Here is the question I am trying to prove:
Let $q\ne1$ be a root of unity of order $d > 1.$ Prove that $yx = qxy$ in a noncommutative algebra implies $$(x + y)^d = x^d + y^d.$$
I do know how to tackle it, could someone help me please?
Some attempts:
1-$(x + y)^2 = (x + y)(x + y) = x^2 + xy + yx + y^2 = x^2 + xy + qxy + y^2$ but I do not know how the expression $xy + qxy$ should be removed in this case ..... can anyone explain this to me please?
2- Also, $$(x + y)^3$$$$ = (x + y)^2(x + y) $$$$= (x + y)(x + y)(x + y)$$ $$ = (x^2 + xy + yx + y^2) (x + y) $$$$= (x^3 + qx^2y + q^2x^2y + q^2xy^2 + x^2y + xy^2 + qxy^2 + y^3)$$
but I do not know how the expression between $x^3$ and $y^3$ should be removed in this case … can anyone explain this to me please?
We have $$(x+y)^n=\sum_{k=0}^n {n\choose k}_qx^{n-k}y^k \quad(*)$$ for some coefficients ${n\choose k}_q.$ Clearly ${n\choose 0}_q={n\choose n}_q=1.$ We will derive a recurrence relation for these quantities. By definition we have $$(x+y)^{n+1}=\sum_{k=0}^{n+1}{n+1\choose k}_qx^{n+1-k}y^k$$ On the other hand $$(x+y)^{n+1}=(x+y)^n(x+y)=\sum_{k=0}^n {n\choose k}_qx^{n-k}y^k(x+y)\\ = \sum_{k=0}^n q^k{n\choose k}_qx^{n+1-k}y^k+\sum_{k=0}^n {n\choose k}_qx^{n-k}y^{k+1}\\ =\sum_{k=0}^n q^k{n\choose k}_qx^{n+1-k}y^k+\sum_{k=1}^{n+1} {n\choose k-1}_qx^{n+1-k}y^{k}\\ =x^{n+1}+\sum_{k=1}^{n}\left [{n\choose k-1}_q+q^k{n\choose k}_q\right ]x^{n+1-k}y^k+y^{n+1} $$ Therefore we obtain $${n+1\choose k}_q={n\choose k-1}_q+q^k{n\choose k}_q\quad (**)$$ Basing on $(**)$ it can be shown, by induction on $n,$ that if $q^k\neq 1$ for $1\le k\le n-1$ then $${n\choose k}_q={(1-q^n)(1-q^{n-1})\ldots (1-q^{n-k+1})\over (1-q^k)(1-q^{k-1})\ldots (1-q)}\quad (***)$$ If $q$ is a primitive root of unity of order $d,$ i.e. $q^d=1$ and $q^k\neq 1$ for $1\le k<d,$ then $(***) $ implies $${d\choose k}_q=0,\quad 1\le k\le d-1$$ as $1-q^k\neq 0$ for $0<k<d$ and $1-q^d=0.$ Hence putting $n:=d$ in $(*)$ gives $$(x+y)^d=x^d+y^d$$ If $q$ is a non-primitive root of order $d$, the conclusion does not hold. For example if $d=2d_0$ and $q$ is a primitive root of unity of order $d_0,$ then $$(x+y)^{d}=[(x+y)^{d_0}]^2=(x^{d_0}+y^{d_0})^2 \\ = x^d+y^d+x^{d_0}y^{d_0}+y^{d_0}x^{d_0}\\ =x^d+y^d+x^{d_0}y^{d_0}+(q^{d_0})^{d_0}x^{d_0}y^{d_0}\\ = x^d+y^d+2x^{d_0}y^{d_0}$$ Concerning $q$-binomial coefficients see here