TITLE: An exercise involving characters
QUESTION [8 upvotes]: Suppose $p$ is a prime, $\chi$ and $\lambda$ are characters on $\mathbb{F}_p$.
How can I show that $\sum_{t\in\mathbb{F}_p}\chi(1-t^m)=\sum_{\lambda}J(\chi,\lambda)$ where $\lambda$ varies over all characters such that $\lambda^m=id$?
($J$ is the Jacobi sum, defined as $J(\chi,\lambda)=\sum_{a+b=1}\chi(a)\lambda(b)$ )
The exercise is taken from A Classical Introduction to Modern Number Theory by Ireland and Rosen - page 105, ex. 8.

REPLY [5 votes]: Write $J(\chi,\lambda) = \sum_{t \in \mathbb F_p} \chi(1-t)\lambda(t),$
so that 
$$\sum_{\lambda} J(\chi,\lambda) = \sum_t \chi(1-t)\sum_{\lambda} 
\lambda(t).$$
Now ask your self: what is the value of $\sum_{\lambda}\lambda(t)$, when $\lambda$ ranges
over all characters of order dividing $m$?  If you sort this out, you will
have answered your question.
(One way to think about is is that the group of $\lambda$ with $\lambda^m = id$ is a subgroup of the full character group, so it is dual to a quotient of $\mathbb F_p^{\times}$.  What is this quotient group explicitly?)