Duadic Codes over Finite Commutative Rings - غزةlibrary.iugaza.edu.ps/thesis/112745.pdf ·...

The Islamic University of Gaza

Faculty of Science

Department of Mathematics

Duadic Codes over Finite Commutative Rings

PRESENTED BY

Ikhlas Ibraheem Diab Al-Awar

SUPERVISED BY

Prof. Mohammed Mahmoud AL-Ashker

A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENT

FOR THE DEGREE OF MASTER OF MATHEMATICS

March 4, 2014

Contents

Acknowledgments iii

Abstract iv

Introduction v

1 Preliminaries 1

1.1 Block codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Properties of codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Cyclic codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.2 Idempotent generators of cyclic codes . . . . . . . . . . . . . . . . . . 11

2 Duadic codes over finite fields 14

2.1 Fundamental properties of duadic codes . . . . . . . . . . . . . . . . . . . . . 14

2.2 Fundamental properties of quadratic residue codes . . . . . . . . . . . . . . . 18

2.3 Quadratic residue codes over fields of characteristic 2 . . . . . . . . . . . . . 20

2.4 Quadratic residue codes over fields of characteristic 3 . . . . . . . . . . . . . 22

3 Quadratic residue codes over finite commutative rings 25

3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2 Quadratic residue codes over Z4 . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.3 Quadratic residue codes over Z9 . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.4 Quadratic residue codes over F3 + vF3, v2 = 1 . . . . . . . . . . . . . . . . . 32

3.5 Quadratic residue codes over F3 + uF3, u2 = 0 . . . . . . . . . . . . . . . . . 35

4 Duadic codes over finite commutative chain rings 58

4.1 Duadic group algebra codes . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.2 Duadic codes over Z4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

i

4.3 Duadic codes over F2 + uF2, u2 = 0 . . . . . . . . . . . . . . . . . . . . . . . 78

4.4 Duadic codes over some other finite commutative chain rings . . . . . . . . . 79

Conclusion 83

Appendix 84

References 93

ii

Acknowledgment

Firstly, I offer my thank to Almighty Allah who always help and guide me to bring forth

to light this search.

Foremost, I would like to express my sincere gratitude to Prof. Mohammed Mahmoud

Al-ashker who helped me in all the time of thesis.

Besides my supervisor, I would like to thank the rest of my thesis committee for their

insightful comments.

Also, I would like to express my sincere thanks to the staff members of mathematics

department and the faculty of science of the Islamic University.

At the last but not least, I would like to thank my family: my parents Ibraheem and

Zainab for supporting me spiritually throughout my study and taught me to come to this

stage of learning and to my brothers Diab and Loay and my sisters Marwa, Sarah and Hadeel.

iii

Abstract

Duadic codes are a generalization of quadratic residue codes. We obtain in this thesis

quadratic residue codes over the chain ring F3+uF3, u2 = 0 and then obtain duadic codes for

this ring. Also, we give a construction of duadic codes over the rings Z4 and F2+uF2, u2 = 0.

We give specific constructions of these codes over finite fields and extend this construction

to finite commutative chain rings Fq + uFq + u2Fq + ... + us−1Fq with us = 0 where q = pr,

p is a prime number and r ∈ Z.

iv

Introduction

Coding theory is concerned with successfully transmitting data through a noisy channel and

correcting errors in corrupted messages. It is of central importance for many applications

in computer science or engineering [21]. Coding theory originated with the 1948 publication

of the paper A mathematical theory of communication by Claude Shannon [29]. For the

past half century, coding theory has grown into a discipline intersecting mathematics and

engineering with applications to almost every area of communication such as satellite and

cellular telephone transmission, compact disc recording, and data storage [14]. Legend goes

that Hamming was so frustrated the computer halted every time it detected an error after

he handed in a stack of punch cards, he thought about a way the computer would be able

not only to detect the error but also to correct it automatically. He came with his nowadays

famous code named after him [25].

The idea of channel coding as in the following diagram is to encode the message again after

the source coding by introducing some form of redundancy so that errors can be detected or

even corrected [21].

Protecting digital information with a suitable error-control code enables the efficient

detection and correction of any errors that may have occurred [24].

I hope that this thesis, with its numerous theorems and examples, will serve as a lucid

introduction that will enable readers to pursue some of the many themes of coding theory.

In this thesis, I briefly describe the contents of four chapters, with appendix.

Chapter 1 covers the main of block codes and give types of codes; in particular linear

codes and dual codes with general properties of codes. On the other hand, gives the basic

theory of cyclic codes. My presentation interrelates the concepts of description and idem-

potent generator of cyclic codes. Continuing with the theory of cyclic codes, Chapter 2

presents the fundamentals properties of duadic codes, which include the family of quadratic

v

residue codes which is delineates QR codes over fields of characteristic 2 and 3 and then

in the pages that follow, In chapter 3, we study the quadratic residue codes over finite

commutative rings. Finally, after the basics have been covered, chapter 4 concludes the

thesis with calculations of duadic codes with examples over the finite commutative chain

ring, F3 + uF3, u2 = 0.

Appendix is certainly essential for the understanding all of the parts of the thesis; Ap-

pendix says about basic concepts on algebra and is broken into three addresses, the first

address discusses some definitions about the fields and rings, the second address examines

some definitions on modules and the third one gives an important part called a polynomials

over fields.

vi

Chapter 1

Preliminaries

In the first section we discuss the basic definitions of codes, linear codes and the crucial

concept of the dual of a code, in the second section, we give some properties of codes. In

particular, we introduce in the third section a type of codes called cyclic codes.

1.1 Block codes

Let us display a few simple definitions.

Definition 1.1.1. [21] Let A = {a1, a2, ..., aq} be a set of size q, which we refer to as a code

alphabet and whose elements are called code symbols.

Let Fq be a field of order q, where q is a power of prime p. [21] The set Fnq =

{(x1, ..., xn)|xi ∈ Fq} of all n-letter words with components from Fq is an n-dimensional

vector space, with addition of vectors and multiplication of vectors by a scalar performed in

Fq:

(x1, ..., xn) + (y1, ..., yn) = (x1 + y1, ..., xn + yn),

α(x1, ..., xn) = (αx1, ..., αxn), α ∈ Fq.

Definition 1.1.2. [21] A code C of length n over a finite field Fq of size q is a subset C of

the set Fnq of all n-letter words with components from Fq.

Remark 1.1.1. [21] 1. An element of C is called a codeword in C.

2. The number of codewords in C, denoted by |C|, is called the size of C.

1

3. A code of length n and size M is called an (n,M)− code.

Example 1.1.1. [21] 1. A code over the code alphabet F2 = {0, 1} is called a binary code;

i.e., the code symbols for a binary code are 0 and 1 (Example 5, Appendix A).

Some examples of binary codes are:

(i) C1 = {00, 01, 10, 11} is a (2, 4)− code,

(ii) C2 = {000, 011, 101, 110} is a (3, 4)− code, and

(iii) C3 = {0011, 0101, 1010, 1100, 1001, 0110} is a (4, 6)− code.

2. A code over the code alphabet F3 = {0, 1, 2} is called a ternary code (Example 6,

Appendix A).

3. The term quaternary code is sometimes used for a code over the code alphabet F4

(Example 7, Appendix A).

Remark 1.1.2. [14] The term quaternary has also been used to refer to codes over the ring

Z4 of integers modulo 4.

Definition 1.1.3. [21] A linear code over Fq is a linear subspace of the n-dimensional vector

space Fnq .

Remark 1.1.3. [14] The linear code C has qk codewords.

Example 1.1.2. [21] The following are linear codes:

1. (q = 3)C = {0000, 1100, 2200, 0001, 0002, 1101, 1102, 2201, 2202}.

2. (q = 2)C = {000, 001, 010, 011, 100, 101, 110, 111}.

Definition 1.1.4. [21] Let x and y be words of length n over C. The (Hamming) distance

from x to y, denoted by d(x,y), is defined to be the number of places at which x and y

differ, i.e., If x = x1...xn and y = y1...yn, then d(x, y) = d(x1, y1) + ...+ d(xn, yn),

where xi and yi are regarded as words of length 1, and

d(xi, yi) =

1, xi 6= yi;

0, xi = yi.

Example 1.1.3. [24] If x = (0011010) and y = (1011100), then d(x, y) = 3.

A part from the length and size of a code, another important and useful characteristic of

a code is its distance.

2

Definition 1.1.5. [21] For a code C containing at least two words, the (minimum) distance

of C, denoted by d(C), is

d(C) = min{d(x, y) : x, y ∈ C, x 6= y}.

Remark 1.1.4. [14] We denote a linear code C over Fq of length n and dimension k by [n, k].

If furthermore the minimum distance of the code is d, then we call by [n, k, d]-code.

Example 1.1.4. [21] Let C = {00000, 00111, 11111} be a binary code. Then d(C) = 2, since

d(00000, 00111) = 3,

d(00000, 11111) = 5,

and

d(00111, 11111) = 2.

Hence, C is a binary [5, 3, 2]− code.

Example 1.1.5. [21] Let C = {000000, 000111, 111222} be a ternary code. Then d(C) = 3,

since

d(000000, 000111) = 3,

d(000000, 111222) = 6,

and

d(000111, 111222) = 6.

Hence, C is a ternary [6, 3, 3]− code.

Definition 1.1.6. [25] For an [n, k] code C we define the dual or orthogonal code C⊥ as

C⊥ = {x ∈ Fnq | c · x = 0 for all c ∈ C}.

Example 1.1.6. [21] 1. Let q = 2 and let n = 4.

If u = (1, 1, 1, 1), v = (1, 1, 1, 0) and w = (1, 0, 0, 1), then

u · v = 1 · 1 + 1 · 1 + 1 · 1 + 1 · 0 = 1,

u ·w = 1 · 1 + 1 · 0 + 1 · 0 + 1 · 1 = 0,

v ·w = 1 · 1 + 1 · 0 + 1 · 0 + 0 · 1 = 1.

Hence, u and w are orthogonal.

3

Remark 1.1.5. [25] A code C is self -orthogonal provided C ⊆ C⊥ and self -dual provided

C = C⊥. The length n of a self-dual code is even and the dimension is n/2.

[14] When studying quaternary codes over the field F4, it is often useful to consider

another inner product, called the Hermitian inner product, given by

〈x,y〉 = x · y =n∑i=1

xiyi,

where − called conjugation, is given by 0 = 0, 1 = 1, and ω = ω.

Using this inner product, we can define the Hermitian dual of a quaternary code C to

be,

C⊥H = {x ∈ Fnq |< x, c >= 0 for all c ∈ C}.

Define the conjugate of C to be

C = {c|c ∈ C},

where c = c1 c2...cn when c = c1c2...cn.

Remark 1.1.6. [14] 1. C⊥H = C⊥, where C is a codeword over F4.

2. C is Hermitian self -orthogonal if C ⊆ C⊥H and Hermitian self -dual if C = C⊥H .

1.2 Properties of codes

We put some examples and facts about vector spaces over finite fields. While the proofs of

most of the facts stated in this section are omitted, it should be noted that many of them

are practically identical to those in the case of vector spaces over R or C (see ref. [6] and

[23] for some definitions).

Example 1.2.1. We want to see which of the following sets of vectors are linearly indepen-

dent?

1. (000), (101) ∈ F32.

Note that 1(000) + 0(101) = (000), so (000) and (101) are not linearly independent.

2.The vectors (0212), (0010), (2212) ∈ F43 are linearly independent.

4

Example 1.2.2. [21] 1. Any set S which contains 0 is linearly dependent.

2. For any Fq, {(0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0)} is linearly independent.

3. For any Fq, {(0, 0, 0, 1), (1, 0, 0, 0), (1, 0, 0, 1)} is linearly dependent.

Example 1.2.3. [21] 1. If q = 2 and S = {0001, 1000, 1001}, then the span of S is

〈S〉 = {0000, 0001, 1000, 1001}.

2. If q = 3 and S = {0001, 1000, 1001}, then the span of S is

〈S〉 = {0000, 0001, 0002, 1000, 2000, 1001, 1002, 2001, 2002}.

Theorem 1.2.1. [21] Let V be a vector space over Fq. If dim(V ) = k, then V has qk

elements.

Example 1.2.4. [21] Let q = 2, S = {0001, 0010, 0100} and V = 〈S〉, then

V = {0000, 0001, 0010, 0100, 0011, 0101, 0110, 0111}.

Since S is linearly independent, so dim(V ) = 3. We see that |V | = 8 = 23.

The two most common ways to present the basis of a linear code are with either a

generator matrix or a parity check matrix.

Definition 1.2.1. [14] A k × n matrix G whose rows form a basis for an [n, k] linear code

C is called a generator matrix of the code C.

Definition 1.2.2. [14] For any set of k independent columns of a generator matrix G, the

corresponding set of coordinates forms an information set for C.

Remark 1.2.1. [14] In general, there are many generator matrices for a code. But if the first

k coordinates form an information set, the code has a unique generator matrix of the form

[Ik|A] where Ik is the k × k identity matrix. Such a generator matrix is in standard form

or say systematic.

Definition 1.2.3. [14] The remaining r = n − k coordinates are termed a redundancy set

and r is called the redundancy of C.

5

Definition 1.2.4. [25] An (n− k)× n matrix of rank n− k is called a parity check matrix

of an [n, k]-code C if C is the null space of this matrix.

Theorem 1.2.2. [14] If G = [Ik|A] is a generator matrix for the [n, k] code C in standard

form, then H = [−AT |In−k] is a parity check matrix for C, where Ik is the k × k identity

matrix.

Example 1.2.5. [14] The matrix G = [I4|A], where

G =

1 0 0 0 0 1 1

0 1 0 0 1 0 1

0 0 1 0 1 1 0

0 0 0 1 1 1 1

is a generator matrix in standard form for a [7, 4] binary code that we denote by H3. By

Theorem 1.2.2, a parity check matrix for H3 is

H = [−AT |I3] =

0 1 1 1 1 0 0

1 0 1 1 0 1 0

1 1 0 1 0 0 1

Remark 1.2.2. [14] The code in Example 1.2.5 is called the [7, 4] Hamming code.

We have seen that there are several relationships among the generator matrix G of a

linear code C, the parity check matrix H of a linear code C and the dual code C⊥ in the

following corollary:

Corollary 1.2.1. [25] Let C be a linear code. Then:

1. G is a generator matrix of C if and only if G is a parity check matrix of C⊥.

2. H is a parity check matrix of C if and only if H is a generator matrix of C⊥.

Example 1.2.6. [14] The [4, 2] ternary code H3,2, often called the tetracode, has generator

matrix G, in standard form, given by

G =

1 0 1 1

0 1 1 −1

,

This code is also self-dual.

6

Definition 1.2.5. [25] Consider the situation of two Fq-linear codes C and D of length n.

If D ⊆ C, then D is called a subcode of C, and C is a supercode of D.

Example 1.2.7. 1. [14] If the code C is the [n, 1] binary linear code consisting of the two

codewords 0 = 00...0 and 1 = 11...1, then it is called the binary repetition code of length n.

2. [25] The repetition code over Fq of length n consists of all words c = (c, c, ..., c) with

c ∈ Fq. This is a linear code of dimension 1 and minimum distance n.

Example 1.2.8. [25] The repetition code of length n has generator matrix

G =(

1 1 · · · 1).

Example 1.2.9. [14] The [6, 3] quaternary code over F4, G6 has generator matrix in standard

form given by

G6 =

1 0 0 1 ω ω

0 1 0 ω 1 ω

0 0 1 ω ω 1

This code is often called the hexacode. It is Hermitian self-dual.

Definition 1.2.6. [14] The (Hamming) weight wt(x) of a vector x ∈ Fnq is the number of

nonzero coordinates in x.

Definition 1.2.7. [25] The minimum weight of a code C, denoted by wt(C), is defined as

the minimal value of the weights of the nonzero codewords:

wt(C) = min{wt(c) | c ∈ C, c 6= 0}.

Proposition 1.2.3. [25] The minimum distance of a linear code C is equal to its minimum

weight.

Definition 1.2.8. [14] Let Ai, also denoted Ai(C), be the number of codewords of weight i

in C. The list Ai for 0 ≤ i ≤ n is called the weight distribution.

Definition 1.2.9. [14] Two linear codes C1 and C2 are permutation equivalent provided

that there is a permutation of coordinates which sends C1 to C2.

7

Remark 1.2.3. [14] 1. This permutation in Definition 1.2.9, can be described using a

permutation matrix, which is a square matrix with exactly one (1) in each row and column

and 0’s elsewhere.

2. If P is a permutation matrix sending C1 to C2, we will write C1P = C2, where

C1P = {y|y = xP for x ∈ C1}.

Example 1.2.10. The two codes C1 and C2 below are equivalent since the second is merely

the permutation of the first two positions in the first code.

G1 =

0 0 0

1 0 1

0 1 0

1 1 1

→ G2 =

0 0 0

0 1 1

1 0 0

1 1 1

.

Definition 1.2.10. [14] A monomial matrix is a square matrix with exactly one nonzero

entry in each row and column.

Remark 1.2.4. [14] A monomial matrix M can be written either in the form DP or the form

PD1, where D and D1 are diagonal matrices and P is a permutation matrix.

Example 1.2.11. [14] The monomial matrix

M =

0 a 0

0 0 b

c 0 0

equals

DP =

a 0 0

0 b 0

0 0 c

0 1 0

0 0 1

1 0 0

= PD1 =

0 1 0

0 0 1

1 0 0

c 0 0

0 a 0

0 0 b

.

1.3 Cyclic codes

Cyclic codes are an important class of linear block codes, the mathematical structure theory

of cyclic codes suggest the study of cyclic invariance in the context of linear codes.

In this section, we study the description of cyclic codes and their generators and idem-

potent.

8

1.3.1 Description

Definition 1.3.1. [14] An [n, k, d] linear code C is cyclic if whenever

(c0, c1, ..., cn−1)

is a codeword in C, then

(cn−1, c0, ....., cn−2)

is also a codeword in C.

Example 1.3.1. The binary code C = {000, 110, 011, 101} is a cyclic code.

Remark 1.3.1. [14] The cyclic codes over Fq are precisely the ideals of Rn = Fq[x]/〈xn − 1〉.

We now use the language of polynomials to characterize cyclic codes:

Definition 1.3.2. [25] Let C be a cyclic code and let g(x) be the monic polynomial of

minimal degree such that g(x) ∈ C, then g(x) is called the generator polynomial of C.

We will give some properties of a generator polynomial in the following theorem.

Theorem 1.3.1. [14] Let C be a nonzero cyclic code in Rn. There exists a polynomial

g(x) ∈ C with the following properties:

(i) g(x) is the unique monic polynomial of minimum degree in C,

(ii) C = 〈g(x)〉 and

(iii) g(x) divides xn − 1.

Example 1.3.2. [25] The polynomial x3 + x+ 1 divides x8 − 1 in F3[x], since

(x3 + x+ 1)(x5 − x3 − x2 + x− 1) = x8 − 1.

Hence 1 + x+ x3 is a generator polynomial of a ternary cyclic code of length 8.

Theorem 1.3.2. [21] Let g(x) = g0 + g1x+ ...+ gn−kxn−k be the generator polynomial of a

cyclic code C in Fqn with deg(g(x)) = n− k. Then the matrix

g0 g1 . . . gn−k 0 0 . . . 0

0 g0 g1 . . . gn−k 0 · · · 0

0 0 g0 g1 . . . gn−k. . .

......

.... . . . . . . . . . . .

. . . 0

0 0 . . . 0 g0 g1 . . . gn−k

9

is a generator matrix of C.

Example 1.3.3. [25] The ternary cyclic code of length 8 with generator polynomial 1+x+x3

of Example 1.3.2 has dimension 5.

Theorem 1.3.3. [14] The dual code of a cyclic code is cyclic.

We will find the generator polynomial of the dual of a cyclic code as in the following

theorem.

Theorem 1.3.4. [14] Let C be an [n, k] cyclic code with generator polynomial g(x). Let

h(x) = (xn − 1)/g(x) =∑k

i=0 hixi. Then the generator polynomial of C⊥ is g⊥(x) =

xkh(x−1)/h(0).

Definition 1.3.3. [21] An element α in a finite field Fq is called a primitive element (or

generator) of Fq if

Fq = {0, α, α2, ..., αq−1}.

Example 1.3.4. [21] Consider the field F4 = F2[α], where α is a root of the irreducible

polynomial 1 + x+ x2 ∈ F2[x]. Then we have

α2 = −(1 + α) = 1 + α and α3 = α(α2) = α(1 + α) = α + α2 = α + 1 + α = 1.

Thus,

F4 = {0, α, 1 + α, 1} = {0, α, α2, α3},

So α is a primitive element.

Proposition 1.3.5. [21] Every finite field has at least one primitive element.

Definition 1.3.4. [14] The order of q modulo n (ordn(q)) is the smallest positive integer a

such that qa ≡ 1 (mod n).

Definition 1.3.5. [28] An element α of a field is an nth root of unity if αn = 1.

Definition 1.3.6. [28] If α is an nth root of unity and if n is the smallest positive integer

for which αn = 1, then α is called primitive nth root of unity.

Remark 1.3.2. [14] If t = ordn(q), then Fqt contains a primitive nth root of unity α, but no

smaller extension field of Fq contains such a primitive root.

10

Definition 1.3.7. [14] The q − cyclotomic coset of s modulo n is the set

Cs = {s, sq, ..., sqr−1}(mod n),

where r is the smallest positive integer such that sqr ≡ s (mod n).

Remark 1.3.3. [14]

The distinct q-cyclotomic cosets modulo n partition the set of integers {0, 1, ..., n− 1}.

Example 1.3.5. The 3-cyclotomic cosets modulo 11 are C0 = {0}, C1 = {1, 3, 4, 5, 9} and

C2 = {2, 6, 7, 8, 10}.

So ord11(3) = 5(= the size of C1) and the primitive 11th roots of unity lie in F243 but in

no smaller extension field of F3.

Theorem 1.3.6. [14] The size of each q-cyclotomic coset is a divisor of ordn(q). Further-

more, the size of C1 is ordn(q).

Corollary 1.3.1. [14] The number of cyclic codes in Rn equals 2m, where m is the number

of q-cyclotomic cosets modulo n. Moreover, the dimensions of cyclic codes in Rn are all

possible sums of the sizes of the q-cyclotomic cosets modulo n.

Example 1.3.6. [14] Over F2, x9 − 1 = (1 + x)(1 + x + x2)(1 + x3 + x6), and so there are

(23) eight binary cyclic codes Ci of length 9.

Definition 1.3.8. [14] Let T = ∪sCs be the union of the q-cyclotomic cosets, then the set

T is called the defining set of C.

1.3.2 Idempotent generators of cyclic codes

After we say about a generator polynomial, we next introduce idempotent generators of

cyclic codes which is an important part of these codes.

Definition 1.3.9. [14] An element e of the ring Rn = Fq[x]/〈xn−1〉 is called an idempotent

if e2 = e.

In particular case, take e = 0 and e = 1.

11

Theorem 1.3.7. [14] Let q be a prime power and n be a positive integer such that gcd(n, q) =

1, and let C be a linear cyclic code of length n over GF (q). There exists an idempotent

e(x) ∈ C such that

C = 〈e(x)〉 = {a(x)e(x)|a(x) ∈ Rn}.

Remark 1.3.4. [14] An idempotent e(x) of a cyclic code C such that C = 〈e(x)〉 is called an

idempotent generator of C.

Example 1.3.7. [14] The generating idempotent for the zero cyclic code {0} is 0, while that

for the cyclic code Rn is 1.

Theorem 1.3.8. [14] Let C be a cyclic code in Rn. Then:

i. There exists a unique idempotent e(x) ∈ C such that C = 〈e(x)〉, and

ii. If e(x) is a nonzero idempotent in C, then C = 〈e(x)〉 if and only if e(x) is a unity

of C.

Remark 1.3.5. [14] If e1(x) and e2(x) are idempotents, so are e1(x)e2(x), e1(x) + e2(x) −

e1(x)e2(x), and 1− e1(x).

Theorem 1.3.9. [14] Let C be a cyclic code over Fq with generating idempotent e(x). Then

the generator polynomial of C is g(x) = gcd(e(x), xn − 1) computed in Fq[x].

Example 1.3.8. The generator polynomial of the binary cyclic code of length 7 with idem-

potent generator x3 + x6 + x5 is

gcd(x3 +x6 +x5, x7−1) = gcd(x3(1+x2 +x3), (x−1)(x3 +x+1)(x3 +x2 +1)) = x3 +x2 +1.

Definition 1.3.10. [14] If C1 and C2 are codes of length n over Fq, then

C1 + C2 = {c1 + c2 | c1 ∈ C1 and c2 ∈ C2}

is the sum of C1 and C2.

Remark 1.3.6. [14] The intersection and the sum of two cyclic codes are cyclic.

Theorem 1.3.10. [14] Let Ci be a cyclic code of length n over Fq with generator polynomial

gi(x) and generating idempotent ei(x) for i = 1 and 2. Then:

12

i. C1∩C2 has generator polynomial lcm(g1(x), g2(x)) and generating idempotent e1(x)e2(x),

and

ii. C1 +C2 has generator polynomial gcd(g1(x), g2(x)) and generating idempotent e1(x)+

e2(x)− e1(x)e2(x).

Meggitt decoding of cyclic codes is presented as are extended cyclic codes.

Let C be an [n, k, d] cyclic code over Fq with generator polynomial g(x) of degree n− k;

C will correct t = b(d − 1)/2c errors. Suppose that c(x) ∈ C is transmitted and y(x) =

c(x) + e(x) is received, where e(x) = e0 + e1x + ... + en−1xn−1 is the error vector with

wt(e(x)) ≤ t. The Meggitt decoder stores syndromes of error patterns with coordinate n− 1

in error. The two versions of the Meggitt Decoding Algorithm that we present can briefly

be described as follows:

In the first version, by shifting y(x) at most n times, the decoder finds the error vector

e(x) from the list and corrects the errors. In the second version, by shifting y(x) until an

error appears in coordinate n − 1, the decoder finds the error in that coordinate, corrects

only that error, and then corrects errors in coordinates n− 2, n− 3, ..., 1, 0 in that order by

further shifting [14].

