## 69.21 Noetherian valuative criterion

We have already proved some results in Cohomology of Spaces, Section 68.19. The corresponding section for schemes is Limits, Section 32.15.

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 69.21.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. 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 point $x \in |X|$ corresponding to $\mathop{\mathrm{Spec}}(K) \to X$ is a codimension $0$ point1 and that $K$ is the residue field of a point on a scheme étale over $X$.

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 65.4.3 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. $\square$

First we state the result concerning separation. We will often use solid commutative diagrams of morphisms of algebraic spaces over a base scheme $S$ having the following shape

69.21.1.1
$$\label{spaces-limits-equation-valuative} \vcenter { \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & Y } }$$

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

Lemma 69.21.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-separated and locally of finite type and $Y$ is locally Noetherian. The following are equivalent:

1. The morphism $f$ is separated.

2. For any diagram (69.21.1.1) there is at most one dotted arrow.

3. For all diagrams (69.21.1.1) with $A$ a discrete valuation ring there is at most one dotted arrow.

4. For all diagrams (69.21.1.1) where $A$ is a discrete valuation ring and where the image of $\mathop{\mathrm{Spec}}(K) \to X$ is a point of codimension $0$ on $X$ there is at most one dotted arrow.

Proof. We have (1) $\Rightarrow$ (2) by Morphisms of Spaces, Lemma 66.43.1. 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 66.4.1. 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 66.12.1). Since $X \times _ Y X$ is locally of finite type over $Y$ we see that $U$ is Noetherian (use Morphisms of Spaces, Lemmas 66.23.2, 66.23.3, and 66.23.5). 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

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & V \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & U }$

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 65.10. 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. $\square$

Lemma 69.21.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-separated and of finite type and $Y$ is locally Noetherian. The following are equivalent:

1. $f$ is proper,

2. $f$ satisfies the valuative criterion, see Morphisms of Spaces, Definition 66.41.1,

3. for any diagram (69.21.1.1) there exists exactly one dotted arrow,

4. for all diagrams (69.21.1.1) with $A$ a discrete valuation ring there exists exactly one dotted arrow, and

5. for all diagrams (69.21.1.1) where $A$ is a discrete valuation ring and where the image of $\mathop{\mathrm{Spec}}(K) \to X$ is a point of codimension $0$ on $X$ there exists exactly one dotted arrow2.

Proof. We have (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3) by Morphisms of Spaces, Lemma 66.44.1. It is clear that (3) $\Rightarrow$ (4) $\Rightarrow$ (5). To finish the proof we will now show (5) implies (1).

Assume (5). By Lemma 69.21.2 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 66.9.5. Let

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & V \times _ Y X \ar[d] \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] \ar@{..>}[rru] & V \ar[r] & Y }$

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 65.13.3 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 68.18.1). 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 66.40.7. To prove this, according to Limits, Lemma 32.15.3 it suffices to prove that in every solid commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X' \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[rr] \ar@{-->}[ru]^ a \ar@{-->}[rru]_ b & & Y }$

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 $a$. $\square$

Lemma 69.21.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $Y$ is locally Noetherian and $f$ is of finite type. Then the following are equivalent

1. $f$ is universally closed,

2. $f$ satisfies the existence part of the valuative criterion,

3. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $|\mathbf{A}^ n \times X \times _ Y V| \to |\mathbf{A}^ n \times V|$ is closed for all $n \geq 0$,

4. for all diagrams (69.21.1.1) with $A$ a discrete valuation ring there there exists a finite separable extension $K'/K$ of fields, a discrete valuation ring $A' \subset K'$ dominating $A$, and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ such that the following diagram commutes

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$
5. for all diagrams (69.21.1.1) with $A$ a discrete valuation ring there there exists a field extension $K'/K$, a valuation ring $A' \subset K'$ dominating $A$, and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ such that the following diagram commutes

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[rru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y }$

Proof. Parts (1), (2), and (3) are equivalent by Lemma 69.20.2 and Morphisms of Spaces, Lemma 66.42.1. These equivalent conditions imply part (4) as Morphisms of Spaces, Lemma 66.41.3 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 66.9.5. Exactly as in the proof of Lemma 69.21.3 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 69.21.1 we may choose 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 $y$. It follows immediately from property (5) that $y$ is in the image of $|X| \to |Y|$ and the proof is complete. $\square$

[1] See discussion in Properties of Spaces, Section 65.11.
[2] There is a sharper formulation where in the existence part one only requires the dotted arrow exists after an extension of discrete valuation rings.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CMB. Beware of the difference between the letter 'O' and the digit '0'.