International Scholarly Research Notices

International Scholarly Research Notices / 2012 / Article

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

Yonglin Cao, "Cogredient Standard Forms of Symmetric Matrices over Galois Rings of Odd Characteristic", International Scholarly Research Notices, vol. 2012, Article ID 520148, 9 pages, 2012. https://doi.org/10.5402/2012/520148

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

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 𝑝𝑠[𝑥]/((𝑥)) for any basic irreducible polynomial (𝑥) of degree 𝑚 over 𝑝𝑠. It is clear that 𝑅=𝔽𝑝𝑚, that is, a finite field of 𝑝𝑚 elements, if 𝑠=1, and 𝑅=𝑝𝑠, that is the ring of residue classes of modulo its ideal 𝑝𝑠, 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)(𝑝𝑠1)(𝑝)(𝑝0)=𝑅. By [1, Theorem 14.8], there exists an element 𝜉𝑅 of multiplicative order 𝑝𝑚1, which is a root of a basic primitive polynomial (𝑥) of degree 𝑚 over 𝑝𝑠 and dividing 𝑥𝑝𝑚11 in 𝑝𝑠[𝑥], and every element 𝑎𝑅 can be written uniquely as 𝑎=𝑎0+𝑎1𝑝++𝑎𝑛1𝑝𝑛1,𝑎0,𝑎1,,𝑎𝑛1𝒯,(1.1) where 𝒯={0,1,𝜉,,𝜉𝑝𝑚2}. Moreover, 𝑎 is a unit if and only if 𝑎00, 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++𝑎𝑛1𝑝𝑛1 with 𝑎𝑖0. By [1, Corollary 14.9], 𝑅𝜉×[1+(𝑝)], where 𝜉 is the cyclic group of order 𝑝𝑚1, and 1+(𝑝)={1+𝑥𝑥(𝑝)} is the one group of Galois ring 𝑅, so |𝑅|=(𝑝𝑚1)𝑝(𝑠1)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 𝐴𝐵=𝐴00𝐵 which is a (𝑙+𝑞)×(𝑛+𝑟) matrix over 𝑅.

For any matrix 𝐴M𝑛(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,𝑆2M𝑛(𝑅), if there exists matrix 𝑃GL𝑛(𝑅) such that 𝑃𝑆1𝑃𝑇=𝑆2, we say that 𝑆1 is cogredient to 𝑆2 over 𝑅. It is clear that 𝑆1𝒮(𝑛,𝑅) if and only if 𝑆2𝒮(𝑛,𝑅). So cogredience of matrices over 𝑅 is an equivalent relation on 𝒮(𝑛,𝑅). If 𝑆1𝒮(𝑛,𝑅), we call {𝑃𝑆1𝑃𝑇𝑃GL𝑛(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,𝐵𝒮(𝑛,𝑅),𝐴𝐵𝒮𝑖,𝑖=0,1,,𝑑.(1.2) Then the system (𝒮(𝑛,𝑅),{Γ𝑖}0𝑖𝑑) 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 𝑝𝑠 are given in [3, 4]. If 𝑛2, 𝑠>1 and 𝑝1 (mod 4), a complex formula to count the number of all distinct cogredient classes of 𝒮(𝑛,𝑝𝑠) is given in [3], which shows that, for example,

if 𝑚 is odd and 𝑠 is odd, then 𝑑+1=𝑚12+1+𝑠10,or𝑠𝑖,𝑖𝑚12𝑠1𝑠2+𝑠3+𝑠4+𝑠5+𝜀2+1×𝑠11+𝑠12𝑠111++𝑠1𝑠1×𝑠12𝑠2𝑠+12𝑠3𝑠12𝑠4𝑠+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,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=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 𝑝𝑚1, and 𝐺2=1+𝑝 is a group of order 𝑝(𝑠1)𝑚.

