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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 Q0 be a finite field extension of Q and let W 0 be the integral closure 0 0 of W in Q . Suppose that there exists a prime ideal P1 of W and a prime p of W such 0 0 ≈ 0 p that WP1 /pWP1 W/p. Let R = WP1 .ThenRis -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). X

The hypothesis for Example 7.1 will hold, for instance, if the prime p “splits 0 0 completely” in W , i.e. pW = P1 ...Pn where P1,...,Pn are distinct prime ideals 0 0 0 0 0 1 ··· n ≈ i in W and n =[W :W]. In that case LpWPi = P P WLPi P WPi for all 0 0 0 0 0 0 i W /pW ≈ W /pW P ≈ W /pW and so by PropositionP 0.* ( ) i Pi Pi ,sothat 0 n 0 0 p-rank W = n = 1 p-rank WPi and necessarily p-rank WPi =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. 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 7→ rx. proof: It is well known that the map r 7→ r˜ is an isomorphism from R to EndR R. But EndR R ⊆ EndW R and in fact EndR R C EndW R. Indeed, if ϕ ∈ End R and wϕ ∈ EndR R for some w =06 ∈W,thenwϕ is R-linear and clearly ϕ itself must be W -linear so ϕ ∈ EndR R.Thusr7→ r˜ maps R onto a pure subring of End R. X

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 2.15. (2) By (1)TIT(R)andIT(R/N) are idempotent.T Thus by Proposition 2.3 IT(R)=t( XWp)andIT(R/N)=t( Y Wp)whereXis the set of prime ideals p such that pR =6 R and Y is the set of p with p(R/N) =6 R/N .Butp(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). X

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)]. X 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. X

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 7.5. X

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 rad R. Then by Proposition 1.31 N C 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. X

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 ⊃ ··· ⊃ Nr = 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. X 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. X

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 7.10. 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 QQ 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.   00 N = . Q0

2 Then it is easy to see that N is an ideal in R and N = 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 00 00 00 w 0 00 = , = . qq0 x0 q0x0 x0 qq0 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 Proposition 7.8 R is left noetherian but not right noetherian. X 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 (w1,g1)·(w2,g2)=(w1w2,w1g2+w2g1). One can note that G is the nil radical of this ring.

example 7.11. 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 =0.6 On the other hand, if one lets R = W × Q then as seen in Example 7.10, nil rad End R =6 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.) z 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 7.12. 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 radQR = 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

proposition 7.13. (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. X

We will need the following well known lemma from the theory of finite-dimensional algebras. lemma 7.14. 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 =6 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. X

proposition 7.15. 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 C 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 rad QR =0andQR 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 =0then 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

(4) ⇒ (3): If N = nil rad R then by Proposition 1.31 N is an ideal and N n =0, where n =rankR.NowifN=0then06

Condition (4) In Proposition 7.15 is the usual definition of a prime ring in general non-commutative ring theory. proposition 7.16. (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.ThuseG 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.ThusifGhas 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). X

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 7.17. 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

(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

corollary 7.18. 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. X

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 7.19. Let R be a finite rank torsion free ring, let Z be its center and N = nil rad R.ThenQR = QR1 ×···×QRk where each Ri is a subring of QR which is quasi-pure in R and Ri/(Ri ∩ QN) is a prime ring. Furthermore

(1) R is quasi-equal to R1 ×···×Rk. (2) If R is commutative and integrally closed then R = R1 ×···×Rn. (3) Z is quasi-equal to Z1 ×···×Zk,whereZi =CenterRi, with equality if Z is integrally closed. (4) N is quasi-equal to N1 ⊕···⊕Nk,whereNi =RiN. (5) Ni is quasi-equal to N ∩ Ri . (6) Ni = nil rad Ri . 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. Lete ¯1,...,e¯k be the central idempotents corresponding to this decomposition of QR/QN . By Lemma 1.34 thee ¯i are images of central idempotents ei ∈ QR. Write Ri = Rei .Notethat R⊆R1×···×Rk since every r ∈ R can be written as r = r1=re1 + ···+rek and note also that QR = Q(R1 ×···×Rk). Since R1 ×···×Rk is generated as an R-module by the finite set e1,...,ek it follows from Lemma 3.13 that R is quasi-equal to R1 ×···×Rk.In particular each Ri is a quasi-summand of R and hence by Proposition 3.21 is quasi-pure in R. z (2) (3) The center of R is quasi-equal to the center of R1 ×···×Rk,whichisZ1×···×Zk. (4) The same reasoning as in (1) shows that N is quasi-equal to e1N ⊕···⊕ekN. Furthermore eiN = eiRN = RiN = Ni . (5) Since ei ∈ QR there exists w =06 ∈Wsuch that wei ∈ R and so 2 wNi = weiN ⊆ N ∩ eiR = N ∩ Ri . On the other hand, since ei = ei , multiplication by ei restricts to the identity on eiR = Ri and hence also on N ∩ Ri .Thus N∩Ri=ei(N∩Ri)⊆eiN=Ni⊆Ri.ThusN∩Ri is quasi-equal to Ni . (6) By (3) Ni is a quasi-summand of R and hence is quasi-pure in R,sothatNi is quasi-equal to R ∩ QNi = R ∩ eiQN . By Proposition 1.32 QN = nil rad QR and so it is well known from the theory of artinian rings that eiQN = nil rad QRi .ThusNi is quasi-equal to R ∩ nil rad QRi ⊆ Ri ∩ nil rad QRi = nil rad Ri .SinceNi CRi (why?)it follows that Ni = nil rad Ri .

In particular, this shows that each Ri/(Ri ∩ QN)isaprimering. X

lemma 7.20. [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 Si such that Si/(nil rad Si) is a simple ring. Thus z no generality is lost in supposing that R/N is simple. 2) Suppose now that R is commutative of characteristic c =0.Then6 QR/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 QRc denote the set of all elements xc for x ∈ QR.Sincec=charR,Rc is a subring of R. Furthermore, since c ≥ 2andN2 =0,ifxc =0then6 x/∈QN and so x is invertible since QN is the unique maximal ideal of QR.ThusQRc is a subfield of QR.SinceQR is finite dimensional, there is a maximal subfield S of QR containing Rc . Since elements of QN are nilpotent, S ∩ QN = 0. We claim that S ⊕ QN = QR. If not, let x ∈ QR with x/∈S⊕QR.Thenxc ∈Rc ⊆S. Now the polynomial Xc − xc =(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

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 .NowletTbe the subring of QR such that QN r−1 ⊆ T and T/QN r−1 = S¯ ≈ QR/QN r−1 .ThenQN r−1 = nil rad T . Furthermore (QN r−1)2 = 0. Thus by the case r = 2 it follows that T contains a subring r−1 z S such that T = S ⊕ QN .ButthenQR = S ⊕ QN since **** X

proposition 7.21. 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 7.28.

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.ThusRis 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 7.22. 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 D0 is an integral domain containing D such that QD = QD0 then D0 is a finite integral extension of D if and only if D and D0 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 dimQD 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 D0 and QD = QD0 and D0 is a finite integral extension of D, i.e. a finitely generated D-module, then by Lemma 3.13 D0 is quasi-equal to D. ( ⇐ ): If D is a subring of D0 and D0 is quasi-equal to D then since D is a finitely generated D-module, by Lemma 7.5 D0 is also a finitely generated D-module, i.e. a finite integral extension of D. X

lemma 7.23. Let F be a finite separable extension of Q and let W 0 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 0R. (2) If D is an integral domain with quotient field F then W 0D is the integral closure of D. In particular, W 0D is a dedekind domain. proof: (1) If F is a separable extension of Q, then by Proposition 0.* W 0 is a finitely generated W -module. It follows that W 0R is a finitely generated R-module since any finite set of generators for W 0 as a W -module will also generate W 0R as a R-module. Since R ⊆ W 0R it follows from Lemma 3.13 that R is quasi-equal to W 0R. (2) Let D0 be the integral closure of D (in its quotient field F ). Then W ⊆ D0 and so W 0 ⊆ D0 and thus W 0D ⊆ D0 .ItiswellknownthatW0 is a dedekind domain (Proposition 0.*). Since W 0 ⊆ W 0D ⊆ QD = QW 0 , W 0D is a dedekind domain by Proposition 1.*. In particular W 0D is integrally closed, so since D ⊆ W 0D it follows that D0 ⊆ W 0D.ThusD0=W0D. X

theorem 7.24. 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 =EndF 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 Q0 of Q. Let D0 be the integral closure of D in Q0 .Then (7) F is the fixed field of {σ ∈ Gal(Q0/Q) | σ(D0)=D0}. 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 .Andif ϕ∈Center End D then in particular ϕ commutes with right multiplication by elements of D.Thusifforr∈Dwe write ρr for the map x 7→ xr then ϕ(r)=ϕ(1r)=ϕρr(1) = ρrϕ(1) = ϕ(1)r.Thusϕ=d˜where d = ϕ(1), i.e. given 13

