98
5 Quantum Code Constructions
= q
2 j q
2m q
−2
+ q
2 j q
2m q
−4
+ · · · + q
2 j q
2m q
−2 j+2
+
+ q
2 j q
2m q
−2 j
+ q
2 j q
2m q
−2 j−2
+ · · · + q
2 j q
2
+ q
2 j
≡ (mod n)q
2 j q
−2
+ q
2 j q
−4
+ · · · + q
2 j q
−2 j+2
+
+ q
2 j q
−2 j
+ q
2m−2
+ q
2m−4
+ · · · + q
2 j q
2
+ q
2 j
=
= (q
2
)
m−1 + (q
2
)
m−2 + · · · + (q
2
)
j+1 + (q
2
)
j +
+ (q
2
)
j−1 + (q
2
)
j−2 + · · · + q
2
+ 1 =
=
m−1
i=0
(q
2
)
i = s
Lemma 5.3.6 Let q = 2 be a prime power and n = q
2m
− 1, where m = ord n (q
2
) ≥
3. Let s =
m−1
i=0
(q
2
)
i . Then the following results hold:
(a) the q
2 -ary cosets of the form C [s+i] are mutually disjoints, where 1 ≤ i ≤ q
2
− 1;
(b) the q
2 -ary cosets of the form C [s− j] are mutually disjoints, where 1 ≤ j ≤ q
2
−
1;
(c) the q
2 -ary cosets of the form C [s+i] are mutually disjoints from the q
2 -ary cosets
of the form C [s− j] , where 1 ≤ i, j ≤ q
2
− 1.
Proof We only show Item (a). Items (b) and (c) are left to the reader.
(a) Assume that there exist i = j, where 1 ≤ i, j ≤ q
2
− 1 such that C [s+i] =
C [s+ j] . Then there exists 0 ≤ t ≤ m − 1 such that s + i ≡ (s + j)q
2t mod n. From
Lemma 5.3.5, we know that sq
2t
≡ s mod n. Since gcd(q
2
, n) = 1 and q
2m
≡
1 mod n, then one has
s + i ≡ (s + j)q
2t
≡ s + jq
2t mod n
=⇒ i ≡ jq
2t mod n.
Because 1 ≤ i, j ≤ q
2
− 1, it follows that
i ≡ jq
2t mod n =⇒ i = jq
2t
.
If t = 0, then i = j, a contradiction; if t ≥ 1, the equality i = jq
2t does not hold.
Therefore, it follows that the cosets C [s+i] and C [s+ j] are disjoint. The proof is
complete.
Exercise 5.3.1 Show Items (b) and (c) of Lemma 5.3.6.
Lemma 5.3.7 Let q ≥ 4 be a prime power and n = q
2m
− 1, where m = ord n (q
2
) ≥
3. Let s =
m−1
i=0
(q
2
)
i . Then the following hold: