Bourbaki

Theorem 1 (Jacobson’s density theorem)

Tag 00SD, permanent. Cite it as [Bourbaki, Tag 00SD].

Algebra, VIII, p. 442, Appendix 1. Algebras without Unit Element

Let M be a semisimple left A-pseudomodule. Let bb be an element of the bicommutant of the A~\widetilde{A}-module M. Let F={x1,...,xn}F =\{x_1, . . . , x_n\} be a finite subset of M. Then there exists an element a∈Aa\in A such that bxi=axibx_i=ax_i for every i∈[1,n]i\in [1, n].

Let B be the bicommutant of the A~\widetilde{A}-module M. The A-pseudomodule MnM^n is semisimple. Let x=(x1,...,xn)∈Mn\boldsymbol{x}= (x_1, . . . , x_n)\in M^n. It follows from Lemma 2 that x∈\boldsymbol{x}\in AxA\boldsymbol{x}. The bicommutant of the A~\widetilde{A}-module MnM^n coincides with the homotheties of the B-module MnM^n (VIII, p. 79, Proposition 2). By Lemma 1, the A-sub-pseudomodule AxA\boldsymbol{x} of MnM^n is therefore a B-submodule of MnM^n. We therefore have the inclusion Bx⊂AxB\boldsymbol{x}\subset A\boldsymbol{x}. So there exists an a∈Aa\in A such that bx=axb\boldsymbol{x}=a\boldsymbol{x}. The result follows.

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.