# Cogredient Standard Forms of Symmetric Matrices over Galois Rings of Odd Characteristic

Authors: Yonglin Cao
Journal: ISRN Algebra
 (2012)

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## Abstract

LetR=GR(ps,psm) be a Galois ring of characteristic ps and cardinality psm, where s and m are positive integers and p is an odd prime number. Two kinds of cogredient standard forms of symmetric matrices over R are given, and an explicit formula to count the number of all distinct cogredient classes of symmetric matrices over R is obtained.


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## Body

## 1. Introduction and Preliminaries

Letp be a prime number, s and m be positive integers, and R=GR(ps,psm) a Galois ring of characteristic ps and cardinality psm. Then GR(ps,psm) is isomorphic to the ring ℤps[x]/(h(x)) for any basic irreducible polynomial h(x) of degree m over ℤps. It is clear that R=𝔽pm, that is, a finite field of pm elements, if s=1, and R=ℤps, that is the ring of residue classes of ℤ modulo its ideal psℤ, if m=1.We denote byR* the group of units of R. R is a local ring with the maximal ideal (p)=pR, and all ideals of R are given by (0)=(ps)⊂(ps-1)⊂⋯⊂(p)⊂(p0)=R. By [1, Theorem 14.8], there exists an element ξ∈R* of multiplicative order pm-1, which is a root of a basic primitive polynomial h(x) of degree m over ℤps and dividing xpm-1-1 in ℤps[x], and every element a∈R can be written uniquely as
(1.1)a=a0+a1p+⋯+an-1pn-1,a0,a1,…,an-1∈T,
where 𝒯={0,1,ξ,…,ξpm-2}. Moreover, a is a unit if and only if a0≠0, and a is a zero divisor or 0 if and only if a0=0. Define the p-exponent of a by τ(0)=s and τ(a)=i if a=aipi+⋯+an-1pn-1 with ai≠0. By [1, Corollary 14.9], R*≅〈ξ〉×[1+(p)], where 〈ξ〉 is the cyclic group of order pm-1, and 1+(p)={1+x∣x∈(p)} is the one group of Galois ring R, so |R*|=(pm-1)p(s-1)m.For a fixed positive integern, let Mn(R) and GLn(R) be the set of all n×n matrices and the multiplicative group of all n×n invertible matrices over R, and denote by I(n) and 0(n) the n×n identity matrix and zero matrix, respectively. In this paper, for l×n matrix A and q×r matrix B over R, we adopt the notation A⊕B:=(A00B) which is a (l+q)×(n+r) matrix over R.For any matrixA∈Mn(R), A is said to be symmetric if AT=A, where AT is the transposed matrix of A. We denote the set of all n×n symmetric matrices over R by 𝒮(n,R). Then (𝒮(n,R),+) is a group under the addition of matrices. For any matrices S1,S2∈Mn(R), if there exists matrix P∈GLn(R) such that PS1PT=S2, we say that S1 is cogredient to S2 over R. It is clear that S1∈𝒮(n,R) if and only if S2∈𝒮(n,R). So cogredience of matrices over R is an equivalent relation on 𝒮(n,R). If S1∈𝒮(n,R), we call {PS1PT∣P∈GLn(R)} the cogredient classes of 𝒮(n,R) containing S1 over R. Let 𝒮0={0}, 𝒮1,…,𝒮d be all distinct cogredient classes of 𝒮(n,R). As in [2] we define relations on 𝒮(n,R) by
(1.2)Γi:={(A,B)∣A,B∈S(n,R),A-B∈Si},i=0,1,…,d.
