Galois invariant basisPosted on May 29, 2019
Suppose that is a Galois extension with Galois group . Let be a vector space with a semi-linear Galois action, i.e., a group homomorphism denoted such that for all and . We define the -vector space
Theorem. The natural homomorphism given by is an isomorphism.
To show injectivity it suffices to show that if are -linearly independent, then they are also -linearly independent. Suppose is the least integer such that there exists a -linearly independent set and such that By dividing by we may assume that . Moreover, since is -linearly independent, we must have for some . Thus there exists such that . Observe that Therefore is a smaller -linearly independent set that is not -linearly independent, a contradiction to our selection of as the smallest number of vectors needed for such an occurrence.
It remains to show that the map is surjective. Let us pick a -basis for and write with . Define the matrix Observe that each column defines a character of with values in given by . Thus the columns are linearly independent by linear independence of characters. So is an invertible matrix with inverse . For any let us also define Now let us pick a fixed . Observe we have the equation given explicitly by Multiplying both sides on the right by gives us . In particular, we have We finally observe that for each as each is invariant under the action of . Thus the map is surjective.
The inspiration for the above proof comes from (1, Proposition 1.2.2). Different proofs of the statement can be found in (2, Lemma II.5.8.1) or (3, Lemma 2.3.8).
- Nicolas Garrel, An introduction to Galois cohomology through central simple algebras, 2014
- Joseph Silverman, The arithmetic of elliptic curves, 2nd ed., Springer Graduate Texts in Mathematics, 1992
- Gille Szamuely, Central simple algebras and galois cohomology, Cambridge Studies in Advanced Mathematics, 2006.