The Stacks project

Lemma 10.47.12. Let $K/k$ be a field extension. The following are equivalent

  1. $K/k$ is geometrically irreducible, and

  2. every element $\alpha \in K$ separably algebraic over $k$ is in $k$.

Proof. Assume (1) and let $\alpha \in K$ be separably algebraic over $k$. Then $k' = k(\alpha )$ is a finite separable extension of $k$ contained in $K$. By Lemma 10.47.6 the extension $k'/k$ is geometrically irreducible. In particular, we see that the spectrum of $k' \otimes _ k \overline{k}$ is irreducible (and hence if it is a product of fields, then there is exactly one factor). By Fields, Lemma 9.13.4 it follows that $\mathop{\mathrm{Hom}}\nolimits _ k(k', \overline{k})$ has one element which in turn implies that $k' = k$ by Fields, Lemma 9.12.11. Thus (2) holds.

Assume (2). Let $k' \subset K$ be the subfield consisting of elements algebraic over $k$. By Lemma 10.47.8 the extension $K/k'$ is geometrically irreducible. By assumption $k'/k$ is a purely inseparable extension. By Lemma 10.46.7 the extension $k'/k$ is geometrically irreducible. Hence by Lemma 10.47.9 we see that $K/k$ is geometrically irreducible. $\square$


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Your email address will not be published.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 0G33. Beware of the difference between the letter 'O' and the digit '0'.