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. $\square$

There are also:

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

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).