Let $p$ be a prime number, $G=\textrm{Gal}(\overline{\mathbb{Q}}\_p/\mathbb{Q}\_p)$, and $\chi:G\rightarrow\mathbb{Z}\_p^\times$ the cyclotomic character. Let $\mathbb{C}\_p$ denote the completion of the algebraic closure of $\mathbb{Q}\_p$ and $\mathcal{O}\_{\mathbb{C}\_p}$ denote its ring of integers.


The Tate-Sen theorem implies that (among many other things) $H^0(G,\mathbb{C}\_p(\chi))=0$.


**Question:** Is $H^0(G,\mathcal{O}\_{\mathbb{C}\_p}(\chi)\otimes \mathbb{F}\_p)=0$?


If the answer is no, do the Galois invariants generate (over $\mathcal{O}\_{\mathbb{C}\_p}$), or are they killed by some power of $p$ less than 1?


**Motivation:** Basically, I am trying to understand if there is some remannt of Hodge-Tate theory modulo $p$.


Thanks!



A Tate-Sen theorem mod $p$
$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\cO}{\mathcal{O}}\newcommand{\bC}{\mathbb{C}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\bF}{\mathbb{F}}$The open subgroup $Gal(\overline{\bQ}\_p/\bQ\_p(\mu\_p))$ acts trivially on the mod $p$ reduction of the cyclotomic character so the invariants in question can be computed as $$H^0(G\_{\bQ\_p},\cO\_{\bC\_p}/p(1))=((\cO\_{\bC\_p}/p)^{G\_{\bQ\_p(\mu\_p)}}(1))^{Gal(\bQ\_p(\mu\_p)/ \bQ\_p)}$$


The group $\cO\_{\bC\_p}^{G\_{\bQ\_p(\mu\_p)}}/p=\bZ\_p[\mu\_p]/p$ injects into $(\cO\_{\bC\_p}/p)^{G\_{\bQ\_p(\mu\_p)}}$ so inside the group that we want to compute there is a subgroup given by $(\bF\_p[\mu\_p](1))^{\bF\_p^{\times}}$. The module $\bF\_p[\mu\_p]$ over the group $(\bF\_p)^{\times}$ decomposes as a direct sum of the trivial character and the regular representation. As the tensor product of the regular representation with any character is still the regular representation, we have $(\bF\_p[\mu\_p](1))^{\bF\_p^{\times}}=\bF\_p$ so $$H^0(G\_{\bQ\_p},\cO\_{\bC\_p}/p(1))\neq 0$$


**Edit** I've just noticed that the question is also answered [here](https://mathoverflow.net/a/315487/39304) by David Hansen: computing $H^0(G\_{\bQ\_p},\cO\_{\bC\_p}/p(1))$ is equivalent to computing the $p$-torsion in $H^1(G\_{\bQ\_p},\cO\_{\bC\_p}(1))$.


