The Stacks project

Proof. First we reduce (1) to (2). Namely, let $Z \subset X$ be the scheme theoretic support of $\mathcal{F}$. Let $i : Z \to X$ be the corresponding closed immersion and write $\mathcal{F} = i_*\mathcal{G}$ for some finite type quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$, see Morphisms, Section 29.5. In case (1) $Z \to S$ is proper by assumption. Thus by Lemma 38.16.7 case (1) reduces to case (2).

Assume $f$ is proper. Let $(g : T \to S, t' \leadsto t, \xi )$ be an impurity of $\mathcal{F}$ above $s \in S$. Since $f$ is proper, it is universally closed. Hence $f_ T : X_ T \to T$ is closed. Since $f_ T(\xi ) = t'$ this implies that $t \in f(\overline{\{ \xi \} })$ which is a contradiction. $\square$

Proof. Let $Z \subset X$ be the scheme theoretic support of $\mathcal{F}$, see Morphisms, Definition 29.5.5. Then $Z \to S$ is a separated, finite type morphism of schemes with finite fibres. Hence it is separated and quasi-finite, see Morphisms, Lemma 29.20.10. By Lemma 38.16.7 it suffices to prove the lemma for $Z \to S$ and the sheaf $\mathcal{F}$ viewed as a finite type quasi-coherent module on $Z$. Hence we may assume that $X \to S$ is separated and quasi-finite and that $\text{Supp}(\mathcal{F}) = X$.

It follows from Lemma 38.17.1 and Morphisms, Lemma 29.44.11 that (2) implies (3). Trivially (3) implies (1). Assume (1) holds. We will prove that (2) holds. It is clear that we may assume $S$ is affine. By More on Morphisms, Lemma 37.43.3 we can find a diagram

with $\pi $ finite and $j$ a quasi-compact open immersion. If we show that $j$ is closed, then $j$ is a closed immersion and we conclude that $f = \pi \circ j$ is finite. To show that $j$ is closed it suffices to show that specializations lift along $j$, see Schemes, Lemma 26.19.8. Let $x \in X$, set $t' = j(x)$ and let $t' \leadsto t$ be a specialization. We have to show $t \in j(X)$. Set $s' = f(x)$ and $s = \pi (t)$ so $s' \leadsto s$. By More on Morphisms, Lemma 37.41.4 we can find an elementary étale neighbourhood $(U, u) \to (S, s)$ and a decomposition

into open and closed subschemes, such that $V \to U$ is finite and there exists a unique point $v$ of $V$ mapping to $u$, and such that $v$ maps to $t$ in $T$. As $V \to T$ is étale, we can lift generalizations, see Morphisms, Lemmas 29.25.9 and 29.36.12. Hence there exists a specialization $v' \leadsto v$ such that $v'$ maps to $t' \in T$. In particular we see that $v' \in X_ U \subset T_ U$. Denote $u' \in U$ the image of $t'$. Note that $v' \in \text{Ass}_{X_ U/U}(\mathcal{F})$ because $X_{u'}$ is a finite discrete set and $X_{u'} = \text{Supp}(\mathcal{F}_{u'})$. As $\mathcal{F}$ is pure relative to $S$ we see that $v'$ must specialize to a point in $X_ u$. Since $v$ is the only point of $V$ lying over $u$ (and since no point of $W$ can be a specialization of $v'$) we see that $v \in X_ u$. Hence $t \in X$. $\square$

Proof. Let $\xi \in X$ with $s' = f(\xi )$ and $s' \leadsto s$ a specialization of $S$. If $\xi $ is an associated point of $X_{s'}$, then $\xi $ is the unique generic point because $X_{s'}$ is an integral scheme. Let $\xi _0$ be the unique generic point of $X_ s$. As $X \to S$ is flat we can lift $s' \leadsto s$ to a specialization $\xi ' \leadsto \xi _0$ in $X$, see Morphisms, Lemma 29.25.9. The $\xi \leadsto \xi '$ because $\xi $ is the generic point of $X_{s'}$ hence $\xi \leadsto \xi _0$. This means that $(\text{id}_ S, s' \to s, \xi )$ is not an impurity of $\mathcal{O}_ X$ above $s$. Since the assumption that $f$ is finite type, flat with geometrically integral fibres is preserved under base change, we see that there doesn't exist an impurity after any base change. In this way we see that $X$ is universally $S$-pure. $\square$

Proof. After reducing to the case where $S$ is the spectrum of a henselian local ring this follows from Lemma 38.14.1. $\square$