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Math Help - Galois Theory Question

  1. #1
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    Galois Theory Question

    Let f(x)=x^4+6x^2+3x-3 and let L be a splitting field of f over \mathbb{Q}.

    I can show that 73 has a cube root in L. Now I'm asked to "deduce" that the Galois group G of f over \mathbb{Q} has a normal subgroup H such that the quotient group G/H is isomorphic to S_3, and then deduce that G is isomorphic to S_4.

    My first thought was that we can show G is isomorphic to S_4 easily enough anyway - it's not difficult to narrow down the possibilities to either A_4 or S_4, then we can use the discriminant to rule out the first case. Once we know G is isomorphic to S_4 then we can show the "normal subgroup" part.

    But this is obviously not what the question is getting at...the idea is to presumably use the fact that 73 has a cube root. Can anyone help out? Thanks.
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  2. #2
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    Quote Originally Posted by Boysilver View Post
    Let f(x)=x^4+6x^2+3x-3 and let L be a splitting field of f over \mathbb{Q}.

    I can show that 73 has a cube root in L. Now I'm asked to "deduce" that the Galois group G of f over \mathbb{Q} has a normal subgroup H such that the quotient group G/H is isomorphic to S_3, and then deduce that G is isomorphic to S_4.

    My first thought was that we can show G is isomorphic to S_4 easily enough anyway - it's not difficult to narrow down the possibilities to either A_4 or S_4, then we can use the discriminant to rule out the first case. Once we know G is isomorphic to S_4 then we can show the "normal subgroup" part.

    But this is obviously not what the question is getting at...the idea is to presumably use the fact that 73 has a cube root. Can anyone help out? Thanks.

    73 has a root in L\Longrightarrow g(x)=x^3-73 is an irreducible pol. over \mathbb{Q} and it thus splits in linear factors over L (why?).

    We thus have the fields tower \mathbb{Q}\le \mathbb{Q}(\alpha)\le L , with \alpha=\sqrt[3]{73}\in L.

    Use now the Fund. Theorem of Galois Theory with the group of \mathbb{Q}-automorphisms of L that leave \mathbb{Q}(\alpha) fixed...

    Of course, you STILL have to prove that the resulting group is the non-abelian group of order 6, but if you succeed with the above I think this already will be easier.

    Tonio
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