Let $E$ be an extension field of $F$. If $a \in E$ has a minimal polynomial of odd degree over $F$, show that $F(a)=F(a^2)$.
let $n$ be the degree of the minimal polynomial $p(x)$ of $a$ over $F$ and $k$ be the degree of the minimal polynomial $q(x)$ of $a^2$ over $F$.
Since $a^2 \in F(a)$, We have $F(a^2) \subset F(a)$, then $k\le n$
In order to prove the converse:
$q(a^2)=b_0+b_1a^2+b_2(a^2)^2\ldots+b_k(a^2)^k=0$
implies
$q(a)=b_0+b_1a^2+b_2a^4\ldots+b_ka^{2k}=0$ Then
$p(x)|q(x)$, because $p(x)$ is the minimal polynomial of $a$ over $F$.
If I prove that $n|2k$ we done, since $k$ is odd, we have $n|k$ and $n\le k$ and finally $n=k$.
So I almost finished the question I only need to know how to prove that $n|2k$ It should be only a detail, but I can't see, someone can help me please?
Thanks