The lengths of Hermitian Self-Dual Extended Duadic Codes

arxiv:math/0595v [math.co] 9 May 006 The lengths of Hermitian Self-Dual Extended Duadic Codes Lilibeth Dicuangco, Pieter Moree, Patrick Solé Abstract Duadic codes are a class of cyclic codes that generalizes quadratic residue codes from prime to composite lengths. For every prime power q, we characterize integers n such that there is a duadic code of length n over F q with an Hermitian self-dual paritycheck extension. We derive asymptotic estimates for the number of such n as well as for the number of lengths for which duadic codes exist. Keywords: Duadic codes, Extended cyclic codes, Hermitian self-dual codes, Splittings. Mathematics Subject Classification: N64, 94B5, N37 Introduction Duadic codes are a family of cyclic codes over fields that generalize quadratic residue codes to composite lengths. For a general introduction, see [], [6] and [5]. It can be determined when an extended duadic code is self-dual for the Euclidean scalar product ([]). In this work, we study for which n there exist duadic codes over F q of length n the extension of which by a suitable parity-check is self-dual for the Hermitian scalar product n+ i= x iy q i. First, we characterize the Hermitian self-orthogonal cyclic codes by their defining sets (Proposition 3.6), then the duadic codes (Proposition 4.4). Next, we study under what conditions the extension by a parity-check of a duadic code is Hermitian self-dual (Proposition 4.8). Finally, we derive by elementary means an arithmetic condition bearing on the divisors of n (Theorem 5.7) for the previous situation. This condition was arrived at in [9] using Mathematics Department, University of the Philippines, Diliman, Quezon City, 0 Philippines, ldicuangco@math.upd.edu.ph Max-Planck-Institut, Vivatsgasse 7, D-53 Bonn, Germany, moree@mpim-bonn.mpg.de CNRS, I3S ESSI, BP 45, Route des Colles, 06 903 Sophia Antipolis, France, sole@essi.fr

representation theory of groups. In an appendix, we derive asymptotic estimates for x large on A q (x), the number of integers x that are split by the multiplier µ q, and on D q (x), the number of possible lengths x of a duadic code. The proofs are based on analytic number theory. Preliminaries We assume the reader is familiar with the theory of cyclic codes (see e.g., [], []). Let q be a power of a prime p and let F q denote the Galois field with q elements. Let n be a positive integer such that gcd(n, q) =. Let R n = F q [x]/(x n ). We view a cyclic code over F q of length n as an ideal in R n. Let 0 < s < n be a nonnegative integer. Let C s = {s, sq, sq,...,sq rs }, where r s is the smallest positive integer such that sq rs s (mod n). The coset C s is called the q-cyclotomic coset of s modulo n. The subscript of C s is usually taken to be the smallest number in the set and is also taken as the coset representative. The distinct q-cyclotomic cosets modulo n partition the set {0,,,..., n }. Let α be a primitive nth root of unity in some extension field of F q. A set T {0,,,..., n } is called the defining set (relative to α) of a cyclic code C whenever c(x) C if and only if c(α i ) = 0 for all i T. In this paper, we assume implicitly that an nth root of unity has been fixed when talking of defining sets. A ring element e such that e = e is called an idempotent. Since gcd(n, q) =, the ring R n is semi-simple. Thus, by invoking the Wedderburn Structure theorems, we can say that each cyclic code in R n contains a unique idempotent element which generates the ideal. Alternatively, this fact has also been proven directly in [, Theorem 4.3.]. We call this idempotent element the generating idempotent (or idempotent generator) of the cyclic code. Let a be an integer such that gcd(a, n) =. We define the function µ a, called a multiplier, on {0,,,, n } by iµ a ia (mod n). Clearly, µ a gives a permutation of the coordinate positions of a cyclic code of length n. Note that this is equivalent to the action of µ a on R n by f(x)µ a f(x a ) (mod x n ). If C is a code of length n over F q, we define a complement of C as a code C c such that C + C c = F n q and C Cc = {0}. In general, a complement of a code is not unique. But it is easy to show that if C is cyclic, then C c is unique and that it is also cyclic (see e.g., Exercise 43, []). In this case, we call C c the cyclic complement of C. 3 Cyclic Codes over F q We now consider cyclic codes over the Galois field F q, where q is a power of a prime p. In this case, we note that R n = F q [x]/(x n ).

3. Idempotents in R n Consider the involution : z z q defined on F q. We extend this map component-wise to F n q. For an element a(x) = a 0 + a x + + a n x n in R n, we set a(x) = a q 0 + a q x + + a q n xn. Let C be a code of length n over F q. We define the conjugate of C to be the code C = {c c C}. It can easily be shown that if C is a cyclic code with generating idempotent e(x), then C is also cyclic and its generating idempotent is e(x). Suppose we list all the distinct q -cyclotomic cosets modulo n in the following way: such that C, C,...,C k, D, D,..., D l, E, E,...,E l, C i = qc i for i k and E i = qd i for i l. By Corollary 4.3.5 of [], an idempotent in R n has the form k l l e(x) = a j x i + b j x i + c j x i. () i C j i D j i E j Thus, Hence, which implies Thus, e(x) = e(x) q = = e(x) = e(x) = = k k a q j a q j i C j x qi + k a j k a q j i C j x i + l l b q j b q j i D j x qi + i E j x i + a q j = a j j k ; b q j = c j j l, i C j x i + x i + i C j k a j x qi + i C j = e(x)µ q. l b j i D j x i + l c q j x qi i E j l l c q j b q j i D j x i. l l b q j x i + b q j i D j l l c j x qi + b j i E j 3 i D j x qi. () i D j x qi i D j x qi

