Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!


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.

share|cite|improve this question
What paper? ${}{}$ – anon Jun 20 '12 at 5:51

$$\Bigl|\sum_{x_i\in D}\chi(1-x_ix)\Bigr|=\Bigl|\sum_{x_i\in D}\chi(x)\chi(x^{-1}-x_i)\Bigr|=\Bigl|\chi(x)\sum_{x_i\in D}\chi(x^{-1}-x_i)\Bigr|$$ $$=|\chi(x)|\Bigl|\sum_{x_i\in D}\chi(x^{-1}-x_i)\Bigr|=\Bigl|\sum_{x_i\in D}\chi(x^{-1}-x_i)\Bigr|=\Bigl|\sum_{a\in D}\chi(x^{-1}-a)\Bigr|$$ 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?

share|cite|improve this answer
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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.