a class of minimal cyclic codes over finite fields

Des. Codes Cryptogr.DOI 10.1007/s10623-013-9857-9

A class of minimal cyclic codes over finite fields

Bocong Chen · Hongwei Liu · Guanghui Zhang

Received: 21 September 2012 / Revised: 23 June 2013 / Accepted: 24 June 2013© Springer Science+Business Media New York 2013

Abstract In this paper, we study minimal cyclic codes of length �m over a finite field Fq ,where � is a prime divisor of q − 1 and m is a positive integer. Explicit expressions for theprimitive idempotents, check polynomials, minimum Hamming distances and the dimensionsof these codes are obtained.

Keywords Finite field · Minimal cyclic code · Primitive idempotent

Mathematics Subject Classification 94B05 · 94B15

1 Introduction

Cyclic codes over finite fields constitute a remarkable class of linear codes. In particular,some important codes, such as Hamming codes and BCH codes, are cyclic codes. Cycliccodes also have practical applications, as they can be efficiently encoded by shift registers.

Let Fq be the finite field of order q . Any cyclic code of length n over Fq is identified withexactly one ideal of the quotient algebra Fq [X ]/〈Xn − 1〉. A cyclic code is called minimalif the corresponding ideal is minimal. The quotient algebra Fq [X ]/〈Xn − 1〉 is semi-simpleif the code length n is relatively prime to the characteristic of Fq . In this case, every cyclic

Communicated by C. Mitchell.

B. Chen · H. Liu (B)School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, Hubei, Chinae-mail: [email protected]

B. Chene-mail: [email protected]

G. ZhangSchool of Mathematical Sciences, Luoyang Normal University, Luoyang 471022, Henan, Chinae-mail: [email protected]

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B. Chen et al.

code turns out to be a direct sum of some minimal cyclic codes. Observe that, minimal cycliccodes include the important family of simplex codes (see [15], Chap. 8).

Recently, minimal cyclic codes have been an attractive topic of study; see e.g. [2,4,5,7,8,10,17,18]. It is well known that there is a one-to-one correspondence between minimalcyclic codes and primitive idempotents, as every minimal cyclic code is generated by exactlyone primitive idempotent (see [11, Theorem 4.3.8]). In general, it is a challenge to determineexplicitly the primitive idempotents in Fq [X ]/〈Xn − 1〉. Assume that p is an odd primecoprime to the characteristic of Fq . In [16], Pruthi and Arora studied minimal cyclic codes oflength pn over Fq , where q is a primitive root modulo pn . The authors explicitly describedthe q-cyclotomic cosets modulo pn and the primitive idempotents in Fq [X ]/〈X pn − 1〉. In[1], the authors obtained the primitive idempotents in Fq [X ]/〈X2pn −1〉, where q is odd and

ord2pn (q) = φ(pn). Primitive idempotents in Fq [X ]/〈X pn −1〉 with ordpn (q) = φ(pn)2 were

also determined (see [6] or [2]). Batra and Arora (see [8]) used the method presented in [2] toobtain the primitive idempotents in Fq [X ]/〈X2pn −1〉, where q is odd and ord2pn (q) = φ(pn)

2 .

We mention that, either ordpn (q) = φ(pn) or ordpn (q) = φ(pn)2 except when p = 3, leads

to gcd(p, q − 1) = 1.In this paper, we study minimal cyclic codes of length �m over Fq , where � is a prime

divisor of q − 1 and m is a positive integer; we further assume that 4 | (q − 1) if � = 2.Write q − 1 = �vc with v > 0, gcd(�, c) = 1, and m = bv + r , where b, r are non-negativeintegers with r < v. We propose a new approach to obtain the primitive idempotents inFq [X ]/〈X�m − 1〉. We mention that the key step to obtain the primitive idempotents in

Fq [X ]/〈X�m − 1〉 is to get the primitive idempotents in Fq [X ]/〈X�v+r − 1〉 as one can obtain

the primitive idempotents in Fq [X ]/〈X�m − 1〉 from Fq [X ]/〈X�v+r − 1〉 recursively (seeTheorem 3.6). The check polynomials, minimum Hamming distances, and the dimensionsof minimal cyclic codes generated by the primitive idempotents in Fq [X ]/〈X�m −1〉 are alsodiscussed.

This paper is organized as follows. Necessary notations and known results are providedin the next section. In Sect. 3, we give the irreducible factorization of X�m − 1 in Fq [X ] and

obtain the primitive idempotents in Fq [X ]/〈X�v+r −1〉. Then we get the primitive idempotentsin Fq [X ]/〈X�m − 1〉 recursively. We end this section with a remark to give new proofs to themain results of [1,8] by using the method proposed in Lemma 3.3. In Sect. 4, we give thecheck polynomials, minimum Hamming distances and the dimensions of the codes generatedby the primitive idempotents in Fq [X ]/〈X�v+r − 1〉. We indicate that the parameters of theminimal cyclic codes in Fq [X ]/〈X�m − 1〉 also can be obtained from the minimal cyclic

codes in Fq [X ]/〈X�v+r − 1〉 recursively. In Sect. 5, we give two examples to illustrate ourmain results.

2 Preliminaries

Throughout this paper Fq denotes the finite field of order q . We denote by F∗q the multiplicative

group of non-zero elements of Fq . For β ∈ F∗q we denote by ord(β) the order of β in the group

F∗q and β is called a primitive ord(β)th root of unity. Let F

∗q = 〈ξ 〉, where ξ is a primitive

(q − 1)th root of unity.Let also n be a positive integer. Any element of the quotient algebra Fq [X ]/〈Xn − 1〉 is

uniquely represented by a polynomial a0 + a1 X + · · · + an−1 Xn−1 of degree less thann, hence it can be identified with a word (a0, a1, . . . , an−1) of length n over Fq ; so we

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have the corresponding Hamming weight and Hamming distance on the quotient algebraFq [X ]/〈Xn − 1〉. A cyclic code C of length n over Fq is called a [ n, k, dH (C) ] code if itsdimension is k and its minimum Hamming distance is dH (C). For more details on cycliccodes, the readers may refer to [11] or [15].

