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 n⩾2,m⩾2. 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 n≥3 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 n⩾2,m⩾2. 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 n≥3 and m=p is a prime.
Let R be a finite ring with 1∈R, and let R∗ denote the multiplicative group of units in R. Let k be an integer with k≥2, and let ♯S denote the cardinality of any finite set S. For any c∈R, we define
Sk(R,c):={(x1,x2,…,xk)∈(R∗)k | k∑i=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(t⩾2) and c∈Z, let
Sn;k1,k2,…,kt(c):={(x1,x2,…,xt) |1≤xi≤n/ki,(xi,n/ki)=1,i=1,2,…,t,∑ti=1kixi≡c(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)=M∗n,m. For any matrix A∈Mn,m, we define
Sn,m,k(A):={(x1,x2,…,xk)∈Gkn,m | k∑i=1xi=A}, |
and
Nn,m,k(A):=♯Sn,m,k(A). |
We also define
Mn,m,r={A∈Mn,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×n∈Mn,p and l∈{1,2,…,n}. Define
tl(A):=l∑i=1gii, |
cn,p,r(l):=♯{A∈Mn,p,r|tl(A)=0}−♯{A∈Mn,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,C∈M2,p with rank(B)=1, rank(C)=2, set
αk:=N2,p,k(O), βk:=N2,p,k(B), γk:=N2,p,k(C), k≥2. |
Let
T=[p000(p−1)2p(p+1)000−p(p−1)], S=[(p−1)2(p+1)1(p+1)2(p−1)1−p1p2−p−111−p−1]. |
Then we have
α2=(p−1)2p(p+1), β2=(p2−p−1)(p−1)p, γ2=p4−2p3−p2+3p, |
and
(αk,βk,γk)t=STk−2S−1(α2,β2,γ2)t, k≥2. |
Theorem 1.2. Let p be a prime. For any A∈Mn,p with rank(A)=r, we have
Nn,p,k(A)=(♯Gn,p)k♯Mn,pn∑l=0♯Mn,p,l⋅cn,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 m∈N∗ and m≥2 with m=pe11pe22⋅⋅⋅pett, where p1,p2,…,pt are different primes, ej≥1, j=1,2,…,t. For any A∈Mn,m, let Aj∈Mn,pejj be the reduction of A module pejj, j=1,2,...,t. Then we have
Nn,m,k(A)=t∏j=1Nn,pejj,k(Aj). |
Proof. Let (B1,B2,…,Bk)∈Sn,m,k(A). Then Bij∈Gn,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,m≅t⨁j=1Mn,pejj, Gn,m≅t⨁j=1Gn,pjii, |
So there is a unique Bi∈Gn,m such that Bij are the reduction of Bi module pejj, i=1,2,…,k, j=1,2,…,t. We have
k∑i=1Bk=A, |
i.e., (B1,B2,…,Bk)∈Sn,m,k(A).
For any prime p and e≥2, 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 e≥2. For any A∈Gn,pe, let ˜A∈Mn,p be the reduction of A module p. Then we have
Nn,pe,k(A)=p(e−1)⋅n2⋅(k−1)Nn,p,k(˜A). |
Proof. Let (B1,B2,…,Bk)∈Sn,pe,k(A). Then ˜Bi∈Gn,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×n∈Gn,p with bst∈{0,1,…,p−1}, then B is a lift of ˜B in Gn,pe if and only if B is of the form as
(kstp+bst)n×n, 0≤kst≤pe−1−1, s,t=1,2,…,n. |
So the number of lifts of ˜B in Gn,pe is p(e−1)⋅n2. So if we choose
(˜B1,˜B2,…,˜Bk)∈Sn,p,k(˜A), |
fix an lift (B1,B2,…,Bk−1) of (˜B1,˜B2,…,˜Bk−1), there is only one lift Bk of ˜Bk such that
k∑i=1Bi=A. |
So we have
Nn,pe,k(A)=p(e−1)⋅n2⋅(k−1)Nn,p,k(˜A). |
Next, we start to consider the case m=p, where p is a prime.
