Construction and number of self-dual skew codes over F _p 2

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1 Constructon and number of self-dual skew codes over F _p 2 Delphne Boucher To cte ths verson: Delphne Boucher. Constructon and number of self-dual skew codes over F _p 2. Advances n Mathematcs of Communcatons, AIMS, 206, Volume 0 (Issue 4), pp <0.3934/amc >. <hal v2> HAL Id: hal Submtted on Feb 206 HAL s a mult-dscplnary open access archve for the depost and dssemnaton of scentfc research documents, whether they are publshed or not. The documents may come from teachng and research nsttutons n France or abroad, or from publc or prvate research centers. L archve ouverte plurdscplnare HAL, est destnée au dépôt et à la dffuson de documents scentfques de nveau recherche, publés ou non, émanant des établssements d ensegnement et de recherche franças ou étrangers, des laboratores publcs ou prvés.

2 Constructon and number of self-dual skew codes over IF p 2. February, 206 Delphne Boucher IRMAR (UMR 6625) Unversté de Rennes, Campus de Beauleu F Rennes Abstract The am of ths text s to construct and to enumerate self-dual θ-cyclc and θ-negacyclc codes over IF p 2 where p s a prme number and θ s the Frobenus automorphsm. Introducton A lnear code over a fnte feld IF q s a k-dmensonal subspace of IF n q. Cyclc codes over IF q form a class of lnear codes who are nvarant under a cyclc shft of coordnates. Ths cyclcty condton enables to descrbe a cyclc code as an deal of IF q [X]/(X n ). A self-dual lnear code s a code who s equal to ts annhlator (wth respect to the scalar product). One reason of the nterest n self-dual codes s that they have strong connectons wth combnatorcs. In 983, N. J. A. Sloane and J. G. Thompson nvestgated the constructon and the enumeraton of self-dual cyclc bnary codes wth a gven length n ([9]). These codes are determned by a polynomal equaton whose solutons can be descrbed thanks to some factorzaton propertes of X n + n IF 2 [X]. Later ths study was generalzed to self-dual cyclc codes over fnte felds of characterstc 2 ([, 0]) and to self-dual negacyclc codes over fnte felds of odd characterstc ([5], [7]). For θ automorphsm of a fnte feld IF q, θ-cyclc codes (also called skew cyclc codes) of length n were defned n [2]. These codes are such that a rght crcular shft of each codeword gves another word who belongs to the code after applcaton of θ to each of ts n coordnates. If θ s the dentty, θ-cyclc codes are cyclc codes; f q s the square of a prme number and θ s the Frobenus automorphsm (who therefore has order 2), θ-cyclc codes form a subclass of the class of quas-cyclc codes of ndex 2 ([8]). Self-dual quas-cyclc codes have been also studed n [8], [3], [4]. Skew cyclc codes have an nterpretaton n the Ore rng R = IF q [X; θ] of skew polynomals where multplcaton s defned by the rule X a = θ(a)x for a n IF q. Lke self-dual cyclc codes, self-dual θ-cyclc codes over IF q are characterzed by an equaton, called self-dual skew equaton and defned n the Ore rng IF q [X; θ]. When q s the square of a prme number and θ s the Frobenus automorphsm over IF q, propertes specfc to the rng IF q [X; θ] wll enable to extend N. J. A. Sloane and J. G. Thompson orgnal approach to solve the self-dual skew equaton.

3 The text s organzed as follows. In Secton 2, some defntons and facts about θ-cyclc codes, θ-negacyclc codes and self-dual codes are recalled. The self-dual skew equaton characterzng self-dual θ-cyclc or θ-negacyclc codes s recalled. Its solutons are least common rght multples of skew polynomals who satsfy ntermedate skew equatons n IF q [X; θ] ([3]). The man goal of ths paper conssts n constructng and enumeratng the solutons of these ntermedate skew equatons when q s the square of a prme number p and θ s the Frobenus automorphsm over IF p 2. In Secton 3, self-dual θ-cyclc and θ-negacyclc codes whose dmenson s a power of p are consdered over IF p 2. In ths case, the self-dual skew equaton splts nto one sngle ntermedate skew equaton. When p s equal to 2, the complete descrpton of ts solutons was obtaned n [3] thanks to some factorzaton propertes (recalled n Proposton 3) specfc to IF p 2[X; θ]. Usng the same arguments, one can also descrbe the solutons of the self-dual skew equaton when p s an odd prme number (Proposton 4). The results are summed up n Table. In Secton 4, self-dual θ-cyclc and θ-negacyclc codes whose dmenson s prme to p are consdered over IF p 2 (Proposton 8). A resoluton of the ntermedate skew equatons based on Cauchy nterpolatons over IF p 2 (Propostons 6 and 7) enables to provde a parametrzaton of the solutons. In Secton 5, self-dual θ-cyclc and θ-negacyclc codes of any dmenson over IF p 2 are constructed and enumerated (Theorem ). The steps of the resolutons of the ntermedate skew equatons are summed up n Tables 4 and 5. Proposton 4 (Secton 3) and Proposton 8 (Secton 4) can be seen as partcular cases of Theorem. The text ends n Secton 6 wth some concludng remarks and perspectves. 2 Generaltes on self-dual skew constacyclc codes For a fnte feld IF q and θ an automorphsm of IF q one consders the rng R = IF q [X; θ] where addton s defned to be the usual addton of polynomals and where multplcaton s defned by the rule : for a n IF q X a = θ(a) X. () The rng R s called a skew polynomal rng or Ore rng (cf. [6]) and ts elements are skew polynomals. When θ s not the dentty, the rng R s not commutatve, t s a left and rght Eucldean rng whose left and rght deals are prncpal. Left and rght gcd and lcm exst n R and can be computed usng the left and rght Eucldean algorthms. The center of R s the commutatve polynomal rng Z(R) = IF θ q[x m ] where IF θ q s the fxed feld of θ and m s the order of θ. The bound B(h) of a skew polynomal h wth a nonzero constant term s the monc skew polynomal f wth a nonzero constant term belongng to Z(R) of mnmal degree such that h dvdes f on the rght n R ([9]). Defnton (defnton of [3]) Consder an element a of IF q and two ntegers n, k such that 0 k n. A (θ, a)-constacyclc code or skew constacyclc code C of length n s a left R-submodule Rg/R(X n a) R/R(X n a) n the bass, X,..., X n where g s a monc skew polynomal dvdng X n a on the rght n R wth degree n k. If a =, the code s θ-cyclc and f a =, t s θ-negacyclc. The skew polynomal g s called skew generator polynomal of C. 2

4 If θ s the dentty then θ-cyclc and θ-negacylc codes are respectvely cyclc and negacyclc codes. Example Consder p a prme number, θ : x x p the Frobenus automorphsm over IF p 2 and α n IF p 2. The remander n the rght dvson of X 2 by X + α n IF p 2[X; θ] s equal to α p+ : X 2 = (X θ(α)) (X + α) + αθ(α). Therefore, there are p + θ-cyclc codes of length 2 and dmenson over IF p 2; ther skew generator polynomals are the skew polynomals X + α where α p+ =. Defnton 2 ([3], Defnton 2) Consder an nteger d and h = d. The skew recprocal polynomal of h s h = d X d h = =0 d h X n R of degree =0 d θ (h d ) X. If m s the degree of the tralng term of h, the left monc skew recprocal polynomal of h s h := θ d m (h m) h. The skew polynomal h s self-recprocal f h = h. Remark For f, g n R, (f g) = Θ deg(f) (g ) f (Lemma 4 of [3]). In partcular, for f, h n R f f dvdes h on the left then f dvdes h on the rght. The (Eucldean) dual of a lnear code C of length n over IF q s defned as C = {x IF n q y C, < x, y >= 0} where for x, y n IF n q, < x, y >:= n = x y s the (Eucldean) scalar product of x and y. The code C s self-dual f C s equal to C. Accordng to [3], self-dual θ-constacyclc codes are necessarly θ-cyclc or θ-negacyclc. They can be characterzed by a skew polynomal equaton who s recalled below. Proposton (Corollary of [3]) Consder ε n {, }, two ntegers k, n wth k n and C a (θ, ε)-constacyclc code wth length n, dmenson k. Consder g the skew generator polynomal of C and h the skew check polynomal of C defned by g h = X n ε. The Eucldean dual C of C s a (θ, ε)-constacyclc code generated by h. The code C s Eucldean self-dual f, and only f, h h = X 2k ε. (2) The equaton (2) s called self-dual skew equaton. When k s fxed, a frst approach to solve the self-dual skew equaton conssts n constructng the polynomal system satsfed by the unknown coeffcents of a soluton : Example 2 Consder p a prme number and θ : x x p the Frobenus automorphsm over IF p 2. The self-dual θ-cyclc codes of dmenson over IF p 2 are the θ-cyclc codes whose skew check polynomals h satsfy the self-dual skew equaton h h = X 2. The monc skew solutons of the self-dual skew equaton are the monc skew polynomals h = X + α where α s n IF p 2 and ( X + ) (X + α) = X 2. θ(α) 3 =0

