IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 41, NO.2, MARCH 1995 387
The q-ary Image of a qm-ary Cyclic Code Gerald E. Seguin
Abstract-For (n, q) = 1 V a qm_ary cyclic code of length n and with generator polynomial g(.r I, we show that there exists a basis for Fq= over Fq with respect to which the q-ary image of V is cyclic, if and only if: i) g(J!) is over Fq; or iiJ g(:;) =
go (.r)(." - ;-'11'1, go (;r) is over Fq, Fq # Fqk = F'1(-y) C Fqm, jI an integer modulo k, and w'" - r has a dhisor over Fq' of
degree f = III / k; or iii) g(.e i = flo (.r) TI"Es (.1' - -, -q" i. flo (J!) is over Fq, Fq # Fqk = Fqb) c Fq=, S a set or integers modulo k of cardinality �. - 1 andle'" - ; has a divisor over Fqk of degree (' = TIl / k. In aU of the above cases, we determine all of the bases with respect to which the q-ary image of F is cyclic.
LET Q = (ao, 0:]," ', Om- I ) be a basis (ordered) for Fq� over Fq and define the mapping da from Fqm [z]/(z" -1)
into Fq[z]/(znm - 1) by -
where
and
m-ln-l d�(a(z) = L Lai jZmHi
i=O j=O
n-l a(z) = Lajzj
i=O
",-1 nj = LILi,jO;,
i=O
( 1)
(2)
It then follows that d� is a bijective q-ary linear (i .e., linear over F q) map and moreover
(3)
If V is a q"'-ary [n, k] cyclic code, then its q-ary image with respect to the basis Q'. is d�(V) where
d�(V) = {d�(IL(z»la(z) E V}. (4)
Manuscript received April 25, 1993; revised April 2, 1994. Part of this work was done in the Summer uf 1991 while the author was at the Universite de Toulon, Toulon, France, supported by a research fellowship from the French government. This work was also partly funded by the Natural
It follows that d� (V) is a q-ary [nm, km]linear code invariant under multiplication by zm; hence d�(V) is a q-ary quasicyclic code [1]-[3]. The problem we address in this paper is that of determining all pairs (Q, V), V a qTn_ary cyclic code and Q a basis for Fqm over Fq, for which d�(V) is cyclic. This problem was first addressed by Hanan and Palermo in 1964[4J and then in 1970 by MacWilliams [51. These authors restricted themselves to the ca'>e q = 2. The theorems they obtained are difficult to apply and so do not lead, in the general case, to a simple characterization of the qm_ary cyclic codes which have a cyclic q-ary image. However, for a polynomial basis (�i = (Wi, 0 :::; i < m, MacWilliams completely characterized all the 2m._ary cyclic codes which have a cyclic binary image. This latter result was subsequently generalized to qm_ary cyclic codes by Rabizzoni [6], For the special case m = 2, and a fixed basis Q, Mouaha [7] determined all the qm_ary cyclic codes having a cyclic q-ary image with respect to Q. See also Wolfmann [81 and [91, [10].
More recently, Leonard [18] has presented a solution to the problem being studied in this paper. As it turns out, this solution is incorrect [ 191; Examples 1 and 2 of the present paper being counterexamples to Leonard's theorem.
In this paper, under the only restriction (n, q) = 1, we give a very simple characterization of all the cylic codes V for which there exists a basis Q such that do (V) is cyclic. Our method involves two essential steps: first. decompsoing d�(V) as a direct sum of primary modules and second, using Jensen's concatenation map [ 11] to represent the primary components of d.Q:.(V)'
II. THE PRIMARY COMPONENTS OF A CYCLIC CODE
Mathematically, it will be more convenient to work, not with da(V), but with a permuted version of do(V). To this end, we introduce the permutation 'if considered-as a function from Fq[z]/(zmn - 1) into the external direct product Am where A = Fq[x]/(xn -1). The elements of Am are m-tuples over A, i.e., objects of the form (no(x), (l1(X),"', (lm-l(X»), where o,i(:];) E A. The elements of Am may also be written polynomially as
m-l m-l n-l L(],i(:D):t/ = L Lai,jxjyi i=O i=O j=O
(5)
Science and Engineering Research Council uf Canada under a Research Grant OGPD006288, held by the author at L 'Ecole Poly technique de Montreal. The where material in this paper was presented at FUROCODE, Udine, Italy, October 1992.
The author is with the Department uf Eleclrical and Computer Engineering, Royal Military College, Kingston, Ont. K7K 5LO, Canada.
388 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 41, :.10. 2, MARCH 1995
We define 7f by setting as
lnj+i j i (m-1 n-1 ) Tn-1 n-1 7f t; f;ai,jz = t; f;ai,jX y.
For example, if q = 2, m = 2, and n = 3, then
11"(100011) = (101001).
Now Am is an A-module [21 if we define m-l m-l
a(x) L a; (X)yi = L a(x)ai(:r;)yi i=O
(7)
(8)
where in (8), a(x)a.i(x) is computed in A. Similarly, Fq[zJl(zmn - 1) is an A-module if we set
is a submodule of Am then its order ord (M) is the nonzero monic polynomial f(x) of least degree such that f(x)g = Q for every gEM. It readily follows that if
(16)
is a direct sum of the submodules Mi, I � .i, � s, then
. (\7)
If h(x) = ord(M) and hex) = h1(:I;)hdx) .. ·ht(x) is its prime factorization (here and hereafter we assume that (n, q) = 1 so that hex) has no repeated factors), then
M - t hex) M - EEl1 hi(x) (18)
a(x)b(z) = a(zm)b(z) (9) and
for a(x) E A, b(z) E Fq[zl/(zrrm - 1). Since
7f(xb(z» = 7f(zmb(z» = x7f(b(z)) (10)
it follows that 7f is a bijective module homomorphism. Moreover, we observe that if
then
7f(z a (z» = (Xam_1(X), ao(x),"', am-2(x» = r(g). (12)
where (12) defines r. Consequently, d9'.(V) is cyclic if and only if 7f(rl,,(V» is invariant under the action of r. Finally, we define the mapping D9'. from Fqm [zl/(zn - 1) into Am as D 9'. = 7f 0 d9'. so that
(13)
where the ai, J are defined by (2). Our problem then is to determine the pairs (g, V) for which the q-ary linear code D,,(V) is r-invariant. Since D,,(V) is a submodule of Am we shall first consider the more general problem of determining when a submodule of Am is T-invariant. The solution to this problem will be in such a form as to be readily applicable to D9'.(V), We now introduce a few elementary notions from module theory.
If g E Am, then the order of g is the nonzero monic polynomial f(x) of least degree such that f(x)g = Q. The order of Q is 1. The order of a(x) E A is defined similarly and from the theory of cyclic codes, we know that
xn - 1 ord(a(x» =
(a(:r;), :I;n -1) (14)
where (a(:r;), :I;n - 1) denotes the greatest common divisor between a(x) and xn - 1. It then immediately follows that if g = (ao(x), aj (x), .. " am-1 (x», then
ord(g) = lcmo�i<mord(ai('T,». (15)
Since (x" - l)Q = Q for every Q in Am, it follows that oId (g,)lxn - 1 (this follows equally well from (15». If M
h ex) Mi = -h'( )1'1'1, t x
are called the primary components [12l. [13l of M. The order of Mi is h.;(x) and it easily follows that Mi is the collection of all elements in M whose order di vides hi (x) (hence is I or hi(x) since h i (x) is irreducible). These primary components are important in our study because of the following elementary results [2l:
Lemma 1: If Me Am is submodule. then Mis r-invariant if and only if its primary components are r-invariant.
Proof' The condition is clearly sufficient. Suppose therefore that ]Iv! = EEl i Mi .• the NI,' s being its primary components. is r-invariant. If g E Mi, then ord(g,) = h;(x) (or g = D. in which case r(g) E Mi automatically). But r(g) = (Xam-l(X), ao(x),"" am-2(x) and it follows from (14) and (15) that ordr(g) = ordg = h;(x) hence r(g) E Mi since r(g) E 1\1. •
If ord (J'lvf) is irreducible we shall say that M is primary
(more generally. a submodule is said to be primary if its order is a power of an irreducible polynomial [12l. [13]). We have therefore reduced our problem to that of determining when a primary submodule of Am is r-invariant. To do this we shall exploit the concatenated structure of such submodules. Let xn -1 = h1(x)h2(x)··· hs(x) be the prime factorization of xn - 1, then
(xn -1) Pi = h;(x) Am, (19)
are the primary components of Am. Hence if M c Am is a primary submodule of order hi(x). then M C Pi. Let hex) be an irreducible factor of xn - 1 and let P be the primary component of Am of order hex). Let 1'-1 be a zero of hex), Fqk = Fq(ry), k = degh(x). Following Jensen [ill. [20] we define the function rf; from Fqk [wl/( WID - 1'-1). where (wm - 1'-1) is the ideal in F qk [wl generated by w'" - 1'-1. into P by setting
(20)
SEGUIN: THE q-ARY IMAGE OF A q= -ARY CYCLIC CODE
where
m-1 a(w) = LaiWi
i=O
and where Tr1 is the trace function from F qk into F q (in the sequel Tr� shall denote the trace function from Fqt into Fq,). Since ry-l is a zero of h(:r;), it follows that
is an element of the q-ary cyclic code of block length n with parity check polynomial hex) and so it follows that ord(.p(a(w»)lh(x); hence .p(a(w» E P. Next, we define the function 'I/J from Pinto Fqk [w]/(w'" - ry-1) by setting
m-l
389
and referring to (\2), we recognize the right-hand side of (24) as T( .p(b( w»); hence
¢(wb(w» = T(.p(b(w»). (25)
Equation (25) establishes the theorem. _ In the language of Jensen [20J, Theorem I says that the T
invariant submodules of P (the primary component of Am of order hex»� are obtained as the concatenation of conslacyclic codes (Le., ideals in Fqk[wl/(wm - .,.-1» and an irreducible cyclic code. So there is nothing new in this theorem. One of the reasons why Theorem I is useful is that the ideal structure of Fqk [wl/(w1n - ry-1) is very simple. The only fact that we shall need is that if I cf (0) is an ideal in Fqk [wl/(w'" _ry-1), then there exists a unique monic divisor a(w) of wTn - ry-l such that
1 = {b(w)a(w)lb(w) E Fqk[WJ, 'I/J(g,) = Lakr-1)wi (21) degb(w) < m - dega(w)}. (26)
i=O
where g, (ao (x), a1(x),· .. , am-1(X» and as always (n, q) = 1. It may be verified by direct calculation that ¢ and 1/1 are inverses of each other. Finally, we make Fqk[W]j(Wm _ry-1) into an A-module by setting
u�-l m-l a(x) Lbiwi = La(ry-1)biWi (22)
for a(x) E A and
m-1
o 0
Lbiwi E Fqk[W]j(Wm _ry-1). o
It is then verified that ¢ (and consequently 'I/J) is a bijective A-module homomorphism. But by (22), the A-submodules of Fqk[W]/(WTn - ry-1) are simply its qk_ary subspaces. Hence cp sets up a one-to-one correspondence between the qk_ary subspaces of Fqk [w]j(w'" - ry-1) and the A-submodules of P ( this fact was used in [2]). More precisely, if M C P is a submodulc. then 'I/J(M) C Fqk[W]/(wm _ry-1) is a subspace and
(23)
We have the following fundamental result [14], [20J: Theorem 1: Let (n, q) = 1, hC .'];) a q-ary irreducible factor
of xn - 1 of degree k, and ry-l E Fqk a zero of hex). Let M be a primary submodule of ATn of order hex) and U = 'I/J(M) the corresponding qk_ary subspace of Fqk[WJlCWTn - ry-1); then M is T-invariant if and only if U is an ideal.
Proof If b(w) E Fq.[wl/Cwm - ,}-1), then
</J(wb(w» = cp(ry-1bm_1 + bow + . .. + bm_2wm-l)
(24)
The polynomial a( w) is called ·the generator polynomial of I and m-deg 11.('IlI ) is the qk_ary dimension of I. The polynomial (wm - ry-1 )Ia( w) is called the parity check polynomial of I. Conversely, if a(w) is a monic divisor of wm - ry-1, then I given by (26) is an ideal in Fqk [wl/( wm _,}-1) with generator polynomial a(w).
Let y' be a qm_ary cyclic code of length n. To make use of Lemma 1 and Theorem 1, we must first detennine when D,,(V) is primary. Suppose first that V is irreducible with parity check polynomial hex) of degree k. Let ,}-1 E Fq=k =
Fq=C'"Y) be a zero of h(:c), then as is well known [15J
(27)
Next, let f3 = ((30, fh, . .. , f3m -1) be the trace-dual basis of g, then [16J
rn-l Tr;;;k (pry]) = L Tr;" (f3; Tr;;:k (P'Yj) )Qi
Consequently
(28)
'm-ln-l = � L .L Tr;"k (pf3d )xj yi (29)
;=0 j=O
so that the q-ary code D�(V) is given as
390 IEEE TRANSACTIONS ON INFORMATION THEORY. VOL. 41. NO. 2. MARCH 1995
Let ho(x) = irr(--y-l, Fq), i.e., the q-ary minimal polynomial of 'i-I, then Fqko = Fq(--y) where ko = deg ho(:D). If N is the order of " then ko is the order of IJ modulo Nand k is the order of qm modulo N, hence it follows that
k = ko/(m, ko). (31)
Decomposing TrIk as Tr�o Trrok, (30) become's
D�(V) � {��% .Tr�o(fjTrrok(p8i))xjyilp E Fq=k } . (32)
Comparing (32) with the defining expression (20) for ¢J, it becomes apparent that
U = 'I/J(DQJV)) = {Trrok (pe)lp E Fq=k} C F',;ku (33)
and
D�(V) = ¢J(U). (34)
It is clear that
(35)
so that D",(V) is primary; i.e., if V is irreducible with parity check polYnomial h(:1;) = irr(--y-I,Fq=), then Do(V) is primary with order ho(x) = irr (,-1, Fq) .