Proposition 1.2. (i)𝑅2 is a subgroup of 𝑅 with index [𝑅𝑅2]=2.
(ii) For any 𝑧𝑅𝑅2, 𝑅𝑅2=𝑧𝑅2, and |𝑅2|=|𝑧𝑅2|=(1/2)|𝑅|.
(iii) For any 𝑢𝑅 and 𝑎𝑝, there exists 𝑐𝑅 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 𝑝𝑚1. Since 𝑝 is odd and 𝑝𝑚1 is even, 𝜉2 is of order (1/2)(𝑝𝑚1) and 𝐺21=𝜉2. Since 𝑝(𝑠1)𝑚 is odd and 𝐺2 is a commutative group of order 𝑝(𝑠1)𝑚 by Lemma 1.1, for every 𝑎𝐺2, there exists a unique 𝑏𝐺2 such that 𝑎=𝑏2, so 𝐺22=𝐺2. Moreover, by Lemma 1.1 each 𝑢𝑅 can be uniquely expressed as 𝑢=𝑔 where 𝑔𝐺1 and 𝐺2. (i) For every 𝑢=𝑔𝑅 where 𝑔𝐺1 and 𝐺2, 𝑢𝑅2 if and only if there exist 𝑔1𝐺1 and 1𝐺2 such that 𝑔=(𝑔11)2=𝑔2121, which is then equivalent to that 𝑔=𝑔21 and =21. So 𝑢𝑅2 if and only if 𝑢𝐺21×𝐺2 by Lemma 1.1. Then 𝑅2=𝐺21×𝐺2 and so |𝑅2|=|𝐺21||𝐺2|=(1/2)(𝑝𝑚1)𝑝(𝑠1)𝑚=(1/2)|𝑅|. Hence, [𝑅𝑅2]=2 by group theory. (ii) Since [𝑅𝑅2]=2, for any 𝑧𝑅𝑅2, we have 𝑅=𝑅2𝑧𝑅2 and 𝑅2𝑧𝑅2= by group theory. So |𝑧𝑅2|=|𝑅||𝑅2|=(1/2)|𝑅| by the proof of (i). (iii) Let 𝑢𝑅 and 𝑎𝑝. Then 𝑢1(𝑢+𝑎)=1+𝑢1𝑎1+𝑝=𝐺2. From this and by Lemma 1.1, there exists a unique element 𝑏𝐺2𝑅 such that 𝑢1(𝑢+𝑎)=𝑏2. Now, let 𝑐=𝑏1. Then 𝑐𝑅 satisfying 𝑐2(𝑢+𝑎)=𝑢.

Proposition 1.3. Let 1𝑅2. Then for any 𝑧𝑅𝑅2, there exist 𝑥,𝑦𝑅 such that 𝑧=(1+𝑥2)𝑦2.

Proof. Let 𝑢𝑅. Suppose that 1+𝑢2𝑅. Then there exists 𝑎𝑅 such that 1+𝑢2=𝑎𝑝. So 𝑢2=(1𝑎𝑝). Since 𝑝 is odd and 𝑝𝑠=0 in 𝑅, there exists 𝑏𝑅 such that (𝑢𝑝𝑠)2=(1𝑎𝑝)𝑝𝑠=(1𝑝𝑝𝑠𝑏)=1. From 𝑢𝑝𝑠𝑅 we deduce 1𝑅2, which is a contradiction. Hence 1+𝑢2𝑅. Therefore, 𝜎𝑤1+𝑤 (forall𝑤𝑅2) is a mapping from 𝑅2 to 𝑅. Suppose that 𝜎(𝑅2)𝑅2. Then for 1𝑅2, there exists 𝑤0𝑅2 such that 𝜎(𝑤0)=1+𝑤0=1, which implies that 𝑤0=0, and we get a contradiction. So there exists 𝑥𝑅 such that 1+𝑥2𝑅2, that is, 1+𝑥2𝑅𝑅2=𝑧𝑅2 by Proposition 1.2. Then there exists 𝑐𝑅 such that 1+𝑥2=𝑧𝑐2, so (1+𝑥2)𝑦2=𝑧, where 𝑦=𝑐1𝑅.

