On the weight structure of cyclic codes over $GF(q)$, $q>2$

TL;DR

This paper explores the weight structure of cyclic codes over GF(q) for q>2, establishing relations with their proportional elements, and determines conditions for equidistance and minimum distance in certain nonprimitive codes.

Contribution

It introduces a new approach linking cyclic code structure with proportional elements and provides new results on equidistance and minimum distance for nonprimitive codes.

Findings

Derived conditions for equidistance of irreducible nonprimitive codes

Identified minimum distances for specific classes of nonprimitive cyclic codes

Established the relation between cyclic structure and weight distribution

Abstract

The interrelation between the cyclic structure of an ideal, i.e., a cyclic code over Galois field $GF(q)$, $q>2$, and its classes of proportional elements is considered. This relation is used in order to define the code's weight structure. The equidistance conditions of irreducible nonprimitive codes over GF(q) are given. Besides that, the minimum distance for some class of nonprimitive cyclic codes is found.

Equations

$${\alpha }_{1}z(x),{\alpha }_{2}z(x),\dots ,{\alpha }_{q-1}z(x)$$

$$J=\sum _{i=1}^t{J}_i,$$

$$h(x)=\prod _{i=1}^t{h}_i(x),$$

$$c(x)=\prod _{j=1}^k{h}_{i_j}(x),1\le k\le t.$$

$$n_c\cdot s_c=R_c\cdot (q-1),$$

$$x^{r_z}\cdot z(x)=\alpha \cdot z(x)mod(x^{n_z}-1),$$

$$z(x),x\cdot z(x),...,x^{r^z-1}\cdot z(x)$$

$$P_{z(x)},P_{xz(x)},\dots ,...,P_{x^{r_z-1}z(x)}.$$

$$d=gcd(q-1,n),r=n/d.$$

$$x^{r_f}\cdot f(x)=\gamma \cdot f(x),$$

$$r_c=gcd(R_c,n_c),d_c=gcd(q-1,n_c).$$

$$n_c=r_1{d}_1\cdot r_2{d}_2/gcd(r_1{d}_1\cdot r_2{d}_2).$$

$$n_c=(r_1{r}_2/gcd(r_1,r_2))\cdot (d_1{d}_2/gcd(d_1,d_2)).$$

$$x^{r_a}\cdot a(x)=\theta a(x),$$

$$d_c=gcd(q-1,n_c),r_c=n_c/d_c.$$

$$x^{r_z}\cdot z(x)=\beta z(x),$$

$$n_c=lcm(n_1,n_2)>(q-1).$$

$$h(x)=\prod _{i=1}^t{h}_i(x),$$

$$d_{min}=q^{m-1}-\sum _{j=1}^{C_t^1}{q}^{m_{j,1}-1}--\sum _{j=1}^{C_t^2}{q}^{m_{j,2}-1}-...--\sum _{j=1}^{C_t^{t-1}}{q}^{m_{j,t-1}-1},t\ge 2,$$

$$d_{min}=q^{m-1}(q-1)/b,t=1,1\le b\le q-1,$$

$$d_{min}=q^{m-1}(q-1),t=1,b=1.$$

Full text

The relation of proportionality for elements of algebra $A_{n}$, consisting of polynomials in $x$ over Galois field $GF(q)$, modulo polynomial ${x^{n}-1}$, is the equivalence relation  [1]. Therefore $A_{n}$ falls into several disjoint subsets and every such subset contains all elements which are proportional to each other. These subsets will be called the classes of proportionality. Let ${z(x)\neq 0}$ be some vector of $A_{n}$. If $\alpha_{1}=1,\alpha_{2},\ldots,\alpha_{q-1}$ are all different elements of the multiplicative group $GF(q)^{*}$ of the field $GF(q)$, then the following $q-1$ vectors

[TABLE]

are some different and proportional to each other elements of $A_{n}$. The set of vectors $(1)$ is closed under the multiplication by the elements of the group $GF(q)^{*}$. Hence the set $(1)$ represents some class of proportional elements, which will be denoted by $P_{z(x)}$. Because of an arbitrary choice for $z(x)$, every nonzero class consists of $q-1$ elements of the form $(1)$. Consequently $A_{n}$ contains $(q^{n}-1)/(q-1)$ different nonzero classes. Evidently all elements of one class have the same period  [2, 3] or the same order [8]. Clearly, the supporting sets [2] of vectors, entering into the same class of proportionality, are similar too. Hence, the Hamming weight is also the same for all vectors of one class. Thus, we can say that any proportionality class  $P_{z(x)}$, $P_{z(x)}\subset A_{n}$, has its order, its supporting set and its Hamming weight. Obviously, any proportionality class of $A_{n}$ is characterizied by its unique monic polynomial.

