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'/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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