Let f (X) be a polynomial in Fq [X ]. Write

f (X) = f1(X)e1 f2(X)e2 · · · fr (X)er ,

where f1(X), f2(X), . . . , fr (X) are distinct irreducible divisors of f (X) in Fq [X ] ande1, e2, . . . , er are positive integers. By the Chinese Remainder Theorem (e.g. see [9, Theorem3.5.15]), we have the Fq -algebra isomorphism:

ϕ : Fq [X ]/〈 f (X)〉 −→ Fq [X ]/〈 f1(X)e1〉 × · · · × Fq [X ]/〈 fr (X)er 〉, (2.1)

p(X) + 〈 f (X)〉 �→ (p(X) + 〈 f1(X)e1〉, . . . , p(X) + 〈 fr (X)er 〉 )

.

The following lemma contains a criterion on irreducible non-linear binomials over Fq ,which was given by Serret in 1866 (see [13, Theorem 3.75] or [19, Theorem 10.7]).

Lemma 2.1 Assume that n ≥ 2 is a positive integer. For any γ ∈ F∗q with ord(γ ) = e,

the binomial Xn − γ is irreducible over Fq if and only if both the following conditions aresatisfied:

(i) Every prime divisor of n divides e, but does not divide (q − 1)/e;(ii) If 4 | n, then 4 | (q − 1).

3 Primitive idempotents in Fq[X]/〈X�m − 1〉

Let Fq be the finite field of order q and F∗q = 〈ξ 〉 as before. Let m be a positive integer.

Throughout this paper, we always assume that � is a prime divisor of q − 1, and furtherassume that 4 | (q −1) if � = 2. Hence, in order to check that a non-linear binomial of degree�m in Fq [X ] is irreducible it is enough to verify Condition (i) in Lemma 2.1.

Write q − 1 = �vc, where gcd(�, c) = 1 and v is a positive integer. Let η = ξ c, then η

is a primitive �v th root of unity in Fq . Write also m = bv + r , where b, r are non-negativeintegers with r < v. In this section, we shall explicitly determine the irreducible factors ofX�m − 1 in Fq [X ]; then we go on to obtain the primitive idempotents in Fq [X ]/〈X�v+r − 1〉.As we shall see, this is a key step to obtain the primitive idempotents in Fq [X ]/〈X�m − 1〉.Theorem 3.1 Let � be a prime divisor of q − 1, and further assume that 4 | (q − 1) if � = 2.Write q − 1 = �vc with gcd(�, c) = 1 and v a positive integer. Then

X�m − 1 ={∏�v−1

k=0

(X − ηk

) · ∏m−vj=1

∏i=1

�v−1

� � i

(X� j − ηi

), i f m > v;

∏�m−1

k=0

(X − δk

), i f m ≤ v,

where δ is a primitive �mth root of unity in Fq for m ≤ v. All the factors on the right handside of the equation above are irreducible over Fq . In particular, the number of primitiveidempotents in Fq [X ]/〈X�m − 1〉 is equal to �v + (m − v)(�v − �v−1) for m > v.

Proof If m ≤ v, then there exists a primitive �m th root of unity δ in Fq . The result thenfollows trivially.

Now we investigate the case m > v. Note that η is a primitive �v th root of unity in Fq .For 1 ≤ i ≤ �v − 1 and gcd(�, i) = 1, we have ord(ηi ) = �v

gcd(i,�v)= �v , it follows that

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� | ord(ηi ) but gcd(�,

q−1ord(ηi )

)= 1. By Lemma 2.1, we get that X� j − ηi is irreducible in

Fq [X ]. We deduce that X − ηk and X� j − ηi are divisors of X�m − 1, where 0 ≤ k ≤ �v − 1,1 ≤ j ≤ m − v and 1 ≤ i ≤ �v − 1 with gcd(�, i) = 1. Since the irreducible factors X − ηk

and X� j − ηi are distinct from each other, it follows that

�v−1∏

k=0

(X − ηk) ·

m−v∏

j=1

�v−1∏

i=1� � i

(X� j − ηi )

∣∣∣ (X�m − 1).

On the other hand, the degree of∏

i=1�v−1� � i

(X� j − ηi

)is � j (�v − �v−1), hence the degree

of∏�v−1

k=0

(X − ηk

) · ∏m−vj=1

�v−1∏i=1

� � i

(X� j − ηi

)is

�v + (� + �2 + · · · + �m−v)(�v − �v−1) = �v + �m−v+1 − �

� − 1(�v − �v−1) = �m .

To conclude, we note that the number of primitive idempotents in Fq [X ]/〈X�m − 1〉 andthe number of irreducible factors of X�m − 1 in Fq [X ] coincide. �

Recall that q − 1 = �vc and m = bv + r . If b = 0, i.e. m = r < v, then the primitiveidempotents in Fq [X ]/〈X�m − 1〉 are easy to obtain (see [16, Theorem 2.1] or [14, Corollary2.1.7]). Hence, in this paper, we assume b to be a positive integer. In the following, we shallexplicitly determine the primitive idempotents in Fq [X ]/〈X�v+r − 1〉. Observe that

X�v+r − 1 = (X�r − η)(X�r − η2) · · · (X�r − η�v−1)(X�r − 1),

is a decomposition of X�v+r − 1 in Fq [X ]. By (2.1), we have the algebra isomorphism:

ϕ : Fq [X ]/〈X�v+r − 1〉 −→ Fq [X ]/〈X�r − η〉 × · · · × Fq [X ]/〈X�r − η�v−1〉 × Fq [X ]/〈X�r − 1〉.(3.1)

We first calculate the primitive idempotents in Fq [X ]/〈X�r − ηk〉 for 1 ≤ k ≤ �v . Note thatk can be written in a unique way as k = �z y such that z ≥ 0 and gcd(y, �) = 1. If z = 0, i.e.gcd(�, k) = 1, then by Lemma 2.1, X�r −ηk is irreducible in Fq [X ], hence Fq [X ]/〈X�r −ηk〉is a finite field and it contains the only primitive idempotent 1. In the following, we give theprimitive idempotents in Fq [X ]/〈X�r − η�z y〉 with z ≥ 1.

