The Artin-Rees Lemma

November 2, 2009 by hilbertthm90 | 2 Comments

We have a somewhat bumpy road to traverse today. I’ll start with the Artin-Rees lemma and see if we can get to a use of it to continue our set of inequalities we’re trying to prove.

First we’ll need some new ideas. Suppose $R$ is any old ring (in particular, we are dropping graded and Noetherian assumptions). Then if $\frak{a}$ is an ideal, we can form a new ring $R^*=\bigoplus_{n=0}^\infty \frak{a}^n$ which by construction is graded. Now for any $R$ -module, say $M$ and an $\frak{a}$ -filtration $M_n$ we can form a graded $R^*$ -module, $M^*=\bigoplus M_n$ .

Note that if $R$ is Noetherian in the situation above, then $\frak{a}=(x_1, \ldots, x_r)$ , so $R^*=R[x_1, \ldots , x_r]$ , so by Hilbert Basis Theorem, we get $R^*$ is Noetherian.

We’ll need that in the situation above the following two statements are equivalent: $M^*$ is finitely generated as an $R^*$ -module, and that the filtration $M_n$ is stable.

Proof: Each $M_n$ is finitely generated, so $Q_n=\bigoplus_{r=0}^n M_r$ is finitely generated for all $n$ . Let’s form $M_n^*=Q_n\oplus\left(\bigoplus_{k=1}^\infty \frak{a}^kM_n\right)$ . We have that each $Q_n$ is finitely generated as an $R$ -module, so we get that $M_n^*$ is finitely generated as an $R^*$ -module.

Clearly, $M_0^*\subset M_1^*\subset \cdots$ , so since $R^*$ is Noetherian we get that $M^*$ is finitely generated iff the ascending chain terminates iff $M_{n_0+r}=\frak{a}^r M_{n_0}$ for some $n_0$ and for all $r\geq 0$ iff the filtration is stable.

Now we can prove the Artin-Rees Lemma which says that if $R$ is a Noetherian ring, $\frak{a}$ an ideal, $M$ a finitely generated $R$ -module, $M_n$ a stable $\frak{a}$ -filtration and $M'$ a submodule of $M$ , then $M'\cap M_n$ is a stable $\frak{a}$ -filtration of $M'$ .

The situation is fairly simple from the previous fact. Note that $\frak{a}(M'\cap M_n)\subset \frak{a}M'\cap \frak{a}M_n\subset M'\cap M_{n+1}$ . So we do indeed get a filtration. But $M'^*$ is a graded $A^*$ -submodule of $M^*$ , so it is finitely generated. Now by the equivalence of finitely generated and stable we are done.

There are two important corollaries (both get referred to as the Artin-Rees Lemma as well). In the special case $M_n=\frak{a}^nM$ we get that the stable filtration condition says that there is some integer $N$ such that $(\frak{a}^nM)\cap M'=\frak{a}^{n-N}((\frak{a}^NM)\cap M')$ for all $n\geq N$ .

The other result uses the bounded difference result from last time. Since $\frak{a}^nM'$ and $(\frak{a}^nM)\cap M'$ are both stable $\frak{a}$ -filtrations, they have bounded difference, so the $\frak{a}$ -topology of $M'$ coincides with induced topology from the $\frak{a}$ -topology on $M$ .

I think that is sufficient for today. Next time I’ll go ahead and knock off the next step of the inequalities: $d(R)\geq \dim R$ .

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

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The Artin-Rees Lemma

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