82
5 Quantum Code Constructions
Lemma 5.2.3 Let n = q
m
− 1, where q ≥ 3 is a prime power and consider c be a
positive integer. Then the c cosets given by
{C q+1 , C 2q+1 , C 3q+1 , . . . , C cq+1 }
are disjoint and each of them has m elements provided cq + 1 < q
m/2
− 1. Moreover, each of these cosets are distinct of the cosets C 1 , C 2 , . . . , C c .
Proof Apply Lemma 5.2.2.
Theorem 5.2.4 Let n = q
m
− 1, where q ≥ 3, and assume that the inequality cq +
1 < q
m/2
− 1 holds. Then the last elements in the c cosets given by
{C q+1 , C 2q+1 , C 3q+1 , . . . , C cq+1 },
form a sequence of c consecutive integers.
Proof Let us consider the q-ary cosets modulo n = q
m
− 1 given by
{C q+1 , C 2q+1 , C 3q+1 , . . . , C cq+1 }.
We show that the last elements in these cosets form a sequence of c consecutive
positive integers. In fact, from Lemma 5.2.3, each of these cosets has cardinality m.
Let C s and C s+q be two of them. Let u and v be the last elements in the cosets C s and
C s+q , respectively, where u and v are integers considered without using the modulo
n operation. Then it follows that
u = sq
m−1
= sq
t
; v = (s + q)q
m−1
= (s + q)q
t
.
Therefore, we know that v = sq
t
+ q
t+1 ; so, v ≡ sq
t
+ 1 mod (q
m
− 1), that is,
v ≡ u + 1 mod (q
m
− 1).
Let n = q
m
− 1. Applying the division algorithm for v and n and for u + 1 and
n, there exist integers a, b, r 1 and r 2 , where 0 ≤ r 1 , r 2 < n such that
v = an + r 1 u + 1 = bn + r 2 .
Since v ≡ u + 1 mod n, it follows that r 1 = r 2 . Denote this common number by
v
∗ . We thus have
v = an + v
∗
; u + 1 = bn + v
∗
,
where 0 ≤ v
∗
< n. Therefore,
u = bn + v
∗
− 1.