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 , ap-reduced module G with rankG > 1 and p-rankG = 1. Example 1.47, the Pontryaginmodule, was an example of such a construction, and it ultimately led us to the concept ofsplitting 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 closureof W in Q′ . Suppose that there exists a prime ideal P1 of W ′ and a prime p of W suchthat W ′

P1/pW ′

P1≈W/p . Let R = W ′

P1. Then R is p-reduced and p-rankR = 1.

proof: By hypothesis, p-rankR = 1. Furthermore R is an integral domain and p∞R isa pure ideal in R and so Qp∞R is an ideal in the field QR , which forces p∞R = 0 (seeCorollary 7.17 below). X

The hypothesis for Example 7.1 will hold, for instance, if the prime p “splitscompletely” in W ′ , i.e. pW ′ = P1 . . . Pn where P1, . . . , Pn are distinct prime idealsin W ′ and n = [W ′ : W ] . In that case pW ′

Pi= P1 · · ·PnW

′Pi

≈ PiW′Pi

for alli and so by Proposition 0.* W ′/pW ′ ≈ ⊕

(W ′/pW ′)Pi≈ ⊕

W ′Pi/pW ′

Pi, so that

p-rankW ′ = n =∑n

1 p-rankW ′Pi

and necessarily p-rankW ′Pi

= 1 for all i . From algebraicnumber 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 classof W -modules . Although we will see that Example 7.1 can be described, at least inprinciple, by the splitting ring construction (see Proposition 7.43), the above constructionhas the advantage of being simple and relatively uncontrived.

There are two questions that come up when we think about finite rank torsion freemodules 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 aW -module? And second, what does the W -module structure of a ring R tell us aboutthe ring-theoretic properties of R?

We have already seen several instances of answers to the second question in thediscussion 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 thatring must be an integral domain.

3

GENERAL RESULTS. A few elementary results of general interest are the following:

proposition 7.2. A ring R can be identified as a pure subring of EndR by the mapthat 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 EndRR .But EndRR ⊆ EndW R and in fact EndRR C EndW R . Indeed, if ϕ ∈ EndR andwϕ ∈ EndRR for some w 6= 0 ∈ W , then wϕ is R-linear and clearly ϕ itself must beW -linear so ϕ ∈ EndRR . Thus r 7→ r maps R onto a pure subring of EndR . X

proposition 7.3. If R is a finite rank torsion free ring and N = nil radR 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 , so Q ∩ R is a pure rank-one subring of R . Thust(Q ∩ R) ∈ T(R) , so IT(R) = inf T(R) ≤ t(Q ∩ R) . On the other hand, R is aQ∩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 byProposition 2.15.

(2) By (1) IT(R) and IT(R/N) are idempotent. Thus by Proposition 2.3IT(R) = t (

⋂X Wp) and IT(R/N) = t (

⋂Y Wp) where X is 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 . But p(R/N) = R/N ifand only if

R = pR +N = pR +RN = pR+ (pR +N)N = pR+ pN +N2 = pR +N2

and inductively we see that

p(R/N) = R/N ⇐⇒ R = pR +Nk for every k.

Since by Proposition 1.32 Nk = 0 for large k it follows that p(R/N) = R/N if and onlyif pR = R , so that X = Y and IT(R) = IT(R/N) . X

corollary 7.4. For any finite rank torsion free module G , IT(EndG) =[IT(G) : IT(G)] .proof: By Proposition 7.3 IT(EndG) = t(Q ∩ EndG) and clearly Q ∩ EndGis 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 ifG is t(A)-saturated, which is the same as saying that t(A) ≤ IT(G) . ThereforeIT(EndG) = t(Q ∩ EndG) is the largest idempotent type t such that t ≤ IT(G) . Butby 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 Mand N be R-submodules of U which are quasi-equal as W -modules. Then M is finitelygenerated 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 andN 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 Mtogether with a set of pre-images for generators of the W -module N/M . Therefore N isalso a finitely generated R-module. X

proposition 7.6. If R is a finite rank torsion free ring with nil radR = 0 then every leftideal 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 radR = 0then QR is a finite dimensional algebra with trivial radical, hence is semi-simple,so QL = QRe for some idempotent e ∈ QR . Now e is a right identity on QL soL = Le ⊆ Re . Then since L is an R-submodule of the finitely generated R-module Reand QL = QRe , Lemma 3.13 shows that L is quasi-equal to Re . But right multiplicationby e is an idempotent quasi-endomorphism of R , so Re is a quasi-summand of R , andthus L is a quasi-summand of R . Furthermore since L is quasi-equal to the principal leftideal 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 thatL ⊇ nil radR then L is quasi-pure in R .

proof: Let N = nil radR . Then by Proposition 1.31 N C R . Furthermore byProposition 7.6 L/N is a quasi-summand of R/N , hence by Proposition 3.21 L/N isquasi-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 ifnil radR is a finitely generated left ideal.

proof: ( ⇒ ): Clear since if R is noetherian then all left ideals are finitely generated.( ⇐ ): Let N = nil radR . Since nil radR/N = 0, it follows from Proposition7.6

that all left ideals of R/N are finitely generated so R/N is left noetherian. Now byProposition 1.31 there exists a filtration R ⊃ N ⊃ N2 ⊃ · · · ⊃ Nr = 0 for some finiter , where N = nil radR . The factors R/N and Nk/Nk+1 are all modules over thenoetherian 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 isnoetherian. X

5

corollary 7.9. An integral domain (over W ) is integrally closed if and only if it is adedekind 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 thenQP is a non-trivial ideal in the field QD , so QP = QD . Since P is a D-submodule ofD 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 inaddition 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 bealgebras 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 “leftnoetherian.” The following well known example shows that these two concepts are notequivalent.

example 7.10. The ring R = End(W ⊕Q) is left noetherian but not right noetherian.

proof: R is isomorphic to the matrix ring(W 0Q Q

), i.e. the ring of 2 × 2 matrices

such that the entries in the second row belong to Q , the upper left hand corner belongsto W and the upper right hand entry is 0. Now let N be the subset of R consisting ofthose matrices whose diagonal elements vanish, i.e.

N =(

0 0Q 0

).

Then it is easy to see that N is an ideal in R and N2 = 0. Furthermore

R/N ≈(W 00 Q

)≈ W × Q , a ring with trivial nil radical. Thus N = nil radR . Now

multiplication between elements of R and N operates as follows:

(w 0q q′

) (0 0x 0

)=

(0 0q′x 0

),

(0 0x 0

) (w 0q q′

)=

(0 0wx 0

).

Thus as a left ideal N is essentially just a Q-module, and since N ≈ Q it is finitelygenerated. On the other hand, as a right ideal it is essentially just a W -module , and sinceQ 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. X

6

PRIME & SEMI-PRIME RINGS. We are primarily interested in torsion free ringsbecause those W -modules which are the underlying modules of certain torsion free ringsare 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 Qplay a special role. Hence it is plausible to suspect that the underlying modules of ringsin general might be special.

Unfortunately, though, the mere fact that a finite rank torsion free module Gcan be given a ring structure imposes only mild restrictions on the structure of theW -module G . Indeed any G can be given the structure of a commutative “ring withoutidentity,” simply by making the multiplication trivial. If one now uses the standardconstruction to “unitize” G one gets a commutative W -algebra with the underlyingW -module isomorphic to W ⊕ G and containing G as a prime ideal. Explicitly, themultiplication on W ⊕ G is given by (w1, g1) · (w2, g2) = (w1w2, w1g2 + w2g1) . One cannote that G is the nil radical of this ring.

example 7.11. By applying the construction in the preceding paragraph in the caseG = W one gets a ring R such that R = W ⊕W . Thus EndR = End(W ⊕W ) so thatnil radEndR = 0. And yet nil radR = 0 ⊕W 6= 0.

On the other hand, if one lets R = W × Q then as seen in Example 7.10,nil radEndR 6= 0, but clearly nil radR = 0.

Thus there is in general no nice relationship between the nil radical of a ring and thenil radical of its endomorphism ring. (However see Proposition 7.16 below.)

Delete? As soon as one imposes the condition that nil radR = 0, the restrictionszon R as a W -module become much more stringent. For instance, we have seen inProposition 7.8 that in this case R is noetherian. Another important observation is thatif H is a fully invariant submodule of any ring R then in particular H is invariant underboth left and right multiplication by elements of R so that H must be an ideal in R andQH must also be an ideal in QR . If nil radR = 0 then QR is semi-simple and sincesemi-simple rings don’t have many (two-sided) ideals, then R can have few pure fullyinvariant submodules. (See Proposition 7.16 below.) Delete?

We shall see that the study of finite rank torsion free modules which can be given thestructure of rings with trivial nil radical reduces to the study of those with the structureof (commutative) integral domains. It turns out that these integral domains play a keyrole 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 radR = 0and prime if QR is simple (i.e. has no proper two-sided ideals). Note that since byProposition 1.32, nil radQR = Q(nil radR) , a ring R is semi-prime if and only if QR issemi-simple. (The reader should not confuse this usage with the use of the term “primering” by a few other (misguided!) authors to denote the rings Z and Z/pZ for primenumbers 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 withoutzero 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. Buta commutative ring without non-trivial proper ideals is a field, hence R is an integraldomain. Since QR is a field and is the localization of R with respect to a multiplicativeset in R , namely the set of non-trivial elements of W , it follows that QR is the quotientfield of R . X

We will need the following well known lemma from the theory of finite-dimensionalalgebras.

lemma 7.14. A finite dimensional algebra S over a field is a simple algebra if and only ifS has trivial radical and no proper central idempotents.

proof: ( ⇒ ): If e is a central idempotent in S then eS is a (two-sided) ideal. Likewisenil radS is an ideal. Thus if S is a simple ring then eS and nil radS must be eitherall of S or trivial. This implies that e = 1 or e = 0, and that nil radS = 0 (sincenil radS 6= S .)

( ⇐ ): If a finite dimensional algebra S has trivial radical then it is semi-simple, hencea product of simple algebras. If it has no non-trivial central idempotents then there canbe 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 areequivalent:

(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 ifand 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 properideals, 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 7.14QR if simple if and only if nil radQR = 0 and QR has no proper idempotents. But byProposition 1.32 nil radQR = Q(nil radR) and by definition nil radR = 0 if and only if Ris semi-prime.

(2) ⇒ (4): If IJ = 0 then I∗J∗ = 0. Thus without loss of generality we may supposethat I and J are pure. Now IJ = 0 is not possible if I = J = R , so one of I and Jmust be a proper pure ideal. Thus if R has no non-trivial proper pure ideals then eitherI or J must be trivial.

8

(4) ⇒ (3): If N = nil radR then by Proposition 1.31 N is an ideal and Nn = 0,where n = rankR . Now if N 6= 0 then 0 < rankN < n = rankR so n ≥ 2 and ifwe set I = J = Nn−1 then IJ = 0. Likewise if e is a central idempotent in QR andI = R ∩ QRe = R ∩ eQR , J = R ∩ QR(1 − e) = R ∩ (1 − e)QR , then I and J areideals in R and IJ = 0. Thus if there are no proper non-trivial ideals I and J withIJ = 0 then e = 0 or e = 1 so that R has no proper central idempotents and likewiseN = nil radR = 0. X

Condition (4) In Proposition 7.15 is the usual definition of a prime ring in generalnon-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 EndG is a prime ring.(4) If R is a prime ring then EndR is a prime ring.

proof: (1) If M is a fully invariant submodule of R then in particular M is invariantunder 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 = Ror M = 0.

(3) Let R = EndG . If N = nil radR then (NG)∗ is a fully invariant pure submoduleof G and by Proposition 1.32 rank(NG)∗ = rankNG < rankG . Thus if G has nonon-trivial proper pure fully invariant submodules except G then NG = 0. SinceN ⊆ EndG this implies that N = 0. Thus R is semi-prime.

Now if e is a central idempotent in QR = QEndG then by Proposition 3.16 G isquasi-equal to eG⊕ (1−e)G . Thus eG and (1−e)G are quasi-pure submodules of G andthey are fully invariant since if ϕ ∈ EndG = R then ϕ(eG) = ϕe(G) = eϕ(G) ⊆ eG andlikewise ϕ(1 − e)G ⊆ (1 − e)G . Thus if G has no non-trivial proper pure fully invariantsubmodules 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 7.15R 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-trivialproper pure fully invariant subgroups is called irreducible, the point being that QG isan irreducible (QEndG)-module. In module theory the term “irreducible” has beenalready 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 inR .

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 Proposition7.16 since by Proposition 1.4,Proposition 1.15 and Proposition 4.2, d(R) , p∞R , R(t) and R[t] are fully invariant puresubmodules of R .

(3) & (4) It follows from (1) and (2) that if R is prime then d(R) = 0 or d(R) = Rso that R is either reduced or divisible. Likewise p∞R = 0 or p∞R = R so R is eitherp-reduced or p-divisible. And for any type t , either R(t) = R or R(t) = 0. Thus ift > IT(R) then R(t) = 0, so we see that all elements of R have type IT(R) and R ishomogeneous. Furthermore by Proposition 7.3 IT(R) is idempotent. Likewise R[t] = Ror R[t] = 0, so R[t] = 0 if t < OT(R) and we see that R is cohomogeneous. X

corollary 7.18. If a prime ring is a Butler module then it is a homogeneous completelydecomposable module.

proof: By Corollary 7.17 a prime ring is homogeneous. By Proposition 5.* ahomogeneous Butler module is completely decomposable. X

QUASI-PRODUCTS OF RINGS. If R is semi-prime, then QR is semi-simple andso the center of QR is a finite product of fields. We say that R is quasi-separableif these fields are all separable extensions of Q . (Quasi-separability is defined only forsemi-prime rings.)

If R is a ring, then by Proposition 7.2 we may identify R as a subring of EndR . Itfollows that if e is an idempotent in R then eR is a summand of the W -module R . Andif e is an idempotent in QR then eR is a quasi-summand of R . This makes the followingresult not terribly surprising:

proposition 7.19. Let R be a finite rank torsion free ring, let Z be its center andN = nil radR . Then QR = QR1 × · · · ×QRk where each Ri is a subring of QR which isquasi-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 , where Zi = CenterRi , with equality if Z is

integrally closed.(4) N is quasi-equal to N1 ⊕ · · · ⊕Nk , where Ni = RiN .(5) Ni is quasi-equal to N ∩Ri .(6) Ni = nil radRi .

10

proof: (1) By Proposition 1.32 QN is the nil radical of QR . Therefore QR/QN is asemi-simple algebra and hence is a finite product of simple rings. Let e1, . . . , ek be thecentral idempotents corresponding to this decomposition of QR/QN . By Lemma 1.34the ei are images of central idempotents ei ∈ QR . Write Ri = Rei . Note thatR ⊆ R1 × · · · ×Rk since every r ∈ R can be written as r = r1 = re1 + · · · + rek and notealso that QR = Q(R1×· · ·×Rk) . Since R1×· · ·×Rk is generated as an R-module by thefinite set e1, . . . , ek it follows from Lemma 3.13 that R is quasi-equal to R1 × · · ·×Rk . Inparticular each Ri is a quasi-summand of R and hence by Proposition 3.21 is quasi-purein R .

(2)z(3) The center of R is quasi-equal to the center of R1×· · ·×Rk , which is Z1×· · ·×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 6= 0 ∈ W such that wei ∈ R and so

wNi = weiN ⊆ N ∩ eiR = N ∩ Ri . On the other hand, since e2i = ei , multiplicationby ei restricts to the identity on eiR = Ri and hence also on N ∩ Ri . ThusN ∩Ri = ei(N ∩Ri) ⊆ eiN = Ni ⊆ Ri . Thus N ∩Ri is quasi-equal to Ni .

(6) By (3) Ni is a quasi-summand of R and hence is quasi-pure in R , so that Ni isquasi-equal to R ∩ QNi = R ∩ eiQN . By Proposition 1.32 QN = nil radQR and so itis well known from the theory of artinian rings that eiQN = nil radQRi . Thus Ni isquasi-equal to R ∩ nil radQRi ⊆ Ri ∩ nil radQRi = nil radRi . Since Ni C Ri (why?) itfollows that Ni = nil radRi .

In particular, this shows that each Ri/(Ri ∩QN) is a prime ring. X

lemma 7.20. [Wedderburn Principal Theorem] Let N = nil radR . If R/N isquasi-separable or R is commutative then there exists a semi-simple subring S ⊆ QRsuch that QR = S ⊕QN .

proof: QR is a finite product of rings Si such that Si/(nil radSi) is a simple ring. Thusno generality is lost in supposing that R/N is simple.z

2) Suppose now that R is commutative of characteristic c 6= 0. Then QR/QN is acommutative simple ring, i.e. a field. By Proposition 1.* there exists r ≥ 1 such thatN r = 0. We use induction on r , this case r = 1 being trivial. Suppose now that N2 = 0.Let QRc denote the set of all elements xc for x ∈ QR . Since c = charR , Rc is asubring of R . Furthermore, since c ≥ 2 and N2 = 0, if xc 6= 0 then x /∈ QN and sox is invertible since QN is the unique maximal ideal of QR . Thus QRc is a subfield ofQR . Since QR is finite dimensional, there is a maximal subfield S of QR containingRc . Since elements of QN are nilpotent, S ∩ QN = 0. We claim that S ⊕ QN = QR .If not, let x ∈ QR with x /∈ S ⊕ QR . Then xc ∈ Rc ⊆ S . Now the polynomialXc − xc = (X − x)c must be irreducible over S since c = charS and x /∈ S . But thismeans that x is algebraic over S so that S[x] is a proper extension field of S containedin QR , contradicting the maximality of S . Thus S ⊕QN = QR , as desired.

11

Now suppose that r > 2. By induction we may assume that QR/QN r−1 contains asubfield S such that QR/QNr−1 = S ⊕ QN/QNr−1 . Now let T be the subring of QRsuch that QNr−1 ⊆ T and T/QNr−1 = S ≈ QR/QNr−1 . Then QNr−1 = nil radT .Furthermore (QNr−1)2 = 0. Thus by the case r = 2 it follows that T contains a subringS such that T = S ⊕QNr−1 . But then QR = S ⊕QN since **** Xz

proposition 7.21. Let N = nil radR and suppose R/N is quasi-separable. Let S bea semi-simple subalgebra of QR such that QR = S ⊕ QN . Then R ∩ S is semi-primeand R is quasi-equal to (R ∩ S) ⊕ N . Furthermore if R is commutative and R ∩ S isintegrally 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 ringis quasi-equal to a product of prime rings so that the study of semi-prime rings prettymuch comes down to the study of prime rings. In particular, a strongly indecomposablesemi-prime ring must in fact be a prime ring.

Now there exist prime rings which are not integral domains, for instance the ring R ofn × n matrices over W is prime since QR is the ring of n × n matrices over the fieldQ and hence a well known simple ring. However it turns out that every prime ring Rhas a subring D which is an integral domain and such that R is quasi-isomorphic to adirect sum of copies of D . Thus R is a finitely generated D-module . This result wasfirst proved in [Pierce], using steps from [B & P]. The proof is a fascinating amalgamof Wedderburn Theory, Galois Theory, commutative ring theory, as well as torsion freemodule theory.

