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Math Help - Galois Group of some polynomials of low degree

  1. #1
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    Galois Group of some polynomials of low degree

    Okay, I still having some trouble with finding Galois group.

    Let's say that I wish to find the Galois group of f(x)=x^4+1 over \mathbb{Q}. Now the roots of f(x) are \omega, \omega^3, \omega^5, \omega^7 where \omega = \sqrt{i} and so the splitting fields of f(x) over \mathbb{Q} is \mathbb{Q}(\omega). Also the minimal polynomial of \omega is f(x) hence we have [\mathbb{Q}(\omega) : \mathbb{Q}]=4.

    Since the characteristics of \mathbb{Q} = 0 we know that |Gal(\mathbb{Q}(\omega)/\mathbb{Q})| = [\mathbb{Q}(\omega) : \mathbb{Q}]=4. There are only two unique groups (up to isomorphism) of order 4, hence either Gal(\mathbb{Q}(\omega)/\mathbb{Q}) \cong \mathbb{Z}_4 or Gal(\mathbb{Q}(\omega)/\mathbb{Q}) \cong \mathbb{Z}_{2} \times \mathbb{Z}_{2}.

    I feel like it should be \mathbb{Z}_{2} \times \mathbb{Z}_{2}. This is my reason (which I know is not right) but I think I am on the right track. Since we have \omega^2 = i we can write f(x)=(x^2+\omega^2)(x^2-\omega^2). Now if x^2+\omega^2, x^2-\omega^2 \in \mathbb{Q}[x] I would be able to assert that for \phi \in Gal(\mathbb{Q}(\omega)/\mathbb{Q}) we must have \phi(\omega^2) = \pm \omega^2 and \phi(\omega^3) = \pm \omega^3.

    So in essence I need some like the above that will enable me to eliminate some of the 24 permuatations (unless my argument is totally wrong and it's acutally isomorphic to S_{4} ).

    Also, the second problem I was doing is this: find the Galois group of f(x) = x^8 -1 . Since f(x)=(x^4-1)(x^4+1)=(x^2-1)(x^2+1)(x^4+1)=(x-1)(x+1)(x^2+1)(x^4+1) the spltting field of f(x) over \mathbb{Q} is \mathbb{Q}(\omega) (because i = \omega^2 \in \mathbb{Q}(\omega)). Hence once I solve the above problem I have solved this one as well since they have the same splitting fields, right?

    Thanks in advance.
    Last edited by mpetnuch; May 2nd 2008 at 10:18 PM. Reason: Forgot the second polynomial.
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  2. #2
    Junior Member
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    When I woke up this morning I realized what an idiot I am. The answer was staring me right in the face. The group is cyclic!!!

    Let \phi \in Gal(\mathbb{Q}(\omega)/\mathbb{Q}), then \phi(\omega) = \omega, \omega^3, \omega^5, or \omega^7. Once we choose the value for \phi(\omega), we have defined \phi entirely because it is an automorphism! That is \phi(\omega^{n}) = \phi^{n}(\omega). Thus we must have Gal(\mathbb{Q}(\omega)/\mathbb{Q}) \cong \mathbb{Z}_{4}.

    And for the second problem, since the splitting field of f(x)=x^8-1 is \mathbb{Q}(\omega) as well, the Galois group of f(x) over \mathbb{Q} is \mathbb{Z}_{4} as well.
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