It is Exercise 4.3.2 in the book An Invitation to Algebraic Geometry.
Show that the ring $\mathcal{O}_V(U)$ of regular functions on the punctured plane $U=\mathbb{A}^2\backslash\{(0,0)\}$ is the polynomial ring $\mathbb{C}[x,y]$ (where I think $V=\mathbb{A}^2$). Conclude that this quasi-projective variety is not affine.
My try is as follows.
Obviously we have $\mathbb{C}[x,y]\subset \mathcal{O}_V(U)$.
Now let $f\in\mathcal{O}_V(U)$, by definition we have for any $p$ in $U$, we can choose a neighborhood $U_p$ of $p$ such that
$$
\left.f\right|_{U_p}=\frac{h_p}{k_p}
$$
for some polynomial $h_p$ and $k_p$ and $k_p(p)\neq 0$.
Because Zariski topology is compact, we can select finitely many such neighborhood, indexed by $i=1,\cdots,n$.
Because $k_i$c cannot simultaneously vanish on $U$, we must have
$$
\mathbb{V}(k_1,\cdots,k_n)\subset U^C=\{(0,0)\}
$$
If $\mathbb{V}(k_1,\cdots,k_n)=\varnothing$, then $(k_1,\cdots,k_n)=(1)$, then we can select
$$
1=\sum_{j}l_jk_j
$$
for some $l_j$, and the rest is easy.
My problems lies in the case $\mathbb{V}(k_1,\cdots,k_n)=\{(0,0)\}$ which follows
$$
(k_1,\cdots,k_n)=(x,y)
$$
and I don't know how to move on…
Best Answer
So, your method is interesting. It is not how I approached the problem when I did it first. $\def\O{\mathcal{O}}$ $\def\Spec{\mathop{Spec}}$
I think the main ingredient you are missing is imposing the compatibility condition that on $U_{i} \cap U_{j}$
$$\frac{h_{i}}{k_{i}} = \frac{h_{j}}{k_{j}}.$$
But it is hard to get a handle on the compatibility condition if you don't impose some controls on the open sets $U_{i}, U_{j}.$ So instead, maybe you can approach the problem like this.
Suppose $f \in \O_{V}(U)$. Then, we can cover $U$ by the affines $$U_{1} = (V)_{x} = \Spec \mathbb{C}[x, y]_{x}, U_{2} = (V)_{y} = \Spec \mathbb{C}[x, y]_{y}.$$
These are the open sets where $x$ and $y$ are respectively nonzero. Restricting $f$ to each of these open affines give $g_{1} \in k[x, y]_{x}, g_{2} \in k[x, y]_{y}.$ Hence, for some integer $m$, we have $$f|_{U_{1}} = \frac{p_{1}(x, y)}{x^{m}}, f|U_{2} = \frac{p_{2}(x, y)}{y^{m}}.$$
Now, on $U_{1} \cap U_{2} = \Spec \mathbb{C}[x, y]_{x, y}$, these expressions must agree. So, we must have $$\frac{p_{1}}{x^{m}} = \frac{p_{2}}{y^{m}} \Rightarrow y^{m}p_{1} = x^{m}p_{2}.$$ Since $k[x, y]$ is a unique factorization domain, this implies that $x^{m}$ must divide $p_{1}$ and $y^{m}$ must divide $p_{2}$. Hence, $f = \frac{p_{1}}{x^{m}} \in \mathbb{C}[x, y]$