Then the system (𝒮(n,R),{Γi}0≤i≤d) is an association scheme of class d on the set 𝒮(n,R) and denoted by Sym(n,R).Letp stand for an odd prime number in the following. When s=1, we know that the class number of Sym(n,𝔽pm) is given by d=2n and the association scheme Sym(n,𝔽pm) has been investigated in [2]. When m=1, two kinds of cogredient standard forms of symmetric matrices over ℤps are given in [3, 4]. If n≥2, s>1 and p≡1 (mod 4), a complex formula to count the number of all distinct cogredient classes of 𝒮(n,ℤps) is given in [3], which shows that, for example,ifm′ is odd and s is odd, then
(1.3)d+1=(m′-12+1)+∑s1≠0,orsi′,∃i(m′-12-s1-s2′+s3′+s4′+s5′+ε2+1)×[(s-11)+(s-12)(s1-11)+⋯+(s-1s1)]×(s-12s2′)(s+12s3′)(s-12s4′)(s+12s5′),
where the meanings of m′,s1,s2′,s3′,s4′,s5′,ε and formulas for other cases are referred to [3].Then two problems arise.(1) Is there a simple and explicit formula to count the number of all distinct cogredient classes of 𝒮(n,ℤps)? (2) For arbitrary Galois ring R, in order to determine precisely the class number d of the association scheme Sym(n,R), we have to count the number of all distinct cogredient classes of 𝒮(n,R).In Section2 we give two kinds of cogredient standard forms for every symmetric matrix over arbitrary Galois ring R of odd characteristic. In Section 3 we obtain an explicit formula to count the number of all distinct cogredient classes of 𝒮(n,R), which is simpler than that of [3] for the special case R=ℤps.Now, we list some properties for the Galois ringR which will be needed in the following sections. For general theory of Galois rings, one can refer to [1].Lemma 1.1 (see [1, Theorem 14.11]).
R
*
=
G
1
×
G
2 where G1 is a cyclic group of order pm-1, and G2=1+〈p〉 is a group of order p(s-1)m.Proposition 1.2.
(
i
)
R
*
2 is a subgroup of R* with index [R*:R*2]=2.
(
ii
)
For any z∈R*∖R*2, R*∖R*2=zR*2, and |R*2|=|zR*2|=(1/2)|R*|.
(
iii
)
For any u∈R* and a∈〈p〉, there exists c∈R* such that c2(u+a)=u.Proof .
In the notation of Lemma1.1. Let ξ be a generator of the cyclic group G1. Then ξ is of order pm-1. Since p is odd and pm-1 is even, ξ2 is of order (1/2)(pm-1) and G12=〈ξ2〉. Since p(s-1)m is odd and G2 is a commutative group of order p(s-1)m by Lemma 1.1, for every a∈G2, there exists a unique b∈G2 such that a=b2, so G22=G2. Moreover, by Lemma 1.1 each u∈R* can be uniquely expressed as u=gh where g∈G1 and h∈G2. (i)
For everyu=gh∈R* where g∈G1 and h∈G2, u∈R*2 if and only if there exist g1∈G1 and h1∈G2 such that gh=(g1h1)2=g12h12, which is then equivalent to that g=g12 and h=h12. So u∈R*2 if and only if u∈G12×G2 by Lemma 1.1. Then R*2=G12×G2 and so |R*2|=|G12|·|G2|=(1/2)(pm-1)·p(s-1)m=(1/2)|R*|. Hence, [R*:R*2]=2 by group theory.(ii)
Since[R*:R*2]=2, for any z∈R*∖R*2, we have R*=R*2∪zR*2 and R*2∩zR*2=∅ by group theory. So |zR*2|=|R*|-|R*2|=(1/2)|R*| by the proof of (i).(iii)
Letu∈R* and a∈〈p〉. Then u-1(u+a)=1+u-1a∈1+〈p〉=G2. From this and by Lemma 1.1, there exists a unique element b∈G2⊆R* such that u-1(u+a)=b2. Now, let c=b-1. Then c∈R* satisfying c2(u+a)=u.Proposition 1.3.
Let-1∉R*2. Then for any z∈R*∖R*2, there exist x,y∈R* such that z=(1+x2)y2.Proof.
