# The Stacks Project

## Tag 03KI

Example 58.40.6. Consider the algebraic space $X$ constructed in Spaces, Example 56.14.2. 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 8317–8349 (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}

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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