Cfrgtkky

I really need help with this question.

Assume you have a morphism $varphi: mathbb{A}^2 rightarrow
> mathbb{A}^1$ such that $varphi (x,y)= x^2-y^4$.

1) Find all points in $mathbb{C}$ such that $varphi^{-1}(a)$ is
singular. And its type of singularity.

2) If $Y_{a}$ is the closure of curve $varphi^{-1}(a)$ in
$mathbb{P}^{2}$ and L is line at $infty$ (z=0), find the point
$Y_{a}cap L$.

3) Let $psi: mathbb{P}^{2}rightarrow mathbb{P}^{1}$ be rational
map that extends $varphi$, find the domain of $psi$ and the fiber
$psi^{-1}(infty)$.

This is a reducible curve that I've never seen before. I do not know what can I do!

I found $(0,0)$ is a singular point by using Jacobian. For the second one, I took homogenization of curve but I couldn't how to find the intersection with the line at infinity. For the last one, I have no idea.

1) As you said, by the Jacobian criterion indeed $varphi^{-1}(0)$ is the unique singular fiber and the singularity type is two tangent parabolas $(x-y^2)(x+y^2) = 0$.

2) The fiber has homogenous equation $x^2z^2 = y^4 + az^4 $. Its intersection with the line at infinity given by $z=0$ and the previous equation, i.e this is the point $(1,0,0)$ (with multiplicity $4$).

3) We take $$psi(x,y,z) = frac{x^2z^2 - y^4}{z^4}$$

and the domain of $psi$ is exactly when $z^4 neq 0$ or $x^2z^2 - y^4 neq 0$. This means that the domain is $Bbb P^2 backslash {(1,0,0)}$. This is not surprising since by the previous question, $(1,0,0)$ was in the closure of each fibers !

Finally, the fiber over $infty$ is simply given by the line $z=0$, minus the point $[1:0:0]$. For completness, the other projective fibers over $[a:1] in Bbb P^1 $ are given by ${ (x,y,1) : x^2 = y^4 + a }$. Notice that it does not include the point $(1,0,0)$ by what we said, even though this point is in the closure of all the fibers.

A final remark : this is possible to find a surface $X$ and a morphism $f : X to Bbb P^2$ so that $psi$ becomes defined everywhere, this morphism is called a blow-up and making your map defined everywhere is typically why algebraic geometers introduce blow-up.

About your final remark, If I make blow-up, can I get a such a morphism $f:Xrightarrow mathbb{P}^{1}$ which an extension of $varphi$ ? Does it have to be $mathbb{P}^{2}$