The Stacks project

Proof. (1) implies (2) implies (3) implies (4). We will now show (4) implies (1). As in the proof of Lemma 32.15.2, we can reduce to the case $S$ is affine, since properness is local on the base, and if $X \to S$ satisfies (4), then $X_\alpha \to S_\alpha$ does as well for open $S_\alpha \subset S$ and $X_\alpha = f^{-1}(S_\alpha )$.

Now $S$ is a Noetherian scheme, and so $X$ is as well, since $X \to S$ is of finite type. Now we may use Chow's lemma (Cohomology of Schemes, Lemma 30.18.1) to get a surjective, proper, birational $X' \to X$ and an immersion $X' \to \mathbf{P}^ n_ S$. We wish to show $X \to S$ is universally closed. As in the proof of Lemma 32.14.3, it is enough to check that $X' \to \mathbf{P}^ n_ S$ is a closed immersion. For the sake of contradiction, assume that $X' \to \mathbf{P}^ n_ S$ is not a closed immersion. Then there is some $y \in \mathbf{P}^ n_ S$ that is in the closure of the image of $X'$, but is not in the image. So $y$ is in the closure of the image of an irreducible component $X_0'$ of $X'$, but not in the image. Let $\bar X_0' \subset \mathbf{P}^ n_ S$ be the closure of the image of $X_0'$. As $X' \to \mathbf{P}^ n_ S$ is an immersion of Noetherian schemes, the morphism $X'_0 \to \bar X_0'$ is open and dense. By Algebra, Lemma 10.119.13 or Properties, Lemma 28.5.10 we can find a discrete valuation ring $A$ dominating $\mathcal{O}_{\bar X_0', y}$ and with identical field of fractions $K$. It is clear that $K$ is the residue field at the generic point of $X_0'$. Thus the solid commutative diagram

Note that the closed point of $A$ maps to $y \in \mathbf{P}^ n_ S$. By construction, there does not exist a set theoretic lift to $X'$. As $X' \to X$ is birational, the image of $X'_0$ in $X$ is an irreducible component $X_0$ of $X$ and $K$ is also identified with the function field of $X_0$. Hence, as $X \to S$ is assumed to satisfy (4), the dotted arrow $\mathop{\mathrm{Spec}}(A) \to X$ exists. Since $X' \to X$ is proper, the dotted arrow lifts to the dotted arrow $\mathop{\mathrm{Spec}}(A) \to X'$ (use Schemes, Proposition 26.20.6). We can compose this with the immersion $X' \to \mathbf{P}^ n_ S$ to obtain another morphism (not depicted in the diagram) from $\mathop{\mathrm{Spec}}(A) \to \mathbf{P}^ n_ S$. Since $\mathbf{P}^ n_ S$ is proper over $S$, it satisfies (2), and so these two morphisms agree. This is a contradiction, for we have constructed the forbidden lift of our original map $\mathop{\mathrm{Spec}}(A) \to \mathbf{P}^ n_ S$ to $X'$. $\square$