Modular Representation Theory of Finite Groups

Home , Character theory, Grothendieck group, Indecomposable module, Simple module

Modular representation theory of ﬁnite groups

Peter Schneider

Course at M¨unster,2010/11, version 19.08.2011

Contents

I Module theory continued 1

1 Chain conditions and more 1

2 Radicals 3

3 I-adic completeness 4

4 Unique decomposition 10

5 Idempotents and blocks 17

6 Projective modules 29

7 Grothendieck groups 36

II The Cartan-Brauer triangle 45

8 The setting 45

9 The triangle 48

10 The ring structure of RF (G), and induction 57

11 The Burnside ring 62

12 Cliﬀord theory 70

13 Brauer’s induction theorem 75

14 Splitting ﬁelds 79

15 Properties of the Cartan-Brauer triangle 83

III The Brauer character 92

16 Deﬁnitions 92

i 17 Properties 96

IV Green’s theory of indecomposable modules 102

18 Relatively projective modules 102

19 Vertices and sources 110

20 The Green correspondence 115

21 An example: The group SL2(Fp) 125

22 Green’s indecomposability theorem 148

V Blocks 155

23 Blocks and simple modules 155

24 Central characters 159

25 Defect groups 161

26 The Brauer correspondence 168

27 Brauer homomorphisms 175

References 185

ii iii Chapter I Module theory continued

Let R be an arbitrary (not necessarily commutative) ring (with unit). By an R-module we always will mean a left R-module. All ring homomorphisms respect the unit element. But a subring may have a diﬀerent unit element.

1 Chain conditions and more

For an R-module M we have the notions of being

ﬁnitely generated, artinian, noetherian, simple, and semisimple.

The ring R is called left artinian, resp. left noetherian, resp. semisimple, if it has this property as a left module over itself.

Proposition 1.1. i. The R-module M is noetherian if and only if any submodule of M is ﬁnitely generated.

ii. Let L ⊆ M be a submodule; then M is artinian, resp. noetherian, if and only if L and M/L are artinian, resp. noetherian.

iii. If R is left artinian, resp. left noetherian, then every ﬁnitely generated R-module M is artinian, resp. noetherian.

iv. If R is left noetherian then an R-module M is noetherian if and only if it is ﬁnitely generated.

Proposition 1.2. (Jordan-H¨older)For any R-module M the following conditions are equivalent:

i. M is artinian and noetherian;

ii. M has a composition series {0} = M0 ⊆ M1 ⊆ ... ⊆ Mn = M such that all Mi/Mi−1 are simple R-modules.

In this case two composition series {0} = M0 ⊆ M1 ⊆ ... ⊆ Mn = M ∼ and {0} = L0 ⊆ L1 ⊆ ... ⊆ Lm = M satisfy n = m and Li/Li−1 = M(i)/M(i)−1, for any 1 ≤ i ≤ m, where is an appropriate permutation of {1, . . . , n}.

1 An R-module M which satisﬁes the conditions of Prop. 1.2 is called of ﬁnite length and the integer l(M) := n is its length. Let

Rˆ := set of all isomorphism classes of simple R-modules.

For ∈ Rˆ and an R-module M the -isotypic component of M is

M() := sum of all simple submodules of M in .

Lemma 1.3. For any R-module homomorphism f : L −→ M we have f(L()) ⊆ M().

Proposition 1.4. i. For any R-module M the following conditions are equivalent:

a. M is semisimple, i. e., isomorphic to a direct sum of simple R- modules; b. M is the sum of its simple submodules; c. every submodule of M has a complement.

ii. Submodules and factor modules of semisimple modules are semisimple.

iii. If R is semisimple then any R-module is semisimple.

iv. Any -isotypic component M() of any R-module M is semisimple.

v. If the R-module M is semisimple then M = ⊕∈RˆM(). Q vi. If R is semisimple then R = ∈Rˆ R() as rings. Lemma 1.5. Any R-module M contains a unique maximal submodule which is semisimple (and which is called the socle soc(M) of M). P Proof. We deﬁne soc(M) := ∈Rˆ M(). By Prop. 1.4.i the submodule soc(M) is semisimple. On the other hand if L ⊆ M is any semisimple sub- P module then L = L() by Prop. 1.4.v. But L() ⊆ M() by Lemma 1.3 and hence L ⊆ soc(M).

Deﬁnition. An R-module M is called decomposable if there exist nonzero submodules M1,M2 ⊆ M such that M = M1 ⊕ M2; correspondingly, M is called indecomposable if it is nonzero and not decomposable.

Lemma 1.6. If M is artinian or noetherian then M is the direct sum of ﬁnitely many indecomposable submodules.

2 Proof. We may assume that M ∕= {0}. Step 1: We claim that M has a nonzero indecomposable direct summand N. If M is artinian take a minimal element N of the set of all nonzero direct summands of M. If M is noetherian let N ′ be a maximal element of the set of all direct summands ∕= M of M, ′ ′ and take N such that M = N ⊕N. Step 2: By Step 1 we ﬁnd M = M1 ⊕M1 ′ ′ ′ with M1 ∕= {0} indecomposable. If M1 ∕= {0} then similarly M1 = M2 ⊕ M2 with M2 ∕= {0} indecomposable. Inductively we obtain in this way a strictly increasing sequence

{0} ⫋ M1 ⫋ M1 ⊕ M2 ⫋ ... as well as a strictly decreasing sequence

′ ′ M1 ⫌ M2 ⫌ ... of submodules of M. Since one of the two (depending on M being artinian ′ or noetherian) stops after ﬁnitely many steps we must have Mn = {0} for some n. Then M = M1 ⊕ ... ⊕ Mn is the direct sum of the ﬁnitely many indecomposable submodules M1,...,Mn.

n Exercise. The ℤ-modules ℤ and ℤ/p ℤ, for any prime number p and any n ≥ 1, are indecomposable.

2 Radicals

The radical of an R-module M is the submodule

rad(M) := intersection of all maximal submodules of M.

The Jacobson radical of R is Jac(R) := rad(R). Proposition 2.1. i. If M ∕= {0} is ﬁnitely generated then rad(M) ∕= M.

ii. If M is semisimple then rad(M) = {0}.

iii. If M is artinian with rad(M) = {0} then M is semisimple.

iv. The Jacobson radial

Jac(R) = intersection of all maximal left ideals of R = intersection of all maximal right ideals of R = {a ∈ R : 1 + Ra ⊆ R×}

is a two-sided ideal of R.

3 v. Any left ideal which consists of nilpotent elements is contained in Jac(R).

vi. If R is left artinian then the ideal Jac(R) is nilpotent and the factor ring R/ Jac(R) is semisimple.

vii. If R is left artinian then rad(M) = Jac(R)M for any finitely generated R-module M. Corollary 2.2. If R is left artinian then any artinian R-module is noetherian and, in particular, R is left noetherian and any finitely generated R- module is of finite length. Lemma 2.3. (Nakayama) If L ⊆ M is a submodule of an R-module M such that M/L is finitely generated, then L + Jac(R)M = M implies that L = M. Lemma 2.4. If R is left noetherian and R/ Jac(R) is left artinian then R/ Jac(R)m is left noetherian and left artinian for any m ≥ 1. Proof. It suffices to show that R/ Jac(R)m is noetherian and artinian as an R-module. By Prop. 1.1.ii it further suffices to prove that the R-module Jac(R)m/ Jac(R)m+1, for any m ≥ 0, is artinian. Since R is left noetherian it certainly is a finitely generated R-module and hence a finitely generated R/ Jac(R)-module. The claim therefore follows from Prop. 1.1.iii.

Lemma 2.5. The Jacobson ideal of the matrix ring Mn×n(R), for any n ∈ ℕ, is the ideal Jac(Mn×n(R)) = Mn×n(Jac(R)) of all matrices with entries in Jac(R).

3 I-adic completeness

We begin by introducing the following general construction. Let

fn (Mn+1 −−→ Mn)n∈ℕ be a sequence of R-module homomorphisms, usually visualized by the diagram fn+1 fn fn−1 f1 ... −−−−→ Mn+1 −−→ Mn −−−−→ ... −−→ M1 . In the direct product module Q M we then have the submodule n∈ℕ n Y lim M := {(x ) ∈ M : f (x ) = x for any n ∈ }, ←− n n n n n n+1 n ℕ n

4 which is called the projective limit of the above sequence. Although sup- pressed in the notation the construction depends crucially, of course, on the homomorphisms fn and not only on the modules Mn. Let us ﬁx now a two-sided ideal I ⊆ R. Its powers form a descending sequence R ⊇ I ⊇ I2 ⊇ ... ⊇ In ⊇ ... of two-sided ideals. More generally, for any R-module M we have the descending sequence of submodules

M ⊇ IM ⊇ I2M ⊇ ... ⊇ InM ⊇ ... and correspondingly the sequence of residue class projections pr pr ... −→ M/In+1M −−→ M/InM −→ ... −→ M/I2M −−→ M/IM .

We form the projective limit of the latter

lim M/InM = ←− n Y n n {(xn + I M)n ∈ M/I M : xn+1 − xn ∈ I M for any n ∈ ℕ}. n There is the obvious R-module homomorphism

I : M −→ lim M/InM M ←− n x 7−→ (x + I M)n . Deﬁnition. The R-module M is called I-adically separated, resp. I-adically I T n complete, if M is injective (i. e., if n I M = {0}), resp. is bijective. n Exercise. If I 0 M = {0} for some n0 ∈ ℕ then M is I-adically complete. Lemma 3.1. Let J ⊆ R be another two-sided ideal such that In0 ⊆ J ⊆ I for some n0 ∈ ℕ; then M is J-adically separated, resp. complete, if and only if it is I-adically separated, resp. complete. Proof. We have

nn n n I 0 M ⊆ J M ⊆ I M for any n ∈ ℕ. This makes obvious the separatedness part of the assertion. Furthermore, by the right hand inclusions we have the well deﬁned map

: lim M/J nM −→ lim M/InM ←− ←− n n (xn + J M)n 7−→ (xn + I M)n .

5 J I Since ∘ M = M it suﬃces for the completeness part of the assertion to show that is bijective. That () = 0 for (x + J nM) ∈ lim M/J nM n n ←− n nn0 n means that xn ∈ I M for any n ∈ ℕ. Hence xnn0 ∈ I M ⊆ J M. On the n n other hand xnn0 − xn ∈ J M. It follows that xn ∈ J M for any n, i. e., that = 0. For the surjectivity of let (y + InM) ∈ lim M/InM be any n n ←− nn0 element. We deﬁne xn := ynn0 . Then xn+1 −xn = ynn0+n0 −ynn0 ∈ I M ⊆ n n n n J M and hence (xn +J M)n ∈ lim M/J M. Moreover, xn +I M = ynn0 + n n ←− n n I M = yn + I M which shows that ((xn + J M)n) = (yn + I M)n. Lemma 3.2. If R is I-adically complete then I ⊆ Jac(R).

Proof. Let a ∈ I be any element. Then

(1 + a + ... + an−1 + In) ∈ lim R/In . n ←− Therefore, by assumption, there is an element c ∈ R such that

n n−1 n c + I = 1 + a + ... + a + I for any n ∈ ℕ.

It follows that \ \ (1 − a)c ∈ (1 − a)(1 + a + ... + an−1) + In = 1 − an + In = {1} n n and similarly that c(1 − a) = 1. This proves that 1 + I ⊆ R×. The assertion now follows from Prop. 2.1.iv.

Lemma 3.3. Let f : M −→ N be a surjective R-module homomorphism and suppose that M is I-adically complete; if N is I-adically separated then it already is I-adically complete.

Proof. The R-module homomorphism

: lim M/InM −→ lim N/InN ←− ←− n n (xn + I M)n 7−→ (f(xn) + I N)n

ﬁts into the commutative diagram

f / / M N _ I ∼ I M = N lim M/InM lim N/InN. ←− / ←−

6 It therefore suﬃces to show that is surjective. Let (y +InN) ∈ lim N/InN n n ←− be an arbitrary element. By the surjectivity of f we ﬁnd an x1 ∈ M such that f(x1) = y1. We now proceed by induction and assume that elements x1, . . . , xn ∈ M have been found such that

j xj+1 − xj ∈ I M for 1 ≤ j < n and j j f(xj) + I N = yj + I N for 1 ≤ j ≤ n . ′ ′ We choose any xn+1 ∈ M such that f(xn+1) = yn+1. Then

′ n n f(xn+1 − xn) ∈ yn+1 − yn + I N = I N.

f Since the restricted map InM −→ InN still is surjective we ﬁnd an element z ∈ ker(f) such that ′ n xn+1 − xn − z ∈ I M. ′ n Therefore, if we put xn+1 := xn+1 −z then xn+1 −xn ∈ I M and f(xn+1) = ′ f(xn+1) = yn+1. Proposition 3.4. Suppose that R is commutative and noetherian and that I ⊆ Jac(R); then any ﬁnitely generated R-module M is I-adically separated.

T n Proof. We have to show that the submodule M0 := n I M is zero. Since R is noetherian M0 is finitely generated by Prop. 1.1. We therefore may apply the Nakayama lemma 2.3 to M0 (and its submodule {0}) and see that it suffices to show that IM0 = M0. Obviously IM0 ⊆ M0. Again since R is noetherian we find a submodule IM0 ⊆ L ⊆ M which is maximal with respect to the property that L ∩ M0 = IM0. In an intermediate step we establish that for any a ∈ I there is an integer n0(a) ≥ 1 such that an0(a)M ⊆ L. j Fixing a we put Mj := {x ∈ M : a x ∈ L} for j ≥ 1. Since R is commutative the Mj ⊆ Mj+1 form an increasing sequence of submodules of M. Since R is noetherian this sequence

M1 ⊆ ... ⊆ Mn0(a) = Mn0(a)+1 = ...

n (a) has to stabilize. We trivially have IM0 ⊆ (a 0 M + L) ∩ M0. Consider any n (a) element x0 = a 0 x + y, with x ∈ M and y ∈ L, in the right hand side. n (a)+1 We have ax0 ∈ IM0 ⊆ L and ay ∈ L, hence a 0 x = ax0 − ay ∈ L or

7 n0(a) equivalently x ∈ Mn0(a)+1. But Mn0(a)+1 = Mn0(a) so that a x ∈ L and n (a) consequently x0 ∈ L∩M0 = IM0. This shows that IM0 = (a 0 M +L)∩M0 holds true. The maximality of L then implies that an0(a)M ⊆ L. The ideal I in the noetherian ring R can be generated by ﬁnitely many elements a1, . . . , ar. We put n0 := maxi n0(ai) and n1 := rn0. Then

n1 rn0 n0 n0 n0(a1) n0(ar) I = (Ra1 + ... + Rar) ⊆ Ra1 + ... + Rar ⊆ Rar + ... + Rar and hence In1 M ⊆ L which implies

\ n n1 M0 = I M ⊆ I M ⊆ L and therefore M0 = L ∩ M0 = IM0 . n

Let R0 be a commutative ring. If : R0 −→ Z(R) is a ring homomorphism into the center Z(R) of R then we call R an R0-algebra (with respect to ). In particular, R then is an R0-module. More generally, by restriction of scalars, any R-module also is an R0-module. For any R-module M we have the two endomorphism rings EndR(M) ⊆

EndR0 (M). Both are R0-algebras with respect to the homomorphism

R0 −→ EndR(M)

a0 7−→ a0 ⋅ idM .

Lemma 3.5. Suppose that R is an R0-algebra which as an R0-module is ﬁnitely generated, and assume R0 to be noetherian; we than have: i. R is left and right noetherian;

ii. for any ﬁnitely generated R-module M its ring EndR(M) of endomorphisms is left and right noetherian and is ﬁnitely generated as an R0- module;

iii. Jac(R0)R ⊆ Jac(R).

Proof. i. Any left or right ideal of R is a submodule of the noetherian R0- module R and hence is ﬁnitely generated (cf. Prop. 1.1). ii. Step 1: We claim that, for any ﬁnitely generated R0-module M0, the

R0-module EndR0 (M0) is ﬁnitely generated. Let x1, . . . , xr be generators of the R0-module M0. Then

EndR0 (M0) −→ M0 ⊕ ... ⊕ M0 f 7−→ (f(x1), . . . , f(xr))

8 is an injective R0-module homomorphism. The right hand side is finitely generated by assumption and so is then the left hand side since R0 is noetherian. Step 2: By assumption M is finitely generated over R, and R is finitely generated over R0. Hence M is finitely generated over R0, Step 1 applies, and EndR0 (M) is a finitely generated R0-module. Since R0 is noetherian the submodule EndR(M) is finitely generated as well. Furthermore, since any left or right ideal of EndR(M) is an R0-submodule the ring EndR(M) is left and right noetherian. iii. Set L ⊆ R be a maximal left ideal. Then M := R/L is a simple R-module. By ii. the R0-module EndR(M) is finitely generated. It is nonzero since M is nonzero. The Nakayama lemma 2.3 therefore implies that Jac(R0) EndR(M) ∕= EndR(M). But EndR(M), by Schur’s lemma, is a skew field. It follows that Jac(R0) EndR(M) = {0}. Hence

Jac(R0)M = idM (Jac(R0)M) = Jac(R0) idM (M) = {0} or equivalently Jac(R0)R ⊆ L. Since L was arbitrary we obtain the assertion.

For simplicity we call an R-module complete if it is Jac(R)-adically complete.

Proposition 3.6. Suppose that R is an R0-algebra which as an R0-module is ﬁnitely generated, and assume that R0 is noetherian and complete and that R0/ Jac(R0) is artinian; for any of the rings S = R or S = EndR(M), where M is a ﬁnitely generated R-module, we then have:

i. S is left and right noetherian;

ii. S/ Jac(S) is left and right artinian;

iii. any ﬁnitely generated S-module is complete.

Proof. By Lemma 3.5.ii the case S = R contains the case S = EndR(M). Lemma 3.5.i says that R is left and right noetherian. Since R0 maps to the center of R the right ideal Jac(R0)R in fact is two-sided. Clearly R/ Jac(R0)R is ﬁnitely generated as an R0/ Jac(R0)-module. Hence R/ Jac(R0)R is left and right artinian by Prop. 1.1.iii. According to Lemma 3.5.iii the ring R/ Jac(R) is a factor ring of R/ Jac(R0)R and therefore, by Prop. 1.1.ii, is left and right artinian as well. Using Prop. 2.1.vi it also follows that

n0 Jac(R) ⊆ Jac(R0)R ⊆ Jac(R)

9 for some n0 ∈ ℕ. Because of Lemma 3.1 it therefore remains to show that any ﬁnitely generated R-module N is Jac(R0)R-adically complete. Since

n n (Jac(R0)R) N = Jac(R0) N for any n ≥ 1 this further reduces to the statement that any finitely generated R0-module N is complete. We know from Prop. 3.4 that the R0-module N is Jac(R0)- adically separated. On the other hand, by finite generation we find, for some m m m ∈ ℕ, a surjective R0-module homomorphism R0 ↠ N. With R0 also R0 is complete by assumption. We therefore may apply Lemma 3.3 and obtain that N is complete.

4 Unique decomposition

We ﬁrst introduce the following concept.

Deﬁnition. A ring A is called local if A ∖ A× is a two-sided ideal of A.

We note that a local ring A is nonzero since 1 ∈ A× whereas 0 must lie in the ideal A ∖ A×.

Proposition 4.1. For any nonzero ring A the following conditions are equivalent:

i. A is local;

ii. A ∖ A× is additively closed;

iii. A ∖ Jac(A) ⊆ A×;

iv. A/ Jac(A) is a skew ﬁeld;

v. A contains a unique maximal left ideal;

vi. A contains a unique maximal right ideal.

Proof. We note that A ∕= {0} implies that Jac(A) ∕= A is a proper ideal. i. =⇒ ii. This is obvious. ii. =⇒ iii. Let b ∈ A ∖ A× be any element. Then also −b ∕∈ A×. Suppose that 1 + ab ∕∈ A× for some a ∈ A. Using that A ∖ A× is additively closed we ﬁrst obtain that −ab ∈ A× and in particular that −a, a ∕∈ A×. It then follows that a + b ∕∈ A× and that 1 + a, 1 + b ∈ A×. But in the identity (1 + ab) + (a + b) = (1 + a)(1 + b)

10 now the right hand side is contained in A× whereas the left hand side is not. This contradiction proves that 1 + Ab ⊆ A× and therefore, by Prop. 2.1.iv, that b ∈ Jac(A). We conclude that A ⊆ A× ∪ Jac(A). iii. =⇒ iv. It immediately follows from iii. that any nonzero element in A/ Jac(A) is a unit. iv. =⇒ v.,vi. Let L ⊆ A be any maximal left, resp. right, ideal. Then Jac(A) ⊆ L ⫋ A by Prop. 2.1.iv. Since the only proper left, resp. right, ideal of a skew field is the zero ideal we obtain L = Jac(A). v., vi. =⇒ i. The unique maximal left (right) ideal, by Prop. 2.1.iv, is necessarily equal to Jac(A). Let b ∈ A ∖ Jac(A) be any element. Then Ab (bA) is not contained in any maximal left (right) ideal and hence Ab = A (bA = A). Let a ∈ A such that ab = 1 (ba = 1). Since a ∕∈ Jac(A) we may repeat this reasoning and find an element c ∈ A such that ca = 1 (ac = 1). But c = c(ab) = (ca)b = b (c = (ba)c = b(ac) = b). It follows that a ∈ A× and then also that b ∈ A×. This shows that A = Jac(A)∪A×. But the union is disjoint. We finally obtain that A ∖ A× = Jac(A) is a two-sided ideal.

We see that in a local ring A the Jacobson radical Jac(A) is the unique maximal left (right, two-sided) ideal and that A ∖ Jac(A) = A×.

Lemma 4.2. If A is nonzero and any element in A ∖ A× is nilpotent then the ring A is local.

Proof. Let b ∈ A ∖ A× and let n ≥ 1 be minimal such that bn = 0. For any a ∈ A we then have (ab)bn−1 = abn = 0. Since bn−1 ∕= 0 the element ab cannot be a unit in A. It follows that Ab ⊆ A ∖ A× and hence, by Prop. 2.1.v, that Ab ⊆ Jac(A). We thus have shown that A ∖ A× ⊆ Jac(A) which, by Prop. 4.1.iii, implies that A is local.

Lemma 4.3. (Fitting) For any R-module M and any f ∈ EndR(M) we have:

i. If M is noetherian and f is injective then f is bijective;

ii. if M is noetherian and f is surjective then f is bijective;

iii. if M is of ﬁnite length then there exists an integer n ≥ 1 such that

a. ker(f n) = ker(f n+j) for any j ≥ 0, b. im(f n) = im(f n+j) for any j ≥ 0, c. M = ker(f n) ⊕ im(f n), and

11 ∼ ∼ n = n n = d. the induced maps idM −f : ker(f ) −→ ker(f ) and f : im(f ) −→ im(f n) are bijective.

Proof. We have the increasing sequence of submodules

ker(f) ⊆ ker(f 2) ⊆ ... ⊆ ker(f n) ⊆ ... as well as the decreasing sequence of submodules

im(f) ⊇ im(f 2) ⊇ ... ⊇ im(f n) ⊇ ...

If M is artinian there must exist an n ≥ 1 such that

im(f n) = im(f n+1) = ... = im(f n+j) = ...

For any x ∈ M we then ﬁnd a y ∈ M such that f n(x) = f n+1(y). Hence f n(x − f(y)) = 0. If f is injective it follows that x = f(y). This proves M = f(M) under the assumptions in i. If M is noetherian there exists an n ≥ 1 such that

ker(f n) = ker(f n+1) = ... = ker(f n+j) = ...

Let x ∈ ker(f). If f, and hence f n, is surjective we find a y ∈ M such that x = f n(y). Then f n+1(y) = f(x) = 0, i. e., y ∈ ker(f n+1) = ker(f n). It follows that x = f n(y) = 0. This proves ker(f) = {0} under the assumptions in ii. Assuming that M is of finite length, i. e., artinian and noetherian, we at least know the existence of an n ≥ 1 satisfying a. and b. To establish c. (for any such n) we first consider any x ∈ ker(f n) ∩ im(f n). Then f n(x) = 0 and x = f n(y) for some y ∈ M. Hence y ∈ ker(f 2n) = ker(f n) which implies x = 0. This shows that

ker(f n) ∩ im(f n) = {0} .

Secondly let x ∈ M be arbitrary. Then f n(x) ∈ im(f n) = im(f 2n), i. e., f n(x) = f 2n(y) for some y ∈ M. We obtain f n(x−f n(y)) = f n(x)−f 2n(y) = 0. Hence x = (x − f n(y)) + f n(y) ∈ ker(f n) + im(f n) . n n For d. we note that idM −f : ker(f ) −→ ker(f ) has the inverse idm +f + ... + f n−1. Since ker(f) ∩ im(f n) = {0} by c., the restriction f∣ im(f n) is injective. Let x ∈ im(f n). Because of b. we ﬁnd a y ∈ M such that x = f n+1(y) = f(f n(y)) ∈ f(im(f n)).

12 Proposition 4.4. For any indecomposable R-module M of ﬁnite length the ring EndR(M) is local.

Proof. Since M ∕= {0} we have idM ∕= 0 and hence EndR(M) ∕= {0}. Let f ∈ EndR(M) be any element which is not a unit, i. e., is not bijective. According to Lemma 4.3.iii we have

M = ker(f n) ⊕ im(f n) for some n ≥ 1.

But M is indecomposable. Hence ker(f n) = M or im(f n) = M. In the latter case f would be bijective by Lemma 4.3.iii.d which leads to a contradiction. It follows that f n = 0. This shows that the assumption in Lemma 4.2 is satisfied so that EndR(M) is local. Proposition 4.5. Suppose that R is left noetherian, R/ Jac(R) is left artinian, and any finitely generated R-module is complete; then EndR(M), for any finitely generated indecomposable R-module M, is a local ring. Proof. We abbreviate J := Jac(R). By assumption the map

∼ J : M −−→= lim M/J mM M ←− is an isomorphism. Each M/J mM is a ﬁnitely generated module over the factor ring R/J m which, by Lemma 2.4, is left noetherian and left artinian. Hence M/J mM is a module of ﬁnite length by Prop. 1.1.iii. On the other hand for any f ∈ EndR(M) the R-module homomorphisms

m m fm : M/J M −→ M/J M x + J mM 7−→ f(x) + J mM are well deﬁned. The diagrams

fm+1 (1) M/J m+1M / M/J m+1M

pr pr

fm M/J mM / M/J mM obviously are commutative so that in the projective limit we obtain the R-module homomorphism

f : lim M/J mM −→ lim M/J mM ∞ ←− ←− m m (xm + J M)m 7−→ (f(xm) + J M)m .

13 Clearly the diagram

f M / M

J ∼ ∼ J M = = M f lim M/J mM ∞ lim M/J mM ←− / ←− is commutative. If follows, for example, that f is bijective if and only if f∞ is bijective. We now apply Fitting’s lemma 4.3.iii to each module M/J mM and obtain an increasing sequence of integers 1 ≤ n(1) ≤ ... ≤ n(m) ≤ ... such m that the triple (M/J M, fm, n(m)) satisﬁes the conditions a. – d. in that lemma. In particular, we have

m n(m) n(m) M/J M = ker(fm ) ⊕ im(fm ) for any m ≥ 1. The commutativity of (1) easily implies that we have the sequences of surjective (!) R-module homomorphisms

pr n(m+1) pr n(m) pr pr n(1) ... −−→ ker(fm+1 ) −−→ ker(fm ) −−→ ... −−→ ker(f1 ) and

pr n(m+1) pr n(m) pr pr n(1) ... −−→ im(fm+1 ) −−→ im(fm ) −−→ ... −−→ im(f1 ) . Be deﬁning

X := lim ker(f n(m)) and Y := lim im(f n(m)) ←− m ←− m we obtain the decomposition into R-submodules

M ∼ lim M/J mM = X ⊕ Y. = ←− But M is indecomposable. Hence X = {0} or Y = {0}. Suppose ﬁrst that X = {0}. By the surjectivity of the maps in the corresponding sequence this n(m) implies ker(fm ) = {0} for any m ≥ 1. The condition d. then says that all the fm, hence f∞, and therefore f are bijective. If, on the other hand, n(m) Y = {0} then analogously im(fm ) = {0} for all m ≥ 1, and conditions d. implies that idM −f is bijective. So far we have shown that for any f ∈ EndR(M) either f or idM −f is a unit. To prove our assertion it suﬃces, by Prop. 4.1.ii, to verify that the nonunits in EndR(M) are additively closed. Suppose therefore that

14 × × f, g ∈ EndR(M) ∖ EndR(M) are such that ℎ := f + g ∈ EndR(M) . −1 By multiplying by ℎ we reduce to the case that ℎ = idM . Then the left hand side in the identity g = idM −f is not a unit, but the right hand side is by what we have shown above. This × is a contradiction, and hence EndR(M)∖EndR(M) is additively closed. Proposition 4.6. Let

M = M1 ⊕ ... ⊕ Mr = N1 ⊕ ... ⊕ Ns be two decompositions of the R-module M into indecomposable R-modules Mi and Nj; if EndR(Mi), for any 1 ≤ i ≤ r, is a local ring we have: i. r = s; ∼ ii. there is a permutation of {1, . . . , r} such that Nj = M(j) for any 1 ≤ j ≤ r.

Proof. The proof is by induction with respect to r. The case r = 1 is trivial since M then is indecomposable. In general we have the R-module homomorphisms

pr prN Mi ⊆ j ⊆ fi : M −−−−→ Mi −−→ M and gj : M −−−−→ Nj −−→ M in EndR(M) satisfying the equation

idM = f1 + ... + fr = g1 + ... + gs .

In particular, f1 = f1g1 + ... + f1gs, which restricts to the equation

idM1 = (prM1 g1)∣M1 + ... + (prM1 gs)∣M1 in EndR(M1). But EndR(M1) is local. Hence at least one of the summands must be a unit. By renumbering we may assume that the composed map

pr pr ⊆ N1 ⊆ M1 M1 −−→ M −−−−→ N1 −−→ M −−−−→ M1 is an automorphism of M1. This implies that ∼ N1 = M1 ⊕ ker(prM1 ∣N1) .

