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4
votes
Accepted
Deformations of Galois cohomology
The answer to your example question is "No".
Let $G$ be isomorphic to $\widehat{\mathbf{Z}}$, and $\phi$ the topological generator of $G$ (corresponding to $1 \in \widehat{\mathbf{Z}}$). Then one comp …
6
votes
Accepted
Galois cohomology of $\mathbf{Z}_\ell(m)$ over finite fields
The Galois cohomology of finite fields is pretty straightforward: a $G_k$-module $M$ is entirely determined by the action of Frobenius $\varphi$, and we have $H^0(k, M) = M^{\varphi = 1}$, $H^1(k, M) …
33
votes
1
answer
3k
views
How is etale cohomology of integer rings related to Galois cohomology?
In the paper of Bloch and Kato in the Grothendieck Festschrift, and some other papers relating to the Bloch-Kato conjecture and the ETNC, the cohomology groups
$H^i_{\mathrm{et}}(\operatorname{Spec} …
8
votes
Why does $H^1(G_p/I_p,\mathbb{F}(\delta\epsilon^{-1}))$ vanish?
Cohomology of $\widehat{\mathbb{Z}}$-modules is something you can just write down. If $V$ is a (finite) module with a (continuous) action of $\widehat{\mathbb{Z}}$, so the generator acts by some endom …
9
votes
Accepted
Selmer Group versus Selmer Variety
You should read the Bloch--Kato paper in the Grothendieck Festschrift. This was, I believe, the first paper to consider Selmer groups of Galois representations defined by local conditions coming from …
8
votes
Accepted
Weight filtration on certain Galois representations
No, life is not so easy I'm afraid. For instance, the group $\mathrm{Ext}^1_{G_{\mathbf{Q}}}(\mathbf{Z}_\ell, \mathbf{Z}_\ell(n)) = H^1(\mathbf{Q}, \mathbf{Z}_\ell(n))$ has positive rank for all odd i …
15
votes
Accepted
Forms of ${\rm SL}(2)$
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Br{Br}\DeclareMathOperator\U{U}\DeclareMathOperator\disc{disc}\DeclareMathOperator\Nm{Nm}\DeclareMathOperator\diag{diag}\D …
8
votes
Which cases of Beilinson-Bloch-Kato for elliptic motives are known?
There are three approaches I know of to studying $H^1_{\mathrm{f}}(K, V)$, where $V = Sym^k(h^1(E))(n))$. All rely on $E$ being modular, so let me assume this henceforth (of course, this is no assumpt …