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Theorem. Let $R$ be a left (resp. right) s-unital ring. If $R$ satisfy $(P_1)$ (resp. $(P_2)$). Then R is commutative(and conversely).

$(P_1)$ $y^{s}[x,\, y]=\pm x^{p}[x^{m},y^{n}]^{r}y^{q}$ where $m>1,r>0,n\geq0, s\geq0, p\geq0 , q\geq0$

$(P_2)$ $[x,\, y]y^{t}=\pm x^{p}[x^{m},y^{n}]^{r}y^{q}$ where $m>1,r>0,n\geq0, t\geq0, p\geq0 , q\geq0$

A ring is called left (resp. right) s-unital if $x\in Rx$ (resp. $x\in xR$) for all $x$ in $R$. A ring is called s-unital if and only if $x\in xR\cap Rx$ for all $x$.

Two examples: $R_1=\left\{ \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix},\begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 1 \end{pmatrix},\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\right\} $, $R_2=\left\{ \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix},\begin{pmatrix}1 & 0\\ 1 & 0 \end{pmatrix},\begin{pmatrix}0 & 1\\ 0 & 1 \end{pmatrix},\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\right\} $ which are subrings of $2×2$ matrices over $GF(2)$.

The ring $R_1$ (resp. $R_2)$ are left (resp. right) s-unital ring and it satisfies $(P_1)$ (resp. $(P_2)$) for some integers. Hovewer, they are not commutative. Could you elaborate why it is like this?

The theorem is true and there is a proof in this article: http://www.m-hikari.com/ija/ija-2010/ija-25-28-2010/khanIJA25-28-2010-1.pdf

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Am I reading this right? It looks like you say there is a proof that a left $s$-unital ring satisfying $P_1$ is commutative, but you are also saying $R_1$ is a counterexample to that statement. In that case, either the proof is wrong or the counterexample doesn't satisfy all the requirements. –  rschwieb Aug 16 '13 at 11:51
Dear Monika: Can't you be more specific about where you think the problem is? Questions like this, which include rather involved computations and a link to a paper, accompanied by the question "can someone explain this?" will probably not get a lot of responses because it feels like all the work is dumped on the reader. –  rschwieb Aug 16 '13 at 11:53
I quickly perused the article, and it seems to me that commutativity is claimed to follow from $(P_1)$ or $(P_2)$, if the rng is semiprime (whatever that means) or if it has a unit, i.e. is a ring. May be your examples $R_1$ and $R_2$ fail at that point? At least the latter case does not apply as they don't have a multiplicative neutral element. –  Jyrki Lahtonen Aug 16 '13 at 11:53

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