hcju/stacksqa huggingface dataset:
The dualizing module $\omega_A$ is a finite $A$-module whose stalk at the generic point is invertible. Namely, $\omega_A \otimes_A K$ is a dualizing module for the fraction field $K$ of $A$, hence has rank $1$. Thus there exists a blowup $b : Y \to \Spec(A)$ such that the strict transform of $\omega_A$ with respect to $b$ is an invertible $\mathcal{O}_Y$-module, see Divisors, Lemma \ref{divisors-lemma-blowup-fitting-ideal}. By Lemma \ref{lemma-dominate-by-normalized-blowing-up} we can choose a sequence of normalized blowups $$ X_n \to X_{n - 1} \to \ldots \to X_1 \to \Spec(A) $$ such that $X_n$ dominates $Y$. By Lemma \ref{lemma-blow-up-normal-rational} and arguing by induction each $X_i \to X_{i - 1}$ is simply a blowing up. \medskip\noindent We claim that $\omega_{X_n}$ is invertible. Since $\omega_{X_n}$ is a coherent $\mathcal{O}_{X_n}$-module, it suffices to see its stalks are invertible modules. If $x \in X_n$ is a regular point, then this is clear from the fact that regular schemes are Gorenstein (Dualizing Complexes, Lemma \ref{dualizing-lemma-regular-gorenstein}). If $x$ is a singular point of $X_n$, then each of the images $x_i \in X_i$ of $x$ is a singular point (because the blowup of a regular point is regular by Lemma \ref{lemma-blowup-regular}). Consider the canonical map $f_n^*\omega_A \to \omega_{X_n}$ of Lemma \ref{lemma-dualizing-rational}. For each $i$ the morphism $X_{i + 1} \to X_i$ is either a blowup of $x_i$ or an isomorphism at $x_i$. Since $x_i$ is always a singular point, it follows from Lemma \ref{lemma-dualizing-blow-up-rational} and induction that the maps $f_i^*\omega_A \to \omega_{X_i}$ is always surjective on stalks at $x_i$. Hence $$ (f_n^*\omega_A)_x \longrightarrow \omega_{X_n, x} $$ is surjective. On the other hand, by our choice of $b$ the quotient of $f_n^*\omega_A$ by its torsion submodule is an invertible module $\mathcal{L}$. Moreover, the dualizing module is torsion free (Duality for Schemes, Lemma \ref{duality-lemma-dualizing-module}). It follows that $\mathcal{L}_x \cong  \omega_{X_n, x}$ and the proof is complete.
Let us write $\mathcal{O}_n = \mathcal{O}_{X_n, p_n}$ and $\mathcal{O} = \mathcal{O}_{X, p}$. Denote $\mathfrak m \subset \mathcal{O}$ and $\mathfrak m_n \subset \mathcal{O}_n$ the maximal ideals. \medskip\noindent We claim that $x_1^t \not \in \mathfrak m_n^{t + 1}$. Namely, if this were the case, then in the local ring $\mathcal{O}_{n + 1}$ the element $x_1^t$ would be in the ideal of $(t + 1)E_{n + 1}$. This contradicts the assumption that $x_1$ defines $E_{n + 1}$. \medskip\noindent For every $n$ choose generators $y_{n, 1}, \ldots, y_{n, t_n}$ for $\mathfrak m_n$. As $\mathfrak m_n \mathcal{O}_{n + 1} = x_1\mathcal{O}_{n + 1}$ by assumption (2), we can write $y_{n, i} = a_{n, i} x_1$ for some $a_{n, i} \in \mathcal{O}_{n + 1}$. Since the map $\mathcal{O}_n \to \mathcal{O}_{n + 1}$ defines an isomorphism on residue fields by (1) we can choose $c_{n, i} \in \mathcal{O}_n$ having the same residue class as $a_{n, i}$. Then we see that $$ \mathfrak m_n = (x_1, z_{n, 1}, \ldots, z_{n, t_n}), \quad z_{n, i} = y_{n, i} - c_{n, i} x_1 $$ and the elements $z_{n, i}$ map to elements of $\mathfrak m_{n + 1}^2$ in $\mathcal{O}_{n + 1}$. \medskip\noindent Let us consider $$ J_n = \Ker(\mathcal{O} \to \mathcal{O}_n/\mathfrak m_n^{n + 1}) $$ We claim that $\mathcal{O}/J_n$ has length $n + 1$ and that $\mathcal{O}/(x_1) + J_n$ equals the residue field. For $n = 0$ this is immediate. Assume the statement holds for $n$. Let $f \in J_n$. Then in $\mathcal{O}_n$ we have $$ f = a x_1^{n + 1} + x_1^n A_1(z_{n, i}) + x_1^{n - 1} A_2(z_{n, i}) + \ldots + A_{n + 1}(z_{n, i}) $$ for some $a \in \mathcal{O}_n$ and some $A_i$ homogeneous of degree $i$ with coefficients in $\mathcal{O}_n$. Since $\mathcal{O} \to \mathcal{O}_n$ identifies residue fields, we may choose $a \in \mathcal{O}$ (argue as in the construction of $z_{n, i}$ above). Taking the image in $\mathcal{O}_{n + 1}$ we see that $f$ and $a x_1^{n + 1}$ have the same image modulo $\mathfrak m_{n + 1}^{n + 2}$. Since $x_n^{n + 1} \not \in \mathfrak m_{n + 1}^{n + 2}$ it follows that $J_n/J_{n + 1}$ has length $1$ and the claim is true. \medskip\noindent Consider $R = \lim \mathcal{O}/J_n$. This is a quotient of the $\mathfrak m$-adic completion of $\mathcal{O}$ hence it is a complete Noetherian local ring. On the other hand, it is not finite length and $x_1$ generates the maximal ideal. Thus $R$ is a complete discrete valuation ring. The map $\mathcal{O} \to R$ lifts to a local homomorphism $\mathcal{O}_n \to R$ for every $n$. There are two ways to show this: (1) for every $n$ one can use a similar procedure to construct $\mathcal{O}_n \to R_n$ and then one can show that $\mathcal{O} \to \mathcal{O}_n \to R_n$ factors through an isomorphism $R \to R_n$, or (2) one can use Divisors, Lemma \ref{divisors-lemma-characterize-affine-blowup} to show that $\mathcal{O}_n$ is a localization of a repeated affine blowup algebra to explicitly construct a map $\mathcal{O}_n \to R$. Having said this it is clear that our sequence of blowups comes from the nonsingular arc $a : T = \Spec(R) \to X$.
We may assume $X$ is affine. Then we may write $X = \Spec(B)$. Let $\mathfrak q \subset B' = B \otimes_A A^\wedge$ be the prime corresponding to $q$ and let $\mathfrak p \subset B$ be the prime ideal corresponding to $p$. By Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated} we have $$ B'/(\mathfrak m^\wedge)^n B' = A^\wedge/(\mathfrak m^\wedge)^n \otimes_A B = A/\mathfrak m^n \otimes_A B = B/\mathfrak m^n B $$ for all $n$. Since $\mathfrak m B \subset \mathfrak p$ and $\mathfrak m^\wedge B' \subset \mathfrak q$ we see that $B/\mathfrak p^n$ and $B'/\mathfrak q^n$ are both quotients of the ring displayed above by the $n$th power of the same prime ideal. The lemma follows.
Since $A \to A^\wedge$ is faithfully flat (Algebra, Lemma \ref{algebra-lemma-completion-faithfully-flat}), we see that $Y \to X$ is flat. Hence (1) by Algebra, Lemma \ref{algebra-lemma-descent-regular}. Lemma \ref{lemma-iso-completions} shows the morphism $Y \to X$ induces an isomorphism on complete local rings at points of the special fibres. Thus (2) by More on Algebra, Lemma \ref{more-algebra-lemma-completion-regular}. If $X$ is proper over $A$, then $Y$ is proper over $A^\wedge$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-proper}) and we see every closed point of $X$ and $Y$ lies in the closed fibre. Thus we see that $Y$ is a regular scheme if and only if $X$ is so by Properties, Lemma \ref{properties-lemma-characterize-regular}.
