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I have $\alpha= 2^{1/3}+\sqrt{2}$ and $E(\alpha)=\mathbb{Q}$, I want to prove that $E=\mathbb{Q}(2^{1/3}, \sqrt{2})$.

I tried to compute the irreducible polynomial of $\alpha$, but I didn't get anywhere. Any hint?

qwerty89
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    You mean $E=\Bbb Q[\alpha]$? – Hagen von Eitzen Apr 19 '21 at 19:09
  • Yes, sorry @HagenvonEitzen – qwerty89 Apr 19 '21 at 20:38
  • Thank you @DietrichBurde, but I am looking for a an "easier" answer, maybe without computing the minimal polynomial – qwerty89 Apr 19 '21 at 20:39
  • Notice that $\mathbb Q(2^{1/3},\sqrt{2})$ contains $\alpha$, so it's free $\mathbb Q(\alpha)\subseteq\mathbb Q(2^{1/3},\sqrt{2})$. On the other hand you need to prove that $2^{1/3},\sqrt{2}\in\mathbb Q(\alpha)$, which implies $$\mathbb Q(2^{1/3},\sqrt{2})\subseteq\mathbb Q(\alpha)$$To do so take $\alpha=2^{1/3}+\sqrt{2}\rightarrow (\alpha-\sqrt{2})^3=2$, develop the cubic power and separate $\sqrt{2}$, and you prove that $\sqrt2\in\mathbb Q(\alpha)$, then clearly $2^{1/3}=\alpha-\sqrt{2}\in\mathbb Q(\alpha)$, thus the conclusion – Alessandro Apr 19 '21 at 20:45
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    @qwerty89 If you look at the first answer in the linked question, it shows the equality without computing the minimal polynomial, exactly as suggested by Alessandro. – Viktor Vaughn Apr 19 '21 at 20:51

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