Writing this up as a question seemed to clarify things, but I just want to make sure that I am understanding things correctly. Let $\mathbb{F}_{p} = \mathbb{Z}/p\mathbb{Z}$, $\chi$ a nontrivial multiplicative character on $\mathbb{F}_{p}$, and $\chi^{-1}$ denote the inverse of $\chi$. I am trying to follow the proof that $J(\chi, \chi^{-1}) = -\chi(-1)$ from Ireland and Rosen's A Classical Introduction to Modern Number Theory. Now we have that $$ J(\chi, \chi^{-1}) = \sum_{a + b = 1} \chi(a) \chi^{-1}(b) = \sum_{\substack{a + b = 1\\b \neq 0 }} \chi\left(\frac{a}{b}\right) = \sum_{a \neq 0} \chi \left( \frac{a}{1- a} \right).$$ Now if we set $a/(1 - a) = c$ and $c \neq -1$, then $a = c/(1 + c)$. Now comes the part of the proof that I was struggling with, they go on to say:
It follows that as $a$ varies over $\mathbb{F}_{p}$, less the element 1, that $c$ varies over $\mathbb{F}_{p}$, less the element $-1$. Thus $$ J(\chi, \chi^{-1}) = \sum_{c \neq -1} \chi(c) = - \chi(-1).$$
Is this because $$ J(\chi, \chi^{-1}) = \sum_{a \neq 0} \chi\left(\frac{a}{1 -a} \right) = \sum_{c \neq -1}\chi(c) = - \chi(-1) + \sum_{c \in \mathbb{F}_{p}} \chi(c) = - \chi(-1)?$$
EDIT: Forgot the $-$ sign in the line $J(\chi, \chi^{-1}) = - \chi(-1)$
