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

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

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 $$\mathbb{Q }(\sqrt{a})$$. For example, $$\alpha = 1 + \sqrt{3}$$.

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

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 $$\Bbb Q(\sqrt{a})$$ 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).
• For example, the roots of $x^3 - 3x - 1$? Thank you, sir. Oct 2, 2018 at 20:12
• As long as you know that $x^3-3x-1$ is irreducible, yes! Oct 3, 2018 at 0:15