Let $C$ be the cubic curve $ y^2 = x^3 + ax^2 + bx + c $ with singular point $S$. Show that $S$ must have rational coordinates.

Question

This is exercise 3.14 of Silverman's book Rational points on elliptic curves.

Let $C$ be the cubic curve given by the equation $$ y^2 = x^3 + ax^2 + bx + c $$

with $a,b,c \in \mathbb{Q}$. Suppose that $C$ is singular and let $S=(x_0, y_0)$ be the singular point.

(a) Prove that $x_0$ and $y_0$ are in $\mathbb{Q}$.

Attempt:

By definition of singular point, the singular point means that $f'(x_0)=0$ and $2y_0=0$. So $x_0$ is a root of $3x^2+2ax+b=0$ and also a double root of $x^3+ax^2+bx+c=0$. Then I factor $x^3+ax^2+bx+c=(x-x_0)^2(x-x_1)$. But then I am stuck here, because I don't know how to show that $x_0$ must be rational.

Answer 1

$x_0$ is a zero of both $f(x)=x^3+ax^2+bx+c$ and of $f'(x)$. So $x_0$ is a zero of $g(x)=\gcd(f(x),f'(x))$. One can do Euclidean algorithm for polynomials over $\Bbb Q$ to see that $g(x)$ has coefficients in $\Bbb Q$.

Either $g(x)=x-x_0$ (when $x_0\ne x_1$) or $g(x)=(x-x_0)^2$ (when $x_0=x_1$). Each way, we get $x_0\in\Bbb Q$.