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My book states that an integral domain is a commutative ring with unity in which $1 \neq 0$ and there are no nonzero zero-divisors.

So if $R$ is not an integral domain, does that mean $R$ is a commutative ring with unity in which $1 = 0$ and there are nonzero zero-divisors?

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    $\begingroup$ Well, $R$ could not be an integral domain if the ring $R$ has no unity for example (since your definition requires it). But usually $R$ is not an integral domain if there is some non trivial zero divisor. $\endgroup$ – Sigur Jan 27 '13 at 20:57
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    $\begingroup$ Well, in principle $R$ could be a rat. But the tacit universe here is probably commutative rings with unit, and probably with $0\ne 1$. So what is intended is probably only that there exist $a$, $b$, neither $0$, such that $ab=0$. $\endgroup$ – André Nicolas Jan 27 '13 at 20:58
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No. By de Morgan's laws, the negation of a ring being an integral domain is: either $0=1$ or there is a nonzero zero divisor (or both).

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  • $\begingroup$ Ah I see what I missed. But if $R$ is not an integral domain, $R$ is still possibly a commutative ring with unity right? $\endgroup$ – Student Jan 27 '13 at 20:58
  • $\begingroup$ @Student: That's right! $\endgroup$ – Clive Newstead Jan 27 '13 at 20:59

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