ON THE GROWTH OF NOETHERIAN FILTERED RINGS ^{∗}

Andrei Jaikin-Zapirain

Departamento de Matemáticas, Facultad de Ciencias Canto Blanco Ciudad Universitaria, 28049 Madrid, Spain e-mail: ajaikin@uam.es

Dmitri Piontkovski

Central Institute of Economics and Mathematics Nakhimovsky prosp. 47, Moscow 117418, Russia e-mail: piont@mccme.ru

May 21, 2002

Abstract The goal of this note is to show that associated graded ring of a Noetherian ring with a descending filtration, grows subexponentially. The same is true for completed group algebras of Noetherian pro-p groups and for group algebras of Noetherian groups which are residually a finite p-group.

1 Introduction

It is proved in [8] that Noetherian associative graded algebras have at most subexponential growth. It seems very reasonable that the condition on graded algebras to be Noetherian should imply some strong consequences. For example, it has been conjectured in [8] the following problem:

Conjecture 1. Let A be a Noetherian graded algebra. Then A has finite Gelfand-Kirillov dimension, i.e. A has polynomial growth.

The condition “Noetherian graded algebra” seems to be too strong for subexponential growth; however, if we omit the assumption that the algebra is graded, then the conclusion of the Stephenson–Zhang theorem does not hold. It is pointed out in [8] that the algebra S constructed in [6] is a Noetherian domain of exponential growth; in particular, for every ascending filtration on S, its associated graded algebra has the same growth and hence is not Noetherian.

∗The work of the first author has been partially supported by the Grant BFM2001-0180 of the Ministry of Science and Technology. The work of the second author has been partially supported by the grant 02-01-00468 of the Russian Basic Research Foundation.

Now, let us deal with descending filtrations. Our first result is the following generalization of the Stephenson–Zhang theorem:

Theorem 1.1. Let R be a right Noetherian ring with a descending ℤ–filtration

R = R_{−q} ≥ ⋯ ≥ R_{0} ≥ R_{1} ≥ …

such that the associated graded ring R̄ is an algebra over a field k with finitedimensional components. Then R̄ has subexponential growth.

Notice that the condition “R is Noetherian” does not imply the condition “R̄ is Noetherian”: there are examples of Noetherian complete local algebras R such that, for every descending ℕ–filtration on R, the algebra R̄ is not right Noetherian [7, Example 4.5]. Also, the conclusion of Theorem 1.1 can not be replaced by the statement “R itself has subexponential growth”: for example, if G is a polycyclic but not virtually nilpotent group and is residually a finite p-group, then, by [4, Theorem 11.9], G, and so, its group ring R = 𝔽_{p}[G], grows exponentially. However, R is Noetherian and it is separably filtered by the powers of its augmentation ideal.

In group theory, there are some conjectures similar to Conjecture 1. Let us consider a pro-p group G and let Λ = 𝔽_{p}[[G]] be its completed group algebra over the field 𝔽_{p}. It is known that if G has finite rank (i.e. there exists a number d such that every closed subgroup of G can be generated topologically by d elements) then the algebra Λ is Noetherian (see, for example, [11, Theorem 8.7.8]). Moreover, it has been conjectured in [3, p. 4983] that the result is also true in the reverse direction:

Conjecture 2. Let G be a pro-p group. If Λ is a Noetherian ring then G is of finite rank.

Recall that Λ has the unique maximal ideal Δ. The powers of this ideal form a separable filtration. Let Λ̄ be the graded ring associated with this filtration. This algebra plays an important role in the investigation of G. By a theorem of Quillen, we know that Λ̄ is isomorphic to the universal p-enveloping algebra of the Lie algebra L(G) = ⊕_{i≥1}D_{i}/D_{i+1}, where D_{i} = {g ∈ G | g − 1 ∈ Δ^{i}} is the i-th dimension subgroup of G. Theorem 12.1 of [1] says that G has finite rank if and only if Λ̄ grows polynomially. Hence Conjecture 2 is equivalent to

Conjecture 2′. Let G be a pro-p group. If Λ is a Noetherian ring then Λ̄ has finite Gelfand-Kirillov dimension.

Now let H be a group and R = 𝔽_{p}[H] its group ring over 𝔽_{p}. It is well-known that if H is polycyclic-by-finite then R is Noetherian. The converse implication is an open question. In the case when H is residually a finite p-group, it is equivalent to the following conjecture:

Conjecture 3. Let H be a group which is residually a finite p-group and suppose that R, its group ring over 𝔽_{p}, is Noetherian. Let R̄ = ⊕_{i≥0}Δ^{i}/Δ^{i+1}, where Δ = id {1 − h|h ∈ H} is the augmentation ideal of R. Then R̄ has finite Gelfand-Kirillov dimension.

