On the radical idealizer chain of symmetric orders.

Gabriele Nebe ^{∗}

Abstract: If Λ is an indecomposable, non maximal, symmetric order, then the idealizer of the radical Γ := Id(J(Λ)) = J(Λ)^{#} is the dual of the radical. If Γ is hereditary then Λ has a Brauer tree (under modest additional assumptions). Otherwise Δ := Id(J(Γ)) = (J(Γ)^{2})^{#}. If Λ = ℤ_{p}G for a p-group G ≠ 1, then Γ is hereditary iff G ≅ C_{p} and otherwise [Δ : Λ] = p^{2}|G/(G′G^{p})|. For Abelian groups G, the length of the radical idealizer chain of ℤ_{p}G is (n−a)(p^{a} −p^{a−1})+p^{a−1}, where p^{n} is the order and p^{a} the exponent of the Sylow p-subgroup of G.

1 Introduction

Throughout this paper let R be a discrete valuation ring with maximal ideal πR and residue class field R/πR =: k. Let K be the field of fractions of R and A a separable finite dimensional K-algebra. If Λ is an R-order in A, then there is a canonical process, the so called radical idealizer process, that constructs an ascending chain of over-orders of Λ ending in a hereditary order Λ_{N}, called the head order of Λ (see Remark 2.7). We call the length of this chain the radical idealizer length l_{rad}(Λ) of Λ. Hereditary orders are well understood, see [Jac], [Rei, Chapter 9]. They are direct sums of hereditary orders in the simple components of A. So one might hope to classify R-orders according to the length of the radical idealizer chain and the head order.

An important class of R-orders are the symmetric orders, that are self dual with respect to some trace bilinear form on A. Examples of symmetric orders are provided by blocks of group rings RG for finite groups G. The main tool to deal with symmetric orders is Jacobinski’s conductor formula (see Theorem 4.3) stating that for any overorder Γ of a symmetric order Λ the conductor F_{Γ}(Λ) (which is the largest Γ ideal in Λ) is the dual of Γ. If Γ = Id(J(Λ)) is the idealizer of J(Λ), then a converse of this formula holds: Theorem 5.1 shows that for indecomposable, non hereditary, symmetric orders Λ the dual of J(Λ) is the idealizer of J(Λ). Using his conductor formula, Jacobinski shows that the indecomposable symmetric orders Λ with l_{rad}(Λ) = 0 are maximal orders (Theorem 4.6). If l_{rad}(Λ) = 1, then one may derive the Brauer tree of Λ using the idealizer of J(Λ) (see Proposition 7.2 and [Jac, Section 11]). In the present paper the first two steps of the radical idealizer chain for symmetric orders are investigated and properties of symmetric orders Λ with l_{rad}(Λ) = 2 are determined. We apply the theorems to p-groups G showing that l_{rad}(ℤ_{p}G) = 1 if and only if G ≅ C_{p}, l_{rad}(ℤ_{p}G) = 2 if and only if G ≅ C_{4} or G ≅ C_{2} × C_{2}. Moreover we calculate l_{rad}(ℤ_{p}G) for Abelian groups G.

∗Abteilung Reine Mathematik, Universität Ulm, 89069 Ulm, Germany, nebe@mathematik.uniulm.de

2 The radical idealizer chain

Let Λ be an R-order in A. Then the Jacobson radical J(Λ) is the intersection of all maximal right ideals of Λ. It is a 2-sided ideal of Λ, in fact the smallest ideal I of Λ, such that Λ/I is a semi-simple k-algebra. One other important characterization of J(Λ) is that J(Λ) is the biggest Λ-ideal I in Λ that is pro-nilpotent, i.e. for which there is m ∈ ℕ such that I^{m} ⊂ πΛ (cf. [Jac, Lemma 8.5]).

Definition 2.1. (see [Rei, Section 39]) Let Λ, Λ′ be R-orders in A. Then Λ radically covers Λ′, Λ ≻ Λ′, if Λ ⊇ Λ′ and J(Λ) ⊇ J(Λ′). If Λ is maximal with respect to ≻, then Λ is called extremal.

Lemma 2.2. Let Γ ≻ Λ be two R-orders in A. Then J(Λ) = J(Γ) ∩ Λ and Λ/J(Λ) is isomorphic to a sub-algebra of Γ/J(Γ). Moreover every simple Γ-module is semi-simple as a Λ-module.

\underline{Proof.} Since J(Γ) ∩ Λ is a pro nilpotent ideal of Λ, it is contained in J(Λ). On the other hand J(Λ) ⊆ J(Γ), because Γ ≻ Λ. Therefore J(Γ) ∩ Λ = J(Λ) and Λ/J(Λ) ≅ (Λ + J(Γ))/J(Γ) is naturally embedded in Γ/J(Γ). The second assertion follows from the fact that ΓJ(Λ) = J(Λ) ⊆ J(Γ) which implies that Γ/J(Γ) is a semi-simple Λ-module. □

Recall that an order Γ is called hereditary, if every left ideal of Γ is projective (see [Rei, Section 10]).

Theorem 2.3. ([Jac, Satz 8.12], [Rei, Theorem 39.14]) An R-order Λ in A is extremal, if and only if Λ is hereditary.

Definition 2.4. Let L be a full R-lattice in A. The left order of L is O_{l}(L) := {a ∈ A | aL ⊆ L}. Analogously one defines the right order O_{r}(L) of L. Id(L) := O_{l}(L) ∩ O_{r}(L) is called the idealizer of L.

Remark 2.5. Let Λ be an order, Γ := Id(J(Λ)). Then Γ ≻ Λ.

The following characterization of hereditary orders is shown in [Rei, Theorem 39.11] for O_{l}(J(Λ)) instead of Id(J(Λ)). With a completely analogous proof (see [Neb]) one shows

Theorem 2.6. Let Λ be an R-order in A. Then Λ = Id(J(Λ)) if and only if Λ is hereditary.

