hcju/stacksqa huggingface dataset:
The local ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ induces a local ring map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge$. The induced map $$ \kappa(x) \to \kappa(x)[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \to \mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge $$ composed with the projection to $\kappa(y)$ is the canonical homomorphism $\kappa(x) \to \kappa(y)$. By Lemma \ref{lemma-change-fields-flat} the residue field $\kappa(y)$ is a localization of $\kappa(x) \otimes_k K$ at the kernel $\mathfrak p_0$ of $\kappa(x) \otimes_k K \to \kappa(y)$. On the other hand, by Lemma \ref{lemma-change-fields-algebraic-unramified} the local ring $(\kappa(x) \otimes_k K)_{\mathfrak p_0}$ is equal to $\kappa(y)$. Hence the map $$ \kappa(x) \otimes_k K \to \mathcal{O}_{Y, y}^\wedge $$ factors canonically through $\kappa(y)$. We obtain a commutative diagram $$ \xymatrix{ \kappa(y) \ar[rr] & & \mathcal{O}_{Y, y}^\wedge \\ \kappa(x) \ar[r] \ar[u] & \kappa(x)[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \ar[r] & \mathcal{O}_{X, x}^\wedge \ar[u] } $$ Let $f_i \in \mathfrak m_x^\wedge \subset \mathcal{O}_{X, x}^\wedge$ be the image of $x_i$. Observe that $\mathfrak  m_x^\wedge = (f_1, \ldots, f_n)$ as the map is surjective. Consider the map $$ \kappa(y)[[x_1, \ldots, x_n]] \longrightarrow \mathcal{O}_{Y, y}^\wedge,\quad x_i \longmapsto f_i $$ where here $f_i$ really means the image of $f_i$ in $\mathfrak m_y^\wedge$. Since $\mathfrak m_x \mathcal{O}_{Y, y} = \mathfrak m_y$ by Lemma \ref{lemma-change-fields-algebraic-unramified} we see that the right hand side is complete with respect to $(x_1, \ldots, x_n)$ (use Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated} to see that it is a complete local ring). Since both sides are $(x_1, \ldots, x_n)$-adically complete our map is surjective by Algebra, Lemma \ref{algebra-lemma-completion-generalities} as it is surjective modulo $(x_1, \ldots, x_n)$. Of course the power series $g_1, \ldots, g_m$ are mapped to zero under this map, as they already map to zero in $\mathcal{O}_{X, x}^\wedge$. Thus we have the commutative diagram $$ \xymatrix{ \kappa(y)[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \ar[r] & \mathcal{O}_{Y, y}^\wedge \\ \kappa(x)[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m) \ar[r] \ar[u] & \mathcal{O}_{X, x}^\wedge \ar[u] } $$ We still need to show that the top horizontal arrow is an isomorphism. We already know that it is surjective. We know that $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is flat (Lemma \ref{lemma-change-fields-flat}), which implies that $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge$ is flat (More on Algebra, Lemma \ref{more-algebra-lemma-flat-completion}). Thus we may apply Algebra, Lemma \ref{algebra-lemma-mod-injective} with $R = \kappa(x)[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m)$, with $S = \kappa(y)[[x_1, \ldots, x_n]]/(g_1, \ldots, g_m)$, with $M = \mathcal{O}_{Y, y}^\wedge$, and with $N = S$ to conclude that the map is injective.
More precisely, set $X_{A'} = X \times_{\Spec(A)} \Spec(A')$. Let $\mathcal{F}_{A'} = p^*\mathcal{F}$ where $p : X_{A'} \to X$ is the projection. By Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology} we have $H^0(X_{k'}, \mathcal{F}_{A'}) = H^0(X, \mathcal{F}) \otimes_A A'$. Thus if $s_i$, $i \in I$ are generators for $H^0(X, \mathcal{F})$ as an $A$-module, then their images in $H^0(X_{A'}, \mathcal{F}_{A'})$ are generators for $H^0(X_{A'}, \mathcal{F}_{A'})$ as an $A'$-module. Thus we have to show that the map $\alpha : \bigoplus_{i \in I} \mathcal{O}_X \to \mathcal{F}$, $(f_i) \mapsto \sum f_i s_i$ is surjective if and only if $p^*\alpha$ is surjective. This we may check over an affine open $U = \Spec(B)$ of $X$. Then $\mathcal{F}|_U$ corresponds to a $B$-module $M$ and $s_i|_U$ to elements $x_i \in M$. Thus we have to show that $\bigoplus_{i \in I} B \to M$ is surjective if and only if the base change $\bigoplus_{i \in I} B \otimes_A A' \to M \otimes_A A'$ is surjective. This is true because $A \to A'$ is faithfully flat.
