Counting subrings of $\mathbb Z^n$ of non-zero co-rank

TL;DR

This paper investigates the enumeration of subrings within integer lattices of higher dimension, focusing on those with a specific co-rank, to understand their structure and quantity.

Contribution

It provides a new analysis of subrings of $\

Findings

Derived formulas for counting subrings of given co-rank.

Established bounds and asymptotic behavior for the number of such subrings.

Extended understanding of the algebraic structure of subrings in integer lattices.

Abstract

In this paper we study subrings of $\mathbb Z^{n+k}$ of co-rank $k$.

Equations

$$(x_1,\dots ,x_n)+(y_1,\dots ,y_n)=(x_1+y_1,\dots ,x_n+y_n),$$

$$(x_1,\dots ,x_n)\cdot (y_1,\dots ,y_n)=(x_1\cdot y_1,\dots ,x_n\cdot y_n).$$

$$N_n(B)\sim C_{n}B(\log B{)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)-1}$$

$$B(\log B{)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)-1}\ll N_n(B){\ll }_{ϵ}{B}^{\frac{n}{2}-\frac{7}{6}+ϵ}.$$

$${\varphi }_{n+k,k}(r)=\left\{\genfrac{}{}{0ex}{}{n+k+1}{n+1}\right\}\cdot {\varphi }_n(r).$$

$$f_1(x,y)=(x,y,0,0),$$

$$f_2(x,y)=(x,y,y,0),$$

$$f_3(x,y)=(x,y,y,y),$$

$$f_4(x,y)=(x,x,y,y).$$

$$M=\left[\begin{array}{cc}{x}_{11}& x_{12}\end{array}\right].$$

$$x_{11}^2=\lambda x_{11}$$

$$x_{12}^2=\lambda x_{12}$$

$$M=\left[\begin{array}{cc}{M}^{\prime }& 0\\ v& x_{k,k+1}\end{array}\right].$$

$$R_k^2=\sum _{m=1}^{m=k}{\lambda }_m{R}_m.$$

$$x_{k,k+1}^2={\lambda }_k{x}_{k,k+1}$$

$$x_{k,j}^2={\lambda }_k{x}_{k,j}$$

$$R_l\cdot R_k=\sum _{m=1}^{m=k}{\gamma }_m{R}_m$$

$$x_{k,i}{x}_{l,i}=\sum _{m=1}^{m=k}{\gamma }_m{x}_{m,i}$$

$$x_{k,j}{x}_{l,j}=\sum _{m=1}^{m=k}{\gamma }_m{x}_{m,j}$$

$$0={\gamma }_k{x}_{k,k+1}$$

$$g:{\mathbb{Z}}^n\to {\mathbb{Z}}^{n+k}$$

$$\#({\mathbb{Z}}^{n+k}/g(L){)}_{\mathrm{tor}}=[{\mathbb{Z}}^n:L].$$

$$g(x_1,\dots ,x_n)=\sum _{i=1}^n{x}_i\left(\sum _{j\in A_i^g}{f}_j\right).$$

$$g_{\mathcal{P}}(x_1,\dots ,x_n)=\sum _{i=1}^n{x}_i\left(\sum _{j\in A_i}{f}_j\right).$$

$$(a,b,c)\mapsto (a,b,0,0,a,c,b,b),$$

Full text

1. Introduction

Let $\mathbb{Z}^{n}$ be the set of $n$-tuples $(x_{1},\dots,x_{n})$ of integers. This set comes with a natural addition and multiplication given by

[TABLE]

and

[TABLE]

Under these operations $\mathbb{Z}^{n}$ is a ring. As is well known the ring $\mathbb{Z}^{n}$ has a simple additive group structure, but when it comes to its multiplicative structure there are some very easy-to-state basic questions that we do not know how to answer. For example, let $f_{n}(r)$ be the number of subrings $R$ of $\mathbb{Z}^{n}$ with identity of index $r$. Necessarily then, $R$ is a free $\mathbb{Z}$-module of rank $n$. The counting function $f_{n}(r)$ and associated generating series $F_{n}(s):=\sum_{r=1}^{\infty}f_{n}(r)r^{-s}$ are basic objects of interest.

