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

Abstract

Let 𝑅=GR(𝑝^𝑠,𝑝^𝑠𝑚) be a Galois ring of characteristic 𝑝^𝑠 and cardinality 𝑝^𝑠𝑚, 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(𝑝^𝑠,𝑝^𝑠𝑚) a Galois ring of characteristic 𝑝^𝑠 and cardinality 𝑝^𝑠𝑚. Then GR(𝑝^𝑠,𝑝^𝑠𝑚) is isomorphic to the ring ℤ_𝑝𝑠[𝑥]/(ℎ(𝑥)) for any basic irreducible polynomial ℎ(𝑥) of degree 𝑚 over ℤ_𝑝𝑠. It is clear that 𝑅=𝔽_𝑝𝑚, that is, a finite field of 𝑝^𝑚 elements, if 𝑠=1, and 𝑅=ℤ_𝑝𝑠, that is the ring of residue classes of ℤ modulo its ideal 𝑝^𝑠ℤ, 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)=(𝑝^𝑠)⊂(𝑝^𝑠−1)⊂⋯⊂(𝑝)⊂(𝑝⁰)=𝑅. By [1, Theorem 14.8], there exists an element 𝜉∈𝑅^∗ of multiplicative order 𝑝^𝑚−1, which is a root of a basic primitive polynomial ℎ(𝑥) of degree 𝑚 over ℤ_𝑝𝑠 and dividing 𝑥^𝑝𝑚−1−1 in ℤ_𝑝𝑠[𝑥], and every element 𝑎∈𝑅 can be written uniquely as 𝑎=𝑎₀+𝑎₁𝑝+⋯+𝑎_𝑛−1𝑝^𝑛−1,𝑎₀,𝑎₁,…,𝑎_𝑛−1∈𝒯,(1.1) where 𝒯={0,1,𝜉,…,𝜉^𝑝𝑚−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=𝑎_𝑖𝑝^𝑖+⋯+𝑎_𝑛−1𝑝^𝑛−1 with 𝑎_𝑖≠0. By [1, Corollary 14.9], 𝑅^∗≅⟨𝜉⟩×[1+(𝑝)], where ⟨𝜉⟩ is the cyclic group of order 𝑝^𝑚−1, and 1+(𝑝)={1+𝑥∣𝑥∈(𝑝)} is the one group of Galois ring 𝑅, so |𝑅^∗|=(𝑝^𝑚−1)𝑝^{(𝑠−1)𝑚}.

For a fixed positive integer 𝑛, let M_𝑛(𝑅) and GL_𝑛(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∶=^𝐴0_0𝐵 which is a (𝑙+𝑞)×(𝑛+𝑟) matrix over 𝑅.

For any matrix 𝐴∈M_𝑛(𝑅), 𝐴 is said to be symmetric if 𝐴^𝑇=𝐴, where 𝐴^𝑇 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∈M_𝑛(𝑅), if there exists matrix 𝑃∈GL_𝑛(𝑅) such that 𝑃𝑆₁𝑃^𝑇=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 {𝑃𝑆₁𝑃^𝑇∣𝑃∈GL_𝑛(𝑅)} the cogredient classes of 𝒮(𝑛,𝑅) containing 𝑆₁ over 𝑅. Let 𝒮₀={0}, 𝒮₁,…,𝒮_𝑑 be all distinct cogredient classes of 𝒮(𝑛,𝑅). As in [2] we define relations on 𝒮(𝑛,𝑅) by Γ_𝑖∶=(𝐴,𝐵)∣𝐴,𝐵∈𝒮(𝑛,𝑅),𝐴−𝐵∈𝒮_𝑖,𝑖=0,1,…,𝑑.(1.2) Then the system (𝒮(𝑛,𝑅),{Γ_𝑖}_{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,𝔽_𝑝𝑚) is given by 𝑑=2𝑛 and the association scheme Sym(𝑛,𝔽_𝑝𝑚) has been investigated in [2]. When 𝑚=1, two kinds of cogredient standard forms of symmetric matrices over ℤ_𝑝𝑠 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,ℤ_𝑝𝑠) is given in [3], which shows that, for example,

