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10.77. Finite projective modules

Definition 10.77.1. Let $R$ be a ring and $M$ an $R$-module.

  1. We say that $M$ is locally free if we can cover $\mathop{\mathrm{Spec}}(R)$ by standard opens $D(f_i)$, $i \in I$ such that $M_{f_i}$ is a free $R_{f_i}$-module for all $i \in I$.
  2. We say that $M$ is finite locally free if we can choose the covering such that each $M_{f_i}$ is finite free.
  3. We say that $M$ is finite locally free of rank $r$ if we can choose the covering such that each $M_{f_i}$ is isomorphic to $R_{f_i}^{\oplus r}$.

Note that a finite locally free $R$-module is automatically finitely presented by Lemma 10.23.2.

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 10.23.2, 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 $\mathop{\mathrm{Hom}}\nolimits_R(M, -)$ 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 10.19.1 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.) $\square$

Remark 10.77.3. It is not true that a finite $R$-module which is $R$-flat is automatically projective. A counter example is where $R = \mathcal{C}^\infty(\mathbf{R})$ is the ring of infinitely differentiable functions on $\mathbf{R}$, and $M = R_{\mathfrak m} = R/I$ where $\mathfrak m = \{f \in R \mid f(0) = 0\}$ and $I = \{f \in R \mid \exists \epsilon, \epsilon > 0 : f(x) = 0 \forall x, |x| < \epsilon\}$.

Lemma 10.77.4. (Warning: see Remark 10.77.3.) Suppose $R$ is a local ring, and $M$ is a finite flat $R$-module. Then $M$ is finite free.

Proof. Follows from the equational criterion of flatness, see Lemma 10.38.11. Namely, suppose that $x_1, \ldots, x_r \in M$ map to a basis of $M/\mathfrak mM$. By Nakayama's Lemma 10.19.1 these elements generate $M$. We want to show there is no relation among the $x_i$. Instead, we will show by induction on $n$ that if $x_1, \ldots, x_n \in M$ are linearly independent in the vector space $M/\mathfrak mM$ then they are independent over $R$.

The base case of the induction is where we have $x \in M$, $x \not\in \mathfrak mM$ and a relation $fx = 0$. By the equational criterion there exist $y_j \in M$ and $a_j \in R$ such that $x = \sum a_j y_j$ and $fa_j = 0$ for all $j$. Since $x \not\in \mathfrak mM$ we see that at least one $a_j$ is a unit and hence $f = 0 $.

Suppose that $\sum f_i x_i$ is a relation among $x_1, \ldots, x_n$. By our choice of $x_i$ we have $f_i \in \mathfrak m$. According to the equational criterion of flatness there exist $a_{ij} \in R$ and $y_j \in M$ such that $x_i = \sum a_{ij} y_j$ and $\sum f_i a_{ij} = 0$. Since $x_n \not \in \mathfrak mM$ we see that $a_{nj}\not\in \mathfrak m$ for at least one $j$. Since $\sum f_i a_{ij} = 0$ we get $f_n = \sum_{i = 1}^{n-1} (-a_{ij}/a_{nj}) f_i$. The relation $\sum f_i x_i = 0$ now can be rewritten as $\sum_{i = 1}^{n-1} f_i( x_i + (-a_{ij}/a_{nj}) x_n) = 0$. Note that the elements $x_i + (-a_{ij}/a_{nj}) x_n$ map to $n-1$ linearly independent elements of $M/\mathfrak mM$. By induction assumption we get that all the $f_i$, $i \leq n-1$ have to be zero, and also $f_n = \sum_{i = 1}^{n-1} (-a_{ij}/a_{nj}) f_i$. This proves the induction step. $\square$

Lemma 10.77.5. Let $R \to S$ be a flat local homomorphism of local rings. Let $M$ be a finite $R$-module. Then $M$ is finite projective over $R$ if and only if $M \otimes_R S$ is finite projective over $S$.

Proof. By Lemma 10.77.2 being finite projective over a local ring is the same thing as being finite free. Suppose that $M \otimes_R S$ is a finite free $S$-module. Pick $x_1, \ldots, x_r \in M$ whose images in $M/\mathfrak m_RM$ form a basis over $\kappa(\mathfrak m)$. Then we see that $x_1 \otimes 1, \ldots, x_r \otimes 1$ are a basis for $M \otimes_R S$. This implies that the map $R^{\oplus r} \to M, (a_i) \mapsto \sum a_i x_i$ becomes an isomorphism after tensoring with $S$. By faithful flatness of $R \to S$, see Lemma 10.38.17 we see that it is an isomorphism. $\square$

Lemma 10.77.6. Let $R$ be a semi-local ring. Let $M$ be a finite locally free module. If $M$ has constant rank, then $M$ is free. In particular, if $R$ has connected spectrum, then $M$ is free.

