Product of two left zero divisors
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)
- The multiplication operator of a ring is by definition associative. Hence, the product of any x with a left zero divisor a (in that order) y = xa will always be a left zero divisor (here b is a right zero divisor):
- — Quondum☏ 16:21, 10 August 2012 (UTC)
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 184.108.40.206 17:12, 4 January 2007 (UTC)
The correct definition of zero divisor is the one given on page 98 of Bourbaki, Algebra 1-3. Namely, a is a left zero divisor if there exists a nonzero b such that ab=0. This applies even when a=0! It follows that 0 is a zero divisor in every ring except for the zero ring (the ring in which 0=1). It has to be this way if you want the set of non-zero-divisors in a commutative ring to form a submonoid of the multiplicative monoid of the ring, and this in turn is essential if you want to be able to define the total quotient ring of any commutative ring by localizing at the set of non-zero-divisors. If there are no objections, I will fix the article. Ebony Jackson (talk) 06:54, 15 November 2013 (UTC)
- Having such a notable reference means that this definition must be included in the article. If the other definition (that excludes zero as a zero divisor) is sufficiently notable, you'll have objections to excluding it. So, as in Divisor, a clear statement of two competing definitions may be appropriate. You may also be interested in Divisibility (ring theory), which not only gives only one definition, but is internally inconsistent as noted in Talk:Divisibility (ring theory)#Fixing internal inconsistency in article: which definitions are notable?. —Quondum 09:25, 15 November 2013 (UTC)
- Bourbaki discusses this too, explaining that the usual definition of "a is a divisor of b" should not be specialized to the case b=0 to give a definition of "zero divisor" since it would make every element a zero divisor (I know you understand this already). Another advantage of Bourbaki's definition of zero divisor is that it makes the set of zero divisors equal to the union of the associated primes in any commutative ring. It is true that there are other textbooks that use different conventions for 0 being a zero divisor, but I have not seen any such books that argue convincingly why their convention is better, and I suspect that the authors in most cases have simply not thought things through. If someone knows an advantage of a different convention (in terms of elegant theorems that become true with it but false with Bourbaki's definition, say), then it would be good to know what they are. Ebony Jackson (talk) 14:12, 15 November 2013 (UTC)
- I don't fully follow you, when you say "Bourbaki discusses this too" (my emphasis), as Bourbaki is merely saying that "zero divisor" and "divisor of zero" are not synonymous, and does not relate the constraint a≠0, which is what has been mentioned and is hence what I'd think you meant by "this".
- I too am of the opinion that the exclusion of zero as a candidate as a divisor and as a zero divisor is simply not thought through, and may originate from a desire to avoid dealing with slightly confusing trivial cases. Or worse, it may come from the argument that you cannot divide by zero, hence zero cannot be a divisor of any type (which argument, when applied to the definition of zero divisors, would result in there being no zero divisors). However, as one will see from Talk:Divisor, this kind of question tends to be hotly debated. I suspect that of a term such as "factor" had been used instead of "divisor", fewer people (including textbook authors) would become confused about what is being defined. However, here we should be steered by notable sources, preferably Bourbaki and equivalent rather than the average textbook. —Quondum 16:03, 15 November 2013 (UTC)
- Sorry, my fault, when I wrote "Bourbaki discusses this too", I was confused about what we were talking about; in that sentence, I meant to be referring to the standard abuse of terminology in which one uses "zero divisor" to mean something different from what one gets by taking b=0 in the definition of "divisor of b". In any case, thank you for the advice to look at Divisibility (ring theory) and the Talk:Divisor page. Ebony Jackson (talk) 05:44, 16 November 2013 (UTC)
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 220.127.116.11 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 18.104.22.168 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)
- But if xy = 1, then x and y are units, and hence not zero divisors. (The proof in the article that zero divisors cannot be units did not rely on commutativity, so it should hold for non-commutative rings as well.) Your counter-example cannot exist. —Preceding unsigned comment added by 22.214.171.124 (talk) 14:01, 20 March 2011 (UTC)
- The author says "Set of zero divisors is a union of prime ideals." I guess there should be "minimal prime ideals" instead of "prime ideals". Minimal prime (commutative algebra) --126.96.36.199 (talk) 14:22, 7 March 2010 (UTC)
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)
Bug on the definition of "zero divisor" from the definition of lateral zero divisors
In its present version, the article reads: "An element that is both a left and a right zero divisor is simply called a zero divisor". This is indeed coherent with _some_ sources (since we are Saturday, I have no library at hand but Google Books is my friends), e.g.  (A course in algebra by Yun Fan, Q. Y. Xiong, Y. L. Zheng) but certainly not all e.g. Concepts in abstract algebra by Charles Lanski  which defines a zero divisor as an element which is either a right zero divisor or a left zero divisor, and uses two-sided zero divisor for elements which are both.
The present version seems odd to me, and it seems to have propagated to most if not all Wikipedias in major languages. I have enough to do to correct the numerous mistakes in Wikipedia in French, so I just post the problem here on the discussion page, ready to have a look at the replies - but I don't believe things are rightly done in the current version. French Tourist (talk) 13:24, 5 February 2011 (UTC)