Prove that $R$ has exactly one nontrivial ideal

Question

Let $R$ be a commutative ring with identity and with the following properties:

(a) The intersection of all of its nonzero ideals is nontrivial.

(b) If $x$ and $y$ are zero divisors in $R$, then $xy=0$.

Prove that $R$ has exactly one nontrivial ideal.

I was trying to prove that $I$, the intersection of all ideals is a maximal ideal. I can see that $I^2=0$ but cannot go any further. I'd much prefer hints than complete answers. Thank you.

Answer 1

Since $I$ (as you denoted in your original post) is clearly a minimal ideal, $R/M\cong I$ for some maximal ideal $M$. We have $MI=0$ so that $M$ consists entirely of zero divisors, and by a) we have that $M^2=\{0\}$.

I will leave it to you to prove that $I=ann(M)$, and that $R$ is a local ring with maximal ideal $M$.

Finally, putting the things together, $M\subseteq ann(M)=I$.

So $I$ is the unique minimal and unique maximal ideal, therefore it is the unique nontrivial ideal.