DEFECT THEORY FOR PRIME IDEALS AND DRESS’S INDUCTION THEOREM

Hubert Fottner

Friedrich-List-Str. 7, 86153 Augsburg, Germany

§1. INTRODUCTION

It was first observed by E. C. Dade [8] that defect groups can be defined for arbitrary rings with finite group action as long as one replaces points (resp. primitive idempotents) by maximal ideals, which, in general, are no longer in bijection. J. Thévenaz has shown in [24] (cf. also [25] chapter 8) that also Puig’s theory of pointed groups [18] (for text book accounts cf. [16] and [25]) carries over to this more general setup. So one can also define “sources” of maximal ideals (in the fixed subring of the group), which are then maximal ideals in the fixed subring of “the” defect group (of the original maximal ideal). Moreover, Thévenaz demonstrates that this cannot only be done for a ring with group action, but also for the more general notion of a Green functor. This refines and extends earlier work of Thévenaz [23], where he already points out the close relation between primordial groups and Green’s defect bases [13], i.e. the relevance of primordial groups for induction theorems. This approach in turn has its origin in the generalization of the Brauer homomorphism to arbitrary G-algebras introduced by M. Broué and L. Puig in [5]. (Of course, Thévenaz’s paper [23] has also many other objectives.) However, Thévenaz needs some additional assumptions to guarentee the existence of defect pointed groups, and to establish the Puig resp. Green correspondences.

The objective of this paper is to show that the results carry over to a situation where one replaces maximal ideals by prime ideals, and, then, no additional assumptions are necessary to prove the existence of defect pointed groups (or to establish the above correspondences). Although, Thévenaz’s assumptions in [24] are harmless in a sense that they are satisfied in most (if not in all interesting) examples, for instance, if the involved rings are finitely generated as modules over their centers (and, as we will show, one of these assumptions is superfluous anyway), the fact that we can do without any additional hypotheses might be an evidence that focusing on prime ideals rather than maximal ideals is the more natural point of view. Of course, at some places the results can be improved under additional hypotheses.

Apart from that, by focusing on prime ideals rather than maximal ideals, new structures come into game, namely (Zariski) topologies and partial orders, and it will turn out that the Puig and Green correspondences become homeomorphisms and isomorphisms of posets, respectively. We will use the latter fact to show that, for certain Green functors A, the Krull dimension of A(G) can be bounded in terms of the Krull dimensions of \overline{A}(P), where P ranges over all primordial subgroups of G for A (cf. corollary M for a precise statement, and section §2 for the notation, cf. also remark (6.2)). Moreover, we show that Puig’s version of Sylow’s first theorem for local pointed groups has generalizations to the situation we are concerned with here (cf. corollaries P and R). (Results of this type are neither contained in [24] nor in [25].) As an application, we demonstrate that Dress’s induction theorem [11] (cf. also [28]) is a consequence of this version of Sylow’s first theorem (cf. theorem U). (In fact, Dress’s induction theorem [11] is more precise than the result we obtain using Sylow’s first theorem. One might say that this result is just “qualitatively (not quantitatively) equivalent” to Dress’s induction theorem.)

In Appendix A we show that this version of Sylow’s first theorem is correlated to a theorem of Dress [10], which asserts that the integral Brunside ring of a finite group G has no non-trivial idempotents if and only if G is solvable. (Of course, in view of the above paragraph, this might not be suprising as the techniques and ideas of Dress’s paper [10] served as a basis for the proof of Dress’s induction theorem.) In Appendix B we consider the Grothendieck group G_{0}(𝕂G) of a group algebra 𝕂G over a field 𝕂 of characteristic 0. In order to illustrate the theory, we demonstrate how some classical results around this Grothendieck group can be deduced using defect theory, including, for instance, character theory, and Artin’s resp. Brauer’s induction theorem. Much of this could as well be done using defect theory for maximal ideals. The contribute of the new theory shows up in our proof of Brauer’s induction theorem on the grounds of Sylow’s first theorem, which allows to relate defects of maximal ideals with defects of certain minimal prime ideals. Moreover, we give a proof of a result of Berman [3] asserting that G_{0}(𝕂G) does not have non-trivial idempotents, which is based on this version of Sylow’s first theorem, as well.

So this generalized approach to Puig’s theory of pointed groups might contribute further to standardization of different branches of representation theory of finite groups.

Finally, note that another objective of Thévenaz’s paper [24] is to describe the defect theory in terms of functorial ideals, and, as a consequence, to classify the simple Green functors. Although it seems likely that at least parts of this theory carry over to the situation we are concerned with here, we do not follow this approach in this paper. Quite recently, S. Bouc [4] and, independently, F. Luca in his Ph. D. thesis classified also the simple modules over Green functors. This result extends the classification of simple Mackey functors given by J. Thévenaz and P. Webb in [26]. Moreover, Bouc shows in [4] that also Thévenaz’s classification of simple Green functors can be recovered from this result.

§2. NOTATION AND MAIN RESULTS

§2. NOTATION AND MAIN RESULTS 3

Throughout, we fix a finite group G, denote by k a commutative (ground) ring with identity 1, and let A be a Green functor for G over k. We refer to [25] for the definition of Green functors. However, for our purpose it is not necessary to require the rings A(H), H a subgroup of G, to possess a multiplicative identity. Moreover, we prefer to refer to the transfer maps of A as induction maps, and denote them by Ind_{K}^{H} : A(K) → A(H) for subgroups K and H of G such that K is contained in H; in this situation, A_{K}^{H} denotes the image of the induction map Ind_{K}^{H}, which is an ideal in A(H), by the Frobenius axiom. Besides, we denote the restriction maps of A by Res_{K}^{H} : A(H) → A(K) (H and K as above). We refer to Green functors as in the definition of [25] on page 501 as unitary Green functors, i.e. Green functors A for G such that A(H) has a multiplicative identity for all subgroups H of G, and such that the restriction maps are unitary homomorphisms of k-algebras. (In this case the conjugation maps are necessarily unitary since they are bijective.)

Note that we will use the symbol ⊆ to denote set-theoretical inclusion, whereas ⊂ will always mean proper inclusion. Unless stated otherwise, algebra resp. Green functor is supposed to mean algebra resp. Green functor over k, and all our algebras are associative, and unitary as k-modules, but we do not require the existence of a multiplicative identity. (The notion of a G-algebra is to be understood in this way.) Moreover, ideal in an algebra will always mean two-sided k-algebra ideal.

Let B be a G-algebra. Then B may be regarded as Green functor where B(H) := B^{H}, the subring {b ∈ B : ^{h}b = b for all h ∈ H} of fixed points of H on B, for any subgroup H of G, and the induction maps are the relative trace maps

Tr_{K}^{H} : B^{K} → B^{H}, b ↦ ∑_{hK∈H/K} ^{h}b

for subgroups K, H of G such that K is contained in H; and we keep the above notation B_{K}^{H} to denote the image Tr_{K}^{H}(B^{K}) of Tr_{K}^{H}. The restriction maps are just the canonical inclusions, and the conjugation maps come from the given G-action on B.

Note that, for any subgroup H of G, N_{G}(H) acts on A(H) via the conjugation maps; so A(H) is a N_{G}(H)-algebra, and thus the notation A(H)_{L}^{K} makes sense for any subgroup K of N_{G}(H), and any subgroup L of K. Note also that, as H acts trivially on A(H), A(H) may be regarded as \overline{N}_{G}(H)-algebra where \overline{N}_{G}(H) := N_{G}(H)/H.

We refer to section §3 for the definition and basic properties of prime ideals in noncommutative rings.

DEFINITION A. Let G be a finite group, and let A be a Green functor for G. A pointed group on A is a pair (H, 𝔭) consisting of a subgroup H of G and a prime ideal 𝔭 of A(H). We shall write H_{𝔭} instead of (H, 𝔭).

The group G acts on the set of pointed groups on A, where, for a pointed group H_{𝔭} on A,

g(H_{𝔭}) := ^{g}H_{^{g}𝔭} = (gHg^{−1})_{c_{g,H}(𝔭)};

we denote the stabilizer of H_{𝔭} in G by N_{G}(H_{𝔭}). Note that, obviously H is contained, and hence normal, in N_{G}(H_{𝔭}); we set \overline{N}_{G}(H_{𝔭}) := N_{G}(H_{𝔭})/H. This action generalizes the conjugation action of G on subgroups of G, and we refer to it as conjugation, as well.

The following definition yields a partial order on the set of pointed groups on A:

DEFINITION B. Let G be a finite group, let A be a Green functor, and let H_{𝔭} and K_{𝔮} be pointed groups on A. We say that K_{𝔮} is a pointed subgroup of H_{𝔭}, and write K_{𝔮} ≤ H_{𝔭}, provided that K is a subgroup of H and

(Res_{K}^{H})^{−1}(𝔮) ⊆ 𝔭,

i.e. any element x of A(H) such that Res_{K}^{H}(x) ∈ 𝔮 is contained in 𝔭. As usual, we write K_{𝔮} < H_{𝔭} to indicate that K_{𝔮} ≤ H_{𝔭}, but K_{𝔮} ≠ H_{𝔭}.

Obviously, the relation ≤ on the set of pointed groups on A is in fact a partial order. So the poset of pointed groups on A refines the subgroup lattice of G.

Moreover, it is straight forward to check that this relation is G-equivariant, i.e. K_{𝔮} ≤ H_{𝔭}, for pointed groups K_{𝔮} and H_{𝔭} on A, implies ^{g}(K_{𝔮}) ≤ ^{g}(H_{𝔭}) for all g ∈ G.

Note that if A is a G-algebra, and K_{𝔮} and H_{𝔭} are pointed groups on A such that K ⊆ H, then K_{𝔮} ≤ H_{𝔭} means nothing but A^{H} ∩ 𝔮 ⊆ 𝔭. So, in some weaker sense as in commutative algebra, “𝔮 lies over 𝔭”. Although, in this situation one is rather tempted to say that “𝔭 lies over 𝔮”, because of the containment ⊆ and the fact that one usually thinks of restriction maps as going down.

There is also another partial order on the set of pointed groups on A which is defined via the induction maps; so it can be regarded as a kind of a dual of the first one.

DEFINITION C. Let G be a finite group, let A be a Green functor for G, let H_{𝔭} be a pointed group on A, and let K be a subgroup of H. We say that H_{𝔭} is relatively K-projective (or relatively projective to K) provided that A_{K}^{H} ⊈ 𝔭. A relatively 1-projective pointed group is just called projective. Moreover, for a prime ideal 𝔮 of A(K), we call H_{𝔭} relatively K_{𝔮}-projective, and write H_{𝔭} pr K_{𝔮}, if any ideal I of A(K), such that Ind_{K}^{H}(I) ⊆ 𝔭, is contained in 𝔮.

Note that, in case H_{𝔭} is relatively K_{𝔮}-projective, H_{𝔭} is necessarily relatively projective to K, as A(K) is not contained in 𝔮. We will see that, conversely, if H_{𝔭} is relatively K-projective then there exists a prime ideal 𝔮 in A(K) such that H_{𝔭} is relatively K_{𝔮}-projective, and 𝔮 can be chosen in such a way that K_{𝔮} ≤ H_{𝔭} (cf. lemma (4.1)). Moreover, note that also the relation pr is G-equivariant, as one easily verifies.

Note also that, for any prime ideal 𝔯 of A(K) containing 𝔮, H_{𝔭} is plainly relatively K_{𝔯}-projective, if it is relatively K_{𝔮}-projective; and similarly, for any prime ideal 𝔱 of A(H) contained in 𝔭, H_{𝔱} is relatively K_{𝔮}-projective, in case H_{𝔭} is.

Since, for any subgroup H of G, Ind_{H}^{H} is the identity on A(H), it is plain that the relation pr on the set of pointed groups on A is reflexive and antisymmetric. Moreover, since induction is transitive, so is the relation pr, and hence pr is a partial order on the set of pointed groups on A.

Recall the definition of the Brauer homomorphism. For a subgroup H of G, we define

\overline{A}(H) := A(H)/∑_{K} A_{K}^{H}

where the sum is taken over proper subgroups K of H, and let Br_{H}^{A} be the canonical epimorphism A(H) → \overline{A}(H). Br_{H}^{A} is called the Brauer homomorphism of H (with respect to A). We abbreviate Br_{H}^{A} to Br_{H} when it is clear from context which Green functor we refer to.

LEMMA D. Let G be a finite group, let A be a Green functor for G, and let P_{𝔮} be a pointed group on A. Then the following assertions are equivalent:

(a) P_{𝔮} is not relatively projective to a proper subgroup of P.

(b) If P_{𝔮} is relatively Q_{𝔭}-projective for some pointed group Q_{𝔭} on A then P = Q and 𝔮 ⊆ 𝔭.

(c) If P_{𝔮} is relatively Q_{𝔭}-projective for some pointed group Q_{𝔭} on A then P = Q.

(d) ker Br_{P} ⊆ 𝔮.

A pointed group P_{𝔮} on A satisfying the equivalent assertions of lemma D is called primordial. Moreover, a subgroup P of G is called primordial (for A) provided that Br_{P}^{A} ≠ 0, i.e. ∑_{Q<P} A_{Q}^{P} ⊂ A(P). Note that if P_{𝔮} is a primordial pointed group on A then P is necessarily a primordial subgroup of G since, in this case, ker Br_{P} ⊆ 𝔮 ⊂ A(P). Conversely, if P is a primordial subgroup of G and A(P) has an identity then there exists a prime ideal 𝔮 in A(P) such that P_{𝔮} is primordial. (Just take a maximal ideal of A(P) containing ker Br_{P}.) However, this does not hold, in general, without the assumption that A(P) has an identity. (For instance, let A be a non-zero nilpotent algebra over a field of positive characteristic p, and regard it as P -algebra with trivial action, where P is some p-group. Then P is primordial. However, A^{P} does not contain any prime ideals.) Clearly, any Gconjugate of a primordial pointed group on A is still primordial.

Recall that a Green functor A for G is called cohomological, if

Ind_{K}^{H}(Res_{K}^{H}(x)) = |H : K| x

for all subgroups H and K of G such that K is contained in H, and for all x ∈ A(H). Important examples of cohomological Green functors are G-algebras and (giving rise to the terminology) the cohomology ring functor.

Assume that A is a cohomological Green functor, and let P_{𝔮} be a primordial pointed group on A. Denote by c the characteristic of A(P)/𝔮, i.e. c is the minimum of all non-zeronatural numbers d such that dA(P) ⊆ 𝔮 in case at least one such d exists, and otherwise c = 0. Then c is either 0 or a prime number as 𝔮 is a prime ideal in A(P). (Note that if nmA(P) ⊆ 𝔮 for certain n, m ∈ ℕ then (nA(P))(mA(P)) ⊆ 𝔮, and so nA(P) ⊆ 𝔮 or mA(P) ⊆ 𝔮.)

Let Q be a proper subgroup of P. Then

|P : Q|A(P) = Ind_{Q}^{P}(Res_{Q}^{P}(A(P)) ⊆ ker Br_{P} ⊆ 𝔮

as P_{𝔮} is primordial. Thus c divides |P : Q| (by the division algorithm). This shows on the one hand that P = 1 in case c = 0. On the other hand, in case c is the prime number p, we can take Q to be a Sylow p-subgroup of P, and obtain that we must have P = Q, i.e. P is a p-group.

So the primordial pointed groups may be regarded as an analogue of the psubgroups of G (p any prime). Note that in the context of G-algebras primordial pointed groups (with prime ideals replaced by maximal ideals) are called local pointed groups; we kept the convention of J. Thévenaz in [25] (cf. [25] page 519) of choosing a different terminology in the more general setup of Green functors since the word “local” is usually associated with p-subgroups of G; and, in general, primordial groups need not be p-subgroups of G. For instance, for the integral Green ring functor defined with respect to a field 𝔽 of positive characteristic p, the primordial subgroups are well-known to be the subgroups of G which are elementary modulo p (i.e. subgroups H of G such that H/ O_{p}(H) is elementary in the sense of Brauer).

The following definition provides also an analogue of the Sylow subgroups of G, as will become clear later on:

DEFINITION E. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} be a pointed group on A. A primordial pointed subgroup P_{𝔮} of H_{𝔭} is called a defect pointed subgroup of H_{𝔭} (or simply a defect of H_{𝔭}) provided that H_{𝔭} is relatively P_{𝔮}-projective. In this case, P is called a defect group of H_{𝔭}.

In other words, a pointed group P_{𝔮} on A is a defect of a pointed group H_{𝔭} on A, if and only if

P_{𝔭} is primordial, P_{𝔮} ≤ H_{𝔭} and H_{𝔭} pr P_{𝔮}.

Since the relations ≤ and pr, as well as the property of being primordial behave well with respect to conjugation, ^{g}(P_{𝔮}) is a defect of ^{g}(H_{𝔭}), for all g ∈ G, in case P_{𝔮} is a defect of H_{𝔭}. In particular, any N_{G}(H_{𝔭})-conjugate of a defect of H_{𝔭} is a defect of H_{𝔭}, as well.

Note that, if a pointed group on a unitary Green functor with maximal second component has a defect in Thévenaz’s sense then this is a defect of the pointed group in the sense of definition E, as well; for our definitions of the relations ≤ and pr coincide with the ones of [25], except that we defined them on a possibly bigger set. Moreover, a pointed group is primordial in the sense of [25], if and only if it is primordial in our sense.

The following lemma asserts that defects always exist:

LEMMA F. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} be a pointed group on A. Then a defect of H_{𝔭} exists.

The next result can be regarded as an analogue of Sylow’s second theorem in the following version: If P is a p-subgroup of a subgroup H of G, and K is a subgroup of H with index coprime to p in H then P is H-conjugate to a subgroup of K.

THEOREM G. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} be a pointed group on A. Moreover, let P_{𝔮} be a primordial pointed subgroup of H_{𝔭}, and let K_{𝔯} be a pointed group on A such that H_{𝔭} is relatively K_{𝔯}-projective. Then there exists h ∈ H such that ^{h}(P_{𝔮}) ≤ K_{𝔯}. In particular, any two defects of H_{𝔭} are conjugate by an element of H.

The following result provides several necessary and sufficient criteria for a pointed group to be a defect of some given other pointed group. Moreover, it illustrates even more why defect pointed subgroups are analogues of Sylow subgroups: As Sylow p-subgroups are maximal p-subgroups, defect pointed subgroups are maximal primordial pointed subgroups. At this, and in the following, the words “maximal” or “minimal” in connection with pointed groups always refer to the partial order relation ≤.

THEOREM H. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} and P_{𝔮} be pointed groups on A. Then the following assertions are equivalent:

(a) P_{𝔮} is a defect of H_{𝔭}.

(b) P_{𝔮} is minimal among the pointed groups K_{𝔯} on A satisfying H_{𝔭} pr K_{𝔯}.

(c) P_{𝔭} is a maximal primordial pointed subgroup of H_{𝔭}.

(d) H_{𝔭} is relatively P_{𝔮}-projective, ker(Br_{P} ∘ Res_{P}^{H}) ⊆ 𝔭, and 𝔮 = 𝔱 for any prime ideal 𝔱 in A(P) which is contained in 𝔮 such that H_{𝔭} is relatively P_{𝔱}-projective.

Note that theorem H “(a) ⇔ (c)” and lemma F imply the existence of maximal primordial pointed subgroups, which, as far as we can see, is not at all obvious.

Next we state the Puig and Green correspondences for our generalized notion of pointed groups. In this more general setup, new structures come into game, namely Zariski topologies and partial orders. Before we state the results, let us recall the definition of the Zariski topology. Let A be an algebra. Then the set of prime ideals in A, denoted by Spec(A), is called the spectrum of A. This set is, on the one hand, a partially ordered set (poset) just by set-theoretical inclusion, and, on the other hand, it is a topological space. This topology, called the Zariski topology, is defined as follows: The closed subsets of Spec(A) are, by definition, the sets of the form

V (X) := {𝔭 ∈ Spec(A) : X ⊆ 𝔭}

where X is some subset of A. We shall abbreviate V ({x}) to V (x) for x ∈ X. Clearly, we have V (X) = V (kX + AXA) for any subset X of A, where kX + AXA denotes the ideal of A generated by X. Thus any closed subset of Spec(A) is of the form V (I) for some ideal I in A, and any such subset is closed. We have V (A) = ∅, V (0) = Spec(A), and, for ideals I_{y} in A (y ∈ Y, Y some index set),

⋂_{y∈Y} V (I_{y}) = V (∑_{y∈Y} I_{y}).

Moreover, V (I)∪V (J) = V (IJ)(= V (JI)) for all ideals I, J in A (although A need not be commutative!), thanks to the fact that Spec(A) consists only of prime ideals. Hence this defines in fact a topology on Spec(A).

Now let us describe the open subsets. Let X be a subset of A. Then

Spec(A) ∖ V (X) = {𝔭 ∈ Spec(A) : X ⊈ 𝔭}.

Hence the open subsets of Spec(A) are precisely the sets of the form U(X) := {𝔭 ∈ Spec(A) : X ⊈ 𝔭} where X is some subset of A. Cleary, U(X) = ⋃_{x∈X} U(x) for any subset X of A. (Where U(x) := U({x}) for x ∈ A.) Thus the sets U(x) (x ∈ A) form a basis of the topology. Moreover, U(X) = U(kX + AXA) for any subset X of A, and therefore the open subsets of Spec(A) are precisely the sets of the form U(I), where I is some ideal of A.

Let 𝔭 and 𝔮 be prime ideals in A such that 𝔮 ⊆ 𝔭, and denote by C the connected component of Spec(A) containing 𝔮. As C is closed, we must have C = V (I) for some ideal I in A. Hence I ⊆ 𝔮 ⊆ 𝔭, and thus 𝔭 ∈ C. This shows that any path component of Spec(A), as poset, is contained in some connected component of Spec(A) as topological space. Hence any connected component of Spec(A) is a disjoint union of path components. In case A possesses only finitely many minimal prime ideals, these notions in fact coincide; for if the path component P of Spec(A) is contained in the connected component C, 𝔭_{1}, …, 𝔭_{n} (n ∈ ℕ) denote the minimal prime ideals of A contained in P, and 𝔮_{1}, …, 𝔮_{m} (m ∈ ℕ) denote the minimal prime ideals of A contained in C ∖ P, then we obviously have P = V (𝔭_{1}) ∪ … ∪ V (𝔭_{n}), which is closed, and

P = C ∩ ⋂_{i=1}^{m} U(𝔮_{i}),

which is open. (Recall that any prime ideal contains some minimal prime ideal by Zorn’s lemma (cf. the remark subsequent to corollary (3.7)).) Consequently, P = C as P ≠ ∅.

In this more general setup the Puig correspondence reads as follows:

THEOREM I (Puig correspondence). Let H be a subgroup of a finite group G, let A be a Green functor for G, and let P_{𝔮} be a primordial pointed group on A such that P is contained in H. Set S := A(P)/𝔮, and denote by π_{𝔮} : A(P) → S the canonical epimorphism. Then the homomorphism π_{𝔮} ∘ Res_{P}^{H} induces a bijection

Γ : {𝔭 ∈ Spec(A(H)) : P_{𝔮} is a defect of H_{𝔭}} ⟶
\{\overline{𝔭} ∈ Spec(S^{\overline{N}_{H}(P_{𝔮})}) : 1_{0} is a defect of \overline{N}_{H}(P_{𝔮})\overline{𝔭} \}

which is both, a homeomorphism, as well as an isomorphism of posets. The inverse of Γ is given by the assignment \overline{𝔭} ↦ (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}). Moreover, in case A is unitary, a prime ideal 𝔭 contained in the domain of Γ is a maximal ideal in A(H), if and only if \overline{𝔭} := Γ(𝔭) is a maximal ideal in S^{\overline{N}_{H}(P_{𝔮})}; and in this case, the homomorphism π_{𝔮} ∘ Res_{P}^{H} induces an isomorphism A(H)/𝔭 →^{∼} S^{\overline{N}_{H}(P_{𝔮})}/\overline{𝔭}.

We adopt the notation of theorem I. First of all, note that, as N_{H}(P_{𝔮}) fixes 𝔮, \overline{N}_{H}(P_{𝔮}) acts on S via the conjugation maps. Hence S is a \overline{N}_{H}(P_{𝔮})-algebra which may be regarded as Green functor for \overline{N}_{H}(P_{𝔮}). Therefore, we may speak of pointed groups on S and defects of such. If prime ideals 𝔭 and \overline{𝔭} in the domain respectively codomain of Γ correspond under the bijection Γ then 𝔭 is called the Puig correspondent of \overline{𝔭}, and conversely \overline{𝔭} is called the Puig correspondent of 𝔭. We also say that H_{𝔭} is the Puig correspondent of \overline{N}_{H}(P_{𝔮})\overline{𝔭}, and vice versa.

Note also that theorem I shows that the version of the Puig correspondence stated in [25] (theorem (58.7)) is equally valid without the additional hypothesis on A ([25] assumption (58.1)).

The Puig correspondence has several consequences; we start with an immediate one:

COROLLARY J. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} be a pointed group on A with defect P_{𝔮}. Assume that the residue algebra A(P)/𝔮 is commutative. Then 𝔭 = (Res_{P}^{H})^{−1}(𝔮).

Thus, in the commutative case, the second component 𝔭 of some pointed group H_{𝔭} on A is uniquely determined by a defect of H_{𝔭}. Therefore, the sets involved in the Puig correspondence are either empty or singletons in this case. Note that this does not hold in general without the commutativity hypothesis; for it is well-known that distinct pointed groups on a G-algebra might very well have the same defect.

COROLLARY K. Let G be a finite group, let A be a unitary Green functor for G, and let P_{𝔮} be a defect of some pointed group H_{𝔭} on A. Denote by S the residue algebra A(P)/𝔮, and assume that S is finitely generated as module over its center. Then 𝔭 is a maximal ideal in A(H) if and only if 𝔮 is a maximal ideal in A(P).

A similar result is the following. It demonstrates in particular that, for Green functors which are “sufficiently close to commutative”, there are no proper inclusions in the posets involved in the Puig correspondence.

COROLLARY L. Let G be a finite group, let A be a unitary Green functor for G, let P_{𝔮} be a primordial pointed group on A, and let H_{𝔭_{1}} ≤ H_{𝔭_{2}} be pointed groups on A which both have defect P_{𝔮}. Assume further that S := A(P)/𝔮 is finitely generated as module over its center. Then H_{𝔭_{1}} = H_{𝔭_{2}}.

Corollary L immediately implies the following result:

COROLLARY M. Let G be a finite group, let A be a unitary Green functor for G, and let R be a primordial subgroup of G for A of maximal order. Assume that there is a natural number N such that, for any primordial pointed group P_{𝔮} on A, A(P)/𝔮 is finitely generated as module over its center, and the Krull dimension of A(P)/𝔮 is less or equal than N. Then the Krull dimension of A(G) is less or equal than N + log_{2}(|R|)N + log_{2}(|R|).

Note that for a unitary Green functor A the hypotheses of corollary M are in particular satisfied in case \overline{A}(P) is finitely generated as module over its center and has finite Krull dimension for all primordial subgroups P of G for A. We can then take R as above and N to be the maximum of all these Krull dimensions.

We will see in §6 that, using Dress’s induction theorem, the assertion of corollary M can be refined (cf. remark (6.2)).

