# When are MCM ideals principal?

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?

-
For hypersurfaces in say a powerseries rings, a theorem of Eisenbud says that all MCM rank one ideals are principal if and only if the hyersurface is not determinantal-that is it can not be written as the determinant of a matrix of size at least 2 with entries in the maximal ideal. –  Mohan Jul 14 '13 at 16:54
Interesting, and it answers my question for certain types of domains (since you mention rank one, I assume the hypersurface is not even assumed to be irreducible). I guess this must follow from the theory of matrix factorizations. You should post this as an answer. –  Hans Schoutens Jul 14 '13 at 17:45
The mentioned result of Eisenbud holds in 3-dimensional regular local rings for a prime element $f$. The reference is: (page 124) Eisenbud, D.: Recent progress in commutative algebra, Algebraic geometry - Arcata 1974, AMS Proc. of Pure Math. XXIX (1975), 111-128, available online at: msri.org/~de/papers/pdfs/1975-002.pdf . –  Mahdi Majidi-Zolbanin Jul 15 '13 at 12:54
In your two-dimensional example, if the conductor must be principal (and $R$ is Gorenstein) then we must have $R = \overline{R}$. Since any height-one prime is MCM, in fact $R$ is factorial. It begins to look to me like the answer is that every MCM ideal is principal iff $R$ is a Gorenstein UFD. Is it possible that's true in higher dimensions? –  Graham Leuschke Jul 16 '13 at 17:13
By the way, Bruns' bound from PAMS 81 implies that a hypersurface with singular locus of codimension $c$ has no non-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. –  Graham Leuschke Jul 16 '13 at 17:56

## 1 Answer

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.

-