7. TORSION FREE RINGS

E. L. Lady

June 27, 1998

In Chapter 1 we considered the problem of constructing, for a given prime p, a p-reduced module G with rank G > 1 and p-rank G = 1. Example 1.47, the Pontryagin module, was an example of such a construction, and it ultimately led us to the concept of splitting fields and splitting rings.

Another approach is as follows:

example 7.1. Let Q′ be a finite field extension of Q and let W′ be the integral closure of W in Q′. Suppose that there exists a prime ideal P_{1} of W′ and a prime p of W such that W′_{P_{1}}/pW′_{P_{1}} ≈ W/p. Let R = W′_{P_{1}} . Then R is p-reduced and p-rank R = 1. proof: By hypothesis, p-rank R = 1. Furthermore R is an integral domain and p^{∞}R is a pure ideal in R and so Qp^{∞}R is an ideal in the field QR, which forces p^{∞}R = 0 (see Corollary 7.17 below). ✓

The hypothesis for Example 7.1 will hold, for instance, if the prime p “splits completely” in W′, i.e. pW′ = P_{1} … P_{n} where P_{1}, …, P_{n} are distinct prime ideals in W′ and n = [W′ : W]. In that case pW′_{P_{i}} = P_{1} ⋯ P_{n}W′_{P_{i}} ≈ P_{i}W′_{P_{i}} for all i and so by Proposition 0.* W′/pW′ ≈ ⨁(W′/pW′)_{P_{i}} ≈ ⨁ W′_{P_{i}}/pW′_{P_{i}}, so that p-rank W′ = n = ∑_{1}^{n} p-rank W′_{P_{i}} and necessarily p-rank W′_{P_{i}} = 1 for all 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 instance, that if the underlying W -module of a ring is strongly indecomposable then that ring must be an integral domain.

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 End R by the map that associates to r ∈ R the left multiplication r̃: x ↦ rx. proof: It is well known that the map r ↦ r̃ is an isomorphism from R to End_{R} R. But End_{R} R ⊆ End_{W} R and in fact End_{R} R ◁ End_{W} R. Indeed, if φ ∈ End R and wφ ∈ End_{R} R for some w ≠ 0 ∈ W, then wφ is R-linear and clearly φ itself must be W -linear so φ ∈ End_{R} R. Thus r ↦ r̃ maps R onto a pure subring of End R. ✓

proposition 7.3. If R is a finite rank torsion free ring and N = nil rad R then (1) IT(R) = t(Q ∩ R), in particular IT(R) is idempotent.

(2) IT(R/N) = IT(R).

proof: (1) Q is a subring of QR, so Q ∩ R is 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) IT(R) and IT(R/N) are idempotent. Thus by Proposition 2.3 IT(R) = t (⋂_{X} W_{p}) and IT(R/N) = t (⋂_{Y} W_{p}) where X is the set of prime ideals p such that pR ≠ R and Y is the set of p with p(R/N) ≠ R/N. But p(R/N) = R/N if and only if

R = pR + N = pR + RN = pR + (pR + N)N = pR + pN + N^{2} = pR + N^{2}

and inductively we see that

p(R/N) = R/N ⟺ R = pR + N^{k} for every k.

Since by Proposition 1.32 N^{k} = 0 for large k it follows that p(R/N) = R/N if and only if pR = R, so that X = Y and IT(R) = IT(R/N). ✓

corollary 7.4. For any finite rank torsion free module G, IT(End G) = [IT(G): IT(G)]. proof: By Proposition 7.3 IT(End G) = t(Q ∩ End G) and clearly Q ∩ End G is the largest subring A of Q such that G is an A-module. On the other hand, by Proposition 4.11 if A is a subring of Q then G is an A-module if and only if G is t(A)-saturated, which is the same as saying that t(A) ≤ IT(G). Therefore IT(End G) = t(Q ∩ End G) is the largest idempotent type t such that t ≤ IT(G). But by Proposition 2.* this is [IT(G): IT(G)]. ✓ lemma 7.5. Let R be a finite rank torsion free ring, let U be a QR-module, and let M and N be R-submodules of U which are quasi-equal as W -modules. Then M is finitely generated as an R-module if and only if N is.

proof: Suppose that M is a finitely generated R-module. By Proposition 3.1 if M and N are quasi-equal we may as well assume that M ⊆ N and N/M is finitely generated (as a W -module). Then N is generated as an R-module by a set of generators for M together with a set of pre-images for generators of the W -module N/M. Therefore N is also a finitely generated R-module. ✓

proposition 7.6. If R is a finite rank torsion free ring with nil rad R = 0 then every left ideal in R is a quasi-summand and is finitely generated as a left ideal.

proof: If L is a left ideal in R then QL is a left ideal in QR. But if nil rad R = 0 then QR is a finite dimensional algebra with trivial radical, hence is semi-simple, so QL = QRe for some idempotent e ∈ QR. Now e is a right identity on QL so L = Le ⊆ Re. Then since L is 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, so Re 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. ✓

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 ◁ R. Furthermore by Proposition 7.6 L/N is a quasi-summand of R/N, hence by Proposition 3.21 L/N is quasi-pure in R/N. Thus by Proposition 3.14 L is quasi-pure in R. ✓

proposition 7.8. A finite rank torsion free ring R is left noetherian if and only if nil rad R is a finitely generated left ideal.

proof: ( ⇒ ): Clear since if R is noetherian then all left ideals are finitely generated.

( ⇐ ): Let N = nil rad R. Since nil rad R/N = 0, it follows from Proposition 7.6 that all left ideals of R/N are finitely generated so R/N is left noetherian. Now by Proposition 1.31 there exists a filtration R ⊃ N ⊃ N^{2} ⊃ ⋯ ⊃ N^{r} = 0 for some finite r, where N = 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. ✓ 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, so QP = QD. Since P is a D-submodule of D and D is a cyclic D-module it follows from Lemma 3.13 that P is quasi-equal to D. Thus by Proposition 3.11 D/P has finite length, so D/P is an artinian integral domain, hence a field. Thus P is a maximal ideal.

Thus D is a noetherian integral domain such that all prime ideals are maximal. If in addition D is integrally closed then by Proposition 0.* D is a dedekind domain. ✓

caution: Corollary 7.9 depends on our convention that all rings are assumed to be algebras over W. It is not a valid theorem in general commutative ring theory.

Obviously Proposition 7.8 is equally valid if we substitute “right noetherian” for “left noetherian.” The following well known example shows that these two concepts are not equivalent.

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

proof: R is isomorphic to the matrix ring ( \begin{matrix}W & 0 \\ Q & Q \end{matrix} ), i.e. the ring of 2 × 2 matrices

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

N = ( \begin{matrix}0 & 0 \\ Q & 0 \end{matrix} ) .

Then it is easy to see that N is an ideal in R and N^{2} = 0. Furthermore

R/N ≈ ( \begin{matrix}W & 0 \\ 0 & Q \end{matrix} ) ≈ W × Q, a ring with trivial nil radical. Thus N = nil rad R. Now

multiplication between elements of R and N operates as follows:

( \begin{matrix}w & 0 \\ q & q′ \end{matrix} ) ( \begin{matrix}0 & 0 \\ x & 0 \end{matrix} ) = ( \begin{matrix}0 & 0 \\ q′x & 0 \end{matrix} ) , ( \begin{matrix}0 & 0 \\ x & 0 \end{matrix} ) ( \begin{matrix}w & 0 \\ q & q′ \end{matrix} ) = ( \begin{matrix}0 & 0 \\ wx & 0 \end{matrix} ) .

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. ✓ PRIME & SEMI-PRIME RINGS. We are primarily interested in torsion free rings because those W -modules which are the underlying modules of certain torsion free rings are very special for the general structure theory of finite rank torsion free W -modules. For instance in Chapter 2 we noticed that among rank-one modules the subrings of Q play a special role. Hence it is plausible to suspect that the underlying modules of rings in general might be special.

Unfortunately, though, the mere fact that a finite rank torsion free module G can be given a ring structure imposes only mild restrictions on the structure of the W -module G. Indeed any G can be given the structure of a commutative “ring without identity,” simply by making the multiplication trivial. If one now uses the standard construction to “unitize” G one gets a commutative W -algebra with the underlying W -module isomorphic to W ⊕ G and containing G as a prime ideal. Explicitly, the multiplication on W ⊕ G is given by (w_{1}, g_{1}) · (w_{2}, g_{2}) = (w_{1}w_{2}, w_{1}g_{2} + w_{2}g_{1}). One can note that G is the nil radical of this ring.

example 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. Thus End R = End(W ⊕ W) so that nil rad End R = 0. And yet nil rad R = 0 ⊕ W ≠ 0.