Remark 2.7. (cf. [BeZ]) Letting Λ_{0} := Λ and Λ_{n+1} := Id(J(Λ_{n})) for n = 0, 1, 2, … defines a canonical process, the so called radical idealizer process that constructs from an R-order Λ in A successively bigger R-orders Λ_{0} ⊂ Λ_{1} ⊂ … ⊂ Λ_{N} = Λ_{N+1}, the so called radical idealizer chain. The order Λ_{N} is hereditary and called the head order of Λ. If N is minimal such that Λ_{N} = Λ_{N+1}, then N is called the radical idealizer length l_{rad}(Λ) of Λ.

The next remark gives a lower bound on the length of the radical idealizer chain and is also useful for the explicit calculation of Id(J(Λ)), since one may calculate modulo the maximal ideal πR:

Remark 2.8. Let Λ be an R-order in A and Γ := Id(J(Λ)). Then J(Z(Λ))Γ ⊆ Λ, in particular πΓ ⊆ Λ.

\underline{Proof.} J(Z(Λ))Λ ⊂ J(Λ) ⊂ Λ, since J(Z(Λ))Λ is a pro-nilpotent ideal of Λ. Therefore

J(Z(Λ))Γ = J(Z(Λ))ΛΓ ⊆ J(Λ)Γ = J(Λ) ⊆ Λ.

□

Definition 2.9. Let Λ be an R-order in A and let ϵ_{1}, …, ϵ_{s} be the central primitive idempotents of A. Then the defect of Λ is the minimal d such that π^{d}ϵ_{t} ∈ Λ for all 1 ≤ t ≤ s. Note that this coincides with the usual definition of defect for blocks of group rings, if K is an unramified extension of ℚ_{p}.

Since hereditary orders contain the central primitive idempotents of A, one gets the following corollary:

Corollary 2.10. The radical idealizer length l_{rad}(Λ) is greater or equal than the defect of Λ.

Definition 2.11. For two R-orders Λ, Γ in A, the conductor of Γ in Λ is the biggest 2-sided Γ-ideal F_{Γ}(Λ) that is contained in Λ. Analogously one defines the left conductor F_{Γ}^{(l)}(Λ) and the right conductor F_{Γ}^{(r)}(Λ).

The following lemma is a straightforward generalization of [CPW, Theorem 2.2]:

Lemma 2.12. Assume that A is Abelian, let ϵ_{1}, …, ϵ_{s} be the primitive idempotents in A and assume that Γ := ⊕_{i=1}^{s}ϵ_{i}Λ is the maximal order in A. For i ∈ {1, …, s} let π_{i} be a prime element in ϵ_{i}Λand put π := (π_{1}, …, π_{s}) ∈ Γ. Let

Λ = Λ_{0} ⊂ Λ_{1} ⊂ … ⊂ Λ_{N}

be the radical idealizer chain of Λ. Then for n = 0, …, N

F_{Γ}(Λ_{n}) = π^{−n}F_{Γ}(Λ) ∩ Γ.

Proof. We argue by induction on n, the case n = 0 being trivial. Assume that

F_{Γ}(Λ_{n}) = π^{−n}F_{Γ}(Λ) ∩ Γ = ⨁_{i=1}^{s} π_{i}^{a_{i}}Λϵ_{i}.

Splitting off the direct summands of Λ_{n} that are maximal orders, we may assume that a_{i} > 0 for all i. Then

F_{Γ}(Λ_{n}) ⊆ J(Λ_{n}) = πΓ ∩ Λ_{n}.

Since F_{Γ}(Λ_{n}) is a Γ-ideal one gets

π^{−1}F_{Γ}(Λ_{n})J(Λ_{n}) ⊆ π^{−1}F_{Γ}(Λ_{n})πΓ ⊆ F_{Γ}(Λ_{n}) ⊆ J(Λ_{n}).

Hence

π^{−1}F_{Γ}(Λ_{n}) ⊆ Id(J(Λ_{n})) = Λ_{n+1}

and therefore

F_{Γ}(Λ_{n+1}) ⊇ π^{−(n+1)}F_{Γ}(Λ) ∩ Γ.

The opposite inclusion follows from Remark 2.8. □

Corollary 2.13. In the notation of Lemma 2.12 let

F_{Γ}(Λ) = ⨁_{i=1}^{s} π_{i}^{a_{i}}Λϵ_{i}.

Then l_{rad}(Λ) = max_{i=1,...,s} a_{i}.

3 Idealizers and bilinear forms

Idealizers can be calculated using a symmetric non degenerate bilinear form

ϕ : A × A → K that is associative, i.e. ϕ(ab, c) = ϕ(a, bc) for all a, b, c ∈ A.

It is easy to see that such an associative bilinear form ϕ is of the form

Tr_{z} : A × A → K, (a, b) ↦ tr_{red}(zab),

where z ∈ Z(A)^{∗} is an invertible element of the center of A and tr_{red} denotes the reduced trace of A. Fix such an associative symmetric bilinear form ϕ = Tr_{z}. For a full R-lattice L in A let

L^{#} := {a ∈ A | ϕ(L, a) ⊂ R}

be the dual lattice with respect to ϕ. It is frequently used that dualizing is an inclusion reversing bijection of the set of full R-lattices in A and that (L^{#})^{#} = L for all full R-lattices L in A.

Lemma 3.1. Let Γ be an R-order in A. Then Γ^{#} is a 2-sided Γ-ideal with

Γ = O_{l}(Γ^{#}) = O_{r}(Γ^{#}) = Id(Γ^{#}).

\underline{Proof.} Let x, γ ∈ Γ and y ∈ Γ^{#}. Then ϕ(x, γy) = ϕ(xγ, y) ∈ R and ϕ(yγ, x) = ϕ(y, γx) ∈ R and therefore Γ ⊂ Id(Γ^{#}). On the other hand let λ ∈ O_{l}(Γ^{#}). Then for all y ∈ Γ^{#} and x ∈ Γ = (Γ^{#})^{#}

ϕ(xλ, y) = ϕ(x, λy) ∈ R.

