Bourbaki

Exercise 15, Appendix 1

Tag 00Q0, permanent. Cite it as [Bourbaki, Tag 00Q0].

Algebra, VIII, p. 445

Let A be a kk-algebra. A multiplier of A is a multiplier of the (A,A)(A,A)-pseudobimodule A. We denote the kk-algebra of multipliers of A by M(A)\mathscr{M}(A). a) Prove that the mapping μ:a↦ma\mu:a\mapsto m_a is a kk-algebra homomorphism and that we have am=mamam=m_am and ma=mmama=mm_a for a∈A,m∈M(A)a\in A, m\in \mathscr{M}(A). The image of the homomorphism μ\mu is therefore a two-sided ideal of the ring M(A)\mathscr{M}(A).

b) A two-sided ideal b\mathfrak{b} of a kk-algebra A is called faithful if its left and right annihilators are zero. Show that if A is faithful in A, then the homomorphism μ\mu is injective and the ideal μ(A)\mu(A) is faithful in M(A)\mathscr{M}(A).

c) Let B be a unital kk-algebra containing A as a two-sided ideal. For b∈Bb\in B, let σb\sigma_b and δb\delta_b be the mappings from A to A such that for a∈Aa\in A, we have σb(a)=ba\sigma_b(a) =ba and δb(a)=ab\delta_b(a) =ab, and set π(b)=(σb,δb)\pi (b) = (\sigma_b, \delta_b). Show that π(b)∈M(A)\pi (b)\in \mathscr{M}(A) for every b∈Bb\in B and that the resulting mapping π:B→M(A)\pi : B\rightarrow \mathscr{M}(A) is a ring homomorphism whose restriction to A is the mapping a↦maa\mapsto m_a. Show that π\pi is injective if the ideal A is faithful in B.

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Transcribed from the printed volume. Not translated.