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Chapter 10: Commutative Algebra > Section 10.77: Finite projective modules

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 $\mathfrak p \in \mathop{\rm Spec}(R)$ 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 $$ \rho_M : \mathop{\rm Spec}(R) \to \mathbf{Z}, \quad \mathfrak p \longmapsto \dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p) $$ 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) $\mathop{\rm Ker}(\varphi)$ is a finite $R_g$-module (see Lemma 10.5.3) and (b) $\mathop{\rm Ker}(\varphi) \otimes \kappa(\mathfrak p) = 0$ 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 $\mathop{\rm Ker}(\varphi)_{g'} = 0$. 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 10.38.19) 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 $\mathop{\rm Hom}\nolimits_R(M, -)$ 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 $0 \to \mathop{\rm Hom}\nolimits_R(M, N'') \to \mathop{\rm Hom}\nolimits_R(M, N) \to \mathop{\rm Hom}\nolimits_R(M, N') \to 0$ is exact. As $M$ is finite locally free there exist a covering $\mathop{\rm Spec}(R) = \bigcup D(f_i)$ such that $M_{f_i}$ is finite free. By Lemma 10.10.2 we see that $$ 0 \to \mathop{\rm Hom}\nolimits_R(M, N'')_{f_i} \to \mathop{\rm Hom}\nolimits_R(M, N)_{f_i} \to \mathop{\rm Hom}\nolimits_R(M, N')_{f_i} \to 0 $$ is equal to $0 \to \mathop{\rm Hom}\nolimits_{R_{f_i}}(M_{f_i}, N''_{f_i}) \to \mathop{\rm Hom}\nolimits_{R_{f_i}}(M_{f_i}, N_{f_i}) \to \mathop{\rm Hom}\nolimits_{R_{f_i}}(M_{f_i}, N'_{f_i}) \to 0$ 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 $0 \to \mathop{\rm Hom}\nolimits_R(M, N'') \to \mathop{\rm Hom}\nolimits_R(M, N) \to \mathop{\rm Hom}\nolimits_R(M, N') \to 0$ 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 10.19.1 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 10.23.1. 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 10.15.4. (Of course this last fact follows from a simple matrix argument also.) $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 18074–18100 (see updates for more information).

    \begin{lemma}
    \label{lemma-finite-projective}
    Let $R$ be a ring and let $M$ be an $R$-module.
    The following are equivalent
    \begin{enumerate}
    \item $M$ is finitely presented and $R$-flat,
    \item $M$ is finite projective,
    \item $M$ is a direct summand of a finite free $R$-module,
    \item $M$ is finitely presented and
    for all $\mathfrak p \in \Spec(R)$ the
    localization $M_{\mathfrak p}$ is free,
    \item $M$ is finitely presented and
    for all maximal ideals $\mathfrak m \subset R$ the
    localization $M_{\mathfrak m}$ is free,
    \item $M$ is finite and locally free,
    \item $M$ is finite locally free, and
    \item $M$ is finite, for every prime $\mathfrak p$ the module
    $M_{\mathfrak p}$ is free, and the function
    $$
    \rho_M : \Spec(R) \to \mathbf{Z}, \quad
    \mathfrak p
    \longmapsto
    \dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p)
    $$
    is locally constant in the Zariski topology.
    \end{enumerate}
    \end{lemma}
    
    \begin{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 \ref{lemma-characterize-projective}.
    
    \medskip\noindent
    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).
    
    \medskip\noindent
    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 \ref{lemma-NAK}
    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) $\Ker(\varphi)$ is a finite
    $R_g$-module (see Lemma \ref{lemma-extension})
    and (b) $\Ker(\varphi) \otimes \kappa(\mathfrak p) = 0$
    by flatness of $M_g$ over $R_g$ (see
    Lemma \ref{lemma-flat-tor-zero}).
    Hence by Nakayama's lemma again there exists a $g' \in R_g$ such that
    $\Ker(\varphi)_{g'} = 0$. In other words, $M_{gg'}$ is free.
    
    \medskip\noindent
    A finite locally free module is a finite module, see
    Lemma \ref{lemma-cover},
    hence (7) $\Rightarrow$ (6).
    It is clear that (6) $\Rightarrow$ (7) and that (7) $\Rightarrow$ (8).
    
    \medskip\noindent
    A finite locally free module is a finitely presented module, see
    Lemma \ref{lemma-cover},
    hence (7) $\Rightarrow$ (4).
    Of course (4) implies (5).
    Since we may check flatness locally (see
    Lemma \ref{lemma-flat-localization})
    we conclude that (5) implies (1).
    At this point we have
    $$
    \xymatrix{
    (2) \ar@{<=>}[r] & (3) \ar@{=>}[r] & (1) \ar@{=>}[r] &
    (7)  \ar@{<=>}[r] \ar@{=>}[rd] \ar@{=>}[d] & (6) \\
    & & (5) \ar@{=>}[u] & (4) \ar@{=>}[l] & (8)
    }
    $$
    
    \medskip\noindent
    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 $\Hom_R(M, -)$
    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
    $0 \to \Hom_R(M, N'') \to \Hom_R(M, N) \to
    \Hom_R(M, N') \to 0$ is exact.
    As $M$ is finite locally free there exist a covering
    $\Spec(R) = \bigcup D(f_i)$ such that $M_{f_i}$ is finite free.
    By
    Lemma \ref{lemma-hom-from-finitely-presented}
    we see that
    $$
    0 \to \Hom_R(M, N'')_{f_i} \to \Hom_R(M, N)_{f_i} \to
    \Hom_R(M, N')_{f_i} \to 0
    $$
    is equal to
    $0 \to \Hom_{R_{f_i}}(M_{f_i}, N''_{f_i}) \to
    \Hom_{R_{f_i}}(M_{f_i}, N_{f_i}) \to
    \Hom_{R_{f_i}}(M_{f_i}, N'_{f_i}) \to 0$
    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
    $0 \to \Hom_R(M, N'') \to \Hom_R(M, N) \to
    \Hom_R(M, N') \to 0$ is exact by
    Lemma \ref{lemma-cover}.
    
    \medskip\noindent
    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 \ref{lemma-NAK}
    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 \ref{lemma-characterize-zero-local}.
    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 \ref{lemma-fun}.
    (Of course this last fact follows from a simple matrix argument also.)
    \end{proof}

    Comments (2)

    Comment #1191 by Lenny Taelman on December 5, 2014 a 10:47 am UTC

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

    Comment #1205 by Johan (site) on December 8, 2014 a 2:14 pm UTC

    Thanks! See here.

    There are also 2 comments on Section 10.77: Commutative Algebra.

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