13

Chapter 2

Duadic codes over finite fields

In this chapter we study duadic codes over finite fields.

Binary duadic codes were introduced by Leon, Masley and Pless in [18] are a generaliza-

tion of quadratic residue codes, for further studies see [31].

2.1 Fundamental Properties of duadic codes

Definition 2.1.1. [14] A vector x = x1x2...xn in Fnq is an even-like if

n∑i=1

xi = 0

and is an odd-like otherwise.

Remark 2.1.1. [14] 1. The even-like vectors in a code form a subcode of a code over Fq.

2. We say that a code is even-like if it has only even-like codewords; a code is odd-like

if it is not even-like.

In defining the duadic codes, we will obtain two pairs of codes; one pair will be even-like

codes, which are thus subcodes of a code, and the other pair will be odd-like codes.

Duadic codes are a class of cyclic codes that generalizes quadratic residue codes from

prime to composite lengths.

We first define duadic codes in terms of their idempotents. Duadic codes come in two

pairs as we say previously, one even-like pair, which we usually denoted C1 and C2, and one

odd-like pair, usually denoted D1 and D2.

14

Definition 2.1.2. [8] Let a be an integer such that gcd(a, n) = 1. We define the function

µa, called a multiplier, on {0, 1, 2, ..., n− 1} by iµa ≡ ia (mod n).

Remark 2.1.2. [8]

1. µa gives a permutation of the coordinate positions of a cyclic code of length n.

2. Note that this is equivalent to the action of µa on Rn by f(x)µa ≡ f(xa)(mod xn − 1).

Definition 2.1.3. [30] A binary cyclic code of length n is called a duadic code if its idem-

potent is one of the following:

e1(x), e2(x), 1 + e1(x) or 1 + e2(x).

Definition 2.1.4. [13] Let n be an odd positive integer with n > 1. A pair (S1, S2) of two

sets S1 and S2 is called a (generalized duadic) splitting of n if the following two conditions

are satisfied:

(i) S1 and S2 satisfy

S1 ∪ S2 = {1, 2, ....., n− 1} and S1 ∩ S2 = ∅,

(ii) There is a multiplier µa such that

S1µa = S2 and S2µa = S1.

Definition 2.1.5. [13] Suppose that there is a splitting (S1, S2) of n and let q be a power

of a prime with gcd(q, n) = 1. Furthermore, if S1 and S2 are unions of nonzero q-cyclotomic

cosets, then the cyclic codes with defining sets S1 and S2 are called the odd-like duadic codes

of length n over Fq.

On the other hand, the cyclic codes with defining sets S1 ∪ {0} and S2 ∪ {0} are called

the even-like duadic codes of length n over Fq.

Definition 2.1.6. [16] Let m be a positive integer and a any integer such that gcd(a,m) = 1.

Then a is a quadratic residue of m if the quadratic congruence x2 ≡ a (mod m) has a

solution; otherwise, it is a quadratic nonresidue of m.

Example 2.1.1. The quadratic residues of 7 are: {1, 2, 4} and the non residues of 7 are:

{3, 5, 6}, since

12 ≡ 62 ≡ 1 (mod 7),

15

22 ≡ 52 ≡ 4 (mod 7),

and

32 ≡ 42 ≡ 2 (mod 7).

Lemma 2.1.1. [14] Let n = n1n2 where gcd(n1, n2) = 1. There is a splitting of n given by µa

if and only if there are splittings of n1 and n2 given by µamod(n1) and µamod(n2), respectively.

Furthermore, q is a square modulo n if and only if q is a square modulo n1 and a square

modulo n2.

Now we recall the criteria for the existence of duadic codes of length n over Fq.

Theorem 2.1.2. [14] Duadic codes of length n over Fq exist if and only if q is a square

modulo n.

Proof. By Lemma (2.1.1), we may assume that n = pm, where p is an odd prime. We first

show that if a splitting of n = pm exists, then q is a square modulo n. Let U be the group

of units in Zn. This group is cyclic of order φ(pm) = (p− 1)pm−1, which is even as p is odd.

Since q is relatively prime to n, q ∈ U. Let R be the subgroup of U consisting of the squares

in U. We only need to show that q ∈ R. If u generates U, then u2 generates R. As U has

even order, R has index 2 in U. Since q ∈ U, define Q to be the subgroup of U generated by

q. Notice that if a ∈ U, then aq ∈ U and hence U is a union of q-cyclotomic cosets modulo

n; in fact, the q-cyclotomic cosets contained in U are precisely the cosets of Q in U. The

number of q-cyclotomic cosets in U is then the index |U : Q| of Q in U. Let S1 and S2 form a

splitting of n given by µb. Each q-cyclotomic coset of U is in precisely one Si as U ⊆ S1 ∪S2

and S1 ∩ S2 = φ Because b and n are relatively prime, b ∈ U and so Uµb = U implying that

(U ∩ S1)µb = U ∩ S2. In particular, this says that U has an even number of q-cyclotomic

cosets. Thus |U : Q| is even; as |U : R| = 2 and U is cyclic, Q ⊆ R. Thus q ∈ R as desired.

Now assume that q is a square modulo n. We show how to construct a splitting of n. For

1 ≤ t ≤ m, let Ut be the group of units in Zpt . Let Rt be the subgroup of Ut consisting of

the squares of elements in Ut and let Qt be the subgroup of Ut generated by q. As in the

previous paragraph, Ut is cyclic of even order, and Rt has index 2 in Ut. As q is a square

modulo pm, then q is a square modulo pt implying that Qt ⊆ Rt. Finally, Ut is a union of

q-cyclotomic cosets modulo pt and these are precisely the cosets of Qt in Ut. The nonzero

16

elements of Zn are the set

∪mt=1pm−tUt. (2.1.1)

We are now ready to construct the splitting of n. Since Ut is cyclic and Qt ⊆ Rt ⊆ Ut

with |Ut : Rt| = 2, there is a unique subgroup Kt of Ut containing Qt such that |Kt : Qt| = 2.

Note that Ut, Rt, Qt, and Kt can be obtained from Um, Rm, Qm, and Km by reducing the

latter modulo pt. Let b ∈ Km\Qm. Then Km = Qm ∪ bQm, and hence, by reducing modulo

pt, Kt = Qt∪bQt for 1 ≤ t ≤ m. Also, b2 ∈ Qt modulo pt. Let g(t)1 , g

(t)2 , ..., g

(t)it

be distinct coset

representatives of Kt in Ut. Then the q-cyclotomic cosets modulo pt in Ut are precisely the

cosets g(t)1 Qt, g

(t)2 Qt, ..., g

(t)itQt, bg

(t)1 Qt, bg

(t)2 Qt, ..., bg

(t)itQt. Let S1,t = g

(t)1 Qt∪g(t)2 Qt∪ ...∪g(t)it Qt

and S2,t = bg(t)1 Qt ∪ bg(t)2 Qt ∪ ... ∪ bg(t)it Qt. Then S1,tµb = S2,t and S2,tµb = S1,t as b2 ∈ Qt

modulo pt, S1,t ∩ S2,t = ∅, and S1,t ∪ S2,t = Ut. Note that pm−tg(t)j Qt and pm−tbg

(t)j Qt are

q-cyclotomic cosets modulo pm. Thus by (2.1.1), S1 = ∪mt=1pm−tS1,t and S2 = ∪mt=1p

m−tS2,t

form a splitting of n given by µb.

The following theorem necessary and sufficient conditions on the length n for the existence

of binary, ternary, and quaternary duadic codes as in the following theorem:

Theorem 2.1.3. [14] Let n = pa11 pa22 ...p

arr where p1, p2, ..., pr are distinct odd primes. The

following assertions hold:

i. Duadic codes of length n over F2 exist if and only if pi ≡ ±1 (mod 8) for 1 ≤ i ≤ r.

ii. Duadic codes of length n over F3 exist if and only if pi ≡ ±1 (mod 12) for 1 ≤ i ≤ r.

iii. Duadic codes of length n over the quaternary field F4 exist for all (odd) n.

Proof. Duadic codes of length n over Fq exist if and only if q is a square modulo n by

Theorem (2.1.2). Since q is a square modulo n if and only if q is a square modulo paii for

1 ≤ i ≤ r and q is a square modulo paii if and only if q is a square modulo pi. Part (i) now

follows if p is an odd prime, then 2 is a square modulo p if and only if p ≡ ±1 (mod 8), where

p is an odd prime. (ii) from 3 is a square modulo p if and only if p ≡ ±1 (mod 12), where

p is an odd prime and p 6= 3. Finally, part (iii) follows from the simple fact that 4 = 22 is

always a square modulo n.

17

Example 2.1.2. We want to find the integers n, 3 < n ≤ 24, for which binary, ternary and

quaternary duadic codes exist.

By Theorem 2.1.3, the binary duadic codes exist for n = {7, 17, 23},

The ternary duadic codes exist for n = {11, 13, 23} and

The quatrnary duadic codes exist for n = {5, 7, 9, 11, 13, 15, 17, 19, 21, 23}.

Example 2.1.3. [14] We construct the generating idempotents of the duadic codes of length

11 over F3. The 3-cyclotomic cosets modulo 11 are C0 = {0}, C1 = {1, 3, 4, 5, 9}, and

C2 = {2, 6, 7, 8, 10}, then there is only one pair of even-like duadic codes: e1(x) = 1 + x +

x3 + x4 + x5 + x9 and e2(x) = 1 + x2 + x6 + x7 + x8 + x10. The corresponding generating

idempotents for the odd-like duadic codes are 1− e2(x) and 1− e1(x). These odd-like codes

are called the ternary Golay codes.

We want to say about codeword weights in duadic codes.

Theorem 2.1.4. [14] Let D1 and D2 be odd-like binary duadic codes of length n with splitting

given by µ−1. Then for i = 1 and 2, the weight of every even weight codeword of Di is 0 mod

4, and the weight of every odd weight codeword of Di is n mod 4.

2.2 Fundamental Properties of quadratic residue codes

In this section we study more closely the family of quadratic residue codes, which, as we

have seen, are special cases of duadic codes.

Definition 2.2.1. [14] Let Qp denote the set of nonzero squares modulo p, and let Np be the

set of nonsquares modulo p. The sets Qp and Np are called the nonzero quadratic residues

and the quadratic nonresidues modulo p, respectively.

Proposition 2.2.1. [21] Let p be an odd prime. Denote by Qp and Np the sets of nonzero

quadratic residues and quadratic nonresidues modulo p, respectively. Then we have the fol-

lowing:

i. The product of two quadratic residues modulo p is a quadratic residue modulo p.

ii. The product of two quadratic nonresidues modulo p is a quadratic residue modulo p.

iii. The product of a nonzero quadratic residue modulo p with a quadratic nonresidue

modulo p is a quadratic nonresidue modulo p.

18

Lemma 2.2.2. [14] Let p be an odd prime. The following hold:

i. |Qp| = |Np| = (p−1)2.

ii. Modulo p, we have Qpa = Qp, Npa = Np, Qpb = Np, and Npb = Qp where a ∈ Qp and

b ∈ Np.

Proof. The nonzero elements of the field Fp form a cyclic group F∗p of even order p− 1 with

generator α. Qp is the set of even order elements, that is, Qp = {α2i|0 ≤ i < p−12}; this set

forms a subgroup of index 2 in F∗p. Furthermore Np is the coset Qpα. Then the results hold.

Definition 2.2.2. [14] The pair of sets Qp and Np is a splitting of p given by the multiplier

µb for any b ∈ Np, that is, this splitting determines the defining sets for a pair of even-like

duadic codes and a pair of odd-like duadic codes is called the quadratic residue codes or

QR codes, of length p over Fq.

Remark 2.2.1. [14] The odd-like QR codes have defining sets Qp and Np and dimension (p+1)2

,

while the even-like QR codes have defining sets Qp ∪ {0} and Np ∪ {0} and dimension (p−1)2

.

Example 2.2.1. Consider the finite field F11. The set of nonzero quadratic residues mod-

ulo 11 is Q11 = {1, 4, 5, 9, 3}, and the set of quadratic nonresidues modulo 11 is N11 =

{2, 8, 10, 7, 6}.

We have |Q11| = |N11| = 5 = (11− 1)/2. Furthermore, by choosing 3 ∈ Q11 and 7 ∈ N11,

we have:

3Q11 = {3, 1, 4, 5, 9} = Q11,

3N11 = {6, 2, 8, 10, 7} = N11,

7Q11 = {7, 6, 2, 8, 10} = N11,

and

7N11 = {3, 1, 4, 5, 9} = Q11.

Remark 2.2.2. [14] 1. Quadratic residue codes is a type of cyclic code and exist only for

prime lengths, but there are duadic codes of composite length.

2. At prime lengths there may be duadic codes that are not quadratic residue codes.

Duadic codes possess many of the properties of quadratic residue codes.

19

Examples of binary quadratic residue codes [22]:

1. The [7, 4, 3] Hamming code (Example 1.2.5) with generator polynomial x3 + x+ 1.

2. The [17, 9, 5] code with generator polynomial x8 + x5 + x4 + x3 + 1 and idempotent

x16 + x15 + x13 + x11 + x8 + x4 + x2 + x+ 1.

3. The [31, 16, 7] code with generator polynomial x15 + x12 + x7 + x6 + x2 + x + 1. The

quadratic residues mod 31 are C1 ∪ C5 ∪ C7.

4. The [47, 24, 11] code with generator polynomial x23 + x19 + x18 + x14 + x13 + x12 +

x10 + x9 + x7 + x6 + x5 + x3 + x2 + x+ 1.

Theorem 2.2.3. [14] Quadratic residue codes of odd prime length p exist over Fq if and

only if q ∈ Qp.

2.3 Quadratic residue codes over fields of characteris-

tic 2

We study the generating idempotents of all the QR codes over any field of characteristic 2.

We recall that we only have to look at the generating idempotents of QR codes over F2 and

F4.

Theorem 2.3.1. [14] Let p be an odd prime. The following hold:

i. Binary quadratic residue codes of length p exist if and only if p ≡ ±1 (mod 8).

ii. The even-like binary quadratic residue codes have generating idempotents:

δ +∑

j∈Qpxj and δ +

∑j∈Np

xj, where δ =

1, p ≡ -1(mod 8);

0, p ≡ 1(mod 8).

iii. The odd-like binary quadratic residue codes have generating idempotents:

ε+∑

j∈Qpxj and ε+

∑j∈Np

xj, where ε =

0, p ≡ -1(mod 8);

1, p ≡ 1(mod 8).

Proof. i. Binary QR codes of length p exist if and only if 2 ∈ Qp, which is equivalent to

p ≡ ±1 (mod 8).


20

i. If p ≡ ±1 (mod 8), the generating idempotents of the quadratic residue codes over F4

are the same as those over F2 given in Theorem 2.3.1.

ii. The even-like quadratic residue codes over F4 have generating idempotents:

δ+ω∑

j∈Qpxj+ω

∑j∈Np

xj and δ+ω∑

j∈Qpxj+ω

∑j∈Np

xj, where δ =

1, p ≡ 3(mod 8);

0, p ≡ -3(mod 8).

iii. The odd-like quadratic residue codes over F4 have generating idempotents:

ε+ω∑

j∈Qpxj+ω

∑j∈Np

xj and ε+ω∑

j∈Qpxj+ω

∑j∈Np

xj, where ε =

1, p ≡ -3(mod 8);

0, p ≡ 3(mod 8).

Example 2.3.1. We consider the binary QR codes of length 17. In that case, Q17 =

{1, 2, 4, 8, 9, 13, 15, 16} and N17 = {3, 5, 6, 7, 10, 11, 12, 14}.

Using Theorem 2.3.1, we take ε = 1 and δ = 0 since 17 ≡ 1(mod 8).

The generating idempotents of the odd-like QR codes are:

1 +∑

j∈Q17xj = 1 + x + x2 + x4 + x8 + x9 + x13 + x15 + x16 and 1 +

∑j∈N17

xj =

1 + x3 + x5 + x6 + x7 + x10 + x11 + x12 + x14,

and the generating idempotents of the even-like QR codes are:

0 +∑

j∈Q17xj = x+ x2 + x4 + x8 + x9 + x13 + x15 + x16 and 0 +

∑j∈N17

xj = x3 + x5 +

x6 + x7 + x10 + x11 + x12 + x14.

Example 2.3.2. Now we want to find QR codes of length 11 over F4,

Q11 = {1, 3, 4, 5, 9} and N11 = {2, 6, 7, 8, 10}.

By Theorem 2.3.2, we take ε = 0 and δ = 1 since 11 ≡ 3(mod 8).


0 + ω∑

j∈Q11xj + ω

∑j∈N11

xj = ω(x + x3 + x4 + x5 + x9) + ω(x2 + x6 + x7 + x8 + x10)

and 0 + ω∑

j∈Q11xj + ω

∑j∈N11

xj = ω(x+ x3 + x4 + x5 + x9) + ω(x2 + x6 + x7 + x8 + x10),

and the generating idempotents of the even-like QR codes are:

1 +ω∑

j∈Q11xj + ω

∑j∈N11

xj = 1 +ω(x+x3 +x4 +x5 +x9) + ω(x2 +x6 +x7 +x8 +x10)

and 1+ ω∑

j∈Q17xj +ω

∑j∈N17

xj = 1+ ω(x+x3 +x4 +x5 +x9)+ω(x2 +x6 +x7 +x8 +x10).

The idempotents that arise in Theorems 2.3.1 and 2.3.2 are the generating idempotents

for QR codes over any field of characteristic 2 as the next result shows:


i. Quadratic residue codes of length p over F2t , where t is odd, exist if and only if p ≡ ±1

(mod 8), and the generating idempotents are those given in Theorem 2.3.1.

21

ii. Quadratic residue codes of length p over F2t, where t is even, exist for all p, and the

generating idempotents are those given in Theorems 2.3.1 and 2.3.2.

2.4 Quadratic residue codes over fields of characteris-

tic 3

Analogous results hold for fields of characteristic 3. We assume our QR codes have length p

an odd prime that cannot equal 3. We first examine quadratic residue codes over F3.

Theorem 2.4.1. [14] Let C be a cyclic code of odd prime length p over Fq, where q is a

square modulo p. Let e(x) be the generating idempotent of C. Then e(x) = a0+a1∑

i∈Qpxi+

a2∑

i∈Npxi, for some a0, a1, and a2 in Fq.

Theorem 2.4.2. [14] Let p > 3 be prime. The following hold:

i. Quadratic residue codes over F3 of length p exist if and only if p ≡ ±1 (mod 12).


−∑

j∈Qpxj and −

∑j∈Np

xj, if p ≡ 1 (mod 12), and

1 +∑

j∈Qpxj and 1 +

∑j∈Np

xj, if p ≡ −1 (mod 12).


1 +∑

j∈Qpxj and 1 +

∑j∈Np


−∑

j∈Qpxj and −

∑j∈Np

xj, if p ≡ −1 (mod 12).

Example 2.4.1. [14] We find the generating idempotents of the QR codes of length 11 over

F3 by Theorem 2.4.2. Here

Q11 = {1, 3, 4, 5, 9} and N11 = {2, 6, 7, 8, 10}. Since 11 ≡ −1 mod 12, then:

The generating idempotents of the odd-like QR codes are

−(x+ x3 + x4 + x5 + x9) and −(x2 + x6 + x7 + x8 + x10),

and the generating idempotents of the even-like QR codes are

1 + x+ x3 + x4 + x5 + x9 and 1 + x2 + x6 + x7 + x8 + x10.

We now turn to QR codes over F9. Because 9 = 32 is a square modulo any odd prime p,

by Theorem 2.4.2, QR codes over F9 exist for any odd prime length greater than 3.

The field F9 can be constructed by adjoining an element ρ to F3, where ρ2 = 1 + ρ.

22

So F9 = {a+ bρ|a, b ∈ F3}. Multiplication in F9 is described in Table 2.1; note that ρ is

a primitive 8th root of unity.

ρi a+ bρ

0 0

1 1

ρ ρ

ρ2 1 + ρ

ρ3 1− ρ

ρ4 −1

ρ5 −ρ

ρ6 −1− ρ

ρ7 −1 + ρ

Table 2.1: The field F9


i. If p ≡ ±1 (mod 12), the generating idempotents of the quadratic residue codes over F9

are the same as those over F3 given in Theorem 2.4.2.


1 + ρ∑

j∈Qpxj + ρ3

∑j∈Np

xj and 1 + ρ3∑

j∈Qpxj + ρ

∑j∈Np


−ρ∑

j∈Qpxj − ρ3

∑j∈Np

xj and −ρ3∑

j∈Qpxj − ρ

∑j∈Np

xj, if p ≡ −5 (mod 12).


−ρ∑

j∈Qpxj − ρ3

∑j∈Np

xj and −ρ3∑

j∈Qpxj − ρ

∑j∈Np


1 + ρ∑

j∈Qpxj + ρ3

∑j∈Np

xj and 1 + ρ3∑

j∈Qpxj + ρ

∑j∈Np

xj, if p ≡ −5(mod 12).

Example 2.4.2. We want to find the generating idempotents of the QR codes of length 17

over F9.

Then Q17 = {1, 2, 4, 8, 9, 13, 15, 16} and N17 = {3, 5, 6, 7, 10, 11, 12, 14}.

Since 17 ≡ 5 mod 12, then by Theorem 2.4.3:


−ρ∑

j∈Q17xj − ρ3

∑j∈N17

xj = ρ5(x+ x2 + x4 + x8 + x9 + x13 + x15 + x16) + ρ7(x3 + x5 +

x6 + x7 + x10 + x11 + x12 + x14) and −ρ3∑

j∈Q17xj − ρ

∑j∈N17

xj = −ρ3(x+ x2 + x4 + x8 +

23

x9 + x13 + x15 + x16)− ρ(x3 + x5 + x6 + x7 + x10 + x11 + x12 + x14).

The generating idempotents of the even-like QR codes are:

1 + ρ∑

j∈Q17xj + ρ3

∑j∈N17

xj = 1 + ρ(x+ x2 + x4 + x8 + x9 + x13 + x15 + x16) + ρ3(x3 +

x5 + x6 + x7 + x10 + x11 + x12 + x14) and 1 + ρ3∑

j∈Q17xj + ρ

∑j∈N17

xj = 1 + ρ3(x + x2 +

x4 + x8 + x9 + x13 + x15 + x16) + ρ(x3 + x5 + x6 + x7 + x10 + x11 + x12 + x14).

Theorem 2.4.4. [14] Let p be an odd prime with p 6= 3. The following hold:

i. Quadratic residue codes of length p over F3t , where t is odd, exist if and only if p ≡ ±1

(mod 12), and the generating idempotents are those given in Theorem 2.4.2.

ii. Quadratic residue codes of length p over F3t, where t is even, exist for all p, and the

generating idempotents are those given in Theorems 2.4.2 and 2.4.3.

24

Chapter 3

Quadratic residue codes over finite

commutative rings

Codes over finite rings have been studied for many years. More recently, codes over a wide

variety of rings have been studied.

Pless and Qian [26] defined quadratic residue codes over Z4 in terms of their idempotent

generators.

Taeri [32], defined quadratic residue codes over Z9 in terms of their idempotent generators

and show that these codes also have many good properties which are analogous in many

respects to properties of quadratic residue codes over finite fields.

In [1], the author studied idempotent generators of cyclic codes and quadratic residue

codes over the non-chain ring F3 + vF3, where v2 = 1.

Finally, we extend this study to obtain quadratic residue codes over the chain ring F3 +

uF3, where u2 = 0.

3.1 Preliminaries

Definition 3.1.1. [32] A subset of n tuples of elements of Zn is said to be a code over Znif it is a Zn-module.

Remark 3.1.1. [1] A linear code C of length n over R is an R-submodule of Rn.

Definition 3.1.2. [9] A finite ring is called a chain ring if its ideals are linearly ordered by

25

inclusion.

In particular, this means that any finite chain ring has a unique maximal ideal.

Definition 3.1.3. [4] A ring R with a unique maximal ideal M is called a local ring. The

quotient R/M is called the residue field.

Example 3.1.1. 1. All fields are local rings, since {0} is the only maximal ideal in these

rings (So, Q is a local ring).

2. The integers Z do not form a local ring, but for any prime ideal P we find a local ring

in Zp = {fg

: f, g ∈ Z, g /∈ P}.

3. R[x] and C[x] are not local ring.

Proposition 3.1.1. [4] 1. Let R be a ring and M 6= (0) be an ideal such that each x ∈ R \

M is a unit. Then R is a local ring and M is its maximal ideal.

2. Let R be a ring and M a maximal ideal such that each element of the set 1 + M :=

{1 + x|x ∈M} is a unit in R. Then R is a local ring.

Definition 3.1.4. [4] A ring which contains only a finite number of maximal ideals is called

semi-local ring.

Example 3.1.2. Z6 is a semi-local ring with two maximal ideals 2Z6 and 3Z6.

Definition 3.1.5. [32] h = 1 + e1 + e2 is the all one vector over any ring.

Definition 3.1.6. [1] The extended code of C over R denoted by C, is the code obtained

by adding an over all parity check to each codeword of C, that is, if C = (c0, c1, ..., cn−1),

then C = (c0, c1, ..., cn−1, cn) iff∑n

i=o ci = 0.

3.2 Quadratic residue codes over Z4

We divide this section into two parts; one is a cyclic codes with generating polynomials and

idempotents and another is quadratic residue codes over Z4.

Notation: In describing codes over Z4, we denote the elements of Z4 in either of the

natural forms {0, 1, 2, 3} or {0, 1, 2,−1}, whichever is most convenient.

• Cyclic codes over Z4

26

As with usual cyclic codes over Fq, we view codewords c = c0c1...cn−1 in a cyclic Z4-linear

code of length n as polynomials c(x) = c0 + c1x+ ...+ cn−1xn−1 ∈ Z4[x].

