Bull. Soc. Math. Bel. (series A) 45 (1993), 29 - 37.

Cancellation over Rings of Dimension ≦ 1

ANTHONY BAK Department of Mathematics University of Bielefeld W-4800 Bielefeld, Germany

MASAHARU MORIMOTO Department of Mathematics Okayama University Okayama 700, Japan

Abstract Let A be a module finite R-algebra such that the Bass-Serre dimension (R) ≦ 1. Let M, M′ and P be A-modules. Then M ⊕A⊕P ≅ M′⊕A⊕P implies M ⊕ A ≅ M′ ⊕ A, providing the following holds: (1) P is finitely generated and projective. (2) M is finitely presented. (3) There is a 2-sided ideal I in A such that the general linear group GL_{2}(A) acts transitively on the (A/I)-unimodular vectors in A/I ⊕ A/I and for almost all maximal ideals 𝔪 of R there is locally an A_{𝔪}-homomorphism f^{𝔪} : M_{𝔪} ⟶ A_{𝔪} such that modulo the Jacobson radical(A_{𝔪}), image (f^{𝔪}) ⊇ I_{𝔪}.

1. Introduction The purpose of this note is to extend a recent cancellation result of Hambleton and Kreck [H-K, Theorem A] for modules over a separable order to modules over a module finite R-algebra where dimension (R) ≤ 1. We define the dimension dim(R) of a commutative ring R to be 0 (resp. 1) if it is semilocal (resp. there is a finite set 𝔐 of maximal ideals of R such that for each element s ∈ R\ ∪_{𝔪∈𝔐}𝔪, the quotient ring R/Rs is semilocal.) This notion of dimension is weaker than that of Bass-Serre dimension which was used by H. Bass in his fundamental work on cancellation, cf. [B, IV].

Our main result is the following. □ THEOREM 1.1 Let A be a module finite R-algebra such that dim(R) ≦ 1. Let M, M′, and P be A-modules. Suppose the following conditions hold.

(1.1.1) There is a 2-sided ideal I in A such that the general linear group GL_{2}(A) acts transitively on the (A/I)-unimodular vectors in A/I ⊕ A/I. (1.1.2) M is finitely presented (this is automatic if A is Noetherian and M is finitely generated) and for all but a finite number of maximal ideals 𝔪 of R, there is locally an A_{𝔪}-homomorphism f^{𝔪} : M_{𝔪} ⟶ A_{𝔪} such that modulo the Jacobson radical(A_{𝔪}) the image (f^{𝔪}) ⊇ I_{𝔪}. If P is finitely generated and projective then P ⊕ A ⊕ M ≅ P ⊕ A ⊕ M′ implies A ⊕ M ≅ A ⊕ M′. □

It is very likely that there are appropriate generalizations of (1.1) to module finite R-algebras A where the only condition imposed on A is that R is finite dimensional. As Hambleton and Kreck [H-K] have shown, one can expect such results to find applications in the classification of 2-dimensional C.W. complexes. □ COROLLARY 1.2 Let A be a module finite R-algebra such that dim(R) ≦ 1. Let M, M′ and P be A-modules. Suppose the following conditions hold. (1.2.1) There is a 2-sided ideal I in A such that A/I is commutative and each element of the special linear group SL_{2}(A/I) lifts to GL_{2}(A); e.g., SL_{2}(A/I) is equal to the elementary group E_{2}(A/I). (1.2.2) Condition (1.1.2) above. If P is finitely generated and projective then P ⊕ A ⊕ M ≅ P ⊕ A ⊕ M′ implies A ⊕ M ≅ A ⊕ M′. □ PROOF The conclusion above will follow from (1.1), once we show that condition (1.1.1) is satisfied. It suffices to show that SL_{2}(A/I) acts transitively on unimodular vectors of A/I ⊕ A/I. Let a, c ∈ A/I such that (a, c) ∈ A/I ⊕ A/I is unimodular. Choose elements b, d ∈ A/I such that

ad + bc = 1. The matrix (\begin{matrix}a & b \\ c & d \end{matrix}) has determinant 1, i.e. ∈ SL_{2}(A/I),

and (\begin{matrix}a & b \\ c & d \end{matrix}) (\begin{matrix}1 \\ 0 \end{matrix}) = (\begin{matrix}a \\ c \end{matrix}). It follows that SL_{2}(A/I) acts transitively on the

unimodular vectors in A/I ⊕ A/I. Q.E.D.

