11. DIRECT SUM DECOMPOSITIONS

E. L. Lady

June 27, 1998

The examples of pathological direct sum decompositions given in previous chapters are worth going over very carefully, to understand exactly what makes them work and how to do variations on them. Such careful study makes one appreciate how artfully contrived they are. This is in turn suggests that pathological decompositions are in fact far from typical.

We now turn to a sequence of theorems showing various conditions under which direct sum decompositions are nice.

This chapter is an anomaly in this book, because many of the major results really have nothing to do with finite rank torsion free modules. In this chapter, unless otherwise indicated, the symbols G, H, M, etc. will denote modules over an arbitrary ring R, which need not even be commutative.

The material in this chapter is nonetheless an essential part of the theory of finite rank torsion free modules. For one thing, it is essential that anyone working in the field know this information, and he will have a difficult time finding it anywhere else in readily accessible form, except in [Arnold]. Secondly, even though many of these results are valid over an arbitrary ring R, and were originally published as such, the original discoverers were for the most part abelian group theorists, or at least part-time abelian group theorists. (A partial exception is the material on stable range, which was originally due to ring theorists working on the Serre Conjecture. But even here, the results were extended to their present scope by Warfield, motivated by the study of finite rank torsion free abelian groups.)

These results are intimately associated with finite rank torsion free modules not because that is the environment required to nurture them, but because that is the environment in which they are needed.

We begin with an old chestnut. lemma 11.1. If H ⊆ M ⊆ H ⊕ K then M = H ⊕ (M ∩ K). proof: Let π be the projection from H ⊕ K onto H with kernel K. Then since H = π(G) ⊆ M, π restricts to a projection π′ of M onto H, so

M = H ⊕ Ker π′ = H ⊕ (M ∩ Ker π) = H ⊕ (M ∩ K). ✓

The holy grail of the study of direct sum decompositions is the Krull-Schmidt Property:

definition 11.2. An R-module G has the Krull-Schmidt Property if whenever G = M_{1} ⊕ ⋯ ⊕ M_{s} = N_{1} ⊕ ⋯ ⊕ M_{t} where all M_{i} and N_{j} are indecomposable, then t = s and the N_{i} can be renumbered as N′_{1}, …, N′_{s} such that for all i, N′_{i} ≈ M_{i}.

It is worth noting that the classical Krull-Schmidt Theorem, as well as Azumaya’s refinement of it, actually has a stronger conclusion than the Krull-Schmidt Property given above. Namely, if all the endomorphism rings End_{R} M_{i} are local rings, then the indecomposable summands M_{i} and N′_{i} are interchangeable with each other. In other words, for any k, G = M_{1} ⊕ ⋯ ⊕ M_{k} ⊕ N′_{k+1} ⋯ ⊕ N′_{s}. In fact, if X_{1}, …, X_{s} are modules such that for each i, X_{i} = M_{i} or X_{i} = N′_{i}, then G = X_{1} ⊕ ⋯ ⊕ X_{s}. We will call summands M_{i} with this property exchangeable.

We will essentially give Azumaya’s proof of the Krull-Schmidt Theorem as Proposition 11.28.

Even this is not the most one can ask for, though. The strongest possible uniqueness property one could ask for is that for each i, M_{i} and N′_{i} are actually identical as submodules of G: M_{i} = N′_{i}. This actually happens if each M_{i} is a fully invariant submodule of G.

The preceding paragraph brings up the need to distinguish between the two notations H = K and H ≈ K for submodules H and K of G. Often in algebra it is convenient to use an equal sign to indicate isomorphism, especially when the isomorphism in question is natural. In discussing direct sum decompositions, however, it is vital to reserve the notation H = K to mean that H and K are the same submodule of G. For instance, the hypothesis G = H ⊕ K = H ⊕ L implies K ≈ L since both K and L are isomorphic to G/H, whereas Example 1.49 shows that from H ⊕ K ≈ H ⊕ L one cannot necessarily conclude that K ≈ L.

One the other hand, we will seldom need to be equally punctillious in distinguishing between internal direct sums and external ones. Thus when forming the external direct sum H ⊕ K of two modules H and K, we will as usual denote the submodules {(h, 0) | h ∈ H} and {(0, k) | k ∈ K} simply by H and K. We will also not try to be totally consistent about whether we denote a typical element of this external direct sum as (h, k) or h + k.

Of course this notational convenience fails us for H ⊕ H, in which case we shall denote the two submodules in question by H ⊕ 0 and 0 ⊕ H.

FULLY INVARIANT SUMMANDS. Recall that a submodule H of G is called fully invariant in G if φ(H) ⊆ H for all φ ∈ End_{R} G. This is equivalent to the assertion that H is an (End_{R} G)-submodule of G.

proposition 11.3. Let G = H ⊕ K. Then H is fully invariant in G if and only if Hom_{R}(H, K) = 0.

proof: ( ⇐ ): Let κ be the projection of G = H ⊕ K onto K with kernel H and let φ ∈ End_{R} G. If Hom(H, K) = 0 then the restriction of κφ to H must be trivial. But this means that φ(H) ⊆ H. Thus H is fully invariant.

( ⇒ ): Let φ′ ∈ Hom_{R}(H, K). Then φ′ extends to a map φ ∈ End_{R} G defined by φ(h + k) = φ′(h). If H is fully invariant in K, then for all h ∈ H, φ′(h) = φ(h) ∈ H. But φ′(h) ∈ K and H ∩ K = 0. Thus φ′ = 0 for all φ′ ∈ Hom_{R}(H, K), so

Hom_{R}(H, K) = 0. ✓

proposition 11.4. If G = H ⊕ K = L ⊕ M and H is fully invariant in G there there exist summands L′′ of L and M′′ of M such that

H = (H ∩ L) ⊕ (H ∩ M) , L = (H ∩ L) ⊕ L′′, M = (H ∩ M) ⊕ M′′

and L′′ ⊕ M′′ ≈ K.

proof: Let λ and μ be the idempotents in End_{R} G with L = λ(G), M = μ(G) and λ + μ = 1. If H is fully invariant in G then λ and μ restrict to endomorphisms λ′ and μ′ of H, which must necessarily still be idempotent. Furthermore λ(H) = λ(λ(H)) = λ′(λ(H)) = λ′(H) since λ(H) ⊆ H and likewise μ(H) = μ′(H). Thus H = λ′(H) ⊕ μ′(H) = λ(H) ⊕ μ(H). Now λ(H) ⊆ H ∩ L and on the other hand H ∩ L = λ(H ∩ L) ⊆ λ(H) because λ restricted to H ∩ L is the identity map. Thus λ(H) = H ∩ L and likewise μ(H) = H ∩ M, so that H = (H ∩ L) ⊕ (H ∩ M).

Since L ∩ H and M ∩ H are summands of H and H is a summand of G, they are thus summands of G. But H ∩ L ⊆ L so by Proposition 11.1, L = (H ∩ L) ⊕ L′′ for some L′′. Likewise M = (H ∩ M) ⊕ M′′ for some M′′. Furthermore G = H ⊕ K = L ⊕ M = (H ∩ L) ⊕ (H ∩ M) ⊕ L′′ ⊕ M′′ = H ⊕ L′′ ⊕ M′′, so

L′′ ⊕ M′′ ≈ G/H ≈ K. ✓

corollary 11.5. Let H be an indecomposable summand of G. Then H is fully invariant in G if and only if it has the property that whenever G = L ⊕ M then H ⊆ L or H ⊆ M. proof: ( ⇒ ): By Proposition 11.4, H = (H ∩ L) ⊕ (H ∩ M). Since H is indecomposable, either H = H ∩ L or H = H ∩ M. Thus H ⊆ L or H ⊆ M.

( ⇐ ): Let G = H ⊕ K. We will show that if G has the stated property then Hom_{R}(H, K) = 0. In fact, let φ ∈ Hom_{R}(H, K) and let H′ = {h+φ(h) ∈ H⊕K | h ∈ H}. One checks that G = H′ ⊕ K (see Lemma 11.10 below). By hypothesis, H ⊆ H′ or H ⊆ K. Since H ∩ K = 0, thus H ⊆ H′. Thus for every h ∈ H, h + φ(h) ∈ H′ and h ∈ H′, so φ(h) ∈ H′. But φ(h) ∈ K and H′ ∩ K = 0. Thus φ(h) = 0 for all h ∈ H, so φ = 0. Thus by Proposition 11.3, H is fully invariant in G. ✓

corollary 11.6. (1) If G = M_{1} ⊕ ⋯ ⊕ M_{s} where the M_{i} are indecomposable modules and H is a fully invariant indecomposable summand of G then H = M_{i} for some i.

(2) If G = M_{1} ⊕ ⋯ ⊕ M_{s} = N_{1} ⊕ ⋯ ⊕ N_{t}, where the M_{i} and N_{j} are all indecomposable and the N_{i} are fully invariant in G, then s = t and the set {N_{1}, …, N_{s}} is the same as the set {M_{1}, …, M_{s}}.

proof: (1) It is immediate from Corollary 11.5 by induction that H ⊆ M_{i} for some i. By Proposition 11.1, H is a summand of M_{i}. Since M_{i} is indecomposable, it follows that H = M_{i}.

(2) By (1), for each j, N_{j} = M_{i} for some i and the result follows. ✓

The following proposition can be seen as a variation on Proposition 11.4.

proposition 11.7. Let G = H ⊕ K = L ⊕ M. If Hom_{R}(H, M) = 0 then there exists a summand K_{1} of K such that

G = H ⊕ (L ∩ K) ⊕ M = H ⊕ (L ∩ K) ⊕ K_{1}.

proof: Let μ be the projection onto M. Since Hom_{R}(H, M) = 0, μ(H) = 0 so that H ⊆ Ker μ = L. Thus by Proposition 11.1, L = H ⊕ (L ∩ K). Furthermore, since L ∩ K ⊆ K ⊆ G = H ⊕ (L ∩ K) ⊕ M, by Proposition 11.1 K = (L ∩ K) ⊕ K_{1} , where

K_{1} = K ∩(H ⊕M). Thus G = L⊕M = H ⊕(L∩K)⊕M = H ⊕K = H ⊕(L∩K)⊕K_{1} . ✓

the traditional proof of Baer’s Theorem (Theorem 4.33): Let G be a completely decomposable W -module and let G = L ⊕ M. We prove by induction on rank G (or on the cardinality of T(G)) that L is also completely decomposable. If G is t-homogeneous for some type t then this is Proposition 4.19. Otherwise note that if G = ⨁ A_{i} where the A_{i} are rank-one W -modules and if t is any type, then by Proposition 2.* G(t) = ⨁{A_{i} | t(A_{i}) ≥ t}. In particular, G(t) is completely decomposable and a summand of G. Choose t maximal in T(G) (which is finite by Proposition 2.*). Then G(t) is t-projective and by Proposition 4.4 it is fully invariant in G. And G = G(t) ⊕ K, where K = ⨁{A_{i} | t(A_{i}) ≠ t}. Thus by Proposition 11.4, L = (G(t) ∩ L) ⊕ L′′. Furthermore G(t) ∩ L is a summand of G(t) and L′′ is isomorphic to a summand of the completely decomposable module K. Then G(t) ∩ L is completely decomposable by Proposition 4.19 and L′′ is completely decomposable by the induction hypothesis. Thus L is completely decomposable. ✓

definition 11.8. Let G = H ⊕ K. We say that H is a cancellable summand of G if whenever G ≈ H ⊕ M then M ≈ K.

it is often more convenient to rephrase the definition as follows: H is a cancellable summand of H ⊕ K if whenever H ⊕ K = H′ ⊕ M with H′ ≈ H then K ≈ M.

proposition 11.9. A fully invariant summand H of an R-module G is cancellable if and only if H is not isomorphic to any proper summand of itself. proof: ( ⇐ ): Suppose G = H ⊕K = H_{1} ⊕M with H_{1} ≈ H. An isomorphism from H to H_{1} is given by an element φ ∈ Hom_{R}(H, G), and φ can be extended to an endomorphism of G by setting φ(K) = 0. Hence if H is fully invariant in G then H_{1} = φ(H) ⊆ H. Since by assumption H_{1} is a summand of G, it follows from Lemma 11.1 that H_{1} is a summand of H. Thus if H is not isomorphic to any proper summand then H_{1} = H and M ≈ G/H ≈ K. Thus H is cancellable.

( ⇒ ): Suppose that H ≈ H ⊕ P. If G = H ⊕ K then G ≈ H ⊕ P ⊕ K. If H is a cancellable summand of G then K ≈ P ⊕ K. But if H is fully invariant in H ⊕ K then by Proposition 11.3, Hom_{R}(H, K) = 0 and it follows that Hom_{R}(H, P) = 0. Since P is a summand of H, this forces P = 0. This shows that H is not isomorphic to a proper summand of itself. ✓

Fully invariant summands of a module are like uniformly convergent sequences in analysis: essentially any plausible assertion about them will be valid. Unfortunately they are few and far between. A large part of the remainder of this chapter will be devoted to finding conditions weaker than full invariance which still yield nice conditions for direct sum decompositions.

The following example from Chapter 1 shows that not all direct summands are cancellable.

Example 1.49. Let W be a dedekind domain having an infinite number of prime ideals. Let P_{1} and P_{2} be disjoint infinite sets of prime ideals not containing a certain non-unit w ≠ 0 ∈ W. Let a be another non-trivial non-unit in W such that a ∉ p for all p ∈ P_{1} ∪ P_{2} . Let H_{1} be the submodule of Q ⊕ Q generated by all (p^{−1}, 0) for p ∈ P_{1} and (0, p^{−1}) for p ∈ P_{2} together with (1, 1)/w. Let H_{2} be generated by the same set of elements except with (1, 1)/w replaced with (1, a)/w. Let A be the submodule of Q generated by all p^{−1} for p ∈ P_{2} . Then

(1) H_{1} and H_{2} are not isomorphic.

(2) A ⊕ H_{1} ≈ A ⊕ H_{2} .

definition 11.10. We say that an R-module H has the cancellation property if H is a cancellable summand of H ⊕ K for every K.

We note the following triviality.

proposition 11.11. If H is a module with the cancellation property then H is not isomorphic to a proper summand of itself.

proof: If H = H ⊕ 0 ≈ H ⊕ K and H has the cancellation property then it follows that

K = 0. ✓

proposition 11.12. The class of modules with the cancellation property is closed under (finite) direct sums and direct summands.

proof: Suppose that H_{1} and H_{2} have the cancellation property and H_{1} ⊕ H_{2} ⊕ K ≈ H_{1} ⊕ H_{2} ⊕ M. Then since H_{1} has the cancellation property, H_{2} ⊕ K ≈ H_{2} ⊕ M. Since H_{2} has the cancellation property, K ≈ M. Thus H_{1} ⊕ H_{2} is a cancellable summand of H_{1} ⊕ H_{2} ⊕ K for every K, hence H_{1} ⊕ H_{2} has the cancellation property.

Conversely, if H_{1} ⊕ H_{2} has the cancellation property and H_{1} ⊕ K ≈ H_{1} ⊕ M, then H_{1} ⊕ H_{2} ⊕ K ≈ H_{1} ⊕ H_{2} ⊕ M, so K ≈ M. Thus H_{1} is a cancellable summand of H_{1} ⊕ K for every K, so H_{1} has the cancellation property. ✓

THE EXCHANGE PROPERTY. Let G = H ⊕ K and let H′ ⊆ G. We say that H′ replaces H (relative to K) if G = H′ ⊕ K.

Note that if G = H ⊕ K = H′ ⊕ K′ then to say that H′ replaces H is equivalent to saying that K replaces K′ (relative to H′). lemma 11.13. Let H′ be a submodule of G = H ⊕ K. Let π be the projection of G onto H with Ker π = K.

(1) H′ replaces H if and only if π restricts to an isomorphism from H′ onto H.

(2) If H = H_{1} ⊕ H_{2} and H′_{1} is a submodule of G mapped isomorphically onto H_{1} by π, then G = H′_{1} ⊕ H_{2} ⊕ K.

(3) If H′ replaces H and H = H_{1} ⊕ H_{2} then there is a direct decomposition H′ = H′_{1} ⊕ H′_{2} such that H′_{1} replaces H_{1} (with respect to H_{2} ⊕ K.)

(4) The submodules H′ of G which replace H are precisely those of the form

H′ = {(h, φ(h)) | h ∈ H}

for φ ∈ Hom_{R}(H, K).

proof: (1) Since K = Ker π, G = H′ ⊕ K if and only if there exists a monomorphism σ: H → G such that σ(H) = H′ and σ splits π, i.e. πσ = 1_{H}. If π′ is the restriction of π to H′, then such a σ will exist if and only if π′ is an isomorphism (in which case σ = π^{′−1} ).

