What are the automorphisms acting on $Q\sqrt[3]{2}$ that fixes Q?

Question

For the field $Q(\sqrt[3]{2})$, the elements in the field are in the form of $a + b\sqrt[3]{2}+c\sqrt[3]{4}$, where a,b,c, are elements of Q.

Is the automorphism group that fixes Q $C_{3}$?

multiplying all elements in $Q(\sqrt[3]{2})$ by $\sqrt[3]{2}$, will result a cyclic cycle of period 3, but what about mapping $\sqrt[3]{2}$ to $-\sqrt[3]{2}$? For extension fields like $Q(\sqrt{2},\sqrt{3})$, the mapping towards a radical's negative counterpart is part of the automorphism group. Should this also be part of the automorphism group that fixes Q?

Answer 1

Hint: note that such an automorphism is determined by the image of $\sqrt[3]{2}$. The latter has to get mapped to a root of the minimal polynomial $m(\sqrt[3]{2}, \Bbb Q)$ that additionally lies in $\Bbb Q(\sqrt[3]{2})$.

The only such root is $\sqrt[3]{2}$ itself, hence the group is trivial.