5 Developng the left hand sde of ths relaton thanks to the commutaton law () and equatng the terms of both sdes, one gets the condtons α 2 + = 0 and α p =. If p = 2 then α = and f p s an odd prme number then α 2 = and ( ) p 2 =. Therefore f p = 2 there s one self-dual θ-cyclc code of dmenson over IF 4 ; f p 3 (mod 4) there are two self-dual θ-cyclc codes of dmenson over IF p 2; f p (mod 4) then there s no self-dual θ-cyclc code of dmenson over IF p 2. When k s not fxed, a second approach s based on the factorzaton propertes of the monc solutons of the self-dual skew equaton. The startng pont of the study s nspred from Sloane and Thompson constructon of self-dual bnary cyclc codes ([9]) who s extended to fnte felds wth characterstc 2 n [0]. Let us recall ther strategy (and therefore assume that IF q has characterstc 2 and that θ s the dentty). Consder two ntegers s and t such that k = 2 s t wth t odd. The polynomal X n + = X 2k + s factorzed n IF q [X] as the product of r polynomals f (X) 2s+ where f (X) s a self-recprocal polynomal whch s ether rreducble or product of two dstnct rreducble polynomals g (X) and g (X) n IF q [X]. Consder h n IF q [X] such that h h = X 2k +. Necessarly, h s the product of polynomals f (X) α, g (X) β and g (X)γ, where α, β and γ are ntegers of {0,..., 2 s+ }. The relaton h h = X 2k + s satsfed f and only f α = 2 s and β +γ = 2 s+, therefore there are (2 s+ + ) m self-dual cyclc codes of dmenson k where m s the number of polynomals f (X) = g (X)g (X) dvdng Xn + n IF q [X]. Lastly one can notce that the polynomals h who satsfy the relaton h h = X 2k + are least common multples of polynomals h who are defned by the ntermedate equatons h h = f (X) 2s+ : h h = X 2k + h = lcm(h,..., h r ), h h = f (X) 2s+. In [3], ths lcm decomposton was generalzed to a lcrm decomposton over R = IF q [X; θ] n the partcular case when q s the square of a prme number and θ s the Frobenus automorphsm (Proposton 28 of [3]). Ths decomposton enables to derve a frst formula for the number of (θ, ε)-constacyclc codes of dmenson k (Proposton 2 below). Frst one ntroduces some notatons that wll be useful later : Notaton For F = F (X 2 ) n IF p [X 2 ], k n IN and ε n {, }, H F := {h R h s monc and h h = F (X 2 )} H F := {h H F no non constant dvsor of F (X 2 ) n IF p [X 2 ] dvdes h n R} D F := {f = f(x 2 ) IF p [X 2 ] f s monc and f dvdes F (X 2 )} F := {f = f(x 2 ) IF p [X 2 ] f s rreducble n IF p [X 2 ] and deg X 2(f) > } G := {f = f(x 2 ) IF p [X 2 ] f = gg wth g g rreducble n IF p [X 2 ]} F k,ε := D X 2k ε F G k,ε := D X 2k ε G Followng ths notaton, the monc solutons of the self-dual skew equaton are the elements of H X 2k ε. 4

6 Proposton 2 Consder p a prme number, θ the Frobenus automorphsm over IF p 2, R = IF p 2[X; θ], k a postve nteger, s, t two ntegers such that k = p s t and p does not dvde t. The number of self-dual (θ, ε)-constacyclc codes of dmenson k over IF p 2 s where and #H X 2k ε = N ε f F k,ε #H f ps f G k,ε #H f ps #H (X 2 +) ps f p = 2 N = #H (X 2 ) ps f k (mod 2) and p odd #H (X 2 ) ps #H (X 2 +) ps f k 0 (mod 2) and p odd N = { #H(X 2 +) ps f k (mod 2) and p odd f k 0 (mod 2) and p odd. Proof. Consder the factorzaton of X 2t ε over IF p [X 2 ] nto the product of dstnct rreducble polynomals of IF p [X 2 ] and splt ths product nto two sub-products, the product of self-recprocal rreducble factors and the product of non self-recprocal rreducble factors. In ths second product, factors appear by pars (g, g g) therefore X 2k ε = (X 2t ε) ps = r = f ps where f = f (X 2 ) s self-recprocal, ether rreducble n IF p [X 2 ] or product of two dstnct rreducble polynomals g (X 2 ) and g (X2 ) of IF p [X 2 ]. Followng [3], one has. H X 2k ε = {lcrm(h,..., h r ) h H f p s } ([3], Proposton 28); 2. If h belongs to H X 2k ε, then h = lcrm(h,..., h r ) where h ([3], Proposton 28, pont (2)). H f p s Therefore, the followng applcaton φ s well defned and s njectve : { HX φ : 2k ε H f p s H f p s r h (h,..., h r ), h ps = gcrd(f, h ). ps = gcrd(f, h ) and h Let us prove that φ s surjectve. Consder (h,..., h r ) n H f p s H f p s and h = r lcrm(h,..., h r ). Accordng to pont., the skew polynomal h belongs to H X 2k ε. It remans to prove that for all n {,..., r}, h = gcrd(f ps, h ). Accordng to pont 2., h = lcrm( h,..., h r ) where h = gcrd(f ps, h ) and h H f p s. Consder n {,..., r}, one has h h = f ps and f s central, therefore h dvdes f ps on the rght. Furthermore h = lcrm(h,..., h r ) therefore h dvdes h on the left and h dvdes h on the rght (see Remark ). As h s the greatest common rght dvsor of f ps and h, h dvdes h on the rght. Furthermore h h = h h = f ps so h and h have the same degree and h = h ps = gcrd(f, h ). To conclude φ s bjectve and #H X 2k ε = r = where N ε = deg(f )= #H f ps. #H f p s = N ε f F k,ε #H f ps f G k,ε #H f ps 5

