2024-09-13

Theorem

Let be a ring. Then

is an (adjoint) equivalence of categories.

Corollary

Let be a scheme. Then is an abelian subcategory, whose inclusion preserves all finite limits and all colimits.

Infinite products don’t agree.

Proof

Let be a functor, where is small.

Claim: , or where is finite in -modules is quasi-coherent.

Let be an affine open. Then

because restriction is a left adjoint to pushfoward , and so preserves all colimits. Further, when is finite, we have that

because restriction is exact (it preserves terminal objects, finite products, and equalisers). This means we can reduce to the case where . But is exact, fully faithful, preserves finite limits and all colimits with image (this is Theorem and also Lemma 2), and therefore we have

So we’re done.

Theorem (Serre)

Let be a ring. Let . Consider

be an exact sequence of -modules. If is quasi-coherent, then

is exact. Note that is always left exact because it’s a right adjoint:

If all of , , were quasi-coherent, then we would be done because is exact.

Proof

We know that

is exact. So we just need to show that is surjective. Let . The exact sequence of sheaves (the fact that is surjective, i.e., surjective on stalks) implies that there exist such that (i.e. an open cover ) and such that for all . #TODO diagram. To glue these together, we need to know they agree on overlaps. Let . We don’t know that these are . However, we certainly have that

so . Let where is an -module (by Lemma 3). Then . We have by the cocycle condition

Now, we can clear denominators: there exists large such that for all ,

Now write (since ). Set

Observe that still since come from . Then on ,

So since is a sheaf, there exists a unique such that . But since , then , and so we’re done.

Remark

This result tells us that there is no higher sheaf cohomology on an affine scheme. There is also a converse. If is a quasi-compact scheme, i.e., it admits a finite cover by affine schemes and it has no higher cohomology, then is actually an affine scheme.

Quasi-compact and quasi-separated morphisms

Let be a map of schemes. Then . Can we do ?

Definition (Quasi-compact, quasi-separated)

Let be a topological space.

  • is quasi-compact if every open cover admits a finite subcover.
  • is quasi-separated if the intersection of any two quasi-compact opens is quasi-compact.

Definition (Quasi-compact, quasi-separated map)

Relative version of the above.

Let be continuous. We say that

  • is quasi-compact if for all quasi-compact open, is quasi-compact open.
  • is quasi-separated if for all quasi-separated open, is quasi-separated open.

Examples

  1. is quasi-compact and quasi-separated. In fact, is quasi-compact if and only if there exists a finitely-generated ideal such that .
  2. If is Noetherian, then every open in is quasi-compact.
  3. is not quasi-compact.

Lemma

Let be a topological space and an open cover. If are quasi-compact, then is quasi-compact.

Proof

Omitted.

Proposition

Let be a scheme. Then the following are equivalent:

  1. is quasi-compact.
  2. is quasi-compact.
  3. has a finite cover by affine opens.

Proof

Since is quasi-compact, then we immediately have 1. 2. 3.. It remains to show that 3. 1.. Let be open. Then for some (because is a PID). Let an affine open form the fibre squares:

Actually, the outside square is Cartesian as well, and so quasi-compact. If , then , which is quasi-compact by the Lemma.