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I'm trying to figure out a solution to the question:

If a commutative ring $R$ has a nontrivial proper ideal $I$ that contains no nontrivial zero divisor of $R$, is $R$ an integral domain?

I haven't been able to find a counterexample, and it is simple to answer this in the affirmative if $I$ is prime. However, I have found it very difficult to make progress in the general case. I considered the contrapositive, and obviously in a ring that is not an integral domain there is an ideal that contains zero divisors, but I can't figure out how to show that every ideal contains zero divisors.

Does anyone have any suggestions for a proof? Or a counterexample? Thanks so much, and I apologize that I don't have more progress to show.

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Let $R$ be a ring containing a zero divisor $r \ne 0$, and let $I$ be any non-trivial ideal of $R$.

Since $I$ is an ideal, $rI \subset I$, but every multiple of $r$ is also a zero divisor. Either $rI \ne 0$, and therefore $I \supset rI$ contains a non-zero zero divisor, or $rI = 0$, which also implies that all elements of $I$ are zero divisors (at least one of which is non-zero, since $I$ is non-trivial).

Thus, every non-trivial ideal of $R$ contains a zero divisor.

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  • $\begingroup$ So you mean that if there exists a non-trivial ideal $I$ of $R$ which contains no zero divisor then $R$ itself contains no zero divisor. Am I right @M.Vinay? But then it proves the result exactly what OP needs. $\endgroup$ – Dbchatto67 Mar 20 at 15:35
  • $\begingroup$ @Dbchatto67 Yes, exactly. It seems too easy, so I'm still not entirely confident about what I've myself written here (despite not being able to see any flaws in the reasoning). $\endgroup$ – M. Vinay Mar 20 at 16:22
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    $\begingroup$ @M.Vinay: Don't worry. It's correct. $\endgroup$ – tomasz Mar 20 at 16:31
  • $\begingroup$ Thanks! Super clean proof. $\endgroup$ – ReallyBigDog Mar 20 at 17:21

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