nt.number theory - Subfields joining an algebraic element to another

EDIT: Okay, wrong. Sorry for spamming.

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 $betain Kleft[alpharight]$, then $betainleft(Kcapmathbb{Q}left(alpha,betaright)right)left[alpharight]$.

Adding to this the trivial observation that

(b) whenever a subfield $K$ of $mathbb{C}$ satisfies $betanotin K$, then $betanotin Kcapmathbb{Q}left(alpha,betaright)$,

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

So let's prove (a) now: Since $betain Kleft[alpharight]$, there exists a polynomial $Pin Kleft[Xright]$ such that $beta=Pleft(alpharight)$. Since $Kcapmathbb{Q}left(alpha,betaright)$ is a vector subspace of $K$ (where "vector space" means $mathbb{Q}$-vector space), there exists a linear map $phi:Kto Kcapmathbb{Q}left(alpha,betaright)$ such that $phileft(vright)=v$ for every $vin Kcapmathbb{Q}left(alpha,betaright)$ (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 $Pin Kleft[Xright]$, we get another polynomial $Qin left(Kcapmathbb{Q}left(alpha,betaright)right)left[Xright]$ which also satisfies $beta=Qleft(alpharight)$ (since all powers of $alpha$ and $beta$ lie in $Kcapmathbb{Q}left(alpha,betaright)$ and thus are invariant under $phi$). Thus, $betainleft(Kcapmathbb{Q}left(alpha,betaright)right)left[alpharight]$, qed.

Can anyone check this for nonsense?

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