The question is quite long since I give some background but really I am interested in some very concrete fact about matrix representations. Scroll all the way to the bottom for a self contained question, feel free to ignore the rest.
Let $L/K$ be a Galois field extension with Galois group $\Gamma$. Let $G$ be a finite group and $V$ a representation of $G$ over $L$ (aka, a module over $L[G])$.
Suppose there is a semi linear action of $\Gamma$ on $V$ with respect to $L(G)$. That is, for all $a \in L[G]$,$\gamma \in \Gamma$ and $v \in V$, we have $(av)^\gamma = a^\gamma v^\gamma$.
Does there necessarily exist a $k[G]$ submodule $W \subset L$ such that the natural map $W\otimes_{k[G]} L[G] \to V$ is an isomorphism?
I believe I can prove this using general ideas from faithfully flat descent (extend to a separably closed base field...) but I am looking for a more concrete way of proving this along the following lines:
Let us decompose $V = \bigoplus_k V_k$ as a sum of irreducible representations. Now, note that we can define conjugate representations $g^\gamma$ on $V$ since $g^\gamma (v) = (g(v^{\gamma^{-1}}))^\gamma \in V.$ Moreover, this conjugate action preserves irreducibility and so the representation $(G^\gamma,V_k)$ corresponds to some $(G,V_{\gamma(k)})$.
In concrete terms, if we fix a basis of vectors fixed under the semilinear action of $\Gamma$ and $g$ is given by a matrix $[a_{ij}]$ on $V_k$, then $g^\gamma$ is given by the matrix $[\gamma(a_{ij})]$.
Therefore, we can group the conjugate representations together and now suppose $W_k = \bigoplus_{\gamma \in \Gamma}V_k^\gamma$. (In general, there might be repeated terms in this sum but let me ignore that for now). This is closed under both the $G$ action and the $\Gamma$ action. Concretely, for any $g$ with matrix $[a_{ij}]$ for $W_k$, the matrix for $W$ is given by a block matrix with $|\Gamma|$ blocks and the block corresponding to $\gamma$ is equal to $[a_{ij}^\gamma]$.
In this case, I want to show that the matrix for each $g$ is similar to a matrix over $k$ and moreover that this similarity is realized by the same matrix (over $L$) for all $g \in G$.
The second condition is what is tripping me up. Since the characteristic polynomial for any $g$ is a polynomial over $k$, the rational canonical form realizes an element in the conjugacy class over the base field. But how do I make sure that I can do just one base change to get all the matrices corresponding to all the $g \in G$ into rational form?
Concrete question:
Let $L/K$ be a Galois extension with Galois group $\Gamma$. Let $G$ be a finite group. Let $\rho: G \in M_n(L)$ be a representation and suppose that for each $g, \rho(g)$ is a block matrix with the blocks indexed by $\gamma \in \Gamma$ such that if the block corresponding to the identity in $\Gamma$ is $[a_{ij}]$, then the block corresponding to $\gamma$ is $[\gamma(a_{ij})]$.
Can I find an invertible matrix $P$ over $L$ such that for all $g \in G$, $P\rho(g)P^{-1}$ is a matrix over $K$? Note that $\rho(g)$ always has characteristic polynomial defined over $K$ so that the rational canonical form gives us some $P_g$ for each $g \in G$ with the above property.
The question is eliminate the dependence of $P_g$ on $g$.
