Lemma 10.165.4. Let $k$ be a field. Let $K/k$ be a separable field extension. Then $K$ is geometrically normal over $k$.

**Proof.**
This is true because $k^{perf} \otimes _ k K$ is a field. Namely, it is reduced by Lemma 10.43.6. By Lemma 10.45.4 (or by Definition 10.45.5) the field extension $k^{perf}/k$ is purely inseparable. Hence by Lemma 10.46.10 the ring $k^{perf} \otimes _ k K$ has a unique prime ideal. A reduced ring with a unique prime ideal is a field.
$\square$

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## Comments (2)

Comment #8096 by Rubén Muñoz--Bertrand on

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