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Let $a,b$ be elements of a non-commutative ring $R$ with $\operatorname{char}(R) =p > 0$ and suppose that $ab-ba=[a,b]=1$. My question is simply:

Could you give a formula for the element $(a^n b^m)^p$ in the form $\sum\limits_{i,j} c_{ij} a^i b^j$?

For example, $(ab)^2=abab=a(ab-1)b=a^2b^2-ab$. I've tried to find some patterns in order to use induction, but without much success. May be someone more familiar with manipulations like this could help. Thank you.

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No doubt this is some nifty combinatorial problem or other with a well-known solution, but brute force can get us some of the way. Using $[A,BC]=[A,B]C+B[A,C]$, it is easy to check that $$ [b,a^n] = [b,a^{n-1}]a+a^{n-1}[b,a] = [b,a^{n-1}]a-a^{n-1}, $$ and iterating gives us inductively $$ [b,a^n] = -na^{n-1}. $$

Now, supposing initially that $n\geqslant m$, I believe one can similarly derive the following: $$ [b^m,a^n] = \sum_{k=0}^{m-1} (-1)^{k+1} \binom{m}{k} \frac{n!}{(n-m+k)!} a^{n-m+k} b^k. $$ (if there's a more symmetric way of writing this, I'm happy to hear it...)

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  • $\begingroup$ Thank you for your answer, @Chappers! But how can we use the expression for $[b^m,a^n]$ to evaluate $(a^n b^m)^p$? $\endgroup$ – u1571372 Mar 10 '15 at 21:44
  • $\begingroup$ At this point, we have to start writing $(a^n b^m)^p = a^n b^m a^n b^m (a^n b^m)^{p-2} = a^n ([b^m, a^n]+a^n b^m) b^m (a^n b^m)^{p-2} $. This looks pretty grim. As an alternative, we could look at $a^n([b^m,(a^n b^m)^{p-1}]+(a^n b^m)^{p-1}b^m)$, and hope that some sort of induction works there as it did on $a^n b^m$. $\endgroup$ – Chappers Mar 10 '15 at 22:54

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