# [Galois theory] Dimension and automorphisms

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• Mar 25th 2013, 03:07 PM
Siron
[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!
• Mar 25th 2013, 03:35 PM
Gusbob
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.