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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.

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    $\begingroup$ Take the zero ideal in $\mathbb Z$. $\endgroup$ Commented Dec 1, 2011 at 17:01
  • $\begingroup$ 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"). $\endgroup$ Commented Dec 1, 2011 at 17:07
  • $\begingroup$ 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? $\endgroup$
    – Adam
    Commented Dec 1, 2011 at 17:11
  • $\begingroup$ 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. $\endgroup$
    – Adam
    Commented Dec 1, 2011 at 17:12
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    $\begingroup$ @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. $\endgroup$
    – Ted
    Commented Dec 1, 2011 at 17:28

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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.

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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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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.

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