if 𝑚^ is odd and 𝑠 is odd, then 𝑑+1=𝑚^−12+1+_{𝑠1≠0,or𝑠𝑖,∃𝑖}⎛⎜⎝𝑚^−12−𝑠₁−𝑠^₂+𝑠^₃+𝑠^₄+𝑠^₅+𝜀2+1⎞⎟⎠×⎡⎢⎢⎣⎛⎜⎜⎝𝑠−11⎞⎟⎟⎠+⎛⎜⎜⎝𝑠−12⎞⎟⎟⎠⎛⎜⎜⎝𝑠₁−11⎞⎟⎟⎠+⋯+⎛⎜⎜⎝𝑠−1𝑠₁⎞⎟⎟⎠⎤⎥⎥⎦×⎛⎜⎜⎜⎝𝑠−12𝑠^₂⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝𝑠+12𝑠^₃⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝𝑠−12𝑠^₄⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝𝑠+12𝑠^₅⎞⎟⎟⎟⎠,(1.3) where the meanings of 𝑚^,𝑠₁,𝑠^₂,𝑠^₃,𝑠^₄,𝑠^₅,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,ℤ_𝑝𝑠)? (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=ℤ_𝑝𝑠.

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 [1, Theorem 14.11]). 𝑅^∗=𝐺₁×𝐺₂ where 𝐺₁ is a cyclic group of order 𝑝^𝑚−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 𝑝^𝑚−1. Since 𝑝 is odd and 𝑝^𝑚−1 is even, 𝜉² is of order (1/2)(𝑝^𝑚−1) and 𝐺²₁=⟨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 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=𝑔²₁ℎ²₁, which is then equivalent to that 𝑔=𝑔²₁ and ℎ=ℎ²₁. So 𝑢∈𝑅^∗2 if and only if 𝑢∈𝐺²₁×𝐺₂ by Lemma 1.1. Then 𝑅^∗2=𝐺²₁×𝐺₂ and so |𝑅^∗2|=|𝐺²₁|⋅|𝐺₂|=(1/2)(𝑝^𝑚−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) 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 𝑝^𝑠=0 in 𝑅, there exists 𝑏∈𝑅 such that (𝑢^𝑝𝑠)²=−(1−𝑎𝑝)^𝑝𝑠=−(1−𝑝^𝑝𝑠𝑏)=−1. From 𝑢^𝑝𝑠∈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 𝐻_2𝜈=⎛⎜⎜⎝0𝐼^(𝜈)𝐼^(𝜈)0⎞⎟⎟⎠,𝐻_2𝜈+2,Δ=𝐻_2𝜈⊕Δ,whereΔ=⎛⎜⎜⎝100−𝑧⎞⎟⎟⎠,𝐻_2𝜈+1,(1)=𝐻_2𝜈⊕(1),𝐻_{2𝜈+1,(𝑧)}=𝐻_2𝜈⊕(𝑧).(2.1)

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,𝑝^𝑠)=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 𝑃𝐼⁽²⁾𝑃^𝑇=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⁾,(2.2) 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⁾𝑄^𝑇=𝑄𝑄^𝑇=𝑧𝐼⁽²⁾. Hence, 𝑧𝐼⁽²⁾ is cogredient to 𝐼⁽²⁾ as well.
Then 𝑧𝐼^(2𝜈)=𝑧𝐼⁽²⁾⊕⋯⊕𝑧𝐼^{𝜈𝑠(2)} is cogredient to 𝐼^(2𝜈)=𝐼⁽²⁾⊕⋯⊕𝐼^{𝜈𝑠(2)}.

Lemma 2.2. Let 𝑧∈𝑅^∗⧵𝑅^∗2 and 𝜈∈ℤ⁺. (i) If −1∈𝑅^∗2, then 𝐼^(2𝜈) is cogredient to 𝐻_2𝜈. (ii) If −1∉𝑅^∗2, then 𝐼^(𝜈)⊕𝑧𝐼^(𝜈) is cogredient to 𝐻_2𝜈.

