The Stacks project

Proof. By the non-Noetherian version of this lemma (Morphisms, Lemma 29.6.5) there exists a point $x \in X$ such that $f(x)$ specializes to $y$. We may replace $x$ by any point specializing to $x$, hence we may assume that $x$ is a generic point of an irreducible component of $X$. This produces a ring map $\mathcal{O}_{Y, y} \to \kappa (x)$ (see Schemes, Section 26.13). Let $R \subset \kappa (x)$ be the image. Then $R$ is Noetherian as a quotient of the Noetherian local ring $\mathcal{O}_{Y, y}$. On the other hand, the extension $\kappa (x)$ is a finitely generated extension of the fraction field of $R$ as $f$ is of finite type. Thus there exists a discrete valuation ring $A \subset \kappa (x)$ with fraction field $\kappa (x)$ dominating $R$ by Algebra, Lemma 10.119.13. Then

gives the desired diagram. $\square$

Proof. Clearly (1) implies (2), (2) implies (3), and (3) implies (4). It remains to show (4) implies (1). Assume (4). We begin by reducing to $S$ affine. Being separated is a local on the base (see Schemes, Lemma 26.21.7). Hence, if we can show that whenever $X \to S$ has (4) that the restriction $X_\alpha \to S_\alpha $ has (4) where $S_\alpha \subset S$ is an (affine) open subset and $X_\alpha := f^{-1}(S_\alpha )$, then we will be done. The generic points of the irreducible components of $X_\alpha $ will be the generic points of irreducible components of $X$, since $X_\alpha $ is open in $X$. Therefore, any two distinct dotted arrows in the diagram

would then give two distinct arrows in diagram (32.15.1.1) via the maps $X_\alpha \to X$ and $S_\alpha \to S$, which is a contradiction. Thus we have reduced to the case $S$ is affine. We remark that in the course of this reduction, we prove that if $X \to S$ has (4) then the restriction $U \to V$ has (4) for opens $U \subset X$ and $V \subset S$ with $f(U) \subset V$.

We next wish to reduce to the case $X \to S$ is finite type. Assume that we know (4) implies (1) when $X$ is finite type. Since $S$ is Noetherian and $X$ is locally of finite type over $S$ we see $X$ is locally Noetherian as well (see Morphisms, Lemma 29.15.6). Thus, $X \to S$ is quasi-separated (see Properties, Lemma 28.5.4), and therefore we may apply the valuative criterion to check whether $X$ is separated (see Schemes, Lemma 26.22.2). Let $X = \bigcup _\alpha X_\alpha $ be an affine open cover of $X$. Given any two dotted arrows, in a diagram (32.15.1.1), the image of the closed points of $\mathop{\mathrm{Spec}}A$ will fall in two sets $X_\alpha $ and $X_\beta $. Since $X_\alpha \cup X_\beta $ is open, for topological reasons it must contain the image of $\mathop{\mathrm{Spec}}(A)$ under both maps. Therefore, the two dotted arrows factor through $X_\alpha \cup X_\beta \to X$, which is a scheme of finite type over $S$. Since $X_\alpha \cup X_\beta $ is an open subset of $X$, by our previous remark, $X_\alpha \cup X_\beta $ satisfies (4), so by assumption, is separated. This implies the two given dotted arrows are the same. Therefore, we have reduced to $X \to S$ is finite type.

Assume $X \to S$ of finite type and assume (4). Since $X \to S$ is finite type, and $S$ is an affine Noetherian scheme, $X$ is also Noetherian (see Morphisms, Lemma 29.15.6). Therefore, $X \to X \times _ S X$ will be a quasi-compact immersion of Noetherian schemes. We proceed by contradiction. Assume that $X \to X \times _ S X$ is not closed. Then, there is some $y \in X \times _ S X$ in the closure of the image that is not in the image. As $X$ is Noetherian it has finitely many irreducible components. Therefore, $y$ is in the closure of the image of one of the irreducible components $X_0 \subset X$. Give $X_0$ the reduced induced structure. The composition $X_0 \to X \to X \times _ S X$ factors through the closed subscheme $X_0 \times _ S X_0 \subset X \times _ S X$. Denote the closure of $\Delta (X_0)$ in $X_0 \times _ S X_0$ by $\bar X_0$ (again as a reduced closed subscheme). Thus $y \in \bar X_0$. Since $X_0 \to X_0 \times _ S X_0$ is an immersion, the image of $X_0$ will be open in $\bar X_0$. Hence $X_0$ and $\bar X_0$ are birational. Since $\bar{X}_0$ is a closed subscheme of a Noetherian scheme, it is Noetherian. Thus, the local ring $\mathcal O_{{\bar X_0, y}}$ is a local Noetherian domain with fraction field $K$ equal to the function field of $X_0$. By the Krull-Akizuki theorem (see Algebra, Lemma 10.119.13), there exists a discrete valuation ring $A$ dominating $\mathcal O_{{\bar X_0, y}}$ with fraction field $K$. This allows to construct a diagram:

which sends $\mathop{\mathrm{Spec}}K$ to the generic point of $\Delta (X_0)$ and the closed point of $A$ to $y \in X_0 \times _ S X_0$ (use the material in Schemes, Section 26.13 to construct the arrows). There cannot even exist a set theoretic dotted arrow, since $y$ is not in the image of $\Delta $ by our choice of $y$. By categorical means, the existence of the dotted arrow in the above diagram is equivalent to the uniqueness of the dotted arrow in the following diagram:

Therefore, we have non-uniqueness in this latter diagram by the nonexistence in the first. Therefore, $X_0$ does not satisfy uniqueness for discrete valuation rings, and since $X_0$ is an irreducible component of $X$, we have that $X \to S$ does not satisfy (4). Therefore, we have shown (4) implies (1). $\square$

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 $X'$. $\square$

Proof. The equivalence of (1) and (2) is a special case of Lemma 32.14.2. The equivalence of (1) and (3) is a special case of Schemes, Proposition 26.20.6. Trivially (3) implies (4). Thus all we have to do is prove that (4) implies (2). We will prove that $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is closed by the criterion of Schemes, Lemma 26.19.8. Pick $n$ and a specialization $z \leadsto z'$ of points in $\mathbf{A}^ n \times S$ and a point $y \in \mathbf{A}^ n \times X$ lying over $z$. Note that $\kappa (y)$ is a finitely generated field extension of $\kappa (z)$ as $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is of finite type. Hence by Properties, Lemma 28.5.10 or Algebra, Lemma 10.119.13 implies that there exists a discrete valuation ring $A \subset \kappa (y)$ with fraction field $\kappa (z)$ dominating the image of $\mathcal{O}_{\mathbf{A}^ n \times S, z'}$ in $\kappa (z)$. This gives a commutative diagram

Now property (4) implies that there exists a morphism $\mathop{\mathrm{Spec}}(A) \to X$ which fits into this diagram. Since we already have the morphism $\mathop{\mathrm{Spec}}(A) \to \mathbf{A}^ n$ from the left lower horizontal arrow we also get a morphism $\mathop{\mathrm{Spec}}(A) \to \mathbf{A}^ n \times X$ fitting into the left square. Thus the image $y' \in \mathbf{A}^ n \times X$ of the closed point is a specialization of $y$ lying over $z'$. This proves that specializations lift along $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ and we win. $\square$