The Stacks project

Proof. Choose an affine scheme $V$, a point $v \in V$ and an étale morphism $V \to Y$ mapping $v$ to $y$. The map $|V| \to |Y|$ is open and by Properties of Spaces, Lemma 66.4.3 the image of $|X \times _ Y V| \to |V|$ is the inverse image of the image of $|f|$. We conclude that the point $v$ is in the closure of the image of $|X \times _ Y V| \to |V|$. If we prove the lemma for $X \times _ Y V \to V$ and the point $v$, then the lemma follows for $f$ and $y$. In this way we reduce to the situation described in the next paragraph.

Assume we have $f : X \to Y$ and $y \in |Y|$ as in the lemma where $Y$ is an affine scheme. Since $f$ is quasi-compact, we conclude that $X$ is quasi-compact. Hence we can choose an affine scheme $W$ and a surjective étale morphism $W \to X$. Then the image of $|f|$ is the same as the image of $W \to Y$. In this way we reduce to the case of schemes which is Limits, Lemma 32.15.1. $\square$

Proof. We have (1) $\Rightarrow $ (2) by Morphisms of Spaces, Lemma 67.43.1. The implications (2) $\Rightarrow $ (3) and (3) $\Rightarrow $ (4) are immediate. It remains to show (4) implies (1).

Assume (4). We have to show that the diagonal $\Delta : X \to X \times _ Y X$ is a closed immersion. We already know $\Delta $ is representable, separated, a monomorphism, and locally of finite type, see Morphisms of Spaces, Lemma 67.4.1. Choose an affine scheme $U$ and an étale morphism $U \to X \times _ Y X$. Set $V = X \times _{\Delta , X \times _ Y X} U$. It suffices to show that $V \to U$ is a closed immersion (Morphisms of Spaces, Lemma 67.12.1). Since $X \times _ Y X$ is locally of finite type over $Y$ we see that $U$ is Noetherian (use Morphisms of Spaces, Lemmas 67.23.2, 67.23.3, and 67.23.5). Note that $V$ is a scheme as $\Delta $ is representable. Also, $V$ is quasi-compact because $f$ is quasi-separated. Hence $V \to U$ is separated and of finite type. Consider a commutative diagram

of morphisms of schemes where $A$ is a discrete valuation ring with fraction field $K$ and where $K$ is the residue field of a generic point of the Noetherian scheme $V$. Since $V \to X$ is étale (as a base change of the étale morphism $U \to X \times _ Y X$) we see that the image of $\mathop{\mathrm{Spec}}(K) \to V \to X$ is a point of codimension $0$, see Properties of Spaces, Section 66.10. We can interpret the composition $\mathop{\mathrm{Spec}}(A) \to U \to X \times _ Y X$ as a pair of morphisms $a, b : \mathop{\mathrm{Spec}}(A) \to X$ agreeing as morphisms into $Y$ and equal when restricted to $\mathop{\mathrm{Spec}}(K)$ and that this restriction maps to a point of codimension $0$. Hence our assumption (4) guarantees $a = b$ and we find the dotted arrow in the diagram. By Limits, Lemma 32.15.3 we conclude that $V \to U$ is proper. In other words, $\Delta $ is proper. Since $\Delta $ is a monomorphism, we find that $\Delta $ is a closed immersion (Étale Morphisms, Lemma 41.7.2) as desired. $\square$

Proof. We have (1) $\Leftrightarrow $ (2) $\Leftrightarrow $ (3) by Morphisms of Spaces, Lemma 67.44.1. It is clear that (3) $\Rightarrow $ (4) $\Rightarrow $ (5). To finish the proof we will now show (5) implies (1).

