Let $F$ be a finite field of characteristic a prime $p$ with $q$ elements and let $E/F$ be a finite extension of degree $n > 1$ over $F$. Let $\chi$ be the additive canonical character on $F$ defined by $\chi(x) = e^{2\pi \sqrt{-1}T(x)/p}$, where $T : F \to \mathbb{F}_p$ is the ($\mathbb{F}_p$-linear surjective) trace map $T(x) = \sum_{i=0}^{n-1} x^{p^i}$.

The class of polynomials $f : E \to F$ that I'm interested in have the following two properties:

(1) $f$ is surjective;

(2) There exists an integer $m$ such that $f(cx) = c^m f(x)$ for every $c \in F$.

Question: Is it known a good upper bound for $$ S = \left| \sum_{x \in E} \chi(f(x))\right|? $$

Remark: Of course we have a general upper bound for these sums, the so called Weil bound $S \leq (\deg(f) - 1)q^{n/2}$. However from some computer experiments I have tried it seems possible this should be a lot lower. In particular when $\gcd(m, q-1) = 1$, it seems that $S = 0$ always. This should have to do with the fact that the mapping $x \mapsto x^m$ permutes $F$ if and only if $\gcd(m, q-1) = 1$. In any case I think there should be some dependency of $S$ on $\gcd(m, q-1)$. Hopefully someone has seen these things before! Thanks!

  • 1
    $\begingroup$ if $f(x)=\sum_{i=1}^{Q-1} f_ix^i$, $Q=q^n$, then the condition (2) implies that $i\equiv m\pmod{q-1}$ whenever $f_i\neq0$. The condition $f:E\to F$ implies that $f_i=f_{qi}$, where $qi$ is calculated modulo $Q-1$. Don't know about condition (1). Anyway, we can typically then write $T(f(x))=T^E_F(g(x)$, where $T^E_F:E\to F$ is the relative trace, and $g(x)$ is a polynomial consisting of those terms of $f(x)$, where we select the lowest degree term from each $q$-cyclotomic coset. Typically $\deg g$ is much less than $\deg f$. There may be problems with cyclotomic cosets having $<n$ numbers. $\endgroup$ – Jyrki Lahtonen Jul 31 '15 at 13:21
  • $\begingroup$ Instead of writing $T(f(x)) = T_F^E(g(x))$, don't you mean that we must write $f(x) = T_F^E(g(x))$ instead? This way $S$ becomes $\sum_{x \in E} \psi(g(x))$ where $\psi$ is canonical additive character of $E$ (by the transitivity of the trace function). Then we could hopefully apply Weil's bound to a better version. This I think is called the trace representation of a function, the coefficients of $g(x)$ being Fourier coefficients of $f(x)$. $\endgroup$ – user152169 Aug 2 '15 at 20:21
  • $\begingroup$ Still I'm not sure how this could prove a better bound, although, as you say, the degree of $g(x)$ obtained through this method could be often low. $\endgroup$ – user152169 Aug 2 '15 at 20:27
  • $\begingroup$ Oops. Yes, that's what I was thinking, but writing... Thinking about the rest. What kind of polynomials did you check? $\endgroup$ – Jyrki Lahtonen Aug 2 '15 at 20:49
  • 1
    $\begingroup$ Another possibility is that when $\gcd(m,q-1)$, and $c$ ranges over $F^*$ the elements $f(cx)=c^mf(x)$ do the same unless they are all zero. This gives us a lot of cancellation in the sum, but there may be problems with those zeros (also because the sum over $F^*$ gives a non-cancelled $-1$. $\endgroup$ – Jyrki Lahtonen Aug 2 '15 at 20:52

I have some doubts now that $$ S_m := \sum_{x \in E} \chi(f_m(x)) = 0 $$ whenever $(m, q-1) = 1$ in general, but nevertheless here are some calculations that might be of help in certain cases.

For now let $m$ be any integer. Clearly for any $k \in E^*$, $x \mapsto kx$ is a permutation of $E$. Let $e(z) := e^{2\pi \sqrt{-1}z}$. Thus \begin{align} (p-1) S_m &= \sum_{c \in \mathbb{F}_p^*}\sum_{x \in E} \chi(f_m(x)) = \sum_{c \in \mathbb{F}_p^*}\sum_{x \in E} \chi(f_m(cx)) = \sum_{x \in E}\sum_{c \in \mathbb{F}_p^*}\chi(f_m(cx))\\ &= \sum_{x \in E}\sum_{c \in \mathbb{F}_p^*}\chi(c^m f_m(x)) = \sum_{x \in E} \sum_{c \in \mathbb{F}_p^*} \chi(f_m(x))^{c^m} = \sum_{x \in E} \left( -1 + \sum_{c \in \mathbb{F}_p} e\left( \dfrac{T(f_m(x)) c^m)}{p} \right) \right)\\ &= -q^n + p N_m + \sum_{x \in E \setminus K_m} \sum_{c \in \mathbb{F}_p} e\left( \dfrac{T(f_m(x)) c^m)}{p} \right) \end{align} where $$K_m := \{x \in E \mid f_m(x) \in \ker(T)\}$$ and $N_m := \#K_m$.

Some things to notice now.

(1) Note that
$$ N_m = \dfrac{1}{p}\#\{ (x, y) \in E \times F \mid f_m(x) = y^p - y \}. $$

(2) In the case that $(m, p-1) = 1$ (which happens if $(m, q-1) = 1$) we have $$ G_m := \sum_{c \in \mathbb{F}_p} e\left( \dfrac{T(f_m(x)) c^m)}{p} \right) = 0. $$ Thus $S_m \in \mathbb{Q}$ here.

Moreover, when $(m, p-1) = 1$, we have $S_m = 0$ if and only if $$ \#\{ (x, y) \in E \times F \mid f_m(x) = y^p - y \} = \#E := q^n. $$ I believe $G_m$ is known also when $m = 2$ or $m = 3$ (quadratic and cubic Gauss sums) but it seems that a good upper bound is not known in the general case of $m$. See Section 6.1 here:


(3) Assume $(m, q-1) = 1$. Further assume that $p \nmid n$. It follows that \begin{align} (q - 1)S_m &= \sum_{c \in F^*} \sum_{x \in E} \chi(f_m((cn)^{1/m} x)) = \sum_{c \in F^*} \sum_{x \in E}\chi(c nf_m(x)) = \sum_{c \in F^*} \sum_{x \in E}\chi( T_F^E (cf_m(x))) \\ &= \sum_{x \in E} \sum_{c \in F^*} \psi(cf_m(x)), \end{align} where $\psi$ is the canonical additive character of $E$. By the orthogonality relations it follows that $$ (q-1)S_m = -q^n + q \cdot \#\{x \in E \mid f_m(x) = 0\}. $$ Thus we get $$ (q - p)S_m = q \cdot \#\{x \in E \mid f_m(x) = 0\} - p \cdot \#\{x \in E \mid f_m(x) \in \ker(T_{q|p})\}. $$ If $q \neq p$, then $S_m = 0$ if and only if $$ \#\{x \in E \mid f_m(x) \in \ker(T_{q|p})\} = \dfrac{q}{p} \cdot \#\{x \in E \mid f_m(x) = 0\}. $$


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.