When are MCM ideals principal?

Question

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?

Answer 1

On question 1, for what rings all MCM ideals are principal, we can say quite a bit more if one knows that $R$ is parafactorial (that is, the Picard group of the punctured spectrum $Spec^o(R):=Spec(R)-\{m\}$ is trivial). For instance:

If R is a local complete intersection which is locally a UFD in codimension $3$, then any MCM ideal $I$ is free.

Proof: Suppose it is not, then there is a prime $P$ such that $I_P$ is not $R_P$- free. Then the height of $P$ is at least $4$, and $I_P$ is locally free on $Spec^o(R_P)$, so is an element in the Picard group of $Spec^o(R_P)$. But $R_P$ is a complete intersection of dimension at least $4$, so by SGA, it is parafactorial, and $I_P$ is free.

The last question, whether there are finitely many MCM module of rank one, is well-known among certain people, and has resurfaced from time to time. The most ambitious conjecture is perhaps

If $R$ is a normal local domain with an algebraically closed residue field, and the class group $Cl(R)$ finitely generated, then it contains finitely many MCM elements.

For references and some positive results in this direction, see Section 4 of my paper with Kurano and the subsection 6 (on page 3) in this paper by Kollár.

Answer 2

I am not clear whether we need to assume irreducibility, but certainly it works in that case. If $R$ is the power series ring, $0\neq f\in R$ defines a hypersurface (if necessary, assume irreducible) and $f\in I\subset R$ is an MCM ideal, then since the height of $I$ in $R$ is two, we have a minimal resolution $0\to R^{k-1}\stackrel{M}{\to} R^k\to I\to 0$. Further, $I$ is generated by the $(k-1)\times (k-1)$ minors $a_i, 1\leq i\leq k$ of $M$. Since $f\in I$, we can write $f=\sum x_ia_i$. $I$ is principal modulo $f$ if and only if one of the $x_i$'s is a unit and if not it is clear that $f$ is determinantal of a $k\times k$ matrix with all entries in the maximal ideal.