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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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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
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
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 $I$

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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.]

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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 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

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