The Stacks project

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 $A_{red}$. $\square$

Comments (3)

Comment #8693 by on

There is a fourth way to write , which is a consequence of the integrality of for open affine. Namely, if is an open affine neighborhood of , is the prime in associated with and is the integral closure of in the total ring of fractions of , then

(4) , where .

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 some domain and , , a multiplicatively closed subset, ,

one gets: if is irreducible (hence and are integral), then where are subrings of the function field of and thus we can intersect them. (Intuitively, identifies with the rational functions on defined on all points of the fibre .)

Comment #8694 by on

In the general case, if is any scheme locally with finitely many irreducible components, then , where are the irreducible components of , Tag 29.54.6, so that , where . Thus, where we have used that “taking stalks” is exact, and the fact that the product is locally finite. Using the facts (i) equals the composite (second proof idea of Tag 29.54.6), (ii) for a closed immersion of topological spaces and a sheaf over , the natural map is an isomorphism (Tag 6.32.1) and (iii) the last equation in #8693, we get where the -th intersection is taken inside the function field of . (Slogan: the stalk is rational functions on defined at all points of .)

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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  • 2 comment(s) on Section 29.54: Normalization

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