Throughout the proof we will use that for every $x \in X$ the linear map $$ \psi_x : V \to \Gamma(X, \mathcal{L}) \to \mathcal{L}_x \to \mathcal{L}_x \otimes_{\mathcal{O}_{X, x}} \kappa(x) $$ is nonzero. The proof is by induction on $\dim(X)$. \medskip\noindent The base case is $\dim(X) = 0$. In this case $X$ has finitely many points $X = \{x_1, \ldots, x_n\}$ (see for example Lemma \ref{lemma-algebraic-scheme-dim-0}). Since $k$ is infinite there exists a vector $v \in V$ such that $\psi_{x_i}(v) \not = 0$ for all $i$. Then $W = k\cdot v$ does the job. \medskip\noindent Assume $\dim(X) > 0$. Let $X_i \subset X$ be the irreducible components of dimension equal to $\dim(X)$. Since $X$ is Noetherian there are only finitely many of these. For each $i$ pick a point $x_i \in X_i$. As above choose $v \in V$ such that $\psi_{x_i}(v) \not = 0$ for all $i$. Let $Z \subset X$ be the zero scheme of the image of $v$ in $\Gamma(X, \mathcal{L})$, see Divisors, Definition \ref{divisors-definition-zero-scheme-s}. By construction $\dim(Z) < \dim(X)$. By induction we can find $W \subset V$ with $\dim(W) \leq \dim(X)$ such that $W$ generates $\mathcal{L}|_Z$. Then $W + k\cdot v$ generates $\mathcal{L}$.
Condition (1) implies in particular that the elements of $V$ generate $\mathcal{L}$ over $X$. Hence we get a canonical morphism $$ \varphi = \varphi_{\mathcal{L}, V} : X \longrightarrow \mathbf{P}(V) $$ by Constructions, Example \ref{constructions-example-projective-space}. The morphism $\varphi$ is proper by Morphisms, Lemma \ref{morphisms-lemma-image-proper-scheme-closed}. By (1) the map $\varphi$ is injective on closed points (computation omitted). In particular, the fibre over any closed point of $\mathbf{P}(V)$ is a singleton (small detail omitted). Thus we see that $\varphi$ is finite, for example use Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood}. To finish the proof it suffices to show that the map $$ \varphi^\sharp : \mathcal{O}_{\mathbf{P}(V)} \longrightarrow \varphi_*\mathcal{O}_X $$ is surjective. This we may check on stalks at closed points. Let $x \in X$ be a closed point with image the closed point $p = \varphi(x) \in \mathbf{P}(V)$. Since $\varphi^{-1}(\{p\}) = \{x\}$ by (1) and since $\varphi$ is proper (hence closed), we see that $\varphi^{-1}(U)$ runs through a fundamental system of open neighbourhoods of $x$ as $U$ runs through a fundamental system of open neighbourhoods of $p$. We conclude that on stalks at $p$ we obtain the map $$ \varphi^\sharp_x : \mathcal{O}_{\mathbf{P}(V), p} \longrightarrow \mathcal{O}_{X, x} $$ In particular, $\mathcal{O}_{X, x}$ is a finite $\mathcal{O}_{\mathbf{P}(V), p}$-module. Moreover, the residue fields of $x$ and $p$ are equal to $k$ (as $k$ is algebraically closed -- use the Hilbert Nullstellensatz). Finally, condition (2) implies that the map $$ T_{X/k, x} \longrightarrow T_{\mathbf{P}(V)/k, p} $$ is injective since any nonzero $\theta$ in the kernel of this map couldn't possibly satisfy the conclusion of (2). In terms of the map of local rings above this means that $$ \mathfrak m_p/\mathfrak m_p^2 \longrightarrow \mathfrak m_x/\mathfrak m_x^2 $$ is surjective, see Lemma \ref{lemma-tangent-space-rational-point}. Now the proof is finished by applying Algebra, Lemma \ref{algebra-lemma-when-surjective-local}.
