Reduced ring are SI? A ring $R$ is called an SI-ring if for any $a\in R$ the right annihilator of $a$ is an ideal of $R$. It is equivalent to the following statement: "if $ab=0$ for $a,b\in R$ then $aRb=0$". 
Is it true that any reduced ring (a ring without non-zero nilpotent elements) is an SI-ring?
I tried to prove that any $axb$ is nilpotent for $x\in R$, but to no avail. Any suggestion would be thanked!
 A: A ring $R$ is called symmetric if $abc=0$ implies $acb=0$  for $a,b,c\in R$. Any symmetric ring is obviously an SI-ring. We prove that "reduced gives symmetric".
Let $abc=0$. We have to show that $acb=0$. Firstly, $c(abc)ab=0$ hence $cab=0$ (since it is nilpotent). Now, $aba(cab)ac=0$, which gives $abac=0$. Then we write $bacb(abac)ba=0$, yielding $bacba=0$ and $ac(bacba)cb=0$, whence $acb$ is  nilpotent and is equal to zero.
A: Yes, and the place you probably want to look is any of Greg Marks' articles involving $2$-primal rings. The one I have in mind is 

G. Marks, A taxonomy of $2$-primal rings
  Journal of Algebra
  Volume 266, Issue 2, 15 August 2003, Pages 494–520

which contains a chart in the first paragraph in the introduction that says reduced rings are $SI$ (and says a whole lot more.)
I don't recall if a proof appears in that particular paper, but you can just follow the citation and check 

G. Marks, Direct product and power series formations over 2-primal rings, in: S.K. Jain, S.T. Rizvi (Eds.),
  Advances in Ring Theory, in: Trends in Math., Birkhäuser, Boston, 1997, pp. 239–245, MR 98i:16020.

A: yes, since $R$ is reduced $ab=0$ implies that $ba=0$ so for any $x \in R$, $(axb)^2=0$ so $axb=0$