-
In the more general case, we can decompose V as V EEl 1 v, , the Vi'S being the irreducible cyclic subcodes of V. Let hi(:r:) be the parity check polynomial of Vi and let ,;1 be a . zero of hi(X) in some extension of Fqm. Thcn
and D�(vd is a primary module with order hi , o(x) = irr (--Yi-I, F q). Based on the fact that the order of a direct sum is the lea,t common multiple of the orders of its summands, DQ.(V) will be primary if and only if irr (,i-1, Fq) = irr('l\ Fq), 1 ::; i ::; r, i.e., if and only if the nonzeros of V are Fq-conjugates (two elements Ct, (3 arc Fq-eonjugates if irr(Ct, Fq) = irr(tJ, Fq). Setting, = ,I, this will happen if and only if there exists integers /li, 1 ::; i ::; r such that
(36)
and /11 = O. If Fqko = Fq(!) (all the ,i'S generate the same Fqko) then
the integers l1i may be reduced modulo ko. Since the hi(x)'s must be distinct (since (n, q) = 1), then the 'iq"' 's must not be Fqm-conjugates, hence the /l;'S must be incongruent modulo (m, ko). In other words, if Zko denotes the ring of integers modulo ko, m,Zko = (m, ka)Zko' the ideal generated by m. then /li, 1 .:; i ::; r, must belong to distinct cosets of (rn, ko)Zko in Zko' Since the number of such cosets is (rn. ko), then we always have that r ::; (m, ko). In the extreme case when r = (m, ko), then ,q'" is a zero of the parity check polynomial hex) = TI� hi(x) of V for every 11, i.e.,
hex) = irr(!-l, Fq).
Since deg irr (,q"' ,Fqm) = k, 1::; i::; r, then deg h(x) = rk. If V is a qm_ary cyclic code of length n, we can decompose
V = CDfVi, Vi C V cyclic
in such a way that D",(V;), 1 ::; i ::; ,'; are the distinct primary components of D Q. (V). From the above analysis we see that this decomposition does not depend on Q and so we call the Vi's the primary components of v. If the parity check polynomial of Vi is over F q' we call Vi a trivial
primary component and we call it nontrivial otherwise. We can group together the trivial primary components obtaining the decomposition
V = Va EEl (EEl1Vi) (37)
where in (37), Vo is the sum of the trivial primary con:.ponents of V and Vi, 1 ::; 'i ::; '(', are the distinct nontrivial primary components of V. We shall soon see that Va plays no role in determining whether or not dQ.(V) is cyclic. If hex) is the parity check polynomial of V, then the parity check polynomial of Vo is what MacWilliams [5] called the interlaced part of hex) and the parity check polynomial of EEljVi is what she called the noninterlaced part of hex).
We now illustrate the above decomposition by means of a simple example.
.
Example 1: Let q = 2, 'In = 3, and n = 7; then (33 = 1+(3 generates the field Fq� = Fs and x7 - 1 = TI�(x - (3j). Let V be the 8-ary cyclic code of length 7 with parity check polynomial hex) = (x -l)(x - (3)(x -(32)(x - (33), then V = Va EEl VI ED V2 where Va has parity check polynomial x-I, VI has parity check polynomial (x - 8)(x - (32), and V2 has parity check polynomial x - (33.
We now come to the first application of Theorem I to the study of d�(V ).
Theorem 2: Let Fqmk = Fq=(--y), Fqku = Fq(ry), S a set of r ::; (m, ko) integers modulo ko which are incongruent modulo (rn, ko). Let V be the qm_ary primary cyclic code of length n with parity check polynomial hex) =
TII'ES irr(,-q�, Fqm).lfr = (m" ka), then d�(V) is cyclic for every basis Q for Fqm over F q and the parity check polynomial of d,!.(V) is ha(ym) where holY) = irr CI-I, Fq). If r < (m, ka) and if d�(V ) is cyclic for some basis Q then , E Fqm, i.e., k = 1.
Proof' As we have already seen Da(V) is primary with order ho(x). Hence by appealing to the functions ¢J and 'If; previously introduced, there exists a subspace U of F';ko such that DQ.(V) = ¢J(U). Since deg hex) = rk, then the dimension of D �JY) is mrk so that the F q<o -dimension of U is mrk/ko = er where e = mk/ko = m/(rn, ka) is the dimension of F q=k over F qko' First, suppose that r = (m, ko), then U has dimension elm, ko ) = m so that U = Fqko[w]/(Wffi _,-1) is an ideal in that algebra. Consequently, by Theorem I, d�(V) is cyclic irrespective of ,g. Notice that in this case DQ.(Y) = ¢;(U) = ¢;(F;�u) is the primary component of Am of order holY). Since holY) is the order of D",(V), then (see (10» ho(ym) annihilates dQ.(V), Since degho(ym) = mko = mrk is the dimension of dQ.(V), it follows that ho(ym) is the parity check polynomial of dQ.(V),
SEGCIN: THE q-ARY IMAGE OF A qm ARY CYCLIC CODE
On the other hand, if r < (Tn, ko) and do (V) is cyclic for some basis g, then U is an ideal in F qko [�/( wm _,),-1) of dimension er < m. Let a(w) = wt -alwt-I -... -atlWm,-I, t = m - er, be the generator polynomial of U. Referring to the defining equation (33) of U, it becomes apparent that U is invariant under the transformation
m,-l rn-l 2:: biwi --+ 2:: b(' wi o 0
since fJ( = fJi, () :-:; i < m. Hence wt - af wt-1 - ... - a'{' E U and consequently
(ar -al)wt-I+ . .. +(a(' -at) E U. Since t is the minimum degree of the nonzero polynomials in U, we must have that
1 :-:; i :-:; t.
Hence a(w) E Fqm [w]. If p is a zero of a(w), then so in pq'" and since a(w)111Jm - ,-I we must also have that pm = f!mqn.
= ,),-1 Consequently, (,_I)qm = ,),-1 which implies that ,),-1 E Fqm, hence: E Fqm. •
According to Theorem 2, we see that Vo in the decomposition given by (37) plays no role in determining for which g the q-ary image do(V) is cyclic. This was already known to MacWilliams [5). Moreover, Theorem 2 says that if tiQ.(V) is cyclic for some basis. then the nontrivial primary components of V are of a very special type. In the next section, we shall completely solve the problem under study for V a qm_ ary primary cyclic code. In Section IV, we shall show that if dQ.(V) is cyclic for some basis, then V has at most one nontrivial primary component. This will complete the solution of the problem.
III. THE CASE V: A PRIMARY CYCLIC CODE
Let Fq cJ Fqk = Fqh) c Fqm, 5 c;:: Zb h(x) IlI'Es(x - ,-q"), and V the qm_ary primary cyclic code of length n with parity check polynomial h(x). By Theorem 2, these are the only types of primary cyclic codes we need consider. We have imposed the conditions, rf. Fq and 05 cJ Zk so that V is not a trivial primary cyclic code. We have also used k instead of ko since what was called k in Section II is now 1.
