II.4.2

4/28/2022

4.2

0.1 The set maps

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, p1 = p2. If z Z, let y = Δ1(z) then

 p1(z) = p1(Δ(y)) = i(y) = y = i(y) = p2(Δ(y)) = p2(z)

Now, let’s bring X into the picture. To say that f,g are ”S-morphisms”, 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 ×SY with Z. Then I’d be able to draw this:

In particular,

 f(x) = p1(h(x)) = p2(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 ×SY ? 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 T2-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 h1(Z) is closed. h1(Z) is a closed set containing the open dense subset U. Hence h1(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 hbe 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.

0.2 The sheaf maps

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 h0). Then the restrictions

are equal. And thus so are the corresponding ring maps

But these maps can be factored as

 f′′ : B A → Ah g′′ : B A → Ah

Hence,

 f′′(b) = g′′(b) ∃n ≥ 0 : hn(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 zero-product-property 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.

Where that comes from is a complete mystery to me. But I gotta move, and I am not delaying this post another week to figure it out.

Now that we have introduced the annihilator, it’s time for a lemma:

0.2.1 Lemma

For any n > 0, D(h) V (Ann(hn))

Proof

Let P D(h), so h P. Now let s Ann(hn). Then hn s = 0. Since P is prime, hn P or s P. But hn P h P, a contradiction, so we must have s P. Since s was arbitrary, Ann(hn) P i.e. P V (Ann(hn)).

QED

Suppose n > 0. Then we can apply the lemma: Let α = Ann(hn), 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(hn). Hence

 hn ⋅ 1 = 0 hn = 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

 h0(x − y) = 0 1 ⋅ (x − y) = 0 x − y = 0 x = y

Done.