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Math Help - galois group of a polynomial of degree 5

  1. #1
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    galois group of a polynomial of degree 5

    We've done degree 3 and 4, but not 5.

    Find Galois groups of the following polynomials over \mathbb{Q}:

    x^5+x^3-3x^2-3

    (x^2-8)(x^3-7)

    Please help me out, again.
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  2. #2
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    Quote Originally Posted by marianne View Post
    x^5+x^3-3x^2-3
    This polynomial factors as (x^2 + 1)(x^3 - 3).
    Over the complex numbers this factors as (x+i)(x-i)(x - \sqrt[3]{3})(x-\zeta\sqrt[3]{3})(x - \zeta^2 \sqrt[3]{3}).
    Where, \zeta = -\tfrac{1}{2} + i \tfrac{\sqrt{3}}{2}.

    Thus, the splitting field is given by \mathbb{Q}(i,\sqrt[3]{3},\zeta) = \mathbb{Q}(i,\sqrt[3]{3},\sqrt{3}).
    Now, [\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3}) : \mathbb{Q}(\sqrt[3]{3},\sqrt{3})] = 2 since i \not \in \mathbb{Q}(\sqrt[3]{3},\sqrt{3}).

    The polynomial x^2 - 3 is irreducible over \mathbb{Q} so it must be irreducible over \mathbb{Q}(\sqrt[3]{3}) because [\mathbb{Q}(\sqrt[3]{3}): \mathbb{Q}] = 3 and \gcd(2,3) = 1. This means that [\mathbb{Q}(\sqrt[3]{3},\sqrt{3}) : \mathbb{Q}(\sqrt[3]{3}) ] = 2.
    And of course, [\mathbb{Q}(\sqrt[3]{3}) : \mathbb{Q}] = 3.

    Thus, [\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3}):\mathbb{Q}] = 2\cdot 2\cdot 3 = 12.

    The Galois group therefore must have order 12 and that already gives us a few ideas of this group might be.
    There are five possibilites: \mathbb{Z}_{12},\mathbb{Z}_4\times \mathbb{Z}_3, A_4, D_6, T.

    I will let you think about which one out of the five it must be. But before you do that it would be helpful to make the following realizations. First, if \sigma is an automorphism then it is completely determined by the values \sigma(i),\sigma(\sqrt[3]{3}),\sigma(\sqrt{3}). Now, \sigma must permute the roots of the minimal polynomial of each element, so, \sigma(i) = \pm i, \sigma(\sqrt[3]{3}) = \sqrt[3]{3},\zeta\sqrt[3]{3},\zeta^2\sqrt[3]{3},\sigma(\sqrt{3}) = \pm \sqrt{3}. Thus, there are a total of at most 12 such permutations but because the Galois group has order twelve it means for each one of these possibilities we have a permuation. Let \alpha be the permutation: \alpha(i) = -i,\alpha(\sqrt[3]{3}) = \sqrt[3]{3},\alpha(\sqrt{3}) = \sqrt{3} i.e. \alpha is complex conjugation. Let \beta be the permutation: \beta(i) = i,\beta(\sqrt[3]{3}) = \zeta\sqrt[3]{3},\beta(\sqrt{3}) = \sqrt{3}. Let \gamma be the permutation: \gamma(i) = i,\gamma(\sqrt[3]{3}) = \sqrt[3]{3},\gamma(\sqrt{3}) = -\sqrt{3}. Notice that \alpha,\beta,\gamma are non-trivial automorphisms with \alpha^2 = \beta^3 = \gamma^2 = \text{id}.

    Remember that \overline{\zeta} = \zeta^2, so, \alpha\beta (\sqrt[3]{3}) = \alpha ( \zeta \sqrt[3]{3}) = \alpha(\zeta)\alpha(\sqrt[3]{3}) = \overline{\zeta}\sqrt[3]{3} = \zeta^2 \sqrt[3]{3}. While \beta \alpha(\sqrt[3]{3}) = \beta (\sqrt[3]{3}) = \zeta\sqrt[3]{3}. We see that \alpha \circ \beta \not = \beta \circ \alpha so the Galois group is not abelian. This eliminates the first two abelian cases and leaves us with only 3. Can you argue from here?
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    This polynomial factors as (x^2 + 1)(x^3 - 3).
    Could this be done like this?

    Gal_\mathbb{Q}f=\mathbb{Z}_2\otimes A_3 or Gal_\mathbb{Q}f=\mathbb{Z}_2\otimes \mathbb{Z}_3, since
    Gal_\mathbb{Q}(x^3-3)=\mathbb{Z}_3 or A_3
    and then we just check, using discriminant, what is exactly Gal_\mathbb{Q}(x^3-3).
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  4. #4
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    Quote Originally Posted by georgel View Post
    Could this be done like this?

    Gal_\mathbb{Q}f=\mathbb{Z}_2\otimes A_3 or Gal_\mathbb{Q}f=\mathbb{Z}_2\otimes \mathbb{Z}_3, since
    Gal_\mathbb{Q}(x^3-3)=\mathbb{Z}_3 or A_3
    and then we just check, using discriminant, what is exactly Gal_\mathbb{Q}(x^3-3).
    It is not true in general that the Galois group of a product of two polynomials is the direct product of the Galois groups of each of the polynomials. Thus, you do not know if this is one of those cases if it works or does not without further argument.
    ----

    There is something further we can do to determine the Galois group. Notice that \mathbb{Q}(\sqrt{3}),\mathbb{Q}(i) are two distinct subfields of \mathbb{Q}(i,\sqrt[3]{3},\sqrt{3}). Now \mathbb{Q}(\sqrt{3})/\mathbb{Q} and \mathbb{Q}(i)/\mathbb{Q} are normal extensions of degree 2. Therefore, by the fundamental theorem of Galois theory it means \text{Gal}(\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})/\mathbb{Q}(i)) and \text{Gal}(\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})/\mathbb{Q}(\sqrt{3})) are two normal subgroups of index two in \text{Gal}(\mathbb{Q}(i,\sqrt[3]{3},\sqrt{3})/\mathbb{Q}). However A_4 has no subgroup of order 6 while D_6 has only one normal subgroup of order 6. Thus, it must be the remaining possibility and that is T.
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