Hello. Not an impressive turnout for March, but you know what they say: Never trust a Hartshorner. Our elk graze
in bursts, the irregularity governed by whim. Actually, I'm a woodpecker. I peck out my prospective home in bursts,
the irregularity governed by whim. Actually, I don't understand how a compound interest works, let alone a
mortgage. You'll have to forgive me, reader: I'm currently busy getting buttfucked by conics. Oh, this part of the
exercise has nothing to do with conics, sure. But the next part is buttfucking me. It's buttfucking me so hard that I
don't even care anymore. I'm starting to selfinsert as the girl in Fatalpulse doujinshi, it's great. I'm just living
through the stupidity, baby. Yes, we're still on part (b), and the hiatus is slated to continue. I either have to figure it
out or look at the solution. Enough time has passed to use the latter, but you know... sunk cost and
stubbornness. We've all been there. But hey you know what I think, motherfuckers (bad advice incoming),
STAY TRUE TO YOURSELF AND JUST LIVE THROUGH THE STUPIDITY HAHAHAHA.
If it wasn't clear already: This is an educational blog. I'm teaching you how not to live your life. It's actually very
noble. Whenever I think about holeinmyheart's sacrifice for the enrichment his avid readers, it brings
a tear to my eye. holeinmyheart is such an inspiration to me. The way he doesn't update his birds
page, the bad CSS... He's doing it all for our sake. He's such a lovely site admin. And, honestly, he's
just such an amazing person in general. I admire him so much. Sometimes I
just wish... he'd look at me, you know? That he'd pay attention to me a bit. But I know I'm nothing
special. *Sigh*... I really do admire him though. Can I be honest for a second? Okay.... it's kind of
embarassing but... sometimes I masturbate to holeinmyheart. Ughhhh, I can't believe I said that out
loud! I always feel so guilty after I do it, but sometimes I can't help it. When I lie down in bed, I
often feel kinda lonely, so I just start thinking of him and.... Uggggghhhh! I'm such a bad person. If
holeinmyheart ever found out about this, he'd be so disgusted. He should be. Someone like me, having those
sorts of thoughts about him... Haha. Sometimes I feel so ashamed when I'm talking to him, like I
can't face him directly. He's probably thinking, like, "what's wrong with this weirdo, looking away like
someone who committed a crime" haha. I just feel like he's reading my mind or something. I feel so
dumb. But I still find so much inspiration in him. I know it sounds silly, but he's the thing that keeps
me going, sometimes. I sometimes wish I could thank him somehow, but idk how. I feel like i'd just
stammer all over the place like i always do... haha. maybe one day, i can get closer to him. One day...
Anyway, part (c) notwithstanding, I do have the goods for this exercise. Remember Proposition 3.5?
Last time, I was uneasy about whether an isomorphism in Hom(A(Y ),
(X)) necessarily induced an
isomorphism in Hom(X,Y ). After trying to prove it, my guess is that the answer is NO.
However: I figured out that it works in the reverse (in a slightly more general form, even!). Here's a lemma for that:
LEMMA 1:
Given any two varieties X,Y where ϕ : X → Y is an isomorphism, then the induced map ψ :
(Y ) →
(X) is
an isomorphism.
PROOF:
You see, the proof of 3.5 tells us that ψ is given by

(we know that the image is in
(X) because ϕ "carries regular functions to regular functions", by virtue of being a
morphism) On the other hand, let me define a function
δ :
(X)  →
(Y )  

