4
$\begingroup$

So I'm curious about how the Cohen-Lenstra heuristics on ideal class groups might work over non-abelian extensions. Specifically:

Fix a finite group $G$ and suppose I have a family of Galois extensions of $\mathbf Q$ with Galois group $G$. Fix a prime $p$ not dividing #$G$. If $K$ is one of these extensions, then the $p$-torsion $C_K[p]$ in the ideal class group of $K$ is a semisimple ${\mathbf F}_p[G]$-module. How would one expect these to be distributed as $K$ varies?

An obvious guess is that for any finite ${\mathbf F}_p[G]$-module $M$, the (inverse of the) size of the ${\mathbf F}_p[G]$-automorphisms of $M$ would control the frequency with which $M$ occurs. However, I don't feel comfortable enough with the usual Cohen-Lenstra heuristics to have a feeling if there should be more to it than that.

Any suggestions and/or references? I've had no luck finding anything so far.

$\endgroup$

1 Answer 1

4
$\begingroup$

The original work is by Henri Cohen and Jacques Martinet. The references are

Class groups of number fields: Numerical heuristics. Mathematics of Computation, 48(177):123–137, 1987.

and

Étude heuristique des groupes de classes des corps de nombres. Journal für die Reine und Angewandte Mathematik, 404:39–76, 1990.

For what may be the latest work on this topic, see Conjectures for distributions of class groups of extensions of number fields containing roots of unity by Will Sawin and Melanie Matchett Wood https://arxiv.org/abs/2301.00791

One can't just take the inverse of the automorphism of the module as that would predict the same distribution for real and imaginary quadratic fields but these are different. A better approach is to take the inverse of the product of the order of the automorphism group and the order of the invariants under the complex conjugation conjugacy class in $G$. This will give the right answer for $p>2$ when $|G|$ is prime to $p$, and then for $p=2$ the formula is more complicated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.