If every prime ideal is maximal then the ring is a PID

Question

I was thinking if it's true that in a ring R every prime ideal is maximal if and only if R is a PID. If R is a PID then it's trivial, but I don't know how to prove or disprove the other direction. Personally I think it is false, but I cannot find a counterexample.

Answer 1

If $R$ is a PID , then any nonzero prime ideal is maximal. The zero ideal is prime, but not maximal if $R$ is not a field.

So, a more reasonable version of your question would be: is is true that in a ring $R$ (that i suppose to be commutative since you are talking about prime ideals) every nonzero prime ideal is maximal if and only if $R$ is a PID.

Unfortunately, the answer is NO. You have infinitely many examples of rings satisfying your condition without being PIDs.

For example, any subring $R$ of $\mathbb{C}$ which is also a finitely generated abelian group satisfies your condition.

Sketch of proof. if $\mathfrak{p}$ is a nonzero prime ideal, one may show ny intersecting with $\mathbb{Z}$ that it contains a prime number $p$. Then $R/\mathbb{p}$ is an $\mathbb{F}_p$-vector algebra for the law $\bar{m}\cdot (x+\mathfrak{p})=mx+\mathfrak{p}$, which is finite dimensional since $R$ is a finitely generated abelian group. Hence, $R/\mathbb{p}$ is a finite dimensional algebra over a field which is also an integral domain. It is well known that it implies that it is a field.

In particular, any quadratic ring $\mathbb{Z}[\sqrt{d}]$ satisfies that any nonzero prime ideal is maximal. However, we know lot of examples where these rings are not UFDs, and thus not PIDs: any $d\leq -3$ or any $d\equiv 1 \mod 4$ will do.