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Thread: 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 $\displaystyle f(x) \in F[x]$ with splitting field E such that $\displaystyle [E : F] \not = |Gal(E/F)|$. 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 $\displaystyle \mathbb{Z}_p(x)$ (the field of rational functions with elements of $\displaystyle \mathbb{Z}_p$ 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 $\displaystyle [E : F]$ or $\displaystyle |Gal(E/F)|$. 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 $\displaystyle f(x) \in F[x]$ with splitting field E such that $\displaystyle [E : F] \not = |Gal(E/F)|$. 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 $\displaystyle \mathbb{Z}_p(x)$ (the field of rational functions with elements of $\displaystyle \mathbb{Z}_p$ 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 $\displaystyle [E : F]$ or $\displaystyle |Gal(E/F)|$. Any help or hints would be much appreciated.
    If you are looking for $\displaystyle E|F$ which is inseparable I cannot help you but if you are looking for $\displaystyle E|F$ such that $\displaystyle [E:F] \neq Gal(E|F)$ I have a solution.

    If you take $\displaystyle \mathbb Q (\alpha )| \mathbb Q$, where $\displaystyle \alpha \in \mathbb R, \alpha ^3 = 2$ then you have an extension with $\displaystyle [E:F] = 3$ and $\displaystyle Gal(E|F) = 1$. 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 $\displaystyle f(x) \in F[x]$ is a separable polynomial and if $\displaystyle E/F$ is its splitting field, then $\displaystyle |Gal(E/F)|=[E:F]$.

    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 $\displaystyle |Gal(E/F)|=1$ and $\displaystyle [E:F]=3$? 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 $\displaystyle |Gal(E/F)|>1$.

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

    therefore $\displaystyle |Gal(\mathbb Q(\alpha)|\mathbb Q)|= 1$.
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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 $\displaystyle E = \mathbb Q(\alpha)$ and this field is not a splitting field for the polynomial $\displaystyle f(x) = x^3 - 2$ (because the other two roots does not belong to $\displaystyle \mathbb Q(\alpha)$)
    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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