Hence Γλ ⊆ (Γ^{#})^{#} = Γ and therefore λ = 1λ ∈ Γ. Analogously one gets O_{r}(Γ^{#}) ⊆ Γ.

□

Proposition 3.2. If L is a full R-lattice in A, then O_{l}(L) = (LL^{#})^{#}, O_{r}(L) = (L^{#}L)^{#}, and hence

Id(L) = (LL^{#})^{#} ∩ (L^{#}L)^{#}.

\underline{Proof.} We only show O_{l}(L) = (LL^{#})^{#}. Let γ ∈ A. Then γ ∈ (LL^{#})^{#} if and only if for all x ∈ L, y ∈ L^{#}

ϕ(γ, xy) = ϕ(γx, y) ∈ R

which is equivalent to γL ⊆ (L^{#})^{#} = L, i.e. γ ∈ O_{l}(L). Analogously O_{r}(L) = (L^{#}L)^{#}.

□

From this one gets an interesting direct description of the idealizer of the radical.

Corollary 3.3. Let Λ be an R-order in A and Γ := Id(J(Λ)). Then Γ is the biggest Λ-ideal I ⊂ \frac{1}{π}Λ such that I/J(Λ) is a semi-simple Λ-Λ-bimodule.

\underline{Proof.} By Proposition 3.2

Γ^{#} = (J(Λ)J(Λ)^{#}) + (J(Λ)^{#}J(Λ))

is the smallest Λ-ideal J in J(Λ)^{#} for which J(Λ)^{#}/J is a semi-simple Λ-Λ-bimodule. Since the dual of a bimodule is semi-simple if and only if the module is semi-simple, the corollary follows. □

4 Symmetric orders.

Definition 4.1. An R-order Λ in A is called symmetric, if there is a non degenerated, symmetric, associative, bilinear form ϕ = Tr_{z} : A × A → K with Λ = Λ^{#}.

Lemma 4.2. (see e.g. [The, Proposition (1.6.2)]) Let e, f ∈ Λ be two idempotents in the symmetric order Λ. Then ϕ_{|eΛf×fΛe} is a regular R-bilinear pairing. In particular eΛe is a symmetric order in eAe.

An important tool to deal with symmetric orders is Jacobinski’s conductor formula:

Theorem 4.3. ([Jac, Satz 10.6]) Let Λ be a symmetric R-order in A and Γ ⊇ Λ an over-order. Then left and right conductor coincide and are equal to the dual of Γ:

F_{Γ}(Λ) = F_{Γ}^{(l)}(Λ) = F_{Γ}^{(r)}(Λ) = Γ^{#}.

\underline{Proof.} Since Γ^{#} ⊆ Λ is a Γ-ideal by Lemma 3.1 one has Γ^{#} ⊆ F_{Γ}(Λ). On the other hand if x ∈ F_{Γ}(Λ) and γ ∈ Γ then ϕ(x, γ) = ϕ(xγ, 1) ∈ R is integral, since xγ ∈ Λ = Λ^{#}.

□

From this proof one even gets that Γ^{#} is the largest R-lattice L in Λ with ΓL ⊆ Λ. [Jac, Satz 10.7] and [Ple, Theorem III.8] describe the conductor of F_{Γ}(Λ) for hered-

itary and (more general) graduated over-orders Γ of the symmetric order Λ. To apply this precise version of the conductor formula, we need the following (technical) notation:

Notation 4.4. Let ϵ_{1}, …, ϵ_{s} be the central primitive idempotents in A. Let z ∈ Z(A) be such that Λ = Λ^{#} with respect to T r_{z} and z_{i} = ϵ_{i}z ∈ Z(ϵ_{i}A) = K_{i} (1 ≤ i ≤ s). Let R_{i} be the maximal order in K_{i} with maximal ideal ℘_{i}, ℘_{i}^{−d_{i}} the inverse different of R_{i} over R and n_{i} ∈ ℤ with z_{i}R_{i} = ℘_{i}^{−n_{i}} (1 ≤ i ≤ s). The simple algebra Aϵ_{i} is isomorphic to a matrix ring over a central K_{i}-division algebra D_{i}. Let Ω_{i} be the maximal order in D_{i} and m_{i}^{2} = dim_{K_{i}}(D_{i}).

Theorem 4.5. ( [Jac, Satz 10.7], [Ple, Theorem III.8]) With the notation above let Δ be a hereditary order in A. Then

Δ^{#} = ⊕_{i=1}^{s}℘_{i}^{m_{i}(n_{i}−d_{i}−1)}J(Δ)ϵ_{i}.

Theorem 4.6. Let Λ be an indecomposable symmetric R-order in A and 0 ≠ e^{2} = e ∈ Λ be an idempotent such that eΛe is hereditary. Then Λ is a maximal order.

\underline{Proof.} ϵ_{i}eΛe =: Λ_{i} is either \{0\} or a symmetric hereditary order in eAe for all 1 ≤ i ≤ s. Let i be fixed such that Λ_{i} ≠ {0}. Then the conductor formula 4.5 yields that

Λ_{i} = Λ_{i}^{#} = ℘_{i}^{m_{i}(n_{i}−d_{i}−1)}J(Λ_{i}).

In particular Λ_{i} is isomorphic to J(Λ_{i}) as a bimodule and therefore Λ_{i} a maximal order in ϵ_{i}eAe, m_{i} = 1 and n_{i} = d_{i}. But then the conductor of every maximal over-order Γ of Λ in Λ is of the form F_{Γ}(Λ) = Γ′ ⊕ ϵ_{i}Γ for a suitable order Γ′. In particular ϵ_{i} ∈ Λ. Since Λ is indecomposable Λ = ϵ_{i}Λ = ϵ_{i}Γ and Λ is a maximal order in the simple K-algebra A = ϵ_{i}A. □

Putting e = 1 in Theorem 4.6 this characterizes the symmetric orders Λ with l_{rad}(Λ) = 0 as maximal orders. In particular if Λ is a block of a group ring RG, then l_{rad}(Λ) = 0 if and only if the defect of Λ is 0 (see [Jac, Satz 11.1]).