7 Let us determne N ε n the three followng cases : p = 2, ε = ; p odd prme, ε = and p odd prme, ε =. For p = 2, the self-recprocal polynomal of degree n X 2 dvdng X 2k s X 2 + therefore N = #H (X 2 +) ps. For p odd prme, the self-recprocal polynomals of degree n X 2 dvdng X 2k are X 2 f k s odd; X 2 and X 2 + f k s even therefore, { N = #H (X 2 ) ps f k (mod 2) #H (X 2 ) ps #H (X 2 +) ps f k 0 (mod 2). For p odd prme and k even number, X 2k + has no self-recprocal factor of degree n X 2. If k s odd, X 2 + s the only self-recprocal polynomal of degree n X 2 dvdng X 2k +. Therefore, { #H(X N = 2 +) ps f k (mod 2) f k 0 (mod 2). The rest of the paper wll be devoted to the enumeraton of the elements of the set H X 2k ε when k s a power of p (Secton 3), k s coprme wth p (Secton 4) and k s any nteger (Secton 5). Followng Proposton 2, the man task wll consst n constructng H f p s for f = X 2 ±, f n F and f n G. The man dffculty comes from the non uncty of the factorzaton of skew polynomals n the Ore rng R. In Secton 3, one assumes that k s a power of p, therefore X 2k ε factorzes over IF p [X 2 ] as X 2k ε = (X 2 ε) ps and the self-dual skew equaton splts nto one sngle ntermedate skew equaton. For s > 0, t s solved by usng a partton and factorzaton propertes specfc to IF p 2[X; θ]. 3 Self-dual θ-cyclc and θ-negacyclc codes wth dmenson p s over IF p 2. The am of ths secton s to construct and to enumerate self-dual θ-cyclc and θ-negacyclc codes over IF p 2 whose dmenson s p s where θ s the Frobenus automorphsm. Recall that over IF 4, there s one sngle self-dual cyclc code of dmenson 2 s. When p s an odd prme number there s no self-dual cyclc code over IF p 2 and there are p s + self-dual negacyclc codes of dmenson p s (Corollary 3.3 of [5]). Lastly, there are only three self-dual θ-cyclc codes of dmenson 2 s > over IF 4 (Corollary 26 of [3]). In what follows one proves that the number of self-dual θ-cyclc and θ-negacyclc codes of dmenson p s over IF p 2 s exponental n the dmenson p s when p s an odd prme number (Proposton 4 and Table ). In order to construct the set H X 2k ε = H (X 2 ε) ps, factorzaton propertes specfc to IF p 2[X; θ] wll be useful. The followng proposton enables to characterze the skew polynomals that have a unque factorzaton nto the product of monc lnear skew polynomals dvdng X 2 ε (see also Proposton 6 of [3]). Proposton 3 Consder p a prme number, θ the Frobenus automorphsm over IF p 2, R = IF p 2[X; θ], m a nonnegatve nteger, f(x 2 ) n IF p [X 2 ] rreducble and h = h h m n R where h s rreducble n R, monc and dvdes f(x 2 ). The followng assertons are equvalent : 6

8 () The above factorzaton of h s not unque. () f(x 2 ) dvdes h. () There exsts n {,..., m } such that h h + = f(x 2 ). Proof. Consder f(x 2 ) IF p [X 2 ] rreducble wth degree d > such that f (X 2 ) = f(x 2 ). Accordng to [5], page 6 (or Lemma.4. of [4] wth e = 2), as f(x 2 ) s rreducble n the center of R, the skew polynomal f(x 2 ) has ((p 2 ) d )/(p d ) = p d + rreducble monc rght factors of degree d n R, n partcular t s reducble n R. Accordng to Proposton 6 of [3] the ponts (), () and () are therefore equvalent. Corollary Consder p a prme number, θ the Frobenus automorphsm over IF p 2, R = IF p 2[X; θ], m a nonnegatve nteger, ε n {, } and h = (X + λ ) (X + λ m ) n R where λ p+ = ε. The followng assertons are equvalent : () The above factorzaton of h s not unque. () X 2 ε dvdes h. () There exsts n {,..., m } such that (X +λ ) (X +λ + ) = X 2 ε.e. λ λ + = ε. Proof. Ths a consequence of Proposton 3 wth f(x 2 ) = X 2 ε. It suffces to notce that X + λ dvdes X 2 ε f and only f λ p+ = ε. In ths case (X + λ ) (X + λ + ) = X 2 + (λ + ε λ + )X + λ λ + and (X + λ ) (X + λ + ) = X 2 ε λ λ + = ε. The elements of H (X 2 ε) ps who have a unque factorzaton n R nto the product of monc rreducble skew polynomals are therefore not dvsble by X 2 ε. In what follows one constructs for m n IN the set of elements of H (X 2 ε) m who are not dvsble by X2 ε. Recall that one denotes H (X 2 ε) m ths set of elements (see notatons n Secton 2) : H (X 2 ε) m := {h H (X 2 ε) m X2 ε does not dvde h}. Lemma Consder p a prme number, θ the Frobenus automorphsm, R = IF p 2[X; θ], m a nonnegatve nteger and ε n {, }. Assume that p s odd and m s odd, then the number of elements of H (X 2 ε) m s #H (X 2 ε) m = { 0 f ε =, p (mod 4) or ε =, p 3 (mod 4) 2p m 2 f ε =, p 3 (mod 4) or ε =, p (mod 4). Assume that p s equal to 2, then the number of elements of H (X 2 +) m 0 f m > 2 #H (X 2 +) m = 2 f m = 2 f m =. Proof. s 7

9 One frst proves that the elements h of H (X 2 ε) m are where h = (X + λ ) (X + λ m ) {,..., m}, λ p+ = ε {,..., m }, λ λ + ε λ 2 = j {,..., m 2 }, (λ 2jλ 2j+ ) 2 =. (3) Namely, consder h n H (X 2 ε) m. As h dvdes (X2 ε) m and as X 2 ε s rreducble wth degree n IF p [X 2 ], h s a (non necessarly commutatve) product of lnear monc skew polynomals dvdng X 2 ε (Lemma 3 (2) of [3] or [5] page 6). Furthermore, the degree of h s equal to m (because deg(h h) = 2m) therefore one has : h = (X + λ ) (X + λ m ) where λ IF p 2, λ p+ = ε. In partcular, the frst relaton of (3) s satsfed. As X 2 ε does not dvde h, accordng to Corollary : therefore {,..., m }, (X + λ ) (X + λ + ) X 2 ε (4) {,..., m }, λ λ + ε whch s the second relaton of (3). The followng expresson of h can be obtaned usng an nducton argument (left to the reader) : where for n {,..., m}, λ s defned by : h = (X + λ m ) (X + λ ) { /λ ε (λ λ ) 2 f (mod 2) λ := ε /λ (λ λ ) 2 f 0 (mod 2). (5) Furthermore, X 2 ε does not dvde h, otherwse X 2 ε would dvde h, therefore {,..., m }, (X + λ + ) (X + λ ) X 2 ε. (6) The relaton h h = (X 2 ε) m can be wrtten (X + λ m ) (X + λ ) (X + λ ) (X + λ m ) = (X 2 ε) m. (7) As X 2 ε s central, the factorzaton of the skew polynomal (X 2 ε) m nto the product of monc skew polynomals dvdng X 2 ε s not unque, therefore, accordng 8

10 to Corollary, X 2 ε s necessarly the product of two consecutve monc lnear factors of the left hand sde of (7). Accordng to (4) and (6), the only possblty s (X + λ ) (X + λ ) = X 2 ε. As X 2 ε s central, the relaton (7) can be smplfed and one gets (X + λ m ) (X + λ 2 ) (X + λ 2 ) (X + λ m ) = (X 2 ε) m. Usng the same argument as before, one gets (X + λ 2 ) (X + λ 2 ) = X 2 ε. (X + λ m ) (X + λ m ) = X 2 ε. From the equaltes above, one deduces that {,..., m}, λ λ = ε and usng the defnton of λ gven n (5), one gets λ 2 = (thrd relaton of (3)) and for odd, (λ λ + ) 2 = (fourth relaton of (3)). Conversely, consder h = (X + λ ) (X + λ m ) where λ,..., λ m are defned by (3). Accordng to the frst relaton of (3), the monc skew polynomals X + λ dvde X 2 ε. Accordng to the second relaton of (3) and to Corollary, X 2 ε does not dvde h. Lke prevously the skew polynomal h s equal to (X + λ m ) (X + λ ) where λ s defned by the relatons (5). Furthermore, accordng to the thrd and fourth relatons of (3), f s odd, (λ λ ) 2 =, so for all n {,..., m}, λ λ = ε and X 2 ε = (X + λ ) (X + λ ). The product h h can be smplfed as follows : h h = (X + λ m ) (X + λ ) (X + λ ) (X + λ m ) = (X 2 ε) (X + λ m ) (X + λ 2 ) (X + λ 2 ) (X + λ m ) (because X 2 ε s central). = (X 2 ε) m (X + λ m ) (X + λ m ) = (X 2 ε) m and one concludes that h belongs to H (X 2 ε) m. The relatons (3) enable to count the number of elements of H (X 2 ε) m. Namely accordng to Corollary, the elements of H (X 2 ε) m have a unque factorzaton nto the product of monc skew lnear polynomals dvdng X 2 ε. Therefore the number of elements of the set H (X 2 ε) m s the number of m-tuples (λ,..., λ m ) of (IF p 2) m satsfyng the condtons (3). Assume that p = 2 and that m s an nteger greater than 2. Then the condtons λ 2 λ 3 and (λ 2 λ 3 ) 2 = are not compatble, therefore the set H (X 2 ) m s empty. If m =, t s reduced to {X + } (see Example ). If m = 2, the set H (X 2 ) m s equal 9

