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Tag 03KI

Chapter 55: Morphisms of Algebraic Spaces > Section 55.40: Valuative criteria

Example 55.40.6. Consider the algebraic space $X$ constructed in Spaces, Example 53.14.2. Recall that it is Galois twist of the affine line with zero doubled. The Galois twist is with respect to a degree two Galois extension $k'/k$ of fields. As such it comes with a morphism $$ \pi : X \longrightarrow S = \mathbf{A}^1_k $$ which is quasi-compact. We claim that $\pi$ is universally closed. Namely, after base change by $\mathop{\rm Spec}(k') \to \mathop{\rm Spec}(k)$ the morphism $\pi$ is identified with the morphism $$ \text{affine line with zero doubled} \longrightarrow \text{affine line} $$ which is universally closed (some details omitted). Since the morphism $\mathop{\rm Spec}(k') \to \mathop{\rm Spec}(k)$ is universally closed and surjective, a diagram chase shows that $\pi$ is universally closed. On the other hand, consider the diagram $$ \xymatrix{ \mathop{\rm Spec}(k((x))) \ar[r] \ar[d] & X \ar[d]^\pi \\ \mathop{\rm Spec}(k[[x]]) \ar[r] \ar@{..>}[ru] & \mathbf{A}^1_k } $$ Since the unique point of $X$ above $0 \in \mathbf{A}^1_k$ corresponds to a monomorphism $\mathop{\rm Spec}(k') \to X$ it is clear there cannot exist a dotted arrow! This shows that a finite separable field extension is needed in general.

    The code snippet corresponding to this tag is a part of the file spaces-morphisms.tex and is located in lines 8294–8326 (see updates for more information).

    \begin{example}
    \label{example-finite-separable-needed}
    Consider the algebraic space $X$ constructed in
    Spaces, Example \ref{spaces-example-non-representable-descent}.
    Recall that it is Galois twist of the affine line with zero doubled.
    The Galois twist is with respect to a degree two Galois extension
    $k'/k$ of fields. As such it comes with a morphism
    $$
    \pi : X \longrightarrow S = \mathbf{A}^1_k
    $$
    which is quasi-compact. We claim that $\pi$ is universally closed.
    Namely, after base change by $\Spec(k') \to \Spec(k)$
    the morphism $\pi$ is identified with the morphism
    $$
    \text{affine line with zero doubled}
    \longrightarrow
    \text{affine line}
    $$
    which is universally closed (some details omitted). Since the morphism
    $\Spec(k') \to \Spec(k)$ is universally closed and
    surjective, a diagram chase shows that $\pi$ is universally closed.
    On the other hand, consider the diagram
    $$
    \xymatrix{
    \Spec(k((x))) \ar[r] \ar[d] & X \ar[d]^\pi \\
    \Spec(k[[x]]) \ar[r] \ar@{..>}[ru] & \mathbf{A}^1_k
    }
    $$
    Since the unique point of $X$ above $0 \in \mathbf{A}^1_k$
    corresponds to a monomorphism $\Spec(k') \to X$
    it is clear there cannot exist a dotted arrow! This shows that
    a finite separable field extension is needed in general.
    \end{example}

    Comments (2)

    Comment #2099 by Matthew Emerton on June 22, 2016 a 2:40 pm UTC

    At the end of line two, in the phrase ``in a Galois twisted relative'', the language is slightly garbled.

    Comment #2127 by Johan (site) on July 21, 2016 a 1:03 pm UTC

    Thanks, fixed here.

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