## 32.15 Noetherian valuative criterion

If the base is Noetherian we can show that the valuative criterion holds using only discrete valuation rings.

Many of the results in this section can (and perhaps should) be proved by appealing to the following lemma, although we have not always done so.

Lemma 32.15.1. Let $f : X \to Y$ be a morphism of schemes. Assume $f$ finite type and $Y$ locally Noetherian. Let $y \in Y$ be a point in the closure of the image of $f$. Then there exists a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y } \]

where $A$ is a discrete valuation ring and $K$ is its field of fractions mapping the closed point of $\mathop{\mathrm{Spec}}(A)$ to $y$. Moreover, we can assume that the image point of $\mathop{\mathrm{Spec}}(K) \to X$ is a generic point $\eta $ of an irreducible component of $X$ and that $K = \kappa (\eta )$.

**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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\kappa (x)) \ar[d] \ar[rrr] & & & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathop{\mathrm{Spec}}(R) \ar[r] & \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \ar[r] & Y } \]

gives the desired diagram.
$\square$

First we state the result concerning separation. We will often use solid commutative diagrams of morphisms of schemes having the following shape

32.15.1.1
\begin{equation} \label{limits-equation-valuative} \vcenter { \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & S } } \end{equation}

with $A$ a valuation ring and $K$ its field of fractions.

Lemma 32.15.2. Let $S$ be a locally Noetherian scheme. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is locally of finite type. The following are equivalent:

The morphism $f$ is separated.

For any diagram (32.15.1.1) there is at most one dotted arrow.

For all diagrams (32.15.1.1) with $A$ a discrete valuation ring there is at most one dotted arrow.

For any irreducible component $X_0$ of $X$ with generic point $\eta \in X_0$, for any discrete valuation ring $A \subset K = \kappa (\eta )$ with fraction field $K$ and any diagram (32.15.1.1) such that the morphism $\mathop{\mathrm{Spec}}(K) \to X$ is the canonical one (see Schemes, Section 26.13) there is at most one dotted arrow.

**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

32.15.2.1
\begin{equation} \label{limits-equation-valuative-alpha} \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X_\alpha \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & S_\alpha } \end{equation}

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:

32.15.2.2
\begin{equation} \label{limits-equation-valuative-generic} \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X_0 \ar[d]^{\Delta } \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ur]& X_0 \times _ S X_0 \\ } \end{equation}

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:

32.15.2.3
\begin{equation} \label{limits-equation-valuative-nonexistent} \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X_0 \ar[d]\\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ur] & S \\ } \end{equation}

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).
$\square$

Lemma 32.15.3. Let $S$ be a locally Noetherian scheme. Let $f : X \to S$ be a morphism of finite type. The following are equivalent:

The morphism $f$ is proper.

For any diagram (32.15.1.1) there exists exactly one dotted arrow.

For all diagrams (32.15.1.1) with $A$ a discrete valuation ring there exists exactly one dotted arrow.

For any irreducible component $X_0$ of $X$ with generic point $\eta \in X_0$, for any discrete valuation ring $A \subset K = \kappa (\eta )$ with fraction field $K$ and any diagram (32.15.1.1) such that the morphism $\mathop{\mathrm{Spec}}(K) \to X$ is the canonical one (see Schemes, Section 26.13) there exists exactly one dotted arrow.

**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

32.15.3.1
\begin{equation} \label{limits-equation-solid} \xymatrix{ \mathop{\mathrm{Spec}}K \ar[r] \ar[d] & X' \ar[r] \ar[d] & \mathbf{P}^ n_ S \ar[d] \\ \mathop{\mathrm{Spec}}A \ar@{-->}[r] \ar@{-->}[ru] \ar[urr] & X \ar[r] & S\\ } \end{equation}

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 $X'$.
$\square$

Lemma 32.15.4. Let $f : X \to S$ be a finite type morphism of schemes. Assume $S$ is locally Noetherian. Then the following are equivalent

$f$ is universally closed,

for every $n$ the morphism $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is closed,

for any diagram (32.15.1.1) there exists some dotted arrow,

for all diagrams (32.15.1.1) with $A$ a discrete valuation ring there exists some dotted arrow.

**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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\kappa (y)) \ar[r] \ar[d] & \mathbf{A}^ n \times X \ar[d] \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathbf{A}^ n \times S \ar[r] & S } \]

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.
$\square$

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