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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 $R$.

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    @wxu: yeah, the intersection is taken over minimal primes of $I$2011-12-13

2 Answers 2

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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 $I$

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    I got it thanks.2011-12-13
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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.]

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    @Gene: Dear Gene, No worries. Cheers,2011-12-13