Gauss like sum evaluation / estimate Let $\mathbb{F}_{p^2}$ be the finite field of cardinality $p^2$, $\chi$ be a (multiplicative) character of $\mathbb{F}_{p^2}$ and $e(.)=e^{2\pi i .}$. What is the evaluation of the following two sums,
$$\sum_{b\in\mathbb{F}_{p^2}}e\left(\frac{a Norm(b)}{p}\right)$$
and
$$\sum_{b\in\mathbb{F}_{p^2}}\chi(b)e\left(\frac{a Norm(b)}{p}\right),$$
where $a\in \mathbb{F}_p$. In general, what is the evaluation/estimate if $\mathbb{F}_{p^2}$ is replaced by another finite field?
 A: The first sum is easy. We need the following basic properties of the norm map
$$
N:\Bbb{F}_{p^2}\to\Bbb{F}_p, N(z)=z^{p+1}.
$$


*

*$N(0)=0$ and $N(z)\neq0$ whenever $z\neq0$.

*Every element $a\in\Bbb{F}_p,a\neq0$, is the norm of exactly $p+1$ elements of $\Bbb{F}_{p^2}.$


I write $e(\cdot)$ in place of $e(\frac{\cdot}p)$ because that $p$ is always the same. It is well known that $\sum_{z\in\Bbb{F}_p}e(z)=0$, so the sum
$\sum_{z\in\Bbb{F}_p,z\neq0}e(z)=-1$. Using these
we get
$$
\begin{aligned}
\sum_{b\in \Bbb{F}_{p^2}}e(aN(b))
&=e(0)+(p+1)\sum_{z\in\Bbb{F}_p,z\neq0}e(az)\\
&=\begin{cases}1+(p+1)\cdot(-1)=-p,&\ \text{if $a\neq0$, and}\\
1+(p+1)(p-1)=p^2,&\ \text{if a=0.}
\end{cases}\\
\end{aligned}
$$

By the same process
$$
\sum_{b\in\Bbb{F}_{p^2}}\chi(b)e(a N(b))
=\chi(0)+\sum_{z\in\Bbb{F}_p,z\neq0}e(az)\sum_{b\in\Bbb{F}_{p^2},N(b)=z}\chi(b).
$$
Here the inner sum
$$
S(z):=\sum_{b\in\Bbb{F}_{p^2},N(b)=z}\chi(b)
$$
can be analyzed. The set of, call it $G_z$, of elements $b\in\Bbb{F}_{p^2}$ such that $N(b)=z$ is a coset of the subgroup $G_1=\operatorname{Ker}(N)$ (cyclic of order $p+1$). If the restriction of $\chi$ to $G_1$ is not constant, then all those inner sums are automatically zero. On the other hand, if $\chi(G_1)=\{1\}$, then $S(z)=(q+1)\chi(b_z)$, where $b_z$ is any element of $\Bbb{F}_{p^2}$ such that $N(b_z)=z$. 
In this remaining case the mapping $z\mapsto \chi(b_z)$ is a multiplicative character, call it $\tilde\chi$, of $\Bbb{F}_p$. This is because for any elements $z,z'\in\Bbb{F}_p$ we can use $b_zb_{z'}$ is place of $b_{zz'}$. Here the quotient
$b_{zz'}/(b_zb_{z'})$ is in the subgroup $G_1$, and we now assume that $\chi(G_1)=\{1\}$. Therefore $\chi(b_{zz'})=\chi(b_z)\chi(b_{z'})$.
Hence we can conclude that your sum
$$
\sum_{b\in\Bbb{F}_{p^2}}\chi(b)e(a N(b))
=\begin{cases}\chi(0),&\ \text{if $\chi_{\vert G_1}$ is non-trivial, and}\\
\chi(0)+(q+1)\sum_{z\in\Bbb{F}_p}e(az)\tilde{\chi}(z),&\ \text{if $\chi(b)=1$ whenever $N(b)=1$.}
\end{cases}
$$
In the last form we clearly have a Gauss sum of the prime field with known absolute value $\sqrt p$ except in the trivial cases $a=0$ or $\tilde{\chi}=\chi_0$.
