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.


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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