The Stacks project

Lemma 66.42.5. Let $S$ be a scheme. Let $f : X \to Y$ and $h : U \to X$ be morphisms of algebraic spaces over $S$. If

  1. $f$ and $h$ are quasi-compact,

  2. $|h|(|U|)$ is dense in $|X|$, and

given any commutative solid diagram

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

where $A$ is a valuation ring with field of fractions $K$

  1. there exists at most one dotted arrow making the diagram commute, and

  2. there exists an extension $K'/K$ of fields, a valuation ring $A' \subset K'$ dominating $A$ and a morphism $\mathop{\mathrm{Spec}}(A') \to X$ such that the following diagram commutes

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

then $f$ is universally closed. If moreover

  1. $f$ is quasi-separated

then $f$ is separated and universally closed.

Proof. Assume (1), (2), (3), and (4). We will verify the existence part of the valuative criterion for $f$ which will imply $f$ is universally closed by Lemma 66.42.1. To do this, consider a commutative diagram
\begin{equation} \label{spaces-morphisms-equation-start-with} \vcenter { \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y } } \end{equation}

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 Properties of Spaces, Lemma 65.12.4. 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 66.16.4.

Reduction to the case $Y$ affine. Choose an étale morphism $\mathop{\mathrm{Spec}}(R) \to Y$ such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to an element of $\mathop{\mathrm{Im}}(|\mathop{\mathrm{Spec}}(R)| \to |Y|)$. By Lemma 66.42.4 we can find a local ring map $A \to A'$ of valuation rings and a morphism $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(R)$ fitting into a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(R) \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y } \]

Since in Definition 66.41.1 we allow for extensions of valuation rings it is clear that we may replace $A$ by $A'$, $Y$ by $\mathop{\mathrm{Spec}}(R)$, $X$ by $X \times _ Y \mathop{\mathrm{Spec}}(R)$ and $U$ by $U \times _ Y \mathop{\mathrm{Spec}}(R)$.

From now on we assume that $Y = \mathop{\mathrm{Spec}}(R)$ is an affine scheme. Let $\mathop{\mathrm{Spec}}(B) \to X$ be an étale morphism from an affine scheme such that the morphism $\mathop{\mathrm{Spec}}(K) \to X$ is in the image of $|\mathop{\mathrm{Spec}}(B)| \to |X|$. Since we may replace $K$ by an extension $K' \supset K$ and $A$ by a valuation ring $A' \subset K'$ dominating $A$ (which exists by Algebra, Lemma 10.50.2), we may assume the morphism $\mathop{\mathrm{Spec}}(K) \to X$ factors through $\mathop{\mathrm{Spec}}(B)$ (by definition of $|X|$). In other words, we may think of $K$ as a $B$-algebra. Choose a polynomial algebra $P$ over $B$ and a $B$-algebra surjection $P \to K$. Then $\mathop{\mathrm{Spec}}(P) \to X$ is flat as a composition $\mathop{\mathrm{Spec}}(P) \to \mathop{\mathrm{Spec}}(B) \to X$. 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 66.30.12. By Lemma 66.16.5 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 solid commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K'_1) \ar@{-->}[r] \ar@{-->}[d] & \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & U \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(C_1) \ar@{-->}[r] \ar@{-->}[rrru] & \mathop{\mathrm{Spec}}(C) \ar[rr] & & Y } \]

as in the statement of the lemma. Thus assumption (4) produces $C \to C_1$ and the dotted arrows making the diagram commute. Let $A_1' = (C_1)_\mathfrak p$ be the localization of $C_1$ at a prime $\mathfrak p \subset C_1$ lying over $\mathfrak m_{A'} \subset C$. Since $C \to C_1$ is flat by More on Algebra, Lemma 15.22.10 such a prime $\mathfrak p$ exists by Algebra, Lemmas 10.39.17 and 10.39.16. Note that $A'$ is the localization of $C$ at $\mathfrak m_{A'}$ and that $A'_1$ is a valuation ring (Algebra, Lemma 10.50.8). In other words, $A' \to A'_1$ is a local ring map of valuation rings. Assumption (3) implies

\[ \xymatrix{ \mathop{\mathrm{Spec}}(A'_1) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(C_1) \ar[r] & X \\ \mathop{\mathrm{Spec}}(A') \ar[r] & \mathop{\mathrm{Spec}}(P) \ar[r] & \mathop{\mathrm{Spec}}(B) \ar[u] } \]

commutes. Hence the restriction of the morphism $\mathop{\mathrm{Spec}}(C_1) \to X$ to $\mathop{\mathrm{Spec}}(C_1/\mathfrak p)$ restricts to the composition

\[ \mathop{\mathrm{Spec}}(\kappa (\mathfrak p)) \to \mathop{\mathrm{Spec}}(A'/\mathfrak m_{A'}) = \mathop{\mathrm{Spec}}(K'') \to \mathop{\mathrm{Spec}}(K) \to X \]

on the generic point of $\mathop{\mathrm{Spec}}(C_1/\mathfrak p)$. Moreover, $C_1/\mathfrak p$ is a valuation ring (Algebra, Lemma 10.50.8) dominating $A''$ which dominates $A$. Thus the morphism $\mathop{\mathrm{Spec}}(C_1/\mathfrak p) \to X$ witnesses the existence part of the valuative criterion for the diagram ( as desired.

Next, suppose that (5) is satisfied as well, i.e., the morphism $\Delta : X \to X \times _ S X$ is quasi-compact. In this case assumptions (1) – (4) hold for $h$ and $\Delta $. Hence the first part of the proof shows that $\Delta $ is universally closed. By Lemma 66.40.9 we conclude that $f$ is separated. $\square$

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