15 =∼ Since N1 is indecomposable we obtain that g1 : M1 −−→ N1 is an isomorphism. In particular

M1 ∩ ker(g1) = M1 ∩ (N2 ⊕ ... ⊕ Ns) = {0} .

On the other hand let x ∈ N1 and write x = g1(y) with y ∈ M1. Then

g1(x − y) = x − g1(y) = 0 and hence

x = y + (x − y) ∈ M1 + ker(g1) = M1 + (N2 ⊕ ... ⊕ Ns) .

This shows that N1 ⊆ M1 + (N2 ⊕ ... ⊕ Ns), and hence

M = N1 + ... + Ns = M1 + (N2 ⊕ ... ⊕ Ns) . Together we obtain M = M1 ⊕ N2 ⊕ ... ⊕ Ns and therefore ∼ ∼ M/M1 = M2 ⊕ ... ⊕ Mr = N2 ⊕ ... ⊕ Ns . We now apply the induction hypothesis to these two decompositions of the R-module M/M1. Theorem 4.7. (Krull-Remak-Schmidt) The assumptions of Prop. 4.6 are satisfied in any of the following cases: i. M is of finite length; ii. R is left artinian and M is finitely generated; iii. R is left noetherian, R/ Jac(R) is left artinian, any finitely generated R-module is complete, and M is finitely generated;

iv. R is an R0-algebra, which is finitely generated as an R0-module, over a noetherian complete commutative ring R0 such that R0/ Jac(R0) is artinian, and M is finitely generated. Proof. i. Use Prop. 4.4. ii. This reduces to i. by Prop. 1.1.iii. iii. Use Prop. 4.5. iv. This reduces to iii. by Prop. 3.6. Example. Let K be a field and G be a finite group. The group ring K[G] is left artinian. Hence Prop. 4.6 applies: Any finitely generated K[G]-module has a ”unique” decomposition into indecomposable modules.

16 5 Idempotents and blocks

Of primary interest to us is the decomposition of the ring R itself into indecomposable submodules. This is closely connected to the existence of idempotents in R. Deﬁnition. i. An element e ∈ R is called an idempotent if e2 = e ∕= 0.

ii. Two idempotents e1, e2 ∈ R are called orthogonal if e1e2 = 0 = e2e1. ii. An idempotent e ∈ R is called primitive if e is not equal to the sum of two orthogonal idempotents. iv. The idempotents in the center Z(R) of R are called central idempotents in R. We note that eRe, for any idempotent e ∈ R, is a subring of R with unit element e. We also note that for any idempotent 1 ∕= e ∈ R the element 1−e is another idempotent, and e, 1 − e are orthogonal.

Exercise. If e1, . . . , er ∈ R are idempotents which are pairwise orthogonal then e1 + ... + er is an idempotent as well. Proposition 5.1. Let L = Re ⊆ R be a left ideal generated by an idempotent e; the map

set of all sets {e1, . . . , er} of pair- set of all decompositions ∼ wise orthogonal idempotents ei ∈ R −−→ L = L1 ⊕ ... ⊕ Lr of L

such that e1 + ... + er = e into nonzero left ideals Li

{e1, . . . , er} 7−→ L = Re1 ⊕ ... ⊕ Rer is bijective.

Proof. Suppose given a set {e1, . . . , er} in the left hand side. We have L = Re = R(e1 + ... + er) ⊆ Re1 + ... + Rer. On the other hand L ⊇ Reie = R(eie1 + ... + eier) = Rei. Hence L = Re1 + ... + Rer. To see that the sum is direct let X X aej = aiei ∈ Rej ∩ ( Rei) with a, ai ∈ R i∕=j i∕=j be an arbitrary element. Then X X aej = (aej)ej = ( aiei)ej = aieiej = 0 . i∕=j i∕=j

17 It follows that the asserted map is well deﬁned. ′ ′ To establish its injectivity let {e1, . . . , er} be another set in the left hand side such that the two decompositions

′ ′ Re1 ⊕ ... ⊕ Rer = L = Re1 ⊕ ... ⊕ Rer coincide. This means that there is a permutation of {1, , . . . , r} such that ′ Rei = Re(i) for any i. The identity ′ ′ e(1) + ... + e(r) = e = e1 + ... + er ′ ′ ′ then implies that ei = e(i) for any i or, equivalently, that {e1, . . . , er} = {e1, . . . , er}. We prove the surjectivity of the asserted map in two steps. Step 1: We assume that e = 1 and hence L = R. Suppose that R = L1 ⊕ ... ⊕ Lr is a decomposition as a direct sum of nonzero left ideals Li. We then have 1 = e1 +...+er for appropriate elements ei ∈ Li and, in particular, Rei ⊆ Li and R = R ⋅ 1 = R(e1 + ... + er) ⊆ Re1 + ... + Rer. It follows that Rei = Li. Since Li ∕= {0} we must have ei ∕= 0. Furthermore,

ei = ei ⋅ 1 = ei(e1 + ... + er) = eie1 + ... + eier for any 1 ≤ i ≤ r.

Since eiej ∈ Lj we obtain

2 ei = ei and eiej = 0 for j ∕= i .

We conclude that the set {e1, . . . , er} is a preimage of the given decomposition under the map in the assertion. Step 2: For a general e ∕= 1 we ﬁrst observe that the elements 1 − e and e are orthogonal idempotents in R. Hence, by what we have shown already, we have the decomposition R = R(1 − e) ⊕ Re = R(1 − e) ⊕ L. Suppose now that L = L1 ⊕ ... ⊕ Lr is a direct sum decomposition into nonzero left ideals Li. Then

R = R(1 − e) ⊕ L1 ⊕ ... ⊕ Lr is a decomposition into nonzero left ideals as well. By Step 1 we ﬁnd pairwise orthogonal idempotents e0, e1, . . . , er in R such that

1 = e0 + e1 + ... + er, Re0 = R(1 − e), and Rei = Li for 1 ≤ i ≤ r.

Comparing this with the identity 1 = (1 − e) + e where 1 − e ∈ Re0 and e ∈ L = L1 ⊕ ... ⊕ Lr we see that 1 − e = e0 and e = e1 + ... + er. Therefore the set {e1, . . . , er} is a preimage of the decomposition L = L1 ⊕ ... ⊕ Lr under the asserted map.

18 There is a completely analogous right ideal version, sending {e1, . . . , er} to e1R ⊕ ... ⊕ erR, of the above proposition. Corollary 5.2. For any idempotent e ∈ R the following conditions are equivalent:

i. The R-module Re is indecomposable;

ii. e is primitive;

iii. the right R-module eR is indecomposable;

iv. the ring eRe contains no idempotent ∕= e.

Proof. The equivalence of i., ii., and iii. follows immediately from Prop. 5.1 and its right ideal version. ii. =⇒ iv. Suppose that e ∕= f ∈ eRe is an idempotent. Then ef = fe = f, and e = (e − f) + f is the sum of the orthogonal idempotents e − f and f. This is a contradiction to the primitivity of e. iv. =⇒ ii. Suppose that e = e1 + e2 is the sum of the orthogonal idem- 2 2 potents e1, e2 ∈ R. Then ee1 = e1 + e2e1 = e1 and e1e = e1 + e1e2 = e1 and hence e1 = ee1e ∈ eRe. Since e1 ∕= e this again is a contradiction. Proposition 5.3. Let I = Re = eR be a two-sided ideal generated by a central idempotent e; we then have:

i. I is a subring of R with unit element e;

ii. the map

set of all sets {e1, . . . , er} set of all decompositions

of pairwise orthogonal I = I1 ⊕ ... ⊕ Ir −−→∼ idempotents ei ∈ Z(R) such that of I into nonzero two-sided

e1 + ... + er = e ideals Ii ⊆ R

{e1, . . . , er} 7−→ I = Re1 ⊕ ... ⊕ Rer

is bijective; moreover, the multiplication in I = Re1 ⊕ ... ⊕ Rer can be carried out componentwise.

Proof. i. We have I = eRe. ii. Obviously, since ei is central the ideal Rei = eiR is two-sided. Taking Prop. 5.1 into account it therefore remains to show that, if in the decomposition I = Re1 ⊕ ... ⊕ Rer with e = e1 + ... + er

19 the Rei are two-sided ideals, then the ei necessarily are central. But for any a ∈ R we have

ae1 + ... + aer = ae = ea = a(e1 + ... + er)

= (e1 + ... + er)a

= e1a + ... + era where aei and eia both lie in Rei. Hence aei = eia since the summands are uniquely determined in Rei. For the second part of the assertion let a = a1 + ... + ar and b = b1 + . . . br with ai, bi ∈ Ii be arbitrary elements. Then ai = aiei and bi = eibi and hence

ab = (a1 + ... + ar)(b1 + ... + br)

= (a1e1 + ... + arer)(e1b1 + ... + erbr) X X = aieiejbj = aieibi i,j i X = aibi . i

Corollary 5.4. A central idempotent e ∈ R is primitive in Z(R) if and only if Re is not the direct sum of two nonzero two-sided ideals of R.

Proposition 5.5. If R is left noetherian then we have:

i. 1 ∈ R can be written as a sum of pairwise orthogonal primitive idempotents;

ii. R contains only ﬁnitely many central idempotents;

iii. any two diﬀerent central idempotents which are primitive in Z(R) are orthogonal;

iv. if {e1, . . . , en} is the set of all central idempotents which are primitive in Z(R) then e1 + ... + en = 1.

Proof. i. By Lemma 1.6 we have a direct sum decomposition R = L1 ⊕ ... ⊕ Lr of R into indecomposable left ideals Li. According to Prop. 5.1 this corresponds to a decomposition of the unit element

(2) 1 = f1 + ... + fr

20 as a sum of pairwise orthogonal idempotents fi such that Li = Rfi. More- over, Cor. 5.2 says that each fi is primitive. ii. We keep the decomposition (2). Let e ∈ Z(R) be any idempotent. Then e = ef1 + ... + efr and fi = efi + (1 − e)fi. We have

2 2 2 2 2 2 (efi) = e fi = efi, ((1 − e)fi) = (1 − e) fi = (1 − e)fi, 2 2 efi(1 − e)fi = e(1 − e)fi = 0, and (1 − e)fiefi = (1 − e)efi = 0 .

But fi is primitive. Hence either efi = 0 or (1 − e)fi = 0, i. e., efi = fi. This shows that there is a subset S ⊆ {1, . . . r} such that X e = fi , i∈S and we see that there are only finitely many possibilities for e. iii. Let e1 ∕= e2 be two primitive idempotents in the ring Z(R). We then have e1 = e1e2 + e1(1 − e2) 2 2 where the summands satisfy (e1e2) = e1e2, (e1(1 − e2)) = e1(1 − e2), and e1e2e1(1−e2) = 0. Hence e1e2 = 0 or e1 = e1e2. By symmetry we also obtain e2e1 = 0 or e2 = e2e1. It follows that e1e2 = 0 or e1 = e1e2 = e2. The latter case being excluded by assumption we conclude that e1e2 = 0. iv. First we consider any idempotent f ∈ Z(R). If f is not primitive in Z(R) then we can write it as the sum of two orthogonal idempotents in Z(R). Any of the two summands either is primitive or again can be written as the sum of two new (exercise!) orthogonal idempotents in Z(R). Proceeding in this way we must arrive, because of ii., after finitely many steps at an expression of f as a sum of pairwise orthogonal idempotents which are primitive in Z(R). This, first of all, shows that the set {e1, . . . , en} is nonempty. By iii. the sum e := e1 +...+en is a central idempotent. Suppose that e ∕= 1. Then 1 − e is a central idempotent. By the initial observation we find a subset S ⊆ {1, . . . , n} such that X 1 − e = ei . i∈S For i ∈ S we then compute X ei = ei( ej) = ei(1 − e) = ei − ei(e1 + ... + en) = ei − ei = 0 j∈S which is a contradiction.

21 Let e ∈ R be a central idempotent which is primitive in Z(R). An R- module M is said to belong to the e-block of R if eM = M holds true. Exercise. If M belongs to the e-block then we have:

a. ex = x for any x ∈ M;

b. every submodule and every factor module of M also belongs to the e-block.

We now suppose that R is left noetherian. Let {e1, . . . , en} be the set of all central idempotents which are primitive in Z(R). By Prop. 5.5.iii/iv the ei are pairwise orthogonal and satisfy e1 + ... + en = 1. Let M be any R- module. Since ei is central eiM ⊆ M is a submodule which obviously belongs to the ei-block. We also have M = 1⋅M = (e1+...+en)M ⊆ e1M +...+enM and hence M = e1M + ... + enM. For X X eix = ejxj ∈ eiM ∩ ( ejM) with x, xj ∈ M j∕=i j∕=i we compute X X eix = eieix = ei ejxj = eiejxj = 0 . j∕=i j∕=i

Hence the decomposition

M = e1M ⊕ ... ⊕ enM is direct. It is called the block decomposition of M.

Remark 5.6. i. For any R-module homomorphism f : M −→ N we have f(eiM) ⊆ eiN.

ii. If the submodule N ⊆ M lies in the ei-block then N ⊆ eiM. iii. If M is indecomposable then there is a unique 1 ≤ i ≤ n such that M lies in the ei-block.

Proof. i. Since f is R-linear we have f(eiM) = eif(M) ⊆ eiN. ii. Apply i. to the inclusion eiN = N ⊆ M. iii. In this case the block decomposition can have only a single nonzero summand.

22 As a consequence of Prop. 5.3 the map

n =∼ Y R −−→ Rei i=1

a 7−→ (ae1, . . . , aen) is an isomorphism of rings. If M lies in the ei-block then

ax = a(eix) = (aei)x for any a ∈ R and x ∈ M. This means that M comes, by restriction of scalars along the projection map R −→ Rei, from an Rei-module (with the same underlying additive group). In this sense the ei-block conincides with the class of all Rei-modules. We next discuss, for arbitrary R, the relationship between idempotents in the ring R and in a factor ring R/I. Proposition 5.7. Let I ⊆ R be a two-sided ideal and suppose that either every element in I is nilpotent or R is I-adically complete; then for any idempotent " ∈ R/I there is an idempotent e ∈ R such that e + I = ". Proof. Case 1: We assume that every element in I is nilpotent. Let " = a+I and put b := 1 − a. Then ab = ba = a − a2 ∈ I, and hence (ab)m = 0 for some m ≥ 1. Since a and b commute the binomial theorem gives

2m X 2m 1 = (a + b)2m = a2m−ibi = e + f i i=0 with m 2m X 2m X 2m e := a2m−ibi ∈ aR and f := a2m−jbj . i j i=0 j=m+1 For any 0 ≤ i ≤ m and m < j ≤ 2m we have

a2m−ibia2m−jbj = ambma3m−i−jbi+j−m = (ab)ma3m−i−jbi+j−m = 0 and hence ef = 0. It follows that e = e(e+f) = e2. Moreover, ab ∈ I implies

m X 2m e + I = a2m + ( a2m−i−1bi−1)ab + I = a2m + I = "2m = ". i i=1 ∼ Case 2: We assume that R is I-adically complete. Since R −−→= lim R/In it n ←− suﬃces to construct a sequence of idempotents "n ∈ R/I , for n ≥ 2, such

23 2n n+1 n n+1 that pr("n+1) = "n and "1 = ". Because of I ⊆ I the ideal I /I in the ring R/In+1 is nilpotent. Hence we may, inductively, apply the ﬁrst case n n+1 to the idempotent "n in the factor ring R/I of the ring R/I in order to obtain "n+1. We point out that, by Prop. 2.1.v and Lemma 3.2, the assumptions in Prop. 5.7 imply that I ⊆ Jac(R).

Remark 5.8. i. Jac(R) does not contain any idempotent.

ii. Let I ⊆ Jac(R) be a two-sided ideal and e ∈ R be an idempotent; we then have:

a. e + I is an idempotent in R/I; b. if e + I ∈ R/I is primitive then e is primitive.

Proof. i. Suppose that e ∈ Jac(R) is an idempotent. Then 1 − e is an idempotent as well as a unit (cf. Prop. 2.1.iv). Multiplying the equation (1 − e)2 = 1 − e by the inverse of 1 − e shows that 1 − e = 1. This leads to the contradiction that e = 0. ii. The assertion a. is immediate from i. For b. let e = e1 + e2 with orthogonal idempotents e1, e2 ∈ R. Then e + I = (e1 + I) + (e2 + I) with (e1 + I)(e2 + I) = e1e2 + I = I. Therefore, by a., e1 + I and e2 + I are orthogonal idempotents in R/I. This, again, is a contradiction.

Lemma 5.9. i. Let e, f ∈ R be two idempotents such that e + Jac(R) = f + Jac(R); then Re =∼ Rf as R-modules.

ii. Let I ⊆ Jac(R) be a two-sided ideal and e, f ∈ R be idempotents such that the idempotents e+I, f +I ∈ R/I are orthogonal; then there exists an idempotent f ′ ∈ R such that f ′ +I = f +I and e, f ′ are orthogonal.

Proof. i. We consider the pair of R-modules L := Rfe ⊆ M := Re. Since f − e ∈ Jac(R) we have

L + Jac(R)M = Rfe + Jac(R)e = R(e + (f − e))e + Jac(R)e = Re + Jac(R)e = Re = M.

The factor module M/L being generated by a single element e + L we may apply the Nakayama lemma 2.3 and obtain Rfe = Re. On the other hand the decomposition R = Rf ⊕ R(1 − f) leads to Re = Rfe ⊕ R(1 − f)e =

24 Re⊕R(1−f)e. It follows that R(1−f)e = {0} and, in particular, (1−f)e = 0. We obtain that e = fe and, by symmetry, also f = ef . This easily implies that the R-module homomorphisms of right multiplication by f and e, respectively,

Re ⇄ Rf re 7→ ref r′fe ← r′f [ are inverse to each other. ii. We have fe ∈ I ⊆ Jac(R), hence 1 − fe ∈ R×, and we may introduce the idempotent −1 f0 := (1 − fe) f(1 − fe) . −1 −1 Obviously f0+I = f+I and f0e = (1−fe) f(e−fe) = (1−fe) (fe−fe) = 0. We put ′ f := (1 − e)f0 . Then

′ f + I = f0 − ef0 + I = (f + I) − (e + I)(f + I) = f + I.

Moreover,

′ ′ f e = (1 − e)f0e = 0 and ef = e(1 − e)f0 = 0 .

Finally

′2 2 ′ f = (1 − e)f0(1 − e)f0 = (1 − e)(f0 − f0ef0) = (1 − e)f0 = f .

Proposition 5.10. Under the assumptions of Prop. 5.7 we have: i. If e ∈ R is a primitive idempotent then the idempotent e + I ∈ R/I is primitive as well;

ii. if "1,...,"r ∈ R/I are pairwise orthogonal idempotents then there are pairwise orthogonal idempotents e1, . . . , er ∈ R such that "i = ei + I for any 1 ≤ i ≤ r.

25 Proof. i. Let e + I = "1 + "2 with orthogonal idempotents "1,"2 ∈ R/I. By Prop. 5.7 we find idempotents ei ∈ R such that "i = ei + I, for i = 1, 2. By ′ ′ Lemma 5.9.ii there is an idempotent e2 ∈ R such that e2+I = e2+I = "2 and ′ ′ e1, e2 are orthogonal. The latter implies that f := e1 + e2 is an idempotent ∼ as well. It satisfies f + I = "1 + "2 = e + I. Hence Rf = Re by Lemma 5.9.i. ′ ∼ Applying Prop. 5.1 we obtain that Re1 ⊕ Re2 = Re and we see that e is not primitive. This is a contradiction. ii. The proof is by induction with respect to r. We assume that the idempotents e1, . . . , er−1 have been constructed already. On the one hand we then have the idempotent e := e1 + ... + er−1. On the other hand we find, by Prop. 5.7, an idempotent f ∈ R such that f + I = "r. Since e + I = "1 + ... + "r−1 and "r are orthogonal there exists, by Lemma 5.9.ii, an idempotent er ∈ R such that er + I = f + I = "r and e, er are orthogonal. It remains to observe that

erei = er(eei) = (ere)ei = 0 and eier = eieer = 0 for any 1 ≤ i < r.

Proposition 5.11. Suppose that R is complete and that R/ Jac(R) is left artinian; then R is local if and only if 1 is the only idempotent in R.

Proof. We ﬁrst assume that R is local. Let e ∈ R be any idempotent. Since e ∕∈ Jac(R) by Remark 5.8.i we must have e ∈ R×. Multiplying the identity e2 = e by e−1 gives e = 1. Now let us assume, vice versa, that 1 is the only idempotent in R. As a consequence of Prop. 5.7, the factor ring R := R/ Jac(R) also has no other idempotent than 1. Therefore, by Prop. 5.1, the R-module L := R is indecomposable. On the other hand, the ring R being left artinian the module L, by Cor. 2.2, is of ﬁnite length. Hence Prop. 4.4 implies that EndR(L) is a local ring. But the map

op =∼ R −−→ EndR(R) c 7−→ [a 7→ ac]

op is an isomorphism of rings (exercise!). We obtain that R and R are local rings. Since Jac(R) = Jac(R/ Jac(R)) = {0} the ring R in fact is a skew ﬁeld. Now Prop. 4.1 implies that R is local.

Proposition 5.12. Suppose that R is an R0-algebra, which is ﬁnitely generated as an R0-module, over a noetherian complete commutative ring R0

26 such that R0/ Jac(R0) is artinian; then the map

set of all central set of all central idem- −−→≃ idempotents in R potents in R/ Jac(R0)R

e 7−→ e := e + Jac(R0)R is bijective; moreover, this bijection satisﬁes: – e, f are orthogonal if and only if e, f are orthogonal;

– e is primitive in Z(R) if and only if e is primitive in Z(R/ Jac(R0)R).

Proof. In Lemma 3.5.iii we have seen that Jac(R0)R ⊆ Jac(R). Hence Jac(R0)R, by Remark 5.8.i, does not contain any idempotent. Therefore e ∕= 0 which says that the map in the assertion is well deﬁned. To establish its injectivity let us assume that e1 = e2. Then

ei − e1 e2 = 0 and hence ei − e1e2 ∈ Jac(R0)R.

But since e1 and e2 commute we have

2 2 2 2 (ei − e1e2) = ei − 2eie1e2 + e1e2 = ei − 2e1e2 + e1e2 = ei − e1e2 .

It follows that ei − e1e2 = 0 which implies e1 = e1e2 = e2. For the surjectivity let " ∈ R := R/ Jac(R0)R be any central idempotent. In the proof of Prop. 3.6 we have seen that R is Jac(R0)R-adically complete. Hence we may apply Prop. 5.7 and obtain an idempotent e ∈ R such that e = ". We, in fact, claim the stronger statement that any such e necessarily lies in the center Z(R) of R. We have

R = Re + R(1 − e) = eRe + (1 − e)Re + eR(1 − e) + (1 − e)R(1 − e) and correspondingly

R = "R" + (1 − ")R" + "R(1 − ") + (1 − ")R(1 − ") .

But

(1 − ")R" = (1 − ")"R = {0} and "R(1 − ") = R"(1 − ") = {0} since " is central in R. It follows that (1 − e)Re and eR(1 − e) both are contained in Jac(R0)R. We see that

2 2 (1 − e)Re = (1 − e) Re ⊆ (1 − e) Jac(R0)Re = Jac(R0)(1 − e)Re

27 and similarly eR(1 − e) ⊆ Jac(R0)eR(1 − e). This means that for the two ﬁnitely generated (as submodules of R) R0-modules M := (1 − e)Re and M := eR(1−e) we have Jac(R0)M = M. The Nakayama lemma 2.3 therefore implies (1 − e)Re = eR(1 − e) = {0} and consequently that

R = eRe + (1 − e)R(1 − e) .

If we write an arbitrary element a ∈ R as a = ebe + (1 − e)c(1 − e) with b, c ∈ R then we obtain

ea = e(ebe) = ebe = (ebe)e = ae .

This shows that e ∈ Z(R). For the second half of the assertion we first note that with e, f also e, f are orthogonal for trivial reasons. Let us suppose, vice versa, that e, f are orthogonal. By Lemma 5.9.ii we find an idempotent f ′ ∈ R such that f ′ = f and e, f ′ are orthogonal. According to what we have established above f ′ necessarily is central. Hence the injectivity of our map forces f ′ = f. So e, f are orthogonal. Finally, if e = e1 + e2 with orthogonal central idempotents e1, e2 then obviously e = e1 + e2 with orthogonal idempotents e1, e2 ∈ Z(R). Vice versa, let e = "1 + "2 with orthogonal idempotents "1,"2 ∈ Z(R). By the surjectivity of our map we find idempotents ei ∈ Z(R) such that ei = "i. We have shown already that e1, e2 necessarily are orthogonal. Since e1 + e2 then is a central idempotent with e1 + e2 = e1 + e2 = "1 + "2 = e the injectivity of our map forces e1 + e2 = e. Remark 5.13. Let e ∈ R be any idempotent, and let L ⊆ M be R-modules; then eM/eL =∼ e(M/L) as Z(R)-modules.

Proof. We have the obviously well deﬁned and surjective Z(R)-module homomorphism

eM/eL −→ e(M/L) ex + eL 7−→ e(x + L) = ex + L.

If ex + L = L then ex ∈ L and hence ex = e(ex) ∈ eL. This shows that the map is injective as well.

Keeping the assumptions of Prop. 5.12 we consider the block decomposition M = e1M ⊕ ... ⊕ enM

28 of any R-module M. It follows that

M/ Jac(R0)M = e1M/ Jac(R0)e1M ⊕ ... ⊕ enM/ Jac(R0)enM

= e1M/e1 Jac(R0)M ⊕ ... ⊕ enM/en Jac(R0)M

= e1(M/ Jac(R0)M) ⊕ ... ⊕ en(M/ Jac(R0)M) is the block decomposition of the R/ Jac(R0)R-module M/ Jac(R0)M. If eiM = {0} then obviously ei(M/ Jac(R0)M) = {0}. Vice versa, let us suppose that ei(M/ Jac(R0)M) = {0}. Then eiM ⊆ Jac(R0)M and hence 2 eiM = ei M ⊆ Jac(R0)eiM.

If M is finitely generated as an R-module then eiM is finitely generated as an R0-module and the Nakayama lemma 2.3 implies that eiM = {0}. We, in particular, obtain that a finitely generated R-module M belongs to the ei-block if and only if M/ Jac(R0)M belongs to the ei-block.

6 Projective modules

We ﬁx an R-module X. For any R-module M, resp. for any R-module homomorphism g : L −→ M, we have the Z(R)-module

HomR(X,M) , resp. the Z(R)-module homomorphism

HomR(X, g) : HomR(X,L) −→ HomR(X,M) f 7−→ g ∘ f .

g Lemma 6.1. For any exact sequence 0 −→ L −→ℎ M −→ N of R-modules the sequence

HomR(X,ℎ) HomR(X,g) 0 −→ HomR(X,L) −−−−−−−−→ HomR(X,M) −−−−−−−−→ HomR(X,N) is exact as well. Proof. Whenever a composite

f X −→ L −→ℎ M is the zero map we must have f = 0 since ℎ is injective. This shows the injectivity of HomR(X, ℎ). We have

HomR(X, g) ∘ HomR(X, ℎ) = HomR(X, g ∘ ℎ) = HomR(X, 0) = 0 .

29 Hence the image of HomR(X, ℎ) is contained in the kernel of HomR(X, g). Let now f : X −→ M be an R-module homomorphism such that g ∘ f = 0. Then im(f) ⊆ ker(g). Hence for any x ∈ X we ﬁnd, by the exactness of the original sequence, a unique fL(x) ∈ L such that

f(x) = ℎ(fL(x)) .

This fL : X −→ L is an R-module homomorphism such that ℎ ∘ fL = f. Hence image of HomR(X, ℎ) = kernel of HomR(X, g) .

Example. Let R = ℤ,X = ℤ/nℤ for some n ≥ 2, and g : ℤ −→ ℤ/nℤ be the surjective projection map. Then Homℤ(X, ℤ) = {0} but Homℤ(X, ℤ/nℤ) ∋ idX ∕= 0. Hence the map Homℤ(X, g) : Homℤ(X, ℤ) −→ Homℤ(X, ℤ/nℤ) cannot be surjective.

Deﬁnition. An R-module P is called projective if, for any surjective R- module homomorphism g : M −→ N, the map

HomR(P, g) : HomR(P,M) −→ HomR(P,N) is surjective.

In slightly more explicit terms, an R-module P is projective if and only if any exact diagram of the form

P f | | f ′ | ~| g M / N / 0 can be completed, by an oblique arrow f, to a commutative diagram.

Lemma 6.2. For an R-module P the following conditions are equivalent:

i. P is projective;

ii. for any surjective R-module homomorphism ℎ : M −→ P there exists an R-module homomorphism s : P −→ M such that ℎ ∘ s = idP .

30 Proof. i. =⇒ ii. We obtain s by contemplating the diagram

P s } } id } P ~} ℎ M / P / 0 . ii. =⇒ i. Let P

f ′ g M / N / 0 be any exact “test diagram”. In the direct sum M ⊕P we have the submodule

M ′ := {(x, y) ∈ M ⊕ P : g(x) = f ′(y)} .

The diagram ℎ((x,y)):=y M ′ / P

f ′′((x,y)):=x f ′ g M / N is commutative. We claim that the map ℎ is surjective. Let y ∈ P be an arbitrary element. Since g is surjective we ﬁnd an x ∈ M such that g(x) = f ′(y). Then (x, y) ∈ M ′ with ℎ((x, y)) = y. Hence, by assumption, there is ′ ′′ an s : P −→ M such that ℎ ∘ s = idP . We deﬁne f := f ∘ s and have

′′ ′ ′ ′ g ∘ f = g ∘ f ∘ s = f ∘ ℎ ∘ s = f ∘ idP = f .

In the situation of Lemma 6.2.ii we see that P is isomorphic to a direct summand of M via the R-module isomorphism

∼ ker(ℎ) ⊕ P −−→= M (x, y) 7−→ x + s(y)

(exercise!). Deﬁnition. An R-module F is called free if there exists an R-module isomorphism ∼ F = ⊕i∈I R for some index set I.

