Augmentation modules for affine groups

Christopher J. B. Brookes

Corpus Christi College, Cambridge CB2 1RH email: brookes@dpmms.cam.ac.uk and

David M. Evans

School of Mathematics, University of East Anglia, Norwich NR4 7TJ email: d.evans@uea.ac.uk

Abstract: We prove that if k, F are fields of different characteristics then the permutation module kF^{n} for the affine group AGL(n, F) has simple augmentation submodule. This proves a conjecture of A. R. Camina.

1. Introduction

Suppose k and F are (commutative) fields of characteristics l and p respectively. Write A for the additive group of the n-dimensional vector space F^{n} and Q for its automorphism group, and consider the affine group G = A.Q acting on F^{n} (as x ↦ qx + a for x ∈ F^{n}, q ∈ Q and a ∈ A). We are interested in the submodule structure of the kG-permutation module kF^{n}. Equivalently, consider the group algebra kA with Q acting on it (via conjugation in G). Then the kG-submodules of kF^{n} are exactly the Q-invariant ideals in kA. We write A multiplicatively from now on, and denote the Q-action on the right as a superscript.

For any group B denote by I(B) the augmentation ideal of the group algebra kB. Thus I(B) is the kernel of the augmentation map kB → k (with ∑_{b∈B} α_{b}b ↦ ∑_{b∈B} α_{b}). Our main result is the following, which proves a conjecture of Alan Camina (personal communication):

Theorem 1 Let the characteristics of k and F be different. Then the zero ideal is the only Qinvariant ideal of kA strictly contained in I(A). When F is infinite, the Q-invariant ideals of kA are 0, I(A) and kA. When F is finite, there is an additional Q-invariant ideal, that generated by ∑_{a∈A} a.

Expressed in terms of permutation modules, this says that the augmentation submodule of the kG-permutation module kF is simple when the characteristics of F and k are different. Special cases of this result were known previously. The case n = 1 is handled when F = ℚ by A. R. Camina and D.M. Evans in [2] and when F is a finite extension of ℚ by D. R. Farkas and R. L. Snider in [3]. If F is finite the result is ‘well-known’ and the locally finite case follows easily from this. Note that, as any element of kA is contained in kA_{1} where A_{1} is the additive group of F_{1}^{n} for some countable subfield F_{1} of F, it is enough to prove Theorem 1 in the case where F is countable.

The proof of Theorem 1 varies according to whether the characteristic of F is zero or not, but the strategy of the proof is the same in all cases. Let Q_{0} denote the multiplicative group of the prime subfield F_{0} of F. So we may consider A as an F_{0}-vector space. We consider semiprime ideals of kA which are contained in I(A) and which are Q_{0}-invariant (recall that an ideal J in a commutative ring R is semiprime if R/J has no non-zero nilpotent elements). Suppose B is a subgroup of A. The natural homomorphism A → A/B induces a k-algebra homomorphism kA → k(A/B) whose kernel consists of all ∑_{x∈A} α_{x}x with the property that ∑_{x∈T} α_{x} = 0 for all cosets T of B in A. It is easy to see that this equals I(B)kA.

Proposition 2 Assume either that F is of characteristic zero, or that the characteristics of k and F are different. Suppose J < I(A) is a Q_{0}-invariant semiprime ideal. Then there exists a proper F_{0}-subspace B of A such that J ≤ I(B)kA.

This is proved in Section 2. Most work is needed when the characteristic of F is zero and valuation-theoretic techniques come into play. The other main ingredient in the proof of Theorem 1 is Proposition 3, which we prove in Section 3.

Proposition 3 Suppose either that F is of characteristic 0, or of different characteristic to k. Let B be a proper F_{0}-subspace of A. Then

⋂_{q∈Q} I(B^{q})kA = {0}.