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

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

Delete? As soon as one imposes the condition that nil rad R = 0, the restrictions

✠ on R as a W -module become much more stringent. For instance, we have seen in Proposition 7.8 that in this case R is noetherian. Another important observation is that if H is a fully invariant submodule of any ring R then in particular H is invariant under both left and right multiplication by elements of R so that H must be an ideal in R and QH must also be an ideal in QR. If nil rad R = 0 then QR 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 rad QR = Q(nil rad R), a ring R is semi-prime if and only if QR is semi-simple. (The reader should not confuse this usage with the use of the term “prime ring” by a few other (misguided!) authors to denote the rings ℤ and ℤ/pℤ for prime numbers p.) 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. ✓

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 = 1 or e = 0, and that nil rad S = 0 (since nil rad S ≠ S.)

( ⇐ ): If a finite dimensional algebra S has trivial radical then it is semi-simple, hence a product of simple algebras. If it has no non-trivial central idempotents then there can be only one factor in this product, so S is a simple algebra. ✓

proposition 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 ◁ 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 = 0 and QR has no proper idempotents. But by Proposition 1.32 nil rad QR = Q(nil rad R) and by definition nil rad R = 0 if and only if R is semi-prime.

(2) ⇒ (4): If IJ = 0 then I_{∗}J_{∗} = 0. Thus without loss of generality we may suppose that I and J are pure. Now IJ = 0 is not possible if I = J = R, so one of I 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.

(4) ⇒ (3): If N = nil rad R then by Proposition 1.31 N is an ideal and N^{n} = 0, where n = rank R. Now if N ≠ 0 then 0 < rank N < n = rank R so n ≥ 2 and if we set I = J = N^{n−1} then IJ = 0. Likewise if e is a central idempotent in QR and I = R ∩ QRe = R ∩ eQR, J = R ∩ QR(1 − e) = R ∩ (1 − e)QR, then I and J are ideals in R and IJ = 0. Thus if there are no proper non-trivial ideals I and J with IJ = 0 then e = 0 or e = 1 so that R has no proper central idempotents and likewise

N = nil rad R = 0. ✓

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

proposition 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 = End G. If N = nil rad R then (NG)_{∗} is a fully invariant pure submodule of G and by Proposition 1.32 rank(NG)_{∗} = rank NG < rank G. Thus if G has 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 = QEnd G then by Proposition 3.16 G is quasi-equal to eG ⊕ (1 − e)G. Thus eG and (1 − e)G are quasi-pure submodules of G and they are fully invariant since if φ ∈ End G = R then φ(eG) = φe(G) = eφ(G) ⊆ eG and likewise φ(1 − e)G ⊆ (1 − e)G. Thus if G has no non-trivial proper pure fully invariant submodules then G ∩ eQG = 0 or G ∩ (1 − e)QG = 0 and it follows that e = 0 or e = 1. Thus R is semi-prime and has no proper central idempotents, hence by Proposition 7.15 R is a prime ring.

(4) This follows immediately from (2) and (3). ✓

In abelian group theory, a finite rank torsion free group G with no non-trivial proper pure fully invariant subgroups is called irreducible, the point being that QG is an irreducible (QEnd G)-module. In module theory the term “irreducible” has been already preempted so, in line with prevailing fashion, your author would suggest the term “E-irreducible.”

corollary 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.

(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) and R[t] are fully invariant pure submodules of R.

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

corollary 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. ✓

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. Then QR = QR_{1} × ⋯ × QR_{k} where each R_{i} is a subring of QR which is quasi-pure in R and R_{i}/(R_{i} ∩ QN) is a prime ring. Furthermore

(1) R is quasi-equal to R_{1} × ⋯ × R_{k}.

(2) If R is commutative and integrally closed then R = R_{1} × ⋯ × R_{n}.

(3) Z is quasi-equal to Z_{1} × ⋯ × Z_{k}, where Z_{i} = Center R_{i}, with equality if Z is integrally closed.

(4) N is quasi-equal to N_{1} ⊕ ⋯ ⊕ N_{k}, where N_{i} = R_{i}N.

(5) N_{i} is quasi-equal to N ∩ R_{i}.

(6) N_{i} = nil rad R_{i}. proof: (1) By Proposition 1.32 QN is the nil radical of QR. Therefore QR/QN is a semi-simple algebra and hence is a finite product of simple rings. Let ē_{1}, …, ē_{k} be the central idempotents corresponding to this decomposition of QR/QN. By Lemma 1.34 the ē_{i} are images of central idempotents e_{i} ∈ QR. Write R_{i} = Re_{i}. Note that R ⊆ R_{1} × ⋯ × R_{k} since every r ∈ R can be written as r = r1 = re_{1} + ⋯ + re_{k} and note also that QR = Q(R_{1} ×⋯×R_{k}). Since R_{1} ×⋯×R_{k} is generated as an R-module by the finite set e_{1}, …, e_{k} it follows from Lemma 3.13 that R is quasi-equal to R_{1} × ⋯ × R_{k}. In particular each R_{i} is a quasi-summand of R and hence by Proposition 3.21 is quasi-pure in R.

(2) ✠

(3) The center of R is quasi-equal to the center of R_{1} ×⋯×R_{k}, which is Z_{1} ×⋯×Z_{k}.

(4) The same reasoning as in (1) shows that N is quasi-equal to e_{1}N ⊕ ⋯ ⊕ e_{k}N. Furthermore e_{i}N = e_{i}RN = R_{i}N = N_{i}.

(5) Since e_{i} ∈ QR there exists w ≠ 0 ∈ W such that we_{i} ∈ R and so wN_{i} = we_{i}N ⊆ N ∩ e_{i}R = N ∩ R_{i}. On the other hand, since e_{i}^{2} = e_{i}, multiplication by e_{i} restricts to the identity on e_{i}R = R_{i} and hence also on N ∩ R_{i}. Thus N ∩ R_{i} = e_{i}(N ∩ R_{i}) ⊆ e_{i}N = N_{i} ⊆ R_{i}. Thus N ∩ R_{i} is quasi-equal to N_{i}.

(6) By (3) N_{i} is a quasi-summand of R and hence is quasi-pure in R, so that N_{i} is quasi-equal to R ∩ QN_{i} = R ∩ e_{i}QN. By Proposition 1.32 QN = nil rad QR and so it is well known from the theory of artinian rings that e_{i}QN = nil rad QR_{i}. Thus N_{i} is quasi-equal to R ∩ nil rad QR_{i} ⊆ R_{i} ∩ nil rad QR_{i} = nil rad R_{i}. Since N_{i} ◁ R_{i} (why?) it follows that N_{i} = nil rad R_{i}.

In particular, this shows that each R_{i}/(R_{i} ∩ QN) is a prime ring. ✓

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

proof: QR is a finite product of rings S_{i} such that S_{i}/(nil rad S_{i}) is a simple ring. Thus no generality is lost in supposing that R/N is simple.

✠

2) Suppose now that R is commutative of characteristic c ≠ 0. Then 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, this case r = 1 being trivial. Suppose now that N^{2} = 0. Let QR^{c} denote the set of all elements x^{c} for x ∈ QR. Since c = char R, R^{c} is a subring of R. Furthermore, since c ≥ 2 and N^{2} = 0, if x^{c} ≠ 0 then x ∉ QN and so x is invertible since QN is the unique maximal ideal of QR. Thus QR^{c} is a subfield of QR. Since QR is finite dimensional, there is a maximal subfield S of QR containing R^{c}. Since elements of QN are nilpotent, S ∩ QN = 0. We claim that S ⊕ QN = QR. If not, let x ∈ QR with x ∉ S ⊕ QR. Then x^{c} ∈ R^{c} ⊆ S. Now the polynomial X^{c} − x^{c} = (X − x)^{c} must be irreducible over S since c = char S 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. Thus S ⊕ QN = QR, as desired.

Now suppose that r > 2. By induction we may assume that QR/QN^{r−1} contains a subfield S̄ such that QR/QN^{r−1} = S̄ ⊕ QN/QN^{r−1} . Now let T be the subring of QR such that QN^{r−1} ⊆ T and T/QN^{r−1} = S̄ ≈ QR/QN^{r−1} . Then QN^{r−1} = nil rad T. Furthermore (QN^{r−1})^{2} = 0. Thus by the case r = 2 it follows that T contains a subring S such that T = S ⊕ QN^{r−1} . But then QR = S ⊕ QN since **** ✓

✠

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. Then R ∩ S is 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. Thus R is a finitely generated D-module. This result was first proved in [Pierce], using steps from [B & P]. The proof is a fascinating amalgam of Wedderburn Theory, Galois Theory, commutative ring theory, as well as torsion free module theory.

