In the solution to my homework that $F(a) = F(a^2)$, we are trying to use the assumption that $a \notin F(a^2)$ and arrive at a contradiction. We will show that assuming $a \notin F(a^2)$, then $[F(a):F]$ would be even. We know that $[F(a):F] = [F(a):F(a^2)][F(a^2):F]$ We know that $f(x) = x^2-a^2 \in F(a^2)[x]$ has $f(a) = 0$, i.e. a is a root of $f(x)$. Then the professor claims that $[F(a):F(a^2)]$ must divide $deg(f(x)) = 2$, i.e. $[F(a):F(a^2)] =1 \text{ or } 2$. This is the statement I'm not sure about, why must $[F(a):F(a^2)]$ divide $deg(f(x)) = 2$ and why can't we just directly say that $[F(a):F(a^2)] = 2$?
THanks for your help!
The fact you are trying to prove is not generally true. There must be some assumptions you haven't told us.
For instance: Let $F = \mathbb{Q}$ and $a = \sqrt{2}$. Then $$F(a^2) = \mathbb{Q}$$ but $$F(a) = \mathbb{Q}[\sqrt{2}].$$
To address your concern in the problem: The minimal polynomial of $\alpha$ over $F(\alpha^2)$ divides $f(x)$. That is, the degree of the minimal polynomial of $\alpha$ over $F(\alpha^2)$ is $1$ or $2$. But the degree of the minimal polynomial of $\alpha$ over $F(\alpha^2)$ is $[F(\alpha) : F(\alpha^2)]$.
There are instances where $F(\alpha) = F(\alpha^2)$. Check out my answer to another field theory question which gives examples of such situations.