Proof of Theorem 1: Let I be a non-zero Q-invariant ideal in kA. If its intersection with I(A) is zero then it is one dimensional as a k-vector space and A acts trivially upon it. This can only happen when A is finite and I is the ideal generated by ∑_{a∈A} a. So suppose I ∩ I(A) is non-zero. We shall show that this implies that I(A) ≤ I. It follows that I is I(A) or kA as required. So it is enough to consider the case where I ≤ I(A).

Let J be the radical of I (that is, the smallest semiprime ideal of kA containing I). Then J ≤ I(A) is a Q-invariant semiprime ideal of kA. It follows from Propositions 2 and 3 that J = I(A).

Let u = a − 1 ∈ I(A) (for some a ∈ A \ {1}). Then there exists a natural number m such that u^{m} ∈ I. Choose m as small as possible. It is enough to prove that m = 1. Case l ≠ 0. Let s be the non-negative integer such that l(s − 1) < m ≤ ls. Then (a^{l} − 1)^{s} = (a − 1)^{ls} ∈ I, and so (a − 1)^{s} ∈ I, by Q-invariance. Hence, s ≥ m. If s = 1, we are done. If s > 1 then \frac{s}{s−1} > l, which contradicts l ≥ 2. Case l = 0. In the polynomial algebra k[X], the greatest common divisor of (X −1)^{m} and (X^{p}−1) is X − 1. Since a^{p} = 1 we see that (a − 1)^{m} ∈ I forces a − 1 ∈ I. □

The Propositions 2 and 3 also give information when the characteristics of k and F are both zero. We are grateful to J.E. Roseblade for providing us with the following argument, depending crucially on certain binomial coefficients being non-zero in k; the strategy is similar to that of the proof of the Nagata-Higman theorem to be found on p.274 of [4]. On the other hand, when the characteristics are both equal to p > 0 we do not know whether all non-zero Q-invariant ideals of kA necessarily contain a positive power of I(A).

Theorem 4 Suppose k and F are of characteristic zero. Then any non-zero Q-invariant ideal of kA contains a positive power of the augmentation ideal I(A) of A.

Proof. Let J be the radical of a non-zero, proper, Q-invariant ideal I. By Propositions 2 and 3 we again know that J is I(A), and so (a − 1)^{m} lies in I for some m and some non-identity a ∈ A. Since Q acts transitively and I is invariant this implies (a − 1)^{m} ∈ I for all a ∈ A.

Let I_{r} be the ideal generated by the set of (a − 1)^{r} with a ∈ A. Thus the ideals I_{r} are Q-invariant and I_{m} ≤ I. We prove by induction that I_{r} ≥ I(A)^{2^{r−1}} by showing that I_{r−1}^{2} ≤ I_{r} for each r and observing that I_{1} = I(A).

First note that ab − 1 = (a − 1)b + (b − 1). Set α = (a − 1)b and β = b − 1. Thus (α + β)^{r} ≡ 0 (modI_{r}). We show that α^{r−1}β^{r−1} ≡ 0.

If r is even, set r = 2s and observe that α^{s−1}(α + β)^{r}β^{s−1} ≡ 0. Expanding and using that α^{2s} ≡ 0 ≡ β^{2s} we find that

α^{s−1}\binom{2s}{s}α^{s}β^{s}β^{s−1} ≡ 0.

Since the characteristic of k is zero this implies that α^{r−1}β^{r−1} ≡ 0. Similarly if r = 2s + 1 the expansion of α^{s}(α + β)^{r}β^{s−1} ≡ 0 yields α^{r−1}β^{r−1} ≡ 0.

Since b is a unit we deduce that (a − 1)^{r−1}(b − 1)^{r−1} ∈ I_{r} for all a and b in A, and therefore that I_{r−1}^{2} ≤ I_{r}. □

2. Proof of Proposition 2

2.1. The case char F = 0.

Lemma 5 Let A ≤ B be finitely generated abelian groups and k be a field. Let J be a semiprime ideal of kB. Denote the primes of kB minimal over J by P_{1}, P_{2}, …, P_{m} and those of kA minimal over J ∩ kA by Q_{1}, Q_{2}, …, Q_{r}.

