Timeline for A question on the injectivity of a canonical map between galois cohomology groups
Current License: CC BY-SA 4.0
11 events
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| May 26, 2018 at 3:16 | comment | added | gualterio | @FilippoAlbertoEdoardo: First of all, thank you very much for answering the question. I think you're right that K is not closed with repsect to p-extensions. For example if k is a local field not containing p-th roots of unity and if the characteristic of the residue class field is not p, then K is the maximal unramified p-extension of k. In this case, we can find nontrivial p-extensions of K. I think by "with respect to p-extensions" Koch meant "with respect to (normal) p-extensions" | |
| May 25, 2018 at 8:03 | comment | added | Filippo Alberto Edoardo | @Yonatan Harpaz: After two days of thinking I found the mistake in my (deleted) answer, thanks! You were right with your example, thank you. I have one question: Koch says in his book (p. 93) that $K$ is closed with respect to $p$-extensions because if $L/K$ is a $p$-extensions, than all conjugates of $L$ over $k$ are such (I agree) and hence their compositum is $K$ by maximality. But why is the compositum of all conjugates of $L$ a $p$-extension of $k$? This is in general false, take $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(j\sqrt[3]{2})$ which compose to a $\mathfrak{S}_3$-extension. | |
| May 23, 2018 at 11:35 | comment | added | Yonatan Harpaz | But this means that any automorphism of $K'$ over $k$ maps $K_f$ into itself, so $K_f$ would be a normal $p$-extension of $k$ which is bigger than $K$, contradicting the maximality of $K$. | |
| May 23, 2018 at 11:34 | comment | added | Yonatan Harpaz | --> There is a natural action of $G(K/k)$ on the set of cyclic $p$-extensions of $K$ inside $K'$ (because there is a natural action of $G(K'/k)$ on this set, but any automorphism of $K'$ which fixes $K$ must map any normal extension of $K$ to itself). There is also a natural action of $G(K/k)$ on $H^1(G(K',K),\mathbb{F}_p)$, and the association $f \mapsto K_f$ is $G(K/k)$-equivariant. This means that if $f$ is a non-zero $G(K/k)$-invariant element of $H^1(G(K',K),\mathbb{F}_p)$ then $K_f$ is a non-trivial cyclic $p$-extension of $K$ which invariant under $G(K/k)$.--> | |
| May 23, 2018 at 11:27 | comment | added | Yonatan Harpaz | Well, $H^1(G(K',K), \mathbb{F}_p)$ is the group of homomorphisms $G(K'/K) \to \mathbb{F}_p$. To each such homomorphism $f: G(K'/K) \to \mathbb{F}_p$ we can associate the fixed field $K \subseteq K_f \subseteq K'$ of its kernel $Ker(f) \subseteq G(K'/K)$. Then $K_f$ is either $K$ (if $f=0$) or a cyclic p-extension of $K$ (if $f \neq 0$).--> | |
| May 23, 2018 at 10:38 | vote | accept | gualterio | ||
| May 23, 2018 at 10:36 | vote | accept | gualterio | ||
| May 23, 2018 at 10:36 | |||||
| May 23, 2018 at 9:40 | comment | added | gualterio | Thanks again. So the last sentence is the one that I needed the most. Can you please suggest me any references on the theorem?, because I haven't seen them before and maybe in someday I might need those kinds of theorems. | |
| May 23, 2018 at 8:49 | comment | added | Yonatan Harpaz | If $K \subseteq L \subseteq K'$ is a cyclic $p$-extension of $K$ in $K'$ which is normal over $k$, then $L$ is in particular a normal $p$-extension of $k$. Since $K$ is the maximal normal $p$-extension of $k$, it would have to contain $L$, and so $L$ must be trivial. In cohomological terms, cyclic $p$-extensions of $K$ in $K'$ which are normal over $k$ are classified by the group $H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)}$, and so this group must be trivial. | |
| May 23, 2018 at 8:23 | comment | added | gualterio | Thank you for answering the question. I guess your intention is to show that H^1(G(K'/K),F_p)=0 by showing that K is the maximal elementary abelian p-subextension of K'/K. Can you explain a little bit more about how to use the maximality of K to show that K'/K does not admit nontrivial cyclic p-extension of K? (For example there can be non-trivial cyclic p-extension say L of K contained in K' such that L/k is not normal.) | |
| May 22, 2018 at 19:14 | history | answered | Yonatan Harpaz | CC BY-SA 4.0 |