Let $k$ be a field and $K/k$ be a field extension. For a scheme $X$ of finite type over $k$, denote $X_K:=X\times_k \text{Spec}K$. Let $x\in X$ be a closed point and $x'\in X_K$ be a point lying over $x$. In this situation, is $x'$ also a closed point? (This is true for $K/k$ is purely inseparable extension since two schemes are homeomorphic.)


After the usual reductions (work locally on $X$ and mod out the prime ideal corresponding to $x$) the question is equivalent to: Let $L$ be a finite field extension of $k$, is every prime ideal of $L \otimes_k K$ maximal, i.e. do we have $\dim(L \otimes_k K)=0$? The answer is yes, because $K \hookrightarrow L \otimes_k K$ is an integral extension.

More generally Grothendieck (EGA IV, Quatrième partie, page 349, Remarque ( has proven the formula $\dim(L \otimes_k K) = \min(\mathrm{tr.deg}_k(L),\mathrm{tr.deg}_k(K))$ for arbitrary field extensions $L/k$, $K/k$.

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    $\begingroup$ For OP: If $B$ is integral over $A$, then $\dim B = \dim A$. $\endgroup$ – user38268 Aug 8 '13 at 10:28

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