7. TORSION FREE RINGS. E. L. Lady. June 27, 1998

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1 1 7. TORSION FREE RINGS E. L. Lady June 27, 1998 In Chapter 1 we considered the problem of constructing, for a given prime p, a p-reduced module G with rank G>1 and p-rank G = 1. Example 1.47, the Pontryagin module, was an example of such a construction, and it ultimately led us to the concept of splitting fields and splitting rings. Another approach is as follows: example 7.1. Let Q be a finite field extension of Q and let W be the integral closure of W in Q. Suppose that there exists a prime ideal P 1 of W and a prime p of W such that W P 1 /pw P 1 W/p. Let R = W P 1.ThenRis p-reduced and p-rank R =1. proof: By hypothesis, p-rank R = 1. Furthermore R is an integral domain and p R is a pure ideal in R and so Qp R is an ideal in the field QR, which forces p R = 0 (see Corollary 7.17 below). The hypothesis for Example 7.1 will hold, for instance, if the prime p splits completely in W, i.e. pw = P 1...P n where P 1,...,P n are distinct prime ideals in W and n =[W :W]. In that case pw P i = P 1 P n W P i P i W P i for all i and so by Proposition 0.* W /pw (W /pw ) Pi W P i /pw P i,sothat p-rank W = n = n 1 p-rank W P i and necessarily p-rank W P i =1 forall i. From algebraic number theory it is known that this is a condition that occurs reasonably often. This example suggests that finite rank torsion free W -algebras can be a useful class of W -modules. Although we will see that Example 7.1 can be described, at least in principle, by the splitting ring construction (see Proposition 7.43), the above construction has the advantage of being simple and relatively uncontrived. There are two questions that come up when we think about finite rank torsion free modules which are in fact rings. (As always, we use the term ring to mean W -algebra.) First, if a W -module is also a ring, what does that tell us about its structure as a W -module? And second, what does the W -module structure of a ring R tell us about the ring-theoretic properties of R? We have already seen several instances of answers to the second question in the discussion of Murley rings in Chapter 3. Later in this chapter, we will also see, for

2 instance, that if the underlying W -module of a ring is strongly indecomposable then that ring must be an integral domain. 2

3 3 GENERAL RESULTS. A few elementary results of general interest are the following: proposition 7.2. AringRcan be identified as a pure subring of End R by the map that associates to r R the left multiplication r : x rx. proof: It is well known that the map r r is an isomorphism from R to End R R. But End R R End W R and in fact End R R End W R. Indeed, if ϕ End R and wϕ End R R for some w 0 W,thenwϕ is R-linear and clearly ϕ itself must be W -linear so ϕ End R R.Thusr r maps R onto a pure subring of End R. proposition 7.3. If R is a finite rank torsion free ring and N = nil rad R then (1) IT(R) =t(q R), in particular IT(R) is idempotent. (2) IT(R/N) =IT(R). proof: (1) Q is a subring of QR, soq Ris a pure rank-one subring of R. Thus t(q R) T(R), so IT(R) =inf T(R) t(q R). On the other hand, R is a Q R-module, hence by Proposition 4.11 R is t(q R)-saturated, i.e. IT(R) t(q R). Thus IT(R) =t(q R). Since Q R is a subring of Q, t(q R) is idempotent by Proposition (2) By (1) IT(R) andit(r/n) are idempotent. Thus by Proposition 2.3 IT(R) =t( X W p)andit(r/n) =t( Y W p)wherexis the set of prime ideals p such that pr R and Y is the set of p with p(r/n) R/N. But p(r/n) =R/N if and only if R = pr + N = pr + RN = pr +(pr + N)N = pr + pn + N 2 = pr + N 2 and inductively we see that p(r/n) =R/N R = pr + N k for every k. Since by Proposition 1.32 N k = 0 for large k it follows that p(r/n) =R/N if and only if pr = R, sothatx=y and IT(R) =IT(R/N). corollary 7.4. For any finite rank torsion free module G, IT(End G) = [IT(G): IT(G)]. proof: By Proposition 7.3 IT(End G) =t(q End G) and clearly Q End G is the largest subring A of Q such that G is an A-module. On the other hand, by Proposition 4.11 if A is a subring of Q then G is an A-module if and only if G is t(a)-saturated, which is the same as saying that t(a) IT(G). Therefore IT(End G) =t(q End G) is the largest idempotent type t such that t IT(G). But by Proposition 2.* this is [IT(G): IT(G)].

4 lemma 7.5. Let R be a finite rank torsion free ring, let U be a QR-module, and let M and N be R-submodules of U which are quasi-equal as W -modules. Then M is finitely generated as an R-module if and only if N is. proof: Suppose that M is a finitely generated R-module. By Proposition 3.1 if M and N are quasi-equal we may as well assume that M N and N/M is finitely generated (as a W -module). Then N is generated as an R-module by a set of generators for M together with a set of pre-images for generators of the W -module N/M. Therefore N is also a finitely generated R-module. proposition 7.6. If R is a finite rank torsion free ring with nil rad R =0 then every left ideal in R is a quasi-summand and is finitely generated as a left ideal. proof: If L is a left ideal in R then QL is a left ideal in QR. But if nil rad R =0 then QR is a finite dimensional algebra with trivial radical, hence is semi-simple, so QL = QRe for some idempotent e QR. Noweis a right identity on QL so L = Le Re. ThensinceLis an R-submodule of the finitely generated R-module Re and QL = QRe, Lemma 3.13 shows that L is quasi-equal to Re. But right multiplication by e is an idempotent quasi-endomorphism of R, sore is a quasi-summand of R, and thus L is a quasi-summand of R. Furthermore since L is quasi-equal to the principal left ideal Re, L is a finitely generated left ideal by Lemma corollary 7.7. If L is a left ideal in a finite rank torsion free ring R such that L nil rad R then L is quasi-pure in R. proof: Let N = nil radr. Then by Proposition 1.31 N R. Furthermore by Proposition 7.6 L/N is a quasi-summand of R/N, hence by Proposition 3.21 L/N is quasi-pure in R/N. Thus by Proposition 3.14 L is quasi-pure in R. proposition 7.8. A finite rank torsion free ring R is left noetherian if and only if nil rad R is a finitely generated left ideal. proof: ( ): Clear since if R is noetherian then all left ideals are finitely generated. ( ): Let N = nil rad R. SincenilradR/N = 0, it follows from Proposition 7.6 that all left ideals of R/N are finitely generated so R/N is left noetherian. Now by Proposition 1.31 there exists a filtration R N N 2 N r = 0 for some finite r,wheren = nil rad R. The factors R/N and N k /N k+1 are all modules over the noetherian ring R/N and if N is finitely generated over R they are all finitely generated, and hence notherian. It follows that if N is a finitely generated left ideal then R is noetherian.

5 5 corollary 7.9. An integral domain (over W ) is integrally closed if and only if it is a dedekind domain. proof: ( ): By definition dedekind domains are integrally closed. ( ): Since an integral domain has trivial nil radical, by Proposition 7.8 it is noetherian. Furthermore if P is a non-trivial prime ideal in an integral domain D then QP is a non-trivial ideal in the field QD, soqp = QD. SincePis a D-submodule of D and D is a cyclic D-module it follows from Lemma 3.13 that P is quasi-equal to D. Thus by Proposition 3.11 D/P has finite length, so D/P is an artinian integral domain, hence a field. Thus P is a maximal ideal. Thus D is a noetherian integral domain such that all prime ideals are maximal. If in addition D is integrally closed then by Proposition 0.* D is a dedekind domain. caution: Corollary 7.9 depends on our convention that all rings are assumed to be algebras over W. It is not a valid theorem in general commutative ring theory. Obviously Proposition 7.8 is equally valid if we substitute right noetherian for left noetherian. The following well known example shows that these two concepts are not equivalent. example The ring R = End(W Q) is left noetherian but not right noetherian. ( ) W 0 proof: R is isomorphic to the matrix ring, i.e. the ring of 2 2 matrices Q Q such that the entries in the second row belong to Q, the upper left hand corner belongs to W and the upper right hand entry is 0. Now let N be the subset of R consisting of those matrices whose diagonal elements vanish, i.e. N = ( ) 0 0. Q 0 Then it ( is easy to) see that N is an ideal in R and N 2 = 0. Furthermore W 0 R/N W Q, a ring with trivial nil radical. Thus N = nil rad R. Now 0 Q multiplication between elements of R and N operates as follows: ( w 0 q q )( ) 0 0 = x 0 ( ) 0 0 q, x 0 ( )( ) 0 0 w 0 x 0 q q = ( ) 0 0. wx 0 Thus as a left ideal N is essentially just a Q-module, and since N Q it is finitely generated. On the other hand, as a right ideal it is essentially just a W -module, and since Q is not finitely generated as a W -module, N is not finitely generated as a right ideal. Thus by Proposition7.8 R is left noetherian but not right noetherian.

6 6 PRIME & SEMI-PRIME RINGS. We are primarily interested in torsion free rings because those W -modules which are the underlying modules of certain torsion free rings are very special for the general structure theory of finite rank torsion free W -modules. For instance in Chapter 2 we noticed that among rank-one modules the subrings of Q play a special role. Hence it is plausible to suspect that the underlying modules of rings in general might be special. Unfortunately, though, the mere fact that a finite rank torsion free module G can be given a ring structure imposes only mild restrictions on the structure of the W -module G. Indeed any G can be given the structure of a commutative ring without identity, simply by making the multiplication trivial. If one now uses the standard construction to unitize G one gets a commutative W -algebra with the underlying W -module isomorphic to W G and containing G as a prime ideal. Explicitly, the multiplication on W G is given by (w 1,g 1 ) (w 2,g 2 )=(w 1 w 2,w 1 g 2 +w 2 g 1 ). One can note that G is the nil radical of this ring. example By applying the construction in the preceding paragraph in the case G = W one gets a ring R such that R = W W.ThusEndR= End(W W )sothat nil rad End R = 0. And yet nil rad R =0 W 0. On the other hand, if one lets R = W Q then as seen in Example 7.10, nil rad End R 0, but clearly nil rad R =0. Thus there is in general no nice relationship between the nil radical of a ring and the nil radical of its endomorphism ring. (However see Proposition 7.16 below.) Delete? As soon as one imposes the condition that nil rad R = 0, the restrictions on R as a W -module become much more stringent. For instance, we have seen in Proposition 7.8 that in this case R is noetherian. Another important observation is that if H is a fully invariant submodule of any ring R then in particular H is invariant under both left and right multiplication by elements of R so that H must be an ideal in R and QH must also be an ideal in QR. If nil rad R =0thenQR is semi-simple and since semi-simple rings don t have many (two-sided) ideals, then R can have few pure fully invariant submodules. (See Proposition 7.16 below.) Delete? We shall see that the study of finite rank torsion free modules which can be given the structure of rings with trivial nil radical reduces to the study of those with the structure of (commutative) integral domains. It turns out that these integral domains play a key role in the structure theory of finite rank torsion free W -modules. definition A finite rank torsion free ring R is called semi-prime if nil rad R =0 and prime if QR is simple (i.e. has no proper two-sided ideals). Note that since by Proposition 1.32, nil rad QR = Q(nil rad R), a ring R is semi-prime if and only if QR is semi-simple. (The reader should not confuse this usage with the use of the term prime ring by a few other (misguided!) authors to denote the rings Z and Z/pZ for prime numbers p.)

7 7 proposition (1) A ring without zero divisors is a prime ring. (2) A commutative ring R is a prime ring if and only if it is an integral domain. In this case QR is the quotient field of R. proof: (1) If R has no zero divisors then QR is a finite dimensional Q-algebra without zero divisors, hence is a skew field, which is a simple ring. Thus R is a prime ring. (2) If R is a commutative prime ring, then QR is a commutative simple ring. But a commutative ring without non-trivial proper ideals is a field, hence R is an integral domain. Since QR is a field and is the localization of R with respect to a multiplicative set in R, namely the set of non-trivial elements of W, it follows that QR is the quotient field of R. We will need the following well known lemma from the theory of finite-dimensional algebras. lemma A finite dimensional algebra S over a field is a simple algebra if and only if S has trivial radical and no proper central idempotents. proof: ( ): If e is a central idempotent in S then es is a (two-sided) ideal. Likewise nil rad S is an ideal. Thus if S is a simple ring then es and nil rad S must be either all of S or trivial. This implies that e =1ore= 0, and that nil rad S = 0 (since nil rad S S.) ( ): If a finite dimensional algebra S has trivial radical then it is semi-simple, hence a product of simple algebras. If it has no non-trivial central idempotents then there can be only one factor in this product, so S is a simple algebra. proposition Let R be a finite rank torsion free ring. The following conditions are equivalent: (1) R is a prime ring. (2) R has no proper non-trivial pure (two-sided) ideals. (3) R is semi-prime and QR has no proper central idempotents. (4) If I and J are ideals in R such that IJ =0 then I =0 or J =0. proof: (1) (2): If I is a pure ideal in R then QI is an ideal in QR and QI = QR if and only if I is essential in R, i.e. if and only if I = R, since by assumption I R. Thus R has no non-trivial proper pure ideals if and only if QR has no non-trivial proper ideals, i.e. if and only if QR is simple, i.e. if and only if R is prime. (1) (3): By definition, R is prime if and only if QR is simple. And by Lemma 7.14 QR if simple if and only if nil radqr = 0 and QR has no proper idempotents. But by Proposition 1.32 nil rad QR = Q(nil rad R) and by definition nil rad R =0 ifandonlyif R is semi-prime. (2) (4): If IJ =0 then I J = 0. Thus without loss of generality we may suppose that I and J are pure. Now IJ = 0 is not possible if I = J = R, sooneofi and J must be a proper pure ideal. Thus if R has no non-trivial proper pure ideals then either I or J must be trivial.