Then for each i there is some j with P_{i} ∩ kA ≥ Q_{j}, and each Q_{j} is of the form P_{i} ∩ kA for some i. Thus each Q_{j} extends, not necessarily uniquely, to a prime P_{i}.

Proof. Because P_{i} ≥ J we have P_{i} ∩ kA ≥ J ∩ kA. But P_{i} ∩ kA is prime in kA and so contains some prime Q_{j} minimal over J ∩ kA.

A semiprime ideal is the intersection of the primes minimal over it and so the product P_{1} … P_{m} ≤ ⋂_{i=1}^{m} P_{i} = J. Hence ∏_{i}(P_{i} ∩ kA) ≤ J ∩ kA = ⋂_{j} Q_{j} ≤ Q_{j} for each j. But Q_{j} is prime and so some P_{i} ∩ kA ≤ Q_{j}. But P_{i} ∩ kA is prime in kA and so the minimality of Q_{j} ensures P_{i} ∩ kA = Q_{j}. □

Proposition 6 Let A be isomorphic to the additive group ℚ^{n} and k be a field. Consider the multiplicative group ℚ^{×} acting on A, via multiplication, and hence on kA. Let J be a semiprime ideal of kA invariant under this action.

Then there are finitely many primes of kA minimal over J and they are of the form I(B_{i})kA where I(B_{i}) is the augmentation ideal of the subgroup B_{i} of A corresponding to a rational subspace of ℚ^{n} (for i = 1, …, m). Thus J = ⋂_{i=1}^{m} I(B_{i})kA.

Remark. Since each ℚ^{×}-invariant semiprime ideal of kA is the intersection of finitely many primes, of the form I(B)kA, the set of ℚ^{×}-invariant semiprime ideals satisfies the maximum condition.

Before starting the proof, we recall a definition. Suppose A is an abelian group (written multiplicatively). We say that an ideal P in its group ring kA is faithful if (P + 1) ∩ A = {1}. Equivalently, the natural homomorphism embeds A into the group of units of kA/P.

Proof of Proposition 6. We choose finitely generated subgroups A_{j} with A_{1} < A_{2} < … and ⋃_{j} A_{j} = A so that A_{j} = A_{1}^{q_{j}} for some q_{j} ∈ ℚ^{×}. For example within ℚ^{n} we may pick

ℤ^{n} < \frac{1}{2}ℤ^{n} < \frac{1}{2^{2}.3}ℤ^{n} < \frac{1}{2^{3}.3^{2}.5}ℤ^{n} < ….

Since J is semiprime, J ∩ kA_{j} is also semiprime and therefore the intersection of the finitely many primes minimal over it. But J = J^{q_{j}} and so J ∩ kA_{j} = (J ∩ kA_{1})^{q_{j}}. Denoting the minimal primes over J ∩ kA_{1} by Q_{1}, …, Q_{m} the minimal primes over J ∩ kA_{j} are Q_{1}^{q_{j}}, …, Q_{m}^{q_{j}}. Thus the number of primes minimal over J ∩ kA_{j} is independent of j. Lemma 5 says that each prime minimal over J ∩ kA_{1} extends to one minimal over J ∩ kA_{j}, and, because the numbers of such primes are the same, this extension is unique. Taking the union over j of the extensions of Q_{i} in kA_{j} minimal over J ∩ kA_{j} gives a prime P_{i} of kA with P_{i} ∩ kA_{j} minimal over J ∩ kA_{j} for all j and P_{i} ∩ kA_{1} = Q_{i}. Thus P_{i} ≥ J and, since (⋂_{i} P_{i}) ∩ kA_{j} = ⋂_{i}(P_{i} ∩ kA_{j}) = J ∩ kA_{j} for all j, we have ⋂_{i} P_{i} = J. Also, no P_{i} contains any other. Indeed the P_{i} are precisely the primes of kA minimal over J. To see this consider any prime P of kA containing J. Then the product P_{1} … P_{m} ≤ ⋂_{i} P_{i} = J ≤ P and so P_{i} ≤ P for some i using the primality of P.

