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Why is the affine space $\mathbb{A}^{2}$ not isomorphic to $\mathbb{A}^{2}$ minus the origin?

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Lots of reasons. One is that the latter isn't the spectrum of a ring. –  Hurkyl Mar 21 '12 at 7:03
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3 Answers

up vote 32 down vote accepted

It is enough to show that $X=\mathbb A^2_k\setminus \lbrace 0\rbrace$ is not affine since $\mathbb A^2_k \:$ is affine.

First proof of non-affineness
The key point is that the restriction map $\Gamma(\mathbb A^2_k,\mathcal O_{\mathbb A^2_k})=k[T_1,T_2] \to \Gamma(X,\mathcal O_X )$ is bijective.
This is the analogue of the Hartogs phenomenon in several complex variables and in algebraic geometry results from the fact that for a normal ring $A$ we have $A=\cap_{\mathfrak p} A_{\mathfrak p}$, where the intersection is over primes of height $1$.

Now if $X$ were affine we would have the canonical isomorphism of schemes
$X\stackrel {\cong}{\to} Spec(\Gamma(X,\mathcal O_X ))=Spec (k[T_1,T_2])=\mathbb A^2_k$
which is false since the origin of $\mathbb A^2_k$ is not in $X$.

Edit: Second proof of non-affineness
Consider the open covering $\mathcal U$ of $X$ by the two open affine subsets $U_1=D(T_1)$ and $U_2=D(T_2)$, i.e. those open subsets determined by $T_1\neq0$ and $T_2\neq0$.
The covering is a Leray covering for $\mathcal O$ (acyclic opens with acyclic intersection), since $U_1, U_2$ and $U_1\cap U_2$ are affine.
Hence by Leray's theorem we have $H^1(X,\mathcal O)=\check {H}^1(\mathcal U,\mathcal O)$ and thus $H^1(X,\mathcal O)$ is the cohomology of the complex $\Gamma(U_1,\mathcal O)\times \Gamma(U_1,\mathcal O)\to \Gamma(U_1\cap U_1,\mathcal O)\to 0$ where the non-trivial map is $$k[T^1,T^{-1}]\times k[T^2,T^{-2}] \to k[T^1,T^{-1},T^2,T^{-2}]]:(f,g)\mapsto g-f$$

Thus $H^1(X,\mathcal O)=\oplus _{i,j\lt 0} \;\; k \cdot T^{i}T^{j}$, an infinite-dimensional $k$-vector space, in sharp contrast to Serre's theorem stating that positive dimensional cohomology groups of coherent sheaves on affine schemes are zero.

Other Edit: Third proof of non-affineness
If $k=\mathbb C $ let us show that $X$ is not affine by proving that its underlying holomorphic manifold $X_{hol}$ is not Stein.
Indeed suppose it were and consider the discrete closed subset $D=\lbrace (1/n,0): n=1,2,3,...\rbrace\subset X$.
Since $D$ is a $0$-dimensional submanifold the restriction map $\Gamma(X_{hol}, \mathcal O_{X_{hol}})\to \Gamma(D, \mathcal O_D)$ would be surjective.
On the other hand, by Hartogs's theorem the restriction map $\Gamma(\mathbb C^2,\mathcal O_{\mathbb C^2}) \to \Gamma(X_{hol},\mathcal O_{X_{hol}}) $ is also surjective so that by composition we would conclude if $X_{hol}$ were Stein to the surjectivity of the restriction map $$ \Gamma(\mathbb C^2,\mathcal O_{\mathbb C^2}) \to \Gamma(D, \mathcal O_D): f\mapsto f_0=f\mid D $$

But this is clearly false because a holomorphic function $f_0:D\to \mathbb C$ is an arbitrary function, and if $f_0$ is not bounded [ for example $f_0(1/n,0)=n$ ] it cannot be the restriction of a holomorphic function $f:\mathbb C^2\to \mathbb C$.

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You give wonderful explanations! –  Bruno Joyal Mar 21 '12 at 7:50
    
@Georges Elencwajg: Thanks. I'm actually trying to show that $\mathbb{A}^{2}\setminus \{(0,0)\}$ is not affine so it suffices to show (because the ring of regular functions coincides with the coordinate ring in the affine case) that there is no isomorphism between $\mathbb{A}^{2}$ and $\mathbb{A}^{2} \setminus \{(0,0)\}$ without an affine argument. (Otherwise the argument is circular), do you know another argument? –  user6495 Mar 21 '12 at 7:55
    
@user6495: As Georges wrote, the core of the argument amounts to showing that a regular function on the punctured affine plane always extends to a unique function on the whole affine plane. Once you show that, you can prove that the punctured plane is not affine by invoking the universal property of maps to affine varieties (see, e.g. [Hartshorne, Ch. I, Proposition 3.5]). –  Zhen Lin Mar 21 '12 at 8:16
    
Thanks a lot to @Zhen for his illuminating comment. Although the proof I posted is logically sufficient, I have posted a second proof, for the sheer pleasure of computing a cohomology group (but that second proof is a bit more advanced). –  Georges Elencwajg Mar 21 '12 at 8:32
    
@GeorgesElencwajg Can you explain your first argument a little bit more? I don't understand how the intersection identity results the isomorphism of global section rings? –  mqx Oct 25 '13 at 0:48
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I'll call your punctured plane $X$, as Georges does. As he says, the key is to prove a Hartogs lemma for the inclusion $X \to \mathbf A^2$. What follows is (I think) a special case of the result in commutative algebra that he mentions.

Let $f$ be a regular function on $X$. Then $X - Z(x)$ is affine, isomorphic to $Z(xz - 1) \subset \mathbf A^3$, and hence the restriction of $f$ to this open set of agrees with $g(x, y)/x^n$ for some $g \in k[x, y]$ and $n \geq 0$. Similarly, $f$ restricted to $X - Z(y)$ looks like $h(x, y)/y^m$.

Now, regular functions on $\mathbf A^2$ correspond exactly to elements of $k[x, y]$, and you have two such elements \[ y^mg(x, y) \quad \text{and} \quad x^nh(x, y) \] which agree on the dense subset $\mathbf A^2 - Z(xy)$. Can you argue that, after putting some conditions on $g, h, n, m$, we must have $n = m = 0$ and $f = g$?

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One of the two is an affine variety while the other isn't.

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