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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 $R$.

Community
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Gene Simmons
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2 Answers2

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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 $I$

wxu
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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.]

Matt E
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  • 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 05:30
  • @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 05:37
  • Sorry I meant to say associated primes. – Gene Simmons Dec 13 '11 at 05:40
  • @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 05:41
  • 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 05:42
  • @wxu: Dear wxu, I realize that. Obviously, given your answer, the examples where the $n$th power and the $n$th symbolic power disagree are precisely those with embedded primes. Regards, – Matt E Dec 13 '11 at 05:43
  • @Gene: Dear Gene, I'm a bit confused again; my answer was targeted at the general form of your question, in which the ideal $I$ is not necessarily assumed to have no embedded primes. Since the concept of symbolic power is directly related to your question in this case, I thought it might be of interest to you to point this out. In any case, I wrote my answer at the same time wxu was writing theirs; if I'd seen wxu's answer before writing mine, I probably wouldn't have bothered answering. Regards, – Matt E Dec 13 '11 at 05:47
  • Dear Matt: In my first comment I meant to say associated instead of minimal. Later I realized, that was not a valid comment given my question. Actually I was thinking of a slightly different question from the one I posted when I wrote the first comment. I understand your answer and it does correctly answer the question as posted without the additional assumption used in the second part. To sum up - my bad! – Gene Simmons Dec 13 '11 at 14:39
  • @Gene: Dear Gene, No worries. Cheers, – Matt E Dec 13 '11 at 15:35