Theorem 3: Let Fq cJ Fqk = Fq(!) C FqnS c;:: Zk. and V the qm_ary cyclic code of length n with parity check polynomial h(x) = Il,'oEs(.r -,-q"). Let g be a basis for F q"" over F q with trace-dual basis {1, then do (V) is cyclic if and only if
- -
{ I m-I
1(5) = Q E F�i 2:: a.;/3;"'-" = 0, i=O
for every /1, E 5 } (38)
391
Proof' Let VI' be the qm_ary cyclic code with parity check polynomial x - ,-q", /-l E 5; then
(39)
If UI' C F�'1 is such that DQ.(V/L) = cP(U/L)' then. following a procedure identical to that used to obtain (33), we find that
(40)
If now we set
(41)
then
and by theorem I, Da(V) is T-invariant if and only if U is an ideal in Fqk[1lJ]/(1;7m _,-I). If (g, fJ) denotes the usual dot product between two m-tuples, then the dual U � of U is
U� = {Q E F01(Q, 12) = 0, for every 12 E U}. (43)
It is easy to verify (or we may invoke Delsarte's theorem, [IS, p. 208)) that
I(S) = U� { I m-l
Tn-f/ = Q E F0 � aifJ? = 0, for every It E 05 }.
(44)
Now U is an ideal in Fqk [1ll]/( 111m - �!-I), if and only if U � is an ideal in Fq' [w]/(wm - ,). This proves the theorem . •
For I(S) defined by (38), we define In(s) by setting (see (45) at bottom of this page) We shall shortly need the following lemma, whose proof may be found in the Appendix.
Lemma 2: Things being as in Theorem 3, 1(05) is an ideal in Fqk[W]/(Wm - ,) if and only if In(s) is an ideal in Fqk[W]/(lllmn - 1). When this is the case, 1(05) and In(s) have the same generator polynomial and In (s) n F;;n is the dual of dQ.(V),
Lemma 2 will allow us to describe the parity check polynomial of d",(V) when the latter is cyclic. Again let Fq cJ Fqk = Fq(!) C Fqm, S s: Zk. and V the qm_ary cyclic code with parity check polynomial h(x) = Il/LEs(x - ,-q" ) . We now show that if d",(V) is cyclic for some basis g, then the cardinality #(8) of8 is either I or k - 1.
From the description of U given by (40) and (41) it follows that U contains the polynomials
m-I f-I,(w) = 2:: Trk' (fJ,(-")wi
m-I = 2:: (Trk' (fJiWm "wi (46)
i=O
(45)
392 IEEE TRANS ACTIONS ON INFORMATION THEORY. VOL. 41, NO. 2, MARCH 1995
for every p, E 8. Let F q' be the splitting field of WW -, over F q� and define the automorphism 71", on F q' by setting
(47)
The inverse of TI", is TI_"" where -p, is reduced modulo t, We
that irr (pi, Fqk) = irr (P.i' F qk) and so i = j, Consequently. from (51) we deduce that
d
n n irr (pr , Fqk )Ei Ifo(w),
extend TJJ.t to an automorphism on F q' [w] by setting and so
(48)
It now follows that
(49)
Moreover. for any fi E F q" we have that
71", (irr (p, Fqk)) = irr(pq", Fqk). (50)
We shall need the following simple lemma. Lemma 3: Let p E F q' be a zero of wm - " then i) if figt is a zero of wm - ,. then kit. ii) the polynomials irr (pq" , Fqk). P, E Zk, are distinct. The same is true if we replace , by ,-1.
Proof' i) Recall that Fqk = Fq(!). if pg' is a zero of wm -,.
then pm = , = pmqt = ,q' implying that, E Fqt. hence Fqk C Fqf. hence kit.
ii) irr (pq" , F qk) = irr (pqV, F qk) implies that pqV = pq"+" for some integer s. Hence pq"-'4"k = P and from part i) we have that kip, - lJ + sk. hence �;I/J, - v. But this implies that It = IJ since IL, v E Zk. •
We can now prove the following theorem: Theorem 5: Let Fq i- Fqk = Fq(!) C Fqm, 8 � Zk
and Y the qm_ary cyclic code of length n with parity check polynomial 11,(:1:) = ITI'ES(x -,-q" ) , If there exists a basis g for Fq= over Fq for which duty) is cyclic then r = #(8) = 1 or r = #(8) = k - 1. -
Proof: If d" (V) is cyclic then U, as described by (40) and (41), is an ideal with generator a( w) of degree m -er, e = m/k. Then a(w)lf-",(w) = 71-1'(10(W)) for every p, E 8, Let
d a(w) = n irr(pi' Fqkr'
i=l where irr (pi, F qk)' 1 < z ::; d, are the distinct irreducible factors of a ( w ). Then
d ;Q irr(pi' Fqk)ti 17I-",(10(W») from which it follows that
d d TI irr (p(, Fqk)'" = TITJI,(irr(pi' Fgk)ti) I fo(w), (51)
j.l �, I.' �+Tk If irr (p'f ,F qk) = irr (P) , F qk ), then PJ = pr for some �-I.'+tk
integer t and so Pj = p'f and by Lemma 3, kilL - v
which implies that p, = v. Hence Pj = pr'k which implies
rem, - eT)::; degfo (w ) < m. (52)
From (52) and the fact that m = ek, we get that k
T < k _ T'
(53)
If k = 2 or 3 then forcibly T = 1 or k -1. If k 2': 4, then a simple analysis of (53) shows that r = 1 or T = k - 1. •
We remark that MacWilliams arrived at the result of Theorem 5 for the very special case when q = 2 and m is a prime for which 2 is primitive. This is the content of her corollary 6,5 in [5],
We shall now completely solve our problem for V, a primary cyclic code for which the cardinality of the defining set S C Zk is I.
The case when the cardinality of S is k - 1 will then be solved by appealing to the dual of V. We shall require two auxilliary results; one of which deals with the subfield subcodes of certain codes [15],
Let F qk he an extension of F q and let a( w) = irr (p, F qk ) be an irreducible polynomial over F q" We then set a( w) = irr (p, F q). More generally, if a( w) = IT� ai( w)", the ai ( w),s
, being distinct irreducible polynomials over F g" then a( w) = 1Cml<i<sai(W)f,. We then have the following result:
Le;;;w 4: Let a( w) E F qk [w] be a polynomial of degree less than m and let
J = {b(w)a(w)lb(w) E Fqk[W], degb(w) < m - dega(w)}.
Then the subfield subcode J n F'; is given by
JnF'; = {b(w)a(w)lb(w) E Fq [w], degb(w) < m -dega(w) } ,
Hence J n F;' = to} if and only if dega(w) 2': m, Proof' Let a(w) = IT� ai(w)" be the prime factorization
of a(w) in Fqk [w], d, = deg lLi(W), and {Ii be a zero of tZi(W), 1 ::; " ::; s; then
d.t-l
(},i(llJ)= ncw-p;'j),
j=O
If di = dega;('IlI) then di = (k, d;)di and d,-l
a,(w) = n (w - pf), j=O
Let c( w) E .JnF7, then (},(m) Ic(w) which implies that Pi is a zero of c(w) with multiplicity at least ti, hence a;(w)" Ic(w) and so a(wllc(w).
