Timeline for zeros of a homogeneous polynomial

6 events

when toggle format	what		by	license	comment
Jun 1, 2013 at 21:13	vote	accept	user34548
May 31, 2013 at 14:05	vote	accept	user34548
May 31, 2013 at 14:05
May 30, 2013 at 22:55	comment	added	François Brunault		(Then $4\lambda_0^3+27$ is automatically nonzero, since it is the discriminant of an irreducible cubic polynomial.)
May 30, 2013 at 22:50	comment	added	François Brunault		The number of $(t,\lambda)$ such that $t^3+\lambda t+1=0$ is at most $q-1$, hence there must exist $\lambda_0 \in \mathbf{F}_q$ such that the polynomial is irreducible.
May 30, 2013 at 22:35	comment	added	Noam D. Elkies		... and that has to happen once $p$ is large enough: the discriminant $-(4\lambda^3+27)$ must be a nonzero square, which happens for $p/2+O(p^{1/2})$ choices of $\lambda$, and then the polynomial must not factor completely, which happens for $p/6+O(p^{1/2})$ choices (corresponding to points on the elliptic curve $t_1+t_2+t_3=0$, $t_1 t_2 t_3 = -1$ where the $t_i$ are the roots, up to permutation and minus the few points where the $t_i$ are not pairwise distinct).
May 30, 2013 at 22:08	history	answered	Abhinav Kumar	CC BY-SA 3.0