Set of elements of degree $2^n$ over a base field is itself a field

Let $F \subset L$ be two fields, and define $K = \{\alpha \in L\mid [F(\alpha): F] \text{ is a power of 2} \}$. Our problem is to prove that $K$ is a field.

Closure under reciprocation is easy (because $F(\alpha) = F(\alpha^{-1})$). We ran into trouble proving closure under addition and multiplication. Our initial attempt was to prove that for any $\alpha, \beta \in L$, $[F(\alpha, \beta): F]$ divides $[F(\alpha): F] [F(\beta): F]$, but this relies on a result that $[F(\alpha, \beta): F(\alpha)] = \deg_{F(\alpha)} (\beta)$ divides $[F(\beta), F] = \deg_F (\beta)$, which (as far as we can tell) is not necessarily true.

EDIT: As pointed out below, the problem is unsolvable; the professor who assigned it agrees.

Solution 1:

Let $\alpha$ and $\beta$ be two of the roots of an $S_4$ quartic; then they both have degree 4, but their sum has degree 6.

A thorough discussion of degrees of sums, and products, of algebraic numbers, can be found here.

EDIT: With a little help from Wolfram Alpha, and possibly with many mistakes of my own, I present this example. The $S_4$-quartic $x^4-x-1$ has two real roots, which add up to $$y=\sqrt{\left({9+\sqrt{849}\over18}\right)^{1/3}-4\left({2\over3(9+\sqrt{849})}\right)^{1/3}}$$ and the minimal polynomial for $y$ is $y^6+4y^2-1$.

The two real roots of the quartic are of the form $(1/2)y\pm z$, where $z$ is a radical expression about twice as complicated as the expression for $y$. Go to Wolfram, and see for yourself.

EDIT: The dead link pointed to this paper: Paulius Drungilas, Arturas Dubickas, Chris Smyth, A degree problem for two algebraic numbers and their sum, Publ. Mat. 56 (2012), 413--448. This link should work (for now).