Question about affine plane minus origin not being affine I am confused about the example given in Vakil's algebraic geometry notes that states $U:=\mathbb{A}_k^2-\{(x,y)\}$ is not an affine scheme. Questions about this example have been asked on here before, but none seem to have the same confusion as I do (nor is my confusion relieved by the answers provided).
We start by assuming that $U$ is affine. Say $(U,\mathcal{O}_{\mathbb{A}_k^2|U})=(\operatorname{Spec}A,\mathcal{O}_{\operatorname{Spec}A})$ for some ring $A$. Then we can recover $A$ by global setions: $A=\Gamma(U,\mathcal{O}_{\mathbb{A}_2|U})$, which we have previously identified to be $k[x,y]$. So by assuming $U$ is affine, we have that $U\cong\mathbb{A}^2_k$.
Now, the next part is what I have been struggling with. Vakil writes "But this bijection between prime ideals in a ring and points of the spectrum is more constructive than that: given the prime ideal $I$, you can recover the point as the geometric point of the closed subset cut out by $I$, i.e., $V(I)$, and given a point $p$, you can recover the ideal as those functions vanishing at p, i.e., $I(p)$. In particular, the prime ideal $(x,y)$ of $A$ should cut out a point of $\operatorname{Spec}A$".
I am not sure what he means here. Don't both sides of $U\cong\mathbb{A}^2_k$ consist of prime ideals (which can also be interpreted as point on $\operatorname{Spec}A$), except $U$ has one fewer point (prime ideal), namely $(x,y)$? Grammatically, this paragraph has pronouns for which the antecedent is unclear. (For example, does "this bijection" refer to $U\cong\mathbb{A}^2_k$ or to the general association of prime ideals with points on a spectrum?)Perhaps if I had more intuition, I'd be able to figure it out from context. Unfortunately, I don't. Could someone please explain more explicitly what is going on here?
 A: I always find this example to be annoyingly and confusingly explained.
(NB: I assume you have typo and that $U=\mathbb{A}^2_k-\{0\}$ and $x$ and $y$ are the coordinates on $\mathbb{A}^2_k$)
The point is this. We have the natural open embedding $j:U\hookrightarrow \mathbb{A}^2_k$ and not only is $\mathcal{O}(U)\cong \mathcal{O}(\mathbb{A}^2_k)$ as abstract $k$-algebras but, in fact (as the 'Algebraic Hartog's lemma' in Vakil shows) the induced map
$$j^\sharp:\mathcal{O}(\mathbb{A}^2_k)\to \mathcal{O}(U)$$
is an isomorphism. In particular, if $U$ were affine then this would imply that $j$ is an isomorphism (since $j^\sharp$) is which, in particular, would imply that $j$ is bijective. But, of course, this is false.
What Vakil is saying then is that since it's $j^\sharp$ that is an isomorphism one would have that the 'point' of $U$ corresponding to $0$ would be a point $p$ of $U$ such that $j(p)$ agrees with $0$. Indeed, by $0$ in $U$ he really means
$$\ker(\mathcal{O}(U)\to \mathcal{O}(\mathbb{A}^2_k)=k[x,y]\twoheadrightarrow k[x,y]/(x,y)\cong k)$$
but this just means that $j(p)$ in $\mathbb{A}^2_k$ is
$$\ker(\mathcal{O}(\mathbb{A}^2_k)=k[x,y]\twoheadrightarrow k[x,y]/(x,y)=k)$$
which is just $0$. But, of course, no point $p$ can exist since $j^{-1}(0)$ is empty.