Now consider the action of ℚ^{×}. If q ∈ ℚ^{×} then P_{1}^{q}, …, P_{m}^{q} are the primes minimal over J^{q} = J. Thus ℚ^{×} permutes the primes P_{i}. Some subgroup S of finite index in ℚ^{×} fixes each P_{i}. Let B_{i} = (P_{i} + 1) ∩ A. For each i this is a subgroup of A invariant under S, and I(B_{i})kA ≤ P_{i}. The only subgroups of ℚ^{n} invariant under a subgroup of finite index in ℚ^{×} are the additive groups of vector subspaces and so B_{i} is such a subspace.

Consider the image of P_{i} in k(A/B_{i}). It is faithful and invariant under the inherited action of S on k(A/B_{i}). In addition, S contains a positive integer which, when k has positive characteristic, may be chosen coprime to the characteristic. Thus the following proposition ensures that the image of P_{i} in k(A/B_{i}) is zero. The kernel of the natural map kA → kA/B_{i} is I(B_{i})kA, and so we obtain P_{i} = I(B_{i})kA. □

Proposition 7 Let A be isomorphic to the additive group of a finite dimensional rational vector space (written multiplicatively), and k be a field of characteristic p. Let P be a prime ideal of kA with (1 + P) ∩ A = 1. Let m be a positive integer greater than 1, coprime to p in the case when p > 0. Let γ be the ring automorphism of kA induced by the group automorphism of A obtained by taking mth roots. Suppose that P is invariant under γ. Then P is zero.

Proof. Suppose that P is non-zero. We may assume, by passing to the intersection of P with the group algebra of a subgroup corresponding to a rational subspace, that P ∩ kB = 0 for any subgroup B of A of smaller torsion-free rank.

For subgroups B of A we shall write k[B] for the integral domain kB/(P ∩ kB) and k(B) for its field of fractions. Note that as P is non-zero, the transcendence degree of k(A) over k is less than the torsion free rank of A. The minimality assumption of the previous paragraph ensures that k(A) is algebraic over k(B) whenever the torsion-free rank of A/B is equal to 1. Since (1 + P) ∩ A = 1 there is a natural embedding of A into k(A).

Suppose first that A has torsion-free rank 1. Then kA/P is an algebraic field extension of k. If a ∈ A is not 1 then k[< a >] is a field extension of finite degree d say. Applying the automorphism γ we discover that k[< a^{1/m} >] is also of degree d over k. But this field contains k[< a >] and so is equal to it. Thus k[< a >] is invariant under the k-automorphism γ. So some non-zero power γ^{s} of γ fixes k[< a >] and hence a^{1/m^{s}} − a ∈ P. Hence a^{1/m^{s}}a^{−1} ∈ (1 + P) ∩ A, contradicting the faithfulness of P.

Thus the torsion-free rank of A is greater than 1, and so the minimality assumption of the first paragraph together with the above argument, imply that only the trivial element of A yields an image in k[A] algebraic over k. Thus k(A) has transcendence degree at least 1 over k. Let A_{0} be a free Abelian subgroup of A of maximal rank. Then k(A) has the same transcendence degree over k as k(A_{0}) and so P ∩ kA_{0} is non-zero, but P ∩ kB_{0} = 0 for any subgroup B_{0} of A_{0} of smaller rank. Thus k(A_{0}) is algebraic over k(B_{0}) whenever A_{0}/B_{0} is infinite cyclic.

