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Research article

On the sumsets of units in a ring of matrices over Z/mZ

  • Received: 23 December 2024 Revised: 24 February 2025 Accepted: 25 February 2025 Published: 05 March 2025
  • Let Mn,m:=Matn(Z/mZ) be the ring of matrices of n×n over Z/mZ and Gn,m:=Gln(Z/mZ) be the multiplicative group of units of Mn,m with n2,m2. In this paper, we obtain an exact formula for the number of representations of any element of M2,m as the sum of k units in M2,m. Furthermore, by using the technique of Fourier transformation, we also give a formula for the case n3 and m=p is a prime.

    Citation: Yifan Luo, Kaisheng Lei, Qingzhong Ji. On the sumsets of units in a ring of matrices over Z/mZ[J]. Electronic Research Archive, 2025, 33(3): 1323-1332. doi: 10.3934/era.2025059

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  • Let Mn,m:=Matn(Z/mZ) be the ring of matrices of n×n over Z/mZ and Gn,m:=Gln(Z/mZ) be the multiplicative group of units of Mn,m with n2,m2. In this paper, we obtain an exact formula for the number of representations of any element of M2,m as the sum of k units in M2,m. Furthermore, by using the technique of Fourier transformation, we also give a formula for the case n3 and m=p is a prime.



    Let R be a finite ring with 1R, and let R denote the multiplicative group of units in R. Let k be an integer with k2, and let S denote the cardinality of any finite set S. For any cR, we define

    Sk(R,c):={(x1,x2,,xk)(R)k | ki=1xi=c},

    and

    Nk(R,c):=Sk(R,c).

    For a positive integer n, let Z/nZ be the ring of residue classes modulo n. In 2000, Deaconescu [1] obtained a formula for N2(Z/nZ,c). In 2009, Sander [2] gave a generalization of the above result. In fact, for any integer c, he determined the number of representations of c as a sum of two units (two nonunits, a unit, and a nonunit, respectively) in Z/nZ.

    For a positive integer n with divisors k1,k2,...,kt(t2) and cZ, let

    Sn;k1,k2,,kt(c):={(x1,x2,,xt) |1xin/ki,(xi,n/ki)=1,i=1,2,,t,ti=1kixic(modn)}.

    We define Nn;k1,k2,,kt(c):=Sn;k1,k2,,kt(c).

    In 2013, Sander and Sander [3] gave a formula for Nn;k1,k2(c). In 2014, Sun and Yang [4] obtained a formula for Nn;k1,k2,,kt(c). In 2015, Yang and Tang [5] extended Sander's results to the quadratic case. In 2017, Ji and Zhang [6] extended Sander's results to the residue ring of a Dedekind ring.

    In this paper, we shall extend the above results to the ring of matrices over Z/mZ. Let Mn,m:=Matn(Z/mZ), Gn,m:=Gln(Z/mZ)=Mn,m. For any matrix AMn,m, we define

    Sn,m,k(A):={(x1,x2,,xk)Gkn,m | ki=1xi=A},

    and

    Nn,m,k(A):=Sn,m,k(A).

    We also define

    Mn,m,r={AMn,m | rank(A)=r},   r=0,1,,n.

    Clearly, Mn,m,0={O}, Mn,p,n=Gn,p where p is a prime.

    By Lemmas 2.1 and 2.2, it is sufficient to compute Nn,p,k(A), where p is a prime. Let A=(gij)n×nMn,p and l{1,2,,n}. Define

    tl(A):=li=1gii,
     cn,p,r(l):={AMn,p,r|tl(A)=0}{AMn,p,r|tl(A)=1}Mn,p,r.

    In this paper, our main results are the followings:

    Theorem 1.1. Let p be a prime. For any B,CM2,p with rank(B)=1, rank(C)=2, set

    αk:=N2,p,k(O), βk:=N2,p,k(B), γk:=N2,p,k(C),   k2.

    Let

    T=[p000(p1)2p(p+1)000p(p1)], S=[(p1)2(p+1)1(p+1)2(p1)1p1p2p111p1].