Now consider an ideal $J$, $J\subset A_{n}$, i.e., some cyclic code over  $GF(q)$, having the following generator $g(x)=(x^{n}-1)/h(x)$ [7], where $h(x)$ is some parity-check polynomial of degree $m$, having some order $n$, $n=ord(h(x))$ [8]. Below we suppose that $q>2$ and $gcd(n,q)=1$.

It’s also known [3, 4, 10] that any ideal is partitioned into several disjoint subsets, that is cycles, under the multiplication of ideal’s vectors by $x$. On the other hand, some ideal $J$, as a subspace of $A_{n}$, consists of $(q^{m}-1)/(q-1)$ nonzero proportionality classes. Obviously, the existence of two different partitions into some disjoint subsets of any ideal assumes a certain dependence between proportionality classes and cycles of ideal.

Further, any ideal $J\subset A_{n}$ is the direct sum of minimal ideals  [2, 8]

[TABLE]

where $J_{i}$ is some minimal ideal, having an irreducible parity-check polynomial $h_{i}(x)$ of degree $m_{i}$ and of order $n_{i}$, $n_{i}=ord(h_{i}(x))$, ${1\leq i\leq t}$. This implies that the following polynomial

[TABLE]

is the parity-check polynomial of $J$ and $n$, $n=lcm(n_{1},n_{2},\ldots,n_{t})$, is the order of $h(x)$ [8]. It should be stressed that under the condition $gcd(n,q)=1$ the polynomial $(3)$ has no repeated factors [8].

Remark $1$. It is worth mentioning that the number $n$ can be some number of either the primitive form $n=q^{m}-1$ or of the nonprimitive form $n\neq q^{m}-1$. In the first case, we have some cyclic primitive code and the second case corresponds to a certain cyclic nonprimitive code [11]. Let us stress that $n\neq q^{m}-1$ if and only if $n_{i}\neq q^{m_{i}}-1$, $1\leq i\leq t$. But if there is at least one primitive polynomial among the polynomials $h_{i}(x)$, $1\leq i\leq t$, then $n=q^{m}-1$.

Furthermore, applying the theory of linear recurring sequences [8, 12] to elements of an ideal, we obtain that every element of some ideal $J\subset A_{n}$ is characterized by its unique minimal polynomial. Denote by  $C$ the set of all elements of  $J$, having the same minimal polynomial $c(x)$. The set $C$ is either some minimal ideal $J_{i}$, $1\leq i\leq t$, or a certain subset of all such elements of $J$, whose characteristic polynomial of the smallest degree coincides with $c(x)$. In the general case, the polynomial $c(x)$ is equal to the product of some $k$, $1\leq k\leq t$, polynomials from $t$ different prime divisors of $h(x)$. Thus,

[TABLE]

This means that any element of $C$ has the same period or the same order $n_{c}=ord(c(x))$, $n_{c}\leq n$, $n_{c}|n$.

Lemma $1$. For the set $C$, $C\subset J$, having some minimal polynomial $c(x)$ in terms of $(4)$, the following equality takes place

[TABLE]

where $s_{c}$ is the number of all cycles, and $R_{c}$ is the number of all proportionality classes of $C$.

Proof. The set $C$ is closed under two different operations. The first operation is the cyclic shift of vector and the second one is the multiplication of vectors by elements of group $GF(q)^{*}$. Hence equality $(5)$ can be obtained by the counting of the number of all elements, belonging to $C$, via the two different ways. The lemma is proved.

Theorem $1$. Any cycle $\{z(x)\}$ of ideal $(2)$, having some period $n_{z}$, $n_{z}|n$, consists of  $d_{z}$ subsets. The first element of each such subset is proportional to $z(x)$. Every such subset contains $r_{z}$ nonproportional to each other vectors, i.e., $n_{z}=r_{z}\cdot d_{z}$, $d_{z}|(q-1)$. And the number $r_{z}$ is the index of the subgroup, belonging to $GF(q)^{*}$, *of order * $d_{z}$ in the group of the roots of unity, having the least possible order.