(2) Let π_{1} be the projection of G onto H_{1} with kernel H_{2} ⊕ K. Suppose that π maps H′ isomorphically onto H_{1} . Then π_{1} = π_{1}π and π(H′_{1}) = H_{1} so that π_{1} and π agree on H′_{1} . Thus π_{1} also maps H′_{1} isomorphically onto H_{1} . Therefore by (1), H′_{1} replaces H_{1} .

(3) Let π_{1} and π_{2} be the projections of G onto H_{1} and H_{2} associated with the direct decomposition G = H_{1} ⊕ H_{2} ⊕ K and let π = π_{1} + π_{2} , so that π is the projection of G onto H with kernel K. If H′ replaces H_{1} ⊕ H_{2} then by (1), π maps H′ isomorphically onto H_{1} ⊕ H_{2} . Thus there is a decomposition H′ = H′_{1} ⊕ H′_{2} such that H_{1} = π(H′_{1}). Thus by (2) H′_{1} replaces H_{1} .

(4) Any submodule H′ of H ⊕ K is a set of ordered pairs (h, k) for h ∈ H, k ∈ K, and is thus (the graph of) a relation from H to K. This is necessarily an additive and R-linear relation. It is a function φ if and only if no two ordered pairs in H′ have the same first coordinate, i.e. if and only if the restriction of the projection π to H′ is monic, and if that is so then

H′ = {(h, φ(h)) | h ∈ Domain(φ)}.

The domain of φ is H if and only if every element of H occurs as a first coordinate in H′, i.e. if and only if the restriction of π to H′ is surjective. Thus H′ has the form indicated in (4) if and only if the restriction of π to H′ is an isomorphism, and by (1) this is equivalent to the condition that H′ replaces H. ✓

It is a truism in mathematics that good theorems become definitions. We now use a weakened version of the conclusion of Proposition 11.4 to define an important property for direct summands. definition 11.14. We will say that that the summand H of G = H ⊕ K is exchangeable if whenever G = L ⊕ M then there exist direct sum decompositions

L = L′ ⊕ L′′, M = M′ ⊕ M′′

such that H replaces L′ ⊕ M′, i.e. G = H ⊕ L′′ ⊕ M′′.

lemma 11.15. If H is an exchangeable summand of H ⊕ K and H′ replaces H then H′ is also an exchangeable summand of H ⊕ K.

proof: This is clear since H′ ≈ H and there is thus an automorphism of H⊕K = H′⊕K taking H to H′. ✓

lemma 11.16. Let G = H ⊕ K = L ⊕ M. Suppose that there exist submodules L′′ ⊆ L and M′′ ⊆ M such that G = H ⊕ L′′ ⊕ M′′. Let L′ = L ∩ (H ⊕ M′′) and M′ = M ∩ (H ⊕ L′′). Then L = L′ ⊕ L′′, M = M′ ⊕ M′′ and there is a decomposition H = H_{1} ⊕ H_{2} such that

G = L ⊕ M = H_{1} ⊕ L′′ ⊕ M = L ⊕ H_{2} ⊕ M′′.

Furthermore, L′ ≈ H_{1} and M′ ≈ H_{2} .

proof: Since L′′ ⊆ L ⊆ G = L′′ ⊕ H ⊕ M′′, it follows from Lemma 11.1 that L = (L ∩ (H ⊕ M′′)) ⊕ L′′ = L′ ⊕ L′′. Likewise M = M′ ⊕ M′′. And since G = L ⊕ M = L′ ⊕ L′′ ⊕ M′ ⊕ M′′ = H ⊕ L′′ ⊕ M′′, H replaces L′ ⊕ M′. Therefore by Lemma 11.13 there is a decomposition H = H_{1} ⊕ H_{2} such that G = L ⊕ M = H_{1} ⊕ L′′ ⊕ M = L ⊕ H_{2} ⊕ M′′.

Since H_{1} replaces L′ and H_{2} replaces M′, then L′ ≈ H_{1} and M′ ≈ H_{2} by

Lemma 11.13. ✓

proposition 11.17. If H is fully invariant in G = H ⊕ K then H and K are exchangeable summands of G.

proof: If G = H ⊕ K = L ⊕ M then by Proposition 11.4 H = (H ∩ L) ⊕ (H ∩ M) and there exist L′′ ⊆ L and M′′ ⊆ M such that L = (L ∩ H) ⊕ L′′ and M = (M ∩ H) ⊕ M′′. Thus G = L ⊕ M = (H ∩ L) ⊕ L′′ ⊕ (H ∩ M) ⊕ M′′ = H ⊕ L′′ ⊕ M′′. This shows that H is an exchangeable summand of G. Furthermore G = H ⊕ K = (H ∩ L) ⊕ (H ∩ M) ⊕ K, so K is an exchangeable summand of G. ✓ lemma 11.18. If H_{1} ⊕ H_{2} is an exchangeable summand of H_{1} ⊕ H_{2} ⊕ K, then H_{2} is an exchangeable summand of H_{2} ⊕ K. proof: Let H_{2} ⊕ K = L ⊕ M. Then G = H_{1} ⊕ H_{2} ⊕ K = H_{1} ⊕ L ⊕ M. If H_{1} ⊕ H_{2} is an exchangeable summand of G then there exist decompositions

H_{1} ⊕ L = X′ ⊕ X′′ and M = M′ ⊕ M′′

such that

G = H_{1} ⊕ H_{2} ⊕ X′′ ⊕ M′′.

Now since

H_{2} ⊕ M′′ ⊆ H_{2} + M ⊆ H_{2} ⊕ K ⊆ G = H_{2} ⊕ M′′ ⊕ H_{1} ⊕ X′′,

it follows from Lemma 11.1, H_{2} ⊕ K = H_{2} ⊕ L′′ ⊕ M′′, where

L′′ = (H_{2} ⊕ K) ∩ (H_{1} ⊕ X′′) = (L ⊕ M) ∩ (H_{1} ⊕ X′′) ⊆ (L ⊕ M) ∩ (H_{1} ⊕ L) = L.

Then by Lemma 11.1 L′′ is a summand of L. This shows that H_{2} is a exchangeable summand of H_{2} ⊕ K. ✓

proposition 11.19. If H_{2} is an exchangeable summand of H_{2} ⊕ K and H_{1} is an exchangeable summand of H_{1} ⊕ H_{2} ⊕ K, then H_{1} ⊕ H_{2} is an exchangeable summand of H_{1} ⊕ H_{2} ⊕ K. proof: Let G = H_{1} ⊕ H_{2} ⊕ K and suppose G = L ⊕ M. Since H_{1} is an exchangeable summand of G, there exist decompositions L = L′ ⊕ L_{1} , M = M′ ⊕ M_{1} such that

G = H_{1} ⊕ H_{2} ⊕ K = H_{1} ⊕ L_{1} ⊕ M_{1}.

By Lemma 11.13 the projection ω of G onto L_{1} ⊕ M_{1} with kernel H_{1} maps H_{2} ⊕ K isomorphically onto L_{1} ⊕ M_{1} , so that L_{1} ⊕ M_{1} = ω(H_{2}) ⊕ ω(K) and since H_{2} is an exchangeable summand of H_{2} ⊕ K, ω(H_{2}) is an exchangeable summand of ω(H_{2}) ⊕ ω(K) = L_{1} ⊕ M_{1} . Thus there exist decompositions L_{1} = L′_{1} ⊕ L′′ and M_{1} = M′_{1} ⊕ M′′ such that

L_{1} ⊕ M_{1} = ω(H_{2}) ⊕ L′′ ⊕ M′′.

Then L = L′ ⊕ L_{1} = L′ ⊕ L′_{1} ⊕ L′′ and M = M′ ⊕ M′_{1} ⊕ M′′ and

G = H_{1} ⊕ L_{1} ⊕ M_{1} = H_{1} ⊕ ω(H_{2}) ⊕ L′′ ⊕ M′′.

Since ω maps H_{2} isomorphically on ω(H_{2}), by Lemma 11.13 H_{2} replaces ω(H_{2}) and G = H_{1} ⊕ ω(H_{2}) ⊕ L′′ ⊕ M′′ = H_{1} ⊕ H_{2} ⊕ L′′ ⊕ M′′. Thus H_{1} ⊕ H_{2} is an exchangeable summand of G. ✓ definition 11.20. We say that H has the exchange property if whenever H is a summand of an R-module G then H is an exchangeable summand of G.

(It is more customary to call this property the finite exchange property. Since we will have no interest in infinite direct sums, we have abbreviated the term.)

proposition 11.21. The class of R-modules with the exchange property is closed under direct sums and direct summands.

proof: If H_{1} and H_{2} have the exchange property and G = H_{1} ⊕ H_{2} ⊕ K then H_{2} is an exchangeable summand of H_{2} ⊕ K and H_{1} is an exchangeable summand of H_{1} ⊕ H_{2} ⊕ K. Therefore by Proposition 11.19 H_{1} ⊕ H_{2} is an exchangeable summand of G = H_{1} ⊕ H_{2} ⊕ K. Since this is true for any K, this shows that H_{1} ⊕ H_{2} has the exchange property.

Conversely if H_{1} ⊕ H_{2} has the exchange property then for any K, H_{1} ⊕ H_{2} is an exchangeable summand of H_{1} ⊕ H_{2} ⊕ K. Therefore by Lemma 11.18 H_{2} is an exchangeable summand of H_{2} ⊕ K. Since this is true for any K, this shows that H_{2} has the exchange property. ✓

The modules of interest in this book are always direct sums of indecomposable modules. Thus in light of Proposition 11.21 in order to identify those modules having the exchange property it suffices to identify the indecomposable ones. This turns out to be very straightforward.

lemma 11.22. Let φ, ψ ∈ End_{R} H with φ + ψ = 1. Let L and M be the submodules of H ⊕ H defined as

L = {(h, φ(h)) | h ∈ H}
M = {(h, −ψ(h) | h ∈ H}.

Then H ⊕ H = L ⊕ M.

proof: If (h, φ(h)) ∈ L ∩ M then φ(h) = −ψ(h). Since φ + ψ = 1, h = φ(h) + ψ(h) = −ψ(h) + ψ(h) = 0. Thus L ∩ M = 0. Now for every h ∈ H, (0, h) = (h, φ(h)) − (h, −ψ(h)) ∈ L ⊕ M and (h, 0) = (ψ(h), φψ(h)) + (φ(h), −ψφ(h)) ∈ L ⊕ M. (Note that φψ = ψφ since

ψ = 1 − φ.) Thus H ⊕ H = L ⊕ M. ✓

proposition 11.23. Let H be an indecomposable R-module . The following conditions are equivalent:

(1) H has the exchange property.

(2) H ⊕ 0 is an exchangeable summand of H ⊕ H.

(3) End_{R} H is a local ring.

proof: (1) ⇒ (2): Clear.

(2) ⇒ (3): Suppose that H ⊕ 0 is an exchangeable summand of H ⊕ H. In order to prove that End_{R} H is local, it suffices to prove that if φ ∈ End_{R} H then either φ or 1 − φ is an automorphism. Let ψ = 1 − φ and let

L = {(h, φ(h)) | h ∈ H} and
M = {(h, −ψ(h)) | h ∈ H} ⊆ H ⊕ H.

By Lemma 11.22, H ⊕ H = L ⊕ M. There is an isomorphism H → L given by h ↦ (h, φ(h)) and an analogous isomorphism H → M. In particular L and M are indecomposable. Write H_{1} = H ⊕ 0 and H_{2} = 0 ⊕ H. Since H_{1} is exchangeable in H_{1} ⊕ H_{2} = L ⊕ M and L and M are indecomposable, it follows that either H_{1} ⊕ H_{2} = H_{1} ⊕ L or H_{1} ⊕ H_{2} = H_{1} ⊕ M. If H_{1} ⊕ H_{2} = H_{1} ⊕ L it then follows from Proposition 11.13 that the projection π_{2} from H_{1} ⊕ H_{2} onto H_{2} restricts to an isomorphism from L onto H_{2} . But then the composition h ↦ (h, φ(h)) ↦ π_{2}(h, φ(h)) = φ(h) is an automorphism of H, i.e. φ is invertible. Likewise if H_{1} ⊕ H_{2} = H_{1} ⊕ M then ψ is invertible. Thus for any φ ∈ End_{R} H, either φ or 1 − φ is invertible, so End_{R} H is a local ring.

(3) ⇒ (1): Suppose that End_{R} H is local and G = H ⊕ K = L ⊕ M. Let λ and μ be the idempotents in End_{R} G projecting G onto L and M and let π be the projection from G onto H with Ker π = K. Then λ + μ = 1 so that π = πλ + πμ. Since π restricts to the identity map in End_{R} H and End_{R} H is local, either πλ or πμ restricts to an invertible map on H. If, say, the restriction of πλ to H is invertible, then λ restricts to a split monomorphism from H into L and L = λ(H) ⊕ L′′ for some L′′ so that G = λ(H) ⊕ L′′ ⊕ M. And since λ maps H isomorphically onto λ(H), by Lemma 11.13 G = H ⊕ L′′ ⊕ M. This shows that H is an exchangeable summand of G. Since this is true for all G containing H as a summand, H has the exchange property. ✓

proposition 11.24. If H has the exchange property then H has the cancellation property if and only if every summand of H is a cancellable summand. proof: ( ⇒ ): If H has the cancellation property then by Proposition 11.12 every summand of H also has the cancellation property and therefore is a cancellable summand of H.

( ⇐ ): Suppose that H has the exchange property and G = H ⊕ K = H′ ⊕ M with H′ ≈ H. Then by Lemma 11.16 there exist decompositions H = H_{1} ⊕ H_{2} , H′ = H′_{1} ⊕ H′′, and M = M′ ⊕ M′′ such that

G = H ⊕ H′′ ⊕ M′′

with H_{2} ⊕ M′′ ≈ M and

H_{1} ⊕ H′′ ≈ H′ ≈ H ≈ H_{1} ⊕ H_{2}.

If H_{1} is a cancellable summand of H this implies H′′ ≈ H_{2} . Then

K ≈ G/H ≈ H′′ ⊕ M′′ ≈ H_{2} ⊕ M′′ ≈ M.

This shows that H has the cancellation property. ✓

The following proposition will follow as a consequence of later results on the substitution property. But we can give a very simple proof right now.

proposition 11.25. If an R-module is a finite direct sum of indecomposable modules and has the exchange property then it has the cancellation property. proof: If H is a finite direct sum of indecomposable modules then by Proposition 11.12 H has the cancellation property if each indecomposable summand does. Furthermore by Proposition 11.21 if H has the exchange property then each summand of H does as well. Therefore it suffices to prove that an indecomposable module with the exchange property has the cancellation property. But if H is indecomposable then trivially every summand of H is an cancellable summand. Therefore the result follows from Proposition 11.24. ✓

proposition 11.26. Injective R-modules have the exchange property. proof: Let G = H ⊕ K = L ⊕ M, where H is injective. Let λ: G → L and μ: G → M be the idempotents in End_{R} G such that λ(G) = H, μ(G) = K, and λ + μ = 1. Since H is injective it contains injective envelopes H_{1} and H′_{2} for H ∩ L and H ∩ M. Furthermore, since H ∩L is an essential submodule of H_{1} and (H ∩L)∩(H ∩M) = 0, then H_{1} ∩ (H ∩ M) = 0 and since H ∩ M is an essential submodule of H′_{2} , thus H_{1} ∩ H′_{2} = 0. Thus H_{1} ⊕ H′_{2} is an injective submodule of H and hence H = H_{1} ⊕ H′_{2} ⊕ H_{3} for some H_{3} . Let H_{2} = H′_{2} ⊕ H_{3} , so that H = H_{1} ⊕ H_{2} . Now H ∩ Ker λ = H ∩ M ⊆ H_{2} and H_{1} ∩ H_{2} = 0, so H_{1} ∩ Ker λ = 0 and so λ(H_{1}) ≈ H_{1} and λ(H_{1}) is an injective envelope for λ(H ∩ L) = H ∩ L. Since λ(H_{1}) is an injective submodule of L, L = λ(H_{1}) ⊕ L′′ for some L′′. Likewise μ(H_{2}) ≈ H_{2} , so μ(H_{2}) is injective and M = μ(H_{2}) ⊕ M′′ for some M′′. Thus

G = λ(H_{1}) ⊕ L′′ ⊕ μ(H_{2}) ⊕ M′′.

Since μ maps H_{2} isomorphically onto μ(H_{2}), by Lemma 11.13

G = λ(H_{1}) ⊕ L′′ ⊕ H_{2} ⊕ M′′.

