For a finite field $\mathbb{F}$ of char $p$ $\geq 2$, let $f$ be a normalised eigenform of weight $k\geq 2$ that is ordinary at $p$ and $\overline{\rho}\_f$ be the mod $p$ Gal representation attached to it. Let $G\_p$ and $I\_p$ be the decomposition and inertia subgroups. We know that the restriction of $\overline{\rho}\_f$ to $G\_p$ is upper-triangular with characters $\delta$ and $\epsilon$ (unramified) on the diagonal. Under the assumption that $\delta$ and $\epsilon$ are distinct mod $p$ ($p$-distinguishedness) why is it true that $H^1(G\_p/I\_p,\mathbb{F}(\delta\epsilon^{-1})) = 0$? I know that $G\_p/I\_p \simeq \hat{\mathbb{Z}}$ has cohomological dimension $1$ but this only shows that the cohomology groups vanish from $H^2$ onwards . Is it to do with the fact that $G\_p/I\_p$ is topologically generated? Could someone please explain this to me?



Why does  $H^1(G_p/I_p,\mathbb{F}(\delta\epsilon^{-1}))$ vanish?
A variant of David Loeffler's answer...


If more generally $V$ is a $G\_{\mathbb Q\_{p}}$-representation with coefficients in a field, then the dimension of $H^{1}(G\_{p}/I\_{p},V)$ (EDIT : or rather $H^{1}(G\_{p}/I\_{p},V^{I\_{p}})$) is equal to the dimension of $H^{0}(G\_{p},V)$ because the former is the cokernel of the map $Fr\_{p}-1$ acting on $V^{I\_{p}}$ and the latter is the kernel of this map. 


In the situation you described, $\mathbb F(\delta\epsilon^{-1})$ has no non-trivial invariants under $G\_{p}$ because $\delta≠\epsilon$ (and this is why you need your form to be $p$-distinguished) so its $H^{1}$ is trivial as well.


