Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $L/K$ be a finite Galois extension with Galois group $G$ and $V$ a $L$-vector space, on which $G$ acts by $K$-automorphisms satisfying $g(\lambda v)=g(\lambda) g(v)$. It is known that the canonical map

$V^G \otimes_K L \to V$

is an isomorphism. However, I can't find any short and nice proof for that. Actually I'm wondering if it is possible to construct an explicit inverse map, by choosing a basis of $L/K$ and taking some average with respect to $G$. Any ideas? I'm interested in the case of positive characteristic.

share|improve this question
    
Dear Martin: You can take a look at Theorem 5.6.3 p. 305 in Algèbre et théories galoisiennes by Douady and Douady. –  Pierre-Yves Gaillard Jun 14 '10 at 5:43
add comment

4 Answers

up vote 6 down vote accepted

Martin, http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf is a handout on this kind of stuff and Theorem 2.14 there gives a proof of the bijection between $K$-forms of $V$ and $G$-structures on $V$. It doesn't qualify as "short", and whether it's "nice" or not is too subjective. I wrote it for a target audience that knows only Galois theory and tensor products.

As for an explicit inverse map, see the top of page 6. Let $Tr_G \colon V \rightarrow V^G$ by $Tr_G(v) = \sum_{\sigma \in G} \sigma(v)$. If $d = [L:K]$ and $\alpha_1,\dots,\alpha_d$ is a $K$-basis of $L$, there exist $\beta_1,\dots,\beta_d$ in $L$ such that $$ v = \sum_{j=1}^d \alpha_jTr_G(\beta_j v) $$ for all $v$ in $V$. The right side provides a decomposition coming from $L \otimes_K V^G$.

share|improve this answer
add comment

I dunno about the explicit inverse, but there are two simple ways I know of showing the map is an isomorpism. The first is just to apply Grothendieck's faithfully flat descent theory to L/K -- one identifies the descent data on an L-vector space as exactly the kind of Galois action you describe. The other, maybe more down-to-earth, is by considering twisted group K-algebra L{G} built so that modules over it are exactly L-vector spaces with the kind of G-action you describe. Here is the key:

Claim: Every finite-dimensional L{G}-module is a direct sum of copies of the module L with its Galois action.

Proof: The natural map L{G} --> End_{K-vect}(L) is injective by linear independence of automorphsims hence an isomorphism by dimension count, and the corresponding fact for the matrix algebra End_{K-vect}(L) is well-known (e.g. by Morita).

Given this the reason that your map is an isomorphism is that by the claim we can reduce to V=L where it's just Galois theory (a direct limit argument reduces to the finite-dimensional case, or we could remove "finite-dimensional" from the above claim using Choice).

share|improve this answer
add comment

Isn't it Hilbert 90? Choose a basis for $L$, and denote by $c_g$ the matrix s.t. $g(e_j) = \sum_i (c_g)_{i,j} e_i$ Then $g \mapsto c_g$ is a cocycle with values in $GL_n (L)$, i.e. $c_{gh} = c_g g(c_h)$.

Hilbert 90 tells you that $H^1(G,GL_n(L))=0$, so that $c_g=b g(b)^{-1}$ for some invertible $b$, which is the matrix of an invariant basis. You can take $b= \sum_g c_g g(a)$ for some well-chosen matrix $a$ (so that $b$ is invertible).

See Serre's Local fields, chapter X for the proof (which relies on linear independance of the elements of $G$).

share|improve this answer
add comment

A natural idea to try would be to try to express a given $v\in V$ in the form $$v=\sum_{g\in G} g(a)v_g$$ where each $v_g\in V^G$ and $a\in L$ is a normal basis: the $g(a)$ for $g\in G$ form a $K$-basis of $L$. There is a unique representation of $v$ in this form. From the nonsingularity of the trace pairing then there is a unique $b\in L$ with $T(ba)=1$ but $T(b g(a))=0$ for all $\in G$ apart from the identity. (Here $T$ denotes the trace). Then $$\sum_{h\in G}h(b)h(v) =\sum_{g,h\in G}h(b)h(g(a))v_g=\sum_{g\in G}T(b g(a))v_g=v_1.$$ Similarly for any $k\in G$ $$\sum_{h\in G}h(b)h(k^{-1}(v))=v_k$$ so we get a formula for the $v_k$.

This relies on finding suitable $a$ and $b$. Somehow I think there must be some Hopf algebra formalism that does the job instantly :-)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.