31 Example. If R = K is a ﬁeld then, by the existence of bases for vector spaces, any K-module is free. On the other hand for R = ℤ the modules ℤ/nℤ, for any n ≥ 2, are neither free nor projective.

Lemma 6.3. Any free R-module F is projective. ∼ Proof. If F = ⊕i∈I R then let ei ∈ F be the element which corresponds to the tuple (..., 0, 1, 0,...) with 1 in the i-th place and zeros elsewhere. The set {ei}i∈I is an “R-basis” of the module F . In particular, for any R-module M, the map

=∼ Y HomR(F,M) −−→ M i∈I

f 7−→ (f(ei))i

g is bijective. For any surjective R-module homomorphism M −→ N the lower horizontal map in the commutative diagram

HomR(F,g) HomR(F,M) / HomR(F,N)

=∼ =∼ Q (xi)i7−→(g(xi))i Q i∈I M / i∈I N obviously is surjective. Hence the upper one is surjective, too.

Proposition 6.4. An R-module P is projective if and only if it isomorphic to a direct summand of a free module.

Proof. First we suppose that P is projective. For a suﬃciently large index set I we ﬁnd a surjective R-module homomorphism

ℎ : F := ⊕i∈I R −→ P.

By Lemma 6.2 there exists an R-module homomorphism s : P −→ F such that ℎ ∘ s = idP , and ker(ℎ) ⊕ P =∼ F. Vice versa, let P be isomorphic to a direct summand of a free R-module F . Any module isomorphic to a projective module itself is projective (exercise!). Hence we may assume that

F = P ⊕ P ′

32 ⊆ for some submodule P ′ ⊆ F . We then have the inclusion map i : P −−→ F as well as the projection map pr : F −→ P . We consider any exact “test diagram” P

f ′ g M / N / 0 . By Lemma 6.3 the extended diagram

F pr f˜ P ′ f Ö g M / N / 0 can be completed to a commutative diagram by an oblique arrow f˜. Then

g ∘ (f˜∘ i) = (g ∘ f˜) ∘ i = f ′ ∘ pr ∘i = f ′ which means that the diagram

P f:=f˜∘i || || f ′ || ~|| g M / N is commutative. Hence P is projective.

Example. Let e ∈ R be an idempotent. Then R = Re ⊕ R(1 − e). Hence Re is a projective R-module.

Corollary 6.5. If P1,P2 are two R-modules then the direct sum P1 ⊕ P2 is projective if and only if P1 and P2 are projective. Proof. If ′ ∼ (P1 ⊕ P2) ⊕ P = F for some free R-module F then visibly P1 and P2 both are isomorphic to direct summands of F as well and hence are projective. If on the other hand

′ ∼ ′ ∼ P1 ⊕ P1 = ⊕i∈I1 R and P2 ⊕ P2 = ⊕i∈I2 R

33 then ′ ′ ∼ (P1 ⊕ P2) ⊕ (P1 ⊕ P2) = ⊕i∈I1∪I2 R.

Lemma 6.6. (Schanuel) Let

ℎ1 g1 ℎ2 g2 0 −→ L1 −−→ P1 −−→ N −→ 0 and 0 −→ L2 −−→ P2 −−→ N −→ 0 be two short exact sequences of R-modules with the same right hand term ∼ N; if P1 and P2 are projective then L2 ⊕ P1 = L1 ⊕ P2. Proof. The R-module

M := {(x1, x2) ∈ P1 ⊕ P2 : g1(x1) = g2(x2)} sets in the two short exact sequences

y7−→(0,ℎ2(y)) (x1,x2)7−→x1 0 −→ L2 −−−−−−−−−→ M −−−−−−−−−→ P1 −→ 0 and y7−→(ℎ1(y),0) (x1,x2)7−→x2 0 −→ L1 −−−−−−−−−→ M −−−−−−−−−→ P2 −→ 0 . By applying Lemma 6.2 we obtain ∼ ∼ L2 ⊕ P1 = M = L1 ⊕ P2 .

Lemma 6.7. Let R −→ R′ be any ring homomorphism; if P is a projective ′ ′ R-module then R ⊗R P is a projective R -module. ∼ Proof. Using Prop. 6.4 we write P ⊕ Q = ⊕i∈I R and obtain ′ ′ ′ ∼ ′ (R ⊗R P ) ⊕ (R ⊗R Q) = R ⊗R (P ⊕ Q) = R ⊗R (⊕i∈I R) ′ ′ = ⊕i∈I (R ⊗R R) = ⊕i∈I R .

Deﬁnition. i. An R-module homomorphism f : M −→ N is called essential if it is surjective but f(L) ∕= N for any proper submodule L ⫋ M. ii. A projective cover of an R-module M is an essential R-module homomorphism f : P −→ M where P is projective.

34 Lemma 6.8. Let f : P → M and f ′ : P ′ → M be two projective covers; ∼ then there exists an R-module isomorphism g : P ′ −−→= P such that f ′ = f ∘g.

Proof. The “test diagram”

P ′ g } } f ′ } ~} P / M / 0 f shows the existence of a homomorphism g such that f ′ = f ∘ g. Since f ′ is surjective we have f(g(P ′)) = M, and since f is essential we deduce that g(P ′) = P . This shows that g is surjective. Then, by Lemma 6.2, there exists ′ an R-module homomorphism s : P −→ P such that g ∘ s = idP . We have

f ′(s(P )) = f(g(s(P ))) = f(P ) = M.

Since f ′ is essential this implies s(P ) = P ′. Hence s and g are isomorphisms.

Remark 6.9. Let f : M −→ N be a surjective R-module homomorphism between ﬁnitely generated R-modules; if

ker(f) ⊆ Jac(R)M then f is essential.

Proof. Let L ⊆ M be a submodule such that f(L) = N. Then

M = L + ker(f) = L + Jac(R)M.

Hence L = M by the Nakayama lemma 2.3.

Proposition 6.10. Suppose that R is complete and that R/ Jac(R) is left artinian; then any ﬁnitely generated R-module M has a projective cover; more precisely, there exists a projective cover f : P −→ M such that the ∼ induced map P/ Jac(R)P −−→= M/ Jac(R)M is an isomorphism.

Proof. By Cor. 2.2 and Prop. 5.5.i we may write 1 + Jac(R) = "1 + ... + "r as a sum of pairwise orthogonal primitive idempotents "i ∈ R := R/ Jac(R). According to Prop. 5.1 and Cor. 5.2 we then have

R = R"1 ⊕ ... ⊕ R"r

35 where the R-modules R"i are indecomposable. But R is semisimple by Prop. 2.1.vi. Hence the indecomposable R-modules R"i in fact are simple, and all simple R-modules occur, up to isomorphism, among the R"i. On the other hand, M/ Jac(R)M is a semisimple R-module by Prop. 1.4.iii and as such is a direct sum M/ Jac(R)M = L1 ⊕ ... ⊕ Ls of simple submodules Lj. For any 1 ≤ j ≤ s we ﬁnd an 1 ≤ i(j) ≤ r such that ∼ Lj = R"i(j) .

By Prop. 5.7 there exist idempotents e1, . . . , er ∈ R such that ei + Jac(R) = "i for any 1 ≤ i ≤ r. Using Remark 5.13 we now consider the R-module isomorphism

s s ∼ s ∼ s ∼ f :(⊕j=1Rei(j))/ Jac(R)(⊕j=1Rei(j)) = ⊕j=1R"i(j) = ⊕j=1Lj = M/ Jac(R)M. It sits in the “test diagram”

s P := ⊕j=1Rei(j) o o o pr o o f o o (⊕s Re )/ Jac(R)(⊕s Re ) o o j=1 i(j) j=1 i(j) o o o =∼ f o o wo o M pr / M/ Jac(R)M / 0 .

Each Rei is a projective R-module. Hence P is a projective R-module by Cor. 6.5. We therefore ﬁnd an R-module homomorphism f : P −→ M which makes the above diagram commutative. By construction it induces the isomorphism f modulo Jac(R). It remains to show that f is essential. Since f ∘ pr is surjective we have f(P ) + Jac(R)M = M, and the Nakayama lemma 2.3 implies that f is surjective. Moreover, ker(f) ⊆ Jac(R)P by construction. Hence f is essential by Remark 6.9.

7 Grothendieck groups

We ﬁrst recall that over any set S one has the free abelian group ℤ[S] with basis S given by X ℤ[S] = { mss : ms ∈ ℤ, all but ﬁnitely many ms are equal to zero} s∈S

36 and X X X ( mss) + ( nss) = (ms + ns)s . s∈S s∈S s∈S Its universal property is the following: For any map of sets : S −→ B from S into some abelian group B there is a unique homomorphism of abelian groups ˜ : ℤ[S] −→ B such that ˜∣S = . Let now A be any ring and let ℳ be some class of A-modules. It is partitioned into isomorphism classes where the isomorphism class {M} of a module M in ℳ consists of all modules in ℳ which are isomorphic to M. We let ℳ/ =∼ denote the set of all isomorphism classes of modules in ℳ, ∼ and we form the free abelian group ℤ[ℳ] := ℤ[ℳ/ =]. In ℤ[ℳ] we consider the subgroup Rel generated by all elements of the form

{M} − {L} − {N} whenever there is a short exact sequence of A-module homomorphisms

0 −→ L −→ M −→ N −→ 0 with L, M, N in ℳ .

The corresponding factor group

G0(ℳ) := ℤ[ℳ]/ Rel is called the Grothendieck group of ℳ. We deﬁne [M] ∈ G0(ℳ) to be the image of {M}. The elements [M] are generators of the abelian group G0(ℳ), and for any short exact sequence as above one has the identity

[M] = [L] + [N] in G0(ℳ). Remark. For any A-modules L, N we have the short exact sequence 0 → L → L ⊕ N → N → 0 and therefore the identity [L ⊕ N] = [L] + [N] in G0(ℳ) provided L, N, and L ⊕ N lie in the class ℳ. For us two particular cases of this construction will be most important. In the first case we take MA to be the class of all A-modules of finite length, and we define R(A) := G0(MA) . The set Aˆ of isomorphism classes of simple A-modules obviously is a subset ∼ ˆ of MA/ =. Hence ℤ[A] ⊆ ℤ[MA] is a subgroup, and we have the composed map ˆ ⊆ pr ℤ[A] −−→ ℤ[MA] −−→ R(A) .

37 =∼ Proposition 7.1. The above map ℤ[Aˆ] −−→ R(A) is an isomorphism.

Proof. We define an endomorphism of ℤ[MA] as follows. By the universal property of ℤ[MA] we only need to define ({M}) ∈ ℤ[MA] for any A- module M of finite length. Let {0} = M0 ⫋ M1 ⫋ ... ⫋ Mn = M be a composition series of M. According to the Jordan-HölderProp. 1.2 the isomorphism classes {M1}, {M2/M1},..., {M/Mn−1} do not depend on the choice of the series. We put

({M}) := {M1} + {M2/M1} + ... + {M/Mn−1} , and we observe:

– The modules M1,M2/M1, . . . , M/Mn−1 are simple. Hence we have im() ⊆ ℤ[Aˆ]. – If M is simple then obviously ({M}) = {M}. It follows that the ˆ endomorphism of ℤ[MA] is an idempotent with image equal to ℤ[A], and therefore ˆ (3) ℤ[MA] = im() ⊕ ker() = ℤ[A] ⊕ ker() .

– The exact sequences

0 −→ M1 −→ M −→ M/M1 −→ 0

0 −→ M2/M1 −→ M/M1 −→ M/M2 −→ 0 . . .

0 −→ Mn−2/Mn−1 −→ M/Mn−2 −→ M/Mn−1 −→ 0 imply the identities

[M] = [M1] + [M/M1], [M/M1] = [M2/M1] + [M/M2],...,

[M/Mn−2] = [Mn−1/Mn−2] + [M/Mn−1] in R(A). It follows that

[M] = [M1] + [M2/M1] + ... + [M/Mn−1] and therefore {M} − ({M}) ∈ Rel. We obtain ˆ (4) ℤ[MA] = im() + Rel = ℤ[A] + Rel .

38 f g – Finally, let 0 −→ L −→ M −→ N −→ 0 be a short exact sequence with L, M, N in MA. Let {0} = L0 ⫋ ... ⫋ Lr = L and {0} = N0 ⫋ ... ⫋ Ns = N be composition series. Then

{0} = M0 ⫋ M1 := f(L1) ⫋ ... ⫋ Mr := f(L) −1 −1 ⫋ Mr+1 := g (N1) ⫋ ... ⫋ Mr+s−1 := g (Ns−1) ⫋ Mr+s := M

is a composition series of M with ( ∼ Li/Li−1 for 1 ≤ i ≤ r, Mi/Mi−1 = Ni−r/Ni−r−1 for r < i ≤ r + s.

Hence

r+s r s X X X ({M}) = {Mi/Mi−1} = {Li/Li−1} + {Nj/Nj−1} i=1 i=1 j=1 = ({L}) + ({N})

which shows that ({M} − {L} − {N}) = 0. It follows that

(5) Rel ⊆ ker() .

The formulae (3) - (5) together imply ˆ ℤ[MA] = ℤ[A] ⊕ Rel which is our assertion.

In the second case we take ℳA to be the class of all ﬁnitely generated projective A-modules, and we deﬁne

K0(A) := G0(ℳA) .

Remark. As a consequence of Lemma 6.2.ii the subgroup Rel ⊆ ℤ[ℳA] in this case is generated by all elements of the form

{P ⊕ Q} − {P } − {Q} where P and Q are arbitrary ﬁnitely generated projective A-modules. Let A˜ denote the set of isomorphism classes of ﬁnitely generated inde- ˜ composable projective A-modules. Then ℤ[A] is a subgroup of ℤ[ℳA].

39 Lemma 7.2. If A is left noetherian then the classes [P ] for {P } ∈ A˜ are generators of the abelian group K0(A). Proof. Let P be any ﬁnitely generated projective A-module. By Lemma 1.6 we have P = P1 ⊕ ... ⊕ Ps with ﬁnitely generated indecomposable A-modules Pi. The Prop. 6.4 implies that the Pi are projective. Hence {Pi} ∈ A˜. As remarked earlier we have

[P ] = [P1] + ... + [Ps] in K0(A). Remark. Under the assumptions of the Krull-Remak-Schmidt Thm. 4.7 (e. g., if A is left artinian) an argument completely analogous to the proof of Prop. 7.1 shows that the map

˜ =∼ ℤ[A] −−→ K0(A) {P } 7−→ [P ] is an isomorphism. But we will see later that, since the unique decomposition into indecomposable modules is needed only for projective modules, weaker assumptions suﬃce for this isomorphism.

Remark 7.3. If A is semisimple we have:

i. A˜ = Aˆ;

ii. any A-module is projective;

iii. K0(A) = R(A). Proof. By Prop. 1.4.iii any A-module is semisimple. In particular, any indecomposable A-module is simple which proves i. Furthermore, any simple A-module is isomorphic to a module Ae for some idempotent e ∈ A and hence is projective (compare the proof of Prop. 6.10). It follows that any A-module is a direct sum of projective A-modules and therefore is projective (extend the proof of Cor. 6.5 to arbitrarily many summands!). This establishes ii. For iii. it remains to note that by Cor. 2.2 the class of all ﬁnitely generated (projective) A-modules coincides with the class of all A-modules of ﬁnite length.

40 Suppose that A is left artinian. Then, by Cor. 2.2, any ﬁnitely generated A-module is of ﬁnite length. Hence the homomorphism

K0(A) −→ R(A) [P ] 7−→ [P ] is well deﬁned. Using the above Remark as well as Prop. 7.1 we may introduce the composed homomorphism

˜ =∼ =∼ ˆ cA : ℤ[A] −−→ K0(A) −→ R(A) −−→ ℤ[A] .

It is called the Cartan homomorphism of the left artinian ring A. If X cA({P }) = n{M}{M} {M}∈Aˆ then the integer n{M} is the multiplicity with which the simple A-module M occurs, up to isomorphism, as a subquotient in any composition series of the ﬁnitely generated indecomposable projective module P . Let A −→ B be a ring homomorphism between arbitrary rings A and B. By Lemma 6.7 the map ∼ ∼ ℳA/ = −→ ℳB/ =

{P } 7−→ {B ⊗A P } and hence the homomorphism

ℤ[ℳA] −→ ℤ[ℳB] {P } 7−→ {B ⊗A P } ∼ are well deﬁned. Because of B ⊗A (P ⊕ Q) = (B ⊗A P ) ⊕ (B ⊗A Q) the latter map respects the subgroups Rel in both sides. We therefore obtain a well deﬁned homomorphism

K0(A) −→ K0(B)

[P ] 7−→ [B ⊗A P ] .

Exercise. Let I ⊆ A be a two-sided ideal. For the projection homomorphism A −→ A/I and any A-module M we have

A/I ⊗A M = M/IM .

41 In particular, the homomorphism

K0(A) −→ K0(A/I) [P ] 7−→ [P/IP ] is well deﬁned.

Proposition 7.4. Suppose that A is complete and that A := A/ Jac(A) is left artinian; we then have:

i. The maps

ˆ =∼ A˜ −→ A and K0(A) −−→ K0(A) = R(A) {P } 7−→ {P/ Jac(A)P } [P ] 7−→ [P/ Jac(A)P ]

are bijective;

ii. the inverses of the maps in i. are given by sending the isomorphism class {M} of an A-module M of ﬁnite length to the isomorphism class of a projective cover of M as an A-module; ˜ =∼ iii. ℤ[A] −−→ K0(A). Proof. First of all we note that A is semisimple by Prop. 2.1.iii. We already have seen that the map

: K0(A) −→ K0(A) = R(A) [P ] 7−→ [P/ Jac(A)P ] is well defined. If M is an arbitrary A-module of finite length then by Prop. f 6.10 we find a projective cover PM −→ M of M as an A-module such that

=∼ (6) PM / Jac(A)PM −−→ M.

The proof of Prop. 6.10 shows that PM in fact is a ﬁnitely generated A- module. According to Lemma 6.8 the isomorphism class {PM } only depends on the isomorphism class {M}. We conclude that

ℤ[ℳA] −→ ℤ[ℳA]

{M} 7−→ {PM }

42 is a well deﬁned homomorphism. If N is a second A-module of ﬁnite length g with projective cover PN −→ N as above then

f⊕g PM ⊕ PN −−−→ M ⊕ N is surjective with

(PM ⊕ PN )/ Jac(A)(PM ⊕ PN ) = PM / Jac(A)PM ⊕ PN / Jac(A)PN =∼ M ⊕ N.

It follows that ker(f ⊕ g) = Jac(A)(PM ⊕ PN ). This, by Remark 6.9, implies that f ⊕ g is essential. Using Cor. 6.5 we see that f ⊕ g is a projective cover of M ⊕ N as an A-module. Hence we have

{PM ⊕ PN } = {PM⊕N } . This means that the above map respects the subgroups Rel in both sides and consequently induces a homomorphism

: K0(A) −→ K0(A)

[M] 7−→ [PM ] .

But it also shows that M is a simple A-module if PM is an indecomposable A-module. The isomorphism (6) says that

∘ = id . K0(A) On the other hand, let P be any ﬁnitely generated projective A-module. As a consequence of Remark 6.9 the projection map P −→ M := P/ Jac(A)P is essential and hence a projective cover. We then deduce from Lemma 6.8 that {P } = {PM } which means that

∘ = idK0(A) . It follows that is an isomorphism with inverse . We also see that if P = P1 ⊕ P2 is decomposable then M = P1/ Jac(A)P1 ⊕ P2/ Jac(A)P2 is decomposable as well. This establishes the assertions i. and ii. For iii. we consider the commutative diagram

=∼ K0(A) / R(A) O O =∼ ∼ ˜ = ˆ ℤ[A] / ℤ[A]

43 where the horizontal isomorphisms come from i. and the right vertical isomorphism was shown in Prop. 7.1. Hence the left vertical arrow is an isomorphism as well.

Corollary 7.5. Suppose that A is complete and that A/ Jac(A) is left artinian; then A = P1 ⊕ ... ⊕ Pr decomposes into a direct sum of finitely many finitely generated indecomposable projective A-modules Pj, and any finitely generated indecomposable projective A-module is isomorphic to one of the Pj. Proof. The projection map A −→ A/ Jac(A) is a projective cover. We now decompose the semisimple ring

A/ Jac(A) = M1 ⊕ ... ⊕ Mr as a direct sum of ﬁnitely many simple modules Mj, and we choose projective covers Pj −→ Mj. Then P1 ⊕ ... ⊕ Pr is a projective cover of A/ Jac(A) and consequently is isomorphic to A. If P is an arbitrary ﬁnitely generated indecomposable projective A-module then P is a projective cover of the simple module P/ Jac(A)P . The latter has to be isomorphic to some Mj. ∼ Hence P = Pj. Assuming that A is left artinian let us go back to the Cartan homomorphism ˆ ∼ ∼ ˆ cA : ℤ[A] = K0(A) −→ R(A) = ℤ[A] .

By Cor. 7.5 the set A˜ = {{P1},..., {Pt}} is ﬁnite. We put

Mj := Pj/ Jac(A)Pj .

Then, by Prop. 7.4.i, {M1},..., {Mt} are exactly the isomorphism classes of simple A/ Jac(A)-modules. But due to the deﬁnition of the Jacobson radical the simple A/ Jac(A)-modules coincide with the simple A-modules, i. e.,

Aˆ = {{M1},..., {Mt}} . The Cartan homomorphism therefore is given by an integral matrix

CA = (cij)1≤i,j≤t ∈ Mt×t(ℤ) deﬁned by the equations

cA({Pj}) = c1j{M1} + ... + ctj{Mt} .

The matrix CA is called the Cartan matrix of A.

44 Chapter II The Cartan-Brauer triangle

Let G be a ﬁnite group. Over any commutative ring R we have the group ring X R[G] = { agg : ag ∈ R} g∈G with addition X X X ( agg) + ( bgg) = (ag + bg)g g∈G g∈G g∈G and multiplication X X X X ( agg)( bgg) = ( aℎbℎ−1g)g . g∈G g∈G g∈G ℎ∈G

We ﬁx an algebraically closed ﬁeld k of characteristic p > 0. The modular representation theory of G over k is the module theory of the group ring k[G]. This is our primary object of study in the following.

8 The setting

The main technical tool of our investigation will be a (0, p)-ring R for k which is a complete local commutative integral domain R such that

– the maximal ideal mR ⊆ R is principal,

– R/mR = k, and – the ﬁeld of fractions of R has characteristic zero.

j Exercise. The only ideals of R are mR for j ≥ 0 and {0}. We note that there must exist an integer e ≥ 1 – the ramification index of e R – such that Rp = mR. There is, in fact, a canonical (0, p)-ring W (k) for k – its ring of Witt vectors – with the additional property that mW (k) = W (k)p. Let K/K0 be any finite extension of the field of fractions K0 of W (k). Then

R := {a ∈ K : NormK/K0 (a) ∈ W (k)} is a (0, p)-ring for k with ramiﬁcation index equal to [K : K0]. Proofs for all of this can be found in §3 - 6 of [TdA].

45 In the following we ﬁx a (0, p)-ring R for k. We denote by K the ﬁeld of fractions of R and by R a choice of generator of mR, i. e., mR = RR. The following three group rings are now at our disposal:

K[G] O ⊆ pr R[G] / k[G] such that

K[G] = K ⊗R R[G] and k[G] = R[G]/RR[G] . As explained before Prop. 7.4 there are the corresponding homomorphisms between Grothendieck groups

K0(K[G]) O [P ]7−→[K⊗RP ] K0(R[G]) / K0(k[G]) . [P ]7−→[P/RP ] For the vertical arrow observe that, quite generally for any R[G]-module M, we have

K[G] ⊗R[G] M = K ⊗R R[G] ⊗R[G] M = K ⊗R M. We put RK (G) := R(K[G]) and Rk(G) := R(k[G]) . Since K has characteristic zero the group ring K[G] is semisimple, and we have RK (G) = K0(K[G]) by Remark 7.3.iii. On the other hand, as a ﬁnite dimensional k-vector space the group ring k[G] of course is left and right artinian. In particular we have the Cartan homomorphism

cG : K0(k[G]) −→ Rk(G) [P ] 7−→ [P ] . Hence, so far, there is the diagram of homomorphisms

RK (G) Rk(G) O O cG K0(R[G]) / K0(k[G]) .

46 Clearly, R[G] is an R-algebra which is ﬁnitely generated as an R-module. Let us collect some of what we know in this situation.

– (Prop. 3.6) R[G] is left and right noetherian, and any ﬁnitely generated R[G]-module is complete as well as R[G]R-adically complete. – (Thm. 4.7) The Krull-Remak-Schmidt theorem holds for any ﬁnitely generated R[G]-module.

– (Prop. 5.5) 1 ∈ R[G] can be written as a sum of pairwise orthogonal primitive idempotents; the set of all central idempotents in R[G] is ﬁnite; 1 is equal to the sum of all primitive idempotents in Z(R[G]); any R[G]-module has a block decomposition.

– (Prop. 5.7) For any idempotent " ∈ k[G] there is an idempotent e ∈ R[G] such that " = e + R[G]R. – (Prop. 5.12) The projection map R[G] −→ k[G] restricts to a bijection between the set of all central idempotents in R[G] and the set of all central idempotents in k[G]; in particular, the block decomposition of an R[G]-module M reduces modulo R[G]R to the block decomposition of the k[G]-module M/RM. – (Prop. 6.10) Any ﬁnitely generated R[G]-module M has a projective ∼ cover P −→ M such that P/ Jac(R[G])P −−→= M/ Jac(R[G])M is an isomorphism.

We emphasize that R[G]R ⊆ Jac(R[G]) by Lemma 3.5.iii. Moreover, the ideal Jac(k[G]) = Jac(R[G])/R[G]R in the left artinian ring k[G] is nilpotent by Prop. 2.1.vi. We apply Prop. 7.4 to the rings R[G] and k[G] and we see that the maps {P } 7−→ {P/RP } 7−→ {P/ Jac(R[G])P } induce the commutative diagram of bijections between ﬁnite sets

≃ R][G] / kg[G] K KK tt KK tt KK tt ≃ KK tt ≃ K% ytt R[[G] = kd[G]

47 as well as the commutative diagram of isomorphisms

=∼ ℤ[R][G]] / ℤ[kg[G]]

=∼ =∼ =∼ K0(R[G]) / K0(k[G]) .

For purposes of reference we state the last fact as a proposition. =∼ Proposition 8.1. The map : K0(R[G]) −−→ K0(k[G]) is an isomorphism; its inverse is given by sending [M] to the class of a projective cover of M as an R[G]-module. We deﬁne the composed map

−1 eG : K0(k[G]) −−−→ K0(R[G]) −−→ K0(K[G]) = RK (G) . Remark 8.2. Any ﬁnitely generated projective R-module is free. Proof. Since R is an integral domain 1 is the only idempotent in R. Hence the free R-module R is indecomposable by Cor. 5.2. On the other hand, according to Prop. 7.4.i the map R˜ −−→≃ kˆ

{P } 7−→ {P/RP } is bijective. Obviously, k is up to isomorphism the only simple k-module. Hence R is up to isomorphism the only finitely generated indecomposable projective R-module. An arbitrary finitely generated projective R-module P , by Lemma 1.6, is a finite direct sum of indecomposable ones. It follows that P must be isomorphic to some Rn.

9 The triangle

We already have the two sides

RK (G) Rk(G) eLLL ss9 LLL sss eG LL sscG LL sss K0(k[G]) of the triangle. To construct the third side we ﬁrst introduce the following notion.

48 Deﬁnition. Let V be a ﬁnite dimensional K-vector space; a lattice L in V is an R-submodule L ⊆ V for which there exists a K-basis e1, . . . , ed of V such that L = Re1 + ... + Red . Obviously, any lattice is free as an R-module. Furthermore, with L also aL, for any a ∈ K×, is a lattice in V .

Lemma 9.1. i. Let L be an R-submodule of a K-vector space V ; if L is ﬁnitely generated then L is free.

ii. Let L ⊆ V be an R-submodule of a ﬁnite dimensional K-vector space V ; if L is ﬁnitely generated as an R-module and L generates V as a K-vector space then L is a lattice in V .

iii. For any two lattices L and L′ in V there is an integer m ≥ 0 such that m ′ R L ⊆ L . Proof. i. and ii. Let d ≥ 0 be the smallest integer such that the R-module L has d generators e1, . . . , ed. The R-module homomorphism

Rd −→ L

(a1, . . . , ad) 7−→ a1e1 + ... + aded is surjective. Suppose that (a1, . . . , ad) ∕= 0 is an element in its kernel. Since at least one ai is nonzero the integer

j ℓ := max{j ≥ 0 : a1, . . . , ad ∈ mR}

ℓ × is deﬁned. Then ai = Rbi with bi ∈ R, and bi0 ∈ R for at least one index 1 ≤ i0 ≤ d. Computing in the vector space V we have

ℓ 0 = a1e1 + ... + aded = R(b1e1 + ... + bded) and hence b1e1 + ... + bded = 0 . But the latter equation implies e = − P b−1b e ∈ P Re which is i0 i∕=i0 i0 i i i∕=i0 i a contradiction to the minimality of d. It follows that the above map is an isomorphism. This proves i. and, in particular, that

L = Re1 + ... + Red .

49 For ii. it therefore suﬃces to show that e1, . . . , ed, under the additional assumption that L generates V , is a K-basis of V . This assumption immediately guarantees that the e1, . . . , ed generate the K-vector space V . To show that they are K-linearly independent let

c1e1 + ... + cded = 0 with c1, . . . , cd ∈ K.

j We ﬁnd a suﬃciently large j ≥ 0 such that ai := Rci ∈ R for any 1 ≤ i ≤ d. Then ℓ ℓ 0 = R ⋅ 0 = R(c1e1 + ... + cded) = a1e1 + ... + aded .

