Serre's Criterion On 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'':
*$R_k: A_$ is a regular local ring for any prime ideal $\mathfrak$ of height ≤ ''k''.
*$S_k: \operatorname A_ \ge \inf \$ for any prime ideal $\mathfrak$.
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. 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 $\mathfrak_i, \, 1 \le i \le r$ are the minimal prime ideals of ''A'', then the total ring of fractions ''K'' of ''A'' is the direct product of the residue fields $\kappa(\mathfrak_i) = Q(A/\mathfrak_i)$: 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 $\kappa(\mathfrak_i)$ and such that $e_i e_j = 0, \, i \ne j$. 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
:$(f/g)^n + a_1 (f/g)^ + \dots + a_n = 0$
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 $\mathfrak$ of $A/gA$ has height one. This is because if $\mathfrak$ has height greater than one, then $\mathfrak$ would contain a non zero divisor in $A/gA$. However, $\mathfrak$ is associated to the zero ideal in $A/gA$ so it can only contain zero divisors, se
here
By the condition ''R''₁, the localization $A_$ is integrally closed and so $\phi(f) \in \phi(g)A_$, where $\phi: A \to A_$ is the localization map, since the integral equation persists after localization. If $gA = \cap_i \mathfrak_i$ is the primary decomposition, then, for any ''i'', the radical of $\mathfrak_i$ is an associated prime $\mathfrak$ of $A/gA$ and so $f \in \phi^(\mathfrak_i A_) = \mathfrak_i$; the equality here is because $\mathfrak_i$ is a $\mathfrak$- primary ideal. Hence, the assertion holds.

Necessity

Suppose ''A'' is a normal ring. For ''S''₂, let $\mathfrak$ 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 $\mathfrak$. By definition, there is an element ''g'' in ''A'' such that $\mathfrak = \$ and $g \not\in fA$. Put ''y'' = ''g''/''f'' in the total ring of fractions. If $y \mathfrak \subset \mathfrak$, then $\mathfrak$ is a faithful A /math>-module and is a finitely generated ''A''-module; consequently, $y$ is integral over ''A'' and thus in ''A'', a contradiction. Hence, $y \mathfrak = A$ or $\mathfrak = f/g A$, which implies $\mathfrak$ has height one ( Krull's principal ideal theorem).

For ''R''₁, we argue in the same way: let $\mathfrak$ be a prime ideal of height one. Localizing at $\mathfrak$ we assume $\mathfrak$ is a maximal ideal and the similar argument as above shows that $\mathfrak$ is in fact principal. Thus, ''A'' is a regular local ring. $\square$

Notes

References

*{{EGA , book=IV-2
*H. Matsumura, ''Commutative algebra'', 1970.

Theorems in ring theory