Exercise 10.14

Proof. Let $𝒞$ be the curve defined by $f(x,y)=y^2-x^3-\mathit{ax}-b$. The homogeneous equation of the projective closure $\overline{𝒞}$ of $𝒞$ is

The only point at infinity is given by $t=0,-x^3=0$, thus is the point $p=[0,0,1]$. Since

then $\frac{\partial \bar{f}}{\mathit{\partial t}}(0,0,1)=1$, thus the point at infinity $p$ is not singular.

For some other points $a=(u,v)$ on $\bar{C}$ not at infinity, it is sufficient by Exercise 10 to verify $(\mathit{\partial f}∕\mathit{\partial x}(u,v),\mathit{\partial f}∕\mathit{\partial y}(u,v))\ne (0,0)$. Since

if $a$ is singular, then

Therefore

If $a=0$, then $u=v=0$, thus $b=0$, so that $4a^3+27b^2=0$.

If $a\ne 0$, we eliminate $u$ between these two equations to obtain,

thus $u=-\frac{3b}{2a}$, and $u^2=\frac{9b^2}{4a^2}=-\frac{a}{3}$, which gives $4a^3+27b^2=0$. To conclude, if $4a^4+27b^2\ne 0$, then the curve defined by $y^2=x^3+\mathit{ax}+b$ has no singular points, finite or infinite. □