Lemma 3.2 Let y, z be positive integers with r ≤ z, gcd(y, �) = 1 and �z y ≤ �v . Thenthere are �r primitive idempotents in Fq [X ]/〈X�r − η�z y〉; they are 1

�r

∑�r −1i=0

(ρ−iuai

�z y Xi),

where 0 ≤ u ≤ �r − 1, ρ = η�v−rand a�z y = η−�z−r y .

Proof Take a�z y = η−�z−r y . It is easy to see that the following map

ϑ : Fq [X ]/〈X�r − 1〉 −→ Fq [X ]/〈X�r − η�z y〉f (X) + 〈X�r − 1〉 �→ f (a�z y X) + 〈X�r − η�z y〉,

is an Fq -algebra isomorphism. Now the result is a direct consequence of [16, Theorem 2.1]or [14, Corollary 2.1.7]. �

The next lemma gives the primitive idempotents in Fq [X ]/〈X�r − η�z y〉 with z, y beingpositive integers, gcd(y, �) = 1, �z y ≤ �v − 1 and r > z.

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Lemma 3.3 Let a, y, z be positive integers with a > z, gcd(y, �) = 1 and �z y ≤ �v − 1.Then there are �z primitive idempotents in Fq [X ]/〈X�a − η�z y〉; they are

Ds(X) = 1

�z

�z−1∑

j=0

(εsηy)− j X j�a−z

,

where ε = η�v−zis a primitive �z th root of unity in Fq and 1 ≤ s ≤ �z .

Proof Since z ≥ 1 and �z y ≤ �v − 1, we have 1 ≤ z < v. Let ε = η�v−zbe a primitive �z th

root of unity in Fq . By Lemma 2.1, we have the following irreducible factorization in Fq [X ]:X�a − η�z y = (X�a−z − εηy) · · · (X�a−z − ε�z−1ηy)(X�a−z − ηy). (3.2)

By (2.1), we have the algebra isomorphism:

ϕ : Fq [X ]/〈X�a − η�z y〉 −→ Fq [X ]/〈X�a−z − εηy〉 × · · · × Fq [X ]/〈X�a−z − ηy〉. (3.3)

We represent the polynomial p(X) = ∑�a−1i=0 pi Xi ∈ Fq [X ]/〈X�a − η�z y〉 by the follow-

ing matrix

P =

⎛

⎜⎜⎜⎝

p0 p1 · · · p�a−z−1p�a−z p�a−z+1 · · · p2�a−z−1

· · · · · · . . . · · ·p(�z−1)�a−z p(�z−1)�a−z+1 · · · p�a−1

⎞

⎟⎟⎟⎠

�z×�a−z

.

Accordingly, any element( ∑�a−z−1

i=0 qi Xi ,∑�a−z−1

i=0 qi+�a−z Xi , . . . ,∑�a−z−1

i=0 qi+(�z−1)�a−z Xi )

on the right hand side in (3.3) is represented by

Q =

⎛

⎜⎜⎜⎝

q0 q1 · · · q�a−z−1q�a−z q�a−z+1 · · · q2�a−z−1

· · · · · · . . . · · ·q(�z−1)�a−z q(�z−1)�a−z+1 · · · q�a−1

⎞

⎟⎟⎟⎠

�z×�a−z

,

where the first row of Q represents the first polynomial∑�a−z−1

i=0 qi Xi , . . ., and the last

row of Q represents the last polynomial∑�a−z−1

i=0 qi+(�z−1)�a−z X i . Suppose that ϕ(p(X)) =( ∑�a−z−1

i=0 qi Xi ,∑�a−z−1

i=0 qi+�a−z Xi , . . . ,∑�a−z−1

i=0 qi+(�z−1)�a−z Xi ), then ϕ can be expressed as

V P = Q, where V is the Vandermonde matrix

V =

⎛

⎜⎜⎜⎜⎝

1 εηy (εηy)2 · · · (εηy)�z−1

1 ε2ηy (ε2ηy)2 · · · (ε2ηy)�z−1

· · · · · · · · · · · · · · ·1 ε�z−1ηy (ε�z−1ηy)2 · · · (ε�z−1ηy)�

z−1

1 ηy (ηy)2 · · · (ηy)�z−1

⎞

⎟⎟⎟⎟⎠

�z×�z

.

We can obtain the preimage of every vector on the right hand side in (3.3) under ϕ using theabove matrix form. In more detail, given an arbitrary vector on the right hand side in (3.3), wewrite it as an �z × �a−z matrix Q such that the entries in its i th row consist of the coefficientof the i th polynomial. Then the preimage of the given vector under ϕ can be obtained byrewriting the matrix V −1 Q as a polynomial of degree �a .

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It is easy to check that

V −1 = 1

�z

⎛

⎜⎜⎜⎜⎝

1 · · · 1 1ε−1η−y · · · ε−(�z−1)η−y η−y

· · · · · · · · · · · ·(ε−1η−y)�

z−2 · · · (ε−(�z−1)η−y)�z−2 (η−y)�

z−2

(ε−1η−y)�z−1 · · · (ε−(�z−1)η−y)�

z−1 (η−y)�z−1

⎞

⎟⎟⎟⎟⎠

�z×�z

.

Since (3.2) gives the irreducible factorization of X�a − η�z y in Fq [X ], each quotient algebra

Fq [X ]/〈X�a−z − εiηy〉 is a finite field for 1 ≤ i ≤ �z and its primitive idempotent is just 1.We compute the primitive idempotents Ds(X) in Fq [X ]/〈X�a − η�z y〉 by the above matrixform. For 1 ≤ s ≤ �z ,

Ds(X) = V −1 ·

⎛

⎜⎜⎜⎜⎜⎜⎝

0 0 0 · · · 00 0 0 · · · 0· · · · · · · · · · · · · · ·1 0 0 · · · 0· · · · · · · · · · · · · · ·0 0 0 · · · 0

⎞

⎟⎟⎟⎟⎟⎟⎠

�z×�a−z

= 1�z

⎛

⎜⎜⎜⎜⎜⎜⎝

1 0 0 · · · 0ε−sη−y 0 0 · · · 0

(ε−sη−y)2 0 0 · · · 0· · · · · · · · · · · · · · ·· · · · · · · · · · · · · · ·

(ε−sη−y)(�z−1) 0 0 · · · 0

⎞

⎟⎟⎟⎟⎟⎟⎠

�z×�a−z

= 1�z

�z−1∑

j=0

(εsηy)− j X j�a−z

.

�

The expression V P = Q of ϕ used in the proof of Lemma 3.3 is originally due to Hughesin [12].

Theorem 3.4 Assume that the notation as given above. Then there are �v + r(�v − �v−1)

primitive idempotents in Fq [X ]/〈X�v+r − 1〉 given as follows:

(i)

Ek(X) = 1

�v

�v−1∑

i=0

η−ki Xi�r,

where 1 ≤ k ≤ �v and gcd(�, k) = 1;(ii)

G(u)

�z1 y1(X) = 1

�v+r

�v−1∑

j=0

�r −1∑

i=0

(η− j�z1 y1(a�z1 y1ρ

−u)i X i+ j�r),

where z1, y1 are positive integers with r ≤ z1, �z1 y1 ≤ �v , gcd(�, y1) = 1, 0 ≤ u ≤

�r − 1, a�z1 y1 = η−�z1−r y1 and ρ = η�v−r;

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(iii)

F (s)�z0 y0

(X) = 1

�v+z0

�v−1∑

i=0

�z0 −1∑

j=0

(η−i�z0 y0(εsηy0)− j X i�r + j�r−z0

),

where z0, y0 are positive integers with r > z0 ≥ 1, �z0 y0 ≤ �v − 1, gcd(�, y0) = 1,1 ≤ s ≤ �z0 and ε = η�v−z0 .

Proof The number of primitive idempotents in Fq [X ]/〈X�v+r − 1〉 is a direct consequenceof Theorem 3.1. Recall that η is a primitive �vth root of unity, and

X�v+r − 1 = (X�r − η)(X�r − η2) · · · (X�r − η�v−1)(X�r − 1),

is a decomposition of X�v+r − 1 in Fq [X ].As we did in the proof of Lemma 3.3, we can express ϕ defined in (3.1) with matrix form:

V P = Q, where

P =

⎛

⎜⎜⎜⎝

p0 p1 · · · p�r −1

p�r p�r +1 · · · p2�r −1

· · · · · · . . . · · ·p(�v−1)�r p(�v−1)�r +1 · · · p�v+r −1

⎞

⎟⎟⎟⎠

�v×�r

,

Q =

⎛

⎜⎜⎜⎝

q0 q1 · · · q�r −1

q�r q�r +1 · · · q2�r −1

· · · · · · . . . · · ·q(�v−1)�r q(�v−1)�r +1 · · · q�v+r −1

⎞

⎟⎟⎟⎠

�v×�r

,

and V is the Vandermonde matrix

V =

⎛

⎜⎜⎜⎜⎝

1 η η2 · · · η�v−1

1 η2 (η2)2 · · · (η2)�v−1

· · · · · · · · · · · · · · ·1 η�v−1 (η�v−1)2 · · · (η�v−1)�

v−1

1 1 1 · · · 1

⎞

⎟⎟⎟⎟⎠

�v×�v

.

It is easy to verify that

V −1 = 1

�v

⎛

⎜⎜⎝

1 1 · · · 1 1η−1 η−2 · · · η−�v−1 1· · · · · · · · · · · · · · ·

η−(�v−1) (η−2)�v−1 · · · (η−(�v−1))�

v−1 1

⎞

⎟⎟⎠

�v×�v

.

Using arguments similar to those of Lemma 3.3, we can get the primitive idempotents inFq [X ]/〈X�v+r − 1〉. They are as follows:

(i) For 1 ≤ k ≤ �v and gcd(�, k) = 1, recall that Fq [X ]/〈X�r − ηk〉 is a finite fieldin this case. By the isomorphism ϕ, we get the following primitive idempotents inFq [X ]/〈X�v+r −1〉,

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Ek(X) = V −1 ·

⎛

⎜⎜⎜⎜⎜⎜⎝

0 0 0 · · · 00 0 0 · · · 0· · · · · · · · · · · · · · ·1 0 0 · · · 0

· · · · · · · · · · · · · · ·0 0 0 · · · 0

⎞

⎟⎟⎟⎟⎟⎟⎠

�v×�r

= 1�v

⎛

⎜⎜⎜⎜⎜⎜⎝

1 0 0 · · · 0η−k 0 0 · · · 0

(η−k)2 0 0 · · · 0· · · · · · · · · · · · · · ·· · · · · · · · · · · · · · ·

(η−k)�v−1 0 0 · · · 0

⎞

⎟⎟⎟⎟⎟⎟⎠

�v×�r

= 1�v

�v−1∑

i=0η−ki Xi�r

.