□ COROLLARY 1.3 Let R be a Dedekind ring with field of fractions F. Let A be an R-order on a finite separable semisimple F -algebra. Let M, M′, and N be finitely generated A-modules. Let I be a 2-sided ideal of A such that conditions (1.1.1) and (1.1.2) are satisfied. Suppose that N is Rtorsion free and that there is a natural number r such that for each maximal ideal 𝔪 of R, N_{𝔪} is a direct summand of (A_{𝔪} ⊕ M_{𝔪})^{r}. Then N ⊕ A ⊕ M ≅ N ⊕ A ⊕ M′ implies A ⊕ M ≅ A ⊕ M′. □

PROOF Clearly, N ⊕ A ⊕ A ⊕ M ≅ N ⊕ A ⊕ A ⊕ M′. By Swan’s cancellation theorem [S, (9.4) and (9.7)], A ⊕ A ⊕ M ≅ A ⊕ A ⊕ M′. Since dim(R) ≦ 1, it follows now from Theorem (1.1) that A ⊕ M ≅ A ⊕ M′. Q.E.D. □ COROLLARY 1.4 (Hambleton - Kreck [H-K, Theorem A]) Let A be a separable R-order as in (1.3). Let M, M′ and N be finitely generated A-modules where N is as in (1.3). Suppose there is a 2-sided ideal I in A such that the ring A/I is also a separable R-order and the following conditions hold. (1.4.1) GL_{2}(A) acts transitively on the (A/I)-unimodular vectors of A/I ⊕ A/I. (1.4.2) There is a natural number k such that for all but a finite number of maximal ideals 𝔪 of R, ((A/I)^{k} ⊕ M)_{𝔪} has a direct summand isomorphic to A_{𝔪}. Then N ⊕ A ⊕ M ≅ N ⊕ A ⊕ M′ implies A ⊕ M ≅ A ⊕ M′. □

PROOF The conclusion of Hambleton - Kreck will follow from (1.3), once we show that condition (1.1.2) is satified.

Let B = A/I. For almost all maximal ideals 𝔪 of R, there are by hypothesis A_{𝔪}-homomorphisms f : A_{𝔪} ⟶ B_{𝔪}^{k} ⊕ M_{𝔪} and g : B_{𝔪}^{k} ⊕ M_{𝔪} ⟶ A_{𝔪} such that gf = 1_{A_{𝔪}}. Write f = (f_{1}, f_{2}) where f_{1} : A_{𝔪} ⟶ B_{𝔪}^{k} and f_{2} : A_{𝔪} ⟶ M_{𝔪} and write g = (g_{1}, g_{2}) where g_{1} : B_{𝔪}^{k} ⟶ A_{𝔪} and g_{2} : M_{𝔪} ⟶ A_{𝔪}. For a ∈ I_{𝔪}, f_{1}(a) = f_{1}(1)a = 0; thus, a = gf(a) = g_{2}f_{2}(a). Thus, g_{2}f_{2}|_{I_{𝔪}} = 1_{I_{𝔪}}. Thus, M_{𝔪} contains a direct summand isomorphic to I_{𝔪}. Q.E.D.

In the next section, we recall a few definitions and then prove Theorem (1.1). Our methods are elementary and require little beyond a familiarity with semilocal rings, Nakayama’s lemma, and localization.

