QUADRATIC RESIDUE CODES OVER Z 9

Transcription

1 J. Korean Math. Soc. 46 (009), No. 1, pp QUADRATIC RESIDUE CODES OVER Z 9 Bijan Taeri Abstract. A subset of n tuples of elements of Z 9 is said to be a code over Z 9 if it is a Z 9 -module. In this paper we consider an special family of cyclic codes over Z 9, namely quadratic residue codes. We define these codes in term of their idempotent generators and show that these codes also have many good properties which are analogous in many respects to properties of quadratic residue codes over finite fields. 1. Introduction Let R be a finite commutative ring with identity. A subset C of n tuples of elements of R is called an R-code (code over R) if it is an R-module. Recall that a code C over R is called cyclic if cyclic shift of every element of C is an element of C, or equivalently C is an ideal of R[x]/ x n 1. Hammons, Kumar, Calderbank, Sloane, and Solé in a seminal paper [3], discuss the Z 4 -linearity of Kerdock, Preparata, Goethals, and other codes. The structure of cyclic Z 4 - codes is considered by Pless and Qian [10], and Pless, Solé, and Qian [9]. They found generator polynomials as well as idempotent generators for cyclic Z 4 - codes. They have also outlined the necessary and sufficient conditions for these codes to be self dual. An interesting family of cyclic codes is quadratic residue codes. Quadratic residue codes firstly defined and investigated by Andrew Gleason. The minimum weights of many modest quadratic residue codes are quite high, making this class of codes promising. Pless and Qian [10] defined quadratic residue codes over Z 4 in terms of their idempotent generators. They showed that these codes have many good properties which are analogous in many respect to properties of quadratic residue codes over finite fields. The same results were obtained for quadratic residue codes over Z 8 (see []). In this paper we consider Z 9 and define quadratic residue codes over Z 9. We prove that the results of [10] and [] remain valid over Z 9. Our method suggests that if one has the idempotent generators of quadratic residue codes Received April 1, Mathematics Subject Classification. 11T71, 94B05, 94B15, 94B99. Key words and phrases. cyclic codes, quadratic residue codes, extended codes, automorphism group of a code. This research was in part supported by a grant from IPM (No ). 13 c 009 The Korean Mathematical Society

2 14 BIJAN TAERI over Z q m, one can obtain idempotents over Z q m+1 and define quadratic residue codes over Z q m+1. We show that these quadratic residue codes over Z 9 have large automorphism group which will be useful in decoding these codes by using the permutation decoding methods described in [5, Chapter 16]. We also define a distance preserving map from Z N 9 (Lee distance) to Z 4N (Hamming distance). Our notation and terminology are standard and can be found in [5]. Throughout the paper we will assume that R is a finite commutative ring with identity. We begin with some elementary facts (see [, pp ], where we can replace Z p m by R). Theorem 1. Let e 1, e be idempotents of R[x]/ x n 1 and let C 1 = e 1, C = e. Then C 1 C and C 1 + C have idempotent generators e 1 e and e 1 + e e 1 e, respectively. Theorem. Let e 1 (x) be the idempotent generator of an R-cyclic code C. Then 1 e(x 1 ) is the idempotent generator of the dual code C. Recall that f, g R[x] are called coprime if fu+gv = 1 for some u, v R[x]. If x n 1 = fg, where f and g are coprime, then it is easy to see that C = f in R[x]/ x n 1 has an idempotent generator, namely e = fu, which is the identity of C so it is unique (see [, Theorem.0.1]). Let e i, i = 1,,..., r be elements of R. If e i e j = 0 for all i j, and r i=1 e i = 1, we say that e i, i = 1,,..., r are primitive idempotents. In this case R is the direct sum of minimal ideals Re i, i = 1,,..., r (see, for example, [6, p. 95]). Theorem 3. Let S and R be finite commutative rings with identity and characteristic q m and q m+1, respectively, where q is a prime. Let f : R S be an epimorphism, with ker f = q m R. (a) If f(e) = e 1 is an idempotent of S. Then e q is an idempotent of R. (b) If e i, i = 1,,..., r are primitive idempotents of S, and f(θ i ) = e i, i = 1,,..., r, then θ q i, i = 1,,..., r, are primitive idempotents of R. Proof. (a) Since e 1 = e 1, we have e e ker f. Thus e e = q m b 1 for some b 1 R. Therefore, since q divides ( q i ), 1 i q 1, and qmi+1 q m+1, q mq q m+1, we have e q = (e + q m b 1 ) q = q 1 ( ) e q q + e q i (q m b i 1 ) i + (q m b 1 ) q i=1 q 1 = e q + n i qq mi e q i b i 1 + q mq b q 1 (n i N) = e q. i=1

