A Tate-Sen theorem mod $p$ 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: $\newcommand{\bQ}{\mathbb{Q}}\newcommand{\cO}{\mathcal{O}}\newcommand{\bC}{\mathbb{C}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\bF}{\mathbb{F}}$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 by David Hansen: computing $H^0(G_{\bQ_p},\cO_{\bC_p}/p(1))$ is equivalent to computing the $p$-torsion in $H^1(G_{\bQ_p},\cO_{\bC_p}(1))$.
