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Hi, I am reading about p-adic representations from Fontaine's book which can be found at On page 145 where they prove Proposition 5.24 which is essentially the theorem of Tate-Sen, they show $H^{n}(Gal(L/K_{\infty},C(i)^{G_L})=0$ and the argument is essentially same as in the proof of hilbert Thm 90. But then they are concluding that implies $H^{n}(H_K,C(i))=0$ by passing to the limit. I am confused because I thought that you can only pass through the limit in case of discrete modules. I think this same argument will also show that $H^{1}(G_K,C_K)=0$ which is not true. I am sure I am missing something obvious. I will greatly appreciate any kind of clarification.

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up vote 7 down vote accepted

The whole thing is done with more details in Tate's original article "p-divisible groups", section 3.2. Tate proves that one can approximate a cocyle in $C_p(i)$ by cocyles with values in $Q_p^{alg}(i)$ and this is how he reduces the computation to the "discrete case".

I would suggest that it's better to prove the result by $p$-adic approximation. This way, you can basically work with cocycles with values in $O_{C_p}(i)/p^n$, also a discrete space.

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Thanks a lot, Prof. Berger for the reply.I can understand Tate's proof but somehow I did not understand the proof from the book. I think I can prove it by working in O$_{\bar{K}}/p^{n}$. Thanks again. – Arijit Nov 27 '11 at 15:05

They are using continuous cohomology, so that $$ H^n(G,M) = \varinjlim H^n(G/H,M^H) $$ if $G$ is topological and $M$ is discrete (thanks Arjit) $G$-module (the limit runs over open compact subgroups $H$ of $G$). Look in p. 38 for the definition.

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But this is only true if M is discrete. But this is not the case here. – Arijit Nov 11 '11 at 0:44

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