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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.

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up vote 5 down vote accepted

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?)

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