The Stacks project

Proof. By Lemma 15.45.3 the map $A^ h \to A^\wedge$ is flat and injective. Combining going down (Algebra, Lemma 10.39.19) and Algebra, Lemma 10.30.5 we see that part (1) holds. Part (2) follows from this, Definition 15.106.6, and the fact that $A^\wedge$ is henselian (Algebra, Lemma 10.153.9). By Lemma 15.45.3 we have $(A^\wedge )^{sh} = A^{sh} \otimes _{A^ h} A^\wedge$. Thus we can repeat the arguments above using the flat injective map $A^{sh} \to (A^\wedge )^{sh}$ to prove (3). $\square$

Proof. Follows from Lemma 15.108.1 and the fact that there are only a finite number of branches for both $A$ and $A^\wedge$ by Algebra, Lemma 10.31.6 and the fact that $A^ h$ and $A^\wedge$ are Noetherian (Lemma 15.45.3). $\square$

Proof. Choose generators $f_1, \ldots , f_ r$ of $I$. Write

for some idempotent $e_ i \in A_{f_ i}$. Write $e_ i = a_ i/f_ i^ n$ for some $a_ i \in A$ and $n \geq 0$; we may use the same $n$ for all $i = 1, \ldots , r$. After replacing $a_ i$ by $f_ i^ ma_ i$ and $n$ by $n + m$ for a suitable $m \gg 0$, we may assume $a_ i^2 = f_ i^ n a_ i$ for all $i$. Since $e_ i$ maps to $1$ in $C_{f_ if_ j} = (A_{f_ if_ j})_{e_ j} = A_{f_ if_ ja_ j}$ we see that

for some $N$ (we can pick the same $N$ for all pairs $i, j$). Using $a_ j^2 = f_ j^ na_ j$ this gives

After increasing $n$ to $n + N + nN$ and replacing $a_ i$ by $f_ i^{N + nN}a_ i$ we see that $f_ i^ n a_ j$ is in the ideal of $a_ i$ for all pairs $i, j$. Let $C' = A/(a_1, \ldots , a_ r)$. Then

because $a_ j$ is in the ideal generated by $a_ i$ after inverting $f_ i$. Since for an idempotent $e$ of a ring $B$ we have $B = B_ e \times B/(e)$ we see that the conclusion of the lemma holds for $f$ equal to one of $f_1, \ldots , f_ r$. Using glueing of functions, in the form of Algebra, Lemma 10.23.2, we conclude that the result holds for all $f \in I$. Namely, for $f \in I$ the elements $f_1, \ldots , f_ r$ generate the unit ideal in $A_ f$ so $A_ f \to (C \times C')_ f$ is an isomorphism if and only if this is the case after localizing at $f_1, \ldots , f_ r$. $\square$

Proof. Let $f \in I$. Since $B^\wedge$ is Noetherian (Algebra, Lemma 10.97.6), we see that $J$ is a finitely generated ideal. Hence we conclude from Algebra, Lemma 10.21.5 that

for some idempotent $e \in (B^\wedge )_ f$. By Lemma 15.108.3 we can find a surjection $B^\wedge \to C'$ such that $B^\wedge \to C \times C'$ becomes an isomorphism after inverting any $f \in I$. Observe that $C \times C'$ is a finite $B^\wedge$-algebra.

Choose generators $f_1, \ldots , f_ r \in I$. Denote $\alpha _ i : (C \times C')_{f_ i} \to B_{f_ i} \otimes _ B B^\wedge$ the inverse of the isomorphism of $(B^\wedge )_{f_ i}$-algebras we obtained above. Denote $\alpha _{ij} : (B_{f_ i})_{f_ j} \to (B_{f_ j})_{f_ i}$ the obvious $B$-algebra isomorphism. Consider the object

