Let $K\subseteq F,L \subseteq \Omega$ be fields such that $F/K$ and $L/K$ are finite extensions (don't know if this is necessary). Prove that $$\alpha \in LF \Longrightarrow \alpha = \sum _{i=1}^nf_il_i \qquad f_i \in F,\ l_i \in L$$
I don't understand why this is true (and also don't know if the finite extensions requirement is necessary). My attempt was to observe that $LF=L(F)$ by definition, and so $\alpha \in LF=L(F) \Longrightarrow \alpha \in L(f_1,\dots,f_n)$ for some $f_1,\dots,f_n \in F$, and this implies that $$\alpha=\frac{p(f_1,\dots,f_n)}{q(f_1,\dots,f_n)} \qquad \text{for some }p,q \in L[x_1,\dots,x_n],\ q \neq 0$$ But I don't know how to arrive at the linear combination from this rational function. Is it the right way? How can I finish it?