Lemma 10.78.2. Let $R$ be a ring and let $M$ be an $R$-module. The following are equivalent

1. $M$ is finitely presented and $R$-flat,

2. $M$ is finite projective,

3. $M$ is a direct summand of a finite free $R$-module,

4. $M$ is finitely presented and for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$ the localization $M_{\mathfrak p}$ is free,

5. $M$ is finitely presented and for all maximal ideals $\mathfrak m \subset R$ the localization $M_{\mathfrak m}$ is free,

6. $M$ is finite and locally free,

7. $M$ is finite locally free, and

8. $M$ is finite, for every prime $\mathfrak p$ the module $M_{\mathfrak p}$ is free, and the function

$\rho _ M : \mathop{\mathrm{Spec}}(R) \to \mathbf{Z}, \quad \mathfrak p \longmapsto \dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p)$

is locally constant in the Zariski topology.

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

$\xymatrix{ (2) \ar@{<=>}[r] & (3) \ar@{=>}[r] & (1) \ar@{=>}[r] & (7) \ar@{<=>}[r] \ar@{=>}[rd] \ar@{=>}[d] & (6) \\ & & (5) \ar@{=>}[u] & (4) \ar@{=>}[l] & (8) }$

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

$0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'')_{f_ i} \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N)_{f_ i} \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N')_{f_ i} \to 0$

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

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

is an isomorphism. This claim will show that $M$ is finite locally free, i.e., that (7) holds. 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.) $\square$

Comment #1191 by Lenny Taelman on

In (1) ==> (7): "The corresponding surjection" is to M_g, not M_g^r.

Comment #6248 by Rein Janssen Groesbeek on

$\def\sb{_}$ In (1) $\Rightarrow$ (7), in the first application of Nakayama, I think it is applied on $M\sb{\mathfrak{p}}$ instead of $M$, as $M \otimes\sb{R} \kappa(\mathfrak{p})$ is a quotient of $M\sb{\mathfrak{p}}$ and not of $M$ if $\mathfrak{p}$ is not maximal. So Nakayama's Lemma gives $\operatorname{Coker}(R^{\oplus r} \xrightarrow{x_i} M)\sb{\mathfrak{p}} = 0$, and the existence of $g \in R, g \not \in \mathfrak{p}$ such that $x_1, \dots, x_r$ generate $M_g$ follows from finiteness of $\operatorname{Coker}(R^{\oplus r} \xrightarrow{x_i} M)$.
Similarly in the second application, Nakayama gives $\operatorname{Ker}(\varphi)\sb{\mathfrak{p}} =0$ and the existence of $g' \in R\sb{g}$ follows from finiteness of $\operatorname{Ker}(\varphi)$.

Comment #6382 by Laurent Moret-Bailly on

Condition (7) looks incomplete.

Comment #6392 by Laurent Moret-Bailly on

@#6384: I had not realized that "and" might announce the last condition, so I read it as an unfinished sentence. Another thing: the difference between (6) and (7) will be unclear to those reading this lemma without seeing Definition 00NW first.

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