5.3 BCH Codes—Part III
97
q + l + q
3 r + q
2
+ j < q
3
(q − 1) + 2q + 2q
2
= q
4
− q
3
+ 2q + 2q
2
.
Since q
3
> 2q
2
+ 2q + 1, it follows that q + l + q
3 r + q
2
+ j < q
4
− 1; hence,
q + l + q
3 r = −q
2
− j, a contradiction. Consequently, C is Hermitian
dual-containing.
We next compute the minimum distance and the dimension of C. Since the defining
set of C contains the sequence q
2
+ 1, q
2
+ 2, . . . , 2q
2
− 1, it follows from the BCH
bound that C has minimum distance greater than or equal to q
2 . On the other hand,
from Lemma 5.3.4, the defining set of C has 2(q
2
− 2) + 1 elements. Hence, g(x)
has degree deg(g(x)) = 2(q
2
− 2) + 1, so C has dimension n − 2(q
2
− 2) − 1, i.e.,
C is an [n, n − 2(q
2
− 2) − 1, d ≥ q
2
] q 2 code. Applying Lemma 5.1.5, there exists
an [[n, n − 4(q
2
− 2) − 2, d ≥ q
2
[] q quantum code. The proof is complete.
Corollary 5.3.1 Let q ≥ 3 be a prime power and n = q
4
− 1. Then there exists an
[[n, n − 4(c − 2) − 2, d ≥ c[] q , where 3 ≤ c ≤ q
2
− 1
Proof It suffices to consider C as the cyclic code generated by
M
(q
2 +1)
(x)M
(q
2 +2)
(x) · . . . · M
(q
2 +c−1)
(x),
after proceeding similarly to the proof of Theorem 5.3.1.
Example 5.3.1 As an example, let us consider m = 2 and q = 3. Let C be generated
by M
(10)
(x)M
(11)
(x). Applying Theorem 5.3.1 we have an [[80, 74, d ≥ 3]] 3 code.
5.3.2 Construction II
In this subsection, we construct suitable Hermitian dual-containing non-narrow-sense
BCH codes with good parameters in order to obtain good quantum codes derived
from them.
We start with the following result.
Lemma 5.3.5 Let q = 2 be a prime power and n = q
2m
− 1, where m = ord n (q
2
) ≥
3. If s =
m−1
i=0
(q
2
)
i , then the q
2 -coset C [s] has only one element.
Proof We know that gcd(q
2
, n) = 1 and q
2m
≡ 1 mod n. The result follows from
direct computation.
sq
2 j
=
m−1
i=0
(q
2
)
i
q
2 j
= q
2 j
(q
2
)
m−1 + q
2 j
(q
2
)
m−2 + · · · + q
2 j q
2
+ q
2 j
=
97
q + l + q
3 r + q
2
+ j < q
3
(q − 1) + 2q + 2q
2
= q
4
− q
3
+ 2q + 2q
2
.
Since q
3
> 2q
2
+ 2q + 1, it follows that q + l + q
3 r + q
2
+ j < q
4
− 1; hence,
q + l + q
3 r = −q
2
− j, a contradiction. Consequently, C is Hermitian
dual-containing.
We next compute the minimum distance and the dimension of C. Since the defining
set of C contains the sequence q
2
+ 1, q
2
+ 2, . . . , 2q
2
− 1, it follows from the BCH
bound that C has minimum distance greater than or equal to q
2 . On the other hand,
from Lemma 5.3.4, the defining set of C has 2(q
2
− 2) + 1 elements. Hence, g(x)
has degree deg(g(x)) = 2(q
2
− 2) + 1, so C has dimension n − 2(q
2
− 2) − 1, i.e.,
C is an [n, n − 2(q
2
− 2) − 1, d ≥ q
2
] q 2 code. Applying Lemma 5.1.5, there exists
an [[n, n − 4(q
2
− 2) − 2, d ≥ q
2
[] q quantum code. The proof is complete.
Corollary 5.3.1 Let q ≥ 3 be a prime power and n = q
4
− 1. Then there exists an
[[n, n − 4(c − 2) − 2, d ≥ c[] q , where 3 ≤ c ≤ q
2
− 1
Proof It suffices to consider C as the cyclic code generated by
M
(q
2 +1)
(x)M
(q
2 +2)
(x) · . . . · M
(q
2 +c−1)
(x),
after proceeding similarly to the proof of Theorem 5.3.1.
Example 5.3.1 As an example, let us consider m = 2 and q = 3. Let C be generated
by M
(10)
(x)M
(11)
(x). Applying Theorem 5.3.1 we have an [[80, 74, d ≥ 3]] 3 code.
5.3.2 Construction II
In this subsection, we construct suitable Hermitian dual-containing non-narrow-sense
BCH codes with good parameters in order to obtain good quantum codes derived
from them.
We start with the following result.
Lemma 5.3.5 Let q = 2 be a prime power and n = q
2m
− 1, where m = ord n (q
2
) ≥
3. If s =
m−1
i=0
(q
2
)
i , then the q
2 -coset C [s] has only one element.
Proof We know that gcd(q
2
, n) = 1 and q
2m
≡ 1 mod n. The result follows from
direct computation.
sq
2 j
=
m−1
i=0
(q
2
)
i
q
2 j
= q
2 j
(q
2
)
m−1 + q
2 j
(q
2
)
m−2 + · · · + q
2 j q
2
+ q
2 j
=