Example of ring such that for an element $s \in R$, the set $sR$ does not contain $s$

Question

I'm looking for a ring $R$, such that for an element $s \in R$, the set $sR$ does not contain $s$.

My first instinct was to turn to non-commutative rings, such as the quaternions or 2x2 matrices, but I could not find an element where this was the case. Would appreciate any guidance into the right direction - thank you!

Answer 1

You need a ring without a unit element, otherwise $s=s1\in sR$.

Consider $R=2\mathbb{Z}$ and $s=2$.

Answer 2

Consider the following very synthetic example:

Let $(A,+)$ be a non-trivial abelian group (with identity element $0$) and let $\cdot: A \times A \rightarrow A$ via $x\cdot y = 0$. This makes $(A, +, \cdot)$ into a commutative ring (without identity!) and for any element $s \in A$ s.t. $s \neq 0$ we have $sA = \{0\}$, which does not contain $s$.