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Tag 0CL9

91.38. Valuative criteria

We need to be careful when stating the valuative criterion. Namely, in the formulation we need to speak about commutative diagrams but we are working in a $2$-category and we need to make sure the $2$-morphisms compose correctly as well!

Definition 91.38.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider a $2$-commutative solid diagram \begin{equation} \tag{91.38.1.1} \vcenter{ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\ \mathop{\rm Spec}(A) \ar[r]^-y \ar@{..>}[ru] & \mathcal{Y} } } \end{equation} where $A$ is a valuation ring with field of fractions $K$. Let $$ \gamma : y \circ j \longrightarrow f \circ x $$ be a $2$-morphism witnessing the $2$-commutativity of the diagram. (Notation as in Categories, Sections 4.27 and 4.28.) Given (91.38.1.1) and $\gamma$ a dotted arrow is a triple $(a, \alpha, \beta)$ consisting of a morphism $a : \mathop{\rm Spec}(A) \to \mathcal{X}$ and $2$-arrows $\alpha : a \circ j \to x$, $\beta : y \to f \circ a$ such that $\gamma = (\text{id}_f \star \alpha) \circ (\beta \star \text{id}_j)$, in other words such that $$ \xymatrix{ & f \circ a \circ j \ar[rd]^{\text{id}_f \star \alpha} \\ y \circ j \ar[ru]^{\beta \star \text{id}_j} \ar[rr]^\gamma & & f \circ x } $$ is commutative. A morphism of dotted arrows $(a, \alpha, \beta) \to (a', \alpha', \beta')$ is a $2$-arrow $\theta : a \to a'$ such that $\alpha = \alpha' \circ (\theta \star \text{id}_j)$ and $\beta' = (\text{id}_f \star \theta) \circ \beta$.

The category of dotted arrows depends on $\gamma$ in general. If $\mathcal{Y}$ is representable by an algebraic space (or if automorphism groups of objects over fields are trivial), then of course there is at most one $\gamma$ and one does not need to check the commutativity of the triangle. More generally, we have Lemma 91.38.3. The commutativity of the triangle is important in the proof of compatibility with base change, see proof of Lemma 91.38.4.

Lemma 91.38.2. In the situation of Definition 91.38.1 the category of dotted arrows is a groupoid. If $\Delta_f$ is separated, then it is a setoid.

Proof. Since $2$-arrows are invertible it is clear that the category of dotted arrows is a groupoid. Given a dotted arrow $(a, \alpha, \beta)$ an automorphism of $(a, \alpha, \beta)$ is a $2$-morphism $\theta : a \to a$ satisfying two conditions. The first condition $\beta = (\text{id}_f \star \theta) \circ \beta$ signifies that $\theta$ defines a morphism $(a, \theta) : \mathop{\rm Spec}(A) \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$. The second condition $\alpha = \alpha \circ (\theta \star \text{id}_j)$ implies that the restriction of $(a, \theta)$ to $\mathop{\rm Spec}(K)$ is the identity. Picture $$ \xymatrix{ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] & & \mathop{\rm Spec}(K) \ar[d]^j \ar[ll]_{(a \circ j, \text{id})} \\ \mathcal{X} & & \mathop{\rm Spec}(A) \ar[ll]_a \ar[llu]_{(a, \theta)} } $$ In other words, if $G \to \mathop{\rm Spec}(A)$ is the group algebraic space we get by pulling back the relative inertia by $a$, then $\theta$ defines a point $\theta \in G(A)$ whose image in $G(K)$ is trivial. Certainly, if the identity $e : \mathop{\rm Spec}(A) \to G$ is a closed immersion, then this can happen only if $\theta$ is the identity. Looking at Lemma 91.6.1 we obtain the result we want. $\square$

Lemma 91.38.3. In Definition 91.38.1 assume $\mathcal{I}_\mathcal{Y} \to \mathcal{Y}$ is proper (for example if $\mathcal{Y}$ is separated or if $\mathcal{Y}$ is separated over an algebraic space). Then the category of dotted arrows is independent (up to noncanonical equivalence) of the choice of $\gamma$ and the existence of a dotted arrow (for some and hence equivalently all $\gamma$) is equivalent to the existence of a diagram $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\ \mathop{\rm Spec}(A) \ar[r]^-y \ar[ru]_a & \mathcal{Y} } $$ with $2$-commutative triangles (without checking the $2$-morphisms compose correctly).

