## 15.105 Branches of the completion

Let $(A, \mathfrak m)$ be a Noetherian local ring. Consider the maps $A \to A^ h \to A^\wedge $. In general the map $A^ h \to A^\wedge $ need not induce a bijection on minimal primes, see Examples, Section 108.19. In other words, the number of branches of $A$ (as defined in Definition 15.104.6) may be different from the number of branches of $A^\wedge $. However, under some conditions the number of branches is the same, for example if the dimension of $A$ is $1$.

Lemma 15.105.1. Let $(A, \mathfrak m)$ be a Noetherian local ring.

The map $A^ h \to A^\wedge $ defines a surjective map from minimal primes of $A^\wedge $ to minimal primes of $A^ h$.

The number of branches of $A$ is at most the number of branches of $A^\wedge $.

The number of geometric branches of $A$ is at most the number of geometric branches of $A^\wedge $.

**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 15.104.6, 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).
$\square$

Lemma 15.105.2. Let $(A, \mathfrak m)$ be a Noetherian local ring. The number of branches of $A$ is the same as the number of branches of $A^\wedge $ if and only if $\sqrt{\mathfrak qA^\wedge }$ is prime for every minimal prime $\mathfrak q \subset A^ h$ of the henselization.

**Proof.**
Follows from Lemma 15.105.1 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).
$\square$

A simple glueing lemma.

Lemma 15.105.3. Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $A \to C$ be a ring map such that for all $f \in I$ the ring map $A_ f \to C_ f$ is localization at an idempotent. Then there exists a surjection $A \to C'$ such that $A_ f \to (C \times C')_ f$ is an isomorphism for all $f \in I$.

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

\[ C_{f_ i} = (A_{f_ i})_{e_ i} \]

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

\[ (f_ if_ ja_ j)^ N(f_ j^ n a_ i - f_ i^ na_ j) = 0 \]

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

\[ f_ i^{N + n} f_ j^{N + nN} a_ j = f_ i^ N f_ j^{N + n} a_ ia_ j^ N \]

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

\[ C'_{f_ i} = A_{f_ i}/(a_ i) = A_{f_ i}/(e_ i) \]

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 $f_1, \ldots , f_ r$.
$\square$

Lemma 15.105.4 can be used to construct finite type extensions from given finite type extensions of the formal completion. We will generalize this lemma in Algebraization of Formal Spaces, Lemma 86.10.3.

Lemma 15.105.4. Let $A$ be a Noetherian ring and $I$ an ideal. Let $B$ be a finite type $A$-algebra. Let $B^\wedge \to C$ be a surjective ring map with kernel $J$ where $B^\wedge $ is the $I$-adic completion. If $J/J^2$ is annihilated by $I^ c$ for some $c \geq 0$, then $C$ is isomorphic to the completion of a finite type $A$-algebra.

**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

\[ C_ f = ((B^\wedge )_ f)_ e \]

for some idempotent $e \in (B^\wedge )_ f$. By Lemma 15.105.3 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

\[ (C \times C', B_{f_ i}, \alpha _ i, \alpha _{ij}) \]

of the category $\text{Glue}(B \to B^\wedge , f_1, \ldots , f_ r)$ introduced in Remark 15.88.10. 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 15.88.15 and Remark 15.88.19. 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 $C'$.
$\square$

Lemma 15.105.5. Let $(A, \mathfrak m)$ be a Noetherian local ring with henselization $A^ h$. Let $\mathfrak q \subset A^\wedge $ be a minimal prime with $\dim (A^\wedge /\mathfrak q) = 1$. Then there exists a minimal prime $\mathfrak q^ h$ of $A^ h$ such that $\mathfrak q = \sqrt{\mathfrak q^ hA^\wedge }$.

**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 15.105.4 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.
$\square$

Lemma 15.105.6. Let $(A, \mathfrak m)$ be a Noetherian local ring. The punctured spectrum of $A^\wedge $ is disconnected if and only if the punctured spectrum of $A^ h$ is disconnected.

**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

\[ \mathop{\mathrm{Spec}}(A^\wedge ) \setminus \{ \mathfrak m^\wedge \} = Z \amalg Z' \]

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 15.105.4 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 15.105.5; 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 $A$.
$\square$

Lemma 15.105.7. Let $(A, \mathfrak m)$ be a Noetherian local ring of dimension $1$. Then the number of (geometric) branches of $A$ and $A^\wedge $ is the same.

**Proof.**
To see this for the number of branches, combine Lemmas 15.105.1, 15.105.2, and 15.105.5 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.
$\square$

reference
Lemma 15.105.8. Let $(A, \mathfrak m)$ be a Noetherian local ring. If the formal fibres of $A$ are geometrically normal (for example if $A$ is excellent or quasi-excellent), then $A$ is Nagata and the number of (geometric) branches of $A$ and $A^\wedge $ is the same.

**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 10.155.10). 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 15.104.2). Also, recall that

\[ (A')^\wedge = A' \otimes _ A A^\wedge = \prod \nolimits _{\mathfrak m' \subset A'} (A'_{\mathfrak m'})^\wedge \]

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

\[ A^\wedge \subset A' \otimes _ A A^\wedge \subset K \otimes _ A A^\wedge \]

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.
$\square$

## Comments (2)

Comment #5045 by anonymous on

Comment #5267 by Johan on