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How to see why the affine curves $V(4x^{2}+x+y^{2})$ and $V(4x^{2}+xy+1) \subseteq \mathbb{A}^{2}$ are isomorphic? I was thinking in using the fact that every quasi-affine variety is birational to its projective closure, now their projective closures are given by $V(4x^{2}+xz+y^{2})$ and $V(4x^{2}+xy+z^{2})$ so the map $[x : y : z] \mapsto [x : y: z]$ gives an isomorphism. How can we check the first claim without using this fact?

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The map you say that gives an isomorphism is the identity, so there must be a typo somewhere. In any case, your argument seems only to show that the two affine curves are birational, not isomorphic. –  Mariano Suárez-Alvarez May 20 '12 at 4:34
    
@Mariano Suárez-Alvarez: yeah I meant $[x:y:z] \mapsto [x:z:y]$ –  user31509 May 21 '12 at 7:20

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up vote 7 down vote accepted

If you make the change of variables $x\leadsto x/8-1/8$, $y\leadsto y/4$, the equation $$4x^2+x+y^2=0$$ turns into $$x^2+y^2=1.$$

Similarly, if you change $x\leadsto\sqrt{-1}x- y$ and $y\leadsto -3\sqrt{-1}x+5y$ on the equation $$4x^2+xy+1=0$$ you also get $$x^2+y^2=1.$$

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There is no magic involved in doing this by the way: it is just the well-known affine classification of conics that I used. Classifying conics by hand on the plane is something that everyone should do sometime before even learning what «quasi-affine variety» means! :( –  Mariano Suárez-Alvarez May 20 '12 at 4:53
    
I agree with that! Unfortunately, nowadays, this doesn't appear in any lectures. The same for other classical, but still very important topics. –  Martin Brandenburg May 21 '12 at 8:24

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