3 QUADRATIC RESIDUE CODES 15 (b) In S we have r i=1 e i = 1 and e i e j = 0, i j. Thus r i=1 θ i = 1 + q m a 0 and θ i θ j = q m b 0 for some a 0, b 0 R. Now ( r ) q r θ i = θ q i + qu, i=1 where u is the sum of terms containing a factor of the form θ i θ j. Thus, in R, r θ q i = (1 + q m a 0 ) q q m+1 c 0 (where c 0 R) i=1 = 1 + = 1 + q i=1 i=1 ( q i) (q m a 0 ) i q n i qq mi a i 0 (where n i N) i=1 = 1 (since q mi+1 q m+1 ). Finally θ q i θq j = (θ iθ j ) q = (q m b 0 ) q = q mq b q 0 = 0, since qmq q m+1. Let R j, j = 1,,..., m + 1 be finite commutative rings with identity and characteristic q j, where q is prime. Let f j : R j+1 R j, j = 1,,..., m, be epimorphisms, with ker f j = q j R j+1. Suppose that e 1 is an idempotent of R 1. Since f 1 is epimorphism, there exists e in R such that f 1 (e ) = e 1. By Theorem 3, e q is an idempotent of R. Now since f is epimorphism, there exists e 3 in R 3 such that f(e 3 ) = e. Thus f (e q 3 ) = eq is an idempotent of R. So, by Theorem 3, e q 3 is an idempotent of R 3. Continuing in this way we obtain elements e 1, e,..., e m+1 such that f j (e j+1 ) = e j such that e qj j+1 is an idempotent of R j. In particular e qm m+1 is an idempotent of R m. Now suppose that f 1 (e 1 ) = e 11, f 1 (e ) = e 1,..., f 1 (e r ) = e r1 are primitive elements of R 1. Then, by Theorem 3, e q 1, eq,..., eq r are primitive elements of R. Now there exist elements e i3, i = 1,,..., r, such that f (e i3 ) = e i. Thus f (e q i3 ) = eq i, i = 1,,..., r, are primitive elements of R and so, by Theorem 3, e q i3, i = 1,,..., r are primitive elements of R 3. Continuing in this way we obtain elements e 1j, e j,..., e r,m+1 such that f j (e i,j+1 ) = e ij and e qj 1,j+1, eqj,j+1,..., eqj r,j+1 are primitive elements of R j. e qm 1,m+1, eqm,m+1,..., eqm r,m+1 are primitive elements of R m. In particular Let p and q be primes and m a positive integer. Put R m = Z q m[x]/ x p 1. The epimorphism f m : Z q m+1 Z q m a + q m+1 a + q m can be extended to an epimorphism R m+1 R m, which is again denoted by f m. Thus if e R m is an idempotent of R m and viewed as an element of R m+1,

4 16 BIJAN TAERI then e q is an idempotent of R m+1. In this paper we deal with some special idempotents of Z 9 [x]/ x p 1, so if e is an idempotent of Z 3 [x]/ x p 1, then by Theorem 3, e 3 is an idempotent of Z 9 [x]/ x p 1. Let Q be the set of quadratic residues and N be the set of non-residues for p (a prime). Let e 1 = i Q xi and e = i N xi. By [5, Problem (5), p. 50], 3 is quadratic residue (mod p) if and only if p = 1r ± 1. Therefore for considering quadratic residue codes over Z 3 (and hence over Z 3 m) we must assume that p = 1r ± 1. It is well known e i, 1 + e i, i = 1,, are idempotents of Z 3 [x]/ x p 1 (see [5, p. 486]). A Z 3 -cyclic code is a Z 3 -quadratic residue (QR) code if it is generated by one of the idempotents e i, 1 + e i, i = 1,. If p = 1r 1 put Q 1 = e 1, Q = e, Q 1 = 1 + e and Q = 1 + e 1. If p = 1r +1 put Q 1 = 1+e, Q = 1+e 1, Q 1 = e 1 and Q = e. Now, as we have seen, (e i ) 3, (1 + e i ) 3, i = 1,, are idempotents over Z 9 [x]/ x p 1. In order to define quadratic residue codes over Z 9 in terms of idempotent generators, we must compute these elements modulo 9. The following theorem is needed for such computations. Theorem 4 ([7], or [5, p. 519]). (i) Suppose that p = 4k 1 and a is a number prime to p. Then in the set a + (Q {0}), there are k elements in Q {0} and k elements in N. In the set a + N, there are k elements in Q {0} and k 1 elements in N. (ii) Suppose that p = 4k + 1 and a is a number prime to p. Then in the set a + (Q {0}), if a Q, there are k + 1 elements in Q {0} (including 0) and k elements in N; and if a N, there are k elements in Q and k + 1 elements in N. In the set a + N, if a Q, there are k elements in Q and k elements in N; and if a N, there are k + 1 elements in Q {0} (including 0) and k 1 elements in N. By a routine application of Theorem 4, we obtain the following result. Theorem 5. If p = 4k 1, then If p = 4k + 1, then e 1 = (k 1)e 1 + ke, e = ke 1 + (k 1)e, e 1 e = (k 1) + (k 1)e 1 + (k 1)e. e 1 = (k 1)e 1 + ke + k, e = ke 1 + (k 1)e + k, e 1 e = ke 1 + ke. By the above theorem and a little computations, when p = 4k 1 we have e 3 1 = (3k 3k + 1)e 1 + k(k 1)e + k k, e 3 = k(k 1)e 1 + (3k 3k + 1)e + k k,

