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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

algeom
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    Relevant: (http://math.stackexchange.com/questions/101262/projective-varieties-basics) and (http://math.stackexchange.com/questions/122821/mathbba2-not-isomorphic-to-affine-space-minus-the-origin/122826#122826) – Georges Elencwajg Apr 26 '12 at 18:52

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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!

Mariano Suárez-Álvarez
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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-Álvarez 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
  • How can we show that there are no nonconstant regular functions on $Y$? – user302934 Oct 23 '20 at 03:02