Proof. Let $\gamma, \gamma' : y \circ j \longrightarrow f \circ x$ be two $2$-morphisms. Then $\gamma^{-1} \circ \gamma'$ is an automorphism of $y$ over $\mathop{\rm Spec}(K)$. Hence if $\mathit{Isom}_\mathcal{Y}(y, y) \to \mathop{\rm Spec}(A)$ is proper, then by the valuative criterion of properness (Morphisms of Spaces, Lemma 58.43.1) we can find $\delta : y \to y$ whose restriction to $\mathop{\rm Spec}(K)$ is $\gamma^{-1} \circ \gamma'$. Then we can use $\delta$ to define an equivalence between the category of dotted arrows for $\gamma$ to the category of dotted arrows for $\gamma'$ by sending $(a, \alpha, \beta)$ to $(a, \alpha, \beta \circ \delta)$. The final statement is clear. $\square$

Lemma 91.38.4. Assume given a $2$-commutative diagram $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_-{x'} \ar[d]_j & \mathcal{X}' \ar[d]^p \ar[r]_q & \mathcal{X} \ar[d]^f \\ \mathop{\rm Spec}(A) \ar[r]^-{y'} & \mathcal{Y}' \ar[r]^g & \mathcal{Y} } $$ with the right square $2$-cartesian. Choose a $2$-arrow $\gamma' : y' \circ j \to p \circ x'$. Set $x = q \circ x'$, $y = g \circ y'$ and let $\gamma : y \circ j \to f \circ x$ be the composition of $\gamma'$ with the $2$-arrow implicit in the $2$-commutativity of the right square. Then the category of dotted arrows for the left square and $\gamma'$ is equivalent to the category of dotted arrows for the outer rectangle and $\gamma$.

Proof. This lemma, although a bit of a brain teaser, is straightforward. (We do not know how to prove the analogue of this lemma if instead of the category of dotted arrows we look at the set of isomorphism classes of morphisms producing two $2$-commutative triangles as in Lemma 91.38.3; in fact this analogue may very well be wrong.) To prove the lemma we are allowed to replace $\mathcal{X}'$ by the $2$-fibre product $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$ as described in Categories, Lemma 4.31.3. Then the object $x'$ becomes the triple $(y' \circ j, x, \gamma)$. Then we can go from a dotted arrow $(a, \alpha, \beta)$ for the outer rectangle to a dotted arrow $(a', \alpha', \beta')$ for the left square by taking $a' = (y', a, \beta)$ and $\alpha' = (\text{id}_{y' \circ j}, \alpha)$ and $\beta' = \text{id}_{y'}$. Details omitted. $\square$

Lemma 91.38.5. Assume given a $2$-commutative diagram $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_-x \ar[dd]_j & \mathcal{X} \ar[d]^f \\ & \mathcal{Y} \ar[d]^g \\ \mathop{\rm Spec}(A) \ar[r]^-z & \mathcal{Z} } $$ Choose a $2$-arrow $\gamma : z \circ j \to g \circ f \circ x$. Let $\mathcal{C}$ be the category of dotted arrows for the outer rectangle and $\gamma$. Let $\mathcal{C}'$ be the category of dotted arrows for the square $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_-{f \circ x} \ar[d]_j & \mathcal{Y} \ar[d]^g \\ \mathop{\rm Spec}(A) \ar[r]^-z & \mathcal{Z} } $$ and $\gamma$. There is a canonical functor $\mathcal{C} \to \mathcal{C}'$ which turns $\mathcal{C}$ into a category fibred in groupoids over $\mathcal{C}'$ and whose fibre categories are categories of dotted arrows for certain squares of the form $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\ \mathop{\rm Spec}(A) \ar[r]^-y & \mathcal{Y} } $$ and some choice of $y \circ j \to f \circ x$.

Proof. Omitted. Hint: If $(a, \alpha, \beta)$ is an object of $\mathcal{C}$, then $(f \circ a, \text{id}_f \star \alpha, \beta)$ is an object of $\mathcal{C}'$. Conversely, if $(y, \delta, \epsilon)$ is an object of $\mathcal{C}'$ and $(a, \alpha, \beta)$ is an object of the category of dotted arrows of the last displayed diagram with $y \circ j \to f \circ x$ given by $\delta$, then $(a, \alpha, (\text{id}_g \star \beta) \circ \epsilon)$ is an object of $\mathcal{C}$. $\square$

Definition 91.38.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ satisfies the uniqueness part of the valuative criterion if for every diagram (91.38.1.1) and $\gamma$ as in Definition 91.38.1 the category of dotted arrows is either empty or a setoid with exactly one isomorphism class.

