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Topologies and the Artin-Rees lemma August 19, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
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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.

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Comments»

1. Completions of rings and modules « Delta Epsilons - August 25, 2009

[...] of all, I claim that . This follows from the Artin-Rees lemma: the projective limit is the completion of with respect to the topology from the filtration . But [...]

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[...] nonzero as for a unit, because for some —this is the Krull intersection theorem (Cor. 6 here). Thus from this representation is a domain, and the result is then [...]


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