CYCLIC CODES OF LENGTH 2 p n OVER FINITE CHAIN RINGS

. We use group algebra methods to study cyclic codes over ﬁnite chain rings and under some restrictive hypotheses, described in section 2, for codes of length 2 p n , p a prime, we are able to compute the minimum weights of all possible cyclic codes of that length.


Introduction
Let R be a finite commutative ring with unity. A linear code of length n over R is a submodule C of R n . A linear code C is cyclic if, for every codeword x = (x 0 , . . . , x n−1 ) ∈ C, its cyclic shift (x n−1 , x 0 , . . . , x n−2 ) is also in C and many of the important codes in use are of this type. It is well-known that cyclic codes of length n over a ring R can be identified with the ideals of the group ring RA, where A is the cyclcic group of order n.
The Hamming weight of an element in R n is the number of its non-zero coordinates. We denote by w(C) the minimum weight of the code, which is the smallest Hamming weight of non-zero elements in the code.
A commutative ring R is a chain ring, or a uniserial ring, if the set of all ideals of R is a chain under set-theoretic inclusion. A ring R is local if it has a unique maximal ideal which is actually two-sided. We recall the following. Notice that, since M is maximal, then R = R/M is a field and M is the set of non units of R and its Jacobson radical; hence, it is nilpotent. Let a be a (fixed) generator of the maximal ideal M . Then a is a nilpotent element and we denote its nilpotency index by t. (a) For some prime q and positive integers k and l with k ≥ l, we have |R| = q k , |R| = q l and the characteristic of R and R are powers of q.
Minimum weights and dimensions are very important parameters of codes, as they determine the error-correcting capacity and the amount of information the code is capable of transmitting, respectively. In what follows, we compute the minimum Hamming weights and dimensions of an extensive family of cyclic codes of length 2p n , extending results from [1]. It should be noted that, to this end, we are using the Hamming weight while others weights are possible for codes over rings, such as the homogeneous weight (see [11]). Under some restrictive hypothesis described below, the family we consider is the family of all minimal codes.

Minimum weights
In what follows, R will denote a finite commutative chain ring with unity of order |R| = q k , where q denotes a prime rational integer, and maximal ideal M = a . Let t denote the nilpotency index of a. Set R = R/M . Then |R| = q l with = k/t. Also, we shall denote by A a cyclic group of order 2p n , such that gcd(q, p) = 1, and write A = B × G where G is the p-Sylow subgroup of A and B = {1, d} is its 2-Sylow subgroup.
We denote by ϕ Euler's ϕ-function; i.e., for a positive integer m, ϕ(m) denotes the number of positive integers smaller that m that are relatively prime to m. We denote by o(q) the multiplicative order of q in the group of units U (Z 2p n ). Notice that o(q) = ϕ(p n ) if and only if q is a generator in U (Z 2p n ).
Given a field K and a subgroup H of a group G such that gcd(char(K), |H|) = 1, the element is an idempotent of KG known as the idempotent determined by H. Ferraz and Polcino Milies proved the following.  Since G is a cyclic group of order p n its chain of subgroups is: where G i is the unique subgroup of G of order p n−i and the primitive idempotents of KG are In what follows, we shall always assume that o(q) = ϕ(p n ) in U (Z 2p n ), just to be able to consider these codes as minimal. In the general case, they are not necessarily minimal but are still a significant family of codes.
The proof of the following result is straightforward. The minimum weight of a code of the form KA 1±d 2 e i was computed in [5]. A very similar result holds over chain rings, but requires new arguments. Then Proof. We consider first the case of an index i with 0 < i ≤ n. Let T = {τ 1 , . . . , τ n } be a transversal of G i in G; i.e., a full set of representatives of cosets of G i in G. As we have that C ⊂ RA a k 1±d 2 G i . Thus, an element α = 0 ∈ C, is of the form Finally, consider a code of the form C = RA a k 1±d 2 e 0 and take 0 = α ∈ C. We can write α in the form As a k r g ∈ R and w(e 0 ) = |A|, it follows that w(C) = 2|G| = |A|.