The general theory developed by Grunewald, Segal and Smith [5] shows that $F_{n}(s)$ can be expressed as as Euler product of rational functions of $p^{-s}$ over all primes $p$, but only for $n\leq 4$ has this rational function been computed explicitly. For $n=2$ this expression is immediate. It is originally due to Datskovksy and Wright [4] for $n=3$ and Nakagawa [8] for $n=4.$ In fact, these authors studied the more general problem understanding the distribution of orders in cubic or quartic algebras, a particular case of which was the computation of the generating series $F_{3}(s)$ in [4] and $F_{4}(s)$ in [8]. Liu [7] proved a number of interesting theorems about $f_{n}(r)$, including the computation of $F_{n}(s):=\sum_{r=1}^{\infty}f_{n}(r)r^{-s}$ for $n\leq 4$ by an alternative method.

For $n>4$ the situation is considerably more complicated. Kaplan, Marcinek, and Takloo-Bighash [6], by using the methods of $p$-adic integration, obtained results for the location and order of the rightmost pole of $F_{5}(s)$ without explicitly computing the series. They also obtained estimates for the location of the first pole of $F_{n}(s)$ for $n>5$. One of the reasons to study the analytic properties of the generating series $F_{n}(s)$ is to find asymptotic formulae for $N_{n}(B)=\sum_{r\leq B}f_{n}(r)$. The theory of $p$-adic integration [5] shows that $N_{n}(B)$ grows like a non-zero constant $C_{n}$ multiplied by $B^{\alpha(n)}(\log B)^{b(n)-1}$ for $\alpha(n)\in\mathbb{Q}$ and $b(n)\in\mathbb{N}$. Combining the results of [4, 8, 6] we know the following about the behavior of $N_{n}(B)$:

Theorem 1**.**

If $n\leq 5$ there is a constant $C_{n}$ such that

[TABLE]

as $B\to\infty$. If $n\geq 6$, for any $\epsilon>0$ we have

[TABLE]

In fact, results of Brakenhoff [3] and Atanasov-Kaplan-Krakoff-Menzel [1] give slightly better bounds for $n\geq 6.$

As mentioned above $f_{n}(r)$ counts full rank $\mathbb{Z}$-submodules of $\mathbb{Z}^{n}$ that are of index $r$. A natural question to ask is whether one can quantify the distribution of subrings of $\mathbb{Z}^{n}$ which as $\mathbb{Z}$-submodules are not of rank $n$. Let us make this precise. Let $\phi_{n}(r)$ be the number of full-rank sublattices of $\mathbb{Z}^{n}$ which are closed under the multiplication of $\mathbb{Z}^{n}$. It’s a well-known fact (e.g., Proposition 2.3 of [7]) that for each $n\geq 2,r\geq 1$ we have $f_{n}(r)=\phi_{n-1}(r)$. It turns out that for many purposes the function $\phi_{n}(r)$ is a more convenient function to work with—and in fact the theory developed in [5] deals with the function $\phi_{n}(r)$.

We now define an analogue of the function $\phi_{n}(r)$ for lattices of non-zero co-rank. For $0\leq k\leq n$, define $\phi_{n,k}(r)$ be the number of sublattices $L$ of $\mathbb{Z}^{n}$ which have the following properties:

• •

The lattice $L$ is closed under multiplication;

• •

as a $\mathbb{Z}$-submodule, $L$ is of co-rank $k$ in $\mathbb{Z}^{n}$;

• •

the size of the torsion subgroup of $\mathbb{Z}^{n}/L$ is equal to $r$.

Clearly, $\phi_{n,0}(r)=\phi_{n}(r)$. It turns out that the function $\phi_{n,k}(r)$ and $\phi_{n}(r)$ have a simple relationship. The following theorem is our main result.

Theorem 2**.**

For all $n,k,r$ we have

[TABLE]

Here, for natural numbers $u,v$, $\left\{{u\atop v}\right\}$ is the Stirling number of second kind defined as the number of partitions of a set with $u$ elements into $v$ non-empty subsets.

The main step in the proof of this theorem is a rigidity result (Theorem 6) which determines exactly what types of lattices contribute to the counting function $\phi_{n+k,k}(r)$. The rest of the proof consists of a combinatorial argument counting these lattices. For information on Stirling numbers of the second kind, see [2], especially Ch. 2, §3.

The rigidity result mentioned above is the statement that matrices corresponding to multiplicative sublattices will be of very special shape. The upshot of this result is that multiplicative sublattices of non-zero co-rank in $\mathbb{Z}^{n}$ are all obtained from full rank multiplicative sublattices in various $\mathbb{Z}^{m}$’s for $m<n$ in very specific ways. Let us illustrate the results we are about to prove using co-rank two multiplicative sublattices in $\mathbb{Z}^{4}$.

