# If every prime ideal is maximal, what can we say about the ring?

Suppose $R$ is a ring and every prime ideal of $R$ is also a maximal ideal of $R$. Then what can we say about the ring $R$?

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It's a principal ideal domain? –  Jeremy Dec 2 '12 at 5:52
@Jeremy A PID is integral, right? Then $(0)$ is prime, so it's maximal, and $R$ is actually a field. –  jathd Dec 2 '12 at 6:14

If we assume $R$ is commutative and Noetherian, then this property is equivalent to $R$ being an Artinian ring (i.e., satisfying the descending chain condition). Such rings are finite products of Artin local rings.

Reduced Artin local rings are fields. Some non-reduced examples include $k[x]/(x^n)$, $k$ a field, and more generally $k[x_1,\ldots,x_n]/I$, where Rad$(I)=(x_1,\ldots,x_n)$. There are also examples that don't contain a field, like $\mathbb{Z}/(p^n)$, $p$ a prime.

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I assume $R$ is commutative. Such a ring is said to have Krull dimension $0$ or to be zero-dimensional.

• Every field is zero-dimensional. More generally, every Artinian local ring is zero-dimensional.
• A product of zero-dimensional rings is zero-dimensional. In particular, every product of Artinian local rings is zero-dimensional.
• Every Boolean ring is zero-dimensional. This gives a supply of examples that are in general neither Noetherian nor products of Artinian local rings.
• According to Wikipedia, zero-dimensional and reduced is equivalent to von Neumann regular.

I don't think there is a nice classification of arbitrary rings of Krull dimension $0$ (and I have no idea what happens in the noncommutative case).

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If $R$ is an integral domain, then $(0)$ is prime, so it's maximal, and $R$ only has two ideals, $(0)$ and $R$. In other words, it's a field.
If not, but it's Noetherian, then it's still Artinian (because its Krull dimension is $0$).
I'm not sure what can be said if $R$ is not Noetherian.