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

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.

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 $$x\in N$$, $$x = x(1-r) \in xI$$.
• I would have thought that $IM=M$ is the hard part. The only proof I know is to use Artin-Rees. – Mohan Nov 13 at 14:17
• @Mohan Yes, my carelessness and you're right. Proving $M\subset IM$ is indeed non-trivial. – pisco Nov 13 at 16:26