This is a generalization of this question. Let $k$ be a field. Let $k'$ be an extension field of $k$. Let $X$ be a $k$-scheme of finite type. Suppose $X\times_k k'$ is separated over $k'$. Is $X$ separated over $k$? If yes, how do you prove it?
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I think same argument I posted in that other answer should work for any field extension as long as the projection $X_{k'} \to X$ is closed and surjective. I'm still not sure if/when this will true but it seems true for e.g. algebraic extensions in characteristic $0$. – Dori Bejleri Dec 24 '13 at 18:32
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@DoriBejleri Please see this question. – Makoto Kato Dec 25 '13 at 17:36
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Ok so it is true when the extension is algebraic. Cool, thanks! – Dori Bejleri Dec 25 '13 at 17:49
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Formation of the diagonal is compatible with base-change; i.e. the base-change to $k'$ of the diagonal morphism $X \to X\times_k X$ is canonically identified with the diagonal morphism $X' \to X' \times_{k'} X'$. (Here I have written $X' := X\times_k k'$.)
Now $X$ is separated over $k$ iff the diagonal is a closed immersion, and so what we want to check is that $Y \to X$ is a closed immersion iff the base-changed morphism $Y' \to X'$ is closed immersion.
This latter statement is true, and not too hard to check. Indeed, the property of being a closed immersion can be checked after making any faithfully flat quasi-compact base-change (one says that the property of being a closed immersion satisfies fpqc descent).
Matt E
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