Ytukyg

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

It is
enough to show that $X=mathbb A^2_ksetminus lbrace 0rbrace$ 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

$Xstackrel {cong}{to} Spec(Gamma(X,mathcal O_X ))=Spec (k[T_1,T_2])=mathbb A^2_k$

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 $T_1neq0$ and $T_2neq0$.

The covering is a Leray covering for $mathcal O$ (acyclic opens with acyclic intersection), since $U_1, U_2$ and $U_1cap U_2$ 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
$Gamma(U_1,mathcal O)times Gamma(U_2,mathcal O)to Gamma(U_1cap U_2,mathcal O)to 0$ where the non-trivial map is
$$k[T_1,T_1^{-1},T_2]times k[T_1,T_2,T_2^{-1}] to k[T_1,T_1^{-1},T_2,T_2^{-1}]:(f,g)mapsto g-f$$

Thus $H^1(X,mathcal O)=oplus _{i,jlt 0} ;; k cdot T^{i}T^{j}$, 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 $D=lbrace (1/n,0): n=1,2,3,...rbracesubset X$.

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
$$ Gamma(mathbb C^2,mathcal O_{mathbb C^2}) to Gamma(D, mathcal O_D): fmapsto f_0=fmid D $$

But this is clearly false because a holomorphic function $f_0:Dto mathbb C$ 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 $f:mathbb C^2to mathbb C$.

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 $f = g$?

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