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I want to prove that $X = \mathbb{A}^2\setminus (0,0)$ is not affine.

My attempt: If $\Bbbk[X] = \Bbbk[x,y]$ then $X$ is not affine since $(x,y) \subset \Bbbk[x,y]$ is a proper ideal, but $V(x,y) \cap X = \emptyset$. $X = \{x \ne 0\} \cup \{y \ne 0\}$ therefore i want to say that $\Bbbk[X] = \Bbbk[x,y,y^{-1}] \cap \Bbbk[x,y,x^{-1}] = \Bbbk[x,y]$.

But i can't prove rigorously both steps.

user26857
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qwenty
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  • Just for the last equality $k[X] = k[x,y]$ -- what's unclear here? – Hoot May 31 '15 at 20:35
  • 1)$M$ - algebraic manifold. $(U,\varphi), (W,\psi)$ - charts and $M = U \cup W$. Why $\Bbbk[M] = \Bbbk[U] \cap \Bbbk[W]$? Moreover i think that $\Bbbk[M]$ is not well defined(if $M$ is not affine)(??). So all steps are obvious, but i'm new in this area and i want to know how to prove it rigorously like proving that $S^1 \cong [0,1]/(0\sim 1)$ – qwenty May 31 '15 at 20:45
  • $k[M]$ is very much defined, but maybe not by whatever source you're using. This question has been asked many times on this site but maybe it's worth going through the motions one more time with your specific concerns. What are you using for a reference, anyway? – Hoot May 31 '15 at 21:08
  • http://math.stackexchange.com/a/122826/3217 – Georges Elencwajg Jun 01 '15 at 07:01

1 Answers1

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Suppose that $\mathbb{A}^2 \setminus \{\left(0,0\right)\}$ is affine. Let $i \colon \mathbb{A}^2 \setminus \{\left(0,0\right)\} \hookrightarrow \mathbb{A}^2 $ be the inclusion map. Then $i$ induces $i^* \colon \mathcal{O}_{\mathbb{A}^2}\left( \mathbb{A}^2\right) \to \mathcal{O}_{\mathbb{A}^2 \setminus \{\left(0,0\right)\}}\left( \mathbb{A}^2 \setminus \{\left(0,0\right)\}\right)$. We have the following identifications. $\mathcal{O}_{\mathbb{A}^2}\left( \mathbb{A}^2\right)$ is just the ring of polynomials in two variables $k\left[X,Y\right]$, and we know that $\mathcal{O}_{\mathbb{A}^2 \setminus \{\left(0,0\right)\}}\left( \mathbb{A}^2 \setminus \{\left(0,0\right)\}\right) = \mathcal{O}_{\mathbb{A}^2 }\left( \mathbb{A}^2 \setminus \{\left(0,0\right)\}\right) = k\left[X,Y\right]$. So $i^*$ is just the identity map, an isomorphism. This says that $i \colon \mathbb{A}^2 \setminus \{\left(0,0\right)\} \hookrightarrow \mathbb{A}^2 $ is an isomorphism. This is a contradiction since $i$ is not onto.

aGer
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    Why we know that $\mathcal{O}{\mathbb{A}^2 \setminus {\left(0,0\right)}}\left( \mathbb{A}^2 \setminus {\left(0,0\right)}\right) = \mathcal{O}{\mathbb{A}^2 }\left( \mathbb{A}^2 \setminus {\left(0,0\right)}\right) = k\left[X,Y\right]$ And what is $\mathcal{O_X}(Y)$? Rational functions on $X$ which regular on $Y$? – qwenty Jun 01 '15 at 10:05
  • I'm sorry. I thought you were familiar with those kind of definitions. Nevermind, take a look at my further post. http://math.stackexchange.com/questions/1298508/a-morphism-which-is-not-a-comorphism-of-a-regular-map – aGer Jun 01 '15 at 10:29
  • To see that $\mathcal{O}{X}\left( X \right) = \mathcal{O}{\mathbb{A}^2 }\left( X \right)$ (where $X=\mathbb{A}^2 \setminus {\left(0,0\right)}$), don't you use the fact that $X$ and $\mathbb{A}^2$ are birationally isomorphic, so they have the same function field? – Watson Apr 21 '17 at 20:49