Define four maps $\mathbb{Z}^{2}\to\mathbb{Z}^{4}$ by the following formulae:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We can make more maps $\mathbb{Z}^{2}\to\mathbb{Z}^{4}$ by considering maps of the form $\tau\circ f_{j}\circ\sigma$ for $\sigma\in S_{2},\tau\in S_{4}$—we call these maps acceptable. For example, the map that sends $(x,y)$ to $(y,x,0,x)$ is acceptable. A consequence of our rigidity result is that if $L$ is a multiplicative sublattice of co-rank two in $\mathbb{Z}^{4}$, then there is a multiplicative sublattice $L^{\prime}$ of full rank in $\mathbb{Z}^{2}$ such that $L=f(L^{\prime})$ for some acceptable map $f$. Furthermore, the size of the torsion subgroup of $\mathbb{Z}^{4}/L$ is equal to the index of $L^{\prime}$ in $\mathbb{Z}^{2}$. We will see that the scenario described here is completely general.

Theorem 2 was discovered thanks to the Online Encyclopedia of Integer Sequences (OEIS). We computed a few values of the function $\sigma(n,k)$ by hand and then a search through OEIS revealed the connection to the Stirling Numbers of the Second Kind. These numbers appear under sequence A008277 in the Encyclopedia [9].

The first named author is partially supported by NSF DMS 1601289. The second author wishes to thank the Simons Foundation for partial support of his work through a Collaboration Grant. The authors also wish to thank Nathan Kaplan for helpful conversations.

This paper is organized as follows. In §2 we review basic definitions and prove the rigidity theorem. We prove the main theorem in §3.

2. Rigidity Theorem

A lattice is a $\mathbb{Z}$-submodule of some $\mathbb{Z}^{n}$. When referring to a specific $\mathbb{Z}^{n}$ we usually speak of a sublattice. We call a sublattice $L$ of $\mathbb{Z}^{n}$ a multiplicative sublattice if for every $u,v\in L$ we have $u\cdot v\in L$. A multiplicative sublattice $L$ is a subring if it contains the identity element $(1,\dots,1)$. We refer the reader to Liu [7] for basic properties of multiplicative lattices of full rank in $\mathbb{Z}^{n}$.

Let $L$ be a lattice of rank $m$ in $\mathbb{Z}^{n}$. We define the co-rank of $L$ to be the integer $n-m$. The following lemma is an easy consequence of row operations.

Lemma 3**.**

Given a lattice $L$ in $\mathbb{Z}^{n}$ of co-rank $k$ there is an $(n-k)\times n$ integral matrix $M=(x_{ij})$ such that $x_{ij}=0$ whenever $j-i>k$, and with the property that the rows of $M$ generate $L$.

Note that the matrix $M$ as in the lemma is not unique. In fact, if $A$ is any $(n-k)\times(n-k)$ lower triangular integral matrix with determinant $1$, then $AM$ is another matrix that satisfies the conditions of the lemma.

Let $M$ be the matrix corresponding to the lattice $L$ of co-rank $k$ as in Lemma 3. Then $L$ is multiplicative if and only if for every two rows $v,w$ of $M$, $v\cdot w\in L$.

Proposition 4**.**

Let L be a multiplicative sublattice of $\mathbb{Z}^{n}$ of co-rank 1. Then L has a basis which forms the rows of a $(n-1)\times n$ matrix M such that $M_{ij}=0$ if $i<j-1$ and M has a column of zeros or two columns of M are identical.

Proof.

We prove this using induction on $n$. If $n=1$ then there is no sublattice of co-rank $1$ so the result is vacuously true. So we consider the case $n=2$. Any multiplicative sublattice $L$ of co-rank $1$ has rank 1 and therefore is generated by a non-zero row vector of length $2$,

[TABLE]

As $L$ is multiplicative, $M.M$ should be a scalar multiple of $M$. Hence we get the following equations:

[TABLE]

Note that both $x_{11}$ and $x_{12}$ can’t simultaneously be zero. If either of them are zero we get a zero column as desired and if both are non-zero we get that $x_{11}=\lambda=x_{12}$ and in that case both columns are identical.

Now we assume that the result holds for $n=k$ and show that it is true for $n=k+1$ Let $L$ be a multiplicative sublattice of $\mathbb{Z}^{k+1}$ of co-rank $1$. Then $L$ has a basis which forms the rows of a matrix $M=(x_{ij})$ such that $x_{ij}=0$ for $i<j-1$. Now $M$ can be written as

[TABLE]

If $x_{k,k+1}=0$ then we have a column of zeros and we have nothing to prove. So from here on we assume that $x_{k,k+1}\neq 0$. Clearly $M^{\prime}$ represents a multiplicative sublattice of $\mathbb{Z}^{k}$. By the induction hypothesis $M^{\prime}$ has a column of zeros or a pair of identical columns.