By what we have shown above we must have ai = 0 and hence ci = 0 for any 1 ≤ i ≤ d. iii. Let e1, . . . , ed and f1, . . . , fd be K-bases of V such that

′ L = Re1 + ... + Red and L = Rf1 + ... + Rfd .

We write ej = c1jf1 + ... + cdjfd with cij ∈ K, and we choose an integer m ≥ 0 such that

m R cij ∈ R for any 1 ≤ i, j ≤ d .

m ′ It follows that R ej ∈ Rf1 + ... + Rfd = L for any 1 ≤ j ≤ d and hence m ′ that R L ⊆ L . Suppose that V is a finitely generated K[G]-module. Then V is finite dimensional as a K-vector space. A lattice L in V is called G-invariant if we have gL ⊆ L for any g ∈ G. In particular, L is a finitely generated R[G]-submodule of V , and L/RL is a k[G]-module of finite length. Lemma 9.2. Any finitely generated K[G]-module V contains a G-invariant lattice.

′ Proof. We choose a basis e1, . . . , ed of the K-vector space V . Then L := Re1 + ... + Red is a lattice in V . We deﬁne the R[G]-submodule X L := gL′ g∈G of V . With L′ also L generates V as a K-vector space. Moreover, L is ﬁnitely generated by the set {gei : 1 ≤ i ≤ d, g ∈ G} as an R-module. Therefore, by Lemma 9.1.ii, L is a G-invariant lattice in V .

50 Whereas the K[G]-module V always is projective by Remark 7.3 a G- invariant lattice L in V need not to be projective as an R[G]-module. We will encounter an example of this later on. Theorem 9.3. Let L and L′ be two G-invariant lattices in the ﬁnitely generated K[G]-module V ; we then have ′ ′ [L/RL] = [L /RL ] in Rk(G) . Proof. We begin by observing that, for any a ∈ K×, the map

=∼ L/RL −−→ (aL)/R(aL)

x + RL 7−→ ax + R(aL) is an isomorphism of k[G]-modules, and hence

[L/RL] = [(aL)/R(aL)] in Rk(G) . m By applying Lemma 9.1.iii (and replacing L by R L for some suﬃciently large m ≥ 0) we therefore may assume that L ⊆ L′. By applying Lemma 9.1.iii again to L and L′ (while interchanging their roles) we ﬁnd an integer n ≥ 0 such that n ′ ′ RL ⊆ L ⊆ L . We now proceed by induction with respect to n. If n = 1 we have the two exact sequences of k[G]-modules ′ ′ ′ ′ 0 −→ L/RL −→ L /RL −→ L /L −→ 0 and ′ ′ 0 −→ RL /RL −→ L/RL −→ L/RL −→ 0 . It follows that ′ ′ ′ ′ ′ ′ [L /RL ] = [L/RL ] + [L /L] = [L/RL ] + [RL /RL] ′ ′ = [L/RL ] + [L/RL] − [L/RL ]

= [L/RL] in Rk(G). For n ≥ 2 we consider the R[G]-submodule n−1 ′ M := R L + L. It is a G-invariant lattice in V by Lemma 9.1.ii and satisﬁes n−1 ′ ′ R L ⊆ M ⊆ L and RM ⊆ L ⊆ M.

Applying the case n = 1 to L and M we obtain [M/RM] = [L/RL]. The ′ ′ ′ induction hypothesis for M and L gives [L /RL ] = [M/RM].

51 The above lemma and theorem imply that

ℤ[ℳK[G]] −→ Rk(G)

{V } 7−→ [L/RL] , where L is any G-invariant lattice in V , is a well deﬁned homomorphism. If V1 and V2 are two ﬁnitely generated K[G]-modules and L1 ⊆ V1 and L2 ⊆ V2 are G-invariant lattices then L1 ⊕ L2 is a G-invariant lattice in V1 ⊕ V2 and

[(L1 ⊕ L2)/R(L1 ⊕ L2)] = [L1/RL1 ⊕ L2/RL2]

= [L1/RL1] + [L2/RL2] .

It follows that the subgroup Rel ⊆ ℤ[ℳK[G]] lies in the kernel of the above map so that we obtain the homomorphism

dG : RK (G) −→ Rk(G)

[V ] 7−→ [L/RL] .

It is called the decomposition homomorphism of G. The Cartan-Brauer triangle is the diagram

dG RK (G) / Rk(G) fMMM rr8 MMM rrr eG MM rrcG MM rrr K0(k[G]) .

Lemma 9.4. The Cartan-Brauer triangle is commutative.

Proof. Let P be a finitely generated projective R[G]-module. We have to show that dG(([P ])) = cG(([P ])) holds true. By definition the right hand side is equal to [P/RP ] ∈ Rk(G). Moreover, ([P ]) = [K ⊗R P ] ∈ RK (G). According to Prop. 6.4 the R(G)- module P is a direct summand of a free R[G]-module. But R[G] and hence any free R[G]-module also is free as an R-module. We see that P as an R-module is finitely generated projective and hence free by Remark 8.2. We ∼ d ∼ conclude that P = R is a G-invariant lattice in the K[G]-module K ⊗R P = d d d K ⊗R R = (K ⊗R R) = K , and we obtain dG(([P ])) = dG([K ⊗R P ]) = [P/RP ].

52 Let us consider two “extreme” situations where the maps in the Cartan- Brauer triangle can be determined completely. First we look at the case where p does not divide the order ∣G∣ of the group G. Then k[G] is semisimple. Hence we have kg[G] = kd[G] and K0(k[G]) = Rk(G) by Remark 7.3. The map cG, in particular, is the identity. Proposition 9.5. If p ∤ ∣G∣ then any R[G]-module M which is projective as an R-module also is projective as an R[G]-module. Proof. We consider any “test diagram” of R[G]-modules M

L / N / 0 . Viewing this as a “test diagram” of R-modules our second assumption en- sures the existence of an R-module homomorphism 0 : M −→ L such that ∘ 0 = . Since ∣G∣ is a unit in R by our ﬁrst assumption, we may deﬁne a new R-module homomorphism ˜ : M −→ L by

−1 X −1 ˜(x) := ∣G∣ g 0(g x) for any x ∈ M. g∈G One easily checks that ˜ satisﬁes ˜(ℎx) = ℎ ˜(x) for any ℎ ∈ G and any x ∈ M. This means that ˜ is, in fact, an R[G]-module homomorphism. Moreover, we compute

−1 X −1 −1 X −1 (˜ (x)) = ∣G∣ g ( 0(g x)) = ∣G∣ g (g x) g∈G g∈G X = ∣G∣−1 gg−1 (x) = (x) . g∈G

Corollary 9.6. If p ∤ ∣G∣ then all three maps in the Cartan-Brauer triangle are isomorphisms; more precisely, we have the triangle of bijections

≃ K[[G] / kd[G] bF < FF ≃ ≃ yy FF yy {P }7−→{K⊗ P }FF y{yP }7−→{P/ P } R F yy R R][G] .

53 Proof. We already have remarked that cG is the identity. Hence it suffices to show that the map : K0(R[G]) −→ RK (G) is surjective. Let V be any finitely generated K[G]-module. By Lemma 9.2 we find a G-invariant lattice L in V . It satisfies V = K ⊗R L by definition. Prop. 9.5 implies that L is a finitely generated projective R[G]-module. We conclude that [L] ∈ K0(R[G]) with ([L]) = [V ]. This argument in fact shows that the map ∼ ∼ ℳR[G]/ = −→ ℳK[G]/ =

{P } 7−→ {K ⊗R P } is surjective. Let P and Q be two ﬁnitely generated projective R[G]-modules ∼ such that K ⊗R P = K ⊗R Q as K[G]-modules. The commutativity of the Cartan-Brauer triangle then implies that

[P/RP ] = dG([K ⊗R P ]) = dG([K ⊗R Q]) = [Q/RQ] .

Let P = P1 ⊕ ... ⊕ Ps and Q = Q1 ⊕ ... ⊕ Qt be decompositions into indecomposable submodules. Then

s t P/RP = ⊕i=1Pi/RPi and Q/RQ = ⊕j=1Qj/RQj are decompositions into simple submodules. Using Prop. 7.1 the identity

s t X X [Pi/RPi] = [P/RP ] = [Q/RQ] = [Qj/RQj] i=1 j=1 implies that s = t and that there is a permutation of {1, . . . , s} such that ∼ Qj/RQj = P(j)/RP(j) for any 1 ≤ j ≤ s as k[G]-modules. Applying Prop. 7.4.i we obtain ∼ Qj = P(j) for any 1 ≤ j ≤ s as R[G]-modules. It follows that P =∼ Q as R[G]-modules. Hence the above map between sets of isomorphism classes is bijective. Obviously, if K ⊗R P is indecomposable (i. e., simple) then P was indecomposable. Vice versa, if P is indecomposable then the above reasoning says that P/RP is simple. Because of [P/RP ] = dG([K ⊗R P ]) it follows that K ⊗R P must be indecomposable.

54 The second case is where G is a p-group. For a general group G we have the ring homomorphism

k[G] −→ k X X agg 7−→ ag g∈G g∈G which is called the augmentation of k[G]. It makes k into a simple k[G]- module which is called the trivial k[G]-module. Its kernel is the augmentation ideal X X Ik[G] := { agg ∈ k[G]: ag = 0} . g∈G g∈G

Proposition 9.7. If G is a p-group then we have Jac(k[G]) = Ik[G]; in particular, k[G] is a local ring and the trivial k[G]-module is, up to isomorphism, the only simple k[G]-module.

Proof. We will prove by induction with respect to the order ∣G∣ = pn of G that the trivial module, up to isomorphism, is the only simple k[G]-module. There is nothing to prove if n = 0. We therefore suppose that n ≥ 1. Note that we have

n n n gp = 1 and hence (g − 1)p = gp − 1 = 1 − 1 = 0 for any g ∈ G. The center of a nontrivial p-group is nontrivial. Let g0 ∕= 1 be a central element in G. We now consider any simple k[G]-module M and we denote by : k[G] −→ Endk(M) the corresponding ring homomorphism. Then

pn pn pn ((g0) − idM ) = (g0 − 1) = ((g0 − 1) ) = (0) = 0 .

Since g0 is central ((g0) − idM )(M) is a k[G]-submodule of M. But M is simple. Hence ((g0)−idM )(M) = {0} or = M. The latter would inductively pn imply that ((g0)−idM ) (M) = M ∕= {0} which is contradiction. We obtain (g0) = idM which means that the cyclic subgroup < g0 > is contained in the kernel of the group homomorphism : G −→ Autk(M). Hence we have a commutative diagram of group homomorphisms

Aut (M) G J / k JJJ o7 JJ ooo pr J oo JJJ oo J% ooo G/ < g0 > .

55 We conclude that M already is a simple k[G/ < g0 >]-module and therefore is the trivial module by the induction hypothesis. The identity Jac(k[G]) = Ik[G] now follows from the deﬁnition of the Jacobson radical, and Prop. 4.1 implies that k[G] is a local ring.

Suppose that G is a p-group. Then Prop. 9.7 and Prop. 7.1 together imply that the map

=∼ ℤ −−→ Rk(G) m 7−→ m[k] is an isomorphism. To compute the inverse map let M be a k[G]-module of ﬁnite length, and let {0} = M0 ⫋ M1 ⫋ ... ⫋ Mt = M be a composition ∼ series. We must have Mi/Mi−1 = k for any 1 ≤ i ≤ t. It follows that

t t X X [M] = [Mi/Mi−1] = t ⋅ [k] and dimk M = dimk Mi/Mi−1 = t . i=1 i=1 Hence the inverse map is given by

[M] 7−→ dimk M.

Furthermore, from Prop. 7.4 we obtain that

=∼ ℤ −−→ K0(R[G]) m 7−→ m[R[G]] is an isomorphism. Because of dimk k[G] = ∣G∣ the Cartan homomorphism, under these identiﬁcations, becomes the map

cG : ℤ −→ ℤ m 7−→ m ⋅ ∣G∣ .

For any ﬁnitely generated K[G]-module V and any G-invariant lattice L in V we have dimK V = dimk L/RL. Hence the decomposition homomorphism becomes

dG : RK (G) −→ ℤ [V ] 7−→ dimK V.

56 Altogether the Cartan-Brauer triangle of a p-group G is of the form

[V ]7−→dimK V RK (G) / ℤ cHH ~> HH ~~ HH ~ 17−→[K[G]] HH ~~ ⋅∣G∣ H ~~ ℤ . We also see that the trivial K[G]-module K has the G-invariant lattice R which cannot be projective as an R[G]-module if G ∕= {1}. Before we can establish the ﬁner properties of the Cartan-Brauer triangle we need to develop the theory of induction.

10 The ring structure of RF (G), and induction

In this section we let F be an arbitrary ﬁeld, and we consider the group ring F [G] and its Grothendieck group RF (G) := R(F [G]). Let V and W be two (ﬁnitely generated) F [G]-modules. The group G acts on the tensor product V ⊗F W by g(v ⊗ w) := gv ⊗ gw for v ∈ V and w ∈ W.

In this way V ⊗F W becomes a (ﬁnitely generated) F [G]-module, and we obtain the multiplication map

ℤ[MF [G]] × ℤ[MF [G]] −→ ℤ[MF [G]]

({V }, {W }) 7−→ {V ⊗F W } . Since the tensor product, up to isomorphism, is associative and commutative this multiplication makes ℤ[MF [G]] into a commutative ring. Its unit element is the isomorphism class {F } of the trivial F [G]-module.

Remark 10.1. The subgroup Rel is an ideal in the ring ℤ[MF [G]]. Proof. Let V be a (ﬁnitely generated) F [G]-module and let

0 −→ L −→ M −→ N −→ 0 be a short exact sequence of F [G]-modules. We claim that the sequence

idV ⊗ idV ⊗ 0 −→ V ⊗F L −−−−−→ V ⊗F M −−−−−→ V ⊗F N −→ 0 is exact as well. This shows that the subgroup Rel is preserved under multiplication by {V }. The exactness in question is purely a problem about F -vector spaces. But as vector spaces we have M =∼ L ⊕ N and hence ∼ V ⊗F M = (V ⊗F L) ⊕ (V ⊗F N).

57 It follows that RF (G) naturally is a commutative ring with unit element [F ] such that [V ] ⋅ [W ] = [V ⊗F W ] . Let H ⊆ G be a subgroup. Then F [H] ⊆ F [G] is a subring (with the same unit element). Any F [G]-module V , by restriction of scalars, can be viewed as an F [H]-module. If V is finitely generated as an F [G]-module then V is a finite dimensional F -vector space and, in particular, is finitely generated as an F [H]-module. Hence we have the ring homomorphism

G resH : RF (G) −→ RF (H) [V ] 7−→ [V ] .

On the other hand, for any F [H]-module W we have, by base extension, the F [G]-module F [G] ⊗F [H] W . Obviously, the latter is finitely generated over F [G] if the former was finitely generated over F [H]. The first Frobenius reciprocity says that

=∼ HomF [G](F [G] ⊗F [H] W, V ) −−→ HomF [H](W, V ) 7−→ [w 7−→ (1 ⊗ w)] is an F -linear isomorphism for any F [H]-module W and any F [G]-module V .

Remark 10.2. For any short exact sequence 0 −→ L −→ M −→ N −→ 0 of F [H]-modules the sequence of F [G]-modules

idF [G] ⊗ idF [G] ⊗ 0 → F [G] ⊗F [H] L −−−−−−−→ F [G] ⊗F [H] M −−−−−−−→ F [G] ⊗F [H] N → 0 is exact as well.

Proof. Let g1, . . . , gr ∈ G be a set of representatives for the left cosets of H in G. Then g1, . . . , gr also is a basis of the free right F [H]-module F [G]. It follows that, for any F [H]-module W , the map

r =∼ W −−→ F [G] ⊗F [H] W

(w1, . . . , wr) 7−→ g1 ⊗ w1 + ... + gr ⊗ wr is an F -linear isomorphism. We see that, as a sequence of F -vector spaces, the sequence in question is just the r-fold direct sum of the original exact sequence with itself.

58 Remark 10.3. For any F [H]-module W , if the F [G]-module F [G] ⊗F [H] W is simple then W is a simple F [H]-module.

′ ′ Proof. Let W ⊆ W be any F [H]-submodule. Then F [G] ⊗F [H] W is an F [G]-submodule of F [G] ⊗F [H] W by Remark 10.2. Since the latter is sim- ′ ple we must have F [G] ⊗F [H] W = {0} or = F [G] ⊗F [H] W . Comparing dimensions using the argument in the proof of Remark 10.2 we obtain

′ ′ [G : H] ⋅ dimF W = dimF F [G] ⊗F [H] W = 0 or

= dimF F [G] ⊗F [H] W = [G : H] ⋅ dimF W.

′ ′ We see that dimF W = 0 or = dimF W and therefore that W = {0} or = W . This proves that W is a simple F [H]-module.

It follows that the map

ℤ[MF [H]] −→ ℤ[MF [G]]

{W } 7−→ {F [G] ⊗F [H] W } preserves the subgroups Rel in both sides and therefore induces an additive homomorphism

G indH : RF (H) −→ RF (G) [W ] 7−→ F [G] ⊗F [H] W (which is not multiplicative!). Proposition 10.4. We have

G G G indH (y) ⋅ x = indH (y ⋅ resH (x)) for any x ∈ RF (G) and y ∈ RF (H) . Proof. It suﬃces to show that, for any F [G]-module V and any F [H]-module W , we have an isomorphism of F [G]-modules ∼ (F [G] ⊗F [H] W ) ⊗F V = F [G] ⊗F [H] (W ⊗F V ) . One checks (exercise!) that such an isomorphism is given by

(g ⊗ w) ⊗ v 7−→ g ⊗ (w ⊗ g−1v) .

G Corollary 10.5. The image of indH : RF (H) −→ RF (G) is an ideal in RF (G).

59 We also mention the obvious transitivity relations

H G G G H G resH′ ∘ resH = resH′ and indH ∘ indH′ = indH′ for any chain of subgroups H′ ⊆ H ⊆ G. An alternative way to look at induction is the following. Let W be any F [H]-module. Then

G −1 IndH (W ) := { : G −→ W : (gℎ) = ℎ (g) for any g ∈ G, ℎ ∈ H} equipped with the left translation action of G given by

g(g′) := (g−1g′) is an F [G]-module called the module induced from W . But, in fact, the map

=∼ G F [G] ⊗F [H] W −−→ IndH (W ) X ′ X ( agg) ⊗ w 7−→ (g ) := ag′ℎℎw g∈G ℎ∈H is an isomorphism of F [G]-modules. This leads to the second Frobenius reciprocity isomorphism

G =∼ HomF [G](V, IndH (W )) −−→ HomF [H](V,W ) 7−→ [v 7−→ (v)(1)] .

We also need to recall the character theory of G in the semisimple case. For this we assume for the rest of this section that the order of G is prime to the characteristic of the field F . Any finitely generated F [G]-module V is a finite dimensional F -vector space. Hence we may introduce the function

V : G −→ F g g 7−→ tr(g; V ) = trace of V −→ V which is called the character of V . It depends only on the isomorphism class of V . Characters are class functions on G, i. e., they are constant on each conjugacy class of G. For any two ﬁnitely generated F [G]-modules V1 and V2 we have

V1⊕V2 = V1 + V2 and V1⊗F V2 = V1 ⋅ V2 .

60 Let Cl(G, F ) denote the F -vector space of all class functions G −→ F . By pointwise multiplication of functions it is a commutative F -algebra. The above identities imply that the map

Tr : RF (G) −→ Cl(G, F )

[V ] 7−→ V is a ring homomorphism.

Definition. The field F is called a splitting field for G if, for any simple F [G]-module V , we have EndF [G](V ) = F . If F is algebraically closed then it is a splitting field for G.

Theorem 10.6. i. If F has characteristic zero then the characters { V : {V } ∈ F[[G]} are F -linearly independent.

ii. If F is a splitting ﬁeld for G then the characters { V : {V } ∈ F[[G]} form a basis of the F -vector space Cl(G, F ).

iii. If F has characteristic zero then two ﬁnitely generated F [G]-modules

V1 and V2 are isomorphic if and only if V1 = V2 holds true. Corollary 10.7. i. If F has characteristic zero then the map Tr is injective.

ii. If F is a splitting ﬁeld for G then the map Tr induces an isomorphism of F -algebras =∼ F ⊗ℤ RF (G) −−→ Cl(G, F ) . iii. If F has characteristic zero then the map ∼ MF [G]/ = −→ Cl(G, F )

{V } 7−→ V

is injective.

Proof. For i. and ii use Prop. 7.1.

61 11 The Burnside ring

A G-set X is a set equipped with a G-action

G × X −→ X (g, x) 7−→ gx such that

1x = x and g(ℎx) = (gℎ)x for any g, ℎ ∈ G and any x ∈ X.

Let X and Y be two G-sets. Their disjoint union X ∪ Y is a G-set in an obvious way. But also their cartesian product X × Y is a G-set with respect to g(x, y) := (gx, gy) for (x, y) ∈ X × Y. We will call X and Y isomorphic if there is a bijective map : X −−→≃ Y such that (gx) = g (x) for any g ∈ G and x ∈ X. Let SG denote the set of all isomorphism classes {X} of ﬁnite G-sets X. In the free abelian group ℤ[SG] we consider the subgroup Rel generated by all elements of the form

{X ∪ Y } − {X} − {Y } for any two ﬁnite G-sets X and Y.

We deﬁne the factor group

B(G) := ℤ[SG]/ Rel , and we let [X] ∈ B(G) denote the image of the isomorphism class {X}. In fact, the map

ℤ[SG] × ℤ[SG] −→ ℤ[SG] ({X}, {Y }) 7−→ {X × Y } makes ℤ[SG] into a commutative ring in which the unit element is the isomorphism class of the G-set with one point. Because of (X1 ∪ X2) × Y = X1 × Y ∪ X2 × Y the subgroup Rel is an ideal in ℤ[SG]. We see that B(G) is a commutative ring. It is called the Burnside ring of G. Two elements x, y in a G-set X are called equivalent if there is a g ∈ G such that y = gx. This deﬁnes an equivalence relation on X. The equivalence classes are called G-orbits. They are of the form Gx = {gx : g ∈ G} for some x ∈ X. A nonempty G-set which consists of a single G-orbit is called simple

62 (or transitive or a principal homogeneous space). The decomposition of an arbitrary G-set X into its G-orbits is the unique decomposition of X into simple G-sets. In particular, the only G-subsets of a simple G-set Y are Y and ∅. We let SG denote the set of isomorphism classes of simple G-sets.

=∼ Lemma 11.1. ℤ[SG] −−→ B(G).

Proof. If X = Y1 ∪ ... ∪ Yn is the decomposition of X into its G-orbits then we put ({X}) := {Y1} + ... + {Yn} .

This deﬁnes an endomorphism of ℤ[SG] which is idempotent with im() = ℤ[SG]. It is rather clear that Rel ⊆ ker(). Moreover, using the identities

[Y1] + [Y2] = [Y1 ∪ Y2], [Y1 ∪ Y2] + [Y3] = [Y1 ∪ Y2 ∪ Y3],...,

[Y1 ∪ ... ∪ Yn−1] + [Yn] = [X] we see that [X] = [Y1] + ... + [Yn] and hence that {X}−({X}) ∈ Rel. As in the proof of Prop. 7.1 these three facts together imply ℤ[SG] = ℤ[SG] ⊕ Rel .

For any subgroup H ⊆ G the coset space G/H is a simple G-set with respect to

G × G/H −→ G/H (g, g′H) 7−→ gg′H.

Simple G-sets of this form are called standard G-sets.

Remark 11.2. Each simple G-set X is isomorphic to some standard G-set G/H.

Proof. We ﬁx a point x ∈ X. Let Gx := {g ∈ G : gx = x} be the stabilizer of x in G. Then

≃ G/Gx −−→ X

gGx 7−→ gx is an isomorphism.

63 It follows that the set SG is ﬁnite.

Lemma 11.3. Two standard G-sets G/H1 and G/H2 are isomorphic if and −1 only if there is a g0 ∈ G such that g0 H1g0 = H2. −1 Proof. If g0 H1g0 = H2 then

≃ G/H1 −−→ G/H2

gH1 7−→ gg0H2 is an isomorphism of G-sets. Vice versa, let

≃ : G/H1 −−→ G/H2 be an isomorphism of G-sets. We have (1H1) = g0H2 for some g0 ∈ G and then g0H2 = (1H1) = (ℎ1H1) = ℎ1 (1H1) = ℎ1g0H2 −1 for any ℎ1 ∈ H1. This implies g0 H1g0 ⊆ H2. On the other hand

−1 −1 −1 −1 −1 g0 H1 = (1H2) = (ℎ2H2) = ℎ2 (1H2) = ℎ2g0 H1

−1 for any ℎ2 ∈ H2 which implies g0H2g0 ⊆ H1.

Exercise 11.4. Let G/H1 and G/H2 be two standard G-sets; we then have

X −1 [G/H1] ⋅ [G/H2] = [G/H1 ∩ gH2g ] in B(G)

g∈H1∖G/H2 where H1∖G/H2 denotes the space of double cosets H1gH2 in G. Let F again be an arbitrary field. For any finite set X we have the finite dimensional F -vector space X F [X] := { axx : ax ∈ F } x∈X “with basis X”. Suppose that X is a finite G-set. Then the group G acts on F [X] by X X X g( axx) := axgx = ag−1xx . x∈X x∈X x∈X

64 In this way F [X] becomes a ﬁnitely generated F [G]-module (called a permutation module). If : X −−→≃ Y is an isomorphism of ﬁnite G-sets then

∼ ˜ : F [X] −−→= F [Y ] X X X axx 7−→ ax (x) = a −1(y)y x∈X x∈X y∈Y is an isomorphism of F [G]-modules. It follows that the map ∼ SG −→ MF [G]/ = {X} 7−→ {F [X]} is well deﬁned. We obviously have

F [X1 ∪ X2] = F [X1] ⊕ F [X2] for any two ﬁnite G-sets X1 and X2. Hence the above map respects the subgroups Rel in both sides and induces a group homomorphism

b : B(G) −→ RF (G) [X] 7−→ F [X] .

Remark. There is a third interesting Grothendieck group for the ring F [G] which is the factor group

AF (G) := ℤ[MF [G]]/ Rel⊕ with respect to the subgroup Rel⊕ generated by all elements of the form

{M ⊕ N} − {M} − {N} where M and N are arbitrary ﬁnitely generated F [G]-modules. We note that Rel⊕ ⊆ Rel. The above map b is the composite of the maps

pr B(G) −→ AF (G) −−→ RF (G) [X] 7−→ F [X] 7−→ F [X].

Remark 11.5. For any two ﬁnite G-sets X1 and X2 we have ∼ F [X1 × X2] = F [X1] ⊗F F [X2] as F [G]-modules.

65 Proof. The vectors (x1, x2), resp. x1 ⊗ x2, for x1 ∈ X1 and x2 ∈ X2, form an F -basis of the left, resp. right, hand side. Hence there is a unique F -linear isomorphism mapping (x1, x2) to x1 ⊗ x2. Because of

g(x1, x2) = (gx1, gx2) 7−→ gx1 ⊗ gx2 = g(x1 ⊗ x2) , for any g ∈ G, this map is an F [G]-module isomorphism.

It follows that the map

b : B(G) −→ RF (G) is a ring homomorphism. Note that the unit element [G/G] in B(G) is mapped to the class [F ] of the trivial module F which is the unit element in RF (G). Lemma 11.6. i. For any standard G-set G/H we have

G b([G/H]) = indH (1)

where 1 on the right hand side denotes the unit element of RF (H). ii. For any ﬁnite G-set X we have

tr(g; F [X]) = ∣{x ∈ X : gx = x}∣ ∈ F for any g ∈ G.

Proof. i. Let F be the trivial F [H]-module. It suﬃces to establish an isomorphism ∼ F [G/H] = F [G] ⊗F [H] F. For this purpose we consider the F -bilinear map

: F [G] × F −→ F [G/H] X X ( agg, a) 7−→ aaggH . g∈G g∈G Because of (gℎ, a) = gℎH = gH = (g, a) = (g, ℎa) for any ℎ ∈ H the map is F [H]-balanced and therefore induces a well deﬁned F -linear map ˜ : F [G] ⊗F [H] F −→ F [G/H] X X ( agg) ⊗ a 7−→ aaggH . g∈G g∈G

66 ∼ [G:H] As discussed in the proof of Remark 10.2 we have F [G] ⊗F [H] F = F as F -vector spaces. Hence ˜ is a map between F -vector spaces of the same dimension. It obviously is surjective and therefore bijective. Finally the identity

′ X X ′ X ′ ˜(g (( agg) ⊗ a)) = ˜(( agg g) ⊗ a) = aagg gH g∈G g∈G g∈G ′ X ′ X = g ( aaggH) = g ˜(( agg) ⊗ a) g∈G g∈G for any g′ ∈ G shows that ˜ is an isomorphism of F [G]-modules. g⋅ ii. The matrix (ax,y)x,y of the F -linear map F [X] −−→ F [X] with respect to the basis X is given by the equations X gy = ax,yx . x∈X But gy ∈ X and hence ( 1 if x = gy ax,y = 0 otherwise.

It follows that X X tr(g; F [X]) = ax,x = 1 = ∣{x ∈ X : gx = x}∣ ∈ F. x∈X gx=x

Remark. 1. The map b rarely is injective. Let G = S3 be the symmetric group on three letters. It has four conjugacy classes of subgroups. Using Lemma 11.1, Remark 11.2, and Lemma 11.3 it therefore follows that ∼ 4 B(S3) = ℤ . On the other hand, S3 has only three conjugacy classes of ∼ 3 elements. Hence Prop. 7.1 and Thm. 10.6.ii imply that Rℂ(S3) = ℤ . 2. In general the map b is not surjective either. But there are many structural results about its cokernel. For example, the Artin induction theorem implies that

∣G∣ ⋅ Rℚ(G) ⊆ im(b) .

67 We therefore introduce the subring

PF (G) := im(b) ⊆ RF (G) .

