1
$\begingroup$

I'm interested in the quantity $|\sum_{x \in \mathbb{F}_p} \omega^{ax^3+bx^2+cx}|$ where $a \in \mathbb{F}_p^*,b,c \in \mathbb{F}_p$ and $\omega$ is a primitive $p$-th root of unity i.e. $\omega=e^{\frac{2 \pi i}{p}}$.

I know that the case $a=0$ is well understood and evaluates to $\sqrt{p}$ (quadratic Gauss sum).

When $a\neq 0$ I have observed numerically that, for fixed $a$, the set $$ \mathcal{S}_b:= \{|\sum_{x \in \mathbb{F}_p} \omega^{ax^3+bx^2+cx}|, c \in \mathbb{F}_p \} $$ obeys $$ \mathcal{S}_b = \mathcal{S}_{b^\prime\neq b}$$

In other words, it appears that the quadratic coefficient is somehow unimportant for my purposes.

Question: (i) Is there some way of manipulating my expression to see why this property holds? or, (ii) Is there a relevant reference that states something along these lines?

$\endgroup$
1
$\begingroup$

Let us assume that $\Bbb{F}_p$ does not have characteristic three (so $p>3$ if you are, as seems to be the case, only interested in prime fields). Denote $$ T(a,b,c)=\sum_{x\in\Bbb{F}_p}\omega^{ax^3+bx^2+cx}. $$ Assume that $a\neq0$. Let $z\in\Bbb{F}_p$ be a parameter. For a fixed $z$ as $x$ ranges over $\Bbb{F}_p$ so does $x+z$. Thus $$ \begin{aligned} T(a,b,c)&=\sum_{x\in\Bbb{F}_p}\omega^{a(x+z)^3+b(x+z)^2+c(x+z)}\\ &=\sum_{x\in\Bbb{F}_p}\omega^{[ax^3+(b+3az)x^2+(c+2bz+3az^2)x]+[az^3+bz^2+cz]}\\ &=\omega^{az^3+bz^2+cz}T(a,b+3az,c+2bz+3az^2). \end{aligned} $$ In particular (any power of $\omega$ has absolute value $=1$) $$ |T(a,b,c)|=|T(a,b+3az,c+2bz+3az^2)|. $$ Here $3a$ is an invertible element of $\Bbb{F}_p$, so by varying $z$, the second parameter on the right hand side, $b+3az$, ranges over the entire field. The particular choice $z_0=-b/3a$ makes it zero giving us $$ |T(a,b,c)|=|T(a,0,c+2bz_0+3az_0^2)|. $$ Fixing $a,b$ (and thus also $z_0$) and varying $c$ shows then that $$ S(b)=S(0), $$ because $c$ ranges over the elements of $\Bbb{F}_p$ as $c+2bz_0+3az_0^2$ does.


This kind of linear substitution tricks are common in manipulating character sums over finite fields. A more general linear substitution would allow you to replace $a$ with any other element $a'$ such that $a'/a$ is a non-zero cube.

$\endgroup$
  • $\begingroup$ Thank you for an extremely helpful answer. The only thing I don't understand is the very last sentence. When you use the variable $a$, am I correct that this is just a generic choice and not related to the above problem? In that case your criterion applied to the original problem is that $\frac{x+z}{x}$ is a non-zero cube, which I don't think makes sense as cubes form a subset of $\mathbb{F}_p$ for $p=1 \bmod 3$. I have another question on mathSE (related to this) for which a clear understanding of such "tricks" would be very helpful. $\endgroup$ – Mark Aug 23 '14 at 17:25
  • $\begingroup$ @Mark: What I was suggesting is that if we denote $$S(a,b)=\{|T(a,b,c)|\mid c\in\Bbb{F}_p\},$$ then in addition to the above result $S(a,b)=S(a,0)$ we should also get $S(a,0)=S(ag^3,0)$ for any non-zero $g\in\Bbb{F}_p$. As you pointed out, if $3\nmid p-1$ then the end result should be that $S(a,b)=S(1,0)$ for all $a\neq0,b$, because every element is a cube. But if $p\equiv 1\pmod3$ it may be more complicated. $\endgroup$ – Jyrki Lahtonen Aug 23 '14 at 17:49

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.