"Umm, teacher, when am I going to use the Frobenius morphism in real life?" Ummm, you're not.
You came here to waste your time, not become a better person. You're not going to go out in the
park and see a couple feeding each other icecream and be like, "Ohhh. That's just like the Frobenius
morphism!". If you can think of a situation where the Frobenius morphism saves someone's life, let me know.
Yea, I don't know why I'm here either. I could spend all this free time on selfimprovement, attending to
some urgent life things, and finally getting myself together and pulling myself out of the impending
doom that is coming in at many angles towards my near future. BUT remember, reader, as I've said
before: The purpose of this blog is not to get better, but to get worse. So here we are. At least there's a
little "antilesson" to be had at the end of this exercise. First we have to handle part (a), though.
PART (a):
Let's start by showing that ϕ is bijective onto Z(y^{2}  x^{3}).
Given (X,Y ) ∈ Z(y^{2}  x^{3}), we know that
Now, let
T,T be the square roots of
X (since we're in an algebraically closed field), so that
(T^{2})^{3}  Y ^{2}  = 0  

(T^{3})^{2}  Y ^{2}  = 0  

(T^{3}  Y )(T^{3} + Y )  = 0   
so that
Y = T^{3} or
Y = T^{3} = (T)^{3}. In the former case, let
t = T, and in the latter case let
t = T. Then
clearly
ϕ(t) = (X,Y ). Hence,
ϕ is surjective.
So is
ϕ injective? Let's see:
ϕ(t_{1})  = ϕ(t_{2})  

(t_{1}^{2},t_{1}^{3})  = (t_{2}^{2},t_{2}^{3})  

t_{1}^{2} = t_{2}^{2}  and t_{1}^{3} = t_{2}^{3}  

t_{1}t_{2}^{2}  = t_{2}^{3}  

t_{1}t_{2}^{2}  t_{2}^{3}  = 0  

t_{2}^{2}(t_{1}  t_{2})  = 0   
The case
t_{1} t_{2} = 0 is exactly what we want. And the case
t_{2}^{2} = 0 implies that
t_{2} = 0 and
t_{1} = 0, so
t_{1} = t_{2}
again. Hence
ϕ is injective.
Great! So
ϕ is bijective onto
Z(y^{2}  x^{3}). Now we have to show that it is a homeomorphism.
Let
f(x,y) ∈ k[x,y] so
Z(f) is a closed set in
A^{2}. Then
T  ∈ ϕ^{1}(Z(f))  

⇐⇒ϕ(T)  ∈ Z(f)  

⇐⇒f(ϕ(T))  = 0  

⇐⇒f(T,T^{2})  = 0   
Hence, letting
g(t) = f(t,t^{2}) ∈ k[t], we have that
Z(g) = ϕ^{1}(Z(f)). So
ϕ is indeed continuous.
IMPORTANT: Note how in this direction, given
f in the
A^{2} side, there is an easy way to get a corresponding
polynomial
g on the
A^{1} side. But what happens when we try the other direction? We need to show that
ϕ is a
closed map to prove that it's a homeomorphism. But if you give me an element
f ∈ k[t], what is
ϕ(Z(f))? What
polynomial will I get for the
A^{2} side?
I struggled trying to construct one for a good hour or so, and then I realized: wait, if there were a way to get a nice,
single polynomial from
f, that would actually be a problem: that would indicate that there's an isomorphism of
coordinate rings, which would imply that
ϕ is an isomorphism, which we know according to the exercise
can't be
true.
Z(f) isn't gonna map to some simple
Z(g), it's gonna map to
Z(somethingmorecomplicated
). But still: what
would that be?
Then I realized, oh wait,
f comes from a singleindeterminate coordinate ring (
k[t]): we can fully factor it into
linear factors:
f = (t  t_{0})
(t  t_{n}). But wait.... that just means that
Z(f) is a finite set. So
ϕ(Z(f)) is a
finite set, which is trivially closed in
A^{2}. (unless
f = 0, in which case
Z(f) = A^{1}, so
ϕ(Z(f)) = A^{2}, since
ϕ
is surjective). Oh, right...... I just rederived the fact that
A^{1} has the finite complement topology....
YEP. Well,
ϕ is a homeomorphism. Good. One consequence of this, by the way, is that
Z(y^{2}  x^{3}) is
irreducible (since
A^{1} is), so it's actually a variety, with coordinate ring
k[x,y]∕(y^{2}  x^{3}). In fact, we can
view
ϕ : A^{1} → Z(y^{2}  x^{3}) as the morphism induced by the following map of coordinate rings
ψ : k[x,y]∕(x^{3}  y^{2})  → k[t]  

x 
t^{2}  

y 
t^{3}   
So
ϕ is therefore a bicontinuous
morphism. Finally, since we're talking about affine varieties, the map of coordinate
rings tells all: This induced map is clearly not a
kalg isomorphism (since, e.g.
ψ is not surjective: it doesn't reach
t), hence
ϕ is not an isomorphism either. DONE.
PART (b):
Here's the
Frobenius morphism! Proof is basically identical to (a), so I'll skip most of it. The most difficult part is
showing that
ϕ is injective. So let's say we had
ϕ(x) = ϕ(y). I.e.
I got stuck here for like an hour or so as well....
Here's the problem: Fixing
y, the polynomial
x^{p}  y^{p} = 0 could in general have
p many solutions for
x. One of
them is the one we want
x = y, but how do we guarantee that will occur? I mean, for that to be the only solution,
this polynomial would have to factor into...
(xy)^{p}. Yeah, fucking right. What crazy ass fucking nonsense is that?
In your dreams, motherfucke–oh, right.
THERE U GO, my _______ vision goggle wearers. Do not work hard to make your dreams come true. Work hard to set up a context exotic enough that
they have to come true.