Non minimal codes
A non minimal code is the direct sum of codes of the form RA a ki 1±d 2 e i . We shall consider first the case when all the idempotents involved are of the form a ki ( 1+d 2 )e i . The case when e 0 is not involved is simpler. To simplify notations, we shall denote an ideal of the form RA a ki 1±d 2 e i as a ki 1±d 2 e i . Theorem 3.1. Let C be a code of the form Then If there exist just one coefficient x such that xa k = 0 we write α in the form On the other hand, as Thus, multiplying the equation above by G i1−1 we get When the code involves e 0 there are two different cases. We consider first the case when all the idempotents from e 0 to idempotents e are involved.
Proof. As a k 1+d Then, C = a k ( 1+d 2 ) G . As a k G and a k d G have disjoint supports, we have that: Now we consider codes involving e 0 but different from the above.
Proof. First, note that w(C) ≤ w(a ki ( 1+d 2 )e i l ) = 4|G i |. As G i ⊂ G j , for j < i , we have G i .e j = e j . Thus, as above, Let T = {τ 1 , . . . , τ n } be a transversal of G i in G and take α ∈ C, α = 0. We can write α = ( Assume now, by way of contradiction, that there exists only one coefficiente x of α such that xa k = 0. Then α = xτ a k ( 1+d 2 ) G i , with τ ∈ T . As there exist β 0 , β 1 , . . . , β ∈ RA, such that As {i 1 , . . . , i } {1, . . . , i }, there exists an index i r ∈ {1, . . . , i } such that the idempotent ( 1+d 2 )e ir is not in the initial sum. Let i r be the smallest such index. Multiplying both sides of the equality above by G ir , we get Notice that the ideal is as in Theorem 3.3 so its minimum weight is w(J) = 2|G ir−1 | = 2p n−ir+1 and α G ir ∈ J.
Thus there exist at least two coefficients x r , x s such that x r a k and x s a k are non zero, so w(C) ≥ 4|G i | and thus w(C) = 4|G i |.
The case when all the idempotents involved are of the form a ki ( 1−d 2 )e i is not different. Now, we will compute the minimum weight of codes that involve ideals of both forms. As before, we study first codes not involving e 0 .
Theorem 3.4. Let C be a code of the form where 0 ≤ k ij < t, 1 ≤ j ≤ , 0 < i 1 ≤ i 2 ≤ . . . ≤ i −1 < i , and assume that at least one idempotent of the form ( 1+d 2 )e i and one idempotent of the form ( 1−d 2 )e j are involved in C.
Proof. Note that w(C) ≤ w( a ki l ( 1±d 2 )e i ) = 4|G i |. Also, as in the previous results, we have that C ⊂ a k G i , where k = min {k i1 , . . . , k i }. Let T = {τ 1 , . . . , τ n } be a transversal of G i in G and set α ∈ C, α = 0. Then we can write Assume first, by way of contradiction, that there exists only one coefficient x in the expression of α such that xa k = 0, so that α = xa k d δ τ G i , where δ is equal to either 0 or 1, and τ ∈ T .
As α ∈ C, there exist β i1 , . . . , β i ∈ RA such that and we can write Multiplying the equality above by G i1−1 , we get Thus, xa k = 0, a contradiction. Assume now that there exist precisely two coefficients x 1 , x 2 in the expression of α such that x 1 a k = 0 and x 2 a k = 0. Then, where both of δ and δ are equal to either 0 or 1, and τ, τ ∈ T . Suppose that the last idempotent in this expression is ( 1+d 2 )e i , the other possibility being similar. As ( 1+d 2 )( 1−d 2 ) = 0, multiplying by ( 1−d 2 ) we get where j s < i . This shows that α( 1−d 2 ) belongs to code which is as in the previous theorem.
A very similar argument shows that if there are precisely three coefficients x 1 , x 2 , x 3 such that x 1 a k , x 2 a k and x 1 a k are non zero, we get again a contradiction.
To complete the study of ideals not involving e 0 we need to consider first a particular case.
We are now ready to complete this case.
Theorem 3.6. Let C be a code of the form Then w(C) = 2|G i |.
Proof. By the previous lemma, we have Let T = {τ 1 , . . . , τ n } be a transversal of G i in G and α ∈ C, α = 0.
We can write Suppose, again, that there exist only one coefficient x in the expression above such that xa k = 0.
As α ∈ C, there exist β i1 , . . . , β 1 , β 2 ∈ RA such that Then Multiplying the above equality by G i1−1 , we get xa k d δ τ G i1−1 = 0. Thus xa k = 0, a contradiction. Therefore w(C) ≥ 2|G i | and thus w(C) = 2|G i | Now we consider ideals involving e 0 . As a first step, we will assume that if j is the greatest subindex such that e j is involved, then 1+d 2 e j is involved, say, but 1−d 2 e j is not, the symmetric case being identical.
Theorem 3.7. Let C be a code of the form where 0 ≤ k i , h i ≤ t, for 0 < i < j, k j < t and k 0 or h 0 < t. Then w(C) = 4|G j |.
Proof. Note that w(C) ≤ 4|G j | and C ⊂ a k G j , where k is the minimal non zero exponent of a. Let T = {τ 1 , . . . , τ n } a transversal of G j in G and α ∈ C, α = 0. We can write α = (x 1 τ 1 + . . . + x n τ n + x 1 dτ 1 + . . . + x n dτ n )a k G j Using exactly the same technique as in Theorem 4.4, we can show that there are at least four coefficients whose product with a k is non zero.
Finally, we consider the case when the code C involves e 0 and both 1+d 2 e j and Theorem 3.8. Let C be a code of the form where at least either k 0 or h 0 is less than t and k j < t and h j < t. Then w(C) = |G j | or w(C) = 2|G j |.
Proof. We shall assume first that all idempotents of the form 1±d 2 e i are involved in C, for 0 ≤ i ≤ j; i.e., that k i , h i < t, for 0 ≤ i ≤ j.
As before, we have that C ⊂ G j . Therefore, w(C) ≥ w( G j ) = |G j |.
As in the proof of Theorem 4.2, one can show that a k G j ⊂ C so actually w(C) = |G j |. Now, assume that there exists an index r such that either k r or h r is equal to t. Note that a kj 1+d 2 e j ⊕ a hj 1−d 2 e j ⊂ C and thus, by Lemma 3.5 we have w(C) ≤ 2|G j |.
If k denotes the minimum of all exponents of a, we have as always that C ⊂< a k G j > and, taking a transversal T = {τ 1 , . . . , τ n } of G j in G, and element α ∈ C, α = 0, we can write One can show, as in the proof of Theorem 4.3, that there exist at least two coefficients in this expression above, whose product by a k is non zero. Hence w(α) ≥ 2|G j | and, in this case, w(C) = 2|G j |.

