Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

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!

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.