The Stacks project

Lemma 38.27.1. In Situation 38.20.7. For each $p \geq 0$ the functor $H_ p$ ( is representable by a locally closed immersion $S_ p \to S$. If $\mathcal{F}$ is of finite presentation, then $S_ p \to S$ is of finite presentation.

Proof. For each $S$ we will prove the statement for all $p \geq 0$ concurrently. The functor $H_ p$ is a sheaf for the fppf topology by Lemma 38.20.8. Hence combining Descent, Lemma 35.39.1, More on Morphisms, Lemma 37.57.1 , and Descent, Lemma 35.24.1 we see that the question is local for the ├ętale topology on $S$. In particular, the question is Zariski local on $S$.

For $s \in S$ denote $\xi _ s$ the unique generic point of the fibre $X_ s$. Note that for every $s \in S$ the restriction $\mathcal{F}_ s$ of $\mathcal{F}$ is locally free of some rank $p(s) \geq 0$ in some neighbourhood of $\xi _ s$. (As $X_ s$ is irreducible and smooth this follows from generic flatness for $\mathcal{F}_ s$ over $X_ s$, see Algebra, Lemma 10.118.1 although this is overkill.) For future reference we note that

\[ p(s) = \dim _{\kappa (\xi _ s)}( \mathcal{F}_{\xi _ s} \otimes _{\mathcal{O}_{X, \xi _ s}} \kappa (\xi _ s) ). \]

In particular $H_{p(s)}(s)$ is nonempty and $H_ q(s)$ is empty if $q \not= p(s)$.

Let $U \subset X$ be an open subscheme. As $f : X \to S$ is smooth, it is open. It is immediate from ( that the functor $H_ p$ for the pair $(f|_ U : U \to f(U), \mathcal{F}|_ U)$ and the functor $H_ p$ for the pair $(f|_{f^{-1}(f(U))}, \mathcal{F}|_{f^{-1}(f(U))})$ are the same. Hence to prove the existence of $S_ p$ over $f(U)$ we may always replace $X$ by $U$.

Pick $s \in S$. There exists an affine open neighbourhood $U$ of $\xi _ s$ such that $\mathcal{F}|_ U$ can be generated by at most $p(s)$ elements. By the arguments above we see that in order to prove the statement for $H_{p(s)}$ in an neighbourhood of $s$ we may assume that $\mathcal{F}$ is generated by $p(s)$ elements, i.e., that there exists a surjection

\[ u : \mathcal{O}_ X^{\oplus p(s)} \longrightarrow \mathcal{F} \]

In this case it is clear that $H_{p(s)}$ is equal to $F_{iso}$ ( for the map $u$ (this follows immediately from Lemma 38.19.1 but also from Lemma 38.12.1 after shrinking a bit more so that both $S$ and $X$ are affine.) Thus we may apply Theorem 38.23.3 to see that $H_{p(s)}$ is representable by a closed immersion in a neighbourhood of $s$.

The result follows formally from the above. Namely, the arguments above show that locally on $S$ the function $s \mapsto p(s)$ is bounded. Hence we may use induction on $p = \max _{s \in S} p(s)$. The functor $H_ p$ is representable by a closed immersion $S_ p \to S$ by the above. Replace $S$ by $S \setminus S_ p$ which drops the maximum by at least one and we win by induction hypothesis.

Assume $\mathcal{F}$ is of finite presentation. Then $S_ p \to S$ is locally of finite presentation by Lemma 38.20.8 part (2) combined with Limits, Remark 32.6.2. Then we redo the induction argument in the paragraph to see that each $S_ p$ is quasi-compact when $S$ is affine: first if $p = \max _{s \in S} p(s)$, then $S_ p \subset S$ is closed (see above) hence quasi-compact. Then $U = S \setminus S_ p$ is quasi-compact open in $S$ because $S_ p \to S$ is a closed immersion of finite presentation (see discussion in Morphisms, Section 29.22 for example). Then $S_{p - 1} \to U$ is a closed immersion of finite presentation, and so $S_{p - 1}$ is quasi-compact and $U' = S \setminus (S_ p \cup S_{p - 1})$ is quasi-compact. And so on. $\square$

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 05UC. Beware of the difference between the letter 'O' and the digit '0'.