Take C_{0} ≤ A_{0} such that the transcendence degree of k(A_{0}) over k(C_{0}) is 1 and let x ∈ A_{0} be transcendental over k_{1} = k(C_{0}). Let v_{1} be the valuation on k_{1}(x) obtained by localising the polynomial ring k_{1}[x] at x. This valuation is discrete with x taking value 1 and the characteristics of k and the residue field of the valuation ring are the same. As k(A_{0}) is finitely generated over k it is a finite extension of k_{1}(x), and so v_{1} extends to a discrete valuation v on k(A_{0}). Now let B_{0} be the intersection of A_{0} with the kernel of v. Thus A_{0}/B_{0} is an infinite cyclic group and there is some a ∈ A with A_{0} = B_{0} < a >.

The discrete valuation v extends to one on the field k(B_{0}^{1/m} < a >). We need to establish that no ramification takes place. We refer the reader to VI.8.5 of Bourbaki [1] for the standard theory of the behaviour of valuations under field extension. Our field extension is obtained by adjoining mth roots of elements of value 0. It is therefore enough to show that such a valuation v on a field K does not ramify on extending to a valuation w on L = K(α^{1/m}) when α has value 0. This is probably well-known, but we include a proof, for lack of a suitable reference. Write K̂ and L̂ for the completions with respect to v and w. Since L is a separable extension of K, the characteristic being coprime to m, we know that K̂ ⊗ L is isomorphic to a direct product of fields L̂_{i}, one of which is isomorphic to L̂. Moreover, if f(X) is the minimal polynomial of α^{1/m} over K, then L̂ is isomorphic to K̂[X]/(f_{i}(X)) where f_{i}(X) is one of the irreducible factors of f(X) in K̂[X], and the degree of f_{i}(X) is equal to the product of the ramification index e of w over v with the degree t of the residue field L_{w} over the residue field K_{v}. However f(X) divides X^{m} − α and all its coefficients lie in the valuation ring of v. Since m is coprime to the characteristic of the residue field K_{v}, we know that f(X) does not have repeated roots modulo the unique maximal ideal P of the valuation ring of v. Hensel’s lemma says that in such circumstances any factorisation of f(X) modulo P induces a factorisation of f(X) over K̂. Thus the irreducibility of f_{i}(X) in K̂[X] ensures that f_{i}(X) is irreducible modulo P. However f_{i}(X) modulo P does have a root in the residue field L_{w}, namely a root of f_{i}(X) in L̂ modulo the maximal ideal of the valuation ring, and so the degree of f_{i}(X) is at most equal to the degree t of the extension L_{w} over K_{v}. Thus et ≤ t and so e = 1.

The value group of the extended valuation on k(B_{0}^{1/m} < a >) is therefore precisely that of v. Repeated extension in this way yields a discrete valuation w_{1} on k(D < a >), where D is the closure of B_{0} under taking mth roots. Let d be the degree of the algebraic extension k(D < a >) of k(D). Applying the automorphism γ, we get that k(D < a^{1/m} >) also has degree d over k(D). Since one extension lies in the other they must be the same field and it is closed under the automorphism induced by γ. It is therefore k(E) where E is the closure of A_{0} under < γ >. The valuation w_{1} therefore induces a non-trivial group homomorphism from the m-divisible group E to the value group which is infinite cyclic, a contradiction. □

Proposition 8 Let A be isomorphic to the additive group of a countably infinite dimensional ℚ-vector space and k be a field. Suppose J is a semiprime ideal of kA invariant under the action of ℚ^{×}.

Then J is the intersection of some primes of the form I(B)kA where the subgroups B correspond to the additive groups of vector subspaces. In particular J is contained in some I(B)kA.