Letu∈R*. Suppose that 1+u2∉R*. Then there exists a∈R such that 1+u2=ap. So u2=-(1-ap). Since p is odd and ps=0 in R, there exists b∈R such that (ups)2=-(1-ap)ps=-(1-ppsb)=-1. From ups∈R* we deduce -1∈R*2, which is a contradiction. Hence 1+u2∈R*. Therefore, σ:w↦1+w (forallw∈R*2) is a mapping from R*2 to R*. Suppose that σ(R*2)⊆R*2. Then for 1∈R*2, there exists w0∈R*2 such that σ(w0)=1+w0=1, which implies that w0=0, and we get a contradiction. So there exists x∈R* such that 1+x2∉R*2, that is, 1+x2∈R*∖R*2=zR*2 by Proposition 1.2. Then there exists c∈R* such that 1+x2=zc2, so (1+x2)y2=z, where y=c-1∈R*.


## 2. Cogredient Standard Forms of Symmetric Matrices

In this section, we give two kinds of cogredient standard forms of symmetric matrices overR corresponding to that of cogredient standard forms of symmetric matrices over finite fields (see [5], or [6], Theorems 1.22 and 1.25).Notation 1.
For any nonnegative integerν and z∈R*∖R*2, define
(2.1)H2ν=(0I(ν)I(ν)0),H2ν+2,Δ=H2ν⊕Δ,whereΔ=(100-z),H2ν+1,(1)=H2ν⊕(1),H2ν+1,(z)=H2ν⊕(z).Lemma 2.1.
For anyν∈ℤ+ and z∈R*∖R*2, zI(2ν) is cogredient to I(2ν).Proof.
Let-1∈R*2. Then there exists u∈R* such that u2=-1, that is, 1+u2=0. Since p is an odd prime number, we have gcd(2,ps)=1 and so 2∈R*. Let P=2-1((1+z)u-1(1-z)u(1-z)(1+z)). Since R is a commutative ring, we have detP=(2-1)2[(1+z)(1+z)-u-1(1-z)u(1-z)]=(2-1)2·2·2z=z∈R*. Hence, P∈GL2(R). Then by (u-1)2=(u2)-1=-1 and u(1-z2)+u-1(1-z2)=u-1(u2+1)(1-z2)=0, we get
(2.2)PI(2)PT=(2-1)2((1+z)u-1(1-z)u(1-z)(1+z))((1+z)u(1-z)u-1(1-z)(1+z))=(2-1)2(2⋅2z002⋅2z)=zI(2),
so zI(2) is cogredient to I(2).
Let-1∉R*2. Then by Proposition 1.3 there exist x,y∈R* such that (1+x2)y2=z. Let Q=(xyy-yxy). Then detQ=(1+x2)y2=z∈R* and so Q∈GL2(R). By (1+x2)y2=z, a matrix computation shows that QI(2)QT=QQT=zI(2). Hence, zI(2) is cogredient to I(2) as well.
ThenzI(2ν)=zI(2)⊕⋯⊕zI(2)⏞ν′s is cogredient to I(2ν)=I(2)⊕⋯⊕I(2)⏞ν′s.Lemma 2.2.
Letz∈R*∖R*2 and ν∈ℤ+. (i)
If-1∈R*2, then I(2ν) is cogredient to H2ν.(ii)
If-1∉R*2, then I(ν)⊕zI(ν) is cogredient to H2ν.Proof.
