For anyone unfamiliar with algebraic geometry, let me just define a couple of things. (This is fairly basic geometry, as far as that goes; no scheme theory or anything here.) All of our points will come from some algebraically closed field $\displaystyle k$. Let $\displaystyle X,Y$ be two varieties (affine, quasi-affine, projective, quasi-projective; doesn't matter). Aregular functionon $\displaystyle Y$ is a mapping $\displaystyle \frac{g}{h}:Y\longrightarrow k$ such that $\displaystyle \frac{g}{h}$ is defined at every point in $\displaystyle Y$. Amorphismis a continuous (with respect to the Zariski topology) function $\displaystyle \phi:X\longrightarrow Y$ with the property that, for any regular function $\displaystyle f:Y\longrightarrow k$, the composition $\displaystyle f\circ \phi:X\longrightarrow k$ is a regular function on $\displaystyle X$. Anisomorphismof varieties is just a morphism for which there exists an inverse morphism.

The rings of regular functions on $\displaystyle X,Y$ are denoted $\displaystyle \mathcal{O}(X),\mathcal{O}(Y)$ respectively.

Now, let $\displaystyle X=\mathbb{A}^2\setminus\{(0,0)\}$ (which is a quasi-affine variety).

The problem: show that $\displaystyle X\ncong \mathbb{A}^2$. (in which case we would say $\displaystyle X$ is not affine).

First, just a couple of facts:

1. I have shown that $\displaystyle \mathcal{O}(X)=\mathcal{O}(\mathbb{A}^2)=k[x,y]$.

2. We have this theorem: if $\displaystyle X$ is any variety and $\displaystyle Y$ an affine variety, then there is a bijection between the set of morphisms from $\displaystyle X$ to $\displaystyle Y$ and $\displaystyle k$-algebra homomorphisms from $\displaystyle \mathcal{O}(Y)$ to $\displaystyle \mathcal{O}(X)$. This correspondence is given by $\displaystyle \phi\leftrightarrow\tilde{\phi}$ where, if $\displaystyle f\in \mathcal{O}(Y)$, we have $\displaystyle \tilde{\varphi}(f)=f\circ\varphi$. Under this, isomorphisms of varieties correspond to $\displaystyle k$-algebra isomorphisms.

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So, number (2) says that if there were an isomorphism of varieties $\displaystyle \varphi:X\rightarrow\mathbb{A}^2$, there would be a corresponding $\displaystyle k$-algebra isomorphism $\displaystyle \tilde\varphi$, or equivalently, a corresponding $\displaystyle k-$algebra automorphism of $\displaystyle k[x,y]$. This is where I'm stuck; showing that there is no such $\displaystyle k$-algebra automorphism which would give rise to an isomorphism on the varieties. I found something online which says

"...so isomorphisms would have to correspond to automorphisms of $\displaystyle k[x,y]$ , but this is just the set of invertible linear transformations of $\displaystyle x$ and $\displaystyle y$; none of these yield an isomorphism."

I guess the first thing I don't quite get is what is meant by an invertible linear transformation... a function of the form $\displaystyle x\mapsto p(x,y),y\mapsto q(x,y)$? And, either way, how to prove that each invertible linear transformation (whatever they are) cannot possibly give a corresponding isomorphism on the varieties.