Some other results of the above type can be found in §5 (cf. lemmas (5.6) and (5.7)). The most important application of theorem I is the Green correspondence which we state next:

THEOREM N (Green correspondence). Let G be a finite group, let A be a Green functor for G, let P_{𝔮} be a primordial pointed group on A, and let H be a subgroup of G containing N_{G}(P_{𝔮}). Moreover, let 𝔭 be a prime ideal in A(G) such that P_{𝔮} is a defect of G_{𝔭}, and denote by 𝔖_{𝔭} the subposet of Spec(A(H)) consisting of prime ideals 𝔱 in A(H) such that P_{𝔮} ≤ H_{𝔱} ≤ G_{𝔭}. Then P_{𝔮} is a defect of H_{𝔱} for all 𝔱 ∈ 𝔖_{𝔭}. Furthermore, 𝔖_{𝔭} has a unique maximal element 𝔯_{𝔭} (in particular, 𝔖_{𝔭} ≠ ∅), and the assignment 𝔭 ↦ 𝔯_{𝔭} defines a map

Γ : {𝔭 ∈ Spec(A(G)) : P_{𝔮} is a defect of G_{𝔭}} ⟶
{𝔯 ∈ Spec(A(H)) : P_{𝔮} is a defect of H_{𝔯}}

which is both, a homeomorphism, as well as an isomorphism of posets. Moreover, if 𝔭 is a prime ideal contained in the domain of Γ and 𝔯 := Γ(𝔭), then the following assertions hold:

(a) 𝔭 = (Res_{H}^{G})^{−1}(𝔯).

(b) G_{𝔭} is relatively H_{𝔯}-projective.

(c) In case A is unitary, 𝔭 is a maximal ideal of A(G), if and only if 𝔯 is a maximal ideal of A(H); and, in this case, Res_{H}^{G} induces an isomorphism

A(G)/𝔭 →^{∼} A(H)/𝔯.

The bijection Γ of theorem N is called the Green correspondence. If prime ideals 𝔭 and 𝔯 correspond under this bijection then 𝔭 is called the Green correspondent of 𝔯, and conversely 𝔯 is called the Green correspondent of 𝔭. We also say that G_{𝔭} is the Green correspondent of H_{𝔯}, and vice versa. Note that the Green correspondence depends on the choice of the primordial subgroup P_{𝔮} (cf. [25] remark (20.2)).

We adopt the notation of theorem N. Note that in case A(P)/𝔮 is finitely generated as module over its center, corollary L tells us that, for any primordial pointed group G_{𝔭} on A with defect P_{𝔮}, the set 𝔖_{𝔭} of theorem O consists just of a single element, namely the Green correspondent of 𝔭 (with respect to P_{𝔮}).

An important consequence of the Green correspondence is the Burry-Carlson-Puig theorem which we state next. (Actually, the result is more or less included in the version of the Green correspondence we stated above.)

THEOREM O. Let G be a finite group, let A be a Green functor for G, let P_{𝔮} be a primordial pointed group on A, and let H be a subgroup of G containing N_{G}(P_{𝔮}). Moreover, let G_{𝔭} and H_{𝔯} be pointed groups on A such that P_{𝔮} ≤ H_{𝔯} ≤ G_{𝔭}. Then the following assertions hold:

(a) If P_{𝔮} is a defect of G_{𝔭} then P_{𝔮} is a defect of H_{𝔯}. If, in addition, 𝔯 = 𝔱 for any prime ideal 𝔱 in A(H) such that H_{𝔯} ≤ H_{𝔱} ≤ G_{𝔭} then H_{𝔯} is the Green correspondent of G_{𝔭} (with respect to P_{𝔮}).

(b) If P_{𝔮} is a defect of H_{𝔯} and 𝔭′ = 𝔭 for any prime ideal 𝔭′ in A(G) such that H_{𝔯} ≤ G_{𝔭′} ≤ G_{𝔭} then P_{𝔮} is a defect of G_{𝔭} and G_{𝔭} is the Green correspondent of H_{𝔯} (with respect to G_{𝔭}).

(c) Assume that A is unitary, and that 𝔯 is a maximal ideal in A(H). Then P_{𝔮} is a defect of H_{𝔯}, if and only if P_{𝔮} is a defect of G_{𝔭}; and, in this case, H_{𝔯} is the Green correspondent of G_{𝔭} (with respect to P_{𝔮}).

The next result, which is a corollary of the Burry-Carlson-Puig theorem, may be regarded as an analogue of Sylow’s first theorem in the following version: If Q is a proper subgroup of a Sylow p-subgroup P of G then there exists a p-subgroup R of G normalizing Q such that Q ⊂ R ⊆ P (e.g. take R = N_{P} (Q)).

COROLLARY P. Let G be a finite group, let A be a unitary Green functor for G, and let Q_{𝔮} < P_{𝔭} be primordial pointed groups on A. Moreover, assume that 𝔮 is a maximal ideal in A(Q), and that A(Q)/𝔮 is finitely generated as module over its center. Then there exists a primordial pointed group R_{𝔯} on A such that Q_{𝔮} < R_{𝔯} ≤ P_{𝔭}, and R ⊆ N_{P} (Q_{𝔮}). In particular, Q ⊂ N_{P} (Q_{𝔮}).

We adopt the notation of corollary P. As far as we can see, it is not clear whether we may choose R_{𝔯} such that 𝔯 is a maximal ideal in A(R), not even if we assume that A(T)/𝔱 is finitely generated as module over its center for all primordial pointed groups T_{𝔱} on A.

The following fact demonstrates, in particular, that one can say even more about the situation of corollary P.

LEMMA Q. Let G be a finite group, let A be a Green functor for G, and let Q_{𝔮} ≤ P_{𝔭} be primordial pointed groups on A such that P ⊆ N_{G}(Q_{𝔮}). Then P/Q is a p-group for some prime number p. Moreover, if c_{𝔮} and c_{𝔭} denote the characteristics of A(Q)/𝔮 and A(P)/𝔭, respectively, then the following assertions hold:

(a) In case c_{𝔭} = 0, we have c_{𝔮} = 0 and P = Q.

(b) If Q ≠ P then c_{𝔭} = p and c_{𝔮} ∈ {0, p}.

Under the additional hypothesis of all the involved algebras being finitely generated as modules over the ground ring, Sylow’s first theorem for pointed groups can be improved essentially:

COROLLARY R. Let G be a finite group, let A be a unitary Green functor for G, and let Q_{𝔮} ≤ P_{𝔭} be primordial pointed groups on A. Assume further that, for any primordial pointed group T_{𝔱} on A such that Q_{𝔮} ≤ T_{𝔱} ≤ P_{𝔭}, A(T)/𝔱 is finitely generated as module over the ground ring k. Then there is a chain

Q_{𝔮} = (R_{0})_{𝔯_{0}} ≤ (R_{1})_{𝔯_{1}} ≤ … ≤ (R_{s})_{𝔯_{s}} = P_{𝔭}

(s ∈ ℕ) of primordial pointed groups on A such that R_{i+1} ⊆ N_{P} ((R_{i})_{𝔯_{i}}) for all i = 0, …, s − 1, and |P : Q| is a power of some prime number p. In particular, Q is subnormal in P. Moreover, if c_{𝔮} and c_{𝔭} denote the characteristics of A(Q)/𝔮 and A(P)/𝔭, respectively, then the following assertions hold:

(a) In case c_{𝔭} = 0, we have c_{𝔮} = 0 and Q = P.

(b) If Q ≠ P then c_{𝔭} = p and c_{𝔮} ∈ {0, p}.

At a first sight, it might be a little bit suprising that also this result of Puig’s theory carries over to the more general setup we are concerned with here; for primordial groups need not be p-groups in general. For instance, if B is the Burnside ring functor for G then any subgroup of G is the first component of some primordial pointed group on B. The reader might imagine that corollary R, in view of this fact, is correlated to a theorem of Dress [10] which asserts that the integral Burnside ring of a finite group G has non-trivial idempotents if and only if the group G is not solvable. (Recall that a commutative ring R has non-trivial idempotents if and only if Spec(R) is not connected (cf. [15] corollary (7.3)).) We will give a proof of this result using corollary R in Appendix A in order to illustrate the theory.

We will show in section §5 (cf. corollary (6.1)) that, using Dress’s induction theorem, some of the assertions of corollary R carry over to arbitrary Green functors.

The following immediate consequence of Sylow’s first theorem for primordial pointed groups will serve as a basis for our proof of Dress’s induction theorem [11] (cf. also [28]). Apart from corollary R, the proof of this result relies on the ideas of the concept of hyperminimal pointed groups introduced in [24] by J. Thévenaz. Before we state the result, let us recall that, for a prime p, O^{p}(G), the p-residue of G, denotes the unique minimal normal subgroup of G the index of which in G is a power of p.

COROLLARY S. Let G be a finite group, let A be a unitary Green functor for G, let H_{𝔭} be a pointed group on A with defect group P, and let C be a set of subgroups of H such that

⋂_{K∈C} ker Res_{K}^{H} ⊆ 𝔭.

Assume further that, for any primordial pointed group T_{𝔱} on A, A(T)/𝔱 is finitely generated as module over the ground ring k. Denote by c the characteristic of A(H)/𝔭. Then the following assertions hold:

(a) In case c = 0, P is conjugate in H to a subgroup of K for some K ∈ C.

(b) Otherwise (i.e. in case c ≠ 0) c is some prime number p, and, for some K ∈ C, O^{p}(P) is conjugate in H to a subgroup of K.

REMARK T. We adopt the notation of corollary S. Note that

( ⋂_{K∈C} ker Res_{K}^{H})( ∑_{K∈C} A_{K}^{H}) = 0,

by the Frobenius axiom. Thus, for any prime ideal 𝔭 in A(H), we either have

⋂_{K∈C} ker Res_{K}^{H} ⊆ 𝔭,

and corollary S applies (yielding information on the defect group of H_{𝔭}), or

∑_{K∈C} A_{K}^{H} ⊆ 𝔭

(or both).

Using corollary S one can easily deduce Dress’s induction theorem [11] in the following form. (We refer to [25] for the definition of Mackey functors.)

THEOREM U (Dress’s Induction Theorem). Let G be a finite group, let M be a Mackey functor for G over k, and let C be a set of subgroups of G. Then the following assertions hold:

(a) Assume that |G| is invertible in k. Then

M(G) = ( ⋂_{H∈C} ker Res_{H}^{G}) ⊕ ( ∑_{H∈C} M_{H}^{G}).

(b) Let π ⊆ ℤ be a non-empty set of primes, and let |G| = |G|_{π}|G|_{π′} be the decomposition of |G| into its π-part and π′-part, respectively (i.e. any prime divisor of |G|_{π} lies in π, whereas no prime divisor of |G|_{π′} is contained in π). Assume that |G|_{π′} is invertible in k. Moreover, denote by C^{π} the set of subgroups H of G such that, for some p ∈ π and some K ∈ C, O^{p}(H) ⊆ K. Then

M(G) = ( ⋂_{H∈C} ker Res_{H}^{G}) + ( ∑_{H∈C^{π}} M_{H}^{G})

= ( ⋂_{H∈C^{π}} ker Res_{H}^{G}) ⊕ ( ∑_{H∈C^{π}} M_{H}^{G}).

Of course, with notation of theorem U, M_{H}^{G} denotes the image of Ind_{H}^{G} : M(H) → M(G) for any subgroup H of G.

Note that Dress’s induction theorem in [11] is more precise; for it asserts even that

|G| M(G) ⊆ ( ⋂_{H∈C} ker Res_{H}^{G}) + ( ∑_{H∈C} M_{H}^{G})

for any Mackey functor M over ℤ, and analogously for the second assertion. This makes it more explicit which denominators are actually needed in order to decompose an element of M(G) as above. However, theorem U probably covers the most important applications of Dress’s induction theorem.

The next result deals with Puig’s approach to induction theorems for the Green ring introduced in [19], which replaces kernels of restriction maps by kernels of Brauer homomorphisms. In [23], J. Thévenaz extended this theory to arbitrary Green functors (or more generally, Mackey functors) and also proved several new results in this context.

Let M be a Mackey functor for G over k, and let H be a subgroup of G. As for Green functors, \overline{M}(H) denotes the residue k-module

M(H)/ ∑_{K} M_{K}^{H},

where K ranges over all proper subgroups of G. Moreover, Br_{H}^{M} (or just Br_{H}), called the Brauer homomorphism, denotes the canonical epimorphism M(H) → \overline{M}(H). Of course, H is called primordial for M in case \overline{M}(H) ≠ 0. More generally, Br_{H}^{G} : M(G) → \overline{M}(H) denotes the composition of the restriction map Res_{H}^{G} : M(G) → M(H) followed by the Brauer homomorphism Br_{H} : M(H) → \overline{M}(H). (Thus Br_{G} = Br_{G}^{G}.) Also Br_{H}^{G} is called a Brauer homomorphism.

Moreover, we set \underline{M}(H) := ⋂_{K} ker Res_{K}^{H}, where K ranges over all proper subgroups K of H. This submodule of M(H) is a kind of dual of the kernel of the Brauer homomorphism of H. Also, for a set of primes π ⊆ ℤ, set

\underline{M}_{π}(H) := ⋂_{K} ker Res_{K}^{H},

where K ranges over all proper subgroups of H such that O^{p}(U) ≠ O^{p}(H) for any subgroup U of K and any p ∈ π. Now we are in a position to formulate an analogue of Dress’s induction theorem:

THEOREM V (Dress’s Induction Theorem à la Puig). Let G be a finite group, let M be a Mackey functor for G over k, and let 𝔓 be a set of primordial subgroups of G for M. Then the following assertions hold:

(a) Assume that |G| is invertible in k. Then

M(G) = ( ⋂_{Q∈𝔓} ker Br_{Q}^{G}) ⊕ ( ∑_{Q∈𝔓} Ind_{Q}^{P}(\underline{M}(Q))).

(b) Let π ⊆ ℤ be a non-empty set of primes, and let |G| = |G|_{π}|G|_{π′} be the decomposition of |G| into its π-part and π′-part, respectively. Assume that |G|_{π′} is invertible in k. Moreover, denote by 𝔓^{π} the set of primordial subgroups R of G for M such that, for some p ∈ π and some Q ∈ 𝔓, O^{p}(R) = O^{p}(Q). Then

M(G) = ( ⋂_{Q∈𝔓} ker Br_{Q}^{G}) + ( ∑_{Q∈𝔓^{π}} Ind_{Q}^{G}(\underline{M}_{π}(H))).

Assertion (a) of theorem V was proved by L. Puig [19] for the Green ring over ℚ (cf. the remarks subsequent to corollary W, below).

An immediate consequence of Dress’s induction theorem and theorem V (a) is the following special case of a result of Thévenaz in [23]:

COROLLARY W. Let G be a finite group, let M be a Mackey functor for G over k, and let 𝔓 be the set of primordial subgroups of G for M. Assume that |G| is invertible in k. Then

⋂_{Q∈𝔓} ker Br_{Q}^{G} = 0.

This result was proved by L. Puig [19] for the Green ring a_{ℚ}(OG) over ℚ (with respect to a complete discrete valuation ring O of characteristic 0 and algebraically closed residue field of characteristic p ≠ 0). Let a_{ℚ} denote the corresponding Green functor for G over ℚ. Puig points out in [19] that the product of the Brauer homomorphisms induces an isomorphism

∏_{Q} Br_{Q}^{G} : a_{ℚ}(OG) → ∏_{Q} \overline{a}_{ℚ}(Q)^{N_{G}(Q)},

where Q ranges over a transversal for the conjugacy classes of primordial subgroups of G for a_{ℚ} (which are well-known to be the subgroups H of G such that H/ O^{p}(H) is cyclic, in this case). This result was generalized in [23] to arbitrary Mackey functors by J. Thévenaz. Thévenaz shows that, in general, both, the kernel as well as the cokernel of the above map are |G|-torsion modules. Moreover, Thévenaz demonstrates that, for an arbitrary unitary Green functor, the kernel of the above map is a nilpotent ideal.

§3. SOME REMARKS ON PRIME IDEALS IN NONCOMMUTATIVE RINGS

The aim of this section is to recall the required facts on prime ideals. Some results on prime ideals in noncommutative rings can be found in [14] or [17]. For some of the results we need, we don’t know any reference; so we sate the proofs, although some of these facts are certainly and some others might be well-known. Anyway, this might be convenient for the reader.

Throughout this section, we fix an algebra A (over k). Of course, ideal in A will always mean k-algebra ideal. Recall that a prime ideal in A is a proper ideal 𝔭 in A such that

IJ ⊆ 𝔭 ⇒ I ⊆ 𝔭 or J ⊆ 𝔭

for all ideals I and J in A. Recall also that, if A is noncommutative, this does not necessarily imply that

ab ∈ 𝔭 ⇒ a ∈ 𝔭 or b ∈ 𝔭

for all a, b ∈ A. However, any proper ideal 𝔭 in A having this property is a prime ideal. Moreover, if A has an identity (or, more generally, in case A^{2} = A), any maximal ideal of A is prime.

Recall that an m-system (generalized multiplicative system) (in A) is a nonempty subset S of A such that for all s, t ∈ S either st ∈ S or there exists some a ∈ A such that sat ∈ S (or both). (Not that, as our algebras do not necessarily have identities, one needs to include the possibility st ∈ S to make things, namely propositions (3.1) and (3.2), work.)

Like in the commutative case, this notion yields an alternative description of prime ideals. The following result can be found in [17] (corollary (10.4)) in the case of rings with identity; we leave it to the reader to verify that its proof carries over to this more general setup.

(3.1) PROPOSITION. Let 𝔭 be an ideal in an algebra A. Then 𝔭 is a prime ideal in A, if and only if A ∖ 𝔭 is an m-system.

Also the following proposition is an analogue of a well-known result of commutative algebra. Again, its proof can be found in [17] (proposition (10.5)) in the case of rings with identity; and we leave it to the reader to check that it carries over to this more general situation.

§3. PRIME IDEALS IN NONCOMMUTATIVE RINGS 15

(3.2) PROPOSITION. Let S be an m-system in an algebra A. Then any maximal element (with respect to set-theoretic inclusion) in the set 𝔖 of ideals I in A satisfying I ∩ S = ∅ is a prime ideal in A.

We adopt the notation of proposition (3.2). Note that, if I is an ideal in A such that I ∩ S = ∅ (thus 𝔖 ≠ ∅) then, by Zorn’s lemma, 𝔖 has a maximal element containing I. (If any member of an ascending chain of ideals intersects trivially with S and contains I then so does the union of all these ideals.)

As an application, we show that the easy part of the Lying-over Theorem carries over to a noncommutative situation (and an arbitrary extension). The proof of this result follows the ideas of the proof of the Lying-over Theorem stated in [14] (theorem VIII.5.9).

(3.3) COROLLARY (“Weak Lying-over Theorem”). Let A be a subalgebra of an algebra B, and let 𝔭 be a prime ideal in A. Then there exists a prime ideal 𝔮 in B such that A ∩ 𝔮 ⊆ 𝔭. Moreover, we may choose 𝔮 maximal subject to this condition.

PROOF. By proposition (3.2), A∖𝔭 is an m-system in A; but then it is, plainly, also an m-system in B. Moreover, 0 is an ideal in B such that 0 ∩ (A ∖ 𝔭) = ∅. Therefore, there exists a prime ideal 𝔮 in B such that 𝔮∩(A∖𝔭) = ∅, by proposition (3.2). Hence A ∩ 𝔮 ⊆ 𝔭. Moreover, proposition (3.2) shows that we may choose 𝔮 maximal subject to the condition (A ∖ 𝔭) ∩ 𝔮 = ∅, i.e. A ∩ 𝔮 ⊆ 𝔭. □

For later use, we state a slightly modified version of corollary (3.3):

(3.4) COROLLARY. Let A, B be algebras, and let ψ : A → B be a homomorphism of algebras. Moreover, let I be an ideal in B, and let 𝔭 be a prime ideal in A such that ψ^{−1}(I) ⊆ 𝔭. Then there exists a prime ideal 𝔮 in B containing I such that ψ^{−1}(𝔮) ⊆ 𝔭. Moreover, we may choose 𝔮 maximal subject to this condition.

PROOF. Clearly, ψ induces an isomorphism of algebras

A/ψ^{−1}(I) → ψ(A) + I/I

(where ψ(A) + I/I stands for (ψ(A) + I)/I). As 𝔭 contains ψ^{−1}(I), 𝔭/ψ^{−1}(I) is a prime ideal in A/ψ^{−1}(I), and so ψ(𝔭) + I/I is a prime ideal in ψ(A) + I/I ⊆ B/I. By corollary (3.3), there exists a prime ideal 𝔮 in B containing I such that

(𝔮 ∩ ψ(A)) + I = 𝔮 ∩ (ψ(A) + I) ⊆ ψ(𝔭) + I;

and we may assume that 𝔮 is maximal subject to this condition. Let x ∈ ψ^{−1}(𝔮). Then ψ(x) ∈ 𝔮 ∩ ψ(A) ⊆ ψ(𝔭) + I. Consequently, ψ(x) = ψ(p) + y for certain p ∈ 𝔭 and y ∈ I. Thus ψ(x − p) = y ∈ I, hence x − p ∈ ψ^{−1}(I) ⊆ 𝔭, and so x ∈ 𝔭. Therefore, ψ^{−1}(𝔮) ⊆ 𝔭, as desired. Finally, let 𝔮′ be a prime ideal in B containing 𝔮 such that ψ^{−1}(𝔮′) ⊆ 𝔭. Then, for ψ(x) ∈ 𝔮′ ∩ ψ(A), we have x ∈ ψ^{−1}(𝔮′) ⊆ 𝔭, and thus (𝔮′ ∩ ψ(A)) + I ⊆ ψ(𝔭) + I. Therefore, 𝔮 = 𝔮′ by maximality of 𝔮. □

The next application is a generalization of [25] lemma (58.4), which serves as a basis for Thévenaz’s proof of the Puig correspondence for maximal ideals ([25] theorem (58.7)); and the following result will be the tool for the proof of the Puig correspondence for prime ideals:

(3.5) COROLLARY. Let A be a subalgebra of an algebra B, and let I be an ideal in B which is contained in A. Then the map

{𝔮 ∈ Spec(B) : I ⊈ 𝔮} → {𝔭 ∈ Spec(A) : I ⊈ 𝔭}, 𝔮 ↦ A ∩ 𝔮

is both, a homeomorphism, as well as an isomorphism of posets. Moreover, for a prime ideal 𝔮 in B such that I ⊈ 𝔮, the following assertions are satisfied:

(a) If 𝔮 is a maximal ideal in B, then A ∩ 𝔮 is a maximal ideal in A; and, in this case, the inclusion A ↪ B induces an isomorphism A/A ∩ 𝔮 →^{∼} B/𝔮.

(b) In case A and B both have (not necessarily the same) multiplicative identities, the converse holds, as well, i.e. 𝔮 is a maximal ideal in B, if A∩𝔮 is a maximal ideal in A. Thus, in this case, the above map restricts to a bijection

{𝔪 ∈ Max(B) : I ⊈ 𝔪} → {𝔫 ∈ Max(A) : I ⊈ 𝔫}

where Max(B) and Max(A) denote the set of maximal ideals in B and A, respectively.

PROOF. Denote the map of corollary (3.5) by Γ. Let 𝔮 be a prime ideal of B such that I ⊈ 𝔮. We show that A ∩ 𝔮 is a prime ideal in A. Let J, K be ideals in A such that JK ⊆ A ∩ 𝔮. Clearly, IJI and IKI are ideals in B. Moreover, (IJI)(IKI) ⊆ JK ⊆ A ∩ 𝔮 ⊆ 𝔮 since I ⊆ A. As 𝔮 is prime, we must have IJI ⊆ 𝔮 or IKI ⊆ 𝔮. Hence I(J +BJB)I ⊆ 𝔮 or I(K +BKB)I ⊆ 𝔮 since I is an ideal in B. Therefore, J ⊆ J + BJB ⊆ 𝔮 or K ⊆ K + BKB ⊆ 𝔮 since 𝔮 is a prime ideal in B which is not containing I. Consequently, J = A ∩ J ⊆ A ∩ 𝔮 or K = A ∩ K ⊆ A ∩ 𝔮. Clearly, I ⊈ A ∩ 𝔮, as I ⊈ 𝔮. In particular, A ∩ 𝔮 is a proper ideal in A, and hence a prime ideal. This shows that Γ is in fact well-defined.

Let 𝔭 be a prime ideal in A such that I ⊈ 𝔭. By corollary (3.3), there exists a prime ideal 𝔮 in B such that A ∩ 𝔮 ⊆ 𝔭. The proof of corollary (3.3) shows that we may assume that 𝔮 is a maximal element in the poset of ideals J in B satisfying J ∩ A ⊆ 𝔭. (Note that the latter inclusion is equivalent to J ∩ (A ∖ 𝔭) = ∅.) We must have I ⊈ A ∩ 𝔮, as I ⊈ 𝔭, and thus I = A ∩ I ⊈ 𝔮. Hence A ∩ 𝔮 is a prime ideal in A, by the first part of the proof. Moreover, I𝔭I is an ideal in B since I is. Besides,

A ∩ (𝔮 + I𝔭I) = (A ∩ 𝔮) + I𝔭I ⊆ 𝔭

since I ⊆ A. Hence I𝔭I ⊆ 𝔮 + I𝔭I = 𝔮, by maximality of 𝔮. Thus I𝔭I ⊆ A ∩ 𝔮, as I ⊆ A. Therefore, 𝔭 ⊆ A ∩ 𝔮 since A ∩ 𝔮 is prime, and hence 𝔭 = A ∩ 𝔮. This shows that Γ is surjective.

Let 𝔮, 𝔮′ be prime ideals in B such that I ⊈ 𝔮, I ⊈ 𝔮′, and A ∩ 𝔮 ⊆ A ∩ 𝔮′. Suppose 𝔮 ⊈ 𝔮′. Then I𝔮 ⊈ 𝔮′. However, I𝔮 ⊆ I ∩ (I𝔮) ⊆ A ∩ (I𝔮) ⊆ I𝔮, and thus I𝔮 = A ∩ (I𝔮) ⊈ A ∩ 𝔮′. On the other hand, I𝔮 = A ∩ (I𝔮) ⊆ A ∩ 𝔮, which contradicts the assumption A ∩ 𝔮 ⊆ A ∩ 𝔮′. Therefore, 𝔮 ⊆ 𝔮′. On the one hand, this shows that Γ is bijective. On the other hand, this implies that Γ^{−1} preserves order. Since Γ plainly preserves order, Γ is an isomorphism of posets.

We show that Γ is an open and continuous map. First of all, note that the domain and the codomain of Γ are open subsets of Spec(B) and Spec(A), respectively. Thus a subset of the domain (resp. codomain) is open in the induced topology, if and only if it is open in Spec(B) (resp. Spec(A)). Let U be an open subset of the domain of Γ. Then

U = {𝔮 ∈ Spec(B) : J ⊈ 𝔮}

for some ideal J in B. But, as U is contained in the domain of Γ, we also have

U = {𝔮 ∈ Spec(B) : IJ ⊈ 𝔮}.

Let 𝔮 ∈ U. Then IJ ⊈ A ∩ 𝔮. Conversely, let 𝔮 be a prime ideal in B such that IJ ⊈ A ∩ 𝔮. Then IJ ⊈ 𝔮 since IJ ⊆ I ⊆ A. This shows that, for 𝔮 in the domain of Γ, we have 𝔮 ∈ U, if and only if

Γ(𝔮) = A ∩ 𝔮 ∈ Ũ := {𝔭 ∈ Spec(A) : IJ ⊈ 𝔭},

and thus Γ(U) = Ũ, which is open since IJ ⊆ A.

