II.4.2
4/28/2022
With a stunning dearth of anything resembling intuition regarding the notion of
”separation,” let’s meditate a bit... Drops of dew on our fingers. Lick em.
Young man, I do not mean to clear your mind, but to fill ’er up. You, me,
we aint engineers of ye olde, trained neither theoretically to a T, nor in
practice to a P. Is there a bigger picture to all this? This week, you are
merely in love with the construction of this battery. A construction in a
more ambitious location than your peers who broke the virginity: From
here, you can show off your gallantry. We pick out locations that look
like death zones, but are actually relatively safe, so long as you do well
under stress. You do well under stress boy? You like the sound of fire?
You link the beat of the bang? Look visible, it’s all for show: Give em
something to engage with, but not hit. That’s the rule of coastal artillery: Buy
high, sell low–I mean, build high, shoot low. Be a temptation–the fleur of
their eyes. The more roundshot you swallow, the more you lure them in.
Being under fire means they’re in range. Being under fire means they’re
under fire. If you’re under fire and still standing, it means they’re stuck at
armslength–the sweet spot for red hot shot. Look at me. I’m not hiding behind
the rampart, I’m standing on it. Is there something special about me? That I’m a
short? A little thin, yet thickskinned? I’m in plain sight, and I’m not
dodging–they’re just missing. You see? My figure serves the purpose to
confirm that measurement: That’s the location I picked out for you. And for
them: My figure serves the purpose of imprinting my serious, young face
into their minds. The face that cannot be hit, not because it’s special,
but because it was intentionally placed there. And for you: Think of me
as a scarecrow. A temporary decoration to your construction. I am to
you as this carnation on my lapel is to me. Weak stemmed, onioning
at the bloom. Your uniforms are no longer adorned with the feminine
perfection of delis, and neither is mine with speculative euphoria of tulips
(BUY HIGH, SELL LOW). My hands are dirty; I’m down to earthwork.
A carnation: A bullseye that is actually a vortex. Every shot of theirs
should peel back a layer, only to reveal more, while your brilliant red
continues to seduce them: Into spilling their own blood. And when I alight
from the rampart to construct another, who will take my place? Who
would like to act as the measuring stick for who has who by the balls?
We’ll begin by establishing that the set maps of f and g are the same. So in this
subsection, we’ll only consider things in the set level.
Let’s start by looking into ”separation” a bit. The main diagram of reference, of
course, is this one:
Note the identities i. I would like to break this up a bit. Let’s set Z = Δ(Y ).
Then the fact that those i’s are identities, means that I can draw
Now, surprisingly, I haven’t said anything about separatedness yet. All that is just
how the diagonal works. All separatedness gives us is the fact that Z is closed.
Hmm...
Now here’s something important (again, thanks to the identities): on Z, p_{1} = p_{2}.
If z ∈ Z, let y = Δ^{−1}(z) then
p_{1}(z)  = p_{1}(Δ(y))  
= i(y)  
= y  
= i(y)  
= p_{2}(Δ(y))  
= p_{2}(z) 
Now, let’s bring X into the picture. To say that f,g are ”Smorphisms”, is to
say that the following diagram commutes:
Where ∙ can refer to either f and g. Although we’ve guaranteed nothing about f
and g being identical, when composed with Y → S, they terminate in
the same manner. Which means that the following diagram commutes:
And I earn that dashed line, h, via the universal property of the fiber product.
Now what? Well, let’s imagine that we could replace the Y ×_{S}Y with Z. Then
I’d be able to draw this:
In particular,
f(x)  = p_{1}(h(x))  
= p_{2}(h(x))  
= g(x) 
So we’d have f = g (as set maps).
READY FOR SOME TARGET×PRACTICE? TO A T, TO A P? Can
you fit h(X) inside Z ⊂ Y ×_{S}Y ? Because if you can, we’re done.
Ahh, all this abstract talk, and it feels like we’ve done such a huge amount of
work, but we’ve done nothing. We haven’t used any property provided to us:
Density, reducedness, separatedness. Now, I have hindsight, so I’ll tell
you this. Separatedness is meant to be a generalization of T_{2}ity. These
spaces have the property that a two maps into them that agree on an
open dense subset agree everywhere. We are proving this holds on set
maps. Ideally, we’d invoke density and separatedness. It’d make even
more sense to invoke separatedness, given that we’re already working with
the diagonal. And it’d make sense to invoke density because it would be
frightening if we proved the equality of two set maps without doing so.
Target practice? Let’s refine this even more: Let’s refine the tip of our arrow.
Would it not suffice to merely fit h(U) into Z? Because in that case,
Z  ⊃ h(U)  
h^{−1}(Z)  ⊃ h^{−1}(h(U))  
⊃ U  cheatsheet 
Now HERE’S where separation would come in: Z is closed, so h^{−1}(Z) is
closed. h^{−1}(Z) is a closed set containing the open dense subset U. Hence
h^{−1}(Z) = X. This argument is inspired by an old one back in varietyland, from
Hartshorne himself:
So
h(X)  = h(h^{−1}(Z))  
⊂ Z  cheatsheet 
AND WE’RE DONE... Or we would be done, provided that we prove the
following inclusion

