$F(u) = F(u^2)$ if $u$ is algebraic of odd degree
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 $F(u) = F(u^2)$.