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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.

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    $\begingroup$ What is your motivation for asking exactly this? $\endgroup$ – quid Jan 27 '15 at 12:07
  • $\begingroup$ @quid If there exist atleast one left zero divisor there exist atleast one right zero divisor. How much we can go further $\endgroup$ – Sushil Jan 29 '15 at 8:26
  • $\begingroup$ I included your reply, slightly expanded, in the post and voted to reopen. Thank you for following up on this. $\endgroup$ – quid Jan 29 '15 at 9:47
  • $\begingroup$ @quid thanks for reopening the question. I think i am not able to ask questions properly n words. Next time I better remember this $\endgroup$ – Sushil Jan 29 '15 at 12:17
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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 $x+r=\left(\begin{array}{} 1 & 0\\ 1 & 0 \end{array}\right)$.

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    $\begingroup$ This part doesn't make sense to me: "$(x+y)r=xr+yr=r+r=0$. Therefore $x+y=0$ or $x+y=r$. Clearly, knowing that $(x+y)$ is a left zero-divisor doesn't prove that it's a right zero divisor (indeed, all the ring elements are left zero divisors, and only $0$ and $r$ are right zero divisors!). So I don't see how you can conclude that either $x+y = 0$ or $x+y = r$. (Also, I think you mean $(x+y)r = xr + yr = 0 + 0 = 0$, but that's just a minor comment.) $\endgroup$ – mathmandan Feb 14 '15 at 19:49
  • $\begingroup$ @mathmandan It is a misprint. Here should be "$r(x+y)=rx+ry=r+r=0$" instead. Corrected. $\endgroup$ – Alex Ravsky Feb 14 '15 at 20:10
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    $\begingroup$ Great, thanks! Note that this proof uses the hypothesis that $R$ has at least three elements; it remains to consider rings with two elements (since the ring of one element has no nonzero zero divisors). The only possibility would be $\mathbb{Z}_2$ under the usual addition, with multiplication $x*y =0$ for all $x, y$. This could be considered another (degenerate) example--there is only one nonzero element, which is both a right and left zero-divisor. Anyway, nice work! $\endgroup$ – mathmandan Feb 14 '15 at 20:26
  • $\begingroup$ @mathmandan Thanks. I assumed that $R$ has at least three elements in order to exclude trivial rings in which there is only one non-zero (right or left) divisor of zero. $\endgroup$ – Alex Ravsky Feb 14 '15 at 20:30
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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.

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  • $\begingroup$ Sorry I didn't get the hint $\endgroup$ – Sushil Jan 29 '15 at 8:25
  • $\begingroup$ group of order 4 with generators a,b what it mean? There are only teo groups of order 4 right one cyclic and other one Klein 4 $\endgroup$ – Sushil Jan 30 '15 at 13:13
  • $\begingroup$ Right, and the non-cyclic group has two generators. I'm calling them $a$ and $b$ so it is explicit how the ring multiplication is defined. $\endgroup$ – Jason Jan 31 '15 at 2:14
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    $\begingroup$ What would be a.b would it be o or b?? $\endgroup$ – Sushil Feb 2 '15 at 17:51

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