Lately, I have been using Steve Kleiman and Allen Altman lecture notes on commutative algebra, A Term of Commutative Algebra, that are available for free on internet, to study the subject. In those, the authors state the following result, which I adapt for the given purposes
Krull intersection theorem: let $R$ be a Noetherian ring, $\mathfrak{a}\subset R$ an ideal, and $M$ a finitely generated $R$-module. Set $N=\bigcap_{n\geq 0}\mathfrak{a}^nM$. Then there exists $x\in\mathfrak{a}$ such that $(1+x)N=0$
Proof: We know, by the characterisation of Noetherian modules over Noetherian rings, that $N$ should be finitely generated. So the desired $x\in\mathfrak{a}$ exists provided $N=\mathfrak{a}N$. Clearly, $\mathfrak{a}N\subset N$. To prove $N\subset\mathfrak{a}N$, we know from Lasker-Noether theorem that there should exist an irredundant primary decomposition of $\mathfrak{a}N$, that is, $\mathfrak{a}N=\bigcap_{i=1}^rQ_i$ with $Q_i$ $\mathfrak{p}_i$-primary. Fix some index $i$. If there's $a\in\mathfrak{a}\setminus\mathfrak{p}_i$, then $aN\subset Q_i$ (?), and so, by the characterization of the $\mathfrak{p}$-primary submodules of $M$, yields $N\subset Q_i$. (...)
The proof goes on, and everithing is undestandable, except by the inclusion $aN\subset Q_i$ I highlighted on the penultimate line of the proof. How can justify such inclusion? I guess we should be applying the definition of the given set $N$ somewhere, and somehow, but I'm not seeing it. Can anyone help me with this point?
Thanks in advance for your answers.
Say $a \in \mathfrak{a}$ and $n \in N$. First note that $an \in N$.
Indeed, we need to show $an \in \mathfrak{a}^j N$ for $j \geq 0$; the result is obvious when $j = 0$ and otherwise we use $n \in \mathfrak{a}^{j-1}N$ to get $an \in \mathfrak{a}^jN$.
So $aN \subset N$. Since $N \subset Q_i$, the result you want follows.