Yes, we're doing 3 parts in one post. Strap in.
Half of the challenge is figuring out what the flying fuck is going on, and one thing we've learned in this blog is that
notation is one way to help with that.
Some lemmas, to help me out:
LEMMA 1:
If g is a degree r monomial in k[y_{0},…,y_{N}], then θ(g) is a degree r ⋅ d monomial in k[x_{0},…,x_{n}]
COROLLARY:
If g is a degree r homogenous polynomial in k[y_{0},…,y_{N}], then θ(g) is either a degree r ⋅ d homogenous
polynomial in k[x_{0},…,x_{n}], or 0
Alrighty. Let's begin.
PART (a):
So I want to show that α is homogenous, eh? First instinct is to find a finite set of homogenous generators for α,
but that isn't a great idea because it's hard, especially so early in the exercise where you're not sure what's going
on. Easier: Let's just show that α is generated by its homogenous elements (i.e. instead of a finite set of homogenous
generators, take all the homogenous elements as a generating set). Indeed, given some f ∈ α, let's decompose f
into homogenous components:
 (1) 
where deg Y _{i} = i
And all we have to do is show that each of these homogenous components are in α in the first place.
Taking θ of both sides, we get
θ(f)  = θ(∑
_{i=0}^{l}Y _{i})    

0  = θ(∑
_{i=0}^{l}Y _{i})  (since f ∈ ker θ)    

0  = ∑
_{i=0}^{l}θ(Y _{i})     
Now note that we've split
θ(f) into homogenous components, with some components possibly being
0: Indeed, by
the COROLLARY, each
θ(Y _{i}) is either
0 or has degree
i ⋅ d. BUT it can't be the latter because then
θ(f) would
be nonzero (because it'd have nonzero homogenous components). So
Y _{i} ∈ ker θ for each
i. Thus,
α is
homogenous.
And how about showing it's prime? Another lesson we've learned through this blog is that it's often easier to use the
"QUOTIENT INTEGRAL DOMAIN" thingy instead of showing primeness directly. Now, I did not construct a
finite set of generators so I can't do the usual Euclidean division trick like in
1.2 etc. But it's still very easy. What is
the IMAGE of
θ? Well, it's a subring of
k[x_{0},…,x_{n}]. And subrings of integral domains are themselves
integral domains (see e.g.
here). So we're DONE. (actually, I now just realized that I could have used
this "subring" argument numerous times in the past and skipped a bunch of extra work. WOOPS)
PART (b):
I'll repaste the exercise here for reference!
One inclusion is, in fact, easier than the other. It's the im
ρ
Z(α) one.
Give me an element
P ∈ P^{n} and pick an affine representative
a = (a_{0},…,a_{n}) for it (BTW, if you're confused
about the phrasing of the exercise where it says "
M_{0}(a)" etc., that's what they mean).
Now give me an element
f ∈ α, and break up
f into monomials:
Write
Y _{i} = c_{i}y_{0}^{bi,0}y_{
1}^{bi,1}
y_{
N}^{bi,N}. And note that
θ(Y _{
i}) = c_{i}M_{0}^{bi,0}M_{
1}^{bi,1}
M_{
N}^{bi,N} So we can then
write
But the right hand side is just the point
ρ(P) plugged into
f, so
f(ρ(P)) = 0. Since
f ∈ α was arbitrary,
ρ(P) ∈ Z(α). And since
P was arbitrary, im
ρ ⊂ Z(α) indeed.
Yeah.... "easy". Now you want to see "hard" (aka "some calculation")? Get fucking ready.
To make everything easier to see, I'm actually going to work for the specific case where
n = 2 and
d = 3. DON'T
WORRY. It does generalize, and the generalization is straightforward (except for one part, as you'll see, lol). We've
been ignoring the actual format of the monomials
M_{i} up to now. What I'm going to do is actually show you what
they look like. For instance, we know that
θ maps
y_{0} to some
ddegree monomial in
k[x_{0},…,x_{n}]. Which
one? Well, that depends on how you order the monomials. I'll order them like this, without loss of
generality:
y_{0} 
x_{0}^{3}  

