10. NEAR ISOMORPHISM

E. L. Lady

April 17, 1998

We have seen that quasi-isomorphic modules look very much like each other, since only a finite number of elements prevents them from being isomorphic. Nonetheless, this finite number of elements can make for some important differences.

example 10.1. Let p and q be distinct prime ideals and let A and B be rank-one modules with Hom(A, B) = Hom(B, A) = 0 such that A and B are neither p-divisible nor q-divisible. Let a ∈ A, b ∈ B be such that a ∉ pA ∪ qA, b ∉ pB ∪ qB. Let G = A ⊕ B, let H = ⟨G, p^{−1}(a + b)⟩, K = ⟨G, q^{−1}(a + b)⟩. Then

(1) G, H, and K are all quasi-isomorphic, but H and K are indecomposable whereas G is not.

(2) Let s = t(A) and t = t(B). Then H/(H(s) + H(t)) ≈ W/p and K/(K(s) + K(t)) ≈ W/q.

(3) If φ is a monomorphism from H to K then ann(K/φ(H)) ⊆ pq.

(4) If p and q are principal ideals then there exists a monomorphism φ: H → K such that ann(K/φ(H)) = pq.

proof:

(4) If we assume QH = QK = Q ⊕ Q, and that the prime ideals p and q are principal, say p = (p_{1}), q = (q_{1}), then h ↦ q_{1}h is a monomorphism from H to K with

ann(K/φ(H)) = pq. ✓

Assertions (3) and (4) of Example 10.1 say that the ideal pq is in some sense a measure of how far H and K are from being isomorphic. (The case where one of the ideals p or q is not principal requires only a slight adjustment to assertion (4).) Assertion (2) shows the existence of an intrinsic quality distinguishing H and K.

On the other hand, there are examples of quasi-isomorphic modules where the problem of distinguishing between them is much more subtle. Although one may be able to prove that no isomorphism between the two exists, it is difficult to find any intrinsic quality distinguishing them. example 10.2. Let H = ⟨G, p^{−1}(a + b)⟩ be as in Example 10.1 and let H_{2} = ⟨G, p^{−1}(a + 2b)⟩. (Assume char W ≠ 2 and 2 ∉ p and 2 is not congruent modulo p to an invertible element of W.)

(1) If w ∈ W is such that 2w ≡ 1 (mod p) and if either 2B = B or wA = A, then H ≈ H_{2} .

(2) if End A = End B = W then H_{2} ≉ H.

(3) If p is principal then there exist monomorphisms φ and φ′ from H to H_{2} such that H_{2}/φ(H) ≈ (W/p)^{2} and H_{2}/φ′(H) ≈ W/2W. proof: Note that there always exists w with 2w ≡ 1 (mod p) since W/p is a field. Now if 2B = B then define φ: QH → QH by φ(a) = a, φ(b) = 2b. Then φ(A) = A, φ(B) = B and φ(p^{−1}(a + b)) = p^{−1}(a + 2b), so φ maps H isomorphically onto H_{2} .

Likewise if wA = A then define φ′ by φ′(a) = wa, φ′(b) = b. Then φ′(A) = A, φ′(B) = B and p^{−1}(a + 2b) = p^{−1}(2(wa + b)) = p^{−1}φ′(2(a + b)) ⊆ φ′(H), so φ′ maps H onto H_{2} in this case as well.

(2) If s = t(A) then A = H(s) = H_{2}(s). Thus an isomorphism φ from H to H_{2} must restrict to an automorphism of A, and this restriction will be given by multiplication by some invertible element u ∈ W. Likewise φ restricts to an automorphism on B given by multiplication by some invertible v ∈ W. Multiplying by u^{−1}, we may assume that φ restricts to the identity map on A. Since a + b ∈ pH, φ(a + b) = a + vb ∈ pH_{2} . But also a + 2b ∈ pH_{2} , so (v − 2)b ∈ B ∩ pH_{2} = pB. Since End B = W, it follows that v ≡ 2 (mod p). But this is a contradiction to the hypothesis that 2 is not congruent modul p to a unit of W. Therefore there cannot exist an isomorphism from H to H_{2} .

