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 Feb 12 accepted Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)] Feb 12 awarded Nice Question Feb 12 revised Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)] edited title 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 $\operatorname{Br}(\mathbb{Q}_p)\cong \mathbb{Q}/\mathbb{Z}$