y_{1} 
x_{0}^{2}x_{1}  

y_{2} 
x_{0}^{2}x_{2}  

y_{3} 
x_{1}^{3}  

y_{4} 
x_{2}^{3}  

y_{5} 
x_{0}x_{1}^{2}  

y_{6} 
x_{0}x_{2}^{2}  

y_{7} 
x_{1}^{2}x_{2}  

y_{8} 
x_{2}^{2}x_{1}  

y_{9} 
x_{0}x_{1}x_{2}   
"WTF is with that unnatural ordering". Trust me, reader. It'll help. Yes, in the more general case, I'm putting the
x_{0}^{d} term in front, and then setting
y_{i} = x_{0}^{d1}x_{i} all the way up to
i = n.
If you give me a point
c = (Y _{0},…,Y _{N}) ∈ Z(α), what do I want? I want
c to be in the image of
ρ. I want
(Y _{0},…,Y _{N}) = (M_{0}(a),…,M_{N}(a)) for some
a = (X_{0},…,X_{n}) ∈ P^{n}. I.E. I want a solution to the following
system of equations:
Y _{0}  = X_{0}^{3}  

Y _{1}  = X_{0}^{2}X_{1}  

Y _{2}  = X_{0}^{2}X_{2}  

Y _{3}  = X_{1}^{3}  

Y _{4}  = X_{2}^{3}  

Y _{5}  = X_{0}X_{1}^{2}  

Y _{6}  = X_{0}X_{2}^{2}  

Y _{7}  = X_{1}^{2}X_{2}  

Y _{8}  = X_{2}^{2}X_{1}  

Y _{9}  = X_{0}X_{1}X_{2}   
(the
Y _{i}s are the "given" scalars, and the
X_{i}s are the "unknowns") Erm...well that's a complicated system
of equations. It's not even
linear (...okay, now I'm actually starting to miss linear algebra). What
do?
It's pretty easy to show that at least one of
Y _{0},Y _{3},Y _{4} must be nonzero (EXERCISE LEFT TO READER LOL).
Well, you've already pointed out my funny ordering of the monomials: Yes, I'm assuming without loss of
generality that
Y _{0}≠0. Now take a look at that first equation:
Y _{0}  X_{0}^{3} = 0. Since we're in an
algebraically closed field , we can actually narrow down the possible solutions of X_{0} to a set of at most three
elements. Let's call them X_{0}′,X_{0}′′,X_{0}′′′. I know that X_{0} ∈{X_{0}′,X_{0}′′,X_{0}′′′}. Okay. Cool. Which
one?
This is where I got stuck for a considerable amount of time, playing with the other equations and
trying to eliminate some of the options... It didn't turn out well. In fact, it turned out so badly that I
got angry and decided, "FUCK IT. LET ME JUST PICK ANY ONE OF THEM AND SEE WHAT
HAPPENS."
YES. LET'S JUST SET X_{0} = X_{0}′ AND SEE WHAT GOES WRONG. Well, here's something interesting: once
I've arbitrarily set X_{0} in that manner, that FULLY DETERMINES the other variables. Look at the next 2
equations (in general, the next n equations). I can divide by X_{0}^{2} on both sides (since it's nonzero, heh) and I
get
X_{1}  = Y _{1}∕(X_{0}^{2})  

X_{2}  = Y _{2}∕(X_{0}^{2})  

  
So now we have a possible solution for X_{0},X_{1},X_{2}. We have something that at least satisfies the first 3 equations
(in general, the first n equations). Does it satisfy the rest? Well, let's try the 4th equation: Y _{3}
X_{
1}^{3}. Suppose
equality didn't hold. Then we'd have:
Y _{3}  ≠X_{1}^{3}  