By definition there exists an ideal $J \subset A^\wedge$ such that $V(J) = \{\mathfrak m A^\wedge\}$ and such that $Y$ is the blowup of $S^\wedge$ in the closed subscheme defined by $J$, see Divisors, Definition \ref{divisors-definition-admissible-blowup}. Since $A^\wedge$ is Noetherian this implies $\mathfrak m^n A^\wedge \subset J$ for some $n$. Since $A^\wedge/\mathfrak m^n A^\wedge = A/\mathfrak m^n$ we find an ideal $\mathfrak m^n \subset I \subset A$ such that $J = I A^\wedge$. Let $X \to S$ be the blowup in $I$. Since $A \to A^\wedge$ is flat we conclude that the base change of $X$ is $Y$ by Divisors, Lemma \ref{divisors-lemma-flat-base-change-blowing-up}.
Let $Y \to \Spec(A^\wedge)$ be a modification with $Y$ normal. We have to show that $H^1(Y, \mathcal{O}_Y) = 0$. By Varieties, Lemma \ref{varieties-lemma-modification-normal-iso-over-codimension-1} $Y \to \Spec(A^\wedge)$ is an isomorphism over the punctured spectrum $U^\wedge = \Spec(A^\wedge) \setminus \{\mathfrak m^\wedge\}$. By Lemma \ref{lemma-dominate-by-scheme-modification} there exists a $U^\wedge$-admissible blowup $Y' \to \Spec(A^\wedge)$ dominating $Y$. By Lemma \ref{lemma-descend-admissible-blowup} we find there exists a $U$-admissible blowup $X \to \Spec(A)$ whose base change to $A^\wedge$ dominates $Y$. Since $A$ is Nagata, we can replace $X$ by its normalization after which $X \to \Spec(A)$ is a normal modification (but possibly no longer a $U$-admissible blowup). Then $H^1(X, \mathcal{O}_X) = 0$ as $A$ defines a rational singularity. It follows that $H^1(X \times_{\Spec(A)} \Spec(A^\wedge), \mathcal{O}_{X \times_{\Spec(A)} \Spec(A^\wedge)}) = 0$ by flat base change (Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology} and flatness of $A \to A^\wedge$ by Algebra, Lemma \ref{algebra-lemma-completion-flat}). We find that $H^1(Y, \mathcal{O}_Y) = 0$ by Lemma \ref{lemma-exact-sequence}. \medskip\noindent Finally, let $X \to \Spec(A)$ be the blowing up of $\Spec(A)$ in $\mathfrak m$. Then $Y = X \times_{\Spec(A)} \Spec(A^\wedge)$ is the blowing up of $\Spec(A^\wedge)$ in $\mathfrak m^\wedge$. By Lemma \ref{lemma-blow-up-normal-rational} we see that both $Y$ and $X$ are normal. On the other hand, $A^\wedge$ is excellent (More on Algebra, Proposition \ref{more-algebra-proposition-ubiquity-excellent}) hence every affine open in $Y$ is the spectrum of an excellent normal domain (More on Algebra, Lemma \ref{more-algebra-lemma-finite-type-over-excellent}). Thus for $y \in Y$ the ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{Y, y}^\wedge$ is regular and by More on Algebra, Lemma \ref{more-algebra-lemma-normal-goes-up} we find that $\mathcal{O}_{Y, y}^\wedge$ is normal. If $x \in X$ is a closed point of the special fibre, then there is a unique closed point $y \in Y$ lying over $x$. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ induces an isomorphism on completions (Lemma \ref{lemma-iso-completions}) we conclude.
