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.



Galois descent, explicit inverse map
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 $L \otimes\_K V^G$.


