# Galois descent, explicit inverse map

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.

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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

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 $L \otimes_K V^G$.

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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).

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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 $G$).

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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 :-)

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