Summing up the elements of a finite field.

Question

I want to show that $\sum_{x\in \mathbb F_{q}}x^i=0$ if $q-1$ does not divide $i$. Can someone give me a hint?

Answer 1

Let $F$ be the field with $q$ elements and let $g$ be a generator of the multiplicative group. Note that for any integer $n$ we have $g^n=1$ if and only if $q-1$ divides $n$ because the order of $g$ is $q-1$. Following Daniel Fischer's suggestion, note that $$g^{i}\sum_{k=1}^{q-1}{g^{ik}}=\sum_{k=1}^{q-1}{g^{i(k+1)}}=\sum_{k=2}^{q}{g^{ik}}=\sum_{k=1}^{q-1}{g^{ik}}$$ The equality holds because the group is cyclic and all we've done is permuted the elements in the sum. But $F$ is a field, and $g^i\neq 1$ because $q-1$ does not divide $i$. There is only one element $x$ such that for some element $y\neq 1$ we have $yx=x$, and that is the only element we can't divide by.