Proof. Let $r_{z}$, $1\leq r_{z}\leq n_{z}$, is the smallest natural number such that the following equality holds

[TABLE]

where $\alpha$ is some element of $GF(q)^{*}$. Then the following $r_{z}$ vectors of cycle $\{z(x)\}$

[TABLE]

are some non–proportional to each others vectors because, assuming the inverse, we should be able to decrease the number $r_{z}$, but it is impossible. Hence the set of elements $(7)$ belongs to the following $r_{z}$ classes of proportionality

[TABLE]

Since $x^{n_{z}}z(x)=z(x)$ in the ring $A_{n}$ and also, considering  $(6)$, we see that $P_{z(x)}=P_{x^{r_{z}}z(x)}=P_{x^{n_{z}}z(x)}$. This means that the cycle $\{z(x)\}$ belongs to the classes $(8)$ and every such class contains $d_{z}$ vectors, $d_{z}=n_{z}/r_{z}$, $1<d_{z}\leq q-1$. In terms of equality  $(6)$ the following different vectors $z(x),x^{r_{z}}z(x),x^{2r_{z}}z(x),\ldots,x^{(d_{z}-1)r_{z}}z(x)$ of class $P_{z(x)}$ can be represented as ${\alpha^{0}z(x),\alpha z(x),\ldots,\alpha^{d_{z}-1}z(x)}$. This implies that $\alpha^{0}=1,\alpha,\alpha^{2},\ldots,\alpha^{d_{z}-1}$ are the different elements of group $GF(q)^{*}$. Since $x^{n_{z}}z(x)=x^{r_{z}d_{z}}z(x)=\alpha^{d_{z}}z(x)=z(x)$, we see that $\alpha^{d_{z}}=1$. This yields that $d_{z}$ is the order of element $\alpha$ in $GF(q)^{*}$. Consequently, $d_{z}|q-1$, $1\leq d_{z}\leq q-1$.

Finally, under the condition $gcd(n,q)=1$ the polynomial $x^{n}-1$ has  $n$ different roots in $GF(q^{h})$ field, where $h$ is the multiplicative order of $q$ modulo $n$ [8]. Denote by $E(n)$ the multiplicative group of $n$- th roots of unity over $GF(q)$. Let $\xi\in E(n)$ be some  $n$- th primitive root of unity. Then the following set of elements $\xi^{0}=1,\xi,\xi^{2},\ldots,\xi^{n_{z}-1},\ldots,\xi^{n-1}$ represents the group $E(n)$. Since $n_{z}|n$, we have $E(n_{z})\subset E(n)$, where $E(n_{z})$ is the multiplicative group of $n_{z}$-th roots of unity. Moreover, taking into account the isomorphism of the groups, having the same order  [9], we can state that the subgroup $\{\alpha\}$, $\{\alpha\}\subset GF(q)^{*}$, of order $d_{z}$ belongs to  $E(n_{z})$ because $d_{z}|n_{z}$.

As mentioned above, $n_{z}$ is the period of $z(x)$, so that $n_{z}$ is the smallest divisor of  $n$ such that the following congruence $x^{r_{z}}\equiv\alpha\,{mod}(x^{n_{z}}-1)$ takes place. Hence $E(n_{z})$ is the smallest group of  $n$- th roots of unity, which contains $\{\alpha\}$. Since $\alpha=\xi^{r_{z}}$, we see that the following elements $\xi^{r_{z}d_{z}}=1,\xi^{r_{z}},\xi^{2r_{z}},\ldots,\xi^{(d_{z}-1)r_{z}}$ represent the subgroup  $\{\alpha\}$ in the group $E(n_{z})$. Besides that, the decomposition of $E(n_{z})$ relative to the subgroup $\{\alpha\}$ consists of $r_{z}$ different cosets. Thus, the number $r_{z}$ is the index of subgroup $\{\alpha\}$ in the group of roots of unity, having the smallest possible order. The theorem is proved.

Remark $2$. Notice that when a parity-check polynomial of code is some primitive polynomial of degree $m$ and of order $n=q^{m}-1$, $m>1$, then $r=R={(q^{m}-1)/(q-1)}$ and $d=q-1$. (see [2], [7]).