(3) If p = (p_{1}) then define φ on H by φ(x) = p_{1}x. Then φ(H) = pH ⊆ A ⊕ B ⊆ H_{2} and H_{2}/φ(H) ≈ (W/p)^{2} .

Now define φ′: QH → QH by φ′(a) = a, φ′(b) = 2b. Then φ′(A) ⊆ A, φ′(B) ⊆ B and φ′(p^{−1}(a + b)) = p^{−1}(a + 2b), so φ(H) ⊆ H_{2} . Furthermore H_{2}/φ(H) is cyclic generated by the image of b, and so H_{2}/φ(H) ≈ W/2W since 2b ∈ φ(H). ✓

Note that assertions (1) and (2) in Example 10.2 are not sufficient to determine in every case whether H_{2} ≈ H. The main point of the example is that this is a much subtler question than the non-isomorphism of the modules H and K in Example 10.1.

The relevant point in assertion (3) of Example 10.2 is that 2 and p are relatively prime, thus offering a sharp contrast to Example 10.1

definition 10.3. We say that modules G and H are nearly isomorphic if they are quasi-isomorphic and for every non-trivial ideal 𝔞 of W there exists a monomorphism φ: G → H such that 𝔞 + ann(H/φ(G)) = W.

(Note that if G and H are quasi-isomorphic and φ: G → H is a monomorphism, then H/φ(G) has finite length by Proposition 3.4.) proposition 10.4. Near isomorphism is an equivalence relation. proof: First suppose φ: G → H is a monomorphism and wH ⊆ φ(G) for w ≠ 0 ∈ W. Then ψ = wφ^{−1} is a monomorphism from H to G. Furthermore, since φ(G) ⊆ H, then wG ⊆ wφ^{−1}(H) = ψ(H). From this, it can be seen that if G is nearly isomorphic to H then H is nearly isomorphic to G.

Now notice that if φ: G → H and ψ: H → K are monomorphisms and wH ⊆ φ(G) and w′K ⊆ ψ(H), then ww′K ⊆ ψ(wH) ⊆ ψφ(G). Furthermore if (w) + 𝔞 = W and (w′) + 𝔞 = W, then (ww′) + 𝔞 = W by Proposition 0.*. From this it can be seen that if G is nearly isomorphic to H and H nearly isomorphic to K, then G is nearly isomorphic to K.

Since every G is clearly nearly isomorphic to itself, it follows that near isomorphism is an equivalence relation. ✓

proposition 10.5. If W is semi-local, then nearly isomorphic modules are isomorphic. proof: Let 𝔞 be the Jacobson radical of W. Then 𝔞 ≠ 0 since W is semi-local. If G and H are nearly isomorphic, then there exists a monomorphism φ: G → H such that 𝔞 + ann(H/φ(G)) = W. By Nakayama’s Lemma, ann H/φ(G) = W, i. e. φ(G) = H. Thus φ is an isomorphism. ✓

lemma 10.6. If G is a reduced finite rank torsion free W -module then there exists a non-trivial ideal 𝔞 such that 𝔞K ≠ K for all non-trivial submodules K of G. proof: Since G is reduced, there is at least one prime ideal p such that p^{∞}G ≠ G. If p^{∞}G = 0 then no non-trivial submodule K of G is p-divisible so we may choose 𝔞 = p. Otherwise, by induction on rank we may assume that there exists a non-trivial ideal 𝔞′ such that 𝔞′K ≠ K for all non-trivial submodules K of p^{∞}G. Now let 𝔞 = p𝔞′. Then if 0 ≠ K ⊆ p^{∞}G then 𝔞K = 𝔞′K ≠ K. And if K ⊈ p^{∞}G then K is not p-divisible, so

𝔞K = p𝔞′K ⊆ pK ⊊ K. ✓

proposition 10.7. Let G and H be reduced finite rank torsion free W -modules. The following conditions are equivalent:

(1) G and H are nearly isomorphic.

(2) G and H are quasi-isomorphic and for every prime ideal p there exists a monomorphism φ: G → H such that p ∉ Ass H/φ(G).

(3) For every non-trivial ideal 𝔞 there exist φ ∈ Hom(G, H) and ψ ∈ Hom(H, G) such that the induced maps G/𝔞G → H/𝔞H and H/𝔞H → G/𝔞G are isomorphisms.

(4) G and H are quasi-isomorphic and there exist monomorphisms φ_{1} and φ_{2} from G to H such that ann H/φ_{1}(G) + ann H/φ_{2}(G) = W. proof: (1) ⇒ (2): Recall from Proposition 0.* that since H/φ(G) is finitely generated, p ∈ Ass H/φ(G) if only if ann(H/φ(G)) ⊆ p. Now if G and H are nearly isomorphic there exists a monomorphism φ: G → H such that p + ann(H/φ(G)) = W. Thus p ∉ Ass H/φ(G).

(2) ⇒ (3): Consider first the case where 𝔞 = p, a prime. By assumption there exists a monomorphism φ: G → H such that p ∉ Ass H/φ(G). Thus ann H/φ(G) ⊈ p and so ann H_{p}/φ(G_{p}) = W_{p}, i. e. H_{p} = φ(G_{p}). It follows that H/pH = (φ(G) + pH)/pH, i. e. the induced map G/pG → H/pH is surjective. Since G and H are quasi-isomorphic, the finite length modules G/pG and H/pH are isomorphic, so the induced map must also be monic.

By induction on the number of prime factors of 𝔞 it now suffices to prove that if there exist ideals 𝔞 and 𝔟 such that 𝔞 + 𝔟 = W and maps α, β ∈ Hom(G, H) such that the induced maps ᾱ: G/𝔞G → H/𝔞H and β̄: G/𝔟G → H/𝔟H are isomorphisms, then there exists φ ∈ Hom(G, H) inducing an isomorphism G/𝔞𝔟G → H/𝔞𝔟H. In fact, define φ = uα + vβ, where u ∈ 𝔟, v ∈ 𝔞, and u + v = 1. Now notice that by the Chinese Remainder Theorem, G/𝔞𝔟G ≈ G/𝔞G ⊕ G/𝔟G and H/𝔞𝔟H ≈ H/𝔞H ⊕ H/𝔟H. Furthermore, if φ̄: G/𝔞𝔟G → H/𝔞𝔟H is the induced map, then the restriction of φ̄ to G/𝔞G agrees with ᾱ and the restriction to G/𝔟G agrees with β̄. Therefore φ̄ is an isomorphism.

(3) ⇒ (1): Let 𝔞 be a non-trivial ideal. It suffices to prove that there exists a monomorphism φ: G → H such that H/φ(G) has finite length and 𝔞′+ann H/φ(G) = W for some ideal 𝔞′ contained in 𝔞. Therefore, using Lemma 10.6, we may assume that 𝔞K ≠ K for every non-trivial submodule K of G. If (3) holds then there exists φ: G → H which induces an isomorphism G/𝔞G → H/𝔞H. It follows that Ker φ ⊆ 𝔞G. Since Ker φ is necessarily pure, it follows that Ker φ = 𝔞 Ker φ. Thus by assumption Ker φ = 0 so φ is monic. Likewise there exists a monomorphism H → G. By Proposition 3.*, G and H are quasi-isomorphic. Therefore by Proposition 3.*, H/φ(G) has finite length.

Since φ induces an isomorphism G/𝔞G ⟶^{≈} H/𝔞H then H = φ(G) + 𝔞H so that 𝔞(H/φ(G)) = H/φ(G) and thus a fortiori p(H/φ(G)) = H/φ(G) for any prime p containing 𝔞. It then follows from Proposition 0.* that if 𝔞 ⊆ p then p ∉ Supp H/φ(G) or, equivalently, ann H/φ(G) ⊈ p (c. f. Proposition 0.*). Thus no prime ideal p contains both 𝔞 and ann H/φ(G). Therefor 𝔞 + ann H/φ(G) = W.

(1) ⇒ (4): If G and H are nearly isomorphic then in particular there exists a monomorphism φ_{1} : G → H and H/φ_{1}(G) has finite length, so ann(H/φ_{1}(G)) is a non-trivial ideal. Then by definition of near isomorphism there exists a monomorphism φ_{2} : G → H such that ann H/φ_{1}(G) + ann H/φ_{2}(G) = W.

