---
tag: W27Z60Q1PF
kind: statement
role: theorem
name: Serre's criterion for normality
msc: [13B22, 13H10, 13C15]
uses: [0670T59HGE, KR7SWBYHRQ, 63XXTF6B5M, M2RADBFW99, X6JD4AG973, 87WH65FJB8, MVGBS3J1P7, 00MHHZ9SNE, H20P600HQS, N1QJQEGBFC]
about: [KR7SWBYHRQ]
sources:
  - url: https://en.wikipedia.org/wiki/Serre%27s_criterion_for_normality
same_as:
  wikidata: Q25098875
  wikipedia: Serre's criterion for normality
---

## Statement

Let $A$ be a [[Noetherian ring|0670T59HGE]].
For a [[prime ideal|00MHHZ9SNE]] $\mathfrak{p}$, write $A_{\mathfrak{p}}$ for the [[localization|X6JD4AG973]] at $\mathfrak{p}$, $\operatorname{ht} \mathfrak{p}$ for its [[height|87WH65FJB8]], and $\operatorname{depth} A_{\mathfrak{p}}$ for the [[depth|M2RADBFW99]] of $A_{\mathfrak{p}}$.
Consider the conditions:

- $(R_1)$: $A_{\mathfrak{p}}$ is a [[regular local ring|63XXTF6B5M]] for every prime $\mathfrak{p}$ with $\operatorname{ht} \mathfrak{p} \le 1$.
- $(S_2)$: $\operatorname{depth} A_{\mathfrak{p}} \ge \min(2, \dim A_{\mathfrak{p}})$ for every prime $\mathfrak{p}$.

Call $A$ a [[normal ring|KR7SWBYHRQ]] if $A_{\mathfrak{p}}$ is an [[integrally closed domain|H20P600HQS]] for every prime $\mathfrak{p}$.
Then $A$ is normal if and only if $A$ satisfies $(R_1)$ and $(S_2)$.

## Remarks

The proof works locally.
A normal local ring of dimension $1$ is a discrete valuation ring, hence regular, which gives $(R_1)$.
For $(S_2)$, let $\dim A_{\mathfrak{p}} \ge 2$ and let $x \in \mathfrak{p}A_{\mathfrak{p}}$ be a nonzerodivisor.
Suppose the maximal ideal were associated to $A_{\mathfrak{p}}/(x)$.
Then some quotient $y/x$ would be integral but not in the ring, or it would make the maximal ideal principal and the dimension $1$.
Conversely, $(S_2)$ says that $A$ is the intersection of its localizations at height one primes, and $(R_1)$ says those localizations are discrete valuation rings, hence integrally closed.
The companion statement says that $A$ is reduced if and only if it satisfies $(R_0)$ and $(S_1)$.

Example: $A = k[x, y, z]/(z^2 - xy)$ over a field $k$ of characteristic not $2$ is normal.
It is a hypersurface, hence [[Cohen–Macaulay|N1QJQEGBFC]], so $(S_2)$ holds; its only singular point is the origin, whose prime has height $2$, so $(R_1)$ holds.

Both conditions are needed.
The cusp $k[x, y]/(y^2 - x^3)$ is Cohen–Macaulay but fails $(R_1)$ at the origin, a prime of height $1$; indeed $t = y/x$ is integral, since $t^2 = x$, but not in the ring.
Two planes meeting in one point, $k[x, y, u, v]/((x, y) \cap (u, v))$, satisfy $(R_1)$.
But they have depth $1$ at the origin, where the dimension is $2$, so $(S_2)$ fails.
The ring is not normal, because its local ring at the origin is not a domain.

## Explanations

### L1 human

A normal ring is one with no "missing fractions", which in geometry means mild singularities.
Serre's criterion splits normality into two simpler checks.
The first check says the space has no singularities in codimension one, so it is smooth away from a small set.
The second says that the ring has enough depth, so functions defined away from that small set extend over it.

### L2 human

In practice, $(S_2)$ is often automatic: it holds for every [[Cohen–Macaulay ring|N1QJQEGBFC]], in particular for complete intersections.
Then normality reduces to $(R_1)$, which one checks with the Jacobian criterion by computing the codimension of the singular locus.
This is the standard way to prove that a hypersurface with isolated singularities in dimension at least $2$ is normal.
The criterion also explains the algebraic Hartogs property of normal rings: a normal Noetherian domain is the intersection of its localizations at height one primes.