TITLE: $F(u) = F(u^2)$ if $u$ is algebraic of odd degree
QUESTION [11 upvotes]: This has been asked before, but I have a different solution and I would like to check it.

Let $F$ be a field and $u$ be algebraic over $F$ of odd degree. Prove that $F(u) = F(u^2)$.

Of course $F(u^2) \subset F(u)$. If we can show that $u \in F(u^2)$, we are through. 
Let $$f(x) = x^{2n+1} + a_{2n}x^{2n} + \cdots + a_1 x + a_0$$ be the minimum polynomial of $u$ over $F$. Then:
$$u g(u) + h(u) = 0$$
where
$$g(x) = x^{2n} + a_{2n-1}bx^{2n-2} + a_{2n-3} x^{n-2} + \cdots + a_3 x^2 + a_1$$
and 
$$h(x) = a_{2n}x^{2n} + a_{2n-2} x^{2n-2} + \cdots + a_2 x^2 + a_0$$
We can't have $g(u) = 0$ because the degree of $u$ is $2n+1$. So $g(u)$ is a unit in $F(u^2)$. Hence $u = - h(u)g(u)^{-1} \in F(u^2)$.
Does this look good?

REPLY [11 votes]: Not sure why you want a confirmation.

Your proof 

Let $f(x) = g(x^2)+xh(x^2)$ the minimal polynomial of $u$ over $F$, with $deg(f) = 2n+1,deg(h)=n, deg(g) \le n$. 
$h(u^2) \ne 0$ since otherwise $u$ would be a root of $h(x^2)$ whose degree is $< 2n+1$. 
Thus $f(u) = 0 \implies u = \frac{-g(u^2)}{h(u^2)} \in F(u^2)$.

The traditional one

$[F(u):F] =[F(u):F(u^2)][F(u^2):F]$. Since $[F(u):F(u^2)] \le 2$ and $2 \nmid [F(u):F]$ we obtain $[F(u):F(u^2)]=1$ and hence $F(u) = F(u^2)$.