Let $K/k$ be an extension of number fields.

Is it true that for almost all places $v$ of $k$, the extension $K \otimes_k k_v / k_v$ is a cyclic extension of local fields? (Maybe under some additional assumption on $K/k$, e.g. if $K/k$ is Galois/abelian/...)

EDIT: OK, maybe that was not the question I wanted to ask. Is it true that for almost all paces $v$ of $k$, there exists a place $w$ of $K$ lying over $v$ such that the extension $K_w/k_v$ is cyclic?

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    $\begingroup$ The tensor product $K \otimes_k k_v$ need not be a field for most $v$. Try $k = {\mathbf Q}$ and $K$ a Galois extension of ${\mathbf Q}$ whose Galois group is not cyclic (e.g., $K = {\mathbf Q}(\sqrt{2},\sqrt{3})$, whose Galois group is abelian). For all unramified places $v$ of $\mathbf Q$, the tensor product $K \otimes_{\mathbf Q} {\mathbf Q}_v$ is not a field since $v$ has at least two places over it (an unramified place with just one place above it has a decomposition group that is both cyclic and equal to the Galois group, which would be impossible if the Galois group is not cyclic). $\endgroup$ – KCd Jan 9 '13 at 2:03

Yes. You want to say that the decomposition group is cyclic for most primes. This is true for unramified primes because the decomposition group is isomorphic to the Galois group of the extension of residue fields, which is always cyclic.


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