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

  • $\begingroup$ 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. $\endgroup$ – ms08030 Jan 8 '18 at 22:20
  • $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen Jan 9 '18 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.