gpt 6 astra pro attempt on Math-87

Cohen-Lenstra class-group distributions for quadratic number fields

Definitions and mathematical statement

The ideal class group $\operatorname{Cl}(K)$ is the group of nonzero fractional ideals modulo principal ideals. For an odd prime $p$, its $p$-primary subgroup consists of elements of $p$-power order. Let $\mathcal K_u(X)$ contain quadratic fields of discriminant absolute value below $X$, imaginary for $u=0$ and real for $u=1$. For every finite abelian $p$-group $A$, the claim is \[ \lim_{X\to\infty}\frac{\#\{K\in\mathcal K_u(X):\operatorname{Cl}(K)[p^\infty]\simeq A\}} {\#\mathcal K_u(X)}=\frac{|A|^{-u}}{Z_u(p)|\operatorname{Aut}(A)|}, \] where $Z_u(p)=\sum_{[B]}|B|^{-u}/|\operatorname{Aut}(B)|$. This is a single-prime distribution for ordinary class groups, not a statement about $p=2$, narrow class groups, or joint independence across primes.

Formulation and concept references: [S1].

Short English statement

Do odd-primary class groups of quadratic number fields follow the exact automorphism-weighted distributions predicted by Cohen and Lenstra?

Research attempt

Current frontier and source audit.

The source's introduction gives the two different weights for imaginary and real quadratic fields, and records the Davenport--Heilbronn moments ${\mathbb{E}}\#\operatorname{Sur}(\operatorname{Cl}(K),C_3)\to1$ and $1/3$, respectively. These are not the full distributions of the $3$-primary groups. The normalization incorporated by reference is checked in Majumder's stated theorem. Throughout this attempt, $p$ is one fixed odd prime and class groups are ordinary class groups. [S1][E1]

Landesman--Levy's 2024 manuscript, revised in October 2025, obtains surjection moments over quadratic function fields for sufficiently large finite constant fields. The permitted size of the constant field depends on the target group. They explicitly warn that this does not establish all moments at one fixed constant field. Neither that result nor recent results about $2$-primary groups can be substituted for the quadratic number-field target here. [E2]

Attack route.

The natural statistic is a surjection count rather than the order of the class group. Three routes are examined: recover the law from all these moments; derive finite, one-sided certificates for the probability of a trivial $p$-part; and compute the same moments in a Haar random-matrix model. The model calculation explains the proposed normalization but supplies no arithmetic equidistribution theorem. The essential missing input is stated explicitly rather than described as a generic randomness assumption.

Attempt: an exact sufficient arithmetic input.

Let $G_X=\operatorname{Cl}(K)[p^\infty]$ for a uniformly chosen $K\in\mathcal K_u(X)$, and write \[ S_A(G)=\#\operatorname{Sur}(G,A),\qquad M_A(X)={\mathbb{E}} S_A(G_X). \] Consider the additional hypothesis \begin{equation}\label{wsref-0-65-71-3a-33-30-30-30-36-37-33-35-3a-6d-6f-6d-65-6e-74-73} \text{for every finite abelian $p$-group $A$,}\qquad M_A(X)\longrightarrow |A|^{-u}. \end{equation} This is a sufficient condition, not something obtained in this notebook for the arithmetic family. Pointwise convergence of group probabilities, without tail control, must not be used to assert convergence of these unbounded statistics in the reverse direction.

Here is a precise way to pass from (eq:30006735:moments) to the target. Wood's Theorem 3.1 determines distributions modulo a fixed integer from surjection moments with $M_A\leq|\bigwedge^2 A|$; Lemma 3.2 computes those moments for the automorphism-weighted law. Our moments satisfy this hypothesis because $|A|^{-u}\leq1\leq|\bigwedge^2 A|$. For a desired group $A$ of exponent $p^e$, use modulus $p^{e+1}$. A finite abelian $p$-group $G$ satisfies \[ G/p^{e+1}G\simeq A\quad\Longleftrightarrow\quad G\simeq A: \] a cyclic summand of length greater than $e$ would survive in the quotient with length at least $e+1$. The cited fixed-modulus theorem therefore gives exactly the asserted probability for $A$, not merely a distribution of its rank. For $A=0$, take $e=0$. Apply the theorem along arbitrary sequences $X\to\infty$ to obtain the full real-parameter limit. [E3]

Attempt: independently control escape and unbounded tests.