8 8 (4) (3): If N = nil rad R then by Proposition 1.31 N is an ideal and N n =0, where n =rankr.nowifn 0then0<rank N < n =rankrso n 2andif we set I = J = N n 1 then IJ = 0. Likewise if e is a central idempotent in QR and I = R QRe = R eqr, J = R QR(1 e) =R (1 e)qr, theni and J are ideals in R and IJ = 0. Thus if there are no proper non-trivial ideals I and J with IJ =0thene=0ore=1sothatRhas no proper central idempotents and likewise N = nil rad R =0. Condition (4) In Proposition 7.15 is the usual definition of a prime ring in general non-commutative ring theory. proposition (1) A fully invariant W -submodule of a ring is a (two-sided) ideal. (2) If R is a prime ring then R has no fully invariant pure submodules except 0 and R. (3) If G is a W -module with no fully invariant pure submodules except 0 and G then End G is a prime ring. (4) If R is a prime ring then End R is a prime ring. proof: (1) If M is a fully invariant submodule of R then in particular M is invariant under left and right multiplication by elements of R so M is an ideal. (2) If M is a fully invariant pure submodule of R then by (1) M is a pure ideal in R. If R is prime then by Proposition 7.15 R has no non-trivial proper pure ideals, so M = R or M =0. (3) Let R =EndG.IfN= nil rad R then (NG) is a fully invariant pure submodule of G and by Proposition 1.32 rank(ng) =rankng < rank G. ThusifGhas no non-trivial proper pure fully invariant submodules except G then NG = 0.Since N End G this implies that N =0. Thus R is semi-prime. Now if e is a central idempotent in QR =QEndGthen by Proposition 3.16 G is quasi-equal to eg (1 e)g. Thus eg and (1 e)g are quasi-pure submodules of G and they are fully invariant since if ϕ End G = R then ϕ(eg) =ϕe(g) =eϕ(g) eg and likewise ϕ(1 e)g (1 e)g. ThusifG has no non-trivial proper pure fully invariant submodules then G eqg =0 or G (1 e)qg = 0 and it follows that e =0 or e=1. Thus R is semi-prime and has no proper central idempotents, hence by Proposition 7.15 R is a prime ring. (4) This follows immediately from (2) and (3). In abelian group theory, a finite rank torsion free group G with no non-trivial proper pure fully invariant subgroups is called irreducible, the point being that QG is an irreducible (QEnd G)-module. In module theory the term irreducible has been already preempted so, in line with prevailing fashion, your author would suggest the term E-irreducible. corollary Let R be a ring. (1) For any prime p, p R is a pure ideal in R and likewise d(r) is a pure ideal in R.

9 9 (2) For every type t, R(t) and R[t] are pure ideals. (3) If R is a prime ring then R is either divisible or reduced, and for every prime p, R is either p-divisible or p-reduced. (4) If R is a prime ring then R is homogeneous of idempotent type and cohomogeneous (of idempotent type?). proof: (1) & (2) These follow from Proposition 7.16 since by Proposition 1.4, Proposition 1.15 and Proposition 4.2, d(r), p R, R(t)andR[t] are fully invariant pure submodules of R. (3) & (4) It follows from (1) and (2) that if R is prime then d(r) =0ord(R)=R so that R is either reduced or divisible. Likewise p R =0orp R=Rso R is either p-reduced or p-divisible. And for any type t, either R(t) = Ror R(t) = 0.Thusif t>it(r) thenr(t) = 0, so we see that all elements of R have type IT(R) andris homogeneous. Furthermore by Proposition 7.3 IT(R) is idempotent. Likewise R[t] =R or R[t] =0,so R[t]=0 if t<ot(r) and we see that R is cohomogeneous. corollary If a prime ring is a Butler module then it is a homogeneous completely decomposable module. proof: By Corollary 7.17 a prime ring is homogeneous. By Proposition 5.* a homogeneous Butler module is completely decomposable. QUASI-PRODUCTS OF RINGS. If R is semi-prime, then QR is semi-simple and so the center of QR is a finite product of fields. We say that R is quasi-separable if these fields are all separable extensions of Q. (Quasi-separability is defined only for semi-prime rings.) If R is a ring, then by Proposition 7.2 we may identify R as a subring of End R. It follows that if e is an idempotent in R then er is a summand of the W -module R. And if e is an idempotent in QR then er is a quasi-summand of R. This makes the following result not terribly surprising: proposition Let R be a finite rank torsion free ring, let Z be its center and N = nil rad R. ThenQR = QR 1 QR k where each R i is a subring of QR which is quasi-pure in R and R i /(R i QN) is a prime ring. Furthermore (1) R is quasi-equal to R 1 R k. (2) If R is commutative and integrally closed then R = R 1 R n. (3) Z is quasi-equal to Z 1 Z k,wherez i =CenterR i, with equality if Z is integrally closed. (4) N is quasi-equal to N 1 N k,wheren i =R i N. (5) N i is quasi-equal to N R i. (6) N i = nil rad R i.

