**Proof.**
Pick a maximal ideal $\mathfrak m \subset R$. Pick $x_1, \ldots , x_ r \in M$ which map to a $\kappa (\mathfrak m)$-basis of $M \otimes _ R \kappa (\mathfrak m) = M/\mathfrak mM$. In particular $\rho _ M(\mathfrak m) = r$. By Nakayama's Lemma 10.20.1 there exists an $f \in R$, $f \not\in \mathfrak m$ such that $x_1, \ldots , x_ r$ generate $M_ f$ over $R_ f$. By the assumption that $\rho _ M$ is locally constant there exists a $g \in R$, $g \not\in \mathfrak m$ such that $\rho _ M$ is constant equal to $r$ on $D(g)$. We claim that

\[ \Psi : R_{fg}^{\oplus r} \longrightarrow M_{fg}, \quad (a_1, \ldots , a_ r) \longmapsto \sum a_ i x_ i \]

is an isomorphism. This claim will show that $M$ is finite locally free, i.e., that (7) holds. Since $\Psi $ is surjective, it suffices to show that $\Psi $ is injective. Since $R_{fg}$ is reduced, it suffices to show that $\Psi $ is injective after localization at all minimal primes $\mathfrak p$ of $R_{fg}$, see Lemma 10.25.2. However, we know that $R_\mathfrak p = \kappa (\mathfrak p)$ by Lemma 10.25.1 and $\rho _ M(\mathfrak p) = r$ hence $\Psi _\mathfrak p : R_\mathfrak p^{\oplus r} \to M \otimes _ R \kappa (\mathfrak p)$ is an isomorphism as a surjective map of finite dimensional vector spaces of the same dimension.
$\square$

## Comments (0)

There are also: