# 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?

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 $\tilde{\chi}=\chi_0$.

• Although I don't see why restriction of $\chi$ to $G_1$ not constant implies the inner sum is zero. Note that the inner sum is over $b\in\mathbb{F}_{p^2}, N(b)=z$, so orthagonality does not apply. – ms08030 Jan 8 '18 at 22:20
• @ms08030 The idea is that $G_z$ is the coset $b_zG_1$. Therefore $$S(z)=\sum_{b\in\Bbb{F}_{p^2},N(b)=z}\chi(b)=\sum_{x\in G_1}\chi(b_zx)= \chi(b_z)\sum_{x\in G_1}\chi(x).$$ That last sum then vanishes whenever the restriction of $\chi$ to $G_1$ is non-trivial. – Jyrki Lahtonen Jan 9 '18 at 4:54