Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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
show 2 more comments

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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