# Ideal with no zero divisors implies integral domain?

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.

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.
• 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. – Dbchatto67 Mar 20 at 15:35