2. Proof of Theorem 1.1 We begin by recalling a few definitions.

Let A be an associative ring with identity and 𝔮 a 2-sided ideal in A. Let M = M_{1}⊕⋯⊕M_{n} be a direct sum of right A-modules. If f : M_{i} ⟶ M_{j}(i ≠ j) is an A-homomorphism and \overline{f} its unique extension to an A-endomorphism of M such that \overline{f}(M_{k}) = 0 for all k ≠ i, we set ϵ(f) = 1_{M} +\overline{f}. Clearly ϵ(f) is an A-automorphism of M with inverse ϵ(−f). ϵ(f)is called the elementary transformation defined by f. If image (f) ⊆ M_{j}𝔮 then ϵ(f) is called a 𝔮-elementary transformation. Let E(M_{1} …, M_{n}) denote the subgroup of Aut_{A}(M) generated by all elementary transformations ϵ(f) where f ranges over all A-homomorphisms f : M_{i} ⟶ M_{j} such that i ≠ j, 1 ≤ i ≤ n, 1 ≤ j ≤ n. Let E(M_{1}, …, M_{n}; 𝔮) denote the normal subgroup of E(M_{1}, …, M_{n}) generated by the 𝔮-elementary transformations. If M_{1} = ⋯ = M_{n} = A then by definition E_{n}(A) = E_{n}(M_{1}, …, M_{n}) and E_{n}(A, 𝔮) = E(M_{1}, …, M_{n}; 𝔮).

Let M be a right A-module. An element m ∈ M is called unimodular if there is an A-homomorphism f : M ⟶ A such that f(m) = 1. It follows that m = (m_{1}, …, m_{n}) ∈ M_{1} ⊕ ⋯ ⊕ M_{n} is unimodular ⇔ there

are A-homomorphisms f_{i} : M_{i} ⟶ A(i = 1, …, n) such that ∑_{i=1}^{n} f_{i}(m) =

1 ⇔ there are A-homomorphisms f_{i} : M_{i} ⟶ A(i = 1, …, n) such that (f_{1}(m_{1}), …, f_{n}(m_{n})) ∈ A^{n} = A ⊕ ⋯ ⊕ A (n times) is unimodular. A vector (a_{1}, …, a_{n}) ∈ A^{n} is unimodular ⇔ there are elements b_{1}, …, b_{n} ∈ A such

that ∑_{i=1}^{n} b_{i}a_{i} = 1.

If M is a right A-module and m ∈ M, one defines o_{M}(m) = {f(m)|f ∈

Hom_{A}(M, A)}. Clearly, o_{M}(m) is a left ideal in A and m is unimodular ⇔ o_{M}(m) = A.

□ LEMMA 2.1 Let A be an associative ring with identity. Let (a_{1}, …, a_{n}) ∈ A^{n} be unimodular. Then there is an element b ∈ A such that (a_{1}, …, a_{n−1}, (ba_{n})^{2}) is unimodular. □ PROOF By definition, there are elements c_{1}, …, c_{n} ∈ A such that 1 = c_{1}a_{1} + ⋯ + c_{n}a_{n}. Thus, a_{n} = a_{n}(c_{1}a_{1} + ⋯ + c_{n}a_{n}). Thus, 1 = c_{1}a_{1} ⋯ + c_{n}a_{n} = c_{1}a_{1}+⋯+c_{n−1}a_{n−1}+c_{n}[a_{n}(c_{1}a_{1}+⋯+c_{n}a_{n})] = (c_{n}a_{n}c_{1}+c_{1})a_{1}+⋯+ (c_{n}a_{n}c_{n−1} + c_{n−1})a_{n−1} + (c_{n}a_{n})^{2}. Thus, (a_{1}, …, a_{n−1}, (c_{n}a_{n})^{2}) is unimodular. Q.E.D.

□ LEMMA 2.2 Let A be a semilocal ring and let 𝔮 be a 2-sided ideal in A. Let 𝔞 be a left ideal in A. Let M be a right A-module. Let (a, m) ∈ A ⊕ M such that m ∈ M𝔮 and 𝔞 + o_{A⊕M}(a, m) = A. Then there is an A-homomorphism f : M ⟶ A𝔮 such that 𝔞 + o_{A}(a + f(m)) = A. □

PROOF By definition, there is an A-homomorphism g : M ⟶ A such that 𝔞 + o_{A⊕A}(a, g(m)) = A. Since m ∈ M𝔮, g(m) ∈ 𝔮. It follows from (2.1) that 𝔞 + o_{A⊕A}(a, (bg(m))^{2}) = A for some b ∈ A. By [B, III (2.8)], there is an element c ∈ A such that 𝔞 + o_{A}(a + c(bg(m))^{2}) = A. Define f to be

the compositon of the A-homomorphisms M ⟶^{g} A ⟶^{cbg(m)b} A where cbg(m)b denotes left multiplication by cbg(m)b. Clearly, f has the desired properties. Q.E.D.

