## Topologies and the Artin-Rees lemma August 19, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
Tags: , , , ,

Today I’ll continue the series on graded rings and filtrations by discussing the resulting topologies and the Artin-Rees lemma.

All filtrations henceforth are descending.

Topologies

Recall that a topological group is a topological space with a group structure in which the group operations of composition and inversion are continuous—in other words, a group object in the category of topological spaces.

So, let’s now consider a filtered module ${M}$ over some filtered ring ${A}$ with a descending filtration. Recall that this means that we have descending subgroups ${M_i, A_i}$ such that for all ${i,j}$, ${A_i M_j \subset M_{i+j}}$.

We can actually generalize this as follows:

Definition 1 A filtered abelian group is an abelian group ${G}$ with a sequence of subgroups ${G_i}$ such that ${G_0=G}$, ${G_i \supset G_j}$ for ${i>j}$.

(Alternatively this is a filtered module over ${A=\mathbb{Z}}$, with the filtration ${A_0 = \mathbb{Z}, A_i=0}$ for ${i>0}$.)

Given a filtered additive commutative group ${G}$, we can give it a topology by taking a base at ${x \in G}$ to be ${x + G_i}$. It can directly be checked that addition and inversion are continuous operations since the ${G_i}$ are subgroups: for instance, the inverse image of ${x+y+G_i}$ under the addition map ${G \times G \rightarrow G}$ contains the open set ${(x+G_i) \times (y + G_i)}$.

In general, however, the topology will be Hausdorff if and only if

$\displaystyle \bigcap_i G_i = 0;$

if this is true, then given ${x \neq 0}$, we can find some ${G_N}$ with ${x \notin G_N}$ so ${G}$ is Hausdorff, and the argument can be reversed.

Now return to the case of a filtered module ${M}$ over a filtered ring ${A}$. As above, ${M}$ and ${A}$ have topologies. Moreover, the condition ${A_i M_j \subset M_{i+j}}$ implies for instance that the map

$\displaystyle A \times M \rightarrow M, \ (a,m) \rightarrow am$

is continuous as well.

The result I am aiming for today is the Artin-Rees lemma, which states under suitable conditions, if you pick an ideal ${I}$ in a ring ${A}$, ${A}$-modules ${N \subset M}$, then the ${I}$-adic topology on ${N}$ (from the filtration ${I^iN}$) is induced by the ${I}$-adic topology on ${M}$. But first, we need to talk more about a variant of ${\mathrm{gr} }$.

The Artin-Rees Lemma

The variant of ${\mathrm{gr} }$ I was talking about is the blowup algebra or Rees algebra defined as

$\displaystyle B(A) := A_0 \oplus A_1 \oplus \dots ,$

where multiplication of ${a \in A_n, b \in A_m}$ is ${ab \in A_{n+m}}$. This is now a graded algebra. Similarly, given a filtered module ${M}$, we can make ${B(M) := \bigoplus M_n}$ into a graded module over the blowup algebra. Then ${B}$ is functorial, from the category of filtered rings to the category of graded rings (resp. from the category of filtered modules to the category of graded modules over ${B(A)}$).

The case we care about most is with the filtration ${I^j}$ or ${I^jM}$. In this case we denote ${B(A)}$ by ${B_I(A)}$, following Eisenbud.

Proposition 2 If ${A}$ is Noetherian and ${I}$ an ideal, the blowup algebra ${B_I(A)}$ is Noetherian.

Indeed, the algebra is ${B_I(A) = A \oplus I \oplus I^2 \oplus \dots}$; then if ${x_1, \dots, x_n}$ generate ${I}$ as an ideal, it follows that ${x_1, \dots, x_n}$ generate ${B_I(A)}$ as an ${A}$-algebra, so the blowup algebra is Noetherian by Hilbert’s basis theorem.

Now let ${M}$ be a filtered ${A}$-module, where ${A}$ is given the ${I}$-adic filtration.

