The Stacks project

Proof. First suppose $M$ is finite projective, i.e., (2) holds. Take a surjection $R^ n \to M$ and let $K$ be the kernel. Since $M$ is projective, $0 \to K \to R^ n \to M \to 0$ splits. Hence (2) $\Rightarrow$ (3). The implication (3) $\Rightarrow$ (2) follows from the fact that a direct summand of a projective is projective, see Lemma 10.77.2.

Assume (3), so we can write $K \oplus M \cong R^{\oplus n}$. So $K$ is a direct summand of $R^ n$ and thus finitely generated. This shows $M = R^{\oplus n}/K$ is finitely presented. In other words, (3) $\Rightarrow$ (1).

Assume $M$ is finitely presented and flat, i.e., (1) holds. We will prove that (7) holds. Pick any prime $\mathfrak p$ and $x_1, \ldots , x_ r \in M$ which map to a basis of $M \otimes _ R \kappa (\mathfrak p)$. By Nakayama's lemma (in the form of Lemma 10.20.2) these elements generate $M_ g$ for some $g \in R$, $g \not\in \mathfrak p$. The corresponding surjection $\varphi : R_ g^{\oplus r} \to M_ g$ has the following two properties: (a) $\mathop{\mathrm{Ker}}(\varphi )$ is a finite $R_ g$-module (see Lemma 10.5.3) and (b) $\mathop{\mathrm{Ker}}(\varphi ) \otimes \kappa (\mathfrak p) = 0$ by flatness of $M_ g$ over $R_ g$ (see Lemma 10.39.12). Hence by Nakayama's lemma again there exists a $g' \in R_ g$ such that $\mathop{\mathrm{Ker}}(\varphi )_{g'} = 0$. In other words, $M_{gg'}$ is free.

A finite locally free module is a finite module, see Lemma 10.23.2, hence (7) $\Rightarrow$ (6). It is clear that (6) $\Rightarrow$ (7) and that (7) $\Rightarrow$ (8).

A finite locally free module is a finitely presented module, see Lemma 10.23.2, hence (7) $\Rightarrow$ (4). Of course (4) implies (5). Since we may check flatness locally (see Lemma 10.39.18) we conclude that (5) implies (1). At this point we have

Suppose that $M$ satisfies (1), (4), (5), (6), and (7). We will prove that (3) holds. It suffices to show that $M$ is projective. We have to show that $\mathop{\mathrm{Hom}}\nolimits _ R(M, -)$ is exact. Let $0 \to N'' \to N \to N'\to 0$ be a short exact sequence of $R$-module. We have to show that $0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'') \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N') \to 0$ is exact. As $M$ is finite locally free there exist a covering $\mathop{\mathrm{Spec}}(R) = \bigcup D(f_ i)$ such that $M_{f_ i}$ is finite free. By Lemma 10.10.2 we see that

is equal to $0 \to \mathop{\mathrm{Hom}}\nolimits _{R_{f_ i}}(M_{f_ i}, N''_{f_ i}) \to \mathop{\mathrm{Hom}}\nolimits _{R_{f_ i}}(M_{f_ i}, N_{f_ i}) \to \mathop{\mathrm{Hom}}\nolimits _{R_{f_ i}}(M_{f_ i}, N'_{f_ i}) \to 0$ which is exact as $M_{f_ i}$ is free and as the localization $0 \to N''_{f_ i} \to N_{f_ i} \to N'_{f_ i} \to 0$ is exact (as localization is exact). Whence we see that $0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'') \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N') \to 0$ is exact by Lemma 10.23.2.

Finally, assume that (8) holds. 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

is an isomorphism. This claim will show that $M$ is finite locally free, i.e., that (7) holds. To see the claim it suffices to show that the induced map on localizations $\Psi _{\mathfrak p} : R_{\mathfrak p}^{\oplus r} \to M_{\mathfrak p}$ is an isomorphism for all $\mathfrak p \in D(fg)$, see Lemma 10.23.1. By our choice of $f$ the map $\Psi _{\mathfrak p}$ is surjective. By assumption (8) we have $M_{\mathfrak p} \cong R_{\mathfrak p}^{\oplus \rho _ M(\mathfrak p)}$ and by our choice of $g$ we have $\rho _ M(\mathfrak p) = r$. Hence $\Psi _{\mathfrak p}$ determines a surjection $R_{\mathfrak p}^{\oplus r} \to M_{\mathfrak p} \cong R_{\mathfrak p}^{\oplus r}$ whence is an isomorphism by Lemma 10.16.4. (Of course this last fact follows from a simple matrix argument also.) $\square$