I've been scouring google and asking friend about something I was certain must be absolutely the easiest thing to people who do homological algebra, and none seem to know the answer to this, so if it's something really easy, I apologize.

The setup is relatively simple, I have a profinite group, $G$, and a sequence of $G$-modules--that is to say $\mathbb{Z}[G]$ modules--

$$1\to X_0\to X_1\to X_2\to 1$$

Given a trivial $G$ module, $R$ which is injective as an abelian group, I'd like to analyze the Galois cohomology groups, $H^*(G, R\otimes_{\mathbb{Z}} X_i)$ given knowledge of $H^*(G, X_i)$. If it helps at all, one can assume I'm talking about $R=\mathbb{R}$, as even that case would be of great use. As far as I can tell, with a trivial $G$ action, $R$ is a flat $G$ module, so the snake lemma still applies to the modified sequence. The main problem I have is finding any result which indicates how the new cohomology groups relate to the old ones.

Furthermore, if I have a $\mathbb{Q}[G]$ module, $V$--that is, a vector space with a continuous $G$ action--how can I compute $H^*(G, \mathbb{F}\widehat{\otimes}_{\mathbb{Q}}V)$ where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, and the hat means I intend to *complete* my vector space $\mathbb{F}\otimes V$ with respect to some topology (my case is a norm topology if that helps at all). Is it just the same as $H^*(G,\mathbb{F}\otimes_{\mathbb{Q}}V)$--i.e. does completion affect anything? (I'm inclined to think not, as the $\mathbb{F}$ are endowed with a trivial $G$ action, but I'm not enough of an expert to be confident of that.) If so, is then $H^*(G, \mathbb{F}\otimes_{\mathbb{Q}}V)$ the same as $H^*(G, V)$ or a simple operation away from $H^*(G,V)$?

**Edit**: The answer posted by user 43326 makes me realize that I must have gotten the wrong tensor, it should be over $\mathbb{Q}$ (or even $\mathbb{Z}$) and not, $\mathbb{Q}[G]$ since the latter trvializes the $G$ action, which should not happen.

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