Valuation lifts detect universal closedness

Assume first that $f$ is universally closed and fix a valuative diagram. By [F3] and [F4] the generic map and the identity of $\mathrm{Spec}R$ give a map $\mathrm{Spec}K\to X_R:=X{\times }_S\mathrm{Spec}R$. The projection $f_R:X_R\to \mathrm{Spec}R$ is closed by [F2]. The closure of the image point of $\mathrm{Spec}K$ is closed in $X_R$; its image is closed, contains the generic point of $\mathrm{Spec}R$, and therefore is all of $\mathrm{Spec}R$. Choose a point $z$ in this closure over the closed point of $\mathrm{Spec}R$.

Conversely, assume every valuative diagram for $f$ has a lift. Fix an arbitrary base change $T\to S$ and a closed subset $Z\subseteq X_T$. Give $Z$ its reduced closed-subscheme structure. On an affine chart $U=\mathrm{Spec}C$ of $X_T$, write $Z\cap U=V(I_U)$ and use the quotient chart $\mathrm{Spec}(C\mathrm{/}\sqrt{I_U})$. If $a\in C$, localization gives $(\sqrt{I_U}{)}_a=\sqrt{(I_U{)}_a}$: if $(x\mathrm{/}1{)}^n$ belongs to $(I_U{)}_a$, some power $a^m{x}^n$ lies in $I_U$, whence $(a^{m}x{)}^n\in I_U$ and $x\mathrm{/}1$ lies in $(\sqrt{I_U}{)}_a$. The reverse inclusion follows directly from the definition of radical. Therefore the quotient charts restrict to reduced closed subschemes on principal opens. Given two affine charts $U,U^{\mathrm{\prime }}$ and a point of $U\cap U^{\mathrm{\prime }}$, choose a principal open $D_U(a)$ around it contained in $U^{\mathrm{\prime }}$, then a principal open $D_{U^{\mathrm{\prime }}}(b)$ around it contained in $D_U(a)$. By [F27], the restriction of $b$ to $D_U(a)$ is $c\mathrm{/}{a}^n$ for some $c\in C$; hence this common neighborhood is $D_U(ac)=D_{U^{\mathrm{\prime }}}(b)$, principal in both charts. On it the two quotient restrictions define radical ideals with the same vanishing set $Z\cap D_U(ac)$. By [F25] the two ideals are equal, so the identity on this quotient gives their canonical overlap isomorphism. These isomorphisms satisfy the cocycle condition because they are induced by the same restrictions of ${\mathcal{O}}_{X_T}$. The quotient charts glue by [F18] to a reduced closed subscheme with underlying set $Z$. Let $g:Z\to T$ be its composite map.

To verify the DVR clause, fix a field $k$. Let $F=\mathrm{Frac}(k[\mathbb{Q}])$, where $k[\mathbb{Q}]$ is the group algebra with monomials $t^q$ for $q\in \mathbb{Q}$. The product of the least-exponent monomials is the unique least term of the product and has nonzero coefficient, so $k[\mathbb{Q}]$ is a domain. The least exponent of a nonzero Laurent polynomial defines a valuation $w:F^{\times }\to \mathbb{Q}$; for a fraction $P\mathrm{/}Q$, its value is the least exponent of $P$ minus that of $Q$; additivity of least exponents under multiplication makes this independent of the representative. The least exponent of a sum of two nonzero Laurent polynomials is at least the minimum of their least exponents, with strict increase possible when leading terms cancel; after clearing denominators this gives the valuation inequality for fractions. Let $W=\{0\}\cup \{a\in F^{\times }:w(a)\ge 0\}$, and put $K=F((u))$. For $h\in K^{\times }$, define $v(h)=({\mathrm{ord}}_u(h),w({\mathrm{lc}}_u(h)))$ in $\mathbb{Z}\times \mathbb{Q}$ with lexicographic order. Multiplication adds the $u$-orders and multiplies leading coefficients; for a sum with unequal $u$-orders, the lower-order term determines the leading term, while for equal orders the leading coefficients add and the valuation inequality for $w$ handles their sum. Leading terms show $v(h{h}^{\mathrm{\prime }})=v(h)+v(h^{\mathrm{\prime }})$ and $v(h+h^{\mathrm{\prime }})\ge \min (v(h),v(h^{\mathrm{\prime }}))$ when $h+h^{\mathrm{\prime }}\mathrm{\ne }0$. Set $v(0)=\mathrm{\infty }$. Thus $R=\{0\}\cup \{h\in K^{\times }:v(h)\ge (0,0)\}$ is a valuation ring: if $v(h)<(0,0)$ then $v(h^{-1})=-v(h)>(0,0)$. Every element of $K^{\times }$ is therefore a quotient of elements of $R$, so $\mathrm{Frac}(R)=K$.

