# Solution: Kaplansky’s Commutative Rings 4.1.01

If $R$ is a commutative ring and $M$ an $R$-module, a regular sequence on $M$ is a sequence $x_1,\dots,x_n\in R$ such that $(x_1,\dots,x_n)M \not=M$ and for each $i$, the element $x_{i+1}$ is not a zero divisor on the module $M/(x_1,\dots,x_{i})$. The term regular sequence in $R$ just refers to a regular sequence on $R$ as an $R$-module over itself. The length of any regular sequence is the number of elements in the sequence.

The projective dimension of an $R$-module $M$ is the infimum over all lengths of projective resolutions of $M$, and hence is either a nonnegative integer or infinity. We write ${\rm pd}_R(M)$ for the projective dimension of an $R$-module $M$. (Clearly, this definition also makes sense for noncommutative rings.)

In particular, any ideal of $R$ is an $R$-module by definition, so we can look at the projective dimension of ideals. For instance, if $x\in R$ is a nonzerodivisor (=an element that is not a zero divisor), then the ideal $Rx$ is a left $R$-module that is free because $x$ is not a zero divisor. The following is a generalisation of this remark:

Problem. If $I$ is an ideal in a commutative ring generated by a regular sequence of length $n > 0$ then ${\rm pd}_R(I) = n-1$.

# A Case of No Positive Finite Projective Dimension

A commutative Noetherian local ring $R$ with maximal ideal $M$ is called a regular local ring if the Krull dimension of $R$ is the same as the dimension of $M/M^2$ as a $R/M$-vector space.

In studying regular local rings one often uses the following lemma in inductive arguments: if $R$ is an arbitrary commutative Noetherian local ring with maximal ideal $M$, and $M$ consists entirely of zero divisors, then the projective dimension $\mathrm{pd}_R(A)$ is either zero or infinity for any finitely-generated $R$-module $A$. In other words, the only $R$-modules of finite projective dimension are the projective (hence free) modules.

This is a neat little result that has a fun More »