(ii) For k = �z1 y1, where z1, y1 are positive integers with r ≤ z1, �z1 y1 ≤ �v and

gcd(�, y1) = 1, by Lemma 3.2, taking z = z1, y = y1, the primitive idempo-tents in Fq [X ]/〈X�r − ηk〉 are 1

�r

∑�r −1i=0

(ρ−iuai

�z1 y1Xi

), where 0 ≤ u ≤ �r − 1,

a�z1 y1 = η−�z1−r y1 and ρ = η�v−r.

We have the following primitive idempotents in Fq [X ]/〈X�v+r −1〉,

G(u)

�z1 y1(X) = V −1 · 1

�r

⎛

⎜⎜⎜⎜⎜⎜⎝

0 0 · · · · · · 00 0 · · · · · · 0· · · · · · · · · · · · · · ·1 a�z1 y1ρ

−u a2�z1 y1

ρ−2u · · · a�r −1�z1 y1

ρ−(�r −1)u

· · · · · · · · · · · · · · ·0 0 · · · · · · 0

⎞

⎟⎟⎟⎟⎟⎟⎠

�v×�r

= 1�v+r

⎛

⎜⎜⎜⎜⎜⎝

1 · · · a�r −1�z1 y1

ρ−(�r −1)u

η−�z1 y1 · · · η−�z1 y1 a�r −1�z1 y1

ρ−(�r −1)u

· · · · · · · · ·· · · · · · · · ·

(η−�z1 y1)�v−1 · · · (η−�z1 y1)�

v−1a�r −1�z1 y1

ρ−(�r −1)u

⎞

⎟⎟⎟⎟⎟⎠

�v×�r

= 1�v+r

�v−1∑

j=0

�r −1∑

i=0


−u)i X i+ j�r).

(iii) For k = �z0 y0 , where z0, y0 are positive integers with r > z0, �z0 y0 ≤ �v − 1 and

gcd(�, y0) = 1, by Lemma 3.3, taking a = r, z = z0, y = y0, then the primitiveidempotents in Fq [X ]/〈X�r −ηk〉 are 1

�z0

∑�z0 −1j=0

(εsηy0)− j X j�r−z0 where 1 ≤ s ≤ �z0

and ε = η�v−z0 is a primitive �z0 th root of unity in Fq . We have the following primitive

idempotents in Fq [X ]/〈X�v+r −1〉,

123


F (s)�z0 y0

(X) = V −1 · 1�z0

⎛

⎜⎜⎜⎜⎜⎜⎝

0 0 · · · 0 · · · 00 0 · · · 0 · · · 0· · · · · · · · · · · · · · · · · ·1 · · · ε−sη−y0 · · · (ε−sη−y0)(�

z0 −1) · · ·· · · · · · · · · · · · · · · · · ·0 0 · · · 0 · · · 0

⎞

⎟⎟⎟⎟⎟⎟⎠

�v×�r

= 1�v+z0

⎛

⎜⎜⎜⎜⎝

1 · · · (ε−sη−y0)�z0 −1 · · ·

η−�z0 y0 · · · η−�z0 y0(ε−sη−y0)�z0 −1 · · ·

· · · · · · · · · · · ·· · · · · · · · · · · ·

(η−�z0 y0)�v−1 · · · (η−�z0 y0)�

v−1(ε−sη−y0)�z0 −1 · · ·

⎞

⎟⎟⎟⎟⎠

�v×�r

= 1�v+z0

�v−1∑

i=0

�z0 −1∑

j=0

(η−i�z0 y0(εsηy0)− j X i�r + j�r−z0

).

�As we shall see, Theorem 3.4 is a crucial step for obtaining the primitive idempotents in

Fq [X ]/〈X�m − 1〉, where m = bv + r with b > 0 and 0 ≤ r < v.

Corollary 3.5 Assume that the notation as given above. Then Fq [X ]/〈X�2v+r − 1〉 has �v +(v + r)(�v − �v−1) primitive idempotents given as follows:

(i)

H2,k(X) = 1

�v

�v−1∑

i=0

η−ki Xi�v+r,


H (s)2,�z y(X) = 1

�v+z

�v−1∑

i=0

�z−1∑

j=0

(η−i�z y(εsηy)− j X i�v+r + j�v+r−z )

,

where z, y are positive integers with �z y ≤ �v − 1, gcd(�, y) = 1, ε = η�v−zand

1 ≤ s ≤ �z;(iii)

E2,k(X) = 1

�vEk(X)

�v−1∑

i=0

Xi�v+r,

G(u)

2,�z1 y1(X) = 1

�vG(u)

�z1 y1(X)

�v−1∑

i=0

Xi�v+r,

F (s)2,�z0 y0

(X) = 1

�vF (s)

�z0 y0(X)

�v−1∑

i=0

Xi�v+r,

where Ek(X), G(u)

�z1 y1(X) and F (s)

�z0 y0(X) were introduced in Theorem 3.4.

Proof By (2.1), we have the algebra isomorphism:

Fq [X ]/〈X�2v+r − 1〉 −→ Fq [X ]/〈X�v+r − η〉 × · · · × Fq [X ]/〈X�v+r − 1〉. (3.4)

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B. Chen et al.

Using arguments similar to Theorem 3.4 we obtain

V =

⎛

⎜⎜⎜⎜⎝

1 η η2 · · · η�v−1

1 η2 (η2)2 · · · (η2)�v−1

· · · · · · · · · · · · · · ·1 η�v−1 (η�v−1)2 · · · (η�v−1)�

v−1

1 1 1 · · · 1

⎞

⎟⎟⎟⎟⎠

�v×�v

.