For the converse we shall assume that deg a( w 1 < m, for otherwise there is nothing to prove. Let c( w) E F q [w], degc(w) < m be a multiple of a(w) . Since ai(wllai(w) (in Fqk [w]) and ail w lE, Ic( w), it follows that ail w)E' Ic( w) and so a(w)lc(w). Hence c(w) E .J n F;; and we are done. •
SEGUIN: THE q-ARY IMAGE OF A q=-ARY CYCLIC CODE
Lemma 5: Let Fqk = Fqh) then i) if a(w) = rr� a;(w)'i divides wm - " thc a; (w) 's
being the distinct irreducible factors of a(w) in Fq,[w], then a(w) = rr�a,(w)",
Proof" i) Let ai(w) = irr(p" Fqk), then by Lemma 3 the Pi'S are
not F q-conjugates, hence the a, (w) = irr (pi , F q), 1 s: i s: 8, arc distinct and so a(w) = 1cmlSiSsai(W)" = rr�ai(w)",
ii) Since Pi is a zero of w7ri -" then pi = , so that Fqk = Fkll = Fq(pil C Fq(p;) = Fqdi' where di = degai (w). This implies that Fqkd, = Fq'(Pi) = Fq(pi) = Fqd" d, = degai(w), hence ki = kdi. Consequently,
dega(w) = LEi degai(w)
= k LEi deg ai (w) 1
= kdega(w ) .
iii) wm - , = rr� irr (p." F qk ) " where irr (pi, F qk), 1 ::; i ::; t are the distinct irreducible factors of wm -,. By (i)
t wm - �I = II irr (p" Fq)". 1
By ii), deg irr (pi, F q) = k deg irr (pi, F qk ) and so it follows that
k-l irr(pi' Fq) = II 171'(irr(pi, Fq,))
p,=o
where T}p, is defined by (48). Consequently,
t t k-l wm - , = II irr (p;, Fqt' = II II 7}p.(irr(pi' Fqk ))"
1 1 /,=0
But
k-l f(w)=irrh,Fq)= II 7}p.(w-,)
1'=0
hence
•
We now come to the first conclusive theorem of the paper: Theorem 6: Let Fq =P Fqk = Fqrr) c Fqm and let V be
the (r-ary primary cyclic code uf length n with parity check polynomial x - ,-q�, 11 E Zk. Then there exists a basis .IT for F qm over F q for which do (V) is cyclic if and only if w Tn - , has a divisor over Fqk �)] of degree e = m I k. If this is the case, then the only bases Q for which dg,(V) is cyclic
393
are the trace-duals of (3 where (30, (31, . . . ,{3e-l is a basis for Fqm over Fqk, and -
e
{3j = La;" {3j-i, e::;j < m
i=l where
a(w) = we -alwc-1 - ... - ae E Fq..['II!]
divides wm - ,. In this latter case the reciprocal a_leW) = wma(w-I) of a(w) is the parity check polynomial of d",(V).
Prooj- If d�(V) is cyclic then by Theorem 3 -
1= {12 E F;� I �lbi{3,!m-" = o} (54)
is an ideal modulo (wm - ,) of dimension m - e. Let a(w) = uf -aIWe-l- . . . -ae be the generator polynomial of I. Since e is the minimum degree of the nonzero polynomials in I, then (3(-", 0 ::; i < e must be independent over Fqk and similarly for (ii, 0 ::; i < e. Hence 8i, 0 ::; i < e form a basis lor Fqn. uver Fqk . Since wja(w) E I, 0 ::; j < m - e, it follows that
e {3j"'-u = Lai{3J:i-"
;=1 and so
8j = La( /3j-i, for e::; j < m. i=l
Conversely, let {3o, 81,' . . ,{3e-l be a basis for Fq", over Fqk and let
fij = Lai" (3j-i, e::; j < m i=l
where a(w) = we - al wc-l - ... - ae is a divisor of wm -, in Fqk [w]. Cunsider I as defined by (54). Clearly, 1 is a vector space over Fq,. of dimension m - c. We proceed to show that every polynomial in I is a multiple of a( 1ll) , hence establishing that I is an ideal with generator a( w). Let
then
b(w) = Lb ;wi E I
e
� =-� o = �bi{3? o
';=0
= �1)i{3rm-1' + be (tar (3,-i) qm-"
e-l e-l - � q=-e � ,q=" - �bi(3i + be�ae-i(3i
i=O i=O c-I
= L(bi + beae-i)(3'{'-". i=O
394 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 41, NO.2. MARCH 1995
By the independence of (J'(-M, 0 :S i < e, we conclude that bi = -beae-i, 0 :S i < e, from which we get that b( w) = bear w). Assume that every polynomial in T of degree less than or equal to .s, (� < .S < In - 1, is divisible by a( w) and consider
Then
s+1 b(w) = 2..)iwi E I.
o
8+1 0= I)i(3'(-M
o S s
= Lbi(3;m-M + bs+1 LaifJ;:;�i o 1 s s
= LbifJ'(-M + b'+l L as+l_ifJ(-e o .+1-e s
= L(bi + b.,+las+l_d{-J'(-M o
where in the above expressions ai = 0 if i > e, hence
L(bi + b.+las+l_i)Wi E I o
and so by induction hypothesis this latter polynomial is c(w)(}.(w) for some polynomial c(w). Hence
0+1 s LiJ(Wi = c(w)a (w) -bB+lLaO+l-iWi + bs+1Ws+l
and we are done. Next we must show that fJi, 0 :S i < m (or equivalently
fJ,(-M, 0 :S i < m) are F q-independent, i.e., that In F:;' = {O}. To accomplish this we use Lemmas 5 and 6. By Lemma 4
InF�' = {b(w)a(w)lb(w) E Fq[w], deg b( w) < m - deg a( w )}
and by Lemma 5
dega(w) = kdega(w) = ke = m.
Hence I n F�' = to}. To complete the proof, we show that a_I (w) = wma( w-1)
is the parity check polynomial of d,,(V). Now In, as defined by (45) with 5 = {J.t}, is by
-Lemma 2 an ideal in
Fqk[W]/(Wmn - 1 ) with generator polynomial a(w). Moreover, by tbe same lemma, In nF;;n is the dual of d.9:,(V), But
by Lemma 4, a(w) is the generator polynomial of In nF�nn; hence a_l eW) is the parity check polynomial of d.9:,(V)' •
We shall illustrate the contents of Theorem 6 by means of the following example:
Example 2: Let q = 2, m = 4, V the [3, 1] irreducible l6-ary cyclic code with parity check polynomial x - 1-1 where 1<'4 = F2C'Y). Hence, e = rn/k = 4/2 = 2 and w'" - 1 = '1JJ4 -1 = (w -7)4. Using the recursion p4 = 1 + P to generate F16, we find that
F16 = to, 1, p, /, p3, p4 = 1 + p, pO = P + l, p6=p2+p3, p7=I+p+p3, p8=I+p2, p9 = P + p3, plO = 1 + P + l, pll = P + / + i, /2 = 1 + Ii + / + l, /3 = 1 + p2 + p3, p14 = 1 + l}.
