The Artin-Rees Lemma

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.

2 thoughts on “The Artin-Rees Lemma

  1. There is a typo in the proof of equivalence of finitely generatedness and stable: the index in the direct sum starts from k = 1.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s