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 $k$. $\square$

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