11 to {(X + λ ) (X + λ 2 ) λ =, λ 2 } = {(X + ) (X + a), (X + ) (X + a 2 )} where a 2 + a + = 0. Assume that p and m are odd, then the condtons (3) can be smplfed as follows : λ 2 = λ 2 ελ {,..., m}, λ p+ = ε j {,..., (m )/2}, λ 2j+ = ε/λ 2j j {,..., (m 3)/2}, λ 2j+2 λ 2j Frst, the condtons λ 2 = and λp+ = ε mply ( ) (p+)/2 = ε so H (X 2 ε) m s empty f p 3 (mod 4) and ε = or p (mod 4) and ε =. If p 3 (mod 4) and ε = or p (mod 4) and ε =, then there are two possbltes for λ, p possbltes for λ 2, one possblty for λ 3, p possbltes for λ 4, one for λ 5, and m so on, thererefore H (X 2 ε) m has 2p 2 elements. Remark 2 If m s odd, one can smplfy the relatons (3) by takng α 0 = λ, α = λ 2 and for n {2,..., (m )/2}, α = λ 2. Therefore one gets : H (X 2 ε) m = {(X + α 0) (X 2 + 2α X + ε) (X 2 + 2α (m )/2 X + ε) α0 2 =, α εα 0, {0,..., (m )/2}, α p+ = ε, {2,..., (m )/2}, α α }. To descrbe the set H (X 2 ε) ps one uses the followng partton : Lemma 2 Consder p a prme number, θ the Frobenus automorphsm over IF p 2, R = IF p 2[X; θ], s n IN and f = f(x 2 ) {X 2 ± } F. One has the followng partton : H f p s = ps 2 f H f p s 2. (8) =0 Proof. Consder M = ps 2, h = h(x) n H f ps and the bggest nteger n {0,..., M} such that f dvdes h. Consder H = H(X) n R such that h = f H and f does not dvde H. As f s central, h = f H therefore H H = f ps 2 and H belongs to H f p s 2. Conversely, f H n H f p s 2, then f H belongs to H f p s. Furthermore consder >, H n H f p s 2 and H n H f p s 2 such that f H = f H then f dvdes H, whch s mpossble as f does not dvde H. Therefore, for, the sets f H f p s 2 and f H f p s 2 are dsjont. Remark 3 If p = 2 and f(x 2 ) = X 2 +, accordng to Lemma 2, one gets the followng partton : 0

12 2 s H (X 2 +) 2s = (X 2 + ) H (X 2 +) 2s 2. =0 Accordng to Lemma, the sets H (X 2 +) 2s 2 are empty when 2 s 2 > 2 and H (X 2 +) 2 = {(X + ) (X + a), (X + ) (X + a 2 )} where a 2 + a + = 0. Therefore : H (X 2 +) 2s = (X 2 + ) 2s H (X 2 +) 0 (X 2 + ) 2s H (X 2 +) 2 = {(X + ) 2s, (X + ) 2s (X + a), (X + ) 2s (X + a 2 )} One gets that for s > 0 there are only three self-dual θ-cyclc codes of dmenson 2 s over IF 4 (see also Corollary 26 of [3]). Proposton 4 below gves a formula for the number of self-dual θ-cyclc and θ-negacyclc codes whose dmenson s a power of p when p s an odd prme number. The results are also summed up n Table. Proposton 4 Consder p an odd prme number, s an nteger, ε n {, } and θ the Frobenus automorphsm over IF p 2. The number of self-dual (θ, ε)-constacyclc codes of dmenson p s over IF p 2 s 0 f ε =, p (mod 4) or ε =, p 3 (mod 4) 2 p(ps +)/2 f ε =, p 3 (mod 4) or ε =, p (mod 4). p Proof. Consder R = IF p 2[X; θ]. The number of self-dual (θ, ε)-constacyclc codes of dmenson p s over IF p 2 s equal to #H X 2p s ε. Accordng to Lemma 2, one has the followng partton : M H X 2p s ε = (X 2 ε) H (X 2 ε) ps 2 =0 where M = ps 2. Accordng to Lemma, each set H (X 2 ε) ps 2 s empty f ε ( ) p+ 2 and has 2p M elements f ε = ( ) p+ 2. Therefore, f ε ( ) p+ 2, H X 2p s ε s empty and otherwse t has M =0 2p M = 2 pm+ p = 2 p(ps +)/2 p elements. Example 3 Accordng to Corollary 3.3 of [5], there are 4 self-dual negacyclc codes of dmenson 3 over IF 9. The correspondng skew check polynomals are the polynomals (X γ) (X + γ) 3 IF 9 [X] where s n {0,, 2, 3} and γ 2 =. Accordng to Proposton 4, for θ : x x 3 Frobenus automorphsm over IF 9, there are 2 (3 (3+)/2 )/(3 ) = 8 self-dual θ-cyclc codes of dmenson 3 over IF 9. Ther skew check polynomals are the elements of H X 6 and accordng to Lemma 2, H X 6 = H (X 2 ) 3 (X 2 ) H (X 2 ). The sets H (X 2 ) (wth cardnal 2) and H (X 2 ) 3 (wth cardnal 6) are constructed wth Lemma and Remark 2 : H X 2 = {X + α 0 α 2 0 =, α 4 0 = } = {X + γ, X γ}

13 p negacyclc θ-cyclc θ-negacyclc p 3 (mod 4) p s + 2 p(ps +)/2 p p (mod 4) p s p(ps +)2 0 p Table : Numbers of self-dual negacyclc (Corollary 3.3 of [5]), θ-cyclc (Proposton 4) and θ-negacyclc (Proposton 4) codes over IF p 2 of dmenson p s wth p odd prme number and θ : x x p. and H (X 2 ) 3 = {(X + α 0) (X 2 + 2α X + ) α 0 = ±γ, α α 0, α {±γ, ±}}. The 2 3 = 6 elements of H (X 2 ) 3 are lsted below : (X + γ) (X 2 + 2X + ) = X 3 + (γ )X 2 + ( γ)x + γ (X + γ) (X 2 + X + ) = X 3 + (γ + )X 2 + (γ + )X + γ (X + γ) (X 2 2γX + ) = X 3 + γ (X γ) (X 2 + 2X + ) = X 3 + ( γ )X 2 + ( + γ)x γ (X γ) (X 2 + X + ) = X 3 + ( γ + )X 2 + ( γ + )X γ (X γ) (X 2 + 2γX + ) = X 3 γ. Proposton 4 enables also to smplfy Proposton 2 as follows. It wll be useful n the two next sectons. Proposton 5 Consder p a prme number, θ the Frobenus automorphsm over IF p 2, R = IF p 2[X; θ], k a postve nteger, s, t two ntegers such that k = p s t and p does not dvde t. The number of self-dual (θ, ε)-constacyclc codes over IF p 2 wth dmenson k s where N = #H X 2k ε = N ε f F k,ε #H f ps f G k,ε #H f ps 0 f k (mod 2) and p (mod 4) or k 0 (mod 2) and p odd f s = 0 and p = 2 3 f s > 0 and p = 2 2 p(ps +)/2 p f k (mod 2) and p 3 (mod 4) and 0 f k (mod 2) and p 3 (mod 4) f k 0 (mod 2) and p odd N = 2 p(ps +)/2 f k (mod 2) and p (mod 4). p Proof. One starts wth Proposton 2 where the expresson of N ε s gven n functon of #H (X 2 ±) ps. One smplfes N ε thanks to Proposton 4 (for p odd prme) and Remark 3 (for p = 2). 2

