Let $G$ be a finite abelian group (written multiplicatively), $R$ a commutative ring and let $R [G]$ denote the set of all formal linear combinations of elements of $G$ with coefficients in $R$. Then $R[G]$ is an $R$-algebra and in particular a ring with multiplication of elements defined in the obvious way:
$(\sum_{g\in G} a_g g)\cdot (\sum_{h\in G} b_h h) = \sum_{g\in G} \sum_{h\in G} a_g b_h (g h)$
where the product of group elements occurs in $G$. This object is called the group ring or the group algebra of $G$ (over $R$).
This is just an idle speculative question which occurred to me during my representation theory class, but given that we've got a natural $R$-algebra $R[G]$ here, does the affine scheme $\text{Spec}R[G]$ carry any information about $G$? Particular cases of interest to me are when $R = \mathbb{C}$ (this most closely ties in with my representation theory course) and when $R$ is somehow "arithmetic" e.g. $\mathbb{Z}$ or $\mathbb{F}_p$ for a prime $p$.
This is not an answer, since it doesn't address general connections between $G$ and $\operatorname{Spec} R[G]$, but it's a little long for a comment. My goal is to make some basic observations about the spectrum of a group ring.
First, observe that $R[G\times H] \cong R[G]\otimes_R R[H]$, so $\operatorname{Spec} R[G\times H] \cong \operatorname{Spec} R[G] \times_{\operatorname{Spec} R} \operatorname{Spec} R[H]$. Thus we can restrict our attention to indecomposable groups.
In particular, we can compute this spectrum geometrically for finitely generated abelian groups. For example, we have $R[\mathbb{Z}] \cong R[X,X^{-1}]$, so $\operatorname{Spec} R[\mathbb{Z}^n] \cong \mathbb{G}_m^n$ is the $n$-torus over $R$.
Similarly, $\operatorname{Spec} R[\mathbb{Z}/k] \cong \operatorname{Spec} R[X]/(X^k-1)$ is the closed subscheme of $\mathbb{G}_m$ of $k$-th roots of unity (in characteristic relatively prime to $k$, anyway).
To connect these affine schemes with our original group $G$, we should of course keep track of the action of $G$. Or perhaps coaction? I don't know very much about this.