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Let $F$ be a global field without real places (that is, a function field or a totally imaginary number field). Let $K/F$ be a cyclic extension of degree $n$. Let $S$ be a ${\rm Gal}(K/F)$-invariant finite set of non-archimedean places of $K$.

Let $r\ge 1$ be an integer. By Theorem X.6 in the book "Class field theory" by Artin and Tate, there exists a cyclic extension $L/K$ of degree $n^r$ such that all local extensions $L_w/K_v$ for $v\in S$ are of degree $n^r$.

Question. Can one construct such an extension $L/K$ with the additional property that $L$ is normal over $F$?

Note that $L$ is normal over $F$ if and only if the norm subgroup $N_{L/K}(C_L)\subset C_K$ is ${\rm Gal}(K/F)$-invariant, where $C_L$ and $C_K$ are the corresponding idèle class groups.

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    $\begingroup$ 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. $\endgroup$
    – user491858
    Commented Mar 2 at 13:16
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    $\begingroup$ 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 $F$. $\endgroup$
    – user491858
    Commented Mar 2 at 13:18
  • $\begingroup$ Many thanks indeed! $\endgroup$
    – Mikhail Borovoi
    Commented Mar 2 at 14:38

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