Talk:Zero divisor

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Mathematics rating:	Start Class	Low Priority	Field: Algebra
Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. Needs references. Geometry guy 20:48, 31 May 2007 (UTC)

I am translating math articles from the english Wikipedia into the german one, and I think the last example of this article is not correct, and that the given definition is only applicable to commutative rings.

In a noncommutative ring, a can be a left zero divisor (ab=0), right zero devisor (ba=0) or (both sided) zero divisor (ab=ca=0).

The given infinite matrix A is a left zero divisor, as A * (1&0&0&...\\ 0&0&0...\\...) = 0, but it is not a right zero divisor, because C*A=0 implies C=0. In contrast to that, B is not a left zero divisor, but a right zero divisor (using the same matrix as for A as the left factor).

Also it is only proven there that a left zero divisor cannot have a left inverse (a'a=1), but it may still have a right inverse (aa'=1). (As the example shows, the matrix A has such an inverse B.)

Can the product of two left zero divisors be not a left zero divisor? -- SirJective 10:38, 12 Aug 2003 (UTC)

[edit] Is zero a zero divisor?

The ring Z of integers does not have any zero divisors,

What about... zero?

By definition, zero divisors are non-zero.

Opinion differs as to whether zero is in fact a zero divisor. I would contend that it is simpler to make zero a zero divisor, otherwise one has to say that the zero divisors together with zero is a prime ideal of a ring, rather than just saying that the set of zero divisors of a ring is a prime ideal. I have made the necessary change. Xantharius 20:53, 30 December 2006 (UTC)

In his famous algebra book, vanderWaerden considers zero as a zero divisor. This is useful, because then the complement of the set of all zero divisors is multiplicatively closed and does not contain zero. By using this set, one can easily form the total ring of fractions. The claim, that the set of zero divisors is a prime ideal, is wrong. See my talk section below, in particular the counterexample. ASlateff 128.131.37.74 17:12, 4 January 2007 (UTC)

[edit] In general, zero divisors do not form a prime ideal!

The article claims that the set of zero divisors of a commutative ring is a prime ideal. This is wrong. Neither the product nor the sum of zero divisors need be zero divisors. In fact, if a,b are zero divisors different from zero, then there are r,s different from zero such that ar=0 and bs=0. Maybe the author meant that (ab)(rs)=0 and hence ab should be a zero divisor. However, it may happen that rs=0. Therefore the conclusion, that ab is a zero divisor, does not follow from this.

Anyway, it is true that the set Z of zero divisors is a union of prime ideals, and it contains every minimal prime ideal. This is due to the fact that the complement of Z is a saturated multiplicatively closed set (and contains all units). ASlateff 128.131.37.74 16:24, 4 January 2007 (UTC)

A counterexample is Z/6Z, where Z denotes the integers. 2 x 3 = 4 x 3 = 0 (mod 6), the zero divisors are 2,3, and 4. But 2+3=5, which is not a zero divisor. Therefore the set of zero divisors is not closed under addition and in particular is not an ideal, let alone a prime ideal. ASlateff 128.131.37.74 16:58, 4 January 2007 (UTC)

If we use the definition of zero divisor that allows 0 to be a zero divisor, then any multiple of a zero divisor in a commutative ring is also a zero divisor. The obvious non-commutative counter-example are zero divisors x and y such that xy = 1. Albmont (talk) 18:22, 22 January 2008 (UTC)

[edit] Cross product in R²

the first example:

Seems to overlook a small detail that there is no cross product in R², in fact the "cross product" as we know it is only defined in R³, being a particular case of the wedge product. I took the liberty of changing the cross product to a dot product, either way, in the current example the wedge product would yield one (scalar). —Preceding unsigned comment added by Carvasf (talk • contribs) 23:18, 7 June 2008 (UTC)

The cited example doesn't involve cross products or dot products. Rather, it involves componentwise multiplication, which is the standard product in the ring Z×Z. I've edited the example a bit in an attempt to clarify this. Michael Slone (talk) 01:35, 8 June 2008 (UTC)