# Complete intersections

Let us say that a Noetherian local ring is a strong complete intersection if it is of the form S/(f_1, …, f_r) where S is a regular local ring and f_1, …, f_r is a regular sequence. It turns out that if R = S/I = S’/I’ where S, S’ are regular local rings, then I is generated by a regular sequence if and only if I’ is (this is not a triviality!). But, as there exist Noetherian local rings which are not the quotient of any regular local rings, this definition of a strong complete intersection does not make a whole lot of sense. Note that it is clear that if R is a strong complete intersection, then so is R_p for any prime ideal p of R.

The Cohen structure theorem tells us we can write the completion of any Noetherian local ring as the quotient of a regular local ring. Thus we say a Noetherian local ring is a complete intersection if its completion is a strong complete intersection. By the Cohen structure theorem we can write the completion as a quotient of a regular local ring, so this definition makes sense.

The problem: Why is the localization of a complete intersection at a prime a complete intersection?

The solution to this conundrum comes from a theorem of Avramov: If R —> R’ is a flat local homomorphism of Noetherian local rings, and if R’ is a complete intersection, then R is a complete intersection. How do we use this? Suppose that R is a complete intersection and p a prime ideal of R. Let R’ be the completion of R. As R —> R’ is faithfully flat, we can find a prime p’ of R’ lying over p. By assumption R’ is a strong complete intersection, hence R’_{p’} is a strong complete intersection. Since the local ring map R_p —> R’_{p’} is flat, Avramov’s theorem kicks in and we see that R_p is a complete intersection!

Cool, no?

## 6 thoughts on “Complete intersections”

1. student on said:

Thanks for the post, a small question: Why R’ being complete intersection implies the same for R?

• Johan on said:

That is a hard theorem by Avramov which is discussed in later posts.

• student on said:

I am sorry, I asked nonsense, I wanted to ask how does R’ being CI implies CI of R’_{p’}:

“By assumption R’ is a strong complete intersection, hence R’_{p’} is a strong complete intersection”

Thanks!

• student on said:

You say that that fact is clear at the beggining but anyway/ To add some other question (all of them quite dummies!): doesn’t localization commute with completion? I suppose it does not since that would make the question trivial without using Avramov’s theorem

• Johan on said:

Completion does not commute with localization. Suppose R is a local ring, R’ is its completion, p is a prime in R and p’ is a prime of R’ over p. The problem is to relate properties of R_p to properties of R’_p’. Avramov’s theorem does this for “complete intersection”. But for other properties (e.g., being reduced) this fails. For a general Noetherian local ring R there isn’t much you can say about the local ring map R_p —> R’_p’ beyond the fact that it is flat and local. Hope this helps!