Now let U′ be an open subset of the codomain of Γ. Then, as above,

U′ = {𝔭 ∈ Spec(A) : IK ⊈ 𝔭}

for some ideal K in A. Let 𝔭 be a prime ideal in A, and let 𝔮 be a prime ideal in B such that 𝔭 = A ∩ 𝔮. Plainly, IK ⊆ 𝔭 = A ∩ 𝔮, if and only if IK ⊆ 𝔮, as IK ⊆ A. Thus the inverse image of U′ under Γ is

{𝔮 ∈ Spec(B) : IK ⊈ 𝔮},

which is an open subset of Spec(B).

It remains to prove the additional assertion. Let 𝔮 be a prime ideal in B such that I ⊈ 𝔮. Assume that 𝔮 is a maximal ideal in B. We have

A/A ∩ 𝔮 ≅ A + 𝔮/𝔮 ⊆ B/𝔮.

Moreover, B = I + 𝔮 ⊆ A + 𝔮, as 𝔮 ⊂ I + 𝔮 and 𝔮 is maximal. Thus A/A ∩ 𝔮 ≅ B/𝔮 is a simple algebra, and hence A ∩ 𝔮 is a maximal ideal in A.

Assertion (b) follows from the above and the fact that, in case A and B have multiplicative identities, the sets Max(A) resp. Max(B) coincide with the sets of maximal elements in Spec(A) resp. Spec(B). Thus Γ restricts to a bijection of the desired form in this case. □

Note that corollary (3.5) shows that lemma (58.4) in [25] is equally valid for an arbitrary algebra with identity. This implies that the additional assumption (58.1) in [25] is superfluous for the proof of the Puig correspondence for maximal ideals ([25] theorem (58.7)).

The remaining results of this section are not needed for the general defect theory; we will use them at some places to improve the results under additional hypotheses.

Let S be a multiplicative subset of k, and let A be a k-algebra with identity 1_{A} which is S-torsion-free as k-module (i.e. sa ≠ 0 for all s ∈ S and a ∈ A ∖ {0}). By [20] corollary (3.4), the localization S^{−1}A may be identified with S^{−1}k ⊗_{k} A where, for a ∈ A and s ∈ S, \frac{a}{s} corresponds to s^{−1} ⊗ a. As A is S-torsion-free, by assumption, the canonical map A → S^{−1}A, a ↦ \frac{a}{1}, is injective; so we may regard A as unitary k-subalgebra of S^{−1}A. Moreover, for any k-submodule U of A, S^{−1}U may be regarded as S^{−1}k-submodule of S^{−1}A. Then S^{−1}U is the S^{−1}k-submodule of S^{−1}A generated by U. The following result is well-known for commutative rings.

(3.6) PROPOSITION. Let S be a multiplicative subset of k, and let A be a kalgebra with identity 1_{A} which is S-torsion-free as k-module. Moreover, let 𝔭 be a prime ideal in A such that 𝔭 ∩ S1_{A} = ∅, and let 𝔮 be a prime ideal in S^{−1}A. Then S^{−1}𝔭 is a prime ideal in S^{−1}A such that 𝔭 = A ∩ (S^{−1}𝔭), and A ∩ 𝔮 is a prime ideal in A such that 𝔮 = S^{−1}(A ∩ 𝔮). Thus the assignment 𝔭 ↦ S^{−1}𝔭 induces an isomorphism of posets between the poset of prime ideals 𝔭 in A satisfying 𝔭∩S1_{A} = ∅ and the poset of prime ideals in S^{−1}A.

PROOF. At first, we show that 𝔭 = A ∩ (S^{−1}𝔭). Clearly, 𝔭 ⊆ A ∩ S^{−1}𝔭. Conversely, let x ∈ A ∩ (S^{−1}𝔭). Then sx ∈ 𝔭 for some s ∈ S. Thus (sA)(AxA) ⊆ 𝔭. Hence s1_{A} ∈ sA ⊆ 𝔭 or x ∈ AxA ⊆ 𝔭, as x ∈ A and 𝔭 is prime in A. However, s ∈ S and 𝔭 ∩ S1_{A} = ∅, by assumption; so we are in the latter case, as we wanted to show. In particular, S^{−1}𝔭 is a proper ideal of S^{−1}A.

Next we show that S^{−1}𝔭 is a prime ideal in S^{−1}A. Suppose not. Then there are ideals I and J in S^{−1}A such that IJ ⊆ S^{−1}𝔭, but I ⊈ S^{−1}𝔭 and J ⊈ S^{−1}𝔭. Thus there is x ∈ I ∖(S^{−1}𝔭). We have sx ∈ A for some s ∈ S, and sx ∉ 𝔭 since otherwise x = \frac{1}{s}sx ∈ S^{−1}𝔭. Therefore, sx ∈ (A ∩ I) ∖ 𝔭; and so A ∩ I ⊈ 𝔭. Analogously, one shows that A ∩ J ⊈ 𝔭. However,

(A ∩ I)(A ∩ J) ⊆ A ∩ IJ ⊆ A ∩ (S^{−1}𝔭) = 𝔭,

by the first part of the proof. As 𝔭 is prime, this implies that A∩I ⊆ 𝔭 or A∩J ⊆ 𝔭. Anyway, this is a contradiction. Thus S^{−1}𝔭 is a prime ideal in S^{−1}A.

Now we show that A ∩ 𝔮 is a prime ideal in A. Note first that we must have (A ∩ 𝔮) ∩ S1_{A} = ∅ since otherwise 𝔮 would contain a unit of S^{−1}A. In particular, A ∩ 𝔮 is a proper ideal of A. Let X and Y be ideals of A such that XY ⊆ A ∩ 𝔮. Then

(S^{−1}X)(S^{−1}Y) ⊆ S^{−1}(A ∩ 𝔮) ⊆ 𝔮.

Thus S^{−1}X ⊆ 𝔮 or S^{−1}Y ⊆ 𝔮, and so X ⊆ A ∩ 𝔮 or Y ⊆ A ∩ 𝔮. Therfore, A ∩ 𝔮 is in fact a prime ideal in A. Let x ∈ 𝔮. Then sx ∈ A ∩ 𝔮 for some s ∈ S, and so 𝔮 = S^{−1}(A ∩ 𝔮). The additional assertion is then immediate. □

We adopt the notation of proposition (3.6). Note that it is not really necessary to assume that A is S-torsion-free: Denote by T_{S}(A) the S-torsion submodule of A (i.e. the set of all elements a ∈ A such that sa = 0 for some s ∈ S). Plainly, T_{S}(A) is an ideal in A, and A/T_{S}(A) is S-torsion-free. Moreover, it is straight forward to check that T_{S}(A) ⊆ 𝔭 for any prime ideal 𝔭 of A such that 𝔭∩S1_{A} = ∅; and, by [20] section 3b, there is a canonical isomorphism S^{−1}A → S^{−1}(A/T_{S}(A)). However, the assertion of proposition (3.6) suffices for our purpose, and this assertion is easier to formulate than the more general fact.

Let R be an integral domain with quotient field 𝕂, i.e. 𝕂 = (R ∖ 0)^{−1}R. We then write 𝕂M instead of (R ∖ 0)^{−1}M for any torsion-free R-module M.

(3.7) COROLLARY. Let R be an integral domain with quotient field 𝕂, and let A be an R-algebra with identity 1_{A} which is torsion-free as R-module. Assume that 𝕂A is finite dimensional over 𝕂. Then a prime ideal 𝔭 in A is a minimal prime ideal in A, if and only if R ∩ 𝔭 = 0. Moreover, A has only finitely many minimal prime ideals, and the prime radical of A is nilpotent.

Note that, as A is torsion-free, the canonical map R → R1_{A} is an isomorphism; so we may identify R with R1_{A}. Recall also that, in general, any prime ideal 𝔭 of an algebra A contains some minimal prime ideal of A by Zorn’s lemma; for if 𝔔 is a chain of prime ideals in A, which are all contained in 𝔭 then ⋂_{𝔮∈𝔔} 𝔮 is an ideal in A contained in 𝔭, which is easily checked to be prime, as well.

PROOF of Corollary (3.7). By proposition (3.6), there is an isomorphism of posets from the poset of prime ideals 𝔭 in A satisfying R ∩ 𝔭 = 0 and the set of prime ideals in 𝕂A, which is given by 𝔭 ↦ 𝕂𝔭. However, 𝕂A is an Artinian algebra, by assumption; and so any prime ideal in 𝕂A is a maximal ideal in 𝕂A.

Let 𝔮 ⊆ 𝔭 be prime ideals in A such that R∩𝔭 = 0. Then we also have R∩𝔮 = 0, and so 𝕂𝔮 = 𝕂𝔭. Hence 𝔮 = 𝔭, and thus 𝔭 is a minimal prime ideal in A.

Since 𝕂A has only finitely many maximal ideals, there are only finitely many prime ideals 𝔭_{1}, …, 𝔭_{n} in A which intersect trivially with R. Moreover,

I := 𝔭_{1} · . . . · 𝔭_{n} ⊆ 𝔭_{1} ∩ … ∩ 𝔭_{n} ⊆ 𝕂𝔭_{1} ∩ … ∩ 𝕂𝔭_{n} = J(𝕂A),

where J(𝕂A) denotes the Jacobson radical of A. However, J(𝕂A) is nilpotent, and so is I. Hence I ⊆ 𝔮 for any prime ideal 𝔮 in A; and thus 𝔭_{i} ⊆ 𝔮 for some i. This shows that the 𝔭_{i}’s are precisely the minimal prime ideals of A. So the prime radical of A equals 𝔭_{1} ∩ … ∩ 𝔭_{n} which is contained in the nilpotent ideal J(𝕂A) of 𝕂A, and we are done. □

As far as we can see, it is not clear whether lemma (3.7) holds also without the hypothesis of A being torsion-free.

Note that the assumption of 𝕂A being finite dimensional over 𝕂 is in particular satisfied if A is an R-submodule of some finitely generated R-module, since 𝕂 is flat as R-module by [20] section 3c. Note also that this is less than requiring A to be finitely generated over R, as R may not be Noetherian.

The following consequence of corollary (3.7) is an analogue of a well-known result of commutative algebra.

(3.8) COROLLARY. Let A be a k-algebra with identity 1_{A}, and let 𝔮 ⊆ 𝔮′ be prime ideals in A such that k1_{A} ∩ 𝔮 = k1_{A} ∩ 𝔮′. Assume further that A/𝔮 is finitely generated as module over k. Then 𝔮 = 𝔮′.

PROOF. Replacing k by k1_{A} + 𝔮/𝔮 ≅ k1_{A}/k1_{A} ∩ 𝔮, A by A/𝔮, 𝔮 by 0 and 𝔮′ by 𝔮′/𝔮, we may assume that k is an integral domain which is contained in A, A is a finitely generated prime algebra, and k1_{A} ∩ 𝔮′ = 0. (Note that (k1_{A} + 𝔮) ∩ 𝔮′ = (k1_{A} ∩ 𝔮′) + 𝔮 ⊆ 𝔮, by assumption.) In particular, A is torsion-free as k-module. (Note that αa = 0 for α ∈ k and a ∈ A implies that (αA)(AaA) ⊆ 0.) Thus 𝔮′ is a minimal prime ideal in A, by corollary (3.7), and we are done. □

Let B an algebra with identity, and let A be a unitary subalgebra of B. Assume that A is contained in Z(B), the center of B. Note that, in this situation, it makes sense to define when an element of B is integral over A like in commutative algebra. We leave it to the reader to verify that the proof of [14] Theorem VIII.5.3 carries over verbatim to this more general situation, thanks to the fact that, for b ∈ B, A[b], the subalgebra of B generated by A and b, is commutative. This implies that if B is finitely generated as module over A then any element of B is integral over A. We will apply this fact to A = k1_{B}.

The following result may be regarded as an analogue of the Lying-over Theorem in the noncommutative case; it will serve as a basis for the proof of our best version of Sylow’s first theorem for primordial pointed groups under the additional hypothesis that all the involved algebras are finitely generated over the ground ring (cf. corollary R).

(3.9) THEOREM. Let B be a k-algebra with identity 1_{B}, and let A be a unitary subalgebra of B. Moreover, let 𝔭 be a prime ideal in A, and let 𝔮 be a prime ideal in B such that A ∩ 𝔮 ⊆ 𝔭. Assume that A/𝔭 and B/𝔮 are finitely generated as modules over k. Then there is a prime ideal 𝔮′ in B containing 𝔮 such that A ∩ 𝔮′ ⊆ 𝔭 and k1_{B} ∩ 𝔮′ = k1_{B} ∩ 𝔭.

We adopt the notation of theorem (3.9). Note that, given 𝔭, there exists a prime ideal 𝔮 in B as in theorem (3.9), by corollary (3.3). Moreover, we may choose 𝔮 maximal subject to the property that A ∩ 𝔮 ⊆ 𝔭. In this case, k1_{B} ∩ 𝔮 = k1_{B} ∩ 𝔭, by theorem (3.9).

PROOF of Theorem (3.9). Set \overline{k} := k1_{B} + 𝔮/𝔮 ≅ k1_{B}/k1_{B} ∩ 𝔮. Then B/𝔮 is a finitely generated \overline{k}-algebra, and 0 is a prime ideal in B/𝔮. Moreover, A + 𝔮/𝔮 is a unitary \overline{k}-subalgebra of B/𝔮. We have A+𝔮/𝔮 ≅ A/A∩𝔮, and 𝔭+𝔮/𝔮 is mapped to 𝔭 + (A ∩ 𝔮)/A ∩ 𝔮 = 𝔭/A ∩ 𝔮 under this isomorphism, by assumption. Hence 𝔭 + 𝔮/𝔮 is a prime ideal in A + 𝔮/𝔮. Moreover,

(A + 𝔮/𝔮)/(𝔭 + 𝔮/𝔮) ≅ A/𝔭;

so this algebra is finitely generated as module over k (hence also over \overline{k}).

Assume that we have already found a prime ideal 𝔮′/𝔮 in B/𝔮 such that (A + 𝔮) ∩ 𝔮′ ⊆ 𝔭 + 𝔮, and such that (k1_{B} + 𝔮) ∩ 𝔮′ = (k1_{B} + 𝔮) ∩ (𝔭 + 𝔮). Then 𝔮′ is a prime ideal in B containing 𝔮, and

A ∩ 𝔮′ ⊆ A ∩ ((A + 𝔮) ∩ 𝔮′) ⊆ A ∩ (𝔭 + 𝔮) = 𝔭 + (A ∩ 𝔮) ⊆ 𝔭,

by assumption; in particular, k1_{B} ∩ 𝔮′ ⊆ k1_{B} ∩ 𝔭. Moreover,

k1_{B} ∩ 𝔭 ⊆ k1_{B} ∩ ((k1_{B} + 𝔮) ∩ (𝔭 + 𝔮)) = k1_{B} ∩ ((k1_{B} + 𝔮) ∩ 𝔮′) ⊆ k1_{B} ∩ 𝔮′.

Hence k1_{B} ∩ 𝔮′ = k1_{B} ∩ 𝔭. Therefore, replacing k by \overline{k}, B by B/𝔮, A by A + 𝔮/𝔮, 𝔮 by 0, and 𝔭 by 𝔭 + 𝔮/𝔮, we may assume that k is an integral domain contained in A, B is a prime algebra, finitely generated as k-module, and 𝔮 = 0. In particular, B is torsion-free as k-module.

Then 𝔭̃ := k ∩ 𝔭 is a prime ideal in k, and so k ∖ 𝔭̃ is a multiplicative subset S of k. For a k-module M, we write M𝔭̃ instead of S^{−1}M. Then B𝔭̃ is a k𝔭̃-algebra, finitely generated as k𝔭̃-module. A𝔭̃ may be regarded as unitary subalgebra of B𝔭̃. Moreover, 𝔭∩S = 𝔭∩(k ∖𝔭) = ∅. So 𝔭𝔭̃ is a prime ideal in A𝔭̃, by proposition (3.6). Similarly, 0 is a prime ideal in B𝔭̃. Clearly, B𝔭̃ is also torsion-free as k𝔭̃-module.

We claim that 𝔭𝔭̃ is a maximal ideal in A𝔭̃. By [20] section 3c, k𝔭̃ is a flat kmodule. Hence A𝔭̃/𝔭𝔭̃ is isomorphic to (A/𝔭)𝔭̃ which is finitely generated as module over k𝔭̃ as A/𝔭 is finitely generated as module over k, by assumption. Let 𝔪 be a maximal ideal of A𝔭̃ containing 𝔭𝔭̃. (Note that 𝔪 exists and is a prime ideal in A𝔭̃ as A𝔭̃ possesses an identity.) Then k𝔭̃ ∩ 𝔭𝔭̃ ⊆ k𝔭̃ ∩ 𝔪. As any element of k𝔭̃ which is not contained in 𝔭𝔭̃ is invertible in k𝔭̃, this implies that k𝔭̃ ∩ 𝔭𝔭̃ = k𝔭̃ ∩ 𝔪. Thus 𝔭𝔭̃ = 𝔪, by corollary (3.8).

Assume that we have already found a prime ideal 𝔮′ in B𝔭̃ such that A𝔭̃ ∩𝔮′ ⊆ 𝔭𝔭̃ and k𝔭̃ ∩ 𝔮′ = k𝔭̃ ∩ 𝔭𝔭̃. Then, by proposition (3.6), B ∩ 𝔮′ is a prime ideal in B,

A ∩ (B ∩ 𝔮′) = A ∩ 𝔮′ = A ∩ (A𝔭̃ ∩ 𝔮′) ⊆ A ∩ 𝔭𝔭̃ = 𝔭,

and

k ∩ 𝔭 ⊆ k ∩ 𝔭𝔭̃ = k ∩ 𝔮′ ⊆ k ∩ (B ∩ 𝔮′);

so k ∩ 𝔭 = k ∩ (B ∩ 𝔮′). Therefore, replacing k by k𝔭̃, B by B𝔭̃, A by A𝔭̃ and 𝔭 by 𝔭𝔭̃, we may assume, in addition, that k ∩ 𝔭 is a maximal ideal in k and 𝔭 is a maximal ideal in A.

Set \overline{k} := k/k ∩ 𝔭, \overline{A} := A + (k ∩ 𝔭)B/(k ∩ 𝔭)B and \overline{B} := B/(k ∩ 𝔭)B. Then \overline{k} is a field, \overline{B} is a finite dimensional \overline{k}-algebra, and \overline{A} is a unitary subalgebra of \overline{B}. In particular, \overline{B} is Artinian.

We claim that A ∩ (k ∩ 𝔭)B ⊆ 𝔭. Suppose not. Then (A ∩ (k ∩ 𝔭)B) + 𝔭 = A as 𝔭 is maximal in A. Thus there are α ∈ k ∩ 𝔭 and b ∈ B such that a := αb ∈ A and 1_{A} ≡ a (mod 𝔭). By the observations preceeding theorem (3.9), there are n ∈ ℕ and α_{0}, …, α_{n−1} ∈ k such that

b^{n} + α_{n−1}b^{n−1} + … α_{1}b + α_{0} = 0.

Multiplying this equation with α^{n} yields

a^{n} + αα_{n−1}a^{n−1} + … + α^{n−1}α_{1}a + α^{n}α_{0} = 0.

As α ∈ 𝔭 and a ∈ A, this implies that a^{n} ∈ 𝔭. However,

1_{A} = 1_{A}^{n} ≡ a^{n} ≡ 0 (mod 𝔭)

is a contradiction as 𝔭 is a proper ideal of A, proving our claim.

Therefore, \overline{𝔭} := 𝔭 + (k ∩ 𝔭)B/(k ∩ 𝔭)B is mapped to 𝔭/A ∩ (k ∩ 𝔭)B under the canonical isomorphism \overline{A} →^{∼} A/A ∩ (k ∩ 𝔭)B. Hence \overline{𝔭} is a prime ideal in \overline{A}. Let 𝔪_{1}, …, 𝔪_{n} (n ∈ ℕ) be the maximal ideals of \overline{B}. Then

(\overline{A} ∩ 𝔪_{1}) · . . . · (\overline{A} ∩ 𝔪_{n}) ⊆ 𝔪_{1} ∩ … ∩ 𝔪_{n} = J(\overline{B}).

As J(\overline{B}) is nilpotent, this implies

(\overline{A} ∩ 𝔪_{1}) · . . . · (\overline{A} ∩ 𝔪_{n}) ⊆ \overline{𝔭},

and so \overline{A} ∩ 𝔪_{i} ⊆ \overline{𝔭} for some i ∈ \{1, …, n\}. We have 𝔪_{i} = 𝔮′/(k ∩ 𝔭)B for some prime ideal 𝔮′ in B. Then

A ∩ 𝔮′ ⊆ A ∩ (𝔭 + (k ∩ 𝔭)B) ⊆ 𝔭 + (A ∩ (k ∩ 𝔭)B) = 𝔭;

in particular, k ∩ 𝔮′ ⊆ k ∩ 𝔭. On the other hand, k ∩ 𝔭 ⊆ (k ∩ 𝔭)B ⊆ 𝔮′, and so k ∩ 𝔮′ = k ∩ 𝔭. This completes the proof. □

Note that theorem (3.9) can be in particular applied to A := k1_{B}; then it is in fact a Lying-over Theorem. As far as we can see, it is not clear whether it is really necessary to assume that also A/𝔭 is finitely generated as module over k.

For later use, we state a slightly modified version of theorem (3.9). Before we state the result, let us introduce some convenient, though sloppy notation: Let A be a k-algebra with identity 1_{A}, and let I be an ideal in A. Then k ∩ I denotes the full preimage of I under the canonical epimorphism k → k1_{A}.

(3.10) COROLLARY. Let A and B be k-algebras with identities, let ψ : A → B be a unitary homomorphism of algebras, let 𝔭 be a prime ideal in A, and let 𝔮 be a prime ideal in B such that ψ^{−1}(𝔮) ⊆ 𝔭. Assume further that A/𝔭 and B/𝔮 are finitely generated as modules over k. Then there is a prime ideal 𝔮′ in B containing 𝔮 such that ψ^{−1}(𝔮′) ⊆ 𝔭 and k ∩ 𝔮′ = k ∩ 𝔭.

PROOF. As in the proof of corollary (3.4), ψ induces an isomorphism of algebras

A/ψ^{−1}(𝔮) →^{∼} ψ(A) + 𝔮/𝔮,

and ψ(𝔭) + 𝔮/𝔮 is a prime ideal in ψ(A) + 𝔮/𝔮. Moreover,

(ψ(A) + 𝔮/𝔮)/(ψ(𝔭) + 𝔮/𝔮) ≅ A/𝔭

which is finitely generated as module over k, by assumption.

By theorem (3.9) applied to ψ(A) + 𝔮/𝔮 ⊆ B/𝔮, there exists a prime ideal 𝔮′ in B containing 𝔮 such that

(𝔮′ ∩ ψ(A)) + 𝔮 = 𝔮′ ∩ (ψ(A) + 𝔮) ⊆ ψ(𝔭) + 𝔮,

and k ∩ 𝔮′/𝔮 = k ∩ (ψ(𝔭) + 𝔮/𝔮). As in the proof of corollary (3.4), one shows that this implies ψ^{−1}(𝔮′) ⊆ 𝔭. Hence k ∩ 𝔮′ ⊆ k ∩ 𝔭. Let α ∈ k such that α1_{A} ∈ 𝔭. Thus α1_{B/𝔮} = ψ(α1_{A}) + 𝔮 ∈ ψ(𝔭) + 𝔮. Therefore,

α ∈ k ∩ (ψ(𝔭) + 𝔮/𝔮) = k ∩ (𝔮′/𝔮),

and thus α1_{B} ∈ 𝔮′. This shows k ∩ 𝔮′ = k ∩ 𝔭, and we are done. □

§4. PROOF OF THE SYLOW THEOREMS AND FURTHER RESULTS

In this section we prove the existence of defects and their uniqueness up to conjugacy. Throughout, we fix a finite group G, and let A be a Green functor for G. The following result will be crucial for the existence of defects:

(4.1) LEMMA. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} be a pointed group on A. Moreover, assume that H_{𝔭} is relatively K-projective for some subgroup K of H. Then there exists a prime ideal 𝔮 in A(K), such that K_{𝔮} is a pointed subgroup of H_{𝔭}, and such that H_{𝔭} is relatively K_{𝔮}-projective (i.e. K_{𝔮} ≤ H_{𝔭} and H_{𝔭} pr K_{𝔮}).

PROOF. Set J := ∑_{I} I where the sum is taken over all ideals I in A(K) satisfying Ind_{K}^{H}(I) ⊆ 𝔭. Then J is an ideal in A(K), and, as Ind_{K}^{H} is additive, we have Ind_{K}^{H}(J) ⊆ 𝔭. Clearly, J is the uniquely determined maximal element in the set of ideals I in A(K) satisfying Ind_{K}^{H}(I) ⊆ 𝔭.

By the Frobenius axiom, we have

(Res_{K}^{H})^{−1}(J)A_{K}^{H} ⊆ Ind_{K}^{H}(J) ⊆ 𝔭.

Consequently, (Res_{K}^{H})^{−1}(J) ⊆ 𝔭 since A_{K}^{H} ⊈ 𝔭 and 𝔭 is prime. By corollary (3.4) applied to the homomorphism Res_{K}^{H} : A(H) → A(K), there exists a prime ideal 𝔮 in A(K) containing J such that (Res_{K}^{H})^{−1}(𝔮) ⊆ 𝔭, i.e. K_{𝔮} ≤ H_{𝔭}. Moreover, if I is an ideal in A(K) such that Ind_{K}^{H}(I) ⊆ 𝔭 then I ⊆ J ⊆ 𝔮, by definition of J. Hence H_{𝔭} pr K_{𝔮}, and we are done. □

PROOF of Lemma D. “(a) ⇒ (b)”. Assume that P_{𝔮} is not relatively projective to a proper subgroup of P, and let Q_{𝔭} be a pointed group on A such that P_{𝔮} is relatively Q_{𝔭}-projective. Thus Q ⊆ P, and Ind_{Q}^{P}(A(Q)) = A_{Q}^{P} ⊈ 𝔮, since otherwise A(Q) ⊆ 𝔭, by definition of relative projectivity. Hence P_{𝔮} is relatively Q-projective, and therefore P = Q, by assumption. But then Ind_{Q}^{P}(𝔮) = 𝔮, and thus 𝔮 ⊆ 𝔭 since P_{𝔮} pr Q_{𝔭}, as was to be shown.

The implication “(b) ⇒ (c)” is trivial.

§4. PROOF OF THE SYLOW THEOREMS AND FURTHER RESULTS 23

“(c) ⇒ (d)”. Assume that if P_{𝔮} is relatively Q_{𝔭}-projective for some pointed group Q_{𝔭} on A then Q = P. Suppose ker Br_{P} ⊈ 𝔮. Then A_{Q}^{P} ⊈ 𝔮, i.e. P_{𝔮} is relatively Q-projective, for some proper subgroup Q of P. Thus, by lemma (4.1), there exists a prime ideal 𝔭 in A(Q) such that P_{𝔮} is relatively Q_{𝔭}-projective. By assumption, this yields the contradiction Q = P.

