Question on the set of zero-divisors of a ring

[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.

 

[Landau]

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

 

[xixi]

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

 

[Landau]

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

 

[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?

 

[xixi]

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?

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 .

 

[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.

 

[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)

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.

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.

 

[Landau]

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.

How so?

let R be a non-commutative ring.

 

[rasmhop]

How so?

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