u1n for some U E K, which shows that K is the center of D. Finally, the equation (n~)n = n1n proves that n~ is a prime element for D. Thus D is a skewfie1d with Hasse invariant rln, as claimed. We shall conclude this section by calculating the discriminant d(AIR) of the R-order A (see § 10) as well as the different :D = :D(AI R), defined below. To begin with, let tr:D ~ K be the reduced trace map, and let r::D x D ~ K be the trace form given by r:(a, b) = tr(ab), a, bED. By (9.8), r: is a symmetric nondegenerate associative bilinear form. We may then define ~

=

{xED: r:(X, A) c R}

=

{xED: tr(xA) c R},

By Exercise 4.12,'& is a full R-Iattice in D, and is clearly a two-sided A-module. Since n;~ c A for some r, it follows from (13.2) that m . A ,& = nD

for some integer m, and m ~ 0 since ,& define the different as

:=J

A. Call ,& the inverse different, and

:D = :D(AIR) =

n~

. A.

For a nonzero ideal L = n~A of A, define the norm N D/KL

= ordR (AIL),

where ordR is the R-order ideal (see §4a). We have, as in (4.31 iii), ordR(Alnb A) = {ordR(AlnDAW = pnl, since (~:1~) = n in this case. Therefore N DIK(n~A)

(14.8)

LEMMA.

= (N D/K nDY . R,

d(AIR) = N D/K(:D(AIR)).

Proof. Keeping the above notation, we observe that left multiplication by gives an R-isomorphism

n~

Therefore N D/K(:D) = ordR (A/:D) = ordR (~/A). Now we imitate the proof of (4.35). We may write A=

L' Rx i ,'& = L' Ry j ,

where

tr(xiy)

= (jij'

(14.9)

FINITE RESIDUE CLASS FIELD CASE

and the subscripts i, j range from 1 to n 2 . Set a .. = tr(x,x.), and so by Exercise 4.2 we have 'J

'

Xj

151

= Lajjy j , ajjEK. Then

J

ordR(~/,1)

=

R· det(a jj )

=

d(,1IR),

as claimed. Let us now calculate !'J(,1I R) and d(,11 R) explicitly. (14.9) Theorem. We have !'J(,1IR)

=

~-1

d(,1IR)

A,

= p n(n-1).

Proof. The formula for the discriminant follows from that for the different, by taking norms. Let us write ~ = n;-m,1; then m is maximal such that tr n;- mAc R. Hence we need only show that tr(n;-(n-1),1) c R,

tr(n;-n,1)

Suppose n D chosen as in (14.5), so that

n~

=

cf R.

n. Since

n-1

A=

L'

ow' (nDY,

j=O

we have n-(n-1),1

=

n- 1 . nD,1

=

n-1 L' (n- 1ow )'(nD EB ow'

Y

j= 1

Let a E D; to compute tr a, we may use the isomorphism W ®K D ~ Mn(W) to represent 1 ® a as an element of Mn(W), and then take the trace of the matrix thus obtained. But the isomorphism W ®KD ~ Mn(W) has been exhibited explicitly in (14.7), and hence

tr(J~ aj(nDy)

TW/Ka O '

ajE W.

Hence we have at once tr (n D - (n -1) A)

= T.W/K (0) W c R.

On the other hand, if tr (n;-n A) c R, then the above argument gives C R, whence T W/K(OW) c nR. If 'f denote~the trace map from Ow to R, then since the galois groups of WI K and owlR are isomorphic, it follows that for each A E oW' T(A) is the image of TW/K(A). We may thus conclude that 'f is the zero map, which contradicts the fact that Ow is separable over R. This proves that tr n;-n A cf R, and establishes the theorem. TW/K (n-=: lOW)

152

MAXIMAL ORDERS IN SKEWFIELDS (LOCAL CASE)

EXERCISES 1. Let D be a skewfield whose center K is infinite. Let a E D - K, f(X) = min. pol. Ka. Show that there are infinitely many elements a' ED such that f(a') = o. [Hint: For each nonzero xED, f(xax- 1) = o. Since a ¢ K, there exists an element y E D which does not commute with a; the same holds for each y + ex, ex E K. If

(y

+ ex)a(y + ex)-l

=

(y

+

f3)a(y

+ (3)-1,

ex, (3 E K,

ex 1= fl,

then setting z = y + fl, it follows that Z-l(Z + ex - fl) commutes with a. Hence z commutes with a, a contradiction. Thus {(y + a)a(y + ex)-l: aE K} is an infinite set of zeros off(X)]. 2. Keep the notation of (14.4) and the discussion preceding it. Show that for each i, ~ f - 1, w q ' is conjugate to win D.

o~ i

3. Let K be the 2-adic completion of the rational field, R its ring of 2-adic integers, and V K the valuation on K. Let D = K EB Ki EB Kj EB Kk (quaternions over K), and set A

=

R

+ Ri + Rj + Rk + Ra,

a = (1

+ i + j + k)/2.

Show how to extend vK to a valuation on D, and prove that A is its valuation ring. Find the discriminant of A, and compute e(D/K), f(D/K), a prime element 1[D' the residue class field K, an R-basis for ~, and the inertia fields of Dover K. 4. Let K be the 3-adic completion of Q. Show that there exists an element a E K with a 2 = -2. Let

=

(X2 - aX - 1)(X2

+ aX

- 1)

in

K[X],

and that each quadratic factor is irreducible in K[ Xl Let w be a zero of the first factor in some extension field of K. Prove that w is a primitive 8th root of unity, and that X 2 - aX - 1

=

(X - w)(X - w 3 ),

X 2 + aX - 1 = (X - WS)(X - w 7). 5. Keeping the notation of (14.5), let W be an inertia field for D. Prove that

nr D1K (1[D) = (_1)"-11[, [Hint: Since X" - 1[ is irreducible min. pol.K1[D = X" - 1[. But then

nrD1K(ex) = NWIK(ex),

exE

W

K[ X] by Exercise 5.1, it follows that

III

red. char. pol. DIK 1[ D = min. pol. K1[ D' by Exercise 9.1. This gives the desired formula for nr(1[D). Next, let ex E Wand let f(X) = min. pol.Kex. By Exercise 9.1, red. char. POI.DIK ex

=

U(x)},w:K(a)).

But also char. pol.wlK ex = U(X)}(W:K(a))

153

EXERCISES

(compare degrees !l, whence red. char. pol. D/K IX

=

char.

IX E

POI.W/K IX,

W.

This yields the formula for nr(IX). Another proof can be given by using the matrix representation of n D and IX (IX E W) occurring in the proof of (14.6). Then

n

"-1

nrD/K IX = det IX* =

B'(IX) = N W/K IX.]

i=O

6. Let D be a skewfield whose center K is complete with respect to a discrete valuation, with a finite residue class field R (as in §14). Let nr D

Prove that nr D Then

=

=

{nrD/K(d): d ED}.

K. [Hint: Use the notation and results of the preceding exercise. nr D

:::>

NW/K(W)

=

{xEK: nlvK(x)},

by (14.1). But nr D also contains (_1)"-l n, whence nr D

=

K.]

4. Morita Equivalence We shall need to study the relationship between A-modules and M.(A)_ modules, where as usual Mn(A) is the ring of all n x n matrices over the ring A. The generalization of this relationship leads to the concept of Morita equivalence, and we sketch here some of the main results of this theory. As references, we cite Bass [1], P. M. Cohn [1], Faith [1]. While the general theory holds for abelian categories, we shall make the simplifying assumption that our categories are in fact categories of modules. Denote by .ff" the category of right A-modules, by j..//I/\ the category of (~, A)-bimodules, and so on.

15. PROGENERATORS We shall use the notation of § 2b. Let .91 be a category with objects A, A', ... , and f!J a category with objects B, B', .... We shall write A Ed to indicate that A is an object in d. All functors F:d ---+f!J considered below are assumed to be additive. Unless otherwise stated, functors are taken to be covariant. Let F: .91 ---+ f!J and G: .91 ---+ f!J be a pair of functors. A natural transformation t:F ---+ G is a family wherefor each A E .91, tA E

HomM(FA, GA),

and where for each a E Hom(A, A'), the diagram FA~GA

Fal

Get!

FA'~GA'

is commutative. If furthermore each t A is an isomorphism, we call the functors F and G naturall y equivalent, and write F ~ G. Example. Let .91 = f!J = /\.//1, F the identity functor, G the functor A ®/\ .. For each A E .91, let t A: A ---+ A ®/\ A be the left A-isomorphism given in (2.8). Then t is a natural equivalence from F to G.

154

(15.1)

PROGENERATORS

155

Let id" be the identity functor on .91. Two categories .91 and []I are called equivalent if there exist functors F, G:

such that GF ~ id<>f and FG ~ id!M' For example, if N denotes the opposite ring of A (same elements, but multiplication is reversed), then the categories ".A and .A,," are equivalent. N ow let F: .91 ---+ []I be any functor. Call F faithful if the map (15.1)

a E Hom(A, A')

---+

Fa E Hom(FA, FA')

is injective, that is, if Fa = 0 implies that a = O. For example, if M is a faithfully flat right A-module, then the functor M ®" . : ".A ---+ .916 is faithful (see (2.21)). On the other hand, the functor F:.91 ---+[]I is full if the map in (15.1) is surjective, that is, if each (3 E Hom(F A, FA') equals some Fa. For example, let A = AI I where I is a two-sided ideal of A Let .91 be the category of projective left A-modules, []I the corresponding category for A, and let F: .91 ---+ []I be given by F(M) = M/IM. Then F is a full functor, since each A-homomorphism {3:M/IM---+N/IN, where M,NE.91, can be lifted to a A-homomorphism a:M ---+ N, and thus (3 = Fa. We recall from (2.14) that an object XEd is projective if the functor Hom(X, .) : .91

---+

.916

is exact. This functor carries A Ed onto the abelian group Hom(X, A), and takes each a E Hom(A, A') onto a* : Hom(X, A)

---+

Hom(X, A'),

where a,,.! = af, f E Hom(X, A). Whether or not X is projective, let us consider the functor Hom(X,·). We call X a generator of the category .91 if Hom(X, . ) is a faithful functor, that is, if a* = 0 implies that a = O. In other words, X is a generator if and only if for each nonzero a: A ---+ A' there exists an f: X ---+ A such that af i= O. Dually, XEd is injective if the contravariant functor Hom( " X) is exact, while X is a cogenerator ford if Hom(', X) is faithful. (15.2) THEOREM. An object X in the category .91 is a generator ford only if every A E .91 is a quotient of a direct sum of copies of X.

if and

Proof. Let X be a generator, and let A Ed. For each f E Hom(X, A), let X J be a copy of X, and set Y = X J' Each X J maps into A (by means of f),

I" J

156

(15.3)

MORITA EQUIVALENCE

so there is a map '1: Y -+ A, defined by '1 = 1J(Ixf

}

= If(x f

I'-I Thus

),

XfEX f ·

Clearly (15.3)

f(X) c A.

im1J = fEHom(X,A)

Let us show that im 1J = A; if not, there exists a nonzero map IX: A -4 A/im 1J. Since X is a generator, there must be an f: X -4 A such that IXf i= 0. But (rxf)X = CJ:(f(X)) = 0, since f(X) c im1J. This gives a contradiction, and shows that '1 is an epimorphism of Yonto A. Therefore A is a quotient of a direct sum of copies of X, as claimed. Conversely, suppose that for each A E d there is an epimorphism ~:L" Xi -4 A, whereeachXiisacopyofX. WemustshowthatXisagenerator ford. Let IX: A -4 A' be nonzero; then also IX~ i= 0, since ~ is epic. If we write ~ = L" ~i' where ~i:Xi -4 A, then IX~ = lX~i· Therefore lX~i i= for some i. But Xi ~ X, and so there exists an f: X -4 A such that IXf i= 0. This proves that X is a generator for d, as desired.

°

I"

The construction in (15.3) is of special importance in the following circumstance. Let X E A~It, and define the trace ideal of X as trace X =

(15.4)

f(X) c A.

It is easily verified that trace X is a two-sided ideal of A.

(15.5)

COROLLARY. The left A-module X is a generator for A~

if trace X =

if

and only

A.

= A, then the proof of the preceding theorem shows that A is a quotient module of Xi' where each Xi ~ X. But every A-module A is a quotient of a direct sum I" Aj , with each Aj ~ A. Hence A is also a quotient of a direct sum of copies of X, whence X is a generator for Aolt by (15.2). Suppose conversely that X is a generator. The proof of (15.2) shows that im 1J = A in (15.3), where A is any A-module. Taking A = AA, we see that the trace ideal of X equals im 1J, and thus coincides with A as claimed. Proof. If trace X

I"

It is clear from (15.2), or directly from the definition of generator, that A is itself a generator for the category A~It. Indeed, every nonzero free module is a generator. Given a possibly infinite family {X;.} of objects in d, we may form their

( 15.6)

157

PROGENERA TORS

direct sum L' X ... There are the usual injection maps {i..} and projection maps {p..}, with

We shall say that a functor F:d --4 ffI preserves direct sums if for each family {XJ, there is an isomorphism F(L' Xv) ~ L' FX v '

induced by the maps {Fi .. , Fp .. }. For example, tensor product functors preserve direct sums. So also does the functor HomA(P, .), where P is any finitely generated projective A-module (see Exercise 15.3). (15.6) THEOREM. Let F: d --4 ffI and G: d --4 ffI be a pair of right exact functors which preserve direct sums. Let t: F --4 G be a natural transformation, and let X be a generator for d. If tx:FX ~ GX is an isomorphism in the category ffI, then t is a natural equivalence of functors. Proof. Since t = {tA:AEd} is a natural transformation, each ccA --4 A' gives rise to a commutative diagram FA ~ FA' tAl

tAJ

GA ~ GA'.

We are assuming that tx is an isomorphism, and are trying to prove that each t A is an isomorphism. . . ' . Since X is a generator of d, each A E d IS a quotIent of a dIrect sum L X of copies of X. Hence there is an d-exact sequence U --4 V --4 A --40,

where

U =

L' X,

V =

L' X.

But F is right exact, and so the sequence FU --4 FV --4 FA --4 0

is ffI-exact. Hence there is a commutative diagram with exact rows: FU-FV-FA-O tv t tv t tA t GU GV - - GA - - O. Now FU ~ L' FX, GU ~ L' GX, because F and G preserve dire:t su~s by hypothesis. Since tx: FX ~ GX, it follows that tu: FU ~ GU. LIkewIse tv is an isomorphism, and the commutativity of the above dIagram shows that t A is also an isomorphism, as desired.

158

(15.7)

MORITA EQUIVALENCE

An object P E.sd is fait~fully projective if it satisfies the following three conditions: (i) P is a projective object in ,91. (ii) P is a generator for ,91. (iii) Hom(P,') preserves direct sums. In particular, considering the category AJIt, we call a left A-module P a progenerator for A.A if (i) P is a finitely generated projective A-module, and (ii) P is a generator for A.A. Note each such P is faithfully projective, since by Exercise 15.3 Hom(P,') necessarily preserves direct sums. (15.7) let

THEOREM.

Let P be a faithfully projective object in the category ,91, and

A = Homsd(P, P),

fJi =

A.A.

View P as right A-module. Then the functor Hom(P, . ) : ,91

-+

fJi

gives an equivalence of categories. Proof. It is clear that A is a ring, and that P can be viewed as a right A-module. Then for each A E ,91, Homsd(P, A) is a left A-module. Thus the functor F: ,91 -+!!4, given by F(A) = Hom(P, A), is well defined. Clearly FP = A Since P is a projective object, the functor F is exact. In order to show that F is a category equivalence, we use the result of Exercise 15.2. We show first that F is full and faithful, that is, for each pair A, A' E ,91, the map

(15.8)

F: Hom(A, A')

-+

Hom(F A, FA')

given by IX -+ FIX, is an isomorphism. Let us first verify this when A = where each P). ~ P. By (2.7) Hom(A, A') ~

f1 Hom(P)., A')

~

I' P).,

f1 FA', ).

the latter because Hom(P)., A') ~ Hom(P, A') = FA'. On the other hand, F preserves direct sums since P is faithfully projective, and thus Hom A (FA, F A') ~ HomA

(I' FP)., F A') ~ f1 HomA (A, F A') ~ f1 FA'. ).

).

We have now shown that the map F in (15.8) is an isomorphism whenever A is a direct sum of copies of P.

(15.9)

159

PROGENERA TORS

Now let A and A' be arbitrary, and let V-+V-+A-+O

be an ,r;1-exact sequence in which V and V are direct sums of copies of P. Then the sequence FV -+ FV -.... FA -+ 0

is [Ji-exact, since F is an exact functor. Since Hom is a left exact functor, we obtain a commutative diagram with exact rows

o -+ Hom (A, A')

--+

Hom (V, A') -----. Hom (V, A')

tr2

1- F,

1- F ,

0-+ Hom (FA, FA') -+ Hom (FV, FA') -+ Hom (FV, FA'). By the preceding paragraph, both F 2 and F 3 are isomorphisms. Hence so is F 1 , and this completes the proof that F: ,91 -+ [Ji is a full and faithful functor. (For another proof see Exercise 15.6.) In order to deduce that F gives an equivalence of categories, it remains for us to show that every left A-module M is of the form FA for some A Ed. Let us first write a A-exact sequence X1

X 0 -+ M -+ 0,

X O ' Xl

A-free.

Since A = F(P), we may write Xi = F(V i ), with Vi a direct sum of copies of P, i = 0, 1. By the preceding paragraphs,

so we have ~ = Frx for some rx: V 1 -+ V o. Let A = cok rx, and apply F to the ,r;1-exact sequence V 1 7 V 0 -+ A -+ 0,

thus obtaining a A-exact sequence Xl Therefore M

~

"t X 0

-+ FA -+ O.

FA, and the proof of the theorem is complete.

(I5.9) Corollary. Let M be a progenerator for the category />..A, where L1 is a ring, and let A = Hom/>. (M, M) be the endomorphism ring of M. View M as a bimodule />.MA. Then there is an equivalence of categories Hom/>.(M, . ) : />..A -+ A.A.

160

MORITA EQUIVALENCE

EXERCISES 1. Let F: d ~91 be an equivalence of categories. Show that A is projective if and only if FA is projective.

2. Let F: d --> 91 be a full faithful functor such that every BE 14 is the image FA of some A Ed. Prove that F is a category equivalence. 3. Let P be a finitely generated projective left A-module. Show that the functor HomA(P, .) : A.;/(

dt

-->

preserves direct sums. [Hint: First check the case where P HomA(P ..j.. p".)

~

=

1\ then use the fact that

HomA(P, . ) ..j.. HomA(P',' ).]

4. Let A.;/( --> I>.;/( be an equivalence of categories, where A and /',. are rings. Show that A is left hereditary if a~d only if /',. is left hereditary. [Hint: A is left hereditary if and only if subobjects of projective objects in A.;/( are projective.] 5. Let dtO denote the category which is opposite to dt, that is, dtO has the same objects as dt, but all arrows are reversed. (i) Prove that direct products in dt correspond to direct sums in dt 0. (ii) Given an object A' in a category d, define a functor G: d --> dtOby setting G(A) = {Hom"" (A, A')t E dt~

for each A E d.

(There is an obvious definition of the action of the functor G on maps in d.) Prove that G is a right exact covariant functor which preserves direct sums! 6. Use Theorem 15.6 to prove that the map F in (15.8) is an isomorphism. [Hint: Keep the notation of the proof of (15.7), so F(A') = Hom.of(P, A')

Keeping A' fixed, define functors G, H: d G(A) = {Hom.of (A, A'no,

-->

for each A' Ed. d (10 by

H(A) = {Hom!j(F A, F A

'no,

A E sil,

with obvious definitions of G and H on maps. Since P is a faithfully projective object in d, the functor F is exact and preserves direct sums. Use this to prove that H is a right exact covariant functor which preserves direct sums. The functor F: d --> 91 determines homomorphisms FA. A' : Hom.of (A, A')

-->

Hom1ll (F A, FA'),

A,A'Ed.

For each AEd, define tA:HA --> GA by tA = {FA,A,r Prove that t:H--> G is a natural transformation. In order to prove that the map F in (15.8) is an isomorphism, it suffices to establish that each t A is an isomorphism in ..;;1{10. By (15.6), this will follow once we know that tp is an isomorphism in d{l°. But this is true since GP

= Hom.of (P, A') = FA'

HP = HomS\j(FP, FA') = HomA(A, FA') ~ FA'.]

161

MORITA CORRESPONDENCE

(16.1)

7. Let P;1 be a progenerator for functors defined by G(L) = Hom;1(P' L),

vi{ tl'

and let G, H: vI{;1

--->

d6 be the covariant

H(L) = L0" P*,

where P* = Hom;1(P'~) (viewed as left ~-module). Show that the functors G, Hare naturally equivalent. [Hint: Prove (i) P* is a finitely generated projective left ~-module. (ii) The functors G and H are exact, and preserve direct sums. (iii) There is a natural transformation t: H ---> G, given by

where tL is defined by the formula

{t L (l0 rp)}x (iv) t;1: H(~) ~ Then use (15.6).]

=

I· rp(x),

1E L,

rp

E

P*,

X E

P.

G(~).

16. MORITA CORRESPONDENCE Let M,1 be a nonzero right A-module, and set A = Hom,1 (M, M),

(16.1)

Let A act on the left on M, so there is a bimodule structure AM,1. By definition, M* = Hom,1(... M,1' ,1A,1)' the set of all right A-homomorphisms from M to A. The left action of A on the bimodule ,1A,1 enables us to make M* into a left A-module. The left action of A on the bimodule AM,1 likewise makes M* a right A-module (see §2). It is easily vedfied that M* is a bimodule ,1M* A' We define a map (16.2)

f.l: M <8l,1 M* ~ A, m <8l f ~ (m, f), where (m, f)m'

for m, m' E M, (16.3)

f

E

= m(fm')

M*. Likewise, let

r:M* <8lAM

-4

A,

f <8l m' ~ [j, m'], where [j, m'] = fin'·

It is easily checked that f.l and r are well defined, f.l is a two-sided A-homomorphism, and r a two-sided A-homomorphism. Further, one obtains

(16.4)

(m, f)m' = m[j, m'J,

gem, f) = [g, m]f,

for m, m' E M, f, g E M*. Finally, we note that (16.5)

im r = trace M,1 ,

where "trace" is defined as in (15.4).

162

(16.6)

MORITA EQUIVALENCE

(16.6) Definition. A Morita context consists of a pair of bimodules AM,1,' ,1,NA relative to rings t;, A, and bimodule maps 11 : M ®,1, N

-4

A,

' : N ® AM

-4

t;,

given symbolically by l1(m ® n) = (m, n), ,(n ® m) = en, m], interrelated by the formulas (m, n)m' = men, m'], n(m, n') = en, m]n', for all m, m' EM, n, n' EN. If both 11 and, are isomorphisms, we call the rings t;, AMorita equivalent. We have just seen how to construct a Morita context by starting with M,1,' and defining A, M* by (16.1). We call it the Morita context derived from the right t;-module M. We now proceed to investigate this derived context more fully, keeping the notation of (16.1)-(16.5). (16.7) THEOREM. Let M be a nonzero right t;-module, and let A = Hom,1,(M, M). Let 11 : M ®,1, Hom,1,(M, t;) -4 Hom,1,(M, M), , : Hom,1,(M, t;) ® AM

-4

t;,

be the maps defined in (16.2) and (16.3). Then 11 is epic if and only if M is a finitely generated projective right t;-module; in this case, 11 is also monic. On the other hand, ' is epic if and only if M is a generator of the category .it,1,; in this case, ' is also monic. FinallY,11 is epic if and only if 1A E im 11, and, is epic if and only if 1,1 E im ,. Proof The last statement in the theorem is obvious, since im 11 is a two-sided ideal in A, and im, a two-sided ideal in t;. The identity map 1M on M is the identity element 1A. Hence if 11 is epic, we may write

J; E Hom,1,(M, t;). i=1

Define right t;-homomorphisms t; (r)

7

M by

r

a(a 1 ,

••• ,

ar ) =

L miai , 1

It is easily seen that a{3 = 1M' whence MIt; (r), and thus M,1, is finitely generated and projective. The argument can be reversed, and thus 11 is epic if and only if M,1, is finitely generated and projective. Next, im, = trace M,1,' Hence by (15.5), ' is epic if and only if M is a

MORITA CORRESPONDENCE

(16.9)

163

generator for All.' Finally, 11 epic implies 11 monic, by Exercise 16.1, and likewise for T. This completes the proof. The same argument shows (16.8)

COROLLARY.

For any Morita context (16.6), we have lA E im 11

~

11 epic

Ill. EimT

~

T

epic

~

11 monic,

~T

monic.

Recall that Mil. is a progenerator of .Ad if M is a finitely generated projective which is a generator of the category All.' The preceding results give at once ~-module

(16.9) Corollary. The module Mil. is a pro generator of All. if and only if both of the maps 11, T in (16.7) are isomorphisms. If Mil. is a pro generator, then the Morita context derived from M gives aMorita equivalence between the rings ~ and Hom/l.(M, M). Remark. Let Mil. be a progenerator of All.' Then M is a projective object in the category All.' and by (15.9) there is an equivalence of categories (16.10) where A = Hom/l.(M, M), and M is viewed as a bimodule AM /l.' It is sometimes useful to exhibit this category equivalence in terms of tensor products, rather than Hom. Define M* as in (16.1); by Exercise 2.5, there is an isomorphism and it is easily verified that this is a right A-isomorphism which is natural in L. In other words, there is an equivalence of functors (16.11) (This also follows from Exercise 15.7.) Hence the equivalence in (16.10) can also be given as

We shall show below that every Morita equivalence of rings ~,A is obtained in the manner described above, by means of the Morita context derived from a progenerator Mil. of category .All.' As a first step in establishing this fact, let us start with an arbitrary Morita context (16.6), not necessarily derived from a ~-module M, and let us see what conclusions may be drawn if we assume that 11 is epic.

164

MORITA EQUIVALENCE

(16.12)

(16.12) THEOREM. Suppose that the map /l occurring in the Morita context (16.6) is epic. Then both M!J. and !J.N arejinitely generated projective modules, and A acts faithfullyt on N. Furthermore, M

~ Hom!J.(N,~),

The jirst of these is a (A, ~)-bimodule isomorphism, and the second is an isomorphism as rings of right operators on N. Proof. Let us form the Morita context derived from !J.N, by setting

Then A' is a ring, and there are bimodule structures !J. NA" A' N* !J.' and . blmodule homomorphisms /l':N* ®!J.N T': N

®A' N*

->

A',

f ®n

->

{f, n},

n® f

->

(n, f)

->~,

=

f(n),

where

{f, n}g = f (n, g). We may now define a right

tjJ:M

~-homomorphism ->

N*,

m->[·,m],

where we use ( , ) and [ ,] for the maps /l, ring homomorphism cp:A

->

tjJ by

T

occurring in (16.6). Also define a

A',

where AR is right multiplication by A on N. Let us verify that the following diagram is commutative: ""01

---.N*®!J.N

(16.13) rp

til' A'.

To prove equality of two elements of A', we compute their action on an arbitrary element n' E N. We have n'{(cp/l)' (m ® n)} = n'· (m, n) = en', m]n,

t This means that for ,\ E A, jf N,\ = 0 then ,\ = O.

(16.13)

165

MORITA CORRESPONDENCE

whereas

n'· [(/l'(1/1 ® 1)' (m ® n)} = n'· {11'(l/Im ® n)} = n' . {I/Im, n}

= (n',l/Im)·n

=

{(I/Im)n'}'n

=

[n',mln.

This shows that (16.13) is commutative. Up to this point, we have not used our hypothesis that /l is epic. Under this hypothesis, we know from (16.8) that /l is in fact an isomorphism. We shall show that the other maps cP, /l' and 1/1 ® 1 in (16.13) are also isomorphisms. We begin by proving that 1/1 itself is an isomorphism. Choose Z EM ® N so that /l(z) = l A ; then If mE ker 1/1, then [N, m] = 0, whence

m=

I

(m i , ni)m =

I

m;[ ni , m]

=

0,

and thus 1/1 is monic. To prove 1/1 epic, let f E N* and set m = Im i ' f(n i ).

We show that I/I(m) = f by checking that for each n EN, {I/I(m)}n = f(n). We have {I/I(m)}n = [n, I mi' f(n i )] = I [n, mJf(nJ But f is a <"1-homomorphism, and [n, mJ E <"1, whence

{I/I(m)}n

fCI en, m,]nJ = f(I n(mp nJ)

= f(n).

This completes the verification that 1/1 is an isomorphism. It follows at once that 1/1 ® 1 is also an isomorphism. Now we note that IN = CP/l(Z) = /l'{(1/1 ® l)z}, whence by (16.8) /l' is an isomorphism. Since (16.13) commutes, it follows that cP is an isomorphism, as claimed. This in turn implies that A acts faithfully on N. Finally, since /l' is epic, it follows by (16.7) that N is a finitely generated projective left <"1-module. But then N* is a finitely generated projective right <"1-module, since (6<"1)* = <"16 ' We have shown above that M ~ N*, whence we may conclude that M6 is finitely generated and projective. This completes the proof. We called the rings <"1, A Morita equivalent if there exists a Morita context (16.6) in which both /l and Tare epimorphisms (and hence isomorphisms). We saw in (16.9) that the Morita context derived from a progenerator M6

166

(16.l4)

MORIT A EQUIVALENCE

of .4t ~ always gives a Morita equivalence, and in that special case we constructed an equivalence of categories ./It~ ~ .4t A' where A = Hom~(M, M). We now prove that every Morita equivalence of rings is derived from a progenerator, and gives an equivalence of categories. (16.14) T~eorem. Let the rings ~ and A be Morita equivalent relative to a Morita context AM~, ~NA as in (16.6). Then (i) M is a progenerator for both .4t ~ and A.4t. N is a progenerator for both ~.4t and .4tA • (ii) There are bimodule isomorphisms N

~ Hom~(M,~) ~

HomA(M, A),

(iii) There are ring isomorphisms A

~

Hom,..(M, M)

~ Hom~(N,

N),

where Hom~(N,N) acts as a ring of right operators on N, and likewise HomA(M, M) on M. (iv) There are inverse pairs of equivalences of categories A Jt

where

s=

. Q9 A M

S' =N Q9A'

~

HomA(N,'),

T

= . Q9~N

~

HomA(M,'),

T

=

- Hom~(M,'),

M Q9~' ~ Hom~(N, ').

(v) The correspondence J -+ JM gives an isomorphism of the lattice of right ideals of A and the lattice of submodules of M ~ , and the two-sided ideals correspond to sub-bimodules of AM ~. The correspondence I -+ MI gives an isomorphism of the lattice of left ideals of ~ and the lattice of sub modules of AM, and the two-sided ideals correspond to sub-bimodules of AM~. The correspondence I -+ Hom~(M, MI) gives an isomorphism between the lattice of two-sided ideals of ~ and the lattice of two-sided ideals of A, once we identify A with Hom~(M, M). Proof. By hypothesis both /1 and M, n's in N such that

L

in (16.6) are epic, so there exist m's in

(16.15)

It is easily checked that for n E N, (', n)EHomA(M,A),

MORITA CORRESPONDENCE

(16.l6)

167

The first equation in (16.15) then asserts that /\M has trace ideal A, whence /\M is a generator of /\At by (15.5). Likewise, the second equation in (16.15) implies that MIJ. is a generator of AtIJ.' Further, since Jl is epic, it follows from (16.12) that MIJ. is finitely generated and projective; the same holds for /\M since L is epic. This proves the first part of (i), and the second part follows by symmetry. The assertions in (ii) and (iii) are immediate consequences of (16.12), using symmetry arguments. (See also Exercise 16.7). To prove (iv) we observe that f or each X E At/\ ' {(. 0N)(' 0M)}X = (X 0/\M)0IJ.N ~ X0/\(M0IJ.N) ~

X 0/\ A

~

X,

since M 0IJ. N ~ A as A-A-bimodules. All of these isomorphisms are natural in X, and thus ( . 0 N)( . 0 M) is naturally equivalent to the identity functor on .,,$//\. Analogous arguments hold for the other functors in (iv), so we obtain category equivalences, as claimed. The natural equivalences of the type' 0/\ M ~ Hom/\(N,') follow from (ii) and (16.l1). Finally S(A) = M, so the right ideal J of A maps onto S(1) = J 0" M. But J 0/\ M ~ J M as ~-modules, since /\M is projective. This implies the first statement in (v). The second statement follows by symmetry. Finally, if I is a two-sided ideal of ~, then I corresponds to the sub-bimodule MI of /\MIJ.' and this in turn corresponds (via T) to the two-sided ideal HomlJ.(M, MI). This completes the proof. Let us briefly consider the special case where MIJ. is free on r generators, and where (16.16) Clearly MIJ. is a pro generator for At", and there is a category equivalence HomlJ.(M,·): AtIJ. ~ At/\'

The isomorphism in (16.16) can be described explicitly, once we start with a free ~-basis {m1 ' ... , mr} of M. Each f E A determines a matrix TJ E M.(~), such that f(m 1 , · · · , mr) = (m 1 , · · · , mr)TJ . This equation means that r

(16.17)

f(m)

=

L miL

ij ,

l;:;:j;:;: r,

;=1

For any two-sided ideal I of the ring matrices with entries in 1.

~,

let M.(I) denote the set of all r x r

168

(16.18)

MORITA EQUIVALENCE

(16.18) COROLLARY. The map 1--+ M,(I) gives an isomorphism between the lattice of two-sided ideals of ~ and the lattice of two-sided ideals of M.(~). Proof By (16.14( v)), the map I --+ Hom l1(M, M I) gives the desired isomorphism, once we identify A with M,(~) as in (16.16). We have

where {ml , ... , mJ is a free ~-basis for M. Let f E A correspond to Tf E M.(A), as in (16.17). It is clear that f E Hom l1 (M, MI) if and only if each tij in (16.17) lies in I. Hence the image of Hom l1 (M, MI) in M.(~) is precisely M,(I), which proves the corollary.

EXERCISES 1. Keep the notation in (16.1)-(16.5). Prove that if /1 is epic, then fl is monic, and likewise for r. [Hint: It suffices to prove the result for /1, by symmetry. Let /1 be epic,

and write 1,1. =

f (m,,!;). Let Lnj (8) gj

E

ker fl, so L(n j , g)

=

o. Then

1

L nj (8) gj = L (m" f.)n j (8) gj = L mi (8) [f., nJgj j

i,j

2. Consider an equivalence of categories .At A +}.At", where A, ~ are rings. Show that this equivalence comes from a Morita context. [Hint: Set M = T(~), N = SeA). Then there are bimodule structures "M A' AN" which give a Morita equivalence between A and ~. Further, the functors Sand . (8) N are naturally equivalent, as are T and· (8) M.J 3. Let M be a free right ~-module on r generators, and set A = Hom,,(M, M). Then M is a progenerator for.At", and there is an equivalence of categories .At

·eM*

A

.~ ~~

.®M

,.At

A

where M* = Hom,,(M, ~). What is the two-sided ideal in A which corresponds to a given two-sided ideal J in !l? 4. Show that Morita equivalent rings have isomorphic centers. [Hint: Identify the center of A with the ring of bimodule endomorphisms ofAM" .J 5. Let R be a commutative ring, A a left noetherian R-algebra, and M a finitely generated left A-module. Suppose that for each maximal ideal P of R, the localization M p is a generator for the category of left Ap-modules. Prove that M is a generator for A.At. Also prove the corresponding statement when "generator" is replaced by "progenerator". 6. Let

R be the

completion of the commutative local noetherian ring R, let A be a

169

EXERCISES

left noetherian R-algebra. and let M be a finitely generated left A-module. Show that M is a generator of I\JIt if and only if R (8) R M is a generator of 'AId, where A = R (8) R A Prove the corresponding result for progenerators. 7. Show that the isomorphisms in (16.14(ii)) are given as follows: N

~ Hom~(M, ~),

M

~

n -> [n,· J;

Homl\(N, A), m

->

(m,');

N

~

Homl\(M, A), n

-> (.,

n).

M ~ Homa(N, ~), m

-> [.,

mJ.

Further, each isomorphism in (16.14(iii)) is given by letting the elements of A and act as left or right multiplications on the modules M or N.

~

8. Show that Morita equivalence of rings is transitive. 9, Let M be a minimal left ideal of the simple left art in ian ring A. Prove that M is a progenerator for I\JIt. Set ~ = Homl\(M, M), and view M as a bimodule I\M a' Prove (i) ~ is a skewfield. (ii) M is a finite dimensional ~-space. (iii) A ~ Homa(M, M). [Hint: Use (15.5) to show that M is a generator forl\JIt. By Schur's Lemma, ~ is a skewfield. Assertions (ii) and (iii) then follow from (16.14). This provides another proof of the harder part of Wedderburn's Theorem (7.4).J 10. Let ~ be a skewfield, Va = ~(r) a right ~-space, and let A = Homa(V, V). Prove that A is a simple left artinian ring, and that ~ = Homl\(V, V). [Hint: View Vas a bimodule 1\ Va' By (16.18), the ring A is simple. Its left ideals satisfy the D.Ce., since they are in bijection with the set of ~-subspaces ofVby (16.14). Finally, ~ ~ Homl\(V, V) by (16.14). This proves the easier part of Wedderburn's Theorem (7.4).J

5. Maximal Orders Over Discrete Valuation Rings Throughout this chapter, R denotes a discrete valuation ring with quotient field K, maximal ideal P = nR ¥- 0, and residue class field R = RIP. We have already seen in (10.5) that the study of maximal R-orders in separable K-algebras can be reduced to the case of central simple algebras. We shall investigate this case in detail here, by using the results from Chapter IlIon maximal orders in skewfields, and then applying the Morita equivalence between skewfields and full matrix algebras over skewfields. We assume throughout that A is a central simple K-algebra, and that V is a minimal left ideal of A. Set D = HomA(V, V), and view AVD as a bimodule. If {v 1 ' ... , Vr} is any right D-basis for V, then for each a E A we may write (17.1) and the map defined by a --> (ct. i ) permits us to identify A with Mr(D). Likewise, if ,1 is an R-order in D, we may choose a free ,1-lattice in V, say M = (17.2)

t; m ,1· For each a Hom~(M, M) we may write

;= 1

E

i

a(m 1 ,

••• ,

mJ = (m 1 ' ••. , mr)(ct.i ),

(ct.;)

E

MJ,1).

The map defined by a --> (ct.;) then gives an identification of Hom~Ct\1, M) with Mr(,1), and this identification is consistent with that obtained from (17.1). Finally, we set (D:K)

17. MAXIMAL ORDERS (COMPLETE LOCAL CASE) In this section we assume always that R is a complete discrete valuation ring. We know by §12 that the skewfield D contains a unique ~aximal R-or~er ,1 which is the integral closure of R in D. Denote by n D a prIme element of ,1, a~d let p = n D ,1. We have seen in (13.2) that the powers (pm} give ~ll n~nzero one-sided ideals of ,1, and that these ideals are necessarIly tw~sIded Ideals. We set ~ = ,1/p, a skewfield of finite dimension over ~he field.R. Let A = M (D) be a central simple K-algebra. Our aIm here ~s to show that M (,1), and its' conjugates, give all possible maximal R-orders III A. We shall al;o study the one-sided and two-sided ideals of such orders. 170

(17.3)

MAXIMAL ORDERS (COMPLETE LOCAL CASE)

171

(17.3) Theorem. (i) Let A = M/,1). Then A is a maximal R-order in A, and has a unique maximal two-sided ideal nDA The powers m

= 0,1,2, ... ,

give all of the nonzero two-sided ideals of A. (ii) Every maximal R-order in A is of the form uAu- 1 for some u E u(A), and each such ring is a maximal order. (iii) Every maximal order A' is left and right hereditary, and each of its onesided ideals is principal. The unique maximal two-sided ideal of uAu - 1 is 1 1 U' nDA· u-l, that is, un u- . uAu- . D Proof We know by (8.7) that A is a maximal order. Keeping the notam i,1; then we may identify A with Hom,., (M, M). tion (17.2), we set M = Since M,., is a progenerator for the category .A,." it follows from (16.9) that A is Morita equivalent to ,1, relative to the Morita context derived from M,.,. By (16.18), the two-sided ideals of A are given by MP), with I ranging over the two-sided ideals of ,1. But each nonzero I is some n~,1, by (13.2). This proves (i), once we observe that

IO

M/n~,1) = n~

. M,(,1)

=

(nDA)m.

Furthermore, everyone-sided ideal of,1 is isomorphic to ,1 by (13.2), whence ,1 is left and right hereditary. By Exercise 15.4, we may conclude that also A is left and right hereditary. Now let us prove that every right ideal of A is principal,t that is, has the form xA for some x E A By (16.14), every right ideal of A can be written as Hom iM, L) for some ,1-submodule L of M. However, M is ,1-free, and every right ideal of,1 is principal, and is isomorphic to ,1. Hence by (2.44) we may write L =

s

IO /,,1, a free ,1-module on s generators, where necessarily s !!( r.

j= 1 J

Let ,1sx, denote the set of s x r matrices with entries in ,1. There is then a bijection Hom,., (M, L) +-+ ,1s x" given by f +-+ U J' where f(m 1 ' ... , m)

= (11" .. , I.) U J"

On the other hand, there is a unique BE ,1'xs such that (11' ... , I.)

= (m 1 ' ... , m)B,

and thus we have f(m 1 ,···,m)

=

(m 1

, ...

,m)·BUr

Comparing this with (17.2), it follows that Hom,., (M, L) can be identified with the right ideal B·,1Sx, of M,(,1). But for each U E,1Sx " we have t Caution: A principal ideal xA is not isomorphic to A, except when x E u(A).

172

MAXIMAL ORDERS OVER DISCRETE VALVA nON RINGS

(17.4 )

where the right-hand expression is a product of two r x r matrices. Therefore

and so we have shown that the given right ideal of A is principal, with generator [B 0]. A corresponding statement holds for left ideals of A, by symmetry. (A rather different proof of these facts is given in (18.7).) Turning next to the proof of (ii), let A' be another maximal order in A. Then A' A is a full right A-lattice in A, so there exists a nonzero a E R such that aA'A c A. Then aA'A is a right ideal of A, hence is principal by the above, and so A'A = uA for some UEA. Clearly A = uA, whence uEu(A) by (6.4). Therefore uA =A'A = A" A'A = A'. uA =:> A'u, and thus A' ~ uAu- 1 . But uAu- 1 is also an order, and hence A' = uAu- 1 since A' is maximal, Finally, the inner automorphism of A given by x -4 uxu-l, X E A, carries A onto A'. The assertions about A' follow at once from those about A, and the proof is complete. (17.4) COROLLARY. Let A = HomD (V, V), where D is a skewjield with center K, and where V is a simple left A-module which is a right vector space over D. Let Il be the maximal R-orderinD, and letA beany maximal R-order in A. Then there exists a full free Il-lattice M in V such that A = Hom ~ (M, M). Conversely, each such A is maximal. Proof Let N be any free Il-Iattice spanning V, and set A' = Hom ~(N, N). By the above theorem, there exists u E u(A) such that A = uA'u- 1 • But then choose M = uN, and we have Hom~ (uN, uN)

= U'

Hom~ (N, N)' u-

1

= A,

as desired. Note that since Nil. is free, so is M,1' Thefactthateach Hom,1 (M, M) is maximal has already been observed in (17.3(i)). Another important consequence of (17.3) is (17.5) COROLLARY. For every maximal R-order A in A, we have A/radA

~

M/Il/rad Il),

where "rad" denotes Jacobson radical, and where Il/rad Il is a skewjleld.

(17.6)

MAXIMAL ORDERS (COMPLETE LOCAL CASE)

173

Proof By (17.4) we may choose A = M,(Ll). We may write rad A = n~A for some m > O. If m > 1, then A/rad A contains the nonzero nilpotent ideal nDA/rad J\, which is impossible since A/rad A is semisimple. Thus m = 1, and A/rad A ~ Mr(A/nDLl) = Mr(Ll/rad Ll). This finishes·the proof.

Now let X be any full R-lattice in A, and define its left and right orders as in (8.1), (8.2). (17.6) THEOREM. 0lX)isa maximal order if and only if O/X) isa maximal order. Proof Let A = O/X) be maximal, and set;\' = O/(X). Then X is a right A-lattice in A, so as in (17.3) we may write X = uA for some u E u(A). Then uAu- l . X C X, l whence uAu- c N. Since uAu- 1 is a maximal order, we obtain uAu- l = N, and so N is maximal, as claimed.

Keeping the notation of (17.3), let us determine all one-sided ideals of A, and especially the maximal such ideals. We begin with an easy result. (17.7) THEOREM. Every matrix with entries in Ll can be diagonalized by means of elementary row and column operations. Given any x u E u(M)Ll)), v E u(M,(Ll)), such that uxv

E

Ll s x r, there exist matrices

= diag Crr~" ... , n~'),

The integers a 1 , ••• ,as are uniquely determined by x. (Possibly some of the {aJ are injinite, and we interpret n'; as 0.) Proof One can imitate the usual prooffor the case where Ll is a (commutative) principal ideal domain. Indeed, the argument is even simpler in this case, since Ll is a local ring without zero divisors, and the fact that multiplication in Ll is non-commutative does not cause any difficulties. We shall not give the details of the proof. For more general theorems of this type, see Knebusch [1].

= A/rad A ~ MJ3). There is a one-to-one correspondence J -> J between maximal right ideals of A and those of A. Here, J is the image of J in 1\, and A/J ~ 1\/1. The module AIJ is a simple A-module. (ii) Let Xl = diag (n D, 1, ... , 1) E Mr(Ll). Then xlA is a maximal right ideal of A, and its conjugates

(17.8) THEOREM. (i) Set A

{u'xlA'u- l : uEu(A)} give all of the maximal right ideals of A.

174

MAXIMAL ORDERS OVER DISCRETE VALUATION RINGS

(iii) For x E A, the right ideal xA is maximal maximal.

(17.8)

if and only if the left ideal Ax is

Proof Each maximal right ideal J of A contains rad A, by (6.3) and (6.6).

Therefore

A/ J ~ (A/rad A)/(J/rad A) = A/1.

Since J is maximal, A/J is a simple A-module, whence A/J is a simple Amodule. Therefore J is also maximal. Conversely for each maximal right ideal Y of A, the kernel of the epimorphism A ~ A/Y is a maximal right ideal J of A such that J = Y This proves (i). Next, we have 1 - Xl = ell = diag (1, 0, ... ,0) EA, whence A/XIA

= A/(1

- ell)A ~ ellA.

But ell is a primitive idempotent in A, so ellA is a simple A-module. Thus xlA is a maximal right ideal of A, as claimed. The same argument shows that Ax! is a maximal left ideal of A. By (17.3), every maximal right ideal of A has the form xA for some x E A. As in (17.7), choose u, v E u(A) so that uxv

But if 0 ::;; b i uXV' A

::;;

= diag (tr;', ... , n:;),

a p I ::;; i ::;; r, then

= diag (n~', ... , n:;)' A

c:

diag (n~, . .. , n;;)' A c A.

Since xA is maximal, so is uxv . A, because left multiplication by u carries A onto itself, and carries xA onto uxv . A. Hence we have a l = " . = ar _ 1 = 0, ar = 1, and thus uxv

= diag (1, ... , 1, n r ) = x"

say. But x, = txlt' for some t, t' Eu(A), and thus uxv = txlt' , so xA

= u- 1 t· x 1 A = (u-lt) 'xlA' (U-lt)-l.

This completes the proof of (ii). Finally, let xA be maximal, By the above, we may write x = U'Xl v' for some u', v' E u(A). Then Ax = AXl . v', whence Ax is a maximal left ideal of A, as claimed. This proves the theorem. 18. MAXIMAL ORDERS (LOCAL CASE)

Keeping the notation introduced at the beginning of this chapter, let R

(18.1)

175

MAXIMAL ORDERS (LOCAL CASE)

be a discrete valuation ring, not necessarily complete. Let R denote the Padic completion of R, and K the quotient field of R. Given any separable K-algebra B, we may form the K-algebra B = K ®K B, which is a separable K-algebra by Exercise 7.13. Of course, when B is a central simple K-algebra, we know from (7.8) that B is a central simple K-algebra. However, if B is merely assumed separable over K, then it may well happen that B will have more simple components than B. This may occur even when B is a field (see Exercise 7.7, for example). Even in the case where B is central simple over K, the full matrix algebras B, B may consist of different size matrices. Despite these precautionary remarks, however, we can obtain results on R-orders in B by using the theorems in §17 on R-orders in B. If r is any R-order in B, then setting = R ®R r, we obtain an R-order tin B. If M is any r -module, then M = R ®R M is a t-module. We shall use such notation hereafter without further comment. By (11.5), r is a maximal order in B if and only if t is a maximal order in B.

t

(18.1) THEOREM. Every maximal R-order r in a separable K-algebra B is left and right hereditary.

t

Proof By (10.5), the maximal R-order is expressible as a ring direct sum L'A; of maximal R(orders in central simple algebras, where each R; is a complete discrete valuation ring in the center of some simple component of B. Each of these direct summands A; is hereditary by (17.3), and thus itself is hereditary. Therefore also r is hereditary, by §3d.

t

(18.2) THEOREM. Let A be any R-algebra, finitely generated as R-module, and let A = A/nA. (i) For each s :;, 1, there is a ring isomorphism

A/nsA

~

A/nsA.,

(ii) There are ring isomorphisms A/rad A ~ A/rad A ~ A/rad A. (iii) Let M, N be fInitely generated left A-modules. Then M modules if and only if M ~ N as A-modules.

~

N as A-

Proof Assertion (i) follows immediately from Exercise 5.7. Taking s = 1 in (i), we have A ~ A/n'A. We then obtain (ii) by applying (6.15) twice, once for the R-order A and once for the R-order A. To prove (iii), we note that if M ~ N, then obviously M ~ N. Conversely, let q>: M ~ N, and let t/I = q> -1. Since R is a flat R-module (see (2.23) or Exercise 5.3), it follows at once by (2.38) that R ®R HomA(M, N) ~ HOll1A(M, N).

176

MAXIMAL ORDERS OVER DISCRETE VALUATION RINGS

(18.3)

Thus we may view HomA(.M', N) as the P-adic completion of HomA(M, N). But HomA(M, N) is dense in its completion (see §5, or (6.16)), and so we can find anf E HomA(M, N) such that

f == q; (mod n' HomA(M, N)). Likewise, we may choose 9 E Hom A(N, M) such that 9 == 1/1 (mod n). Then

gf E HomA(M, M) is such that

m - (gf)m E nM,

mEM.

However, M n nM = nM by Exercise 5.7, and thus m - (gf)m mE M. This gives M = (gf)M

E

nM for all

+ nM.

But rad R = nR, and M is a finitely generated R-module, so by Nakayama's Lemma it follows from the above that M = (gf)M. Therefore gfis a A-automorphism of M, by (6.3a). Likewise,fg is a A-automorphism of N. Thus both fandg are isomorphisms, which shows that M ~ N, and completes the proof of the theorem. (18.3) Th.eorem. Let A be a maximal R-order in a central simple K-algebra A. Then A has a unique maximal two-sided ideal ~, given by ~ = An radA. Then rad A = ~, and every nonzero two-sided ideal of A is a power of ~. Further, radA. is the P-adic completion ofrad A. Proof Since A is a maximal ~-order in the central simple 1?-algebra .4, the ring A/radA. is simple. Hence also A/rad A is simple, by (18.2). But every maximal two-sided ideal of A contains rad A by (6.13), and thus rad A is the unique maximal two-sided ideal of A. Furthermore, (18.2(ii)) implies at once that rad A = A n rad A, ~ ® R rad A = rad A.

Next, since A is simple, every nonzero two-sided ideal M of A is such that KM = A. Thus M ::::> nSA for some s, and hence M is the inverse image of a two-sided ideal of A/nsA. But A/nsA ~ AinsA, and all two-sided ideals of Ainst... are images of powers of rad A. It follows at once that M itself is a power of

~,

as claimed. This completes the proof of the theorem.

We next give a criterion, due to Auslander-Goldman [1J, for an order to be maximaL (18.4) THEOREM. Let A be an R-order in the central simple K-algebra A. Then A is maximal if and only if A is hereditary, and rad A is its unique maximal two-sided ideal.

(18.5)

MAXIMAL ORDERS (LOCAL CASE)

177

Proof Maximal orders have the indicated properties, by (18.1) and (18.3). Conversely, let A be hereditary, with unique maximal two-sided ideal rad A, and let r be an order properly containing A. Then r is a left A-lattice, hence is a finitely generated projective left A-module by (10.7). It follows then from (16.7) that the map fi. : Hom", (r, A) ® '" r -> Hom", (r, r) is epic, where

,1(1 ® y) =

y'(f, y) = [y',f]y,

(f, y),

[y', fJ is I evaluated at y'. Iy(.t;, y) = y for all y Er, that is,

and

I

Suppose that ft{I.t; ® yJ = 1. Then

[y,fJy; = y,

Y E r.

N ow consider the trace ideal T of the left A-module r (see (15.4)). Then T is the two-sided ideal of A spanned by {[y,f]: y E r,fE Hom",(r, An, so the above equation implies that Tr = r. Let us show that T is a proper ideal of A. If T = A, then r is a progenerator of N;/{. This gives (by (16.14)) A = Homn(r, r),

where

n=

Hom",(r, r),

and where ",rn is viewed as a bimodule. But then n = r, since every A-endomorphism of ",r is given by right multiplication by an element of r. Therefore we obtain A = Homn(r, r) = r, a contradiction. We have now shown that T is a proper ideal of A, and thus T is contained in the unique maximal two-sided ideal rad A of A. Since Tr = r, it follows that (rad A)r = r, which is impossible by Nakayama's Lemma. This shows that A is a maximal order, and completes the proof of the theorem. We are going to show that maximal R-orders in separable K-algebras are principal ideal rings, that is, everyone-sided ideal is principal. Theorems 10.5 and 18.2 will enable us to reduce this question to the case of maximal orders in central simple algebras. The key to the proof is the fact that such an order A is a primary ring, that is, Ajrad A is a simple artinian ring. We begin with a result on primary algebras. (18.5)

THEOREM. Let B be a primary linite dimensional algebra over a field k, and let M, N be finitely generated projective left B-modules. Then M ~ N if and only if (M:k) = (N:k).

Proof Let 13

= B/rad B, a simple artinian ring, and let b E B map onto li E 13.

178

MAXIMAL ORDERS OVER DISCRETE VALVA TION RINGS

(18.6)

Su ppose that

is a decomposition of B into a direct sum of indecomposable left ideals, wh!re th,: {eJ are primitive idempotents in B. By (6.21), Be i ~ Be j if and only if Be.t ~ Be., J and the _modules {Be.} are minimal left ideals of B. Since B is simple artinian, the {BeJ are mutually isomorphic. Therefore Be ~ Bel for i all i. But M and N are direct summands of B(s) for some s, whence by the Krull-Schmidt Theorem (Exercise 6.6), both M and N are direct sums of copies of Bel' Hence M ~ N if and only if M and N have the same k-dimension. l

We need one more preliminary result: (18.6) THEOREM. Let A bf any R-order in a finite dimensional K-algebra KA, and let M, N be projective left A-lattices. Set A = A/nA, M = M/nM, N = N/nN. Then M ~ N as A-modules if and only if M ~ N as A-modules. Proof Clearly M ~ N implies that M ~ N. Conversely, let cp: M ~ N be a A-isomorphism. Since both M and N are A-projective, we may find A-homomorphisms a, [J lifting cp, cp- \ respectively, making the following diagram commute:

'P

1

M~N.

For m EM, we have ([Ja)m - mE nM. Hence M = {JaM + nM, so M .= [J~M by Nakayama's Lemma. Then [Ja is an automorphism of.M by (6.3a). LIkewIse, r:xf3 is an automorphism of N, and so a: M ~ N, as deSIred. There is a generalization of the above theorem, in which the hypothesis on M and N is dropped. The general version states: If KA is K-separable, ~hen there exists a positive integer q depending on A, such that for any paIr of A-lattices M N we have M ~ N if and only if M/nqM ~ N/~N. Indeed, q may be cho;en'so that n q - 1 R is the Higman ideal of A (see Curtis-Reiner [1, Th. 75.11], where the Higman ideal is denoted by i(A)). .. We are now ready to generalize (17.3)-(17.6) to the present case, III WhICh R need not be complete.

(18.7)

MAXIMAL ORDERS (LOCAL CASE)

179

(18.7) Theorem. Let A be a maximal R-order in a central simple K-algebra A. (i) Let M and N be left A-lattices. Then M ~ N if and only if M and N have the same R -rank. (ii) Everyone-sided ideal of A is principal. (iii) Every maximal R -order in A is a conjugate uAu -1 of A, where u E u(A). (iv) Let A = K @K A ~ M/E), where E is a skewfield with center K, and let Q be the unique maximal R-order in E. Then

A/rad A

~

M/Q/rad Q),

and Q/rad Q is a skewfield. Proof Since A is hereditary, M is A-projective by (10.7). Thus M = M/nM is A-projective, since if MIN') then M IA('). Now let M, N be any left A-lattices. Then by (18.6), M ~ N if and only if M ~ N as A-modules. Since A/rad A is a simple artinian ring by (17.5). it follows from (18.2) that A is a primary ring. Hence by (18.5), M ~ N if and only if (M: R) = (N:R). Finally, we note that (M: R) = rank R M. This proves that M ~ N if and only if M and N have the same R-rank. Suppose next that M is a left ideal of A. Then KM is a left ideal of the simple artinian ring A, whence KM = Ae = K· Ae for some idempotent e EA. Hence Ae ~ M by (i); if this isomorphism maps e onto m, then M = Am. This completes the proof of (ii). It should be pointed out that this proof is completely independent of that given for the corresponding statement in (17.3(iii)), and so constitutes another proof of that statement. The argument in (17.3) which proves that every maximal order is conjugate to A, carries over unchanged to the present case. It remains for us to prove (iv). Keeping the notation of (iv), we showed in (17.5) that

A/rad A ~ Mt(Q/rad Q). Thus the desired formula for A/rad A holds by virtue of (18.2). This completes the proof of the theorem. (18.8) Remarks. (i) The integer t occurring in (18.7(iv)) is called the capacity of rad A. If A ~ Mr(D), where D is a skewfield, and if

D = K @KD

~ Ms(E)

for some skewfield E, then

A~

M,(D) ~ MriE).

This E is the same as that which occurs in (18.7(iv)), and t = rs. (ii) The statements and proofs of (17.4) and (17.6) carryover unchanged to the present case where R need not be complete.

180

MAXIMAL ORDERS OVER DISCRETE VALVA TION RINGS

(18.9)

(iii) If A is a maximal order in a skewfield D with center K, (18.2) shows that L'1/rad L'1 ~ ~/rad~. Here, ~ is a maximal R-order in the central simple K-algebra D. But if D is not a skew field, then ~/rad ~ is certainly not a skewfield. Indeed, in terms of the notation of the first remark above, we have ~/rad ~ ~ Ms(n/rad n),

where n is the maximal R-order in E. Generalizing (17.8), we have (18.9) THEOREM. Let A be a maximal R-order in a central simple K-algebra A. (i) There is a one-to-one correspondence J -+ J between the set of maximal right ideals J of A, and the set of maximal right ideals J of A/rad A. (ii) All maximal right ideals of A are mutually conjugate under inner automorphisms by units of A. (iii) If xA is a maximal right ideal of A, then Ax is a maximal left ideal of A. Proof The proof of (i) is exactly the same as in (17.8). Next, let xA and yA be maximal right ideals of A. Since A/xA is an R-torsion module, it follows from Exercise 5.7 that there is a right A-isomorphism A/xA ~ A/xA, whence xA is a maximal right ideal in A. But then Ax is a maximal left ideal in A by (17.8), and so Ax is a maximal left ideal in A. This proves (iii). Next, by (17.8) there exists u E u(A) such that yA = u . xA. Choose v E A with v == u (mod nA); since v, u have the same image in A/nA, and since nA c rad A, it follows from Exercise 6.2 that v E u(A). But then yA and vxA are a pair of maximal right ideals of A, having the same image in A/rad A. It follows from (i) that yA = vxA = v· xA . V-I as desired, which completes the proof.

(18.10) THEOREM Let A be a maximal R-order in a separable K-algebra A. Then everyone-sided ideal of A is principal. If M and N are left A-lattices, then M ~ N if and only if KM ~ KN as left A-modules. Proof If M

KM

~

~

N, then clearly KM

~

KN. Conversely if KM

~

KN then

KN as left A-modules. If we decompose A into simple components,

then there are corresponding decompositions

and each Ai is a maximal Ri-order in a central simple algebra, as in the proof of(18.1). From the isomorphism KM ~ KN it follows that K· AiM ~ K·AiN

IDEALS

181

for each i, whence by (18.7(i)) we may conclude that AJU ~ A/J for each i. Hence £II ~ N, and so M ~ N by (18.2(iii)). Finally, if M is a left ideal of A, then KM = Ae = K· Ae for some idempotent e in the semisimple ring A. Thus M ~ Ae by the first part of this proof, and the theorem is established. EXERCISES Unless otherwise stated, R denotes a discrete valuation ring, not necessarily complete. 1. Let ~ be a maximal R-order in a skewfield D with center K. Show that every left ~-lattice is free. [Hint: Since ~ is hereditary, each left ~-lattice M is isomorphic to a direct sum of ideals of ~. Each such ideal is principal, hence is isomorphic to ~.]

2. Prove (18.9(ii), (iii)) without using the results for the case where R is complete. [Hint: Let 7\ = A/rad A, and let xII., yA be maximal right ideals of A. Then xA, yA are maximal right ideals of the simple ring A. By Exercise 7.12, there exists u E u(7\) such that U· xA . u- 1 = yA. Choose v E A with v = u; then v E u(A) by Exercise 6.2. Therefore v . xII.' v- 1 = yA, since these maximal right ideals have the same image in

A. Further, Ax is a maximal left ideal of A, by Exercise 7.12. Therefore Ax is a maximal left ideal of A.] 3. In this exercise, R denotes a semilocal Dedekind domain, that is, a Dedekind domain with onlya finite number of maxim ali deals {P1 , ... , Pn}' Let A be an R-order in a separable K-algebra A, and let M, N be left A-lattices. Show that M ~ N if and only if M p, ~ NN 1 ~ i ~ n. [Hint: Let f.:Mp, ~ N p" 1 ~ i ~ n. Replacing f. by ai;, a E R - Pi' we may assume that each f. E HomA(M, N). Choose P1 " .. , Pn E R such that Pi == 1 (mod PJ, Pi == 0 (mod P), j i=- i. Set f = 'L.,PJ" and show that f:M -> N is an epimorphism, by verifying that f Pi : M Pi -> N Pi is epicfor 1 ~ i ~ n. Deduce from this that f is an isomorphism.] 4. Which of the results of this section remain true when R is assumed to be a semilocal Dedekind domain, rather than a discrete valuation ring?

19. IDEALS Throughout this section, let A be a maximal R-order in a central simple K-algebra A, where R is a discrete valuation ring, not necessarily complete. We shall study here the factorization properties of one-sided ideals of A, and begin with a number of definitions. For a pair of full R-lattices M, N in A, define their product M' N in the usual way, as the full R-lattice consisting of all finite sums Lmin i , mi EM, ni E N. Call M a normal ideal in A if both 0l(M) and O/M) are maximal orders. By (17.6) and (18.8(ii)), if one of these orders is maximal, then so is the other. Call these orders similar; by (18.7), they are necessarily conjugate in A. The normal ideal M is two-sided if O,(M) = O/M}. In the special case where

182

MAXIMAL ORDERS OVER DISCRETE VALUATION RINGS

(19.1 )

A = K, the normal ideals are precisely the fractional R-ideals of K (see §4a), and are of course two-sided. An integral ideal in A is a normal ideal M such that M c O,(M).

(19.1) THEOREM. If M isanormalidealsuchthatM cO,(M), then also M c O,(M), and conversely. Proof Let A = O,(M), so M is a left ideal of A. Hence M = Ax for some x E u(A), and clearly (19.2)

O,(M)

= 0r(Ax) = x-lAx.

Since x E A, we have xA c A, whence M = x - 1 . xA . x

c

X - 1

Ax =

°

JM).

This completes the proof. If M is a two-sided normal ideal with A = O,(M) = 0r(M), we shall also refer to M as a two-sided A-ideal in A.

(19.3)

THEOREM. The set of two-sided A-ideals in A is an il'!finite cyclic group with generator rad A and identity element A, relative to the usual multiplication of lattices.

Proof Let Il3 = rad A, so by (18.3) every nonzero two-sided ideal of A is a power of 1l3. In particular we have nA = Il3m for some m. Define 1l3- 1 = n- 1 ll3 m- \ a two-sided A-ideal in A. Then Il3 .1l3 -1 = Il3 -1 '1l3 = A. If X is any two-sided A-ideal in A, then for some s, n S X is a two-sided ideal in A, and so X = n- s ll3 k for some k. Thus X is a power of 1l3, and X- 1 = nS (Il3- 1 )k, This completes the proof.

An integral ideal M will be called a maximal integral ideal if M is a maximal left ideal in O,(M). (Some authors refer to such ideals as indecomposable ideals, but we shall not use this terminology here.) (19.4) THEOREM. If the normal ideal M is a maximal left ideal of O,(M), then M is a maximal right ideal in r(M), and conversely.

°

Proof Let M = Ax be a maximal left ideal in A, where A = O,(M). By (19.2),OJM) = x- 1 Ax = N, say. By (18.9), xA is a maximal right ideal in A.

Therefore x- 1 . xA . x is a maximal right ideal in x-lAx. This proves that M is a maximal right ideal in N, as claimed. We also remark for future use that conjugation by x- 1 yields an isomorphism of R-modules:

(19.6) (19.5)

183

IDEALS

A x-lAx N Ax = x - . . x - A'x'

We have thus proved that the property of an integral ideal being maximal does not depend on whether we consider it as a left ideal in its left order, or a right ideal in its right order. We may also conclude that all maximal integral ideals in A are mutually conjugate, since each pair of maximal orders are conjugate, and for a given maximal order A; any two maximal left ideals of A are conjugate by (18.9) Let M, N be any pair of full R-lattices in A. We say that their product M . N is proper if O,(M) = Oz{N). Likewise, a product M 1 . M 2' •• Mk is proper if 0r(MJ = O/(M i + 1), 1 :( i :( k - 1. (19.6) THEOREM. I"et M be a proper left ideal of the maximal order A, such that the A-module AIM has a composition series of length k. Then M is expressible as a proper product of k maximal integral ideals M

= M1M2 "'M k ,

with

Proof Use induction on k, the result being trivial when k = 1. Assume that k > 1, and that the theorem is known for the case k - 1. We can choose a maximal left ideal Ax of A such that M c: Ax c: A, and then AxlM has composi tion length k - 1. Let A' = x - 1 Ax; then N is a maximal order, and there is a bijection W -. x - 1 W between the set of A-modules Wand the set of A' -modules x- 1 W. Therefore x- 1 M is a left ideal of N, and x- 1 Axlx- 1 M has composition length k - 1 as left A' -module. It follows from the induction hypothesis that we may write x- 1M = M 2 '" M k , a proper product of maximal integral ideals, with O/(M 2) = O/(x- 1M) = x-lAx = N.

°

If we set M 1 = Ax, then M 1 is a maximal integral ideal such that r( M 1) = N. Therefore the factorization

M = Ax . x - 1 M = M 1 M 2 .•. M k expresses M as a proper product of maximal integral ideals. Once such a factorization is known, then there are obvious inclusions

Since all of these orders are maximal, equality must hold. This completes the proof.

184

MAXIMAL ORDERS OVER DISCRETE VALVA nON RINGS

(20.1)

(19.7) COROLLARY. Let rad A have capacity t, that is, Ajrad A ~ M/S) for some skewjield S. Then rad A is expressible as a proper product of t maximal integral ideals. Proof Mr(S) is a direct sum of t simple left A-modules, and hence A/rad A has composition length t.

EXERCISES 1. Show that in a proper product of maximal integral ideals, the factors need not commute with one another. [Hint: Suppose that M 1 . M 2 = M 2 . M l' where M l' M 2 .. . are maXImal mtegral Ideals. If the algebra A is M neD), where n > 1, then M 1 is not a two-sided D,(M i)-ideal. Therefore

2. Show that in some cases an integral ideal may be expressed in more than one way as a proper product of maximal integral ideals. [Hint: Let A = M 2(K), A = M 2(R), Xl =

(~ ~),

X2

=

(~ ~), where n is a prime element of R. Then

Xl' X 2

commute, and

nA = AXI . X1- 1(Ax 2)X I = Ax 2 ' x; 1(AxI)X2

gives two different factorizations of nA.J

20. DIFFERENT, DISCRIMINANT

Throughout this section, R denotes a discrete valuation ring, not necessarily complete, and A a maximal R-order in a central simple K-algebra A. Let tr A/K denote the reduced trace map, and, A/K: A x A ---+ K the bilinear reduced trace form defined by ,(a, b) = tr(ab). We shall drop the subscript A/K when there is no danger of confusion. We define

A=

{xEA: ,(x, A) c R}

=

{xEA: tr(xA) c R}.

By Exercise 4.12, A is a full R-Iattice in A, and is clearly a two-sided A-module containing A (by (10.1)). Let ~ = rad A; then by (19.3) we have A = ~-m for some m ;;;:: O. Call A the inverse different, and define the different

!:l = !:l(A/R) =

~m.

(20.1) THEOREM. If R is the P-adic completion of R, then

!:l(f.../R) = R ®R !:l(A/R). Proof By (9.29), tr A/R extends the map tr A/K' Hence for each x

E

A we have

(20.2)

185

DIFFERENT, DISCRIMINANT

tr A/K(xA) c R

=

tr J/kCxA) c

R=

X E A.

This proves that & = An X. If '-l! = rad A, then '-l! = A n~, where $ = radA. Thus if A = $-\ then f... = '-l!-k. This gives the desired result. For later use, we remark that the proof is equally valid for the more general case where A is a separable K-algebra. (20.2) THEOREM. Let A = M,(D), where D is a skewfield with center K, and let A = M,(fl), where fl is a maximal R-order in D. Then

=

':n(A/R) Proof Let x = (d i )

E

':n(fl/R) . A

M,(D), and let A = (P ij )

E

M,(fl). Then by Exercise 9.5,

trA/K(xA) = ItrD/K(dijPji)' I. )

Hence if each dij E Li, then x E f.... On the other hand, let A have entry P at position (k, I), and zeros elsewhere, with P variable in fl. Then tr A/K(XA) = tr D/K(dlkP), so if x E f... then dLk E X. This proves that f... = M,(Li). If Li = (rad f1)-m, then

f... =

M,((radfl)-m) = {M,(radfl)}-m = {radA}-m,

which proves the theorem. Together, these results yield (20.3) THEOREM. Let 'k ® K A ~ Mt(E), where E is a skewfield with center K and index m, that is, m:Z = (E: 'k). Suppose that the residue class field R is finite. Then ':n(A/R) = '-l!m-l. ~roof By (20.1) it suffices to prove that ':n(A/R) = $m-t. Now Ais a maximal ~-order in Mt(E), so by (17.4) we may write A = M/O) for some maximal R-order 0 in E. By (20.2) we obtain

':n(A/R) = ':nCO/R)' A. But ':nCO/R) = (rad o)m-l by (14.9), which completes the proof.

For any left ideal L of A, define its norm as ~!K(L)

(20.4)

THEOREM.

= ord R A/L.

The discriminant dCA/R) is given by dCA/R)

=

NA/KC':nCA/R)).

186

MAXIMAL ORDERS OVER DISCRETE VALVA nON RINGS

(20.4)

Proof Both the discriminant and the different behave properly under the passage from R to R, by (20.1) and Exercise 10.6. So likewise does the norm, and thus it suffices to prove the result for the case where R is complete. But then A ~ Mt(Q) as in the proof of (20.3), and 'JJ(AjR) = 7r~ . A for some k. Therefore N('JJ(AjR)) = ordRAj~A = ordR 7r;kAjA = ordRA/A.

The rest of the proof is exactly like that of (14.8).

6. Maximal Orders over Dedekind Domains Throughout this chapter, R denotes a Dedekind domain with quotient field K, R i= K, and P ranges over the maximal ideals of R. Let Rp be the localization of R at P, and Rp the P-adic completion of R. Let Kp be the quotient field of Rp. For A any R-order in the separable K-algebra A, and M any A-module, set Then A. p is an Rp-order in Ap, and !VIp is a A.p-module. It follows from Exercise 7.13 that Ap is a separable Kp-algebra. In this chapter we shall generalize the theorems obtained in Chapters III and V. There are two ways of accomplishing this. In the first approach, followed in §§ 21-22, we use the local results from Chapter V, together with the techniques of § Sa, to obtain global results. The second approach, given in § 23, is independent of the first, and also yields the basic theorems for the global case. In this second method, we give some straightforward generalizations of some of the familiar proofs in algebraic number theory. Indeed, this same dichotomy already occurs in algebraic number theory. One can develop the ideal theory of Dedekind domains either from the standpoint of valuation theory (first approach), or by manipulations with the ideals themselves (second approach). Just as in algebraic number theory, however, we cannot in the long run avoid the use of P-adic completions. Many of the deeper results depend on the interrelations between the global and local theories. Indeed, we have already seen in (20.3) that when R is a discrete valuation ring, we may calculate the different 'JJ(A/R) by passing to completions.

21. BASIC RESULTS Let A be an R-order in the separable K-algebra A, and suppose that there are decompositions (21.1)

Ap =

I' Ai (simple components),

where Ai is an Rp-order in Ai' Let Ki be the center of Ai' and Ri the integral closure of Rp in K i . By (10.5), the maximal Rp-orders in Ai coincide with the maximal Ri -orders in Ai' In particular, if A is a maximal R-order in A, then 187

188

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(21.2)

by (11.6) Ap is a maximal Rp-order in Ap , and so by (10.5) there is a decomposition Ap = I" Ai' with each Ai a maximal Ri -order in the central simple Ki -algebra Ai' We may then apply the results of Chapter V, on maximal orders in central simple algebras over complete fields, to the orders Ai' Conversely, if the order Ap is known to have a decomposition as in (21.1) for each P, and if each Ai is a maximal order in Ai' then we may conclude from (10.5) that A is necessarily a maximal order in A. As an illustration of these remarks, we prove a number of basic results. (21.2) Theorem. Let M be a full R-lattice in A. Then 01(M) is a maximal R-order in A if and only if OJM) is a maximal R-order in A. Proof. We know that Rp is R-flat, by (2.22) or Exercise 5.4. As in the proof of (8.5), it follows that

Or (Ai p) = Rp 0

(21.3)

R

0r(M),

and likewise for left orders. Now suppose that A = 0/ (M) is maximal, and let r = Or (M). Keeping the notation of (21.1), there is also a decomposition A1p = Mp where Mi is a full Rp-lattice in Ai' and where Ai = O/(M;), But then there is a decomposition

I"

Since A is maximal, so is each Ai' whence so is each r i by (17.6). Therefore r is maximal, by the remarks preceding the theorem, and the proof is complete. (21.4) Theorem. Maximal orders are left and right hereditary. Proof Let A be a maximal R-order in A, and use the notation of (21.1). Each Ai is hereditary by (17.3), whence A. p is hereditary for each P. Thus A is itself

hereditary, by (3.24) and (3.30). (21.5) COROLLARY. Let A be a maximal order. Then every left A-lattice M is A-projective. Further, M is indecomposable A-module.

if

and only

if

KM is a simple

Proof. Since A is hereditary, M is projective by (10.7). If M decomposes, so does KM. Conversely, let W be any A-submodule of KM, and set N = M n W. Then N is a A-submodule of M, and by (4.0) the A-module MIN is an Rlattice. Hence MIN is A-projective, and so the A-exact sequence

O->N->M->MIN->O

(21.6)

189

BASIC RESULTS

is split. This gives a decomposition of M, unless W completes the proof.

= 0 or W = KM,

which

We are now ready to determine all maximal R-orders in A. We have seen that it suffices to handle the case of central simple algebras. We prove (21.6) Theorem. Let A = HomDCV, V) be a simple algebra, where V is a right vector space of dimension r over the skewjield D with center K. Let ~ be some jixed maximal R-order in D, and let M be any full right ~-lattice in V. Then Hom", (M, M) is a maximal R-order in A. If i\.' is any maximal R-order in A, then there exists afull right ~-lattice N in V such that i\.' = Hom", (N, N).

Proof Letting A = Hom", (M, M), we have Ap = Rp ®RA = Hom",p(Mp,M p) for each P. By (18.7) every right ideal of~p is principal, whence M p is ~p-free. Hence by (8.7) Ap is a maximal Rp-order. This holds for each P, and so A is maximal by (11.2). Now let i\.' be any maximal R-order in A. Since A and i\.' are a pair of full R-lattices in A, we have A~ = Ap a.e., say except at P l , · · · , Pn • By (18.7), for each P there exists an element up E u(A) such that A~ = upApU; 1, and we may take up = 1 for P i= P 1'" ., Pn . For each p) set X(P) = upM p; then we

A~ = Hom",)X(P), X(P)).

have If we define N =

nX(P),

then by (4.22) N is a full right ~-lattice in A,

p

such that N p = X(P) for all P. But then i\.' and Hom",(N, N) are a pair of full R-lattices in A, with the same localizations for each P. Therefore i\.' = Hom",(N, N) by Exercise 3.2, and the theorem is proved. (21. 7) COROLLARY. Keeping the above notation, every maximal order in A is Morita equivalent to ~.

Proof. Every maximal order A in A is of the form A = Hom", (M, M), where M is some full right ~-lattice in V. Since ~ is hereditary, it is clear that M is a finitely generated projective ~-module. We wish to show that M is a progenerator for the category At"" and so we must verify that Hom", (M, .) is a faithful functor. But for each ~-module L, we known that {Hom", (M, L)}p ~ Hom",p (M P' Lp) for each P. Since M is ~ -free by Exercise 18.1, it follows from §15 that Hom (M .) is a f:lthfu{ functor. Therefore Hom", (M,' ) is also faithful, .1P

p'

190

(22.1)

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

and so we have proved that M,.. is a progenerator for At,... Hence A and t1 are Morita equivalent, by (16.9), and the result is established.

EXERCISES 1. Let A be a maximal R-order in a separable K-algebra A, and let M be any left A-lattice. Show that the endomorphism ring End" M is a maximal R-order in the separable K-algebra End A KM. [Hint: It suffices to prove the result when A is a central simple K-a\gebra, and where R is a discrete valuation ring. Set r = End" M, B = End A KM, and let V be a simple left A-module. Then D = End A V is the skewfield part of A. We have KM :;::; V(t) for some t, whence B :;::; End A V(t) :;::; Mt(D). Thus B is also a central simple K-algebra. Now choose a left A-lattice L in V such that KL = V. Then KM :;::; (KL)(t), whence M :;::; L(t) by (18.10). Therefore

r

=

End" M :;::; End" LCt) :;::; Mt(~)'

r

where ~ = End" L is a maximal R-order in D. Hence as claimed.]

is a maximal R-order in B,

22. IDEAL THEORY In this section we shall develop the theory of one-sided and two-sided ideals in maximal orders over a Dedekind ring R with quotient field K, by using the results of §§18, 19 for the local case. Let P, Q, ... denote prime ideals of R, always assumed to be nonzero. To begin with, let A be any R-order in a separable K-algebra A. A prime ideal of A is a proper two-sided ideal ~ in A, such that K· ~ = A, and such that for every pair 0 f two-sided ideals S, T in A, S· T c ~ ==> S c ~

(22.1)

or

T c ~.

There is no loss of generality in imposing this condition only for the special case where both Sand Tcontain ~; for if (22. 1) holds in this special case, then in the general case we have S' T c

~

--=!>

(S

+ ~) (T +

~) c

=S c

~ ~

= or

S

+~

c ~

or

T

+

~ c ~

Tc~.

Hence (22.1) is equivalent to the following condition: for each pair of two -sided ideals J, f in A/~, (22.2)

J . J' = 0

~--=>

J = 0

or

J'

= O.

(22.3) THEOREM. The prime ideals of A coincide with the maximal two-sided ideals of A. If~ is a prime ideal of A, then P = ~ n R is a (nonzero) prime ideal

(22.4) of R, and

191

IDEAL THEORY

A = A/~ is ajinite dimensional simple algebra over

the jield RIP.

Proof Let ~ be a prime ideal of A, and set P = ~ n R, A = AI~. Since ~ is a full R-Iattice in A, we have P #- O. On the other hand, P < R since 1 ~ ~. If a, fJ E R, then

af3

E

P

= aA' f3A c ~ ==> aA c: == a E P or f3 E P.

~

or

f3A c:

~

Thus P is a prime ideal of R, and A is a finite dimensional algebra over the field RIP. Next we observe that rad A is nilpotent, since A is artinian. Thus rad A = 0 by (22.2), and so A is semisimple. However, the simple components of A are two-sided ideals of A which annihilate one another. Hence by (22.2) there is only one simple component, that is, A is a simple algebra. Therefore X has no nontrivial two-sided ideals, whence ~ is a maximal two-sided ideal of A. Conversely, let ~ be any maximal two-sided ideal of A. Then (22.1) automatically holds true when Sand Tcontain ~, since the only possibilities for Sand Tare'll and A. This shows that ~ is a prime ideal, and establishes the theorem. Keeping the above notation, we prove. (22.4) THEOREM. Let A be a maximal R-order in a central simple K -algebra A. There is a one-to-one correspondence ~ +4 P between the set of prime ideals ~ of A, and the set of prime ideals P of R, given by

P = R n~, ~ = A n rad Ap. F urther, ~. Ap = rad Ap' the unique prime ideal of Ap' For each prime ideal Q of R different from P, we have ~Q = A Q • Proof Let ~ be a prime ideal of A, and set X = Ai'll, P = R n ~. Then ann R A = P, so by (3.6) it follows that

A;;; Rp ® R A;;; Ap/~· Ap. But A is a simple ring, and hence 'l3 . Ap is a maximal two-sided ideal of Ap . By (18.3), all two-sided ideals of Ap are powers of rad Ap. Therefore ~p = ~'Ap

= radAp,

as claimed. Furthermore, AI~ is torsionfree relative to the multiplicative set R - P, so by (3.10) we obtain ~ = A n ~p = A n rad Ap.

192

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(22.5)

Finally, for Q ¥- P we have RQ ®R (A!'lJ) = 0, and thus 'lJ Q = A Q . Conversely, given a prime ideal P of R, we may form the two-sided ideal PA of A. Then PA ¥- A, since A is an R-Iattice, and so there exists a maximal two-sided ideal 'lJ of A with PA c 'lJ. Clearly P = R n 'lJ, since the intersection is a prime ideal of R containing P. The preceding discussion shows that'lJ = An rad Ap, so P uniquely determines 'lJ. This completes the proof of the theorem. Now let L be any full R-Iattice in the separable K-algebra A, and define (22.5)

D

1

=

{X E

A : L· x . L

c

L}.

Clearly

(22.6)

D1 = {xEA: L·x

c

O,(L)} = {xEA: x'Lc O/Ln.

If A = O,(L), then rxA c L c [3A for some nonzero rx, [3 E R. Therefore rx- 1A :J D 1 :J [3- 1A, since rx- 1 A

= {xEA: rxi\·x

c A}:J

{xEA: L·x

c A}

= Dl,

and likewise for the other inclusion. Thus D 1 is also a full R-lattice in A. We shall call La full left A-lattice in A, to indicate that L is a left A-module which is a full R-Iattice in A. Then D 1 is a full right A-lattice in A. We may identify D 1 with HomA (L, A), where L, A are viewed as left A-modules. For each prime ideal P of R, Lp is a full left Ap-Iattice in A. By (3.18) we have (D l)p

= Rp ® R HomA (L, A)

~ HomAp

(Lp, Ap) = (Lp )-1.

Hereafter, we shall identify (D l)p with (Lp )-1. Likewise, we identify

Rp®RL-1

and

(L p )-l,

where Rp, Lp are P-adic completions of R, L, respectively. (22.7) THEOREM. Let L be a full left A-lattice in A, where A is a maximal R-order in a separable K-algebra A. Then

L· L- 1

= A,

Proof In order to prove that L· D

1

= A, it suffices by Exercise 3.2 to prove

that

Lp' (L p)-l = Ap for each P. By (18.10) we may write Lp = ApY for some YEA. Then A which implies by (6.3a) that Y E u(A). Therefore

(L p)-l = {xEA: ApY'x cAp} = y-1Ap.

= Ay,

(22.8)

193

IDEAL THEORY

This gives Lp' L; 1 = Ap y . y - 1 Ap = Ap , as desired, and shows that L· L -1 = A.. The argument also yields the formula (Lp)-l 'L p = y-1Apy = O,(L p) for each P, whence C l . L = O/L). Likewise (Cl)-l = L, since this holds locally at each P, and the theorem is established. (22.8)

COROLLARY.

Keeping the above notation, we have OI(C l ) = O.(L),

Or(C

l

)

= OlL).

Proof The first equality holds locally: 0lL;l) = 0iy-1Ap) = y-1Apy = 0r(Lp).

Thus the first assertion is correct, and the second follows in the same way. We now repeat several definitions, which were first given in the local case in §19. All ideals considered below are full R-Iattices in A. Call a full R-Iattice M a normal ideal if both 01(M) and OJM) are maximal orders. An integral ideal of A is a normal ideal M such that M c 01(M). (22.9)

THEOREM. If M is a normal ideal such that Me 01(M), then also M c OJM), and conversely.

Proof It suffices to establish the result after localizing at an arbitrary prime ideal P of R. But the local result has already been proved in (19.1). (See also Exercise 22.9.) If M is a normal ideal with O,(M) = O,(M) = A, we shall also refer to M as a two-sided A-ideal in A.

(22.10) T~eorem. Let A be a maximal R-order in the central simple K-algebra A. Then the set of two-sided A-ideals in A is the free abelian group generated

by the prime ideals of A.. The group operation is the usual multiplication of lattices, the identity element is A, and inverses are given by (22.5). Proof Let 'l3 t , 'l3 2 be distinct prime ideals of A, and let Pi P l # P 2• For Q any prime ideal of R, we have AQ'

(22.11)

=

'l3 i

n R, so

Q # Pi'

('l3 i )Q = { ('l3;)Q' Q = Pi'

But then 'l3 1 . 'l3 2 = 'l3 2 . 'l31' since this is true locally at every Q. Now let M be a two-sided A-ideal in A. Then for each Q, MQ is a two-sided

194

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(22.12)

AQ-ideal in A, and so by (19.3) we have MQ =

OQmQ

for some

mQ E

Z,

where 0 is the prime ideal of A corresponding to Q. It should be pointed out t~at the definition of 0; 1 given in (19.3) agrees with the definition of inverse gIven III (22.6). Furthermore, mQ = 0 a.e., since M Q = AQ a.e. It follows at once that M =TIn'" Q , Q

since this result holds everywhere locally. Indeed, for any P we have Mp = m +'p mp

--

TI""" mQ "-"p, Q

by (22.11). Finally, the ideal group is free abelian on the generators {'P}, since the ideal M uniquely determines the exponents in the factorization of M. This completes the proof. (22.12) COROLLARY. The preceding theorem is valid for the more general case where A is a maximal R-order in a separable K-algebra A. Proof Let A and A be decomposed into components as in (10.5). Then every two-sided A-ideal M in A decomposes into a direct sum M i , where Mi is a two-sided Ai-ideal in the central simple Ki-algebra Ai' In particular, each maximal two-sided ideal of A is of the form

I·

(22.13) for some j, 1 ~ j ~ t, and some prime ideal 'Pj in Aj . Conversely, each such expression is a prime ideal of A It follows at once that if we express M j as a power product of prime ideals of Aj , then (22.13a) is a power product of the corresponding prime ideals of A of the form (22.13). Since M is the product (for 1 ~ j ~ t) of the expressions in (22.13a), it is clear that M is a power product of prime ideals of A It is then also obvious that the prime ideals of A generate a free abelian group, and so the result is proved. We have at our disposal all of the information necessary to describe how a prime ideal P of R behaves in a maximal R-order in a simple K-algebra: (22.14) THEOREM. Let A be a central simple L-algebra, and let L be a finite separable extension of K. Let S denote the integral closure of R in L. Given a prime ideal P of R, let PI' ... , Pd be the distinct primes of S dividing P, and set

(22.15)

195

IDEAL THEORY

ei = e(Pi , L/K). Each Pi determines a prime ideal ~i of A by (22.4), and we may write Pi A = ~t', 1 ~ i ~ d, for some positive integers {mJ. (i) The factorization of PA into prime ideals of A is given by d

PA =

TI~ieim,. i= 1

(ii) For each i, mi is the ram!fication index of the skewfield part of the central simple I-algebra .4, where ~ denotes Pi-adic completion. (iii) If the residue class field R/P isfinite, thenf{N" each i, m. is the index of the skewfield part of A. I

TI Pt', whence = TI (Pi A)e = TI ~/'m,

Proof From (4.27) we have PS = PA

i

as cIai!lled. Now keep i fixed, and let I, l\, A denote P(adic completions. Since A is a maximal ~-order in.4, by (17.3) we may write

A = Mr(A), where D is a skewfield with center I, and A is the maximal ~-order in D. Um denotes the ramification index of D over I, then by §13 we have PiA = (nDAt,

where n D is a prime element of A. Therefore

Pi A = Mr(P i A) =

n; A = (rad At·

On the other hand, rad A is the P(adic completion of ~i by (18.3). Therefore when we take completions on both sides of the equation Pi A = ~ti we obtain pi, = (rad At'· This shows that mi = m, and establishes (ii). Assertion (iii) follows from (i), by virtue of (14.3). Let us turn next to the theory of one-sided ideals in a separable K-algebra A. A normal ideal M is called a maximal integral ideal in A if M is a maximal left ideal in its left order O/M). (22.l 5) THEOREM. For each maximal left ideal M of a maximal R-order A in a separable K-algebra A, there is a unique prime ideal ~ of A such that (22.16)

~ c MeA,

~

= ann" AIM

= {x EA:xA eM}.

We say that M belongs to~. Then AIM is a simple left module over the simple ring AI~. Conversely, each ~ determines a maximal left ideal M of A which belongs to ~.

196

MAXIMAL ORDERS OVER DE DE KIND DOMAINS

(22.17)

Proof Given a maximal ideal M of A, we can decompose M and A according to the decomposition of A into simple components. It is then sufficient to prove the result when A is a central simple K -algebra. Now let ~ = ann" AIM, a two-sided ideal of A. Choose a nonzero rx E R such that rxA eM; then rx E ~, so rxA c ~, and thus A/~ is an artinian ring. But AIM is a faithful left (AI~)-module, and is a simple module. Therefore the ring AI~ is a simple artinian ring, and so ~ is a prime ideal of A, as claimed. Clearly ~ = ~A c M. Furthermore, if n is any two-sided ideal of A contained in M, then n c ann" AIM = ~. Thus there is a unique prime ideal of A contained in M. Conversely, given a prime ideal ~ of A, by Zorn's Lemma there exists a maximal left ideal M of A such that ~ c MeA. Then ~ c ann" AIM, so equality must hold, and thus M belongs to ~. This completes the proof. (22.17) THEOREM. Let M be a normal ideal in the separable K-algebra A. If M is a maximal left ideal of O/(M), then M is a maximal right ideal of 0r(M), and conversely.

Proof As remarked above, it suffices to prove the result for the special case where A is a central simple K-algebra. Let A = 0z{M), and let ~ be the prime ideal of A to which M belongs. Setting P = R II~, it is clear that annRAIM = P, whence by(3.6) we have Ap/Mp ~ AIM. ThusMpisamaximal left ideal of Apo If we set N = O/M), it then follows from (19.4) that Mp is a maximal right ideal of A~. On the other hand, let Q be any prime ideal of R different from P. Then ~Q = MQ = AQ, and so

°

~ = /M Q) = AQ = M Q. Now let MeN c N, where N is any right ideal of N. Then MQ = N Q for Q # P, whereas N p is either M p or A~. In the former case we see that N = M, while in the latter case N = N. Thus M is a maximal right ideal of N, and the theorem is proved. Let M, N be any pair of full R-Iattices in the separable K-algebra A. Their product M . N is proper if r(M) = z{N). Likewise, a product M 1 M 2 ... M k is called proper if O/M i ) = O/(M i + 1 ), 1 ~ i ~ k - 1. Generalizing (19.6), we prove

°

°

(22.18) Theorem. Let M be a full left ideal of the maximal R-order A in the

separable K-algebra A, and suppose that the A-module AIM has composition length k. Then M is expressible as a proper product of k maximal integral ideals M 1 ... M k' such that O/(M) = O/(M 1 ), 0r(M) = 0r(M k ).

(22.19)

197

IDEAL THEORY

Proof Use induction on k, the result being trivial when k = 1. Assume that k > 1, and that the theorem is known for the case k - 1. We can choose a maximal left ideal N of A such that MeN c A, and then N/M has composition length k - I as left A-module. Both M and N are normal ideals, with left order A. By (22.7) we have N· N- 1 = A,

N- 1 . N

= OreN) = A'(say).

1

Now there is a bijection W --+ N- W between the set of left A-submodules W of N, and the set of left A' -submodules N- 1W of A', with the inverse mapping given by N -1 W --+ N(N- 1 W). Therefore N- 1 M is a left ideal of A', and the quotient A'/ N -1 M has composition length k - 1 as left A' -module. It follows from the induction hypothesis that we may write N- 1 M=M 2 "·M k,

a proper product of k - 1 maximal integral ideals, with Bu t then M = N M 7. ••• M k is the desired proper product of k maximal integral ideals, and the theorem is proved. We shall next prove the non-commutative analogue of the algebraic number theory dictum "To contain is to divide" (see H. Cohn [1]). In the proof, it will be convenient to use the notation N = N ij to indicate that the normal ideal N has left order A, and right order Ai" Thus a proper product might be written as N 12 N 23 , where A1 ,A 2 and A3 need not be distinct. From (22.8) we obtain N-1·N=A .. J

(22.19) THEOREM. (i) A proper product of integral ideals is integral. (ii) Let M 12' N 14 be normal ideals with the same left order A 1. Then MeN if and only if M 12 = N 14' C 42 for some integral ideal C. (iii) Let M 12' N 34 be normalideals. Then MeN !fand only if M is a proper product (22.20)

for some integral ideals B, C. Proof To prove (i), we show by induction on k that M12 M 23

'"

Mk,k+l

C

Ak+l'

if the M's are integral ideals. The result is clear for k = 1, so let k > 1, and assume the result for k - 1 factors. Then

198

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(22.21)

. M I2 ··· Mk-l.kMk.k+l C A k M k. k+1 = M k,k+l c A k + 1 , as desIred. We next prove (iii), since (ii) will then fo11O':" as a special case. If (22.20) holds with B, C integral, then M 12 C A3 N 34 A4 = N 34 ' as claimed. Conversely, let M 12 C N 34' and set

B 13 = M 12 (A 3 M 12 )-I,

C42 = (N 34 )-1 M . 12 It is easily ve.rified that B, C are normal ideals with the indicated orders. Next, the normal Ideals M ~nd A3M have the same right order, and Me A3 M , so by (22.6) we obtam M- 1 :::J (A 3 M)-I. Therefore Be M'M- 1 = A ~nd hence. B is integral. Further, C = N -1 M C N- 1 N = A 4 , so C is al~~ mtegral. Fmally, B13N34C42 = M(A 3 M)-I'N'N- 1 M = M(A 3 M)-I'A M 3 = M'A 2 = M, which completes the proof of (iii). To prove (ii), note that when Al = A3 , we obtain B = M(A 1 M)-1 = Al in the above. This establishes the theorem. Of course, the analogue of (ii) holds for a pair of normal ideals with the same right order. Let us now investigate the uniqueness properties of the factorization of an integral ideal into a proper product of maximal integral ideals. As we have already seen in Exercises 19.1 and 19.2, the factors need not commute, and there may be more than one possible factorization. To some extent, the difficulty arises from the fact that we must consider not just one maximal order, but the sequence of maximal orders associated with the successive factors. Let us prove some results on change of orders. (22.21) THEOREM. Let AI' A2 be a pair of maximal R-orders in the separable K-algebra A. Then there exists a normal ideal M = M 12' If I(A) denotes the group of two-sided A{ideals in A (j = 1,2), there is an isomorphism
The map


where


XEI(A ). 1

is independent of the choice of M 12'

Proof The product M = AI' A2 is a normal ideal with left order Al and right order A2 . It is clear that
(22.22)

199

IDEAL THEORY

M 12 used to define ({)12' Finally, ({)12 is an isomorphism since it has an inverse ({)21' given by (()21(Y) = MY M-l, Y E I(A 2 ). This completes the proof.

(22.22)

COROLLARY. Let M 12 be a maximal integral ideal which belongs to the prime ideal '-PI of AI' and to the prime ideal '-P 2 of A 2 • Then '-P 2 = ({)d'-P 1).

Proof It is clear from (22.21) that ({)d'-P 1) is a prime ideal of A 2 , and that ({)d'-P 1 ) = M-1'-P 1 M. We have '-PI c: M => M- 1 c '-P;l => M-l~lM c: '-P~l'-PIM = M.

Thus ({)12('-P 1 ) is a prime ideal of A2 contained in M, whence M belongs to ({)12('-P I ). This completes the proof.

(22.23)

COROLLARY. Let M 23 be a maximal integral ideal belonging to the prime ideal '-P 2 of A 2 . Then for each normal ideal B = B 12 , the quotient B12 /B12 M 23 is a simple left AI-module, and

annAl B/BM = {x

E

Al :xB c BM}

= ({)21('-P 2).

Proof If BM c X c B is a chain of left AI-modules, then M c: B-1X C A is a chain of left A 2-modules. Thus B- 1 X is either M or A 2 , whence X i~ either BM or B. This shows that B/BM is a simple left AI-module. As in the proof of (22.15), annAl B/BM is a prime ideal of AI' But ({)21 ('-P 2) = B'-P 2 B- 1 is a prime ideal of Al which annihilates B/BM. Therefore annAl B/BM = ({)21 ('-P 2 ), and the corollary is established.

Let (()12:I(Al) ~ I(A 2 ) be the isomorphism described in (22.21). For XE I(A l ), we call the ideals X and ({)12 (X) similar. Thus, by (22.22), each maximal integral ideal determines a pair of similar prime ideals. We may caution that the converse need not hold, that is, given a pair of similar prime ideals '-PI and '-P 2 , it is not necessarily true (even in the complete local case) that there exists a maximal integral ideal M 12 belonging to both '-PI and '-P 2 (see Exercise 22.1). We are now ready to deal with the uniqueness properties of factorizations into proper products of maximal integral ideals, still assuming that A is any separable K-algebra.

(22.24) THEOREM. In any factorization of an integral ideal L into a proper product of maximal integral ideals, the number of factors is uniquely determined by L, and the prime ideals to which these factors belong are uniquely determined up to similarity and order of occurrence. Proof Let L = L lr be an integral ideal, and suppose that

200

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(22.25)

(22.25)

where each M i,i+1 is a maximal integral ideal, and M k ,k+1 is interpreted to be M kr' By (22.23), the chain of left A 1-modules (22.26)

A1

:::J

M12

:::J M12 M 23 :::J .•• :::J M 12 M 23 ··· M kr =

L 1r

is strictly decreasing, and is unrefinable. The factor modules (22.27)

are therefore the composition factors of the left A 1-module A1/L. By the Jordan-Holder Theorem, these cpmposition factors are uniquely determined by L, up to A 1-isomorphism and order of occurrence. There are k such factor modules, so L determines k uniquely. Furthermore, let ~i denote the prime ideal of Ai to which M i, i+ 1 belongs, 1 :( i :( k. Then by (22.23) we have

Thus L uniquely determines the set of prime ideals of A1 {~1'

([J21

(~2)" .. , ([Jk1(~k))'

This completes the proof of the theorem. As already remarked in Exercises 19.1 and 19.2, the individual factors M i, i+1 occurring in (22.25) are not uniquely determined (except in special cases). Indeed, even the set of maximal orders {A 2 ' •.. , Ak } is not an invariant of L. We now prove, however, that the order of occurrence of the composition factors of A1 IL in (22.27) may be specified in advance. (22.28) THEoREM. Let L be a left ideal of the maximal R-order A in a separable K-algebra A, and let {Sl"'" Sk} be the composition factors of the A-module AI L, arranged in any preassigned order. Then there is afactorization (22.25) of

L into a product of maximal integral ideals, such that the factor modules in (22.27) are precisely S 1 ' ... , Sk' in that order. Proof Any composition series for the left A-module AIL lifts to an unrefinable chain of left ideals of A, and this chain can be expressed in the form (22.26) for some choice of M's (by (22.19)). Hence it suffices to prove that AIL has a composition series in which the composition factors occur in the preassigned order S 1 ' ... , Sk' Furthermore, a decomposition of A into simple components yields corresponding decompositions of A and L, and hence of AIL. It is therefore sufficient to deal with the case where A is a central simple K-algebra. Let us set T = AIL, an R-torsion finitely generated left A-module, and let

(22.28)

201

IDEAL THEORY n

~

= ann"T. Then ~ is a two-sided ideal of A, and so we may write ~ =

f1 ~t; 1

where the But

{IPJ are distinct prime ideals of A. Then

A/~ ~ whence there is a decomposition T

=

t 1

T is a left (A/~)-module.

A/~ti,

±" 7;, with 1

7; = {x E T:

~~i x

= O} = {x E T:

~~ x

= 0 for some m}.

(Call I; the ~(primary component of T.) It suffices to show that the left A-module I; has a composition series in which the factors occur in a preassigned order. Each composition factor X of I; is a simple left A-module, hence as in (22.15), ann" X is a prime ideal of A. But then ann" X = ~i' since we already know that ~ti X = O. Hence X is a simple left (A/~i )-module. Since A/~i is a simple artinian ring, there is only one isomorphism class of such modules X. Thus all of the composition factors of 7; are mutually isomorphic. This proves that I; has a composition series with preassigned order of composition factors (indeed, any composition series of 7; will do), and thus the same holds true for AIL. The theorem is thus established. To conclude this section, we observe that the collection of normal ideals in a separable K-algebra forms a groupoid, according to the following definition. A groupoid G is a collection of elements, certain of whose products are defined and lie in G, such that (i) For each aij EG, there exist unique elements ep ej EG such that eia ij = alj = ail j , where all indicated products are defined. Further, eie i = ei , eje j = ej" Call ei the left unit of aij' and ej the right unit of air (ii) aij bk1 is defined if and only if j = k, that is, if and only if the right unit of alj equals the left unit of bk1 • (iii) If ab and bc are defined, so are (ab)c and a(bc), and these are equal. (iv) For each aij E G, there exists an ail 1 E G with left unit ej , right unit ei , such that (v) Given any pair of units e, e' E G, there is an element au E G with left unit e, right unit e'.

The collection of all normal ideals in A, relative to proper multiplication, is the Brandt groupoid associated with A. The units are the maximal orders in A; the inverse of a normal ideal M is the normal ideal M - 1. If A and N are maximal Qrders, the normal ideal A . N has left unit A, right unit N.

202

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

EXERCISES Unless otherwise stated, A denotes a maximal R-order in a separable K-algebra A.

1. Let R be a complete discrete valuation ring with prime element n, and let A1 = M3 (R), Show that no maximal left ideal of A1 can be a maximal right ideal of A z . Deduce from this that the prime ideals nA 1 , nA 2 are similar, but that there is no maximal integral ideal M 12 which belongs to both nA 1 and nA 2 . [Hint: Let A1x be a maximal left ideal of A1 ' and suppose that 0,(A 1 x) = A 2 . Then A2 = x- 1 A1 x, so (xY)'A 1 '(xy)-1 =A 1

·

Therefore (see Exercise 38.5) xy = nku for some k ~ 1 and some u Eu(A 1 ). But then

x has elementary divisors {n k - \ nk, n k },yvhich contradicts (17.8).J

2. Let M, N be normal ideals in A. Prove that M· N is a proper product if and only if replacing either factor by a larger R-Iattice (in A) increases the product. [Hint: Let M = M 12' N = N 34' in the notation introduced just before (22.19). Suppose that increasing either factor increases the product MN. Then

=

MN = MA 2 'N = M·A 2 N

N

=

A2N

=

A2

=

A3 •

Conversely, let Az = A3 and let M'N = MN, where M' is an R-Iattice properly containing M. Then M' c M'A z = M"NN- 1 = M'NN- 1 = M, a contradiction.J 3. Let 1JJ 1 ' ... , IJJn be distinct prime ideals of A. Let n

W=

IT lJJ,a" 1

n

IT IJJ/',

~ =

1

Prove (i) \!I

::J

~ if and only if a, 0;;: b" 1 0;;: i 0;;: n.

(ii) W

+

~ =

n

IT lJJ;nin

(a"

bil.

1 n

(iii) W n ~ =

IT lJJ;nax(a"bd, 1

(iv) If each a, ~ 0, then there is a ring isomorphism n

Aj\!I ~

.I' AjIJJ,"i,

1=1

4. Show that each finitely generated R-torsion left A-module X is expressible as a finite direct sum I'APi' where each J i is a left ideal of A of the same R-rank as A [Hint: As in the proof of (22.28), the problem reduces to the case where R is a complete discrete valuation ring, and where A = Mk(Ll) for some maximal R-order Ll in a skewfield. By means of the Morita correspondence, the problem further reduces to that of finding R-torsion Ll-modules. Now use (17.7).]

203

EXERCISES

5. Any simple left A-module is of the form A./M, where M is a maximal left ideal of A. Let P = R n M. Show that P is a prime ideal of R, and that P = annR A./ M. [Hint: Use (22.3).] 6. Let M 23 be a maximal integral ideal, and B12 any normal ideal. Prove that ann R A2 IM 23

ann R B12IB12 M 23 ·

=

[Hint: If P = annRA21M, then PB-IB = PA 2 C M 23 , whence PB c BM.] 7. Keeping the notation of (22.28), let l.Jli = annA Si' Pi = l.Jli n R = ann R Si' 1",: i",: k. Let PI' ... , Pn be the distinct prime ideals of R among {PI' ... ,Pk }. Show that we can express L as a proper product L =

N(I). N(2) .. 'N(n),

where each N(i) is an integral ideal such that (N(i»)Pi = L pi ,

(N(i»)p

=

maximal order

(P

'"I

PJ,

for 1 ",: i ",: n. 8. Let L, M be left ideals in A such that ann R A./ L

+ ann R A./ M

= R.

Show that there exists a left ideal N such that A./ N ~ A./L

+-

A./M.

[Hint: Define a left A-homomorphism

f:A--->A./L+- AIM by f(A) = (A + L, A + M), )_ E A.. For each prime ideal P of R, either (A./L)p = 0 or (A./M)p = O. Thus fp is epicfor each P, whence also f is epic. Now choose N = ker f.] 9. Prove (22.9) directly. [Hint: M c O,(M)

= M .M

c M

=M

c O,(M).]

10. Verify that the definition of inverses in (19.3) agrees with the definition in (22.6). 11. Show that any two maximal orders AI' A2 in A are Morita equivalent. [Hint: (1) Reduce to central simple case, then use (21. 7) and transitivity of Morita equivalence; or (2) Let M = MI2 be a normal ideal, and identify HomAl (M, AI) with the module of right multiplications by elements of M- I . The formula M· M- I = Al is then equivalent to the assertion that f1:

M ®A2 HomAl (M, AI)

--->

Al

is an epimorphism. Likewise, HomAl (M, AI) ®Al M ---> A2 is epic. Hence Al and A2 are Morita equivalent by § 16.] !:

12. Let r be an arbitrary R-order in a central simple K-algebra A. Show that r is a maximal order if and only if r is hereditary, and for each prime ideal P of R there is a unique prime ideal of r containing P. [Hint: To prove r maximal, it suffices to prove rp maximal for each P. There is a one-to-one correspondence between the

204

(23.1 )

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

prime ideals of r containing P, and the prime ideals of r

p.

Now use (18.4).]

13. Let P be a prime ideal of R, and let PA = I1 'P/', where the 'Pi are prime ideals of the maximal order A. Let fi = (A/'Pi: RIP). Prove that

Ledi

= (A:K).

[Hint: Consider the R-algebra AI PA, where R

=

RIP, and imitate the proof of (4.30).]

23. ALTERNATE APPROACH TO GLOBAL IDEAL THEORY

In §§21-22, we derived results on ideals and orders in the global case by making use of the local theorems from Chapter V. In this section we shall give an alternate approach to some of the global theorems obtained in §§21-22. The direct global approach given below generalizes some of the standard arguments of algebraic number theory. We shall not use any of those results from the preceding sections, whose proofs were based on localization techniques. This section may be skipped if desired without affecting the continuity of the exposition. As usual, let A be a separable K-algebra, where K is the quotient field of the Dedekind domain R. In §22, we defined prime ideals of an R-order A in A, and showed that these coincide with the maximal two-sided ideals of A. We now recall some other definitions from §22. All ideals M considered below are assumed to be full R-lattices in A. Call M a normal ideal if 01(M) is a maximal R-order. An integral ideal is a normal ideal M such that M c I(M). A maximal integral ideal is an integral ideal M which is a maximal left ideal in 01(M). (We shall eventually remedy this asymmetry in the definitions (see (23.2), (23.10), and (23.11).) Let us write M = Mij to indicate that M is an ideal with 0l(M ').. ) = A., 0 r (M t).. ) = A ).. r.

°

Our aim is to derive the factorization properties of one-sided and two-sided ideals. We begin with

(23.1) LEMMA. Let A be any R-order in A. Then every two-sided ideal in A contains a product of prime ideals. Proof If not, then the set of two-sided ideals which do not contain such products is non-empty, hence has a maximal element J since A is noetherian. Clearly J is not a prime ideal, so by (22.2) there exist two-sided ideals B, C of A, properly containing J, such that J ~ Be. But both Band C contain products of prime ideals, whence so does J. This yields a contradiction, and proves the lemma.

(23.2) Mij

c

LEMMA. Aj"

Let

Mij

be a full R-lattice in A. Then

Mij c

Ai

if and

only

if

(23.3)

ALTERNATE APPROACH TO GLOBAL IDEAL THEORY

Proof Let Mij

c

205

Ai; then

Mij'Mij

C

Ai' Mij

= Mij ==- Mij

C

0 reM,)

= Aj"

This completes the proof. For any full R-Iattice L

Lx c Ai

= L,j in A, define D 1 as in (22.5). Then for x E L- 1, =?

L·xA i c Ai

=

XAi c L- l ,

whence L- 1 is a right A(module. Thus we obtain (23.3) Our next step is to develop the theory of two-sided ideals, and we now prove (23.4) LEMMA. Let M be a two-sided ideal in the maximal order A. If M < A, then M- l > A.

Proof It is clear from (23.3) that M- l is a two-sided A-ideal in A, and obviously M- l ::> A. Let us assume that M- l = A, and obtain a contradiction. Since M < A, we can choose a prime ideal ~ of A with M c ~ c A. Let ex be a nonzero element of R n ~ (such exist, since ~ is a full R-Iattice in A). Then by (23.1) IXA contains a product of prime ideals, and so we may write ~::> IXA ::> ~l ~2'" ~r'

where the ~i are prime ideals, and where r is minimal. Since ~ ::> ~l ... ~r and ~ is a prime ideal, it follows that ~ contains some factor ~j' and then of course ~ = ~ j ' Thus we may write ~ ::>

IXA

::> B~C,

where B, C are two-sided ideals of A. We then obtain ex- l B~C c A ==> B'ex-l~C'B c B ==> ex-1CB c ~-l c

=

ex-l~CB c Oy(B)

M- = A l

=

CB c etA.

This shows that exA contains a product CB of r - 1 prime ideals, which contradicts the minimality of r, and completes the rather mystifying proof. (23.5) THEOREM. Let M be a full R-lattice in A such that OtCM) is a maximal

order. Then

Proof Let A in A. Then

= 0/(M) be maximal, and set B = M· M- 1, a two-sided ideal

206

BB- 1 C A

MAXI¥AL ORDERS OVER DEDEKIND DOMAINS

=

M'M- 1B- 1 C A

=

(23.6)

M- 1B- 1 C M- 1

=

1

But 0r(M- ):=J O,(M) = A by (23.3), whence 0r(M- 1) B- 1 c A, so B = A by (23.4). This completes the proof.

B- 1

= A.

C

0r(M- 1 ). Therefore

The preceding result is the key step in this chain of arguments. Using it, we prove (23.6) THEOREM. Let A be a maximal order. For each two-sided ideal M in A, we have (23.7)

M' M- 1

= .ft;r 1 . M = A,

(M- 1)-1 =M.

Furthermore, every two-sided ideal in A is uniquely expressible as a product of prime ideals of A, and multiplication of prime ideals is commutative. Proof The first two equalities in (23.7) are immediate consequences of(23.5). To establish the last formula, choose a nonzero ex E R such that exM- 1 c A. Then exM -1 is a two-sided ideal of A, whence (exM- 1)(exM- 1)-1 = A. But (exM- 1 1 = ex- 1L, where L = (M-1)-1.

t

Therefore M

-1 .

L

= A, whence L

= AL = MM- 1 . L = MA = M.

We shall now show that every two-sided ideal N of A is a product of prime ideals (an "empty" product, if N = A.) If the result is false, let N be a largest counterexample. Surely N is not prime, and so there exists a prime ideal \.l3 with N < \.l3 < A, and hence

N c N\.l3-1 cA. If N = N\.l3-l, then \.l3-1 c O.(N) = A, which is impossible by (23.4). Thus N < N\.l3 - 1, whence N\.l3 -1 is a product of prime ideals \.l3 . . . \.l3r. This 1 gives N = \.l31·· ·\.l3r \.l3, a contradiction. We have thus established that every two-sided ideal of A is a product of prime ideals. Now let \.l3, \.l3' be distinct prime ideals of A. Then

\.l3-1\.l3'\.l3

c

\.l3-1A\.l3 = A,

and Thus \.l3' contains the product \.l3(\.l3-1\.l3'\.l3) ofa pair of two-sided ideals of A, and consequently \.l3' :=J \.l3 -1 \.l3'\.l3. Therefore \.l3\.l3' :=J \.l3'\.l3. Since the reverse

(23.8)

ALTERNATE APPROACH TO GLOBAL IDEAL THEORY

207

inclusion follows by symmetry, we obtain \13\13' = \13'\13, as desired. It remains to prove uniqueness (up to order of occurrence of the factors), so let (23.8) where the \13's and Q's are prime ideals. Then \13 1 ::::J nQ; implies that \13 1 = .ok for some k. Multiplying (23.8) by \13~ 1 and repeating the argument, we see that the {\13;} coincide wIth the {.oj}, apart from order of occurrence. This establishes the theorem. It is easily deduced from the above that the set of two-sided A-ideals in A is the free abelian group generated by the prime ideals of A. We leave the obvious proofs to the reader. Remark. The method of proof of (23.5) can be applied to more general situations, and leads to the theory of Asano orders in arbitrary rings. As in Jacobson [1, Ch. 6], one first generalizes the concepts of orders and fractional ideals. Then one defines an Asano order to be an order whose fractional twosided ideals form a multiplicative group. For the theory of Asano orders, the reader may consult Jacobson [1] as basic reference. Further developments are given in Asano [1,2], Harada [6], Michler [1], and Robson [1,2]'

(23.9)

COROLLARY.

If M is a proper left ideal of the maximal order A, then

M- 1 > A.

Proof Let MeN c A, where N is a maximal left ideal of A. Then l'vr 1 ::::J N- 1 ::::J A, and so it suffices to prove that N- 1 > A. Let us set

\13 = ann/\. A/N. As in the proof of (22.15), we find that \13 is a prime ideal of A, and AjN is a simple left module over the simple ring A = A/\13. Hence N is the inverse image, under the map A -+ A, of some maximal left ideal of A. Therefore (see Exercise 7.12) we may write N = A(1 - e) + \13, where e E A is such that its image e is a primitive idempotent in A. Set L = eA + \13, a right ideal in A. Since (1 - e)e E \13, it follows that N· L c \13. On the other hand, N· L is a two-sided ideal of A containing \13 2 • It follows from (23.6) that NLis either \13 or \13 2 • If NL = \13 2 , then \13e c \13 2 ; multiplying by \13 -1, we deduce that e E \13, which is impossible. This shows that NL = \13. Now choose a nonzero rt. E R !l \13, and write rt.A as a product of prime ideals of A. Since rt.A c \13, one of the factors must equal \13, and so we may write rt.A = \13 Q= NL.Q,

208

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

= A, then == LA c rxA· S) - 1 =

(23.10)

where Q is some two-sided ideal of A If N- 1 rx - 1 LS)

c

==

N-1 = A

LS)

c

rxA

~

==Lc~.

But L = eA + \.l3, so the last incl usion is impossible. This shows that N - 1 > A, and completes the proof of the theorem. Weare finally ready to prove (23.10) THEOREM. Let M be afull R-lattice in A. Then 0/ (M) is a maximal order

if and only if O,(M) is a maximal order. Proof Let A = O/(M) be maximal. Replacing M by rxM if need be, where rx E R,rx '10, we may hereafter assume that MeA When M = A, then O/M) = A, so we may restrict ourselves to the case where M < A. If the theorem is false, let M be a maximal counterexample, and choose a left ideal L of A such that M < LeA,

LIM = simple left A-module.

(Possibly L = A) Then O/(L) = A, and since M is assumed to be a maximal counterexample, it follows that O/L) = N is a maximal order. Then L· L- 1 = A by (23.5). But the analogue of(23.5), obtained by reversing "left" and "right", is also true. This analogue yields the equality L- 1 . L = N, since O,(L) is a maximal order. Each left ideal X of A' maps onto a A-submodule LX of L, and the correspondence X +-. LX is one-to-one, since X = L- 1 . LX. Set N = L- 1 M; then

It follows that N is a maximal left ideal of A', since M is a maximal A-sub-

module of L. Consequently we obtain N·N- 1 = A',

by (23.9) and (23.5). Next, for x E A we have

== LNx c LN == xEO,(M), == L-1Mx c L-1M == xEO/N).

Nx c N Mx eM

Therefore O,(N) = 0r(M). Since M is assumed to be a counterexample to the theorem, it follows that O,(N) is not a maximal order. Thus O,(N) < A" for some maximal order A". We observe next that NAil N- 1 is an order, since

(23.11)

ALTERNATE APPROACH TO GLOBAL IDEAL THEORY NA"N- 1 'NA"N-

1

c: NA"'O/N)'A"N-

1

c: NA"N-

209 1

.

But also NA"N- 1

::l

N· N- 1

=

N,

and hence NAil N- 1 = N, since N is a maximal order. Consequently there is a chain ofleft N-ideals N c: NAil c: NA"N- 1 = A'.

Since N is a maximal left ideal in N, it follows that NAil is either N or N. If = N, then A" c: O/N), which is impossible. Hence NAil = N, and thus

NAil

A' = NA"· N- 1 = A ' N- 1 . Therefore N-1 = N, which is also impossible. This proves that there can be no counterexample to the theorem, and so the result is established. We have now shown, without the use of local results, that for a normal ideal M, both 01(M) and Oy(M) are maximal orders. We have also shown that

M· M- 1 = O,(M),

M- 1 . M

= O/M)

for any normal idealM,and furthermore, M c: O,(M) ifand only if M c: O/M). The proofs of Theorems 22.18 and 22.19 carryover unchanged, since they depend only upon the above-stated results, rather than any localization arguments. We are left with the task of proving (22.17) without using results from Chapter V. (23.11) THEOREM. Let M = M12 be a normal ideal. Then M is a maximal left ideal in A1 if and only if M is a maximal right ideal in A2. Proof Let M bea maximal left ideal in A 1. Then M c: A2 , by (23.2). If M = A2 ' then A1 = O,(M) = A2 = M, which is impossible. Suppose that M is not a maximal right ideal of A 2 . Then there exists a normal ideal N = N 32 such

that

M12 <

N32

< A2 •

Surely N 32 < A 3 , since otherwise A2 = A3 = N.1t follows from (22.19) that we may write M 12 = B 13 . N 32 for some integral ideal B 13' The product B . N is a proper product, whence by Exercise 22.2 we obtain M = B . N < B . A3 = B c: A 1 . Therefore B = A1 ' and so A3 = 0r(B) = Ai' which implies that M = B . N = A3 . N 32 = N. This is a contradiction and the theorem is proved. At this point we have established all of the major results of §22 without the use oflocalization arguments. We shall not proceed further with the program

210

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(24.1J

of obtaining global results directly however, and merely observe that the remaining theorems in §22 can be derived readily from the results so far established in this section.

24. NORMS OF IDEALS Throughout this section let A be a separable K-algebra. As in the preceding sections, M 12 denotes an ideal in A with left order A1 and right order A 2 . We shall define norms and reduced norms of normal ideals of A, so as to generalize the concept of relative norm in algebraic number theory described in (4.31 iii). To begin with, let M = M 12 be an integral ideal of A, and define its norm by (24.1 )

When there is no danger of confusion, we shall omit the subscripts A/K and R. Since M12 is a full R-lattice in its left order Ai' the quotient Al /M12 is a finitely generated torsion R-module. Thus ord Al / M is a nonzero ideal of R, and equals R if and only if M = AI' We shall soon remove the asymmetry in the definition of norm, and we begin with some easy results. (24.2) THEOREM.

(i) Let M be an integral ideal of A. Then N(Mp) = {N(M)}p

for each prime ideal P of R. (ii) For each maximal R-order A in A, N(Aa) Proof. (i) Let A

= R· N a =

a EA n u(A).

N(aA),

= O/(M), so Ap = O/(Mp ) by (8.5). Therefore

N(Mp) = ordRpAp/Mp = Rp@R ordRA/M = {N(M))}p,

using (4.20 ii). (ii) We have N(Aa) = ord A/Aa = R· Na,

by Exercise 10.7. In order to compute N(aA), we se.t A'

=

O/(aA)

= aAa- I .

Then N(aA) = ordA'/aA = ordaAa- 1 /aA.

However, there is an R-isomorphism aAa- 1 faA ~ A/ Aa,

given by),

-+

a- 1 ),a,

),

E

aAa- 1.

(24.3)

NORMS OF IDEALS

Thus by Exercise 10.7, ordA'/aA

211

= ordAjAa = R·Na.

This completes the proof. Using this result, we obtain a number of important consequences. (24.3)

THEOREM.

