Lemma 33.27.4. Let $k$ be a field. Let $X$ be an algebraic $k$-scheme. Then there exists a finite purely inseparable extension $k \subset k'$ such that the normalization $Y$ of $X_{k'}$ is geometrically normal over $k'$.

**Proof.**
Let $K = k^{perf}$ be the perfect closure. Let $Y_ K$ be the normalization of $X_ K$, see Lemma 33.27.1. By Limits, Lemma 32.10.1 there exists a finite sub extension $K/k'/k$ and a morphism $\nu : Y \to X_{k'}$ of finite presentation whose base change to $K$ is the normalization morphism $\nu _ K : Y_ K \to X_ K$. Observe that $Y$ is geometrically normal over $k'$ (Lemma 33.10.3). After increasing $k'$ we may assume $Y \to X_{k'}$ is finite (Limits, Lemma 32.8.3). Since $\nu _ K : Y_ K \to X_ K$ is the normalization morphism, it induces a birational morphism $Y_ K \to (X_ K)_{red}$. Hence there is a dense open $V_ K \subset X_ K$ such that $\nu _ K^{-1}(V_ K) \to V_ K$ is a closed immersion (inducing an isomorphism of $\nu _ K^{-1}(V_ K)$ with $V_{K, red}$, see for example Morphisms, Lemma 29.51.6). After increasing $k'$ we find $V_ K$ is the base change of a dense open $V \subset Y$ and the morphism $\nu ^{-1}(V) \to V$ is a closed immersion (Limits, Lemmas 32.4.11 and 32.8.5). It follows readily from this that $\nu $ is the normalization morphism and the proof is complete.
$\square$

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