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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Ok. Thank you for your help @user26857 $\endgroup$ – chí trung châu Jan 12 '17 at 15:44

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 $sr=0\}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.