“(a) ⇐ (d)”. Assume ker Br_{P} ⊆ 𝔮. Hence A_{Q}^{P} ⊆ ker Br_{P} ⊆ 𝔮 for any proper subgroup Q of P, i.e. P_{𝔮} is not relatively projective to a proper subgroup of P, and we are done. □

Lemmas (4.1) and D immediately imply the existence of defects:

PROOF of Lemma F. Let H_{𝔭} be a pointed group on A. First of all, note that H_{𝔭} is certainly H-projective; for A_{H}^{H} = Ind_{H}^{H}(A(H)) = A(H) ⊈ 𝔭. Thus, as G is finite, we may choose a subgroup P of H which is minimal subject to the condition of H_{𝔭} being P -projective. By lemma (4.1), there exists a prime ideal 𝔮 in A(P) such that P_{𝔮} ≤ H_{𝔭} and H_{𝔭} pr P_{𝔮}. Let Q_{𝔯} be a pointed group on A such that P_{𝔮} pr Q_{𝔯}. Then, by transitivity of pr, we have H_{𝔭} pr Q_{𝔯}. In particular, H_{𝔭} is Q-projective. Therefore, Q = P, by choice of P. Hence P_{𝔮} satisfies assertion (c) of lemma D, i.e. P_{𝔮} is primordial, by definition. This means that P_{𝔮} is a defect of H_{𝔭}. □

REMARK. Recall that we defined a defect group of a pointed group H_{𝔭} on A to be the first component of some defect pointed subgroup of H_{𝔭}. But usually one would define a defect group of H_{𝔭} to be a minimal subgroup P of H subject to the condition of H_{𝔭} being P -projective. However, these two definitions are in fact equivalent, which we will be able to show as soon as we have proved theorem G; for in the proof of lemma F we started with an arbitrary such minimal subgroup P and constructed a prime ideal 𝔮 in A(P) such that P_{𝔮} is a defect pointed subgroup of H_{𝔭}. Conversely, if P_{𝔮} is a defect of H_{𝔭}, and Q is a subgroup of P such that H_{𝔭} is relatively Q-projective then there exists a prime ideal 𝔯 in A(Q) such that H_{𝔭} is relatively Q_{𝔯}-projective, by lemma (4.1). Thus, by theorem G, there is h ∈ H such that ^{h}(P_{𝔮}) ≤ Q_{𝔯}. In particular, ^{h}P ⊆ Q ⊆ P, and hence P = Q.

The reader is reminded of our notation introduced in connection with corollary (3.10): For a k-algebra A with identity 1_{A} and an ideal I in A, k ∩ I denotes the full preimage of I under the canonical epimorphism k → k1_{A}. Then we have the following:

(4.2) LEMMA. Let G be a finite group, let A be a unitary Green functor for G, and let H_{𝔭} be a pointed group on A with defect P_{𝔮}. Then k ∩ 𝔭 = k ∩ 𝔮.

PROOF. Let α ∈ k ∩ 𝔮, i.e. α ∈ k and α1_{A(P)} ∈ 𝔮. Hence Res_{P}^{H}(α1_{A(H)}) = α1_{A(P)} ∈ 𝔮, and therefore α1_{A(H)} ∈ (Res_{P}^{H})^{−1}(𝔮) ⊆ 𝔭 as P_{𝔮} ≤ H_{𝔭}. Thus α ∈ k ∩ 𝔭.

Conversely, let α ∈ k ∩ 𝔭. Then αA(P) is an ideal in A(P), and

Ind_{P}^{H}(αA(P)) = αA_{P}^{H} ⊆ 𝔭.

As H_{𝔭} is relatively P_{𝔮}-projective, this implies that

α1_{A(P)} ∈ αA(P) ⊆ 𝔮;

so α ∈ k ∩ 𝔮, completing the proof. □

The next fact will be frequently used in the sequel. We already observed that the relation ≤ on pointed groups is antisymmetric, but we have even more:

(4.3) LEMMA. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} and K_{𝔯} be pointed groups on A. Moreover, assume that there are x, y ∈ G such that ^{x}(K_{𝔯}) ≤ H_{𝔭} ≤ ^{y}(K_{𝔯}). Then ^{x}(K_{𝔯}) = H_{𝔭} = ^{y}(K_{𝔯}).

PROOF. First of all, we have^{x}K ⊆ H ⊆ ^{y}K, and thus^{x}K = H = ^{y}K. Therefore,

x𝔯 = (Res_{^{x}K}^{H})^{−1}(^{x}𝔯) ⊆ 𝔭 = (Res_{H}^{^{y}K})^{−1}(𝔭) ⊆ ^{y}𝔯

since ^{x}(K_{𝔯}) ≤ H_{𝔭} ≤ ^{y}(K_{𝔯}). Thus, for z := yx^{−1} ∈ G, we have

𝔭 ⊆ ^{y}𝔯 = ^{z}(^{x}𝔯) ⊆ ^{z}𝔭,

and hence

𝔭 ⊆ ^{z}𝔭 ⊆ ^{z^{2}}𝔭 ⊆ ^{z^{3}}𝔭 ⊆ … ⊆ 𝔭,

as G is finite. Therefore, 𝔭 = ^{z}𝔭, hence 𝔭 = ^{y}𝔯, and thus 𝔭 = ^{z^{−1}}𝔭 = ^{z^{−1}y}𝔯 = ^{x}𝔯, completing the proof. □

The basic tool for our proof of Sylow’s second theorem (theorem G) is the following result:

(4.4) LEMMA. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} be a pointed group on A. Moreover, let P_{𝔮} be a primordial pointed subgroup of H_{𝔭}, and let K be a subgroup of H such that H_{𝔭} is relatively K-projective. Then there exists h ∈ H such that ^{h}P := hP h^{−1} ⊆ K.

PROOF. We have (Res_{P}^{H})^{−1}(𝔮) ⊆ 𝔭, by definition of P_{𝔮} ≤ H_{𝔭}. Moreover, A_{K}^{H} ⊈ 𝔭 since H_{𝔭} is relatively K-projective. By the Mackey axiom, we have

Res_{P}^{H}(A_{K}^{H}) ⊆ ∑_{P hK∈P \H/K} A_{P ∩^{h}K}^{P}.

This implies that we must have A_{P ∩^{h}K}^{P} ⊈ 𝔮 for some h ∈ H since otherwise

A_{K}^{H} ⊆ (Res_{K}^{H})^{−1}(𝔮) ⊆ 𝔭.

Hence P_{𝔮} is relatively P ∩^{h}K-projective. However, as P_{𝔮} is primordial, this implies P = P ∩^{h}K ⊆ ^{h}K, and thus ^{h^{−1}}P ⊆ K. □

PROOF of Theorem G. Recall that the situation is as in lemma (4.4) except that we are also given a prime ideal 𝔯 in A(K) such that H_{𝔭} is relatively K_{𝔯}-projective; and we want to show that ^{h}(P_{𝔮}) ≤ K_{𝔯} for some h ∈ H.

In view of lemma (4.4), we may assume that we already have P ⊆ K; for any H-conjugate of P_{𝔮} is obviously a primordial pointed subgroup of H_{𝔭}, as well.

Set 𝔖 := {h ∈ H : P ⊆ ^{h}K}. Then 𝔖 ≠ ∅, as 1 ∈ 𝔖. We write 𝔖 = {h_{1}, …, h_{n}}, i.e. we choose a linear order of 𝔖. Note that, for h ∈ 𝔖, ^{h^{−1}}𝔮 is a prime ideal in A(^{h^{−1}}P), and ^{h^{−1}}P is a subgroup of K. Now define

I := (Res_{^{h{−1 \atop 1}}P}^{K})^{−1}(^{h_{1}^{−1}}𝔮) · . . . · (Res_{^{h{−1 \atop n}}P}^{K})^{−1}(^{h_{n}^{−1}}𝔮).

Then I is an ideal in A(K). Moreover, for h ∈ 𝔖, we have

hI ⊆ ^{h}((Res_{^{h^{−1}}P}^{K})^{−1}(^{h^{−1}}𝔮)) = (Res_{P}^{^{h}K})^{−1}(𝔮),

i.e. Res_{P}^{^{h}K}(^{h}I) ⊆ 𝔮.

Suppose I ⊈ 𝔯. Then Ind_{K}^{H}(I) ⊈ 𝔭 since H_{𝔭} pr K_{𝔯}. By Mackey formula, we have

Res_{P}^{H}(Ind_{K}^{H}(I)) = ∑_{P hK∈P \H/K} Ind_{P ∩^{h}K}^{P}(Res_{P ∩^{h}K}^{^{h}K}(^{h}I)).

Therefore, Ind_{P ∩^{h}K}^{P}(Res_{P ∩^{h}K}^{^{h}K}(^{h}I)) ⊈ 𝔮 for some h ∈ H since otherwise

Ind_{K}^{H}(I) ⊆ (Res_{P}^{H})^{−1}(𝔮) ⊆ 𝔭,

as P_{𝔮} ≤ H_{𝔭}. In particular, P_{𝔮} is relatively P ∩^{h}K-projective. Thus P = P ∩^{h}K ⊆ hK, i.e. h ∈ 𝔖, as P_{𝔮} is primordial. Hence

Res_{P}^{^{h}K}(^{h}I) = Ind_{P ∩^{h}K}^{P}(Res_{P ∩^{h}K}^{^{h}K}(^{h}I)) ⊈ 𝔮,

which is a contradiction.

Therefore, I ⊆ 𝔯. Thus (Res_{^{h^{−1}}P}^{K})^{−1}(^{h^{−1}}𝔮) ⊆ 𝔯 for some h ∈ 𝔖 since 𝔯 is a prime ideal, i.e. ^{h^{−1}}(P_{𝔮}) ≤ K_{𝔯}, completing the proof of the first assertion.

Now we deduce that any two defects of H_{𝔭} are H-conjugate. Let P_{𝔮} and Q_{𝔯} be defects of H_{𝔭}. Since P_{𝔮} is a primordial subgroup of H_{𝔭} and H_{𝔭} is relatively Q_{𝔯}- projective, the first part of the proof yields an element h ∈ H such that ^{h}(P_{𝔮}) ≤ Q_{𝔯}. Symmetrically, there is k ∈ H such that Q_{𝔯} ≤ ^{k}(P_{𝔮}). Thus ^{h}(P_{𝔮}) = Q_{𝔯} = ^{k}(P_{𝔮}), by lemma (4.3). □

Now we are in a position to prove theorem H.

PROOF of Theorem H. “(a) ⇒ (b)”. Assume that P_{𝔮} is a defect of H_{𝔭}. In particular, H_{𝔭} is relatively P_{𝔮}-projective. Let Q_{r} be a pointed subgroup of P_{𝔮} such that H_{𝔭} is relatively Q_{𝔯}-projective. Since P_{𝔮} is, in particular, a primordial pointed subgroup of H_{𝔭}, there exists h ∈ H such that ^{h}(P_{𝔮}) ≤ Q_{𝔯} ≤ P_{𝔮}. Thus Q_{𝔯} = P_{𝔮}, by lemma (4.3), as was to be shown.

“(b) ⇒ (c)”. Assume that P_{𝔮} is minimal among the pointed groups K_{𝔯} on A satisfying H_{𝔭} pr K_{𝔯}. In particular, H_{𝔭} is relatively P_{𝔮}-projective. Firstly, we show that P_{𝔮} is a primordial subgroup of H_{𝔭}. By lemma F, H_{𝔭} has a defect Q_{𝔱}. Thus, by theorem G, there exists h ∈ H such that ^{h}(Q_{𝔱}) ≤ P_{𝔮} since Q_{𝔱} is in particular a primordial subgroup of H_{𝔭}. Moreover, ^{h}(Q_{𝔱}) is a defect of H_{𝔭} since Q_{𝔱} is and h ∈ H. In particular, H_{𝔭} pr^{h}(Q_{𝔱}). Thus P_{𝔮} = ^{h}(Q_{𝔱}), by assumption. In particular, P_{𝔮} is a primordial pointed subgroup of H_{𝔭}.

Secondly, we show maximality of P_{𝔮}. Let R_{𝔰} be a primordial pointed subgroup of H_{𝔭} such that P_{𝔮} ≤ R_{𝔰}. Then there exists h ∈ H such that ^{h}(R_{𝔰}) ≤ P_{𝔮}, by theorem G, as H_{𝔭} pr P_{𝔮}. Consequently, P_{𝔮} = R_{𝔰}, by lemma (4.3).

“(c) ⇒ (d)”. Assume that P_{𝔮} is a maximal primordial pointed subgroup of H_{𝔭}. By lemma F, H_{𝔭} has a defect Q_{𝔯}. Then, in particular, H_{𝔭} pr Q_{𝔯}. Thus there exists h ∈ H such that P_{𝔮} ≤ ^{h}(Q_{𝔯}), by theorem G. Morever, ^{h}(Q_{𝔯}) is a defect of H_{𝔭} since Q_{𝔯} is and h ∈ H. In particular, ^{h}(Q_{𝔯}) is a primordial pointed subgroup of H_{𝔭}, and thus assumption implies that P_{𝔮} = ^{h}(Q_{𝔯}). Thus P_{𝔮} is a defect of H_{𝔭}, and, in particular, H_{𝔭} is relatively P_{𝔮}-projective. Since P_{𝔮} is primordial, we have ker Br_{P} ⊆ 𝔭. Moreover, (Res_{P}^{H})^{−1}(𝔮) ⊆ 𝔭, as P_{𝔮} ≤ H_{𝔭}. Hence, for x ∈ ker(Br_{P} ∘ Res_{P}^{H}), we have Res_{P}^{H}(x) ∈ ker Br_{P} ⊆ 𝔮, and thus x ∈ (Res_{P}^{H})^{−1}(𝔮) ⊆ 𝔭. Finally, let 𝔱 be a prime ideal in A(K) contained in 𝔮 such that H_{𝔭} pr P_{𝔱}. Then P_{𝔱} ≤ P_{𝔮}, and the implication (a) ⇒ (b) (which is already proved) yields that P_{𝔮} = P_{𝔱}, as P_{𝔮} is a defect of H_{𝔭}, completing the proof of this implication.

“(a) ⇐ (d)”. Assume that H_{𝔭} is relatively P_{𝔮}-projective, that ker(Br_{P} ∘ Res_{P}^{H}) ⊆ 𝔭, and that 𝔮 = 𝔱 for any prime ideal in A(P) contained in 𝔮 such that H_{𝔭} is relatively P_{𝔱}-projective. By lemma F, H_{𝔭} has a defect Q_{𝔯}. Then, by theorem G, there exists h ∈ H such that ^{h}(Q_{𝔯}) ≤ P_{𝔮}, as Q_{𝔯} is a primordial pointed subgroup of H_{𝔭}. Replacing Q_{𝔯} by ^{h}(Q_{𝔯}) (which is also a defect of H_{𝔭}), we may assume that Q_{𝔯} ≤ P_{𝔮}. We have A_{Q}^{H} ⊈ 𝔭, as H_{𝔭} is relatively Q_{𝔯}-projective, and hence relatively Q-projective. Thus A_{Q}^{H} ⊈ ker(Br_{P} ∘ Res_{P}^{H}), i.e. Res_{P}^{H}(A_{Q}^{H}) ⊈ ker Br_{P}, by assumption. However, the Mackey axiom yields

Res_{P}^{H}(A_{Q}^{H}) ⊆ ∑_{P hQ∈P \H/Q} A_{P ∩^{h}Q}^{P},

and so A_{P ∩^{h}Q}^{P} ⊈ ker Br_{P}, i.e. P = P ∩^{h}Q ⊆ ^{h}Q, for some h ∈ H. Hence P = Q, as Q ⊆ P. Thus

𝔯 = (Res_{Q}^{P})^{−1}(𝔯) ⊆ 𝔮.

Therefore, 𝔮 = 𝔯, by assumption, as H_{𝔭} pr Q_{𝔯}; so P_{𝔮} = Q_{𝔯} is a defect of H_{𝔭}. □

We will prove Sylow’s first theorem (corollaries P and R) in the following section as an application of the Green correspondence.

REMARK. We have not checked this in detail, but it seems to be the case that a big part of this theory carries over to infinite groups as long as one restricts attention to subgroups of finite index (resp. pointed subgroups where the first component has finite index). For the definition of a Green functor for infinite groups one would anyway require the existence of induction maps Ind_{K}^{H} : A(K) → A(H) only if K has finite index in H. Note that the proof of lemma (4.3) carries over to this more general setup; for, given subgroups H and K of finite index in a group G such that H ⊆ K ⊆ ^{g}H for some g ∈ G, one has g^{n} ∈ H for some n ∈ ℕ, so

H ⊆ ^{g}H ⊆ ^{g^{2}}H ⊆ … ⊆ ^{g^{n}}H = H,

and thus H = K = ^{g}H; and similarly for the prime ideals. The only problem here is the proof of the existence of a defect of a pointed group H_{𝔭}: Recall that we started with a minimal subgroup P such that H_{𝔭} is relatively P -projective. However, in this more general situation it is not all clear whether such a subgroup exists (even though we look only at subgroups of finite index in H). This fact shows the difference between the above proof of “Sylow’s theorems” (following [25]) and the ordinary proofs. Essentially, what we did is starting with a subgroup P which is minimal subject to the condition that the index of P in G is not divisible by the given prime p, and after this we have shown that P is in fact a p-group. But usually one starts with a maximal p-subgroup P, and shows then that the index of P in G is not divisible by p. So the question is whether there is a direct proof of the fact that maximal primordial pointed subgroups are defects. Although, in this more general setup of infinite groups it is also not clear, as far as we can see, whether a pointed group H_{𝔭} has a maximal primordial pointed subgroup P_{𝔮} such that |H : P | is finite. But it seems that, in case H_{𝔭} has a primordial pointed subgroup Q_{𝔯}, at all, such that |H : Q| is finite, it also has a maximal one such by corollary (3.4).

§5. THE PUIG AND GREEN CORRESPONDENCES 27

§5. THE PUIG AND GREEN CORRESPONDENCES

The main objective of this section is to prove theorems I, N and O, and, as a corollary, Sylow’s first theorem (corollaries P and R), as well as some applications of the Puig correspondence and Sylow’s first theorem. An alternative proof of Dress’s induction theorem using this version of Sylow’s first theorem will be given in the following section.

We will need the following result stated in [25] (proposition (54.2)) which is an analogue of a well-known result of G-algebra theory:

(5.1) PROPOSITION. Let G be a finite group, let A be Green functor for G, let P be a subgroup of G for A, and let H be a subgroup of G containing P. Then

Br_{P} (Res_{P}^{H}(Ind_{P}^{H}(a))) = Tr_{1}^{\overline{N}_{H}(P)}(Br_{P} (a))

for all a ∈ A(P).

Note that, in case P is not primordial for A, proposition (5.1) holds trivially. Recall also that A(P) is a \overline{N}_{G}(P)-algebra, and hence a \overline{N}_{H}(P)-algebra; so we have in fact relative trace maps for A(P).

The next result is an analogue of proposition (5.1) where one replaces the Brauer homomorphism by the canonical epimorphism π_{𝔮} : A(P) → A(P)/𝔮 associated to a primordial pointed group P_{𝔮} on A. Recall that π_{𝔮} factors through Br_{P} by definition of primordiality. A proof of this proposition can be found in [25] (proposition (55.10)) for maximal ideals; we leave it to the reader to verify that the proof carries over verbatim to the situation below:

(5.2) PROPOSITION. Let H be a subgroup of a finite group G, let A be a Green functor for G, and let P_{𝔮} be a primordial pointed group on A such that P is contained in H. Denote by π_{𝔮} the canonical epimorphism A(P) → A(P)/𝔮. Then

π_{𝔮}(Res_{P}^{H}(Ind_{P}^{H}(a))) = Tr_{1}^{\overline{N}_{H}(P_{𝔮})}(π_{q}(a))

for all a ∈ A(P) satisfying a ∈ ^{h}𝔮 for any h ∈ N_{H}(P) ∖ N_{H}(P_{𝔮}).

We also need the following generalization of proposition (5.2) the proof of which is left in [25] as an exercise for the reader (cf. exercise (55.6)). Again, this result is an analogue of a well-known fact of G-algebra theory (cf. [25] corollary (14.8)).

(5.3) PROPOSITION. Let G be a finite group, let A be a Green functor for G, and let P_{𝔮} be a primordial pointed group on A. Moreover, let H, K be subgroups of G such that P ⊆ K ⊆ H, and let b ∈ A(P) such that b ∈ ^{h}𝔮 for all h ∈ N_{H}(P) ∖ N_{H}(P_{𝔮}). Denote by π_{𝔮} the canonical epimorphism A(P) → A(P)/𝔮. Then, for a := Ind_{P}^{K}(b), we have

π_{𝔮}(Res_{P}^{H}(Ind_{K}^{H}(a))) = Tr_{\overline{N}_{K}(P_{𝔮})}^{\overline{N}_{H}(P_{𝔮})}(π_{𝔮}(Res_{P}^{K}(a))).

The following result is a further characterization of defects under additional hypotheses:

(5.4) LEMMA. Let G be a finite group, let A be a Green functor for G, let H_{𝔭} be a pointed group on A, and let P_{𝔮} be a primordial pointed subgroup of H_{𝔭}. Moreover, let I be an ideal in A(P) which is not contained in 𝔮. Then P_{𝔮} is a defect of H_{𝔭}, if and only if the following assertions are satisfied:

(a) Ind_{P}^{H}(I) ⊈ 𝔭.

(b) If 𝔱 is a prime ideal in A(P) such that P_{𝔮} ≤ P_{𝔱} ≤ H_{𝔭} then 𝔮 = 𝔱.

PROOF. “⇒”. Assume that P_{𝔮} is a defect of H_{𝔭}. Since H_{𝔭} is relatively P_{𝔮}- projective, Ind_{P}^{H}(I) ⊈ 𝔭 as I ⊈ 𝔮. Let 𝔱 be a prime ideal in A(P) such that P_{𝔮} ≤ P_{𝔱} ≤ H_{𝔭}. Then P_{𝔱} is a primordial subgroup of H_{𝔭} since ker Br_{P} ⊆ 𝔮 ⊆ 𝔱 as P_{𝔮} is primordial, by assumption. Therefore, P_{𝔮} = P_{𝔱}, i.e. 𝔮 = 𝔱, by theorem H “(a) ⇔ (c)”.

“⇐”. Assume that assertions (a) and (b) are satisfied. In particular, H_{𝔭} is relatively P -projective since Ind_{P}^{H}(I) ⊆ A_{P}^{H}. Thus, by lemma (4.1), there exists a prime ideal 𝔱 in A(P) such that P_{𝔱} ≤ H_{𝔭} and H_{𝔭} pr P_{𝔱}. Hence there is h ∈ H such that P_{𝔮} ⊆ ^{h}(P_{𝔱}), by theorem G. In particular, P ⊆ ^{h}P, thus h ∈ N_{H}(P), and hence P_{𝔮} ≤ ^{h}(P_{𝔱}) = P_{^{h}𝔱}, i.e. 𝔮 ⊆ ^{h}𝔱. On the other hand, ^{h}(P_{𝔱}) is a pointed subgroup of H_{𝔭}, as P_{𝔱} is and h ∈ H. Therefore, 𝔮 = ^{h}𝔱, by assumption, and so P_{𝔮} = P_{^{h}𝔱} = ^{h}(P_{𝔱}) satisfies H_{𝔭} pr P_{𝔮}, as P_{𝔱} does. Hence P_{𝔮} is a defect of H_{𝔭}. □

Now we are in a position to prove the Puig correspondence (theorem I); we will prove a stronger version as stated in §2. In order to state the result, the following terminology will be convenient: Let P_{𝔮} be a pointed group on A, and let H be a subgroup of G containing P. We denote by 𝔓_{H}(𝔮) the set of prime ideals 𝔯 in A(P) such that, for any h ∈ N_{H}(P),

h𝔮 ⊆ 𝔯 if and only if h ∈ N_{H}(P_{𝔮}).

In particular, 𝔮 ⊆ 𝔯 for any 𝔯 ∈ 𝔓_{H}(𝔮). Cleary, 𝔮 ∈ 𝔓_{H}(𝔮) as ^{h}𝔮 ⊆ 𝔮 implies ^{h}𝔮 = 𝔮, i.e. h ∈ N_{H}(P_{𝔮}), for all h ∈ N_{H}(P), by the argument of the proof of lemma (4.3). Moreover, for 𝔯 ∈ 𝔓_{H}(𝔮), we must have N_{H}(P_{𝔯}) ⊂ N_{H}(P_{𝔮}) since otherwise there is h ∈ N_{H}(P_{𝔯}) ∖ N_{H}(P_{𝔮}), and thus ^{h}𝔮 ⊆ ^{h}𝔯 = 𝔯 contradicting the choice of 𝔯. The version of the Puig correspondence we prove reads as follows:

(5.5) THEOREM. Let G be a finite group, let A be a Green functor for G, let P_{𝔮} be a primordial pointed group on A, and let H be a subgroup of G containing P. Set \overline{N} := \overline{N}_{H}(P_{𝔮}). Moreover, set S := A(P)/𝔮, and denote by π_{𝔮} the canonical epimorphism A(P) → S. Then the homomorphism π_{𝔮} ∘ Res_{P}^{H} induces a bijection

Δ : {𝔭 ∈ Spec(A(H)) : ∃ 𝔯 ∈ 𝔓_{H}(𝔮) : P_{𝔯} is a defect of H_{𝔭}} ⟶

\{\overline{𝔭} ∈ Spec(S\overline{N}) : ∃ 𝔯 ∈ 𝔓_{H}(𝔮) : 1_{𝔯/𝔮} is a defect of \overline{N}\overline{𝔭}\}

which is both, a homeomorphism, as well as an isomorphism of posets. Moreover, for a prime ideal 𝔭 contained in the domain of Δ, and \overline{𝔭} := Δ(𝔭), the following assertions hold:

(a) 𝔭 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}).

(b) For any 𝔯 ∈ 𝔓_{H}(𝔮), P_{𝔯} is a defect of H_{𝔭}, if and only if 1_{𝔯/𝔮} is a defect of \overline{N}\overline{𝔭}.

(c) In case A is unitary, 𝔭 is a maximal ideal in A(H), if and only if \overline{𝔭} is a maximal ideal in S\overline{N}; and in this case the homomorphism π_{𝔮} ∘ Res_{P}^{H} induces an isomorphism A(H)/𝔭 →^{∼} S\overline{N}/\overline{𝔭}.

PROOF. We denote by T the image of the homomorphism π_{𝔮} ∘ Res_{P}^{H} : A(H) → S. As any element in A(H) is fixed by H, and hence by N_{H}(P_{𝔮}), T is a subalgebra of S\overline{N}. Write N_{H}(P) ∖ N_{H}(P_{𝔮}) = \{h_{1}, …, h_{n}\} (n ∈ ℕ), and set

J := (^{h_{1}}𝔮)(^{h_{2}}𝔮) · . . . · (^{h_{n}}𝔮).

(Note that J = A(P), in case N_{H}(P) = N_{H}(P_{𝔮}), i.e. n = 0.) Then J ⊈ 𝔯 for all 𝔯 ∈ 𝔓_{H}(𝔮), by definition of 𝔓_{H}(𝔮) and the fact that any such 𝔯 is a prime ideal in A(P). Moreover, J ⊆ ^{h}𝔮 for all h ∈ N_{H}(P) ∖ N_{H}(P_{𝔮}). Therefore, proposition (5.2) applies to all elements of J, and thus yields

T ⊇ π_{𝔮}(Res_{P}^{H}(Ind_{P}^{H}(J))) = Tr\overline{N} ; 1(π_{𝔮}(J)).

Besides, I := Tr\overline{N} ; 1(π_{𝔮}(J)) is an ideal in S\overline{N}, by the Frobenius axiom for relative trace maps and the fact that π_{𝔮}(J) is an ideal in S.

