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.

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    $\begingroup$ $F=\Bbb{Q},u=e^{2i\pi/3}$ $\endgroup$ – reuns May 16 at 16:05
  • $\begingroup$ @reuns, your $u$ is of degree $2$ over $\mathbf{Q}$, so the only possible $k$ is $k=1$. $\endgroup$ – Stephen May 16 at 16:17
  • $\begingroup$ @Stephen There is some obvious $k$ contradicting OP's claim $\endgroup$ – reuns May 16 at 16:31
  • $\begingroup$ The OP probably wants $k<m$. $\endgroup$ – Dionel Jaime May 16 at 16:37
  • $\begingroup$ @DionelJaime .... $u=e^{2i\pi/3} \sqrt2$ $\endgroup$ – 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 $F((\sqrt3+\sqrt5)^3 )$.

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    $\begingroup$ 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 $. $\endgroup$ – Dionel Jaime May 16 at 18:09
  • $\begingroup$ @Dionel Jaime, Ofcourse here $u = \sqrt3+\sqrt5$ , $m=4$ and $k =3$. $\endgroup$ – Infinity_hunter May 17 at 0:45
  • $\begingroup$ The OP is not asking for a specific $u, m,$ and $k$. They even gave one. They asked whether it was true more generally. $\endgroup$ – Dionel Jaime May 17 at 1:09

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