I just wanted to confirm the following: when a text says a ring has a "unique prime ideal", does it really mean "unique nontrivial prime ideal", because 0 is a prime ideal, correct? I am writing this after seeing the following the exercise:
Let $R$ be a commutative ring with identity, let $M$ be a maximal ideal of $R$, and let $n$ a positive integer. Prove $R/M^{n}$ has a unique prime ideal, so that, in particular, $R/M^n$ is a local ring.
I believe "nontrivial" was intended, because a local ring is, by definition, one that has a unique maximal ideal. Since any proper ideal of a ring is contained in a maximal ideal and any maximal ideal is prime, this is how we arrive at "unique prime ideal $\Rightarrow$ unique maximal ideal $\Rightarrow$ the ring is local" Is this right?
Best Answer
Collecting the comments, given a commutative ring $R$ with unity, $(0)$ is a prime ideal if and only if the ring $R$ is actually an integral domain. This given that $$ \text{Then an ideal }p\text{ is prime iff }R/p\text{ is an integral domain.} $$ In your case, $(0)$ is prime iff $R/(0)\cong R$ is an integral domain. But despite this, your final conclusion is still true since every maximal ideal is prime, if you suppose that your ring has two maximal ideals (at least) then you have two prime ideals which is a contradiction, therefore your ring is local