10 10 proof: (1) By Proposition 1.32 QN is the nil radical of QR. Therefore QR/QN is a semi-simple algebra and hence is a finite product of simple rings. Let ē 1,...,ē k be the central idempotents corresponding to this decomposition of QR/QN. By Lemma 1.34 the ē i are images of central idempotents e i QR. Write R i = Re i. Note that R R 1 R k since every r R can be written as r = r1 =re 1 + +re k and note also that QR = Q(R 1 R k ). Since R 1 R k is generated as an R-module by the finite set e 1,...,e k it follows from Lemma 3.13 that R is quasi-equal to R 1 R k. In particular each R i is a quasi-summand of R and hence by Proposition 3.21 is quasi-pure in R. (2) (3) The center of R is quasi-equal to the center of R 1 R k,whichisz 1 Z k. (4) The same reasoning as in (1) shows that N is quasi-equal to e 1 N e k N. Furthermore e i N = e i RN = R i N = N i. (5) Since e i QR there exists w 0 W such that we i R and so wn i = we i N N e i R = N R i. On the other hand, since e 2 i = e i, multiplication by e i restricts to the identity on e i R = R i and hence also on N R i. Thus N R i = e i (N R i ) e i N = N i R i.thusn R i is quasi-equal to N i. (6) By (3) N i is a quasi-summand of R and hence is quasi-pure in R, sothatn i is quasi-equal to R QN i = R e i QN. By Proposition 1.32 QN = nil rad QR and so it is well known from the theory of artinian rings that e i QN = nil rad QR i.thusn i is quasi-equal to R nil rad QR i R i nil rad QR i = nil rad R i.sincen i R i (why?) it follows that N i = nil rad R i. In particular, this shows that each R i /(R i QN) isaprimering. lemma [Wedderburn Principal Theorem] Let N = nil rad R. IfR/N is quasi-separable or R is commutative then there exists a semi-simple subring S QR such that QR = S QN. proof: QR is a finite product of rings S i such that S i /(nil rad S i ) is a simple ring. Thus no generality is lost in supposing that R/N is simple. 2) Suppose now that R is commutative of characteristic c 0. ThenQR/QN is a commutative simple ring, i.e. a field. By Proposition 1.* there exists r 1 such that N r = 0. We use induction on r,thiscaser= 1 being trivial. Suppose now that N 2 =0. Let QR c denote the set of all elements x c for x QR. Sincec=charR,R c is a subring of R. Furthermore, since c 2andN 2 =0,ifx c 0thenx/ QN and so x is invertible since QN is the unique maximal ideal of QR. ThusQR c is a subfield of QR. SinceQR is finite dimensional, there is a maximal subfield S of QR containing R c. Since elements of QN are nilpotent, S QN = 0. We claim that S QN = QR. If not, let x QR with x / S QR. Thenx c R c S. Now the polynomial X c x c =(X x) c must be irreducible over S since c =chars and x/ S. But this means that x is algebraic over S so that S[x] is a proper extension field of S contained in QR, contradicting the maximality of S. ThusS QN = QR, as desired.

11 11 Now suppose that r>2. By induction we may assume that QR/QN r 1 contains a subfield S such that QR/QN r 1 = S QN/QN r 1. Now let T be the subring of QR such that QN r 1 T and T/QN r 1 = S QR/QN r 1. Then QN r 1 = nil rad T. Furthermore (QN r 1 ) 2 = 0. Thus by the case r = 2 it follows that T contains a subring S such that T = S QN r 1.ButthenQR = S QN since **** proposition Let N = nil rad R and suppose R/N is quasi-separable. Let S be a semi-simple subalgebra of QR such that QR = S QN.ThenR Sis semi-prime and R is quasi-equal to (R S) N. Furthermore if R is commutative and R S is integrally closed, then R =(R S) N. proof: Deferred until after Corollary FIELD OF DEFINITION. It follows from Proposition 7.19 that a semi-prime ring is quasi-equal to a product of prime rings so that the study of semi-prime rings pretty much comes down to the study of prime rings. In particular, a strongly indecomposable semi-prime ring must in fact be a prime ring. Now there exist prime rings which are not integral domains, for instance the ring R of n n matrices over W is prime since QR is the ring of n n matrices over the field Q and hence a well known simple ring. However it turns out that every prime ring R has a subring D which is an integral domain and such that R is quasi-isomorphic to a direct sum of copies of D. Thus R is a finitely generated D-module. This result was first proved in [Pierce], using steps from [B & P]. The proof is a fascinating amalgam of Wedderburn Theory, Galois Theory, commutative ring theory, as well as torsion free module theory. Recall that if an an integral domain D is a subring of the center of a ring R then we say that R is a finite integral extension of D if R is a finitely generated D-module. lemma Let D be an integral domain (1) If D is a subring of the center of a ring R and QD is separable over Q, then Ris a finite integral extension of D if and only if R is quasi-equal to a free D-submodule of QR. (2) If D is an integral domain containing D such that QD = QD then D is a finite integral extension of D if and only if D and D are quasi-equal. proof: (1) ( ): If R is quasi-equal to a free D-submodule of QR then R is a finite integral extension of D since by Lemma 7.5 a module quasi-equal to a finitely generated D-module is itself finitely generated as a D-module. ( ): Since R is a D-module, QR is a vector space over the field QD and dim QD QR rank R<. Let M be the D-module generated by a basis for QR over QD contained in R. Then since this basis is linearly independent over D, M is a free D-module. Furthermore M R and QM = QR. ThusifRis finitely generated over D then R is quasi-equal to M by Lemma 3.13.

12 (2) ( ): If D is a subring of D and QD = QD and D is a finite integral extension of D, i.e. a finitely generated D-module, then by Lemma 3.13 D is quasi-equal to D. ( ): If D is a subring of D and D is quasi-equal to D then since D is a finitely generated D-module, by Lemma 7.5 D is also a finitely generated D-module, i.e. a finite integral extension of D. 12 lemma Let F be a finite separable extension of Q and let W be the integral closure of W in F. Then (1) If R is a ring such that F is a subring of QR then R is quasi-equal to W R. (2) If D is an integral domain with quotient field F then W D is the integral closure of D. In particular, W D is a dedekind domain. proof: (1) If F is a separable extension of Q, then by Proposition 0.* W is a finitely generated W -module. It follows that W R is a finitely generated R-module since any finite set of generators for W as a W -module will also generate W R as a R-module. Since R W R it follows from Lemma 3.13 that R is quasi-equal to W R. (2) Let D be the integral closure of D (in its quotient field F ). Then W D and so W D and thus W D D.ItiswellknownthatW is a dedekind domain (Proposition 0.*). Since W W D QD = QW, W D is a dedekind domain by Proposition 1.*. In particular W D is integrally closed, so since D W D it follows that D W D. Thus D = W D. theorem Let D be a finite rank torsion free prime ring. Let F =CenterQEndD. Then (0) QD can be naturally identified as a subring of QEnd D. (1) F Center QD. (2) F D =CenterEndD. (3) QEnd D =End F QD. (4) D is a finite integral extension of the commutative ring F D and is quasi-isomorphic to a free F D -module. (5) End(F D) F D. (6) F D is strongly indecomposable. Suppose in addition that D is commutative and contained in a galois extension Q of Q. Let D be the integral closure of D in Q. Then (7) F is the fixed field of {σ Gal(Q /Q) σ(d )=D }. proof: (0) By Proposition 7.2. (2) Since D End D and F =CenterQEndD it follows that D F Center End D. On the other hand, clearly Center End D Center QEnd D = F. And if ϕ Center End D then in particular ϕ commutes with right multiplication by elements of D. Thus if for r D we write ρ r for the map x xr then ϕ(r) =ϕ(1r) =ϕρ r (1) = ρ r ϕ(1) = ϕ(1)r.thusϕ= dwhere d = ϕ(1), i.e. given

