I was under the impression that for all field extensions the degree would be $2^n$ for some $n \in \mathbb{Z}$. Am I mistaken?
I am wondering as I was faced with the problem of showing that: for $[F(\alpha):F]$ is odd, it implies $F(\alpha) = F$.
In advance, thank you.
I suggest you read through all the definitions and the formal process of constructing simple algebraic extensions. $\Bbb{Q}(\sqrt[3]{2})\cong \frac{\Bbb{Q}[X]}{(X^{3}-2)}$. An explicit basis for this vector space over $F$ will be given by $\{1,\sqrt[3]{2},(\sqrt[3]{2})^{2}\}$.
That is any element in $\Bbb{Q}(\sqrt[3]{2})$ can be written as $a+b\sqrt[3]{2}+c(\sqrt[3]{2})^{2}$ such that $a,b,c\in F$.