When is a torus the closure of the cyclic subgroup generated by one of its elements?

Question

$K$ is an algebraically closed field. And $T$ is an algebraic group conatined in $GL(n,K)$.

Assume $K$ is not the algebraic closure of a finite field. If $T$ is a torus, show that $T$ is the closure of the cyclic subgroup generated by one of its elements. [Note that $K^*$ contains free abelian groups of arbitrarily large finite rank.]

Is this right?

1. Suppose that $\dim T = m$. Then there exsits $d = \mathrm{diag}(d_1, \cdots, d_n) \in T$ such that the subgroup of $K^*$ generated by $d_1, \cdots, d_n$ is free of rank $m$.
2. Then $T$ is the closure of the subgroup generated by $d$.

I have no idea whether this is right or not. If it is, would you please help me with the detail? Or, if it is wrong, sincere thanks for any answer, hint, or reference.

Answer 1

In fact, we have the following wonderful theorem of Tits:

Theorem (Tits): Let $S$ be a torus (not necessarily split) over an infinite field $F$ (not necessarily algebraically closed). Then, there exists an element $s$ in $S(F)$ such that $s^{\mathbb{Z}}$ is dense in $S$.