I wanna to solve this problem dealing with a polynomial map between two affine curves.
First curve is $A \subset \mathbb{C}^2$ defined by equation $s(1+t^2)-1=0$ and the second curve is $B \subset \mathbb{C}^2$ defined by equation $(x^2+y^2)^2+3x^2y-y^3 = 0$
I have a polynomial map $f: A \rightarrow B$ defined as $f(s,t)= ( s^2t(1-3t^2), s^2(1-3t^2) )$
The problem is to prove that $f$ is surjective and also one-to-one (except the point $(0,0)$
I have no idea how to approach this. Can someone give me a little nudge what to look for? I only see that in the image of $f$ this holds $x=yt$, but I dont see how it can help. Maybe looking for inverse might be the way?
Thanks in advance
We can approach this by solving the polynomial system $$ \left\{\begin{matrix} x = ts^2(1-3t^2)\\ y = s^2(1-3t^2) \end{matrix} \right. $$ with $(x,y) \neq (0,0)$ satisfying $(x^2+y^2)^2+3x^2y-y^3=0$ and $(s,t)$ satisfying $s(1+t^2)=1$. First assume that $y\neq 0$, then $t = \frac{x}{y}$. From the equation of $A$ we see that $s= \frac{y^2}{y^2+x^2}$ (Note that the equation of $B$ implies that $x^2+y^2 \neq 0$) and you can verify that they solve the system above. Hence for each $(x,y)$ with $y\neq 0$ we have a unique point $$ (s,t) = \left(\frac{y^2}{y^2+x^2},\frac{x}{y} \right) $$ such that $f(s,t)=(x,y)$.
Now we turn to the case $y=0$. The equation of $B$ becomes $x^4 =0$ whence $x=0$ (we are at the origin). In this case our system becomes $$ \left\{\begin{matrix} 0 = ts^2(1-3t^2)\\ 0 = s^2(1-3t^2) \end{matrix} \right. . $$ The equation of $A$ says that $s \neq 0$ hence $1-3t^2 =0$. Therefore $ t = \pm \sqrt{\frac{1}{3}}$ and $s = \frac{3}{4}$ give the two possible solutions. This means that the inverse image of the origin by $f$ is composed by two points.