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

We now return to the proof of Theorem 4.

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

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