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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.

Pont
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  • Does 'contain' mean 'properly contain'? If it does not so, the proposition cannot be used here. – Pont Apr 02 '20 at 16:46
  • $p$ should be maximal so there shouldn't be any proper ideals that properly contain $p$. On the other hand there is some maximal ideal that contains $p$ and if that's $p$ then you're good and if that's not $p$ then the Krull dimension can't be $0$. – Trevor Gunn Apr 02 '20 at 16:47
  • I understood, thank you very much! – Pont Apr 02 '20 at 16:59

1 Answers1

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  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.

  1. Every maximal ideal is prime.

Which should be familiar.

Trevor Gunn
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