# Descent for vector spaces

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 ${k\subset K}$ be a field extension with (finite) Galois group ${G}$. Suppose ${W}$ is a vector space over ${K}$.

Definition 1 A ${k}$-form of ${W}$ is a vector space ${V}$ over ${k}$ and a ${k}$-linear map ${V\rightarrow W}$ such that the extension ${V\otimes_k K\rightarrow W}$ is an isomorphism.

Note that on ${V\otimes_k K}$ there is a natural ${G}$-action on the second factor. On ${W}$, this translates as follows.

Definition 2 A ${G}$-semilinear structure on ${W}$ is a ${k}$-linear representation of ${\rho:G\rightarrow End_k(W)}$, required to satisfy

$\displaystyle \rho(\sigma)(aw)=\sigma(a)w ,\forall a\in K$

The action of ${G}$ on ${K}$ is the natural Galois action.

Definition 3 Consider the algebra

$\displaystyle C(G) = \bigoplus_{\sigma\in G} K\cdot e_\sigma$

with the operations

$\displaystyle e_\sigma \cdot e_\tau = e_{\sigma\cdot \tau}\\$

$\displaystyle e_\sigma \cdot \lambda e_\tau = \sigma(\lambda) e_{\sigma \cdot \tau}$

This is a ${k}$-algebra, but not a ${K}$-algebra.

Example 1 For the situation ${{\mathbb R}\subset {\mathbb C}}$, ${G={\mathbb Z}/2}$ and then ${C(G)=M_2({\mathbb R})}$.

Remark 1 A ${G}$-semilinear structure on ${W}$ is the same as giving it a ${C(G)}$-module structure compatible with the ${K}$-structure. Also, note that ${K}$ has a canonical ${G}$-semilinear structure.

— 2. The main theorem for vector spaces —

The following result is basic for understanding more general situations.

Theorem 4 We have an equivalence of categories

$\displaystyle C(G)-mod \rightleftarrows k-vect$

The functor to ${k}$-vector spaces is taking ${C(G)}$-invariants, and the inverse one is given by tensoring with ${K}$:

$\displaystyle W\rightarrow W^{C(G)}$

$\displaystyle V \otimes_k K \leftarrow V$

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

Before proving this, we need some preparations.

Lemma 5 Suppose ${W}$ is a ${C(G)}$-module and consider the map ${Tr_G:W\rightarrow W}$ given by

$\displaystyle Tr_G(w) = \sum_{\sigma \in G} \sigma w$

Then we have

• (i) The image of ${Tr_G}$ is in ${W^{C(G)}}$.
• (ii) For all ${w\in W}$, there exists ${a\in K}$ such that ${Tr_G(aw)\neq 0}$.

Proof: The first claim is immediate from the definition. For the second part, suppose by contradiction there exists ${w\in W}$ such that

$\displaystyle \sum_{\sigma\in G} \sigma(a) \sigma(w)=0$

holds for all ${a\in K}$. We have that ${\sigma:K^\times\rightarrow K^\times}$ is a character for all ${\sigma\in G}$. We obtain a contradiction from the result below which states that distinct characters must be linearly independent. $\Box$

Lemma 6 Suppose ${A}$ is an abelian group and ${\sigma_i:A\rightarrow K^\times}$ are distinct characters. Then there cannot exist ${c_i\in K}$ not all zero such that

$\displaystyle \sum_i c_i \sigma_i = 0$

This also applies under if ${c_i}$ 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

$\displaystyle \sigma_{n+1} = \sum_{i=1}^{n} c_i \sigma_i$

This implies that for all ${a,b\in A}$ we have

$\displaystyle \sigma_{n+1}(a) = \sum_i c_i \sigma_i(a)$

$\displaystyle \sigma_{n+1}(a)\sigma_{n+1}(b) =\sigma_{n+1}(ab) = \sum_i c_i \sigma_i(ab) =\sum_i c_i \sigma_i(a) \sigma_i (b)$

Because of the minimality assumption on ${n}$, multiplying the first line by ${\sigma_{n+1}(b)}$ and subtracting from the second we find that the rhs must be zero for all ${a}$. This gives ${\sigma_i(b)=\sigma_{n+1}(b)}$ for all ${b}$, which is a contradiction. $\Box$

Proof: From Lemma 5 we know that ${W^{C(G)}}$ is always non-trivial. However, if we consider ${W^{C(G)}\otimes_k K\rightarrow W}$ it is an injective map. If it is not surjective, we can take the quotient and get a non-trivial ${C(G)}$-module. But the lemma allows us to find non-trivial invariants again, which is a contradiction. $\Box$

Remark 2 The above arguments apply when ${G}$ 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 ${K}$-algebra, one needs to make it a ${C(G)}$-module in a way compatible with the multiplication. This applies to not necessarily associative algebras or Lie algebras.

Example 2 The algebra ${M_2({\mathbb C})}$ of ${2\times 2}$ complex matrices has two real forms. These are ${M_2({\mathbb R})}$ and the quaternions ${{\mathbb H}}$. The interesting conjugation (i.e. action of ${{\mathbb Z}/2}$) which yields ${{\mathbb H}}$ is

$\displaystyle \sigma\begin{bmatrix} a & b\\ c & d \end{bmatrix} =\begin{bmatrix} \overline{d} & \overline{-d}\\ \overline{-b} & \overline{a} \end{bmatrix}$

The following is also useful.

Theorem 7 Let ${I\subset K[x_1,\ldots,x_n]}$ be an ideal which is invariant under the Galois action of ${G}$. Then there exists a generating set ${f_1,\ldots,f_N}$ for ${I}$ with ${f_i\in k[x_1,\ldots,x_n]}$.

Proof: Consider the ideal ${I'=I\cap k[x_1,\ldots,x_n]}$. It is finitely generated by some elements ${f_1,\ldots,f_N}$. If we consider all polynomials in ${I}$ up to some degree ${d}$, the ones that are ${G}$-invariant are in ${I'}$. However, by Theorem 4 they also span the corresponding space after tensoring ${\otimes_k K}$. $\Box$

Example 3 We work out what is ${K\otimes_k K}$ as an algebra. First, let ${G}$ act on the first factor only:

$\displaystyle \sigma(x\otimes y) = \sigma(x)\otimes y$

This gives a ${C(G)}$-module structure which gives as invariants ${K}$. Now consider the algebra

$\displaystyle A=\prod_{\sigma \in G} K_\sigma$

Endow this with a ${K}$-action twisted on each factor by the corresponding automorphism:

$\displaystyle \lambda(x_\sigma)_\sigma = (\sigma(\lambda) x_\sigma)_\sigma$

Let ${G}$ act by permutation on the right of the indices. Namely

$\displaystyle \tau (x_\sigma)_\sigma = (x_{\sigma \tau})_\sigma$

The algebra of invariants is again ${K}$, so it must be that

$\displaystyle A \cong K\otimes_k K$