Since $A$ is analytically unramified it is reduced by Algebra, Lemma \ref{algebra-lemma-analytically-unramified-easy}. Since the normalization of $X$ depends only on the reduction of $X$, we may replace $X$ by its reduction $X_{red}$; note that $X_{red} \to X$ is an isomorphism over the open $U$ where $X \to \Spec(A)$ is \'etale because $U$ is reduced (Descent, Lemma \ref{descent-lemma-reduced-local-smooth}) hence condition (3) remains true after this replacement. In addition we may and do assume that $X = \Spec(B)$ is affine. \medskip\noindent The map $$ K = \prod\nolimits_{\mathfrak p \subset A\text{ minimal}} \kappa(\mathfrak p) \longrightarrow K^\wedge = \prod\nolimits_{\mathfrak p^\wedge \subset A^\wedge\text{ minimal}} \kappa(\mathfrak p^\wedge) $$ is injective because $A \to A^\wedge$ is faithfully flat (Algebra, Lemma \ref{algebra-lemma-completion-faithfully-flat}) hence induces a surjective map between sets of minimal primes (by going down for flat ring maps, see Algebra, Section \ref{algebra-section-going-up}). Both sides are finite products of fields as our rings are Noetherian. Let $L = \prod_{\mathfrak q \subset B\text{ minimal}} \kappa(\mathfrak q)$. Our assumption (3) implies that $L = B \otimes_A K$ and that $K \to L$ is a finite \'etale ring map (this is true because $A \to B$ is generically finite, for example use Algebra, Lemma \ref{algebra-lemma-generically-finite} or the more detailed results in Morphisms, Section \ref{morphisms-section-generically-finite}). Since $B$ is reduced we see that $B \subset L$. This implies that $$ C = B \otimes_A A^\wedge \subset L \otimes_A A^\wedge = L \otimes_K K^\wedge = M $$ Then $M$ is the total ring of fractions of $C$ and is a finite product of fields as a finite separable algebra over $K^\wedge$. It follows that $C$ is reduced and that its normalization $C'$ is the integral closure of $C$ in $M$. The normalization $B'$ of $B$ is the integral closure of $B$ in $L$. By flatness of $A \to A^\wedge$ we obtain an injective map $B' \otimes_A A^\wedge \to M$ whose image is contained in $C'$. Picture $$ B' \otimes_A A^\wedge \longrightarrow C' $$ As $A^\wedge$ is Nagata (by Algebra, Lemma \ref{algebra-lemma-Noetherian-complete-local-Nagata}), we see that $C'$ is finite over $C = B \otimes_A A^\wedge$ (see Algebra, Lemmas \ref{algebra-lemma-Noetherian-complete-local-Nagata} and \ref{algebra-lemma-nagata-in-reduced-finite-type-finite-integral-closure}). As $C$ is Noetherian, we conclude that $B' \otimes_A A^\wedge$ is finite over $C = B \otimes_A A^\wedge$. Therefore by faithfully flat descent (Algebra, Lemma \ref{algebra-lemma-descend-properties-modules}) we see that $B'$ is finite over $B$ which is what we had to show.
There is an immediate reduction to the case where $X = \Spec(B)$ is affine with $B$ a finite type $A$-algebra. Set $C = B \otimes_A A^\wedge$ so that $Y = \Spec(C)$. Since $A \to A^\wedge$ is faithfully flat, for any prime $\mathfrak q \subset B$ there exists a prime $\mathfrak r \subset C$ lying over $\mathfrak q$. Then $B_\mathfrak q \to C_\mathfrak r$ is faithfully flat. Hence if $\mathfrak q$ does not lie over $\mathfrak m$, then $C_\mathfrak r$ is normal by assumption on $Y$ and we conclude that $B_\mathfrak q$ is normal by Algebra, Lemma \ref{algebra-lemma-descent-normal}. In this way we see that $X$ is normal away from the special fibre. \medskip\noindent Recall that the complete Noetherian local ring $A^\wedge$ is Nagata (Algebra, Lemma \ref{algebra-lemma-Noetherian-complete-local-Nagata}). Hence the normalization $Y^\nu \to Y$ is finite (Morphisms, Lemma \ref{morphisms-lemma-nagata-normalization}) and an isomorphism away from the special fibre. Say $Y^\nu = \Spec(C')$. Then $C \to C'$ is finite and an isomorphism away from $V(\mathfrak m C)$. Since $B \to C$ is flat and induces an isomorphism $B/\mathfrak m B \to C/\mathfrak m C$ there exists a finite ring map $B \to B'$ whose base change to $C$ recovers $C \to C'$. See More on Algebra, Lemma \ref{more-algebra-lemma-application-formal-glueing} and Remark \ref{more-algebra-remark-formal-glueing-algebras}. Thus we find a finite morphism $X' \to X$ which is an isomorphism away from the special fibre and whose base change recovers $Y^\nu \to Y$. By the discussion in the first paragraph we see that $X'$ is normal at points not on the special fibre. For a point $x \in X'$ on the special fibre we have a corresponding point $y \in Y^\nu$ and a flat map $\mathcal{O}_{X', x} \to \mathcal{O}_{Y^\nu, y}$. Since $\mathcal{O}_{Y^\nu, y}$ is normal, so is $\mathcal{O}_{X', x}$, see Algebra, Lemma \ref{algebra-lemma-descent-normal}. Thus $X'$ is normal and it follows that it is the normalization of $X$.