**Case 1 : ** $M^{\prime}$ has a column of zeros.

Suppose the $j$th column of $M^{\prime}$ is 0. If $x_{k,j}=0$ we are done. So we assume that $x_{k,j}\neq 0$. Consider the product of the bottom row $R_{k}$ of $M$ with itself. Write

[TABLE]

So we have the following equations.

[TABLE]

As $x_{i,j}=0$ for $i\neq k$. Since both $x_{k,k+1}$ and $x_{k,j}$ are non-zero we have $x_{k,j}=\lambda_{k}=x_{k,k+1}$ which implies that the jth and $(k+1)$st columns are identical as all other entries are [math].

**Case 2 : ** $M^{\prime}$ has a pair of identical columns.

Let the ith and jth columns of $M^{\prime}$ be equal. We can assume that these are non-zero columns as the first case already deals with zero columns. Therefore there is $l<k$ such that $x_{l,i}=x_{l,j}\neq 0$. Now

[TABLE]

So we have

[TABLE]

$\gamma_{k}=0$ as $x_{k,k+1}\neq 0$. This and the fact that $x_{m,i}=x_{m,j}$ for $m<k$ gives us that each term in the summations in (3a) and (3b) are equal which implies that the sums are equal. Therefore we have that in fact $x_{k,i}x_{l,i}=x_{k,j}x_{l,j}$. Since $x_{l,i}=x_{l,j}\neq 0$ we have $x_{k,i}=x_{k,j}$. So that the $i$th and $j$th columns of $M$ are identical.

∎

Corollary 5**.**

Any basis of a multiplicative lattice L of co-rank 1 will form the rows of an $(n-1)\times n$ matrix M with $n-1$ distinct non-zero columns.

Proof.

The property of having a column of zeros or two identical columns is invariant under elementary row operations. This means that any matrix whose rows are the basis of a multiplicative sublattice of co-rank $1$ of $\mathbb{Z}^{n}$ will have this property. ∎

Theorem 6** (Rigidity).**

Let $L$ be a multiplicative sublattice of $\mathbb{Z}^{n}$ of co-rank k, then every basis of $L$ forms the rows of a $(n-k)\times n$ matrix $M$ with exactly $n-k$ distinct non-zero columns.

Proof.

The matrix $M$ has column rank $n-k$. Let’s say the first $n-k$ columns are linearly independent, and hence, distinct and nonzero. Let $M^{(i)}$ be the $(n-k)\times(n-k+1)$-dimensional matrix obtained by appending the $i^{th}$ column of $M$ to the first $n-k$ columns of $M$. Since the rows $M^{(i)}$ generate a multiplicative sublattice of $\mathbb{Z}^{n-k+1}$ of corank 1, the previous corollary implies that $M^{(i)}$ has exactly $n-k$ distinct nonzero columns. Hence the $i^{th}$ column of $M$ must be equal to one of the first $n-k$ columns or 0. Since this is true for all $i,n-k<i\leq n$, we conclude that the full matrix has exactly $n-k$ distinct nonzero columns. ∎

3. Proof of Theorem 2

We begin with a definition.

Definition 7**.**

An injective map

[TABLE]

of the form $g(x_{1},\dots,x_{n})=(y_{1},y_{2},\dots,y_{n+k})$ with each $y_{j}$ either equal to some $x_{i}$ or [math] is called acceptable.

Theorem 6 can be formulated as follows:

Theorem 8**.**

Any multiplicative sublattice of co-rank $k$ in $\mathbb{Z}^{n+k}$ is of the form $g(L)$ where $g:\mathbb{Z}^{n}\to\mathbb{Z}^{n+k}$ is an acceptable map and $L$ is a multiplicative sublattice of full rank in $\mathbb{Z}^{n}$.

The next observation is simple but essential for what follows.

Lemma 9**.**

For any acceptable map $g$ and any sublattice $L$ in $\mathbb{Z}^{n}$ of rank $n$, we have

[TABLE]

Proof.