13 the identification in (0) we can write ϕ = ϕ(1) D. Thus Center End D D so that Center End D = F D. (1) Since by (2) Center End D D it follows that F =CenterQEndD QD. Nowif d QD and d Center QEnd D = F then d commutes with (left multiplication by) the other elements of QD, so d Center QD. Thus F Center QD. (3) By Proposition 7.16 End D is a prime ring, so QEnd D is a simple algebra. We claim that QD is a simple (QEnd D)-module. In fact, if M is a (QEnd D)-submodule of QD then in particular M must be invariant under left and right multiplication by elements of QD so that M is a two-sided ideal in QD and since QD is a simple ring it follows that either M = QD or M =0. It now follows from Jacobson s version of the Wedderburn Theorem that QEnd D End QD where = End QEnd D QD, a skew field. Now if ϕ, then in particular ϕ is QD-linear on the right, so, as seen in the proof of (2), QD QEnd D. It follows that consists of those mappings in QEnd D which are (QEnd D)-linear, i.e. = Center QEnd D = F. Thus QEndD End F QD. (4) Since QEnd D =End F QD, the quasi-direct decompositions of D are the same as the F -linear decompositions of QD. But as an F -space, QD is a direct sum of a finite number of copies of F. Thus D is a quasi-direct sum of a finite number of copies of F D, so D is quasi-isomorphic to a free F D-module, and thus by Lemma 7.22 is a finite integral extension of F D. (5) Since D is quasi-isomorphic to a free F D-module, any endomorphism of F D can be extended to a quasi-endomorphism of D, hence by (3) is F -linear, and so a fortiori is F D-linear. Thus End(F D) =End F D F D=F D. (6) By (5) QEnd(F D) Q(F D), a field. Thus QEnd(F D) has no non-trivial idempotents, so F D is strongly indecomposable by Proposition (7) If D is commutative and contained in a galois extension Q of Q, letd be the integral closure of D in Q. Then since Q is separable over QD, by Proposition 0.* D is a finite integral extension of D and hence by Lemma 7.22 is quasi-isomorphic to a direct sum of copies of D. Thus QEndD is isomorphic to a full matrix ring over QEnd D and so Center QEnd D =CenterQEndD=F and so F is also the field of definition for D. In order to show that F isthefixedfieldof{σ Gal(Q /Q) σ(d ) D },it suffices by Galois Theory to show that for σ Gal(Q /Q), σ(d ) D if and only if σ fixes F. Now σ(d ) is isomorphic to D and therefore is also a dedekind domain, and in particular is integrally closed in Q. On the other hand since σ is monic, if σ(d ) D then by Proposition 3.9 σ(d )isquasi-equaltod and therefore by Lemma 7.22 D is an integral extension of σ(d ). It follows that if σ(d ) D then σ(d )=D. Now the field of definition for σ(d )isσ(f), so since the field of definition is determined uniquely by the ring D,ifσ(D )=D then σ(f )=F andthusalso σ(f D )=F D,sothatσ restricts to an endomorphism σ F End F D.But by (5) End F D = F D,sothatσ F is given by multiplication by some f D F. Since σ F (1) = 1 we conclude that f =1andσ F =1 Gal(Q /Q). In other words, if σ(d ) D then σ restricts to the identity map on D F. Conversely if σ restricts to 13

14 14 the identity map on D F then σ(d )=D since by (4) D is the integral closure in Q of D F. Since Q(D F)=F, this shows that σ(d ) D if and only if σ fixed F. definition If D is a prime ring then Center(QEnd D), identified as above with a subfield of QD, is called the field of definition for D. corollary Let D be a prime ring. The following conditions are equivalent: (1) End D D. (2) QEnd D QD. (3) D is strongly indecomposable. (4) The field of definition for D is QD. Furthermore in this case D is an integral domain. proof: (1) (2): Clear. (2) (4): If F is the field of definition for D then by Theorem 7.24 QEnd D End F QD, sodim F QEnd D = (dim F QD) 2. This means that if QEnd D QD then dim F QD =1 so F =QD. (4) (3): Immediate from part (6) of Theorem (3) (4): By part (4) of Theorem 7.24, if F is the field of definition for D then D is a finite integral extension of F D and hence by Lemma 7.22 D is quasi-isomorphic to a free F D-module. If D is strongly indecomposable this implies that D is quasi-isorphic to F D. Thus QD = Q(F D) =F,soQD is the field of definition for D. (4) (1): By part (5) of Theorem Furthermore, if QD is the field of definition for D then QD is a field, so D is an integral domain. corollary (1) Let D be an integral domain and let D be a finite integral extension of D. Then D and D have the same field of definition. In particular, the field of definition for D is contained in QD. (2) Let D be a prime ring and K a subfield of the Q-algebra QD. Then K contains the field of definition for D if and only if D is a finite integral extension of K D. proof: (2) Let F be the field of definition for D. ( ): By Theorem 7.24 D is a finite integral extension of F D. If F K then a fortiori D is a finite integral extension of K D. ( ): If D is a finite integral extension of K D then by Lemma 7.22 D is quasi-equal to a direct sum of k copies of K D for some k 1. Thus QEnd D is isomorphic to the ring of k k matrices over QEnd(K D). Therefore the center F of QEnd D is the center of QEnd(K D), and thus by part (2) of Theorem 7.24 is contained in K. (1) Let F be the field of definition for D and F the field of definition for D.By Theorem 7.24 D is a finite integral extension of F D. Thus if D is a finite integral