Y _{3}  ≠(Y _{1}∕(X_{0}^{2}))^{3}  

Y _{3}X_{0}^{6}  ≠Y _{1}^{3}  

Y _{3}Y _{0}^{2}  ≠Y _{1}^{3}  

Y _{3}Y _{0}^{2}  Y _{1}^{3}  ≠0   
What I'm saying is that the point c = (Y _{0},Y _{1},Y _{2}) is not zero at the polynomial g = y_{3}y_{0}^{2} y_{1}^{3} ∈ k[y_{0},…,y_{N}].
However, what happens when we map this under θ?
y_{3}y_{0}^{2}  y_{1}^{3} 
(x_{1})^{3}(x_{0}^{3})^{2}  (x_{0}^{2}x_{1})^{3}  

 = x_{1}^{3}x_{0}^{6}  x_{0}^{6}x_{1}^{3}  

 = 0   
Yep, g ∈ ker θ. And we assumed that c ∈ Z(ker θ) in the first place, so that would be a contradiction. So our
(X_{0},X_{1},X_{2}) has to satisfy that 4th equation.
.....Now we just have to verify it satisfies the other 5 equations. Fuck. That's a lot of work. And this is just the case
of d = 3,n = 2. What if we had like, d = 100,n = 1000? This is one place where I guess I need to take a detour
from the specific case and show it works for the more general case...
Okay, so in the more general case, what's going on?
The relationships given by θ are
y_{0} 
x_{0}^{d}  

y_{1} 
x_{0}^{d1}x_{1}  

y_{2} 
x_{0}^{d1}x_{2}  

 

y_{n} 
x_{0}^{d1}x_{n}  

 

y_{k} 
x_{0}^{b0}x_{1}b_{1}
x_{n}^{bn}  

 

  
The last case is the most general case, where the right hand side is an arbitrary d degree polynomial
(b_{0} + b_{1} +
+ b_{n} = d).
Given a point c = (Y _{0},…,Y _{N}) ∈ Z(α), we need a point a = (X_{0},…,X_{n}) ∈ P^{n} to satisfy
Y _{0}  = X_{0}^{d}  

Y _{1}  = X_{0}^{d1}X_{1}  

Y _{2}  = X_{0}^{d1}X_{2}  

 

Y _{n}  = X_{n}^{d1}X_{n}  

 

Y _{k}  = X_{0}^{b0}X_{1}b_{1}
X_{n}^{bn}  

 

  
Again, picking an arbitrary solution for the first equation, we use the next n equations to set
X_{1}  = Y _{1}∕(X_{0}^{d1})  

X_{2}  = Y _{2}∕(X_{0}^{d1})  

 

X_{n}  = Y _{n}∕(X_{n}^{d1})  

  
And now we need to show that our choice satisfies the arbitrary equation
Y _{k} 
X_{
0}^{b0}X_{1}b_{1}
X_{n}^{bn}  

  
So again, we assume for the sake of contradiction that it doesn't, so
Y _{k}  ≠X_{0}^{b0}X_{1}^{b1}
X_{n}^{bn}  