Proof. We select 𝑃₁=2⁻¹^{𝐼(𝜈)−𝐼(𝜈)}_𝐼^(𝜈)_𝐼^(𝜈) and denote that 𝑀=2^𝐼(𝜈)0_0−𝐼^(𝜈). From 𝑃₁^{𝐼(𝜈)𝐼(𝜈)}_0𝐼^(𝜈)=^{2−1𝐼(𝜈)0}₂⁻¹_𝐼^(𝜈)_𝐼^(𝜈) we deduce det𝑃₁=det(2⁻¹𝐼^(𝜈))=(2⁻¹)^𝜈∈𝑅^∗. Hence 𝑃₁∈GL_2𝜈(𝑅). Then by 𝑃₁𝑀𝑃^𝑇₁=2⁻¹^{𝐼(𝜈)−𝐼(𝜈)}_𝐼^(𝜈)_𝐼^(𝜈)^{𝐼(𝜈)𝐼(𝜈)}_𝐼^(𝜈)_−𝐼^(𝜈)=𝐻_2𝜈, we see that 𝑀 is cogredient to 𝐻_2𝜈.(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 𝐻_2𝜈 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)0_0−𝐼^(𝜈). If 2∈𝑅^∗2, there exists 𝑤∈𝑅^∗ such that 2=𝑤², so ^𝐼(𝜈)0_0−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))𝐻_2𝜈^𝐼(𝜈)0_0−𝐼^(𝜈)𝐻^𝑇_2𝜈(𝑎𝐼^(2𝜈))=𝑀. Hence, ^𝐼(𝜈)0_0−𝐼^(𝜈) is cogredient to 𝑀 as well. Therefore, 𝐼^(𝜈)⊕𝑧𝐼^(𝜈) is cogredient to 𝐻_2𝜈.

Lemma 2.3. Let 𝑧∈𝑅^∗⧵𝑅^∗2 and 𝐷=diag(𝑢₁,…,𝑢_𝑟), where 𝑢_𝑖∈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) If 𝑟=2𝜈+1 is odd, then D is necessarily cogredient to either 𝐻_2𝜈+1,(1) or 𝐻_{2𝜈+1,(𝑧)}. Moreover, these two matrices are not cogredient. If 𝑟=2𝜈 is even, then D is necessarily cogredient to either 𝐻_2𝜈 or 𝐻_{2(𝜈−1)+2,Δ}. Moreover, these two matrices are not cogredient.

Proof. (i) We may assume that 𝑢₁,…,𝑢_𝑡∈𝑅^∗2 and 𝑢_𝑡+1,…,𝑢_𝑟∈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 that 𝐼^(𝑟) is cogredient to 𝐼^(𝑟−1)⊕(𝑧) over 𝑅. Then there exists 𝑄∈GL_𝑟(𝑅) such that 𝑄𝐼^(𝑟)𝑄^𝑇=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)Let 𝑟=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 𝐻_2𝜈 by Lemma 2.2(i). From this and by (i) we deduce that 𝐷 is cogredient to 𝐻_2𝜈+1,(1) when 𝐷 is cogredient to 𝐼^(𝑟), or 𝐷 is cogredient to 𝐻_{2𝜈+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 𝐻_2𝜈 by Lemma 2.2(ii), by (i) we see that: 𝐷 is cogredient to 𝐻_2𝜈+1,(1) when 𝐷 is cogredient to 𝐼^(𝑟), or 𝐷 is cogredient to 𝐻_{2𝜈+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 𝐻_2(2𝜔+1)=𝐻_2𝜈 by Lemma 2.2(ii), 𝐼^(𝑟) is cogredient to 𝐻_{2𝜈+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 𝐻_2𝜈 by Lemma 2.2(ii), 𝐼^(𝑟−1)⊕(𝑧) is cogredient to 𝐻_2𝜈+1,(1). Therefore, 𝐷 is necessarily cogredient to either 𝐻_2𝜈+1,(1) or 𝐻_{2𝜈+1,(𝑧)} by (i).(ii-2) 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.(ii-2-1) Let −1∈𝑅^∗2. Then −1=𝑢² for some 𝑢∈𝑅^∗ and so ¹⁰_0𝑧 is cogredient to ¹⁰_0−𝑧=Δ. By Lemma 2.2(i) 𝐷 is cogredient to 𝐻_2𝜈 when 𝐷 is cogredient to 𝐼^(𝑟), or 𝐷 is cogredient to 𝐻_{2(𝜈−1)+2,Δ} when 𝐷 is cogedient to 𝐼^(𝑟−1)⊕(𝑧)=𝐼^{(2(𝜈−1))}⊕¹⁰_0𝑧.(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) 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 𝐻_2𝜈. 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=𝐻_2⋅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 𝐻_{2(𝜈−1)+2,Δ} by Lemma 2.2(ii). Therefore, 𝐷 is necessarily cogredient to either 𝐻_2𝜈 or 𝐻_{2(𝜈−1)+2,Δ} by (i).(𝛽) 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 𝐻_{2(2𝜔)+2,Δ}=𝐻_{2(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 𝐻_2(2𝜔+1)=𝐻_2𝜈 by Lemma 2.2(ii). Therefore, 𝐷 is necessarily cogredient to either 𝐻_2𝜈 or 𝐻_{2(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: 𝐷_{(𝑛,𝑘,𝑡;𝑘1,…,𝑘𝑡;𝑟1,…,𝑟𝑡)}∶=diag𝑝^𝑟1𝐷₁,𝑝^𝑟2𝐷₂,…,𝑝^𝑟𝑡𝐷_𝑡,0^{(𝑛−𝑘)},(2.3) where 0≤𝑡≤𝑘≤𝑛, 𝐷_𝑖=𝐼^(𝑘𝑖) or 𝐼^{(𝑘𝑖−1)}⊕(𝑧) for all 𝑖=1,…,𝑡, 0≤𝑟₁<𝑟₂<⋯<𝑟_𝑡≤𝑠−1, and 𝑘_𝑖∈ℤ⁺ satisfy Σ^𝑡_𝑖=1𝑘_𝑖=𝑘.

