Regular functions on the plane minus a point Let $k$ be an algebraically closed field. Let $X$ be the quasi-afine variety $X = A^2 - \{0,0\}$. I know from complex-analytic arguments that the ring of regular functions on $X$ is equal to $k[x,y]$. How can I prove this algebraically?
Thanks
 A: The ring of regular functions on $A^2 \setminus \{ x=0 \}$ is $k[x,y,x^{-1}]$ and the ring of regular functions on $A^2 \setminus \{ y=0 \}$ is $k[x,y,y^{-1}]$, and $A^2 \setminus \{0\}$ is the union of these open subsets, so its ring of regular functions is the intersection of the two, which is equal to $k[x,y]$.
A: The answer of Plop is correct, but I'll try to explain it a little bit more. I know how confused you are. I'm assuming you are using Scheme Theory.
By definition, $\mathbb{A}^2_k=\text{Spec }k[x,y]$. The origin of the plane  is given by the maximal ideal $(x,y)$. Let $X=\text{Spec }k[x,y] \setminus (x,y)$. Then $X=D(x) \cup D(y)$. We got an affine cover for $X$. By the definition of the sheaf, a function on $\mathcal{O}_{\mathbb{A}^2_k}(X)$ can be represented as an ordered pair of a function in  $\mathcal{O}_{\mathbb{A}^2_k}(D(x))$ and the other in $\mathcal{O}_{\mathbb{A}^2_k}(D(y))$ wich coincides in the intersection $\mathcal{O}_{\mathbb{A}^2_k}(D(x)\cap D(y))$. The 'coincides' means that they are equal after applying the restriction map.
Now, we know that $\mathcal{O}_{\mathbb{A}^2_k}(D(x))=k[x,y]_x=k[x,y,\frac1x]$ and $\mathcal{O}_{\mathbb{A}^2_k}(D(y))=k[x,y]_y=k[x,y,\frac1y]$. Since $D(x)\cap D(y)=D(xy)$, we get $\mathcal{O}_{\mathbb{A}^2_k}(D(x)\cap D(y))=k[x,y]_{xy}=k[x,y,\frac{1}{xy}]$.
What you have to do now is to show that if you have a function $f\in k[x,y,\frac1x]$ and $g\in k[x,y,\frac1y]$ which coincides on $k[x,y,\frac{1}{xy}]$, they must be equal and belong to $k[x,y]$. Remember that, in this case, the restriction map is a localization map.
This is an important example because it's one of the first (and maybe the simpler) examples of a non-affine scheme! Can you see why?
