The groups whose subgroups are totally ordered by inclusion are easy to classify; they are subgroups of $\mathbb{Z}/p^{\infty} = \text{colim } \mathbb{Z}/p^k$ for some prime $p$, and thus $\mathbb{Z}/p^{\infty}$ or finite cyclic of prime power order. What about fields?

Is it possible to classify (or at least, give some properties and examples) the field extensions $E/F$, whose intermediate fields are totally ordered by inclusion?

If $E/F$ is a Galois extension, then we may rephrase the condition: The *closed* subgroups of $Gal(E/F)$ are totally ordered. It can be shown that then there is a prime $p$ such that $Gal(E/F)$ is pro-$p$-cyclic and thus isomorphic to $\mathbb{Z}/p^k$ for some $k \geq 0$ or to $\mathbb{Z}_p = \lim_k \mathbb{Z}/p^k$. Thus $E/F$ is built up out of cyclic Galois extensions $F_{i+1} / F_i$ degree $p$ (for example $E = \mathbb{F}_{q^{p^\infty}}, F = \mathbb{F}_q$). In characteristic $p$, cyclic Galois extensions of degree $p$ are characterized by a Theorem of Artin-Schreier. In characteristic $q \neq p$ ($q=0$ allowed), there is a characterization if $F_i$ contains a primitive $p$th root of unity. What can be said if this is not the case?

Now do not assume that $E/F$ is Galois. Here is a simple observation:

The intermediate fields of $E/F$ are totally ordered iff $E/F$ is algebraic and the finite intermediate fields of $E/F$ are totally ordered.

*Proof*: $\Rightarrow:$ If $t$ is a variable, then the intermediate fields of $F(t)/F$ are not totally ordered, consider $F(t^2)$ and $F(t^3)$. $\Leftarrow:$ Let $K,L$ be intermediate fields which are not compatible. Choose $a \in K - L, b \in L - K$. Then $F(a), F(b)$ are finite extensions, which are not compatible, contradiction.

An immediate consequence is, that we may first restrict to the finite case: Namely every $E/F$ as above is a directed union of finite subextensions, whose intermediate fields are totally ordered by inclusion; and vice versa.

Note that we cannot restrict the degree of $E/F$. Namely, $S_{n-1}$ is a maximal subgroup of $S_n$. Since $S_n=Gal(E/F)$ for some Galois extension $E/F$, if $K$ is the fixed field of $S_{n-1}$, the extension $K/F$ has degree $n$ and no nontrivial intermediate fields at all.

What about inseparable extensions? And what happens if we take the normal closure? Perhaps we can reduce everything to the Galois case?