178
7 Constructions of QCCs
Lemma 7.4.5 Suppose that n = q
2m
− 1, where q ≥ 4 is a prime power and m =
ord n (q
2
) ≥ 3. Let s =
m−1
i=0
(q
2
)
i . Then the following hold:
(a) the q
2 -coset C [s] has only one element;
(b) the q
2 -cosets C [s+i] are mutually disjoints, where 1 ≤ i ≤ q
2
− 1;
(c) the q
2 -cosets C [s− j] are mutually disjoints, where 1 ≤ j ≤ q
2
− 1;
(d) the q
2 -cosets of the forms C [s+i] and C [s− j] are mutually disjoints, where 1 ≤
i, j ≤ q
2
− 1;
(e) the cosets of the form C [s+i] , where 1 ≤ i ≤ q
2
− 1, contain m elements;
(f) the cosets of the form C [s− j] contain m elements, where 1 ≤ j ≤ q
2
− 1.
Proof All these results can be found in [89, Lemmas III.3, III.4, and III.5].
Lemma 7.4.6 Let q ≥ 4 be a prime power and n = q
2m
− 1. Assume also that
gcd(q
2
, n) = 1 and m = or d n (q
2
) ≥ 3. Let s =
m−1
i=0
(q
2
)
i . If C is the cyclic code
generated by
M
(s)
(x)M
(s+1)
(x) . . . M
(s+i)
(x) · M
(s−1)
(x) . . . M
(s− j)
(x),
for all 1 ≤ i, j ≤ q
2
− 1, then C is Hermitian self-orthogonal.
Proof See [89, Lemma III.6].
Keeping these results in mind we are able to prove Theorems 7.4.3 and 7.4.4 and
their respective corollaries.
Theorem 7.4.3 Let n = q
2m
− 1, where q ≥ 4 is a prime power and m = ord n (q
2
)≥
3. Then there exists an [(n, n − 2m(2q
2
− 3) − 2, 1; m, d f ≥ 2q
2
+ 2)] q QCC.
Proof Clearly one has gcd(q, n) = 1. Consider first that C is the BCH code of length
n = q
2m
− 1 over F q 2 , generated by the product of the minimal polynomials
M
(s)
(x)M
(s+1)
(x) . . . M
(s+q
2 −2)
(x)M
(s+q
2 −1)
(x) ·
·M
(s−1)
(x) . . . M
(s−q
2 +1)
(x),
where s =
m−1
i=0
(q
2
)
i . A parity check matrix of C is obtained from the matrix