Galois theory and degree of extension Let $E = F (u),$ where $u$ is algebraic over $F$, of degree $m.$ If $k$ is a natural with $(k, m) = 1,$ it is true that $E = F(u^{k})$? I tried but I couldn't.
I did another exercise that is reasonably similar. Let $E = F (u),$ where $u$ is algebraic over $F$, of odd degree. So $E = F(u^{2}).$  I tried to imitate the demonstration of this exercise, but I couldn't.
 A: Sometime it fails,

*

*with $F=\Bbb{Q}, u=e^{2i\pi/3}, m=2,k=3$


*with $F=\Bbb{Q}, u=e^{2i\pi/3}\sqrt2, m=4,k=3$
When $m$ is coprime with every integer $\le k$ then it works. Same when $x^k-1$ splits completely in $F(u^k)$.
A: Let $F = \mathbb{Q}$ and consider the extension $K = F(\sqrt3, \sqrt5)$. It is fairly easy to show that degree of $K$ over $F$ is $4$. Also it is easy to see that $K = F(\sqrt3 + \sqrt5)$. Now observe that $(\sqrt3+\sqrt5)^3 = 18\sqrt3 + 14\sqrt5$ and  $(\sqrt3-\sqrt5)^3 = 18\sqrt3 - 14 \sqrt5$. Now we claim that $K = F((\sqrt3+\sqrt5)^3 )$.
Observe that $F$ is subset of $F((\sqrt3+\sqrt5)^3 )$ by closure. So it is enough to show that $\sqrt3$ and $\sqrt5$ are in $F((\sqrt3+\sqrt5)^3 )$. First observe that $(\sqrt3-\sqrt5)^3$ is in $F((\sqrt3+\sqrt5)^3 )$ since $\sqrt3-\sqrt5 = \frac{-2}{\sqrt3+\sqrt5}$.
Now again by using closure we can show that $\sqrt3$ and $\sqrt5$ are in $F((\sqrt3+\sqrt5)^3 )$. With this now it is straightforward to show that $K$ is indeed $F((\sqrt3+\sqrt5)^3 )$.
