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?
Isomorphism between affine curves
2
$\begingroup$
algebraic-geometry
-
0@Mariano Suárez-Alvarez: yeah I meant $[x:y:z] \mapsto [x:z:y]$ – 2012-05-21
1 Answers
7
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 $x^2+y^2=1.$
-
0I agree with that! Unfortunately, nowadays, this doesn't appear in any lectures. The same for other classical, but still very important topics. – 2012-05-21