Recall that if an an integral domain D is a subring of the center of a ring R then wesay 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 , thenR is a finite integral extension of D if and only if R is quasi-equal to a freeD-submodule of QR .

(2) If D′ is an integral domain containing D such that QD = QD′ then D′ is a finiteintegral 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 finiteintegral extension of D since by Lemma 7.5 a module quasi-equal to a finitely generatedD-module is itself finitely generated as a D-module.

( ⇒ ): Since R is a D-module, QR is a vector space over the field QD anddimQD QR ≤ rankR < ∞ . Let M be the D-module generated by a basis for QR overQD contained in R . Then since this basis is linearly independent over D , M is a freeD-module . Furthermore M ⊆ R and QM = QR . Thus if R is finitely generated over Dthen 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 extensionof 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 finitelygenerated D-module , by Lemma 7.5 D′ is also a finitely generated D-module, i.e. a finiteintegral extension of D . X

lemma 7.23. Let F be a finite separable extension of Q and let W ′ be the integralclosure 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 finitelygenerated W -module. It follows that W ′R is a finitely generated R-module since anyfinite 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′ andso W ′ ⊆ D′ and thus W ′D ⊆ D′ . It is well known that W ′ is a dedekind domain(Proposition 0.*). Since W ′ ⊆ W ′D ⊆ QD = QW ′ , W ′D is a dedekind domain byProposition 1.*. In particular W ′D is integrally closed, so since D ⊆ W ′D it follows thatD′ ⊆W ′D . Thus D′ = W ′D . X

theorem7.24. Let D be a finite rank torsion free prime ring.Let F = Center QEndD . Then

(0) QD can be naturally identified as a subring of QEndD .(1) F ⊆ CenterQD .(2) F ∩D = Center EndD .(3) QEndD = 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 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 ⊆ EndD and F = Center QEndD it follows that D ∩ F ⊆ Center EndD .

On the other hand, clearly Center EndD ⊆ Center QEndD = F . And ifϕ ∈ Center EndD then in particular ϕ commutes with right multiplicationby elements of D . Thus if for r ∈ D we write ρr for the map x 7→ xr thenϕ(r) = ϕ(1r) = ϕρr(1) = ρrϕ(1) = ϕ(1)r . Thus ϕ = d where d = ϕ(1), i.e. given

13

the identification in (0) we can write ϕ = ϕ(1) ∈ D . Thus Center EndD ⊆ D so thatCenter EndD = F ∩D .

(1) Since by (2) Center EndD ⊆ D it follows that F = Center QEndD ⊆ QD . Now ifd ∈ QD and d ∈ Center QEndD = F then d commutes with (left multiplication by) theother elements of QD , so d ∈ CenterQD . Thus F ⊆ CenterQD .

(3) By Proposition 7.16 EndD is a prime ring, so QEndD is a simple algebra. Weclaim that QD is a simple (QEndD)-module. In fact, if M is a (QEndD)-submoduleof QD then in particular M must be invariant under left and right multiplication byelements of QD so that M is a two-sided ideal in QD and since QD is a simple ring itfollows that either M = QD or M = 0.

It now follows from Jacobson’s version of the Wedderburn Theorem thatQEndD ≈ 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 ⊆ QEndD . It follows that ∆ consists of those mappings in QEndD which are(QEndD)-linear, i.e. ∆ = Center QEndD = F . Thus QEndD ≈ EndF QD .

(4) Since QEndD = EndF QD , the quasi-direct decompositions of D are the sameas the F -linear decompositions of QD . But as an F -space, QD is a direct sum of afinite number of copies of F . Thus D is a quasi-direct sum of a finite number of copies ofF ∩D , so D is quasi-isomorphic to a free F ∩D-module , and thus by Lemma 7.22 is afinite integral extension of F ∩D .

(5) Since D is quasi-isomorphic to a free F ∩D -module, any endomorphism of F ∩Dcan be extended to a quasi-endomorphism of D , hence by (3) is F -linear, and so afortiori is F ∩D -linear. Thus End(F ∩D) = EndF∩D F ∩D = F ∩D .

(6) By (5) QEnd(F ∩D) ≈ Q(F ∩D) , a field. Thus QEnd(F ∩D) has no non-trivialidempotents, so F ∩D is strongly indecomposable by Proposition 3.16.

(7) If D is commutative and contained in a galois extension Q′ of Q , let D′ be theintegral 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 toa direct sum of copies of D . Thus QEndD′ is isomorphic to a full matrix ring overQEndD and so Center QEndD′ = Center QEndD = F and so F is also the field ofdefinition for D′ .

In order to show that F is the fixed field of {σ ∈ Gal(Q′/Q) | σ(D′) ⊆ D′} , itsuffices by Galois Theory to show that for σ ∈ Gal(Q′/Q) , σ(D′) ⊆ D′ if and only ifσ 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, ifσ(D′) ⊆ D′ then by Proposition 3.9 σ(D′) is quasi-equal to D′ and therefore byLemma 7.22 D′ is an integral extension of σ(D′) . It follows that if σ(D′) ⊆ D′ thenσ(D′) = D′ . Now the field of definition for σ(D′) is σ(F ) , so since the field of definitionis determined uniquely by the ring D′ , if σ(D′) = D′ then σ(F ) = F and thus alsoσ(F ∩ D′) = F ∩ D′ , so that σ restricts to an endomorphism σF ∈ EndF ∩D′ . Butby (5) EndF ∩D′ = F ∩D′ , so that σF is given by multiplication by some f ∈ D ∩ F .Since σF (1) = 1 we conclude that f = 1 and σF = 1 ∈ Gal(Q′/Q) . In other words, ifσ(D′) ⊆ D′ then σ restricts to the identity map on D′ ∩ F . Conversely if σ restricts to

14

the identity map on D′ ∩ F then σ(D′) = D′ since by (4) D′ is the integral closure inQ′ of D′ ∩ F . Since Q(D′ ∩ F ) = F , this shows that σ(D′) ⊆ D′ if and only if σ fixedF . X

definition 7.25. If D is a prime ring then Center(QEndD) , identified as above with asubfield 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) EndD ≈ D .(2) QEndD ≈ 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

QEndD ≈ EndF QD , so dimF QEndD = (dimF QD)2 . This means that ifQEndD ≈ 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 afree F ∩D-module . If D is strongly indecomposable this implies that D is quasi-isorphicto F ∩D . Thus QD = Q(F ∩D) = F , so QD 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 anintegral domain. X

corollary 7.27. (1) Let D be an integral domain and let D′ be a finite integralextension of D . Then D′ and D have the same field of definition. In particular, the fieldof 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 containsthe 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 QEndD is isomorphic to thering of k × k matrices over QEnd(K ∩ D) . Therefore the center F of QEndD is thecenter 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′ . ByTheorem 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 . Then since D′ is a finite integral extension of F ′ ∩D andF ′ ∩D ⊆ D ⊆ D′ , a fortiori D is a finite integral extension of F ′ ∩D and thus by (2)F ′ ⊇ F . X

corollary 7.28. A quasi-separable semi-prime ring R is a finitely generated moduleover its center.

proof: By Proposition 7.19 R is quasi-equal to R1 × · · · × Rn , where the Ri are primerings and the center Z of R is quasi-equal to Z1 × · · · × Zn , where Zi is the center ofRi . Thus by Lemma 7.5 Z1 × · · · × Zn is a finitely generated Z-module and furthermoreR is a finitely generated Z-module if and only if R1 × · · · × Rn is a finitely generatedZ-module . It follows that it suffices to prove that each Ri is a finitely generatedZi-module , for if Xi is a finite set of generators for Ri over Zi and Y is a finite set ofgenerators for Z1 × · · · ×Zn over Z , then R1 × · · · ×Rn is generated over Z by the set ofall elements xy for y ∈ Y and x ∈ Xi for some i . Thus we may assume without loss ofgenerality that R is a prime ring. Let Z be the center of R . If F is the field of definitionfor R then by Theorem 7.24 F ∩R ⊆ Z and R is a finite integral extension of F ∩R , soa 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 isquasi-separable. Then by Lemma 7.20, QR = S ⊕ QN where N = nil radR and S isa semi-simple Q-algebra. We wish to prove that R is quasi-equal to (R ∩ S) ⊕N , withequality 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/QNonto S . Since Q(R ∩ S) = QR ∩ S = S , it follows that R ∩ S is semi-prime. Since(R ∩ S) ∩ Ker π = R ∩ S ∩ N = 0, to prove that R is quasi-equal to (R ∩ S) ⊕ Nit suffices by Proposition 3.18 to show that R ∩ S is quasi-equal to π(R) . (Note thatR ∩ S = π(R ∩ S) since the restriction of π to S is the identity map.) Furthermore wewill show that if R is commutative and R ∩ S is integrally closed then R ∩ S = π(R) , sothat 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 thefield S . Thus W ′ ⊆ R ∩ S if 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 . But thenwe may as well replace the ground ring W by W ′ , hence it suffices to consider the caseS = 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 typeare identical. Thus S ∩R = π(R) , so R = (S ∩R) ⊕ Kerπ = (S ∩R) ⊕N , as required.

16

Now consider the more general case that S is commutative. Then by Proposition7.19R 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 andall the Ri ∩ S are dedekind domains. Furthermore N is quasi-equal to N1 ⊕ · · · ⊕ Nn

where Ni = nil radRi . Note also that R ∩ S is quasi-equal to∏

(Ri ∩ S) (why?). It thussuffices to prove that for each i , Ri is quasi-equal to (Ri ∩ S) ⊕Ni , with equality if Ri isintegrally closed. This was handled in the preceding paragraph.

