## 29.42 Valuative criteria

We have already discussed the valuative criterion for universal closedness and for separatedness in Schemes, Sections 26.20 and 26.22. In this section we will discuss some consequences and variants. In Limits, Section 32.15 we will show that it suffices to consider discrete valuation rings when working with locally Noetherian schemes and morphisms of finite type.

Lemma 29.42.1 (Valuative criterion for properness). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Assume $f$ is of finite type and quasi-separated. Then the following are equivalent

1. $f$ is proper,

2. $f$ satisfies the valuative criterion (Schemes, Definition 26.20.3),

3. given any commutative solid diagram

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

where $A$ is a valuation ring with field of fractions $K$, there exists a unique dotted arrow making the diagram commute.

Proof. Part (3) is a reformulation of (2). Thus the lemma is a formal consequence of Schemes, Proposition 26.20.6 and Lemma 26.22.2 and the definitions. $\square$

One usually does not have to consider all possible diagrams when testing the valuative criterion.

Lemma 29.42.2. Let $f : X \to S$ and $h : U \to X$ be morphisms of schemes. Assume that $f$ and $h$ are quasi-compact and that $h(U)$ is dense in $X$. If given any commutative solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & U \ar[r]^ h & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[rr] \ar@{-->}[rru] & & S }$

where $A$ is a valuation ring with field of fractions $K$, there exists a unique dotted arrow making the diagram commute, then $f$ is universally closed. If moreover $f$ is quasi-separated, then $f$ is separated.

Proof. To prove $f$ is universally closed we will verify the existence part of the valuative criterion for $f$ which suffices by Schemes, Proposition 26.20.6. To do this, consider a commutative diagram

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

where $A$ is a valuation ring and $K$ is the fraction field of $A$. Note that since valuation rings and fields are reduced, we may replace $U$, $X$, and $S$ by their respective reductions by Schemes, Lemma 26.12.7. In this case the assumption that $h(U)$ is dense means that the scheme theoretic image of $h : U \to X$ is $X$, see Lemma 29.6.7. We may also replace $S$ by an affine open through which the morphism $\mathop{\mathrm{Spec}}(A) \to S$ factors. Thus we may assume that $S = \mathop{\mathrm{Spec}}(R)$.

Let $\mathop{\mathrm{Spec}}(B) \subset X$ be an affine open through which the morphism $\mathop{\mathrm{Spec}}(K) \to X$ factors. Choose a polynomial algebra $P$ over $B$ and a $B$-algebra surjection $P \to K$. Then $\mathop{\mathrm{Spec}}(P) \to X$ is flat. Hence the scheme theoretic image of the morphism $U \times _ X \mathop{\mathrm{Spec}}(P) \to \mathop{\mathrm{Spec}}(P)$ is $\mathop{\mathrm{Spec}}(P)$ by Lemma 29.25.16. By Lemma 29.6.5 we can find a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & U \times _ X \mathop{\mathrm{Spec}}(P) \ar[d] \\ \mathop{\mathrm{Spec}}(A') \ar[r] & \mathop{\mathrm{Spec}}(P) }$

where $A'$ is a valuation ring and $K'$ is the fraction field of $A'$ such that the closed point of $\mathop{\mathrm{Spec}}(A')$ maps to $\mathop{\mathrm{Spec}}(K) \subset \mathop{\mathrm{Spec}}(P)$. In other words, there is a $B$-algebra map $\varphi : K \to A'/\mathfrak m_{A'}$. Choose a valuation ring $A'' \subset A'/\mathfrak m_{A'}$ dominating $\varphi (A)$ with field of fractions $K'' = A'/\mathfrak m_{A'}$ (Algebra, Lemma 10.50.2). We set