Indeed, suppose that R̄ has finite Gelfand-Kirillov dimension and let G be a pro-p completion of H. Then R̄ ≅ Λ̄. Hence, as we have said before, G is of finite rank. From [1] we know that G is linear over the ring of p-adic numbers ℤ_{p}. Since H is residually a finite p-group, H is a subgroup of G and, so, it is also linear over ℤ_{p}. Using the Tits alternative [9], we obtain that either H is virtually soluble or contains a free non-abelian subgroup. The second case is impossible because, by [5, Lemma 10.2.2], 𝔽_{p}[N] is right Noetherian for every subgroup N of H. Hence H is virtually soluble. But in this case we can apply [5, Lemma 10.2.3] and conclude that H is polycyclic-by-finite.

On the other hand, when H is polycyclic-by-finite, its pro-p completion is of finite rank and we have that R̄ has polynomial growth. Notice that to show Conjecture 3 it is sufficient to prove that the growth of R̄ is less than [2^{\sqrt{n}}], see [2].

The parallels between Conjecture 1, Conjecture 2′ and Conjecture 3 have lead us to think about an analogous result of Stephenson-Zhang’s for Noetherian rings with descending filtrations (Theorem 1.1). This also leads us to grouptheoretic results which are analogous to Theorem 1.1.

Now, let G be a pro-p group and H a dense subgroup of G. Let N be a subgroup of H and N̄ its closure in G. We say that N is closed in H if N̄ ∩ H = N, that is if N is closed in the topology of H inherited from G.

Theorem 1.2. Let G be a pro-p group and H a dense subgroup of G. Suppose H satisfies the a.c.c. on closed subgroups in H. Then the growth of Λ̄ is subexponential.

Putting H = G, we obtain the following corollary:

Corollary 1.3. Let G be a Noetherian pro-p group (i.e. G satisfies a.c.c. on closed subgroups). Then the growth of Λ̄ is subexponential.

Since every group which is residually a finite p-group, is a dense subgroup of its pro-p completion, we have:

Corollary 1.4. Let H be a Noetherian group which is residually a finite p-group and R = 𝔽_{p}[H] its group ring over 𝔽_{p}. Then the growth of R̄ is subexponential.

Remark 1.5. Recall that if 𝔽_{p}[H] is right Noetherian then H is Noetherian, and if G is a pro-p group and 𝔽_{p}[[G]] is right Noetherian then G is Noetherian (as a pro-p group).

Remark 1.6. The conclusion of Theorem 1.2 is equivalent to the subexponential growth of the dimensions of D_{i}/D_{i−1} (see [1, Lemma D3]).

In Section 2 we give a technical lemma, which is a criterion for a graded ring associated to a ring filtration to have subexponential growth and in Sections 3 and 4 we prove our theorems. The notation is standard. We refer the reader to [1, 11] for background on pro-p groups and to [4, 10] for background on growth of algebras.

2 A criterion for subexponential growth of associated graded rings

Let R be an associative ring (with unit or without it) with a ℤ–filtration

R = R_{−q} ≥ ⋯ ≥ R_{0} ≥ R_{1} ≥ …

We assume that the associated graded ring R̄ admits a structure of algebra over a field k such that all the graded components are finite-dimensional kspaces. Our main examples are filtered k-algebras and noncommutative local rings filtered by the powers of the radical. Let us say that an element e ∈ R has a degree d if and only if e ∈ R_{d} but e ∉ R_{d+1}. For the zero element, we admit the equality deg 0 = i for every integer i. The following lemma describes those rings whose associated graded algebra has subexponential growth.

Lemma 2.1. For a ring R as described above, the following conditions are equivalent:

(i) the algebra R̄ grows subexponentially;

(ii) there exists a sequence of integers α = (α_{0}, α_{1}, …) such that

1) lim_{i→∞} α_{i} = ∞,

2) the sequence of coefficients of the generating function α(z) = ∑_{i} A_{i}z^{i}, where A_{k} = |{i | α_{i} = k}|, grows subexponentially, and

3) for every d ≥ 0, there is a sequence b^{(d)} = (b_{0}, b_{1}, …) of elements of R with sequence of degrees (α_{d}, α_{d+1}, …) such that the image Ī in R̄ of the right ideal I generated by b^{(d)} has finite codimension in R̄.

Proof. For the implication (i) ⟹ (ii), it is sufficient to take as α the sequence of degrees of the elements of any homogeneous k-linear basis of R̄. Then we can take as b_{i} ∈ b^{(d)} any pre-image in R of the (i + d)-th element of this basis.

