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I would like to calculate the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$. I know that the elements of $\mathbb{Q}(\sqrt[3]{2})$ have the form of ${a+b\sqrt[3]{2}+c\sqrt[3]{4}}$, where a,b,c $\in \mathbb{Q}$ and the elements of $\mathbb{Q}(\sqrt{3})$ have the form of $a+b\sqrt{3}$, where a,b $\in \mathbb{Q}$. I also calculated the minimal polynomial of $(\sqrt[3]{2}+\sqrt{3})$ over $\mathbb{Q}$ , which is: $x^6−9x^4−4x^3+27x^2−36x−23$.

Can you help me to calculate the form of the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$? I have to find the elements of the linear combinations which form the higher powers of $(\sqrt[3]{2}+\sqrt{3})$ which are also powers of $(\sqrt[3]{2}+\sqrt{3})$? Could you give me a proper method to find the solution? Also, could you write down your calculation in your answer? Thank you for helping me!

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Let $\beta=\sqrt[3]{2}+\sqrt{3}$.

Your problem becomes easy if $\mathbb{Q}(\beta)=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$. $\mathbb{Q}(\beta)\subset\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$ ($\beta\in\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$). Let's see that $\sqrt{3}\in\mathbb{Q}(\beta)$:

$$\beta-\sqrt{3}=\sqrt[3]{2}$$ $$(\beta-\sqrt{3})^3=2$$ $$\beta^3-3\sqrt{3}\beta^2-9\beta+3\sqrt{3}=2$$ $$\sqrt{3}=(\beta^3-9\beta^2+1)/3(\beta^2-1)\in\mathbb{Q}(\beta)$$

Thus, $\sqrt[3]{2}\in\mathbb{Q}(\beta)=\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$ too, so $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3}) =\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$.

Lancet S.
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Given that $\theta=\sqrt[3]{2}+\sqrt{3}$ has degree $6$ over $\mathbb Q$, $\mathbb Q(\theta)$ is the set of polynomial expressions in $\theta$ of degree at most $5$.

Another take: $Q(\theta)=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$.

lhf
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  • See also https://math.stackexchange.com/a/3922059/589 – lhf Feb 25 '21 at 13:21
  • Thank you for your answer! Can you give me a similar form as in the case for example $\mathbb{Q}(\sqrt(2))$? How can you prove that $\mathbb{Q}(\sqrt[3]{2},\sqrt{3})=\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$? –  Feb 25 '21 at 15:12
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    @mathfairy, adapt the answer linked above. – lhf Feb 25 '21 at 15:48
  • @mathfairy: You may also use primitive element theorem to prove those two fields equal. – Paramanand Singh Feb 27 '21 at 09:39