Bourbaki

Proposition 4

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

Algebra, VIII, p. 381, § 20. Linear Representations of Algebras

a) There exists a group homomorphism λS\lambda_S from S⊗DS∗S\otimes_DS^* to ΘS(A)\Theta_S(A), characterized by

λS(x⊗x∗)=cS(x,x∗)(6) \lambda_S(x\otimes x^*) =c_S(x, x^*) \tag{6}

for x∈Sx\in S and x∗∈S∗x^*\in S^*. This mapping is an isomorphism of (A,A)(A,A)-bimodules.

b) The (A,A)(A,A)-bimodule ΘS(A)\Theta_S(A) is simple.

The mapping γS:S⊗KS∗→ΘS(A)\gamma_S: S\otimes_KS^*\rightarrow \Theta_S(A) characterized by the formula γS(x⊗x∗)=cS(x,x∗)\gamma_S(x\otimes x^*) =c_S(x, x^*) is (A(A, A)-linear and surjective and satisfies γS(xd⊗x∗)\gamma_S(xd\otimes x^*) =γS(x⊗dx∗)=\gamma_S(x\otimes dx^*) for x∈S,x∗∈S∗x\in S,x^*\in S^*, and d∈Dd\in D. It therefore defines a surjective (A(A, A)-linear mapping λS:S⊗DS∗→ΘS(A)\lambda_S: S\otimes_DS^*\rightarrow \Theta_S(A) characterized by formula (6).

Let us prove that S⊗DS∗S\otimes_DS^* is a simple (A(A, A)-bimodule. By Corollary 2 of VIII, p. 63, every (A(A, A)-sub-bimodule of S⊗DS∗S\otimes_DS^* is of the form S⊗DHS\otimes_DH, where H is a (D(D, A)-sub-bimodule of S∗S^*. Since S∗S^* is a simple right A-module (VIII, p. 380, Theorem 1, c)), we have H = 0 or H=S∗H = S^*; the assertion follows.

The homomorphism λS\lambda_S is (A(A, A)-linear and nonzero; since S⊗DS∗S\otimes_DS^* is a simple bimodule, λS\lambda_S is injective (VIII, p. 47, Proposition 2, a)).

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.