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Observation 2.6.2.
1. Define in the polynomial algebra Z[q] for n ∈ N:
Define
(n) q := 1 + q + . . . + q n−1 .
(n)! q := (n) q ∙ ∙ ∙ (2) q (1) q ∈ Z[q] .
Finally, define for 0 ≤ i ≤ n in the field of fractions
( ) n (n)! q
:=
.
i (n − i)! q (i)! q
2. We note that
and thus deduce
q k ( n
k
)
q
+
( n
k − 1
q
q k (n + 1 − k) q + (k) q = (n + 1) q
)
q
=
=
(n)! q
(n+1−k)! q(k)! q
∙ ( )
q k (n + 1 − k) q + (k) q
( n + 1
k
( ) n
from which we conclude by induction on n that
k
)
q
q
∈ Z[q].
For any associative algebra A over a field K, we can then specialize for q( ∈ K)
the values of
n
(n) q , (n)! q and the of the q-binomials and denote them by (n) q , (n)! q and . Note that
k
q
(n) 1 = n. If q is an N-th root of unity different from 1, then
(N) q = 1 + q + . . . + q N−1 = 1 − qN
1 − q = 0 .
In a field of characteristic p with p a divisor of q, the quantity N = 1+. . .+1 also vanishes. There
are similarities between q-deformed situations and situations in fields of prime characteristic.
As a further consequence,
( ) N
= 0 for all 0 < k < N .
k
q
Lemma 2.6.3.
Let A be an associative algebra over a field K and q ∈ K. Let x, y ∈ A be two elements that
q-commute, i.e. xy = qyx. Then the quantum binomial formula holds for all n ∈ N:
(x + y) n =
n∑
( ) n
i
i=0
q
y i x n−i .
Proof.
By induction on n, using the relation we just proved.
✷
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