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I was looking at a proof in one of my course notes and I saw the following statement, where we assume that $I$ is an ideal in a Noetherian ring $R$:

If $x \in \cap_{n=1}^\infty I^n$, then $x \in xI$.

We of course know that $I$ is finitely generated and hence, so are all the $I^n$ but I don't know how you can then conclude that statement.

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Let $N = \cap_{i=1}^\infty I^n$, since $R$ is Noetherian, $N$ is a finitely generated $R$-module, then Artin-Rees lemma says $IN=N$ (set $M=R$ in Wikipedia notations).

Therefore by Nakayama's lemma (Statement 1 in Wikipedia) there exists $r\in R$ such that $r-1 \in I$ and $rN=0$. In particular for $x\in N$, $x = x(1-r) \in xI$.

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    $\begingroup$ Is it trivial to assume that IM = M? $\endgroup$ – Mee98 Nov 13 at 11:55
  • $\begingroup$ I would have thought that $IM=M$ is the hard part. The only proof I know is to use Artin-Rees. $\endgroup$ – Mohan Nov 13 at 14:17
  • $\begingroup$ @Mohan Yes, my carelessness and you're right. Proving $M\subset IM$ is indeed non-trivial. $\endgroup$ – pisco Nov 13 at 16:26

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