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I am trying to find a irreducible polynomial over $\mathbb{Q}[x]$ and over $\mathbb{Q}[\sqrt[3]{2}][x]$ that have $\alpha = \sqrt[3]{2} + i$ as zero.

I realized that $p(x) = x^6 + 3x^4 - 4x^3 + 3x^2 + 12x + 5$ is a polynomial in $\mathbb{Q}[x]$ which have $\alpha$ as root. But I failed to prove that $p(x)$ is irreducible over $\mathbb{Q}$.

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  • $\begingroup$ Thanks @DietrichBurde, I corrected that $\endgroup$ – rrsc Jun 27 at 15:42
  • $\begingroup$ How did you find your $p(x)$ polynomial? $\endgroup$ – Peter Phipps Jun 27 at 16:09
  • $\begingroup$ Well, $\alpha = \sqrt[3]{2} + i \implies (\alpha - i)^3 = (\sqrt[3]{2})^3 \implies \alpha^3 - 3\alpha^2i + 3\alpha i^2 - i^3 = 2 \implies \alpha^3 - 3\alpha - 2 = i(3\alpha^2 - 1) \implies (\alpha^3 - 3\alpha - 2)^2 = [i(3\alpha^2 - 1)]^2 \implies \alpha^6 + 3\alpha^4 - 4\alpha^3 - 6\alpha^2 + 12\alpha + 5 = 0$ $\endgroup$ – rrsc Jun 27 at 18:10
  • $\begingroup$ I think your $-6\alpha^2$ should be $+3\alpha^2$. I solved your polynomial on WolframAlpha and $\sqrt[3] 2 \pm i$ wasn't one of the roots. $\endgroup$ – Peter Phipps Jun 27 at 18:28
  • $\begingroup$ You are right, @PeterPhipps. I edited the question $\endgroup$ – rrsc Jun 27 at 18:35
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To prove that the polynomial you found is irreducible over $\mathbb Q$, it s true that you can't use any simple criterion like Eisenstein, but you can work with degrees and dimensions. Let $b=2^{1/3}$. First, prove that $\mathbb Q[a]= \mathbb Q[b,\omega b]=\mathbb Q[b,\omega]$ where $\omega$ is third root of unity. Then, you can see that $[\mathbb Q[b,\omega] : \mathbb Q]=6$ by tower law. But $[\mathbb Q[a],\mathbb Q]= deg irr_{\mathbb Q, a}(x)$= the degree of the minimal polynomial over $\mathbb Q$ with $a$ as its root. However, you ve already found a polynomial of degree 6 with $a$ as a root so it must be irreducible.

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Another method to show that $p(x)$ is irreducible over $\Bbb Q$ is to show that it is already irreducible over $\Bbb F_7$. This is easier to show.

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