Let $\mathbb{Q}$ be the field of rational numbers and $\mathbb{G}_m$ the multiplicative group. Let $S \cong \mathbb{G}_m^n$ be a split torus.
How to show that the coordinate ring $\mathbb{Q}[S]$ is isomorphic to the group algebra $\mathbb{Q} Hom(S, \mathbb{G}_m)$?
That is, every polynomial $f$ in $\mathbb{Q}[S]$ can be written as $$ f = \sum_{\chi} c_{\chi} \chi, $$ for some $c_{\chi} \in \mathbb{Q}$.
Any hint will be greatly appreciated.
This is an instance of a more general phenomenon: the representing Hopf algebra of any diagonalizable group scheme is naturally isomorphic to the group algebra on its group of characters.
Let $F$ be a field. For any abelian group $H$, the group algebra $F[H]$ can be given the structure of a Hopf algebra as follows, where the following maps are given on elements of $H$:
The co-multiplication map $c \colon F[H] \to F[H] \otimes_{F} F[H]$ is given by $c(h) = h \otimes h$.
The co-inversion map $i \colon F[H] \to F[H]$ is given by $c(h) = h^{-1}$.
The co-unit map $u \colon F[H] \to F$ is given by $u(h) = 1$ (i.e., the usual augmentation morphism).
An affine group scheme $G$ over $F$ is said to be diagonalizable if $G$ is represented by $F[H]$ for some abelian group $H$; for an abelian group $H$, we denote the diagonalizable group scheme associated to $H$ by $H_{\mathrm{diag}}$. One can see that
$$(\mathbb{Z})_{\mathrm{diag}} = \mathbb{G}_{m, F}; \quad (\mathbb{Z}^{n})_{\mathrm{diag}} = \mathbb{G}_{m, F}^{n}$$
Now, let $G$ be an affine group scheme, with representing Hopf algebra $A$ (and structural morphisms denoted $c, i, u$). We say an element $x \in A^{\times}$ is group-like if $c(x) = x \otimes x$. One can see that the group-like elements of $A^{\times}$ form a subgroup of $A^{\times}$; I claim that this subgroup is isomorphic to $G^{\ast} := \mathrm{Hom}(G, \mathbb{G}_{m, F})$.
Indeed, any element $\rho \in G^{\ast}$ is determined by its comorphism $\rho^{\sharp} \colon F[t, t^{-1}] \to A$. It is well-known that the $F$-algebra homomorphisms $F[t, t^{-1}] \to A$ are in bijection with the elements of $A^{\times}$, where the bijection takes an $F$-algebra homomorphism $\varphi \colon F[t, t^{-1}] \to A$ to $\varphi(t)$. Since $\rho^{\sharp}$ is a morphism of Hopf algebras over $F$, it must further commmute with the co-mutliplication, co-inversion and co-unit maps. Commuting with the comultiplication on $F[t, t^{-1}]$ forces that we must have $c(\rho^{\sharp}(t)) = \rho^{\sharp}(t) \otimes \rho^{\sharp}(t)$, i.e. $\rho^{\sharp}(t)$ must be a group-like element of $A^{\times}$. Somewhat surprisingly, this condition is not only necessary but sufficient; every Hopf algebra homomorphism $F[t, t^{-1}] \to A$ is given by some group-like element of $A^{\times}$, which specifies the image of $t$.
Back to our original problem, one can see that for any abelian group $H$, the group-like elements of $F[H]$ (with the Hopf algebra structure described above) is just the copy of $H$ inside $F[H]^{\times}$. Hence, given a diagonalizable group scheme $H_{\mathrm{diag}}$, we can recover the group of the representing group algebra via its group of characters $(H_{\mathrm{diag}})^{\ast}$. In fact, we can say more. We have two functors: there is the functor from abelian groups to diagonalizable group schemes which sends an abelian group $H$ to $H_{\mathrm{diag}}$; and there is a functor from diagonalizable group schemes to abelian groups which sends a diagonalizable group $G$ to its group of characters $G^{\ast}$. One can check that these two functors define an equivalence of categories.
In particular, we get $F[((G_{m, F})^{n})^{\ast}] \cong F[\mathbb{Z}^{n}]$ as Hopf algebras. Explicitly, the isomorphism takes $\rho \in ((G_{m, F})^{n})^{\ast}$ to $\rho^{\sharp}(t)$.