K)$?

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Dec 15, 2021 at 3:05	comment	added	MAS		I don't find reason why to close this question. When I said there exists $\sigma \in \text{Gal}(K(S_n)/K)$ such that $\sigma(\alpha)=g(\alpha)$ for all $\alpha \in S_n$ is quite possible. I find no reason to disagree. In fact this is an established fact in a published paper in a reputed journal.
Dec 14, 2021 at 12:02	review	Close votes
Dec 20, 2021 at 3:02
Dec 14, 2021 at 4:39	comment	added	MAS		@WillSawin, no, perhaps I made mistake in the language. I mean there exists `at least one` $\sigma \in \text{Gal}(K(S_n)/K)$ such that $\sigma(\alpha)=g(\alpha)$ for all $\alpha \in S_n$. I didn't mean all $\sigma$ will do it. Now I want to show whether that particular $\sigma$ correspond to some $\tau \in \text{Gal}(\bar K/K)$ satisfying relation like $(3)$. All I confirm is that the relation $(3)$ is valid.
Dec 14, 2021 at 4:33	history	edited	MAS	CC BY-SA 4.0	added 24 characters in body
Dec 14, 2021 at 3:57	comment	added	Will Sawin		Equation (3) is also certainly false, as to be true for all $\sigma$s would imply that $\sigma(\alpha)$ is independent of $\sigma$, which, since this is also true for all $\alpha \in S_n$, only holds if the Galois group is trivial.
Dec 14, 2021 at 3:35	history	edited	MAS	CC BY-SA 4.0	deleted 212 characters in body
Dec 14, 2021 at 3:28	comment	added	MAS		@WillSawin, well. Suppose $g(x) \in x\mathcal{O}[[x]]$ be another power series so that $\sigma(\alpha)=g(\alpha)$ for all $ \alpha \in S_n$ and $\sigma \in \text{Gal}(K(S_n)/K)$. I want to show this relation can be extended to $\sigma(\alpha)=g(\alpha)$ for all $\alpha \in S$ and $ \sigma \in \text{Gal}(\bar K/K)$. This is what I want. When does it follow ?
Dec 14, 2021 at 3:09	comment	added	Will Sawin		Can you simply edit this question to ask only what you intend to ask, rather than leaving the old material up, which makes it a bit confusing? If you ask for all roots of all iterates of all power series, that would be enough, since every finite extension is generated by the roots of some polynomial, thus of some power series.
Dec 14, 2021 at 2:35	history	edited	MAS	CC BY-SA 4.0	added 466 characters in body
Dec 14, 2021 at 2:19	comment	added	MAS		I admit I should not have written the left side of $(2)$. In other words, following the above nice comments, $\text{Gal}(\bar K/K) \neq \text{Aut}(K(S)/K)$. So now the question is- when would we get such equality ? Does adjoining `all roots of all iterates of all power series` $f(x) \in x \mathcal{O}[[x]]$ with $K$ produce $\bar K$ or the $\text{Gal}(\bar K/K)$ ?
Dec 13, 2021 at 20:50	comment	added	KConrad		Why do you write ${\rm Gal}(\overline{K}/K)$ on the left side of (2)? It would amount to saying $\overline{K} = K(S)$.
Dec 13, 2021 at 19:10	comment	added	LSpice		Am I missing something obvious? Isn't the answer that the fibres of $\operatorname{Gal}(\bar K/K) \to \operatorname{Gal}(K(S)/K)$ are $\operatorname{Gal}(\bar K/K(S))$-torsors, so (regardless of what $K$ and $S$ are) cannot be trivial unless $K(S) = \bar K$? Or do you mean whether there is some isomorphism other than the one given by restriction?
Dec 13, 2021 at 19:08	history	edited	LSpice	CC BY-SA 4.0	Minor TeXing and proofreading
Dec 13, 2021 at 18:51	comment	added	Will Sawin		Under this assumption it seems that the answer is no since the Galois group of $\overline{K}$ is nonabelian but the Galois group of $K(S)$ is (by assumption) abelian.
Dec 13, 2021 at 17:02	history	edited	MAS	CC BY-SA 4.0	added 140 characters in body
Dec 13, 2021 at 16:53	history	asked	MAS	CC BY-SA 4.0

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