Galois invariant basis

Suppose that $K/k$ is a Galois extension with Galois group $G$ . Let $V$ be a $K$ vector space with a semi-linear Galois action, i.e., a group homomorphism $G\times V\to V$ denoted $(\sigma,v)\mapsto \sigma(v)$ such that $\sigma(\lambda v)=\sigma(\lambda)\sigma(v)$ for all $\lambda\in K$ and $v\in V$ . We define the $k$ -vector space $V^G=\{v\in V : \sigma(v)=v \ \text{for all} \ \sigma\in G\}.$

Theorem. The natural homomorphism $V^G\otimes_k K\to V$ given by $v\otimes\lambda\mapsto \lambda v$ is an isomorphism.

To show injectivity it suffices to show that if $\{v_1,\ldots,v_n\}$ are $k$ -linearly independent, then they are also $K$ -linearly independent. Suppose $n$ is the least integer such that there exists a $k$ -linearly independent set $\{v_1,\ldots,v_n\}$ and $\lambda_1,\ldots, \lambda_n\in K$ such that $\sum_{i=1}^n\lambda_i v_i =0.$ By dividing by $\lambda_1$ we may assume that $\lambda_1=1$ . Moreover, since $\{v_1,\ldots,v_n\}$ is $k$ -linearly independent, we must have $\lambda_j\not\in k$ for some $j\ge 2$ . Thus there exists $\sigma\in G$ such that $\sigma(\lambda_j)\ne\lambda_j$ . Observe that $\begin{aligned}[t] \sum_{i=2}^n(\sigma(\lambda_i)-\lambda_i)v_i &= \sigma\left(\sum_{i=1}^n\lambda_iv_i\right) -\sum_{i=1}^n\lambda_iv_i\\ &= 0. \end{aligned}$ Therefore $\{v_2,\ldots,v_n\}$ is a smaller $k$ -linearly independent set that is not $K$ -linearly independent, a contradiction to our selection of $n$ as the smallest number of vectors needed for such an occurrence.

It remains to show that the map is surjective. Let us pick a $k$ -basis $\lambda_1,\ldots,\lambda_n$ for $K$ and write $G=\{\sigma_1,\ldots,\sigma_n\}$ with $\sigma_1=\text{id}_{K}$ . Define the matrix $A= \begin{bmatrix} \sigma_1(\lambda_1) & \sigma_1(\lambda_2) & \ldots & \sigma_1(\lambda_n)\\ \sigma_2(\lambda_1) & \sigma_2(\lambda_2) & \ldots & \sigma_2(\lambda_n)\\ \vdots & \vdots & \vdots & \vdots\\ \sigma_n(\lambda_1) & \sigma_n(\lambda_2) & \ldots & \sigma_n(\lambda_n) \end{bmatrix}.$ Observe that each column defines a character of $G$ with values in $K$ given by $\sigma\mapsto\sigma(\lambda_i)$ . Thus the columns are linearly independent by linear independence of characters. So $A$ is an invertible matrix with inverse $A^{-1}=\left( a_{ij}\right)_{i,j}$ . For any $v\in V$ let us also define $a(v)= \begin{bmatrix} \sigma_1(v)\\ \sigma_2(v)\\ \vdots\\ \sigma_n(v) \end{bmatrix}, \ b(v)= \begin{bmatrix} \sum_{i=1}^n\sigma_i(\lambda_1 v)\\ \sum_{i=1}^n\sigma_i(\lambda_2 v)\\ \vdots\\ \sum_{i=1}^n\sigma_i(\lambda_n v) \end{bmatrix}.$ Now let us pick a fixed $v\in V$ . Observe we have the equation $Aa(v)=b(v)$ given explicitly by $\begin{bmatrix} \sigma_1(\lambda_1) & \sigma_2(\lambda_1) & \ldots & \sigma_n(\lambda_1)\\ \sigma_1(\lambda_2) & \sigma_2(\lambda_2) & \ldots & \sigma_n(\lambda_2)\\ \vdots & \vdots & \vdots & \vdots\\ \sigma_1(\lambda_n) & \sigma_2(\lambda_n) & \ldots & \sigma_n(\lambda_n) \end{bmatrix} \begin{bmatrix} \sigma_1(v)\\ \sigma_2(v)\\ \vdots\\ \sigma_n(v) \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^n\sigma_i(\lambda_1 v)\\ \sum_{i=1}^n\sigma_i(\lambda_2 v)\\ \vdots\\ \sum_{i=1}^n\sigma_i(\lambda_n v) \end{bmatrix}.$ Multiplying both sides on the right by $A^{-1}$ gives us $a(v)=A^{-1}b(v)$ . In particular, we have $v=\sigma_1(v)=\sum_{j=1}^n a_{1j}b(v)_j.$ We finally observe that $b(v)_j\in V^G$ for each $j$ as each is invariant under the action of $G$ . 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).

References

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.