I've been thinking about subfield lattices. Textbook always ask about a extension $F/\mathbb{Q}$ and ask to draw the lattice and ask questions based on that. But what if we go the other way? Say $F/\mathbb{Q}$ is an extension of degree $n$. We can draw it's lattice of subfields. Say it has a subfield $F'$. $F'$ can appear as a subfield of other fields $L$. So being told that you have $F'$ means you can't know that this came from a field $F$.
But then I thought about counterexamples. It wasn't too hard to show that you can create infinitely many fields $L/F'/\mathbb{Q}$ such that $L/\mathbb{Q}$ has degree $d$ so that this proves knowing $F'$ appears in the lattice does not tell you "anything" about $F$. [I mean, you can get some information about $F$, either degree or discriminant, etc.] But can you take this farther?
Q: So given an extension $F'/\mathbb{Q}$ of degree $k$, can $F'$ be a subfield of all fields $F/\mathbb{Q}$ of degree $n$, where $n$ is divisible by $k$?
The answer should definitely be no. I feel like this should be easy to show but I couldn't think of an easy argument. I know you have to be careful because from the stuff I did in the other direction, there are infinitely many fields $F$ with degree $n$ that do contain $F'$ as a subfield. But this should not be all fields of degree $d$. There should still be infinitely many left. Is there a proof for this?
I also couldn't figure out what happens if you specified more than one subfield in the lattice. Like if you have $F'$ and $F''$ instead. Or generally, a whole finite collection of fields in the lattice, are there still infinitely many fields with degree $n$ that do not contain these fields in their lattice of subfields?