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.