The Stacks project

Proof. We may assume $X$ is affine. Then we may write $X = \mathop{\mathrm{Spec}}(B)$. 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 10.96.3 we have

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. $\square$

Proof. Since $A \to A^\wedge $ is faithfully flat (Algebra, Lemma 10.97.3), we see that $Y \to X$ is flat. Hence (1) by Algebra, Lemma 10.164.4. Lemma 54.11.1 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 15.43.4. If $X$ is proper over $A$, then $Y$ is proper over $A^\wedge $ (Morphisms, Lemma 29.41.5) 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 28.9.2. $\square$

Proof. 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 31.34.1. 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 31.32.3. $\square$

Proof. Let $Y \to \mathop{\mathrm{Spec}}(A^\wedge )$ be a modification with $Y$ normal. We have to show that $H^1(Y, \mathcal{O}_ Y) = 0$. By Varieties, Lemma 33.17.3 $Y \to \mathop{\mathrm{Spec}}(A^\wedge )$ is an isomorphism over the punctured spectrum $U^\wedge = \mathop{\mathrm{Spec}}(A^\wedge ) \setminus \{ \mathfrak m^\wedge \} $. By Lemma 54.7.2 there exists a $U^\wedge $-admissible blowup $Y' \to \mathop{\mathrm{Spec}}(A^\wedge )$ dominating $Y$. By Lemma 54.11.3 we find there exists a $U$-admissible blowup $X \to \mathop{\mathrm{Spec}}(A)$ whose base change to $A^\wedge $ dominates $Y$. Since $A$ is Nagata, we can replace $X$ by its normalization after which $X \to \mathop{\mathrm{Spec}}(A)$ 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 $H^1(X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge ), \mathcal{O}_{X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )}) = 0$ by flat base change (Cohomology of Schemes, Lemma 30.5.2 and flatness of $A \to A^\wedge $ by Algebra, Lemma 10.97.2). We find that $H^1(Y, \mathcal{O}_ Y) = 0$ by Lemma 54.8.1.

Finally, let $X \to \mathop{\mathrm{Spec}}(A)$ be the blowing up of $\mathop{\mathrm{Spec}}(A)$ in $\mathfrak m$. Then $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$ is the blowing up of $\mathop{\mathrm{Spec}}(A^\wedge )$ in $\mathfrak m^\wedge $. By Lemma 54.9.4 we see that both $Y$ and $X$ are normal. On the other hand, $A^\wedge $ is excellent (More on Algebra, Proposition 15.52.3) hence every affine open in $Y$ is the spectrum of an excellent normal domain (More on Algebra, Lemma 15.52.2). 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 15.42.2 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 54.11.1) we conclude. $\square$

Proof. Since $A$ is analytically unramified it is reduced by Algebra, Lemma 10.162.10. 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 $X \to \mathop{\mathrm{Spec}}(A)$ is étale because $U$ is reduced (Descent, Lemma 35.18.1) hence condition (3) remains true after this replacement. In addition we may and do assume that $X = \mathop{\mathrm{Spec}}(B)$ is affine.

The map

is injective because $A \to A^\wedge $ is faithfully flat (Algebra, Lemma 10.97.3) hence induces a surjective map between sets of minimal primes (by going down for flat ring maps, see Algebra, Section 10.41). 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 étale ring map (this is true because $A \to B$ is generically finite, for example use Algebra, Lemma 10.122.10 or the more detailed results in Morphisms, Section 29.51). Since $B$ is reduced we see that $B \subset L$. This implies that

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

As $A^\wedge $ is Nagata (by Algebra, Lemma 10.162.8), we see that $C'$ is finite over $C = B \otimes _ A A^\wedge $ (see Algebra, Lemmas 10.162.8 and 10.162.2). 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 10.83.2) we see that $B'$ is finite over $B$ which is what we had to show. $\square$

Proof. There is an immediate reduction to the case where $X = \mathop{\mathrm{Spec}}(B)$ is affine with $B$ a finite type $A$-algebra. Set $C = B \otimes _ A A^\wedge $ so that $Y = \mathop{\mathrm{Spec}}(C)$. 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 10.164.3. In this way we see that $X$ is normal away from the special fibre.

Recall that the complete Noetherian local ring $A^\wedge $ is Nagata (Algebra, Lemma 10.162.8). Hence the normalization $Y^\nu \to Y$ is finite (Morphisms, Lemma 29.54.11) and an isomorphism away from the special fibre. Say $Y^\nu = \mathop{\mathrm{Spec}}(C')$. 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 15.89.17 and Remark 15.89.20. 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 10.164.3. Thus $X'$ is normal and it follows that it is the normalization of $X$. $\square$

Proof. Given the sequence $Y_ n \to \ldots \to Y_1 \to Y_0 = \mathop{\mathrm{Spec}}(A^\wedge )$ we inductively construct $X_ n \to \ldots \to X_1 \to X_0 = \mathop{\mathrm{Spec}}(A)$. 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 $\mathop{\mathrm{Spec}}(A^\wedge )$ is isomorphic to $Y'_ i$. By Lemma 54.11.6 the normalization $X_{i + 1} \to X'_ i$ is finite and its base change to $\mathop{\mathrm{Spec}}(A^\wedge )$ is isomorphic to $Y_{i + 1}$. $\square$