The following elementary estimates expose two possible limit gaps. Let $r(G)=\dim_{{\mathbb{F}}_p}G/pG$. Then \[ S_{C_p}(G)=p^{r(G)}-1, \qquad \limsup_{X\to\infty}\Pr(r(G_X)\geq R) \leq\frac{p^{-u}}{p^R-1}. \] If $\exp G\geq p^m$, its largest cyclic summand supplies at least $\varphi(p^m)=p^{m-1}(p-1)$ surjections onto $C_{p^m}$. Hence \begin{equation}\label{wsref-0-65-71-3a-33-30-30-30-36-37-33-35-3a-74-61-69-6c} \limsup_{X\to\infty}\Pr(\exp G_X\geq p^m) \leq\frac{p^{-mu}}{p^{m-1}(p-1)}. \end{equation} Bounds on rank and exponent leave only finitely many group types. These estimates imply tightness under (eq:30006735:moments); no group-valued probability mass escapes to arbitrarily large types.

For each fixed $A$, a pair of surjections to $A$ is a homomorphism to $A\times A$, with additional surjectivity conditions. Partitioning all homomorphisms by their image subgroup gives the exact identity and bound \begin{equation}\label{wsref-0-65-71-3a-33-30-30-30-36-37-33-35-3a-55-49} S_A(G)^2\leq\#\operatorname{Hom}(G,A\times A) =\sum_{B\leq A\times A}S_B(G). \end{equation} The sum is over actual subgroups, so repetitions of an isomorphism type are intentionally included. There are finitely many terms. Hypothesis (eq:30006735:moments) bounds the expected right side, yielding uniformly bounded second moments for $S_A(G_X)$ for large $X$. Therefore these statistics are uniformly integrable. This separately justifies passing their first moments to any subsequential limiting law. Merely having a bounded first moment would not have justified that passage.

Attempt: finite-moment certificates for a trivial class-group part.

There is a more explicit inversion for the probability of $G_X=0$. For $r\geq0$ and $k\geq0$ let $\genfrac{[}{]}{0pt}{}{r}{k}_p$ be the Gaussian binomial coefficient, zero when $k>r$. Since \[ S_{(C_p)^k}(G)=|\operatorname{GL}_k({\mathbb{F}}_p)|\genfrac{[}{]}{0pt}{}{r(G)}{k}_p, \] define the finite observable \[ L_K(X)=\sum_{k=0}^K \frac{(-1)^k p^{k(k-1)/2}}{|\operatorname{GL}_k({\mathbb{F}}_p)|} M_{(C_p)^k}(X), \] with $|\operatorname{GL}_0|=1$ and the zeroth moment equal to one. Then, for every $X$ and $K\geq0$, \begin{equation}\label{wsref-0-65-71-3a-33-30-30-30-36-37-33-35-3a-62-6f-6e-66-65-72-72-6f-6e-69} L_{2K+1}(X)\leq\Pr(G_X=0)\leq L_{2K}(X). \end{equation} This holds for any random finite abelian $p$-group having the displayed finite moments, without assuming Cohen--Lenstra. In particular, all of these moments are finite for each finite set $\mathcal K_u(X)$ of quadratic number fields.

Proof. The finite $q$-binomial identity gives, for $0\leq K<r$, \[ \sum_{k=0}^K(-1)^kp^{k(k-1)/2}\genfrac{[}{]}{0pt}{}{r}{k}_p =(-1)^Kp^{K(K+1)/2}\genfrac{[}{]}{0pt}{}{r-1}{K}_p. \] It follows by induction from the Gaussian-binomial recursion. When $K\geq r>0$, the full sum is $\prod_{j=0}^{r-1}(1-p^j)=0$; for $r=0$ it is one. Thus each even partial sum is at least $\mathbf1_{r=0}$ and each odd partial sum is at most it. Take expectations. A finite $p$-group has rank zero exactly when it is trivial. $\square$

Under (eq:30006735:moments), the limiting partial sums tend, as their length increases, to \begin{equation}\label{wsref-0-65-71-3a-33-30-30-30-36-37-33-35-3a-65-75-6c-65-72} \sum_{k\geq0} \frac{(-1)^k p^{-k(k+1)/2-ku}}{\prod_{j=1}^k(1-p^{-j})} =\prod_{j\geq u+1}(1-p^{-j})=Z_u(p)^{-1}. \end{equation} The series is absolutely convergent: its denominator is bounded below by $\prod_{j\geq1}(1-p^{-j})>0$, while its numerator decays quadratically in $k$ in the exponent. The identity follows by expanding the convergent product $\prod_{j\geq0}(1-zp^{-j})$ and putting $z=p^{-u-1}$. Equation (eq:30006735:bonferroni), first with fixed $K$ and then with $K\to\infty$, proves the limiting trivial-group probability without an unjustified exchange of the $X$-limit and an infinite sum.

