What are the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$?

Question

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!

Answer 1

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

Answer 2

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 $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3}) =\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$.