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The question I am working on is

"Show that $\mathbb{A}^2-\{(0,0)\}$ is not affine".

There are already many questions and proofs here for this, for example see: (1) or (2).

I only have very basic knowledge of algebraic geometry and I couldn't understand some of the arguments of these proofs. I came up with the basic proof (?), but it is probably wrong seeing that the other proofs are much more involved. Can you tell me where I am wrong?

My short attempt is:

Assume $\mathbb{A}^2-\{(0,0)\}$ is an affine variety. This means $\{(0,0)\}$ is an open set in the Zariski topology in $\mathbb{A}^2$. Open sets must be dense in Zariski topology but a single point cannot be dense, so we have a contradiction.

user666150
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  • $\mathbb{A}^2-{(0,0)}$ being affine does not imply that ${(0,0)}$ is open. – Douglas Molin Apr 13 '22 at 14:47
  • Isn't an affine variety $(\mathbb{A}^2-{(0,0)})$ a closed set and so its complement $({(0,0)})$ an open set? – user666150 Apr 13 '22 at 14:55
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    No, that would be a closed subvariety. For example, $\mathbb{A}^1-{0}$ is affine because it is isomorphic to the subvariety of $\mathbb{A}^2$ defined by $XY-1$. – Douglas Molin Apr 13 '22 at 14:58
  • Thanks, now I see that my attempt does not work for $\mathbb{A}^1-{0}$. I am probably confused with the basic definitions and will study them again. – user666150 Apr 13 '22 at 15:10

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