I really couldn't figure out the answer to the following question: Let $X$ be a scheme of finite type over a field $k$ and let $K$ be an extension field of $k$. Let $X_K := K \times_k X$ be the base extension and let $p: X_K \rightarrow X$ be the projection. Is it true that $p$ maps closed points to closed points?

This should be true if $K$ is of finite type over $k$ (so not necessarily finite!). Proof: $X$ is Jacobson, the projection $p$ is of finite type and now it follows for example from the Stacks Project Lemma 27.17.8 (see http://stacks.math.columbia.edu/tag/01TB) that $p$ maps closed points to closed points. But I would like to have the same for any $K$. Is this true?


1 Answer 1


No, it is not ! For instance, if $k=\bar{\mathbb{Q}}$, $K=\mathbb{C}$ and $X$ has dimension $\geq 1$, there is exactly one closed point of $X_K$ above each closed point of $X$. But there is only countably many closed points of $X$, whereas $X_K$ has uncountably many closed points.

In general, a point of $X$ will be the image of a closed point of $X_K$ if and only if its residue field may be embedded in a finite extension of $K$ (in a way that respects $k$).

In particular, your statement is not true in general, even when $K$ is of finite type over $k$ : for instance, if $K$ is the residue field of a generic point of $X$. The problem with your argument is that $K$ is of finite type over $k$ as a field, but not as an algebra.


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