Lemma 91.38.7. The base change of a morphism of algebraic stacks which satisfies the uniqueness part of the valuative criterion by any morphism of algebraic stacks is a morphism of algebraic stacks which satisfies the uniqueness part of the valuative criterion.

Proof. Follows from Lemma 91.38.4 and the definition. $\square$

Lemma 91.38.8. The composition of morphisms of algebraic stacks which satisfy the uniqueness part of the valuative criterion is another morphism of algebraic stacks which satisfies the uniqueness part of the valuative criterion.

Proof. Follows from Lemma 91.38.5 and the definition. $\square$

Lemma 91.38.9. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. Then the following are equivalent

  1. $f$ satisfies the uniqueness part of the valuative criterion,
  2. for every scheme $T$ and morphism $T \to \mathcal{Y}$ the morphism $\mathcal{X} \times_\mathcal{Y} T \to T$ satisfies the uniqueness part of the valuative criterion as a morphism of algebraic spaces.

Proof. Omitted. $\square$

Definition 91.38.10. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ satisfies the existence part of the valuative criterion if for every diagram (91.38.1.1) and $\gamma$ as in Definition 91.38.1 there exists an extension $K'/K$ of fields, a valuation ring $A' \subset K'$ dominating $A$ such that the category of dotted arrows for the outer rectangle of the diagram $$ \xymatrix{ \mathop{\rm Spec}(K') \ar[r] \ar@/^2em/[rr]_{x'} \ar[d]_{j'} & \mathop{\rm Spec}(K) \ar[d]_j \ar[r]_-x & \mathcal{X} \ar[d]^f \\ \mathop{\rm Spec}(A') \ar[r] \ar@/_2em/[rr]^{y'} & \mathop{\rm Spec}(A) \ar[r]^-y & \mathcal{Y} } $$ with induced $2$-arrow $\gamma' : y' \circ j' \to f \circ x'$ is nonempty.

We have already seen in the case of morphisms of algebraic spaces, that it is necessary to allow extensions of the fractions fields in order to get the correct notion of the valuative criterion. See Morphisms of Spaces, Example 58.40.6. Still, for morphisms between separated algebraic spaces, such an extension is not needed (Morphisms of Spaces, Lemma 58.40.5). However, for morphisms between algebraic stacks, an extension may be needed even if $\mathcal{X}$ and $\mathcal{Y}$ are both separated. For example consider the morphism of algebraic stacks $$ [\mathop{\rm Spec}(\mathbf{C})/G] \to \mathop{\rm Spec}(\mathbf{C}) $$ over the base scheme $\mathop{\rm Spec}(\mathbf{C})$ where $G$ is a group of order $2$. Both source and target are separated algebraic stacks and the morphism is proper. Whence it satisfies the uniqueness and existence parts of the valuative criterion (see Lemma 91.42.1). But on the other hand, there is a diagram $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r] \ar[d] & [\mathop{\rm Spec}(\mathbf{C})/G] \ar[d] \\ \mathop{\rm Spec}(A) \ar[r] & \mathop{\rm Spec}(\mathbf{C}) } $$ where no dotted arrow exists with $A = \mathbf{C}[[t]]$ and $K = \mathbf{C}((t))$. Namely, the top horizontal arrow is given by a $G$-torsor over the spectrum of $K = \mathbf{C}((t))$. Since any $G$-torsor over the strictly henselian local ring $A = \mathbf{C}[[t]]$ is trivial, we see that if a dotted arrow always exists, then every $G$-torsor over $K$ is trivial. This is not true because $G = \{+1, -1\}$ and by Kummer theory $G$-torsors over $K$ are classified by $K^*/(K^*)^2$ which is nontrivial.

Lemma 91.38.11. The base change of a morphism of algebraic stacks which satisfies the existence part of the valuative criterion by any morphism of algebraic stacks is a morphism of algebraic stacks which satisfies the existence part of the valuative criterion.

Proof. Follows from Lemma 91.38.4 and the definition. $\square$

Lemma 91.38.12. The composition of morphisms of algebraic stacks which satisfy the existence part of the valuative criterion is another morphism of algebraic stacks which satisfies the existence part of the valuative criterion.