This gives C = e(x) = e(x)µ q = Cµ q by Theorem 4.3.3 of []. The discussion above is summarized in the following proposition. Proposition 3. Let C be a cyclic code over F q with generating idempotent e(x). The following hold:. e(x) has the form given in ().. C is cyclic with generating idempotent e(x). 3. e(x) = e(x)µ q. 4. C = Cµ q. 3. Euclidean and Hermitian Duals Let x = (x 0, x,...,x n ) and y = (y 0, y,...,y n ) be any vectors in F n q. Consider the involution : z z q defined on F q. The Hermitian scalar product of x and y is given by x y = n i=0 x iy i. If C is a linear code over F q, the Euclidean dual of C is denoted C E. The Hermitian dual of C is C H = {u F n q u w = 0 for all w C}. We say that a code C is Euclidean self-orthogonal if C C E, and that C is Euclidean self-dual if C = C E. Similarly, C is said to be Hermitian self-orthogonal if C C H, and that C is Hermitian self-dual if C = C H. Let f(x) = f 0 + f x + + f r x r F q [x]. The reciprocal polynomial of f(x) is the polynomial f (x) = x r f(x ) = x r (f(x)µ ) = f r + f r x + + f 0 x r. Lemma 3. Let a = (a 0, a,...,a n ), b = (b 0, b,...,b n ) be vectors in F n q with associated polynomials a(x) and b(x). Then a is Hermitian orthogonal (similarly Euclidean orthogonal) to b and all its cyclic shifts if and only if a(x)b (x) = 0 (similarly a(x)b (x) = 0) in R n. Proof: See Lemma 4.4.8 of [] for the Euclidean case. The proof for the case of Hermitian orthogonality follows a similar argument and is omitted. Recall that a vector a = (a 0, a,...,a n ) of F n q is called an even-like vector if n i=0 a i = 0. A code C is called an even-like code if all its codewords are even-like, otherwise it is called odd-like. The following lemma appears as Exercise 38 of [] and its proof is left to the reader. Lemma 3.3 Let C be a cyclic code over F q with defining set T and generator polynomial g(x). Let C e be the subcode of C consisting of all the even-like vectors in C. Then:. C e is cyclic and has defining set T {0}. 4

. C = C e if and only if 0 T if and only if g() = 0. 3. If C C e, then the generator polynomial of C e is (x )g(x). The following propositions generalize some results on Euclidean duals of cyclic codes over an arbitrary finite field to Hermitian duals of cyclic codes over F q. Proposition 3.4 Let C be a cyclic code of length n over F q with generating idempotent e(x) and defining set T. The following hold:. C H is a cyclic code and C H = C c µ q.. C H has generating idempotent e(x)µ q. 3. If N = {0,,,..., n }, then N \ ( q)t mod n is the defining set for C H. 4. Precisely one of C and C H is odd-like and the other is even-like. Proof: Let a = (a 0, a,...,a n ) C. Denote by a (i) the i th cyclic shift of a. By assumption a (i) C for all i. Let b = (b 0, b,..., b n ) C H. For i = 0,,..., n, we have b (i) a = b a n i = 0. Thus C H is cyclic. Note that C H = C E and C E = C c µ (Theorem 4.4.9 of []). It can easily be shown that C c = C c and so using Proposition 3. we have C c = C c µ q. Hence C H = C c µ = C c µ q µ = C c µ q, proving part. Using Theorem 4.4.6 and Theorem 4.3.3 of [], the generating idempotent for C H = C c µ q is ( e(x))µ q = e(x)µ q. Thus part holds. The defining set for C c is N \ T (Theorem 4.4.6 of []) and hence applying Corollary 4.4.5 of [] the defining set for C H is ( q) (N \ T) = N \ ( q) T mod n. Since µ q = µ ( q) = µ q fixes each q -cyclotomic coset and ( q) T is a union of q -cyclotomic coset (using Theorem 4.4. of []), it follows that ( q) T = ( q) ( q) T = ( q)t mod n. Thus the defining set for C H is N \ ( q)t mod n. This proves part 3. Lastly, since exactly one of T and N \ ( q)t contains 0, part 4 follows from part 3 and Lemma 3.3. The following lemma is from Exercise 39 of []. Lemma 3.5 Let C i be a cyclic code of length n over F q with defining sets T i for i =,. Then:. C C has defining set T T.. C + C has defining set T T. 3. C C T T. 5