Given the sequence $Y_n \to \ldots \to Y_1 \to Y_0 = \Spec(A^\wedge)$ we inductively construct $X_n \to \ldots \to X_1 \to X_0 = \Spec(A)$. The base case is $i = 0$. Given $X_i$ whose base change is $Y_i$, let $Y'_i \to Y_i$ be the blowing up in the closed point $y_i \in Y_i$ such that $Y_{i + 1}$ is the normalization of $Y_i$. Since the closed fibres of $Y_i$ and $X_i$ are isomorphic, the point $y_i$ corresponds to a closed point $x_i$ on the special fibre of $X_i$. Let $X'_i \to X_i$ be the blowup of $X_i$ in $x_i$. Then the base change of $X'_i$ to $\Spec(A^\wedge)$ is isomorphic to $Y'_i$.  By Lemma \ref{lemma-normalization-completion} the normalization $X_{i + 1} \to X'_i$ is finite and its base change to $\Spec(A^\wedge)$ is isomorphic to $Y_{i + 1}$.
Consider the partial derivatives $b + 2dx + ey$ and $c + ex + 2fy$. By the Leibniz rules these are contained in $I$. If one of these is nonzero, then after a linear change of coordinates, i.e., of the form $x \mapsto \alpha + \beta x + \gamma y$ and $y \mapsto \delta + \epsilon x + \zeta y$, we may assume that $x \in I$. Then we see that $I = (x)$ or $I = (x, F)$ with $F$ a monic polynomial of degree $\geq 2$ in $y$. In the first case the statement is clear. In the second case observe that we can write any element in $I^2$ in the form $$ A(x, y) x^2 + B(y) x F + C(y) F^2 $$ for some $A(x, y) \in \kappa[x, y]$ and $B, C \in \kappa[y]$. Thus $$ a + b x + c y + d x^2 + exy + f y^2 = A(x, y) x^2 + B(y) x F + C(y) F^2 $$ and by degree reasons we see that $B = C = 0$ and $A$ is a constant. \medskip\noindent To finish the proof we need to deal with the case that both partial derivatives are zero. This can only happen in characteristic $2$ and then we get $$ a + d x^2 + f y^2 \in I^2 $$ We may assume $f$ is nonzero (if not, then switch the roles of $x$ and $y$). After dividing by $f$ we obtain the case where the characteristic of $\kappa$ is $2$ and $$ a + d x^2 + y^2 \in I^2 $$ If $a$ and $d$ are squares in $\kappa$, then we are done. If not, then there exists a derivation $\theta : \kappa \to \kappa$ with $\theta(a) \not = 0$ or $\theta(d) \not = 0$, see Algebra, Lemma \ref{algebra-lemma-derivative-zero-pth-power}. We can extend this to a derivation of $\kappa[x, y]$ by setting $\theta(x) = \theta(y) = 0$. Then we find that $$ \theta(a) + \theta(d) x^2 \in I $$ The case $\theta(d) = 0$ is absurd. Thus we may assume that $\alpha + x^2 \in I$ for some $\alpha \in \kappa$. Combining with the above we find that $a + \alpha d + y^2 \in I$. Hence $$ J = (\alpha + x^2, a + \alpha d + y^2) \subset I $$ with codimension at most $2$. Observe that $J/J^2$ is free over $\kappa[x, y]/J$ with basis $\alpha + x^2$ and $a + \alpha d + y^2$. Thus $a + d x^2 + y^2 = 1 \cdot (a + \alpha d + y^2) + d \cdot (\alpha + x^2) \in I^2$ implies that the inclusion $J \subset I$ is strict. Thus we find a nonzero element of the form $g + hx + iy + jxy$ in $I$. If $j = 0$, then $I$ contains a linear form and we can conclude as in the first paragraph. Thus $j \not = 0$ and $\dim_\kappa(I/J) = 1$ (otherwise we could find an element as above in $I$ with $j = 0$). We conclude that $I$ has the form $(\alpha + x^2, \beta + y^2, g + hx + iy + jxy)$ with $j \not = 0$ and has colength $3$. In this case $a + dx^2 + y^2 \in I^2$ is impossible. This can be shown by a direct computation, but we prefer to argue as follows. Namely, to prove this statement we may assume that $\kappa$ is algebraically closed. Then we can do a coordinate change $x \mapsto \sqrt{\alpha} + x$ and $y \mapsto \sqrt{\beta} + y$ and assume that $I = (x^2, y^2, g' + h'x + i'y + jxy)$ with the same $j$. Then $g' = h' = i' = 0$ otherwise the colength of $I$ is not $3$. Thus we get $I = (x^2, y^2, xy)$ and the result is clear.