the identification in (0) we can write ϕ = ϕ(1) ∈ D.ThusCenterEndD⊆Dso 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,sod∈Center QD.ThusF⊆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 ∆ = EndQEnd 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 .ThusQEndD≈EndF QD. (4) Since QEnd D =EndFQD, 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 .ThusDis a quasi-direct sum of a finite number of copies of F ∩ D,soDis 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)=EndF∩DF∩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 3.16. (7) If D is commutative and contained in a galois extension Q0 of Q,letD0 be the integral closure of D in Q0 .ThensinceQ0 is separable over QD, by Proposition 0.* D0 is a finite integral extension of D and hence by Lemma 7.22 is quasi-isomorphic to a direct sum of copies of D.ThusQEndD0 is isomorphic to a full matrix ring over QEnd D and so Center QEnd D0 =CenterQEndD=F and so F is also the field of definition for D0 . In order to show that F isthefixedfieldof{σ∈Gal(Q0/Q) | σ(D0) ⊆ D0},it suffices by Galois Theory to show that for σ ∈ Gal(Q0/Q), σ(D0) ⊆ D0 if and only if σ fixes F .Nowσ(D0) is isomorphic to D0 and therefore is also a dedekind domain, and in particular is integrally closed in Q0 . On the other hand since σ is monic, if σ(D0) ⊆ D0 then by Proposition 3.9 σ(D0)isquasi-equaltoD0 and therefore by Lemma 7.22 D0 is an integral extension of σ(D0). It follows that if σ(D0) ⊆ D0 then σ(D0)=D0. Now the field of definition for σ(D0)isσ(F) , so since the field of definition is determined uniquely by the ring D0 ,ifσ(D0)=D0 then σ(F )=F andthusalso 0 0 0 σ(F∩D)=F∩D,sothatσ restricts to an endomorphism σF ∈ End F ∩ D .But 0 0 by (5) End F ∩ D = F ∩ D ,sothatσF is given by multiplication by some f ∈ D ∩ F . 0 Since σF (1) = 1 we conclude that f =1andσF =1∈Gal(Q /Q). In other words, if σ(D0) ⊆ D0 then σ restricts to the identity map on D0 ∩ F . Conversely if σ restricts to 14

the identity map on D0 ∩ F then σ(D0)=D0 since by (4) D0 is the integral closure in Q0 of D0 ∩ F .SinceQ(D0∩F)=F, this shows that σ(D0) ⊆ D0 if and only if σ fixed F . X

definition 7.25. 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 7.26. 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 2 QEnd D ≈ EndF QD,sodimF QEnd D =(dimFQD) . This means that if QEnd D ≈ QD then dimF QD =1 so F =QD. (4) ⇒ (3): Immediate from part (6) of Theorem 7.24. (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.ThusQD = Q(F ∩ D)=F,soQD is the field of definition for D. (4) ⇒ (1): By part (5) of Theorem 7.24. Furthermore, if QD is the field of definition for D then QD is a field, so D is an integral domain. X

corollary 7.27. (1) Let D be an integral domain and let D0 be a finite integral extension of D.ThenD0 and D have the same field of definition. In particular, the field of definition for D0 is contained in QD. (2) Let D be a prime ring and K a subfield of the Q-algebra QD.ThenKcontains 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.IfF⊆Kthen 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 0 the field of definition for D0 .By Theorem 7.24 D is a finite integral extension of F ∩ D.ThusifD0 is a finite integral 15

extension of D then D0 is also a finite integral extension of F ∩ D and so by (2) F ⊇ F 0 . Thus in particular F 0 ⊆ QD.ThensinceD0 is a finite integral extension of F 0 ∩ D and F 0 ∩ D ⊆ D ⊆ D0 , a fortiori D is a finite integral extension of F 0 ∩ D and thus by (2) F 0 ⊇ F . X

corollary 7.28. 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 R1 ×···×Rn, where the Ri are prime rings and the center Z of R is quasi-equal to Z1 ×···×Zn,whereZi is the center of Ri . Thus by Lemma 7.5 Z1 ×···×Zn is a finitely generated Z-module and furthermore R is a finitely generated Z-module if and only if R1 ×···×Rn is a finitely generated Z-module. It follows that it suffices to prove that each Ri is a finitely generated Zi-module, for if Xi is a finite set of generators for Ri over Zi and Y is a finite set of generators for Z1 ×···×Zn over Z ,thenR1×···×Rn is generated over Z by the set of all elements xy for y ∈ Y and x ∈ Xi 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 . X

We can now prove Theorem 7.21. 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 rad R 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 .NotethatsinceQN 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 0 be the integral closure of W in the field S .ThusW0⊆R∩Sif R ∩ S is integrally closed, and in general W 0R ⊆ S ⊆ QR. Since by assumption S is separable over Q, by Lemma 7.23 W 0R is quasi-equal to R and (W 0R) ∩ S is quasi-equal to R ∩ S . It thus suffices to prove that W 0R is quasi-equal to (W 0R ∩ S) ⊕ W 0N . I.e. without loss of generality we may suppose W 0 ⊆ R.Butthen we may as well replace the ground ring W by W 0 , 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 R1 ×···×Rn, where for each i, Ri is a commutative prime ring, i.e. an integral domain, and if R ∩ S is integrally closed then R = R1 ×···×Rn and all the Ri ∩ S are dedekind domains. Furthermore N is quasi-equalQ to N1 ⊕···⊕Nn where Ni = nil rad Ri .NotealsothatR∩Sis quasi-equal to (Ri ∩ S)(why?). It thus suffices to prove that for each i, Ri is quasi-equal to (Ri ∩ S) ⊕ Ni , with equality if Ri 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 . X

corollary 7.29. 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 = nil rad R. By Theorem 7.21 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=0since16 ∈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. X

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 Q0 of Q.ButifDis integrally closed and W 0 is the integral closure of W in Q0 ,thenW0 ⊆Dso that D is a rank-one W 0-module. Thus we get the following proposition. 17

proposition 7.30. (1) A finite rank torsion free quasi-separable dedekind domain D is determined up to a quasi-equality by a finite field extension Q0 of Q and a subset Y of Spec W 0 ,whereW0 is the integral closure of W in Q0 . (Recall that according to our 0 0 convention,T Spec W denotes the set of non-trivial prime ideals of W ). Specifically, 0 0 0 D = Y WP where Y is the set of prime ideals P of W such that DP =6 Q . (2) If Q0 is a separable extension of Q then a quasi-equality class of essential subrings of Q0 contains exactly one integrally closed subring D. Furthermore every subring of Q0 quasi-equal to D is a subring of D. (3) Let Q0 be a galois extension of Q and let Y be a subset of Spec W 0 . Let G be the largest subgroup of Gal(Q0/Q) under which Y is invariant, i.e.

G = {σ ∈ Gal(Q0/Q) | (∀P ∈ Y ) σ(P ) ∈ Y }. T 0 Then the field of definition for Y WP is the fixed field of G . proof: (1) Let D be a finite rank torsion free quasi-separable integral domain and let Q0 = QD.ThenQ0 is a finite separable extension of Q. Let W 0 be the integral closure of W in Q. Then as indicated in the remarks preceding the Proposition, the integral closure of D0 in Q is a rank-one W 0-module of idempotent type and by Proposition 7.23 is quasi-equal to D. Furthermore by Proposition 2.2 D0 is uniquely determined by its type over W 0 and hence no two integrally closed subrings of Q0 can be quasi-equal as W 0-modules. But this in fact means that they cannot be quasi-equal as W -modules, since if w =06 ∈W is such that wD1 ⊆ D2 and wD2 ⊆ D1 ,whereD1 and D2 are 0 0 0 W -submodules of Q ,thenD1 and D2 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 0 , where a domain D corresponds to the set of prime ideals P of W 0 0 such that D is not P -divisible, i.e. DP =6 Q . (2) If D is an essential subring of Q0 and D0 is its integral closure, then D ⊆ D0 and by Proposition 7.22 D is quasi-equal to D0 . Furthermore if two integrally closed essential subrings D1 and D2 are quasi-equal then by Lemma 3.13 they are quasi-equal to D1D2 . Thus by Proposition 7.22 D1D2 is an integral extension of D1 and D2 . Since these are integrally closed,T D1 = D2 = D1D2 . 0 0 (3) If D = P ∈Y WP 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 7.24. X

We say that two rings R1 and R2 are quasi-isomorphic as rings if there is a ring isomorphism from R1 ontoasubringofQR2 whichisquasi-equaltoR2.

