# Zero-divisors

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

1. ### Felix GoldbergGuest

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?

THanks,
Felix.

Felix Goldberg, Apr 30, 2004

2. ### Holger WalliserGuest

Hi Felix

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

Holger Walliser, Apr 30, 2004

3. ### HagenGuest

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.

H

Hagen, Apr 30, 2004
4. ### Herman RubinGuest

The same holds for Z/p^2Z, p prime. K = p-1.

Herman Rubin, Apr 30, 2004
5. ### Doug ChathamGuest

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
degree?"

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. ### Edwin ClarkGuest

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

Edwin Clark, May 6, 2004
7. ### Felix GoldbergGuest

Actually, that's the application I had in mind...
So I'm not sure it helps much but thanks anyway...

Regards,
Felix.

Felix Goldberg, May 8, 2004