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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R$ be a commutative, Noetherian ring with unity. I know that the following is true:

For any ideal $I\subset R$, there are prime ideals $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$ such that $\mathfrak{p}_1\cdots\mathfrak{p}_n\subset I$.

If we only consider proper ideals, can we always find prime ideals $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$ such that $\mathfrak{p}_1\cdots\mathfrak{p}_n=I$?

I came across this problem on an old qualifying exam and it has me puzzled. I'm beginning to think the problem is stated incorrectly. Are there any counterexamples, or is the result true?

share|cite|improve this question
up vote 4 down vote accepted

For example in the ring $R=k[x,y]$, the ideal $I=(x,y^2)$ is $P$-primary, where $P = (x, y)$, but is not a power of $P$, so the answer to your question is negative.

share|cite|improve this answer
    
This is helpful, but if we know $I$ is not a power of $P$, why can we conclude it is not the product of any prime ideals? I'm not too familiar with properties of $P$-primary ideals. – Jared Aug 28 '13 at 5:43
    
In our case $\sqrt I=P$. If $I=\mathfrak{p}_1\cdots\mathfrak{p}_n$ what about $\sqrt I$? – user26857 Aug 28 '13 at 5:45
    
@Jared But why does $P = \mathfrak{p}_1 \cap \ldots \mathfrak{p}_n$ force $\mathfrak{p}_i = P$ for all $i$? Couldn't you have a single $\mathfrak{p}_i = P$ and all the other $\mathfrak{p}$'s being prime ideals strictly bigger than $P$? You have to use the fact that $I$ is primary somehow, not just that $\sqrt{I}$ is prime. It is not true that if $\sqrt{I}$ is prime, and $I$ is expressed as the product of prime ideals, then all those prime ideals are equal. For example, $I=(x^2,xy)$ has prime radical $(x)$ and can be written as the product of distinct primes: $I=(x)(x,y)$. – Ted Aug 28 '13 at 6:57
    
@Ted In our case $P$ is maximal and this makes things very simple. – user26857 Aug 28 '13 at 22:40
    
Yes that's true. But is the original claim correct, that if $I$ is primary and expressed as a product of prime ideals, then all the primes are equal? – Ted Aug 29 '13 at 4:25

A commutative ring $R$ with unity is said to be a general zpi ring if every ideal of $R$ is a finite product of prime ideals.

Examples: 1) $\Bbb Z$ is a general zpi ring; 2) $\Bbb Z[[X]]$ is noetherian but not a general zpi ring.

Therefore the answer to your question is negative.

share|cite|improve this answer
    
Why 'zpi'? Do you have a reference for these facts? – Michael Albanese Jan 19 at 14:28
    
@ Michael Albanese you can see this paper "ON GENERAL Z.P.I.-RINGS" CRAIG A. WOOD – M.A Jan 19 at 14:58

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.