Timeline for What is the ring $A_{\Gamma}$ in the Cohen-Lenstra Heuristics?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2020 at 1:27 | comment | added | Alex B. | @Melanka, that just shows that $\mathbb{Z}[G]$ projects surjectively onto the maximal order of $\mathbb{Q}[\Gamma]/\sum g$, not that it itself is maximal. You could inspect the case $p=2$ more closely, perhaps draw a picture to see what is going on. I am not sure what the hang-up is, but one possibility is that you are confusing subrings and quotient rings. | |
| Nov 3, 2020 at 20:48 | comment | added | Melanka | @AlexB. Isn't it the case that under the isomorphism $\mathbb{Q}[\zeta_p] \cong \mathbb{Q}[\Gamma]/\langle\sum g \rangle$, $\zeta_p$ is identified with $\sigma$ a generator of $\Gamma$? So then wouldn't it be that the maximal order with basis $g \in \Gamma$ be the same as with the basis of $\zeta_p$. | |
| Nov 3, 2020 at 18:31 | comment | added | Alex B. | ...needs to be modified to take into account genus theory. This was done by Frank Gerth III some years after the original paper of C-L. | |
| Nov 3, 2020 at 18:30 | comment | added | Alex B. | @Melanka: if $\Gamma$ is cyclic of order $p$, then a maximal order in $\mathbb{Q}[\Gamma]$ will have denominators, when expressed in terms of the standard basis $g\in \Gamma$. When you pass to a quotient, there is no well-defined meaning of "having denominators", because it depends on the basis. Of course you can pick a $\mathbb{Z}$-basis for your maximal order, and then by definition the elements of that order will have no denominators w.r.t. that basis. Having said that, there is a well-defined action of $\mathbb{Z}[\zeta_p]$ on the $p$-class group in your example, but the C-L heuristic... | |
| Nov 3, 2020 at 2:25 | comment | added | Melanka | Correction: $\Gamma = \mathbb{Z}/p\mathbb{Z}$ ($\Gamma$ instead of $G$) | |
| Nov 3, 2020 at 2:17 | comment | added | Melanka | @AlexB. in the original paper by Cohen - Lenstra they define $A_{\Gamma}$ to be the maximal order of $\mathbb{Q}[\Gamma]/⟨\sum_{g \in \Gamma}g⟩$. But when $G = \mathbb{Z}/p\mathbb{Z}$, $\mathbb{Q}[\Gamma]/⟨\sum_{g \in \Gamma}g⟩ \cong \mathbb{Q}[\zeta_p]$ and it's maximal order is $\mathbb{Z}[\zeta_p]$. So as per your explanation (since there are no denominators in this case) it seems as if they could have conjectured for the whole class group (instead of the prime to p part of the class group) in this case. I think I am missing something in this. Would be grateful if you could help. | |
| Dec 1, 2017 at 11:37 | history | edited | Alex B. | CC BY-SA 3.0 |
Fixed several typos
|
| Oct 16, 2014 at 9:53 | comment | added | Alex B. | Sorry Jan, I was being sloppy. Of course, an arbitrary element of $\mathbb{Q}$ has no way of acting on the class groups. But the maximal order in $\mathbb{Z}[\Gamma]$ does act on the coprime-to-$|\Gamma|$ part of the class group. I have expanded the answer to explain this. | |
| Oct 16, 2014 at 9:52 | history | edited | Alex B. | CC BY-SA 3.0 |
Substential rewrite, mainly to fix sloppy notation and to explain how a maximal order acts on the class group
|
| Oct 16, 2014 at 0:48 | comment | added | JAN | One more question: What is the action of $\mathbb{Q}[\Gamma]$ on $\textrm{Cl}(K)$? I.e how do each of the matrix algebras act on $\textrm{Cl}(K)$? | |
| Oct 15, 2014 at 16:07 | comment | added | JAN | Perfect. This is exactly what I needed. | |
| Oct 15, 2014 at 13:14 | vote | accept | JAN | ||
| Oct 15, 2014 at 8:40 | history | edited | Alex B. | CC BY-SA 3.0 |
added 126 characters in body
|
| Oct 15, 2014 at 8:24 | history | answered | Alex B. | CC BY-SA 3.0 |