Let's follow the hint. For now let's just assume that
(X) ≃ k[x,y]. Then if X
were isomorphic to an affine variety Y , then we'd have that
(Y ) ≃
(X) (see
Lemma 1, here. Hence
(Y ) ≃ k[x,y], which means that we must have that
Y ≃ A^{2} (corollary 3.7). Hence, X ≃ A^{2}. Let's call this isomorphism ϕ:
Now, the question being begged is if
ϕ maps
A^{2} to
A^{2}  0, where is the origin
being mapped to? It has to be mapped
somewhere, so let's suppose that
ϕ(0, 0) = (R,S) Now, nothing immediately wrong with that persay. There are isomorphisms of
affines that map the origin to nonzero points. Like translations for example, like
I've shown
before. But WAIT:
This map couldn't be a translation. Because then
what would
(R,S) map to? HMMMMMMM.
Something is geometrically wrong with this map. And since this subject is called
"algebraic geometry", let's look at what's happening with the induced map of
coordinate rings:
ϕ* : k[x,y]  → k[x,y]  

f 
f ∘ ϕ   
This map is an isomorphism (again, thanks to Lemma 1 from
here). And how is it
defined? Well, figuring out how a
kalgebra morphism means figuring out where
x
and
y go. Let's start with
x.
x 
ϕ * (x)  

 = x ∘ ϕ  

 ∈ k[x,y]   
Errr... okay.... Let's try plugging in the special point
(0, 0), in which case we get
that
(x ∘ ϕ)(0, 0)  = x(ϕ(0, 0))  

 = x(R,S)  

 = R   
Ah, so
ϕ * (x) has to be a polynomial that evaluates to
R on
(0, 0).
Which means that it must be a polynomial with a constant term of
R:
 (1) 
Okay, that thing on the right is kind of complicated. But I believe! I believe that
it's simpler than that. Remember: I had a strong suspicion that this map would
end up representing a translation. could the right hand side, possibly, be simply
x + R?
Well: ϕ* is an isomorphism. Okay. And the fact that ϕ maps (0, 0) to (R,S) is
equivalent to saying that ϕ* maps (x,y) to (x  R,y  S). In particular, it
has to map the generators of the former ideal, x,y to generators of the latter
ideal.
NOW: are the generators of the latter ideal necessarily x  R and y  S? Not
necessarily. But it's pretty clear to see that they have to be at least
linear. So I can write (1) as
Similarly, I can write
OKAY: sooooooo this isn't quite a translation..... BUT, this does tell us that ϕ
must look like this:
(x,y) 
(R + ax + by,S + cx + dy)   
And remember: The contradiction I'm looking for is to produce a 0 in the image.
Is that possible? Well, I would like.
ax + by  = R  

cx + dy  = S   
BUAHAHAHAHAHA. A system of linear equations. And it's clear to see that it
must have full rank (remember we need to generate the maximal ideal
(x  R,y  S), and thus has a solution.
SO WE'RE DONE....
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Hi. So I decided to check for myself
just how "pretty clear" it was that the images must be linear. Of course, this
resulted in me spending
2 days trying to "check" this. I spent fucking hours.
Hours spent messing around with summations, exponents, indices.......... and
then discovered that
it's not even true. Which means that, as you can
tell from the mega strikethrough, we have to go all the way back to (
1):
Let's call this f. And let's write an analogous polynomial for ϕ * (y)
Let's call this g. So I can't necessarily claim that f and g are linear (thus
screwing up the "translation" dream). In fact, this is sort of the best I can do
unless I want to use the advanced "automorphism theorem of k[x,y]" linked
above. Fine then I'll have to do with using that as my base map. If that's what
ϕ* is, then ϕ is given by
(x,y) 
(R + ∑
γ_{i}x^{ai}y^{bi},S + ∑
δ_{i}x^{ai}y^{bi})  

 = (f(x,y),g(x,y))   
Now, why did I want translation or linearity in the first place? I wanted to
produce
(0, 0) in the image, to yield a contradiction. Indeed, I would like to find
a point
(x,y) that is zero on both polynomials. I.e. I want
Z(f) ∩ Z(g) to be
nonempty. Here's the thing. we know that
ϕ* maps
x
f and
y
g. So
ϕ must
map
(Z(x) ∩ Z(y)) to
Z(f) ∩ Z(g). But the former set is nonempty (it
contains
(0, 0)), so since
ϕ is an isomorphism, the latter set must be nonempty
too.
So there you fucking go. FUCKING DONE.
actually, we still need to show that
(X) ≃ k[x,y].... Fuck.
Alright. Fine. So
(X) = S^{1}k[x,y] where
S = {ff≠0 on
X}. Now
let's suppose that
f ∈ S. If
f is reducible, then it can be factored into
f = g ⋅ h where both
g and
h must be in
S. So let's assume without
loss of generality that
f is
irreducible. Then
f must be nonzero on all of
A^{2}, possibly excluding the origin. Well, if it has
no zeros at all, then it
would have to be a constant function (
f ∈ k). What if it was
0 at the
origin, and only the origin? Then
f(x, 1) would have to be a constant
polynomial (otherwise
f(x, 1) would have zeros. Say it had a zero at
a. Then
(a, 1) would be a zero of
f, contradicting nonzeroness outside the origin).
Actually, this would apply for any
f(x,b) where
b ∈ k. And I'm PRETTY
SURE that this means that
f has no
x terms. I know we JUST saw the
consequences of trusting my "pretty sure"ness, but.... I've already spent way too
much time on this. Exercise left to you, dear reader, lol (I'll hoist a hint
for you: plug in
1 and then a large value
a >> 1 for
y and see what
happens to the like degree
x terms in either case.). If you apply this same
"reasoning" to show that
f has no
y terms either, that means that
f ∈ k yet
again. Hence,
S = k. Which means that
S^{1}k[x,y] = k[x,y]. DONE.