Lemma 29.54.4. Let $X$ be a scheme such that every quasi-compact open has a finite number of irreducible components. Let $\nu : X^\nu \to X$ be the normalization of $X$. Let $x \in X$. Then the following are canonically isomorphic as $\mathcal{O}_{X, x}$-algebras

the stalk $(\nu _*\mathcal{O}_{X^\nu })_ x$,

the integral closure of $\mathcal{O}_{X, x}$ in the total ring of fractions of $(\mathcal{O}_{X, x})_{red}$,

the integral closure of $\mathcal{O}_{X, x}$ in the product of the residue fields of the minimal primes of $\mathcal{O}_{X, x}$ (and there are finitely many of these).

**Proof.**
After replacing $X$ by an affine open neighbourhood of $x$ we may assume that $X$ has finitely many irreducible components and that $x$ is contained in each of them. Then the stalk $(\nu _*\mathcal{O}_{X^\nu })_ x$ is the integral closure of $A = \mathcal{O}_{X, x}$ in the product $L$ of the residue fields of the minimal primes of $A$. This follows from the construction of the normalization and Lemma 29.53.1. Alternatively, you can use Lemma 29.54.3 and the fact that normalization commutes with localization (Algebra, Lemma 10.36.11). Since $A_{red}$ has finitely many minimal primes (because these correspond exactly to the generic points of the irreducible components of $X$ passing through $x$) we see that $L$ is the total ring of fractions of $A_{red}$ (Algebra, Lemma 10.25.4). Thus our ring is also the integral closure of $A$ in the total ring of fractions of $A_{red}$.
$\square$

## Comments (2)

Comment #8693 by Elías Guisado on

Comment #8694 by Elías Guisado on

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