We just note that by Lemma 3.3, taking a = v + r , we can get the primitive idempotentsH (s)

2,�z y(X). �Recall that m = bv + r with b > 0 and 0 ≤ r < v. Generalizing (3.1) and (3.4), for any

positive integer t , we have an algebra isomorphism:

Fq [X ]/〈X�tv+r − 1〉 −→ Fq [X ]/〈X�(t−1)v+r − η〉 × · · · × Fq [X ]/〈X�(t−1)v+r − 1〉.

All the primitive idempotents in Fq [X ]/〈X�tv+r − 1〉 with t = 1, 2 have been obtainedusing Theorem 3.4 and Corollary 3.5. We now move to the general case of obtaining all theprimitive idempotents in Fq [X ]/〈X�m − 1〉.Theorem 3.6 Let t be a positive integer and let S be the set of the primitive idempotents inFq [X ]/〈X�tv+r − 1〉. Then the primitive idempotents of Fq [X ]/〈X�(t+1)v+r − 1〉 are given asfollows:

(i)

1

�v

�v−1∑

i=0

η−ki Xi�tv+r,


1

�v+z

�v−1∑

i=0

�z−1∑

j=0

(η−i�z y(εsηy)− j X i�tv+r + j�tv+r−z )

,

where z, y are positive integers with �z y ≤ �v − 1 and gcd(�, y) = 1;(iii)

1

�ve(X)

�v−1∑

i=0

Xi�tv+r,

where e(X) ranges over S.

In Sect. 5, a concrete example will be given to illustrate Theorem 3.6.At the end of this section, we shall show that, using the method proposed in Lemma 3.3,

we can give a new proof to the main results of [1,8]. More generally, we have the followingtheorem.

Theorem 3.7 Suppose that q is odd and let n be an odd positive integer with gcd(n, q) = 1.Assume that there are h primitive idempotents in Fq [X ]/〈Xn −1〉. Then there are 2h primitiveidempotents in Fq [X ]/〈X2n −1〉 given by 1

2 e(X)(1+ Xn) and 12 e(−X)(1− Xn), where e(X)

ranges over the primitive idempotents in Fq [X ]/〈Xn − 1〉.

123


Proof Since n is an odd integer, it is easy to check that the following map is an Fq -algebraisomorphism:

Fq [X ]/〈Xn − 1〉 −→ Fq [X ]/〈Xn + 1〉f (X) + 〈Xn − 1〉 �→ f (−X) + 〈Xn + 1〉.

By this isomorphism, we get the primitive idempotents in Fq [X ]/〈Xn + 1〉, i.e. e(−X),where e(X) ranges over the primitive idempotents in Fq [X ]/〈Xn − 1〉. Now we compute theprimitive idempotents in Fq [X ]/〈X2n − 1〉.

By (2.1), we have the algebra isomorphism

Fq [X ]/〈X2n − 1〉 � Fq [X ]/〈Xn + 1〉 × Fq [X ]/〈Xn − 1〉.Using arguments similar to Lemma 3.3, we have AP = Q, where

P =(

f0 f1 · · · fn−1

fn fn+1 · · · f2n−1

)

2×n,

Q =(

q0 q1 · · · qn−1

qn qn+1 · · · q2n−1

)

2×n,

and A is the matrix

A =(

1 −11 1

)

2×2.

Suppose e(X) = ∑n−1k=0 qk Xk is a primitive idempotent in Fq [X ]/〈Xn − 1〉. We have

A−1 ·(

0 0 · · · 0q0 q1 · · · qn−1

)

2×n= 1

2

(1 1

−1 1

)·(

0 0 · · · 0q0 q1 · · · qn−1

)

2×n

= 12

(q0 q1 · · · qn−1q0 q1 · · · qn−1

)

2×n= 1

2 e(X)(1 + Xn);and

A−1 ·(

q0 −q1 · · · qn−10 0 · · · 0

)

2×n= 1

2

(1 1

−1 1

)·(

q0 −q1 · · · qn−10 0 · · · 0

)

2×n

= 12

(q0 −q1 · · · qn−1

−q0 q1 · · · −qn−1

)

2×n= 1

2 e(−X)(1 − Xn).

�

Remark 3.8 As the primitive idempotents of Fq [X ]/〈X pn −1〉 have been determined in [16,2]

for ordpn (q) = φ(pn) and ordpn (q) = φ(pn)2 respectively, it is possible to use Theorem 3.7

in order to give new proofs of the main results of [1,8] for the primitive idempotents inFq [X ]/〈X2pn − 1〉, respectively.

4 Minimal cyclic codes of length �m

We use the notation I to denote the minimal cyclic code generated by the primitive idempotentI (x) in Fq [X ]/〈X�v+r − 1〉. In this section, we first obtain the explicit expressions for thecheck polynomials, dimensions and the minimum Hamming distances of minimal cycliccodes in Fq [X ]/〈X�v+r − 1〉.

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B. Chen et al.

Theorem 4.1 With the notation of Theorem 3.4, we have the following:

(i) Ek is a [ �v+r , �r , �v ] cyclic code, where 1 ≤ k ≤ �v and gcd(�, k) = 1; its checkpolynomial is X�r − ηk;

(ii) F (s)�z0 y0

is a [ �v+r , �r−z0 , �v+z0 ] cyclic code, where r > z0 ≥ 1, 1 ≤ s ≤ �z0 , 1 ≤�z0 y0 ≤ �v and gcd(�, y0) = 1; its check polynomial is X�r−z0 − εsηy0 ;

(iii) G(u)

�z1 y1is a [ �v+r , 1, �v+r ] cyclic code, where r ≤ z1 ≤ v, 1 ≤ �z1 y1 ≤ �v , 0 ≤ u ≤

�r − 1 and gcd(�, y1) = 1; its check polynomial is X − a−1�z1 y1

ρu.