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 𝑧𝑅𝑅2, define 𝐻2𝜈=0𝐼(𝜈)𝐼(𝜈)0,𝐻2𝜈+2,Δ=𝐻2𝜈Δ,whereΔ=100𝑧,𝐻2𝜈+1,(1)=𝐻2𝜈(1),𝐻2𝜈+1,(𝑧)=𝐻2𝜈(𝑧).(2.1)

Lemma 2.1. For any 𝜈+ and 𝑧𝑅𝑅2, 𝑧𝐼(2𝜈) is cogredient to 𝐼(2𝜈).

Proof. Let 1𝑅2. Then there exists 𝑢𝑅 such that 𝑢2=1, that is, 1+𝑢2=0. Since 𝑝 is an odd prime number, we have gcd(2,𝑝𝑠)=1 and so 2𝑅. Let 𝑃=21(1+𝑧)𝑢1(1𝑧)𝑢(1𝑧)(1+𝑧). Since 𝑅 is a commutative ring, we have det𝑃=(21)2[(1+𝑧)(1+𝑧)𝑢1(1𝑧)𝑢(1𝑧)]=(21)222𝑧=𝑧𝑅. Hence, 𝑃GL2(𝑅). Then by (𝑢1)2=(𝑢2)1=1 and 𝑢(1𝑧2)+𝑢1(1𝑧2)=𝑢1(𝑢2+1)(1𝑧2)=0, we get 𝑃𝐼(2)𝑃𝑇=212(1+𝑧)𝑢1(1𝑧)𝑢(1𝑧)(1+𝑧)(1+𝑧)𝑢(1𝑧)𝑢1(1𝑧)(1+𝑧)=21222𝑧0022𝑧=𝑧𝐼(2),(2.2) so 𝑧𝐼(2) is cogredient to 𝐼(2).
Let 1𝑅2. Then by Proposition 1.3 there exist 𝑥,𝑦𝑅 such that (1+𝑥2)𝑦2=𝑧. Let 𝑄=𝑥𝑦𝑦𝑦𝑥𝑦. Then det𝑄=(1+𝑥2)𝑦2=𝑧𝑅 and so 𝑄GL2(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 𝑧𝑅𝑅2 and 𝜈+. (i) If 1𝑅2, then 𝐼(2𝜈) is cogredient to 𝐻2𝜈. (ii) If 1𝑅2, then 𝐼(𝜈)𝑧𝐼(𝜈) is cogredient to 𝐻2𝜈.

Proof. We select 𝑃1=21𝐼(𝜈)𝐼(𝜈)𝐼(𝜈)𝐼(𝜈) and denote that 𝑀=2𝐼(𝜈)00𝐼(𝜈). From 𝑃1𝐼(𝜈)𝐼(𝜈)0𝐼(𝜈)=21𝐼(𝜈)021𝐼(𝜈)𝐼(𝜈) we deduce det𝑃1=det(21𝐼(𝜈))=(21)𝜈𝑅. Hence 𝑃1GL2𝜈(𝑅). Then by 𝑃1𝑀𝑃𝑇1=21𝐼(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 1𝑅2 there exists 𝑢𝑅 such that 1=𝑢2. Then 𝑀 is cogredient to 2𝐼(2𝜈). If 2𝑅2, 2𝐼(2𝜈) is cogredient to 𝐼(2𝜈) by Lemma 2.1. If 2𝑅2, there exists 𝑤𝑅 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 1𝑅2. Then by Proposition 1.2 there exists 𝑐𝑅 such that 1=𝑧𝑐2. Hence 𝐼(𝜈)𝑧𝐼(𝜈) is cogredient to 𝐼(𝜈)00𝐼(𝜈). If 2𝑅2, there exists 𝑤𝑅 such that 2=𝑤2, so 𝐼(𝜈)00𝐼(𝜈) is cogredient to 𝑀. If 2𝑅2, then 2=(1)2𝑅2, and hence there exists 𝑎𝑅 such that 2=𝑎2, so (𝑎𝐼(2𝜈))𝐻2𝜈𝐼(𝜈)00𝐼(𝜈)𝐻𝑇2𝜈(𝑎𝐼(2𝜈))=𝑀. Hence, 𝐼(𝜈)00𝐼(𝜈) is cogredient to 𝑀 as well. Therefore, 𝐼(𝜈)𝑧𝐼(𝜈) is cogredient to 𝐻2𝜈.

