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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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1 Answer 1

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.

Of course, one has to prove all this!

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Sorry if I'm being stupid, but I don't see how this helps. What do you mean by "one (of) the two"? –  algeom Apr 26 '12 at 18:16
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
Please edit the question and make it clearer. –  Mariano Suárez-Alvarez Apr 26 '12 at 18:24
To clarify: when Mariano writes "it has non-trivial cohomology", he means specifically that it has nontrivial coherent cohomology. There are for instance plenty of affine varieties with non-trivial singular cohomology. –  Dan Petersen Apr 26 '12 at 18:59

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