5 QUADRATIC RESIDUE CODES 17 and when p = 4k + 1 we have e 3 1 = (k + 1)e 1 + (k k)e + k k, e 3 = (k k)e 1 + (k + 1)e + k k.. Quadratic residue codes over Z 9 Throughout the paper we assume that p is a prime and h = 1 + e 1 + e is the all one vector. As mentioned in introduction, for considering quadratic residue codes over Z 3 we must assume that p = 1r ± 1. First of all we find idempotents of Z 9 [x]/ x p 1 from idempotent generators of quadratic residue codes over Z 3 [x]/ x p 1. Theorem 6. I. Suppose that p = 1r 1. (a) If r = 3k, then 8e i, 1+e i, 8h, 1+h are idempotents over Z 9 / x p 1, where i = 1,. (b) If r = 3k +1, then 3+6e i +8e j, 7+e i +3e j, 5h, 1+5h are idempotents over Z 9 / x p 1, where 1 i j. (c) If r = 3k +, then 6+3e i +8e j, 4+e i +6e j, h, 8+8h are idempotents over Z 9 / x p 1, where 1 i j. II. Let p = 1r + 1. (a) If r = 3k, then 1 + e i, 8e i, h, 1 + h are idempotents over Z 9 / x p 1, where i = 1,. (b) If r = 3k +1, then 4+e i +6e j, 6+3e i +8e j, 7h, 1+h are idempotents over Z 9 / x p 1, where 1 i j. (c) If r = 3k +, then 7+e i +3e j, 3+6e i +8e j, 4h, 1+5h are idempotents over Z 9 / x p 1, where 1 i j. Proof. I. Let p = 1r 1. Since e 1 is an idempotent of Z 3 [x]/ x p 1, (e 1 ) 3 is an idempotent Z 9 [x]/ x p 1. By Theorem 5, in Z 9 [x]/ x p 1 we have (e 1 ) 3 = 8e 3 1 = [(18r 9r + 1)e 1 + 6r(3r 1)e + 18r 3r] Similarly, = 3r + 8e 1 + 6re 8e 1 r = 3k = 3 + 8e 1 + 6e r = 3k e 1 + 3e r = 3k +. (1 + e 1 ) 3 = 1 + 3e 1 + 3e 1 + e 3 1 = 1 + 3e 1 + 3[(3r 1)e 1 + 3re ] + [e 1 6re 3r] = 1 3r + e 1 6re 1 + e 1 r = 3k = 7 + e 1 + 3e r = 3k e 1 + 6e r = 3k +.

6 18 BIJAN TAERI Other cases are proved similarly. A Z 9 -cyclic code is called a Z 9 -quadratic residue (QR) code if it is generated by one of the idempotents in the above theorem. Let a be a nonzero element of Z p. The map µ a is defined as µ a (i) = ai (mod p). It is easy to see that µ a (fg) = µ a (f)µ a (g) for polynomials f and g in Z 9 [x]/ x p 1. In the following theorem we investigate some properties of QR-codes over Z 9. Theorem 7. Suppose that p = 1r 1. If r = 3k, let Q 1 = 8e 1, Q = 8e and Q 1 = 1 + e, Q = 1 + e 1. If r = 3k + 1, let Q 1 = 3 + 6e 1 + 8e, Q = 3+8e 1 +6e and Q 1 = 7+e 1 +3e, Q = 7+3e 1 +e. If r = 3k +, let Q 1 = 6 + 3e 1 + 8e, Q = 6 + 8e 1 + 3e and Q 1 = 4 + e 1 + 6e, Q = 4 + 6e 1 + e. Then the following hold for Z 9 -QR codes Q 1, Q, Q 1, Q : (a) Q 1 and Q are equivalent and Q 1 and Q are equivalent; (b) Q 1 Q = h and Q 1 + Q = Z 9 [x]/ x p 1, where h is a suitable element in {8h, 5h, h} listed in the Theorem 6; (c) Q 1 = 9 (p+1)/ = Q ; (d) Q 1 = Q 1 + h, Q = Q + h ; (e) Q 1 = 9 (p 1)/ = Q ; (f) Q 1 and Q are self-orthogonal and Q 1 = Q 1 and Q = Q ; (g) Q 1 Q = {0} and Q 1 + Q = 1 h ; also Q i Q j = {0} and Q i + Q j = Z 9[x]/ x p 1, where 1 i j. Proof. Since p = 4(3r) 1, 1 N, by [8, Theorem 65]. We prove the case r = 3k +. The proof of other cases are similar. (a) Let a be an element of N. Then µ a e 1 = e and µ a e = e 1. Thus µ a 6 + 3e 1 + 8e = 6 + 8e 1 + 3e and µ a 4 + 6e 1 + e = 4 + e 1 + 6e. (b) Since (6 + 3e 1 + 8e ) + (6 + 8e 1 + 3e ) = 1 + h, we have (6 + 3e 1 + 8e )(5h) = (6 + 3e 1 + 8e )[8 + (6 + 3e 1 + 8e ) + (6 + 8e 1 + 3e )] = 8(6 + 3e 1 + 8e ) + (6 + 3e 1 + 8e ) + (6 + 3e 1 + 8e )(6 + 8e 1 + 3e ) = (6 + 3e 1 + 8e )(6 + 8e 1 + 3e ). Now since p = 1(3k + ) 1 = 36k + 3 and p 1 mod 9, we have (6+3e 1 +8e )(h) = 6(h)+3 p 1 (h)+8p 1 1 (h) = (6+p )(h) = h. Therefore (6 + 3e 1 + 8e )(6 + 8e 1 + 3e ) = h. By Theorem 1, Q 1 Q has idempotent generator h. Hence Q 1 Q = h = 9.

