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I need to find the minimal polynomial of $\sqrt[5]{3} \sqrt{2} i$ over $\mathbb{Q}$ and prove that this is the minimal polynomial. I let $x =\sqrt[5]{3} \sqrt{2} i$. Then $x^5 = 3 \sqrt{32}i$ and so $x^{10} + 288 = 0$. I want to prove this is the minimal polynomial now, by showing it is irreducible over $\mathbb{Q}$.

I consider the well known ring morphism $\phi : \mathbb{Q}[X] \to \mathbb{Z}_5[X]$ and consider the polynomial $x^{10} + 288 = x^{10} + 3 \in \mathbb{Z}_5[X]$. How can I conclude that this polynomial is irreducible in $\mathbb{Z}_5[X]$?

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  • $\begingroup$ Why did you decide to go to $\;\Bbb F_5\;$ precisely? $\endgroup$
    – DonAntonio
    Dec 2, 2016 at 7:33
  • $\begingroup$ Because I noticed that $288$ is not divisible by the prime $5$. $\endgroup$
    – Kamil
    Dec 2, 2016 at 7:40
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    $\begingroup$ There is no well-known map $\mathbf Q[X]\to \mathbf F_5[X]$, but a map $\mathbf Z[X]\to \mathbf F_5[X]$ and it is enough to prove irreducibility over $\mathbf Z$. $\endgroup$
    – martini
    Dec 2, 2016 at 7:42

2 Answers 2

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I would suggest an alternate approach. Instead, consider the extension $K = \mathbb{Q}(\sqrt[5]{3}\sqrt{2}i)$. It is enough to show that $[K:\mathbb{Q}] = 10$. Indeed, $(\sqrt[5]{3}\sqrt{2}i)^{4} = 4(3)^{4/5}$, so $3^{4/5} \in K$, whence $(3^{4/5})^{4} = 27 \cdot 3^{1/5} \in K$, and thus finally $\sqrt[5]{3} \in K$. Likewise, $3^{4/5} \cdot \sqrt[5]{3}\sqrt{2}i = 3\sqrt{2}i \in K$, so $\sqrt{2}i \in K$. Thus, $K$ contains the extensions $\mathbb{Q}(\sqrt[5]{3})$ and $\mathbb{Q}(\sqrt{2}i)$ as subfields. The degree of the former is $5$ (the minimal poylnomial of $\sqrt[5]{3}$ over $\mathbb{Q}$ is $X^{5}-3$) and the degree of the latter is $2$ (the minimal poylnomial of $\sqrt{2}i$ over $\mathbb{Q}$ is $X^{2}+2$). This establishes the claim.

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  • $\begingroup$ Thank you for the reply. But why is it enough to show that $[K: \mathbb{Q}] = 10 $ ? Is there some theorem regarding this? $\endgroup$
    – Kamil
    Dec 2, 2016 at 10:06
  • $\begingroup$ I mean, I understand that $[K: \mathbb{Q}] = 10$. But how exactly does it follow from this that $x^{10} + 288 $ is irreducible over $\mathbb{Q}$? $\endgroup$
    – Kamil
    Dec 2, 2016 at 10:23
  • $\begingroup$ @Kamil: yes. Namely, if $K = F(\alpha)$, then $[K:F]$ is the degree of the minimal polynomial of $\alpha$ over $F$. $\endgroup$
    – Alex Wertheim
    Dec 2, 2016 at 17:50
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You won't succeed to prove that $x^{10} + 3 \in \mathbb{Z}_5[X]$ is irreducible... It is reducible as $3 = 3^5 \in \mathbb{Z}_5$ and therefore: $$\begin{aligned} x^{10}+3 &= x^{10}+3^5=(x^2)^5-(-3)^5\\ &=(x^2+3)(x^8+(-3)x^6+(-3)^2 x^4+(-3)^3 x^2+ (-3)^4)\\ &=(x^2+3)(x^8+2 x^6 +4 x^4 +3 x^2 +1) \end{aligned}$$

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