We selectP1=2-1(I(ν)-I(ν)I(ν)I(ν)) and denote that M=2(I(ν)00-I(ν)). From P1(I(ν)I(ν)0I(ν))=(2-1I(ν)02-1I(ν)I(ν)) we deduce detP1=det(2-1I(ν))=(2-1)ν∈R*. Hence P1∈GL2ν(R). Then by P1MP1T=2-1(I(ν)-I(ν)I(ν)I(ν))(I(ν)I(ν)I(ν)-I(ν))=H2ν, we see that M is cogredient to H2ν.(i)
By-1∈R*2 there exists u∈R* such that -1=u2. Then M is cogredient to 2I(2ν). If 2∉R*2, 2I(2ν) is cogredient to I(2ν) by Lemma 2.1. If 2∈R*2, there exists w∈R* such that 2=w2, so 2I(2ν) is cogredient to I(2ν) as well. Therefore, I(2ν) is cogredient to H2ν in this case.(ii)
Let-1∉R*2. Then by Proposition 1.2 there exists c∈R* such that -1=zc2. Hence I(ν)⊕zI(ν) is cogredient to (I(ν)00-I(ν)). If 2∈R*2, there exists w∈R* such that 2=w2, so (I(ν)00-I(ν)) is cogredient to M. If 2∉R*2, then -2=(-1)2∈R*2, and hence there exists a∈R* such that -2=a2, so (aI(2ν))H2ν(I(ν)00-I(ν))H2νT(aI(2ν))=M. Hence, (I(ν)00-I(ν)) is cogredient to M as well. Therefore, I(ν)⊕zI(ν) is cogredient to H2ν.Lemma 2.3.
Letz∈R*∖R*2 and D=diag(u1,…,ur), where ui∈R*, i=1,…,r and r∈ℤ+. Then, One has the following. (i)
D is necessarily cogredient to either I(r) or I(r-1)⊕(z). Moreover, these two matrices are not cogredient over R.(ii)
Ifr=2ν+1 is odd, then D is necessarily cogredient to either H2ν+1,(1) or H2ν+1,(z). Moreover, these two matrices are not cogredient. If r=2ν is even, then D is necessarily cogredient to either H2ν or H2(ν-1)+2,Δ. Moreover, these two matrices are not cogredient.Proof.
(i) We may assume thatu1,…,ut∈R*2 and ut+1,…,ur∈zR*2, where 0≤t≤r. Then D is cogredient to I(t)⊕zI(r-t). If r-t is even, by Lemma 2.1  zI(r-t) is cogredient to I(r-t) and hence D is cogredient to I(t)⊕I(r-t)=I(r). Now, let r-t be odd. If r-t=1, D is obviously cogredient to I(1)⊕(z). If r-t≥3, by Lemma 2.1  zI(r-t-1) is cogredient to I(r-t-1), and hence D is cogredient to I(t)⊕I(r-t-1)⊕(z)=I(r-1)⊕(z).
Suppose thatI(r) is cogredient to I(r-1)⊕(z) over R. Then there exists Q∈GLr(R) such that QI(r)QT=I(r-1)⊕(z). From this and by detQ∈R*, we obtain that z=(detQ)2∈R*2, which is a contradiction. So I(r) and I(r-1)⊕(z) are not cogredient over R.
(ii) We have one of the following two cases.(ii-1)
Letr=2ν+1 be an odd number. Then r-1=2ν is even and we have one of the following two cases.(ii-1-1)
Let-1∈R*2. Then I(2ν) is cogredient to H2ν by Lemma 2.2(i). From this and by (i) we deduce that D is cogredient to H2ν+1,(1) when D is cogredient to I(r), or D is cogredient to H2ν+1,(z) when D is cogredient to I(r-1)⊕(z).(ii-1-2)
Let-1∈zR*2. Then we have one of the following two cases.(α)
