I'm trying to find a basis for the field extension $\mathbb{Q}(\zeta , \sqrt[3]{2})/\mathbb{Q}$, where $\zeta$ is the cube root of unity. I attempted this with starting with a set of elements I know which span the field, i.e $1, \zeta, \zeta^2,\sqrt[3]{2}, \zeta\sqrt[3]{2}, \zeta^2\sqrt[3]{2}, \sqrt[3]{2^2}, \zeta\sqrt[3]{2^2}, \zeta^2\sqrt[3]{2^2}$
From this spanning set I need to sift any elements which are a linear combination of other elements in order to end up with a basis. I decided to remove $\zeta^2, \zeta^2\sqrt[3]{2}, \zeta^2\sqrt[3]{2^2}$, because $\zeta^2=-1-\zeta$. Have I removed the dependant elements correctly?
But then the next part of the question asks to find the 3 automorphisms of the field. If $\sigma$ is an automorphism. but then what does $\sigma (\zeta)$ and $\sigma(\sqrt[3]{2})$ equal
Let $\;F\;$ be a field, $\;a,b\;$ two algebraic elements (in some extension field) over $\;F\;$ . Assume they are no conjugate to avoid trivial cases.
Prove that if $\;A\,,\,\,B\;$ are basis for $\;F(a)/F\;,\;\;F(b)/F\;$ , resp. , then $\;C:=\{ab\;;\;a\in A\;,\;\;b\in B\}\;$ is a basis for $\;F(a,b)/F\;$