The following question is from Carlos Moreno's "Algebraic Curves over Finite Fields":
I am trying to solve part (i), but I am keep getting stuck at roughly the same place. Write $C: y=x^d$ a curve over $\mathbb{F}_p$, the sum can be written as (with $e(x)=e^{2\pi ix}$)
$$ \sum_{x,y\in\mathbb{F}_q} e(ay/p)e(bx/p)1_{x,y\in C} = (*) $$
Now write $1_{x,y\in C}=1_{y\in\mathbb{F}_p^{\times d}}1_{x\in\text{roots}_d(y)}$. We get
$$ (*) = \sum_y e(ay/p)1_{y\in\mathbb{F}_p^{\times d}}\sum_x e(bx/p)1_{x\in\text{roots}_d(y)} $$
Denote the inner sum by $\phi(y)$, and write $H$ the group of characters orthogonal to the $d$-powers. We have then $1_{y\in\mathbb{F}_p^{\times d}} = \frac{1}{\# H}\sum_{\chi\in H}\chi(y)$. Then we can rewrite the sum as
$$ \frac{1}{\# H}\sum_{\chi\in H}\sum_y \chi(y)e(ay/p)\phi(y) $$
Now $phi(y)$ is of size at most $d$, and for each $\chi$, $|\sum_y \chi(y)e(ay/p)|=\sqrt{p}$ (it is a Gauss sum). But there might be some cancellations, so we can't just apply the triangle inequality. I tried to bypass it by factoring father $\phi(y)=\sum_{k=1}^p \hat{\phi}(k)e(ky/p)$, and then for each $\chi$ the sum is
$$ \sum_y \chi(y)e(ay/p)\sum_{k=1}^p \hat{\phi}(k)e(ky/p)=\sum_{k=1}^p\hat{\phi}(k)\sum_y e((a+k)y/p)\chi(y) $$
The inner sum is of size $p$, and we can bound the outer sum by the $L_1$ norm of $\hat{\phi}$, which is sadly not helpful. If we could somehow replace it with the $L_2$ norm of it we and square on the left-hand side of the equation this would solve it, but I don't know if it is possible.
Is there hope for this direction? Can anyone suggest a proof? Thanks in advance!