Discussion in 'Math Research' started by Felix Goldberg, Apr 30, 2004.

  1. Hi,

    I'm looking for examples of rings R (finite or otherwise, commutative
    or otherwise) that have the following property:

    There exists a constant K so that for every zero-divisor x in R there
    are exactly K zero-divisors y's in R such that xy=0.

    Has anybody met such rings or has any thoughts how to construct them?

    Felix Goldberg, Apr 30, 2004
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  2. Hi Felix

    How about Z/9Z
    The only zero divisors are 3 and 6 and
    for x=3 you have y in {3,6}
    for x=6 you have y in {3,6}
    so K is 2
    I don't know if this is what you are interested in

    Greetings from
    Holger Walliser, Apr 30, 2004
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  3. Felix Goldberg

    Hagen Guest

    A simple example seems to be the following: take a finite field F
    and consider the ring R=FxF with componentwise addition and
    multiplication. The zero-divisors of R are the elements of the
    form (a,0) and (0,b) for a,b non-zero. Now its easy to verify
    the required property.

    Hagen, Apr 30, 2004
  4. Felix Goldberg

    Herman Rubin Guest

    The same holds for Z/p^2Z, p prime. K = p-1.
    Herman Rubin, Apr 30, 2004
  5. Felix Goldberg

    Doug Chatham Guest

    The zero-divisor graph of a commutative ring is the graph whose
    vertices are the nonzero zero-divisors and for which distinct vertices
    x and y are adjacent if and only if xy=0. So (for commutative rings)
    the original poster's question can be translated to "What rings have
    zero-divisor graphs such that every vertex has the same finite

    If the zero-divisor graph is finite and complete (every vertex
    connected to every other vertex), then the ring will be one of the
    examples the original poster is looking for.

    The following example is Example 2.11(a) in Anderson and Livingston's
    paper "The Zero-Divisor Graph of a Commutative Ring", J. Algebra
    (217), 434-447 (1999): Let p be a prime, X={X_\alpha}_{\alpha \in B} a
    family of indeterminates, A= Z_{p} [X], and I=({X_\alpha X_\beta |
    \alpha \beta \in B}). Then R=A/I has a complete zero-divisor graph.
    The zero-divisor graph of R is finite iff the index set B is finite.

    I hope this helps.
    Doug Chatham, May 1, 2004
  6. Felix Goldberg

    Edwin Clark Guest

    The ring of 2x2 matrices over a field with q elements has this property with
    K = q^2-1. (I assume you don't allow x or y to be 0).

    Edwin Clark, May 6, 2004
  7. Actually, that's the application I had in mind... :)
    So I'm not sure it helps much but thanks anyway...

    Felix Goldberg, May 8, 2004
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