Bourbaki

Theorem 3

Tag 004N, permanent. Cite it as [Bourbaki, Tag 004N].

Algebra, VIII, p. 64, § 4. Semisimple Modules

Let M and M′M' be isotypical A-modules of type S. Let (V,α)(V, \alpha ) and (V′,α′)(V', \alpha ') be descriptions of M and M′M', respectively. For any D-linear mapping f:V→V′f: V\rightarrow V', denote by f~:M→M′\widetilde{f}: M\rightarrow M' the unique A-linear mapping that makes the following diagram commute:

S⊗DVαS\otimes_DV^{\alpha} // M

1S⊗ff~(9) 1_{_S\otimes f}\widetilde{f} \tag{9}

S⊗DV′α′S\otimes_D{V'}^{\alpha'} // M′M'.

The mapping f↦f~f\mapsto \widetilde{f} from HomD(V,V′)_D(V,V') to HomA(M,M′)_A(M,M') is a group isomorphism.

It suffices to prove that the Z\mathbf{Z}-linear mapping u↦1S⊗uu\mapsto 1_S\otimes u from HomD(V,V′)_D(V,V') to HomA(S⊗DV,S⊗DV′)_A(S\otimes_DV,S\otimes_DV') is bijective. In the notation of No.3 applied to the (A(A, D)-bimodule S, this corresponds to showing that the mapping

T:\mathscr{T}: HomD(V,V′)⟶_D(V,V')\longrightarrow HomA(T(V),T(V′))_A(\mathscr{T}(V),\mathscr{T}(V'))

is bijective. But by Remark 1 of VIII, p. 60, since the adjunction isomorphism (VIII, p. 59) is bijective, this corresponds to showing that the mapping that sends uu to βV′∘u\beta_{V'}\circ u is bijective, which follows from the fact that the D-linear mapping βV′\beta_{V'} is bijective (VIII, p. 62, Proposition 3, b)).

We keep the notation of Theorem 3. Let M′′M'' be an isotypical A-module of type S, and let (V′′,α′′)(V'', \alpha '') be a description of M′′M''. For every f∈f\in HomD(V,V′)_D(V,V') and every g∈g\in HomD(V′,V′′)_D(V',V''), we have g]∘f=g~∘f~g]\circ f=\widetilde{g}\circ \widetilde{f}. In particular, for M=M′M = M', V=V′V = V', and α=α′\alpha =\alpha ', the mapping f↦f~f\mapsto \widetilde{f} from EndD(V)_D(V) to EndA(M)_A(M) is a ring isomorphism.

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.