Proposition 3.6 Let C be a Hermitian self-orthogonal cyclic code over F q of length n with defining set T. Let C, C,...,C k, D, D,...,D l, E, E,..., E l be all the distinct q - cyclotomic cosets modulo n partitioned such that C i = C i µ q for i k and D i = E i µ q for i l. Then the following hold:. C i T for i k, and at least one of D i or E i is contained in T for each i l.. C is even-like. 3. C Cµ q = {0}. Conversely, if C is a cyclic code with defining set T that satisfies part, then C is an Hermitian self-orthogonal code. Proof: Let N = {0,,,..., n }. By Proposition 3.4, T = N \ ( q)t mod n is the defining set for C H. By assumption, C C H and so N \ ( q)t T by Lemma 3.5. If C i T for some i, then C i µ q ( q)t. Since C i = C i µ q, it follows that C i N \( q)t T, a contradiction. Thus C i T for all i. If D i T, then E i = D i µ q ( q)t mod n. Thus E i N \ ( q)t T. Hence part holds. To prove part, note that {0} = C i for some i. Hence 0 T by part. By Lemma 3.3, C is even-like. As noted in the proof of Proposition 3.4, ( q) T = ( q)t mod n. Using Corollary 4.4.5 of [], Cµ q has defining set ( q) T = ( q)t. Since N \ ( q)t T, it follows that T ( q)t = N. By Lemma 3.5, T ( q)t is the defining set for C Cµ q. Thus C Cµ q = {0}, which proves part 3. For the converse, assume T satisfies part. We will show that T T which will imply that C is Hermitian self-orthogonal. By Proposition 3.4, T = N \ ( q)t mod n. Note that C i T = C i = C i µ q ( q)t = C i T. Hence T is a union of some E i s and D i s. If D i T = N \ ( q)t, then D i ( q)t mod n, implying that ( q)d i T. Since ( q)d i = E i, it follows that E i T. By part, D i T. By a similar argument, it can be shown that if E i T, then E i T. 4 Duadic Codes Let n be an odd positive integer. We let j(x) = n ( + x + x + + x n ), the generating idempotent for the repetition code of length n over F q. We first define duadic codes over arbitrary finite fields. Then we proceed to examine duadic codes over finite fields of square order. The goal of this section is to present some results concerning Hermitian orthogonality of duadic codes over such finite fields. 6

4. Definitions and Basic Properties Definition 4. Let e (x) and e (x) be a pair of even-like idempotents and let C = e (x) and C = e (x). The codes C and C form a pair of even-like duadic codes if the following properties are satisfied: a) the idempotents satisfy e (x) + e (x) = j(x), b) there is a multiplier µ a such that C µ a = C and C µ a = C. To the pair of even-like codes C and C, we associate a pair of odd-like duadic codes D = e (x) and D = e (x). We say that the multiplier µ a gives a splitting for the even-like duadic codes or for the odd-like duadic codes. Theorem 4. ([].) Let C and C be cyclic codes over F q with defining sets T = {0} S and T = {0} S, respectively, where 0 S and 0 S. Then C and C form a pair of even-like duadic codes if and only if the following conditions are satisfied: a) S and S satisfy S S = {,,..., n } and S S =, b) there is a multiplier µ b such that S µ b = S and S µ b = S. If the conditions in the preceding theorem are satisfied, we say that S and S gives a splitting of n by µ b over F q. This gives us another way of describing duadic codes. Note that for a fixed pair of duadic codes over F q of length n, we can use the same multiplier for the splitting in Definition 4. and the splitting of n in Theorem 4.. Theorem 4.3 ([].) Duadic codes of length n over F q exist if and only if q is a square mod n. 4. Hermitian Orthogonality of Duadic Codes over F q From this point onwards, we consider codes over the Galois field F q, where q is a power of some prime p. Again we assume that n is an odd positive integer and gcd(n, q) =. Thus duadic codes of length n over F q always exist by Theorem 4.3. The following theorem is the Hermitian analogue of Theorem 6.4. of [], where the Euclidean self-orthogonality of duadic codes over F q are considered. Proposition 4.4 Let C be any [n, n ] cyclic code of length n over F q. Then C is Hermitian self-orthogonal if and only if C is an even-like duadic code whose splitting is given by µ q. Proof: ( ) Suppose C = C is an even-like duadic code whose splitting is given by µ q. Let e(x) be the generating idempotent for C. By Theorem 6..3 (vi) of [], C = C D = e(x)µ q. By Proposition 3.4, the generating idempotent for C H is also e(x)µ q. Thus D = C H and so C is Hermitian self-orthogonal. 7

