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I know that in a principal ideal DOMAIN every $\neq 0$ prime ideal is maximal. is this also true for just a commutative principal ideal ring? It seems to be true for $\mathbf{Z}/n\mathbf{Z}$ ($>1$, is always a PIR) innit? as every finite integral domain is a field.

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I know that in a principal ideal DOMAIN every ≠0 prime ideal is maximal.

Sure...

is this also true for just a commutative principal ideal ring?

Mariano's already provided the fine example of $\Bbb Z\times \Bbb Z$.

It seems to be true for $\Bbb Z/n\Bbb Z$ ($n>1$, is always a PIR) innit?

Yes, that is true. The main reason this is the case is that $\Bbb Z/n\Bbb Z$ is an Artinian ring, and prime ideals of Artinian rings are maximal. (This is a sort of generalization of Wedderburn's little theorem that you mentioned.) The proof is easy: for a prime ideal $P$ in a commutative Artinian ring $R$, $R/P$ is an Artinian domain. But it's easy to show that an Artinian domain is a field. Thus $P$ is maximal.

But these aren't the only examples.

Take an infinite direct product of fields: the prime ideals are again maximal. This time, the reason is that the ring is a von Neumann regular ring. This example is not even Noetherian!

The best you can say is this: for a commutative ring $R$, all prime ideals of $R$ are maximal iff $R/J(R)$ is von Neumann regular and $J(R)$ is nil, where $J(R)$ is the Jacobson radical of $R$.

This absorbs almost all the previous cases mentioned. For Artinian rings, $R/J(R)$ is semisimple (hence VNR) and $J(R)$ is nilpotent (hence nil.) For the product of fields case, $J(R)=\{0\}$ and $R$ is VNR. Finally, $\Bbb Z\times \Bbb Z$ has zero Jacobson radical, but it isn't VNR, so it is consistent with the statement I gave.

Notice though that the PID case is not captured, namely because we had been excusing the $\{0\}$ ideal from being maximal! If we go back to thinking about domains for which nonzero primes are maximal, we have what are called $1$-dimensional domains. The class of $1$-dimensional domains includes the class of Dedekind domains, so there are more than just PID's there.

So, to summarize and return to commutative PIRs whose nonzero prime ideals are maximal, two things can happen:

  1. $\{0\}$ is prime, so you have a PID which is $1$-dimensional.
  2. $\{0\}$ isn't prime, $R$ is Noetherian with nil radical, hence nilpotent radical, so by Hopkins-Levitzki, the ring is an Artinian PIR.
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The ring $\mathbb Z\times\mathbb Z$ is a principal ideal ring.

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  • $\begingroup$ But its prime ideals are maximal $\endgroup$ – Sergio Nov 19 '16 at 1:34

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