Let(1/2)(r-1)=ν be even. Then I(ν) is cogredient to zI(ν) by Lemma 2.1, so I(2ν) is cogredient to I(ν)⊕zI(ν). Since I(ν)⊕zI(ν) is cogredient to H2ν by Lemma 2.2(ii), by (i) we see that: D is cogredient to H2ν+1,(1) when D is cogredient to I(r), or D is cogredient to H2ν+1,(z) when D is cogredient to I(r-1)⊕(z).(β)
Let(1/2)(r-1)=ν be odd. Then ν=2ω+1 for some nonnegative integer ω and so r-1=4ω+2. By Lemma 2.1 we see that I(2ω) is cogredient to zI(2ω), and I(2) is cogredient to zI(2). Hence I(r)=I(2ω)⊕I(2ω)⊕I(2)⊕(1) is cogredient to I(2ω)⊕zI(2ω)⊕zI(2)⊕(1), which is then cogredient to I(2ω+1)⊕zI(2ω+1)⊕(z). Since I(2ω+1)⊕zI(2ω+1) is cogredient to H2(2ω+1)=H2ν by Lemma 2.2(ii), I(r) is cogredient to H2ν+1,(z). Moreover, I(r-1)⊕(z)=I(2ω)⊕I(2ω)⊕I(2)⊕(z) is cogredient to I(2ω)⊕zI(2ω)⊕I(2)⊕(z), which is then cogredient to I(2ω+1)⊕zI(2ω+1)⊕(1). Since I(ν)⊕zI(ν) is cogredient to H2ν by Lemma 2.2(ii), I(r-1)⊕(z) is cogredient to H2ν+1,(1). Therefore, D is necessarily cogredient to either H2ν+1,(1) or H2ν+1,(z) by (i).(ii-2)
Letr=2ν be an even number. Then r-2=2(ν-1) is also even and we have one of the following two cases.(ii-2-1)
Let-1∈R*2. Then -1=u2 for some u∈R* and so (100z) is cogredient to (100-z)=Δ. By Lemma 2.2(i)  D is cogredient to H2ν when D is cogredient to I(r), or D is cogredient to H2(ν-1)+2,Δ when D is cogedient to I(r-1)⊕(z)=I(2(ν-1))⊕(100z).(ii-2-2)
Let-1∈zR*2. Then -1=zc2 for some c∈R*. By 1=(-z)c2, we see that I(2) is cogredient to Δ. Now, we have one of the following two cases.(α)
Letν be even. Then I(ν) is cogredient to zI(ν) by Lemma 2.1 and so I(r)=I(ν)⊕I(ν) is cogredient to I(ν)⊕zI(ν). From this and by Lemma 2.2(ii), we see that I(r) is cogredient to H2ν. Let ν=2. Since I(2) is cogredient to Δ and I(1)⊕(z) is cogredient to H2 by Lemma 2.2(ii), I(3)⊕(z)=I(2)⊕I(1)⊕(z) is cogredient to H2⊕Δ=H2·1+2,Δ. Now, let ν≥4. Since ν-2 is even, I(ν-2) is cogredient to zI(ν-2) by Lemma 2.1, so I(ν-2)⊕I(ν-2) is cogredient to I(ν-2)⊕zI(ν-2). Hence, I(r-1)⊕(z)=I(ν-2)⊕I(ν-2)⊕I(3)⊕(z) is cogredient to I(ν-2)⊕zI(ν-2)⊕I(3)⊕(z), which is then cogredient to I(ν-1)⊕zI(ν-1)⊕I(2). Since I(2) is cogredient to Δ, we see that I(r-1)⊕(z) is cogredient to H2(ν-1)+2,Δ by Lemma 2.2(ii). Therefore, D is necessarily cogredient to either H2ν or H2(ν-1)+2,Δ by (i).(β)
Letν be odd. Then there exists nonnegative integer ω such that ν=2ω+1 and so r=4ω+2. Since I(2ω) is cogredient to zI(2ω) by Lemma 2.1, I(r)=I(2ω)⊕I(2ω)⊕I(2) is cogredient to I(2ω)⊕zI(2ω)⊕Δ, that is then cogredient to H2(2ω)+2,Δ=H2(ν-1)+2,Δ by Lemma 2.2(ii). Now, I(r-1)⊕(z)=I(2ω)⊕I(2ω)⊕(1)⊕(z) is cogredient to I(2ω)⊕zI(2ω)⊕(1)⊕(z) by Lemma 2.1, which is then cogredient to I(2ω+1)⊕zI(2ω+1). Hence I(r-1)⊕(z) is cogredient to H2(2ω+1)=H2ν by Lemma 2.2(ii). Therefore, D is necessarily cogredient to either H2ν or H2(ν-1)+2,Δ by (i).Theorem 2.4.
