Projective Line as quotient ringed space

Question

Let $\mathbb{k}$ be an infinite field, $X=\mathbb{A}^2 \setminus \{(0,0)\}$ and $G$ the group of automorphisms $(x,y) \mapsto (\alpha x, \alpha y)$ for all $\alpha \in \mathbb{k}^{*}$. Seeing $Y=X/G$ as topological space, it's obvious that it is $\mathbb{P}^1$. In order to show that $Y$ is a ringed space, I have to proof that the sheaf $\mathcal{O}_Y$ such that $\mathcal{O}_Y(U)=\mathcal{O}_X(p^{-1}(U))^G$ is isomorphic to the structure sheaf of $\mathbb{P}^1$.

1) How can proof this result?

2) Moreover, if $G$ is the group of automorphism $(x,y) \mapsto (\alpha x, \alpha^{-1} y)$, is $Y/X/G$ a scheme?

3) Is it true even if $X=\mathbb{A}^2$?

Answer 1

Question 1: You have a $G=k^*$ invariant map of varieties $\mathbb{A}^2 -(0,0) \rightarrow \mathbb{P}^1$ sending $(x,y)$ to $(x:y)$. Topologically, this is your quotient map. It restricts to a $G$ invariant map $\mathbb{A}^2 - y$ axis $\rightarrow \mathbb{P}^1-\infty$. Both schemes are affine in this case, so the restricted map corresponds to a map of $k$-algebras $k[Y/X] \rightarrow k[X,X^{-1},Y]$ which does the obvious thing. The induced $G$ action on the right hand side is such that $\alpha$ sends $p(X,X^{-1},Y)$ to $p(\alpha X,\alpha^{-1}X^{-1}, \alpha Y)$. It is clear that the image of $k[Y/X]$ is invariant under $G$.

I guess the hard part is to see that you get every invariant polynomial in the image. But it doesn't seem too bad; $p$ is invariant if and only if every monomial in $p$ of the form $X^mY^n$ (where $m$ is allowed to be negative) is invariant. But clearly such a monomial is invariant if and only if $m=-n$, i.e. is $Y/X$ raised to a power. So this proves that $O_{\mathbb{A}^2-(0,0)}|_{x\neq0}$ is isomorphic under the given map to $O_{\mathbb{P}^1}|_{\mathbb{P}^1-\infty}$. Play the same argument with $\mathbb{A}^2 - x$ axis to get the other isomorphism you need.

Question 3: The standard way for affine varieties, in this case $Speck[X,Y]$, to get a quotient with decent functorial properties is to just take $Spec$ of the $G$ invariants. That will give you a point $Spec(k)$ in this case, which isn't the topological quotient. This is because the topological quotient is a bad space. The only neighborhood of the origin is the whole space, so it's not even Hausdorff.

Answer 2

1. Do the computation. What does an open set in $\mathbb{P}^1$ look like? Can you describe it's inverse image under $p$? Can you compute the sections of the sheaves? This computation is relatively straightforwards- just start writing and eventually you'll finish.

2. This is an example of weighted projective space. You can construct it as a gluing of two copies of $\mathbb{A}^1$ (where either $x$ or $y$ is nonzero) along the copy of $\mathbb{A}^1\setminus 0$ where they're both nonzero. Weighted projective spaces are toric varieties, and therefore schemes.

3. You can take the quotient, but the quotient object is fairly badly behaved- every point is any any neighborhood of the origin, for example. GIT (geometric invariant theory) can tell you more about these types of strangely behaved quotients.