I wish to understand a remark made by Rubin in his book "Euler Systems" (just after definition 5.1) about the first cohomology group of an unramified Galois representation.

Let $K$ be a number field, $v$ a finite place of $K$ above the rational prime $l$. Denote by $K_v^{unr}$ the maximal unramified Galois extension of $K_v$, and by $I_v=\mathrm{Gal}(\overline{K_v}/K_v^{unr})$ the inertia group. Let $p\neq l$ be another prime and $T$ a $\mathbb{Z}_p$-module (which we can assume free of finite rank) together with a $\mathbb{Z}_p$-linear continuous $G_K=\mathrm{Gal}(\overline{K}/K)$-action. Let finally assume that $T$ is unramified at $v$ (i.e., $I_v$ acts trivially).

I want to understand why $$H^1(K_v,T)=\ker\left(H^1(K_v,T)\xrightarrow{\text{res}} H^1(I_v,T)^{\mathrm{Gal}(K_v^{unr}/K_v)}\right)$$

or equivalently, by considering the associated inflation-restriction sequence, that the inflation morphism $$\text{inf}: H^1(\mathrm{Gal}(K_v^{unr}/K_v),T)\rightarrow H^1(K_v,T)$$ is in fact an isomorphism.

What I did so far was to show that, given any $c\in H^1(I_v,T)^{\mathrm{Gal}(K_v^{unr}/K_v)}$, then $|k_v|\mathrm{Fr}_v\cdot{}c_{\sigma}=c_{\sigma}$ (modulo a switch between $|k_v|$ and $|k_v|^{-1}$, with $k_v$ the residue field of $K_v$ and $\mathrm{Fr}_v$ the Frobenius at $v$) for all $\sigma\in I_v$. But then I'm stuck when trying to show that this implies $c_{\sigma}=0$.

Edit: (just an obvious reformulation of the preceeding) if the (geometric) Frobenius has eigenvalues different from $|k_v|^{+/-1}$ then we're done. But I guess there must be easy cases where $|k_v|$ is an eigenvalue of the Frobenius.

Edit 2: The problem is solved, thanks to peter a g for providing this nice counter-example. In fact, denoting $H^1_f(K_v,T):=\ker\left(H^1(K_v,T)\xrightarrow{\text{res}} H^1(I_v,T)^{\mathrm{Gal}(K_v^{unr}/K_v)}\right)$, we have $$H^1(K_v,T)/H^1_f(K_v,T)\simeq T(-1)^{\mathrm{Fr}_v}$$ which is $0$ iff $|k_v|$ isn't an eigenvalue of $\mathrm{Fr}_v$ acting on $T$.

Thanks for your help !


  • $\begingroup$ Anyone, any idea ? $\endgroup$ – Yoël Jan 2 '17 at 16:40
  • $\begingroup$ I don't have a hard copy: could you provide a link to a web version please? I didn't find your reference. $\endgroup$ – peter a g Jan 2 '17 at 22:33
  • $\begingroup$ namely, I didn't see the remark after 5.1, so I think what I was looking at did not correspond to what you have. $\endgroup$ – peter a g Jan 3 '17 at 0:49
  • $\begingroup$ It's not namely a remark, just a sentence written between parenthesis. In this link (citeseerx.ist.psu.edu/viewdoc/…), this is the fist sentence of page 12. $\endgroup$ – Yoël Jan 7 '17 at 13:22

I think the statement, as given above, is not correct: there are (continuous) cocycles, with the above hypotheses on $T$, that do not vanish on inertia. Here is one - unless I am making a mistake - I hope I'm not! Tell me what you think.


  • Write (for simplicity of notation) $K= \mathbb Q_l$. Fix an algebraic closure $\bar K$ once and for all.
  • Write $G = \text{Gal}(\bar K/ K)$, and $K^u$ for the maximal unramified (sub)-extension of $\bar K/K$. The field $K^u$ is the fixed field of $I$, the inertial group.

  • Denote ${\mathfrak g} = \text {Gal}(K^u/K)$, which is canonically isomorphic to $ \text {Gal }(\bar{\mathbb F}_l/{\mathbb F}_l)$.

  • Denote now by $K_p$ the subfield of $\bar K/ K^u$, defined by $$K_p = \bigcup_{n}\,K^u[l^{1/p^n}].$$ We have that $I_p=\text{Gal} (K_p/K^u)$ is isomorphic (as topological groups) to $\mathbb Z_p$, that $K_p$ is Galois over $K$, with Galois group we denote here $G_p$, and an extension $$ 1 \to I_p \to G_p\to {\mathfrak g} \to 1.$$

  • The previous extension is split: choose as topological generator $\tilde F \in {\mathfrak g}$, characterized by
    $$\tilde F x \equiv x^l \pmod l,$$ for all $x$ in the ring of integers of $K^u$. (The choice of $\tilde F$ identifies an isomorphism of ${\mathfrak g} \simeq \hat{\mathbb Z}$.) Choose a lifting $F\in G_p$ of $\tilde F$. Then $F$ acts on $I_p$ by conjugation on $I_p$ as multiplication by $l$, and this lifts extends by continuity to an action by $\mathfrak g$. In any case, we may identify, as $G$-modules, $I_p$ with
    $${\mathbb Z}_p(1) = \lim_{\leftarrow}\mu_{p^n}(\bar K) = \lim_{\leftarrow}\mu_{p^n}(K^u).$$ (So, of course, as inertia acts trivially on ${\mathbb Z}_p(1)$, we may also view it as a $G_p$-module.) [Remark: I see I chose the opposite $F$ from the one you did. Oh well.]

