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

Question

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.

Answer 1

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 $x\in N$, $x = x(1-r) \in xI$.