DVR uniqueness need not detect nonseparatedness

For each $n\ge 1$, write $u=t^{1\mathrm{/}n}$ and let ${\mathrm{ord}}_{u=0}$ be the usual order-of-vanishing discrete valuation on $k(u)$. Scale it by $1\mathrm{/}n$. These valuations are compatible under the transition maps indexed by divisibility, since $t^{1\mathrm{/}n}=(t^{1\mathrm{/}m}{)}^{m\mathrm{/}n}$ when $n\mid m$; multiplicativity and the ultrametric inequality for ${\mathrm{ord}}_{u=0}$ are preserved by this positive rescaling, so they define a valuation $v:K\to \mathbb{Q}\cup \{\mathrm{\infty }\}$ with $v(0)=\mathrm{\infty }$ [F2]. Its value group is $\mathbb{Q}$, since $v(t^{1\mathrm{/}n})=1\mathrm{/}n$. At each stage, the elements of nonnegative value are exactly $k[t^{1\mathrm{/}n}{]}_{(t^{1\mathrm{/}n})}$, so $V=\{z\in K:v(z)\ge 0\}$ is a valuation ring by [F1], with maximal ideal ${\mathfrak{m}}_V=\{z:v(z)>0\}\cup \{0\}$ and residue field $k$ (taking residues at any stage gives the same element of $k$). It is not a discrete valuation ring by [F2]. It is not Noetherian: if ${\mathfrak{m}}_V$ were generated by finitely many nonzero elements, the least positive value $q$ of those generators would bound below the value of every element they generate, whereas $t^{1\mathrm{/}n}\in {\mathfrak{m}}_V$ has value $1\mathrm{/}n<q$ for all sufficiently large $n$.

We show that every valuative diagram for $f$ over a discrete valuation ring $A$ has at most one lift. Let $A$ be a discrete valuation ring by [F2], with fraction field $L$, closed point $s$ and generic point ${\eta }_A$, and let $u,v:\mathrm{Spec}A\to X$ be two lifts of one diagram; then $u$ and $v$ agree on $\mathrm{Spec}L$, so $u({\eta }_A)=v({\eta }_A)=x$.

Every nonzero prime ideal $\mathfrak{p}$ of $V$ is ${\mathfrak{m}}_V$. Indeed, choose $0\mathrm{\ne }g\in \mathfrak{p}$; since $\mathfrak{p}$ is proper, $v(g)>0$. For any nonzero $h\in {\mathfrak{m}}_V$, choose $N$ with $Nv(h)\ge v(g)$. Then $h^N\mathrm{/}g\in V$, so $h^N\in gV\subseteq \mathfrak{p}$, and primality gives $h\in \mathfrak{p}$. Thus ${\mathfrak{m}}_V\subseteq \mathfrak{p}$, so equality holds. The only primes of $V$ are therefore $(0)$ and ${\mathfrak{m}}_V$. Fix $0\mathrm{\ne }g\in {\mathfrak{m}}_V$. For any $z\in K$, if $v(z)\ge 0$ then $z\in V$; if $v(z)<0$, choose $N$ with $v(z)+Nv(g)\ge 0$, so $z{g}^N\in V$ and $z=(z{g}^N)\mathrm{/}{g}^N\in V_g$. Hence $V_g=K$, and [F7] identifies $\mathrm{Spec}K$ with $D(g)\subseteq \mathrm{Spec}V$. Now [F3] glues $U$ and $W$ along this open to a scheme $X$ with $U\cap W=\mathrm{Spec}K$ and $U\cup W=X$, and the identity $V\to V$ on each chart induces $f:X\to S=\mathrm{Spec}V$.

Apply [F10] to the one-member affine cover $S=\{S\}$ and the affine open cover $f^{-1}(S)=X=U\cup W$. The pairwise intersections in this cover are $U$, $W$, and $U\cap W=\mathrm{Spec}K$; all are affine by construction in step 2.1, hence each is a finite union of affine opens. Therefore $f$ is quasi-separated.

