102
5 Quantum Code Constructions
Lemma 5.3.12 Let q ≥ 4 be a prime power and n = q
m
− 1, where m = ord n (q) ≥
3. Let s =
m−1
i=0
q
i . Let C be the cyclic code generated by
M
(s)
(x)M
(s+1)
(x) · . . . · M
(s+ j)
(x)M
(s−1)
(x) · . . . · M
(s− j)
(x),
where 1 ≤ j ≤ q − 1. Then C is Euclidean dual-containing.
Proof From Lemma 5.1.3, it is sufficient to prove that Z
Z
−1
= ∅. Forcing a
contradiction, we assume the Z
Z
−1
= ∅. The cases concerning the coset C [s] are
trivial.
(1) Assume first that C [s+i] = C [−(s+ j)] , where 1 ≤ i, j ≤ q − 1. Then there
exists 0 ≤ t ≤ m − 1 such that s + i ≡ −(s + j) mod n. Since gcd(q, n) = 1,
q
m
≡ 1 mod n and sq
t
≡ s mod n, 0 ≤ t ≤ m − 1, we have
s + i ≡ −s − jq
t mod n =⇒ 2s ≡ −(i + jq
t
) mod n.
If 0 ≤ t ≤ m − 2 and because q ≥ 4, it follows that
2s + i + jq
t
≤
2q
m
− 2
q − 1
+ (q − 1)(1 + q
m−2
)
< q
m
− 1
=⇒ 2s + i + jq
t
< q
m
− 1.
Hence s + i = −s − jq
t , a contradiction.
Let us next consider the case t = m − 1. We know that for each 1 ≤ i, j ≤ q −
3 we have 2s + i + jq
m−1
< q
m
− 1; since s + i = −s − jq
m−1 does not hold,
this implies in a contradiction. Analogously, if j = q − 3 and 1 ≤ i ≤ q − 1, it
follows that 2s + i + jq
m−1
< q
m
− 1, and because s + i = −s − jq
m−1 , we have
a contradiction.
If j ≥ q − 2, we have 2s + i + jq
t
> q
m
− 1. Let us compute the equivalence
2s ≡ −(i + jq
m−1
) mod n for j = q − 2 and 1 ≤ i ≤ q − 1.
2s ≡ −[i + (q − 2)q
m−1
] mod n =⇒ 2s ≡ −i − 1 + 2q
m−1 mod n.
As 0 < 2s + i + 1 − 2q
m−1
< q
m
− 1 and also 2s = −i − 1 + 2q
m−1 are true, one
has a contradiction.
Let j = q − 1 and 1 ≤ i ≤ q − 1. Computing the equivalence 2s ≡ −(i +
jq
m−1
) mod n we obtain
2s ≡ −[i + (q − 1)q
m−1
] mod n =⇒ 2s ≡ −i − 1 + q
m−1 mod n.