Classification of subfields of $K(S)$

Question

Let $F:K$ be a field extension and $S \subset F$. Is there any classification of subfields of $K(S)$ In case this isn't standard, $K(S)$ is the intersection of every subfield containing $K$ and $S$, i.e.

\begin{equation*} K(S) = \bigcap_{\substack{L \supset K \cup S \\ L \text{ is a subfield of } F}} L. \end{equation*}

I couldn't find anything, but my intuition is telling me that the only subfields of $K(S)$ are either

1. subfields of $K$;
2. subfields of the form $K(T)$, where $T$ is "built" from $S$.

I'm not exactly sure what "built" should be, but my thoughts so far are that elements of $T$ should be linear combinations of elements of $K(S)$ with coefficients in $K$. However, I am at a loss of how to start proving this.

If not, can we say something along these lines if $S$ is a finite set, say $S = \{\alpha_1, \dotsc, \alpha_n\}$?

Answer 1

EDIT: Apologies, I made a silly error in my construction of $K(S)$. What I wrote is correct if you assume that $K(S)/K$ is algebraic. If not, then you must instead consider quotients of these sums. For instance, if we look at $K(x)/K$ then the set I wrote out is just $K[x, x^{-1}]$, which is not equal to $K(x)$. The rest of the argument holds the same way, but with a broader notion of $T$ being "built" from $S$.

Your conjecture is essentially correct, and here's a way to prove it. We have an expression for a generic element of $K(S)$ as sums of the form $\sum a_i s_i^{n_i}$ where $a_i \in K$, $s_i \in S$, and $n_i \in \mathbb Z$ (I'm assuming here that $0 \notin S$). This will indeed be a field containing $K$ and $S$, and clearly any field containing $K$ and $S$ must contain this. Now, let $K(S)/E/K$ be an intermediate extension. Then every element of $E$ is of the form of the sum I wrote above, so any generating set $T$ of $E$ over $K$ will necessarily consist of such elements as well. In that sense, $T$ is wholly "built" from $S$.

For completion, I want to emphasize that you cannot strengthen this to say that $T \subseteq S$ or even that elements of $T$ are $K$-linear combinations of elements of $S$. Take for instance $K(x)/K(x^2)$. You do really need these to be algebraic combinations, not just linear ones.

Also, I'd like to point out that any discussion of the set of intermediate fields in an extension $F/K$ warrants some mention of Galois theory. In certain nice cases, subgroups of the group of automorphisms of $F$ that fix $K$ (called the Galois group of $F/K$) completely determine the intermediate fields in a very elegant fashion.