# Serre's remark on group algebras and related questions

I've recently heard about an idea of Serre that for each finite group $G$ there exists a group scheme $X$ such that for each field $K$ the group $X(K)$ is naturally isomorphic to the unit group of $K[G]$. Unfortunately, the article where this fact was mentioned gave no reference, so I ask you if you know how to construct such a scheme. Of course, an interesting question would be: what about a set of $R$-points of $X$, where $R$ is a ring, how is it related to $R[G]$?

And can this be generalized somehow to arbitrary groups? In the form given above it sounds not really possible as for $K[\mathbb Z]$ the group of units is isomorphic to $K^*\times \mathbb Z$ and one can hardly imagine a group scheme whose group of $K$-points is isomorphic to $\mathbb Z$.

By the way, why there is no group scheme whose group of points is isomorphic to $\mathbb Z$? Or it exists?

## 1 Answer

It's fairly easy to do this for finite groups. In fact, the functor $R \mapsto R[G]$ is naturally representable by a ring scheme: the underlying set functor is represented by $\mathbb A^n$ where $n = |G|$, and the ring structure comes from the functor of points $R \mapsto R[G]$. Write $Y$ for this ring scheme (say over $\operatorname{Spec} \mathbb Z$).

Now the unit group can be constructed as the closed subset $V \subseteq Y \times Y$ of pairs $(x,y)$ such that $xy = 1$. It is closed because it is the pullback of the diagram $$\begin{array}{ccc}V & \to & Y \times Y\\\downarrow & & \downarrow \\ 1 & \hookrightarrow & Y\end{array},$$ where the right vertical map is the multiplication morphism on $Y$. This shows that $R \mapsto R[G]^\times$ is representable. It naturally becomes a group scheme, again by the functor of points point of view. $\square$

In the infinite case, this construction doesn't work, because the functor $R \mapsto R[G]$ is not represented by $\mathbb A^G$ (the latter represents the infinite direct product $R \mapsto R^G$, not the direct sum $R \mapsto R^{(G)}$). I have no idea whether the functor $R \mapsto R^{(G)}$ (equivalently, the sheaf $\mathcal O^{(G)}$) is representable, but I think it might not be.

On the other hand, in the example you give of $G = \mathbb Z$, the functor on fields $$K \mapsto K[x,x^{-1}]^\times = K^\times \times \mathbb Z$$ is representable by $\coprod_{i \in \mathbb Z} \mathbb G_m$, but this does not represent the functor $R \mapsto R[x,x^{-1}]^\times$ on rings for multiple reasons. Indeed, it is no longer true that $R[x,x^{-1}]^\times = R^\times \times \mathbb Z$ if $R$ is non-reduced, nor does $\coprod \mathbb G_m$ represent $R \mapsto R^\times \times \mathbb Z$ if $\operatorname{Spec} R$ is disconnected. These problems do not cancel out, as can already be seen by taking $R = k[\varepsilon]/(\varepsilon^2)$.