In a commutative ring with unit every maximal ideal is prime. Under what conditions does the converse occur?
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2You 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 05 '12 at 21:13
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If the ring R is a finite integral domain, then R is a field. – DYBnor May 20 '20 at 02:38
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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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2Possibly the questioner intended to exclude the $0$-ideal in case it is prime. – paul garrett Jan 05 '12 at 21:28
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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. – Jan 05 '12 at 21:30
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2@JackManey: what do you mean exactly with "in the contenxt of ID the concepts of UFD and PID coincide"? – Jan 05 '12 at 23:50
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@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. – Jan 06 '12 at 02:08
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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.
Eric Towers
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1I 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. – Sampah Oct 02 '20 at 12:06
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@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. – Eric Towers Oct 02 '20 at 20:11
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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.
Steven Gamer
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