We will show that G = H ⊕ L′′ ⊕ M′′. Let φ: G → λ(H_{1}) be the projection map with Ker φ = L′′ ⊕ H_{2} ⊕ M′′. Note that since H ∩ L ⊆ L,

H ∩ L = λ(H ∩ L) ⊆ λ(H_{1})

so that H ∩ L = φ(H ∩ L) and also (H ∩ L) ∩ (L′′ ⊕ H_{2} ⊕ M′′) = 0. But H ∩ L is an essential submodule of its injective envelope H_{1} , so it follows that H_{1} ∩ (L′′ ⊕ H_{2} ⊕ M′′) = H_{1} ∩ Ker φ = 0. Thus φ(H_{1}) ≈ H_{1} and φ(H_{1}) is an injective envelope of φ(H ∩ L) = H ∩ L. But φ(H_{1}) ⊆ λ(H_{1}) and λ(H_{1}) is an injective envelope for H ∩ L. Therefore φ(H_{1}) = λ(H_{1}). It then follows from Lemma 11.13 that

G = H_{1} ⊕ L′′ ⊕ H_{2} ⊕ M′′ = H ⊕ L′′ ⊕ M′′.

Thus H is an exchangeable summand of G. Since this is true for every G containing H as a summand, H has the exchange property. ✓

Since over most rings there exist injective modules which are isomorphic to proper summands of themselves, Proposition 11.26 and Proposition 11.11 show that for a module which is not a finite direct sum of indecomposable modules, the exchange property need not imply the cancellation property.

theorem 11.27. Suppose that H has the exchange property and H⊕K = M_{1}⊕⋯⊕M_{n}. Then there exist direct sum decompositions H = H_{1} ⊕ ⋯ ⊕ H_{n} and M_{i} = M′_{1} ⊕ M′′_{i} for all i such that

M_{1} ⊕ ⋯ ⊕ M_{n} = H ⊕ M′′_{1} ⊕ ⋯ ⊕ M′′_{n}
= H_{1} ⊕ M′′_{1} ⊕ M_{2} ⊕ ⋯ ⊕ M_{n} = ⋯
= M_{1} ⊕ ⋯ ⊕ M_{n−1} ⊕ H_{n} ⊕ M′′_{n}.

proof: If, for fixed n, the assertion of the theorem is true for all direct sums H ⊕ K = M_{1} ⊕ ⋯ ⊕ M_{n}, then for convenience (and the sake of tradition) we will say that H has the n-exchange property. If H has the exchange property then by definition it has the 2-exchange property, and by Proposition 11.21 the same is true of all the summands of H. We now assume inductively that H and its summands have the (n − 1)-exchange property. Let G = H ⊕ K = M_{1} ⊕ ⋯ ⊕ M_{n}.

Since H has the exchange property it is an exchangeable summand of G and so there exist direct sum decompositions M_{1} = M′_{1} ⊕ M′′_{1} and M_{2} ⊕ ⋯ ⊕ M_{n} = X ⊕ Y such that

G = M_{1} ⊕ ⋯ ⊕ M_{n} = H ⊕ M′′_{1} ⊕ Y.

Furthermore by Lemma 11.16 we may choose

X = (M_{2} ⊕ ⋯ ⊕ M_{n}) ∩ (H ⊕ M′′_{1}) ⊆ H ⊕ M′′_{1}.

Furthermore H ≈ G/(M′′_{1} ⊕ Y) ≈ M′_{1} ⊕ X. Since X is thus isomorphic to a summand of H, by the induction hypothesis it has the (n − 1)-exchange property, so since X ⊕ Y = M_{2} ⊕ ⋯ ⊕ M_{n} it follows that there exist direct sum decompositions

M_{i} = M′_{i} ⊕ M′′_{i}

for i ≥ 2 such that

X ⊕ Y = M_{2} ⊕ ⋯ ⊕ M_{n} = X ⊕ M′′_{2} ⊕ ⋯ ⊕ M′′_{n}.

Since

X ⊆ H ⊕ M′′_{1} ⊆ G = X ⊕ Y ⊕ M_{1},

by Lemma 11.1, H ⊕ M′′_{1} = X ⊕ Z for some Z. Then

G = H ⊕ M′′_{1} ⊕ Y
= X ⊕ Z ⊕ Y  (because H ⊕ M′′_{1} = X ⊕ Z)
= X ⊕ Z ⊕ M′′_{2} ⊕ ⋯ ⊕ M′′_{n}  (because X ⊕ Y = X ⊕ M′′_{2} ⊕ …)
= H ⊕ M′′_{1} ⊕ M′′_{2} ⊕ ⋯ ⊕ M′′_{n}  (because X ⊕ Z = H ⊕ M′′_{1}).

Thus H has the n-exchange property. ✓

In light of Proposition 11.23, the following proposition is essentially Azumaya’s refinement of the Krull-Schmidt Theorem. We include it to show how it fits into the framework here.

proposition 11.28. Let G = H_{1} ⊕ ⋯ ⊕ H_{r} where the H_{i} are indecomposable R-modules and each H_{i} has the exchange property. Then G has the Krull-Schmidt Property. More precisely, whenever G = K_{1} ⊕ ⋯ ⊕ K_{s} and the K_{i} are also indecomposable, then r = s and the subscripts can be renumbered so that for each i the summands H_{i} and K_{i} are interchangeable, i.e. if for each i we set X_{i} = H_{i} or X_{i} = K_{i}, then G = X_{1} ⊕ ⋯ ⊕ X_{r}. proof: Since H_{1} has the exchange property, by Theorem 11.27 there exist decompositions K_{i} = K′_{i} ⊕ K′′_{i} for each i such that G = H_{1} ⊕ K′′_{1} ⊕ ⋯ ⊕ K′′_{s}. And furthermore H_{1} ≈ K′_{1} ⊕ ⋯ ⊕ K′_{s}. But since H_{1} is indecomposable, this implies that H_{1} ≈ K′_{i} for some i and K′_{j} = 0 for j ≠ i, so that K′′_{j} = K_{j} . Renumbering, we may suppose that i = 1, so H_{1} ≈ K′_{1} . Furthermore, since K_{1} is indecomposable, this implies that K′_{1} = K_{1} . Thus G = K_{1} ⊕ H_{2} ⊕ ⋯ ⊕ H_{r}. Continuing in this fashion completes the

proof. ✓

C-PROJECTIVE MODULES. We can notice that the traditional proof of Baer’s Theorem, as given above using Proposition 11.4, falls into two parts: the induction step, which is just general nonsense about fully invariant summands, and the case where G is t-projective for some t, which is where the real finite rank torsion free theory (namely Baer’s Lemma) comes into play.

By definition, a t-projective W -module is simply a module isomorphic to a summand of A^{n}, for some n ≥ 1 and a rank-one W -module A with t = t(A). In general, given a module C over some ring R, we will see that the decompositions of C^{n} play an essential role in the study of direct sum decompositions involving C. In particular, the properties of that C has as a summand of C ⊕ C will mostly determine the universal properties C will have as a summand of arbitrary R-module G (see Proposition 11.23 and Proposition 11.57 below).

It turns out that the study of the direct sum decompositions of C^{n} reduces to the study of finitely generated projective modules over End_{R} C.

It is a truism in category theory that the objects in a category do not play any significant role, but can best be thought of as mere placeholders for morphisms. In fact, one can actually dispense with the objects altogether, simply replacing them by the corresponding identity morphisms. The practical significance of this is that if one can match up the objects in two categories in such a way that the corresponding morphism groups are always the same, then the two categories are essentially the same. It is customary to formalize this by the use of functors and natural transformations, but such formalism may obscure the essential triviality of the result.

The categories of interest in this book, such as the category of finite rank torsion free W -modules under homomorphisms or under quasi-homomorphisms, or the category of left (right) R-modules over some ring R, are additive categories. This means that they have finite direct sums and a zero object (which can be thought of as the empty direct sum) and the morphism sets are abelian groups such that composition of morphisms is distributive over addition: α(β + γ) = αβ + αγ and (β + γ)α = βα + γα. Furthermore the Hom functors for the categories respect direct sums. Moreover, all these categories (at least up to now) have the properties that idempotents split, which means that if φ is an idempotent in the endomorphism ring of an object X, then φ(X) also belongs to the category. It then follows from basic principles that X = φ(X) ⊕ (1 − φ)(X). (Certainly this is clear for the specific categories of interest in this book.) We will call a subcategory of one of our categories a (full) additive subcategory if the group of morphisms between two objects in the subcategory is the same as the morphism group in the original category and the subcategory is closed under finite direct sums. proposition 11.29. Let X and Y be objects in two additive categories C and C′ such that idempotents split and such that the endomorphism rings of X and Y are isomorphic. Let P be the full additive subcategory of C consisting of all direct summands of finite direct sums of copies of X, and P′ the category of summands of finite direct sums of Y . Then P and P′ are equivalent. proof: We need to see that the objects in the two categories match up, and that if U and V are objects in P and U′ and V′ are the corresponding objects in P′, then Hom_{P}(U, V) is isomorphic to Hom_{P′}(U′, V′) in a natural way. By hypothesis, Hom_{C}(X, X) ≈ Hom_{C′}(Y, Y), and it then follows immediately that for m, n ≥ 1, Hom_{C}(X^{m}, X^{n}) ≈ Hom_{C′}(Y^{m}, Y^{n}). Finally, if U is a direct summand of X^{m} and V a direct summand of X^{n} in C, then there exist idempotents φ ∈ End_{C}(X^{m}) and ψ ∈ End_{C}(X^{n}) such that U = φ(X^{m}), V = ψ(X^{n}). These idempotents correspond to morphisms φ′ ∈ End_{C′}(Y^{m}) and ψ′ ∈ End_{C′}(Y^{n}) which must also be idempotent. Then the objects U and V in P will correspond to the objects φ′(Y^{m}) and ψ′(Y^{n}) in P′. Furthermore, the homomorphisms from U to V have the form ψαφ, where α ∈ Hom_{C}(X^{m}, X^{n}). Such a homomorphism will correspond to ψ′α′φ′ ∈ Hom_{C′}(φ′(Y^{m}), ψ′(Y^{n})), where of course α′ ∈ Hom_{C′}(Y^{m}, Y^{n}) is the morphism corresponding to α ∈ Hom_{C}(X^{m}, X^{n}). It is routine to verify that this correspondance between objects and morphisms of P to objects and morphisms of P′ is in fact a functor and has an obvious inverse functor. ✓

definition 11.30. Let C be an R-module. We say that an R-module X is C-projective if X is a summand of C^{n} for some n. (It is more traditional to call a module of this form finitely C-projective.)

Since a lot is known about projective modules over certain rings, the following proposition is useful in a variety of situations, and cuts through some routine but somewhat lengthy traditional proofs in the theory of finite rank torsion free W -modules.

proposition 11.31. The category of C-projective R-modules is equivalent to the category of projective right modules over the ring End_{R} C. In this equivalence, the summands of C correspond to the right ideals of End_{R} C generated by idempotents. Furthermore, this equivalence may be afforded by the functor Hom_{R}(C, ). proof: Let E = End_{R} C. It is a well known triviality in ring theory that if we think of E as a right E-module then End_{R} C = E ≈ End_{E}(E). (In fact, an isomorphism End_{E}(E) → E is given by φ ↦ φ(1).) Therefore by Proposition 11.29 the category of summands of finite direct sums of C, i.e. the category of C-projective modules, is isomorphic to the category of summands of finite direct sums of E, i. e. to the category of projective right E-modules. Furthermore, if we look at the isomorphism End_{R} C = E → End_{E}(E), we see that it takes an element φ ∈ E to the mapping e ↦ φe = φ_{∗}(e), i.e. this isomorphism is obtained simply by applying the functor Hom_{R}(C, ), yielding Hom_{R}(C, C) → Hom_{E}(Hom_{R}(C, C), Hom_{R}(C, C)). It now follows by additivity that the map

Hom_{R}(U, V) → Hom_{E}(Hom_{R}(C, U), Hom_{R}(C, V))

where U and V are direct summands of C^{m} and C^{n}, is also obtained by applying

Hom_{R}(C, ). ✓

In particular, when interested in direct sum decompositions of a fixed R-module G one can apply Proposition 11.31 for C = G. This is in a way remarkable, for it says that in studying direct sum decompositions there is no loss of generality in restricting attention to finitely generated projective modules, provided that one is willing to work over a fairly complicated ring, as End_{R} G will often be. Although the study of direct sum decompositions of projective modules is by no means simple—as was shown by the work on the Serre Conjecture—nonetheless, as we will see below, the assumption that modules are projective can be very useful.

example 11.32.

Using Corollary 11.32 we will later be able to identify those dedekind domains W such that finite rank torsion free W -modules cannot have an infinite number of non-isomorphic summands.

lemma 11.33. If G is an C-projective R-module then the natural map σ: Hom_{R}(C, G) ⊗_{E} C → G given by φ ⊗ c ↦ φ(c) is an isomorphism. proof: Any C-projective R-module is a direct summand of C^{n} for some finite n. Therefore by Proposition 0.* it suffices to prove that result for the case G = C. But then σ: E ⊗_{E} C → C is simply the canonical isomorphism e ⊗ c ↦ ec. ✓

proposition 11.34. If G is an C-projective R-module and H is a summand of G then H is fully invariant in G if and only if Hom_{R}(C, H) is a fully invariant (End_{R} C)-submodule of Hom_{R}(C, G). proof: If G = H ⊕ K then by Proposition 11.3, H is fully invariant in G if and only if Hom_{R}(H, K) = 0. But since H and K are C-projective modules, by Proposition 11.31 this will be true if and only if Hom_{E}(Hom_{R}(C, H), Hom_{R}(C, K)) = 0, where E = End_{R} C. Again by Proposition 11.3, this will hold if and only if Hom_{R}(C, H) is a fully invariant E-submodule of Hom_{R}(C, G). ✓ corollary 11.35. If H is a summand of G then H is fully invariant in G if and only if the functor Hom_{R}(G, ) takes H to a two-sided ideal of End_{R} G. proof: By Proposition 11.34 H is fully invariant if and only if Hom_{R}(C, H) is a fully invariant left ideal in the ring End_{R} G. But the End_{R} G-linear endomorphisms of End_{R} G are given by right multiplication by elements of End_{R} G. Thus H is fully invariant in G if and only if Hom_{R}(G, H) is invariant under right multiplication by elements of End_{R} G, i. e. if and only if Hom_{R}(G, H) is a right ideal. Since it is a left ideal in any case, this is equivalent to its being a two-sided ideal. ✓

REDUCTION MODULO THE NIL RADICAL.

lemma 11.36. Let E be a ring and let N be an ideal in E such that N^{r} = 0 for some r ≥ 1. Let M and P be right E-modules.

(1) If P is projective then the standard map Hom_{E}(P, M) → Hom_{E}(P/P N , M/MN) is surjective.

(2) If φ ∈ End_{E} P then φ is an automorphism of P if and only if the induced map φ̄ ∈ End_{E} P/P N is an automorphism of P/P N.

(3) If X̄ is a summand of P/P N then there exists a summand X of P such that X̄ = (X + P N)/P N.

proof: (1) Let φ̄ ∈ Hom_{E}(P/P N , M/MN). Consider the diagram

P  M
↓  ↓
P/P N ⟶^{φ̄} M/MN ,

where the vertical maps are the canonical surjections. Since P is projective and M → M/MN is a surjection, there exists a map φ: P → M making this diagram commute. Then φ induces φ̄. This shows that Hom_{E}(P, M) → Hom_{E}(P/P N , M/MN) is surjective.

(2) Clearly if φ is an automorphism of P then φ̄ is an automorphism of P/P N. Now consider the ring morphism ρ: End_{E} P → End_{E}(P/P N) such that ρ(φ) = φ̄. If φ ∈ Ker ρ then φ(P) ⊆ P N and by induction we see that φ^{k}(P) ⊆ P N^{k}. But by hypothesis N^{k} = 0 for sufficiently large k. Thus Ker ρ is an ideal in End_{E} P consisting of nilpotent elements. Thus by Lemma 1.*, if φ̄ = ρ(φ) is invertible in End_{E}(P/P N) then φ is invertible in End_{E} P, i. e. an automorphism of P.

(3) If X̄ is a summand of P/P N then X̄ = π̄(P/P N), where π̄ is an idempotent in End_{E}(P/P N). As noted above, the kernel of the surjection ρ: End_{E} P → End_{E}(P/P N) consists of nilpotent elements. Therefore by Lemma 1.36 there is an idempotent pre-image π ∈ End_{E} P for π̄. Let X = π(P). Then X is a summand of P since π is idempotent

and X̄ = π̄(P/P N) = (π(P) + P N)/P N = (X + P N)/P N. ✓

proposition 11.37. Let C be an R-module and let N be an ideal in E = End_{R} C such that N^{r} = 0 for some r ≥ 1. Let G, H, and K be C-projective R-modules. Let F be the functor from R-modules to right E/N -modules defined by F(G) = Hom_{R}(C, G)/ Hom_{R}(C, G)N. Then the following assertions are true:

(1) The canonical map Hom_{R}(H, K) → Hom_{E}(F(H), F(K)) is surjective.