Corrollary $1$. The period $n_{z}$, $n_{z}=r_{z}\cdot d_{z}$,of element $z(x)$, ${z(x)\in J}$, equals $d_{z}$, $d_{z}|(q-1)$, $1\leq d_{z}\leq(q-1)$,if and only if the cycle $\{z(x)\}$ is contained in one class of proportionality.

Corollary $2$. All code words of any cyclic code, havinq some length $n$ over $GF(q)$, fall into some equal-weight subsets and every such subset includes all proportional to each other cycles.

Besides that, consider some minimal ideal $J$, $J\subset A_{n}$, having an irreducible nonprimitive parity-check polynomial $h(x)$ of degree $m$ and of order $n$, $n\neq q^{m}-1$, i.e., some irreducible code of nonprimitive length.

Remark $3$. The degree $m$ of the polynomial $h(x)$ coincides with the multiplicative order $h$ of the number $q$ modulo $n$ [8]. Also, the order $n$ of the polynomial $h(x)$ is some divisor of $q^{h}-1$. This means that the order $n$ can change in the following limits $1<n<q^{h}-1$, ${n\neq q-1}$. If ${1<n<q-1}$, i.e., ${n\neq q-1}$, then some minimal ideal $J$, ${J\subset A_{n}}$, of dimention one contains only one nonzero class of proportyonality. Consequently, $n=d$, $1<d<q-1$, $d|q-1$.

Theorem $2$. Any cycle of minimal ideal $J$, $J\subset A_{n}$, having some parity-check polynomial $h(x)$ of degree $m$, $m>1$, and of order $n$, $n\neq q^{m}-1$, is contained in $r$ proportionality classes, $1\leq r\leq R$, $r|R$, $R=(q^{m}-1)/(q-1)$. Every such class consists of $d$, $1\leq d\leq(q-1)$, different vectors of cycle, and

[TABLE]

Proof. All elements of minimal ideal $J$, $J\subset A_{n}$, have the same order $n$, $n=ord(h(x))$. Applying the theorem $1$ to some element $f(x)$, $f(x)\in J$, we have $n=r_{f}\cdot d_{f}$, $d_{f}|(q-1)$, $1\leq r_{f}\leq n$, $1\leq d_{f}\leq(q-1)$, and also the following equality

[TABLE]

where $\gamma$ is some element of $GF(q)^{*}$. Evidently, if either $r_{f}=1$, i.e., $d_{f}=n$, or $r_{f}=n$ and $d_{f}=1$, then equalities (9) take place. Hence, below we suppose that $1<r_{f}<n$, $1<d_{f}<q-1$, and therefore $q-1<n<q^{m}-1$.

Taking into account $(9)$, we have $d_{f}\leq d$. Let us show that the strong unequality $d_{f}<d$ is impossible. Indeed, if $d_{f}<d$, then the subgroup $\{\gamma\}$, where $\gamma$ is the element from equality $(10)$, belongs to some group of $n$-th roots of unity, having the order $d$, because $d_{f}|d$. Since $d|(q-1)$, we see that the subgroup $\{\gamma\}$ belongs to $GF(q)^{*}$. This means that the cycle $\{f(x)\}$ is contained in one class of proportionality, i. e., $r_{f}=1$. But this fact contradicts to the condition $r_{f}>1$. This implies that the strong unequality $d_{f}<d$ is impossible. Hence $d_{f}=d$. Because of an arbitrary choice of $f(x)$ we can conclude that equalities $(9)$ take place for any element of $J$. The theorem is proved [6].

Remark $4$. It is necessary to note that the theorem $2$ is valid only for irreducible codes of non–primitive length except Reed-Solomon codes of length $n=q-1$ as it was shown above. In the case of irreducible codes of primitive length $n$, $n=q^{m}-1$, $m>1$, the theorem $2$ will be valid if and only if $gcd(m,q-1)=1$. Indeed, when the last condition takes place, then ${gcd((q^{m}-1)/(q-1),q-1)=1}$ [7]. Thus, $d=q-1={gcd(q^{m}-1,q-1)}$ that is $d=gcd(n,q-1)$. It follows that the theorem $2$ holds.