Proof. Omitted. Hints: First show that $M/\mathfrak m_iM$ has the same dimension $d$ for all maximal ideal $\mathfrak m_1, \ldots, \mathfrak m_n$ of $R$ using the rank is constant. Next, show that there exist elements $x_1, \ldots, x_d \in M$ which form a basis for each $M/\mathfrak m_iM$ by the Chinese remainder theorem. Finally show that $x_1, \ldots, x_d$ is a basis for $M$. $\square$

Here is a technical lemma that is used in the chapter on groupoids.

Lemma 10.77.7. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and infinite residue field. Let $R \to S$ be a ring map. Let $M$ be an $S$-module and let $N \subset M$ be an $R$-submodule. Assume

  1. $S$ is semi-local and $\mathfrak mS$ is contained in the radical of $S$,
  2. $M$ is a finite free $S$-module, and
  3. $N$ generates $M$ as an $S$-module.

Then $N$ contains an $S$-basis of $M$.

Proof. Assume $M$ is free of rank $n$. Let $I = \text{rad}(S)$. By Nakayama's Lemma 10.19.1 a sequence of elements $m_1, \ldots, m_n$ is a basis for $M$ if and only if $\overline{m}_i \in M/IM$ generate $M/IM$. Hence we may replace $M$ by $M/IM$, $N$ by $N/(N \cap IM)$, $R$ by $R/\mathfrak m$, and $S$ by $S/IS$. In this case we see that $S$ is a finite product of fields $S = k_1 \times \ldots \times k_r$ and $M = k_1^{\oplus n} \times \ldots \times k_r^{\oplus n}$. The fact that $N \subset M$ generates $M$ as an $S$-module means that there exist $x_j \in N$ such that a linear combination $\sum a_j x_j$ with $a_j \in S$ has a nonzero component in each factor $k_i^{\oplus n}$. Because $R = k$ is an infinite field, this means that also some linear combination $y = \sum c_j x_j$ with $c_j \in k$ has a nonzero component in each factor. Hence $y \in N$ generates a free direct summand $Sy \subset M$. By induction on $n$ the result holds for $M/Sy$ and the submodule $\overline{N} = N/(N \cap Sy)$. In other words there exist $\overline{y}_2, \ldots, \overline{y}_n$ in $\overline{N}$ which (freely) generate $M/Sy$. Then $y, y_2, \ldots, y_n$ (freely) generate $M$ and we win. $\square$

Lemma 10.77.8. Let $R$ be ring. Let $L$, $M$, $N$ be $R$-modules. The canonical map $$ \mathop{\mathrm{Hom}}\nolimits_R(M, N) \otimes_R L \to \mathop{\mathrm{Hom}}\nolimits_R(M, N \otimes_R L) $$ is an isomorphism if $M$ is finite projective.

Proof. By Lemma 10.77.2 we see that $M$ is finitely presented as well as finite locally free. By Lemmas 10.10.2 and 10.11.16 formation of the left and right hand side of the arrow commutes with localization. We may check that our map is an isomorphism after localization, see Lemma 10.23.2. Thus we may assume $M$ is finite free. In this case the lemma is immediate. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 18118–18445 (see updates for more information).

    \section{Finite projective modules}
    \label{section-finite-projective-modules}
    
    \begin{definition}
    \label{definition-locally-free}
    Let $R$ be a ring and $M$ an $R$-module.
    \begin{enumerate}
    \item We say that $M$ is {\it locally free} if we can cover $\Spec(R)$ by
    standard opens $D(f_i)$, $i \in I$ such that $M_{f_i}$ is a free
    $R_{f_i}$-module for all $i \in I$.
    \item We say that $M$ is {\it finite locally free} if we can choose
    the covering such that each $M_{f_i}$ is finite free.
    \item We say that $M$ is {\it finite locally free of rank $r$}
    if we can choose the covering such that each $M_{f_i}$ is isomorphic
    to $R_{f_i}^{\oplus r}$.
    \end{enumerate}
    \end{definition}
    
    \noindent
    Note that a finite locally free $R$-module is
    automatically finitely presented by
    Lemma \ref{lemma-cover}.
    