Recall that if an an integral domain D is a subring of the center of a ring R then we say that R is a finite integral extension of D if R is a finitely generated D-module.

lemma 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 R is a finite integral extension of D if and only if R is quasi-equal to a free D-submodule of QR.

(2) If D′ is an integral domain containing D such that QD = QD′ then D′ is a finite integral extension of D if and only if D and D′ are quasi-equal.

proof: (1) ( ⇐ ): If R is quasi-equal to a free D-submodule of QR then R is a finite integral extension of D since by Lemma 7.5 a module quasi-equal to a finitely generated D-module is itself finitely generated as a D-module.

( ⇒ ): Since R is a D-module, QR is a vector space over the field QD and dim_{QD} QR ≤ rank R < ∞. Let M be the D-module generated by a basis for QR over QD contained in R. Then since this basis is linearly independent over D, M is a free D-module. Furthermore M ⊆ R and QM = QR. Thus if R is finitely generated over D then R is quasi-equal to M by Lemma 3.13.

(2) ( ⇒ ): If D is a subring of D′ and QD = QD′ and D′ is a finite integral extension of D, i.e. a finitely generated D-module, then by Lemma 3.13 D′ is quasi-equal to D.

( ⇐ ): If D is a subring of D′ and D′ is quasi-equal to D then since D is a finitely generated D-module, by Lemma 7.5 D′ is also a finitely generated D-module, i.e. a finite integral extension of D. ✓

lemma 7.23. Let F be a finite separable extension of Q and let W′ be the integral closure of W in F. Then

(1) If R is a ring such that F is a subring of QR then R is quasi-equal to W′R.

(2) If D is an integral domain with quotient field F then W′D is the integral closure of D. In particular, W′D is a dedekind domain.

proof: (1) If F is a separable extension of Q, then by Proposition 0.* W′ is a finitely generated W -module. It follows that W′R is a finitely generated R-module since any finite set of generators for W′ as a W -module will also generate W′R as a R-module. Since R ⊆ W′R it follows from Lemma 3.13 that R is quasi-equal to W′R.

(2) Let D′ be the integral closure of D (in its quotient field F). Then W ⊆ D′ and so W′ ⊆ D′ and thus W′D ⊆ D′. It is well known that W′ is a dedekind domain (Proposition 0.*). Since W′ ⊆ W′D ⊆ QD = QW′, W′D is a dedekind domain by Proposition 1.*. In particular W′D is integrally closed, so since D ⊆ W′D it follows that

D′ ⊆ W′D. Thus D′ = W′D. ✓

theorem 7.24. Let D be a finite rank torsion free prime ring. Let F = Center QEnd D. Then

(0) QD can be naturally identified as a subring of QEnd D.

(1) F ⊆ Center QD.

(2) F ∩ D = Center End D.

(3) QEnd D = End_{F} QD.

(4) D is a finite integral extension of the commutative ring F ∩ D and is quasi-isomorphic to a free F ∩ D-module.

(5) End(F ∩ D) ≈ F ∩ D.

(6) F ∩ D is strongly indecomposable. Suppose in addition that D is commutative and contained in a galois extension Q′ of Q. Let D′ be the integral closure of D in Q′. Then

(7) F is the fixed field of {σ ∈ Gal(Q′/Q) | σ(D′) = D′}.

proof: (0) By Proposition 7.2.

(2) Since D ⊆ End D and F = Center QEnd D it follows that D ∩ F ⊆ Center End D. On the other hand, clearly Center End D ⊆ Center QEnd D = F. And if φ ∈ Center End D then in particular φ commutes with right multiplication by elements of D. Thus if for r ∈ D we write ρ_{r} for the map x ↦ xr then φ(r) = φ(1r) = φρ_{r}(1) = ρ_{r}φ(1) = φ(1)r. Thus φ = d̃ where d = φ(1), i.e. given the identification in (0) we can write φ = φ(1) ∈ D. Thus Center End D ⊆ D so that Center End D = F ∩ D.

(1) Since by (2) Center End D ⊆ D it follows that F = Center QEnd D ⊆ QD. Now if d ∈ QD and d ∈ Center QEnd D = F then d commutes with (left multiplication by) the other elements of QD, so d ∈ Center QD. Thus F ⊆ Center QD.

(3) By Proposition 7.16 End D is a prime ring, so QEnd D is a simple algebra. We claim that QD is a simple (QEnd D)-module. In fact, if M is a (QEnd D)-submodule of QD then in particular M must be invariant under left and right multiplication by elements of QD so that M is a two-sided ideal in QD and since QD is a simple ring it follows that either M = QD or M = 0.

It now follows from Jacobson’s version of the Wedderburn Theorem that QEnd D ≈ End_{Δ} QD where Δ = End_{QEnd D} QD, a skew field. Now if φ ∈ Δ, then in particular φ is QD-linear on the right, so, as seen in the proof of (2), Δ ⊆ QD ⊆ QEnd D. It follows that Δ consists of those mappings in QEnd D which are (QEnd D)-linear, i.e. Δ = Center QEnd D = F. Thus QEnd D ≈ End_{F} QD.

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

(5) Since D is quasi-isomorphic to a free F ∩ D-module, any endomorphism of F ∩ D can be extended to a quasi-endomorphism of D, hence by (3) is F -linear, and so a fortiori is F ∩ D-linear. Thus End(F ∩ D) = End_{F ∩D} F ∩ D = F ∩ D.

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

(7) If D is commutative and contained in a galois extension Q′ of Q, let D′ be the integral closure of D in Q′. Then since Q′ is separable over QD, by Proposition 0.* D′ is a finite integral extension of D and hence by Lemma 7.22 is quasi-isomorphic to a direct sum of copies of D. Thus QEnd D′ is isomorphic to a full matrix ring over QEnd D and so Center QEnd D′ = Center QEnd D = F and so F is also the field of definition for D′.

In order to show that F is the fixed field of {σ ∈ Gal(Q′/Q) | σ(D′) ⊆ D′}, it suffices by Galois Theory to show that for σ ∈ Gal(Q′/Q), σ(D′) ⊆ D′ if and only if σ fixes F. Now σ(D′) is isomorphic to D′ and therefore is also a dedekind domain, and in particular is integrally closed in Q′. On the other hand since σ is monic, if σ(D′) ⊆ D′ then by Proposition 3.9 σ(D′) is quasi-equal to D′ and therefore by Lemma 7.22 D′ is an integral extension of σ(D′). It follows that if σ(D′) ⊆ D′ then σ(D′) = D′. Now the field of definition for σ(D′) is σ(F) , so since the field of definition is determined uniquely by the ring D′, if σ(D′) = D′ then σ(F) = F and thus also σ(F ∩ D′) = F ∩ D′, so that σ restricts to an endomorphism σ_{F} ∈ End F ∩ D′. But by (5) End F ∩ D′ = F ∩ D′, so that σ_{F} is given by multiplication by some f ∈ D ∩ F. Since σ_{F} (1) = 1 we conclude that f = 1 and σ_{F} = 1 ∈ Gal(Q′/Q). In other words, if σ(D′) ⊆ D′ then σ restricts to the identity map on D′ ∩ F. Conversely if σ restricts to the identity map on D′ ∩ F then σ(D′) = D′ since by (4) D′ is the integral closure in Q′ of D′ ∩ F. Since Q(D′ ∩ F) = F, this shows that σ(D′) ⊆ D′ if and only if σ fixed

F. ✓

definition 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 QEnd D ≈ End_{F} QD, so dim_{F} QEnd D = (dim_{F} QD)^{2} . This means that if QEnd D ≈ QD then dim_{F} QD = 1 so F = QD.

(4) ⇒ (3): Immediate from part (6) of Theorem 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. 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 an integral domain. ✓

corollary 7.27. (1) Let D be an integral domain and let D′ be a finite integral extension of D. Then D′ and D have the same field of definition. In particular, the field of definition for D′ is contained in QD.

(2) Let D be a prime ring and K a subfield of the Q-algebra QD. Then K contains the field of definition for D if and only if D is a finite integral extension of K ∩ D. proof: (2) Let F be the field of definition for D.

( ⇒ ): By Theorem 7.24 D is a finite integral extension of F ∩ D. If F ⊆ K then a fortiori D is a finite integral extension of K ∩ D.

