How do I prove $F(a)=F(a^2)?$ 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
 A: I am writing the converse part $F(a)\subset F(a^2).$ Enough to show that $a \in F(a^2).$ Let $a\notin  F(a^2)$ then $ F(a^2)\subsetneq F(a). $   Since $a^2$ satisfies the polynomial $x^2-a^2\in F(a^2)$ so $[F(a):F(a^2)]=2$. Now 
$$  [F(a): F] = [F(a) : F(a^2)][F(a^2):F]=2[F(a^2):F].$$ which contradicts the fact that [F(a) : F] is odd. Thus $a \in F(a) $ and therefore $F(a) = F(a^2)$.
A: You know that  $[F(a):F]=[F(a):F(a^2)][F(a^2):F]$.  The minimal polynomial of $a$ over $F(a^2)$, assuming that $a\notin F(a^2)$, is $x^2-a^2$, so we have $[F(a):F(a^2)]=2$.  Now, what is the problem with that?
A: let $n$ be the degree of the minimal polynomial $p(x)$ of a over $F$ and $L$ be the degree of the minimal polynomial $q(x)$ of $a^2$ over $F$ .
$[F(a):F]=n$, $n$ is odd then it is clear $F⊂ F(a^2) ⊂ F(a)$ (because $a^2∈F(a)$).
So (by law, or Theorem): 
$$ [F(a):F]=[F(a): F(a^2)] [F(a^2) : F] $$
If prove $[F(a): F(a^2)]=1$ then $F(a)= F(a^2)$. 
Conversely, suppose $[F(a): F(a^2)]≠1$.  Since $[F(a): F(a^2)]$ can’t be even (because $n$ is odd) then $[F(a): F(a^2)]≥3$.
$[F(a^2):F]=L$, then $$ n≥3L \tag{$*$} $$ in other hand $[F(a^2):F]=L$ implies the polynomial of degree $L$  that $a^2$ is its root.
in other way: it exist the polynomial $q(x)=b_0+b_1x+\cdots +b_L x^L$ such that $b_0+b_1a^2+\cdots +b_L a^{2L}=0$ .
it’s mean $a$ is root of the polynomial $g(x)$ of degree $2L$  and since the minimaml polynomial degree of
the $a$ is $n$ we must have $$ n≤2L \tag{${*}{*}$}. $$ (because we have $p(x)|g(x)$ by theorem. so $p(x)=g(x)r(x)$ its mean : degree $p(x)$=degree $g(x)$ +degree $r(x)$ .so $n≤2L$ ) 
So we must have (by attention to $(*)$ and $(**)$), $3L≤n≤2L$ that is a contradiction.
In general case we have : if $[F(a):F]=n$ and $m>0$ , $(n,m!)=1$ then $F(a)=F(a^m)$
Hint: if $m<n$ then $ (m!,n)\neq1$ so $m\ge n$. if suppose $[F(a^m):F]=k$ then $n=kl$.
If $k=1$ then $F(a)=F(a^m)$.                                                         since $a^m$ is root of the polynomial of degree $ml$ ,we have $n\le ml$
or $kl\le ml$ so $k\le m$ . so we have $k|m!$ and since $k|n$ , $(n,m!)=1$ we must have :$k=1$.
A: Since $a^2 \in F(a)$ then $F(a^2) \subseteq F(a).$ Since $a$ is algebraic over $F$ of odd degree, then $[F(a):F]$ is odd (finite)   ( and so $F(a)$ is algebraic over $F$). So,
$$[F(a):F] = [F(a):F(a^2)] x [F(a^2):F].$$
But since $f(x) = x^2 - a^2 \in F(a^2)[x]$ has $f(a) = 0,$ i.e., $a$ is a root,
then $[F(a):F(a^2)]$ must divide deg$(f(x)) = 2,$ i.e., $[F(a):F(a^2)] = 1$ or $2.$
But since $[F(a):F]$ is odd, $[F(a):F(a^2)]$ cannot be $2.$ Therefore, $[F(a):F(a^2)] = 1$ implying that $F(a) = F(a^2).$ qed
A: To show that $a\in F(a^2)$: The case deg $(p)=1$ is trivial as then $a$ and $a^2$ belong to $F.$ For the case deg $(p) >1$, we have $m\geq 1$, where the minimal polynomial of $a$ over $F$ is$$p(x)=\sum_{j=0}^{2 m +1}x^jf_j .$$  $$ \text {Let } \quad r(x)=\sum_{j=0}^mx^j f_{2 j+1}.$$  $$ \text { Let }\quad s(x)=\sum_{j=0}^mx^jf_{2 j}.$$ We have $0<$deg $r(x^2)=2 m<$ deg $(p)$. So $ r( a^2)\ne 0$ by the minimality of deg$(p)$. Therefore $$a=-s(a^2)/r(a^2)\in F(a^2).$$
