A local ring not a quotient of a regular local ring

In his book Commutative Ring Theory, Matsumura proves that if a local ring is equidimensional, and a quotient of a regular local ring, then its completion is equidimensional.

What is an example of a local ring which does not admit such a presentation?

-

1 Answer

A source available online is this paper "Examples of bad Noetherian rings" by Mariani (example 2.1).

The reason many of these types of construction work is because of the following vague and counter-intuitive phenomemon:

It is usually easier than we think for a complete local ring to be a completion of a Noetherian ring with certain properties.

For example, there is this amazing theorem by Heitmann that most complete local ring of depth at least $2$ is a completion of a UFD !

So back to Mariani's paper, the example is as follows: start with some local Artinian ring $(Q,m)$ such that $Q$ is not Gorenstein. Then $Q[[X]]$ is complete, and one can find a local domain $R$ such that $\hat R=Q[[X]]$. Now if $R$ is a quotient of a regular local ring, then the comletion of $R$ is generically a complete intersection. But $Q[[X]]$ is not even generically Gorenstein, since $Q$ is not.

-
The same general idea was used by Ferrand and Raynaud in '70: numdam.org/item?id=ASENS_1970_4_3_3_295_0 – Graham Leuschke Apr 14 '10 at 10:46
To fill in one detail in Hailong's answer, the fact that one can find a local domain $R$ with completion $Q[[x]]$ follows from a theorem of Lech (springerlink.com/content/y435pxn32m8q0703): a complete local ring is the completion of a local domain if and only if it has depth at least one and no integer is a zerodivisor. These are the obvious necessary conditions; that they are sufficient is (to me, still) very surprising. – Graham Leuschke Apr 14 '10 at 12:58
That is an amazing paper title, by the way. – Ben Webster Apr 14 '10 at 14:34
@Graham: thanks a lot for the details. – Hailong Dao Apr 14 '10 at 14:54