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The problem is as follows:

Let $I\subseteq J$ be ideals in a Noetherian ring. Show that if $I_{p}=J_{p}$ for every associated prime $p$ of $I$,then $I=J$.

It seems reasonable to consider $J/I\subseteq R/I$ but I couldn't go on.

Let M be an $R-\mathrm{module}$. A prime ideal $p$ is an associated prime of $M$ if there exists nonzero $x\in M$ such that $p=\mathrm{ann}_{R}(x)$.

user26857
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user12580
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    Please say what you have tried. Also, if this is a problem from textbook, please give a proper reference. – Manos May 11 '16 at 14:15
  • The use of `associated prime of $I$' can be confusing. You really mean associated prime of $R/I$ where $R$ is your Noetherian ring. What can you say about the associated primes of $J/I$? – Mohan May 11 '16 at 14:16
  • @ Manos The problem is from handouts, so I cannot give the reference. – user12580 May 11 '16 at 14:19
  • Do not vandalise your post, we are alerted to it in Charcoal HQ and have to clean it up. This may result in a question ban. see question bans –  May 26 '16 at 07:41

1 Answers1

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Hint: We have that $\operatorname{Ass}(J/I) \subset \operatorname{Ass}(R/I)$ (why?). Now, use this together with the hypothesis $I_P = J_P, \, \forall P \in \operatorname{Ass}(R/I)$ to conclude that $J/I = 0$. In doing so, recall that for a finite $R$-module $M$ it is always the case that $\operatorname{Ass}(M) \subset \operatorname{Supp}(M)$.

Remark: The definition of associated prime that you gave is correct, but recall that it is a convention that when one talks about associated primes of an ideal $I$, they really mean the associated primes of the quotient $R/I$. As Mohan says, this can be quite confusing and takes some time to get used to.

Manos
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  • Thank you for your reminding me of the definition! I was thinking of the ideal I being an $R-module $ but forgot the definition of the associated prime of an ideal. It's my fault. – user12580 May 11 '16 at 14:55
  • @user1: Let $P$ be an associated prime of $J/I$. Then $P$ is in the support of $J/I$ and so $(J/I)_P \neq 0$. On the other hand, $P$ is an associated prime of $R/I$ and so $J_P = I_P$ by hypothesis, i.e., $(J/I)_P = 0$. This contradiction shows that $Ass(J/I)$ is empty, which implies that $J/I=0$. In this last deduction, i used the fact that for any non-zero module over a Noetherian ring, its set of associated primes is non-empty. Is that clear now? – Manos May 11 '16 at 18:50
  • yes thanks. maybe its better to add the comment to answer(?) – user 1 May 12 '16 at 06:16