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Math Help - Funadamental theorem of Galois Theory

  1. #1
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    Cool Funadamental theorem of Galois Theory

    k(x) is an automorphism defined by \sigma_1(x)=x, \sigma_2(x)=1-x, and \sigma_3(x)=1/x Let I(x)=\frac{(x^2-x+1)^3}{x^2(x-1)^2}

    Assuming the fundamental theorem of Galois Theory, exhibit all intermediate fields between E=k(x) and F=k(I).
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  2. #2
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    Quote Originally Posted by dabien View Post
    k(x) is an automorphism defined by \sigma_1(x)=x, \sigma_2(x)=1-x, and \sigma_3(x)=1/x Let I(x)=\frac{(x^2-x+1)^3}{x^2(x-1)^2}

    Assuming the fundamental theorem of Galois Theory, exhibit all intermediate fields between E=k(x) and F=k(I).
    I'll assume that k is a field and k(x) is a field of rational functions in the variable x. Unfortunately, \sigma_1(x)=x, \sigma_2(x)=1-x, and \sigma_3(x)=1/x do not form a group of automorphisms (it is not even closed), so you can't find the corresponding field. However, it seems like \sigma_2(x)=1-x, and \sigma_3(x)=1/x generate a group A of automorphisms (isomorphic to a symmetric group S_3), A=\{\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5, \sigma_6\}, where \sigma_4(x)=1-1/x, \sigma_5(x)=1/(1-x), and \sigma_6(x)=x/(x-1).

    So your base field is F=k(I) where I=I(x)=\frac{(x^2-x+1)^3}{x^2(x-1)^2}. Now consider the splitting field E=k(x) for an irreducible polynomial f in k(I) such that f(X) ={(X^2 - X + 1 )}^3 -IX^2{(X-1)}^2, where f(X) \in k(I)[X] (verify this). We see that Gal(E/F)=S_3.

    Now we shall find an intermediate field corresponding a subgroup of Gal(E/F). Let H=\{\sigma_1, \sigma_3\}. It is easy to check that H is a subgroup of Gal(E/F). The field corresponding H is S=k(x+\frac{1}{x}), which is fixed by elements of H.

    I'll leave it to you to find all other intermediate fields, because it seems like a lengthy computation is required to find and verify it.
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