# Two commutator relations.

$$\newcommand{\ad}{\operatorname{ad}}$$Let $$R$$ be an associative ring. Set $$[x, y] = xy - yx$$ and $$\ad_x(-) = [x, -]: R \to R$$.

1. Is there a formula for $$(ab)^n$$ in general? I found one formula for the group theoretic commutator, i.e., $$(xy)^n = x^{\binom{n}1} y^{\binom{n}1} [y,x]^{\binom{n}{2}}[[y,x],x]^{\binom{n}{3}} \cdots [\cdots[[[ y, x],x],x]\cdots,x]^{\binom{n}{n}},$$ I am wondering if there is any nice formula for the ring theoretic commutator. BTW, I found https://mathoverflow.net/q/78813/19222, which is a for $$(x+y)^n$$.
2. In characteristic $$p$$, suppose that $$r$$ commutes with $$[x, r]$$, we should have $$\ad_{rx}^{p-1}(r) = -r\ad_x^{p-1}(r^{p-1})$$. But now I have no idea how to prove it.

Any hints? Thank you!

• I suspect the answer to question 1 is negative. Hall's formula for groups relies on the fact that the elements are invertible, which allows the right hand side to have huge exponents compared to the left. But in a ring, with $a$ and $b$ not being invertible, you are limited to degree $2n$ -- any higher degree cannot be "balanced out". – darij grinberg Apr 13 at 14:53
• What kind of formula are you even looking for in 1? What would it be for $n=2$? -- For no. 2, writing out both sides with the commutator for $p=2,3$ should give an idea what's going on. – Torsten Schoeneberg Apr 13 at 16:01