Lemma 2.3. Let A,B∈Mn,p with rank(A)=rank(B). Then we have
Nn,p,k(A)=Nn,p,k(B). |
Proof. By assumption, there exist C,D∈Gn,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 1≤r<n, we have
♯Gn,p=n−1∏i=0(pn−pi), ♯Mn,p,r=r−1∏i=0(pn−pi)2pr−pi. |
Next we consider the case n=2,k=2.
Theorem 2.5. Let p be a prime and A∈M2,p. Then we have
N2,p,2(A)={(p−1)2p(p+1),if rank(A)=0,(p2−p−1)(p−1)p,if rank(A)=1,p4−2p3−p2+3p,if rank(A)=2. |
Proof. Case 1. rank(A)=0, i.e., A=O. For any x1∈G2,p, O−x1=−x1∈G2,p. Hence we have
N2,p,2(A)=♯G2,p=(p−1)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 x1∈G2,p such that A−x1 is not in G2,p, i.e., rank(A−x1)=1. Assume
x1=[abcd], a,b,c,d∈Z/pZ. |
then
A−x1=[1−a−b−c−d]. |
For x1∈G2,p, we have (−c,−d)≠(0,0). Then there exists k∈Z/pZ such that
A−x1=[−kc−kd−c−d],x1=[kc+1kdcd]. |
We have
det(x1)=(kc+1)d−kcd=d≠0. |
For x1 to be uniquely determined by k,c,d, then the number of such x1 is p2(p−1). So
N2,p,2(A)=♯G2,p−p2(p−1)=(p2−p−1)(p−1)p. |
Case 3. rank(A)=2. We know that
♯M2,p,1=♯M2,p−♯M2,p,0−♯M2,p,2=p4−1−(p−1)2p(p+1)=(p+1)2(p−1). |
Choose B∈M2,p,0, C∈M2,p,1, then we have
♯G22,p=∑x∈M2,pN2,p,2(x)=♯M2,p,0⋅N2,p,2(B)+♯M2,p,1⋅N2,p,2(C)+♯M2,p,2⋅N2,p,2(A)=(p−1)2p(p+1)+(p+1)2(p−1)⋅(p2−p−1)(p−1)p+(p−1)2p(p+1)⋅N2,p,2(A). |
So we have
N2,p,2(A)=(p−1)4p2(p+1)2−(p−1)2p(p+1)−(p−1)2(p+1)⋅(p2−p−1)(p−1)p(p−1)2p(p+1)=(p−1)2p(p+1)⋅((p−1)2p(p+1)−1−(p2−p−1)(p+1))(p−1)2p(p+1)=p4−2p3−p2+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, H≅ˆH. For any function f:H→C, the function
ˆf:ˆH→C, χ↦∑x∈Hf(x)¯χ(x), ∀χ∈ˆH |
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:H→C. Then we have
f=∑χ∈ˆH1♯Hˆf(¯χ)χ. |
Consider the equation
x1+x2+⋯+xk+1=A, x1,x2,…,xk+1∈G2,p, A∈M2,p. |
Case 1. rank(A)=0, i.e., A=O. Fix an xk+1, then O−xk+1=−xk+1∈G2,p. So the number αk+1 of solutions of the equation
x1+x2+⋯+xk=−xk+1, x1,x2,…,xk+1∈G2,p, |
is ♯G2,p⋅γk=(p−1)2p(p+1)γk.
Case 2. rank(A)=1. By Theorem 2.5, the number of xk+1 such that A−xk+1∈M2,p,2 is β2, the number of xk+1 such that A−xk+1∈M2,p,1 is ♯G2,p−β2. So we have
βk+1=(♯G2,p−β2)βk+β2γk=p2(p−1)βk+(p2−p−1)(p−1)pγk |
Case 3. rank(A)=2. Use the same way as Case 2; we have
γk+1=αk+(♯G2,p−γ2−1)βk+γ2γk=αk+(p3−2p−1)βk+(p4−2p3−p2+3p)γk. |
Let
P=[00(p−1)2p(p+1)0p2(p−1)(p2−p−1)(p−1)p1p3−2p−1p4−2p3−p2+3p]. |
Then (αk,βk,γk)t=P(αk−1,βk−1,γk−1)t=⋯=Pk−2(α2,β2,γ2)t. The characteristic polynomial of P is
det(λE−P)=det[λ0−(p−1)2p(p+1)0λ−p2(p−1)−(p2−p−1)(p−1)p−1−(p3−2p−1)λ−(p4−2p3−p2+3p)]=λ(λ−p2(p−1))(λ−p(p3−2p2−p+3))−(p−1)2p(p+1)(λ−p2(p−1))−(p2−p−1)(p−1)p(p2−p−1)(p+1)λ=(λ−p)(λ−(p−1)2p(p+1))(λ+p(p−1)). |
Hence, P is similar to
T:=[p000(p−1)2p(p+1)000−p(p−1)]. |
The eigenvectors of p,(p−1)2p(p+1),−p(p−1) are respectively
e1=[(p−1)2(p+1)1−p1], e2=[111], e3=[(p+1)2(p−1)p2−p−1−p−1]. |
Define S:=(e1,e2,e3). Then we have P=STS−1.
