2
$\begingroup$

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.)

$\endgroup$
3
$\begingroup$

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 (4.2.1.4)) has proven the formula $\dim(L \otimes_k K) = \min(\mathrm{tr.deg}_k(L),\mathrm{tr.deg}_k(K))$ for arbitrary field extensions $L/k$, $K/k$.

$\endgroup$
  • 1
    $\begingroup$ For OP: If $B$ is integral over $A$, then $\dim B = \dim A$. $\endgroup$ – user38268 Aug 8 '13 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.