Timeline for Computing $H^1$ with coefficients in a torsion-free abelian group
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 12, 2021 at 8:18 | comment | added | Daniel Loughran | This cohomology group is both torsion and divisible, hence trivial. | |
| Jul 12, 2021 at 3:10 | comment | added | oleout | So I apply the functor $H^i(k,-)$ to your exact sequence and obtained $$0 \rightarrow M^{G_k} \rightarrow (M \otimes \mathbb{Q})^{G_k} \rightarrow (M \otimes \mathbb{Q}/\mathbb{Z})^{G_k} \rightarrow H^1(k,M) \rightarrow H^1(k, M \otimes \mathbb{Q}).$$ Is the second last map surjective, i.e., is the last term zero? Otherwise I don't see how $H^1(k,M)$ can be realised as a quotient of $(M \otimes \mathbb{Q}/\mathbb{Z})^{G_k}$. | |
| Jul 11, 2021 at 16:13 | comment | added | Chris Wuthrich | It just follows from the short exact sequence $0\to M \to M\otimes\mathbb{Q} \to M\otimes\mathbb{Q}/\mathbb{Z}\to 0$. | |
| Jul 11, 2021 at 16:09 | comment | added | oleout | @ChrisWuthrich I'm just curious here... Could you explain how does $\mathbb{Q}/\mathbb{Z}$ come about in this instance? This abelian additive group appears quite a lot in what I'm studying, for example, it contains the Brauer group of any local field, or that this is the colimit of all the $n$-th roots of unity. Your result looks neat, are there any suitable references? | |
| Jul 11, 2021 at 16:02 | vote | accept | oleout | ||
| Jul 11, 2021 at 11:14 | comment | added | Chris Wuthrich | To answer the general question in the title: If $M$ is a $G_k$-module that is free as a $\mathbb{Z}$-module, then $H^1(k,M)$ is the quotient of $(M\otimes \mathbb{Q}/\mathbb{Z})^{G_k}$ by $M^{G_k}\otimes\mathbb{Q}/\mathbb{Z}$. That is what computer algebra systems like magma do behind the scene, | |
| Jul 11, 2021 at 7:59 | answer | added | Daniel Loughran | timeline score: 4 | |
| Jul 11, 2021 at 3:10 | history | asked | oleout | CC BY-SA 4.0 |