Example 67.41.6 (03KI)—The Stacks project

Example 67.41.6. Consider the algebraic space $X$ constructed in Spaces, Example 65.14.2. 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.

Comments (2)

Comment #2099 by Matthew Emerton on June 22, 2016 at 14:40

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

Comment #2127 by Johan on July 21, 2016 at 13:03

Thanks, fixed here.

Comments (2)

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