If we consider our polynomials as elements of the quotient ring Rn = Z4[x]/〈xn−1〉, then

xc(x) modulo xn − 1 represents the cyclic shift of c [14].

Generating polynomials of cyclic codes over Z4

Theorem 3.2.1. [26] Let C be a cyclic code over Z4 of odd length n. Then there exist

unique monic polynomials f(x), g(x) and h(x) such that xn − 1 = f(x)g(x)h(x) and |C| =

4deg(h)2deg(g).

Corollary 3.2.1. [14] Let n be odd. Assume that xn − 1 is a product of k irreducible

polynomials in Z4[x]. Then there are 3k cyclic codes over Z4 of length n.

Example 3.2.1. [14] x7−1 = g1(x)g2(x)g3(x), where g1(x) = x−1, g2(x) = x3 +2x2 +x−1

and g3(x) = x3−x2 +2x−1 are the monic irreducible factors of x7−1. By Corollary (3.2.1),

there are 33 = 27 cyclic codes over Z4 of length 7.

Generating idempotents of cyclic codes over Z4

Lemma 3.2.2. [14] The following hold:

i. Let C = 〈e(x)〉 be a Z4-linear cyclic code with generating idempotent e(x). Then the

generating idempotent for C⊥ is 1− e(x)µ−1.

ii. For i = 1 and 2 let Ci = 〈ei(x)〉 be Z4-linear cyclic codes with generating idempo-

tents ei(x). Then the generating idempotent for C1 ∩ C2 is e1(x)e2(x), and the generating

idempotent for C1 + C2 is e1(x) + e2(x)− e1(x)e2(x).

•• Quadratic residue codes over Z4

Let Q(x) =∑

i∈Qpxi and N(x) =

∑i∈Np

xi. Note that 1, Q(x) and N(x) are idempotents

in Rp = F2[x]/〈xp + 1〉. A combination of these will be idempotents in Rp that lead to the

definition of Z4-quadratic residue codes [14].

Lemma 3.2.3. [14] Define r by p = 8r ± 1.

If r is odd, then Q(x) + 2N(x), N(x) + 2Q(x), 1−Q(x) + 2N(x) and 1−N(x) + 2Q(x)

are idempotents in Rp.

If r is even, then −Q(x),−N(x), 1 +Q(x) and 1 +N(x) are idempotents in Rp.

27

We now define the Z4-quadratic residue codes using the idempotents of Lemma 3.2.3.

The definitions depend upon the value of p modulo 8.

1. Z4-quadratic residue codes: p ≡ −1 (mod 8)

We first look at the case where p ≡ −1 (mod 8), Let p+ 1 = 8r.

If r is odd, define D1 = 〈Q(x) + 2N(x)〉, D2 = 〈N(x) + 2Q(x)〉, C1 = 〈1−N(x) + 2Q(x)〉

and C2 = 〈1−Q(x) + 2N(x)〉.

If r is even, define D1 = 〈−Q(x)〉, D2 = 〈−N(x)〉, C1 = 〈1 +N(x)〉 and C2 = 〈1 +Q(x)〉.

These codes are called Z4-quadratic residue codes when p ≡ −1 (mod 8).

2. Z4-quadratic residue codes: p ≡ 1 (mod 8)

When p ≡ 1 (mod 8), we simply reverse the C ′is and D′is from the codes defined in the

case p ≡ −1 (mod 8). Again let p− 1 = 8r.

If r is odd, define D1 = 〈1−N(x)+2Q(x)〉, D2 = 〈1−Q(x)+2N(x)〉, C1 = 〈Q(x)+2N(x)〉

and C2 = 〈N(x) + 2Q(x)〉.

If r is even, define D1 = 〈1 +N(x)〉, D2 = 〈1 +Q(x)〉, C1 = 〈−Q(x)〉 and C2 = 〈−N(x)〉.

These codes are called Z4-quadratic residue codes when p ≡ 1 (mod 8).

3.3 Quadratic residue codes over Z9

In section (2.4), we study QR codes over fields of characteristic 3. Now we will discuss the

quadratic residue codes and an extended codes over Z9 as in [32].

Remark 3.3.1. [32] If we say about QR codes over Z9, we will consider the following:

Let e1 =∑

i∈Q xi and e2 =

∑i∈N x

i, where Q is the set of quadratic residues and N is

the set of non-residues for p (a prime).

Since 3 is quadratic residue (mod p) if and only if p = 12r± 1. Therefore for considering

quadratic residue codes over Z3, we must assume that p = 12r ± 1. It is well known that

2ei, 1 + ei, i = 1, 2, are idempotents over Z3[x]/〈xp − 1〉.

A Z3-cyclic codes is a Z3-quadratic residue code if it is generated by one of the idempo-

tents 2ei, 1 + ei, i = 1, 2.

Therefore, we find idempotents of Z9[x]/〈xp−1〉 from idempotent generators of quadratic

residue codes over Z3[x]/〈xp − 1〉.

28

If p = 12r−1, put Q1 = 〈2e1〉, Q2 = 〈2e2〉, Q′1 = 〈1+e2〉 and Q′2 = 〈1+e1〉. If p = 12r+1,

put Q1 = 〈1 + e2〉, Q2 = 〈1 + e1〉, Q′1 = 〈2e1〉 and Q′2 = 〈2e2〉.

Theorem 3.3.1. [32] Let S and R be finite commutative rings with identity and characteristic

qm and qm+1, respectively, where q is a prime. Let f : R → S be an epimorphism, with

kerf = qmR. Then if f(e) = e1 is an idempotent of S. Then eq is an idempotent of R.

Now, by Theorem (3.3.1), (2ei)3, (1 + ei)3, i = 1, 2, are idempotents over Z9[x]/〈xp − 1〉.

In order to define quadratic residue codes over Z9 in terms of idempotent generators, we

must compute these elements modulo 9. The following theorem is needed for such compu-

tations:

Theorem 3.3.2. [22] (i) Suppose that p = 4k − 1 and a is a number relatively prime to p.

Then in the set a + (Q ∪ {0}), there are k elements in (Q ∪ {0}) and k elements in N . In

the set a+N , there are k elements in (Q ∪ {0}) and k − 1 elements in N .

(ii) Suppose that p = 4k + 1 and a is a number relatively prime to p. Then in the set

a+ (Q ∪ {0}), if a ∈ Q, there are k + 1 elements in (Q ∪ {0})(including 0) and k elements

in N ; and if a ∈ N , there are k elements in Q and k + 1 elements in N . In the set a + N ,

if a ∈ Q, there are k elements in Q and k elements in N ; and if a ∈ N , there are k + 1

elements in (Q ∪ {0})(including 0) and k − 1 elements in N .

By a routine application of Theorem (3.3.2), we obtain the following result:

Theorem 3.3.3. [32] If p = 4k − 1, then:

e21 = (k − 1)e1 + ke2,

e22 = ke1 + (k − 1)e2,

e1e2 = (2k − 1) + (k − 1)e1 + (k − 1)e2,

e31 = (3k2 − 3k + 1)e1 + 2k(k − 1)e2 + 2k2 − k,

e32 = 2k(k − 1)e1 + (3k2 − 3k + 1)e2 + 2k2 − k.

If p = 4k + 1, then:

e21 = (k − 1)e1 + ke2 + 2k,

29

e22 = ke1 + (k − 1)e2 + 2k,

e1e2 = ke1 + ke2,

e31 = (2k2 + 1)e1 + (2k2 − k)e2 + 2k2 − 2k,

e32 = (2k2 − k)e1 + (2k2 + 1)e2 + 2k2 − 2k.

• Quadratic residue codes over Z9

A Z9-cyclic code is called a Z9-quadratic residue (QR) code if it is generated by one of

the idempotents in the following theorem:

Theorem 3.3.4. [32] I- Suppose that p = 12r − 1, then:

(a) If r = 3k, then 8ei, 1 + ei, 8h, 1 + 2h are idempotents over Z9[x]/〈xp − 1〉, where

i = 1, 2.

(b) If r = 3k+1, then 6ei+8ej+3, ei+3ej+7, 5h, 1+5h are idempotents over Z9[x]/〈xp−1〉,

where 1 ≤ i 6= j ≤ 2.

(c) If r = 3k+2, then 3ei+8ej+6, ei+6ej+4, 2h, 8+8h are idempotents over Z9[x]/〈xp−1〉,

where 1 ≤ i 6= j ≤ 2.

II- Suppose that p = 12r + 1, then:

(a) If r = 3k, then 1+ei, 8ei, h, 1+h are idempotents over Z9[x]/〈xp−1〉, where i = 1, 2.

(b) If r = 3k+1, then ei+6ej+4, 3ei+8ej+6, 7h, 1+2h are idempotents over Z9[x]/〈xp−1〉,

where 1 ≤ i 6= j ≤ 2.

(c) If r = 3k+2, then ei+3ej+7, 6ei+8ej+3, 4h, 1+5h are idempotents over Z9[x]/〈xp−1〉,

where 1 ≤ i 6= j ≤ 2.

In the following theorem we investigate some properties of QR-codes over Z9.

Theorem 3.3.5. [32] Suppose that p = 12r − 1.

(1)If r = 3k, let Q1 = 〈8e1〉, Q2 = 〈8e2〉 and Q′1 = 〈1 + e1〉, Q′2 = 〈1 + e2〉.

(2)If r = 3k+1, Let Q1 = 〈6e1+8e2+3〉, Q2 = 〈8e1+6e2+3〉 and Q′1 = 〈e1+3e2+7〉, Q′2 =

〈3e1 + e2 + 7〉.

(3)If r = 3k+2, Let Q1 = 〈3e1+8e2+6〉, Q2 = 〈8e1+3e2+6〉 and Q′1 = 〈e1+6e2+4〉, Q′2 =

〈6e1 + e2 + 4〉.

Then the following hold for Z9-QR codes Q1, Q2, Q′1 and Q′2:

30

(a) Q1 and Q2 are equivalent and Q′1 and Q′2 are equivalent.

(b) Q1 ∩ Q2 = 〈h〉 and Q1 + Q2 = Z9[x]/〈xp − 1〉, where h is a suitable element in

{8h, 5h, 2h} listed in Theorem (3.3.4(I)).

(c) |Q1| = 9p+12 = |Q2|.

(d) Q1 = Q′1 + 〈h〉, Q2 = Q′2 + 〈h〉.

(e) |Q′1| = 9p−12 = |Q′2|.

(f) Q′1 and Q′2 are self-orthogonal and Q⊥1 = Q′1 and Q⊥2 = Q′2.

(g) Q′1∩Q′2 = {0} and Q′1+Q′2 = 〈1−h〉; also Q′i∩Q′j = {0} and Qi+Qj = Z9[x]/〈xp−1〉,

where 1 ≤ i 6= j ≤ 2.

Theorem 3.3.6. [32] Suppose that p = 12r + 1.

(1)If r = 3k, let Q1 = 〈1 + e1〉, Q2 = 〈1 + e2〉 and Q′1 = 〈8e2〉, Q′2 = 〈8e1〉.

(2)If r = 3k+1, Let Q1 = 〈e1+6e2+4〉, Q2 = 〈6e1+e2+4〉 and Q′1 = 〈3e1+8e2+6〉, Q′2 =

〈8e1 + 3e2 + 6〉.

(3)If r = 3k+2, Let Q1 = 〈e1+3e2+7〉, Q2 = 〈e1+3e2+7〉 and Q′1 = 〈6e1+8e2+3〉, Q′2 =

〈8e1 + 6e2 + 3〉.

Then the following hold for Z9-QR codes Q1, Q2, Q′1 and Q′2:


(b) Q1 ∩ Q2 = 〈h〉 and Q1 + Q2 = Z9[x]/〈xp − 1〉, where h is a suitable element in

{h, 4h, 7h} listed in Theorem (3.3.4(II)).

(c) |Q1| = 9p+12 = |Q2|.

(d) Q1 = Q′1 + 〈h〉, Q2 = Q′2 + 〈h〉.

(e) |Q′1| = 9p−12 = |Q′2|.

(f) Q⊥1 = Q′2 and Q⊥2 = Q′1.

(g) Q′1 ∩ Q′2 = {0} and Q′1 + Q′2 = 〈1 − h〉; also Q′i ∩ Q′j = {0} and Qi + Qj =< u >,

where 1 ≤ i 6= j ≤ 2, and u is a suitable element of {1 + 2h, 1 + 5h, 1 + h} listed in the

Theorem (3.3.4(II)).

•• Extended codes over Z9

Let Q1 and Q2 denoted an extended codes of Q1 and Q2 respectively.

Theorem 3.3.7. [32] Suppose p = 12r−1 and Q1 and Q2 are the Z9-QR codes in Theorems

3.3.5. Then Q1 and Q2 are self-dual.

31

When p = 12r + 1, we define Q1 to be the Z9-code generated by the matrix,

∞ 0 1 2 ... p− 1

0

0 G′1...

1 1 1 1 ... 1

,

where each row of G′1 is a cyclic shift of 8e1 when r = 3k, a cyclic shift of 6 + 3e1 + 8e2

when r = 3k + 1, a cyclic shift of 3 + 6e1 + 8e2 when r = 3k + 2. We define Q2 similarly.

Note that these are not extended codes of Q1 and Q2, since the sum of the components of

the all one vector is not 0 (mod 9).

Theorem 3.3.8. [32] Suppose p = 12r+1 and Q1 and Q2 are the Z9-QR codes in Theorems

3.3.6. Then the dual of Q1 is Q2 and the dual of Q2 is Q1.

Remark 3.3.2. [32] For p = 12r ± 1, the extended codes Q1 and Q2 are equivalent, since Q1

and Q2 are equivalent. They are also equivalent to Q1⊥

and Q2⊥

.

3.4 Quadratic residue codes over F3 + vF3, v2 = 1

Definition 3.4.1. [1] The alphabet R = F3 + vF3 = {0, 1, 2, v, 2v, a = 1 + v, b = 2 + v, c =

1 + 2v, d = 2 + 2v}, where v2 = 1,F3 = {0, 1, 2}, is a non chain commutative ring with nine

elements.

Remark 3.4.1. [1] 1. The elements {1, 2, v, 2v} are units.

2. This ring is a semi-local ring, since it has two maximal ideals (v − 1) = (b) and

(1 + v) = (a).

Addition and multiplication over R are given in the following tables:

32

⊕0 1 2 v 2v a b c d

0 0 1 2 v 2v a b c d

1 1 2 0 a c b v d 2v

2 2 0 1 b d v a 2v c

v v a b 2v 0 c d 1 2

2v 2v c d 0 v 1 2 a b

a a b v c 1 d 2v 2 0

b b v a d 2 2v c 0 1

c c d 2v 1 a 2 0 b v

d d 2v c 2 b 0 1 v a

⊗0 1 2 v 2v a b c d

0 0 0 0 0 0 0 0 0 0

1 0 1 2 v 2v a b c d

2 0 2 1 2v v d c b a

v 0 v 2v 1 2 a c b d

2v 0 2v v 2 1 d b c a

a 0 a d a d d 0 0 a

b 0 b c c b 0 b c 0

c 0 c b b c 0 c b 0

d 0 d a d a a 0 0 d

Theorem 3.4.1. [1] If n is odd, then every cyclic codes C of length n over R contains a

unique idempotent e(x) ∈ C, such that C = 〈e(x)〉.

• Quadratic residue codes over F3 + vF3

Theorem 3.4.2. [1] Let p ≡ ±1(mod 12). If 2ei, 1+ei are idempotent generators of quadratic

residue codes over F3, then cei + aej and b(1 + ei) + d(1 + ej) = 1 + bei + dej where a =

1 + v, b = 2 + v, c = 1 + 2v, d = 2 + 2v, v2 = 1 are idempotent generators over R[x]/〈xp− 1〉,

where i, j = 1, 2.

In the following theorem we investigate some properties of QR-codes over R = F3 + vF3.

Theorem 3.4.3. [1] Let p be a prime with p = 12r − 1, Q1 = 〈ce1 + ae2〉, Q2 = 〈ce2 + ae1〉

and Q′1 = 〈1 + be1 + de2〉, Q′2 = 〈1 + be2 + de1〉, then the following holds for Q1, Q2, Q′1 and

Q′2:


(b) Q1 ∩Q2 = 〈2h〉 and Q1 +Q2 = R[x]/〈xp − 1〉.

(c) Q1 = Q′1 + 〈2h〉, Q2 = Q′2 + 〈2h〉.

33

(d) |Q1| = |Q2| = 9p+12 , |Q′1| = |Q′2| = 9

p−12 .

(e) Q⊥1 = Q′2 and Q⊥2 = Q′1.

(f) Q′1 ∩Q′2 = {0} and Q′1 +Q′2 = 〈1 + h〉.

Theorem 3.4.4. [1] Let p be a prime with p = 12r+1, Q1 = 〈1+be1+de2〉, Q2 = 〈1+be2+de1〉

and Q′1 = 〈ce1 + ae2〉, Q′2 = 〈ce2 + ae1〉, then the following holds for Q1, Q2, Q′1 and Q′2:


(b) Q1 ∩Q2 = 〈h〉 and Q1 +Q2 = R[x]/〈xp − 1〉.

(c) Q1 = Q′2 + 〈h〉, Q2 = Q′1 + 〈h〉.

(d) |Q1| = |Q2| = 9p+12 , |Q′1| = |Q′2| = 9

p−12 .

(e) Q⊥1 = Q′1 and Q⊥2 = Q′2.

(f) Q′1 ∩Q′2 = {0} and Q′1 +Q′2 = 〈1− h〉.

•• Extended codes over F3 + vF3

When p ≡ −1(mod 12), we define Q1 to be the R-code generated by the matrix

∞ 0 1 2 ... p− 1

0

0 G′1...

2 2 2 2 ... 2

,

where each row of G′1 is a cyclic shift of the vector 1 + be1 + de2. We define Q2 similarly.

Note that these are extended codes of Q1 and Q2, since the sum of components of all one

vector is 0 (mod 3).

Theorem 3.4.5. [1] Let Q1, Q2, Q′1 and Q′2 be the quadratic residue codes over R in Theorem

3.4.3. Let Q1 and Q2 denoted their extended codes. When p ≡ −1(mod 12), then the dual of

Q1 is Q2 and the dual of Q2 is Q1.

When p ≡ 1(mod 12), we define Q1 to be the R-code generated by the matrix

∞ 0 1 2 ... p− 1

0

0 G′2...

1 1 1 1 ... 1

,

34

where each row of G′2 is a cyclic shift of the vector ce2 + ae1. We define Q2 similarly.

Note that these are not extended codes of Q1 and Q2, since the sum of components of all

one vector is not 0 (mod 3).

Theorem 3.4.6. [1] Let Q1, Q2, Q′1 and Q′2 be the quadratic residue codes over R in Theorem

3.4.4. When p ≡ 1(mod 12), then the dual of Q1 is Q1 and the dual of Q2 is Q2.

3.5 Quadratic residue codes over F3 + uF3, u2 = 0

In this section, we study generator idempotent of cyclic and quadratic residue codes over

the commutative chain ring F3 + uF3, where u2 = 0.

• Definition and properties

We will give in this subsection definition and some properties of the ring F3 + uF3.

Definition 3.5.1. [11] The ring R = F3 + uF3 = {a + ub : a, b ∈ F3} is a commutative

chain ring with nine elements {0, 1, 2, u, 1 + u, 2 + u, 2u, 1 + 2u, 2 + 2u}, where u2 = 0 and

F3 = {0, 1, 2}.

Remark 3.5.1. The ring F3 + uF3 is of characteristic 3 and is analogous to Z9, since u plays

the role of 3.

Addition and multiplication over R are given in the following tables:

⊕0 1 2 u 1 + u 2 + u 2u 1 + 2u 2 + 2u

0 0 1 2 u 1 + u 2 + u 2u 1 + 2u 2 + 2u

1 1 2 0 1 + u 2 + u u 1 + 2u 2 + 2u 2u

2 2 0 1 2 + u u 1 + u 2 + 2u 2u 1 + 2u

u u 1 + u 2 + u 2u 1 + 2u 2 + 2u 0 1 2

1 + u 1 + u 2 + u u 1 + 2u 2 + 2u 2u 1 2 0

2 + u 2 + u u 1 + u 2 + 2u 2u 1 + 2u 2 0 1

2u 2u 1 + 2u 2 + 2u 0 1 2 u 1 + u 2 + u

1 + 2u 1 + 2u 2 + 2u 2u 1 2 0 1 + u 2 + u u

2 + 2u 2 + 2u 2u 1 + 2u 2 0 1 2 + u u 1 + u

35

⊗0 1 2 u 1 + u 2 + u 2u 1 + 2u 2 + 2u

0 0 0 0 0 0 0 0 0 0

1 0 1 2 u 1 + u 2 + u 2u 1 + 2u 2 + 2u

2 0 2 1 2u 2 + 2u 1 + 2u u 2 + u 1 + u

u 0 u 2u 0 u 2u 0 u 2u

1 + u 0 1 + u 2 + 2u u 1 + 2u 2 2u 1 2 + u

2 + u 0 2 + u 1 + 2u 2u 2 1 + u u 2 + 2u 1

2u 0 2u u 0 2u u 0 2u u

1 + 2u 0 1 + 2u 2 + u u 1 2 + 2u 2u 1 + u 2

2 + 2u 0 2 + 2u 1 + u 2u 2 + u 1 u 2 1 + 2u

In [11], There is a new weight function for codes over R. These are the Gray weight,

Gray weight enumerator and Gray distance.

The Gray weight, denoted by Gw(x) of a codeword x = (x1, x2, ..., xn) is defined as∑ni=1Gw(xi), where

Gw(xi) =

0, xi=0;

1, xi = 1, 2, u and 2u;

2, otherwise.

The Gray weight enumerator of C is a polynomial∑

c∈C yGw(c).

And the Gray distance, denoted by Gd(x, y) between two codewords x and y is the Gray

weight of x− y.

We want to define a Gray map α from (F3 + uF3) to F23 as:

xi α(xi)

0 00

1 01

2 02

u 10

1 + u 11

2 + u 12

2u 20

1 + 2u 21

2 + 2u 22

then the Gray map φ : (F3 + uF3)n → F2n

3 is defined by

φ(x) = (α(x1), α(x2), ..., α(xn)),

where x = (x1, x2, ..., xn).

36

Lemma 3.5.1. [11] The Gray map φ is a distance-preserving map from

((F3 + uF3)n, Gray distance) to (F2n

3 , Hamming distance).

• • Quadratic residue codes over F3 + uF3, u2 = 0

Remark 3.5.2. According to Theorem(3.3.3) and if we replace 3 by u, we have the following

calculations:

Let k = 3r and let h = 1 + e1 + e2 then:

(I)If p = 4k − 1, then:

e21 = (ur − 1)e1 + ure2,

e22 = ure1 + (ur − 1)e2,

e1e2 = (2ur − 1) + (ur − 1)e1 + (ur − 1)e2,

e31 = (3(ur)2 − 3ur + 1)e1 + 2ur(ur − 1)e2 + 2(ur)2 − ur = (u(ur) + 1)e1 − 2ure2 − ur =

e1 + ure2 + 2ur,

e32 = 2ur(ur− 1)e1 + (3(ur)2− 3ur+ 1)e2 + 2(ur)2−ur = −2ure1 + (u(ur) + 1)e2−ur =

ure1 + e2 + 2ur.

h2 = (1+e1+e2)2 = (1+e1+e2)(1+e1+e2) = 1+e1+e2+e1+e21+e1e2+e2+e1e2+e22 =

1 + 2e1 + 2e2 + (ur−1)e1 +ure2 + 2((2ur−1) + (ur−1)e1 + (ur−1)e2) +ure1 + (ur−1)e2 =

1 − e1 − e2 + ure1 + ure2 + (ur + 1) = 2 − e1 − e2 + ure1 + ure2 + (ur) = 2 + 2e1 + 2e2 +

ure1 + ure2 + (ur) = (2 + ur)(1 + e1 + e2) = (2 + ur)h.

(II)If p = 4k + 1, then:

e21 = (ur − 1)e1 + ure2 + 2ur,

e22 = ure1 + (ur − 1)e2 + 2ur,

e1e2 = ure1 + ure2,

e31 = (2(ur)2 + 1)e1 + (2(ur)2−ur)e2 + 2(ur)2− 2ur = e1−ure2− 2ur = e1 + 2ure2 +ur,

e32 = (2(ur)2−ur)e1 +(2(ur)2 +1)e2 +2(ur)2−2ur = −ure1 +e2−2ur = 2ure1 +e2 +ur.

h2 = (1+e1+e2)2 = (1+e1+e2)(1+e1+e2) = 1+e1+e2+e1+e21+e1e2+e2+e1e2+e22 =

1 + 2e1 + 2e2 + (ur − 1)e1 + ure2 + 2ur + 2(ure1 + ure2) + ure1 + (ur − 1)e2 + 2ur =

1 + e1 + e2 + ure1 + ure2 + ur = (1 + ur)(1 + e1 + e2) = (1 + ur)h.

Theorem 3.5.2. [32] Let e(x) be the idempotent generator of an R-cyclic code C. Then

1− e(x−1) is the idempotent generator of the dual code C⊥.

37

Theorem 3.5.3. I- Suppose that p = 12r − 1, then:

(a) If r = 3k, then (2+2u)ei, 1+ei, (2+u)h, 1+h are idempotents over (F3+uF3)[x]/〈xp−

1〉, where i=1,2.

(b) If r = 3k + 1, then (2 + 2u)ei + 2uej + u, ei + uej + (1 + 2u), 2h, 1 + (1 + u)h are

idempotents over (F3 + uF3)[x]/〈xp − 1〉, where 1 ≤ i 6= j ≤ 2.

(c) If r = 3k + 2, then (2 + 2u)ei + uej + 2u, ei + 2uej + (1 + u), (2 + 2u)h, 1 + (1 + 2u)h

are idempotents over (F3 + uF3)[x]/〈xp − 1〉, where 1 ≤ i 6= j ≤ 2.

II- Suppose that p = 12r + 1, then:

(a) If r = 3k, then 1 + ei, (2 + 2u)ei, h, (1 + u) + (2 + 2u)h are idempotents over (F3 +

uF3)[x]/〈xp − 1〉, where i=1,2.