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 7.31. Let D1 and D2 be integral domains. (1) If D1 and D2 are strongly indecomposable then they are isomorphic as W -modules if and only if they are isomorphic as rings. 18

(2) If D1 and D2 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 D1 and D2 are quasi-isomorphic as W -modules if and only if they have the same rank, their fields of definition F1 and F2 are isomorphic fields, and F1 ∩ D1 and F2 ∩ D2 are quasi-isomorphic as rings.

proof: (1) Clearly if D1 and D2 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 D1 ≈ End D1 ≈ End D2 ≈ D2 since isomorphic modules have isomorphic endomorphism rings. (3) ( ⇒ ): If D1 ∼ D2 then clearly rank D1 =rankD2.AndifFi is the field of definition for Di then by Theorem 7.24 Fi is the center of QEnd Di and QEnd D1 ≈ QEnd D2 by Proposition 3.8, so F1 ≈ F2 . Furthermore F1 ∩ D1 and F2 ∩ D2 are strongly indecomposable by Theorem 7.24 and each Di is quasi-isomorphic to a direct sum of copies of Fi ∩ Di .SinceD1∼D2 it follows from J´onsson’s Theorem (Theorem 3.24) that F1 ∩ D1 ∼ F2 ∩ D2 . But then as rings

F1 ∩ D1 ≈ End(F1 ∩ D1) ∼ End(F2 ∩ D2) ≈ F2 ∩ D2

by Theorem 7.24 and Proposition 3.8. ( ⇐ ): If F1 ∩ D1 and F2 ∩ D2 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 Di is quasi-isomorphic to a direct sum of copies of Fi ∩ Di it follows that if rank D1 =rankD2 then D1 ∼ D2 . (2) If D1 and D2 are strongly indecomposable then by Corollary 7.26 their fields of definition are QD1 and QD2 . Therefore by (3) if they are quasi-isomorphic as W -modules then they are quasi-isomorphic as rings. Therefore we may suppose that D1 and D2 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 D1 and D2 are integrally closed then they are equal. X

proposition 7.32. 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.SinceDis 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. X 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 0 is the integral closure of W in some proper separable extension field of Q then W 0 is a projective W -module and hence quasi-isomorphism to a free W -module W n ,forn>1. Then QEndW0 is the ring of n × n matrices over Q so that End W 0 is not commutative, and hence certainly is much larger than W 0 . 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 7.33. (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 R1 × R2 is quasi-equal to an E-ring then R1 and R2 are E-rings. Furthermore Hom(R1,R2)=0. proof: (1) Recall that by Proposition 7.2 R C End R.ThusifQR =QEndRthen R = QR ∩ End R =QEndR∩End R =EndR,soRis an E-ring. (2) If R and R0 are quasi-isomorphic then we may suppose them quasi-equal. Then QR = QR0 and QEnd R =QEndR0. 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)wemayassumethatR1×R2 is an E-ring. Now any endomorphism ϕ of R1 extends (non-uniquely) to an endomorphism of R1 × R2 and hence since R1 × R2 is an E-ring is given by multiplication by some (r1,r2)∈R1 ×R2. It follows that ϕ is given by multiplication by r1 . Thus the canonical embedding R1 → End R1 is surjective, so R1 is an E-ring. Furthermore if ψ ∈ Hom(R1,R2)thenψextends to an endomorphism of R1 × R2 , which must be given by multiplication by some r ∈ R1 × R2 . But since R1 is an ideal in R1 × R2 , rR1 ⊆ R1 and it follows that ψ =0.ThusHom(R1,R2)=0. X

Note that by Proposition 3.43 Murley rings are E-rings. For Murley rings, the following proposition was part of Proposition 3.43. proposition 7.34. 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 =1yieldsr=1r=s1=s.Thusxr = rx for all x ∈ R.Since this holds for all r ∈ R, R is commutative. X 20

proposition 7.35. Let G be quasi-equal to G1 ⊕···⊕Gn, where the Gi are strongly indecomposable. Then End G is commutative if and only if End Gi is commutative for each i and Hom(Gi,Gj)=0 for i =6 j . In this case, End G ∼ End G1 ×···×End Gn . proof: ( ⇐ ): Easy since if G is quasi-equal to G1 ⊕···⊕Gn and Hom(Gi,Gj)=0for i=6 j then End G ∼ End G1 ×···×End Gn . ( ⇒ ): If End G is commutative, let πi be the the quasi-projection of G onto Gi and letπ ˜i be the corresponding idempotent in QEnd G.Ifϕ∈Hom(Gi,Gj) with j =6 i then ϕπi ∈ QHom(G, Gj) and thus ϕπi corresponds to a mapϕ ˜ ∈ QHom(G, G)=QEndG given byϕ ˜(x)=ϕπi(x). Sinceϕ ˜(G) ⊆ Gj and j =6 i,˜πiϕ˜= 0. But since by hypothesis QEnd G is commutative andϕ ˜ =˜ϕπ˜i,weget0=˜πiϕ˜=˜ϕπ˜i=˜ϕ. It follows that ϕ =0. Thus Hom(Gi,Gj)=0fori=6 j. Likewise if ϕ, ψ ∈ QEnd Gi thenϕ ˜ψ˜ = ψ˜ϕ˜ and it follows that ϕψ = ψϕ. X

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.

theorem 7.36. AringRis an E-ring if and only if R is quasi-isomorphic to a finite product of strongly indecomposable integral domains D1 ×···× Dn such that Hom(Di,Dj)=0 for i =6 j . In particular, a strongly indecomposable E-ring is an integral domain. proof: ( ⇐ ): By Lemma 7.33 it suffices to prove that if D1 ×···×Dn are strongly indecomposable domains with Hom(Di,Dj)=0fori=6 jthen D1 ×···×Dn is an E-ring. Now since the Di are strongly indecomposable, End Di ≈ Di by Corollary 7.26. Thus D1 ×···×Dn ≈ End D1 ×···×End Dn ≈ End(D1 ⊕···⊕Dn), where the latter isomorphism holds by the assumption that Hom(Di,Dj)=0 for i=6 j. Furthermore this isomorphism takes an element (d1,...,dn)∈D1 ×···×Dn to the endomorphism given by left multiplication by (d1,...,dn). Thus D1 ×···×Dn 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 R1 ⊕···⊕Rn where the Ri are strongly indecomposable pure W -submodules of R such that Hom(Ri,Rj)=0 for i =6 j and R ≈ End R ∼ End R1 ×···×End Rn . It further follows from the condition Hom(Ri,Rj) = 0 that then the subspaces QRi 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 QRi .Nowletei be the identity element of QRi and let Di be the subring of QRi generated by Ri together with ei .Thenby Lemma 3.13 Di is quasi-equal to Ri and is therefore also strongly indecomposable, and R is quasi-equal to D1 ×···×Dn. Furthermore the Di are integral domains by z Corollary 7.29 [This requires separability!]. X 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 7.37. Let D1 and D2 be integral domains, at least one of which is quasi-separable. By a compositum of D1 and D2 is meant a non-trivial torsion free (D1 ⊗ D2)-algebra C such that C is an integral domain and the structural map D1 ⊗ D2 → C is surjective. Note that the restrictions of the structural map to D1 ⊗ 1 and 1 ⊗ D2 must be monic, since these maps extend to maps from the fields QD1 ⊗ 1and 1⊗QD2 to QC . We say that two composita are equivalent if they are isomorphic as (D1 ⊗ D2)-algebras. proposition 7.38. Let D1 and D2 be integral domains, at least one of which is Then quasi-separable. Q (1) D1 ⊗ D2 ∼ Ci , where the Ci form a maximal set of non-equivalent composita of D1 and D2 . Q0 (2) D1 ?D2 ∼ Ci where the product is taken over the set of Ci which are not fields. (3) If Q¯ is an algebraically closed extension of Q containing D1 and D2 then we may choose Ci = D1σ(D2) ,whereσ:QD2 → Q¯ is a morphism of Q-algebras. 0 0 (4) If D2 ⊆ Q Q⊆ QD1 where Q is a normal extension of Q,then D1 ⊗D2 ∼ D1σ(D2),whereσ ranges over the set of all Q-algebra 0 morphisms from QD2 into Q . proof: (1) Partly this has already been covered in Proposition 7.19 since the result is well known for fields. Since either QD1 or QD2 is separable over Q, QD1 ⊗ QD2 has trivial nil radical [Reference], hence isQ a product of commutative simple algebras, i.e. a product of fields, say QD1 ⊗ QD2 = Fi . Then the Fi are composita of QQD1 and QD2 and are clearly non-equivalent since the structuralQ maps QD1 ⊗ QD2 = Fi → Fi have different kernels. By Proposition 7.19 D1 ⊗ D2 ∼ Ci where Ci is the (D1 ⊗ D2)-submodule of Fi generated by its identity element. Now clearly each Ci is a compositum of D1 ⊗ D2 and they are all non-equivalent. Furthermore if CQ is any compositum then the surjection D1 ⊗ D2 → C induces a ring morphism ϕ: QCi → QC and since C is a domain Ker ϕ is a prime ideal, meaning that there is a unique i with ϕ(Ci) =6 0. Then there exists a commutative diagram ⊆ Ci −−−−→ QCi     D1 ⊗ D2 y y

