If we have $\mathscr{O}(\mathbf{A}^2\backslash\{(0,0)\})=k[x,y]$, now how can we show that $\mathbf{A}^2\backslash\{(0,0)\}$ is not isomorphic to $\mathbf{A}^2$?
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Did you bother searching? This question is a very common one. Here is an example of someone else asking this question: http://math.stackexchange.com/questions/424067/mathbba2-backslash-0-0-is-not-affine-variety?rq=1 – KReiser Nov 24 '16 at 07:22
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The punctured plane isn't even an affine variety. It's a result (probably in Shafarevich) that if an affine variety contains a line segment, it must contain the entire line. This is obviously not true for the punctured plane.
A. Thomas Yerger
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You're dealing with polynomials. The only way a polynomial (or family thereof) can vanish along a line segment is if its' actually linear. – A. Thomas Yerger Nov 24 '16 at 07:25