The Stacks project

Proof. Since $X$ is locally of finite type over a field, we see that $X$ is locally Noetherian (Morphisms, Lemma 29.15.6) hence every quasi-compact open has finitely many irreducible components (Properties, Lemma 28.5.7). Thus Morphisms, Definition 29.54.1 applies. The normalization $X^\nu $ is always a disjoint union of normal integral schemes and the normalization morphism $\nu $ is always dominant, see Morphisms, Lemma 29.54.5. Since $X$ is universally Nagata (Morphisms, Lemma 29.18.2) we see that $\nu $ is finite (Morphisms, Lemma 29.54.11). Hence $X^\nu $ is locally algebraic too. At this point we have proved (1).

Part (2) is Morphisms, Lemma 29.54.2.

Part (3) is Morphisms, Lemma 29.54.7.

Part (4) follows from (1), (2), (3), and the fact that $X^\nu $ is separated as a scheme finite over a separated scheme. $\square$

Proof. By Lemma 33.27.1 the morphism $\nu $ is finite. Hence $X^\nu $ is proper over $k$ by Morphisms, Lemmas 29.44.11 and 29.41.4. The morphism $\nu $ is projective by Morphisms, Lemma 29.44.16 and hence if $X$ is projective over $k$, then $X^\nu $ is projective over $k$ by Morphisms, Lemma 29.43.14. $\square$

Proof. Since $Y$ is quasi-separated (by Properties, Lemma 28.5.4 and Morphisms, Lemma 29.15.6) the morphism $f$ is quasi-separated (Schemes, Lemma 26.21.13). Hence Morphisms, Definition 29.53.3 applies. The result follows from Morphisms, Lemma 29.53.14. This uses that locally algebraic schemes are locally Noetherian (hence have locally finitely many irreducible components) and that locally algebraic schemes are Nagata (Morphisms, Lemma 29.18.2). Some small details omitted. $\square$

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. $\square$

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

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. $\square$

Proof. We will use the results of Lemma 33.27.1 without further mention. Let $x' \in X^\nu $ be a point over $x$. By dimension theory (Section 33.20) we have $\dim (\mathcal{O}_{X^\nu , x'}) = 1$. Since $X^\nu $ is normal, we see that $\mathcal{O}_{X^\nu , x'}$ is a discrete valuation ring (Properties, Lemma 28.12.5). Thus $\mathcal{O}_{X^\nu , x'}$ is a regular local $k$-algebra whose residue field is separable over $k$. Hence $k \to \mathcal{O}_{X^\nu , x'}$ is formally smooth in the $\mathfrak m_{x'}$-adic topology, see More on Algebra, Lemma 15.38.5. Then $\mathcal{O}_{X^\nu , x'}$ is geometrically regular over $k$ by More on Algebra, Theorem 15.40.1. Thus $X^\nu $ is geometrically regular at $x'$ by Lemma 33.12.2.

Since $\mathcal{O}_{X, x}$ is reduced, the family of maps $\mathcal{O}_{X, x} \to \mathcal{O}_{X^\nu , x'}$ is injective. Since $\mathcal{O}_{X^\nu , x'}$ is a geometrically reduced $k$-algebra, it follows immediately that $\mathcal{O}_{X, x}$ is a geometrically reduced $k$-algebra. Hence $X$ is geometrically reduced at $x$ by Lemma 33.6.2. $\square$