66
5 Quantum Code Constructions
cyclic code generated by M
(4)
(x)M
(6)
(x)M
(8)
(x). It is easy to see that C is Euclidean
dual-containing and has parameters [31, 22, d ≥ 5] 5 . Let C
be the cyclic code generated by M
(4)
(x)M
(8)
(x). The code C
has parameters [31, 25, d ≥ 4] 5 . Thus there
exists an [[31, 16, d ≥ 5]] 5 quantum code.
We next establish Theorem 5.1.7, an analogous to Theorem 5.1.1.
Theorem 5.1.7 Suppose that q ≥ 5 is a prime power and n > q is an integer such
that gcd(q, n) = 1. Assume also that (q − 1) | n and m = ord n (q) = 2 hold. Then
there exists a quantum code with parameters [[n, n − 4c, d ≥ c + 2]] q , where 1 ≤
c ≤ r − 3 and r > 3 is such that n = r (q − 1).
Proof We only prove the existence of an [[n, n − 4(r − 3), d ≥ r − 1]] q code, since
the constructions of the other codes are quite similar.
Let C be the cyclic code generated by
M
(r )
(x)M
(r +1)
(x) · . . . · M
(2r −3)
(x).
From Lemma 5.1.1 and from the proof of Theorem 5.1.1, we know that the q-cosets
given by C [r ] = {r }, C [r +1] = {r + 1, r + q}, C [r +2] = {r + 2, r + 2q}, . . . ,
C [2r −3] = {2r − 3, r + (r − 3)q} are mutually disjoint and each of them has two
elements. Therefore, C has dimension k = n − 2(r − 3) − 1 and minimum distance
d ≥ r − 1.
Let us prove that C is Euclidean dual-containing. In fact, if (r + i) ≡ −(r + j)
mod n, where 0 ≤ i, j ≤ r − 3, it follows that 2r + i + j ≡ 0 mod n. Since the
inequality 2r + i + j < n holds because q ≥ 5, one has a contradiction. On the
other hand, if (r + i)q ≡ −(r + j) mod n holds then
(iq + j)(q − 1) ≡ 0 mod n =⇒
i(q
2
− q) + j (q − 1) ≡ 0 mod n =⇒
j (q − 1) ≡ i(q − 1) mod n,
where the latter congruence holds because ord n (q) = 2. Then the unique solution is
when i = j. Let us investigate this case. Seeking a contradiction, we assume that the
congruence (r + i)q ≡ −(r + i) mod n is true. Then we obtain
(r + i)q ≡ −(r + i) mod n =⇒
2r + i(q + 1) ≡ 0 mod n =⇒
r (q − 3) ≡ i(q + 1) mod n.
If 0 ≤ i ≤ r − 4, then
r (q − 3) − i(q + 1) ≥ r (q − 3) − (r − 4)(q + 1) = 4q − 4r + 4 > 0,
where the latter inequality holds because r < q since we only consider nonprimitive
BCH codes. Moreover, the inequality r (q − 3) − i(q + 1) < n also holds, which is a
5 Quantum Code Constructions
cyclic code generated by M
(4)
(x)M
(6)
(x)M
(8)
(x). It is easy to see that C is Euclidean
dual-containing and has parameters [31, 22, d ≥ 5] 5 . Let C
be the cyclic code generated by M
(4)
(x)M
(8)
(x). The code C
has parameters [31, 25, d ≥ 4] 5 . Thus there
exists an [[31, 16, d ≥ 5]] 5 quantum code.
We next establish Theorem 5.1.7, an analogous to Theorem 5.1.1.
Theorem 5.1.7 Suppose that q ≥ 5 is a prime power and n > q is an integer such
that gcd(q, n) = 1. Assume also that (q − 1) | n and m = ord n (q) = 2 hold. Then
there exists a quantum code with parameters [[n, n − 4c, d ≥ c + 2]] q , where 1 ≤
c ≤ r − 3 and r > 3 is such that n = r (q − 1).
Proof We only prove the existence of an [[n, n − 4(r − 3), d ≥ r − 1]] q code, since
the constructions of the other codes are quite similar.
Let C be the cyclic code generated by
M
(r )
(x)M
(r +1)
(x) · . . . · M
(2r −3)
(x).
From Lemma 5.1.1 and from the proof of Theorem 5.1.1, we know that the q-cosets
given by C [r ] = {r }, C [r +1] = {r + 1, r + q}, C [r +2] = {r + 2, r + 2q}, . . . ,
C [2r −3] = {2r − 3, r + (r − 3)q} are mutually disjoint and each of them has two
elements. Therefore, C has dimension k = n − 2(r − 3) − 1 and minimum distance
d ≥ r − 1.
Let us prove that C is Euclidean dual-containing. In fact, if (r + i) ≡ −(r + j)
mod n, where 0 ≤ i, j ≤ r − 3, it follows that 2r + i + j ≡ 0 mod n. Since the
inequality 2r + i + j < n holds because q ≥ 5, one has a contradiction. On the
other hand, if (r + i)q ≡ −(r + j) mod n holds then
(iq + j)(q − 1) ≡ 0 mod n =⇒
i(q
2
− q) + j (q − 1) ≡ 0 mod n =⇒
j (q − 1) ≡ i(q − 1) mod n,
where the latter congruence holds because ord n (q) = 2. Then the unique solution is
when i = j. Let us investigate this case. Seeking a contradiction, we assume that the
congruence (r + i)q ≡ −(r + i) mod n is true. Then we obtain
(r + i)q ≡ −(r + i) mod n =⇒
2r + i(q + 1) ≡ 0 mod n =⇒
r (q − 3) ≡ i(q + 1) mod n.
If 0 ≤ i ≤ r − 4, then
r (q − 3) − i(q + 1) ≥ r (q − 3) − (r − 4)(q + 1) = 4q − 4r + 4 > 0,
where the latter inequality holds because r < q since we only consider nonprimitive
BCH codes. Moreover, the inequality r (q − 3) − i(q + 1) < n also holds, which is a