( ) Let C = C be a Hermitian self-orthogonal cyclic code. Let e (x) be the generating idempotent for C and T its defining set. Since C is Hermitian self-orthogonal and j(x) is not orthogonal to itself, j(x) C. Hence by Lemma 6.. (iii) of [], C is even-like. Let e (x) = e (x)µ q and let C = e (x). By Theorem 4.3.3 of [], C = C µ q. Let (a 0, a,..., a n ) C. Since C is even-like, it follows that n i=0 a i = 0. Thus (,,..., ) (a 0, a,...,a n ) = (,,..., ) (a q 0, a q,...,a q n ) = n i=0 aq i = ( n i=0 a q i) = 0 which implies that j(x) C H. Since C H has dimension n+ and C C H, we have C H = C + j(x). Using Theorem 4.3.7 of [] and Lemma 6.. (i) of [], C H has generating idempotent e (x) + j(x). By Proposition 3.4, the generating idempotent for C H is e (x)µ q. By the uniqueness of the idempotent generator, we must have e (x)µ q = e (x) + j(x) which implies j(x) = e (x) + e (x)µ q = e (x) + e (x). Clearly e (x) = e (x)(µ q ) = e (x)(µ q ). Therefore C and C form a pair of even-like codes whose splitting is given by µ q. Lemma 4.5 Let C be a cyclic code. Then (Cµ a ) H = C H µ a. Proof: Use Proposition 3.4 above and Theorem 4.3.3 of [] to show that (Cµ a ) H and C H µ a have the same idempotent generator. Proposition 4.6 Suppose that C and C are a pair of even-like duadic codes over F q, having D and D as their associated odd-like duadic codes. Then the following are equivalent.. C H = D. C H = D 3. C µ q = C 4. C µ q = C Proof: From the definition of duadic codes and Theorem 6..3 (vii) of [], we obtain C µ a = C, C µ a = C, D µ a = D and D µ a = D for some a. Hence by Lemma 4.5, if part holds, then C H = (C µ a ) H = C H µ a = D µ a = D and if part holds, then C H = (C µ a ) H = C H µ a = D µ a = D. Hence parts and are equivalent. Part 3 is equivalent to part 4 since (µ q ) = µ q. If part holds, then by Theorem 6..3 (vi) of [], C is Hermitian self-orthogonal. Hence by Proposition 4.4, part 3 holds. 8

If part 3 holds, then µ q gives a splitting for C and C. Let e i (x) be the generating idempotent for C i. By Theorem 4.3.3 of [], e (x)µ q = e (x). Hence by Proposition 3.4, the generating idempotent for C H is e (x)µ q = e (x). Thus, part holds, completing the proof. Proposition 4.7 Suppose that C and C are a pair of even-like duadic codes over F q, having D and D as their associated odd-like duadic codes. Then the following are equivalent.. C H = D. C H = D 3. C µ q = C 4. C µ q = C Proof: From the definition of duadic codes and Theorem 6..3 (vii) of [], we obtain C µ a = C, C µ a = C, D µ a = D and D µ a = D for some a. Hence, by Lemma 4.5, if part holds, then C H = (C µ a ) H = C H µ a = D µ a = D and if part holds, then C H = (C µ a ) H = C H µ a = D µ a = D. Hence parts and are equivalent. Let e i (x) be the generating idempotent for C i. By Proposition 3.4, C H has generating idempotent e (x)µ q. Thus C H = D if and only if e (x)µ q = e (x) if and only if e (x)µ q = e (x) if and only if C µ q = C by Theorem 4.3.3 of []. Hence parts and 3 are equivalent. It can be shown by an analogous argument that parts and 4 are equivalent. 4.3 Extensions of Odd-like Duadic Codes Odd-like duadic codes have parameters [n, n+ ]. Hence it is interesting to consider extending such codes because such extensions could possibly be Hermitian self-dual codes. The goal of this section is to give a way of extending odd-like duadic codes and to give conditions under which these extensions are Hermitian self-dual. We also prove that any cyclic code whose extended code is Hermitian self-dual must be an odd-like duadic code. Let D be an odd-like duadic code. The code D can be obtained from its even-like subcode C by adding j(x) to a basis of C (Theorem 6..3 (ix), [] ). Hence it is natural to define an extension for which the all-one vector is Hermitian orthogonal to itself. 9

In F q consider the equation + γ q+ n = 0. (3) Since q is a power of a prime p and n F p F q, we have n q = n, or n q+ = n in F q. So + γ q+ n = 0 n + γ q+ n = 0 n + γ q+ n q+ = 0 n + (γn) q+ = 0. Thus Equation (3) is equivalent to n + γ q+ = 0. (4) Note that {a q+ a F q } = F q. Thus (4) will always have a solution in F q, which implies that (3) is solvable in F q. We are now ready to describe the extension. Let γ be a solution to (3). Let c = (c 0, c,...,c n ) D. Define the extended codeword c = (c 0, c,...,c n, c ), where n c = γ c i. Let D = { c c D} be the extended code of D. Proposition 4.8 Let D and D be a pair of odd-like duadic codes of length n over F q. The following hold:. If µ q gives the splitting for D and D, then D and D are Hermitian self-dual.. If D µ q = D, then D and D are Hermitian duals of each other. i=0 Proof: Let C and C be the even-like duadic codes associated to D and D. Note that j(x) j(x) ( = n, n,..., ) ( n, γ n, n,..., ) n, γ ( = n, n,..., ) ( n, γ n, n,..., ) n, ( γ)q = n + γq+ = n ( + γq+ n) = 0, by our choice of γ. This shows that j(x) is Hermitian orthogonal to itself. Since C i is even-like, Ci is obtained by adding a zero coordinate to C i and so j(x) is also orthogonal to C i. We first prove part. Proposition 4.4 ensures that C is Hermitian self-orthogonal, and so C is Hermitian self-orthogonal. Since j(x) is orthogonal to C, the code spanned 0