Let M 12 be an integral ideal of A. Then

ord R A 1/M 12 = ordR A2 /M 12 , so that the norm of M can be computed by using either the left order of M or the right order of M. Proof. In order to prove the equality of a pair of order ideals, it suffices to

establish the equality after replacing R by each of its localizations R p , with P an arbitrary prime ideal of R (see Exercise 3.2). Changing notation,

assume now that R is a discrete valuation ring. By (18.10) we may write M = A1 a for some a E u(A), and then A2 = O.(M) = a- 1A1 a. The formula, which we are trying to prove, thus becomes ord A1 /A1 a = ord a- 1A1 a/A 1 a. But we have already established this result in the proof of (24.2), and so the theorem is proved. (24.4)

THEOREM.

Let L· M be a proper product of integral ideals. Then N(L· M)

= N(L)· N(M).

Proof. Let L = L 12 , M = M 23. By (4.17), the exact sequence of left A 1modules 0----> L/LM ----> A1 /LM ----> A1 /L ----> 0 gives rise to the equality N(LM) = N(L)· ord L/ LM.

Thus it suffices to show that ord L/LM = N(M). As in the proof of (24.2), it is enough to prove this when R is a discrete valuation ring. But then L = xA2 for some x E u(A), and so L/LM = xAz/xM c::: Az/M,

where the indicated isomorphism is an R-isomorphism. This implies at once that ord L/ LM = N(M), and the theorem is proved. We are now in a position to extend our definition of norm from integral

212

(24.5)

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

ideals to normal ideals. For each normal ideal I, choose a nonzero rx such that rxl is integral, and let N*(J) = rx-"N(rxJ),

where

n

E

R

= (A: K).

Let us show at once that N*(J) does not depend on the choice of rx. If also [31 c A = OP), then rx[31 = rxA' [31 = [3A' rxl, and both of the indicated products are proper products. Hence N(rxA) N([3J) = N(f3A) N(rxJ)

by (24.4). But N(rxA) = R . rt,", N([3A)

= R' [3n, whence

rx-nN(rxJ) = r"N([3J),

as claimed. In particular, if I is an integral ideal, we may choose thus N*(J) = N(J). (24.5)

THEOREM.

rt,

=

1, and

The above-dejined norm N* satisjies

N*(I' 1') = N*(J)' N*(I') for each proper product 1· I' of normal ideals. Further

N*(F 1) = N*(J)-

1.

Proof Choose a nonzero rx E R such that both rxl and rxl' are integral. Then rxl' rxl' is a proper product of integral ideals, and so N(rxJ) N(rxl') = N(rx z II') = rt, Zn N*(II').

This implies at once that N*(IJ) = N*(J) N*(I'). Further, if A = OP) then 1· F 1 = A, whence R = N(A) = N*(J)· N*(F 1). This completes the proof of the theorem. Hereafter we shall write N instead of N*, so we have now defined the norm N(J) of each normal ideal I of A. If 11 is a two-sided A1-ideal in A, where A1 is a maximal order, then for each normal ideal L = L12 ' the proper product L - 111 L is a two-sided A z -ideal l z . We have called 11 and l z similar (see § 22). It follows at once that N(l z ) = N(L -1 11 L) = N(1 1 ),

so similar ideals have the same norm. Let '1J be any prime ideal of the maximal order A, and set P = R n '-l3, R = R/ P. We define the inertial degree of'-l3 as

213

NORMS OF IDEAlS

(24.6)

f = f('+1, AIR) = (A/'+1: R). Since A/'+1 ~

R(f)

as R-modules, we obtain

N('+1) = ord R A/'+1 = pr.

(24.6)

Define the capacity

K

of '+1 by the condition

A/'+1

~

M K( S),

S = skewfield.

(24.7) THEOREM. Similar prime ideals have the same inertial degree, capacity,

and norm. Proof. Let '+1 1 be a prime ideal ofthe maximal order A 1 , and let '+1 2 = L- 1'+1 1L be a similar prime ideal in A 2 , where L = L12 is a normal ideal. Clearly '+1 1 n R = '+1 2 n R = P, and by the proof of (22.4) we have A;/'+1i ~ (Ai )p/('+1 i )p,

i

= 1,2.

To prove the theorem, it suffices to show that there is an isomorphism of RI P-algebras A1 1'+11 ~ A2 1'+1 2 , under the assumption that R is a discrete valuation ring. But in this case, L = A1 x for some x E u(A), so

A2/'+12 = x-tAl XIX- 1 '+1 1 X ~ Ad'+1 1 · This completes the proof. (24.8) COROLLARY. Let M be a maximal integral ideal, belonging to a prime

ideal of inertial degree f and capacity N(M) = pfI\

K.

Then

where

P = R n M.

Proof. Let A = 0, (M), and let M belong to the prime ideal '+1 of A. Then Af'+1 ~ (AI M)("l as left A-modules, whence pI

= N('+1) = ord R Af'+1 = {ord R AI My = N(M)".

This proves the theorem. It should be pointed out that if A' = Or (M), then M also belongs to a prime ideal '+1' of A'. However, '+1 and '+1' are similar by (24.22), and they have the same P, f, K by (24.7), so the assertion of the theorem holds equally well when we work with '+1' rather than '+1. We shall next give another interpretation of norms, which relates norms of ideals with norms of elements. (24.9) THEOREM. Let J be a normal ideal of A, and iet N denote N A/K' Then

N(J) = R-ideal generated by {N(x): x

E

J}.

214

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(24.10)

Proof. It suffices to prove the result when R is a discrete valuation ring, and an obvious argument shows that we may restrict ourselves to the case where I is an integral ideaL Then I = Aa for some a E u(A), where A = 0, (1), and so N(1) = R' Na by Exercise 10.7. On the other hand.

L

L

R· N(x) =

XEJ

R· N(),) N(a) = R' N(a),

lEA

since N(),) E R for each ), EA. This completes the proof. (24.10) COROLLARY. Let K be a finite separable extension of the field E, and let R be the integral closure in K of the Dedekind domain S with quotient field E. Then N K/E(N A/K1) = NA/El for each normal ideal 10f A. Proof. Let Ibea normal ideal of A, and set A = 0,(1). Then A is a maximal S-order in A, as well as a maximal R-order in A. To prove the stated equality of S-ideals in E, we need only prove the result when S is replaced by its localizations at prime ideals of S. Changing notation, we assume that S is a discrete valuation ring. Then I = Ax for some XE u(A), and so NA/El = S' NA/E X,

But by (24.2), for each nonzero

0: E

NA/Kl = R·NA/KX.

R we have

NK/E(Rex) = S·NK/Eo:·

Therefore N K/E(N A/K 1) = S'NK/E(NA/KX) = S'NA/EX

by Exercise 1.5. This completes the proof. We shall now define reduced norms of normal ideals, and in order to avoid cumbersome notation, we shall restrict our attention to the central simple case. For the remainder of this section, let A denote a central simple K-algebra, and let (A,' K) = n 2 hereafter. Denote by "nr" the reduced norm map nr: A -+ K, so by (9.7) we have N A/K a

=

(nr a)",

aEA.

(24.11) THEOREM. For each normal ideal I in the central simple K-algebra A, the norm N(1) is the n-th power of an R-ideal nr I, called the reduced norm of 1. For each proper product 1· J' of normal ideals in A, we have nr (1'1') = (nr 1) (nr I').

(24.12)

NORMS OF IDEALS

215

Proof. To prove that N(1) is an nth power, it suffices to show that for each prime ideal P of R, N(lp) is an nth power. Changing notation, we may assume that R is a discrete valuation ring. But then I = Aa for some a E' u(A), where A = 01 (1), and thus N(1) = R·NA/Ka = (R·nra)n.

Finally, N is multiplicative on proper products by (24.5), whence so is "nr", This completes the proof. As a direct consequence of (24.9) and (24.11), we obtain (24.12) COROLLARY. For each normal ideal I in A, nr 1= R-ideal generated by {nr x:x E'l}. We shall now establish the following important result: (24.13) T'-eorem. Let M be a maximal integral ideal of the .central simple K-algebra A, and set P = M n R, R = R/P. If R is a finite field, then N(M) = pn,

nr M = P, Proof. Let A = By (24.8)

01 (M),

and let

~

be the prime ideal of A to which M belongs.

nr M = pfl Kn , where f is the inertial degree of ~ and K is the capacity of ~. Letting denote the P-adic completion of R, we have

A/$ ~

R

Ap /~p ~ A/~

by (18.2) and (22.4). Further, A is a central simple K-algebra and (A: K) = (A: K) = n2 . Thus we can compute f, K, n by working with the case where the underlying ring is R, rather than R. Let us write A ~ Mr (D), where D is a skew field with center K and index m. Then A ~ Mr (~), where ~ is the maximal R-order in D, and we have

A/~ ~ Mr (~/rad ~). Further, ~/rad ~ is a skewfield of dimension mover R, by (14.3). Hence we obtain K =

This gives and completes the proof.

r,

216

(24.14)

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(24.14) C<~roUary. Let M be an integral ideal of the central simple K-algebra A, and assume that RIP is finite for each maximal ideal P of R. Then nr M is the product of the R-annihilators of the A-composition factors of the left A-module AIM, where A = 01(M). Proof. Let {S;} be the set of A-composition factors of AIM, and write Si ~ A/Mi , with Mi a maximal left ideal of A. Then ann R Si = Mi n R = Pi (say), and N(M) = P7 by (24.13). Hence by (4.17) we have

TI ord R Si = TI N(Mi) = (TI Pi)". = (nr M)", we obtain nr M = TI Pi' as desired.

N(M) Since N(M)

= ord R AI M =

EXERCISES In Exercises 1-3, A is a separable K-algebra, M is a normal ideal in A, and A

=

O,(M).

1. Let R be the P-adic completion of R, where P is a prime ideal of R. Show that R ® R N(M) = N(M). [Hint: It suffices to prove the result when M is integral. But

R ®RordRA/M

=

ord R A.llQ

by Exercise 4.4.]

2. If M is integral, prove that ord R A/M [Hint: After localizing, write M

=

=

ordRM-1/A.

Ax. Then

M-I/A

=

x-lA/A ~ A/Ax.

Now use (24.2).] 3. Let M be integral, and let K

=

Q, R

N(M)

=

=

Z. Prove that

(card AIM)' Z.

Call card AIM the counting norm or absolute norm of M. 4. Let A be a maximal order in a central simple K-algebra A, and let J be a proper two-sided ideal of A. Defined a J-adic valuation w on A as follows: for a E A, a =I 0, let w(a) be the least integer such that A·a·A

=

r(Q)BC-l,

where Band C are two-sided ideals of A such that B B. Set w(O) = + 00. Prove that

w(a

+ b)

~ min

(w(a), w(b)),

w(ab)

~

+C

w(a)

=

A and J does not divide

+ w(b),

a, b

E

A.

Now give A the J-adic topology, in which two elements a, b E A are near each other if w(a - b) is large. Let AJ denote the J-adic completion of A, and AJ the completion of A. Show that each element of AJ is representable by a sequence {aJ from A, with

(25.1)

an+ 1 == an (mod J"), Let J

217

DIFFERENT, DISCRIMINANT

=

n

=

1,2, ...

IT \lJ:', where the {\lJi} are distinct prime ideals

of A, and each ei

;.'

l.

i= 1

Using Exercise 22.3 (iv), show that

Prove further that if Pi

=

R n \lJi' then

AIPi ~ Ap , ~ Rpi ®R A. Calculate the completion

Aui\' where a

is any ideal of R.

5. Let A be a maximal order in a central simple K-algebra A, and let M be a left ideal in A, and J a two-sided ideal in A. Call M and J relatively prime if M + J = A. Prove that the following are equivalent: (i) M and J are relatively prime. (ii) For ea ch prime \lJ of A dividing J, M'jJ = AlP' (iii) For each prime P of R, either J p = Ap or Mp = Ap. (iv) NAIKM + NA1KJ = R. (See also EX.ercise 27.S.)

25. DIFFERENT, DISCRIMINANT

Through this section let A be a maximal R-order in the separable K -algebra A, and let tr: A ~ K be the reduced trace map. By "ideal" we mean, as usual, a full R-Iattice in A. Given a two-sided A-ideal J in A, define its complementary ideal as j

= {xEA: trxJ

c

R}.

Then j is also a two-sided A-ideal in A. In particular, we call different of A, and define the different of A as !'J = !'J(A/R) =

A the

inverse

A-1,

Since A ::::> A, it is clear that !'J is a two-sided ideal of A As in the proofs of (4.35) and (20.1), it is easily verified that the process of forming complementary ideals commutes with localization or completion at prime ideals of R. Furthermore, as explained in Exercise 25.1, the calculation of !'J(A/R) can always be reduced to the central simple case. (25.1)

THEOREM.

Keeping the above notation, we have j

=kr1.

Proof It suffices to prove the result locally, when R is a discrete valuation

218

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(25.2)

ring. But then I = yA for some y E u(A), and so tr xl = tr xyA. Thus x E j if and only if xy E A. Therefore j = Ay-1 = krl, which completes the proof. The discriminant d(AIR) is the ideal of R generated by {det(trx i x j )1";;i,j",": x 1 ,···,Xn EA}, where n = (A : K). As in Exercises 4.13 and 10,6, forming discriminants commutes with localization or completion at a prime ideal of R. (25.2)

THEOREM,

d(AjR) = N A/K(:D(AjR)).

Proof. It suffices to prove the result locally, since the norm behaves properly under localization by (24.2). We have N(:D) = ordR Aj:D = ordR A/A

by Exercise 24,2. Now let A =

L'" RXi ; as

in the proof of (4.35), we have

i= 1

d(AjR)

= R 'det(tr X,X')1<' 1

)

-...:;:1,

'<

)~n

.

On the other hand, choose a dual basis {y. } of A so that tr x. y. = i5 ... Then - ' . J '} 'J A= RYj' and If we set Xi = C1. ij Yj' C1. ij E K, then

L

L

ordR AI A = R . det (C1. i J by Exercise 4.2. But (25.3)

THEOREM.

C1. ij

= tr Xi x j ' so the proof is completed.

The discriminant d(AjR) is independent of the choice of the

maximal order A. Proof. Let r be another maximal R-order in A, and choose a normal ideal L in A with left order r and right order A Then r = LAL - 1, and :D(r) = L· :D(A)' L - 1. Taking norms, we obtain d(r) = d(A), as claimed.

We next prove the non-commutative analogue of (4.37). ~ be a prime ideal of the maximal R-order A in a separable K-algebra A. Set P = ~ n R, R = RIP, and write PA = ~e,Q, where e ): 1 and Q is a two-sided ideal of A not divisible by ~. TIJen ~e-1 I :D(AjR). If e = 1, then ~ I :D(AjR) if and only if the center of the simple R-algebra A/~ is inseparable over It

(25.4). Theorem. Let

Proof Step 1. We show first that the pro'of can be reduced to the case where

(25.5)

219

DIFFERENT, DISCRIMINANT

A is simple, and R is a complete discrete valuation ring. Indeed, starting with the general case, let R be the P-adic completion of R. If we replace each ideal

by its P-adic completion, then the hypotheses are unchanged. (In this connection, note that from the relation ~ + Q = A it follows that ~ + Q = A, so ~ does not divide Q.) Since n(AIR) is the P-adic completion of n(AIR), we know that ~kln(AlR) if and only if ~kln(A/R). Hence it suffices to prove the desired result for the case where R is complete. Changing notation, we assume for the remainder of this proof that R t

is a complete discrete valuation ring. If we write A as a direct sum

r

L . Ai of i= 1

simple components, then correspondingly A = Ai' with Ai a maximal R -order in Ai' Renumbering the {AJ if need be, we may take ~

where form

~1 =

=

~1 EEl

A2 EEl ... EEl At'

rad A1 is the unique prime ideal of A1 . Then 0 must have the

0= A1 EEl O 2 EEl ... EEl 0t' OJ = two-sided ideal in Aj' and likewise there is a decomposition Then we have PAl = ~~, and ~kl n(AlR) if and only if ~; I n 1 . Changing notation once more, we may thus assume for the remainder of the proof that A is a simple separable K-algebra and that 0 = A, that is, PA = ~e. Step 2. We now show that if PA = ~e then ~e - 11 n, where n = n(AIR). Consider the R-algebra l\. = AlPA; since an R-basis of A maps onto an R-basis of X, we may conclude (as in the proof of formula (4.38)) that

aEA,

(25.5)

where bars denote reduction mod P. For each a E ~ we have ae = 0 in X, and so char. pol. a is a power of X. But then by (25.5), char. pol. a (taken mod P) is a power of X. Since char. pol. a is a power of red. char. pol. a by (9.24), we conclude that also red, char. pol. a (taken mod P) is a power of X. But we have red. char. pol. a = xn - (tr a) X n- 1 + ... + (-1)" nr a, and so we have now shown that

tr A / K Therefore ~ c p.

~ c P.

n-\ whence ~e-1In as claimed.

220

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(25.6)

Step 3. For the rest of the proof let e = 1, so that PA = \.l5 is the prime ideal

of A. Let A = M, (D), where D is a skewfield with center L, and let S be the integral closure of R in L. We set (D:L) = n 2 , (L:K) = k. Then A is also a maximal S-order in A, and by (17.4) we may identify A with a full matrix ring M, (~) over the unique maximal S-order ~ in D. From (17.5) and the formula \.l5 = PA = M,(P~) = M,(PS'~), it follows that PS is the prime ideal of S, and P~ is the prime ideal of ~. Therefore .1 = MP~ is a skewfield; since A/\.l5 ~ M,(.1), the skewfield ~ has the same center C as does the R-algebra A/\.l5. Furthermore, !)(A/ R) = M, (!)(~/ R))

by (20.2). Thus we need to prove that P~ I!)(~/R) if and only if C is inseparable over R, so we have now reduced the problem to the case of skewfields. Step 4. Keeping the notation of the preceding step, let a E ~ map onto

a E~,

and let f(X) = red. char. pol. D/K a,

m(X) = min. pol. R a.

Then f(X) has degree nk, whereas degree ofm(X) = (R(a):R) = d, say. Since ~ is a skewfield, m(X) must be irreducible over R, and hence char. pol. X/R a is a power of m(X). It then follows from (25.5) that J(X) is a power of m(X), whence dink. Furthermore, we obtain (25.6)

frD/K a

= (nk/d)· T(a),

a E~,

where T is the ordinary trace from the field R(a) to R. We shall now show that if C (= the center of~) is inseparable over R, then trD/Ka = 0 for each aE~. This will imply that trD/K~ c P, whence p-l~ c !)(~/R)-l, and so P~I!)(~/R) as asserted. Let p = charR, so pEP. If a is inseparable over R then T(a) = 0, so by (25.6) we obtain fr a = 0 as desired. On the other hand, suppose that ais separable over R. Since Cis inseparable over R, there exists an element b E ~ such that 5 E C and 0' is inseparable over R(a). Then R(a, 5) is a finite algebraic extension of R, so ·by the "Theorem of the primitive element" (see van der Waerden [1]), there exists an element C E ~ such that R(a, 0) = R(c). Therefore (R(c): R) = (R(a, 5): R(a)) (R(a): R) = (pSd')' d

for some s ~ I, where (R(a, 5): R(a)) = pS d' and where d = (R(a): R) as before. However, (R(c):R) is the degree of min. pol.jiC, and thus divides nk by the remarks in the preceding paragraph. This proves that pS. dink, and so

(25.6)

DIFFERENT, DISCRIMINANT

221

pSI(nkld). Therefore frDlKa = 0 whenever a is separable over R. We have thus established that PI). I'!l(I).IR) whenever C is inseparable over R. Step 5. Suppose finally that C is separable over R, and that e = 1 as before. Since PS is the prime ideal of S, it follows that S = SIPS is a field, and (S:R) = (L:K) = k by (5.6). ~

D-I).

1

m

'

C

It'

t

'1-1

K-R

Further, (.~:S) = (I).IPI).:S) Set (~: C) = m 2 ; then

=

(D:L)

n2 =

R

=

(~:S)

n

2

.

= m2 (C:S),

so (C: S) = t 2 for some integer t. We show next that C = S, that is, t = 1. By (7.15) there exists a maximal sub field E of the skew field ~, with the property that E is a separable extension of C. Then E is a separable field extension of R, and so E = R(c) for some eEl).. Since (E: C) = m by (7.15), we obtain (R(c):R) = (E:C) (C:S)(S:R) = mt 2 k. But on the other hand (R(c): R) is a divisor of nk, by the first paragraph of Step 4. Thus mt 2 k Imtk, whence t = 1 and C = S. It now follows that for each a E 1)., both of the polynomials red. char. pol. DIK a, red. char. po1.lolR a have degree nk. Both of them are powers of the irreducible polynomial min. pol.:R a, and hence they are equal. This shows that frDIKa = trlolRa,

aEI)..

But by hypothesis X is a separable R-algebra, and so there exists an element a E I). such that trX/.R a i= O. This shows that trDIl( I). P, and hence it follows that PI). '!l(1).1 R). This completes the proof of the theorem.

cf

t

If A is a central simple K-algebra, and (A: K) = n 2 , then we have d(AjR)

=

NAIl( ('!l(AjR))

=

(nr '!l)".

222

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(25.7)

We call nr '!l the ground ideal of A. It is an ideal of R, and is independent of the choice of the maximal order A. For each prime ideal P of R, the P-adic completion ~ is isomorphic to a full matrix algebra MKP(S), where S is a skewfield with center Kp and index mp. We shall call mp the local index of A at P, and Kp the local capacity of A at P. Now suppose that K is a global field. Then of course each residue class field R/P is a finite field, and has no inseparable extensions. Hence for each prime ideal ~ of the maximal order A, we have ~I '!lCAjR)

if and only if

~21 PA,

where

P = ~

(1

R.

In fact, we have the more precise statement: (25.7) T~eorem. Let A be a central simple K -algebra, where K is a global field, and let A be a maximal R-order in A. For each P, let mp be the local index of A at P, and let ~ be the prime ideal of A corresponding to P. Then mp = 1 a.e., and (25.8)

'!l(AjR)

IT ~mp

=

-1.

p

Proof. Since K is a global field, the first of the above equalities follows from (22.14). As remarked at the start of this section, for each P we have

(25.9)

"Y here R, A denote P-adic completions. The left hand expression equals ~mp- 1 by (20.3), since K is a global field. On the other hand, '!l(AjR) is an Ideal in A, so '!lp = Ap a.e., whence '!l(A/R) = A a.e. by (25.9). This shows that mp = 1 a.e. It also establishes the second equality in (25.8) since that equality holds at each P-adic completion (see (5.2)). W,e may remark that in the above situation, the power of ~ in '!l(AjR) is precIsely ~e- \ where PA = ~e. Thus (25.4) is in a sense best possible. (25.10)

COROLLARY.

Keeping the above notation, we have

nr '!l(AjR) =

IT p(mp

-l)Kp ,

d(AjR)

p

= {nr '!l(AjRW

where n 2 = (A: K), and Kp denotes the local capacity of A at P. Proof. Let P be any prime ideal of R, ~ the corresponding prime ideal of A and M a maximal left ideal of A belonging to ~. Then nr M = P by (24.l3): On the other hand, the proof of (24.8) shows that N(~)

= N(Mt

p

•

223

EXERCISES

Since the ordinary norm is the nth power of the reduced norm, this gives (25.11) The formulas in (25.10) for the ground ideal nr Tl, and for the discriminant d(A/R), now follow directly from (25.8) by taking reduced norms.

EXERCISES Let A be a maximal R-order in the separable K-algebra A. 1. Using the notation of (10.5), show that

'j)(AjR)

=

I" 'j)(AJR),

and that 'j)(AJR)

=

'j)(AJR i ) ' 'j)(RJR).

Here, 'j)(RJR) is the different defined as in §4d. [Hint: Imitate the proof of Exercise 4.11, using (9.15).J 1a. Keeping the above notation, show that d(A/R) =

TI d(AJR),

and that d(AJR) = {IVK,/K deA, /R i )} {d(RJR)}(Ai:Kf),

where d(Ri / R) is the discriminant of Ri with respect to R (see §4d). [Hint: By (25.2) and Exercise 1,

Further, d(AJR)

= IV A
By Exercise 1.5, IV AdK 'j)(AJR i )

= IV K,/K {IV A,/K, 'j)(AJR i )} = IV K,/K

d(AjRJ

Likewise

This yields the desired formulas for d(AjR) and deA, /R).J 2. Let M = M 12 be a normal ideal of A, and define its complementary ideal by

fir

= {xEA: trxM c R}.

Show that l~1 = (Alb is also a normal ideal of A, and that 4.12). Prove further that

M= M (compare Exercise

224

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(26.1)

[Remark: Manipulations of this type can be used to give another proof of the fact that 01(M) is maximal if and only if O,(M) is maximal; see Deuring [1, p. 84].J 3. An clement a E u(A) is called an R-unit if both a and a- 1 are integral over R. Prove that if a is integral over R, then a is an R-unit if and only if N A/K a E u(R). [Hint: Let a be integral over R, and let II. be a maximal R-order in A containing R[ a]. If Na E u(R), then ordR 1I./lI.a =.R, so II. = lI.a. Therefore 1 = xa for some x Ell., and clearly x = a- 1.] 4. Let r c r' be R-orders in A, and let x E r. Show that x E u(r) if and only if x E u(r'). [Remark: The result is clear from Exercise 3. Alternate proof (A. Dress): if x Ern u(r'), then x -1 is integral over R, whence x -1 E R [x J c r; thus x E u(r).J

26. IDEAL CLASSES; JORDAN-ZASSENHAUS THEOREM Throughout this section let R denote a Dedekind domain whose quotient field K is agIo bal field. As shown in the references listed in § 4, the ring R/ Ra is finite for each nonzero a E R. Let us set lal cardR/Ra, a"# 0; 101 = O. Let A be an R-order in a K-algebra A. By a left A-ideal in A we shall mean, as usual, a left A-lattice M in A such that KM = A. We may partition the set ofleft A-ideals in A into ideal classes, by placing two such ideals M, N in the same class if M ~ N as left A-modules. Each such isomorphism extends to a left A-isomorphism KM ~ KN, hence is given by right multiplication by some x E u(A). Thus the ideal class containing M consists of all ideals {Mx : x E u(A)}. We shall show that under suitable hypotheses, the number h(A) of ideal classes of A is finite. This result will be a special case of the more general 10rdan-Zassenhaus Theorem proved below, and will be valid for all orders, not just for maximal orders. As we shall see, the general case reduces quickly to the case of ideals. We shall follow the treatment in Swan-Evans [1]; another approach is given in Curtis-Reiner [1 J. Given an R-order A, we shall say that the Jordan-Zassenhaus condition JZ(A) holds true if for each positive integer t, there are only finitely many A-isomorphism classes of left A-lattices of R-rank at most t. We intend to prove that JZ holds for all R-orders in semisimple K-algebras. Exercise 26.6 shows that the restriction to semis imp Ie algebras cannot be omitted. Our first lemma shows that in order to test whether JZ(A) holds true, it suffices to test for JZ(Ll), with Ll ranging over orders in skewfields. (26.1) LEMMA. Let A be an R-order in the semisimple K-algebra A, let Di be the skewjield part of the i-th simple component of A, and let Ll; be an R-order in D,. If JZ(Ll,) holds for each i, then JZ(A) also holds true.

Proof. To begin with, let

r

be another R-order in A such that

rcA Any

(26.1)

IDEAL CLASSES; JORDAN-ZASSENHAUS THEOREM

225

two left A-lattices M, N are also left r-Iattices, and

K ® R Hom r (M, N)

~

HomA(KM, KN)

~

K ® R HomA (M, N).

Thus M ~ N as A-lattices if and only if there exists an A-isomorphism
Clearly R-rank M(a) = L. R-rank M\a) , ,

for each rJ..

~ence the se.t of positive integers {R-rank M(a)} is bounded above if and only If for ~ach I,. the set {R-rank ~\~)} is bounded above. We shall usually abbrevIate thIS statement by omIttmg the superscript rJ., and shall say "The

226

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(26.2)

R-rank of M is bounded if and only if for each i, the R-rank of Mi is bounded". Thus JZ(A) holds true if and only if JZ(A i ) holds for each i. By § 16, Ai is Morita equivalent to Ai' and there exists a (Ai' Ai )-bimodule L such that the correspondence M -+ L ® A, M gives a one-to-one isomorphism-preserving correspondence between the set of left Ai-modules M and the set of left A/modules L ® M. This bimodule L is a projective right Ai-module, and thus if M is an R-Iattice, so is L® M. Further, rank R M is bounded if and only if rank R L ® M is bounded. It follows at once that JZ(A i ) holds true if and only if JZ(A i ) holds true. This completes the proof of the theorem. (26.2) LEMMA. Let A be an R-order in a skewjield Dover K. Then JZ(A) holds if and only if the number h(A) of ideal classes of A is jinite. Proof If JZ(A) holds, then clearly h(A) is finite. Conversely, suppose h(A) is finite. If D is separable over K, then we can show quickly that JZ(A) holds true. Namely, by the proof of (26.1), we may assume that A is a maximal R-order in D. Then A is hereditary. so every left A-lattice M is isomorphic to a direct sum ofleft ideals of A. If rank RM is bounded, so is the number of summands. Up to isomorphism, there are only h(A) possible choices for each ·summand, whence the number of non-isomorphic M's of bounded Rrank is finite. Thus JZ(A) holds in this case. Still assuming that h(A) is finite, let us drop the assumption that D is separable over K. By (10.6), each left A-lattice M may be embedded in a free A-lattice A(n), where KM = D(n), and then

rank R M = (KM:K) = n· (D:K). Let d = (D: K). We show by induction on t that there are only finitely many non-isomorphic left A-lattices of R-rank td; this of course implies that JZ(A) holds true. The result is clear for t = 1, since we have assumed that h(A) is finite. Let t > 1, and assume the result holds for lattices of R-rank (t - l)d. If M is any left A-lattice of R-rank td, then by the method of proof of (2.44), there is a A-exact sequence

o -+ N

-+

M

-+

J

-+

0,

where J is a left ideal of A, and where rankR N = (t - l)d. Up to isomorphism, there are finitely many choices for Nand J. Hence (see §2d) we need only show that for each fixed choice of Nand J, the group Ext~(J, N) is finite. This group is a finitely generated R-module by (2.34), so in view of our assumptions on R, we need only prove that it is an R-torsion module, that is, K ®R Ext~(J, N) =

o.

(26.3)

IDEAL CLASSES; JORDAN-ZASSENHAUS THEOREM

227

The left-hand expression equals Ext1 (KJ, KN) by (2.43), and hence is zero by (2.28 v) since KJ is free as left D-module. This completes the proof of the lemma.

(26.3) LEMMA. Suppose that R is either Z or a polynomial domain k[X] over a jinite jield k. Let D be a skewjield over the quotient jield K of R. Then JZ(d) holds true for every R-order d in D.

Proof By the preceding lemma, it suffices to prove the finiteness of the number of isomorphism classes of left ideals of d. Since R is a principal ideal domain, we may write d =

" Rx;, L"

;; I

XjX = L )"; IX;]! Xl' ;,1

Hence if X -=f. 0, then by Exercise 26.3 we obtain cardd/dX = INv/KXI = Idet(LA;IX;jl)l".j,l'snl, i

where for nonzero IX E R, IIXI = card R/RIX. But the indicated determinant is a homogeneous polynomial of degree n in the A's, with coefficients in R. It follows from Exercise 26.1 that there exists a positive constant c such that IAII ~ t,···,IAnl ~ t

= INv/K(LAix;)1

~ ctn.

N ow let M be any proper left ideal of d. Then diM is a finite group, since it is a finitely generated torsion R-module. Let s be the greatest integer such that s" ~ card diM, and let )01"'" An range independently over all elements A E R with IAI ~ s + 1. By Exercise 26.1 there are at least (s + 1)" such n-tuples (AI' ... , An)' and thus L )"i Xi ranges over more than card diM distinct elements of d. Hence two such sums must be congruent mod M, and therefore M contains a nonzero element X

" = ;L A;x" ;I

where A; E R, and I Ail ~ 2(s + 1) for each i. Therefore dx c M c d, and card d/dX while card diM

~ sO.

= INv/K(LIl;X;)1

~ c· 2"(s

+

1)",

Hence

card M/dx = {card d/dX}/{ card diM} ~ c· 2n(s

+ l)n/sn =

c· 2"(1

+ S-I) ~

c· 4".

228

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(26.4)

We have thus shown that every proper left ideal M of d contains a nonzero element x such that card M/dx ~ c· 4n. The same holds when M = d, just choosing x = 1. Consider now all R-modules Thaving at most c· 4n elements. Since R is a principal ideal domain, we may express T as a finite direct sum R/Rrxi where card R/Rrxi .::s c· 4n for each i. Then each rx i oF 0 since R is infinite, and so Irxil.::s c· 4n for each i. The number of possible choices for the {rx i } is therefore finite, by Exercise 26.1, and thus there exists a nonzero 13 E R such that f3T = 0 for each T. Note that 13 does not depend on T. In particular, we have 13M c dx always, and so

IO

13 . dx c 13M c dx. Therefore /3d eM, /3x- 1 c d. But 13 was chosen independently of M, and Ll//3d is finite. Hence the number of choices for M· /3x- 1 is also finite. Since M ~ M· /3x- 1 as left d-modules, it follows that the number of isomorphism classes of left ideals of d is finite, and the proof is complete. We are now ready to prove the fundamental result (26.4) T~eorem (Jordan-Zassenhaus). Let R be any Dedekind domain whose quotient field K is a global field. Then for each R-order A in a semisimple K-algebra A, and for each positive integer t, there are only finitely many isomorphism classes of left A-lattices of R-rank at most t. Proof We are trying to prove that JZ holds for R-orders in semisimple

K-algebras. We shall first establish the existence of a Dedekind domain Ro with quotient field K, such that Ro c R, and such that JZ holds for Ro-orders. When K is an algebraic number field, we choose Ro = alg. int.{ K}, the integral closure of Z in K. Since Z c R, and R is integrally closed, it follows that Ro cR. By (26.1) and (26.3), JZ holds for Z-orders. Hence it also holds for Ro-orders, by Exercise 26.4. On the other hand, suppose that K is a function field. By definition, we may write K = k(X1 ' ... , x n ) for some elements Xl"'" x n ' where k is a finite field, and where K has transcendence degree 1 over k. Each Xi is expressible as Xi = Y, /Zi with Yi' Zi E R, and therefore K = k(y l' Zl' ... , Yn , Zn)' By a theorem of F. K. Schmidt (see Zariski-Samuel [1, Th. 31, p. 105]), we may choose one of the elements Yl' Zl' ... , Yn' zn (call the chosen element X), such that k(X) is purely transcendental over k and K is a finite separable extension of k(X). Of course X E R, from the manner in which X was chosen. Furthermore, the prime field F of k lies in R, and every element of k is integral over F. Therefore R :=J k[ X], and hence also R :=J Ro' where Ro is the integral

(26.5)

IDEAL CLASSES; JORDAN-ZASSENHAUS THEOREM

229

closure of k[XJ in K. By (26.1) an<;l (26.3), JZ holds for k[X]-orders. Hence by Exercise 26.4, JZ also holds for Ro-orders. Now let A be any R-order in a semisimple K-algebra, where K is either an algebraic number field or a function field. We have shown above that R contains a Dedekind domain Ro with quotient field K, such that JZ holds for Ro-orders. By (8.8), there exists an Ro-order Ao in A such that A = RAo. Further, each left A-lattice may be expressed as RM 0' for some left Ao-lattice Mo. Clearly, the R-rank of RMo equals the Ro-rank of Mo. Since JZ(Ao) holds true, it follows at once that JZ(A) is also true. This completes the proof of the theorem. It is worth pointing out another approach to the proof of the JordanZassenhaus Theorem. Once the theorem is established for Ro-orders, it follows for arbitrary R-orders by Exercise 26.5, since it can be shown (see Swan-Evans [1, pp. 226-227J) that R = S-l Ro for some mUltiplicative subset S of Ro. Finally, keeping the notation of (26.4), let L be any finite separable extension of K, and let R' be any R-order in L. Even if R' is not a Dedekind domain, the condition JZ(A') is meaningful for R'-orders A' in semisimple L-algebras. The discussion in Exercise 26.4 remains valid, and shows that JZ(A') holds for each such A'. We shall now consider the special case

A = Q EB Qi EB Qj EB Qk,

(26.5)

A

=Z

EB Zi EB Zj EB Za, a

= (1 + i + j + k)/2,

where A is a maximal Z-order in the rational quaternion algebra A. By Exercise 25.3, u(A) consists of the 24 elements ±1,

±i,

±j,

±k, (±1 ±i ±j ±k)/2.

It follows readily that for each x E A, there exists a unit u E u(A) such that ux EZ EB Zi EB Zj EB Zk. We shall prove that h(A) = 1, and indeed that A is a euclidean domain though not commutative. Let x, YEA, Y # 0, and write '

= q + r, q = CX o + cxli + cxzj + cx 3 k, r = Po + f3li + f3zj + f3 3 k, where CX EZ, f3 v E Q. We may choose q so that each 113.1 ~ 1/2. Further, if it

xy-l

V

turns out that each If3v! = 1/2, we may change our choice of q and r so that q E A and r = O. Hence we can always pick q E A, rEA, such that nr r = f3~ + f3i + f3~ + 13; < 1. Therefore we may write x = qy + t, where q, tEA, and 0 ~ nr t < nr y.

230

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(26.6)

This shows that A has a euclidean algorithm, and consequently every left ideal in A is principal, and is generated by an element of smallest reduced norm. Therefore h(A) = 1, as claimed. For orders of class number 1, we can obtain a factorization theory for elements, rather than for ideals. Let A be any maximal order in a separable K -algebra, with h(A) = 1. Let x E A 11 u(A); we call x indecomposable if x is a non-unit of A which cannot be expressed as a product of non-units. We claim that x is indecomposable if and only if Ax is a maximal left ideal of A. If Ax is not maximal, then Ax < Am < A for some m, and then there is a factorization x = ym into non-units. The procedure can be reversed, and the claim is established. Now let x E All u(A) be arbitrary, and let Ax

= Ax 1 <

Ax2

< ... < Ax. = A

be an unrefinable chain of left ideals of A, where x = Xl and x. = 1. Then Xi = u i x i+ 1 for some U i E A, and AXi+l/Ax i ~ A/Aui ,

so each u i is indecomposable. The equation expresses x as a product of n - 1 indecomposable elements. The prime ideals of A associated with these elements are uniquely determined by x, up to similarity and order of occurrence (see (22.24)). Returning to the case of quaternions, we use the preceding discussion to prove (26.6) THEOREM. Every positive rational integer is expressible as a sum of four squares. Proof Keep the notation of (26.5), and let p be any rational prime. Then there exists a maximal left ideal M = Am such that pA c MeA. Since A is a central simple Q-algebra, it follows from (24.13) that (nr m)Z = nr M = pZ.

Therefore p = nr m, since nr m > O. Now set m = (Xo + (Xli + (X2j + (X3 k; replacing m by u E u(A), we may assume that each (Xv E Z. But then

urn

if need be, where

p = (X~ + (Xi + (X~ + (X~, a sum of four squares. Finally, let r = Il p~v be any positive rational integer, where the {pJ

231

EXERCISES

are primes, and let mv EZ EB Zi EB Zj EB Zk.

Pv = nrm.,

Then r

=

Il p~v = Il nr (m~v) =

nr

Il m~v.

This proves that r is a sum of four squares of rational integers, and the theorem is established. EXERCISES 1. Let R = Z or R = k[ XJ, where card k = q. For nonzero rx E R. let Irx I = card R/Rrx, and set 101 = O. Prove that Irxl is finite for all rxE R, and

(i) IrxPI = Irxllpl, Irx + pi ~ Irxl + IPI, rx, PER. (ii) For each positive integer n, the number of elements rx E R with I rxl ,,;: n is finite, and is at least n. [Hin t: Let deg rx denote the degree of a polynomial rx E k[ XJ, with deg 0 = - 00. Then Irx I = qde g ., which yields (i). Further, Irx I ,,;: n if and only if deg rx ,,;: logq n. The number of such rx's equals q 1 +., where r is the greatest integer such that r ,,;: logq n.J 2. Keeping the above notation, let N c M be free R-modules, where d

d

M =

L' Rmp i=

N=L"Rn., j~ I

1

}

Prove that card M/N Idet (a I. [Hint: R is a principal ideal domain. After basis changes in M and N, which leave Idet (aj)1 unchanged, we may assume that (a j ) = diag (b l ' .•. , bdl. Then i)

cardM/N = I1cardR/Rbj

=

I1lbj l

=

Idet(aij)l.

Compare with Exercise 4.2.J 3. Keep the above notation, and let A be an R-order in a K-algebra A. Then for each x E A such that x E u(A), we have cardA/Ax

=

INAIRxl,

where in this case the norm NAIR must be computed by letting x act by right multiplication on an R-basis of A. 4. Let L be a finite separable extension of K, and let R' be the integral closure of R in L. Show that if JZ(A) holds for R-orders A in semisimple K-algebras, then JZ(N) holds for R'-orders N in semisimple L-algebras. [Hint: Since R' is a finitely generated R-module, each such N is also an R-order. Further, for each N-Iattice M, rank R , M = (L: K)' rankR M.J 5. Let R' = S- I R be a ring of quotients of R. Show that if JZ holds for R-orders (in semisimple K-algebras), then JZ also holds true for R'-orders. [Hint: Let the R'order N have R' -generators {x j }, and let Xj x j = L rxijl xi' rx's in R'. Choose S E S so

232

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(27.1)

that each SCI.;jl E R, and set I\. = R + I R(sx;). Then I\. isan R-order such that R' I\. = 1\'. Likewise, if M' = Il\'mj is a I\'-Iattice, then M = Il\.mi is a I\.-Iattice such that I\'M = M', and rankR M = rank R • M'. Deduce that if JZ(I\.) holds, then SG does JZ(I\'). See also Exercise 3.5.J 6. Let A = Q EB Qv, where v 2 = 0, and let I\. = Z EB Zv be a Z-order in A. For r ;;. 1, let M, be the unitall\.-Iattice Z Z, where

+

v(a, b)

Show that if M, '" M s ' then r this case.

=

=

(0, rb),

(a, b) EM,.

s. (Consider M, jvM,.) Thus JZ(I\.) cannot hold in

7. Suppose that JZ(I\.) holds true for the R-order I\. in the semisimple K-algebra A, and let h(l\.) be the number of classes of left I\.-ideals M in A such that KM = A. Prove (i) If I\. c I\' = R-order, then h(l\.) ;;. h(I\'). (ii) If I\. and N are maximal orders, then h(l\.) = h(I\'). (iii) If I\. is maximal, then h(l\.) equals the number of classes of right I\.-ideals in A. (iv) h(l\') < h(l\.) if N is a localization or completion of I\. at a prime ideal of R. 8. Let I\. be a maximal R-order in a separable K-algebra A. For x E u(A), we call the maximal order x I\. x- 1 conjugate to 1\.. Show that if JZ(I\.) holds true, then there are finitely many conjugacy classes of maximal R-orders in A. 9. Let L = Q(.J2), R = Z[.J2], S = Z[2.J2J. Show that R = alg. into {L}, and that R, S are Z-orders in L. Prove that h(S) > h(R) = 1. [Hint: h(R) = 1 since R has a euclidean algorithm (see Exercise 30.4). Show that the ideal 2S + 2.J2S is not a principal ideal of S, so h(S) ;;. 2.J

27. GENUS Throughout this section let R denote a Dedekind domain with quotient field K. Let A be an R-order in a K-algebra A. Two left A-lattices M, N are in the same genus (notation: M v N) if for each prime ideal P of R, there is a Ap-isomorphism M p ~ N p . (27.1) THEOREM. Let M, N be left A-lattices. Then Mv N if and only iffor each nonzero ideal 1 of R, the condition

(27.2)

o -> M -> N

->

T

->

0,

1

+ ann R T

=

R,

holds for some A-exact sequence for some A-module T Proof If the condition (27.2) holds when 1 = P, then P whence Tp = 0 by (4.20). Since o ->Mp->Np-> Tp-> 0

+ ann R T =

R,

is Ap-exact, it follows that M p ~ N p. Thus if (27.2) holds when 1 ranges over the prime ideals of R, then M v N.

(27.3)

233

GENUS

Conversely, assume that M v N, and let I be a preassigned nonzero ideal of R. Let P l ' . . . , P n be the distinct prime ideals dividing I. By the method of proof of Exercise 18.3, there exists cP E Hom" (M, N) such that CPPi: M Pi ~ N P, ' 1 :0:;: ; :0:;: n. Since (ker CP)Pi = ker CPP i = 0, and ker cP is R-torsionfree, it follows that ker cP = O. Consider the A-exact sequence o -> M !4 N -> T -> 0, where T = cok cpo Then TPi = 0 for 1 ~ i :0:;: n, whence Pi each i. Therefore I + annR T = R, as desired.

+ ann R T =

R for

It should be remarked that when I = R, the preceding proof is valid if we just choose P 1 to be any prime ideal of R. (27.3) COROLLARY. Let L, M, N be left A-lattices in the same genus. Then there exists a left A-lattice L in the genus, such that M+N~L+L.

Proof By (27.1), there exist A-exact sequences

o -> M ~ L -> T -> 0, such that T, U are R-torsion A-modules for which

annR T

+ annR

U = R.

Thus for each prime ideal P of R, either Tp = 0 or Up = 0, that is, either cpp:Mp ~ Lp or l/Ip:Np ~ Lp. Now consider the A-exact sequence

o -> L -> M + N ~ L -> 0, where L is defined to be the kernel of the map (cp, l/I). The map (cp, l/I) is epic . and split, since for each P, the epimorphism (cp, l/I)p:Mp Np -> Lp is split. N ~ L ~, and it remains for us to prove that ~ v L. Therefore M Let R be the P-adic completion of R, where P is a prime ideal of R, and set Nt = R ®R M, etc. Then

+

+

+

Nt

+ N ~ f + 1'"

Nt

~

f.

Since the Krull-Schmidt Theorem holds for finitely generated A-modules by Exercise 6.6, the above implies (by Exercise 6.7) that N ~ f. Therefore N p ~ Lp by (18.2), and so Lv N. This completes the proof. (27.4) THEOREM. Let A be a maximal R-order in a separable K-algebra, and let M be any left A-ideal in A (always subject to the condition that KM = A).

234

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(27.5)

Then M is in the same genus as A. Furthermore, given any set of left A-ideals M l ' ... , Mr in A, there exists a left ideal M in A such that (27.5)

M1

-i- ... -i-

Mr ~ Alr-1)

-i-

M.

Proof For each P, M p is a principal Ap-ideal by (18.10), and hence M v A. By (27.3), we have M1 -i- M2 ~ A -i- N for some left A-lattice N such that N v A. But by (27.1) we may embed N in A, so replacing N by an isomorphic copy, we may assume that N is a left ideal of A. Continuing in this way, we find after r - 1 steps that (27.5) holds for some left ideal M of A. (27.6) COROLLARY. ~et A = Mr (D), where D is a skewjield separable over K, and let ~ be any maxImal R-order in D For each right ideal J of~, let ~' = 0 (1) and let _ I , ~

~;-1

denote the ring ofall r x r matrices (Xi), where X 11 ranges over all elements of~, ... , X1r ranges over all elements of J-1, and so on. Then A is a maximal R-order in A, and every maximal R-order is of this form, for some right ideal J of ~. !,roo.( By (21.6~, each maximal R-order A is of the form Hom,1 (L, L), where L a ng~t ~-lattIce such that KL ~ D(r). Since ~ is hereditary, L can be written as a dIrect. sum of r right ideals of ~. Hence by (27.4), we may take L = ~(r-l) + J, where J is a right ideal of ~. But we may identify IS

Hom,1(~(r-l)

-i-

J,

~(r-l)

-i-

1)

with the indicated ring of matrices, since there are identifications Hom,1 (~,~) =~, Hom,1 (~, 1) = J, Hom,1 (J,~)

= ;- 1, Hom,1 (J, 1) = K

This completes the proof. Let A be a maximal R-order in a separable K-algebra A. We shall say that the left ideals M, N of A are relatively prime if M + N = A. (27.7) COROLLARY. Let N be any left ideal of A. Then every class of left A-ideals in A contains an integral ideal relatively prime to N.

Proof Choose a nonzero

('f.

ERn N, and let M be any left ideal of A. Since

235

EXERCISES

M v A, there is an exact sequence O~M.!4A~T~O

with

+ ann R T = R, + {J, with pER, {JEann R T

rxR

by (27.1). We may write 1 = rxp Then (JA c cp(M), and rxp EN, whence cp(M) + N = A. But cp(M) = Mx for some x E u(A), and thus cp(M) lies in the ideal class of M. This completes the proof. To conclude this section, we prove (27.8) THEOREM. Let A be a maximal R-order in a separable K-algebra A, and let X be a left A-lattice such that KX ~ A(r) for some positive integer r. Then there exists a left A-ideal M in A such that X

~

A(r-l)

-i-

M

as A-modules.

Proof By hypothesis, there exists an A-isomorphism cp: KX ~ A(r). This isomorphism carries X onto a full R-Iattice cp(X) in AIr). Since AIr) is also a full R-Iattice in A(r), there exists a nonzero rx E R such that rx . cp(X) c Nr). Replacing X by the isomorphic A-lattice rx . cp(X), we may hereafter assume that Xc A(r),

KX = K· A(r) = A(r).

(The above reasoning has also been used in the proof of (10.6).) By (21.4), the maximal order A is hereditary. It follows from (2.44) and (2.45 ii) that we may write X=M 1 +.·.+Mr' where each M! is a left ideal of A. Since KX = L' KMi = A(r), we see that KMi = A for each i, so each M j is a left A-ideal in A. Hence by (27.4) there exists a left A-ideal M in A such that Ml

-i- ... -i-

Mr ~ A(r-l)

-i-

M.

This completes the proof of the theorem.

EXERCISES In the following, A denotes an R-order in a separable K-algebra A. Let {P 1 " ' " P n } be the set of all prime ideals P of R such that Ap is not a maximal Rp-order. 1. Let M, N be left A-lattices such that KM ~ KN. Prove that M v N if and only if Rp®RM~Rp®RN

236

MAXIMAL ORDERS OVER DEDEKIND DOMAINS

(27.8)

for each P, whereR p denotes the P-adic completion of R. [Remark: If this isomorphism holds for even one P, then necessarily KM ~ KN. Indeed, the isomorphism implies that whence

Kp ®x KM This implies that KM [1, § 29J).J

~

~

Kp ®K KN.

KN by the Noether-Deuring Theorem (see Curtis-Reiner ~

2. Let M, N be left A-lattices such that KM M p, ~ N p"

KN _Prove that M v N if and only if 1 ~ i ~ n.

3. Let 0 .... L .... M .... N .... 0 be an exact sequence of left A-lattices. Prove that the sequence splits if and only if for i = I, ... , n, the sequence is split. 4. Let M be a left ideal of A such that KM = A. Show that ord R A/M c M. [Hint: ordRA/M c annRA/M.] 5. Let M l ' M 2 be left ideals of A with KM j ordRA/M 1

= A.

Show that if

+ ord RA/M 2

=

R,

then M 1 + M 2 = A. In particular, if M 1 and M 2 are left ideals in a maximal order A, then M 1 and M 2 are "relatively prime" whenever their norms are relatively prime. Is the converse true? 6. Let L, M, N be left A-lattices such that Lv (M +- N). Show that L ~ X +- Y, for some A-lattices X, Y such that X v M, Yv N. [Hint: By (27.1), there exists an fE HomA(L,M +- N) such that fp:Lp ~ Mp N p, PE{Pl""'P.}' Let n: M N .... M be the projection map, and set X = nf(L), Y = ker nf. Then

+

+

04Y .... L .... X .... O is an exact sequence of left A-lattices. For P

E

{p 1'" .,Pn }, the sequence

0 .... Yp .... Lp .... X p''''' 0

is Ap-split, since Lp ~ Mp -i- N p,

Xp = nf(Lp) = n(Mp

Therefore, by Exercise 3, L ~ X

+ Y, and X v M,

+Np ) =

Mp.

Y v N.]

7. Let L, M be left A-lattices such that Lv M(n). Then there exist A-lattices M 1 ,··· ,Mn v M such that L~Ml+-···+-Mn·

Further, given any M J

, .•.

,M n v M, there exists an M' v M such that

M J +- ... +- Mn ~

M(n-l)

-+

M'.

[Hint: The first assertion follows from Exercise 6, the second by repeated use of (27.3).J

7. Crossed-product Algebras In this chapter we collect a number of important results on Brauer groups, crossed-product algebras, and cyclic algebras. We have included proofs in order to make the subject matter more accessible to the reader who is encountering these topics for the first time. As will be seen below, we work directly with the simple algebras themselves, taking advantage of our earlier results from § 7 and § 14. This enables us to avoid introducing the machinery of cohomology groups. All of the topics considered below may be found in standard references. Two excellent introductions are Artin-Nesbitt-Thrall [1] and Herstein [1]. For more sophisticated approaches, the reader may consult Albert [1], Bourbaki [1], Cassels-Frohlich [1], Deuring [1], Jacobson [1, Ch. 5], Weil [1]. The book by Albert has a detailed bibliography of all work on these subjects up to 1938. As additional references on cohomology groups, we cite Babakhanian [1], Gruenberg [1], Neukirch [1], Rotman [1], and Weiss [2]. 28. BRAUER GROUPS Throughout this section let K, L, and E denote fields, with K c L, and let D denote a skewfield. The symbols A and B will always denote central simple K-algebras. If A ~ Mr (D), we shall refer to D as the skewpeld part of A. We shall call A and B similar (notation: A "'" B) if their skewfield parts are K-isomorphict. By Wedderburn's Theorem (7.4), A"", B if and only if there is an isomorphism of K-algebrast A ®K Mr(K) ~ B ®K Ms(K) for some integers r, s. Let [A] denote the similarity class of A. For each A and B (always assumed to be central simple K-algebras), the tensor product A ®K B is also a central simple K-algebra by (7.6). Let us define multiplication of classes by the formula (28.1) [A] [B] = [A ®K B].

Then, by (7.7), multiplication of classes is well defined. Clearly

t Two K -algebras X, rp(rtx)

=

rtrp(x), rt

E

Yare K -isomorphic if there exists a ring isomorphism rp: X ~ Y such that K, x E X.

237

238

CROSSED-PRODUCT ALGEBRAS

[A][B] = [B][A],

[A][K]

(28.2)

= [A].

Finally, we remark that if A is a central simple K-algebra, then so is its opposite ring AO. THEOREM. The classes of central simple K -algebras /o.rm ~n abe!ian group B(K), called the Brauer group of K, r~lative to the m~ltlphDcatwn defmed in (28.1). The identity class is [K], and the mverse of [A] IS [A J.

(28.2)

Proof. Since the associative law holds for tensor prod~cts, ~t also holds for multiplication of similarity classes. Clearly [K] acts as IdentIty, and we need only prove that [A] [AD] = [KJ. We choose B = A in (7.14), so then B' = K. By (7.14), we obtain A ®K AD ~ Mn(K),

and thus

n = (A:K),

This completes the proof. (28.3)

THEOREM. Let

B(K)

->

K c L. There is a homomorphism of groups

B(L),

[A]

->

[A]

[L ®K A],

E

B(K).

Proof. As usual, let A and B denote central simple K-algebras. Then L®K A is a central simple L-algebra by (7.8). It is easily verified that A "-'B=L®KA ~L®KB, L ®K(A ®K B ) ~ (L ®KA) ®L(L ®KB),

L ®KAo ~ (L ®KAt,

which imply the desired result. Keeping the above notation, let B(L/K) be the kernel of the homomorphism -> B(L). Thus [A] E B(L/K) if and only if L®K A ~ M.(L) for some r. (In this case we say that L splits A, or is a splitting field for A.) There is an exact sequence of groups B(K)

(28.4)

1 -> B(L/K) -> B(K) -> B(L) whenever K c L. The next set of theorems will show that each class [A'] E B(L/ K) contains a representative A in which L is embedded as a maximal subfield. In § 29, we shall see that such algebras A can be described explicitly as crossed-products.

(28.5) Theorem. Let D be a skewfield with center K, and let m = J(D:K) be the index of D. Let E be a finite extension of K.

(28.6)

239

BRAUER GROUPS

(i) If E splits D, then m I(E: K). (ii) There exists a smallest positive integer r for which there is an embedding E c M, (D) as K-algebras. With this choice of r, E splits D if and only if E is a maximal subjleld of M. (D). Furthermore, the centralizer E' of E in M. (D) is a skewjleld, and E is a maximal subfield of M. (D)

if

and only

if

E = E'.

Proof. Let S = E ®K D, a central simple E-algebra, and let V be a simple right S-module. Then V is a right vector space over the skewfield D, say (V:D) = r, and there is an embedding of K-algebras (28.6) E c HomD (V, V) ~ M, (D), since E and D commute elementwise in S. Let us show that this gives an embedding with minimal r. Indeed, given any embedding E c M t (D), there is a right D-space W of dimension t over D, such that E c Hom D (W, W). By Exercise 28.3, W may be viewed as a right (E ®K D)-module, and hence W is isomorphic to a direct sum of copies of V. Hence t is a multiple of r. This proves that the embedding given in (28.6) has minimal r. Keeping this notation, let E' be the centralizert of E in the central simple K-algebra B = HomD (V, V). Each E E' is then an (E ® K D)-endomorphism of V, that is, f E Horns (V, V). Conversely, it is clear that Horns (V, V) c E'. This shows that E' = Horns (V, V), and the latter is a skewfield by Schur's Lemma (see Exercise 7.1). We have now proved that E' is a skewfield containing E. Let us deduce at once that E' = E if and only if E is a maximal subfield of B. Obviously, if E' = E then E must be a maximal subfield of B. On the other hand, if E' > E, choose any x E E' - E. Since E' centralizes E, the element x commutes with each element of E. Therefore E(x) is a field (since it is contained in the skewfield E'), and E(x) > E. Thus if E' > E, then E is not a maximal sub field of B. We also observe that by (7.13) we have

.r

(28.7) whether or not E' = E. After these preliminaries, let us turn to the proof of (i). Suppose that E splits D; then S = E ®K D ~ Mm (E), so S ~ v(m) as right S-modules. Comparing dimensions as right D-spaces, we obtain (28.8)

mr = m(V: D) = (S: D) = (E: K).

Therefore ml (E: K), which proves (i). Furthermore, since E' ::::l E, it follows from (28.7) that E' = E. Thus E is a maximal subfield of B, which proves part of (ii). We now drop the assumption that E splits D, and we show finally that if E is a maximal subfield of B, then necessarily E splits D. Indeed, if E is a maximal t See pp. 94-95.

240

CROSSED-PRODUCT ALGEBRAS

(28.9)

subfield of B, then we have seen that E' = E. On the other hand, by (7.14) we have B®KEO~Ms(E'), s=(E:K). Since E = EO = E', this shows that E splits B. But B splits D. This completes the proof of the theorem.

~

Mr (D), and so E also

(28.9) Remarks. If D is a skewfield with center K and index m, then we know from (7.15) that every maximal subfield of D splits D, and has K-dimension m. Part (i) of the preceding theorem asserts that we cannot find any smaller extensions of K, whether contained in D or not, which split D. Thus (7.15) is a "best possible" result. The preceding theorem shows that if E splits D, then E can be embedded in a full matrix algebra B over D, in such a way that (B: K)

= (E: K)2,

and that E coincides with its centralizer in B. We shall called E a selfcentralizing maximal subfield of B. We conclude with some easy consequences of the precedIng discussion. (28.10) COROLLARY. Let A be a central simple K-algebra split by E, where E :::J K, and where (A:K) = (E:K)2. Then E can be embedded in A as a se?fcentralizing maximal subfield of A. Proof. Let D be the skewfield part of A, so E splits D, and let E c Mr (D) = B with r minimal, as in (28.5). By (28.9) we have (B:K) = (E:Kf = (A:K), whence A ~ B. The desired result is now obvious, since E is a self-centralizing maximal subfield of B. (28.11) COROLLARY. For each central simple K-algebra A', there exists a finite galois extension L of K which splits A'. Further, A' ~ A for some central simple K-algebra A containing L as a self-centralizing maximal subfield. Finally, (28.12)

B(K)

= UB(L/K),

where L ranges over all finite galois extensions of K. Proof. Given A', by (7.15) there exists a finite separable extension E of K such that E splits A'. Let L be the normal closure of E over K, that is, the

(28.12)

241

CROSSED-PRODUCT ALGEBRAS

smallest galois extension of K containing E. Then L is also a splitting field for A', and L is a finite galois extension of K. The above shows that [A'] E B(LI K), which implies at once that (28.12) holds true. Finally, if D is the skewfield part of A', then L splits D. As in (28.5), let L c Mr(D) be an embedding with minimal r, and set A = Mr(D). By (28.9), L is a self-centralizing maximal subfield of A. Clearly A' ~ A, and the proof is complete.

EXERCISES 1. Let D be a skewfield with center K, index m. Let E be a finite extension of K which splits D. Then the least value ofr for which there is an embedding of K-algebras E c:: Mr(D) is given by r = (E:K)/m.

2. Let K be a field. Show that the set of sub fields of M, (K) which contain the center K of M, (K) is in one-to-one correspondence with the set of extension fields E of K for which (E: K) divides r. What are the maximal subfields of M, (K) containing K? Show that such maximal subfields are not necessarily self-centralizing. 3. Let A and B be K-algebras, and let M be simultaneously a right A-module and a right B-module. We wish to make M into a right A @K B-module, by defining m(a @ b)

=

mab,

aEA,

bEB,

mEM.

