CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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Two DVR lifts of one diagram over the doubled-origin line

Statement refuted

For the affine line $D$ with doubled origin over a field $k$ , every valuative diagram for $D \to Spec k$ whose valuation ring is a discrete valuation ring has at most one lift.

Facts & Assumptions

Given: A field $k$ , the doubled-origin line $D$ with charts $U = Spec k [x]$ , $V = Spec k [y]$ glued by the identity on $D (x) ≅ D (y)$ , and the ring $R = k [t]_{(t)}$ with fraction field $K = k (t)$ .

[F1]

$D$ is obtained by gluing $U$ and $V$ by the identity on the complement of the origin: the two copies of every nonzero point are identified and the two closed points $0_{1}, 0_{2}$ remain distinct. The chart inclusions agree on the identified open $D (x) ≅ D (y)$ ; the maps $k [x] \to K$ , $x \mapsto t$ , and $k [y] \to K$ , $y \mapsto t$ , therefore define the same morphism $Spec K \to D$ . The two origins remain distinct. (The affine line with doubled origin is not separated)

[F2]

A valuative diagram for $f : X \to S$ is a valuation ring $R \subseteq K$ with fraction field $K$ together with morphisms $Spec K \to X$ and $Spec R \to S$ forming a commutative square; a lift is a morphism $Spec R \to X$ making both triangles commute. (Valuative uniqueness diagram)

[F3]

A discrete valuation on a field $K$ is a valuation $v : K \to Z \cup {\infty}$ such that $v : K^{\times} \to Z$ is surjective; its valuation ring is $V_{v} = {x \in K : v (x) \geq 0}$ , and a discrete valuation ring is a subring of this form, so it is not a field. (Discrete valuations, Discrete valuation rings)

[F4]

A valuation on a field $K$ is a function $v : K \to Γ \cup {\infty}$ with values in an ordered abelian group, satisfying $v (x) = \infty$ if and only if $x = 0$ , $v (x y) = v (x) + v (y)$ and $v (x + y) \geq \min (v (x), v (y))$ . (Valuations on a field)

Counterexample

1.1

Define $v (f / g) : = {ord}_{0} (f) - {ord}_{0} (g)$ for nonzero $f, g \in k [t]$ and $v (0) : = \infty$ , where ${ord}_{0}$ is the order of vanishing at $0$ . By [F4] this is a valuation: multiplicativity is clear from additivity of ${ord}_{0}$ and the ultrametric inequality follows from the Taylor expansion of $f$ and $g$ at $0$ ; it is surjective onto $Z$ since $v (t) = 1$ , so by [F3] it is a discrete valuation and $R = {h \in K : v (h) \geq 0} = k [t]_{(t)}$ is a discrete valuation ring with fraction field $K$ .

F3F4algebra

2.1

Let $Spec K \to D$ be the morphism with image the generic point of the shared $G_{m}$ , obtained by composing $k [x] \to K$ , $x \mapsto t$ , with the chart inclusion $U ↪ D$ , and let $Spec R \to Spec k$ be the structure morphism. The square commutes, so this is a valuative diagram for $D \to Spec k$ in the sense of [F2].

F1F2step 1.1

3.1

The generic point of $Spec R$ lies in the shared overlap, so the composite $Spec K \to U ↪ D$ coincides with $Spec K \to V ↪ D$ : both are given by the inclusion $k [t] \to k (t)$ read in the two charts, and $t$ is a unit in the glued $G_{m}$ .

F1step 2.1

4.1

The ring maps $k [x] \to R$ , $x \mapsto t$ , and $k [y] \to R$ , $y \mapsto t$ , define morphisms $u_{R} : Spec R \to U ↪ D$ and $w_{R} : Spec R \to V ↪ D$ . Their composites to $Spec k$ are the structure morphism. After restricting to $Spec K$ , they agree with the generic map of step 2.1 because the chart identifications on $D (x) ≅ D (y)$ identify $x$ and $y$ . Hence $u_{R}$ and $w_{R}$ are two lifts of the valuative diagram.

F1F2step 2.1step 3.1

5.1

The lifts $u_{R} \neq w_{R}$ are distinct: the closed point of $Spec R$ , corresponding to the maximal ideal $(t)$ , is sent by $u_{R}$ to the origin $0_{1}$ of the chart $U$ and by $w_{R}$ to the origin $0_{2}$ of the chart $V$ , and $0_{1} \neq 0_{2}$ by [F1].

F1step 4.1

6.1

Hence the displayed valuative diagram has two distinct lifts, refuting the claimed uniqueness for $D$ . In this example the single discrete valuation ring $k [t]_{(t)}$ already detects the failure of uniqueness; this witness does not establish that DVR tests are insufficient for other morphisms.

F2step 4.1step 5.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ravi Vakil, The Rising Sea, Exercise 13.7.C, printed p.382 (standard reference, not scraped)
The Stacks Project, Schemes, Lemma 26.22.2 and Example 26.22.2, printed p.44 (standard reference, not scraped)