15 extension of D then D is also a finite integral extension of F D and so by (2) F F. Thus in particular F QD. ThensinceD is a finite integral extension of F D and F D D D, a fortiori D is a finite integral extension of F D and thus by (2) F F. 15 corollary A quasi-separable semi-prime ring R is a finitely generated module over its center. proof: By Proposition 7.19 R is quasi-equal to R 1 R n, where the R i are prime rings and the center Z of R is quasi-equal to Z 1 Z n,wherez i is the center of R i. Thus by Lemma 7.5 Z 1 Z n is a finitely generated Z-module and furthermore R is a finitely generated Z-module if and only if R 1 R n is a finitely generated Z-module. It follows that it suffices to prove that each R i is a finitely generated Z i -module, for if X i is a finite set of generators for R i over Z i and Y is a finite set of generators for Z 1 Z n over Z,thenR 1 R n is generated over Z by the set of all elements xy for y Y and x X i for some i. Thus we may assume without loss of generality that R is a prime ring. Let Z be the center of R. IfFis the field of definition for R then by Theorem 7.24 F R Z and R is a finite integral extension of F R, so a fortiori R is a finite integral extension of Z. We can now prove Theorem proof of theorem 7.21: Suppose that S is a Q-subalgebra of QR such that R/N is quasi-separable. Then by Lemma 7.20, QR = S QN where N = nil radr and S is a semi-simple Q-algebra. We wish to prove that R is quasi-equal to (R S) N, with equality if R is commutative and R S is integrally closed. Let π : QR S be the projection with kernel QN. Note that since QN is an ideal, π is a ring morphism and induces an isomorphism from the semi-simple ring QR/QN onto S.SinceQ(R S)=QR S = S, it follows that R S is semi-prime. Since (R S) Ker π = R S N =0,toprovethatRis quasi-equal to (R S) N it suffices by Proposition 3.18 to show that R S is quasi-equal to π(r). (Note that R S = π(r S) since the restriction of π to S is the identity map.) Furthermore we will show that if R is commutative and R S is integrally closed then R S = π(r), so that π restricts to a split surjection from R onto R S and so R =(R S) N. Consider first the case where S is a field. Let W be the integral closure of W in the field S.ThusW R Sif R S is integrally closed, and in general W R S QR. Since by assumption S is separable over Q, by Lemma 7.23 W R is quasi-equal to R and (W R) S is quasi-equal to R S. It thus suffices to prove that W R is quasi-equal to (W R S) W N. I.e. without loss of generality we may suppose W R. Butthen we may as well replace the ground ring W by W, hence it suffices to consider the case S = Q. Then by Proposition 7.3 t(s R) =t(q R)=IT(R) =IT(R/N) =t(π(r)), since R/N π(r) Q. But by Proposition 2.15 two subrings of Q with the same type are identical. Thus S R = π(r), so R =(S R) Ker π =(S R) N, as required.

16 Now consider the more general case that S is commutative. Then by Proposition 7.19 R is quasi-equal to R 1 R n, where for each i, R i is a commutative prime ring, i.e. an integral domain, and if R S is integrally closed then R = R 1 R n and all the R i S are dedekind domains. Furthermore N is quasi-equal to N 1 N n where N i = nil rad R i. Note also that R S is quasi-equal to (R i S) (why?). It thus suffices to prove that for each i, R i is quasi-equal to (R i S) N i, with equality if R i is integrally closed. This was handled in the preceding paragraph. Finally, consider a general ring R such that R/N is quasi-separable and consequently QR = S QN.Asabove,letπ:QR S. By Corollary 7.28 R is a finitely generated module over its center Z and likewise R S is a finitely generated module over Z S.It further follows that π(r) is a finitely generated module over π(z). Now by applying the previous paragraph to the commutative ring π 1 (Z) we see that π(z) isquasi-equalto Z S.Now Z S=π(Z S) π(z) π(r) and since π(r) is a finitely generated module over π(z) andπ(z)isquasi-equaltoz S it follows that π(r) is a finitely generated module over Z S. (In fact, by Lemma 7.5 π(z) is finitely generated over Z S.Thusπ(R) is a finite integral extension of π(z), which is a finite integral extension of Z S.) Furthermore R S = π(r S) isan essential submodule of π(r). Thus by Lemma 3.13 π(r) isquasi-equaltor S. Thus πrestricts to a quasi-split surjection from R onto π(r), so by Proposition 3.18 R is quasi-equal to (R S) N. 16 corollary If R is a strongly indecomposable ring such that R/ nil rad R is quasi-separable then R is an integral domain. proof: Let N = nilradr. By Theorem7.21 if R/N is quasi-separable then R is quasi-equal to (S R) N.ThusifRis strongly indecomposable, either N =0or S R= 0. ButS R 0since1 S R.ThusN= 0soRis semi-prime. Then by Proposition 7.19 R is quasi-equal to a product of prime rings. Since R is strongly indecomposable, there can be only one factor in this product, so R is prime. Then by Corollary 7.26, R is an integral domain. STRONGLY INDECOMPOSABLE DOMAINS. We saw in Proposition 2.2 and Proposition 2.3 that the subrings of Q are the localizations of W with respect to multiplicative sets and correspond in one-to-one fashion to the subsets of Spec W.Nowif Dis any finite rank integral domain then D is a essential subring of a finite dimensional extension field Q of Q. ButifDis integrally closed and W is the integral closure of W in Q,thenW Dso that D is a rank-one W -module. Thus we get the following proposition.

