## Tag `03KI`

Chapter 58: Morphisms of Algebraic Spaces > Section 58.41: Valuative criteria

Example 58.41.6. 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 $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ 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 $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ is universally closed and surjective, a diagram chase shows that $\pi$ is universally closed. On the other hand, consider the diagram $$ \xymatrix{ \mathop{\mathrm{Spec}}(k((x))) \ar[r] \ar[d] & X \ar[d]^\pi \\ \mathop{\mathrm{Spec}}(k[[x]]) \ar[r] \ar@{..>}[ru] & \mathbf{A}^1_k } $$ Since the unique point of $X$ above $0 \in \mathbf{A}^1_k$ corresponds to a monomorphism $\mathop{\mathrm{Spec}}(k') \to X$ 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 8412–8444 (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}
```

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