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

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

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

3. 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 $A_{red}$. $\square$

Comment #8693 by on

There is a fourth way to write $(\nu_*\mathcal{O}_{X^\nu})_x$, which is a consequence of the integrality of $\mathcal{O}_X(U)\to\nu_*\mathcal{O}_{X^\nu}(U)$ for $U\subset X$ open affine. Namely, if $U=\operatorname{Spec} A$ is an open affine neighborhood of $x$, $\mathfrak{q}$ is the prime in $A$ associated with $x$ and $B$ is the integral closure of $A$ in the total ring of fractions of $A_{red}$, then

(4) $S^{-1}B$, where $\displaystyle S=\bigcap_{\substack{\mathfrak{p}\subset B\\\text{lying over }\mathfrak{q}}} B-\mathfrak{p}$.

This is Bourbaki, Commutative Algebra, Ch. V, § 2, no. 1, Proposition 2.

As far as I can tell, (4) is independent of (1)-(3) (the latter do not prove (4) nor are proven by it).

In particular, using the fact:

for $B$ some domain and $S_i\subset B$, $i\in I$, a multiplicatively closed subset, $(\bigcap S_i)^{-1}B=\bigcap S_i^{-1}B$,

one gets: if $X$ is irreducible (hence $X_{red}$ and $X^\nu$ are integral), then where $\mathcal{O}_{X^\nu,y}$ are subrings of the function field of $X^\nu$ and thus we can intersect them. (Intuitively, $(\nu_*\mathcal{O}_{X^\nu})_x$ identifies with the rational functions on $X^\nu$ defined on all points of the fibre $\nu^{-1}(x)$.)

Comment #8694 by on

In the general case, if $X$ is any scheme locally with finitely many irreducible components, then $X^\nu=\coprod Z_i^\nu$, where $Z_i\subset X$ are the irreducible components of $X$, Tag 29.54.6, so that $\nu_*\mathcal{O}_{X^\nu}=\prod\nu_{i,*}\mathcal{O}_{Z_i^\nu}$, where $\nu_i:Z_i^\nu\to X^\nu\to X$. Thus, $(\nu_*\mathcal{O}_{X^\nu})_x = \prod(\nu_{i,*}\mathcal{O}_{Z_i^\nu})_x,$ where we have used that “taking stalks” is exact, and the fact that the product $\prod\nu_{i,*}\mathcal{O}_{Z_i^\nu}$ is locally finite. Using the facts (i) $\nu_i$ equals the composite $Z_i^\nu\to Z_i\to X$ (second proof idea of Tag 29.54.6), (ii) for a closed immersion of topological spaces $f:Z\to T$ and a sheaf $\mathcal{F}$ over $Z$, the natural map $(f_*\mathcal{F})_{z}\to\mathcal{F}_z$ is an isomorphism (Tag 6.32.1) and (iii) the last equation in #8693, we get where the $i$-th intersection is taken inside the function field of $X_i^\nu$. (Slogan: the stalk $(\nu_*\mathcal{O}_{X^\nu})_x$ is rational functions on $X^\nu$ defined at all points of $\nu^{-1}(x)$.)

Comment #9377 by on

Sounds good (I didn't check the Bourbaki reference). The intersections are potentially over infinite sets of points, so difficult to work with in general. I am going to leave this alone until we need it for another statement in the Stacks project.

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