# Why are $\mathbb A_k^2 \backslash \{(0,0) \}$ and $\mathbb P_k^2 \backslash \{(0,0) \}$ not isomorphic to affine nor projective varieties?

Why are $\mathbb A_k^2 \backslash \{(0,0) \}$ and $\mathbb P_k^2 \backslash \{(0,0) \}$ isomorphic to neither affine nor projective varieties?

I've seen this question in several different places, but haven't been able to do it. Any hints/explanations appreciated. Thanks

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– Georges Elencwajg Apr 26 '12 at 18:52

Hint: There are non-constant regular functions on $X=A^2\setminus\{\text{point}\}$, so it is not a projective variety. On the other hand, it has non-trivial cohomology, so it is also not affine (See the link provided by Georges in a comment to the question for a non-cohomological argument)
On $Y=P^2\setminus\{\text{point}\}$ there are no non-constant regular functions, so it is not affine, and it is not complete so it is not projective.
Ok, I now realise I didn't state the question very clearly. I meant to ask a) Why is $\mathbb A^2 \backslash 0$ not isomorphic to any affine or projective varieties? and b) Why is $\mathbb P^2 \backslash 0$ not isomorphic to any affine or projective varieties? – algeom Apr 26 '12 at 18:20