Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
7  
Take the zero ideal in $\mathbb Z$. –  Pierre-Yves Gaillard Dec 1 '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 1 '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 1 '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 1 '11 at 17:12
2  
@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 1 '11 at 17:28
show 8 more comments

3 Answers

up vote 5 down vote accepted

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.

share|improve this answer
add comment

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).

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.