The Stacks project

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. $\square$