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Let $\sigma $ act on the $m$th roots of unity by $\sigma \zeta = \zeta^r$ for $1\le r <m, (r,m)=1.$ I'm looking to evaluate $\displaystyle\sum_{i=0}^{t(\zeta )-1} \zeta^{r^i},$ where $t(\zeta )$ is the size of the orbit of $\zeta $ under $\sigma .$

I keep running into this pesky little sum and was wondering if anyone on this forum might be able to point me in the direction of converting it to something simple (or at least more amenable for further computation.) It wouldn't surprise me if I'm overlooking something elementary, but it's already frustrated me to the point of posting here, so any guidance whatsoever would be greatly appreciated.

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Where does $\sigma$ come from? Isn't $r$ fixed, but arbitrary? –  user21436 Jan 19 '12 at 18:17
    
$\sigma $ is just a name I gave to the map $\zeta \mapsto \zeta^r.$ $r$ is fixed but arbitrary. –  Tim Duff Jan 19 '12 at 18:22
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1 Answer

up vote 2 down vote accepted

You're looking at a sum consisting of values of an additive character on ${\mathbb Z}_m$, where the inputs run over a multiplicative subgroup of ${\mathbb Z}_m^\times$. You can use a linear combination of Dirichlet characters to detect that multiplicative subgroup, and this converts your problem to a linear combination of Gauss sums. The relevant Dirichlet characters will be those $\chi$ such that $\chi(r)=1$.

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PS: Welcome back Wikipedia :) –  Greg Martin Jan 19 '12 at 18:25
    
Looks like I'll be busy, thank you! –  Tim Duff Jan 19 '12 at 18:36
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