Let $x$ be the image of the generic point. Choose an affine open $U=\mathrm{Spec}B$ around $z$; it also contains $x$, since $x$ is a generalization of $z$. Write $x=\mathfrak{p}$ and $z=\mathfrak{q}$. By [F12] and [F13], $\mathfrak{p}\subseteq \mathfrak{q}$; [F24] gives the generalization relation used here. The map $B\to K$ from the given field-valued point has kernel $\mathfrak{p}$ and induces the specified embedding $\mathrm{Frac}(B\mathrm{/}\mathfrak{p})=\kappa (x)\hookrightarrow K$ by [F21], [F28], and [F30]. Every element outside $\mathfrak{q}$ is outside $\mathfrak{p}$, so the map extends to $B_{\mathfrak{q}}={\mathcal{O}}_{X_R,z}\to K$ by [F14] and [F29]. Its kernel is $\mathfrak{p}{B}_{\mathfrak{q}}$, contained in the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$; hence its image $A\cong B_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{B}_{\mathfrak{q}}$ is local with maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{B}_{\mathfrak{q}}$ by [F6]. The structural local map $R\to {\mathcal{O}}_{X_R,z}$ followed by this quotient map is local, and its composite with $A\hookrightarrow K$ is the original inclusion $R\hookrightarrow K$. Thus it identifies $R$ with a subring of $A$ and contracts the maximal ideal of $A$ to that of $R$, so $A$ dominates $R$.

Suppose $t^{\mathrm{\prime }}\in g(Z)$ and $t$ is a specialization of $t^{\mathrm{\prime }}$. Fix $z^{\mathrm{\prime }}\in Z$ over $t^{\mathrm{\prime }}$ and put $K=\kappa (z^{\mathrm{\prime }})$. By [F21], the residue-field point gives $\mathrm{Spec}K\to Z$, with its induced field embedding $\kappa (t^{\mathrm{\prime }})\hookrightarrow K$. Choose an affine open $\mathrm{Spec}C$ around $t$; since it is open and contains the specialization $t$, it also contains $t^{\mathrm{\prime }}$. Write $t^{\mathrm{\prime }}=\mathfrak{p}$ and $t=\mathfrak{q}$, so [F13] gives $\mathfrak{p}\subseteq \mathfrak{q}$. By [F28] and [F30], $\kappa (t^{\mathrm{\prime }})=\mathrm{Frac}(C\mathrm{/}\mathfrak{p})$. Every denominator outside $\mathfrak{q}$ is outside $\mathfrak{p}$, so the stalk map ${\mathcal{O}}_{T,t}=C_{\mathfrak{q}}\to \kappa (t^{\mathrm{\prime }})$ has image exactly $(C\mathrm{/}\mathfrak{p}{)}_{\mathfrak{q}\mathrm{/}\mathfrak{p}}$ by [F14] and [F29]. This is a local domain, and its embedding in $K=\kappa (z^{\mathrm{\prime }})$ preserves that local-domain structure. By [F8], choose a valuation ring $V\subseteq K$ dominating that image. The resulting local map ${\mathcal{O}}_{T,t}\to V$ gives $\mathrm{Spec}V\to T$ by [F21], with closed point over $t$ by [F19] and generic point over $t^{\mathrm{\prime }}$.

