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An open immersion has valuative uniqueness but not existence

Example

Let $k$ be a field and let $j : D (t) = Spec k [t, t^{- 1}] ↪ Spec k [t]$ be the inclusion of the complement of the origin. Then $j$ is an open immersion, hence separated, so every valuative diagram for $j$ has at most one lift. Existence can fail: the diagram with $R = k [t]_{(t)} \subseteq k (t) = K$ , base map $Spec R \to Spec k [t]$ the localization $k [t] \to k [t]_{(t)}$ , and generic map $Spec K \to D (t)$ corresponding to $k [t, t^{- 1}] \to K$ , $t \mapsto t$ , has no lift to $D (t)$ . Thus uniqueness is strictly weaker than existence, exactly as the criterion of separatedness asserts.

Facts & Assumptions

Given: A field $k$ , the open immersion $j : D (t) ↪ Spec k [t]$ of the complement of the origin, and the ring $R = k [t]_{(t)}$ with fraction field $K = k (t)$ .

[F1]

Every open immersion, every closed immersion and every immersion of schemes is separated as a morphism. (Open and closed immersions are separated)

[F2]

If $f : X \to S$ is separated, then every valuative diagram for $f$ has at most one lift. (Separatedness implies valuative uniqueness)

[F3]

A valuative diagram for $f : X \to S$ consists of a valuation ring $R \subseteq K$ with fraction field $K$ , a morphism $Spec K \to X$ and a morphism $Spec R \to S$ forming a commutative square; a lift is a compatible $Spec R \to X$ . (Valuative uniqueness diagram)

[F4]

The order of vanishing at $0$ defines a discrete valuation $v$ on $k (t)$ with $v (k (t)^{\times}) = Z$ , and $k [t]_{(t)}$ is its valuation ring, so it is a discrete valuation ring with fraction field $k (t)$ , not a field. (Discrete valuations, Discrete valuation rings)

Verification

1.1F1F2

The morphism $j$ is an open immersion, so by [F1] it is separated; hence [F2] gives at most one lift for every valuative diagram for $j$ .

1.2F3F4

By [F4] the ring $R = k [t]_{(t)}$ is a discrete valuation ring with fraction field $K = k (t)$ , so $R \subseteq K$ is a valuation ring for the purposes of [F3].

2.1F3step 1.2given

Let $Spec K \to D (t)$ correspond to the ring map $k [t, t^{- 1}] \to K$ with $t \mapsto t$ ; its image is the generic point, which lies in $D (t)$ . Let $Spec R \to Spec k [t]$ correspond to the localization $k [t] \to R$ . The two composites $Spec K \to Spec k [t]$ agree, so this is a valuative diagram for $j$ in the sense of [F3].

3.1F3step 2.1algebra

Suppose there were a lift $u : Spec R \to D (t)$ . Then $u$ corresponds to a ring homomorphism $ψ : k [t, t^{- 1}] \to R$ with $ψ (t) = t$ , since composing $u$ with $j$ must give the base map, whose corresponding ring map is the localization $k [t] \to R$ .

4.1step 1.2step 3.1algebra

Here $t$ is a unit of $k [t, t^{- 1}]$ , so $ψ (t)$ must be a unit of $R$ , every ring homomorphism sending units to units. But the image of $t$ under the localization $k [t] \to R$ is the element $t \in R$ , which is not a unit: the maximal ideal of $R$ is $(t)$ , so $t$ lies in the maximal ideal of the local ring $R$ and cannot be invertible there.

5.1step 1.1step 3.1step 4.1∎

Steps 3.1 and 4.1 contradict each other, so no lift exists; combined with step 1.1, the displayed valuative diagram has exactly zero lifts although every valuative diagram for $j$ has at most one. This shows that the uniqueness part of the valuative criterion carries no existence assertion.

Depends on

Used by

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Sources

Ravi Vakil, The Rising Sea, Theorem 13.7.4 and Exercise 13.7.A, printed p.383 (standard reference, not scraped)
The Stacks Project, Schemes, Lemma 26.22.1 and Section 26.23, printed pp.44-45 (standard reference, not scraped)