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A normal Noetherian domain is the intersection of its height-one localizations

Statement

Assume AC. Let $A$ be a Noetherian normal domain (normal noetherian ring) with fraction field $K$ , so that $A$ is integrally closed in $K$ . Then, inside $K$ , $A = ⋂_{ht p = 1} A_{p},$ the intersection running over all height-one prime ideals $p$ of $A$ . Equivalently, a rational function which is regular at every height-one point of $Spec A$ is regular.

Facts & Assumptions

Given: AC, a Noetherian normal domain $A$ with fraction field $K$ , and an element $z = b / a \in K$ with $a, b \in A$ , $a \neq 0$ .

[F1]

A Noetherian ring is normal when every prime localization is an integrally closed domain; for a domain this means integrally closed in its fraction field (normal noetherian ring). A normal Noetherian domain satisfies Serre's condition $(S_{2})$ : $depth A_{p} \geq \min (2, ht p)$ for every prime $p$ (normal domain implies s two, assuming AC).

[F2]

A nonzero module over a Noetherian ring has an associated prime (A nonzero module over a Noetherian ring has an associated prime, assuming DC, hence in particular under AC); a prime $p$ is associated to $M$ exactly when $p = Ann (m)$ for some $m \in M$ , equivalently when $A / p$ embeds in $M$ (Associated primes are exactly primes of embedded cyclic residue modules).

[F3]

For a Noetherian local ring $(R, m)$ and a nonzero finite module $M$ , $depth (M) = 0$ if and only if $m \in Ass (M)$ (The local depth-zero associated-prime criterion, assuming AC).

[F4]

If $R$ is Noetherian, $M$ finite, $I \subseteq J (R)$ and $x \in I$ is $M$ -regular, then ${depth}_{I} (M / x M) = {depth}_{I} (M) - 1$ (Depth drops by one after quotienting by a regular element, assuming AC).

[F5]

A height-one prime localization of a Noetherian integrally closed domain is a discrete valuation ring (Height-one localizations of normal Noetherian domains are DVRs).

Proof

technique · direct, by contraposition of the nontrivial inclusion

1.1givenalgebra

The inclusion $A \subseteq ⋂_{ht p = 1} A_{p}$ holds because every $A_{p}$ contains $A$ , compatibly with the common fraction field $K$ ; all rings involved are subrings of $K$ .

2.1F2step 1.1algebra

Suppose $z = b / a \notin A$ . Then $b \notin a A$ , so the class $\overset{ˉ}{b}$ of $b$ in the finite nonzero $A$ -module $A / a A$ generates a nonzero cyclic submodule $C = A \overset{ˉ}{b}$ . By [F2] $C$ has an associated prime $p$ , say $p = Ann (c \overset{ˉ}{b})$ for some $c \in A$ ; then $0 \neq a \in p$ because $a$ annihilates $A / a A$ .

3.1F2F3step 2.1algebra

The localized module $C_{p}$ is nonzero, since $Ann (c \overset{ˉ}{b}) = p$ : if $s \in A ∖ p$ satisfied $s (c \overset{ˉ}{b}) = 0$ , then $s \in p$ , a contradiction. As $C_{p} \subseteq (A / a A)_{p} = A_{p} / a A_{p}$ and the annihilator of the image of $c \overset{ˉ}{b}$ in this localization is $p A_{p}$ , the associated-prime depth criterion [F3] gives ${depth}_{A_{p}} (A_{p} / a A_{p}) = 0$ .

4.1F4step 3.1algebra

Since $A$ is a domain and $a \neq 0$ , the element $a$ is $A_{p}$ -regular and lies in the maximal ideal $p A_{p} \subseteq J (A_{p})$ ; the regular-element depth formula [F4] applied to $M = A_{p}$ gives $depth (A_{p} / a A_{p}) = depth (A_{p}) - 1$ , so $depth A_{p} = 1$ .

5.1F1F5step 2.1step 4.1algebra

By the $(S_{2})$ condition of [F1], $1 = depth A_{p} \geq \min (2, ht p)$ , so $ht p \leq 1$ ; since $0 \neq a \in p$ we also have $ht p \geq 1$ , hence $ht p = 1$ , and $A_{p}$ is a discrete valuation ring by [F5].

6.1F2step 2.1step 5.1algebra∎

Finally $z \notin A_{p}$ . Indeed, for the annihilator $Ann (\overset{ˉ}{b}) \subseteq A$ of $\overset{ˉ}{b} \in A / a A$ one has $Ann (\overset{ˉ}{b}) \subseteq p$ , since $s \overset{ˉ}{b} = 0$ implies $s (c \overset{ˉ}{b}) = c (s \overset{ˉ}{b}) = 0$ and hence $s \in Ann (c \overset{ˉ}{b}) = p$ . If $b / a \in A_{p}$ , write $b = a u$ with $u = r / s$ , $r \in A$ , $s \in A ∖ p$ ; then $s b = a r \in a A$ , so $s \in Ann (\overset{ˉ}{b}) \subseteq p$ , contradicting $s \notin p$ . Thus every $z \in K$ outside $A$ lies outside $A_{p}$ for some height-one prime $p$ , and with step 1.1 the intersection equals $A$ .

The last step also yields the standard Hartogs form: an element of $K$ contained in $A_{p}$ for every height-one prime $p$ lies in $A$ , so on a normal Noetherian scheme a rational function regular in codimension one is regular.

Depends on

Used by

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Tag 031T and Tag 0AVK (Hartogs for normal domains) (standard reference, not scraped)
Bosch, Lutkebohmert, Raynaud, Neron Models (1990), 2.5/1 and scheme Hartogs in owner-arithmetic-models/neron-source/closure-supplement.md (standard reference, not scraped)