Elliptic Curves and "roots"

Question

Given elliptic curve $\omega$ in $\mathbb{R}^2$ such that $y^2 = x^3 + ax + b$, how can you find how many solutions (and what they are) of $x^3+ax+b$ have a $y$ value of $0$; or as they call it, a homogeneous equation?

Answer 1

If you're asking for the number of solutions of $x^3 + ax + b = 0$, where $y^2 = x^3+ax+b$ describes an elliptic curve, then the number of complex solutions is necessarily 3, and the number of real solutions is either 1 or 3.

The fact that the number of complex solutions is 3 follows from the fact that the curve is nonsingular iff the discriminant of $x^3+ax+b$ is nonzero (ie, all 3 roots are distinct). This is described in Silverman's "Arithmetic of Elliptic Curves" in the chapter describing weierstrass equations.

Whether there are 1 or 3 real roots can be deduced by analyzing the famously complicated cubic formula.

Answer 2

If you are given curve $\omega$ defined with $y^2 = x^3+ax+b$, and you want to find the roots, recall this equation is cubic (degree 3). First we must find the discriminant, which in a cubic equation $ax^3+bx^2+cx+d$ works out to $b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$. When plugged in with an elliptic curve, you get $4b^3-27a^2$. Now, in a cubic equation, if the discriminant is less than 0, there is one solution, but if it is greater than, there is three (because the discriminant of an elliptic curve is non-zero). Now we may define how many solutions there are, $s$.

If $4b^3-27a^2<0$ then $s = 1$

If $4b^3-27a^2>0$ then $s = 3$