I am trying to solve the following problem:
Let $D_n=(\mathbb{C}^{\times})^n$ (an algebraic torus of rank $n$). If $D_k\cong D_n$ as an algebraic group, show that $k=n$.
(HINT: The given isomorphism induces a surjective algebra homomorphism from $\mathcal{O}[D_k]$ onto $\mathcal{O}[D_n]$; clear denominators to obtain a polynomials relation of the form $x_nf(x_1,\ldots,x_n)=g(x_1,\ldots, x_k)$, wich implies $n\leq k$.)
I have proved the first part of the HINT, concluding that $$ f (x_1, \ldots, x_k,\frac{1}{\det}) = g (x_1, \ldots,x_k, \frac{1}{\det}) $$
where the two determinants have different ranges, but I do not see how to determine the relationship $x_nf(x_1,\ldots,x_k)=g(x_1,\ldots, x_k)$.
