Lemma 33.6.11. Let $k$ be a field. Let $X \to \mathop{\mathrm{Spec}}(k)$ be locally of finite type. Assume $X$ has finitely many irreducible components. Then there exists a finite purely inseparable extension $k'/k$ such that $(X_{k'})_{red}$ is geometrically reduced over $k'$.

Proof. To prove this lemma we may replace $X$ by its reduction $X_{red}$. Hence we may assume that $X$ is reduced and locally of finite type over $k$. Let $x_1, \ldots , x_ n \in X$ be the generic points of the irreducible components of $X$. Note that for every purely inseparable algebraic extension $k'/k$ the morphism $(X_{k'})_{red} \to X$ is a homeomorphism, see Algebra, Lemma 10.46.7. Hence the points $x'_1, \ldots , x'_ n$ lying over $x_1, \ldots , x_ n$ are the generic points of the irreducible components of $(X_{k'})_{red}$. As $X$ is reduced the local rings $K_ i = \mathcal{O}_{X, x_ i}$ are fields, see Algebra, Lemma 10.25.1. As $X$ is locally of finite type over $k$ the field extensions $K_ i/k$ are finitely generated field extensions. Finally, the local rings $\mathcal{O}_{(X_{k'})_{red}, x'_ i}$ are the fields $(K_ i \otimes _ k k')_{red}$. By Algebra, Lemma 10.45.3 we can find a finite purely inseparable extension $k'/k$ such that $(K_ i \otimes _ k k')_{red}$ are separable field extensions of $k'$. In particular each $(K_ i \otimes _ k k')_{red}$ is geometrically reduced over $k'$ by Algebra, Lemma 10.44.1. At this point Lemma 33.6.8 part (3) implies that $(X_{k'})_{red}$ is geometrically reduced. $\square$

Comment #767 by Keenan Kidwell on

The local rings $\mathscr{O}_{X_{k^\prime},x_i^\prime}$ should be $\mathscr{O}_{(X_{k^\prime})_{\mathrm{red}},x_i^\prime}$.

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