# Is there an adjective for rings whose every non-zero prime ideal is maximal?

(All my rings are commutative and unital.)

Question. Is there an adjective for rings whose every non-zero prime ideal is maximal?

Remarks:

• Every PID has this property; more generally, every Dedekind domain does.

• Every semiprimitive ring with this property is trivially a Jacobson ring.

• These rings are exactly zero-dimensional rings which are not domains, and one-dimensional domains. – Crostul Apr 26 '16 at 15:35
• @goblin No, I am not going to explain how to read the definition of Krull dimension to you, since I know you are perfectly capable of reading it on the wiki. These imprecisions you see are the result of working on a mobile phone. Perhaps you can take this into consideration next time before writing responses with such tones. Regards – rschwieb Apr 27 '16 at 2:55
• @Crostul nearly exact: that omits fields, which vacuously satisfy the condition – rschwieb Apr 27 '16 at 2:59
• Don't criticize Wikipedia, edit Wikipedia. Bad pages are less the result of poor authors, and more the result of potentially good authors deciding it's someone else's problem. In any case, the Krull dimension is properly defined, on the current edit of Wikipedia, as the supremum of the lengths of the chains of primes $\mathfrak{p}_0 \subsetneq \cdots \subsetneq \mathfrak{p}_n$, where by "length" we mean simply $n$. The supremum here is taken in $[0,\infty]$. In other words, a ring is said to have Krull dimension $\infty$ exactly when there exist chains of arbitrary (finite) length. – Slade Apr 27 '16 at 3:09
• In particular, $\mathbb{Z}$ has Krull dimension $1$. The primes do not have to be "adjacent"—your example of a ring with prime spectrum isomorphic as a poset to $\mathbb{R}$ still has arbitrarily long chains, e.g. $0<1<\cdots < n$. – Slade Apr 27 '16 at 3:11