I.6.7
9/10/2021
I'm starting to self-insert as the girl in Fatalpulse doujinshi, it's
great. LEMMA 1: LEMMA 2 (Regular funcs on open sets of A1 are rational):
Hello, reader! Welcome to SEEZON FEENALAY. Indeed, this section only turned
out to have 7 exercises! Well, I skipped a few of them, but you'll have to give me a
break. I've generally been doing more exercises per section than an actual
math class would, rite? .... which is why I'm taking so long to get through
these sections.... So there you go. gimme some leeway here, aight, yo?
This is one of those times I'm thankful I made this blog. I would have struggled a
lot with this exercise, maybe not even been able to finish it, but I was able to
largely thanks to my hard work back in 3.1b
......................................................................
Anyway, I did create two very crafty lemmas during that exercise, which will help
me out here. Actually, turns out I'm doing this exercise in a completely different
way than our Queen intended, but whatevs. IF IT WORKS, IT WORKS. Here
are the lemmas, btw:
Given any two varieties X,Y where ϕ : X → Y is an isomorphism, then the
induced map ψ :
(Y ) →
(X) is an isomorphism.
If f ∈
(X) where X ⊂ A1 is open in A1, 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}.
And don't worry: I maintain a very careful mental androgyny. It's a delicate
balance–not too fem, not too masc–like walking on a tightrope. Whichever you
side you fall, your death is waiting for you. Ever play Super Mario 64? Imagine
Cool, Cool Mountain on one side and Lethal Lava Land on the other. Fall one
way, you freeze to death, fall the other way, you burn.... Or, will you walk the
rope? "What about F.L.U.D.D?" Heh. You clever bastard. This is why you're my
disciple. Well, what about it, then? You could choose to say "Wahoo!" and jump
off the tightrope, and hover nozzle your way forward to the age of 30, but
remember: Your tank's only 1/4 full. You think you'll make it with that much
water? Suppose you run out before reaching the platform, and you miss
the rope on the way down, your flapping skirt clipping right through
it. That "Wahoo!" might just be your last words. And let me tell you,
that would be an embarassment. An utter embarassment. You know who
else died before 30? Evariste Galois, in a duel of passion. You think he
said "Wahoo!" when the bullet pierced his chest? Lili Boulanger died
before 30. You think she said "Wahoo!" in the throes of intestinal pain?
Janis Joplin died before 30. You really FUCKING think her last words
were, of all FUCKING things—actually, it was a heroin overdose, so it
might have been... but you get my point, right? You damn well don't want
"Wahoo!" on your epitaph. When you're on the rope, pick your words
wisely. "Then what? How do I achieve solemn dignity? What should I
say on my leap of faith?" Well, well, now. Have you learned nothing?
That's right. Say it with me. Say it. SAY IT, YOU BASTARD. YOU
FAGGOT BASTARD, SAY IT. *Grabs you by the shoulders* LOOK INTO
MY EYES AND SAY IT. MOTHERFUCKER *Shakes you* YOU'RE
ALWAYS FUCKING WITH ME! YOU'RE ALWAYS FUCKING WITH ME!
ARRRRRRRRRRRRRRRRRRRRRRRRRGGGGGGHHHHHHH. I KNOW
YOUR TRICKS. I KNOW YOUR GOD DAMN FUCKING TRICKS! I KNOW
THE LENGTHS OF YOUR SELF-DENIAL. DON'T PLAY FUCKING DUMB
WITH YOURSELF. YOU'VE ALWAYS KNOWN. YOU'VE KNOWN FROM
THE BEGINNING. *Shakes you* SAY IT. COME ON NOW. SAY ITTTTTT.
THAT'S RIGHT..... That's right. You know what I'm talking about, don't you?
You've always known. It's been inside you. The sacred rites have been passed
down from generation to generation, but only some have the power to conjure
them with sincerity. Can you hear it, see it? The running organ fantasia. The
stadium of ominous green black flagstone marble, erected high into the dark
skies, studded on its circumference with spiked spheres. The dark, fey
hour. The crimson horizon. The rising incantation. So long, gay Bowser.
