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

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

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

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

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

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 $.

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