5 The radical idealizer of symmetric orders.

In this and the next section the first two steps of the radical idealizer chain of symmetric orders are made precise. The first theorem is a sort of converse of the conductor formula.

Theorem 5.1. Let Λ be a non hereditary, indecomposable, symmetric R-order in A and Γ := Id(J(Λ)) the idealizer of the radical of Λ. Then Γ = J(Λ)^{#}.

\underline{Proof.} Γ^{#} ⊆ Λ is the largest Γ-ideal in Λ by the conductor formula. Since J(Λ) ⊆ Λ is a Γ-ideal, one has J(Λ) ⊆ Γ^{#} and therefore Γ ⊆ J(Λ)^{#}. On the other hand let e_{1}, …, e_{h} ∈ Λ be orthogonal idempotents that map onto the central primitive idempotents of Λ/J(Λ) with 1 = e_{1} + … + e_{h}.

Then Λ = ⊕_{i,j=1}^{h}e_{i}Λe_{j}, (e_{i}Λe_{j})^{#} = e_{j}Λe_{i} and e_{i}Λe_{i} is a symmetric R-order in e_{i}Ae_{i}, which is not hereditary because of Theorem 4.6. Now

J(Λ) = ⨁_{i≠j=1}^{h} e_{i}Λe_{j} ⊕ ⨁_{i=1}^{h} J(e_{i}Λe_{i})

and with Lemma 4.2

J(Λ)^{#} = ⨁_{i≠j=1}^{h} e_{i}Λe_{j} ⊕ ⨁_{i=1}^{h} J(e_{i}Λe_{i})^{#}.

Assume first that h = 1 so Λ/J(Λ) is a simple k-algebra. Then J(Λ) is a maximal 2-sided ideal in Λ. Therefore either Γ^{#} = Λ or Γ^{#} = J(Λ). In the first case Γ = (Γ^{#})^{#} = Λ^{#} = Λ and therefore Λ is hereditary contradicting the assumption. In the second case Γ = J(Λ)^{#} and the theorem follows.

Now let h be arbitrary and Γ_{j} := J(e_{j}Λe_{j})^{#}. From above Γ_{j} = Id(J(e_{j}Λe_{j})) is an order, so it remains to show that for i ≠ j the summand e_{i}Λe_{j} of J(Λ)^{#} is a Γ_{i}-Γ_{j}-bimodule. The inclusion (e_{i}Λe_{j})Γ_{j} ⊆ e_{i}Λe_{j} is equivalent to

e_{j}Λe_{i} = (e_{i}Λe_{j})^{#} ⊆ (e_{i}Λe_{j}Γ_{j})^{#}.

So let e_{j}ae_{j} ∈ Γ_{j} with a ∈ J(Λ)^{#} and γ, λ ∈ Λ. Then

ϕ(e_{i}λe_{j}e_{j}ae_{j}, e_{j}γe_{i}) = ϕ(e_{j}γe_{i}, e_{i}λe_{j}ae_{j}) = ϕ(e_{j}γe_{i}λe_{j}, e_{j}ae_{j}) ∈ R,

because e_{j}γe_{i}λe_{j} ∈ J(Λ) (note that i ≠ j) and a ∈ J(Λ)^{#}. Analogously Γ_{i}(e_{i}Λe_{j}) ⊂ e_{i}Λe_{j}.

Since J(e_{j}Λe_{j}) is a Γ_{j}-bimodule (1 ≤ j ≤ s), one gets J(Λ)^{#}J(Λ)J(Λ)^{#} ⊆ J(Λ) and hence J(Λ)^{#} ⊆ Γ. □

The orders eΛe, where e^{2} = e ∈ Λ is an idempotent in Λ mapping onto a central primitive idempotent of Λ/J(Λ) play an important role in the above proof. These orders are 2-sided local orders, i.e. they have a unique maximal 2-sided ideal. These orders have a unique simple module, or equivalently eΛe/eJ(Λ)e is a simple k-algebra.

Lemma 5.2. Let Λ be a 2-sided local, symmetric R-order and Γ := Id(J(Λ)). Then either Γ/J(Γ) ≅ Λ/J(Λ) as Λ-Λ-bimodule and J(Γ) = J(Γ)^{#} or J(Γ) = J(Λ) and Γ is hereditary.

\underline{Proof.} Since Λ/J(Λ) is a simple Λ-Λ-bimodule, also its dual Γ/Λ is simple. Now Γ ⊇ J(Γ) + Λ ⊇ Λ and J(Λ) = J(Γ) ∩ Λ. Therefore either Γ = J(Γ) + Λ ≠ Λ and Γ/J(Γ) ≅ Λ/J(Λ) or J(Γ) ⊆ Λ whence J(Γ) = J(Λ). In the latter case Γ = Id(J(Γ)) is hereditary. In the first case Λ is not hereditary and J(Γ) is the unique maximal 2-sided Γ-ideal in Γ. Since J(Λ) ⊂ J(Γ) ⊂ J(Λ)^{#} = Γ (by Theorem 5.1), one also has J(Λ) ⊂ J(Γ)^{#} ⊂ Γ. Here all inclusions are proper. So J(Γ)^{#} is also a maximal 2-sided Γ-ideal in Γ and hence J(Γ)^{#} = J(Γ). □

Following the lines of [Jac, 11.4] one gets the following remark:

Remark 5.3. With the assumptions of Lemma 5.2 assume that J(Λ) = J(Γ). Then Γ is hereditary and Γ/J(Γ) has two (isomorphic) composition factors as a Λ−Λ-bimodule, namely the submodule Λ/J(Λ) and its dual, the factor module Γ/Λ = (Λ/J(Λ))^{#}.

