International Scholarly Research Notices

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International Scholarly Research Notices / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 520148 | 9 pages | https://doi.org/10.5402/2012/520148

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

Yonglin Cao 1

1School of Sciences, Shandong University of Technology, Shandong, Zibo 255091, China

Academic Editor: D. Kressner
Received20 Mar 2012
Accepted13 May 2012
Published19 Jul 2012

Abstract

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 𝑎=𝑎0+𝑎1𝑝+⋯+𝑎𝑛−1𝑝𝑛−1,𝑎0,𝑎1,…,𝑎𝑛−1∈𝒯,(1.1) 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 Γ𝑖∶=(𝐴,𝐵)∣𝐴,𝐵∈𝒮(𝑛,𝑅),𝐴−𝐵∈𝒮𝑖,𝑖=0,1,…,𝑑.(1.2) 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 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⎞⎟⎟⎟⎠,(1.3) 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 [1, 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 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)(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) 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 𝐻2𝜈=⎛⎜⎜⎝0𝐼(𝜈)𝐼(𝜈)0⎞⎟⎟⎠,𝐻2𝜈+2,Δ=𝐻2𝜈⊕Δ,whereΔ=⎛⎜⎜⎝100−𝑧⎞⎟⎟⎠,𝐻2𝜈+1,(1)=𝐻2𝜈⊕(1),𝐻2𝜈+1,(𝑧)=𝐻2𝜈⊕(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,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 𝑃𝐼(2)𝑃𝑇=2−12⎛⎜⎜⎝(1+𝑧)𝑢−1(1−𝑧)𝑢(1−𝑧)(1+𝑧)⎞⎟⎟⎠⎛⎜⎜⎝(1+𝑧)𝑢(1−𝑧)𝑢−1(1−𝑧)(1+𝑧)⎞⎟⎟⎠=2−12⎛⎜⎜⎝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)𝑄𝑇=𝑄𝑄𝑇=𝑧𝐼(2). Hence, 𝑧𝐼(2) is cogredient to 𝐼(2) as well.
Then 𝑧𝐼(2𝜈)=𝜈𝑠𝑧𝐼(2)⊕⋯⊕𝑧𝐼(2) is cogredient to 𝐼(2𝜈)=𝜈𝑠𝐼(2)⊕⋯⊕𝐼(2).

Lemma 2.2. Let 𝑧∈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 select 𝑃1=2−1𝐼(𝜈)−𝐼(𝜈)𝐼(𝜈)𝐼(𝜈) and denote that 𝑀=2𝐼(𝜈)00−𝐼(𝜈). From 𝑃1𝐼(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 𝑃1𝑀𝑃𝑇1=2−1𝐼(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))𝐻2𝜈𝐼(𝜈)00−𝐼(𝜈)𝐻𝑇2𝜈(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. (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 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 2.1  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∈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)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 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) 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 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) 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).(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 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: 𝐷(𝑛,𝑘,𝑡;𝑘1,…,𝑘𝑡;𝑟1,…,𝑟𝑡)∶=diag𝑝𝑟1𝐷1,𝑝𝑟2𝐷2,…,𝑝𝑟𝑡𝐷𝑡,0(𝑛−𝑘),(2.3) 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 Σ𝑡𝑖=1𝑘𝑖=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 𝑃1𝐴𝑃𝑇1=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 Σ𝑡𝑖=1𝑘𝑖=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: 𝐻(𝑛,𝑘,𝑡;𝑘1,…,𝑘𝑡;𝑟1,…,𝑟𝑡)∶=diag𝑝𝑟1𝐻1,𝑝𝑟2𝐻2,…,𝑝𝑟𝑡𝐻𝑡,0(𝑛−𝑘),(2.4) 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 Σ𝑡𝑖=1𝑘𝑖=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: (i) If 𝑛≤𝑠, then 𝒞𝑠,𝑛=1+∑𝑛−1𝑗=0∑𝑛−1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1; (ii) If 𝑛≥𝑠+1, then 𝒞𝑠,𝑛=1+∑𝑠−1𝑗=0∑𝑛−1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1.

Proof. Let 𝐷∶=diag(𝑝̂𝑟1𝐷1,𝑝̂𝑟2𝐷2,…,𝑝̂𝑟̂‌𝑡𝐷̂‌𝑡,0(𝑛−̂‌𝑘)), where 𝐷𝑖=𝐼(̂‌𝑘𝑖) or 𝐼(̂‌𝑘𝑖−1)⊕(𝑧) for all 𝑖=1,…,̂‌𝑡, 0≤̂‌𝑡≤̂‌𝑘≤𝑛, 0≤̂𝑟1<̂𝑟2<⋯<̂𝑟̂‌𝑡≤𝑠−1, and ̂‌𝑘𝑖∈ℤ+ satisfy Σ̂‌𝑡𝑖=1̂‌𝑘𝑖=̂‌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+,Σ𝑡𝑖=1=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+∑𝑛𝑘=1(∑𝑘𝑡=1|𝑆1|⋅|𝑆2|⋅2𝑡). Therefore, 𝒞𝑠,𝑛=1+∑𝑛−1𝑗=0∑𝑛−1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1 if 𝑛≤𝑠 and, 𝒞𝑠,𝑛=1+∑𝑠−1𝑗=0∑𝑛−1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+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.(i) If 𝑛≤𝑠, then 𝑑=∑𝑛−1𝑗=0∑𝑛−1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1; (ii) If 𝑛≥𝑠+1, then 𝑑=∑𝑠−1𝑗=0∑𝑛−1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+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+∑1𝑗=0∑1𝑖=𝑗𝑖𝑗2𝑗+12𝑗+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: ⎛⎜⎜⎝0000⎞⎟⎟⎠,⎛⎜⎜⎝1000⎞⎟⎟⎠,⎛⎜⎜⎝𝑧000⎞⎟⎟⎠,⎛⎜⎜⎝𝑝000⎞⎟⎟⎠,⎛⎜⎜⎝𝑧𝑝000⎞⎟⎟⎠,⎛⎜⎜⎝1001⎞⎟⎟⎠,⎛⎜⎜⎝100𝑧⎞⎟⎟⎠,⎛⎜⎜⎝𝑝00𝑝⎞⎟⎟⎠,⎛⎜⎜⎝𝑝00𝑧𝑝⎞⎟⎟⎠,⎛⎜⎜⎝100𝑝⎞⎟⎟⎠,⎛⎜⎜⎝𝑧00𝑝⎞⎟⎟⎠,⎛⎜⎜⎝100𝑧𝑝⎞⎟⎟⎠,⎛⎜⎜⎝𝑧00𝑧𝑝⎞⎟⎟⎠.(3.1) The number 𝒞2,3 of all cogredient classes of 3×3 symmetric matrices over GR(𝑝2,𝑝2𝑚) is given by 𝒞2,3=1+∑1𝑗=0∑2𝑖=𝑗𝑖𝑗2𝑗+12𝑗+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 ⎛⎜⎜⎜⎜⎝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 𝒞5,4 of all cogredient classes of 4×4 symmetric matrices over Galois ring 𝐺𝑅(𝑝5,𝑝5𝑚) is given by 𝒞5,4=1+∑3𝑗=0∑3𝑖=𝑗𝑖𝑗5𝑗+12𝑗+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+∑4𝑗=0∑6𝑖=𝑗𝑖𝑗5𝑗+12𝑗+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).

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Copyright © 2012 Yonglin Cao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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