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)
 * Edit conflict -- maybe you're just about to change this. --Linus44 (talk) 23:00, 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)


 * Huh? In that sense, $0_R$ divides $0_R$ in every ring, no? --Dfeuer (talk) 00:40, 20 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

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)