Remark $5$. Notice that under the condition ${gcd(m,q-1)=1}$ the number $r$ from $(9)$ has no divisors of $(q-1)$ except $1$, so ${gcd(r,q-1)=1}$. This means that $gcd(r,d)=1$. Hence, considering the fact that $r|R$ and also, taking into account $(9)$ and the following equality $s=R(q-1)/n$ , we have $r=gcd(R,n)$. Besides that, if either one from the two numbers $n$ and $q-1$ does not contain multiple prime divisors or the same prime divisors of these numbers have the same degrees under the decomposition of both $n$ and $q-1$, then the following equalities $r=gcd(R,n)$, $gcd(r,d)=1$ also take place.

Corollary $3$. Both the number $r$ and the number $d$ are the same numbers of all irreducible divisors of polynomial $x^{n}-1$ over $GF(q)$, having the same order.

Corollary $4$. The number $R$, $R=(q^{m}-1)/(q-1)$, of proportionality classes, of some irreducible code $K$,having some length $n$, $n\neq(q^{m}-1)$, $n=d\cdot r$, over $GF(q)$ field, consists of some $v$ different subsets. And every such subset contains $r$ equal-weight proportionality classes, i.e., $R={v\cdot r}$. Besides that, every subset includes $b$ equal-weight cycles, $b=(q-1)/d$, $1<b\leq(q-1)$. So that the number of all cycles for $K$ equals $s=v\cdot b$ and $gcd(r,b)=1$.

Proof. According to the theorem $2$ any cycle of code $K$ is contained in $r$, $r|R$, proportionality classes. Therefore the number $v=R/r$ gives us the common quantity of different subsets of $J$, each of which consists of $r$ classes, i.e., $R=v\cdot r$. The number $b$, $b=(q-1)/d$, is the number of all different equal-weight cycles, contained in every such subset, which consists of some $r$ classes. Hence the number of all cycles for $K$ is equal to $s=v\cdot b$. Since $d=gcd(q-1,n)$ we see that $gcd(r,b)=1$. Actually, assuming the inverse, we would have been able to decrease the number $d$, but it’s impossible. The corollary is proved.

Corollary $5$. The irreducible nonprimitive code $K$ is some equidistant code if $s=b$.Besides that, the last equation is equivalent to the following ones: $r=R$ or $gcd(s,R)=1$.

Remark $6$. Note that the condition $s=b$ was obtained in  [14, 15], but only for some subclass of irreducible nonprimitive codes and under the following additional restriction $gcd(b,m)=1$.

Corollary $6$. The weight of any element, belonging to some irreducible nonprimitive code $K$ of length $n$ over $GF(q)$, is multiple of the number $d$, $d=gcd(q-1,n)$.

Proof. The weight of any element $z(x)$, $z(x)\in K$, of order $n$, $n=ord(h(x))$, is equal to the number of such $j$, $0\leq j\leq n-1$, for which the polynomial ${x^{j}\cdot z(x)}$ has the following degree $n-1$. According to the theorem $2$, the number of such polynomials for the cycle ${z(x)}$, having degree $n-1$, is equal to $w_{r}\cdot d$, where $w_{r}$ is the number of polynomials, having degree $n-1$, among the first $r$ cyclic shifts of $z(x)$, and $d=gcd(q-1,n)$. The corollary is proved.

In addition, consider some ideal $J$, $J\subset A_{n}$, of the form $(2)$, having the parity-check polynomial $h(x)$ in terms of $(3)$.

Theorem $3$.If the following condition $gcd(h,q-1)=1$, where $h$ is the multiplicative order of number $q~{}\mod~{}n$, takes place, then any cycle of set $C$, $C\subset J$, having some minimal polynomial of the form $(4)$, is contained in $r_{c}$ proportionality classes, $r_{c}|R_{c}$, and every such class includes $d_{c}$, $d_{c}|q-1$, elements of cycle, that is $n_{c}=r_{c}\cdot d_{c}$, $n_{c}={ord}(c(x))$, where $R_{c}$  is the number of all proportionality classes of set $C$, and

[TABLE]