(2) H ≈ K if and only if F(H) ≈ F(K).

(3) G = 0 if and only if F(G) = 0.

(4) If H ⊆ G then H = G if and only if F(H) = F(G).

(5) If H, K ⊆ G then G = H ⊕ K if and only if F(G) = F(H) ⊕ F(K).

(6) If X is a summand of F(G) then there exists a summand H of G such that X is the image of F(H) in F(G).

proof: (1) By Theorem 11.31, the canonical map

Hom_{R}(H, K) → Hom_{E}(Hom_{R}(C, H), Hom_{R}(C, K))

is an isomorphism and by Lemma 11.36 the map

Hom_{E}(Hom_{R}(C, H), Hom_{R}(C, K)) → Hom_{E}(F(H), F(K))

is surjective. Since Hom_{R}(H, K) → Hom_{E}(F(H), F(K)) is the composition of these, it is also surjective.

(2) If H ≈ K then F(H) ≈ F(K) since functors preserve isomorphisms. Conversely, if F(H) ≈ F(K) then by (1) there exist φ ∈ Hom_{R}(H, K) and ψ ∈ Hom_{R}(K, H) such that F(ψφ) = 1_{F(H)} and F(φψ) = 1_{F(K)} . But it then follows from Lemma 11.36 and Theorem 11.31 that ψφ and φψ are automorphisms. Therefore φ and ψ are isomorphisms and H ≈ K.

(3) and (4): If H and G are C-projective and H ⊆ G then by Lemma 11.33 there is a commutative diagram

Hom_{R}(C, H) ⊗_{E} C ⟶ Hom_{R}(C, G) ⊗_{E} C
σ_{H} ↓  ^{σ_{G}}↓
H  ⟶  G ,

where σ_{H} and σ_{G} are isomorphisms. Thus G = 0 if and only if Hom_{R}(C, G) = 0. Furthermore if H = G then obviously Hom_{R}(C, H) = Hom_{R}(C, G) and conversely if the natural embedding Hom_{R}(C, H) ↪ Hom_{R}(C, G) is an equality then from the diagram it follows that H = G. Thus H = G if and only if Hom_{R}(C, H) = Hom_{R}(C, G). Since by Theorem 11.31 Hom_{R}(C, H) and Hom_{R}(C, G) are finitely generated projective E-modules and N is contained in the Jacobson radical of E, it then follows from Nakayama’s Lemma that G = 0 if and only if F(G) = 0 and H = G if and only if F(H) = F(G).

(5) ( ⇒ ): Clear.

( ⇐ ): If F(G) = F(H) ⊕ F(K) then the inclusions H ⊆ G and K ⊆ G induce a map H ⊕ K → G such that F(H ⊕ K) = F(H) ⊕ F(K) → F(G) is an isomorphism. Thus by (2) H ⊕ K → G is an isomorphism. This implies that G = H ⊕ K.

(6) If X̄ is a summand of F(G) then applying Lemma 11.36 to End_{E}(Hom(C, G))

shows that X̄ = \frac{X̃ + Hom_{R}(C, G)N}{Hom_{R}(C, G)N} for some summand X̃ of End_{R} G. By

Theorem 11.31, X̃ = Hom_{R}(C, X) for some summand X of G. Then

X̄ = \frac{Hom_{R}(C, X) + Hom_{R}(C, G)N}{Hom_{R}(C, G)N} ,

which is the image of F(X) in F(G) under the map F(ι), where ι: X → G is the inclusion. ✓

corollary 11.38. Let N and F be as above. If G = H ⊕K then H is an exchangeable summand of G if and only if F(H) is an exchangeable summand of F(G). proof: ( ⇒ ): If G = H ⊕ K and F(G) = F(H) ⊕ F(K) = L̄ ⊕ M̄, then by Proposition 11.37, L̄ = F(L) and M̄ = F(M) for some L, M ⊆ G such that G = L ⊕ M. If H is an exchangeable summand of G then there exist summands L′′ of L and M′′ of M such that G = H ⊕ L′′ ⊕ M′′. Then F(L′′) and F(M′′) are summands of F(L) and F(M) and F(G) = F(H) ⊕ F(L′′) ⊕ F(M′′). Thus F(H) is an exchangeable summand of F(G).

( ⇐ ): If G = H ⊕ K = L ⊕ M, then F(G) = F(H) ⊕ F(K) = F(L) ⊕ F(M). If F(H) is an exchangeable summand of F(G) then there are summands L̄′′ and M̄′′ of F(L) and F(M) such that F(G) = F(H) ⊕ L̄′′ ⊕ M̄′′. By Proposition 11.37, there exist summands L′′ of L and M′′ of M such that L̄′′ = F(L′′), M̄′′ = F(M′′) and G = H ⊕ L′′ ⊕ M′′. Thus H is an exchangeable summand of G. ✓

corollary 11.39. Let G be an R-module and Ē = (End_{R} G)/N, where N = nil rad End_{R} G. Then the maximum number of non-isomorphic summands of G equals the maximum number of non-isomorphic summands of the Ē as a right Ē-module. proof: Let N = nil rad End_{R} G and let F be as in Proposition 11.37 with C = G. Then the right Ē-module Ē is F(G). Therefore by Proposition 11.37 non-isomorphic summands of G correspond to non-isomorphic summands of Ē and every summand of Ē has the form F(H) for some summand H of G. Thus the maximum number of G equals the maximum number of non-isomorphic summands of the Ē as a right Ē-module. ✓

If C is a finite rank torsion free module over a dedekind domain W, we can prove an analog of Proposition 11.31 for the category of quasi-summands of the modules C^{k} and the category of projective modules over QEnd C. This can be used as an alternate path to a proof of Jónsson’s Theorem (Theorem 3.24). More important, though, is the following variation.

proposition 11.40. Let G be a finite rank torsion free module over a dedekind domain W and let E = End G and N = nil rad E.

(1) The functor H = QHom(G, )/ QHom(G, )N gives a one-to-one correspondance between the quasi-summands of G and the right ideals in QE/QN.

(2) Under this correspondance the strongly indecomposable quasi-summands correspond to the simple right ideals,

(3) If H and K are non-quasi-isomorphic strongly indecomposable quasi-summands of G then Hom_{QE}(H(H), H(K)) = 0.

proof: (1) The proof is essentially identical to that of Proposition 11.37. One needs the fact that the analog of Proposition 11.31 gives a one-to-one correspondence between the quasi-summands of G and the summands of the right QE-module QE. But this follows from Proposition 11.29 since QEnd G = QE ≈ End_{QE} QE.

(2) By Proposition 1.* QE/QN is a semi-simple ring, therefore indecomposable submodules of QE/QN are the same as simple right ideals.

(3) By Schur’s Lemma the two simple right ideals H(H) and H(K) are nonisomorphic if and only if Hom_{QE}(H(H), H(K)) = 0. ✓

corollary 11.41. Let G be a finite rank torsion free module over a dedekind domain W. Let E = End G, N = nil rad E, and let H be the functor QHom(G, )/ QHom(G, )N. Let G = H ⊕ K, where no non-trivial quasisummand of H is isomorphic to a quasi-summand of K. Then H(H) and H(K) are fully invariant QE-submodules of H(G). proof: By Proposition 11.3 if suffices to prove that Hom_{QE}(H(H), H(K)) = 0 = Hom_{QE}(H(K), H(H)). Since H(H) and H(K) are modules over the semi-simple ring QE/QN, it suffices to prove that Hom_{QE}(S, S′) = 0 whenever S and S′ are simple summands of H(H) and H(K) or vice-versa. But by Proposition 11.40, in this case S and S′ have the form H(H′) and H(K′) where H′ and K′ are strongly indecomposable quasi-summands of H and K (or vice-versa). The result then follows from Proposition 11.40. ✓

Now putting various things together yields a pair of major results on direct sum decompositions of finite rank torsion free modules over dedekind domains. theorem 11.42. Let G be a finite rank torsion free module over a dedekind domain W and G = H ⊕ K, where no non-trivial quasi-summand of H is isomorphic to a quasi-summand of K. Then H and K are exchangeable summands of G.

proof: Let N = nil rad End G, F = Hom(G, )/ Hom(G, )N, and H = QF = QHom(G, )/ QHom(G, )N. By Proposition 11.41 the hypothesis that no quasi-summand of H is isomorphic to a quasi-summand of K implies that H(H) = QF(H) is a fully invariant (End G)-submodule of H(G) = QF(G). Since F(H) ◁ F(G) it follows that F(H) is a fully invariant (End G)-submodule of F(G). (In fact, if γ is an (End G)-linear endomorphism of F(G) then γ(F(H)) ⊆ F(G) ∩ QF(H) = F(H).) Therefore by Proposition 11.17, F(H) is an exchangeable summand of (End G)/N (as an (End G)-module). Thus by Proposition 11.38, H is an exchangeable summand of G. ✓

As a variation on this we have

proposition 11.43. Let G be a finite rank torsion free module over a dedekind domain W and suppose that G = H ⊕ K = L ⊕ M, where no non-trivial quasi-summand of H is isomorphic to a quasi-summand of M. Then there exist summands K_{1} of K and L_{1} of L such that G = H ⊕ K_{1} ⊕ M = H ⊕ L_{1} ⊕ M.

proof: As before, let N = nil rad End G and let H(X) = Hom(G, X)/ Hom(G, X)N. Then by Proposition 11.40 the assumption on H and M implies that Hom(H(H), H(M)) = 0 = Hom(H(M), H(H)). Thus by Proposition 11.7 there exists a summand K̄_{1} of the (End G)-module H(K) such that H(G) = H(H) ⊕ K̄_{1} ⊕ H(M). By Proposition 11.37, K̄_{1} = H(K_{1}) for some summand K_{1} of K and G = H ⊕ K_{1} ⊕ M. Analogously there exists a summand L_{1} of L such that G = H ⊕ L_{1} ⊕ M. ✓

definition 11.44. We say that a dedekind domain W has the Jordan-Zassenhaus Property if every finite integral extension W′ of W has a finite ideal class group [Picard group]. What this means is that there are, up to isomorphism, only finitely many rank-one W′-modules quasi-isomorphic to W′.

We then get the following theorem. In the case where W is the ring of integers, this was proved independently and virtually simultaneously by Chang Mo Bang and your author:

theorem 11.45. A dedekind domain W has the Jordan-Zassenhaus Property if and only if there does not exist a finite rank torsion free W -module G having an infinite number of non-isomorphic direct summands. proof: ( ⇐ ): This was proved in a joint paper by Fuchs and one of his students.

( ⇒ ): Let E = End G. By Corollary 11.39 it suffices to prove that Ē = E/N does not have an infinite number of non-isomorphic direct summands (as an E-module). (As usual, N = nil rad E.) **** ✓

✠

THE BAER SPLITTING PROPERTY. As seen in Lemma 11.33 above, for any R-modules C and G, there exists an “evaluation map” σ: Hom_{R}(C, G) ⊗_{E} C → G given by σ(φ ⊗ c) = φ(c), where here E = End_{R} C. We will denote the image of this map by S_{C}(G). In other words, S_{C}(G) is the R-submodule of G generated by all elements φ(c) with φ ∈ Hom_{R}(C, G) and c ∈ C. S_{C}(G) is sometimes called the C-socle of G.

If W is a dedekind domain and C = W (p^{n}), then for any W -module G, S_{C}(G) = G[p^{n}]. On the other hand if C is a rank-one W -module and t = t(C), then for any torsion free G, S_{C}(G) is simply the submodule G(t) of G defined in Chapter 4.

definition 11.46. We say that an R-module C has the Baer Splitting Property if whenever H is a submodule of an R-module G such that G/H is C-projective and G = H + S_{C}(G), then H is a direct summand of G.

theorem 11.47. If IC ⊊ C for every proper right ideal I of End_{R} C then C has the Baer splitting property.

proof: We claim that if IC ≠ C for every proper right ideal I of E then M ⊗_{E} C ≠ 0 for every non-trivial finitely generated E-module M. In fact, if IC ≠ C for a right ideal I of E then E/I ⊗_{E} C ≠ 0. Thus if this holds for every proper right ideal I then L ⊗_{E} C ≠ 0 for every non-trivial cyclic E-module L. Now if M is a non-trivial finitely generated right E-module then there exists a proper submodule N of M such that M is generated by N together with a single additional element. Thus M/N is a non-trivial cyclic right E-module so the hypothesis implies that M/N ⊗_{E} C ≠ 0. It follows that M ⊗_{E} C ≠ 0.

Now consider a submodule H of a module G. Let K = G/H and let φ: G → K be the quotient map. Then H is a summand of G if and only if φ splits, i. e. if and only if there exists σ: K → G such that φ_{∗}(σ) = φσ = 1_{K}, where φ_{∗}: Hom_{R}(K, G) → Hom_{R}(K, K) is the usual induced map.

We will show that if K is C-projective and G = H + S_{C}(G) then the map φ_{∗} above is a surjection, so that 1_{K} ∈ φ_{∗}(Hom_{R}(K, G)) and φ splits. Let M be the cokernel of φ_{∗}, so that the sequence of right E-modules

0 → Hom_{R}(C, H) → Hom_{R}(C, G) ⟶^{φ_{∗}} Hom_{R}(C, K) ⟶^{β} M → 0

is exact. Since K is C-projective, it follows from Theorem 11.31 that Hom_{R}(C, K) is a finitely generated projective E-module, and thus M is a finitely generated E-module. There is the following commutative diagram:

Hom_{R}(C, G) ⊗_{E} C ⟶^{φ_{∗}⊗_{E}C} Hom_{R}(C, K) ⊗_{E} C ⟶^{β⊗_{E}C} M ⊗_{E} C → 0
σ_{G}↓  ^{σ_{K}} ↓^{≈}  ↓
0 → H ⟶  G  ⟶^{φ}  K  ⟶  0 .

The top row is exact by the right exactness of the tensor product. The image of σ_{G} is S_{C}(G). The map σ_{K} is an isomorphism by Lemma 11.33 since K is C-projective. Thus if G = H + S_{C}(G) = H + Image(σ_{G}) then φσ_{G} is surjective and so φ_{∗} ⊗_{E} C must be surjective. Thus M ⊗_{E} C = 0. Hence by the first paragraph of the proof M = 0, so that φ_{∗} is surjective and thus φ is a split surjection and H is a summand of G. ✓

proposition 11.48. If C has the Baer Splitting Property and I is a proper right ideal in End_{R} C, and the canonical monomorphism ⨁_{S} Hom_{R}(C, C) → Hom_{R}(C, ⨁_{S} C) is surjective for an index set S with the same cardinality as that of a set of generators for I, then IC ⊊ C. In particular, this is the case if either I is a finitely generated right ideal or C is a finitely generated R-module or R is a an integral domain and C is torsion free with finite rank.

proof: The canonical embedding ⨁_{S} Hom_{R}(C, C) → Hom_{R}(C, ⨁_{S} C) is an isomorphism if and only if every map φ: C → ⨁_{S} C has the property that φ(C) ⊆ ⨁_{S′} C for some finite subset S′ of S. This will be true if I is finitely generated (in which case we may choose a finite set for S) or C is a finitely generated R-module or R is an integral domain and C is torsion free with finite rank.

Note now that for any r ∈ R, the endomorphism of C given by c ↦ rc is E-linear and thus induces for any right E-module M a map M ⊗_{E} C → M ⊗_{E} C. This shows that M ⊗_{E} C is a left R-module, where r(m ⊗ c) = m ⊗ rc.

Now suppose that C has the Baer splitting property and let I be a right ideal in E = End_{R} C such that IC = C. Then the obvious map I ⊗_{E} C → C is surjective. Now let F = ⨁_{S} E be a free right E-module such that there exists a surjection μ: F → I. This yields a surjection F ⊗_{E} C → I ⊗_{E} C and therefore a surjection γ: ⨁_{S} C ≈ F ⊗_{E} C → IC = C of R-modules. Since C has the Baer Splitting Property, γ splits. Therefore the induced map γ_{∗}: Hom_{R}(C, F ⊗_{E} C) → Hom_{R}(C, C) = E is also a split surjection. Now there is an E-linear map θ: F → Hom_{R}(C, F ⊗_{E} C) where θ(x)(c) = x ⊗ c. But this is essentially just the map ⨁_{S} C → Hom_{R}(C, ⨁_{S} C) and so by hypothesis, θ is a surjection. There is thus a commutative diagram

⨁_{S} Hom_{R}(C, C)  F  ⟶^{μ} I
≈↓  ^{θ}↓  ↓^{⊆}
Hom_{R}(C, ⨁_{S} C) ⟶^{≈} Hom_{R}(C, F ⊗_{E} C) ⟶^{γ_{∗}} E = End_{R} C,

where γ_{∗} is a surjection. Therefore the inclusion map I → E is surjective,

i. e. I = E. ✓

In the following corollary, the assumption that surjective endomorphisms of C are automorphisms holds in case C is a noetherian R-module or in case R is a finite rank torsion free algebra over a dedekind domain W and C is torsion free with finite rank.

corollary 11.49. If every right ideal of End_{R} C is principal and every surjection φ ∈ End_{R} C is an automorphism then C has the Baer Splitting Property. proof: By Theorem 11.47 it suffices to prove that if IC = C for some right ideal I of End_{R} C then I = End_{R} C. But if End_{R} C is right principal then I = φ End_{R} C for some φ and IC = φ(C). Therefore if IC = C then φ is a surjection and hence by assumption is an automorphism of C and thus an invertible element of End_{R} C and so

I = φ End_{R} C = End_{R} C. ✓

THE SUBSTITUTION PROPERTY.

definition 11.50. Let G = H ⊕ K. We say that H is a substitutable summand of G if whenever G = H′ ⊕ L with H′ ≈ H then there exists H′′ ⊆ G which simultaneously replaces H with respect to K and replaces H′ with respect to L, i.e. G = H′′ ⊕ K = H′′ ⊕ L.

proposition 11.51. A substitutable summand is cancellable. proof: Let G = H ⊕ K = H′ ⊕ L with H′ ≈ H. If H is a substitutable summand of G then there exists H′′ such that G = H′′ ⊕ K = H′′ ⊕ L. Then K ≈ G/H′′ ≈ L. Thus H is a cancellable summand of G. ✓

It seems at first that the definition of a substitutable summand depends on a particular direct sum decomposition G = H ⊕ K. However not only is the definition independent of the complement K but the following stronger assertion holds. proposition 11.52. If H is a substitutable summand of G then any summand of G isomorphic to H is also substitutable. proof: Let G = H ⊕ K = H′ ⊕ L, where H′ ≈ H. If H is a substitutable summand of G then by Proposition 11.51, K ≈ L. It follows that there is an automorphism of G taking H to H′ and K to L. Thus H′ is also a substitutable summand of G. ✓

proposition 11.53. If H is a substitutable summand of G then H is not isomorphic to any proper summand of itself. proof: In fact, if G = H ⊕ K and H is a substitutable summand of G and H = H′ ⊕ P with H′ ≈ H , then G = H′ ⊕ P ⊕ K and there must exist H′′ ⊆ G such that G = H′′ ⊕ K = H′′ ⊕ P ⊕ K. This is only possible if P = 0, so that H′ = H. ✓

We now note the following:

lemma 11.54. Let H be a any R-module. The following conditions are equivalent: (1) H is not isomorphic to any proper summand of itself.

(2) Every left invertible element of the ring End_{R} H is in fact an automorphism of H.

(3) Ever right invertible element of End_{R} H is an automorphism of H.

proof: (2) ⇒ (1) and (3) ⇒ (1): : Let H = H′ ⊕ P with H′ ≈ H. Then an isomorphism from H′ to H extends to an endomorphism α of H with α(P) = 0 and its inverse can be thought of as a map β ∈ End_{R} H such that αβ = 1_{H}. Then α is right invertible and β is left invertible. Thus if either (2) or (3) holds then either α or β is an automorphism of H, and in either case it follows that H′ = H. Thus no proper summand of H is isomorphic to H.

(1) ⇒ (2) & (3): Let α, β ∈ End_{R} H with αβ = 1_{H}. Then β is a split monomorphism and α a split surjection so that β(H) is a summand of H isomorphic to H. Thus if H is not isomorphic to any proper summand then β(H) = H, such that β is an automorphism of H and it follows that α must also be an automorphism. Thus any left or right invertible element of End_{R} H is in fact invertible. ✓

A fully invariant submodule need not be substitutable. However it follows from the next proposition that a fully invariant summand of a finite rank torsion free module will be substitutable.

proposition 11.55. A fully invariant summand H of G is substitutable if and only if H is not isomorphic to any proper submodule of itself. proof: ( ⇒ ): It follows from Proposition 11.53 that if H is a substitutable summand of G then H is not isomorphic to any proper summand of itself.

( ⇐ ): If G = H ⊕ K = H′ ⊕ L and H′ ≈ H and H is fully invariant and not isomorphic to any proper submodule of itself, then as in the proof of Proposition 11.9 one sees that H′ = H, so that G = H ⊕ K = H ⊕ L. Thus H is a substitutable summand

of G. ✓

proposition 11.56. If G = H_{1} ⊕ H_{2} ⊕ K and H_{1} and H_{2} are substitutable summands of G, then so is H_{1} ⊕ H_{2} .

proof: Let G = H_{1} ⊕ H_{2} ⊕ K = H′ ⊕ L where H′ ≈ H_{1} ⊕ H_{2} . Then H′ = H′_{1} ⊕ H′_{2} with H′_{i} ≈ H_{i}, so that

G = H_{1} ⊕ H_{2} ⊕ K = H′_{1} ⊕ H′_{2} ⊕ L.

Since H_{1} is a substitutable summand of G, there exists H′′_{1} ⊆ G such that

G = H′′_{1} ⊕ H_{2} ⊕ K = H′′_{1} ⊕ H′_{2} ⊕ L = H_{2} ⊕ (H′′_{1} ⊕ K) = H′_{2} ⊕ (H′′_{1} ⊕ L).

Since H_{2} is a substitutable summand of G there exists H′′_{2} such that G = H′′_{2} ⊕ H′′_{1} ⊕ K = H′′_{2} ⊕ H′′_{1} ⊕ L. Thus H_{1} ⊕ H_{2} is a substitutable summand of G. ✓

proposition 11.57. If H_{1} ⊕ H_{2} is a substitutable summand of H_{1} ⊕ H_{2} ⊕ K, then H_{2} is a substitutable summand of H_{2} ⊕ K.

proof: Suppose that H_{2} ⊕ K = H′_{2} ⊕ L with H′_{2} ≈ H_{2} . Then G = H_{1} ⊕ H_{2} ⊕ K = H_{1} ⊕ H′_{2} ⊕ L with H_{1} ⊕ H′_{2} ≈ H_{1} ⊕ H_{2} . By hypothesis, then, there exists H′′ ⊆ G such that G = H′′ ⊕ K = H′′ ⊕ L. Since K, L ⊆ H_{2} ⊕ K = H′_{2} ⊕ L, we can apply Lemma 11.1 to get H_{2} ⊕ K = ((H_{2} ⊕ K) ∩ H′′) ⊕ K = ((H_{2} ⊕ K) ∩ H′′) ⊕ L. Setting H′′_{2} = (H_{2} ⊕ K) ∩ H′′, we have

H_{2} ⊕ K = H′_{2} ⊕ L = H′′_{2} ⊕ K = H′′_{2} ⊕ L,

so that H_{2} is a substitutable summand of H_{2} ⊕ K. ✓

definition 11.58. We say that H has the substitution property if H is a substitutable summand of H ⊕ K for every K. theorem 11.59. The following conditions are equivalent:

(1) H has the substitution property.

(2) H ⊕ 0 is a substitutable summand of H ⊕ H.

(3) Whenever β_{1}, β_{2}, α_{1}, α_{2} ∈ End_{R} H have the property that β_{1}α_{1} + β_{2}α_{2} = 1 then there exists φ ∈ End_{R} H such that β_{1} + β_{2}φ is an automorphism of H.

Furthermore any of these conditions applies that every left or right invertible element of End H is in fact an automorphism of H.

proof: (1) ⇒ (2): Clear.

(2) ⇒ (3): Suppose that β_{1}, β_{2}, α_{1}, α_{2} ∈ End_{R} H have the property that β_{1}α_{1} + β_{2}α_{2} = 1. Define σ: H → H ⊕ H and μ: H ⊕ H → H by σ(h) = (α_{1}(h), α_{2}(h)) and μ(h_{1}, h_{2}) = β_{1}(h_{1}) + β_{2}(h_{2}). Then μσ = 1_{H} so σ is a monomorphism split by μ and H ⊕ H = σ(H) ⊕ Ker μ and σ(H) ≈ H. Thus if H ⊕ 0 is a substitutable summand of H ⊕ H then there exists H′′ ⊆ H ⊕ H such that H ⊕ H = H′′ ⊕ Ker μ = H′′ ⊕ H. Since H′′ replaces H ⊕ 0, by Lemma 11.13 there exists φ ∈ End H such that H′′ = {(h, φ(h)) | h ∈ H}. Since H ⊕ H = H′′ ⊕ Ker μ, it follows that μ maps H′′ isomorphically onto H, so h ↦ (h, φ(h)) ↦ μ(h, φ(h)) = β_{1}(h) + β_{2}φ(h) is an automorphism of H.

(3) ⇒ (1): Let G = H ⊕ K = H′ ⊕ L with H′ ≈ H. Let σ: H → H ⊕ K and μ: H ⊕ K → H be such that σ(H) = H′, Ker μ = L, and μσ = 1_{H}. Then σ and μ must be given by σ(h) = (α_{1}(h), α_{2}(h)) and μ(h, k) = β_{1}(h) + β_{2}(k) where β_{1}, α_{1} ∈ End_{R} H, α_{2} ∈ Hom_{R}(H, K), and β_{2} ∈ Hom_{R}(K, H). Furthermore β_{1}α_{1} + β_{2}α_{2}1_{H} = 1_{H}. If condition (3) holds then there exists φ ∈ End_{R} H such that

δ =_{def} β_{1} + β_{2}α_{2}φ ∈ End_{R} H

is invertible, i.e. an automorphism of H. Define η: H → H ⊕ K by η(h) = (δ^{−1}(h), α_{2}φδ^{−1}(h)). Then

μη(h) = β_{1}δ^{−1}(h) + β_{2}α_{2}φδ^{−1}(h) = δδ^{−1}(h) = h

so that μη = 1_{H}. Thus G = η(H) ⊕ Ker μ = η(H) ⊕ L. On the other hand let π: H ⊕ K → H be the projection onto the first cooordinate. Then δπη = 1_{H} so that G = η(H) ⊕ Ker δπ = η(H) ⊕ K. Thus η(H) replaces both H and H′, showing that H is a substitutable summand of H ⊕ K. Since this is true for any K, H has the substitution property. ✓ proposition 11.60. If End_{R} H is a local ring or a commutative semi-local ring then H has the substitution property.

proof: Let M_{1}, …, M_{n} be the finite number of maximal ideals in End_{R} H. Suppose that α_{1}, α_{2}, β_{1}, β_{2} ∈ End_{R} H and β_{1}α_{1} + β_{2}α_{2} = 1. For given i, we claim that either β_{1} or β_{1} + β_{2} is not in M_{i}. In fact, if β_{1} ∈ M_{i} and β_{1} + β_{2} ∈ M_{i}, then β_{2} ∈ M_{i}, so that 1 = β_{1}α_{1} + β_{2}α_{2} ∈ M_{i}, a contradiction.

If End_{R} H is local then M_{1} consists of all the non-units in End_{R} H so the above means that either β_{1} + β_{2}0 or β_{1} + β_{2}1_{H} is invertible, so H has the substitution property by Theorem 11.59.

Now suppose that End_{R} H is commutative. If β_{1} ∉ M_{i} then since M_{i} is prime we can choose ξ_{i} ∈ M_{i} with ξ_{i} ∉ M_{j} for j ≠ i. And if β_{1} ∈ M_{i} choose ξ_{i} = 1. Let φ = ξ_{1} ⋯ ξ_{n}. Then φ ∈ M_{i} ⟺ β_{1} ∉ M_{i}. Thus β_{1} ∉ M_{i} ⇒ β_{2}φ ∈ M_{i} and β_{1} ∈ M_{i} ⇒ β_{2}, φ ∉ M_{i} ⇒ β_{2}φ ∉ M_{i} since M_{i} is prime. Thus for all i, β_{1} + β_{2}φ ∉ M_{i}. Thus β_{1} + β_{2}φ is not contained in any maximal ideal of the commutative ring End_{R} H and hence is invertible. Once again, Theorem 11.59 shows that H has the substitution property. ✓

proposition 11.61. The class of R-modules with the substitution property is closed under direct sums and direct summands.

proof: If H_{1} and H_{2} have the substitution property then for any K, H_{1} and H_{2} are substitutable summands of H_{1} ⊕ H_{2} ⊕ K. Therefore by Proposition 11.56, H_{1} ⊕ H_{2} is a substitutable summand of H_{1} ⊕ H_{2} ⊕ K. Since this is true for any K, H_{1} ⊕ H_{2} has the substitution property.

Conversely, if H_{1} ⊕ H_{2} has the substitution property then for any K, H_{1} ⊕ H_{2} is a substitutable summand of H_{1} ⊕ H_{2} ⊕ K. Therefore by Proposition 11.57 H_{2} is a substitutable summand of H_{2} ⊕ K. Since this is true for any K, H_{2} has the substitution property. ✓

corollary 11.62. If H is a finite direct sum of indecomposable modules and H has the exchange property, then H has the substitution property.

proof: By Proposition 11.61 and Proposition 11.21 it suffices to suppose H indecomposable. Then by Proposition 11.23, End_{R} H is a local ring. Thus H has the substitution property by Proposition 11.60. ✓ proposition 11.63. An R-module with finite length has the substitution property. proof: If H has finite length it is a direct sum of indecomposables. Since by Proposition 11.61 a finite direct sum has the substitution property if each of the summands does, it suffices to suppose that H is indecomposable with finite length. But then End_{R} H is a local ring so by Proposition 11.60 H has the substitution property. ✓

STABLE RANGE. We now proceed to an important (although highly technical) generalization of condition (3) of Theorem 11.59.

definition 11.64. Let H and K be R-modules. We say that H^{n} is an n-substitutable summand of H^{n} ⊕ K if whenever H^{n} ⊕ K = H′ ⊕ L with H′ ≈ H, then there exists H′′ ⊆ H^{n} ⊕ K and P ⊆ H^{n} such that H^{n} ⊕ K = H′′ ⊕ L = H′′ ⊕ P ⊕ K.

For given n, we say that H has the n-substitution property if H^{n} is an n-substitutable summand of H^{n} ⊕ K for every K.

note 11.65. If H^{n} ⊕ K = H′′ ⊕ L = H′′ ⊕ P ⊕ K where P ⊆ H^{n}, then L ≈ P ⊕ K and H ⊕ P ≈ H^{n}. proof: In fact, L ≈ (H^{n} ⊕ K)/H′′ ≈ P ⊕ K and H ⊕ P ≈ H′′ ⊕ P ≈ (H^{n} ⊕ K)/K ≈

H^{n}. ✓

The following rather curious proposition will prove essential in proving the main theorem on stable range below.

proposition 11.66. Let G = H^{n} ⊕ K = H′ ⊕ L, where H′ ≈ H. The following conditions are equivalent:

(1) There exist submodules H′′ ⊆ G and P ⊆ H^{n} such that

G = H′′ ⊕ P ⊕ K = H′′ ⊕ L.

(2) There exist submodules M ⊆ G and K̃ ⊆ L such that

G = M ⊕ K̃ = M ⊕ K.

proof: (1) ⇒ (2): Suppose that there exists P ⊆ H^{n} and H′′ ⊆ G such that

G = H′′ ⊕ P ⊕ K = H′′ ⊕ L.

Let M = H′′ ⊕ P so that

G = M ⊕ K.

Since H′′ ⊆ H′′ ⊕ K ⊆ G = H′′ ⊕ L, by Lemma 11.1 there exists K̃ ⊆ L such that H′′ ⊕ K = H′′ ⊕ K̃ and thus

G = H′′ ⊕ P ⊕ K = H′′ ⊕ P ⊕ K̃ = M ⊕ K̃.

(2) ⇒ (1): Suppose that there exist K̃ ⊆ L and M ⊆ G such that

G = M ⊕ K̃ = M ⊕ K.

Since K̃ ⊆ L ⊆ K̃ ⊕ M, by Lemma 11.1 L = S ⊕ K̃, where S = L ∩ M. Then S is a summand of G and S ⊆ M so

M = H′′ ⊕ S

for some H′′ ⊆ M, and thus

G = M ⊕ K̃ = H′′ ⊕ S ⊕ K̃ = H′′ ⊕ L.

Now G = M ⊕ K = H′′ ⊕ S ⊕ K and K ⊆ S ⊕ K ⊆ H^{n} ⊕ K so by Lemma 11.1

S ⊕ K = P ⊕ K,

where P ⊆ H^{n}. Then

G = M ⊕ K = H′′ ⊕ S ⊕ K = H′′ ⊕ P ⊕ K. ✓

proposition 11.67. Suppose that H ⊕ K ≈ H ⊕ L and that K has a summand K_{1} isomorphic to H^{n} which is an n-substitutable summand of H ⊕ K. Then L ≈ K.

