I came across the following question -
Let $u$ be algebraic over $\mathbb Q$ of degree which is not divisible by $3$. Does necessarily $\mathbb Q(u)=\mathbb Q(u^3)$?
What I have so far:
Since $x^3-u^3$ is reducible over $\mathbb Q$ with $u$ as a root, then the dimension of $\mathbb Q(u)$ over $\mathbb Q(u^3)$ is at most $3$. From the dimension theorem:
$[\mathbb Q(u):\mathbb Q]=[\mathbb Q(u):\mathbb Q(u^3)]\cdot [\mathbb Q(u^3):\mathbb Q]$
The LHS is not divisible by $3$, and so since $[\mathbb Q(u):\mathbb Q(u^3)]$ is either $1$ or $2$, and cannot be $3$. If it is $1$, then we know that the two fields are equal. But I'm not sure why it can't be $2$. and can't come up with a counter example which fits the problem.
Can anyone help me with proving that the two fields are equal, or coming up with a counter example?
Thanks in advance.
Let $u$ be a root of $x^2+x+1$. This polynomial is irreducible over $\Bbb{Q}$. Also note that
$$(x-1)\cdot(x^2+x+1)=x^3-1.$$
Since $u$ is a root of $x^2+x+1$, it is also a root of $x^3-1$. So we have that $u^3=1$ and $u\notin\Bbb{Q}$.
It follows that $[\Bbb{Q}(u):\Bbb{Q}(u^3)]=2$, since $\Bbb{Q}(u^3)=\Bbb{Q}$, and $\left[\Bbb{Q}(u):\Bbb{Q}\right]=2$,