Problem about intermediate fields in the extension Let $E,K$ be intermediate fields in the extension $L/F$
(a) If $[EK:F]$ is finite, then
$$[EK:F] \leq [E:F][K:F] $$
(b) If  $E$ and $K$ are algebraic over $F$, then so is $EK$
For (a), I try two ways:
(1) I try to prove that $[EK:K]\leq [E:F]$ and $[EK:E]\leq [K:F]$.
(2) $x_1,\dotsc,x_m$ are base of $E/F$, $y_1,\dotsc,y_n$ are base of $K/F$, but how to prove that $\{x_iy_j\}$ span  $EK$?
 A: Hints:
A. Pick bases $S$ (resp. $T$) for $E/F$ (resp. $K/F$(. Show that the products $st,s\in S, t\in T$ span all of $EK$.
B. Review what you know about products and sums of two algebraic elements.

Extending... In both parts, A and B, you know that $E$ and $K$ are algebraic. This implies that the compositum is
$$
EK=\{\sum_{i=1}^n e_ik_i\mid n\in\mathbb{N};  e_i\in E, k_i\in K\ \text{for all $i$}\},
$$
i.e. the elements of the compositum are finite sums of products of elements of $E$ and $K$.
For part A you can pick bases $S=\{s_1,s_2,\ldots,s_m\}$ and $T=\{t_1,t_2,\ldots,t_n\}$ as in the hint. Take a sum $\sum_i e_ik_i$ as above. For all $i$, you can write $e_i=\sum_j a_{ij}s_j$ and $k_i=\sum_\ell b_{i\ell}t_\ell$ with all the coefficients $a_{ij},b_{i\ell}\in F$. Plugging these in you see that
$$
\sum_ie_ik_i=\sum_{j=1}^m\sum_{\ell=1}^n\left(\sum_{p,q}a_{pj}b_{q\ell}\right)s_jt_\ell.
$$
So all the elements of $EK$ can be written as $F$-linear combinations of the elements $s_jt_\ell$. Therefore the elements $s_jt_\ell$ span $EK$. There are
$mn=[E:F][K:F]$ of them.
In part B you are supposed to recall results saying that if all the elements $e_i$ and $k_i$ are algebraic over $F$, then so are the products $e_ik_i$. And also that the sum
$$
e_1k_1+e_2k_2+e_3k_3+\cdots+e_nk_n
$$
is algebraic as a sum of algebraic elements.
