Given a finite Galois fiedl extension $L/K$ with Galoisgroup $G$ i.e. the fixpoints of $L$ under G are $K$ ($L^G=K$). Let $G$ act on $L^n$ semilinearly (i.e. $g(ax)=g(a)g(x)$ where $a\in L$ and $x\in L^n$)
This induces a action $\coprod_{k=1}^n Spec L\rightarrow \coprod_{k=1}^n Spec L$. I want to describe the $G$ action. First of all $\coprod_{k=1}^n Spec L$ is a collection of points and each $g\in G$ acts by permuting them. But also $g$ acts "internally" on the sheaf. Will it act by $g$ or $g^{-1}$?
Another way to phrase this question: Assume we started on the geometry side $\coprod_{k=1}^n Spec L\rightarrow \coprod_{k=1}^n Spec L$ where $G$ acts by a covering action. When I go to the algebra side now, and I want to compute $(L^n)^G$: Will the condition be
- $(z_1,\ldots,z_n)\in (L^n)^G$ iff $\forall g\in G: g(z_{g(i)})=z_i$ or
- $(z_1,\ldots,z_n)\in (L^n)^G$ iff $\forall g\in G: g^{-1}(z_{g(i)})=z_i$
I hope I made my question clear.