by C, j(x) is Hermitian self-orthogonal. However, by Theorem 6..3 (ix) of [], D = C, j(x). Clearly D = C, j(x). Thus D is Hermitian self-orthogonal. Since the dimension of D is n+, D is Hermitian self dual. Analogous arguments will prove that D is Hermitian self-dual. We now prove part. Suppose D µ q = D. It follows that C µ q = C. By Proposition 4.7, C H = D and so by Theorem 6..3 (vi) of [], C C H. Therefore C and C are orthogonal to each other and consequently the codes spanned by C, j(x) and C, j(x) are orthogonal. By Theorem 6..3 (v) & (vi) of [], these codes must be D and D of dimension n+. Therefore D and D are duals of each other. Corollary 4.9 Let C be a cyclic code over F q. The extended code C is Hermitian self-dual if and only if C is an odd-like duadic code whose splitting is given by µ q. Proof: ( ) This follows directly from the preceding proposition. ( ) Since the extended code C has length n +, the dimension of C is n+ and therefore C cannot be Hermitian self-orthogonal. The assumption that C is self-dual implies that the even-like subcode of C is necessarily Hermitian self-orthogonal. Since C is not Hermitian self-orthogonal, C cannot be even-like. Let C e be the even-like subcode of C. The code C e is an [n, n ] Hermitian self-orthogonal cyclic code and so by Proposition 4.4, C e is an even-like duadic code with splitting by µ q. Thus C is an odd-like duadic code with a splitting by µ q. 5 Lengths with Splittings By µ q All throughout this section, we let q be a power of a prime p and we assume that n is an odd integer with gcd(n, q) =. Define ord r (q) to be the smallest positive integer t such that q t (mod r). In view of Proposition 4.4 and Corollary 4.9, it is natural to ask under what conditions do we get a splitting of n by µ q. We note that the study of the feasibility of an integer in [7] becomes a special case of this with q =. The main result of this section is the following theorem. Theorem 5. The permutation map µ q gives a splitting of n if and only if ord r (q) (mod 4) for every prime r dividing n. Our proof of this theorem will be based on several lemmas. Lemma 5. is a well-known fact from elementary number theory, see e.g. Proposition 3 in [0], and we leave its proof as an exercise to the reader. Lemma 5. Let r be a prime distinct from p. Then r divides q k + for some positive integer k if and only if ord r (q) is even.

Lemma 5.3 Let r be a prime distinct from p. Then r divides q i + for some integer i if and only if ord r (q) (mod 4). Proof: By Lemma 5., r divides q k + for some positive integer k if and only if ord r (q) is even. If ord r (q) is even, then r q k + q k (mod r) k ord r(q) (mod ord r (q)). Thus r divides q i + if and only if ord r (q) is even and i ord r(q) (mod ord r (q)). (5) But (5) has a solution i if and only if ord r (q) (mod 4). Proposition 5.4 Assume gcd(n, q) =. Then gcd(n, q i + ) = for every integer i if and only if ord r (q) (mod 4) for every prime r dividing n. Proof: Write n = r e re res s. Then, using Lemma 5.3, ord r j (q) (mod 4) for all j =,...,s if and only if for all j =,..., s, r j does not divide q i + for every i if and only if for all j =,,..., s, gcd(r j, q i + ) = for every i if and only if gcd(n, q i + ) = for every i. Proposition 5.5 Let t be an integer such that t (q ) j (mod n) and t (q ) j (mod n) for some non-negative integer j. Suppose gcd(t, n) =. Then µ t gives a splitting of n if and only if gcd(n, q i t) = for every integer i. Proof: Clearly by the assumptions on t, (µ t ) (C s ) = C s for every q -cyclotomic coset C s. Thus µ t gives a splitting of n if and only if it does not fix any q -cyclotomic coset. Let C a be a q -cyclotomic coset. Then µ t fixes C a if and only if ta (q ) i a (mod n) for some positive integer i. Thus µ t gives a splitting of n if and only if ta (q ) i a (mod n) for every i if and only if gcd(n, q i t) = for every i. Theorem 9 of [4] is a special case of Proposition 5.5 with q =. Corollary 5.6 The permutation map µ q gives a splitting of n if and only if gcd(n, q i + ) = for every integer i. Proof: This follows immediately from Proposition 5.5 since gcd(n, q) = by assumption. We are now ready to prove the main theorem of this section.

Proof of Theorem 5.: By Corollary 5.6, the permutation map µ q gives a splitting of n if and only if gcd(n, q i + ) = for every integer i. By Proposition 5.4, gcd(n, q i + ) = for every integer i if and only if ord r (q) (mod 4) for every prime r dividing n. We remark that Theorem 5. says that µ q gives a splitting of n if and only if for all prime r dividing n, either ord r (q) is odd or ord r (q) is doubly even. However, it is easy to show that ord r (q) is doubly even if and only if ord r (q ) is even. Thus we can restate Theorem 5. as: Theorem 5.7 The permutation map µ q gives a splitting of n if and only if for every prime r dividing n, either ord r (q) is odd or ord r (q ) is even. Lastly, we arrive at the following result which gives sufficient and necessary conditions for the existence of Hermitian self-dual extended cyclic code. We note that the same result was obtained in [9] for the more general case of group codes. Theorem 5.8 Cyclic codes of length n over F q whose extended code is Hermitian self-dual exist if and only if for every prime r dividing n, either ord r (q) is odd or ord r (q ) is even. Proof: This follows directly from Corollary 4.9 and Theorem 5.7. The table below enumerates all the splittings (up to symmetry between S and S ) of n by µ q over F q for 5 n 45 and q = 3, q = 4 and q = 5 by listing all the possible sets for the S in Theorem 4.. The C i s are q -cyclotomic cosets modulo n. We omit those n for which no such splitting exists for all values of q. A Quantitative Aspects A. Counting integers that are split by µ q Theorem 5. raises the question of counting the number of integers n x such that µ q gives a splitting of n. In other words, we are interested in counting those integers n such that n is coprime with the sequence S(q) := {q i + } i=. We let A q(x) denote the associated counting function. We are interested in sharp estimates for A q (x) as x gets large. We use the shorthand GRH to denote the Generalized Riemann Hypothesis. The best we can do in this respect is stated in the following theorem: Theorem A. Let q = p t be a prime power. Put λ = ν (t).. For some positive constant c q we have ( ) x x(log log x) 5 A q (x) = c q log δ(q) x + O q, log +δ(q) x where the implicit constant depends at most on q. 3

