Consider the scheme $X:=$ Spec $ \mathbb{C}[x,y]$ and delete the closed point $(x,y)$ (i.e. the ideal generated by $x,y$). This gives an open set $U \subset$ Spec $\mathbb{C}[x,y]$. I'm trying to prove that $\mathcal{O}(U)$ is isomorphic to $\mathbb{C}[x,y]$. My strategy was to take the basic-open cover $X_x \cup X_y = U$, since I know that by definition $\mathcal{O}(X_x) = \mathbb{C}[x,y]_{(x)}$, $\mathcal{O}(X_y) = \mathbb{C}[x,y]_{(y)}$ and I believe that then we have
$$\mathcal{O}(U) = \mathcal{O}(X_x)\times_{\mathcal{O}(X_{xy})} \mathcal{O}(X_y) = \mathbb{C}[x,y]_{(x)}\times_{\mathbb{C}[x,y]_{(xy)}} \mathbb{C}[x,y]_{(y)}$$
where by this notation I mean the set of pairs which restrict to the same thing in $\mathcal{O}(X_{xy})$. Please let me know if this is incorrect. Now I certainly seem to have an injective ring hom $$\mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y]_{(x)}\times_{\mathbb{C}[x,y]_{(xy)}} \mathbb{C}[x,y]_{(y)}.$$
So I need to prove its surjective. The only thing I know to do is to take a pair $$\big(\frac{p(x,y)}{q(y)+ x\cdot r(x,y)}, \frac{p'(x,y)}{q'(x)+ y\cdot r'(x,y)}\big)\in \mathbb{C}[x,y]_{(x)}\times_{\mathbb{C}[x,y]_{(xy)}} \mathbb{C}[x,y]_{(y)}$$ and use the fact that the first entry is equal to the second entry (in the fraction field, say) to prove that in fact the denominators are in fact just elements of $\mathbb{C}$. I tried playing around with degrees but to no avail. So Am I on the right/easiest track? and is there a simple way to finish the argument?
We show that the restriction $\rho_{X,U}: \Bbb C[x,y]= \mathcal{O}_X(X)\to \mathcal{O}_X(U) $ is an isomorphism of rings.
Surjectivity: Let $s\in \mathcal{O}_X(U); U=X_x \cup X_y$ we can find representations $s=\frac{\alpha}{x^m}$ on $X_x$ and $s=\frac{\beta}{y^n}$ on $X_y$ for some $\alpha, \beta \in \Bbb C[x,y]$ and $m,n \in \Bbb N$.
On the intersection $X_x\cap X_y=X_{xy}$ we have $\frac{\alpha}{x^m}=\frac{\beta}{y^n}$ and hence the equality $\alpha y^n=\beta x^m $. But $ \Bbb C[x,y]$ us a UFD and we find that $x^m|\alpha$ and $y^n|\beta$. Therefore there exsits some $\gamma \in \Bbb C[x,y]$ with $\gamma= \frac{\alpha}{x^m}=\frac{\beta}{y^n}$ and $\rho_{X,U}=s$.
Injectivity: The restriction maps on an integral scheme are injective.