□ LEMMA 2.3 Let A be a module finite R-algebra. Let 𝔮 be a 2-sided ideal in A. Let 𝔐 be a finite set of maximal ideals in R. Let M be a right Amodule. If (a, m) ∈ A ⊕ M is a unimodular element such that m ∈ M𝔮 then there is an A-homomorphism f : M ⟶ A𝔮 such that A(a + f(m)) ⫆ As for

some s ∈ R \ ∪_{𝔪∈𝔐}𝔪. □

PROOF Let 𝔭 = ∩_{𝔪∈𝔐}𝔪. R/𝔭 is a semilocal ring (with maximal ideals

{𝔪/𝔭|𝔪 ∈ 𝔐}). Since A is module finite over R, A/A𝔭 is module finite over R/R𝔭 and hence semilocal. By hypothesis, there is an A-homomorphism g : M ⟶ A such that (a, g(m)) is unimodular. Since m ∈ M𝔮, g(m) ∈ 𝔮. By (2.1), (a, (bg(m))^{2}) is unimodular for some b ∈ A. By [B, III (2.8)], there is an element c ∈ A such that a+c(bg(m))^{2} is a unit mod A𝔭. Let f denote the

composition of M ⟶^{g} A ⟶^{cbg(m)b} A where cbg(m)b denotes left multiplication by cbg(m)b. Then a+f(m) is a unit in A/A𝔭. Let S denote the multiplicative set R \ ∪_{𝔪∈𝔐}𝔪. Since the ideal S^{−1}A𝔭 in S^{−1}A is contained in the Jacobson

radical(S^{−1}A) and S^{−1}A/S^{−1}A𝔭 = A/𝔭, it follows from Nakayama’s lemma [B, III (2.2)] that a+f(m) is a unit in S^{−1}A. Thus, there are elements d ∈ A and s ∈ S such that s^{−1}d(a + f(m)) = 1 in S^{−1}A. Thus, there is an element t ∈ S such that the equality td(a + f(m)) = ts holds in A. Q.E.D.

□ PROPOSITION 2.4 Let A be a module finite R-algebra such that dim(R) ≦ 1. Let M be a finitely presented right A-module. Let I be a 2-sided ideal in A with the following properties. (2.4.1) There is a subgroup G of the general linear group GL_{2}(A), which acts transitively on the (A/I)-unimodular vectors in A/I ⊕ A/I. (2.4.2) there is a finite set 𝔐 of maximal ideals of R such that for each maximal ideal 𝔪 ∉ 𝔐, there is locally an A_{𝔪}-homomorphism f^{𝔪} : M_{𝔪} ⟶ A_{𝔪} such that modulo the Jacobson radical(A_{𝔪}), the image (f^{𝔪}) ⊇ I_{𝔪}. Let 𝔮 be a 2-sided ideal in A and let G(𝔮) be a subgroup of the 𝔮-relative general linear group GL_{2}(A, 𝔮), which acts transitively on the (A/I)-unimodular vectors u of A/I ⊕ A/I such that u ≡ (1, 0) mod (𝔮 + I)/I. (If 𝔮 = A, one can take G(𝔮) = G.) If v, w ∈ A ⊕ A ⊕ M are unimodular elements such that v ≡ w mod 𝔮, i.e v − w ∈ (A ⊕ A ⊕ M)𝔮, then there is an automorphism σ in the normal closure of < E(A, A, M; 𝔮), G(𝔮) > by < E(A, A, M), G > such that σv = w. □

