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I am attempting to prove the converse to $I$ maximal implies $I$ prime is not true.

$I$ prime $\iff A/I$ is an integral domain and $I$ maximal $\iff A/I$ is a field so I'm looking for a prime ideal that yields an integral domain but not a field.

Obviously the integers fail (as the only ideals are $d\mathbb{Z}$ for some $d$) and these are prime iff $d$ is prime and the quotient ring modulo $p$ is a field.

I have proved $\langle p(x)\rangle$ is maximal iff $p(x)$ is irreducible in $K[x]$ where $K$ is a field, again tried looking for prime ideals that aren't generated by an irreducible polynomial but did not get far.

I imagine I should be considering the ring of diagonal matrices as this has zero divisors and I think zero divisors are important in the underlying examples.

Davide Giraudo
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Adam
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    Take the zero ideal in $\mathbb Z$. – Pierre-Yves Gaillard Dec 01 '11 at 17:01
  • It's not quite right to say the prime ideal "generates an integral domain but not a field"; 'generating' is a term of art, so what you write is probably not what you meant (perhaps, "yields", or "gives"). – Arturo Magidin Dec 01 '11 at 17:07
  • Thanks for the response. I understand how your example answers the question but now how 0 is 'prime' in a sense we could use. It's clear {0} is a prime ideal in an integral domain but pZ is a prime ideal iff p is prime would surely be a definition we want? I realized I skipped d being zero in dZ <-> d prime which I shouldn't have done but why would we allow this? – Adam Dec 01 '11 at 17:11
  • I agree, yield etc would have been wiser words to use I'll edit. It is quite annoying when you have specific meanings for words that seem almost natural to describe a process. – Adam Dec 01 '11 at 17:12
  • @Adam: As it turns out, no, the definition we want is not "$p\mathbb{Z}$ is a prime ideal if and only if $p$ is a prime". Even in the integers, the definition of that says "$p$ is not a unit and if $p|ab$ then $p|a$ or $p|b$" includes $p=0$. We don't want it because it makes things bad in general. What is correct is that "$p\mathbb{Z}$ is a nonzero prime ideal if and only if $p$ is a prime number." – Arturo Magidin Dec 01 '11 at 17:16
  • Why do we want to include {0} as a prime Ideal though? I know it satisfies the definitions but we specify I =/= R (I understand that's required to produce at least two elements of the quotient ring) why not force I =/= {0} as well? Is there a fundamental reason for including it? – Adam Dec 01 '11 at 17:20
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    @Adam: If we want the equivalence "$I$ is prime iff $R/I$ is an integral domain" to hold, then we need to include 0 and exclude $R$. I'm sure someone else can come up with a better answer. – Ted Dec 01 '11 at 17:28
  • @Adam: It makes for far fewer exceptions. If you exclude $(0)$, then "$I$ is prime if and only if $R/I$ is an integral domain" is false. "$I$ maximal implies $I$ prime" is false. And when you move away from $\mathbb{Z}$, it makes sense to include the $0$ ideal when the definition applies. – Arturo Magidin Dec 01 '11 at 17:29
  • Maybe you'd also like another example: take $\mathbb{Z}[X]$, the ring of polynomials with integer coefficients. Then $\langle X \rangle$ is prime, since $\mathbb{Z}[X]/\langle X \rangle \cong \mathbb{Z}$, but obviously it is not maximal, since $\mathbb{Z}$ is not a field. Also, we can fit $\langle 2,X \rangle$ in between $\langle X \rangle$ and $\mathbb{Z}[X]$. – Oliver Braun Dec 01 '11 at 17:53
  • @Adam The method of modular reduction exhibited in my answer (and esp. its link) yield quite compelling reasons for adopting the standard convention that $0$ is prime (but $1$ is not). – Bill Dubuque Dec 01 '11 at 18:27
  • @BillDubuque: I have been giving the concept some thought. I think I understand modulo ideal slightly more, does this sound okay? Suppose we have an equivalence relation on a ring R. Then if $x \equiv y$ and $v \equiv w$ implies x+v is equivalent to x + w and xv is equivalent to yw then [0] - the equivalence class of 0 is an ideal. Also, we see for r in R [r] = r + [0]. That is, if the equivalence relations satisfies these properties then R/[0] is just a set of sets of elements of R and these elements of R are uniquely determined up to addition of multiples of [0]. – Adam Dec 02 '11 at 02:45
  • So elements of [0] are what we need to be think of as a kind of zero generally in R (I don't think I worded that correctly)? Conversely, take an ideal. Then elements which are in the same coset produces the equivalence relation. – Adam Dec 02 '11 at 02:49
  • And the set of Ideals partitions R and keeps elements from the same Ideal in the same cosets under + and . in R, in the same cosets? – Adam Dec 02 '11 at 02:59
  • See also: Examples of prime ideals that are not maximal – Martin Sleziak Apr 24 '16 at 07:56

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Your assertion about the integers is not quite correct.

It is true that $d\mathbb{Z}$ is a prime ideal if $d$ is a prime integer, and that these ideals are also maximal.

But you forgot one very important prime ideal of $\mathbb{Z}$: $0\mathbb{Z}$! In a commutative ring with unity, $(0)$ is a prime ideal if and only if the ring is an integral domain, and $\mathbb{Z}$ is definitely an integral domain (one might say it's the integral domain; it's why they are called "integral domains", after all...). So you can get lots of examples just by taking integral domains that are not fields.

For examples with zero divisors, consider a product of two rings: $R\times S$. Notice that if $I$ is an ideal of $R$ and $J$ is an ideal of $S$, then $I\times J$ is an ideal of $R\times S$, and $(R\times S)/(I\times J) \cong (R/I)\times (S/J)$. So you can get further examples this way. You can think of the case of diagonal matrices that way, since the ring of diagonal $n\times n$ matrix with coefficients in $R$ is isomorphic to the direct product of $R$ with itself, $n$ times.

Arturo Magidin
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HINT $\rm\ \ A\:$ is a field $\iff\: (0)\:$ is maximal; $\rm\ A\:$ is a domain $ \iff\: (0)\:$ is prime.

Therefore $(0)$ is a non-maximal prime in any domain that is not a field.

As I mentioned in your question yesterday on the converse, factoring out by the prime $\rm\:P\:$ to reduce to the domain case $\rm\:P = 0\:$ is a technique that is very useful in algebra. Again I highly recommend reading the masters (Kaplansky).

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Bill Dubuque
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Another class of examples: Take any integral domain $R$ that isn't a field. Then the ideal of the polynomial ring $R[x]$ generated by $x$ is prime but not maximal, since $R[x]/(x)\cong R$ is an integral domain but not a field.