Proof. Follows from Lemma 91.38.5 and the definition. $\square$

Lemma 91.38.13. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. Then the following are equivalent

  1. $f$ satisfies the existence part of the valuative criterion,
  2. for every scheme $T$ and morphism $T \to \mathcal{Y}$ the morphism $\mathcal{X} \times_\mathcal{Y} T \to T$ satisfies the existence part of the valuative criterion as a morphism of algebraic spaces.

Proof. Omitted. $\square$

Lemma 91.38.14. A closed immersion of algebraic stacks satisfies both the existence and uniqueness part of the valuative criterion.

Proof. Omitted. Hint: reduce to the case of a closed immersion of schemes by Lemmas 91.38.9 and 91.38.13. $\square$

    The code snippet corresponding to this tag is a part of the file stacks-morphisms.tex and is located in lines 8576–8965 (see updates for more information).

    \section{Valuative criteria}
    \label{section-valuative}
    
    \noindent
    We need to be careful when stating the valuative criterion. Namely, in
    the formulation we need to speak about commutative diagrams but we are
    working in a $2$-category and we need to make sure the $2$-morphisms
    compose correctly as well!
    
    \begin{definition}
    \label{definition-fill-in-diagram}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
    Consider a $2$-commutative solid diagram
    \begin{equation}
    \label{equation-diagram}
    \vcenter{
    \xymatrix{
    \Spec(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\
    \Spec(A) \ar[r]^-y \ar@{..>}[ru] & \mathcal{Y}
    }
    }
    \end{equation}
    where $A$ is a valuation ring with field of fractions $K$. Let
    $$
    \gamma : y \circ j \longrightarrow f \circ x
    $$
    be a $2$-morphism witnessing the $2$-commutativity of the diagram.
    (Notation as in Categories, Sections \ref{categories-section-formal-cat-cat}
    and \ref{categories-section-2-categories}.)
    Given (\ref{equation-diagram}) and $\gamma$
    a {\it dotted arrow} is a triple $(a, \alpha, \beta)$ consisting of a
    morphism $a : \Spec(A) \to \mathcal{X}$ and $2$-arrows
    $\alpha : a \circ j \to x$, $\beta : y \to f \circ a$
    such that
    $\gamma = (\text{id}_f \star \alpha) \circ (\beta \star \text{id}_j)$,
    in other words such that
    $$
    \xymatrix{
    & f \circ a \circ j \ar[rd]^{\text{id}_f \star \alpha} \\
    y \circ j \ar[ru]^{\beta \star \text{id}_j} \ar[rr]^\gamma & &
    f \circ x
    }
    $$
    is commutative. A {\it morphism of dotted arrows}
    $(a, \alpha, \beta) \to (a', \alpha', \beta')$ is a
    $2$-arrow $\theta : a \to a'$ such that
    $\alpha = \alpha' \circ (\theta \star \text{id}_j)$ and
    $\beta' = (\text{id}_f \star \theta) \circ \beta$.
    \end{definition}
    
    \noindent
    The category of dotted arrows depends on $\gamma$ in general.
    If $\mathcal{Y}$ is representable by an algebraic space
    (or if automorphism groups of objects over fields are trivial), then
    of course there is at most one $\gamma$ and one does not need
    to check the commutativity of the triangle. More generally, we
    have Lemma \ref{lemma-cat-dotted-arrows-independent}.
    The commutativity of the triangle is important in the proof
    of compatibility with base change, see
    proof of Lemma \ref{lemma-cat-dotted-arrows-base-change}.
    
    \begin{lemma}
    \label{lemma-cat-dotted-arrows}
    In the situation of Definition \ref{definition-fill-in-diagram}
    the category of dotted arrows is a groupoid. If $\Delta_f$
    is separated, then it is a setoid.
    \end{lemma}
    