Finally, consider a general ring R such that R/N is quasi-separable and consequentlyQR = S ⊕QN . As above, let π : QR → S . By Corollary 7.28 R is a finitely generatedmodule over its center Z and likewise R ∩ S is a finitely generated module over Z ∩ S . Itfurther follows that π(R) is a finitely generated module over π(Z) . Now by applying theprevious paragraph to the commutative ring π−1(Z) we see that π(Z) is quasi-equal toZ ∩ S . Now

Z ∩ S = π(Z ∩ S) ⊆ π(Z) ⊆ π(R)

and since π(R) is a finitely generated module over π(Z) and π(Z) is quasi-equal to Z ∩Sit follows that π(R) is a finitely generated module over Z ∩ S . (In fact, by Lemma 7.5π(Z) is finitely generated over Z ∩ S . Thus π(R) is a finite integral extension of π(Z) ,which is a finite integral extension of Z ∩ S .) Furthermore R ∩ S = π(R ∩ S) is anessential submodule of π(R) . Thus by Lemma 3.13 π(R) is quasi-equal to R ∩ S . Thusπ restricts to a quasi-split surjection from R onto π(R) , so by Proposition 3.18 R isquasi-equal to (R ∩ S) ⊕N . X

corollary 7.29. If R is a strongly indecomposable ring such that R/ nil radR isquasi-separable then R is an integral domain.

proof: Let N = nil radR . By Theorem 7.21 if R/N is quasi-separable then R isquasi-equal to (S ∩ R) ⊕ N . Thus if R is strongly indecomposable, either N = 0 orS ∩ R = 0. But S ∩ R 6= 0 since 1 ∈ S ∩ R . Thus N = 0 so R is semi-prime. Thenby Proposition 7.19 R is quasi-equal to a product of prime rings. Since R is stronglyindecomposable, there can be only one factor in this product, so R is prime. Then byCorollary 7.26, R is an integral domain. X

STRONGLY INDECOMPOSABLE DOMAINS. We saw in Proposition 2.2and Proposition 2.3 that the subrings of Q are the localizations of W with respect tomultiplicative sets and correspond in one-to-one fashion to the subsets of SpecW . Now ifD is any finite rank integral domain then D is a essential subring of a finite dimensionalextension field Q′ of Q . But if D is integrally closed and W ′ is the integral closure ofW in Q′ , then W ′ ⊆ D so that D is a rank-one W ′-module. Thus we get the followingproposition.

17

proposition 7.30. (1) A finite rank torsion free quasi-separable dedekind domain Dis determined up to a quasi-equality by a finite field extension Q′ of Q and a subset Yof SpecW ′ , where W ′ is the integral closure of W in Q′ . (Recall that according to ourconvention, SpecW ′ 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 DP 6= Q′ .

(2) If Q′ is a separable extension of Q then a quasi-equality class of essential subringsof 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 SpecW ′ . 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 letQ′ = QD . Then Q′ is a finite separable extension of Q . Let W ′ be the integral closureof W in Q . Then as indicated in the remarks preceding the Proposition, the integralclosure of D′ in Q is a rank-one W ′-module of idempotent type and by Proposition 7.23is quasi-equal to D . Furthermore by Proposition 2.2 D′ is uniquely determined by itstype over W ′ and hence no two integrally closed subrings of Q′ can be quasi-equal asW ′-modules . But this in fact means that they cannot be quasi-equal as W -modules ,since if w 6= 0 ∈ W is such that wD1 ⊆ D2 and wD2 ⊆ D1 , where D1 and D2 areW ′-submodules of Q′ , then D1 and D2 are actually quasi-equal as W ′-modules . Thusby Proposition 2.3 the set of such D is in one-to-one correspondence with the family ofsubsets of SpecW ′ , where a domain D corresponds to the set of prime ideals P of W ′

such that D is not P -divisible, i.e. DP 6= 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 essentialsubrings D1 and D2 are quasi-equal then by Lemma 3.13 they are quasi-equal to D1D2 .Thus by Proposition7.22 D1D2 is an integral extension of D1 and D2 . Since these areintegrally closed, D1 = D2 = D1D2 .

(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 7.24. X

We say that two rings R1 and R2 are quasi-isomorphic as rings if there is a ringisomorphism from R1 onto a subring of QR2 which is quasi-equal to R2 .

The class of strongly indecomposable integrally closed integral domains has theimportant property that if two modules in this class are quasi-isomorphic they are in factisomorphic, 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 asW -modules if and only if they are isomorphic as rings.

18

(2) If D1 and D2 are integrally closed and strongly indecomposable then they arequasi-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 asW -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 asrings.

proof: (1) Clearly if D1 and D2 are isomorphic as rings [i.e. as W -algebras] thenthey are isomorphic as W -modules . Conversely if they are strongly indecomposable andisomorphic as W -modules then by Corollary 7.26 D1 ≈ EndD1 ≈ EndD2 ≈ D2 sinceisomorphic modules have isomorphic endomorphism rings.

(3) ( ⇒ ): If D1 ∼ D2 then clearly rankD1 = rankD2 . And if Fi is thefield of definition for Di then by Theorem 7.24 Fi is the center of QEndDi andQEndD1 ≈ QEndD2 by Proposition 3.8, so F1 ≈ F2 . Furthermore F1 ∩ D1 andF2 ∩D2 are strongly indecomposable by Theorem 7.24 and each Di is quasi-isomorphicto a direct sum of copies of Fi ∩Di . Since D1 ∼ D2 it follows from Jonsson’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 isquasi-isomorphic to a direct sum of copies of Fi ∩Di it follows that if rankD1 = 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 asW -modules then they are quasi-isomorphic as rings. Therefore we may suppose thatD1 and D2 have the same quotient field and are quasi-equal. But by Proposition 7.30 aquasi-equality class of integral domains in a field contains only a single integrally closeddomain. 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 aW -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 isomorphicto a direct sum of ideals of F ∩D . Since D is indecomposable, there can be only oneterm in this direct sum decomposition, so D is isomorphic to an ideal of F ∩D . ThusrankD = rankF . Since F ⊆ QD , it follows that F = QD . Therefore by Corollary 7.26D is strongly indecomposable. X

19

EARRINGS. We have seen in Corollary 7.26 that if D is a strongly indecomposableintegral domain then D ≈ EndD . 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 EndR .One can then wonder just how large a subring R is of EndR . For instance one mightconjecture that R = EndR if 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 insome proper separable extension field of Q then W ′ is a projective W -module and hencequasi-isomorphism to a free W -module Wn , for n > 1. Then QEndW ′ is the ring ofn× n matrices over Q so that EndW ′ is not commutative, and hence certainly is muchlarger than W ′ .

Rings R such that the canonical embedding R ⊆ EndR is an equality have come tobe called E-rings [Schultz]. (In other words, R is an E-ring if every endomorphism of Ris given by left multiplication by some element r ∈ R .) We shall see that he structure ofE-rings is quite simple.

lemma 7.33. (1) R is an earring if and only if the canonical embedding QR → QEndRis an equality.

(2) A ring quasi-isomorphic to an E-ring is an E-ring.(3) If EndR ≈ 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 EndR . Thus if QR = QEndR thenR = QR ∩ EndR = QEndR ∩ EndR = EndR , so R is an E-ring.

(2) If R and R′ are quasi-isomorphic then we may suppose them quasi-equal. ThenQR = QR′ and QEndR = QEndR′ . Thus the result follows from (1).

(3) By Proposition 7.2 the canonical map λ : QR → QEndR is monic. If EndR ≈ Rthen rankEndR = rankR so λ is an isomorphism. Thus R is an E-ring by (1).

(4) By (2) we may assume that R1 × R2 is an E-ring. Now any endomorphism ϕ ofR1 extends (non-uniquely) to an endomorphism of R1 × R2 and hence since R1 × R2 isan E-ring is given by multiplication by some (r1, r2) ∈ R1 ×R2 . It follows that ϕ is givenby multiplication by r1 . Thus the canonical embedding R1 → EndR1 is surjective, so R1

is an E-ring.Furthermore if ψ ∈ Hom(R1, R2) then ψ extends to an endomorphism of R1 × R2 ,

which must be given by multiplication by some r ∈ R1 ×R2 . But since R1 is an ideal inR1 ×R2 , rR1 ⊆ R1 and it follows that ψ = 0. Thus Hom(R1, R2) = 0. X

Note that by Proposition 3.43 Murley rings are E-rings. For Murley rings, the followingproposition 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 , henceby hypothesis must be given by left multiplication by some s ∈ R , i.e. for all x ∈ R ,xr = sx . Substituting x = 1 yields r = 1r = s1 = s . Thus xr = rx for all x ∈ R . Sincethis 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 stronglyindecomposable. Then EndG is commutative if and only if EndGi is commutative foreach i and Hom(Gi, Gj) = 0 for i 6= j . In this case, EndG ∼ EndG1 × · · · × EndGn .

proof: ( ⇐ ): Easy since if G is quasi-equal to G1 ⊕ · · · ⊕Gn and Hom(Gi, Gj) = 0 fori 6= j then EndG ∼ EndG1 × · · · × EndGn .