    \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}
    
    \begin{remark}
    \label{remark-warning}
    It is not true that a finite $R$-module which is
    $R$-flat is automatically projective. A counter
    example is where $R = \mathcal{C}^\infty(\mathbf{R})$
    is the ring of infinitely differentiable functions on
    $\mathbf{R}$, and $M = R_{\mathfrak m} = R/I$ where
    $\mathfrak m = \{f \in R \mid f(0) = 0\}$ and
    $I = \{f \in R \mid \exists \epsilon, \epsilon > 0 :
    f(x) = 0\ \forall x, |x| < \epsilon\}$.
    \end{remark}
    
    \begin{lemma}
    \label{lemma-finite-flat-local}
    (Warning: see Remark \ref{remark-warning}.)
    Suppose $R$ is a local ring, and $M$ is a finite
    flat $R$-module. Then $M$ is finite free.
    \end{lemma}
    
    \begin{proof}
    Follows from the equational criterion of flatness, see
    Lemma \ref{lemma-flat-eq}. Namely, suppose that
    $x_1, \ldots, x_r \in M$ map to a basis of
    $M/\mathfrak mM$. By Nakayama's Lemma \ref{lemma-NAK}
    these elements generate $M$. We want to show there
    is no relation among the $x_i$. Instead, we will show
    by induction on $n$ that if $x_1, \ldots, x_n \in M$
    are linearly independent in the vector space
    $M/\mathfrak mM$ then they are independent over $R$.
    
    \medskip\noindent
    The base case of the induction is where we have
    $x \in M$, $x \not\in \mathfrak mM$ and a relation
    $fx = 0$. By the equational criterion there
    exist $y_j \in M$ and $a_j \in R$ such that
    $x = \sum a_j y_j$ and $fa_j = 0$ for all $j$.
    Since $x \not\in \mathfrak mM$ we see that
    at least one $a_j$ is a unit and hence $f = 0 $.
    
    \medskip\noindent
    Suppose that $\sum f_i x_i$ is a relation among $x_1, \ldots, x_n$.
    By our choice of $x_i$ we have $f_i \in \mathfrak m$.
    According to the equational criterion of flatness there exist
    $a_{ij} \in R$ and $y_j \in M$ such that
    $x_i = \sum a_{ij} y_j$ and $\sum f_i a_{ij} = 0$.
    Since $x_n \not \in \mathfrak mM$ we see that
    $a_{nj}\not\in \mathfrak m$ for at least one $j$.
    Since $\sum f_i a_{ij} = 0$ we get
    $f_n = \sum_{i = 1}^{n-1} (-a_{ij}/a_{nj}) f_i$.
    The relation $\sum f_i x_i = 0$ now can be rewritten
    as $\sum_{i = 1}^{n-1} f_i( x_i + (-a_{ij}/a_{nj}) x_n) = 0$.
    Note that the elements $x_i + (-a_{ij}/a_{nj}) x_n$ map
    to $n-1$ linearly independent elements of $M/\mathfrak mM$.
    By induction assumption we get that all the $f_i$, $i \leq n-1$
    have to be zero, and also $f_n = \sum_{i = 1}^{n-1} (-a_{ij}/a_{nj}) f_i$.
    This proves the induction step.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-finite-projective-descends}
    Let $R \to S$ be a flat local homomorphism of local rings.
    Let $M$ be a finite $R$-module. Then $M$ is finite projective
    over $R$ if and only if $M \otimes_R S$ is finite projective
    over $S$.
    \end{lemma}
    
    \begin{proof}
    By Lemma \ref{lemma-finite-projective} being finite projective
    over a local ring is the same thing as being finite free.
    Suppose that $M \otimes_R S$ is a finite free $S$-module.
    Pick $x_1, \ldots, x_r \in M$ whose images in $M/\mathfrak m_RM$
    form a basis over $\kappa(\mathfrak m)$. Then
    we see that $x_1 \otimes 1, \ldots, x_r \otimes 1$
    are a basis for $M \otimes_R S$. This implies that
    the map $R^{\oplus r} \to M, (a_i) \mapsto \sum a_i x_i$
    becomes an isomorphism after tensoring with $S$.
    By faithful flatness of $R \to S$, see Lemma \ref{lemma-local-flat-ff}
    we see that it is an isomorphism.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-locally-free-semi-local-free}
    Let $R$ be a semi-local ring.
    Let $M$ be a finite locally free module.
    If $M$ has constant rank, then $M$ is free. In particular, if $R$ has
    connected spectrum, then $M$ is free.
    \end{lemma}
    
    \begin{proof}
    Omitted. Hints: First show that $M/\mathfrak m_iM$ has the
    same dimension $d$ for all maximal ideal $\mathfrak m_1, \ldots, \mathfrak m_n$
    of $R$ using the rank is constant.
    Next, show that there exist elements $x_1, \ldots, x_d \in M$
    which form a basis for each $M/\mathfrak m_iM$ by the Chinese
    remainder theorem. Finally show that $x_1, \ldots, x_d$ is a basis for $M$.
    \end{proof}
    
