Bourbaki

Theorem 3 (Jacobson)

Tag 00S1, permanent. Cite it as [Bourbaki, Tag 00S1].

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

Let M be a semisimple A-module, and let cc be an endomorphism of the additive group M. Then cc belongs to the bicommutant AM′′A''_M of M if and only if it satisfies the following condition:

(D) For every finite subset F of M, there exists an element aa of A such that cc coincides with aMa_M on F.

First, suppose that cc satisfies condition (D). Let uu be an element of AM′A'_M. Let xx be an element of M, and apply condition (D) to the subset F={x,u(x)}F =\{x, u(x)\}. There exists an element aa of A such that c(x)=axc(x) =ax and c(u(x))=au(x)c(u(x)) =au(x), so that u(c(x))=u(ax)=au(x)=c(u(x))u(c(x)) =u(ax) =au(x) =c(u(x)). Since xx is arbitrary, we have cu=uccu=uc; as this holds for all uu, we have c∈AM′′c\in A''_M.

For the converse, we use the following lemma.

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.