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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?

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up vote 3 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.

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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

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