Let ℋ be a family of subgroups of G with the property that if H′ ⊆ H is a subgroup of some H ∈ ℋ then also H′ ∈ ℋ. We introduce the subgroup B(G, ℋ) ⊆ B(G) generated by all [G/H] for H ∈ ℋ as well as its image PF (G, ℋ) ⊆ PF (G) under the map b.

Lemma 11.7. B(G, ℋ) is an ideal in B(G), and hence PF (G, ℋ) is an ideal in PF (G).

Proof. We have to show that, for any H1 ∈ ℋ and any subgroup H2 ⊆ G, the element [G/H1] ⋅ [G/H2] lies in B(G, ℋ). This is immediately clear from Exercise 11.4. But a less detailed argument suﬃces. Obviously [G/H1]⋅ [G/H2] is the sum of the classes of the G-orbits in G/H1 × G/H2. Let −1 ′ G(g1H1, g2H2) = G(H1, g1 g2H2) be such a G-orbit. The stabilizer H ⊆ G −1 of the element (H1, g1 g2H2) is contained in H1 and therefore belongs to ℋ. It follows that

′ [G(g1H1, g2H2)] = [G/H ] ∈ B(G, ℋ) .

Definition. i. Let ℓ be a prime number. A finite group H is called ℓ- hyper-elementary if it contains a cyclic normal subgroup C such that ℓ ∤ ∣C∣ and H/C is an ℓ-group. ii. A finite group is called hyper-elementary if it is ℓ-hyper-elementary for some prime number ℓ. Exercise. Let H be an ℓ-hyper-elementary group. Then: i. Any subgroup of H is ℓ-hyper-elementary;

ii. let C ⊆ H be a cyclic normal subgroup as in the deﬁnition, and let L ⊆ H be any ℓ-Sylow subgroup; then the map C × L −−→≃ H sending (c, g) to cg is a bijection of sets.

Let ℋℎe denote the family of hyper-elementary subgroups of G. By the exercise Lemma 11.7 is applicable to ℋℎe. Theorem 11.8. (Solomon) Suppose that F has characteristic zero; then

PF (G, ℋℎe) = PF (G) .

68 Proof. Because of Lemma 11.7 it suﬃces to show that the unit element 1 ∈ PF (G) already lies in PF (G, ℋℎe). According to Lemma 11.6.ii the characters

F [X](g) = ∣{x ∈ X : gx = x}∣ ∈ ℤ , for any g ∈ G and any ﬁnite G-set X, have integral values. Hence we have the well deﬁned ring homomorphisms

tg : PF (G) −→ ℤ z 7−→ Tr(z)(g) for g ∈ G. On the one hand, by Cor. 10.7.i, they satisfy \ (7) ker(tg) = ker(Tr ∣PF (G)) = {0} . g∈G

On the other hand, we claim that

(8) tg(PF (G, ℋℎe)) = ℤ holds true for any g ∈ G. We fix a g0 ∈ G in the following. Since the image tg0 (PF (G, ℋℎe)) is an additive subgroup of ℤ and hence is of the form nℤ for some n ≥ 0 it suffices to find, for any prime number ℓ, an ℓ-hyper-elementary subgroup H ⊆ G such that the integer tg0 ( F [G/H] ) = F [G/H](g0) = ∣{x ∈ G/H : g0x = x}∣ −1 = ∣{gH ∈ G/H : g g0g ∈ H}∣ is not contained in ℓℤ. We also fix ℓ. The wanted ℓ-hyper-elementary subgroup H will be found in a chain of subgroups C ⊆ < g0 > ⊆ H ⊆ N with C being normal in N which is constructed as follows. Let n ≥ 1 be the s order of g0, and write n = ℓ m with l ∤ m. The cyclic subgroup < g0 > ⊆ G generated by g0 then is the direct product

ℓs m < g0 > = < g0 > × < g0 >

m ℓs where < g0 > is an ℓ-group and C := < g0 > is a cyclic group of order prime to ℓ. We deﬁne N := {g ∈ G : gCg−1 = C} to be the normalizer of C in G. It contains < g0 >, of course. Finally, we choose H ⊆ N in such a way

69 that H/C is an ℓ-Sylow subgroup of N/C which contains the ℓ-subgroup < g0 > /C. By construction H is ℓ-hyper-elementary. In the next step we study the cardinality of the set

−1 {gH ∈ G/H : g0gH = gH} = {gH ∈ G/H : g g0g ∈ H} .

−1 −1 −1 −1 Suppose that g g0g ∈ H. Then g Cg ⊆ g < g0 > g = < g g0g > ⊆ H. But, the two sides having coprime orders, the projection map g−1Cg −→ H/C has to be the trivial map. It follows that g−1Cg = C which means that g ∈ N. This shows that

{gH ∈ G/H : g0gH = gH} = {gH ∈ N/H : g0gH = gH} .

The cardinality of the right hand side is the number of < g0 >-orbits in N/H which consist of one point only. We note that the subgroup C, being normal in N and contained in H, acts trivially on N/H. Hence the < g0 >-orbits coincide with the orbits of the ℓ-group < g0 > /C. But, quite generally, the cardinality of an orbit, being the index of the stabilizer of any point in the orbit, divides the order of the acting group. It follows that the cardinality of any < g0 >-orbit in N/H is a power of ℓ. We conclude that

∣{gH ∈ N/H : g0gH = gH}∣ ≡ ∣N/H∣ = [N : H] mod ℓ .

But by the choice of H we have ℓ ∤ [N : H]. This establishes our claim. By (8) we now may choose, for any g ∈ G, an element zg ∈ PF (G, ℋℎe) such that tg(zg) = 1. We then have Y tg( (zg′ − 1)) = 0 for any g ∈ G, g′∈G and (7) implies that Y (zg′ − 1) = 0 . g′∈G

Multiplying out the left hand side and using that PF (G, ℋℎe) is additively and multiplicatively closed easily shows that 1 ∈ PF (G, ℋℎe).

12 Cliﬀord theory

As before F is an arbitrary field. We fix a normal subgroup N in our finite group G. Let W be an F [N]-module. It is given by a homomorphism of F -algebras : F [N] −→ EndF (W ) .

70 For any g ∈ G we now deﬁne a new F [N]-module g∗(W ) by the composite homomorphism F [N] −→ F [N] −→ EndF (W ) ℎ 7−→ gℎg−1 or equivalently by

F [N] × g∗(W ) −→ g∗(W ) (ℎ, w) 7−→ gℎg−1w .

∗ Remark 12.1. i. dimF g (W ) = dimF W . ii. The map U 7−→ g∗(U) is a bijection between the set of F [N]-submodules of W and the set of F [N]-submodules of g∗(W ).

iii. W is simple if and only if g∗(W ) is simple.

∗ ∗ ∗ iv. g1(g2(W )) = (g2g1) (W ) for any g1, g2 ∈ G.

v. Any F [N]-module homomorphism : W1 −→ W2 also is a homomor- ∗ ∗ phism of F [N]-modules : g (W1) −→ g (W2). Proof. Trivial or straightforward.

Suppose that g ∈ N. One checks that then

∼ W −−→= g∗(W ) w 7−→ gw is an isomorphism of F [N]-modules. Together with Remark 12.1.iv/v this implies that ∼ ∼ G/N × (MF [N]/ =) −→ MF [N]/ = (gN, {W }) 7−→ {g∗(W )} ∼ is a well deﬁned action of the group G/N on the set MF [N]/ =. By Remark 12.1.iii this action respects the subset F[[N]. For any {W } in F[[N] we put

∗ IG(W ) := {g ∈ G : {g (W )} = {W }} .

As a consequence of Remark 12.1.iv this is a subgroup of G, which contains N of course.

71 Remark 12.2. Let V be an F [G]-module, and let g ∈ G be any element; then the map

set of all F [N]- set of all F [N]- −−→≃ submodules of V submodules of V W 7−→ gW is an inclusion preserving bijection; moreover, for any F [N]-submodule W ⊆ V we have:

i. The map

∼ g∗(W ) −−→= g−1W w 7−→ g−1w

is an isomorphism of F [N]-modules;

ii. gW is a simple F [N]-module if and only if W is a simple F [N]-module; ∼ ∼ iii. if W1 = W2 are isomorphic F [N]-submodules of V then also gW1 = gW2 are isomorphic as F [N]-modules. Proof. For ℎ ∈ N we have

ℎ(gW ) = g(g−1ℎg)W = gW since g−1ℎg ∈ N. Hence gW indeed is an F [N]-submodule, and the map in the assertion is well deﬁned. It obviously is inclusion preserving. Its bi- jectivity is immediate from g−1(gW ) = W = g(g−1W ). The assertion ii. is a direct consequence. The map in i. clearly is an F -linear isomorphism. Because of

g−1(gℎg−1w) = ℎ(g−1w) for any ℎ ∈ N and w ∈ W it is an F [N]-module isomorphism. The last assertion iii. follows from i. and Remark 12.1.v.

Theorem 12.3. (Cliﬀord) Let V be a simple F [G]-module; we then have:

i. V is semisimple as an F [N]-module;

ii. let W ⊆ V be a simple F [N]-submodule, and let W˜ ⊆ V be the {W }- isotypic component; then

72 a. W˜ is a simple F [IG(W )]-module, and b. V =∼ IndG (W˜ ) as F [G]-modules. IG(W ) Proof. Since V is of finite length as an F [N]-module we find a simple F [N]- submodule W ⊆ V . Then gW , for any g ∈ G, is another simple F [N]- P submodule by Remark 12.2.ii. Therefore V0 := g∈G gW is, on the one hand, a semisimple F [N]-module by Prop. 1.4. On the other hand it is, by definition, a nonzero F [G]-submodule of V . Since V is simple we must have V0 = V which proves the assertion i. As an F [N]-submodule the {W }- isotypic component W˜ of V is of the form W˜ = W1 ⊕ ... ⊕ Wm with simple ∼ F [N]-submodules Wi = W . Let first g be an element in IG(W ). Using ∼ ∼ Remark 12.2.i/iii we obtain gWi = gW = W for any 1 ≤ i ≤ m. It follows that gWi ⊆ W˜ for any 1 ≤ i ≤ m and hence gW˜ ⊆ W˜ . We see that W˜ is an F [IG(W )]-submodule of V . For a general g ∈ G we conclude from Remark 12.2 that gW˜ is the {gW }-isotypic component of V . We certainly have X V = gW.˜

g∈G/IG(W )

Two such submodules g1W˜ and g2W˜ , being isotypic components, either are ˜ ˜ −1 ˜ ˜ equal or have zero intersection. If g1W = g2W then g2 g1W = W , hence −1 ∼ −1 g2 g1W = W , and therefore g2 g1 ∈ IG(W ). We see that in fact ˜ (9) V = ⊕g∈G/IG(W ) gW.

The inclusion W˜ ⊆ V induces, by the ﬁrst Frobenius reciprocity, the F [G]- module homomorphism

IndG (W˜ ) =∼ F [G] ⊗ W˜ −→ V IG(W ) F [IG(W )] X X ( agg) ⊗ w˜ 7−→ aggw˜ . g∈G g∈G

Since V is simple it must be surjective. But both sides have the same dimension as F -vector spaces [G : IG(W )] ⋅ dimF W˜ , the left hand side by the argument in the proof of Remark 10.2 and the right hand side by (9). Hence this map is an isomorphism which proves ii.b. Finally, since ˜ ∼ F [G] ⊗F [IG(W )] W = V is a simple F [G]-module it follows from Remark 10.3 that W˜ is a simple F [IG(W )]-module. In the next section we will need the following particular consequence of this result. But ﬁrst we point out that for an F [N]-module W of dimension

73 dimF W = 1 the describing algebra homomorphism is of the form

: F [N] −→ F.

The corresponding homomorphism for g∗(W ) then is (g.g−1). Since an endomorphism of a one dimensional F -vector space is given by multiplication × by a scalar we have g ∈ IG(W ) if and only if there is a scalar a ∈ F such that a(ℎ)w = (gℎg−1)aw for any ℎ ∈ N and w ∈ W. It follows that

−1 (10) IG(W ) = {g ∈ G : (gℎg ) = (ℎ) for any ℎ ∈ N} if dimF W = 1. Remark 12.4. Suppose that N is abelian and that F is a splitting ﬁeld for N; then any simple F [N]-module W has dimension dimF W = 1.

Proof. Let : F [N] −→ EndF (W ) be the algebra homomorphism describing W . Since N is abelian we have im() ⊆ EndF [N](W ). By our assumption on the ﬁeld F the latter is equal to F . This means that any element in F [N] acts on W by multiplication by a scalar in F . Since W is simple this forces it to be one dimensional.

Proposition 12.5. Let H be an ℓ-hyper-elementary group with cyclic normal subgroup C such that ℓ ∤ ∣C∣ and H/C is an ℓ-group, and let V be a simple F [H]-module; we suppose that

a. F is a splitting ﬁeld for C,

b. V does not contain the trivial F [C]-module, and

c. the subgroup C0 := {c ∈ C : cg = gc for any g ∈ H} acts trivially on V ;

′ ′ ′ then there exists a proper subgroup H ⫋ H and an F [H ]-module V such ∼ H ′ that V = IndH′ (V ) as F [H]-modules. Proof. We pick any simple F [C]-submodule W ⊆ V . By applying Cliﬀord’s Thm. 12.3 to the normal subgroup C and the module W it suﬃces to show that IH (W ) ∕= H.

74 According to Remark 12.4 the module W is one dimensional and given by an algebra homomorphism : F [C] −→ F , and by (10) we have

−1 IH (W ) = {g ∈ H : (gcg ) = (c) for any c ∈ C} .

The assumption c. means that

C0 ⊆ C1 := ker(∣C) .

We immediately note that any subgroup of the cyclic normal subgroup C also is normal in H. By assumption b. we ﬁnd an element c2 ∈ C ∖ C1 so that (c2) ∕= 1. Let L ⊆ H be an ℓ-Sylow subgroup. We claim that we ﬁnd −1 an element g0 ∈ L such that (g0c2g0 ) ∕= (c2). Then g0 ∕∈ IH (W ) which establishes what we wanted. We point out that, since H = C × L as sets, we have C0 = {c ∈ C : cg = gc for any g ∈ L} . −1 Arguing by contradiction we assume that (gc2g ) = (c2) for any g ∈ L. Then −1 −1 gc2C1g = gc2g C1 = c2C1 for any g ∈ L.

This means that we may consider c2C1 as an L-set with respect to the conjugation action. Since L is an ℓ-group the cardinality of any L-orbit in c2C1 is a power of ℓ. On the other hand we have ℓ ∤ ∣C1∣ = ∣c2C1∣. There must therefore exist an element c0 ∈ c2C1 such that

−1 gc0g = c0 for any g ∈ L

(i. e., an L-orbit of cardinality one). We conclude that c0 ∈ C0 ⊆ C1 and hence c2C1 = c0C1 = C1. This is in contradiction to c2 ∕∈ C1.

13 Brauer’s induction theorem

In this section F is a ﬁeld of characteristic zero, and G continuous to be any ﬁnite group.

Deﬁnition. i. Let ℓ be a prime number. A ﬁnite group H is called ℓ- elementary if it is a direct product H = C × L of a cyclic group C and an ℓ-group L.

ii. A ﬁnite group is called elementary if it is ℓ-elementary for some prime number ℓ.

75 Exercise. i. If H is ℓ-elementary then H = C × L is the direct product of a cyclic group C of order prime to ℓ and an ℓ-group L.

ii. Any ℓ-elementary group is ℓ-hyper-elementary.

iii. Any subgroup of an ℓ-elementary group is ℓ-elementary.

Let ℋe denote the family of elementary subgroups of G. Theorem 13.1. (Brauer) Suppose that F is a splitting ﬁeld for every subgroup of G; then X G indH (RF (H)) = RF (G) . H∈ℋe

G Proof. By Cor. 10.5 each indH (RF (H)) is an ideal in the ring RF (G). Hence the left hand side of the asserted identity is an ideal in the right hand side. To obtain equality we therefore need only to show that the unit element 1G ∈ RF (G) lies in the left hand side. According to Solomon’s Thm. 11.8 together with Lemma 11.6.i we have

X X X G 1G ∈ ℤ F [G/H] = ℤb([G/H]) = ℤ indH (1H ) H∈ℋℎe H∈ℋℎe H∈ℋℎe X G ⊆ indH (RF (H)) . H∈ℋℎe By the transitivity of induction this reduces us to the case that G is ℓ- hyper-elementary for some prime number ℓ. We now proceed by induction with respect to the order of G and assume that our assertion holds for all ′ proper subgroups H ⫋ G. We also may assume, of course, that G is not elementary. Using the transitivity of induction again it then suﬃces to show that X G ′ 1G ∈ indH′ (RF (H )) . H′⫋G Let C ⊆ G be the cyclic normal subgroup of order prime to ℓ such that G/C is an ℓ-group. We ﬁx an ℓ-Sylow subgroup L ⊆ G. Then G = C × L as sets. In C we have the (cyclic) subgroup

C0 := {c ∈ C : cg = gc for any g ∈ G} .

Then H0 := C0 × L

76 is an ℓ-elementary subgroup of G. Since G is not elementary we must have G H0 ⫋ G. We consider the induction IndH0 (F ) of the trivial F [H0]-module F . By semisimplicity it decomposes into a direct sum

G IndH0 (F ) = V0 ⊕ V1 ⊕ ... ⊕ Vr

G of simple F [G]-modules Vi. We recall that IndH0 (F ) is the F -vector space of all functions : G/H0 −→ F with the G-action given by

g ′ −1 ′ (g H0) = (g g H0) .

′ ′ This G-action ﬁxes a function if and only if (gg H0) = (g H0) for any g, g′ ∈ G, i. e., if and only if is constant. It follows that the one dimensional subspace of constant functions is the only simple F [G]-submodule in G IndH0 (F ) which is isomorphic to the trivial module. We may assume that V0 is this trivial submodule. In RF (G) we then have the equation

G 1G = indH0 (1H0 ) − [V1] − ... − [Vr] . This reduces us further to showing that, for any 1 ≤ i ≤ r, we have

G [Vi] ∈ indHi (RF (Hi)) for some proper subgroup Hi ⫋ G. This will be achieved by applying the criterion in Prop. 12.5. By our assumption on F it remains to verify the conditions b. and c. in that proposition for each V1,...,Vr. Since C0 is central G in G and is contained in H0 it acts trivially on IndH0 (F ) and a fortiori on any Vi. This is condition c. For b. we note that CH0 = G and C ∩ H0 = C0. ≃ Hence the inclusion C ⊆ G induces a bijection C/C0 −−→ G/H0. It follows that the map

G =∼ C IndH0 (F ) −−→ IndC0 (F ) 7−→ ∣(C/C0) is an isomorphism of F [C]-modules. It maps constant functions to constant functions and hence the unique trivial F [G]-submodule V0 to the unique trivial F [C]-submodule. Therefore Vi, for 1 ≤ i ≤ r, cannot contain any trivial F [C]-submodule.

Lemma 13.2. Let H be an elementary group, and let N0 ⊆ H be a normal subgroup such that H/N0 is not abelian; then there exists a normal subgroup N0 ⊆ N ⊆ H such that N/N0 is abelian but is not contained in the center Z(H/N0) of H/N0.

77 Proof. With H also H/N0 is elementary (if H = C × L with C and L ∼ having coprime orders then H/N0 = C/C ∩ N0 × L/L ∩ N0). We therefore may assume without loss of generality that N0 = {1}. Step 1: We assume that H is an ℓ-group for some prime number ℓ. By assumption we have Z(H) ∕= H so that H/Z(H) is an ℓ-group ∕= {1}. We pick a cyclic normal subgroup {1}= ∕ N = < g > ⊆ H/Z(H). Let Z(H) ⊆ N ⊆ H be the normal subgroup such that N/Z(H) = N and let g ∈ N be a preimage of g. Clearly N = < Z(H), g > is abelian. But Z(H) ⫋ N since g ∕= 1. Step 2: In general let H = C × L where C is cyclic and L is an ℓ-group. With H also L is not abelian. Applying Step 1 to L we ﬁnd a normal abelian subgroup NL ⊆ L such that NL ⊈ Z(L). Then N := C × NL is a normal abelian subgroup of H such that N ⊈ Z(H) = C × Z(L). Lemma 13.3. Let H be an elementary group, and let W be a simple F [H]- module; we suppose that F is a splitting ﬁeld for all subgroups of H; then there exists a subgroup H′ ⊆ H and a one dimensional F [H′]-module W ′ such that ∼ H ′ W = IndH′ (W ) as F [H]-modules.

Proof. We choose H′ ⊆ H to be a minimal subgroup (possibly equal to H) ′ ′ ∼ H ′ such that there exists an F [H ]-module W with W = IndH′ (W ), and we observe that W ′ necessarily is a simple F [H′]-module by Remark 10.3. Let

′ ′ ′ : H −→ EndF (W )

′ be the corresponding algebra homomorphism, and put N0 := ker( ). We ′ claim that H /N0 is abelian. Suppose otherwise. Then there exists, by Lemma ′ 13.2, a normal subgroup N0 ⊆ N ⊆ H such that N/N0 is abelian but is ′ ′ not contained in Z(H /N0). Let W¯ ⊆ W be a simple F [N]-submodule. By Cliﬀord’s Thm. 12.3 we have

′ ∼ H′ ¯˜ W = Ind ¯ (W ) IH′ (W ) where W¯˜ denotes the {W¯ }-isotypic component of W ′. Transitivity of induction implies

∼ H H′ ¯˜ ∼ H ¯˜ W = Ind ′ (Ind ¯ (W )) = Ind ¯ (W ) . H IH′ (W ) IH′ (W ) ′ ¯ ′ By the minimality of H we therefore must have IH′ (W ) = H which means that W ′ = W¯˜ is {W¯ }-isotypic.

78 ′ ′ On the other hand, W is an F [H /N0]-module. Hence W¯ is a simple F [N/N0]-module for the abelian group N/N0. Remark 12.4 then implies ¯ ¯ ¯ (note that EndF [N/N0](W ) = EndF [N](W ) = F ) that W is one dimensional given by an algebra homomorphism

: F [N/N0] −→ F.

It follows that any ℎ ∈ N acts on the {W¯ }-isotypic module W ′ by multiplication by the scalar (ℎN0). In other words the injective homomorphism

′ ′ H /N0 −→ EndF (W ) ′ ℎN0 7−→ (ℎ) satisﬁes ′ (ℎ) = (ℎN0) ⋅ idW ′ for any ℎ ∈ N. ′ But (ℎN0) ⋅ idW ′ lies in the center of EndF (W ). The injectivity of the ′ homomorphism therefore implies that N/N0 lies in the center of H /N0. This is a contradiction. ′ We thus have established that H /N0 is abelian. Applying Remark 12.4 ′ ′ ′ to W viewed as a simple F [H /N0]-module we conclude that W is one dimensional.

Theorem 13.4. (Brauer) Suppose that F is a splitting ﬁeld for any subgroup of G, and let x ∈ RF (G) be any element; then there exist integers m1, . . . , mr, elementary subgroups H1,...,Hr, and one dimensional F [Hi]- modules Wi such that r X G x = mi indHi ([Wi]) . i=1 Proof. Combine Thm. 13.1, Prop. 7.1, Lemma 13.3, and the transitivity of induction.

14 Splitting ﬁelds

Again F is a field of characteristic zero. Lemma 14.1. Let E/F be any extension field, and let V and W be two finitely generated F [G]-modules; we then have

HomE[G](E ⊗F V,E ⊗F W ) = E ⊗F HomF [G](V,W ) .

79 Proof. First of all we observe, by comparing dimensions, that

HomE(E ⊗F V,E ⊗F W ) = E ⊗F HomF (V,W ) holds true. We now consider U := HomF (V,W ) as an F [G]-module via

G × U −→ U (g, f) 7−→ gf := gf(g−1.) .

G g Then HomF [G](V,W ) = U := {f ∈ U : f = f for any g ∈ G} is the {F }- isotypic component of U for the trivial F [G]-module F . Correspondingly we obtain

G G HomE[G](E ⊗F V,E ⊗F W ) = HomE(E ⊗F V,E ⊗F W ) = (E ⊗F U) .

This reduces us to proving that

G G (E ⊗F U) = E ⊗F U for any F [G]-module U. The element

1 X " := g ∈ F [G] ⊆ E[G] G ∣G∣ g∈G is an idempotent with the property that

G U = "G ⋅ U, and hence

G G (E ⊗F U) = "G ⋅ (E ⊗F U) = E ⊗F "G ⋅ U = E ⊗F U .

Theorem 14.2. (Brauer) Let e be the exponent of G, and suppose that F contains a primitive e-th root of unity; then F is a splitting ﬁeld for any subgroup of G.

Proof. We ﬁx an algebraic closure F¯ of F . Step 1: We show that, for any ﬁnitely generated F¯[G]-module V¯ , there is an F [G]-module V such that ∼ V¯ = F¯ ⊗F V as F¯[G]-modules .

80 According to Brauer’s Thm. 13.4 we ﬁnd integers m1, . . . , mr, subgroups H1,...,Hr of G, and one dimensional F¯[Hi]-modules W¯i such that r X [V¯ ] = m F¯[G] ⊗ W¯ i F¯[Hi] i i=1 X m X −m = F¯[G] ⊗ W¯ i − F¯[G] ⊗ W¯ i F¯[Hi] i F¯[Hi] i mi>0 mi<0 M m M −m = (F¯[G] ⊗ W¯ i ) − (F¯[G] ⊗ W¯ i ) . F¯[Hi] i F¯[Hi] i mi>0 mi<0

Let i : F¯[Hi] −→ F¯ denote the F¯-algebra homomorphism describing W¯i. e e We have i(ℎ) = i(ℎ ) = i(1) = 1 for any ℎ ∈ Hi. Our assumption on F therefore implies that i(F [Hi]) ⊆ F . Hence the restriction i∣F [Hi] describes a one dimensional F [Hi]-module Wi such that ∼ W¯i = F¯ ⊗F Wi as F¯[Hi]-modules . We deﬁne the F [G]-modules

M mi M −mi V+ := (F [G] ⊗F [Hi] Wi ) and V− := (F [G] ⊗F [Hi] Wi ) . mi>0 mi<0 Then

¯ M ¯ mi M ¯ mi F ⊗F V+ = (F ⊗F F [G] ⊗F [Hi] Wi ) = (F [G] ⊗F [Hi] Wi ) mi>0 mi>0 M = (F¯[G] ⊗ F¯[H ] ⊗ W mi ) F¯[Hi] i F [Hi] i mi>0 M = (F¯[G] ⊗ (F¯ ⊗ W mi )) F¯[Hi] F i mi>0 M m ∼ (F¯[G] ⊗ W¯ i ) = F¯[Hi] i mi>0 and similarly M −m F¯ ⊗ V ∼ (F¯[G] ⊗ W¯ i ) . F − = F¯[Hi] i mi<0 It follows that [V¯ ] = [F¯ ⊗F V+] − [F¯ ⊗F V−] or, equivalently, that

[V¯ ⊕ (F¯ ⊗F V−)] = [F¯ ⊗F V+] .

81 Using Cor. 10.7.i/iii we deduce that ∼ V¯ ⊕ (F¯ ⊗F V−) = F¯ ⊗F V+ as F¯[G]-modules .

′ If V− is nonzero then V− = U ⊕V− is a direct sum of F [G]-modules where U is simple. On the other hand let V+ = U1 ⊕...⊕Um be a decomposition into simple F [G]-modules. Then F¯ ⊗F U is a direct summand of F¯ ⊗F V− and hence is isomorphic to a direct summand of F¯ ⊗F V+. We therefore must ¯ ¯ have HomF¯[G](F ⊗F U, F ⊗F Uj) ∕= {0} for some 1 ≤ j ≤ m. Lemma 14.1 ∼ implies that HomF [G](U, Uj) ∕= {0}. Hence U = Uj as F [G]-modules. We ∼ ′ ′ conclude that V+ = U ⊕ V+ with V+ := ⊕i∕=jUi, and we obtain ¯ ¯ ′ ¯ ∼ ¯ ′ ¯ V ⊕ (F ⊗F V−) ⊕ (F ⊗F U) = (F ⊗F V+) ⊕ (F ⊗F U) .

The Jordan-H¨olderProp. 1.2 then implies that

¯ ¯ ′ ∼ ¯ ′ V ⊕ (F ⊗F V−) = F ⊗F V+ .

By repeating this argument we arrive after ﬁnitely many steps at an F [G]- module V such that ∼ V¯ = F¯ ⊗F V.

Step 2: Let now V be a simple F [G]-module, and let F¯⊗F V = V¯1⊕...⊕V¯m be a decomposition into simple F¯[G]-modules. By Step 1 we ﬁnd F [G]-modules Vi such that ∼ V¯i = F¯ ⊗F Vi .

For any 1 ≤ i ≤ m, the F [G]-module Vi necessarily is simple, and we have ¯ ¯ HomF¯[G](F ⊗F Vi, F ⊗F V ) ∕= {0}. Hence HomF [G](Vi,V ) ∕= {0} by Lemma ∼ 14.1. It follows that Vi = V for any 1 ≤ i ≤ m. By comparing dimensions we conclude that m = 1. This means that F¯ ⊗F V is a simple F¯[G]-module. Using Lemma 14.1 again we obtain ¯ ¯ ¯ F ⊗F EndF [G](V ) = EndF¯[G](F ⊗F V ) = F.

Hence EndF [G](V ) = F must be one dimensional. This shows that F is a splitting field for G. Step 3: Let H ⊆ G be any subgroup with exponent eH . Since eH divides e the field F a fortiori contains a primitive eH -th root of unity. Hence F also is a splitting field for H.

82 15 Properties of the Cartan-Brauer triangle

We go back to the setting from the beginning of this chapter: k is an algebraically closed field of characteristic p > 0, R is a (0, p)-ring for k with maximal ideal mR = RR, and K denotes the field of fractions of R. The ring R will be called splitting for our finite group G if K contains a primitive e-th root of unity where e is the exponent of G. By Thm. 14.2 the field K, in this case, is a splitting field for any subgroup of G. This additional condition can easily be achieved by defining K′ := K() and ′ ′ ′ R := {a ∈ K : NormK′/K (a) ∈ R}; then R is a (0, p)-ring for k which is splitting for G. It is our goal in this section to establish the deeper properties of the Cartan-Brauer triangle

dG RK (G) / Rk(G) M 8 MMM rrr MMM rr eG MM rrcG M& rrr K0(k[G]) .