Show that this procedure is justified if and only if A c:: Hom B (M, M).

29. CROSSED-PRODUCT ALGEBRAS Throughout this section, L denotes a finite galois extension of the field K, with galois group Gal (LI K), and E denotes an arbitrary field containing K. Let D be a skewfield with center K. We set E* = E - {O}, the multiplicative group of nonzero elements of E. In § 14 we were concerned with the special case where K is a complete field with a finite residue class field. We showed how to construct a skewfield D with center K and given index n. Recall that we start with a field W ::J K such that Gal (WIK) is a cyclic group of order n, with generator (J (say). Then n-1

D =

L' Wzj,

IX E

W,

0

~

j ~ n - 1,

i=O

and z" E W*. Here, W is a maximal subfield of D, and D has a W-basis whose elements correspond to those of Gal (WI K). This correspondence is such that IX E W, 0 ~ j ~ n - 1. Further, the set of basis elements {zi} is closed under multiplication, apart

242

(29.1)

CROSSED-PRODUCT ALGEBRAS

from constant factors from W*. This construction is a special case of the extremely important concept of crossed-product algebras, which we proceed to define. Returning to the case of arbitrary K, let G = Gal (L/ K). We shall define an algebra A = I" Lu", having as L-basis a set of symbols {u,,:O' E G}. ueG

These symbols are to be manipulated according to the formulas (29.1)

u,,' x

= O'(x)' u",

X E

L,

0', T E

G,

where each f" r E L *. Note that K is contained in the center of A, but that L is not (if L # iq, since the u,,'s need not centralize L. The K -algebra A will be associative if and only if u" (u" u,) = (u" u,,) u, for all p, 0', T E G, or equivalently, if and only if (29.2)

pUu.Jf",,,, =f",,,'f,,,,,r'

p,O',TEG.

A map f: G x G ---+ L* satisfying (29.2) is a factor set from G to L *. Given such an f, the algebra A constructed above is called a crossed-product algebra, and is denoted by (L/ K, f). If f and 9 are factor sets, so is fg, where by definition O',TEG.

I"

Now let A = Lu" as above, let {c,,: 0' E G} be any set of elements of L *, and put v" = c"u", 0' E G. Then A = Lv", and

I"

XEL,

O',TEG,

where gu"=cu'O'(c,)'c;,l'f,,,r'

O',TEG.

The map Oc: G x G ---+ L *, given by (29.3)

(c5c)", r = c,,' O'(c,)' c;/,

0', T E

G,

is easily found to be a factor set, hereafter called a principal factor set. We have thus shown that 9 = (c5c) 'I, and that (29.4)

(L/K,f) ~ (L/K,(&)'f)

for each c:G ---+ L*. A few definitions are in order. The trivial factor set has all values equal to 1. The collection of all factor sets from G to L * is a multiplicative group Z2(G, L*), and the collection of all principal factor sets is a subgroup B2( G, L *). We set H 2(G,L*)

= Z2(G,L*)/B 2 (G,L*),

the second cohomology groupt ofG with coefficients inL*. Call f, 9 E Z2(G, L *)

t

See Exercise 29.11.

(29.5)

CROSSED-PRODUCT ALGEBRAS

243

equivalent if f, 9 have the same image [J] in H2(G, L *), that is, if 9 = (bc) f for some c. It is easily shown that every factor set 9 is equivalent to a normalized factor set f satisfying f(f,l = f 1 ,(f = 1,

(29.5)

(J E G.

(See Exercise 29.1). By (29.4), equivalent factor sets yield isomorphic crossedproduct algebras, In particular, if (L/ K, f) = I'Lu(f is a crossed-product algebra in which f is normalized, then ul is the unity element of the ring (L/ K, f), and each u(f is a unit in this ring. We shall always identify L with the subring LU l of (L/K, f). Even when f is not necessarily normalized, the preceding remarks show that each u(f must be a unit in (L/K, f). (29.6) THEOREM. For each factor set f: G x G -> L *, the crossed-product algebra A = (L/K, f) is a central simple K-algebra. The field L is its own centralizer in A, and is a maximal subfield of A. If 9 is another factor set, then there exists a K -isomorphism

(L/ K, f)

~

(L/ K, g)

if and only if f is equivalent tc! g. Proof. Without loss of generality, we may assume that both f and 9 are normalized. Identify L with LU l c A. We leave it as an exercise to check that K is the center of A, and that every element of A which commutes with each x E L must lie in L. Thus L is its own centralizer in A, and is therefore a maximal subfield of A. To prove that A is simple, let X be a nonzero two-sided ideal of A, and let x

= a(flu(fl + .,. + a(fru(fr EX ,

x # 0,

with r minimal.Ifr > 1, choose bEL with (J 1 (b) # (J2 (b). Then x - (J 1 (b)-l xb is a shorter nonzero element of X. This is impossible, so necessarily r = 1, and X contains a unit aC11 U 0'1 of A. Therefore X = A, which shows that A is a central simple K -algebra, as claimed. Now let B = Lv cr , where the {vO"} multiply according to the factor set g. Any K -algebra isomorphism ¢: A ':'0 B must preserve identity elements, so ¢(u 1 ) = v l . Therefore ¢(Lu l ) = L'v l , where L' is a field K-isomorphic to L. Let u~ = ¢(U(f)' (J E G; then ¢(Lu(f) = L'v l 'u~ = L'u~. Therefore

I'

(29.7)

B = " L'U~, cr~

u~'¢(x)

= ¢((Jx)'u~,

u~u~ = ¢(f",<)u~t'

for x E L, (J, rEG. On the other hand, L'Vl and LVl are K-isomorphic simple subalgebras of B, so by the Skolem-Noether Theorem (7.21), there is an

244

CROSSED-PRODUCT ALGEBRAS

(29.8)

inner automorphism e of B such that e(rjJ(x)) = x, x E L. Applying e to equa tions (29.7) and setting w
wr ==

/(1,1'

wITt'

But for each (J E G, W
~

Mn (K)

if and only if

Proof. We need only show that (L/K, 1) ~ Mn (K), where 1 denotes the trivial factor set. Now (L/K,l) = L'Lu.,., u.,.·x = (J(x)u.,., U.,.U = u.,.t' t


for x E L, (J, rEG. For x E L, let x' denote left multiplication by x acting on L. Then there is an algebra homomorphism

t/!:(L/K,l)--> HomK(L,L),

where

xU.,.-->x'o(J.

The kernel is a two-sided ideal of the simple algebra (L/K, 1), hence is zero. Then t/! is monic, and must be an isomorphism since both (L/K, 1) and Hom K (L, L) have K-dimension n2 •

Example. Let K = Q, L = Q(i), i2 = -1. Then G = Gal (L/ K) is cyclic of order 2; with generator (J, where (J(a + bi) = a - bi, a, bE Q. Let f: G x G --> L * be the factor set given by f

1 ,1

= f",l = fl . .,. = 1,

f.,.,.,. = -1.

Then A = (L/K, f) = LU 1 EB Lu.,., where u1 is the unity element of A, and where xEL. U~ = -1, Thus A is isomorphic to the quaternion algebra Q EB Qi EB Qj EB Qk over Q, by identifying Q EB Qi with L, and j with u.,., On the other hand, if we had chosen f normalized, with f.,.,
(L/K,f) ®K(L/K,g)

~

(L/K,fg).

(29.10)

245

CROSSED-PRODUCT ALGEBRAS

Proof. Let C = A ®K B, where A

=

(L/K, f)

=

L" LU(J'

B = (L/K, g) =

L" LV(J'

Then C is a central simple K-algebra, by (7.6). We shall find an idempotent eEL ®K L c C such that eCe ~ (L/K, fg). This will yield the desired result, since C ~ eCe by Exercise 29.3. To begin witp., we observe that the subfields L® 1 and 1 ® L of L ® K L commute elementwise. Since L is separable over K, there exists an a E L such that L = K(a); then f(x) = min. pol.K a has degree n, where n = (L :K). Now define e=

TI (a ® 1 -

1 ® O'a)

ITI (a -

O'a) ® 1,

where these products are taken over all 0' E G - {1}. Note that the denominator is not zero, since a =1= O'a for each 0' E G - {1}. The numerator is also different from zero in L ® K L, since the elements {a r ® 1 : 0 :( r :( n - I} are linearly independent over 1 ® L. Thus e =1= 0 in L ®K L. Next we note that f(X) = (X - a)' TI (X - O'a), where 0' ranges over G - {I}. Therefore (a ® 1 - 1 ® a) 'TI (a ® 1 - 1 ® O'a) = f(a ® 1) = O. This shows that (1 ® a) e = (a ® 1) e in L ® L. Therefore by induction on r, it follows that

(a' ® 1) e = (1 ® a') e, Since L =

~

r

O.

"-1

L" Ka

r

,

and multiplication in L ® L is commutative, we obtain

,=0

(29.10)

(x ® 1) e = e(x ® 1) = e(l ® x) = (1 ® x) e,

XEL.

Consequently

ITI (a - O'a) ® 1 O'a) ® 1)/TI (a - O'a) ® 1

e2 = e' TI (a ® 1 - 1 ® O'a)

= e' TI (a -

= e. Thus e is an idempotent in L ®K L. It remains for us to prove that eCe eCe =

L

~

(L/K,fg). We have

eeL ® L)'(U a ®

Vt

)

e

1'T,t'EG

=

L eeL ® 1) e'e(1

® L) e' e(uq ® v,) e.

But eO ® L) e = eeL ® l)e by (29.10), and eeL ® l)e = 1:

IS

a field

246

(29.11)

CROSSED-PRODUCT ALGEBRAS

K-isomorphic to L. Let us compute e(u u ® ut)e, by using the formulas Vt'X=T(X)'V t ,

X E

L,

(Y,

T E

G.

Then e(u u ® vt ) e = (u u ®

Vr)'

{TI (mx ® 1 -

1 ® Tpa) /

I1

(CTa -

(Ypa) ® l}' e;

where p ranges over all elements of G - {1}. From (29.10), we obtain (29.11)

e(u u ® vt ) e = (uu ® vr )· e'

TI ((Ya -

Tpa) ®

I/O (cm

- O'fla)

® I.

If (Y # T, then (Ya - Tpa = 0 for p = T- 10'E G - {1}, and hence e(uu ® vt)e = O. On the other hand, when (Y = T we obtain e(uu ® v,,) e = (u" ® v,,) e. An analogous argument proves that e(u" ® v.,.) e = e(uO' ® v.,.). We have now shown that

eCe =

L'

ueG

where

L'wu '

L' = e(L ® 1)e,

(Y E G,

and have also shown that Wu = (uu ® vu ) e = e(uu ® vu)'

Clearly Gal (L'/ K)

~

(yE

G.

G, and we have for x E L,

wu' e(x ® 1) e = e(uu ® vu)(x ® 1) e = e((Yx ® 1) e'w u '

Thus, conjugation by Wu acts as (Y on L'. Further, for (Y,

TE

G,

wu'wt=e(uuut®v_vt)e=e(fu, ucrt ®g u, v)e crt v

t"

T

= e(fu, r gu. r ® 1) e . W ut .

Thus the w's multiply according to the factor set fg. This proves that eCe ~ (L'/K, fg), and establishes the theorem. We are now in a position to relate crossed-product algebras with the Brauer group. We have denoted by B(L/ K) the subgroup of the Brauer group B(K) consisting of all classes of central simple K-algebras split by L. (29.12) Theorem. Let L be a finite galois extension of K, with galois group G. Then H2( G, L *) ~ B(L/ K), and the isomorphism is given by mapping [f] [(L/ K,f)] E B(L/ K).

E

H2(G, L *) onto the class

(29.12)

247

CROSSED· PRODUCT ALGEBRAS

Proof. The indicated map is a well-defined monomorphism of groups, by (29.6) and (29.9). To show that the map is epic, let A' be any central simple K-algebra which is split by L. Then by (28.9) there exists a central simple K-algebra A ~ A', such that L is a self-centralizing maximal subfield of A, and

(A:K) = n Z ,

where

n = (L:K).

Of course [A'] = [AJ in B(L/ K). For each (JEG, the map (J:L-+ L is a K-isomorphism of simple subalgebras of A. Hence by the Skolem-Noether Theorem (7.21), there exists u" E u(A) such that

u" xu:; 1 = (J(x), In order to prove that A =

XEL,

(JEG.

L' Lu", it suffices to show that the right hand "EG

expression is a direct sum, for then both sides will have K-dimension n Z• If the sum is not direct, let

be a relation with minimal k. Surely k > 1, since each u" bEL* such that (Jl(b) i= (Jz(b), one easily checks that

E

u(A). Choosing

(J 1 (b) s - sb = 0 gives a shorter nontrivial relation connecting U"2"" diction, and establishes the formula A = L' Lu". We already know that

•

(J

E

u,,' x

(J(x)' u",

XEL,

uak . This is a contra-

G.

Suppose now that (J, T E G. Then the preceding equation shows that u" ut u~ 1 commutes with each x E L. Since L is its own centralizer in A, we obtain

f If," EL*. The associativity of multiplication in A implies that f: G x G -+ L * is a factor set, and thus we have shown that A ~ (L/ K, f) for some f. This completes the proof of the theorem. The preceding theorem shows that every class in the Brauer group B(K) is represented by a crossed-product algebra over K. Hence every skew field D with center K, such that (D: K) is finite, is similar to some crossed-product algebra (L/K,f). However, as shown by Amitsur [1], there exist skew fields

248

(29.l3)

CROSSED-PRODUCT ALGEBRAS

D which are not isomorphic to crossed-product algebras. (See also Schacher-

Small [lJ). Let us next establish some results on change of fields in crossed-product algebras. (29.13) THEOREM. Let LIK be a finite galois extension, and ElK an arbitrary extension. Then E Q!h (LI K, f) ~ (ELI E, 1'),

I'

where

is obtained from f by restriction, as described below.

Proof. Let EL be the composite of the fields E andL, in some extension field of K containing both E and L. Let F = E n L. By galois theory, we know that ELIE is a finite galois extension, and we have a diagram EL

/~

L

E

H

~/

= Gal (ELI E)

~

Gal (LI F)

c:

Gal (LI K) = G.

F

I

K

Th us each factor set f: G x G ~ J} restricts to a factor set f': H x H ~ L*, and so we may construct crossed-product algebras (LI F, f') and (ELI E, 1'). We shall prove (29.14)

F 0 K (LIK,f) ~ (LIF,f'),

(29.15)

E 0 F (LIF,f') ~ (ELIE,f').

Together these yield the assertion in the theorem, since E 0

K

(LIK,f) ~ E 0 F (F 0

K

(LIK,f)).

Let F' be the centralizer of F in the central simple K-algebra A = (LIK, f). By (7.14) we have F 0

K

A

~

F',

so (29.14) will be established as soon as we ,show that F' Let

L

~

(LIF,f').

a" u" E F', where each a" E L. Then for all x E F,

'(1EG

x'La"U"

= (La"u,,)'x = La"a(x)u".

Hence whenever a" =P 0, we must have a(x) = x for all x

E

F, that

IS,

(29.16) (J

E

249

CROSSED-PRODUCT ALGEBRAS

H = Gal (LI F). This shows that F' = L· LU(T = (LIF,f'), (TEH

so (29.14) is proved.

n. The F-algebra homomorphism

Now let B = (LIF,

E ® FL

->

EL,

x ® y

->

xy,

is an isomorphism (since the map is epic, and both sides have E-dimension (L:F)). Therefore

E®FB= L'(E®FL)(1®u(T)~ L' EL·v(T' (TEH (TEGal(ELIE)

where the {v(T} multiply according to the factor set 1'. This proves (29.15), and establishes the theorem. (29.16) THEOREM. Let K c: L c: E, where LIK and ElK are finite galois extensions. Then (LIK,f) ~ (EIK,g), as K-algebras, where 9 is the inflation of f; as described below. Proof By galois theory, there is a diagram E H

I L

r

G

I

s

G

K

G

= Gal (ElK), H

~

Gal (ElL), HtlG,

G = GIH ~ Gal (LIK), r = (E :L), s = (L:K).

Given a factor set f : G x G -> L*, we define a factor set g: G x G -> L* c: E* by the formula (J, rEG, 9 a. t = J+-U, -, t where (j denotes the image of (J in G. Call 9 the inflation of f Let us set B = (LIK,f) ®KMr(K), a central simple K-algebra split by L (and hence also split by the field E which contains L). Then (LlK,!)~' B, {B:K) = (rs)2, (E:K) = rs. By (28.10), we know that E can be embedded inB as a self-centralizing maximal subfield of B. We shall give a specific such embedding, and shall use it to prove that B ~ (ElK, g), where 9 is the inflation of f This will complete the proof of the theorem.

250

(29.17)

CROSSED-PRODUCT ALGEBRAS

It follows from (29.8) that there is an L-isomorphism

(ElL, 1) ~ Mr(L}. We shall need to know an explicit form of this isomorphism. Let E =

I

r.

Lei'

i= 1

and for each x E L, let T(x) E Mr (L) describe the action of x by left multiplication on the {e i }. Thus x(el' '" ,e r ) = (e 1 ' ... ,er ) T(x), x E L. On the other hand, each pEG may be viewed as an element of HomL (E, E), and hence determines a matrix PI' E Mr (L) such that (pel"'" pe)

Let us put e briefly as (29.17)

= (e 1 , ... ,e.).

=

(el"'" e.)PI"

pEG.

Then the preceding formulas may be written

xe = eT(x),

pe

= eP p '

XEE,

pEG.

The isomorphism (ElL, 1) ~ Mr(L), given by (29.8), is obtained by the mapping x -+ T(x), p -+ PI" x EE, P Ell. Let T(E)

=

{T(x): x E E},

a field L-isomorphic to E. Then we have (29.18)

L' T(E)PI"

Mr(L) =

I'EH

For each matrix X with entries in L, and each a E G, let a(X) be the matrix obtained from X by applying a to each of its entries. Then (pa)e

=

p(eP,,)

=

eP p ' p(P,,),

p, a

E

G,

whence (29.19)

p, aE G.

On the other hand, for x

E

a(x) a(e)

E and a E G we obtain

=

a(xe)

=

=

ae' a(T(x))

a(eT(x))

=

eP u . a(T(x)).

But also a(x) a(e)

= a(x)' eP u = eT(a(x))P u'

Therefore (29.20)

T(a(x))· P u = P u ' a(T(x)),

X E

E,

a E G.

We are now ready to return to the K-algebra B defined above. We may write B = (LIK,f) @KMr(K) = L' Luu @ Mr(K) ~ L' Mr(L)uij' aEG aEG

251

CROSSED-PRODUCT ALGEBRAS

(29.21)

In the above, the symbols {Uij} multiply according to the normalized factor set j; and in the last summation we have written U;j in place of Uij @ 1. From the definition of the crossed-product algebra (L/K, f), we have Uif'

Y = a(y)' Uij'

a E G.

Y E L,

Therefore Uff ' X

=

a(X)' Uij,

I' MJL)u

Let us identify B with Then we may write

ii ,

X

E

Mr(L),

aE

G.

and let us use formula (29.18) for M (L). r

B

I' _T(E)P Uij .

=

p

pEH,aEG

now exhibited an embedding of E in B, since E ~ T(E). Of course identity matrix in Mr(L), and UI is the unity element of B. We are prove that B ~ (E/K,g). write G = Hal U ... u Has' Each a EGis uniquely expressible as a = pa , for some p E H and some i between 1 and s. Then a = a p and of i course G = {a I " ' " as}' Now define

We have PI is the trying to Let us

aE G. We shall show that

B

(29.21)

=

I' T(E)v".

nEG

Let a = pap p E H, 1 ~ i ~ s. By (29.19), PeT = P p ' p(P,,) Since P", has entries in L, and since p E H, it follows that P u = Pp . P Ui' Furthermore, P Ui is a unit in the ring T(E)P p ' Therefore

I'

pEH

B

=

t {I'

i= 1

pEH

T(E)P p }u ii , =

L' {I' T(E)P

p

i

}P",uij

P

I

which establishes formula (29.21). We are going to prove that B ~ (T(E)/K, g), and so shall first compute G = Gal (T(E)/K). Since E ~ T(E), it follows that G = {a: a E G}, where a(T(x))

= T(ax),

x E E,

a

E

G,

Using this formula, we may now compute the action of v" on T(x), for a E G and x E E. We have v,,' T(x)

But a(Y)

=

=

PeT uij' T(x)

= p,,' a(T(x))' u;;.

a(Y)for any Y E Mr(L) and any a E G, since a acts as a on elements

252

(29.21)

CROSSED-PRODUCT ALGEBRAS

of L. Hence by (29.20) we obtain vO"· T(x) = p p • a(T(x))· u~ = T(a(x))· PO"· u&

= T(a(x))· VO" = O'(T(x))· vO"'

X E

E,

a E G.

This proves that VO" transforms T(E) according to the action of 0' on T(E). Finally, we obtain

a, rEG. This establishes that B of the theorem.

~

(T(E)/K, g) as claimed, and completes the proof

Remarks. (i) Let L/K be a finite galois extension, F an intermediate field, and set G

= Gal (L/K), H = Gal (L/F).

Since H is a subgroup of G, each factor set from G to L* restricts to a factor set from H to I!, there by ind ucing a homomorphism res: H2(G, L*)

-+

H2(H, L*).

On the other hand, by (28.3) there is a homomorphsim

F ®K·: B(L/K)

-+

B(L/F).

Formula (29.14) is equivalent to the assertion that the following diagram commutes:

where the vertical arrows denote the isomorphisms given by (29.12). (ii) Keeping the above notation, assume further that H {;. G, and set G = G/H = Gal (F/K). As in the proof of (29.16), each factor set from G to F* inflates to a factor set from G to L*, thereby inducing a homomorphism inf: H2(G, F*)

-+

H2(G, L*).

On the other hand, each central simple K-algebra split by Fis also split by L, so there is a monomorphism B(F/K) -+ B(L/K). Theorem 29.16 is equivalent to the assertion that the following diagram commutes:

(29.22)

CROSSED-PRODUCT ALGEBRAS

H'T

F') --"'L.

B(F/K)

H'(f'

253

I:')

- - B(L/K) ,

where the vertical arrows are the isomorphisms of (29.12). We conclude this section with a discussion of the relation between the exponent and the index of an element of the Brauer group B(K). For [A] E B(K), let exp [A] denote the exponent of [A] in B(K), that is, the least positive integer t such that [AJ = 1 in B(K). On the other hand, define the index of [A] to be the index of the skew field part of A. This means that for [A] E B(K), index [A] = index [D] = J(D:K), where D is a skewfield such that A (29.22) Theorem. For each [A]

E

~

D.

B(K), we have

[A]m = 1 in B(K), where m = index [AJ. Therefore exp [A] divides index [Al Proof Let A ~ D, where D is a skewfield of index m, so [A] = [D] in B(K). By (28.11) there exists a finite galois extension L of K such that L splits A. Then L also splits D. As in the proof of (28.5), using L in place of the field E occurring in that proof, we let V be a simple right (L ® K D)-module, and set r = (V: D). Then there is an embedding B

~

M,(D),

and (L: K) = mr by (28.8). On the other hand, we may view Vas left B-module, and then B ~ 0') as left B-modules. Therefore r(V:L)

= (B:L)

= (B:K)/(L:K) = mr,

and so (V:L) = m. Since L is a self-centralizing maximal subfield of the central simple Kalgebra B, it follows as in (29.12) that there exists a factor set f: G x G -+ L* such that B ~ (L/K,j). Hence by (29.9) we obtain

[A]m = [B]m = [(L/K,r)] It therefore suffices to prove that

r

in

B(K).

is a principal factor set. This will be done by using a technique of Schur, who first proved this theorem. Each (J E G = Gal (L/K) determines an element U E (L/K,f). We shall identify B u with (L/K, j), and then the element Uu of B acts on the left B-module V. If

254

(29.23)

CROSSED-PRODUCT ALGEBRAS

pcu) describes the action of u.,. on some specific L-basis of V, then we have obtained a mapping (J E G -4 pC") E Mm (L). We shall show that this mapping satisfies the identity (29.23)

(J,

TE

G,

where (J{ pIt)} is obtained from pCt) by applying (J to each of its entries. To establish the above formula, let us recall that (V: L) = m, so we may write V

=

f'

;~

LVi'

For

(J

G, let us set

E

1

U . V.

"

)

Then ( U" ) U Vj t

=

'\' (t).} U" { L., PkjV k

k

=

'\'

L,

(J

(It)). Pkj

(
Pik

i'

i, k

On the other hand,

Comparing this formula with the preceding one, we deduce (29.23) at once. Now take determinants in (29.23), and let c" = det pC,,) E L*. Then we obtain

Thereforef m = &, a principal factor set, and so [A]m = [B]m = [(L/Kr)] = 1 in B(K). This completes the proof. As we shall see in Chapter 8 (see (32.19)), exp [A] = index [A],

[A]

E

B(K),

when K is any algebraic number field. For the moment, we content ourselves with a somewhat weaker result, which holds for general K. (29.24) THEOREM. For each [A]EB(K), the integers exp[A] and index [A] have the same prime factors, apart from multiplicities. Proof We may assume that A is a skewfield D with center K and index m. Since exp [D] divides m by (29.22), it suffices for us to prove that each rational primep dividing m also divides exp [D]. We shall suppose that plm, exp [D], and shall obtain a contradiction. There exists a finite galois extension L/K such that L splits D, by (28.11). Let G = Gal (L/K), and let H be a Sylow p-subgroup of G. Then

PI

(29.24)

255

EXERCISES

IHI

= pr, r ~ U,

[G:H] = q, ptq.

Let E be the subfield of L fixed by H; then (E: K) = [G: H] = q. Since m t q, it follows from (28.5i) that E does not split D. Then S = E ® K D is a central simple E-algebra, such that [S] i= 1 in B(E). Consider the diagram LI H

S

p'

171

G

'i;yD K

The homomorphism B(K) - B(E) defined by E ® K', carries [D] onto [SJ. exp [D], we conclude that Therefore exp [S] divides exp [D]. Since ptexp [S]. On the other hand, we have

PI

since L splits D. Therefore L also splits S, and so n I(L: E) by (28.5i), where n is the index of S. Hence n is a power of p. However, by (29.22) we know that exp [S] divides n, so exp [S] is also a power of p. But we have shown above that pt exp [S], and so it follows that [S] = 1 in B(E). This is a contradiction, and thus the theorem must be true. EXERCISES 1. Show that every factor set f: G x G ---> L* is equivalent to a normalized factor set. [Hint: Let c 1 = fl. L Ca = 1 for 6 E G - {1}. Then (tk)q, 1 = 6(e 1 ), 6 E G. Set 9 = Uk)}; and show first that g1.1 = 1, then that gu.l = gl.a = 1 for all 6 E G.] 2. Verify that (LI K,f) has center K, and that L is its own centralizer in (LI K, f). 3. Let e be an idempotent of the central simple K-algebra C. Prove that C ~ eCe. [Hint: We may take C = Mr (D), D = skewfield with cenlt:r K. After a suitable change of basis (see Exercise 7.11), we may assume that e = diag (1, ... , 1,0, ... ,0), where s I's occur. But then eCe ~ Ms (D) ~ D ~ c.]

4. Let D be a skewfield with center K, index m. Let E be a finite field extension of K, and let E iElK D ~ Mr (S), where S is a skewfield of index n. Show that S has c~nter E, and that min is an integer which divides (E: K). [Hint: This is the Index Reduction Theorem of Albert [1 J. To prove it, note first that E iEl K D is a central simple E-algebra, so S has center E. Next, m2 so r

= min.

=

(E iElK D: E)

=

(Mr(S):E) = r2 n2 ,

Finally, if Vis a simple right Mr (S)-module, then Mr (S)

~

v(r).

Comparing

256

(29.25)

CROSSED-PRODUCT ALGEBRAS

right D-dimensions, we obtain (E: K) = r(V: D),

so rl(E: K).] 5. Keep the notation of the preceding exercise. Prove that if (E: K) is relatively prime to the index m of D, then E ®KD is a skewfield with center E and index m. [Hint: U sing the nota tion of Exercise 29.4, we see that r = 1, since r divides both m and (E: K).] 6. Let D j be a skew field with center K, index mp i = 1,2, and suppose that (m l , m 2 ) =1. Prove that Dl ®KD2 is a skewfield with centerK and index ml m2 . [Hint: Let Dl ®K D2 ~ M,(S), where S is a skewfield with center K, and let E be a maximal subfield of S. Then E splits S, so [E ®KD1J [E ®KD2] = 1 in B(E).

But under the homomorphism B(K) --+ B(E), [D;] maps onto [E ®KDjJ. Since [DJm, = 1, this shows that [E ®KDj] and [E ®KDJ have relatively prime orders. Therefore [E ®KDj] = 1 in B(E), i = 1,2, so E splits both Dl and D 2. Then (E:K) is divisible by m 1 and m 2 , so (E: K) ~ ml m 2 • Therefore (S:K) = (E:K)2 ~ m~m~ = (Dl:K) (D2:K) = (Dl ®KD2 :K) = (M,(S):K),

so S

=

= m~ m~.]

M,(S). This proves that t = 1, and that (S: K)

7. Let D be a skew field with center K and index m, where m

=

,

IT pt', with the {Pi} i= 1

distinct primes and each ej ~ 1. Prove that there exist skewfields D l center K, such that D j has index pt', and

, ... ,

D, with

D~Dl®K"·®KD,.

(29.25)

[Hint: exp [D] divides m, and each pjlexp [D]. We may then w,ne [D] = [AJ··· [A,]

where exp [Ai]

=

p/', with 1

~

in

B(K),

b j ~ ej . Let D j be the skewfieldpart of A j. Then

[D] = [Dl] ... [D,] = [Dl ®K .,. ®KD,]

in

B(K).

But index [Dj] = p/' for some cj ~ 1, by (29.22). Hence D 1 ® K ' .. ® K D, is a skewfield, by Exercise 29.6, and so (29.25) is valid.] No hints will be given for Exercises 29.8-29.13. The reader may consult the references on cohomology of groups listed at the start of this chapter. 8. Let G be a finite group, and let ZG be the integral group ring consisting of all sums 0'.9 g : r1. g E Z} (see beginning of §8, example (iv)). Every additive group M, on

{L

geG

which G acts as left operator domain, can be made into a left ZG-module. Let Z be the trivial ZG-module, that is, the additive group Z on which G acts trivially: gz = z, g E G, Z E Z. Let Q. be the free left ZG-module having as free basis the collection of all n-tuples [Xl' _.. , X.] of elements of G. For n = 0, interpret Qo as the free module on

257

EXERCISES

(29.26)

one generator, with this generator denoted by the empty bracket [ a ZG-exact sequence (29.26)

~

J. Show that there is

d, Q d, Q d, Z 1'-'+ o~ ......... 0 , Q 2~

where doe ]

=

1,

d 1 [x]

=

x[ ] - [ ],

d 2 [x,y] = x[y] - [xy]

+ [x],

d3 [x,y,z] = x[y,z] - [xy,z]

+

[x,yz] - [x,y],

x,y,zEG,

and so on. Each map di is a left ZG-homomorphism, that is, dig[xl'" " xJ} = ~d;[xl' ... , Xi]' ~ E ZG, x. E G. 9. Given a left ZG-module M, we apply HomZG (., M) to (29.26), obtaining a sequence of additive groups (29.27)

0

dO

d'

->

do

Hom/.'G (Z, M) -4 Hom ZG (Qo, M)-4 Hom ZG (Q1' M)~ ... ,

where di induces di as in §2a. Thus

Prove first that im di c ker di+ l' Then define l/o(G, M) = ker

d1,

W(G, M) = ker d:+ dim d:, n ~ I.

Call W(G, M) the nth cohomology group of G with coefficients in M. Prove that (29.28)

W(G, M)

~ Ext~G

n

(Z, M),

~

0,

where on the right, Z denotes the trivial ZG-module. 10. Keep the above notation. EachfEHomZG(Qn,M) is determined by its values {f[x 1, ... , x n]} in M, so we may identify HomZG(Qn, M) with the additive group of all

maps

f:G x ... xG

--->

M.

"-----y--~

n factors

For n = 0, identify Hom ZG (Go, M) with M, letting to f[ ] E M. Prove that

f

E Homza(G o, M) correspond

(d!f) [x] =f(d 1 [x]) =xf[] -f[], wheref:[] --->M, (d!fII [x,yJ

=

xf(y) - f(xy)

+ f(x), wheref:G

(d~f) [x, y, z] = xf(y, z) - f(xy, z)

+ f(x, yz)

--->

M,

- f(x, y), where f: G x G ---> M,

and so on. Prove that HO(G, M) ~ MG as additive groups, where MG={mEM:gm=m forallgEG}.

258

(29.29)

CROSSED-PRODUCT ALGEBRAS

II. Keep the above notation, and let G = Gal (L/ K), where L/ K is a finite galois extension. Choose M to be the multiplicative group L* on which G acts, and view M as ZG-module. Show that ker di consists of all factor sets from G to L *, and that im di consists of all principal factor sets. Deduce that the second cohomology group H2( G, L*), defined at the beginning of §29, is a special case of the cohomology groups given in Exercise 29.8.

12. Let G = (0"> be a finite cyclic group of order n,andlet M be aleft ZG-module. Let D=6-1,

N=I+6+6 2 + ... +O"n-l E ZG.

Prove that there is a ZG-exact sequence

.,. fi. ZG IJ. ZG t!" ZG lJ. ZG .4 Z

->

0,

where ~ means left multiplication by N, lJ. by D, and where map defined by

I

Ii(

rtg

g) =

geG

I

rtg , rtg E

Ii

is the augmentation

z.

geG

Using this projective resolution of the trivial ZG-module Z, deduce that HO(G, M) ~ M G , and that

Here,

M'

=

{mEM:Nm

= O} =

{mEM:(1

+ 0" + '" + O"n-l)m

= O}.

13. Let L/K be a finite galois extension, with galois group G = (6) cyclic of degree n. Choosing M = L* in the preceding exercise, show that MG = {xEL*:6X = x}

=

K*, (l

+ 0" + ... + 6 n -

1

)M

=

NL1K(L*),

where N UK (L*) = {N L/K X: x E L*}. Deduce that H2( G, L *) ~ K* / N LIK(L *).

(29.30)

14. Let L/ K be a finite galois extension, with galois group G. Prove the Normal Basis Theorem: There exists an x E L such that {a(x):6 E G} is a K-basis of L. [Hint (Berger-Reiner [IJ): Let A = (L/K, 1) ~ HomK(L, L), .as in the pr.oof of (29.8), so L is a left module over the simple ring A. Companng K-dlmenSlOns, It follows that A ~ Dol as len A-modules. Let A = I'Lu",

B

where

Uu .

x =

6(X)Uu' 0" E

L'KI/ :leG

(tEG

G, x

E

L.

q

,

U~UT

li q

,.

6,rEG,

(30.1)

259

CYCLIC ALGEBRAS n

I' Kx" then I' L = I' {I' Ku

Then B is a subring of A. If L

=

i= 1

A

=

UU

i

(J

u } Xi

(J

~

B(n)

-

as left B-modules. Hence L(n) ~ B(n) as left B-modules, whence L ~ B by the Krull-Schmidt Theorem (Exercise 6.6). Therefore L = Bx for some X E L, and Bx =

(I

I

Kuu)x =

t7EG

Ka(x).]

t7EG

30. CYCLIC ALGEBRAS Throughout this section let L denote a finite galois extension of the field K, and E any extension field of K. The symbol D will always stand for a skewfield. Call L a cyclic extension of Kif Gal (L/K) is cyclic; we write Gal (L/K) =
(30.1) A

= (L/K, a, a) = I'Lu i ,

u .x

=

a(x) . U,

=

un

a,

x E L,

i=O

where we identify UO with the unity element of A. We shall call A a cyclic algebra. Obviously A is a crossed-product algebra, with the elements {u i } playing the role of elements {va i : 0 ~ j ~ II - I} such that x EL.

0:( j :( n - 1,

Thus A

~

(L/K, f) where the factor set f from
(30.2)

jUi,(Ji =

{

i

+j <

.' (I,

r +.J ;,

n ..

11

0:( I.] ~ n -

1.

By §29, we know that A is a central simple K-algebra split by L, and that L is a self-centralizing maximal subfield of A. Conversely, if L/K is cyclic, we show that every crossed-product algebra is a cyclic algebra.

= Gal (L/K) =
nO.3) THEOREM. Let G

n-1

(L/K, g)

~

(L/K, a, a),

a

=

f1

j=O

gai a E '

K*.

260

CROSSED-PRODUCT ALGEBRAS

(30.4)

Proof We may write "-1

B

=

2::'

Lv aj ,

j=O

Uaj'X

= U.i(X)·U"j'

XEL,

Va,V aj

=

g~i.ajVai+j' v

for 0 ~ i,j ~ n - 1. Then we obtain

v,2 ,--

V " • V , , -- -

v: = a' v,," Therefore B

=

g0',0' V,,2 ,

= a.

"-1

~' Lv j L. ",

j=O

Since v::. lies in the center of B, it follows that a

E

K*, and the theorem is proved.

Using this result, we may carryover to cyclic algebras the results on crossedproduct algebras established in §29. Let us set NL/K(L*)

= {NL/KX:XEL"'}.

To begin with, we prove

= Gal (L/K) = (u) be cyclic C?forder n, and let a, bE K*. Then (i) (L/K, u, a) ~ (L/K, cr, a') for each s E Z such that (s, n) = 1. (ii) (L/K, u, 1) ~ M" (K). (iii) (L/K, a, a) ~ (L/K, a, b) if and only if

(30.4) THEOREM. Let G

b

=

(N L/K c)a for some c E L*.

In particular, (L/K, a, a) ~ K if and only if a E N L/K (L*). (iv) (L/ K, a, a) ® K (L/ K, a, b) '" (L/ K, a, ab). Proof Let (30.5)

A

= (L/K, a, a) = 2::' Luj,

B

= (L/K, a, b) =

2::' Lv

L,

where for all x

E

(30.6)

u .x

= a(x)u, u"

=

a;

v . x = a(x)u,

If (s, n) = 1, then G = (as) and "-1

A

= L'Lw j ,

where

w

= US.

j=O

It is easily checked that W' X

:-=

as (x) . w,

w" = as,

xEL,

u" = b.

j

,

(30.7)

CYCLIC ALGEBRAS

261

so (i) is established. Assertion (ii) is a special case of (29.8). To prove (iii), note first that for any c E L* we have "-1

A

= L'L(cuY, j=O

and

cu . x = O'(x)' cu,

X E

L,

(cu)" = cu' cu ... cu = c· O'(c)· . , O'"-l(C) uri = (NL/K

c)a.

Then A ~ (L/K,O', a' Nc). Conversely, given any K-isomorphism A ~ B, we may assume that A = B by using the argument in the proof of (29.6). But then vu- 1 centralizes L, so v = cu for some c E L*. Therefore b = v" = (cu)" = (N L/K c)a, as claimed. Analogously, assertion (iv) follows from (29.9). We leave the details as exercise for the reader. (30.7) COROLLARY. Let A = (L/ K, 0', a) as above. Then exp [A] is the least positive integer t such that dEN LiK(L *), If exp [A] = (L: K), then A is a skewfield, Proof We have [AJ = [(L/ K, 0', at)] in B(K). Thus [AJ = 1 if and only if at isanorm.Forthesecondassertion,letn = (L:K),so(A:K) = n 2 ,IfA ~ Mr(D), where D has index m, then n = mr. But [A]m = 1 in B(K) by (29.22), whence exp [A] divides m. Hence if exp [A] = n, it follows that m = nand r = 1. Thus A is a skewfield, and the proof is completed. Let us carryover to the case of cyclic algebras, two further results from §29. (30.8) THEOREM. Let G = Gal (L/K) = (0') be cyclic of order n, and let a E K*. Let E be anyfield containing K, and let EL be the composite of E and L in some larger field containing both E and L. We may write (30.9)

H = (O'k) = Gal (L/L n E) ~ Gal (EL/E),

where k is the least positive integer such that ~ fixes L n E. Then E ® K (L/ K,

0',

a) - (EL/E, O'k, a).

Proof The isomorphism given in (30.9) is well known from galois theory,

262

CROSSED-PRODUCT ALGEBRAS

(30.10)

and we treat it as an identification. Let A = (LIK, (J, a) = (LIK, f)

where f is normalized and is given by (30.2). If f' denotes the restriction off to H x H, then by (29.13) we have E ®KA ~ (ELIE,!,).

Since ELIE is a cyclic extension with galois group H from (30.3) that

=

«(Jk), it follows

(ELIE, f') = (ELIE, (Jk, b),

where m-1

b

IT

=

7=1

f~'k,ak'

From (30.2) we find immediately that b the theorem.

m =

IHI =

= a,

which completes the proof of

nlk.

(30.10) THEOREM. Let K c LeE, where G = Gal (ElK) = «(J) is cyclic of finite order t. Let H = Gal (ElL), G = GIH = Gal (LIK) = (ii),

where ii is the image of (J in G. Then for any a E K*, (LIK,ii, a)

~

(ElK, (J, a(E:L»).

Proof Let A = (LIK, ii, a) = (LIK,f), where f-,

_j

=

{1'

i

a,

i

a ,a

+j < +j ~

n n,

0 ~ i, j ~ n - 1,

and where n = (L: K). By (29.16), we have A ~ (ElK, g), where g is the inflation off But from (30.3) we obtain (ElK,

g) =?=

(ElK, (J,b), whereb

=

<-1

IT

gaJ,u'

j=O

Now

o~ j

~

t

= (E :K) = (E:L)(L:K) = sn, t - 1 we have .

_

gal a ,

where

Further,

_fa, j =n -1,2n -1, ... ,sn -1, L1, otherwise.

J,j - a ,a

s = (E:L).

Therefore b = as, and the theorem is proved.

for

(30.10)

CYCLIC ALGEBRAS OVER LOCAL FIELDS

263

EXERCISES 1. Let L be a cyclic extension of K. Prove that

Hl(G, L*) ~ B(L/K) ~ K*/N LIK(L*) where G = Gal (L/K). [Hint: Let G = <0"), and consider the map K* -> B(L/K) defined by a E K* -> [(L/ K, 0", a)] E B(L/ K). Show that this gives an epimorphism of groups, with kernel N LIK (L*). For another proof, see Exercise 29.13.J 2. Let A be a central simple K -algebra split by E, where E is a cyclic extension of K such that (A: K) = (E: K)l. Prove that A ~ (E/ K, 0", a) for some a E K*, where G = Gal (E/ K) = <0"). [Hint: By (28.10), we may embed E in A as a sel f-centralizing maximal sub field of A. The proof of (29.12) then shows that A is isomorphic to a crossed-product algebra (E/K,f), for some [fJ E Hl(G, E*). Finally, (E/K,f)

~

(E/K, 0", a)

for some a E K*, by (30.3).] 3. Let A' be a central simple K-algebra split by E, where E is a finite cyclic extension of K. Prove that A' ~ (E/K, 0", a) for some a EK*, where Gal (E/K) = <0"). [Hint: The proof of(29 .12) shows that A' ~ A, for some central simple K-algebra A split by E, such that (A: K) = (E: Kf Now use the preceding exercise.J 4. Let K = Q, L = Q(J2), and let 0" E Gal (L/ K) be defined by 0"(.j2) = -.j2. For p an odd prime, let A = (L/K, 0", pl. Show that A ~ Ml (K) if 2 is a quadratic residue mod p, while A is a skewfield if 2 is not a quadratic residue mod p. [Hint: A is a skewfield if and only if p ¢ N(L*), where N means N LIK. Let R = Z[ J2:], and show that R = alg. int. {L}. Imitating the proof in §26 that the quaternion order has a euclidean algorithm, show that R is also euclidean, and hence every ideal of R is principal. Ifp E N(L*), then there exist integers a, b, c E Z, not all zero, such that a l - 2b l = pc 2 • If pia, then also plb and pic; after cancellation and repetition of the process, we may assume that pia. Since a l == 2b l (mod p), it follows that 2 is a quadratic residue mod p. Conversely, if 2 is a quadratic residue mod p, then there exist a, bE Z with pta, p{b, such that pl(a 2 - 2b l ). Then pl(a + bJ2) (a - bJ2), so pR is not a prime ideal of R. Further, pis unramified in L/K, since p does not divide the discriminant of R over Z. Hence pR = PI Pl , where the Pi are distinct ideals of R. Therefore pZ = N LIK(P1 ), and if P 1 = Rrt, then p = ±Nrt. But u = 1 + E l/(R), and N(l/) = -1. Hence p is either Nrt or N(urt), and so p E N(L*).]

31. CYCLIC ALGEBRAS OVER LOCAL FIELDS Throughout this section let R be a complete discrete valuation ring, with maximal ideal P = nR f:. 0, and let R = R/ P. Let K be the quotient field of R, and vK the exponental valuation on K. We assume throughout that the residue class field R is finite, and set card R = q. Let D be a skewfield with center K and index m, where by definition

264

(31.1 )

CROSSED-PRODUCT ALGEBRAS

m = J(D: K). Let us recall some details from §14, where we showed that each such skewfield D is isomorphic to a cyclic algebra. Let W be the unique

unramified extension of K of degree m. Then W = K(m), wheremis a primitive (qm _ l)th root of unity over K. The galois group Gal (WIK) is cyclic of

order m, and has as canonical generator the Frobenius automorphism (J of WIK. Recall that (J is defined by the equation (J(m) = mq • We shall denote the Frobenius automorphism of the extension WIK by (JW/K; it is defined only for unramified extensions. As in §5b, we use the notation Ow for the valuation ring of l-f~ Pw for the maximal ideal of ow' and Ow for the residue class field owlpw' We showed in §14 that W may be embedded in D, and that there exists a prime element ZED such that m-l

D =

L:'

Wzi, where zrtz- 1 = a'(rt), rt

E

l-f~

and zm = n.

i=O

The integer r is relatively prime to the index m, and D determines r mod m uniquely. Further, each pair r, m with (r, m) = 1 arises from some D. In terms of the notation for cyclic algebras introduced in §30, we have D ~ (WIK, (Jr, n). Choose s EZ so that rs == 1 (mod m). Then also (s, m) = 1, and by (30.4 i), D ~ (WIK, a', n) ~ (WIK, (Jrs, n S) = (WIK, (J, nS). Furthermore, we could have restricted s to lie in the range 1 cp(m) denote the number of such integers s. We now prove

~

s

~

m. Let

(31.1) THEOREM. Let WIK be an unramified extension of degree m, and let (J be

the Frobenius automorphism of WIK. Let {as} be any set of cp(m) elements of K*, such that the values {v K(as)} are relatively prime to m, and are incongruent mod m. Then the cp(m) cyclic algebras {(WIK, (J, as)} give a full set of nonisomorphic skewfields with center K and index m. Proof We have already seen that the cyclic algebras {(WIK, (J, n5): 1

~

s

~

m, (s, m) = I},

give a full set of non-isomorphic skewfields with center K and index m. Now let a E K*, and let NW/K(W*) = {NW/K(x):XEW*}.

We showed in (14.1) that a E N W/K (W*) if and only if m IvK(a). On the other hand, from (30.4) we know that for a, bE K*, we have (31.2)

(WIK,(J,a)~(WIK,(J,b)

ifandonlyif ba-1ENw/K(W*),

(31.3)

CYCLIC ALGEBRAS OVER LOCAL FIELDS

265

It then follows that

(31.3) (W/K, a, a) ~ (W/K, a, b) if and only if

UK (a) ~ uK(b) (mod m).

The assertion in the theorem is now obvious. For each [A] E B(K), we have denoted by exp [A] the exponent of [A] in the Brauer group B(K), and by index [A] the index of the skew field part of A. An important consequence of the preceding discussion is the following fundamental result: (31.4) THEOREM. Let D be a skewfield with center K and index m. Then m = exp [D]. Hencefor each [A] E B(K), exp [A] = index [A].

Proof We may take D = (W/K, a, rr') as in (31.1), with (s, m) = 1. Then

[DJ = [(W/K, a, 7Ost)] EB(K) by (30.4). But (W/K, a, 1) ~ Kby(30.4),andso it follows that (W/K, a, 7Ost) if and only if there is a K-isomorphism

(W/K, a, nst)

~

~

K

(W/K, a, 1).

But by (31.3), such an isomorphism exists if and only if m Ist. This proves that [DJ = 1 in B(K) if and only if m Ist. Since (s, m) = 1, it is clear that mist if and only if mit. This proves that exp [D] = m. The second assertion in the theorem is immediate. Whether or not (s, m) = 1, we may still form the cyclic algebra A = (W /K, a, nS). By (31.3), the isomorphism class of A depends only on s(mod m), that is, on the fraction s/m viewed as an element of the additive group Q/ Z. Let us find the skewfield part of A; of course, we already know that A is a

skewfield whenever (s, m) = 1. (31.5) THEOREM. Let W/K be an unram~fled extension of degree m, with Frobenius automorphism a, and let s E Z. Write s/m = s' /m', where (s', m') = 1. Then (W/K, a, n ~ (W'/K, a', n = skewjield of index m', S

S

)

')

where W'/K is an unramUied extension of degree m', with Frobenius automorphism a'. Proof Let W' be the subfield of Wfixed by
266

CROSSED-PRODUCT ALGEBRAS

(31.6)

(W'/K, a', 71:") ~ (W/K, a, (71:")d) = (W/K, a, nS). Since (s', m') is proved.

= 1, we know that (W'/K, a', nS') is a skew field, and the theorem

(31.6) COROLLARY. Keep the above notation, and let a E K*. Then the cyclic algebra (W/K, a, a) is a skewjzeld if and only if(m, vK(a)) = 1.

Proof Let s from (31.5).

= vK (a), so (W/ K, a, a)

~

(W/ K, a, nS). The result now follows

Let W/K be an unramified extension of degree m, with Frobenius automorphism aW / K ' Given an integer s, not necessarily prime to m, let us form the cyclic algebra A = (W/K, a W / K' 71:'). We now define the Hasse invariant of A, denoted by inv A, by the formula (31.7)

inv (W/K, a W / K ' nS ) = s/mEQ/Z.

The skewfield part of A can be calculated by use of (31.5), and it has the same Hasse invariant as A. Therefore inv A depends only upon the class [A] E B(K), and we shall write inv [A] rather than inv A hereafter. By (31.1), every class in B(K) is represented by some cyclic algebra (W/K, a W / K' n S ) with W/K unramified, and hence there is a well defined map inv: B(K) --> Q/Z. Caution. Let L/K be a cyclic extension with galois group (a) cyclic of order n, and let a E K*. Then the cyclic algebra B = (L/K, a, a) determines a class [B] in B(K). However, it is not necessarily true that inv [B] = vK(a)/n. Indeed, even when L/K is unramified, the formula is valid only when a equals the Frobenius automorphism a L/K' Further, in the ramified case the Frobenius automorphism a L/K is not even defined. In order to compute inv [B] when L/K is ramified, we must first write B ~ (W/K, a W / K' n S ) = A for some unramified extension W/K, and then we have inv [B] = inv [A] = s/m. We are now ready to prove the important result: (31.8) THEOREM. inv: B(K)

~

Q/Z.

Proof. The map inv is epic, since each element of Q/Z is of the form s/m, with (s, m) = 1, m;:: 1. Now let [A] E B(K) be represented by the cyclic algebra A = (W/K, a,n S), where W/K isunramified, a = a W / K' and (W:K) = m. Then inv [A] = s/m E Q/Z, so inv [A] = 0 if and only if mis, that is, if and only if A ~ K. This proves that inv is monic.

(31.9)

CYCLIC ALGEBRAS OVER LOCAL FIELDS

267

Finally, given [A], [B] EB(K), we may choose W/K unramified so that A = (W/K, a, rf),

B

= (W/K, a, nt),

a = a WIK ·

Then by (30.4). Therefore inv [A] [B] = inv [A]

+ inv [B],

which completes the proof that inv is an isomorphism of groups. We shall denote inv by inv K when we need to specify the underlying field K. Our next result, of great importance, describes the effect on inv of a change in ground fields. (31.9) THEOREM. Let E be any finite extension of K. The following diagram commutes:

B(K)~K E®K

1

~Q/Z

1

(E:K)

B(E) - ~~Q/Z, where the horizontal maps are isomorphisms, and where the second vertical map is defined to be multiplication by (E :K). Proof· Let A

= (W/K, a, n S ) represent a class [A] E B(K), where W/K is unramified, a = aWIK ' and (W:K) = m. We may view Wand E as embedded in some algebraic closure of K, and then we may form their intersection F = E n Wand their composite EW. If k is the least positive integer such

= k.

that d' fixes F, then F/K is unramified and (F:K) E

®K

A ~ (EW/E, a , n k

By (30.8) we have

S ),

so it remains for us to compute the Hasse invariant of the right hand expression. As in § 5, let 0E denote the valuation ring of E, 0E its residue class field, and vE the exponential valuation on E. Then (E: K) = ef,

Note that m

e = e(E/K),

f = f(E/K),

= (W:K) = (OW:OK). Let us put d = (m, f), m = dm',

f = dr,

where (m', f') = 1. Since d divides both m and f, there is a field (5 cow n 0E with (8:o K ) By (5.9) it follows that there is a field FeW n E = F, such that

= d.

268

(31.10)

CROSSED-PRODUCT ALGEBRAS

P/K is unramified, On the other hand, the field OF is contained in both Ow and 0E' whence f, hence divides d. But

(OF:OK) divides both m and

d = (P:K), Since we have just shown that kid, it follows that F = quently we obtain

P and k = d.

Conse-

f(E/F) = f(E/K)If(F/K) = f/d = f'. Diagrammatically, we have

E·W

/~.

Gal (EW/E) ~ Gal (W/F)

m'

W~/E

=

(Uk),

= (W: F) = (EW:E).

F=EnW

kl

K

Now let r be the Frobenius automorphism of EW/E, and view cI' as element of Gal (EW/E). By Exercise 31.1 (iv) we have r

= (crk)f(EIF)

= Ukf'.

But (f', m') = 1, and m' = (EW:E), so by (30.4 i) we obtain (EW/E, crk, n') ~ (EW/E, crkf', n'!'). Therefore

and so invE [E ®K A]

= vE(ns!')jm' = sf" vE(n)/m' = sf'e/m' = sfe/m = (E:K)·inv[AJ.

This completes the proof of the theorem. (31.10) Corollary. Let D be a skewfield with center K and index m, and let E be any finite extension of K. Then E splits D if and only if ml(E:K).

Proof Consider the element [DJ E B(K); then inv [DJ = s/m for some s relatively prime to m. Clearly E splits D if and only if invE [E ®K DJ = O.

269

EXERCISES

(31.11 ) On the other hand,

Since (s, m) = 1, we conclude that invE [E ®K D] = 0 if and only if m I(E: K). This gives the desired result. As a consequence of the preceding corollary, we obtain the following amazing result: (31.11) THEOREM. Let D be a skewjield with center K and index m. Then every irreducible m-th degree equation over K has a solution in D.

Proof. Let f(X) E K[X] be irreducible, and of degree m. Let E = K(a), where min. pol.K a = f(X). Then (E: K) = m, so E splits D by (31.10). It follows from (28.10) that there is a K-embedding of E into D. The image of a is then an element of D which satisfies the equation f(X) = O. We conclude with a further consequence of (31.8) and (31.9). (31.12) Theorem. Let E/K be any finite extension of degree n. Then B(E/K) is cyclic of order n.

Proof. By definition, B(E/K) consists of all classes [A] E B(K) which are split by E. Hence by (31.9) there is a commutative diagram 1 -> B(E/K)

->

B(K) -B(E)

1

1

Q/Z

---,,--+ Q/ z,

mVE

mVK

where the top row is exact, the bottom map is given by multiplication by n, and the vertical maps are isomorphisms. It follows at once that B(E/ K) is isomorphic to the kernel of the bottom map, that is,

B(E/K) ~ -~ Z;Z. This completes the proof. (31.13)

COROLLARY.

If (E: K) = n, then B(E/K) = {[A] EB(K):[A]n =1}.

270

CROSSED-PRODUCT ALGEBRAS

EXERCISES The hypotheses and notation given at the beginning of this section remain in force throughout these exercises. 1. Consider the diagram E·W

W/~ ~F/ I

K

in which all of the fields indicated are finite extensions of K. We assume that W/K is unramified, and denote its Frobenius automorphism by a WIK' (i) Prove that a W1K E Gal (W/K) is characterized by the condition

(ii) Prove that F / K is unramified, and that the epimorphism Gal (W/K) --> Gal (F/K) carries a W1K onto a F1K . If k is the least positive integer such that (aWIK)k

E

Gal (W/F),

prove that (F:K) = k,

aW/F = (aWIKt

(iii) Let E· W be the composite of E and W in some extension field of K. Show that EW/E is an unramified extension, and that a EWIE(restricted to W) = (aW/F'f[EIF)= (cfwIK'fiEfI'l,

where f(E/F) = (OE:OF)' (iv) When F = E n W, there is an isomorphism \f': Gal (W/F)

~

GaJ(EW/E),

obtained by extending each F-automorphism of W to an E-automorphism of EW For each a E Gal (ElIf1E), \P-l a is just the restriction ofa to W Deduce from (iii) that a EW1E = 'l'{(~WIK)f(EIF)}.

[Hint: The results are straightforward consequences of the definition of the Frobenius a utomorphism of an unramifiedextension. Detailed proofs area vaila blein the references listed at the beginning of §4.J 2. Let L/K be a cyclic extension of degree n, possibly ramified. Prove that K* /N LIK(L*) is cyclic of order n. [Hint: Use Exercise 30.1 and Theorem 31.12.J 3. Let K c E c F, where (F: E) = m, (E: K) = n. We make no assumptions as to ramification, nor do we assume that the extensions are galois extensions. Prove that the sequence of groups 1--> B(E/K)i;0 B(F/K)e®; B(F/E) --> 1

271

EXERCISES

is exact, where incl. is the inclusion map. [Hint: Consider the diagram

1

1-->B(EjK)----+B(FjK)

t I

lnYK

1·-+~ZjZ

lDVK

-;'~nZIZ

Show that the left hand square commutes (use (31.12)), and so does the right hand square (use (31.9)). Then use exactness of the bottom row.] 4. Let LjK be any cyclic extension, with galois group inv K [(Lj K,

CT,

ItS)]

=

s· inv K [(Lj K,

(CT>.

CT,

Prove that for

S E

Z,

It)].

5. Let R be the real field, C the complex field, R+ = {xER:x;;.O}.

Let N:C -;. R be the norm map. Prove that N(C)

=

R+.

6. Let D be a skew field with center C, such that (D: C) is finite. Prove that D = C. [Hint: Let a E D, and set min. pol. c(a) = f(X) E C[X]. Then there exist elements {IX,} in C such that f(X) = [I(X - IX,). Since [I(a - IX,) = 0, and D is a skewfield, some factor a - lXi must be zero.] 7. Let

CT

be complex conjugation on C, and let H be the cyclic algebra H

=

(CjR,

CT,-1).

Prove that H is the skewfield consisting of quaternions over R. [Hint: The least positive integer t such that ( -1)1 E NC/R(C*) is given by t = 2. Thus H is a skewfield, by (30.7). As in the example following (29.8), we have

H

~

REfJ Ri EfJ Rj EfJ Rk,

=

k 2 = -1, ij = - ji = k.]

where i2

=/

8. Prove that if D is a skewfield whose center contains R, and for which (D: R) is finite, then D = C, R, or H. [Hint: If the center of D is C, then D = C by Exercise 31.6. Suppose now that D has center R, and that the index of Dis m, where m > 1. Then D has a maximal subfield E, with (E: R) = m. Hence E = C, m = 2, and by §30 we may write D ~ (CjR,

CT,

a), a E R*,

where CT is complex conjugation. The isomorphism class of the cyclic algebra A = (CjR, CT, a) depends only on the coset of a in R* jN(C*), where N is the norm map from C to R. By Exercise 31.5, it follows that A is determined up to isomorphism by the sign of the element a. But we have (CjR,

CT,

1) ~ MiR),

so the result follows.] 9. Prove that [H]

E

B(R) has exponent 2.

(CjR,

CT,

-1) = H,

8. Simple Algebras Over Global Fields Throughout this chapter let K denote a global field (see §4e). Our aim is to study central simple K-algebras, and ideal class groups of maximal orders in such algebras. As general references for the topics considered below, the reader may consult Albert [1], Cassels-Frohlich [1 J, Deuring [1], Roggenkamp [IJ, Roggenl<;amp, Huber-Dyson [IJ, Swan-Evans [IJ, Weil [1]. We shall state without proof two basic results from class field theory, namely the Hasse Norm Theorem and the Grunwald-Wang Theorem. To include the proofs of these results in this book would require several additional chapters, and would adversely affect the proper balance of the subject matter covered in this book. Apart from these two theorems, however, the rest of the exposition is self-contained. 32. SPLITTING OF SIMPLE ALGEBRAS Let A be a central simple K-algebra, where K is a global field, and let P range over the primes of K (see §4e). In order to simplify the notation, we shall use Kp (rather than Kp) to denote the P-adic completion of K. Then put Ap

=

Kp ®K A

= P-adic completion of A.

