Timeline for Given a number field $K$, when is its Hilbert class field an abelian extension of $\mathbb{Q}$?
Current License: CC BY-SA 2.5
13 events
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| Feb 23, 2010 at 14:42 | comment | added | Franz Lemmermeyer | The quartic subfield of Q(zeta_p), p = 1 mod 4, is real if and only if p = 1 mod 8. So in order to avoid having a complex extension of a real field, you have to make sure that either the base field is complex or the class field is real. | |
| Feb 22, 2010 at 11:25 | comment | added | Kevin Buzzard | "There's no guarantee that this is the whole Hilbert class field, however.". No, of course not---but I just wanted a toy example of a class of unramified (everywhere) extensions to carry around---I wanted to better my Q(zeta_{pq})/Q(zeta_{pq})^+ example above. Your idea is not to use more primes but to go deeper into the 2-part of the cyclo field---there are six extensions of Q unramified outside pq.infty with Galois group cyclic of order 4 and this gives you the extra space you need. Thanks! Don't you just need p,q to be 1 mod 4 by the way? | |
| Feb 20, 2010 at 11:32 | comment | added | Franz Lemmermeyer | It's not so much a question about the number of primes as it is of getting rid of ramification at infinity. Here's a simple example that generalizes easily: take primes p = 1 mod 8 and q = 5 mod 8, and let K be the cyclic quartic extension of conductor pq*\infty. This complex field must have a class group containing a cyclic group of order 4, and the extension you get by adjoining the quartic subfield of Q(\zeta_p) is unramified and abelian over K. There's no guarantee that this is the whole Hilbert class field, however. | |
| Feb 19, 2010 at 15:48 | comment | added | Kevin Buzzard | While I'm here---can my example be modified (e.g. by using more primes) to come up with a cheap general class of examples unramified at all places? | |
| Feb 19, 2010 at 15:48 | comment | added | Franz Lemmermeyer | For every statement about Hilbert class fields in the usual sense (unramified everywhere) there is an analogous statement for Hilbert class fields in the strict sense (unramified at all finite places). So your comment is not wrong, it's the question that has two interpretations. That's all I wanted to point out. | |
| Feb 19, 2010 at 15:46 | comment | added | Kevin Buzzard | Aah yes very nice! For years I've carried David's example around in my head as an example unramified at all places, and the Q(zeta_{pq}) example I mentioned above as an example unramified at all finite primes, but I'd never seen your even simpler example of a non-trivial extension unramified at all finite primes until today. Thanks! | |
| Feb 19, 2010 at 15:17 | comment | added | user1073 | @Kevin: The real quadratic field K=Q(sqrt(3)) has class number one, but its pretty obvious that the quadratic extension K(i) of K is unramified at all finite primes. So a Hilbert class field has to be everywhere unramified (infinite primes included). | |
| Feb 19, 2010 at 15:05 | history | edited | Felipe Voloch | CC BY-SA 2.5 |
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| Feb 19, 2010 at 14:36 | comment | added | Kevin Buzzard | @Franz: I've never known whether Hilbert class Fields are allowed to be ramified at infinity! From what you say, they aren't, so my comment is wrong, but David's still stands as an explicit counterexample. David's example is the reason that you can use binary quadratic forms and the usual tricks to figure out precisely which primes are of the form 2x^2+2xy+3y^2 (it's a congruence condition: 3 or 7 mod 20), but you can't figure out which primes are of the form 2x^2+xy+3y^2 (because now the Hilbert class field isn't abelian). Of course you know this already :-) | |
| Feb 19, 2010 at 9:25 | comment | added | Franz Lemmermeyer | @Kevin: this is correct if you neglect infinite primes. What this means is that the corresponding class field corresponds to a subgroup of the class group in the strict sense. | |
| Feb 19, 2010 at 8:17 | comment | added | Kevin Buzzard | @Felipe: I think that if N has more than 1 prime divisor then Q(zeta_N) is unramified over its real subfield. | |
| Feb 19, 2010 at 4:01 | comment | added | David E Speyer | That's not true. The class filed of Q(sqrt(-5)) is Q(sqrt(-5), sqrt(-1)). | |
| Feb 19, 2010 at 3:59 | history | answered | Felipe Voloch | CC BY-SA 2.5 |