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

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.

• This is the standard primitive element theorem where we take $\alpha =\alpha_1,\dots,\alpha_r$ as roots of minimal polynomials for $\alpha$ and $\beta=\beta_1,\dots,\beta_s$ as roots of minimal polynomial for $\beta$ and choose $n$ such that $\alpha+n\beta$ doesn't equal any of $\alpha_i+n\beta_j$ except when $i=j=1$. This is possible because we have infinitely many integers $n$ and only finitely many combinations of $\alpha_i, \beta_j$. Commented Sep 2, 2023 at 9:45
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 $$F$$.