Proof. We may pick subgroups A_{j} isomorphic to ℚ^{j} so that A_{1} < A_{2} < … and ⋃_{j} A_{j} = A. By Lemma 5, each prime minimal over J ∩ kA_{j} extends to a prime minimal over J ∩ kA_{j+1}. Taking the union of such a chain of minimal primes gives a prime P of kA containing J, since P ∩ kA_{j} contains J ∩ kA_{j} for each j. But each P ∩ kA_{j} is a minimal prime over J ∩ kA_{j}, which is a semiprime ideal of kA_{j} invariant under the action of ℚ^{×}, and so is of the form I(B_{j})kA_{j} for some additive group B_{j} of a vector subspace of A_{j}, by Proposition 6. Thus B_{1} ≤ B_{2} ≤ … and setting B = ⋃_{j} B_{j} we have B ∩ A_{j} = B_{j} and P = I(B)kA. Thus J ≤ I(B)kA. In fact considering the set X of all primes P obtained in this way by extending primes minimal over some J ∩ kA_{j} one can see that

J ∩ kA_{j} = ⋂_{P ∈X}(P ∩ kA_{j}) = ( ⋂_{P ∈X} P) ∩ kA_{j}

for each j and so J = ⋂_{P ∈X} P. □

In the case where char F = 0, Proposition 2 now follows immediately from Propositions 6 and 8.

2.2. The case char F ≠ 0.

Lemma 9 Let p be a prime, m ∈ ℕ, and A_{m} = ⟨x_{1}, …, x_{m}⟩ an elementary abelian group of order p^{m}. Let Q_{0} consist of the automorphisms of A_{m} given by a ↦ a^{i}, for a ∈ A and 1 ≤ i ≤ p − 1. Let k be a field of characteristic l ≠ p. Suppose J_{m} is a semiprime ideal in the group algebra kA_{m} which is properly contained in I(A_{m}), and which is Q_{0}-invariant. Then there is a proper subgroup B_{m} of A_{m} such that J_{m} ≤ I(B_{m})kA_{m}. In fact, J_{m} can be written as the intersection of ideals of this form.

Proof. There is no loss of generality in assuming that k contains a primitive p-th root of unity, ζ. The group ring kA_{m} is isomorphic to the polynomial ring k[X_{1}, …, X_{m}] factored by the ideal I generated by the polynomials X_{i}^{p} − 1 (for i = 1, …, m). Thus the semiprime ideals in kA_{m} are in one-to-one correspondence with semiprime ideals in k[X_{1}, …, X_{m}] which contain I, and by the Nullstellensatz these are precisely the annihilators of subsets of

{(ζ^{j_{1}}, …, ζ^{j_{m}}) : 0 ≤ j_{i} ≤ p − 1}.

Denoting the field of integers modulo p by 𝔽_{p}, this implies that there exists a non-zero j = (j_{1}, …, j_{m}) ∈ 𝔽_{p}^{m} such that if f = ∑_{i∈𝔽_{p}^{m}} α_{i}x^{i} is in J_{m} then ∑_{i} α_{i}ζ^{i·j} = 0, where x^{i} = x^{i_{1}} ⋯ x^{i_{m}} and i · j denotes the dot product in 𝔽_{p}^{m}. For r ∈ 𝔽_{p} let c_{r} = ∑{α_{i} : i · j = r}. It will be enough to show that each c_{r} is zero, for then f ∈ I(B_{m})kA_{m}, where B_{m} = {x^{i} : i · j = 0}. But invariance of J_{m} under the action of Q_{0} means that for s = 1, …, p − 1 we have

∑_{r} c_{r}ζ^{rs} = 0.

As J_{m} ≤ I(A) we also have

∑_{r} c_{r} = 0.