( ⇒ ): If EndG is commutative, let πi be the the quasi-projection of G onto Gi andlet πi be the corresponding idempotent in QEndG . If ϕ ∈ Hom(Gi, Gj) with j 6= i thenϕπi ∈ QHom(G,Gj) and thus ϕπi corresponds to a map ϕ ∈ QHom(G,G) = QEndGgiven by ϕ(x) = ϕπi(x) . Since ϕ(G) ⊆ Gj and j 6= i , πiϕ = 0. But since by hypothesisQEndG is commutative and ϕ = ϕπi , we get 0 = πiϕ = ϕπi = ϕ . It follows that ϕ = 0.Thus Hom(Gi, Gj) = 0 for i 6= j . Likewise if ϕ, ψ ∈ QEndGi then ϕψ = ψϕ and itfollows that ϕψ = ψϕ . X

We noted in Chapter 3 that a Murley ring is indecomposable if and only if it is adedekind domain. For E-rings in general we have an analogous result.

theorem7.36. A ring R is an E-ring if and only if R is quasi-isomorphic to a finiteproduct of strongly indecomposable integral domains D1 × · · · × Dn such thatHom(Di, Dj) = 0 for i 6= j . In particular, a strongly indecomposable E-ring is an integraldomain.

proof: ( ⇐ ): By Lemma 7.33 it suffices to prove that if D1 × · · · × Dn are stronglyindecomposable domains with Hom(Di, Dj) = 0 for i 6= j then D1 × · · · × Dn is anE-ring. Now since the Di are strongly indecomposable, EndDi ≈ Di by Corollary 7.26.Thus D1 × · · · ×Dn ≈ EndD1 × · · · × EndDn ≈ End(D1 ⊕ · · · ⊕Dn) , where the latterisomorphism holds by the assumption that Hom(Di, Dj) = 0 for i 6= j . Furthermore thisisomorphism takes an element (d1, . . . , dn) ∈ D1 × · · · ×Dn to the endomorphism given byleft multiplication by (d1, . . . , dn) . Thus D1 × · · · ×Dn is an E-ring.

( ⇒ ): If R is an E-ring then EndR ≈ R and by Proposition 7.34 R is commutative.Thus by Proposition 7.35 R has a quasi-direct decomposition R1 ⊕ · · · ⊕ Rn where theRi are strongly indecomposable pure W -submodules of R such that Hom(Ri, Rj) = 0for i 6= j and R ≈ EndR ∼ EndR1 × · · · × EndRn . It further follows from thecondition Hom(Ri, Rj) = 0 that then the subspaces QRi are invariant under leftand right multiplication by elements of QR and hence are ideals in QR and so QRis the ring-theoretic product of the rings QRi . Now let ei be the identity element ofQRi and let Di be the subring of QRi generated by Ri together with ei . Then byLemma 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 byCorollary 7.29 [This requires separability!]. XzCOMPOSITA. Although the question of determining the structure of the tensorproduct of two finite rank torsion free modules is a very difficult one, if the two modulesare rings then their tensor product is also a ring, which narrows down the range ofpossibilities considerably. In the case of the tensor product of two integral domains, theproblem 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 isquasi-separable. By a compositum of D1 and D2 is meant a non-trivial torsionfree (D1 ⊗ D2)-algebra C such that C is an integral domain and the structural mapD1 ⊗D2 → C is surjective. Note that the restrictions of the structural map to D1 ⊗ 1and 1⊗D2 must be monic, since these maps extend to maps from the fields QD1 ⊗ 1 and1 ⊗ 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 isquasi-separable. Then

(1) D1 ⊗D2 ∼ ∏Ci , where the Ci form a maximal set of non-equivalent composita of

D1 and D2 .(2) D1 ? D2 ∼ ∏′

Ci where the product is taken over the set of Ci which are notfields.

(3) If Q is an algebraically closed extension of Q containing D1 and D2 then we maychoose Ci = D1σ(D2) , where σ : QD2 → Q is a morphism of Q-algebras.

(4) If D2 ⊆ Q′ ⊆ QD1 where Q′ is a normal extension of Q , thenD1 ⊗ D2 ∼ ∏

D1σ(D2) , where σ ranges over the set of all Q-algebramorphisms from QD2 into Q′ .

proof: (1) Partly this has already been covered in Proposition 7.19 since the result iswell known for fields.

Since either QD1 or QD2 is separable over Q , QD1 ⊗ QD2 has trivial nil radical[Reference], hence is a product of commutative simple algebras, i.e. a product of fields,say QD1 ⊗QD2 =

∏Fi . Then the Fi are composita of QD1 and QD2 and are clearly

non-equivalent since the structural 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 andthey are all non-equivalent. Furthermore if C is any compositum then the surjectionD1 ⊗D2 → C induces a ring morphism ϕ :

∏QCi → QC and since C is a domain Kerϕ

is a prime ideal, meaning that there is a unique i with ϕ(Ci) 6= 0. Then there exists acommutative diagram

D1 ⊗D2

Ci⊆−−−−→ QCiy

yC

⊆−−−−→ QC.

Now both left-hand (slanted) maps are surjections. And the right-hand vertical map is anon-trivial homomorphism of fields, hence is monic. It thus follows that the map Ci → Cis an isomorphism of (D1 ⊗D2)-algebras, so that C and Ci are equivalent composita ofD1 and D2 .

(2) By Corollary 7.17 each integral domain Ci is either reduced or divisible. And byProposition 7.13 it is divisible if and only if it is a field. Thus D1 ? D2 is the product ofthose Ci which are not fields.

22

(3) Since QC is a finite dimensional extension field of Q we may choose C so thatQC ⊆ Q . Furthermore, since C is a compositum of D1 and D2 , the compositionD1 → D1 ⊗ D2 → C is a monomorphism, hence we may choose C so that it is aninclusion. Then we get a morphism of W -algebras σ : D2 → D1 ⊗ D2 → C such thatC = D1σ(D2) .

(4) If D2 ⊆ Q′ ⊆ D1 and Q′ is a normal extension of Q , then σ(D2) ⊆ QD1 . Aswe have seen, the composita may be chosen in the form D1σ(D2) . Now suppose twocomposita D1σ(D2) and D1τ(D2) are equivalent. We then have a commutative diagram

D1 ⊗D2

1 ⊗ σ

1 ⊗ τ

D1σ(D2)⊆−−−−→ QD1y

∥∥∥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-onecorrespondence with the Q-algebra morphisms from QD2 into Q′ . X

VALUATION RINGS. Recall that by a valuation ring on a field K is meant avaluation ring (in the present context, necessarily a discrete valuation ring) whosequotient field is K . In algebraic number theory, the field of definition for a discretevaluation ring V is called the decomposition field for V . The following twopropositions are restatements of well known results in ramification theory.

proposition 7.39. Let V be a strongly indecomposable quasi-separable discretevaluation 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 . If p′ is aprime of W and p′ 6= p then there exists w ∈ W such that w /∈ p , so w /∈ P and thus wis invertible in V . Therefore V = wV ⊆ p′V ⊆ V , so V is p′-divisible.

(2) Let F = QV . Since V is strongly indecomposable, by Corollary 7.26 F is the fieldof definition for V . Let Q′ be a galois extension of Q containing F and let W ′ and V ′

be the the integral closures of W and V in Q′ . By Proposition 7.38

W ′ ⊗ V ≈∏

W ′σ(V ) =∏

σ(V )′,

where σ ranges over the set of distinct Q-algebra morphisms from F into Q′ , and σ(V )′

denotes the integral closure of σ(V ) in Q′ . We can also think of σ as ranging over aset of representatives for the left coset space Gal(Q′/Q)/Gal(Q′/F ) and we then haveσ(V )′ = σ(V ′) . Now Gal(Q′/Q) acts transitively on the family of σ(V ′) and since Fis the field of definition of V ′ , by part (7) of Theorem 7.24 Gal(Q′/F ) is the stabilizer

23

of V ′ under this action. Thus the σ(V ′) , and also the σ(V ) , are distinct. Now sinceσ(V ) is a maximal proper subring of σ(F ) it follows that if P ′ is a prime ideal of W ′

then σ(V ′) ⊆ W ′P ′ if and only if σ(V ) ⊆ W ′

P ′ if and only if σ(V ) = σ(F ) ∩W ′P ′ .

Therefore for each P ′ there is only one σ with σ(V ′) ⊆ W ′P ′ . But σ(V )′P ′ = W ′

P ′σ(V ′)so σ(V ′)P ′ = W ′

P ′ if σ(V ′) ⊆ W ′P ′ and σ(V ′)P ′ = Q′ otherwise. Therefore, using

Lemma 6.27,

p-rankV = P ′-rankW ′ W ′ ⊗ V = P ′-rankW ′∏

σ(V ′) = 1.

Since q-rankV = 0 for q 6= p ∈ SpecW and since by Corollary 7.17 V is reduced, itfollows that V is a Murley ring. [Rewrite!] Xz

proposition 7.40. Let Q′ be a finite galois extension of Q and let V ′ be a valuationring on Q′ . Let p be the unique prime ideal of W such that pV ′ 6= V ′ and let F be thefield of definition for V ′ . Then

(1) V ′ is the only valuation ring on Q′ containing V ′ ∩ F .(2) If WF is the integral closure of W in F then there exist unique prime ideals

P ∈ SpecWF and P ′ ∈ SpecW ′ such that V ′ = W ′P ′ and V ′ ∩ F = (WF )P .

(3) P ′ is the only prime ideal of W ′ lying above P .(4) If W ′ is the integral closure of W in Q′ then W ′ ⊗ V ≈ ∏

σ(V ′) , where σ rangesover a set of representatives for the left cosets of Gal(Q′/Q)/Gal(Q′/F ) .