Letz∈R*∖R*2. Then every n×n symmetric matrix A over R is necessarily cogredient to one of the following matrices:
(2.3)D(n,k,t;k1,…,kt;r1,…,rt):=diag⁡(pr1D1,pr2D2,…,prtDt,0(n-k)),
where 0≤t≤k≤n, Di=I(ki) or I(ki-1)⊕(z) for all i=1,…,t, 0≤r1<r2<⋯<rt≤s-1, and ki∈ℤ+ satisfy Σi=1tki=k.Proof.
The statement holds obviously ifA=0 (corresponding to the case k=0) or n=1. Now, let n≥2 and A=(aij)n×n≠0. Then, there exist 1≤i0,j0≤n such that ai0j0≠0 and τ(ai0j0)=min{τ(aij)∣aij≠0,1≤i,j≤n}. Let s1=ν(ai0j0). Then 0≤s1≤s-1, and there exists P1∈GLn(R) such that P1AP1T=diag(u1ps1,B) where u1∈R* and B=(bij) is a (n-1)×(n-1) symmetric matrix over R satisfying B=0 or τ(bij)≥s1 for all bij≠0, 1≤i,j≤n-1. By induction there exists X∈GLn-1(R) such that XBXT=diag(u2ps2,…,ukpsk,0(n-k)), where u2,…,uk∈R* and s2≤⋯≤sk≤s-1. Then P=diag(1,X)P1∈GLn(R) satisfies PAPT=diag(u1ps1,…,ukpsk,0(n-k)).
Now, there must existt,ki∈ℤ+, i=1,…,t and 0≤r1<⋯<rt≤s-1 such that s1=⋯=sk1=r1<sk1+1=⋯=sk1+k2=r2<⋯<sk1+k2+⋯+kt-1+1=⋯=sk1+k2+⋯+kt-1+kt=rt. Then Σi=1tki=k and A is cogredient to M=diag(pr1M1,pr2M2,…,prtMt,0(n-k)), where Mi=diag(uk1+⋯+ki-1+1,…,uk1+⋯+ki-1+ki) is a ki×ki matrix over R for all i=1,…,t. Since Mi is cogredient to Di for every 1≤i≤t by Lemma 2.3(i), we deduce that A is cogredient to diag(pr1D1,pr2D2,…,prtDt,0(n-k)).Theorem 2.5.
Letz∈R*∖R*2. Then every n×n symmetric matrix A over R is necessarily cogredient to one of the following matrices:
(2.4)H(n,k,t;k1,…,kt;r1,…,rt):=diag⁡(pr1H1,pr2H2,…,prtHt,0(n-k)),
where Hi is a ki×ki matrix over R such that Hi is equal to either H2νi+1,(1) or H2νi+1,(z) when ki=2νi+1 is odd, and Hi is equal to either H2νi or H2(νi-1)+2,Δ when ki=2νi is even, for all i=1,…,t; 0≤t≤k≤n, 0≤r1<r2<⋯<rt≤s-1, and ki∈ℤ+ satisfy Σi=1tki=k.Proof.
It follows from Theorem2.4 and the proof of Lemma 2.3(ii).
For anyn×n symmetric matrix A, we call D(n,k,t;k1,…,kt;r1,…,rt) the cogredient standard form of kind (I) of A if A is cogredient to D(n,k,t;k1,…,kt;r1,…,rt), and call H(n,k,t;k1,…,kt;r1,…,rt) the cogredient standard form of kind (II) of A if A is cogredient to H(n,k,t;k1,…,kt;r1,…,rt).


## 3. The Number of Cogredient Classes of Symmetric Matrices

In order to count the number of all distinct cogredient classes ofn×n symmetric matrices over R, we show that every n×n symmetric matrix over R has only one cogredient standard form of kind (I) first, then the number of all distinct cogredient classes of n×n symmetric matrices over R is equal to the number of all cogredient standard forms of kind (I) by Theorem 2.4.Theorem 3.1.