(b) If r = 3k + 1, then ei + 2uej + (1 + u), (2 + 2u)ei + uej + 2u, (1 + 2u)h, (1 + u) + 2h


(c) If r = 3k+2, then ei+uej +(1+2u), (2+2u)ei+2uej +u, (1+u)h, (1+u)+(2+u)h


Proof. (I)Let p = 12r − 1. Since 2e1 is an idempotent of (F3)[x]/〈xp − 1〉, (2e1)3 is an

idempotent (F3 + uF3)[x]/〈xp − 1〉.

By Remark(3.5.2), in (F3 + uF3)[x]/〈xp − 1〉 we have:

(2e1)3 = 23e31 = 2e32 = (2 + 2u)(e1 + ure2 + 2ur).

• If r = 3k, then:

(2e1)3 = (2 + 2u)(e1 + u(uk)e2 + 2u(uk)) = (2 + 2u)e1.

((2 + 2u)e1)2 = (2 + 2u)2e21 = (1 + 2u)((u(uk) − 1)e1 + u(uk)e2) = (1 + 2u)(−e1) =

−(1 + 2u)e1 = (2 + u)e1 = (2 + 2u)e1.

((2 + u)h)2 = (2 + u)2h2 = (1 + u)(2 + u(uk))h = 2(1 + u)h = (2 + 2u)h = (2 + u)h.

• If r = 3k + 1, then:

(2e1)3 = (2 + 2u)(e1 +u(uk+ 1)e2 + 2u(uk+ 1)) = (2 + 2u)(e1 +ue2 + 2u) = (2 + 2u)e1 +

2ue2 + u.

((2+2u)e1+2ue2+u)2 = ((2+2u)e1+2ue2+u)((2+2u)e1+2ue2+u) = (1+2u)e21+ue1e2+

2ue1+ue1e2+2ue1 = (1+2u)((u(uk+1)−1)e1+u(uk+1)e2)+ue1+2u((2u(uk+1)−1)+(u(uk+

1)−1)e1+(u(uk+1)−1)e2) = (1+2u)((2+u)e1+ue2)+ue1+2u((2u+2)+(1+u)e1+(2+u)e2) =

(2 + 2u)e1 + ue2 + ue1 + u+ 2ue1 + ue2 = (2 + 2u)e1 + 2ue2 + u.

38

(2h)2 = 22h2 = h2 = (2 + u(uk + 1))h = (2 + u)h = 2h.

• If r = 3k + 2, then:

(2e1)3 = (2 + 2u)(e1 +u(uk+ 2)e2 + 2u(uk+ 2)) = (2 + 2u)(e1 + 2ue2 +u) = (2 + 2u)e1 +

ue2 + 2u.

((2+2u)e1+ue2+2u)2 = ((2+2u)e1+ue2+2u)((2+2u)e1+ue2+2u) = (1+2u)e21+2ue1e2+

ue1 + 2ue1e2 +ue1 = (1 + 2u)((u(uk+ 2)−1)e1 +u(uk+ 2)e2) +u((2u(uk+ 2)−1) + (u(uk+

2)−1)e1+(u(uk+2)−1)e2)+2ue1 = (1+2u)((2+2u)e1+2ue2)+u((2+u)+(2+2u)e1+(2+

2u)e2)+2ue1 = 2e1+2ue2+2u+2ue1+2ue2+2ue1 = (2+u)e1+ue2+2u = (2+2u)e1+ue2+2u.

((2+2u)h)2 = (2+2u)2h2 = (1+2u)(2+u(uk+2))h = (1+2u)(2+2u)h = 2h = (2+2u)h.

Therefore, (2e1)3 =

(2 + 2u)e1, r=3k;

(2 + 2u)e1 + 2ue2 + u, r=3k+1;

(2 + 2u)e1 + ue2 + 2u, r=3k+2.

Also, (1 + e1)3 = 1 + 3e1 + 3e21 + e31 = 1 + ue1 + u((ur− 1)e1 + ure2) + e1 + ure2 + 2ur =

1 + ue1 − ue1 + e1 + ure2 + 2ur = 1 + e1 + ure2 + 2ur.


(1 + e1)3 = 1 + e1 + u(uk)e2 + 2u(uk) = 1 + e1.

(1 + e1)2 = 1 + 2e1 + e21 = 1 + 2e1 + (u(uk)− 1)e1 + u(uk)e2 = 1 + 2e1 − e1 = 1 + e1.

(1 + h)2 = 1 + 2h+ h2 = 1 + 2h+ (2 + u(uk))h = 1 + 2h+ 2h = 1 + h.

• If r = 3k + 1, then:

(1 + e1)3 = 1 + e1 + u(uk + 1)e2 + 2u(uk + 1) = 1 + e1 + ue2 + 2u = e1 + ue2 + (1 + 2u).

(e1 + ue2 + (1 + 2u))2 = (e1 + ue2 + (1 + 2u))(e1 + ue2 + (1 + 2u)) = e21 + ue1e2 +

(1 + 2u)e1 + ue1e2 + ue2 + (1 + 2u)e1 + ue2 + (1 + u) = (u(uk + 1) − 1)e1 + u(uk + 1)e2 +

2u((2u(uk + 1)− 1) + (u(uk + 1)− 1)e1 + (u(uk + 1)− 1)e2) + (2 + u)e1 + ue2 + (1 + u) =

(2 + u)e1 + ue2 + 2u((2 + 2u) + (2 + u)e1 + (2 + u)e2) + (2 + u)e1 + ue2 + (1 + u) =

(2+u)e1 +ue2 +u+ue1 +ue2 +(2+u)e1 +ue2 +(1+u) = e1 +(1+2u) = e1 +ue2 +(1+2u).

(1 + (1 +u)h)2 = 1 + 2(1 +u)h+ (1 +u)2h2 = 1 + (2 + 2u)h+ (1 + 2u)(2 +u(uk+ 2))h =

1 + (2 + 2u)h+ (1 + 2u)(2 + 2u)h = 1 + (2 + 2u)h+ 2h = 1 + (1 + 2u)h = 1 + (1 + u)h.

• If r = 3k + 2, then:

(1 + e1)3 = 1 + e1 + u(uk + 2)e2 + 2u(uk + 2) = 1 + e1 + 2ue2 + u = e1 + 2ue2 + (1 + u).

(e1 + 2ue2 + (1 + u))2 = (e1 + 2ue2 + (1 + u))(e1 + 2ue2 + (1 + u)) = e21 + 2ue1e2 + (1 +

u)e1 + 2ue1e2 + 2u(1 +u)e2 + (1 +u)e1 + 2u(1 +u)e2 + (1 +u)2 = (u(uk+ 2)− 1)e1 +u(uk+

39

2)e2+u((2u(uk+2)−1)+(u(uk+2)−1)e1+(u(uk+2)−1)e2)+(2+2u)e1+ue2+(1+2u) =

(2 + 2u)e1 + 2ue2 + 2u+ 2ue1 + 2ue2 + (2 + 2u)e1 + ue2 + (1 + 2u) = e1 + 2ue2 + (1 + u).

(1 + (1 + 2u)h)2 = 1 + 2(1 + 2u)h+ (1 + 2u)2h2 = 1 + (2 +u)h+ (1 +u)(2 +u(uk+ 2))h =

1 + (2 + u)h+ (1 + u)(2 + 2u)h = 1 + (2 + u)h+ 2h = 1 + (1 + u)h = 1 + (1 + 2u)h.

Therefore, (1 + e1)3 =

1 + e1, r=3k;

e1 + ue2 + (1 + 2u), r=3k+1;

e1 + 2ue2 + (1 + u), r=3k+2.

Similarly for (2e2)3 and (1 + e2)

3.

(II)Let p = 12r + 1.

By remark(3.5.2), in (F3 + uF3)[x]/〈xp − 1〉 we have:

(1 + e2)3 = 1 + 3e2 + 3e22 + e32 = 1 + ue2 + u(ure1 + (ur− 1)e2 + 2ur) + 2ure1 + e2 + ur =

1 + ue2 − ue2 + 2ure1 + e2 + ur = 1 + 2ure1 + e2 + ur.


(1 + e2)3 = 1 + 2u(uk)e1 + e2 + u(uk) = 1 + e2.

(1+e2)2 = 1+2e2+e22 = 1+2e2+u(uk)e1+(u(uk)−1)e2+2u(uk) = 1+2e2−e2 = 1+e2.

h2 = (1 + u(uk))h = (1 + 0)h = h.

• If r = 3k + 1, then:

(1 + e2)3 = 1 + 2u(uk + 1)e1 + e2 + u(uk + 1) = 1 + 2ue1 + e2 + u = 2ue1 + e2 + (1 + u).

(2ue1 +e2 +(1+u))2 = (2ue1 +e2 +(1+u))(2ue1 +e2 +(1+u)) = 2ue1e2 +2u(1+u)e1 +

2ue1e2 + e22 + (1 + u)e2 + 2u(1 + u)e1 + (1 + u)e2 + (1 + u)2 = u(u(uk+ 1)e1 + u(uk+ 1)e2) +

ue1 +u(uk+1)e1 +(u(uk+1)−1)e2 +2u(uk+1)+(2+2u)e2 +(1+2u) = 2ue1 +e2 +(1+u).

((1 + 2u)h)2 = (1 + 2u)2h2 = (1 + u)(1 + u(uk + 1))h = (1 + u)(1 + u)h = (1 + 2u)h.

• If r = 3k + 2, then:

(1 + e2)3 = 1 + 2u(uk + 2)e1 + e2 + u(uk + 2) = 1 + ue1 + e2 + 2u = ue1 + e2 + (1 + 2u).

(ue1 + e2 + (1 + 2u))2 = (ue1 + e2 + (1 + 2u))(ue1 + e2 + (1 + 2u)) = ue1e2 +u(1 + 2u)e1 +

ue1e2 +e22 +(1+2u)e2 +u(1+2u)e1 +(1+2u)e2 +(1+2u)2 = 2u(u(uk+2)e1 +u(uk+2)e2)+

2ue1 +u(uk+2)e1 +(u(uk+2)−1)e2 +2u(uk+2)+(2+u)e2 +(1+u) = ue1 +e2 +(1+2u).

((1 + u)h)2 = (1 + u)2h2 = (1 + 2u)(1 + u(uk + 2))h = (1 + 2u)(1 + 2u)h = (1 + u)h.

Therefore, (1 + e2)3 =

1 + e2, r=3k;

2ue1 + e2 + (1 + u), r=3k+1;

ue1 + e2 + (1 + 2u), r=3k+2.

40

Also, (2e2)3 = 23e32 = 2e32 = (2 + 2u)(2ure1 + e2 + ur) = ure1 + (2 + 2u)e2 + 2ur.


(2e2)3 = u(uk)e1 + (2 + 2u)e2 + 2u(uk) = (2 + 2u)e2.

((2+2u)e2)2 = (2+2u)2e22 = (1+2u)(u(uk)e1+(u(uk)−1)e2+2u(uk)) = (1+2u)(2e2) =

(2 + u)e2 = (2 + 2u)e2.

((1 + u) + (2 + 2u)h)2 = (1 + u)2 + 2(1 + u)(2 + 2u)h + (2 + 2u)2h2 = (1 + 2u) + (1 +

2u)h+ (1 + 2u)(1 + u(uk))h = (1 + u) + (2 + u)h = (1 + u) + (2 + 2u)h.

• If r = 3k + 1, then:

(2e2)3 = u(uk + 1)e1 + (2 + 2u)e2 + 2u(uk + 1) = ue1 + (2 + 2u)e2 + 2u

(ue1 +(2+2u)e2 +2u)2 = (ue1 +(2+2u)e2 +2u)(ue1 +(2+2u)e2 +2u) = u(2+2u)e1e2 +

u(2 + 2u)e1e2 + (2 + 2u)2e22 + 2u(2 + 2u)e2 + 2u(2 + 2u)e2 = u(u(uk + 1)e1 + u(uk + 1)e2) +

(1+2u)(u(uk+1)e1 +(u(uk+1)−1)e2 +2u(uk+1))+2ue2 = ue1 +(2+2u)e2 +2u+2ue2 =

ue1 + (2 + u)e2 + 2u = ue1 + (2 + 2u)e2 + 2u.

((1+u)+2h)2 = (1+u)2 +2(2(1+u))h+22h2 = (1+2u)+(1+u)h+(1+u(uk+1))h =

(1 + 2u) + (1 + u)h+ (1 + u)h = (1 + 2u) + (2 + 2u)h = (1 + u) + 2h.

• If r = 3k + 2, then:

(2e2)3 = u(uk + 2)e1 + (2 + 2u)e2 + 2u(uk + 2) = 2ue1 + (2 + 2u)e2 + u.

(2ue1 + (2 + 2u)e2 + u)2 = (2ue1 + (2 + 2u)e2 + u)(2ue1 + (2 + 2u)e2 + u) = 2u(2 +

2u)e1e2 + 2u(2 + 2u)e1e2 + (2 + 2u)2e22 +u(2 + 2u)e2 +u(2 + 2u)e2 = 2u(u(uk+ 2)e1 +u(uk+

2)e2) + (1 + 2u)(u(uk+ 2)e1 + (u(uk+ 2)− 1)e2 + 2u(uk+ 2)) +ue2 = 2ue1 + 2e2 +u+ue2 =

2ue1 + (2 + u)e2 + u = 2ue1 + (2 + 2u)e2 + u.

((1+u)+(2+u)h)2 = (1+u)2 +2(1+u)(2+u)h+(2+u)2h2 = (1+2u)+h+(1+u)(1+

u(uk+2))h = (1+u)+h+(1+u)(1+2u)h = (1+u)+h+h = (1+u)+2h = (1+u)+(2+u)h.

Therefore, (2e2)3 =

(2 + 2u)e2, r=3k;

ue1 + (2 + 2u)e2 + 2u, r=3k+1;

2ue1 + (2 + 2u)e2 + u, r=3k+2.

Similarly for (2e1)3 and (1 + e1)

3.

Theorem 3.5.4. Suppose that p = 12r − 1.

(1)If r = 3k, let Q1 = 〈(2 + 2u)e1〉, Q2 = 〈(2 + 2u)e2〉, and Q′1 = 〈1 + e1〉, Q′2 = 〈1 + e2〉.

(2)If r = 3k + 1, Let Q1 = 〈(2 + 2u)e1 + 2ue2 + u〉, Q2 = 〈2ue1 + (2 + 2u)e2 + u〉, and

Q′1 = 〈e1 + ue2 + (1 + 2u)〉, Q′2 = 〈ue1 + e2 + (1 + 2u)〉.

41

(3)If r = 3k + 2, Let Q1 = 〈(2 + 2u)e1 + ue2 + 2u〉, Q2 = 〈ue1 + (2 + 2u)e2 + 2u〉, and

Q′1 = 〈e1 + 2ue2 + (1 + u)〉, Q′2 = 〈2ue1 + e2 + (1 + u)〉.

Then the following hold for (F3 + uF3)-QR codes Q1, Q2, Q′1, and Q′2:


(b) Q1 ∩Q2 = 〈h〉 and Q1 + Q2 = (F3 + uF3)[x]/〈xp − 1〉, where h is a suitable element

in {(2 + u)h, 2h, (2 + 2u)h}.

(c) |Q1| = 9p+12 = |Q2|.

(d) Q1 = Q′2 + 〈h〉, Q2 = Q′1 + 〈h〉.

(e) |Q′1| = 9p−12 = |Q′2|.

(f) Q⊥1 = Q′2 and Q⊥2 = Q′1.

(g) Q′1 ∩Q′2 = {0} and Q′1 +Q′2 = 〈1− h〉.

Proof. (a) Since p = 12r − 1, -1∈ N . Let a∈ N , then µae1 = e2 and µae2 = e1.

(1)If r = 3k, then:

µa((2 + 2u)e1) = ((2 + 2u)e2) and µa((2 + 2u)e2) = ((2 + 2u)e1).

µa(1 + e1) = (1 + e2) and µa(1 + e2) = (1 + e1).

(2)If r = 3k + 1, then:

µa((2 + 2u)e1 + 2ue2 + u) = (2 + 2u)e2 + 2ue1 + u and µa(2ue1 + (2 + 2u)e2 + u) =

2ue2 + (2 + 2u)e1 + u.

µa(e1+ue2+(1+2u)) = e2+ue1+(1+2u) and µa(ue1+e2+(1+2u)) = ue2+e1+(1+2u).

(3)If r = 3k + 2, then:

µa((2 + 2u)e1 + ue2 + 2u) = (2 + 2u)e2 + ue1 + 2u and µa(ue1 + (2 + 2u)e2 + 2u) =

ue2 + (2 + 2u)e1 + 2u.

µa(e1+2ue2+(1+u)) = e2+2ue1+(1+u) and µa(2ue1+e2+(1+u)) = 2ue2+e1+(1+u).

Therefore, Q1 and Q2 are equivalent and Q′1 and Q′2 are equivalent.

(b) By Theorem 1.3.10;

(1)If r = 3k, then:

Q1 ∩Q2 has an idempotent generator,

((2+2u)e1)((2+2u)e2) = (2+2u)2e1e2 = (1+2u)((2u(uk)−1)+(u(uk)−1)e1+(u(uk)−

1)e2) = (1 + 2u)(2 + 2e1 + 2e2) = 2(1 + 2u)(1 + e1 + e2) = (2 + u)h.

So, Q1 ∩Q2 = 〈(2 + u)h〉.

42

Q1 +Q2 has an idempotent generator,

(2 + 2u)e1 + (2 + 2u)e2 − ((2 + 2u)e1)((2 + 2u)e2) = (2 + 2u)e1 + (2 + 2u)e2 − (2 + u)−

(2 + u)e1 − (2 + u)e2 = ue1 + ue2 + (1 + 2u) = 1.

So, Q1 +Q2 = (F3 + uF3)[x]/〈xp − 1〉.

(2)If r = 3k + 1, then:


((2+2u)e1+2ue2+u)(2ue1+(2+2u)e2+u) = 2u(2+2u)e21+(2+2u)2e1e2+u(2+2u)e1+

2u(2 + 2u)e22 +u(2 + 2u)e2 = u((u(uk+ 1)−1)e1 +u(uk+ 1)e2) + (1 + 2u)((2u(uk+ 1)−1) +

(u(uk+ 1)− 1)e1 + (u(uk+ 1)− 1)e2) + 2ue1 + u(u(uk+ 1)e1 + (u(uk+ 1)− 1)e2) + 2ue2 =

u((2 + u)e1 + ue2) + (1 + 2u)((2 + 2u) + (2 + u)e1 + (2 + u)e2) + 2ue1 + u(2 + u)e2 + 2ue2 =

2ue1 + 2 + (2 + 2u)e1 + (2 + 2u)e2 + 2ue1 + ue2 = 2 + 2e1 + 2e2 = 2(1 + e1 + e2) = 2h.

So, Q1 ∩Q2 = 〈2h〉.

Q1 +Q2 has an idempotent generator, ((2 + 2u)e1 + 2ue2 +u) + (2ue1 + (2 + 2u)e2 +u)−

((2 + 2u)e1 + 2ue2 + u)(2ue1 + (2 + 2u)e2 + u) = (2 + u)e1 + (2 + u)e2 + 2u− 2− 2e1− 2e2 =

ue1 + ue2 + 1 + 2u = 1.

So, Q1 +Q2 = (F3 + uF3)[x]/〈xp − 1〉.

(3)If r = 3k + 2, then:


((2+2u)e1+ue2+2u)(ue1+(2+2u)e2+2u) = u(2+2u)e21+(2+2u)2e1e2+2u(2+2u)e1+

u(2 + 2u)e22 + 2u(2 + 2u)e2 = 2u((u(uk + 2)− 1)e1 + u(uk + 2)e2) + (1 + 2u)((2u(uk + 2)−

1)+(u(uk+2)−1)e1 +(u(uk+1)−1)e2)+ue1 +2u(u(uk+2)e1 +(u(uk+2)−1)e2)+ue2 =

2u(2+2u)e1+(1+2u)((2+u)+(2+2u)e1+(2+2u)e2)+ue1+2u(2+2u)e2+ue2 = ue1+(2+2u)+

2e1+2e2+ue1+ue2+ue2 = (2+2u)+(2+2u)e1+(2+2u)e2 = (2+2u)(1+e1+e2) = (2+2u)h.

So, Q1 ∩Q2 = 〈(2 + 2u)h〉.


((2+2u)e1+ue2+2u)+(ue1+(2+2u)e2+2u)−((2+2u)e1+ue2+2u)(ue1+(2+2u)e2+2u) =

2e1+2e2+u−(2+2u)−(2+2u)e1−(2+2u)e2 = −2−2u−2ue1−2ue2 = 1+u+ue1+ue2 = 1.

So, Q1 +Q2 = (F3 + uF3)[x]/〈xp − 1〉.

(c)By (a) and (b) for all cases of r we have:

9p = |Q1 +Q2| = |Q1||Q2||Q1∩Q2| = |Q1|2

9,

So, |Q1| = 9p+12 = |Q2|.

43

(d) using Theorem 1.3.10;

(1)If r = 3k, then:

Q′1 ∩ 〈(2 + u)h〉 has an idempotent generator,

(1+e1)((2+u)h) = (1+e1)((2+u)+(2+u)e1+(2+u)e2) = (2+u)+(2+u)e1+(2+u)e2+(2+

u)e1+(2+u)e21+(2+u)e1e2 = (2+u)+(1+2u)e1+(2+u)e2+(2+u)((u(uk)−1)e1+u(uk)e2)+

(2+u)((2u(uk)−1)+(u(uk)−1)e1 +(u(uk)−1)e2) = (2+u)+(1+2u)e1 +(2+u)e2 +2(2+

u)e1+(2+u)(2+2e1+2e2) = (2+u)+(2+u)e1+(2+u)e2+(1+2u)+(1+2u)e1+(1+2u)e2 = 0.

So, Q′1 ∩ 〈(2 + u)h〉 = {0}.

Q′1 + 〈(2 + u)h〉 has an idempotent generator,

(1 + e1) + ((2 + u)h)− (1 + e1)((2 + u)h) = 1 + e1 + (2 + u) + (2 + u)e1 + (2 + u)e2− 0 =

2e2 = (2 + 2u)e2.

So, Q′1 + 〈(2 + u)h〉 = Q2.

(2)If r = 3k + 1, then:

Q′1 ∩ 〈2h〉 has an idempotent generator,

(e1+ue2+(1+2u))(2h) = (e1+ue2+(1+2u))(2e1+2e2+2) = 2e21+2e1e2+2e1+2ue1e2+

2ue22+2ue2+(2+u)e1+(2+u)e2+(2+u) = 2((u(uk+1)−1)e1+u(uk+1)e2)+(2+2u)((2u(uk+

1)−1)+(u(uk+1)−1)e1+(u(uk+1)−1)e2)+(1+u)e1+2u(u(uk+1)e1+(u(uk+1)−1)e2)+

2e2+(2+u) = 2((2+u)e1+ue2)+(2+2u)((2+2u)+(2+u)e1+(2+u)e2)+(1+u)e1+2u(ue1+

(2+u)e2)+2e2+(2+u) = (1+2u)e1+2ue2+(1+2u)+e1+e2+(1+u)e1+ue2+2e2+(2+u) = 0.

So, Q′1 ∩ 〈2h〉 = {0}.

Q′1 + 〈2h〉 has an idempotent generator,

(e1+ue2+(1+2u))+(2h)−(e1+ue2+(1+2u))(2h) = e1+ue2+(1+2u)+2e1+2e2+2−0 =

(2 + u)e2 + 2u = 2ue1 + (2 + 2u)e2 + u.

So, Q′1 + 〈2h〉 = Q2.

(3)If r = 3k + 2, then:

Q′1 ∩ 〈(2 + 2u)h〉 has an idempotent generator,

(e1+2ue2+(1+u))((2+2u)h) = (e1+2ue2+(1+u))((2+2u)e1+(2+2u)e2+(2+2u)) =

(2+2u)e21+(2+2u)e1e2+(2+2u)e1+2u(2+2u)e1e2+2u(2+2u)e22+2u(2+2u)e2+(1+u)(2+

2u)e1+(1+u)(2+2u)e2+(1+u)(2+2u) = (2+2u)((u(uk+2)−1)e1+u(uk+2)e2)+2((2u(uk+

2)−1)+(u(uk+2)−1)e1+(u(uk+2)−1)e2)+e1+u(u(uk+2)e1+(u(uk+2)−1)e2)+(2+u)e2+

(2+u) = (2+2u)((2+2u)e1+2ue2)+2((2+u)+(2+2u)e1+(2+2u)e2)+e1+u(2+2u)e2+(2+

44

u)e2+(2+u) = (1+2u)e1+ue2+(1+2u)+(1+u)e1+(1+u)e2+e1+2e2+(2+u) = 2ue2 = 0.

So, Q′1 ∩ 〈(2 + 2u)h〉 = {0}.

Q′1 + 〈(2 + 2u)h〉 has an idempotent generator,

(e1 + 2ue2 + (1 + u)) + ((2 + 2u)h)− (e1 + 2ue2 + (1 + u))((2 + 2u)h) = e1 + 2ue2 + (1 +

u) + (2 + 2u)e1 + (2 + 2u)e2 + (2 + 2u)− 0 = 2ue1 + (2 + u)e2 = ue1 + (2 + 2u)e2 + 2u.

So, Q′1 + 〈(2 + 2u)h〉 = Q2.

(e)By (a) and (b) for all cases of r we have:

9p+12 = |Q2| = |Q′1 + 〈h〉| = |Q′1||〈h〉| = 9|Q′1|

Thus |Q′1| = 9p−12 .

(f)Since −1 ∈ Q and by Theorem 3.5.2, we have:

(1)If r = 3k, then:

Q⊥1 has an idempotent generator,

1− ((2 + 2u)e1(x−1)) = 1 + (1 + u)e1(x

−1) = 1 + (e2).

So, Q⊥1 = Q′2.