PROOF The proof will be divided into two steps. Step 1: There is an element ρ ∈< E(A, A, M), G > such that ρw = (1, 0, 0) ∈ A ⊕ A ⊕ M. Step 2: If w = (1, 0, 0) then there is an element τ ∈< E(A, A, M; 𝔮), G(𝔮) > such that τv = (1, 0, 0). Assume Steps 1 and 2 have been established. The proof is then completed as follows. By Step 1, there is a ρ such that ρw = (1, 0, 0). Clearly, ρv ≡ ρw mod 𝔮. Thus, according to Step 2, there is a τ ∈< E(A, A, M; 𝔮), G(𝔮) > such that τρv = ρw. Clearly, (ρ^{−1}τρ)v = w and ρ^{−1}τρ is in the normal closure of < E(A, A, M; 𝔮), G(𝔮) > under < E(A, A, M), G >.

Step 1 is the special case of Step 2 where 𝔮 = A. Thus, it suffices to prove Step 2.

Let v = (1 + a, b, m) be a unimodular element in A ⊕ A ⊕ M such that a, b ∈ 𝔮 and m ∈ M𝔮. Enlarge 𝔐 to a finite set, denoted again by 𝔐, such that if m is a maximal ideal ∉ 𝔐 and s ∈ R \ 𝔪 then A/As is semilocal. This can be done, since dim(R) ≦ 1. By Lemma (2.3), there is an A-homomorphism f : A ⊕ M ⟶ A𝔮 such that A(a + f(b, m)) ⫆ As for some s ∈ R \ ∪_{𝔪∈𝔐}𝔪. Clearly, ϵ(f)v = (1 + a + f(b, m), bm). Thus, we can

assume right from the start that A(1 + a) ⫆ As for some s ∈ R \ ∪_{𝔪∈𝔐}𝔪.

Since (1 + a, b, m) is unimodular, there is an A-homomorphism f : M ⟶ A such that (1 + a, b, f(m)) ∈ A ⊕ A ⊕ A is unimodular. Applying Lemma (2.2) to the vector (1 + a, b, f(m)) over the semilocal ring A/As, we can find an A-homomorphism g : A ⟶ A𝔮 such that (1 + a, b + gf(m)) is unimodular over A/As. But, since A(1 + a) ⫆ As, it follows that (1 + a, b + gf(m)) is unimodular over A. Clearly, ϵ(gf)v = (1 + a, b + gf(m), m). Thus, we can assume right from the start that v = (1 + a, b, m) where (1 + a, b) is unimodular. By hypothesis, there is an element τ ∈ G(𝔮) such that τ ⊕ 1_{M}(v) = (1 + a′, b′, m) where a′, b′ ∈ I ∩ 𝔮. Thus, we can assume v = (1 + a, b, m) where a, b ∈ I ∩𝔮 and (1+a, b) is unimodular. Moreover, by applying if necessary an elementary transformation ϵ(f) to v, where f : A ⟶ A(I ∩𝔮) has the property that A(1 + a + f(b)) ⊇ As for some s ∈ R \ ∪_{𝔪∈𝔐}𝔪, we can

assume that A(1 + a) ⊇ As.

Let V (Rs) = {𝔪|𝔪 a maximal ideal of R, 𝔪 ⊇ Rs}. Evidently, V (Rs)∩ 𝔐 = ∅. Thus, R/Rs is semilocal and V (Rs) is finite. Let 𝔪 ∈ V (Rs). Let f^{𝔪} : M_{𝔪} ⟶ A_{𝔪} be as in the hypothesis of the proposition. Since M is finitely presented, we can apply [B, III (4.5)] to find an A-homomorphism f^{[𝔪]} : M ⟶ A and an element s^{[𝔪]} ∈ R \ 𝔪 such that (s^{[𝔪]})^{−1}f^{[𝔪]} = f^{𝔪}. Let t^{[𝔪]} ∈ R\𝔪 such that t^{[𝔪]} ≡ (s^{[𝔪]})^{−1} mod (R_{𝔪}𝔪). Let g^{[𝔪]} = t^{[𝔪]}f^{[𝔪]}. Let J^{[𝔪]} denote the inverse image in A of the Jacobson radical (A_{𝔪}/A_{𝔪}s). Each g^{[𝔪]} has the property that mod J^{[𝔪]}, image (g^{[𝔪]}) ⊇ I. Let x^{[𝔪]} ∈ M such that mod J^{[𝔪]}, g^{[𝔪]}x^{[𝔪]} = b − g^{[𝔪]}(m). Let r^{[𝔪]} ∈ R such that r^{[𝔪]} ≡ 1 mod 𝔪

and r^{[𝔪]} ≡ 0 mod 𝔪′ for each 𝔪′ ≠ 𝔪 ∈ V (Rs). Let x = ∑_{𝔪∈V (Rs)}x^{[𝔪]}r^{[𝔪]}.

