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?
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2Hint: the group of units is cyclic. Let $g$ be a generator and write everything in terms of powers of $g$. – Daniel Fischer Aug 16 '15 at 21:03
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By "$\mathbb{Z}_{q-1}$", do you mean "$\mathbb{Z}_q \backslash {0}$", where $q$ is a power of a prime? – 727 Aug 16 '15 at 21:12
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1@LJL q ist just a prime here. But I know that this theorem is even valid if q is a power of a prime. – Masoun Alhachmi Aug 16 '15 at 21:18
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@DanielFischer I already have written: $\sum_{j=1}^k {g^{n_j}}^i$ but I dont know how to continue from here, especially how to use the assumption that (q-1) does not divide i. – Masoun Alhachmi Aug 16 '15 at 21:20
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I've tried to address the part you're having trouble with in my edited answer. – Matt Samuel Aug 16 '15 at 21:25
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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.
Matt Samuel
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Thanks alot for helping me. Probably a shame but I still dont understand why we have $g^{i}\sum_{k=1}^{q-1}{g^{ik}}=\sum_{k=1}^{q-1}{g^{ik}}$. Where does this come from? – Masoun Alhachmi Aug 16 '15 at 21:29
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Well done. But there's no need to use powers of $g$ in the inner sum for the trick to be apparent. – Jyrki Lahtonen Aug 16 '15 at 22:31
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@JyrkiLahtonen How would you write it? I'm not always known to explain things in the best way (I was a terrible lecturer). – Matt Samuel Aug 16 '15 at 23:37
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Yeah. Sorry about blowing my own trumpet. I'm a bit allergic to duplicate questions in general. And I've been around this place, and I pay special attention to questions about finite fields in particular :-) The upvote was there, of course. After all, there's no doubt whatsoever that you solved this under your own steam! – Jyrki Lahtonen Aug 17 '15 at 07:55
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Great answer! Let me ask you a question please? When we can consider the sum $\sum\limits_{\alpha \in \mathbb{F}_q} \alpha^i$ where $i\in \mathbb{Z}$, then how do we define $\alpha^i$ if $i\leq 0$ and $\alpha=0$? – RFZ Jan 16 '23 at 18:34
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@ZFR If $i=0$, you define it as $1$. If $i<0$ that would be division by $0$. That doesn't occur anywhere in the answer. All the powers are nonnegative. You don't need negative powers anyway for nonzero elements because they occur as positive powers. – Matt Samuel Jan 17 '23 at 00:27
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@MattSamuel, so you mean if $i=0$, then we define $\alpha^i=1$ even if $\alpha=0$, right? – RFZ Jan 17 '23 at 01:11
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@MattSamuel, but it does not agree with the formula $\sum\limits_{\alpha\in \mathbb{F}q}\alpha^i=-1$ if $q-1\mid i$. However, $q-1\mid 0$. If we follow what you've suggested then we would get that $\sum\limits{\alpha\in \mathbb{F}_q}\alpha^i=0$. – RFZ Jan 17 '23 at 01:14
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@MattSamuel, i did not get your last comment. Why 0 is not an option? I explained in the above comment that it does not agree with the formula. What is wrong? – RFZ Jan 21 '23 at 00:30
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