g
g ∘ ϕ^{1}   
Both of these are
kalg homomorphisms and clearly they are inverses, so we're dun.
LEMMA 1 FIN
Now, here's why this is relevant to this exercise. Give me
Y = A^{1} and an open set
X ⊂ Y . Then if there
were an
isomorphism
X ≃ Y , then there would have to be an isomorphism
A(Y ) ≃
(X). If I can show the condition of
the contrapositive, I'm good. I.e. if I'm able to show
A(Y )
(X), I'm good (So I've reduced a "geometric"
problem to an "algebraic" problem).
Now obviously
A(Y ) = k[x]. What about
(X)? Well to figure that out, I'm gonna need another lemma.
I know, there's an assload of bookkeeping here. Here's the thing, folks.
(X) are all the regular functions on
X,
right? Let's review the definition of "regular":
Don't you see how complicated these are to classify? A regular function on
X isn't just a "rational function". It's a
locally rational function. If
f ∈
(X), I can't just say that
f = g∕h. I have to say, oh,
f = g∕h in some subset
U of
X, and maybe
f = g′∕h′ in some other subset
U′ of
X, etc. until I cover
X with open subsets.
That is so inconvenient. The fact that regular functions are in general not globally rational but locally rational
makes the algebra hard. It makes it difficult to calculate what
(X) actually
is.
However, I found out that in the single dimensional affine case (in our case), regular functions
are (globally)
rational. Check this out:
LEMMA 2 (Regular funcs on open sets of
A^{1} are rational):
If
f ∈
(X) where
X ⊂ A^{1} is open in
A^{1}, then as a function on
X,
f = g∕h on
X for
g,h ∈ k[x] where
h
nowhere zero. Hence
(X) = {rational functions on X
}.
PROOF:
We know that
f is regular, so it's locally rational. I.e. we may have
f = g∕h on
U, but
f = g′∕h′ on
U′. How are
these two representations of
f related? I want to show that they're the same, or at least can be represented as the
same rational func.
I know that
g∕h = g′∕h′ on
U ∩U′, so
gh′hg′ = 0 on
U ∩U′. I.e.
Z(gh′hg′) ⊃ U ∩U′. Now since
X
is open, by
1.6 I know that
X is irreducible. Which means that
U ∩ U′ must be nonempty. And being open
sets of not just
X, but
A^{1}, I know that they're dense in
A^{1}. Thus
Z(gh′ hg′) is a closed sed
containing a dense sets of
A^{1}, meaning that we must have
Z(gh′ hg′) = A^{1}. Hence we can write
gh′ = hg′ on all of
A^{1}, which is equivalent to saying that
gh′ = hg′ in the coordinate ring
k[x].
Whew. So that was another "convert the geometry problem into an algebra problem" procedure. Now, readers: How
do we compare two polynomials? That's right: Factor em. Since, we're in an algebraically closed field, I
can factor g,
h′, h, g′ and write
A(x  g_{1})
(x  g_{a})B(x  h_{1}′)
(x  h_{b}′)  = C(x  g_{1}′)
(x  g_{c}′)D(x  h_{1})
(x  h_{d})   
And since this factorization is unique, I know that AB = CD, and the linear factors on the left must match up
with the linear factors on the right, perhaps permuted.
Now, in the case that a = c and (after a possible repermutation) the g_{i}s match up with the g_{i}′s, I can actually
conclude that g = g′ and therefore h = h′, and thus g∕h = g′∕h′, which is what I want. But what if the g_{i}s and
g_{i}′s don't match up? What if, say, there's an (x  g_{1}) on the left hand side but there's no i such that g_{1} = g_{i}′.
Since the factorization is unique, I then know that then g_{1} has to match up with one of the h_{i}s. Repermuting
it so that it's i = 1, I can say that (x  g_{1}) = (x  h_{1})... But this just means that (x  g_{1})
appears in the numerator of g∕h and (x  h_{1}) appears in the denominator, so I can cancel them out to
get a new rational expression g∕h, where certainly h≠0 on U (since I'm essentially getting rid of a
hole) and f = g∕h on U. So what do I do? Just keep cancelling out elements until g∕h = g′∕h′.
Hence, if f has a rational representaion on U and a separate rational representation on U′, I can assume
without loss of generality that it has a rational representation on U ∪ U′: f = g∕h on U ∪ U′.
Alright, that's good. But I want f to be globally rational. I want to write f = g∕h on all of X.
If I start with U, I can glob up another U′ so f is rational on U ∪ U′. And then maybe from there, I can glob up
another U′′, and then after that another U′′′, so f is rational on U ∪ U′∪ U′′∪ U′′′. But that still might not
constitute all of X, like I want. Uh oh.
Well, by the definition of regularity, I know that there's a U_{P} at least for each point P in X, so these U_{P}s form an
open cover of X, but again, that maybe infinitely many open sets to glob up, so we're still stuck at the same
problem.
Can I, perhaps, make a finite subcover then? If X were just quasicompact, then I would be done, (since I can
certainly glob up a finite amount of open sets). So, is X quasicompact? Well, from 1.7c, I know that since A^{1} is
Noetherian, X is at least Noetherian. But then 1.7b tells me that X then has to be quasicompact.
SO LEMMA 2 DUN.
Actually, while writing this, I realized that pretty much the exact same argument works for any open set X ⊂ A^{n}
(since k[x_{1},…,x_{n}] is a UFD). n doesn't have to be 1. So if X is an open set of A^{n}, regular functions are rational
functions. NOTE: This is not the same thing as saying "if X is quasi affine, regular functions are rational
functions". Quasiaffine varieties are sets open with respect to any variety. So if, e.g., all we knew
about X were that it was an open set of some variety Z(f), we wouldn't be able to make the same
argument (the part in the proof where I claim the zero set of the polynomials are dense in A^{n} is where
the proof would fail). BTW: I honestly forgot for a while that quasiaffine meant not just open in
A^{n} but open in any variety, which caused some major confusions while thinking about this lemma.
OKAY. Back to the main thread of the proof. Errr... what were we doing? Right. I wanted to show that
A(Y )
(X) and I was trying to compute what
(X) was. Now I know that it's simply all the rational
functions whose denominators are nonzero on X. I.e.
(X) = S^{1}k[x], the localization of a ring where
S = {hh≠0onX}.
So our problem reduces to showing that, in general, k[x]
S^{1}k[x], for S≠k {0} (because this is just the case
where S^{1}k[x] = k[x] i.e. X = A^{1})
Alright. Let's say we had a kalgebra isomorphism
ψ : k[x]  → S^{1}k[w]  

