Bourbaki

Proposition 6

Tag 005Y, permanent. Cite it as [Bourbaki, Tag 005Y].

Algebra, VIII, p. 86, § 5. Commutation

a) Viewed as an (A,Bo)(A,B^o)-bimodule, M is semisimple.

b) For every λ∈SM,Mλ\lambda \in \mathscr{S}_M,M_{\lambda} is a simple (A,Bo)(A,B^o)-sub-bimodule of M.

c) For every (A,Bo)(A,B^o)-sub-bimodule N of M, there exists a unique subset Λ of SM\mathscr{S}_M such that N is equal to ⊕λ∈ΛMλ\oplus_{\lambda\in\Lambda}M_{\lambda}.

Let λ\lambda be in SM\mathscr{S}_M. The left A-module SλS_{\lambda} and the right B-module VλV_{\lambda} are simple, and DλD_{\lambda} is the opposite ring of the field of endomorphisms of SλS_{\lambda}. By Corollary 2 of VIII, p. 63, the (A,Bo(A,B^o)-bimodule Sλ⊗DλVλS_{\lambda}\otimes_{D_{\lambda}}V_{\lambda} is simple, and the same holds for MλM_{\lambda}, which is isomorphic to it. This proves b), and a) follows.

If λ\lambda and μ\mu are distinct in SM\mathscr{S}_M, then MλM_{\lambda} and MμM_\mu are not isomorphic as A-modules, nor a fortiori as (A,Bo(A,B^o)-bimodules. Assertion c) follows by Corollary 2 of VIII, p. 68.

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.