17 proposition (1) A finite rank torsion free quasi-separable dedekind domain D is determined up to a quasi-equality by a finite field extension Q of Q and a subset Y of Spec W,whereW is the integral closure of W in Q. (Recall that according to our convention, Spec W denotes the set of non-trivial prime ideals of W ). Specifically, D = Y W P where Y is the set of prime ideals P of W such that D P Q. (2) If Q is a separable extension of Q then a quasi-equality class of essential subrings of Q contains exactly one integrally closed subring D. Furthermore every subring of Q quasi-equal to D is a subring of D. (3) Let Q be a galois extension of Q and let Y be a subset of Spec W. Let G be the largest subgroup of Gal(Q /Q) under which Y is invariant, i.e. G = {σ Gal(Q /Q) ( P Y ) σ(p ) Y }. Then the field of definition for Y W P is the fixed field of G. proof: (1) Let D be a finite rank torsion free quasi-separable integral domain and let Q = QD. Then Q is a finite separable extension of Q. Let W be the integral closure of W in Q. Then as indicated in the remarks preceding the Proposition, the integral closure of D in Q is a rank-one W -module of idempotent type and by Proposition 7.23 is quasi-equal to D. Furthermore by Proposition 2.2 D is uniquely determined by its type over W and hence no two integrally closed subrings of Q can be quasi-equal as W -modules. But this in fact means that they cannot be quasi-equal as W -modules, since if w 0 W is such that wd 1 D 2 and wd 2 D 1,where D 1 and D 2 are W -submodules of Q,thenD 1 and D 2 are actually quasi-equal as W -modules. Thus by Proposition 2.3 the set of such D is in one-to-one correspondence with the family of subsets of Spec W, where a domain D corresponds to the set of prime ideals P of W such that D is not P -divisible, i.e. D P Q. (2) If D is an essential subring of Q and D is its integral closure, then D D and by Proposition 7.22 D is quasi-equal to D. Furthermore if two integrally closed essential subrings D 1 and D 2 are quasi-equal then by Lemma 3.13 they are quasi-equal to D 1 D 2. Thus by Proposition 7.22 D 1 D 2 is an integral extension of D 1 and D 2. Since these are integrally closed, D 1 = D 2 = D 1 D 2. (3) If D = P Y W P and σ Gal(Q /Q), then σ(d) D if and only if σ(p ) Y whenever P Y. Therefore the claim follows from part (7) of Theorem We say that two rings R 1 and R 2 are quasi-isomorphic as rings if there is a ring isomorphism from R 1 ontoasubringofqr 2 whichisquasi-equaltor 2. The class of strongly indecomposable integrally closed integral domains has the important property that if two modules in this class are quasi-isomorphic they are in fact isomorphic, as the following proposition shows. proposition Let D 1 and D 2 be integral domains. (1) If D 1 and D 2 are strongly indecomposable then they are isomorphic as W -modules if and only if they are isomorphic as rings. 17

18 (2) If D 1 and D 2 are integrally closed and strongly indecomposable then they are quasi-isomorphic as W -modules if and only if they are isomorphic as rings. (3) In general, two integral domains D 1 and D 2 are quasi-isomorphic as W -modules if and only if they have the same rank, their fields of definition F 1 and F 2 are isomorphic fields, and F 1 D 1 and F 2 D 2 are quasi-isomorphic as rings. proof: (1) Clearly if D 1 and D 2 are isomorphic as rings [i.e. as W -algebras] then they are isomorphic as W -modules. Conversely if they are strongly indecomposable and isomorphic as W -modules then by Corollary 7.26 D 1 End D 1 End D 2 D 2 since isomorphic modules have isomorphic endomorphism rings. (3) ( ): If D 1 D 2 then clearly rank D 1 = rankd 2. And if F i is the field of definition for D i then by Theorem 7.24 F i is the center of QEnd D i and QEnd D 1 QEnd D 2 by Proposition 3.8, so F 1 F 2. Furthermore F 1 D 1 and F 2 D 2 are strongly indecomposable by Theorem 7.24 and each D i is quasi-isomorphic to a direct sum of copies of F i D i. Since D 1 D 2 it follows from Jónsson s Theorem (Theorem 3.24) that F 1 D 1 F 2 D 2. But then as rings 18 F 1 D 1 End(F 1 D 1 ) End(F 2 D 2 ) F 2 D 2 by Theorem 7.24 and Proposition 3.8. ( ): If F 1 D 1 and F 2 D 2 are quasi-isomorphic as rings (i.e. as W -algebras) then a fortiori they are quasi-isomorphic as W -modules. Since by Theorem 7.24 each D i is quasi-isomorphic to a direct sum of copies of F i D i it follows that if rank D 1 =rankd 2 then D 1 D 2. (2) If D 1 and D 2 are strongly indecomposable then by Corollary 7.26 their fields of definition are QD 1 and QD 2. Therefore by (3) if they are quasi-isomorphic as W -modules then they are quasi-isomorphic as rings. Therefore we may suppose that D 1 and D 2 have the same quotient field and are quasi-equal. But by Proposition 7.30 a quasi-equality class of integral domains in a field contains only a single integrally closed domain. Thus if D 1 and D 2 are integrally closed then they are equal. proposition If an integrally closed integral domain is indecomposable as a W -module then it is strongly indecomposable. proof: Let D be an integrally closed domain and let F be the field of definition for D. Then D F is also integrally closed, therefore by Corollary 7.9 is a dedekind domain. Since D is a finite rank torsion free F D-module, by Proposition 1.19 D is isomorphic to a direct sum of ideals of F D. Since D is indecomposable, there can be only one term in this direct sum decomposition, so D is isomorphic to an ideal of F D. Thus rank D =rankf.sincef QD, it follows that F = QD. Therefore by Corollary 7.26 D is strongly indecomposable.

19 EARRINGS. We have seen in Corollary 7.26 that if D is a strongly indecomposable integral domain then D End D. This result suggests an interesting line of thought. If R is a ring, then by Proposition 7.2 R is embedded as a pure subring of End R. One can then wonder just how large a subring R is of End R. For instance one might conjecture that R =EndRif R is a ring with very nice properties. This fails, however, even when R is an integral domain. For instance if W is the integral closure of W in some proper separable extension field of Q then W is a projective W -module and hence quasi-isomorphism to a free W -module W n,forn>1. Then QEndW is the ring of n n matrices over Q so that End W is not commutative, and hence certainly is much larger than W. Rings R such that the canonical embedding R End R is an equality have come to be called E-rings [Schultz]. (In other words, R is an E-ring if every endomorphism of R is given by left multiplication by some element r R.) We shall see that he structure of E-rings is quite simple. lemma (1) R is an earring if and only if the canonical embedding QR QEnd R is an equality. (2) Aringquasi-isomorphictoanE-ringisanE-ring. (3) If End R R then R is an E-ring. (4) If R 1 R 2 is quasi-equal to an E-ring then R 1 and R 2 are E-rings. Furthermore Hom(R 1,R 2 )=0. proof: (1) Recall that by Proposition 7.2 R End R. ThusifQR =QEndRthen R = QR End R =QEndR End R =EndR,soRis an E-ring. (2) If R and R are quasi-isomorphic then we may suppose them quasi-equal. Then QR = QR and QEnd R =QEndR. Thus the result follows from (1). (3) By Proposition 7.2 the canonical map λ: QR QEnd R is monic. If End R R then rank End R =rankr so λ is an isomorphism. Thus R is an E-ring by (1). (4)By(2)wemayassumethatR 1 R 2 is an E-ring. Now any endomorphism ϕ of R 1 extends (non-uniquely) to an endomorphism of R 1 R 2 and hence since R 1 R 2 is an E-ring is given by multiplication by some (r 1,r 2 ) R 1 R 2. It follows that ϕ is given by multiplication by r 1. Thus the canonical embedding R 1 End R 1 is surjective, so R 1 is an E-ring. Furthermore if ψ Hom(R 1,R 2 )thenψextends to an endomorphism of R 1 R 2, which must be given by multiplication by some r R 1 R 2. But since R 1 is an ideal in R 1 R 2, rr 1 R 1 and it follows that ψ =0. Thus Hom(R 1,R 2 )=0. Note that by Proposition 3.43 Murley rings are E-rings. For Murley rings, the following proposition was part of Proposition proposition If R is an E-ring then R is commutative. proof: For any r R, right multiplication by r is an endomorphism of R, hence by hypothesis must be given by left multiplication by some s R, i.e. for all x R, xr = sx. Substituting x =1 yieldsr=1r=s1=s.thusxr = rx for all x R. Since this holds for all r R, R is commutative. 19

