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Let $\mathbb{Q} \subset \mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i)$ be a field extension. I computed that $[\mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i) : \mathbb{Q}] = 10$.

I now want to show that $\mathbb{Q}(\sqrt[5]{3} \sqrt{2}i) = \mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i)$.

The first inclusion seems trivial to me. Since $\mathbb{Q}(\sqrt[5]{3}, \sqrt{2}i)$ is the smallest subfield of $\mathbb{C}$ that contains $\mathbb{Q}, \sqrt[5]{3}$ and $\sqrt{2}i$, it must also contain their product $\sqrt[5]{3}\sqrt{2}i$. This shows that $\mathbb{Q}(\sqrt[5]{3} \sqrt{2}i) \subseteq \mathbb{Q}(\sqrt[5]{3} ,\sqrt{2}i)$ (is this argument correct)?

But the other inclusion is harder. I want to take the product $\sqrt[5]{3}\sqrt{2}i$ and then by applying field operations on this, show that this field must also contain the separate factors. Obviously it must also contain the $5$th power, that is $(\sqrt[5]{3}\sqrt{2}i)^5 = 3 \sqrt[5]{2} i.$ Now, can I conclude from this that it must also contain $\sqrt{2}i$? and how to show it contains $\sqrt[5]{3}$?

Also, Let's say I would want to find the minimal polynomial of $\sqrt[5]{3}\sqrt{2}i$ over $\mathbb{Q}$. I let $x = \sqrt[5]{3}\sqrt{2}i$. Then $x^5 = 3 \sqrt{32}i$ and so $x^{10} + 288 = 0$. This is the minimal polynomial I think. I want to prove this, by proving it is irreducible over $\mathbb{Q}$. I wanted to show this using Eisenstein criterion, but couldn't find the right prime number. So how would I prove it is the minimal polynomial? Should I assume it is not, and then derive a contradiction somehow?

Help/suggestions are appreciated.

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    $\begingroup$ You got your inclusions mixed up, if $\sqrt[5]3\sqrt 2 i\in \Bbb Q(\sqrt[5]3,\sqrt 2 i)$, why would you conclude that $\Bbb Q(\sqrt[5]3,\sqrt 2 i)\subseteq \Bbb Q(\sqrt[5]3\sqrt 2 i)$? $\endgroup$
    – Ennar
    Nov 30, 2016 at 19:10
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    $\begingroup$ I think Eisentine's criterion won't work here, so it is better to show a contradition $\endgroup$
    – Qwerty
    Nov 30, 2016 at 19:10
  • $\begingroup$ @Ennar Sure the OP made a typo. I corrected it $\endgroup$
    – Qwerty
    Nov 30, 2016 at 19:13
  • $\begingroup$ @Qwerty, most likely, but I still wanted to check with them. $\endgroup$
    – Ennar
    Nov 30, 2016 at 19:15
  • $\begingroup$ You're right—the first inclusion is trivial (I assume you are taking for granted the fact that every subfield of $\Bbb C$ contains $\Bbb Q$). Your first approach to the opposite inclusion is good as well. Note that you made a computational error: $(\sqrt[5]{3}\sqrt{2}i)^5 = 12 \sqrt{2} i$, not $3 \sqrt[5]{2} i$. Does that help you? What happens if you take the $6$th power instead of the $5$th power? $\endgroup$
    – Greg Martin
    Nov 30, 2016 at 19:27

2 Answers 2

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To show that $\Bbb{Q}(\sqrt[5]{3},\sqrt{2}i)$ is a subset of $\Bbb{Q}(\sqrt[5]{3}\sqrt{2}i)$ it suffices to show that both $\sqrt[5]{3}$ and $\sqrt{2}i$ can be formed as a rational expression involving only rational numbers and $\sqrt[5]{3}\sqrt{2}i$.

Well, $$ \frac{(\sqrt[5]{3}\sqrt{2}i)^5}{12} = \sqrt{2}i \\ \frac{(\sqrt[5]{3}\sqrt{2}i)^6}{-72} = \sqrt[5]{3} $$

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  • $\begingroup$ It should be -24 instead of -72 I believe $\endgroup$
    – NoDisplayName
    Nov 22, 2020 at 18:55
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Hint: $\Bbb Q(\sqrt[5]3\sqrt 2 i)\subseteq\Bbb Q(\sqrt[5]3,\sqrt 2 i)$ implies that $[\Bbb Q(\sqrt[5]3\sqrt 2 i):\Bbb Q]\mid 10$. Now, try to eliminate cases $[\Bbb Q(\sqrt[5]3\sqrt 2 i):\Bbb Q]\in\{1,2,5\}$.

Also, Eisenstein criterion can't work on $x^{10}+288$ since $288 = 2^5\cdot 3^2$.

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