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seen Dec 27 '13 at 2:36

Feb
12
comment Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)]
The field of Laurent series. $\mathbb{Q}((x))=Quot(\mathbb{Q}[[x]])$, the quotient field of the ring of formal power series with coefficients in $\mathbb{Q}$.
Feb
12
asked Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)]
Jan
23
comment For a $G$-module $A$ is there a maximal subgroup $H$ of $G$ such that the image $H^2(G,A)\rightarrow H^2(H,A)=0$?
Just out of curiosity, what would you have said under the assumption that $G$ is finite?
Jan
23
comment For a $G$-module $A$ is there a maximal subgroup $H$ of $G$ such that the image $H^2(G,A)\rightarrow H^2(H,A)=0$?
No, not necessarily. Although in the cases that I'm thinking of A is finite.
Jan
23
asked For a $G$-module $A$ is there a maximal subgroup $H$ of $G$ such that the image $H^2(G,A)\rightarrow H^2(H,A)=0$?
Dec
28
comment Interpretations of the first cohomology group
Aha! You're quite correct. The problem is that the "automorphism" I gave is not a homomorphism in general. This means that the action is not transitive, but is free. And the action is free and transitive if we restrict to the set of group-theoretic sections (which may be an empty set, but is not empty iff it's a semi-direct product). Good, I'm happy with this conclusion!
Dec
28
accepted Interpretations of the first cohomology group
Dec
28
comment Interpretations of the first cohomology group
P.S. You can find "torsor" on wiki. The idea is that $H^1(X,G)$, where $X$ is a "space" (e.g. scheme, manifold, topological space, ...) and $G$ is a "group object" (e.g. group scheme, group variety, finite group, ...), classifies $G$-torsors over $X$. Think of this naively as principal $G$-bundles over a topological space $X$.
Dec
28
comment Interpretations of the first cohomology group
Aha... So what you're saying is that I can't just choose any group extension in this interpretation - I have to work with the trivial extension (the semi-direct product). I see... Can you tell me where I was wrong with my argument in the question? It seemed like I defined an action of $H^1(G,A)$ on the set-theoretic sections of any given extension (not nec. the semi-direct product extension) which is free and transitive. Where did I go wrong?
Dec
27
revised Interpretations of the first cohomology group
edited body
Dec
27
comment Interpretations of the first cohomology group
Ah! I will correct this.
Dec
27
comment Interpretations of the first cohomology group
$E$ is some (any) specific group extension of $G$ by $A$. As for your second comment, I'm not at all sure what you mean. The first and second group cohomologies are sometimes trivial, but usually not.
Dec
27
revised Interpretations of the first cohomology group
added 11 characters in body; deleted 38 characters in body
Dec
27
asked Interpretations of the first cohomology group
Dec
20
awarded  Yearling
Oct
31
revised About the isomorphism $\operatorname{Br}(\mathbb{Q}_p)\cong \mathbb{Q}/\mathbb{Z}$
added 11 characters in body
Oct
31
asked About the isomorphism $\operatorname{Br}(\mathbb{Q}_p)\cong \mathbb{Q}/\mathbb{Z}$
Oct
30
comment How can one deduce quadratic reciprocity from Hilbert reciprocity?
Thanks! I thought it might be something with $a=p$ and $b=q$. I can't quite convince myself of all of the facts above... Is there a proof of these identities in Cassels?
Oct
30
accepted How can one deduce quadratic reciprocity from Hilbert reciprocity?
Oct
30
asked How can one deduce quadratic reciprocity from Hilbert reciprocity?