# When a prime ideal is a maximal ideal

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

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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. –  Dylan Moreland Jan 5 '12 at 21:13

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).

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Possibly the questioner intended to exclude the $0$-ideal in case it is prime. –  paul garrett Jan 5 '12 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 Jan 5 '12 at 21:30
@JackManey: what do you mean exactly with "in the contenxt of ID the concepts of UFD and PID coincide"? –  user14174 Jan 5 '12 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 Jan 6 '12 at 2:08

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.

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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.

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