We have defined and used the associated graded ring . Now we want to see how it behaves under completions.
By the last post, we have , so we immediately get that .
A great theorem that I’ll skip proving is that if is Noetherian, and is any ideal, then the completion with respect to the -adic topology is Noetherian. As a corollary we get that for any Noetherian ring , is Noetherian by noting that the completion of the Noetherian ring with respect to the -adic topology is .
After this brief excursion, we’ll come back to the dimension theory we left off from. The next natural place to go is to regular local rings. A local ring is regular if . (Recall that it is always true that ).
Suppose we have a Noetherian local ring such that . Then the following are equivalent definitions of regular: or can be generated by elements.
The first condition implies that , so it implies regular. Regular implies that can be generated by elements, by this post. Lastly, if can be generated by (if you’ve seen the term, this is a system of parameters), then we have a surjective map of graded rings with kernel . So it is an iso. This finishes up the equivalences.
Last time we saw without proof that (for Noetherian local rings) is regular if and only if is regular. Now we can prove it.
By the equivalent definition of regular, is regular iff , but we proved that , so but this happens iff is regular.
We’ll wrap up today with trying to keeping the geometric picture in mind. Regular means non-singular geometrically. So we see that passing to the completion doesn’t introduce any singularities. But since the dimension of the local ring at a point equals the dimension of the variety we actually get that completion of where is non-singular is where is the dimension of the variety.
So if we interpret completion of as the “analytically local” picture, then we see that locally all non-singular points on a variety are analytically isomorphic.