Only A and B today, folks. C shall be done another day, maybe. Also speaking of A and B and B and A, AND NOW YOU SEE ANOTHER ME I'VE BEEN RELOADED YEA-AH I'M FIRED UP DON'T SHUT ME DOWN. I'M LIKE A DREAM WITHIN A DREAM THAT'S BEEN DECODED. I'M FIRED UP I'M HOT DON'T SHUT ME DOWNNNNN. Omg, guys. It's ABBA: Reloaded. Catchy AF. ABBAsed ABBA.

Now, reader. Will you leave me standing in the hall, or let me enter? ...Oh, the hall.... FUCJ. Reader...? Knock knock. KNOCK KNOCK. Oi, oy, lemme in ye little punk. Come onn. Come on y'little... Knock knock... Ah! Don't ABBAndon me! Don't ABBAnish me! Let me hear your voice, in any form! ABBAlk at my doorstep ABBArgain, and in this ABBArren hall I'd hold siege–ABBAttalion of one! ABBArk at me like an grunting, untuned ABBAssoon, and I'll hark it were a whistling ABBAmboo. The ABBArbs of love have ripped your voice ABBAre–a beach flooded into ABBAy. I used to ABBAwl at your ABBArbaric ABBArrages, but now it's like ABBAnter to me! Ahhhh, your slurry ABBArley, your hoarse ABBAsil, mixed into ABBAtter and ABBAked with lye into aural ABBAlm. Delicious, like ABBAnana split for my ears. Lemme hear it again!

......... Reader? Reeeeeeeeaaaaaaader. Knock knock. I know you're in there. I see the lights on. Knocky knockyyyyy~. Hmmm, reader must be shy. Well, as I always say, if they don't respond to knocky knocky, then you gotta picky picky. My back pocket is always stashed for picky picky. Here we are–Snake Rake, tension wrench. Kneel, penetrate, slide. Slide in, slide out. In, out... In... out... in... out. Scrubby, scrubby, scrub the hole. torquey torquey, turn the plug–oh there's a spin—! .... Eh? ...It's.. stuck. Oi, oi, reader, don't tell me you have a spool! Let's see here. Pin 1, rattle. Pin 2, rattle. Pin 3.... is a cockblock. Oh, reader! A 3 to separate us 2! A pin twixt kin! How could you! Okie dokie, let's lift, lift, put your arms in the air, counter-rotate.... WEEEEEEEEE THERE'S MY SPIN. *Barges in* HELLO, MY SWEE–Oh, they escaped out the window.

Well, well, the picker becomes the scavenger, eh? Hmm, let's see... A yellowing, toothmarked spooled apple core, standing up on a small wood round table. A recent dine, eh? Yellowing, but not brown–ripped off like raw meat, wounds wound fresh from a ravenous rampage. Muahaha, reader always used to cut their apples with a knife in my presence. Yes, one eats differently in private. Eating alone is a specially intimate experience–selfcestual–and I'm here for the postmortem. What mess took place at this table for two? Sticky liquid painting their face, skin stuck between their teeth, unashamedly picked out by their bare fingers? Lick, lick. Mmuuuahhh! There it is, my indirect kiss. Oh–oh, what's this? A fallen sock. Sniff, sniff. Ahhhhhhhh, what a fragrance! I'll pocket this one–I am the sock gnome for this apartment room, nay, the sock puff. Yes, I've huffed and puffed my way in, and I'll huff and puff my way through. Skip, skip. What a lovely playhouse! I could frolick here all the way till their return. Hm, what's this? A diary?

A

SO, we're identifying P¹ with A¹ ∪{∞}: given a point (X,Y ) ∈ P¹, let X∕Y ∈ A¹ ∪{∞} ((1, 0) getting mapped to ∞ ofc)

Our "fractional linear transformation" is given by

ϕ : A1 ∪{∞}	→ A1 ∪{∞}			(1)
x	(ax + b)∕(cx + d)			(2)
ad - bc	≠0			(3)

A few things to note here.

FIRSTLY: Implicit here is that we send -d∕c to ∞. Singularity=infinity. Ez.

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

FUG.

THIRDLY, I'm kinda wondering why we had to make the "affine identification" in order to define the map. I mean, couldn't we just have said this?

