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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?

Ofir
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1 Answers1

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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 $F=0$.

  • Thanks. I've known this lemma (and its improvements such as the Combinatorial Nullstellensatz), but somehow, when I approach abstract algebra problems I sometimes blank out and get scared. Thank you for a clear solution again. – Ofir Mar 18 '13 at 20:30