⊆ 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 Ci → C is an isomorphism of (D1 ⊗ D2)-algebras, so that C and Ci are equivalent composita of D1 and D2 . (2) By Corollary 7.17 each integral domain Ci is either reduced or divisible. And by Proposition 7.13 it is divisible if and only if it is a field. Thus D1 ?D2 is the product of those Ci which are not fields. 22

(3) Since QC is a finite dimensional extension field of Q we may choose C so that QC ⊆ Q¯ . Furthermore, since C is a compositum of D1 and D2 , the composition D1 → D1 ⊗ D2 → C is a monomorphism, hence we may choose C so that it is an inclusion. Then we get a morphism of W -algebras σ : D2 → D1 ⊗ D2 → C such that C = D1σ(D2). 0 0 (4) If D2 ⊆ Q ⊆ D1 and Q is a normal extension of Q,thenσ(D2)⊆QD1 .As we have seen, the composita may be chosen in the form D1σ(D2). Now suppose two composita D1σ(D2)andD1τ(D2) are equivalent. We then have a commutative diagram

⊆ 1 ⊗ σ D1σ(D2) −−−−→ QD1   D1 ⊗ D2 y

⊆ 1 ⊗ τ D1τ(D2) −−−−→ QD1.

Since these maps are all D1-linear, the vertical maps must be identity maps. Thus σ = τ so that the non-equivalent composita of D1 and D2 are in one-to-one 0 correspondence with the Q-algebra morphisms from QD2 into Q . X

VALUATION RINGS. Recall that by a valuation ring on a field K is meant a valuation ring (in the present context, necessarily a discrete valuation ring) whose quotient field is K . In algebraic number theory, the field of definition for a discrete valuation ring V is called the decomposition field for V . The following two propositions are restatements of well known results in ramification theory. proposition 7.39. Let V be a strongly indecomposable quasi-separable discrete valuation ring. (1) There is exactly one prime ideal p of W such that V is not p-divisible. (2) V is a Murley ring. proof: (1) Let P be the unique (non-trivial) prime of V and let p = P ∩ W .Ifp0 is a prime of W and p0 =6 p then there exists w ∈ W such that w/∈p,sow/∈P and thus w is invertible in V . Therefore V = wV ⊆ p0V ⊆ V ,soV is p0-divisible. (2) Let F = QV .SinceV is strongly indecomposable, by Corollary 7.26 F is the field of definition for V . Let Q0 be a galois extension of Q containing F and let W 0 and V 0 be the the integral closures of W and V in Q0 . By Proposition 7.38 Y Y W 0 ⊗ V ≈ W 0σ(V )= σ(V)0,

where σ ranges over the set of distinct Q-algebra morphisms from F into Q0 ,andσ(V)0 denotes the integral closure of σ(V )inQ0. We can also think of σ as ranging over a set of representatives for the left coset space Gal(Q0/Q)/ Gal(Q0/F ) and we then have σ(V )0 = σ(V 0). Now Gal(Q0/Q) acts transitively on the family of σ(V 0) and since F is the field of definition of V 0 , by part (7) of Theorem 7.24 Gal(Q0/F ) is the stabilizer 23

of V 0 under this action. Thus the σ(V 0), and also the σ(V), are distinct. Now since σ(V ) is a maximal proper subring of σ(F ) it follows that if P 0 is a prime ideal of W 0 0 0 0 0 then σ(V ) ⊆ WP 0 if and only if σ(V ) ⊆ WP 0 if and only if σ(V )=σ(F)∩WP0. 0 0 0 0 0 0 Therefore for each P there is only one σ with σ(V ) ⊆ WP 0 .Butσ(V)P0 =WP0σ(V) 0 0 0 0 0 0 so σ(V )P 0 = WP 0 if σ(V ) ⊆ WP 0 and σ(V )P 0 = Q otherwise. Therefore, using Lemma 6.27, Y 0 0 0 0 p-rank V = P -rankW 0 W ⊗ V = P -rankW 0 σ(V )=1.

Since q-rank V =0forq=6 p∈Spec W and since by Corollary 7.17 V is reduced, it z follows that V is a Murley ring. [Rewrite!] X

proposition 7.40. Let Q0 be a finite galois extension of Q and let V 0 be a valuation ring on Q0 . Let p be the unique prime ideal of W such that pV 0 =6 V 0 and let F be the field of definition for V 0 .Then (1) V 0 is the only valuation ring on Q0 containing V 0 ∩ F . (2) If WF is the integral closure of W in F then there exist unique prime ideals 0 0 0 0 0 P ∈ Spec WF and P ∈ Spec W such that V = WP 0 and V ∩ F =(WF)P. 0 0 (3) P is the only prime ideal of W lying above P . Q (4) If W 0 is the integral closure of W in Q0 then W 0 ⊗ V ≈ σ(V 0),whereσ ranges over a set of representatives for the left cosets of Gal(Q0/Q)/ Gal(Q0/F ). (5) The valuation rings σ(V 0) in the above product are the only valuation rings on Q0 containing Wp . (6) If K is any subfield of Q0 containing Q then K ∩ V is indecomposable if and only if K ⊆ F . proof: (1) By Theorem 7.24 V 0 is integral over V 0 ∩ F and therefore is the integral 0 0 0 0 0 closure of V ∩ F in Q .ThusifV1 is any valuation ring on Q containing V ∩ F then 0 0 0 0 0 0 V1 ⊇ V . But by Proposition 0.* V is a maximal subring of Q . Therefore V1 = V so that V 0 is the unique valuation ring on Q0 containing V 0 ∩ F . (2) This follows from Proposition 2.*. 0 0 0 (3) If P1 is a prime ideal of W lying above P then W 0 is a discrete valuation ring P1 0 0 0 0 0 0 0 0 on Q and V ∩ F =(WF)P ⊆W 0. Therefore by (1) W 0 = V = WP 0 and so P1 = P . P1 P1 (4) Immediate from Proposition 7.38. T 0 0 0 0 0 (5) If V1 is a valuation ring on Q containing Wp then Wp ⊆ V1 .Now σ(V) 0 is invariant under Gal(T Q /Q), so by TheoremT 7.24 its field of definition is Q, 0 0 0 0 and it follows that σ(V )=Wp,sothat σ(V ) ⊆ V1.Nowforeachσ,if 0 0 0 0 0 0 0 0 V1 =6 σ(V )thenV1σ(V)=Q since V1 and σ(V ) are maximal subrings of Q . 0 Furthermore by Proposition 2.3 the set of Wp-subalgebras of Q is a distributive 0 0 0 lattice, so the assumptionT thatT V1 σ(V )=Q for all σ leads to the contradiction 0 0 0 0 0 0 0 0 0 0 V1 = V1 Wp = V1 σ(V )= V1σ(V )=Q. Therefore V1 = σ(V )forsomeσ. (6) ( ⇐ ): If K ⊆ F then K ∩ V C F ∩ V . Therefore by Proposition 7.39 K ∩ V is a Murley ring as well as an integral domain, hence is indecomposable by Proposition 3.46. 24

( ⇒ ): Since K ∩ V is a discrete valuation ring, if it is indecomposable then it has p-rank one by Proposition 7.39. Now the compositum (K ∩ V )(F ∩ V ) is a homomorphic image of (K ∩ V ) ? (F ∩ V ), hence also has p-rank one, hence is indecomposable, so that FK is its field of definition. But since F ∩ V ⊆ (K ∩ V )(F ∩ V ) ⊆ V , the field of z definition for (K ∩ V )(F ∩ V )isF (why?). Therefore if K ∩ V is indecomposable then FK = F, i.e. K ⊆ F . X

QUASI-SEPARABLE INTEGRAL DOMAINS ARE R-SPLIT. We will now show that every finite rank quasi-separable domain D is R-split for some finite rank quasi-separable splitting ring R. Since the proof is a bit intricate we begin with a special case.

