$\alpha, \beta\in\mathbb C$ algebraic over $F \subseteq \mathbb{C}$. Prove: $\exists n\in \mathbb{N}$ such that $F(\alpha, \beta) = F(\alpha+n\beta)$

Question

I am trying to prove the following claim:

$F \subseteq \mathbb{C}$. $\alpha, \beta \in \mathbb{C}$ algebraic over $F$. Prove: $\exists n\in \mathbb{N}$ such that $F(\alpha, \beta) = F(\alpha+n\beta)$.

The inclusion $F(\alpha + n\beta) \subseteq F(\alpha, \beta)$ is immediate since $F(\alpha, \beta)$ is closed to addition and multiplication of elements in $F\cup \{\alpha, \beta\}$ and $F$ must include $\mathbb{Z}$ since $1_\mathbb{C} \in F$ and the addition operation in $F$ is the same as that in $\mathbb{C}$.

The other direction I am unable to crack.

I've tried the following:

1. Playing around with the minial polynimials of $\alpha$ and $\beta$: $m_\alpha(x)$ and $m_\beta(x)$ (which must exists since they are algebraic) to obtain a representation for $\alpha +n\beta$ and then subtracting some power of $\alpha + n\beta$ to get $\alpha$ and $\beta$ (then claiming that all of these operations were in $F(\alpha +n\beta)$, which would then prove that both are in $F(\alpha +n\beta)$ and using the fact that $F(\alpha, \beta)$ is the smallest one containing both and $F$). This got very messy.

2. Using the fact that $G((F(\alpha, \beta), F)$ must be finite and therefore the number of $F(\alpha, \beta)$'s proper subfields must be finite, and then trying to show that if, for contradiction, for all $n$ $F(\alpha+n\beta)$ is a proper subfield of $F(\alpha, \beta)$ then either from some $n_0$ these are all equal or there must be an $n$ that satisfies equality. Haven't been able to convince myself that infinitely many of these must be different though.

I do realize that I am missing the use of $\mathbb{C}$ somehow, since in the only place that I used it it could have been easily replaced by $\mathbb{Q}$. I just don't see yet how this might be helpful.

Any advice?

Answer 1

Because there are only finitely many intermediate extensions between $F$ and $F(\alpha,\beta)$ (you don't need Galois theory for this result, and you have to transfer to the Galois closure of $F(\alpha,\beta)$ to apply Galois here), there exists distinct $n, m\in\mathbb N$ such that $F(\alpha+n\beta)=F(\alpha+m\beta)=:K$, that is $K$ contains both of $\alpha+n\beta$ and $\alpha+m\beta$, hence also contains their difference $(n-m)\beta$ and then $\beta$ and finally $\alpha=(\alpha+m\beta)-m\beta$.

All we need is $F$ has characteristic $0$ (and $\alpha,\beta$ are algebraic), therefore $F(\alpha,\beta)$ is separable over $F$, and $n-m\not=0$ in $F$.