The Stacks project

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

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

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

which is a valuation ring by Algebra, Lemma 10.50.10. As $C$ is an $R$-algebra with fraction field $K'$, we obtain a commutative diagram

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

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. $\square$