Proof. The statement holds obviously if 𝐴=0 (corresponding to the case 𝑘=0) or 𝑛=1. Now, let 𝑛≥2 and 𝐴=(𝑎_𝑖𝑗)_𝑛×𝑛≠0. Then, there exist 1≤𝑖₀,𝑗₀≤𝑛 such that 𝑎_𝑖0𝑗0≠0 and 𝜏(𝑎_𝑖0𝑗0)=min{𝜏(𝑎_𝑖𝑗)∣𝑎_𝑖𝑗≠0,1≤𝑖,𝑗≤𝑛}. Let 𝑠₁=𝜈(𝑎_𝑖0𝑗0). Then 0≤𝑠₁≤𝑠−1, and there exists 𝑃₁∈GL_𝑛(𝑅) such that 𝑃₁𝐴𝑃^𝑇₁=diag(𝑢₁𝑝^𝑠1,𝐵) where 𝑢₁∈𝑅^∗ and 𝐵=(𝑏_𝑖𝑗) is a (𝑛−1)×(𝑛−1) symmetric matrix over 𝑅 satisfying 𝐵=0 or 𝜏(𝑏_𝑖𝑗)≥𝑠₁ for all 𝑏_𝑖𝑗≠0, 1≤𝑖,𝑗≤𝑛−1. By induction there exists 𝑋∈GL_𝑛−1(𝑅) such that 𝑋𝐵𝑋^𝑇=diag(𝑢₂𝑝^𝑠2,…,𝑢_𝑘𝑝^𝑠𝑘,0^{(𝑛−𝑘)}), where 𝑢₂,…,𝑢_𝑘∈𝑅^∗ and 𝑠₂≤⋯≤𝑠_𝑘≤𝑠−1. Then 𝑃=diag(1,𝑋)𝑃₁∈GL_𝑛(𝑅) satisfies 𝑃𝐴𝑃^𝑇=diag(𝑢₁𝑝^𝑠1,…,𝑢_𝑘𝑝^𝑠𝑘,0^{(𝑛−𝑘)}).
Now, there must exist 𝑡,𝑘_𝑖∈ℤ⁺, 𝑖=1,…,𝑡 and 0≤𝑟₁<⋯<𝑟_𝑡≤𝑠−1 such that 𝑠₁=⋯=𝑠_𝑘1=𝑟₁<𝑠_𝑘1+1=⋯=𝑠_𝑘1+𝑘2=𝑟₂<⋯<𝑠_{𝑘1+𝑘2+⋯+𝑘𝑡−1+1}=⋯=𝑠_{𝑘1+𝑘2+⋯+𝑘𝑡−1+𝑘𝑡}=𝑟_𝑡. Then Σ^𝑡_𝑖=1𝑘_𝑖=𝑘 and 𝐴 is cogredient to 𝑀=diag(𝑝^𝑟1𝑀₁,𝑝^𝑟2𝑀₂,…,𝑝^𝑟𝑡𝑀_𝑡,0^{(𝑛−𝑘)}), where 𝑀_𝑖=diag(𝑢_{𝑘1+⋯+𝑘𝑖−1+1},…,𝑢_{𝑘1+⋯+𝑘𝑖−1+𝑘𝑖}) is a 𝑘_𝑖×𝑘_𝑖 matrix over 𝑅 for all 𝑖=1,…,𝑡. Since 𝑀_𝑖 is cogredient to 𝐷_𝑖 for every 1≤𝑖≤𝑡 by Lemma 2.3(i), we deduce that 𝐴 is cogredient to diag(𝑝^𝑟1𝐷₁,𝑝^𝑟2𝐷₂,…,𝑝^𝑟𝑡𝐷_𝑡,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: 𝐻_{(𝑛,𝑘,𝑡;𝑘1,…,𝑘𝑡;𝑟1,…,𝑟𝑡)}∶=diag𝑝^𝑟1𝐻₁,𝑝^𝑟2𝐻₂,…,𝑝^𝑟𝑡𝐻_𝑡,0^{(𝑛−𝑘)},(2.4) where 𝐻_𝑖 is a 𝑘_𝑖×𝑘_𝑖 matrix over R such that 𝐻_𝑖 is equal to either 𝐻_{2𝜈𝑖+1,(1)} or 𝐻_{2𝜈𝑖+1,(𝑧)} when 𝑘_𝑖=2𝜈_𝑖+1 is odd, and 𝐻_𝑖 is equal to either 𝐻_2𝜈𝑖 or 𝐻_{2(𝜈𝑖−1)+2,Δ} when 𝑘_𝑖=2𝜈_𝑖 is even, for all 𝑖=1,…,𝑡; 0≤𝑡≤𝑘≤𝑛, 0≤𝑟₁<𝑟₂<⋯<𝑟_𝑡≤𝑠−1, and 𝑘_𝑖∈ℤ⁺ satisfy Σ^𝑡_𝑖=1𝑘_𝑖=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 𝒞_𝑠,𝑛 of all distinct cogredient classes of 𝑛×𝑛 symmetric matrices over 𝑅 is given by the following: (i) If 𝑛≤𝑠, then 𝒞_𝑠,𝑛=1+∑^𝑛−1_𝑗=0∑^𝑛−1_𝑖=𝑗^𝑖𝑗^𝑠𝑗+12^𝑗+1; (ii) If 𝑛≥𝑠+1, then 𝒞_𝑠,𝑛=1+∑^𝑠−1_𝑗=0∑^𝑛−1_𝑖=𝑗^𝑖𝑗^𝑠𝑗+12^𝑗+1.