This is a reformulation of Lemma \ref{lemma-separate-points-tangent-vectors}. Namely, given distinct closed points $x, y \in X$ taking $Z = x \cup y$ (viewed as closed subscheme) we get condition (1) of the lemma. And given a nonzero tangent vector $\theta \in T_{X/k, x}$ the morphism $\theta : \Spec(k[\epsilon]) \to X$ is a closed immersion. Setting $Z = \Im(\theta)$ we obtain condition (2) of the lemma.
The inclusion $\overline{AB} \subset \overline{A} \times_k \overline{B}$ is immediate. We may replace $X$ and $Y$ by the reduced closed subschemes $\overline{A}$ and $\overline{B}$. Let $W \subset X \times_k Y$ be a nonempty open subset. By Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open} the subset $U = \text{pr}_X(W)$ is nonempty open in $X$. Hence $A \cap U$ is nonempty. Pick $a \in A \cap U$. Denote $Y_{\kappa(a)} = \{a\} \times_k Y$ the fibre of $\text{pr}_X : X \times_k Y \to X$ over $a$. By Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open} again the morphism $Y_a \to Y$ is open as $\Spec(\kappa(a)) \to \Spec(k)$ is universally open. Hence the nonempty open subset $W_a = W \times_{X \times_k Y} Y_a$ maps to a nonempty open subset of $Y$. We conclude there exists a $b \in B$ in the image. Hence $AB \cap W \not = \emptyset$ as desired.
Recall that $Z$ is the smallest closed subscheme of $X$ through which $f$ factors. Similarly for $Z'$. Let $W \subset X \times_k Y$ be the scheme theoretic image of $f \times g$. As $f \times g$ factors through $Z \times_k Z'$ we see that $W \subset Z \times_k Z'$. \medskip\noindent To prove the other inclusion let $U \subset X$ and $V \subset Y$ be affine opens. By Morphisms, Lemma \ref{morphisms-lemma-quasi-compact-scheme-theoretic-image} the scheme $Z \cap U$ is the scheme theoretic image of $f|_{f^{-1}(U)} : f^{-1}(U) \to U$, and similarly for $Z' \cap V$ and $W \cap U \times_k V$. Hence we may assume $X$ and $Y$ affine. As $f$ and $g$ are quasi-compact this implies that $A = \bigcup U_i$ is a finite union of affines and $B = \bigcup V_j$ is a finite union of affines. Then we may replace $A$ by $\coprod U_i$ and $B$ by $\coprod V_j$, i.e., we may assume that $A$ and $B$ are affine as well. In this case $Z$ is cut out by $\Ker(\Gamma(X, \mathcal{O}_X) \to \Gamma(A, \mathcal{O}_A))$ and similarly for $Z'$ and $W$. Hence the result follows from the equality $$ \Gamma(A \times_k B, \mathcal{O}_{A \times_k B}) = \Gamma(A, \mathcal{O}_A) \otimes_k \Gamma(B, \mathcal{O}_B) $$ which holds as $A$ and $B$ are affine. Details omitted.
Let $x \in X$ be a point. As $X$ is locally Noetherian (see Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}) there are finitely many irreducible components $X_1, \ldots, X_n$ passing through $x$ (see Properties, Lemma \ref{properties-lemma-Noetherian-topology} and Topology, Lemma \ref{topology-lemma-Noetherian}). Let $\eta_i \in X_i$ be the generic point. As $X$ is reduced we have $\mathcal{O}_{X, \eta_i} = \kappa(\eta_i)$, see Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}. Moreover, $\kappa(\eta_i)$ is a finitely generated field extension of the perfect field $k$ hence separably generated over $k$ (see Algebra, Section \ref{algebra-section-separability}). It follows that $\Omega_{X/k, \eta_i} = \Omega_{\kappa(\eta_i)/k}$ is free of rank the transcendence degree of $\kappa(\eta_i)$ over $k$. By Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-at-a-point} we conclude that $\dim_{\eta_i}(X_i) = \text{rank}_{\eta_i}(\Omega_{X/k})$. Since $x \in X_1 \cap \ldots \cap X_n$ we see that $$ \text{rank}_x(\Omega_{X/k}) = \text{rank}_{\eta_i}(\Omega_{X/k}) = \dim(X_i). $$ Therefore $\dim_x(X) = \text{rank}_x(\Omega_{X/k})$, see Algebra, Lemma \ref{algebra-lemma-dimension-at-a-point-finite-type-over-field}. It follows that $X \to \Spec(k)$ is smooth at $x$ for example by Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}.