$C = \{ \lambda \in A' \mid \lambda \bmod \mathfrak m_{A'} \in A''\} .$

which is a valuation ring by Algebra, Lemma 10.50.9. As $C$ is an $R$-algebra with fraction field $K'$, we obtain a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & U \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(C) \ar[rr] \ar@{-->}[rru] & & S }$

as in the statement of the lemma. Thus a dotted arrow fitting into the diagram as indicated. By the uniqueness assumption of the lemma the composition $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(C) \to X$ agrees with the given morphism $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(P) \to \mathop{\mathrm{Spec}}(B) \subset X$. Hence the restriction of the morphism to the spectrum of $C/\mathfrak m_{A'} = A''$ induces the given morphism $\mathop{\mathrm{Spec}}(K'') = \mathop{\mathrm{Spec}}(A'/\mathfrak m_{A'}) \to \mathop{\mathrm{Spec}}(K) \to X$. Let $x \in X$ be the image of the closed point of $\mathop{\mathrm{Spec}}(A'') \to X$. The image of the induced ring map $\mathcal{O}_{X, x} \to A''$ is a local subring which is contained in $K \subset K''$. Since $A$ is maximal for the relation of domination in $K$ and since $A \subset A''$, we have $A = K \cap A''$. We conclude that $\mathcal{O}_{X, x} \to A''$ factors through $A \subset A''$. In this way we obtain our desired arrow $\mathop{\mathrm{Spec}}(A) \to X$.

Finally, assume $f$ is quasi-separated. Then $\Delta : X \to X \times _ S X$ is quasi-compact. Given a solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & U \ar[r]^ h & X \ar[d]^\Delta \\ \mathop{\mathrm{Spec}}(A) \ar[rr] \ar@{-->}[rru] & & X \times _ S X }$

where $A$ is a valuation ring with field of fractions $K$, there exists a unique dotted arrow making the diagram commute. Namely, the lower horizontal arrow is the same thing as a pair of morphisms $\mathop{\mathrm{Spec}}(A) \to X$ which can serve as the dotted arrow in the diagram of the lemma. Thus the required uniqueness shows that the lower horizontal arrow factors through $\Delta$. Hence we can apply the result we just proved to $\Delta : X \to X \times _ S X$ and $h : U \to X$ and conclude that $\Delta$ is universally closed. Clearly this means that $f$ is separated. $\square$

Remark 29.42.3. The assumption on uniqueness of the dotted arrows in Lemma 29.42.2 is necessary (details omitted). Of course, uniqueness is guaranteed if $f$ is separated (Schemes, Lemma 26.22.1).

Lemma 29.42.4. Let $S$ be a scheme. Let $X$, $Y$ be schemes over $S$. Let $s \in S$ and $x \in X$, $y \in Y$ points over $s$.

1. Let $f, g : X \to Y$ be morphisms over $S$ such that $f(x) = g(x) = y$ and $f^\sharp _ x = g^\sharp _ x : \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$. Then there is an open neighbourhood $U \subset X$ with $f|_ U = g|_ U$ in the following cases

1. $Y$ is locally of finite type over $S$,

2. $X$ is integral,

3. $X$ is locally Noetherian, or

4. $X$ is reduced with finitely many irreducible components.

2. Let $\varphi : \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ be a local $\mathcal{O}_{S, s}$-algebra map. Then there exists an open neighbourhood $U \subset X$ of $x$ and a morphism $f : U \to Y$ mapping $x$ to $y$ with $f^\sharp _ x = \varphi$ in the following cases

1. $Y$ is locally of finite presentation over $S$,

2. $Y$ is locally of finite type and $X$ is integral,

3. $Y$ is locally of finite type and $X$ is locally Noetherian, or

4. $Y$ is locally of finite type and $X$ is reduced with finitely many irreducible components.

Proof. Proof of (1). We may replace $X$, $Y$, $S$ by suitable affine open neighbourhoods of $x$, $y$, $s$ and reduce to the following algebra problem: given a ring $R$, two $R$-algebra maps $\varphi , \psi : B \to A$ such that

1. $R \to B$ is of finite type, or $A$ is a domain, or $A$ is Noetherian, or $A$ is reduced and has finitely many minimal primes,

2. the two maps $B \to A_\mathfrak p$ are the same for some prime $\mathfrak p \subset A$,

show that $\varphi , \psi$ define the same map $B \to A_ g$ for a suitable $g \in A$, $g \not\in \mathfrak p$. If $R \to B$ is of finite type, let $t_1, \ldots , t_ m \in B$ be generators of $B$ as an $R$-algebra. For each $j$ we can find $g_ j \in A$, $g_ j \not\in \mathfrak p$ such that $\varphi (t_ j)$ and $\psi (t_ j)$ have the same image in $A_{g_ j}$. Then we set $g = \prod g_ j$. In the other cases (if $A$ is a domain, Noetherian, or reduced with finitely many minimal primes), we can find a $g \in A$, $g \not\in \mathfrak p$ such that $A_ g \subset A_\mathfrak p$. See Algebra, Lemma 10.31.9. Thus the maps $B \to A_ g$ are equal as desired.

