Talk:Nilpotent Element is Zero Divisor
Good call. $R \ne 0$ is necessary, since as defined the null ring has no zerodivisors. --Linus44 (talk) 22:52, 19 March 2013 (UTC)
- "Zero divisors" or "proper zero divisors"? $0_R$ is itself a Definition:Zero Divisor as we have it defined here. --prime mover (talk) 22:57, 19 March 2013 (UTC)
- Hang on bear with me, we seem to have a confusion over definitions. Needs sorting out. --prime mover (talk) 22:58, 19 March 2013 (UTC)
- In all other rings, zero is a zerodivisor. However for the null ring:
- A zero divisor (in $R$) is an element $x \in R$ such that:
- $\exists y \in R^*: x \circ y = 0_R$
- where $R^*$ is defined as $R \setminus \left\{{0_R}\right\}$.
- In the null ring, $\exists y \in R^*: x \circ y = 0_R$ is false, since $R^* = \emptyset$. --Linus44 (talk) 22:59, 19 March 2013 (UTC)
- It's all right, I've got it now, I've cleared my confusion.
- My point still stands. This theorem does hold for the null ring - there are no nilpotent elements in a null ring so (vacuously) all nilpotent elements are zero divisors. No problem, null ring need not be excluded. --prime mover (talk) 23:07, 19 March 2013 (UTC)
But we have:
- An element $x \in R$ is nilpotent if $x^n = 0_R$ for some (strictly) positive integer $n$.
and in the null ring $0_R^n = 0_R$ for all $n \geq 1$, e.g. because the null ring is closed under multiplication. --Linus44 (talk) 23:10, 19 March 2013 (UTC)
- Oh yeah okay. --prime mover (talk) 23:12, 19 March 2013 (UTC)
- Worth making that very observation on the proof page itself, to explain why the result specifically does not hold for a null ring, because (as you see) it does cause confusion. --prime mover (talk) 23:14, 19 March 2013 (UTC)
I think it's worth considering allowing $0_R \in \{0_R\}$ to be a zero-divisor. It's a trivial point, but a zerodivisor is named thus because it is "something that divides zero". We have the definition:
- We define the term $x$ divides $y$ in $D$ as follows:
- $x \mathop {\backslash_D} y \iff \exists t \in D: y = t \circ x$
Since we have $0_R = 0_R \circ 0_R$ in the null ring, $0_R$ divides $0_R$ in this ring. --Linus44 (talk) 23:19, 19 March 2013 (UTC)
- Oh no not this again. We've been through this before. "We as amateur trainee wet-behind-the-ears undergrads don't like the general definition of something so we're going to change it to match what we do like." No we are not going to redefine the concept of "zero divisor" just so it makes one of our results a little bit more pretty. --prime mover (talk) 06:22, 20 March 2013 (UTC)
It's nothing to do with making this result simpler. It's an observation that something called a zero-divisor ought to be defined to be a divisor of zero. --Linus44 (talk) 10:57, 20 March 2013 (UTC)
Actually, I disagree with myself. A more "natural" definition is that $x \in R$ is a zerodivisor if the map
| \(\ds x\) | \(:\) | \(\ds R \to R\) | ||||||||||||
| \(\ds \) | \(:\) | \(\ds y \to xy\) |
is injective, and this is true in the zero ring. I propose also adding a note on the page Def:Zero Divisor to clarify that in the zero ring, zero is not a zero-divisor --Linus44 (talk) 11:05, 20 March 2013 (UTC)
While I should probably leave it there; here's something non-trivial that should be considered in the definition of zero-divisors: at present, we have a zero-divisor if
- $\exists y \in R^*: x \circ y = 0_R$
but we could take the definition
- $\exists y \in R^*: y \circ x = 0_R$
In a non-commutative ring these two definitions are different. See wikipedia for an example. --Linus44 (talk) 11:15, 20 March 2013 (UTC)
- Hadn't thought of that. We need therefore to add "left zero divisor" and "right zero divisor" in order to be consistent. I will go back to the sources and see what they say on the matter. It is possible that in those sources the definition is made only for commutative rings. That exercise won't be immediate. --prime mover (talk) 11:43, 20 March 2013 (UTC)
- Amended the definition of Definition:Zero Divisor accordingly. --prime mover (talk) 20:12, 20 March 2013 (UTC)
- If it offends your sensibilities, don't contribute to it. --prime mover (talk) 19:01, 22 March 2013 (UTC)
Notation
I personally like the $x^n$ notation used here, but I should mention I have seen (and even written, following their example) pages using a less readable but perhaps more formal notation ${\circ}^n (x)$. I don't personally know if the latter is even standard. I'm guessing the mover of primes may care one way or another. --Dfeuer (talk) 15:21, 22 March 2013 (UTC)
- That notation comes from 1965: Seth Warner: Modern Algebra, a big, complicated and extremely thorough work covering most aspects of AA. The notation he uses for general unspecified operations is $\triangle$ and the same upside down. $\triangle^n \left({x}\right)$ is a logical notation, as it encapsulates both the operation and its multiplicity, and therefore can be useful when establishing the basic truths. Generally one would not use it in the context where there is but one operation under discussion, but when establishing e.g. proofs of distributivity between two operations where either or both is non-commutative it has a lot in its favour.
- I understand it's fashionable for children to diss their elders and betters, but there are occasions where the latter actually know what they are talking about, and a particular course of action may be embarked upon for good reasons which may not always be immediately apparent.
- Why do you bring the subject up here? Your language suggests that you're being deliberately provocative. --prime mover (talk) 19:00, 22 March 2013 (UTC)
- In this context I think it's more common to put the operation in the power, i.e. $x^{\circ n}$. This is used a lot in the context of direct sums, tensor products etc.: $A^{\oplus n}$, $A^{\otimes n}$.
- Composition of functions is also sometimes written $\circ_{i =1}^n f_i$, though I've only seen this used informally (the order has to be understood). I'm not bothered personally. --Linus44 (talk) 15:41, 22 March 2013 (UTC)