Given a galois extension of number fields $L/K$ of even degree, set $n_0=\text{lcm} (\{[L_v:K_v] : v \in M_K \})$ ($L_v$ is any completion corresponding to a place dividing $v$).
Does $2$ divide $n_0$?
This comes up in this question.
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asked Apr 10, 2011 at 15:48
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3$\begingroup$ Is this not a trivial consequence of Cebotarev density?? What am I missing? $\endgroup$– Kevin BuzzardCommented Apr 10, 2011 at 16:36
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1$\begingroup$ It is! If no one answer had come up I would have deleted the question. $\endgroup$– Dror SpeiserCommented Apr 10, 2011 at 18:55
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2$\begingroup$ The question of the title does not agree with the question in the body (the latter is easy if you know Cebotarev density, whereas the former is trivial). $\endgroup$– Pete L. ClarkCommented Apr 10, 2011 at 22:01
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1 Answer
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If I'm not mistaken, the slightly stronger result is true, that the lcm of orders of Frobenius elements must be even (forget ramified primes, that is). Isn't that a corollary of the Chebotaryov density theorem (there is a Frobenius in every conjugacy class, and some such class contains elements of even order)?
answered Apr 10, 2011 at 16:38
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2$\begingroup$ Why does everyone spell Chebotarev differently? (Both than me and each other.) $\endgroup$– KimballCommented Apr 10, 2011 at 21:17
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$\begingroup$ The sound yo is written ë in Cyrillic, and often therefore just e in Romanised text. Khrushchev is the well known example, so the final syllable is actually sounded shchoff (though there the ch sound swallows the y). $\endgroup$ Commented Apr 10, 2011 at 21:40