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Is it true that the extension $\mathbb{Q}(\sqrt [n]{3})/\mathbb{Q}$ is the splitting field of some polynomial over $\mathbb{Q}$? My guess is no. But I can not prove it. Some observations I made are as $[\mathbb{Q}(\sqrt [n]{3}):\mathbb{Q}]=n$ any element of $\mathbb{Q}(\sqrt [n]{3})$ will satisfy a polynomial (irreducible over $\mathbb{Q}$) of degree atmost $n$. Splitting field of $x^{n}-3$ is $\mathbb{Q}(\sqrt [n]{3},\zeta_{n})$ where $\zeta_{n}$ is a primitive $n$th root of unity. I don't know whether these observations are of any use.

Any idea/help is most welcome. Thanks. Can this problem be solved without using Galois theory?

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It is not the splitting field since it is not Galois. Such fields contain all the conjugate roots, $\zeta_n\sqrt[n]{3}$

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Hint: If $K$ is a splitting field of some polynomial over $F$, and if some other irreducible polynomial over $F$ has a zero in $K$, then this other polynomial splits over $K$.

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  • $\begingroup$ Can your hint be proved without using Galois theory? $\endgroup$ – Bingo Sep 17 '14 at 19:58
  • $\begingroup$ Hm... I don't think so. $\endgroup$ – Amitai Yuval Sep 17 '14 at 20:30
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If F is a splitting field over K, then that is normal over K. But the constructed field is not normal over the base field, because x^n-3 is irreducible (by Eisenstein's Criterion); and the field has a root, and does not have all roots of x^n-3. (n is greater than 2)

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