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Oct 23, 2015 at 2:20 comment added LSpice Of course this example requires $\zeta^n = 1$ (or else $A \not\in \operatorname{SL}_n(k)$), and so I think doesn't work when $n$ is the characteristic of $k$.
Mar 21, 2013 at 21:26 comment added Salvo Tringali Clear and very nice. I really wonder if one can cluster all the matrices in ${\rm SL}_n(\mathbb C)$ which don't have a $p$-th root for some prime $p \le n$ in a finite number of conjugacy classes: Your example is still very particular, which tempts me to think that there may be only "few" exceptions.
Mar 21, 2013 at 20:06 history edited Ralph CC BY-SA 3.0
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Mar 21, 2013 at 19:59 history answered Ralph CC BY-SA 3.0