In particular, if H has the n-substitution property then H^{r} is a cancellable summand of H^{r} ⊕ K for r ≥ 1 and any R-module K having a summand isomorphic to to H^{n}. proof: We may suppose that K = H^{n} ⊕ K_{2} . Suppose that

G = H′ ⊕ L = H ⊕ K = H^{n} ⊕ H ⊕ K_{2}

where H′ ≈ H. If H^{n} is an n-substitutable summand of G then there exists H′′ ⊆ G and P ⊆ H^{n} with H′′ ≈ H and G = H′′ ⊕ P ⊕ H ⊕ K_{2} = H′′ ⊕ L. Then by Note 11.65, H′′ ⊕ P ≈ H^{n} and

L ≈ G/H′′ ≈ P ⊕ H ⊕ K_{2} ≈ P ⊕ H′′ ⊕ K_{2} ≈ H′′ ⊕ P ⊕ K_{2} ≈ H^{n} ⊕ K_{2} = K.

Thus H is a cancellable summand of H ⊕ K.

Now suppose that H has the n-substitution property and H^{r} ⊕ K ≈ H^{r} ⊕ M for r ≥ 1, where K has a summand isomorphic to H^{n}. Then H^{n} is an n-substitutable summand of H^{r} ⊕ K and H^{r−1} ⊕ K has a summand isomorphic to H^{n}, so by (1), H^{r−1} ⊕ K ≈ H^{r−1} ⊕ M. Continuing inductively, we prove that K ≈ M. Thus H^{r} is a cancellable summand of H^{r} ⊕ K. ✓ proposition 11.68. If H has the n-substitution property and H^{r} ⊕ K ≈ H^{r} ⊕ L, where r ≥ 1 and K ≈ H^{n−1} ⊕ K_{1} for some K_{1} , then L ≈ P ⊕ K_{1} for some P such that H ⊕ P ≈ H^{n}. proof: H^{r−1} ⊕ H ⊕ K ≈ H^{r−1} ⊕ H ⊕ L and H ⊕ K has a summand isomorphic to H^{n}. Thus by Proposition 11.67, H ⊕ K ≈ H ⊕ L. We may suppose that L ⊆ H ⊕ K and

H′ ⊕ L = H ⊕ K = H ⊕ H^{n−1} ⊕ K_{1},

with H′ ≈ H. If H has the n-substitution property then there exists P ⊆ H ⊕ H^{n−1} and H′′ ⊆ H ⊕ K such that

H ⊕ H^{n−1} ⊕ K_{1} = H ⊕ K = H′′ ⊕ L = H′′ ⊕ P ⊕ K_{1}.

Then L ≈ P ⊕ K_{1} and H ⊕ P ≈ H′′ ⊕ P ≈ H ⊕ H^{n−1} = H^{n}. ✓

corollary 11.69. Let H and H_{1} be R-modules such that H has the n-substitution property and let r ≥ 1.

(1) If m ≥ n and H^{m+r} ≈ H_{1}^{m} ⊕ H^{r}, then H^{m} ≈ H_{1}^{m}.

(2) If m ≥ 1 and H^{m+r} ≈ H_{1}^{m} ⊕ H^{r}, then H^{sm} ≈ H_{1}^{sm} for any s such that sm ≥ n.

proof: (1) This follows directly from Proposition 11.67, with K = H^{m} and L = H_{1}^{m}. (2) If H^{m+r} ≈ H_{1}^{m} ⊕ H^{r} then

H^{sm+r} ≈ H^{(s−1)m} ⊕ H^{m} ⊕ H^{r}
≈ H^{(s−1)m} ⊕ H_{1}^{m} ⊕ H^{r}
≈ … ≈ H_{1}^{sm} ⊕ H^{r}.

Thus if sm ≥ n then H^{sm} ≈ H_{1}^{sm} by (1). ✓

It seems only reasonable to think that a 1-substitutable summand is the same as a substitutable summand. However this is not quite true.

remark 11.70. If H is isomorphic to a proper summand of itself then H is a 1-substitutable summand of itself but not a substitutable summand. proof: Every R-module H is a 1-substitutable summand of itself since if H = H ⊕ 0 = H′ ⊕ L with H′ ≈ H then H = H′′ ⊕ L = H′′ ⊕ P ⊕ 0 with H′′ = H′ and P = L ⊆ H. On the other hand, by Proposition 11.53, if H is isomorphic to a proper summand of itself then it cannot be a substitutable summand of itself. ✓ proposition 11.71. Let H be a 1-substitutable summand of H ⊕ K. Then H is a substitutable summand of H ⊕ K if and only if H is not isomorphic to any proper summand of itself. proof: ( ⇒ ): By Proposition 11.53, if H is a substitutable summand of H ⊕ K then H is not isomorphic to any proper summand of itself.

( ⇐ ): Suppose that H ⊕ K = H′ ⊕ L where H′ ≈ H. If H is a 1-substitutable summand of H ⊕ K then there exist H′′ ⊆ H ⊕ K and P ⊆ H^{1} = H such that

H ⊕ K = H′′ ⊕ L = H′′ ⊕ P ⊕ K.

Then H′′ ≈ H′ ≈ H ≈ H′′ ⊕ P. If H is not isomorphic to a proper summand of itself, this implies that P = 0, so that H ⊕ K = H′′ ⊕ L = H′′ ⊕ K, showing that H is a substitutable summand of H ⊕ K. ✓

In light of Proposition 11.71, the following seems surprising.

proposition 11.72. The 1-substitution property is the same as the substitution property. proof: If H has the substitution property and H^{1} ⊕ K = H′ ⊕ L with H′ ≈ H then there exists H′′ ⊆ H ⊕ K such that H ⊕ K = H′′ ⊕ L = H′′ ⊕ K = H′′ ⊕ P ⊕ K, where P = 0. Thus H is a 1-substitutable summand of H ⊕ K.

Suppose now that H has the 1-substitution property. We claim that H cannot be isomorphic to a proper summand of itself. In fact, if H is isomorphic to a proper summand of itself then there exists a non-trivial module K such that H ⊕ K ≈ H. Let H′ = H ⊕ K. Since H has the 1-substitution property then there exists H′′ ⊆ H ⊕ K and P ⊆ H such that H ⊕ K = H′′ ⊕ 0 = H′′ ⊕ P ⊕ K. This forces K = 0, a contradiction. Thus H is not isomorphic to a proper summand of itself, so that by Proposition 11.71 H is a substitutable summand of H ⊕ K for every K and thus has the substitution property. ✓

proposition 11.73. If H_{1} and H_{2} have the n-substitution property then so does H_{1} ⊕ H_{2} . proof: Let G = (H_{1} ⊕ H_{2})^{n} ⊕ K = H′ ⊕ L with H′ ≈ H_{1} ⊕ H_{2} . Then H′ = H′_{1} ⊕ H′_{2} with H′_{1} ≈ H_{1} and H′_{2} ≈ H_{2} and

G = H_{1}^{n} ⊕ H_{2}^{n} ⊕ K = H′_{1} ⊕ H′_{2} ⊕ L.

Now if H_{1}^{n} is an n-substitutable summand of G then there exist H′′_{1} ⊆ G and P_{1} ⊆ H_{1}^{n} with

G = (H′′_{1} ⊕ P_{1}) ⊕ H_{2}^{n} ⊕ K = H′′_{1} ⊕ H′_{2} ⊕ L.

Now if H_{2}^{n} is an n-substitutable summand of G then there exist H′′_{2} ⊆ G and P_{2} ⊆ H_{2}^{n} such that

G = H′′_{2} ⊕ P_{2} ⊕ (H′′_{1} ⊕ P_{1} ⊕ K) = H′′_{2} ⊕ (H′′_{1} ⊕ L)
= (H′′_{1} ⊕ H′′_{2}) ⊕ (P_{1} ⊕ P_{2}) ⊕ K = (H′′_{1} ⊕ H′′_{2}) ⊕ L,

with P_{1} ⊕ P_{2} ⊆ H_{1}^{n} ⊕ H_{2}^{n} = (H_{1} ⊕ H_{2})^{n}, which shows that (H_{1} ⊕ H_{2})^{n} is an n-substitutable summand of (H_{1} ⊕ H_{2})^{n} ⊕ K. Since this is true for every K, H_{1} ⊕ H_{2} has the n-substitution property. ✓

There are examples, however, showing that if n > 1 then a direct summand of a module with the n-substitution property need not have the n-substitution property.

proposition 11.74. If H has the n-substitution property then H has the m-substitution property for every m > n. proof: Let m > n and suppose that H^{m} ⊕ K = H′ ⊕ L with H′ ≈ H. Write H^{m} ⊕ K = H^{n} ⊕ H^{r} ⊕ K, where r = m − n. If H has the n-substitution property there exists H′′ ⊆ H^{n} ⊕H^{r} ⊕K and P ⊆ H^{n} such that H^{m} ⊕K = H′′ ⊕L = H′′ ⊕P ⊕H^{r} ⊕K. Since P ⊕ H^{r} ⊆ H^{m}, this shows that H has the m-substitution property. ✓

A more technical description of the n-substitution property will lay the ground for the stable range conditions. First we need a lemma.

lemma 11.75. Let α: S → G = X ⊕ K be a split monomorphism and β: X ⊕ K → S a surjection such that βα = 1_{S}.

(1) If π: X ⊕ K → X is the projection onto X with kernel K then πα is a split monomorphism from S to X if and only if there exists P ⊆ X such that

G = α(S) ⊕ P ⊕ K.

(2) Let M = Ker β. Then the restriction of β to X is a split surjection if and only if there exists S̃ ⊆ X such that G = S̃ ⊕ Ker β = S̃ ⊕ M.

proof: (1) ( ⇒ ): If πα: S → X is split monic then X = πα(S) ⊕ P for some P ⊆ X and π maps α(S) isomorphically onto πα(S). Therefore by Lemma 11.13, G = πα(S) ⊕ P ⊕ K = α(S) ⊕ P ⊕ K.

( ⇐ ): If G = α(S) ⊕ P ⊕ K for some P ⊆ X then by Lemma 11.13, π maps α(S) ⊕ P isomorphically onto X. Therefore πα is monic and πα(S) is a summand of X, showing that πα is split.

(2) ( ⇒ ): If the restriction β_{1} of β to X is a split surjection then there exists S̃ ⊆ X such that X = S̃ ⊕ Ker β_{1} = S̃ ⊕ (X ∩ M). But then S̃ ∩ M ⊆ S̃ ∩ X ∩ M = 0 and β(S̃) = β_{1}(S̃) = S, so G = S̃ ⊕ Ker β = S̃ ⊕ M.

( ⇐ ): If there exists S̃ ⊆ X such that G = S̃ ⊕ M, then S̃ ∩ Ker β_{1} ⊆ S̃ ∩ M = 0 and β_{1}(S̃) = β(S̃) = S, showing that β_{1} maps S̃ isomorphically onto S and so β_{1} is a split surjection from X onto S. ✓

We now easily get the following.

proposition 11.76. Let π: H^{n} ⊕ K → H^{n} be the projection with kernel K. Then H^{n} is an n-substitutable summand of H^{n} ⊕ K if and only if for every split surjection β: H^{n} ⊕ K → H there exists a splitting map α: H → H^{n} ⊕ K such that πα is a split monomorphism. proof: The definition of an n-substitutable summand can be restated as saying that H^{n} is an n-substitutable summand of H^{n} ⊕ K if and only if whenever β: H^{n} ⊕ K → H is a split surjection and L = Ker β, then there exists a map α: H → H^{n} ⊕ K splitting β such that H^{n} ⊕ K = α(H) ⊕ P ⊕ K for some P ⊆ H^{n}. By Lemma 11.75 this holds if and only if πα is a split monomorphism. ✓

For the case K = H, the condition given in Proposition 11.76 can be stated in a more computational (and obscure!) form. We need some definitions first.

definition 11.77. Let E be a ring. If (β_{1}, …, β_{n}), (α_{1}, …, α_{n}) ∈ E^{n} and ∑ β_{i}α_{i} = 1, we say that (β_{1}, …, β_{n}) is a right unimodular row over E and that (α_{1}, …, α_{n}) is a left unimodular row over E.

As suggested by the use of Greek letters, we will be interested here in the case where E is the endomorphism ring of some R-module. The definition amounts to saying that (β_{1}, …, β_{n}) is a right unimodular row if β_{1}, …, β_{n} generate E as a right ideal and that (α_{1}, …, α_{n}) is a left unimodular row if α_{1}, …, α_{n} generate E as a left ideal. More significant, though, is the fact that if E = End H then a left unimodular row corresponds to a split monomorphism from H to H^{n} and a right unimodular row corresponds to a split surjection from H^{n} to H.

lemma 11.78. Let H be an R-module and E = End H. Let (β_{1}, …, β_{n}) ∈ E^{n} and let β: H^{n} → H be given by β(h_{1}, …, h_{n}) = ∑ β_{i}(h_{i}). Let (α_{1}, …, α_{n}) ∈ E^{n} and let α: H → H^{n} be the map given by α(h) = (α_{1}(h), …, α_{n}(h)).

(1) α maps H isomorphically onto a summand of H^{n} if and only if (α_{1}, …, α_{n}) is a left unimodular row.

(2) β is a split surjection from H^{n} onto H if and only if (β_{1}, …, β_{n}) is a right unimodular row. proof: Note that any map β: H^{n} → H must have the indicated form for some n-tuple (β_{1}, …, β_{n}) with β_{i} ∈ E, and likewise any map H → H^{n} must correspond to some n-tuple (α_{1}, …, α_{n}). Furthermore, in this case βα = ∑ β_{i}α_{i} ∈ End H. Thus the assertions of the Lemma simply reduce to the definition of a split monomorphism and split surjection. ✓

If α: H → H^{n} is given by α(h) = (α_{1}(h), …, α_{n}(h)), where α_{i} ∈ End H, we will refer to the n-tuple (α_{1}, …, α_{n}) as the coordinates of α. Similarly, if β: H^{n} → H is given by β(h_{1}, …, h_{n}) = β_{1}(h_{1}) + ⋯ + β_{n}(h_{n}), with β_{i} ∈ End H, we will call the n-tuple (β_{1}, …, β_{n}) the coordinates of β.

definition 11.79. We say that a positive integer n is in the stable range for a ring E if for every right unimodular row (β_{1}, …, β_{n+1}) over E there exists (α_{1}, …, α_{n+1}) ∈ E^{n+1} such that ∑_{1}^{n+1} β_{i}α_{i} = 1 and (α_{1}, …, α_{n}) ∈ E^{n} is a left unimodular row. (Obviously if ∑_{1}^{n+1} β_{i}α_{i} = 1 then (α_{1}, …, α_{n+1}) has to be a left unimodular row. The point of the definition is that one can choose (α_{1}, …, α_{n+1}) so that one still has a left unimodular row when the last coordinate is dropped off.)

The following theorem restates Proposition 11.76 for the case K = H in the language of stable range.

theorem 11.80. Let H be an R-module and E = End H. The following conditions are equivalent:

(1) H^{n} is an n-substitutable summand of H^{n} ⊕ H.

(2) H has the n-substitution property.

(3) Whenever H′ is a summand of H^{n} ⊕ H with H′ ≈ H, there exists H̃ ⊆ H^{n} and M ⊆ H^{n} ⊕ H such that H^{n} ⊕ H = H′ ⊕ M = H̃ ⊕ M.

(4) E has n in its stable range, i.e. for every right unimodular row (β_{1}, …, β_{n+1}) ∈ E^{n+1} there exists (α_{1}, …, α_{n+1}) ∈ E^{n+1} such that ∑ β_{i}α_{i} = 1 and (α_{1}, …, α_{n}) is a left unimodular row in E^{n}.

(5) For every left unimodular row (α_{1}, …, α_{n+1}) over E there exists a row (β_{1}, …, β_{n+1}) ∈ E^{n+1} such that ∑_{1}^{n+1} β_{i}α_{i} = 1 and (β_{1}, …, β_{n}) is a right unimodular row over E.

(6) For every right unimodular row (β_{1}, …, β_{n+1}) ∈ E^{n+1} there exist γ_{1}, …, γ_{n} ∈ E such that (β_{1}, …, β_{n}) + β_{n+1}(γ_{1}, …, γ_{n}) is a right unimodular row over E.

(7) for every left unimodular row (α_{1}, …, α_{n+1}) ∈ E^{n+1} there exist γ_{1}, …, γ_{n+1} ∈ E such that (α_{1}, …, α_{n}) + (γ_{1}, …, γ_{n})α_{n+1} is a left unimodular row over E. proof: (1) ⇒ (3): Let G = H^{n} ⊕ H and suppose that G = H′ ⊕ L with H′ ≈ H. If H^{n} is an n-substitutable summand of G then by Proposition 11.67, L ≈ H^{n}. Thus there exists an automorphism of

G = L ⊕ H′ = H ⊕ H^{n}

taking H^{n} to L and H to H′. (The order of the summands here has been reversed to make it easier for the reader to see the pattern that will be used.) Thus we can interchange the roles of H^{n} and L (as well as H and H′). In other words, L = H_{1}^{n}, where H_{1} ≈ H, and from the fact that H^{n} is an n-substitutable summand of G it follows L is also an n-substitutable summand of G, so that that there exists H′′ ⊆ G and P ⊆ L such that G = H′′ ⊕ H^{n} = H′′ ⊕ P ⊕ H′. Thus by Proposition 11.66 (with the roles of H^{n} and L reversed and H′ taking the role of K), there exist H̃ ⊆ H^{n} and M ⊆ G such that G = H̃ ⊕ M = H̃ ⊕ H′, as required.

