Cohomology of Severi-Brauer varieties What can be said about Galois-module structure of $l$-adic cohomology of a Severi-Brauer variety over a local field?
In particular, I'm interested in the proof of the proposition given at the top of the 4th page in this paper Good reduction, bad reduction (arXiv link)



 A: Let $X$ be defined over a finite extension $K$ of $\mathbb Q_p$.  Recall that when we say "the $\ell$-adic cohom. of $X$" we actually mean the etale cohom., with $\mathbb Q_{\ell}$-coeffs., of the base-change $\overline{X}$ of $X$ to $\overline{K} = \overline{\mathbb Q}_p$.  (And actually, we take etale cohom. of this base-change with $\mathbb Z/\mathbb l^n$ coeffs., then take a limit in $n$, and then invert $\ell$; but I won't dwell on this aspect of things, since it is not the key to what follows.) The $\mathrm{Gal}(\overline{K}/K)$-action on cohom. is then induced by the $\mathrm{Gal}(\overline{K}/K)$-action on the base-change $\overline{X}$.
The cohom. ring of $\mathbf P_n$ is isomorphic to $\mathbb Q_{\ell}[h]/h^{n+1}),$ where $h$ spans $H^2$, and the Galois action on $h$ is via the inverse cyclotomic character.
If $X$ is a Brauer--Severi variety, then there is a finite extension $L$ of $K$ so that $X_{L}$ is isomorphic to $\mathbf P_{n/L}$.  In particular, $X$ and $\mathbf P_n$ have the same base-change to $\overline{K}$, and hence there cohom. is isomorphic as rings.  Thus the cohom. ring of $X$ is also isomorphic to $\mathbb Q_{\ell}[h]/(h^{n+1})$, with $h$ spanning $H^2$.
Furthermore, $\mathrm{Gal}(\overline{K}/L)$ acts on $h$ through the inverse cyclotomic char.
Since $h$ spans $H^2$ of $X$, we see that
$\mathrm{GaL}(\overline{K}/K)$ acts on $h$ through some char.; and we need to check that this is an unram. char.   What we know is that its restriction to $L$ is inverse cyclotomic, and hence unram.  (Here I am assuming $\ell \neq p$.)  
But any central simple alg. over $K$, and hence any Brauer--Severi, has an an unram. splitting field; in other words, we can choose $L$ to be an unram. extension of $K$.  Thus the action on $h$ is through a char. which becomes unram. after restricting to an unram. extension; thus it was unram.  to begin with.
In the case $\ell = p$, the argument is the same, using the fact that a char. which becomes crystalline (as the inverse of the $p$-adic cyclo. char. is) after restricting to an unram. extension is already crystalline.
