MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am interested in computing direct sum decomposition of modules over some quotients of polynomial rings over a field (do not care much about the field at this point). Does any one know a package which does that? I tried searching Macaulay 2 but could not find any. Thanks.

EDIT: Perhaps the question was a bit vague, so I will add a specific example. Let $R=k[X_{11},...X_{33}]/I$, with $k$ say, $\mathbb Q$ or $\mathbb Z/(p)$ with $p>3$ and $I$ generated by the 2x2 minors. Let $M=(x_{11},x_{12},x_{13})$. I would like to be able to understand the direct summands of syzygies of $M$. They are all maximal Cohen-Macaulay, but that's all I know.

share|cite|improve this question
Polynomials over what kind of object? – Ryan Budney Dec 18 '09 at 7:40
Any field will do. – Hailong Dao Dec 18 '09 at 7:54
I've given this a little thought over the years, and I think it might be very hard even to tell algorithmically whether a given module (given as a cokernel of some matrix) is indecomposable. Can Singular tell whether a (mildly) non-commutative ring such as $\operatorname{End}(M)$ is local or not? I know that M2 cannot. – Graham Leuschke Dec 18 '09 at 21:52
Good point, Graham! May be there should be a project! – Hailong Dao Dec 18 '09 at 21:56
May I draw you attention to some recent criterion of decomposability: arXiv:1305.2256 . Probably it does not help for the syzygies of this particular $M$, but it seems quite useful in many cases. – Dmitry Kerner Jul 19 '15 at 18:07

Suppose A = C[[x,y,z]]/f(x,y,z) is one of the ADE singularities, where there are finitely many indecomposables P_1,...,P_n. In analogy with character theory of finite groups, we want to set up a situation where Hom(P_i,P_j) = delta _ij. That will allow us to decompose a reflexive A-module into a direct sum of indecomposables (in the same way one decomposes a representation into a direct sum of irreducible representations).

The triangulated category \underline{CM}(A) = CM(A)/A has a t-structure with heart CM(A), in which the finitely many indecomposables are the simple objects. The simples satisfy Hom^0(S_i, S_j) = delta ij. To compute Hom^0 in this category use the equation \underline{Hom}(M,N) \simeq Ext^2_A(M,N). (See Burban and Drozd's survey paper, especially page 46.)

So basically you just have to compute Ext^2 in the complete local ring A. Singular can do this directly.

If you don't want to download Singular, Macaulay2 can do it, although it takes some care, because Macaulay naturally works with graded modules over polynomial rings; (one has to be careful with grading shifts.) For more information on the graded case, see the papers by [Kajiura Saito and Takahashi].

share|cite|improve this answer
I believe you're taking the analogy with representation theory too literally. The ADE singularities have finitely many indecomposable MCM modules, not indecomposable modules. Also, if $M_i$ and $M_j$ are indecomposable MCM modules, $Hom_R(M_i,M_j)$ is going to be another reflexive module, and almost certainly not zero. (Specifically, there are certainly examples where $i\neq j$ and the $Hom$ is not zero.) The later comments in the triangulated situation are correct, but not helpful in general. – Graham Leuschke Dec 18 '09 at 14:56
Thanks, Graham, for the feedback. you are correct there are only finitely many indecomposable maximal cm modules. the method described above allows one to decompose a MCM module (or reflexive) module. Also, you are right that Hom_R(M_i,M_j) is not delta_ij; but in the triangulated category it is, that was my point. – user2686 Dec 18 '09 at 19:28
Thanks, Brian. I was thinking of something like $k[x_11,...,x_33]$ modulo the 2x2 minors, so typically there will be quite a few even MCM indecomposables. – Hailong Dao Dec 18 '09 at 20:40

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.