Y _{k}  ≠X_{0}^{b0}(
)^{b1}
(
)^{bn}  

Y _{k}  ≠X_{0}^{b0}
 

Y _{k}  ≠X_{0}^{b0}
 

Y _{k}  ≠
 

Y _{k}  ≠
 

Y _{k}(X_{0}^{d})^{(b1+
+bn)}  ≠(X_{0})^{d}Y _{1}^{b1}
Y _{n}^{bn}  

Y _{k}(Y _{0})^{(b1+
+bn)}  ≠(Y _{0})Y _{1}^{b1}
Y _{n}^{bn}  

Y _{k}(Y _{0})^{(b1+
+bn)}  (Y _{0})Y _{1}^{b1}
Y _{n}^{bn}  ≠ 0  

  
yes, that was a fucking pain to type out. NOW, following what we were doing earlier, Let's consider the
polynomial
g  = y_{k}(y_{0})^{(b1+
+bn)}  (y_{0})y_{1}^{b1}
y_{n}^{bn}   
So we're saying that g(c)≠0 Putting this under θ yields
(x_{0}^{b0}x_{1}^{b1}
x_{n}^{bn})(x_{0}^{d})^{(b1+
+bn)}  (x_{0}^{d})(x_{0}^{d1}x_{1})^{b1}
(x_{0}^{d1}x_{n})^{bn}   
Which is clearly equal to 0, which means that the g ∈ Z(α), which means that we'd need g(c) = 0, a
contradiction.
SO THAT'S THE FUCKING GENERAL CASE. *Pant pant* BACK TO d = 3,n = 2.
So, let's review: From Y _{0} = X_{0}^{3}, we had three choices: X_{0} ∈{X_{0}′,X_{0}′′,X_{0}′′′}. What did I do? I just
daredeviled it and arbitrarily picked one of them. And what was the result? It just werkd. Yes, turns out, we can
pick any one of them and that yields a solution.
When you think of the equations
Y _{0}  = X_{0}^{3}  

Y _{1}  = X_{1}^{3}  

Y _{2}  = X_{2}^{3}  

  
They each yield 3 possibilities
X_{0} ∈{X_{0}′,X_{0}′′,X_{0}′′′}  

X_{1} ∈{X_{1}′,X_{1}′′,X_{1}′′′}  

X_{2} ∈{X_{2}′,X_{2}′′,X_{2}′′′}  

  
If you pick X_{0} = X_{0}′, then that forces the choice X_{1} = X_{1}′,X_{2} = X_{2}′ (assuming I've ordered them in the
right way), so we can set a = (X_{0}′,X_{1}′,X_{2}′). Similarly, picking X_{0}′′ forces a = (X_{0}′′,X_{1}′′,X_{2}′′), and
picking X_{0}′′′ forces a = (X_{0}′′′,X_{1}′′′,X_{2}′′′). Any of these 3 choices work (in general, d choices); i.e. any of
them map to c under ρ. So we're done.
PART (c):
Alright, to show that it's a homeomorphism, we need to show that ρ is bijective. And that both it and its inverse is
continuous.
"Wait... bijective? Didn't you just say there 3 possible points that map to a point in Z(α)? That contradicts
injectivity."
.........FUCK. So.... uhh......... did I completely fuck up part (b)?
Okay, are you ready for some high IQ shit. We actually do have some leeway, because we're working in projective
space. Indeed: There is a way that (X_{0}′,X_{1}′,X_{2}′), (X_{0}′′,X_{1}′′,X_{2}′′), (X_{0}′′′,X_{1}′′′,X_{2}′′′) can all actually
be the same point: If they are scalar multiples of each other.
Well, are they?
Let me start by setting s = X_{0}′∕X_{0}′′, so that X_{0}′ = sX_{0}′′.
Okay, so given that, how does X_{1}′′ relate to X_{1}′? Let's see...
X_{1}′′  = Y _{1}∕(X_{0}′′)^{2}  

