FLAT MODULES OVER GROUP RINGS OF FINITE GROUPS

D. J. BENSON

Abstract. Let k be a commutative ring of coefficients and G be a finite group. Does there exist a flat kG-module which is projective as a kmodule but not as a kG-module? We relate this question to the question of existence of a k-module which is flat and periodic but not projective. For either question to have a positive answer, it is at least necessary to have |k| ≥ ℵ_{ω}. There can be no such example if k is Noetherian of finite Krull dimension, or if k is perfect.

1. Introduction

In the course of work on categories of modules for Kropholler’s class of LH𝔉-groups [2, 3], the author was led to the following question. If k is a commutative ring of coefficients and G is a finite group, can there exist a nonprojective flat kG-module which is projective as a k-module? This paper relates this question to the question of finding periodic flat k-modules which are not projective. We investigate some conditions on k which ensure that no such examples exist. Some of the results in Sections 3 and 4 of this paper have been obtained independently by Ed Alcock as part of his Ph. D. thesis under the supervision of Peter Kropholler.

2. Cyclic groups

We begin by reducing the question to the case where G is cyclic. By Chouinard’s theorem [5], a kG-module is projective if and only if its restriction to every elementary abelian subgroup is projective. So the question reduces immediately to the case where G is an elementary abelian group (ℤ/p)^{r}. Moreover, by replacing the coefficient ring k by k(ℤ/p)^{r−1}, it reduces to the case where G ≅ ℤ/p.

In this case, we have the following criterion for flatness. We deal with the case p = 2 first because it is easier.

Theorem 2.1. Suppose that k is a commutative ring of coefficients and let ℤ/2 = ⟨T | T^{2} = 1⟩. A k(ℤ/2)-module M is flat if and only if:

(i) Ker(T − 1) = Im(T + 1) and Ker(T + 1) = Im(T − 1), and

(ii) M/(T − 1)M and M/(T + 1)M are flat as k-modules.

The author is partly supported by a grant from the NSF.

Proof. First suppose that M is a flat k(ℤ/2)-module. Let k_{+} and k_{−} denote the k(ℤ/2)-modules whose underlying k-module is k itself, and with T acting via multiplication by 1 and −1 respectively. The standard resolution of k_{+} as a k(ℤ/2)-module is an exact sequence

⋯ ⟶^{T −1} k(ℤ/2) ⟶^{T +1} k(ℤ/2) ⟶^{T −1} k(ℤ/2) → k_{+} → 0

where the kernels are alternately copies of k_{−} and k_{+}. Tensoring with M, we obtain an exact sequence

⋯ ⟶^{T −1} M ⟶^{T +1} M ⟶^{T −1} M → M/(T − 1)M → 0,

and so (i) holds. Since M/(T − 1)M ≅ M ⊗_{k(ℤ/2)} k_{+} and M/(T + 1)M ≅ M ⊗_{k(ℤ/2)} k_{−}, the associativity of tensor product shows that (ii) holds.

Conversely, suppose that M is a k(ℤ/2)-module which is projective as a kmodule and satisfies (i) and (ii). We must prove that for all k(ℤ/2)-modules N we have Tor_{1}^{k(ℤ/2)}(M, N) = 0. We proceed in a number of stages. First suppose that N = k_{+}. Then tensoring M with the standard resolution of k_{+} as above we obtain the sequence

⋯ ⟶^{T +1} M ⟶^{T −1} M → M/(T − 1)M → 0.

By condition (i), this sequence is exact, and so Tor_{1}^{k(ℤ/2)}(M, k_{+}) = 0. Similarly, tensoring M with the standard resolution

⋯ ⟶^{T −1} k(ℤ/2) ⟶^{T +1} k(ℤ/2) → k_{−} → 0

of k_{−}, we obtain

⋯ ⟶^{T −1} M ⟶^{T +1} M → M/(T + 1)M → 0.

Again applying condition (i) we obtain Tor_{1}^{k(ℤ/2)}(M, k_{−}) = 0.

Next, by taking direct sums and direct summands, we find that

Tor_{1}^{k(ℤ/2)}(M, N) = 0

for any k(ℤ/2)-module N which is projective as a k-module and on which T acts via multiplication by ±1.

If N is a k(ℤ/2)-module which is not necessarily projective as a k-module, but on which T acts via multiplication by ±1, then we begin a projective resolution

0 → N′ → P → N → 0

as a k-module, and let T act in the same way on P and N′ to obtain a short exact sequence of k(ℤ/2)-modules. Tensoring M with this sequence, we obtain a sequence

[figure: Tor_{1}^{k(ℤ/2)}(M, P); Tor_{1}^{k(ℤ/2)}(M, N); M ⊗_{k(ℤ/2)} N′; M ⊗_{k(ℤ/2)} P; 0; \frac{M}{(1∓T)M} ⊗_{k} N′; \frac{M}{(1∓T)M} ⊗_{k} P]

By condition (ii), the right hand map is injective and so again we have Tor_{1}^{k(ℤ/2)}(M, N) = 0.

