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

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    [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.
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    [edit: sorry for the error]
    Last edited by hatsoff; March 19th 2010 at 02:16 PM.
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    Quote Originally Posted by xixi View Post
    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.
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    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.)
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