The Stacks project

Proof. Let $s = g(t)$. During the proof we may always replace $T$ by an open neighbourhood of $t$. Hence we may also replace $S$ by an open neighbourhood of $s$. Thus we may and do assume that $T$ and $S$ are affine. Say $S = \mathop{\mathrm{Spec}}(A)$, $T = \mathop{\mathrm{Spec}}(B)$, $g$ is given by the ring map $A \to B$, and $t$ correspond to the prime ideal $\mathfrak q \subset B$.

As $X \to S$ is quasi-compact and $S$ is affine we may write $X = \bigcup _{i = 1, \ldots , n} U_ i$ as a finite union of affine opens. Write $U_ i = \mathop{\mathrm{Spec}}(C_ i)$. In particular we have $X_ T = \bigcup _{i = 1, \ldots , n} U_{i, T} = \bigcup _{i = 1, \ldots n} \mathop{\mathrm{Spec}}(C_ i \otimes _ A B)$. Let $I_ i \subset C_ i \otimes _ A B$ be the ideal corresponding to the closed subscheme $Z \cap U_{i, T}$. The condition that $Z \cap X_ t = \emptyset$ signifies that $I_ i$ generates the unit ideal in the ring

Since $I_ i (B \setminus \mathfrak q)^{-1}(C_ i \otimes _ A B) = (B \setminus \mathfrak q)^{-1} I_ i$ this means that $1 = x_ i/g_ i$ for some $x_ i \in I_ i$ and $g_ i \in B$, $g_ i \not\in \mathfrak q$. Thus, clearing denominators we can find a relation of the form

with $x_ i \in I_ i$, $f_{i, j} \in \mathfrak q$, $c_{i, j} \in C_ i \otimes _ A B$, and $g_ i \in B$, $g_ i \not\in \mathfrak q$. After replacing $B$ by $B_{g_1 \ldots g_ n}$, i.e., after replacing $T$ by a smaller affine neighbourhood of $t$, we may assume the equations read

with $x_ i \in I_ i$, $f_{i, j} \in \mathfrak q$, $c_{i, j} \in C_ i \otimes _ A B$.

To finish the argument write $B$ as a colimit of finitely presented $A$-algebras $B_\lambda$ over a directed set $\Lambda$. For each $\lambda$ set $\mathfrak q_\lambda = (B_\lambda \to B)^{-1}(\mathfrak q)$. For sufficiently large $\lambda \in \Lambda$ we can find

1. an element $x_{i, \lambda } \in C_ i \otimes _ A B_\lambda$ which maps to $x_ i$,

2. elements $f_{i, j, \lambda } \in \mathfrak q_{i, \lambda }$ mapping to $f_{i, j}$, and

3. elements $c_{i, j, \lambda } \in C_ i \otimes _ A B_\lambda$ mapping to $c_{i, j}$.

After increasing $\lambda$ a bit more the equation

will hold. Fix such a $\lambda$ and set $T' = \mathop{\mathrm{Spec}}(B_\lambda )$. Then $t' \in T'$ is the point corresponding to the prime $\mathfrak q_\lambda$. Finally, let $Z' \subset X_{T'}$ be the scheme theoretic image of $Z \to X_ T \to X_{T'}$. As $X_ T \to X_{T'}$ is affine, we can compute $Z'$ on the affine open pieces $U_{i, T'}$ as the closed subscheme associated to $\mathop{\mathrm{Ker}}(C_ i \otimes _ A B_\lambda \to C_ i \otimes _ A B/I_ i)$, see Morphisms, Example 29.6.4. Hence $x_{i, \lambda }$ is in the ideal defining $Z'$. Thus the last displayed equation shows that $Z' \cap X_{t'}$ is empty. $\square$

Proof. It is clear that (1) implies (2). Let us prove that (2) implies (1). Suppose that the base change $X_ T \to T$ is not closed for some scheme $T$ over $S$. By Schemes, Lemma 26.19.8 this means that there exists some specialization $t_1 \leadsto t$ in $T$ and a point $\xi \in X_ T$ mapping to $t_1$ such that $\xi$ does not specialize to a point in the fibre over $t$. Set $Z = \overline{\{ \xi \} } \subset X_ T$. Then $Z \cap X_ t = \emptyset$. Apply Lemma 32.14.1. We find an open neighbourhood $V \subset T$ of $t$, a commutative diagram

and a closed subscheme $Z' \subset X_{T'}$ such that

1. the morphism $b : T' \to S$ is locally of finite presentation,

2. with $t' = a(t)$ we have $Z' \cap X_{t'} = \emptyset$, and

3. $Z \cap X_ V$ maps into $Z'$ via the morphism $X_ V \to X_{T'}$.

Clearly this means that $X_{T'} \to T'$ maps the closed subset $Z'$ to a subset of $T'$ which contains $a(t_1)$ but not $t' = a(t)$. Since $a(t_1) \leadsto a(t) = t'$ we conclude that $X_{T'} \to T'$ is not closed. Hence we have shown that $X \to S$ not universally closed implies that $X_{T'} \to T'$ is not closed for some $T' \to S$ which is locally of finite presentation. In order words (2) implies (1).

Assume that $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is closed for every integer $n$. We want to prove that $X_ T \to T$ is closed for every scheme $T$ which is locally of finite presentation over $S$. We may of course assume that $T$ is affine and maps into an affine open $V$ of $S$ (since $X_ T \to T$ being a closed is local on $T$). In this case there exists a closed immersion $T \to \mathbf{A}^ n \times V$ because $\mathcal{O}_ T(T)$ is a finitely presented $\mathcal{O}_ S(V)$-algebra, see Morphisms, Lemma 29.21.2. Then $T \to \mathbf{A}^ n \times S$ is a locally closed immersion. Hence we get a cartesian diagram

of schemes where the horizontal arrows are locally closed immersions. Hence any closed subset $Z \subset X_ T$ can be written as $X_ T \cap Z'$ for some closed subset $Z' \subset \mathbf{A}^ n \times X$. Then $f_ T(Z) = T \cap f_ n(Z')$ and we see that if $f_ n$ is closed, then also $f_ T$ is closed. $\square$

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,

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:

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

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

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

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