Why $F(\alpha_1)∩F(\alpha_2)=F$ is false $\def\Q{\mathbb Q}$Let $\alpha_1$ and $\alpha_2$ be conjugates over a field $F$ such that $\alpha_2 \not\in F(\alpha_1)$. Is true that  $F(\alpha_2)\cap F(\alpha_2)=F$.
Here is my attempt:
Is false. Let $F= \Q(\sqrt 2)$. We know that $\sqrt[3]2$ , $\omega\sqrt[3]2$, $\omega^2\sqrt[3]2$ are roots of the polynomial $x^3-2$. But my doubt is how to prove that $\Q(\sqrt2,\sqrt[3]2)∩\Q(\sqrt2,\omega \sqrt[3]2)≠\Q(\sqrt2)$?
 A: I'm not sure your example works, but here's one that does:
Consider any field $F$ and any polynomial $f(x)\in F[x]$ which has Galois group with at least one pair of elements $g,h$ which do not generate the group or each other. Then the fixed fields of $g$ and $h$ intersect in a proper extension of $F$. By the Primitive Element Theorem, these fixed fields are simple, i.e. we have $\alpha_1,\alpha_2$ such that they are $F(\alpha_1)$ and $F(\alpha_2)$ respectively.
For a concrete example, let $F=\mathbb Q$ and $f(x)=(x^2-2)(x^2-3)(x^2-5)$. This has Galois group $C_2\times C_2\times C_2$, so any pair of distinct nonzero elements satisfies the desired condition. Choosing the automorphisms
$$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5$$
$$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a+b\sqrt2-c\sqrt3+d\sqrt5$$
gives fixed fields $\mathbb Q(\sqrt3,\sqrt5)$ and $\mathbb Q(\sqrt2,\sqrt5)$ which intersect in $\mathbb Q(\sqrt5)$. If you'd like, you can show that $\mathbb Q(\sqrt3,\sqrt5)=\mathbb Q(\sqrt3+\sqrt5)$ and $\mathbb Q(\sqrt2,\sqrt5)=\mathbb Q(\sqrt2+\sqrt5)$.
