Possible Duplicate:
$\mathbb{A}^{2}$ not isomorphic to affine space minus the origin
Hello I'm using the book Basic algebraic geometry I of Shafarevich.
Definitions: I'm very complicated with the computation of the coordinate ring of a quasiprojective variety. In the affine case, this ring it´s easier to compute, because (using the notation of Shafarevich) if X it´s my closed set, the it´s ring of coordinates it's $k[T]/U_X$ ( at least could be something explicit). But in the projective case. Given $X \subset P^n$ a quasiprojective variety, the ring of coordinates denoted also by $k[X]$ is defined as the set of all functions of the form $ f=P/Q $ where P,Q homogeneous polynomials of the same degree , and $ Q(x) \ne 0 $ for every x $\in X$.
Problem: For example, I have to prove that the quasiprojective variety $ {\rm A}^2 - \left\{ {\left( {0,0} \right)} \right\} $ is not isomorphic to an affine variety. And the hint of the problem it's to compute $k[X]$ and showing that it has a proper ideal $ J \subseteq k\left[ X \right] $, such that defines a nonempty set.
But I have no idea , what can I do with this abstract (at least for me now) $k[X]$. Sorry for this stupid question