    Then we have

    α2=(p1)2p(p+1), β2=(p2p1)(p1)p, γ2=p42p3p2+3p,

    and

    (αk,βk,γk)t=STk2S1(α2,β2,γ2)t,   k2.

    Theorem 1.2. Let p be a prime. For any AMn,p with rank(A)=r, we have

    Nn,p,k(A)=(Gn,p)kMn,pnl=0Mn,p,lcn,p,n(l)kcn,p,l(r).

    This paper is organized as follows: In Section 2, we shall prove some lemmas that will be used in the proofs of our main results. In Sections 3 and 4, we shall give the proofs of Theorems 1.1 and 1.2, respectively.

    Lemma 2.1. Let mN and m2 with m=pe11pe22pett, where p1,p2,,pt are different primes, ej1, j=1,2,,t. For any AMn,m, let AjMn,pejj be the reduction of A module pejj, j=1,2,...,t. Then we have

    Nn,m,k(A)=tj=1Nn,pejj,k(Aj).

    Proof. Let (B1,B2,,Bk)Sn,m,k(A). Then BijGn,pejj, i=1,2,,k, j=1,2,,t, where Bij is the reduction of Bi module pejj. It is clear that (B1j,B2j,,Bkj)Sn,pejj,k(Aj), j=1,2,,t. Conversely, let (B1j,B2j,,Bkj)Sn,pejj,k(Aj),j=1,2,,t. By the Chinese remainder theorem, the reduction induces two isomorphisms:

    Mn,mtj=1Mn,pejj,  Gn,mtj=1Gn,pjii,

    So there is a unique BiGn,m such that Bij are the reduction of Bi module pejj, i=1,2,,k, j=1,2,,t. We have

    ki=1Bk=A,

    i.e., (B1,B2,,Bk)Sn,m,k(A).

    For any prime p and e2, the next lemma shows the relation between Nn,pe,k(A) and Nn,p,k(A).

    Lemma 2.2. Let p be a prime and e2. For any AGn,pe, let ˜AMn,p be the reduction of A module p. Then we have

    Nn,pe,k(A)=p(e1)n2(k1)Nn,p,k(˜A).

    Proof. Let (B1,B2,,Bk)Sn,pe,k(A). Then ˜BiGn,p, where ˜Bi are the reduction of Bi module p, i=1,2,,k. It is clear that (˜B1,˜B2,,˜Bk)Sn,p,k(˜A). Conversely, let ˜B=(bst)n×nGn,p with bst{0,1,,p1}, then B is a lift of ˜B in Gn,pe if and only if B is of the form as

    (kstp+bst)n×n,   0kstpe11, s,t=1,2,,n.

    So the number of lifts of ˜B in Gn,pe is p(e1)n2. So if we choose

    (˜B1,˜B2,,˜Bk)Sn,p,k(˜A),

    fix an lift (B1,B2,,Bk1) of (˜B1,˜B2,,˜Bk1), there is only one lift Bk of ˜Bk such that

    ki=1Bi=A.

    So we have

    Nn,pe,k(A)=p(e1)n2(k1)Nn,p,k(˜A).

    Next, we start to consider the case m=p, where p is a prime.

    Lemma 2.3. Let A,BMn,p with rank(A)=rank(B). Then we have

    Nn,p,k(A)=Nn,p,k(B).

    Proof. By assumption, there exist C,DGn,p such that CAD=B. It is obvious that the map

    Sn,p,k(A)Sn,p,k(B), (x1,x2,,xk)(Cx1D,Cx2D,,CxkD)

    is bijective. Hence Nn,p,k(A)=Nn,p,k(B).

    It is well known that we have the following results.

    Lemma 2.4. [7] For any 1r<n, we have

    Gn,p=n1i=0(pnpi), Mn,p,r=r1i=0(pnpi)2prpi.

    Next we consider the case n=2,k=2.

    Theorem 2.5. Let p be a prime and AM2,p. Then we have

    N2,p,2(A)={(p1)2p(p+1),if rank(A)=0,(p2p1)(p1)p,if rank(A)=1,p42p3p2+3p,if rank(A)=2.