    \begin{proof}
    Since $2$-arrows are invertible it is clear that the category of
    dotted arrows is a groupoid. Given a dotted arrow $(a, \alpha, \beta)$
    an automorphism of $(a, \alpha, \beta)$ is a $2$-morphism
    $\theta : a \to a$ satisfying two conditions. The first condition
    $\beta = (\text{id}_f \star \theta) \circ \beta$ signifies that
    $\theta$ defines a morphism
    $(a, \theta) : \Spec(A) \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$.
    The second condition
    $\alpha = \alpha \circ (\theta \star \text{id}_j)$
    implies that the restriction of $(a, \theta)$ to $\Spec(K)$
    is the identity. Picture
    $$
    \xymatrix{
    \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] & &
    \Spec(K) \ar[d]^j \ar[ll]_{(a \circ j, \text{id})} \\
    \mathcal{X} & & \Spec(A) \ar[ll]_a \ar[llu]_{(a, \theta)}
    }
    $$
    In other words, if $G \to \Spec(A)$ is the group algebraic space
    we get by pulling back the relative inertia by $a$, then
    $\theta$ defines a point $\theta \in G(A)$ whose image
    in $G(K)$ is trivial. Certainly, if the identity $e : \Spec(A) \to G$
    is a closed immersion, then this can happen only if
    $\theta$ is the identity.
    Looking at Lemma \ref{lemma-diagonal-diagonal}
    we obtain the result we want.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-cat-dotted-arrows-independent}
    In Definition \ref{definition-fill-in-diagram}
    assume $\mathcal{I}_\mathcal{Y} \to \mathcal{Y}$ is proper
    (for example if $\mathcal{Y}$ is separated or if $\mathcal{Y}$
    is separated over an algebraic space). Then the category of dotted arrows
    is independent (up to noncanonical equivalence) of the choice of $\gamma$
    and the existence of a dotted arrow
    (for some and hence equivalently all $\gamma$)
    is equivalent to the existence of a diagram
    $$
    \xymatrix{
    \Spec(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\
    \Spec(A) \ar[r]^-y \ar[ru]_a & \mathcal{Y}
    }
    $$
    with $2$-commutative triangles
    (without checking the $2$-morphisms compose correctly).
    \end{lemma}
    
    \begin{proof}
    Let $\gamma, \gamma' : y \circ j \longrightarrow f \circ x$
    be two $2$-morphisms. Then $\gamma^{-1} \circ \gamma'$
    is an automorphism of $y$ over $\Spec(K)$.
    Hence if $\mathit{Isom}_\mathcal{Y}(y, y) \to \Spec(A)$
    is proper, then by the valuative criterion of properness
    (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-characterize-proper})
    we can find $\delta : y \to y$ whose restriction to
    $\Spec(K)$ is $\gamma^{-1} \circ \gamma'$.
    Then we can use $\delta$ to define an equivalence
    between the category of dotted arrows for $\gamma$
    to the category of dotted arrows for $\gamma'$ by
    sending $(a, \alpha, \beta)$ to $(a, \alpha, \beta \circ \delta)$.
    The final statement is clear.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-cat-dotted-arrows-base-change}
    Assume given a $2$-commutative diagram
    $$
    \xymatrix{
    \Spec(K) \ar[r]_-{x'} \ar[d]_j &
    \mathcal{X}' \ar[d]^p \ar[r]_q &
    \mathcal{X} \ar[d]^f \\
    \Spec(A) \ar[r]^-{y'} &
    \mathcal{Y}' \ar[r]^g &
    \mathcal{Y}
    }
    $$
    with the right square $2$-cartesian. Choose a $2$-arrow
    $\gamma' : y' \circ j \to p \circ x'$. Set
    $x = q \circ x'$, $y = g \circ y'$ and let
    $\gamma : y \circ j \to f \circ x$ be the composition of
    $\gamma'$ with the $2$-arrow implicit in the $2$-commutativity
    of the right square. Then the category of dotted arrows
    for the left square and $\gamma'$ is equivalent to the category of dotted
    arrows for the outer rectangle and $\gamma$.
    \end{lemma}
    
