0
$\begingroup$

Let $R$ be a commutative ring without unity and $n \in R\setminus\{0\}$. Prove that $n\mid n$ implies that $n$ is a zero divisor.

$\endgroup$

closed as off-topic by Watson, user26857, Zelos Malum, Namaste, user223391 Nov 5 '16 at 7:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Watson, user26857, Zelos Malum, Namaste, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Is this a commutative ring? $\endgroup$ – Arthur Nov 3 '16 at 16:40
  • 1
    $\begingroup$ Yes, it is. I forget to mention that. $\endgroup$ – Taufi Nov 3 '16 at 16:41
  • $\begingroup$ We can prove it only for commutative ring. For the not commutative ring: Let n = kn nn = knn nn = nkn 0 = (kn - nk)*n => n is the zero divisor or kn = n*k and we can go to subring, created by n and k $\endgroup$ – kotomord Nov 3 '16 at 16:43
  • $\begingroup$ Can we extend R to a ring with unity? $\endgroup$ – Jacob Wakem Nov 3 '16 at 17:33
12
$\begingroup$

Assume that $n=nk$ for some $k\in R$. Because $R$ has no unity, there exists an element $r\in R$ such that $kr\neq r$. In other words $kr-r\neq0$.

But by standard applications of rng axioms $$ n(kr-r)=n(kr)-nr=(nk)r-nr=nr-nr=0. $$ Therefore $n$ is a zero-divisor.

$\endgroup$
  • $\begingroup$ Thanks so much, Jyrki Lahtonen. You saw the trick! $\endgroup$ – Taufi Nov 3 '16 at 17:07
  • 2
    $\begingroup$ For an example of a rng where this can happen consider the direct sum of infinitely many copies of $\Bbb{Z}$. In other words $$R=\{(n_1,n_2,\ldots)\mid n_i\in\Bbb{Z}\ \text{such that $n_i\neq0$ for only finitely many $i$}\}.$$ This is a rng with componentwise operations. Clearly $n=(1,0,0,0,\ldots)$ is divisible by itself. And also a zero divisor. $\endgroup$ – Jyrki Lahtonen Nov 3 '16 at 17:25
  • $\begingroup$ Where is commutativity used? $\endgroup$ – Serge Seredenko Nov 3 '16 at 20:47
  • $\begingroup$ @Serge I use commutativity in concluding that there exists an $r$ such that $kr\neq r$ specifically. Otherwise I would need to worry about the possibility that $k$ might be a one-sided unity but not a two-sided one. That is, I avoid the headache possibility that $kr=r$ for all $r$, but may be $rk\neq r$ for some $r$, when we still could not conclude that $k$ is a multiplicative neutral element. My exposure to rngs is kinda lacking, so I can't tell right away, whether it is possible that a rng could have a one-sided identity :-) $\endgroup$ – Jyrki Lahtonen Nov 3 '16 at 20:51
  • 1
    $\begingroup$ @JyrkiLahtonen Moreover, if $R$ is not commutative, $n\mid n$ is not well defined: one should distinguishing between left divisor and right divisor. $\endgroup$ – egreg Nov 3 '16 at 21:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.