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This is Exercise 8.37 in R.Y.Sharp's Steps in Commutative Algebra:

Let $P$ be a prime ideal of the commutative Noetherian ring $R$. Prove that $$\bigcap_{n=1}^{\infty} P^{(n)}=\{ r\in R \mid \exists s\in R \setminus P ,sr=0 \}$$ in which $P^{(n)}=(P^n)^{\text{ec}}$ with extension and contraction notation in conjunction with the natural ring homomorphism $R\to R_P$.

There's 2 reasons I got stuck. First, I'm confused with $\bigcap_1^{\infty}$, I don't know how to use this notation. Second, the only theorem I know involving with $\bigcap_1^{\infty}$ is Krull's Intersection Theorem, stating "If $I\subseteq\mathrm{Jac}(R)$ then $\bigcap_{n=1}^{\infty}I^n=0$", which I believe is useless in this situation.

So help me with this problem. THank you.

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  • $\begingroup$ Your guess that the KIT is useless is wrong. If $x\in P^{(n)}$ then $x\in P^nR_P=(PR_P)^n$. Can you take it from here? $\endgroup$ – user26857 Jan 12 '17 at 15:18
  • $\begingroup$ Okay, thanks to your hint, I solved the inclusion. One more thing, now that I recheck, I was wrong about the converse. Now I solved it using your hint, too. $\endgroup$ – chí trung châu Jan 12 '17 at 15:38
  • $\begingroup$ like this, $r/1=0\in (P^n)^e$, so $r\in (P^n)^{ec}$. It's true, right? $\endgroup$ – chí trung châu Jan 12 '17 at 15:39
  • $\begingroup$ It sounds right. $\endgroup$ – user26857 Jan 12 '17 at 15:43
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    $\begingroup$ Ok. Thank you for your help @user26857 $\endgroup$ – chí trung châu Jan 12 '17 at 15:44
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We have:

  • $\bigcap_{n=1}^{\infty} P^{(n)}=\bigcap_{n=1}^{\infty} (P^{n})^{ec}= (\bigcap_{n=1}^{\infty} (P^{e})^{n})^{c}=(\frac{0}{1})^{c}$

  • $(\frac{0}{1})^{c}=\{r\in R$: $\exists s\notin P $ such that $sr=0\}$

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