    \begin{proof}
    This lemma, although a bit of a brain teaser, is straightforward.
    (We do not know how to prove the analogue of this lemma if instead
    of the category of dotted arrows we look at the set of isomorphism
    classes of morphisms producing two $2$-commutative
    triangles as in Lemma \ref{lemma-cat-dotted-arrows-independent};
    in fact this analogue may very well be wrong.)
    To prove the lemma we are allowed to replace
    $\mathcal{X}'$ by the $2$-fibre product
    $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$
    as described in Categories, Lemma
    \ref{categories-lemma-2-product-categories-over-C}.
    Then the object $x'$ becomes the triple $(y' \circ j, x, \gamma)$.
    Then we can go from a dotted arrow $(a, \alpha, \beta)$ for the
    outer rectangle to a dotted arrow $(a', \alpha', \beta')$
    for the left square by taking $a' = (y', a, \beta)$ and
    $\alpha' = (\text{id}_{y' \circ j}, \alpha)$ and
    $\beta' = \text{id}_{y'}$. Details omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-cat-dotted-arrows-composition}
    Assume given a $2$-commutative diagram
    $$
    \xymatrix{
    \Spec(K) \ar[r]_-x \ar[dd]_j & \mathcal{X} \ar[d]^f \\
    & \mathcal{Y} \ar[d]^g \\
    \Spec(A) \ar[r]^-z & \mathcal{Z}
    }
    $$
    Choose a $2$-arrow $\gamma : z \circ j \to g \circ f \circ x$.
    Let $\mathcal{C}$ be the category of dotted arrows for
    the outer rectangle and $\gamma$. Let $\mathcal{C}'$ be the
    category of dotted arrows for the square
    $$
    \xymatrix{
    \Spec(K) \ar[r]_-{f \circ x} \ar[d]_j & \mathcal{Y} \ar[d]^g \\
    \Spec(A) \ar[r]^-z & \mathcal{Z}
    }
    $$
    and $\gamma$. There is a canonical functor $\mathcal{C} \to \mathcal{C}'$
    which turns $\mathcal{C}$ into a category fibred in groupoids over
    $\mathcal{C}'$ and whose fibre categories are categories of dotted arrows
    for certain squares of the form
    $$
    \xymatrix{
    \Spec(K) \ar[r]_-x \ar[d]_j & \mathcal{X} \ar[d]^f \\
    \Spec(A) \ar[r]^-y & \mathcal{Y}
    }
    $$
    and some choice of $y \circ j \to f \circ x$.
    \end{lemma}
    
    \begin{proof}
    Omitted. Hint: If $(a, \alpha, \beta)$ is an object of $\mathcal{C}$,
    then $(f \circ a, \text{id}_f \star \alpha, \beta)$ is an object
    of $\mathcal{C}'$. Conversely, if $(y, \delta, \epsilon)$ is an
    object of $\mathcal{C}'$ and $(a, \alpha, \beta)$ is an object
    of the category of dotted arrows of the last displayed diagram
    with $y \circ j \to f \circ x$ given by $\delta$, then
    $(a, \alpha, (\text{id}_g \star \beta) \circ \epsilon)$ is an
    object of $\mathcal{C}$.
    \end{proof}
    
    \begin{definition}
    \label{definition-uniqueness}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
    We say $f$ satisfies the {\it uniqueness part of the valuative criterion}
    if for every diagram (\ref{equation-diagram}) and $\gamma$
    as in Definition \ref{definition-fill-in-diagram}
    the category of dotted arrows is either empty or
    a setoid with exactly one isomorphism class.
    \end{definition}
    
    \begin{lemma}
    \label{lemma-base-change-uniqueness}
    The base change of a morphism of algebraic stacks which satisfies the
    uniqueness part of the valuative criterion by any morphism of
    algebraic stacks is a morphism of algebraic stacks which satisfies the
    uniqueness part of the valuative criterion.
    \end{lemma}
    
    \begin{proof}
    Follows from Lemma \ref{lemma-cat-dotted-arrows-base-change}
    and the definition.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-composition-uniqueness}
    The composition of morphisms of algebraic stacks which satisfy the
    uniqueness part of the valuative criterion is another
    morphism of algebraic stacks which satisfies the
    uniqueness part of the valuative criterion.
    \end{lemma}
    
    \begin{proof}
    Follows from Lemma \ref{lemma-cat-dotted-arrows-composition}
    and the definition.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-uniqueness-representable}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks
    which is representable by algebraic spaces. Then the following are equivalent
    \begin{enumerate}
    \item $f$ satisfies the uniqueness part of the valuative criterion,
    \item for every scheme $T$ and morphism $T \to \mathcal{Y}$
    the morphism $\mathcal{X} \times_\mathcal{Y} T \to T$ satisfies
    the uniqueness part of the valuative criterion as a morphism
    of algebraic spaces.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \begin{definition}
    \label{definition-existence}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
    We say $f$ satisfies the {\it existence part of the valuative criterion}
    if for every diagram (\ref{equation-diagram}) and $\gamma$
    as in Definition \ref{definition-fill-in-diagram}
    there exists an extension $K'/K$ of fields, a valuation ring $A' \subset K'$
    dominating $A$ such that the category of dotted arrows for the
    outer rectangle of the diagram
    $$
    \xymatrix{
    \Spec(K') \ar[r] \ar@/^2em/[rr]_{x'} \ar[d]_{j'} &
    \Spec(K) \ar[d]_j \ar[r]_-x &
    \mathcal{X} \ar[d]^f \\
    \Spec(A') \ar[r] \ar@/_2em/[rr]^{y'} &
    \Spec(A) \ar[r]^-y &
    \mathcal{Y}
    }
    $$
    with induced $2$-arrow $\gamma' : y' \circ j' \to f \circ x'$ is nonempty.
    \end{definition}
    
