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.


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    $\begingroup$ Is this a commutative ring? $\endgroup$ – Arthur Nov 3 '16 at 16:40
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    $\begingroup$ 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

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.

  • $\begingroup$ Thanks so much, Jyrki Lahtonen. You saw the trick! $\endgroup$ – Taufi Nov 3 '16 at 17:07
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    $\begingroup$ 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
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    $\begingroup$ @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

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