Posted by Jason Polak on 21. September 2016 · Write a comment · Categories: commutative-algebra · Tags:

Suppose $R$ is a Noetherian local ring with unique maximal ideal $m\subset R$. We say that $R$ is regular if the dimension of $R$ is equal to the dimension of $m/m^2$ as an $R/m$-vector space. Regular local rings arise as the local rings of varieties over a field corresponding to smooth points, and this gives an abundant supply of them: for a field $k$, the rings $k[x,y]_{(x,y)}$ and $k[x,y]_{(x-a,y-a^2)}/(y – x^2)$ for instance are regular local rings.

Recall the notion of a regular sequence: if $R$ is a commutative Noetherian ring and $M$ a finitely-generated $R$-module, then a regular sequence in $R$ on $M$ is a sequence of elements $x_1,\dots,x_n\in R$ such that $(x_1,\dots,x_n)M\not= M$ and $x_i$ is not a zero divisor on $M/(x_1,\dots,x_{i-1})$ for all $i$. If we fix an ideal $I\subseteq R$ then we denote by $G(I,M)$ the maximum length of a regular sequence of $R$ on $M$ contained in $I$. This is sometimes called the grade of $I$ on $M$. For a Noetherian local ring $R$, it is convention to denote $G(M,R)$ by $G(R)$, and to call this quantity the grade of $R$. The prototypical example of a regular sequence is $x_1,\dots,x_n$ in the polynomial ring $R[x_1,\dots,x_n]$ on the module $R[x_1,\dots,x_n]$, considered as a module over itself. Regular sequences can be thought of as a generalization of this example, in the sense that if $R$ contains a field $k$ and $x_1,\dots,x_n$ is a regular sequence in $R$ on $R$, then $k[x_1,\dots,x_n]$ actually is a polynomial ring in $n$-variables; this is a theorem of Kaplansky. Here is a classic:

Theorem.The $n$-dimensional regular local rings are precisely those Noetherian local rings whose maximal ideals can be generated by a regular sequence $x_1,\dots,x_n\in m – m^2$ of length $n$.

There is a related notion of regular local ring: the Cohen-Macaulay ring. A Noetherian local ring $R$ with maximal ideal $m$ is called Cohen-Macaulay if $G(R) = \dim(R)$, and of course in this case the dimension $\dim(R)$ is just the rank (a.k.a. codimension) of $m$. Of course, regular local rings are Cohen-Macaulay. However, the converse is false, for otherwise, why name them?

Producing a counterexample is not difficult, and that is the point of this blog post after all. In fact, the machine for producing such counterexamples comes from the following small result:

Theorem. A reduced local ring of dimension one is Cohen-Macaulay.
Proof. Let $R$ be the ring. It is a general fact that $rank(M)\geq G(R)$. Therefore, it suffices to produce a nonunit nonzerodivisor in $R$. If every element of $R$ were a zero divisor, then the maximal ideal $m$ would be the set of zero divisors of $R$, which by the theory of associated primes, would mean that $m$ is the annihilator of a single nonzero $x\in R$. Therefore $x^2 = 0$, and hence $R$ would not be reduced. Therefore, there exists a nonzerodivisor in $m$. QED

Therefore, in order to produce a Cohen-Macaulay ring that is not a regular local ring, we should look at the local rings of varieties of dimension one at the singular points. For example, $k[x,y]/(xy)$ represents the union of the two axes in $k^2$ and thus localizing at $(x,y)$ gives the desired example $A = k[x,y]_{(x,y)}/(xy)$. In this case, $x + y$ is the requisite nonzerodivisor. As the residue field at the point $(0,0)$ is $k$, we see that the ring cannot be regular because this would imply the corresponding affine variety is smooth there. Of course, another way to see that $A$ is not regular is simply by observing that $A$ has zero divisors – a regular local ring cannot have these. In fact, regular local rings are unique factorization domains.

It is a famous result that the Noetherian regular local rings are precisely the Noetherian local rings with finite global dimension, so our example $k[x,y]_{(x,y)}/(xy)$ has infinite global dimension as well!

Leave a Reply

Your email address will not be published. Required fields are marked *