Let us now prove (ii) ⟹ (i). We have to show that the radius of convergence r of the Hilbert series R̄(z) is equal to 1 [10, 4]. Fix ε > 0; we will see that r ≥ 1 − ε . Let b^{(d)} be the sequence described in (ii), I the right ideal in R generated by b^{(d)} and Ī its image in R̄. Without loss of generality we can assume that the zero component R_{0} includes the unit of R. For the dimension of the i-th component of Ī (here i > 0), we have the inequalities

dim Ī_{i} = dim(I ∩ R_{i}/I ∩ R_{i+1})
≤ dim ((b_{0}R + b_{1}R + …)/(b_{0}R_{i+1−α_{d}} + b_{1}R_{i+1−α_{d+1}} + …))
≤ dim ((b_{0}R + b_{1}R_{i+1−α_{d+1}} + …)/(b_{0}R_{i+1−α_{d}} + b_{1}R_{i+1−α_{d+1}} + …)) +

dim ((b_{0}R_{i+1−α_{d}} + b_{1}R + …)/(b_{0}R_{i+1−α_{d}} + b_{1}R_{i+1−α_{d+1}} + …)) + …

≤ dim(b_{0}R/b_{0}R_{i+1−α_{d}}) + dim(b_{1}R/b_{1}R_{i+1−α_{d+1}}) + …
≤ dim(R/R_{i+1−α_{d}}) + dim(R/R_{i+1−α_{d+1}}) + …

(here the last sum is finite since q < ∞). Denoting a_{i} = dim R_{i}/R_{i+1}, we have

dim Ī_{i} ≤ ∑_{j≤i−α_{d}} a_{j} + ∑_{j≤i−α_{d+1}} a_{j} + ⋯ = ∑_{s≥0} ∑_{j≤i−α_{d+s}} a_{j}.

Thus we have the coefficient-wise inequality

Ī(z) ≤ ∑_{i∈ℤ} z^{i} ∑_{s≥0} ∑_{j≤i−α_{d+s}} a_{j} = ∑_{j} a_{j} ∑_{s≥0} ∑_{i≥j+α_{d+s}} z^{i}
= ∑_{j} a_{j}z^{j} ∑_{s≥0} z^{α_{d+s}} ∑_{i≥0} z^{i} = R̄(z)b^{(d)}(z)/(1 − z).

(We use the obvious notation b^{(d)}(z) = ∑_{k}(|{i ≥ d | α_{i} = k}|)z^{k}.) Since

R̄(z) = Ī(z) + p(z),

where p(z) = (R̄/Ī)(z) is a polynomial, we obtain the coefficient-wise inequality

R̄(z)f_{d}(z) ≤ p(z),

where f_{d}(z) = 1 − b^{(d)}(z)/(1 − z). Now, the coefficients of α(z) grow subexponentially, whence the series α(z) converges for all z for |z| ≤ 1 − ε. It follows that for sufficiently large d we have b^{(d)}(z) < 1 − z, or, equivalently, f_{d}(z) > 0 for all 0 < z ≤ 1 − ε . For such d, we have R̄(z) ≤ p(z)/f_{d}(z) < ∞ for all 0 < z ≤ 1 − ε, so r ≥ 1 − ε.

3 Noetherian filtered rings

Proof of Theorem 1.1. In the notation of Lemma 2.1, let α = (0, 1, 2, …). For every d > 0, let b_{0} ∈ b^{(d)} be any element of degree d (nonzero, if possible). Recursively, let b_{i} ∈ b^{(d)} be any element of degree d + i which does not lie in b_{0}R + ⋯ + b_{i−1}R (if such an element does not exists, put b_{i} = 0). By the ascending chain condition for the sequence of right ideals

b_{0}R ⊆ b_{0}R + b_{1}R ⊆ …,

we have that, for n >> 0, all elements of degree n lie in ∑_{i≥0} b_{i}R. This means that dim R/(∑_{i≥0} b_{i}R) < ∞. Now, it only remains to apply Lemma 2.1.

4 Noetherian groups and pro-p groups

In this section we use the same notation as in Introduction. So, if G is a pro-p group, then Λ = 𝔽_{p}[[G]] is its completed group algebra over the field 𝔽_{p} and Δ is the maximal ideal of Λ.

Proof of Theorem 1.2. Recall that the filtration on Λ induces a filtration on H by the subgroups H_{n} = {h ∈ H|h − 1 ∈ Δ^{n}}. In order to apply Lemma 2.1, put α = (1, 2, …). For d > 0, let h_{0} = 1 ∈ H and let b_{0} = 0. Recursively, for i > 0 let h_{i} be any element of H_{d+i} \ H_{d+i+1} such that h_{i} − 1 does not lie in the right ideal I_{i−1} of Λ generated by b_{0}, …, b_{i−1} (if such an element does not exist, set h_{i} = 1) and let b_{i} = h_{i} − 1. Since I_{i} is finitely generated, I_{i} is closed in Λ and, so, T_{i} = {h ∈ H|h − 1 ∈ I_{i}} is closed in H. By the a.c.c. for the chain of closed subgroups in H

T_{0} ⊆ T_{1} ⊆ …,

we obtain that there is n > 0 such that for i > n, h_{i} ∈ T_{n}. It follows that T_{n} is a subgroup of finite index in H and so, the right ideal I_{n} has finite codimension in Λ.

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