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.
 A: 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 $L \otimes_K V^G$.
A: 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).
A: 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 $G$).
A: 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 :-)