Now, certainly it must have something to do with the fact that f and g are equal
on U. Here is my argument. Define:
 (1) 
which is the same whether I use f or g for X → Y . That gives me this guy:
However, you can check that the following, more complicated diagram also
commutes:
Now d (1), the way we defined it, is made up of f and g, so letting h′ be the
restriction of h to U, I can even draw:
But the map obtained by universal property of the fiber product is unique, hence
h′  = Δ ∘ d  
h′(U)  = Δ(d(U))  
h(U)  = Δ(d(U))  
⊂ Δ(Y )  
= Z 
And we are done.
Now we have to show that the sheaf maps are equal, and it suffices to do this
locally. So let

be maps of affine schemes that agree as set maps, and agree fully on an open
dense subset U. Assume X is reduced. I need to show that the corresponding
maps

are equal. I.e. given b ∈ B, I would like to show that f ∗ (b) = g ∗ (b). Now, let
D(h) ⊂ U (assume h≠0). Then the restrictions

are equal. And thus so are the corresponding ring maps

But these maps can be factored as
f′′ : B  A → A_{h}  
g′′ : B  A → A_{h}  
Hence,
f′′(b)  = g′′(b)  
∃n ≥ 0 : h^{n}(f ∗ (b) − g ∗ (b))  = 0  
Let x = f ∗ (b),y = g ∗ (b). I want to show x = y. Again, what I do have is
 (2) 
(Remember, we might not be an integral domain, so no zeroproductproperty here). This
is far as a I got without looking at the solution (+1).
The most blatant issue was not being aware of the construction that is the
annihiliator:

In particular, this is an ideal.
What’s worse, however, is that there is another crucial assumption that the
solution makes:
We can assume without loss of generality that D(h) is an open dense set.
For any n > 0, D(h) ⊂ V (Ann(h^{n}))
Let P ∈ D(h), so h P. Now let s ∈ Ann(h^{n}). Then h^{n} ⋅ s = 0. Since P is prime, h^{n} ∈ P or s ∈ P. But h^{n} ∈ P h ∈ P, a contradiction, so we must have s ∈ P. Since s was arbitrary, Ann(h^{n}) ⊂ P i.e. P ∈ V (Ann(h^{n})).
Suppose n > 0. Then we can apply the lemma: Let α = Ann(h^{n}), so
D(h) ⊂ V (α). Since D(h) is dense, V (α) = X. This is again inspired by
Remark I.3.1.1
So α = (1), which means that 1 ∈ Ann(h^{n}). Hence
h^{n} ⋅ 1  = 0  
h^{n}  = 0 
But we assumed n > 0, making h nilpotent. But X is reduced (THERE SHE
IS), so that would mean h = 0, a contradiction. Hence n = 0, and (2) becomes
h^{0}(x − y)  = 0  
1 ⋅ (x − y)  = 0  
x − y  = 0  
x  = y 
Done.