Q Recall first that by Proposition 3.43 any splitting ring R can be written as a product Ri , where the Ri are dedekind domains. We will use this notation consistently in this sense in the remainder of the chapter.

lemma 7.41. Let K be a finite separable extension of Q and let {Vi}i∈I be a family of valuation rings on K all having the same field of definition F . Suppose further that for each prime p thereisatmostoneT T i∈I such that Vi is not p-divisible. Then F is the field of definition for Vi and Vi ∩ F is a Murley ring. proof: If F is the field of definitionT for the valuation ring Vi then by Proposition 7.39 ∩ ∩ ∩ Vi F is a Murley ring. NowT ( Vi F )p =(VTj F)p,whereVj is the unique Vj with 6 ∩ ∩ ∩ T(Vj)p = K .Thusp-rank( Vi F) = p-rank ( Vi F )p =p-rank(Vj F)p =1.Thus Vi∩F is a Murley ring and is an integral domain, hence by Proposition 3.44 is strongly indecomposable.T Furthermore by Theorem 7.24 Vi is an integral extension of Vi ∩ F . Thus if x ∈ Vi then for any i, x ∈ Vi , so the coefficients for the minimal polynomial for Tx over W lie in Vi ∩ F .T Hence the minimal polynomialT for x over F hascoefficientsin Vi∩F, showing that Vi is an integral extension of Vi ∩ F , and by theT separability ofT K is a finite integral extension (Proposition 0.*).T Hence by Corollary 7.27 Vi and Vi ∩ F have the same field of definition. Since Vi ∩ F is strongly indecomposable, by Corollary 7.26 this field of definition is F . X

lemma 7.42. Let D be a quasi-separable domain and suppose that all the valuation rings on QD containing D have the sameQ field of definition F . Then there exists a finite rank quasi-separable splitting ring R = Ri such that Ri ⊆ F for all i andsuchthatDis R-split. proof: By Lemma 7.23 D is quasi-equal to its integral closure so there is no loss of generality in supposing D integrally closed. Then by Proposition 0.* D is the intersection of the valuation rings containing it. By hypothesis, these all have the same field of definition F . We claim that F contains the field of definition for D.Infact, each valuation ring V containingT D is a finite integral extension of a valuation ring VF with QVF = F .IfDF = VF then D is an integral extension of DF , since if d ∈ D then d ∈ V for each valuation ring V and so since V is an integral extension 25 T of VF the minimal polynomial for d hascoefficientsin VF =DF. Furthermore since QD is separable, D is a finite integral extension of DF by Proposition 0.* and so by Corollary 7.27 F contains the field of definition for D. Now since by Lemma 7.22 D is quasi-equal to a direct sum of copies of DF it suffices to prove that DF is R-split for some R. Thus there is no loss of generality in supposing that F = QD. Thus by Proposition 7.39 if V is a valuation ring on F containing D then V is a Murley ring. Now for each prime p with p-rank D =6 0 choose a valuation ring Vp on F containingTD with p-rank Vp =6 0. (The subscript p here does not denote localization.) Let R1 = p Vp .ThenR1⊆Fand in particular R1 is quasi-separable and by Lemma 7.41 R1 is a Murley ring. Now repeat the same process using valuation rings on F containing D which were not utilized in the first step, thus constructing a second Murley ring R2 contained in F and containing D. Since for each prime p there can be at most [F : Q] valuation rings V on F with pV =6 V (why?), in this way we finally obtain a set of Murley rings R1,...,Rn with n ≤ [F : Q] such that for each valuation ring V containing D , D ⊆ Ri ⊆ V for some i.SinceDisT integrally closed, it is the intersection of the valuation rings containingL it. Thus D = Ri , so by Proposition 1.10 D is isomorphic to a pure submodule of Ri . If, as in the proof of Proposition 6.21, we let R˜i = Ri ⊕ Ai ,whereAi is an appropriate subring of Q,thenR˜i is a quasi-separable splitting ring and Ri is R˜i-split. Thus by Proposition 6.20 D is R-split for some finite rank quasi-separable R. X

proposition 7.43. Let D be an integral domain and let Q0 be a finite galois extension 0 ofQ Q.ThenQ contains the field of definition for D if and only if there is a splitting ring 0 Ri such that D is R-split and Ri ⊆ Q for all i. proof: ( ⇒ ): Let F be the field of definition for D. Since by Theorem 7.24 D is quasi-isomorphic to a directQ sum of copies of F ∩ D, it suffices to prove that there exists a splitting ring R = Ri with all Ri ⊆ Q such that F ∩ D is R-split. In other words, there is no loss of generality in supposing that F = QD. Since by Lemma 7.22 D is quasi-equal to its integral closure, we may suppose D integrally closed. For each subfield E of QD let DE be the intersection of the valuation rings V on QD whose field of definitionQ is E . By Lemma 7.42 DE is R-split for some finite rank splitting 0 ring R = Ri with all RTi ⊆ Q . And since D is the intersection of the valuation rings containingL it, D = DE , and so by Proposition 1.10 D is isomorphic to a pure submodule of E DE .SinceQD is a separable extension of Q, there are only finitely many subfieldsQ E . Thus by Proposition 6.20 D is R-split for some finite rank splitting 0 ring R = Ri with Ri ⊆ Q for all i. Q 0 ( ⇐ ): Suppose that D is R-split for some splitting ring Ri with all Ri ⊆ Q . Since the field Q0D is separable over QD, by Proposition 0.* the integral closure D0 of D in Q0D is a finite integral extension of D. Hence by Proposition 7.27 the field of definition for D0 is the same as the field of definition for D, so that, replacing D 0 0 with D , noL generality is lost in assuming that Q ⊆ QD. Then by Proposition 7.38 0 Ri ?D ∼ σ(Ri)D, the product being taken over certain σ ∈ Gal(Q /Q). 26

(The Ri-module structure on σ(Ri)D is given via σ , i.e.Q (r, x) 7→ σ(r)x.) If 0 ϕ ∈ EndQ0 QDL then since σ(Ri) ⊆ Q , ϕ extends to a Ri-linear endomorphism of Q(R?D)= Qσ(Ri)D. Therefore by Proposition 6.4 EndQ0 QD ⊆ QEnd D =EndF QD where F is the field of definition for D and the latter equality is by Theorem 7.24. In particular, a Q0-linear projection QD → Q0 is F -linear, hence restricts to the identity on F .ThusF⊆Q0. X

WHEN IS A MODULE ISOMORPHIC TO AN INTEGRAL DOMAIN? We can now give two characterizations of those W -modules which are quasi-isomorphic to integral domains.

theorem 7.44. Let F be a finite separable extension of Q and G a W -module. Then G is quasi-isomorphic to an essential subring of F if and only if the following three conditions hold: (1) rank G =[F :Q]. (2) There is an embedding of rings F → QEnd G. (3) G is R-split for some finite rank splitting ring R.

proof: ( ⇒ ): If G is an essential subring of F then rank G =dimF =[F:Q]. Furthermore by Proposition 7.2 G ⊆ End G and F = QG ⊆ QEnd G. Finally, by Proposition 7.43 G is R-split for some finite rank splitting ring R. Thus conditions (1), (2) and (3) hold. ( ⇐ ): If conditions (1) and (2) hold then QG is a one-dimensional vector space over F andwemaysupposeQG = F . Let W 0 be the integral closure of W in F .By separability W 0 is a finitely generated W -module and since G is an essential submodule of W 0G, W 0G is quasi-equal to G by Lemma 3.13. Therefore without loss of generality we may suppose that G = W 0G, i.e. that G is a W 0-module. Hence if G is R-split then by Proposition 6.28 W 0 ?R is a splitting ring over W 0 and G is (W 0 ?R)-split. In other words, without loss of generality we may replace the ground ring W by W 0 and hence suppose that F = QQand G ⊆ Q.ThenGis a rank-one module and G C R?G so t(G) ∈ T(R?G)=T( Ri?G). But Ri ?G is a finitely generated Ri-module, henceQ is quasi-isomorphic to a direct sum of copies of Ri , and hence by Proposition 2.24 T( Ri ?G) is contained in the lattice generated by the sets T(Ri). But since Ri is an integral domain, by Corollary 7.17 T(Ri) consists of a single idempotent type.Q Since by Proposition 2.15 the set of idempotent types is a lattice, it follows that T( Ri ?G) consists of idempotent types, so t(G) is idempotent and thus G is quasi-isomorphic to a subring of Q by Proposition 2.15. X

note: In the proof of the sufficiency above it is not necessary to assume that R has finite rank.

