Suppose $(R,\mathfrak m, k)$ is a $d$-dimensional Cohen-Macaulay local ring with canonical module $\omega_R$ and $d>1$. Suppose $I\subset R$ is an ideal which is MCM (=maximal Cohen-Macaulay, i.e., its depth as a module over $R$ is $d$). My main question is:

Under which assumptions on $R$ can we conclude that every MCM ideal $I$ is principal?

Of course, if $d=1$ there are too many MCMs, so we must take $d>1$. Clearly, this is true for regular local rings (as any MCM is then free). However, if $R$ is a domain (or more generally, generically Gorenstein), then $\omega_R$ is isomorphic to a MCM ideal of $R$, and this ideal will be principal precisely when $R$ is Gorenstein. So we must at least impose that $R$ is Gorenstein. It is not too hard to see that any MCM ideal is principal if $R$ is a unique factorization domain (for $I$ must be unmixed of height one by the depth lemma, whence principal). Using Graham's observation below, we see that in a two-dimensional local Gorenstein ring, every MCM ideal is principal if and only if the ring is a unique factorization domain. Is this true in higher dimensions?

Sometimes we can prove that a certain ideal is MCM: for instance, if $R$ is two-dimensional and $\bar R$ is its integral closure, then the conductor ideal $I=\text{Hom}(\bar R,R)$ is MCM by the depth lemma.

More generally, one could ask when are there only finitely many different isomorphism types of MCM ideals. Do we have Brauer-Thrall-like behavior?

nonon-free MCM ideals as soon as $c > 3$. There's a conjectural lower bound on the rank of MCMs over hypersurfaces due to Buchweitz-Greuel-Schreyer: $r \geq 2^{\mathrm{dim}R -2}$, which would rule out non-free MCM ideals in dim 3 as well. $\endgroup$ – Graham Leuschke Jul 16 '13 at 17:562more comments