Exercise 5.6.7 (Double points on the real curve $y^2 = ax^3 + bx^2 + cx + d$)

Question

Let $p(x) = ax^3 + bx^2 + cx+d$, where $a,b,c,d$ are real numbers, not all $0$. Show that if the curve $y^2 = p(x)$ has a double point, then it must be of the form $(r,0)$ where $r$ is a double root of $p(x)$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Put $f(x,y) = y^2 - p(x)$. Then

$$\frac{\partial f}{\partial x}(x,y) = p'(x),\qquad \frac{\partial f}{\partial x}(x,y)  = 2y,$$
and
$$\frac{\partial^2 f}{\partial x^2}(x,y) = p''(x), \qquad \frac{\partial^2 f}{\partial y^2}(x,y) =2,\qquad  \frac{\partial^2 f}{\partial x \partial y}(x,y) = 0.$$
If the curve $y^2 = p(x)$ has a double point $(r,s)$, then $ f(r,s) =\frac{\partial f}{\partial x}(r,s)  = \frac{\partial f}{\partial y}(r,s) = 0$ (and these condition are sufficient, because $\frac{\partial^2 f}{\partial y^2}(r,s) =2 
e 0$).

Therefore $0 = s^2  - p(r), p'(r) = 0, 2s = 0$, so $s = 0$ and $p(r) = p'(r) =0$. This shows that the order of multiplicity of the root $r$ is at least $2$.

If the curve $y^2 = p(x)$ has a double point, then it must be of the form $(r,0)$ where $r$ is a double root (at least) of $p(x)$.
\end{proof}

Note: The curve $y^2 = (x-1)^3$, where $p(x) = (x-1)^3$ has point double at $(1,0)$, not triple because $\frac{\partial^2 f}{\partial y^2}(1,0) =2 
e 0$ (and $p(x) = f(x,m(x-1)) = m^2(x-1)^2 -(x-1)^3$ shows that $M = 2$). But $1$ is a triple root of $p(x)$.

Exercise 5.6.7 (Double points on the real curve $y^2 = ax^3 + bx^2 + cx + d$)

Answers

Comments

Exercise 5.6.7 (Double points on the real curve $y^2 = ax^3 + bx^2 + cx + d$)

Answers

Comments

Add answer