( ⇐ ): If D is a finite integral extension of K ∩D then by Lemma 7.22 D is quasi-equal to a direct sum of k copies of K ∩ D for some k ≥ 1. Thus QEnd D is isomorphic to the ring of k × k matrices over QEnd(K ∩ D). Therefore the center F of QEnd D is the center of QEnd(K ∩ D), and thus by part (2) of Theorem 7.24 is contained in K.

(1) Let F be the field of definition for D and F′ the field of definition for D′. By Theorem 7.24 D is a finite integral extension of F ∩ D. Thus if D′ is a finite integral 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 and F′ ∩ D ⊆ D ⊆ D′, a fortiori D is a finite integral extension of F′ ∩ D and thus by (2)

F′ ⊇ F. ✓

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 R_{1} × ⋯ × R_{n}, where the R_{i} are prime rings and the center Z of R is quasi-equal to Z_{1} × ⋯ × Z_{n}, where Z_{i} is the center of R_{i}. Thus by Lemma 7.5 Z_{1} × ⋯ × Z_{n} is a finitely generated Z-module and furthermore R is a finitely generated Z-module if and only if R_{1} × ⋯ × R_{n} is a finitely generated Z-module. It follows that it suffices to prove that each R_{i} is a finitely generated Z_{i}-module, for if X_{i} is a finite set of generators for R_{i} over Z_{i} and Y is a finite set of generators for Z_{1} × ⋯ × Z_{n} over Z, then R_{1} × ⋯ × R_{n} is generated over Z by the set of all elements xy for y ∈ Y and x ∈ X_{i} for some i. Thus we may assume without loss of generality that R is a prime ring. Let Z be the center of R. If F is the field of definition for R then by Theorem 7.24 F ∩ R ⊆ Z and R is a finite integral extension of F ∩ R, so a fortiori R is a finite integral extension of Z. ✓

We can now prove Theorem 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. Note that since QN is an ideal, π is a ring morphism and induces an isomorphism from the semi-simple ring QR/QN onto S. 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) ⊕ N it suffices by Proposition 3.18 to show that R ∩ S is quasi-equal to π(R). (Note that R ∩ S = π(R ∩ S) since the restriction of π to S is the identity map.) Furthermore we will show that if R is commutative and R ∩ S is integrally closed then R ∩ S = π(R), so that π restricts to a split surjection from R onto R ∩ S and so R = (R ∩ S) ⊕ N.

Consider first the case where S is a field. Let W′ be the integral closure of W in the field S. 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 then we may as well replace the ground ring W by W′, hence it suffices to consider the case S = Q. Then by Proposition 7.3 t(S ∩ R) = t(Q ∩ R) = IT(R) = IT(R/N) = t(π(R)), since R/N ≈ π(R) ⊆ Q. But by Proposition 2.15 two subrings of Q with the same type are identical. Thus S ∩ R = π(R), so R = (S ∩ R) ⊕ Ker π = (S ∩ R) ⊕ N, as required.

Now consider the more general case that S is commutative. Then by Proposition 7.19 R is quasi-equal to R_{1} × ⋯ × R_{n}, where for each i, R_{i} is a commutative prime ring, i.e. an integral domain, and if R ∩ S is integrally closed then R = R_{1} × ⋯ × R_{n} and all the R_{i} ∩ S are dedekind domains. Furthermore N is quasi-equal to N_{1} ⊕ ⋯ ⊕ N_{n} where N_{i} = nil rad R_{i}. Note also that R ∩ S is quasi-equal to ∏(R_{i} ∩ S) (why?). It thus suffices to prove that for each i, R_{i} is quasi-equal to (R_{i} ∩ S) ⊕ N_{i}, with equality if R_{i} is integrally closed. This was handled in the preceding paragraph.

Finally, consider a general ring R such that R/N is quasi-separable and consequently QR = S ⊕ QN. As above, let π: 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) is quasi-equal to Z ∩ 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 ∩ 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. 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 an essential submodule of π(R). Thus by Lemma 3.13 π(R) is quasi-equal to R ∩ S. Thus π restricts to a quasi-split surjection from R onto π(R), so by Proposition 3.18 R is quasi-equal to (R ∩ S) ⊕ N. ✓

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. Thus if R is strongly indecomposable, either N = 0 or S ∩ R = 0. But S ∩ R ≠ 0 since 1 ∈ S ∩ R. Thus N = 0 so R is semi-prime. Then by Proposition 7.19 R is quasi-equal to a product of prime rings. Since R is strongly indecomposable, there can be only one factor in this product, so R is prime. Then by Corollary 7.26, R is an integral domain. ✓

STRONGLY INDECOMPOSABLE DOMAINS. We saw in Proposition 2.2 and Proposition 2.3 that the subrings of Q are the localizations of W with respect to multiplicative sets and correspond in one-to-one fashion to the subsets of Spec W. Now if D is any finite rank integral domain then D is a essential subring of a finite dimensional extension field Q′ of Q. But if D is integrally closed and W′ is the integral closure of W in Q′, then W′ ⊆ D so that D is a rank-one W′-module. Thus we get the following proposition. 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 Q′ of Q and a subset Y of Spec W′, where W′ is the integral closure of W in Q′. (Recall that according to our convention, Spec W′ denotes the set of non-trivial prime ideals of W′). Specifically, D = ⋂_{Y} W′_{P} where Y is the set of prime ideals P of W′ such that D_{P} ≠ Q′.

(2) If Q′ is a separable extension of Q then a quasi-equality class of essential subrings of Q′ contains exactly one integrally closed subring D. Furthermore every subring of Q′ quasi-equal to D is a subring of D.

(3) Let Q′ be a galois extension of Q and let Y be a subset of Spec W′. Let G be the largest subgroup of Gal(Q′/Q) under which Y is invariant, i.e.

G = {σ ∈ Gal(Q′/Q) | (∀P ∈ Y) σ(P) ∈ Y }.

Then the field of definition for ⋂_{Y} W′_{P} is the fixed field of G. proof: (1) Let D be a finite rank torsion free quasi-separable integral domain and let Q′ = QD. Then Q′ is a finite separable extension of Q. Let W′ be the integral closure of W in Q. Then as indicated in the remarks preceding the Proposition, the integral closure of D′ in Q is a rank-one W′-module of idempotent type and by Proposition 7.23 is quasi-equal to D. Furthermore by Proposition 2.2 D′ is uniquely determined by its type over W′ and hence no two integrally closed subrings of Q′ can be quasi-equal as W′-modules. But this in fact means that they cannot be quasi-equal as W -modules, since if w ≠ 0 ∈ W is such that wD_{1} ⊆ D_{2} and wD_{2} ⊆ D_{1} , where D_{1} and D_{2} are W′-submodules of Q′, then D_{1} and D_{2} are actually quasi-equal as W′-modules. Thus by Proposition 2.3 the set of such D is in one-to-one correspondence with the family of subsets of Spec W′, where a domain D corresponds to the set of prime ideals P of W′ such that D is not P -divisible, i.e. D_{P} ≠ Q′.

(2) If D is an essential subring of Q′ and D′ is its integral closure, then D ⊆ D′ and by Proposition 7.22 D is quasi-equal to D′. Furthermore if two integrally closed essential subrings D_{1} and D_{2} are quasi-equal then by Lemma 3.13 they are quasi-equal to D_{1}D_{2} . Thus by Proposition 7.22 D_{1}D_{2} is an integral extension of D_{1} and D_{2} . Since these are integrally closed, D_{1} = D_{2} = D_{1}D_{2} .

(3) If D = ⋂_{P ∈Y} W′_{P} and σ ∈ Gal(Q′/Q), then σ(D) ⊆ D if and only if σ(P) ∈ Y whenever P ∈ Y . Therefore the claim follows from part (7) of Theorem 7.24. ✓

We say that two rings R_{1} and R_{2} are quasi-isomorphic as rings if there is a ring isomorphism from R_{1} onto a subring of QR_{2} which is quasi-equal to R_{2} .

The class of strongly indecomposable integrally closed integral domains has the important property that if two modules in this class are quasi-isomorphic they are in fact isomorphic, as the following proposition shows.

proposition 7.31. Let D_{1} and D_{2} be integral domains.

(1) If D_{1} and D_{2} are strongly indecomposable then they are isomorphic as W -modules if and only if they are isomorphic as rings.

(2) If D_{1} and D_{2} are integrally closed and strongly indecomposable then they are quasi-isomorphic as W -modules if and only if they are isomorphic as rings.