14 4 Self-dual θ-cyclc and θ-negacyclc codes wth dmenson prme to p over IF p 2. Over IF p 2 wth p odd prme number, there s no self-dual cyclc code and the number of selfdual negacyclc codes wth dmenson k prme to p s gven n Theorem 2 of [7]. Self-dual cyclc codes over IF 4 wth odd dmenson are studed n [0], The am of ths secton s to construct and to enumerate self-dual θ-cyclc and θ-negacyclc codes over IF p 2 whose dmenson s prme to p when p s a prme number and θ s the Frobenus automorphsm (Proposton 8). The startng pont of the study s Proposton 5 appled n the partcular case when the dmenson k of the code s prme to p (.e. k = p s t, s = 0 and p t). One wants to determne now the set H f for f n F G. Consder f = f(x 2 ) n F G. Note that f f s n F then the degree d of f n X 2 s even (see exercse 3.4 page 4 of [2]). Consder δ n IN such that d = 2δ where δ s n IN. Let h n R monc wth degree 2δ : h = X 2δ + 2δ =0 h X = (X 2δ + ( δ =0 h δ ) 2X 2 ) + X =0 θ(h 2+)X 2. The skew recprocal polynomal h of h s h = + δ = h 2δ 2X 2 + ( δ =0 θ(h 2δ 2 )X 2 ) X. One can assocate to h the two polynomals defned n IF p 2[Z] by δ δ A(Z) := Z δ + h 2 Z and B(Z) := θ(h 2+ )Z. (9) =0 Usng the commutaton law (), one gets that h h = f(x 2 ) f and only f the followng polynomal relatons n IF p 2[Z] are satsfed : ( ) ( ) Z δ A A(Z) + Z δ B B(Z) h 0 f(z) = 0 ( Z ) Z ( ) (0) Z δ A Θ(B)(Z) + Z δ B Θ(A)(Z) = 0 Z Z where Θ : a Z a p Z. In the rest of the secton, the followng notaton wll be useful : Notaton 2 Consder P (X 2 ) = P X 2 n IF p [X 2 ], one denotes P (Z) the polynomal n IF p 2[Z] defned by P (Z) = P Z. For a n IF p 2 and P (X 2 ) n IF p [X 2 ], P (a) s P a. The Frobenus automorphsm θ defned over IF p 2 s extended to IF p 2 and s denoted wth the same letter θ. Fndng (A, B) n IF p 2[Z] IF p 2[Z] satsfyng (0) wth A monc, deg(a) = δ and deg(b) δ enables to construct the elements h of H f. One frst consders the resoluton of (0) when B(Z) = 0. Ths amounts to fnd the elements of H f IF p 2[X 2 ]. =0 3

15 Lemma 3. Consder f = f(x 2 ) n F wth degree 2δ n X 2 and f(x 2 ) = f(x 2 ) Θ( f)(x 2 ) the factorzaton of f(x 2 ) n IF p 2[X 2 ]. { H f IF p 2[X 2 ] = f δ 0 (mod 2) { f(x 2 ), Θ( f)(x 2 )} f δ (mod 2) 2. Consder f = f(x 2 ) n G wth degree 2δ n X 2 and g(x 2 ) such that f(x 2 ) = g(x 2 )g (X 2 ). When δ s even, consder the factorzaton of g(x 2 ) n IF p 2[X 2 ] : g(x 2 ) = g(x 2 ) Θ( g)(x 2 ). H f IF p 2[X 2 ] = { {g(x 2 ), g (X 2 ), g(x 2 )Θ( g )(X 2 ), g (X 2 )Θ( g)(x 2 )} f δ 0 (mod 2) {g(x 2 ), g (X 2 )} f δ (mod 2) Proof. Recall that h s n H f f and only f (A(Z), B(Z)) defned by (9) satsfes the relaton (0). Furthermore h s n IF p 2[X 2 ] f and only f B(Z) = 0. The elements of H f IF p 2[X 2 ] are therefore characterzed by the relatons B(Z) = 0 and Z δ A ( ) Z A(Z) = h0 f(z) where h 0 s the constant term of A. Here are now necessary condtons for h belongng to H f \ IF p 2[X 2 ]. Lemma 4 Consder f = f(x 2 ) n F G wth degree 2δ n X 2, h n R monc wth degree 2δ and (A(Z), B(Z)) defned n (9). If h H f \ IF p 2[X 2 ] then () gcd(a(z), B(Z)) = () gcd(b(z), f(z)) =. Proof. () Assume that A(Z) and B(Z) have a common factor n IF p 2[Z] then accordng to the frst relaton of (0), ths factor must dvde f(z). Furthermore, B(Z) 0 and the degree of B(Z) s δ, therefore f(z) must have a nontrval factor n IF p 2[Z] wth degree δ. Necessarly δ s even, f = gg wth g = gθ( g) product of two rreducble polynomals of degree δ/2 n IF p 2[Z]. Wthout loss of generalty one can assume that g(z) s the common factor of A(Z) and B(Z) n IF p 2[Z]. Consder β such that g(β) = 0, a(z) and b(z) n IF p 2[Z] such that A(Z) = g(z)a(z) and B(Z) = g(z)b(z). From relatons (0), one gets that { Z δ/2 a (/Z) a(z) + Z δ/2 b (/Z) b(z) = λθ( g)(z) g (Z) Z δ/2 a (/Z) Θ(b)(Z) + Z δ/2 b (/Z) Θ(a)(Z) = 0 where λ s a nonzero constant. Consder u n IF p δ \ {0} such that a(γ) = u b(γ). Accordng to () evaluated at γ, { a(γ)a(/γ) + b(γ)b(/γ) = 0 γθ(b)(γ)a(/γ) + Θ(a)(γ)b(/γ) = 0. From the frst relaton, one deduces that a(/γ) = /u b(/γ) and from the second relaton, one deduces γ/uθ(b)(γ) + Θ(a)(γ) = 0 so ( γ/u) p b(γ p ) + a(γ p ) = 0. As a(γ p )a(/γ p ) + b(γ p )b(/γ p ) = 0, one gets a(/γ p ) = (u/γ) p b(/γ p ). Therefore () 4

16 a(γ) = u b(γ) a(/γ) = /u b(/γ) a(γ p ) = γ p /u p b(γ p ) a(/γ p ) = u p /γ p b(/γ p ). In partcular, the polynomal Z δ/2 a (/Z) a(z) + Z δ/2 b (/Z) b(z) cancels at γ, /γ, γ p, /γ p, therefore t s dvsble by f(z), whch s mpossble because of the frst relaton of (). () Assume that B(Z) and f(z) are not coprme n IF p 2[Z]. Necessarly δ s even, f = gg wth g = gθ( g) product of two rreducble polynomals of degree δ/2 n IF p 2[Z]. Wthout loss of generalty one can assume that g(z) s the common factor of f(z) and B(Z) n IF p 2[Z]. Consder β such that g(β) = 0, one has B(β) = 0, B(β ), B(β p ), B(β p ) 0. Furthermore Θ(B)(β p ) = 0 so accordng to the second relaton of (0), Θ(A)(β p )B(/β p ) = 0 and A(β) = 0. Therefore A(Z) and B(Z) have a common factor n IF p 2[Z], whch s mpossble accordng to (). To characterze the elements h of H f such that h does not belong to IF p 2[X 2 ], one wll use the followng ratonal nterpolaton problem or Cauchy nterpolaton problem (Secton 5.8 of [20]): gven 2δ dstnct ponts x 0,..., x 2δ n IF p 2δ and 2δ values y 0,..., y 2δ n IF p 2δ, fnd a ratonal functon r/t IF p 2δ(Z) such that (RI) : t(x ) 0, r(x ) t(x ) = y for 0 < 2δ, deg(r) < δ +, deg(t) δ Note that ths problem can be rewrtten as gcd(t, f) =, r P t (mod f), deg(r) < δ +, deg(t) δ where f = 2δ =0 (Z x ), P has degree 2δ and P (x ) = y for 0 < 2δ. Ths problem can be solved usng extended Eucldean algorthm ([20]). 4. Constructon of H f for f n F For f n F, one frst gves a characterzaton of the elements h of H f \ IF p 2[X 2 ]. Lemma 5 Consder f = f(x 2 ) n F wth degree d = 2δ n X 2 and α n IF p 2δ f(α) = 0. such that. Consder h n R monc wth degree d = 2δ and (A(Z), B(Z)) defned n (9). Then h H f \ IF p 2[X 2 ] f and only f there exsts u n IF p d such that A(α) = u B(α) A (α p ) = α p /u p B(α p ) gcd(a(z), B(Z)) = gcd(b(z), f(z)) = (2) 5