Clearly, the restriction A(H) → T of π_{𝔮} ∘ Res_{P}^{H} induces a bijection

{𝔭 ∈ Spec(A(H)) : ker(π_{𝔮} ∘ Res_{P}^{H}) ⊆ 𝔭} → Spec(T)

the inverse of which is given by the assignment \overline{𝔭} ↦ (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}). Moreover, it is obvious that this bijection is both, an isomorphism of posets, as well as a homeomorphism. Besides, if A is unitary, then this bijection restricts to a bijection between the respective sets of maximal ideals since then A(H) has an identity, and so does T, so maximal ideals in A(H) resp. T are in fact prime. Apart from that, for any prime ideal \overline{𝔭} in T, π_{𝔮} ∘ Res_{P}^{H} induces trivially an isomorphism

A(H)/𝔭 →^{∼} T/\overline{𝔭}

where 𝔭 := (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}).

Let 𝔭 be a prime ideal in A(H) containing ker(π_{𝔮} ∘ Res_{P}^{H}), denote by \overline{𝔭} the corresponding prime ideal in T, i.e. 𝔭 = (π_{𝔮} ∘Res_{P}^{H})^{−1}(\overline{𝔭}), and let 𝔯 be a prime ideal in A(P) containing 𝔮. Then, by definition, P_{𝔯} ≤ H_{𝔭}, if and only if (Res_{P}^{H})^{−1}(𝔯) ⊆ 𝔭; and this in turn is equivalent to T ∩ 𝔯/𝔮 ⊆ \overline{𝔭}; for if (Res_{P}^{H})^{−1}(𝔯) ⊆ 𝔭 then, for any \overline{x} ∈ T ∩ 𝔯/𝔮, we have

π_{𝔮}(𝔯) = 𝔯/𝔮 ∋ \overline{x} = π_{𝔮}(Res_{P}^{H}(x))

for some x ∈ A(H); thus Res_{P}^{H}(x) ∈ 𝔯 + ker π_{q} = 𝔯, as 𝔯 contains 𝔮; hence x ∈ (Res_{P}^{H})^{−1}(𝔯) ⊆ 𝔭 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}), and therefore \overline{x} = π_{𝔮}(Res_{P}^{H}(x)) ∈ \overline{𝔭}. Conversely, if T ∩ 𝔯/𝔮 ⊆ \overline{𝔭} then, for any x ∈ (Res_{P}^{H})^{−1}(𝔯), we have

π_{𝔮}(Res_{P}^{H}(x)) ∈ T ∩ π_{𝔮}(𝔯) = T ∩ 𝔯/𝔮 ⊆ \overline{𝔭},

and hence x ∈ (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}) = 𝔭. Moreover, (Res_{P}^{H})^{−1}(𝔯) ⊆ 𝔭 implies in particular

ker(π_{𝔮} ∘ Res_{P}^{H}) = (Res_{P}^{H})^{−1}(𝔮) ⊆ 𝔭,

as 𝔯 contains 𝔮. Thus the above bijection restricts to a bijection

{𝔭 ∈ Spec(A(H)) : ∃ 𝔯 ∈ Spec(A(P)) : 𝔮 ⊆ 𝔯 and P_{𝔯} ≤ H_{𝔭}} ⟶
\{\overline{𝔭} ∈ Spec(T) : ∃ 𝔯 ∈ Spec(T) : 𝔮 ⊆ 𝔯 and T ∩ 𝔯/𝔮 ⊆ \overline{𝔭}\}

which, of course, is still a homeomorphism and an isomorphism of posets.

Let 𝔯 ∈ 𝔓_{H}(𝔮), let 𝔭 be a prime ideal in A(H) such that P_{𝔯} ≤ H_{𝔭}, and denote by \overline{𝔭} the corresponding prime ideal in T. Thus 𝔭 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}) and T ∩ 𝔯/𝔮 ⊆ \overline{𝔭}.

Note that P_{𝔯} is a primordial pointed group as P_{𝔮} is and 𝔮 ⊆ 𝔯. Thus, by lemma (5.4), P_{𝔯} is a defect of H_{𝔭}, if and only if the following assertions are satisfied:

(a) Ind_{P}^{H}(J) ⊈ 𝔭,

(b) 𝔯 = 𝔱 for any prime ideal 𝔱 in A(P) such that P_{𝔯} ≤ P_{𝔱} ≤ H_{𝔭}, as J ⊈ 𝔯. Moreover, assertion (a) is equivalent to I ⊈ \overline{𝔭}; for if Ind_{P}^{H}(J) ⊆ 𝔭 then

I = Tr\overline{N} ; 1(π_{𝔮}(J)) = π_{𝔮}(Res_{P}^{H}(Ind_{P}^{H}(J))) ⊆ π_{𝔮}(Res_{P}^{H}(𝔭)) = \overline{𝔭}.

Conversely, if I ⊆ \overline{𝔭} then π_{𝔮}(Res_{P}^{H}(Ind_{P}^{H}(J))) ⊆ \overline{𝔭}, and thus Ind_{P}^{H}(J) ⊆ (π_{q} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}) = 𝔭. Moreover, assertion (b) is equivalent to:

(c) 𝔯 = 𝔱 for any prime ideal 𝔱 in A(P) containing 𝔯 such that T ∩ 𝔱/𝔮 ⊆ \overline{𝔭},

by the above observations. (Note that such a prime ideal 𝔱 necessarily contains 𝔮, as 𝔯 does.) Therefore, restricting the above bijection yields a bijection

{𝔭 ∈ Spec(A(H)) : ∃ 𝔯 ∈ 𝔓_{H}(𝔮) : P_{𝔯} is a defect of H_{𝔭}} ⟶
\{\overline{𝔭} ∈ Spec(T) : ∃ 𝔯 ∈ 𝔓_{H}(𝔮) : I ⊈ \overline{𝔭} and 𝔯 satisfies assertion (c) \}

Moreover, if 𝔭 and \overline{𝔭} correspond under this bijection, and 𝔯 ∈ 𝔓_{H}(𝔮) then P_{𝔯} is a defect of H_{𝔭}, if and only if I ⊈ \overline{𝔭} and 𝔯 satisfies assertion (c). So it remains to pass from T to S\overline{N}.

Since I is an ideal in S\overline{N} contained in T, corollary (3.5) yields a homeomorphism and isomorphism of posets

\{𝔭̃ ∈ Spec(S\overline{N}) : I ⊈ 𝔭̃\} → \{\overline{𝔭} ∈ Spec(T) : I ⊈ \overline{𝔭}\}, 𝔭̃ ↦ T ∩ 𝔭̃

Let 𝔭̃ be a prime ideal in S\overline{N} and let 𝔯 ∈ 𝔓_{H}(𝔮). Then J ⊈ 𝔯, and thus π_{𝔮}(J) ⊈ π_{𝔮}(𝔯) = 𝔯/𝔮 as 𝔯 contains 𝔮. Thus, by lemma (5.4), 1_{𝔯/𝔮} is a defect of \overline{N} 𝔭̃, if and only if the following assertions are satisfied:

(a) I = Tr\overline{N} ; 1(π_{𝔮}(J)) ⊈ 𝔭̃.

(b) 𝔯 = 𝔱 for any prime ideal 𝔱 in A(P) containing 𝔯 such that S\overline{N} ∩ 𝔱/𝔮 ⊆ 𝔭̃.

Let 𝔱 be a prime ideal in A(P) containing 𝔮, and let \overline{𝔭} = T ∩ 𝔭̃. Then S\overline{N} ∩ 𝔱/𝔮 ⊆ 𝔭̃ implies T ∩ 𝔱/𝔮 ⊆ T ∩ 𝔭̃ = \overline{𝔭}. Conversely, assume that T ∩ 𝔱/𝔮 ⊆ \overline{𝔭}. Then

I(S\overline{N} ∩ 𝔱/𝔮) ⊆ I ∩ (S\overline{N} ∩ 𝔱/𝔮) ⊆ T ∩ 𝔱/𝔮 ⊆ \overline{𝔭} = T ∩ 𝔭̃ ⊆ 𝔭̃

since I is an ideal in S\overline{N} containing T. As 𝔭̃ is a prime ideal in S\overline{N} which does not contain I, this implies S\overline{N} ∩ 𝔱/𝔮 ⊆ 𝔭̃.

Therefore, the above bijection restricts to a bijection

\{𝔭̃ ∈ Spec(S\overline{N}) : ∃ 𝔯 ∈ 𝔓_{H}(𝔮) : 1_{𝔯/𝔮} is a defect of \overline{N} 𝔭̃\} ⟶
\{\overline{𝔭} ∈ Spec(T) : ∃ 𝔯 ∈ 𝔓_{H}(𝔮) : I ⊈ \overline{𝔭} and 𝔯 satisfies assertion (c) \}
𝔭̃ ↦ T ∩ 𝔭̃,

which is still both, a homeomorphism, as well as an isomorphism of posets. (Note that, by construction, if 𝔭̃ and \overline{𝔭} correspond under this bijection then an 𝔯 ∈ 𝔓_{H}(𝔮) “belongs” to 𝔭̃, if and only if it “belongs” to \overline{𝔭}.) So composing these bijections, respectively one with the inverse of the other, we obtain a bijection of the desired form. Note that if 𝔭 ∈ Spec(A(H)), \overline{𝔭} ∈ Spec(T) and 𝔭̃ ∈ Spec(S\overline{N}) correspond under these bijections then

𝔭 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}) and \overline{𝔭} = T ∩ 𝔭̃;

as T is the image of π_{𝔮} ∘ Res_{P}^{H} this implies 𝔭 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(𝔭̃). The additional assertion follows from corollary (3.5) and the observations in the beginning of this proof. □

PROOF of Theorem I. The restriction of the bijection Δ of theorem (5.5) to the sets we are concerned with in theorem I is a bijection Γ having the desired properties. □

Next we prove some applications of the Puig correspondence; we start with a proof of corollary J:

PROOF of Corollary J. Set \overline{N} := N_{H}(P_{𝔮})/P, and denote by \overline{N}\overline{𝔭} the Puig correspondent of H_{𝔭} with respect to P_{𝔮}. Thus \overline{N}\overline{𝔭} is a pointed group on the \overline{N}- algebra S := A(P)/𝔮 with defect 1_{0}, and

𝔭 := (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}),

by theorem I, where π_{𝔮} : A(P) → S denotes the canonical epimorphism.

We claim that \overline{𝔭} = 0. In fact, for s ∈ \overline{𝔭} ⊆ S\overline{N}, ks + sS is an ideal in S as S is commutative, by assumption, and we have

Tr\overline{N} ; 1(s + sS) = |\overline{N}|ks + sS\overline{N} ; 1 ⊆ \overline{𝔭}

as s ∈ S\overline{N}; thus s ∈ ks + sS = 0 since \overline{N}\overline{𝔭} is relatively 1_{0}-projective, proving our claim.

Therefore, we even have

𝔭 = ker(π_{𝔮} ∘ Res_{P}^{H}) = (Res_{P}^{H})^{−1}(𝔮).
□

A similar result is the following:

(5.6) LEMMA. Let G be a finite group, let A be a Green functor for G, let P_{𝔮} be a primordial pointed group on A, and let H be a subgroup of G containing P. Assume further that the residue algebra S := A(P)/𝔮 has a multiplicative identity and is finitely generated as module over its center, and that the characteristic c of S does not divide the order of N_{H}(P_{𝔮})/P. Then there is a prime ideal 𝔭 in A(H) such that H_{𝔭} has defect P_{𝔮}. Moreover, 𝔭 = (Res_{P}^{H})^{−1}(𝔮) in case S is commutative.

PROOF. Set \overline{N} := \overline{N}(P_{𝔮}), and denote by Z the center of S. As S is a prime ring, by assumption, Z is an integral domain and S is torsion-free as Z-module. Thus Z\overline{N} is an integral domain, as well, i.e. 0 is a prime ideal in Z\overline{N}. By corollary (3.3), there is a prime ideal \overline{𝔭} in S\overline{N} such that Z\overline{N} ∩ \overline{𝔭} = 0.

Let R_{𝔯} be a defect of \overline{N}\overline{𝔭}. Then

Z\overline{N} ∩ 𝔯 = Z\overline{N} ∩ \overline{𝔭} = 0,

by lemma (4.2), since S may be regarded as \overline{N}-algebra over Z\overline{N} and, as S has a multiplicative identitiy, by assumption, the k-liner ideals are the same as the Z\overline{N}- linear ones, so that defect theory over k and over Z\overline{N} coincide. Moreover, R_{𝔯} is primordial, and so R = 1, by the observations subsequent to lemma D, as c does not divide |R|, by assumption.

We show that 𝔯 = 0. Suppose not. Then Z ∩ 𝔯 ≠ 0, by corollary (3.7), as 0 is a minimal prime ideal in S (and S is finitely generated as module over Z, by assumption). Thus there exists z ∈ (Z ∩ 𝔯) ∖ 0. Therefore,

w := ∏_{x∈\overline{N}} ^{x}z ∈ Z\overline{N} ∖ 0

as Z is an integral domain. However, as z divides w and z ∈ 𝔯, this yields the contradiction

0 ≠ w ∈ Z\overline{N} ∩ 𝔯.

Thus 𝔯 = 0, and so \overline{N}\overline{𝔭} has defect 1_{0}. Set 𝔭 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(\overline{𝔭}) where π_{𝔮} : A(P) → S denotes the canonical epimorphism. By theorem I, 𝔭 is a prime ideal in A(H), and H_{𝔭} has defect P_{𝔮}. The additional assertion follows from corollary J. □

The following result can be regarded as a kind of converse of lemma (5.6).

(5.7) LEMMA. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} be a pointed group on A with defect P_{𝔮}. Moreover, assume that the characteristic c of the residue algebra S := A(P)/𝔮 is non-zero, i.e. c = q for some prime number q. Then the kernel K of the action of \overline{N}_{H}(P_{𝔮}) on S is a q′-group.

PROOF. Set \overline{N} := \overline{N}_{H}(P_{𝔮}), and denote by \overline{N}\overline{𝔭} the Puig correspondent of H_{𝔭} with respect to P_{𝔮}. Then \overline{N}\overline{𝔭} is a pointed group on the \overline{N}-algebra S with defect 1_{0}.

Suppose that q divides the order of K. Thus K contains an element x of order q. Denote by C the cyclic subgroup of K generated by x. By corollary (3.3), there is a prime ideal 𝔯 in S^{C} such that S\overline{N} ∩ 𝔯 ⊆ \overline{𝔭}, i.e. C_{𝔯} ≤ \overline{N}\overline{𝔭}. Trivially, S^{C} ∩ 0 ⊆ 𝔯, and so

1_{0} ≤ C_{𝔯} ≤ \overline{N}\overline{𝔭}.

Moreover, 1 is the only proper subgroup of C, and

S_{1}^{C} = Tr_{1}^{C}(S) ⊆ |C|S = qS = 0 ⊆ 𝔯

as C acts trivially on S. This shows that C_{𝔯} is primordial. However, as 1_{0} is a defect of \overline{N}\overline{𝔭}, theorem H yields C_{𝔯} = 1_{0}, and thus x ∈ C = 1, which is a contradiction. □

Next we prove corollary K:

PROOF of Corollary K. Set \overline{N} := \overline{N}_{H}(P_{𝔮}), and denote by \overline{𝔭} the Puig correspondent of 𝔭 (with respect to P_{𝔮}). Thus, by theorem I, 1_{0} is a defect of \overline{N}\overline{𝔭} (as pointed groups on the \overline{N}-algebra S), and 𝔭 is a maximal ideal in A(H) if and only if \overline{𝔭} is a maximal ideal in S\overline{N}. Moreover, 𝔮 is a maximal ideal in A(P) if and only if 0 is a maximal ideal in S. Therefore, it suffices to show that 0 is maximal in S if and only if \overline{𝔭} is maximal in S\overline{N}.

“⇒”. Assume that 0 is a maximal ideal in S, i.e. S is a simple algebra, and so Z(S) is a field. Thus S is a finite dimensional Z(S)-algebra, by assumption. Set 𝔽 := Z(S)\overline{N}, a subfield of Z(S). By Artin’s theorem, Z(S) is a Galois extension of 𝔽 with Galois group (isomorphic to) \overline{N}/K, where K denotes the kernel of the action of \overline{N} on Z(S); and, in particular, Z(S) is finite dimensional as 𝔽-vector space (cf. [14] theorem V.2.15). Hence S is also a finite dimensional 𝔽-algebra. On the other hand, S\overline{N} is an 𝔽-subalgebra of S. Thus S\overline{N} is finite dimensional over 𝔽, as well; in partiuclar, S\overline{N} is an Artinian ring. Therefore, \overline{𝔭} is a maximal ideal in S\overline{N}, and we are done.

“⇐”. Assume now that \overline{𝔭} is a maximal ideal in S\overline{N}. Set Z := Z(S). Then S is a Z-algebra which is finitely generated as module over Z, by assumption. Moreover, Z is an integral domain as S is a prime ring, by definition.

Let z ∈ Z\overline{N} ∩ \overline{𝔭}. Then zS is an ideal in S and

Tr\overline{N} ; 1(zS) = zA\overline{N} ; 1 ⊆ \overline{𝔭},

by the Frobenius axiom. Thus z ∈ zS = 0 as \overline{N}\overline{𝔭} is relatively 1_{0}-projective; so Z\overline{N} ∩ \overline{𝔭} = 0. Hence

Z\overline{N} ≅ Z\overline{N}/Z\overline{N} ∩ \overline{𝔭} ≅ Z\overline{N} + \overline{𝔭}/\overline{𝔭} ⊆ Z(S\overline{N}/\overline{𝔭}).

As S\overline{N}/\overline{𝔭} is a simple algebra, by assumption, its center is a field, and so is Z\overline{N}.

We claim that Z is a field, as well. Let z ∈ Z ∖ 0. Then w := ∏_{x∈\overline{N}} ^{x}z ∈ Z\overline{N} ∖ 0 as Z has no zero divisors. Moreover, z divides w in Z. Hence w is invertible, and so is z.

Therefore, S is an Artinian algebra. As 0 is a prime ideal in S, 0 is thus even maximal in S, and we are done. □

The proof of corollary L is similar:

PROOF of Corollary L. Set \overline{N} := N_{H}(P_{𝔮})/P, and denote by \overline{N}_{\overline{𝔭}_{1}} and \overline{N}_{\overline{𝔭}_{2}} the Puig correspondents of H_{𝔭_{1}} and H_{𝔭_{2}}, respectively, with respect ot P_{𝔮} (which are pointed groups on the \overline{N}-algebra S). By theorem I, we have \overline{N}_{\overline{𝔭}_{1}} ≤ \overline{N}_{\overline{𝔭}_{2}}, and it suffices to show that \overline{𝔭}_{1} = \overline{𝔭}_{2}; moreover, \overline{N}_{\overline{𝔭}_{1}} and \overline{N}_{\overline{𝔭}_{2}} both have defect 1_{0}.

Set Z := Z(S), and let z ∈ Z\overline{N} ∩ \overline{𝔭}_{2}. Then zS is an ideal in S, and

Tr\overline{N} ; 1(zS) = z Tr\overline{N} ; 1(S) ⊆ \overline{𝔭}_{2}.

Therefore, z ∈ zS ⊆ 0 as \overline{N}_{\overline{𝔭}_{2}} pr 1_{0}. This shows that Z\overline{N} ∩ \overline{𝔭}_{2} = 0.

As S is a prime ring, by construction, Z is an integral domain; moreover, S is torsion-free as Z-module. Denote by 𝕂 the quotient field of Z. For any Z-module M, write 𝕂M instead of (Z ∖ 0)^{−1}M. By assumption, S is finitely generated over Z; in particular, 𝕂S is finite dimensional over 𝕂. Moreover, S may be regarded as unitary Z-subalgebra of 𝕂S, and 𝕂S is spanned by S as 𝕂-vector space (cf. the observations preceding proposition (3.6)).

Clearly, the action of \overline{N} on S extends to an action of \overline{N} on 𝕂S. Let a ∈ (𝕂S)\overline{N}. Then za ∈ S for some z ∈ Z ∖ 0. Set w := ∏_{x∈\overline{N}} ^{x}z. Plainly, w ∈ Z\overline{N} ∖ 0. As w = uz for some u ∈ Z, we also have wa ∈ S. Moreover, wa is fixed by \overline{N}; so wa ∈ S\overline{N}. This argument applied to Z instead of S yields in particular that 𝕂\overline{N} is a quotient field of Z\overline{N}. Since S is torsion-free as Z-module, 𝕂\overline{N}S\overline{N} may be regarded as unitary 𝕂\overline{N} subalgebra of 𝕂S. (Then the above argument implies that 𝕂\overline{N}S\overline{N} = (𝕂S)\overline{N}.)

By Artin’s Theorem, 𝕂 is finite dimensional over 𝕂\overline{N} (cf. the proof of corollary K). Therefore, 𝕂S is also a finite dimensional 𝕂\overline{N}-algebra. In particular, 𝕂\overline{N}S\overline{N} is finite dimensional over 𝕂\overline{N}. Therefore, by corollary (3.7), \overline{𝔭}_{2} is a minimal prime ideal in S\overline{N}; and so \overline{𝔭}_{1} = \overline{𝔭}_{2}, as we wanted to show. □

Now we can easily deduce corollary M.

PROOF of Corollary M. Let

𝔭_{0} ⊂ 𝔭_{1} ⊂ … ⊂ 𝔭_{d}

(d ∈ ℕ) be a chain of prime ideals in A(G). For i = 0, …, d, denote by (P_{i})_{𝔮_{i}} a defect of G_{𝔭_{i}}. Then (P_{0})_{𝔮_{0}} is a primordial pointed subgroup of G_{𝔭_{1}}, and G_{𝔭_{1}} is relatively (P_{1})_{𝔮_{1}}-projective. Therefore, there is g ∈ G such that (P_{0})_{𝔮_{0}} ≤ ^{g}((P_{1})_{𝔮_{1}}), by theorem G. Replacing (P_{1})_{𝔮_{1}} by ^{g}((P_{1})_{𝔮_{1}}) (which is a defect of G_{𝔭_{1}}, as well), we may assume that (P_{0})_{𝔮_{0}} ≤ (P_{1})_{𝔮_{1}}. Arguing inductively on d, we may assume that

(P_{0})_{𝔮_{0}} ≤ (P_{1})_{𝔮_{1}} ≤ … ≤ (P_{d})_{𝔮_{d}}.

By corollary L, we even have

(P_{0})_{𝔮_{0}} < (P_{1})_{𝔮_{1}} < … < (P_{d})_{𝔮_{d}}.

Let i_{1} < i_{2} < … < i_{m−1} be the elements of {1, …, d} such that P_{i_{j}} ≠ P_{i_{j}−1} for all j. Moreover, set i_{0} := 0 and i_{m} = d + 1 Thus, for any j = 0, …, m − 1,

𝔮_{i_{j}}/𝔮_{i_{j}} ⊂ 𝔮_{i_{j}+1}/𝔮_{i_{j}} ⊂ … ⊂ 𝔮_{i_{j+1}−1}/𝔮_{i_{j}}

are prime ideals in A(P_{i_{j}})/𝔮_{i_{j}}; hence i_{j+1} −1−i_{j} ≤ N, by assumption. This implies that

d + 1 = (i_{m} − i_{m−1}) + (i_{m−1} − i_{m−2}) + … + (i_{1} − i_{0})
≤ m(N + 1).

On the other hand, by choice of the i_{j}’s, we have

P_{0} = P_{i_{0}} ⊂ P_{i_{1}} ⊂ … ⊂ P_{i_{m−1}},

and so 2^{m−1} ≤ |P_{i_{m}}|. However, P_{i_{m}} is primordial for A, and thus |P_{i_{m}}| ≤ |R|, by choice of R. This yields that d ≤ (log_{2}(|R|) + 1)(N + 1) − 1 = N + log_{2}(|R|)N + log_{2}(|R|), completing the proof. □

Note that the proof of corollary M demonstrates that corollary M can be refined if A is, in addition, a cohomological Green functor. Then any primordial subgroup of G for A is a p-group for some rational prime p. Let p be a prime number. Denote by R_{p} a primordial p-subgroup of G for A of maximal order, and let N_{p} be an upper bound for the Krull dimension of A(P)/𝔮, where P_{𝔮} ranges over all primordial pointed groups P_{𝔮} on A such that P is a p-group. Then the Krull dimension of A(G) is less or equal than the maximum of all numbers N_{p} + log_{p}(|R_{p}|)N_{p} + log_{p}(|R_{p}|), where p ranges over all prime divisors of |G|. We will demonstrate below (cf. remark (6.2)) that, using Dress’s induction theorem, a similar assertion holds in the general situation.

Our next aim is a proof of the Green correspondence:

PROOF of Theorem N. As H contains N_{G}(P_{𝔮}), by assumption, we have

N_{H}(P_{𝔮}) = H ∩ N_{G}(P_{𝔮}) = N_{G}(P_{𝔮}).

We set \overline{N} := \overline{N}_{H}(P_{𝔮}) = \overline{N}_{G}(P_{𝔮}). Moreover, set S := A(P)/𝔮 and denote by π_{𝔮} : A(P) → S the canonical the canonical epimorphism. Consider the following topological spaces (resp. posets):

𝔛 := {𝔭 ∈ Spec(A(G)) : P_{𝔮} is a defect of G_{𝔭} },
𝔜 := {𝔯 ∈ Spec(A(H)) : P_{𝔮} is a defect of H_{𝔯} },
ℨ := \{𝔰 ∈ Spec(S\overline{N}) : 1_{0} is a defect of \overline{N}_{𝔰} \}.

Then the Puig correspondence (theorem I) yields bijections 𝔛 → ℨ and 𝔜 → ℨ which are both, homeomorphisms, as well as isomorphism of posets. This way we obtain an homeomorphism (and an isomorphism of posets) Γ : 𝔛 → 𝔜. At first, we show that this map Γ satisfies assertions (a)-(c), and later on we will see that this bijection, in fact, coincides with the map Γ of theorem N.

First of all, recall that if 𝔭 ∈ 𝔛, 𝔰 ∈ ℨ and 𝔯 ∈ 𝔜 such that 𝔭 corresponds to 𝔰, which in turn corresponds to 𝔯 under the above bijections then

𝔭 = (π_{𝔮} ∘ Res_{P}^{G})^{−1}(𝔰) and 𝔯 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(𝔰).

Thus

𝔭 = ((π_{𝔮} ∘ Res_{P}^{H}) ∘ Res_{H}^{G})^{−1}(𝔰) = (Res_{H}^{G})^{−1}((π_{𝔮} ∘ Res_{P}^{H})^{−1}(𝔰)) = (Res_{H}^{H})^{−1}(𝔯)

Therefore, Γ satisfies assertion (a) of theorem N, and in particular H_{𝔯} ≤ G_{𝔭}.

Let 𝔭 ∈ 𝔛, and set 𝔯 := Γ(𝔭). Thus P_{𝔮} ≤ H_{𝔯} ≤ G_{𝔭}. Denote by 𝔰 ∈ ℨ the Puig correspondent of 𝔯. Then 𝔯 = (π_{𝔮} ∘ Res_{P}^{H})^{−1}(𝔰), by theorem I. Moreover, 𝔰 is also the Puig correspondent of 𝔭, by definition of Γ, and thus 𝔭 = (π_{𝔮} ∘ Res_{P}^{G})^{−1}(𝔰), by theorem I. We show that G_{𝔭} pr H_{𝔯}. Write N_{G}(P)∖N_{G}(P_{𝔮}) = {g_{1}, …, g_{n}} (n ∈ ℕ), and set

J := (^{g_{1}}𝔮)(^{g_{2}}𝔮) · . . . · (^{g_{n}}𝔮).

