When a prime ideal is a maximal ideal [duplicate]

In a commutative ring with unit every maximal ideal is prime. Under what conditions does the converse occur?

• You want to look at the literature on rings of "dimension zero". If the ring is Noetherian, then it is dimension zero iff it is Artinian. Commented Jan 5, 2012 at 21:13
• If the ring R is a finite integral domain, then R is a field. Commented May 20, 2020 at 2:38

When the ring has Krull dimension equal to zero. If we're talking about integral domains then every prime ideal of $R$ is maximal if and only if $R$ is a field (since $0$ is a prime ideal in any integral domain).

• Possibly the questioner intended to exclude the $0$-ideal in case it is prime. Commented Jan 5, 2012 at 21:28
• Possibly. In that case, @Lmn6, you'll want to look at one-dimensional rings. In the context of integral domains, the concepts of UFD and PID coincide.
– user5137
Commented Jan 5, 2012 at 21:30
• @JackManey: what do you mean exactly with "in the contenxt of ID the concepts of UFD and PID coincide"?
– user14174
Commented Jan 5, 2012 at 23:50
• @Lmn6 - If $R$ is an integral domain of Krull dimension at most 1, then $R$ is a UFD if and only if $R$ is a PID.
– user5137
Commented Jan 6, 2012 at 2:08

When the ring contains no elements that are neither units nor zero divisors. Irritatingly, I know of no short name for this third category of elements of rings.

Polynomials are a typical example: Let $R$ be an integral domain and consider $R[x]$. Polynomials (of positive degree) are not units (they don't have multiplicative inverses), nor are they zero divisors. In such a ring, there are prime ideals that are not maximal ideals. For instance $(x)$, which has $R[x]/(x) \cong R$. Since $R$ had no zero divisors, neither does this quotient, so $(x)$ is prime. However, unless $R$ was already a field, this quotient is not a field so $(x)$ is not maximal.

• I am sorry to have revived this old thread, but I am curious as to how to prove such a ring necessarily have all its prime ideals being maximal. My own google search seems (to the best of my effort) to yield nothing on this. Although a chat in Discord claims it can be proved provided it has finite number of minimal prime ideals. Commented Oct 2, 2020 at 12:06
• @Sampah : Quoting from the Answer: "In such a ring, there are prime ideals that are not maximal ideals.". Quoting from your comment "such a ring necessarily have all its prime ideals being maximal." You ask about an occurrence that is not described in this Answer. Commented Oct 2, 2020 at 20:11

A ring $R$ with identity is artinian if and only if it is noetherian and every prime ideal is a maximal ideal by Hopkins theorem.

• Z is PID and noetherian but not artin Commented Dec 20, 2014 at 19:10