2 1. THE KRULL-REMAK-SCHMIDT THEOREM and a cokernel, and in which every monomorphism (respectively epi- morphism) is a kernel (respectively cokernel). Over any ring R the category R-Mod of all left R-modules is abelian if R is left Noether- ian, then the category R-mod of finitely generated left R-modules is abelian. It is easy to see that idempotents split in an abelian category. Indeed, suppose e: M −→ M is an idempotent, and let u: K −→ M be the kernel of 1M − e. Since (1M − e)e = 0, the map e factors through u that is, there is a map p: M −→ K satisfying up = e. Then upu = eu = eu + (1M − e)u = u = u1K. Since u is a monomorphism (as kernels are always monomorphisms), it follows that pu = 1K. A non-zero object M in the additive category A is said to be decomposable if there exist non-zero objects M1 and M2 such that M ∼ M1 ⊕ M2. Otherwise, M is indecomposable. We leave the proof of the next result as an exercise: 1.1. Proposition. Let M be a non-zero object in an additive cat- egory A, and let E = EndA(M). (i) If 0 and 1 are the only idempotents of E, then M is indecom- posable. (ii) Conversely, suppose e = e2 ∈ E, with e = 0, 1. If both e and 1 − e split, then M is decomposable. We say that the Krull-Remak-Schmidt Theorem (KRS for short) holds in the additive category A provided (i) every object in A is a biproduct of indecomposable objects, and (ii) if M1 ⊕ · · · ⊕ Mm ∼ N1 ⊕ · · · ⊕ Nn, with each Mi and each Nj an indecomposable object in A, then m = n and, after renumbering, Mi ∼ Ni for each i. It is easy to see that every Noetherian object is a biproduct of finitely many indecomposable objects (cf. Exercise 1.19), but there are easy examples to show that (ii) can fail. For perhaps the simplest example, let R = k[x, y], the polynomial ring in two variables over a field. Letting m = Rx + Ry and n = R(x − 1) + Ry, we get a short exact sequence 0 −→ m ∩ n −→ m ⊕ n −→ R −→ 0 , since m+n = R. This splits, so m⊕n ∼ R⊕(m∩n). Since neither m nor n is isomorphic to R as an R-module, KRS fails for finitely generated R-modules. Alternatively, let D be a Dedekind domain with a non-principal ideal I. We have an isomorphism (see Exercise 1.20) (1.1) R ⊕ R ∼ I ⊕ I−1 ,

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