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Question from chapter on 'Integral Domains' from Gallian .which said an integral domain does not contains zero-divisor and is a commutative ring.

Does there exist any other commutative ring which is not a integral domain and also does not contain zero-divisor...

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  • $\begingroup$ Does the definition of a ring in Gallian require a multiplicative identity? It seems this makes a difference. $\endgroup$ – mdp Oct 2 '14 at 9:10
  • $\begingroup$ @MattPressland in the book ring does not requires to have a multiplicative identity.. $\endgroup$ – spectraa Oct 2 '14 at 9:25
  • $\begingroup$ OK - for completeness and future reference, because lots of people use lots of different definitions, it might be good to say this in the question, and to say whether or not an integral domain is required to have an identity. $\endgroup$ – mdp Oct 2 '14 at 10:20
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    $\begingroup$ This question is very enlightening. I learned that definitions have not been standardized apparently. Perhaps NIST should get involved. It can be quite confusing. Wikipedia says one thing, my book says another, etc, etc. $\endgroup$ – user172428 Oct 2 '14 at 11:08
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Disclaimer: In this answer, the definition of a ring requires a multiplicative identity. If you use another definition of ring that does not make this requirement, then you might get a different answer - indeed, other people have provided such different answers.

We know that an integral domain doesn't contain (non-zero) zero-divisors because that (as well as the ring being non-zero itself) is the definition of an integral domain! So apart from the vacuous example of the zero ring, which has no non-zero zero-divisors but is not an integral domain by definition, there are no examples.

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  • $\begingroup$ It depends on your definition of integral domain, for me integral domains may be non-unital, but for others not (see answer below). $\endgroup$ – AIM_BLB Oct 1 '14 at 22:07
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    $\begingroup$ @CSA I should have guessed something like this would happen! Thanks for catching, I'll add a disclaimer. $\endgroup$ – mdp Oct 2 '14 at 9:07
  • $\begingroup$ No problem, glad i could help :) $\endgroup$ – AIM_BLB Oct 3 '14 at 22:29
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Yes. From my understanding, the ring $nZ$ for $n > 0$, with the usual multiplication and division, counts as a commutative ring without zero divisors but is not an integral domain because it doesn't contain unity (aka, multiplicative identity). I think your question stems from not noting that an integral domain is distinguished by containing unity.

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  • $\begingroup$ $n\mathbb{Z}$ is an ideal and not a ring (unless by ring you mean non-unital ring,but i don't expect that's what he means by ring). $\endgroup$ – AIM_BLB Oct 1 '14 at 21:27
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    $\begingroup$ nZ together with its additive operation qualifies as a group. It multiplicative operation is associative and the left and right distributive laws apply so from my understanding nZ is definitely a commutative ring. $\endgroup$ – user172428 Oct 1 '14 at 21:34
  • $\begingroup$ Then what would the unit be? since $nZ$ has the multiplicative structure of $\mathbb{Z}$ (according to your definition) and yet $\mathbb{Z}$'s unit (unique) is $1$, but $1\not\in n\mathbb{Z}$. $\endgroup$ – AIM_BLB Oct 1 '14 at 21:36
  • $\begingroup$ There is no unity. Hence it is a what I believe is a good example of a commutative ring that doesn't contain unity. Not all rings contain unity. $\endgroup$ – user172428 Oct 1 '14 at 21:39
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    $\begingroup$ @TheOldHag my definition of integral domain requires to have a unity... $\endgroup$ – spectraa Oct 2 '14 at 1:57

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