    Proof. Case 1. rank(A)=0, i.e., A=O. For any x1G2,p, Ox1=x1G2,p. Hence we have

    N2,p,2(A)=G2,p=(p1)2p(p+1).

    Case 2. rank(A)=1. By Lemma 2.3, it is sufficient to compute N2,p,2(A), where A=[1000]. To compute N2,p,2(A), we only need to compute the number of x1G2,p such that Ax1 is not in G2,p, i.e., rank(Ax1)=1. Assume

    x1=[abcd],   a,b,c,dZ/pZ.

    then

    Ax1=[1abcd].

    For x1G2,p, we have (c,d)(0,0). Then there exists kZ/pZ such that

    Ax1=[kckdcd],x1=[kc+1kdcd].

    We have

    det(x1)=(kc+1)dkcd=d0.

    For x1 to be uniquely determined by k,c,d, then the number of such x1 is p2(p1). So

    N2,p,2(A)=G2,pp2(p1)=(p2p1)(p1)p.

    Case 3. rank(A)=2. We know that

    M2,p,1=M2,pM2,p,0M2,p,2=p41(p1)2p(p+1)=(p+1)2(p1).

    Choose BM2,p,0, CM2,p,1, then we have

    G22,p=xM2,pN2,p,2(x)=M2,p,0N2,p,2(B)+M2,p,1N2,p,2(C)+M2,p,2N2,p,2(A)=(p1)2p(p+1)+(p+1)2(p1)(p2p1)(p1)p+(p1)2p(p+1)N2,p,2(A).

    So we have

    N2,p,2(A)=(p1)4p2(p+1)2(p1)2p(p+1)(p1)2(p+1)(p2p1)(p1)p(p1)2p(p+1)=(p1)2p(p+1)((p1)2p(p+1)1(p2p1)(p+1))(p1)2p(p+1)=p42p3p2+3p.

    Next, we introduce the Fourier Transformation. Let H be a finite abelian group, and let ˆH=:HomH(H,C) be the character group of H. Clearly, HˆH. For any function f:HC, the function

    ˆf:ˆHC, χxHf(x)¯χ(x),   χˆH

    is called the Fourier Transformation of f. The transformation can be inverted. We have

    Lemma 2.6. [8] Let ˆf be the Fourier Transformation of f:HC. Then we have

    f=χˆH1Hˆf(¯χ)χ.

    Consider the equation

    x1+x2++xk+1=A,   x1,x2,,xk+1G2,p, AM2,p.

    Case 1. rank(A)=0, i.e., A=O. Fix an xk+1, then Oxk+1=xk+1G2,p. So the number αk+1 of solutions of the equation

    x1+x2++xk=xk+1,   x1,x2,,xk+1G2,p,

    is G2,pγk=(p1)2p(p+1)γk.

    Case 2. rank(A)=1. By Theorem 2.5, the number of xk+1 such that Axk+1M2,p,2 is β2, the number of xk+1 such that Axk+1M2,p,1 is G2,pβ2. So we have

    βk+1=(G2,pβ2)βk+β2γk=p2(p1)βk+(p2p1)(p1)pγk

    Case 3. rank(A)=2. Use the same way as Case 2; we have

    γk+1=αk+(G2,pγ21)βk+γ2γk=αk+(p32p1)βk+(p42p3p2+3p)γk.

    Let

    P=[00(p1)2p(p+1)0p2(p1)(p2p1)(p1)p1p32p1p42p3p2+3p].

    Then (αk,βk,γk)t=P(αk1,βk1,γk1)t==Pk2(α2,β2,γ2)t. The characteristic polynomial of P is

    det(λEP)=det[λ0(p1)2p(p+1)0λp2(p1)(p2p1)(p1)p1(p32p1)λ(p42p3p2+3p)]=λ(λp2(p1))(λp(p32p2p+3))(p1)2p(p+1)(λp2(p1))(p2p1)(p1)p(p2p1)(p+1)λ=(λp)(λ(p1)2p(p+1))(λ+p(p1)).