The number𝒞s,n of all distinct cogredient classes of n×n symmetric matrices over R is given by the following: (i)
Ifn≤s, then 𝒞s,n=1+∑j=0n-1∑i=jn-1(ij)(sj+1)2j+1;(ii)
Ifn≥s+1, then 𝒞s,n=1+∑j=0s-1∑i=jn-1(ij)(sj+1)2j+1.Proof.
LetD^:=diag(pr^1D^1,pr^2D^2,…,pr^t^D^t^,0(n-k^)), where D^i=I(k^i) or I(k^i-1)⊕(z) for all i=1,…,t^, 0≤t^≤k^≤n, 0≤r^1<r^2<⋯<r^t^≤s-1, and k^i∈ℤ+ satisfy Σi=1t^k^i=k^. In the notation of Theorem 2.4, by [7, Theorem D], it follows that D=D^ if D:=D(n,k,t;k1,…,kt;r1,…,rt) is cogredient to D^ over R. Hence, every n×n symmetric matrix over R has only one cogredient standard form of kind (I).
For any1≤t≤k≤n, denote that S1={(k1,…,kt)∣ki∈ℤ+,Σi=1t=k} and S2={(r1,…,rt)∣ri∈ℤ,0≤r1<r2<⋯<rt≤s-1}. Then |S1|=(k-1t-1), |S2|=(st) if t≤s and, |S2|=0 if t≥s. From this and by Theorem 2.4 it follows that 𝒞s,n=1+∑k=1n(∑t=1k|S1|·|S2|·2t). Therefore, 𝒞s,n=1+∑j=0n-1∑i=jn-1(ij)(sj+1)2j+1 if n≤s and, 𝒞s,n=1+∑j=0s-1∑i=jn-1(ij)(sj+1)2j+1 if n≥s+1.
In the notations of Section1, we see that the class number d of the association scheme Sym(n,R) is determined by d+1=𝒞s,n. Then by  Theorem 3.1, we have the following corollary.Corollary 3.2.
The class number of the association schemeSym(n,R) is given by the following.(i)
Ifn≤s, then d=∑j=0n-1∑i=jn-1(ij)(sj+1)2j+1;(ii)
Ifn≥s+1, then d=∑j=0s-1∑i=jn-1(ij)(sj+1)2j+1.Example 3.3.
Letp be an odd prime number and s=2. Then by  Theorem 3.1 the number 𝒞2,2 of all cogredient classes of 2×2 symmetric matrices over Galois ring GR(p2,p2m) is given by 𝒞2,2=1+∑j=01∑i=j1(ij)(2j+1)2j+1=13. In fact, for a fixed element z∈R*∖R*2, all cogredient standard forms of kind (I) of 2×2 symmetric matrices over GR(p2,p2m) are given by the following:
(3.1)(0000),(1000),(z000),(p000),(zp000),(1001),(100z),(p00p),(p00zp),(100p),(z00p),(100zp),(z00zp).
The number 𝒞2,3 of all cogredient classes of 3×3 symmetric matrices over GR(p2,p2m) is given by 𝒞2,3=1+∑j=01∑i=j2(ij)(2j+1)2j+1=25. In fact, all cogredient standard forms of kind (I) of 3×3 symmetric matrices over GR(p2,p2m) are given by the following: (J000) where J is one of matrices in (3.1), and
(3.2)(100010001),(10001000z),(p000p000p),(p000p000zp),(10001000p),(1000z000p),(10001000zp),(1000z000zp),(1000p000p),(z000p000p),(1000p000zp),(z000p000zp).Example 3.4.
Letp be an odd prime number and s=5. Then by  Theorem 3.1 the number 𝒞5,4 of all cogredient classes of 4×4 symmetric matrices over Galois ring GR(p5,p5m) is given by 𝒞5,4=1+∑j=03∑i=j3(ij)(5j+1)2j+1=681; the number 𝒞5,7 of all cogredient classes of 7×7 symmetric matrices over GR(p5,p5m) is given by 𝒞5,7=1+∑j=04∑i=j6(ij)(5j+1)2j+1=6943.



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*Source: 520148-2012-07-19.xml*