Proof Note that there is a one-to-one correspondence between the irreducible factors ofX�v+r − 1 over Fq and the check polynomials of minimal cyclic codes of length �v+r overFq . To show that X�r − ηk is a check polynomial of Ek , it suffices to prove that the roots ofX�r − ηk do not satisfy Ek(X) = 0.

Set α to be any root of X�r − ηk in some extension field of Fq . Then

Ek(α) = 1

�v

�v−1∑

i=0

η−kiαi�r = 1

�v

�v−1∑

i=0

η−kiηki = 1 = 0.

Therefore, X�r − ηk is a check polynomial of Ek .Now we determine the minimum Hamming distance dH (Ek) of Ek . Obviously the

Hamming weight of Ek(X) is �v ; hence, we get dH (Ek) ≤ �v . For the reverse inequal-ity, let c(X) + 〈X�v+r − 1〉 be any non-zero element in Fq [X ]/〈X�v+r − 1〉. The degreeof c(X) is less than �v+r . In Fq [X ], we have polynomials q(X) and r(X) such thatc(X) = q(X)(X�r − ηk) + r(X), where r(X) = 0 or deg r(X) < �r .

Note that (X�r − ηk) · Ek(X) = 0 in Fq [X ]/〈X�v+r − 1〉, hence we have

(c(X) + 〈X�v+r − 1〉

)·(

Ek(X) + 〈X�v+r − 1〉)

= r(X)Ek(X) + 〈X�v+r − 1〉.

If r(X) = 0, then we get the codeword 0 in Ek ; hence, we may further assume r(X) = 0.Observe that deg r(X) + deg Ek(X) < �v+r ; then the Hamming weight of r(X) · Ek(X) is

wH (r(X) · Ek(X)) = wH (r(X)) · wH (Ek(X)) ≥ wH (Ek(X)) = �v.

The first equality holds since Ek(X) = 1�v

∑�v−1i=0 η−ki Xi�r

and deg(r(X)) < �r . Therefore,we get dH (Ek) = �v . This completes the proof of the first statement of the theorem.

123


Similarly, for the check polynomial of F (s)�z0 y0

, set β to be any root of X�r−z0 − εsηy0 insome extension field of Fq . Then

F (s)�z0 y0

(β) = 1

�v+z0

�v−1∑

i=0

�z0 −1∑

j=0

(η−i�z0 y0(εsηy0)− jβ i�r + j�r−z0 )

= 1

�v+z0

�v−1∑

i=0

�z0 −1∑

j=0

(η−i�z0 y0ε− jsη− j y0β i�r

β j�r−z0 )

= 1

�v+z0

�v−1∑

i=0

�z0 −1∑

j=0

(η−i�z0 y0ε− jsη− j y0β i�r

(εsηy0) j )

= 1

�v+z0

�v−1∑

i=0

�z0 −1∑

j=0

(η−i�z0 y0β i�r )

= 1

�v+z0

�v−1∑

i=0

�z0 −1∑

j=0

η−i�z0 y0(εsηy0)i�z0 = 1 = 0.

The last equality holds because ε is a primitive �z0 th root of unity. Therefore, X�r−z0 − εsηy0

is a check polynomial of F (s)�z0 y0

.

For the minimum Hamming distance of F (s)�z0 y0

, using approach similar to the previous

paragraph, we get dH (F (s)�z0 y0

) = �v+z0 .

Finally, it is obvious that the root of X − a−1�z1 y1

ρu = 0 is a−1�z1 y1

ρu . Then

G(u)

�z1 y1(a−1

�z1 y1ρu) = 1

�v+r

�v−1∑

j=0

�r −1∑

i=0


−u)i (a−1�z1 y1

ρu)i+ j�r )

= 1

�v+r

�v−1∑

j=0

�r −1∑

i=0

(η− j�z1 y1 a− j�r

�z1 y1

) = 1 = 0.

The second equality holds because ρ is a primitive �r th root of unity in Fq and the

last equality holds because a�r

�z1 y1η�z1 y1 = 1. Hence, the check polynomial of G(u)

�z1 y1is

X − a−1�z1 y1

ρu . It is easy to see that dH (G(u)

�z1 y1) = �v+r . �

Using arguments similar to Theorem 4.1, we can easily get the check polynomials andthe parameters of the minimal cyclic codes in Fq X ]/〈X�2v+r − 1〉. Note that the minimal

cyclic codes generated by the primitive idempotents E2,k(X), G(u)

2,�z1 y1(X) and F (s)

2,�z0 y0(X)

in Fq [X ]/〈X�2v+r − 1〉, as presented in Corollary 3.5, are just the �v times repetition of the

codes Ek , F (s)�z0 y0

and G(u)

�z1 y1. We omit the proof of the following corollary.

Corollary 4.2 With the notation of Corollary 3.5, we have the following:

(i) H2,k is a [ �2v+r , �v+r , �v ] cyclic code, where 1 ≤ k ≤ �v and gcd(�, k) = 1; its checkpolynomial is X�v+r − ηk;

(ii) H (s)2,�z y is a [ �2v+r , �v+r−z, �v+z ] cyclic code, where z, y being positive integers with

�z y ≤ �v − 1, gcd(�, y) = 1; its check polynomial is X�v+r−z − εsηy;

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B. Chen et al.

(iii) E2,k is a [ �2v+r , �r , �2v ] cyclic code; F (s)2,�z0 y0

is a [ �2v+r , �r−z0 , �2v+z0 ] cyclic code

and G(u)

2,�z1 y1is a [ �2v+r , 1, �2v+r ] cyclic code.