n S q = 3 q = 4 q = 5 5 C 7 C 9 C C 3, C C 6 C C C 3 C C, C C 7 C C C 4, C C C 6, C C 3 C 4, C C 3 C 6 7 C C C C 3 C 6, C C C 3 C 7, C C C C 5 C 6, C C C 5 C 7, C C 3 C 6 C 8, C C 3 C 7 C 8, C C 5 C 6 C 8, C C 5 C 7 C 8, 9 C C C C C 3 C 7, C C C 3 C 4, C C C 7 C 9, C C C 9 C 4, C C 3 C 7 C 0, C C 3 C 0 C 4, C C 7 C 9 C 0, C C 9 C 0 C 4 3 C C 5 C C 5, C C 0 7 C C 3 C 9, C C 3 C 8, C C 6 C 9, C C 6 C 8 9 C 3 C C 3 C 5, C C 3 C, C C C 3 C 4 C 8, C C 5 C 7, C C 7 C, C C C 3 C 4 C 7, C C C 3 C 8 C, C C C 3 C C 7, C C C 4 C 8 C 6, C C C 4 C 6 C 7, C C C 8 C C 6, C C C C 6 C 7, C C 3 C 4 C 8 C, C C 3 C 4 C C 7, C C 3 C 8 C C, C C 3 C C C 7, C C 4 C 8 C C 6, C C 4 C C 6 C 7, C C 8 C C C 6, C C C C 6 C 7, 33 C C 3 C 5 C, C C 3 C 5 C, C C 3 C 7 C, C C 3 C 7 C, C C 5 C 6 C, C C 5 C 6 C, C C 6 C 7 C, C C 6 C 7 C, 37 C 4 C C C 4 C 7 C 8, C C 3, C C 6 C C C 4 C 7 C, C C C 4 C 8 C 6, C C C 4 C C 6, C C C 7 C 8 C, C C C 7 C C, C C C 8 C C 6, C C C C C 6, C C 4 C 6 C 7 C 8, C C 4 C 6 C 7 C, C C 4 C 6 C 8 C 6, C C 4 C 6 C C 6, C C 6 C 7 C 8 C, C C 6 C 7 C C, C C 6 C 8 C C 6, C C 6 C C C 6, 43 C C 3 C 7, C C 3 C 9, C C 6 C 7, C C 6 C 9 Table : Splittings of n by µ q ( denotes splittings of Quadratic Residue codes) 4

. Let ǫ > 0 and v be arbitrary. Assuming GRH we have that A q (x) = ( ) b j x log δ(q)+j x + O x ǫ,q, log δ(q)+v ǫ x 0 j<v where the implied constant depends at most on ǫ and q, and b 0 (= c q ),..., b v are constants that depend at most on q. The constant δ(q) is the natural density of primes r such that ord r (q) (mod 4) and is given as follows: 7/4 if p = and λ = 0; δ(p t /3 if p = and λ = ; ) = λ /3 if p = and λ ; λ /3 if p. Our proof of Theorem A. rests on various lemmas. Let χ q (n) be the characteristic function of the integers n that are coprime with the sequence S(q), i.e. { if (n, S(q)) = ; χ q (n) = 0 otherwise. Clearly A q (x) = n x χ q(n). Note that χ q (n) is a completely multiplicative function in n, i.e., χ q (nm) = χ q (n)χ q (m) for all natural numbers n and m. This observation reduces the study of χ q (n) to that of χ q (r) with r a prime. Using Lemma 5.3 we infer the following lemma. Lemma A. We have χ q (r) = if and only if r = p or ord r (q) (mod 4) in case r p. This result allows one to count the number of primes r x such that (r, S(q)) =. Recall that Li(x), the logarithmic integral, is defined as x dt/ log t. Lemma A.3 Write q = p t. Let λ = ν (t).. We have r x, (r,s(q))= = r x χ q (r) = ( δ(q))li(x) + O q ( x(log log x) 4 log 3 x ). (6). Assuming GRH the estimate (6) holds with error term O q ( x log x), where the index q indicates that the implied constant depends at most on q. 5

