Pick your favorite prime $p$, as well as three positive integers $e,f,g$.
For each such choice, does there exist at least one Galois number field $K/\mathbf{Q}$ of degree $n=efg$ in which $p$ has ramification index $e$ and inertial degree $f$?
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$\begingroup$ I think there’s a “trivial” local obstruction due to ramification theory: the prime-to-$p$ part of $e$ must divide $p^f-1$. $\endgroup$– AphelliCommented Oct 5, 2023 at 13:57
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1$\begingroup$ @Aphelli Good point; I'm most interested in the case where $e$ is 1. $\endgroup$– Jeff HCommented Oct 5, 2023 at 15:20
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1$\begingroup$ This well-known to be false for $p=2$ and $(e,f,g) = (1,8,1)$. en.wikipedia.org/wiki/… $\endgroup$– user491858Commented Oct 5, 2023 at 19:22
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1 Answer
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Yes for odd primes in the case $e=1$, and you can take $K$ to be cyclic, and this is also true relatively to any base number field and for any finite collection of primes. It follows from the Grunwald-Wang theorem, according to which every finite collection of local cyclic extensions at distinct primes can be globalised into a single cyclic extension of degree the LCM of the local degrees (See Artin-Tate, Class field theory, Chapter X, Section 2, Theorem 5).
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3$\begingroup$ This feels like a result I should have known, but definitely did not know. Thanks so much! $\endgroup$– Jeff HCommented Oct 5, 2023 at 15:59
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1$\begingroup$ degree the LCM of the local degrees, possibly times 2. $\endgroup$ Commented Oct 5, 2023 at 19:22
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1$\begingroup$ @user491858 No, the times 2 occurs only when you want to globalise characters, but you don't need it to globalise field extensions (see Theorem 6 in Artin-Tate). $\endgroup$– AurelCommented Oct 5, 2023 at 19:49
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1$\begingroup$ @user491858 Oh sorry, that's not true. You can ensure that the local extension has the right degree, but not necessarily that it is unramified. So there is indeed the times 2 if you insist on unramifiedness. I've edited to add "odd", so that the bad case is avoided. $\endgroup$– AurelCommented Oct 5, 2023 at 19:52