Maybe this title isn’t exactly what the post is about, but today will mostly be a hodgepodge attempt to get some more out there. I’m not sure what else to do with regular local rings (other than systems of parameters which I’m not too excited to post on), so I’ll move on. The next set of theorems in Hartshorne (that are not proved in the text) has to do with Noetherian local domains of dimension 1.

Before this is stated we need quite a bit of terminology. Let be a field and a totally ordered abelian group. A valuation is a map such that and .

We form a subring of by taking the set which is called the valuation ring of . This ring is local with maximal ideal .

Mostly we care about discrete valuation rings (DVR). These are the ones whose value group is the integers. Now we can state and prove the Theorem stated in Hartshorne:

Let be a Noetherian local domain of dimension 1. Then the following are equivalent:

1) is a discrete valuation ring

2) is integrally closed

3) is principal

4)

5) Every non-zero ideal is a power of

6) There exists such that every non-zero ideal is of the form , .

Proof: We’ll just go in the standard cyclic order for proof. If is a DVR, then we consider an integral element . If then we are done. If , then the claim is that . This is simply because . Since , and $v(x)=-v(x^{-1})$, we get that . Thus if we get by multiplying by $x^{1-n}$ that and every integral element is in the ring.

For the next, assume is integrally closed. Let and . Since is the only non-zero prime ideal, . Thus there is some integer such that such that . Now let . Let . Now since , we cannot reduce to a form , so . By integrally closed, we get that is not integral over . If , then would be a finitely generated (as an module) faithful -module, and hence would be integral. Thus . Clearly, , so . Thus is principal.

Now if is principal, we have . But since by the Noetherian condition, we get that .

Suppose . Thus is principal. Let be any non-zero ideal. Since all ideals are -primary we again get that for some n. Since is Artinian we get that for some r. Thus .

Suppose every non-zero ideal is a power of . By Noetherian we have , so we can pick . So there is some such that . Thus or else we’d have a prime chain longer than 1. Now given any ideal, .

Suppose such that every non-zero ideal has the form . Again, we must have . So by Noetherian . Thus if is non-zero, there is a well-defined such that . Naturally we get a discrete valuation and extend in the obvious way to the rest of the field by . By putting everything in reduced form, we see that something in that is not in has negative valuation, and hence is the valuation ring of .

January 11, 2010 at 10:59 am

What does it mean when there is a overline over an ideal? as used in the 3rd last paragraph.

Please help. Thank you.

January 11, 2010 at 3:48 pm

It is the image of the ideal in the quotient ring.