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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.

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1 Answer 1

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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).

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  • $\begingroup$ For example, the roots of $x^3 - 3x - 1$? Thank you, sir. $\endgroup$
    – ElementX
    Oct 2, 2018 at 20:12
  • $\begingroup$ As long as you know that $x^3-3x-1$ is irreducible, yes! $\endgroup$
    – Lubin
    Oct 3, 2018 at 0:15

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