In Atiyah-Macdonald, in the proof of Theorem 10.17 (Krull's intersection theorem), the authors go through a 4-line chain of arguments to show that that the kernel $E=\bigcap_{n=1}^{\infty}\mathfrak a^nM$ of $M\to\hat M$ satisfies $\mathfrak aE=E$. When checking some other sources (for example Prof. May's notes), I see that the approach is similar: use some form of Artin-Rees lemma to conclude $\mathfrak aE=E$.
Question: Doesn't it follow trivially from the equality $E=\bigcap_{n\ge1}\mathfrak a^nM$ that $\mathfrak aE=\mathfrak a\left(\bigcap_{n\ge1}\mathfrak a^nM\right)=\bigcap_{n\ge2}\mathfrak a^nM=E$? (basically the first term $\mathfrak aM$ of the intersection can be ignored, as it contains every $\mathfrak a^nM$ anyway). Am I missing something trivial here?
The equality $\mathfrak a\left(\bigcap_{n\ge1}\mathfrak a^nM\right)=\bigcap_{n\ge2}\mathfrak a^nM$ assumes that multiplication by ideals and intersection of submodules commute. This is unfortunately not the case.
In general we only have $\mathfrak{a} \left(\bigcap_{n\ge1} M_n\right) \subset \left(\bigcap_{n\ge1}\mathfrak{a} M_n\right)$, but not equality.
The Artin-Rees lemma tells you that you have equality in the special case where $M_n = \mathfrak a^nM$.