of the category $\text{Glue}(B \to B^\wedge , f_1, \ldots , f_ r)$ introduced in Remark 15.89.10. We omit the verification of conditions (1)(a) and (1)(b). Since $B \to B^\wedge$ is a flat map (Algebra, Lemma 10.97.2) inducing an isomorphism $B/IB \to B^\wedge /IB^\wedge$ we may apply Proposition 15.89.16 and Remark 15.89.20. We conclude that $C \times C'$ is isomorphic to $D \otimes _ B B^\wedge$ for some finite $B$-algebra $D$. Then $D/ID \cong C/IC \times C'/IC'$. Let $\overline{e} \in D/ID$ be the idempotent corresponding to the factor $C/IC$. By Lemma 15.9.10 there exists an étale ring map $B \to B'$ which induces an isomorphism $B/IB \to B'/IB'$ such that $D' = D \otimes _ B B'$ contains an idempotent $e$ lifting $\overline{e}$. Since $C \times C'$ is $I$-adically complete the pair $(C \times C', IC \times IC')$ is henselian (Lemma 15.11.4). Thus we can factor the map $B \to C \times C'$ through $B'$. Doing so we may replace $B$ by $B'$ and $D$ by $D'$. Then we find that $D = D_ e \times D_{1 - e} = D/(1 - e) \times D/(e)$ is a product of finite type $A$-algebras and the completion of the first part is $C$ and the completion of the second part is $C'$. $\square$

Proof. Since the completion of $A$ and $A^ h$ are the same, we may assume that $A$ is henselian (Lemma 15.45.3). We will apply Lemma 15.108.4 to $A^\wedge \to A^\wedge /J$ where $J = \mathop{\mathrm{Ker}}(A^\wedge \to (A^\wedge )_{\mathfrak q})$. Since $\dim ((A^\wedge )_\mathfrak q) = 0$ we see that $\mathfrak q^ n \subset J$ for some $n$. Hence $J/J^2$ is annihilated by $\mathfrak q^ n$. On the other hand $(J/J^2)_\mathfrak q = 0$ because $J_\mathfrak q = 0$. Hence $\mathfrak m$ is the only associated prime of $J/J^2$ and we find that a power of $\mathfrak m$ annihilates $J/J^2$. Thus the lemma applies and we find that $A^\wedge /J = C^\wedge$ for some finite type $A$-algebra $C$.

Then $C/\mathfrak m C = A/\mathfrak m$ because $A^\wedge /J$ has the same property. Hence $\mathfrak m_ C = \mathfrak m C$ is a maximal ideal and $A \to C$ is unramified at $\mathfrak m_ C$ (Algebra, Lemma 10.151.7). After replacing $C$ by a principal localization we may assume that $C$ is a quotient of an étale $A$-algebra $B$, see Algebra, Proposition 10.152.1. However, since the residue field extension of $A \to C_{\mathfrak m_ C}$ is trivial and $A$ is henselian, we conclude that $B = A$ again after a localization. Thus $C = A/I$ for some ideal $I \subset A$ and it follows that $J = IA^\wedge$ (because completion is exact in our situation by Algebra, Lemma 10.97.2) and $I = J \cap A$ (by flatness of $A \to A^\wedge$). Since $\mathfrak q^ n \subset J \subset \mathfrak q$ we see that $\mathfrak p = \mathfrak q \cap A$ satisfies $\mathfrak p^ n \subset I \subset \mathfrak p$. Then $\sqrt{\mathfrak p A^\wedge } = \mathfrak q$ and the proof is complete. $\square$

Proof. Since the completion of $A$ and $A^ h$ are the same, we may assume that $A$ is henselian (Lemma 15.45.3).

Since $A \to A^\wedge$ is faithfully flat (see reference just given) the map from the punctured spectrum of $A^\wedge$ to the punctured spectrum of $A$ is surjective (see Algebra, Lemma 10.39.16). Hence if the punctured spectrum of $A$ is disconnected, then the same is true for $A^\wedge$.

Assume the punctured spectrum of $A^\wedge$ is disconnected. This means that

with $Z$ and $Z'$ closed. Let $\overline{Z}, \overline{Z}' \subset \mathop{\mathrm{Spec}}(A^\wedge )$ be the closures. Say $\overline{Z} = V(J)$, $\overline{Z}' = V(J')$ for some ideals $J, J' \subset A^\wedge$. Then $V(J + J') = \{ \mathfrak m^\wedge \}$ and $V(JJ') = \mathop{\mathrm{Spec}}(A^\wedge )$. The first equality means that $\mathfrak m^\wedge = \sqrt{J + J'}$ which implies $(\mathfrak m^\wedge )^ e \subset J + J'$ for some $e \geq 1$. The second equality implies every element of $JJ'$ is nilpotent hence $(JJ')^ n = 0$ for some $n \geq 1$. Combined this means that $J^ n/J^{2n}$ is annihilated by $J^ n$ and $(J')^ n$ and hence by $(\mathfrak m^\wedge )^{2en}$. Thus we may apply Lemma 15.108.4 to see that there is a finite type $A$-algebra $C$ and an isomorphism $A^\wedge /J^ n = C^\wedge$.

