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.
Best Answer
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.