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 ${\rm pd}_R(I) = n-1$.

Solution. We proceed by induction on $n$, the case of $n=1$ being the observation we made above that $Rx\cong R$ as $R$-modules for any $x\in R$ a nonzerodivisor.

Consider the image $I/x_1$ of the ideal $I$ in the ring $R/x_1$. There is a short exact sequence $0\to I/x_1\to R/x_1\to R/I\to 0$ of $R/x_1$-modules, and $I/x_1$ in $R/x_1$ is generated by a regular sequence of length $n-1$, so by the induction hypothesis, ${\rm pd}_{R/x_1}(I) = n-2$.

If $n > 2$ then we see that ${\rm pd}_{R/x_1}(R/I) = n -1$. If $n = 2$, then $I/x_1$ is $R/x_1$-projective, so that ${\rm pd}_{R/x_1}(R/I) \leq 1$. But we cannot have $R/I$ being $R/x_1$ projective since $x_2$ is a zero divisor on $R/I$ but not on $R/x_1$. Hence ${\rm pd}_{R/x_1}(R/I) = n-1$, which implies that ${\rm pd}_R(R/I)= n$ since $x_1$ is not a zero divisor in $R$.

Now consider the short exact sequence $0\to I\to R\to R/I\to 0$. If $I$ were projective then we would have ${\rm pd}_R(R/I)\leq 1$, which contradicts ${\rm pd}_R(R/I) = n\geq 2$. Hence $I$ is not projective. Hence $n = {\rm pd}_R(R/I) = {\rm pd}_R(I) + 1$, showing that ${\rm pd}_R(I) = n-1$ as desired. .

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