7 QUADRATIC RESIDUE CODES 19 Also, by Theorem 1, Q 1 + Q has idempotent generator (6 + 3e 1 + 8e ) + (6 + 8e 1 + 3e ) (6 + 3e 1 + 8e )(6 + 8e 1 + 3e ) = 3 + e 1 + e ( + e 1 + e ) = 1. Thus Q 1 + Q = Z 9 [x]/ x p 1. (c) By (a) and (b) we have 9 p = Q 1 + Q = Q 1 Q Q 1 Q = Q 1 9, so Q 1 = Q = 9 (p+1)/. (d) By Theorem 1, Q 1 h has idempotent generator (4 + 6e 1 + e )(h) = 4(h) + 6 p 1 (h) + p 1 1 (h) = (4 + 7p )(h) = 0. Thus Q 1 h = {0}. By Theorem 1, Q 1 + h has idempotent generator (4 + 6e 1 + e ) + h (4 + 6e 1 + e )(h) = 6 + 8e 1 + 3e. Hence Q 1 + h = 6 + 8e 1 + 3e = Q 1. Similarly Q + h = Q. (e) 9 (p+1)/ = Q 1 = Q 1+ h = Q 1 h = 9 Q 1. Thus Q 1 = 9 (p 1)/. (f) By Theorem and the fact that 1 N, Q 1 has idempotent generator 1 [6 + 3e 1 (x 1 ) + 8e (x 1 )] = 4 + 6e 1 (x 1 ) + e (x 1 ) = 4 + e 1 + 6e. Hence Q 1 = Q 1. Similarly Q = Q. By (d), Q 1 Q = Q 1 and Q 1 Q = Q 1, so Q 1 and Q are self-orthogonal. (g) Since (4 + e 1 + 6e ) + (4 + 6e 1 + e ) = 1 + 7h, we have (4 + e 1 + 6e )(7h) = (4 + e 1 + 6e )[8 + (4 + e 1 + 6e ) + (4 + 6e 1 + e )] = 8(4 + e 1 + 6e ) + (4 + e 1 + 6e ) + (4 + e 1 + 6e )(4 + 6e 1 + e ) = (4 + e 1 + 6e )(4 + 6e 1 + e ). Now since p 1 mod 9, we have (4 + e 1 + 6e )(7h) = 4(7h) + p (7h) + 6p (7h) = (4 + 7p )(h) = 0. Therefore (6 + 3e 1 + 8e )(6 + 8e 1 + 3e ) = 0. By Theorem 1, Q 1 Q = {0}, and Q 1 + Q has idempotent generator (4 + e 1 + 6e ) + (4 + 6e 1 + e ) = 1 + 7h, so Q 1 + Q = 1 + 7h. Theorem 8. Let p = 1r + 1 be a prime. If r = 3k, let Q 1 = 1 + e 1, Q = 1+e and Q 1 = 8e, Q = 8e 1. If r = 3k+1, let Q 1 = 4+e 1 +6e, Q = 4+6e 1 +e and Q 1 = 6+3e 1 +8e, Q = 6+8e 1 +3e. If r = 3k+, let Q 1 = 7 + e 1 + 3e, Q 1 = 7 + 3e + e 1 and Q 1 = 3 + 6e 1 + 8e, Q = 3 + 8e 1 + 6e. Then the following hold for Z 9 -QR codes Q 1, Q, Q 1, Q :

8 0 BIJAN TAERI (a) Q 1 and Q are equivalent and Q 1 and Q are equivalent: (b) Q 1 Q = h and Q 1 + Q = Z 9 [x]/ x p 1, where h is a suitable element in {h, 4h, 7h} listed in the Theorem 6; (c) Q 1 = 9 (p+1)/ = Q ; (d) Q 1 = Q 1 + h, Q = Q + h ; (e) Q 1 = 9 (p 1)/ = Q ; (f) Q 1 = Q and Q = Q 1; (g) Q 1 Q = {0} and Q 1 + Q = 1 h ; also Q i Q j = {0} and Q i + Q j = u, where 1 i j, and u is a suitable element of {1 + h, 1 + 5h, 1 + h} listed in the Theorem 6. Proof. We prove the case r = 3k + 1. The proof of other cases are similar. (a) Let a be an element of N. Then µ a e 1 = e and µ a e = e 1. Thus µ a 4 + 6e 1 + e = 4 + e 1 + 6e and µ a 3 + 8e 1 + 6e = 3 + 6e 1 + 8e. (b) Since (4 + 6e 1 + e ) + (4 + e 1 + 6e ) = 1 + 7h, we have (4 + 6e 1 + e )(7h) = (4 + 6e 1 + e )[8 + (4 + 6e 1 + e ) + (4 + e 1 + 6e )] = 8(4 + 6e 1 + e ) + (4 + 6e 1 + e ) + (4 + 6e 1 + e )(4 + e 1 + 6e ) = (4 + 6e 1 + e )(4 + e 1 + 6e ). Now since p = 1(3k + 1) + 1 = 36k + 13 and p 1 6 mod 9, we have (4 + 6e 1 + e )(7h) = 4(7h) + 6 p 1 (7h) + p 1 1 (7h) = (4 + 7p )(7h) = 7h. Therefore (4 + 6e 1 + e )(4 + e 1 + 6e ) = 7h. idempotent generator 7h. Hence Q 1 Q = 7h = 9. Also, by Theorem 1, Q 1 + Q has idempotent generator By Theorem 1, Q 1 Q has (4 + 6e 1 + e ) + (4 + e 1 + 6e ) (4 + 6e 1 + e )(4 + e 1 + 6e ) = 8 + 7e 1 + 7e (7 + 7e 1 + 7e ) = 1. Thus Q 1 + Q = Z 9 [x]/ x p 1. (c) By (a) and (b) we have 9 p = Q 1 + Q = Q 1 Q Q 1 Q = Q 1 9, so Q 1 = Q = 9 (p+1)/. (d) By Theorem 1, Q 1 7h has idempotent generator (6 + 3e 1 + 8e )(7h) = 6(7h) + 3 p 1 (7h) + 8p 1 1 (7h) = (6 + p )(7h) = 0.