17 and { u pδ + = f δ even u pδ = /α f δ odd. (3) 2. Consder u n IF p d such that the condton (3) s satsfed. There exsts a unque soluton (A, B) n IF p 2[Z] IF p 2[Z] to (2) wth A monc, deg(a) = δ, deg(b) δ. 3. The set H f \ IF p 2[X 2 ] has p δ + elements f δ s even and p δ elements f δ s odd. Proof. As f belongs to F, f(α ) = 0 so there exsts n {0,..., d } such that α = α p. As α p2 = (α ) p = α, and as f(x 2 ) s rreducble n IF p [X 2 ] wth degree d = 2δ, necessarly = δ and α = α pδ.. Consder h n R wth degree d = 2δ and (A(Z), B(Z)) defned by (9). If h belongs to H f \ IF p 2[X 2 ] then the relatons (0) are satsfed by (A(Z), B(Z)) wth { A(α) = u B(α) B(Z) 0. Consder u, v n IF p d such that A(α p ) = v B(α p. Accordng to ) Lemma 4, gcd(b, f) = so B(α), B(α p ) 0. If δ s even, then A(α ) = A(α) pδ and B(α ) = B(α) pδ. Evaluatng (0) at α one gets : { u pδ u + = 0 α u pδ + θ (v) = 0 therefore v = α p /u p and u pδ + =. If δ s odd, then A(α ) = A(α p ) pδ and B(α ) = B(α p ) pδ. Evaluatng (0) at α, one gets : { v pδ u + = 0 α v pδ + θ (v) = 0 therefore v = α p /u p and u pδ = /α. Conversely, f there exsts u n IF p d such that (2) and (3) are satsfed, then one can ( ) ( ) check that the polynomals Z δ A A(Z) + Z δ B B(Z) h 0 f(z) and Z Z ( ) ( ) Z δ A Θ(B)(Z) + Z δ B Θ(A)(Z) cancel at α and α p. Therefore the relatons Z Z (0) are satsfed and h belongs to H f. As gcd(b, f) =, B 0 so h belongs to H f \ IF p 2[X 2 ]. 2. Consder u n IF p 2δ such that the condton (3) s satsfed. Consder the 2δ ponts (x, y ) 0 2δ defned by (x, y ) = { (θ (α), θ (u)) f 0 (mod 2) (θ (α), θ (α/u)) f (mod 2). Accordng to Corollary 5.8 of [20] there exsts two nonzero polynomals A and B n IF p 2δ[Z] such that deg(a) < δ +, deg(b) δ and A(x ) = y B(x ). Wthout loss of 6

18 generalty, one can assume that A and B are coprme and that A s monc. Furthermore, the set (x, y ) 0 2δ s stable under the acton of θ 2, therefore (Θ 2 (A), Θ 2 (B)) satsfes the relatons Θ 2 (A)(x ) = y Θ 2 (B)(x ). As A and B are coprme and A s monc, Θ 2 (A) = A, Θ 2 (B) = B. Therefore A and B are polynomals of IF p 2[Z]. Consderng the two frst relatons A(x 0 ) = y 0 B(x 0 ) and A(x ) = y B(x ) one gets the relaton (2), so the relaton (0) s satsfed and the skew polynomal h assocated to A and B belongs to H f. As A and B are coprme, B and f are also coprme (see Lemma 4). Assume that deg(a) δ, then Z δ A(/Z)A(Z) + Z δ B(/Z)B(Z) would be the zero polynomal and h would satsfy h h = 0 whch s mpossble. Therefore for u n IF p 2δ such that the condton (3) s satsfed, there exsts (A, B) n IF p 2[Z] IF p 2[Z] satsfyng (2) wth A monc, deg(a) = δ and deg(b) δ. The uncty of (A, B) follows from the fact that A/B s the unque soluton to the ratonal nterpolaton problem (RI) wth A and B coprme (Corollary 5.8 of [20] ). 3. Accordng to., H f = u {h R h monc, deg(h) = d, (A, B) defned n (9) soluton of (2)} where u satsfes (3). Accordng to 2., for each u satsfyng (3), there s a unque h n R monc of degree d such that (A, B) defned n (9) s soluton of (2). Therefore, the number of elements of H f s the number of u n IF p d satsfyng u pδ + = f δ s even and u pδ = /α s δ s odd. Proposton 6 Consder p a prme number, m a postve nteger, θ the Frobenus automorphsm over IF p 2 and R = IF p 2[X; θ]. Let f = f(x 2 ) n F and d = 2δ ts degree n X 2, then the set H f = H f has + p δ elements. Proof. The elements of H f IF p 2[X 2 ] are gven n pont. of Lemma 3: there are two elements f δ s odd and no element f δ s even. The elements of H f who do not belong to IF p 2[X 2 ] are gven n pont 3. of Lemma 5. There are p δ elements f δ s odd and p δ + elements f δ s even. Example 4 Consder p = 2, θ the Frobenus automorphsm over IF 4 = IF 2 (a) where a 2 + a + = 0, R = IF 4 [X; θ] and f(x 2 ) = X 4 + X 2 + n F. Consder h = X 2 + h X + h 0 n R, A(Z) = Z + h 0 and B(Z) = θ(h ) n IF 4 [Z]. One has h h = X 4 + X 2 + { ZA (/Z) A(Z) + ZB (/Z) B(Z) = h0 (Z 2 + Z + ) ZA (/Z) Θ(B)(Z) + B (/Z) Θ(A)(Z) = 0. If h = 0, one gets h h = X 4 + X 2 + f and only f ZA ( Z ) A(Z) = h0 (Z 2 + Z + ). As Z 2 + Z + = (Z + a)(z + a 2 ) and a 2 = /a, one gets A(Z) = Z + a or A(Z) = Z + a 2 (see Lemma 3), therefore f h = 0, h = X 2 + a or h = X 2 + a 2. Followng Lemma 5, f h 0, then h h = X 4 + X 2 + f and only f there exsts u n IF 4 such that u = /a and { A(a) = u B(a) = /a B(a) A(a 2 ) = a2 B(a 2 ) = a B(a 2 ). u 2 7

