Minimal polynomial of odd degree I'm stuck on trying to prove this:
Let $K\supset F$ and let $u$ be an algebraic element of $K$ with a minimal polynomial of odd degree. Prove that $F(u)=F(u^2)$.
I know that in general, $F(u^2)\subseteq F(u)$.
I'm trying to figure out which theorems I could use from here. I could, perhaps, somehow show that $u$ and $u^2$ have the same minimal polynomial, implying that $F(u)\cong F(u^2)$. Is it possible to do this and would it help me?
 A: So, since $u$ is an element with minimal polynomial of odd degree, then the degree of $F(u)$ over $F$ is odd. 
Now, $u$ satisfies the polynomial $x^{2} - u^{2} \in F(u^{2})[x]$. Thus, the minimal polynomial of $u$ over $F(u^{2})$ has degree $1$ or $2$. 
If the degree is 1, then this implies that $u \in F(u^{2}) \Rightarrow F(u) \subseteq F(u^{2}).$ As you mentioned, the reverse containment is always true. Hence, the degree being 1 would imply $F(u^{2}) = F(u)$. 
If the degree is 2, then we have the following tower of field extensions
$$F(u) \supset F(u^{2}) \supseteq F.$$
Now, the degree of $F(u)$ over $F(u^{2})$ is $2.$ Then, if $n$ denotes the degree of $F(u^{2})$ over $F$, we have degree of $F(u)$ over $F$ is $2n$, because whenever
$$L \supseteq K \supseteq F$$
we have
$$[L : K][K: F] = [L : F].$$
So we see that the degree of the minimal polynomial of $u$ over $F(u^{2})$ being $2$ implies that the degree of $F(u)$ over $F$ is even, a contradiction. Hence, the degree of the minimal polynomial of $u$ over $F(u^{2})$ is 1 and hence $F(u) = F(u^{2}).$
