Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 6 / b

A stable $I$-filtration of $M$ is a descending sequence $M=M_0\supseteq M_1\supseteq \cdots$ with $I{M}_n\subseteq M_{n+1}$ for all $n$ and equality for all sufficiently large $n$. The Rees ring and associated Rees module areThe filtration condition makes multiplication by $I^r{t}^r$ send $M_n{t}^n$ into $M_{n+r}{t}^{n+r}$, so $M^{\ast }$ is a graded $R^{\ast }$-module.

If the filtration is stable from degree $\ell$, then $M^{\ast }$ is generated over $R^{\ast }$ by finite generating sets for $M_0,\dots ,M_{\ell }$. Conversely, let homogeneous elements of degrees at most $\ell$ generate $M^{\ast }$. In every degree $n>\ell$, each expression for an element of $M_n$ uses a positive-degree coefficient from $R^{\ast }$, so $M_n=I{M}_{n-1}$. Hence finite generation is equivalent to stability.

For $M^{\prime }\subseteq M$, takeThen $M^{\ast }$ is a graded submodule of the finite Rees module $⨁I^{n}M$. Since $R$ is Noetherian and $I$ is finitely generated, $R^{\ast }$ is Noetherian; hence $M^{\ast }$ is finite and the filtration is stable. Therefore, for some $\ell$ and all $n\ge \ell$,This is the Artin-Rees lemma.

Let $N={\bigcap }_{n\ge 0}{I}^{n}M$. Apply Artin--Rees to $N\subseteq M$. Since $N\subseteq I^{n}M$ for every $n$, stability givesfor all sufficiently large $n$, and in particular $N=IN$. The module $N$ is finitely generated because $M$ is Noetherian. The determinant trick applied to a finite generating set of $N=IN$ produces $a\in I$ withTaking $x=-a\in I$ yieldswhich is Krull intersection theorem.