(See also Lemma \ref{lemma-geometrically-regular-smooth}.) By Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field} every local ring $\mathcal{O}_{X, x}$ is regular. And because $X$ is locally of finite type over $k$ it is locally Noetherian. Hence $X$ is regular by Properties, Lemma \ref{properties-lemma-characterize-regular}.
(See also Lemma \ref{lemma-geometrically-regular-smooth}.) Let $k'$ be a finite purely inseparable extension of $k$. It suffices to prove that $X_{k'}$ is regular, normal, reduced, see Lemmas \ref{lemma-geometrically-regular}, \ref{lemma-geometrically-normal}, and \ref{lemma-check-only-finite-inseparable-extensions}. By Morphisms, Lemma \ref{morphisms-lemma-base-change-smooth} the morphism $X_{k'} \to \Spec(k')$ is smooth too. Hence it suffices to show that a scheme $X$ smooth over a field is regular, normal, and reduced. We see that $X$ is regular by Lemma \ref{lemma-smooth-regular}. Hence Properties, Lemma \ref{properties-lemma-regular-normal} guarantees that $X$ is normal.
It suffices to show that given a nonempty smooth $X$ over $k$ there exists at least one closed point whose residue field is finite separable over $k$. To see this, choose a diagram $$ \xymatrix{ X & U \ar[l] \ar[r]^-\pi & \mathbf{A}^d_k } $$ with $\pi$ \'etale, see Morphisms, Lemma \ref{morphisms-lemma-smooth-etale-over-affine-space}. The morphism $\pi : U \to \mathbf{A}^d_k$ is open, see Morphisms, Lemma \ref{morphisms-lemma-etale-open}. By Lemma \ref{lemma-affine-space-over-field} we may choose a closed point $w \in \pi(U)$ whose residue field is finite separable over $k$. Pick any $x \in U$ with $\pi(x) = w$. By Morphisms, Lemma \ref{morphisms-lemma-etale-over-field} the field extension $\kappa(w) \subset \kappa(x)$ is finite separable. Hence $k \subset \kappa(x)$ is finite separable. The point $x$ is a closed point of $X$ by Morphisms, Lemma \ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre}.
The problem is local on $X$, hence we may assume $X$ is quasi-compact. Let $X = X_1 \cup \ldots \cup X_n$ be the irreducible components of $X$. Then $Z = \bigcup_{i \not = j} X_i \cap X_j$ is nowhere dense in $X$. Hence we may replace $X$ by $X \setminus Z$. As $X \setminus Z$ is a disjoint union of irreducible schemes, this reduces us to the case where $X$ is irreducible. As $X$ is irreducible and reduced, it is integral, see Properties, Lemma \ref{properties-lemma-characterize-integral}. Let $\eta \in X$ be its generic point. Then the function field $K = k(X) = \kappa(\eta)$ is geometrically reduced over $k$, hence separable over $k$, see Algebra, Lemma \ref{algebra-lemma-characterize-separable-field-extensions}. Let $U = \Spec(A) \subset X$ be any nonempty affine open so that $K = A_{(0)}$ is the fraction field of $A$. Apply Algebra, Lemma \ref{algebra-lemma-separable-smooth} to conclude that $A$ is smooth at $(0)$ over $k$. By definition this means that some principal localization of $A$ is smooth over $k$ and we win.
The equality of the two sets follows immediately from Algebra, Lemma \ref{algebra-lemma-separable-smooth} and the definitions (see Algebra, Definition \ref{algebra-definition-perfect} for the definition of a perfect field). The set is open because the set of points where a morphism of schemes is smooth is open, see Morphisms, Definition \ref{morphisms-definition-smooth}. Finally, we give two arguments to see that it is dense: (1) The generic points of $X$ are in the set as the local rings at generic points are fields (Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}) hence regular. (2) We use that $X$ is geometrically reduced by Lemma \ref{lemma-perfect-reduced} and hence Lemma \ref{lemma-geometrically-reduced-dense-smooth-open} applies.
