This looks just too easy, so I guess I'm doing something stupid.

I'll prove that

**(a)** whenever a subfield $K$ of $\mathbb{C}$ satisfies $\beta\in K\left[\alpha\right]$, then $\beta\in\left(K\cap\mathbb{Q}\left(\alpha,\beta\right)\right)\left[\alpha\right]$.

Adding to this the trivial observation that

**(b)** whenever a subfield $K$ of $\mathbb{C}$ satisfies $\beta\not\in K$, then $\beta\not\in K\cap\mathbb{Q}\left(\alpha,\beta\right)$,

we see that whenever a subfield $K$ of $\mathbb{C}$ joins $\alpha$ to $\beta$, its subfield $K\cap\mathbb{Q}\left(\alpha,\beta\right)$ does the same. This obviously settles 1) (even without the normal closure) and therefore 2).

So let's prove **(a)** now: Since $\beta\in K\left[\alpha\right]$, there exists a polynomial $P\in K\left[X\right]$ such that $\beta=P\left(\alpha\right)$. Since $K\cap\mathbb{Q}\left(\alpha,\beta\right)$ is a vector subspace of $K$ (where "vector space" means $\mathbb{Q}$-vector space), there exists a linear map $\phi:K\to K\cap\mathbb{Q}\left(\alpha,\beta\right)$ such that $\phi\left(v\right)=v$ for every $v\in K\cap\mathbb{Q}\left(\alpha,\beta\right)$ (this may require the axiom of choice for infinite $K$, but if you want to consider infinite field extensions of $\mathbb{Q}$ I think you can't help but use the axiom of choice). Applying the linear map $\phi$ to every coefficient of the polynomial $P\in K\left[X\right]$, we get another polynomial $Q\in \left(K\cap\mathbb{Q}\left(\alpha,\beta\right)\right)\left[X\right]$ which also satisfies $\beta=Q\left(\alpha\right)$ (since all powers of $\alpha$ and $\beta$ lie in $K\cap\mathbb{Q}\left(\alpha,\beta\right)$ and thus are invariant under $\phi$). Thus, $\beta\in\left(K\cap\mathbb{Q}\left(\alpha,\beta\right)\right)\left[\alpha\right]$, qed.

Can anyone check this for nonsense?