Also F4 = to, 1, p5, plO} and so we may take 1 = p5. The code V is
V = {oCI, 1, 12)10 E F16} = {(pi, /+5, /+10)10 :S i < 15, i = -Do}
where in the above description of V, i = -00 gives the all-O codeword.
To find a basis Q for F16 over F" for which d,,(V) is cyclic we proceed according to Theorem 6. For a( w f we have the unique choice a(w) = (w -1)2 = w2 -12 = w2 - plO. Next we pick a basis for F 16 over F 4, say (30 = 1, (31 = p. Then
2 fJj = Lai(3j-i = plUfJj_2,
1 2:S.i<4
so that fJ2 = pIO, fJ3 = pll. We then compute the trace-dual basis Q of fJ to be Q = (p9, p5, p4, 1). This gives rise to the following iahle:
It is now easily verified that the 16 binary 12-tuples which appear in the above table form a [ 1 2, 4, 4] cyclic code with generator polynomial 1 + x2 + x6 + x8. The polynomial a( x) is ( 1 + x + x2)2 and so 1 + x2 + x4 is the parity check polynomial of d.9:,(V), Since '/lJ4 - I h<L� only one monic divisor over /<'4
SEGUIN: THE q-ARY IMAGE OF A qm ARY CYCLIC CODE
of degree 2, then the total number of bases g for which d� (V)
is cyclic is simply the number of bases for F 16 over F 4. This number is (24 - 1)(24 - 22) = 180. Since there are only 3 binary [12, 4] cyclic codes, then clearly there are many g's for which d�(V) is the same binary cyclic code.
To settle the case where the defining set S C Z k has cardinality k - 1, we shall need the following simple lemma whose proof is left to the reader.
LRmma 6: Let V be a qm_ary linear code with dual V �.
Let g be a basis for F q= over F q with trace-dual g� then
(55)
Let R be the reciprocal operator, Le., if
then
R(g,) = (as-l, as-2, - " , ao) . We use the same generic symbol R for various values of s. Now if V is a cyclic code of length n and parity check polynomial h( x), then R(V �) is the cyclic code with parity check polynomial fl(:r.) , where g(:I:)h(:I:) = :I:n - 1 [ 15]. Moreover, it is easy to see that
(56)
We now come to the second conclusive theorem of the paper and which is a sort of dual of Theorem 6.
Theorem 7: Let Fq i- Fqk = Fqb) c Fqm, S a subset of Zk of cardinality k - 1, and V the qm_ary primary cyclic code of length n and parity check polynomial hex) = TIIlES(X -'1-q"). Then there exists a basis g for F qm over F q for which d�(V) is cyclic if and only if wm - '"'I has a divisor over Fqk of degree e = m/k. If this is the case, then d�(V) is cyclic if and only if Om-I, Om-2, - - -, O:m-e form a basis for Fqm over Fqk and
O-::;'.i<m-c
where a(w) = we - (J,l'/IIe-1 -... - nc E Fqk.[W] divides wm -'"'I and where v E Zk. V � S. Moreover, the parity check polynomial of d�(V) is the reciprocal b_l(y) of bey) where b(w) = (wm - '1l/a(w).
Proof: Now 11",(V) is cyclic if and only if d" ('l)� =
daJ.(V�) is cyclic-and daJ.(V�) is cyclic if and only if Rlda�(V�)) = dR(oJ.)(R(V�)) is cyclic_ R(V�) is the cyclic code with parity check polynomial (:1:" -1 )/11.(:/:). Now )-1 is a zero of xn - 1 and
k-I I-I(x) = irrh-l, Fq) = II (x -'1-q")
1'=0
so
xn - 1 = I-I(x)c(x) (57)
and consequently
395
where v E Zk. V � S. If Vo is the cyclic code with parity check polynomial c(x) E Fq[x] and if V' is the cyclic code with parity check polynomial x - 'Y-,-qV, then
(59)
It follows that Vo is the direct sum of the trivial primary components of R(V �) and V' is the unique nontrivial primary component ofR(V�)_ By the theory developed in Section II, dR(Q. ') (R(V �)) is cyclic if and only if dR(Q.�) (V*) is cyclic and by Theorem 6, the latter is cyclic if and only if wm - 'Y has a divisor over Fqk of degree e = m/k. Moreover, when this is the case, dR(!!J.)(V*) is cyclic if and only if the trace-dual basis of R(g�), Le., R(g) = (am-I, a",-2,'" , ao), is such that am-l, . - - ,O:m-< are independent over Fqk and
O-::;' j<m-e
where n( w) = we -alWe-1 - ... - nc divides w11l - 'Y over Fqk.
We must now identify the parity check polynomial of da(V)_ By Theorem 6, the parity check polynomial of dR(oJ.)(R(V*)) is a-ley) of degree m. Since dR(n-J.)(R(V�)) = dRC'!J.)(Vo) 61 dR(�J.)(V*) and the parity check polynomial of d R(Q. I ) (Vo) is, by Theorem 2, c(ym), then the parity check polynomial of dR(�� )(R(V �)) is
(60)
of degree men - k) + m = men - k + 1). Now d,,(V) = R(dR�J.)(R(V�)�) (by (55) and (56) so that the parity check polynomial of d,,(V) is the generator polynomial of dRt,!J.)(R(V�)), i.e., itis
(61)
of degree mn - men -k + 1) = m(k - 1) as expected. Now
ym - 'Y = n(y)b(y) (62)
and, by Lemma 5, ym - "y = I(y""') where f(y) = irrb, Fq) and notice that I-I (y) = iff b -1, F q) is the reciprocal of I(y) up to a nonzero constant in Fq• Hence from (62), we have that
(63)
From (57) and (63), we have that
and using (64) in (61) yields
(xn - l)/h(x) = (x -'1-q")c(:I:) (58) as claimed. •
396 IEEE TRANSACTIONS ON INFORMATION T HEORY, VOL. 41. NO. 2, MARCH 1995
We illustratc Theorem 7 by means of an example: Example 3: Let q = 2, k = 3, m = 6, Fs = F 2h),
and V the 64-ary cyclic code of length 7 and with parity check polynomial h(x) = (x - �/-2)(.r _ ,-4). We generate FfH by means of the recursion p6 = 1 + p (for a list of the elements of F 64 sec [17, p. 562 J). For , we choose plH and so w'" - �I = w6 _ p 18 = (w3 - p9)2 and w3 _ p9 = (w _ p:J)(w _ p24)(w - p45). We have that (w - p3)(w - p2"*-) = w 2 + p45w + p27 E Fs['IlI] is a divisor of w6 - , of degrcc 2 = E. Setting 0:5 = 1, 0:4 = p, and a(w ) = w2 + p4Gw + p27 we have that
no = p2 9 is a basis for F64 over F2 for which d,,(V) is cyclic. Now /i(m) = (w6 - 'Y)/a(w) = (w -l)(w -p24)(w _ p4.'»2 = (w2 + p4Gw + p27)(w _ p45? so that b(w) = irr(p3, F2)(irr(p"*-5, F2»2 = (1 + w + w2 + w 4 + w6)(1 + w + 1113)2 = 1 + w + 1113 + w 6 + w7 + 11110 + w12. Hence, according to Theorem 7, the parity check polynomial of do(V) is b_1(y) = 1 + y 2 + yG + y 6 + y9 + y ll + y12 . Thc generator polynomial of V is g(.1;) = (:1;7 - 1)/h(x) = (x-,-I)(I+x)(1+:r2+x3) since irrh-1, F2) = 1+.r+x3 and (1+X+l;3)/(x-,-2)(X-,,-4) = X-,-l. Consequently
g(x) = pl5 + p9x + p9x2 + :1;3 + p4·'ix4 + xG
and so dgJnig(x»), 0 � i < 6, are, respectively:
(10101 L 001011. 001011. 100000, 101011, 100000, 0(0000)
It may now be easily verified that these six 42-tuples, considcred as polynomials, when multiplied by b_1(y) modulo y 42 + 1 give O. Note that only the first, second, and fourth need be verified since the others arc cyclic shifts of these. This shows that dg,(V) is indeed the [42, 121 binary cyclic code with parity check polynomial b_1 (y) (the latter does divide y12+ 1). The above six 42-tuples and the six 42-tuples obtained by cyclically shifting these by six places will form the rows of a generator matrix for d,,(V).
