Galois descent for semilinear endomorphisms

Question

Let $K \subset L$ be a finite Galois extension, $\sigma$ an automorphism of $L$ (not necessarily fixing $K$) and let $E$ be a finite-dimensional vector space over $L$ together with an $\sigma$-linear endomorphism $\phi : E \rightarrow E$. I would like to do Galois descent. So my question is "Under what conditions does $\phi$ ascent from a $\sigma$-linear endomorphism defined over $K$?".

Thanks.

EDIT: I guess I should describe my problem more precisely.

$k$ a perfect field of char. $p>0$, $K$ the fraction field of the Witt ring $W(k)$, $F$ finite extension of $\mathbb{Q}_p$, $L$ the compositum of $K$ and $F$ in $\bar{K}$, $\tau$ the Frobenius automorphism of $L$ over $F$ and $\sigma$ the Frobenius automorphism of $K$.

A $\tau$-$L$-space is a finite dimensional vector space $V$ over $L$ together with a $\tau$-semilinear bijection $\Phi: V\rightarrow V$.

If I'm given an $\sigma^f$-isocrystal $V$ over $k$, then I can associate to it an $\tau$-$L$-space as follows. Let $F$ be the unique unramified extension of $\mathbb{Q}_p$ of degree $f$, $L=KF$ then $\tau|_K = \sigma^f$ and tensoring with $L$ gives a $\tau$-$L$-space.

I'm interested in the other direction. Let $F$ be an unramified extension of $\mathbb{Q}_p$ of degree $f$ and $E$ an $\tau$-$L$-space. Does $E$ come from a $\sigma^f$-isocrystal over $k$?

Answer 1

If $E$ is of dimension $1$, then $\phi$ is given in some basis by a scalar $x \in L$ and you are asking whether there is an element $y \in L^\times$ such that $x \cdot \sigma(y)/y$ belongs to $K$. I don't see why there should be a simple general criteria to decide whether such a $y$ exists. If $\sigma$ fixes $K$, then the norm of $x$ should be the norm of an element of $K$, and conversely this would be enough if in addition $Gal(L/K)$ was generated by $\sigma$.

More generally, you can at least rewrite your problem in terms of cohomology: the group $\mathbb{Z}$ acts on $L$ by $n \cdot x = \sigma^n(x)$, and you're asking about the image of $H^1(\mathbb{Z},GL_d(K)) \to H^1(\mathbb{Z},GL_d(L))$. But without more specific information on $K$, $L$ and $\sigma$, I don't see what one could say in general.