Lemma 2.3. Let 𝑧𝑅𝑅2 and 𝐷=diag(𝑢1,,𝑢𝑟), where 𝑢𝑖𝑅, 𝑖=1,,𝑟 and 𝑟+. Then, One has the following. (i)𝐷 is necessarily cogredient to either 𝐼(𝑟) or 𝐼(𝑟1)(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(𝜈1)+2,Δ. Moreover, these two matrices are not cogredient.

Proof. (i) We may assume that 𝑢1,,𝑢𝑡𝑅2 and 𝑢𝑡+1,,𝑢𝑟𝑧𝑅2, where 0𝑡𝑟. Then 𝐷 is cogredient to 𝐼(𝑡)𝑧𝐼(𝑟𝑡). If 𝑟𝑡 is even, by Lemma 2.1  𝑧𝐼(𝑟𝑡) is cogredient to 𝐼(𝑟𝑡) and hence 𝐷 is cogredient to 𝐼(𝑡)𝐼(𝑟𝑡)=𝐼(𝑟). Now, let 𝑟𝑡 be odd. If 𝑟𝑡=1, 𝐷 is obviously cogredient to 𝐼(1)(𝑧). If 𝑟𝑡3, by Lemma 2.1  𝑧𝐼(𝑟𝑡1) is cogredient to 𝐼(𝑟𝑡1), and hence 𝐷 is cogredient to 𝐼(𝑡)𝐼(𝑟𝑡1)(𝑧)=𝐼(𝑟1)(𝑧).
Suppose that 𝐼(𝑟) is cogredient to 𝐼(𝑟1)(𝑧) over 𝑅. Then there exists 𝑄GL𝑟(𝑅) such that 𝑄𝐼(𝑟)𝑄𝑇=𝐼(𝑟1)(𝑧). From this and by det𝑄𝑅, we obtain that 𝑧=(det𝑄)2𝑅2, which is a contradiction. So 𝐼(𝑟) and 𝐼(𝑟1)(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 𝑟1=2𝜈 is even and we have one of the following two cases. (ii-1-1)Let 1𝑅2. 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(𝑟1)(𝑧). (ii-1-2) Let 1𝑧𝑅2. Then we have one of the following two cases.(𝛼) Let (1/2)(𝑟1)=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(𝑟1)(𝑧). (𝛽) Let (1/2)(𝑟1)=𝜈 be odd. Then 𝜈=2𝜔+1 for some nonnegative integer 𝜔 and so 𝑟1=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(𝑟1)(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(𝑟1)(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 𝑟2=2(𝜈1) is also even and we have one of the following two cases.(ii-2-1) Let 1𝑅2. Then 1=𝑢2 for some 𝑢𝑅 and so 100𝑧 is cogredient to 100𝑧=Δ. By Lemma 2.2(i) 𝐷 is cogredient to 𝐻2𝜈 when 𝐷 is cogredient to 𝐼(𝑟), or 𝐷 is cogredient to 𝐻2(𝜈1)+2,Δ when 𝐷 is cogedient to 𝐼(𝑟1)(𝑧)=𝐼(2(𝜈1))100𝑧.(ii-2-2) Let 1𝑧𝑅2. Then 1=𝑧𝑐2 for some 𝑐𝑅. 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 𝐻2Δ=𝐻21+2,Δ. Now, let 𝜈4. Since 𝜈2 is even, 𝐼(𝜈2) is cogredient to 𝑧𝐼(𝜈2) by Lemma 2.1, so 𝐼(𝜈2)𝐼(𝜈2) is cogredient to 𝐼(𝜈2)𝑧𝐼(𝜈2). Hence, 𝐼(𝑟1)(𝑧)=𝐼(𝜈2)𝐼(𝜈2)𝐼(3)(𝑧) is cogredient to 𝐼(𝜈2)𝑧𝐼(𝜈2)𝐼(3)(𝑧), which is then cogredient to 𝐼(𝜈1)𝑧𝐼(𝜈1)𝐼(2). Since 𝐼(2) is cogredient to Δ, we see that 𝐼(𝑟1)(𝑧) is cogredient to 𝐻2(𝜈1)+2,Δ by Lemma 2.2(ii). Therefore, 𝐷 is necessarily cogredient to either 𝐻2𝜈 or 𝐻2(𝜈1)+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(𝜈1)+2,Δ by Lemma 2.2(ii). Now, 𝐼(𝑟1)(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(𝑟1)(𝑧) is cogredient to 𝐻2(2𝜔+1)=𝐻2𝜈 by Lemma 2.2(ii). Therefore, 𝐷 is necessarily cogredient to either 𝐻2𝜈 or 𝐻2(𝜈1)+2,Δ by (i).

