I understand the work in Cohen and Lenstra's paper that leads up to the heuristics themselves, where they count weighted averages of functions defined over isomorphism classes of $A$-modules, where $A$ is a number ring. When they get to the heuristics themselves, they start by saying: "Let $\Gamma$ be an abelian group of order $N$ and $r_1, r_2$ chosen such that $r_1+2r_2=N$. Finally we let $A=A_{\Gamma}$ be the maximal order in the ring $\mathbb{Q}[\Gamma]/\sum_{g\in\Gamma}g$. It is well known that $A_{\Gamma}$ is unique and that it is a product of rings of integers of number fields".

I don't really understand their notation. Is this supposed to be a quotient of the group ring $\mathbb{Q}[\Gamma]$? Also, what do they mean by "maximal order"? Isn't an order just a subring of a number ring? And for a number ring $\mathcal{O}$, $\mathcal{O}$ itself is the maximal order contained in $\mathcal{O}$. But $\mathbb{Q}[\Gamma]/\sum_{g\in\Gamma}g$ is not a number ring? Finally, what is the intuition behind the ring $A_{\Gamma}$? I assume it is supposed to be related in some way to an abelian extension of $\mathbb{Q}$ with Galois group $\Gamma$.

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    $\begingroup$ If $\Gamma$ is cyclic of order $n$ then their ring $\mathbf Q[\Gamma]/\sum_{g\in \Gamma} g$ is basically $\mathbf Q[X]/(1+X+...+X^{n-1})$, which is not a field unless $n$ is prime. In general this ring is a product of fields, namely the fields $\mathbf Q[X]/(\Phi_d(X))$ where $d$ runs over the factors of $n$ greater than $1$. $\endgroup$
    – KConrad
    Oct 15, 2014 at 3:17

1 Answer 1


The group $\Gamma$ is indeed isomorphic to the Galois groups of the fields in the family, whose class groups one studies. The class groups come with a natural action of $\Gamma$, but under this action, the primitive central idempotent $\sum_{g\in \Gamma}g$, which corresponds to the trivial representation, acts as zero on the coprime-to-$|\Gamma|$ part (in other words, the coprime-to-$|\Gamma|$ part has no fixed points under the action of $\Gamma$), so this part is naturally a module under the quotient ring $\mathbb{Z}[\Gamma]/\langle\sum_{g\in \Gamma}g\rangle$ (and more, see below).

In general, for arbitrary finite groups $\Gamma$, $\mathbb{Q}[\Gamma]$ is a semisimple $\mathbb{Q}$-algebra, i.e. a direct product of matrix algebras over division algebras. It always has a direct summand isomorphic to $\mathbb{Q}$, generated by the above-mentioned idempotent, and Cohen–Lenstra pass to the quotient, since that idempotent is not interfering with their Galois action.

If $R$ is any integrally closed Noetherian domain (such as $\mathbb{Z}$, or the localisation $\mathbb{Z}_{(p)}$ at a prime $p$), and $\mathbb{K}$ is the field of fractions, then an $R$-order in a semisimple $\mathbb{K}$-algebra is a subring that is a finitely generated $R$-module and that generates the algebra over $\mathbb{K}$. For example $\mathbb{Z}[\Gamma]$ is a $\mathbb{Z}$-order (or just "order" when $R=\mathbb{Z}$) in $\mathbb{Q}[\Gamma]$. But also, $\mathbb{Z}_{(p)}[\Gamma]$ is a $\mathbb{Z}_{(p)}$-order in the same algebra. A maximal $R$-order is an $R$-order that is not properly contained in any other $R$-order. It requires proof that such a thing exists. In the particular case that Cohen–Lenstra are talking about, the maximal order is unique, and contains $\mathbb{Z}[\Gamma]$.

One can show by localising that for any $\Gamma$ (not just abelian), if $A_\Gamma$ is a maximal order in $\mathbb{Q}[\Gamma]$ that contains $\mathbb{Z}[\Gamma]$, then the coefficients of the elements in $A_{\Gamma}$ have denominators supported only at prime divisors of $|\Gamma|$. In other words, if $p$ is a prime that does not divide $|\Gamma|$, then $\mathbb{Z}_{(p)}[\Gamma]$ is a maximal $\mathbb{Z}_{(p)}$-order in $\mathbb{Q}[\Gamma]$. It follows that if $A$ is a $\Gamma$-module of order coprime to $|\Gamma|$, then $A$ has a well-defined action of $A_{\Gamma}$ (because if $p\nmid |A|$, then you can "divide by $p$" in $A$).

For this and much more on maximal orders, see I. Reiner's book "Maximal Orders".

When $\Gamma$ is not assumed to be commutative, a maximal order in $\mathbb{Q}[G]$ is no longer unique, in general, and the situation is more complicated. This case was treated a few years later by Cohen–Martinet.

  • $\begingroup$ Perfect. This is exactly what I needed. $\endgroup$
    – JAN
    Oct 15, 2014 at 16:07
  • $\begingroup$ 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 $\textrm{Cl}(K)$? $\endgroup$
    – JAN
    Oct 16, 2014 at 0:48
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    $\begingroup$ 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. $\endgroup$
    – Alex B.
    Oct 16, 2014 at 9:53
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    $\begingroup$ @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... $\endgroup$
    – Alex B.
    Nov 3, 2020 at 18:30
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    $\begingroup$ ...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. $\endgroup$
    – Alex B.
    Nov 3, 2020 at 18:31

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