(Note that J = A(P) in case n = 0.) Then J ⊆ ^{g}𝔮 for all g ∈ N_{G}(P) ∖ N_{G}(P_{𝔮}). Moreover, by definition, J ⊈ 𝔮, as 𝔮 is prime. Thus Ind_{P}^{H}(J) ⊈ 𝔯 since H_{𝔯} is relatively P_{𝔮}-projective. Let I be an ideal in A(H) such that Ind_{H}^{G}(I) ⊆ 𝔭. Set Ĩ = I(Ind_{P}^{H}(J)) ⊆ I ∩ (Ind_{P}^{H}(J)). Then, by proposition (5.3), we have

π_{𝔮}(Res_{P}^{G}(Ind_{H}^{G}(Ĩ))) = Tr_{\overline{N}_{H}(P_{𝔮})}^{\overline{N}_{G}(P_{𝔮})}(π_{𝔮}(Res_{P}^{H}(Ĩ)) = π_{𝔮}(Res_{P}^{H}(Ĩ)),

and thus

π_{𝔮}(Res_{P}^{H}(Ĩ)) = π_{𝔮}(Res_{P}^{G}(Ind_{H}^{G}(Ĩ))) ⊆ π_{𝔮}(Res_{P}^{G}(Ind_{H}^{G}(I))) ⊆ π_{𝔮}(Res_{P}^{G}(𝔭)) ⊆ 𝔰.

Therefore, I(Ind_{P}^{H}(J)) = Ĩ ⊆ (π_{𝔮} ∘ Res_{P}^{H})^{−1}(𝔰) = 𝔯. Consequently, I ⊆ 𝔯, as Ind_{P}^{H}(J) ⊈ 𝔯 and 𝔯 is prime. This shows that G_{𝔭} pr H_{𝔯}, and (b) is settled.

Let H_{𝔱} be a pointed group on A such that P_{𝔮} ≤ H_{𝔱} ≤ G_{𝔭}. We show that P_{𝔮} is a defect of H_{𝔱}. Suppose not. Then, by lemma (5.4), one of the following assertions holds:

(i) Ind_{P}^{H}(J) ⊆ 𝔱. (ii) There is a prime ideal 𝔲 in A(P) which is properly containing 𝔮 such that

P_{𝔲} ≤ H_{𝔱}; as J ⊈ 𝔮. However, as P_{𝔮} is a defect of G_{𝔭} and H_{𝔱} ≤ G_{𝔭}, lemma (5.4) tells us that any prime ideal 𝔲 in A(P) such that P_{𝔮} ≤ P_{𝔲} ≤ H_{𝔱} equals 𝔮. Therefore, assertion (i) holds. By proposition (5.2), we have

π_{𝔮}(Res_{P}^{H}(Res_{H}^{G}(Ind_{P}^{G}(J)))) = π_{𝔮}(Res_{P}^{G}(Ind_{P}^{G}(J))) = Tr\overline{N} ; 1(π_{𝔮}(J))) and

π_{𝔮}(Res_{P}^{H}(Ind_{P}^{H}(J))) = Tr\overline{N} ; 1(π_{𝔮}(J)).

Therefore,

Res_{H}^{G}(Ind_{P}^{G}(J)) ⊆ Ind_{P}^{H}(J) + ker(π_{𝔮} ∘ Res_{P}^{H}) ⊆ 𝔱 + ker(π_{𝔮} ∘ Res_{P}^{H}).

As ker(π_{𝔮} ∘ Res_{P}^{H}) ⊆ 𝔱 since P_{𝔮} ≤ H_{𝔱}, this implies Ind_{H}^{G}(J) ⊆ (Res_{H}^{G})^{−1}(𝔱) ⊆ 𝔭. However, this contradicts G_{𝔭} pr P_{𝔮} as J is not contained in 𝔮. Therefore, P_{𝔮} is a defect of H_{𝔱}, i.e. t ∈ 𝔜.

Next we show that 𝔱 ⊆ 𝔯. Let 𝔭′ := Γ^{−1}(𝔱) ∈ 𝔛. Then 𝔭′ = (Res_{H}^{G})^{−1}(𝔱), by the first part of the proof. However, 𝔭′ = (Res_{H}^{G})^{−1}(𝔱) ⊆ 𝔭, as H_{𝔱} ≤ G_{𝔭}, and thus Γ^{−1}(𝔱) = 𝔭′ ⊆ 𝔭 = Γ^{−1}(𝔯). Since Γ preserves order, this implies 𝔱 ⊆ 𝔯. Thus, 𝔖_{𝔭} has a unique maximal element which is precisely Γ(𝔭).

It remains to prove assertion (c). Assume that A is unitary. Let 𝔭 ∈ 𝔛, and set 𝔯 := Γ(𝔭). Moreover, let 𝔰 ∈ ℨ be the Puig correspondent of 𝔭 and 𝔯, respectively. By theorem I, 𝔭 is a maximal ideal in A(G), if and only if 𝔰 is a maximal ideal in S\overline{N} which in turn is equivalent to 𝔯 being a maximal ideal in A(H). We assume that this is the case. Then, as 𝔭 = (Res_{H}^{G})^{−1}(𝔯), Res_{H}^{G} induces a monomorphism φ : A(G)/𝔭 → A(H)/𝔯. Besides, theorem I yields isomorphisms ψ_{G} : A(G)/𝔭 → S\overline{N}/𝔰 and ψ_{H} : A(H)/𝔯 → S\overline{N}/𝔰 which are induced by π_{𝔮} ∘ Res_{P}^{G} and π_{𝔮} ∘ Res_{P}^{H}, respectively. This plainly implies

φ = ψ_{H}^{−1} ∘ ψ_{G},

and forces φ to be an isomorphism. □

We can now easily prove theorem O.

PROOF of Theorem O. ad (a). Assume that P_{𝔮} is a defect of G_{𝔭}. Let H_{𝔯′} be the Green correspondent of G_{𝔭} with respect to P_{𝔮}. Then, by theorem N, P_{𝔮} is a defect of H_{𝔯} and 𝔯 ⊆ 𝔯′. So, in case 𝔯 satisfies the additional hypothesis, H_{𝔯} = H_{𝔯′} is the Green correspondent of G_{𝔭} (with respect to P_{𝔮}).

ad (b). Assume that P_{𝔮} is a defect of H_{𝔯} and that 𝔭 is minimal subject to the condition H_{𝔯} ≤ G_{𝔭}. Let G_{𝔭′} be the Green correspondent of H_{𝔯} with respect to P_{𝔮}. By theorem N, 𝔭′ = (Res_{H}^{G})^{−1}(𝔯) ⊆ 𝔭 as H_{𝔯} ≤ G_{𝔭}; in particular, H_{𝔯} ≤ G_{𝔭′} ≤ G_{𝔭}. So 𝔭 = 𝔭′, by assumption, and thus G_{𝔭} = G_{𝔭′} is the Green correspondent of H_{𝔯} (with respect to P_{𝔮}), which has defect P_{𝔮}, by theorem N.

ad (c). Assume that A is unitary, and that 𝔯 is a maximal ideal of A(H). In particular, 𝔯 satisfies the additional hypothesis of assertion (a); so the implication “⇐” is settled (including the additional statement). In order to prove the implication “⇒”, assume that P_{𝔮} is a defect of H_{𝔯}, and proceed as in the proof of (b). Then, here, 𝔭′ is a maximal ideal in A(G), by theorem N (c). Thus 𝔭′ = 𝔭 as 𝔭′ ⊆ 𝔭, and so G_{𝔭} = G_{𝔭′}, completing the proof. □

A further consequence of the Green correspondence is corollary P, which we prove next.

PROOF of Corollary P. Set N := N_{P} (Q_{𝔮}) ⊇ Q and I := (Res_{Q}^{N})^{−1}(𝔮). Then (Res_{N}^{P})^{−1}(I) = (Res_{Q}^{P})^{−1}(𝔮) ⊆ 𝔭 as restriction is transitive. By corollary (3.4) applied to the homomorphism Res_{N}^{P} : A(P) → A(N), there exists a prime ideal 𝔰 in A(N) such that (Res_{Q}^{N})^{−1}(𝔮) = I ⊆ 𝔰 and (Res_{N}^{P})^{−1}(𝔰) ⊆ 𝔭, i.e. Q_{𝔮} ≤ N_{𝔰} ≤ P_{𝔭}.

Suppose that Q_{𝔮} is a defect of N_{𝔰}. Then 𝔰 is a maximal ideal in A(N), by corollary K. Denote by P_{𝔭′} the Green correspondent of N_{𝔰} with respect to Q_{𝔮}.

Then, by theorem N, 𝔭′ is a maximal ideal in A(P) and

𝔭′ = (Res_{N}^{P})^{−1}(𝔰) ⊆ 𝔭,

as N_{𝔰} ≤ P_{𝔭}. Thus 𝔭′ = 𝔭 as 𝔭 is a proper ideal of A(P). Hence P_{𝔭} = P_{𝔭′} is the Green correspondent of N_{𝔰} (with respect to Q_{𝔮}), which has defect Q_{𝔮}, as well, by theorem N. In particular, P_{𝔭} is relatively Q-projective, contradicting the assumption that P_{𝔭} is primordial. (Note that, as 𝔮 is maximal in A(Q), Q_{𝔮} < P_{𝔭} implies that Q is a proper subgroup of P.)

Therefore, Q_{𝔮} is not a defect of N_{𝔰}. Let R_{𝔯} be a defect of N_{𝔰}. In particular, R_{𝔯} is primordial. By theorem G, there exists x ∈ N such that Q_{𝔮} ≤ ^{x}(R_{𝔯}) as Q_{𝔮} is primordial, by assumption, and N_{𝔰} pr R_{𝔯}. Replacing R_{𝔯} by ^{x}(R_{𝔯}) (which is a defect of N_{𝔰}, as well), we may assume that Q_{𝔮} < R_{𝔯} ≤ N_{𝔰} ≤ P_{𝔭}. In particular, Q ⊂ N = N_{P} (Q_{𝔮}). □

Next we prove Lemma Q which does not rely on the Puig resp. Green correspondences, but merely on elementary properties of primordial pointed groups.

PROOF of Lemma Q. We associate a Green functor S for P/Q with the given data, as follows: For a subgroup R/Q of P/Q set S(R/Q) := Res_{Q}^{R}(A(R)) + 𝔮/𝔮, with its k-algebra structure as subalgebra of A(Q)/𝔮. The restriction maps on S are just the canonical inclusions and conjugation on S is induced from conjugation on A in the obvious way. (Note that P fixes 𝔮, by assumption.) Let R/Q ⊆ T/Q be subgroups of P/Q. Define

Ind_{R/Q}^{T/Q} : S(R/Q) → S(T/Q), a + 𝔮 ↦ Tr_{R}^{T}(a) + 𝔮,

where a ∈ Res_{Q}^{R}(A(R)). We check that this map is well-defined. (In case it is, it is plainly k-linear.) Let a = Res_{Q}^{R}(b) ∈ Res_{Q}^{R}(A(R)). In particular, a ∈ A(Q)^{R}; so that Tr_{R}^{T}(a) makes sense. Moreover, we have

Res_{Q}^{T}(Ind_{R}^{T}(b)) = ∑_{QxR∈Q\T/R} Ind_{Q∩^{x}R}^{Q}(Res_{Q∩^{x}R}^{^{x}R}(^{x}b))

= ∑_{xR∈T/R} Res_{Q}^{^{x}R}(^{x}b) = Tr_{R}^{T}(a),

as Q is normal in T. This shows that Tr_{R}^{T}(a) ∈ Res_{Q}^{T}(A(T)). Apart from that, for a ∈ Res_{Q}^{R}(A(R)) ∩ 𝔮, we have Tr_{R}^{T}(a) ∈ 𝔮 as T fixes 𝔮. Thus Ind_{R/Q}^{T/Q} is in fact well-defined.

It is easily checked that S is a well-defined Green functor for P/Q. Moreover, S is plainly cohomological.

Let ε be the composition of Res_{Q}^{P} : A(P) → Res_{Q}^{P}(A(P)) followed by the canonical epimorphism

Res_{Q}^{P}(A(P)) → Res_{Q}^{P}(A(P)) + 𝔮/𝔮 = S(P/Q).

Then ε is an epimorphism, as well, and

ker ε = (Res_{Q}^{P})^{−1}(𝔮) ⊆ 𝔭

as Q_{𝔮} ≤ P_{𝔭}. Therefore, ε(𝔭) is a prime ideal in S(P/Q), and (P/Q)_{ε(𝔭)} is a pointed group on S.

We show that this pointed group is primordial. Let R/Q be a proper subgroup of P/Q, and let a = Res_{Q}^{R}(b) ∈ Res_{Q}^{R}(A(R)). Then the first part of the proof shows that

Ind_{R/Q}^{P/Q}(a+𝔮) = Res_{Q}^{P}(Ind_{R}^{P}(b))+𝔮 ⊆ Res_{Q}^{P}(ker Br_{P}^{A})+𝔮/𝔮 ⊆ Res_{Q}^{P}(𝔭)+𝔮/𝔮 = ε(𝔭),

as P_{𝔭} is primordial. This proves our claim. Hence P/Q is a p-group for some rational prime p, by the observations subsequent to lemma D.

We now prove the additional assertion. Recall first that c_{𝔭} is either 0 or a prime numbers as 𝔭 is a prime ideal in A(P). Analogously, c_{𝔮} is either 0 or a prime number. As, by definition, S(P/Q) is a subalgebra of A(Q)/𝔮, the characteristic of S(P/Q) divides c_{𝔮}. Thus the characteristic c of S(P/Q)/ε(𝔭) divides c_{𝔮}, as well.

We show that c = c_{𝔭}. Since c_{𝔭}A(P) ⊆ 𝔭, we have

c_{𝔭}(Res_{Q}^{P}(A(P)) + 𝔮) = Res_{Q}^{P}(c_{𝔭}A(P)) + 𝔮 ∈ Res_{Q}^{P}(𝔭) + 𝔮/𝔮 = ε(𝔭),

and so c divides c_{𝔭}. Conversely, let a ∈ A(P), and we wish to show that ca ∈ 𝔭. We have c(Res_{Q}^{P}(A(P)) + 𝔮) ⊆ Res_{Q}^{P}(𝔭) + 𝔮, by definition of c. Thus c Res_{Q}^{P}(a) = Res_{Q}^{P}(x) + y for some x ∈ 𝔭 and some y ∈ 𝔮. Hence

Res_{Q}^{P}(ca − x) = c Res_{Q}^{P}(a) − Res_{Q}^{P}(x) = y ∈ 𝔮,

therefore ca − x ∈ (Res_{Q}^{P})^{−1}(𝔮) ⊆ 𝔭, and so ca ∈ 𝔭 as x ∈ 𝔭. Hence cA(P) ⊆ 𝔭. This shows that c_{𝔭} divides c. Thus c = c_{𝔭}, and so c_{𝔭} divides c_{𝔮}.

By the observations subsequent to lemma D, we have the following:

(a) P/Q = 1, ie. P = Q, in case c_{𝔭} = 0. Moreover, c_{𝔮} = 0 in this case as c_{𝔭} divides c_{𝔮}.

(b) If Q ≠ P then c_{𝔭} = p, and thus c_{𝔮} ∈ {0, p}. This completes the proof. □

Our proof of corollary R is based on corollary (3.10) and the ideas of the proof of corollary P.

PROOF of Corollary R. We argue by induction on |P : Q|, and proceed as in the proof of corollary P. Assume first that P = Q. Then the first assertions of corollary R are plainly satisfed. The additional statement follows from the fact that, in this case, A(P)/𝔭 is isomorphic to a residue algebra of A(Q)/𝔮, and so c_{𝔭} divides c_{𝔮}.

So we assume that Q ⊂ P. By corollary (3.4), there is a prime ideal 𝔮′ in A(Q) containing 𝔮 such that (Res_{Q}^{P})^{−1}(𝔮′) ⊆ 𝔭 (i.e. Q_{𝔮} ≤ Q_{𝔮′} ≤ P_{𝔭}); and we may assume that 𝔮′ is maximal subject to this condition. As Q_{𝔮} is primordial, i.e. ker Br_{Q} ⊆ 𝔮 ⊆ 𝔮′, Q_{𝔮′} is primordial, as well. By corollary (3.10), we have k ∩ 𝔮′ = k ∩ 𝔭 as A(P)/𝔭 and A(Q)/𝔮′ are finitely generated as modules over k, by assumption.

As in the proof of corollary P, set N := N_{P} (Q_{𝔮′}) ⊇ Q, and I := (Res_{Q}^{N})^{−1}(𝔮′). Then (Res_{N}^{P})^{−1}(I) = (Res_{Q}^{P})^{−1}(𝔮′) ⊆ 𝔭 as restriction is transitive. By corollary (3.4) applied to the homomorphism Res_{N}^{P} : A(P) → A(N), there is a prime ideal 𝔰 in A(N) such that (Res_{Q}^{N})^{−1}(𝔮′) = I ⊆ 𝔰 and (Res_{N}^{P})^{−1}(𝔰) ⊆ 𝔭, i.e. Q_{𝔮′} ≤ N_{𝔰} ≤ P_{𝔭}.

Suppose that Q_{𝔮′} is a defect of N_{𝔰}. Denote by P_{𝔭′} the Green correspondent of N_{𝔰} with respect to Q_{𝔮′}. As in the proof of corollary P, we have 𝔭′ ⊆ 𝔭 and Q_{𝔮′} is a defect of P_{𝔭′}. Hence

k ∩ 𝔭′ = k ∩ 𝔮′ = k ∩ 𝔭,

§6. A PROOF OF DRESS’S INDUCTION THEOREM 39

by lemma (4.2), and so k1_{A(P)} ∩ 𝔭′ = k1_{A(P)} ∩ 𝔭. Thus 𝔭 = 𝔭′, by corollary (3.8), as A(P)/𝔭′ is finitely generated as module over k, by assumption. Therefore, P_{𝔭} = P_{𝔭′} has defect Q_{𝔮′}. In particular, P_{𝔭} is relatively Q-projective. As P_{𝔭} is primordial, by assumption, and Q ⊂ P, this is a contradiction.

Thus Q_{𝔮′} is not a defect of N_{𝔰}. Thus there exists a primordial subgroup R_{𝔯} of N_{𝔰} such that Q_{𝔮′} < R_{𝔯} ≤ N_{𝔰} ≤ P_{𝔭}, by theorem H “(a) ⇔ (c)”.

Suppose that Q = R. Thus 𝔮′ ⊂ 𝔯. In particular, k ∩ 𝔮′ ⊆ k ∩ 𝔯. Let α ∈ k ∩ 𝔯, i.e. α ∈ k and α1_{A(R)} ∈ 𝔯. Then

Res_{R}^{P}(α1_{A(P)}) = α1_{A(R)} ∈ 𝔯,

and so α1_{A(P)} ∈ (Res_{R}^{P})^{−1}(𝔯) ⊆ 𝔭 as R_{𝔯} ≤ P_{𝔭}. Hence α ∈ k ∩ 𝔭. This shows that

k ∩ 𝔮′ ⊆ k ∩ 𝔯 ⊆ k ∩ 𝔭 = k ∩ 𝔮′,

and so we have equality everywhere. Hence k1_{A(Q)} ∩ 𝔮′ = k1_{A(Q)} ∩ 𝔯, and corollary (3.8) yields the contradiction 𝔮′ = 𝔯.

Therefore, Q ⊂ R, and thus |P : R| < |P : Q|. By induction, there is a chain

R_{𝔯} = (R_{2})_{𝔯_{2}} ≤ (R_{3})_{𝔯_{3}} ≤ … ≤ (R_{s})_{𝔯_{s}} = P_{𝔭}

(s ∈ ℕ) of primordial pointed groups on A such that R_{i+1} ⊆ N_{P} ((R_{i})_{𝔯_{i}}) for all i = 2, … s − 1, and |P : R| is a power of some prime number p.

Denote by c_{𝔮}, c_{𝔮′}, c_{𝔯} and c_{𝔭} the characteristics of A(Q)/𝔮, A(Q)/𝔮′, A(R)/𝔯 and A(P)/𝔭, respectively. Then c_{𝔮′} divides c_{𝔮} as q ⊆ 𝔮′.

Suppose that c_{𝔭} = 0. Then R = P and c_{𝔯} = 0, by induction. Thus lemma Q yields the contradiction Q = R.

Hence c_{𝔭} = p and c_{𝔯} ∈ {0, p}, again by induction. Then c_{𝔯} = p since otherwise lemma Q would imply Q = R. Thus R/Q is a p-group and c_{𝔮′} ∈ {0, p}, by lemma Q. Hence also c_{𝔮} ∈ {0, p} as c_{𝔮′} divides c_{𝔮}. (Note that c_{𝔮} is either 0 or a prime number as A(Q)/𝔮 is a prime ring.) Moreover, |P : Q| is a p-power.

Set (R_{0})_{𝔯_{0}} := Q_{𝔮} and (R_{1})_{𝔯_{1}} := Q_{𝔮′}. Then the chain of primordial pointed groups on A

Q_{𝔮} = (R_{0})_{𝔯_{0}} ≤ (R_{1})_{𝔯_{1}} ≤ (R_{2})_{𝔯_{2}} ≤ … ≤ (R_{s})_{𝔯_{s}} = P_{𝔭}

has the desired property, completing the proof. □

§6. A PROOF OF DRESS’S INDUCTION THEOREM

The main objective of this section is to prove Dress’s induction theorem as presented in theorem U. We begin with deducing corollary S from Sylow’s first theorem for primordial pointed groups.

PROOF of Corollary S. Note first that C ≠ ∅ as 𝔭 is a proper ideal of A(H). Let K_{1}, …, K_{n} (n ∈ ℕ) be the elements of C. By assumption,

(ker Res_{K_{1}}^{H}) · . . . · (ker Res_{K_{n}}^{H}) ⊆ ⋂_{K∈C} ker Res_{K}^{H} ⊆ 𝔭.

As 𝔭 is a prime ideal in A(H), there exists K ∈ C such that ker Res_{K}^{H} ⊆ 𝔭. Thus corollary (3.4), applied to the homomorphism Res_{K}^{H} : A(H) → A(K), yields a prime ideal 𝔯 in A(K) such that

(Res_{K}^{H})^{−1}(𝔯) ⊆ 𝔭,

i.e. K_{𝔯} ≤ H_{𝔭}. Let Q_{𝔰} be a defect of K_{𝔯}. In particular, Q_{𝔰} is a primordial pointed subgroup of H_{𝔭}.

By definition of a defect group, there is a prime ideal 𝔮 in A(P) such that H_{𝔭} has defect P_{𝔮}. In particular, H_{𝔭} is relatively P_{𝔮}-projective. Therefore, there is h ∈ H such that

Q′_{𝔰′} := (^{h}Q)_{^{h}𝔰} = ^{h}(Q_{𝔰}) ≤ P_{𝔮}

(and Q′_{𝔰′} is a primordial pointed group on A, as well).

Moreover,

k/k ∩ 𝔭 ≅ k1_{A(H)}/k1_{A(H)} ∩ 𝔭 ≅ k1_{A(H)} + 𝔭/𝔭,

by definition of k ∩ 𝔭 (cf. the paragraph preceding corollary (3.10)), and the latter algebra is a unitary subalgebra of A(H)/𝔭. Hence c is the characteristic of k/k ∩ 𝔭, as well. By lemma (4.2), k ∩ 𝔭 = k ∩ 𝔮, and so, by a similar argument, c is also the characteristic of A(P)/𝔮.

Assume first that c = 0. Then P = Q′ = ^{h}Q, by corollary R, and Q ⊆ K ∈ C. Second, assume that c is some prime number p. Again by corollary R, there is a chain of primordial pointed groups

Q′_{𝔰′} = (T_{0})_{𝔱_{0}} ≤ … ≤ (T_{m})_{𝔱_{m}} = P_{𝔮}

(m ∈ ℕ) on A such that, for all i = 0, …, m − 1, T_{i+1} ⊆ N_{P} (T_{i}) and T_{i+1}/T_{i} is a p-group. As O^{p}(Q′) is characteristic in Q′ = T_{0}, it is normal in T_{1}. Moreover, T_{1}/ O^{p}(Q′) is a p-group. Thus O^{p}(T_{1}) ⊆ O^{p}(Q′). In particular, O^{p}(T_{1}) is a normal subgroup of Q′. Moreover, Q′/ O^{p}(T_{1}) ⊆ T_{1}/ O^{p}(T_{1}) is a p-group. Thus we also have O^{p}(Q′) ⊆ O^{p}(T_{1}), and so O^{p}(Q′) = O^{p}(T_{1}). Arguing inductively on m, we obtain that

O^{p}(P) = O^{p}(T_{m}) = O^{p}(T_{m−1}) = … = O^{p}(T_{1}) = O^{p}(Q′),

and so O^{p}(P) = O^{p}(^{h}Q) = ^{h}(O^{p}(Q)). As O^{p}(Q) ⊆ Q ⊆ K ∈ C, this completes the proof. □

Now we are in a position to prove theorem U.

PROOF of Theorem U. It suffices to prove assertion (b); for if π = {p}, and p is a prime which does not divide the order of G, then |G| = |G|_{π′} and C^{π} is the set of subgroups of elements of C; thus, as restriction and induction are transitive, the summands in question do not change if we replace C by C^{π}, in this case.

Denote by B the Burnside ring functor of G over k (cf. appendix A for the notation), a unitary Green functor for G over k. At first, we show that the assertion of theorem U holds for B, and afterwards we deduce the general case.

Plainly, B satisfies the hypotheses of corollary S. Set

I := ( ⋂_{H∈C} ker Res_{H}^{G}) + ( ∑_{H∈C^{π}} B_{H}^{G}),

which is an ideal in B(G).

Suppose that I ⊂ B(G). As B(G) has an identity, there is a maximal (and hence prime) ideal 𝔪 in B(G) containing I. Let P be a defect group of G_{𝔪}, and denote by c the characteristic of B(G)/𝔪. We show that P ∈ C^{π}. In case c = 0, we have P = ^{g}H for some g ∈ G and some H ∈ C, by corollary S appliued to A = B and H_{𝔭} = G_{m}; replacing P by ^{g^{−1}}P (which is a defect group of G_{𝔪}, as well), we may assume that P = H ∈ C, in particular O^{p}(P) = O^{p}(H) for any p ∈ π, and so P ∈ C^{π}. So we assume that c = p for some prime p(≠ 0). As |G|_{π′} is invertible in k, by assumption, p = c cannot divide |G|_{π′}. Hence p ∈ π, and so P is conjugate in G to an element of C^{π}, by corollary S. Again, replacing P by a suitable conjugate, we may assume that P ∈ C^{π}.

Thus B_{P}^{G} ⊆ 𝔪, by definition of I. However, this contradicts the fact that G_{𝔪} is relatively P -projective.

Therefore, I = B(G). It is well-known (and straightforward to check) that M may be regarded as module of the Green functor B (cf. [27] for the notation), where for subgroups K ⊆ H of G and m ∈ M(H), [H/K] · m is defined by

Ind_{K}^{H}(Res_{K}^{H}(m)).