Finally, for an arbitrary k(ℤ/2)-module N we have a short exact sequence

0 → (T − 1)N → N → \frac{N}{(T −1)N} → 0.

The action of T on the right hand term is trivial, and the action on the left hand term is via multiplication by −1. So the sequence

Tor_{1}^{k(ℤ/2)}(M, (T − 1)N) → Tor_{1}^{k(ℤ/2)}(M, N) → Tor_{1}^{k(ℤ/2)}(M, \frac{N}{(T −1)N})

shows that Tor_{1}^{k(ℤ/2)}(M, N) = 0, and so M is flat.

The reason why the case p = 2 is easier than p odd is that the equation x^{2} − 1 = 0 already splits into linear factors in k. For p odd, we have to work under the assumption that the equation

x^{p−1} + x^{p−2} + ⋯ + 1 = 0

has a solution ε in k. We do not preclude here the case where p = 0 in k so that ε = 1 is a solution. If such an ε exists, then x^{p} − 1 splits into linear factors as follows:

x^{p} − 1 = (x − 1)(x − ε)(x − ε^{2}) … (x − ε^{p−1}).

Theorem 2.2. Suppose that k is a commutative ring of coefficients in which the equation

x^{p−1} + x^{p−2} + ⋯ + x + 1 = 0

has a solution ε, and let ℤ/p = ⟨T | T^{p} = 1⟩. A k(ℤ/p)-module M is flat if and only if for 0 ≤ i ≤ p − 1 we have:

(i) Ker(T − ε^{i}) = Im(T^{p−1} + ε^{i}T^{p−2} + ε^{2i}T^{p−3} + ⋯ + ε^{(p−1)i}) and Im(T − ε^{i}) = Ker(T^{p−1} + ε^{i}T^{p−2} + ε^{2i}T^{p−3} + ⋯ + ε^{(p−1)i}), and

(ii) M/(T − ε^{i})M is flat as a k-module.

Proof. Let k_{i} denote the k(ℤ/p)-module whose underlying k-module is k itself, and where T acts via multiplication by ε^{i}. Then the standard resolution of k as a k(ℤ/p)-module is

… ⟶^{T −ε^{i}} k(ℤ/p) ⟶^{T^{p−1}+ε^{i}T^{p−2}+···+1} k(ℤ/p) ⟶^{T −ε^{i}} k(ℤ/p) → k_{i} → 0.

So the first part of the proof works just as in the case p = 2.

For the last part of the proof, we have the following equation in k(ℤ/p):

(T − 1)(T − ε)(T − ε^{2}) … (T − ε^{p−1}) = 0.

So if N is any k(ℤ/p)-module, then there is a filtration

F^{0}N = N
F^{1}N = (T − 1)N
F^{i}N = (T − 1)(T − ε) … (T − ε^{i−1})N  (2 ≤ i ≤ p − 1)
F^{p}N = 0.

The generator T acts as multiplication by ε^{i} on the filtered quotient

F^{i}N/F^{i+1}N 0 ≤ i ≤ p − 1.

So we inductively prove that Tor_{1}^{k(ℤ/p)}(M, N/F^{i}N) = 0.

We can remove the assumtion on k by passing to the extension ring k′ = k[ε]/(ε^{p−1} + ε^{p−2} + ⋯ + 1), which is free as a k-module of rank p − 1 with k-basis 1, ε, …, ε^{p−2}. So tensoring over k with k′ is exact, and the k′-module k′ ⊗_{k} M is flat if and only if the k-module M is flat.

Theorem 2.3. Suppose that k is a commutative ring of coefficients and let ℤ/p = ⟨T | T^{p} = 1⟩. A k(ℤ/p)-module M is flat if and only if

(i) Ker(T − 1) = Im(T^{p−1} + T^{p−2} + ⋯ + 1) and Im(T − 1) = Ker(T^{p−1} + T^{p−2} + ⋯ + 1),

(ii) M/(T − 1)M is flat as a k-module, and

(iii) M/(T^{p−1} + ⋯ + 1)M is flat as a k[ε]/(ε^{p−1} + ε^{p−2} + ⋯ + 1)-module (with multiplication by ε acting via T).

