Lemma 33.27.5. Let $k$ be a field. Let $X$ be a locally algebraic $k$-scheme. Let $K/k$ be an extension of fields. Let $\nu : X^\nu \to X$ be the normalization of $X$ and let $Y^\nu \to X_ K$ be the normalization of the base change. Then the canonical morphism

$Y^\nu \longrightarrow X^\nu \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(K)$

is an isomorphism if $K/k$ is separable and a universal homeomorphism in general.

Proof. Set $Y = X_ K$. Let $X^{(0)}$, resp. $Y^{(0)}$ be the set of generic points of irreducible components of $X$, resp. $Y$. Then the projection morphism $\pi : Y \to X$ satisfies $\pi (Y^{(0)}) = X^{(0)}$. This is true because $\pi$ is surjective, open, and generizing, see Morphisms, Lemmas 29.23.4 and 29.23.5. If we view $X^{(0)}$, resp. $Y^{(0)}$ as (reduced) schemes, then $X^\nu$, resp. $Y^\nu$ is the normalization of $X$, resp. $Y$ in $X^{(0)}$, resp. $Y^{(0})$. Thus Morphisms, Lemma 29.53.5 gives a canonical morphism $Y^\nu \to X^\nu$ over $Y \to X$ which in turn gives the canonical morphism of the lemma by the universal property of the fibre product.

To prove this morphism has the properties stated in the lemma we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Let $Q(A_{red})$ be the total ring of fractions of $A_{red}$. Then $X^\nu$ is the spectrum of the integral closure $A'$ of $A$ in $Q(A_{red})$, see Morphisms, Lemmas 29.54.2 and 29.54.3. Similarly, $Y^\nu$ is the spectrum of the integral closure $B'$ of $A \otimes _ k K$ in $Q((A \otimes _ k K)_{red})$. There is a canonical map $Q(A_{red}) \to Q((A \otimes _ k K)_{red})$, a canonical map $A' \to B'$, and the morphism of the lemma corresponds to the induced map

$A' \otimes _ k K \longrightarrow B'$

of $K$-algebras. The kernel consists of nilpotent elements as the kernel of $Q(A_{red}) \otimes _ k K \to Q((A \otimes _ k K)_{red})$ is the set of nilpotent elements.

If $K/k$ is separable, then $A' \otimes _ k K$ is normal by Lemma 33.10.6. In particular it is reduced, whence $Q((A \otimes _ k K)_{red}) = Q(A' \otimes _ k K)$ and $B' = A' \otimes _ k K$ by Algebra, Lemma 10.37.16.

Assume $K/k$ is not separable. Then the characteristic of $k$ is $p > 0$. We will show that for every $b \in B'$ there is a power $q$ of $p$ such that $b^ q$ is in the image of $A' \otimes _ k K$. This will prove that the displayed map is a universal homeomorphism by Algebra, Lemma 10.46.7. For a given $b$ there is a subfield $F \subset K$ with $F/k$ finitely generated such that $b$ is contained in $Q((A \otimes _ k F)_{red})$ and is integral over $A \otimes _ k F$. Choose a monic polynomial $P = T^ d + \alpha _1 T^{d - 1} + \ldots + \alpha _ d$ with $P(b) = 0$ and $\alpha _ i \in A \otimes _ k F$. Choose a transcendence basis $t_1, \ldots , t_ r$ for $F$ over $k$. Let $F/F'/k(t_1, \ldots , t_ r)$ be the maximal separable subextension (Fields, Lemma 9.14.6). Since $F/F'$ is finite purely inseparable, there is a $q$ such that $\lambda ^ q \in F'$ for all $\lambda \in F$. Then $b^ q$ is in $Q((A \otimes _ k F')_{red})$ and satisfies the polynomial $T^ d + \alpha _1^ q T^{d - 1} + \ldots + \alpha _ d^ q$ with $\alpha _ i^ q \in A \otimes _ k F'$. By the separable case we see that $b^ q \in A' \otimes _ k F'$ and the proof is complete. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).