Since (1 + a, b) is unimodular, we can find an A-homomorphism h : A ⊕ A ⟶ M such that h(1 + a, b) = x. Clearly, ϵ(h)(1 + a, b, m) = (1 + a, b, m + h(1 + a, b)) = (1 + a, b, m + x). Since (1 + a, b) is unimodular and g^{[𝔪]}(m+x) ≡ g^{[𝔪]}(m)+g^{[𝔪]}x^{[𝔪]} ≡ g^{[𝔪]}(m)+(b−g^{[𝔪]}(m) = b mod J^{[𝔪]}, we see that (1 + a, m + x) ∈ A ⊕ M is unimodular mod J^{[𝔪]}. Thus, (1 + a, m + x) is unimodular over A_{𝔪}/A_{𝔪}s = (A/As)_{𝔪} for each 𝔪 ∈ V (Rs). Thus, by Nakayama, (1 + a, m + x) is unimodular over A/As. (More specifically, one can argue as follows. Choose c, d ∈ A such that c(1 + a) + db = 1 and let g = ∑_{𝔪∈V (Rs)}x^{[𝔪]}g^{[𝔪]}. It suffices to show that A(ca + dg(m + x)) ≡ A

mod As. By the local-global principle, it suffices to show that for all maximal ideals of R/Rs, equivalently for all 𝔪 ∈ V (Rs), A_{𝔪}(ca + dg(m + x)) ≡ A_{𝔪} mod (A_{𝔪}s). By Nakayama’s lemma [B, III (2.2)], it suffices to show that A_{𝔪}(ca + dg(m + x)) ≡ A_{𝔪} mod A_{𝔪}J^{[𝔪]}. But ca + dg(m + x)) ≡ ca + db = 1 mod A_{𝔪} mod J^{[𝔪]}.) Since A(1 + a) ⊇ As, it follows that (1 + a, m + x) is unimodular over A. Choose h′ : A ⊕ M ⟶ A such that h′(1 + a, m + x) = 1 − b. Clearly, ϵ(h′)ϵ(h)(1 + a, b, m) = (1 + a, 1, m + x). Choose h′′ : A ⟶ A such that h′′(1) = −a. If τ = ϵ(h)^{−1}ϵ(h′)^{−1}ϵ(h′′)ϵ(h′)ϵ(h) then τ(1 + a, b, m) = (1, b′, m′) for suitable b′ and m′. Furthermore, since image (h′′) ⊆ I ∩𝔮, τ ∈ E(A, A, , M; 𝔮). Thus, b′ ≡ 0 mod 𝔮 and m′ ≡ 0 mod M𝔮. Letting h_{1} : A ⟶ A such that h_{1}(1) = −b′ and h_{2} : A ⟶ M such that h_{2}(1) = −m′, we obtain that ϵ(h_{2})ϵ(h_{1})τ(1 + a, b, m) = (1, 0, 0). Q.E.D.

□ THEOREM 2.5 Let A be a module finite R-algebra such that dim(R) ≦ 1. Let M be a finitely presented right A-module. Let I be a 2-sided ideal in A satisfying (2.4.1) and (2.4.2). If M′ and P are right A-modules and P is finitely generated and projective then P ⊕ A ⊕ M ≅ P ⊕ A ⊕ M′ implies A ⊕ M ≅ A ⊕ M′. □

