Let $p \neq 2$ be a prime, let $a,b,c,d \in \mathbb{F}_p$ satisfy $acd \neq 0$, and let $C$ be the conic given by the homogeneous equation $$ C : ax^2 + bxy + cy^2 = dz^2. $$

a) If $b^2 \neq 4ac$, prove that $\#C(\mathbb{F}_p)=p+1$.

b) If $b^2 = 4ac$, prove either $\#C(\mathbb{F}_p) = 1$ or $2p+1$. Give examples for $p = 3$ to show that both possibilities can occur. More generally, show that both possibilities occur for all odd primes.

I don't really know where to begin with a question like this. I have tried a few examples where $p =3$ and $a=b=c=d = 1$ but I don't really have any intuition. I'm using Silverman and Tate's Rational Points on Elliptic Curves.

  • 1
    $\begingroup$ Hint $b^2=4ac$ means that $ax^2+bxy+cy^2=A(Bx+Cy)^2$, If you need another hint, do ping! $\endgroup$ – Jyrki Lahtonen Mar 14 at 20:42

I would try to aim for the following intuition: Your conic is either non-degenerate or degenerate, which is what your discriminant decides. If it is non-degenerate, then over the reals you'd get something that is topologically equivalent to a projective line. You can define a projective basis on it and describe every point in a unique fashion with respect to that basis. All of this should carry over to finite fields just fine.

General idea is that if you take a point on your conic then there are $p+1$ lines through that point. Each of them will intersect the conic in one well-defined other point, except for one line, the tangent, where that second point is in fact the same one you started from. So the $p+1$ lines through the point correspond $1:1$ to the $p+1$ points on the conic.

For the degenerate case over the reals, you might want to distinguish three cases. Either your conic is a pair of real projective lines, or a single line with algebraic multiplicity $2$, or a pair of complex conjugate lines which intersect in a single real point. Transported to finite fields, the first of these cases would yield $2(p+1)-1=2p+1$ points since the point of intersection is only counted once. The third case yields $1$ point. The second case again yields $p+1$ points, so at first glance I would expect that to be an option as well. But since your formula is not fully general due to the absence of mixed terms for $xz$ and $yz$, it can't express duplicate lines while maintaining $d\neq 0$.


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.