$\newcommand{\C}{\mathcal{C}}$ This is an exercise in Silverman and Tate's Rational Points on Elliptic Curves:

Let $\C$ be the conic given by the equation $$ F(x,y)=ax^2+bxy+cy^2+dx+ey+f=0.$$ Let $\delta$ be the determinant $$ \delta:=\begin{vmatrix}2a&b&d\\ b&2c&e\\ d&e&2f\end{vmatrix}.$$ Show that if $\delta\neq0$, then $\C$ has no singular points; i.e. no $(x,y)$ satisfy $F(x,y)=F_x(x,y)=F_y(x,y)=0$.

Here, $F_x$ and $F_y$ are the partial derivatives of $F$. I think I've already solved the problem, but am quite unsure of the proof, and hence I would appreciate if somone can check it for me. Here's my attempted proof. Because there's a $3\times 3$ whose determinant is being taken, this motivates the introduction of a third variable $z$, so I thought to define: $$ G(x,y,z):=ax^2+bxy+cy^2+dxz+eyz+fz^2. $$ Note that $G(x,y,1)\equiv F(x,y)$ under this definition. Then consider the sequence of equations given by setting each partial derivative of $G$ to $0$: $$\begin{cases} G_x=2ax+by+dz=0,\\ G_y=bx+2cy+ez=0,\\G_z=dx+ey+2fz=0. \end{cases}$$ Now if $\mathbf M$ is the matrix in the question, then $\delta=\det \mathbf M$, and we now have an equation of the form $$\mathbf{Mx}=\mathbf 0,$$ which looks right! So we use the condition $\delta\neq 0$ we get $\mathbf x=\mathbf M^{-1}\mathbf 0=\mathbf 0$, hence $z=0$, which is a contradiction, because we need $z=1$ for $(x,y)$ to be a singular point of $F(x,y)=G(x,y,1)$.

I'm not confident of this last part in italics. Can I really just impose the restriction $z=1$ at the end? Also, maybe a bigger issue is, how can I just introduce the extra condition $G_z=0$, when that's not required (I think) for $F(x,y)$ to have singular point $(x,y)$? I never used the condition $F(x,y)=0$ either, which feels really fishy. Can someone help explaining if my steps were valid, and if not, what the right idea is?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.