A Faithfully flat morphisms
In this appendix, we document some important facts about flat and faithfully flat morphisms.
Let be a ring homomorphism. We say is flat if the functor
A morphism is flat if for all and , the map is flat. This in particular implies the pullback functor
is quasi-compact and quasi-separated, then
being exact implies
Compositions and pullbacks of flat maps are flat.
We only have to show that the pullback of flat maps is flat. Since flatness is local, it suffices to show that if
is a flat map of rings and
is any map of rings, then
is flat. But if
is a exact sequence of chain complexes, then
where we think of as an -module via . So this is exact.
Let be flat. Then the following are equivalent:
is faithful, i.e. if is a morphism of quasi-coherent sheaves over , and , then .
If , then .
reflects exactness, i.e. if a sequence is such that is exact, then so is .
When these hold, we say is faithfully flat.
(1) (2): Take .
(2) (3): Apply (2) to the homology groups of .
(3) (1): iff is exact.
(4) (2): Take . We may assume is in fact coherent, for contains a coherent subsheaf and preserves subsheaves by flatness. So if , then .
Pick such that . By surjectivity, there is a field and a map that sends the unique point to and has a lift to (e.g. by first picking a map to that hits a preimage of ). This means the pullback is non-empty. Moreover, by Nakayama, and is free since is a field. So the pullback of to is non-zero. Hence .
(2) (4): Let , and an affine open containing . Set and extend by zero. Then implies there is some affine open such that . Then a prime of is a prime of that gets mapped to under .
Using (4), it is clear that
Compositions and pullbacks of faithfully flat maps are faithfully flat.
We say a property P of morphisms satisfies fpqc descent if whenever we have a pullback diagram
with faithfully flat, then has property P iff does.
Flat morphisms satisfy fpqc descent.
is flat. If we have a sequence
of quasi-coherent sheaves on
is exact since
are flat, and since
, we know
is exact by faithfulness.
Finite morphisms satisfy fpqc descent.
[Proof sketch] The pullback of a finite morphism is clearly finite. For the other direction, We will prove the affine version. The gluing step part (which is the hard step) is annoying and will be omitted.
Suppose that is faithfully flat and is an -module. We want to show that being finitely-generated implies is finitely generated.
Suppose generate , and . Then the generate , since they generate as an -module and is faithfully flat, hence reflects surjectivity.