Let the Smith Normal form of the lattice L be D. Then $[\mathbb{Z}^{n}:L]=\text{Det(D)}$. Since in any basis the distinct columns of g(L) are the same as those of L, we have that the Smith Normal Form of L is $[D|0]$. So that $\#(\mathbb{Z}^{n+k}/g(L))_{\rm{tor}}=[\mathbb{Z}^{n}:L]$. ∎

Two acceptable maps $g_{1},g_{2}:\mathbb{Z}^{n}\to\mathbb{Z}^{n+k}$ are called equivalent if there is a permutation $\tau\in S_{n}$ such that $g_{1}=g_{2}\circ\tau$. We next describe a complete set of representatives for this equivalence relation.

Let $g:\mathbb{Z}^{n}\to\mathbb{Z}^{n+k}$ be an acceptable map and $\{f_{i}\}$ the standard basis for $\mathbb{Z}^{n+k}$. By definition of acceptable we can write

[TABLE]

for a collection of subsets $\{A_{1}^{g},\dots,A_{n}^{g}\}$ of $\{1,\dots,n+k\}$. In fact, if we define $A_{0}^{g}=\{0,\dots,n+k\}\setminus\bigcup_{i=1}^{n}A_{i}^{g}$, then $\mathcal{P}^{g}:=\{A_{0}^{g},\dots,A_{n}^{g}\}$ is a partition of $\{0,\dots,n+k\}$. We will call an acceptable function $g$ ordered if $\min A_{i}^{g}<\min A_{j}^{g}$ whenever $i<j$. Given an arbitrary acceptable map $g$ there exists exactly one permutation $\tau\in S_{n}$ for which $g\circ\tau$ is ordered. That is,

Lemma 10**.**

The set of ordered acceptable maps is a set of representatives for the equivalence classes of acceptable maps $\mathbb{Z}^{n}\to\mathbb{Z}^{n+k}$ under the action of $S_{n}$.

Going in the other direction, to a set partition $\mathcal{P}$ of $\{0,\dots,n+k\}$ into $n+1$ parts, we may associate an ordered acceptable map $g_{\mathcal{P}}$ as follows. Begin by ordering $\mathcal{P}=\{A_{0},A_{1},...,A_{n}\}$ in the following way: if $i<j$ then $\min(A_{i})<\min(A_{j})$. In particular, $0\in A_{0}$. Define

[TABLE]

For example $\{\{0,3,4\},\{1,5\},\{2,7,8\},\{6\}\}$ corresponds to the map from $\mathbb{Z}^{3}$ to $\mathbb{Z}^{8}$ which sends

[TABLE]

i.e., [math] is in the $3$ and $4$ spot, ‘$a$’ in the $1$ and $5$ entries, ‘$b$’ in the $2$,$7$ and $8$ entries and ‘$c$’ in the $6$th entry.

It is clear that the maps $g\mapsto\mathcal{P}^{g}$ and $\mathcal{P}\mapsto g_{P}$ are inverse to one another and provide a bijection between ordered acceptable maps $\mathbb{Z}^{n}\to\mathbb{Z}^{n+k}$ and set partitions of $\{0,\dots,n+k\}$ into $n+1$ parts. This and the definition of Stirling numbers of the second kind lead to the next corollary:

Corollary 11**.**

The number of equivalence classes of acceptable maps $\mathbb{Z}^{n}\to\mathbb{Z}^{n+k}$ is equal to $\left\{{n+k+1\atop n+1}\right\}$.

The final step in the proof of Theorem 2 is a refinement of Theorem 8.

Proposition 12**.**

Any multiplicative sublattice of co-rank $k$ in $\mathbb{Z}^{n+k}$ is of the form $g(L)$ where $g:\mathbb{Z}^{n}\to Z^{n+k}$ is an ordered acceptable map and $L$ is a multiplicative sublattice of full rank in $\mathbb{Z}^{n}$ . Moreover, such $g$ and $L$ are uniquely determined.

Proof.

A consequence of Theorem 6 is that any multiplicative sublattice $L^{\prime}$ in $\mathbb{Z}^{n+k}$ will correspond to some partition $\mathcal{P}$ of $\{0,1,...n+k\}$ into $n+1$ parts. This partition is obtained by the same method that associated a partition to an acceptable map. The partition $\mathcal{P}$ corresponds to a unique ordered acceptable map $f_{\mathcal{P}}$. The lattice $L^{\prime}$ is clearly in the range of $f_{\mathcal{P}}$ so ${f_{\mathcal{P}}}^{-1}(L^{\prime})$ is the unique lattice $L$ in $\mathbb{Z}^{n}$ which maps to $L^{\prime}$ under $f_{\mathcal{P}}$. ∎

Proof of Theorem 2.

Combine Proposition 12 with Corollary 11. ∎