# Coproduct on coordinate ring of finite algebraic group

I'm reading Mukai's book "An introduction to invariants and moduli", and I am having trouble understanding one of his examples. It is example 3.49 on page 101.

The setup is as follows. Let $G$ be a finite group, considered as an algebraic group over a field $k$. The coordinate ring of $G$ is then just the set of functions $G \rightarrow k$ with the usual pointwise addition and multiplication. This can be identified with the group ring $k[G]$ in the obvious way (an element $[g] \in k[G]$ corresponds to the function $G \rightarrow k$ that takes $g$ to $1$ and $h$ to $0$ for $h \neq g$). Under this identification, it seems to me that the coproduct is the function

$$\phi : k[G] \rightarrow k[G] \otimes k[G]$$

$$\phi([g]) = \sum_{h \in G} [h] \otimes [h^{-1} g]$$

However, Mukai asserts that if $G$ is the finite cyclic group of order $n$, then the coordinate ring of $G$ is $k[t]/(t^n-1)$ with the coproduct $t \mapsto t \otimes t$. These do not seem like the same thing to me -- what am I doing wrong?

• What Mukai has written down here is the co-ordinate ring of the finite group scheme $\mu_n$ that parametrizes $n^\text{th}$-roots of unity. As people have noted below, this is the dual to the group scheme $\mathbb{Z}/n\mathbb{Z}$, whose co-ordinate ring will have the co-multiplication you describe. – Keerthi Madapusi Pera Feb 13 '12 at 2:03

I think the author accidentally described the dual of the Hopf algebra you're thinking of. Finite group rings are usually endowed with multiplication $(g,h)\mapsto gh$ and comultiplication $g \mapsto g\otimes g$ (see here).

The coordinate ring $k[G]$ is obtained by dualizing. Then $g \mapsto g\otimes g$ becomes $e_g^2 = e_g$, where $e_g$ is the function on $G$ that maps $g$ to $1$ and all other group elements to $0$. Comultiplication will look exactly the way you described it (i.e. $e_g \mapsto \sum_h e_{gh^{-1}}\otimes e_h$).

The book presumably assumes the field $k$ contains $n$ distinct roots of unity (in particular, that the characteristic of $k$ is coprime to $n$). Then you get a $k$-algebra isomorphism between $k[x]/(x^n-1)$ (isomorphic to the group ring $k[G]$ by sending $x$ to a generator) and the coordinate ring $\bigoplus_{g \in G} k$ by a finite Fourier transform. This reflects the fact that finite abelian groups are Pontryagin self-dual.

The Hopf algebra structure here involves a coproduct taking the function $f$ on $G$ to $\sum_i g_i \otimes h_i$, where $f(xy) = \sum_i f_i(x) g_i(y)$ when $x,y \in G$. Whatever Mukai is doing for a cyclic group should be consistent with this formulation of the coproduct, but I'm unfamiliar with his book.

More generally, this kind of formalism occurs when you consider a finite group scheme as in Jantzen Representations of Algebraic Groups (AMS, 2003), I.2.3.

It might be that the author accidentally described the dual of the Hopf algebra as Florian Eisele suggested, however in this case the Hopf algebra is self dual so k[G] is actualy isomorphic to $k[t]/(t^n−1)$. The isomorphism is not compliantly canonical, it becomes canonical if $G$ is the group of n-s roots of $1$. Then it is given by $[g] \mapsto \sum_1^n g^i t^i$.

So may be this is what the author meant.

• Are you sure $k[\mathbb Z/n\mathbb Z]$ is self-dual? If $char(k)=p$ and $n=p$, then $k[t](t^p-1)$ is not semisimple as an algebra, but its dual (the coordinate ring $k\oplus \ldots\oplus k$) is. – Florian Eisele Feb 13 '12 at 2:08