By definition of a module over a Green functor, we plainly have

( ∑_{H∈C^{π}} B_{H}^{G})M(G) ⊆ ∑_{H∈C^{π}} M_{H}^{G} and

( ⋂_{H∈C} ker(Res_{H}^{G} : B(G) → B(H)))M(G) ⊆ ⋂_{H∈C} ker(Res_{H}^{G} : M(G) → M(H))

As M(G) is a unitary B(G)-module, this implies that

M(G) = B(G)M(G) ⊆ ( ⋂_{H∈C} ker Res_{H}^{G}) + ( ∑_{H∈C^{π}} M_{H}^{G})

⊆ ( ⋂_{H∈C^{π}} ker Res_{H}^{G}) + ( ∑_{H∈C^{π}} M_{H}^{G}) ⊆ M(G),

as C ⊆ C^{π}, and it remains to prove that the latter sum is direct. But this follows from the fact that

( ⋂_{H∈C^{π}} ker(Res_{H}^{G} : B(G) → B(H)))( ∑_{H∈C^{π}} M_{H}^{G}) = 0 and

( ∑_{H∈C^{π}} B_{H}^{G})( ⋂_{H∈C^{π}} ker(Res_{H}^{G} : M(G) → M(H))) = 0,

which in turn are immediate consequences of the Frobenius axioms. □

Using Dress’s induction theorem, one can show that, in parts, Sylow’s first theorem for primordial pointed groups holds for arbitrary Green functors:

(6.1) COROLLARY. Let G be a finite group, let A be a Green functor for G, and let Q_{𝔮} ≤ P_{𝔭} be primordial pointed groups on A. Denote by c_{𝔮} and c_{𝔭} the characteristics of A(Q)/𝔮 and A(P)/𝔭, respectively. Then the following assertions hold:

(a) If c_{𝔭} = 0 then c_{𝔮} = 0 and Q = P.

(b) In case Q ≠ P, c_{𝔭} is some (non-zero) prime p, O^{p}(Q) = O^{p}(P) and c_{𝔮} ∈ {0, p}. Thus Q is subnormal in P and |P : Q| is a power of p.

Note that corollary (6.1) is correlated to results of Thévenaz in [23] on primordial subgroups for Green functors, which are based on Dress’s induction theorem.

PROOF of Corollary (6.1). Recall that c_{𝔭} (resp. c_{𝔮}) is either 0 or a prime number as A(P)/𝔭 (resp. A(Q)/𝔮) is a prime algebra. We have

Res_{Q}^{P}(c_{𝔮}A(P)) ⊆ c_{𝔮}A(Q) ⊆ 𝔮.

Hence c_{𝔮}A(P) ∈ (Res_{Q}^{P})^{−1}(𝔮) ⊆ 𝔭 as Q_{𝔮} ≤ P_{𝔭}. Thus c_{𝔭} divides c_{𝔮}. Therefore, c_{𝔮} = 0 in case c_{𝔭} = 0, and c_{𝔮} ∈ {0, p} if c_{𝔭} = p for some prime p.

We need a generalization of some of the assertions of proposition (3.6). Set B := A(P), and regard B as ℤ-algebra. Moreover, set S := ℤ ∖ c_{𝔭}ℤ, which is a multiplicative subset of ℤ as c_{𝔭}ℤ is a prime ideal in ℤ, and denote by φ : B → S^{−1}B the canonical homomorphism. We claim that S^{−1}𝔭 := φ(𝔭) is a prime ideal in S^{−1}B. First of all, we show that 𝔭 = φ^{−1}(S^{−1}𝔭). Plainly, 𝔭 ⊆ φ^{−1}(S^{−1}𝔭). Conversely, let x ∈ B, y ∈ 𝔭 and s ∈ S such that φ(x) = \frac{y}{s}. Thus sx − y ∈ ker φ. By [20] section 3b, ker φ = T_{S}(B), the S-torsion submodule of B. Hence tsx−ty = 0 ∈ 𝔭 for some t ∈ S, and so tsx ∈ 𝔭. Therefore,

(tsB)(kx + BxB) ⊆ 𝔭.

Thus tsB ⊆ 𝔭 or x ∈ kx + BxB ⊆ 𝔭 as 𝔭 is a prime ideal in B. Suppose we are in the first case. Then c_{𝔭} divides ts ∈ S as c_{𝔭} is the characteristic of B/𝔭, which contradicts the choice of S. Hence we are in the second case, proving our claim. In particular, S^{−1}𝔭 is a proper ideal of S^{−1}B. Now one can proceed completely analogous to the proof of proposition (3.6) in order to show that S^{−1}𝔭 is a prime ideal in S^{−1}B. Similarly, S^{−1}𝔮 is a prime ideal in S^{−1}A(Q) as c_{𝔮} does not divide any element of S.

For a subgroup H of G, set (S^{−1}A)(H) := S^{−1}A(H). Moreover, the restriction, induction and conjugation maps on A induce maps Ind_{K}^{H} : S^{−1}A(K) → S^{−1}A(H), Res_{K}^{H} : S^{−1}A(H) → S^{−1}A(K) and c_{g,H} : S^{−1}A(H) → S^{−1}A(^{g}H), for subgroups K ⊆ H of G and g ∈ G. It is clear that the collection (S^{−1}A(H)), where H ranges over all subgroups of G, together with the above induction, restriction and conjugation maps, is a well-defined Green functor S^{−1}A for G over S^{−1}ℤ = ℚ, in case c_{𝔭} = 0, or over S^{−1}ℤ = ℤ_{(p)}, in case c_{𝔭} is the prime p. Moreover, for subgroups K ⊆ H of G, we clearly have

(S^{−1}A)_{K}^{H} = S^{−1}A_{K}^{H},

so that Q_{S^{−1}𝔮} and P_{S^{−1}𝔭} are primordial pointed groups on S^{−1}A. Besides, it is plain that Q_{S^{−1}𝔮} ≤ P_{S^{−1}𝔭}. Thus

ker Res_{Q}^{P} ⊆ (Res_{Q}^{P})^{−1}(S^{−1}𝔮) ⊆ S^{−1}𝔭.

Set C := {Q}, and set π := {q}, where q is some prime number which does not divide the order of G in case c_{𝔭} = 0, and q = c_{𝔭} = p otherwise. Then |G|_{π′} is invertible in S^{−1}ℤ, in any case. Thus theorem U, applied to the restricted Green functor Res_{P}^{G} A, yields that

(S^{−1}A)_{H}^{P} ⊈ S^{−1}𝔭

for some H ∈ C^{π} as S^{−1}𝔭 is a proper ideal in S^{−1}A(P). Therefore, P = H, by lemma D, as P_{S^{−1}𝔭} is primordial. In case c_{𝔮} = 0, i.e. q does not divide |G|, this implies P = Q. So we assume that c_{𝔭} = p, i.e. π = {p}. Then O^{p}(P) ⊆ Q. Thus Q is subnormal in P and |P : Q| is a power of p. As in the proof of corollary S, one shows that this implies O^{p}(P) = O^{p}(Q). □

(6.2) REMARK. Note that, using corollary (6.1), the assertion of corollary M can be refined. We adopt the notation of corollary M and its proof. Recall that, given a chain

𝔭_{0} ⊂ 𝔭_{1} ⊂ … ⊂ 𝔭_{d}

of prime ideals in A(G), we constructed a chain of primordial pointed groups on A:

(P_{0})_{𝔮_{0}} < (P_{1})_{𝔮_{1}} < … < (P_{d})_{𝔮_{d}}.

Moreover, we defined elements 0 = i_{0} < i_{1} < … < i_{m−1} of {0, 1, …, d} such that

P_{0} = P_{i_{0}} ⊂ P_{i_{1}} ⊂ … ⊂ P_{i_{m−1}} = P_{d}.

As in the proof of corollary M, one shows that this implies that d ≤ (N_{0} + 1) + (m − 1)(N + 1) − 1, where N_{0} is the Krull dimension of A(P_{0})/𝔮_{0} (which is finite, by assumption), and N is an an upper bound for the Krull dimensions of A(P_{j})/𝔮_{j} (j = i_{1}, …, d) (which exists, by assumption).

Denote by c the characteristic of A(P_{d})/𝔮_{d}. Assume first that c = 0. Then

P_{0} = P_{i_{0}} = … P_{i_{m−1}} = P_{d},

and also the characteristic of A(P_{j})/𝔮_{j} is 0 for all i, by corollary (6.1). Second, assume that c = p for some prime p (≠ 0). Then O^{p}(P_{0}) = O^{p}(P_{d}), p is the characteristic of A(P_{j})/𝔮_{j} for all j = i_{1}, …, d, and, for j = 0, …, i_{1} − 1, the characteristic of A(P_{j})/𝔮_{j} is contained in {0, p}, again by corollary (6.1). Thus m − 1 ≤ log_{p}(r), where r = |P_{d} : O^{p}(P_{d})|. Apart from that, we may assume, without loss of generality, that (P_{0})_{𝔮_{0}} is a minimal primordial pointed group on A as P_{0} is finite and any prime ideal contains some minimal prime ideal (cf. the remark subsequent to corollary (3.7)). Similarly, we may assume that (P_{d})_{𝔮_{d}} is a maximal primordial pointed group on A as P_{d} is finite and A is unitary.

So summarizing, we obtain the following result: Let N_{0} be an upper bound for the Krull dimensions of A(Q)/𝔮, where Q_{𝔮} ranges over all minimal primordial pointed groups on A. Moreover, for any prime divisor p of G, let (R_{p})_{𝔭} be a maximal primordial pointed group on A such that A(R_{p})/𝔭 has characteristic p, and choose (R_{p})_{𝔭} in such a way that among all such maximal primordial pointed groups r_{p} := |R_{p} : O^{p}(R_{p})| is greatest possible. Finally, let N_{p} be an upper bound for the Krull dimensions of A(P)/𝔰, where P_{𝔰} ranges over all primordial pointed groups P_{𝔰} on A such that A(P)/𝔰 has characteristic p. Then the Krull dimension of A(G) is less or equal than the maximum of the numbers N_{0} + log_{p}(r_{p})(N_{p} + 1), where p ranges over all prime divisors of G.

Of course, in the above proof we still used corollary L, i.e. we still rely on the assumption that A is unitary and that A(P)/𝔭 is finitely generated as module over its center for any primordial pointed group P_{𝔭} on A.

The following result (together with corollary (6.1)) is an analogue of corollary S.

(6.3) LEMMA. Let G be a finite group, let A be a Green functor for G, and let H_{𝔭} and K_{𝔮} be a pointed groups on A such that H_{𝔭} is relatively K_{𝔮}-projective. Moreover, let 𝔓 be a set of subgroups of H such that

⋂_{Q∈𝔓} ker Br_{Q}^{H} ⊆ 𝔭.

Then there exist h ∈ H, Q ∈ 𝔓 and a prime ideal 𝔯 in A(Q) such that ^{h}(Q_{r}) is a primordial pointed subgroup of K_{𝔮}.

PROOF. Note first that 𝔓 ≠ ∅ as 𝔭 is a proper ideal of A(H). Let Q_{1}, …, Q_{n} (n ∈ ℕ) be the elements of 𝔓. By assumption,

(ker Br_{Q_{1}}^{H}) · . . . · (ker Br_{Q_{n}}^{H}) ⊆ ⋂_{Q∈𝔓} ker Br_{Q}^{H} ⊆ 𝔭.

Thus there is Q ∈ 𝔓 such that

(Res_{Q}^{H})^{−1}(ker Br_{Q}) = ker Br_{Q}^{H} ⊆ 𝔭,

as 𝔭 is a prime ideal in A(H). Hence corollary (3.4), applied to the homomorphism Res_{Q}^{H} : A(H) → A(Q), yields a prime ideal 𝔯 in A(Q) containing ker Br_{Q} such that

(Res_{Q}^{H})^{−1}(𝔯) ⊆ 𝔭,

i.e. Q_{𝔯} is a primordial pointed subgroup of H_{𝔭}. As H_{𝔭} is relatively K_{𝔮}-projective, by assumption, the result follows from theorem G. □

Now we are in a position to prove theorem V; its proof is completely analogous to the proof of theorem U.

PROOF of Theorem V. Except for the proof of the fact that the sum of assertion (a) is direct, it suffices to prove assertion (b); for if p is a prime which does not divide the order of G, and π := \{p\}, then 𝔓^{π} = 𝔓, |G| = |G|_{π′} and \underline{M}_{π}(H) = \underline{M}(H) for any subgroup H of G as, in this case, O^{p}(U) = U ≠ H = O^{p}(H) for any proper subgroup U of H.

As in the proof of theorem U, denote by B the Burnside ring functor for G over k. Set

I := ( ⋂_{Q∈𝔓} ker Br_{Q}^{G}) + ( ∑_{Q∈𝔓^{π}} Ind_{Q}^{G}(\underline{B}_{π}(Q))),

which is an ideal in B(G), by the Frobenius axiom.

Suppose that I ⊂ B(G). Then there is a prime ideal 𝔪 in B(G) such that I ⊆ 𝔪 (e.g. a maximal ideal). Thus

( ⋂_{Q∈𝔓} ker Br_{Q}^{G}) ⊆ I ⊆ 𝔪.

Let P_{𝔮} be a defect of G_{𝔪}. By lemma (6.3), there is g ∈ G, Q ∈ 𝔓 and a prime ideal 𝔯 in B(Q) such that Q_{𝔯} is a primordial pointed subgroup of ^{g}(P_{𝔮}). Replacing ^{g}(P_{𝔮}) by P_{𝔮} (which is a defect of G_{𝔪}, as well), we may assume that

Q_{𝔯} ≤ P_{𝔮}.

We claim that P ∈ 𝔓^{π}. Denote by c the characteristic of B(P)/𝔮. In case c = 0, we have P = Q ∈ 𝔓, by corollary (6.1), in particular O^{p}(P) = O^{p}(Q) for any p ∈ π, and so P ∈ 𝔓^{π}, by definition of 𝔓^{π}. So we assume that c = p for some prime p(≠ 0). As |G|_{π′} is invertible in k, by assumption, we must have p ∈ π. Moreover, O^{p}(P) = O^{p}(Q), by corollary (6.1); and so P ∈ 𝔓^{π} also in this case.

Therefore,

Ind_{P}^{G}(\underline{B}_{π}(P)) ⊆ I ⊆ 𝔪,

and thus \underline{B}_{π}(P) ⊆ 𝔮 as G_{𝔪} is relatively P_{𝔮}-projective.

Let C be the set of proper subgroups Q of P such that, for any p ∈ π and any subgroup U of Q, O^{p}(U) ≠ O^{p}(Q). Thus

\underline{B}_{π}(P) = ⋂_{Q∈C} ker Res_{Q}^{P},

by definition of \underline{B}_{π}(P). As P_{𝔮} is its own defect, by theorem H, we either have P ∈ C or O^{p}(P) ⊆ Q for some p ∈ π and some Q ∈ C (or both), by corollary S. As C consists only of proper subgroups of P, we are in the latter case. Thus O^{p}(O^{p}(P)) = O^{p}(P) ⊆ Q ∈ C, which contradicts the definition of C.

Therefore, I = B(G). As in the proof of theorem U, M is a module of the Green functor B. Let H be a subgroup of G. Clearly,

\underline{B}_{π}(H)M(H) ⊆ \underline{M}_{π}(H),

and so

(Ind_{H}^{G}(\underline{B}_{π}(H)))M(G) = Ind_{H}^{G}(\underline{M}_{π}),

by the Frobenius axiom. Similarly,

(∑_{K} Ind_{K}^{H}(B(K)))M(H) ⊆ ∑_{K} Ind_{K}^{H}(M(K)),

where K ranges over all proper subgroups of H, and so

(ker Br_{H}^{G})M(G) = ((Res_{H}^{G})^{−1}(ker Br_{H}^{B}))M(G)
⊆ (Res_{H}^{G})^{−1}(ker Br_{H}^{M})
= ker Br_{H}^{G} .

As M(G) is a unitary B(G)-module, this implies that

M(G) = ( ⋂_{Q∈𝔓} ker Br_{Q}^{G}) + ( ∑_{Q∈𝔓^{π}} Ind_{Q}^{G}(\underline{M}_{π}(Q))),

and it remains to prove that, in the situation of asserion (a), the corresponding sum is direct. However, this follows from the fact that, for any subgroup H of G,

(ker Br_{H}^{B})\underline{M}(H) = 0 and
\underline{B}(H)(ker Br_{H}^{M}) = 0,

by the Frobenius axioms, and so

( ⋂_{Q∈𝔓} ker Br_{Q}^{G})( ∑_{Q∈𝔓} Ind_{Q}^{G}(\underline{M}(Q))) = 0 and

( ∑_{Q∈𝔓} Ind_{Q}^{G}(\underline{B}(Q)))( ⋂_{Q∈𝔓} ker Br_{Q}^{G}) = 0,

again by the Frobenius axioms. □

It remains to prove corollary W.

PROOF of Corollary W. In view of theorem V (a), it suffices to show that

M(G) = ∑_{Q∈𝔓} Ind_{Q}^{G}(\underline{M}(Q)).

We argue by induction on |G|.

Let Q ∈ 𝔓, and denote by C the set of proper subgroups of Q. Thus \underline{M}(Q) = ⋂_{R∈C} ker Res_{R}^{Q}. By Dress’s induction theorem (theorem U), we have

M(Q) = \underline{M}(Q) ⊕ ∑_{R∈C} Ind_{R}^{Q}(M(R)).

Let R ∈ C. By induction (applied to the restricted Mackey functor Res_{R}^{G} M), we have

M(R) = ∑_{{S∈𝔓 \atop S⊆R}} Ind_{S}^{R}(\underline{M}(S)),

and, therefore,

M(Q) = ∑_{{R∈𝔓 \atop R⊆Q}} Ind_{R}^{Q}(\underline{M}(R)).

Thus the assertion follows as

M(G) = ∑_{Q∈𝔓} M_{Q}^{G},

by [25] proposition (54.3). □

APPENDIX A. THE SPECTRUM OF THE BURNSIDE RING

The aim of this appendix is to demonstrate that some (very well-known) facts around the prime spectrum of the Burnside ring of a finite group can be easily obtained using defect theory. For references on work on the Burnside ring let us mention, for instance, the papers [10], [22] and [28] by Dress, Solomon and Yoshida, respectively.

(A.1) Throughout we fix a finite group G, let B(G) denote the integral Burnside ring of G, and denote by B the Burnside ring functor of G (i.e. the associated Green functor for G). We refer to [2] for the definition of B(G).

For a G-set S, we denote by [S] its isomorphism class. Let H be a subgroup of G. Then the set G/H of cosets gH (g ∈ G) is a transitive G-set, where G acts by left multiplication. Recall that the set {[G/H]}, where H runs through a transversal for the conjugacy classes of subgroups of G, is a ℤ-basis of B(G). Recall also that [G/G] is a multiplicative identity of B(G), and that, for any subgroup H of G, the induction map Ind_{H}^{G} : B(H) → B(G) can be easily described on the canonical basis by Ind_{H}^{G}([H/K]) = [G/K] for any subgroup K of H. Therefore, \overline{B}(G), the image of the Brauer homomorphism of G, is spanned by the coset of [G/G], i.e. we have

\overline{B}(G) ≅ ℤ[G/G] ≅ ℤ.

Thus the primordial pointed groups on B are precisely the pointed groups H_{(H,p)} where H is any subgroup of G, p is either 0 or a prime number, and (H, p) stands for the prime ideal pℤ + ker Br_{H} in B(H). It is plain that two such pointed groups, H_{(H,p)} and K_{(K,q)}, are conjugate in G if and only if p = q and H and K are conjugate in G.

(A.2) We give an alternative description of the Brauer homomorphism. Let H be a subgroup of G. For a finite G-set S, S^{H} denotes the set of fixed points of H on S and |S^{H}| denotes the number of its elements. The assignment [S] ↦ |S^{H}| yields a homomorphism from the abelian monoid of isomorphism classes of finite G-sets into the integers, and, therefore, gives rise to a homomorphism of groups B(G) → ℤ. We denote this homomorphism by | ^{H}|_{G}, and we omit the subscript when it is clear from context the Burnside ring of which group we refer to. It is easily checked that, for any subgroup K of G, |[G/K]^{H}| is the number of cosets gK ∈ G/K such that H ⊆ ^{g}K. In particular, |[G/G]^{G}| = 1 and |[G/K]^{G}| = 0 for any proper subgroup K of G. Thus the map | ^{G}| : B(G) → ℤ is precisely the Brauer homomorphism of G. Hence, for p as above, the prime ideal (H, p) in B(H) is the full preimage of pℤ under the homomorphism | ^{H}| : B(H) → ℤ, i.e. the collection of all elements x ∈ B(H) such that |x^{H}| is divisible by p.

(A.3) We want to find all prime ideals of B(G), i.e. all pointed groups on B of type G_{𝔭}. We do this by looking for all primordial pointed groups on B which are defects of a pointed group of type G_{𝔭}. Let H be a subgroup of G. Then H_{(H,0)} is a primordial pointed group on B, and \overline{N}_{G}(H_{(H,0)}) = N_{G}(H)/H (the order of which is certainly not divisible by 0). Thus,

𝔭 := (Res_{H}^{G})^{−1}((H, 0))

is a prime ideal in B(G) such that G_{𝔭} has defect H_{(H,0)}, by lemma (5.6). As counting fixed points commutes with restriction, 𝔭 is the kernel of the homomorphism

| |^{H} : B(G) → ℤ.

Moreover, by lemma (4.2), we have ℤ ∩ 𝔭 = ℤ ∩ (H, 0) = 0. We denote this prime ideal 𝔭 by (H, 0)_{G}, and the corresponding pointed group G_{𝔭} by G_{(H,0)}. Thus, if T denotes a transversal for the conjugacy classes of subgroups of G, the (H, 0)_{G} (H ∈ T) are pairwise distinct minimal prime ideals in B(G), by corollary (3.7). Hence the ℚ(H, 0)_{G} (H ∈ T) are pairwise distinct maximal ideals in ℚB(G), by proposition (3.6) (cf. also the proof of corollary (3.7)). However, ℚB(G) is Artinian, and thus

dim_{ℚ} ℚB(G)/J(ℚB(G)) ≥ |{maximal ideals of ℚB(G)}| ≥ |T |.

On the other hand, ℚB(G) has dimension T. Thus ℚB(G) is semisimple and split, and |T | is precisely the number of maximal ideals of ℚB(G). This implies that the induced homomorphisms

| |^{H} : ℚB(G) → ℚ

(H ∈ T) are precisely the species, i.e. the irreducible representations of ℚB(G). Moreover, the (H, 0)_{G} (H ∈ T) are precisely the minimal prime ideals of B(G).

(A.4) Next we determine the prime ideals of B(G) which lie over some non-zero prime ideal of ℤ. Let p ∈ ℤ be a prime number, and let H be subgroup of G such that the order of N_{G}(H)/H is not divisible by p. As above

(H, p)_{G} := (Res_{H}^{G})^{−1}((H, p))

is a prime ideal in B(G) such that G_{(H,p)} := G_{(H,p)_{G}} has defect H_{(H,p)} and

ℤ ∩ (H, p)_{G} = pℤ,

using lemmas (5.6) and (4.2). Moreover, (H, p)_{G} is the full preimage of pℤ under the homomorphism | ^{H}| : B(G) → ℤ.

Conversely, let G_{𝔭} a pointed group on B such that 𝔭 lies over p. Let H_{𝔮} be a defect of G_{𝔭}. Then also 𝔮 lies over p by lemma (4.2). Thus H_{𝔮} = H_{(H,p)} as H_{𝔮} is primordial. Moreover, 𝔭 = (Res_{H}^{G})^{−1}((H, p)), by corollary J, as B(G) is commutative. Finally, as N_{G}(H)/H acts trivially on \overline{B}(H), N_{G}(H)/H is a p′- group, by lemma (5.7).

Thus if T_{p} denotes the set of elements H of T such that the order of N_{G}(H)/H is not divisible by p then the (H, p)_{G} (H ∈ T_{p}) are precisely the (pairwise distinct) prime ideals of B(G) which lie over p.

(A.5) As we have determined Spec(B(G)) as set, we investigate its structure as poset. Let G_{(K,q)} < G_{(H,p)} (with the notation above) be pointed groups on B. Thus K_{(K,q)} and H_{(H,p)} are defects of G_{(K,q)} and G_{(H,p)}, respectively. By theorem G (and corollary L), there exists g ∈ G such that ^{g}K_{(^{g}K,q)} = ^{g}(K_{(K,q)}) < H_{(H,p)}.

We must have p ≠ 0 since otherwise ^{g}K = H and q = 0, by corollary R, and so G_{(K,q)} = G_{(H,p)}.

Suppose that q ≠ 0. Then we have q = p, by corollary R. Thus ^{g}K ⊂ H since otherwise G_{(K,q)} = G_{(H,p)}. Hence there is a subgroup R of G such that ^{g}K ⊂ R ⊆ N_{G}(^{g}K) and R/^{g}K is a p-group, again by corollary R. (With notation of corollary R take R to be the minimal R_{i} such that^{g}K ⊂ R_{i}. Then R_{i−1} = ^{g}K.) This contradicts the fact that q = p does not divide the order of N_{G}(K)/K ≅ N_{G}(^{g}K)/^{g}K.

Thus q = 0. Moreover, ^{g}K is subnormal in H and |H : ^{g}K| is a power of p, by corollary R (i.e. O^{p}(H) = O^{p}(^{g}K), cf. the proof of corollary S).

Conversely, let p be a prime number, let H be subgroup of G such that N_{G}(H)/H is p′-group, and let K be a subnormal subgroup of H the index of which in H is a power of p. Thus there is a chain of subgroups of H

K = N_{0} ⊴ N_{1} ⊴ … ⊴ N_{s} = H

(s ∈ ℕ) such that N_{i}/N_{i−1} is a p-group for i = 1, …, s. Let S be a finite G-set. Then S^{N_{0}} is a N_{1}/N_{0}-set, and S^{N_{1}} = (S^{N_{0}})^{N_{1}/N_{0}}. As N_{1}/N_{0} is a p-group, this yields |S^{N_{1}}| ≡ |S^{N_{0}}| (mod p). Arguing inductively on s, we obtain |S^{K}| ≡ |S^{H}| (mod p). This implies that, for any x ∈ B(G), we have |x^{K}| ≡ |x^{H}| (mod p), and so

G_{(K,0)} < G_{(H,p)}.

(A.6) Using the results of (A.5), we can now easily complete the proof of Dress’s theorem asserting that the connected components of Spec(B(G)), and hence the primitive idempotents of B(G) (cf. [15] Theorem 7.3), are parametrized by the conjugacy classes of perfect subgroups of G. Recall that a finite group H is called perfect provided that H has no non-trivial solvable quotient groups. So that, in particular, B(G) has no non-trivial idempotents if and only if G is solvable.

For a finite group H denote by O^{s}(H) the unique minimal normal subgroup of H such that the corresponding quotient group H/ O^{s}(H) is solvable; thus H is perfect if and only if H = O^{s}(H). By the observations preceding theorem I, it suffices to consider path components of Spec(B(G)) as poset instead of connected components of Spec(B(G)) as topological space.

Let G_{(K,q)} and G_{(H,p)} be pointed groups on B. We claim that the corresponding prime ideals 𝔮 and 𝔭 of B(G) belong to the same path component of Spec(B(G)) as poset if and only if O^{s}(K) and O^{s}(H) are conjugate in G.

Assume first that 𝔮 and 𝔭 belong to the same path component. Clearly, it suffices to assume that 𝔮 ⊆ 𝔭, i.e. G_{(K,q)} < G_{(H,p)}. Thus q = 0, p > 0 and there is g ∈ G such that ^{g}K is a subnormal subgroup of H and |H : ^{g}K| is a power of p, by (A.5). As in the proof of corollary S (with O^{p} replaced by O^{s}), one shows that this implies

g(O^{s}(K)) = O^{s}(^{g}K) = O^{s}(H).

Conversely, assume that O^{s}(H) and O^{s}(K) are conjugate in G. Thus

G_{(O^{s}(H),0)} = G_{(O^{s}(K),0)}.