(3) In general, two integral domains D_{1} and D_{2} are quasi-isomorphic as W -modules if and only if they have the same rank, their fields of definition F_{1} and F_{2} are isomorphic fields, and F_{1} ∩ D_{1} and F_{2} ∩ D_{2} are quasi-isomorphic as rings.

proof: (1) Clearly if D_{1} and D_{2} are isomorphic as rings [i.e. as W -algebras] then they are isomorphic as W -modules. Conversely if they are strongly indecomposable and isomorphic as W -modules then by Corollary 7.26 D_{1} ≈ End D_{1} ≈ End D_{2} ≈ D_{2} since isomorphic modules have isomorphic endomorphism rings.

(3) ( ⇒ ): If D_{1} ∼ D_{2} then clearly rank D_{1} = rank D_{2} . And if F_{i} is the field of definition for D_{i} then by Theorem 7.24 F_{i} is the center of QEnd D_{i} and QEnd D_{1} ≈ QEnd D_{2} by Proposition 3.8, so F_{1} ≈ F_{2} . Furthermore F_{1} ∩ D_{1} and F_{2} ∩ D_{2} are strongly indecomposable by Theorem 7.24 and each D_{i} is quasi-isomorphic to a direct sum of copies of F_{i} ∩ D_{i}. Since D_{1} ∼ D_{2} it follows from Jónsson’s Theorem (Theorem 3.24) that F_{1} ∩ D_{1} ∼ F_{2} ∩ D_{2} . But then as rings

F_{1} ∩ D_{1} ≈ End(F_{1} ∩ D_{1}) ∼ End(F_{2} ∩ D_{2}) ≈ F_{2} ∩ D_{2}

by Theorem 7.24 and Proposition 3.8.

( ⇐ ): If F_{1} ∩ D_{1} and F_{2} ∩ D_{2} are quasi-isomorphic as rings (i.e. as W -algebras) then a fortiori they are quasi-isomorphic as W -modules. Since by Theorem 7.24 each D_{i} is quasi-isomorphic to a direct sum of copies of F_{i} ∩ D_{i} it follows that if rank D_{1} = rank D_{2} then D_{1} ∼ D_{2} .

(2) If D_{1} and D_{2} are strongly indecomposable then by Corollary 7.26 their fields of definition are QD_{1} and QD_{2} . Therefore by (3) if they are quasi-isomorphic as W -modules then they are quasi-isomorphic as rings. Therefore we may suppose that D_{1} and D_{2} have the same quotient field and are quasi-equal. But by Proposition 7.30 a quasi-equality class of integral domains in a field contains only a single integrally closed domain. Thus if D_{1} and D_{2} are integrally closed then they are equal. ✓

proposition 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. Since D is indecomposable, there can be only one term in this direct sum decomposition, so D is isomorphic to an ideal of F ∩ D. Thus rank D = rank F. Since F ⊆ QD, it follows that F = QD. Therefore by Corollary 7.26 D is strongly indecomposable. ✓ 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 = End R 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 in some proper separable extension field of Q then W′ is a projective W -module and hence quasi-isomorphism to a free W -module W^{n}, for n > 1. Then QEnd W′ is the ring of n × n matrices over Q so that End W′ is not commutative, and hence certainly is much larger than W′.

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

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

(2) A ring quasi-isomorphic to an E-ring is an E-ring.

(3) If End R ≈ R then R is an E-ring.

(4) If R_{1} × R_{2} is quasi-equal to an E-ring then R_{1} and R_{2} are E-rings. Furthermore Hom(R_{1}, R_{2}) = 0.

proof: (1) Recall that by Proposition 7.2 R ◁ End R. Thus if QR = QEnd R then R = QR ∩ End R = QEnd R ∩ End R = End R, so R is an E-ring.

(2) If R and R′ are quasi-isomorphic then we may suppose them quasi-equal. Then QR = QR′ and QEnd R = QEnd R′. 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 = rank R so λ is an isomorphism. Thus R is an E-ring by (1).

(4) By (2) we may assume that R_{1} × R_{2} is an E-ring. Now any endomorphism φ of R_{1} extends (non-uniquely) to an endomorphism of R_{1} × R_{2} and hence since R_{1} × R_{2} is an E-ring is given by multiplication by some (r_{1}, r_{2}) ∈ R_{1} × R_{2} . It follows that φ is given by multiplication by r_{1} . Thus the canonical embedding R_{1} → End R_{1} is surjective, so R_{1} is an E-ring.

Furthermore if ψ ∈ Hom(R_{1}, R_{2}) then ψ extends to an endomorphism of R_{1} × R_{2} , which must be given by multiplication by some r ∈ R_{1} × R_{2} . But since R_{1} is an ideal in R_{1} × R_{2} , rR_{1} ⊆ R_{1} and it follows that ψ = 0. Thus Hom(R_{1}, R_{2}) = 0. ✓

Note that by Proposition 3.43 Murley rings are E-rings. For Murley rings, the following proposition was part of Proposition 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 = 1 yields r = 1r = s1 = s. Thus xr = rx for all x ∈ R. Since this holds for all r ∈ R, R is commutative. ✓ proposition 7.35. Let G be quasi-equal to G_{1} ⊕ ⋯ ⊕ G_{n}, where the G_{i} are strongly indecomposable. Then End G is commutative if and only if End G_{i} is commutative for each i and Hom(G_{i}, G_{j}) = 0 for i ≠ j. In this case, End G ∼ End G_{1} × ⋯ × End G_{n}. proof: ( ⇐ ): Easy since if G is quasi-equal to G_{1} ⊕ ⋯ ⊕ G_{n} and Hom(G_{i}, G_{j}) = 0 for i ≠ j then End G ∼ End G_{1} × ⋯ × End G_{n}.

( ⇒ ): If End G is commutative, let π_{i} be the the quasi-projection of G onto G_{i} and let π̃_{i} be the corresponding idempotent in QEnd G. If φ ∈ Hom(G_{i}, G_{j}) with j ≠ i then φπ_{i} ∈ QHom(G, G_{j}) and thus φπ_{i} corresponds to a map φ̃ ∈ QHom(G, G) = QEnd G given by φ̃(x) = φπ_{i}(x). Since φ̃(G) ⊆ G_{j} and j ≠ i, π̃_{i}φ̃ = 0. But since by hypothesis QEnd G is commutative and φ̃ = φ̃π̃_{i}, we get 0 = π̃_{i}φ̃ = φ̃π̃_{i} = φ̃. It follows that φ = 0. Thus Hom(G_{i}, G_{j}) = 0 for i ≠ j. Likewise if φ, ψ ∈ QEnd G_{i} then φ̃ψ̃ = ψ̃φ̃ and it follows that φψ = ψφ. ✓

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

theorem 7.36. A ring R is an E-ring if and only if R is quasi-isomorphic to a finite product of strongly indecomposable integral domains D_{1} × ⋯ × D_{n} such that Hom(D_{i}, D_{j}) = 0 for i ≠ j. In particular, a strongly indecomposable E-ring is an integral domain. proof: ( ⇐ ): By Lemma 7.33 it suffices to prove that if D_{1} × ⋯ × D_{n} are strongly indecomposable domains with Hom(D_{i}, D_{j}) = 0 for i ≠ j then D_{1} × ⋯ × D_{n} is an E-ring. Now since the D_{i} are strongly indecomposable, End D_{i} ≈ D_{i} by Corollary 7.26. Thus D_{1} × ⋯ × D_{n} ≈ End D_{1} × ⋯ × End D_{n} ≈ End(D_{1} ⊕ ⋯ ⊕ D_{n}), where the latter isomorphism holds by the assumption that Hom(D_{i}, D_{j}) = 0 for i ≠ j. Furthermore this isomorphism takes an element (d_{1}, …, d_{n}) ∈ D_{1} × ⋯ × D_{n} to the endomorphism given by left multiplication by (d_{1}, …, d_{n}). Thus D_{1} × ⋯ × D_{n} is an E-ring.