ϕ : P1	→ P1			(4)
(x,y)	(ax + by)∕(cx + dy)			(5)
i.e.(x,y)	(ax + by,cx + dy)			(6)
				(7)

Well, admittedly, that does make some parts of the exercise harder (e.g. finding an inverse). So the affine indentification is maybe not necessary but helpful. OK!

FOURTHLY, I'm going to assume that c≠0 (which means thanks to ad-bc≠0, we also have d≠0. Obviously the case c = 0 actually makes things easier, so no need to worry about that.

OKAY: So the goal of part A is to show that ϕ is a morphism. Been a while since we've done one of these morphism checks. Let's start by finding an inverse set map. To do that, we'll jump back to good old high school PreCalc. How do we do this? Set it equal to y and solve? OK.

y	= (ax + b)∕(cx + d)
y(cx + d)	= ax + b
ycx + yd	= ax + b
ycx - ax	= b - yd
x(yc - a)	= b - yd
x(yc - a)	= b - yd
x	= (-dy + b)∕(cy - a)

Holy jesus mother of god. In my notes I fucked this up so badly that I got x = (cy -a)∕(dy -b). WHAT THE FUCKING FUCK LOL. I was sleepy, ok?

In any case, the above gives us an inverse for ϕ... and it happens to be another "fractional linear transformation"! Let's call this map δ:

δ : A1 ∪{∞}	→ A1 ∪{∞}
y	(-dy + b)∕(cy - a)

This map sends a∕c to ∞. And, btw, you may be wondering "where do either of these maps send ∞?" Well, in order to make these inverses of each other, we're forced to set

ϕ(∞)	= a∕c
δ(∞)	= -d∕c

And those choices are pretty sensible if you look at the original definitions and think of it from a limits standpoint. You guys see it? HELP ME, burypink!

FUCK.

Well, anyway, IT WORKS. ϕ is a bijection. NOW: All I have to do is show that ϕ is continuous and... morphic...?, and I'm done. "Ummm what about δ?" Psshh, δ is a fractional linear transformation just like ϕ, so the same arguments that apply to ϕ will apply to delta.

OK: I sort of lied. I'm actually going to show that ϕ is a closed map (which translates to δ is continuous), but that's equivalent in the end. Okay: Let's say that f ∈ k[x,y]. I'd like to make a polynomial f′ so that the image of ϕ satisfies f′ exactly when the preimage satisfies f. This is sufficient to show that ϕ is closed (ϕ(Z(f₁,…,f_t) = Z(f₁′,…,f₂′)). Well, I LIED AGAIN, but let's go forth with this as our motivation.

Let's let (X, 1) ∈ P¹. Hence, from the A¹ ∪{∞} perspective, we're assuming that X≠∞. Let Y = ϕ(X). Thanks to X≠∞, we can assume that Y ≠a∕c. OK. Let's set

g(x)

= f(x, 1)

So ofc

f(X, 1)	= 0
⇔g(X)	= 0

Now, letting d = deg g I'm gonna define

f′(y)

= g( )(cy - a)d

So, f′ is an element in k[y]. And note that

f′(Y )	= 0
⇔g( ) = 0	or (cY - a)d = 0
⇔g( )	= 0	(since Y ≠a∕c)
⇔g(X)	= 0

yaaay, we can homogenize the polynomial f′ and thus we have a "counterpart" for f.... For all the points that are NOT ∞. What about infinity THOUGH. What if (1, 0) = ∞ satisfies f? Would its image (a∕c) satisfy our homogenized f′? I almost went on a struggle to figure this out, until I realized: WHO CARES. Singletons are closed in P¹, so we can just union in a∕c and the resulting set still remains closed. CLOSED SET MAP TO CLOSED SETS. ϕ IS CLOSED (δ is cont). WEEEEEEEE. (By the same argument δ is closed so ϕ is cont).

So now I just have to show that our ϕ is a morphism. It's been a while, so here's the definition of morphism:


As for regular functions, here's the definition for quasi-projective varieties:


well, this is super tricky if you try and use the original definition (2) (WTF do you do with ∞ and the singularity). HOWEVER, I think this is where my alternate definition (6) comes in and, it's pretty easy to see that it trivially works (I won't even go through the proof). DUN.

B

errrrr..... Look at Corollary 6.12:


yea.... It's just (i) ⇔ (iii), rite?. THERE'S NO WORK TO DO HERE LOL.