Timeline for Galois cohomology of finite fields
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 27, 2017 at 5:56 | comment | added | user1225 | When $N$ is finite, $\text{Ext}_\mathcal{C}^k (N, \mathbb{Z})=0$ for $k \neq 1$ by Example 1.10 of chapter I of Milne's book on Arithmetic Duality theorem (2nd edition). And as already answered below, $\text{Ext}_\mathcal{C}^1 (N, \mathbb{Z})=\text{Ext}_\mathcal{C}^0 (N, \mathbb{Q}/\mathbb{Z}) = \text{Hom}_{G}(N, \mathbb{Q}/\mathbb{Z})=\text{Hom}_{G}(\mathbb{Z}, N^D)=H^0(G, N^D)$. | |
| Mar 27, 2017 at 5:48 | comment | added | user1225 | @Chris I wonder that $\text{Ext}_\mathcal{C}^k (\mathbb{Z}, M)$ is isomorphic to $\text{Ext}_{\mathcal{C}}^k (M^D, \mathbb{Z}^D)$, by taking Pontryagin duality. I saw a reference that this is true for $k=0, 1$ though | |
| Mar 24, 2017 at 12:15 | comment | added | Chris Wuthrich | Take the Pontryagin dual of the extension given in the above comment. This is an element of $\operatorname{Ext}^k_{\mathcal{C}}(M^D,\mathbb{Q}/\mathbb{Z})$. Now use the isomorphism to $\operatorname{Ext}^{k+1}_{\mathcal{C}}(M^D,\mathbb{Z})$ in R. van Dobben de Bruyn's answer. Or do I miss something? | |
| Mar 24, 2017 at 6:15 | vote | accept | user1225 | ||
| Mar 24, 2017 at 5:30 | answer | added | nfdc23 | timeline score: 6 | |
| Mar 24, 2017 at 4:50 | answer | added | R. van Dobben de Bruyn | timeline score: 9 | |
| Mar 23, 2017 at 18:28 | comment | added | მამუკა ჯიბლაძე | Note that $H^k(G;M)=\operatorname{Ext}^k_{\mathcal C}(\mathbb Z,M)$ | |
| Mar 23, 2017 at 16:54 | history | edited | user1225 | CC BY-SA 3.0 |
added 124 characters in body
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| Mar 23, 2017 at 16:07 | history | asked | user1225 | CC BY-SA 3.0 |