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Suppose that $\Gamma=\{$exp $kX\mid k\in\mathbb{Z}\}$ where $X\in\mathfrak{gl}(n,\mathbb{R})$ is a diagonal matrix. How do we prove that the Zariski closure of $\Gamma$ must contain exp $tX$ for all real number $t$?

Rupert
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Hint: The group $\mathbb{Z}$ is Zariski dense in $\mathbb{R}$, since any nonzero polynomial $f(x)$ has only finitely many roots, so any polynomial $f(x)$ such that $f(k) = 0$ for all $k\in \mathbb{Z}$ must vanish identically. Similarly $\mathbb{Z}^n$ is Zariski dense in $\mathbb{R}^n$.

Dietrich Burde
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    Fair enough. But there is no guarantee that ${$exp $tX\mid t\in\mathbb{R}}$ is the $\mathbb{R}$-rational part of a Zariski closed set. We can say that the Zariski closure of the cyclic group $\Gamma$ must have dimension at least one because it's infinite. But it's not clear to me how we can be sure that it contains ${$exp $tX\mid t\in\mathbb{R}}$. There's probably some simpe argument that I'm missing somehow. – Rupert May 30 '15 at 06:58