(4) ⇒ (2): If H/φ_{1}(G) and H/φ_{2}(G) have finite length and p is a prime ideal then by Proposition 0.*, p ∈ Ass H/φ_{i}(G) if and only if ann H/φ_{i}(G) ⊆ p. But if ann H/φ_{1}(G) + ann H/φ_{2}(G) = W, then p cannot contain both these annihilators, so p ∉ Ass H/φ_{i}(G) for either i = 1 or i = 2. ✓ proposition 10.8. If G is nearly isomorphic to H and quasi-isomorphic to K, then G is isomorphic to a direct summand of H ⊕ K. proof: If G is quasi-isomorphic to K then by Proposition 3.* we may assume that K ⊆ G and ann G/K ≠ 0. If H is nearly isomorphic to G then we may assume that H ⊆ G with ann G/H + ann G/K = 1. Choose u ∈ ann G/H and v ∈ ann G/K with u + v = 1. Then uG ⊆ H and vG ⊆ K. Consider the short exact sequence

0 ⟶ H ∩ K ⟶^{α} H ⊕ K ⟶^{σ} G ⟶ 0,

where α(x) = (x, −x) and σ(h, k) = h + k. (The map σ is surjective since W = ann G/H + ann G/K ⊆ ann G/(H + K) = ann G/σ(H ⊕ K). ) Then σ is split by the

map ζ: G → H ⊕ K given by ζ(g) = (ug, vg) ∈ H ⊕ K. Thus H ⊕ K ≈ G ⊕ (H ∩ K). ✓

corollary 10.9. If H_{1}, H_{2}, and H_{3} are nearly isomorphic then H_{1} ⊕ H_{2} ≈ H_{3} ⊕ H_{4} , for some H_{4} nearly isomorphc to H_{1} . proof: It is immediate from Proposition 10.8 that H_{1} ⊕ H_{2} ≈ H_{3} ⊕ H_{4} for some H_{4} . It follows from Proposition *** that H_{4} must be isomorphic to H_{1} . ✓

corollary 10.10. G is projective if and only if G is nearly isomorphic to a free module. proof: ( ⇐ ): If G is nearly isomorphic to a free module F then by Corollary 10.9 G is isomorphic to a summand of the free module F ⊕ F, hence is projective.

( ⇒ ): If G is projective then by Proposition 1.* it is isomorphic to a direct sum of ideals. Therefore by Proposition 10.* it suffices to proof that every non-trivial ideal is nearly isomorphic to W. In fact, by Proposition 0.* if 0 ≠ I ⊆ W then there exists J ⊆ W with J ≈ I such that I + J = W. Therefore I is nearly isomorphic to W by Proposition 10.*. ✓

One of the major historical landmarks in the theory of of finite rank torsion free W -modules was that direct sum decompositions have no nice uniqueness properties. In fact, not only can a small W -module have many different non-isomorphic direct sum decompositions, but it is possible that G and H be non-isomorphic and yet G ⊕ L ≈ H ⊕ L or G^{r} ≈ H^{r} for some r ≥ 2.

The original interest in quasi-isomorphism was the direct sum decompositions are in fact unique up to quasi-isomorphism. As we have seen in Chapter 3, this is a consequence of the fact that if G has no non-trivial quasi-direct decomposition then QEnd G is a local ring.

The concept of near isomorphism is not quite as nice in this respect. Even in the case when W is local there will generally exist modules having several decompositions which are not unique up to near isomorphism. example 10.11. Let A_{1}, …, A_{n} be rank-one modules with incomparable types and let B_{i} ≈ A_{i} for i = 1, …, n. Let p and q be distinct prime ideals such that no A_{i} (and thus no B_{i}) is either p-divisible or q-divisible. In each A_{i}, choose a_{i} ∉ pA_{i} and likewise choose b_{i} ∈ B_{i} with b_{i} ∉ qB_{i}. Let G be the submodule of A_{1} ⊕ ⋯ ⊕ A_{n} generated by p(A_{1} ⊕ ⋯ ⊕ A_{n}) together with a_{1} + ⋯ + a_{n}. Let H be the submodule of B_{1} ⊕ ⋯ ⊕ B_{n} generated by q(B_{1} ⊕ ⋯ ⊕ B_{n}) together with b_{1} + ⋯ + b_{n}. Then G and H are indecomposable. Furthermore, there exists an indecomposable module K with rank n such that G ⊕ H ≈ K ⊕ A_{1} ⊕ ⋯ ⊕ A_{n}. proof:

