Is it always true that $\deg[F(\alpha): F(\alpha^2)] = 2$, given $\alpha$ has degree 2 over $F$?

Question

where $\alpha$ is algebraic over $F$ and $F$ is characteristic 0.

It seems obvious but I can't seem to think of a good enough proof for it. I am given $\deg[F(\alpha):F] = 2$. And I want to show $\alpha^2 \in F$. So I assumed the negation and tried to prove by contradiction.

Assuming $\alpha^2 \notin F$, we know $\deg[F(\alpha^2):F] \geq 2$ and we know $$\deg[F(\alpha):F] = \deg[F(\alpha):F(\alpha^2)] \cdot \deg [F(\alpha^2):F]$$ then using this and the fact $\deg[F(\alpha^2):F] \geq 2$, we get $\deg[F(\alpha^2):F] = 2$.

This immediately tells us $F(\alpha) = F(\alpha^2)$. But I don't know how to proceed from here. If we can somehow show that $\deg[F(\alpha):F(\alpha^2)] = 2$, then we will arrive at the contradiction to show $\alpha^2 \in F$.

Answer 1

It is not true in general. Consider for example $\alpha =1+\sqrt 2, F=\mathbb Q$.

Minimal polynomial of $\alpha$ in this case is $p(x)= x^2-2x-1$.

So degree of $F(\alpha)/F=2$ but $\alpha^2= 3+2\sqrt 2\not\in F$

Answer 2

You always have $[F(\alpha):F(\alpha^2)]\leq 2$, since $X^2-\alpha^2 \in F(\alpha^2)$ annihilates $\alpha$, and you have $[F(\alpha):F(\alpha^2)]=1$ if and only if $\alpha^2$ has a root in $F(\alpha^2)$.

This can in fact happen, as the counterexample $[\mathbb{Q}(\zeta_6):\mathbb{Q}(\zeta_3)]=1$ provided by J.W.Tanner shows (which uses the fact that $\zeta_6 = \zeta_3+1$)

Answer 3

There is a general result to this. Suppose $[L : K] = p$ where $p$ is prime. Show that if $M$ is an intermediate field, then $[L:K] = p$ or $1$. From here it is easy to show that if $[K(\alpha):K] = p$ is odd, you will always get $K(\alpha) = K(\alpha^2)$ but if you have that if $p$ is even, you could have that $[K(\alpha^2):K] = 1$