Let $\alpha$ and $\beta$ be two algebraic numbers over $\mathbb Q$. Say that a subfield $\mathbb K$ of $\mathbb C$ joins $\alpha$ to $\beta$ iff $\beta \in {\mathbb K}[\alpha]$ but $\beta \not\in {\mathbb K}$. Now, if $\mathbb K$ joins $\alpha$ to $\beta$ and we add a completely unrelated algebraic number to $\mathbb K$, we still have a join from $\alpha$ to $\beta$. So it is natural to consider the minimal joins from $\alpha$ to $\beta$, i.e. the joins that are minimal with respect to field inclusion. Let ${\cal M}(\alpha,\beta)$ denote the set of all minimal joins from $\alpha$ to $\beta$. My guesses are that :

1) Any field in ${\cal M}(\alpha,\beta)$ is always contained in the normal (Galois) closure
of ${\mathbb Q}(\alpha,\beta)$

2) ${\cal M}(\alpha,\beta)$ is always finite

3) The two facts above should be provable using Galois theory.

Note that 2) follows from 1).

Can anyone confirm this ?

A simple example : ${\cal M}(\sqrt{2},\sqrt{3})$ consists of ${\mathbb Q}(\sqrt{6})$. Indeed, suppose $\mathbb K$ joins $\sqrt{2}$ to $\sqrt{3}$ and $x$ and $y$ are numbers in $\mathbb K$ such that $x+y\sqrt{2}=\sqrt{3}$. If $x \neq 0$ then $\sqrt{3}=\frac{3+x^2-2y^2}{2x} \in \mathbb K$ which is absurd. So $x=0$ and $y=\frac{\sqrt{6}}{2}$.