Next: Test ideals Up: Integral closure of rings Previous: Integral closure of rings

## Ring normalization

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:
• reduced, the total quotient ring of , and . In this case and is the normalization of .

• and arbitrary. Then is the integral closure of in .
Let us consider the normalization first.

Lemma 1.1 (Key-lemma)   Let be a reduced Noetherian ring and an ideal containing a non-zerodivisor of . Then there are natural inclusions of rings

and inclusions of -modules. Here .

Proof. If , then is independent of the non-zerodivisor 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 Cayley-Hamilton) 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 non-normal 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 non-zerodivisor 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.

Corollary 1.4   Let be as in Proposition 1.2 and define . Then .

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 non-zerodivisor 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
2001-03-21