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
p
is a regular local ring for any prime ideal
p
of height ≤ k.
S
k
:
depth
A
p
≥
inf
{
k
,
ht
(
p
)
}
for any prime ideal
p
.
The statement is:
A is a reduced ring
⇔
R
0
,
S
1
hold.
A is a normal ring
⇔
R
1
,
S
2
hold.
A is a Cohen–Macaulay ring
⇔
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.
(After EGA IV. Theorem 5.8.6.)
Suppose A satisfies S2 and R1. Then A in particular satisfies S1 and R0; hence, it is reduced. If
p
i
,
1
≤
i
≤
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
κ
(
p
i
)
=
Q
(
A
/
p
i
)
: see total ring of fractions of a reduced ring. That means we can write
1
=
e
1
+
⋯
+
e
r
where
e
i
are idempotents in
κ
(
p
i
)
and such that
e
i
e
j
=
0
,
i
≠
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 Aei'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
+
⋯
+
a
n
=
0
where all f, g, ai's are in A and g is moreover a non-zerodivisor. We want to show:
f
∈
g
A
.
Now, the condition S2 says that
g
A
is unmixed of height one; i.e., each associated primes
p
of
A
/
g
A
has height one. By the condition R1, the localization
A
p
is integrally closed and so
ϕ
(
f
)
∈
ϕ
(
g
)
A
p
, where
ϕ
:
A
→
A
p
is the localization map, since the integral equation persists after localization. If
g
A
=
∩
i
q
i
is the primary decomposition, then, for any i, the radical of
q
i
is an associated prime
p
of
A
/
g
A
and so
f
∈
ϕ
−
1
(
q
i
A
p
)
=
q
i
; the equality here is because
q
i
is a
p
-primary ideal. Hence, the assertion holds.
Suppose A is a normal ring. For S2, let
p
be an associated prime of
A
/
f
A
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
p
. By definition, there is an element g in A such that
p
=
{
x
∈
A
|
x
g
≡
0
mod
f
A
}
and
g
∉
f
A
. Put y = g/f in the total ring of fractions. If
y
p
⊂
p
, then
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
p
=
A
or
p
=
f
/
g
A
, which implies
p
has height one (Krull's principal ideal theorem).
For R1, we argue in the same way: let
p
be a prime ideal of height one. Localizing at
p
we assume
p
is a maximal ideal and the similar argument as above shows that
p
is in fact principal. Thus, A is a regular local ring.
◻
