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Thread: 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 $\displaystyle f(x)=x^4+1$ over $\displaystyle \mathbb{Q}$. Now the roots of $\displaystyle f(x)$ are $\displaystyle \omega, \omega^3, \omega^5, \omega^7$ where $\displaystyle \omega = \sqrt{i}$ and so the splitting fields of $\displaystyle f(x)$ over $\displaystyle \mathbb{Q}$ is $\displaystyle \mathbb{Q}(\omega)$. Also the minimal polynomial of $\displaystyle \omega$ is $\displaystyle f(x)$ hence we have $\displaystyle [\mathbb{Q}(\omega) : \mathbb{Q}]=4$.

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

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

    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 $\displaystyle S_{4}$ ).

    Also, the second problem I was doing is this: find the Galois group of $\displaystyle f(x) = x^8 -1 $. Since $\displaystyle f(x)=(x^4-1)(x^4+1)=(x^2-1)(x^2+1)(x^4+1)=(x-1)(x+1)(x^2+1)(x^4+1)$ the spltting field of $\displaystyle f(x)$ over $\displaystyle \mathbb{Q}$ is $\displaystyle \mathbb{Q}(\omega)$ (because $\displaystyle i = \omega^2 \in \mathbb{Q}(\omega)$). 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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    May 2007
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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 $\displaystyle \phi \in Gal(\mathbb{Q}(\omega)/\mathbb{Q})$, then $\displaystyle \phi(\omega) = \omega, \omega^3, \omega^5,$ or $\displaystyle \omega^7$. Once we choose the value for $\displaystyle \phi(\omega)$, we have defined $\displaystyle \phi$ entirely because it is an automorphism! That is $\displaystyle \phi(\omega^{n}) = \phi^{n}(\omega)$. Thus we must have $\displaystyle Gal(\mathbb{Q}(\omega)/\mathbb{Q}) \cong \mathbb{Z}_{4}$.

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