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By Kurinczuk

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1. In both cases M = R and we use that EndR (R)op ∼ = R. • R = kG with k = F2 , G = C2 = C , so R = {0, 1, c, 1 + c}. We can check directly that this has no non-trivial idempotents; c2 = 1 = c, and (1 + c)2 = 0. Hence R is indecomposable as an R-module. • R = Mn (k), we can write 1 = e1 + · · ·en as a decomposition into primitive idempotents, where   0  ...      0     1 ei =     0     .  ..  0 is the matrix with a 1 in the i-th row i-th column. Then   0          .

A unit in EndR (M ). If λ ∈EndR (M ) is not a unit then it is nilpotent. So N = {non-units in EndR (M )}, and provided it is an ideal it is certainly the unique maximal ideal. (1) If ν ∈ N , λ ∈EndR (M ) then λν, νλ are not injective so are not units and λν, νλ ∈ N . (2) If ν ∈ N , λ ∈ N then λ + ν is invertible with inverse λ−1 (1 + λ−1 ν)−1 = λ−1 1 − (λ−1 ν) + ... + (−1)n (λ−1 ν)n as λ−1 ν ∈ N . e. it is in N . 1. This says that EndR (M ) is a local ring. Azumaya (1950) proved that the Krull-Schmidt theorem holds under the condition that all indecompasable components have local endomorphism rings (so drop the finite length condition).

0 is an idempotent, and all primitive idempotents are  1  0   ..  . conjugate to    . 1. What does this have to do with Krull-Schmidt? Let R be a ring and M an R-module. Let e be an idempotent in EndR (M ); then eM is a submodule of M and M = eM ⊕ (1 − e)M is a direct sum decomposition of M into submodules. So M is indecomposable if and only if EndR (M ) has no idempotents apart from 0, 1. eM is an indecomposable submodule if and only if e is primitive in EndR (M ).