PROOF The proof follows the pattern of that in Bass [B, IV (3.5)]. Choose Q such that P ⊕ Q ≅ A^{n} for some n. If n = 0 then P = 0 and we are done. Thus, we can assume n > 0. It suffices now to show that A^{n+1} ⊕ M ≅ A^{n+1} ⊕ M′ implies A^{n} ⊕ M ≅ A^{n} ⊕ M′ for any n > 0. Let v = (1, 0, …, 0) ∈ A^{n+1} ⊕ M, w = (1, 0, …, 0) ∈ A^{n+1} ⊕ M′, and identify A^{n+1} ⊕ M with A^{n+1} ⊕ M′. By Proposition (2.4), there is a transformation σ ∈< E(A, …, A, M), G > such that σv = w. σ induces an isomorphism

A ⊕ A^{n} ⊕ M/vA ⟶^{≅} A ⊕ A^{n} ⊕ M′/wA. But A^{n} ⊕ M ≅ A ⊕ A^{n} ⊕ M/vA and

A^{n} ⊕ M′ ≅ A ⊕ A^{n} ⊕ M′/wA. Q.E.D.

□ THEOREM 2.6 Let A be a module finite R-algebra such that dim(R) ≤ 1 and R is Noetherian. Let M, M′, and N be finitely generated right A-modules (and therefore finitely presented, because A is Noetherian). Let B denote the A-endomorphism ring End_{A}(N) of N and suppose that the canonical A-homomorphisms Hom _{A}(N, M) ⊗_{B} N ⟶ M, f ⊗ n ↦ f(n), and Hom _{A}(N, M′) ⊗_{B} N ⟶ M′, f ⊗ n ↦ f(n) are isomorphisms; e.g., M and M′ are direct summands of a direct sum of N’s. Let ℑ be a 2-sided ideal in B such that ℑ and the right B-module Hom_{A}(N, M) satisfy conditions (2.4.1) and (2.4.2). Let Q be a right A-module which is a direct summand of a direct sum of finitely many copies of N. Then Q ⊕ N ⊕ M ≅ Q ⊕ N ⊕ M′ implies N ⊕ M ≅ N ⊕ M′. □

PROOF Since N finitely generated over A and A is module finite over R with R Noetherian, it follows that B is module finite over R. Consider the functor ((right A-modules)) ⟶ ((right B-modules)), X ↦ Hom _{A}(N, X). Applying the functor to the isomorphism Q⊕N ⊕M ≅ Q⊕N ⊕M′, we obtain an isomorphism Hom _{A}(N, Q) ⊕ B ⊕ Hom _{A}(N, M) ≅ Hom _{A}(N, Q) ⊕ B ⊕ Hom _{A}(N, M′). Since Q is a direct summand of a direct sum of finitely many copies of N, it follows that Hom _{A}(N, Q) is finitely generated and projective over B. Hom _{A}(N, M) is finitely presented over B, since it is finitely generated already over R and B is Noetherian. Thus, we can apply Theorem (2.5). By the conclusion of that theorem, B⊕ Hom _{A}(N, M) ≅ B⊕ Hom _{A}(N, M′). Applying the functor −⊗_{B} N to the isomorphism above, we obtain an isomorphism N ⊕ Hom _{A}(N, M)⊗_{B}N ≅ N ⊕ Hom _{A}(N, M′)⊗_{B}N. But by hypothesis, Hom _{A}(N, M) ⊗_{B} N ≅ M and Hom _{A}(N, M′) ⊗_{B} N ≅ M′. Q.E.D.

REMARK 2.7 One can replace in (2.4) and (2.5) (resp. (2.6)) the hypothesis that M is finitely presented by the weaker hypothesis that M contains a direct summand M_{0} such that M_{0} is finitely presented and the ideal I (resp. ℑ) in A (resp. B) satisfies (2.4.2) with respect to the submodule M_{0} (rersp. Hom _{A}(N, M_{0})). The details are a little tedious, but not difficult. We shall skip them.

References

B. H. Bass, Algebraic K-Theory, W.A. Benjamin (1968), New York

H-K. I. Hambleton and M. Kreck, Cancellation of Lattices and Finite Two-Colexes, preprint

S. R. Swan, K-Theory of Finite Groups and Orders, Lecture Notes in Math. 149 (1970), Springer-Verlag, Berlin