Lemma 15.117.2. Let $R$ be a ring. Let $M$ be an $R$-module. Equivalent are

1. $M$ is finite locally free module of rank $1$,

2. $M$ is invertible, and

3. there exists an $R$-module $N$ such that $M \otimes _ R N \cong R$.

Moreover, in this case the module $N$ in (3) is isomorphic to $\mathop{\mathrm{Hom}}\nolimits _ R(M, R)$.

Proof. Assume (1). Consider the module $N = \mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ and the evaluation map $M \otimes _ R N = M \otimes _ R \mathop{\mathrm{Hom}}\nolimits _ R(M, R) \to R$. If $f \in R$ such that $M_ f \cong R_ f$, then the evaluation map becomes an isomorphism after localization at $f$ (details omitted). Thus we see the evaluation map is an isomorphism by Algebra, Lemma 10.23.2. Thus (1) $\Rightarrow$ (3).

Assume (3). Then the functor $K \mapsto K \otimes _ R N$ is a quasi-inverse to the functor $K \mapsto K \otimes _ R M$. Thus (3) $\Rightarrow$ (2). Conversely, if (2) holds, then $K \mapsto K \otimes _ R M$ is essentially surjective and we see that (3) holds.

Assume the equivalent conditions (2) and (3) hold. Denote $\psi : M \otimes _ R N \to R$ the isomorphism from (3). Choose an element $\xi = \sum _{i = 1, \ldots , n} x_ i \otimes y_ i$ such that $\psi (\xi ) = 1$. Consider the isomorphisms

$M \to M \otimes _ R M \otimes _ R N \to M$

where the first arrow sends $x$ to $\sum x_ i \otimes x \otimes y_ i$ and the second arrow sends $x \otimes x' \otimes y$ to $\psi (x' \otimes y)x$. We conclude that $x \mapsto \sum \psi (x \otimes y_ i)x_ i$ is an automorphism of $M$. This automorphism factors as

$M \to R^{\oplus n} \to M$

where the first arrow is given by $x \mapsto (\psi (x \otimes y_1), \ldots , \psi (x \otimes y_ n))$ and the second arrow by $(a_1, \ldots , a_ n) \mapsto \sum a_ i x_ i$. In this way we conclude that $M$ is a direct summand of a finite free $R$-module. This means that $M$ is finite locally free (Algebra, Lemma 10.78.2). Since the same is true for $N$ by symmetry and since $M \otimes _ R N \cong R$, we see that $M$ and $N$ both have to have rank $1$. $\square$

Comment #1837 by Keenan Kidwell on

In the first sentence of the third text block, "equivalence conditions" should be "equivalent conditions."

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).