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Is there any condition over a number field $K$ for an unramified quadratic extension of $K$ to admit an embedding into an unramified cyclic extension of degree 4 of $K$?

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    $\begingroup$ Is this not answered by Hilbert classfield stuff? $\endgroup$
    – paul garrett
    Commented Oct 19, 2022 at 16:42
  • $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Commented Oct 19, 2022 at 19:19
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    $\begingroup$ As Paul Garrett is saying, there is an answer in terms of class field theory: let $\chi$ be the quadratic class group character corresponding to your extension $L/K$, then there is an embedding as you ask if and only if $\chi$ is the square of another character (necessarily of order $4$) of the class group. Is this the type of answer you are looking for? $\endgroup$
    – Aurel
    Commented Oct 19, 2022 at 19:25
  • $\begingroup$ i didn't know this result by Hilbert, thank u so much for this information, does this result exist on his book "The Theory of Algebraic Number Fields "? $\endgroup$
    – ayoub-chess
    Commented Oct 19, 2022 at 20:55

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Ah... perhaps an amplification of my comment and @Aurel's would be useful to you:

First, this kind of thing is not really in Hilbert's "Zahlbericht", because he only treats a sub-class of extensions... which was already a novelty, etc.

But by the 1920's, Takagi and Artin had clarified/proved the reasonable general case of Hilbert's earlier examples, namely, that the Galois group of the maximal unramified abelian extension of a number field was naturally isomorphic to the absolute ideal class group of the ring of integers of that number field. This is a little bit more delicate than just the general assertions of classfield theory, since it requires further attention to ramification...

The general theorems of classfield theory are proven many places, but, as far as I know, the subtler versions, about ramification and "Hilbert classfields", are not reliably proven. (E.g., I think Lang's otherwise very useful "Alg No Th" does not actually prove that, though remarks upon it.)

Anyway, the facts are fairly straightforward! :)

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