Elements of Group Rings, Are they Reducible?

Question

In general, elements of group rings are written as sums of the group elements multiplied by scalars from the ring, correct? What is the utility of that, if we don't know how to reduce the elements of the group ring? In the case of Linear Algebra, we know that the module the group ring acts on is over the same field as the group ring, but in a situation where the module is an arbitrary module, and the group ring is formed via an arbitrary ring and group, what is the utility? It's not reducible, is it?

Answer 1

You question is quite vague and the topic of group rings is very broad. So I'll just mention some well-known facts and some examples and hopefully it will satisfy you a little.

Let's recall that group rings ale related to free modules. In particular, any group $G$ has its underlying set of elements and we can consider free $R$-modules over that set. In case $R$ is some arithmetic ring, this lets us count how many of each element of $G$ we have in our pockets. The quintessential example is that of $R = {\bf Z}$ for which free $R$-modules are nothing else than free abelian groups. These are used all over the place to count stuff (for example divisors in algebraic geometry or singular chains in algebraic topology). So if you believe that free modules are fantastic, group rings must be twice as good because besides addition they also allow you to do multiplication.

Now let's look at this from another angle. For the moment I'll assume $G$ is finite and $R$ is commutative (think of $\bf C$ for concreteness). Write $\sum_g c_g \cdot g$ for an element of $R[G]$. This element carries no more information than a function $G \to R$ that sends $g$ to $c_g$. Viewed this way, the group ring is nothing else than just set of all functions from $G \to R$ with point-wise addition. But the multiplication is a bit weird, it's given by something called convolution. The rings $R[G]$ and $R^G$ (functions with point-wise additions and multiplication) are dual to each other. This is actuality nothing else than Fourier transform (in the finite, discrete setting) and the restating of the familiar fact that Fourier transform of the convolution of functions is the product of their Fourier duals.

Let's remove the assumption that $G$ be finite. A pretty well-known case is when $G = {\bf Z}$. In this case, ${\bf R}[{\bf Z}]$ is nothing else than the Laurent polynomials in $R$ (that is polynomials, where you also allow negative powers) such as $z^2 + z^{-3}$. What $\bf Z$ keep track of in this setting is the degree of a monomial. Remember how I said in the first paragraph that group rings and more generally free modules are used to count things? Here we have one monomial of degree $2$ and one one degree $-3$.

I believe the above two view-points: keeping track of counting of things and relating the group ring to some ring of functions are the most important. But there is obviously much more to be said since the theory of group rings can't be much easier than the theory of groups or rings themselves (that is, very hard). For example, representation theory of finite groups over a field $K$ is nothing else than the study of $K[G]$-modules.

Finally, all of this stuff can be spiced up with a little topology. That is, $G$ can be something like a Lie group and we can consider a ring of continuous functions (or $L^1$, $L^2$, etc.). There are again notions of convolution and Fourier transform (this is related to the Pontryagin duality) in this setting as well. This stuff is mostly studied in the fields of $C^*$-algebras and harmonic analysis.