Consider $x^3 - 2$ with roots $2^{1/3}$, $2^{1/3}w$, $2^{1/3}w^{2}$ over $\mathbf{Q}$. where $\mathbf{Q}$ is the set of rationals and $w$ is the cube root of unity. Then which of the following are true
(a) $\mathbf{Q}[2^{1/3}w]$ is a normal extension of $\mathbf{Q}$ over $x^3 - 2$.
(b) $\mathbf{Q}[2^{1/3},w]$ is a normal extension of $\mathbf{Q}$ over $x^3 - 2$.
Definitely (b) is true, since it is the splitting field of Q of this polynomial but what about (a)? Since $\mathbf{Q}[2^{1/3}w]$ is a field closure holds so if $2^{1/3}w$ is in a field then $2^{1/3}w, 2^{1/3}w^2, 2^{1/3}$ will be in the field. Am I right in my argument? If not please say where I am wrong.
$(a)$ is false. This is because $x^3-2$ is an irreducible polynomial over $\mathbf{Q}$ that has one root in $\mathbf{Q}(2^{1/3}w)$, but doesn't split into linear factors.
For instance start with the root $2^{1/3}w^2$. Since $2^{1/3}w$ is in $\mathbf{Q}(2^{1/3}w)$, then if we suppose that $2^{1/3}w^2$ lies in this field, then so should $w=\frac{2^{1/3}w^2}{2^{1/3}w}.$ This implies that $\mathbf{Q}(w)$ is a subfield of $\mathbf{Q}(2^{1/3}w)$.
However, [$\mathbf{Q}(w):\mathbf{Q}]=2$, while [$\mathbf{Q}(2^{1/3}w):\mathbf{Q}]=3$, which isn't possible by the tower law.