\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

The Stacks project

Lemma 10.77.2. 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 10.76.2.

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 10.19.1 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 10.38.12). 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 10.22.2, 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 10.22.2, hence (7) $\Rightarrow $ (4). Of course (4) implies (5). Since we may check flatness locally (see Lemma 10.38.19) 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 10.22.2.

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 10.19.1 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 10.22.1. 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 10.15.4. (Of course this last fact follows from a simple matrix argument also.) $\square$


Comments (2)

Comment #1191 by Lenny Taelman on

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

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  • 2 comment(s) on Section 10.77: Finite projective modules

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