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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 $R$?

fan
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  • It's a principal ideal domain? – Jemmy Dec 02 '12 at 05:52
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    @Jeremy A PID is integral, right? Then $(0)$ is prime, so it's maximal, and $R$ is actually a field. – jathd Dec 02 '12 at 06:14
  • @jathd I believe that's incorrect. For instance, $\mathbb{Z}$ is a domain such that every prime ideal is maximal, but of course, isn't a field. The problem in your argument, I believe, is that 0 is not considered as a prime element. – D777 Jan 12 '19 at 18:00
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    @D777 (0) is a prime ideal in Z which is not maximal – Tim kinsella Aug 05 '19 at 08:33

5 Answers5

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For a commutative ring, having all primes maximal has a simple characterization: $R/J(R)$ is von Neumann regular and $J(R)$ is a nil ideal, where $J(R)$ is the Jacobson radical.

This recovers everything previously mentioned:

  • When $R$ is Artinian, $R/J(R)$ is semisimple (hence VNR) and $J(R)$ is nilpotent (hence nil.)

  • When $R$ is VNR, then $J(R)=\{0\}$ (hence nil) and $R/J(R)=R$ is obviously VNR.

  • When $R$ is Noetherian, a nil $J(R)$ becomes a nilpotent $J(R)$, and a Noetherian $R/J(R)$ is necessarily semisimple, so Hopkins-Levitzki says that $R$ is Artinian.

I'm not aware of a definitive answer for noncommutative rings. Things are different there because simple rings are a lot more diverse than fields, prime ideals are less nice, localization is not nice, and nilpotent elements don't necessarily live in $J(R)$ anymore.

Here's a fitting generalization I found. The above theorem for commutative rings is "$R/J(R)$ is VNR and $J(R)$ is nil iff $R/P$ is a field for every prime ideal $P$," and the following is also true: "$R/P$ is a division ring for all prime ideals $P$ iff $R/J(R)$ is strongly regular and $J(R)$ is nil."

rschwieb
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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 (edit: finite) product of zero-dimensional rings is zero-dimensional. In particular, every (edit: finite) 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).

Qiaochu Yuan
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    A product of finitely many zero-dimensional rings is zero-dimensional. $\dim(\prod_n \mathbb{Z}/p^n)=\infty$. – Martin Brandenburg Jan 12 '14 at 17:56
  • @Martin: True. But also many infinite products as well, e.g. any product of fields. The issue of when a product of zero-dimenional rings stays zero-dimensional seems to be missing from the commutative algebra texts (which mostly seem to think that we will all rather need to know that Cohen-Macaulay rings are Gorenstein). I saw some treatment of this in several research papers, but I would be very interested to see a unified discussion in a textbook. – Pete L. Clark Jan 12 '14 at 18:15
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    Getting back to Qiaochu's answer: I honestly think that $\operatorname{dim} R = 0 \iff$ $R/\operatorname{nil} R$ is absolutely flat (= von Neumann regular) is not bad. – Pete L. Clark Jan 12 '14 at 18:16
  • @Martin: Oh, I just saw your answer. It addresses everything I said in my comments. Thanks! – Pete L. Clark Jan 12 '14 at 18:22
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    Infinite products of von Neumann regular rings are von Neumann regular. But in the non-reduced setting, usually strange things happen. – Martin Brandenburg Jan 12 '14 at 18:33
  • @Martin: whoops. I'm not entirely sure why I wrote that. Thanks for the correction! – Qiaochu Yuan Jan 12 '14 at 18:47
  • @Martin: "usually strange things happen". Yes, usually but not always -- the full truth is a bit complicated. – Pete L. Clark Jan 12 '14 at 20:01
  • @PeteL.Clark: If Cohen-Macaulay rings are Gorenstein, then I've been reading the wrong books! Also, Qiaochu: why are all your Artinian rings local? Considering (nonlocal) Artinian rings would divide your points neatly into the Noetherian and non-Noetherian cases. – zcn Jan 13 '14 at 01:37
  • @user115654: Yes, I had it the wrong way around. Maybe there are not enough commutative algebra books containing this material after all... – Pete L. Clark Jan 13 '14 at 05:18
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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.

Julian Rosen
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8

For a commutative ring $R$, the following are equivalent:

  • Every prime ideal is maximal.
  • $\dim(R)=0$
  • $R/\sqrt{0}$ is von Neumann regular
  • For all $x$ there is some $n \in \mathbb{N}$ such that $x^{n+1}$ divides $x^n$.

There is a big theory about $0$-dimensional commutative rings, see for example David F. Anderson, David Dobbs, "Zero-Dimensional Commutative Rings" (Lecture Notes in Pure and Applied Mathematics).

Martin Brandenburg
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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.

jathd
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