Let $G$ be the abolute Galois group of $\mathbb Q\_p$, let $\delta\_1, \delta\_2: G\rightarrow L^{\times}$ be continuous characters, where $L$ is a finite extension of $\mathbb Q\_p$. Assume that $\delta\_1\delta\_2^{-1}$ is neither trivial nor the cyclotomic character then 
$Ext^1\_{G}(\delta\_2, \delta\_1)$ is one dimensional. Hence there exists a unique non-split extension: 


$0\rightarrow \delta\_1\rightarrow V\rightarrow \delta\_2\rightarrow 0$. 


When is $V$ de Rham?


I believe that the answer is if and only if both $\delta\_1$ and $\delta\_2$ are de Rham and the Hodge-Tate weight of $\delta\_1\delta\_2^{-1}$ is $\ge 1$ (at least if the Hodge-Tate weights of $\delta\_1$ and $\delta\_2$ are distinct) and I guess I could x it out by using Bloch-Kato's paper in Grothendieck Festschrift, bu the answer must be well 
known and maybe even written down somewhere. 


Ideally, I would like to be able to quote a reference, where this has been worked out. 



When is an extension of characters de Rham? 
Dear Vytas,


lemma 6.5 of my 2002 inventiones paper says that if $V$ is any de Rham representation all of whose HT weights are at least 1, then any extension of $Q\_p$ by $V$ is itself de Rham. This holds for reps of $G\_K$ where the residue field $k$ of $K$ can be any perfect field (not merely finite).


If $k$ is finite, then this was well-known before and follows from the results of Bloch and Kato (which I think you should quote). See proposition 1.28 of Nekovar's "On $p$-adic height pairings" where this is stated explicitly and proved using BK's computations.


EDIT: see also the "Proposition" on page 196 of Perrin-Riou's "Représentations $p$-adiques ordinaires". It predates Nekovar's paper, and although the result is less strong, it's enough for what you need. 