Proof. Let 𝐷∶=diag(𝑝𝑟₁̂𝐷₁,𝑝𝑟₂̂𝐷₂,…,𝑝𝑟𝑡̂𝐷̂‌̂‌𝑡,0^(𝑛−𝑘̂‌₎), where 𝐷_𝑖=𝐼⁽𝑘̂‌_𝑖) or 𝐼⁽𝑘̂‌_𝑖−1)⊕(𝑧) for all 𝑖=1,…,𝑡̂‌, 0≤𝑡̂‌≤𝑘̂‌≤𝑛, 0≤𝑟₁̂<𝑟₂̂<⋯<𝑟𝑡̂̂‌≤𝑠−1, and 𝑘̂‌_𝑖∈ℤ⁺ satisfy Σ𝑡̂‌_𝑖=1̂‌𝑘_𝑖=𝑘̂‌. 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,…,𝑘_𝑡)∣𝑘_𝑖∈ℤ⁺,Σ^𝑡_𝑖=1=𝑘} and 𝑆₂={(𝑟₁,…,𝑟_𝑡)∣𝑟_𝑖∈ℤ,0≤𝑟₁<𝑟₂<⋯<𝑟_𝑡≤𝑠−1}. Then |𝑆₁|=^𝑘−1_𝑡−1, |𝑆₂|=(^𝑠_𝑡) if 𝑡≤𝑠 and, |𝑆₂|=0 if 𝑡≥𝑠. From this and by Theorem 2.4 it follows that 𝒞_𝑠,𝑛=1+∑^𝑛_𝑘=1(∑^𝑘_𝑡=1|𝑆₁|⋅|𝑆₂|⋅2^𝑡). Therefore, 𝒞_𝑠,𝑛=1+∑^𝑛−1_𝑗=0∑^𝑛−1_𝑖=𝑗^𝑖𝑗^𝑠𝑗+12^𝑗+1 if 𝑛≤𝑠 and, 𝒞_𝑠,𝑛=1+∑^𝑠−1_𝑗=0∑^𝑛−1_𝑖=𝑗^𝑖𝑗^𝑠𝑗+12^𝑗+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=𝒞_𝑠,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.(i) If 𝑛≤𝑠, then 𝑑=∑^𝑛−1_𝑗=0∑^𝑛−1_𝑖=𝑗^𝑖𝑗^𝑠𝑗+12^𝑗+1; (ii) If 𝑛≥𝑠+1, then 𝑑=∑^𝑠−1_𝑗=0∑^𝑛−1_𝑖=𝑗^𝑖𝑗^𝑠𝑗+12^𝑗+1.