    \noindent
    We have already seen in the case of morphisms of algebraic spaces,
    that it is necessary to allow extensions of the fractions fields
    in order to get the correct notion of the valuative criterion.
    See Morphisms of Spaces, Example
    \ref{spaces-morphisms-example-finite-separable-needed}.
    Still, for morphisms between separated algebraic spaces, such
    an extension is not needed
    (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-usual-enough}).
    However, for morphisms between algebraic stacks, an extension may
    be needed even if $\mathcal{X}$ and $\mathcal{Y}$ are both separated.
    For example consider the morphism of algebraic stacks
    $$
    [\Spec(\mathbf{C})/G] \to \Spec(\mathbf{C})
    $$
    over the base scheme $\Spec(\mathbf{C})$
    where $G$ is a group of order $2$. Both source and target are separated
    algebraic stacks and the morphism is proper. Whence it
    satisfies the uniqueness and existence parts of the valuative criterion
    (see Lemma \ref{lemma-criterion-proper}).
    But on the other hand, there is a diagram
    $$
    \xymatrix{
    \Spec(K) \ar[r] \ar[d] & [\Spec(\mathbf{C})/G] \ar[d] \\
    \Spec(A) \ar[r] & \Spec(\mathbf{C})
    }
    $$
    where no dotted arrow exists with $A = \mathbf{C}[[t]]$ and
    $K = \mathbf{C}((t))$. Namely, the top horizontal arrow is given
    by a $G$-torsor over the spectrum of $K = \mathbf{C}((t))$. Since any $G$-torsor
    over the strictly henselian local ring $A = \mathbf{C}[[t]]$ is trivial, we see
    that if a dotted arrow always exists, then every $G$-torsor over
    $K$ is trivial. This is not true because $G = \{+1, -1\}$
    and by Kummer theory $G$-torsors over $K$ are classified by
    $K^*/(K^*)^2$ which is nontrivial.
    
    \begin{lemma}
    \label{lemma-base-change-existence}
    The base change of a morphism of algebraic stacks which satisfies the
    existence part of the valuative criterion by any morphism of
    algebraic stacks is a morphism of algebraic stacks which satisfies the
    existence part of the valuative criterion.
    \end{lemma}
    
    \begin{proof}
    Follows from Lemma \ref{lemma-cat-dotted-arrows-base-change}
    and the definition.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-composition-existence}
    The composition of morphisms of algebraic stacks which satisfy the
    existence part of the valuative criterion is another
    morphism of algebraic stacks which satisfies the
    existence part of the valuative criterion.
    \end{lemma}
    
    \begin{proof}
    Follows from Lemma \ref{lemma-cat-dotted-arrows-composition}
    and the definition.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-existence-representable}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks
    which is representable by algebraic spaces. Then the following are equivalent
    \begin{enumerate}
    \item $f$ satisfies the existence part of the valuative criterion,
    \item for every scheme $T$ and morphism $T \to \mathcal{Y}$
    the morphism $\mathcal{X} \times_\mathcal{Y} T \to T$ satisfies
    the existence part of the valuative criterion as a morphism
    of algebraic spaces.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Omitted.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-closed-immersion-valuative-criteria}
    A closed immersion of algebraic stacks satisfies both
    the existence and uniqueness part of the valuative criterion.
    \end{lemma}
    
    \begin{proof}
    Omitted. Hint: reduce to the case of a closed immersion of
    schemes by Lemmas \ref{lemma-uniqueness-representable} and
    \ref{lemma-existence-representable}.
    \end{proof}

    Comments (2)

    Comment #2449 by Andrea on March 9, 2017 a 11:25 pm UTC

    A little typo: twelve lines after definition 85.33.10, it should be "top horizontal arrow" instead of "top vertical arrow", right?

    Comment #2491 by Johan (site) on April 13, 2017 a 11:04 pm UTC

    Yep! Fixed here. If you want your name in the list of contributors, leave a full one.

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