Proof. Let k′ = k[ε]/(ε^{p−1} +ε^{p−2} +⋯+1) as above. On k′ ⊗_{k} M, Ker(T −ε^{i}) (1 ≤ i ≤ p − 1) consists of elements of the form

x + (1 − T^{−i})(x)ε + (1 + T^{−i} + T^{−2i})(x)ε^{2} +

⋯ + (1 + T^{−i} + ⋯ + T^{−(p−2)i})(x)ε^{p−2}

where x is an element of M satisfying (T^{p−1} + T^{p−2} + ⋯ + 1)(x) = 0. On the other hand, Im(T^{p−1} + ε^{i}T^{p−2} + ⋯ + ε^{(p−1)i}) consists of elements of the form

(1 − T^{−i})(y) + (1 − T^{−2i})(y)ε + ⋯ + (1 − T^{i})(x)ε^{p−2}.

So the first part of condition (i) of Theorem 2.2 on k′ ⊗_{k} M can be rewritten as

Im(T^{i} − 1) = Ker(T^{p−1} + T^{p−2} + ⋯ + 1)

on M. Since T and T^{i} are powers of each other, we have Im(T^{i} − 1) = Im(T − 1).

Similarly, the other part of condition (i) of Theorem 2.2 translates to

Ker(T − 1) = Ker(T^{i} − 1) = Im(T^{p−1} + T^{p−2} + ⋯ + 1),

and condition (ii) of Theorem 2.2 translates to conditions (ii) and (iii).

Theorem 2.4. Let k be a commutative ring of coefficients of characteristic p, in the sense that p = 0 in k, and let ℤ/p = ⟨T | T^{p} = 1⟩. Then the following are equivalent:

(i) There exists a nonprojective flat k(ℤ/p)-module N which is projective as a k-module.

(ii) There exists a nonprojective flat k-module M and a short exact sequence

0 → M → P → M → 0

with P a projective k-module.

Proof. We begin with the case p = 2 since this is simpler than the odd case. In this case, if N is a k(ℤ/2)-module satisfying (i) then let P be the underlying k-vector space of N and let M = (T +1)N. Then by Theorem 2.1, M is a flat k-module, and by the first isomorphism theorem and the fact that +1 = −1 in k, we have

P/M = N/(T + 1)N = N/Ker(T + 1) ≅ Im(T + 1) = M.

So we have a short exact sequence of the type described in (ii). If M were a projective k-module then the sequence would split and we would have N ≅ k(ℤ/2) ⊗_{k} M, and N would be a projective k(ℤ/2)-module. So M is a nonprojective k-module.

Conversely, given a sequence of the type described in (ii), P has an endomorphism θ given by composing P → M → P. We have θ^{2} = 0, and so we may make a k(ℤ/2)-module N with underlying k-module P and where T acts via 1 + θ. It is easy to check that the conditions of Theorem 2.1 hold, and so N is a module satisfying (i).

For the odd case, we have to work harder. First, note that since we are in characteristic p, the identity element ε = 1 of k satisfies the equation

x^{p−1} + x^{p−2} + ⋯ + x + 1 = 0

and so we are in a position to apply Theorem 2.2. Moreover, we have

T^{p−1} + T^{p−2} + ⋯ + T + 1 = (T − 1)^{p−1}.

If N is a k(ℤ/p)-module satisfying (i), then we have

(T − 1)N = Ker(T − 1)^{p−1}
(T − 1)^{p−1}N = Ker(T − 1).

So we set M = (T − 1)N ⊕ (T − 1)^{p−1}N and P equal to a direct sum of two copies of the underlying k-module of N to produce a sequence of the type described in (ii). By part (ii) of Theorem 2.2, N is flat as a k-module. If (T −1)N were a projective k-module, then so would (T −1)^{p−1}N be. Then P would be isomorphic to a direct sum of p copies of (T −1)^{p−1}N as a k-module, and N would be isomorphic to k(ℤ/p) ⊗_{k} (T − 1)^{p−1}N as a k(ℤ/p)-module, which is projective. So (T − 1)N is flat but not projective as a k-module.

Conversely, given a sequence of the type described in (ii), let θ be the endomorphism of P described above. Then we let N be the k(ℤ/p)-module whose underlying k-module is a direct sum of p copies of P, and where T acts through the matrix

(\begin{matrix}1 & & 1 & & 0 & & … & & & 0 & & 0 \\ 0 & & 1 & & 1 & & & & & 0 & & 0 \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 0 & & 0 & & 0 & & … & & & 1 & & 0 \\ 0 & & 0 & & 0 & & & & & 1 & & θ \\ θ & & 0 & & 0 & & … & & & 0 & & 1 \end{matrix})

The kernel of (T − 1)^{i} and the image of (T − 1)^{p−i} on N both consist of the column vectors in which the ith and pth coordinates lie in M, and the ones between them are zero. In particular, we have Ker(T − 1) = Im(T − 1)^{p−1} and Ker(T − 1)^{p−1} = Im(T − 1), and the quotient N/(T − 1)N ≅ M ⊕ M is a flat k-module. So by Theorem 2.2, N is a flat k(ℤ/p)-module. If it were projective then M ⊕ M ≅ k ⊗_{k(ℤ/p)}N would be a projective k-module, which it is not.

