Generalised Gauss sums

Question

Let $\chi$ be a non-trivial Dirichlet character modulo an odd prime $p$ and let $f(x) \in \mathbb{Z}[x]$ be a polynomial. We define the generalised Gauss sum $$ G(\chi, f):=\sum_{y \in \mathbb{F}_p^*} \chi(y) \left(\frac{f(y)}{p}\right).$$ Under which conditions for $f$ can we prove that $$ G(\chi,f)\ll \deg(f) \sqrt{p},$$ with an absolute implied constant ?

Answer 1

Let $\eta$ be a generator of the character group $\widehat{\Bbb{Z}_p^*}$. Therefore $\chi=\eta^k$ for some integer $k$, $0<k<p-1$ and the Legendre symbol is equal to $\eta^{(p-1)/2}$. Therefore $$ G(\chi,f)=\sum_{y\in\Bbb{F}_p}\eta(y^kf(y)^{(p-1)/2}). $$

The general Weil bound for multiplicative character sums says that the sum $$ S(g)=\sum_{y\in\Bbb{F}_p}\eta(g(y)) $$ is in the non-trivial cases bounded by $$ |S(g)|\le (d-1)\sqrt p, $$ where $d$ is the number of zeros of $g(y)$ in its splitting field over $\Bbb{F}_p$. The sum is trivial, if $g(y)$ is of the form $g(y)=c h(y)^{p-1}$ for some $c\in\Bbb{F}_p$ and $h(y)\in\Bbb{F}_p[y]$.

Obviously the number of zeros of $g(y)$ is bounded from above by $\deg g$. In your case the number of zeros of $y^kf(y)^{(p-1)/2}$ is bounded from above by $\deg f+1$ (add one for the power of $y$ factor, unless $f(0)=0$). Therefore

When your sums are non-trivial they are bounded as guessed. With absolute constant $1$ and without the logarithmic factor.

The sum is trivial, if $y^kf(y)^{(p-1)/2}$ is a constant times a $(p-1)$th power of a polynomial (modulo $p$). This happens for example when $f(y)=y$ and $\chi$ is also the Legendre character.