In this note, I would like to record mainly for my own benefit some basic things about how to descend algebraic data along Galois extensions of fields. I have benefited from some notes of K. Conrad available online. The exposition below is a shorter version of those.

** — 1. Setup — **

Let be a field extension with (finite) Galois group . Suppose is a vector space over .

Definition 1A -form of is a vector space over and a -linear map such that the extension is an isomorphism.

Note that on there is a natural -action on the second factor. On , this translates as follows.

Definition 2A -semilinear structure on is a -linear representation of , required to satisfyThe action of on is the natural Galois action.

Definition 3Consider the algebrawith the operations

This is a -algebra, but not a -algebra.

Example 1For the situation , and then .

Remark 1A -semilinear structure on is the same as giving it a -module structure compatible with the -structure. Also, note that has a canonical -semilinear structure.

** — 2. The main theorem for vector spaces — **

The following result is basic for understanding more general situations.

Theorem 4We have an equivalence of categoriesThe functor to -vector spaces is taking -invariants, and the inverse one is given by tensoring with :

This in particular implies that to give a -form of is the same as to give it a -module structure, and all -forms arise so uniquely (up to the right notion of isomorphism).

Before proving this, we need some preparations.

Lemma 5Suppose is a -module and consider the map given byThen we have

- (i) The image of is in .
- (ii) For all , there exists such that .

*Proof:* The first claim is immediate from the definition. For the second part, suppose by contradiction there exists such that

holds for all . We have that is a character for all . We obtain a contradiction from the result below which states that distinct characters must be linearly independent.

Lemma 6Suppose is an abelian group and are distinct characters. Then there cannot exist not all zero such thatThis also applies under if are vectors in a vector space, by looking in a basis. In this form, it yields the contradiction in the lemma above.

*Proof:* Pick a minimal number of distinct characters which are not linearly independent. We thus have

This implies that for all we have

Because of the minimality assumption on , multiplying the first line by and subtracting from the second we find that the rhs must be zero for all . This gives for all , which is a contradiction.

We now return to the proof of Theorem 4.

*Proof:* From Lemma 5 we know that is always non-trivial. However, if we consider it is an injective map. If it is not surjective, we can take the quotient and get a non-trivial -module. But the lemma allows us to find non-trivial invariants again, which is a contradiction.

Remark 2The above arguments apply when is profinite, provided that the modules are continuous. One can also allow for infinite-dimensional spaces, under the same assumption.

** — 3. Applications — **

The above considerations imply that in order to descend for example a -algebra, one needs to make it a -module in a way compatible with the multiplication. This applies to not necessarily associative algebras or Lie algebras.

Example 2The algebra of complex matrices has two real forms. These are and the quaternions . The interesting conjugation (i.e. action of ) which yields is

The following is also useful.

Theorem 7Let be an ideal which is invariant under the Galois action of . Then there exists a generating set for with .

*Proof:* Consider the ideal . It is finitely generated by some elements . If we consider all polynomials in up to some degree , the ones that are -invariant are in . However, by Theorem 4 they also span the corresponding space after tensoring .

Example 3We work out what is as an algebra. First, let act on the first factor only:This gives a -module structure which gives as invariants . Now consider the algebra

Endow this with a -action twisted on each factor by the corresponding automorphism:

Let act by permutation on the right of the indices. Namely

The algebra of invariants is again , so it must be that