Structure theorem for finitely generated Z[G] modules

Question

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 ?

Answer 1

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.

Answer 2

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/

Answer 3

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.