Notation 5.4. We fix the following notation: Λ denotes an indecomposable non hereditary symmetric R-order in A, Γ := Id(J(Λ)) = J(Λ)^{#} the idealizer of the radical of Λ, ϵ_{1}, …, ϵ_{s} are the central primitive idempotents of A, and e_{1}, …, e_{h} ∈ Λ are orthogonal lifts of the central primitive idempotents of Λ/J(Λ). We order the e_{i} such that e_{i}Λe_{i}/J(e_{i}Λe_{i}) ≅ e_{i}Γe_{i}/J(e_{i}Γe_{i}) for 1 ≤ i ≤ t ≤ h and J(e_{i}Γe_{i}) = J(e_{i}Λe_{i}) for t < i ≤ h and put

e := ∑_{i=1}^{t} e_{i} and f := ∑_{i=t+1}^{h} e_{i}.

From Lemma 5.2 one now gets

Corollary 5.5. For 1 ≤ i ≤ h the idempotent e_{i} + J(Γ) is a central idempotent of Γ/J(Γ). If 1 ≤ i ≤ t, then e_{i} + J(Γ) is a central primitive idempotent. If t < i ≤ h then e_{i}Γe_{i} is hereditary.

6 The second step of the radical idealizer chain

We keep the Notation 5.4. Moreover let Δ := Id(J(Γ)) and let Λ_{N} be the head order of Λ.

Theorem 6.1.

Δ = (J(eΓe)^{2} + eΓfΓe)^{#} ⊕ fΓf ⊕ fΓe ⊕ eΓf.

\underline{Proof.} Δ = ((J(Γ)^{#}J(Γ) + J(Γ)J(Γ)^{#})^{#} by Proposition 3.2. Now

J(Γ)^{#} = (eJ(Γ)e)^{#} ⊕ fΓf ⊕ eΓf ⊕ fΓe.

Lemma 5.2 says (eJ(Γ)e)^{#} = eJ(Γ)e and therefore

J(Γ)^{#}J(Γ) = (eJ(Γ)e ⊕ fΓf ⊕ eΓf ⊕ fΓe)J(Γ) =

((eJ(Γ)e)^{2} + eΓfΓe) ⊕ (fJ(Γ)f + fΓeΓf) ⊕ (fΓe) ⊕ (eJ(Γ)eΓf + eΓfJ(Γ)f).

Since

fΓeΓf = fΛeΛf ⊆ J(fΛf) = J(fΓf)

one gets

Δ^{#} = ((eJ(Γ)e)^{2} + eΛfΛe) ⊕ fJ(Λ)f ⊕ fΛe ⊕ eΛf

and therefore the theorem follows. □

Corollary 6.2. For the head order Λ_{N} one finds

fΛ_{N}f = fΓf, eΛ_{N}f = eΛf, fΛ_{N}e = fΛe

and e + J(Λ_{N}) and f + J(Λ_{N}) are central idempotents of Λ_{N}/J(Λ_{N}).

\underline{Proof.} Let Λ_{0} := Λ and Λ_{i} := Id(J(Λ_{i−1})) (1 ≤ i ≤ N).

Using induction we show that fΛ_{i}f = fΓf, eΛ_{i}f = eΛf, fΛ_{i}e = fΛe and that f + J(Λ_{i}) (hence also e + J(Λ_{i})) lies in the center of Λ_{i}/J(Λ_{i}) for all 1 ≤ i ≤ N.

For i = 1 this is trivial. Now let i ≥ 1 and assume that the statement is true for i. Then Λ_{i+1} = ((J(Λ_{i})^{#}J(Λ_{i}) + J(Λ_{i})J(Λ_{i})^{#})^{#}. By assumption

J(Λ_{i}) = J(eΛ_{i}e) ⊕ J(fΛf) ⊕ eΛf ⊕ fΛe

and therefore

J(Λ_{i})^{#} = J(eΛ_{i}e)^{#} ⊕ fΓf ⊕ eΛf ⊕ fΛe.

Since Λ_{i} radically covers Λ_{i−1}, it holds that J(Λ_{i−1}) ⊆ J(Λ_{i}). In particular J(Λ_{0}) ⊆ J(Λ_{i}). Therefore J(Λ_{i})^{#} ⊆ J(Λ_{0})^{#} = Γ. One calculates J(Λ_{i})^{#}J(Λ_{i}) + J(Λ_{i})J(Λ_{i})^{#}

= (J(eΛ_{i}e)^{#}J(eΛ_{i}e) + J(eΛ_{i}e)J(eΛ_{i}e)^{#} + eΛfΛe) ⊕ J(fΛf) ⊕ eΛf ⊕ fΛe.

After dualizing, one gets the desired form of Λ_{i+1}.

Let λ ∈ Λ_{i+1}. Then

fλ − λf = fλe − eλf ∈ eΛ_{i+1}f ⊕ fΛ_{i+1}e = eΛf ⊕ fΛe ⊆ J(Λ) ⊆ J(Λ_{i+1}).

Therefore f +J(Λ_{i+1}) ∈ Z(Λ_{i+1}/J(Λ_{i+1})) and hence also e+J(Λ_{i+1}) = (1−f)+J(Λ_{i+1}) is central. □

Corollary 6.3. The conductor of Λ_{N} in Λ is

Λ_{N}^{#} = (eΛ_{N}e)^{#} ⊕ fJ(Λ)f ⊕ eΛf ⊕ fΛe.

Theorem 6.4. Let Λ, Γ = Id(J(Λ)), Δ = Id(J(Γ)) and f be as in Notation 5.4. Then f = 0 or f = 1. If f = 1, then J(Γ) = J(Λ) and Γ = Δ is hereditary. If f = 0, then Γ = J(Γ) + Λ, Γ/J(Γ) ≅ Λ/J(Λ), J(Γ) = J(Γ)^{#} and Δ = (J(Γ)^{2})^{#}.

\underline{Proof.} With Notation 4.4 the conductor formula 4.5 gives

Λ_{N}^{#} = (eΛ_{N}e)^{#}⊕fJ(Λ)f⊕eΛf⊕fΛe = ⊕_{i=1}^{s}℘_{i}^{m_{i}(n_{i}−d_{i}−1)}(eJ(Λ_{N})e⊕fJ(Λ)f⊕eΛf⊕fΛe).

