Lemma 32.14.3. Let $S$ be a scheme. Let $f : X \to S$ be a separated morphism of finite type. The following are equivalent:

1. The morphism $f$ is proper.

2. For any morphism $S' \to S$ which is locally of finite type the base change $X_{S'} \to S'$ is closed.

3. For every $n \geq 0$ the morphism $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is closed.

First proof. In view of the fact that a proper morphism is the same thing as a separated, finite type, and universally closed morphism, this lemma is a special case of Lemma 32.14.2. $\square$

Second proof. Clearly (1) implies (2), and (2) implies (3), so we just need to show (3) implies (1). First we reduce to the case when $S$ is affine. Assume that (3) implies (1) when the base is affine. Now let $f: X \to S$ be a separated morphism of finite type. Being proper is local on the base (see Morphisms, Lemma 29.41.3), so if $S = \bigcup _\alpha S_\alpha$ is an open affine cover, and if we denote $X_\alpha := f^{-1}(S_\alpha )$, then it is enough to show that $f|_{X_\alpha }: X_\alpha \to S_\alpha$ is proper for all $\alpha$. Since $S_\alpha$ is affine, if the map $f|_{X_\alpha }$ satisfies (3), then it will satisfy (1) by assumption, and will be proper. To finish the reduction to the case $S$ is affine, we must show that if $f: X \to S$ is separated of finite type satisfying (3), then $f|_{X_\alpha } : X_\alpha \to S_\alpha$ is separated of finite type satisfying (3). Separatedness and finite type are clear. To see (3), notice that $\mathbf{A}^ n \times X_\alpha$ is the open preimage of $\mathbf{A}^ n \times S_\alpha$ under the map $1 \times f$. Fix a closed set $Z \subset \mathbf A^ n \times X_\alpha$. Let $\bar Z$ denote the closure of $Z$ in $\mathbf{A}^ n \times X$. Then for topological reasons,

$1 \times f(\bar Z) \cap \mathbf{A}^ n \times S_\alpha = 1 \times f(Z).$

Hence $1 \times f(Z)$ is closed, and we have reduced the proof of (3) $\Rightarrow$ (1) to the affine case.

Assume $S$ affine, and $f : X \to S$ separated of finite type. We can apply Chow's Lemma 32.12.1 to get $\pi : X' \to X$ proper surjective and $X' \to \mathbf{P}^ n_ S$ an immersion. If $X$ is proper over $S$, then $X' \to S$ is proper (Morphisms, Lemma 29.41.4). Since $\mathbf{P}^ n_ S \to S$ is separated, we conclude that $X' \to \mathbf{P}^ n_ S$ is proper (Morphisms, Lemma 29.41.7) and hence a closed immersion (Schemes, Lemma 26.10.4). Conversely, assume $X' \to \mathbf{P}^ n_ S$ is a closed immersion. Consider the diagram:

32.14.3.1
\begin{equation} \label{limits-equation-check-proper} \xymatrix{ X' \ar[r] \ar@{->>}[d]_{\pi } & \mathbf{P}^ n_ S \ar[d] \\ X \ar[r]^ f & S } \end{equation}

All maps are a priori proper except for $X \to S$. Hence we conclude that $X \to S$ is proper by Morphisms, Lemma 29.41.9. Therefore, we have shown that $X \to S$ is proper if and only if $X' \to \mathbf{P}^ n_ S$ is a closed immersion.

Assume $S$ is affine and (3) holds, and let $n, X', \pi$ be as above. Since being a closed morphism is local on the base, the map $X \times \mathbf{P}^ n \to S \times \mathbf{P}^ n$ is closed since by (3) $X \times \mathbf{A}^ n \to S \times \mathbf{A}^ n$ is closed and since projective space is covered by copies of affine $n$-space, see Constructions, Lemma 27.13.3. By Morphisms, Lemma 29.41.5 the morphism

$X' \times _ S \mathbf{P}^ n_ S \to X \times _ S \mathbf{P}^ n_ S = X \times \mathbf{P}^ n$

is proper. Since $\mathbf{P}^ n$ is separated, the projection

$X' \times _ S \mathbf{P}^ n_ S = \mathbf{P}^ n_{X'} \to X'$

will be separated as it is just a base change of a separated morphism. Therefore, the map $X' \to X' \times _ S \mathbf{P}^ n_ S$ is proper, since it is a section to a separated map (see Schemes, Lemma 26.21.11). Composing these morphisms

$X' \to X' \times _ S \mathbf{P}^ n_ S \to X \times _ S \mathbf{P}^ n_ S = X \times \mathbf{P}^ n \to S \times \mathbf{P}^ n = \mathbf{P}^ n_ S$

we find that the immersion $X' \to \mathbf{P}^ n_ S$ is closed, and hence a closed immersion. $\square$

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