( ⇒ ): If R is an E-ring then End R ≈ R and by Proposition 7.34 R is commutative. Thus by Proposition 7.35 R has a quasi-direct decomposition R_{1} ⊕ ⋯ ⊕ R_{n} where the R_{i} are strongly indecomposable pure W -submodules of R such that Hom(R_{i}, R_{j}) = 0 for i ≠ j and R ≈ End R ∼ End R_{1} × ⋯ × End R_{n}. It further follows from the condition Hom(R_{i}, R_{j}) = 0 that then the subspaces QR_{i} are invariant under left and right multiplication by elements of QR and hence are ideals in QR and so QR is the ring-theoretic product of the rings QR_{i}. Now let e_{i} be the identity element of QR_{i} and let D_{i} be the subring of QR_{i} generated by R_{i} together with e_{i}. Then by Lemma 3.13 D_{i} is quasi-equal to R_{i} and is therefore also strongly indecomposable, and R is quasi-equal to D_{1} × ⋯ × D_{n}. Furthermore the D_{i} are integral domains by Corollary 7.29 [This requires separability!]. ✓

✠

COMPOSITA. Although the question of determining the structure of the tensor product of two finite rank torsion free modules is a very difficult one, if the two modules are rings then their tensor product is also a ring, which narrows down the range of possibilities considerably. In the case of the tensor product of two integral domains, the problem reduces to a variation of a standard result in field theory. definition 7.37. Let D_{1} and D_{2} be integral domains, at least one of which is quasi-separable. By a compositum of D_{1} and D_{2} is meant a non-trivial torsion free (D_{1} ⊗ D_{2})-algebra C such that C is an integral domain and the structural map D_{1} ⊗ D_{2} → C is surjective. Note that the restrictions of the structural map to D_{1} ⊗ 1 and 1 ⊗ D_{2} must be monic, since these maps extend to maps from the fields QD_{1} ⊗ 1 and 1 ⊗ QD_{2} to QC. We say that two composita are equivalent if they are isomorphic as (D_{1} ⊗ D_{2})-algebras.

proposition 7.38. Let D_{1} and D_{2} be integral domains, at least one of which is quasi-separable. Then

(1) D_{1} ⊗ D_{2} ∼ ∏ C_{i}, where the C_{i} form a maximal set of non-equivalent composita of D_{1} and D_{2} .

(2) D_{1} ⋆ D_{2} ∼ ∏′ C_{i} where the product is taken over the set of C_{i} which are not fields.

(3) If Q̄ is an algebraically closed extension of Q containing D_{1} and D_{2} then we may choose C_{i} = D_{1}σ(D_{2}), where σ: QD_{2} → Q̄ is a morphism of Q-algebras.

(4) If D_{2} ⊆ Q′ ⊆ QD_{1} where Q′ is a normal extension of Q, then D_{1} ⊗ D_{2} ∼ ∏ D_{1}σ(D_{2}), where σ ranges over the set of all Q-algebra morphisms from QD_{2} into Q′.

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

Since either QD_{1} or QD_{2} is separable over Q, QD_{1} ⊗ QD_{2} has trivial nil radical [Reference], hence is a product of commutative simple algebras, i.e. a product of fields, say QD_{1} ⊗ QD_{2} = ∏ F_{i}. Then the F_{i} are composita of QD_{1} and QD_{2} and are clearly non-equivalent since the structural maps QD_{1} ⊗ QD_{2} = ∏ F_{i} → F_{i} have different kernels. By Proposition 7.19 D_{1} ⊗ D_{2} ∼ ∏ C_{i} where C_{i} is the (D_{1} ⊗ D_{2})-submodule of F_{i} generated by its identity element. Now clearly each C_{i} is a compositum of D_{1} ⊗ D_{2} and they are all non-equivalent. Furthermore if C is any compositum then the surjection D_{1} ⊗ D_{2} → C induces a ring morphism φ: ∏ QC_{i} → QC and since C is a domain Ker φ is a prime ideal, meaning that there is a unique i with φ(C_{i}) ≠ 0. Then there exists a commutative diagram

D_{1} ⊗ D_{2} \begin{aligned}C_{i} & ⟶^{⊆} & QC_{i} \\ ↓ & & ↓ \\ C & ⟶^{⊆} & QC. \end{aligned}

Now both left-hand (slanted) maps are surjections. And the right-hand vertical map is a non-trivial homomorphism of fields, hence is monic. It thus follows that the map C_{i} → C is an isomorphism of (D_{1} ⊗ D_{2})-algebras, so that C and C_{i} are equivalent composita of D_{1} and D_{2} .

(2) By Corollary 7.17 each integral domain C_{i} is either reduced or divisible. And by Proposition 7.13 it is divisible if and only if it is a field. Thus D_{1} ⋆ D_{2} is the product of those C_{i} which are not fields.

(3) Since QC is a finite dimensional extension field of Q we may choose C so that QC ⊆ Q̄. Furthermore, since C is a compositum of D_{1} and D_{2} , the composition D_{1} → D_{1} ⊗ D_{2} → C is a monomorphism, hence we may choose C so that it is an inclusion. Then we get a morphism of W -algebras σ: D_{2} → D_{1} ⊗ D_{2} → C such that C = D_{1}σ(D_{2}).

(4) If D_{2} ⊆ Q′ ⊆ D_{1} and Q′ is a normal extension of Q, then σ(D_{2}) ⊆ QD_{1} . As we have seen, the composita may be chosen in the form D_{1}σ(D_{2}). Now suppose two composita D_{1}σ(D_{2}) and D_{1}τ(D_{2}) are equivalent. We then have a commutative diagram

D_{1} ⊗ D_{2} {1 ⊗ σ \atop 1 ⊗ τ} \begin{aligned}D_{1}σ(D_{2}) & ⟶^{⊆} & QD_{1} \\ ↓ & & ∥ \\ D_{1}τ(D_{2}) & ⟶^{⊆} & QD_{1}. \end{aligned}

Since these maps are all D_{1}-linear, the vertical maps must be identity maps. Thus σ = τ so that the non-equivalent composita of D_{1} and D_{2} are in one-to-one correspondence with the Q-algebra morphisms from QD_{2} into Q′. ✓

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. If p′ is a prime of W and p′ ≠ p then there exists w ∈ W such that w ∉ p, so w ∉ P and thus w is 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 field of 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 a set of representatives for the left coset space Gal(Q′/Q)/ Gal(Q′/F) and we then have σ(V)′ = σ(V′). Now Gal(Q′/Q) acts transitively on the family of σ(V′) and since F is the field of definition of V′, by part (7) of Theorem 7.24 Gal(Q′/F) is the stabilizer of V′ under this action. Thus the σ(V′), and also the σ(V), are distinct. Now since σ(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-rank V = P′-rank_{W′} W′ ⊗ V = P′-rank_{W′} ∏ σ(V′) = 1.

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

✠

proposition 7.40. Let Q′ be a finite galois extension of Q and let V′ be a valuation ring on Q′. Let p be the unique prime ideal of W such that pV′ ≠ V′ and let F be the field of definition for V′. Then

(1) V′ is the only valuation ring on Q′ containing V′ ∩ F.

(2) If W_{F} is the integral closure of W in F then there exist unique prime ideals P ∈ Spec W_{F} and P′ ∈ Spec W′ such that V′ = W′_{P′} and V′ ∩ F = (W_{F})_{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 ranges over 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 W_{p}.

(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 integral closure of V′ ∩ F in Q′. Thus if V′_{1} is any valuation ring on Q′ containing V′ ∩ F then V′_{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′_{1}} is a discrete valuation ring on Q′ and V′ ∩ F = (W_{F})_{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 W_{p} 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, if V′_{1} ≠ σ(V′) then V′_{1}σ(V′) = Q′ since V′_{1} and σ(V′) are maximal subrings of Q′. Furthermore by Proposition 2.3 the set of W_{p}-subalgebras of Q′ is a distributive lattice, so the assumption that V′_{1}σ(V′) = Q′ for all σ leads to the contradiction V′_{1} = V′_{1}W′_{p} = V′_{1} ⋂ σ(V′) = ⋂ V′_{1}σ(V′) = Q′. Therefore V′_{1} = σ(V′) for some σ.

(6) ( ⇐ ): If K ⊆ F then K ∩ V ◁ 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.

( ⇒ ): 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 F K is its field of definition. But since F ∩ V ⊆ (K ∩ V)(F ∩ V) ⊆ V , the field of definition for (K ∩ V)(F ∩ V) is F (why?). Therefore if K ∩ V is indecomposable then

✠

F K = F, i.e. K ⊆ F. ✓

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.

