We have , and satisfying the cocycle condition

is a -cocycle. From this we get and .

The setup is we have a finite Galois extension, and (so ). Last timeLast time, we found that there is an injection

Consider the vector space . Tensoring with , we get . The automorphism group is

Suppose we have is a -cocycle. We can use this to define a new action of on :

where is the “old” action (a cocycle is a “difference” between two actions). To see that this is an action, we check:

This action satisfies:

We axiomatise this:

Let be an -vector space. A semilinear -action on is an action of on satisfying 1.1. and 2.2. above.

Let be a basis of . We define for

Let be a -vector space. Let . Then for , we have

A morphism between two vector spaces with semilinear actions is a map which is -linear such that the following diagram commutes for all :

There is an equivalence of categories

Whatever the definition is, we want an equivalence:

For two functors and

we want there to be natural isomorphisms (the counit) and (the unit), i.e.:

is an equivalence of categories if and only if it is both fully faithful and essentially surjective.

A functor is full if is a surjection for all and .

A functor is faithful if is an injection for all and .

We say that is fully faithful if it is full and faithful.

A functor is essentially surjective if for all , there exists such that .

Given an adjoint pair , the unit is and the counit is , both natural transformations.

Let’s prove that is fully faithful. We consider the case in Definition (Full)Definition (Full) and Definition (Faithful)Definition (Faithful). We need to show that

Let . Where can go? By -linearity, we have a commutative diagram

So we must have for all , i.e., . So , and we have the composition

So the middle map must be an isomorphism. The -dimensional case generalises to the general case due to the fact that vector spaces have bases.

We define:

Suppose is a subgroup fixed by the -action. There exists an exact sequence (of pointed sets)

Here, and are still groups.

Suppose is a normal subgroup. Then there exists an exact sequence

(because now is a group). Furthermore, if is abelian, then the map

is a group homomorphism (because now it makes sense for to be a group).

If is a central subgroup, then we further have an exact sequence

where is defined because is abelian.

and . We have and so the last sequence above applies.

Let , , be pointed sets and and be pointed maps. We say that the sequence

is exact at if .