The Stacks project

Lemma 10.47.9. Let $K/k$ be a geometrically irreducible field extension. Let $S$ be a geometrically irreducible $K$-algebra. Then $S$ is geometrically irreducible over $k$.

Proof. By Definition 10.47.4 and Lemma 10.47.3 it suffices to show that the spectrum of $S \otimes _ k k'$ is irreducible for every finite separable extension $k'/k$. Since $K$ is geometrically irreducible over $k$ we see that $K' = K \otimes _ k k'$ is a finite, separable field extension of $K$. Hence the spectrum of $S \otimes _ k k' = S \otimes _ K K'$ is irreducible as $S$ is assumed geometrically irreducible over $K$. $\square$


Comments (0)

There are also:

  • 6 comment(s) on Section 10.47: Geometrically irreducible algebras

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Your email address will not be published.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 0G30. Beware of the difference between the letter 'O' and the digit '0'.