However what turns out to be true is that if G ⊕ L and H ⊕ L or G^{r} and H^{r} are nearly isomorphic, then G and H must be nearly isomorphic.

Thus in particular, finite rank torsion free W -modules have over a semi-local dedekind domain has the “cancellation” and “n^{th}-root” properies, since in this case near isomorphism is the same as isomorphism (Proposition ***).

For the next lemma, we anticipate some ideas from Chapter 11. Now let 𝔞 be a prime idea define Ē_{𝔞} (G) = (End G)/𝔞(End G) and H̄_{𝔞} (G, H) = Hom(G, H)/𝔞Hom(G, H). Then H̄_{𝔞} (G, ) is a functor from W -modules to right Ē_{𝔞} (G)-modules.

lemma 10.12. Suppose that H is isomorphic to a summand of G ⊕ G. Then G and H are nearly isomorphic if and only if for every non-trivial idea 𝔞, H̄_{𝔞} (G, G) and H̄_{𝔞} (G, H) are isomorphic Ē_{𝔞} (G)-modules.

proof: ( ⇐ ): For convenience, let Λ = Ē_{𝔞} (G) and let H(X) = H̄_{𝔞} (G, X). If φ: X → Y then φ induces a map φ̄: H(X) → H(Y). We claim first that if X and Y are summands of G ⊕ G then the map Hom(X, Y) → Hom_{Λ}(H(X), H(Y)) which takes φ to φ̄ is surjective. In fact, let F̄(X, Y) be the set of maps in Hom_{Λ}(H(X), H(Y)) induced by maps in Hom(X, Y). Using the additivity of the functors, it suffices to prove that F̄(X, Y) = Hom_{Λ}(H(X), H(Y)) for X = Y = G (Proposition 0.*), in other words to prove that every Λ-linear endomorphism of the right Λ-module H(G) is induced by a map in End G. But H(G) is the cyclic right Λ-module Λ/𝔞Λ, so every endomorphism λ̄ of H(G) is given by left multiplication by φ̄ = λ̄(1), and this is the same as saying that it is the map induced by any pre-image φ for φ̄.

Thus if H(X) and H(Y) are isomorphic as right Λ-modules then there exist maps φ: X → Y and ψ: Y → X such that φ̄: H(X) → H(Y) and ψ̄: H(Y) → H(X) are isomorphisms. Then φ and ψ in turn induce isomorphisms G/𝔞G ⟶^{≈} H/𝔞H and H/𝔞H ⟶^{≈} G/𝔞G. Thus by Proposition 10.7, G and H are nearly isomorphic.

( ⇒ ): **** ✓

proposition 10.13. (1) If G ⊕ L and H ⊕ L are nearly isomorphic, then G is nearly isomorphic to H.

(2) If G^{r} and H^{r} are nearly isomorphic for some r ≥ 1, then G is nearly isomorphic to H.

proof:

In Chapter 11 we will define the Jordan-Zassenhaus Property for a dedekind domain W and show that W has the Jordan-Zassenhaus Property if and only if no finite rank torsion free W -module has an infinite number of mutually non-isomorphic direct summands.

theorem 10.14. Suppose that W has the Jordan-Zassenhaus Property and let H and K be W -modules. The following conditions are equivalent:

(1) K is nearly isomorphic to H.

(2) There exist r, s ≥ 1 such that H^{r+s} ≈ K^{r} ⊕ H^{s}.

(3) There exist r ≥ 1 and a finite rank torsion free module L such that H^{r} ⊕ L ≈ K^{r} ⊕ L.

