Theorem (Artin-Rees): Let $R$ be a Noetherian ring, $I \subseteq R$ an ideal, $M$ a finitely generated $R$-module, and $N \subseteq M$ a submodule. Then there exists $c \geq 0$ such that $I^n M \cap N = I^{n-c}(I^c M \cap N)$ for all $n \geq c$.

Step 1: The Rees algebra.

Define the Rees algebra of $I$ as the graded ring $\mathcal{R} = \mathcal{R}(I) = \bigoplus_{n \geq 0} I^n = R \oplus I \oplus I^2 \oplus \cdots \subseteq R[t]$ where we identify $I^n$ with $I^n t^n \subseteq R[t]$ to make the grading explicit. If $I = (a_1, \ldots, a_r)$, then $\mathcal{R}$ is generated as an $R$-algebra by $a_1 t, \ldots, a_r t$ in degree $1$. Since $R$ is Noetherian, the Hilbert basis theorem implies $\mathcal{R}$ is Noetherian.

Step 2: Rees modules.

Define the graded $\mathcal{R}$-module $\mathcal{M} = \bigoplus_{n \geq 0} I^n M$ with the natural $\mathcal{R}$-module structure: $I^j \cdot I^n M \subseteq I^{j+n} M$. Since $M$ is finitely generated over $R$, the module $\mathcal{M}$ is finitely generated over $\mathcal{R}$ (if $M = Rm_1 + \cdots + Rm_s$, then $\mathcal{M}$ is generated by $m_1, \ldots, m_s$ in degree $0$).

Similarly, define the graded $\mathcal{R}$-submodule $\mathcal{N} = \bigoplus_{n \geq 0} (I^n M \cap N)$

Step 3: Finite generation of $\mathcal{N}$.

Since $\mathcal{R}$ is Noetherian and $\mathcal{M}$ is a finitely generated $\mathcal{R}$-module, every submodule of $\mathcal{M}$ is finitely generated. In particular, $\mathcal{N}$ is a finitely generated graded $\mathcal{R}$-module.

Let $\mathcal{N}$ be generated by homogeneous elements $n_1, \ldots, n_k$ of degrees $d_1, \ldots, d_k$ respectively. Set $c = \max(d_1, \ldots, d_k)$.

Step 4: The Artin-Rees conclusion.

For $n \geq c$, the degree-$n$ component of $\mathcal{N}$ is $I^n M \cap N = \mathcal{N}_n = \sum_{i=1}^k I^{n - d_i} \cdot n_i \subseteq I^{n-c} \cdot \mathcal{N}_c = I^{n-c}(I^c M \cap N)$

The reverse inclusion $I^{n-c}(I^c M \cap N) \subseteq I^n M \cap N$ is clear since $I^{n-c} \cdot I^c M \subseteq I^n M$ and $I^{n-c} \cdot N \subseteq N$.

Therefore $I^n M \cap N = I^{n-c}(I^c M \cap N)$ for all $n \geq c$. $\square$

Corollary (Krull Intersection Theorem): Let $L = \bigcap_{n=1}^\infty I^n M$. By the Artin-Rees lemma with $N = L$, there exists $c$ such that for all $n \geq c$: $I^n M \cap L = I^{n-c}(I^c M \cap L)$ Since $L \subseteq I^n M$ for all $n$, the left side is $L$. Also $I^c M \cap L = L$. So $L = I^{n-c} L$ for all $n \geq c$; in particular, $L = IL$. By Nakayama's lemma (applied to the finitely generated module $L$ over $R$): there exists $a \in I$ with $(1-a)L = 0$. $\square$