(3) ⇒ (1): Suppose that

G = H^{n} ⊕ H_{0} = H′ ⊕ L = H′ ⊕ M = H̃ ⊕ M,

where H′ ≈ H, H̃ ⊆ H^{n}, and H_{0} = H. Then by Lemma 11.1, H^{n} = H̃ ⊕ X for some X and G = H̃ ⊕ X ⊕ H_{0} . Then

L ≈ G/H′ ≈ M ≈ G/H̃ ≈ X ⊕ H_{0} ≈ X ⊕ H̃ = H^{n}.

Therefore, analogously to the preceding step, we may apply condition (3) to G = L ⊕ H′ = H_{0} ⊕ H^{n} with the roles of L and H^{n} interchanged (as well as those of H′ and H_{0} ). I.e. there exist H̃_{0} ⊆ L and M_{0} ⊆ G such that G = M_{0} ⊕ H̃_{0} = M_{0} ⊕ H_{0} . Therefore by Proposition 11.66 there exist H′′ ⊆ G and P ⊆ H^{n} such that G = H′′ ⊕ L = H′′ ⊕ P ⊕ H_{0} , showing that H^{n} is an n-substitutable summand of H^{n} ⊕ H_{0} = H^{n} ⊕ H.

(2) ⇒ (1): Clear.

(1) ⇔ (4): Let (β_{1}, …, β_{n+1}) be a right unimodular row and let β: H^{n+1} → H be the split surjection with coordinates (β_{1}, …, β_{n+1}). Let π: H^{n} ⊕ H → H^{n} be the usual projection and let L = Ker β. Note that if (α_{1}, …, α_{n+1}) is a left unimodular row and if α: H → H^{n} ⊕ H is the split monomorphism with coordinates (α_{1}, …, α_{n+1}) then (α_{1}, …, α_{n}) are the coordinates for πα. Therefore (4) can be restated as saying that for every split projection β: H^{n} ⊕ H → H there exists a splitting map α such that πα is a split monomorphism. By Proposition 11.76 this is equivalent to the assertion that H^{n} is a n-substitutable summand of H^{n} ⊕ H.

(4) ⇒ (2): Suppose that for some K, H^{n} ⊕ K = H′ ⊕ L, with H′ ≈ H. Then there exists a split monomorphism γ: H → H^{n} ⊕ K with γ(H) = H′. Let (γ_{1}, …, γ_{n}, φ) denote the coordinates of γ, where γ_{i} ∈ E = End H and φ ∈ Hom(H, K). Then there exists a splitting map β: H^{n} ⊕ K → H withKer β = L and if the coordinates of β are (β_{1}, …, β_{n}, ψ) with β_{i} ∈ End H and ψ ∈ Hom(K, H), then ∑ β_{i}γ_{i} + ψφ = 1. Thus (β_{1}, …, β_{n}, ψφ) is a left unimodular row over E. Now if (4) holds then there exists (α_{1}, …, α_{n+1}) ∈ E^{n+1} such that ∑_{1}^{n} β_{i}α_{i} + ψφα_{n+1} = 1 and (α_{1}, …, α_{n}) is a left unimodular row. This says that if α: H → H^{n} ⊕ K is the map with coordinates (α_{1}, …, α_{n}, φα_{n+1}) then α splits β and by Lemma 11.78 πα is a split monomorphism, where π: H^{n}⊕K → H^{n} is the usual projection. Thus H^{n}⊕K = α(H)⊕Ker β = α(H)⊕L and by Lemma 11.75 there exists P ⊆ H^{n} such that

H^{n} ⊕ K = α(H) ⊕ P ⊕ H = α(H) ⊕ L.

Setting H′′ = α(H), this shows that H^{n} is an n-substitutable summand of H^{n} ⊕ K. Since this holds for every K, H has the n-substitution property.

(3) ⇔ (5): Let (α_{1}, …, α_{n+1}) be a left unimodular row over E and let α: H → H^{n} ⊕ H be the split monomorphism with coordinates α_{i}. Note that if (β_{1}, …, β_{n+1}) is a right unimodular row over E and β: H^{n+1} → H the corresponding split surjection then (β_{1}, …, β_{n}) are the coordinates of the restriction of β to H^{n}. Thus (5) states that for every split monomorphism α: H → H^{n} ⊕ H there exists a splitting map β such that the restriction of β to H^{n} is a split surjection. By Proposition 11.75 this is equivalent to condition (3).

(5) ⇒ (7): Let (α_{1}, …, α_{n+1}) be a left unimodular row over E and let (β_{1}, …, β_{n+1}) be such that ∑_{1}^{n+1} β_{i}α_{i} = 1 and (β_{1}, …, β_{n}) is a right unimodular row. Then there exists (γ_{1}, …, γ_{n}) such that ∑_{1}^{n} β_{i}γ_{i} = 1. Then ∑_{1}^{n} β_{i}(α_{i} + γ_{i}β_{n+1}α_{n+1}) = 1, so that (α_{1}, …, α_{n}) + (γ_{1}β_{n+1}, …, γ_{n}β_{n+1})α_{n+1} is a left unimodular row.

(7) ⇒ (5): Let (α_{1}, …, α_{n+1}) be a left unimodular row and let (γ_{1}, …, γ_{n}) be such that (α_{1}, …, α_{n}) + (γ_{1}, …, γ_{n})α_{n+1} is a left unimodular. Thus there exists (β_{1}, …, β_{n}) ∈ E^{n} such that ∑_{1}^{n} β_{i}(α_{i} + γ_{i}α_{n+1}) = 1. Then (β_{1}, …, β_{n}) is a right unimodular row and β_{1}α_{1} + ⋯ + β_{n}α_{n} + (∑_{1}^{n} β_{i}γ_{i})α_{n+1} = 1.

(4) ⇒ (6) and (6) ⇒ (4): Analogous. ✓

proposition 11.81. If n is in the stable range of E then so is m for every m > n. proof: This is clear from Theorem 11.80 and Proposition 11.74. ✓

lemma 11.82. Let J be an ideal contained in the Jacobson radical of E.

(1) Let a_{1}, … α_{n}, β_{1}, …, β_{n} ∈ E and let ᾱ_{i}, β̄_{i} denote the images of α_{i} and β_{i} in E/J. Then ∑ β_{i}α_{i} is invertible in E if and only if ∑ β̄_{i}ᾱ_{i} is invertible in E/J.

(2) Let (γ_{1}, …, γ_{n}) ∈ E^{n} and let γ̄_{i} denote the image of γ_{i} in E/J. Then (γ_{1}, …, γ_{n}) is a left unimodular row [right unimodular row] over E if and only if (γ̄_{1}, …, γ̄_{n}) is a right unimodular row [left unimodular row] over E/J. proof: (1) This is just Proposition 1.*.

(2) Clear from (1). ✓

proposition 11.83. Let J be contained in the Jacobson radical of E. Then n is in the stable range of E if and only if n is in the stable range of E/J.

proof: ( ⇐ ): Let (β_{1}, …, β_{n+1}) be a right unimodular row in E and let β̄_{i} be the image of β_{i} in E/J. Then by Lemma 11.82, (β̄_{1}, …, β̄_{n+1}) is a right unimodular row in E/J. Thus if n is in the stable range of E/J there exists (α_{1}, …, α_{n+1}) ∈ E^{n+1} such that ∑_{1}^{n+1} β̄_{i}ᾱ_{i} = ¯1 and (ᾱ_{1}, …, ᾱ_{n}) is a left unimodular row over Ē. Then by Lemma 11.82, ∑ β_{i}α_{i} is invertible and (α_{1}, …, α_{n}) is a left unimodular row over E. This shows that n is in the stable range of E.

( ⇒ ): Let (β_{1}, …, β_{n+1}) ∈ E^{n+1} be such that (β̄_{1}, …, β̄_{n+1}) is a right unimodular row over E/J. By Lemma 11.82, (β_{1}, …, β_{n+1}) is a right unimodular row over E. Thus if E has n in its stable range then there exists (α_{1}, …, α_{n+1}) ∈ E^{n+1} such that ∑_{1}^{n+1} β_{i}α_{i} = 1 and (α_{1}, …, α_{n}) is a left unimodular row. Then ∑ β̄_{i}ᾱ_{1} = ¯1 ∈ E/J and by Lemma 11.82, (ᾱ_{1}, …, ᾱ_{n}) is a left unimodular row over E/J. This shows that n is in the stable range of E/J. ✓

proposition 11.84. (1) 1 is in the stable range for a ring E if and only if whenever β_{i}, α_{i} ∈ E are such that β_{1}α_{1} + β_{1}α_{1} = 1 then there exists φ ∈ E such that β_{1} + β_{2}φ is invertible.

(2) If H is an R-module then 1 is in the stable range of End_{R} H if and only if H has the substitution property.

proof: (2) This follows from Theorem 11.80 and Proposition 11.72.

(1) This follows from (2) and Theorem 11.59. Alternatively, it follows from Theorem 11.80 if one notes that (β_{1} + β_{2}φ) is a right unimodular row of length 1 if and only if β_{1} + β_{2}φ is invertible. (Recall that if H has the substitution property then by Proposition 11.53 and Lemma 11.54 a right invertible element of End H is invertible.) ✓

proposition 11.85. If E is artinian, then 1 is in the stable range of E. proof: Since E ≈ End_{E} E, by Proposition 11.84 it suffices to prove that E has the substitution property as an E-module. But if E is artinian then E has finite length so this follows from Proposition 11.63. ✓ FINITE RANK TORSION FREE MODULES. We now return to the environment of this book. Namely, we will consider in this section only finite rank torsion free modules over a dedekind domain W, and rings E which are finite rank torsion free W -algebras. Note that a finite rank torsion free W -module is never isomorphic to a proper summand of itself.

Among other things, we will show that if E if a finite rank torsion free ring over a dedekind domain W then 2 is in the stable range for E.

proposition 11.86. Finite rank projective W -modules have the cancellation property. proof: Let G = H ⊕ K = H′ ⊕ M where H is projective and H′ ≈ H. (We do not assume that K and M are torsion free.) Let π be the projection onto H with kernel K. Then π(M) ⊆ H so π(M) is projective and thus the restriction of π to M splits and so M = M_{1} ⊕ (M ∩ Ker π) = M_{1} ⊕ (K ∩ M), where M_{1} ≈ π(M). Likewise K = K_{1} ⊕ (K ∩ M) where K_{1} is projective. It suffices to prove that M_{1} ≈ K_{1} . Now H ⊕ K_{1} ≈ G/(K ∩ M) ≈ H′ ⊕ M_{1} ≈ H ⊕ M_{1} . Since K_{1} and M_{1} are projective, by Proposition 1.* they are isomorphic to direct sums of ideals, as is H. But in this case it follows from Proposition 1.* that K_{1} ≈ M_{1} . ✓

proposition 11.87. An indecomposable W -module with the exchange property is p-local for some p. proof: If H is indecomposable and has the exchange property then by Proposition 11.23, End H is a local ring. Thus J(End H) contains all the non-automorphisms, so if p is a prime ideal in W and p ⊈ J(End H) then H is p-divisible. But J(End H) cannot contain two distinct primes of W since if p, p′ ⊆ J(End H) and p ≠ p′ then 1 ∈ p + p′ ⊆ J(End H), a contradiction. Thus H is p-divisible for all primes p of W with at most one exception, so H is local. ✓

It need not be the case, however, that an indecomposable module with the exchange property is strongly indecomposable.

example 11.88. Let W be a discrete valuation ring with maximal ideal p. Let

E = {(x, y) ∈ W × W | x ≡ y (mod p)}.

If G is a W -module such that End G = E, then G is indecomposable and has the exchange property but G is not strongly indecomposable. proof: Since QEnd G = QE = Q × Q, a ring with non-trivial idempotents, then G is not strongly indecomposable. Now to see that G is indecomposable and has the exchange property it suffices to see that E is a local ring. In fact, let M = {(x, y) | x, y ∈ p}.

Then M is a proper ideal in E and if (x, y) ∉ M then (x, y) is invertible. In fact, if (x, y) ∉ M then x, y ∉ p (remembering that x ≡ y (mod p)) and so since W is local there exist u, v ∈ W with xu = 1, yv = 1. Necessarily, u ≡ v (mod p), so (u, v) ∈ E. And (x, y)(u, v) = (1, 1) = 1_{R}. It follows that M is the unique maximal ideal of E and so E is local. ✓

proposition 11.89. If a finite rank torsion free W -module H has the exchange property then it has the substitution property. proof: Since H has finite rank, it is a direct sum of indecomposables. Therefore the result follows from Proposition 11.23 and Proposition 11.60. ✓

proposition 11.90. A W -module H has the substitution or exchange property if and only if H/d(H) does. proof: Since H ≈ d(H) ⊕ H/d(H) and d(H) is a direct sum of copies of Q, by Proposition 11.61 and Proposition 11.21 it suffices to show that Q has the substitution and exchange properties. But since End Q ≈ Q and Q is a field, hence a fortiori a local ring, this follows the Proposition 11.60 and Proposition 11.26. ✓

proposition 11.91. Let H_{1} and H_{2} be nearly isomorphic W -modules.

(1) H_{1} has the exchange property if and only if H_{2} does.

(2) H_{1} has the substitution property if and only if H_{2} does.

proof: If H_{2} is nearly isomorphic to H_{1} then by Proposition 10.* H_{2} is isomorphic to a summand of H_{1} ⊕ H_{1} . But it follows from Proposition 11.21 and Proposition 11.61 that if H_{1} has the exchange or substitution property then so do H_{1} ⊕ H_{1} and its summands. ✓

lemma 11.92. Let H be a reduced semi-local W -module and let 𝔞 = ⋂_{pH≠H} p. Then 𝔞 End H ⊆ J(End H). proof: Let w ∈ 𝔞 = ⋂ p. If φ ∈ End H, we will show that 1−wφ is invertible in End H. In fact, if h ∈ Ker(1 − wφ) then h = wφ(h) ∈ wH. Therefore Ker(1 − wφ) ⊆ wH ⊆ pH for all primes p such that pH ≠ H and obviously also for all p with pH = H. Since Ker(1 − wφ) ◁ H it follows that Ker(1 − wφ) is p-divisible for all primes p and hence is divisible. Since H is reduced, Ker(1 − wφ) = 0 and so 1 − wφ is monic. Therefore by Proposition 1.* (1 − wφ)(H) is an essential submodule of H. But note also that if (1 − wφ)(h) = h − wφ(h) ∈ pH for some p then h ∈ pH, since either w ∈ p or pH = H, and thus (1 − wφ)(h) ∈ (1 − wφ)(pH) = p(1 − wφ)(H). Thus we see that (1 − wφ)(H) ◁ H. Since (1 − wφ)(H) is both pure and essential in H it follows that (1 − wφ)(H) = H. Thus 1 − wφ is an automorphism of H for all φ ∈ End H. It follows that w ∈ J(End H). Since this is true for all w ∈ 𝔞, thus 𝔞 ⊆ J(End H). ✓ proposition 11.93. A semi-local W -module H has the substitution property.

proof: By Proposition 11.90 we may suppose that H is reduced. Now by Lemma 11.92, 𝔞 End H ⊆ J = J(End H) for some non-trivial ideal 𝔞 ⊆ W. Then (End H)/J is a homomorphic image of (End H)/𝔞(End H) and is thus artinian by Proposition 1.*. Therefore by Proposition 11.85 and Proposition 11.83, H has the substitution property. ✓

proposition 11.94. If K is a semi-local W -module then for any H, H is a cancellable summand of H ⊕ K.

proof: If K is semi-local then K = S^{−1}K for some multiplicative set S in W such that S^{−1}W is semi-local. If now H ⊕ K ≈ H ⊕ M then by Proposition 3.* M ∼ K and thus also M = S^{−1}M and so S^{−1}H ⊕ K = S^{−1}H ⊕ S^{−1}K ≈ S^{−1}H ⊕ M. But since S^{−1}H is a semi-local module it has the substitution property by Proposition 11.93. Thus by Proposition 11.51, K ≈ M. This shows that H is a cancellable summand of H ⊕ K. ✓

The criterion given in Theorem 11.59 for determining whether a module has the substitution property can certainly not be called easy to use. In the case of a strongly indecomposable W -module, there is the following somewhat more workable condition.

proposition 11.95. If H is strongly indecomposable then H has the substitution property if and only if for each non-trivial ideal 𝔞 of W, every invertible element of (End H)/𝔞(End H) lifts to an invertible element of End H. (It is sufficient for this to be true for principal ideals.)

proof: ( ⇒ ): This direction is true without the assumption that H is strongly indecomposable. If β ∈ End H and the image of β in (End H)/𝔞(End H) is invertible, then βα − 1 ∈ 𝔞 for some α ∈ End H and so βα + wα_{2} = 1 for some α_{2} ∈ End H and w ∈ 𝔞. If H has the substitution property then it follows from Theorem 11.59 that β + wφ′ is an automorphism of H for some φ′ ∈ End H. But β + wφ′ has the same image as β in (End H)/𝔞(End H).