Also, Q⊥2 has an idempotent generator,

1− ((2 + 2u)e2(x−1)) = 1 + (1 + u)e2(x

−1) = 1 + (e1).

So, Q⊥2 = Q′1.

(2)If r = 3k + 1, then:


1− ((2 + 2u)e1(x−1) + 2ue2(x

−1) +u) = 1− (2 + 2u)e1(x−1)−2ue2(x

−1)−u = (1 +u)e2 +

ue1 + (1 + 2u)) = ue1 + (1 + u)e2 + (1 + 2u) = ue1 + e2 + (1 + 2u).

So, Q⊥1 = Q′2.


1− ((2 + 2u)e2(x−1) + 2ue1(x

−1) +u) = 1− (2 + 2u)e2(x−1)−2ue1(x

−1)−u = (1 +u)e1 +

ue2 + (1 + 2u)) = ue2 + (1 + u)e1 + (1 + 2u) = ue2 + e1 + (1 + 2u).

So, Q⊥2 = Q′1.

(3)If r = 3k + 2, then:


1 − ((2 + 2u)e1(x−1) + ue2(x

−1) + 2u) = −(2 + 2u)e1(x−1) + −ue2(x−1) + (1 − 2u) =

(1 + u)e2 + 2ue1 + (1 + u) = 2ue1 + (1 + u)e2 + (1 + u) = 2ue1 + e2 + (1 + u).

So, Q⊥1 = Q′2.

45


1 − ((2 + 2u)e2(x−1) + ue1(x

−1) + 2u) = −(2 + 2u)e2(x−1) + −ue1(x−1) + (1 − 2u) =

(1 + u)e1 + 2ue2 + (1 + u) = 2ue2 + (1 + u)e1 + (1 + u) = 2ue2 + e1 + (1 + u).

So, Q⊥2 = Q′1.

(g) By Theorem 1.3.10;

(1)If r = 3k, then:

Q′1 ∩Q′2 has an idempotent generator,

(1+e1)(1+e2) = 1+e2+e1+e1e2 = 1+e2+e1+(2u(uk)−1)+(u(uk)−1)e1+(u(uk)−1)e2 =

1 + e2 + e1 + 2 + 2e1 + 2e2 = 0.

Q′1 +Q′2 has an idempotent generator,

(1 + e1) + (1 + e2) − (1 + e1)(1 + e2) = 1 + e1 + 1 + e2 − 0 = 1 + (1 + e1 + e2) =

1− 2(1 + e1 + e2) = 1− (2 + u)(1 + e1 + e2) = 1− (2 + u)h.

(2)If r = 3k + 1, then:


(e1 + ue2 + (1 + 2u))(ue1 + e2 + (1 + 2u)) = ue21 + e1e2 + (1 + 2u)e1 + ue22 + u(1 + 2u)e2 +

u(1 + 2u)e1 + (1 + 2u)e2 + (1 + 2u)2 = u((u(uk+ 1)−1)e1 +u(uk+ 1)e2) + (2u(uk+ 1)−1) +

(u(uk+1)−1)e1 +(u(uk+1)−1)e2 +e1 +u(u(uk+1)e1 +(u(uk+1)−1)e2)+e2 +(1+u) =

u((2 + u)e1 + ue2) + (2 + 2u) + (2 + u)e1 + (2 + u)e2 + e1 + u(2 + u)e2 + e2 + (1 + u) =

2ue1 + ue1 + 2ue2 = 2ue2 = 0.


(e1 +ue2 + (1 + 2u)) + (ue1 + e2 + (1 + 2u))− (e1 +ue2 + (1 + 2u))(ue1 + e2 + (1 + 2u)) =

e1 +ue2 +(1+2u)+ue1 +e2 +(1+2u)−0 = (1+u)e1 +(1+u)e2 +(2+u) = 1+(1+u)+(1+

u)e1+(1+u)e2 = 1+(1+u)(1+e1+e2) = 1−(2+2u)(1+e1+e2) = 1−2(1+e1+e2) = 1−2h.

(3)If r = 3k + 2, then:


(e1+2ue2+(1+u))(2ue1+e2+(1+u)) = 2ue21+e1e2+(1+u)e1+2ue22+2u(1+u)e2+2u(1+

u)e1+(1+u)e2+(1+u)2 = 2u((u(uk+2)−1)e1+u(uk+2)e2)+(2u(uk+2)−1)+(u(uk+2)−

1)e1+(u(uk+2)−1)e2+(2+2u)e2+2u(u(uk+2)e1+(u(uk+2)−1)e2)+e2+(1+2u) = 2u((2+

2u)e1+2ue2)+(2+u)+(2+2u)e1+(2+2u)e2+2ue2+2u(2+2u)e2+e2+(1+2u)+e1 = ue1 = 0.


(e1 + 2ue2 + (1 +u)) + (2ue1 + e2 + (1 +u))− (e1 + 2ue2 + (1 +u))(2ue1 + e2 + (1 +u)) =

46

e1 + 2ue2 + (1 + u) + 2ue1 + e2 + (1 + u) − 0 = (2 + 2u) + (1 + 2u)e1 + (1 + 2u)e2 =

1 + (1 + 2u) + (1 + 2u)e1 + (1 + 2u)e2 = 1 + (1 + 2u)(1 + e1 + e2) = 1− (2 + u)(1 + e1 + e2) =

1− (2 + 2u)(1 + e1 + e2) = 1− (2 + 2u)h.

Therefore in all cases, Q′1 ∩Q′2 = {0} and Q′1 +Q′2 = 〈1− h〉.

Theorem 3.5.5. Suppose that p = 12r + 1.

(1)If r = 3k, let Q1 = 〈1 + e1〉, Q2 = 〈1 + e2〉 and Q′1 = 〈(2 + 2u)e1〉, Q′2 = 〈(2 + 2u)e2〉.

(2)If r = 3k + 1, Let Q1 = 〈e1 + 2ue2 + (1 + u)〉, Q2 = 〈2ue1 + e2 + (1 + u)〉 and

Q′1 = 〈(2 + 2u)e1 + ue2 + 2u〉, Q′2 = 〈ue1 + (2 + 2u)e2 + 2u〉, .

(3)If r = 3k + 2, Let Q1 = 〈e1 + ue2 + (1 + 2u)〉, Q2 = 〈ue1 + e2 + (1 + 2u)〉 and

Q′1 = 〈(2 + 2u)e1 + 2ue2 + u〉, Q′2 = 〈2ue1 + (2 + 2u)e2 + u〉,

Then the following hold for (F3 + uF3)-QR codes Q1, Q2, Q′1, and Q′2:


(b) Q1 ∩Q2 = 〈h〉 and Q1 + Q2 = (F3 + uF3)[x]/〈xp − 1〉, where h is a suitable element

in {h, (1 + 2u)h, (1 + u)h}.

(c) |Q1| = 9p+12 = |Q2|.

(d) Q1 = Q′2 + 〈h〉, Q2 = Q′1 + 〈h〉.

(e) |Q′1| = 9p−12 = |Q′2|.

(f) Q⊥1 = Q′2 and Q⊥2 = Q′1.

(g) Q′1 ∩Q′2 = {0} and Q′1 +Q′2 = 〈1− h〉.

Proof. (a) Since p = 12r + 1, -1∈ N . Let a∈ N , then µae1 = e2 and µae2 = e1.

(1)If r = 3k, then:

µa(1 + e1) = (1 + e2) and µa(1 + e2) = (1 + e1).

µa((2 + 2u)e1) = ((2 + 2u)e2) and µa((2 + 2u)e2) = ((2 + 2u)e1).

(3)If r = 3k + 1, then:

µa(e1+2ue2+(1+u)) = e2+2ue1+(1+u) and µa(2ue1+e2+(1+u)) = 2ue2+e1+(1+u).

µa((2 + 2u)e1 + ue2 + 2u) = (2 + 2u)e2 + ue1 + 2u and µa(ue1 + (2 + 2u)e2 + 2u) =

ue2 + (2 + 2u)e1 + 2u.

(3)If r = 3k + 2, then:

µa(e1+ue2+(1+2u)) = e2+ue1+(1+2u) and µa(ue1+e2+(1+2u)) = ue2+e1+(1+2u).

47

µa((2 + 2u)e1 + 2ue2 + u) = (2 + 2u)e2 + 2ue1 + u and µa(2ue1 + (2 + 2u)e2 + u) =

2ue2 + (2 + 2u)e1 + u.

Therefore, Q1 and Q2 are equivalent and Q′1 and Q′2 are equivalent.

(b) By Theorem 1.3.10;

(1)If r = 3k, then:


(1 + e1)(1 + e2) = 1 + e1 + e2 + e1e2 = 1 + e1 + e2 + (ur)e1 + (ur)e2 = 1 + e1 + e2 +

(u(uk))e1 + (u(uk))e2 = 1 + e1 + e2 = h.

So, Q1 ∩Q2 = 〈h〉.


(1 + e1) + (1 + e2)− (1 + e1)(1 + e2) = (1 + e1) + (1 + e2)− (1 + e1 + e2) = 1.

So, Q1 +Q2 = (F3 + uF3)[x]/〈xp − 1〉.

(2)If r = 3k + 1, then:


(e1 + 2ue2 + (1 + u))(2ue1 + e2 + (1 + u)) = 2ue21 + e1e2 + (1 + u)e1 + (2u)2e1e2 + 2ue22 +

(2u)(1 + u)e2 + (2u)(1 + u)e1 + (1 + u)e2 + (1 + u)2 = 2u((u(uk + 1)− 1)e1 + u(uk + 1)e2 +

2u(uk + 1)) + u(uk + 1)e1 + u(uk + 1)e2 + (1 + u)e1 + 2u(u(uk + 1)e1 + (u(uk + 1)− 1)e2 +

2u(uk+ 1)) + 2ue2 = ue1 +ue1 +ue2 + (1 +u)e1 +ue2 + 2ue2 + 2ue1 + (1 +u)e2 + (1 + 2u) =

e1 + ue2 + 2ue1 + e2 + ue2 + (1 + 2u) = (1 + 2u)e1 + (1 + 2u)e2 + (1 + 2u) = (1 + 2u)h.

So, Q1 ∩Q2 = 〈(1 + 2u)h〉.


(e1 + 2ue2 + (1 +u)) + (2ue1 + e2 + (1 +u))− (e1 + 2ue2 + (1 +u))(2ue1 + e2 + (1 +u)) =

(1 + 2u)e1 + (1 + 2u)e2 + (2 + 2u)− (1 + 2u)e1 − (1 + 2u)e2 − (1 + 2u) = 1.

So, Q1 +Q2 = (F3 + uF3)[x]/〈xp − 1〉.

(3)If r = 3k + 2, then:


(e1+ue2+(1+2u))(ue1+e2+(1+2u)) = ue21+e1e2+(1+2u)e1+(u)2e1e2+ue22+(u)(1+

2u)e2+(u)(1+2u)e1+(1+2u)e2+(1+2u)2 = u((u(uk+2)−1)e1+u(uk+2)e2+2u(uk+2))+

u(uk+2)e1+u(uk+2)e2+(1+2u)e1+u(u(uk+2)e1+(u(uk+2)−1)e2+2u(uk+2))+ue2+ue1+

(1+2u)e2+(1+u) = 2ue1+2ue1+2ue2+(1+2u)e1+2ue2+ue2+ue1+(1+2u)e2+(1+u) =

ue1 + ue2 + e1 + e2 + (1 + u) = (1 + u)e1 + (1 + u)e2 + (1 + u) = (1 + u)h.

48

So, Q1 ∩Q2 = 〈(1 + u)h〉.


(e1 +ue2 + (1 + 2u)) + (ue1 + e2 + (1 + 2u))− (e1 +ue2 + (1 + 2u))(ue1 + e2 + (1 + 2u)) =

(1 + u)e1 + (1 + u)e2 + (1 + u)− (1 + u)e1 − (1 + u)e2 − (1 + u) = 1.

So, Q1 +Q2 = (F3 + uF3)[x]/〈xp − 1〉.

(c)By (a) and (b) for all cases of r we have:

9p = |Q1 +Q2| = |Q1||Q2||Q1∩Q2| = |Q1|2

9,

So, |Q1| = 9p+12 = |Q2|.

(d) Using Theorem 1.3.10;

(1)If r = 3k, then:

Q′1 ∩ 〈h〉 has an idempotent generator,

((2 + 2u)e1)(1 + e1 + e2) = (2 + 2u)e1 + (2 + 2u)e21 + (2 + 2u)e1e2 = (2 + 2u)e1 + (2 +

2u)((u(uk)− 1)e1 + u(uk)e2 + 2u(uk)) = (2 + 2u)e1 + (1 + u)e1 = 0.

So, Q′1 ∩ 〈h〉 = {0}.

Q′1 + 〈h〉 has an idempotent generator,

((2 + 2u)e1) + (1 + e1 + e2)− ((2 + 2u)e1)(1 + e1 + e2) = ((2 + 2u)e1) + (1 + e1 + e2) =

1 + 2ue1 + e2 = 1 + e2

So, Q′1 + 〈h〉 = Q2.

(2)If r = 3k + 1, then:

Q′1 ∩ 〈(1 + 2u)h〉 has an idempotent generator,

((2+2u)e1+ue2+2u)((1+2u)e1+(1+2u)e2+(1+2u)) = 2e21+2e1e2+2e1+ue1e2+ue22+

ue2 + 2ue1 + 2ue2 + (2u) = 2((u(uk+ 1)− 1)e1 + u(uk+ 1)e2 + 2u(uk+ 1)) + (2 + u)(u(uk+

1)e1 + u(uk+ 1)e2) + (2 + 2u)e1 + u((u(uk+ 1)− 1)e2 + u(uk+ 1)e1 + 2u(uk+ 1)) + (2u) =

(1 + 2u)e1 + (2u)e2 + u+ (2u)e1 + (2u)e2 + (2 + 2u)e1 + (2u)e2 + (2u) = 0.

So, Q′1 ∩ 〈(1 + 2u)h〉 = {0}.

Q′1 + 〈(1 + 2u)h〉 has an idempotent generator,

((2+2u)e1 +ue2 +2u)+((1+2u)e1 +(1+2u)e2 +(1+2u))− ((2+2u)e1 +ue2 +2u)((1+

2u)e1 +(1+2u)e2 +(1+2u)) = ((2+2u)e1 +ue2 +2u)+((1+2u)e1 +(1+2u)e2 +(1+2u)) =

ue1 + e2 + (1 + u) = 2ue1 + e2 + (1 + u).

So, Q′1 + 〈(1 + 2u)h〉 = Q2.

(3)If r = 3k + 2, then:

49

Q′1 ∩ 〈(1 + u)h〉 has an idempotent generator,

((2+2u)e1+2ue2+(u))((1+u)e1+(1+u)e2+(1+u)) = (2+u)e21+(2+u)e1e2+(2+u)e1+

2ue1e2+2ue22+2ue2+(u)e1+(u)e2+(u) = (2+u)((u(uk+2)−1)e1+u(uk+2)e2+2u(uk+2))+

2(u(uk+2)e1+u(uk+2)e2)+(2u)((u(uk+2)−1)e2+u(uk+2)e1+2u(uk+2))+(u)e1+(u) =

e1 + (u)e2 + 2u+ (u)e1 + (2u)e2 + (2 + 2u)e1 + (u)e2 + (u) = (u)e2 = 0.

So, Q′1 ∩ 〈(1 + u)h〉 = {0}.

Q′1 + 〈(1 + u)h〉 has an idempotent generator,

((2+2u)e1 +2ue2 +(u))+((1+u)e1 +(1+u)e2 +(1+u))− ((2+2u)e1 +2ue2 +(u))((1+

u)e1 + (1 + u)e2 + (1 + u)) = ((2 + 2u)e1 + 2ue2 + (u)) + ((1 + u)e1 + (1 + u)e2 + (1 + u)) =

e2 + (1 + 2u) = (u)e1 + e2 + (1 + 2u)

So, Q′1 + 〈(1 + u)h〉 = Q2.

(e)By (a) and (b) for all cases of r we have:

9p+12 = |Q2| = |Q′1 + 〈h〉| = |Q′1||〈h〉| = 9|Q′1|

Thus |Q′1| = 9p−12 .

(f)Since −1 ∈ Q and by Theorem 3.5.2, we have:

(1)If r = 3k, then:


1− (1 + e1(x−1)) = −e1(x−1) = 2(e2) = (2 + 2u)(e2)

So, Q⊥1 = Q′2.


1− (1 + e2(x−1)) = −e2(x−1) = 2(e1) = (2 + 2u)(e1)

So, Q⊥2 = Q′1.

(2)If r = 3k + 1, then:


1−(e1(x−1)+2ue2(x

−1)+(1+u)) = (−e1(x−1)+−2ue2(x−1)+(−u)) = (2e2+ue1+(2u)) =

(ue1 + 2e2 + (2u)) = (ue1 + (2 + 2u)e2 + (2u))

So, Q⊥1 = Q′2.

Also, 1 − (e2(x−1) + 2ue1(x

−1) + (1 + u)) = (−e2(x−1) + −2ue1(x−1) + (−u)) = (2e1 +

ue2 + (2u)) = (ue2 + 2e1 + (2u)) = (ue2 + (2 + 2u)e1 + (2u))

So, Q⊥2 = Q′1.

(3)If r = 3k + 2, then:

50


1−(e1(x−1)+ue2(x

−1)+(1+2u)) = (−e1(x−1)+−ue2(x−1)+(−2u)) = (2e2+2ue1+(u)) =

(2ue1 + 2e2 + (u)) = (2ue1 + (2 + 2u)e2 + (u))

So, Q⊥1 = Q′2.


1−(e2(x−1)+ue1(x

−1)+(1+2u)) = (−e2(x−1)+−ue1(x−1)+(−2u)) = (2e1+2ue2+(u)) =

(2ue2 + 2e1 + (u)) = (2ue2 + (2 + 2u)e1 + (u))

So, Q⊥2 = Q′1.

(g) By Theorem 1.3.10;

(1)If r = 3k, then:


((2 + 2u)e1)((2 + 2u)e2) = (2 + 2u)2e1e2 = (1 + 2u)u(uk)e1 + u(uk)e2 = 0.


((2 + 2u)e1) + ((2 + 2u)e2) − ((2 + 2u)e1)((2 + 2u)e2) = ((2 + 2u)e1) + ((2 + 2u)e2) =

2e1 + 2e2 = −e1 − e21− (1 + e1 + e2) = 1− h.

(2)If r = 3k + 1, then:


((2+2u)e1+ue2+2u)(ue1+(2+2u)e2+2u) = u(2+2u)e21+(2+2u)2e1e2+2u(2+2u)e1+

u2e1e2 +u(2+2u)e22 +(2u)(u)e2 +(2u)(u)e1 +(2u)(2+2u)e2 +(2u)2 = 2u((u(uk+1)−1)e1 +

u(uk+1)e2+2u(uk+1))+(1+2u)(u(uk+1)e1+u(uk+1)e2)+ue1+2u(u(uk+1)e1+(u(uk+1)−

1)e2+2u(uk+1))+ue2 = 2u(2+u)e1+u(1+2u)e1+u(1+2u)e2+ue1+2u(2+u)e2+ue2 = 0.


((2 + 2u)e1 + ue2 + 2u) + (ue1 + (2 + 2u)e2 + 2u) − ((2 + 2u)e1 + ue2 + 2u)(ue1 + (2 +

2u)e2 + 2u)) = 2e1 + 2e2 + (u) = −e1 − e2 − 2u = −(1 + 2u)e1 − (1 + 2u)e2 − 2u =

1− ((1 + 2u) + (1 + 2u)e1 + (1 + 2u)e2) = 1− (1 + 2u)h.

(3)If r = 3k + 2, then:


((2+2u)e1+2ue2+u)(2ue1+(2+2u)e2+u) = 2u(2+2u)e21+(2+2u)2e1e2+u(2+2u)e1+

(2u)2e1e2+2u(2+2u)e22+(u)(2u)e2+(u)(2u)e1+(u)(2+2u)e2+(u)2 = u((u(uk+2)−1)e1+

u(uk+2)e2+2u(uk+2))+(1+2u)(u(uk+2)e1+u(uk+2)e2)+2ue1+u(u(uk+2)e1+(u(uk+2)−

1)e2+2u(uk+2))+2ue2 = 2u(2+2u)e1+2u(1+2u)e1+2u(1+2u)e2+2ue1+2u(2+u)e2+2ue2 =

51

0.


((2+2u)e1+2ue2+u)+(2ue1+(2+2u)e2+u)−((2+2u)e1+2ue2+u)(2ue1+(2+2u)e2+u)) =

(2 + u)e1 + (2 + u)e2 + (2u) = −(1 + 2u)e1 − (1 + 2u)e2 − u = −(1 + u)e1 − (1 + u)e2 − u =

1− ((1 + u) + (1 + u)e1 + (1 + u)e2) = 1− (1 + u)h.

Therefore in all cases, Q′1 ∩Q′2 = {0} and Q′1 +Q′2 = 〈1− h〉.

By direct computation we have the following example:

Example 3.5.1. The (F3 + uF3) quadratic residue code of length p = 11 ≡ −1(mod 12).

We first find the generating idempotents of quadratic residue codes of length 11 over F3.

Here Q11 = {1, 3, 4, 5, 9} and N11 = {2, 6, 7, 8, 10}.

The generating idempotents of odd-like quadratic codes over F3 are 2e1 = 2∑

j∈Q11xj =

2(x+x3 +x4 +x5 +x9) and 2e2 = 2∑

j∈N11xj = 2(x2 +x6 +x7 +x8 +x10) and the generating

idempotents of even-like quadratic residue codes over F3 are 1 + e1 = 1 +∑

j∈Q11xj =

1 + (x+ x3 + x4 + x5 + x9) and 1 + e2 = 1 +∑

j∈N11xj = 1 + (x2 + x6 + x7 + x8 + x10).

Since 11 = 4(3r)− 1, then r = 1, and so, r = 3k + 1 (in Theorem 3.5.4).

Then the generating idempotent of odd-like quadratic residue codes over (F3 + uF3) are

(2+2u)e1+2ue2+u, and (2+2u)e2+2ue1+u which is equivalent to (2+2u)(x+x3+x4+x5+

x9)+2u(x2+x6+x7+x8+x10)+u and (2+2u)(x2+x6+x7+x8+x10)+2u(x+x3+x4+x5+x9)+u

and the generating idempotent of even-like quadratic residue codes over (F3 + uF3) are

e1+2ue2+(1+u), and e2+ue1+(1+2u) which is equivalent to (x+x3+x4+x5+x9)+2u(x2+

x6 +x7 +x8 +x10)+(1+u) and (x2 +x6 +x7 +x8 +x10)+u(x+x3 +x4 +x5 +x9)+(1+2u).

Example 3.5.2. The (F3 + uF3) quadratic residue code of length p = 13 ≡ 1(mod 12).


Here Q13 = {1, 3, 4, 9, 10, 12} and N13 = {2, 5, 6, 7, 8, 11}.

The generating idempotents of even-like quadratic codes over F3 are 2e1 = 2∑

j∈Q13xj =

2(x+x3 +x4 +x9 +x10 +x12) and 2e2 = 2∑

j∈N13xj = 2(x2 +x5 +x6 +x7 +x8 +x11) and the

generating idempotents of odd-like quadratic residue codes over F3 are 1+e1 = 1+∑

j∈Q13xj =

1+(x+x3+x4+x9+x10+x12) and 1+e2 = 1+∑

j∈N13xj = 1+(x2+x5+x6+x7+x8+x11)

Since 13 = 4(3r) + 1, then r = 1, and so, r = 3k + 1 (in Theorem 3.5.5).

52

Then the generating idempotent of even-like quadratic residue codes over (F3 + uF3) are

(2 + 2u)e1 + ue2 + 2u, and (2 + 2u)e2 + ue1 + 2u which is equivalent to (2 + 2u)(x + x3 +

x4 + x9 + x10 + x12) + u(x2 + x5 + x6 + x7 + x8 + x11) + 2u and (2 + 2u)(x2 + x5 + x6 + x7 +

x8 + x11) + u(x+ x3 + x4 + x9 + x10 + x12) + 2u

and the generating idempotent of odd-like quadratic residue codes over (F3 + uF3) are

e1 + 2ue2 + (1 + u), and e2 + 2ue1 + (1 + u) which is equivalent to (x+ x3 + x4 + x9 + x10 +

x12) + 2u(x2 + x5 + x6 + x7 + x8 + x11) + (1 + u) and (x2 + x5 + x6 + x7 + x8 + x11) + 2u(x+

x3 + x4 + x9 + x10 + x12) + (1 + u).

Example 3.5.3. The (F3 + uF3) quadratic residue code of length p = 23 ≡ −1(mod 12).


Here Q23 = {1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18} and N23 = {5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22}.

The generating idempotents of odd-like quadratic codes over F3 are 2e1 = 2∑

j∈Q23=

2(x+x2+x3+x4+x6+x8+x9+x12+x13+x16+x18) and 2e2 = 2∑

j∈N23= 2(x5+x7+x10+

x11+x14+x15+x17+x19+x20+x21+x22) and the generating idempotents of even-like quadratic

residue codes over F3 are 1+e1 = 1+∑

j∈Q23= 1+(x+x2+x3+x4+x6+x8+x9+x12+x13+

x16+x18) and 1+e2 = 1+∑

j∈N23= 1+(x5+x7+x10+x11+x14+x15+x17+x19+x20+x21+x22)

Since 23 = 4(3r)− 1, then r = 2, and so, r = 3k + 2 (in Theorem 3.5.4).