Proof:.) The number of primes r x such that ord r (q) (mod 4) is counted in Theorem of [0]. On invoking the Prime Number Theorem in the form π(x) = Li(x) + O(x log 3 x), the proof of part is then completed..) The proof of this part follows from Theorem 3 of [] together with the well-known result (von Koch, 90) that the Riemann Hypothesis is equivalent with π(x) = Li(x) + O( x log x). We are now ready to prove Theorem A.. Proof of Theorem A.:.) This is a consequence of part of Lemma A.3, Theorem 4 of [0] and the fact that χ q (n) is multiplicative in n..) By part of Lemma A.3 we have r x χ ( q(r) = ( δ(p t ))Li(x) + O q x log v x ). Now invoke Theorem 6 of [] with f(n) = χ q (n). A. Counting duadic codes Theorem 4.3 allows one to study how many duadic codes of length n x (with (n, q) = ) over F q exist as x gets large. We let D q (x) be the associated counting function. Indeed, we will study the more general function D a (x) which is defined similarly, but where a is an arbitrary integer. The trivial case arises when a is a square and thus we assume henceforth that a is not a square. At first glance it seems that D a (x) = ( + ( a ) n ), n x, (n,a)= with (a/n) the Jacobi symbol. However, it is not true that (a/n) = if and only if a is a square modulo n, e.g., (/5) = (/3)(/5) = ( )( ) =, but is not a square modulo 5. It is possible, however, to develop a criterium for a to be a square modulo n in terms of Legendre symbols. To this effect first note that if a is a square modulo n, then a must be a square modulo all prime powers in the factorisation of n. This is a consequence of the following lemma. Lemma A.4 Let n and m be coprime integers. Then a is a square modulo mn if and only if it is a square modulo m and a square modulo n. Proof: By the Chinese Remainder Theorem Z/mnZ Z/m Z/n is an isomorphism of rings and hence a is a square in the ring on the left if and only if a is is square in the ring on the right. Now note that the multiplication in the second ring is coordinatewise. It is a well-known result from elementary number theory that if p is an odd prime and if x a (mod p) is solvable, so is x a (mod p e ) for all e, see e.g. [4, Proposition 4..3]. Using this observation together with Lemma A.4 one arrives at the following criterium for a to be a square modulo n. 6

Lemma A.5 Let a and n be coprime integers. Put ( + ( a Let e = ν (n). Put Then g a (n) = p n 0 if a 3 (mod 4) and e ; f a (n) = 0 if a 5 (mod 8) and e 3; g a (n) otherwise. { if a is a square modulo n; f a (n) = 0 otherwise. By Lemma A.5 we have that D a (x) = n x, (n,a)= f a(n). Note that g a (n) is a multiplicative function, but that f a (n) is a multiplicative function only on the odd integers n (generically). For this reason let us first consider G a (x) := g a (n). n x, (n,a)= As a consequence of the law of quadratic reciprocity, the primes p for which g a (p) = are precisely the primes p in certain arithmetic progressions with modulus dividing 4q. On using the prime number theorem for arithmetic progressions one then infers that for every v > 0 the following estimate holds true: g a (p) = Li(x) + O x q( log v ), (7) x p x On using this one sees that the conditions of Theorem 6 of [] are satisfied and this yields the truth of the following assertion. Lemma A.6 Let ǫ > 0 and v be arbitrary. Suppose that a is not a square. We have G a (x) = ( ) d j x log /+j x + O x ǫ,q, log /+v ǫ x 0 j<v where the implied constant depends at most on ǫ and a, and d 0 (> 0),..., d v are constants that depend at most on a. Now it is straightforward to derive an asymptotic for D a (x). Using Lemma A.5 one infers that G a (x) + G a (x/) if a 3 (mod 4); D a (x) = G a (x) + G a (x/) + G a (x/4) if a 5 (mod 8); (8) G a (x) otherwise. From this and Lemma A.6 it then follows that we have the following asymptotic for D a (x). 7 p ) ).

Theorem A.7 Let ǫ > 0 and v be arbitrary. Suppose that a is not a square. We have D a (x) = ( ) e j x log /+j x + O x ǫ,q, log /+v ǫ x 0 j<v where the implied constant depends at most on ǫ and a, and e 0 (> 0),...,e v are constants that depend at most on a. In particular we have, as x tends to infinity, x x D a (x) D a and G a (x) G a, log x log x where D a and G a are positive constants. We now consider the explicit evaluation of these constants. Note that by (8) it suffices to find an explicit formula for the constant G a. In case a = D is a negative discriminant of a binary quadratic form this constant can be easily computed using results from the analytic theory of binary quadratic forms. We say an integer D is a discriminant if it arises as the discriminant of a binary quadratic form. This implies that either 4 D or D (mod 4). On the other hand, it can be shown that any number D satisfying 4 D or D (mod 4) arises as the discriminant of a binary quadratic form. Now let D be a discriminant and ξ D be the multiplicative function defined as follows: if ( D ) = ; ξ D (p e p ) = if ( D ) = and e; p 0 otherwise. Let n be any integer coprime to D. Then ξ D (n) = if and only if n is represented by some primitive positive integral binary quadratic form of discriminant D. Let B D (x) denote the number of positive integers n x which are coprime to D and which are represented by some primitive integral form of discriminant D 3. Note that B D (x) = n x ξ D(n). It was proved by James [5] that ( ) x x B D (x) = J(D) + O, log x log x where J(D) is the positive constant given by πj(d) = ϕ( D ) D L(, χ D ) ( D p )= p, (9) and p runs over all primes such that (D/p) =. (Recall that the Dirichlet L-series L(s, χ D ) is defined by L(s, χ D ) = n= χ D(n)n s.) Since the behaviour of ξ D is so similar to that of f D, James result can in fact be used to determine the asymptotic behaviour of G D (x) 8