For convenience, let M:=Mn,p, G:=Gn,p, Mr:=Mn,p,r. Let S be a finite set. For any map f:S→M and x∈M, we define
Pf(x):=♯f−1(x)♯S, |
where f−1(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)=∑x∈MPf(x)¯χ(x)=1♯S∑s∈S¯χ(f(s)),χ∈ˆM. |
By Lemma 2.6, we have
Pf(x)=1♯ˆM∑χ∈ˆMˆPf(¯χ)χ(x). |
Let ϕ:G→M be the inclusion map and
φ:Gk→M,(x1,x2,…,xk)↦x1+x2+⋯+xk. |
Clearly,
Nn,p,k(A)=(♯G)k⋅Pφ(A), ∀A∈M. | (4.1) |
For all χ∈ˆM, 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)∑x1∈G¯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×M→Z/pZ→C∗,(x1,x2)↦tr(x1x2)↦ψ(tr(x1x2)) |
is a non-degenerated symmetric bilinear map. Hence ⟨_,_⟩ induces a group isomorphism:
ρ:M→ˆM,y↦χy:=⟨_,y⟩. |
So we have
ˆPϕ(¯χy)=1♯G∑x∈G¯χy(¯x)=1♯G∑x∈Gχy(x)=1♯G∑x∈G⟨x,y⟩. |
If rank(x)=rank(y), i.e., there exits g1,g2∈G such that x=g1yg2. By the properties of the trace function, we have
∑z∈Mr⟨z,x⟩=∑z∈Mr⟨z,g1yg2⟩=∑z∈Mr⟨g2zg1,y⟩=∑z∈Mr⟨z,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 x∈M. |
For any a∈(Z/pZ)∗, it is obvious that
♯{x∈Mr|tl(x)=a}=♯{x∈Mr|tl(x)=1}. |
Note that ∑a∈Z/pZψ(a)=0, hence we have
1♯Mr∑x∈Mr⟨x,yl⟩=1♯Mr∑x∈Mrψ(tl(x))=1♯Mr∑a∈Z/pZψ(a)∑x∈Mr,tl(x)=a1=1♯Mr∑x∈Mr,tl(x)=01+1♯Mr∑a∈(Z/pZ)∗ψ(a)∑x∈Mr,tl(x)=a1=1♯Mr∑x∈Mr,tl(x)=01−1♯Mr∑x∈Mr,tl(x)=11=cn,p,r(l). |
Especially,
ˆPϕ(¯χl)=1♯G∑x∈G⟨x,yl⟩=cn,p,n(l). | (4.3) |
As rank(A)=r, by Eqs (4.2) and (4.3), we have
Pφ(A)=1♯M∑χ∈ˆMˆPφ(¯χ)χ(A)=1♯Mn∑l=0∑y∈MlˆPϕ(¯χl)k⟨A,y⟩=1♯Mn∑l=0cn,p,n(l)k∑y∈Ml⟨A,y⟩=1♯Mn∑l=0cn,p,n(l)k∑y∈Ml⟨y,A⟩=1♯Mn∑l=0cn,p,n(l)k∑y∈Ml⟨y,yr⟩=1♯Mn∑l=0cn,p,n(l)k⋅♯Ml⋅cn,p,l(r). |
Then by Eq (4.1), we have
Nn,p,k(A)=(♯G)k♯Mn∑l=0♯Ml⋅cn,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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