Remark 4.3 The parameters of the minimal cyclic codes in Fq [X ]/〈X�m −1〉 can be obtained

from the minimal cyclic codes in Fq [X ]/〈X�v+r − 1〉 recursively. Let m = bv + r withb > 0 and 0 ≤ r < v. The parameters of the minimal cyclic codes generated bythe primitive idempotents in (i) and (ii) of Theorem 3.6 are [ �bv+r , �(b−1)v+r , �v ] and[ �bv+r , �(b−1)v+r−z, �v+z ] respectively; the minimal cyclic codes generated by the prim-itive idempotents in (iii) of Theorem 3.6 are just multiple repetitions of minimal cyclic codesin Fq [X ]/〈X�v+r − 1〉.

5 Examples

Let � be a prime divisor of q − 1. Write q − 1 = �vc with v > 0 and gcd(�, c) = 1. Assumethat 0 ≤ r < v. Using Theorem 4.1, we can easily determine the parameters [ �v+r , k, d ]of all minimal cyclic codes of length �v+r over Fq . The following examples illustrate ourresults.

Example 5.1 Take q = 72, � = 2, b = 1 and r = 2; let η be a primitive 24th root of unity inF72 . The primitive idempotents in F72 [X ]/〈X26 − 1〉 can be easily obtained by Theorem 3.4.The 32 primitive idempotents in F72 [X ]/〈X26 − 1〉 given by Theorem 3.4 are the following:

(i) For 1 ≤ k ≤ 16 and k odd,

Ek(X) = 1

24

15∑

i=0

η−ki X4i .

(ii) For 1 ≤ s ≤ 2 and ε = −1,

F (s)2 (X) = 1

25

15∑

i=0

1∑

j=0

(η−2i (εsη)− j X4i+2 j ),

F (s)6 (X) = 1

25

15∑

i=0

1∑

j=0

(η−6i (εsη3)− j X4i+2 j ),

F (s)10 (X) = 1

25

15∑

i=0

1∑

j=0

(η−10i (εsη5)− j X4i+2 j ),

F (s)14 (X) = 1

25

15∑

i=0

1∑

j=0

(η−14i (εsη7)− j X4i+2 j ).

(iii) Since 4 = 22 · 1, we have z1 = 2 and y1 = 1. By Theorem 3.4, for 0 ≤ u ≤ 3, ρ beinga primitive 4th root of unity and a−1

4 = η,

G(u)4 (X) = 1

26

15∑

j=0

3∑

i=0

(η−4 j (a4ρ

−u)i X i+4 j ).

123


Table 1 Minimal cyclic codes oflength 64 over F72

Code Dimension Distance Check polynomial

E1 4 16 X4 − η

E3 4 16 X4 − η3

E5 4 16 X4 − η5

E7 4 16 X4 − η7

E9 4 16 X4 − η9

E11 4 16 X4 − η11

E13 4 16 X4 − η13

E15 4 16 X4 − η15

F(1)2 2 32 X2 + η

F(2)2 2 32 X2 − η

F(1)6 2 32 X2 + η3

F(2)6 2 32 X2 − η3

F(1)10 2 32 X2 + η5

F(2)10 2 32 X2 − η5

F(1)14 2 32 X2 + η7

F(2)14 2 32 X2 − η7

Similarly, taking a−18 = η2,

G(u)8 (X) = 1

26

15∑

j=0

3∑

i=0

(η−8 j (a8ρ

−u)i X i+4 j ).

Taking a−112 = η3,

G(u)12 (X) = 1

26

15∑

j=0

3∑

i=0

(η−12 j (a12ρ

−u)i X i+4 j ).

And taking a−116 = η4,

G(u)16 (X) = 1

26

15∑

j=0

3∑

i=0

(η−16 j (a16ρ

−u)i X i+4 j ).

Table 1 exhibits all minimal cyclic codes of length 64 and dimension greater than 1 overF72 .

Example 5.2 Take q = 7, � = 3, b = 3 and r = 0; let η be a primitive 3-rd root of unity in F7.The 7 primitive idempotents in F7[X ]/〈X33 −1〉 are easy to obtain by Theorem 3.6. With thenotation in Theorem 3.4, we have v = 1 and there are 3 primitive idempotents in F7[X ]/〈X3−1〉; they are E1(X) = 1

3

∑2i=0 η−i X i , E2(X) = 1

3

∑2i=0 η−2i X i , and G(0)

3 (X) = 13

∑2i=0 Xi .

By Corollary 3.5, we have the 5 primitive idempotents in F7[X ]/〈X32 − 1〉; they areH2,1(x) = 1

3

∑2i=0 η−i X3i , H2,2(x) = 1

3

∑2i=0 η−2i X3i , E2,1(X) = 1

3 E1(X)∑2

i=0 X3i ,

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B. Chen et al.

E2,2(X) = 13 E2(X)

∑2i=0 X3i , and G(0)

2,3(X) = 13 G(0)

3 (X)∑2

i=0 X3i . Finally, repeating the

process, we have the 7 primitive idempotents in F7[X ]/〈X33 − 1〉; they are 13

∑2i=0 η−i X9i ,

13

∑2i=0 η−2i X9i , and 1

3 e(X)∑2

i=0 X9i , where e(X) ranges over the primitive idempotents

in F7[X ]/〈X32 − 1〉.Acknowledgments The authors would like to deeply thank the referees for their very careful reading andmany valuable comments that helped us improve this paper. The authors also thank NSFC for the supportthrough Grant No. 11171370, and Research Funds of CCNU, Grant No. 11A02014.

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a class of minimal cyclic codes over finite fields

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