We will see in Chapter 8 that condition (3) above can be replaced by the weaker requirement that G is quotient divisible. 27 Q theorem 7.45. Let R = Ri be a splitting ring, where each Ri is a finite rank quasi-separable dedekind domain, and let Q0 be a finite galois extension of Q such that 0 0 Ri ⊆ Q for all i. Let G be a strongly indecomposable R-split W -module and let W be the integral closure of W in Q0 .ThenGis quasi-isomorphic to an integral domain if and only if W 0 ⊗ G is quasi-isomorphic to a direct sum of rank-one W 0-modules of idempotent type. proof: ( ⇒ ): If G is strongly indecomposable and quasi-isomorphic to an integral domain D then by Proposition 7.26 the field of definition for DQis QD. Hence 0 by Proposition 7.43 if G (and hence D )isR-split and if R = Ri and Q is a 0 galois extensionQ of Q containing all the Ri then D ⊆ Q . By Proposition 7.38 W 0 ⊗ D ∼ W 0σ(D)asW0-algebras, where the σ range over a certain subset of Gal(Q0/Q). Since σ(D) ⊆ Q0 = QW 0 , W 0σ(D) is an essential W 0-subalgebra of Q0 , hence by Proposition 2.15 is a rank-one W 0-module of idempotent type. Since G ∼ D, 0 0 0 0 WL ⊗ G and W ⊗ D are quasi-isomorphic W -modules, so W ⊗ G is quasi-isomorphic to 0 0 W σ(D), a direct sumL of rank-one W -modules of idempotent type. 0 0 ( ⇐ ): If W ⊗ G ∼ Bk where the Bk are rank-one W -modules of idempotent type 0 then by Proposition 2.15 the Bk can be taken to be subrings of Q . Since by Lemma 7.22 W 0 is quasi-isomorphic to a free W -module, as a W -module W 0 ⊗ G is quasi-isomorphic to a direct sum of copies of G. Therefore by J´onsson’s Theorem (Theorem 3.24) it follows that G is quasi-isomorphic to a strongly indecomposable quasi-summand of B1 . But by Theorem 7.24 B1 is quasi-isomorphic a direct sum of copies of the strongly indecomposable module B1 ∩ F ,whereF is the field of definition for the ring B1 .Thus Gis quasi-isomorphic to the integral domain B1 ∩ F . X

STRONGLY HOMOGENEOUS MODULES. Recall from Corollary 7.17 that integral domains are homogeneous. This says, of course, that any two pure rank-one submodules of an integral domain are quasi-isomorphic. The following condition is much stronger:

definition 7.46. A strongly indecomposable module G is called strongly homogeneous if G is homogeneous and every non-trivial homomorphism ϕ: A → G,whereAis a pure rank-one submodule of G, extends to an endomorphism of G. lemma 7.47. If G is strongly homogeneous then End G is a prime ring. proof: Let G be strongly homogeneous. We will show that G has no non-trivial proper pure fully invariant submodules. If H is a pure fully invariant submodule of G and H =0,let6 A be a pure rank-one submodule of H and let B be any rank-one submodule of G.SinceGis homogeneous t(A)=t(B) and so by Proposition 2.4 there exists a non-trivial homomorphism ϕ: A → B .Thenϕcan be thought of as a mapping in Hom(A, G), so by hypothesis it extends to an endomorphism ψ of G.SinceHis fully invariant, ψ(A) ⊆ ψ(H) ⊆ H and thus B ∩ H =6 0 and so by Proposition 2.2 B ⊆ H . Since this is true for every rank-one submodule B of G it follows that H = G.Thus 28

G has no non-trivial proper pure fully invariant submodules and so by Proposition 7.16 End G is prime. X

lemma 7.48. Let G be a strongly indecomposable module such that End G is prime. (1) Let A be a pure rank-one submodule of G and let ρ:EndG→Hom(A, G) be the map such that ρ(ϕ) is the restriction of ϕ to A.Thenρis a monomorphism. (2) G is strongly homogeneous if and only if ρ is an isomorphism. (3) If G = A ⊗ G0 ,whereAis a rank-one module with [t(A): t(A)] ≤ IT(G0),then Gis strongly homogeneous if and only if G0 is strongly homogeneous. proof: (1) If ρ(ϕ) = 0 then the restriction of ϕ to A is trivial so that ϕ is not monic, hence since G is strongly indecomposable ϕ ∈ nil rad End G by Proposition 3.26. Since End G is prime, ϕ =0.Thus ρ is monic. (2) The definition of strongly homogeneous can be stated as saying that G is strongly homogeneous if and only if ρ is surjective. Since ρ ismonicby(1),thisisequivalenttoρ being an isomorphism. (3) If G ≈ A ⊗ G0 where A is a rank-one module and if A0 =EndAand t(A0)=[t(A): t(A)] ≤ IT(G0) then there is a commutative diagram ρ End G −−−−−→ Hom(A, G)=Hom(A⊗A0,A⊗G0)     ≈y ≈y

ρ0 End G0 −−−−−→ Hom(A0,G0)

where ρ0 is the corresponding restriction map and the vertical isomorphisms exist by Corollary 4.35. Thus ρ is an isomorphism if and only if ρ0 is an isomorphism and it follows from (2) that G is strongly homogeneous if and only if G0 is strongly homogeneous. X

lemma 7.49. If G is a strongly indecomposable strongly homogeneous module then End G is an E-ring and an integral domain. proof: Let A be a pure rank-one submodule of G and t = t(A). By Lemma 7.48 the restriction map ρ:EndG→Hom(A, G) is an isomorphism. Now note that if ϕ, ψ ∈ End G then ρ(ϕψ)=ϕρ(ψ). This says that ρ is an (End G)-linear isomorphism from End G to Hom(A, G). Now since G is homogeneous, T(G)={t}and G is t-saturated. Thus by Corollary 4.39 the map η :EndG→End Hom(A, G)given by η(ϕ)=ϕ∗ is an isomorphism. But η(ϕ) is simply multiplication by ϕ on the left (End G)-module Hom(A, G). Since End G and Hom(A, G) are isomorphic left (End G)-modules, it follows that the map λ:EndG→End End G taking ϕ to the map given by left multiplication by ϕ is also an isomorphism, so that End G is an E-ring. By Proposition 7.34 E-rings are commutative. And by Lemma 7.47 End G is a prime ring. But by Proposition 7.13 commutative prime rings are integral domains. Therefore End G is an integral domain. X 29

A (two-sided) ideal P in a not-necessarily-commutative ring R is called a prime ideal if whenever I and J are ideals of R such that IJ ⊆ R then either I ⊆ P or J ⊆ P . Note that by Proposition 7.13 a ring R is a prime ring if and only if the trivial ideal in R is a prime ideal. Maximal ideals are prime, since if P is maximal and I, J 6⊆ P then I + P = J + P = R and so IJ + P = R and thus IJ 6⊆ P .NotealsothatifRis commutative then the definition of a prime ideal above agrees with the usual definition in commutative ring theory. lemma 7.50. Let D be a prime ring. The following conditions are equivalent: (1) Every ideal in D has the form aD for some ideal a of W . (2) For every pure rank-one submodule A of D, DAD = D. (3) For every prime p ∈ Spec W , pD is a prime ideal in D. proof: (1) ⇒ (2): Suppose that every ideal in D has the form aD where a is an ideal of W and let A be a pure rank-one submodule of D.ThenDAD is an ideal in D so DAD = aD for some a ⊆ W .ThensinceA⊆DAD, A = A ∩ DAD = A ∩ aD = aA by purity and therefore also A = a−1A.ThusDAD = Da−1AD = a −1DAD = a−1aD = D. (2) ⇒ (1): Suppose that DAD = D for every pure rank-one submodule of D and let I be a non-trivial ideal in D.ThenIis a D-submodule of D and since D is prime, QD is a simple ring so that QI = QD. Therefore since D is a cyclic D-module, I is quasi-equal to D by Lemma 3.13. Thus if a = {w ∈ W | wD ⊆ I} then by Proposition 3.11 a is a non-trivial ideal in W .NowifAis any pure rank-one submodule of D,thenaA⊆aD∩A⊆I∩Aso that, in the notation of Chapter 2, 0 =6 a ⊆ [I ∩ A: A]. On the other hand, by Proposition 2.11 I ∩ A =[I∩A:A]A.Since by assumption D = DAD it follows that

