The Stacks project

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.) $\square$

Comments (7)

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

In (1) (7), in the first application of Nakayama, I think it is applied on instead of , as is a quotient of and not of if is not maximal. So Nakayama's Lemma gives , and the existence of such that generate follows from finiteness of .
Similarly in the second application, Nakayama gives and the existence of follows from finiteness of .

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.

There are also:

  • 4 comment(s) on Section 10.78: Finite projective modules

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00NX. Beware of the difference between the letter 'O' and the digit '0'.