(4) There exists r ≥ 1 such that H^{r} ≈ K^{r}.

proof: (1) ⇒ (2): We prove first by induction that if H is nearly isomorphic to K then for all s ≥ 2 there exists a module H_{s} nearly isomorphic to H such that H^{s} ≈ K^{s−1} ⊕ H_{s}. In fact, by Corollary 10.9, H^{2} ≈ K ⊕ H_{2} for some H_{2} nearly isomorphic to H. Now assume inductively that H^{s−1} ≈ K^{s−2} ⊕ H_{s−1} with H_{s−1} nearly isomorphic to H. By Corollary 10.9, H_{s−1} ⊕ H ≈ K ⊕ H_{s} for some H_{s} nearly isomorphic to H, and thus

H^{s} = H^{s−1} ⊕ H ≈ K^{s−2} ⊕ H_{s−1} ⊕ H ≈ K^{s−2} ⊕ K ⊕ H_{s} ≈ K^{s−1} ⊕ H_{s}.

Now by Corollary 10.9, H_{s} is isomorphic to a summand of H^{2} . If W has the Jordan-Zassenhaus Property then H^{2} does not have an infinite number of non-isomorphic summands so it must be true that H_{s} ≈ H_{r+s} for some r, s ≥ 1, so that

H^{r+s} ≈ K^{r+s−1} ⊕ H_{r+s} ≈ K^{r+s−1} ⊕ H_{s} = K^{r} ⊕ K^{s−1} ⊕ H_{s} ≈ K^{r} ⊕ H^{s}.

(2) ⇒ (3): Choose L = H^{s} in (2).

(3) ⇒ (1): By Proposition 10.13, if H^{r} ⊕ L ≈ K^{r} ⊕ L then H^{r} and K^{r} are nearly isomorphic. But then Proposition 10.13 implies that H and K are nearly isomorphic.

(2) ⇒ (4): By Corollary 11.69 and Theorem 11.105 (c.f. also Corollary 11.107).

(4) ⇒ (1): By Proposition 10.13. ✓

Recall that a finite rank torsion free module is called almost completely decomposable if it is quasi-isomorphic to a completely decomposable module. For almost completely decomposable modules there is a much stronger version of Theorem 10.13. proposition 10.15. Let G be almost completely decomposable and let H be an finite rank torsion free module..

(1) G and H are nearly isomorphic if and only if G ⊕ L ≈ H ⊕ L for some completely decomposable module L.

(2) If a finite rank torsion free module H is nearly isomorphic to G then H is almost completely decomposable and i(G) = i(H) (where i(G) is the regulating index of G).

proof:

NEAR SUMMANDS. Recall from Proposition 3.16, a submodule H of G is a quasi-summand of G if and only if there exists γ ∈ End G and w ≠ 0 ∈ W such that γ^{2} = wγ and H is quasi-equal to γ(G).

definition 10.16. We say that a module H is a near summand of G if for every non-trivial ideal 𝔞 of W there exists a γ ∈ End G such that γ^{2} = wγ where w ≠ 0 ∈ W and (w) + 𝔞 = W and γ(G) ≈ H. (Here we do not assume that H is a submodule of G.)

proposition 10.17. (1) If H is a near summand of G then H is nearly isomorphic to a direct summand G′ of G.

(2) If G is nearly isomorphic to H_{1} ⊕ H_{2} then H_{1} and H_{2} are near summands of G and G = G_{1} ⊕ G_{2} , where G_{i} is nearly isomorphic to H_{i}.

(3) A module nearly isomorphic to an indecomposable module is itself indecomposable.

(4) If G is nearly isomorphic to H^{k} for k ≥ 2, then G ≈ H^{k−1} ⊕ L, where L is nearly isomorphic to H.

proof: (1) If H is a near summand of G then there exists γ_{i} ∈ End G and w_{i} ≠ 0 ∈ W such that γ_{i}^{2} = w_{i}γ_{i} and w_{1} + w_{2} = 1 and there exist isomorphisms θ_{i}: γ_{i}(G) ⟶^{≈} H.

