Lemma 15.106.7. Let $(A, \mathfrak m, \kappa )$ be a local ring.

1. If $A$ has infinitely many minimal prime ideals, then the number of (geometric) branches of $A$ is $\infty$.

2. The number of branches of $A$ is $1$ if and only if $A$ is unibranch.

3. The number of geometric branches of $A$ is $1$ if and only if $A$ is geometrically unibranch.

Assume $A$ has finitely many minimal primes and let $A'$ be the integral closure of $A$ in the total ring of fractions of $A_{red}$. Then

1. the number of branches of $A$ is the number of maximal ideals $\mathfrak m'$ of $A'$,

2. to get the number of geometric branches of $A$ we have to count each maximal ideal $\mathfrak m'$ of $A'$ with multiplicity given by the separable degree of $\kappa (\mathfrak m')/\kappa$.

Proof. This lemma follows immediately from the definitions, Lemma 15.106.2, Lemma 15.106.4, and Fields, Lemma 9.14.8. $\square$

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