Define the Whitney umbrella as the affine surface $V(z^2 - yx^2) \subset \mathbb{A}^3$. I've come across an exercise that asks me to show that this surface is birational, but not isomorphic, to $\mathbb{A}^2$.
My intuition is that this follows from the fact that the Whitney umbrella is self-intersecting. Am I headed in the right direction?
Edit: I thought I had a rational parametrization to start things off, but it doesn't actually seem to work. It was $\mathbb{A}^2 \rightarrow \mathbb{A}^3 : (a, b) \mapsto (ab, b, a^2)$.
a) Since the umbrella $U= V(z^2 - yx^2)$ is singular (along all points of its stick $V(x,z)$) it is not isomorphic to $\mathbb A^2$, since $\mathbb A^2$ is nonsingular.
b) The umbrella $U$ is however birational to $\mathbb A^2$ because $$\mathbb A^2\setminus V(u) \stackrel {\cong}{\to} U\setminus V(x,z):(u,v)\mapsto (u,v^2/u^2,v)$$ is an isomorphism between an open dense set of $\mathbb A^2$ and an open dense set of $U$.
[The inverse isomorphism being $ U\setminus V(x,z)\to\mathbb A^2\setminus V(u): (x,y,z)\mapsto (x,z)$]