Lemma 15.1. For any subgroup H ⊆ G the diagram

dH RK (H) / Rk(H)

G G indH indH

dG RK (G) / Rk(G) is commutative. Proof. Let W be a ﬁnitely generated K[H]-module. We choose an H-invariant lattice L ⊆ W . Then

G G indH (dH ([W ])) = indH ([L/RL]) = k[G] ⊗k[H] (L/RL) .

∼ [G:H] ∼ Moreover, R[G]⊗R[H] L = L is a G-invariant lattice in K[G]⊗K[H] W = W [G:H] (compare the proof of Remark 10.2). Hence

G dG(indH ([W ])) = dG( K[G] ⊗K[H] W ) = (R[G] ⊗R[H] L)/R(R[G] ⊗R[H] L) = k[G] ⊗k[H] (L/RL) .

83 Lemma 15.2. We have X G Rk(G) = indH (Rk(H)) . H∈ℋe G Proof. Since the indH (Rk(H)) are ideals in Rk(G) it suﬃces to show that the unit element 1k[G] ∈ Rk(G) lies in the right hand side. We choose R to be splitting for G. By Brauer’s induction Thm. 13.1 we have

X G 1K[G] ∈ indH (RK (H)) H∈ℋe where 1K[G] is the unit element in RK (G). Using Lemma 15.1 we obtain

X G X G dG(1K[G]) ∈ dG(indH (RK (H))) = indH (dH (RK (H))) H∈ℋe H∈ℋe X G ⊆ indH (Rk(H)) . H∈ℋe

It is trivial to see that dG(1K[G]) = 1k[G].

Theorem 15.3. The decomposition homomorphism dG : RK (G) −→ Rk(G) is surjective. Proof. By Lemma 15.1 we have

X G X G dG(RK (G)) ⊇ dG(indH (RK (H))) = indH (dH (RK (H))) . H∈ℋe H∈ℋe

Because of Lemma 15.2 it therefore suffices to show that dH (RK (H)) = Rk(H) for any H ∈ ℋe. This means we are reduced to proving our assertion in the case where the group G is elementary. Then G = H × P is the direct product of a group H of order prime to p and a p-group P . By Prop. 7.1 it suffices to show that the class [W ] ∈ Rk(G), for any simple k[G]-module W , lies in the image of dG. Viewed as a k[P ]-module W must contain the trivial k[P ]-module by Prop. 9.7. We deduce that W P := {w ∈ W : gw = w for any g ∈ P }= ∕ {0}. Since P is a normal subgroup of G the k[P ]-submodule W P in fact is a k[G]- submodule of W . But W is simple. Hence W P = W which means that k[G] acts on W through the projection map k[G] −→ k[H]. According to Cor. 9.6 we find a simple K[H]-module V together with a G-invariant lattice L ⊆ V ∼ such that L/RL = W as k[H]-modules. Viewing V as a K[G]-module through the projection map K[G] −→ K[H] we obtain [V ] ∈ RK (G) and dG([V ]) = [W ].

84 Theorem 15.4. Let pm be the largest power of p which divides the order of G; the Cartan homomorphism cG : K0(k[G]) −→ Rk(G) is injective, its m cokernel is ﬁnite, and p Rk(G) ⊆ im(cG). m Proof. Step 1: We show that p Rk(G) ⊆ im(cG) holds true. It is trivial from the deﬁnition of the Cartan homomorphism that, for any subgroup H ⊆ G, the diagram cH K0(k[H]) / Rk(H)

G [P ]7−→ k[G]⊗k[H]P indH

cG K0(k[G]) / Rk(G) is commutative. It follows that

G indH (im(cH )) ⊆ im(cG) . Lemma 15.2 therefore reduces us to the case that G is an elementary group. Let W be any simple k[G]-module. With the notations of the proof of Thm. 15.3 we have seen there that k[G] acts on W through the projection map k[G] −→ k[H]. Viewed as a k[H]-module W is projective by Remark 7.3. Hence k[G]⊗k[H] W is a ﬁnitely generated projective k[G]-module. We claim that cG( k[G] ⊗k[H] W ) = ∣G/H∣ ⋅ [W ] holds true. Using the above commutative diagram as well as Prop. 10.4 we obtain

G cG( k[G] ⊗k[H] W ) = indH ([W ]) = k[G] ⊗k[H] k ⋅ [W ] . ′ In order to analyze the k[G]-module k[G] ⊗k[H] k let ℎ, ℎ ∈ H, g ∈ P , and a ∈ k. Then ℎ(gℎ′ ⊗ a) = gℎℎ′ ⊗ a = g ⊗ a = gℎ′ ⊗ a .

This shows that H acts trivially on k[G] ⊗k[H] k. In other words, k[G] acts on k[G] ⊗k[H] k through the projection map k[G] −→ k[P ]. It then follows from Prop. 9.7 that all simple subquotients in a composition series of the k[G]-module k[G] ⊗k[H] k are trivial k[G]-modules. We conclude that k[G] ⊗k[H] k = dimk(k[G] ⊗k[H] k) ⋅ 1 = ∣G/H∣ ⋅ 1

(where 1 ∈ Rk(G) is the unit element). ∼ r Step 2: We know from Prop. 7.1 that, as an abelian group, Rk(G) = ℤ for some r ≥ 1. It therefore follows from Step 1 that Rk(G)/ im(cG) is r m r isomorphic to a factor group of the ﬁnite group ℤ /p ℤ .

85 Step 3: It is a consequence of Prop. 7.4 that K0(k[G]) and Rk(G) are r isomorphic to ℤ for the same integer r ≥ 1. Hence

id ⊗ cG : ℚ ⊗ℤ K0(k[G]) −→ ℚ ⊗ℤ Rk(G) is a linear map between two ℚ-vector spaces of the same ﬁnite dimension r. Its injectivity is equivalent to its surjectivity. Let a ∈ ℚ and x ∈ Rk(G). By m Step 1 we ﬁnd an element y ∈ K0(k[G]) such that cG(y) = p x. Then a a (id ⊗ c )( ⊗ y) = ⊗ pmx = a ⊗ x . G pm pm

This shows that id ⊗ cG and consequently cG are injective.

In order to discuss the third homomorphism eG we ﬁrst introduce two bilinear forms. We start from the maps ∼ ∼ (MK[G]/ =) × (MK[G]/ =) −→ ℤ

({V }, {W }) 7−→ dimK HomK[G](V,W ) and ∼ ∼ (ℳk[G]/ =) × (Mk[G]/ =) −→ ℤ

({P }, {V }) 7−→ dimk Homk[G](P,V ) .

They extend to ℤ-bilinear maps

ℤ[MK[G]] × ℤ[MK[G]] −→ ℤ and ℤ[ℳk[G]] × ℤ[Mk[G]] −→ ℤ .

Since K[G] is semisimple we have, for any exact sequence 0 → V1 → V → ∼ V2 → 0 in MK[G], that V = V1 ⊕ V2 and hence that dimK HomK[G](V,W )−dimK HomK[G](V1,W )−dimK HomK[G](V2,W ) = 0.

The corresponding fact in the “variable” W holds as well, of course. The ﬁrst map therefore induces a well deﬁned ℤ-bilinear form

< , >K[G] : RK (G) × RK (G) −→ ℤ

([V ], [W ]) 7−→ dimK HomK[G](V,W ) .

86 Even though k[G] might not be semisimple any exact sequence 0 → P1 → ∼ P → P2 → 0 in ℳk[G] still satisﬁes P = P1 ⊕P2 as a consequence of Lemma 6.2.ii. Hence we again have

dimk Homk[G](P,V ) − dimk Homk[G](P1,V ) − dimk Homk[G](P2,V ) = 0 for any V in Mk[G]. Furthermore, for any P in ℳk[G] and any exact sequence 0 → V1 → V → V2 → 0 in Mk[G] we have, by the deﬁnition of projective modules, the exact sequence

0 −→ Homk[G](P,V1) −→ Homk[G](P,V ) −→ Homk[G](P,V2) −→ 0 . Hence once more

dimk Homk[G](P,V ) − dimk Homk[G](P,V1) − dimk Homk[G](P,V2) = 0 .

This shows that the second map also induces a well deﬁned ℤ-bilinear form

< , >k[G] : K0(k[G]) × Rk(G) −→ ℤ

([P ], [V ]) 7−→ dimk Homk[G](P,V ) .

If {V1},..., {Vr} are the isomorphism classes of the simple K[G]-modules then [V1],..., [Vr] is a ℤ-basis of RK (G) by Prop. 7.1. We have ( dimK EndK[G](Vi) if i = j, < [Vi], [Vj] >K[G] = 0 if i ∕= j.

In particular, if K is a splitting ﬁeld for G then ( 1 if i = j, < [Vi], [Vj] >K[G] = 0 if i ∕= j.

Let {P1},..., {Pt} be the isomorphism classes of ﬁnitely generated indecomposable projective k[G]-modules. By Prop. 7.4.iii the [P1],..., [Pt] form a ℤ-basis of K0(k[G]), and the [P1/ Jac(k[G])P1],..., [Pt/ Jac(k[G])Pt] form a ℤ-basis of Rk(G) by Prop. 7.4.i. We have

Homk[G](Pi,Pj/ Jac(k[G])Pj) = Homk[G](Pi/ Jac(k[G])Pi,Pj/ Jac(k[G])Pj) ( End (P / Jac(k[G])P ) if i = j, = k[G] i i {0} if i ∕= j ( k if i = j, = {0} if i ∕= j,

87 where the latter identity comes from the fact that the algebraically closed ﬁeld k is a splitting ﬁeld for G. Hence ( 1 if i = j, < [Pi], [Pj/ Jac(k[G])Pj] >k[G] = 0 if i ∕= j.

Exercise 15.5. i. If K is a splitting ﬁeld for G then the map

=∼ RK (G) −−→ Homℤ(RK (G), ℤ) x 7−→< x, . >K[G]

is an isomorphism of abelian groups.

ii. The maps

=∼ =∼ K0(k[G]) −−→ Homℤ(Rk(G), ℤ) and Rk(G) −−→ Homℤ(K0(k[G]), ℤ) y 7−→< y, . >k[G] z 7−→< ., z >k[G]

are isomorphisms of abelian groups.

Lemma 15.6. We have

< y, dG(x) >k[G] =< eG(y), x >K[G] for any y ∈ K0(k[G]) and x ∈ RK (G).

Proof. It suffices to consider elements of the form y = [P/RP ] for some finitely generated projective R[G]-module P (see Prop. 8.1) and x = [V ] for some finitely generated K[G]-module V . We pick a G-invariant lattice L ⊆ V . The asserted identity then reads

dimk Homk[G](P/RP, L/RL) = dimK HomK[G](K ⊗R P,V ) .

We have

Homk[G](P/RP, L/RL) = HomR[G](P, L/RL)(11)

= HomR[G](P,L)/ HomR[G](P, RL)

= HomR[G](P,L)/R HomR[G](P,L)

= k ⊗R HomR[G](P,L)

88 where the second identity comes from the projectivity of P as an R[G]- module. On the other hand

HomK[G](K ⊗R P,V ) = HomR[G](P,V )(12) [ −i = HomR[G](P, R L) i≥0 [ −i = HomR[G](P, R L) i≥0 [ −i = R HomR[G](P,L) i≥0

= K ⊗R HomR[G](P,L) . For the third identity one has to observe that, since P is finitely generated S −i as an R-module, any R-module homomorphism P −→ V = i≥0 R L has −i to have its image inside R L for some sufficiently large i. Both, P being a direct summand of some R[G]m (Remark 8.2) and L by definition are free R-modules. Hence HomR(P,L) is a finitely generated free R-module. The ring R being noetherian the R-submodule HomR[G](P,L) is ∼ s finitely generated as well. Lemma 9.1.i then implies that HomR[G](P,L) = R is a free R-module. We now deduce from (11) and (12) that

∼ s ∼ s Homk[G](P/RP, L/RL) = k and HomK[G](K ⊗R P,V ) = K , respectively.

Theorem 15.7. The homomorphism eG : K0(k[G]) −→ RK (G) is injective and its image is a direct summand of RK (G). Proof. Step 1: We assume that R is splitting for G. By Thm. 15.3 the map dG : RK (G) −→ Rk(G) is surjective. Since, by Prop. 7.1, Rk(G) is a free abelian group we ﬁnd a homomorphism s : Rk(G) −→ RK (G) such that dG ∘ s = id. It follows that

Hom(s, ℤ) ∘ Hom(dG, ℤ) = Hom(dG ∘ s, ℤ) = Hom(id, ℤ) = id . Hence the map

Hom(dG, ℤ) : Homℤ(Rk(G), ℤ) −→ Homℤ(RK (G), ℤ) is injective and

Homℤ(RK (G), ℤ) = im(Hom(dG, ℤ)) ⊕ ker(Hom(s, ℤ)) .

89 But because of Lemma 15.6 the map Hom(dG, ℤ) corresponds under the isomorphisms in Exercise 15.5 to the homomorphism eG. Step 2: For general R we use, as described at the beginning of this section a larger (0, p)-ring R′ for k which contains R and is splitting for G. Let K′ denote the field of fractions of R′. It P is a finitely generated projective ′ ′ R[G]-module then R ⊗R P = R [G]⊗R[G] P is a finitely generated projective R′[G]-module such that

′ ′ ′ ′ (R ⊗R P )/R′ (R ⊗R P ) = (R /R′ R ) ⊗R P = (R/RR) ⊗R P

= P/RP

′ ′ (recall that k = R/RR = R /R′ R ). This shows that the diagram

RK (G) o K0(R[G]) N NN NNN NNN NN& ′ ′ [V ]7−→[K ⊗K V ] [P ]7−→[R ⊗RP ] K0(k[G]) pp8 ppp pp ppp ′ p RK′ (G) o K0(R [G]) is commutative. Hence

eG Tr K0(k[G]) / RK (G) / Cl(G, K) LLL LL [V ]7−→[K′⊗ V ] ⊆ e LL K G LLL & ′ R ′ (G) / Cl(G, K ) K Tr is commutative. The oblique arrow is injective by Step 1 and so then is the upper left horizontal arrow. The two right horizontal arrows are injective by Cor. 10.7.i. This implies that the middle vertical arrow is injective and therefore induces an injective homomorphism

RK (G)/ im(eG) −→ RK′ (G)/ im(eG) .

By Step 1 the target RK′ (G)/ im(eG) is isomorphic to a direct summand of the free abelian group RK′ (G). It follows that RK (G)/ im(eG) is isomorphic to a subgroup in a ﬁnitely generated free abelian group and hence is a free abelian group by the elementary divisor theorem. We conclude that ∼ RK (G) = im(eG) ⊕ RK (G)/ im(eG) .

90 We choose R to be splitting for G, and we ﬁx the ℤ-bases of the three involved Grothendieck groups as described before Exercise 15.5. Let E, D, and C denote the matrices which describe the homomorphisms eG, dG, and cG, respectively, with respect to these bases. We, of course, have

DE = C.

Lemma 15.6 says that D is the transpose of E. It follows that the quadratic Cartan matrix C of k[G] is symmetric.

91 Chapter III The Brauer character

As in the last chapter we fix an algebraically closed field k of characteristic p > 0, and we let G be a finite group. We also fix a (0, p)-ring R for k which is splitting for G. Let mR = RR denote the maximal ideal and K the field of fractions of R.

16 Deﬁnitions

As in the semisimple case we let Cl(G, k) denote the k-vector space of all k-valued class functions on G, i. e., functions on G which are constant on conjugacy classes. For any V ∈ Mk[G] we introduce its k-character

V : G −→ k g 7−→ tr(g; V ) ,

which is a function in Cl(G, k). Let 0 → V1 −→ V −→ V2 → 0 be an exact g sequence in Mk[G]. For i = 1, 2 we let Ai be the matrix of Vi −→ Vi with respect to some k-basis e(i), . . . , e(i). We put e := (e(1)) for 1 ≤ j ≤ d 1 di j j 1 (2) and we choose ej ∈ V , for d1 < j ≤ d1 + d2, such that (ej−d1 ) = ej . Then g the matrix of V −→ V with respect to the basis e1, . . . , ed1 , . . . , ed1+d2 is of the form A A 1 . 0 A2 Since the trace is the sum of the diagonal entries of the respective matrix we see that

V (g) = V1 (g) + V2 (g) for any g ∈ G. It follows that

Tr : Rk(G) −→ Cl(G, k)

[V ] 7−→ V is a well deﬁned homomorphism.

Proposition 16.1. The k-characters V ∈ Cl(G, k), for {V } ∈ kd[G], are k-linearly independent.

92 Proof. Let kd[G] = {{V1},..., {Vr}} = Aˆ where A := k[G]/ Jac(k[G]). Since k[G] is artinian A is semisimple (cf. Prop. 1.2.vi). By Wedderburn’s structure theory of semisimple k-algebras we have

r Y A = EndDi (Vi) i=1 where Di := Endk[G](Vi). Since k is algebraically closed we, in fact, have Di = k. For any 1 ≤ i ≤ r we pick an element i ∈ Endk(Vi) with tr( i) = 1. By the above product decomposition of A we then ﬁnd elements ai ∈ k[G], for 1 ≤ i ≤ r, such that ( ai i if i = j, (Vj −−→ Vj) = 0 if i ∕= j.

If we extend each by k-linearity to a k-linear form : k[G] −→ k then Vj eVj ( 1 if i = j, V (ai) = tr(ai; Vj) = e j 0 if i ∕= j.

Pr We now suppose that j=1 cj Vj = 0 in Cl(G, k) with cj ∈ k. Then also Pr c = 0 in Hom (k[G], k) and hence 0 = Pr c (a ) = c for j=1 j eVj k j=1 j eVj i i any 1 ≤ i ≤ r.

Since, by Prop. 7.1, the vectors 1 ⊗ [V ], for {V } ∈ kd[G], form a basis of the k-vector space k ⊗ℤ Rk(G) it follows that

Tr : k ⊗ℤ Rk(G) −→ Cl(G, k) c ⊗ [V ] −→ c V is an injective k-linear map. But the canonical map Rk(G) −→ k ⊗ℤ Rk(G) is not injective. Therefore the k-character V does not determine the isomorphism class {V }. This is the reason that in the present setting the k- characters are of limited use.

Deﬁnition. An element g ∈ G is called p-regular, resp. p-unipotent, if the order of g is prime to p, resp. is a power of p.

Lemma 16.2. For any g ∈ G there exist uniquely determined elements greg and guni in G such that

- greg is p-regular, guni is p-unipotent, and

93 - g = gregguni = gunigreg; moreover, greg and guni are powers of g.

s Proof. Let p m with p ∤ m be the order of g. We choose integers a and b s aps bm such that ap + bm = 1, and we deﬁne greg := g and guni := g . Then

aps+bm g = g = gregguni = gunigreg .

m apsm ps bmps Furthermore, greg = g = 1 and guni = g = 1; hence greg is p-regular and guni is p-unipotent. Let g = grgu = gugr be another decomposition with p-regular gr and p-unipotent gu. One checks that gr and gu have the order s aps 1−bm 1−bm aps m and p , respectively. Then greg = g = g = gr gu = gr and hence also guni = gu.

′ s ′ Let e denote the exponent of G and let e = e p with p ∤ e . We consider any ﬁnitely generated k[G]-module V and any element g ∈ G. Let 1(g, V ), . . . , d(g, V ), where d := dimk V , be all eigenvalues of the k-linear g endomorphism V −→ V . We list any eigenvalue as many times as its multi- g plicity as a zero of the characteristic polynomial of V −→ V prescribes. We have V (g) = 1(g, V ) + ... + d(g, V ) . The following are easy facts:

1. The sequence (1(g, V ), . . . , d(g, V )) depends up to its ordering only on the isomorphism class {V }.

g 2. If is an eigenvalue of V −→ V then j, for any j ≥ 0, is an eigenvalue j g order(g) of V −−→ V . It follows that i(g, V ) = 1 for any 1 ≤ i ≤ d. In particular, each i(g, V ) is an e-th root of unity. Is g p-regular then ′ the i(g, V ) are e -th roots of unity.

3. If 0 → V1 → V → V2 → 0 is an exact sequence in Mk[G] then the sequence (1(g, V ), . . . , d(g, V )) is, up to a reordering, the union of the

two sequences (1(g, V1), . . . , d1 (g, V1)) and (1(g, V2), . . . , d2 (g, V2)) (where di := dimk Vi). Lemma 16.3. For any ﬁnitely generated k[G]-module V and any g ∈ G the sequences (1(g, V ), . . . , d(g, V )) and (1(greg,V ), . . . , d(greg,V )) coincide up to a reordering; in particular, we have

V (g) = V (greg) .

94 Proof. Since the order of greg is prime to p the vector space

V = V1 ⊕ ... ⊕ Vt decomposes into the diﬀerent eigenspaces Vj for the linear endomorphism greg V −−−→ V . The elements gregguni = gunigreg commute. Hence guni respects the eigenspaces of greg, i. e., guni(Vj) = Vj for any 1 ≤ j ≤ t. The cyclic group < guni > is a p-group. By Prop. 9.7 the only simple k[< guni >]- module is the trivial module. This implies that 1 is the only eigenvalue of guni V −−−→ V . We therefore ﬁnd a basis of Vj with respect to which the matrix 1 ∗ of guni∣Vj is of the form ∖ . The matrix of greg∣Vj is the diagonal matrix 0 1 j 0 ∖ where j is the corresponding eigenvalue. The matrix of g∣Vj then 0 j j ∗ is ∖ . It follows that g∣Vj has a single eigenvalue which coincides with 0 j the eigenvalue of greg∣Vj. The subset Greg := {g ∈ G : g is p-regular} consists of full conjugacy classes of G. We therefore may introduce the k- vector space Cl(Greg, k) of k-valued class functions on Greg. There is the obvious map Cl(G, k) −→ Cl(Greg, k) which sends a function on G to its restriction to Greg. Prop. 16.1 and Lemma 16.3 together imply that the composed map Tr Trreg : k ⊗ℤ Rk(G) −−→ Cl(G, k) −→ Cl(Greg, k) still is injective. We will see later on that Trreg in fact is an isomorphism. Remark 16.4. Let ∈ K× be any root of unity; then ∈ R×. j × m Proof. We have = aR with a ∈ R and j ∈ ℤ. If = 1 with m ≥ 1 then m jm jm × 1 = a R which implies R ∈ R . It follows that j = 0 and consequently that = a ∈ R×.

× × Let e′ (K) and e′ (k) denote the subgroup of K and k , respectively, ′ ′ of all e -th roots of unity. Both groups are cyclic of order e , e′ (K) since R is splitting for G, and e′ (k) since k is algebraically closed of characteristic ′ × prime to e . Since e′ (K) ⊆ R by Remark 16.4 the homomorphism

e′ (K) −→ e′ (k)

7−→ + mR is well deﬁned.

95 =∼ Lemma 16.5. The map e′ (K) −−→ e′ (k) is an isomorphism.

Proof. The elements in e′ (K) are precisely the roots of the polynomial e′ X − 1 ∈ K[X]. If 1 ∕= 2 in e′ (K) were mapped to the same element in e′ e′ (k) then the same polynomial X − 1 but viewed in k[X] would have a ′ zero of multiplicity > 1. But this polynomial is separable since p ∤ e . Hence the map is injective and then also bijective.

We denote the inverse of the isomorphism in Lemma 16.5 by

e′ (k) −→ e′ (K) ⊆ R 7−→ [] .

The element [] ∈ R is called the Teichm¨uller representative of . The isomorphism in Lemma 16.5 allows us to introduce, for any ﬁnitely generated k[G]-module V (with d := dimk V ), the K-valued class function

V : Greg −→ K

g 7−→ [1(g, V )] + ... + [d(g, V )] on Greg. It is called the Brauer character of V . By construction we have

V (g) ≡ V (g) mod mR for any g ∈ Greg .

The ﬁrst fact in the list before Lemma 16.3 implies that V only depends on the isomorphism class {V }, whereas the last fact implies that

V = V1 + V2 for any exact sequence 0 → V1 → V → V2 → 0 in Mk[G]. Therefore, if we let Cl(Greg,K) denote the K-vector space of all K-valued class functions on Greg then

TrB : Rk(G) −→ Cl(Greg,K)

[V ] 7−→ V is a well deﬁned homomorphism.

17 Properties

Lemma 17.1. The Brauer characters V ∈ Cl(Greg,K), for {V } ∈ kd[G], are K-linearly independent.

96 Proof. Let P c = 0 with c ∈ K. By multiplying the c by a {V }∈kd[G] V V V V high enough power of R we may assume that all cV lie in R. Then X (cV mod mR) V = 0 , {V } and Prop. 16.1 implies that all cV must lie in mR. We therefore may write cV = RdV with dV ∈ R and obtain X X X 0 = cV V = R dV V , hence dV V = 0 . {V } {V } {V }

2 Applying Prop. 16.1 again gives dV ∈ mR and cV ∈ mV . Proceeding in- T i ductively in this way we deduce that cV ∈ i≥0 mR = {0} for any {V } ∈ kd[G]. Proposition 17.2. The diagram

dG RK (G) / Rk(G)

Tr TrB res Cl(G, K) / Cl(Greg,K) , where “res” denotes the map of restricting functions from G to Greg, is commutative. Proof. Let V be any ﬁnitely generated K[G]-module. We choose a G-invariant lattice L ⊆ V and put W := L/RL. Then dG([V ]) = [W ], and our assertion becomes the statement that

W (g) = V (g) for any g ∈ Greg holds true. Let e1, . . . , ed be a K-basis of V such that L = Re1 + ... + Red. Thene ¯1 := e1 + RL, . . . , e¯d := ed + RL is a k-basis of W . We ﬁx g ∈ Greg, and we form the matrices Ag and Ag of the linear endomorphisms g g V −→ V and W −→ W , respectively, with respect to these bases. Ob- viously Ag is obtained from Ag by reducing its entries modulo mR. Let 1(g, V ), . . . , d(g, V ) ∈ K, resp. 1(g, W ), . . . , d(g, W ) ∈ k, be the zeros, counted with multiplicity, of the characteristic polynomial of Ag, resp. Ag. We have X X V (g) = tr(Ag) = i(g, V ) and W (g) = tr(Ag) = i(g, W ) . i i

97 Clearly, the set {i(g, V )}i is mapped by reduction modulo mR to the set {i(g, W )}i. Lemma 16.5 then implies that, up to a reordering, we have

i(g, V ) = [i(g, W )] .

We conclude that X X V (g) = i(g, V ) = [i(g, W )] = W (g) . i i

Corollary 17.3. The homomorphism

=∼ TrB : K ⊗ℤ Rk(G) −−→ Cl(Greg,K) c ⊗ [V ] 7−→ c V is an isomorphism.

Proof. From Prop. 17.2 we obtain the commutative diagram

id ⊗dG K ⊗ℤ RK (G) / K ⊗ℤ Rk(G)

Tr TrB res Cl(G, K) / Cl(Greg,K) of K-linear maps. The left vertical map is an isomorphism by Cor. 10.7.ii. The map “res” clearly is surjective: For example,' ˜(g) := '(greg) is a preimage of ' ∈ Cl(Greg,K). Hence TrB is surjective. Its injectivity follows from Lemma 17.1.

Corollary 17.4. The number of isomorphism classes of simple k[G]-modules coincides with the number of conjugacy classes of p-regular elements in G.

Proof. The two numbers whose equality is asserted are the dimensions of the two K-vector spaces in the isomorphism of Cor. 17.3.

=∼ Corollary 17.5. The homomorphism Trreg : k ⊗ℤ Rk(G) −−→ Cl(Greg, k) is an isomorphism.

Proof. We know already that this k-linear map is injective. But by Cor. 17.4 the two k-vector spaces have the same ﬁnite dimension. Hence the map is bijective.

98 We also may deduce a more conceptual formula for Brauer characters.

Corollary 17.6. For any ﬁnitely generated k[G]-module V and any g ∈ Greg we have −1 V (g) = Tr(d([V ]))(g) .

Proof. By construction the element V (g) only depends on V as a k[< g >]- module where < g > ⊆ G is the cyclic subgroup generated by g. Since the order of < g > is prime to p the decomposition homomorphism d is an isomorphism by Cor. 9.6. The assertion therefore follows from Prop. 17.2 applied to the group < g >. Lemma 17.7. Let M be a ﬁnitely generated projective R[G]-module and put V := K ⊗R M; we then have

V (g) = 0 for any g ∈ G ∖ Greg .

Proof. We fix an element g ∈ G ∖ Greg. It generates a cyclic subgroup < g > ⊆ G whose order is divisible by p. The trace of g on V only depends on M viewed as an R[G]-module. We have R[G] =∼ R[< g >][G:]. Hence M, being isomorphic to a direct summand of some free R[G]-module, also is isomorphic to a direct summand of a free R[< g >]-module. We see that M also is projective as an R[< g >]-module (cf. Prop. 6.4). This observation reduces us (by replacing G by < g >) to the case of a cyclic group G generated by an element whose order is divisible by p. We write G = C × P as the direct product of a cyclic group C of order prime to p and a cyclic p-group P ∕= {1}. First of all we note that by repeating the above observation for the subgroup P ⊆ G we obtain that M is projective also as an R[P ]-module. Let g = gC gP with gC ∈ C and gP ∈ P . We decompose V = V1 ⊕ ... ⊕ Vr into its isotypic components Vi as an R[C]-module. Since K is a splitting field for any subgroup of G the Vi are precisely the different eigenspaces of gC the linear endomorphism V −−→ V . Let i ∈ K be the eigenvalue of gC on Vi. Each Vi, of course, is a K[G]-submodule of V . In particular, we have r X V (g) = i tr(gP ; Vi) . i=1

It therefore suﬃces to show that tr(gP ; Vi) = 0 for any 1 ≤ i ≤ r. The element r 1 X " := −jgj ∈ K[C] i ∣C∣ i C j=1

99 is the idempotent such that Vi = "iV . But since p ∤ ∣G∣ and because of Remark 16.4 we see that "i ∈ R[C]. It follows that we have the decomposition as an R[G]-module

M = M1 ⊕ ... ⊕ Mr with Mi := M ∩ Vi .

