The Stacks project

Lemma 10.108.6. Let $R$ be a ring. The following are equivalent:

  1. every $Z \subset \mathop{\mathrm{Spec}}(R)$ 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 10.108.3.

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 10.40.5. On the other hand, if $\mathfrak p \subset \mathfrak p'$, then by Lemma 10.78.5 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

\[ \mathop{\mathrm{Spec}}(R) = U_0 \amalg U_1 \amalg \ldots \amalg U_ r \]

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 10.20.1 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 $\mathop{\mathrm{Spec}}(R_ f)$. 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 $V(J) = \mathop{\mathrm{Spec}}(R_ f)$. Clearly this means that $J = (0)$, see Lemma 10.108.3. 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. $\square$

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