Minimal polynom for $e^{2\pi i/3} + 2^{1/3}$

Question

I have $\alpha = \exp(\frac{2\pi i}{3}) + 2^{\frac13}$ and I need to find a minimal polynomial for It in $\mathbb Q$, the "obvious" Path is to get the third Power of each term, but I'm struggling with It It would also be sufficient for me to show that $\mathbb Q(\alpha) = \mathbb Q(\exp(\frac{2\pi i}3))(2^{\frac13})$ But i dont even know If this is true

Answer 1

It is clear that $$ \mathbb{Q}(e^{2\pi i/3}+\sqrt[3]{2})\subseteq \mathbb{Q}(\sqrt[3]{2})(e^{2\pi i/3}) $$ Note that $$ [\mathbb{Q}(\sqrt[3]{2})(e^{2\pi i/3}):\mathbb{Q}(\sqrt[3]{2})]=2 $$ because the degree of $e^{2\pi i/3}$ over $\mathbb{Q}$ is $2$ and $e^{2\pi i/3}\notin\mathbb{Q}(\sqrt[3]{2})$. Therefore the degree of $e^{2\pi i/3}+\sqrt[3]{2}$ is a divisor of $6$. Try and prove the degree is indeed $6$.

Let $\beta=e^{2\pi i/3}$, for simplicity. We have $$ \sqrt[3]{2}=\alpha-\beta $$ so $2=\alpha^3-3\alpha^2\beta+3\alpha\beta^2-\beta^3$. Since $\beta^3=1$, we have $$ \alpha^3-3=3\alpha^2\beta-3\alpha\beta^2 $$ However, $\beta^2+\beta+1=0$, so we can write the identity as $$ \alpha^3-3\alpha-3=\beta(3\alpha^2+3\alpha) $$ Hence $$ \beta=\frac{\alpha^3-3\alpha-3}{3\alpha^2+3\alpha} $$ which yields, since $\beta^2+\beta+1=0$, $$ \frac{(\alpha^3-3\alpha-3)^2}{(3\alpha^2+3\alpha)^2} +\frac{\alpha^3-3\alpha-3}{3\alpha^2+3\alpha} +1=0 $$ This can be written as a degree $6$ polynomial expression in $\alpha$.

Answer 2

Hint

if $x=e^{\frac{2\pi i}{3}}+2^{\frac{1}{3}}$

then cubing both sides gives

$$x^3=e^{2\pi i}+2+3\cdot e^{\frac{2\pi i}{3}} 2^{\frac{1}{3}}\left(e^{\frac{2\pi i}{3}}+2^{\frac{1}{3}}\right)$$

$$x^3=e^{2\pi i}+2+3\cdot e^{\frac{2\pi i}{3}} 2^{\frac{1}{3}}\left( x \right)$$

$$x^3-3=3\cdot e^{\frac{2\pi i}{3}} 2^{\frac{1}{3}} x$$

Cubing both sides

$$\left(x^3-3\right)^3=\left(3\cdot e^{\frac{2\pi i}{3}} 2^{\frac{1}{3}}x\right)^3 $$

$$\left(x^3-3\right)^3=-54x^3 \cdot$$

This is a polynomial which has $e^{\frac{2\pi i}{3}}+2^{\frac{1}{3}}$ as one of its zero. You might want to check if there is some other polynomial of smaller degree(obviously a factor of the above polynomial) which satisfies $e^{\frac{2\pi i}{3}}+2^{\frac{1}{3}}$ If not, Then this is a minimal polynomial