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Math Help - [Galois theory] Dimension and automorphisms

  1. #1
    MHF Contributor Siron's Avatar
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    [Galois theory] Dimension and automorphisms

    Hi,
    I need some help with the following exercise.

    Let L= \mathbb{Q}(\sqrt[3]{2}). Show that [L:\mathbb{Q}]=3 and that L has no automorphisms other than the identity (hint: T^3-2 has no root in \mathbb{Q} and a single root in \mathbb{R}).

    I have no idea how I can prove \dim_{L}\mathbb{Q}=3 and also no idea for the automorphism.
    Can someone give me a hint?

    Thanks in advance!
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  2. #2
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    Re: [Galois theory] Dimension and automorphisms

    [L:\mathbb{Q}] is the degree of the minimal polynomial of \sqrt[3]{2}, so the first one should be easy. Any automorphism on L must fix the base field \mathbb{Q} and send \sqrt[3]{2} to one of the roots of its minimal polynomial. However, \mathbb{Q}\subset \mathbb{R}, so \sqrt[3]{2} cannot be sent to a complex root, of which there are two. Hence \sqrt[3]{2} can only be sent to itself, which means any automorphism is the identity.
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