(2) Let 𝔞 be a non-trivial ideal and letθ: G → H be a monomorphism with 𝔞 + ann(H/θ(G)) = W. Then there exists w ∈ ann(H/θ(G)) such that 𝔞 + (w) = W. Let η_{i} ∈ End H be an idempotent with H_{i} = η_{i}(H). Defineγ_{i} = wθ^{−1}η_{i}θ ∈ End G (since for g ∈ G, θ^{−1}(wη_{i}θ(g)) ∈ G). Then γ(G) ≈ η_{i}(H) = H_{i} and γ^{2} = wγ. This shows that H_{i} is a near summand of G. It follows from (1) that G = G_{1} ⊕ G_{2} with G_{1} nearly isomorphic to H_{1} . It then follows from Proposition *** that G_{2} is nearly isomorphic to H_{2} .

(3) In fact, by (2) if G is not indecomposable and is nearly isomorphic to H, then H must also be be decomposable.

(4) If G is nearly isomorphic to H^{k} then by (2) G = G_{1} ⊕ ⋯ ⊕ G_{k}, where G_{i} is nearly isomorphic to H for all i. By Corollary ***, H is isomorphic to a summand of G_{1} ⊕ G_{2} , i. e. G_{1} ⊕ G_{2} ≈ H ⊕ H_{2} where H_{2} is nearly isomorphic to G_{2} and hence to H. Inductively we get G ≈ H^{k−1} ⊕ L with L nearly isomorphic to H. ✓ THE PRIME SPECTRUM. At this point it is entertaining, and possibly even enlightening, to interpret various things in terms of the topology on the maximal ideal spectrum of W, simple-minded though this topology is. Namely, let X be the set of non-trivial prime ideals of W, and give X the coarsest possible T_{1} topology, i.e. the closed sets are the finite sets (and, of course, the whole space). If 𝔞 is an ideal of W, we define V (𝔞) to be the set of prime ideals containing 𝔞. Then V (𝔞) is in fact a closed set, and every every closed set in X has the form V (𝔞) for some ideal 𝔞. (One can take 𝔞 as the intersection of the finite number of prime ideals in the closed set.) We will write U_{𝔞} for the complement of V (𝔞). Thus every open set has the form U_{𝔞} for some ideal 𝔞.

Note that any subset of X containing a non-trivial open set is open and that if X is infinite then any two non-trivial open sets have a non-trivial intersection.

We now use the following terminology, adapted (with modification) from the theory of quasi-coherent sheaves over affine schemes. If φ: QG → QH then we say that φ is a homomorphism on U_{𝔞} if φ(𝔞^{−∞}G) ⊆ 𝔞^{−∞}H and we say that φ is an isomorphism on U_{𝔞} if the induced map 𝔞^{−∞}G → 𝔞^{−∞}H is an isomorphism. Likewise we say that φ is an isomorphism on the closed set V if φ is an isomorphism from G_{p} to H_{p} for every p ∈ V .

If U is an open subset of X, we say that G and H are isomorphic on U if there exists a quasi-isomorphism φ from G to H such that φ is an isomorphism from G_{p} to H_{p} for all p ∈ U. (????) warning: This is much stronger than merely saying that G_{p} and H_{p} are isomorphic for each p ∈ U.

For the most part, we will only be interested in the case where φ ∈ QHom(G, H).

proposition 10.18. (1) A quasi-homomorphism φ from G to H is a homomorphism on U_{𝔞} if and only if φ(G_{p}) ⊆ H_{p} for every prime p such that 𝔞 ⊈ p.

(2) If φ ∈ QHom(G, H) then φ is an isomorphism on U_{𝔞} if and only if rank G = rank H and φ(G_{p}) = φ(H_{p}) for every prime p with 𝔞 ⊈ p.

(3) If φ ∈ QHom(G, H) then φ is an isomorphism on U_{𝔞} if and only if φ is monic and H/φ(G) is torsion and p ∉ Ass H/φ(G) for all primes p with 𝔞 ⊈ p.

