3
$\begingroup$

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

  • 1
    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)2012-04-26

1 Answers 1

7

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!

  • 0
    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.2012-04-26