Note that for p odd, the processes we have described are not mutually inverse, as they were for p = 2.

3. Cardinality

In this section, we make some remarks on cardinality conditions on k and M which ensure that a flat kG-module which is projective as a k-module is projective as a kG-module. These rely on the following proposition.

Proposition 3.1. If M is a kG-module of finite projective dimension which is projective as a k-module then M is projective as a kG-module.

Proof. A proof of this statement may be found in Lemma 2.3 of [4].

Proposition 3.2. Over a ring of cardinality ℵ_{t}, every flat module has projective dimension at most t + 1.

Proof. This follows from Gruson and Jensen [7, 8], see also Theorem 7.47 of [9], using the fact that for flat modules the pure projective dimension is equal to the projective dimension.

Proposition 3.3. Over any ring, every ℵ_{t}-presented flat module has projective dimension at most t + 1.

Proof. This can be found as Theorem 2.45 in Osofsky [10].

Theorem 3.4. Let k be a commutative ring and G be a finite group, and let M be a flat kG-module which is projective as a k-module. If either k has cardinality ℵ_{t} or M is ℵ_{t}-presented with t < ω then M is projective as a kG-module.

Proof. This follows from Propositions 3.1, 3.2 and 3.3.

A good candidate for a ring k providing examples of kG-modules which are flat and k-projective but not kG-projective might be a ring of functions from a set of cardinality at least ℵ_{ω} to a field of characteristic p. Such a ring is Von Neumann regular, and so every module is automatically flat. The construction of a suitable module M would then involve some combinatorics of subsets. Such a construction has eluded the author of this paper. Another possible candidate for k would be the group ring of an elementary abelian p-group of cardinality at least ℵ_{ω}, over a field of characteristic p. In both cases it is known that there are flat modules of infinite projective dimension, which is at least a prerequisite.

4. Other conditions on k

In this section, we investigate some other conditions on k which ensure that a flat kG-module which is projective as a k-module is projective as a kG-module.

Proposition 4.1. If k is a Noetherian ring with Krull dimension d then every flat k-module has projective dimension at most d.

Proof. This is proved in 3.2.7 of Raynaud and Gruson [11]. See also proposition 3.1 of Enochs [6].

Theorem 4.2. Let k be a commutative Noetherian ring with finite Krull dimension and G be a finite group. If M is a flat kG-module which is projective as a k-module then M is projective as a kG-module.

Proof. By Chouinard’s theorem [5], it suffices to consider the case where G = E is elementary abelian. In this case, kE is also a commutative Noetherian ring with the same Krull dimension, by the going-up theorem. The result now follows from Propositions 3.1 and 4.1.

Recall from Bass [1] that a ring k is said to be left perfect if every left k-module has a projective cover. An ideal I in k is said to be left T -nilpotent if given any sequence a_{1}, a_{2}, … of elements of I, there exists n such that a_{1} … a_{n} = 0.

Proposition 4.3. The following conditions on a ring k are equivalent: (a) k is left perfect.

(b) The Jacobson radical J(k) is left T -nilpotent and k/J(k) is a semisimple Artinian ring.

(c) Every flat (left) k-module is projective.

Proof. See Bass [1].

Proposition 4.4. Let k be a left perfect ring and G be a finite group. Then kG is left perfect.

Proof. We embed kG in End_{k}(kG)^{op} ≅ Mat_{n}(k) via right multiplication, where n is the order of G. Since it follows from the definition that the property of being left perfect is Morita invariant, it follows that J(Mat_{n}(k)) = Mat_{n}(J(k)) is left T -nilpotent. So kG ∩ Mat_{n}(J(k)) = J(k)G is left T - nilpotent. Now J(k)G ⊆ J(kG), and kG/J(k)G ≅ (k/J(k))G is left Artinian since k/J(k) is Artinian, and so kG/J(kG) is semisimple Artinian. It follows that J((k/J(k))G) is nilpotent, and so J(kG) is T -nilpotent. Now apply the equivalence of (a) and (b) in Proposition 4.3.

Theorem 4.5. Let k be a left perfect ring and G be a finite group. Then every flat kG-module is projective.

Proof. By Proposition 4.4, kG is a left perfect ring, and the result follows from the equivalence of (a) and (c) in Proposition 4.3.

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Department of Mathematics, University of Georgia, Athens GA 30602, USA

E-mail address: djb@byrd.math.uga.edu