    Hence, P is similar to

    T:=[p000(p1)2p(p+1)000p(p1)].

    The eigenvectors of p,(p1)2p(p+1),p(p1) are respectively

    e1=[(p1)2(p+1)1p1], e2=[111], e3=[(p+1)2(p1)p2p1p1].

    Define S:=(e1,e2,e3). Then we have P=STS1.

    For convenience, let M:=Mn,p, G:=Gn,p, Mr:=Mn,p,r. Let S be a finite set. For any map f:SM and xM, we define

    Pf(x):=f1(x)S,

    where f1(x) is the set of all the inverse images of x. Let ˆM=:HomM(M,C) be the additive character group of M. Then we have

    ˆPf(χ)=xMPf(x)¯χ(x)=1SsS¯χ(f(s)),χˆM.

    By Lemma 2.6, we have

    Pf(x)=1ˆMχˆMˆPf(¯χ)χ(x).

    Let ϕ:GM be the inclusion map and

    φ:GkM,(x1,x2,,xk)x1+x2++xk.

    Clearly,

    Nn,p,k(A)=(G)kPφ(A),   AM. (4.1)

    For all χˆM, we have

    ˆPφ(χ)=1(G)k(x1,x2,,xk)Gk¯χ(x1+x2++xk)=1(G)k(x1,x2,,xk)Gk¯χ(x1)¯χ(x2)¯χ(xk)=(1(G)x1G¯χ(x1))k=ˆPϕ(χ)k.

    Next, we consider ˆPϕ(χ). Let ψ be a nontrivial additive character of Z/pZ. Then the map

    _,_:M×MZ/pZC,(x1,x2)tr(x1x2)ψ(tr(x1x2))

    is a non-degenerated symmetric bilinear map. Hence _,_ induces a group isomorphism:

    ρ:MˆM,yχy:=_,y.

    So we have

    ˆPϕ(¯χy)=1GxG¯χy(¯x)=1GxGχy(x)=1GxGx,y.

    If rank(x)=rank(y), i.e., there exits g1,g2G such that x=g1yg2. By the properties of the trace function, we have

    zMrz,x=zMrz,g1yg2=zMrg2zg1,y=zMrz,y. (4.2)

    Specially, we have

    ˆPϕ(¯χx)=ˆPϕ(¯χy).

    Let l{1,2,,n}. Set yl:=[IlOOO]M and χl:=χyl, where Il is the identity matrix of order l. Then

    χl(x)=ψ(tl(x)),   for all xM.

    For any a(Z/pZ), it is obvious that

    {xMr|tl(x)=a}={xMr|tl(x)=1}.

    Note that aZ/pZψ(a)=0, hence we have

    1MrxMrx,yl=1MrxMrψ(tl(x))=1MraZ/pZψ(a)xMr,tl(x)=a1=1MrxMr,tl(x)=01+1Mra(Z/pZ)ψ(a)xMr,tl(x)=a1=1MrxMr,tl(x)=011MrxMr,tl(x)=11=cn,p,r(l).

    Especially,

    ˆPϕ(¯χl)=1GxGx,yl=cn,p,n(l). (4.3)

    As rank(A)=r, by Eqs (4.2) and (4.3), we have

    Pφ(A)=1MχˆMˆPφ(¯χ)χ(A)=1Mnl=0yMlˆPϕ(¯χl)kA,y=1Mnl=0cn,p,n(l)kyMlA,y=1Mnl=0cn,p,n(l)kyMly,A=1Mnl=0cn,p,n(l)kyMly,yr=1Mnl=0cn,p,n(l)kMlcn,p,l(r).

    Then by Eq (4.1), we have

    Nn,p,k(A)=(G)kMnl=0Mlcn,p,n(l)kcn,p,l(r).

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The article is supported by NSFC (Nos. 12071209, 12231009). We would like to thank the referees for reading the manuscript carefully and providing valuable comments and suggestions.

    The authors declare no conflicts of interest.



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