19 Therefore when h 0, one gets h H X 4 +X 2 + f and only h = X 2 + X +. As a concluson the set H X 4 +X 2 + s H X 4 +X 2 + = H X 4 +X 2 + = {X 2 + a, X 2 + a 2, X 2 + X + }. Example 5 Consder p = 2, θ the Frobenus automorphsm over IF 4 = IF 2 (a) wth a 2 +a+ = 0 and R = IF 4 [X; θ]. The skew polynomal X 2 + X 6 + belongs to F and ts degree n X 2 s 6. Consder α n IF 2 6 such that α 6 + α 3 + = 0. Accordng to Lemma 3 the elements of H X 2 +X 6 + wth no term of odd degree are X 6 + a and X 6 + a 2. Accordng to Lemma 5, the other elements of H X 2 +X 6 + are the monc skew polynomals h of degree 6 such that (A(Z), B(Z)) defned by relatons (9) are solutons of (2) wth u 7 = /α. The table below gves the solutons correspondng to the seven problems (2). u h + α X 6 + a 2 X 5 + ax 4 + ax 2 + a 2 X + a 2 + α + α 5 X 6 + X 5 + a 2 X 4 + ax 2 + X + α + α 3 + α 4 + α 5 X 6 + X 4 + a 2 X 3 + a 2 X 2 + a 2 α 5 X 6 + X α 3 + α 4 X 6 + X 4 + ax 3 + ax 2 + a α + α 3 + α 4 X 6 + X 5 + ax 4 + a 2 X 2 + X + + α 3 + α 4 + α 5 X 6 + ax 5 + a 2 X 4 + a 2 X 2 + ax + a The number of elements of H X 2 +X 6 + s 9 = Constructon of H f for f n G For f n G, one gves a characterzaton of the elements of H f \ IF p 2[X 2 ]. Lemma 6 Consder f = f(x 2 ) n G wth degree 2δ n X 2 and g rreducble n IF p [X 2 ] such that f(x 2 ) = g(x 2 )g (X 2 ). Consder β n IF p δ such that g(β) = 0.. Consder h n R monc wth degree 2δ and (A(Z), B(Z)) defned by relatons (9). Then h H f \ IF p 2[X 2 ] f and only f there exsts u n IF p d such that and A(β) = u B(β) A(/β) = /u B(/β) A(β p ) = β p /u p B(β p ) A(/β p ) = u p /β p B(/β p ) gcd(a(z), B(Z)) = gcd(b(z), f(z)) = { u pδ = f δ even u pδ + = β f δ odd. (4) (5) 2. Consder u n IF p d such that the condton (5) s satsfed. There exsts a unque soluton (A, B) n IF p 2[Z] IF p 2[Z] to (4) wth A monc, deg(a) = δ, deg(b) δ. 8

20 3. The set H f \ IF p 2[X 2 ] has p δ elements f δ s even and p δ + elements f δ s odd. Proof.. Consder h n R wth degree d = 2δ and (A(Z), B(Z)) defned by (9). If h H f \IF p 2[X 2 ] then (A(Z), B(Z)) satsfes the relaton (0) wth B(Z) 0. Accordng to Lemma 4, B s coprme wth f, therefore B(β), B(/β), B(β p ) and B(/β p ) 0. Consder u n IF p δ such that A(β) = u B(β). Accordng to (0) evaluated at β, { A(β)A(/β) + B(β)B(/β) = 0 βθ(b)(β)a(/β) + Θ(A)(β)B(/β) = 0. From the frst relaton, one deduces that A(/β) = /u B(/β) and from the second relaton, one deduces β/uθ(b)(β)+θ(a)(β) = 0 so ( β/u) p B(β p )+A(β p ) = 0. As A(β p )A(/β p ) + B(β p )B(/β p ) = 0, one gets A(/β p ) = (u/β) p B(/β p ). Therefore the relatons (4) are satsfed. Furthermore, f δ s odd, one gets another constrant, namely as β = β pδ one gets Θ(A)(β) = (Θ(A)(β p )) pδ = (u p B(β) p ) pδ = u pδ Θ(B)(β). Furthermore, β/uθ(b)(β)+ Θ(A)(β) = 0, so β/u + u pδ = 0 and u pδ + = β. The relatons (5) are therefore satsfed. Conversely, f there exsts u n IF p δ such that (4) and (5) are satsfed, then one can ( ) ( ) check that the polynomals Z δ A A(Z) + Z δ B B(Z) h 0 f(z) and Z Z ( ) ( ) Z δ A Θ(B)(Z) + Z δ B Θ(A)(Z) cancel at β, /β, β p and /β p. Therefore Z Z the relatons (0) are satsfed and h belongs to H f. As gcd(b, f) =, B s nonzero so h belongs to H \ IF p 2[X 2 ]. 2. Consder u n IF p 2δ such that the condton (5) s satsfed. Consder the 2δ ponts (x, y ) 0 2δ defned by { (θ (x, y ) = (β), θ (u)) f 0 (mod 2), < δ (θ (β), θ (β/u)) f (mod 2), < δ { (θ (x +δ, y +δ ) = (/β), θ (/u)) f 0 (mod 2), < δ (θ (/β), θ (u/β)) f (mod 2), < δ. Accordng to Corollary 5.8 of [20] there exsts two nonzero polynomals A and B n IF p 2δ[Z] such that deg(a) < δ +, deg(b) δ and for n {0,..., 2δ }, A(x ) = y B(x ). Wthout loss of generalty, one can assume that A and B are coprme and that A s monc. Furthermore, as the the set of ponts {(x, y )} s stable under the applcaton of θ 2, A(Z) and B(Z) belong to IF p 2[Z]. Consderng the four relatons A(x 0 ) = y 0 B(x 0 ), A(x ) = y B(x ), A(x δ ) = y δ B(x δ ) and A(x δ+ ) = y δ+ B(x δ+ ) one gets the relaton (4), so the relaton (0) s satsfed and the skew polynomal h assocated to A and B belongs to H f. As A and B are coprme, B and f are also coprme (see Lemma 4). Assume that deg(a) δ, then 9

21 Z δ A(/Z)A(Z) + Z δ B(/Z)B(Z) would be the zero polynomal and h would satsfy h h = 0 whch s mpossble. Therefore for u n IF p 2δ such that the condton (5) s satsfed, there exsts (A, B) n IF p 2[Z] IF p 2[Z] satsfyng (4) wth A monc, deg(a) = δ and deg(b) δ wth A and B coprme. The uncty follows form the fact that A/B s the unque soluton to the ratonal nterpolaton problem (RI) wth A and B coprme ( Corollary 5.8 of [20] ). 3. Lke n Lemma 5, the number of elements of H f s deduced from ponts. and 2. Proposton 7 Consder p a prme number, θ the Frobenus automorphsm over IF p 2 and R = IF p 2[X; θ]. Let f = f(x 2 ) n G and d = 2δ ts degree n X 2, then the set H f has 3 + p δ elements and H f has + p δ elements. Proof. The result s deduced from pont 2. of Lemma 3 and pont 3. of Lemma 6. Furthermore f f(x 2 ) = g(x 2 )g (X 2 ) belongs to G, then H f(x 2 ) = H f(x 2 ) \ {g(x 2 ), g (X 2 )} has + p δ elements. Example 6 Consder IF 4 = IF 2 (a) where a 2 + a + = 0, θ the Frobenus automorphsm, R = IF 4 [X; θ] and f(x 2 ) = (X 6 + X 2 + )(X 6 + X 4 + ) n G wth degree 6 = 2 3 n X 2. Consder β n IF 2 3 such that β 3 + β 2 + = 0. Accordng to Lemma 3, the elements of H f wth no term of odd degree are X 6 + X 2 + and X 6 + X 4 +. Accordng to Lemma 6, the other elements of H f are deduced from the polynomals A(Z) and B(Z) of IF 4 [Z] satsfyng (4) wth u 9 = β, A(Z) monc of degree 3 and B(Z) of degree 2. For example take u = v 3 where v 6 +v 4 +v 3 +v+ = 0, then u 9 = β and the unque soluton (A, B) n IF 4 [Z] IF 4 [Z] of (4) wth A monc of degree 3 and B of degree 2 s (A, B) = (Z 3 + a, az 2 +a 2 Z+). Therefore, h(x) = (X 6 +a)+x (ax 4 +a 2 X 2 +) = X 6 +a 2 X 5 +ax 3 +X+a s an element of H f wth at least one non zero term of odd degree. The entre set H f s {X 6 + X 2 +, X 6 + X 4 +, X 6 + X 5 + ax 3 + a 2 X + a, X 6 + a 2 X 5 + X 4 + X 2 + ax +, X 6 + ax 5 +X 4 +X 2 +a 2 X+, X 6 +X 5 +X 4 +X 3 +X 2 +X+, X 6 +ax 4 +ax 3 +X 2 +a, X 6 +a 2 X 4 + a 2 X 3 +X 2 +a 2, X 6 +ax 5 +a 2 X 3 +X+a 2, X 6 +X 5 +a 2 X 3 +ax+a 2, X 6 +a 2 X 5 +ax 3 +X+a}. It has 2 δ + 3 = elements (Proposton 7). 4.3 Concluson The proposton below gves a formula for the number of self-dual θ-cyclc and θ-negacyclc codes over IF p 2 whose dmenson s prme to p. Tables 2 and 3 llustrate ths proposton over IF 4 and IF 9 and gve some elements of comparson wth cyclc and negacyclc codes. Proposton 8 Consder p a prme number, θ the Frobenus automorphsm over IF p 2, k a postve nteger not dvsble by p and ε n {, }. The number of self-dual (θ, ε)-constacyclc codes wth dmenson k defned over IF p 2 s N ε (p δ + ) (p δ + 3) f F k,ε f G k,ε where 2δ s the degree of f n X 2 and N ε s defned below : 20

