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- Thread starter xixi
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Landau

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

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

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Landau

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

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- #5

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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

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ora right zero divisor, the second that a zero divisor is something that is a left zero divisoranda 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

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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].

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

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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 [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.

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

Landau

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How so?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].

let [tex] R[/tex] be a non-commutative ring.

- #10

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How so?

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

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