# Are the two singular cubic curves isomorphic?

Consider two singular cubic curves $$C_1$$ and $$C_2$$ in $$\mathbb{C}^2$$, defined by $$y^2 = x^3+x^2$$ and $$y^2=x^3+3x^2$$, respectively. So both have a double point at origin. They are isomorphic by adding the point $$(0,\infty)$$. Precisely speaking, their projective closures $$\overline{C}_1$$ and $$\overline{C}_2$$ in $$\mathbb{CP}^2$$ with coordinates $$[x,y,z]$$ are isomorphic by a scaling

$$x \mapsto \frac{1}{\sqrt{3}}x, \quad z \mapsto \frac{1}{3\sqrt{3}}z.$$ But are $$C_1$$ and $$C_2$$ themselves isomorphic as affine varieties?

You can use your isomorphism for the projective closures $$\overline{C_1}$$ and $$\overline{C_2}$$ to obtain an isomorphism of $$C_1$$ and $$C_2$$. For clarity, I'll use $$X, Y, Z$$ for projective coordinates and $$x,y$$ for affine coordinates. Recall that on the affine open set where $$Z \neq 0$$, the standard affine coordinates are $$x = X/Z$$ and $$y = Y/Z$$. Restricting your isomorphism \begin{align*} \overline{C_1} &\to \overline{C_2}\\ [X:Y:Z] &\mapsto \left[\frac{1}{\sqrt{3}} X : Y : \frac{1}{3\sqrt{3}} Z \right] \end{align*} to this affine open patch, we obtain \begin{align*} x = X/Z &\mapsto \frac{\frac{1}{\sqrt{3}} X}{\frac{1}{3\sqrt{3}} Z} = \frac{3\sqrt{3}}{\sqrt{3}} \frac{X}{Z} = 3x\\ y = Y/Z &\mapsto \frac{Y}{\frac{1}{3\sqrt{3}} Z} = 3 \sqrt{3} \frac{Y}{Z} = 3 \sqrt{3} y \, . \end{align*} To verify that this really works, note that \begin{align*} (3 \sqrt{3} y)^2 &= 27 y^2\\ (3x)^3 + 3(3x)^2 &= 27 x^3 + 27 x^2 \, . \end{align*}

• Thanks! I just realized this worked. I didn’t know the restriction would work. It seems restriction won’t give a morphism but only a rational map in general. Sep 23 '19 at 19:17