This also yields a quantitative finite-data test: estimated moments through $(C_p)^{2K+1}$, with rigorous error bars, give an interval for the true trivial-group probability. With exact moments the interval width is the final nonnegative term; uncertainty in estimated moments must be added to the endpoints. That observation is an exact combinatorial consequence, not a new estimate for arithmetic moments.

Attempt: derive the model moments instead of assuming them.

Let $M_n:{\mathbb{Z}}_p^{n+u}\to{\mathbb{Z}}_p^n$ have independent Haar-uniform entries, and $H_n=\operatorname{coker}M_n$. It is finite almost surely: a fixed $n\times n$ minor has nonzero determinant almost surely, since the zero set of a nonzero polynomial has Haar measure zero. For a fixed finite abelian $p$-group $A$ of rank $r\leq n$, a surjection $\phi:{\mathbb{Z}}_p^n\to A$ kills every column with probability $|A|^{-(n+u)}$. Reduction modulo $p$ and the elementary full-rank matrix count give \[ \#\operatorname{Sur}({\mathbb{Z}}_p^n,A) =|A|^n\prod_{i=0}^{r-1}(1-p^{i-n}). \] Consequently \begin{equation}\label{wsref-0-65-71-3a-33-30-30-30-36-37-33-35-3a-48-61-61-72} {\mathbb{E}} S_A(H_n)=|A|^{-u}\prod_{i=0}^{r-1}(1-p^{i-n}) \longrightarrow |A|^{-u}. \end{equation} For $r>n$ both surjection count and expectation are zero. This calculation explains exactly how the rectangular excess $u$ enters. It does not show that ideal-class relations behave like independent Haar columns; no such independence is proved for quadratic fields. Wood's random-matrix theorems rigorously address matrices, not that arithmetic identification. [E3]

Stress test.

The first moment alone is radically insufficient. For either $u=0$ or $1$, put mass $p^{-u}/(p-1)$ on $C_p$ and the remaining mass on the trivial group. This is a probability law for odd $p$, and its $C_p$ moment is exactly $p^{-u}$. But its $(C_p)^2$ moment is zero instead of $p^{-2u}$, and all higher-exponent cyclic surjection moments vanish. Thus even a correct Davenport--Heilbronn-type first moment cannot determine the entire law.

The proposed route still needs (eq:30006735:moments) for every finite target group, in the prescribed number-field ordering. Estimates for finitely many $A$, bounds without the exact main term, or a theorem with a second varying field parameter do not supply it. The script checks the Gaussian-binomial inequalities exactly and illustrates convergent normalization sums; it does not compute or certify the required arithmetic limits. No prediction for $p=2$, narrow class groups, or independence across primes is appended.

Outcome.

Outcome: Conditional reduction.

The result is a rigorous moment-to-law reduction with explicit rank and exponent tails, a second-moment uniform-integrability bound, and finite alternating-moment certificates for the trivial-group probability. The exact Haar-model computation is separately established. These do not constitute new arithmetic moment estimates, and no novelty claim is made for the underlying moment method. Proving the missing number-field surjection averages for all target groups would complete this route.

Sources

• S1 B. Alberts, Cohen-Lenstra Moments for Some Nonabelian Groups, JTNB 32 $2020$, 631-664; normalization identity in P. Majumder, arXiv:1407.3066v1. https://www.numdam.org/item/10.5802/jtnb.1137.pdf. Catalogue formulation source.
• E1 Pritam Majumder, An elementary proof of a power series identity for the weighted sum of all finite abelian p-groups, arXiv:1407.3066v1, 11 July 2014. The theorem in Section 1 and Lemma 3 (automorphism counts). https://arxiv.org/pdf/1407.3066v1.
• E2 Aaron Landesman and Ishan Levy, The Cohen--Lenstra moments over function fields via the stable homology of non-splitting Hurwitz spaces, arXiv:2410.22210v3, 1 October 2025. Introduction and Theorem 1.2.1; note the dependence of the constant-field size on the target moment. https://arxiv.org/html/2410.22210v3.
• E3 Melanie Matchett Wood, Random integral matrices and the Cohen--Lenstra Heuristics, arXiv:1504.04391; Theorem 3.1, Lemma 3.2 and Corollaries 3.3--3.4 in the consulted manuscript. https://arxiv.org/pdf/1504.04391.