Lemma 54.11.6. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $X \to \mathop{\mathrm{Spec}}(A)$ be a morphism which is locally of finite type. Set $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$. If the complement of the special fibre in $Y$ is normal, then the normalization $X^\nu \to X$ is finite and the base change of $X^\nu$ to $\mathop{\mathrm{Spec}}(A^\wedge )$ recovers the normalization of $Y$.

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 29.54.10) 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 15.89.16 and Remark 15.89.19. 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 $X$. $\square$

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