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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?

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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*}

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    $\begingroup$ 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. $\endgroup$ Sep 23 '19 at 19:17

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