Lemma 28.15.5. Let $X$ be a scheme and $x \in X$. Let $X_ i$, $i \in I$ be the irreducible components of $X$ passing through $x$. Then the number of (geometric) branches of $X$ at $x$ is the sum over $i \in I$ of the number of (geometric) branches of $X_ i$ at $x$.

**Proof.**
We view the $X_ i$ as integral closed subschemes of $X$, see Schemes, Definition 26.12.5 and Lemma 28.3.4. Observe that the number of (geometric) branches of $X_ i$ at $x$ is at least $1$ for all $i$ (essentially by definition). Recall that the $X_ i$ correspond $1$-to-$1$ with the minimal prime ideals $\mathfrak p_ i \subset \mathcal{O}_{X, x}$, see Algebra, Lemma 10.26.3. Thus, if $I$ is infinite, then $\mathcal{O}_{X, x}$ has infinitely many minimal primes, whence both $\mathcal{O}_{X, x}^ h$ and $\mathcal{O}_{X, x}^{sh}$ have infinitely many minimal primes (combine Algebra, Lemmas 10.30.5 and 10.30.7 and the injectivity of the maps $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^ h \to \mathcal{O}_{X, x}^{sh}$). In this case the number of (geometric) branches of $X$ at $x$ is defined to be $\infty $ which is also true for the sum. Thus we may assume $I$ is finite. Let $A'$ be the integral closure of $\mathcal{O}_{X, x}$ in the total ring of fractions $Q$ of $(\mathcal{O}_{X, x})_{red}$. Let $A'_ i$ be the integral closure of $\mathcal{O}_{X, x}/\mathfrak p_ i$ in the total ring of fractions $Q_ i$ of $\mathcal{O}_{X, x}/\mathfrak p_ i$. By Algebra, Lemma 10.25.4 we have $Q = \prod _{i \in I} Q_ i$. Thus $A' = \prod A'_ i$. Then the equality of the lemma follows from More on Algebra, Lemma 15.106.7 which expresses the number of (geometric) branches in terms of the maximal ideals of $A'$.
$\square$

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