It suffices to prove this when $Y = \Spec(A)$ where $A$ is a Noetherian domain. Let $B$ be the integral closure of $A$ in its fraction field. Set $C = \Gamma(X, \mathcal{O}_X)$. By Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-over-affine-cohomology-finite} we see that $C$ is a finite $A$-module. As $X$ is normal (Properties, Lemma \ref{properties-lemma-regular-normal}) we see that $C$ is normal domain (Properties, Lemma \ref{properties-lemma-normal-integral-sections}). Thus $B \subset C$ and we conclude that $B$ is finite over $A$ as $A$ is Noetherian. \medskip\noindent There exists a nonempty open $V \subset Y$ such that $f^{-1}V \to V$ is finite, see Morphisms, Definition \ref{morphisms-definition-alteration}. After shrinking $V$ we may assume that $f^{-1}V \to V$ is flat (Morphisms, Proposition \ref{morphisms-proposition-generic-flatness}). Thus $f^{-1}V \to V$ is faithfully flat. Then $V$ is regular by Algebra, Lemma \ref{algebra-lemma-descent-regular}.
Consider the commutative diagram $$ \xymatrix{ B \ar[r] \ar[d] & C \ar[d] \\ B^\wedge \ar[r] & C^\wedge } $$ Recall that $\mathfrak m$-adic completion on the category of finite $A$-modules is exact because it is given by tensoring with the flat $A$-algebra $A^\wedge$ (Algebra, Lemma \ref{algebra-lemma-completion-flat}). We will use Serre's criterion (Algebra, Lemma \ref{algebra-lemma-criterion-normal}) to prove that the Noetherian ring $B^\wedge$ is normal. Let $\mathfrak q \subset B^\wedge$ be a prime lying over $\mathfrak p \subset B$. If $\dim(B_\mathfrak p) \geq 2$, then $\text{depth}(B_\mathfrak p) \geq 2$ and since $B_\mathfrak p \to B^\wedge_\mathfrak q$ is flat we find that $\text{depth}(B^\wedge_\mathfrak q) \geq 2$ (Algebra, Lemma \ref{algebra-lemma-apply-grothendieck}). If $\dim(B_\mathfrak p) \leq 1$, then $B_\mathfrak p$ is either a discrete valuation ring or a field. In that case $C_\mathfrak p$ is faithfully flat over $B_\mathfrak p$ (because it is finite and torsion free). Hence $B^\wedge_\mathfrak p \to C^\wedge_\mathfrak p$ is faithfully flat and the same holds after localizing at $\mathfrak q$. As $C^\wedge$ and hence any localization is $(S_2)$ we conclude that $B^\wedge_\mathfrak p$ is $(S_2)$ by Algebra, Lemma \ref{algebra-lemma-descent-Sk}. All in all we find that $(S_2)$ holds for $B^\wedge$. To prove that $B^\wedge$ is $(R_1)$ we only have to consider primes $\mathfrak q \subset B^\wedge$ with $\dim(B^\wedge_\mathfrak q) \leq 1$. Since $\dim(B^\wedge_\mathfrak q) = \dim(B_\mathfrak p) + \dim(B^\wedge_\mathfrak q/\mathfrak p B^\wedge_\mathfrak q)$ by Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-total} we find that $\dim(B_\mathfrak p) \leq 1$ and we see that $B^\wedge_\mathfrak q \to C^\wedge_\mathfrak q$ is faithfully flat as before. We conclude using Algebra, Lemma \ref{algebra-lemma-descent-Rk}.
