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 1 A -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 2 A -semilinear structure on is a -linear representation of , required to satisfy
The action of on is the natural Galois action.
Definition 3 Consider the algebra
with the operations
This is a -algebra, but not a -algebra.
Example 1 For the situation , and then .
Remark 1 A -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.
The 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.
Then 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 6 Suppose is an abelian group and are distinct characters. Then there cannot exist not all zero such that
This 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 2 The 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 2 The 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 7 Let 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 3 We 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