TITLE: Find the order of $\sqrt[3]{2}(1+\zeta_3)$ and $\sqrt[3]{2}(1-\zeta_3)$ over $\mathbb{Q}$
QUESTION [2 upvotes]: Find the order of $\sqrt[3]{2}(1+\zeta_3)$ and $\sqrt[3]{2}(1-\zeta_3)$ over $\mathbb{Q}$
I am not sure how to tackle this problem.
MY notes:
$\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ has degree 6. The degree of both elements is $2$ over $\mathbb{Q}(\sqrt[3]{2})$. I have never done a problem like this so I am not sure how to proceed. The first one I can do as it is a root of $x^3+2$. How do i deal with the second one

REPLY [1 votes]: While not as quick as the suggestions in the comment, here is another way to deduce that $\mathbb{Q}(\sqrt[3]{2}(1 + \zeta_3)) = \mathbb{Q}(\sqrt[3]{2},\zeta_3)$: Note that
$$[\sqrt[3]{2}(1 + \zeta_3)]^3 =2(1 + \zeta_3)^3 = 2 + 6\zeta_3 + 6 \zeta_3^2 + 2\zeta_3^3 $$
Since $\zeta_3^3 = 1$, we have that $[\sqrt[3]{2}(1 + \zeta_3)]^3 = 4 + 6\zeta_3 + 6\zeta_3^2$. After subtracting $4$ and dividing by $6$, we see that $\zeta_3 + \zeta_3^2 \in \mathbb{Q}(\sqrt[3]{2}(1 + \zeta_3))$. Note that,
$$(\zeta_3 + \zeta_3^2)^2 = \zeta_3^2 + 2\zeta_3^3 + \zeta_3^4 = 2\zeta_3^2 + 2 $$
After dividing by $2$ and subtracting $1$, we see that $\zeta_3^2 \in \mathbb{Q}(\sqrt[3]{2}(1 + \zeta_3))$. After squaring we see that $\zeta_3 \in \mathbb{Q}(\sqrt[3]{2}(1 + \zeta_3))$. You can now deduce that $\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]{2}(1 + \zeta_3))$. Hence,
$$\mathbb{Q}(\sqrt[3]{2}(1 + \zeta_3)) = \mathbb{Q}(\sqrt[3]{2},\zeta_3)$$
A similar technique works with $\sqrt[3]{2}(1 - \zeta_3)$
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