Let R=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.

1. Introduction and Preliminaries

Let p 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 by R* 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 integer n, 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 matrix A∈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).

Let p 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,

if m′ 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 Section 2 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 ring R which will be needed in the following sections. For general theory of Galois rings, one can refer to [1].

Lemma 1.1 (see [<xref ref-type="bibr" rid="B6">1</xref>, Theorem 14.11]).

R*=G1×G2 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 Lemma 1.1. 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.

For every u=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.

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).

Let u∈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.

Let u∈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 over R 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 integer ν 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 any ν∈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.

Then zI(2ν)=zI(2)⊕⋯⊕zI(2)⏞ν′s is cogredient to I(2ν)=I(2)⊕⋯⊕I(2)⏞ν′s.

Lemma 2.2.

Let z∈R*∖R*2 and ν∈ℤ+.

If -1∈R*2, then I(2ν) is cogredient to H2ν.

If -1∉R*2, then I(ν)⊕zI(ν) is cogredient to H2ν.

Proof.

We select P1=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.

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.

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.

Let z∈R*∖R*2 and D=diag(u1,…,ur), where ui∈R*, i=1,…,r and r∈ℤ+. Then, One has the following.

D is necessarily cogredient to either I(r) or I(r-1)⊕(z). Moreover, these two matrices are not cogredient over R.

If r=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 that u1,…,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 2.1zI(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 2.1zI(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 that I(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.

Let r=2ν+1 be an odd number. Then r-1=2ν is even and we have one of the following two cases.

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).

Let -1∈zR*2. Then we have one of the following two cases.

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).

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).

Let r=2ν be an even number. Then r-2=2(ν-1) is also even and we have one of the following two cases.

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).

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.

Let ν 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).

Let ν 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.

Let z∈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 if A=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 exist t,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.

Let z∈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 Theorem 2.4 and the proof of Lemma 2.3(ii).

For any n×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 of n×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 number 𝒞s,n of all distinct cogredient classes of n×n symmetric matrices over R is given by the following:

If n≤s, then 𝒞s,n=1+∑j=0n-1∑i=jn-1(ij)(sj+1)2j+1;

If n≥s+1, then 𝒞s,n=1+∑j=0s-1∑i=jn-1(ij)(sj+1)2j+1.

Proof.

Let D^:=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 any 1≤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 Section 1, 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 scheme Sym(n,R) is given by the following.

If n≤s, then d=∑j=0n-1∑i=jn-1(ij)(sj+1)2j+1;

If n≥s+1, then d=∑j=0s-1∑i=jn-1(ij)(sj+1)2j+1.

Example 3.3.

Let p 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.

Let p 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.

Acknowledgment

This reaserach is supported in part by the National Science Foundation of China (No. 10971160) and Natural Science Foundation of Shandong provence (Grant No. ZR2011AQ004).

WanZ.-X.HuoY. J.WanZ. X.Non-symmetric association schemes of symmetric matricesLiuY.NanJ.-Z.Some Anzahl theorems in symmetric matrices over finite local ringsWuY.DicksonL. E.WanZ. X.CaoY.SzechtmanF.Congruence of symmetric matrices over local rings