26
$\begingroup$

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?

$\endgroup$
25
$\begingroup$

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 $R = k[\varepsilon]/(\varepsilon^2)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.