Question on the set of zero-divisors of a ring

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In summary: 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.
  • #1
xixi
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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].
 
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  • #2
I thought zero-divisors are by definition non-zero, so that 0 cannot be in D(R)?
 
  • #3
of course 0 is a zero-divisor and belongs to D(R).
 
  • #4
Maybe you have a http://planetmath.org/encyclopedia/ZeroDivisor.html .
 
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  • #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?
 
  • #6
Martin Rattigan said:
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 .
 
  • #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].
 
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  • #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)

Martin Rattigan said:
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].

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.
 
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  • #9
rasmhop said:
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].
How so?
xixi said:
let [tex] R[/tex] be a non-commutative ring.
 
  • #10
Landau said:
How so?

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

1. What is a zero-divisor in a ring?

A zero-divisor in a ring is an element that, when multiplied by another element, results in the product being equal to zero. In other words, it is an element that has a non-trivial solution to the equation a*b = 0, where a and b are elements of the ring.

2. Can every ring have zero-divisors?

No, not every ring has zero-divisors. In fact, a ring is called an integral domain if it does not have any zero-divisors. This means that in an integral domain, the only way to get the product of two elements to be zero is if one of the elements is already equal to zero.

3. What is the set of zero-divisors in a ring?

The set of zero-divisors in a ring is the collection of all elements that are considered to be zero-divisors. This set can vary in size and can even be empty, depending on the specific ring in question.

4. How do zero-divisors affect the structure of a ring?

The presence of zero-divisors in a ring can significantly affect its structure. For example, if a ring has a large number of zero-divisors, it may not have a well-defined multiplication operation and may not satisfy certain algebraic properties. On the other hand, if a ring has no zero-divisors, it is considered to be a more "well-behaved" ring with a simpler structure.

5. What is the importance of studying the set of zero-divisors in a ring?

The study of zero-divisors in a ring is essential because it can provide insight into the structure and properties of the ring itself. It can also help in the classification and comparison of different types of rings. Furthermore, understanding zero-divisors is crucial in certain applications, such as coding theory and algebraic geometry.

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