20 proposition Let G be quasi-equal to G 1 G n, where the G i are strongly indecomposable. Then End G is commutative if and only if End G i is commutative for each i and Hom(G i,g j )=0 for i j. In this case, End G End G 1 End G n. proof: ( ): Easy since if G is quasi-equal to G 1 G n and Hom(G i,g j )=0 for i j then End G End G 1 End G n. ( ): If End G is commutative, let π i be the the quasi-projection of G onto G i and let π i be the corresponding idempotent in QEnd G. If ϕ Hom(G i,g j ) with j i then ϕπ i QHom(G, G j ) and thus ϕπ i corresponds to a map ϕ QHom(G, G) =QEndG given by ϕ(x) =ϕπ i (x). Since ϕ(g) G j and j i, π i ϕ= 0. But since by hypothesis QEnd G is commutative and ϕ = ϕ π i,weget0= π i ϕ= ϕ π i = ϕ. It follows that ϕ =0. Thus Hom(G i,g j )=0fori j. Likewise if ϕ, ψ QEnd G i then ϕ ψ = ψ ϕ and it follows that ϕψ = ψϕ. We noted in Chapter 3 that a Murley ring is indecomposable if and only if it is a dedekind domain. For E-rings in general we have an analogous result. 20 theorem AringRis an E-ring if and only if R is quasi-isomorphic to a finite product of strongly indecomposable integral domains D 1 D n such that Hom(D i,d j )=0 for i j. In particular, a strongly indecomposable E-ring is an integral domain. proof: ( ): By Lemma 7.33 it suffices to prove that if D 1 D n are strongly indecomposable domains with Hom(D i,d j )=0fori jthen D 1 D n is an E-ring. Now since the D i are strongly indecomposable, End D i D i by Corollary Thus D 1 D n End D 1 End D n End(D 1 D n ), where the latter isomorphism holds by the assumption that Hom(D i,d j )=0 for i j. Furthermore this isomorphism takes an element (d 1,...,d n ) D 1 D n to the endomorphism given by left multiplication by (d 1,...,d n ). Thus D 1 D n is an E-ring. ( ): If R is an E-ring then End R R and by Proposition 7.34 R is commutative. Thus by Proposition 7.35 R has a quasi-direct decomposition R 1 R n where the R i are strongly indecomposable pure W -submodules of R such that Hom(R i,r j )=0 for i j and R End R End R 1 End R n. It further follows from the condition Hom(R i,r j ) = 0 that then the subspaces QR i are invariant under left and right multiplication by elements of QR and hence are ideals in QR and so QR is the ring-theoretic product of the rings QR i.nowlete i be the identity element of QR i and let D i be the subring of QR i generated by R i together with e i.thenby Lemma 3.13 D i is quasi-equal to R i and is therefore also strongly indecomposable, and R is quasi-equal to D 1 D n. Furthermore the D i are integral domains by Corollary 7.29 [This requires separability!]. COMPOSITA. Although the question of determining the structure of the tensor product of two finite rank torsion free modules is a very difficult one, if the two modules are rings then their tensor product is also a ring, which narrows down the range of possibilities considerably. In the case of the tensor product of two integral domains, the problem reduces to a variation of a standard result in field theory.

21 definition Let D 1 and D 2 be integral domains, at least one of which is quasi-separable. By a compositum of D 1 and D 2 is meant a non-trivial torsion free (D 1 D 2 )-algebra C such that C is an integral domain and the structural map D 1 D 2 C is surjective. Note that the restrictions of the structural map to D 1 1 and 1 D 2 must be monic, since these maps extend to maps from the fields QD 1 1and 1 QD 2 to QC. We say that two composita are equivalent if they are isomorphic as (D 1 D 2 )-algebras. proposition Let D 1 and D 2 be integral domains, at least one of which is quasi-separable. Then (1) D 1 D 2 C i, where the C i form a maximal set of non-equivalent composita of D 1 and D 2. (2) D 1 D 2 C i where the product is taken over the set of C i which are not fields. (3) If Q is an algebraically closed extension of Q containing D1 and D 2 then we may choose C i = D 1 σ(d 2 ),whereσ:qd 2 Q is a morphism of Q-algebras. (4) If D 2 Q QD 1 where Q is a normal extension of Q, then D 1 D 2 D 1 σ(d 2 ),whereσ ranges over the set of all Q-algebra morphisms from QD 2 into Q. proof: (1) Partly this has already been covered in Proposition 7.19 since the result is well known for fields. Since either QD 1 or QD 2 is separable over Q, QD 1 QD 2 has trivial nil radical [Reference], hence is a product of commutative simple algebras, i.e. a product of fields, say QD 1 QD 2 = F i. Then the F i are composita of QD 1 and QD 2 and are clearly non-equivalent since the structural maps QD 1 QD 2 = F i F i have different kernels. By Proposition 7.19 D 1 D 2 C i where C i is the (D 1 D 2 )-submodule of F i generated by its identity element. Now clearly each C i is a compositum of D 1 D 2 and they are all non-equivalent. Furthermore if C is any compositum then the surjection D 1 D 2 C induces a ring morphism ϕ: QC i QC and since C is a domain Ker ϕ is a prime ideal, meaning that there is a unique i with ϕ(c i ) 0. Then there exists a commutative diagram D 1 D 2 C i QC i C QC. Now both left-hand (slanted) maps are surjections. And the right-hand vertical map is a non-trivial homomorphism of fields, hence is monic. It thus follows that the map C i C is an isomorphism of (D 1 D 2 )-algebras, so that C and C i are equivalent composita of D 1 and D 2. (2) By Corollary 7.17 each integral domain C i is either reduced or divisible. And by Proposition 7.13 it is divisible if and only if it is a field. Thus D 1 D 2 is the product of those C i which are not fields. 21