If the cubic polynomial with rational coefficients $x^3+ax^2+bx+c$ has a double root, the root is rational.

Question

This question comes from a problem in Rational Points on Elliptic Curves by Silverman and Tate. The problem asks to show that

For a cubic curve $C: y^2=x^3+ax^2+bx+c$ with $a,b,c\in \mathbb{Q}$, if $S=(x_0, y_0)$ is a singular point, then $(x_0,y_0)$ is rational.

It turns out that a singular point $(x_0, y_0)$ must satisfy $$y_0=0, -3x_0^2-2ax_0-b=0.$$ In other words, a point $(x_0,y_0)$ is singular on this cubic curve if and only if $y_0=0$ and $x_0$ is a multiple root of $x^3+ax^2+bx+c$.

If it is a triple root, it is easy to see $x_0$ has to be rational. It is in fact $-\frac{a}{3}$.

If it is a double root, we can write the above cubic polynomial as $(x-x_0)^2(x-x_1)$ where $x_1$ is another root. We have $$2x_0+x_1=-a, x_0^2x_1=-c, x_0^2+2x_0x_1=b.$$ If we suppose $x_0$ is irrational, then $x_0^2, x_1$ have to be irrational too. But this does not help.

How should I continue? Thank you for any help!

Answer 1

Let $f$ be the polynomial $x^3+ax^2+bx+c$ in $\mathbf Q[x]$ and suppose that it has a double root. Then the gcd $g$ of $f$ and its derivative $f'$ is a nonconstant polynomial in $\mathbf Q[x]$. If $\deg(g)=1$ then $g$ has a root in $\mathbf Q$. This root is a root of $f$ and $f'$ in $\mathbf Q$, i.e., $f$ has a double root in $\mathbf Q$. If $\deg(g)\neq1$ then $\deg(g)=2$ since $\deg(g)\leq\deg(f')=2$. Then $g$ is a multiple of $f'$. It follows that $f'$ divides $f$ in $\mathbf Q[x]$. Then $f'$ cannot have distinct roots, otherwise they would be both multiple roots of $f$ and $\deg(f)\geq4$ which is false. Therefore, $f'$ has only one root wich is necessarily a double root in $\mathbf Q$. It follows that $f$ has a triple root in $\mathbf Q$.

Answer 2

HINT.-Put the equation of the cubic as $$f(x)=y^2-(x-\alpha)(x-\beta)(x-\gamma)=0$$ where $\alpha,\beta,\gamma$ are supposed reals in the affin plan.

We must have $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$.

Since $\frac{\partial f}{\partial y}=2y=2\sqrt{(x-\alpha)(x-\beta)(x-\gamma)}$ then $$(x-\alpha)(x-\beta)(x-\gamma)=0$$ On the other hand it is known that the cubic is non-singular if and only if $\alpha,\beta$ and $\gamma$ are distinct so making $\alpha\le\beta\le\gamma$ we have as possibilities $$\begin{cases}(x-\alpha)(x-\beta)^2: (\beta,0)\text{ nodal point}\\(x-\alpha)^2(x-\beta):(\alpha,0)\text{ isolated point}\\(x-\alpha)^3:(\alpha,0) \text{ cusp point}\end{cases}$$

That these points are rational is obtained from the system $$\begin{cases}\alpha+\beta+\gamma=-a\\\alpha\beta+\alpha\gamma+\beta\gamma=b\\\alpha\beta\gamma=-c\end{cases}$$