By (7.8), Ap is a central simple Kp-algebra; thus the map [AJ ---+ [ApJ yields a homomorphism of Brauer groups B(K) ---+ B(Kp). Suppose that Ap ~ MK)S),

where S is a skewfield of index mp over its center K p' As in §§25, 29, we call the local index of A at P, and we write

mp

mp

= index [Ap].

We call Kp the local capacity of A at P. Clearly Ap ~ Kp

ifandonlyif

mp

= 1.

We shall say that A ram(fles at P, or that P is ramified in A, if mp > 1. The following theorem summarizes some of the main results of Chapter VI (see (22.4), (25.7) and (25.1 0)). 272

(32.1)

SPLITTING OF SIMPLE ALGEBRAS

273

(32.1) TlffioREM. Let R be a Dedekind domain whose quotientfield is a global field K, and assume that R =I K. Let A be a maximal R-order in the central simple K-algebra A. The nonzero prime ideals P of R, and the prime ideals'll of A, are in one-to-one correspondence, given by 'lliPA. Let mp be the local index of A at P and Kp the local capacity of A at P. Then (i) mp = 1 a.e., that is, Ap ~ Kp a.e. (ii) P A = 'll mp for all P. (iii) Let n(A/R) denote the d(fferent of A with respect to R, and d(A/R) the discriminant. Then

n(A/R) =

ITp 'llmp- \

d(A/R) =

{IT p1mp-l)"p}v'fA:Xl. p

(iv) We have m p > 1 <=> pld(A/R)

¢=>

~ln(A/R) <=> ~2ipA.

In the present discussion, the infinite primes of K will play an important role. Such infinite primes occur only when K is an algebraic number field. In this case, an infinite prime P of K corresponds to an archimedean valuation on K which extends the ordinary absolute value on the rational field Q. The P-adic completion Kp is either the real field R (in which case P is called a real prime), or else the complex field C (and P is a complex prime). (32.2) THEOREM. Let A be a central simple K-algebra, and let mp be the local index of A at an in.finite prime P of K. (i) If P is a complex prime, then Ap

~

Kp

= 1.

and

mp

and

mp = 1,

(ii) If P is a real prime, then either Ap ~ Kp or else

Ap ~ Hand

mp

= 2,

where H is the skewfield of real quaternions (see Exercise 31.7). Proof Clearly Ap ~ D, where D is a skew field with center K p, and (D:Kp) is finite. If P is complex, then D = C = Kp by Exercise 31.6. If P is real, then Kp = R, and D = R or H by Exercise 31.8. This completes the proof.

If P is any finite prime of K, then Kp is a complete field relative to a discrete valuation, and has a finite residue class field. We saw in §3l how to define the

274

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(32.3)

Hasse invariant inv [A p] of a central simple K p-algebra, thereby obtaining an isomorphism inv:B(Kp) ~ Q/Z. We showed in fact (see (31.4) and (31.7)) that (32.3)

inv [Ap] { exp [A p]

=

sp/mp

.

= 1, = mp. , where mp = mdex [A P] ' (sp' m) P

We would like to have the same formulas true for the case of infinite primes. First we must define Hasse invariants when P is an infinite prime, and in light of (32.2), it is sufficient to define these invariants for the three cases C, Rand H. We set (32.4)

inv [C]

=

inv [R] = 0,

0,

inv [H] =

±.

Formulas (32.3) then hold equally well when P is infinite, provided we know that exp [H] =·2 when [H] is considered as an element of B(R). We have already observed this fact in Exercise 31.9. Now let A be any central simple K-algebra, and let P be any prime of K (finite or infinite). Clearly,

A

~

K

= Ap ~ Kp for all P.

We shall prove the extremely important converse of this implication, by using the Hasse Norm Theorem (see below). First of all, we must review some facts from galois theory. Some of these results have already been mentioned in §4 and §5c. Proofs are readily available in the references listed at the beginning of §4. Let L be a finite galois extension of K, with galois group G = Gal (L/K). Let P be a prime of K, finite or infinite. Even when P is a finite prime, it will be convenient to think of P as representing a class of valuations onK, rather than an ideal in some valuation ring. From this point of view, the valuation P extends to a finite set of inequivalent valuations on L, denoted by p (= p 1)' p 2' ... , p g. For each 0" E G, there is a valuation pO" on L, defined by the formula pO'(x)

=

p( 0"-1 x),

XEL.

We call pO' a conjugate of p; if p is a finite prime, then 0" carries the valuation ring of p onto the valuation ring of pO'. Whether or not p is finite, each Pi is of the form pO' for some 0" E G. Keeping the above notation, we set (32.5)

G p = {aEG:p" = p},

and call Gp the decomposition group of p relative to the extension L/ K. The groups {Gp ,} are mutually conjugate in G. Each O"E Gp induces a Kp-automorphism ft of the p-adic completion L p , since 0" maps each Cauchy sequence from L (relative to the p-adic valuation) onto another such sequence. The

(32.6)

rna p a

SPLITTING OF SIMPLE ALGEBRAS fj

-4

275

yields an isomorphism Gp ~ Gal (Lp/K p).

(32.6)

We define (32.7)

np

=

(Lp:Kp)

=

local degree of L/K at P.

Then (32.8)

np = 1Gp I,

and npl (L:K) for each P.

It should be remarked that the fields {Lp,: 1 ~ i ~ g} are mutually Kp-isomorphic, so np does not depend on the choice of the prime p of L which extends P. The next theorem is of fundamental importance for the entire theory of simple algebras over global fields. The proof depends on class field theory, and is beyond the scope of this book. Proofs may be found in CasselsFrohlich [1], Janusz [1], Neukirch [1], and Weil [1]. (32.9) THEOREM (Hasse Norm Theorem). Let L be a jinite cyclic extension of the globaljield K, and let aEK. For each prime P of K, we choose a prime p

of L which extends P. Then a EN LIK(L) ~ a E N LpIKp(Lp)jor each P. Remarks. (i) The theorem asserts that a is a global norm (from L to K) if and only if at each P, a is a local norm (from Lp to K ). (ii) If p and p' are primes of L, both of whiCh extend P, then there is a Kp-isomorphism Lp ~ L p" and therefore N LpIKp(L p )

=

N Lp'IKp(L p').

This shows that in determining local norms at P, it does not matter which prime p of L we use, provided only that p is an extension of the valuation P from K to L. (iii) By Exercise 32.3, every global norm is also a local norm at each P. The difficult part of the proof of Hasse's Norm Theorem is the converse: if a E K is a local norm at each P, then a is a global norm. In proving this, it is necessary to know that a is a local norm at every prime P of K, including the infinite primes. (iv) The theorem breaks down if we drop the hypothesis that L/K be cyclic. There are counterexamples even when L/K is abelian (see CasselsFrohlich [1, Exercise 5]). (32.1 0) COROLLARY. Let A = (L/K, a, a) bea cyclic algebra, where Gal (L/K) =
276

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(32.11)

Proof By (30.4), A ~ K <= a E NL/K(D'). On the other hand, we showed in (30.8) that for each P, ~

Ap

(Lp/Kp, (Jk, a),

where cI is some generator of Gal (Lp/Kp). Therefore by (30.4), Ap

~

Kp

<=:>

aENLp/K)Lt).

The desired result now follows from the Hasse Norm Theorem. Clearly, we could have deduced (32.9) from (32.10), and so the n\':xt result may be regarded as a generaliza tion of the Hasse Norm Theorem. Historically, it was first proved by use of (32.1 0). (32.11) Theorem (Hasse-Brauer-Noether-Albert). Let A be a central simple K-algebra. Then A

~

K

<=:>

Ap

~

Kp for each

prime

P ofK.

Proof If A ~ K, then clearly Ap ~ Kp for each P. We shall prove the converse by assuming that Ap ~ Kp for each P, but that m = index [A] > 1, and obtaining a contradiction. By (28.11), there exists a finite galois extension L/K such that L splits A, that is, L @K A ~ L. Let G = Gal(L/K), let p be some rational prime dividing m, and let H be a Sylow p-subgroup of G. Then p t [G: H], whereas IHI is a power of p. It is well known from group theory that H has a descending chain of subgroups .. , :;) H n = 1."

[H:H i i

+1

] = p,

with Hi + 1 D. Hi for each i. Let Ei be the sub field of L fixed by Hi , so En = L, and each Ei+ 1/Ei is a cyclic extension of degree p.

Hn-1 :: H

,,

G

L=E

I

n

F=E n-l I

Eo I K

Further,

(Eo:K) = [G:H]

=1=

0 (mod p).

Set F = En _ l' so L/F is a cyclic extension of degree p.

(32.12)

277

SPLITTING OF SIMPLE ALGEBRAS

Let us set B = F (8) K A, a central simple F -algebra. Then L(8)FB~L(8)KA~L,

so L splits B. Hence by Exercise 30.3, B is similar to a cyclic algebra C = (L/F, cr, b), where Gal (L/F) = (cr) and b E F*. We intend to show that B ~ F, so it suffices for us to prove that C ~ F. By (32.10), C

~

F

~

Cp

~

Fp for every prime p of F.

Given a prime p of F, its restriction to K is a prime P of K, and we have Cp

~

Bp = F p (8) F B ~

~

Fp (8)KpAp

F p (8) K A ~

Fp.

This proves that C splits locally everywhere, whence C ~ F, and so B ~ F. We have thus shown that F splits the algebra A. Repeat the argument, using the cyclic extension F/E n _ 2 in place of L/F. Then we find that E._2 also splits A. Continuing in this manner, we see that Eo splits A. But then m I(Eo: K) by (28.5). This is impossible, since pi m but pt(Eo:K), and so the theorem is established. (32.12) Remarks (i) For each prime P of K, there is a homomorphism B(K) -> B(Kp), defmed by Kp (8)K·' Let [A] E B(K), and let mp be the local index of A at P. Then mp = 1 a.e., which means that [Ap] = 1 a.e. Hence there is a well defined homomorphism

The Hasse-Brauer-Noether-Albert Theorem is precisely the assertion that this map is monic. (ii) A stronger result, due to Hasse, describes the image of B(K) in L'B(Kp) by means of Hasse invariants. It can be shown (see references) that the following sequence is exact: (32.13)

1 -> B(K)

->

L' B(Kp) ~ Q/Z

->

0,

p

where inv denotes the Hasse invariant map, computed locally on each component: inv = L'invKp ' It follows from the exactness of (32.13) that (32.l3a)

Lp inv [Ap]

= 0,

[A] E B(K).

t

Of course, inv [Ap] = 0 if P is a complex prime, while inv [Ap] = 0 or if P is a real prime. The exactness of (32.13) also tells us that, other than (32.l3a), these are the only conditions which the set of local invariants {inv [Ap]} must

278

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(32.14)

satisfy. In other words, suppose that we are given in advance any set of fractions {x p } from Q/ Z, such that

xp = 0 { xp = 0

a.e., Ixp = 0 if P is complex, xl' = 0 or~ if P is real.

Then there exists a unique [A]

E

B(K) such that

inv [Ap] = xp forallP. (iii) Some authors prove the exactness of (32.13) directly (see CasselsFrohlich [1], Neukirch [1]), and then deduce (32.9H32.11) as corollaries. To prove that (32.13) is exact, one shows first that the following sequence is exact: (32.14)

2

1 -+ H (G L1K ,L*)

-+

I" H2(GLpIKJ>' Ln ~ ~ Z/Z -+ o. p

Here, L/ K is any finite cyclic extension, and n = (L: K); the direct sum extends over all primes P of K, and for each such P, the prime p is some prime of L extending P. The isomorphism obtained in (29.12) permits one to replace H 2 (G L1K ,L*) by B(L/K), and likewise for the terms in the direct sum. Finally, one proves (32.14a)

B(K) =

U B(L/ K), L

where L ranges over all finite cyclic extensions of K. In (28.12) we proved a weaker version of (32.14a), in which L ranges over all finite galois extensions. By (32.20) below, it suffices to consider only cyclic extensions. Formula (32.14a) can also be proved directly, without using the difficult Theorem 32.20. We may remark that if L/K is a finite galois extension, not necessarily cyclic, then the image of inv in (32.14) may be a proper subgroup of (l/n)Z/Z. For the proofs of the exactness of (32.13) and (32.14), the reader may consult the references listed above. As a first application of (32.11), we give a simple criterion for deciding whether a finite extension of the global field K splits a given central simple K-algebra. (32.15) THEOREM. Let A be a central simple K-algebra. For each prime P of K, let mp = index [Ap]. Let L be any finite extension of K, not necessarily a galois extension. 1hen L is a splitting field for A if and only if for each prime p of L, (32.16)

mpl(Lp:Kp),

where P is the restriction of p to K.

(32.17)

279

SPLITTING OF SIMPLE ALGEBRAS

Proof If P is the restriction to K of a prime p of L, then LP®KpAp ~ Lp ®L(L®K A ).

Hence if L splits A, then L ®K A ~ L, whence Lp ®K p Ap Lp splits Ap. Then (32.16) holds, by (28.5).

~

L p, and so

Conversely, suppose that (32.16) holds for eachp. Then (for each p) Lp splits A p , by (31.10). It follows that the central simple L-algebra L ®K A is split locally at every prime p of L. Hence by the Hasse--Brauer-Noether-Albert Theorem (32.11), L ®K A ~ L. Therefore L splits A, as claimed, and the theorem is proved. A second important consequence of (32.11) is (32.17) THEOREM. Let A be a central simple K-algebra with local indices {m p}, where P ranges over the primes of K. Set m'

= L.C.M. {m p }.

Then exp [A] = m'. Proof By (32.11), [AJ'

=

1 in B(K)

=-

[A pJ'

=

1 in B(Kp) for each P.

But exp [Ap] = mp by (32.3), so [ApJ' = 1 if and only if mp I t. This proves that [AJ' = 1 if and only if t is divisible by each mp. Hence exp [A] = m', as claimed. We are going to prove below that whenever K is a global field, we have index [A] = exp [A],

[A]

E

B(K).

We shall also prove that each central simple K-algebra A is in fact a cyclic algebra, so that (32.10) and (32.11) say exactly the same thing. The proofs will depend on the following result, whose proof is beyond the scope of this book: (32.18) THEOREM (Grunwald-Wang). Let P l ' ... , P s be any set of distinct primes of the globaljield K. Let {m p ,: 1 ~ i ~ s} be positive integers, subject to the conditions: mp = 1 ~r P is complex, mp = 1 or 2 if P is real, where P ranges over the {PJ Let n be any positive integer divisible by each m p" 1 ~ i ~ s. Then there exists a cyclic extension L/K such that (L: K)

= n, (Lp: Kp) = mp,

where p denotes any prime of L extending P. Remarks. (i) For the proof of the Grunwald-Wang theorem, see Artin-Tate

280

(32.19)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

[1, Ch. 10] or Neukirch [2]. Wang [1] gives the prooffor the case where K is an algebraic number field. (ii) For any galois extension L/K and any prime P of K, we know that (Lp: Kp) must divide (L: K), by (32.8). Hence if L/K has local degree mp at P, for P = P 1> ... ,P s' then necessarily (L: K) must be a multiple of m', where

m' = L.C.M. {m p" ... , mpJ. Hence if we seek a galois extension L/K with prescribed local degrees {mp} at some preassigned finite set of primes {PJ of K, then m' is the smallest possible value for (L: K). The Grunwald-Wang theorem tells us that not only can we find such a galois extension L/K of degree m', but in fact there always exists a cyclic extension of degree m', with the prescribed local degrees. Further, we can force (L:K) to be any preassigned multiple of m', if desired. We are now ready to prove (32.19) Theorem. Let [A]

E

B(K) have local indices {mp}. Then

index [A] = exp [A]

= L.C.M. {mp}.

Proof We know that mp = 1 except possibly at some finite set of primes {P 1"'" Ps} of K. We also know that mp = 1 if P is c'omplex, and that mp = 1 or 2 if P is real. Let m' = L.C.M. {mp}, so m' = exp [A] by (32.17). If m = index [A], then m' Im by (29.22). On the other hand, by the Grunwald-Wang Theorem, there exists a cyclic extension L/K such that (L:K) = m',

where as usual p denotes a prime of L extending P. Since the local degree (Lp: Kp) is the same for each such p, and since mp = 1 for P P 1" .. , Ps,lt follows from (32.15) that L splits A. But then ml (L: K), by (28.5i). Therefore m = m', and the theorem is proved.

"*

(32.20) Theorem. Every central simple K-algebra is cyclic.

Proof Let A be a central simple K-algebra, let (A:K) = n2, and let {mp} be the local indices of A. Since (Ap: Kp) = n2, it follows that mpl n for each P. Further, mp = 1 except (say) at P l' ... , P s ' By (32.18) it follows that we may find a cyclic extension E/ K such that (E:K) = n, Then E splits A, and (A:K) = (E:K)2. Hence, by Exercise 30.2, we may conclude that A is isomorphic to a cyclic algebra (E/ K, a, a), where Gal (E/ K) = (a) and a E K*. This completes the proof.

EXERCISES

(32.20)

281

Example. Let us determine some of the local Hasse invariants of the cyclic algebra A = (L/K, c;, a), where Gal (L/K) = (c;) and a E K*. Let P denote a prime of K, and p an extension of P to L Then by (30.8) we have

Ap

~

(Lp/Kp,(Jk, a),

where k is the least positive integer such that C;k lies in the decomposition group Gp of p relative to L/K. Of course, inv [Ap] = 0 whenever Ap ~ Kp. (i) If P is complex, or if both P and p are real, then Ap ~ Kp. (ii) Suppose that P is real, p complex. By Exercise 31.8 we have

Ap

~

Kp

Ap

~

H

if ap > 0, if ap < 0,

where a p represents the image of a under the embedding K --+ Kp. In the latter case, inv [Ap] = t. (iii) Let P be a finite prime, and assume that Pis unramified in the extension L/K. This is equivalent to assuming that Lp/Kp is unramified. Since Gp = (C;k), we may choose r E Z relatively prime to the local degree np = IGp I, such that c;kr is the Frobenius automorphism of the extension Lp/Kp. By (31.7) we obtain inv [Ap] = r' vp(a)/np, where vp is the exponential P-adic valuation. If we reduce the fraction r' vp(a)/np to lowest terms, then mp is the denominator of the fraction thus obtained. In particular, we observe that mp = 1 whenever vp(a) = O. Thus, mp = 1 for every finite prime P, except possibly those primes P which ramify in L/ K, or which contain a. EXERCISES Throughout these exercises, let A denote a central simple K-algebra, where K is a global field, and let P range over the primes of K. 1. Prove that either A ~ K, or else there are at least two primes of K which ramify in A. [Hint: Use (32.1 3a)].

2. Prove that no infinite primes of K are ramified in A, if index [A] is odd. 3. Let L/K be a galois extension, p a prime of L extending the prime P of K. Let Gp be the decomposition group ofp relative to L/K, and write n

G

=

UG

p . (Jp

i= 1

For x Eo' L, let

n

f(x) =

T1 i=l

(J

i(X) Eo'L.

282

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(33.1 )

Prove that N LpIKp(f(X)) = N LIK(X).

Deduce that if a E K is a global nonn (from L to K), then a is a local norm (from Lp to Kp) at each P. [Hint: n

IT IT

NLPIKp(f(X)) =

po)x)

=

peG p i;:::l

IT o-(x) =

NLIK(x).]

dEG

33. REDUCED NORMS Throughout this section K is assumed to. be a global field, unless otherwise stated. :let nr AIK: A ....... K be the reduced norm map, where A is a central simple K-algebra, and let nr A = nrAIK(A) = {nrAIK a:aEA}. Our aim is to determine the image nr A explicitly. Our first result, true for arbitrary fields K, enables us to reduce the problem to the case where A is a skewfield. (33.1) THEOREM. Let A = M,(D), where D is a skewfield with cencer K (not necessarily a global field). Then (i) nrAIK(A) = nr DIK(D). (ii) For a E A, nr AIK a = 0 if and only if a is a nonunit of A. Proof We shall omit the subscripts A/K and D/K when there is no danger of confusion. As in Exercise 9.5, let E be a splitting field for D, and let

WD ....... E ®KD

~

Ms{E).

Then there is an embedding given by a

= (Ct.;)u;;. j<:;' E M,(D) ....... (j.t(Ct.;)

EM ,iE).

By definition of the reduced norm map, we have

nr a = det f}t(Ct. i ) , In particular, if a = (Ct.;) is an upper triangular matrix, then (33.2)

nr a

=

fI det(I'(Ct.;) = fI nr Ct.;i'

i= 1

;= 1

where nr Ct.;; means nr DIK Ct.;;' The same holds true if a is lower triangular.

)3.3)

283

REDUCED NORMS

Finally, let t E M,(D) be any transposition matrix. From the identity

G~) G-~) C~) (~ -~), =

it follows at once that nr t = nr D/ K ( -1). By an "elementary matrix" in M,(D) we shall mean a matrix e which coincides with the r x r identity matrix, except for having a single nonzero entry off the main diagonal. By (33.2), nr e = 1. Given an arbitrary a E A, we can find matrices p, q E A which are products of elementary matrices and transpositions, such that paq = diag (ct1' ... , ct,), ct; ED. Note that a E u(A) if and only if each ct, E u(D). Since nr A/K is multiplicative, and since nr p and nr q can only have the values nr D/K( ± 1), it follows that nr a = nr (±ct 1 ··· ctJ This implies (i) immediately. It also proves (ii), since nr (± ct j if and only if some ct, = O.

•••

ct,) = 0

Let P be a prime of K, and let the subscript P indicate P-adic completion. For any K-algebra A, the embedding A ....... KpQ9KA = Ap carries each x (33.3)

E

THEOREM.

A onto an element xp of Ap. We have at once Let A be a central simple K-algebra, and let P be a prime of

K. Then nr A/K X = nr Ap/Kp

Xp ,

XEA.

Proof Use (9.28), with F replaced by Kp. We call elements of nr A global reduced norms, and those of nr Ap local reduced norms. The above theorem tells us that every global reduced norm is also a local reduced norm at each P. We shall eventually prove the converse, when K is a global field. As a first step, let us determine which elements are local reduced norms. (33.4) Theorem. Let P be a prime of the global field K, and let B be any central

simple Kp-algebra. Then nr B / Kp B = K p , except for the special case where Kp = Rand B '" H. In that case, nr B = nr H = R+ = {ctE R:ct ~ O}.

284

(33.5)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

Proof By (33.1) we may assume that B is a skewfield S with center Kp. If P is a finite prime, then nr S = K p by Exercise 14.6. If P is an infinite prime, then by (322) we have S = K p (and so also nr S = Kp), except for the special case where K p = R, S = H. It remains for us to prove that nr H = R+. Let a = ao + at i + a2 j + a3 kEH, avER. Then (see (9.4)) we have nrH1R a -_ ao2 + at2 + a22 + a32 E R+ .

Conversely, each a E R+ is expressible in the form a~, nr H = R+, and completes the proof.

:Xo E

R. This shows that

Now let A be a central simple K-algebra, where K is a global field, and let P be a prime of K. For a E K, let ap denote its image in K p' By (33.3), we have

(33.5)

a E nr A

=

ap E nr Ap for each P.

If we are trying to decide whether a given element a E K is a global reduced norm, we must therefore first check whether a is a local reduced norm at each P. By (33.4), the condition that ap E nr Ap is automatically satisfied for every finite prime P, and also for every infinite prime P as well, except for the special case where Kp = Rand Ap ~ H. This special case can only arise when P is a real prime which is ram~(zed in A, using the terminology introduced in §32. Given the central simple K-algebra A, we define

(33.6) U(A) = {aEK:a p > OateachrealprimePofKramifiedinA}. The preceding discussion then shows that for a E K*,

(33.7)

ap E nr Ap

for all P

=-

a E U(A).

We intend to prove that a E nr A if and only if a E U(A), and we begin with two preliminary lemmas which are of interest in themselves. (33.8)

LEMMA.

Let K be any field complete with respect to a valuation f(X) = xn

I I. Let

+ atXn-t + ... + anEK[X]

be a separable irreducible polynomial. Set L = K(~), where min. pol.K ~ = f(X). Then for any polynomial g(X) = btxn-t + ... + bnEK[X], for which the values Ibtl, ... , IbnIare sufficiently small, the polynomial f(X) + g(X) is also irreducible and separable, and has a zero in L. Proof. Step 1. The result is obvious when n = 1, so assume that n > 1 hereafter. Let us first settle the case where the valuation is archimedean, in

(33.9)

285

REDUCED NORMS

which case K = R or C. Since n > 1, the only possibility is that K = R, L = C, and f(X) is a quadratic polynomial with no real zeros. If the coefficients of g(X) are sufficiently small, then also f(X) + g(X) has no real zeros, so f(X) + g(X) is irreducible in K[ X], and has a zero in L. Thus the result holds in this case. Step 2. (Krasner's Lemma). For the rest of the proof, we assume that the valuation is non-archimedean, and that n > 1. Let K be an algebraic closure of K containing L, and extend I I to a valuation on K as in §5b. Two elements 1', y' in K are conjugate (over K) if y' = cp(y) for some K-automorphism cp of K; this occurs if and only if I' and y' have the same minimal polynomial over K. It follows from (5.5) that II' I = 11"1 whenever I' and )1' are conjugate. Now let f(X) = (X - ~1)" '(X - ~n)' and set h = min {I ~i

-

~ 1:2 ~ i ~ n}.

Since f(X) is separable by hypothesis, the elements ~ l' ••• , ~n and so h > 0. We shall prove that (33.9)

fJ

EO

K, IfJ

- ~I< h

i;

==>

EO

EO

K are distinct,

K(fJ)·

Suppose that (33.9) is false, and let fJ be a counterexample, so ~ ¢ K(fJ). Let m(X) = min. POl.Klll)~' Then m(X)lf(X), and m(X) has degree greater than 1. Since f(X) has n distinct zeros, namely ~ 1 ' ... , ~n' it follows that m(O = for some ~i =I ~. Therefore the elements fJ - ~ and fJ - ~i are conjugate over K(fJ), hence over K, and so I fJ - ~ I = IfJ - ~J But then

°

I~ - ~il ~ min(lfJ - ~1,lfJ - ~il) = IfJ - ~I < h,

which contradicts the choice of h. We have thus proved (33.9), which may be rephrased as follows: If fJ EO Kis closer to ~ than any conjugate of ~, then ~ EO K(fJ). Step 3. We are now assuming that n > 1; it follows that ~ =I 0, since otherwise f(X) is reducible. Hence we may choose d EO R such that

keeping the notation of Step 2. We shall show that if the coefficients of g(X) are such that 1 ~j

~

n,

then f(X) + g(X) is irreducible, and has a zero in L. We show first that (33.10)

1~j

~

n.

286

(33.11)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

Indeed, for each such j, the coefficient aj in f(X) is (apart from sign) the elementary symmetric function of degree j in the quantities g 1 " ' " ~.}. Since these elements are mutually conjugate over K, we have I~ 1 I = ... = I~J But then (33.10) follows at once, by using the properties (4.l9i-iv) of the valuation I I. (It should be pointed out that (33.10) is merely a restatement of Lemma 12.9.) Next, for 1 ~ j ~ n we obtain

whence

laj + bjl If fJ E it is any zero of f (X) that

ltl(a +

IfJln =

(33.11)

+

j

~ max(lajl, Ibjl) ~ 1~lj·

g(X), then f(fJ)

b)fJn-jl

+ g(fJ) = 0, and so we conclude

~ lrr;::n {1~lj·lfJln-j}.

If I~ I < IfJl , then j max {1~lj ·lfJln- } < max {lfJln}

l~J~n

=

IfJln,

l~j~n

which contradicts (33.11). Therefore IfJl ~ 1~ I. Furthermore, n

TI (fJ -

= f(f3) = -g(fJ) = -(b1fJ n- 1 + '" + bJ

~)

i= 1

This gives { 111 in

l~i~n

max { Ibj II fJl n- j } ~ max {d j

fJ - ~,I} n ~

l:::;)~n

I

l:"1Sj-~n

·

1~ 1n- j}

=

d'

I.; 1n- \

where the last step follows from the fact that d < 1~ I. Since d . 1~ 1n- 1 < hn, we conclude that I

fJ - ~i 1 < h for some i,

1 ~ i ~ n.

Now let ¢ be a K-automorphism of it such that ¢ (~) 1

= ~; then

¢ (fJ) - ~ 1 = I fJ - ~i 1 < h,

and ¢(fJ) is also a zero of f (X) + 9 (X). Set Y = ¢(fJ), so y is a zero of f(X) + g(X), and ~I < h. By Step 1, it follows that ~ E K(y), Therefore

Iy -

(K(y):K)

~ (K(~):K)

= n.

On the other hand, y is a zero of the nth degree polynomial f(X)

+ g(X).

It

(33.12)

287

REDUCED NORMS

follows that this polynomial is irreducible in K[X], and that K(y) = Thus y E L, and f(X) + g(X) is separable. This completes the proof.

K(~).

Our second preliminary result is (33.12) LEMMA. Let R be a complete discrete valuation ring, with quotient field K and finite residue class field R. Let W be an unramUied extension of K of degree n. Then for each a E u(R), there exists an element YEW such that W

=

K(y).

Proof. The result is obvious when n = 1, so take n > 1 hereafter. Denote NW/K by N, for brevity. By (14.1) we know that a = Nx for some x E W, and we must modify x so that it generates W. By (5.10) we may set W = K(w), where w is a primitive (qn - l)th root of lover K, and card R = q. Then WIK is a (separable) galois extension; its galois group is cyclic of order n, and is generated by the Frobenius automorphism (J, where (J(w) = w q. Thus there is a one-to-one correspondence between the set of divisors of n, and the set of intermediate fields L between K and W. Consequently there are at most n such intermediate fields. Let us put O~j~l1-l.

We have Nw q- 1 =

n (Ji(wq-1) = W(q-1)(1+q+"+q"-1)

n-1

1.

i=O

Therefore N(x' Wi (q-1» = Nx = a,

O~j~n-l.

If L. = W for some j, then the desired element YEW such that Ny = IX, W = K(Y), can be chosen as X' wi(q- 0, and we are through. We shall now assume that each of the n fields Lo, ... , L._1 is a proper subfield of W, and we shall obtain a contradiction. We saw above that there are at most n fields between K and W, including the extreme cases K and W. Since each of the n fields {L i : 0 ~ j ~ n - l} is assumed to be distinct from W, it follows that two of the L's must coincide. Suppose that Li = Lj' where 0 ~ i < j ~ n - 1. Since a E u(R) and N x = IX, surely x "# O. Thus we may form the quotient x . wi(q-1)lx' Wi(q-1) ELi'

and so Li contains an element wt(q -1), where 0 < t < n - 1. Now (W:LJ > 1 by assumption, and therefore Li is fixed by some (Js, with 1 ~ s< n, sl n. Thus s ~ n12. The equation

288

(33.13)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

yields the congruence qS . t(q - 1) == t(q - 1)

(mod(qn - 1)).

Therefore qn - I must divide the positive integer

A. = (qS - 1) t . (q - 1). But

A. :( (qn/2 - l)(q - I)(n - 2). The reader will easily verify that the right hand expression is always less than q" - 1. This gives the desired contradiction, and completes the proof of the lemma. (33.13) COROLLARY. Let R be a complete discrete valuation ring with quotient field K, prime element n, and finite residue class field R. Then for each nonzero {3 E K, and each positive integer n, there exists a monic separable irreducible polynomial f(X) E K[ X] of degree n, with constant term ( -1 )n{3. Furthermore, if {3 E R then f(X) may be chosen in R[ Xl Proof. Given {3 E K* and n ~ 1, we need only show tha t there exist a separable extension LIK of degree n, and an element x E L such that

{3 =

L = K(x).

NL/KX,

Once this is done, we may pick f(X) = min. P01.KX. Then f(X) is irreducible and has degree n, since (L: K) = n. Further, f(X) is a separable polynomial because LI K is separable. We note also that if {3 E R, then f(X) E R[ X] by virtue of(12.2). The rest of the proof is devoted to showing how to find Land x. Set {3 = nrC!., r E Z, C!. E u(R); then vK C{3) = r, where vK is the exponential valuation associated with R (see §5b). If n I r, then put s = rln E Z, and choose Wand y as in (33.12). Then W

= K(nSy),

(W: K)

= n.

Hence we may choose L = W, x = n y, in this case where n I r. We now turn to the case where ntr, and put S

d = (n, r),

n = md,

r = sd,

where

(m, s) = 1.

Since ntr, it follows that m =F 1, and thus that m > 1. Now let ElK be an unramified extension of degree d. Since d I r, we may deduce from the preceding paragraph that there exists an element y E E such that E

= K(y).

Now n is also a prime element for the valuation ring of E, since ElK is

(33.14)

289

REDUCED NORMS

unramified. Therefore by (5.6) we have

r = vK({3) = vE({3) = (E: K)vE(y) which shows that VE(Y) = r/d = s.

= d· vE(y),

Let us choose t = Is I + 1, and consider the separable polynomial

g(X)

= xm + ntX + (_l)my

EE[X].

Let z be a zero of g(X) in some algebraic closure of E, and set L L/E is--a separable extension, and (L: E) :( m. As in (5.6) we set

e so ef

= (L:E)

(33.14)

= e(L/E),

f = f(L/E),

:( m, and vL(n) = e. Since g(z) uL(Y) ~ min {vL(zm), vL(ntz)}

where h = vL(z). If mh ~ et

= E(z). Then

= 0, we have = min {mh, et + h},

+ h, then (m - l)h ~ et > O.

Since m > 1, this would give h > 0, and then from (33.14) we would obtain

es = vL(y)

~

et

+h>

et.

This is impossible, since t = lsi + 1, and so it follows that mh < et + h. But then (see (4.l9a, (v)), the equation g(z) = 0 yields a sharper form of (33.14), namely,

es = vL(Y) = min {mh, et

+ h} =

mho

Therefore m Ies, whence m Ie since (m, s) = 1. This proves that m = e, and therefore L/E is a separable extension of degree m, such that L = E(z). Since Y E K(z) and E = K(y), it follows that

L = K(z), (L: K) = md = n, L/K separable. Finally,

{3 = NE/Ky = NE/K(NL/EZ) = NL/KZ by Exercise 1.5. This completes the proof of the corollary. With these preliminaries settled, we are now ready to prove the following important result: (33.15) Theorem. (Hasse-Schilling-Maass). Let A be a central simple Kalgebra, where K is a global field, and let rt E K*. Then rt is the reduced norm of an element of A if and only if rtp > 0 at every infinite prime P of K ramified in A.

290

(33.15)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

Proof. Step 1. Since rx E K*, it follows from (33.1) that rx is a reduced norm if and only if rx is the reduced norm of an invertible element of A. We are trying to prove that

nrA/K u(A)

=

U(A),

where U(A) is defined by (33.6). Now rx E nr A ==> rxp

E

for all P ==> rx E U(A),

nr Ap

by (33.5) and (33.7). Hence we need only prove that, conversely, U(A) c nr A. In what follows, let [3 E U(A). Throughout this proof, let S denote the set of all infinite primes P of K which are ramified in A. Possibly S is the empty set. Choose S' to be a nonempty finite set of finite primes of K, including each finite P ramified in A. Then P rj: SuS' Step 2. Let (A:K) polynomial Ip(X)

= X"

=n

2

.

---~

Ap

~

Kp.

By (33.13), for each PES' we may find a separable

+ a1.pX"-1 + ... + a"_UX + (-1)"[3

EKp[X]

such that Ip(X) is irreducible over Kp. If the set S is empty, proceed directly to the next paragraph. If S is non-empty, then some infinite prime of K ramifies in A, so n is even by Exercise 32.2. We then set Ip(X)

= X" + (-It[3

E

Kp[X],

PES.

Given any e > 0, by (4.11) we can choose a polynomial I(X)

=

X" + C1xn-l + .. ' +Cn-1X + (-1)"[3

EK[X],

such that qJp(c; - ai • p ) < e, { qJp(C) < e,

1::;: i ::;: n - 1,

PES',

PES.

By (33.8), if we pick a sufficiently small (;, then I(X) will be irreducible in Kp[ X] for each PES'. Hence I(X) is also irreducible over K, since S' is non-empty. Furthermore, since [3 E U(A), we have [3p > for each PES. Hence if <: is small enough, we can be sure that for each PES, the polynomial I(X) has no zeros in the real field Kp. Step 3. Now choose an element x in an algebraic closure of K, such that I(x) = 0, and set L = K(x). Then

°

(L:K)

= n,

min. pol.Kx = I(X),

NL/K

x = [3.

291

EXERCISES

We proceed to calculate the local degrees np = (Lp:Kp),

PES uS',

where for each PES uS', the prime p ranges over all primes of L extending P. For each prime PES, the polynomial f(X} has no real zeros in Kp. From the discussion in §5c, it follows that each prime p of L which extends P must be complex. Therefore np = 2 for each such p. On the other hand, for PES' the polynomial f(X} is irreducible in Kp[X]. Again using §5c, we may conclude that np = n, and indeed that there is only one p extending P. N ow let {mp} be the set of local indices of the central simple K -algebra A. Then surely mp I n for each P, and P¢3S uS'; Hence for every prime P of K, and every prime p of L extending P, it follows that Inpl (Lp: Kp). Therefore L is a splitting field for A, by (32.15). Finally, we note that (A:K) = n 2 = (L:Kf, Since L splits A, it follows from (28.10) that we may embed L in A as a self-centralizing maximal subfield of A. Therefore

f3 =

=

NL/KX

nrA/Kx,

by Exercise 9.1, and the proof is finished. Remark. In the above proof, we used the Strong Approximation Theorem (4.11). It would have been enough to apply a weaker version of(4.11), obtained by deleting from the conclusion of (4.11) the assertion that "b E Rp for all P -# PI'" ., P n"· This weaker version is called the Weak Approximation Theorem, and is true f or any domain R (not necessarily a Dedekind domain). EXERCISES 1. Let K be a field with valuation I

I,

and V =

f: KVi

a finite dimensional

i=l

K -space. Define a norm 0 n V by setting n

II

L: i=

IXivill = max{IIX 1 1, .. ·,IIXn l},

1

Prove the following: (i) V is a metric space relative to this norm, with distance between two elements v, v' e V given by Ilv - v'll. (ii) This topology on V is independent or the choice of K-basis of V. (iii) If K is complete with respect to I I, then V is a complete metric space.

292

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

2. Let A be a central simple K-algebra, where K is a field with a valuation I I, and define a norm on A as in the preceding exercise. Let nr: A ..... K be the reduced norm map. Prove that nr is continuous, that is, if x, Y EO A are such that "x - Y II is small, then also Inr(x) - nr(y) I is small. [Hint: Choose a field E :::> K such that (E:K) is fmite, and E splits A. Extend the valuation I I on K to a norm on E. Then prove that each of the following maps is continuous: A ..... E Q9 K A,

where (A:K)

=

det:Mr{E) ..... E,

r2]

3. Let A be a central simple K -algebra, P a finite prime of K, Rp the P-adic valuation ring of K p' Given a finite set of finite primes {P, P 2' ... , Pr }, show that there exists an element YEA such that nr A/K Y = prime element of Rp, { nrA/KYEu(R ) [or Q = Pl, ... ,P,.. Q

[Hint: Extend the P-adic valuation C{Jp on Kp to a norm II lip on A and Ap. By (33.4), there exists an xp E Ap with nr xp = np, a preassigned prime element of Rp. If YP E A p is chosen so that I Yp - xp lip is sufficiently small, then nr YP

=0

nr xp (mod n;,Rp),

so nr YP is also a prime element of Rp. Now use (4.11) to choose YEA so that

Ily -

xpll p < e,

Ily - lllQ <

dor Q

=

P z ,· '" P r ,

with e small. If e is sufficiently small, then nr Y will have the desired properties.] 4. Let L be a cyclic extension of the algebraic number field K, and let A = (L/K, (J, a) be a cyclic algebra. When is A a totally definite quaternion algebra? (See (34.1).)

34. EICHLER'S THEOREM The following notation remains in force throughout this section: K = global field R = Dedekind domain with quotient field K, R i' K Kp = P-adic completion of K at a prime P of K CfJ p = P-adic valuation on K and Kp (see §4e, 5a) Rp = P-adic valuation ring in K p , if P = finite prime

= {xEKp:CfJp(x) ~ I} central simple K-algebra, (A: K) = n2 = index [Ap] = local index of A at P S = set of all infinite primes P of K for which mp > 1 = {P: P an infinite prime of K ramified in A}

A mp

=

U(A) = {(x E K*: (Xp > 0 for each PES}.

(We remind the reader that the notation K p , R p is different from that of the

(34.1 )

293

EICHLER'S THEOREM

earlier chapters, where we had previously used Rp to ~enote the localization of R at a prime ideal P, and Rp for the P-adic completIOn of R.) According to the Hasse-Schilling-Maass Norm Theorem, we know that nr A/K u(A) = U(A), a fact which we shall use repeatedly. In this section, we shall present Eichler's refinement of this Norm Theorem. The refinement deals with the problem: Given a nonzero element /3 E R, does there exist an element x E A such that nr A/K x = /3, and x is integral over R? In short, we wish to characterize reduced norms of integral elements. Obviously we must assume that /3 E U(A), since otherwise /3 cannot be a reduced norm. We shall find that the discussion below proceeds smoothly as long as A satisfies some mild restriction, called the Eichler condition. In the exceptional cases, where this condition fails to hold, many of the proofs and results given below are no longer valid. Our first task is to single out these exceptional cases, in which the Eichler condition does not hold. We shall begin with a number of definitions, the significance of which will not become apparent until we are in the middle of the proof of the preliminary Lemma 34.5 below. We shall follow the treatment given in Swan-Evans [ll

(34.1) Definition. Let K be an algebraic number field. A central simple Kalgebra A is called a totally definite quaternion algebra if every infinite prime P of K is ramified in A, and if furthermore Ap ~ H for each such P. Here, H denotes the quaternion skewfield over the real field K p' We remark that no complex prime of K can possibly ramify in A. Further, if A is a totally definite quaternion algebra, then (A:K) = (Ap:Kp) = (H:R) = 4,

PES.

Hence, A is a totally definite quaternion algebra if and only if all of the following conditions hold true: (i) (A:K) = 4, (ii) Every infinite prime of K is ~eal, (iii) Ap ~ H for every infinite pnme P of K. . . that (A: K) = 4 and that S is the set of all mfimte . 1 tl these mean E qUlva en y, d f' d' (94) primes of K. It is clear that the quaternion algebra over Q, e me m . ,

is totally definite. . h Totally definite quaternion algebras play the role o~ exceptIOns to t. e theory to be developed in this section. We shall need to fmd a corre~pon~mg class of exceptions for the case where the ground field is a fu.nctIOn fleld. In order to do so, we introduce a bit of non-standard termmol?gy. Let K be any global field, either an algebraic number field or a functIOn field,

294

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(34.2)

and let R be a Dedekind domain with quotient field K. Each nonzero prime ideal P of R gives rise to a prime of K, which we shall also denote by P, and refer to as a "prime of R". Those primes of K, which do not come from prime ideals of R, will be called "non-R" primes of K. For example, when K is an algebraic number field and R = alg. int. {K}, the ring of all algebraic integers in K (see §4), then the "non-R" primes of K are precisely the same as the infinite primes of K. (For the proof of this statement, see the references listed in § 4.) It will turn out (see proof of (34.5)) that the central simple K-algebras A, which are "exceptional" relative to R, are those which occur in the following situations:

(34.2) (i) A ramifies at every "non-R" prime of K, and (A: K) = 4, where K is an algebraic number field, or (ii) A ramifies at every "non-R" prime of K, where K is a function field. This leads to the following definition: (34.3) Definition. The central simple K -algebra A satisfies the Eichler condition relative to R (notation: A = Eichler/R), if either (i) K is an algebraic number field, and (A: K) '" 4 if A ramifies at every "non-R" prime of K, or (ii) K is a function field, and some "non-R" prime of K does not ramify in A. (34.4) Remark. In particular, suppose that K is an algebraic number field, and R = alg. int. {K}. To say that "A satisfies the Eichler condition relative to R" means that either (A: K) '" 4, or else (A: K) = 4 but A is not ramified at some infinite prime of K. In other words, in this case we have A = Eichler / R if and only if A is not a totally definite quaternion algebra. The proofs of the important theorems, given later in this section, will rely heavily on the following rather technical "approximation lemma". The proof of this lemma requires the hypothesis that A be Eichler/R. We have stated the lemma in a fairly general version, since the extra generality does not make the proof any harder, and will be needed later.

(34.5) LEMMA. We assume that (i) A is a central simple K-algebra satisfying the Eichler condition relative to R, with (A :K) = n 2 , (ii) S is the set of all infinite primes of K ramified in A, (iii) S' is some non-empty finite collection of primes of R, including all those which ramify in A, (iv) S" is any finite collection (possible empty) of primes of R, such that S" is disjoint from S',

(34.6)

295

EICHLER'S THEOREM

(v) f3 is a given element of U(A) II R, (vi) For each PES' uS", we are given a polynomial (34.6)fp(X) = xn + b 1,pX n- 1 + .. , + bn _ 1. p X + (_l)nf3

ERp[X],

such that for each PES', j~(X) is separable and irreducible over Kp. Then for each e > 0, there exists a polynomial (34.7)

f(X) = xn

+ a 1 xn-1 + .. , + an -

1X

+ (_l)nf3

E

R[X],

satisjying all of the following conditions: (CI) For each PES' uS",

l:'(i:'(n-1. (C2) For each P E S',f(X) is irreducible over K . (C3) The polynomial f(X) is separable and irreducible over K. (C4) For each PES, f(X) has no zeros in the real jield Kp.

°

Proof. Let e > be given. After replacing e by a smaller positive number if need be, we may assume that e < 1. We may further assume that the conclusion of Lemma 33.8 is valid whenever the coefficients {bJ occurring in (33.8) are subject to the condition that each qJp(b) < e. Our proof now splits in to three cases. Case 1. K = algebraic number field, n #- 2. For each value of i, 1 :'( i :'( n - 1, we are given a collection of elements {b" p E Rp : PES' uS"}, which ~ccur. as coefficients in the polynomials in (34.6). By the Strong ApproxImatIOn Theorem (4.11), we may approximate these elements simultaneously by an element ai E K. Thus we may choose a., E K so that

qJp(a i - b"p) < e for all PES' U S" { qJQ(a):'( 1 for all primes Q of R not in S' uS". Since e < 1, it follows that for each PES' uS" we have a i - bi. p E RJ" and hence also a i E Rp. But also a, E RQ for every prime Q of R not in S' uS". Hence ai is integral at every prime of R, and so at E R. We shall tentatively define f(X) by formula (34.7), so (Cl) surely holds true. By virtue of the restriction placed on e at the beginning of this proof, it follows from (Cl) and (33.8) that f(X) is separable and irreducible over K p' for each PES'. Thus (C2) is true, whence (C3) is also true, since S' is non-empty. Unfortunately, however, the polynomial f(X) may fail to satisfy (C4).

296

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(34.7)

If the set S is empty, (C4) is vacuous, and so the lemma is established in this case. Now suppose that S is not empty; then n is even, and hence n > 2, since we are considering here the case where n i= 2. Let t be a positive rational integer, and consider the graph of the function f(x) + tx 2 , where x is a real variable. Since n is even, it follows that by choosing t sufficiently large, we may be certain that the graph lies entirely above the x-axis. At the same time, we can make t divisible by a high power of each prime PES' uS". Therefore we may choose t so that ((Jp(t) < e for all PES' uS", { f(X) + tX 2 has no zero in the real field K , for each PES. p

Since n > 2, the polynomial f(X) + tX 2 is monic. Its coefficients satisfy the same inequalities (CI) as do those of f(X). Hence if we use f(X) + tX 2 rather than the original f(X), we have obtained a polynomial which satisfies the conditions (CI)-(C4), as desired. Case 2. K

=

algebraic number field, n

=

2.

Since n = 2, and A = Eichler/ R by hypothesis, it follows from (34.3) that there exists at least one "non-R" prime Po of K at which A does not ramify. Thus Po rj: SuS' uS". We shall now imitate the proof given in Case 1, but this time we must use the Very Strong Approximation Theorem (4.40), rather than (4.11). For each value of i, where 1 ~ i ~ n - 1, we may choose an element a i E K such that ({Jp(a; - bi, p) < e for each PES' uS", ({JQ(aJ ~ 1 for every prime Q of R not in S' uS", { rpp(aJ < e for every PES.

This application of (4.40) is permissible, since no condition has been imposed on ai at the prime Po' A" in the previous case, it follows that each a i E R, and we may again define f(X)ER[X] by formula (34.7). Just as before, f(X) satisfies (CI}--(C3). Furthermore, since n = 2, we may write

and by hypothesis {3p > 0 for each PES. Since ((Jp(a 1 ) < e for each PES, it is clear that if e is chosen sufficiently small, then f(X) will have no zero in the real field K p , for each PES. Thus f(X) satisfies (C4), and this case is settled. Case 3. K = function field. Since A = Eichler/R by hypothesis, there exists a "non-R" prime Po of K at which A is not ramified. The set S is empty, and Po rj: S' uS", since S' u S"

(34.8)

EICHLER'S THEOREM

297

is a collection of primes of R. Given polynomials (34.6) with P ranging over S' u S", we use (4.40) to obtain an element ai E K such that
Proof We may choose Sand S' as in (34.5), with S" empty. Let PER II U(A) be given. For each PES', it follows from (33.13) that we may find a separable irreducible polynomial fp(X) = xn + bl,pX n- l + ... + bn_l,pX + (_l)np ERp[X], with constant term ( - l)n fJ. By (34.5), we may then find a separable irreducible polynomial f(X) = xn + a l xn-l + ... + an_ l X + (-l)"fJ ER[X], satisfying conditions (Cl)-(C4) of (34.5). The discussion in Step 3 of the proof of (33.15) carries over unchanged. It shows that there exists an element x E A such that f(x) = 0, fJ = llfAIK X. But then x is integral over R, since f{x)

=

0, and so the theorem is established.

Before proceeding to the statement and proof of Eichler's Theorem below, we shall recall some definitions from §22. A normal ideal in A is a full R-Iattice L in A whose left and right orders are maximal R-orders in A. If L is a normal ideal with left order A, right order N, then we call L a principal ideal if L = Ax for some x E A. In this case, A'

=

x-lAx,

L= xA',

so in speaking about principal ideals, it does not matter whether we deal with the left order or the right order of L. We are now ready to prove the following important result:

298

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(34.9)

(3~.9) Theore.~. (Eichl~r). Let A be a central simple K-algebra satisfying the Eichler condition relatIVe to R. Let L be any normal ideal in A. Then L is principal ideal if and only ~f its reduced norm nrA K L is a principal ideal Rrx f o~ some rx E U(A). / Proof. For brevity, we write nr instead of nr A/K' Let L be a normal ideal in A, and let A = 01(L), a maximal R-order in A. If L is principal, then L = Ax for some x Eu(A). By (24.12) we obtain

nr L = nr Ax = R . nr x, and nr x E U(A) by (33.15). We must prove that, conversely, if nr L = Rrx for some rx E U(A), then L is principal. We shall assume hereafter that n > 1, where (A:K) = n2 , since the result is obvious when n = 1. Step 1. Throughout the proof, let d = d(Aj R) be the discriminant of the maximal order A (see §25). Then d is a nonzero ideal in R, and for each prime P of R we have (by (32.1))

Ap ~'Kp

if and only if

pi d.

By Exercise 26.8, there are only finitely many conjugacy classes of maximal

R -orders in A. Let At (=A), k~, .. . , AI( be a full set of representatives of these conjugacy classes. Then each maximal E u( A) and some i between 1 and K. Choose once and for all a nonzero element r E R such that r ¢; u(R), and

R -order in A is of the form uA;u -1 for some u

(34.1 0)

i

= 1, ... , K.

We temporarily fix a prime P of R such that PI rd, and let M, N be a pair of maximal left ideals of A such that nr M = nr N = P (see (24.13)). In Step 4, we shall prove that necessarily M = N (J for some (J E u(A). The remainder of Step 1, and the material in Steps 2 and 3, will be needed in Step 4. Let us draw some consequences from the hypothesis that Pt rd. As remarked above, it follows that Ap ~ K p, so Ap ~ M,,(Kp). Since Ap is a maximal Rp-order in Ap by (11.5), it follows from (17.3) and (17.5) that Ap ~ Mn(Rp),

rad Ap = P . Ap.

Let us set

A=

AI/PA

C': -

AplPA p' I

R = RiP I

Then (34.11 )

A

~ M n (R)' ~ Hom_(V V) R"

c:::: R p j'PR p' -

(34.12)

299

EICHLER'S THEOREM

where V is the vector space consisting of all n x column vectors with entries in R. We may view Vas a simple left A-module. Since K is a global field, R is a finite field, and hence j\ is a finite ring. Thus we may choose a finite set of elements, 1" .. , Tq E A such that (34.12)

u(A)

= {i1'" ., i 9 }.

Let AlP) denote the localization of A at P. By (18.3) we have rad AlP) = AlP)

rI

rad Ap = AlP)

rI

P . Ap

= p. AlP)'

Since AlP/rad AlP) = AlP/PA lP ) ~ A, it follows from Exercise 6.2 that each ' j is a unit in AlP)" Hence we may choose an element t E R - P such that (34.13)

t·

-1 'j E

A,

j

= 1, .. . ,g.

For later use, we note that pt t. Step 2. We use the notation of Lemma 34.5, choosing Sf to be the set of all primes of R which either ramify in A or which divide the principal ideal rt·R. (Possibly some prime divisors of rt· R may ramify in A.) Since r ¢ u(R), the set Sf is non-empty. We choose S" to consist of the single prime {P}, so S" is disjoint from Sf. We are going to find a collection of polynomials UQ(X):Q E Sf} and a polynomial !p(X), with certain properties. Once this is done, we shall approximate these polynomials by a polynomial in R[X], by using (34.5). To begin with, we claim that for each Q E Sf, we may find a polynomial fQ(X) = xn

+ b1,QX"-1 + ... + bn_1,QX + (-l)"ER Q[X],

such that (34.14)

!Q(X) is separable and irreducible over K Q, { !Q(X) := (X - 1)" mod Qnn',

where Qn' is the power of Q occurring in rt . R. (Congruence of polynomials means coefficientwise congruence.) For each prime Q E Sf, we proceed as follows: let W = KQ(w) be an unramified extension of degree n, with Frobenius automorphism cr, where w is a primitive (qn - 1)-th root of 1, q = card RQ . Let n be a prime element of R Q . By Exercise 34.4, we may choose a positive integer k > nnf so large that (34.15)

u:= 1 mod nkRQ ===> u

=

yn for some y E R Q, Y := 1 mod nnn' R Q.

Having done so, we set g(X)

= min. pol. KQ(l + nkw) ERQ[X].

300

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(34.1S)

Then g(X) is a separable irreducible polynomial of degree n, and its zeros are {l + nkwq' : 0 ~ i ~ n - I}. Therefore g(X) == (X - If mod nk. Let us write g(X)

= xn + C1X n- 1 + ... + cn_1X + (-I)nu.

Since u == 1 mod n\ it follows from (34.1S) that u = yn for some y E u(RQ) such that y == 1 mod nnn'. We now set fQ(X)

= y-ng(yx) = xn + y-1c1xn-l + ... + y-tn-1)cn_1X + (_1)n ERQ[X].

Then the polynomial fQ(X) satisfies conditions (34.14). Finally, we shall choose fp (X) E Rp[X] as in Exercise 34.1, so that fp (X) is a monic nth degree polynomial whose image!p (X) in R[X] has no zeros in the field R = RIP, and such that flAX) has constant term (_I)n. Step 3. We have thus constructed a set of polynomials {fQ(X): Q E Sf} and a polynomial fp(X), all having constant term (-1)". We apply Lemma 34.5 with f3 = 1, and with the sets S, Sf, S" as defined in Step 2. Then there exists a separable irreducible polynomial f(X) = xn + a1X n- 1 + ... + an_IX + (-1)" ER[X], such that f(X) is coefficient wise near fQ(X) for each Q E Sf uS". Now Sf includes all primes of R which divide rt· R, as well as all primes of R ramified in A. If we choose c; in (34.S) sufficiently small, then our polynomial f(X) will satisfy all of the following conditions: (i) (ii) (iii) (iv)

leX) has no zero in R[ X], where R = RIP. f(X) == (X - 1)" mod (rt)". f(X) is irreducible over K0,2,for each Q ESf. f(X) has no zeros in K Q , for each Q E S.

Now let x be a zero off(X) in some algebraic closure of K, and setL = K(x). As in the proof of (34.8), we may embed L in A. Then x is an element of A integral over R, and nr x = 1. By Exercise 2S.3, X-I is also integral over R. We now set y = (x - 1)Irt EA.

We may writef(X) = (X - 1)" + (rt)nh(X), where h(X)ER[X] by virtue of (ii). Then f(x) = 0 implies that yn = -h(X)ER[x], and hence by (1.11) yn is integral over R. Therefore y is also integral over R, and so there exists a maximal R-order in A containing y. Replacing x and y by suitable conjugates if need be, we may assume that y E Ai for some choice of i between 1 and K. Then ryE A by (34.10), so also x = 1 + rty FA.

(34.16)

301

EICHLER'S THEOREM

Since X-I is integral over R, it follows that x E u(A). Let us set . ..... (34 .16) x j = 'L jX'L j- 1 = 1 + 'L j . ry . t7: j- I EAI..... , "" ] "" g, where the fact that Xj EA follows from (34.13) Then nr Xj = nr x = 1, so each x j E u(A). Step 4. We are now ready to prove the assertion: given any pair of maximal left ideals M, N in the maximal R-order A, such that nr M = nr N = P, where Plrd, then M = N(J for some (J E u(A). We keep the notation of (34.11), so V is a left A-module, and hence also a left A-module. By (2.7) there is a natural isomorphism cp

(34.17)

-+

cp(I).

Recall that the action of an element A E A on an element cp E HomA(A, V) is given by (Acp)~ = cp(~A), ~ E A. Since nr M = P, it follows from (24.14) that AIM is a simple left A-module, and likewise for A/N. Hence there exist a pair of A-exact sequences

°

-+

M

A~ V

-+

-+

°

0.

-+

N

-+

A~ V

-+

0,

with M

= ker cp,

N

= kefl/t,

V = Acp(l) = AI/I(l).

Case 1. Suppose that the elements cp(l), 1/1(1) of the R-space V are linearly dependent over R. Then there exists an a E R - P such that cp(l) = a·I/I(I). But then (cp - al/l)(l) = 0, so by (34.17) we conclude that cp - al/l = 0. Therefore ker cp = ker 1/1, since a i= 0. This gives M = N, so the assertion at the start of Step 4 is proved in this case. Case 2. Now suppose that cp(l) and 1/1(1) are linearly independent elements of the R-space v. In Step 3 we constructed an x E A such that f(x) = 0. Let XL denote left multiplication by X on the A-module v. Then J(x L ) = 0, whence

min. pol.1fx L

divides J(X).

Since J(X) has no zeros in R, it follows that the R-linear transformation XL on V has no characteristic root in R. Hence for each nonzero v E V, the vectors v and xv are linearly independent over R. Choose a nonzero v E V. Then the pair {v, xv} are part of an R-basis of V. So also are the pair {cp(I), 1/I(1)}. Hence there exists an invertible linear transformation T on V such that T:cp(I)-+v,

1/1(1)-+ xv.

302

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(34.18)

"!

Since ~ Hom ii(V' V), T must be given by left multiplication by some unit j E u(A). Therefore

i

i j ' ep(1) = v,

for some j between 1 and g. These give

1/1(1) = i:-1Xi.· ep(l) = x.ep(l), }

}

}

where Xj is defined by (34.16). In view of the natural isomorphism (34.17), we obtain 1/1 = xjep. Therefore N

= kerl/l = ker xjep = g =

g:ep(~x) =

EA:(xjep)(~) = O}

O} = g:~xjEM} = Mxjl.

This S?oWS that M = Nxj,with xjEu(A), and completes the proof of the assertwn made at the beginning of this Step. Step 5. Suppose to start with that J is any left ideal in the maximal R-order A. By (22.18) we may express J as a proper product (34.18)

J = M1···M k , Mi = maximal integral ideal.

Then by (24.11), nr J = (nr M 1)" . (nr M k ), and each nr M i is a prime ideal of R (by (24.13)). Hence the number offactors k is uniquely determined by nr J. Furthermore, if P is a prime of R dividing nr J, then nr Mi = P for some i, 1 :( i :( k. Indeed, by (22.28) or Exercise 22.7, we can find a factorization of J in which nr M 1 = P. (This important fact may also be proved directly, as follows: given the ideal J, define J*=AnJ n p

nA

Q*P

Q'

where Q ranges over all primes of R except P. By (5.3) it follows that J* is a left ideal of A such that J*=>J,

J;=J p ,

J~=AQ

foraB

Q=fP.

Then nr J* isa power of P, and J = J*. J** (proper product) for some integral ideal J**, by (22.19). When we express 1* as a proper product of maximal integral ideals, each factor will have reduced norm P. Hence we obtain a factorization (34.18) in which nr M 1 = P.) Now let J and J' be any pair of left ideals of A. We shall prove that (34.19)

nr J = nr J' } nr J + rd = R

=f

= Jx for some x E u(A).

