If $K$ is field extension of $F$ and if $a \in K$ such that $[F(a):F]$ is odd, show that $F(a) = F(a^{2})$. Given an example to show that this can be false if the degree of $F(a)$ over $F$ is even.
I know that $F(a^{2}) \subseteq F(a)$ and $[F(a):F] = [F(a):F(a^{2})][F(a^{2}):F]$. I know too that if $[K:F] = p$ with $p$ prime, so there are no intermediate fields between $K$ and $F$. Any hints?
On
In $F(a^2)[X]$ consider the polynomial of degree $2$, $P(X)=X^2-a^2$.
$a\in F(a)\supset F(a^2)$ is such that $P(a)=0$. So the degree of the extension is at most $2$. In other words $[F(a):F(a^2)]\in \{1,2\}$.
Because $[F(a):F]=[F(a):F(a^2][F(a^2):F]$, if $[F(a):F]$ is odd then $[F(a):F(a^2)]=1$ and this means $F(a)=F(a^2)$.
The counterexample is obvious. Consider $F=\Bbb{Q}$ and $a=\sqrt{2}$
Hint : First prove that $[F(a):F(a^2)]\leq 2$, and then prove that it can't be equal to $2$.