Bourbaki

Corollary 1

Tag 004W, permanent. Cite it as [Bourbaki, Tag 004W].

Algebra, VIII, p. 67, § 4. Semisimple Modules

Let M be a semisimple A-module and N a submodule of M. The following properties are equivalent:

(i) There exists a unique submodule supplementary to N in M.

(ii) We have HomA(M/N,N)=0_A(M/N,N) = 0.

(iii) There exists a subset Λ of S\mathscr{S} such that N=⨁λ∈ΛMλN =\bigoplus_{\lambda\in\Lambda}M_{\lambda}.

Choose a submodule N′N' supplementary to N in M (VIII, p. 56, Corollary 2). If we identify M with N′×NN'\times N, then the submodules of M supplementary to N are the graphs of the A-linear mappings from N′N' to N. Since N′N' is isomorphic to M/NM/N, we have proved the equivalence of properties (i) and (ii). By Proposition 5, b), the group HomA(N′,N)_A(N',N) is isomorphic to the group ∏λ∈S\prod_{\lambda\in\mathscr{S}} HomA(Nλ′,Nλ)_A(N'_{\lambda},N_{\lambda}). It is zero if and only if for every λ∈S\lambda \in \mathscr{S}, we have Nλ=0N_{\lambda}= 0 or Nλ′=0N'_{\lambda}= 0 (VIII, p. 61, Remark), that is, Nλ=0N_{\lambda}= 0 or Nλ=MλN_{\lambda}= M_{\lambda}. This proves the equivalence of properties (ii) and (iii).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.