Each Mi, being a direct summand of the projective R[P ]-module M, is a ﬁnitely generated projective R[P ]-module, and Vi = K ⊗R Mi. This further reduces us (by replacing M by Mi and g by gP ) to the case where G is a cyclic p-group with generator g ∕= 1. We know from Prop. 7.4 and Prop. 9.7 that in this situation the (up to isomorphism) only ﬁnitely generated indecomposable projective R[G]- module is R[G]. Hence M =∼ R[G]m for some m ≥ 0, and

tr(g; V ) = m tr(g; K[G]) .

Since g ∕= 1 we have gℎ ∕= ℎ for any ℎ ∈ G. Using the K-basis {ℎ}ℎ∈G of g K[G] we therefore see that the corresponding matrix of K[G] −→ K[G] has all diagonal entries equal to zero. Hence

tr(g; K[G]) = 0 .

The above lemma can be rephrased by saying that there is a unique homomorphism Trproj : K0(k[G]) −→ Cl(Greg,K) such that the diagram

eG K0(k[G]) / RK (G)

Trproj Tr

ext0 Cl(Greg,K) / Cl(G, K), where “ext0” denotes the homomorphism which extends a function on Greg to G by setting it equal to zero on G ∖ Greg, is commutative.

=∼ Proposition 17.8. i. The map Trproj : K ⊗ℤ K0(k[G]) −−→ Cl(Greg,K) is an isomorphism.

100 eG ii. im K0(k[G]) −−→ RK (G) = {x ∈ RK (G) : Tr(x)∣G ∖ Greg = 0}. Proof. i. We have the commutative diagram

id ⊗eG K ⊗ℤ K0(k[G]) / K ⊗ℤ RK (G)

Trproj Tr

ext0 Cl(Greg,K) / Cl(G, K).

The maps id ⊗eG and Tr are injective by Thm. 15.7 and Cor. 10.7.ii, respectively. Hence Trproj is injective. Using Prop. 7.4 and Cor. 17.3 we obtain

dimK K ⊗ℤ K0(k[G]) = dimK K ⊗ℤ Rk(G) = dimK Cl(Greg,K) .

It follows that Trproj is bijective. ii. As a consequence of Lemma 17.7 the left hand side im(eG) of the asserted identity is contained in the right hand side. According to Thm. 15.7 there exists a subgroup Z ⊆ RK (G) such that RK (G) = im(eG) ⊕ Z and a fortiori

K ⊗ℤ RK (G) = im(id ⊗eG) ⊕ (K ⊗ℤ Z) .

Suppose that z ∈ Z is such that Tr(z)∣G ∖ Greg = 0. By i. we then ﬁnd an element x ∈ im(id ⊗eG) such that Tr(z) = Tr(x). It follows that

z = x ∈ Z ∩ im(id ⊗eG) ⊆ (K ⊗ℤ Z) ∩ im(id ⊗eG) = {0} .

Hence z = 0, and we see that any x ∈ RK (G) such that Tr(x)∣G ∖ Greg = 0 must be contained in im(eG).

101 Chapter IV Green’s theory of indecomposable modules

18 Relatively projective modules

Let f : A −→ B be any ring homomorphism. Any B-module M may be considered, via restriction of scalars, as an A-module. Any homomorphism : M −→ N of B-modules automatically is a homomorphism of A-modules as well.

Deﬁnition. A B-module P is called relatively projective (with respect to f) if for any pair of B-module homomorphisms

M / N for which there is an A-module homomorphism 0 : P −→ M such that ∘ 0 = there also exists a B-module homomorphism : P −→ M satisfying ∘ = .

We start with some very simple observations.

Remark. 1. The existence of 0 in the above definition implies that the image of contains the image of . Hence we may replace N by im( ). This means that in the above definition it suffices to consider pairs ( , ) with surjective .

2. Any projective B-module is relatively projective.

3. If P is relatively projective and is projective as an A-module then P is a projective B-module.

4. Every B-module is relatively projective with respect to the identity homomorphism idB. Lemma 18.1. For any B-module P the following conditions are equivalent:

i. P is relatively projective.

102 ii. for any (surjective) B-module homomorphism : M −→ P for which there is an A-module homomorphism 0 : P −→ M such that ∘ 0 = idP there also exists a B-module homomorphism : P −→ M satisfying ∘ = idP . Proof. i. =⇒ ii. We apply the deﬁnition to the pair

idP M / P. ii. =⇒ i. Let P

M / N be any pair of B-module homomorphisms such that there is an A-module homomorphism 0 : P −→ M with ∘ 0 = . By introducing the B-module

M ′ := {(x, y) ∈ M ⊕ P : (x) = (y)} we obtain the commutative diagram

pr2 M ′ / P

pr1 M / N where pr1((x, y)) := x and pr2((x, y)) := y. We observe that

′ 0 : P −→ M

y 7−→ ( 0(y), y) is a well deﬁned A-module homomorphism such that pr2 ∘0 = idP . There exists therefore, by assumption, a B-module homomorphism : P −→ M ′ satisfying pr2 ∘ = idP . It follows that the B-module homomorphism := pr1 ∘ satisﬁes

∘ = ( ∘ pr1) ∘ = ( ∘ pr2) ∘ = ∘ (pr2 ∘) = .

103 Lemma 18.2. For any two B-modules P1 and P2 the direct sum P := P1 ⊕ P2 is relatively projective if and only if P1 and P2 both are relatively projective.

Proof. Let pr1 and pr2 denote, quite generally, the projection map from a direct sum onto its first and second summand, respectively. Similarly, let i1, resp. i2, denote the inclusion map from the first, resp. second, summand into their direct sum. We first suppose that P is relatively projective. By symmetry it suffices to show that P1 is relatively projective. We are going to use Lemma 18.1. Let therefore 1 : M1 −→ P1 be any B-module homomorphism and 0 :

P1 −→ M1 be any A-module homomorphism such that 1 ∘ 0 = idP1 . Then

1⊕idP2 : M := M1 ⊕ P2 −−−−−−→ P1 ⊕ P2 = P is a B-module homomorphism and

⊕id ′ 0 P2 0 : P = P1 ⊕ P2 −−−−−−→ M1 ⊕ P2 = M is an A-module homomorphism and the two satisfy

′ ∘ 0 = ( 1 ∘ 0) ⊕ idP2 = idP1 ⊕ idP2 = idP . By assumption we ﬁnd a B-module homomorphism ′ : P −→ M such that ′ ∘ = idP . We now deﬁne the B-module homomorphism : P1 −→ M1 as the composite

′ i1 pr1 P1 −−→ P1 ⊕ P2 −−→ M1 ⊕ P2 −−−→ M1 , and we compute

′ ′ 1 ∘ = 1 ∘ pr1 ∘ ∘ i1 = pr1 ∘( 1 ⊕ idP2 ) ∘ ∘ i1 ′ = pr1 ∘ ∘ ∘ i1 = pr1 ∘i1

= idP1 .

We now assume, vice versa, that P1 and P2 are relatively projective. Again using Lemma 18.1 we let : M −→ P and 0 : P −→ M be a B- module and an A-module homomorphism, respectively, such that ∘ 0 = idP . This latter relation implies that 0,j := 0∣Pj can be viewed as an A-module homomorphism

−1 0,j : Pj −→ (Pj) for j = 1, 2.

104 −1 −1 We also deﬁne the B-module homomorphism j := ∣ (Pj): (Pj) −→ Pj. We obviously have

j ∘ 0,j = idPj for j = 1, 2.

By assumption we therefore ﬁnd B-module homomorphisms j : Pj −→ −1 (Pj) such that j ∘ j = idPj for j = 1, 2. The B-module homomorphism

: P = P1 ⊕ P2 −→ M

y1 + y2 7−→ 1(y1) + 2(y2) then satisﬁes

∘ (y1 + y2) = (1(y1) + 2(y2)) = ∘ 1(y1) + ∘ 2(y2)

= 1 ∘ 1(y1) + 2 ∘ 2(y2)

= y1 + y2 for any yj ∈ Pj.

Lemma 18.3. For any A-module L0 the B-module B ⊗A L0 is relatively projective.

Proof. Consider any pair

B ⊗A L0

M / N of B-module homomorphisms together with an A-module homomorphism 0 : B ⊗A L0 −→ M such that ⊗ 0 = . Then

: B ⊗A L0 −→ M

b ⊗ y0 7−→ b 0(1 ⊗ y0) is a B-module homomorphism satisfying ∘ = since

∘ (b ⊗ y0) = (b 0(1 ⊗ y0))

= b( ∘ 0)(1 ⊗ y0) = b (1 ⊗ y0)

= (b ⊗ y0) .

105 Proposition 18.4. For any B-module P the following conditions are equivalent: i. P is relatively projective;

ii. P is isomorphic to a direct summand of B ⊗A P ;

iii. P is isomorphic to a direct summand of B ⊗A L0 for some A-module L0. Proof. For the implication from i. to ii. we observe that we have the B- module homomorphism

: B ⊗A P −→ P b ⊗ y 7−→ by as well as the A-module homomorphism

0 : P −→ B ⊗A P y 7−→ 1 ⊗ y which satisfy ∘ 0 = idP . Hence by Lemma 18.1 there exists a B-module homomorphism : P −→ B ⊗A P such that ∘ = idP . Then B ⊗A P = im() ⊕ ker( ), and P −→ im() is an isomorphism of B-modules. The implication from ii. to iii. is trivial. That iii. implies i. follows from Lemmas 18.2 and 18.3.

We are primarily interested in the case of an inclusion homomorphism R[H] ⊆ R[G], where R is any commutative ring and H is a subgroup of a ﬁnite group G. An R[G]-module which is relatively projective with respective to this inclusion will be called relative R[H]-projective. Example. If R = k is a ﬁeld then a k[G]-module is relatively k[{1}]- projective if and only if it is projective. We recall the useful fact that R[G] is free as an R[H]-module with basis any set of representatives of the cosets of H in G. Lemma 18.5. An R[G]-module is projective if and only if it is relatively R[H]-projective and it is projective as an R[H]-module. Proof. By our initial Remark it remains to show that any projective R[G]- module P also is projective as an R[H]-module. But P is a direct summand ∼ of a free R[G]-module F = ⊕i∈I R[G] by Prop. 6.4. Since R[G] is free as an R[H]-module F also is free as an R[H]-module. Hence the reverse implication in Prop. 6.4 says that P is projective as an R[H]-module.

106 As we have discussed in section 10 the R[G]-module R[G] ⊗R[H] L0, for any R[H]-module L0, is naturally isomorphic to the induced module G IndH (L0). Hence we may restate Prop. 18.4 as follows. Proposition 18.6. For any R[G]-module P the following conditions are equivalent:

i. P is relatively R[H]-projective;

G ii. P is isomorphic to a direct summand of IndH (P ); G iii. P is isomorphic to a direct summand of IndH (L0) for some R[H]- module L0. Next we give a useful technical criterion.

Lemma 18.7. Let {g1, . . . , gm} ⊆ G be a set of representatives for the left cosets of H in G; a left R[G]-module P is relatively R[H]-projective if and only if there exists an R[H]-module endomorphism : P −→ P such that

m X −1 gi gi = idP . i=1 Proof. We introduce the following notation. For any g ∈ G there are uniquely determined elements ℎ1(g), . . . , ℎm(g) ∈ H and a uniquely determined permutation (g, .) of {1, . . . , m} such that

ggi = g(g,i)ℎi(g) for any 1 ≤ i ≤ m.

Step 1: Let 0 : M −→ N be any R[H]-module homomorphism between any two R[G]-modules. We claim that

:= N ( 0): M −→ N m X −1 x 7−→ gi 0(gi x) i=1

107 is an R[G]-module homomorphism. Let g ∈ G. We compute

m m −1 X −1 −1 X −1 (g x) = gi 0(gi g x) = gi 0((ggi) x) i=1 i=1 m m X −1 X −1 −1 = gi 0((g(g,i)ℎi(g)) x) = gi 0(ℎi(g) g(g,i)x) i=1 i=1 m m X −1 −1 X −1 −1 = giℎi(g) 0(g(g,i)x) = g g(g,i) 0(g(g,i)x) i=1 i=1 = g−1 (x) for any x ∈ M which proves the claim. We also observe that, for any R[G]- module homomorphisms : M ′ −→ M and : N −→ N ′, we have

(13) ∘ N ( 0) ∘ = N ( ∘ 0 ∘ ) .

Step 2: Let us temporarily say that an R[H]-module endomorphism : M −→ M of an R[G]-module M is good if N ( ) = idM holds true. The assertion of the present lemma then says that an R[G]-module is relatively R[H]-projective if and only if it has a good endomorphism. In this step we establish that any R[G]-module P with a good endomorphism is relatively R[H]-projective. Let P

M / N be a pair of R[G]-module homomorphisms and 0 : P −→ M be an R[H]- module homomorphism such that ∘ 0 = . Using (13) we see that := N ( 0 ∘ ) satisﬁes

∘ = ∘ N ( 0 ∘ ) = N ( ∘ 0 ∘ ) = N ( ∘ ) = ∘ N ( ) = idP .

Step 3: If the direct sum M = M1 ⊕ M2 of R[G]-modules has a good endomorphism then each summand Mj has a good endomorphism as ij prj well. Let Mj −−→ M −−−→ Mj be the inclusion and the projection map, respectively. Then j := prj ∘ ∘ ij satisﬁes

N ( j) = N (prj ∘ ∘ ij) = prj ∘N ( ) ∘ ij = prj ∘ij = idMj where, for the second equality, we again have used (13).

108 Step 4: In order to show that any relatively R[H]-projective R[G]-module P has a good endomorphism it remains, by Step 3 and Prop. 18.4, to see that any R[G]-module of the form R[G]⊗R[H] L0, for some R[H]-module L0, has a good endomorphism. We have

R[G] ⊗R[H] L0 = g1 ⊗ L0 + ... + gm ⊗ L0 and, assuming that g1 ∈ H, we deﬁne the map

: R[G] ⊗R[H] L0 −→ R[G] ⊗R[H] L0 m X gi ⊗ xi 7−→ g1 ⊗ x1 . i=1 Let ℎ ∈ H and observe that the permutation (ℎ, .) has the property that (ℎ, 1) = 1. We compute

m m m X X X ℎ gi ⊗ xi = ℎgi ⊗ xi = g(ℎ,i)ℎi(ℎ) ⊗ xi i=1 i=1 i=1 m X = g(ℎ,i) ⊗ ℎi(ℎ)xi = g(ℎ,1) ⊗ ℎ1x1 i=1

= g(ℎ,1)ℎ1(ℎ) ⊗ x1 = ℎg1 ⊗ x1 m X = ℎ gi ⊗ xi . i=1 This shows that is an R[H]-module endomorphism. Moreover, using that −1 (gj , i) = 1 if and only if i = j, we obtain

m m m m X −1 X X X −1 gj (gj gi ⊗ xi) = gj gj gi ⊗ xi j=1 i=1 j=1 i=1 m m m m X X −1 X X −1 = gj g −1 ℎi(g ) ⊗ xi = gj g −1 ⊗ ℎi(g )xi (gj ,i) j (gj ,i) j j=1 i=1 j=1 i=1 m m X −1 X −1 = gjg −1 ⊗ ℎj(g )xj = gjg −1 ℎj(g ) ⊗ xj (gj ,j) j (gj ,j) j j=1 j=1 m m X −1 X = gjgj gj ⊗ xj = gj ⊗ xj . j=1 j=1 Hence is good.

109 If k is a ﬁeld of characteristic p > 0 and the order of the group G is prime to p then we know the group ring k[G] to be semisimple. Hence all k[G]-modules are projective (cf. Remark 7.3). This fact generalizes as follows to the relative situation. Proposition 18.8. If the integer [G : H] is invertible in R then any R[G]- module M is relatively R[H]-projective.

1 Proof. The endomorphism [G:H] idM of M is good.

19 Vertices and sources

For any R[G]-module M we introduce the set

V(M) := set of subgroups H ⊆ G such that M is relatively R[H]-projective.

For trivial reasons G lies in V(M). Lemma 19.1. The set V(M) is closed under conjugation, i. e., for any H ∈ V(M) and g ∈ G we have gHg−1 ∈ V(M). Proof. Let M

L / N be any pair of R[G]-module homomorphisms and 0 : M −→ L be an −1 R[gHg ]-module homomorphism such that ∘ 0 = . We consider the −1 map 1 := g 0g : M −→ L. For ℎ ∈ H we have

−1 −1 −1 −1 −1 1(ℎy) = g 0(gℎy) = g 0(gℎg gy) = g gℎg 0(gy) = ℎ 1(y) for any y ∈ M. This shows that 1 is an R[H]-module homomorphism. It satisﬁes

−1 −1 ∘ 1(y) = (g 0(gy)) = g ( ∘ 0(gy)) = g−1 (gy) = (g−1gy) = (y) for any y ∈ M. Since, by assumption, M is relatively R[H]-projective there exists an R[G]-module hommorphism : M −→ L such that ∘ = .

110 We let V0(M) ⊆ V(M) denote the subset of those subgroups H ∈ V(M) which are minimal with respect to inclusion.

Exercise 19.2. i. V0(M) is nonempty.

ii. V0(M) is closed under conjugation. iii. V(M) is the set of all subgroups of G which contain some subgroup in V0(M).

Definition. A subgroup H ∈ V0(M) is called a vertex of M. Lemma 19.3. Let p be a fixed prime number and suppose that all prime numbers ∕= p are invertible in R (e.g., let R = k be a field of characteristic p or let R be a (0, p)-ring for such a field k); then all vertices are p-groups.

Proof. Let H1 ∈ V0(M) for some R[G]-module M and let H0 ⊆ H1 be a p-Sylow subgroup of H1. We claim that M is relatively R[H0]-projective which, by the minimality of H1, implies that H0 = H1 is a p-group. Let

L / N be a pair of R[G]-module homomorphisms together with an R[H0]-module homomorphism 0 : M −→ L such that ∘ 0 = . Since [H1 : H0] is invertible in R it follows from Prop. 18.8 that M viewed as an R[H1]-module is relatively R[H0]-projective. Hence there exists an R[H1]-module homomorphism 1 : M −→ L such that ∘ 1 = . But H1 ∈ V(M). So there further must exist an R[G]-module homomorphism : M −→ L satisfying ∘ = .

In order to investigate vertices more closely we need a general property of induction. To formulate it we ﬁrst generalize some notation from section 12. Let H ⊆ G be a subgroup and g ∈ G be an element. For any R[gHg−1]- module M, corresponding to the ring homomorphism : R[gHg−1] −→ ∗ EndR(M), we introduce the R[H]-module g M deﬁned by the composite ring homomorphism

−1 R[H] −→ R[gHg ] −→ EndR(M) . ℎ 7−→ gℎg−1

111 Proposition 19.4. (Mackey) Let H0,H1 ⊆ G be two subgroups and ﬁx a set {g1, . . . , gm} ⊆ G of representatives for the double cosets in H0∖G/H1. For any R[H1]-module L we have an isomorphism of R[H0]-modules

G ∼ H0 −1 ∗ H0 −1 ∗ IndH1 (L) = Ind −1 ((g1 ) L) ⊕ ... ⊕ Ind −1 ((gm ) L) . H0∩g1H1g1 H0∩gmH1gm

Proof. As an H0-set through left multiplication and simultaneously as a (right) H1-set through right multiplication G decomposes disjointly into

G = H0g1H1 ∪ ... ∪ H0gmH1 . G Therefore the induced module IndH1 (L), viewed as an R[H0]-module, decomposes into the direct sum of R[H0]-modules IndG (L) = IndH0g1H1 (L) ⊕ ... ⊕ IndH0gmH1 (L) H1 H1 H1 where, for any g ∈ G, we put IndH0gH1 (L) := { ∈ IndG (L): (g′) = 0 for any g′ ∈/ H gH }. H1 H1 0 1 We note that IndH0gH1 (L) is the R[H ]-module of all functions : H gH −→ H1 0 0 1 −1 L such that (ℎ0gℎ1) = ℎ1 (ℎ0g) for any ℎ0 ∈ H0 and ℎ1 ∈ H1. It remains to check that the map

H0 −1 ∗ H0gH1 Ind −1 ((g ) L) −→ Ind (L) H0∩gH1g H1 ♯ −1 7−→ (ℎ0gℎ1) := ℎ1 (ℎ0) , where ℎ0 ∈ H0 and ℎ1 ∈ H1, is an isomorphism of R[H0]-modules. In order ♯ to check that is well deﬁned let ℎ0gℎ1 = ℎ˜0gℎ˜1 with ℎ˜i ∈ Hi. Then ˜ ˜ −1 −1 ˜ −1 −1 −1 ℎ0 = ℎ0(gℎ1ℎ1 g ) with gℎ1ℎ1 g ∈ H0 ∩ gH1g and hence ˜ ˜ −1 −1 ˜ −1 −1 ˜ (ℎ0) = (ℎ0(gℎ1ℎ1 g )) = (ℎ1ℎ1 ) (ℎ0) which implies ♯ −1 ˜−1 ˜ ♯ ˜ ˜ (ℎ0gℎ1) = ℎ1 (ℎ0) = ℎ1 (ℎ0) = (ℎ0gℎ1) . It is clear that the map ♯ : H gH −→ L lies in IndH0gH1 (L). For ℎ ∈ H 0 1 H1 0 we compute

ℎ ♯ −1 ℎ −1 −1 ( ) (ℎ0gℎ1) = ℎ1 (ℎ0) = ℎ1 (ℎ ℎ0) ♯ −1 = (ℎ ℎ0gℎ1) ℎ ♯ = ( )(ℎ0gℎ1) .

112 ♯ This shows that 7−→ is an R[H0]-module homomorphism. It is visibly injective. To establish its surjectivity we let ∈ IndH0gH1 (L) and we deﬁne H1 the map

: H0 −→ L

ℎ0 7−→ (ℎ0g) .

−1 For ℎ ∈ H0 ∩ gH1g we compute

−1 −1 −1 (ℎ0ℎ) = (ℎ0ℎg) = (ℎ0gg ℎg) = (g ℎg) (ℎ0g) −1 −1 = g ℎ g(ℎ0)

H0 −1 ∗ which means that ∈ Ind −1 ((g ) L). We obviously have H0∩gH1g

♯ −1 −1 (ℎ0gℎ1) = ℎ1 (ℎ0) = ℎ1 (ℎ0g) = (ℎ0gℎ1) , i. e., ♯ = .

Proposition 19.5. Suppose that R is noetherian and complete and that R/ Jac(R) is artinian. For any ﬁnitely generated and indecomposable R[G]- module M its set of vertices V0(M) consists of a single conjugacy class of subgroups.

Proof. We know from Ex. 19.2.ii that V0(M) is a union of conjugacy classes. In the following we show that any two H0,H1 ∈ V0(M) are conjugate. By Prop. 18.6 the R[G]-module M is isomorphic to a direct summand of G G G IndH0 (M) as well as of IndH1 (M). The latter implies that IndH0 (M) is iso- G G morphic to a direct summand of IndH0 (IndH1 (M)). Together with the former G G we obtain that M is isomorphic to a direct summand of IndH0 (IndH1 (M)). At this point we induce the R[H0]-module decomposition

m G ∼ M H0 −1 ∗ IndH1 (M) = Ind −1 ((gi ) M) H0∩giH1gi i=1 from Mackey’s Prop. 19.4 to G and get, by transitivity of induction, that

m G G ∼ M G −1 ∗ IndH0 (IndH1 (M)) = Ind −1 ((gi ) M) H0∩giH1g1 i=1 as R[G]-modules. Our assumptions on R guarantee the validity of the Krull- Remak-Schmidt Thm. 4.7 for R[G], i. e., the “unicity” of the decomposition into indecomposable modules of the ﬁnitely generated R[G]-modules

113 G G G −1 ∗ IndH0 (IndH1 (M)) and Ind −1 ((gi ) M) . It follows that any inde- H0∩giH1gi G G composable direct summand of IndH0 (IndH0 (M)), for example our M, must G −1 ∗ be a direct summand of at least one of the Ind −1 ((gi ) M). This H0∩giH1gi −1 implies, by Prop. 18.6, that M is relatively R[H0 ∩ gH1g ]-projective for −1 some g ∈ G. Hence H0 ∩ gH1g ∈ V(M). Since V(M) is closed under con- −1 jugation according to Lemma 19.1 we also have g H0g ∩ H1 ∈ V(M). Since −1 H0 and H1 both are minimal in V(M) we must have H0 = gH1g . Definition. Suppose that M is finitely generated and indecomposable, and let V ∈ V0(M). Any finitely generated indecomposable R[V ]-module Q such G that M is isomorphic to a direct summand of IndV (Q) is called a V -source of M. A source of M is a V -source for some V ∈ V0(M). −1 We remind the reader that NG(H) = {g ∈ G : gHg = H} denotes the normalizer of the subgroup H of G. Since G is finite any of the inclusions −1 −1 gHg ⊆ H or gHg ⊇ H already implies g ∈ NG(H). Proposition 19.6. Suppose that R is noetherian and complete and that R/ Jac(R) is artinian. Let M be a finitely generated indecomposable R[G]- module and let V ∈ V0(M). We then have: i. M has a V -source Q which is a direct summand of M as an R[V ]- module;

ii. if Q is a V -source of M then (g−1)∗Q, for any g ∈ G, is a gV g−1- source of M;

iii. for any two V -sources Q0 and Q1 of M there exists a g ∈ NG(V ) such ∼ ∗ that Q1 = g Q0. Proof. i. We decompose M, as an R[V ]-module, into a direct sum

M = M1 ⊕ ... ⊕ Mr of indecomposable R[V ]-modules Mi. Then

r G M G IndV (M) = IndV (Mi) . i=1 By Prop. 18.6 the indecomposable R[G]-module M is isomorphic to a direct G summand of IndV (M). As noted already in the proof of Prop. 19.5 the Krull- Remak-Schmidt Thm. 4.7 implies that M then has to be isomorphic to a G direct summand of some IndV (Mi0 ). Put Q := Mi0 .

114 ii. That Q is a ﬁnitely generated indecomposable R[V ]-module implies that (g−1)∗Q is a ﬁnitely generated indecomposable R[gV g−1]-module. The map

G =∼ G −1 ∗ IndV (Q) −→ IndgV g−1 ((g ) Q) 7−→ (.g) is an isomorphism of R[G]-modules. Hence if M is isomorphic to a direct summand of the former it also is isomorphic to a direct summand of the latter. iii. By i. we may assume that Q0 is a direct summand of M as an R[V ]- module. But, by assumption, M, as an R[G]-module, and therefore also Q0, G as an R[V ]-module, is isomorphic to a direct summand of IndV (Q1) as well G as of IndV (Q0). This implies that Q0 is isomorphic to a direct summand of G G IndV (IndV (Q1)). By the same reasoning with Mackey’s Prop. 19.4 as in the proof of Prop. 19.5 we deduce that V −1 ∗ (14) Q0 is isomorphic to a direct summand of IndV ∩gV g−1 ((g ) Q1) G for some g ∈ G. Then IndV (Q0) and hence also M are isomorphic to a G −1 ∗ direct summand of IndV ∩gV g−1 ((g ) Q1). Since V is a vertex of M we −1 −1 must have ∣V ∣ ≤ ∣V ∩ gV g ∣, hence V ⊆ gV g , and therefore g ∈ NG(V ). By inserting this information into (14) it follows that Q0 is isomorphic to a −1 ∗ −1 ∗ direct summand of (g ) Q1. But with Q1 also (g ) Q1 is indecomposable. ∼ −1 ∗ We ﬁnally obtain Q0 = (g ) Q1.

Exercise 19.7. If Q is a V -source of M then V(Q) = V0(Q) = {V }.

20 The Green correspondence

Throughout this section we assume the commutative ring R to be noetherian and complete and R/ Jac(R) to be artinian. We ﬁx a subgroup H ⊆ G. We consider any ﬁnitely generated indecomposable R[G]-module M such that H ∈ V(M). Let M = L1 ⊕ ... ⊕ Lr be a decomposition of M as an R[H]-module into indecomposable R[H]- modules Li. The Krull-Remak-Schmidt Thm. 4.7 says that the set

INDH ({M}) := {{L1},..., {Lr}} of isomorphism classes of the R[H]-modules Li only depends on the isomorphism class {M} of the R[G]-module M.

115 Lemma 20.1. Let V ∈ V0(M) and Vi ∈ V0(Li) for 1 ≤ i ≤ r; we then have:

−1 i. For any 1 ≤ i ≤ r there is a gi ∈ G such that Vi ⊆ giV gi ; G ii. M is isomorphic to a direct summand of IndH (Li) for some 1 ≤ i ≤ r; G −1 iii. if M is isomorphic to a direct summand of IndH (Li) then Vi = giV gi and, in particular, V0(Li) ⊆ V0(M);

iv. if V0(Li) ⊆ V0(M) then M and Li have a common Vi-source. (The assertions i. and iv. do not require the assumption that H ∈ V(M).)

Proof. We ﬁx a V -source Q of M. Then M, as an R[G]-module, is isomorphic G to a direct summand of IndV (Q). Using Mackey’s Prop. 19.4 we see that L1 ⊕ ... ⊕ Lr is isomorphic to a direct summand of

M H −1 ∗ IndH∩xV x−1 ((x ) Q) x∈ℛ where ℛ ⊆ G is a ﬁxed set of representatives for the double cosets in H∖G/V . We conclude from the Krull-Remak-Schmidt Thm. 4.7 that, for any 1 ≤ i ≤ r, there exists an xi ∈ ℛ such that Li is isomorphic to a direct summand of H −1 ∗ −1 Ind −1 ((xi ) Q). Hence H ∩ xiV xi ∈ V(Li) by Prop. 18.6 and, since H∩xiV xi V0(Li) is a single conjugacy class with respect to H by Prop. 19.5, we have −1 −1 −1 ℎViℎ ⊆ H ∩ xiV xi for some ℎ ∈ H. We put gi := ℎ xi and obtain

−1 Vi ⊆ giV gi which proves i. Suppose that Vi ∈ V0(M). Then ∣Vi∣ = ∣V ∣ (cf. Prop. 19.5) and hence

−1 −1 −1 Vi = giV gi = ℎ xiV xi ℎ ⊆ H.

