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: The statement is: Items 1, 3 trivially follow from the definitions.
For an integral domain, the criterion is due to Krull.
The general case is due to Serre.
Suppose A satisfies S2 and R1.
Then A in particular satisfies S1 and R0; hence, it is reduced.
are the minimal prime ideals of A, then the total ring of fractions K of A is the direct product of the residue fields
: see total ring of fractions of a reduced ring.
That means we can write
Now, if A is integrally closed in K, then each
is integral over A and so is in A; consequently, A is a direct product of integrally closed domains Aei's and we are done.
Thus, it is enough to show that A is integrally closed in K. For this end, suppose where all f, g, ai's are in A and g is moreover a non-zerodivisor.
We want to show: Now, the condition S2 says that
is unmixed of height one; i.e., each associated primes
By the condition R1, the localization
is the localization map, since the integral equation persists after localization.
is the primary decomposition, then, for any i, the radical of
Hence, the assertion holds.
Suppose A is a normal ring.
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
By definition, there is an element g in A such that
Put y = g/f in the total ring of fractions.
-module and is a finitely generated A-module; consequently,
is integral over A and thus in A, a contradiction.
has height one (Krull's principal ideal theorem).
be a prime ideal of height one.
is a maximal ideal and the similar argument as above shows that
is in fact principal.
Thus, A is a regular local ring.