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Barre
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I've been doing some exercises in introductory Galois theory (self-study hence PF is the only avaliable validator :) ) and a side-result of some of them is surprising to me, hence I would like you to set me straight on this one if I'm wrong.
Let [itex]K(x)[/itex] be the field of rational functions with coefficients in the field [itex]K[/itex] of characteristic zero. Then [itex]K(x)[/itex] can be seen as an extension of [itex]K[/itex] by a transcendent indeterminate [itex]x[/itex], and [itex]K(x)/K[/itex] is actually Galois. Let [itex]G = Gal(K(x)/K)[/itex], which is infinite (for example [itex]\phi(f(x)) = f(x+1)[/itex] generates a cyclic subgroup of infinite order) and let [itex]H[/itex] be any infinite subgroup of [itex]G[/itex]. Then the field fixed by [itex]H[/itex] is [itex]K[/itex].
Here are a couple of results from previous exercises that I've used.
1. If [itex]E \neq K[/itex] is an intermediate field of the extension [itex]K \subset K(x)[/itex], then [itex][K(x):E][/itex] has finite degree.
2. Let [itex]H'[/itex] denote the intermediate field of the extension [itex]K(x)/K[/itex] fixed by automorphisms in the subgroup [itex]H[/itex]. Equivalently, let [itex]E'[/itex] denote the subgroup of [itex]G[/itex] such that all automorphisms in this group fix [itex]E[/itex]. If [itex]H' = E[/itex], then [itex]H \subset E'[/itex].
Let [itex]H[/itex] be any subgroup of [itex]G = Gal(K(x)/K)[/itex] of infinite order and let [itex]H' = E[/itex], and assume that [itex]E[/itex] is not [itex]K[/itex], hence it is a proper extension of K. Because [itex]K(x)[/itex] is finite-dimensional over [itex]E[/itex] and an E-automorphism of [itex]K(x)[/itex] is determined by its action on the basis of [itex]K(x)[/itex] over [itex]E[/itex], and basis is finite, then there can only be a finite amount of E-automorphisms of [itex]K(x)[/itex]. Hence [itex]|E'|[/itex] is finite. By 2, [itex]H \subset E'[/itex], but the first one was infinite which is a contradiction. Then the assumption that [itex]E[/itex] is not [itex]K[/itex] was wrong, and it would follow that [itex]H' = E = K[/itex].Could this be correct? The follow-up exercise asked to prove that no infinite subgroup [itex]H[/itex] has the property that [itex]H = H'' = (H')'[/itex] but I've arrived at a slightly stronger result, hence my question.
Homework Statement
Let [itex]K(x)[/itex] be the field of rational functions with coefficients in the field [itex]K[/itex] of characteristic zero. Then [itex]K(x)[/itex] can be seen as an extension of [itex]K[/itex] by a transcendent indeterminate [itex]x[/itex], and [itex]K(x)/K[/itex] is actually Galois. Let [itex]G = Gal(K(x)/K)[/itex], which is infinite (for example [itex]\phi(f(x)) = f(x+1)[/itex] generates a cyclic subgroup of infinite order) and let [itex]H[/itex] be any infinite subgroup of [itex]G[/itex]. Then the field fixed by [itex]H[/itex] is [itex]K[/itex].
Homework Equations
Here are a couple of results from previous exercises that I've used.
1. If [itex]E \neq K[/itex] is an intermediate field of the extension [itex]K \subset K(x)[/itex], then [itex][K(x):E][/itex] has finite degree.
2. Let [itex]H'[/itex] denote the intermediate field of the extension [itex]K(x)/K[/itex] fixed by automorphisms in the subgroup [itex]H[/itex]. Equivalently, let [itex]E'[/itex] denote the subgroup of [itex]G[/itex] such that all automorphisms in this group fix [itex]E[/itex]. If [itex]H' = E[/itex], then [itex]H \subset E'[/itex].
The Attempt at a Solution
Let [itex]H[/itex] be any subgroup of [itex]G = Gal(K(x)/K)[/itex] of infinite order and let [itex]H' = E[/itex], and assume that [itex]E[/itex] is not [itex]K[/itex], hence it is a proper extension of K. Because [itex]K(x)[/itex] is finite-dimensional over [itex]E[/itex] and an E-automorphism of [itex]K(x)[/itex] is determined by its action on the basis of [itex]K(x)[/itex] over [itex]E[/itex], and basis is finite, then there can only be a finite amount of E-automorphisms of [itex]K(x)[/itex]. Hence [itex]|E'|[/itex] is finite. By 2, [itex]H \subset E'[/itex], but the first one was infinite which is a contradiction. Then the assumption that [itex]E[/itex] is not [itex]K[/itex] was wrong, and it would follow that [itex]H' = E = K[/itex].Could this be correct? The follow-up exercise asked to prove that no infinite subgroup [itex]H[/itex] has the property that [itex]H = H'' = (H')'[/itex] but I've arrived at a slightly stronger result, hence my question.
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