Let $R$ be a commutative ring with 1. If $R$ is a PID, then every prime ideal is either zero or maximal.

Question

My proof: Let $I = (p)$ be a non-zero prime ideal of $R$. Note that $p$ is prime. Since $R$ is an integral domain, $p$ is also irreducible. Suppose $I \subseteq J = (m)$, then we have $m | p$, and so $p = mk$. Since $p$ is irreducible, $m$ is a unit or $k$ is a unit. If $m$ is a unit, then $J = R$. If $k$ is a unit, then $p \sim m$, and so $I = J$. Therefore, I is maximal. Hence, if $R$ is a PID, then every prime ideal is either zero or maximal.

However, I don't quite get where I need to use PID. It seems that it works on integral domain too. Where am I getting it wrong?

You used it when you wrote $J=(m)$. – Nishant Apr 19 '15 at 02:17 — Nishant, Apr 19 '15 at 02:17

score 1 · Answer 1 · answered Apr 19 '15 at 02:17

You used it right in the first sentence:

Let $I=(p)$ be a non-zero prime ideal of $R$.

If $R$ were not a PID, you very well could have a non-zero prime ideal that was not generated by any single element $p$. (Similarly, you used it when you assumed that some other ideal $J$ would be equal to $(m)$ for some $m$.)

As an example, if $R=\mathbb{C}[x,y]$, then the ideal $I=(x,y)$ is not equal to $(f)$ for any $f\in R$ (that is, it's not a principal ideal).

Let $R$ be a commutative ring with 1. If $R$ is a PID, then every prime ideal is either zero or maximal.

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