Assume (5). By Lemma 70.21.2 we see that $f$ is separated. To finish the proof it suffices to show that $f$ is universally closed. Let $V \to Y$ be an étale morphism where $V$ is an affine scheme. It suffices to show that the base change $V \times _ Y X \to V$ is universally closed, see Morphisms of Spaces, Lemma 67.9.5. Let

of algebraic spaces over $S$ be a commutative diagram where $A$ is a discrete valuation ring with fraction field $K$ and where $\mathop{\mathrm{Spec}}(K) \to V \times _ Y X$ maps to a point of codimension $0$ of the algebraic space $V \times _ Y X$. Since $V \times _ Y X \to X$ is étale it follows that the image of $\mathop{\mathrm{Spec}}(K) \to X$ is a point of codimension $0$ of $X$. Thus by (5) we obtain the longer of the two dotted arrows fitting into the diagram. Then of course we obtain the shorter one as well. It follows that our assumptions hold for the morphism $V \times _ Y X \to V$ and we reduce to the case discussed in the next paragraph.

Aassume $Y$ is a Noetherian affine scheme. In this case $X$ is a separated Noetherian algebraic space (we already know $f$ is separated) of finite type over $Y$. (In particular, the algebraic space $X$ has a dense open subspace which is a scheme by Properties of Spaces, Proposition 66.13.3 although strictly speaking we will not need this.) Choose a quasi-projective scheme $X'$ over $Y$ and a proper surjective morphism $X' \to X$ as in the weak form of Chow's lemma (Cohomology of Spaces, Lemma 69.18.1). We may replace $X'$ by the disjoint union of the irreducible components which dominate an irreducible component of $X$; details omitted. In particular, we may assume that generic points of the scheme $X'$ map to points of codimension $0$ of $X$ (in this case these are exactly the generic points of $X$). We claim that $X' \to Y$ is proper. The claim implies $X$ is proper over $Y$ by Morphisms of Spaces, Lemma 67.40.7. To prove this, according to Limits, Lemma 32.15.3 it suffices to prove that in every solid commutative diagram

where $A$ is a dvr with fraction field $K$ and where $K$ is the residue field of a generic point of $X'$ we can find the dotted arrow $a$ (we already know uniqueness as $X'$ is separated). By assumption (5) we can find the dotted arrow $b$. Then the morphism $X' \times _{X, b} \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A)$ is a proper morphism of schemes and by the valuative criterion for morphisms of schemes we can lift $b$ to the desired morphism $a$. $\square$

Proof. Parts (1), (2), and (3) are equivalent by Lemma 70.20.2 and Morphisms of Spaces, Lemma 67.42.1. These equivalent conditions imply part (4) as Morphisms of Spaces, Lemma 67.41.3 tells us that we may always choose $K'/K$ finite separable in the existence part of the valuative criterion and this automatically forces $A'$ to be a discrete valuation ring by Krull-Akizuki (Algebra, Lemma 10.119.12). The implication (4) $\Rightarrow $ (5) is immediate. In the rest of the proof we show that (5) implies (1).

Assume (5). Chose an affine scheme $V$ and an étale morphism $V \to Y$. It suffices to show that the base change of $f$ to $V$ is universally closed, see Morphisms of Spaces, Lemma 67.9.5. Exactly as in the proof of Lemma 70.21.3 we see that assumption (5) is inherited by this base change; details omitted. This reduces us to the case discussed in the next paragraph.

Assume $Y$ is a Noetherian affine scheme and we have (5). To prove that $f$ is universally closed it suffices to show that $|X \times \mathbf{A}^ n| \to |Y \times \mathbf{A}^ n|$ is closed for all $n$ (by the discussion above). Since assumption (5) is inherited by the product morphism $X \times \mathbf{A}^ n \to Y \times \mathbf{A}^ n$ (details omitted) we reduce to proving that $|X| \to |Y|$ is closed.

Assume $Y$ is a Noetherian affine scheme and we have (5). Let $T \subset |X|$ be a closed subset. We have to show that the image of $T$ in $|Y|$ is closed. We may replace $X$ by the reduced induced closed subspace structure on $T$; we omit the verification that property (5) is preserved by this replacement. Thus we reduce to proving that the image of $|X| \to |Y|$ is closed.

Let $y \in |Y|$ be a point in the closure of the image of $|X| \to |Y|$. By Lemma 70.21.1 we may choose a commutative diagram

where $A$ is a discrete valuation ring and $K$ is its field of fractions mapping the closed point of $\mathop{\mathrm{Spec}}(A)$ to $y$. It follows immediately from property (5) that $y$ is in the image of $|X| \to |Y|$ and the proof is complete. $\square$