I.6.7

9/10/2021

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

I'm starting to self-insert as the girl in Fatalpulse doujinshi, it's great.

......................................................................

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:

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.

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}.

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 = A

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-R_{1} has to map to one of the x-Q_{i}. BUT: of course we
assumed r > s, so at some point we'll run out of x - Q_{i}s 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 A^{1} -{P_{1},…,P_{r}}≃ A^{1} -{Q_{1},…,Q_{s}}
(where the P_{i},Q_{j} are distinct)

Or a more concise way of stating it is

A^{1} -{P_{1},…,P_{r}}≃ A^{1} -{Q_{1},…,Q_{r}} (where the
P_{i},Q_{j} are distinct)

(again, I'll let the former be X and the latter be Y ) Now, let's do a little reading between the lines, shall we? The answer to this is "No". How do I know this? Because if the answer was "yes", then Queen Hartshorne wouldn't have asked it separately. Our Queen would have simply phrased the exercise as and if and only if statement. HENCE, the answer is "No". The converse is not true.

Now, can I make up an example to show the converse is not true? NOPE, LOL, BECAUSE I SUCK. FUUUUUUUUUUUUUUUUUUUUCK MYYYYYYYYYYYYYY ASSSSSSSSSSSSSSSSSSSSSS AAAAAAAAAAA THIS SECTION FUCKED ME IN THE ASS SO HARD. I ANSWERED "NO" AND ITS CORRECT I'M CONSIDERING THIS EXERCISE DONE LOLOLOLOL.

.

.

.

.

.

.

.

.

.

.

.

UPDATE: I... actually figured it out. This is crazy, but I actually did. I felt SO SUCKY when writing the paragraph above. Like, wow, I'm a failure and can't do anything. Even when I looked online, the solution just told me "the converse is not true when r > 3", with an explanation that didn't make any sense. Seriously, WTF? WHERE DID YOU GET 3 FROM?????? HOW AM I SUPPOSED TO KNOW SOMETHING BREAKS DOWN AT 3???? honestly i couldn't fathom why 3 would be such a special number here, and somewhere in the next few hours I basically gave up on this exercise. Obviously there's some advanced bs i'm missing or smething right?

Well, before going to bed I basically thought of it again, and then I thought of "3" again, AND THEN I FUCKING REALIZED SOMETHING. here's the thing: as I mentioned, I did the first half of this exercise in a different way you're supposed to. Remember this corollary from last time?

The way you're supposed to do this exercise is by using (i) ⇔ (ii). The isomorphism of X and Y (ii) yields an isomorphism of P

well, last time was all about automorphisms of P

SECONDLY, for complex analysis fans, if k = ℂ, these are Mobius transformations! .... idk what else to say about that. HELP ME, starsofcurtains!

GUESS WHAT: THIS IS ACTUALLY THE KEY POINT (for my proof). ysee, I'm not quite a complex analysis fan. But one of the few things I learned in the past from my disastrous, embarassing attempt at ""self-study"" of the subject is this:

Given a set of three distinct points z_{1},z_{2},z_{3} on the Riemann
sphere and a second set of distinct points w_{1},w_{2},w_{3}, there exists precisely one
Mobius transformation f(z) with f(z_{i}) = w_{i} for i = 1, 2, 3.

YOU GUYS SEE IT? SEE THOSE 3s? OK: Letting k = ℂ If we were extending an isomorphism

ϕ : A^{1} -{P_{
1},…,P_{r}} | → A^{1} -{Q_{
1},…,Q_{r}} |

to get an automorphism

ϕ : P^{1} = ℂ ∪∞ | → P^{1} = ℂ ∪∞ |

We know this has to be a Mobius transformation, thanks to 6.6. And, in
particular, it's a Mobius transformation where the P_{i}s map to the Q_{i}s. 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, P_{4},Q_{4}, 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?