Bourbaki

Lemma 3

Tag 0063, permanent. Cite it as [Bourbaki, Tag 0063].

Algebra, VIII, p. 88, § 5. Commutation

Let M be a semisimple A-module. Let B be the bicommutant of the A-module M. Then every A-submodule of M is a B-submodule of M.

Let N be an A-submodule of M. By Corollary 2 of VIII, p. 56, there exists a projector pp of the A-module M with image N; it belongs to AM′A'_M. Since we have the relation pb=bppb=bp for every b∈Bb\in B, we obtain that N is a B-submodule of M.

Let us conclude the proof of Theorem 3. Suppose that cc belongs to AM′′A''_M, and let F = {x1,...,xn}\{x_1, . . . , x_n\} be a finite subset of M. Denote the element (x1,...,xn)(x_1, . . . , x_n) of MnM^n by xx. The A-module MnM^n is semisimple and, by Proposition 2 of VIII, p. 79, its bicommutant coincides with the homotheties of the AM′′A''_M-module MnM^n. By Lemma 3, the A-submodule AxAx of MnM^n is an AM′′A''_M-submodule of MnM^n. Let a∈Aa\in A be such that (cx1,...,cxn)(cx_1, . . . , cx_n) is equal to axax. Then cc coincides with aMa_M on {x1,...,xn}\{x_1, . . . , x_n\}, which implies condition (D).

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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Cites

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