The rest of the proof is exactly the same as the second part of the proof of Lemma 15.108.5; of course that lemma is a special case of this one! We have $C/\mathfrak m C = A/\mathfrak m$ because $A^\wedge /J^ n$ has the same property. Hence $\mathfrak m_ C = \mathfrak m C$ is a maximal ideal and $A \to C$ is unramified at $\mathfrak m_ C$ (Algebra, Lemma 10.151.7). After replacing $C$ by a principal localization we may assume that $C$ is a quotient of an étale $A$-algebra $B$, see Algebra, Proposition 10.152.1. However, since the residue field extension of $A \to C_{\mathfrak m_ C}$ is trivial and $A$ is henselian, we conclude that $B = A$ again after a localization. Thus $C = A/I$ for some ideal $I \subset A$ and it follows that $J^ n = IA^\wedge$ (because completion is exact in our situation by Algebra, Lemma 10.97.2) and $I = J^ n \cap A$ (by flatness of $A \to A^\wedge$). By symmetry $I' = (J')^ n \cap A$ satisfies $(J')^ n = I'A^\wedge$. Then $\mathfrak m^ e \subset I + I'$ and $II' = 0$ and we conclude that $V(I)$ and $V(I')$ are closed subschemes which give the desired disjoint union decomposition of the punctured spectrum of $A$. $\square$

Proof. To see this for the number of branches, combine Lemmas 15.108.1, 15.108.2, and 15.108.5 and use that the dimension of $A^\wedge$ is one, see Lemma 15.43.1. To see this is true for the number of geometric branches we use the result for branches, the fact that the dimension does not change under strict henselization (Lemma 15.45.7), and the fact that $(A^{sh})^\wedge = ((A^\wedge )^{sh})^\wedge$ by Lemma 15.45.3. $\square$

Proof. Since a normal ring is reduced, we see that $A$ is Nagata by Lemma 15.52.4. In the rest of the proof we will use Lemma 15.51.10, Proposition 15.51.5, and Lemma 15.51.4. This tells us that $A$ is a P-ring where $P(k \to R) =$“$R$ is geometrically normal over $k$” and the same is true for any (essentially of) finite type $A$-algebra.

Let $\mathfrak q \subset A$ be a minimal prime. Then $A^\wedge /\mathfrak q A^\wedge = (A/\mathfrak q)^\wedge$ and $A^ h/\mathfrak qA^ h = (A/\mathfrak q)^ h$ (Algebra, Lemma 10.156.2). Hence the number of branches of $A$ is the sum of the number of branches of the rings $A/\mathfrak q$ and similarly for $A^\wedge$. In this way we reduce to the case that $A$ is a domain.

Assume $A$ is a domain. Let $A'$ be the integral closure of $A$ in the fraction field $K$ of $A$. Since $A$ is Nagata, we see that $A \to A'$ is finite. Recall that the number of branches of $A$ is the number of maximal ideals $\mathfrak m'$ of $A'$ (Lemma 15.106.2). Also, recall that

by Algebra, Lemma 10.97.8. Because $A'_{\mathfrak m'}$ is a local ring whose formal fibres are geometrically normal, we see that $(A'_{\mathfrak m'})^\wedge$ is normal (Lemma 15.52.6). Hence the minimal primes of $A' \otimes _ A A^\wedge$ are in $1$-to-$1$ correspondence with the factors in the decomposition above. By flatness of $A \to A^\wedge$ we have

Since the left and the right ring have the same set of minimal primes, the same is true for the ring in the middle (small detail omitted) and this finishes the proof.

To see this is true for the number of geometric branches we use the result for branches, the fact that the formal fibres of $A^{sh}$ are geometrically normal (Lemmas 15.51.10 and 15.51.8) and the fact that $(A^{sh})^\wedge = ((A^\wedge )^{sh})^\wedge$ by Lemma 15.45.3. $\square$