Finding a line $L\subset V(y-xz)\subset\mathbb A^3_k$

Question

I want to find lines $L\subset V(y-xz)$ and $M\subset\mathbb A_k^2$ such that $$ V(y-xz)\setminus L \simeq \mathbb A_k^2\setminus M\ . $$ Hint suggests that I use the projection $(x,y,z)\mapsto(x,y)$. I tried to find nice inverse rational map plugging some lines but failed. How can I get such $L$ and $M$?

Answer 1

Take $M=\{ax+by=0\} \subset\mathbb{A}_k^2$. The pre-image under the projection of $V(y-xz)$ to the affine plane is contained in the union of two lines $L:\{x=0,y=0\}\cup \{a+bz=0,y-xz=0\}$ (convince yourself that these are lines). On the complement, you have $x\neq 0$ and thus $z=y/x$ makes sense. So the complement looks like the affine plane minus the line $x= 0$ minus the line $x= -b/ay$. You want the second line to be contained in the first line, so $b$ must be zero. Then you see that for lines $M=\{x=0\}\subset \mathbb{A}_k^2$ and $L=\{x=0,y=0\}\subset V(y-xz)$ the complements are isomorphic.