In this section, we have identified all pairs (g, V), Q a basis for Fqm over Fq, Va qm_ary primary cyclic code, for which do< is cyclic. In thc next section, we will show that if V is a qm_ ary cyclic code and if d�(V) is cyclic for some basis Q then V has at most one nontrivial primary component. This will then completely solve the problem set forth in the introduction.
IV. V CAN HAVE AT MOST ONE
NONTRIVIAL PRIMARY COMPONENT
We shall need the following lemma whose proof may be found in the Appendix.
Lemma 7.' Let Fq # Fqk = Fqh) c Fq� , .9 <;:; �k and let V be the qm_ary cyclic code with parity check polynomial TIILES(X - ,,-q�) of degree r = #(S). If Q is a basis for Fq� over Fq for which tl",(V ) is cyclic and if (3 is the tracedual basis of Q, then (3(;;;) =
'L�I-l (3iwi is divisible by a polynomial of the form DILES IJIIL(w) where IJIIL(w) divides w m -,ql' in Fq�[w] and has degree m - er, e = m/k.
We can now prove the following key result: Theurem 8.- Let Fq # Fqk; = Fq("!;) C Fq=, and
Si S Zki for i = 1, 2 and where ,I and "12 are not F q
conjugates. Let Vi be the qm-ary primary cyclic codc of length n with parity check polynomial DILES, (x - ,;q"), i = 1, 2. Then thcre does not exist any basis Q for Fqm over Fq for which both dg. (Vl) and d-,-, (V2) are cyclic.
Proof Suppose, contrary to the statement of the theorem, that there exists a basis for which both de> (VI) and do (V2) are cyclic. Let (3 be the trace-dual basis ;f Q and f:I(1.;;) = 2:;;'-1 (3,1IIi. Then by Lemma 7, there exists polynomials lJI�i)(W) E Fq=[w] such that \[I�)(w)lwm - ,'( , fL E Si, such that
n lJI�i)(W) I fi(w) /LES;
and
deg (n W�)(W)) = ri(m - Eir;) , /LES,
i = 1, 2.
But because ,I and ,2 are not F q-conjugates, then all the polynomials w1n - ,t, fL E Sl and 111m - ,f, V E S2 are pairwise relatively prime. Consequently
f:r II \[I� ) (w) I f:I(w) t=l/LES,
and from the latter we deduce that
Replacing the first m by elkl and the second by e2k2 we get
Moreover, because eilm and Hi # Tn, we have that e i � m/2 and so
(67)
Now by Theorem 5, ri = 1 or Ti = ki - 1. But in either case, T'iEi(ki - Ti) = ei(ki - 1), hence (66) reduces to
from which we get (since e1k1 + e2k2 = 2m) that
contradicting (67). Hence no such Q exists. • We immediately have the following:
SEGUIN: THE q-ARY IMAGE OF A qm ARY CYCLIC CODE
Corollary: If V is a qm_ary cyclic code of length n, (n, q) = 1 , and if there exists a basis for F qm over F q for which d�(V) is cyclic, then V has at most one nontrivial primary component.
The results of Theorems 2, 6, 7 , and thc above corollary lead to the formulation of the first form of the central result of the papcr.
Theorem 9: Let (n , q) = 1 and let V be a qm_ary cyclic code of length n with parity check polynomial h(x) . Then there exists a basis Q for F q� over F q for which d�(V) is cyclic if and only if:
i) hex) E Fq [x] , in which case d�(V) is cyclic for every basis Q and the parity check polynomial of dn (V) is h(ym) ; or
ii) h(:c) = ho(x)(x - ,-q� ) , ho (x) E Fq[x] . Fq -# Fqk = Fq h) c Fq� , M E Zk and wm - , has a divisor over Fqk of degree e = rnjk. In this case, da lY) is cyclic if and only if Q is the trace-dual basis of fi-: (30 , (31 , " . , ,8e-1 are Fq,. -independent and
e
'\"' ql' {lj = �ai (3j-i , e $. j < rn
1
and I1( W) = we - 111We-1 + . . . + l1e E Fqk [W] divides wffi - ,. Moreover, the parity check polynomial of dalY) is ho(ym )a_l (y) where a- l ey) is the reciprocal of a(y); or
iii) hex) = ho (x ) I1/lES(:I: - ,-q� ) . ho (x) E Fq[x] . Fq -#
Fq' = Fq(r) C Fq� , S C Zk. #(S) = k - 1, and wm - , has a divisor over F q' of degree e = rn / k. In this case, d9:(V) is cyclic if and only if am-I , am-2 , ' " , am-c are F qk -indcpendent and
e
'\"' q' {XJ == Lai (tj+i , O $. .j < rn - e I
where I1(W) = we - ([. IWe- 1 - . . . - l1e E Fqk [W] divides '(/JTn - , and v E Zb V !f: S. Moreover, the parity check polynomial of d�(V) is ho(ymj"Ll (Y) where b_I (y) is the reciprocal of bey) and b(w) = (wm - ,)ja(w).
We can restate Theorem " in terms of generator polynomials obtaining the second version of the central result of the paper.
