The Stacks project

Proof. There is an immediate reduction to the case where $S$ is affine. Then $U$ is quasi-compact. Let $U = U_1 \cup \ldots \cup U_ n$ be an affine open covering. We may replace $U$ by $U_1 \amalg \ldots \amalg U_ n$ without changing the assumptions, hence we may assume $U$ is affine. Thus we can find an open immersion $U \to Y$ over $X$ with $Y$ proper over $X$. (First put $U$ inside $\mathbf{A}^ n_ X$ using Morphisms, Lemma 29.39.2 and then take the closure inside $\mathbf{P}^ n_ X$, or you can directly use Morphisms, Lemma 29.43.12.) We can assume $U$ is dense in $Y$ (replace $Y$ by the scheme theoretic closure of $U$ if necessary, see Morphisms, Section 29.7). Note that $g : Y \to X$ is surjective as the image is closed and contains the dense subset $h(U)$. We will show that $Y \to S$ is proper. This will imply that $X \to S$ is proper by Morphisms, Lemma 29.41.9 thereby finishing the proof. To show that $Y \to S$ is proper we will use part (4) of Lemma 32.15.3. To do this consider a diagram

where $A$ is a discrete valuation ring with fraction field $K$ and where $y : \mathop{\mathrm{Spec}}(K) \to Y$ is the inclusion of a generic point. We have to show there exists a unique dotted arrow. Uniqueness holds by the converse to the valuative criterion for separatedness (Schemes, Lemma 26.22.1) since $Y \to S$ is separated as the composition of the separated morphisms $Y \to X$ and $X \to S$ (Schemes, Lemma 26.21.12). Existence can be seen as follows. As $y$ is a generic point of $Y$, it is contained in $U$. By assumption of the lemma there exists a morphism $a : \mathop{\mathrm{Spec}}(A) \to X$ such that

is commutative. Then since $Y \to X$ is proper, we can apply the valuative criterion for properness (Morphisms, Lemma 29.42.1) to find a morphism $b : \mathop{\mathrm{Spec}}(A) \to Y$ such that

is commutative. This finishes the proof since $b$ can serve as the dotted arrow above. $\square$