9 QUADRATIC RESIDUE CODES 1 Thus Q 1 7h = {0}. By Theorem 1, Q 1 + 7h has idempotent generator (6 + 3e 1 + 8e ) + 7h (6 + 3e 1 + 8e )(7h) = 4 + e 1 + 6e. Hence Q 1 + 7h = 4 + e 1 + 6e = Q 1. Similarly Q + 7h = Q. (e) 9 (p+1)/ = Q 1 = Q 1+ 7h = Q 1 7h = 9 Q 1. Thus Q 1 = 9 (p 1)/. (f) By Theorem and the fact that 1 Q, Q 1 has idempotent generator 1 [4 + e 1 (x 1 ) + 6e (x 1 )] = 6 + 8e 1 (x 1 ) + 3e (x 1 ) = 6 + 8e 1 + 3e. Hence Q 1 = Q. (g) Since (6 + 3e 1 + 8e ) + (6 + 8e 1 + 3e ) = 1 + h, we have (6 + 3e 1 + 8e )(h) = (6 + 3e 1 + 8e )[8 + (6 + 3e 1 + 8e ) + (6 + 8e 1 + 3e )] = 8(6 + 3e 1 + 8e ) + (6 + 3e 1 + 8e ) + (6 + 3e 1 + 8e )(6 + 8e 1 + 3e ) = (6 + 3e 1 + 8e )(6 + 8e 1 + 3e ). Now since p 1 6 mod 9, we have (6+3e 1 +8e )(h) = 6(h)+3 p 1 (h)+8p 1 1 (h) = (6+11p )(h) = 0. Therefore (6+3e 1 +8e )(6+8e 1 +3e ) = 0. By Theorem 1, Q 1 Q = {0}, and Q 1 +Q has idempotent generator (6+3e 1 +8e )+(6+8e 1 +3e ) = 1+h. The extended code of an R code C will be denoted by C, which is the code obtained by adding an overall parity check to each codeword of C. Theorem 9. Suppose p = 1r 1 and Q 1, Q are the Z 9 -QR codes in Theorem 7. Then Q 1 and Q are self-dual. Proof. We prove the case p = 3k +. The proof of other cases are similar. We know, by Theorem 7, that Q 1 = Q 1 + h), and Q 1 has the p+1 (p + 1) generator matrix p G 1, where each row of G 1 is a cyclic shift of the vector 4 + e 1 + 6e. We know that G 1 generates Q 1. Since Q 1 is self-orthogonal (Theorem 7(f)), the rows of G 1 are orthogonal to each other and obviously also orthogonal to h. Since the vector (8, h) is orthogonal to itself and Q 1 = Q 1 = 9 (p+1)/, by comparing the order of Q 1 and Q 1, Q 1 is self-dual. Similarly, Q is self-dual.

10 BIJAN TAERI When p = 1r + 1, we define Q 1 to be the Z 9 -code generated by the matrix p G 1, where each row of G 1 is a cyclic shift of 8e 1 when r = 3k, a cyclic shift of 6 + 3e 1 + 8e when r = 3k + 1, a cyclic shift of 3 + 6e 1 + 8e when r = 3k +. We define Q similarly. Note that these are not extended codes of Q 1 and Q, since the sum of the components of the all one vector is not 0 (mod 9). Theorem 10. Suppose p = 1r + 1 and Q 1, Q are the Z 9 -QR codes in Theorem 8. Then the dual of Q 1 is Q and the dual of Q is Q 1. Proof. We prove the case p = 3k +. The proof of other cases are similar. In this case, since Q 1 = Q 1 + 4h (Theorem 8(d)), Q 1 has the p+1 (p + 1) generator matrix p G 1, where each row of G 1 is a cyclic shift of the vector 3 + 6e 1 + 8e. Since G 1 generates Q 1 and Q = Q 1, by Theorem 8(f), any row in the above matrix is orthogonal to any row in the matrix which defines Q. By comparing the order of the dual of Q 1 and the order of Q, we find that Q 1 = Q. For p = 1r ± 1, the extended codes Q 1 and Q are equivalent, since Q 1 and Q are equivalent. They are also equivalent to Q 1 and Q. Therefore the group of extended codes, which will investigated in the next section, is the group of either one of the extended codes. 3. Extended quadratic residue codes over Z 9 Let a Q. Define permutations σ and µ a on {, 0, 1,..., p 1} by σ : i i + 1 mod p, ; µ a : i ai mod p,. Clearly, the extended QR-codes are fixed by σ and µ a. So the group generated by σ and µ a, a Q, is contained in the group of extended QR-codes. Let χ be

11 QUADRATIC RESIDUE CODES 3 the Legendre symbol on GF (p) which is defined by 1 i Q χ(i) = 1 i N 0 otherwise. Now we define a permutation ρ on extended quadratic residue codes as follows: ρ : i 1 i mod p, followed by multiplication by χ(i), that is, ρ : x i χ(i)x 1/i, and when p = 1r 1, r = 3k 0 followed by multiplication by 8 r = 3k 0 followed by multiplication by 1 r = 3k followed by multiplication by 7 r = 3k followed by multiplication by r = 3k + 0 followed by multiplication by 4 r = 3k + 0 followed by multiplication by 5, and when p = 1r + 1, r = 3k 0 followed by multiplication by 1 r = 3k 0 followed by multiplication by 8 r = 3k followed by multiplication by 4 r = 3k followed by multiplication by 5 r = 3k + 0 followed by multiplication by 7 r = 3k + 0 followed by multiplication by. Let G be the group generated by elements σ, ρ, µ a, a Q. Note that when p = 1r 1, G/(±I) P SL (p), and when p = 1r 1, G P SL (p). In the following theorem we prove that G is contained in the group of the extended QR-code. Theorem 11. Let G be as above. Then G is contained in the group of the extended QR-code. Proof. Suppose that p = 1r 1 and r = 3k + 1. Since p = 4(3r) 1, 1 N, by [8, Theorem 65]. Since Q 1 = Q 1 + 5h, (Theorem 7(d)), the extended code Q 1 is generated by p+1 rows of the (p + 1) (p + 1) matrix r G 1..,.. r where each row of G 1 is a cyclic shift of the vector 7 + e 1 + 3e. Since 1 is a nonresidue, ρ(e 1 ) = e and ρ(e ) = e 1.

