Bourbaki

Exercise 23, § 2

Algebra, II, p. 188

Bourbaki marks this one as harder than the rest.

Soit M an A-module, direct sum of a family (finite or not) (Mλ)λ∈L(M_\lambda)_{\lambda \in L} of submodules such that for every λ∈L\lambda \in L, the non-invertible elements of End(Mλ)\mathrm{End}(M_\lambda) form a two-sided ideal in this ring (which implies that MλM_\lambda is indecomposable). (a) Let f,gf, g be two endomorphisms of M such that f+g=1Mf + g = 1_M. Now prove that for every finite sequence (λk)1≤k≤s(\lambda_k)_{1 \leq k \leq s} of distinct indices in L, there exists a family of submodules NkN_k (1≤k≤s1 \leq k \leq s) of M such that for every k, one of the endomorphisms f,gf, g, restricted to MλkM_{\lambda_k}, is an isomorphism of this submodule onto NkN_k, and that M is the direct sum of the NkN_k (1≤k≤s1 \leq k \leq s) and the MλM_\lambda for the λ\lambda distinct from the λk\lambda_k. (Reduce to the case s=1s = 1; if πλ\pi_\lambda is the canonical projection of M onto MλM_\lambda corresponding to the decomposition into direct sum of MλM_\lambda and ⨁μ≠λMμ\bigoplus_{\mu \neq \lambda} M_\mu, note that either πλ∘f\pi_\lambda \circ f, or πλ∘g\pi_\lambda \circ g, restricted to MλM_\lambda, is an automorphism of MλM_\lambda.) (b) Let ff be an idempotent endomorphism of M. Now prove that there exists at least one index λ∈L\lambda \in L such that the restriction of ff to MλM_\lambda is an isomorphism of MλM_\lambda onto f(Mλ)f(M_\lambda); moreover f(Mλ)f(M_\lambda) is a direct factor of M. (If g=1M−fg = 1_M - f, note that there exists a finite sequence of indices (λk)1≤k≤s(\lambda_k)_{1 \leq k \leq s} such that the intersection of Ker(g)\mathrm{Ker}(g) and the direct sum of the MλkM_{\lambda_k} is ≠{0}\neq \{0\}, and use a).) (c) Deduce from b) that every indecomposable direct factor of M is isomorphic to one of the MλM_\lambda. (d) Let (Nκ)κ∈K(N_\kappa)_{\kappa \in K} be a family of indecomposable submodules of M of which M is the direct sum; every NκN_\kappa is therefore isomorphic to an MλM_\lambda, and conversely, by virtue of c). Let T be the set of classes of indecomposable submodules of M (for the relation of isomorphism) such that an MλM_\lambda (or an NκN_\kappa) belongs to one of these classes. For every t∈Tt \in T, let R(t)R(t) (resp. S(t)S(t)) be the set of λ∈L\lambda \in L (resp. κ∈K\kappa \in K) such that Mλ∈tM_\lambda \in t (resp. Nκ∈tN_\kappa \in t). Now prove that for every t∈Tt \in T, one has Card(R(t))=Card(S(t))\mathrm{Card}(R(t)) = \mathrm{Card}(S(t)). With the notation of a), let J(κ)J(\kappa) be the set of λ∈L\lambda \in L such that the restriction of πλ\pi_\lambda to NκN_\kappa is an isomorphism of NκN_\kappa onto MλM_\lambda; prove that J(κ)J(\kappa) is finite and that the J(κ)J(\kappa) cover R(t)R(t) when κ\kappa runs through S(t)S(t). Deduce that Card(S(t))≤Card(R(t))\mathrm{Card}(S(t)) \leq \mathrm{Card}(R(t)) when R(t)R(t) is infinite. When R(t)R(t) is finite, prove with the help of a) that M is the direct sum of the MλM_\lambda for λ∉R(t)\lambda \notin R(t) and of a subfamily of (Nκ)κ∈S(t)(N_\kappa)_{\kappa \in S(t)} of cardinal equal to that of R(t)R(t). (e) Deduce from d) and Exercises 22 and 21 d) that if (Eλ)λ∈L(E_\lambda)_{\lambda \in L} and (Eκ′)κ∈K(E'_\kappa)_{\kappa \in K} are two reduced indecomposable decompositions of {0} in a module F, L and K are equipotent.

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.