Next: Test ideals
Up: Integral closure of rings
Previous: Integral closure of rings
Let
be a ring extension and
an ideal. Recall that is called (strongly)
integral over if satisfies a relation
with
The set
is integral over is called the integral closure of in , it is a
module, where , the integral closure of in ,
is a ring.
We are interested in the two most interesting cases:
Let us consider the normalization first.
Lemma 1.1 (Keylemma)
Let
be a reduced Noetherian ring and
an
ideal containing a nonzerodivisor
of
. Then there are
natural inclusions of rings
and inclusions
of
modules.
Here
.
Proof.
If
, then
is
independent of the nonzerodivisor and, hence,
defines an embedding
The inclusion
is given by the multiplication with elements of . To see that the
image is contained in
, consider
. The characteristic polynomial of
defines (by CayleyHamilton) an integral relation for
. To see the last inclusion, consider
such that
,
and let
, , be an
integral relation for . For a given multiply the
relation for with . This shows that
, hence
.
Since is normal if and only if the localization is normal
for all
, we define the nonnormal locus of
as
is not normal
It is easy to see that
where
is the conductor of in
, in particular, is closed in
. However, since we cannot yet compute
, we
cannot compute either.
The following proposition is the basis for the algorithm to compute
the normalization as well as for an algorithm to compute an ideal with
zero set . It is basically due to Grauert and Remmert,
[12].
Proposition 1.2 (Criterion for normality)
Let
be a reduced Noetherian ring and
an ideal
satisfying
 1.
 contains a nonzerodivisor of ,
 2.

,
 3.

.
Then
if and only if
.
An ideal
satisfying 1.3. is called a test
ideal for the normalization.
Proof.
If
then
, by Lemma
1.1. For the converse, notice that 3. implies
, hence there exists a minimal
such that , that is,
. If , choose
and
such that
. Since
and
we
have
and hence, using
2.,
, by Lemma 1.1. By
assumption
and, hence, .
This is a contradiction, and we conclude and
.
Remark 1.3
As the proof shows, condition 2. can be weakened to
 2'.

.
However, this
cannot be used in practice, since we do not know
,
while, on the other hand, we can always
pass from
to
without violating the conditions 1 and 3,
and
is computable.
Proof.
This follows from 1.2, 1.1 and the fact that the
operations which define the annihilator are compatible with
localization.
Having a test ideal and a nonzerodivisor of
, we can compute module generators of
and module generators for
,
since we can compute ideal quotients using Gröbner
basis methods, cf. [15].
Let us describe the ring structure of
.
For this, let ,
be generators of
as module, and let
be generators of the module of
syzygies of
. Since
is a ring,
we have
,
for certain
, the quadratic relations between the
. Define
as the ideal
generated by the linear and quadratic relations,
We get an isomorphism
of algebras, by sending to .
This presentation is needed to continue the normalization algorithm.
To compute
, we only need the module
structure of .
Next: Test ideals
Up: Integral closure of rings
Previous: Integral closure of rings
Christoph Lossen
20010321