# An example of s-unital rings

I am just studying s-unital rings.

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 $x$.

Do you know any books which expand this subject?

I have an example of right s-unital: $R=\left\{ \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix},\begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 1 \end{pmatrix},\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\right\}$

So if I take an element $x$ from this ring and multiply by another element it should be equal to $x$. Do I understand this definition properly? However, if I multiply $\begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix}\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}$ then is not equal to \begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix}

Could you give me another example of s-unital ring which is at the same time rigt and left s-unital?

Thank you.

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Note that $x\in xR$ doesn't imply that for all $y\in R$ it holds that $xy=x$, but rather that there exists some $y\in R$ such that $xy=x$. Also, did you intend to take $R$ as a subring of $\mathbb{F}_2^{2\times 2}$? – Jonathan Y. Aug 14 '13 at 11:47
@Jonathan Y.: Yes, I forgot to write it. – Monika Aug 14 '13 at 11:50
And I guess you are mainly interested in examples of rings without identity (since every ring with identity obviously has this property.) – rschwieb Aug 14 '13 at 12:49

Any boolean rng (meaning possibly without identity) is clearly left-right $s$-unital.
For a concrete example, consider this boolean ring which lacks an identity: $\oplus _{i\in \Bbb N}\Bbb F_2$.