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

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!

• To say "elliptic curve" discard the possibility of singular points. Edit your question. Regards. – Piquito May 10 '16 at 14:52
• @Piquito: what does this mean? The meaning of "singular" is adequately expressed in the question... – abiessu May 10 '16 at 14:53
• @abiessu: Thanks! I just checked the definition, it indeed says "assuming that the roots of $f(x)$ are distinct, such a curve is called an elliptic curve". Sorry my fault. – KittyL May 10 '16 at 15:03

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 $\mathbf Q$.

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 $$\begin{cases}\alpha+\beta+\gamma=-a\\\alpha\beta+\alpha\gamma+\beta\gamma=b\\\alpha\beta\gamma=-c\end{cases}$$

• Thanks for the reply! But how does the system tell us that those points are rational? – KittyL May 10 '16 at 16:33
• Because of the double or triple point. (See at the last words of Johannes Huisman´s answer). If the system don't gives rational then it is not matter of singularity. – Piquito May 10 '16 at 17:02

To prove that a double root $x_0$ of $x^3 + ax^2 + bx + c$ with rational coefficients is rational, first substitute $$x = z - \frac{a}{3}$$ to get a cubic polynomial of the form $$z^3 + dz +e$$ with rational $d$ and $e$, a double root $$z_0 = x_0 + \frac{a}{3}, \tag1$$ and a another root $z_1$.

Next, relate those roots to the coefficients: \begin{align} 2z_0 + z_1 & = 0 \tag2\\ z_0^2 + 2z_0z_1 & = d \tag3\\ z_0^2z_1 & = -e \tag4 \end{align} From $(2)$, $z_1 = -2z_0$. Substitute that into $(3)$ and $(4)$, solve for $z_0^2$ and $z_0^3$, and divide to get $$z_0 = -\frac{3e}{2d},$$ a rational number. From $(1)$ we see that $x_0$ is also rational. If $z_0 = 0$, we cannot divide, but in that case, $x_0$ is rational by $(1)$ again.