Krull intersection theorem

Remarks

Let $N = \bigcap_n I^n M$. By the Artin–Rees lemma, $N = I^n M \cap N = I^{n-k}(I^k M \cap N) \subseteq I N$ for $n > k$, so $N = IN$. As $N$ is finitely generated, the determinant trick gives $a \in I$ with $(1 + a) N = 0$. Conversely, if $(1 + a) x = 0$ then $x = -a x = (-a)^n x \in I^n M$ for every $n$. Example: in $R = \Z/6\Z$ with $I = (2)$, one has $I^2 = I$, so the intersection is $I = \set{0, 2, 4}$, and indeed $3 = 1 + 2$ kills $2$ and $4$. The Noetherian hypothesis is needed. In the local ring of germs at $0$ of smooth functions $\R \to \R$, consider the germ of $e^{-1/x^2}$, extended by $0$ at $0$. It is nonzero, but it lies in every power of the maximal ideal.

Going deeper

For a student

Clause 1 shows that a Noetherian local ring embeds into its $\mathfrak m$-adic completion. It also shows that two elements of a Noetherian local ring are equal if they agree modulo every power of $\mathfrak m$. This is the basis of many arguments by successive approximation.