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For a finite abelian group $G$ is there an analogue of structure theorem for finitely generated modules like for P.I.D. rings but with $Z[G]$ group ring over integers instead ?

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Z[G=Z/nZ] has zero-divisors, so I wouldn't bet on it. –  Harry Gindi Nov 11 '10 at 0:08
A finitely generated $Z[Z/2]$-module is the same as a fintely generated abelian group together with an involution. I think these are already hard to classify. –  Martin Brandenburg Nov 11 '10 at 0:31
Harry, note that $\mathbb{C}[G]$ also has zero-divisors, but I think it is fair to say that $\mathbb{C}[G]$-modules are pretty well understood. –  Alex B. Nov 11 '10 at 2:06
Exactly. My point was that the hardness of the problem has nothing to do with zero-divisors. I was merely objecting to the heuristic in your first comment. –  Alex B. Nov 11 '10 at 2:41
Martin says that $\mathbf{Z}[{\mathbf{Z}/2]$-modules are hard to classify. What goes wrong with the following geometric approach: if $R=\mathbf{Z}[\mathbf{Z}/2]$ then $\mathrm{Spec}(R)$ is two copies of $\mathrm{Spec}(\mathbf{Z})$ glued at the point $(2)$, so a f.g. module for $\mathrm{Spec}(R)$ is something like two f.g. abelian groups $M_1$ and $M_2$ (which we understand) equipped with an isomorphism $M_1/2=M_2/2$? This seems to be some sort of a classification, unless I made a slip. –  Kevin Buzzard Nov 11 '10 at 7:40
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3 Answers

up vote 12 down vote accepted

Let me give an answer in the opposite direction of the ones already given. It may that be you can't get a precise classification, but nevertheless, you can say something, and it may be enough for the situation you are considering, or at least helpful.

Let's suppose first that your module $M$ is torsion; then it is a product of its $p$-power torsion parts for each prime $p$, so you an suppose it is $p$-power torsion. It is then a module over $\mathbb Z_p[G]$. You may write $G$ as a product $G_1\times G_2,$ where $G_1$ is of prime-to-$p$ order and $G_2$ is of $p$-power order. The group ring $\mathbb F_p[G_1]$ is a product of finite fields $k$, and the ring $\mathbb Z_p[G_1]$ is isomorphic to a corresponding product of rings of Witt vectors $W(k)$. So one is left with $p$-power torsion modules over $W(k)[G_2]$. This may be something of a mess, but one can say something: for example, any Jordan--H\"older factor of such a module is the trivial representation of $G_2$ over $k$, and so your module is a successive extension of copies of the trivial representation.

Now suppose that you have an $M$ which is not necessarily torsion. Then one can consider the map $M \to \mathbb Q\otimes_{\mathbb Z} M$. The kernel is torsion, and so one can get some handle on its structure as above. The image is a $G$-invariant lattice in the $G$-representation $\mathbb Q \otimes_{\mathbb Z} M.$ The possibilities for this latter representation (of $\mathbb Q$ over $G$) can be described pretty easily, since $\mathbb Q[G]$ is just a product of fields $K$; it will break up into a product of copies of these various fields $K$. The biggest complication with the image of $M$ is that it typically won't break up as a product.

A useful way to think is to consider Spec $\mathbb Z[G]$, which is an affine scheme mapping finitely to Spec $\mathbb Z$, and then to think of $M$ as quasi-coherent sheaf over Spec $\mathbb Z[G]$. Then tensoring with $\mathbb Q$ corresponds to looking at this sheaf at the generic points; looking at the torsion part of $M$ corresponds to looking at sections supported at closed points; and the possible failure of $M$ to decompose as a product corresponds to the fact that although Spec $\mathbb Z[G]$ will have several irreducible components (if $G$ is non-trivial) --- these correspond to the various factors in the decomposition of $\mathbb Q[G]$ into a product of fields --- it will be connected (the components meet at various closed points, corresponding to the primes dividing the order of $G$).

Number theorists frequently work with rings like $\mathbb Z[G]$, and modules over them. Depending on what you want to do, it is not necessary to have a complete classification; a somewhat coarser understanding of the type described above can be quite useful by itself.

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The question came from topology and group actions of abelian group $G$ acting on a product of spheres. We then consider action of group ring $\mathbb{Z}[G]$ on singular cohomology modules. –  B. Naskrecki Nov 11 '10 at 17:58
This has some analogy with the situation in number theory: in number theory, we have Hecke correspondences acting on certain algebraic varieties (Shimura varieties), and we form the Hecke algebra generated by all these correspondences, which as a $\mathbb Z$-module is free of finite rank (just like your $\mathbb Z[G]$), and then we consider the action of this Hecke algebra on the cohomology of the variety, and use techniques of the kind described in my answer to get some handle on the possible module structure that results. –  Emerton Nov 11 '10 at 18:58
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Suppose that $p$ is an odd prime, and suppose that $G = (\mathbf{Z}/p \mathbf{Z})^2$. Then the ("easier") category of $R = \mathbf{Z}_p[G]$-modules is of so called "wild representation type", that is, it contains a full subcategory equivalent to the category of modules over a free associative algebra in two variables over $\mathbf{F}_p$. But classifying modules over such an algebra is hard, because it essentially equivalent to classifying pairs of matrices up to conjugation, which is known to be difficult. So basically you're screwed.

See, for example: http://www.springerlink.com/content/r67475g01n232866/

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A more impressive way of explaining why wild is hard is to observe that the category of modules over the free algebra contains copies of the categories of modules of all finite dimensional algebras. –  Mariano Suárez-Alvarez Nov 11 '10 at 2:36
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The answer to your question is 'no'. Even if you limit yourself to modules that are free over $\mathbb{Z}$, there is no classification known. Indeed, if your abelian group is not cyclic or its order is not cube-free, then there are infinitely many isomorphism classes of indecomposable $\mathbb{Z}$-free modules (this is due to many people). So it is certainly not true, that they are all quotients of the regular module, like in the PID-case.

Of course, it is still true that any finitely generated $\mathbb{Z}[G]$-module is a quotient of a free module, but that is general algebraic non-sense and doesn't tell you anything about their structure.

For a (admittedly slightly aged) survey of known results on $\mathbb{Z}$-free $\mathbb{Z}[G]$-modules, aka integral $G$-representations, see this survey.

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