  • Now set, as Galois-modules, $T =I_p$. In particular, $T$ is an unramified module.

The possible counter-example:

We want to show that the restriction $H^1( G_p, T) \to \hom( I_p, T)$ is non-trivial, i.e, there is a continuous cocycle $c$ that, when restricted to $I_p$, is non-trivial.

Define $c$ on $I_p$ as the identity: $c(\sigma) = \sigma$. Of course, $$ c(\sigma^{l^n)}= l^n\sigma = F^n (\sigma), $$ (using multiplicative notation on the domain $I_p$ and additive on the image $T$), as desired.

If $c$ extends to a continuous cocycle, it is the desired example.

Define $c(F)= 1 \in T$, and proceed inductively, forcing the cocycle condition: for $n\ge 2$, $$ c(F^n) = F c(F^{n-1}) + c(F) = l^{n-1} + \cdots + 1,$$ so that $$ c(F^n) = {l^n -1 \over l - 1}.$$

Likewise $0 = c(F^{-n}F^{n})= l^{-n} c(F^{n}) + c(F^{-n})$ gives $$c(F^{-n}) = -l^{-n} {l^n-1 \over l -1 }.$$

The map $c$ is continuous on $F^{\mathbb Z}\subset F^{\hat{\mathbb Z}}$:
for all $m$ there is an $N$ such that $N| n$ implies that $p^m | (l^n -1) /(l-1)$. Therefore $c$ extends uniquely to (the closure) $F^{\hat{\mathbb Z}}$.

Again, we force the cocycle definition: if $\sigma \in I_p$, define

$$ c(F^n \sigma) = F^n \sigma + c(F^n)= l^n\sigma + c(F^n).$$

A final (unnecessary?) sanity check: suppose also $\tau\in I_p$:

On the one hand, (again, using multiplicative notation on $I_p$ and additive on $T$ - sorry) $$ c( F^n \sigma F^m \tau ) = c(F^{n+m}\sigma^{l^{-m}}\tau) = l^{n+m}(l^{-m}\sigma + \tau) + {l^{n+m} -1\over l -1}.$$

On the other, $$c(F^n\sigma F^m\tau) = F^n c(F^m\tau) + c(F^n\sigma)= l^n \left( l^m \tau + {l^m-1\over l-1}\right) + l^n\sigma + {l^n-1\over l -1},$$ so things work out.

Do you see anything wrong with this?

  • $\begingroup$ Sorry for taking so long to write this - correct or not - up. $\endgroup$ – peter a g Jan 10 '17 at 12:43
  • $\begingroup$ Thank you Peter for this long and detailed answer, and sorry for answering you so late. I think what you wrote is correct, but I'm still not quite convinced that my question is equivalent to finding a cocycle $c\in H^1(G_p,T)$ which doesn't restrict to $0$ on $H^1(I_p, T)$ (because my cocycle $c$ has to be defined in $H^1(G,T)$ and, in case my statement is false, should not vanish in $H^1(I,T)$, keeping your notations): is it obvious that the $c$ you chose can be inflated to $G$ ? $\endgroup$ – Yoël Mar 20 '17 at 10:04
  • $\begingroup$ OK now I understand your idea and why this works, thanks a lot ! However can you explain a bit more the continuity argument (I don't understand what you mean by " for all $m$ there is an $N$ such that $N|n$ implies that $p^m|(l^n−1)/(l−1)$ ? Thanks again. $\endgroup$ – Yoël Mar 20 '17 at 12:08
  • $\begingroup$ I'll try to answer this week - I'm feeling a bit sick and light-headed at the moment so I wouldn't trust myself at the moment to avoid writing nonsense, let alone to write comprehensibly. $\endgroup$ – peter a g Mar 20 '17 at 12:22
  • $\begingroup$ Maybe you can proceed this way: take $N\le N'$ two integers, assume they are positive (for simplicity) and that $N'-N\equiv 0 \mod \phi(n)$, with $\phi(n)=p^{n-1}(p-1)$. Assume (also for simplicity) that $l-1\in \mathbb{Z}_p^{\times}$. Then $c(F^{N'})-c(F^N)=l^N\frac{l^{N'-N}-1}{l-1}\equiv 0\mod p^n$, which means that $c$ can be extended to a continous morphism $c:\widehat{\mathbb{Z}}\rightarrow I_p\simeq\mathbb{Z}_p$. Does that seem OK ? Thanks again and have a good rest. $\endgroup$ – Yoël Mar 20 '17 at 14:06

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