So if ϵ_{i}f ≠ 0 for some 1 ≤ i ≤ s, then n_{i} − d_{i} = 1 and J(Λ_{N})ϵ_{i} ⊆ Λ. Since J(Λ_{N})ϵ_{i} is a pro-nilpotent Λ-ideal in Λ containing J(Λ)ϵ_{i} one gets J(Λ)ϵ_{i} = J(Λ_{N})ϵ_{i} ⊆ J(Λ). But then ϵ_{i} ∈ Γ and Γϵ_{i} = Λ_{N}ϵ_{i} is hereditary. We claim that fϵ_{i} = ϵ_{i}: To see this let 1 ≤ j ≤ t with e_{j}ϵ_{i} ≠ 0. Since e_{j} + J(Γ) is a central primitive idempotent of Γ/J(Γ), one even has e_{j}ϵ_{i} = e_{j}. Now Γϵ_{i} is hereditary, so e_{j}Γe_{j} ≅ Ω_{i}^{x×x} for some x ∈ ℕ. Let P_{i} denote the maximal ideal in Ω_{i}. Since j ≤ t Lemma 5.2 says that P_{i}^{x×x} ≅ J(e_{j}Γe_{j}) is symmetric with respect to the restriction of the form T r_{z} above. But n_{i}−d_{i}−1 = 0, yields together with [Rei, Theorem 14.9] that J(e_{j}Γe_{j})^{#} = ℘_{i}^{−1}e_{j}Γe_{j} which is a contradiction. Therefore eϵ_{i} = 0 and hence fϵ_{i} = ϵ_{i}.

So for all central primitive idempotents ϵ_{i} of A either fϵ_{i} = 0 or fϵ_{i} = ϵ_{i}. Therefore Λ = eΛe ⊕ fΛf. Since Λ is assumed to be indecomposable one has f = 0 or f = 1.

In the latter case J(Γ) = J(Λ) = J(Λ_{N}), hence Γ is hereditary. If f = 0, then Δ = (J(Γ)^{2})^{#} from Theorem 6.1. The fact that J(Γ) is self-dual follows with Lemma 5.2 which also implies that Γ = J(Γ) + Λ and hence Λ/J(Λ) ≅ Γ/J(Γ). □

Summarizing let Λ be an indecomposable, non hereditary, symmetric R-order in A, Γ = Id(J(Λ)) and Δ = Id(J(Γ)). Let f be as in Notation 5.4. Then

[figure: Γ; Λ; J(Λ) = J(Γ); f = 1; Γ hereditary; Γ; Λ; J(Γ) = J(Γ)^{#}; J(Λ); f = 0; Γ/J(Γ) ≅ Λ/J(Λ); Δ = Id(J(Γ)) = (J(Γ)^{2})^{#}; or]

7 Symmetric orders with radical idealizer length 1 or 2.

The case f = 1 in Theorem 6.4 can be dealt with the arguments in [Jac, Satz 11.4]. With a modest additional assumption one gets J(Γ) = J(Λ) and in particular if Λ is a block of a group ring with l_{rad}(Λ) = 1 one can associate a Brauer tree to Λ.

Proposition 7.1. Assume that Γ = Id(J(Λ)) is hereditary. Assume further that Z_{i} := ϵ_{i}Z(Λ) ⊆ ϵ_{i}A is a maximal order for all central primitive idempotents ϵ_{1}, …, ϵ_{s}. Then

J(Γ) = J(Λ).

\underline{Proof.} With the Notation 4.4 and Jacobinski’s conductor formula 4.5

J(Λ) = Γ^{#} = ⨁_{i=1}^{s} ϵ_{i}℘_{i}^{m_{i}(n_{i}−δ_{i}−1)}J(Γ).

By Remark 2.8 and since ϵ_{i}J(Λ) ⊂ J(Λ)

J(Z_{i})J(Γ) ⊊ J(Z_{i})Γ ⊆ J(Λ)

for all i. This implies that n_{i} − δ_{i} = 1 and that J(Λ) = J(Γ). □

Proposition 7.2. Assume that J(Γ) = J(Λ). If moreover the decomposition map from the Grothendieck groups of simple modules G_{0}(A) → G_{0}(k ⊗ Λ) is surjective or k is a splitting field for k ⊗ Γ, then for each idempotent e_{i} ∈ Λ, there are exactly two central primitive idempotents ϵ_{i_{1}} and ϵ_{i_{2}} in A with e_{i}ϵ_{i_{j}} ≠ 0 (j = 1, 2).

\underline{Proof.} We precise the embedding Λ/J(Λ) ↪ Γ/J(Γ) following the lines of the proof of [Jac, Satz 11.4]: Let e_{i}(Λ/J(Λ)) =: S_{i} be the simple algebra summand of Λ/J(Λ) that corresponds to e_{i} (1 ≤ i ≤ h). Then

Λ/J(Λ) ≅ S_{1} ⊕ … ⊕ S_{h} and Γ/Λ = (Λ/J(Λ))^{#} ≅ S_{1}^{∗} ⊕ … ⊕ S_{h}^{∗} ≅ S_{1} ⊕ … ⊕ S_{h} as Λ−Λ-bimodules. Therefore for every simple k-algebra summand S_{i} of Λ/J(Λ) there are either two algebra-summands T_{i′_{1}} and T_{i′_{2}} of Γ/J(Γ) such that S_{i} is diagonally embedded into T_{i′_{1}} ⊕ T_{i′_{2}} or there is a unique summand T_{i} ≅ l_{2}^{n×n} such that S_{i} ≅ l_{1}^{n×n} ⊂ T_{i} for extension fields l_{2}, l_{1} of k with [l_{2} : l_{1}] = 2. The latter is impossible, if k is a splitting field for k ⊗ Γ. Similarly, if the decomposition map of Λ is surjective, the last case cannot happen, since otherwise the simple S_{i}-module occurs with even multiplicity in every simple Γ-module and hence in the reduction of every Γ-lattice modulo π. Therefore

S_{i} ↪ T_{i_{1}} ⊕ T_{i_{2}} ⊂ Γ/J(Γ).

