Let $\mathbb{F}_q$ be a finite field with $q$ elements, let $\chi$ be the canonical additive character of $\mathbb{F}_q$, let $\psi$ be a non-trivial multiplicative character of $\mathbb{F}_q$, and let $f \in \mathbb{F}_q[x]$ be a polynomial. Is an upper bound known for the following general character sum, say in the style of Weil's bound? Thanks! $$ \left| \sum_{x \in \mathbb{F}_q^*} \chi(f(x)) \psi(x) \right| $$

  • 1
    $\begingroup$ Yes, look at the following Springer Lecture Note: W. M. Schmidt, Equations over finite fields: An elementary approach. Springer, Berlin (1976). (I used to have this book in my library until a few years ago....) $\endgroup$ – Dilip Sarwate Dec 19 '14 at 4:14
  • $\begingroup$ Schmidt's book is very useful (our library copy is also mysteriously misplaced). Lidl & Niederreiter don't go into details here even though they explain the use of L-functions IMO a bit better. I benefitted from studying Michael Rosen's book in the sense that there is a more number theoretical account there. Unfortunately he doesn't go into details on the character sums. But you will recognize that those mysterious subgroups of the multiplicative group of $\Bbb{F}_q(x)$ are related to ray classes. $\endgroup$ – Jyrki Lahtonen Dec 19 '14 at 13:27
  • $\begingroup$ Rosen does give Bombieri's version of the Schmidt-Stepanov method of proving the Riemann hypothesis for function fields. (If you are a coding theoretically minded, then a similar account is in Stichtenoth's book). Combining this with what you learn about L-functions from the other listed sources will go a long way. For example, the Shanbhag-Kumar-Helleseth extensions I refer to in my answer become "straightforward" exercises. $\endgroup$ – Jyrki Lahtonen Dec 19 '14 at 13:30

Yes, there is such a bound. Below I list a few variants. To avoid trivial cases we need to assume that $f(x)$ is not of the form $h(x)^p-h(x)+r$ for any polynomial $h(x)$ and any constant $r$. I also include sums where the multiplicative also has a polynomial argument. Then non-triviality condition takes the form that if $\psi$ is of order $d$ the polynomial $g(x)$ should not be of the form $r h(x)^d$, again for any constant $r$ and any polynomial $h(x)$. The definition of a non-trivial multiplicative character is often extended by declaring that $\psi(0)=0$.

A general bound for hybrid sums with polynomial arguments is $$ \left\vert\sum_{x\in\Bbb{F}_q}\chi(f(x))\psi(g(x))\right\vert\le (\deg f+\deg g-1)\sqrt{q}. $$
The sums you specifically asked about are a special case of this with $g(x)=x$ of degree $1$.

Occasionally useful generalizations involve Laurent polynomials and read $$ \left\vert\sum_{x\in\Bbb{F}_q}\chi(f_1(x)+f_2(\frac1x))\right\vert\le (\deg f_1+\deg f_2)\sqrt{q} $$ and $$ \left\vert\sum_{x\in\Bbb{F}_q}\chi(f_1(x)+f_2(\frac1x))\psi(x)\right\vert\le (\deg f_1+\deg f_2)\sqrt{q}. $$

Other generalizations have been developed for analogous sums over Galois rings. Ask if you need to know more. Shanbhag, Kumar and Helleseth extended the calculations from Schmidt's book (see Dilip Sarwate's comment) to cover those cases. Wei-Ching (Winnie) Li gave more class field-theoretical proofs and some generalizations (such as cases where the summation is only over a subfield).


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.