# Question on the set of zero-divisors of a ring

1. Mar 19, 2010

### xixi

let $$R$$ be a non-commutative ring and $$D(R)$$ denotes 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 $$R$$.

2. Mar 19, 2010

### Landau

I thought zero-divisors are by definition non-zero, so that 0 cannot be in D(R)?

3. Mar 21, 2010

### xixi

of course 0 is a zero-divisor and belongs to D(R).

4. Mar 21, 2010

### Landau

Maybe you have a http://planetmath.org/encyclopedia/ZeroDivisor.html [Broken].

Last edited by a moderator: May 4, 2017
5. Mar 27, 2010

### Martin Rattigan

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?

6. Mar 30, 2010

### xixi

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 .

7. Apr 2, 2010

### Martin Rattigan

Denoting $D(R)$ by $D$, $(\forall z\in D)z^2=0$ so every left or right zero divisor is also a two sided zero divisor in $R$.

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

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

But if $zz'z=0$, $(z+z')z'z=0$. Hence $(z+z')\in D$ unless $z'z=0$.

But if $z'z=0$, $(z+z')z=0$. Hence $(z+z')\in D$ unless $z=0$.

But if $z=0$, $(z+z')\in D$.

So in all cases, $(z+z')\in D$.

Also if $z\in D$ then If $(-z)r=-zr=0$, so $-z\in D$.

It follows that $D$ is an ideal of $R$.

Last edited: Apr 2, 2010
8. Apr 2, 2010

### rasmhop

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 $z-z'\not=0$ and:
$$(z+z')(z-z') = z^2 - z'^2 = 0-0=0$$
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
9. Apr 2, 2010

### Landau

How so?

10. Apr 2, 2010

### rasmhop

I didn't think properly. I for some reason assumed the ring was commutative so please ignore my previous post.