Okay, letting X = A1 -{P1,…,Pr}, Y = A1 -{Q1,…,Qs}. Let's assume
wlg that r > s. Then thanks to my lovely LEMMA 1, if I show that
(X)≄
(Y ), I'm done. And I actually know what both of these rings look like
thanks to LEMMA 2! Yaaaaaay. Here's the BIG TRICK: Based on LEMMA 2, we
know that in
(X), the elements x - Pi are invertible (whereas the
elements x - Qi aren't: since we're told these points are distinct). And
it's the opposite situation for
(Y ). Now, if we had an isomorphism:
ϕ : (X) | → (Y ) |
First of all, we know that linear elements have to map to linear elements.
"Ummm, how do you know that?" FUUUUUUUUUUUUUUUUUUUCK
DUUUUUUUUUUUDE I DON'T FUCKIN KNOWWWWWW SO MANY
EXERCISES LATELY HAVE RELIED ON THE ASSUMPTION THAT
ISOMORPHISMS MAP LINEAR ELEMENTS TO LINEAR ELEMENTS
LIKE REMEMBER I.5.14A AND OMGGGGGGG I SKIPPED 6.6C
BUT LIKE THAT'S BASICALLY WHAT 6.6C IS SO LIKE IT'S LIKE
SOOOOOOO COMMON AND I FUCKING SUCK AT PTOVING IT EACH
TIME SO I'M JUST GOING TO ASSUME IT CUZ IT'S INTUITIVE
AND AAAAAAAAHHHHH LEAVE ME ALONE LEAVE ME ALONE I
JUST TOOK A 1.5 HOUR NAP AND IT'S 9PM AND I FEEL LIKE A
ZOMBIE WHY DO I TAKE THESE STUPID IDIOTIC NAPS AND
KILL MY ENERGY IDKKKKKKKKKKKKK HJELP ME SOMEONE..
Second of all, we know that invertible elements have to map to invertible
elements. HENCE, x-R1 has to map to one of the x-Qi. BUT: of course we
assumed r > s, so at some point we'll run out of x - Qis and injectivity will
fail. But injectivity CAN'T fail cuz we need this to be an isomorphism.
Contradiction, DUN.
Now, the rest of the exercise asks us if the converse is true. Lemme state the
converse here:
If r = s, then A1 -{P1,…,Pr}≃ A1 -{Q1,…,Qs} (where the Pi,Qj are distinct)
A1 -{P1,…,Pr}≃ A1 -{Q1,…,Qr} (where the Pi,Qj are distinct)
SECONDLY, for complex analysis fans, if k = ℂ, these are Mobius transformations! .... idk what else to say about that. HELP ME, starsofcurtains!
Given a set of three distinct points z1,z2,z3 on the Riemann sphere and a second set of distinct points w1,w2,w3, there exists precisely one Mobius transformation f(z) with f(zi) = wi for i = 1, 2, 3.
ϕ : A1 -{P 1,…,Pr} | → A1 -{Q 1,…,Qr} |
to get an automorphism
ϕ : P1 = ℂ ∪∞ | → P1 = ℂ ∪∞ |
We know this has to be a Mobius transformation, thanks to 6.6. And, in
particular, it's a Mobius transformation where the Pis map to the Qis. This is
always possible if we only have 3 of each. BUT, as soon as we select 3
distinct points from each set, the resulting Mobius transformation (the
extension) is FIXED (we're using uniqueness: the "precisely one" part
here)–well, OKAY COMBINATORICS FELLAS, I know, there's technically
errr i guess 3! different Mobius transformations, cause le order matters
here, but there's ONLY 3! (!), a finite amount. So if we remove an extra
point from each set, P4,Q4, and if this choice doesn't fit one of the
preestablished Mobius transformations, there's no way to extend this
isomorphism into an automorphism. SO WE ARE NOT GUARANTEED AN
ISOMORPHISM FOR r > 3. THERE IT IS. YAAAAAAAAAAAAAAAAAAY.
Whew. Not bad for a seezon finale eh? I've had worse. I'll see you folks in the next
section! Also i didn't even get myself cake last time... so maybe this time?