The Stacks project

Lemma 38.27.1. In Situation 38.20.7. For each $p \geq 0$ the functor $H_ p$ (38.20.7.2) 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 37.55.1 , 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 (38.20.7.2) 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 $F_{iso}$ (38.20.1.1) 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. $\square$


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