Lemma 1

— Let A and B be k-algebras, P be an invertible $(A, B)_k$-bimodule, M and N be B-modules, and $u : M \to N$ be a B-linear mapping. If the mapping $1_P \otimes u : P \otimes_B M \to P \otimes_B N$ is zero (resp. bijective), then so is u.

Let Q be a bimodule inverse to P and $\theta : Q \otimes_A P \to {}_s B_d$ an isomorphism of $(B, B)_k$-bimodules. The lemma follows from the commutativity of the diagram

Theorem 1. — *Let A and B be k-algebras and P an $(A, B)_k$-bimodule. Denote the $(B, A)_k$-bimodule $\operatorname{Hom}_A(P, A_s)$ by $P^*$. The following properties are equivalent:

(i) The $(A, B)_k$-bimodule P is invertible.

(ii) The A-module P is projective, finitely generated, and generating, and the mapping $b \mapsto b_P$ from B to $\operatorname{End}_A(P)^{\circ}$ is a k-algebra isomorphism.

(iii) The right B-module P is projective, finitely generated, and generating, and the mapping $a \mapsto a_P$ from A to $\operatorname{End}_B(P)$ is a k-algebra isomorphism.

If these properties hold, then the homomorphisms

are isomorphisms, so that the $(B, A)_k$-bimodule $P^*$ is an inverse of P.

If property (ii) holds, then P is an invertible $(A, B)_k$-bimodule with inverse $P^*$ (VIII, p. 98, Proposition 2 and p. 98, Proposition 3). This proves that (ii) implies (i) and the last assertion.

Suppose that the $(A, B)_k$-bimodule P is invertible. Then the A-module P is generating (VIII, p. 98, Proposition 3, (iii) $\Rightarrow$ (i)). It is therefore faithful and balanced (VIII, p. 82, Theorem 2) and, consequently, the mapping $a \mapsto a_p$ from A to $\operatorname{End}_B(P)$ is bijective.

Let us now prove that the mapping $b \mapsto b_P$ from B to $\operatorname{End}_A(P)$ is bijective. Let Q be a $(B, A)_k$-bimodule, inverse to P. Let $u \in \operatorname{End}_A(P)$; then $1_Q \otimes u$ is an endomorphism of the left B-module $Q \otimes_A P$. Since the $(B, B)_k$-bimodule $Q \otimes_A P$ is isomorphic to $_s B_d$, there exists a unique element $b$ of B such that $1_Q\otimes u$ is the homothety with ratio $b$ of the right B-module $Q\otimes_AP$. Consequently, we have $1_Q\otimes (u-b_P) = 0$. Hence, $u=b_P$ by Lemma 1; this proves that the mapping $b\mapsto b_P$ from B to End$_A(P)$ is bijective.

By Proposition 2 of VIII, p. 98, the A-module P is then projective and finitely generated. We have therefore proved the equivalence of (i) and (ii).

By interchanging the roles of A and B, we obtain the equivalence of (i) and (iii), which concludes the proof of the proposition.