By definition of O^{s}(H), there is a chain of subgroups

O^{s}(H) = N_{0} ⊴ N_{1} ⊴ … ⊴ N_{r} = H

(r ∈ ℕ) such that, for any i = 1, …, r, N_{i}/N_{i−1} is a p_{i}-group for some prime number p_{i}. Hence, by (A.5), we have

G_{(O^{s}(H),0)} < G_{(N_{1},p_{1})} > G_{(N_{1},0)} < G_{(N_{2},p_{2})} > … < G_{(N_{s},p_{s})} > G_{(H,0)} ≤ G_{(H,p)}.

By symmetry, this proves our claim that 𝔮 and 𝔭 belong to the same path component of Spec(B(G)) as poset. As a subgroup H of G is perfect if and only if H = O^{s}(L) for some subgroup L of G, this completes the proof of Dress’s theorem.

(A.7) Let p be a prime number, and let H be a subgroup of G. Then the full preimage 𝔭 of pℤ under the epimorphism

| ^{H}| : B(G) → ℤ

is a prime ideal in B(G) (independent of whether or not N_{G}(H)/H is a p′-group). One might ask for a defect of G_{𝔭}. First of all, it is clear that G_{(H,0)} < G_{𝔭}, and so H_{(H,0)} < G_{𝔭}. Therefore, in view of theorem H and corollary L, some defect K_{𝔮} of G_{𝔭} satisfies H_{(H,0)} < K_{𝔮}. Moreover, K_{𝔮} = K_{(K,p)} by (A.1) and lemma (4.2). Thus H is a subnormal subgroup of K such that |K : H| is a power of p. As in the proof of corollary S, this implies that O^{p}(H) = O^{p}(K). Moreover, N_{G}(K) ⊆ N_{G}(O^{p}(H)), and K/ O^{p}(H) is a p-group. Therefore, K/ O^{p}(H) is a Sylow p-subgroup of N_{G}(O^{p}(H))/ O^{p}(H) as p does not divide the order of N_{G}(K)/K.

Conversely, let K be a subgroup of G such that N_{G}(K)/K is a p′-group, and denote by 𝔭 the full preimage of pℤ under | ^{O^{p}(K)}| : B(G) → ℤ. Then, by the above observations, some defect of G_{𝔭} is conjugate to K_{(K,p)}. Hence G_{(K,p)} = G_{𝔭}. Thus the prime ideals of B(G) which lie over some prime p are parametrized by the conjugacy classes of p-perfect subgroups of G. (Recall that a group H is called p-perfect if H = O^{p}(H). Recall also that a subgroup H of G is p-perfect if and only if H equals the p-residue of some subgroup of G.)

Note that most of the above arguments carry over to the Burnside algebra kB(G) over an arbitrary commutative ring k.

APPENDIX B. THE SPECTRUM OF THE CHARACTER RING

Next we determine the prime spectrum of the character ring of a finite group (in a split situation), or more precisely the prime spectrum of the Grothendieck group of a (split) group algebra over a field of characteristic 0. In particular, we give a proof of a theorem of Berman [3] which asserts that the character ring of a finite group does not contain non-trivial idempotents (i.e. the spectrum of the character ring is connected).

The author learnt about the prime spectra of various “representation rings” (e.g. the character ring, the Brauer character ring, the trivial source ring etc.) from the Ph. D. thesis [9] of his former Jena colleague Markus Deiml, which was accepted by the Friedrich-Schiller-Universität in Jena, and supervised by Burkhard Külshammer. The intention of this thesis, which uses different methods, is to understand “modular” versions of these rings. The author of this paper certainly benefited from numerous discussions with Markus Deiml on the subject. We refer to [21] for a description of the prime spectrum of the character ring.

As in appendix A, we will start from scratch, and determine also the species.

(B.1) Throughout, we fix a finite group G, and let R be an integral domain of characteristic 0 with quotient field 𝕂. Moreover, we assume R to contain a primitive |G|th root of unity. For a 𝕂G-module M – we will implicitly assume all such to be finite dimensional over 𝕂 – let [M] denote its isomorphism class. The isomorphism classes of 𝕂G-modules form an abelian monoid where, for 𝕂G-modules M and N, addition is defined by [M]+[N] := [M ⊕N]. Denote by G_{0}(𝕂G) the corresponding Grothendieck group. This group is even a ring with identity where, for 𝕂G-modules M and N, multiplication is given by [M] · [N] := [M ⊗_{𝕂} N]. It is plain (and, of course, very well-known) that the collection of all the rings G_{0}(𝕂H) for all the subgroups H of G form a Green functor for G (over ℤ) where the induction, restriction and conjugation maps are induced from induction, restriction and conjugation of 𝕂H-modules. We denote this Green functor by G_{0}(𝕂, G). Moreover, R G_{0}(𝕂, G) denotes the scalar extension of G_{0}(𝕂, G) to R, i.e. R G_{0}(𝕂, g) := R ⊗_{ℤ} G_{0}(𝕂, G). By Maschke’s Theorem, 𝕂G is semisimple (Artinian); so G_{0}(𝕂G) is spanned by the isomorphism classes of simple 𝕂G-modules. Moreover, by the Jordan-Hölder Theorem, the isomorphism classes of simple 𝕂G-modules even form a ℤ-basis of G_{0}(𝕂G). In particular, G_{0}(𝕂G) is a finitely generated free (hence torsion-free) ℤ-module. Besides, as 𝕂G is semisimple Artinian, there is a canonical bijection between the set of isomorphism classes of simple 𝕂G-modules and the set of primitive idempotents of Z(𝕂G). Since these idempotents are pairwise orthogonal, and hence linearly independent, we obtain

dim_{𝕂} 𝕂 G_{0}(𝕂G) ≤ dim_{𝕂} Z(𝕂G).

The latter dimension in turn equals the number of conjugacy classes of G.

(B.2) We will use the following elementary properties of characters: For a 𝕂Gmodule, denote by χ_{M} the character afforded by M. Then χ_{M} takes values in R since, for g ∈ G, χ_{M}(g) coincides with χ_{Res_{C}^{G} M}(g), where C denotes the cyclic subgroup of G generated by g and Res_{C}^{G} M stands for the restricted 𝕂C-module; however, any simple 𝕂C-module is one-dimensional as C is cyclic and R (and hence 𝕂) contains a primitive |C|th root of unity; hence Res_{C}^{G} M has a decomposition series consisting of one-dimensional modules, and so χ_{M}(g) is a (finite) sum of certain |C|th roots of unity, which are contained in R, by assumption. Moreover, χ_{M} depends only on the isomorphism class of M. It is easily checked that, for any g ∈ G, the assignment [M] ↦ χ_{M}(g) induces a homomorphism of R-algebras φ_{g}^{G} : R G_{0}(𝕂G) → R. We will omit the superscript when it is clear from context the Grothendieck group of which group algebra we refer to.

Let H be a subgroup of G, and let M be a 𝕂H-module. Then the character afforded by the induced 𝕂G-module Ind_{H}^{G} M maps an element g of G to

\frac{1}{|H|} ∑_{{x∈G \atop xgx^{−1}∈H}} χ_{M}(xgx^{−1}).

(B.3) At first, we determine the prime ideals of R G_{0}(𝕂G) which lie over 0. Let g ∈ G, and denote by C the cyclic subgroup of G generated by g. Then the kernel of the epimorphism

φ_{g} : R G_{0}(𝕂C) → R

is a prime ideal in R G_{0}(𝕂C) which we denote by (g, 0). By (B.2), we have ker Br_{C} ⊆ (g, 0) as C is abelian and no proper subgroup of C can contain g. Thus C_{(g,0)} is a primordial pointed group on G_{0}(𝕂, G). Therefore,

(Res_{C}^{G})^{−1}((g, 0))

is a prime ideal (g, 0)_{G} in R G_{0}(𝕂G) such that G_{(g,0)} := G_{(g,0)_{G}} has defect C_{(g,0)}, by lemma (5.7). Moreover, R ∩ (g, 0)_{G} = R ∩ (g, 0) = 0, by lemma (4.2). Clearly, (g, 0) is the kernel of the epimorphism

φ_{g} : R G_{0}(𝕂G) → R.

Let x ∈ C such that (x, 0) = (g, 0). Denote by n the order of C, and let ζ ∈ R be a primitive nth root of unity. Write x = g^{t} with t ∈ {1, …, n − 1}. Let M be the one-dimensional 𝕂G-module on which g acts by multiplication with ζ. Thus χ_{M}(g) = ζ, and so 1 − ζ^{−1}[M] ∈ (g, 0) = (g^{t}, 0). Hence

0 = φ_{g^{t}}(1 − ζ^{−1}[M]) = 1 − ζ^{−1}ζ^{t},

so ζ = ζ^{t}, and thus t = 1. Therefore, x = g.

Moreover, for x ∈ G, we clearly have^{x}(C_{(g,0)}) = (^{x}C)_{(xgx^{−1},0)}. This implies that, for x, y ∈ G, G_{(x,0)} = G_{(y,0)} if and only if x and y are conjugate in G. Denote by C a transversal for the conjugacy classes of G. Then the (g, 0)_{G} (g ∈ C) are pairwise distinct minimal prime ideals in R G_{0}(𝕂G), by corollary (3.7). Therefore, the 𝕂(g, 0)_{G} (g ∈ C) are pairwise distinct maximal ideals in 𝕂 G_{0}(𝕂G), by proposition (3.6) (cf. also the proof of corollary (3.7)). As 𝕂G_{0}(𝕂G) is artinian, we obtain

dim_{𝕂} 𝕂 G_{0}(𝕂G) ≥ dim_{𝕂} 𝕂 G_{0}(𝕂G)/J(𝕂 G_{0}(𝕂G))
≥ |{maximal ideals of 𝕂 G_{0}(𝕂G)}|
≥ |C|.

On the other hand, we already observed that dim_{𝕂} 𝕂 G_{0}(𝕂G) ≤ |C|, so that equality holds everywhere. Hence C is the number of isomorphism classes of simple 𝕂Gmodules. Moreover, 𝕂 G_{0}(𝕂G) is split semisimple, and the 𝕂(g, 0)_{G} (g ∈ C) are precisely the pairwise distinct maximal ideals of 𝕂 G_{0}(𝕂G). Thus the (g, 0)_{G} (g ∈ C) are precisely the pairwise distinct minimal prime ideals of R G_{0}(𝕂G), by corollary (3.7). Moreover, the induced homomorphisms

φ_{g} : 𝕂 G_{0}(𝕂G) → 𝕂, [M] ↦ χ_{M}(g)

(g ∈ C) are precisely (the pairwise distinct) species (i.e. the irreducible representations) of 𝕂 G_{0}(𝕂G). As 𝕂 G_{0}(𝕂G) is semisimple, this implies that the isomorphism class of a 𝕂G-module M is uniquely determined by the character afforded by M.

(B.4) Next we deduce Artin’s induction theorem [7] (15.4) (with coefficients in 𝕂). Let P be a primordial subgroup of G for K G_{0}(𝕂, G). In view of [25] Proposition (54.3), it suffices to show that P is cyclic. Let 𝔭 be a maximal ideal of 𝕂 G_{0}(𝕂P) containing ker Br_{P}^{𝕂 G_{0}(𝕂,G)} (which exists as P is primordial and 𝕂 G_{0}(𝕂P) has an identity). Thus P_{𝔭} is a primordial pointed group on 𝕂 G_{0}(𝕂, G). By proposition (3.6), corollary (3.7) and its proof, there is a minimal prime ideal 𝔮 in R G_{0}(𝕂P) such that 𝔭 = 𝕂𝔮 and 𝔮 = 𝔭 ∩ R G_{0}(𝕂P). Let Q be a proper subgroup of P. Then

(R G_{0}(𝕂, G))_{Q}^{P} ⊆ (𝕂 G_{0}(𝕂, G))_{Q}^{P} ∩ R G_{0}(𝕂P) ⊆ 𝔭 ∩ R G_{0}(𝕂P) = 𝔮

as P_{𝔭} is primordial; and so P_{𝔮} is primordial, as well. Thus P_{𝔮} is its own defect. On the other hand, a defect of P_{𝔮} is of the form C_{(g,0)}, by (B.3) applied to P instead of G, where g ∈ P and C is the cyclic subgroup of P generated by g. Thus P = C is cyclic, and we are done.

(B.5) Now we want to determine the prime ideals of R G_{0}(𝕂G) which lie over some non-zero prime ideal of R. For simplicity, we assume from now on that R = ℤ[ζ] where ζ denotes a primitive |G|th root of unity. (Although this does not really make a difference; we leave it to the reader to consider the general case.) Thus 𝕂 = ℚ(ζ), and, as R is integral over ℤ, R has Krull dimension 1; moreover, for any prime ideal pℤ of ℤ, some prime ideal of R lies over pℤ. Let p(≠ 0) be a prime number, and let 𝔭 be a prime ideal of R which lies over p. Let g be a p-regular element of G (i.e. the order of g is not divisible by p), and denote by C the cyclic subgroup of G generated by g. Moreover, let P be a Sylow p-subgroup of C_{G}(g), the centralizer of g in G. Then E := P C ⊆ C_{G}(g) is a p-elementary subgroup of G.

The full preimage of 𝔭 under the epimorphism

φ_{g} : R G_{0}(𝕂E) → R

is a prime ideal in R G_{0}(𝕂E) which we denote by (g, 𝔭, P). The corresponding pointed group E_{(g,𝔭,P)} is just denoted by E_{(g,𝔭)} (as E determines P uniquely). Clearly, we have

R ∩ (g, 𝔭, P) = 𝔭. (1)

We claim that E_{(g,𝔭)} is primordial. Let H be a proper subgroup of E. Then H is p-elementary, as well, and H = (H ∩ P)(H ∩ C). Assume first that H ∩ C ⊂ C. Thus g ∉ H, and, as g is central in E, we obtain (R G_{0}(𝕂, G))_{H}^{E} ⊆ (g, 𝔭, P), by (B.2).

Secondly, assume that E ∩ C = C. Hence g ∈ H and E ∩ P ⊂ P. Let M be a 𝕂H-module. Then (B.2) yields that

φ_{g}([Ind_{H}^{E} M]) = |E : H| χ_{M}(g) = |P : E ∩ P | χ_{M}(g),

as g is central in E. This shows that (R G_{0}(𝕂, G))_{H}^{E} ⊆ (g, 𝔭, P) as (g, 𝔭, P) lies over p, proving our claim.

It is plain that ^{x}(E_{(g,𝔭)}) = (^{x}E)_{(xgx^{−1},𝔭)} for all x ∈ G. Conversely, let x be a p-singular element of G, let Q be a Sylow p-subgroup of C_{G}(x), denote by D the cyclic subgroup of G generated by x, and let 𝔮 be a prime ideal of R which lies over 𝔭. Set F = QD, and assume that E_{(g,𝔭)} = F_{(x,𝔮)}. Then, first of all, 𝔭 = 𝔮, by (1). Moreover, P is a normal Sylow p-subgroup of E = F, and so is Q. Hence P = Q. Similarly, C is a normal Hall p′-subgroup of E = F, and so is D. Thus C = D. Let n be the order of g. Thus n is not divisible by p, and x = g^{t} for some t ∈ {1, …, n − 1}. As in (B.3) one shows that, for a primitive nth root of unity ξ ∈ R, this implies that

ξ ≡ ξ^{t} mod 𝔭.

However, as n is not divisible by p, the image of ξ under the canonical epimorphism R → R/𝔭 is a primitive nth root of unity in R/𝔭. Thus t = 1, and so x = g. Therefore, N_{G}(E_{(g,𝔭)}) = N_{G}(E) ∩ C_{G}(g). As P is characteristic in E, we have N_{G}(E) ⊆ N_{G}(P). On the other hand, N_{G}(P) ∩ C_{G}(g) ⊆ N_{G}(E), and so

N_{G}(E_{(g,𝔭)}) = N_{G}(E) ∩ C_{G}(g) = N_{G}(P) ∩ C_{G}(g) = C_{N_{G}(P)}(g).

As P ⊆ C_{N_{G}(P)}(g) is a Sylow p-subgroup of C_{G}(g), \overline{N}_{G}(E_{(g,𝔭)}) = C_{N_{G}(P)}(g)/P is a p′-group. Therefore, by lemma (5.6),

(Res_{E}^{G})^{−1}((g, 𝔭, P))

is a prime ideal (g, 𝔭)_{G} in R G_{0}(𝕂G) such that G_{(g,𝔭)_{G}} := G_{(g,𝔭)} has defect E_{(g,𝔭)}. Moreover, (g, 𝔭)_{G} lies over 𝔭, by lemma (4.2). It is plain that one can describe the prime ideal (g, 𝔭)_{G} also as the full preimage of 𝔭 under the epimorphism

φ_{g} : R G_{0}(𝕂G) → R,

as the homomorphisms φ_{g}^{H} (H a subgroup of G) commute with restriction. Denote by C_{p} the set of p-regular elements of C. Then the (g, 𝔭)_{G} (g ∈ C_{p}) are pairwise distinct prime ideals of R G_{0}(𝕂G) which lie over 𝔭. We will show that, conversely, any prime ideal of R G_{0}(𝕂G) which lies over 𝔭 is of the form (g, 𝔭)_{G} for some g ∈ C_{p}.

Note that for g ∈ C_{p} the defect groups of G_{(g,𝔭)} are not precisely the p-defects of the conjugacy class of g, but they are closely related: Namely the defect groups of G_{(g,𝔭)} are precisely the subgroups P D of G, where P is any p-defect of the conjugacy class of g and D is any subgroup of G which is conjugate in G to the cyclic subgroup of G generated by g.

(B.6) Let 𝔪 be a prime ideal of R G_{0}(𝕂G) which lies over 𝔭. Denote by P_{𝔮} a defect of G_{𝔪}. By corollary (3.7), there is a prime ideal 𝔯 in R G_{0}(𝕂P) which is contained in 𝔮 such that R ∩ 𝔯 = 0. Let Q_{𝔰} be a defect of P_{𝔯}. Then also R ∩ 𝔰 = 0, by lemma (4.2), i.e. 𝔰 is a minimal prime ideal of R G_{0}(𝕂Q), by corollary (3.7). By (B.3) applied to Q instead of G, there is g ∈ Q such that C_{(g,0)} is a defect of Q_{𝔰}, where C is the cyclic subgroup of Q generated by g. Then

C_{(g,0)} ≤ Q_{𝔰} ≤ P_{𝔮},

and all these pointed groups are primordial. (As Q_{𝔰} is primordial itself, theorem H yields that in fact C_{(g,0)} = Q_{𝔰}, but we do not need this.)

By corollary R, there is a chain

C_{(g,0)} = (T_{0})_{𝔱_{0}} ≤ (T_{1})_{𝔱_{1}} ≤ … ≤ (T_{m})_{𝔱_{m}} = P_{𝔮}

(m ∈ ℕ) of primordial pointed groups on R G_{0}(𝕂, G) such that T_{i+1} ⊆ N_{P} ((T_{i})_{𝔱_{i}}) and T_{i+1}/T_{i} is a p-group for all i = 0, …, m − 1. Moreover, the proof of corollary R demonstrates that T_{1} = T_{0} = C and R ∩ 𝔱_{1} = R ∩ q = 𝔭. As 𝔱_{0} = (g, 0) is the kernel of the epimorphism

φ_{g} : R G_{0}(𝕂C) → R,

this implies that 𝔱_{1} is the full preimage of 𝔭 under φ_{g}^{C}.

Recall that g can be written uniquely in the form g_{p′}g_{p} where g_{p′} is a p-regular element of G, g_{p} is an element of G the order of which is a power of p, and g_{p′} and g_{p} commute. The elements g_{p′} and g_{p} are called the p′-part respectively p-part of g. Of course, g_{p′} and g_{p} are certain powers of g, i.e. both elements are contained in C. Let M be a simple 𝕂C-module. As C is cyclic (and 𝕂C is split), M is one-dimensional. Thus χ_{M} is the linear representation of M. Hence χ_{M}(g) = χ_{M}(g_{p′})χ_{M}(g_{p}), and χ_{M}(g_{p}) is root of unity in R the order of which is a power of p. As R/𝔭 is a field of characteristic p, this shows that χ_{M}(g_{p}) ≡ 1 (mod 𝔭), and so

φ_{g}([M]) = χ_{M}(g) ≡ χ_{M}(g_{p′}) = φ_{g_{p′}} ([M]) (mod 𝔭)

This yields that

𝔱_{1} = (φ_{g})^{−1}(𝔭) = (φ_{g_{p′}})^{−1}(𝔭),

as R G_{0}(𝕂C) is generated by the isomorphism classes of simple 𝕂C-modules. As g′_{p} is p-regular, one shows as in (B.5) that

C_{N_{G}(T_{1})}(g_{p′}) = N_{G}((T_{1})_{𝔱_{1}}) ⊃ T_{2}.

Let S_{2} be a Sylow p-subgroup of G, and denote by C_{p′} the cyclic subgroup of G generated by g_{p′}. Then T_{2} = S_{2}C_{p′} ≅ S_{2} × C_{p′} as T_{2}/T_{1} is p-group and C_{p′} is a central Hall p′-subgroup of T_{1} = C. In particular, T_{2} is p-elementary. Moreover,

(φ_{g_{p′}}^{T_{2}} )^{−1}(𝔭) = (Res_{T_{1}}^{T_{2}})^{−1}((φ_{g_{p′}}^{T_{1}} )^{−1}(𝔭)) = (Res_{T_{1}}^{T_{2}})^{−1}(𝔱_{1}) ⊆ 𝔱_{2}

as (T_{1})_{𝔱_{1}} ≤ (T_{2})_{𝔱_{2}}. However, as (φ_{g_{p′}}^{T_{2}} )^{−1}(𝔭) plainly is a maximal ideal in R G_{0}(𝕂T_{2}), this yields that

𝔱_{2} = (φ_{g_{p′}}^{T_{2}} )^{−1}(𝔭).

Arguing inductively on m, this implies that P = SC_{p′} ≅ S × C_{p′}, where S is the unique Sylow p-subgroup of P = T_{m}, hence P is p-elementary, and 𝔮 = 𝔱_{m} is the full preimage of 𝔭 under

φ_{g_{p′}} : R G_{0}(𝕂P) → R.

Moreover, one shows as in (B.5) that \overline{N}_{G}(P_{𝔮}) = C_{N_{G}(S)}(g_{p′})/S using the fact that g_{p′} is p-regular. Thus S is a Sylow p-subgroup of C_{N_{G}(P)}(g_{p′}), by lemma (5.7), as \overline{N}_{G}(P_{𝔮}) acts trivially on R G_{0}(𝕂P)/𝔮 ≅ R/𝔭. This implies that S is even a Sylow p-subgroup of C_{G}(G_{p′}), by Sylow’s theorems. Thus P_{𝔮} = (SC_{p′})_{(g_{p′} ,𝔭)} is in fact one of the primordial pointed groups constructed in (B.5), and 𝔪 = (g_{p′}, 𝔭)_{G}.

Therefore, the (x, 𝔭)_{G} (x ∈ C_{p}) are precisely the pairwise distinct prime ideals of R G_{0}(𝕂G) which lie over 𝔭. Hence any primordial pointed group G_{𝔪} on R G_{0}(𝕂, G) has an elementary defect group. As in (B.4) one shows that this implies Brauer’s induction theorem [7] (15.8) (with coefficients in R), using [25] Proposition (54.3).

(B.7) Finally we determine the structure of Spec(R G_{0}(𝕂G)) as poset. With the notation of (B.3) respectively (B.5), let G_{(x,𝔮)} < G_{(y,𝔭)} be pointed groups on R G_{0}(𝕂, G). Thus 𝔮 and 𝔭 are prime ideals in R, and (x, 𝔮)_{G} and (y, 𝔭)_{G} lie over 𝔮 and 𝔭, respectively. Thus 𝔮 ⊂ 𝔭, and so 𝔮 = 0 and 𝔭 lies over some (non-zero) prime number p, as R has Krull dimension 1. Hence y is a p-regular element of G, by (B.5) and (B.6). Denote by C and D the cyclic subgroups of G generated by x and y, respectively. Moreover, let P be a Sylow p-subgroup of C_{G}(y), and set E := P D. Then C_{(x,0)} is a defect of G_{(x,0)}, and E_{(y,𝔭)} is a defect of G_{(y,𝔭)}, by (B.3), (B.5) and (B.6). By theorem G, there is g ∈ G such that C_{(x,0)} ≤ ^{g}(E_{(y,𝔭)}) = (^{g}E)_{(gyg^{−1},𝔭)}. Denote by x_{p′} the p′-part of x. As in (B.6) one shows that x_{p′} generates ^{g}D, ^{g}P is a Sylow p-subgroup of C_{G}(x_{p′}), and (gyg^{−1}, 𝔭, P) is the full preimage of 𝔭 under the epimorphism

φ_{x_{p′}} : R G_{0}(𝕂[^{g}E]) → R,

using corollary R. Hence x_{p′} = gyg^{−1}, by (B.5).

Conversely, let x be an element of G, let 𝔭 be a prime ideal of R which lies over some (non-zero) prime p, and let y be a p-regular element of G such that the p′-part x_{p′} of x is conjugate to y in G. We claim that G_{(x,0)} < G_{(y,𝔭)}. Replacing y be a suitable conjugate in G, we may assume that y = x_{p′}. Denote by C the cyclic subgroup of G generated by x. Then C_{(x,0)} is a defect of G_{(x,0)}. Clearly, (x, 0) is contained in the full preimage of 𝔭 under

φ_{x} : R G_{0}(𝕂C) → R.

Moreover,

(φ_{x}^{C})^{−1}(𝔭) = (φ_{x_{p′}}^{C})^{−1}(𝔭) = (φ_{y}^{C})^{−1}(𝔭)

by (B.6). Therefore,

(x, 0)_{G} = (Res_{C}^{G})^{−1}((x, 0)) ⊆ (Res_{C}^{G})^{−1}((φ_{y}^{C})^{−1}(𝔭))
= (φ_{y}^{G})^{−1}(𝔭) = (y, 𝔭)_{G},

by (B.5), proving our claim.

Using these facts, we can now easily prove Berman’s theorem [3]. It suffices to show that the spectrum of R G_{0}(𝕂G) is connected; and, in order to prove this, it suffices to show that any minimal prime ideal (x, 0)_{G} lies in the same connected component of the poset Spec(R G_{0}(𝕂G)) as (1, 0)_{G}. Let p be a prime divisor of the order of x, and let 𝔭 be a prime ideal of R which lies over p. Then, by the observations above, we have

(x, 0)_{G} ⊂ (x_{p′}, 𝔭)_{G} ⊃ (x_{p′}, 0)_{G}.

Arguing inductively on the order of x, the result follows.

(B.8) Let 𝔭 be a prime ideal of R which lies over some (non-zero) prime number p. Then the full preimage of 𝔭 under the epimorphism

φ_{g} : R G_{0}(𝕂G) → R

is a prime (maximal) ideal 𝔪 in R G_{0}(𝕂G) (independent of whether or not g is pregular). By (B.6), 𝔪 is of the form (y, 𝔭)_{G} for some p-regular element y. Besides, (g, 0)_{G} = ker φ_{g} ⊂ 𝔪 = (y, 𝔭)_{G}. Hence y is conjugate in G to the p′-part of g, by (B.7).

ACKNOWLEDGEMENTS

I wish to thank Robert Boltje for always providing patient answers to questions of mine, and Jacques Thévenaz for bringing some facts on the history of the classification of simple Green functors and related results to my attention. Moreover, I am grateful to Burkhard Külshammer and the referee for spoting some mistakes in an earlier version of this paper and for other helpful remarks and suggestions.

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