Let $k$ be algebraically closed. A character on an algebraic group $G$ over $k$ is a group homomorphism $G \rightarrow k^{\ast}$. If this is also a morphism of varieties, then this is called a rational character.
Let $M$ be a finitely generated additive abelian group, without $p$-torsion if $k$ has prime characteristic. The group algebra $kM$ is the $k$-vector space with basis $e_m : m \in M$, and with multiplication given by $e_m e_{m'} = e_{m+m'}$. Then $kM$ is a finitely generated, reduced $k$-algebra.
One can actually show that $kM$ is a Hopf algebra, with comultiplication given by $\Delta: kM \rightarrow kM \otimes_k kM, e_m \mapsto e_m \otimes e_m$. Therefore, the set $G$ of maximal ideals of $kM$ becomes an algebraic group, with multiplication $m: G \times G \rightarrow G$ given by the following formula: for $(x,y) \in G \times G$, $xy$ is the kernel of the $k$-algebra homomorphism $kM \rightarrow k$ given by $f \mapsto \Delta(f)(x,y)$.
Each $e_m \in kM$ is really a function $G \rightarrow k$, where any maximal ideal $x \in G$ is associated with the unique element of $k$ to which $e_m$ is congruent modulo $kM$. I'm trying to use the structure of the Hopf algebra to show that $e_m(xy) = e_m(x)e_m(y)$ for all $x, y \in G$, but I just don't see it. It should be trivial. Could anyone give me a hint?
Wait I think I have it. Let $\mathfrak m$ be the maximal ideal $xy$. The product $e_m(x)e_m(y)$ is a member of $k$, which we regard as being included in $kM$. So we identify $e_m(x)e_m(y)$ with $$e_m(x)e_m(y) 1_{kM} = e_m(x)e_m(y)e_0$$ Now, we want to show that $e_m - e_m(x)e_m(y)e_0$ lies in $\mathfrak m$. By definition, this means that the evaluation of $$\Delta(e_m - e_m(x)e_m(y)e_0) = e_m \otimes e_m - e_m(x)e_m(y) \cdot e_0 \otimes e_0 $$ at $(x,y)$ is $0$. But this is easy to check. Just use the fact that $f \otimes g(x,y) = f(x)g(y)$, and $e_{m+m'} = e_me_{m'}$.