Proof. It is sufficient to consider the case $k=2$ because the general case can be obtained by the induction. Thus assume that $c(x)=h_{1}(x)\cdot h_{2}(x)$, where $h_{i}(x)$ is of degree $m_{i}$ and of order $n_{i}$, $n_{i}=q^{m_{i}}-1$, $1\leq i\leq 2$, is a certain prime miltiplier of $c(x)$. It is known [8], that the number $m_{i}$, $1\leq i\leq 2$, equals either $h$ or some divisor of this number. Hence, taking into account the theorem $2$, and also remarks $4$ and $5$, we have $n_{i}=r_{i}\cdot d_{i}$, where $r_{i}=gcd(R_{i},n_{i})$, $d_{i}=gcd(q-1,n_{i})$, $gcd(r_{i},q-1)=1$, $1\leq r_{i}\leq R_{i}$, $1\leq d_{i}\leq q-1$, and $R_{i}={(q^{m_{i}}-1)/(q-1)}$, where $R_{i}$ is the number of all proportionality classes of minimal ideal $J_{i}$, $1\leq i\leq 2$. Therefore the order $n_{c}$, $n_{c}=lcm(n_{1},n_{2})=n_{1}\cdot n_{2}/gcd(n_{1},n_{2})$ of the polynomial $c(x)$ can be rewritten as

[TABLE]

Since $gcd(r_{i},q-1)=1$, we have $gcd(r_{i},d_{1}\cdot d_{2})=1$, ${1\leq i\leq 2}$. Thus $gcd(r_{1}r_{2},d_{1}d_{2})=1$. Hence $gcd(gcd(r_{1},r_{2}),gcd(d_{1},d_{2}))=1$ so that $(12)$ may be represented in the following form

[TABLE]

Thus, $n_{c}=lcm(r_{1},r_{2})\cdot lcm(d_{1},d_{2})$. Now by $r_{c}$ and $d_{c}$ we denote $lcm(r_{1},r_{2})$ and $lcm(d_{1},d_{2})$, respectively. Thus $n_{c}=r_{c}\cdot d_{c}$ and $gcd(r_{c},d_{c})=1$. Since $d_{c}|q-1$ and $gcd(r_{c},q-1)=1$, we obtain $d_{c}=gcd(q-1,n_{c})$. Considering $(5)$, it follows that $n_{c}|(R_{c}\cdot(q-1))$. Hence we have $r_{c}=gcd(R_{c},n_{c})$. Consequently the order $n_{c}$ of any element of set $C$ is equal to the product of two relatively prime numbers, i.e., $n_{c}=r_{c}\cdot d_{c}$, where $r_{c}=lcm(r_{1},r_{2})=gcd(R_{c},n_{c})$ and $d_{c}=lcm(d_{1},d_{2})=gcd(q-1,n_{c})$.

Furthermore, applying the theorem $1$ to some element $a(x)\in C$ of period $n_{c}=r_{a}\cdot d_{a}$, we have

[TABLE]

where $\theta\in GF(q)^{*}$ is some element of order $d_{a}$, and $r_{a}$ is the smallest natural number such that equality $(14)$ takes place. Notice that the subgroup $\{\theta\}$ has the order $d_{a}$, ${d_{a}<d_{c}}$. If $d_{c}=1$, then $d_{a}=1$ and $n_{c}=r_{a}=r_{c}=gcd(R_{c},n_{c})$, so that equalities $(11)$ hold. For this reason below we suppose that $d_{c}>1$. If under this condition the number $r_{c}$ is equal to one, then $n_{c}=d_{c}$=$gcd(q-1,n_{c})$ and the theorem is valid. Therefore below we suppose that both $r_{c}>1$ and $d_{c}>1$.

Evidently, $d_{a}\leq d_{c}$. Now let us show that the inequality $d_{a}<d_{c}$ is not possible. Indeed, if $d_{a}<d_{c}$, then we come to the following conclusion. The subgroup $\{\theta\}$, where $\theta$ is the element from equality $(14)$, belongs to some subgroup of $GF(q)^{*}$, having the order $d_{c}$, because $d_{a}|d_{c}$. Since $d_{c}$ is some divisor of $n_{c}$, then, considering the uniqueness of subgroups, having the same order, the subgroup $\{\theta\}$ belongs to some group of $d_{c}$-th roots of unity. This implies that the smallest group of roots of unit, containing $\{\theta\}$, has an order, which either less or equals $d_{c}$. Thus, both the period of $a(x)$ and the order of  $c(x)$ must be either less or equal to $d_{c}$. This yields that the order of $c(x)$ must be some divisor of $q-1$. But this fact contradicts the condition $r_{c}>1$. Hence the inequality $d_{a}<d_{c}$ is impossible so that $d_{a}=d_{c}$ and $r_{a}=r_{c}$. Because of an arbitrary choice of $a(x)$ equalities $(11)$ take place for any element of set $C$. The theorem is proved.

