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)