12 4 BIJAN TAERI By definition of ρ we have x 0 x and x 5x 0. Now r 0 = (0, 7 + e 1 + 3e ) and so ρ(r 0 ) = ( 7, 5 0 e + 3e 1 ) = (5, 3e 1 + 8e ) = 7r 0 + 5h Q 1, where h is the all one vector of length p + 1, i.e., h = (1, 1 + e 1 + e ). Since r = (8, 5 + 5e 1 + 5e ), we have ρ(r ) = ( 5, 7 8 5e + 5e 1 ) = (1, + 5e 1 + 4e ) = 4r 0 + h Q 1. Now we prove that ρ(r s ) Q 1. In all the following proofs we let q Q and n N. We have (0, 7x s + x q+s + 3 x n+s) r s = r 1/s = We consider two cases s Q and s N: ( 0, 7x 1/s + x q 1/s + 3 x n 1/s). Case 1. s Q. We have ρ(r s ) = ( 3, 7χ(s)x 1/s χ(q +s)x 1/(q+s) 3 χ(n + s)x 1/(n+s) ). Since 1 N and s Q, 0 s + N, the position of ρ(r s ) is 3 = 6. We show that ρ(r s ) = 8r 1/s + 7r 0 + 3r. Since 1 N and s Q, the nonresidue position of ρ(r s ) + r 1/s is: 7x 1/s x 1/(q+s) 3 x 1/(n+s) q+s Q +7x 1/s + q 1/s N x q 1/s + 3 n+s Q n 1/s N x n 1/s. By Theorem 4, the set s + Q has 3r 1 elements in Q, 3r elements in N. Thus, since 1 N, the set { 1/(q + s)} has 3r 1 elements in N and 3r elements in Q. Similarly the set { 1/(n + s), n + s 0} has 3r 1 elements in N and 3r 1 elements in Q; the set 1/s + Q has 3r 1 elements in N and 3r 1 elements in Q, and one element is 0; the set 1/s + N has 3r 1 elements in N and 3r elements in Q. Also since for any 1/(q + s) N, there is a q Q such that 1/(q + s) = q 1/s and for any 1/(n + s) N, there is a n N such that 1/(n + s) = n 1/s, the nonresidue position is equal to x q 1/s 3 x n 1/s + x q 1/s +3 x n 1/s = 0. q 1/s N n 1/s N q 1/s N Now the residue position of ρ(r s ) + r 1/s is: x 1/(q+s) + 3 x 1/(n+s) + q+s N n+s N q 1/s Q x q 1/s + 3 n 1/s N n 1/s Q x n 1/s. Since for any 1/(q + s) Q, there is a n N such that 1/(q + s) = n 1/s and for any 1/(n + s) Q, there is a q Q such that 1/(n + s) = q 1/s, there are 3r + 3r 1 = 6r 1 terms appearing, so the residue position is equal

13 QUADRATIC RESIDUE CODES 5 to n 1/s Q x q 1/s +3 q 1/s Q x n 1/s + Thus, since 0 1/s + Q, we have that is, ρ(r s ) = 8r 1/s + 7r 0 + 5h. q 1/s Q x q 1/s +3 ρ(r s ) + r 1/s = (6, 1 + 4e 1 ) = 7r 0 + 3r ; n 1/s Q x n 1/s = 4e 1. Case. s N. We have ρ(r s ) = ( 1, 7χ(s)x 1/s χ(q+s)x 1/(q+s) 3 χ(n + s)x 1/(n+s) ). Since 0 s + Q, the position of ρ(r s ) is 1 =. We show that ρ(r s ) = r 1/s + 7r 0 + 3h. Since 1 N and s N, the residue position of ρ(r s ) r 1/s is 7x 1/s + x 1/(q+s) + 3 x 1/(n+s) q+s N q 1/s Q x q 1/s + 3 n+s Q n 1/s Q x n 1/s By Theorem 4, the set s + Q has 3r 1 elements in Q, 3r 1 elements in N and one element is 0. Thus the set { 1/(q + s), q + s 0} has 3r 1 elements in N and 3r 1 elements in Q. Similarly the set { 1/(n + s)} has 3r elements in N and 3r 1 elements in Q; the set 1/s + Q has 3r elements in N and 3r 1 elements in Q; the set 1/s + N has 3r 1 elements in N and 3r 1 elements in Q, and one element is 0. Also since for any 1/(q + s) Q, there is a q Q such that 1/(q + s) = q 1/s and for any 1/(n + s) Q, there is a n N such that 1/(n + s) = n 1/s, the nonresidue position is equal to x q 1/s + 3 x n 1/s x q 1/s 3 x n 1/s = 0. q 1/s Q q+s Q n 1/s Q n+s Q q 1/s Q Now the nonresidue position of ρ(r s ) r 1/s is: x 1/(q+s) 3 x 1/(n+s) x q 1/s + 3 q 1/s N. n 1/s Q n 1/s N x n 1/s Since for any 1/(q +s) N, there is a n N such that 1/(q +s) = n 1/s and for any 1/(n + s) N, there is a q Q such that 1/(n + s) = q 1/s, and there are 3r + 3r 1 = 6r 1 terms appearing, so the residue position is equal to 4e = 5e. Thus, since 0 1/s + N, we have ρ(r s ) r 1/s = (, 6 + 5e 1 ) = 7r 0 + h; that is, ρ(r s ) = r 1/s + 7r 0 + h Q 1..

