3
$\begingroup$

Let $(R,\frak m)$ be a Cohen-Macaulay local ring. If there exists a maximal Cohen-Macaulay $R$-module $M$ with finite injective dimension, then can we conclude that $R$ has a canonical module?

$\endgroup$
3
$\begingroup$

In short, the answer to your question is no.

Let us recall the following definition due to R. Sharp:

Definition Let $(R,\mathfrak{m},k)$ be a commutative noetherian local ring with $\text{dim}\,R=d$. A finitely generated $R$ module $G$ is said to be Gorenstein of rank (or type) $t$ if $$\mu_{i}(\mathfrak{m},G)= \begin{cases} 0 \mbox{ if }i \neq d \\ t \mbox{ if }i = d \end{cases}$$

Clearly any Gorenstein module is a maximal Cohen-Macaulay $R$-module, as its depth is $d$, and has finite injective dimension (this is clear from Proposition 3.1.14 in Bruns-Herzog). From its definition, the canonical module is just a Gorenstein module of rank 1.

Sharp showed that any local ring admitting a Gorenstein module is a Cohen-Macaulay ring (this is Corollary 3.9 in the paper Gorenstein Modules).

Consequently any local ring admitting a Gorenstein module of rank at least 2 provides a counterexample to your question. I believe the first example of such a ring was described by D. Weston in the paper On Descent in Dimension Two and Non-split Gorenstein Modules, and then examples of Gorenstein modules of arbitrary odd rank were given here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.