Consideration of the Vandermonde matrix with first row 1, ζ, …, ζ^{p−1}, noting that these are all distinct, then shows that each c_{r} is zero, as required. □

Proof of Proposition 2. We may assume that F is countable. Let x_{1}, x_{2}, … be a basis for A a Q_{0}-space, and let A_{m} = ⟨x_{1}, …, x_{m}⟩. Then J_{m} = kA_{m} ∩ J is a Q_{0}-invariant semiprime ideal in kA_{m}, which is properly contained in I(A_{m}) if m is large enough. In particular, there exists a t such that x_{t} −1 ∉ J. Note that for B ≤ A and a ∈ A we have a−1 ∈ I(B)kA if and only if a ∈ B. Thus, by the above lemma, for m ≥ t there is B_{m} ≤ A_{m} with x_{t} ∉ B_{m} and J_{m} ≤ I(B_{m})kA_{m}. Also note that

(I(B_{m+1})kA_{m+1}) ∩ kA_{m} = I(B_{m+1} ∩ A_{m})kA_{m}.

Thus an application of König’s lemma shows that there exists B ≤ A with x_{t} ∉ B and J ≤ I(B)kA, as required. □

3. Proof of Proposition 3

The proof splits into two cases, depending on the characteristic of F, but in both cases we require the following ‘density’ result, which may be of independent interest.

Lemma 10 Let F be a field with prime subfield F_{0}, and let α : F^{n} → F_{0} be a non-zero F_{0}- linear map. For g ∈ GL(n, F) define α^{g} by α^{g}(x) = α(g(x)) (where x ∈ A). Let V be a finite dimensional F_{0}-subspace of F^{n} and α_{0} : V → F_{0} be an F_{0}-linear map, assumed to be non-zero in the case where V = F^{n}. Then there exists g ∈ GL(n, F) such that for all x ∈ V we have α_{0}(x) = α^{g}(x).

Proof. Pick an F -basis e_{1}, …, e_{n} of F^{n} so that α(e_{i}) ≠ 0 for each i, and let α_{i} be a nonzero element of the F_{0}-dual space of a finite dimensional F_{0}-vector space W_{i} ≤ F e_{i}. As the multiplicative group Q_{1} of F is transitive on F \ {0}, it is clear that F e_{i} ∩ ⋂_{q∈Q_{1}} ker(α^{q}) = {0}. Thus ⋂_{q∈Q_{1}} ker(α^{q} | W_{i}) = {0} and so (as W_{i} is finite dimensional) {α^{q} | W_{i} : q ∈ Q_{1}} spans the dual space of W_{i}. Thus there exist q_{1}, …, q_{r} ∈ Q_{1} and μ_{1}, …, μ_{r} ∈ F_{0} such that for all x ∈ W_{i}

α_{i}(x) = ∑_{i=1}^{r} μ_{i}α^{q_{i}}(x) = α((∑_{i=1}^{r} μ_{i}q_{i})x).

Let λ_{i} = ∑_{i} μ_{i}q_{i} and, noting that λ_{i} is non-zero, let g_{1} be the element of GL(n, F) for which each e_{i} is a λ_{i}-eigenvector. Thus α^{g_{1}} coincides with α_{i} on W_{i} for each i.

Now consider the given element α_{0} of the F_{0}-dual space of V. Suppose α_{0} is zero. Then by assumption V ≠ F^{n} and we can extend α_{0} to give a non-zero element α′ in the dual space of a finite dimensional F_{0}-vector space V′ > V. If α^{g} restricts to give α′ on V′ then on V it coincides with α_{0}. Thus, by replacing V by V′, we may assume α_{0} is non-zero.