(5) The valuation rings σ(V ′) in the above product are the only valuation rings on Q′

containing Wp .(6) If K is any subfield of Q′ containing Q then K ∩ V is indecomposable if and only

if K ⊆ F .

proof: (1) By Theorem 7.24 V ′ is integral over V ′ ∩ F and therefore is the integralclosure of V ′ ∩ F in Q′ . Thus if V ′

1 is any valuation ring on Q′ containing V ′ ∩ F thenV ′

1 ⊇ V ′ . But by Proposition 0.* V ′ is a maximal subring of Q′ . Therefore V ′1 = V ′ so

that V ′ is the unique valuation ring on Q′ containing V ′ ∩ F .(2) This follows from Proposition 2.*.(3) If P ′

1 is a prime ideal of W ′ lying above P then W ′P ′

1is a discrete valuation ring

on Q′ and V ′ ∩ F = (WF )P ⊆ W ′P ′

1. Therefore by (1) W ′

P ′1

= V ′ = W ′P ′ and so P ′

1 = P ′ .(4) Immediate from Proposition 7.38.(5) If V ′

1 is a valuation ring on Q′ containing Wp then W ′p ⊆ V ′

1 . Now⋂σ(V ′)

is invariant under Gal(Q′/Q) , so by Theorem 7.24 its field of definition is Q ,and it follows that

⋂σ(V ′) = W ′

p , so that⋂σ(V ′) ⊆ V ′

1 . Now for each σ , ifV ′

1 6= σ(V ′) then V ′1σ(V ′) = Q′ since V ′

1 and σ(V ′) are maximal subrings of Q′ .Furthermore by Proposition 2.3 the set of Wp-subalgebras of Q′ is a distributivelattice, so the assumption that V ′

1σ(V ′) = Q′ for all σ leads to the contradictionV ′

1 = V ′1W

′p = V ′

1

⋂σ(V ′) =

⋂V ′

1σ(V ′) = Q′ . Therefore V ′1 = σ(V ′) for some σ .

(6) ( ⇐ ): If K ⊆ F then K ∩ V C F ∩ V . Therefore by Proposition 7.39 K ∩ V is aMurley 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 hasp-rank one by Proposition 7.39. Now the compositum (K ∩ V )(F ∩ V ) is a homomorphicimage of (K ∩ V ) ? (F ∩ V ) , hence also has p-rank one, hence is indecomposable, sothat FK is its field of definition. But since F ∩ V ⊆ (K ∩ V )(F ∩ V ) ⊆ V , the field ofdefinition for (K ∩ V )(F ∩ V ) is F (why?). Therefore if K ∩ V is indecomposable thenzFK = F , i.e. K ⊆ F . X

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

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 ofvaluation rings on K all having the same field of definition F . Suppose further that foreach prime p there is at most one i ∈ I such that Vi is not p-divisible. Then F is thefield of definition for

⋂Vi and

⋂Vi ∩ F is a Murley ring.

proof: If F is the field of definition for the valuation ring Vi then by Proposition 7.39Vi ∩ F is a Murley ring. Now (

⋂Vi ∩ F )p = (Vj ∩ F )p , where Vj is the unique Vj with

(Vj)p 6= K . Thus p-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 stronglyindecomposable. 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 forx over W lie in Vi ∩ F . Hence the minimal polynomial for x over F has coefficients in⋂Vi ∩ F , showing that

⋂Vi is an integral extension of

⋂Vi ∩ F , and by the separability

of K is a finite integral extension (Proposition 0.*). 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 ringson QD containing D have the same field of definition F . Then there exists a finite rankquasi-separable splitting ring R =

∏Ri such that Ri ⊆ F for all i and such that D is

R-split.

proof: By Lemma 7.23 D is quasi-equal to its integral closure so there is no lossof generality in supposing D integrally closed. Then by Proposition 0.* D is theintersection of the valuation rings containing it. By hypothesis, these all have the samefield of definition F . We claim that F contains the field of definition for D . In fact,each valuation ring V containing D is a finite integral extension of a valuation ringVF with QVF = F . If DF =

⋂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

of VF the minimal polynomial for d has coefficients in⋂VF = DF . Furthermore since

QD is separable, D is a finite integral extension of DF by Proposition 0.* and so byCorollary 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 sufficesto prove that DF is R-split for some R . Thus there is no loss of generality in supposingthat F = QD . Thus by Proposition 7.39 if V is a valuation ring on F containing Dthen V is a Murley ring. Now for each prime p with p-rankD 6= 0 choose a valuationring Vp on F containing D with p-rankVp 6= 0. (The subscript p here does not denotelocalization.) Let R1 =

⋂p Vp . Then R1 ⊆ F and in particular R1 is quasi-separable and

by Lemma 7.41 R1 is a Murley ring. Now repeat the same process using valuation ringson F containing D which were not utilized in the first step, thus constructing a secondMurley ring R2 contained in F and containing D . Since for each prime p there can beat most [F : Q] valuation rings V on F with pV 6= V (why?), in this way we finallyobtain a set of Murley rings R1, . . . , Rn with n ≤ [F : Q] such that for each valuationring V containing D , D ⊆ Ri ⊆ V for some i . Since D is integrally closed, it is theintersection of the valuation rings containing 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 Ri = Ri ⊕ Ai , where Ai is an appropriate subring of Q , then Ri is a quasi-separablesplitting ring and Ri is Ri-split . Thus by Proposition 6.20 D is R-split for some finiterank quasi-separable R . X

proposition 7.43. Let D be an integral domain and let Q′ be a finite galois extensionof Q . Then Q′ contains the field of definition for D if and only if there is a splitting ring∏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 isquasi-isomorphic to a direct sum of copies of F ∩ D , it suffices to prove that thereexists 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 7.22D is quasi-equal to its integral closure, we may suppose D integrally closed. For eachsubfield E of QD let DE be the intersection of the valuation rings V on QD whosefield of definition is E . By Lemma 7.42 DE is R-split for some finite rank splittingring R =

∏Ri with all Ri ⊆ Q′ . And since D is the intersection of the valuation

rings containing it, D =⋂DE , and so by Proposition 1.10 D is isomorphic to a pure

submodule of⊕

E DE . Since QD is a separable extension of Q , there are only finitelymany subfields E . Thus by Proposition 6.20 D is R-split for some finite rank splittingring R =

∏Ri with Ri ⊆ Q′ for all i .

( ⇐ ): Suppose that D is R-split for some splitting ring∏Ri with all Ri ⊆ Q′ .

Since the field Q′D is separable over QD , by Proposition 0.* the integral closure D′

of D in Q′D is a finite integral extension of D . Hence by Proposition 7.27 the fieldof definition for D′ is the same as the field of definition for D , so that, replacing Dwith D′ , no generality is lost in assuming that Q′ ⊆ QD . Then by Proposition7.38Ri ? 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. (r, x) 7→ σ(r)x .) Ifϕ ∈ EndQ′ QD then since σ(Ri) ⊆ Q′ , ϕ extends to a

∏Ri-linear endomorphism of

Q(R?D) =⊕Qσ(Ri)D . Therefore by Proposition 6.4 EndQ′ QD ⊆ QEndD = EndF QD

where F is the field of definition for D and the latter equality is by Theorem 7.24. Inparticular, a Q′-linear projection QD → Q′ is F -linear, hence restricts to the identity onF . Thus F ⊆ Q′ . X

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

theorem7.44. Let F be a finite separable extension of Q and G a W -module. ThenG is quasi-isomorphic to an essential subring of F if and only if the following threeconditions hold:

(1) rankG = [F : Q] .(2) There is an embedding of rings F → QEndG .(3) G is R-split for some finite rank splitting ring R .

proof: ( ⇒ ): If G is an essential subring of F then rankG = dimF = [F : Q] .Furthermore by Proposition 7.2 G ⊆ EndG and F = QG ⊆ QEndG . Finally, byProposition 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 spaceover F and we may suppose QG = F . Let W ′ be the integral closure of W in F . Byseparability W ′ is a finitely generated W -module and since G is an essential submoduleof W ′G , W ′G is quasi-equal to G by Lemma 3.13. Therefore without loss of generalitywe may suppose that G = W ′G , i.e. that G is a W ′-module. Hence if G is R-splitthen by Proposition 6.28 W ′ ? R is a splitting ring over W ′ and G is (W ′ ? R)-split. Inother words, without loss of generality we may replace the ground ring W by W ′ andhence suppose that F = Q and G ⊆ Q . Then G is a rank-one module and G C R ? Gso t(G) ∈ T(R ? G) = T(

∏Ri ? G) . But Ri ? G is a finitely generated Ri-module ,

hence is quasi-isomorphic to a direct sum of copies of Ri , and hence by Proposition 2.24T(

∏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. Sinceby 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 asubring of Q by Proposition 2.15. X

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

We will see in Chapter 8 that condition (3) above can be replaced by the weakerrequirement that G is quotient divisible.

