Lemma 15.115.2. Let $R$ be a ring. Let $M$ be an $R$-module. Equivalent are

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

$M$ is invertible, and

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 $1$.
$\square$

## Comments (2)

Comment #1837 by Keenan Kidwell on

Comment #1874 by Johan on