Proof of (2). To do this we may replace $X$, $Y$, and $S$ by suitable affine opens. Say $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(B)$, and $S = \mathop{\mathrm{Spec}}(R)$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $x$. Let $\mathfrak q \subset B$ be the prime corresponding to $y$. Then $\varphi$ is a local $R$-algebra map $\varphi : B_\mathfrak q \to A_\mathfrak p$. If $R \to B$ is a ring map of finite presentation, then there exists a $g \in A \setminus \mathfrak p$ and an $R$-algebra map $B \to A_ g$ such that

$\xymatrix{ B_\mathfrak q \ar[r]_\varphi & A_\mathfrak p \\ B \ar[u] \ar[r] & A_ g \ar[u] }$

commutes, see Algebra, Lemmas 10.127.3 and 10.9.9. The induced morphism $\mathop{\mathrm{Spec}}(A_ g) \to \mathop{\mathrm{Spec}}(B)$ works. If $B$ is of finite type over $R$, let $t_1, \ldots , t_ m \in B$ be generators of $B$ as an $R$-algebra. Then we can choose $g_ j \in A$, $g_ j \not\in \mathfrak p$ such that $\varphi (t_ j) \in \mathop{\mathrm{Im}}(A_{g_ j} \to A_\mathfrak p)$. Thus after replacing $A$ by $A[1/\prod g_ j]$ we may assume that $B$ maps into the image of $A \to A_\mathfrak p$. If we can find a $g \in A$, $g \not\in \mathfrak p$ such that $A_ g \to A_\mathfrak p$ is injective, then we'll get the desired $R$-algebra map $B \to A_ g$. Thus the proof is finished by another application of See Algebra, Lemma 10.31.9. $\square$

Lemma 29.42.5. Let $S$ be a scheme. Let $X$, $Y$ be schemes over $S$. Let $x \in X$. Let $U \subset X$ be an open and let $f : U \to Y$ be a morphism over $S$. Assume

1. $x$ is in the closure of $U$,

2. $X$ is reduced with finitely many irreducible components or $X$ is Noetherian,

3. $\mathcal{O}_{X, x}$ is a valuation ring,

4. $Y \to S$ is proper

Then there exists an open $U \subset U' \subset X$ containing $x$ and an $S$-morphism $f' : U' \to Y$ extending $f$.

Proof. It is harmless to replace $X$ by an open neighbourhood of $x$ in $X$ (small detail omitted). By Properties, Lemma 28.29.8 we may assume $X$ is affine with $\Gamma (X, \mathcal{O}_ X) \subset \mathcal{O}_{X, x}$. In particular $X$ is integral with a unique generic point $\xi$ whose residue field is the fraction field $K$ of the valuation ring $\mathcal{O}_{X, x}$. Since $x$ is in the closure of $U$ we see that $U$ is not empty, hence $U$ contains $\xi$. Thus by the valuative criterion of properness (Lemma 29.42.1) there is a morphism $t : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to Y$ fitting into a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d]_\xi \ar[r] & \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \ar[d]_ t \\ U \ar[r]^ f & Y }$

of morphisms of schemes over $S$. Applying Lemma 29.42.4 with $y = t(x)$ and $\varphi = t^\sharp _ x$ we obtain an open neighbourhood $V \subset X$ of $x$ and a morphism $g : V \to Y$ over $S$ which sends $x$ to $y$ and such that $g^\sharp _ x = t^\sharp _ x$. As $Y \to S$ is separated, the equalizer $E$ of $f|_{U \cap V}$ and $g|_{U \cap V}$ is a closed subscheme of $U \cap V$, see Schemes, Lemma 26.21.5. Since $f$ and $g$ determine the same morphism $\mathop{\mathrm{Spec}}(K) \to Y$ by construction we see that $E$ contains the generic point of the integral scheme $U \cap V$. Hence $E = U \cap V$ and we conclude that $f$ and $g$ glue to a morphism $U' = U \cup V \to Y$ as desired. $\square$

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).