Ring with many one-sided zero-divisors Does there exist a ring all of whose elements are left zero-divisors but only one element is a right zero-divisor?
The motivation for asking this question is that if there exists atleast one left zero-divisor there exist atleast one-right zero divisor. Now, it seems a natural question to aks if assuming "many" left-zero divisors will result in "more" right-zero divisors. A specific and extremal version if this idea is the question formulated above.
 A: It seems the following. 
Such a non-trivial ring is unique. Indeed, assume that $R$ is a ring all of whose elements are left zero-divisors but only one element $r\ne 0$ is a right zero-divisor and $R$ contains at least three distinct elements. Let $x\in R$ be an arbitrary non-zero element. Since $x$ is a left zero-divisor, there exists a non-zero element $t\in R$ such that $xt=0$. So, $t$ is a right zero-divisor, so $t=r$. Therefore $xr=0$ for each element $x\in R$. In particular, $rr=0$. Since $x(r+r)=xr+xr=0$ for each element $x\in R$, $r+r$ is a right zero-divisor. Since $r\ne 0$, $r+r\ne r$, so $r+r=0$. Again, let $x\in R$ be an arbitrary non-zero element. Then $rx$ is a right zero-divisor, so $rx=0$ or $rx=r$. But if $rx=0$ then $x$ is a right zero-divisor, therefore $x=r$. So $rx=r$ for each element $x\in R$ distinct from $0$ or $r$. Let $x,y$ be elements of the ring $R$ distinct from $0$ or $r$. Then $r(x+y)=rx+ry=r+r=0$. Therefore $x+y=0$ or $x+y=r$. In particular, $x+x=0$ or $x+x=r$. If $x+x=r$ then $0=xr=x(x+x)=$ $xx+xx=(x+x)x=rx=r$, a contradiction. Therefore $x+x=0$. Hence  $y=-x=x$ or $y=r-x=r+x$. Thus the ring $R$ contains exactly four elements: $0$, $r$, $x$ and $x+r$. Since $x$ is not a right zero-divisor, $xx\ne 0$. If $xx=r$ then $0=xr=xxx=rx=r$, a contradiction. If $xx=x+r$ then $xx+r=xx+rx=$ $(x+r)x=xxx=x(x+r)=$ $xx+xr=xx,$ a contradiction. So $xx=x$. Then $(x+r)x=xx+rx=xx+r=x+r$, $x(x+r)=xx+xr=xx=x$,   and $(x+r)(x+r)=xx+xr+rx+rr=xx+r=x+r$. 
The table of addition $a+b$ for the ring $R$:
$$\begin{array}{c}
a\setminus b & 0 & r & x & x+r\\
0 & 0 & r & x & x+r\\
r & r & 0 & x+r & x\\
x & x & x+r & 0 & r\\
x+r & x+r & x & r & 0
\end{array}$$
The table of multiplication $ab$ for the ring $R$:
\begin{array}{c}
a\setminus b & 0 & r & x & x+r\\
0 & 0 & 0 & 0 & 0\\
r & 0 & 0 & 0 & 0\\
x & 0 & 0 & x & x\\
x+r & 0 & 0 & x+r & x+r
\end{array}
Thus the ring $R$ can be realized as the ring of $2\times 2$ matrices of the form 
$\left(
\begin{array}{}
a & 0\\
b & 0
\end{array}
\right)$ over the ring $\Bbb Z_2$, where $0=\left(\begin{array}{}
0 & 0\\
0 & 0
\end{array}\right)$, 
$r=\left(\begin{array}{}
0 & 0\\
1 & 0 \end{array}\right)$,
$x=\left(\begin{array}{}
1 & 0\\
0 & 0 \end{array}\right)$, and 
$x+r=\left(\begin{array}{}
1 & 0\\
1 & 0 \end{array}\right)$.
A: Take $R$ to be the non-cyclic (abelian) group of order $4$ with generators $a,b$. Introduce multiplication on $R$ where $a$ is a right identity and $x\cdot b=0$ for all $x$. I believe this provides an example.
