6

I need to know the value of $\sum_{x \in \mathbb{F}_q} \chi(x^3 - x)$, where $q = p^r$ with $p$ prime and $\chi$ is the Jacobi symbol.

If we take $-x$, we have $\chi((-x)^3 - (-x)) = \chi(-1)\chi(x^3 - x)$ and $\chi(-1) = -1$ since $q \equiv 3 \; (mod \; 4)$. Thus, the terms for $x$ and the terms for $-x$ cancel. Therefore, $\sum_{x \in \mathbb{F}_q} \chi(x^3 - x) = 0$.

But, if $q$ is even or $q \equiv 1 \; (mod \; 4)$, then $\chi(-1) = 1$.

How can I find the value of $\sum_{x \in \mathbb{F}_q} \chi(x^3 - x)$ when $q$ is even or $q \equiv 1 \; (mod \; 4)$?

EvaMGG
  • 506
  • 2
    "$\chi(-1)=-1$ since $x\equiv 3\pmod{4}$"? Anyway, if $q$ is even, every element is a square (Frobenius). – user10354138 Feb 18 '21 at 00:47
  • 2
    The case of an even $q$ is not interesting, because you don't have an elliptic curve in that case. And, as @user10354138 explained, every element of the field has a unique square root, so the number of points is equal to $q+1$. The observations we can immediately make are A) the Hasse-Weil bound, B) the cubic splits modulo any prime so the elliptic curve has full 2-torsion, and hence the number of points is a multiple of four. – Jyrki Lahtonen Feb 18 '21 at 13:02
  • 1
    We also have the Hasse-Davenport relation (explaining how the zeros of the $L$-function behave under extensions of scalars). Implying that it suffices to handle the prime fields. We quickly see that with $p=13,17,29,37$ we have $8,16,40,40$ points respectively. I don't see a pattern. If you can find the rank of the rational variant (surely done in LMFDB), then there may be a tendency related to B-SD conjecture. But that's above my paygrade. – Jyrki Lahtonen Feb 18 '21 at 13:09
  • So if $q=p^n$, $p\equiv-1\pmod4$, $n$ even, then the symmetry you used and Hasse-Davenport settle the matter. – Jyrki Lahtonen Feb 18 '21 at 13:11
  • 2
    For $p\equiv 1\pmod 4$, write $p=\pi\overline{\pi}$ in ${\bf Z}[i]$. Precisely one of $\pm\pi,\pm i\pi$ is congruent to $1\pmod{2+2i}$. Call that one $\pi$ again. Then $\sum_{x\in{\bf F}_p}\chi(x^3-x)$ equal to $-\pi-\overline{\pi}$. – Rene Schoof Feb 18 '21 at 14:37
  • @ReneSchoof Wow! Why are the zeros of the $\zeta$-function of this curve in $\Bbb{Z}[i]$? Is that some CM argument? – Jyrki Lahtonen Feb 19 '21 at 06:07
  • 1
    In general $E(\mathbb{F}_{p^r}) = p^r +1 - \alpha^r - \beta^r$ where $\alpha$ and $\beta$ are the roots of the minimal polynomial of the $p$-power Frobenius endomorphism, which is of the form $T^2 - a T + p$ for some $a$ (the trace of Frobenius). Since $E$ has CM, this trace is related to the trace of the Hecke character of the CM field, as described for instance in this answer. Hopefully someone more familiar with CM can connect the dots! – Viktor Vaughn Feb 19 '21 at 06:54
  • 1
    This is actually done in Ireland and Rosen's Classical Introduction to Number Theory, Ch. 18, $\S4$, Theorem $5$ (p. 307), just using Jacobi sums and reciprocity. – Viktor Vaughn Feb 19 '21 at 07:14

0 Answers0