Proposition 3 Let ${A}$ be a Noetherian ring. Give ${A}$ the ${I}$-adic filtration, and let ${M}$ be a finitely generated and filtered ${A}$-module. The ${B_I(A)}$-module ${B(M)}$ is finitely generated if and only if the filtration ${M_n}$ satisfies ${IM_n = M_{n+1}}$ for ${n}$ sufficiently large (this means the filtration is ${I}$-stable; recall that we already have the inclusion ${\subset}$).

If the filtration ${\{M_i\}}$ is ${I}$-stable, then let ${S}$ be large enough that ${n \geq S}$ implies

$\displaystyle IM_n = M_{n+1}. \ \ \ \ \ (1)$

Then choose generating sets ${G_0 \subset M_0, G_1 \subset M_1, \dots, G_S \subset M_S}$ such that each ${G_i}$ generates ${M_i}$ as an ${A}$-module. Then consider the generating set ${G \subset B(M)}$ consisting of all the ${G_i}$ in the ${i}$-th homogeneous slot. Then ${G}$ generates ${M_0 \oplus \dots \oplus M_S}$ as an ${A}$-module, hence as an ${B_I(A)}$-module. By (1), ${G}$ generates ${M}$ as an ${B_I(A)}$-module.

Conversely, if ${B(M)}$ is finitely generated, then some subset

$\displaystyle M_0 \oplus \dots \oplus M_S \subset B(M)$

generates ${B(M)}$ as an ${B_I(A)}$-module; by the definition of ${B_I(A)}$, it follows that ${M_{S+1} \supset I^S M_0 + I^{S-1} M_1 + \dots + IM_{S} \supset IM_N}$, and similarly it follows more generally that ${M_{n+1} = IM_n}$ for ${n>S}$.

Now we move to proving the Artin-Rees lemma.

Theorem 4 (Artin-Rees) Let ${A}$ be a Noetherian ring, ${I}$ an ideal, ${M}$ a finitely generated ${A}$-module, and ${N}$ a submodule. If ${\{M_i\}}$ is an ${I}$-stable filtration on ${M}$, then ${\{M_i \cap N\}}$ is ${I}$-stable too.

Indeed, considering ${M, N}$ as filtered ${A}$-modules (where ${A}$ has the ${I}$-adic filtration), we find ${B(M)}$ is finitely generated over ${B_I(A)}$; thus ${B(N) \subset B(M)}$ is too, and the filtration on ${N}$ (that is ${\{M_i \cap N\}}$) is thus ${I}$-stable.

Corollary 5 The ${I}$-adic topology on ${N}$ is that induced by the ${I}$-adic topology on ${M}$.

The topology on ${N}$ induced by the ${I}$-adic topology on ${M}$ can be described by intersecting a basis at zero for ${M}$ with ${M}$: that is, we take ${\{I^iM \cap N\}}$ as a basis at ${0}$. We have to show this basis is equivalent to ${\{I^i N\}}$.

By the previous theorem, the filtration ${\{I^iM \cap N\}}$ is ${I}$-stable. Thus for large ${i}$, we have

$\displaystyle I^{2i} \cap N = I^i(I^i M \cap N) \subset I^i N.$

Thus the basis ${\{ I^i M \cap N\}}$ is finer than ${\{I^i N\}}$. Conversely, the basis ${\{I^iN\}}$ is indexwise smaller so finer than ${\{I^i M \cap N\}}$.

There is a nice consequence of this:

Corollary 6 (Krull Intersection Theorem)

Let ${A}$ be a Noetherian local ring with maximal ideal ${\mathfrak{m}}$. Then if ${M}$ is a finitely generated ${A}$-module

$\displaystyle \bigcap_i \mathfrak{m}^i M = 0.$

Let ${N}$ be the submodule given by the intersection. Then the ${\mathfrak{m}}$-adic topology on ${N}$ is given by restriction of the ${\mathfrak{m}}$-adic topology on ${M}$. But each element of the basis at zero in ${M}$ contains ${N}$, so the basis at zero for ${N}$ is just ${N}$ itself, i.e. ${N}$ has the indiscrete ${\mathfrak{m}}$-adic topology. This means just ${N = \mathfrak{m}N}$, so by Nakayama we have ${N = 0}$.

Next I would like to talk about completions. Artin-Rees plays a key role in showing that completion is an exact functor.