14 6 BIJAN TAERI By similar proofs, we obtain the following results: When r = 3k, r 0 = (0, 1 + e ), ρ(r 0 ) = 8r 0 + h, ρ(r s ) = 8r 1/s + r 0, s Q, ρ(r s ) = r 1/s + 8r 0, s N, ρ(r ) = r 0 + 8h. When r = 3k +, r 0 = (0, 4 + e 1 + 6e ), ρ(r 0 ) = 4r 0 + h, ρ(r s ) = 8r 1/s + 4r 0 + 3h, s Q, ρ(r s ) = r 1/s + 4r 0 + 5h, s N, ρ(r ) = r 0 + h. Now suppose that p = 1r + 1 and r = 3k + 1. Since Q 1 = Q 1 + 7h (Theorem 8(d)), the extended code Q 1 is generated by p+1 rows of the (p + 1) (p + 1) matrix r G 1..,. r where each row of G 1 is a cyclic shift of the vector 6 + 3e 1 + 8e. Since p = 4(3r) + 1, 1 is a residue mod p, by [8, Theorem 65]. Thus ρ(e 1 ) = e 1 and ρ(e ) = e. By definition of ρ we have x 0 5x and x 4x 0. Now r 0 = (0, 6+3e 1 + 8e ) and so ρ(r 0 ) = (5 6, 4 0 3e 1 + 8e ) = (3, 6e 1 + 8e ) = 4r 0 + 5h Q 1. Since r = (8, 7 + 7e 1 + 7e ) we have ρ(r ) = (5 7, 4 8 7e 1 + 7e ) = (8, 5 + e 1 + 7e ) = r 0 + 8h Q 1. Now we prove that ρ(r s ) Q 1. In all the following proofs we let q Q and n N. We have ( r s = 0, 6x s + 3 x q+s + 8 x n+s), ( r 1/s = 0, 6x 1/s + 3 x q 1/s + 8 x n 1/s). We consider two cases s Q and s N: Case 1. s Q. We have ρ(r s ) = (5 3, 6χ(s)x 1/s 3 χ(q+s)x 1/(q+s) 8 χ(n + s)x 1/(n+s) ). Since 0 s + Q, the position of ρ(r s ) is 5 3 = 6. We show that ρ(r s ) = 8r 1/s + 4r 0 + 6h. Since 1 Q and s Q, the residue position of ρ(r s ) + r 1/s is: 6x 1/s 3 x 1/(q+s) 8 x 1/(n+s) +6x 1/s + 3 q+s Q q 1/s Q x q 1/s + 8 n+s Q n 1/s Q x n 1/s.

15 QUADRATIC RESIDUE CODES 7 By Theorem 4, the set s + Q has 3r 1 elements in Q, 3r elements in N, and one element is 0. Thus, since 1 Q, the set { 1/(q + s), q + s 0} has 3r 1 elements in Q and 3r elements in N. Similarly the set { 1/(n + s)} has 3r elements in Q and 3r elements in N; the set 1/s + Q has 3r 1 elements in Q and 3r elements in N, and one element is 0; the set 1/s + N has 3r elements in Q and 3r elements in Q. Also since for any 1/(q + s) Q, there is a q Q such that 1/(q + s) = q 1/s and for any 1/(n + s) Q, there is a n N such that 1/(n + s) = n 1/s, the nonresidue position is equal to 3 x q 1/s 8 x n 1/s q 1/s Q + 3 q 1/s Q x q 1/s + 8 n 1/s Q n 1/s Q Now the nonresidue position of ρ(r s ) + r 1/s is: 3 x 1/(q+s) + 8 q+s N + 3 q 1/s N n+s N x q 1/s + 8 x 1/(n+s) n 1/s N x n 1/s = 0. x n 1/s. Since for any 1/(q +s) N, there is a n N such that 1/(q +s) = n 1/s and for any 1/(n + s) N, there is a q Q such that 1/(n + s) = q 1/s, and there are 3r + 3r = 6r terms appearing, so the residue position is equal to 3 x q 1/s + 8 x n 1/s q 1/s N + 3 q 1/s N n 1/s N x q 1/s + 8 Thus, since the 0 1/s + Q, we have that is, ρ(r s ) = 8r 1/s + 4r 0 + 6h. n 1/s N x n 1/s = e. ρ(r s ) + r 1/s = (6, 3 + e 1 ) = 4r 0 + 6h; Case. s N. We have ρ(r s ) = (5 8, 6χ(s)x 1/s 3 χ(q+s)x 1/(q+s) 8 χ(n + s)x 1/(n+s) ). Since 0 s + N, the position of ρ(r s ) is 5 8 = 4. We show that ρ(r s ) = r 1/s +4r 0 +4h. Since 1 Q and s N, the nonresidue position of ρ(r s ) r 1/s is: 6x 1/s + 3 x 1/(q+s) + 8 x 1/(n+s) 6x 1/s + 3 q+s N q 1/s N x q 1/s + 8 n+s N n 1/s N x n 1/s.

