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Let $K$ be a $p$-adic field with Galois group $G$ and inertia subgroup $I\subset G$. Denote $(-)^\ast=\mathrm{Hom}_{cont}(-,\mathbb{Q}/\mathbb{Z})$. Using Tate local duality, we can compute $$H^2(G,\mathbb{Q}/\mathbb{Z})=\varinjlim H^2(G,\mathbb{Z}/n\mathbb{Z})=\varinjlim \mathrm{Hom}_G(\mathbb{Z}/n\mathbb{Z},\mu)^\ast=\mathrm{Hom}_G(\mathbb{Q}/\mathbb{Z},\mu)^\ast=\mathrm{Hom}(\mathbb{Q}/\mathbb{Z},\mu(K))^\ast = 0$$since $\mathbb{Q}/\mathbb{Z}$ has no finite quotient. On the other hand, the Hochschild-Serre spectral sequence for $H^2(G,\mathbb{Q}/\mathbb{Z})$ has $E_2$ page \begin{array}{ll} (I^\ast)^{\phi=1} & I^\ast/1-\phi \\ \mathbb{Q}/\mathbb{Z} & \mathbb{Q}/\mathbb{Z} \end{array} where $\phi$ is the Frobenius in $\widehat{\mathbb{Z}}$. I was under the impression that the action of the Frobenius on the Pontryagin dual of the inertia is trivial by cocycle computations of mine, but this contradicts the first computation as then $I^\ast/1-\phi=I^\ast$ is non-zero. Moreover $H^1(G,\mathbb{Q}/\mathbb{Z})=G^\ast$ is an extension of $\mathbb{Q}/\mathbb{Z}$ by $I^\ast$, which naively suggests that $(I^\ast)^{\phi=1}=I^\ast$ under the canonical inclusion; but this implies that $\phi$ acts trivially.

What is happening here ?

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I don't think the action of Frobenius $\phi$ is trivial. The inf-res five term exact sequence reduces to a short exact sequence $$ 0\to \operatorname{Hom}\bigl(G/I, \mathbb{Q}/\mathbb{Z}\bigr)\to \operatorname{Hom}\bigl(G, \mathbb{Q}/\mathbb{Z}\bigr)\to \operatorname{Hom}\bigl(I, \mathbb{Q}/\mathbb{Z}\bigr)^{G/I}\to 0$$ The first term classifies (by taking the field fixed by the kernel) finite cyclic unramified extensions of $K$, the middle term is all finite cyclic extensions of $K$. Then $\operatorname{Hom}\bigl(I, \mathbb{Q}/\mathbb{Z}\bigr)$ are finite cyclic extensions of $K_{\text{nr}}$. To say that they are fixed by the action of $G/I$ is to ask that they come from totally ramified finite cyclic extensions of $K$. Taking a Kummer extension of degree $n$, where $n$ is not divisible by $p$ and with $\mu_n\not\subset K$, it is clear that $G/I$ does not act trivially on that term. The exactness of the sequence above reflects that every finite cyclic extension of $K$ is a composition of a cyclic unramified and a cyclic totally ramified extension of $K$. That is not true for non-abelian extension and that could be the source of confusion.

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  • $\begingroup$ Now one can go back to the calculation of $H^1\bigl(G/I, \operatorname{Hom}(I,\mathbb{Q}/\mathbb{Z})\bigr)$. But it is too late for me now. $\endgroup$ Sep 22 at 21:11

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