Write $u\in K$ for the series parameter and $t=t^1\in F$ for the coefficient monomial. Every element of $R$ is a power series in $u$ whose constant coefficient lies in $W$. Taking that coefficient is a surjective ring map $R\to W$ with kernel $P=\{0\}\cup \{h\in R:{\mathrm{ord}}_u(h)>0\}$; hence $P$ is prime and $R\mathrm{/}P\cong W$. An element of $R$ is a unit exactly when its value is $(0,0)$: a positive value makes its inverse leave $R$, while value zero puts both it and its inverse in $R$. Hence [F19] identifies the maximal ideal as $\mathfrak{m}=\{h:v(h)>(0,0)\}$. We have $v(u)=(1,0)$ and $v(t)=(0,1)$, so $u\in P$ and $t\in \mathfrak{m}\setminus P$. Moreover $\sqrt{tR}=\mathfrak{m}$: for nonzero $h\in \mathfrak{m}$ with value $(n,q)$ and $n>0$, $v(h\mathrm{/}t)=(n,q-1)>(0,0)$, so $h\mathrm{/}t\in R$ and $h\in tR$. If its value is $(0,q)$ with $q>0$, choose a positive integer $N$ with $Nq\ge 1$; then $v(h^N\mathrm{/}t)=(0,Nq-1)\ge (0,0)$ and $h^N\mathrm{/}t\in R$. Conversely, since the maximal ideal is prime and contains t, it contains every element whose positive power belongs to tR; this proves the reverse inclusion in the radical equality. Thus the only prime containing $t$ is $\mathfrak{m}$. By its definition, every nonzero element of P has positive first value, so P is contained in the maximal ideal. By [F13], $P$ specializes to $\mathfrak{m}$.

Let $D(t)\hookrightarrow \mathrm{Spec}R$ be the principal open immersion. It is quasi-compact: its source is affine by [F15] and hence quasi-compact by [F16], and the single affine cover of the target suffices in [F10]. It is not closed, since it contains $P$ but omits its specialization $\mathfrak{m}$; therefore it is not universally closed by [F2], using the identity base change.

Consider any valuative diagram for this open immersion with a DVR $A\subseteq L$, fraction field $L$, ring map $\varphi :R\to A$, and generic map factoring through $D(t)$. Its kernel $\mathfrak{p}=\ker (R\to L)$ avoids $t$ by [F23]. We claim $\mathfrak{p}\in \{(0),P\}$. If $r\in \mathfrak{p}$ is nonzero and has value $(0,q)$, then $q\ge 0$. If $q=0$, $r$ is a unit, impossible; if $q>0$, $r\mathrm{/}{t}^q$ is a unit, so primality gives $t^q\in \mathfrak{p}$, and a positive integer power of $t^q$ is a positive integer power of $t$, forcing $t\in \mathfrak{p}$, also impossible. Thus every nonzero element of $\mathfrak{p}$ has value $(n,q)$ with $n>0$. For such an element $r$, if $q\le 0$ then $u^n\mathrm{/}r\in R$, so $u^n\in \mathfrak{p}$ and $u\in \mathfrak{p}$. If $q>0$, write $r=u^{n}c$ with $c=r\mathrm{/}{u}^n\in R$ of value $(0,q)$; the previous argument rules out $c\in \mathfrak{p}$, so primality again gives $u\in \mathfrak{p}$. For any $s\in P$ of value $(m,q^{\mathrm{\prime }})$, if $q^{\mathrm{\prime }}\ge 0$, then $s\mathrm{/}{u}^m\in R$, so $s\in (u^m)\subseteq \mathfrak{p}$. If $q^{\mathrm{\prime }}<0$, then $u^m\mathrm{/}s\in R$ and $u^m=s(u^m\mathrm{/}s)$, where $u^m\mathrm{/}s$ has value $(0,-q^{\mathrm{\prime }})>0$; it cannot belong to $\mathfrak{p}$ by the earlier argument, so primality gives $s\in \mathfrak{p}$. Thus $P\subseteq \mathfrak{p}$. Therefore every nonzero prime avoiding $t$ equals $P$, proving the claim.

