# 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.

• $F=\Bbb{Q},u=e^{2i\pi/3}$ – reuns May 16 at 16:05
• @reuns, your $u$ is of degree $2$ over $\mathbf{Q}$, so the only possible $k$ is $k=1$. – Stephen May 16 at 16:17
• @Stephen There is some obvious $k$ contradicting OP's claim – reuns May 16 at 16:31
• The OP probably wants $k<m$. – Dionel Jaime May 16 at 16:37
• @DionelJaime .... $u=e^{2i\pi/3} \sqrt2$ – reuns May 16 at 17:02

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 $$F(u^k)$$.

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 $$F((\sqrt3+\sqrt5)^3 )$$.

• What exactly is this answering? You're giving an example of an $F$, a $u$, and $k$ such that $F(u) = F(u^k)$. But the OP is asking if this is always true for any $F$ and $u$ of degree $m$ where $(m,k) =1$. – Dionel Jaime May 16 at 18:09
• @Dionel Jaime, Ofcourse here $u = \sqrt3+\sqrt5$ , $m=4$ and $k =3$. – Infinity_hunter May 17 at 0:45
• The OP is not asking for a specific $u, m,$ and $k$. They even gave one. They asked whether it was true more generally. – Dionel Jaime May 17 at 1:09