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Math Help - Galois Theory: when does [E : F] \= |Gal(E/F)| ?

  1. #1
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    Galois Theory: when does [E : F] \= |Gal(E/F)| ?

    I'm trying to find a field F and a polynomial f(x) \in F[x] with splitting field E such that [E : F] \not = |Gal(E/F)|. I know that f(x) must be inseparable for this to work, and so F must be infinite with characteristic p > 0. But apart from that, I'm stumped. The fields \mathbb{Z}_p(x) (the field of rational functions with elements of \mathbb{Z}_p as coefficients) for some prime p are the only such fields I can think of, and I can't think of an example in that field that makes it easy for me to compute [E : F] or |Gal(E/F)|. Any help or hints would be much appreciated.
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  2. #2
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    Quote Originally Posted by mtdim View Post
    I'm trying to find a field F and a polynomial f(x) \in F[x] with splitting field E such that [E : F] \not = |Gal(E/F)|. I know that f(x) must be inseparable for this to work, and so F must be infinite with characteristic p > 0. But apart from that, I'm stumped. The fields \mathbb{Z}_p(x) (the field of rational functions with elements of \mathbb{Z}_p as coefficients) for some prime p are the only such fields I can think of, and I can't think of an example in that field that makes it easy for me to compute [E : F] or |Gal(E/F)|. Any help or hints would be much appreciated.
    If you are looking for E|F which is inseparable I cannot help you but if you are looking for E|F such that [E:F] \neq Gal(E|F) I have a solution.

    If you take \mathbb Q (\alpha )| \mathbb Q, where \alpha  \in \mathbb R, \alpha ^3 = 2 then you have an extension with [E:F] = 3 and Gal(E|F) = 1. This happens because the extension is not a normal extension.
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  3. #3
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    Hi zoek,

    This is confusing to me because I have a theorem in my textbook (Rotman, "Galois Theory") which says:

    If f(x) \in F[x] is a separable polynomial and if E/F is its splitting field, then |Gal(E/F)|=[E:F].

    I also have a theorem which says every field of characteristic 0 is perfect i.e. every polynomial is separable. How does this make sense in light of your example? How did you get that |Gal(E/F)|=1 and [E:F]=3? It seems to me that a splitting field for your polynomial would contain both the real cube root of 2 and the two imaginary cube roots of 2, and so |Gal(E/F)|>1.

    Thanks for the help.
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  4. #4
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    in this example E = \mathbb Q(\alpha) and this field is not a splitting field for the polynomial f(x) = x^3 - 2 (because the other two roots does not belong to \mathbb Q(\alpha))
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  5. #5
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    if g \in Gal(\mathbb Q(\alpha)|\mathbb Q) then g(x)=x for all x\in \mathbb Q and 2= g(2)=g(\alpha^3) = g(\alpha)^3, so we have g(\alpha)=\alpha, so finally g(x)=x for all x\in \mathbb Q(\alpha).

    therefore |Gal(\mathbb Q(\alpha)|\mathbb Q)|= 1.
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  6. #6
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    Quote Originally Posted by mtdim View Post
    It seems to me that a splitting field for your polynomial would contain both the real cube root of 2 and the two imaginary cube roots of 2
    this is absolutely right!
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  7. #7
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    one thing that occurs to me is to set K = Z3(u), rational functions in u over Z3, and set f(x) = x^3 - u in K[x]. i believe f is irreducible over K, but if α is a cube root of u, then

    f(x) = (x - α)^3. so E = K(α) is a splitting field of f over K, and [E:K] = 3. since any automorphism of E has to send α-->α, Gal(E/K) = {1}.
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  8. #8
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    Quote Originally Posted by zoek View Post
    in this example E = \mathbb Q(\alpha) and this field is not a splitting field for the polynomial f(x) = x^3 - 2 (because the other two roots does not belong to \mathbb Q(\alpha))
    Ah, I see. That makes sense now; I thought you meant E to be the splitting field for that polynomial.

    Your example isn't quite what I'm looking for; I want the field extension to specifically be the splitting field of the polynomial in question.
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  9. #9
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    Quote Originally Posted by Deveno View Post
    one thing that occurs to me is to set K = Z3(u), rational functions in u over Z3, and set f(x) = x^3 - u in K[x]. i believe f is irreducible over K, but if α is a cube root of u, then

    f(x) = (x - α)^3. so E = K(α) is a splitting field of f over K, and [E:K] = 3. since any automorphism of E has to send α-->α, Gal(E/K) = {1}.
    That's it! I was trying to use the field of rational functions over Z_p, but I couldn't come up with the polynomial that works. I'm quite certain you are correct about f being irreducible over K since there is no cube root of u in K.

    Thanks.
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