Bourbaki

Theorem 1

Tag 0044, permanent. Cite it as [Bourbaki, Tag 0044].

Algebra, VIII, p. 56, § 4. Semisimple Modules

Let M be a module that is the (not necessarily direct) sum of a family (Si)i∈I(S_i)_{i\in I} of simple submodules, and let N be a submodule of M. There exists a subset J of I such that M is the direct sum of the family consisting of N and the modules SjS_j for jj running through J.

Let S\mathscr{S} be the set of subsets I′I' of I such that the sum of the family consisting of the modules N and SiS_i for ii in I′I' is direct. The set S\mathscr{S} is of finite character: a subset J of I belongs to S\mathscr{S} if and only if the same holds for every finite subset of J. Hence, the set S\mathscr{S} has a maximal element J (Set Theory, III, §4, No. 5, p. 171). Set N′=N+∑j∈JSjN'= N +\sum_{j\in J}S_j. Let ii be in I - J. Since J is maximal in S\mathscr{S}, the set J∪{i}J\cup \{i\} does not belong to S\mathscr{S}, so that Si∩N′≠0S_i\cap N'\not= 0. Since SiS_i is a simple module, we have Si∩N′=SiS_i\cap N'= S_i. We therefore have Si⊂N′S_i\subset N' for every i∈Ii\in I, whence N′=MN'= M. This completes the proof.

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.