5.3 BCH Codes—Part III
99
(a) the cosets of the form C [s+i] , where 1 ≤ i ≤ q
2
− 1, contain m elements;
(b) the cosets of the form C [s− j] , 1 ≤ j ≤ q
2
− 1, contain m elements.
Proof We prove Item (a). Item (b) is left as exercise.
(a) The elements of C [s+i] are of the form (s + i)q
2t , where 0 ≤ t ≤ m − 1 for
all 1 ≤ i ≤ q
2
− 1. Since gcd(q
2
, n) = 1, q
2m
≡ 1 mod n and sq
2t
≡ s mod n, it
follows that
(s + i)q
2t
≡ s + iq
2t mod n.
Let us consider that 0 ≤ t ≤ m − 2. We then have
s + iq
2t
< (q
2m
− 1)/(q
2
− 1) + q
2m−2
≤ (q
2m
− 1)/15 + (q
2m
− 1)/15
< q
2m
− 1.
Hence, the first m − 1 elements belonging to C [s+i] are distinct, for all 1 ≤ i ≤
q
2
− 1, i.e., the cosets C [s+i] contain m elements, because m − 1 > m/2.
Lemma 5.3.8 Let q ≥ 4 be a prime power and n = q
2m
− 1, where m = ord n (q
2
) ≥
3. Let s =
m−1
i=0
(q
2
)
i . If C is the cyclic code generated by the product of the minimal
polynomials
M
(s)
(x)M
(s+1)
(x) · · · M
(s+i)
(x)M
(s−1)
(x) · · · M
(s− j)
(x),
where 1 ≤ i, j ≤ q
2
− 1, then C is Hermitian dual-containing.
Proof According to Lemma 5.1.4, we have to show that Z
Z
−q
= ∅. Forcing a
contradiction, we assume that Z
Z
−q
= ∅. The cases concerning the coset C [s]
are immediate. Assume first that C [s+ j] = C [−q(s+i)] , 1 ≤ i, j ≤ q
2
− 1. Then there
exists 0 ≤ h ≤ m − 1 such that
s + j ≡ −q(s + i)q
2h mod n.
Because gcd(q
2
, n) = 1, q
2m
≡ 1 mod n and sq
2t
≡ s mod n for all 0 ≤ t ≤ m − 1,
we obtain
s + j ≡ −qs − qiq
2h mod n,
where 0 ≤ h ≤ m − 1. We now compute the expression s + j + q(s + iq
2h
), 0 ≤
h ≤ m − 1. If h ≤ m − 2, one has