(4) If φ ∈ Hom(G, H) then φ is an isomorphism on U_{𝔞} if and only if φ is a quasi-isomorphism and the induced map G/𝔞G → H/𝔞H is an isomorphism.

proof: ****

In this language, Proposition 3.* can be rephrased as follows:

proposition 10.19. Let φ: QG → QH. Then φ is a quasi-homomorphism from G to H if and only if φ is a quasi-homomorphism at every p ∈ X and φ is a homomorphism on some non-empty open set U. proof: ( ⇒ ): Clearly if φ ∈ QHom(G, H) then φ ∈ QHom(G_{p}, H_{p}) for all prime ideals p. Furthermore there exists w ≠ 0 ∈ W such that wφ(G) ⊆ H. Then the set U of prime ideals with w ∉ p is an open set in X and φ is a homomorphism on U.

( ⇐ ): If φ is a homomorphism on U then φ(G_{p}) ⊆ H_{p} except for the finite set of primes p not in U. Furthermore, if φ is a quasi-homomorphism at p then (φ(G_{p}) + H_{p})/H_{p} is not faithful. Thus we can find w ≠ 0 ∈ W such that w(φ(G_{p}) + H_{p})/H_{p} = 0 for all p ∉ U. Then wφ(G_{p}) ⊆ H_{p} for all primes p so by Proposition 0.*, wφ(G) ⊆ H and thus φ ∈ QHom(G, H). ✓

lemma 10.20. If G and H are quasi-isomorphic and φ ∈ QHom(G, H) then the set of primes p such that φ is an isomorphism at p is an open subset of X. proof: If φ is not monic then the set of p where φ is an isomorphism is the empty set. Now suppose that φ is monic. By Proposition 10.* the set of primes p such that φ is a homomorphism at p is an open set U_{1} . Now φ is an isomorphism at p if and only if p ∈ U_{1} and φ(G_{p}) = H_{p} or, equivalently, (H + φ(G))_{p}/H_{p} = 0. But since by Proposition 3.*, φ is a quasi-isomorphism, Ass((H + φ(G))/H) is finite and (H + φ(G))_{p}/H_{p} = 0 except at this finite set of primes. Thus φ is an isomorphism on the open set {p ∈ U_{1} | p ∉ Ass((H + φ(G))/H)}. ✓

proposition 10.21. (1) If G is quotient divisible and φ: QG → QH, then φ is a quasi-homomorphism if and only if φ is a quasi-homomorphism at each point p ∈ X.

(2) If G is locally free and φ: QG → QH, the φ is a quasi-homomorphism if and only if φ is a homomorphism on some non-empty open set.

proposition 10.22. The following conditions are equivalent:

(1) G and H are nearly isomorphic.

(2) For every prime p there exists a quasi-isomorphism φ: G → H such that φ is an isomorphism at p.

(3) Every finite subset Y of X has an open neighborhood U such that there is a quasi-isomorphism φ: G → H which is an isomorphism on U.

proof: (1) ⇒ (2): Let p be a prime. If G and H are nearly isomorphic then there exists a monomorphism φ: G → H such that H/φ(G) has finite length and p ∉ Ass H/φ(G). Then H_{p}/φ(G_{p}) is a finite length W_{p}-module and p ∉ Ass H_{p}/φ(G_{p}), so Ass H_{p}/φ(G_{p}) = ∅. Thus by Proposition 0.*, H_{p}/φ(G_{p}) = 0, so φ is an isomorphism from G_{p} to H_{p}.

(2) ⇒ (3): Let p_{1}, …, p_{t} be a finite set of distinct primes. It suffices to prove that there exists a quasi-isomorphism φ from G to H such that φ is an isomorphism at each p_{i}, since by Lemma 10.20 φ will then be an isomorphism on an open set containing the p_{i}. By induction we may assume that there exists a quasi-isomorphism φ_{1} which is an isomorphism at p_{1}, …, p_{t−1} . By (2) there exists a quasi-isomorphism φ_{2} : G → H which is an isomorphism at p_{t}. Since p_{t} + ⋂_{1}^{t−1} p_{i} = W there exists v ∈ p_{t} and w ∈ ⋂_{1}^{t−1} p_{i} such that v + w = 1. Now let φ = vφ_{1} + wφ_{2} .

(3) ⇒ (1):