Theorem 2.4. Let 𝑧𝑅𝑅2. 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(𝑛𝑘),(2.3) where 0𝑡𝑘𝑛, 𝐷𝑖=𝐼(𝑘𝑖) or 𝐼(𝑘𝑖1)(𝑧) for all 𝑖=1,,𝑡, 0𝑟1<𝑟2<<𝑟𝑡𝑠1, and 𝑘𝑖+ satisfy Σ𝑡𝑖=1𝑘𝑖=𝑘.

Proof. The statement holds obviously if 𝐴=0 (corresponding to the case 𝑘=0) or 𝑛=1. Now, let 𝑛2 and 𝐴=(𝑎𝑖𝑗)𝑛×𝑛0. Then, there exist 1𝑖0,𝑗0𝑛 such that 𝑎𝑖0𝑗00 and 𝜏(𝑎𝑖0𝑗0)=min{𝜏(𝑎𝑖𝑗)𝑎𝑖𝑗0,1𝑖,𝑗𝑛}. Let 𝑠1=𝜈(𝑎𝑖0𝑗0). Then 0𝑠1𝑠1, and there exists 𝑃1GL𝑛(𝑅) such that 𝑃1𝐴𝑃𝑇1=diag(𝑢1𝑝𝑠1,𝐵) where 𝑢1𝑅 and 𝐵=(𝑏𝑖𝑗) is a (𝑛1)×(𝑛1) symmetric matrix over 𝑅 satisfying 𝐵=0 or 𝜏(𝑏𝑖𝑗)𝑠1 for all 𝑏𝑖𝑗0, 1𝑖,𝑗𝑛1. By induction there exists 𝑋GL𝑛1(𝑅) such that 𝑋𝐵𝑋𝑇=diag(𝑢2𝑝𝑠2,,𝑢𝑘𝑝𝑠𝑘,0(𝑛𝑘)), where 𝑢2,,𝑢𝑘𝑅 and 𝑠2𝑠𝑘𝑠1. Then 𝑃=diag(1,𝑋)𝑃1GL𝑛(𝑅) satisfies 𝑃𝐴𝑃𝑇=diag(𝑢1𝑝𝑠1,,𝑢𝑘𝑝𝑠𝑘,0(𝑛𝑘)).
Now, there must exist 𝑡,𝑘𝑖+, 𝑖=1,,𝑡 and 0𝑟1<<𝑟𝑡𝑠1 such that 𝑠1==𝑠𝑘1=𝑟1<𝑠𝑘1+1==𝑠𝑘1+𝑘2=𝑟2<<𝑠𝑘1+𝑘2++𝑘𝑡1+1==𝑠𝑘1+𝑘2++𝑘𝑡1+𝑘𝑡=𝑟𝑡. Then Σ𝑡𝑖=1𝑘𝑖=𝑘 and 𝐴 is cogredient to 𝑀=diag(𝑝𝑟1𝑀1,𝑝𝑟2𝑀2,,𝑝𝑟𝑡𝑀𝑡,0(𝑛𝑘)), where 𝑀𝑖=diag(𝑢𝑘1++𝑘𝑖1+1,,𝑢𝑘1++𝑘𝑖1+𝑘𝑖) is a 𝑘𝑖×𝑘𝑖 matrix over 𝑅 for all 𝑖=1,,𝑡. Since 𝑀𝑖 is cogredient to 𝐷𝑖 for every 1𝑖𝑡 by Lemma 2.3(i), we deduce that 𝐴 is cogredient to diag(𝑝𝑟1𝐷1,𝑝𝑟2𝐷2,,𝑝𝑟𝑡𝐷𝑡,0(𝑛𝑘)).

