Lemma 10.47.8. Let $K/k$ be a field extension. If $k$ is algebraically closed in $K$, then $K$ is geometrically irreducible over $k$.

**Proof.**
Assume $k$ is algebraically closed in $K$. By Definition 10.47.4 and Lemma 10.47.3 it suffices to show that the spectrum of $K \otimes _ k k'$ is irreducible for every finite separable extension $k'/k$. Say $k'$ is generated by $\alpha \in k'$ over $k$, see Fields, Lemma 9.19.1. Let $P = T^ d + a_1 T^{d - 1} + \ldots + a_ d \in k[T]$ be the minimal polynomial of $\alpha $. Then $K \otimes _ k k' \cong K[T]/(P)$. The only way the spectrum of $K[T]/(P)$ can be reducible is if $P$ is reducible in $K[T]$. Assume $P = P_1 P_2$ is a nontrivial factorization in $K[T]$ to get a contradiction. By Lemma 10.38.5 we see that the coefficients of $P_1$ and $P_2$ are algebraic over $k$. Our assumption implies the coefficients of $P_1$ and $P_2$ are in $k$ which contradicts the fact that $P$ is irreducible over $k$.
$\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: