Question

Why is the Herbrand quotient of the dual $\hat{A}$ equal to the inverse of the Herbrand quotient of $A$ in this situation?

score 503 · Answer 1 · 2011-09-20 04:48:04

Why is the Herbrand quotient of the dual $\hat{A}$ equal to the inverse of the Herbrand quotient of $A$ in this situation?

score 1.5k · Answer 2 · 2011-10-07 01:30:23

1.5k

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442

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score 191 · Answer 5 · 2011-11-05 17:23:48

191

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score 203 · Answer 6 · 2011-11-18 23:18:19

203

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score 494 · Answer 7 · 2011-11-30 00:26:58

494

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score 2.2k · Answer 8 · 2011-12-12 16:12:03

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score 5k · Answer 9 · 2011-12-27 17:48:37

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score 81 · Answer 10 · 2012-01-23 03:35:19

81

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For a $G$-module $A$ is there a maximal subgroup $H$ of $G$ such that the image $H^2(G,A)\rightarrow H^2(H,A)=0$?

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score 889 · Answer 11 · 2012-01-29 11:11:51

889

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score 369 · Answer 12 · 2026-03-26 02:58:26

369

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score 1.5k · Answer 13 · 2012-02-08 22:32:05

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score 241 · Answer 14 · 2012-02-23 11:21:28

241

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score 2.5k · Answer 15 · 2012-03-08 18:27:45

2.5k

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List Question

Why is the Herbrand quotient of the dual $\hat{A}$ equal to the inverse of the Herbrand quotient of $A$ in this situation?

Torsion-free virtually-Z is Z

Cohomology of Locally Indicable groups

Group cohomology: $H^1$ of $\mathbb Z/4 \mathbb Z$

$H^1(G,A)$ is killed by $|G|$ : Proof on the level of cocycles

Why is a class in $H^2(K/k,\mathbb{G}_m)$ locally trivial almost everywhere for $K/k$ a finite Galois extension of number fields?

Group cohomology and Klein four group

Group cohomology and Shapiro's lemma

Interpretations of the first cohomology group

For a $G$-module $A$ is there a maximal subgroup $H$ of $G$ such that the image $H^2(G,A)\rightarrow H^2(H,A)=0$?

How to understand the diagonal approximation?

Explicitly finding a cocycle in $H^3(S_3,\mathbb{Z}_3)$

Cohomology of finite cyclic groups

Find an example about cohomologically trivial

Group cohomology intuition

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