 = Y _{1}∕(sX_{0}′)^{2}  

 =
(Y _{1})∕(X_{0}′)^{2}  

 =
X_{1}′  

  
FUCK. By an entirely analogous argument, we get X_{2}′′ =
X_{2}′. I wanted X_{1}′′ = sX_{1}′ and X_{2}′′ = sX_{2}′.
But instead of s, we have
.
How fucked am I? I am actually not so fucked. You see, s is a very special scalar. Remember: (X_{0}′)^{3} = Y _{0}. But
also, (X_{0}′′)^{3} = Y _{0}. In other words, (X_{0}′)^{3} = Y _{0} and (sX_{0}′)^{3} = Y _{0}. I.e. (X_{0}′)^{3} = (sX_{0}′)^{3}. Isn't that
strange? Slightly unintuitive? What's up with this scalar? If I didn't notice this pecularity, I might have thought I
did something wrong on part (b). But check this out:
The mysterious scalar s is actually a root of unity! In our case, it's a 3rd root of unity (in the general case, a
dth root of unity). In particular, we can divide by 1∕s^{2} on both sides (more generally, 1∕s^{d1}) and we get
 (2) 
Holy fuck. Hence, we actually do have that X_{1}′′ = sX_{1}′, and X_{2}′′ = sX_{2}′′. and a similar argument for the
triple prime case. So we can in fact write (X_{1}′,X_{2}′,X_{3}′) = (X_{1}′′,X_{2}′′,X_{3}′′) = (X_{1}′′′,X_{2}′′′,X_{3}′′′) since
they only differ by scalars, thus making ρ injective, and making us.... SAFE!
Motherfucker. So bijectivity is done. Now we need to show the continuity requirements
Note that a closed set C of Z(α) can be written as C = Z(β) ∩ Z(α).
Alright. Here we go. We already have a map θ : k[y_{0},…,y_{N}] → k[x_{0},…,x_{n}]. And it's pretty easy to see that
ϕ^{1}(Z(α) ∩ Z(β)) = Z(θ(β)), so ϕ^{1} is a closed map (TL note: "pretty easy to see" means "I have
faith").
To show ϕ is closed, I'll do it by way of the d = 3,n = 2 example. Take Z(f_{1},…,f_{r}) is a closed set in P^{2}. Yes,
we can assume a finite set of generators by Noetherianness, and we can in fact assume they're homogenous. We want
to create another set δ ⊂ k[y_{0},…,y_{N}]. such that Z(f_{1},…,f_{r}) is mapped onto Z(δ) ∩ Z(α) Okay. let's look at
f_{1} for example (just call it f for now). If f has degree that is a multiple of d, then just rewrite f_{1} in
terms of the y_{i}s. E.g. if f = x_{0}^{2}x_{1}^{4}  x_{1}^{3}x_{2}^{3}, then set g = y_{1}y_{3}  y_{3}y_{4}. and we know a point
that satisfies f has to satisfy g when mapped under ρ, so we can add g to δ. Well... there are other
ways to write g, aren't there? Fuck it, just include every possible way to write g in δ just in case
lol (IMO, it doesn't matter, because we've already included the requirements of α, s we know that
g = y_{1}y_{3}  y_{3}y_{4} must "evaluate to" f = x_{0}^{2}x_{1}^{4}  x_{1}^{3}x_{2}^{3}. But no harm in adding the extra gs I
guess).
On the other hand, f's degree may not be a multiple of d. E.g. we may have

In this case, we can "force" it to have a degree of d, by multiplying both sides by e.g. x_{0}:
allowing us to set a corresponding "g_{0}" accordingly (in our case g_{0} = y_{9}  y_{0}). But that choice of x_{0} is kind of
arbitrary, isn't it? Yes, Because we're only requiring that our point satisfy x_{0} or f. We included all the points
where x_{0} = 0! i.e. We included an entire hyperplane. Woops. But no worries, because we can also
include
and find a corresponding g_{1} for that. And finally

and grab a corresponding g_{2}, And we include all the g_{i}s in δ. Now a point can't get away with simply being on a
hyperplane, because if it doesn't satisfy f in at least one of the equations, then it would be all 0, which doesn't
count as a point in projective space.
So part (c) is done. That was.... a little handwavy admittedly. Oh well. I think it works.
I also wanted to throw in part (d), but I've spent 2 hours typing this out already, so I'm spent for the night. Which
does beg the question of "is it really worth typing all this out if it takes hours to do so", to which I have
to keep reminding myself that the answer is "yes, because publishing this gives me accountability
and is practically the only thing keeping me from falling off track, and also all this work would get
lost in my scrap paper if I don't type it up leaving me with basically no reference when I need to
look back, and typing it up forces me to reinforce the ideas by thinking about how to organize it as
well as forcing me to catch a shitload of errors in my work because, compared to just putting it in a
notebook, I have higher standards when I have to justify my work to you, dear reader *blows kiss*"