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Let $k$ be a field. Let $k'$ be an extension field of $k$. Let $X$ be a $k$-scheme of finite type. If $k'$ is algebraic over $k$, the morphism $X\times_k k' \rightarrow X$ is integral. Hence it is closed. Suppose $k'$ is not algebraic over $k$. I would like to know examples, if any, that the morphism $X\times_k k' \rightarrow X$ is not closed.

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  • $\begingroup$ @Benja No. I mean a closed morphism. $\endgroup$ – Makoto Kato Dec 25 '13 at 1:52
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Take $\mbox{Spec}(k(z)[x])\to\mbox{Spec}(k[x])$ induced by the inclusion $k[x]\hookrightarrow k(z)[x]$. Then the maximal ideal $(x-z)$ is sent to the generic point of $\mbox{Spec}(k[x])$.

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