Absolute value of a cubic Gauss sum (over Field $\mathbb{F}_p$ )

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?

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.
• 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. – Mark Aug 23 '14 at 17:25
• @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. – Jyrki Lahtonen Aug 23 '14 at 17:49