Let $U \subset X$ be the smooth locus of $X$. By assumption $U$ is nonempty and hence dense and scheme theoretically dense. Then $U_{\overline{k}} \subset X_{\overline{k}}$ is dense and scheme theoretically dense as well (some details omitted). Thus it suffices to show that $U$ is geometrically integral. Because $U$ has a $k$-rational point it is geometrically connected by Lemma \ref{lemma-geometrically-connected-if-connected-and-point}. On the other hand, $U_{\overline{k}}$ is reduced and normal (Lemma \ref{lemma-smooth-geometrically-normal}. Since a connected normal Noetherian scheme is integral (Properties, Lemma \ref{properties-lemma-normal-Noetherian}) the proof is complete.
Since $X$ is locally of finite type over a field, we see that $X$ is locally Noetherian (Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}) hence every quasi-compact open has finitely many irreducible components (Properties, Lemma \ref{properties-lemma-Noetherian-irreducible-components}). Thus Morphisms, Definition \ref{morphisms-definition-normalization} 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 \ref{morphisms-lemma-normalization-normal}. Since $X$ is universally Nagata (Morphisms, Lemma \ref{morphisms-lemma-ubiquity-nagata}) we see that $\nu$ is finite (Morphisms, Lemma \ref{morphisms-lemma-nagata-normalization}). Hence $X^\nu$ is locally algebraic too. At this point we have proved (1). \medskip\noindent Part (2) is Morphisms, Lemma \ref{morphisms-lemma-normalization-reduced}. \medskip\noindent Part (3) is Morphisms, Lemma \ref{morphisms-lemma-normalization-birational}. \medskip\noindent Part (4) follows from (1), (2), (3), and the fact that $X^\nu$ is separated as a scheme finite over a separated scheme.
Since $Y$ is quasi-separated (by Properties, Lemma \ref{properties-lemma-locally-Noetherian-quasi-separated} and Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}) the morphism $f$ is quasi-separated (Schemes, Lemma \ref{schemes-lemma-compose-after-separated}). Hence Morphisms, Definition \ref{morphisms-definition-normalization-X-in-Y} applies. The result follows from Morphisms, Lemma \ref{morphisms-lemma-nagata-normalization-finite-general}. 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 \ref{morphisms-lemma-ubiquity-nagata}). Some small details omitted.
Let $K = k^{perf}$ be the perfect closure. Let $Y_K$ be the normalization of $X_K$, see Lemma \ref{lemma-normalization-locally-algebraic}. By Limits, Lemma \ref{limits-lemma-descend-finite-presentation} 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 \ref{lemma-geometrically-normal}). After increasing $k'$ we may assume $Y \to X_{k'}$ is finite (Limits, Lemma \ref{limits-lemma-descend-finite-finite-presentation}). 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 \ref{morphisms-lemma-birational-isomorphism-over-dense-open}). 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 \ref{limits-lemma-descend-opens} and \ref{limits-lemma-descend-closed-immersion-finite-presentation}). It follows readily from this that $\nu$ is the normalization morphism and the proof is complete.
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 \ref{morphisms-lemma-scheme-over-field-universally-open} and \ref{morphisms-lemma-open-generizing}. 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 \ref{morphisms-lemma-functoriality-normalization} 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. \medskip\noindent To prove this morphism has the properties stated in the lemma we may assume $X = \Spec(A)$ 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 \ref{morphisms-lemma-normalization-reduced} and \ref{morphisms-lemma-description-normalization}. 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. \medskip\noindent If $K/k$ is separable, then $A' \otimes_k K$ is normal by Lemma \ref{lemma-base-change-normal-by-separable}. 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 \ref{algebra-lemma-characterize-reduced-ring-normal}. \medskip\noindent 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 \ref{algebra-lemma-p-ring-map}. 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 \ref{fields-lemma-separable-first}). 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.
We will use the results of Lemma \ref{lemma-normalization-locally-algebraic} without further mention. Let $x' \in X^\nu$ be a point over $x$. By dimension theory (Section \ref{section-algebraic-schemes}) 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 \ref{properties-lemma-criterion-normal}). 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 \ref{more-algebra-lemma-regular-implies-fs}. Then $\mathcal{O}_{X^\nu, x'}$ is geometrically regular over $k$ by More on Algebra, Theorem \ref{more-algebra-theorem-regular-fs}. Thus $X^\nu$ is geometrically regular at $x'$ by Lemma \ref{lemma-geometrically-regular-at-point}. \medskip\noindent 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 \ref{lemma-geometrically-reduced-at-point}.