I was trying the following: Let $R$ be a commutative ring with identity, then $R$ has a unique prime ideal if and only if $R$ has a minimal prime ideal which contains all zero-divisors, and all non-units of $R$ are zero divisors.
While proving necessary part, I stuck at the following place: whether the ideal generated by zero divisors, a proper ideal?
While I am unable to give a rigorous proof for sufficient part. Please somebody explain.
Thank you.
The ideal generated by zero divisors is a proper ideal since it has not $1$.
If $R$ has a unique prime ideal then:
1- It is unique maximal, hence have all non-units;
and
2- It is unique minimal prime, hence it is the set of nilpotent elements (which are zero divisors).
So all non-units of $R$ are zero divisors.
conversely
"$R$ has a minimal prime ideal (say $P$) which contains all zero divisors" and since "all non-units of $R$ are zero divisors", $P$ contains all non-units of $R$. Hence $P$ is the unique maximal ideal of $R$. But $P$ is minimal prime. so $R$ has only one prime ideal, namely $P$.