Let P be a prime of R dividing nr J. Then we may write

(34.19)

303

EICHLER'S THEOREM

J

=

J'

M 1 M z ... AI k'

= N 1 N 2 . . . N k'

where nrM1 = nrN 1 = P, and where the M's and N's are maximal integral ideals. There are as many factors in J' as in J, since nr J' = nr J. We shall establish (34.19) by induction on k. By Step 4, we know that M 1 = N1 Z for some Z E u(A), which proves the result when k = 1. Suppose now that k > 1, and that the result holds (for all maximal orders) for products of k - 1 maximal integral ideals. We may write J' = N 1 z· (Z -1 N Z z)··· (z -1 N k z)· Z-1

= N 1 z· j . z - 1 '

where j

= (Z-1 NzZ)'" (z-lNkz) = z-1(N z '" Nk)z.

It is easily verified that both j and M z ... Mk are left ideals in the maximal order A', where A' = O/(M z ) = 0r(M 1) = 0r(N 1z) = z- 10 r(N 1)z 1 1 = z- 0/(N )z = 01(z- N z z) = ali). z

Furthermore, nrj=nrM z ···Mk'

nr j

+

rd

= R.

It follows from the induction hypothesis that

j=Mz···Mkw

forsome

wEu(A).

Therefore J' -- N 1 Z · j · Z - l

=

M 1 M Z ···M k wz- 1

=

J'wz- 1 '

which completes the proof of (34.19). Step 6. We are now ready to prove the theorem. Let L be any normal ideal in A such that nr L = Rrx, where rx E U(A), and let A = 01(L). We must show that L = Ax for some x E u(A). By (27. 7) we can choose y E u(A) such that Lye A,

Ly

+

rdA

= A.

Set £ = Ly, rx' = rx' nr y. Then also A = 0/(£), and rx' E U(A) by (33.15). If we can prove that £ = Az for some z E u(A), then L = Azy -1 is also principal. Changing notation, we may hereafter assume that L is a left ideal of A such that L

+

rdA

=

A,

nr L = Rrx,

rx ERn U(A).

304

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(34.20)

Let us show that (34.20)

Rrx

+ rd =

R

At each prime P of R dividing rd, we have

Lp

+ rdA p =

Ap

Lp

+ PAp =

Ap.

whence

By Nakayama's Lemma, this implies that Lp = Ar Therefore

= nr Lp =

(nr L)p

nr Ap = R p ,

which proves that pf nr L. Thus (34.20) is established. We next observe that since rx ERn U(A) it follows from (34.8) that rx = nr w for some WE u(A) integral over R. If WE A, then we obtain nr L = nr Aw,

nr L

+ rd

=R,

and hence L = (Aw)w ' for some w' E u(A), by (34.19). Thus the desired result holds when WE A. The manipulations which follow are needed to take care of the possibility that w rI= A. By Exercise 10.5, the ring R[ wJ is contained in some maximal R-order N in A. Consider the normal ideal (A'A)-l in A, with the left order A and right order N. As in the first paragraph of this step, we may choose an element Z E u(A) such that nr {(A'A)-lZ}

+ rd = R.

Let us set M = (A'A)-l Z, Then N(M)

so MeA,

nr M

+ rd =

R

+ rd = R by (24.11), and we have N(M) = ord R AIM c M

by Exercise 27.4. Hence we may choose an element y E N(M) such that

yR

+ rd =

R,

and we note that y ERn M. Therefore

y .Z-

1 WZ

EM·

Z- 1

N Z = M . 0 r(M) c MeA,

and so L = A . y' z-lwz is a left ideal of A. Also Ly is a left ideal of A, since YERWehave nr Ly = (nr L) (nr y) = Rrxy", nr L = R· nr y . nr z-lwz = Ry" . rx,

305

EICHLER'S THEOREM

(34.21)

since nr w = a by the choice of w. Both a and yare prime to rd. It follows from (34.19) that Ly = £'X for some x' E u(A), whence L = Ax for some x E u(A), as desired. This completes the proof of Eichler's Theorem. (34.21) COROLLARY. Let A be a maximal R-order in A, where A = Eichler/R. Let L, £, be left A-ideals in A. Then £,

= Lxfor some xEu(A)<===> nr £, = a(nr L)for some aE U(A).

Proof If £, = Lx, then nr £, = (nr x) (nr El, and nr x E U(A). Conversely, let nr £, = a(nr L) where a E U(A). Let A' = O,.(L), and consider the proper product L- 1 . £. Then

nr(L- 1 £,) = (nr Lr 1 (nr £,) = Ra,

and L- 1 £, has left order A'. By Eichler's Theorem, we may write L- 1 £, = A'x for some x E u(A). Then £,

= LC 1 . £, = L . A' x = Lx,

as desired. This completes the proof. Eichler's Theorem (34.9) requires the hypothesis that A be Eichler/R. Without this hypothesis, Eichler [1] showed that the conclusion of the theorem need not hold true. In his example, A is a totally definite quaternion algebra, and R = alg. int. {K}. He proved that (for suitable choice of A) there may exist a non-principal normal ideal L in A, such that nr A{K L = Ra for some a E U(A). EXERCISES l. Let R = R/ P be a finite field, where P is a prime of R, and let n > 1. Show that there exists a polynomial

such that j(X) has no zeros in R. [Hint: Let card R = q. The number of polynomials Jz(X) = X"

+ C1X"-1 + ... + cn _ 1 X +

(-1)"

ER[X]

equals q" -1. If Jz(X) has a zero A in R, then there is a factorization h(X) = (X

)(X,,-2

+ ... + (_1),,-1 rl).

There are at most q - 1 choice for A, and at most q" - 2 choices for the factor of degree n - 2. Thus, the number of polynomials h(X) with a zero in R is ~ (q - 1)q"-2. Hence there are at least

306

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

polynomials h(X) E R[XJ with no zero in R. Pick one such h(X), and choose f(X) E R[ XJ with 1 = h.] In Exercises 2-4 below, R is a complete discrete valuation ring with exponential valuation v, prime element n, residue class field R. 2. Let n be a positive integer, and set n = n e a, a E u(R). Prove that for each positive integer r, (I + n'X)" = I + ar{+e X + n Z ' g(X), for some g(X) E R[ xl

3. Let a E u(R), {J

E

R, g(X)

E

R[ XJ, and let

h(X)

=

nmg(X)

In

be a positive integer. Set

+ aX

- {J.

Show that h(X) has a zero in R. [Hint: If bars denote passage to Ti(X) =

ax - p =

a(X -

R, then

lJa- I).

By Hensel's Lemma, this factorization of7i(X) lifts to a factorization h(X) = u(X) v(X) in R[XJ, with u(X) monic, u(X) = X - piX-I.J 4. Let In, n be positive integers. Show that there exists a positive integer k such that for u E R, u == I mod n k

==

u=

1'" for some y e R

such that y == 1 (mod nm).

[Hint (Neukirch [IJ): Let n = nea, a E u(R). We may assume that In > e. Choose In + e, and set u = 1 + nk{J, {J E R, and l' = 1 + nmx. We need only show that the equation 1 + nk{J = (1 + nmx)"

k ;;.

has a solution with x

E

R. By Exercise 2, the equation becomes

1+ nk{J = 1 + anm+ex for some g(X)

E

+ nZmg(x)

R[X]. Hence we must solve nm-eg(x)

+ ClX

- {Jnk-(m+ e) = 0,

XER.

N ow use Exercise 3.J

35. IDEAL CLASS GROUPS

Throughout this section, let R be a Dedekind domain with quotient field K, and let A be a maximal R-order in a central simple K-algebra A. We shall eventually restrict our attention to the case where K is a global field, so as to have available Eichler's Theorem and its consequences, but for the moment K may be arbitrary. We shall define the ideal class group of A denoted by CI A, and shall show how to calculate it. The definition will be such that in the special case where A = R, Cl A will coincide with the ideal class group Cl R defined in §4.

(35.1 )

307

IDEAL CLASS GROUPS

By a left A-ideal in A we mean, as usual, a left A-lattice M in A such that KM = A. As in §26, two left A-ideals M, N in A are placed in the same ideal class if M ~ N as left A-modules, or equivalently, if M = Nx for some x E u(A). Let us attempt to define a group structure on this set of ideal classes. Given any two left A-ideals M, N in A, their product M· N is also a left A-ideal in A. However, when A is non-commutative, the ideal class of M . N is not necessarily determined by the ideal classes of M and N. Specifically, for each x E u(A), the ideal Mx is in the same class as M, but possibly Mx' N and M . N are in different classes. Steinitz's Theorem (4.13) suggests one way of overcoming this difficulty. The theorem states (as a special case) that given any two left R-idealsJ, J' in K, then J

(35.1 )

+-

J' ~ R

+-

JJ'

as R-modules,

and furthermore, (35.2)

R

+-

J

c::o:

R

+- J'

¢>

class of J = class of J'

¢>

J ~ J '.

These results show that the ideal class of the product JJ' can be computed unambiguously from the structure of J and J' as R-modules. In (27.4) we obtained a partial generalization of Steinitz's Theorem. A special case of (27.4) asserts that given any pair of left A-ideals M, M' in A, there exists a left A-ideal Mil in A such that (35.3)

M

+-

M' ~ A

+-

Mil as A-modules.

The isomorphism types of M and M' uniquely determine the isomorphism type of the direct sum M +- M', and hence that of A +- Mil. This suggests a way of defining a binary operation on ideal classes. It is customary to write this operation additively, so we tentatively define ideal class of M

+ ideal class of M' =

ideal class of Mil

whenever (35.3) holds true. In order to prove that the operation is well defined, we would need to establish a "cancellation theorem" analogous to (35.2). Specifically, we would need to prove that if M, N are any left A-ideals in A, then (35.4)

A+-M~A+-N=M~N.

Unfortunately, there are examples in which (35.4) is false (see Swan [lJ). The fact that (35.4) is not always true forces us to introduce a new equivalence relation on the set of A-ideals, weaker than isomorphism. It is convenient to define this relation not only for A-ideals, but more generally for A-modules. Given any two left A-modules X, Y, we call X stably isomorphic

308

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(35.5)

to Y if there exists a non-negative integer r such that

+A

X

~ Y

(r)

+A(r)

as left A-modules.

It is easily verified that stable isomorphism is an equivalence relation on the

set of all left A-modules. Clearly, isomorphism implies stable isomorphism, but not conversely (in general). Let [X] denote the stable isomorphism class of the A-module X. (35.5) THEOREM. Let CI A denote the set of stable isomorphism classes of left A-ideals in A. Dejine an additive structure on CI A asfollows: given any pair of left A-ideals M, M' in A, we set [M]

+ [M'] =

[Mil],

where Mil is any left A-ideal in A such that

M

+ M' ~ A -+- Mil.

Then CI A is an abelian additive group, with identity element [A

J.

Proof Given M and M', the existence of an ideal Mil such that (35.3) holds, is guaranteed by (27.4). It is worth pointing out that if Mil is a left A-module satisfying (35.3), then in fact Mil is isomorphic to a left A-ideal in A (see Exercise 35.1). Let us show that addition of stable isomorphism classes is well defined. Let N, N' be left A-ideals in A such that

[M] = [N],

Then there exists an s M

~

[M] = [N'],

N

+-

N' ~ A -+- Nil.

0 such that

+ Ns) ~ N

M' -+- Ns) ~ N'

-+- A(s),

+-

Ns).

But then [Mil] = [Nil], since

Thus addition is well defined. It is easily verified that addition is associative and commutative, and that [A]

+

[M] = [M]

for all

[M]

E

CI A.

It remains to prove the existence of inverses. Any left A-ideal M in A is A-projective, by (21.5). Hence there exists a left A-lattice Y such that

M -+- Y ~

Nt)

for some t.

By Exercise 35.1, Y is isomorphic to a submodule of AU-l). Therefore by (27.8) there exists a left A-ideal N in A such that

(35.6)

309

IDEAL CLASS GROUPS

so we now have Let M -i- N ~ A -i- L for some left A-ideal L in A; then [M] + [N] = [L], by the definition of addition of stable isomorphism classes. However, L

-i-

Nt-I) ~ L

-i-

A

-i-

Nt-2) ~ M

-i-

N

-i-

Nt-2) ~ AU),

so [L] = [A J. Thus [M] + [N] = 0 in CI A, and so inverses exist. This completes the proof that CI A is a group. It follows at once from (35.1) and (35.2) that when A = R, the group CI A coincides with the class group CI R defined in §4. It should also be remarked that when K is a global field and A is any maximal R-order, then the ideal class group CI A is necessarily a finite group, by the lordanZassenhaus Theorem (26.4). For the remainder of this section, we shall restrict our attention to the case where K is a global field. Let us introduce the ray class group CIA R; this is a modified version of CI R, with the modification depending on the given central simple K -algebra A. We shall use the following notation:

feR) = P(R) = CI R = S=

multiplicative group of R-ideals in K, subgroup of principal ideals = {Rcc 0: E K*}, f(R)/P(R) = ideal class group of R, set of all infinite primes of K ramified in A, U(A) = {o: EK*:o:p > 0 for each PES}, P)R) = {Ro:: 0: E U(A)} = ray group (mod S), CIAR = f(R)/ PjR) = ray class group (mod S). The groups PA(R) and peR) may coincide, even when S is non-empty; this certainly occurs when R = Z, for example. In general, P)R) is a subgroup of peR), and there is an epimorphism 0': CIA R ---> CI R. By Exercise 35.2, ker 0' is a finite elementary abelian 2-group. The main result of this section is the fact that the reduced norm map on A-ideals induces an isomorphism CI A ~ CIA R if K is an algebraic number field. This isomorphism also holds for the function field case, if we assume that A satisfies the Eichler condition relative to R. Eichler's Theorem (34.9) plays a key role in the proof of these results. We may restate Corollary 34.21 of Eichler's Theorem in our present notation, as follows: (35.6) THEOREM. Let K be a global lield, and let r be a maximal R-order in a central simple K-algebra B, where B = Eichler/R. For each left r-ideal L in B, let veL) denote the class of the R-ideal nr B/KL in the ray class group CI BR.

310

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(35.7)

Then for each pair of left r -ideals L, L in B, we have L

2;'

L

as

r -modules =

veL) = veL).

A preliminary calculation is necessary before we can apply this result. Given the central simple K-algebra A, we set B = Mr(A), where r ~ l. Then B is also central simple over K, and a prime 0 f K ramifies in B if and only if it ramifies in A. Hence PA(R) = PB(R), and the ray class group CIBR coincides with CIA R. Let r = Mr(A), so by (8.7) r is a maximal R-order in B. Each left r-ideal L in B determines an element veL) E CIA R, where

v(L) = image of nrB/KL

(35.7)

inCIAR.

Let us recall from §24 that by definition,

nrB/KL = {NB/KL}l/S,

(B:K) =

where

S2.

Further, if L c r then

Weare now ready to prove (35.8)

LEMMA.

Let r

~

1, and let B = HomA(A(r), A(r»)

2;'

Mr(A),

where A(r) is viewed as a left A-, right B-bimodule. Let r = HomA(Nrl, Nr»)

2;'

.M)A).

To each lfft A-lattice X in A(r) there corresponds a left r-lattice cp(X) in B, given by cp(X) = HomA(A(r), X). If X = 11 1,., where each 1; is a left A-ideal in A, then

+ ... +

(35.9)

nr B/K cp(X) =

11 nr A/K 1 i .

1=1

Therefore r

(35.10)

v{ cp(X)} =

TI v(li)

in CIA R,

i= 1

where v(l;) denotes the class ofnr A/K li in CIA R. Proof The left A-module Nr) is a progenerator for the category AJIt. Hence by (15.9) or (16.14), there is an equivalence of categories

cp: AJIt

---+

rJlt,

where

cp = HomA(Nr),').

Then cp maps A(r) onto r, and carries each A-submodule X of Atr) onto a

(35.11)

311

IDEAL CLASS GROUPS

left ideal cp(X) in r. By tensoring with the field K, we find readily that cp carries each A-lattice X in A(d onto a r -lattice cp(X) in B. The correspondence X -+ cp(X) is one-to-one, and preserves isomorphism. Now let X be any A-lattice in A(r) such that KX = A(r). By (27.8) we may write X = J 1 J r , where each J., is a left A-ideal in A. Since the reduced norm is multiplicative, it sufflces to prove formulas (35.9) and (35.10) for the case where X C A(r>, and each J, c A We have a commutative diagram

+ ... +

cp(X) = Hom,.. (A (r), X) ~ X

l )

)

cp(Nr) = Hom,.. (Nr, A(r»)

~

+... +X (r copies) .1. A(r) + ... + A(r) (r copies),

where the vertical arrows are inclusions, and the indicated isomorphisms are R-isomorphisms. Therefore

NB/KCP(X) = ordRCP(A(r»)/cp(X) = {ordRA(r)/xy. But

ord R Nr)/x =

r

r

i= 1

i= 1

TI ord R A/li = TI N A/K J

i·

Hence we obtain

~hich gives(35.9) at once. Formula (35.10) follows from (35.9) by taking images CIA R, and the proof is complete.

III

(35.11) COROLLARY. Let K be a global field, and keep the above notation. Let r ~ 1, and let r

X' =

I'J;,

i =1

where each J i and J; is a left ideal in A. Then (i) X ~ X' as A-modules !f and only (f cp(X) (ii) If X ~ X', then vcp(X) = vcp(X'), that is, r

(35.12)

TI v(J

i= 1

(iii) If A

= Eichler/R, then X

~

cp(X') as r-modules.

r

i)

~

=

TI v(JJ

i=l

X' if and only if(35.12) holds.

312

(35.13)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(iv) Let K be an algebraic number field and let r ~ 2. Then X if (35.12) holds true, even when A f= Eichler/ R.

~

X

if and

only

Proof Assertion (i) has already been established at the beginning of the proof of (35.8). From (35.8) we also know that r

wp(X)

= I1 i~

r

Vcp(X) ==

v(lJ,

1

IT v(J;).

i~l

If X ~ X, then cp(X) = cp(X)' b for some bE u(B). Therefore vcp(X) = vcp(X) since (nrB/Kb)R E PB(R), and (ii) is proved. Suppose for the moment that A = Eichler/R; then also B = Eichler/R. Hence by (35.6) we know that cp(X) ~ cp(X) if and only if vcp(X) = vcp(X), and this establishes (iii). Finally, suppose that K is an algebraic number field and r ~ 2. Then B = Eichler/R, whether or not A = Eichler/R. Indeed, if A f= Eichler/R, then (A:K) = 4; but then (B:K) = 4r2 > 4, so necessarily B = Eichler/R. But once we know that B = Eichler/ R, we may use (35.6) to deduce that cp(X) ~ cp(X) if and only if vcp(X) = vcp(X). This gives assertion (iv) and completes the proof of the corollary.

Let us record an important special case of (35.11). (35.13)

COROLLARY. Keep the above notation, and let J, J' be left A-ideals in A. Let [J] denote the stable isomorphism class of J. (i) If A = Eichler/R, then [J] = [J'] if and only if J ~ J'. (ii) Let K be an algebraic number field. Whether or not A = Eichler/R, we have

[J] = [J']

===

J -+- A ~ J' -+- A.

Proof. By definition of stable isomorphism, [J] J -+- Nr) ~ J' -+- N') for some r ~ O. By (35.10),

vcp(J -+- N'»)

= [J']

if and only if

= v(J).

Hence if [J] = [f], then v(J) =. v(J,!. Therefore J ~ f by (35.11 iii), if = Eichler/ R. This proves assertlOn (1). , . Now let K be an algebraic number field. From v(J) = v(J) 1t follows that

A

vcp(J -+- A)

=

vcp(J' -+- A).

Therefore J -+- A ~ J' -+- A by (35.11 iv), which completes the proof of the corollary. We are now ready to prove

(35.14)

313

EXERCISES

(35.14) Theorem (Swan [1]). Keep the above notation. If K is an algebraic number jield, or if K is a function jield and A = Eichler/R, then the reduced norm map induces an isomorphism v : CI A ~ CIA R. Here, CIA is the additive group of stable isomorphism classes of left A-ideals in A, and CIA R the multiplicative ray class group (mod S), where S is the set of all injinite primes of K which ramify in A. Proof. Let J, J' denote left A-ideals in A. We shall define the map v on CIA

by setting v[J] = v(J) = image of nrA/KJ

inCIAR.

It follows from the proof of (35.13) that v is well defined on CIA, since we showed there that if [J] = [J'], then v(J) = v(J'). Now let J -+- J' ~ A -+- J" for some A-ideal J". Then v(J)v(J') = v(J") by (35.11 ii). Therefore

v{[J]

+ [J']} =

v[J"] = v[J] v[J'] ,

so v is a homomorphism. Ifv[J] = 1, then v(J) = v(A). By virtue of the hypotheses of this theorem, we may apply (35.13) to conclude that J -+- A ~ A -i- A Therefore [J] = [A] = 0 in CIA, and thus v is monic. Finally, the group I(R) is the free abelian group generated by the nonzero prime ideals P of R. Given P, we may find a maximal integral ideal J of A such that PAc J c A Then nr A/K J = P by (24.13), and so v[J] equals the image of P in CIA R. This shows that v is epic, and completes the proof of the theorem. Remarks. (i) If A = Eichler/R, then there is a one-to-one correspondence between CIA R and the set of isomorphism classes of left A-ideals in A. (ii) Some of the preceding theorems extend to non-maximal orders. We refer the reader to Frohlich [1], Jacobinski [1,2], Roggenkamp [1], SwanEvans [1], Wilson [1 J. Frohlich's article contains a different proof of (35.14), based on an idele-theoretic version of Eichler's Theorem.

EXERCISES 1. Let L, M be left A-lattices such that KL left A-module such that

~

A(r), KM

~

A(s), and let X be any

L-i-M~A-i-X.

Show that X is isomorphic to a left A-submodule of A (r + s - 1). [Hint: Since X is isomorphic to a submodule of the A-lattice L -+- M, it follows thatXis also a A-lattice. Tensor the given isomorphism with K, to obtain

314

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(36.1 )

A(r) -+- A(s) ~ A -+- KX.

Hence KX ~ A(r+s-1) by Exercise 6.7. Replacing X by an isomorphic copy, we may assume that KX = A(r+s-1) = K· Nr+s-l). We may then choose a nonzero (J. E R such that (XX c 1\(r+s-1), and surely X ~ (XX.] 2. Show that the kernel of the epimorphism CIA R -> CI R is a finite elementary abelian 2-group. [Hint: The kernel is isomorphic to P(R)/ PA(R), which is in turn a homomorphic image of K*/U(A). But there is an exact sequence 1-> UtA)

->

K*.!!.

IT {±1}

->

1,

PES

where for a E K* and PES, the P-th component of eta) is the sign of ap in Kp.].

36. Ko OF MAXIMAL ORDERS We begin by defining the group Ko(i\.) for an arbitrary ring A. Let peA) denote the category of all finitely generated projective left A-modules. Let F be the free abelian group generated by symbols (M), one for each isomorphism class of objects ME peA). Let Fo be the subgroup of F generated by all expressions (L-i- M) - (L) - (M), where L, M E peA). Now set KiA) = F IF o' Then Ko(A) is an abelian additive group, called the Grothendieck group of the category peA). For (M) E F, let [M] denote its image in FIFo. Then every element of Ko(A) is expressible as a difference [M] - [N], with M, N E peA). For any L, M E peA), we have (36.1)

[L -+- M] = [LJ

+ [M]

in KoCA).

Given M, N E peA), we call M stably isomorphic to N if M -+-

Nr) ;:;:

N -+- Nr) for some integer r. By Exercise 36.1, [M] = [N] in KotA) if and only if M is stably isomorphic to N. Hence we may use the symbol [M] to denote both an element of Ko(A) and the stable isomorphism class of M. Each ring A gives rise to an abelian group Ko(A). Further, each homomorphism of rings f{>: A --> A' gives rise to a homomorphism f{> *: Ko(A) --> Ko(A') of additive groups. The map f{>* is defined by setting

f{>*([M] - [N])

= [A'®}\M]

- [A'®}\N],

[MJ - [N]EKoCA).

In the expression A' ®}\ M, the ring A' is made into a right A-module by means of f{>. It is easily verified that f{>* is well defined. In the language of category theory, we may view Ko as a (covariant) functor from the category of rings to the category of abelian groups. n

It follows at once from this fact that if

I . Ai i=l

is a direct sum of rings

("direct product", in category language), then there i s a natural isomorphism

(36.2)

315

Ko OF MAXIMAL ORDERS

(36.2)

Ko

Ct·

Ai)

it·

~

Ko(AJ

We also note that if the rings r and A are Morita equivalent (see § 16), then the equivalence of categories A A --t induces an equivalence peA) --t PCr). This latter equivalence then gives rise to an isomorphism Ko(A) ~ KoCr). We shall use this fact in our discussion below. We are now ready to investigate Ko of a maximal order. By (l0.5), a maximal order in a separable algebra decomposes as a ring direct sum of maximal orders in central simple algebras. In view of (36.2), it therefore suffices to calculate Ko(A) for the case where A is a maximal R-order in a central simple K-algebra. The aim of this section is to prove that

r«

Ko(A) ~ Z -+- CIA, where Cl A is the ideal class group of A defined in § 35. We showed in (35.14) that Cl A is isomorphic to the ray class group ClAR when K is an algebraic number field, or when K is a function field and A = Eichler/R. Hence in these cases the structure of Ko(A) is known explicitly. Our main theorem, which holds whether or not K is a global field, is as follows: (36.3) THEOREM. Let R be a Dedekind domain with quotient field K, and let A be a maximal R-order in the central simple K -algebra A. There is an exact sequence of additive groups (36.4) Furthermore, (36.5)

Proof The inclusion A --t A induces ljJ:Ko(A)

ljJ[M]

--t

Ko(A), where

= [A ®AM] = [K ®RA ®AM] = [K ®RM]

for each M E peA). Since M is a A-lattice, we may identify K ® R M with KM. In order to show that ljJ is epic, it suffices to prove that for each finitely generated left A-module W, there exists an ME peA) such that KM ~ W. By Exercise 3.5 or Exercise 26.5, there exists a left A-lattice M in W such that KM = W. But then ME peA), since by (21.5) every A-lattice is A-projective. Theref ore ljJ is an epimorphism. The ideal class group CI A consists of all stable isomorphism classes of left A-ideals in A, where a "left A-ideal in A" means a left A-lattice M in A such that KM = A. Addition in Cl A is given by [M]

+

[M']

=

[Mil]

if M -+- M' ~ A -+- Mil,

316

SIMPLE ALGEBRAS OVER GLOBAL FIELDS ~

where the M's are left A-ideals in A. We define /1:CIA

/1[M] = [A] - [M],

(36.6)

Ko(A) by setting

[M] ECI A.

Since the element [M] E Ko(A) depends only on the stable isomorphism class of M, it is clear that /1 is monic. Furthermore,

M -+- M' ~ A -+- M":;. [M]

+ [M'] = [A] + [M"] in Ko(A) :;. [A] - [M] + [A] - [M'] = [A] - [M"] :;. /1[M] + /1[M'] = /1[M"] = /1{[M] + [M']}.

Therefore /1 is a homomorphism of groups. For [M] E CI A, we have

1/I/1[M] = [A] - [A] = 0 in

Ko(A),

so 1/1/1 = O. To prove exactness of the sequence (36.4), it remains to verify that ker 1/1 c im/1. Each x E Ko(A) may be written as x = [M] - [N], with M, N E P(A). If we choose M' E P(A) so that M -+- M' ~ A(r) for some r, then x

= [M -+- M'] - [N -+- M'] = [N')] - [L]

for some L E P(A). Suppose now that I/I(x) = 0; then [KL] = [A(')] in Ko(A). Since A is a simple artinian ring, it follows from Exercise 36.3 that KL ~ A(r) as left A-modules. Hence by (27.8) there exists a left A-ideal Lo in A such that L ~ Nr - 1) -+- Lo' Therefore x

= [A(r)]

- [Nr-l) -+- LoJ

= [AJ

= /1[Lol

- [LoJ

This completes the proof that (36.4) is exact. Since A is a simple artinian ring, we have Ko(A) ~ Z by Exercise 36.3. Therefore the sequence (36.4) splits as a sequence of Z-modules. This shows that Ko(A) ~ Z -+- CI A, and completes the proof of the theorem. Let us now compare the sequence (36.4) with that associated with an order which is Morita equivalent to A For simplicity, we treat only the case where r ~ Mn(A).

r

(36.6)

THEOREM.

Keep the notation of (36.3), and let n

~

1. Set

= Atn ),

B = HomA(V, V) ~ Mn(A),

where

V

r = HomA(L, L)

where

L = Nn),

~

Mn(A),

and where the bimodule structures are given by B VA, rLA' Then there is a commutative diagram with exact rows:

(36.7)

o (36.7)

317

Ko OF MAXIMAL ORDERS --+



r

~ Ko(A) ~ Ko(A)

L®'l

--+

0

v®Al

~ Ko(r) ~ Ko(B) ---- 0,

where the map ({J is given by ({J[M] = [L ®A (N"-l) -+-M)],

[M]

E

Cl A.

Each vertical map in (36.7) is an isomorphism. Proof Since VA is a pro generator for j/A' the ring B is Morita equivalent to A, and there is an equivalence of categories V ®A' : AJ!! ~ BJ!!·

Likewise, there is an equivalence L ®A . : AJi -+ r Ji.

The second and third vertical maps in (36.7) are just the isomorphisms induced by these category equivalences. The right hand square in (36.7) is commutative, since both t/J and t/J' are induced by the functor K ® R .. It follows from the Morita equivalence of r and A that there is a left r -isomorphism

(36.8)

p:r~L®AN"),

where rLA is a bimodule, and N") is viewed as left A-module. For M a left ideal in A, we may therefore view L ® A (A (n - 1) -+- M) as a left ideal in r. This readily implies that the stable isomorphism class of M uniquely determines that of L ® A (A (n-1) -+- M), and so ({J is well defined on Cl A. We shall show that the left hand square in (36.7) is commutative. This implies at once that ({J is an isomorphism, as claimed. To prove commutativity, we start with an arbitrary [M] E Cl A, and make the following calculation in Ko(r). (In this calculation, we use (36.8), and write ® in place of ® A for brevity.) We have

.u'({J[M] = [r] - [L ® (Nn-1) -+- M)] = [L ® Nn)] - [L ® N"-l)] - [L®M] = [L®A] - [L®M] = (L®·).u[M]. This shows that the left hand square in (36.7) is commutative, and completes the proof of the theorem.

(36.9) Remarks. (i) Keeping the above notation, let S be a simple left A-module. Then [S] is a free Z-generator for Ko(A), and there is an isomorphism r A: Ko(A) ~ Z, obtained by setting rA[S] = 1. Since V ® A S is a simple

318

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(36.10

left B-module, it follows at once that the following diagram commutes:

Ko(A)~ Z

V@A'l

11

Ko(B)2~ Z.

(ii) Suppose that the hypotheses of (35.14) hold, and keep the above notation. The reduced norm maps induce isomorphisms v:ClA and of course CIA R

~

= CI B R.

v': Clr ~

ClAR,

We 'claim that the diagram ClA~

~1

(36.10)

ClBR,

ClAR

11

Clr~ ClAR

is commutative. Let [MJ E Cl A, and assume without loss of generality that M c: A. We need only show that

where

= nrB1K {L@(N,,-I) -+- M)}, If (A:K) = t 2 , then (B:K) = t 2 n 2 , so

nrA1KM

(36.11) @

means

@A'

NAIK

= (nrAiKY,

NBIK

=

(nrB/KY".

Thus (36.11) holds if and only if

{N A1K M}" = NB1K{L ® (Nil-I) -+- M)}. Using (36.8), the right hand expression equals L ® A(") { A(")}" ord R L @ (Nn--1j-+- M) = ord R N,,-=1) if

r

= {ord R _~

+

= {N AIK M}".

This completes the proof that the diagram (36.10) is commutative. EXERCISES In Exercises 1 and 2, I\. is an arbitrary ring.

1. Let X, Y E P(I\.). Show that [X] = [Y] in Ko(l\.) if and only if X is stably isomorphic to Y [Hint: If X+- 1\.(') ~ Y +- 1\.('), then [X] + r[A] = [Y] + rCA] in Ko(I\.), so [X] = [Y]. Conversely, if [X] = [Y] in Ko(A), then (X) - (Y) lies in the subgroup FoofF. Hence there exist L's, M's in P(I\.) such that in F.

319

PICARD GROUPS

Therefore X

-i- L' Li -i- L'L; -i- L' (M j -i- M;) ~

Choose N

E

Y

-i- L' (Li + £i) -i- L' M j -i- L'M>

P(A) so that

-i- L' Li -i- L'L; -i- L'M j -i- L' M; ~ -i- A(rl ~ Y -i- A('l, as desired.J N

for some r. Then X

Nrl

2. Let Xi' 1'; E P(A). Show that [X IJ - [X 2J = [Y1J - [Y2 J in Ko(A) if and only if -i- Y2 is stably isomorphic to X 2 -i- Y1 .

XI

3. Let A be a semisimple artinian ring with t simple components. Show that (i) P(A) consists of all finitely generated left A-modules. (ii) For X, Y E P(A), we have [XJ = [YJ in Ko(A) if and only if X ~ Y. (iii) Ko(A) ~ Z('l. In fact, if SI' ... ,S, are a full set of non-isomorphic simple left A-modules, show that [SIJ, ... , [SJ form a free Z-basis for Ko(A). In Exercises 4-6, let R be a Dedekind domain with quotient field K, and let A be a maximal R-order in a central simple K-algebra A. 4. Show that every nonzero left A-lattice is a progenerator of the category 1....,11· 5. Show that (36.6) can be generalized, as follows: let L be any nonzero right A-lattice, and set with bimodule structures rLA' A(L*)r' Let V = KL, B = Kr. Show that the diagram (36.7) is commutative, provided we define rp[MJ for [MJ E CI A as follows: rp[MJ = [L ®A

M],

where M is a left A-lattice such that M -i- H ~ A -i- M. [Hint: The equivalence L ® A' : AA -+ rA carries L* onto r. This fact is crucial in proving the commutativity of the left hand square in (36.7).J 6. Keep the notation of the preceding exercise. Show that the analogues of (36.9 i) and (36.9 ii) remain true.

37, PICARD GROUPS Many ofthe results in this section are due to Frohlich [1] (see also Bass [1 ]). Let R be a commutative ring, A any ring (with unity element lA)' and let c(A) denote the center of A We have called A an R-algebra if there is a ring homomorphism R ---> c(A) with 1R ---> 1A' In this case, all A-modules may be viewed as R-modules. Since every ring is a Z-algebra, there is no loss of generality in considering algebras rather than rings. Let A, ~ be R-algebras. We shall say that a bimodule AM~ is over R if rm=mr for all rER, mEM, that is, (rl tJm =

m(r1~)

for all r, m. A bimodule AM~ (over R) is invertible if

320

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.1 )

there exist a bimodule t.N A (over R), and a Morita context (16.6), which give a Morita equivalence (over R) of the rings A, /}. We call N the inverse of M. As shown in §16, the bimodule isomorphism class (M) uniquely determines the inverse class (N); indeed, we have (N) = (M- 1), where M- 1 is the bimodule given by (37.1)

t.(M-1)A

=

HomA(AMt.'AAA) ~ Homt.(AMt.'t./}t.).

We may restate the definition of invertibility in the following explicit form: the bimodule AM t. (over R) is invertible (over R) if there exist a bimodule t. N A (over R) and bimodule isomorphisms M ®t.N

(37.2)

~

A,

N ®AM ~ /},

making the following diagrams commute: M®t.N®AM--+A®A M (37.3)

1

N®AM®t. N --+/}®t. N

1

M®t./}--+

1

M,

N®AA--+

1 N.

Let us recall from §16 that every invertible bimodule AMt. is a progenerator for both of the categories Avit and vltt.. Furthermore, (37.4)

A

~

Homt. (M, M),

/}

~

Hom A(M, M),

where both /} and Hom A(M, M) are viewed as rings ofright operators on M. Conversely, progenerators yield Morita equivalences.

(37.5) Definitions. (i) Let A be an R-algebra. The Picard group of A relative to R, denoted by PicR A, is the multiplicative group consisting of all bimodule isomorphism classes (M) of invertible bimodules AMA over R. Multiplication is defined by the formula (M) (M') = (M ® A M'). The class (A) is the identity element of PicRA, and inverses are given by (M)-l = (M- 1) with M- 1 defined as in (37.1), using A in place of./}. (ii) Let C = c(A), and view A as a C-algebra. We define Picent A = Pice A, so Picent A is the subgroup of Picz A consisting of all classes of invertible bimodules AM A such that (37.6)

cm

= mc

for all c E C, mE M.

We remark that the groups PicR A and Picent A need not be commutative. The group PiczA is called the Picard group of A, and the subscript Z is usually omitted. As we shall see below when we restrict our attention to the case where A is an order in a semisimple algebra A over a field, the group

321

PICARD GROUPS

(37.7)

Picent A is precisely the group of invertible two-sided A-ideals in A, modulo principal ideals generated by units in c(A). From this standpoint, the group Pi cent A is a natural object of investigation. In Exercise 16.4 we have already noted that Morita equivalent rings have isomorphic centers. Let us give this isomorphism explicitly: (37.7) THEOREM. Let A, b.. be R-algebras, and let AM!> be an invertible bimodule over R. Then M determines an R -isomorphism cp: c(A) ~ c(b..) as follows: for each c E c(A), cp(c) is the unique element of c(b..) such that (37.8)

c· m

=

m'

cp(c) for all

mE

M.

Taking R = Z, it follows that Morita equivalent rings have isomorphic centers. Proof Since M is invertible, (37.4) shows that each element of A (or of b..) is completely determined by its action on M. For each c E e, the map m ---+ cm, mE M, is a left A-endomorphism of M. Hence by (37.4) there is a unique cp(c) E b.. such that (37.8) holds true. But then for all x E b.. and mE M, m(cp(c)x) = (cm)x = c(mx)

= (mx)cp(c) =

m(xcp(c)).

Thus cp(c) commutes with each x E b.., so cp(c) E c(b..). It is easily verified that cp is an R-isomorphism of rings, and the theorem is proved. Let us show next that Morita equivalent rings have isomorphic Picard groups. (37.9) THEOREM. If the R-algebras A, b.. are Morita equivalent over R, then Pic R A ~ PicR b... If the rings A, N are Morita equivalent, then Picent A ~ Picent N. Proof. Let AM!> be an invertible bimodule over R. It is easily verified that the map (X) E Pic R A, (X) ---+ (M- 1 ® AX ® AM), gives an isomorphism Pic R A ~ Pic R b... (We should remark that this isomorphism may depend on the choice of M.) Secondly, let the invertible bimodule ANA' determine an isomorphism cp:e ~ C' as in (37.7), where e = c(A), C' = c(N). Then c'n = n'cp(c) for all

CEe,

nEN.

For (X) E Picent A we have cx = xc for all c E e, x E X. Let us deduce from this that (N- 1 ® A X ® A N) E Picent N. Each element of C is of the form cp(c), for some c E C. Then we have

322

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(n'®x®n)cp(c)

(37.10)

= n'®x®cn = n'c®x®n =

cp(c) (n' ® x ® n), for all n' ® x ® n E N- 1 ® X ® N.

The remaining details of the proof are obvious. A supplement to this theorem is given in Exercise 37.4. Now let Aut R A be the group of all R-automorphisms of the R-algebra A. For each u E u(A), let iu be the inner automorphism defined by (37.10)

xEA.

The group of inner automorphisms of A is defined as

In A = {i,,:

U E=

u(A)}.

Since U E

u(A),

it follows that In A is a normal subgroup of Aut R A. We set (37.11)

OutR A = Aut R AjIn A,

the outer automorphism group of A over R. (37.12) Definition. Let A, d be R-algebras, AX~ a bimodule over R. Given any automorphisms f E Aut R A, 9 E Aut R d, let fX 9 be the bimodule (over R) having the same elements as X, but with the action of A "twisted" by f, that of d by g. This means that

Aoxo(j for each x E X,

(infX)=f(A)·x·g((j)

(in X)

2E A, (j Ed. It is easily verified that f'(fX 9)9' ~ f fX 99' as bimodules.

(37.13) LEMMA. Consider the R-algebra A as a (A, A)-bimodule. For each f, 9 E Aut R A. there is a (A, A)-bimodule fAg over R. The following bimodule isomorphisms hold true:

fAg ~ hfAhg ~ lAr 'g ~ g- 'fAp f A 9 ® f,A 9 , ~ fA 9 f' -,9 .,

1

f'(fAg)g' ~ ff'A gg.,

A9 ® 1 A9 . ~ 1 A99 .,

where ® means ® A' Proof Exercise for the reader. (37.14)

THEOREM.

The map WO:AutR A -4 PicR A, defined by

323

PICARD GROUPS

(37.15)

fEAutRA,

is a homomorphism of groups. We have ker roo = In A, the group of inner automorphisms of A. Therefore roo induces a monomorphism ro: Out R A -4 Pic R A, where Out R A is given by (37.11). Proof. Each f E Aut R A gives rise to a (A, A)-bimodule 1 Af over R, and by (37.13), fAl is an inverse for lAf . Thus roo(f) = (lAf)EPic R A. Further, for

f, gE Aut R A,

roo(fg)

=

(lAf)

= (lAf)(lAg)

by (37.13), so roo is a homomorphism. Let us prove that ker 0)0 = In A. We have O)o(f) = 1 if and only if there exists a bimodule isomorphism fJ: A ~ 1 Af' This occurs if and only if there is a bijection fJ: A -4 A such that (37.15)

fJ(axb)

= a' fJ(x) . f(b) for all a, b, x

E

A.

Now let u E u(A), and let f be the inner automorphism iu defined by (37.10). Set e(x) = xu-l, x E A. Condition (37.15) becomes (axb)u- l

a' xu- l . iJb)

=

=

a' xu- l . ubu-l,

which is always true. Therefore A ~ 1 Af whenever f is inner, and thus In Acker 0)0' Conversely, let fJ: A -4 A be a bijection such that (37.15) holds, and set u = fJ(I). Then A = fJ(A) = fJ(A' 1) = AfJ(l) = Au,

and likewise A

= uA,

so u E u(A). Taking x = b = 1 in (37.15), we obtain fJ(a)

=

au for all a E A.

Next choose a = x = 1 in (37.15), whence fJ(b)

=

u' f(b) for all bE A.

Therefore au

=

fJ(a)

=

u . f(a) for all a E A,

whence f = iu - [ E In A. This completes the proof that ker roo = In A, and the remaining assertion in the theorem is now obvious. The preceding discussion is useful in considering the one-sided modul structure of invertible bimodules. The following result is fundamental:

324

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.16)

(37.16) THEOREM. Let (X), (Y) E PicR A. Then AX ~ AY if and only if (Y) E (X) . im OJ, that is, if and only if Y ~ 1 Xf (as bimodules) for some fE Aut R A.

Proof For each f E Aut R A, there is a (A, A)-bimodule isomorphism lX f

=

X ®A lAr

Hence (Y) E (X) . im OJ if and only if Y ~ 1 X{ for some f. It is clear that any bimodule isomorphism Y ~ 1 X f is also a left A-isomorphism AY ~ AX. Conversely, let h: X ~ Y be a left A-isomorphism. Then h induces an isomorphism h* of endomorphism rings: h*:HomA(AY'AY) ~ HomA(AX'A X ), given by h*(cp) = h-1cph, for cpEHomA(AY'AY). Since AXA is invertible, each element in HomA (AX, AX) is given by a right multiplication ar:x -> xa, x E X, for some uniquely determined element a E A. Likewise, every element of Hom A(A Y, AY) is of the form br:y -> yb, y E Y, for some uniquely determined bE A. Therefore each bE A determines a unique a E A such that h*(b r) = a r , that is, This gives

(h-1brh)x = arx = xa

for all

x E X,

that is,

(hx)b = h(xa)

for all x E X.

Setting a = f(b), we obtain the identity

(hx)b = h(x ·f(b))

for all

x E X,

bE A.

It follows readily that f E Aut R A, and that the map h: 1 X f -> Y given by x -> h(x), x E 1 X f' is a bimodule isomorphism. We have thus shown that if AX ~ AY' then Y ~ 1 X f (as bimodules) for some f E Aut R A. This completes the proof of the theorem.

Before proceeding with our investigation of Picent A, let us briefly consider the relation between Picent A and PicR A when A is an R-algebra. Let C = c(A). By (37.7), each invertible AM A over R determines an R-automorphism <1>M of C, according to the condition that for each c E C, <1>M(C) . m = m· c

for all

mE M.

Tt is clear that <1>w depends only upon the bimodule isomorphism class of M. Let us calculate <1> M for the case where M = 1 Af' f EAutRA. We show that

(37.17)

M =

lAf

=

<1>M is the restriction of f to C.

325

PICARD GROUPS

(37.18)

Indeed, let c E C, and let us verify that

f(c)· m = m· c for all mE M. Since M

= 1/1.. f ' this condition becomes f(c)· A= k f(c) for all A E /1..,

which certainly holds true since f(c) E C. This establishes (37.17), which we shall need for the next result. (37.18) THEOREM. For any R-algebra /I.. with center C, there is an exact sequence 1 .... Pi cent /I.. .... Pic R /I.. ~ Aut R C.

(37.19)

If /I.. is commutative, then the sequence 1 .... Picent /I.. .... Pic R /I.. ~ Aut R /I.. .... 1

(37.20) is split exact.

Proof. Let (M), (N) E Pic R /1.., and set f = M' 9 = N. Then for each CE C, we have f(c) . m = mc, g(c)· n = nc, for all mE M, n E N. Therefore

(m ® n)c = m . g(c) ® n = f(g(c))(m ® n) for all m ® n E M ® AN. This shows that M0N = M N' and thus is a homomorphism. Furthermore, M = 1 if and only if cm = mc for all mE M, c E C, that is, if and only if (M) E Picent /1... This completes the proof that the sequence (37.19) is exact, once we observe that the mapping of Picent /I.. into Pic R /I.. is just the inclusion map. Now suppose that /I.. is commutative, and let Wo be the map defined in (37.14). Thus for fEAutR/I.., we put woU) = (I/l.. f ) EPicRI\.. Since /I.. is commutative, it follows from (37.17) that wo(f) = f. This shows that wo is the identity map on Aut R /1.., whence by §2a the map is epic, and the exact sequence (37.20) is split. This completes the proof of the theorem. 0

Remarks. (i) In some cases one can describe explicitly the image of the map t

occurring in (37.19). Suppose for example that A = where for K i . Then Hence f {1, ... , t},

.I· ,=

Ai is semisimple,

I

each i the simple component A. has skew field part Di and center , r } c(A) = Ki' and each f E Aut K c(A) must p~rmute the \Ki . is completely determined by some permutation n of the set together with a collection of K-isomorphisms !;: K, ~ K"(i) ,

I'

326

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.21)

1 ~ i ~ t where f is the restriction of f to K .. Then Pi cent A ~ [1Picent Ai

byEx;;ci~e 37.6, ~nd the exact sequence (37.19) becomest 1 --+ Pi cent A

--+

PicK A ~ AutK c(A).

Each (X) E PicK A yields a Morita equivalence of the K-algebra A with itself. Using this fact, Frohlich [1] proves that the image of cJ>' consists precisely of those elements f E Aut K c(A) such that for 1 ~ i ~ t, the isomorphism f;:Ki ~ K 7t (i) can be extended to an isomorphism Di ~ D7t (i)' We shall see in (37.21) that Picent A = 1, and thus PicK A ~ im cJ>' in the present case. (ii) Now let R be a Dedekind domain with quotient field K, and let A be a maximal R-order in a separable K-algebra A. Keep the above notation. There is a ring isomorphism Aut R c(A)

~

Aut K c(A),

given by extending each f E Aut R c(A) to an 1'. E Aut~ c(A). By using t~e result that any two maximal orders must be Monta eqUIvalent (see ExerCIse 22.11), Frohlich shows that

f E im cJ>

¢>

l' E im cJ>',

where cJ> is the map occurring in (37.19). Thus im cJ> is known also for the case of maximal orders. (iii) Keep the notation of (i), and consider the exact sequence 1 --+ Picent A 1

--+

PiCK A 1 ~ Aut K K 1 .

By (i), im cJ>'1 consists precisely of those K -au tomorphisms f of K 1 which can be extended to a K -automorphism of A l' (37.21) Theorem, Let A be any semisimple artinian ring. Then

,

Proof We may write A

=

Picent A = 1.

I' Ai' where each Ai is a simple artinian ring. Then

i=1

Picent A ~

,

Il1 Picent Ai

;=

by Exercise 37.6, so we need only show that each Picent Ai = 1. Changing notation, let A be a simple artinian ring with center K, that is, A is a central simple K-algebra. Let V be a simple left A-module, D = HomA(V, V) = skewfield part of A, and view V as a bimodule A VD of right D-dimension n. Then A = HomD(V, V) ~ MJD). * To clarify the later discussion, we write
(37.22)

PICARD GROUPS

327

Now let (X) E Picent A. Then AX ~ V(k) for some k, and therefore HomA(AX, AX) ~ HomA(V(kl, V(k)) ~ Mk(D),

where each endomorphism ring is viewed as right operator domain. But Hom A (AX, AX) ~ A since AX A is invertible. Therefore A ~ Mk(D), whence k = n. This proves that X ~ A as left A-modules. Hence by (37.16), there exists an f E Aut K A such that X ~ 1 A f as bimodules. But f is an inner automorphism of A, by the Skolem-Noether Theorem. Hence 1 A f ~ A as bimodules, by (37.14). This shows that (X) = 1 in Pi cent A, and completes the proof of the theorem. One may also prove that Picent A = 1 by using Exercise 37.4, since A is an Azumaya K-algebra, and since PicK K = 1. (37.22) Theorem. Let A be a commutative ring such that A/rad A is a direct sum offields. Then Picent A = 1. This holds in particular when A is a commutative algebra over a local ring R, and A isfinitely generated as R-module. Proof Set N = rad A, A = A/N. Clearly A is a commutative ring such that rad A = O. Thus A is a direct sum of fields if and only if A is an artinian ring. Any left A-module X may be viewed as a bimodule AX A over A, by defining x . A = )ox, )0 E A, x E X. All bimodules over A arise in this way. For each X, let X = X/NX. We claim that there is a homomorphism Picent A ---> Pi cent A, given by (X) ---> (X). Indeed, given a Morita context where X ® Y ~ A, Y ® X ~ A, we obtain a new context X ® Y ---> A, Y ® X ---> A in which both maps are epic, and hence (X) E Picent A. Now Picent A = 1 since A is semisimple artinian, and hence for each (X) EPicent A there exists a A-isomorphism f:X ~ A. In the diagram

x.. ::::~:::~A

1~ 1

-

J X :;:-.=:::::= r- A I

,

we can find A-homomorphisms q>, IjJ which lift f,f -1 respectively, since X and A are A-projective. Then q>1jJ(a) - a EN for all a E A, whence A = q>1jJ(A) + N. It follows that q>1jJ(A) = A, and so q> is epic. Therefore X = ker q> EB Xl' where q>: X 1 ~ A. Consequently ker q> = 0 in X, whence ker q> = O. This proves that X ~ A, as desired. The second assertion in the theorem follows from the first, by use of (6.15). For the remainder of this section let R be a domain with quotient field K, and letA be an R-order ina K-algebra A. If M, N are two-sidedA-submodules of A such that M N = N M = A, we call M an invertible A-ideal in A, and N its inverse. Then KM = KN = A. By Exercise 37.9, M is an invertible

328

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.23)

(A, A)-bimodule, and hence (M) E Picent i\. The same Exercise shows that = (N) in Picent A Let us denote by I(A) the multiplicative group of invertible A-ideals in A, with multiplication performed within A. If C = c(A), then obviously c(A) = KC. We prove

(M)-l

(37.23)

THEOREM.

There is an exact sequence

1 ~ u(C) ~ u(KC) -4 I(A) ~ Picent A -4 Picent A,

where p(u) = Au, a(M) = (M), 'f(X) = (KX). Proof. Exactness at u(KC) is clear. By Exercise 37.8, ker a = im p. Finally, ker'f = im a by Exercise 37.10. (37.24)

COROLLARY.

If A is semisimple, then PicentA

~

I(A)j{Au: uEu(KC)}.

This coroll~ry justifies to some extent our emphasis on the group Picent A, rather than PICR i\. It shows that Pi cent A occurs in a natural way as a class group of invertible A-ideals in A. Let us define Autcent A = AutcA, Outcent A = Outc A = Autc i\jln A, where C = c(A). By (37.14), there is a monomorphism w: Outcent A ~ Picent A,

Now define the normalizer of A in A as

N(A) = {x E u(A): xAx- 1 = A}. (37.25) THEOREM. Let A be a semisimple K-algebra. There is a commutative diagram

N(A)ju(A) u(KC) 4 Outcent A

~

wj Picent A,

where for x

E

N(A), p(x)

= ix'

w'(x)

= (Ax),

and ix is the inner automorphism A ~ XAX -1, A E A The map p is an isomorphism, and Wi a monomorphism. Proof Each f

E

Autc A extends to an

J E AutKc A. By the

Skolem-Noether

(37.26)

PICARD GROUPS

329

Theorem (7.23), J = ix for some x E u(A). This shows that p is epic. The remaining part of the proof is left as exercise for the reader. (37.26) COROLLARY. Let R be a discrete valuation ring. (i) If A is a maximal R-order in a separable K-algebra A, then the maps p, w, w' in (37.25) are isomorphisms. (ii) If A = Mn(R), then N(A) = u(A)' u(K). Proof (i) Let (X) E Picent A By (37.24) we may assume that X is a two-sided A-ideal in A, such that KX = A. Then AX ~ AA by (18.10), and hence (X) E im w by (37.16). This proves that w is epic, and establishes assertion (i) of the corollary. (ii) When A = Mn(R) we have PicentA .~ PicentR = l,andsothesecond assertion of the corollary follows from the first.

One of our ultimate aims is to evaluate Picent of a maximal order. As a first step in this direction, we treat the complete local case. (37.27) THEOREM. Let R be a complete discrete valuation ring, and let A be a maximal R-order in a central simple K-algebra A with skewfield part D. Then Pi cent A

~

Z/eZ,

where e = e(D/K) is the ramijication index of Dover K. If the discriminant d(A/R) equals R, then Picent A = 1. Proof Let 1t be a prime element of R, and let 1tD be a prime element of d, where d is the unique maximal R-order in D (see §12). Then

1td = (1t D d)e,

rad A = 1tDA

by §13 and (17.5). Therefore 1tA = (rad A)e. But by (17.3), I(A) is the infinite cyclic group generated by rad A Since Pi cent A ~ I(A)/{1t k A: k E Z} by (37.24), we obtain Picent A ~ Z/eZ, as desired. If e > 1, then (rad A)e -1 divides the different !)(A/R), by Step 2 of the proof of (25.4). Therefore 1te - 1 R divides the discriminant d(A/ R). Hence if d(A/R) = R, then necessarily e = 1 and Picent A = 1. This completes the proof.

In order to generalize the preceding theorem to the case where R is an arbitrary Dedekind domain, we now prove an important result concerning the relation between global and local Picard groups. This result will be valid for all orders, whether or not they are maximal.

330

(37.28)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.28) Theorem. Let A be any R-order in a separable K-algebra A, where R is a Dedekind domain with quotient field K, and let C = c(A). For each maximal ideal P of R, let Ap denote the P-adic completion of A. Then Picent Ap = 1 almost everywhere, and there is an exact sequence (37.29)

1 --> Picent C -4 Picent A

:s. TI Picent Ap -->

l.

p

The map r is given by r(L) = (L ®c A),

(L)

E

Picent C.

Proof Step 1. For each maximal ideal P of R, let R(p) denote the localization of R at P (as defined in §3), and let Rp be the P-adic completion of R. We may identify Rp with the completion of the discrete valuation ring RIP) relative to its maximal ideal P . R (P). As pointed out in § Sa, for each finitely generated R-module M, we may form its P-adic completion Mp by first forming the localization M(p) = R(p) ® R M, and then passing to completions: Mp = Rp ®R IPi M(p) ~ Rp ®RM. Since A is a separable K-algebra, we know from (lOA) that there exists a maximal R-order in A, say r. By (11.6), rp is a maximal Rp-order in Ap. But A and r are a pair of full R-Iattices in A, and so by Exercise 4.6 we have A(p) = riP) a.e. Therefore Ap = r p a.e., which proves that for almost all P, Ap is a maximal Rp-order in A p. Now let d(AjR) be the discriminant of A with respect to R (see §10, §2S). Then d(A/R) is a nonzero ideal of R, since A is a separable K-algebra, and we have d(Ap/Rp) = {d(AjR)}p by Exercise 10.6. But {d(AjR)}p = Rp whenever Pld(AjR). Since there are only finitely many P's dividing d(Aj R), it follows that d(Ap/ Rp) = R p a.e.

Step 2. We have now shown that for all but a finite number of maximal ideals P of R, we have (37.30)

Ap = maximal Rp-order in Ap ,

and

d(Ap/ Rp) = Rp.

We claim that Pi cent Ap = 1 whenever (37.30) holds true. To verify this, let us write Ap = Ai' where each Ai is a central simple algebra over a field Ki containing K p. Let Ri be the integral closure of R in K i ' so each Ri is a complete discrete valuation ring. By (10.5) we have Ap = Ai' where for each i, Aj is a maximal Rj-order in Aj . Since d(Ap/ Rp) = R p , it follows from Exercise 2S.1a that d(AjRJ = Ri for each i. Hence each Picent Ai = 1, by (37.27). Therefore

IO

Io

Picent Ap ~

TI Picent Ai = 1

331

PICARD GROUPS

(37.31)

by Exercise 37.6. This shows that Pi cent Ap = 1 a.e., as asserted. Step 3. We shall identify the set of left C-modules with the set of (C, C)bimodules over C. For each left C-module L, the bimodule structure AAA permits us to view L ®c A as a (A, A)-bimodule over C. Explicitly we have

x(l ® A)Y

= I ® XAY,

x, A, YEA.

IE L,

This construction may be visualized more easily as follows: given a C-submodule L of KC, we may form the two-sided A-submodule LA in A. Denote by I(C) the group of invertible C-ideals in KC. Then for each L E I(C), there is an isomorphism L ®c A ~ LA as (A, A)-bimodules. Furthermore, if £ E I(C) is such that L£ = C, then (LA)(£A) = (£A)(LA) = A. Hence for each L E I(C) we have LA E I(A). Now let (L) E Pi cent C; by (37.24), we may assume that L preceding discussion shows that r(L) = (L ®c A) = (LA) E Pi cent A.

E

I(C), and the

For Ll' L2 E I(C) we have (L 1 )(L 2 ) = (L 1 L 2 ) in Picent C, and hence r(L l ) r(L 2 ) = (LlA' L2A) = (L 1 L 2 . A) = r(LlLJ Th us r is a homomorphism. We must still show tha t r is monic. For any bimodule AMA' let us set MA = {rnEM:}m = rnA for all

),EA}.

Then MA is a C-module, whose isomorphism class depends only upon the class (M). Obviously

if AN A is another bimodule. We now verify that for each (L) E Picent C, (37.31)

{L ®c AY

= L ®c C

~ L

as

C-modules.

The formula is clearly valid when L = C, and hence also for every projective C-module L by (2.17), and hence it holds true whenever (L) E Picent C. Suppose now that (L) E Picent C is such that r(L) = 1, that is, (L ®c A) = (A). Then there are C-isomorphisms L ~ (L ®c At ~ AA

=

C,

and therefore (L) = 1 in Picent C. This completes the proof that r is monic. Another proof of this fact is given in Exercise 37.12.

332

(37.31)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

Step 4. The map r' in (37.29) is defined by setting

Il (Xp),

r'(X) =

(X)

E

Pi cent A.

p

It is clear that for each P, the (Ap, Ap)-bimodule X p is invertible. Furthermore, cx = xc for all c E C, X E X, since (X) E Picent A The same equality therefore holds for all C E C p , X E X p' But C p = c(Ap) by Exercise 37.14, and hence (X p) E Picent Ap. This shows that im,' c Il Picent Ap , and obviously r' is a homomorphism. We now prove that im r c ker r', that is, (LpAp) = 1 in Picent Ap for each L E I(C) and each P. Since C is an R-submodule of A, C is itself an R-order in KC. Therefore C p is an Rp-order, whence Pi cent C p = 1 by (37.22). Consequently Lp = Cpu for some unit U E (KC)p, and hence (LpAp) = (uAp)

=

I in Picent Ap.

This proves that im r c ker r'. To prove the reverse inclusion, suppose that (X) E Picent A is such that r'(X) = 1, that is, (Xp) = I in Picent Ap for each P. We may assume that X E I(A), and so X p E I(Ap) for each P. It follows from Exercise 37.8 that for each P there exists a unit cp E c(Ap) such that X p = Apcp. Furthermore, since Xp = Ap a.e., we may choose C p = 1 a.e. Now set

By (5.3), L and I.: are C-lattices in KC such that for all P. Therefore LE.

= KC

n{Q Lp[;p} = KC n{QC p} = C,

whence Land [; are invertible C-ideals in KC. Thus (C) need only verify that (X) = r(L). For each P we have Xp

E

Picent C, and we

= Apcp = ApLp,

and therefore X

=

A

n{O

Xp}

=

An {O

(AL)p}

= AL.

This proves that ker r' c im r, and completes the proof that im r

= ker r'.

Step 5. It remains for us to prove that r' is epic. Suppose that for each P we are given an element X(P) E I(Ap), and we wish to find an (X) E Picent A such

(37.32)

333

PICARD GROUPS

that ,'(X) = Il X(P). Since Picent Ap = 1 a.e., we may assume that X(P) Ap a.e. Now set

X

A

n{~X(P)}.

