The Stacks project

54.11 Base change to the completion

The following simple lemma will turn out to be a useful tool in what follows.

Lemma 54.11.1. Let $(A, \mathfrak m, \kappa )$ be a local ring with finitely generated maximal ideal $\mathfrak m$. Let $X$ be a scheme over $A$. Let $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$ where $A^\wedge $ is the $\mathfrak m$-adic completion of $A$. For a point $q \in Y$ with image $p \in X$ lying over the closed point of $\mathop{\mathrm{Spec}}(A)$ the local ring map $\mathcal{O}_{X, p} \to \mathcal{O}_{Y, q}$ induces an isomorphism on completions.

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

\[ B'/(\mathfrak m^\wedge )^ n B' = A^\wedge /(\mathfrak m^\wedge )^ n \otimes _ A B = A/\mathfrak m^ n \otimes _ A B = B/\mathfrak m^ n B \]

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. $\square$

Lemma 54.11.2. 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 $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$. Let $y \in Y$ with image $x \in X$. Then

  1. if $\mathcal{O}_{Y, y}$ is regular, then $\mathcal{O}_{X, x}$ is regular,

  2. if $y$ is in the closed fibre, then $\mathcal{O}_{Y, y}$ is regular $\Leftrightarrow \mathcal{O}_{X, x}$ is regular, and

  3. If $X$ is proper over $A$, then $X$ is regular if and only if $Y$ is regular.

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. $\square$

Lemma 54.11.3. Let $(A, \mathfrak m)$ be a Noetherian local ring with completion $A^\wedge $. Let $U \subset \mathop{\mathrm{Spec}}(A)$ and $U^\wedge \subset \mathop{\mathrm{Spec}}(A^\wedge )$ be the punctured spectra. If $Y \to \mathop{\mathrm{Spec}}(A^\wedge )$ is a $U^\wedge $-admissible blowup, then there exists a $U$-admissible blowup $X \to \mathop{\mathrm{Spec}}(A)$ such that $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$.

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. $\square$

Lemma 54.11.4. Let $(A, \mathfrak m, \kappa )$ be a Nagata local normal domain of dimension $2$. Assume $A$ defines a rational singularity and that the completion $A^\wedge $ of $A$ is normal. Then

  1. $A^\wedge $ defines a rational singularity, and

  2. if $X \to \mathop{\mathrm{Spec}}(A)$ is the blowing up in $\mathfrak m$, then for a closed point $x \in X$ the completion $\mathcal{O}_{X, x}$ is normal.

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. $\square$

Lemma 54.11.5. Let $(A, \mathfrak m)$ be a local Noetherian ring. Let $X$ be a scheme over $A$. Assume

  1. $A$ is analytically unramified (Algebra, Definition 10.162.9),

  2. $X$ is locally of finite type over $A$, and

  3. $X \to \mathop{\mathrm{Spec}}(A)$ is étale at the generic points of irreducible components of $X$.

Then the normalization of $X$ is finite over $X$.

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

\[ K = \prod \nolimits _{\mathfrak p \subset A\text{ minimal}} \kappa (\mathfrak p) \longrightarrow K^\wedge = \prod \nolimits _{\mathfrak p^\wedge \subset A^\wedge \text{ minimal}} \kappa (\mathfrak p^\wedge ) \]

is injective because $A \to A^\wedge $ is faithfully flat (Algebra, Lemma 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

\[ C = B \otimes _ A A^\wedge \subset L \otimes _ A A^\wedge = L \otimes _ K K^\wedge = M \]

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 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. $\square$

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 29.54.11) 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 $X$. $\square$

Lemma 54.11.7. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local domain whose completion $A^\wedge $ is normal. Then given any sequence

\[ Y_ n \to Y_{n - 1} \to \ldots \to Y_1 \to \mathop{\mathrm{Spec}}(A^\wedge ) \]

of normalized blowups, there exists a sequence of (proper) normalized blowups

\[ X_ n \to X_{n - 1} \to \ldots \to X_1 \to \mathop{\mathrm{Spec}}(A) \]

whose base change to $A^\wedge $ recovers the given sequence.

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 $Y_{i + 1}$. $\square$

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