# Commutative ring and its group-algebra, and abelian-group-algebra as a commutative ring.

In course of discussing the algebraic structures, one of my seniors is led quite naturally to considering the $\color{red} {geometric}$ version of following:

Question: Since we could consider the spectrum of an abelian-group-algebra $\color{blue}{as\ a\ commutative\ ring}$, it is natural to ask: what kind of informations of this abelian group could be obtained from the considerations of this spectrum?
On the other hand, we can view the commutative ring $\color{blue} {as\ an\ abelian\ group}$, and thus form its group-algebra. We then ask a further question: what relations are there between the spectrum of the ring, and the spectrum of this group-algebra?
Notice: As I have yet heard of nothing about the subject, any source or reference in this direction is the most appreciated. Thanks in advance.

Also one of my seniors is greatly intrigued in the $\color{green}{geometric}$ point of view of this question, thus it would be quite wonderful if any insight or crucial observations are provided. Thanks again.

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This is pretty interesting. If I understand the questions right: 1) "What does $spec(R[A])$ tell us about abelian group $A$?" 2) If $A$ and $R$ are commutative rings, then how does $Spec(A)$ relate to $Spec(R[A])$? You would think the first question would have answers in representation theory. Superficially the second question seems a little hopeless: it looks like none of the multiplicative monoid information of $A$ is used in the construction of $R[A]$, so hoping for a strong connection seems too optimistic. – rschwieb Feb 20 '13 at 14:23
@rschwieb Yes you understand the question rightly. I see now that the second question is not so hopeful, and that the first question is quite closely related to the representation theory. It should be added that the original ideas were to study group theory via algebraic geometry, a crazy idea of my senior. Thanks in any case for the attention. – awllower Feb 20 '13 at 15:12
Let me know what happens :) It sounds like an interesting program. – rschwieb Feb 20 '13 at 17:00
Duplicate mathoverflow.net/questions/122394 – Martin Brandenburg Feb 20 '13 at 17:24

## 1 Answer

The multiplicative structure of rings is crucial for doing algebraic geometry. In fact, most of it can be done for arbitrary commutative monoids (see the work by Connes, Deitmar, etc.), and more generally, for commutative algebraic monads (see Durov's thesis).

I don't think that the additive structure carries enough information about a ring in order to say something interesting. In particular, the functor $\mathsf{CRing} \to \mathsf{Ring}, (A,+,*) \mapsto \mathbb{Z}[A,+]$ loses lots of information. And in general there are no ring homomorphisms $\mathbb{Z}[A] \to A$ or $A \to \mathbb{Z}[A]$ at all.

EDIT: If $A$ is just an abelian group (not assumed to be the underlying additive group of a ring), then $\mathbb{Z}[A]$ is a quite well-understood commutative ring (in particular all the open problems concerning zero divisors etc. in group rings are solved in this case). It is the directed colimit of the group rings $\mathbb{Z}[A']$, where $A'$ runs through the finitely generated subgroups of $A$. Now, if $A$ is finitely generated, then there is an isomorphism $A \cong \mathbb{Z}^r \oplus \mathbb{Z}/n_1 \oplus \dotsc \oplus \mathbb{Z}/n_s$ for integers $r,s$ and $n_1 | \dotsc | n_s>0$. It follows $\mathbb{Z}[A] \cong \mathbb{Z}[x_1^{\pm 1},\dotsc,x_r^{\pm 1}][y_1,\dotsc,y_s]/(y_i^{n_i}-1)_i$. In particular, we see that if $A$ is torsion-free, then $\mathbb{Z}[A]$ is a regular integral domain.

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Thanks for pointing out all this. I appreciate much. While it could be better if the questin about the spectrum of the group algebra of an abelian group is also addressed. Still thanks in any case. – awllower Feb 20 '13 at 11:45
On the insist of my senior, it would be better if one can give a geometric description. Thanks in advance. – awllower Feb 20 '13 at 14:26