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Let $k$ be a local field with absolute Galois group $\Gamma_k$, inertia subgroup $I_k \subset \Gamma_k$, and residue field $\mathbb{F}$. Let $M$ be a finite Galois module over $k$.

The unramified Galois cohomology of $M$ in degree $i$ is defined to be $$H^i_{\mathrm{un}}(k,M) := \mathrm{Im}(H^i(\mathbb{F},M^{I_k}) \to H^i(k,M)).$$ My question is as follows. Let $$0 \to M_1 \to M_2 \to M_3 \to 0$$ be a short exact sequence of Galois modules. Then does this induce a long exact sequence $$0 \to H^0_{\mathrm{un}}(k,M_1) \to H^0_{\mathrm{un}}(k,M_2) \to H^0_{\mathrm{un}}(k,M_3)\to H^1_{\mathrm{un}}(k,M_1) \to \cdots$$ of unramified Galois cohomology groups?

If the modules $M_i$ are unramified, in the sense that $I_k$ acts trivially on each $M_i$, then this is clear. Otherwise the property is not so clear to me.

Generally I would appreicate any references on the basics of unramified cohomology, as it seems difficult to find the theory developed in a cohesive way in the literature.

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    $\begingroup$ For $k/\mathbf{Q}_p$, the Galois group of the residue field is top. generated by Frobenius $\sigma$ and has cohomological dimension one. It follows (by inf-res for $i=1$) that: $$H^0_{ur}(k,M)=M^{\Gamma};\ H^1_{ur}(k,M)=H^1(\mathbb{F},M^I)\simeq M^I/(\sigma - 1)M^I;\ H^n_{ur}(k,M)=0, n\ge 2.$$ If $k(\zeta_p)/k$ is ramified and you take a non-trivial Kummer extension $$0\rightarrow\mu_p\rightarrow V\rightarrow \mathbf{Z}/p\rightarrow 0$$ Then $$0=H^0_{ur}(k,V)\rightarrow H^0_{ur}(k,\mathbf{Z}/p)\rightarrow H^1_{ur}(k,\mu_p)=0$$ is not exact since $H^0_{ur}(k,\mathbf{Z}/p)=\mathbf{Z}/p$. $\endgroup$
    – user491858
    Commented 4 hours ago
  • $\begingroup$ Here's another example to show it's not an "$\ell = p$" problem: say $k$ is of order $q = p^r$ and let $\ell \neq p$ be prime. Let $M = \mathbf{F}_\ell^2$ and let $\operatorname{Gal}_k$ act tamely on $M$ by sending arithmetic Frobenius to $\mathrm{diag}(q, 1)$ and sending a generator of tame inertia to $\begin{pmatrix} 1 &1 \\ 0 &1 \end{pmatrix}$. Let $N = M^I$ and let $Q = M/N$. Then the map $H^1_{ur}(k, N) \to H^1_{ur}(k, M)$ is surjective but $H^1_{ur}(k, Q)$ is $1$-dimensional (whereas $H^2_{ur}(k, N) = 0$). $\endgroup$
    – SeanC
    Commented 4 hours ago

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