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I'm reading through a paper involving character sums, and I have run into an equality that I am unsure how to justify. Here is the set-up:

Suppose $\chi$ is a multiplicative character of $\mathbb{F}_q[x]/(h(x))$, where $h(x)$ is some irreducible polynomial of degree greater than 1. Let $D$ be a subgroup of the cyclic group $\mathbb{F}_q^*$. The statement to be shown is that

$\displaystyle \left|\sum_{x_i\in D} \chi(1-x_i x)\right|=\left|\sum_{a\in D} \chi(x-a)\right|$

Now it looks to me that we are factoring out the coefficient $x_i$ as in $\chi(1-x_i x)=\chi(-x_i)\chi(x-x_i^{-1})$. Then since the inverse map is bijective and we are summing over all elements of $D$, we can just replace $x_i^{-1}$ with $a$ in the sum. In summary, we can get

$\displaystyle \left|\sum_{x_i\in D} \chi(1-x_i x)\right|=\left|\sum_{a\in D} \chi(-a^{-1})\chi(x-a)\right|$

It isn't exactly clear to me how to eliminate the term $\chi(-a^{-1})$. We can't just pull it out and say that it has norm 1, since it depends on $a$. Is there some kind of reindexing trick that I'm missing here?

I appreciate any suggestions on this problem!

Revision:

I just noticed that there is one other assumption that I missed originally: $\chi$ is a character that is trivial when restricted to $\mathbb{F}_q^*$. If this is the case, the term $\chi(-a^{-1})$ is just $1$, so my equality is proved!

Sorry for the confusion.

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What paper? ${}{}$ –  anon Jun 20 '12 at 5:51

2 Answers 2

$$\Bigl|\sum_{x_i\in D}\chi(1-x_ix)\Bigr|=\Bigl|\sum_{x_i\in D}\chi(x)\chi(x^{-1}-x_i)\Bigr|=\Bigl|\chi(x)\sum_{x_i\in D}\chi(x^{-1}-x_i)\Bigr|$$ $$=|\chi(x)|\Bigl|\sum_{x_i\in D}\chi(x^{-1}-x_i)\Bigr|=\Bigl|\sum_{x_i\in D}\chi(x^{-1}-x_i)\Bigr|=\Bigl|\sum_{a\in D}\chi(x^{-1}-a)\Bigr|$$ is as far as I can get.

I think "The statement to be shown" is false. I'll take a multiplicative character on $F_q$, instead; indeed, let $q=13$. $F_{13}^*=\langle2\rangle$, so define a multiplicative character by $\chi(2^r)=z^r$ where $z=e^{2\pi ir/12}$. Let $D=\{{1,3,9\}}$, a multiplicative subgroup of the units of $F_{13}$.

Let $x=2$.

The left side involves $\chi(1-2),\chi(1-6),\chi(1-18)$ which is $\chi(12),\chi(8),\chi(9)$. Now $12=2^6$, $8=2^3$, $9=2^8$, so it's $|z^6+z^3+z^8|$.

The right side involves $\chi(2-1),\chi(2-3),\chi(2-9)$ which is $\chi(1),\chi(12),\chi(6)$; $1=2^0$, $12=2^6$, $6=2^5$, so it's $|1+z^6+z^5|$. So, is it true that $$|z^6+z^3+z^8|=|1+z^6+z^5|$$ I get $1.506$ (to three decimals) for the left side, while the right side is 1.

So, like anon, I'd like to know, what paper?

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up vote 0 down vote accepted

Just to summarise my revision in my original post, I missed the assumption that $\chi$ is trivial when restricted to $\mathbb{F}_q^*$. If this is the case, $\chi(-a^{-1})=1$, and I get the equality

$\displaystyle \left|\sum_{x_i\in D} \chi(1-x_i x)\right|=\left|\sum_{a\in D} \chi(-a^{-1})\chi(x-a)\right|=\left|\sum_{a\in D} \chi(x-a)\right|$

and I am done!

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