Corollary $7$. The order $n_{c}$ of reducible factor $c(x)$ of polynomial $x^{n}-1$, having some degree $m$ over $GF(q)$, is some divisor of number $q^{m}-1$, if ${gcd(h,q-1)=1}$, where $h$ is the multiplicative order of number $q~{}\mod~{}n$.

Remark $7$. In terms of condition ${gcd(h,q-1)=1}$, where $h$ is the multiplicative order of number $q~{}\mod~{}n$, the theorem $3$ is valid for cyclic codes of both the primitive and the nonprimitive length. Also, taking into consideration the remark $5$, the order of any reducible factor of the polynomial $(x^{n}-1)$ over $GF(q)$ of degree $m$, is some divisor of the number $(q^{m}-1)$, if gcd${(h,q-1)=1}$.

Theorem $4$. Any cycle of set $C$, $C\subset J$, having some minimal polynomial $c(x)$ of the type $(4)$ and of order $n_{c}=lcm(n_{1},n_{2},\ldots,n_{k})$, where $n_{i}\neq q^{m_{i}}-1$, $1\leq i\leq k$, is contained in $r_{c}$, $r_{c}|R_{c}$, $1\leq r_{c}\leq R_{c}$, classes of proportionality and every such class includes $d_{c}$, $1\leq d_{c}\leq q-1$, elements of cycle, where $R_{c}$ is the number of all proportionality classes of $C$, and

[TABLE]

Proof. It is sufficient to assume that $k=2$ because the general case can be obtained by the induction. This implies that $c(x)=h_{1}(x)h_{2}(x)$, where $h_{i}(x)$ is some prime divisor of equality $(3)$, having some degree $m_{i}$, and of order $n_{i}$, ${n_{i}\neq q^{m_{i}}-1}$, $1\leq i\leq 2$. This yields that the theorem $2$ holds for the minimal ideal $J_{i}$, $1\leq i\leq 2$.

Evidently, if one of the two numbers, i.e., either $d_{c}$ or $r_{c}$ is equal to one, then equalities $(15)$ hold. For this reason below we assume that both $r_{c}>1$ and $d_{c}>1$.

Let $z(x)$ be some vector of set $C$. According to the theorem $1$ some cycle $\{z(x)\}\subset C$ is contained in $r_{z}$ classes and every such class includes $d_{z}$ different elements of this cycle. Thus $n_{c}=r_{z}~{}d_{z}$, $1\leq d_{z}\leq q-1$, $d_{z}|q-1$, $1\leq r_{z}\leq n_{c}$. And in addition, the following equality takes place

[TABLE]

where $\beta$ is some element of $GF(q)^{*}$ and the subgroup $\{\beta\}$, $\{\beta\}\subset GF(q)^{*}$, has the order $d_{z}$. Evidently, $d_{z}\leq d_{c}$. Let us show that the number $d_{z}$ can not be smaller than $d_{c}$. Assume the inverse, i.e. let $d_{z}$ be less than $d_{c}$. Since at least one of the two numbers ether $r_{1}$ or $r_{2}$ is not equal to one, we see that at least one of the numbers $n_{i}$, $1\leq i\leq 2$, is more than $q-1$, as was established in the theorem 2. This means that

[TABLE]

Further, since $d_{z}|d_{c}$, we see that the subgroup  $\{\beta\}$ of order $d_{z}$ belongs to some subgroup of $GF(q)^{*}$, having the order $d_{c}$, where $\beta$ is the element from equality $(16)$. Due to the uniqueness of groups, having the same order, the subgroup  $\{\beta\}$ belongs to the group of $d_{c}$-th roots of unity because $d_{c}|n_{c}$. It follows that the smallest group of the roots of unity, which contains the subgroup $\{\beta\}$, has the order less or equals to $d_{c}$. This implies that both the period of $z(x)$ and the order of $c(x)$ is some divisor of $q-1$. But this fact contradicts to $(17)$. It follows that our assumption is not true, so that $d_{z}=d_{c}$, and $r_{z}=r_{c}$.

Besides that, since the number $d_{c}$ is the same number for every cycle of set $C$, we see that every subset, consisting of $r_{c}$ proportionality classes, contains the same number of cycles, which is equal to $b_{c}=q-1/d_{c}$. Moreover, since $d_{c}=gcd(q-1,n_{c})$, we obtain $gcd(b_{c},r_{c})=1$. Hence, taking into account the following equality $s_{c}=R_{c}(q-1)/n_{c}$, which follows from $(5)$, we see that $r_{c}$ is some divisor of $R_{c}$ and $s_{c}=v_{c}b_{c}$, where $v_{c}$ is equal to $R_{c}/r_{c}$. The theorem is proved.

