If K is an extension field of F and if $a \in K$ such that $[F(a):F]$ is odd, show that $F(a)=F(a^2)$.

Question

My approach:

As $[F(a):F]$ is finite and hence an algebraic extension. Also, $F(a)$ can be regarded as a F vectorspace, with basis $1,a,a^2, \dots,a^{n-1}$ where $n$ is odd. I don't then how to proceed..

Any help is appreciated!

Answer 1

Hint: $[F(a):F]=[F(a):F(a^2)]\cdot[F(a^2):F]$.

Answer 2

WE know $F(a^2) \subset F(a)$ and using the tower formula we get: $$ [F(a):F] = [F(a): F(a^2)][F(a^2):F]$$

Clearly $[F(a): F(a^2)] \le 2$ but if it can't be 2 as the left hand side is odd. $[F(a): F(a^2)]=1$ hence we are done.