The valuation ring $R$ has no proper local overring inside $K$ that dominates it. Indeed, if $a\in A\setminus R$, [F7] gives $a^{-1}\in R$. This inverse is not a unit of $R$, since otherwise $a\in R$; hence $a^{-1}\in {\mathfrak{m}}_R$. Domination puts $a^{-1}$ in the maximal ideal of $A$, contradicting $a{a}^{-1}=1$. Thus $A=R$. The quotient map ${\mathcal{O}}_{X_R,z}\to A=R$ is local, so [F21] gives a map $\mathrm{Spec}R\to X_R$. The valuation ring $R$ is local by [F19], and every point of $\mathrm{Spec}R$ generalizes its closed point, whose image is $z$; since $U$ is open and contains $z$, the whole map lands in $U$. Its generic restriction is the original $\mathrm{Spec}K$ map because the two maps into $U=\mathrm{Spec}B$ induce the same ring map $B\to K$. Composing with $X_R\to X$ gives the required lift.

Compose $\mathrm{Spec}K\to Z\to X_T\to X$ and $\mathrm{Spec}V\to T\to S$. They form a valuative diagram for $f$. Its lift $\mathrm{Spec}V\to X$, paired with $\mathrm{Spec}V\to T$, gives a map to $X_T$ by [F4]. By uniqueness in [F4], its generic restriction is the original map through $z^{\mathrm{\prime }}$. The closed point is a specialization of the generic point of $\mathrm{Spec}V$ by [F24], so continuity of the resulting scheme map [F26] and closedness of $Z$ put its image in $Z$ as well. Consequently $t\in g(Z)$; hence $g(Z)$ is stable under specialization.

If $\varphi (t)$ were a nonunit, its contraction $\mathfrak{q}={\varphi }^{-1}({\mathfrak{m}}_A)$ would contain $t$, hence would equal $\mathfrak{m}$ by step 2.3. Thus $\varphi (u),\varphi (t)\in {\mathfrak{m}}_A$. If $\mathfrak{p}=(0)$, the discrete valuation $v_A$ has $v_A(\varphi (u)),v_A(\varphi (t))\ge 1$. For every $n\ge 1$, $u\mathrm{/}{t}^n\in R$ since its value is $(1,-n)>(0,0)$, whence $v_A(\varphi (u))\ge n{v}_A(\varphi (t))\ge n$, impossible for all $n$. If $\mathfrak{p}=P$, the induced map $R\mathrm{/}P\cong W\to A$ is injective. For each $n\ge 1$, the constant-series lifts of $b_n=t^{1\mathrm{/}(2n)}$ and $t$ lie in ${\mathfrak{m}}_R$. Since $\mathfrak{q}=\mathfrak{m}$, their images are nonzero elements of ${\mathfrak{m}}_A$, so have positive integer $v_A$-values. The relation $t=b_n^n{t}^{1\mathrm{/}2}$ gives $v_A(\varphi (t))\ge n$ for every $n$, again impossible. Thus $\varphi (t)$ is a unit. The whole map $\mathrm{Spec}A\to \mathrm{Spec}R$ then factors through $D(t)$, giving the required lift for every DVR diagram.

The morphism $X_T\to T$ is quasi-compact by [F10]. For every affine open $U\subseteq T$, its inverse image is quasi-compact. Its intersection with the closed subset $Z$ is closed in that quasi-compact space and therefore quasi-compact. This intersection is $g^{-1}(U)$, so [F9] shows $g$ is quasi-compact. Now [F11] and step 3.4 show that $g(Z)$ is closed. Since $T$ and $Z$ were arbitrary, every base change of $f$ is closed on every closed subset, which is universal closedness by [F2].

The universally closed-to-lift implication is completed in step 3.3, and the converse in step 4.2. AC is used in step 1.2 through [F25] to identify radical ideals from their vanishing sets, in step 2.2 through [F8] to dominate the local image, and in step 4.2 through [F11] to apply the quasi-compact image criterion; the forward implication and the DVR counterexample use no AC. If $X$ or $Z$ is empty, the relevant image is empty and the closedness assertions are vacuous. A valuation ring that is a field has $\mathrm{Spec}R=\mathrm{Spec}K$ and its generic map itself is the lift. A zero affine chart represents the empty scheme and contributes no points. There is no interval or endpoint parameter in the claim.