Then the generating idempotent of odd-like quadratic residue codes over (F3 + uF3) are

(2+2u)e1+ue2+2u, and (2+2u)e2+ue1+2u which is equivalent to (2+2u)(x+x2+x3+x4+

x6+x8+x9+x12+x13+x16+x18)+u(x5+x7+x10+x11+x14+x15+x17+x19+x20+x21+x22)+2u

and (2 + 2u)(x5 + x7 + x10 + x11 + x14 + x15 + x17 + x19 + x20 + x21 + x22) + u(+x2 + x3 +

x4 + x6 + x8 + x9 + x12 + x13 + x16 + x18) + 2u,

and the generating idempotent of even-like quadratic residue codes over (F3 + uF3) are

e1 +2ue2 +(1+u), and e2 +2ue1 +(1+u) which is equivalent to (x+x2 +x3 +x4 +x6 +x8 +

x9+x12+x13+x16+x18)+2u(x5+x7+x10+x11+x14+x15+x17+x19+x20+x21+x22)+(1+u)

and (x5 + x7 + x10 + x11 + x14 + x15 + x17 + x19 + x20 + x21 + x22) + 2u(+x2 + x3 + x4 + x6 +

x8 + x9 + x12 + x13 + x16 + x18) + (1 + u).

• • • Extended code over F3 + uF3, u2 = 0

Let Q1, Q2 denoted an extended codes of Q1 and Q2 respectively.

53

Theorem 3.5.6. Suppose p = 12r− 1 and Q1, Q2 are the (F3 + uF3)-QR codes in Theorem

3.5.4, then Q1 and Q2 are self-dual.

Proof. (1)If r = 3k,We know by Theorem 3.5.4(d), that Q1 = Q′2+ < (2+u)h > and Q1 has the p+1

2×(p+1)

generator matrix

∞ 0 1 2 ... p− 1

0

0 G′2...

2 + 2u 2 + u 2 + u 2 + u ... 2 + u

where each row of G′2 is a cyclic shift of the vector 1 + e2 (since G′2 generates Q′2).

Since Q′1 is self-orthogonal (Theorem 3.5.4(f)), the rows of G′2 are orthogonal to each

other and obviously also orthogonal to (2 + u)h.

Since the vector ((2 + 2u), (2 + u)h) is orthogonal to itself and | Q1 |= |Q1| = 9p+12 , by

comparing the order of Q1 and Q1⊥, Q1 is self-dual.

(2)If r = 3k + 1,We know by Theorem 3.5.4(d), that Q1 = Q′2+ < 2h > and Q1 has the p+1

2× (p + 1)

generator matrix

∞ 0 1 2 ... p− 1

0

0 G′2...

2 + 2u 2 2 2 ... 2

where each row of G′2 is a cyclic shift of the vector ue1 + e2 + (1 + 2u) (since G′2 generates

Q′2).


other and obviously also orthogonal to 2h.

Since the vector ((2+2u), 2h) is orthogonal to itself and | Q1 |= |Q1| = 9p+12 , by comparing

the order of Q1 and Q1⊥, Q1 is self-dual.

(3)If r = 3k + 2,We know by Theorem 3.5.4(d), that Q1 = Q′2+ < (2+2u)h > and Q1 has the p+1

2×(p+1)

54

generator matrix

∞ 0 1 2 ... p− 1

0

0 G′2...

2 + 2u 2 + 2u 2 + 2u 2 + 2u ... 2 + 2u

where each row of G′2 is a cyclic shift of the vector 2ue1 + e2 + (1 +u) (since G′2 generates

Q′2).


other and obviously also orthogonal to (2 + 2u)h.

Since the vector ((2 + 2u), (2 + 2u)h) is orthogonal to itself and | Q1 |= |Q1| = 9p+12 , by

comparing the order of Q1 and Q1⊥, Q1 is self-dual.

Similarly, Q2 is self-dual.

When p = 12r + 1, we define Q1 to be the F3 + uF3-code generated by the matrix

∞ 0 1 2 ... p− 1

0

0 G′1...

1 1 1 1 ... 1

where each row of G′1 is a cyclic shift of (2 + 2u)e1 when r = 3k, a cyclic shift of 2u+ ue1 +

(2 + 2u)e2 when r = 3k + 1, a cyclic shift of u + 2ue1 + (2 + 2u)e2 when r = 3k + 2. We

define Q2 similarly. Note that these are not extended codes of Q1 and Q2, since the sum of

the components of the all one vector is not 0 (mod 3).

Theorem 3.5.7. Suppose p = 12r + 1 and Q1, Q2 are the F3 + uF3-QR codes in Theorem

3.5.5. Then the dual of Q1 is Q2 and the dual of Q2 is Q1.

Proof. (1)If r = 3k,We know by Theorem 3.5.5(d), that Q1 = Q′2+ < h > and Q1 has the p+1

2× (p + 1)

generator matrix

∞ 0 1 2 ... p− 1

0

0 G′1...

2 + 2u 1 1 1 ... 1

55

where each row of G′1 is a cyclic shift of the vector (2 + 2u)e1.

Since G′1 generates Q′1 and Q⊥2 = Q′1, by Theorem 3.5.5(f), any row in the above matrix

is orthogonal to any row in the matrix which defines Q2, by comparing the order of the dual

of Q1 and the order of Q1, we have Q1⊥

= Q1.

(2)If r = 3k + 1,We know by Theorem 3.5.5(d), that Q1 = Q′2+ < (1+2u)h > and Q1 has the p+1

2×(p+1)

generator matrix

∞ 0 1 2 ... p− 1

0

0 G′1...

2 + 2u 1 + 2u 1 + 2u 1 + 2u ... 1 + 2u

where each row of G′1 is a cyclic shift of the vector 2u+ (2 + 2u)e1 + ue2.




= Q1.

(3)If r = 3k + 2,We know by Theorem 3.5.5(d), that Q1 = Q′2+ < (1+u)h > and Q1 has the p+1

2×(p+1)

generator matrix

∞ 0 1 2 ... p− 1

0

0 G′1...

2 + 2u 1 + u 1 + u 1 + u ... 1 + u

where each row of G′1 is a cyclic shift of the vector u+ (2 + 2u)e1 + 2ue2.




= Q1.

Example 3.5.4. From Example 3.5.1, we want to find the extended code of length 11 (11 =

12(1)− 1). By Theorem (3.5.6), the generator matrix of an extended code is:

56

∞ 0 1 2 3 4 5 6 7 8 9 10

0 1 + 2u 1 u 1 1 1 u u u 1 u

0 u 1 + 2u 1 u 1 1 1 u u u 1

0 1 u 1 + 2u 1 u 1 1 1 u u u

0 u 1 u 1 + 2u 1 u 1 1 1 u u

2 + 2u 2 2 2 2 2 2 2 2 2 2 2

6×12

,

where Q′2 =< ue1 + e2 + (1 + 2u) > and

G′2 =

1 + 2u 1 u 1 1 1 u u u 1 u

u 1 + 2u 1 u 1 1 1 u u u 1

1 u 1 + 2u 1 u 1 1 1 u u u

u 1 u 1 + 2u 1 u 1 1 1 u u

.

57

Chapter 4

Duadic codes over finite commutative

chain rings

We extend the concept of duadic codes over finite fields to finite chain rings. Duadic codes

over finite fields form an important class of linear codes from both the theoretical and prac-

tical perspectives. These codes are a generalization of quadratic residue codes. Duadic codes

share many properties of QR-codes but they are not restricted to prime length. Langevin

and Sole [17] generalized the result to Duadic Z4-Codes. Recently, a duadic codes over the

ring F2 + uF2 has been studied, see [20]. In ref [3], the authors generalized the study of

duadic codes over finite chain rings.

4.1 Duadic group algebra codes

Definition 4.1.1. [19] Let G = {g1, g2, ..., gn}, be a finite group of order n with identity

element 1 and let Fq denote a finite field with q elements such that gcd(q, n) = 1, then the

group algebra R = Fq[G] consists of elements of the form∑n

i=1 aigi, where ai ∈ Fq and

gi ∈ G.

Remark 4.1.1. [2] The set R is a vector space over Fq in which the elements of G form a

basis. Furthermore, R is equipped with a multiplication defined by:

(∑g∈G

agg)(∑g∈G

bgg) =∑g∈G

(∑h∈G

ahbh−1g)g.

58

Definition 4.1.2. [2] A group algebra code in R, or shortly an R− code, is a left ideal I of

R.

Example 4.1.1. [2] If G = Z/nZ is a cyclic group, then Fq[Z/nZ] ∼= Fq[x]/〈xn − 1〉; thus,

in this case a group algebra code is simply a cyclic code.

Definition 4.1.3. [17] R[G] is called the group ring, where G is a finite abelian group of

order n.

We will give some properties of idempotent of group algebra in the following remark:

Remark 4.1.2. [2] 1. An element e in the group algebra R is called an idempotent if and

only if e2 = e, so the elements 0 and 1 are idempotents of R.

2. An idempotent in the center of R is called central idempotent.

3. Since gcd(n, q) = 1, any R-code is generated by an idempotent, that is, for any left

ideal I there exists an idempotent element e in R such that I = Re.

4. Two idempotent elements e and f are called orthogonal if ef = 0 = fe.

5. A nonzero idempotent e in R is called (centrally) primitive if and only if it cannot

be written as the sum of two nonzero orthogonal (central) idempotents in R.

6. If N is a subgroup of G, then N = |N |−1∑

g∈N g is an idempotent element in the

group algebra R.

7. If N is a normal subgroup, then N is a central idempotent.

Notation: The central idempotent G, known as the trivial idempotent, will play a

significant role in this section.

Remark 4.1.3. [2] If we multiply an element b =∑

g∈G bgg by G, then we obtain bG =

(∑

g∈G bgg)G; in particular, we have dim RG = 1.

Definition 4.1.4. [2] An element b =∑

g∈G bgg in R is called even − like if and only if

bG = 0 (i.e.∑

g∈G bg = 0), on the other hand, an element of R that is not even-like is called

odd− like.

Definition 4.1.5. [2] An antiautomorphism on the group algebra R is a bijective map µ

on R that satisfies:

(i) µ(a) + µ(b) = µ(a+ b) and

59

(ii) µ(ab) = µ(b)µ(a) for all a, b in R.

We say that the antiautomorphism µ is isometric if it preserves the Hamming weight.

Remark 4.1.4. 1. [2] An important isometric antiautomorphism on Fq[G] is µ−1 defined as

µ−1(g) = g−1 for g in G.

2. [33] All automorphisms and anti-automorphisms of a group G form a group denoted

by Aut(G).

Definition 4.1.6. [33] Any µ ∈ Aut(G) defines an automorphism or an anti-automorphism

of Fq[G] by

µ(∑

1≤i≤n

aigi) =∑

1≤i≤n

aiµ(gi)

Definition 4.1.7. [2] Let G be a finite group of order n and Fq a finite field such that

gcd(q, n) = 1. If e and f are two even-like idempotents in R = Fq[G] that satisfy the

equations:

A1 e+ f = 1− G and

A2 µ(e) = f and µ(f) = e for some isometric antiautomorphism µ on R,

then the idempotents e and f generate:

C1 a pair of even-like duadic codes Ce := Re and Cf := Rf and

C2 a pair of odd-like duadic codes De := R(1− f) and Df := R(1− e).

Remark 4.1.5. [2] The antiautomorphism µ given in A2 is said to give a splitting and we

also can refer to µ−1 as a splitting.

Lemma 4.1.1. [2] Let G be a group of order n and Fq a finite field such that gcd(n, q) = 1.

If e and f are even-like idempotents in Fq[G] that satisfy A1 and A2 with splitting µ, then

we note that:

(i) The idempotents e, f and G are pairwise orthogonal,

(ii) dim Ce = dim Cf = n−12

and dim De = dim Df = n+12

,

(iii) In particular, the order n of the group G must be odd, and

(iv) The codes satisfy the inclusions Ce ⊆ De and Cf ⊆ Df .

Proof. (i) Since e and f = (1− G− e) are even-like, it follows that the idempotents e, f and

G are pairwise orthogonal; hence, R = Re⊕Rf ⊕RG. For (ii) and (iii) we observe that dim

60

RG = 1 and dim Re = dim Rµ(e) = dim Rf , which implies dim Re = dim Rf = n−12

. The

dimensions for the odd-like duadic codes are an immediate consequence, since Ce ⊕Df = R

and Cf ⊕De = R. For (iv) notice that the orthogonality of e and f yields e(1− f) = e and

f(1− e) = f . Therefore, Ce = Re ⊂ R(1− f) = De and Cf = Rf ⊂ R(1− e) = Df .

Lemma 4.1.2. [2] Let Fq be a finite field of characteristic p. Suppose that µ is an isometric

antiautomorphism of a group algebra Fq[G] that satisfies µ(G) = G. Then there exists a

Galois automorphism σ ∈ Gal(Fq/Fp) and an antiautomorphism µ∗ of the group G such that

µ(αg) = σ(α)µ∗(g) holds for all α ∈ Fq, g ∈ G. (4.1.1)

Proof. The isometry of the antiautomorphism µ implies that the image µ(g) of an element

g in G is of the form µ(g) = cgg′ for some nonzero constant cg ∈ Fq and g′ ∈ G. Since

µ(G) = G, we have cg = 1 for all g ∈ G. Therefore, µ restricts to a antiautomorphism µ∗

on the group G. Since µ preserves addition and multiplication of scalars and µ(Fq1) = Fq1,

we have µ(α1) = σ(α)1 for some automorphism of Fq. The elements of the prime field Fpremain fixed, so σ is an element of Gal(Fq/Fp). The claim is an immediate consequence of

these observations.

We will discuss odd-like weights in duadic group algebra codes:

Lemma 4.1.3. [2] Suppose that e and f are idempotents that determine a pair of even-like

duadic codes in Fq[G] with splitting given by µ. If the group G has order n, then the minimum

weight do of an odd-like element in the odd-like duadic code Df = R(1− e) or De = R(1−f)

satisfies:

(i) d2o ≥ n and

(ii) d2o − do + 1 ≥ n if µ = µ−1.

Proof. (i) Suppose that a is an odd-like element in Df of weight do, so there exists an element

b ∈ Fq[G] such that a = b(1− e). The element b is odd-like, since 0 6= aG = b(1− e)G = bG

holds. A splitting satisfies µ(G) = G; thus, µ(b) is odd-like as well as 0 6= bG implies

0 6= Gµ(b) = µ(b)G. The product of a and µ(a) has Hamming weight ≤ d2o. However, we

61

recall that (1− e)(1− f) = G, so

aµ(a) = b(1− e)(1− f)µ(b) = bGµ(b) = bµ(b)G 6= 0.

If bµ(b) =∑

g∈G cgg, then bµ(b)G = (∑

g∈G cg)G; thus, the product aµ(a) yields an element

of Hamming weight n, which proves the bound n ≤ d2o. For (ii), we note that aµ−1(a) has

at most Hamming weight d2o− do + 1 when a has Hamming weight do. By symmetry similar

results hold for De.

Definition 4.1.8. [2] An Euclidean inner product on Fq[G] is given by

〈∑g∈G

agg|∑g∈G

bgg〉 =∑g∈G

agbg.

Lemma 4.1.4. [2] Let G be a finite group and R = Fq[G].

(i) The product ab = 0 for a, b ∈ R if and only if µ−1(b) ∈ C⊥, where C = Ra.

(ii) If C is an R-code, then C⊥ is also an R-code.

(iii) If e is an idempotent in R and C = Re, then C⊥ = R(1− µ−1(e)).

Proof. (i) We note that the product of a and b can be expressed in the form

ab =∑g∈G

(∑h∈G

aghbh−1)g =∑g∈G

〈g−1a|µ−1(b)〉g,

from which we can directly deduce the claim.

(ii) We note that the inner product satisfies 〈ga|gb〉 = 〈a|b〉 for all g ∈ G and a, b ∈ R.

If a ∈ C and b ∈ C⊥, then for each g ∈ G, we have 〈a|gb〉 = 〈g−1a|b〉 = 0, since g−1a ∈ C.

Extending linearly shows that C⊥ is a left ideal.

(iii) Since e(1 − e) = 0, property (i) shows that the idempotent 1 − µ−1(e) is contained

in C⊥, so R(1− µ−1(e)) ⊆ C⊥ by (ii). Since dim C⊥ = dim R(1− e) = dim R(1− µ−1(e)),

we actually must have equality.

Corollary 4.1.1. [2] If e and f are even-like idempotents that satisfy A1 and A2, then the

following statements hold:

(i) If µ−1(e) = f , then C⊥e = De and C⊥f = Df and

(ii) If µ−1(e) = e, then C⊥e = Df and C⊥f = De.

62

Remark 4.1.6. [2] Cyclic duadic codes exist if and only if q is a quadratic residue modulo n

but this condition is not required for group codes.

Proposition 4.1.5. [2] There exists a splitting µ with even-like (central) idempotents e and

f that satisfy A1 and A2 if and only if each nontrivial (centrally) primitive idempotent h

of Fq[G] satisfies µ(h) 6= h.

Proof. Suppose that e and f are even-like (central) idempotents that satisfy A1 and A2.

These equations imply that e + f + G = 1, where the idempotents e, f and G are pair-

wise orthogonal. Suppose that e = h1 + ... + hm is a decomposition of the idempotent e

into orthogonal (centrally) primitive idempotents. Seeking a contradiction, we assume that

µ(hk) = hk for some k in the range 1 ≤ k ≤ m. However, then e and f cannot be orthogonal

idempotents, contradiction. Conversely, suppose that h 6= µ(h) for all nontrivial primitive

(central) idempotents h of Fq[G]. Partition the nontrivial (central) primitive idempotents

into disjoint pairs {h1, µ(h1)}, ..., {hl, µ(hl)}. Let e = h1 + ... + hl and f = µ(e). Then

e+ f = 1− G and eG = 0 and fG = 0. Further, µ(e+ f + G) = 1 = e+ f + G implies that

e = µ(f). So µ is the desired splitting with (central) even-like idempotents e and f .

Definition 4.1.9. [27] Let (G,+) be a group and let C1 be the multiplicative group of

complex numbers of absolute value 1. A homomorphism χ : G 7→ C1 is called a character

of G and satisfies

χ(g + h) = χ(g)χ(h), ∀ g, h ∈ G,

χ(0) = 1.

The principal character of G is the character

χ(g) = 1g ∀ g ∈ G.

Remark 4.1.7. [2] A centrally primitive idempotent eχ of Fq[G] can be explicitly written in

the form

eχ =nχ|G|

∑g∈G

χ(g)g−1, (4.1.2)

where χ is an irreducible Fq-character and nχ is a positive integer that depends on χ.

63

Notation: The exponent of a group is defined as the least common multiple of the

orders of all elements of the group.

Lemma 4.1.6. [2] Let Fq be a finite field and G a finite group of order n such that gcd(n, q) =

1. Suppose that µ is an antiautomorphism of Fq[G] of the form (4.1.1). Then the action of

µ on a centrally primitive idempotent (4.1.2) is given by

µ(eχ) =nχ|G|

∑g∈G

χ(µ−1∗ (gk))g−1,

where k is a positive integer determined by the Galois automorphism σ and µ−1∗ is the inverse

of the group antiautomorphism µ∗. In particular, k is a power of the characteristic of Fq.

Proof. If the exponent of the group G is m, then the values of the character are contained

in Fq ∩Fp(δ), where δ is a primitive m-th root of unity over the prime field Fp. Suppose that

σ′ is a Galois automorphism of Gal(Fq(δ)/Fp) that restricts to the Galois automorphism

σ on Fq. If σ′(δ) = δk, then σ(χ(g)) = χ(gk) holds for all g in G, and the action of an

antiautomorphism µ is given by

µ(eχ) =nχ|G|

∑g∈G

σ(χ(g))µ∗(g−1) =

nχ|G|

∑g∈G

χ(µ−1∗ (gk))g−1,

where the latter equality is obtained by substituting µ−1∗ (g) for g.

Definition 4.1.10. [2] The Fq-conjugacy class Kq(g) of an element g in a finite group G is

the set

Kq(g) = {h−1gqkh|h ∈ G, k ≥ 0}.

Remark 4.1.8. [2] 1. The two Fq-conjugacy classes are either disjoint or coincide.

2. An irreducible Fq-character is constant on Fq-conjugacy classes.

3. The number of Fq-conjugacy coincides with the number of irreducible Fq-characters.

4. We define an action of an antiautomorphism µ of the form (4.1.1) on an Fq-conjugacy

class Kq(g) by

Kµq (g) = Kq(µ∗(g)l),

where l is a positive integer such that kl ≡ 1 mod m, k is the integer given in Lemma 4.1.6,

and m is the exponent of the group G.

64

Proposition 4.1.7. [2] Let Fq be a finite field and let G be a finite group of order n such that

gcd(n, q) = 1. If µ is an antiautomorphism of Fq[G] of the form (4.1.1), then the number of

Fq-conjugacy classes of G that are fixed by µ coincides with the number of centrally primitive

idempotents of Fq[G] that are fixed by µ.

Proof. Step 1. Suppose that the finite field Fq has characteristic p. There exists a monic

irreducible polynomial f(x) in Fp[x] such that Fp[x]/〈f(x)〉 = Fq. Let Zp denote the ring

of the set {0, 1, ..., p − 1} integers and Qp its quotient field. Then P = pZp is the unique

nonzero prime ideal of Zp and Zp/P ∼= Fp. Choose a monic polynomial g(x) in Zp[x] such

that f(x) ≡ g(x) mod P. Then R = Zp[x]/〈g(x)〉 is a discrete valuation ring with nonzero

prime ideal B = PR.

If K denotes the field Qp[x]/〈g(x)〉. The Galois group Gal(K/Qp) can be identified with

the cyclic group Gal(Fq/Fp). The latter group is generated by the automorphism x −→ xp,

and the generator of Gal(K/Qp) can be characterized by

b ≡ bp mod B for all b ∈ R.

Step 2. The number of centrally primitive idempotents in K[G] and in Fq[G] is the same.

If Y denotes the set of irreducible K-characters of G, then the set of centrally primitive

idempotents {eχ|χ ∈ Y } is bijectively mapped to the centrally primitive idempotents of

Fq[G] by reduction modulo B.

Let k be the integer defined as in Lemma (4.1.6). Then there exists a unique auto-

morphism τ in Gal(K/Qp) such that τ(x) = xk mod B. Therefore, we can define an

antiautomorphism η on K[G] by η(αg) = τ(α)µ∗(g) such that

ηe(χ) mod B = µ(eχ mod B)

holds for all centrally primitive idempotents e(χ) of K[G]. The latter equation guarantees

that the number of idempotents in K[G] fixed by η is the same as the number of idempotents

in Fq[G] fixed by µ.

Step 3. A centrally primitive idempotent in K[G] is of the form

eχ =nχ|G|

∑g∈G

χ(g)g−1.

65

It follows from Lemma (4.1.6) that η(eχ) = eχ if and only if χ(g) = χ(η−1(gk)) holds for all

g in G. Therefore, we define the action of η on an irreducible K-character by

χη(g) = χ(µ−1∗ (gk)),

for all g in G.

An irreducible K-character is constant on K-conjugacy classes. The K-conjugacy classes

coincide with the Fq-conjugacy classes, since the Galois groups are isomorphic. Suppose that

m is the exponent of the group G. There exists a positive integer l such that kl ≡ 1 mod m.

We define the action of η on Fq-conjugacy classes by

Kηq = Kq(µ∗(g

l))

for all g in G. The definitions are carefully chosen such that

χη(Kηq (g)) = χ(Kq(g))

holds for all g in G.

Step 4. Let Kq denote the set of Fq-conjugacy classes. We have |Y | = |Kq|. Therefore,

we can define the square matrix

U = (χ(K))χ∈Y,K∈Kq .

We note that U is nonsingular, since the irreducible K-characters are linearly independent

over K. Let A = (Aχ,ψ)χ,ψ∈Y and B = (BL,K)L,K∈Kq be permutation matrices that are

respectively defined by

Aχ,ψ =

1, if χ = ψη;

0, otherwise.and BL,K =

1, if Kη = L;

0, otherwise.

Since χη(Kη) = χ(K), we have:∑ψ∈Y

Aχ,ψψ(K) = χ(Kη) and∑L∈Kq

χ(L)BL,K = χ(Kη),

So AU = UB. Since U is invertible, we have A = UBU−1. Thus, tr(A) = tr(B). The trace

of A counts the number of characters that remain fixed under the action of η, and the trace

of B counts the number of Fq-conjugacy classes that remain fixed under η. These facts imply

the claim.

66

The goal of the following theorem is to give the existence of duadic group algebra codes.

Theorem 4.1.8. [2] Let G be a finite group of odd order and let Fq be a finite field with

gcd(n, q) = 1. There exists a splitting µ = µ−1 with central idempotents e and f such that

equations A1 and A2 are satisfied if and only if the order of q is odd modulo n.

Proof. Our proof is subdivided into three parts:

Part A. By Proposition (4.1.5), We see a splitting µ−1 with central idempotents e and

f satisfying A1 and A2 exists if and only if no nontrivial centrally primitive idempotent is

fixed by µ−1.

Part B. We can define an action of µ−1 on so-called Fq-conjugacy classes in G. Then by

Proposition (4.1.7), We have no nontrival centrally primitive idempotent is fixed by µ−1 if

and only if no Fq-conjugacy class is fixed by the action of µ−1.

Part C. Finally, K 6= µ−1(K) for all nontrivial Fq-conjugacy classes K if and only the

order of q is odd modulo n.

Then the theorem follows by combining the three parts.

We will discuss existence of duadic group algebra codes when the order of q is even. The

splitting is no longer given by µ−1, but we will show that there exist duadic algebra codes.

We will confine ourselves to the abelian case.

Lemma 4.1.9. [2] Let G = 〈a, b|ap = bp = 1, ab = ba〉, where p is an odd prime and q be

a prime power such that ordp(q) is even and gcd(p, q) = 1. The Fq-conjugacy class of an

element axby in G is given by C(q)x,y = {axqjbyqj |j ∈ Z}. Then the automorphism µ defined as

µ(axby) = aqybx does not fix any of the Fq-conjugacy classes if (x, y) 6= (0, 0). Further, µ−1

fixes each Fq-conjugacy class i.e., µ−1(C(q)x,y) = C

(q)x,y.