for negative discriminants D and, in particular, to determine G D. Using a classical result of Wirsing, see e.g. Theorem 3 of [], one infers that ( p ). G D (x) B D (x) p x ( D p )= From this and the identity (9) it follows that G D is the positive solution of πg D = ϕ( D ) D L(, χ D ) ( D p )= ( p ). (0) For more details on B D (x) and related counting functions the reader is referred to a paper (in preparation) by Moree and Osburn [3]. In [3] it is also pointed out that B D (x) in fact satisfies an asymptotic result similar to the one given for D a (x) in Theorem A.7. The fact that the characteristic functions ξ d and f D are so closely connected, can be exploited to give a criterium for the existence of duadic codes in terms of representability by quadratic forms. Lemma A.8 Let q be an odd prime power, say q = p e with p 3 (mod 4). Let n be an odd squarefree integer satisfying (n, q) = and suppose, moreover, that n can be written as a sum of two integer squares. A duadic code of length n over F q exists if and only if n can be represented by some primitive positive integral binary quadratic form of discriminant p Proof: By assumption p (mod 4) and hence is a discriminant. The assumption that n is odd and squarefree ensures that ξ p (n) = f p (n) = f p e (n). The assumption that n can be represented as a sum of two squares, together with the assumption that n is squarefree ensures that n is a product of primes p satisfying p (mod 4). For every prime p (mod 4) we have ( p e /p) = (pe /p). It thus follows that ξ p (n) = f p e (n) = f p e (n). The result then follows on invoking Theorem 4.3, Lemma A.5 and the fact that, for (n, D) =, ξ D (n) = if and only if n is represented by some primitive positive integral binary quadratic form of discriminant D. It remains, however, to determine G a for a general number a. It is well-known from Tauberian theory that one has G a = Γ(/) lim s F(s), s where F(s) = n= g a(n)n s. An easy computation shows that (s )F(s) = (s )ζ(s) ϕ( a ) a L(s, χ a ) ( a p )= ( p ). 9

On using that the Riemann zeta-function ζ(s) has a simple pole at s = of residue, one obtains that πg a = ϕ( a ) L(, χ a ) ( p ). a Notice that equation (0) is a special case of this. ( a p )= Acknowledgements. The first author gratefully acknowledges financial support from the University of the Philippines and from the Philippine Council for Advanced Science and Technology Research and Development through the Department of Science and Technology. The second author likes to thank I. Shparlinski for suggesting Lemma A.5 and D. Gurevich for some helpful remarks. The three authors would like to thank the anonymous referees for their helpful suggestions that greatly improved the presentation of the material. References [] R. A. Brualdi, W. C. Huffman, V. S. Pless, An Introduction to Algebraic Codes, in Handbook of Coding Theory, V. S. Pless & W. C. Huffman (Editors), Elsevier Science, Amsterdam (998), pp. 3-39. [] W. C. Huffman, V. Pless, Fundamentals of Error-Correcting Codes. Cambridge University Press (003). [3] T. W. Hungerford, Algebra. Springer-Verlag New York (974). [4] K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory (Second Edition). Graduate Texts in Mathematics 84, Springer-Verlag, New York, 990. [5] R.D. James, The Distribution of Integers Represented by Quadratic Forms. American Journal of Mathematics, 60 (938), pp. 737-744. [6] J. S. Leon, J. M. Masley, V. Pless, Duadic Codes. IEEE Transactions on Information Theory, Vol. IT-30, No. 5 (984), pp. 709-74. [7] F. J. MacWilliams, A. M. Odlyzko, N. J. A. Sloane, H. N. Ward, Self-Dual Codes over GF(4). Journal of Combinatorial Theory, Series A 5 (978), pp. 88-38. [8] F. J. MacWilliams, N. J. A. Sloane, The Theory of Error-Correcting Codes. North Holland Publishing Company, Amsterdam (983). [9] C. Martínez-Pérez, W. Willems, Self-Dual Extended Cyclic Codes. Preprint (004) 0

[0] P. Moree, On the divisors of a k +b k. Acta Arithmetica 80, No. 3 (997), pp. 97-. [] P. Moree, On primes p for which d divides ord p (g). Funct. Approx. Comment. Math. 33 (005), pp. 85-95. [] P. Moree and J. Cazaran, On a claim of Ramanujan in his first letter to Hardy. Exposition. Math. 7, No. 4, (999), pp. 89-3. [3] P. Moree and R. Osburn, Two-dimensional lattices with few distances, arxiv:math.nt/060463, submitted for publication. [4] V. Pless, Q-Codes. Journal of Combinatorial Theory, Series A 43 (986), pp. 58-76. [5] M. Smid, Duadic Codes. IEEE Transactions on Information Theory, Vol. IT-33 No. 3 (987), pp. 43-433. [6] K.S. Williams, Note on integers representable by binary quadratic forms. Canad. Math. Bull. 8 (975), pp. 3-5.