Parts (1) and (2) follow from Lemma \ref{lemma-regular-alteration-implies}. We have to redo part of the proof of that lemma in order to set up notation for the proof of (3). Set $C = \Gamma(X, \mathcal{O}_X)$. By Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-over-affine-cohomology-finite} we see that $C$ is a finite $A$-module. As $X$ is normal (Properties, Lemma \ref{properties-lemma-regular-normal}) we see that $C$ is normal domain (Properties, Lemma \ref{properties-lemma-normal-integral-sections}). Thus $B \subset C$ and we conclude that $B$ is finite over $A$ as $A$ is Noetherian. By Lemma \ref{lemma-algebra-helper} in order to prove (3) it suffices to show that the $\mathfrak m$-adic completion $C^\wedge$ is normal. \medskip\noindent By Algebra, Lemma \ref{algebra-lemma-completion-finite-extension} the completion $C^\wedge$ is the product of the completions of $C$ at the prime ideals of $C$ lying over $\mathfrak m$. There are finitely many of these and these are the maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_r$ of $C$. (The corresponding result for $B$ explains the final statement of the lemma.) Thus replacing $A$ by $C_{\mathfrak m_i}$ and $X$ by $X_i = X \times_{\Spec(C)} \Spec(C_{\mathfrak m_i})$ we reduce to the case discussed in the next paragraph. (Note that $\Gamma(X_i, \mathcal{O}) = C_{\mathfrak m_i}$ by Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}.) \medskip\noindent Here $A$ is a Noetherian local normal domain and $f : X \to \Spec(A)$ is a regular alteration with $\Gamma(X, \mathcal{O}_X) = A$. We have to show that the completion $A^\wedge$ of $A$ is a normal domain. By Lemma \ref{lemma-port-regularity-to-completion} $Y = X \times_{\Spec(A)} \Spec(A^\wedge)$ is regular. Since $\Gamma(Y, \mathcal{O}_Y) = A^\wedge$ by Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}, we conclude that $A^\wedge$ is normal as before. Namely, $Y$ is normal by Properties, Lemma \ref{properties-lemma-regular-normal}. It is connected because $\Gamma(Y, \mathcal{O}_Y) = A^\wedge$ is local. Hence $Y$ is normal and integral (as connected and normal implies integral for Noetherian schemes). Thus $\Gamma(Y, \mathcal{O}_Y) = A^\wedge$ is a normal domain by Properties, Lemma \ref{properties-lemma-normal-integral-sections}. This proves (3). \medskip\noindent Proof of (4). Let $\eta \in \Spec(A)$ denote the generic point and denote by a subscript $\eta$ the base change to $\eta$. Since $f$ is an alteration, the scheme $X_\eta$ is finite and faithfully flat over $\eta$. Since $Y = X \times_{\Spec(A)} \Spec(A^\wedge)$ is regular by Lemma \ref{lemma-port-regularity-to-completion} we see that $Y_\eta$ is regular (as a limit of opens in $Y$). Then $Y_\eta \to \Spec(A^\wedge \otimes_A \kappa(\eta))$ is finite faithfully flat onto the generic formal fibre. We conclude by Algebra, Lemma \ref{algebra-lemma-descent-regular}.