The Stacks project

Lemma 33.8.15. Let $k$ be a field. Let $X \to \mathop{\mathrm{Spec}}(k)$ be locally of finite type. Assume $X$ has finitely many irreducible components. Then there exists a finite separable extension $k'/k$ such that every irreducible component of $X_{k'}$ is geometrically irreducible over $k'$.

Proof. Let $\overline{k}$ be a separable algebraic closure of $k$. The assumption that $X$ has finitely many irreducible components combined with Lemma 33.8.14 (3) shows that $X_{\overline{k}}$ has finitely many irreducible components $\overline{T}_1, \ldots , \overline{T}_ n$. By Lemma 33.8.14 (2) there exists a finite extension $\overline{k}/k'/k$ and irreducible components $T_ i \subset X_{k'}$ such that $\overline{T}_ i = T_{i, \overline{k}}$ and we win. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 33.8: Geometrically irreducible schemes

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 054R. Beware of the difference between the letter 'O' and the digit '0'.