Now choose an F_{0}-basis of V with each element taking non-zero value with respect to α_{0}. Taking a maximal F -linearly independent subset e′_{1}, …, e′_{s} of this basis gives a F -basis of F V; extend it to a basis e′_{1}, …, e′_{n} of F^{n}. Let g_{2} ∈ GL(n, F) be such that g_{2}(e′_{i}) = e_{i} for each i. By enlarging V if necessary, we may assume that V = ⨁_{i=1}^{n} V_{i} with V_{i} ≤ F e′_{i}. Let W_{i} = g_{2}(V_{i}) and define α_{i} by α_{i}(g_{2}(x)) = α_{0}(x) for x ∈ V_{i}. Thus W_{i} ≤ F e_{i} and α_{i} is a non-zero element of the F_{0}-dual space of g_{2}(V_{i}). So there is g_{1} ∈ GL(n, F) such that α^{g_{1}}(g_{2}(x)) = α_{i}(g_{2}(x)) = α_{0}(x) for x ∈ V_{i}. Hence α^{g_{1}g_{2}}(x) = α_{0}(x) for x ∈ V_{i} and so α^{g_{1}g_{2}} and α_{0} coincide on V = ⨁_{i=1}^{n} V_{i}. Setting g = g_{1}g_{2} gives the required element of GL(n, F). □

We complete the proof of Proposition 3. First, suppose that the characteristic of F is zero, so F_{0} = ℚ. Let 0 ≠ v = ∑_{a∈A} γ_{a}a ∈ kA. To prove the proposition it will suffice to assume that the proper F_{0}-subspace B is of codimension 1 in A and to find g ∈ Q such that v ∉ I(B^{g})kA.

Let S = {a ∈ A : γ_{a} ≠ 0}. Recall that I(B)kA is the kernel of the natural map kA → k(A/B) and so consists of those ∑_{a∈A} β_{a}a ∈ kA such that Σ_{a∈C}β_{a} = 0 for every coset C ∈ A/B. Thus it is enough to show that we can find g ∈ Q such that no two elements of S lie in the same coset of B^{g} in A. Let Y = {a_{1} − a_{2} : a_{1}, a_{2} ∈ S, distinct}. Then Y is finite, and it will suffice to find g ∈ Q such that Y ∩ B^{g} = ∅. However B is the kernel of a non-zero element α of the F_{0}-dual space of A. We therefore want to find g ∈ Q such that α^{g}(y) ≠ 0 for all y ∈ Y. Let V be the F_{0}-subspace of A generated by Y. As F_{0} is infinite, there exists an F_{0}-linear map α_{0} : V → F_{0} such that α_{0}(y) ≠ 0 for y ∈ Y. An application of Lemma 10 gives the required element of Q.

Finally, suppose the characteristic of F is non-zero, and different from the characteristic of k. Let V be a finite dimensional F_{0}-subspace of F. As (I(B)kA) ∩ kV = I(B ∩ V)kV for any subgroup B of A, it follows from Lemma 10 that it is enough to show

X = ⋂{I(W)kV : W a hyperplane in V } = {0}.

But X is an F_{0}-invariant semiprime ideal of kV, and the required result therefore follows from the Nullstellensatz as in the proof of Lemma 9. □

Remarks. (a) Suppose that k is of characteristic p > 0. Let V be an elementary abelian group of order p^{n} for n ≥ 2, and let X be a subgroup of order at least p^{2}. If W is a subgroup of V of index p, then W ∩ X is non-trivial, and so the number of elements in the intersection of X with any coset of W is divisible by p. Thus ∑_{v∈X} v lies in I(W)kV.

(b) When the characteristic of F is zero, Proposition 3 is also valid for other actions on A by a group Q. The given argument applies as long as the induced orbit of any non-zero element of the ℚ-dual of F^{n} does not lie in a finite union of hyperplanes; one just requires an α^{g} not in the annihilator of any element of Y. However, when the characteristic of F is non-zero, one does need such an orbit to satisfy a density property analogous to that of Lemma 10.

References

[1] BOURBAKI, Commutative algebra (Hermann, 1972).

[2] ALAN R. CAMINA and DAVID M. EVANS, Some infinite permutation modules, Quart. J. Math. Oxford (2), 42 (1991), 15–26.

[3] DANIEL R. FARKAS and ROBERT L. SNIDER, Simple augmentation modules, Quart. J. Math. Oxford (2), 45 (1994), 29–42.

[4] N. JACOBSON, Structure of rings (American Math. Soc., 1964, 2nd edition).