Corollary $7$. (Equidistance signs of the subset $C$, $C\subset J$, having some minimal polynomial of the form $(4)$.)

All vectors of the subset $C$, $C\subset J$, having some minimal polynomial $c(x)$ *of order * $n_{c}=lcm(n_{1},n_{2},...,n_{k})$, $n_{i}\neq q^{m_{i}}-1$, $1\leq i\leq k$, $k>1$, have the same weight if at least one of the following conditions holds: $1.$ $s_{c}=b_{c}$, $1<b_{c}\leq q-1$, $2.$ $r_{c}=R_{c}$, $3.$ $gcd(s_{c},R_{c})=1$.

Corollary $8$. The order of the reducible factor  $c(x)$ of the polynomial $x^{n}-1$ over $GF(q)$, having some degree $m$, is some divisor of the number $q^{m}-1$, if the decomposition of the polynomial $c(x)$ into prime multiples does not contain any primitive polynomial.

Remark $8$. Note that corollaries $(6)$ and $(8)$ show us in what cases corollary $3.4$ from [8] takes place for some reducible polynomial of the degree $m$ over $GF(q)$.

Also, consider some cyclic code, having the following parity-check polynomial

[TABLE]

where $h_{i}(x)$ is an irreducible polynomial over $GF(q)$, $q>2$, of degree $m_{i}$ and of order $n_{i}=(q^{m_{i}}-1)/b_{i}$, $b_{i}|q-1$, $1\leq b_{i}\leq q-1$, $1\leq i\leq t$, $gcd(m_{i},m_{j})=1$ and $gcd(b_{i},b_{j})=1$, provided $i\neq j$, $1\leq i,j\leq t$, so that the order $n$ of $h(x)$ equals $n=lcm(n_{1},n_{2},...,n_{t})$.

It is worth mentioning that in [14] and [15] the following cases of polynomial $(18)$ have been considered. Namely $t=1$ and $t=2$, and besides that, with some additional restrictions, which can be omitted. Also, some particular case of polynomial $(18)$, that is provided $b_{i}=1$ for all $i$, $1\leq i\leq t$, was obtained in [16]. But there are some unnecessary restrictions in this paper too. Also, there are some mistakes in that paper. Namely, the order of the product for some two polynomials from $(18)$ was defined incorrectly in  [16].

Finally, using the results obtained above, we have found the minimal distance of code, having the parity-check polynomial $(18)$, (see [17]).

Denote by $M$ the set of degrees for the polynomials $h_{i}(x)$, $1\leq i\leq t$, from the eq. $(18)$ that is $M=(m_{1},m_{2},...,m_{t})$. Let the number $m$ denote the degree of polynomial $h(x)$, i. e., $m=m_{1}+m_{2}+...+m_{t}$. By $M_{j,k}$, $1\leq j\leq C_{t}^{k}$, $1\leq k\leq t-1$, denote $j^{,}s$ $k$-subset of $M$, where $C_{t}^{k}$ is the binomial coefficient.

Thus $M_{j,k}=(m_{j_{1}},m_{j_{2}},...,m_{j_{k}})$, $1\leq j\leq C_{t}^{k}$, $1\leq k\leq t-1$. At last, denote by $m_{j,k}$ the sum of degrees, from the subset $M_{j,k}$, so that $m_{j,k}=m_{j_{1}}+m_{j_{2}}+...+m_{j_{k}}$, $1\leq j\leq C_{t}^{k}$, $1\leq k\leq t-1$. It is obvious that $1\leq m_{j_{k}}<m$. Let us remark that provided $k=1$ the number $m_{j,k}=m_{j,1}=m_{j}$ and $1\leq j\leq t$ because $C_{t}^{1}=t$.

Theorem $5$.The minimal distance of code, having the parity-check polynomial $(18)$, has the following form

[TABLE]

[TABLE]

[TABLE]

In conclusion I should like to express my sincere gratitude to L.A.Bassaligo, M.I. Boguslavskii and E.T. Akhmedov for helping me to correct some mistakes in the original version of my paper.