Example 3.3. Let 𝑝 be an odd prime number and 𝑠=2. Then by Theorem 3.1 the number 𝒞_2,2 of all cogredient classes of 2×2 symmetric matrices over Galois ring GR(𝑝²,𝑝^2𝑚) is given by 𝒞_2,2=1+∑¹_𝑗=0∑¹_𝑖=𝑗^𝑖𝑗²_𝑗+12^𝑗+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,𝑝^2𝑚) are given by the following: ⎛⎜⎜⎝0000⎞⎟⎟⎠,⎛⎜⎜⎝1000⎞⎟⎟⎠,⎛⎜⎜⎝𝑧000⎞⎟⎟⎠,⎛⎜⎜⎝𝑝000⎞⎟⎟⎠,⎛⎜⎜⎝𝑧𝑝000⎞⎟⎟⎠,⎛⎜⎜⎝1001⎞⎟⎟⎠,⎛⎜⎜⎝100𝑧⎞⎟⎟⎠,⎛⎜⎜⎝𝑝00𝑝⎞⎟⎟⎠,⎛⎜⎜⎝𝑝00𝑧𝑝⎞⎟⎟⎠,⎛⎜⎜⎝100𝑝⎞⎟⎟⎠,⎛⎜⎜⎝𝑧00𝑝⎞⎟⎟⎠,⎛⎜⎜⎝100𝑧𝑝⎞⎟⎟⎠,⎛⎜⎜⎝𝑧00𝑧𝑝⎞⎟⎟⎠.(3.1) The number 𝒞_2,3 of all cogredient classes of 3×3 symmetric matrices over GR(𝑝²,𝑝^2𝑚) is given by 𝒞_2,3=1+∑¹_𝑗=0∑²_𝑖=𝑗^𝑖𝑗²_𝑗+12^𝑗+1=25. In fact, all cogredient standard forms of kind (I) of 3×3 symmetric matrices over GR(𝑝²,𝑝^2𝑚) are given by the following: ^𝐽0₀₀ where 𝐽 is one of matrices in (3.1), and ⎛⎜⎜⎜⎜⎝100010001⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝10001000𝑧⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝𝑝000𝑝000𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝𝑝000𝑝000𝑧𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝10001000𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝1000𝑧000𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝10001000𝑧𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝1000𝑧000𝑧𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝1000𝑝000𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝𝑧000𝑝000𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝1000𝑝000𝑧𝑝⎞⎟⎟⎟⎟⎠,⎛⎜⎜⎜⎜⎝𝑧000𝑝000𝑧𝑝⎞⎟⎟⎟⎟⎠.(3.2)

Example 3.4. Let 𝑝 be an odd prime number and 𝑠=5. Then by Theorem 3.1 the number 𝒞_5,4 of all cogredient classes of 4×4 symmetric matrices over Galois ring 𝐺𝑅(𝑝⁵,𝑝^5𝑚) is given by 𝒞_5,4=1+∑³_𝑗=0∑³_𝑖=𝑗^𝑖𝑗⁵_𝑗+12^𝑗+1=681; the number 𝒞_5,7 of all cogredient classes of 7×7 symmetric matrices over GR(𝑝⁵,𝑝^5𝑚) is given by 𝒞_5,7=1+∑⁴_𝑗=0∑⁶_𝑖=𝑗^𝑖𝑗⁵_𝑗+12^𝑗+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).