Recall first that by Proposition 3.43 any splitting ring R can be written as a product ∏ R_{i}, where the R_{i} 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 {V_{i}}_{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 there is at most one i ∈ I such that V_{i} is not p-divisible. Then F is the field of definition for ⋂ V_{i} and ⋂ V_{i} ∩ F is a Murley ring. proof: If F is the field of definition for the valuation ring V_{i} then by Proposition 7.39 V_{i} ∩ F is a Murley ring. Now (⋂ V_{i} ∩ F)_{p} = (V_{j} ∩ F)_{p}, where V_{j} is the unique V_{j} with (V_{j})_{p} ≠ K. Thus p-rank (⋂ V_{i} ∩ F) = p-rank (⋂ V_{i} ∩ F)_{p} = p-rank(V_{j} ∩ F)_{p} = 1. Thus ⋂ V_{i} ∩ F is a Murley ring and is an integral domain, hence by Proposition 3.44 is strongly indecomposable. Furthermore by Theorem 7.24 V_{i} is an integral extension of V_{i} ∩ F. Thus if x ∈ ⋂ V_{i} then for any i, x ∈ V_{i}, so the coefficients for the minimal polynomial for x over W lie in V_{i} ∩ F. Hence the minimal polynomial for x over F has coefficients in ⋂ V_{i} ∩ F, showing that ⋂ V_{i} is an integral extension of ⋂ V_{i} ∩ F, and by the separability of K is a finite integral extension (Proposition 0.*). Hence by Corollary 7.27 ⋂ V_{i} and ⋂ V_{i} ∩ F have the same field of definition. Since ⋂ V_{i} ∩ F is strongly indecomposable, by Corollary 7.26 this field of definition is F. ✓

lemma 7.42. Let D be a quasi-separable domain and suppose that all the valuation rings on QD containing D have the same field of definition F. Then there exists a finite rank quasi-separable splitting ring R = ∏ R_{i} such that R_{i} ⊆ 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 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. In fact, each valuation ring V containing D is a finite integral extension of a valuation ring V_{F} with QV_{F} = F. If D_{F} = ⋂ V_{F} then D is an integral extension of D_{F} , since if d ∈ D then d ∈ V for each valuation ring V and so since V is an integral extension of V_{F} the minimal polynomial for d has coefficients in ⋂ V_{F} = D_{F} . Furthermore since QD is separable, D is a finite integral extension of D_{F} 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 D_{F} it suffices to prove that D_{F} 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 ≠ 0 choose a valuation ring V_{p} on F containing D with p-rank V_{p} ≠ 0. (The subscript p here does not denote localization.) Let R_{1} = ⋂_{p} V_{p}. Then R_{1} ⊆ F and in particular R_{1} is quasi-separable and by Lemma 7.41 R_{1} 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 R_{2} 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 ≠ V (why?), in this way we finally obtain a set of Murley rings R_{1}, …, R_{n} with n ≤ [F: Q] such that for each valuation ring V containing D, D ⊆ R_{i} ⊆ V for some i. Since D is integrally closed, it is the intersection of the valuation rings containing it. Thus D = ⋂ R_{i}, so by Proposition 1.10 D is isomorphic to a pure submodule of ⨁ R_{i}. If, as in the proof of Proposition 6.21, we let R̃_{i} = R_{i} ⊕ A_{i}, where A_{i} is an appropriate subring of Q, then R̃_{i} is a quasi-separable splitting ring and R_{i} is R̃_{i}-split. Thus by Proposition 6.20 D is R-split for some finite rank quasi-separable R. ✓

proposition 7.43. Let D be an integral domain and let Q′ be a finite galois extension of Q. Then Q′ contains the field of definition for D if and only if there is a splitting ring ∏ R_{i} such that D is R-split and R_{i} ⊆ 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 direct sum of copies of F ∩ D, it suffices to prove that there exists a splitting ring R = ∏ R_{i} with all R_{i} ⊆ 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 D_{E} be the intersection of the valuation rings V on QD whose field of definition is E. By Lemma 7.42 D_{E} is R-split for some finite rank splitting ring R = ∏ R_{i} with all R_{i} ⊆ Q′. And since D is the intersection of the valuation rings containing it, D = ⋂ D_{E}, and so by Proposition 1.10 D is isomorphic to a pure submodule of ⨁_{E} D_{E}. Since QD is a separable extension of Q, there are only finitely many subfields E. Thus by Proposition 6.20 D is R-split for some finite rank splitting ring R = ∏ R_{i} with R_{i} ⊆ Q′ for all i.

( ⇐ ): Suppose that D is R-split for some splitting ring ∏ R_{i} with all R_{i} ⊆ 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 field of definition for D′ is the same as the field of definition for D, so that, replacing D with D′, no generality is lost in assuming that Q′ ⊆ QD. Then by Proposition 7.38 R_{i} ⋆ D ∼ ⨁ σ(R_{i})D, the product being taken over certain σ ∈ Gal(Q′/Q).

(The R_{i}-module structure on σ(R_{i})D is given via σ, i.e. (r, x) ↦ σ(r)x.) If φ ∈ End_{Q′} QD then since σ(R_{i}) ⊆ Q′, φ extends to a ∏ R_{i}-linear endomorphism of Q(R ⋆ D) = ⨁ Qσ(R_{i})D. Therefore by Proposition 6.4 End_{Q′} QD ⊆ QEnd D = End_{F} QD where F is the field of definition for D and the latter equality is by Theorem 7.24. In particular, a Q′-linear projection QD → Q′ is F -linear, hence restricts to the identity on

F. Thus F ⊆ Q′. ✓

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 = dim F = [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 and we may suppose QG = F. Let W′ be the integral closure of W in F. By separability W′ is a finitely generated W -module and since G is an essential submodule of W′G, W′G is quasi-equal to G by Lemma 3.13. Therefore without loss of generality we may suppose that G = W′G, i.e. that G is a W′-module. Hence if G is R-split then by Proposition 6.28 W′ ⋆ R is a splitting ring over W′ and G is (W′ ⋆ R)-split. In other words, without loss of generality we may replace the ground ring W by W′ and hence suppose that F = Q and G ⊆ Q. Then G is a rank-one module and G ◁ R ⋆ G so t(G) ∈ T(R ⋆ G) = T(∏ R_{i} ⋆ G). But R_{i} ⋆ G is a finitely generated R_{i}-module, hence is quasi-isomorphic to a direct sum of copies of R_{i}, and hence by Proposition 2.24 T(∏ R_{i} ⋆ G) is contained in the lattice generated by the sets T(R_{i}). But since R_{i} is an integral domain, by Corollary 7.17 T(R_{i}) consists of a single idempotent type. Since by Proposition 2.15 the set of idempotent types is a lattice, it follows that T(∏ R_{i} ⋆ 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. ✓

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. theorem 7.45. Let R = ∏ R_{i} be a splitting ring, where each R_{i} is a finite rank quasi-separable dedekind domain, and let Q′ be a finite galois extension of Q such that R_{i} ⊆ Q′ for all i. Let G be a strongly indecomposable R-split W -module and let W′ be the integral closure of W in Q′. Then G is quasi-isomorphic to an integral domain if and only if W′ ⊗ G is quasi-isomorphic to a direct sum of rank-one W′-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 D is QD. Hence by Proposition 7.43 if G (and hence D) is R-split and if R = ∏ R_{i} and Q′ is a galois extension of Q containing all the R_{i} then D ⊆ Q′. By Proposition 7.38 W′ ⊗ D ∼ ∏ W′σ(D) as W′-algebras, where the σ range over a certain subset of Gal(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 ∼ ⨁ B_{k} where the B_{k} are rank-one W′-modules of idempotent type then by Proposition 2.15 the B_{k} can be taken to be subrings of Q′. Since by Lemma 7.22 W′ is quasi-isomorphic to a free W -module, as a W -module W′ ⊗ G is quasi-isomorphic to a direct sum of copies of G. Therefore by Jónsson’s Theorem (Theorem 3.24) it follows that G is quasi-isomorphic to a strongly indecomposable quasi-summand of B_{1} . But by Theorem 7.24 B_{1} is quasi-isomorphic a direct sum of copies of the strongly indecomposable module B_{1} ∩ F, where F is the field of definition for the ring B_{1} . Thus G is quasi-isomorphic to the integral domain B_{1} ∩ F. ✓

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, where A is 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, let A be a pure rank-one submodule of H and let B be any rank-one submodule of G. Since G is homogeneous t(A) = t(B) and so by Proposition 2.4 there exists a non-trivial homomorphism φ: A → B. Then φ can be thought of as a mapping in Hom(A, G), so by hypothesis it extends to an endomorphism ψ of G. Since H is fully invariant, ψ(A) ⊆ ψ(H) ⊆ H and thus B ∩ H ≠ 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 G has no non-trivial proper pure fully invariant submodules and so by Proposition 7.16 End G is prime. ✓

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: End G → Hom(A, G) be the map 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 ⊗ G_{0} , where A is a rank-one module with [t(A): t(A)] ≤ IT(G_{0}), then G is strongly homogeneous if and only if G_{0} 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 is monic by (1), this is equivalent to ρ being an isomorphism.

