Let be an algebra and be a right -module. Then is flat if and only if
for all left ideals of .
Special case (PIDs)
If , a principal ideal domain, then for some . Then or is not a zero divisor:
In the case , is always true because is projective over itself.
In the case , then is zero for all . This holds if and only if is torsion-free (by definition).
Proof
)
The direction is true because is the derived functor of .
)
Suppose for all left ideals . We must prove that for all modules . Consider a projective resolution of :
Chopping off and tensoring with ,
Then by definition,
An element is represented by an element , which can be expressed as a sum
At this point, we no longer care about , we only care about the part that we see. To this end, let be the submodule of generated by the . Hence, actually lies in the smaller group and by the inclusion , we get a map
This argument shows that:
Each element actually lives in a smaller Tor group, the corresponding where is finitely-generated.
! Hence, if for all finitely-generated modules , then for all .
Finitely-generation allows us to induct.
It now suffices to prove that for all finitely-generated . Let us induct on the number of generators.
Base case: Number of generators
We just have . So we are done.
Induction step
Now suppose
Take the submodule , which receives a surjective map from given by , . Hence for a left ideal of . Now, is generated by , and there are of these. We have a short exact sequence
By the long exact sequence of , we have
We have:
by assumption;
by induction.
So .
Example: torsion-free flat in general
PIDs are -dimensional. Let us consider something with higher dimension. Let , where is a field. Then is a torsion-free module. We compute . From the assignment, there is a free resolution
Dropping and tensoring with , we have
Note that so . Solving for , we must have . So there is a non-zero element since is not hit by . Hence is not flat, despite being torsion-free.
Grothendieck spectral sequence
Consider two left exact functors and :
How do we compute the right derived functors of the composition (“chain rule”)?
Assume that , , have “enough injectives” (this is true for , and also sheaves).
Theorem (Grothendieck spectral sequence)
Suppose in addition that takes injective objects to -acyclic objects (e.g. injective objects). Then there is a spectral sequence
Q: How do we check the above condition in practice? The answer lies in the PropositionProposition below.
Definition (Acyclic)
An object is -acyclic if for all . Injective objects are -acyclic.
Proposition
Let be an exact functor and be a right adjoint of . Then sends injectives to injectives.
This is the method for checking the aboveabove condition in practice.
Proof
Let be injective. We must prove that is injective, i.e., ie exact. That is, let
be an exact sequence. We are required to prove that
is exact. By adjunction, this is equivalent to the following sequence
being exact. But because is exact,
is exact. Since is injective, is exact.
Example
Let be a group and be a normal subgroup. Consider
If then because there is a well-defined -action: for and
It is clear that the composition . However, if we want to set up the spectral sequence, we should consider the following question: Does the functor have a exact adjoint? The answer is yes. Consider the functor
where receives an action from via . This is not a very exciting functor, but it is an exact functor: it sends an exact sequence to itself.
Claim
is an adjoint pair:
Proof
Let . Since acts trivially on , then acts trivially on the image of , i.e., factors as
Hence, we map to . Conversely, given , we take to be the composition