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Tag 052U

Chapter 10: Commutative Algebra > Section 10.107: Pure ideals

Lemma 10.107.6. Let $R$ be a ring. The following are equivalent:

  1. every $Z \subset \mathop{\rm Spec}(R)$ which is closed and closed under generalizations is also open, and
  2. any finite flat $R$-module is finite locally free.

Proof. If any finite flat $R$-module is finite locally free then the support of $R/I$ where $I$ is a pure ideal is open. Hence the implication (2) $\Rightarrow$ (1) follows from Lemma 10.107.3.

For the converse assume that $R$ satisfies (1). Let $M$ be a finite flat $R$-module. The support $Z = \text{Supp}(M)$ of $M$ is closed, see Lemma 10.39.5. On the other hand, if $\mathfrak p \subset \mathfrak p'$, then by Lemma 10.77.4 the module $M_{\mathfrak p'}$ is free, and $M_{\mathfrak p} = M_{\mathfrak p'} \otimes_{R_{\mathfrak p'}} R_{\mathfrak p}$ Hence $\mathfrak p' \in \text{Supp}(M) \Rightarrow \mathfrak p \in \text{Supp}(M)$, in other words, the support is closed under generalization. As $R$ satisfies (1) we see that the support of $M$ is open and closed. Suppose that $M$ is generated by $r$ elements $m_1, \ldots, m_r$. The modules $\wedge^i(M)$, $i = 1, \ldots, r$ are finite flat $R$-modules also, because $\wedge^i(M)_{\mathfrak p} = \wedge^i(M_{\mathfrak p})$ is free over $R_{\mathfrak p}$. Note that $\text{Supp}(\wedge^{i + 1}(M)) \subset \text{Supp}(\wedge^i(M))$. Thus we see that there exists a decomposition $$ \mathop{\rm Spec}(R) = U_0 \amalg U_1 \amalg \ldots \amalg U_r $$ by open and closed subsets such that the support of $\wedge^i(M)$ is $U_r \cup \ldots \cup U_i$ for all $i = 0, \ldots, r$. Let $\mathfrak p$ be a prime of $R$, and say $\mathfrak p \in U_i$. Note that $\wedge^i(M) \otimes_R \kappa(\mathfrak p) = \wedge^i(M \otimes_R \kappa(\mathfrak p))$. Hence, after possibly renumbering $m_1, \ldots, m_r$ we may assume that $m_1, \ldots, m_i$ generate $M \otimes_R \kappa(\mathfrak p)$. By Nakayama's Lemma 10.19.1 we get a surjection $$ R_f^{\oplus i} \longrightarrow M_f, \quad (a_1, \ldots, a_i) \longmapsto \sum a_im_i $$ for some $f \in R$, $f \not \in \mathfrak p$. We may also assume that $D(f) \subset U_i$. This means that $\wedge^i(M_f) = \wedge^i(M)_f$ is a flat $R_f$ module whose support is all of $\mathop{\rm Spec}(R_f)$. By the above it is generated by a single element, namely $m_1 \wedge \ldots \wedge m_i$. Hence $\wedge^i(M)_f \cong R_f/J$ for some pure ideal $J \subset R_f$ with $V(J) = \mathop{\rm Spec}(R_f)$. Clearly this means that $J = (0)$, see Lemma 10.107.3. Thus $m_1 \wedge \ldots \wedge m_i$ is a basis for $\wedge^i(M_f)$ and it follows that the displayed map is injective as well as surjective. This proves that $M$ is finite locally free as desired. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 25045–25053 (see updates for more information).

    \begin{lemma}
    \label{lemma-finite-flat-module-finitely-presented}
    Let $R$ be a ring. The following are equivalent:
    \begin{enumerate}
    \item every $Z \subset \Spec(R)$ which is closed and closed under
    generalizations is also open, and
    \item any finite flat $R$-module is finite locally free.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    If any finite flat $R$-module is finite locally free then the support
    of $R/I$ where $I$ is a pure ideal is open. Hence the implication
    (2) $\Rightarrow$ (1) follows from
    Lemma \ref{lemma-pure-ideal-determined-by-zero-set}.
    
    \medskip\noindent
    For the converse assume that $R$ satisfies (1).
    Let $M$ be a finite flat $R$-module.
    The support $Z = \text{Supp}(M)$ of $M$ is closed, see
    Lemma \ref{lemma-support-closed}.
    On the other hand, if $\mathfrak p \subset \mathfrak p'$, then by
    Lemma \ref{lemma-finite-flat-local}
    the module $M_{\mathfrak p'}$ is free, and
    $M_{\mathfrak p} = M_{\mathfrak p'} \otimes_{R_{\mathfrak p'}} R_{\mathfrak p}$
    Hence
    $\mathfrak p' \in \text{Supp}(M) \Rightarrow \mathfrak p \in \text{Supp}(M)$,
    in other words, the support is closed under generalization.
    As $R$ satisfies (1) we see that the support of $M$ is open and closed.
    Suppose that $M$ is generated by $r$ elements $m_1, \ldots, m_r$.
    The modules $\wedge^i(M)$, $i = 1, \ldots, r$ are finite flat $R$-modules
    also, because $\wedge^i(M)_{\mathfrak p} = \wedge^i(M_{\mathfrak p})$
    is free over $R_{\mathfrak p}$. Note that
    $\text{Supp}(\wedge^{i + 1}(M)) \subset \text{Supp}(\wedge^i(M))$.
    Thus we see that there exists a decomposition
    $$
    \Spec(R) = U_0 \amalg U_1 \amalg \ldots \amalg U_r
    $$
    by open and closed subsets such that the support of
    $\wedge^i(M)$ is $U_r \cup \ldots \cup U_i$ for all $i = 0, \ldots, r$.
    Let $\mathfrak p$ be a prime of $R$, and say $\mathfrak p \in U_i$.
    Note that
    $\wedge^i(M) \otimes_R \kappa(\mathfrak p) =
    \wedge^i(M \otimes_R \kappa(\mathfrak p))$.
    Hence, after possibly renumbering $m_1, \ldots, m_r$ we may assume that
    $m_1, \ldots, m_i$ generate $M \otimes_R \kappa(\mathfrak p)$. By
    Nakayama's Lemma \ref{lemma-NAK}
    we get a surjection
    $$
    R_f^{\oplus i} \longrightarrow M_f, \quad
    (a_1, \ldots, a_i) \longmapsto \sum a_im_i
    $$
    for some $f \in R$, $f \not \in \mathfrak p$. We may also assume that
    $D(f) \subset U_i$. This means that $\wedge^i(M_f) = \wedge^i(M)_f$
    is a flat $R_f$ module whose support is all of $\Spec(R_f)$.
    By the above it is generated by a single element, namely
    $m_1 \wedge \ldots \wedge m_i$. Hence $\wedge^i(M)_f \cong R_f/J$
    for some pure ideal $J \subset R_f$ with $V(J) = \Spec(R_f)$.
    Clearly this means that $J = (0)$, see
    Lemma \ref{lemma-pure-ideal-determined-by-zero-set}.
    Thus $m_1 \wedge \ldots \wedge m_i$ is a basis for
    $\wedge^i(M_f)$ and it follows that the displayed map is injective
    as well as surjective. This proves that $M$ is finite locally free
    as desired.
    \end{proof}

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