For $\alpha = \sqrt[3]{2} + i$, find the minimal polynomial over $\mathbb{Q}$ and over $\mathbb{Q}[\sqrt[3]{2}]$

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 $$\mathbb{Q}$$.

• Thanks @DietrichBurde, I corrected that – rrsc Jun 27 at 15:42
• How did you find your $p(x)$ polynomial? – Peter Phipps Jun 27 at 16:09
• Well, $\alpha = \sqrt[3]{2} + i \implies (\alpha - i)^3 = (\sqrt[3]{2})^3 \implies \alpha^3 - 3\alpha^2i + 3\alpha i^2 - i^3 = 2 \implies \alpha^3 - 3\alpha - 2 = i(3\alpha^2 - 1) \implies (\alpha^3 - 3\alpha - 2)^2 = [i(3\alpha^2 - 1)]^2 \implies \alpha^6 + 3\alpha^4 - 4\alpha^3 - 6\alpha^2 + 12\alpha + 5 = 0$ – rrsc Jun 27 at 18:10
• 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. – Peter Phipps Jun 27 at 18:28
• You are right, @PeterPhipps. I edited the question – rrsc Jun 27 at 18:35

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.
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.