By (5.3), X is a two-sided A-submodule of A such that X p For each P, let Y(P) be the inverse of X(P), and set

Y = A

=

=

X(P) for all P.

n{r) Y(P)}.

Then, as above, Y is also a two-sided A-submodule of A such that Yp = Y(P) for all P. But then XY = Y X = A, since these equalities hold locally at all P. This shows that (X) E Picent A, and that ,'(X) = Il X(P). Hence " is epic, and the proof of Theorem 37.28 is finished. The exactness of the sequence (37.29) shows that the calculation of Picent A can be carried out in two steps: (i) The determination of the group Picent C for a commutative order C, (ii) The determination of the groups Picent Ap for local orders Ap. There is the further problem of determining the structure of the extension Picent A, once Picent C and TI Picent A p are known. There are as yet no general results concerning this last question. (37.32) COROLLARY. Let R be a Dedekind domain, and let A be a maximal R-order in a central simple K-algebra A. For each maximal ideal P of R dividing the discriminant d(A/ R), let e p be the ramification index of the skewjield part of A p over K p. Then there is an exact sequence of commutative groups (37.33)

1 -+ CI R

-+

Pi cent A

-+

Il

Z/epZ

-+

1.

Pld(A(R)

If K is a global jield, then for each P, ep equals the local index mp of A at P. Proof. In this case A has center R, and Picent R ~ Cl R by (37.24). The group Picent A is abelian, since I(A) is abelian by (22.10). Further, for each P we know that Ap is a maximal R p-order in the central simple K p-algebra A P' and d(Ap/Rp) = {d(A/ R)}p. It follows from (37.27) that Picent Ap = 1 whenever Pfd(A/R), and that Pi cent Ap ~ Z/epZ whenever Pld(A/R). Then (37.33) is just the exact sequence in (37.29). Finally, ep = mp when K is a global field, by (14.3). It is of interest to compare the groups Picent A and CI A when A is a maximal order. Most of the remaining material in this section comes from Frohlich-Reiner-Ullom [1], where non-maximal orders are also considered.

334

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.34)

(37.34) THEOREM. Let A be a maximal R-order, as in (37.32). There is a homo-

morphism

e: Pi cent A --+

Cl A,

given by e(X) = [X], (X) E Picent A If K is a global field and A = Eichler/R (see Remark 37.41), then ker e ~ Outcent A. Proof The map e carries the bimodule isomorphism class ("X,.) onto the stable isomorphism class ["Xl Now let (X), (Y) E Picent A; without loss of generality, wemay assume that X, Y E I(A) and that X c: A Setting T = A/X, we obtain an exact sequence of (A, A)-bimodules

o --+ X

--+

A

--+

T

--+

0,

where T is an R-torsion bimodule. Since AYis projective, the sequence

O--+X®Y--+Y--+T®Y--+O is also exact, where ® means ® A. By (18.10), Yp each P. Therefore

T®Y ~

~

Ap as left Ap-modules, for

L:' (T ® Y)p ~ L:' Tp ® Yp ~ L:' Tp ~ p p. p

T

as left A-modules. Since AY is projective, we may apply Schanuel's Lemma (see Exercise 2.7) to the above pair of exact sequences, so as to deduce that X

-i-

Y ~ A

-i-

X ®Y

as left A-modules.

Hence [X] + [Y] = [X ® Y] in Cl A, which proves that e IS a homomorphism. Now let K be a global field and let A = Eichler/R. If (X) EPicentA is such that e(X) = 1, then [X] = [A], whence X ~ A as left A-modules by (35.13). This latter condition holds if and only if (X) E im w, by (37.16). Therefore ker e = im w ~ Outcent A, as claimed, and the proof is done. (37.35) THEOREM. Keep the above notation, and let K be a global field. If K is a

function field, assume further that A = Eichler / R. Let (A: K) = n2 , and for each maximal ideal P of R, let Kp be the local capacity of A at P, and mp the local index. There is a homomorphism tn:Cl R

--+

CIA R,

induced by mapping each R -idealJ in K onto 1". Then coke

~

CIAR/ Wimt n ,

where W is the subgroup of cl A R generated by the classes of

{r p : pld{A/R)}.

(37.36)

335

PICARD GROUPS

Proof We have CI R = I(R)/ peR), CI A R = I(R)/ PA (R), in terms of the notation preceding (35.6). If no infinite prime of K ramifies in A, then peR) = PA (R) and t. is obviously a homomorphism. On the other hand, if some infinite prime of K does ramify in A, then n must be even, and hence (Ra)" E PA(R) for each a E K*. Thus t. is a well defined homomorphism in any case. By (35.14), the reduced norm map or AIK induces an isomorphism v: CI A ~ CIA R. Hence cok e ~ CIA R/{v[X]:X E I(A)}.

Each P determines a unique prime ideal ~ of A containing P, and these ideals {~} generate I(A) as a free abelian group, as proved in (22.10). Furthermore, for each P we have n = mp/(p, nr AIK ~ = pKp by (25.11). Thus {v[XJ:X EI(A)} is the subgroup of ClAR generated by all classes {P"p}, P arbitrary. But /(p = n except when mp > 1, that is, except when d(AI R). Hence this subgroup is precisely W' im t., and the theorem is proved. (Another approach is given in Exercise 37.15)

pi

e, e', where e: Picent A ..... Cl A, e' :Picent r ..... CI r, for the case where r is a full matrix ring ring M.(A) over a maximal order A We shall now compare the maps

Our main result is (37.36) THEOREM. Let A be a maximal R-order in a central simple K-algebra A, where K is a global field and A = Eichler/R (see Remark 37.41). Let n ? 1, and set E = M .(R), r = M .(A). We may identify r with E ® R A. Then (i) There is a commutative diagram of groups, with exact rows: 1 ..... Outcent A £4 Pi cent A ft. Cl A

III

al

(37.37)

1 ..... Outcent

r

;1

~ Picent

r

~ CI

r,

where CI.(f) = 1 ® f,

P(X) = (E ® R X),

y[Y] = [E ®R Y].

(ii) Let (37.38)

(CIA). =

g ECIA:n~

= O},

n . Cl A

=

{n~: ~

Then ker y = (CI

M.,

cok y

~

CI Aln- Cl A.

E

Cl A}.

336

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.39)

(iii) There is an exact sequence of groups (37.39)

1 ~ Outcent A -"+ Outcent r ~

Cl

.4 (Cl A)n ~

cok 8 ~ cok 8'

A/n- Cl A ~ O.

Proof The top row of (37.37) is exact by (37.34). Next we note that Kr = Mn(A), which is also Eichler/R. Hence we may apply (37.34) to the maximal order r, so as to obtain the exact sequence occurring in the bottom row of (37.37). It is clear that (37.37) is a commutative diagram, since all of the maps IX, /3, yare induced by E ® R '. This establishes assertion (i) of the theorem. Now let cp: Cl A ~ Cl r be the isomorphism which occurs in the commutative diagram (36.7). We claim that y = n' cpo Using the notation of (36.7), it suffices to prove that Il'y = 11' . ncp. For [Y] E Cl A, we have whereas

1l"ncp[YJ = n(L®i\')Il[Y] = n{[L®i\A] - [L®i\ Y]}EKo(r). But in Ko(r) we have

[E ®R A]

= n[L ®i\ A],

[E ®R Y] = n[L ®i\ Y]

by Exercise 37.3. This completes the proof that y = ncp. Since cp is an isomorphism, it follows at once that ker y = (CI A)n' and that

= Cl r/n ·cp(CI A)

~ Cl

A/n· Cl A. Finally let us ~rov~ (iii). B~ Exercise 37.2, the map /3 is an isomorphism. cok y

Hence the f ollowmg dIagram 1

IS

commutative, and has exact rows:

~ Picent A

e1 1 o ~ kery ~ CIA

J!. Picent r

~ 1.

e'l ~

Clr

By the "Snake Lemma" (see Exercise 2.8) there exists an exact sequence 1 ~ ker 8

~

ker 8'

~

ker y ~ cok 8 .!:t. cok 8',

where y* is induced by y. Then cok y* Hence the sequence

=

{Cl r;im 8'}/y * {Cl A/im 8} ~ cok y.

cok 8

l!!,

cok 8'

~

cok y ~ 0

is also exact. This establishes that (37.39) is exact, and completes the proof of the theorem.

(37.40)

EXERCISES

337

(37.40) COROLLARY. Let r = Mn(A) as in (37.36), and keep the notation and hypotheses of that theorem. Then (i) For each f E Aut R r, we have 1" = (1 (8) g)YJ for some 9 E Aut R A and some inner automorphism YJ of r. (ii) Out R Mn(R) ~ (CI R)n' Hence for each f E Aut R Mn(R), f" is an inner automorphism of Mn(R). Proof Let, be the map occuring in (37.39). Since r has center R, we have Outcent r = Out R r. Autcent r = Aut R r, For f E Aut R r, let f denote its image in Out R r. Then 'Un) = n' 'U) = 0, whence fn E im IX. Hence f" = (1 (8) gm for some 9 E AutR A and some YJ E In r, whi' '1 proves (i). Now let A = R. Then ker e = 1, cok e = 0, since e is an isomorphism. The exactness of (37.39) then gives Out R

r

~

(CI R)n'

as claimed. The remaining assertion in (i) is now clear, and the corollary is proved. The result (ii) is due originally to Rosenberg-Zelinsky [1 J. (37.41) Remark. As part of the hypotheses of Theorems 37.34, 37.36 and 37.40, we assumed that K = global field, A = Eichler/R. (37.42) By (35.13), these assumptions imply that (37.43) [X] = [Y] in Cl A ¢> X ~ Y as left A-lattices. It was precisely this result, rather than (37.42) itself, which was needed in the proof of (37.34). The other two theorems, (37.36) and (37.40), depend on (37.34). Thus all three theorems remain valid if, in their hypotheses, we replace condition (37.42) by the weaker condition (37.43). It may well happen that in some cases, (37.43) holds true but (37.42) does not. For example, the quaternion algebra A in (26.5) fails to satisfy the Eichler condition relative to Z; nevertheless, (37.43) holds for the maximal Z-order A given in (26.5), since in fact we showed in §26 that every left A-ideal in A is principal. Even when K is not a global field, there are cases where (37.43) is valid; an obvious example is that obtained by choosing A = K, A = R. In fact, as we shall see in (38.13), stable isomorphism implies isomorphism whenever A is a commutative order.

EXERCISES 1. Verify the assertions of(37.13).

2. Let E = M nCR), where R is a commutative ring and n ~ 1. Let A be an R-algebra, and r = E @R A ~ M)A). Show that the isomorphism Pic R A ~ Pic R r given in

338

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.9), arising from the Morita equivalence of A and rover R, is also given by the map (X) --+ (E ®R X), (X) E Pic R A. [Hint: Let L = Rln) ®RA, viewed as bimodule rL" after identifying r with Hom,,(L, L). Let C 1 = Hom" (L, A), viewed as (A,l)-bimodule. By (37.9), the map defined by (X)E PicRA,

gives an isomorphism Pic R A ~ Pic R r. However, there are two-sided r -isomorphisms

L ®" X ®" L- 1 ~ Hom" (L,L ®"X) ~

~

Hom" (Aln), Xln»)

E ®RX,

Therefore (L ®" X ®" rl) = (E ®RX)in Pic R r.J 3. Keep the above notation. Show that for each [YJ

[E ®R YJ

n[L ®" YJ

=

in

E

Ko(A) (see §36),

Ko(r).

[Hin t: L ®" Y ~ R(n) ® R Y as left r -modules. But E is isomorphic to a direct sum of n copies of the left E-module R(n), and so

[E ®R YJ

=

n[R(n) ®R YJ = n[L ®" YJ,

as desired.J 4. An R-algebra E is an Azumaya R-algebra (or central separable R-algebra) if E has center R, and E is R-separable (see Exercise 7.9). Show that for every R-algebra A and every Azumaya R-algebra E, there is an isomorphism PicRA ~ PicRA ®R E given by (X)

--+

(X ®RE), (X)

E

PicRA.

[Hint: (see DeMeyer-Ingraham [lJ). Let E e = E ®R EO be the enveloping algebra of e E. Since E is an Azumaya R-algebra, the (E e, R)-bimodule E is invertible, and E is Morita equivalent to R. There are maps PicRA

L

PicRA®E

~ PicRA®E®W ~ l'

PicRA

PicRA®E®Eo®E

11

1/11

--

PicRA®E.

The diagram commutes, and 1'/, 1'/' are isomorphisms by (37.9) and Exercise 2. Likewise, the maps ifi 2 ifi 1 and ifi3ifi2 are isomorphisms. Hence also ifi 1 is an isomorphism.J 5. Let A be an R-algebra with center C, and keep the notation of (37.11) and (37.19). Show that the following diagram has exact rows and is commutative, and that w, w' are monic. 1 --+ Outc A

--+

rol I

--+

Picent A

Out R A

ro'!

--+

--+

Aut R C

11

Pic R A ~ Aut R C.

339

EXERCISES

, 2:' Ai be a direct sum of rings. Prove that

6. Let A =

i=l

t

Picent A ~

TI

Picent Ai'

i= 1

LHint: Let e 1 , ... , e, E A be the central idempotents such that Ai = A ej • Each (X) E Picent A is such that ejX = Xe p and thus (ejX) E Picent AI' 1 ~ i ~ t. The desired t

IT (e

isomorphism is given by (X) ->

7. Let A

j

X), (X)

E

Picent A.]

i= 1

t

2: . Ai be a direct su'm of R-algebras.

=

Show that Pic R A need not be

i=l t

isomorphic to

IT

PicR Ai'

i= 1

In Exercises 8-10, let R be a domain with quotient field K, and let A be an R-order in a K-algebra A. We write "M" c A to indicate that M is a two-sided A-submodule of A. 8. Let "M" c A. Show that there is a bimodule isomorphism M ~ A if and only if M = Ae for some e E u(e(A)). [Hint: Surely Ae ~ A for each such e. Conversely, let 8:A ~ M be a bimodule isomorphism, and let e = 8(1). Then 8 extends to a bimodule isomorphism A ~ KM, so A = KM = KA . e, whence e E urAl. Further,

8(1)' a = 8(a) = a' 8(1),

aEA,

so e E e(A).] 9. Let "M", "N" c A be such that MN = NM = A. Show that M is an invertible bimodule, and that (M) E Picent A, and (M) - 1 = (N) in Picent A. [Hint: The maps

11 : M 09" N

--,t

MN = A,

v : N

09"

J\;[ ->

NM

=

A,

give a Morita context in which both 11, v are epic, and hence monic.] 10. Let (X) E Picent A be such that (KX) = I in Picent A. Show that (X) = (M) for some invertible A-ideal M in A. [Hint: Since "X" is invertible, "X is finitely generated and projective as A-module, and thus X -> KX is an embedding. By hypothesis, there is a bimodule isomorphism f: KX ~ A. Then M = f(X) ~ X, and "M" c A. Replacing M by rM with suitably chosen rER, r f= 0, we may assume that MeA. If (X)-l = (Y), then likewise Y ~ N for some "N" cA. We have a Morita context

The map 8:M 09 N -> MN is epic; it is also monic, since each of the maps below is mOlllc: M09N->A09N ....... A09A->A Consequently MN

~

A, NM

M N = Ae, 1

~

A. By Exercise 8, we have N M = Ae '

for some e, e' E u(e(A)).

Then Ne = Ne', and Ne- is an inverse of M.]

340

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(37.44)

11. Complete the proof of (37.25). 12. Keep the hypotheses and notation of (37.28), and let L E I(C) be such that L A ~ A as (A, A)-bimodules. Without using (37.28), show that (L) = 1 in Picent C.

[Hmt: By ExercIse 8, LA = Ac for some c E u(KC). Replacing L by c- 1 L, we may assume that LA = A. Therefore also LA = A, where L E I(C) is such that LL = C. We have L c: A n KC = C, and likewise L c C. Now C is a commutative R-order in a separable K-algebra, so for each maximal ideal P of R we have Picent Cp = 1. HencewemaywnteLp = Cpxp,(L)p = C y forsomex y EC Thenx y C -r . pp' P' P p' pp p-'-'p' so XpEU(C p) and Lp = Cpo ThIS h01.ds for each P, whence L = C.] 13. Let A be a K-algebra, where K is a field, and let E be any extension field of K. Prove that c(E ®K A) = E ®K c(A).

[Hint: Clearly c(E ®K A) :=J E ®K c(A). The reverse inclusion follows from the first paragraph of the proof of (7.6), if we take B = E in that argument.] 14. Let A be an R-order with cenLer C, where R is a Dedekind domain. Prove that Ap has center Cpo [Hint: Let N = A ®R AO be the enveloping algebra of A (see §7), and view A as left N-module. Then there is an identification

C

HomA.(A, A).

=

Therefore by (2.37)

15. Keep the notation and hypotheses of (37.35). Show that the diagram

I ...... Cl R ...... Picent A ......

t,1

01

1 ...... CIA R ......

Cl A

TI Z(mpZ ...... 1 p

L

..... 1

is commutative and has exact rows. Deduce from the "Snake Lemma" that the sequence of groups (37.44)

is exact.

1 ...... ker tn ...... ker 13 ......

TI Z(mp Z ...... cok tn ...... cok 13 ...... 1 p

38. NON-MAXIMAL ORDERS The term "non-maximal order" refers to an order which may be maximal or not. This section is intended to serve as a guide to further study of orders, as well as to indicate some generalizations to non-maximal orders of the material in §§35-37. During the past fifteen years, there have appeared numerous research articles on the theory of orders. There are, however, relatively few books or lecture notes dealing with this topic. The principal ones are as follows:

(38.1)

NON-MAXIMAL ORDERS

341

(i) The lecture notes by Roggenkamp, Huber-Dyson [lJ and Roggenkamp [1 J contain a systematic treatment of orders. The first of these references contains Chapters I-V of the list below, and overlaps the present book considerably; it also contains an extensive bibliography. The second volume (Chapters VI-X) is devoted to proofs of the major recent developments in the theory of orders, up to 1969. It has a small bibliography, supplementing that of the first volume. The chapter headings are these: I. II. III. IV, V, VI. VII, VIII. IX. X.

Preliminaries on rings and modules. Homological algebra. The Morita theorems and separable algebras, Maximal orders, The Higman ideal and extensions of modules, Modules over orders, one-sided ideals over maximal orders, Genera of lattices, Grothendieck groups, Special types of orders, The number of indecomposable lattices over orders,

(ii) The lecture notes by Swan-Evans [lJ provide an elegant and largely self-contained treatment of many topics in the theory of orders, The authors presuppose a basic knowledge of homological algebra, In a few places, they use some results from Swan [2]. Chapter headings: L 2, 3, 4, 5, 6,

Introduction, Frobenius functors, Finiteness theorems, Results on Ko and Go' Maximal orders, Orders, 7, Ko of a maximal order. 8, Kl and G1 , 9, Cancellation theorems, Appendix,

(iii) The survey article by Reiner [1J contains an extensive summary (without proofs) of the principal results on orders, and a large bibliography. Section headings are as follows: 1. Introduction, Notation and definitions,

2, 3. 4, 5,

General remarks, lordan-Zassenhaus theorem, Extensions, Higman ideaL Representations over local domains,

342 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(38.1)

Genus. Maximal orders. Further results on genera. Projective modules and relative projective modules. Grothendieck groups and Whitehead groups. Commutative orders and related results. Divisibility of modules. Hereditary orders. Finiteness of the number of indecomposable representations. Representations of specific groups and orders. Representation rings. Group rings. Algebraic number theory. Krull-Schmidt and Cancellation Theorems.

(iv) Chapter XI of Curtis-Reiner [1], entitled "Integral Representations", is concerned with ~he theory of orders.

(v) There are two major references which deal primarily with maximal orders. The first, by Deuring [1], has an extensive bibliography covering research before 1935. The exposition is self-contained but succinct. Chapter headings are: I. Grundlagen.

II. III. IV. V. VI. VII.

Die Struktursatze. Darstellungen der Algebren durch Matrizes. Einfache Algebren. Faktorensysteme. Theorie der ganzen Grassen. Algebren iiber Zahlkarpern. Zusammenhang mit der Arithmetik der Karper.

The second reference is Weil [1 J. This work is self-contained, except that familiarity with Haar measure is assumed. Chapters are: I. II. III. IV. V. VI. VII.

Locally compact fields. Lattices and duality over local fields. Places of A-fields. Adeles. Algebraic number fields. The theorem of Riemann-Roch. Zeta-functions of A-fields.

(38.1) VIII. IX. X. XI. XII. XIII.

NON-MAXIMAL ORDERS

343

Traces and norms. Simple algebras. Simple algebras over local fields. Simple algebras over A-fields. Local c1assfield theory. Global c1assfield theory.

In the present book, the following sections contain material on nonmaximal orders: §§6, 8, 18, 26, 27, 36-41. We shall now sketch a generalization of the results in §§35-37 to the case of non-maximal orders. For the remainder of this section, let A be an R-order, maximal or not, in a separable K-algebra A. As usual, R denotes a Dedekind domain with quotient field K. Since A need not decompose into a direct sum of orders in simple algebras, we may no longer restrict our attention to the case of central simple algebras. Recall from §27 that two left A-lattices M, N are in the same genus (notation: M v N) if M p ~ N p as left Ap-lattices, for each maximal ideal P of R. By (18.2 iii), in this definition it is immaterial whether the subscript P denotes localization or completion. Call M locally free (of rank n) if M v A(ol. (If A is maximal, then by (18.10) every left A-ideal M in A, such that KM - A, is locally free of rank 1. More generally, any left A-lattice X, such that KX ~ A(n l , is locally free of rank n by (27.8).) Whether or not A is a maximal order, we shall define the locally free class group of A, denoted by CI A, as follows: the elements of CI A are stable isomorphism classes [M] of locally free left A-lattices M in A (see Exercise 38.1). As in (35.3), we define addition of classes thus: given M, M' v A, by (27.3) there exists an M" v A such that M -i- M' ~ A -i- MI'. We then set [M] + [M'] = [M"]. To show that inverses exist in CI A, we need only modify the proof of (35.5) by using Exercises 27.6 and 27.7 in the parts of the proof where we previously used Theorems 27.4 and 27.8. The results in (35.6)--(35.14) need not be true for non-maximal orders. Before stating lacobinski's generalization of these results, we introduce the following definition:

(38.1) Dejinition. Let A be an R-order in a separable K-algebra A, where R is a Dedekind domain and K is a global field. For each simple component A. of A, let R. denote the integral closure of R in the center of Ai. We say that A' satisjies the Eichler condition relative to R (notation: A = Eichler/ R) if for each i, Ai = Eichler/R i . This definition extends that given in (34.3). Remark. Let R = alg. int. {K}, where K is an algebraic number field, and let G be a finite group. Set A = KG, the group algebra of Gover K. Following

344

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(38.2)

the approach in Swan-Evans [1], we may give some sujjicient conditions that A = Eichler/R. Indeed, if A of- Eichler/R, then some simple component Ai of A must be a totally definite quaternion algebra over its center L. Since this can never occur if L has any complex primes, we deduce (i) If K has any complex primes, then A = Eichler/R. Next, suppose that Ai is a totally definite quaternion algebra; then its completion at each real prime of L must be the quaternions H over the real field R. Thus the composition of maps KG

= A

-4

A., -4 H

gives a homomorphism of G onto a subgroup G of u(H), and necessarily H equals RG. Surely G must be nonabelian, since H is a noncommutative ring. The nonabelian finite subgroups of G of u(H) are known explicitly (see Coxeter [1], Coxeter-Moser [1 ]), and are as follows: generalized quaternion group of order 4n, n ~ 2, binary tetrahedral group (2,3,3) of order 24, binary octahedral group (2,3,4) of order 48, { binary icosahedral group (2, 3,5) of order 120. Hence we conclude (ii) A = Eichler/R

if G has no homomorphic

image in the above list.

We now state without proof: (38.2) THEOREM (Jacobinski [2,3]). Let A be an R-order in a separable Kalgebra A, where K is a global field, and where A = Eichler/R. Let Y be the finite set consisting of all maximal ideals P of R such that Ap is not a maximal Rp-order in Ap. Keeping the notation of (38.1), let C = Ri' so KC is the center of A. Denote by I(C, Y) the multiplicative group of all C-lattices ] in KC such that

L"

K']

= KC,

Let PA(C)

= {CnrA1KCx: XEA,XEU(A p) for all PEY},

a subgroup of I(C, Y) consisting of certain principal C-ideals in KC. Then the locally free class group Cl A satisfies

(38.3) Furthermore,

(38.4)

Cl A

if M, N

~

I(C, Y)/PA(C).

are A-lattices in the same genus as A, then

[M] = [N] in CI A

<==>

M ~ N.

(38.5) Remarks. (i) if A is maximal, the set Y may be taken to be the empty

NON-MAXIMAL ORDERS

(38.6)

345

set. The isomorphism occurring in (38.3) can then be described componentwise, according to the simple components of A. At each simple component, we obtain precisely the isomorphism given in (35.14). (ii) In the preceding theorem, the subscript P denotes completion. The result holds equally well if we use localizations rather than completions. (iii) An idele-theoretic version of this theorem, due to Frohlich [2], holds even when A =fo Eichler/ R. See also Wilson [1]. (iv) Cl A is a finite group by virtue of the 10rdan-Zassenhaus Theorem. Let us turn next to the material in §36. The definition of Ko(A) given in §36 holds for all rings A, and hence surely for all orders A, maximal or not. The map ,u:CIA -4 Ko(A) occurring in (36.4) is monic, whether or not A is maximal. However, the sequence (36.4) need not be exact, and (36.5) need not hold true. Now let r = Mn(A), and keep the notation of (36.6). Most of the proof of (36.6) remains valid when A is any R-order in a separable K-algebra. It shows,that, in the notation of (36.7), there is a commutative diagram CIA -4 Ko(A) CPt

(38.6)

L®t; ,

I

~

Cl Mn(A) .4 Ko(Mn(A)), where ,u,,u' are monic, and where cp, L ® A' are isomorphisms. As a consequence of the above, we have (38.7)

A be an R-order in a separable K-algebra A, where K r = M 2(A), B = Kr = M 2 (A). Let C, Y be as in (38.2). Then whether or not A = Eichler/R, we have COROLLARY. Let

is an algebraic number field. Let

Proof Since K is an algebraic number field, it follows that B = Eichler/R even if A =I- Eichler/R. The ring C and the set Yare the same for r as for A.

The formula for CI r is just the one given by (38.3), with r in place of A.

All of the material in §37, up to and including (37.22), is valid for arbitrary R-orders (and indeed for R-algebras). The important results (37.23)-(37.25), and the fundamental (37.28), were established in §37 for all R-orders, maximal or not. Of course, Theorems 37.26, 37.27 and 37.32 hold only for the case of maximal orders. It is possible to salvage (37.34), by introducing the locally free Picard group LFP(A). This is the subgroup of Picent A consisting of all (X) E Picent A such

346

(38.8)

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

that X is locally free as left A-module. (Such X's 'are necessarily of rank 1.) By Exercise 38.4, LF peA) is indeed a group, and consists of all bimodule isomorphism classes (X), where X ranges over all two-sided A-submodules of A such that X v A as left A-modules. (Each such X is necessarily an invertible bimodule, by Exercise 38.4.) If A is a maximal order, then

LF peA) = Pi cent A by (18.10). This also holds true whenever A is commutative, but not necessarily maximal, by virtue of (37.22). Returning to arbitrary orders, consider the map w: Outcent A -- Picent A Clearly its image lies in LF peA), and hence there is a monomorphism

w: Outcent A -- LF peA), with w(f) =

(1 Af).

It follows at once from (37.16) that for (X), (Y)

-=-

AX ~ AY

(38.8)

E

LF peA),

(Y)E(X)·imw.

Furthermore, since LF P(C) = Picent C, we deduce from (37.29) that there is an exact sequence (38.9)

1 -- LFP(C) ~ LFP(A) --

TIp LFP(Ap) -- 1,

with LFP(Ap) = 1 a.e. Finally, we have

LFP(Ap) = {(X) E Pi cent Ap : X ~

~

Ap as left Ap-modules}

Outcent Ap.

Generalizing (37.34), we prove (38.1 0) THEOREM (Frohlich-Reiner-Ullom [1 ]). For any R-order in a separable K-algebra A, there is a homomorphism

e:LFP(A) -- CI A, given by e(X) = [X], (X) E LF peA). Furthermore, suppose thatt for left A-lattices M, N in the genus of A, (38.11 )

[M] = [N]

Then ker

-=

M ~ N

e ~ Outcent i\.

Proof The first half of the proof of (37.34) carries over unchanged, and shows that e is a well defined homomorphism. Now assume that (38.11) holds true, t This hypothesis is satisfied whenever K is a global field and A = Eichler/R, by virtue of (38.4). It also holds for arbitrary K when ;\ is a commutative order. by virtue of (38.13) below.

(38.12)

347

NON-MAXIMAL ORDERS

and let (M) ELF P(A). Then e(M) = 1

-=:>

[M] = [A]

CI A

III

-=:>

{\.~1 ~ (\.A

But AM ~ AA if and only if (M) E im OJ, by (38.8). This shows that ker im OJ, whence ker e ~ Outcent A as claimed.

e=

(38.12) COROLLARY. For any commutative R-order in A in a separable K -algebra A. we have Pi cent A = LF P(A) ~ Cl A Proof We shall identify the set of left A-modules with the set of (A, A)bimodules over A; this can be done since A is commutative. We have already remarked that Picent A = LF P(A), by (37.22). Let us now show that e is an isomorphism. By (38.10), we have ker e ~ Outcent A = 1, so e is monic. Further, each element ofCI A is of the form [M], for some locally free A-lattice Min A. By Exercise 38.4, M is an invertible (A, A)-bimodule, and hence (M) E LF P(A). Clearly e(M) = [M], so e is epic. This completes the proof of the corollary.

In the preceding discussion we made use of the following result, which is of independent interest. (38.13) THEOREM (Kaplansky). Let R be any noetherian domain with quotient field K, let A be any commutative R-order in a K-algebra A (not necessarily separable over K). By a "A-ideal in A" we mean a finitely generated A-submodule M of A such that KM = A. (i) Let M 1 , ... , Mt' N l , ... , Nt be A-ideals in A such that M

+- ... +- M t ~- N 1 +- ... +- N t

1"

as A-modules.

Then M

1'"

M t ~ N 1'" Nt

as A-modules,

where these products are computed within A. (ii) Two A-ideals in A are stably isomorphic

if and only if they are isomorphic.

Proof (i) The hypotheses imply that for all i, j between 1 and t, HomA(KM i , KN) ~ K Q9j{Hom A(M;,N/

Furthermore, the map HomA(M p N) IS

mOllIe.

-+

HomA(KM p KN)

348

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(38.14)

Now let cp: L' Mi ~ L' N j be a A-isomorphism. Then cp can be extended to an A-isomorphism cp*:A(t) ~ A(t). Hence there exists an invertible matrix p = (Ct i ) E Mt(A) such that

cp(ml' ... , mt ) =

(Ct i )

(ml ' ... , mtf,

mi E Mi'

1

~

i

~

t,

where T denotes "transpose". Therefore (38.14)

1

~

i,j

~

t.

Set d = det p; then dE u(A) since p is invertible. Furthermore, d is a sum of products of the form

± Ct,,(l), 1 ., . Ct,,(t).t '

where

11:

is a permutation on {I, ... , t}. It follows from (38.14) that

d· MI'" M t eN 1 ... Nt' On the other band, cp-l also extends to an A-isomorphism (cp-l)*:A(t) ~ A(t), described by the matrix p-l of determinant d- 1 . The above reasoning shows that d- 1 . N 1 ... Nt c MI' .. M t · Therefore

fl N,

= d·

fl M;, whence fl N, ~ fl M; as claimed.

(ii) Now let M, N be A-ideals in A which are stably isomorphic. Then M -i- AIr) ~ N -i- AIr) for some r, whence M ~ N by (i). This completes the proof of the theorem. To conclude this section, we give the following generalization of (37.36): (38.14) THEOREM. Let A be an R-order in a separable K-algebra A, where R is a Dedekind domain. Assumet that condition (38.11) holds, that is, stable isomorphism implies isomorphism for locally free left A-lattices. Let n ~ 1, and set E = Mn(R), r = E 0 R A ~ Mn(A), as in (37.36). Then all of the assertions (i}-(iii) in (37.36) remain true, provided that in (37.37) we replace Pi cent by LFP. Furthermore, (37.40 i) holds true in this case.

Proof. The proofs of (37.37) and (37.40) carryover with only minor modifications. We omit the details. It is worth pointing out a stronger version of (37.40 ii), namely

(38.15) COROLLARY. Let A be a commutative R-order in a separable K-algebra A. Then

for each n

t

~

1.

See footnote to (38.10).

(38.15)

349

EXERCISES

Proor By (38.12) e is an isomorphism, whence cok e = O. Further, Outcent A = 1 since A is commutative. The desired result now follows from the exactness of the sequence (37.39) asserted in (38.14).

EXERCISES In the following, A is an R-order in a separable K-algebra A, where R is a Dedekind domain. 1. Prove that stably isomorphic left A-lattices are in the same genus. [Hint: Let X, Y be A-lattices such that

X

+Nd ~ Y +A(r).

Complete at an arbitrary P, then show that X p ~

Yp

by Exercise 6.7.J

2. Prove that every locally free left A-lattice is A-projective. [Hint: See (3.23).J

3. Let N(A) = {x Eu(A): xAx- 1 = A},

the normalizer of A in A. Show that N(A) = {x

[Hint: Let x

E

E

u(A) : xAx-

1 c

A}.

u(A) be such that xA c Ax. By Exercise 10.7,

ordR A/xA = ordR A/Ax. Hence xA = Ax by (4.17).J 4. Let X be a two-sided A-submodule of A. Show that if X is locally free as left A-module, then X v A and X is an invertible bimodule whose inverse is also locally free. Deduce from this that LFP(A) is a subgroup of Picent A. [Hint: If X V A(n) then KX ~ A(n), so n = 1 and XvA. Now put Y = {aEA:Xa c A}, another bimodule in A. To prove that X is invertible, we need only show that XY = YX = A. It suffices to verify this after passing to completions. Changing nutation, assume that R is complete and that X = Au, where u E u(A). Then Au . A = Au since X is a bimodule, whence uA c Au. Thus uA = Au by Exercise 3, and hence Y = u- I A = AU-I.] 5. Let R be a principal ideal domain with quotient field K, and let r = Mn(R), where n ~ 1. Show that every invertible matrix x E Mn(K), such that x r x- I = r, is of the form x = lXy for some nonzero IX E K and some y E u(r). [Hint: Take A = R in (38.15) to deduce that Outcent r = 1. By (37.25), Outcent r ~ N(r)/u(r) u(K), where N(r) is the normalizer of r in Kr. Hence N(r) = H(r) u(K).J 6. Let G be a finite group, R a Dedekind domain with quotient field K, and let A = RG, A = KA = KG. Suppose that IGI of. 0 in R, and that no prime divisor of I G I is a unit in R. A fundamental theorem due to Swan asserts that, in this case, every projective left A-lattice is locally free. Deduce from this that there is a split exact sequence of additive groups

350

SIMPLE ALGEBRAS OVER GLOBAL FIELDS

(38.16)

(38.16) where fi[M] = [A] - [M], [M]

E

Cl A;

",[XJ

=

[K @RX], [X]

E

Ko(i\).

[Hint: Imitate the relevant parts of the proof of Theorem 36.3. The sequence (38.16) splits, since Ko(A) is a free Z-module by Exercise 36.3.J

9. Hereditary Orders We shall first present the results of Harada [1-5J and Brumer [1J concerning hereditary orders in separable algebras. The approach used is that of Jacobinski [4]. As we shall see, we can give an explicit description of the structure of hereditary orders. We can also determine the relationship between a hereditary order and the maximal orders which contain it. The local theory in § 39 will be applied in §40 to obtain the global theory of hereditary orders. In §41 we consider integral group rings, and prove several theorems which fit in well with the subject matter of this book. Throughout this chapter, R denotes a Dedekind domain with quotient field K, and A is an R-order in a separable K-algebra A. Recall that A is (left) hereditary if every left ideal of A is projective; by (10.7), this occurs if and only if every left A-lattice is projective. Now the ring A is left and right noetherian, since A is finitely generated as R-module. Therefore, as remarked in (2.45 iii), A is left hereditary if and only if A is right hereditary. A direct proof of this fact is given in Theorem 40.1 below. Hence in our discussion of hereditary orders, we may omit the adjectives "left" and "right".

39. LOCAL THEORY OF HEREDITARY ORDERS Throughout this section let R be a complete discrete valuation ring with maximal ideal P = nR and residue class field R. As we shall see, the key to studying hereditary orders is the investigation of their Jacobson radicals. The first result is due to Auslander-Goldman [1]: (39.1) THEOREM. An R-order A is hereditary if and only if rad A is a projective left A-module, or equivalently, if and only if rad A is an invertible (A, A)bimodule. Proof. Step 1. Let I = rad A. Surely I is left A-projective if A is hereditary. Conversely, assume that Al is projective. We shall show that A is hereditary by proving that every left ideal L of A is projective. Given any L, there is a A-exact sequence

351

352

HEREDITARY ORDERS

(39.1)

for some r, where I', is a left A-lattice. We prove below that Ext! (L,1'.) = 0, where Ext means Ext". Once this is known, it follows that the above sequence is A-split. Therefore LI Atr), so L is A-projective, whence A is hereditary as claimed. The A-exact sequence

o -> L

.!4

L -> LlnL -> 0

gives rise to an exact sequence Ext! (L, 1'.) .14 Ext l (L, 1'.) -> Ext2 (LlnL, 1'.) by (2.29), where the arrows labelled "n" are the maps given by multiplication by n. We set M = LlnL, and we show below that Ext2 (M,1'.) = O. This will imply that n . Ext! (L,1'.)

=

Ext! (L, 1'.),

whence Ext! (L, 1'.) = 0 by (2.34) and Nakayama's Lemma. The left A-module M is artinian, since nM = O. We prove by induction on the A-composition length of M that Ext 2 (M, 1'.) = 0 for every left Amodule 1'.. Suppose first that M is a simple A-module; then M is also a simple left A-module, where A is the semisimple artinian ring AjJ. Therefore MIA, whence Ext2 (M, 1'.) is a direct summand of Extl (1\,1'.) by (2.28 iv). However, there is a A-exact sequence

o ->

J -> A -> 1\ -> 0,

and" J is projective by hypothesis. Hence by (2.27) we obtain Ext~ (1\,1'.) = 0 for all 1'., Therefore Ext 2 (M, 1'.) = 0 for all simple A-modules M and all A-modules L Now suppose that the A-module M has composition length at least two, and let S be any simple submodule of M. The A-exact sequence 0-> S -> M -> MIS -> 0 gives an exact sequence Ext 2 (MIS,

I',)

-> Ext 2 (M, 1'.) -> Ext 2 (S, 1'.).

Since the first term vanishes by the induction hypothesis, and since the last term is 0, it follows that Ext 2 (M, 1'.) = 0 for all 1'.. This completes the proof that A is left hereditary whenever "J is projective. Analogously, A is right hereditary whenever J" is projective. Step 2. We now prove that A is hereditary if and only if J is an invertible (A, A)-bimodule. If J is invertible, then "J is projective by (16.7) or §37, whence A is hereditary. For the harder part of the proof, suppose now that

(39.2)

LOCAL THEORY OF HEREDITARY ORDERS

353

A is hereditary, so both /1.1 and J/\ are projective modules. Then JIN') for some r. We shall use this fact in a moment. Let t

A

L' Mi'

= i

[MJ indecomposable left ideals,

1

J

with the ~ M i f numbered so that {M I' ... ,M are a full set of non-isomorphic modules among the {M,}. Setting M, = M;l1M i , it follows from (6.22) that {M I"'" MJ are a full set of non-isomorphic simple left (AjJ)modules. Since N'l, it follows by Exercise 6.8 that

JI

J~ f'M\I!,l -

L.,

i~

,

1

for some non-negative integers {nJ Therefore u

JjJ2 ~

L'

M~nil

as left A-modules,

i=I

If

and so each ni > 0 by Exercise 6.9. Thus A S ) for some s, whence by (15.2) ,/ is a progenerator for the category /\ ..41. Setting ~ = Hom/\ (J, J), it follows from (16.9) that the rings A, ~ are Morita equivalent, so '/6. is an invertible bimodule (see § 37) and A = Homo. (J, J). Using the identification Homo. (1, J) = OP), we conclude at once that A = OP). But then, by symmetry, also A = 0r(J). Therefore J is an invertible (A,A)-bimodule, as claimed. This completes the proof of the theorem. (See also Remark following

(39.18).) As preparation for our later discussion, we introduce the following notation:

(39.2) Definition. Let r be a ring. For each ideal a of r, let (a)m x n denote the set of all m x n matrices with entries in a. If {a ij :l ~ i, j ~ r} is a set of ideals in r, we write (all) (a 21 )

A

(a 12 ) (a 22 )

'"

(a1r)J(." '., Or} (a 2r )

= [ ......................... .. (art)

(a r2 )

.. . (a rr )

to indicate that A is the set of all matrices [TijJ I"; i. j";r' where for each pair (i,j), the matrix Y;. ranges over all elements of (l\.)n ix Now let V be a~ n-dlmensional right vector sp~ce over a skewfield D, and set End a V = Homa (V, V). Thus End a V is the O-endomorphism ring of V, and we shall view V as an (End a V, D)-bimodule. Some ci the calculations

354

HEREDITARY ORDERS

(39.3)

which follow are more easily understood by using matrices instead of linear transformations. Once we pick an O-basis for V, we may identify V with the space on x 1 of all n x 1 column vectors over O. This gives rise to an identification Endn V = M.(O), where the matrices act from the left on the column vectors from V. A chain in V is a strictly decreasing sequence (39.3)

E:V=VO>Vl>···>~=O

of O-subspaces of V. We set 1

and call the ordered r-tuple {n l define the chain ring O(E) by (39.4)

O(E) =

, ... ,

~

i

~

r,

n,} the invariants of the chain E. Now

{x EEndn V : xV;c V;, 0

~

i ~ r}.

We may choose an O-basis of V adapted to the chain E, that is, the first n l basis elements map onto a basis for YaIVl , the next n2 elements onto a basis for Vl/V2, and so on. Clearly n l + ... + n, = n,and the last n, + .,. + n, basis elements of V form a basis for the subspace V; _l ' Relative to this basis, we may identify O(E) with a subring of Mn(O). It is easily seen that O(E) consists of all matrices

~:~~,~~ 1.

(39.5)

r_T,l

1'.2

1'.3 ... 1'.,j

where for 1 ~ j ~ i ~ r, the matrix T,j ranges over all elements of the additive group Homn (~-l/~, V;- l/V;)· Hence in the notation of (39.2), we may write

(39.6)

OrE)

=

[:~;. :~):~; (0)

(0)

(0)

......... ...

:~;

[n,." " nr )

(0)

We collect some facts about chain rings. (39.7) THEOREM, Let the chain E in (39.3) have invariants {n 1 , .. " nr }, and let B = OrE) be its chain ring, Then

(i) rad B = {x

E

B: x V;

c

V; + l' 0

~ i ~

r - I},

(39.8)

355

LOCAL THEORY OF HEREDITARY ORDERS

In terms of the identification in (39.6), rad B is given by (0) (0)

(0) (0)

rad B = ,_ (~) ..

(39.8)

1

(0) (0)

.. .

(O)j (0)

(~) ... (~) ...... : ... (~;

In, .... nr }

,

that is, rad B consists of all matrices in (39.5) in which each 'F;; = 0, 1

~ i ~

r.

(ii) B/rad B ~

r

r

2:' End!) (f!;_,/v) ~ :L'1 Mn,(O). ;= 1 i

The left B-modules {~_ II V; : 1~ i ~ r} are a full set of non-isomorphic simple left B-modules. (iii) For each t, 0 ~ t ~ r, we have

v; = (rad BY V,

(radBY={xEB

xV;cV;+t,

O~i~r},

where we interpret ~ as 0 when s ~ r. Proof Let N={XEB: XV;C~+I'

O~i~r-l}.

Then N is a two-sided ideal of B, and N r = 0, whence N c rad B. On the other hand, N is the kernel of the ring epimorphism r

B ->

L,' End a (f!;- Ill-)·

;= 1

Since the direct sum is a semisimple ring, we conclude from Exercise 6.1 that N ~ rad B. This shows that N = rad B, and establishes (i) and the first statement in (ii). The second assertion in (ii) is an immediate consequence of the first. The easiest way of proving (iii) is by means of matrices. Choosing an O-basis for V adapted to the chain E, we have identifications (0) (0)

(o)nr x 1

356

HEREDITARY ORDERS

(39.9)

for 0 :::; t :::; r. The assertions in (iii) are then immediate consequences of the matrix expression for rad B given in part (i). This completes the proof. By way of illustration, we remark that when the chain E is as fine as possible (that is, each ni = 1), then upon choosing an O-basis of V adapted to E, the chain ring O(E) may be identified with the ring of all lower triangular matrices in Mn(O). Its radical consists of all those lower triangular matrices with U's along the main diagonal. Up to conjugacy, this ring O(E) is the smallest chain ring in M n(O). In this connection, see Exercise 39.5. Now let B be any subring of End n V, not necessarily a chain ring. Setting N = rad B, we see that

EB : V> NV> N 2 V > ... > N'V = 0 is a chain in V. Since B c Endn V, it is clear that s is the least integer such that NS = O. We have

(39.9)

Be O(E B),

and

rad B = N c fad O(EJ.

Weare thus led to partially order the set of subrings of End n V, as follows: given two sub rings B, B' of Endn V, we say that B' radically covers B (notation: B' >- B) if B' =:> B, and rad B' =:> rad B. Call B an extremal subring of Endn V if B' >- B implies that B' = B.

(39.10)

LEMMA. The extremal subrings of End n V are precisely the chain rings.

Proof If B is extremal, then B = O(EJ by (39.9). Conversely, every chain ring O(E) is an extremal sub ring, by Exercise 39.5.

We now turn to the analogues of these concepts for orders. Let A, A' denote R-orders in A. We say that N radically covers A (notation: A' >- A) if N =:> A and rad N =:> rad A. The order A is called extremal if A' >- A implies that N = A. Every maximal order A is obviously extremal, since N>-A = N=:>A = N=A. Our aim is to show that hereditary orders are the same as extremal orders. The problem can be reduced to the central simple case; in this case, we shall show that extremal orders correspond to extremal subrings of full matrix algebras over a skewfield. This correspondence will enable us to determine the structure of hereditary orders in central simple algebras. As a first step in this program, we prove

(39.11) THEOREM. An R-order A in a separable K-algebra A is extremaqr and only if O/(rad A) = A.

LOCAL THEORY OF HER EDIT ARY ORDERS

(39.12)

Proof Let J = rad A; since J A-lattice in A. Set

=:>

357

PA, it is clear that J is a full two-sided

I = 0 P) = {x EO A: x J

e

J}.

Then I is an R-order in A, and J e A e I. For large n, J"e PA e PI; hence J e rad I by Exercise 39.1, which shows that I >- A. Hence I = A if A is an extremal order. Suppose conversely that I = A, and let us show that A is extremal. Let A' :>- A, and set J' = rad A', so J =:> J. Since pr A' e J for some r, it follows that (Jlr e J for some s. Obviously s > 0, since otherwise A' e J e A, which is impossible. If s ~ 2 then (JI)S-l. J e (J'y-l. J' e J, so (J'y-l e 0,(1) = A. Hence (1')S-1 is a two-sided ideal in A. But its sth power lies in J, whence also (Jy-l e J. Continuing in this way, we find eventually that J' e J. Therefore J' = J, and so A

=

OP)

=

O/(JI)

=:>

A',

whence A = A'. This completes the proof. (39.12) COROLLARY. Hereditary orders are extremal.

Proof Let J = rad A, where A is hereditary. The last paragraph of the proof of (39.1) shows that A = 0/(1). Hence A is extremal.

In order to facilitate reduction of the discussion to the central simple case, we now prove the local analogue of (10.5) for extremal orders. t

(39.l3) THEOREM. Let A

=

L:" Ai

be the decomposition of the separable

i= 1

K-algebra A into simple components {AJ, and let Ri be the integral closure of R in the center of Ai' Then every extremal R-order A in A is expressible as a direct t

L:" Ai' where each Ai is an extremal R(order in Ai·Conversely, every 1 such direct sum L:" Ai is an extremal R -order in A.

sum A

=

i=

t

Proof Let C

=

L:" R"an R-order in the center of A. Then CA is an R-order i=1

in A. Let J = rad A; then Jm e PA for some m, whence (CJr e P' CA. Therefore CJ e rad CA by Exercise 39.1, and so CA >- A. Since A is extremal, t

this gives A = CA. But CA =

L:" RiA,

and for each i, RiA is an extremal

i=1

R.-order in A .. This completes the proof that each extremal order A decomI

I

358

(39.14)

HEREDITARY ORDERS

r

poses into a direct sum Ai' as claimed. The last assertion in the theorem is obvious, and the result is established. Since we have shown that every hereditary order is extremal, the above decomposition theorem is valid for all hereditary orders. In (40.7) we shall obtain the global analogue of the above for hereditary orders. The preceding paragraph shows that the classification of hereditary orders can always be reduced to the central simple case. The basic structure theorem for this case is as follows: (39.14) Theorem. Let A be an R-order in the central simple K-algebra A, where center K. Let ~ be the unique maximal A ~ M n(D) and D is a skewjield with _ R-order in D, and set p = rad~, ~ = ~/p. Then (i) A is hereditary if and only if A is an extremal order. (ii) Given any extremal order A in A, there exist positive integers {nl'" ., nr } with sum n, and there exists an identification A = M neD), such that - (~)

(39.15)

A=

(p) (p)

(p)

(~)

(~)

(p)

(p)

(L\)

(~)

(~)

(p)

{n" ... , nr}

1· .. ·,····· ........ ·· ... -

(~)

(~)

.

(~)

,

(~)

"

using the notation in (39.2). Conversely, each such order A is extremal. (iii) For the extremal order A in (39.15), we have (pr (n"

(p) (p) (p) (39.16)

rad A =

(~)

(p) (p)

(p)

(~)

(~)

(p)

(p)

(~)

(~)

(~)

...

.. .

,n.!

(p),

and r

(39.17)

A/rad A ~

L' M

n

,

(L\).

i=l

Proof. Every hereditary order is extremal, by (39.12). To prove the converse, let A be an extremal order in A, and set j = rad A. Let A c: A' where A' is a maximal order in A with radical J'. Then A + j' > A by Exercise 39.2, whence A + j' = A. Therefore j' c: A. so J' c: j by Exercise 39.3. We set

359

LOCAL THEORY OF HEREDITARY ORDERS

(39.17)

A = All',

A' = NIl',

J=J/1'.

Then A is a subring of 1\', and J = rad A by Exercise 39.4. We now use the results of (17.4) and (17.5). We may find a right M such that

~-lattice

II.' = End~ M ~ Mn(~)'

where M

~ ~(n)

J'

is a (11.',

~)-bimodule.

Then

= End~ Mp ~ Mn(p),

A' ~ End;\M ~ Mn(~)'

where M = MIMp = MIJ'M ~ ~(n).

Let p be the canonical map p:N

-+

11.'11'= Mn(~)'

where we have identified NIl' with Mn(~)' Then A = peA), and we shall show that A is an extremal subring of Mn(~)' For suppose that B is any subring of M n(~) which radically covers A, and let Ao = p -l(B), J o = rad Ao' Then Ao is an R-order in A such that 1'c A c Ao c 11.', and so l' c J o by Exercise 39.3. Hence Exercise 39.4 gives Jol1'= radAoll' = radB

::J

radA = JIJ;

whence J o ::J J. Therefore Ao ~ A, so Ao = A si~ce A is an extremal order by hypothesis. This gives B = A, and shows that A is an extremal subnng of Mn(~)' _ . We may now use (39.10) to obtain the structure of A. We form the cham E in M, given by E : M > JM > J2 M > ... > j' M = O. Note that jiM = JiM for each i, since II.' and A act on M via the map p. Let {n 1 , ... , n,} be the invariants of the chain E, so by definition nj

1 ~ i ~ r.

= dim"6, ]i- 1M piM,

Then (39.6) and (39.8) give _

l

(O)j(nl, ... ,n

(~)

(~

(~)

(~)

.. .

(0)

(~)

(~)

.. ,

(~)

A= ................ . ,

r)

-

J=

l

(~

(0)

(O)J

(M (0) ... (0) . ................. (~)

(11)

.. , (0)

Since the map p is precisely "reduction mod p", and since A = P

1

(A),

(nl, " ., n r )

All ~ A/1,

360

(39.18)

HEREDIT AR Y ORDERS

we immediately obtain the asserted formulas (39.15)-(39.17). At this stage we have proved the first two implications in the chain A hereditary

=

A extremal

=

A satisfies (39.15)

=

A hereditary.

To complete the proof of the theorem, we need only show that an order A given by (39.15) is necessarily hereditary. We set A' = M"(~) ~ A, J' = radA' = Mn(p), p:A' -> NIJ', as before. Since p is "reduction mod p", it is clear from (39.15) that A/l is a chain ring in the matrix algebra M n(~)' But then we know rad A/l' from (39.8). Since rad A = p-l {rad AjJ'} by Exercise 39.4, we immediately obtain formula (39.16). The remaining formula (39.17) is then obvious. We are still trying to prove that the order A given by (39.15) is hereditary. Let Ao be the order in A defined by taking n 1 = ... = nr = 1 in (39.15). Since (39.16) gives us rad Ao ' it is easy to check that 000

1 0

0

o o

0

0

...

o

0

1

0

where p = n D ~ = ~ n D . Hence rad Ao is Ao-projective, and thus Ao is hereditary by (39.1). But Ao cAe A, so A is also hereditary by (40.4) below. This completes the proof of the theorem. When A is a hereditary order satisfying (39.15)-(39.17), we shall call r the type of A, and the ordered r-tuple {n 1 , ... , nr } the invariants of A. It is clear from (39.17) that a given order A uniquely determines r and the set of integers {nJ We shall see in (39.24) below that A determines the ordered r-tuple {n 1 , ... , nr } uniquely, up to a cyclic permutation. First, however, we shall obtain some consequences of the structure theorem (39.14). (39.18) COROLLARY. Let A be a hereditary order (ina central simple K-algebra) of type r and invariants {n 1 , ... , nJ Keep the notation of (39.14) and its proof Then (i) r is the unique positive integer such that J"M = Mp. (ii) We have

(39.19) (39.20)

A={XEA:x·JiMcJiM J

=

{xEA : x·JiM c P+1M

forO~i~r},

forO ~ i ~ r -I}.

(iii) Fe = 11:A, where e is the ram(jication index of Dover K. (iv) A is a chain ring in 1\.', corresponding to the chain

(39.21)

LOCAL THEORY OF HEREDITARY ORDERS

(39.21)

E : !VI> JM > PM> '" > PM = 0,

361

with invariants given by ni = dim 6 J i- 1 MjJ iM, 1

1 ::;; i::;; r.

i

(v) The modules {ii- M/J M: 1 ::;; i::;; r} are a full set of non-isomorphic simple left A-modules.

Proof. Step 1. Let r be the canonical map r :

M

-+

M = M/Mp = MjJ'M,

and set Mi =JiM, Mi =

r-

1

(MJ,

0::;; i::;; r.

Since r is a left A-homomorphism, each Mi is a A-submodule of M. We claim that (39.22)

Mi

=

JIM

0::;; i ::;; r,

for

PM

=

Mp = J'M.

The easiest way to prove this is by a matrix calculation, as follows: choose a right ~-basis for M whose images in M form a basis adapted to the chain E in (39.21). We may then write (;i)nlXl

(~)"'Xl

l'vl

= ~ (11) =

, M = ;i (tI) =

Therefore for 0 ::;; i ::;; r, (0)"1

X

1

(pYi-l (~)ni x J

(;i),,' x 1

Since J is given by (39.16), we have

X

1

362

(39.22)

HEREDITARY ORDERS

(p) (~)

(p) (p)

(p) (p)

-(~)

-(p)

(~)

(~)

JM =

=M 1 •

................... (~) (~) ... (p)

(~)

(~)

Continuing in this manner, we easily obtain the formulas in (39.22) by induction on i. This establishes assertions (i) and (iv) of the corollary. Step 2. For the moment let A* denote the order occurring on the right side of (39.19). Obviously A c A* since A· J i = J i for each i. On the other hand, let xEA*; then xM c M, so xEEndL\M = N. Further, for 0 ~ i ~ r, XMi c Mi implies that p(x)Mi c Mi' Therefore p(x) E A, since A is the chain ring associated with the chain E of (39.21). Hence x E A, which proves that A = A* and establishes (39.19). A similar argument proves (39.20). Next, we have have seen in the proof of (39.14) that AjJ ~ A/1, so the simple left A-modules are the same as the simple left A-modules, with A acting via the map p:A ....... A. By (39.7 ii), the modules {M i_ 1 /M i : I ~ i ~ r} are a full set of non-isomorphic simple left A-modules. But '[ induces isomorphisms 1 ~ i ~ r.

and thus we have proved assertion (v) of the corollary. Step 3. Now Jet e be the ramification index of Dover K (see § 13). Then

pe =

n~ ~

=

~ n~

=

n~.

Since J'M = Mp, we obtain J,.eM

= Mpe =

nM.

Therefore T = n- 1 J,e is a two-sided A-ideal in A such that TM = M and T J = JT. Hence TM, - T·JiM =]i. 1M = M i ,

0 ~ i ~ r.

It follows from (39.19) that TeA, and we now show that T = A. Indeed, (T + J)jJ is a two-sided ideal of the semisimpJe artinian ring AjJ. If it is a proper ideal, it is annihilated by some (nonzero) idempotent c; in AjJ. By (6.18) c; is the image of some idempotent a E A, and then obviously aT c 1. Therefore

o~ i

~

r - 1,

whence a E J by (39.20). This is impossible, since rad A contains no idempotents of A. Therefore T + J = A, whence T = A by Nakayama's Lemma.

(39.23)

LOCAL THEORY OF HEREDITARY ORDERS

We have thus proved that n ~ 1 J = A, whence J completes the proof of the corollary. YO

363

= nA as desired. This

YO

Remark. The preceding proof shows that J is an invertible (A, A)-bimodule, with inverse n- 1 ]'e-l. It should be pointed out that the above argument yields another proof, independent of that given for (39.1), of the implication

A hereditary

=-

rad A invertible.

From the preceding results, we deduce (39.23) Theorem. Keeping the hypotheses and notation of (39.18), let

Ii = {x EA : x·JiM

c JiM},

0

~ i ~

r - l.

Then 10 , ... ,1'_1 are precisely the distinct maximal orders of A containing A, and A = 10 n r; n··· n 1'_1 . Furthermore, every indecomposable left A-lattice is isomorphic to exactly one of the A-lattices M, J M, ... , ],"-1 M. Proof. Suppose first that L is any indecomposable left A-lattice. The proof of (21.5) shows that KL must be a simple left A-module, and therefore KL = KM. Replacing L by an isomorphic copy, we may assume that L c M. Choose t so that L c PM, L cf P+lM.

Since J is invertible, it follows that if we set Ll = Ll c M,

Ll

cf

r

t

L, then

J M.

Thus (Ll + JM)/1M is a nonzero submodule of the simple A-module M/1M, whence Ll + JM = M. Therefore Ll = M, and so L = PM. Hence the modules {PM} give all isomorphism classes of indecomposable left A-lattices; each is obviously indecomposable, since K· J'M is a simple A-module. We show now that J'M ~ PM as left A-lattices if and only if s == t(mod r). Any such isomorphism extends to an A-isomorphism KM S; KM, hence must be given by right multiplication by some element dE D. Writing d = U· n~, where u E u(Ll) and n D is a prime element of Ll, it follows that ;S M s; ]':v1 if and only if 1'M = J'M . n~ for some k. Since MnD = 1'M, this occurs if and only if r I (s - t), and the second assertion of the theorem is proved. For each integer t, set r; = {x E A:x' PM c JIM}. Now 1'M is a right Ll-lattice such that K· PM = K' Ll("), whence J'M S; Ll(") by Exercise 18.1.

364

HEREDITARY ORDERS

(39.24)

Since ~ = End,\J'M it follows from.( 17.4) that ~ is a maximal order. Clearly ~ = r; whenever s == t (mod r). Conversely, if rs = r; then ;SM and J'M are a pair of (~, ll)-bimodules; hence by (16.14) ;SM = J'M· pk for some k, whence r I (s - t). This shows that r;;, ... ,~. -1 are distinct orders containing A. Their intersection is A, by (39.19). Finally, let r be any maximal order of A containing A. Every indecomposable left r -lattice is also an indecomposable left A-lattice, hence may be taken to be one of {J iM : 0::::; i::::; r - I}. Therefore r must coincide with one of the orders r 0' . . . , r r _ l ' and the theorem is established. (39.24) COROLLARY. Keep the above notation and hypotheses. Then A uniquely determines the ordered r-tuple {n 1 ' ... , nr } of invariants, up to cyclic permutation.