Since Γ is hereditary, it contains the central primitive idempotents ϵ_{i} of A. Therefore Γ/J(Γ) = ⊕_{i=1}^{s}ϵ_{i}Γ/J(ϵ_{i}Γ) and hence each simple summand of Γ/J(Γ) is a summand of some ϵ_{i}Γ/J(ϵ_{i}Γ). In particular for j = 1, 2 the summand T_{i′_{j}} defines a unique central primitive idempotent ϵ_{i_{j}} with f_{i′_{j}}ϵ_{i_{j}} ≠ 0 for any lift f_{i′_{j}} ∈ Γ of the central primitive idempotent of Γ/J(Γ) that belongs to T_{i′_{j}}. Then ϵ_{i_{j}} (j = 1, 2) are the only central primitive idempotents in A with e_{i}ϵ_{i_{j}} ≠ 0. □

If the decomposition map of Λ is surjective (which is always satisfied when Λ is a block of a group ring) or k is a splitting field for k ⊗ Γ and the other assumptions of Proposition 7.2 hold, we can define a graph G(Λ) whose vertices correspond to ϵ_{1}, …, ϵ_{s} and whose edges correspond to e_{1}, …, e_{h}. Two vertices ϵ_{i} and ϵ_{j} are connected by the edge e_{l}, if e_{l}ϵ_{i} ≠ 0 and e_{l}ϵ_{j} ≠ 0.

As in [Jac, Korollar 11.6] one shows:

Corollary 7.3. If the decomposition map of Λ is surjective then G(Λ) is a tree.

We end this section with a short remark on the length 2 case.

Remark 7.4. With Notation 5.4 assume that f = 0 and Δ := Id(J(Γ)) is hereditary.

(i) J(Λ)^{2}ϵ_{i} ∈ Λ for all 1 ≤ i ≤ s. In particular π^{2}ϵ_{i} ∈ Λ which means that the defect of Λ is ≤ 2.

(ii) ϵ_{i}e_{l}Λe_{j} ⊆ e_{l}Λe_{j} for all 1 ≤ i ≤ s, 1 ≤ l ≠ j ≤ h.

(iii) If s > 1 then ϵ_{i}J(Λ) ⊈ J(Λ) for all 1 ≤ i ≤ s.

\underline{Proof.} (i) Let 1 ≤ i ≤ s. Since ϵ_{i} ∈ Δ one has ϵ_{i}J(Γ) ⊆ J(Γ). Now

J(Λ)^{2} ⊆ J(Γ)^{2} = Δ^{#} = ⨁_{i=1}^{s} J(Γ)^{2}ϵ_{i} ⊆ Λ

implies J(Λ)^{2}ϵ_{i} ∈ Λ. Since π = π1 ∈ πΛ ⊂ J(Λ) one has π^{2}ϵ_{i} ∈ Λ for all i. (ii) Let l ≠ j ∈ {1, …, h}. Then e_{l}Λe_{j} ⊆ J(Γ). Since ϵ_{i}J(Γ) ⊆ J(Γ), the claim follows. (iii) Assume that there is 1 ≤ i ≤ s such that ϵ_{i}J(Λ) ⊆ J(Λ). Then ϵ_{i} ∈ Γ. Since ϵ_{i} ≠ 1 and Λ is indecomposable there is 1 ≤ j ≤ h such that 0 ≠ ϵ_{i}e_{j} ≠ e_{j}. But then e_{j} + J(Γ) ∈ Z(Γ/J(Γ)) is not primitive contradicting Corollary 5.5. □

8 p-groups

Let G ≠ {1} be a p-group, R = ℤ_{p}, and Λ := RG. Then Λ is a symmetric R-order with respect to the associative bilinear form

ϕ(x, y) := \frac{1}{|G|}trace_{reg}(xy) = (xy)_{1} if xy = ∑_{g∈G}(xy)_{g}g

where trace_{reg} is the regular trace of ℚ_{p}G. Moreover

J(Λ) = ⟨pΛ, g − h | g, h ∈ G⟩_{R}

and

Γ := Id(J(Λ)) = J(Λ)^{#} = ⟨Λ, \frac{1}{p} ∑_{g∈G} g⟩_{R}.

because \frac{1}{p} ∑_{g∈G} g idealizes J(Λ) and Γ is an over-order of Λ of index p = |Λ/J(Λ)|. If |G| = p then Γ is hereditary by [Jac, Section 11]. Therefore we assume that |G| ≥ p^{2}. Then the radical of Γ is

J(Γ) = ⟨J(Λ), \frac{1}{p} ∑_{g∈G} g⟩_{R} = J(Γ)^{#}

and is contained in Γ of index p.

Theorem 8.1. Let Δ := Id(J(Γ)). Then

Δ = (J(Γ)^{2})^{#} = ⟨Λ, \frac{1}{p^{2}} ∑_{g∈G} g, \frac{1}{p} ∑_{g∈G} φ(g)g | φ : G → R/pR homomorphism ⟩_{R}.

\underline{Proof.} Clearly λ := \frac{1}{p^{2}} ∑_{g∈G} g ∈ Id(J(Λ)). Let y := ∑_{g∈G} y_{g}g ∈ Id(J(Γ)). Then

p ∑_{g∈G} y_{g}g = a\frac{1}{p} ∑_{g∈G} g + x

with x ∈ J(Λ) ⊂ Λ and a ∈ R. Hence py_{g} ≡ \frac{a}{p} (mod R). Replacing y by y − aλ we may assume that y_{g} = \frac{a_{g}}{p} with a_{g} ∈ R for all g ∈ G. Adding a suitable multiple of pλ to y we can also assume that a_{1} = 0. Now Id(J(Γ)) = (J(Γ)^{2})^{#} and one calculates

J(Γ)^{2} = ⟨p^{2}Λ, p(g − h), (g_{1} − h_{1})(g_{2} − h_{2}), ∑_{g∈G} g | g, h, g_{1}, h_{1}, g_{2}, h_{2} ∈ G⟩_{R}.

