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 . Let be two varieties (affine, quasi-affine, projective, quasi-projective; doesn't matter). A regular function on is a mapping such that is defined at every point in . A morphism is a continuous (with respect to the Zariski topology) function with the property that, for any regular function , the composition is a regular function on . An isomorphism of varieties is just a morphism for which there exists an inverse morphism.
The rings of regular functions on are denoted respectively.
Now, let (which is a quasi-affine variety).
The problem: show that . (in which case we would say is not affine).
First, just a couple of facts:
1. I have shown that .
2. We have this theorem: if is any variety and an affine variety, then there is a bijection between the set of morphisms from to and -algebra homomorphisms from to . This correspondence is given by where, if , we have . Under this, isomorphisms of varieties correspond to -algebra isomorphisms.
So, number (2) says that if there were an isomorphism of varieties , there would be a corresponding -algebra isomorphism , or equivalently, a corresponding algebra automorphism of . This is where I'm stuck; showing that there is no such -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 , but this is just the set of invertible linear transformations of and ; 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 ? And, either way, how to prove that each invertible linear transformation (whatever they are) cannot possibly give a corresponding isomorphism on the varieties.