(3) If G ≈ A ⊗ G_{0} where A is a rank-one module and if A_{0} = End A and t(A_{0}) = [t(A): t(A)] ≤ IT(G_{0}) then there is a commutative diagram

End G ⟶^{ρ} Hom(A, G) = Hom(A ⊗ A_{0}, A ⊗ G_{0})
≈↓  ^{≈}↓
End G_{0} ⟶^{ρ_{0}}  Hom(A_{0}, G_{0})

where ρ_{0} is the corresponding restriction map and the vertical isomorphisms exist by Corollary 4.35. Thus ρ is an isomorphism if and only if ρ_{0} is an isomorphism and it follows from (2) that G is strongly homogeneous if and only if G_{0} is strongly homogeneous. ✓

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: End G → 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: End G → 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: End G → 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. ✓

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 ⊈ P then I + P = J + P = R and so IJ + P = R and thus IJ ⊈ P. Note also that if R is 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. Then DAD is an ideal in D so DAD = 𝔞D for some 𝔞 ⊆ W. Then since A ⊆ DAD, A = A ∩ DAD = A ∩ 𝔞D = 𝔞A by purity and therefore also A = 𝔞^{−1}A. Thus DAD = D𝔞^{−1}AD = 𝔞^{−1}DAD = 𝔞^{−1}𝔞D = 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. Then I is 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. Now if A is any pure rank-one submodule of D, then 𝔞A ⊆ 𝔞D ∩ A ⊆ I ∩ A so that, in the notation of Chapter 2, 0 ≠ 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. Thus 𝔞 = [I ∩ A: A] and I ∩ A = [I ∩ A: A] A = 𝔞A = 𝔞D ∩ A, where the latter equality holds since A ◁ D. Therefore I ∩ A = 𝔞D ∩ A holds for all pure rank-one submodules A of D. Since D is 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, then pD 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. If DAD ≠ D then since by Proposition 7.8 D is noetherian, there exists a maximal ideal I with DAD ⊆ I. Let p = W ∩ I. Then pD ⊆ 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, so since I is prime either xD ⊆ I or yD ⊆ I, which implies x ∈ I ∩ W = p or y ∈ p. Now if pD 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. But then A ⊆ DAD ⊆ I = pD and since A ◁ D, pA = A ∩ pD = A, so A is p-divisible and thus DAD is p-divisible, contradicting the assumption that I = pD is a proper ideal. It thus follows that DAD = D. ✓

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, where 𝔞 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) is an isomorphism. Now if G is strongly homogeneous and A ◁ G, then G is homogeneous and so G(t) = G. Furthermore by Lemma 7.48 the restriction map ρ: End G → 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. Then by (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 = End G. By Lemma 7.49 D is an integral domain. We claim that if A_{0} is any pure rank-one submodule of D then A_{0}D = D. In fact, by (1) D is strongly homogeneous and so by (4) there is an isomorphism μ_{0} : A_{0} ⊗ End D → D given by μ_{0}(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 μ′: A_{0} ⊗ D → D given by μ′(x ⊗ d) = λ(d)(x) = dx = xd. Thus A_{0}D = 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 = End G then by (3) I = 𝔞D for some ideal 𝔞 of W. If I ≠ 0 then 𝔞 ≠ 0 and 𝔞^{−1}D is a D-submodule of QD and 𝔞^{−1}D𝔞D = D. Thus every ideal of D is invertible, so D is a dedekind domain. ✓

theorem 7.52. A strongly indecomposable module G is strongly homogeneous if and only if G ≈ A ⊗ D, where A is 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 ≈ A_{0} ⊗ End G, where A_{0} 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 A_{0} ≈ A ⊗ [A_{0} : A_{0}], for some rank-one module A of locally trivial type. But if t′ = t([A_{0} : A_{0}]) then since G is t(A_{0})-saturated (since G is homogeneous), it follows from Corollary 7.4 that End G is t′-saturated and so by Proposition 4.37 [A_{0} : A_{0}] ⊗ End G ≈ End G. Thus G ≈ A ⊗ [A_{0} : A_{0}] ⊗ 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, then D 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 BD = DBD = D. Now let B′ = φ(B)_{∗}. As with B there is an isomorphism μ′: B′ ⊗ D ⟶^{≈} D. Let φ′ = μ′(φ ⊗ 1_{D})μ^{−1} : D → D. Then the restriction of φ′ to B is given by

b ↦ b ⊗ 1 ↦ φ(b) ⊗ 1 ↦ φ(b),

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

corollary 7.53. Let G and H be strongly indecomposable strongly homogeneous modules such that G ∼ H. If W is a principal ideal domain or if G and H are integral domains, then G ≈ H. proof: By Theorem 7.52 G ≈ A ⊗ G_{0} and H ≈ B ⊗ H_{0} where A and B are rank-one modules with locally trivial types and G_{0} and H_{0} are dedekind domains. Furthermore t(A)IT(G_{0}) = IT(G) = IT(H) = t(B)IT(B_{0}) and since by Corollary 7.3 IT(G_{0}) and IT(H_{0}) 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 G_{0} ≈ Hom(A, A ⊗ G_{0}) ≈ Hom(A, G) and likewise H_{0} ≈ Hom(B, H), so that G_{0} ∼ H_{0} and since G_{0} and H_{0} are strongly indecomposable integrally closed domains, G_{0} ≈ H_{0} by Proposition 7.31. Thus G ≈ H if W is a principal ideal domain or

if G = G_{0} and H = H_{0} . ✓

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 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. Then G is 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. Since G is 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.

( ⇐ ): 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 ≠ α: A → G. Let B = α(A)_{∗}. If φ is an automorphism of G with φ(A) = B then φ^{−1}α ∈ End A ⊆ Q (by Proposition 2.2), so φ^{−1}α is given by multiplication by some w ∈ W. Then the restriction of wφ = φw to A is φφ^{−1}α = 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. ✓

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

✠ 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 − 1 are isomorphic. proof: This has to do with the fact that every element of D can be written as wu, where w ∈ W and u is an invertible element of D. Now if K_{1} and K_{2} are pure submodules of D with corank 1 then K_{i} = Ker φ_{i} for some φ_{i} ∈ Hom(D, Q). But Hom(D, Q) ≈ QD, so it follows that φ_{1} = uφ_{2} for some u ∈ QD. Since Ker φ_{2} = Ker wφ_{2} , we may assume that u is a unit in D. But then

uK_{1} = u Ker uφ_{2} = Ker φ_{2} = K_{2},

so K_{1} ≈ K_{2} . ✓

MULTIPLICATIONS ON A STRONGLY INDECOMPOSABLE MODULE. The question of when a W -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.

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 ≠ 0 ∈ G and consider the map φ: G → G given by φ(x) = gx − xg. Then φ ∈ 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) = g^{2}x − g(gx). Then ψ(g) = (g^{2})g − g(g^{2}) = 0 since multiplication is commutative, so since g ≠ 0, ψ is not monic, and as above ψ = 0. Now consider ζ ∈ End G defined by ζ(x) = (gx)h − g(xh). Then ζ(g) = g^{2}h − 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 ≠ 0 ∈ G and gh = 0 then h is in the kernel of the map x ↦ gx, so this map must be trivial. Then for any g′ ∈ G, g is in the kernel of the map x ↦ xg′, so this 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 ≠ 0 ∈ QG, the map x ↦ gx is monic and hence is an automorphism of QG, so there exists e ∈ QG with ge = g. Then if θ is defined by θ(x) = xe − x, and w ≠ 0 ∈ W is such that we ∈ G, then wθ(G) ⊆ G, so that θ ∈ QEnd G. 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 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 G′ 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 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 G^{2} = G. Since G is an ideal in G′ and by Proposition 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. ✓

The assumption that every non-trivial endomorphism is monic can be weakened considerably under a separability assumption, still assuming that G is strongly indecomposable.

proposition 7.57. Let G be a strongly indecomposable module let N = nil rad End G. Suppose that End G /N is quasi-separable. Let μ: G ⊗ G → G be such that μ(G ⊗ G) ⊈ 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 g_{1} , g_{2} ∈ G such that μ(g_{1} ⊗ g_{2}) ∉ NG. Define π: End G → G by π(φ) = φ(g_{1}). If π(φ) = 0 then φ is 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(g_{2}) = ρ(g_{2})(g_{1}) = μ(g_{1} ⊗ g_{2}) ∉ NG. Thus since G is 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, where S 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, so we conclude 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 7.26 End G ∼ End S ≈ S. Thus End G is an integral domain and N = 0. Since End G has no zero divisors, by Proposition 3.3 every non-trivial endomorphism of G is monic, so the Proposition reduces to Proposition 7.56. ✓

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.

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 and then in a given field extension Q′ finding the set of prime ideals of the integral closure of W in Q′. 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.