Proof. Assume that there is an element axby ∈ G, such that µ(C(q)x,y) = C

(q)x,y. Then there

exists an integer j such that µ(axby) = axqjbyq

j. This implies that (qy, x) = (xqj, yqj) mod p

or qy ≡ xqj mod p and x ≡ yqj mod p. If x = 0, y 6= 0, then we have qy ≡ 0 mod p or y = 0;

a contradiction. If y = 0, x 6= 0, then it follows x = 0, which leads to a contradiction again.

Assuming that both x, y 6= 0, we get qxy ≡ xyq2j mod p. Since q, x, y are all coprime to p

this can be written as 1 ≡ q2j−1 mod p. But as ordp(q) is even, 1 6≡ q2j−1 mod p. Therefore,

67

none of the Fq-conjugacy classes are fixed by µ. Let ordp(q) = 2w, then q2w ≡ 1 mod p,

which implies that qw ≡ −1 mod p. Hence, C(q)x,y = C

(q)−x,−y = µ−1(C

(q)x,y).

Theorem 4.1.10. [2] Let G = 〈a, b|ap = bp = 1, ab = ba〉 with p is an odd prime, q be a

prime power, gcd(p, q) = 1 and ordp(q) is even. Then there exist duadic codes over Fq[G]

with splitting given by µ where µ(axby) = aqybx for any element axby ∈ G. These codes are

fixed by µ−1.

Proof. By Proposition (4.1.5), Proposition (4.1.7) and by Lemma (4.1.9), we know that there

exist a pair of duadic codes over Fq[G] with splitting given by µ such that the codes are fixed

by µ−1.

Example 4.1.2. [33] Let F = F2 and let G = 〈a, b〉, a49 = b7 = 1 and ab = ba8. G is a

non-abelian group of order 343 with gcd(2, |G|) = 1. We shall construct duadic codes in

F2[G]. First, consider the mapping τ : x 7−→ 8x in Z49, so that the residues can be divided

under τ into 13 transitive classes:

T1 = {0},

T2+k = {7.3k}, 0 ≤ k ≤ 5,

T8+j = {7i+ 3j : 0 ≤ i ≤ 6}, 0 ≤ j ≤ 5.

Next, we may compute the conjugacy classes of G as follows:

Ck = {ai : i ∈ Tk}, 1 ≤ k ≤ 13,

C14+t+6j = {b3tai : i ∈ T8+j}, 0 ≤ t ≤ 5, 0 ≤ j ≤ 5.

For each conjugacy class Ci, the sum of its elements σi =∑

g∈Cig is known to be in the

center of F2[G]. Denote σ =∑

0≤i≤48 ai and

e1 = (b+ b2 + b4)σ +∑0≤i≤6

σ2i and e2 = (b6 + b5 + b3)σ +∑0≤i≤6

σ2i+1.

Let µ = µ−1, then µ(e1) = e2 and µ(e2) = e1.

We have also el and e2 are both odd-like vectors satisfying

e1 + e2 = 1 + e0, (∗)

68

where e0 = 1n

∑1≤i≤n gi is the trivial idempotent, which is a central element. We need to

show that e1 and e2 are both idempotents of F2[G]. By (∗) it suffices to show that e1 is an

idempotent of F2[G]. Notice that (b+ b2 + b4)σ is a sum of some σk and then commutes with

σ2i, 1 ≤ i ≤ 6. So,

e21 = [(b+ b2 + b4)σ]2 + (σ2 + σ4 + σ6)2 + (σ8 + σ10 + σ12)

2.

Then (b+ b2 + b4)σ and the two sums σ2 + σ4 + σ6 and σ8 + σ10 + σ12 are all idempotents of

F2[G]. This guarantees that e1 is an idempotent too. By Definition 4.1.7, we have obtained

odd-like duadic codes 〈e1〉 and 〈e2〉. Further, 〈1− e1〉 and 〈1− e2〉 are even-like duadic codes

in F2[G].

We will give an application of duadic group algebra codes which is called quantum error-

correction.

As in ref. [15], The discretization of errors can be found as follows:

Consider errors E = En ⊗ ... ⊗ E1, Ei ∈ {I,X, Y, Z}, where I =

1 0

0 1

, X = 0 1

1 0

, Y =

1 0

0 −1

and Z = XY =

0 −1

1 0

.

The weight of E is the number of Ei 6= I.

Definition 4.1.11. [15] A quantum error control code Q is a K-dimensional subspace of

C2n .

Definition 4.1.12. [15] LetQ ≤ C2n be a quantum error control code, then the stabilizer(S)

of Q is defined to be the set

S = {M ∈ E|Mv = v for all v ∈ Q},

where E = {En ⊗ ...⊗ E1|Ei ∈ {I,X, Y, Z}}.

Remark 4.1.9. [15] S is a group, necessarily abelian if Q 6= {0}.

Definition 4.1.13. [15] Let Q be a quantum error correcting code and let S be the stabilizer

of Q, then the code Q is called a stabilizer code if and only if

The condition {Mv = v for all M ∈ S} implies that v ∈ Q.

69

In ref. [2], Quantum codes can be utilized to protect quantum information over noisy

quantum channels. The CSS (Calderbank-Shor-Steane) construction is particularly trans-

parent method to derive quantum stabilizer codes from a pair of classical codes.

Remark 4.1.10. [2] We derive a family of quantum codes with parameters [[n, 1, d]]q such

that d2 − d+ 1 ≥ n.

Lemma 4.1.11. [2] (CSS Construction)

Suppose that C and D are linear codes over a finite field Fq such that C ⊆ D. If C is

an [n, k1]q code and D is an [n, k2]q code, then there exists an [[n, k2− k1, d]]q stabilizer code

with minimum distance d = min wt{(D\C) ∪ (C⊥\D⊥)}.

Theorem 4.1.12. [2] Let n be an odd positive integer and q a power of a prime such that the

order of q modulo n is odd. Then there exists an [[n, 1, d]]q stabilizer code with d2−d+1 ≥ n.

Proof. There exists a group G of order n. By Theorem 4.1.8, there exist idempotents e

and f in Fq[G] that satisfy A1 and A2 with splitting given by µ = µ−1. By Lemma

4.1.1, we have Ce ⊂ De. The CSS construction shows that there exists an [[n, (n + 1)/2 −

(n − 1)/2, d]]q = [[n, 1, d]]q stabilizer code with minimum distance d = min{(De\Ce) ∪

(C⊥e \D⊥e )} = min{(De\Ce) ∪ (De\Ce)}, where the latter equality follows from Corollary

4.1.1 (i). Since De = RG ⊕ Ce, we observe that De\Ce contains precisely the odd-like

elements of De. By Lemma 4.1.3, the minimum weight of an odd-like element in De satisfies

d2 − d+ 1 ≥ n.

Remark 4.1.11. [2] The distance of the quantum codes does not depend on wt(C) or wt(D)

or even their dual distances, rather the set differences D\C and C⊥\D⊥. This means that

a code that is bad in the classical sense can lead to a good quantum code, if only we can

arrange the low weight codewords of D to be entirely in C and similarly for the low weight

codewords of C⊥ to be in D⊥; this phenomenon is called degeneracy.

Example 4.1.3. [2] We want to construct a degenerate [[81, 1,≥ 9]]2 quantum code which

has many codewords of weight 4. Let Gi = Z3 × Z3 = 〈ai, bi|a3i = b3i = 1, aibi = biai〉,

then from Theorem 4.1.10 we know that there exist duadic group algebra codes over F2[Gi]

with idempotents ei = ai + a2i + aibi + a2i b2i and fi = bi + b2i + aib

2i + a2i bi satisfying A1

70

and A2, under the action of µi(axi byi ) = a2yi b

xi . Further ei, fi are fixed by µ−1. Embedding

Gi into G1 × G2, we get the idempotents of the new code as e = e1 + e2 − e1e2 − f1e2 and

f = f1 + f2 − f1f2 − e1f2. The splitting for this code is given by µ = µ1 × µ2. These

idempotents give us duadic group algebra codes over F2[G1 × G2] that are fixed by µ−1. As

ee1 = e1, wt(Ce) = wt(Cf ) ≤ wt(e1) = 4, while wt(De\Ce) = wt(C⊥e \D⊥e ) = wt(Df\Cf ) ≥ 9

by Corollary 4.1.1 (ii) and Lemma 4.1.3. Thus by Lemma 4.1.11 there exists a [[81, 1,≥ 9]]2

quantum code; this code is degenerate and many errors of weights between 4 and 9 (contained

in Ce or D⊥e ) do not need error-correction.

4.2 Duadic codes over Z4

We will illustrate examples of duadic codes over Z4 with taking into account the duadic

group algebra codes, for more information, see [17].

Definition 4.2.1. [17] A linear code C of length n over Z4 is an additive subgroup of Zn4 .

We will assume all codes over Z4 in this section are linear.

Remark 4.2.1. [14] 1. A generator matrix G of a Z4-linear code C is in standard form if

G =

Ik1 A B1 + 2B2

O 2Ik2 2C

, (4.2.1)

where A,B1, B2, and C are matrices with entries from Z2, and O is the k2 × k1 zero

matrix. The code C is of type 4k12k2 .

2. If C has generator matrix in standard form (4.2.1), then C⊥ has generator matrix

G⊥ =

−(B1 + 2B2)T − CTAT CT In−k1−k2

2AT 2Ik2 O

,

where O is the k2 × (n− k1 − k2) zero matrix. In particular, C⊥ has type 4n−k1−k22k2 .

Example 4.2.1. [14] Let C be the Z4-linear code of length 4 with the 16 codewords: 0000,

1113, 2222, 3331, 0202, 1311, 2020, 3133, 0022, 1131, 2200, 3313, 0220, 1333, 2002, 3111.

Then a generator matrix for C is:

G =

1 1 1 3

0 2 0 2

0 0 2 2

,

71

where I1 =(

1), A =

(1 1

), B1 = B2 =

(1), I2 =

1 0

0 1

, O =

0

0

and

C =

1

1

.

And a generator matrix for C⊥ is:3 1 1 1

2 2 0 0

2 0 2 0

.

Definition 4.2.2. [17] Let C be a Z4-linear code of length n. There are two binary linear

codes of length n associated with C:

The residue code Res(C) = {c(mod 2)|c ∈ C} and the torsion code Tor(C) = {c ∈

Zn2 |2c ∈ C}.

Remark 4.2.2. [14] If C has generator matrix G in standard form (4.2.1), then Res(C) and

Tor(C) have generator matrices GRes =(Ik1 A B1

)and GTor =

Ik1 A B1

O Ik2 C

.

So Res(C) ⊆ Tor(C).

Definition 4.2.3. [14] Let x ∈ Zn4 and suppose that na(x) denotes the number of components

of x equal to a for all a ∈ Z4. The Hamming weight of x is wtH(x) = n1(x)+n2(x)+n3(x),

the Lee weight of x is wtL(x) = n1(x) + 2n2(x) + n3(x), and the Euclidean weight of x is

wtE(x) = n1(x) + 4n2(x) + n3(x).

Remark 4.2.3. [14] The Hamming, Lee, and Euclidean distances between x and y are

dH(x, y) = wtH(x− y), dL(x, y) = wtL(x− y), and dE(x, y) = wtE(x− y), respectively.

Define µ : Z4[x] 7−→ F2[x] by µ(f(x)) = f(x)(mod 2); that is, µ is determined by

µ(0) = µ(2) = 0, µ(1) = µ(3) = 1, and µ(x) = x.

In the following definition, there are a type of weight enumerators of interest with Z4-

linear codes.

Definition 4.2.4. [14] The symmetrized weight enumerator (swe) of a code C is given by

sweC(a, b, c) =∑x∈C

an0(x)bn1(x)+n3(x)cn2(x),

72

and the complete weight enumerator (cwe) of a code C is given by

cweC(a, b, c, d) =∑x∈C

an0(x)bn1(x)cn2(x)dn3(x).

where ni(x) is the number of components of x ∈ C that are congruent to i(mod 4).

Theorem 4.2.1. [17] sweC⊥(a, b, c) = 1|C|sweC(a+ 2b+ c, a− c, a− 2b+ c).

Example 4.2.2. Let C be a Z4-linear code as in Example (4.2.1). Let a = 1, b = 2 and

c = 0, then by Theorem (4.2.1),

sweC⊥(1, 2, 0) =1

16sweC(1 + 4 + 0, 1− 0, 1− 4 + 0) =

1

16sweC(5, 1,−3) =

2046

16= 129.

where sweC(5, 1,−3) =∑

x∈C 5n0(x)1n1(x)+n3(x)(−3)n2(x) =∑

x∈C 5n0(x)(−3)n2(x) = 2064.

Definition 4.2.5. [17] The annihilator (Ann(I)) of an ideal I of the ring R[G] is defined as

Ann(I) := {y ∈ Zpm [G],∀x ∈ I, xy = 0}.,

where Zpm = {0, 1, ..., pm − 1}.

Definition 4.2.6. [17] The hull of a code C is C ∩ C⊥.

Definition 4.2.7. [17] The Z4-code C is isodual if C is equivalent to C⊥.

Definition 4.2.8. [22] For any polynomial P (x) =∑d

k=0 pkxk andN ∈ N, theMacWilliams

transform of order N is defined as:

P (t) = (1+t)NP (1− t1 + t

) =d∑

k=0

pk(1+t)N−k(1−t)k =d∑

k=0

pk

N∑l=0

K(N)l (k)tl =

N∑l=0

d∑k=0

K(N)l (k)pkt

l,

where K(N)l (k) is the Krawtchouk polynomials and defined by

(1 + t)N−k(1− t)k =∞∑l=0

K(N)l (k)tl.

Definition 4.2.9. [17] The Z4-code C is formally self -dual, denoted by fsd, if its swe is

invariant by MacWilliams transform; that is,

sweC⊥(a, b, c) = sweC((a+ 2b+ c)/2, (a− c)/2, (a− 2b+ c)/2).

73

Remark 4.2.4. [17] The Z4-code C is said to be strongly fsd if its cwe is invariant by

MacWilliams transform.

Definition 4.2.10. [17] If C is a Z4-code, then its extended and augmented code C is

defined as follows: Let C denote the extension of C by an overall parity-check and let

C := C ∪ (1 + C).

Remark 4.2.5. [17] 1. If C contains the all-2 word, then C is Z4-linear.

2. If C is self-dual (resp. strongly fsd), so is C.

Definition 4.2.11. [14] A polynomial f(x) ∈ Z4[x] is basic irreducible if µ(f(x)) is irre-

ducible in F2[x] and it is monic if its leading coefficient is 1.

The following, which is a special case of a result called Hensels Lemma, shows how to

get from a factorization of µ(f(x)) to a factorization of f(x).

Theorem 4.2.2. [14] (Hensel’s Lemma) Let f(x) ∈ Z4[x]. Suppose µ(f(x)) = h1(x)h2(x)...hk(x),

where h1(x), h2(x), ..., hk(x) are pairwise coprime polynomials in F2[x]. Then there exist

g1(x), g2(x), ..., gk(x) in Z4[x] such that:

(i) µ(gi(x)) = hi(x) for 1 ≤ i ≤ k,

(ii) g1(x), g2(x), ..., gk(x) are pairwise coprime, and

(iii) f(x) = g1(x)g2(x)...gk(x).

Remark 4.2.6. [14] 1. Let f(x) be a monic basic irreducible polynomial of degree r, then

Z4[x]/(f(x)) is a ring with 4r elements and only one nontrivial ideal; Such a ring is called a

Galois ring.

2. We denote a Galois ring of order 4r by GR(4r).

3. GR(4r) is a ring of characteristic 4 containing the subring Z4.

Definition 4.2.12. [17] The Galois ring GR(pm, d) is the unique Galois extension of Zpm of

degree d.

Remark 4.2.7. [17] 1. When m = 1 in Definition 4.2.12, it is the finite field GF (pd).

2. The Galois ring’s residue field is being GF (pd), which it contains nth roots of unity

for all n dividing pd − 1.

3. The Galois ring is a local ring with maximal ideal (p).

74

Definition 4.2.13. [17] The Fourier transform f of f :=∑

g∈G fgxg in Zpm [G] is defined as∑

g∈G fgxg, where fh :=

∑g∈G fg〈g, h〉.

Theorem 4.2.3. [17] Every ideal I of Zpm [G] can be expressed as

I0 × I1 × ...× Is,

where Ij is one of (0), (1), (p), ..., (pm−1) in GR(pm, dj). In particular there are (m + 1)s+1

such ideals.

Proof. Since the Fourier transform is a ring homomorphism, Ij is an ideal by transport of

structure. It is well known that GR(pm, dj) is a depth m local ring; its only ideals are

therefore (0), (1), (p), ..., (pm−1).

Definition 4.2.14. [10] Let G be a group of permutations of a set S. For each i ∈ S, then

the orbit of x, denoted by O(s) is the subset of S:

O(s) = {gs : g ∈ G}.

Example 4.2.3. [17] If G = Z23 and p=2 are the smallest group of interest for us, we get

s = 4 and the orbits are

{(00), (01, 02), (10, 20), (11, 22), (12, 21)}.

Definition 4.2.15. [17] Let R denote a finite commutative ring with identity. An abelian

code over R of length n is defined to be an ideal in R[G] of a finite abelian group G.

We will discuss a structure of group ring, then we give an example of duadic codes over

Z4[G].

Remark 4.2.8. [17] We denote by O0, O1, ..., Os the orbits of G assumed of order prime to p

under the map x� px. These orbits will play the role of cyclotomic cosets in the theory of

cyclic codes over finite fields.

In ref. [17], the authers enumerate duadic codes of length ≤ 27 over Z4.

75

Example 4.2.4. [17] If the length of a code is n = 9, only G = Z3 × Z3 gives rise to

duadic codes. This group can be realised as the underlying additive group for F9 = F3(i),

with i2 = −1. The orbits are then

{(0), (±1), (±i), (±(1 + i)), (±(1− i))}.

Here Q = {±1,±i} and N = {(±(1 + i)), (±(1− i))}. In Table (4.1), the codes are specified

by their components on each of the five orbits which are labeled in the previous order. We

denote by H1 the code 2−0−0−0−0 of type 4021. We denote by H2 the code 2−0−2−2−0 of

type 4025. There are nine nontrivial duadic codes. Inspection of their hulls and types suggests

that they are only two up to equivalence.

Code C Type dL(C) dE(C) Hull

2− 1− 1− 0− 0 4421 4 4 H1

2− 0− 2− 2− 1 4225 4 6 H2

2− 2− 0− 2− 1 4225 4 6 H2

2− 1− 0− 1− 0 4421 4 4 H1

2− 2− 2− 0− 1 4225 4 6 H2

2− 1− 0− 0− 1 4421 4 4 H1

2− 0− 2− 1− 2 4225 4 6 H2

2− 2− 1− 0− 2 4225 4 6 H2

2− 0− 1− 2− 2 4225 4 6 H2

Table 4.1: Duadic codes of Z4[Z3 × Z3]

Proposition 4.2.4. [17] If C ⊆ Z4[Z3×Z3] is fsd, then it is isodual. Such codes of a given

type are equivalent by a multiplier.

Example 4.2.5. [17] If the length of a code is n = 25, let G = Z5 × Z5 which can be

realised as the underlying additive group of F25 = F5(j) with j2 + j + 1 = 0. There are

six nonzero orbits, all of size 4 and of the shape Ca := (±a,±2a), with a ranging over

{1, j, 1± j, 1± 2j}. With these notations the squares and nonsquares are Q = C1∪Cj ∪C1+j

and N = C1−j∪C1+2j∪C1−2j. There are 71 nontrivial fsd codes up to conjugation in the group

algebra. Table (4.2) classifies them into three classes according to a variety of parameters.

Proposition 4.2.5. [17] If C ⊆ Z4[Z5 × Z5] is fsd of type 41221 or 44217, then it is isodual.

76

Code C Type dL(C) dE(C) dL(C) dE(C) Hull

Class 1 41221 6 6 6 6 H1

Class 2 44217 4 8 6 4 H2

Class 3 4829 8 8 6 6 H3

H1 4021 50 100

H2 40217 8 16

H3 4029 10 20

.


Remark 4.2.9. [17] If G = Z25, since the sizes of the orbits are pairwise distinct, no splitting

can exist in G; and therefore Z5 × Z5 is the only group that gives rise to duadic codes of

length 5.

Example 4.2.6. [17] For n = 27, there are two groups that give rise to duadic codes: Z3×Z9

and Z3 × Z3 × Z3.

In Z3 × Z9, there are seven nonzero orbits. Three have size 6 and four have size 2. The

orbits are by order of representatives:

00, 01, 03, 10, 11, 12, 13, 16.

Table (4.3) list duadic codes of Z4[Z3 × Z9].

But in Z3 × Z3 × Z3, All the 13 nonzero orbits are of the shape Ca = {±a} for some

nonzero a. In Table (4.4), duadic codes of Z4[Z3 × Z3 × Z3] and the a’s are listed in the

order:

000, 001, 010, 011, 012, 100, 101, 102, 110, 111, 112, 120, 121, 122.

Remark 4.2.10. [17] The groups Z27 are of length 27, but since the sizes of the orbits are

pairwise distinct no splitting can exist in this group.

Code C Type dL(C) dE(C)

2− 0− 2− 0− 2− 1− 2− 1 48211 6 6

2− 2− 2− 0− 0− 1− 2− 1 48211 6 6

2− 2− 2− 0− 2− 2− 2− 1 42223 4 8

2− 0− 2− 2− 2− 1− 2− 2 46212 4 6


77

Code C Type dL(C) dE(C)

2− 2− 0− 0− 0− 0− 0− 0− 1− 1− 1− 1− 1− 1 41223 4 4

2− 2− 0− 0− 0− 0− 0− 2− 2− 1− 1− 1− 1− 1 41027 6 6

2− 0− 0− 0− 2− 0− 0− 1− 0− 1− 1− 1− 1− 1 41223 6 6

2− 2− 0− 0− 0− 0− 2− 2− 2− 2− 1− 1− 1− 1 48211 6 6

2− 0− 0− 2− 2− 0− 0− 1− 0− 2− 1− 1− 1− 1 41027 6 6

2− 0− 0− 1− 0− 0− 0− 1− 0− 2− 1− 1− 1− 1 41223 6 6

2− 2− 0− 0− 0− 2− 2− 2− 0− 2− 1− 1− 1− 1 48211 6 6

2− 2− 0− 0− 0− 2− 2− 2− 2− 2− 2− 1− 1− 1 46215 4 6

2− 2− 2− 0− 0− 2− 0− 0− 2− 1− 1− 2− 1− 1 48211 6 8

2− 0− 2− 0− 2− 2− 0− 2− 0− 1− 1− 2− 1− 1 48211 6 8

2− 0− 2− 0− 2− 2− 0− 2− 2− 2− 1− 2− 1− 1 46215 6 12

2− 2− 0− 2− 0− 2− 0− 2− 2− 2− 1− 2− 1− 1 46215 6 12

2− 2− 0− 0− 2− 2− 2− 2− 2− 2− 2− 2− 1− 1 44219 4 8

2− 0− 2− 2− 2− 2− 2− 2− 2− 2− 2− 0− 1− 1 44219 4 8

2− 0− 2− 2− 2− 2− 2− 2− 2− 2− 2− 2− 2− 1 42223 4 8

Table 4.4: Duadic codes of Z4[Z3 × Z3 × Z3]

4.3 Duadic codes over F2 + uF2, u2 = 0

Definition 4.3.1. [20] Let R be the commutative ring F2+uF2 := F2[x]/(x2). The elements

of R may be written as {0, 1, u, u = 1 + u}, where u2 = 0 and F2 = {0, 1}.

The characteristic of this ring is 2.

The ring R = F2 + uF2 shares some good properties of both Z4 and F4. This alphabet is

given by all binary polynomials in indeterminate u of degree less than 2, and is closed under

usual binary polynomial addition and multiplication modulo u2 [5]. Even though F2 + uF2

and Z4 are similar to each other, there are still some significant differences between them;

one of the differences is that x2 + 1 = 0 has a solution in F2 + uF2, but not in Z4 [12].

This ring can also be viewed as a vector space of dimension 2 over F2. Moreover, the

sets {0, 1}, {0, u} and {0, u} form three subspaces in R and the subspace {0, 1}(= F2) is a

subring. Note that the ring R is isomorphic to the quotient ring Z[X]/(2, (X + 1)2), but not

isomorphic to Z4 [5].

The multiplication table coincides with that of Z4, when u and u are replaced by 2 and

3, respectively. In this sense, R is analogous to Z4 and here u plays the role of 2. The

addition table is similar to that of the Galois field F4 = {0, 1, ω, ω = ω + 1}, when u and u

78

are replaced by ω and ω, respectively [5].

So, Addition and multiplication operations for F2 +uF2 are given in the following tables:

+ 0 1 u u

0 0 1 u u

1 1 0 u u

u u u 0 1

u u u 1 0

and

· 0 1 u u

0 0 0 0 0

1 0 1 u u

u 0 u 0 u

u 0 u u 1

.

Remark 4.3.1. 1. [5] R is a local ring with a maximal ideal given by {0, u}.

2. [20] 1 and 1 + u are the only units in R.

3. [20] R has three ideals: (0), (u) and (1).

Duadic codes over F2 + uF2

The method of duadic codes over R in ref. [20] is similar to the duadic codes over Z4

(Section 4.2), when u and u are replaced by 2 and 3 respectively.

4.4 Duadic codes over some other finite commutative

chain rings

Before we give an example for a duadic codes over finite commutative rings, we illustrate

some facts from ref. [3] and [9].

In this section, we assumed that R = Fq + uFq + ...+ uk−1Fq is a finite chain ring, where

uk = 0, q = pr, p is a prime number and r is an integer.

Let M the unique maximal ideal of R, and let u be the generator of the unique maximal

ideal M. Then M = 〈u〉 = Ru, where Ru = 〈u〉 = {βu|β ∈ R}. We have

R = 〈u0〉 ⊇ 〈u1〉 ⊇ ... ⊇ 〈ui〉 ⊇ ...〈uk〉 = {0}.

Let |R| denote the cardinality of R. Let F = R/M = R/〈u〉 be the residue field with

characteristic p, then we have |F| = q = pr.

