# Spectrum and scheme of the commutative group-algebra of an abelian group.

The group-algebra of an abelian group is commutative, so we can consider the spectrum of this algebra. Are there any information about the abelian group that we can obtain from such considerations? That is to say, could we study abelian groups by considering the spectrum and the scheme of its group-algebra?
Since I know nothing about the subject, any reference is mostly welcomed. Thanks in advance.
P.S. I also posted in mathmatics stack exchange Here.

• math.stackexchange.com/questions/308939 Feb 20, 2013 at 13:54
• I also posted in mathstack exchange. Hope this matters not. Maybe I shall delete the repeated part? Feb 20, 2013 at 14:13
• At least you should have informed us. Feb 20, 2013 at 16:29
• It doesn't matter if you cross-post, but it's useful if everyone knows you did. Feb 21, 2013 at 14:15

The spectrum of the group algebra of a commutative group is called a diagonalizable group scheme. This is defined in SGA 3 Exposé VIII Section 1. Several geometric characterizations of group-theoretic properties are given in Proposition 2.1. A lot more is written in later sections, such as material on principal homogeneous spaces, quotients of affine schemes by diagonalizable group schemes, and representability of restriction of scalars.

If that isn't enough for you, Exposés 9-11 are about group schemes that are locally-on-the-base isomorphic to diagonalizable group schemes.

• Thanks! This is very helpful! I am quite surprised to learn this thought has been developped already indeed. :) Feb 20, 2013 at 17:27

What follows does not answer your precise question, but is very related to it and may be of interest to you. I consider the case where $G$ is finite, but not necessarily abelian. Then there are several rings attached to $G$, whose spectrum you might want to consider.

The first one is the ring $R(G)$ of virtual characters of $G$ (or, equivalently, the Grothendieck ring of the category of finite-dimensional complex representations of $G$). When $G$ is abelian, it is exactly the group ring of the dual group $\hat{G}$. This group is defined in [Serre: représentations linéaires des groupes finis, 9.1] and its spectrum is studied in [loc. cit., 11.4], where it is shown to be connected.

The other one is the ring $Burn(G)$ that is the Grothendieck ring of the category of finite $G$-sets. In [Bayer-Fluckiger, Parimala, Serre: Hasse principle for G-trace forms, 4.2] (see also references therein), you will find the precise definition, the statement that the spectrum of $Burn(G)$ is connected if and only if $G$ is solvable, and a very nice example of application of $Burn(G)$.

• One of my seniors is wondering if there is a geometric way of formulating this? In his opinion, this is not so very clear. Thanks in advance in any case. Feb 20, 2013 at 14:48
• The fact that $Spec(Burn(G))$ is connected if and only if $G$ is solvable is certainly a geometrical characterization of a group-theoretic property. Feb 20, 2013 at 14:53

Hi.

Varying the coefficients gives certainly a lot of information about the group. For example the smallest field $K\supseteq\mathbb{Q}$ such that $K[G]$ becomes split semisimple (which means isomorphic to $K^G$ in this case) encodes the exponent of the group (which also can be read of from the Loewy length of the modular group algebras I think).

If you are willing to consider the scheme including the involution $\ast: R[G]\to R[G]$ which is defined by $g\mapsto g^{-1}$, then the group is in fact determined up to isomorphism by $\mathbb{Z}[G]$ since $\lbrace\pm 1\rbrace G$ is the group of "orthogonal" units: $\lbrace x\in\mathbb{Z}[G] \mid xx^\ast=1\rbrace = \lbrace\pm1\rbrace G$.

• One of my seniors is wondering if there is a geometric way of formulating this? In his opinion, this is not so very clear. Thanks in advance in any case. Feb 20, 2013 at 13:36
• I'm not sure. Algebraic geometry is really not my specialty... Maybe one can think of this characteristic zero argument in terms of maximal ideals that cease to be maximal when the the scalars are extended to a bigger field. I don't know how this is best explained geometrically... Feb 20, 2013 at 14:34