16 8 BIJAN TAERI Since 1 Q, by Theorem 4 the set { 1/(q + s)} has 3r elements in Q and 3r elements in N. Similarly, the set { 1/(n + s), n + s 0} has 3r elements in Q and 3r 1 elements in N; the set 1/s + Q has 3r elements in Q and 3r elements in N; the set 1/s + N has 3r elements in Q and 3r 1 elements in N, and one element is 0. Also since for any 1/(q + s) N, there is a q Q such that 1/(q + s) = q 1/s and for any 1/(n + s) N, there is a n N such that 1/(n + s) = n 1/s, the nonresidue position is 0. The residue position of ρ(r s ) r 1/s is: 3 x 1/(q+s) 8 x 1/(n+s) 3 q+s Q q 1/s Q x q 1/s + 8 n+s Q n 1/s Q x n 1/s Since for any 1/(q + s) Q, there is a n N such that 1/(q + s) = n 1/s and for any 1/(n + s) Q, there is a q Q such that 1/(n + s) = q 1/s, there are 3r + 3r = 6r terms appearing, so the residue position is equal to 11e 1 = 7e 1. Thus since 0 1/s +N, ρ(r s ) r 1/s = (4, 1+7e 1 ) = 4r 0 +4h; that is, ρ(r s ) = r 1/s + 4r 0 + 4h Q 1. By similar proofs, we obtain the following results: When r = 3k + 1, r 0 = (0, 6 + 3e 1 + 8e ), ρ(r 0 ) = 4r 0 + 5h, ρ(r s ) = 8r 1/s + 4r 0 + 6h, s Q, ρ(r s ) = r 1/s + 4r 0 + 4h, s N, ρ(r ) = r 0 + 8h. When r = 3k, When r = 3k +,. r 0 = (0, 8e ), ρ(r 0 ) = r 0, ρ(r s ) = 8r 1/s + r 0, s Q, ρ(r s ) = r 1/s + r 0 + 8r, s N, ρ(r ) = 8r. r 0 = (0, 3 + 6e 1 + 8e ), ρ(r 0 ) = 7r 0 + 6h, ρ(r s ) = 8r 1/s + 7r 0 + 3r, s Q, ρ(r s ) = r 1/s + 7r 0 + 7h, s N, ρ(r ) = 4r 0 + 8h. Recall that the Lee weight of a Z m is defined as min{a, m a}, and the Lee wight of a vector is the sum of the Lee wight of its components. Thus in Z 9, the Lee wight of 0 is 0; the Lee wight of 1, 8 is 1; the Lee wight of, 7 is ; the Lee wight of 3, 6 is 3 and the Lee wight of 4, 5 is 4. The Euclidian weight of a Z m is defined as the square of the Lee wight of a, i.e., (min{a, m a}) ; and the Euclidian weight of a vector is the sum of the Euclidian wight of its components. By direct computation with the aid of a computer, we have Theorem 1. The Z 9 -QR code of length 11 has minimum Lee weight 7, minimum Euclidian weight 9, and minimum Hamming weight 5.

17 QUADRATIC RESIDUE CODES 9 We define the maps β i, i = 1,, 3, 4, from Z 9 to Z, by c β 1 (c) β (c) β 3 (c) β 4 (c) The map β i can be extended on N-tuples which is also denoted by β i. The Gray map φ : Z N 9 Z 4N is given by φ(c) = (β 1 (c), β (c), β 3 (c), β 4 (c)). Clearly φ is a distance-preserving map from (Z N 9, Lee distance) to (Z 4N, Hamming distance). The weight distribution of the image of the QR-code of length 11 under the Gray map is given in the following table; i A i i A i , 8, , Acknowledgment. The author thanks the Center of Excellence of Mathematics of Isfahan University of Technology (CEAMA). References [1] A. Bonneaze, P. Solé, and A. R. Calderbank, Quaternary quadratic residue codes and unimodular lattices, IEEE Trans. Inform. Theory 41 (1995), no., [] M. H. Chiu, S. S.-T. Yau, and Y. Yu, Z 8 -cyclic codes and quadratic residue codes, Adv. in Appl. Math. 5 (000), no. 1, 1 33.

18 30 BIJAN TAERI [3] A. R. Hammons, Jr., P. V. Kumar, A. R. Calderbank, N. J. A. Sloane, and P. Solé, The Z 4 -linearity of Kerdock, Preparata, Goethals, and related codes, IEEE Trans. Inform. Theory 40 (1994), no., [4] T. W. Hungerford, Algebra, Springer-Verlag, New York, [5] F. J. MacWilliams and N. J. A. Sloan, The theory of error-correcting codes, North Hoalland, Amsterdam, [6] B. R. McDonald, Finite Rings with Identity, Pure and Applied Mathematics, Vol. 8. Marcel Dekker, Inc., New York, [7] O. Perron, Bemerkungen über die Verteilung der quadratischen Reste, Math. Z. 56 (195), [8] V. Pless, Introduction to The Theory of Error Correcting Codes, John Wiley & Sons, Inc., New York, [9] V. Pless, P. Solé, and Z. Qian, Cyclic self-dual Z 4 -codes, Finite Fields Appl. 3 (1997), no. 1, [10] V. Pless and Z. Qian, Cyclic codes and quadratic residue codes over Z 4, IEEE Trans. Inform. Theory 4 (1996), no. 5, Department of Mathematical Sciences Isfahan University of Technology Isfahan and Institute for Studies in Theoretical Physics and Mathematics Iran address: b.taeri@cc.iut.ac.ir