Lemma 33.40.1. Let $X$ be a scheme. Assume every quasi-compact open of $X$ has finitely many irreducible components. Let $\nu : X^\nu \to X$ be the normalization of $X$. Let $x \in X$.
The number of branches of $X$ at $x$ is the number of inverse images of $x$ in $X^\nu $.
The number of geometric branches of $X$ at $x$ is $\sum _{\nu (x^\nu ) = x} [\kappa (x^\nu ) : \kappa (x)]_ s$.
Proof. First note that the assumption on $X$ exactly means that the normalization is defined, see Morphisms, Definition 29.54.1. Then the stalk $A' = (\nu _*\mathcal{O}_{X^\nu })_ x$ is the integral closure of $A = \mathcal{O}_{X, x}$ in the total ring of fractions of $A_{red}$, see Morphisms, Lemma 29.54.4. Since $\nu $ is an integral morphism, we see that the points of $X^\nu $ lying over $x$ correspond to the primes of $A'$ lying over the maximal ideal $\mathfrak m$ of $A$. As $A \to A'$ is integral, this is the same thing as the maximal ideals of $A'$ (Algebra, Lemmas 10.36.20 and 10.36.22). Thus the lemma now follows from its algebraic counterpart: More on Algebra, Lemma 15.106.7. $\square$
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