Dense Torsion in Torus I am reading Onishchik and Vinberg's "Lie Groups and Algebraic Groups". Upon introducing the torus $T=\mathbb{K}^{\times} \times \cdots \times \mathbb{K}^{\times} = (\mathbb{K}^{\times})^n$ (where $\mathbb{K}$ is algebraically closed field of characteristic zero), the reader is asked to prove that the elements of finite order form a dense subset of $T$.
When $n=1$, this is trivial since closed subsets of $T$ are either all of $T$ or finite sets, but there are infinitely many roots of unity.
For $n>1$ I am not sure how to extend this argument. Any hints?

Let $\mu_{\infty}$ be the torsion of $K^*$. Then the torsion of $T$ is $\mu_{\infty}^n$. Suppose this subset is contained in a proper closed subset of $T$, then it is contained in a proper closed subset $Z(F)$ of $K^n$, with a non-zero polynomial $F\in K[X_1, \dots, X_n]$. We get a contradiction with the following lemma applied to $S=\mu_\infty$. 

Lemma: Let $S$ be a infinite subset of $K$. If $F$ vanishes at $S^n$, then $F=0$. 

Proof: By induction on $n$. The case $n=1$ is trivial. Write 
$$F=g_0(X_1,\dots, X_{n-1})+ \cdots + g_d(X_1, \dots, X_{n-1})X_n^d.$$
Fix $(a_1,\dots, a_{n-1})\in S^{n-1}$. For all $a_n\in S$, we have $F(a_1, \dots, a_{n-1}, a_n)=0$. So $F(a_1, \dots, a_{n-1}, X_n)\in K[X_n]$ has infinitely many roots and is therefore trivial. This implies that 
$g_i(a_1, \dots, a_{n-1})=0$ for all $i\le d$. Now make vary $(a_1,\dots, a_{n-1})\in S^{n-1}$ to conclude (with the induction hypothesis) that $g_i=0$ for all $i\le d$, hence $F=0$. 
