[SOLVED] the set of zero-divisors of a ring

• Mar 19th 2010, 02:26 AM
xixi
[SOLVED] the set of zero-divisors of a ring
let $\displaystyle R$ be a non-commutative ring, and define $\displaystyle D(R)$ to be the set of zero-divisors of the ring . Suppose that $\displaystyle z^2=0$ , for any $\displaystyle z \in D(R)$ . Prove that $\displaystyle D(R)$ is an ideal of $\displaystyle R$.
• Mar 19th 2010, 08:39 AM
hatsoff
[edit: sorry for the error]
• Mar 19th 2010, 01:53 PM
Opalg
Quote:

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

Unfortunately, hatsoff's argument only works in the commutative case. In a noncommutative ring, $\displaystyle a^2-b^2\ne (a+b)(a-b)$ in general.

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

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

That shows that D(R) is closed under addition. For the other property of an ideal, if $\displaystyle a\in D(R)$ and $\displaystyle r\in R$ then clearly $\displaystyle a(ar)=0$. But I cannot see any reason why (in a nocommutative ring) there should exist a nonzero element c such that $\displaystyle arc=0$. 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.
• Mar 21st 2010, 01:21 AM
xixi
thanks
thanks very much , you have already proved it ,as you have showed if $\displaystyle a \in D(R)$ and $\displaystyle r \in R$ then $\displaystyle a(ar)=0$ which means that $\displaystyle ar \in D(R)$ and this shows that $\displaystyle D(R)$ is a right ideal . we also have $\displaystyle ra(a)=0 \Longrightarrow ra \in D(R)$ which shows that $\displaystyle D(R)$ is a left ideal and therefore $\displaystyle D(R)$ is an ideal. (note that here $\displaystyle D(R)$ is the set of left and right zero-divisors together and not just the two-sided ones.)