x 
g∕h   
Let δ = ψ^{1}. So we must have
δ(g∕h)  = x  

δ(g)δ(1∕h)  = x  

  
So between δ(g) and δ(1∕h), one of them has to have degree 0 (but neither have a value of 0) and the other has to
have degree 1.
CASE: deg δ(g) = 0
Let's say deg δ(g) = C where C ∈ k (C≠0). Then
Cδ(1∕h)  = x  

δ(1∕h)  = (1∕C)x  

δ(1∕h)δ(h)  = (1∕C)xδ(h)  

δ(1)  = (1∕C)xδ(h)  

1  = (1∕C)xδ(h)  

  
which is a contradiction because x is not invertible in k[x]. CASE 1 DUN.
CASE 2: deg δ(1∕h) = 0
So this time δ(1∕h) = C for some C≠0 in k, and
δ(g)C  = x  

δ(g)  = (1∕C)x  

  
Now, from this point, note that I can't just multiply by "δ(1∕g)" on both sides because 1∕g might not actually be
an element in
(X)! (g might not be in S).
on the other hand, remember that δ is a kalg isomorphism. And since polynomials over (1∕C)x generate all of
k[x], polynomials over ψ((1∕C)x) = g generate all of S^{1}k[x]. In particular, for a given degree ≥ 1
polynomial in S, say, f (at least one such polynomial exists, since we assumed S≠k {0}, we would have
which, clearly, is impossible. CASE 2 DUN.
And that's the end of this proof. Note that one consequnce of this is that affine parabolas are NOT isomorphic to
hyperbolas/ellipses (since parabolas are A^{1} and hyperbolas are A^{1}  0)
See you NOT soon, as I cannot seem to unbuttfuck myself out of part (c). But I will be back. Keep an eye out, like
those black sheaf symbols.