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

    I have no idea how I can prove $\displaystyle \dim_{L}\mathbb{Q}=3$ 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

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