§1. KRS IN AN ADDITIVE CATEGORY 3 and of course all of the summands in (1.1) are indecomposable. These examples indicate that KRS is likely to fail for modules over rings that aren’t local. It can fail even for finitely generated modules over local rings. An example due to Swan is in Evans’s paper [Eva73]. In Chapter 2 we will see just how badly it can fail. Azumaya [Azu48] observed that the crucial property for guaranteeing KRS is that the endomorphism rings of the summands be local in the non-commutative sense. To avoid a conflict of jargon, we define a ring Λ (not necessarily commutative) to be nc-local provided Λ J (Λ) is a division ring, where J (−) denotes the Jacobson radical. Equivalently (cf. Exercise 1.21) Λ = {0} and J (Λ) is exactly the set of non-units of Λ. It is clear from Proposition 1.1 that any object with nc-local endomorphism ring must be indecomposable. We’ll model our proof of KRS after the proof of unique factorization in the integers, by showing that an object with nc-local endomorphism ring behaves like a prime element in an integral domain. We’ll even use the same notation, writing “M | N”, for objects M and N, to indicate that there is an object Z such that N M Z. Our inductive proof depends on direct-sum cancellation ((ii) below), analogous to the fact that mz = my =⇒ z = y for non-zero elements m, z, y in an integral domain. Later in the chapter (Corollary 1.16) we’ll prove cancellation for arbitrary finitely generated modules over a local ring, but for now we’ll prove only that objects with nc-local endomorphism rings can be cancelled. 1.2. Lemma. Let A be an additive category in which idempotents split. Let M, X, Y , and Z be objects of A, let E = EndA(M), and assume that E is nc-local. (i) If M | X Y , then M | X or M | Y (“primelike”). (ii) If M Z M Y , then Z Y (“cancellation”). Proof. We’ll prove (i) and (ii) simultaneously. In (i) we have an object Z such that M Z X Y . In the proof of (ii) we set X = M and again get an isomorphism M Z X Y . Put J = J (E), the set of non-units of E. Choose reciprocal isomorphisms ϕ: M Z −→ X Y and ψ : X Y −→ M Z. Write ϕ = α β γ δ and ψ = μ ν σ τ , where α: M −→ X, β : Z −→ X, γ : M −→ Y , δ : Z −→ Y , μ: X −→ M, ν : Y −→ M, σ : X −→ Z and τ : Y −→ Z. Since ψϕ = 1M⊕Z = 1M 0 0 1 Z , we have μα + νγ = 1M. Therefore, as E is local, either μα
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