# Serre's criterion for normality

In algebra, Serre's criterion for normality, introduced by Jean-Pierre Serre, gives necessary and sufficient conditions for a commutative Noetherian ring A to be a normal ring. The criterion involves the following two conditions for A:
* $R_{k}:A_{\mathfrak {p}}$ is a regular local ring for any prime ideal ${\mathfrak {p}}$ of height ≤ k.
* $S_{k}:\operatorname {depth} A_{\mathfrak {p}}\geq \inf\{k,\operatorname {ht} ({\mathfrak {p}})\}$ for any prime ideal ${\mathfrak {p}}$.
The statement is:
* A is a reduced ring $\Leftrightarrow R_{0},S_{1}$ hold.
* A is a normal ring $\Leftrightarrow R_{1},S_{2}$ hold.
* A is a Cohen–Macaulay ring $\Leftrightarrow S_{k}$ hold for all k.
Items 1, 3 trivially follow from the definitions. Item 2 is much deeper.
For an integral domain, the criterion is due to Krull. The general case is due to Serre.

## Proof
### Sufficiency

(After EGA IV$_2$. Theorem 5.8.6.)
Suppose A satisfies S$_2$ and R$_1$. Then A in particular satisfies S$_1$ and R$_0$; hence, it is reduced. If ${\mathfrak {p}}_{i},\,1\leq i\leq r$ are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields $\kappa ({\mathfrak {p}}_{i})=Q(A/{\mathfrak {p}}_{i})$: see total ring of fractions of a reduced ring. That means we can write $1=e_{1}+\dots +e_{r}$ where $e_{i}$ are idempotents in $\kappa ({\mathfrak {p}}_{i})$ and such that $e_{i}e_{j}=0,\,i\neq j$. Now, if A is integrally closed in K, then each $e_{i}$ is integral over A and so is in A; consequently, A is a direct product of integrally closed domains Ae$_i$'s and we are done. Thus, it is enough to show that A is integrally closed in K.
For this end, suppose
    $(f/g)^{n}+a_{1}(f/g)^{n-1}+\dots +a_{n}=0$
where all f, g, a$_i$'s are in A and g is moreover a non-zerodivisor. We want to show:
    $f\in gA$.
Now, the condition S$_2$ says that $gA$ is unmixed of height one; i.e., each associated primes ${\mathfrak {p}}$ of $A/gA$ has height one. This is because if ${\mathfrak {p}}$ has height greater than one, then ${\mathfrak {p}}$ would contain a non zero divisor in $A/gA$. However, ${\mathfrak {p}}$ is associated to the zero ideal in $A/gA$ so it can only contain zero divisors, see here. By the condition R$_1$, the localization $A_{\mathfrak {p}}$ is integrally closed and so $\phi (f)\in \phi (g)A_{\mathfrak {p}}$, where $\phi :A\to A_{\mathfrak {p}}$ is the localization map, since the integral equation persists after localization. If $gA=\cap _{i}{\mathfrak {q}}_{i}$ is the primary decomposition, then, for any i, the radical of ${\mathfrak {q}}_{i}$ is an associated prime ${\mathfrak {p}}$ of $A/gA$ and so $f\in \phi ^{-1}({\mathfrak {q}}_{i}A_{\mathfrak {p}})={\mathfrak {q}}_{i}$; the equality here is because ${\mathfrak {q}}_{i}$ is a ${\mathfrak {p}}$-primary ideal. Hence, the assertion holds.

### Necessity

Suppose A is a normal ring. For S$_2$, let ${\mathfrak {p}}$ be an associated prime of $A/fA$ for a non-zerodivisor f; we need to show it has height one. Replacing A by a localization, we can assume A is a local ring with maximal ideal ${\mathfrak {p}}$. By definition, there is an element g in A such that ${\mathfrak {p}}=\{x\in A|xg\equiv 0{\text{ mod }}fA\}$ and $g\not \in fA$. Put y = g/f in the total ring of fractions. If $y{\mathfrak {p}}\subset {\mathfrak {p}}$, then ${\mathfrak {p}}$ is a faithful $A[y]$-module and is a finitely generated A-module; consequently, $y$ is integral over A and thus in A, a contradiction. Hence, $y{\mathfrak {p}}=A$ or ${\mathfrak {p}}=f/gA$, which implies ${\mathfrak {p}}$ has height one (Krull's principal ideal theorem).
For R$_1$, we argue in the same way: let ${\mathfrak {p}}$ be a prime ideal of height one. Localizing at ${\mathfrak {p}}$ we assume ${\mathfrak {p}}$ is a maximal ideal and the similar argument as above shows that ${\mathfrak {p}}$ is in fact principal. Thus, A is a regular local ring. $\square$