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I came across an old exam problem and I wonder if my solution is correct.

Let $L=\mathbb{Q}(\omega)$, where $\omega=e^{\frac{2\pi i}{6}}$ is a primitive sixth root of unity:

a) determine $[L:\mathbb{Q}]$

b) prove that $t^{3}-2$ is irreducible over $L$

c) decide if the field $\mathbb{Q}(\omega, \sqrt[3]{2})$ is normal over $\mathbb{Q}$

Here is what I think

a) consider the polynomial $x^{6}-1\in\mathbb{Q}[x]$ then $\frac{x^{6}-1}{(x-1)(x+1)}=x^4+x^2+1$ which is irreducible over $\mathbb{Q}$ by the rational root theorem, hence $[L:\mathbb{Q}]=4$

b) $t^{3}-2=(x-\sqrt[3]{2})(x-\varepsilon\sqrt[3]{2})(x-\varepsilon^2\sqrt[3]{2})$ where $\varepsilon=e^{\frac{2\pi i}{3}}=-\frac{1}{2}+\frac{\sqrt{3}i}{2}$

I guess I need to show that neither of the above roots is contained in $\mathbb{Q}(\omega)$ but I don't see any good and fast method for proving it. We have
$$\omega^2=e^{\frac{2\pi i}{3}}$$

$$\omega^3=e^{\frac{4\pi i}{3}}$$

Hence $t^{3}-2$ will be irreducible if $\sqrt[3]{2}\not\in \mathbb{Q}(\omega)$

If $\sqrt[3]{2}\in \mathbb{Q}(\omega)$ then $\sqrt[3]{2}=a+b\omega+c\omega^2+c\omega^3$, but then $a=0$ as $a\in\mathbb{Q}$ but how to check the rest?

c) if I can prove b) then I think that $\mathbb{Q}(\omega, \sqrt[3]{2})\supseteq \mathbb{Q}$ is normal as $\mathbb{Q}(\omega, \sqrt[3]{2})\supseteq \mathbb{Q}$ will be the splitting field for $t^{3}-2$.

Any help appreciated! Thank you.

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  • 2
    $\begingroup$ You cannot use the rational root theorem to determine irreducibility of quartics and higher. If anything, you should add a more detailed argument to that part of the proof. $\endgroup$ – Kaj Hansen Aug 19 '14 at 18:10
  • $\begingroup$ So rational root theorem works up to cubics? I tried with Eisenstein criterion applied to f(x+1) but it is not applicable... $\endgroup$ – user124471 Aug 19 '14 at 18:22
  • $\begingroup$ Hint: $$x^4+x^2+1=(x^4+2x^2+1)-x^2$$ is the difference of two squares. $\endgroup$ – Jyrki Lahtonen Aug 19 '14 at 18:58

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