The Stacks project

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

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. $\square$

Proof. The question is local on $X$ and $S$, hence we may assume $X$ and $S$ affine. Choose a closed immersion $i : X \to \mathbf{A}^ n_ S$ and apply Lemma 38.11.1 to $i_*\mathcal{O}_ X$. Some details omitted. $\square$

Proof. Proof in case (1). Let $W \subset X$ be the open subscheme constructed and studied in More on Morphisms, Lemmas 37.22.7 and 37.22.9. Note that every generic point of every fibre is contained in $W$, hence it suffices to prove the result for $W$. Since $W = \bigcup _{d \geq 0} U_ d$, it suffices to prove that $U_ d = \emptyset $ for $d > e$. Since $f$ is flat and locally of finite presentation it is open hence $f(U_ d)$ is open (Morphisms, Lemma 29.25.10). Thus if $U_ d$ is not empty, then $f(U_ d) \cap U \not= \emptyset $ as desired.

Proof in case (2). We may replace $S$ by its reduction. Then $U$ is scheme theoretically dense. Hence $f$ is locally of finite presentation by Lemma 38.11.2. In this way we reduce to case (1). $\square$

Proof. By Lemma 38.11.3 the fibres of $f$ have dimension zero. Hence $f$ is quasi-finite (Morphisms, Lemma 29.29.5) whence has finite fibres (Morphisms, Lemma 29.20.10). Hence $f$ is finite by More on Morphisms, Lemma 37.44.1. $\square$

Proof. By Lemma 38.11.2 the morphism $f$ is locally of finite presentation. The image $f(X) \subset S$ is open (Morphisms, Lemma 29.25.10) hence we may replace $S$ by $f(X)$. Thus we have to prove that $f$ is an isomorphism. We may assume $S$ is affine. We can reduce to the case that $X$ is quasi-compact because it suffices to show that any quasi-compact open $X' \subset X$ whose image is $S$ maps isomorphically to $S$. Thus we may assume $f$ is quasi-compact. All the fibers of $f$ have dimension $0$, see Lemma 38.11.3. Hence $f$ is quasi-finite, see Morphisms, Lemma 29.29.5. Let $s \in S$. Choose an elementary étale neighbourhood $g : (T, t) \to (S, s)$ such that $X \times _ S T = V \amalg W$ with $V \to T$ finite and $W_ t = \emptyset $, see More on Morphisms, Lemma 37.41.6. Denote $\pi : V \amalg W \to T$ the given morphism. Since $\pi $ is flat and locally of finite presentation, we see that $\pi (V)$ is open in $T$ (Morphisms, Lemma 29.25.10). After shrinking $T$ we may assume that $T = \pi (V)$. Since $f$ is an isomorphism over $U$ we see that $\pi $ is an isomorphism over $g^{-1}U$. Since $\pi (V) = T$ this implies that $\pi ^{-1}g^{-1}U$ is contained in $V$. By Morphisms, Lemma 29.25.15 we see that $\pi ^{-1}g^{-1}U \subset V \amalg W$ is scheme theoretically dense. Hence we deduce that $W = \emptyset $. Thus $X \times _ S T = V$ is finite over $T$. This implies that $f$ is finite (after replacing $S$ by an open neighbourhood of $s$), for example by Descent, Lemma 35.23.23. Then $f$ is finite locally free (Morphisms, Lemma 29.48.2) and after shrinking $S$ to a smaller open neighbourhood of $s$ we see that $f$ is finite locally free of some degree $d$ (Morphisms, Lemma 29.48.5). But $d = 1$ as is clear from the fact that the degree is $1$ over the dense open $U$. Hence $f$ is an isomorphism. $\square$