Definition 4.4.1. [9] Let k be the minimal number such that 〈uk〉 = {0}, then the number

k is called the nilpotency index of u.

79

Lemma 4.4.1. [3] Let n be an odd integer such that gcd(p, n) = 1 and q ≡ x2 mod(n). Then

there exists a pair of monic factors of xn − 1, gi(x), i ∈ {1, 2} such that

xn − 1 = (x− 1)g1(x)g2(x)

in R[x].

Proof. Let n be an odd integer such that gcd(p, n) = 1 and q ≡ x2 mod(n), then there exists

a pair of odd-like duadic codes over Fq, generated respectively by f1(x) and f2(x) which

verifies xn − 1 = (x− 1)f1(x)f2(x) over Fq. Since x− 1, f1(x) and f2(x) are monic coprimes

factors of xn − 1 over Fq. Then by Theorem 4.2.2 (Hensel Lemma) there exists unique

monic pairwise coprime polynomials x − a, g1(x), g2(x) factors of xn − 1 in R[x], such that

x− a = x− 1, g1(x) = f1(x) and g2(x) = f2(x). This gives that xn − 1 = (x− a)g1(x)g2(x)

in R[x]. Substituting x = 1 into the above equation we obtain (1 − a)g1(1)g2(1) = 0.

Since gcd(n, q) = 1, hence xn − 1 is with simple roots. Then g1(1) = f1(1) 6= 0 and

g2(1) = f2(1) 6= 0. This gives that g1(1) and g2(1) are both invertible elements of R. Therefor

a = 1 and then in R[x] the following holds xn − 1 = (x− 1)g1(x)g2(x).

Remark 4.4.1. The notation q = x2 mod (n) means that q is a residue quadratic modulo n.

Definition 4.4.2. [3] Let n be an odd integer such that gcd(p, n) = 1 and q ≡ x2 mod(n).

Let gi, i ∈ {1, 2} be the lifted polynomials of fi, where fi are the generator of the duadic

codes over Fq. Then we define the following cyclic codes over R :

1.The free odd-like duadic codes over R are D′1 = 〈g1(x)〉 and D′2 = 〈g2(x)〉 and

2.The free even-like duadic codes over R are C ′1 = 〈(x−1)g1(x)〉 and C ′2 = 〈(x−1)g2(x)〉.

Proposition 4.4.2. [3] Let D′i, and C ′i, i ∈ {1, 2} be the codes given by Definition 4.4.2.

i) If the splitting is given by µ−1, then D′⊥1 = C ′1 and D′⊥2 = C ′2.

ii) If the splitting is not given by µ−1, then D′⊥1 = C ′2 and D′⊥2 = C ′1.

We will use definitions and theorems which is in the introduction of this section and now

we want already to give an example of duadic codes over Z9 and the chain ring F3+uF3, u2 =

0.

80

Example 4.4.1. [3] Let n = 13 be the length of the code over Z9.

Then we have p = 3, and so ord13(9) = 3.

The 9-cyclotomic cosets modulo 13 are: C0 = {0}, C1 = {1, 3, 9}, C2 = {2, 6, 5}, C4 =

{4, 12, 10} and C7 = {11, 7, 8}.

Let β be a primitive 13-th root of unity in Gr(9, 3), but no smaller extension field of F3

contains such a primitive root.

So,

x13 − 1 = Mβ0(x)Mβ(x)Mβ2(x)Mβ4(x)Mβ7(x)

with

• Mβ0(x) = (x− β0) = (x− 1)

• Mβ(x) =∏

i∈c1(x− βi) = x3 + 6x2 + 2x+ 8

• Mβ2(x) =∏

i∈c2(x− βi) = x3 + 7x2 + 3x+ 8

• Mβ4(x) =∏

i∈c4(x− βi) = x3 + 4x2 + 7x+ 8

• Mβ7(x) =∏

i∈c7(x− βi) = x3 + 2x2 + 5x+ 8

We note that (Mβ(x))∗ = −Mβ2(x) and (Mβ4(x))∗ = −Mβ7(x)

So, if we take S1 = C1 ∪ C2 and S2 = C4 ∪ C7, we have µ−1(S1) = S2.

Therefore, a pair of odd-like duadic codes over Z9 are

g1(x) = (x3+6x2+2x+8)(x3+4x2+7x+8) and g2(x) = (x3+7x2+3x+8)(x3+2x2+5x+8).

and a pair of even-like duadic codes over Z9 are

(x− 1)g1(x) and (x− 1)g2(x).

In the same manner of Example (4.4.1), we can give an example of duadic codes over

F3 + uF3, u2 = 0 as follows:

Example 4.4.2. Let n = 13 be the length of the code over F3 + uF3, u2 = 0.

Then we have p = 3, and so ord13(3) = 3.

The 3-cyclotomic cosets modulo 13 are: C0 = {0}, C1 = {1, 3, 9}, C2 = {2, 6, 5}, C4 =

{4, 12, 10} and C7 = {7, 8, 11}.

Let β be a primitive 13-th root of unity in F27(= F33), but no smaller extension field of

F3 contains such a primitive root.

81

So,

x13 − 1 = Mβ0(x)Mβ(x)Mβ2(x)Mβ4(x)Mβ7(x)

with

• Mβ0(x) = (x− β0) = (x− 1)

• Mβ(x) =∏

i∈c1(x− βi) = x3 + 2ux2 + 2x+ (2 + 2u)

• Mβ2(x) =∏

i∈c2(x− βi) = x3 + (1 + 2u)x2 + ux+ (2 + 2u)

• Mβ4(x) =∏

i∈c4(x− βi) = x3 + (1 + u)x2 + (1 + 2u)x+ (2 + 2u)

• Mβ7(x) =∏

i∈c7(x− βi) = x3 + 2x2 + (2 + u)x+ (2 + 2u)

We note that (Mβ(x))∗ = −Mβ2(x) and (Mβ4(x))∗ = −Mβ7(x)

So, if we take S1 = C1 ∪ C2 and S2 = C4 ∪ C7, we have µ−1(S1) = S2.

Therefore, a pair of odd-like duadic codes over F3 + uF3 are

g1(x) = (x3 + 2ux2 + 2x+ (2 + 2u))(x3 + (1 + u)x2 + (1 + 2u)x+ (2 + 2u))

and

g2(x) = (x3 + (1 + 2u)x2 + ux+ (2 + 2u))(x3 + 2x2 + (2 + u)x+ (2 + 2u)).

and a pair of even-like duadic codes over F3 + uF3 are

(x− 1)g1(x) and (x− 1)g2(x).

82

Conclusion

Duadic codes are a generalization of a quadratic residue codes. We benefited in this search

from Batoul,Guenda and Gulliver [3], whose give a duadic codes over finite chain rings. In

this theses, we define a duadic codes over finite commutative chain rings; in particular,

F3 + uF3 where u2 = 0 in terms of their idempotent generators and show that these codes

also have many good properties which are analogous in many respects to properties of duadic

codes over finite fields. Finally, we can generalize this result to any finite commutative chain

rings of the form Fq + uFq + u2Fq + ...+ us−1Fq where us = 0 and q is a power of prime, and

we suggest generalized to any finite non chain rings of the form Fp + vFp where v2 = v and

p is prime.

83

Appendix

• Basic concepts of algebra

All readers are assumed to be familiar with the following notation.

Notation:

N = { natural numbers } = {0, 1, 2, 3, 4, ...}

Z = { integers } = {...,−2,−1, 0, 1, 2, 3, 4, .....}

Q = { rational numbers } = {a/b|a, b ∈ Z, b 6= 0}

R = { real numbers }

C = { complex numbers } = {x+ iy|x, y ∈ R},

So,

N ⊂ Z ⊂ Q ⊂ R ⊂ C.

The field has nice properties with respect to addition and multiplication, properties that

will be extensively discussed in this appendix.

In the context of coding theory fields of finite cardinality play a crucial role. This is

so because most codes consist of vectors taken from a vector space over a finite field and,

secondly, many good codes are constructed by algebraic methods. In this appendix an

introduction will be given to the theory of finite fields. All possible results regarding finite

fields that are needed in the chapters can be found in this appendix.

Definition 1. [21] A field is a nonempty set F of elements with two operations ”+” (called

addition) and ” · ” (called multiplication) satisfying the following axioms:

For all a, b and c ∈ F:

i. (Closure)F is closed under + and · ; i.e., a+ b and a · b are in F.

ii. (Commutativity)a+ b = b+ a, a · b = b · a.

iii. (Associativity)(a+ b) + c = a+ (b+ c), a · (b · c) = (a · b) · c.

iv. (Distributivity)a · (b+ c) = a · b+ a · c and (a+ b)c = a · c+ b · c.

(v - vi)(Existence of identity elements)

84

Furthermore, two distinct identity elements 0 and 1 (called the additive and multiplicative

identities, respectively) must exist in F satisfying the following:

v. a+ 0 = 0 + a = a for all a ∈ F.

vi. a · 1 = 1 · a = a and a · 0 = 0 for all a ∈ F.

vii. (Existence of additive inverses)For any a in F, there exists an additive inverse

element (−a) in F such that a+ (−a) = (−a) + a = 0.

viii. (Existence of multiplicative inverses)For any a 6= 0 in F, there exists a multiplicative

inverse element a−1 in F such that a · a−1 = a−1 · a = 1.

We denote by the set F\{0}, the set F∗ of all nonzero elements of F is also an abelian group

under multiplication with multiplicative identity called one and denoted 1; and multiplication

distributes over addition.

Theorem 1 [21] F∗ is a cyclic group of order p− 1.

In general, we will denote a field with q elements by Fq ; another common notation is

GF (q) and read ”the Galois field with q elements”.

Notation: If p is a prime, the integers modulo p form a field, which is then denoted by

Fp.

Remark 1 1.[14] A field containing only finitely many elements is called a finitefield, and

the number of elements in F is called the order of F.

2. A set F satisfying axioms (i)(vii) in Definition (1) is called a ring.

Example 1 1. The set of all integers Z := {0, 1, 2, ...} forms a ring under the normal addition

and multiplication. It is called the integer ring.

2. The set of all polynomials over a field F,

F[x] := {a0 + a1x+ ...+ anxn : ai ∈ F, n ≥ 0},

forms a ring under the normal addition and multiplication of polynomials.

Theorem 2 [21] Fq is a field if and only if q is a prime.

Definition 2. [21] Let F be a field. The characteristic of F is the least positive integer n

such that n · 1 = 0, where 1 is the multiplicative identity of F. If no such n exists, we define

the characteristic to be 0.

It follows from the following result that the characteristic of a field cannot be composite.

85

Theorem 3[21] The characteristic of a field is either 0 or a prime number.

Remark 2[10] 1. If ab = ba for all a, b in a ring R, then the ring is a commutative ring,

That is, a ring is called commutative if and only if the multiplication is commutative.

2. A commutative ring in which every nonzero element is a unit is a field.

3. If we take a ring R with 0 and with its addition, forgetting the multiplication in R,

then we get an abelian group, called the additive group of R.

Definition 3. [10] Let R be a commutative ring. An ideal in R is a subset I ∈ R such that:

(a) 0 ∈ I

(b) If a, b ∈ I then a+ b ∈ I

(c) If a ∈ R and b ∈ I then ab ∈ I.

For any element x ∈ R, the set Rx = {ax | a ∈ R} is an ideal in R; ideals of this form

are called principal ideals, and we say that x is a generator of Rx.

Example 2 The ideal nZ of the integers is a principle ideal with generator n.

Definition 4. [10] An ideal M in a ring R is said to be maximal if M 6= R and for every

ideal N such that M ⊆ N ⊆ R, either N = M or N = R.

Example 3 1. The ideal 3Z is maximal in Z.

2. The ideal 4Z is not maximal since 4Z ( 2Z ( Z.

Definition 5. [10] An ideal P 6= R is said to be prime if for all a, b ∈ R, ab ∈ P implies

a ∈ P or b ∈ P .

Definition 6. [10] Let R be a commutative ring with unit 1. Let I be an ideal in R. The

quotient ring R/I is the set of cosets

r + I = {r + i : i ∈ I},

with operations of addition and multiplication on R/I by

(r + I) + (s+ I) = (r + s) + I

(r + I) · (s+ I) = (r · s) + I.

86

Remark 3 The zero in the quotient is 0R/I = 0 + I, and the unit is 1R/I = 1 + I.

Proposition 1 [10] For a commutative ring R with unit, and for an ideal I, the quotient

ring R/I is a field if and only if I is a maximal ideal.

Proposition 2 [10] For a commutative ring R with unit, and for an ideal I, the quotient

ring R/I is an integral domain if and only if I is a prime ideal.

Corollary 1 [10] Maximal ideals are prime ideals.

Remark 4 [10] Not all prime ideals are maximal, for example, Let R = Z[x] be the polyno-

mial ring in one variable with integer coefficients. Consider the ideal I = Z[x] · x generated

by x. Then the ideal I is prime, but not maximal. For more example, See [10].

• • Definition and examples of Modules:

The notion of module over a commutative ring is a generalization of the notion of a

vector space over a field, where a scalers belong to a commutative ring. However, many

basic results in the theory of vector spaces are no longer true for module over a commutative

ring, even when the ring is sufficiently nice, say for instance the ring of integers. Historically,

the theory of modules has its origin in Number Theory and Linear Algebra. We shall first

give the formal definitions.

Definition 7. [10] A triple (M,+, ·), where (M,+) is an abelian group and · : R×M −→M

is a map (called scalar multiplication) satisfying the following conditions, is a module over

R :

(1) a · (m+m′) = a ·m+ a ·m′ for all m,m′ ∈M,a ∈ R,

(2) (a+ b) ·m = a ·m+ b ·m for all a, b ∈ R,m ∈M ,

(3) a · (b ·m) = (ab) ·m for all a, b ∈ R,m ∈M ,

(4) 1 ·m = m for all m ∈M .

Definition 8. [10] A subgroup N of M is called a submodule of M , if it is closed under

scalar multiplication induced from M ; i.e., if the following conditions is satisfied:

For all a ∈ R and m ∈ N, am ∈ N .

Example 4 (1) A vector space is a module over a field.

(2) Any ideal in R is a module over R. In particular, R is a module over itself.

Types of fields

87

Three of the fields that are very common in the study of linear codes are the binary

field with two elements, the ternary field with three elements, and the quaternary field with

four elements. One can work with these fields by knowing their addition and multiplication

tables, which we present in the next three examples [14].

Example 5 The binary field F2 with two elements {0, 1} has the following addition and

multiplication tables:

+ 0 1

0 0 1

1 1 0

· 0 1

0 0 0

1 0 1

This is also the ring of integers modulo 2.

Example 6 The ternary field F3 with three elements {0, 1, 2} has addition and multiplication

tables given by addition and multiplication modulo 3:

+ 0 1 2

0 0 1 2

1 1 2 0

2 2 0 1

· 0 1 2

0 0 0 0

1 0 1 2

2 0 2 1

Example 7 The quaternary field F4 with four elements {0, 1, ω, ω} is more complicated. It

has the following addition and multiplication tables; F4 is not the ring of integers modulo 4:

+ 0 1 ω ω

0 0 1 ω ω

1 1 0 ω ω

ω ω ω 0 1

ω ω ω 1 0

· 0 0 ω ω

0 0 0 0 0

1 0 1 ω ω

ω 0 ω ω 1

ω 0 ω 1 ω

Some fundamental equations are observed in these tables. For instance, one notices that

x+ x = 0 for all x ∈ F4. Also ω = ω2 = 1 + ω and ω3 = ω3 = 1.

• • • Polynomials over fields

Definition 9. [21] Let F be a field. The set

F[x] := {n∑i=0

aixi : ai ∈ F, n ≥ 0}

is called the polynomial ring over F. An element of F[x] is called a polynomial over F.

88

Definition 10. [21] Let f(x) =∑r

i=0 aixi be a polynomial of degree r (ar 6= 0) over Fq.

Then the reciprocal polynomial f(x)∗ of f(x) is given by

xrf(x−1).

Remark 5[21] If f(x) is a divisor of xn − 1, then so is f ∗(x).

Proposition 3[3] The cyclic code C over Fq generated by f(x) is equivalent to:

(i) The cyclic code generated by f(−x) if n is even and q is odd, where n is the length

of a code.

(ii) The cyclic code generated by f ∗(x).

Definition 11. [14] Consider a polynomial f(x) =∑n

i=0 aixi = a0 + a1x

1 + .....+ anxn.The

degree of f(x) is the largest d for which an is nonzero. If this coefficient an is equal to one,

we say that f(x) is monic.

Example 8 The polynomial f(x) = 1 + 2x+ 3x3 ∈ Q[x] has degree 3 and is not monic. The

polynomial i+ x6 ∈ C[x] is monic and has degree 6.

Remark 6 [21] A polynomial f(x) of positive degree is said to be reducible over F if there

exist two polynomials g(x) and h(x) over F such that deg(g(x)) < deg(f(x)), deg(h(x)) <

deg(f(x)) and f(x) = g(x)h(x). Otherwise, the polynomial f(x) of positive degree is said

to be irreducible over F.

Example 9 [21] 1. The polynomial f(x) = x4 + 2x6 ∈ Z3[x] is of degree 6. It is reducible as

f(x) = x4(1 + 2x2).

2. The polynomial g(x) = 1 + x+ x2 ∈ Z3[x] is of degree 2. It is irreducible. Otherwise,

it would have a linear factor x or x + 1; i.e., 0 or 1 would be a root of g(x), but g(0) =

g(1) = 1 ∈ Z2.

Definition 12. [14] Let Mα(x) be a monic polynomial in Fq[x] of smallest degree which has

α as a root; this polynomial is called the minimal polynomial of α over Fq.

Example 10 [21] Let α be a root of the polynomial 1 + x+ x2 ∈ F2[x]. Since the two linear

polynomials x and 1+x are not minimal polynomials of α. Therefore, 1+x+x2 is a minimal

polynomial of α. Since 1 + (1 + α) + (1 + α)2 = 1 + 1 + α + 1 + α2 = 1 + α + α2 = 0 and

1 + α is not a root of x or 1 + x, 1 + x+ x2 is also a minimal polynomial of 1 + α.

89

[14] If we begin with an irreducible polynomial f(x) over Fq of degree r, we can adjoin a

root of f(x) to Fq and obtain the field Fqr . Amazingly, all the roots of f(x) lie in Fqr .

Theorem 4 [14] Let f(x) be a monic irreducible polynomial over Fq of degree r. Then:

(i) all the roots of f(x) lie in Fqr and in any field containing Fq along with one root of

f(x),

(ii) f(x) =∏r

i=0(x− α), where αi ∈ Fqr for 1 ≤ i ≤ r, and

(iii) f(x) | xqr − x.

Remark 7 [14] In particular Theorem(4) holds for minimal polynomials Mα(x) over Fq as

such polynomials are monic irreducible.

Theorem 5 [14] Let Fqt be an extension field of Fq and let α be an element of Fqt with

minimal polynomial Mα(x) in Fq[x]. The following hold:

(i) Mα(x) | xqt − x.

(ii) Mα(x) has distinct roots all lying in Fqt .

(iii) The degree of Mα(x) divides t.

(iv) xqt−x =

∏αMα(x), where α runs through some subset of Fqt which enumerates the

minimal polynomials of all elements of Fqt exactly once.

(v) xqt − x =

∏f f(x) , where f runs through all monic irreducible polynomials over Fq

whose degree divides t.

Definition 13. [21] Let f(x) ∈ F[x] be a polynomial of degree n ≥ 1. Then, for any poly-

nomial g(x) ∈ F[x], there exists a unique pair (s(x), r(x)) of polynomials with deg(r(x)) <

deg(f(x)) or r(x) = 0 such that g(x) = s(x)f(x) + r(x). The polynomial r(x) is called the

(principal) remainder of g(x) divided by f(x), denoted by (g(x)(mod f(x))).

Example 11 let f(x) = 1 + x2 and g(x) = x + 2x4 be two polynomials in Z5[x]. Since we

have g(x) = x+ 2x4 = (3 + 2x2)(1 + x) + (2 + x) = (3 + 2x2)f(x) + (2 + x), the remainder

of g(x) divided by f(x) is 2 + x.

Analogous to the integral ring Z, we can introduce the following notions:

Definition 14. [21] i. Let f(x), g(x) ∈ F[x] be two nonzero polynomials.The greatest

common divisor of f(x), g(x), denoted by gcd(f(x), g(x)), is the monic polynomial of the

highest degree which is divisor of both f(x) and g(x).

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ii. The least common multiple of f(x), g(x), denoted by lcm(f(x), g(x)), is the monic

polynomial of the lowest degree which is a multiple of both f(x) and g(x).

Remark 8 1. [21] we say that f(x) is co-prime(or prime) to g(x) if gcd(f(x), g(x)) = 1.

2. [14] Consider a polynomial f(x) =∑n

i=0 aixi ∈ Z[x]. We say that f(x) is primitive

if the greatest common divisor of a0, ..., an is equal to one, or equivalently, there is no prime

that divides all these coefficients.

3. [21] lcm(f(x), g(x)) = f(x)g(x)gcd(f(x),g(x))

.

Theorem 6 [14] Let f(x) and g(x) be in Fq[x] with g(x) nonzero.

i. (Division Algorithm) There exist unique polynomials h(x), r(x) ∈ Fq[x] such that

f(x) = g(x)h(x) + r(x), where deg r(x) < deg g(x) or r(x) = 0.

ii. If f(x) = g(x)h(x) + r(x), then gcd(f(x), g(x)) = gcd(g(x), r(x)).

We can use the Division Algorithm recursively together with part (ii) of Theorem (6) to

find the gcd of two polynomials. This process is known as the Euclidean Algorithm.

The Euclidean Algorithm for polynomials is analogous to the Euclidean Algorithm for

integers. We state it in the next theorem and then illustrate it with an example.

Theorem 7 [14] (Euclidean Algorithm)

Let f(x) and g(x) be polynomials in Fq[x] with g(x) nonzero.

i. Perform the following sequence of steps until rn(x) = 0 for some n:

f(x) = g(x)h1(x) + r1(x), where deg r1(x) < deg g(x),

g(x) = r1(x)h2(x) + r2(x), where deg r2(x) < deg r1(x),

r1(x) = r2(x)h3(x) + r3(x), where deg r3(x) < deg r2(x),

...

rn−3(x) = rn−2(x)hn−1(x) + rn−1(x), where degrn−1(x) < deg rn−2(x),

rn−2(x) = rn−1(x)hn(x) + rn(x), where rn(x) = 0.

Then gcd(f(x), g(x)) = crn−1(x), where c ∈ Fq is chosen so that crn−1(x) is monic.

ii. There exist polynomials a(x), b(x) ∈ Fq[x] such that a(x)f(x)+b(x)g(x) = gcd(f(x), g(x)).

In Example (12) and (13), We want to find gcd((f(x)), (g(x))) between two polynomials,

then we find a(x) and b(x) such that gcd((f(x)), (g(x))) = a(x)f(x) + b(x)g(x) in the field

F[x].

91

Example 12 Let f(x) = x6 + x5 + x4 + x3 + x+ 1, g(x) = x5 + x3 + x2 + x and F[x] = F2[x].

(Note that - and + in F2 are the same).

x6 + x5 + x4 + x3 + x+ 1 = (x5 + x3 + x2 + x)(x+ 1) + (x3 + 1)

x5 + x3 + x2 + x = (x3 + 1)(x2 + 1) + (x+ 1)

x3 + 1 = (x+ 1)(x2 + x+ 1) + 0

Thus

(x+1) = gcd(x3+1, x+1) = gcd(x5+x3+x2+x, x3+1) = gcd(x6+x5+x4+x3+x+1, x5+x3+x2+x).

Now we found a(x) and b(x) such that gcd((f(x)), (g(x))) = a(x)(x6 +x5 +x4 +x3 +x+

1) + b(x)(x5 + x3 + x2 + x) = x+ 1 by reversing the previous steps:

(x+1) = x5+x3+x2+x−(x3+1)(x2+1) = x5+x3+x2+x−(x2+1)((x6+x5+x4+x3+x+1)−

(x5+x3+x2+x)(x+1)) = (x2+1)(x6+x5+x4+x3+x+1)+(1+(x2+1)(x+1))(x5+x3+x2+x) =

(x2 + 1)(x6 + x5 + x4 + x3 + x+ 1) + (x3 + x2 + x)(x5 + x3 + x2 + x).

So, a(x) = x2 + 1 and b(x) = x3 + x2 + x.

Example 13 Let f(x) = x5 − x4 + x+ 1, g(x) = x3 + x and F[x] = F3[x].

x5 − x4 + x+ 1 = (x3 + x)(x2 + 2x+ 2) + (x2 + 2x+ 1)

x3 + x = (x2 + 2x+ 1)(x+ 1) + (x+ 2)

x2 + 2x+ 1 = (x+ 2)(x) + 1

x+ 2 = 1(x+ 2) + 0

Thus

1 = gcd(x2 + 2x+ 1, x+ 2) = gcd(x3 + x, x2 + 2x+ 1) = gcd(x5 − x4 + x+ 1, x3 + x).

Now we found a(x) and b(x) such that gcd((f(x)), (g(x))) = a(x)(x5 − x4 + x + 1) +

b(x)(x3 + x) = 1 by reversing the previous steps.

1 = x2 + 2x + 1 + 2(x + 2)(x) = x2 + 2x + 1 + 2x(x3 + x + 2(x2 + 2x + 1)(x + 1)) =

(x5−x4+x+1)+2(x3+x)(x2+2x+2)+2x(x3+x)+x(x+1)(x5−x4+x+1+2(x3+x)(x2+

2x+2)) = (x5−x4 +x+1)+x(x+1)(x5−x4 +x+1)+2(x3 +x)(x2 +2x+2)+2x(x3 +x)+

2x(x+ 1)(x3 + x)(x2 + 2x+ 2) = (x2 + x+ 1)(x5 − x4 + x+ 1) + (2x4 + x2 + x+ 2)(x3 + x)

So, a(x) = x2 + x+ 1 and b(x) = 2x4 + x2 + x+ 2.

92

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