80
5 Quantum Code Constructions
It is clear that C [n−s] is the complementary coset of C s . We have to show that the
equality m l = m s is true. We first prove that [n − s]q
l
≡ [n − s]q
t mod n holds
for each 0 ≤ t, l ≤ m s − 1, that is, m l ≥ m s . In fact, seeking a contradiction,
we assume that [n − s]q
l
≡ [n − s]q
t mod n holds, where 0 ≤ t, l ≤ m s − 1.
Thus the congruence sq
l
≡ sq
t mod n holds, where 0 ≤ t, l ≤ m s − 1, which
is a contradiction.
On the other hand, seeking a contradiction, we assume that sq
r
≡ sq
t mod
n holds, for each 0 ≤ r, t ≤ m l − 1. Then one has −sq
r
≡ −sq
t mod n, that
is, [n − s]q
r
≡ [n − s]q
t mod n holds for each 0 ≤ r, t ≤ m l − 1, which is a
contradiction. Thus sq
r
≡ sq
t mod n for each 0 ≤ r, t ≤ m l − 1 and so, m l ≤
m s . Therefore, one obtains m l = m s , as required.
(iii) Follows from direct computation.
(iv) Suppose that C s is a q-ary coset with complementary C r and L s =| sq
t 1 − sq
t 2 |,
where 0 ≤ t 1 , t 2 ≤ m s − 1, t 1 = t 2 . We may assume without loss of generality
that
L s = sq
t 1 − sq
t 2 .
(5.1)
From hypothesis, there exists 0 ≤ l ≤ m r − 1 such that s ≡ −rq
l mod n. Replacing the last equivalence in Equation (5.1) one obtains L s ≡ | rq
t 2 +l
− rq
t 1 +l
|
mod n. We may assume that 0 ≤ t 1 + l, t 2 + l ≤ m r − 1 and t 2 > t 1 , so L s =
rq
t 2 +l
− rq
t 1 +l . Therefore L s ≤ L r .
On the other hand, assume L r = rq
a 1 − rq
a 2 , where 0 ≤ a 1 , a 2 ≤ m r − 1, a 1 =
a 2 . Since r ≡ −sq
m r −l mod n it follows that
L r = rq
a 1 − rq
a 2 ≡ | sq
m r −l+a 2 − sq
m r −l+a 1 | mod n.
As above, we can further assume w. l. o. g. that 0 ≤ m r − l + a 1 , m r − l + a 2 ≤
m s − 1 and also a 2 > a 1 . Thus
L r = sq
m r −l+a 2 − sq
m r −l+a 1 ,
and so L r ≤ L s . Therefore, the equality L r = L s holds.
(v) Assume that r ∈ C s . Then there exists 0 ≤ t ≤ m s − 1 such that (s + q
t r ) ≡ 0
mod n since, from Item (ii), m r = m s . Then one obtains (r + q
m s −t s) ≡ 0 mod
n. We can suppose that 0 ≤ m s − t ≤ m s − 1 since for t = 0 the equivalences
(r + q
m s −t s) ≡ (r + s) ≡ (s + q
t r ) mod n hold. From these facts we conclude
that C s = C s since, from Item (i), the complementary coset is unique.
The following lemma already presented (see Lemma 5.1.3) gives necessary and
sufficient condition under which a cyclic code contains its Euclidean dual.
5 Quantum Code Constructions
It is clear that C [n−s] is the complementary coset of C s . We have to show that the
equality m l = m s is true. We first prove that [n − s]q
l
≡ [n − s]q
t mod n holds
for each 0 ≤ t, l ≤ m s − 1, that is, m l ≥ m s . In fact, seeking a contradiction,
we assume that [n − s]q
l
≡ [n − s]q
t mod n holds, where 0 ≤ t, l ≤ m s − 1.
Thus the congruence sq
l
≡ sq
t mod n holds, where 0 ≤ t, l ≤ m s − 1, which
is a contradiction.
On the other hand, seeking a contradiction, we assume that sq
r
≡ sq
t mod
n holds, for each 0 ≤ r, t ≤ m l − 1. Then one has −sq
r
≡ −sq
t mod n, that
is, [n − s]q
r
≡ [n − s]q
t mod n holds for each 0 ≤ r, t ≤ m l − 1, which is a
contradiction. Thus sq
r
≡ sq
t mod n for each 0 ≤ r, t ≤ m l − 1 and so, m l ≤
m s . Therefore, one obtains m l = m s , as required.
(iii) Follows from direct computation.
(iv) Suppose that C s is a q-ary coset with complementary C r and L s =| sq
t 1 − sq
t 2 |,
where 0 ≤ t 1 , t 2 ≤ m s − 1, t 1 = t 2 . We may assume without loss of generality
that
L s = sq
t 1 − sq
t 2 .
(5.1)
From hypothesis, there exists 0 ≤ l ≤ m r − 1 such that s ≡ −rq
l mod n. Replacing the last equivalence in Equation (5.1) one obtains L s ≡ | rq
t 2 +l
− rq
t 1 +l
|
mod n. We may assume that 0 ≤ t 1 + l, t 2 + l ≤ m r − 1 and t 2 > t 1 , so L s =
rq
t 2 +l
− rq
t 1 +l . Therefore L s ≤ L r .
On the other hand, assume L r = rq
a 1 − rq
a 2 , where 0 ≤ a 1 , a 2 ≤ m r − 1, a 1 =
a 2 . Since r ≡ −sq
m r −l mod n it follows that
L r = rq
a 1 − rq
a 2 ≡ | sq
m r −l+a 2 − sq
m r −l+a 1 | mod n.
As above, we can further assume w. l. o. g. that 0 ≤ m r − l + a 1 , m r − l + a 2 ≤
m s − 1 and also a 2 > a 1 . Thus
L r = sq
m r −l+a 2 − sq
m r −l+a 1 ,
and so L r ≤ L s . Therefore, the equality L r = L s holds.
(v) Assume that r ∈ C s . Then there exists 0 ≤ t ≤ m s − 1 such that (s + q
t r ) ≡ 0
mod n since, from Item (ii), m r = m s . Then one obtains (r + q
m s −t s) ≡ 0 mod
n. We can suppose that 0 ≤ m s − t ≤ m s − 1 since for t = 0 the equivalences
(r + q
m s −t s) ≡ (r + s) ≡ (s + q
t r ) mod n hold. From these facts we conclude
that C s = C s since, from Item (i), the complementary coset is unique.
The following lemma already presented (see Lemma 5.1.3) gives necessary and
sufficient condition under which a cyclic code contains its Euclidean dual.