In particular the coefficient of 1 of y(g^{−1} − 1)(1 − h^{−1}) which is \frac{1}{p}(a_{g} + a_{h} − a_{hg}) lies in R. Hence

φ : g ↦ a_{g} + pR ∈ R/pR

is a group homomorphism from G to R/pR, from which the inclusion ⊆ follows. It remains to show that the elements y := \frac{1}{p} ∑_{g∈G} φ(g)g with φ ∈ Hom(G, R/pR) are in the dual of J(Γ)^{2}. Clearly

ϕ(y, p^{2}Λ) ⊂ R and ϕ(y, p(g − h)) ⊂ R for all g, h ∈ G.

For (g_{1} − h_{1})(g_{2} − h_{2}) with g_{1}, g_{2}, h_{1}, h_{2} ∈ G one gets

ϕ(y, (g_{1} − h_{1})(g_{2} − h_{2})) = \frac{1}{p}(φ(g_{2}^{−1}g_{1}^{−1}) − φ(h_{2}^{−1}g_{1}^{−1}) − φ(g_{2}^{−1}h_{1}^{−1}) + φ(h_{2}^{−1}h_{1}^{−1})) ∈ R

since φ is a homomorphism. The last generator is p^{2}λ for which one finds

ϕ(y, ∑_{g∈G} g) = \frac{1}{p} ∑_{g∈G} φ(g) ∈ R.

□

Corollary 8.2. Let Λ := ℤ_{p}G for some p-group G of order |G| ≥ p^{2}. Let Γ := Id(J(Λ)) and Δ := Id(J(Γ)). Then |Γ/Λ| = p and |Δ/Λ| = p^{2}|G/(G′G^{p})|.

Corollary 8.3. Let G be a p-group and Λ := ℤ_{p}G.

1) Γ = Id(J(Λ)) is hereditary if and only if G ≅ C_{p}.

2) Assume that |G| ≥ p^{2}. Then Δ = Id(J(Γ)) is hereditary if and only if |G| = 4 and hence G ≅ C_{2} × C_{2} or G ≅ C_{4}.

\underline{Proof.} 1) Since Γ ⊂ \frac{1}{p}Λ contains all central primitive idempotents of ℚ_{p}G, one gets |G| = p. That Id(J(ℤ_{p}C_{p})) is hereditary follows from [Jac, Abschnitt 11].

2) Since Δ ⊂ \frac{1}{p^{2}} Λ contains all central primitive idempotents of ℚ_{p}G, one gets |G| = p^{2}. Hence G is Abelian, G ≅ C_{p^{2}} or G ≅ C_{p} × C_{p} and by Corollary 8.2, [Δ : Λ] = p^{3} respectively p^{4}. If Δ is hereditary, then Δ is the maximal order in ℚ_{p}G,

Δ ≅ ℤ_{p} ⊕ ℤ_{p}[ζ_{p}] ⊕ ℤ_{p}[ζ_{p^{2}}] respectively Δ ≅ ℤ_{p} ⊕ ℤ_{p}[ζ_{p}]^{p+1}.

The form ϕ above is \frac{1}{|G|}trace_{reg} hence the discriminant of Δ with respect to ϕ is

p^{−2p^{2}} · 1 · p^{p−2} · p^{p(2p−3)} = p^{−2p−2} respectively p^{−2p^{2}} · 1 · (p^{p−2})^{p+1} = p^{−p^{2}−p−2}

(see [Was, Prop. 2.1]). Since Δ contains a symmetric order of index p^{3} respectively p^{4}, one gets −2p − 2 = −6 hence p = 2 respectively −p^{2} − p − 2 = 8 which also implies p = 2. The same argument shows that for G = C_{2} × C_{2} or G = C_{4}, the order Δ has the same discriminant as the maximal order and hence is hereditary (i.e. equal to the maximal order). □

Note that this corollary also follows from Theorem 8.4 below.

For Abelian groups G, the radical idealizer length of ℤ_{p}G can be calculated from the exponent and the order of the Sylow p-subgroup of G:

Theorem 8.4. Let G be an Abelian group with Sylow p-subgroup of order p^{n} and of exponent p^{a} > 1. Then

l_{rad}(ℤ_{p}G) = p^{a−1} + (p^{a} − p^{a−1})(n − a).

Proof. The theorem follows with Corollary 2.13 by calculating the conductor of the maximal order Γ in ℚ_{p}G:

Γ = ⨁_{i=1}^{s} ℤ_{p}Gϵ_{i} = ⨁_{i=1}^{s} R_{i}[ζ_{p^{a_{i}}}]

where R_{i} is an unramified extension of ℤ_{p} and {a_{1}, …, a_{s}} = {0, 1, …, a}. If ^{∗} denotes the different, i.e. the dual with respect to the usual trace bilinear form, then by [Was, Prop. 2.1] R_{i}[ζ_{p^{a_{i}}}]^{∗} = R_{i}[ζ_{p^{a_{i}}}](1 − ζ_{p^{a_{i}}})^{−p^{a_{i}−1}(a_{i}p−a_{i}−1)} and hence the conductor of Γ in ℤ_{p}G is

Γ^{#} = ⨁_{i=1}^{s} R_{i}[ζ_{p^{a_{i}}}]^{∗}p^{n} = ⨁_{i=1}^{s} R_{i}[ζ_{p^{a_{i}}}](1 − ζ_{p}^{a_{i}})^{((n−a_{i})(p−1)+1)p^{a_{i}−1}}.

By Corollary 2.13, the length of the radical idealizer chain is

max_{i=0,...,a}((n − i)(p^{i} − p^{i−1}) + p^{i−1}) = (n − a)(p^{a} − p^{a−1}) + p^{a−1}.

□

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