27

theorem7.45. Let R =∏Ri be a splitting ring, where each Ri is a finite rank

quasi-separable dedekind domain, and let Q′ be a finite galois extension of Q such thatRi ⊆ Q′ for all i . Let G be a strongly indecomposable R-split W -module and let W ′ bethe integral closure of W in Q′ . Then G is quasi-isomorphic to an integral domain if andonly if W ′⊗G is quasi-isomorphic to a direct sum of rank-one W ′-modules of idempotenttype.

proof: ( ⇒ ): If G is strongly indecomposable and quasi-isomorphic to an integraldomain D then by Proposition 7.26 the field of definition for D is QD . Henceby Proposition 7.43 if G (and hence D ) is R-split and if R =

∏Ri and Q′ is a

galois extension of Q containing all the Ri then D ⊆ Q′ . By Proposition 7.38W ′ ⊗ D ∼ ∏

W ′σ(D) as W ′-algebras , where the σ range over a certain subset ofGal(Q′/Q) . Since σ(D) ⊆ Q′ = QW ′ , W ′σ(D) is an essential W ′-subalgebra of Q′ ,hence by Proposition 2.15 is a rank-one W ′-module of idempotent type. Since G ∼ D ,W ′ ⊗G and W ′ ⊗D are quasi-isomorphic W ′-modules , so W ′ ⊗G is quasi-isomorphic to⊕W ′σ(D) , a direct sum of rank-one W ′-modules of idempotent type.( ⇐ ): If W ′ ⊗G ∼ ⊕

Bk where the Bk are rank-one W ′-modules of idempotent typethen by Proposition 2.15 the Bk can be taken to be subrings of Q′ . Since by Lemma 7.22W ′ is quasi-isomorphic to a free W -module, as a W -module W ′ ⊗ G is quasi-isomorphicto a direct sum of copies of G . Therefore by Jonsson’s Theorem (Theorem 3.24) itfollows that G is quasi-isomorphic to a strongly indecomposable quasi-summand ofB1 . But by Theorem 7.24 B1 is quasi-isomorphic a direct sum of copies of the stronglyindecomposable module B1 ∩ F , where F is the field of definition for the ring B1 . ThusG is quasi-isomorphic to the integral domain B1 ∩ F . X

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

definition 7.46. A strongly indecomposable module G is called strongly homogeneousif G is homogeneous and every non-trivial homomorphism ϕ : A → G , where A is a purerank-one submodule of G , extends to an endomorphism of G .

lemma 7.47. If G is strongly homogeneous then EndG is a prime ring.

proof: Let G be strongly homogeneous. We will show that G has no non-trivial properpure fully invariant submodules. If H is a pure fully invariant submodule of G andH 6= 0, let A be a pure rank-one submodule of H and let B be any rank-one submoduleof G . Since G is homogeneous t(A) = t(B) and so by Proposition 2.4 there exists anon-trivial homomorphism ϕ : A → B . Then ϕ can be thought of as a mapping inHom(A,G) , so by hypothesis it extends to an endomorphism ψ of G . Since H is fullyinvariant, 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 7.16EndG is prime. X

lemma 7.48. Let G be a strongly indecomposable module such that EndG is prime.

(1) Let A be a pure rank-one submodule of G and let ρ : EndG→ Hom(A,G) be themap such that ρ(ϕ) is the restriction of ϕ to A . Then ρ is a monomorphism.

(2) G is strongly homogeneous if and only if ρ is an isomorphism.(3) If G = A ⊗G0 , where A is a rank-one module with [t(A) : t(A)] ≤ IT(G0) , then

G is 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 radEndG by Proposition 3.26. SinceEndG is prime, ϕ = 0. Thus ρ is monic.

(2) The definition of strongly homogeneous can be stated as saying that G is stronglyhomogeneous if and only if ρ is surjective. Since ρ is monic by (1), this is equivalent to ρbeing an isomorphism.

(3) If G ≈ A ⊗ G0 where A is a rank-one module and if A0 = EndA andt(A0) = [t(A) : t(A)] ≤ IT(G0) then there is a commutative diagram

EndGρ−−−−−→ Hom(A,G) = Hom(A⊗ A0, A⊗G0)

≈y ≈

yEndG0

ρ0−−−−−→ Hom(A0, G0)

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

lemma 7.49. If G is a strongly indecomposable strongly homogeneous module thenEndG 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 7.48the restriction map ρ : EndG → Hom(A,G) is an isomorphism. Now note that ifϕ, ψ ∈ EndG then ρ(ϕψ) = ϕρ(ψ) . This says that ρ is an (EndG)-linear isomorphismfrom EndG to Hom(A,G) . Now since G is homogeneous, T(G) = {t} and G ist-saturated. Thus by Corollary 4.39 the map η : EndG → End Hom(A,G) givenby η(ϕ) = ϕ∗ is an isomorphism. But η(ϕ) is simply multiplication by ϕ on theleft (EndG)-module Hom(A,G) . Since EndG and Hom(A,G) are isomorphic left(EndG)-modules , it follows that the map λ : EndG → End EndG taking ϕ to the mapgiven by left multiplication by ϕ is also an isomorphism, so that EndG is an E-ring.

By Proposition 7.34 E-rings are commutative. And by Lemma 7.47 EndG is a primering. But by Proposition7.13 commutative prime rings are integral domains. ThereforeEndG is an integral domain. X

29

A (two-sided) ideal P in a not-necessarily-commutative ring R is called a prime idealif whenever I and J are ideals of R such that IJ ⊆ R then either I ⊆ P or J ⊆ P .

Note that by Proposition7.13 a ring R is a prime ring if and only if the trivial ideal inR is a prime ideal.

Maximal ideals are prime, since if P is maximal and I, J 6⊆ P then I+P = J +P = Rand so IJ + P = R and thus IJ 6⊆ P . Note also that if R is commutative then thedefinition of a prime ideal above agrees with the usual definition in commutative ringtheory.

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 ∈ SpecW , pD is a prime ideal in D .

proof: (1) ⇒ (2): Suppose that every ideal in D has the form aD where a is an idealof W and let A be a pure rank-one submodule of D . Then DAD is an ideal in D soDAD = aD for some a ⊆ W . Then since A ⊆ DAD , A = A ∩DAD = A ∩ aD = aA bypurity and therefore also A = a−1A . Thus DAD = Da−1AD = a−1DAD = a−1aD = D .

(2) ⇒ (1): Suppose that DAD = D for every pure rank-one submodule of Dand let I be a non-trivial ideal in D . Then I is a D-submodule of D and sinceD is prime, QD is a simple ring so that QI = QD . Therefore since D is a cyclicD-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 . Now if A is any pure rank-onesubmodule of D , then aA ⊆ aD ∩ A ⊆ I ∩ A so 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 . Sinceby 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 . Thus a = [I ∩ A : A] and I ∩ A = [I ∩ A : A]A = aA = aD ∩ A ,where the latter equality holds since A C D . Therefore I ∩ A = aD ∩ A holdsfor all pure rank-one submodules A of D . Since D is the union of its pure rank-onesubmodules, it follows that I = aD .

(1) ⇒ (3): If p ∈ SpecW and a is a proper ideal of W then pD is not properlycontained in aD since p is maximal. Thus if every ideal in D has the form aD , fora ⊆ W , then pD cannot be properly contained in any proper ideal of D , hence pD ismaximal and therefore prime.

(3) ⇒ (2): Let A be a pure rank-one submodule of D . If DAD 6= D then since byProposition 7.8 D is noetherian, there exists a maximal ideal I with DAD ⊆ I . Letp = W ∩ I . Then pD ⊆ I and p must be a prime ideal of W since if x, y ∈ W andxy ∈ p then xD and yD are ideals in D and xDyD ⊆ pD ⊆ I , so since I is prime eitherxD ⊆ I or yD ⊆ I , which implies x ∈ I ∩W = p or y ∈ p . Now if pD is a prime idealin 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 . But then

30

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

proposition 7.51. If G is a strongly indecomposable strongly homogeneous modulethen

(1) EndG is strongly homogeneous.(2) EndG is a dedekind domain.(3) Every ideal of EndG has the form aEndG , where a is an ideal of W .(4) If A is a pure rank-one submodule of G then the map µ : A ⊗ EndG → 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 mapσ : A ⊗ Hom(A,G) → G(t) is an isomorphism. Now if G is strongly homogeneous andA C G , then G is homogeneous and so G(t) = G . Furthermore by Lemma 7.48 therestriction map ρ : EndG → Hom(A,G) is an isomorphism. Composing σ with A ⊗ ρ

yields an isomorphism µ : A⊗ EndG ≈−→ G given by µ(a⊗ ϕ) = ϕ(a) .(1) Let A be a pure rank-one submodule of G . Then by (4) A ⊗ EndG ≈ G

is strongly homogeneous. Furthermore since T(G) = {t(A)} , by Corollary 7.4[t(A) : t(A)] = IT(EndG) . Thus by part (3) of Lemma 7.48 EndG is stronglyhomogeneous.

(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 stronglyhomogeneous and so by (4) there is an isomorphism µ0 : A0 ⊗ EndD → D given byµ0(x ⊗ ϕ) = ϕ(x) . By by Lemma 7.49 D is an E-ring so the map λ : D → EndDmapping d to left multiplication by d is an isomorphism. Thus there is a surjectionµ′ : A0 ⊗D → D given by µ′(x⊗ d) = λ(d)(x) = dx = xd . Thus A0D = D , as claimed.It thus follows from Lemma 7.50 that every ideal of D has the form DaD = aD for someideal a of W .

(2) By Lemma 7.49 EndG is an integral domain. And if I is an ideal in D = EndGthen by (3) I = aD for some ideal a of W . If I 6= 0 then a 6= 0 and a−1D is aD-submodule of QD and a−1DaD = D . Thus every ideal of D is invertible, so D is adedekind domain. X

theorem7.52. A strongly indecomposable module G is strongly homogeneous if andonly if G ≈ A ⊗D , where A is a rank-one module with locally trivial type and D is astrongly indecomposable dedekind domain such that every ideal in D has the form aDfor some ideal a of W .

proof: ( ⇒ ): By Proposition 7.51 if G is a strongly indecomposable stronglyhomogeneous module then G ≈ A0 ⊗ EndG , where A0 is a pure rank-one submoduleof G and EndG is a dedekind domain such that every ideal in EndG has the forma EndG for some ideal a of W . Now by Proposition 4.35 A0 ≈ A ⊗ [A0 : A0] , for

31

some rank-one module A of locally trivial type. But if t′ = t([A0 : A0]) then sinceG is t(A0)-saturated (since G is homogeneous), it follows from Corollary 7.4 thatEndG is t′-saturated and so by Proposition 4.37 [A0 : A0] ⊗ EndG ≈ EndG . ThusG ≈ A⊗ [A0 : A0] ⊗ EndG ≈ A⊗ EndG .

