The Stacks project

101.41 Valuative criterion for the diagonal

The result is Lemma 101.41.2. We first state and prove a formal helper lemma.

Lemma 101.41.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider a $2$-commutative solid diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[rr]_-x \ar[d]_ j & & \mathcal{X} \ar[d]^{\Delta _ f} \\ \mathop{\mathrm{Spec}}(A) \ar[rr]^{(a_1, a_2, \varphi )} \ar@{..>}[rru] & & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

where $A$ is a valuation ring with field of fractions $K$. Let $\gamma : (a_1, a_2, \varphi ) \circ j \longrightarrow \Delta _ f \circ x$ be a $2$-morphism witnessing the $2$-commutativity of the diagram. Then

  1. Writing $\gamma = (\alpha _1, \alpha _2)$ with $\alpha _ i : a_ i \circ j \to x$ we obtain two dotted arrows $(a_1, \alpha _1, \text{id})$ and $(a_2, \alpha _2, \varphi )$ in the diagram

    \[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d]_ j & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-{f \circ a_1} \ar@{..>}[ru] & \mathcal{Y} } \]
  2. The category of dotted arrows for the original diagram and $\gamma $ is a setoid whose set of isomorphism classes of objects equal to the set of morphisms $(a_1, \alpha _1, \text{id}) \to (a_2, \alpha _2, \varphi )$ in the category of dotted arrows.

Proof. Since $\Delta _ f$ is representable by algebraic spaces (hence the diagonal of $\Delta _ f$ is separated), we see that the category of dotted arrows in the first commutative diagram of the lemma is a setoid by Lemma 101.39.2. All the other statements of the lemma are consequences of $2$-diagramatic computations which we omit. $\square$

Lemma 101.41.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume $f$ is quasi-separated. If $f$ satisfies the uniqueness part of the valuative criterion, then $f$ is separated.

Proof. The assumption on $f$ means that $\Delta _ f$ is quasi-compact and quasi-separated (Definition 101.4.1). We have to show that $\Delta _ f$ is proper. Lemma 101.40.1 says that $\Delta _ f$ is separated. By Lemma 101.3.3 we know that $\Delta _ f$ is locally of finite type. To finish the proof we have to show that $\Delta _ f$ is universally closed. A formal argument (see Lemma 101.41.1) shows that the uniqueness part of the valuative criterion implies that we have the existence of a dotted arrow in any solid diagram like so:

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d] \ar[r] & \mathcal{X} \ar[d]^{\Delta _ f} \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{..>}[ru] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

Using that this property is preserved by any base change we conclude that any base change by $\Delta _ f$ by an algebraic space mapping into $\mathcal{X} \times _\mathcal {Y} \mathcal{X}$ has the existence part of the valuative criterion and we conclude is universally closed by the valuative criterion for morphisms of algebraic spaces, see Morphisms of Spaces, Lemma 67.42.1. $\square$

Here is a converse.

Lemma 101.41.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is separated, then $f$ satisfies the uniqueness part of the valuative criterion.

Proof. Since $f$ is separated we see that all categories of dotted arrows are setoids by Lemma 101.39.2. Consider a diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d]_ j & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-y \ar@{..>}[ru] & \mathcal{Y} } \]

and a $2$-morphism $\gamma : y \circ j \to f \circ x$ as in Definition 101.39.1. Consider two objects $(a, \alpha , \beta )$ and $(a', \beta ', \alpha ')$ of the category of dotted arrows. To finish the proof we have to show these objects are isomorphic. The isomorphism

\[ f \circ a \xrightarrow {\beta ^{-1}} y \xrightarrow {\beta '} f \circ a' \]

means that $(a, a', \beta ' \circ \beta ^{-1})$ is a morphism $\mathop{\mathrm{Spec}}(A) \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. On the other hand, $\alpha $ and $\alpha '$ define a $2$-arrow

\[ (a, a', \beta ' \circ \beta ^{-1}) \circ j = (a \circ j, a' \circ j, (\beta ' \star \text{id}_ j) \circ (\beta \star \text{id}_ j)^{-1}) \xrightarrow {(\alpha , \alpha ')} (x, x, \text{id}) = \Delta _ f \circ x \]

Here we use that both $(a, \alpha , \beta )$ and $(a', \alpha ', \beta ')$ are dotted arrows with respect to $\gamma $. We obtain a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d]_ j \ar[rr]_ x & & \mathcal{X} \ar[d]^{\Delta _ f} \\ \mathop{\mathrm{Spec}}(A) \ar[rr]^{(a, a', \beta ' \circ \beta ^{-1})} & & \mathcal{X} \times _\mathcal {Y} \mathcal{X} } \]

with $2$-commutativity witnessed by $(\alpha , \alpha ')$. Now $\Delta _ f$ is representable by algebraic spaces (Lemma 101.3.3) and proper as $f$ is separated. Hence by Lemma 101.39.13 and the valuative criterion for properness for algebraic spaces (Morphisms of Spaces, Lemma 67.44.1) we see that there exists a dotted arrow. Unwinding the construction, we see that this means $(a, \alpha , \beta )$ and $(a', \alpha ', \beta ')$ are isomorphic in the category of dotted arrows as desired. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CLS. Beware of the difference between the letter 'O' and the digit '0'.