[I ∩ A: A] D =[I∩A:A]DAD = D[I ∩ A: A] AD = D(I ∩ A)D ⊆ I so that [I ∩ A: A] ⊆ a .Thusa=[I∩A:A]andI∩A=[I∩A:A]A=aA=aD∩A, where the latter equality holds since A C D. Therefore I ∩ A = aD ∩ A holds for all pure rank-one submodules A of D.SinceDis the union of its pure rank-one submodules, it follows that I = aD. (1) ⇒ (3): If p ∈ Spec W and a is a proper ideal of W then pD is not properly contained in aD since p is maximal. Thus if every ideal in D has the form aD,for a⊆W,thenpD cannot be properly contained in any proper ideal of D, hence pD is maximal and therefore prime. (3) ⇒ (2): Let A be a pure rank-one submodule of D.IfDAD =6 D then since by Proposition 7.8 D is noetherian, there exists a maximal ideal I with DAD ⊆ I . Let p = W ∩ I .ThenpD ⊆ I and p must be a prime ideal of W since if x, y ∈ W and xy ∈ p then xD and yD are ideals in D and xDyD ⊆ pD ⊆ I ,sosinceI is prime either xD ⊆ I or yD ⊆ I , which implies x ∈ I ∩ W = p or y ∈ p.NowifpD is a prime ideal in D then D/pD is a finite length prime ring and hence is an artinian ring without ****, hence a simple ring. It follows that pD is maximal, and therefore I = pD.Butthen 30

A ⊆ DAD ⊆ I = pD and since A C D, pA = A ∩ pD = A,soAis p-divisible and thus DAD is p-divisible, contradicting the assumption that I = pD is a proper ideal. It thus follows that DAD = D. X

proposition 7.51. If G is a strongly indecomposable strongly homogeneous module then (1) End G is strongly homogeneous. (2) End G is a dedekind domain. (3) Every ideal of End G has the form aEnd G,wherea is an ideal of W . (4) If A is a pure rank-one submodule of G then the map µ: A ⊗ End G → G given by µ(a ⊗ ϕ)=ϕ(a) is an isomorphism. proof: (4) If A is a rank-one module and t = t(A) then by Proposition 4.34 the map σ : A ⊗ Hom(A, G) → G(t)isanisomorphism.NowifGis strongly homogeneous and A C G,thenGis homogeneous and so G(t)=G. Furthermore by Lemma 7.48 the restriction map ρ:EndG→Hom(A, G) is an isomorphism. Composing σ with A ⊗ ρ ≈ yields an isomorphism µ: A ⊗ End G −→ G given by µ(a ⊗ ϕ)=ϕ(a). (1) Let A be a pure rank-one submodule of G.Thenby(4)A⊗End G ≈ G is strongly homogeneous. Furthermore since T(G)={t(A)}, by Corollary 7.4 [t(A): t(A)] = IT(End G). Thus by part (3) of Lemma 7.48 End G is strongly homogeneous. (3) Let D =EndG. By Lemma 7.49 D is an integral domain. We claim that if A0 is any pure rank-one submodule of D then A0D = D. In fact, by (1) D is strongly homogeneous and so by (4) there is an isomorphism µ0 : A0 ⊗ End D → D given by µ0(x ⊗ ϕ)=ϕ(x). By by Lemma 7.49 D is an E-ring so the map λ: D → End D mapping d to left multiplication by d is an isomorphism. Thus there is a surjection 0 0 µ : A0 ⊗ D → D given by µ (x ⊗ d)=λ(d)(x)=dx = xd.ThusA0D=D, as claimed. It thus follows from Lemma 7.50 that every ideal of D has the form DaD = aD for some ideal a of W . (2) By Lemma 7.49 End G is an integral domain. And if I is an ideal in D =EndG then by (3) I = aD for some ideal a of W .IfI=0then6 a=0and6 a−1Dis a D-submodule of QD and a −1DaD = D. Thus every ideal of D is invertible, so D is a dedekind domain. X

theorem 7.52. A strongly indecomposable module G is strongly homogeneous if and only if G ≈ A ⊗ D,whereAis a rank-one module with locally trivial type and D is a strongly indecomposable dedekind domain such that every ideal in D has the form aD for some ideal a of W . proof: ( ⇒ ): By Proposition 7.51 if G is a strongly indecomposable strongly homogeneous module then G ≈ A0 ⊗ End G,whereA0 is a pure rank-one submodule of G and End G is a dedekind domain such that every ideal in End G has the form a End G for some ideal a of W . Now by Proposition 4.35 A0 ≈ A ⊗ [A0 : A0], for 31

0 some rank-one module A of locally trivial type. But if t = t([A0 : A0]) then since G is t(A0)-saturated (since G is homogeneous), it follows from Corollary 7.4 that 0 End G is t -saturated and so by Proposition 4.37 [A0 : A0] ⊗ End G ≈ End G.Thus G≈A⊗[A0:A0]⊗End G ≈ A ⊗ End G. ( ⇐ ): By part (3) of Lemma 7.48 it suffices to prove that if G = D is a domain such that every ideal in D has the form aD for some ideal a of W ,thenD is strongly homogeneous. By Corollary 7.17 domains are homogeneous. Now let B be a pure rank-one submodule of D and ϕ: B → D. By Proposition 2.2 the multiplication map µ: B ⊗ D → BD given by µ(b ⊗ d)=bd is an isomorphism and by Lemma 7.50 0 BD = DBD = D.NowletB =ϕ(B)∗. As with B there is an isomorphism 0 0 ≈ 0 0 −1 0 µ : B ⊗ D −→ D. Let ϕ = µ (ϕ ⊗ 1D)µ : D → D. Then the restriction of ϕ to B is given by b 7→ b ⊗ 1 7→ ϕ(b) ⊗ 1 7→ ϕ(b), so that ϕ0 extends ϕ.ThusDis strongly homogeneous. X

corollary 7.53. Let G and H be strongly indecomposable strongly homogeneous modules such that G ∼ H .IfWis a principal ideal domain or if G and H are integral domains, then G ≈ H . proof: By Theorem 7.52 G ≈ A ⊗ G0 and H ≈ B ⊗ H0 where A and B are rank-one modules with locally trivial types and G0 and H0 are dedekind domains. Furthermore t(A)IT(G0)=IT(G)=IT(H)=t(B)IT(B0) and since by Corollary 7.3 IT(G0)and IT(H0) are idempotent and by Proposition 2.19 the representation of a type as a product of a locally trivial type and an idempotent type is unique, it follows that t(A)=t(B) and so by Corollary 2.10 if W is a principal ideal domain then A ≈ B . Furthermore by Proposition 4.34 G0 ≈ Hom(A, A ⊗ G0) ≈ Hom(A, G) and likewise H0 ≈ Hom(B,H), so that G0 ∼ H0 and since G0 and H0 are strongly indecomposable integrally closed domains, G0 ≈ H0 by Proposition 7.31. Thus G ≈ H if W is a principal ideal domain or if G = G0 and H = H0 . X

proposition 7.54. If W is a principal ideal domain then a strongly indecomposable module G is strongly homogeneous if and only if for every two pure rank-one submodules z A and B of G there exists an automorphism ϕ of G with ϕ(A)=B. proof: ( ⇒ ): Suppose that G is strongly homogeneous and A, B are pure rank-one submodules of G.ThenGis homogeneous, so t(A)=t(B) and hence since W is a ≈ principal ideal domain, by Corollary 2.10 there is an isomorphism α: A −→ B .SinceGis strongly homogeneous, α extends to an endomorphism ϕ of G, and clearly ϕ(A)=B. By Lemma 7.49 End G is an integral domain so ϕ is not a zero divisor in End G and thus by Proposition 3.3 ϕ is monic. It remains to see that ϕ is surjective. It suffices to see that C ⊆ ϕ(G) for every pure rank-one submodule C of G. In fact, by the above there exists ψ ∈ End G with C = ψ(B)=ψϕ(A)=ϕψ(A) ⊆ ϕ(G), using the fact that by Proposition 7.51 End G is commutative. 32

( ⇐ ): Note that if for every two pure rank-one submodules A and B of G there exists an automorphism ϕ of G taking A onto B then every two pure rank-one submodules of G are isomorphic and it follows that G is homogeneous. Now let A be a pure rank-one submodule of G and 0 =6 α: A → G. Let B = α(A)∗ .Ifϕis an automorphism of G with ϕ(A)=Bthen ϕ−1α ∈ End A ⊆ Q (by Proposition 2.2), so ϕ−1α is given by multiplication by some w ∈ W . Then the restriction of wϕ = ϕw to A is ϕϕ−1α = α,so that wϕ is an endomorphism of G extending α. Thus if for every pair of pure rank-one submodules A and B of G there exists an automorphism of G taking A onto B then G is strongly homogeneous. X

By Proposition 7.40 if V 0 is a valuation ring on a galois extension Q0 then there is a unique largest subfield F of Q0 , called the decomposition field for V 0 , such that V 0 ∩ F is indecomposable. Likewise we now see that there is a unique largest subfield K of Q0 such that V 0 ∩ K is strongly homogeneous. K is traditionally called the inertial field of V 0 .

proposition 7.55. Let Q0 be a galois extension of Q and let V 0 be a valuation ring on Q0 . Let P 0 be the unique maximal ideal in V 0 . Let

T = {σ ∈ Gal(Q0/Q) | σ(V 0) ⊆ V 0 and (∀x ∈ V 0) σ(x) ≡ x (mod P 0)}.

