Q:

Are these fields equal?

Let  $\zeta_3$ be the third root of unity.
1) Does it hold that: $\mathbb{Q}(\sqrt{2},\zeta_3)=\mathbb{Q}(\sqrt{2}+\zeta_3)$ ?
2) Does it hold that $\mathbb{Q}(\sqrt[3]{2},\zeta_3)=\mathbb{Q}(\sqrt[3]{2}\zeta_3)$?

My attempt for 1) is to compute the minimal polynomial of $\sqrt{2}+\zeta_3$ as $p(x)=x^4+2x^3-x^2-2x+7$
which is of degree 4. But since $|\mathbb{Q}(\sqrt{2},\zeta_3):\mathbb{Q}|=6$ these can't be equal. Is my proof correct?
For the second one I am not sure.

A:

$\zeta_3$ is a root of $x^2 + x + 1$ since $x^3 - 1 = (x - 1)(x^2 + x + 1)$. So its degree is $2$ not $3$.
For 2) you can do the same kind of thing. The minimal polynomial of $\sqrt[3]2 \zeta_3$ is $x^3 - 2$ so $\mathbb{Q}(\sqrt[3]2 \zeta_3) \ne \mathbb{Q}(\sqrt[3]2, \zeta_3)$. It's a 3-dimensional subspace of a 6-dimensional $\mathbb{Q}$-vector space.

