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F-automorphism group of the field of rational functions

  1. Jun 4, 2012 #1
    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.

    1. The problem statement, all variables and given/known data
    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].


    2. Relevant 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].

    3. 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.
     
    Last edited: Jun 4, 2012
  2. jcsd
  3. Jun 5, 2012 #2
    Hello, I would like to extend my question to this exercise as well. I used some of my work on this to prove assertion 1) in previous post.

    I can see how it is finite, as x (the indeterminate in K(x)) is actually a root of a polynomial in [itex]K(\phi)[/itex], namely [itex](\phi)g(y) - f(y)[/itex]. Hence for sure there exists a minimal polynomial, and by the exercise, I know that it should have degree [itex]max(deg(f), deg(g))[/itex]. But my polynomial has the property that it's degree is the maximum of degrees of f and g, hence it is the minimal polynomial I am looking for. To prove this, I need to prove that it is irreducible.

    I will be honest that I have no idea how that can be done. A hint was given to consider reducibility in [itex]K(\phi)[x][/itex], the ring of polynomials and that the linearity of the equation (in [itex]\phi = f/g[/itex]) together with the fact that f,g are coprime is the way to go. But I barely understand the element [itex]\phi[/itex] here.
     
  4. Jun 12, 2012 #3
    One last bump :)
     
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