Let K bethefixedfieldofT.ThenKis the largest subfield of Q0 such that K ∩ V 0 is strongly homogeneous. z proof: ****

Proposition. Let D be a strongly indecomposable strongly homogeneous principal ideal domain. If rank D = n then any two pure submodules of D with rank n − 1are isomorphic. proof: This has to do with the fact that every element of D can be written as wu,wherew ∈W and u is an invertible element of D.NowifK1 and K2 are pure submodules of D with corank 1 then Ki =Kerϕi for some ϕi ∈ Hom(D, Q). But Hom(D, Q) ≈ QD, so it follows that ϕ1 = uϕ2 for some u ∈ QD.Since Ker ϕ2 =Kerwϕ2 , we may assume that u is a unit in D .Butthen

uK1 = u Ker uϕ2 =Kerϕ2 =K2,

so K1 ≈ K2 . X

MULTIPLICATIONS ON A STRONGLY INDECOMPOSABLE MODULE. ThequestionofwhenaW-module G can be given the structure of an integral domain comes down to asking whether there is a map µ: G ⊗ G → G such that the resulting multiplication is associative and commutative and has an identity element. When Q is a perfect field it follows from Corollary 7.29 that if G is strongly indecomposable and µ is known to be associative and there exists an identity, then commutativity is automatic. We will now see that associativity is also automatic. 33

The most delicate point is actually the existence of an identity. We consider first the hypothesis that every non-trivial endomorphism of G is monic. (This is always the case for a strongly indecomposable integral domain D since by Corollary 7.26 End D is isomorphic to D and thus has no zero divisors.)

proposition 7.56. Let G be such that every non-trivial endomorphism is monic. Then any non-trivial distributive W -bilinear multiplication defined on G is associative and commutative. Furthermore this multiplication extends to a multiplication of QG making QG into a field, and G is quasi-equal to a subring of this field. Finally, G is actually isomorphic to an integral domain under the multiplication µ if and only if µ is surjective. proof: A distributive W -bilinear multiplication is induced by a map µ: G ⊗ G → G. We will write xy = µ(x ⊗ y). Now fix g =06 ∈Gand consider the map ϕ: G → G given by ϕ(x)=gx − xg .Thenϕ∈End G because multiplication is W -bilinear. And g ∈ Ker ϕ so ϕ is not monic. Hence by hypothesis ϕ = 0. This means ϕ(x) = 0, i.e. gx = xg for all x ∈ G. Thus multiplication is commutative. Now let g, h, x, ∈ G. Consider the mapping ψ ∈ End G given by ψ(x)=g2x−g(gx). Then ψ(g)=(g2)g−g(g2) = 0 since multiplication is commutative, so since g =0,6 ψis not monic, and as above ψ = 0. Now consider ζ ∈ End G defined by ζ(x)=(gx)h − g(xh). Then ζ(g)=g2h−g(gh)=ψ(h)=0,soζ is not monic, hence as above ζ = 0, i.e. (gx)h = g(xh) for all x ∈ G. Thus multiplication is associative. Next we claim that G has no zero divisors with respect to the given multiplication. In fact, if g, h =06 ∈Gand gh =0thenhis in the kernel of the map x 7→ gx,sothis map must be trivial. Then for any g0 ∈ G, g is in the kernel of the map x 7→ xg0 ,sothis map must be trivial, and we see that the given multiplication on G is completely trivial, contrary to the hypothesis. Thus G can have no zero divisors. Now the map µ: G ⊗ G → G extends to a map QG ⊗ QG → QG which must also be associative and commutative. For fixed g =06 ∈QG, the map x 7→ gx is monic and hence is an automorphism of QG, so there exists e ∈ QG with ge = g.Thenifθis defined by θ(x)=xe − x,andw=06 ∈W is such that we ∈ G,thenwθ(G) ⊆ G,sothat θ∈QEnd G.Sincegbelongs to the Ker θ , θ is not monic and hence must be trivial, i.e. xe = x for every x ∈ G,soeis an identity in QG.SinceQG is commutative and associative, it is an integral domain and since it is a finite dimensional algebra over the field Q, it must be a field. Now the only possible thing preventing G from being a subring of QG is the possibility that e/∈G. If this in fact occurs, note that the W -submodule G0 of QG generated by G together with e is quasi-equal to G by Proposition 3.1 and is a subring of the field QG, hence an integral domain. Note also that G is an ideal in G0 . Finally, if e ∈ G then for all g ∈ G, g = eg = µ(e ⊗ g), so µ is surjective. Conversely, to say that µ is surjective is to say that G2 = G.SinceGis an ideal in G0 and by Proposition 7.8 G0 is a noetherian integral domain, this implies that G = G0 [Kaplansky, Theorem 76 or Theorem 77, p. 50], so that G is an integral domain. X

The assumption that every non-trivial endomorphism is monic can be weakened 34

considerably under a separability assumption, still assuming that G is strongly indecomposable.

proposition 7.57. Let G be a strongly indecomposable module let N =nilradEndG. Suppose that End G/N is quasi-separable. Let µ: G ⊗ G → G be such that µ(G ⊗ G) 6⊆ NG. Then the multiplication induced on G by µ is associative and commutative without zero divisors so that G is quasi-equal to an integral domain D. Furthermore, G is actually an integral domain if and only if µ is surjective. proof: Choose g1 , g2 ∈ G such that µ(g1 ⊗ g2) ∈/ NG. Define π :EndG→Gby π(ϕ)=ϕ(g1). If π(ϕ)=0thenϕis not monic, hence by Proposition 3.26 ϕ ∈ N since G is strongly indecomposable. Thus Ker π ⊆ N . Now define ρ: G → End G by ρ(g)(x)=µ(x⊗g). Then πρ ∈ End G and πρ∈ / N since πρ(g2)=ρ(g2)(g1)=µ(g1⊗g2)∈/NG.ThussinceGis strongly indecomposable, πρ is a quasi-automorphism of G by Proposition 3.26. Hence π is a quasi-surjection. By Proposition 7.21 End G is quasi-equal to S ⊕ N ,whereS is a semi-prime ring. Since Ker π ⊆ N and G is quasi-equal to π(End G) and hence quasi-equal to π(S ⊕ N), it follows that G ∼ S ⊕ π(N). But G is strongly indecomposable and S =0,sowe6 conclude that π(N)=0andthatSis a strongly indecomposable semi-prime ring. By Proposition 7.29 S is an integral domain. Then G ∼ S and by Corollary 7.26 End G ∼ End S ≈ S .ThusEndGis an integral domain and N =0.SinceEndG has no zero divisors, by Proposition 3.3 every non-trivial endomorphism of G is monic, so the Proposition reduces to Proposition 7.56. X

One of the primary obsessions that has moved the theory of finite rank torsion free modules forward has been the attempt to find classes of modules with properties similar to those of rank-one modules. We have seen, for instance, that the class of indecomposable Murley modules shares some of the important properties of rank-one modules. We now see that the class of strongly indecomposable integral domains is also in some ways a generalization of the class of rank-one modules. In a way, this should not come as a big surprise. For if D is a quasi-separable integral domain then D is quasi-equal to its integral closure. And the integral closure is a rank-one module over the integral closure of W in QD.

The class of W -modules which admit an integral domain structure is a very important class of W -modules. To some extent, we now have a classification of such W -modules. A really satisfactory classification theorem for any type of algebraic structure has three components: (1) A way of constructing of the objects in question. (2) A way of determining whether two such objects are isomorphic in the category in question. (3) Some coherent way of listing (or parametrizing) all objects of the given sort. 35

In the case of the class of underlying W -modules of integral domains, one may add a fourth item: (4) A way of determining whether a given object belongs to the class in question or not. Proposition 7.30 provides us with a solutions to problems (1) and (2), (taking into account Proposition 7.31) inasmuch as it reduces the problem to that of finding all finite dimensional field extensions of Q andtheninagivenfieldextensionQ0 finding the set of prime ideals of the integral closure of W in Q0 . Neither of these two problems is by any means trivial. However since they are in the realm of such things as algebraic number theory and class field theory they are problems that module theorists cannot be expected to address. The fact that we cannot be expected to provide a parametrization for the set of finite dimensional field extensions of Q or the corresponding prime ideal spectra also suggests that there is not much more that can be done with respect to item (3) above. Theorem 7.44 and Theorem 7.45 provide an answer to problem (4). This answer is not the most satisfying that one could imagine, since it involves knowing things about the splitting ring for a module and its endomorphism ring, but it is difficult to see how there could be any more straightforward solution.