Basis of Q$(\sqrt[4]{3})$/Q

Question

I want to show that Q$(\sqrt[4]{3})$/Q is algebraic. I am pretty sure that {1, $(\sqrt[4]{3})$} is a basis of this extension, and I was wondering if in general, for a (some irrational number) and the field extension Q(a)/Q, the basis is {1, a}.

Thanks.

Answer 1

No this is not true. Note you cannot express $(\sqrt[4]{3}))^3$ as a linear combination of $1$ and $\sqrt[4]{3}$.

The degree of the extension is $4$; so you need four elements.