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I want an example to show that if $a,b$ are nilpotent elements of a ring $R$ with 1 and if $c$ is any element of $R$, then $abc=0\Rightarrow acb=0$ but $cab=0$ does not imply $acb=0$.

This is unlike symmetric ring, where we know that if $a,b,c\in R$ and $abc=0$ implies that $acb=0$.

Please help me to find a ring where to search for an example or help me to show that if $abc=0 \Rightarrow acb=0$, then $cab=0 \Rightarrow acb=0$ for above mentioned $a,b,c$.

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  • $\begingroup$ Just to clarify the question: you want to know whether there exists a ring with unit such that whenever $a, b$ are nilpotent and $c$ is any element with $a b c = 0$, then also $a c b = 0$, but also such that there is some $c$ such that $c a b = 0$ and $a c b \neq 0$? Also, I think by "symmetric", you mean commutative. $\endgroup$
    – Todd Trimble
    Oct 10, 2013 at 13:28
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    $\begingroup$ This is crossposted from MSE. math.stackexchange.com/questions/508386/… $\endgroup$ Oct 10, 2013 at 13:31
  • $\begingroup$ @ Todd Trimble: No, symmetric and commutative are not same. Every commutative ring is symmetric but converse is not the truth. $\endgroup$
    – Anupam
    Oct 10, 2013 at 15:16
  • $\begingroup$ Okay, thanks for saying so. I had to look up "symmetric ring": a ring R is called symmetric if $abc=0 \Leftrightarrow acb=0$ for all $a,b,c \in R$. According to this paper by Greg Marks sciencedirect.com/science/article/pii/S0022404902000701 the notion was introduced by Lambek. $\endgroup$
    – Todd Trimble
    Oct 10, 2013 at 16:25
  • $\begingroup$ I want to clarify the second condition: $\forall a,b \ \exists c \ cab=0\ {\rm does\ not\ imply\ }\ acb=0$ or $\exists a,b, c\ cab=0\ {\rm does\ not\ imply\ }\ acb=0$? $\endgroup$ Oct 10, 2013 at 18:25

1 Answer 1

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If I understand correctly, there are two conditions on the ring $R$:

(1) if $a$ and $b$ are nilpotent elements of $R$, and if $c$ is arbitrary, then $abc=0$ implies $acb=0$;

(2) if $a$ and $b$ are nilpotent elements of $R$, and if $c$ is arbitrary, then $cab=0$ implies $acb=0$.

There exists a ring for which condition (1) is satisfied, while condition (2) fails. Indeed, let $k$ be a field, and let $R$ be the path algebra of the quiver $$ 1 \stackrel{x}{\leftarrow} 2 \circlearrowleft^y $$ over $k$, modulo the relation $y^2=0$. Let $e_1$ and $e_2$ be the paths of length $0$ at vertices $1$ and $2$, respectively. I compose arrows from left to right: $xy$ is a non-zero path, while $yx$ is not.

Then any nilpotent element is a linear combination of $x$, $y$ and $xy$. Let $(ax + by+ cxy)$ and $(dx+ey+fxy)$ be two such elements, and let $(ge_1 + he_2 + ix + jy + lxy)$ be an arbitrary element. We have $$(ax + by+ cxy)(dx+ey+fxy)(ge_1 + he_2 + ix + jy + lxy) = (aeh)xy \\ (ax + by+ cxy)(ge_1 + he_2 + ix + jy + lxy)(dx+ey+fxy) = (aeh)xy.$$ Thus condition (1) is satisfied. However, we have that $e_2xy = 0$, while $xe_2y = xy \neq 0$. Thus condition (2) is not satisfied.

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  • $\begingroup$ Well, this looks right. OP has offered a 300 point bounty over at the MSE version of the same question, if you want to claim your prize over there. $\endgroup$
    – Todd Trimble
    Oct 14, 2013 at 23:55
  • $\begingroup$ @ Pierre-Guy Plamondon: Thanks for the answer. Had you provided your affiliation, it would have been esier for me to acknowledge you in all of my future works relating to this. $\endgroup$
    – Anupam
    Oct 15, 2013 at 6:16
  • $\begingroup$ @ToddTrimble Thanks for your suggestion. I wasn't sure if it was considered OK to answer the same question on both sides. I'll copy my answer at MSE, providing a link to this one. $\endgroup$ Oct 15, 2013 at 7:01
  • $\begingroup$ But I see that you've already mentioned this answer in an edit of the MSE question. Maybe reposting the answer is not necessary then. $\endgroup$ Oct 15, 2013 at 8:27
  • $\begingroup$ @Pierre-GuyPlamondon I was the one who made that edit. I felt this was the right thing to do. $\endgroup$
    – Todd Trimble
    Oct 15, 2013 at 11:09

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