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Question on the set of zero-divisors of a ring

  1. Mar 19, 2010 #1
    let [tex] R[/tex] be a non-commutative ring and [tex] D(R)[/tex] denotes the set of zero-divisors of the ring . Suppose that [tex]z^{2} =0 [/tex] for any [tex]z \in D(R)[/tex] . prove that [tex]D(R)[/tex] is an ideal of [tex]R[/tex].
  2. jcsd
  3. Mar 19, 2010 #2


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    I thought zero-divisors are by definition non-zero, so that 0 cannot be in D(R)?
  4. Mar 21, 2010 #3
    of course 0 is a zero-divisor and belongs to D(R).
  5. Mar 21, 2010 #4


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    Maybe you have a http://planetmath.org/encyclopedia/ZeroDivisor.html [Broken].
    Last edited by a moderator: May 4, 2017
  6. Mar 27, 2010 #5
    There are always problems with definitions when it comes to rings. There seem to be two definitions of zero divisor in general use. The first is that a zero divisor is something that is a left zero divisor or a right zero divisor, the second that a zero divisor is something that is a left zero divisor and a right zero divisor.

    Both normally exclude zero, but this seems to be only so its easy to talk about rings "without zero divisors", so it's no hardship to allow zero a status of honorary zero divisor for the purpose of the problem - but apart from zero, which of the above definitions are we supposed to assume?

    Also there were always disgreements about the exact definition of a ring. When I was alive, it was generally an abelian group under + and a groupoid under . with . distributing over + from both sides - and nothing more. Then some (most?) algebra books would state very early that they would only consider associative rings and sometimes only associative rings with a 1. Now Wiki has two slightly different definitions in different articles. One states that a ring should be a monoid under . but carries on to say that it may actually not be.

    In view of the general confusion - what definition of "ring" should we assume for the purposes of this problem?
  7. Mar 30, 2010 #6
    To qualify as a ring the set, together with its two operations, must satisfy certain conditions—namely, the set must be 1. an abelian group under addition; and 2. a monoid (a group without the invertibility property is a monoid) under multiplication; 3. such that multiplication distributes over addition.

    And in here the set of zero-divisors is the set of all zero-divisors of the ring i.e. the left zero-divisors and the right ones and not just the two-sided ones .
  8. Apr 2, 2010 #7
    Denoting [itex]D(R)[/itex] by [itex]D[/itex], [itex](\forall z\in D)z^2=0[/itex] so every left or right zero divisor is also a two sided zero divisor in [itex]R[/itex].

    If [itex]z\in D,r\in R[/itex], since [itex]R[/itex] is associative, [itex]{(rz)z=rz^2=0}[/itex], so [itex]rz[/itex] is a zero divisor. Similarly [itex]zr[/itex].

    If [itex]z,z'\in D[/itex] then because [itex]R[/itex] is associative and [itex]z'z\in D[/itex], [itex]{(z+z')zz'z=z^2z'z+(z'z)^2=0}[/itex]. Hence [itex](z+z')\in D[/itex] unless [itex]zz'z=0[/itex].

    But if [itex]zz'z=0[/itex], [itex](z+z')z'z=0[/itex]. Hence [itex](z+z')\in D[/itex] unless [itex]z'z=0[/itex].

    But if [itex]z'z=0[/itex], [itex](z+z')z=0[/itex]. Hence [itex](z+z')\in D[/itex] unless [itex]z=0[/itex].

    But if [itex]z=0[/itex], [itex](z+z')\in D[/itex].

    So in all cases, [itex](z+z')\in D[/itex].

    Also if [itex]z\in D[/itex] then If [itex](-z)r=-zr=0[/itex], so [itex]-z\in D[/itex].

    It follows that [itex]D[/itex] is an ideal of [itex]R[/itex].
    Last edited: Apr 2, 2010
  9. Apr 2, 2010 #8
    EDIT: This answer is incorrect as pointed out by Landau. Please ignore it. (EDIT2: By "this answer" I'm referring to my own, and not Martin Rattigan's which as far as I can see is correct)

    These cases can be done a little simpler by noting that if z and z' are distinct elements in D, then [itex]z-z'\not=0[/itex] and:
    [tex](z+z')(z-z') = z^2 - z'^2 = 0-0=0[/tex]
    so z+z' is in D. If z=z' is in D, then z+z'=2z is in D by the first condition you proved.
    Last edited: Apr 2, 2010
  10. Apr 2, 2010 #9


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    How so?
  11. Apr 2, 2010 #10
    I didn't think properly. I for some reason assumed the ring was commutative so please ignore my previous post.
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