    \noindent
    Here is a technical lemma that is used in the chapter on groupoids.
    
    \begin{lemma}
    \label{lemma-semi-local-module-basis-in-submodule}
    Let $R$ be a local ring with maximal ideal $\mathfrak m$ and
    infinite residue field.
    Let $R \to S$ be a ring map.
    Let $M$ be an $S$-module and let $N \subset M$ be an $R$-submodule.
    Assume
    \begin{enumerate}
    \item $S$ is semi-local and $\mathfrak mS$ is contained in the radical
    of $S$,
    \item $M$ is a finite free $S$-module, and
    \item $N$ generates $M$ as an $S$-module.
    \end{enumerate}
    Then $N$ contains an $S$-basis of $M$.
    \end{lemma}
    
    \begin{proof}
    Assume $M$ is free of rank $n$. Let $I = \text{rad}(S)$.
    By Nakayama's Lemma \ref{lemma-NAK} a sequence of elements
    $m_1, \ldots, m_n$ is a basis for $M$ if and only if
    $\overline{m}_i \in M/IM$ generate $M/IM$. Hence we may replace
    $M$ by $M/IM$, $N$ by $N/(N \cap IM)$, $R$ by $R/\mathfrak m$,
    and $S$ by $S/IS$. In this case we see that $S$ is a finite product
    of fields $S = k_1 \times \ldots \times k_r$ and
    $M = k_1^{\oplus n} \times \ldots \times k_r^{\oplus n}$.
    The fact that $N \subset M$ generates $M$ as an $S$-module
    means that there exist $x_j \in N$ such that a linear combination
    $\sum a_j x_j$ with $a_j \in S$ has a nonzero component in each
    factor $k_i^{\oplus n}$.
    Because $R = k$ is an infinite field, this means that also
    some linear combination $y = \sum c_j x_j$ with $c_j \in k$ has a
    nonzero component in each factor. Hence $y \in N$ generates a
    free direct summand $Sy \subset M$. By induction on $n$ the result
    holds for $M/Sy$ and the submodule $\overline{N} = N/(N \cap Sy)$.
    In other words there exist $\overline{y}_2, \ldots, \overline{y}_n$
    in $\overline{N}$ which (freely) generate $M/Sy$. Then
    $y, y_2, \ldots, y_n$ (freely) generate $M$ and we win.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-evaluation-map-iso-finite-projective}
    Let $R$ be ring. Let $L$, $M$, $N$ be $R$-modules.
    The canonical map
    $$
    \Hom_R(M, N) \otimes_R L \to \Hom_R(M, N \otimes_R L)
    $$
    is an isomorphism if $M$ is finite projective.
    \end{lemma}
    
    \begin{proof}
    By Lemma \ref{lemma-finite-projective} we see that $M$
    is finitely presented as well as finite locally free.
    By Lemmas \ref{lemma-hom-from-finitely-presented} and
    \ref{lemma-tensor-product-localization} formation of
    the left and right hand side of the arrow commutes with
    localization. We may check that our map is an isomorphism
    after localization, see Lemma \ref{lemma-cover}.
    Thus we may assume $M$ is finite free. In this case
    the lemma is immediate.
    \end{proof}

    Comments (2)

    Comment #2296 by Dario WeiƟmann on November 5, 2016 a 9:43 pm UTC

    In Lemma 10.77.5 we apply Lemma 10.77.2 to get that the module is finite flat. Then we should apply Lemma 10.77.4 to see that it is finite free (this reference is missing). And I think the $x_1,\dots,x_r$ should be chosen such that their residue classes form a basis of $M/\mathfrak{m}_R$ and not just generate it.

    In the hints of Lemma 10.77.6 one should use the rank is constant instead of the spectrum is connected.

    Comment #2322 by Johan (site) on December 11, 2016 a 10:08 pm UTC

    Thanks, fixed here.

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