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.

Let $I$ be an ideal in a Noetherian ring $R$.

Is $I=\cap(IR_P\cap R)$ where the intersection is taken over all minimal primes of $I$?

If not, is it true if we assume $I$ has no embedded primes?

I am motivated to ask this because the statement is true if replace the intersection by the corresponding intersection over the maximal ideals of $R$.

share|improve this question
1  
Certainly not. $(2)\subset\mathbb{Z}$. –  wxu Dec 13 '11 at 4:58
    
@wxu:Sorry, I meant minimal primes of $I$. –  Gene Simmons Dec 13 '11 at 5:05
1  
I mistake your meaning, maybe you mean that the index set is the minimal primes over $I$.. –  wxu Dec 13 '11 at 5:06
    
@wxu: yeah, the intersection is taken over minimal primes of $I$ –  Gene Simmons Dec 13 '11 at 5:20

2 Answers 2

up vote 1 down vote accepted

Right. I think you are done. If $I$ has no embedded primes, the equation holds. Let $I=\cap_i\mathfrak{q}_i$ be a primary decomposition and $\{\mathfrak{p}_i\}_i$ be the corresponding minimal primes over $I$ . then $I_{\mathfrak{p}_i}\cap R=\mathfrak{q}_i$, so RHS contains in LHS, but LHS always contains in RHS. If $I$ has embedded primes, then the associated primes of RHS are the set of minimal primes over $I$ , and it does not equal to the associated primes of $I$

share|improve this answer
    
Thanks for the answer. I am not sure I follow your second equation. Could you please clarify. –  Gene Simmons Dec 13 '11 at 5:31
    
I got it thanks. –  Gene Simmons Dec 13 '11 at 5:35

Consider the case when $I$ is a power $\mathfrak p^n$ of a prime ideal $\mathfrak p$. Then $I R_{\mathfrak p} = (\mathfrak p R_{\mathfrak p})^n$, and the contraction of this ideal back to $R$ is the so-called $n$th symbolic power (see also here) of $\mathfrak p$. If $\mathfrak p$ is maximal this agrees with $\mathfrak p^n$, but not in general. Thus the answer to your question is no, even in this special case.

[But wxu has shown that the answer is yes if one assumes that $I$ has no embedded primes.]

share|improve this answer
    
Thanks for the answer but I don't think this is quite the answer to my question. $p^n$ may have minimal primes other than $p$. –  Gene Simmons Dec 13 '11 at 5:30
    
@Gene: Dear Gene, I'm a bit confused: if $\mathfrak q$ is a prime ideal containing $\mathfrak p^n$, then it contains $\mathfrak p$, so if it is a minimal prime of $\mathfrak p^n$, then it equals $\mathfrak p$. What am I missing? Regards, –  Matt E Dec 13 '11 at 5:37
    
Sorry I meant to say associated primes. –  Gene Simmons Dec 13 '11 at 5:40
    
@Matt E , you are right. The minimal prime over $\mathfrak{p}^n$ is certanly $p$, but it doesnot mean that $\mathfrak{p}^n$ has no embedded primes. –  wxu Dec 13 '11 at 5:41
    
Never mind. I guess I posted the wrong version of the question I had and was thinking about that when I was reading your answer. Anyway both my original and posted questions are clarified as a result of the two answers. –  Gene Simmons Dec 13 '11 at 5:42

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.