# Thread: [SOLVED] the set of zero-divisors of a ring

1. ## [SOLVED] the set of zero-divisors of a ring

let $R$ be a non-commutative ring, and define $D(R)$ to be the set of zero-divisors of the ring . Suppose that $z^2=0$ , for any $z \in D(R)$ . Prove that $D(R)$ is an ideal of $R$.

2. [edit: sorry for the error]

3. Originally Posted by xixi
let $R$ be a non-commutative ring, and define $D(R)$ to be the set of zero-divisors of the ring. Suppose that $z^2=0$ , for any $z \in D(R)$. Prove that $D(R)$ is an ideal of $R$.
Unfortunately, hatsoff's argument only works in the commutative case. In a noncommutative ring, $a^2-b^2\ne (a+b)(a-b)$ in general.

However, it is true that if $a,b\in D(R)$ then $(a+ab)(a-ba) = 0$. If you multiply $a+ab$ on the left by $a$ or on the right by $a-ba$, you get 0. Provided that both $a$ and $a-ba$ are nonzero, that says that $a+ab\in D(R)$ and hence $(a+ab)^2=0$. Therefore $aba+abab=0$ (and that result trivially also holds if $a$ or $a-ba$ is 0).

A similar argument, with $a+ab$ and $a-ba$ interchanged, shows that $aba-baba=0$. Together with the previous result, that gives $abab+baba=0$. But then $(a+b)^4=0$, so $a+b\in D(A)$.

That shows that D(R) is closed under addition. For the other property of an ideal, if $a\in D(R)$ and $r\in R$ then clearly $a(ar)=0$. But I cannot see any reason why (in a nocommutative ring) there should exist a nonzero element c such that $arc=0$. All I can conclude is that in the ring R, the set of left zero-divisors is a right ideal, and the set of right zero-divisors is a left ideal.

4. ## thanks

thanks very much , you have already proved it ,as you have showed if $a \in D(R)$ and $r \in R$ then $a(ar)=0$ which means that $ar \in D(R)$ and this shows that $D(R)$ is a right ideal . we also have $ra(a)=0 \Longrightarrow ra \in D(R)$ which shows that $D(R)$ is a left ideal and therefore $D(R)$ is an ideal. (note that here $D(R)$ is the set of left and right zero-divisors together and not just the two-sided ones.)