20
$\begingroup$

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 $R$?

$\endgroup$
4
  • $\begingroup$ It's a principal ideal domain? $\endgroup$ – Jemmy Dec 2 '12 at 5:52
  • 5
    $\begingroup$ @Jeremy A PID is integral, right? Then $(0)$ is prime, so it's maximal, and $R$ is actually a field. $\endgroup$ – jathd Dec 2 '12 at 6:14
  • $\begingroup$ @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. $\endgroup$ – D777 Jan 12 '19 at 18:00
  • $\begingroup$ @D777 (0) is a prime ideal in Z which is not maximal $\endgroup$ – Tim kinsella Aug 5 '19 at 8:33
21
+50
$\begingroup$

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."

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Finite rings are Artinian. An instructive example is $\mathbb{Z}[x]/(4,2x)$. It has $8$ elements. $\endgroup$ – Martin Brandenburg Jan 12 '14 at 18:40
  • $\begingroup$ Is this really true? I mean, $\mathbb{Z}$ is both commutative and Noetherian, and its maximal ideals are precisely the prime ideals, but $\mathbb{Z}$ is not Artinian. $\endgroup$ – goblin GONE Oct 28 '15 at 5:01
  • 1
    $\begingroup$ @goblin The zero ideal is prime but not maximal. $\endgroup$ – Julian Rosen Oct 28 '15 at 15:16
10
$\begingroup$

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).

$\endgroup$
9
  • 1
    $\begingroup$ A product of finitely many zero-dimensional rings is zero-dimensional. $\dim(\prod_n \mathbb{Z}/p^n)=\infty$. $\endgroup$ – Martin Brandenburg Jan 12 '14 at 17:56
  • $\begingroup$ @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. $\endgroup$ – Pete L. Clark Jan 12 '14 at 18:15
  • 2
    $\begingroup$ 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. $\endgroup$ – Pete L. Clark Jan 12 '14 at 18:16
  • $\begingroup$ @Martin: Oh, I just saw your answer. It addresses everything I said in my comments. Thanks! $\endgroup$ – Pete L. Clark Jan 12 '14 at 18:22
  • 2
    $\begingroup$ Infinite products of von Neumann regular rings are von Neumann regular. But in the non-reduced setting, usually strange things happen. $\endgroup$ – Martin Brandenburg Jan 12 '14 at 18:33
6
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.