I am currently reading this article by Laurent Berger about $p-$adic Galois representations and I have some question regarding its section $4$.
I will go through some of its lines and ask my questions in what follows. Let $K$ be a finite extension of $\mathbb Q_p$ and $G_K$ denote its absolute Galois group.
- We start with a $\mathbb Q_p$-algebra $B$, endowed with a $G_K$ action.
- The author also assumes that
(1) $B$ is a domain such that $\text{Frac}(B)^{G_K}=B^{G_K}$
(2) If $y\in B$ is such that the line $\mathbb Q_p\cdot y$ is stable under $G_K$, then $y\in B^{\times}$.
I am not exactly sure in (1) what $G_K$-action are we assuming on $\text{Frac}(B)$, but I suppose it is expected one by declaring $\sigma\left(\dfrac{a}{b}\right):=\dfrac{\sigma(a)}{\sigma(b)}$. To me it seems that condition $(1)$ means that $B^{G_K}\subset B^{\times}$. The other condition is clear.
- Next, if $V$ is $p-$adic representation of $G_K$, we let $D_B(V):=(B\otimes_{\mathbb Q_p}V)^{G_K}$ and this is a $B^{G_K}$ vector space. Then the author continues:
There is a natural map $$ \alpha: B\otimes_{B^{G_K}} D_B(V)\rightarrow B\otimes_{\mathbb Q_p}V $$
By condition (1) above the map is injective.
I assume this map should be given by $\alpha(b\otimes v)=bv$. But I don't understand how does condition $(1)$ imply its injectivity.
- Finally, the above injectivity implies that $\dim_{B^{G_K}} D_B(V)\le \dim_{\mathbb Q_p}V$ and if equality holds, the author calls $V$ to be $B$-admissible. He claims:
$V$ is $B-$admissible iff $\alpha$ is surjective.
I do understand the converse direction, but I still can't prove the opposite.
Hope someone can help.