Theorem 10: Let (11" q) = 1 and let V be a qm_ary cyclic code of length n with generator polynomial g(x) . Then there exists a basis Q for F qm over F q for which d�(V) is cyclic, if and only if:
i) g(x) E Fq [x] , in which case dQ.(V) is cyclic for every basis Q and the generator polynomial of da W) is g(ym) ; or
ii) .11(:1: ) = go(x) (x - ,-q' ), go (x) E F� [x], Fq -# Fqk = Fq(:) C Fq� , v E Zk and wm - , has a divisor over Fqk of degree e = rn/ k. In this case, d�(V) is cyclic if and only if i:tm-l , am-2 , " ' , O'm-e are Fqk -independent and
O $. j < rn - e
and a(w) = we - alWe-1 - . . . - l1e E Fqk [W] divides wm - ,. Moreover, the generator polynomial of da lY) is a-I (y)gO (ym ) where a-I (y) is the reciprocal of a(yf or
iii) g(:I:) = g(x) I1/lES(x - ,-q" ) , go(x) E Fq [x] . Fq -# Fqk = Fqh) c Fqm , S C Zk, #(S) = k - 1, and wm - ,
397
has a divisor over Fqk of degree e = m/k. When this is the case, d�W) is cyclic if and only if Q is the trace-dual basis of fi; (30 , (31 , . . . , {3e- 1 are F qk -independent and
c
(3j = Lat (3j- " e $. j < rn, /I E Zk, V rf. S 1
and where a(w) = we - alwc- 1 - . . . - l1e E Fqk [w] divides wm - ,. Finally, the generator polynomial of d�(V) is b_1 (y)go(yffi) where b_1 (y) is the reciprocal of bey) and b(w) = (wm - ,)/a(w) .
Remark: In both Theorems 9 and 10, if rn is a power of the characteristic of F q ' then the condition: "wm - , has a divisor of degree e over F qk ." disappears; for then w m -, = (w _ p) m for some p E F q h) = F qk and so has divisors of all degrees between I and rn.
V. CONCLUSIONS
In this paper. under the restriction (n, q) = 1 , we have characterized all pairs (Q, V) . V a qffi_ary cyclic code of length n, Q a basis for Fq� over Fq, for which dQ.(V) is cyclic. This characterization is given i n Theorem 9 or equivalently in Theorem 10. As anticipated by MacWilliams, the class of qm_ary cyclic codes which have a cyclic q-ary image is indeed very restrictive. We note that the necessary condition on the generator polynomial of V given in Theorem 10 was obtained by Leonard [ 18]. However, he does not have the condition that wm - , should have a divisor of a special degree e and his description of the bases Q is incorrect.
We finish by suggesting that a more natural question to ask is "when does a qffi_ary cyclic code have a q-ary image which is equivalent to a cyclic code." Perhaps the tools developed in this paper can be used to answer this more general problem. Finally, the case (n, q) -# 1 should be looked at.
ApPENDIX
In this Appendix we provide proofs to Lemmas 2 and 8. Proof of Lemma 2: From the definition of J" (S) it is clear
that if I1j (w) E 1(8), 0 $. j < n. then
and that
n-l
LWjml1j (w) E r(S) j=O
From these observations, it follows that the monic polynomials of least degrees in J (S) and Tn (S) are the same.
Suppose first that I(S) is an ideal. Since reS) is clearly a vector space over F qk we only need to show that it is closed under cyclic shifting. To this end, let
m-l n-l
L Ll1j, ;Wjm+i E r(S) ,=0 j=O
398 IEEE TRANSACTIONS ON INFORMATION THEORY. VOL. 4 1 , NO. 2. MARCH 1995
for every JL E S. W hence it follows that
'� (�", n} E liS).
Since IrS) is an ideal modulo wm - ;, it follows that
Hence, I"(S) is an ideal modulo wmn - 1. Conversely, suppose that r (S) s an ideal modulo wmn - 1 and let
m-1 b(w) = L biW' E I(S).
i=O Then b(11I) E [" (S) and so wb(w ) E In (s). This implies that
m. - l
L /Ji-dj'(-� + bm-11'{3f-� = 0 £=1
for every /L E 8 and this in turn implies that m-1
IJm-n + L bi_1wi E 1(8). ;=1
Since the latter is 11Ib( w) modulo llr - 1', it follows that 1(8) is an ideal modulo wm - '1'.
Finally, we show that F' (S) n F�m i s the dual of dQ:(V), First, since
then
d�(v')� = n dQ:(VI')� ' p E S
We get the description of da (V) by applying the inverse permutation 7['-1 to D,,(V) given by (30) with k = 1 and with 1'Q" instead of ,.-But the latter is equivalent to using fJ(-1' instead of /3;, hence
Consequently
d"(V,)" � {� '�dj" ",im " E F.iw] ,
LL"j,nir'( ' � n}
Hence (see bottom of this page) and comparing this to the definition of In(s), we see that d�(V)� = 1n(s) n F:;,n . •
Proof of Lemma 7: By Theorem 3, d�(V) is cyclic if and only if
is an ideal in Fqk [W]/ (Wffi - ,) . This ideal has dimension m - cr and so its generator polynomial a( w) has degree fT. Let a(w) = w'r +a1Wer-1 + . . ' + a,r, where we have changed our convention from that used in Sections II and III by dropping the negative signs in n(w). Since a( w) E I, it follows that
'r f3 . - -""'
q� f3 . . J - �ai J-1 , er S j < m 1
or equivalently er
""' q� (3 -�ai j-i. - 0, er s .i < Tf/.. CAl ) o
Define J I' by setting
JI' = {Q E F:;'= I �br (3j-i = 0, er S j < m} (A2)
where J.t E S. Note that we are now working over F qm instead of merely over Fqk . Note also that (3( w) E J/1 .• Clearly, Jf.L is a vector space over Fq= of dimension er. We claim that it is an ideal in Fq= [w]/(wm - 1'q� ) . To show this, let Q E JI' and consider for a moment the generating function
and (wm - ry -q" ) 19J.L('IlI) is a polynomial of degree m - er. Hence the right-hand side of (A8) is a polynomial of degree less than m. From the fact that (wTn - ,-q" )IJ(w) is a polynomial of degree less than m we conclude that
j = m, m + 1 , · · · . (A9)
From (A9) we have that
bo = ,q" bm and this shows that J1, is ' closed under multiplication by !I! modulo ur - ,q" , hence is an ideal in Fq= I( wm - ,q" ). If W/L(w) is the generator polynomial of Jp, then W /l (W) divides wm - ,),q"
in Fq= [w] and has degree m - er. Since f3(w) E Jp , then \[I J.L (w) I,(I( w). But since the polynomials wm - :q" , J.L E S are pairwise relatively prime, it follows that
II wJ.L (w) I f3(w) ,'ES
(A l O)
•
399
ACKNOWLEDGMENT
The author wishes to thank Prof. J . Wolfmann of L'Universite de Toulon, Toulon, France for the many discussions he has had with him on the problem treated in this paper. Without these, and his encouragement, the author would surely have abandoned this problem. The author would also like to thank the reviewers for their many helpful comments (in particular for pointing out the connection between Theorem I and the results of Jensen on the concatenated structure of cyclic codes) and especially referee A for pointing out an error in the original proof of Theorem 8 .
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