Lemma 15.108.8. 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.

[Theorem 2.3, Beddani]

**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. $\square$

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