We considered the ring where is an integer and is a field. Then is an injective -module.

Let be the -dimensional -module with acting by . Let’s compute an injective resolution.

Now, we want to embed into another -module. It turns out that

works. So, an injective resolution is

The injective resolution becomes -periodic.

Let’s apply a left exact functor to this injective resolution. A left exact functor is . Let’s compute its right derived functors at . To do this, we replace the complex by

and then apply . Now we need to know what is . But if and ,

Therefore we are forced to have for some . All such actually work, so as vector spaces. The resulting sequence is

The maps are given by considering

as long as . So we actually have

Taking homology, we find that

At ,

In general, if , the th right derived functor is (definition of )

If is right exact, it has left derived functors , .

A projective resolution of is a long exact sequence

We drop , and apply and take homology ():

The idea is that is also a derived functor of

A projective resolution of is

We drop to get

Now we apply . Now note that for all . So, noting that is a contravariant functor so that it flips all the arrows, we get (for )

Hence .

Suppose

is a short exact sequence.

We can apply . We get a long exact sequence

We have , which maps to some class . This is the obstruction to being the image of an element in , which if it exists would mean that the original short exact sequence splits.

We can instead apply (which is contravariant). We now get a long exact sequence

Now, maps to the same by magic.

Q: When are

the same?
A: If there exists an isomorphism such that the following diagram commutes:

Consider the category . We have for each , the short exact sequences

for if and only if .

classifies short exact sequences

with respect to the equivalence described aboveabove.

Given , how do we produce a short exact sequence? Let

be a projective resolution of . is represented by a map :

We now take . Then there are maps

which we can check sit in a short exact sequence. (Done in Assignment 1 Question 1.)