Proof. Let A c A' = End,\ M, and let us show that the invariants {nl' ... , nr } are unaffected by using N in place of M, where N is another ll-lattice such that A' = End,\ N. By (16.14) we may write N = Mpk = Mn:~ for some integer k. It is then easily verified that the invariants computed using N are the same as those obtained from M. On the other hand, if ~ is one of the maximal orders containing A, then we have seen that ~ = End,\J'M. Thus we obtain invariants {nJ, where n~ t

= dim-6. Ji-1.y Mj.!i .JIM = ni+t' •

1 ::::; i ::::; r,

and where the subscript i + t is read mod r. This shows that all cyclic permutations of {nl ' ... , nr } can arise as invariants, and these are the only possibilities. Returning to the. general case, we have (39.25) THEOREM. Let A be an R-order in a separable K-algebra. Then A is hereditary if and only if A is extremal.

Proof. By (39.12), hereditary orders are extremal. Conversely, any extremal order A can be decomposed as A = Ai by (39.13). But then each Ai is hereditary by (39.14), whence so is A.

I"

If we assume merely that R is a discrete valuation ring, not necessarily complete, then the structure theorem (39.14) and its consequences (39.18)(39.23) remain valid whenever the P-adic completion Dp is a skew field.

EXERCISES 1. Let L be a left ideal of the R-order I\. Show that

Lm c PA

for some m

=

Lc radA.

365

EXERCISES

(39.25)

[Hint: By (6.15), (rad At c PA c rad A for all sufficiently large L c rad A

=

11.

Hence

L n cPA.

Conversely, suppose that Lm c PA for some m. Then LA is a two-sided ideal of A, and (LA)m = (LA)(LA)" . (LA) c LmA c PA c: rad A.

Hence LA c rad A by Exercise 6.3.J 2. Let A c A' be R-orders in A. Show that A + rad A' is an R-order in A whose radical contains rad A. [Hint: Let J = rad A, J' = rad A'. Clearly A + J' is an order, and we must show that J c rad (A + J'). For large m, (J + JT c Jm + J' c PA + J' c PA' + J', and (PA' + JT c PI\' + pm cPA'. Thus (J

+

3

J'r c pmA' c PA c PtA

+ J')

for large m, whence J + J' c rad (A + J') by Exercise 1.J 3. Let A c A' be R-orders in A, and suppose that rad A' c A. Show that rad A' c rad A. [Hint: Let J' = rad A'. As in Exercise 2, we get J" c PA for large r. Since J' is an ideal of A, it follows from Exercise 1 that J' c rad A.] 4. Let N be a two-sided ideal of the R-order A contained in rad A. Prove that (radA)IN = rad(A/N).

[Hint: Let J = rad A, J = JIN, i\ = AjN. Then i\jJ ~ AIJ, a se~isimple ring. Hence J => rad A by Exercise 6.1. On the other hand, I~: A --> A IS epIC, so J = I~(J) c radA by (6.10).J 5. Let E, E' be a pair of chains in the Q-space V, and define the chain rings OrE), OrE') as in (39.4). Prove (i) OrE) ::0 OrE') if and only if E' is a refinement of E. (ii) OrE) OrE') if and only if E = E'. (iii) OrE) is an extremal subring of End n V.

>-

[Hint: To prove (i) and (ii), consider the matrix representations (39.6) and (39.8) of the rings OrE), OrE') and their radicals. To prove (iii), let OrE) be given, and consider all subrings B of Mn(Q) such that B OrE). We may choose a maximal element B' relative to this partial ordering. Then B' is an extremal subring of Mn(Q), whence B' = O(E B,) by(39.10). But then O(E B,) OrE), so by (ii) we have B' = O(EB ,) = O(E).]

>>-

In Exercises 6-11, we keep the hypotheses and notation of(39.18) and (39.23) and their proofs. 6. Show that every invertible (A, A)-bimodule in A is a power of rad A, and deduce that Picent A is cyclic of order reo [Hint: Let T be an invertible bimodule; then T JT- l = rad (TAT-I) = rad A = J, so T J = JT. Let V = J"T with k chosen so that V c A, V q: 1. Then JM < (V + J)M c M, the first inclusion being proper

366

HEREDITARY ORDERS

since otherwise U c J. Hence (U + J)M proof of (39.18) to deduce that U = A.]

(39.26)

M. Now use the argument in Step 3 of the

=

7. Let A be a hereditary order of type r. Prove (i) There are exactly 2' - 1 distinct orders in A containing A, namely the intersections of the orders in each non-empty subset of {r o"" ,r;- J. (ii) The maximal possible length of a chain of orders

A < Al < ... < As in A is r itself. (iii) There are precisely r distinct orders A I with Al > A, such that there is no order properly between Al and A. 8. For the order A in (39.15), describe explicitly those maximal orders containing A. 9. For 0 :s; t :s; r - 1, let J, = {xEA : x·PMc J'+IM}.

Show that {J 0' ... , J, _ I} are the distinct maximal two-sided ideals of A, and that J, = rad r,. Prove that

Ai J

~

,-I

I'

r.l1,.

,~o

For each t, calculate the conductor {x E A :r,' x c A}. to. Prove that FA' = J', where A' is a maximal order containing A, and J' = rad N. [Hint: J'N is a two-sided ideal of N, hence is a power of P. Now use (39.22).]

11. Show that the formula J'

= J'A' = L,·L,_I···L2·Ll' wnere L; = JiA'J-(l-I),

expresses J' as a proper product of r integral ideals L" ... , L I . Prove that (39.26)

A

=

n

01(L),

i=l

and that the chain (39.21) for A is just (39.27)

M > LIM> L2LIM > ... > (L,'" LI)M

=

O.

Show conversely that starting with any maximal order A' and any proper factorization rad A'

=

L, . L, _ I

...

L2 . LI

into integral ideals, the order A given by (39.26) is hereditary and corresponds to the chain (39.27). 12. Give an example of two maximal orders A l' A2 in a central simple K-algebra, for which A I n A2 is not hereditary. 13. Let AI' A2 be hereditary R-orders in the central simple K-algebra A. Show that there exists a full R-Jattice Y in A such that if and only if Al and A2 are of the same type.

(40.1)

GLOBAL THEORY OF HEREDITARY ORDERS

367

40. GLOBAL THEORY OF HEREDITARY ORDERS Keeping the notation introduced at the start of this chapter, we assume now that R is any Dedekind domain. Here we shall develop the theory of hereditary R-orders, by using the local results obtained in § 39. We prove first (40.1)

THEOREM.

A is left hereditary

if and

only!f A is right hereditary.

Proof. Since A is both left and right noetherian, the result is a special case of

(2.45 iii). We give an independent proof here. Each left A-lattice M determines a dual M* = Hom R (M, R), which is a right A-lattice, and there is a natural isomorphism M ~ M** (see Exercise 40.2). Furthermore, any exact sequence of left A-lattices (40.2) must be R-split, since N is R-projective by (4.13). Hence when we apply Hom R (., R) to (40.2), the resulting sequence

o ---> N* ---> 1'v[* ---> L* ---> 0 (40.3) is also R-split, and hence exact. The maps in (40.3) are right A-homomorphisms. Conversely, taking duals in (40.3) gives (40.2) again. Hence the sequence (40.2) is A-split if and only if (40.3) is A-split. Suppose now that A is right hereditary, and let N be any left ideal of A. Then we can find a A-exact sequence (40.2) in which M is a free left A-lattice. Now L* is a right A-lattice, hence is A-projective by (10.7). Therefore the sequence (40.3) is A-split, whence so is (40.2). This proves that every left ideal of A is projective, so A is left hereditary. The theorem now follows by an obvious symmetry argument. (40.4) THEOREM. Let A c A' be R-orders in A. If A is hereditary, then so is A'. Proof. Given any left A' -lattice L, there is an exact sequence ofleft A' -lattices

O--->M--->FY'...L--->O with F N -free. Since L is also a left A-lattice, hence A-projective, the sequence is A-split. Therefore there exists a map l/J E Hom" (L, F) such that cp l/J = 1 on L. But Exercise 40.1 shows that l/J is a N-homomorphism, whence the above sequence is N-split. Therefore L is N-projective, which completes the proof that N is hereditary. It should be pointed out that the preceding theorems and their proofs are valid without the hypothesis that A be a separable K-algebra. For the

368

HEREDITARY ORDERS

(40.5)

remainder of this section, however, we shall keep our hypothesis that A is separable. We intend to investigate R-orders A by considering their P-adic completions Ap , where P ranges over all maximal ideals of R. As a first step in this direction, we show (40.5) THEOREM. A is hereditary if and only if Ap is hereditary for all P, or if and only if Ap is an extremal Rp-order in Ap for all P.

equivalently,

Proof. Note first that since A is K-separable, Ap is Kp-separable for each P, by Exercise 7.13. We showed in (3.24) that A is hereditary if and only if each localization (R - P) -1 A is hereditary. The same argument works if we use completions rather than localizations. Alternatively, we saw in §3 that A is hereditary if and only if dim A ~ 1, where "dim" means "left global dimension". By (3.29) and (3.30),

dim A = sup {dim Ap}. p

This again shows that A is hereditary if and only if each Ap is hereditary. Finally, for each P we know by (39.25) that Ap is hereditary if and only if Ap is extremal. This completes the proof. Let us digress briefly to show that when R contains infinitely many prime ideals, then extremal R-orders are the same as maximal orders. (40.6) THEOREM. For each P, rad A c rad Ap. Further, (rad A)" c (rad R) A, where n = (A: K). If R contains il'!iinitely many prime ideals, then rad R = 0 and rad A = 0; in this case, A is extremal if and only if A is maximal. Proof. Given P, let

1jJ: A ->

By (6.10) we have ljJ(rad A)

c

A

= AIPA ;:; ApIPAp.

rad A. The chain

A > rad A > (rad A)2

> ...

steadily decreases to 0; its length is at most n, since A has dimension n over A)" = 0, and so (rad A)" c ker IjJ = P A. This implies that RIP. Therefore (rad

and therefore rad A as claimed.

c

(rad A)p

rad Ap ,

GLOBAL THEORY OF HEREDITARY ORDERS

(40.7)

369

n

Next we observe that radR = P, where P ranges over all maximal ideals of R. Furthermore PA = P} A = (rad R) A,

n p

{n

since this holds when A = R, and hence also when A is R-projective. Thence

nPA = (rad R) A. If R contains infinitely many prime ideals, then nP = 0 by §4, whence (rad A)" c

(rad A)" = O. But then K· rad A is a nilpotent two-sided ideal of A, and is therefore O. Thus rad A = 0, and the theorem is proved. We are now ready to prove that the decomposition theorem (10.5) holds equally well for hereditary orders. We have (40.7) THEOREM. Let A be a separable K-algebra with simple components {AJ, and let Ri be the integral closure of R in K i . Then (i) For each hereditary R-order A in A, we have A = Ae p where the {eJ are the central idempotents of A such that Ai = Aei . Further, for each i, Ae. is a hereditary R-order in A .. (ii) , If each Ai is a hereditary R -o~der in Ai' then Ai is a hereditary R-order in A. (iii) An R-order Ai in Ai is a hereditary R-order if and only if Ai is a hereditary Ri-order.

I'

r

Proof. As in the proof of (10.5) and (39.13), we need only show that A = CA, where C = R i . For each P, (CA)p is an Rp-order in Ap such that (CA)p>- Ap. Since Ap is extremal, it follows that (CA)p = Ap for all P.

I'

Therefore A = CA, and the rest of the argument is straightforward. We have now shown that, just as in the local case, the study of hereditary orders can be reduced to the situation where A is a central simple K-algebra. Let us now examine this case in some detail. Let A be a hereditary R-order in a central simple K-algebra A oflocal capacity K p , local index m p , at each prime P. This means that Ap ~ MKP(Sp) for some skew field Sp of index mp' Let rp denote the type of the hereditary Rp-order A p , that is, Ap/rad Ap is a direct sum of rp simple components. Call rp the local type of A at P; clearly rp = 1 if and only if Ap is a maximal order, by (39.23). Thus rp = 1 a.e., and we have previously shown that mp = 1 a.e. Whether or not rp = lor mp = 1, we know the structure of Ap from § 39, and we can recover the order A by using the formula Ap}. A = A Ii

{n p

If N is a bigger R-order in A, then Ap c (N)p for all P, with equality a.e.

370

(40.8)

HEREDITARY ORDERS

Conversely, suppose that for each P we are given an Rp-order X(P) in Ap, with X(P) ~ Ap for all P, and X(P) = Ap a.e. Then by (5.3),

X = A n [

nX(P)} p

is an R-order in A containing A, and Xp = X(P) for all P. Hence we may determine all possible orders A' containing A by determining all possible sets of X(P)'s. Our main result is as follows: (40.8) THEOREM. Let A be a hereditary R-order in the central simple K-algebra

A, and let S = {PI' . .. , Pili} be the (finite) set of prime ideals P of R at which Ap is not a maximal order. Let ri denote the type of the hereditary Rp,-order Api' 1 ~ i ~ m. m

(i) There are precisely

TI

ri distinct maximal R-orders in A containing A.

i=1 m

(ii) There are precisely

TI

i~

(2 r ,

-

1) distinct R-orders ill A containing A.

1

m

I

(iii) There are

ri orders in A which minimally containt A.

i= 1

Proof. All three assertions are immediate consequences of (39.23) and Exercise 39.7. 11 is of interest to list the orders mentioned in (i) and (iii). For convenience of notation, we set p¢s We may remark that A is a maximal R-order in A, where R is the ring of elements of K which are integral at each P ¢: S. For each Pi E S, let {~/ 1 ~ j ~ rJ be the distinct maximal orders in Api containing Api. Let

m=

{(al' ... ,am):a;EZ, 1 ~ a; ~ r; for 1 ~ i ~ m}.

For each m = (a j

, ••• ,

am) Em, define

p'=Anr:1, at n···nrm, am . Then 1

~

m

Hence as m ranges over the

TIr;

m-tuples in

i

~

m.

m, the R-order rm ranges over

I

-r-

The order A' minimally contains A if A' > A, but there is no order A" such that A' > A" > A.

GLOBAL THEORY OF HEREDITARY ORDERS

(40.9)

371

all maximal orders in A containing A, and different m's give different rm's. Now put Lis =

n

~i'

1 :( i :( m.

1 ~j:<:;ri.,

iFS

Set 0iS =

Ji. n Lis n {

n Ap},

1:( s :( r j , 1 :( i :( m.

PES PFP, m

Then the L ri distinct orders {Ois} are precisely the orders in A which 1

minimally contain A. To complete our discussion, we observe that every R-order in A containing A is expressible as an intersection of a non-empty set of P"s. This follows at once from Exercise 39.7, which asserts that for 1 :( i :( m, every order in A Pi containing Api is an intersection of some of the orders {~j: 1 :( j :( r i }· It should be pointed out, however, that different sets of rm's may yield the same R-order in A. For instance, suppose that S = {P 1 ,P 2 }, r 1 = r 2 = 2; then

rea, b)

-

-

A n r 1,a n r 2,b'

1 :( a, b :( 2.

We have rO,!) n r(1, 2) n r(2,

1)

= A = r(1, 1) n r(2, 2).

Analogously, for 1 :( i :( m, every order in A p , containing Api is a sum of orders from {LiS: 1 :( s :( rJ. It follows readily that every order A' properly containing A is expressible as a sum of orders chosen from the set {OiS: 1 :( s:( rp 1 :( i :( m}. We leave as an exercise for the reader the proof that each A' is uniquely so expressible. Finally, we caution that not every sum of O's is an order. By the same type of reasoning, we have (40.9) THEOREM. Keep the notation of(40.8). For each P, let

X(P) = A n rad Ap. Then (i) For each P, X(P) is a two-sided ideal in A such that

{X(P)}Q = {rad Ap, Q = P, AQ , Q 1= P, where Q ranges over all prime ideals of R.

372

HEREDTTARY

ORDERS

(40.10)

(ii) Let J(A) be the group of invertible two-sided A-ideals in A (see §37). Then J(A) is thefree abelian group with generators {X(P):P arbitrary}. (iii) There is an exact sequence 1 ---> Cl R ---> Picent A ---> TI Z/eprpZ ---> 1, p

wherefor each P, rp is the local type of A at P, and ep is the ramification index of the skewjield part of Ap' Proof Assertion (i) is clear from the discussion preceding (40.8), and implies (ii) by virtue of Exercise 39.6. Furthermore, (iii) follows from (37.29) and Exercise 39.6. Note that eprp = 1 a.e. Next we shall prove the global version of Exercise 39.11: (40.10) THEOREM. (Jacobinski [3]). Let {1l31"'" Il3J be any set of distinct prime ideals of the maximal R-order N in a central simple K-algebra A. Let (40.11)

113 1

..

'Il3 r =

Ls' ..

Ll

be any factorization of TI 1l3, into a proper product of integral ideals {L J Set s

(40.12)

A

n

= j= 1 OcCL). J

Then A is a hereditary R-order contained in N. Conversely, every hereditary order arises in this way. Proof Starting with Nand (40.11), let A be given by (40.12). All of the formulas behave properly when we pass to completions, and for each prime ideal P of R, we have _ {rad A~ if some 1l3, :::J P, TI (ll3)p - N otherwise. Hence for each P, Ap is hereditary by Exercise 39.11. Therefore A is hereditary by virtue of (40.5). Conversely, let A be a given hereditary order in A, and choose N :::J A, A' maximal. Suppose that Ap is maximal except at {P 1 " ' " PJ, and let 1l3, be the prime ideal of N containing P,. By Exercise 39.11, there is a factori~ zation (for each i) into a proper product of integral ideals, such that (40.12) holds locally at F;. We may assume that the number of factors s is independent of i, by inserting (if need be) additional factors L which are maximal orders. Now choose integral ideals L s , ... , L 1 in A by setting

GLOBAL THEORY OF HEREDITARY ORDERS

(40.13)

~

1

373

j ~ s.

It is then an easy exercise, left to the reader, to verify that (40.11) and (40.12) hold true.

Turning now from abstract properties of hereditary orders, we shall give a concrete example of a hereditary order which arises naturally in representation theory. Let LI K be a finite galois extension (of fields), with galois group G. Let R be a Dedekind domain with quotient field K, and let S be the integral closure of R in L. We shall assume that for each prime ideal P 1 of S, the residue class field SIP 1 is separable over R/(R n P J Call SIR tamely ram(fied if for each P l' the ramification index e(P l' LI K) is not zero in R/(R n PJ Then (see references listed in §4) SIR is tamely ramified if and only if there exists an So E S such that a(so) = 1.

L

aEG

L' LUa be the trivial crossed-product algebra given in Exercise 29.14. We set A = L' SUa' an R-order in the central simple Now let A

= (LI K, 1) =

GEG

K-algebra A. Sometimes A is called a twisted group ring, and is denoted by So G.

(40.13) THEOREM (Rosen). The twisted group ring So G is a hereditary R-order if and only if SIR is tamely ram(fied. Proof Let A = So G; we may regard S as embedded in A via s -> s· up S E S, and then every left A-module is also a left S-module. Suppose that SIR is tamely ramified, and let So E S be such that a(so) = 1. Given any

L

GEG

left ideal X of A, there is an exact sequence of left A-lattices

o -> X'

->

F !4 X

->

0,

with F A-free. Since X is also a left S-lattice, and hence S-projective, the sequence is S-split. Thus there exists a map t/! E Horns (X, F) such that cpt/! = 1 on X. Now set t/!' = L ua ' sot/!· u; 1. (TEG

t/!' E Hom" (X, F), = L cpua ' sot/!· u; lX =

It is easily verified that C(J1jJ'(X)

a

and for x

E

X we have

L uaSO . cpt/!. u; 1X a

= LU,hu;lX = U::a-1(so)}x = X. a

a

Thus the sequence is A-split, whence X is A-projective. This proves that if SIR is tamely ramified, then A is hereditary.

374

HEREDIT AR Y ORDERS

(40.14 )

I

Conversely, suppose that A is hereditary, and set x =

u" EA. Then

"EG

the epimorphism Ax, given by
(I

Therefore

I

(I

(1

"

a(a 1) = 1, whence SIR is tamely ramified. This completes the

" proof. The reader should compare this with the proofs of (7.20) and Exercise 7.9.

Keeping the above terminology and notation, let us prove (40.14) THEOREM (Auslander-Goldman-Rim). The twisted group ring So G isa maximal R-order if and only if SIR is unram(fied, that is, if and only if the discriminant ideal d(SIR) equals R. Proof It suffices to establish the result for the case where R is a discrete valuation ring, with maximal ideal P. Let A = So G, A = LoG = Lu" as above, and set (L: K) = n, so A ~ M n(K). Suppose that A is a maximal order in A; it follows from (17.5) and (18.3) that

I"

rad A = PA =

I" PSu", (J

and that rad A is the unique maximal two-sided ideal of A If d(SIR) 1= R, then wemay write PS = Pie" where the {PJ are distinct prime ideals of S, and where each ei > 1. Setting a = ITPi,it is clear that

IT

radA <

I" au" < A, "

I"

and that au" is a two-sided ideal of A. This gives a contradiction, and shows that SIR must be unramified whenever A is a maximal order. Suppose conversely that SIR is unramified, so the inverse different n-1(SIR) equals S by (4.37). Let r be any R-order in A containing A, and let

y=

I

a"u" E r,

au E L.

" Fix rEG, and let s range over all elements of S. Then tr A/K SYU,-l E R for all s E S.

SYU,-l

E r, whence

(40.15)

375

GLOBAL THEORY OF HEREDITARY ORDERS

By Exercise 40.4, this gives TL/K(sa,)

E

R for all

SE

S.

Therefore a, E n- (S/R) = S for each rEG, whence the proof of the theorem. 1

r =

A. This completes

Also of interest is the following generalization of (40.13): (40.15) THEOREM (Williamson, Harada). Keep the notation and hypotheses introduced before (40.13), and let f: G x G B

=

(L/K,f)

=

-->

u(S) be a factor set. Let

I'Lv a ,
so r is an R-order in the central simple K-algebra B. Then order if and only if S/ R is tamely ram(fied.

J

Proof See Williamson [1 and Harada

r

is a hereditary

[7].

The remainder of this section gives Reiner's simplified version [2J of some results of lacobinski [4]. First of all, we show that when A is hereditary, the calculation of the locally free class group CI A (defined in §38) can be reduced to the case of maximal orders. (40.16) THEOREM. Let A be a hereditary R-order in a separable K-algebra A, and let A' be any R-order in A containing A. Then there is an isomorphism

f3 : CI A given by

~

CI 1\',

f3[XJ = [I\' ®AXJ, [XJ ECIA.

Proof Denote ® A by ® throughout this proof. It is easily checked that the

stable isomorphism class of I\' ® X depends only on that of X, and that (I\' ® X) v I\' whenever X v A. Thus f3 is well defined, and it follows readily from the definition of the additive structure of the class groups that f3 is a homomorphism. We now show that f3 is epic. Given any [X'J E CI 1\', where X' is a locally free left I\'-lattice in A, we may write (for each P)

with xp = 1 a.e. Set

Then X is a locally free left A-lattice in A such that I\' X = X'. Since I\' ® X I\' X, it follows that f3[ XJ = [X'J, whence f3 is epic,

~

376

HEREDITARY ORDERS

(40.17)

Now let f/ = {PI' ... , Pn } be the set of all P's in R such that Ap =1= A'pGiven any [X] E Cl A, we imitate the proof of (37.34). By (27.1) there exists a A-exact sequence (40.17)

0

-->

X

-->

A

-->

T

-->

0,

where Tp = 0 for all P E f/. Since A' is projective as right A-module, applying A' ®. to (40.17) yields a A' -exact sequence

o --> A' ® X

(40.18)

-->

A'

-->

A' ® T

-->

o.

Now we claim that A' ® T ~ T as left A-modules. Indeed, both are R-torsion modules, and for each P we have A~ ® Tp = Tp. (This is clear when P r$ f/, while for P Thus (40.18) yields a A-exact sequence

o --> A' ® X

(40.19)

-->

E

A'

f/ it holds since then -->

T

-->

A~

= Ap.)

o.

Comparing this with (40.17) and using Schanuel's Lemma, we obtain (40.20)

X -+- A'

= A -+- A' ® X

as left A-modules.

Suppose now that [X] E ker {3; then A' ® X is stably isomorphic to A', so there exists a free left A'-lattice F' with A' -+- F' ~ A' ® X Then (40.20) gives

-i-

X -+- (A' -+- F') ~ A -+- (A'

F'.

-i-

F').

But A' -i- F' is a projective A-lattice, whence A' -+- F' -+- L is A-free for some A-lattice L. Adding L to both sides of the preceding equation, we deduce that X is stably isomorphic to A as A-lattices. Thus [X] = 0 in CI A, which shows that {3is monic, and completes the proof of the theorem. The preceding result shows, in particular, that CI A ~ CI A' for every maximal order A' containing A Keeping the notation of (40.7), let us write A' = L' A;, where A; is a maximal Ri-order in Ai. Then CI A ~ CI A' ~

L' CI A;.

If K is an algebraic number field, or if K is a function field and Ai = Eichler/Ri for each i, then by (35.14) we have

Cl A; ~ CI Ali

for each i.

Thus, in these cases the class group CI A is known explicitly as a ray class group of the center of A Generalizing Exercise 21.1, we prove next

(40.21)

GLOBAL THEORY OF HEREDITARY ORDERS

377

(40.21) THEOREM. Let M be any nonzero left A-lattice, where A is a hereditary R-order. Then End AM is a hereditary R-order in End A KM.

= A(n) for some n, and let r = End AA(n). Then r is Morita equivalent to A, whence r is also hereditary by Exercise 15.4. Let e:A(n) ---> M be the projection associated with the direct sum decomposition A(n) = M -+- M'. Then Proof Choose a left A-lattice M' such that M -+- M'

End A M

~

ere,

and so End AM is hereditary by Exercise 40.3. U sing this, we now show (40.22) THEOREM. Let A be a hereditary R-order in the separable K-algebra A, where K is a global field. Let M, N be a pair of left A-lattices such that 1\1 v N and A'M ~ A'N for some order A' containing A If the K-algebra EndAKM satisfies the Eichler condition relative to R (see (38.1)), then M ~ N.

Proof Let us set

r = EndAM,

r' = EndA' A'M,

B = EndAKM.

Then by (40.21), r is a hereditary R-order in the separable K-algebra B. Replacing N by uN for some nonzero (l E R, we may assume that N c M. View M as a bimodule AMr, and set I

= Hom A(M, N).

Then I is a left ideal in r; we claim that Iv r and that N = MI. F or each P, there is an isomorphism M p ~ N p' so we may write Therefore Ip = HomAp(M p , N p) = rpup .

This shows that Iv r. It also establishes that N = MI, since the equality holds locally at all P. Therefore [I] is an element of the locally free class group Clr. Now we have r'I = HomA'(A'M, A' N), since the equality holds locally at each P. On the other hand, A'N = A'M' y for some y E u(B), since A'M ~ A'N by hypothesis. Therefore r'I = r'y, so that [I] maps onto zero under the homomorphism CI r ---> CI r' defined

378

HER EDIT AR Y ORDERS

(40.22)

as in (40.16). Hence by (40.16) it follows that [J] = [rJ. Therefore J ~ r by (38.4), so J = rz for some z E u(B). Hence N = M J = M z ~ M, which completes the proof of the theorem. EXERCISES 1. Let A c II.' be R-orders in a K-algebra A, and let X, Y be left N-modules such that Y is R-torsionfree. Prove that Hom". (X, Y)

=

Hom" (X, Y).

[Hint: The left hand expression is clearly included in the right hand expression. Now let f E Hom" (X, Y). Given A' E 11.', choose a nonzero r E R such that r,1.' EA. Then r ·f(,1.'x) = f(d'· x) = r,1.' f(x),

whence f(A'x)

=

XEX,

,1.'f(x) for all A', x since Y is R-torsionfree.]

2. Let A be an R-order, M and N left A-modules. (i) Let M* = Hom R (M, R), the dual of M. Show that M* is a right A-module. Prove that each f E Hom" (M, N) induces an f* E Hom" (N*, M*). (ii) Show that analogous results hold when "left" and "right" are interchanged. (iii) Let CPM: M -+ M** be the evaluation map defined by

f EM*,

mEM.

Show that CPM is a left A-homomorphism. (iv) ShowthatifM is a A-lattice, thensoareM* andM**. ProvethatCPM:M ~ M**, and show that the isomorphism M ~ M** is natural, that is, for each homomorphism f:M -+ N of left A-lattices, there is a commutative diagram M~N

~M 1

~N 1

M**~N**,

wherefinducesf**. [Hint: Use (4.24).J 3. Let e E r be idempotent, where r is a hereditary R-order. Show that ere is also a hereditary R-order. [Hint: Let L be any left ideal of ere; there exists a left ereepimorphism r

cp(al" .. ,a) =

I. a;li

(say),

i= 1

ai E ere.

Extend cP to a left r -epimorphism

i

where 1/1': r L

rL

:t l'Ji' 1

is a left ideal in r. Since r is hereditary, there exists a left r-map -+ r(r) splitting cp'. Let

379

GROUP RINGS

(41.1) if/(y)

(if; 1 (y), ... , if;,(y)),

=

Then each if;iEHomr(rL,r), and if;,(x) if;(x)

YErL.

eif;i(x), XEL. Define !{t:L

(l/Jl(x)e, ... ,I/J,(x)e),

=

(ere)(r) by

XEL.

Then I/J is an ere-homomorphism, and cpl/J(x) =

i: ifi;Cx)el = i: I/J,(x)li = cp'Ij/(x) = x, i

i=1

x E L.

i=l

Thus L is ere-projective.] 4. Let A = (L/K, 1) =

L' Lu

u

be a trivial crossed-product algebra with u 1 = 1.

I7EG

Prove that for each a ELand each tr

AIK

(J

au

E G, =

{TLIK a,

u

0,

41. GROUP RINGS We shall collect here some miscellaneous results on integral group rings, concerned mostly with their relations with maximal orders. Throughout this section, let G denote a finite group of order n, and let A = KG be the group algebra of Gover K. We assume once and for all that char K In, so A is a separable K-algebra by Exercise 7.8. The integral group ring RG is the R-order in A consisting of all formal sums I rtx x, where the coefficients xeG

lie in R. We set A = RG throughout. Since the study of integral representations of G reduces to the investigation of A-modules, it is desirable to know as much as possible about the R-order A

{rt x }

(41.1) THEOREM. Let r be any R-order in A containing A Then

Furthermore,

A is maximal

=-

A is hereditary

=-

n E u(R).

Proof: Let T = TAlK be the ordinary trace map. For x E G, let x1:a -+ xa, a E A, be left multiplication by x. Relative to the K-basis of A consisting of the elements of G, XI is represented by an n x n permutation matrix. If x -!= 1, then all of the main diagonal entries of this matrix are 0. This gives (41.2)

TAlK

X =

n, { 0,

x = 1, xEG,Xi=1.

380

(41.3)

HEREDITARY ORDERS

Now let y = I axx E r, where each ax E K. Since r contains all elements of G, we have yy-l E r for all y E G. But T(yy-1) E R by Exercise 1.1, and thus T(yy-l)

= Ia x

T(xy-l)

= nayER.

x

Therefore a y En- 1 R for all YEG, whence YEn- 1 ·RG. This proves that r c n - 1 A, as claimed. If n E u(R), the above shows that every order r containing A must coincide with A; thus A is maximal, and also hereditary. Conversely, suppose A is hereditary. Then by (40.7), A contains the central idempotent n- 1 • I x xeG

of A. Therefore n - 1 E R, whence n E u(R). This completes the proof of the theorem. For the rest of this section, let r denote a maximal order in A containing A. We have just shown that nr c A. Jacobinski's refinement of this result, to be given below, is based on using reduced traces rather than ordinary traces. Departing slightly from his terminology, we set (r:A)1 = {x E A:xr

c

A} = left conductor of r into A,

(r:A)r = {x EA:rx c A} = right conductor ofr into A. We shall use the following notation hereafter: t

A=

I' Ai

(simple components).

i= 1

Kj

= center of Aj' (A KJ = j :

nt,

R j = integral closure of R in K j

•

tr i = tr A,IK = reduced trace from Aj to K.

1\ =

n(rJR) = different of r i with respect to R.

Let us recall from §25 that the inverse different ideal of rj with respect to tr i , that is,

n

j-

1

n

i

1

is the complementary

= {x EO Ai : trj(xrJ c R}.

We note that the right conductor (r: A), is a (r, A)-bimodule, and is the largest left r-submodule of A. (41.3) Theorem (Jacobinski [lJ). We have t

(r :A), = (r: A)r =

I" (n/n)n i= 1

j-

1.

381

GROUP RINGS

(41.4)

--> K be the bilinear trace form defined by T(X, y) = T(xy), x, YEA, where T = TAlK as above. Let G = {x!' ... , xnJ, Xl = 1, so that {X!' ... , xn} is a K-basis for A. From (41.2) we obtain

Proof. Let T: A x A

T(X i , n- 1xj1) = b;j

(41.4)

(Kronecker delta),

1:( i, j:( n.

Thus {n-1x~ 1, ... , n- 1X;l} is a dual basis to {Xl' ... ' X.} relative to T, and so the form T is nondegenerate. Every full R-Iattice M in A determines a dual lattice defined by M = {aEA:r(a,M) c R}.

M (see Exercise 4.12),

As shown in Exercise 4.12, the correspondence M --> M is one-to-one inclusion-reversing, and .& = M. If M is a right r-Iattice in A, then M is a left r-lattice in A. If M is R-free with basis {mJ, then if = Rm> where r(m i , mi.) = b .. , 1 :( i,j :( n. From (41.4) we conclude at once that

I'

'J

J

- -A

(41.5)

n

"\"' '-' R n - 1 x.-

j~ 1

1 _

- n

lA .

-

J

Now take M = (r:Alr, the large:t left r-module in A. Then smallest right r-module containing A, that is,

M is

the

M = Ar = n-1A· r = n-1r. Theref ore M = 1~1 = nf, where

f = {x E A: T(xr) c Each

X E

A may be written as

X

=

I

Xi' Xi E

R}.

Ai. By (9.22) we know that

t

T(x)

=

I

i= 1

n i tfiX i ·

This gives at once f = {x E A: ni tr; (xrJ c R =

t "\"'

'-'

i= 1

-l'l'\- 1

ni

Vi

for

1:( i :(

t}

.

Therefore (r: A), = nf =

I' (n/nJDi-l,

as desired. The same argument works equally well for (r: A)/, and the theorem is proved. From the preceding we obtain nr c nf = (r:A)r c A,

382

HER EDIT AR Y ORDERS

(41.6)

so we recover part of(41.1). From the proof 0[(41.3), we may also show directly that if A is hereditary, then A is maximal. If A is hereditary, then r is projective as right A-lattice. Hence by (16.7) we obtain r ® AHomA (r, A)

~

End A r.

After making some obvious identifications, we may rewrite this as r· (r:A)/ = r. But (r:A)/ is a left r-Iattice, since it equals (r:A)r by (41.3). Therefore (r:A)/ = r, which implies that r = A. Turning next to questions of ramification, let us first generalize the definitions given in §§4,32. (41.6) Definition. Let B be a separable K-algebra, and let P be any prime of K. We say that B is unramified at P if the P-adic completion Bp is expressible as Bp ~

I" Mn/E),

where for eachj, E j is a field which is unramified over the P-adic field Kp. (When P is an infinite prime, this latter condition is taken to mean that E j = Kp.) (41.7) THEOREM. Let P be a maximal ideal of R, where R = alg. int. {K} and K is an algebraic number field. Then KG is unramified at P whenever pt nR. Proof Replacing R by its localization (R - P) - 1 R and changing notation, we may assume that R is the P-adic valuation ring in K, and that n E u(R). Keep the notation introduced before (41.3). By Exercise 41.4, it suffices to show that each A; is unramified at P. Keep i fixed; since n; In by Exercise 41.1, we have n/n; E u(R). But n/n; E 1.); by (41.3), whence 1.); = ~. Therefore r; = 1.)(rjRJ . 1.)(Rj R) by Exercise 25.1, that is,

(41.8) Let p range over the primes of R; dividing P. By (25.7) and the first equality in (41.8), we may conclude that Ai has local index 1 at each p. Therefore A; is unramified at each p. On the other hand, from (4.37) and the second equality in (41.8), it follows that each pis unramified in the extension KjK. Hence by §5c, each (K)p is unramified over Kp. Therefore A; is unramified at P, by Exercise 41.6, and the theorem is established. The converse of(41.7) is false, as we shall see below. It may well happen that A is unramified at a prime P dividing n. We shall find examples where this occurs by using the next theorem, which is of importance in its own right.

(41.9)

383

EXERCISES

(41.9) Theorem (Schilling). Let G be a group of order p', where p is prime, and let K be a field of characteristic O. (i) Ifp is odd, then each simple component of KG is afull matrix algebra over a field. (ii) If p = 2, then each simple component of KG is of the form Mk(D), where D is either a field or a skewfleld J( index 2. Proof First let Mk(D) be a simple component of QG, where D is a skew field with center L. Let w be a primitive pr-th root of 1. By Curtis-Reiner [1, (70.8)J, L is a subfield ofQ(w). Since p is completely ramified in the extension Q(w)/Q, there is a unique prime Po of L dividing p (see references listed in §4). Hence by (41.7), Po is the only finite prime of L at which Mk(D) can possibly ramify. Thus D has local Hasse invariant 0 at every finite prime of L except possibly at Po' Suppose now that p is odd. Then D has odd index, since this index divides pr by Exercise 41.1. Therefore no infinite prime of L ramifies in D by Exercise 32.2. It follows from Exercise 32.1 that D = L, which proves (i) for the case K =Q. Now let p = 2, and let D have index m. Some real primes of L may ramify in D; at each such prime, the local Hasse invariant of Dis 1/2. But by (32.13a), the sum of all of the local invariants of D is O. This shows that m = 1 or 2, and establishes (ii) when K = Q. Now let K be arbitrary, and let {BJ range over the simple components of QG. As in the proof of Exercise 41.1, we have KG ~ Fij ®K, B i •

I'

l, j

where K t is the center of Be We have just shown that Bi is of the form M k(D), where D is a skewfield of index 1 or 2. Then each Fij ® K, B; is also of the form Ms(D'), where D' is a skewfield of index I or 2. This completes the proof. This theorem was rediscovered, with different proofs, by Witt and Roquette; see also Feit [1, (14.5)]' EXERCISES In Exercises 1-3, we keep the notation of (41.3) and its proof. 1. Show that ni I n when char K = O. [Hint: First take K = Q, R = Z in (41.3). For each i, we have (n/n,)TI i- 1 c r" and so n/n, E TIi c r i . Hence n/ni is integral over Z, whence ni I n. Now let E be any field of characteristic O. Then E ::::J Q, and EG ~ E Q9 Q QG ~

Fix i, and let E Q9 Q Ki ~

t

L' (E Q9 i=l

Q

K;l Q9K, Ai'

L' Fij be a direct sum of fields

{FiJ. Then

384

HEREDITARY ORDERS

(E ®Q K;l ®Ki Ai ~

L' Fij ®Ki Ai' J

and each summand Fij ® Ai is a simple algebra of dimension

nt over its center Fij.J

2. Let K be an algebraic number field. Prove that for each i, Ri n (n/n;l !f 1 = (n/n;l !l-l(RJR).

Deduce that

n (n/n;l(K n !l-l(RJR)). t

{x€OR:xr c:: A}

=

i= 1

[Hint (Jacobinski [1 J): By Exercise 25.1, !li =

~nB

~ =

where

!l(RJR).

For x €ORi' x€O(n/n;l21-1~-1

<0>

x~ c::

(n/n;l21- 1

<0>

x~ c::

(n/n;l(Ki n21- 1 ).

Write Ki n 21- 1 = (1- \ where (I is an ideal of Ri • If (I oF R" choose a prime ideal P of Ri dividing (I, and let \l3 be the prime ideal of r i corresponding to P. If pr i = \l3 e, then the power of \l3 in 21 is precisely \l3e - 1. But then

P

=>

(I

=>

p- 1 c::

(1-1

=>

p- 1 c:: 21- 1

=>

pri l21,

1

a contradiction. Hence Ki n 21- = Ri, so for x €O Ri we have x€O(n/nJ!li- 1 <0> x~ c:: (n/nJR i <0> x€O(n/nJ~-l. But (n/n) ~ -1 c:: R" since (n/n;l ~ -1 c:: (n/n;l !li- 1 c:: A, so each element of(n/n;l ~-1 is integral over R. This proves the first assertion, and the second is an easy consequence ofthe first.J 3. Let e be a central idempotent in A, and let M be a left A-lattice such that eM = M. Let fe =

n(n/n;l (K n !l-l(RJR)), i

where i ranges over all indices such that ei occurs in e. Prove that fe . Ext~ (M, X) = 0,

fe . Ext~ (X, M) = 0

for all left A-lattices X. [Hint (Jacobinski [1 J): First reduce the problem to the case where eX = X. Changing notation, we may then assume that e = 1. Each r €Ofe is an element of R such that rr c:: i\. Using the fact that r is hereditary, deduce that r annihilates Ext~ (M, .) and Ext~ (., M).J (For generalizations of the above, see T. V. Fossum [1, 2J, Roggenkamp, HuberDyson [1J.) 4. Let B = L' Bi (simple components) be a separable K-algebra, and let P be a prime of K. Show that B is unramified at P if and only if each Bi is unramified at P. 5. Let B be a central simple L-algebra, and let p be a prime of L. Show that B is unramified at p if and only if Bp is a full matrix algebra over Lp, that is, if and only if the local index of Bat p equals 1.

EXERCISES

385

6. Let B be a central simple L-algebra, and let K be a subfield of L such that (L: K) is finite. Let P be a prime of K, and let p range over the primes of L dividing P. Prove that (i) Bp ~

I' B

p,

where Bp = Lp ® L B is a central simple Lp-algebra.

p

(ii) Let P be a finite prime. Then B is unramified at P if and only if for each p, Lp is unramified over Kp and Bp is a full matrix algebra over Lp. (iii) Let P be an infinite prime. Then B is unramified at P if and only if for each p, Lp = Kp and Bp is a full matrix algebra over Lp. [Hint: Write

where p ranges over the distinct primes of L dividing P.]

References ALBERT, A. A.

l. "Structure of Algebras." Amer. Math. Soc., New York, 1939; revised, 1961. S. 1. On central division algebras. Israel J. Math. 12 (1972), 408-420. ARTlN, E. 1. "Questions de Base Minimale dans la Theorie des Nombres Algebriques."

AMITSUR,

Centre Nat. Rech. Sci. XXIV, Colloq. Int., Paris (1950), 19-20. Reprinted in Collected Papers, ed. S. Lang and J. Tate, Addison-Wesley 1965, pp. 229-231. ARTlN, E., NESBITT, C. J., THRALL, R. M. I. "Rings with Minimum Condition." Univ. of Michigan Press, Ann Arbor, 1948. ARTlN, E., TATE, J. T. l. "Class Field Theory." Benjamin. New York, 1967. ASANO,

K.

1. Arithmetische Idealtheorie in nichtkommutativen Ringen. Japan. J. Math. 16 (1936), 1-36. 2. Arithmetik in Schiefringen I. Osaka Math. J. 1 (1949). 98-134. AUSLANDER, M., GOLDMAN, O.

1. Maximal orders. Trans. Amer. Math. Soc. 97 (1960),1-24. 2. The Brauer group of a commutative ring. ibid, 367-409. AZUMAYA,G.

1. On maximally central algebras. Nagoya Math. J. 2 (1951),119-150. BABAKHANIAN, A.

1. "Cohomological Methods in Group Theory." Dekker, New York, 1972. H. 1. "Algebraic K-theory," Math. Lecture Notes. Benjamin, New York, 1968. BERGER, T. R., REINER, I. 1. The normal basis theorem. Amer. Math. Monthly (1975), to appear.

BASS,

BOURBAKI, N.

1. "Algebre: Modules et Anneaux Semi-simples" (Ch. 8). Hermann, Paris, 1958. 2. "Algebre Commutative: Modules Plats, Localisation" (Ch. 1, 2). Hermann, Paris, 1961. 3. "Algebre: "Graduations, ... " (Ch. 3,4). Hermann, Paris, 1961. 4. "Algebre: Entiers, valuations" (Ch. 5, 6). Hermann, Paris. 1964. 5. "Algebre: Diviseurs" (Ch. 7). Hermann, Paris. 1965. BRUMER, A.

1. "Structure of Hereditary Orders." Thesis, Princeton Univ. 1963; Bull. Amer. Math. Soc. 69 (1963), 721-724; addendum, ibid 70 (1964), 185. CARTAN, H., ElLENBERG, S. 1. "Homological Algebra." Princeton Univ. Press, Princeton, 1956.

387

388

REFERENCES

CASSELS, 1. W. S., FROHLICH, A. 1. "Algebraic Number Theory." Academic Press, London and Thompson Publ. Co., Washington, D.C., 1967. COHN,H. 1. "A Second Course in Number Theory." Wiley, New York, 1962. COHN, P. M. 1. "Morita Equivalence and Duality." Queen Mary CoIl. Math. Notes, 1966. COXETER, H. S. M. 1. The binary polyhedral groups and other generalizations of the quaternion group. Duke Math. J. 7 (1940), 367-379. COXETER, H. S. M., MOSER, W. O. 1. 1. "Generators and Relations for Discrete Groups." Springer, Berlin, 1957. CURTIS, C. W., REINER, 1. 1. "Representation Theory of Finite Groups and Associative Algebras." Wiley (Interscience), New York, 1962. DEURING, M. 1. "Algebren." Springer, Berlin, 1935 (revised, 1968). EICHLER, M. J. Uber die Idealklassenzahl total de fin iter Quaternionalgebren. Math. Zeit. 43 (1938), 102-109. 2. Uber die Idealklassenzahl hyperkomplexer Zahlen. Math. Zeit. 43 (1938), 481-494. FADDEEV, D. K. 1. An introduction to the multiplicative theory of integral representations. Proc. Steklov Inst. Math. 80 (1965), A.M.S. translation (1968). FAITH, C. 1. "Algebra: Rings, modules and categories 1." Springer, Berlin-·New York, 1973. FOSSUM, R. I. The Noetherian different of projective orders. J. reine anqew. Math. 224 (1966), 207-218. 2. Maximal orders over Krull domains. J. Algebra 10 (1968), 321-332. 3. Injective modules over Krull orders. Math. Scand. 28 (1971). 233-246. 4. The divisor class group ofa Krull domain. Ergeb. Math., Springer, Berlin, 1973. FOSSUM, T. Y. I. On symmetric orders and separable algebras. Trans. Amer. Math. Soc. (1975), to appear. FROHLICH, A. 1. ThePicardgroupofnon-commutativerings,inparticularoforders. Trans. Amer. Math. Soc. 180 (1973),1-46. 2. Locally free modules over arithmetic orders. J. reine angew. Math. (1975), to appear. FROHLICH, A., REINER, 1., ULLOM, S. J. Class groups and Picard groups of orders. Proc. London Math. Soc. 29 (1974), 405-434. GRUENBERG, K. W. 1. "Cohomological Topics in Group Theory," Springer Lecture notes No. 143. Springer, Berlin, 1970. HARADA, M. 1. Hereditary orders. Trans. Amer. Math. Soc. 107 (1963), 273-290. 2. Structure of hereditary orders. J. Math. Osaka City Univ. 14 (1963), 1-22.

REFERENCES

389

3. Hereditary orders in generalized quaternions. ibid, 71-81. 4. MUltiplicative ideal theory in hereditary orders. ibid, 83-106. 5. Hereditary orders which are dual. ibid, 107-115. 6. On generalization of Asano's maximal orders ina ring. Osaka J. Math. 1 (1964), 61-68.

7. Some criteria for hereditarity of crossed products. ibid, 69-80. HASSE, H.

1. Uber p-adische Schie!1corper und ihre Bedeutung fUr die Arithmetik hyperkomplexer Zahlsysteme. Math. Ann. 104 (1931), 495-534. HERSTElN, 1. N. I. "Noncommutative Rings," Carus Monograph 15. Math. Assoc. America, 1968. HILTON, P. J., STAMMBACH, U. I. "A Course in Homological Algebra?' Springer, Berlin-New York, 1971. JACOBSON, N.

1. "The Theory of Rings." Amer. Math. Soc., New York, 1943. 2. "Structure of Rings." Amer. Math. Soc., New York, 1956. JACOBINSKI, H. 1. On extensions of lattices. Michigan Math. J. 13 (1966), 471-475. 2. Uber die Geschlechter von Gittern iiber Ordnungen. 1. reine angew. Math. 230 (1968), 29-39. 3. Genera and decomposition of lattices over orders. Acta Math. 121 (1968), 1-29. 4. Two remarks about hereditary orders. Proc. Amer. Math. Soc. 28 (1971), 1-8. JANS, J. P. 1. "Rings and Homology". Holt Rinehart Winston, New York, 1964. JANusz, G. J. 1. "Algebraic Number Theory." Academic Press, New York, 1973. 1. 1. "Commutative Rings." Allyn and Bacon, Boston, 1970.

KAPLANSKY,

.

KNEBUSCH, M.

1. Elementarteilertheorie tiber Maximalordnungen. J. reine angew. Math. 226 (1967),175-183. MACLANE, S. 1. "Homology". Springer, Berlin-New York, 1963. MATSUMURA, H.

1. "Commutative Algebra." Benjamin, New York, 1970. G. O. 1. Asano orders. Proc. London Math. Soc. (3) 19 (1969), 421-443. NEUKIRCH, J. 1. "Klassenkorpertheorie." Bibliog. Inst., Mannheim, 1969. 2. Eine Bemerkung zum Existenzsatz von Grunwald-Hasse-Wang. J. reine angew. Math. 268/269 (1974),315-317. NORTHCOTT, D. G. 1. "An Introduction to Homological Algebra." Cambridge Univ Press, Cambridge, 1960. O'MEARA, O. T. 1. "Introduction to Quadratic Forms." Springer, Berlin-New York, 1963. REINER, 1. 1. A survey of integral representation theory. Bull. Amer. Math. Soc. 76 (1970), MICHLER,

159-227.

390

REFERENCES

2. Hereditary orders. Rend. Sem. Mat. Univ. Padova (1975), to appear. P. 1. L'arithmetique des Corps," Hermann, Paris, 1972. ROBSON, 1. C. l. Non-communtative Dedekind rings. J. Algebra 9 (1968), 249-265. 2. Idealizers and hereditary noetherian prime rings. J. Algebra 22 (1972), 45-81. ROGGENKAMP, K. W. I. "Lattices over Orders II," Springer Lecture notes No. 142. Springer, Berlin, 1970. 2. Projective homomorphisms and extensions oflattices. J. reine angew. Math. 246 (1971),41-45. ROGGENKAMP, K. W., HUBER-DYSON, V. I. "Lattices over Orders I," Springer Lecture notes No. 115. Springer, Berlin, 1970. ROSENBERG, A., ZELINSKY, D. I. Automorphisms of separable algebras. Pacific J. Math. 11 (1961). 1109-1117. ROTMAN, 1. 1. 1. "Notes on Homological Algebra." van Nostrand Reinhold, New York, 1970. SAMUEL, P. 1. "Theorie Algebrique des Nombres." Hermann, Paris, 1967. SCHACHER, M. M., SMALL, L. 1. Noncrossed products in characteristic p. J. Algebra 24 (1973), 100-103. SCHILLING, O. F. G. 1. "The theory of valuations." Amer. Math. Soc., New York, 1950. SERRE, 1.-P. 1. "Corps Locaux." Hermann, Paris, 1962. SWAN, R. G. 1. Projective modules over group rings and maximal orders. Ann. of Math. (2) 76 (1962),55-61. 2. "Algebraic K-theory," Springer Lecture notes No. 76. Springer, Berlin, 1968. SWAN, R. G., EVANS, E. G. 1. "K-theory of Finite Groups and Orders," Springer Lecture notes 149. Springer, Berlin, 1':170. RIBENBOIM,

WANG, S.

1. On Grunwald's theorem. Ann. of Math. (2) 51 (1950), 471-484. A. 1. "Basic Number Theory." Springer, Berlin-New York, 1967. WEISS, E. 1. "Algebraic Number Theory." McGraw Hill, New York, 1963. 2. "Cohomology of Groups." Academic Press, New York, 1969.

WElL,

WILLIAMSON, S.

1. Crossed products and hereditary orders. Nagoya Math. J. 23 (1963), 103-120. M. J. I. Reduced norms in the K-theory of orders. J. Algebra (1976), to appear. ZARISKI, 0., SAMUEL, P. 1. "Commutative algebra," Vol. 1. van Nostrand, Princeton, 1958.

WILSON, S.

Subject Index Chinese remainder theorem, 47 class group, 306, 308, 343 cohomology group, 242, 257 cok (cokernel), 8 commutative diagram, 9 complementary ideal, 60, 217,223 complementary lattice, 60 complete (in N-adic topology), 85 complete discrete valuation ring, 135 complete field, 67 completely primary, 82 complex prime, 64 completely ramified, 73 conductor, 366, 380 conjugate, 146, 285 conjugate orders, 232 contravariant (functor), 9, 14 counting norm, 59, 216 covariant (functor), 9, 14 cyclic algebra, 259 cyclic module, 50 crossed-product algebra, 242

A

A.C.C., 13 additive functor 14 a.e. (almost everywhere) algebra Azumaya, 338 central separable, 338 central simple, 92 over a commutative ring, 1 separable, 99, 106 algebraic number field, 46 alg. int., 46 annihilator, 44, 77 archimedean valuation, 52, 71 artinian, 13 Asano order, 207 associative bilinear form, 116 augmentation map, 258 Autcent, 328 Azumaya algebra, 338

B

D

bimodule, 9 bimodule over R, 319 Brandt groupoid, 201 Brauer group, 238

c capacity, 179, 213, 222 centralizer, 95 central simple, 92 chain in a vector space, 354 chain ring, 354 Change of Rings Theorem, 26 characteristic polynomial, 2 characteristic polynomial (reduced), 113, . 119,121

D.C.C., 13 Dedekind domain, 45 deleted projective resolution, 20 dense, 68 decomposition group, 274 derived Morita context, 162 different, 60, 150, 184, 217 dimension shifting, 23 direct product of rings, 90 direct sum of rings, 90 discrete valuation, 51 discrete valuation ring, 52, 135 discriminant (ideal), 61, 66, 126, 218 discriminant (of element), 67 division ring, 91

391

392

SUBJECT INDEX

double centralizer property, 93 dual of lattice, 367, 381 E

Eichler condition, 294, 343 Eichler/R, 294, 343 Eichler's Theorem, 297, 298 Eisenstein extension, 74 Eisenstein polynomial, 73 enveloping algebra, 101 epimorphism (of rings), 77 equivalence of categories, 155 of embeddings, 74 of extensions, 19 of factor sets, 243 of valuations, 51 evaluation map, 55, 378 exact (right, left), 14 exact functor, 14 exact sequence, 8 exp[A],253 exponential valuation, 53 extension of modules, 19 extremal order, 356 extremal subring, 356

F factor set, 242 faithful functor, 155 faithful module, 93 faithfully fiat, 17 faithfully projective, 158 fibre product, 11 finite prime, 63, 64 finitely presented, 24 fiat, 16 free basis, 15 free module, 15 fractional ideal, 47, 143 free resolution, 20 Frobenius automorphism, 73 full functor, 155 full lattice, 44, 108, 192 full matrix algebra, 91 function field, 63 functor, 14

G

Gauss' Lemma, 5 generator (of category), 155 genus, 232 global dimension, 41 global field, 63 global problem, 36 greatest common divisor, 47 Grothendieck group, 314 ground ideal, 222 group algebra, 108 group ring, 256, 373, 379 groupoid, 201 Grunwald-Wang Theorem, 279 H

Hasse-Brauer-Noether-Albert Theorem, 276 Hasse invariant, 148, 266, 274 Hasse Norm Theorem, 275 Hasse-Schilling-Maass Theorem, 289 Hensel's Lemma, 71 hereditary ring, 27, 130 homological dimension, 20, 40 I

ideal class, 48, 224, 307, 343 class group, 48, 306, 308 class number, 224, 226 fractional, 47,143 integral, 47, 182, 193 invertible, 327 maximal, 35 maximal integral, 182, 195 normal, 181, 193 prime, 35, 190 proper, 35 idempotent, 80 im (image), 8 indecomposable element, 230 indecomposable ideal, 182 indecomposable module, 88 index [A], 253 index (of skewfield), 143, 238

393

SUBJECT INDEX

Index Reduction Theorem, 255 inertia field, 72, 145 inertial degree, 140,212 infinite prime, 63 inflation, 249 injective module, 16 inner automorphism, 103, 322 integral closure, 5 integral element, 3 integral group ring, 256, 379 integral ideal, 47, 182, 193 integrally closed, 5, 6 inv [A], 266 Invariant Factor Theorem, 49 invariant (Hasse), 148, 266,274 invariants of chain, 354 invariants of hereditary order, 360 inverse different, 60, 150, 184, 217 invertible action, 33 invertible bimodule, 319-320 invertible ideal, 327 irreducible, 78 isomorphism of categories, 155 isomorphism of K-algebras, 1

J Jacobson radical, 78 J ordan-Zassenhaus condition, 224 J ordan-Zassenhaus Theorem, 228

lifting idempotents, 85, 86 LFP,345 local ring, 82 local capacity, 222 local degree (offield extension), 275 local field, 135 local index, 222 local problem, 36 localization, 36 locally free class group, 343 locally free module, 343 locally free Picard group, 345 M

maximal ideal, 35 maximal integral ideal, 182, 195 maximal order, 110 minimal left ideal, 90 minimal prime, 56 minimally contains, 370 minimum polynomial, 2 module of quotients, 32 monic, 1 Morita context, 162 Morita context, derived, 162 Morita equivalence, 162 morphism, 14 multiplicative subset, 30 N

K

K-isomorphism of K-algebras, 1 ker (kernel), 8 Krasner's Lemma, 285 Krull ring, 56 Krull-Schmidt Theorem, 88 L

Ladder Theorem, 23 lattice, 44, 108, 129 least common multiple, 47 left exact, 14 left hereditary, 27 left order, 109

N -adic topology, 85 Nakayama's Lemma, 81 natural equivalence, 154 natural transformation, 154 nilpotent, 80 noetherian, 13 non-archimedean valuation, 51 nondegenerate form, 46 non-R primes, 294 norm (counting), 59, 216 norm (of element), 3 norm (of ideal), 59, 150,185,210 norm (reduced), 116, 119, 121 norm (reduced, of ideal), 214 normal basis theorem, 258 normal ideal, 181, 193

394

SUBJECT INDEX

normalized factor set, 243 normalized valuation, 64, 139 normalizer of an order, 328

o opposite ring, 91, 101 order, 108 order ideal, 49 order, left, 109 order, maximal, 110 order, right, 109 orthogonal idempotents, 84, 85 Outcent, 328 outer automorphism group, 322 p

P-adic field, 68 P-adic topology, 83 Picard group, 320 Picent,320 preserves direct sums, 157 primary component, 201 primary ring, 177 prime element, 53, 139 prime ideal, 35, 190 primitive idempotent, 85, 90 principal factor set, 242 product formula, 64 progenerator, 158 projective module, 15 projective object, 155 projective resolution, 20 proper ideal, 35 proper product, 183, 196 pullback (diagram), 11 pure, 45 pushout (diagram), 10 Q quaternion skewfield, 114, 244, 271 R

radical, 78

radically cover, 356 ramification (at P), 272 ramification index, 57, 71, 139 ramified prime, 272 rank, 44 ray class group, 309 ray group, 309 real prime, 64 reduced characteristic polynomial, 113, 119, 121 reduced norm, 116, 119, 121 reduced norm of ideal, 214 reduced trace, 116, 119, 121 reflexive, 55 relative norm, 59 relatively prime (ideals), 47, 217,234 residue class degree, 57, 71 restriction map, 248 right exact, 14 right order, 109 ring of quotients, 31

s saturated, 35 Schanuel's Lemma, 30 Schilling's Theorem, 383 Schur's Lemma, 105 self-centralizing maximal subfield, 240 semisimple (ring), 78, 90 separable algebra, 99, 106 short exact sequence, 8 similar algebras, 237 similar ideals, 199 similar orders, 181 simple components, 91 simple (module), 78 simple ring, 90 skewfield, 82, 91 skewfield over K, 92 skew field part, 237 Skolem-Noether Theorem, 103 Snake Lemma, 30 split (sequence), 8 splitting field, 96 stable isomorphism, 307, 314 Steinitz's Theorem, 49 Strong Approximation Theorem, 48

395

SUBJECT INDEX

T tamely ramified, 373 tota~ly definite quaternion algebra, 293 torsIOn element, 31, 32, 44 torsion sub module, 32,44 torsionfree, 17, 44 trace, 3 trace form, 46 trace ideal, 156 trace (reduced), 116, 119, 121 trivial factor set, 242 trivial valuation, 51 trivial ZG-module, 256 twisted group ring, 373 type (of hereditary order), 360

u unit, 79, 224

unital, 1 unramified, 57, 382

v valuation, 51 valuation (exponential), 53 valuation (J-adic) 216 valuation ring, 51, 137 value group (of valuation), 51, 13 5 Very Strong Approximation Theorem, 64

w Weak Approximation Theorem, 291 Wedderburn Structure Theorem 91 Wedderburn Theorem on finit~ skewfields, 104