Timeline for Class groups in dihedral extensions - some sort of Spiegelungssatz?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2018 at 11:24 | vote | accept | Alex B. | ||
| Mar 13, 2018 at 9:19 | comment | added | YCor | Notation: $h(F)$ is the order of the class group of $F$ (it took me some time to find this...) | |
| Mar 13, 2018 at 6:26 | answer | added | Filippo Alberto Edoardo | timeline score: 4 | |
| Aug 22, 2017 at 17:44 | answer | added | Franz Lemmermeyer | timeline score: 5 | |
| Aug 17, 2011 at 11:50 | history | edited | Alex B. | CC BY-SA 3.0 |
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| Aug 17, 2011 at 11:42 | comment | added | Chandan Singh Dalawat | let $K$ ..., and $L$ an intermediate cubic. Do you mean an intermediate degree-$p$ extensions? Also, presumably $p$ is a prime. | |
| Aug 17, 2011 at 7:15 | comment | added | Kevin Ventullo | @Alex: Here's the beginning of an argument: Consider only the p-torsion Hilbert class field over $F$, Hilb$_p$, which is an $\mathbb{F}_p$ vector space. Consider the action of the involution $\sigma\in$Gal$(F/L)$. If there is a +1 eigenvector, then the corresponding $p-$extension of $F$ is abelian over $L$, and so there is a $p-$unramified extension of $L$. Otherwise, $\sigma$ acts by -1 everywhere, hence lies in the center of the action of Gal$(F/Q)$ on Hilb$_p$. Then Gal$(F/K)$ lies in the kernel, and Hilb$_p$ is abelian over $K$. I'm not sure how to proceed from here... | |
| Aug 17, 2011 at 6:30 | history | edited | Alex B. | CC BY-SA 3.0 |
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| Aug 17, 2011 at 6:29 | comment | added | Alex B. | @Kevin: of course, you are right. Thanks! I was getting carried away in my pessimism. But it needn't be abelian over $L$, e.g. it could be dihedral. If we just look at the odd degree bit of the Hilbert class field (which we may wlog), then the Galois group over $L$ is a semi-direct product of an abelian group and $C_2$. I guess, if it's not abelian, then one should be able to say something about the Galois group of $Hilb(F)/K$. I may be beginning to see an argument for the last boxed statement... | |
| Aug 17, 2011 at 6:16 | comment | added | Kevin Ventullo | Actually the Hilbert class field of $F$ is always Galois over $L$. In fact it is Galois over $\mathbb{Q}$: if $\sigma\in$Gal$(\overline{\mathbb{Q}}/\mathbb{Q})$, then $\sigma(Hilb(F))$ is unramified abelian over $\sigma(F)=F$. | |
| Aug 17, 2011 at 5:41 | history | edited | Alex B. | CC BY-SA 3.0 |
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| Aug 17, 2011 at 5:28 | history | asked | Alex B. | CC BY-SA 3.0 |