Say I have a field $F$, and an extension field $L = F(\alpha_1,\cdots,\alpha_n)$. Is it true that every $K$ such that
$$ F \subset K \subset F $$
(all field extensions), $K = F(\alpha_j,\cdots,\alpha_k)$?
That is, $K$ is $F$ adjoined with any number of whichever $\alpha$s.
It seems to be so to me, since $F(\alpha_1)$ is the smallest field containing both $F$ and $\alpha_1$. So there is no field between $F(\alpha_1,\cdots,\alpha_{n-1})$ and $L$, no field between $F(\alpha_1,\cdots,\alpha_{n-1})$ and that last field, etc.
I realize that the alphas could be left out in different orders, to create different fields between $F$ and $L$ not of the form $F(\alpha_1,\cdots,\alpha_k)$. But I think that if we take all of the $F(\alpha_i)$ and build upwards to $L$ in all possible ways, we'll get ever field, because $L$ is the smallest field to contain $F$ and all the $\alpha_i$.
It seems this is true since if there is a field $J$ between $F$ and $L$ and
$$ \forall i,\cdots,j\ni \alpha_i,\cdots,\alpha_j\in\{\alpha_1,\cdots,\alpha_n\}\quad J\ne F(\alpha_i,\cdots,\alpha_j)\\ \implies\exists\beta\in J\ni\beta\not\in F(\alpha_i,\cdots\alpha_j)\text{ and }\beta\in L $$
But I think that would mean that $L$ is not the smallest field containing $\alpha_1,\cdots,\alpha_n$