( ⇐ ): By part (3) of Lemma 7.48 it suffices to prove that if G = D is a domainsuch that every ideal in D has the form aD for some ideal a of W , then D is stronglyhomogeneous. By Corollary 7.17 domains are homogeneous. Now let B be a purerank-one submodule of D and ϕ : B → D . By Proposition 2.2 the multiplicationmap µ : B ⊗D → BD given by µ(b ⊗ d) = bd is an isomorphism and by Lemma 7.50BD = DBD = D . Now let B′ = ϕ(B)∗ . As with B there is an isomorphismµ′ : B′ ⊗D

≈−→ D . Let ϕ′ = µ′(ϕ ⊗ 1D)µ−1 : D → D . Then the restriction of ϕ′ to B isgiven by

b 7→ b⊗ 1 7→ ϕ(b) ⊗ 1 7→ ϕ(b),

so that ϕ′ extends ϕ . Thus D is strongly homogeneous. X

corollary 7.53. Let G and H be strongly indecomposable strongly homogeneousmodules such that G ∼ H . If W is a principal ideal domain or if G and H are integraldomains, then G ≈ H .

proof: By Theorem 7.52 G ≈ A ⊗ G0 and H ≈ B ⊗H0 where A and B are rank-onemodules with locally trivial types and G0 and H0 are dedekind domains. Furthermoret(A)IT(G0) = IT(G) = IT(H) = t(B)IT(B0) and since by Corollary 7.3 IT(G0) andIT(H0) are idempotent and by Proposition 2.19 the representation of a type as a productof 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 byProposition 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 closeddomains, G0 ≈ H0 by Proposition 7.31. Thus G ≈ H if W is a principal ideal domain orif G = G0 and H = H0 . X

proposition 7.54. If W is a principal ideal domain then a strongly indecomposablemodule G is strongly homogeneous if and only if for every two pure rank-one submodulesA and B of G there exists an automorphism ϕ of G with ϕ(A) = B .zproof: ( ⇒ ): Suppose that G is strongly homogeneous and A , B are pure rank-onesubmodules of G . Then G is homogeneous, so t(A) = t(B) and hence since W is aprincipal ideal domain, by Corollary 2.10 there is an isomorphism α : A ≈−→ B . Since G isstrongly homogeneous, α extends to an endomorphism ϕ of G , and clearly ϕ(A) = B .

By Lemma 7.49 EndG is an integral domain so ϕ is not a zero divisor in EndG andthus by Proposition 3.3 ϕ is monic. It remains to see that ϕ is surjective. It suffices tosee that C ⊆ ϕ(G) for every pure rank-one submodule C of G . In fact, by the abovethere exists ψ ∈ EndG with C = ψ(B) = ψϕ(A) = ϕψ(A) ⊆ ϕ(G) , using the fact that byProposition 7.51 EndG is commutative.

32

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

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

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

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

Let K be the fixed field of T . Then K is the largest subfield of Q′ such that K ∩ V ′ isstrongly homogeneous.

proof: ****z

Proposition. Let D be a strongly indecomposable strongly homogeneous principalideal domain. If rankD = n then any two pure submodules of D with rank n − 1 areisomorphic.proof: This has to do with the fact that every element of D can be written aswu , where w ∈ W and u is an invertible element of D . Now if K1 and K2 arepure 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 . SinceKerϕ2 = Kerwϕ2 , we may assume that u is a unit in D . But then

uK1 = uKer uϕ2 = Kerϕ2 = K2,

so K1 ≈ K2 . X

MULTIPLICATIONS ON A STRONGLY INDECOMPOSABLEMODULE. The question of when a W -module G can be given the structure ofan integral domain comes down to asking whether there is a map µ : G ⊗ G → G suchthat the resulting multiplication is associative and commutative and has an identityelement. When Q is a perfect field it follows from Corollary 7.29 that if G is stronglyindecomposable and µ is known to be associative and there exists an identity, thencommutativity 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 thehypothesis that every non-trivial endomorphism of G is monic. (This is always thecase for a strongly indecomposable integral domain D since by Corollary 7.26 EndD isisomorphic to D and thus has no zero divisors.)

proposition 7.56. Let G be such that every non-trivial endomorphism is monic. Thenany non-trivial distributive W -bilinear multiplication defined on G is associative andcommutative. Furthermore this multiplication extends to a multiplication of QG makingQG into a field, and G is quasi-equal to a subring of this field. Finally, G is actuallyisomorphic 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 6= 0 ∈ G and consider the map ϕ : G → Ggiven by ϕ(x) = gx − xg . Then ϕ ∈ EndG because multiplication is W -bilinear.And g ∈ Kerϕ so ϕ is not monic. Hence by hypothesis ϕ = 0. This meansϕ(x) = 0, i.e. gx = xg for all x ∈ G . Thus multiplication is commutative. Now letg, h, x, ∈ G . Consider the mapping ψ ∈ EndG given by ψ(x) = g2x − g(gx) . Thenψ(g) = (g2)g − g(g2) = 0 since multiplication is commutative, so since g 6= 0, ψ is notmonic, and as above ψ = 0. Now consider ζ ∈ EndG defined by ζ(x) = (gx)h− g(xh) .Then ζ(g) = g2h − g(gh) = ψ(h) = 0, so ζ 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 6= 0 ∈ G and gh = 0 then h is in the kernel of the map x 7→ gx , so thismap must be trivial. Then for any g′ ∈ G , g is in the kernel of the map x 7→ xg′ , so thismap 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 alsobe associative and commutative. For fixed g 6= 0 ∈ QG , the map x 7→ gx is monic andhence is an automorphism of QG , so there exists e ∈ QG with ge = g . Then if θ isdefined by θ(x) = xe− x , and w 6= 0 ∈ W is such that we ∈ G , then wθ(G) ⊆ G , so thatθ ∈ QEndG . Since g belongs to the Ker θ , θ is not monic and hence must be trivial,i.e. xe = x for every x ∈ G , so e is an identity in QG . Since QG is commutative andassociative, it is an integral domain and since it is a finite dimensional algebra over thefield Q , it must be a field.

Now the only possible thing preventing G from being a subring of QG is thepossibility that e /∈ G . If this in fact occurs, note that the W -submodule G′ of QGgenerated by G together with e is quasi-equal to G by Proposition 3.1 and is a subringof the field QG , hence an integral domain. Note also that G is an ideal in G′ .

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 . Since G is an ideal in G′ and byProposition 7.8 G′ is a noetherian integral domain, this implies that G = G′ [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 stronglyindecomposable.

proposition 7.57. Let G be a strongly indecomposable module let N = nil radEndG .Suppose that EndG/N is quasi-separable. Let µ : G ⊗ G → G be such thatµ(G ⊗ G) 6⊆ NG . Then the multiplication induced on G by µ is associative andcommutative 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 → G byπ(ϕ) = ϕ(g1) . If π(ϕ) = 0 then ϕ is not monic, hence by Proposition 3.26 ϕ ∈ N sinceG is strongly indecomposable. Thus Ker π ⊆ N .

Now define ρ : G→ EndG by ρ(g)(x) = µ(x⊗ g) . Then πρ ∈ EndG and πρ /∈ N sinceπρ(g2) = ρ(g2)(g1) = µ(g1 ⊗ g2) /∈ NG . Thus since G is strongly indecomposable, πρ is aquasi-automorphism of G by Proposition 3.26. Hence π is a quasi-surjection.

By Proposition 7.21 EndG is quasi-equal to S ⊕ N , where S is a semi-prime ring.Since Ker π ⊆ N and G is quasi-equal to π(EndG) and hence quasi-equal to π(S ⊕N) ,it follows that G ∼ S ⊕ π(N) . But G is strongly indecomposable and S 6= 0, so weconclude that π(N) = 0 and that S is a strongly indecomposable semi-prime ring.By Proposition 7.29 S is an integral domain. Then G ∼ S and by Corollary 7.26EndG ∼ EndS ≈ S . Thus EndG is an integral domain and N = 0. Since EndG has nozero divisors, by Proposition 3.3 every non-trivial endomorphism of G is monic, so theProposition reduces to Proposition 7.56. X

One of the primary obsessions that has moved the theory of finite rank torsionfree modules forward has been the attempt to find classes of modules with propertiessimilar to those of rank-one modules. We have seen, for instance, that the class ofindecomposable Murley modules shares some of the important properties of rank-onemodules. We now see that the class of strongly indecomposable integral domains is alsoin some ways a generalization of the class of rank-one modules. In a way, this shouldnot come as a big surprise. For if D is a quasi-separable integral domain then D isquasi-equal to its integral closure. And the integral closure is a rank-one module over theintegral closure of W in QD .

The class of W -modules which admit an integral domain structure is a very importantclass 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 threecomponents:

(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 afourth item:

(4) A way of determining whether a given object belongs to the class in question ornot.

Proposition7.30 provides us with a solutions to problems (1) and (2), (taking intoaccount Proposition 7.31) inasmuch as it reduces the problem to that of finding all finitedimensional field extensions of Q and then in a given field extension Q′ finding the setof prime ideals of the integral closure of W in Q′ . Neither of these two problems is byany means trivial. However since they are in the realm of such things as algebraic numbertheory and class field theory they are problems that module theorists cannot be expectedto address.

The fact that we cannot be expected to provide a parametrization for the set offinite dimensional field extensions of Q or the corresponding prime ideal spectra alsosuggests 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 notthe most satisfying that one could imagine, since it involves knowing things about thesplitting ring for a module and its endomorphism ring, but it is difficult to see how therecould be any more straightforward solution.