The Stacks project

Lemma 29.26.4. Let $X$ be a scheme. The following are equivalent

  1. every finite flat quasi-coherent $\mathcal{O}_ X$-module is finite locally free, and

  2. every closed subset $Z \subset X$ which is closed under generalizations is open.

Proof. In the affine case this is Algebra, Lemma 10.108.6. The scheme case does not follow directly from the affine case, so we simply repeat the arguments.

Assume (1). Consider a closed immersion $i : Z \to X$ such that $i$ is flat. Then $i_*\mathcal{O}_ Z$ is quasi-coherent and flat, hence finite locally free by (1). Thus $Z = \text{Supp}(i_*\mathcal{O}_ Z)$ is also open and we see that (2) holds. Hence the implication (1) $\Rightarrow $ (2) follows from the characterization of flat closed immersions in Lemma 29.26.1.

For the converse assume that $X$ satisfies (2). Let $\mathcal{F}$ be a finite flat quasi-coherent $\mathcal{O}_ X$-module. The support $Z = \text{Supp}(\mathcal{F})$ of $\mathcal{F}$ is closed, see Modules, Lemma 17.9.6. On the other hand, if $x \leadsto x'$ is a specialization, then by Algebra, Lemma 10.78.5 the module $\mathcal{F}_{x'}$ is free over $\mathcal{O}_{X, x'}$, and

\[ \mathcal{F}_ x = \mathcal{F}_{x'} \otimes _{\mathcal{O}_{X, x'}} \mathcal{O}_{X, x}. \]

Hence $x' \in \text{Supp}(\mathcal{F}) \Rightarrow x \in \text{Supp}(\mathcal{F})$, in other words, the support is closed under generalization. As $X$ satisfies (2) we see that the support of $\mathcal{F}$ is open and closed. The modules $\wedge ^ i(\mathcal{F})$, $i = 1, 2, 3, \ldots $ are finite flat quasi-coherent $\mathcal{O}_ X$-modules also, see Modules, Section 17.21. Note that $\text{Supp}(\wedge ^{i + 1}(\mathcal{F})) \subset \text{Supp}(\wedge ^ i(\mathcal{F}))$. Thus we see that there exists a decomposition

\[ X = U_0 \amalg U_1 \amalg U_2 \amalg \ldots \]

by open and closed subsets such that the support of $\wedge ^ i(\mathcal{F})$ is $U_ i \cup U_{i + 1} \cup \ldots $ for all $i$. Let $x$ be a point of $X$, and say $x \in U_ r$. Note that $\wedge ^ i(\mathcal{F})_ x \otimes \kappa (x) = \wedge ^ i(\mathcal{F}_ x \otimes \kappa (x))$. Hence, $x \in U_ r$ implies that $\mathcal{F}_ x \otimes \kappa (x)$ is a vector space of dimension $r$. By Nakayama's lemma, see Algebra, Lemma 10.20.1 we can choose an affine open neighbourhood $U \subset U_ r \subset X$ of $x$ and sections $s_1, \ldots , s_ r \in \mathcal{F}(U)$ such that the induced map

\[ \mathcal{O}_ U^{\oplus r} \longrightarrow \mathcal{F}|_ U, \quad (f_1, \ldots , f_ r) \longmapsto \sum f_ i s_ i \]

is surjective. This means that $\wedge ^ r(\mathcal{F}|_ U)$ is a finite flat quasi-coherent $\mathcal{O}_ U$-module whose support is all of $U$. By the above it is generated by a single element, namely $s_1 \wedge \ldots \wedge s_ r$. Hence $\wedge ^ r(\mathcal{F}|_ U) \cong \mathcal{O}_ U/\mathcal{I}$ for some quasi-coherent sheaf of ideals $\mathcal{I}$ such that $\mathcal{O}_ U/\mathcal{I}$ is flat over $\mathcal{O}_ U$ and such that $V(\mathcal{I}) = U$. It follows that $\mathcal{I} = 0$ by applying Lemma 29.26.1. Thus $s_1 \wedge \ldots \wedge s_ r$ is a basis for $\wedge ^ r(\mathcal{F}|_ U)$ and it follows that the displayed map is injective as well as surjective. This proves that $\mathcal{F}$ is finite locally free as desired. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 053N. Beware of the difference between the letter 'O' and the digit '0'.