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

Solution 1:

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."

Solution 2:

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).

Solution 3:

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.

Solution 4:

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).

Solution 5:

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.