Let$ K$ be a finite extension of $F$. If $L_1,L_2$ are subfields of $K$ containing $F$ and $L_1⊂ L_2$ then $K$ is simple extension of $F$.

Question

I have seen this problem in field note , but I can not do this. Let $K$ be a finite extension of $F$ such that for every pair $L_1,L_2$ of subfields of $K$ containing $F$, either $L_1⊂ L_2$ or $L_2⊂ L_1$. Show that $K$ is a simple extension of $F$.

Answer 1

The intermediate fields are linearly ordered by inclusion meaning that there exist at most one field of each degree (because if $E_1 \subseteq E_2$ and $[E_1:F]=[E_2:F]$ we get $E_1=E_2$).

Since $K/F$ is finite, there are only finitely many intermediate fields and they are linearly ordered by inclusion, so there is a maximal such field, i. e. a proper subfield $L\subsetneq K$ that contains all the intermediate fields. Then $K=F(\alpha) $ for any $\alpha \in K\setminus L$.