Timeline for Constructing a cyclic extension $L$ with given local behavior of a global field $K$ such that $L$ is normal over a subfield $F$ of $K$
Current License: CC BY-SA 4.0
4 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
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 |