4 1. THE KRULL-REMAK-SCHMIDT THEOREM or νγ must be outside J and hence an automorphism of M. Assuming that μα is an automorphism, we will produce an object W and maps M u −→ X p − → M W v − → X q −→ W satisfying pu = 1M, pv = 0, qv = 1W , qu = 0, and up + vq = 1X. This will show that X = M ⊕ W . (Similarly, the assumption that νγ is an isomorphism forces M to be a direct summand of Y .) Letting u = α, p = (μα)−1μ and e = up ∈ EndA(X), we have pu = 1M and e2 = e. By hypothesis, the idempotent 1 − e splits, so we can write 1 − e = vq, where X q −→ W v − → X and qv = 1W . From e = up and 1 − e = vq, we get the equation up + vq = 1X. Moreover, qu = (qv)(qu)(pu) = q(vq)(up)u = q(1 − e)eu = 0 similarly, pv = pupvqv = pe(1 − e)v = 0. We have verified all of the required equations, so X = M ⊕ W . This proves (i). To prove (ii) we assume that X = M. Suppose first that α is a unit of E. We use α to diagonalize ϕ: 1 0 −γα−1 1 α β γ δ 1 −α−1β 0 1 = α 0 0 −γα−1β + δ Since all the matrices on the left are invertible, so must be the one on the right, and it follows that −γα−1β + δ : Z −→ Y is an isomorphism. Suppose, on the other hand, that α ∈ J. Then νγ / J (as μα+νγ = 1M), and it follows that α + νγ / J. We define a new map ψ = 1M ν σ τ : M ⊕ Y −→ M ⊕ Z , which we claim is an isomorphism. Assuming the claim, we can diago- nalize ψ as we did ϕ, obtaining, in the lower-right corner, an isomor- phism from Y onto Z, and finishing the proof. To prove the claim, we use the equation ψϕ = 1M⊕Z to get ψ ϕ = α + νγ β + νγ 0 1Z . As α + νγ is an automorphism of M, ψ ϕ is clearly an automorphism of M ⊕ Z. Therefore ψ = (ψ ϕ)ϕ−1 is an isomorphism. 1.3. Theorem (Krull-Remak-Schmidt). Let A be an additive cate- gory in which every idempotent splits. Let M1,... , Mm and N1,... , Nn be indecomposable objects of A, with M1 ⊕ · · · ⊕ Mm ∼ N1 ⊕ · · · ⊕ Nn. Assume that EndA(Mi) is nc-local for each i. Then m = n and, after renumbering, Mi ∼ Ni for each i.

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