I am given a non-trivial homomorphism $\chi : \left(\mathbb{Z} / p\mathbb{Z} \right)^\times \rightarrow \mathbb{C}^\times$, p is prime, and $\zeta$ is a primitive p-th root of unity. A generalized Gauss sum is then defined as:
$$\tau_\chi = \sum_{a\in \left(\mathbb{Z} / p\mathbb{Z} \right)^\times} \chi(a) \zeta^a $$
I am asked to show that $|\tau_\chi |^2 = p$.
What I have so far is:
$$|\tau_\chi |^2 = \sum_{a\in \left(\mathbb{Z} / p\mathbb{Z} \right)^\times} \chi(a) \zeta^a \hspace{5pt} \times \sum_{a\in \left(\mathbb{Z} / p\mathbb{Z} \right)^\times} \chi(a^{-1}) \zeta^{-a}$$
$$ = \sum_{a=1}^{p-1} \chi(1) \hspace{5pt} + \hspace{5pt} \sum_{a=1}^{p-1} \hspace{3pt}\sum_{b \in \left(\mathbb{Z} / p\mathbb{Z} \right)^\times, b\neq a} \chi(ab^{-1}) \zeta^{a-b}$$
This is where I get lost. Clearly the sum on the left is $p-1$, then I have:
$$= p-1 + \sum_{a=1}^{p-1}\sum_{b \in \left(\mathbb{Z} / p\mathbb{Z} \right)^\times, b\neq a} \chi(ab^{-1}) \zeta^{a-b}$$
Let $cb = a$, then :
$$= p-1 + \sum_{c=2}^{p-1} \chi(c) \hspace{2pt} \sum_{b \in \left(\mathbb{Z} / p\mathbb{Z} \right)^\times} \zeta^{b\left(c-1\right)} $$
I am unsure if this last step is correct, but if it is, then:
$$\sum_{b \in \left(\mathbb{Z} / p\mathbb{Z} \right)^\times} \zeta^{b\left(c-1\right)} = -1$$
Which gives me:
$$ = p-1 – \sum_{c=2}^{p-1} \chi(c)$$
In this case, I do not know why $\sum_{c=2}^{p-1} \chi(c) = -1$, which would give me the correct answer. In any case, I feel as though I probably made a mistake along the way. Can anyone help me with this? I am not completely comfortable yet with manipulating sums like this. I also lack an understanding in the algebraic concepts behind what this all means. Maybe I am missing a crucial piece of information.
Any help would be very much appreciated!
Best Answer
I figured I'd come back to this and expand my comment into an answer.
Everything you've done so far looks good. You just need to prove that $$\sum_{i=2}^{p-1} \chi(i) = -1,$$ or, equivalently, $$\sum_{i=1}^{p-1} \chi(i) = 0.$$
To prove this last equality, note that the set of integers from $1$ to $p-1$ is a set of representatives for $(\Bbb{Z}/p\Bbb{Z})^\times$. Thus if $a\in(\Bbb{Z}/p\Bbb{Z})^\times$, then $\{ai : 1\le i \le p-1\}=\{i : 1\le i \le p-1\}$, so $$\chi(a)\sum_{i=1}^{p-1} \chi(i) = \sum_{i=1}^{p-1} \chi(a)\chi(i) = \sum_{i=1}^{p-1} \chi(ai) = \sum_{i=1}^{p-1} \chi(i), $$ and since $\chi$ is nontrivial, there is some $a\in (\Bbb{Z}/p\Bbb{Z})^\times$ such that $\chi(a)\ne 1$. Thus, this equality tells us that when $\chi$ is a nontrivial character, $$\sum_{i=1}^{p-1} \chi(i)=0,$$ as desired.