Bourbaki

Proposition 6

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

Algebra, VIII, p. 383, § 20. Linear Representations of Algebras

Let S0\mathscr{S}_0 be the set of classes of simple A-modules of finite dimension over K with nonzero trace. The family of linear forms (TrS_S)S∈S0_{S\in\mathscr{S}_0} is free over K.

Let F be a finite subset of S0\mathscr{S}_0, and let S be an element of F. By assumption, there exists an element a∈Aa\in A such that TrS(a)≠0_S(a)\not= 0. By Corollary 1 of Proposition 4 (VIII, p. 83), there exists an element b∈Ab\in A such that bS=aSb_S=a_S and bT=0b_T= 0 for every T∈F−{S}T\in F-\{S\}. We have TrS(b)≠0_S(b)\not= 0 and TrT(b)=0_T(b) = 0 for T∈F−{S}T\in F-\{S\}. The family (TrS_S)S∈F_{S\in F} is therefore free. Proposition 6 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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Transcribed from the printed edition, 2023, Springer Nature. Not translated.