Algebra representing the group scheme $\mathbb Z/ n\mathbb Z$ as title says, I can not figure out the hopf algebra structure to put on $k^n$ to make it represent the group scheme $\mathbb Z/n\mathbb Z$. With the usual example of group schemes like $G_a$ or $G_m$ it is easy, but I can not use the same reasoning in this case it seems.
Help?
Thanks!
 A: So the scheme $\mathbb{Z}/n\mathbb{Z}$ over a field $k$ is $\text{Spec}(k^{\oplus n})$. I'm going to label the points by some basis, say $e_0,...,e_{n-1}$.
The Hopf-algebra structure is a map $k^{ n}\rightarrow k^{ n} \otimes k^{ n}$ and we need that the preimage of the ideal $(e_i\otimes e_j)$ is the ideal $(e_{{i+j} \pmod n})$.
So define it like that, send the basis vector $e_i$ to the sum of basis vectors $\sum_{j,k} e_j\otimes e_k$ whenever $j+k= i \pmod{n}$.
Example: Label the points of $\mathbb{Z}/2\mathbb{Z}$ by $e_0,e_1$ corresponding to the prime ideals $(1,0)$ and $(0,1)$ in $k^2$. The Hopf algebra map is $k^2\rightarrow k^2\otimes k^2$ by $(0,1)\mapsto (1,0)\otimes (0,1) + (0,1) \otimes (1,0)$ and $(1,0)\mapsto (1,0) \otimes (1,0) + (0,1)\otimes (0,1)$. Now if I want to compute the product of two points like $e_0,e_1$ I need to look at the preimage of the ideal $(1,0)\otimes (0,1)$ under the Hopf algebra map. But this is just $(0,1)$ (which can be seen by observing the ideal generated by $(1,0)\otimes (0,1) + (0,1)\otimes (1,0)$ contains $(1,0)\otimes (0,1)((1,0)\otimes (0,1) + (0,1)\otimes (1,0))=(1,0)\otimes (0,1)$).
Constant group schemes aren't so fun to describe but you can generalize fairly easily to an arbitrary group in a way similar to the above.
For connected spectra $\text{Spec}(R)$ it's easy to see why the $R$-points $\mathbb{Z}/n\mathbb{Z}(R)$ are exactly $\mathbb{Z}/n\mathbb{Z}$.
A: The $n$th roots of unity are well known to be isomorphic to $\mathbb{Z}/n\mathbb{Z}$. The algebra you're looking for should be $\mathbb{C}[x]/(x^n-1)$
