I'm late, but welcome back. This exercise has a smooth start and a rough ending. Begin, reader, by comparing part
(c) and part (a). Turns out: Conics in projective space act more nicely than conics in affine space. All conics are
isomorphic to each other, and to P^{1}.
Now, to start this exercise off, it would be nice if we could start by showing that all conics are isomorphic
to each other, by reducing them to some "standard form". Back in 1.1c, I essentially showed that
translations (LEMMA 2) and invertible linear transformations (LEMMA 3) ("rotations") are isomorphisms.
Now, one big issue with my work there is that I kind of just "assumed" that a series of such translations and
rotations would get us to standard form, without actually showing how. I did link a Wikipedia page that claims this
to be true, but ain't that one hell of a cop out? In fact, if I just assumed this, then it turns out that a
huge segment of my work is actually extraneous. And anyway, that page appears to be discussing
the specific case of when our field is ℝ (whereas we're working over an algebraically closed field k).
...I probably need to make another post just to fix that. HOWEVER: We'll leave that for another day. Today, we're
going to be thorough for the nice, sexy projective case.
Okay, so let's work in P^{2} for now. First thing's first: It appears that, actually, translations are generally not
isomorphisms in projective space. A very simple example is that if you translate all the points of Z(f)
by, say, adding T to all of them, then the new set S can be written as Z(f_{T}) where, as in 1.1c,

Which, in general, is a nonhomogenous polynomial (e.g., think of the case f = x,T = (1, 0, 0)), and I don't see a
good way of constructing a homogenous set of generators for S
HOWEVER: We do not need translations. The power of projective space allows us to get all the conics merely via
means of linear transformation ("rotation").
One folly of this exercise is that it doesn't actually define what a "conic" is for projective space. So I'll give you a
definition: A conic in P^{2} is the zero set of an irreducible homogenous quadratic (definitely has to be irreducible, or
else there's no way the zero set is gonna be isomorphic to P^{1}, lol). So, in general homogenous quadratic's zero set is
going to look like this:
Ax^{2} + Bxy + Cy^{2} + Dxz + Eyz + Fz^{2}  = 0  

 = 0  

x^{T}Ax  = 0   
(sry for reusing the same variable name for different things. YOU CAN FIGURE IT OUT)
The key point here is that we can
diagonalize A, writing it as A = P^{1}DP where D is a diagonal matrix, and P
can be chosen as an orthogonal matrix (since A is symmetric!), so P^{1} = P^{T}, and therefore, we can write
x^{T}Ax  = x^{T}P^{T}DPx  

 = (Px)^{T}D(Px)   
if you look closely, you can think of this diagonalization as converting to a new quadratic by applying the change of
basis P on x (see here). Hence, if we show that invertible linear transformations are isomorphisms, we can assume
that all conics can be represented by the simpler standard form
Ax^{2} + Cy^{2} + Fz^{2}  = 0  

  
Furthermore, with the linear transformation
(note that since we're in an algebraically closed field, such square roots exist, and yes, A,B,C have to all be
nonzero, otherwise the conic would be reducible), we can actually assume that our conic looks like
x^{2} + y^{2} + z^{2}  = 0   
SO: simple linear transformations are sufficient to make all conics isomorphic to each other... provided that
(invertible) linear transformations are isomorphisms. Well, are they?
First of all, note that unlike for translations, linear transformations do map homogenous polynomials to homogenous
polynomials: An arbitrary linear transformation can be represented by
x 
a_{1}x + a_{2}y + a_{3}z  

y 
b_{1}x + b_{2}y + b_{3}z  

z 
c_{1}x + c_{2}y + c_{3}z  

  
and so given a homogenous polynomial γ_{i}x^{ai}y^{bi}z^{ci}, the corresponding polynomial can be written as
γ_{i}(a_{1}x + a_{2}y + a_{3}z)^{ai}(b_{1}x + b_{2}y + b_{3}z)^{ai}(c_{1}x + c_{2}y + c_{3}z)^{ai}   
which is homogenous by the binomial theorem
Now, following my 1.1c notation, given a linear transformation L, I denote the polynomial given by applying L^{1}
on the coordinates as f_{L} (So f_{L}(P) = f(L^{1}P).
Now, it's clear that for a point P ∈ P^{2}, f(P) = 0⇐⇒f_{L}(LP) = 0.
So letting
ϕ_{L} : P^{2}  → P^{2}  

