If krull dimension is $0$, then all prime ideals are maximal?

Question

Let $R$ be a ring, If we assume '$R$'s all prime ideals are maximal',then we can say $R$'s krull dimension is exactly $0$. This is easy because there are two different prime ideal $p$ and $q$ which satisfies $p⊂q⊂R$, and this contradicts the fact that $p$ is maximal.

But is the converse true?

my attempt: If there are prime ideal $p$ which is not maximal, then there are another ideal $q$ which satisfies $p⊂q⊂R$. If I can deduce $q$ is prime ideal, the proof are done.But I cannot show $q$ is a prime ideal.

Any help would be appreciated, thank you.

Answer 1

1. Every ideal is contained in a maximal ideal.

Basically you need to verify that Zorn's lemma applies. If you have a chain of ideals that all contain $p$ and all don't contain $1$ then their union contains $p$ and doesn't contain $1$ (every chain has an upper bound). So the set of proper ideals containing $p$ contains a maximal element and that element is a maximal ideal.

2. Every maximal ideal is prime.

Which should be familiar.