# 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?

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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. $e_g \mapsto \sum_h e_{gh^{-1}}\otimes e_h$).

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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.

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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.

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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.

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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