( ⇐ ): Assume that for every ideal principal ideal 𝔞, every invertible element of (End H)/𝔞(End H) lifts to an invertible element of End H. If H is strongly indecomposable then by Proposition 3.* every endomorphism of H is either monic or belongs to nil rad End H. Now suppose β_{i}, α_{i} ∈ End H and β_{1}α_{1} + β_{2}α_{2} = 1. We will prove that there exists φ ∈ End H such that β_{1} + β_{2}φ is invertible, by considering the two cases where β_{2}α_{2} is nilpotent and where it is monic. If β_{2}α_{2} is nilpotent then β_{1}α_{1} ≡ 1 (mod N), where N = nil rad End H, and so β_{1}α_{1} is invertible by Proposition 1.*. Thus β_{1} = β_{1} + β_{2}0 is also invertible.

If, on the other hand, β_{2}α_{2} is monic then it is invertible in QEnd H and so β_{2} is invertible in QEnd H. Hence there exists β ∈ End H such that

β_{2}β = w

for some w ≠ 0 ∈ W. Now since (End H)/w(End H) is artinian, it has 1 in its stable range by Proposition 11.85. Since β_{1}α_{1} + β_{2}α_{2} = 1, it follows that there exists φ ∈ End H such that the image of β_{1} + β_{2}φ in (End H)/w(End H) is invertible. Thus by hypothesis it is the image of an invertible element of End H, i.e. for some γ ∈ End H,

β_{1} + β_{2}φ − wγ = β_{1} + β_{2}(φ − βγ)

is invertible.

Therefore H has the substitution property by Proposition 11.84. ✓

corollary 11.96. If H is strongly indecomposable and for each non-trivial ideal 𝔞 of W the obvious map End H → End(H/𝔞H) is surjective, then H has the substitution property if and only if for all 𝔞 ≠ 0, every automorphism of H/𝔞H is induced by an automorphism of H. proof: ( ⇒ ): Let φ̄ be an automorphism of H/𝔞H. If End H → End(H/𝔞H) is surjective then φ̄ ∈ (End H)/𝔞(End H). If H has the substitution property then it follows from Proposition 11.95 that there is an automorphism φ of H inducing φ̄.

( ⇐ ): Suppose H is strongly indecomposable and that every automorphism of H/𝔞H is induced by an automorphism of H and let φ̄ be an invertible element of (End H)/𝔞(End H). Then φ̄ induces an automorphism of H/𝔞H and so by assumption this is induced by an automorphism φ of H. Then φ̄ is the image of φ in (End H)/𝔞(End H). Thus by Proposition 11.95, H has the substitution property. ✓

corollary 11.97. Let A be a rank-one W -module and let

S = {w ∈ W | wA = A}.

Let X be the set of ideals 𝔞 of W such that all the associated primes of 𝔞 are disjoint from S. Then A has the substitution property if and only if for each ideal 𝔞 in X and w ∈ W such that w is invertible modulo 𝔞, there exist s, w′ ∈ S such that sw ≡ w′ (mod 𝔞). proof: S is the largest multiplicative set in W such that A = S^{−1}A and an ideal 𝔞 of W belongs to X if and only if 𝔞 = W ∩ 𝔞S^{−1}W. Furthermore End A = S^{−1}W and if 𝔞 ∈ X then End A/𝔞(End A) = S^{−1}W/𝔞S^{−1}W = W/𝔞. Thus End A → (End A)/𝔞(End A) is surjective and so by Proposition 11.95, A has the substitution property if and only if for all w ∈ W such that the image of w in W/𝔞 ≈ (End A)/𝔞(End A) is invertible, there exists w′/s ∈ S^{−1}W such that w′/s is invertible in S^{−1}W and w′/s ≡ w/1 (mod 𝔞). But this means that w′ ∈ S and

sw ≡ w′ (mod 𝔞). ✓

example 11.98. Let p be a prime in W and choose w_{k} ∈ W such that for all w ∉ p there exists k with w ≡ w_{k} (mod p). Let A be a rank-one module such that w_{k}A = A for all k.

lemma 11.99. Let β_{1}, …, β_{n}, δ_{2}, …, δ_{n} ∈ E. Then (β_{1}, …, β_{n}) is a right unimodular row if and only if (β_{1} + ∑_{2}^{n} β_{i}δ_{i}, β_{2}, …, β_{n}) is a right unimodular row. proof: If (β_{1}, …, β_{n}) is a right unimodular row and ∑ β_{i}α_{i} = 1 then

(β_{1} + ∑_{2}^{n} β_{i}δ_{i})α_{1} + ∑_{2}^{n} β_{i}(α_{i} − δ_{i}α_{1}) = 1,

so that (β_{1} + ∑_{2}^{n} β_{i}δ_{i}, β_{2}, …, β_{n}) is a right unimodular row. The converse is analogous. ✓

lemma 11.100. If E is a finite rank torsion free ring with 1 in its stable range and E′ is a subring of E which is quasi-equal to E, then 1 is in the stable range of E′. proof: Since E′ is quasi-equal to E there exists w ≠ 0 ∈ W such that wE ⊆ E′ ⊆ E. Then E′/wE has finite length. For φ ∈ E, let φ̄ denote the image of φ in E/wE. Suppose that α_{i}, β_{i} ∈ E′ and β_{1}α_{1} + β_{2}α_{2} = 1. Then β̄_{1}ᾱ_{1} + β̄_{2}ᾱ_{2} = ¯1. Since E′/wE is an artinian ring it has 1 in its stable range by Proposition 11.85. Thus by Proposition 11.84, β̄_{1} + β̄_{2}ψ̄ is invertible for some ψ̄ ∈ E′/wE. Let ψ be a pre-image in E′ for ψ̄. By Lemma 11.99 there is no loss of generality in replacing β_{1} by β_{1} + β_{2}ψ. Therefore we may assume that β̄_{1} is invertible in E′/wE. Thus there exist λ ∈ E′ and μ ∈ E such that β_{1}λ + wμ = 1. Then

β_{1}(λ + wα_{1}μ) + wβ_{2}α_{2}μ = β_{1}λ + wμ = 1.

There is now no loss of generality in replacing α_{1} by λ + wα_{1}μ ∈ E′ and α_{2} by wα_{2}μ, so that we may suppose that α_{2} ∈ wE.

We now have β_{1}α_{1} + β_{2}α_{2} = 1 with β̄_{1} invertible in E′/wE and α_{2} ∈ wE ⊆ E′. Since 1 is in the stable range of E, thus β_{1} + β_{2}α_{2}φ is invertible in E for some φ ∈ E. Hence there exists δ ∈ E with δ(β_{1} + β_{2}α_{2}φ) = (β_{1} + β_{2}α_{2}φ)δ = 1. Since α_{2} ∈ wE, thus δ̄β̄_{1} = β̄_{1}δ̄ = 1. But now β̄_{1} is an invertible element of E′/wE and so it follows that δ̄ ∈ E′/wE and thus δ ∈ E′ + wE = E′. Therefore β_{1} + β_{2}α_{2}φ is invertible in E′. Furthermore α_{2}φ ∈ wE ⊆ E′. This shows that 1 is in the stable range of E′. ✓

warning. Lemma 11.100 does not say that if 1 is in the stable range of E then 1 is in the stable range of any ring quasi-equal to E. It is essential that E′ ⊆ E. proposition 11.101. Let H′ be a fully invariant submodule of H which is quasi-equal to H. If H′ has the substitution property then so does H. proof: Since H′ is quasi-equal to H, wH ⊆ H′ for some w ≠ 0 ∈ W. Furthermore, QEnd H′ = QEnd H and since H′ is fully invariant in H every endomorphism of H restricts to an endomorphism of H′, so that End H ⊆ End H′. But if φ ∈ End H′ then wφ(H) = φ(wH) ⊆ φ(H′) ⊆ H′ ⊆ H, so that w(End H′) ⊆ End H. Thus End H is quasi-equal to End H′.

Now if H′ has the substitution property then by Theorem 11.59, 1 is in the stable range of End H′ and so by Lemma 11.100 1 is in the stable range of End H. Therefore H has the substitution property. ✓

example 11.102. Let A and B be rank-one modules with the substitution property and incomparable types. Suppose that there exist w ∈ W, a ∈ A and b ∈ B with a ∉ wA, b ∉ wB. G be the submodule of QA ⊕ QB generated by A ⊕ B together with (a + b)/w. Then G has the substitution property. proof: If s = t(A) and t = t(B) then A = G(s) and B = G(t), so A ⊕ B is fully invariant in G by Proposition 4.*. Furthermore by Proposition 11.61 A ⊕ B has the substitution property. Therefore by Proposition 11.101, G has the substitution property. ✓

lemma 11.103. Let E be a finite rank torsion free semi-prime ring and β_{1}, β_{2}, α_{1}, α_{2} ∈ E. If β_{1}α_{1} + β_{2}α_{2} is invertible in QE then there exists φ ∈ E such that β_{1} + β_{2}φ is not a zero divisor in E. proof: Since QE is semi-simple artinian and QE = β_{1}QE + β_{2}QE, there exists a right ideal I ⊆ β_{2}QE with QE = β_{1}QE ⊕ I. Thus there exist γ_{1} ∈ QE and μ ∈ I such that β_{1}γ_{1} + μ = 1.

Now 1 is in the stable range of QE by Proposition 11.85. Thus there exists φ′ ∈ QE such that β_{1} + μφ′ is invertible in QE. Since μφ′ ∈ I ⊆ β_{2}QE, it follows that β_{1} + β_{2}φ = 1 for some φ with β_{2}φ = μφ′ ∈ I. Choose w ≠ 0 ∈ W such that wφ ∈ E. Then wβ_{1} + wβ_{2}φ = w ≠ 0. Furthermore since β_{2}φ ∈ I and QE = β_{1}QE ⊕ I it follows that for ρ ∈ E,

(β_{1} + β_{2}wφ)ρ = 0 ⇒ β_{1}ρ = −β_{2}wφρ ∈ I ∩ β_{1}QE = 0
⇒ wβ_{1}ρ + wβ_{2}φρ = wρ = 0 ⇒ ρ = 0.

Thus β_{1} + β_{2}wφ is not a zero divisor in E. ✓ lemma 11.104. Let I be an ideal in a ring E such that 1 is in the stable range of E/I and suppose that (β_{1}, β_{2}, β_{3}) is a right unimodular row over E such that I ⊆ β_{1}E. Then there exists γ_{2} ∈ E such that (β_{1}, β_{2}) + β_{3}(0, γ_{2}) is a right unimodular row. proof: If (β_{1}, β_{2}, β_{3}) is a right unimodular row then there exists (α′_{1}, α′_{2}, α′_{3}) ∈ E^{3} such that β_{1}α′_{1} + β_{2}α′_{2} + β_{3}α′_{3} = 1. Since I ⊆ β_{1}E, thus β_{2}α′_{2} + β_{3}α′_{3} ≡ 1 (mod I). Since 1 is in the stable range of E/I, by Proposition 11.84 there exists γ_{2} ∈ E such that β_{2} + β_{3}γ_{2} is invertible modulo I. Choose α_{2} ∈ E such that

(β_{2} + β_{3}γ_{2})α_{2} = β_{2}α_{2} + β_{3}γ_{2}α_{2} ≡ 1 (mod I).

Thus there exists μ ∈ I with μ + β_{2}α_{2} + β_{3}γ_{2}α_{2} = 1. But μ ∈ I ⊆ β_{1}E so μ = β_{1}α_{1} for some α_{1} . Thus

1 = β_{1}α_{1} + (β_{2} + β_{3}γ_{2})α_{2} = (β_{1} + β_{3}0)α_{1} + (β_{2} + β_{3}γ_{2})α_{2}.

This shows that (β_{1}, β_{2}) + β_{3}(0, γ_{2}) is a right unimodular row. ✓

theorem 11.105. (1) If E is a finite rank torsion free ring then 2 is in the stable range of E.

(2) Every finite rank torsion free W -module has the 2-substitution property.

proof: (1) By Proposition 11.83, E has 2 in its stable range if and only if E/N does, where N = nil rad E. Thus no generality is lost in supposing E semi-prime. Suppose then that (β_{1}, β_{2}, β_{3}) is a right unimodular row over E. Let (α′_{1}, α′_{2}, α′_{3}) be such that β_{1}α′_{1} + β_{2}α′_{2} + β_{3}α′_{3} = 1 ∈ E. We will show that there exist (α_{1}, α_{2}, α_{3}) ∈ E^{2} such that β_{1}α_{1} + β_{2}α_{2} + β_{3}α_{3} = 1 and (α_{1}, α_{2}) is a left unimodular row.

By Lemma 11.103 there exists φ ∈ E such that β_{1} + (β_{2}α′_{2} + β_{3}α′_{3})φ is not a zero divisor. Let

β′_{1} = β_{1} + (β_{2}α′_{2} + β_{3}α′_{3})φ = β_{1} + β_{2}δ_{2} + β_{3}δ_{3},

where δ_{2} = α′_{2}φ, δ_{3} = α′_{3}φ. By Lemma 11.99, (β′_{1}, β_{2}, β_{3}) is an right unimodular row. Since β′_{1} is not a zero divisor in E, by Proposition 3.* β′_{1}E is quasi-equal to E so there exists w ≠ 0 ∈ W such that wE ⊆ β′_{1}E. Since E/wE is artinian, 1 is in the stable range of E/wE by Proposition 11.85. Thus by Lemma 11.104 with I = wE there exists γ_{2} ∈ E such that (β′_{1}, β_{2}) + β_{3}(0, γ_{2}) is a right unimodular row. Thus there is a left unimodular row (α_{1}, α′′_{2}) such that

1 = β′_{1}α_{1} + (β_{2} + β_{3}γ_{2})α′′_{2}
= (β_{1} + β_{2}δ_{2} + β_{3}δ_{3})α_{1} + (β_{2} + β_{3}γ_{2})α′′_{2}
= β_{1}α_{1} + β_{2}(α′′_{2} + δ_{2}α_{1}) + β_{3}(δ_{3}α_{1} + γ_{2}α′′_{2}).

Since (α_{1}, α′′_{2}) is a left unimodular row, by Lemma 11.99 (α_{1}, α′′_{2} + δ_{2}α_{1}) is a left unimodular row. Thus by Theorem 11.80, 2 is in the stable range of E.

(2) By (1), if G is a finite rank torsion free W -module then 2 is in the stable range for End G. Therefore G has the 2-substitution property by Theorem 11.80. ✓

It seems impossible to state the following result in a way that is not confusing.

corollary 11.106. If H, K, and L are finite rank torsion free W -modules such that H ⊕ K ≈ H ⊕ L and K ≈ H ⊕ X for some X, then L ≈ P ⊕ X for some P such that H ⊕ P ≈ H ⊕ H. proof: Since H has the 2-substitution property by Theorem 11.105, this is just Corollary 11.68 with r = 1. ✓

corollary 11.107. Let H and H_{1} be finite rank torsion free W -modules.

(1) If H^{m+r} ≈ H_{1}^{m} ⊕ H^{r} for some m ≥ 2 and r ≥ 1 then H^{m} ≈ H_{1}^{m}.

(2) If H^{r+1} ≈ H_{1} ⊕ H^{r} for some r ≥ 1 then H^{2} ≈ H_{1}^{2} .

proof: By Theorem 11.105, H has the 2-substitution property, so this is just Corollary 11.69. ✓

theorem 11.108. If W has the Jordan-Zassenhaus property then two finite rank torsion free modules H and H_{1} are nearly isomorphic if and only if H^{n} ≈ H_{1}^{n} for some n ≥ 1. proof: ( ⇐ ): By Proposition 10.*.

( ⇒ ): If H and H_{1} are nearly isomorphic then by Proposition 10.* there exist positive integers m and r such that H^{m} ⊕ H^{r} ≈ H_{1}^{m} ⊕ H^{r}. But then by Corollary 11.107 either

H^{m} ≈ H_{1}^{m} or H^{2} ≈ H_{1}^{2} . ✓