Let $X=\mathbb{A}^2\setminus\{(x,y)\}$ be the affine scheme minus the origin (say over a field $k$). Consider the action of the group $G=k^*$ given by $(x,y)\mapsto (\alpha x,\alpha^{-1}y)$ for any $\alpha\in k^*$. Prove that $X/G$ (which a priori is just a ringed space) is in fact a scheme. What if $X=\mathbb{A}^2$?
I think that the first quotient is $\mathbb{P}^1$, but I'm not sure how to show it for real. My intuitive reasoning is as follows: let's assume that $k$ is algebraically closed. Then the equivalence classes of closed points are the $x$-axis, the $y$-axis, and all the equilateral hyperbolas $xy=c$ ($c\neq 0$). So if I take say the line $y=1$ this intersects exactly once all the classes apart from one (namely, the $x$-axis). The quotient of $X$ minus the $x$-axis is then (or, at least, seems like) $\mathbb{A}^1$, and then the whole quotient is $\mathbb{P}^1=\mathbb{A}^1\cup\{[\text{$x$-axis}]\}$.
Also, if I compute the ring of fixed polynomials of $X$ minus the $x$-axis I get $k[xy]\cong k[t]$ (the problem with this is that I get the same with $X$ minus the $y$-axis, so it doesn't seem to give a $k$ as global sections, as I would expect).
Also this doesn't help me for the second question. Could you please show me how to really prove this (that is: to give a proof in the language of schemes that considers all points equal, and that adapts to the second case).