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Mar 2 at 14:38 comment added Mikhail Borovoi Many thanks indeed!
Mar 2 at 13:18 comment added user491858 Take $n=n'=3$ for convenience. Let $P$ in $F$ be a prime of norm $p \not\equiv 1\bmod 3^2$ which is tamely ramified in $K/F$. This gives a local obstruction to $L/F$ being cyclic. Now suppose $Q$ in $F$ has norm $q\equiv 2\bmod 3$ and is inert in $K$, and take $S=\{Q\}$. The norm of $Q \mathcal{O}_K$ is $q^3\not\equiv 1\bmod 3$, so there is no degree $3$ ramified (at $Q$) extension $L/K$. Your conditions force $L/F$ to be cyclic, a contradiction. An example: $F=\mathbf{Q}(\sqrt{-7})$, $K=\mathbf{Q}(\zeta_7)$, $P$ the (unique) prime in $F$ of norm $7$, and $Q$ any prime of norm $2$ in $F$.
Mar 2 at 13:16 comment added user491858 Dear Mikhail, I don't think so, for local reasons. Consider the case when $n=n'=p$ prime. Then if $L/F$ is Galois, it is abelian, so one can try to set things up where local conditions force $L/F$ to be both cyclic and non-cyclic. For example, if the local conditions imply that a prime $\mathfrak{q}$ in $F$ is inert in $L$, then $L/F$ (if Galois) has to be cyclic, since all unramified extensions are cyclic.
Mar 2 at 9:18 history asked Mikhail Borovoi CC BY-SA 4.0