Consider the fields $F, E,$ and $K$, where $F \subseteq E \subseteq K$. If $E$ is algebraic over $F$, and $K$ is algebraic over $E$, show that $K$ must be algebraic over $F$.
I know this is a well-known, proved theorem, but I'm trying to understand it on my own.
If $E$ is algebraic over $F$, then that means there is some $\alpha_1$ in $E$ s.t. $f(\alpha_1)$ = $0$ for some $f(x) \in F[x]$. Similarly for $K$ and $E$, there is some $\alpha_2$ in $K$ s.t. $e(\alpha_2$) = $0$ for some $e(x) \in E[x]$.
What I'm stuck on is connecting $\alpha_1$ and $\alpha_2$ together. If $\alpha_2$ is algebraic over $E$, how does that translate to it also being algebraic over $F$? Thank you for your help.
We use the characterization that $\alpha \in k'/k$ is an algebraic element of $k'$ over $k$ iff $[k(\alpha):k]<\infty$. Then let $a\in K$. By definition there is $p(x)\in E[x]$ so that $p(a)=0$. Let
$$p(x) = \sum_{i=1}^n a_ix^i$$
and then note that as each $a_i\in E$, we have that
$$[F(a_0, a_1, a_2,\ldots, a_n): F]\le \prod_i [F(a_i):F]<\infty.$$
But then $[F(a_0, a_1,\ldots, a_n, a):F] < \infty$, so $a$ is algebraic over $F$.