I'm working on the following question from Shafarevich's Basic Algebraic Geometry I:
Prove that an irreducible cubic curve $C$ has at most one singular point, and that the multiplicity of a singular point is 2. Prove also that if the singularity is a node then $C$ is projectively equivalent to $y^2=x^3+x^2$; if a cusp then to the curve $y^2=x^3$.
I've managed to do the first part, but can't figure out the projective change of coordinates for the last part. I know that (assuming the singular point is $[1:0:0]$) the curve is given by $$f(x,y,z)=Ay^3+By^2z+Cyz^2+Dz^3+x\Lambda(y,z)$$ where $\Lambda(y,z)=Ey^2+Fyz+Gz^2$, and this gives us two options:
- $\Lambda$ is a perfect square, i.e. $[1:0:0]$ is a cusp;
- $\Lambda$ factors as two distinct linear factors, i.e. $[1:0:0]$ is a node.
But I just can't find a change of coordinates that gives me the required forms.