P 
LP   
be the linear transformation by L, and correspondingly letting
ψ_{L} : k[x,y,z]  → k[x,y,z]  

f  → f_{L}   
be the kalgebra isomorphism of coordinate rings, it's clear that, given a variety X, ψ_{L} gives a homeomorphism
between X and Y = ϕ_{L}(X)
Now, back in 3.1a, I could use the isomorphism of coordinate rings induced by ψ to conclude isomorphism between
X and Y , but of course, Corollary 3.7 only applies to affine varities. So I have to show the isomorphism manually.
No worries: the homeomorphism part is already done, so I just have to show that pesky regularity property.
Suppose V ⊂ Y is open, and f : V → k is regular. Then I need to show that f ∘ϕ_{L} : ϕ_{L}^{1}(V ) → k is regular.
But given an open U ⊂ V where f = g∕h on U, then for any P ∈ ϕ_{L}^{1}(U),
hence it's locally rational, therefore regular. So ϕ_{L} is a morphism. DONE.
And what about showing that the inverse map ϕ_{L}^{1} is a morphism? Muahaha. Guess what. ϕ_{L}^{1} is just ϕ_{L1}.
That's right: the exact same logic proves that it's a morphism. hence ϕ_{L} is an isomorphism. I.e. linear
transformations of varieties in P^{3} are isomorphisms (actually, the same argument applies to P^{n}).
And, recall, since all conics in projective space can be linearly transformed into one another (in particular, into
x^{2} + y^{2} + z^{2} = 0), all projective conics are isomorphic to each other. Great. Now I just have to show that one of
these conics is isomorphic to P^{1}.
This is where I got stuck for 7 days.
You, see, I tried all sorts of nifty tricks trying to show that Z(x^{2} + y^{2} + z^{2}) ≃ P^{1}. I tried covering it with affines
and gluing the maps together, I tried to visualize a cone being mapped into the projective line, I tried to
parametrize the nonreal unit sphere, then the nonreal unit hemisphere.
Then I finally looked up a solution.
Guess what? Trying to construct the isomorphism from Z(x^{2} + y^{2} + z^{2}) is a bad idea. Remember? You can make
the isomorphism from any irreducible conic. Why was I so stubbornly sticking to this one? I mean, you have to
forgive me somewhat for trying. After all, it's the super special (that's why you are here with me (miracle baby
pornography superstar)) diagonalized form–the very form that all conics can "reduce to" via a simple procedure.
But insofar as producing an isomorphism with P^{1}, it's a huge pain: The squares in particular get in the way like a
motherfucker. Of course, there's some way to get an isomorphism directly from Z(x^{2} + y^{2} + z^{2}), but it's a pain.
The better choice of a conic is actually Z(xz  y^{2}). This is what the solution I read used, in any case. Now, this
might seem like an arbitrary choice at first glance. Why would it be any better than the special diagonalized form?
Well, honestly, it's kind of appalling that I didn't try this out. First of all, if you remember 3.1a, I actually used the
affine version of this curve to produce an isomorphism. Why didn't I think of doing the same here in projective
space? And here's another thing: While aimlessly making up crazy martial arts to try and produce a map, for some
reason I never looked back at the exercises in section 2 to see if they would be of any help. This part of the
exercise is about projective space, AND I NEVER THOUGHT TO LOOK BACK AT THE SECTION
THAT WAS ALL ABOUT PROJECTIVE SPACE. Instead, I wasted more and more time trying
to do super hyper cartwheel backflips through this only to repeatedly land on my butt and dance
around the room clenching my buttocks in pain then ending up in a fail compilation which leads to this
post getting licensed by Jukin media and getting bombarded by comments saying, "The stupidity
of Americans today never fails to astound me", written by someone whose profile picture is the old
default Youtube silhouette with a spider on it. My foot is finally in the door to becoming a lolcow.
I am, indeed, butthurt. But, hindsight now tells me that 2.12 exists. And, in fact, now that I've done
3.4, noticing that Z(xz  y^{2}) is simply the 2uple embedding of P^{1} in P^{2} (as it says in the very exposition of 2.12) finishes this exercise.