# Math Help - [Galois theory] Dimension and automorphisms

1. ## [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!

2. ## 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.