$X\to S$ is not separated. The pair $U,W$ consists of affine opens over the affine base $S$, and the map $V{\otimes }_{V}V\to \mathrm{\Gamma }(U\cap W,{\mathcal{O}}_X)=K$ of [F4] is the multiplication $V\to K$, whose image $V$ is a proper subring of $K$: for instance $t^{-1}\in K$ satisfies $v(t^{-1})=-1<0$ and is therefore not in $V$ by step 1.1. So the criterion of [F4] fails and $X\to S$ is not separated.

The point $x\in X$ is either the generic point $\eta$ with residue field $K$ or one of the two closed points $p_1\in U$, $p_2\in W$ with residue field $k$; these three are the only points of $X$ by step 2.1.

Suppose $x=p_1$ (the case $x=p_2$ is symmetric). Since ${\eta }_A$ specializes to $s$, its image $x$ specializes to $u(s)$, and the closure of the closed point $p_1$ is $\{p_1\}$, so $u(s)=p_1$ and likewise $v(s)=p_1$. The stalk map at $p_1$ is a local homomorphism $V\to A$ by [F6], and its localization at the prime $(0)$ of $A$ is the stalk map at ${\eta }_A$, namely the map $V\to L$ induced by the generic map, which lands at the closed point $p_1$ and therefore kills ${\mathfrak{m}}_V$; since $A$ is a domain, an element of $A$ is zero exactly when its image in $L$ is zero, so $V\to A$ kills ${\mathfrak{m}}_V$ and factors through $V\mathrm{/}{\mathfrak{m}}_V=k$. The resulting map $k\to A$ is determined by its composite $k\to A\to L$, which is fixed by the generic map; hence $u=v$.

Suppose $x=\eta$. If some lift $u$ had $u(s)=p_1$ or $u(s)=p_2$, its stalk map $V\to A$ at that closed point would be a local homomorphism by [F6] whose localization at $(0)\subseteq A$ equals the map $V\to K\to L$ induced by the generic map. Set $a=u^{\mathrm{\#}}(t)\in A$. The generic morphism induces a field map $K\to L$, which is injective, so the image of the nonzero element $t\in K$ is nonzero; because $A\hookrightarrow L$, this gives $a\mathrm{\ne }0$. Each $t^{1\mathrm{/}n}$ lies in ${\mathfrak{m}}_V$, so its image $a_n\in A$ lies in the maximal ideal. Since $a_n^n=a\mathrm{\ne }0$, we have $a_n\mathrm{\ne }0$ and $v_A(a_n)\ge 1$. Multiplicativity now gives $v_A(a)=n{v}_A(a_n)\ge n$ for every $n$, contradicting that the nonzero element $a$ has finite discrete valuation. Hence $u(s)=v(s)=\eta$, both lifts factor through the open subscheme $\mathrm{Spec}K\subseteq X$, and each is determined by a ring map $K\to A$ whose composite with $A\to L$ is the fixed map $K\to L$ induced by the generic map; since $K\to L$ is injective, that ring map is unique and $u=v$.

The diagram over the valuation ring $V$ itself does have two lifts: the two chart inclusions $U\hookrightarrow X$ and $W\hookrightarrow X$ are morphisms $\mathrm{Spec}V\to X$ whose composites with $f$ are the identity of $S$ and which agree on $\mathrm{Spec}K$, since the open subscheme $\mathrm{Spec}K$ is glued to itself; they differ at the closed point, which is sent to the distinct points $p_1$ and $p_2$ of step 3.3.

Steps 4.1 and 4.2 cover both cases for the common generic image, so any two lifts of any valuative diagram for $f$ over a discrete valuation ring coincide: every such diagram has at most one lift.

By step 3.2 the morphism $f$ is not separated and by step 3.1 it is quasi-separated, while by step 5.1 every valuative diagram over a discrete valuation ring has at most one lift; step 4.3 exhibits the failure over the non-discrete valuation ring $V$ and completes the refutation.