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Let $k$ be a field. Let $\bar k$ be an algebraic closure of $k$. Let $X$ be a $k$-scheme of finite type. Suppose $X\times_k \bar k$ is separated over $\bar k$. Is $X$ separated over $k$? If yes, how do you prove it?

Makoto Kato
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1 Answers1

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Let $\overline{X}$ denote $X \times_k \bar{k}$. Then this statement is true if $\overline{X} \to X$ is a closed and surjective morphism. My intuition tells me this is true because $X$ should be the quotient of $\overline{X}$ by the Galois action, but I'm not sure yet. I'll edit this post when I figure that out, but assuming it, then we can prove this as follows. You can verify that the diagram

$$ \require{AMScd} \begin{CD} \overline{X} @>>> X \\ @VVV @VVV \\ \overline{X} \times_\bar{k} \overline{X} @>>> X \times_k X \end{CD} $$

is cartesian where the vertical morphisms are the diagonals and the horizontal morphisms are the natural ones induced by base extension. The horizontal morphisms are closed and $\Delta_\overline{X}$ is closed by assumption, and the top morphism is surjective, so the diagonal $\Delta_X$ must be closed.

Dori Bejleri
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