Please help or hints me to solve this question:

Suppose $p>2$ be a prime and $q=p^e$ for some integer $e$ and $f(x) \in \Bbb F_q[x]$.

i) Show that the roots of the equation $1‎\pm‎ f^\frac{q-1}{2} =0$ in $\Bbb F_q$ are multiple root of the equation $R(x) = 2f(x)(1 \pm f^\frac{q-1}{2})+ f^\prime (x) (x^q-x)$.

ii) Then conclude the number $N_q$ that representing the number of $\Bbb F_q$-rational points on the curve $E: y^2= x^3+ax+b$ where $a,b \in \Bbb F_q$, is in the equation $|N_q - q|\leq‎ \dfrac{q+3}{2}‎$ applies. Any suggestion would be appreciated.


Collecting the key steps:

  1. If $R(x)=2f(x)\left(1-f(x)^{(q-1)/2}\right)+f'(x)(x^q-x)$ show that $$R'(x)=2f'(x)\left(1-f(x)^{(q-1)/2}\right)+f''(x)(x^q-x).$$ Conclude that if $x_0\in\Bbb{F}_q$ is such that $f(x_0)^{(q-1)/2}=1$ then $R(x_0)=R'(x_0)=0$.
  2. Warning: Unless I made a mistake you really want to study (with upper/lower signs matching) $$R_{\pm}(x):=2f(x)(1\mp f(x)^{(q-1)/2})\pm f'(x)(x^q-x).$$
  3. Recall that $y^2=z\in\Bbb{F}_q$ has two solutions $y\in\Bbb{F}_q$, when $z^{(q-1)/2}=1$, no solutions when $z^{(q-1)/2}=-1$ and a single solution when $z=0$. Also recall that exactly one of these cases occurs for all $z$.
  4. Apply all of the above with $z=x^3+ax+b=f(x)$. The choices of $x$ that give two solutions for $y$ are double roots of $R_+$ and the choices of $x$ that give no solutions for $y$ are double roots of $R_-$. Give upper bounds to the number of choices resulting in either situation by studying the degrees of $R_{\pm}$.
  • $\begingroup$ @ Jyrki Lahtonen: The degree of $R(x)$ is $ 3/2(q-1)$. Is this true? It is useful? $\endgroup$ – Masoud May 27 '17 at 11:51
  • $\begingroup$ You get (I stopped after that point) by simple counting $$N_q\le \deg f(x)+\deg R(x).$$ Actually I'm no longer sure about the lower bound, nor about the constant! Anyway the $\pm$ variants are kind of complementary to each other. If $g$ is a fixed non-square of $\Bbb{F}_q$, the equations $y^2=f(x)$ and $y^2=gf(x)$ have exactly $2q$ solutions between them. So any upper bound to $N_q-q$ comes together with a loweer bound. $\endgroup$ – Jyrki Lahtonen May 27 '17 at 14:31

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.