Find a field $F$ such that $[F : \mathbb{Q}] = 3$ and $F \ne \mathbb{Q }(\sqrt[3]{a}), \forall a \in \mathbb{Q}$

Question

Find a field $F$ such that $[F : \mathbb{Q}] = 3$ and $F \ne \mathbb{Q }(\sqrt[3]{a}), \forall a \in \mathbb{Q}$

In attempt of solving this problem, I have tried to use many algebraic number $\alpha$ of degree $3$ on $\mathbb{Q}$. But the field $\mathbb{Q} (\alpha )$ always seem to be the same to some $\mathbb{Q }(\sqrt[3]{a})$. For example, $\alpha = 1 + \sqrt[3]{3}$.

Currently I don't have any vision to solve the problem. Please give me a hint. Anything is greatly appreciated.

Answer 1

If you can find an irreducible cubic polynomial $f$ over $\Bbb Q$ which has three real roots, then $F=\Bbb Q(\alpha)$ is a cubic field for any root $\alpha$ of $f$. Moreover $F$ cannot be a $\Bbb Q(\sqrt[3]{a})$ since that field has an embedding into $\Bbb C$ whose image does not lie in $\Bbb R$, but every embedding of $F$ into $\Bbb C$ has image inside $\Bbb R$ (since $f$ has real roots).