Theorem 2.5. Let 𝑧𝑅𝑅2. 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(𝑛𝑘),(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(𝜈𝑖1)+2,Δ when 𝑘𝑖=2𝜈𝑖 is even, for all 𝑖=1,,𝑡; 0𝑡𝑘𝑛, 0𝑟1<𝑟2<<𝑟𝑡𝑠1, and 𝑘𝑖+ 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 𝑛𝑠, then 𝒞𝑠,𝑛=1+𝑛1𝑗=0𝑛1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1; (ii) If 𝑛𝑠+1, then 𝒞𝑠,𝑛=1+𝑠1𝑗=0𝑛1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1.

Proof. Let 𝐷=diag(𝑝̂𝑟1𝐷1,𝑝̂𝑟2𝐷2,,𝑝̂𝑟̂𝑡𝐷̂𝑡,0(𝑛̂𝑘)), where 𝐷𝑖=𝐼(̂𝑘𝑖) or 𝐼(̂𝑘𝑖1)(𝑧) for all 𝑖=1,,̂𝑡, 0̂𝑡̂𝑘𝑛, 0̂𝑟1<̂𝑟2<<̂𝑟̂𝑡𝑠1, and ̂𝑘𝑖+ 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 1𝑡𝑘𝑛, denote that 𝑆1={(𝑘1,,𝑘𝑡)𝑘𝑖+,Σ𝑡𝑖=1=𝑘} and 𝑆2={(𝑟1,,𝑟𝑡)𝑟𝑖,0𝑟1<𝑟2<<𝑟𝑡𝑠1}. Then |𝑆1|=𝑘1𝑡1, |𝑆2|=(𝑠𝑡) if 𝑡𝑠 and, |𝑆2|=0 if 𝑡𝑠. From this and by Theorem 2.4 it follows that 𝒞𝑠,𝑛=1+𝑛𝑘=1(𝑘𝑡=1|𝑆1||𝑆2|2𝑡). Therefore, 𝒞𝑠,𝑛=1+𝑛1𝑗=0𝑛1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1 if 𝑛𝑠 and, 𝒞𝑠,𝑛=1+𝑠1𝑗=0𝑛1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1 if 𝑛𝑠+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 𝑛𝑠, then 𝑑=𝑛1𝑗=0𝑛1𝑖=𝑗𝑖𝑗𝑠𝑗+12𝑗+1; (ii) If 𝑛𝑠+1, then 𝑑=𝑠1𝑗=0𝑛1𝑖=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=01𝑖=𝑗𝑖𝑗2𝑗+12𝑗+1=13. In fact, for a fixed element 𝑧𝑅𝑅2, 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=02𝑖=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=03𝑖=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=06𝑖=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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