Timeline for Unramified Galois cohomology

6 events

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1 hour ago	comment	added	SeanC		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 $H^2_{ur}(k, N) = 0$).
1 hour ago	comment	added	user491858		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 $H^0_{ur}(k,\mathbf{Z}/p)=\mathbf{Z}/p$.
4 hours ago	history	edited	Daniel Loughran	CC BY-SA 4.0	edited body
S 4 hours ago	history	suggested	J. W. Tanner	CC BY-SA 4.0	Corrected spelling in title
4 hours ago	review	Suggested edits
S 4 hours ago
4 hours ago	history	asked	Daniel Loughran	CC BY-SA 4.0