22 0 f k (mod 2) and p (mod 4) or k 0 (mod 2) and p odd N = f p = 2 2 f k (mod 2) and p 3 (mod 4) N = 0 f k (mod 2) and p 3 (mod 4) f k 0 (mod 2) and p odd 2 f k (mod 2) and p (mod 4). Proof. Accordng to Proposton 5, wth s = 0, the number of self-dual (θ, ε)-constacyclc codes over IF p 2 wth dmenson k s #H X 2k ε = N ε f F k,ε #H f f G k,ε #H f where N ε satsfes the above condtons. Proposton 7. The fnal result follows from Proposton 6 and Example 7 Consder θ : x x 2 the Frobenus automorphsm over IF 4 = IF 2 (a) where a 2 + a + = 0. The self-dual θ-cyclc codes of dmenson 9 over IF 4 are characterzed by the monc solutons of the self-dual skew equaton h h = X 8 +. As X 8 + = (X 2 + )(X 4 + X 2 + )(X 2 + X 6 + ) n IF 2 [X 2 ] and as the polynomals X 4 + X 2 + and X 2 + X 6 + are self-recprocal and rreducble n IF 2 [X 2 ], the set F 9, s {X 4 + X 2 +, X 2 + X 6 + } and the set G 9, s empty. Accordng to Proposton 8, the number of self-dual θ-cyclc codes of dmenson 9 over IF 4 s (2 + ) (2 3 + ) = 27. More precsely the set H X 8 + s gven by H X 8 + = {lcrm(h, h 2, h 3 ) h H X 2 +, h 2 H X 4 +X 2 +, h 3 H X 2 +X 6 +} and the sets H X 2 +, H X 4 +X 2 + and H X 2 +X 6 + wth cardnaltes, 3 and 9 were prevously computed n Examples 2, 4 and 5. 5 Self-dual θ-cyclc and θ-negacyclc codes wth any dmenson over IF p 2. In ths secton, one constructs and enumerates all self-dual θ-cyclc and θ-negacyclc codes over IF p 2 where p s a prme number and θ s the Frobenus automorphsm. Lke n Secton 4, the startng pont of the constructon s Proposton 5, who enables to wrte the monc solutons of the self-dual skew equaton as least common rght multples of skew polynomals satsfyng ntermedate skew equatons. The man topc of ths secton s therefore to construct the ntermedate sets H f p s where s > 0 and f = f(x 2 ) belongs to F G. Frst, one assumes that f = f(x 2 ) belongs to F. 2

23 k c θ-c k c θ-c k c θ-c Table 2: Numbers of self-dual cyclc codes (c, Corollary of [0]) and θ-cyclc codes (θ-c, prop. 8) over IF 4 n odd dmenson k < 00 where θ : x x 2. k nc θ-c θ-nc k nc θ-c θ-nc Table 3: Numbers of self-dual negacyclc (nc, Theorem 2 of [7]), self-dual θ-cyclc (θ-c, prop. 8) and self-dual θ-negacyclc (θ-nc, prop. 8) codes over IF 9 n dmenson k 50 coprme wth 3 where θ : x x 3. 22

24 5. Constructon of H f p s for f n F The am of ths subsecton s to compute H f p s for f n F and to compute ts number of elements. The fnal result s gven n Proposton 9 and the man steps are summed up n Table 4. Consder f = f(x 2 ) s n F. Recall that accordng to Lemma 2, one has the partton : where for m n IN, the set H f m H f p s = s defned by ps 2 f H f p s 2 =0 H f m = {h H f m f does not dvde h}. Lemma 7 below generalzes Lemma and uses the same type of arguments lnked to the factorzaton of skew polynomals. Lemma 7 Consder p a prme number, θ the Frobenus automorphsm over IF p 2, R = IF p 2[X; θ], m a nonnegatve nteger and f = f(x 2 ) n F wth degree d = 2δ > n X 2.. The constant coeffcents of the elements of H f are squares n IF p The set H f m has ( + p δ )p δ(m ) elements and s equal to ( ) ( ) m h h m ν j h j H f, νj 2 = (h j ) 0, h j ν j h j ν ν m j= ν j. Proof. To smplfy the presentaton, the followng notatons wll be used n ths proof : h = h(x), f = f(x 2 ). d. Consder h = X d + h X n H f = H f. If p = 2, then h 0 s a nonzero element of IF 4 =0 and therefore s a suqare n IF 4. Assume that p s odd. Accordng to Secton 4, the δ δ polynomals A(Z) = Z δ + h 2 Z and B(Z) = θ(h 2+ )Z defned n (9) satsfy =0 the relatons (0). If f(z) and B(Z) are coprme then f(z) = f (Z) and Z δ B(/Z) are also coprme. Therefore the relatons (0) mply that f(z) = A(Z)Θ(A)(Z) ZB(Z)Θ(B)(Z), Z δ B(/Z) = h 0 Θ(B)(Z) and Z δ A(/Z) = h 0 Θ(A)(Z). In partcular, one has h 0 h p 0 = 0 so h 0 s a square. If f(z) and B(Z) are not coprme, then accordng to Lemma 4, B(Z) = 0 and usng Lemma 3, one gets A(Z) = f(z) or Θ( f)(z) where f(z) = f(z)θ( f)(z) s the factorzaton of f(z) nto rreducble polynomals of IF p 2[Z]. As f = f, the constant coeffcent of f s equal to, so one gets h p+ 0 = and h 0 s a square. =0 23

25 2. Consder h n H f m. As h dvdes f m and f s rreducble n IF p [X 2 ], all the rreducble factors of h dvde f and have the same degree d (Lemma 3 (2) of [3] or [5] page 6) : h = m H, H monc, deg(h ) = d, H f, H rreducble. = Furthermore, f does not dvde h, therefore accordng to Proposton 3, for all j n {... m }, H j H j+ s dstnct of f. Usng an nducton argument (left to the reader), one gets the followng expresson of h : m h = H m µ µ m m =0 where µ = (H H ) 0 s defned as the constant coeffcent of H H. Furthermore, ths factorzaton (nto the product of rreducble monc polynomals of same degree d dvdng f) s unque (because the factorzaton of h s unque). As the factorzaton of f m nto the product of rreducble factors s not unque (because f m s central), accordng to Proposton 3, f m = h h must have two consecutve rreducble monc factors whose product s f. As h and h do not possess two consecutve factors whose product s f, necessarly, µ H µ H = f and proceedng by nducton, one gets j {,..., m }, µ j H j µ j H j = f and H j+ µ j H j µ j. (6) Conversely, consder h = H H m wth µ j H j µ j H j = f, H j+ µ j H j µ j and µ j constant coeffcent of H H j, then h h = f m and H j H j+ f. Furthermore, the skew polynomals H j are all rreducble because they are nontrval factors of f and f s rreducble n IF p [X 2 ], therefore accordng to Proposton 3, the skew polynomal h s not dvsble by f and t belongs to H f m. The concluson follows thanks to the followng equvalence : h = H H m µ j H j µ j H j = f H j+ µ j H j µ j h = ( h ν ) h j h j = f h j+ ν j h j ν j ( ) h m ν m m j= ν j where µ j = (H H j ) 0 s the constant coeffcent of H H j, ν j s defned n IF p 2 by ν 2 j = (H j) 0 = (h j ) 0 and h j = (ν 0 ν j )H j ν 0 ν j. 3. The number of elements of H f m follows from the fact that H f has + p δ elements (Proposton 6). The constructon of the set H f p s for f n F s deduced from Lemma 2 and Lemma 7. The whole constructon s llustrated n Table 4. 24