Theorem 3

Let $A$ be a ring and $M$ ajnitely generated $A$-module.

(i) If there exists an $A$-module $N$ such that $M \otimes_A N$ is isomorphic to $A$, the module $M$ is projective of rank 1.

(ii) Conversely, if $M$ is projective of rank 1 and $M^*$ is the dual of $M$, the canonical homomorphism $u : M \otimes_A M^* \to A$ corresponding to the canonical bilinear form $(x, x^*) \to \langle x, x^* \rangle$ on $M \times M^*$ (Algebra, Chapter II, § 2, no. 3) is bijective.

(i) It is required to prove that, for every maximal ideal $m$ of $A$, the $A_m$-module $M$, is free of rank 1 (Theorem 2 (b)); we are free to replace $A$ by $A_m$ and hence may assume that $A$ is a local ring ($§ 2$, no. 7, Proposition 18). Let $k = A/m$. The isomorphism $v : M \otimes_A N \to A$ defines an isomorphism

as the rank over $k$ of $(M/mM) \otimes_k (N/mN)$ is the product of the ranks of $M/mM$ and $N/mN$, these latter are necessarily equal to 1, in other words $M/mM$ is monogenous. It follows that $M$ is monogenous ($§ 3$, no. 2, Corollary 2 to Proposition 4); on the other hand, the annihilator of $M$ also annihilates $M \otimes_A N$ and hence is zero, which proves that $M$ is isomorphic to $A$.

(ii) It is sufficient to prove that, for every maximal ideal $m$ of $A$, $u_m$ is an isomorphism ($§ 3$, no. 3, Theorem 1). As $M$ is finitely presented (Chapter I, $§ 2$, no. 8, Lemma 8), $(M^*)_m$ is canonically identified with the dual $(M_m)^*$ ($§ 2$, no. 7, Proposition 19) and, as $M_m$ is free of rank 1 like its dual $(M_m)^*$, clearly the canonical homomorphism $u_m : (M_m) \otimes_{A_m} (M_m)^* \to A_m$ is bijective, which completes the proof.

Remark (1). If $M$ is projective of rank 1 and $N$ is such that $M \otimes_A N$ is isomorphic to $A$, then $N$ is isomorphic to $M^*$: there are isomorphisms