The number of words in a code
Since most of the codes over chain rings are not free, we cannot compute dimensions and is then relevant to find the number of words in each code. We begin with a simple case.
Proof. Since e 0 = G we have that RAa k 1±d 2 e 0 = Ra k e 0 , so When i > 0, where e i 1+d 2 Notice that As Then

Free cyclic codes over finite chain rings of lenght 2p n
Let A be a cyclic group of order 2p n , R a chain ring with gcd(| R |, | A |) = 1 and orthogonal primitive idempotents ( 1±d 2 )e i , 0 ≤ i ≤ n. Theorem 5.1. Let γ be a transversal of G i−1 in G and τ a transversal of G i in G i−1 . Then RA( 1±d 2 )e i is a free code with basis over R, where the positive sign in the base elements refers to the ideal RA( 1+d 2 )e i and the negative sign to RA( 1−d 2 )e i .
Proof. We will prove that the code RA( 1+d 2 )e i is free. The proof that RA( 1−d 2 )e i is free being similar. First we show that elements in B belong to the code.
For b ∈ τ \ {1}, we have Now, we show that B is linearly independent. Let x cb ∈ R, where c ∈ γ, and b ∈ τ \ {1} be such that c∈γ b∈τ \{1} The elements of the sum ( c∈γ ( b∈τ x cb c G i )− c∈γ b∈τ x cb cb G i ) have supports that are disjoint with the elements of the sum ( c∈γ ( b∈τ x cb dc G i ) − c∈γ b∈τ x cb dcb G i ) . Thus, We prove now that the elements of the sum ( c∈γ ( b∈τ x cb c G i ) − c∈γ b∈τ x cb cb G i ) has disjoint supports. To this end, we prove that, for every c, b fixed, the element cb G i has disjoint support with any other element in this linear combination. As b ∈ τ , then G i and b G i have disjoint supports. Thus c G i and cb G i have disjoint supports. If c j = c k , then c j G i and c k G i have disjoint supports. As τ is a transversal of G i in G i−1 , we have that for b j = b k ∈ τ , c j b j G i and c k b k G i have disjoint supports. Therefore, x cb = 0, ∀c ∈ γ and b ∈ τ .
By Theorem 4.1, the number of elements of RA( 1+d 2 )e i is given by | R | t(p i −p i−1 ) =| R | (p i −p i−1 ) .
A code C = RAa k 1+d 2 e i , with 0 < k < t, is not free, because α · a t−k · a k 1+d 2 e i = 0, for all α ∈ RA.