Then $\mathcal{F}$ is of finite presentation.
Proof.
The problem is local on $X$ and $S$, hence we may assume $X$ and $S$ affine. Write $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$. Let $N$ be a finite $B$-module such that $\mathcal{F}$ is the quasi-coherent sheaf associated to $N$. We have $U = D(f_1) \cup \ldots \cup D(f_ n)$ for some $f_ i \in A$, see Algebra, Lemma 10.29.1. As $U$ is schematically dense the map $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective. Pick a prime $\mathfrak q \subset B$ lying over $\mathfrak p \subset A$ corresponding to $x \in X$ mapping to $s \in S$. By Lemma 38.10.9 the module $N_\mathfrak q$ is of finite presentation over $B_\mathfrak q$. Choose a surjection $\varphi : B^{\oplus m} \to N$ of $B$-modules. Choose $k_1, \ldots , k_ t \in \mathop{\mathrm{Ker}}(\varphi )$ and set $N' = B^{\oplus m}/\sum Bk_ j$. There is a canonical surjection $N' \to N$ and $N$ is the filtered colimit of the $B$-modules $N'$ constructed in this manner. Thus we see that we can choose $k_1, \ldots , k_ t$ such that (a) $N'_{f_ i} \cong N_{f_ i}$, $i = 1, \ldots , n$ and (b) $N'_\mathfrak q \cong N_\mathfrak q$. This in particular implies that $N'_\mathfrak q$ is flat over $A$. By openness of flatness, see Algebra, Theorem 10.129.4 we conclude that there exists a $g \in B$, $g \not\in \mathfrak q$ such that $N'_ g$ is flat over $A$. Consider the commutative diagram
\[ \xymatrix{ N'_ g \ar[r] \ar[d] & N_ g \ar[d] \\ \prod N'_{gf_ i} \ar[r] & \prod N_{gf_ i} } \]
The bottom arrow is an isomorphism by choice of $k_1, \ldots , k_ t$. The left vertical arrow is an injective map as $A \to \prod A_{f_ i}$ is injective and $N'_ g$ is flat over $A$. Hence the top horizontal arrow is injective, hence an isomorphism. This proves that $N_ g$ is of finite presentation over $B_ g$. We conclude by applying Algebra, Lemma 10.23.2.
$\square$
Comments (1)
Comment #846 by Jason Starr on