−1 −1 ∗ −1 In particular, xiV xi is a vertex of Li and (xi ) Q is an xiV xi -source −1 ∗ of Li. Using Prop. 19.6.ii we deduce that (gi ) Q is a Vi-source of Li. On −1 the other hand we have, of course, that giV gi ∈ V0(M), and Prop. 19.6.ii −1 ∗ again implies that (gi ) Q is a Vi-source of M. This shows iv. Since H ∈ V(M) the R[G]-module M, by Prop. 18.6, also is isomorphic to a direct summand of r G M G IndH (M) = IndH (Li) . i=1

116 Hence ii. follows from the Krull-Remak-Schmidt Thm. 4.7. Now suppose that G M, as an R[G]-module, is isomorphic to a direct summand of IndH (Li) for G some 1 ≤ i ≤ r. Let Qi be a Vi-source of Li. Then IndH (Li), and hence M, is isomorphic to a direct summand of IndG (Q ). It follows that V ∈ V(M) Vi i i −1 and therefore that ∣V ∣ ≤ ∣Vi∣. Together with i. we obtain Vi = giV gi . This proves iii.

We introduce the set

H V0 (M) := {V ∈ V0(M): V ⊆ H} .

H Obviously the subgroup H acts on V0 (M) by conjugation so that we can H speak of the H-orbits in V0 (M). We also introduce

0 INDH ({M}) := {{Li} ∈ INDH ({M}): V0(Li) ⊆ V0(M)} and

1 INDH ({M}) := {{Li} ∈ INDH ({M}): M is isomorphic to a direct G summand of IndH (Li)}. By Lemma 20.1 all three sets are nonempty and

1 0 INDH ({M}) ⊆ INDH ({M}) . Furthermore, using Prop. 19.5 we have the obvious map

M 0 H vH : INDH ({M}) −→ set of H-orbits in V0 (M) {Li} 7−→ V0(Li) .

H Lemma 20.2. For any V ∈ V0 (M) there is a 1 ≤ j ≤ r such that V ∈ V0(Lj); in this case M and Lj have a common V -source.

Proof. By Prop. 19.6.i the R[G]-module M has a V -source M0 which is a direct summand of M as an R[V ]-module. Since V ⊆ H the Krull-Remak- Schmidt Thm. 4.7 implies that M0 is isomorphic to a direct summand of some Lj as R[V ]-modules. This means that we have a decomposition

Lj = X1 ⊕ ... ⊕ Xt ∼ into indecomposable R[V ]-modules Xi such that X1 = M0. We note that V is a vertex of X1 by Ex. 19.7. Let Vj be a vertex of Lj. We now apply Lemma 20.1.i to this decomposition (i. e., with H,Lj,Vj,V instead of G, M, V, V1)

117 −1 and obtain that V ⊆ ℎVjℎ for some ℎ ∈ H. But Lemma 20.1.i applied to the original decomposition M = L1 ⊕ ... ⊕ Lr gives ∣V ∣ ≥ ∣Vj∣. Hence −1 V = ℎVjℎ which implies V ∈ V0(Lj). The latter further implies Vj ∈ V0(Lj) ⊆ V0(M). Therefore M and Lj, by Lemma 20.1.iv, have a common Vj-source and hence, by Prop. 19.6.ii, also a common V -source.

M Lemma 20.2 implies that the map vH is surjective. H Lemma 20.3. For any V ∈ V0 (M) there exists a ﬁnitely generated indecomposable R[H]-module N such that V ∈ V0(N) and M is isomorphic to a G direct summand of IndH (N). Proof. Let Q be a V -source of M so that M is isomorphic to a direct sum- G G H mand of IndV (Q) = IndH (IndV (Q)). By the usual argument there exists an H indecomposable direct summand N of IndV (Q) such that M is isomorphic to G a direct summand of IndH (N). According to Prop. 18.6 we have V ∈ V(N). ′ ′ ′ ′ We choose a vertex V ∈ V0(N) such that V ⊆ V . Let Q be a V -source H ′ of N. Then N is isomorphic to a direct summand of IndV ′ (Q ). Hence M is G H ′ G ′ isomorphic to a direct summand of IndH (IndV ′ (Q )) = IndV ′ (Q ), and consequently V ′ ∈ V(M) by Prop. 18.6. The minimality of V implies V ′ = V which shows that V is a vertex of N.

Lemma 20.4. Let N be a ﬁnitely generated indecomposable R[H]-module, and let U ∈ V0(N); we then have:

G i. The induced module IndH (N), as an R[H]-module, is of the form G ∼ IndH (N) = N ⊕ N1 ⊕ ... ⊕ Ns

with indecomposable R[H]-modules Ni for which there are elements −1 gi ∈ G ∖ H such that H ∩ giUgi ∈ V(Ni); G ′ ′′ ′ ′′ ii. IndH (N) = M ⊕ M with R[G]-submodules M and M such that: – M ′ is indecomposable, – N is isomorphic to a direct summand of M ′, ′ ′ – U ∈ V0(M ), and M and N have a common U-source;

iii. if NG(U) ⊆ H then V0(Ni) ∕= V0(N) for any 1 ≤ i ≤ s and N is not isomorphic to a direct summand of M ′′.

118 Proof. i. Since U ∈ V0(N) we have H ∼ ′ IndU (N) = N ⊕ N for some R[H]-module N ′. On the other hand, let

G ′ IndH (N ) = N0 ⊕ N1 ⊕ ... ⊕ Ns be a decomposition into indecomposable R[H]-modules Ni. Furthermore, it is easy to see (but can also be deduced from Mackey’s Prop. 19.4) that, as R[H]-modules, we have

G ′ ∼ ′ ′′ IndH (N ) = N ⊕ N for some R[H]-module N ′′, and correspondingly that N is isomorphic to a G direct summand of IndH (N). The latter implies that N is isomorphic to ∼ one of the Ni. We may assume that N = N0. Combining these facts with Mackey’s Prop. 19.4 we obtain

′ ′′ N ⊕ N1 ⊕ ... ⊕ Ns ⊕ N ⊕ N ∼ G G ′ = IndH (N) ⊕ IndH (N ) ∼ G H G = IndH (IndU (N)) = IndU (N) ∼ M H −1 ∗ = IndH∩gUg−1 ((g ) N) g∈ℛ H M H −1 ∗ = IndU (N) ⊕ IndH∩gUg−1 ((g ) N) g∈ℛ,g∈ /H ∼ ′ M H −1 ∗ = N ⊕ N ⊕ IndH∩gUg−1 ((g ) N) g∈ℛ,g∈ /H as R[H]-modules, where ℛ is a ﬁxed set of representatives for the double cosets in H∖G/U such that 1 ∈ ℛ. It is a consequence of the Krull-Remak- Schmidt Thm. 4.7 that we may cancel summands which occur on both sides and still have an isomorphism

′′ ∼ M H −1 ∗ N1 ⊕ ... ⊕ Ns ⊕ N = IndH∩gUg−1 ((g ) N) . g∈ℛ,g∈ /H

Moreover, it follows that Ni, for 1 ≤ i ≤ s, is isomorphic to a direct summand H −1 ∗ −1 of Ind −1 ((gi ) N) for some gi ∈/ H. Hence H ∩ giUgi ∈ V(Ni). H∩giUgi ii. Let G ′ ′ IndH (N) = M1 ⊕ ... ⊕ Mt

119 be a decomposition into indecomposable R[G]-modules. By i. and the Krull- Remak-Schmidt Thm. 4.7 there must exist a 1 ≤ l ≤ t such that N is isomor- ′ ′ ′ ′′ L ′ phic to a direct summand of Ml . We deﬁne M := Ml and M := i∕=l Mi . ′ ′ Lemma 20.1.iii/iv (for (M ,N) instead of (M,Li)) implies V0(N) ⊆ V0(M ) and that M ′ and N have a common U-source. iii. Suppose that NG(U) ⊆ H and V0(Ni) = V0(N) for some 1 ≤ i ≤ s. −1 Then H ∩ giUgi ∈ V(Ni) = V(N) and therefore

−1 −1 −1 −1 U ⊆ ℎ(H ∩ giUgi )ℎ = H ∩ ℎgiU(ℎgi) ⊆ ℎgiU(ℎgi) for some ℎ ∈ H. We conclude that ℎgi ∈ NG(U) ⊆ H and hence gi ∈ H which is a contradiction. If N were isomorphic to a direct summand of ′′ G M then N ⊕ N would be isomorphic to a direct summand of IndH (N). ∼ Hence, by i., N = Ni for some 1 ≤ i ≤ s which contradicts the fact that V0(N) ∕= V0(Ni).

H Proposition 20.5. For any V ∈ V0 (M) such that NG(V ) ⊆ H there is a unique index 1 ≤ j ≤ r such that V ∈ V0(Lj), and M is isomorphic to a G direct summand of IndH (Lj). Proof. According to Lemma 20.3 we ﬁnd a ﬁnitely generated indecomposable R[H]-module N such that

– V ∈ V0(N) and

G – M is isomorphic to a direct summand of IndH (N). Lemma 20.4 tells us that

G ∼ IndH (N) = N ⊕ N1 ⊕ ... ⊕ Ns with indecomposable R[H]-modules Ni such that

V0(Ni) ∕= V0(N) for any 1 ≤ i ≤ s.

But by Lemma 20.2 there exists a 1 ≤ j ≤ r such that V ∈ V0(Lj) and hence V0(Lj) = V0(N). If we now apply the Krull-Remak-Schmidt Thm. 4.7 to the fact that

L1 ⊕ ... ⊕ Lr is isomorphic to a direct summand of N ⊕ N1 ⊕ ... ⊕ Ns then we see that, for i ∕= j, we have

′ V0(Li) = V0(Ni′ ) ∕= V0(N) for some 1 ≤ i ≤ s.

120 Hence j is unique with the property that V0(Lj) = V0(N). Since V0(Lj) = ∼ V0(N) ∕= V0(Ni) for any 1 ≤ i ≤ s we furthermore have Lj =∕ N1,...,Ns ∼ and therefore necessarily Lj = N. This shows that M is isomorphic to a G direct summand of IndH (Lj). M In terms of the map vH the Prop. 20.5 says the following. If the H-orbit H V ⊆ V0 (M) has the property that NG(V ) ⊆ H for one (or any) V ∈ V 0 then there is a unique isomorphism class {L} ∈ INDH ({M}) such that M 1 VH ({L}) = V, and in fact {L} ∈ INDH ){M}). We now shift our point of view and ﬁx a subgroup V ⊆ G such that NG(V ) ⊆ H. Then Prop. 20.5 gives the existence of a well deﬁned map

isomorphism classes of ﬁnitely isomorphism classes of ﬁnitely Γ: generated indecomposable −→ generated indecomposable R[G]-modules with vertex V R[H]-modules with vertex V where the image {L} = Γ({M}) is characterized by the condition that L is isomorphic to a direct summand of M.

Theorem 20.6. (Green) The map Γ is a bijection. The image {L} = Γ({M}) is characterized by either of the following two conditions:

a. L is isomorphic to a direct summand of M;

G b. M is isomorphic to a direct summand of IndH (L). Moreover, M and L have a common V -source.

Proof. Let L be any finitely generated indecomposable R[H]-module with vertex V . By Lemma 20.4.ii we find a finitely generated indecomposable R[G]-module M ′ with vertex V such that

– L is isomorphic to a direct summand of M ′;

G ′ ′′ – IndH (L) = M ⊕ M as R[G]-module, and – M ′ and L have a common V -source.

In particular, Γ({M ′}) = {L} which shows the surjectivity of Γ. Let M be any other ﬁnitely generated indecomposable R[G]-module with vertex V . First we consider the case that M is isomorphic to a direct sum- G ∼ ′ mand of IndH (L). Suppose that M =∕ M . Then M must be isomorphic ′′ to a direct summand of M . Let M = X1 ⊕ ... ⊕ Xt be a decomposition into indecomposable R[H]-modules Xi. Then Lemma 20.4 implies that

121 V0(Xi) ∕= V0(L) for any 1 ≤ i ≤ t. But by Lemma 20.2 there exists a 1 ≤ i ≤ t such that V ∈ V0(Xi) which is a contradiction. It follows that M =∼ M ′, hence that L is isomorphic to a direct summand of M, and consequently that Γ({M}) = {L}. This proves that the condition b. characterizes the map Γ. Secondly we consider the case that Γ({M}) = {L}, i. e., that L is isomorphic to a direct summand of M. Then M is isomorphic to a direct summand G of IndH (L) by Prop. 20.5. Hence we are in the ﬁrst case and conclude that M =∼ M ′. This establishes the injectivity of Γ.

The bijection Γ is called the Green correspondence for (G, V, H). We want to establish a useful additional “rigidity” property of the Green correspondence. For this we ﬁrst have to extent the concept of relative projectivity to module homomorphisms.

Deﬁnition. Let ℋ be a ﬁxed set of subgroups of G. ∼ L i. An R[G]-module M is called relatively ℋ-projective if M = i∈I Mi is isomorphic to a direct sum of R[G]-modules Mi each of which is relatively R[Hi]-projective for some Hi ∈ ℋ. ii. An R[G]-module homomorphism : M −→ N is called ℋ-projective if there exist a relatively ℋ-projective R[G]-module X and R[G]-module homomorphisms 0 : M −→ X and 1 : X −→ N such that = 1 ∘ 0. In case ℋ = {H} we simply say that is H-projective. Exercise 20.7. i. An R[G]-module M is relatively R[H]-projective if and only if idM is H-projective. ii. If the R[G]-module homomorphism : M −→ N is ℋ-projective then ∘ ∘ , for any R[G]-module homomorphisms : M ′ −→ M and : N −→ N ′, is ℋ-projective as well.

iii. If : M −→ N is an H-projective R[G]-module homomorphism between two finitely generated R[G]-modules M and N then the relatively R[H]-projective R[G]-module X in the above definition can be chosen to be finitely generated as well. (Hint: Replace X by R[G] ⊗R[H] X.) We keep fixing our subgroups V ⊆ H ⊆ G, and we introduce the family of subgroups h := {H ∩ gV g−1 : g ∈ G ∖ H} .

122 Lemma 20.8. Let : M −→ N be a V -projective R[H]-module homomorphism between two ﬁnitely generated R[G]-modules M and N; then there exists a V -projective R[G]-module homomorphism : M −→ N and an h-projective R[H]-module homomorphism : M −→ N such that = + .

Proof. By assumption we have a commutative diagram

M / N B > BB }} BB }} 0 BB }} 1 B }} X where X is a relatively R[V ]-projective R[H]-module. By Ex. 20.7.iii we may assume that X is a ﬁnitely generated R[H]-module. The idea of the G proof consists in the attempt to replace X by IndH (X). We ﬁrst of all H observe that, X being isomorphic to a direct summand of IndV (X), the G G R[G]-module IndH (X) is isomorphic to a direct summand of IndV (X) and hence is relatively R[V ]-projective as well. We will make use of the two Frobenius reciprocities from section 10. Let

G X −→ IndH (X) −→ X be the R[H]-module homomorphisms given by ( g−1x if g ∈ H, (x)(g) := 0 if g∈ / H and () := (1) . G We obviously have ∘ = idX and therefore IndH (X) = im() ⊕ ker() as an R[H]-module. By applying Lemma 20.4.i to the indecomposable direct summands of X we see that ker() is relatively h-projective. But we have the commutative diagram

G ∘−id G IndH (X) / IndH (X) LLL rr9 LLL rrr − pr LLL rr⊆ L% rrr ker() .

It follows that the R[H]-module homomorphism ∘ − id is h-projective. By the ﬁrst and second Frobenius reciprocity we have the commutative

123 diagrams

˜0 G G ˜1 M / IndH (X) and IndH (X) / N, II O : II uu II uu 0 I uu 1 II uu I$ uu X X respectively, where ˜0 and ˜1 are (uniquely determined) R[G]-module homomorphisms. Then := ˜1 ∘ ˜0 is a V -projective R[G]-module homomorphism, and := ˜1 ∘ ( ∘ − id) ∘ ˜0 is a h-projective R[H]-module homomorphism (cf. Ex. 20.7.ii). We have

= 1 ∘ 0 = ˜1 ∘ ˜0 + ( 1 ∘ 0 − ˜1 ∘ ˜0)

= + (˜ 1 ∘ ∘ ∘ ˜0 − ˜1 ∘ ˜0) = + ˜1 ∘ ( ∘ − id) ∘ ˜0 = + .

Proposition 20.9. Let M be a ﬁnitely generated indecomposable R[G]- module with vertex V such that NG(V ) ⊆ H ⊆ G, and let L be its Green correspondent (i. e., {L} = Γ({M})). For any other ﬁnitely generated R[G]- module M ′ we have:

i. If M ′ is indecomposable such that L is isomorphic to a direct summand of M ′ then M ′ =∼ M;

ii. M is isomorphic to a direct summand of M ′ (as an R[G]-module) if and only if L is isomorphic to a direct summand of M ′ (as an R[H]- module).

Proof. i. By the injectivity of the Green correspondence it suﬃces to show that V is a vertex of M ′. The assumption on M ′ guarantees the existence of R[H]-module homomorphisms : L −→ M ′ and : M ′ −→ L such that ∘ = idL. Since V is a vertex of L the R[H]-module homomorphism := ∘ is V -projective. We therefore, by Lemma 20.8, may write = + with a V -projective R[G]-module endomorphism : M ′ −→ M ′ and an h- projective R[H]-module endomorphism : M ′ −→ M ′. ′ Step 1: We ﬁrst claim that V ∈ V(M ), i. e., that idM ′ is V -projective. For this we consider the inclusion of rings

′ ′ EG := EndR[G](M ) ⊆ EH := EndR[H](M ) ,

124 and we deﬁne

JV := { ∈ EG : is V -projective},

Jh := { ∈ EH : is h-projective}.

Using Lemma 18.2 one checks that JV and Jh are additively closed. More- over, Ex. 20.7.ii implies that JV is a two-sided ideal in EG and Jh is a two-sided ideal in EH . We have ∈ JV and ∈ Jh. Furthermore n n (15) = = ( + ) ≡ mod Jh for any n ∈ ℕ. By Lemma 3.5.ii, on the other hand, EH and hence EH /Jh is ﬁnitely generated as a (left) EG-module (even as an R-module). Prop. 3.6.iii then implies that the EG-module EH /Jh is Jac(EG)-adically complete. This, in particular, means that \ n (16) (Jac(EG) EH + Jh) ⊆ Jh . n∈ℕ ′ We now suppose that JV ∕= EG. Since M is indecomposable as an R[G]- module the ring EG is local by Prop. 3.6 and Prop. 4.5. It follows that ∈ JV ⊆ Jac(EG). We then conclude from (15) and (16) that ∈ Jh. −1 Hence idL = ∘ ∘ ∈ Jh. We obtain that L is relatively R[H ∩ gV g ]- projective for some g ∈ G ∖ H. But as we have seen in the proof of Lemma 20.4.iii the conclusion that H ∩ gV g−1 ∈ V(L) for some g ∈ G ∖ H is in contradiction with V ∈ V0(L). We therefore must have JV = EG which is the assertion that idM ′ is V -projective. ′ ′ ′ Step 2: We now show that V ∈ V0(M ). Let V be a vertex of M . By Step 1 we have ∣V ′∣ ≤ ∣V ∣, and it suﬃces to show that ∣V ′∣ = ∣V ∣. But Lemma 20.1.i applied to M ′ and V ′ says that the order of the vertex V of the indecomposable direct summand (L) of M ′ is ≤ ∣V ′∣. ii. The direct implication is clear since L is isomorphic to a direct sum- ′ ′ ′ mand of M. For the reverse implication let M = M1 ⊕ ... ⊕ Mt be a decomposition into indecomposable R[G]-modules. Then L is isomorphic to ′ a direct summand of Mi for some 1 ≤ i ≤ t, and the assertion i. implies that ∼ ′ M = Mi .

21 An example: The group SL2(Fp)

We ﬁx a prime number p ∕= 2, and we let G := SL2(Fp). There are the following important subgroups 1 0 a 0 U := : c ∈ ⊆ B := ∈ G : ad = 1 c 1 Fp c d

125 of G.

Exercise. i. [G : B] = p + 1 and [B : U] = p − 1. ∼ ii. U = ℤ/pℤ is a p-Sylow subgroup of G.

iii. B = NG(U). A much more lengthy exercise is the following.

Exercise. The group G has exactly p + 4 conjugacy classes which are rep- resented by the elements: 1 0 −1 0 1. , 0 1 0 −1

1 1 −1 1 1 " −1 " 2. , , , (with " ∈ × ∖ ×2 a 0 1 0 −1 0 1 0 −1 Fp Fp ﬁxed element),

a 0 3. where a ∈ × ∖ {±1} (up to replacing a by a−1), 0 a−1 Fp

0 −1 4. where the polynomial X2 − aX + 1 ∈ [X] is irreducible. 1 a Fp The elements in 2. have order divisible by p, all the others an order prime to p.

Let k be an algebraically closed ﬁeld of characteristic p.

Lemma 21.1. There are exactly p isomorphism classed of simple k[G]- modules.

Proof. Since, by the above exercise, there are p conjugacy classes of p-regular elements in G this follows from Cor. 17.4.

We want to construct explicit models for the simple k[G]-modules. Let k[X,Y ] be the polynomial ring in two variables X and Y over k. For any a b g = c d ∈ G we deﬁne the k-algebra homomorphism g : k[X,Y ] −→ k[X,Y ] X 7−→ aX + cY Y 7−→ bX + dY .

126 An easy explicit computation shows that in this way k[X,Y ] becomes a k[G]-module. We have the decomposition M k[X,Y ] = Vn n≥0 where n X i n−i Vn := kX Y i=0 is the k[G]-submodule of all polynomials which are homogeneous of total degree n. We obviously have:

– dimk Vn = n + 1;

– V0 = k is the trivial k[G]-module; ∼ 2 – The G-action on V1 = kX + kY = k is the restriction to SL2(Fp) of 2 the natural GL2(k)-action on the standard k-vector space k .

We will show that V0,V1,...,Vp−1 are simple k[G]-modules. Since they have diﬀerent k-dimensions they then must be, up to isomorphism, all simple k[G]-modules. 1 0 The subgroup U is generated by the element u+ := . Similarly 1 1 1 b 1 1 the subgroup U − := : b ∈ is generated by u− := . In 0 1 Fp 0 1 the following we ﬁx an n ≥ 0, and we consider in Vn the increasing sequence of vector subspaces

{0} = W0 ⊂ W1 ⊂ ... ⊂ Wn ⊂ Wn+1 = Vn deﬁned by i−1 X j n−j Wi := kX Y . j=0 Lemma 21.2. For 0 ≤ i ≤ n we have:

i. Wi+1 is a k[U]-submodule of Vn;

ii. Wi+1/Wi is the trivial k[U]-module;

iii. if i < p then each vector in Wi+1∖Wi generates Wi+1 as a k[U]-module.

127 i n−i + Proof. Obviously Wi+1 = kX Y ⊕ Wi. We have u X = X + Y and u+Y = Y and hence

i X i u+(XiY n−i) = (X + Y )iY n−i = XjY i−jY n−i j j=0 i−1 X i = XiY n−i + XjY n−j j j=0 i n−i ≡ X Y mod Wi .

We now argue by induction with respect to i. The assertions i.–iii. hold n trivially true for W1 = kY . We assume that they hold true for Wi. The above congruence immediately implies i. and ii. for Wi+1. For iii. let v ∈ Wi+1 ∖ Wi be any vector. Then

i n−i × v = aX Y + w with a ∈ k and w ∈ Wi.

+ Using that u w − w ∈ Wi−1 by ii. for Wi, we obtain

u+v − v = a(u+(XiY n−i) − XiY n−i) + (u+w − w) i−1 n−i+1 ≡ aiX Y mod Wi−1 .

Since i < p we have ai ∕= 0 in k, and we conclude that k[U]v contains the + nonzero vector u v − v ∈ Wi ∖ Wi−1. Hence, by iii. for Wi, we get that Wi ⊆ k[U]v ⊆ Wi+1. Since v∈ / Wi we, in fact, must have k[U]v = Wi+1. For convenience we insert the following reminder. Remark. Let H be any ﬁnite p-group. By Prop. 9.7 the trivial k[H]-module is, up to isomorphism, the only simple k[H]-module. It follows that the socle of any k[H]-module M (cf. Lemma 1.5) is equal to

soc(M) = {x ∈ M : ℎx = x for any ℎ ∈ H}.

Suppose that M ∕= {0}. For any 0 ∕= x ∈ M the submodule k[H]x is of ﬁnite k-dimension and nonzero and therefore, by the Jordan-H¨olderProp. 1.2, contains a simple submodule. It follows in particular that soc(M) ∕= {0}.

Lemma 21.3. For 0 ≤ n < p we have:

n i. k[U]X = Vn;

n ii. kY is the socle of the k[U]-module Vn;

128 iii. Vn is indecomposable as a k[U]-module. Proof. i. This is a special cases of Lemma 20.2.iii. ii. The socle in question + n is equal to {v ∈ Vn : u v = v}. It contains W1 = kY by Lemma 21.2.ii. If + u v = v then k[U]v has k-dimension ≤ 1. On the other hand, any 0 ∕= v ∈ Vn is contained in Wi+1 ∖ Wi for a unique 0 ≤ i ≤ n. By Lemma 21.2.iii we then have k[U]v = Wi+1. Since Wi+1 has k-dimension i + 1 we see that, if + u v = v, then necessarily v ∈ W1. iii. Suppose that Vn = M ⊕ N is the direct sum of two nonzero k[U]-submodules M and N. Then the socle of Vn is the direct sum of the nonzero socles of M and N and therefore has k-dimension at least 2. This contradicts ii.

Proposition 21.4. The k[G]-module Vn, for 0 ≤ n < p, is simple. Proof. First of all we observe that

− n k[U ]Y = Vn holds true. This is proved by the same reasoning as for Lemma 21.3.i with only interchanging the roles of X and Y . Let W ⊆ Vn be any nonzero k[G]-submodule. Its socle as a k[U]-module is nonzero and is contained in the corresponding socle of Vn. Hence Lemma n 21.3.ii implies that kY ⊆ W . Our initial observation then gives Vn = k[U −]Y n ⊆ W .

The k[U]-module structure of Vn, for 0 ≤ n < p, can be made even more explicit. Let k[Z] be the polynomial ring in one variable Z over k. Because of (u+ − 1)p = u+p − 1 = 0 we have the k-algebra homomorphism

k[Z]/Zpk[Z] −→ k[U] Z 7−→ u+ − 1 .

Because of (u+)i = ((u+ − 1) + 1)i it is surjective. But both sides have the same k-dimension p. Hence it is an isomorphism. If we view Vn, via this isomorphism, as a k[Z]/Zp[Z]-module then we claim that

∼ n+1 Vn = k[Z]/Z k[Z] holds true. This amounts to the statement that

∼ + n+1 Vn = k[U]/(u − 1) k[U] .

129 By Lemma 21.3.i we have the surjective k[U]-module homomorphism

k[U] −→ Vn 7−→ Xn .

+ On the other hand, Lemma 21.2.ii implies that (u − 1)Wi+1 ⊆ Wi and hence by induction that (u+ − 1)n+1Xn = 0. We deduce that the above homomorphism induces a homomorphism

+ n+1 k[U]/(u − 1) k[U] −→ Vn which has to be an isomorphism since it is surjective and both sides have the same k-dimension n + 1. ∼ Remark 21.5. Vp−1 = k[U] as a k[U]-module. Proof. This is the case n = p − 1 of the above discussion.

This latter fact has an interesting consequence. For this we also need the following general properties. Lemma 21.6. i. If H is a finite p-group then any finitely generated projective k[H]-module is free. ii. If H is a finite group and V ⊆ H is a p-Sylow subgroup then the k- dimension of any finitely generated projective k[H]-module is divisible by ∣V ∣. Proof. i. It follows from Remark 6.9 and Prop. 9.7 that k[H] is a projective cover of the trivial k[H]-module k. Since the trivial module, up to isomorphism, is the only simple k[H]-module it then is a consequence of Prop. 7.4.i that k[H], up to isomorphism, is the only finitely generated indecomposable projective k[H]-module. This implies the assertion by Lemma 1.6. ii. Let M be a finitely generated projective k[H]-module. By Lemma 18.5 it also is projective as an k[V ]-module. The assertion i. therefore says that ∼ m M = k[V ] for some m ≥ 0. It follows that dimk M = m∣V ∣.

Proposition 21.7. Among the simple k[G]-modules V0,...,Vp−1 only Vp−1 is a projective k[G]-module.

Proof. By Lemma 19.3 we have U ∈ V(Vp−1), i. e., Vp−1 is relatively k[U]- projective. But Vp−1 also is projective as a k[U]-module by Remark 21.5. Hence it is projective as a k[G]-module. On the other hand, if some Vi with 0 ≤ i < p is a projective k[G]-module then i + 1 = dimk Vi must be divisible by ∣U∣ = p by Lemma 21.6.ii. It follows that i = p − 1.

130 Let M be a finitely generated indecomposable k[G]-module. The vertices of M, by Lemma 19.3, are p-groups. It follows that either {1} or U is a vertex of M. We have observed earlier that {1} is a vertex of M if and only if M is a projective k[G]-module. We postpone the investigation of this case. In the other case we may apply the Green correspondence Γ with B = NG(U) to obtain a bijection between the isomorphism classes of finitely generated indecomposable nonprojective k[G]-modules and the isomorphism classes of finitely generated indecomposable nonprojective k[B]-modules.

Example. V0,...,Vp−2 of course are indecomposable k[G]-modules which, by Prop. 21.7, are nonprojective. As a consequence of Lemma 21.3.iii they are indecomposable as k[B]-modules. Hence they are their own Green correspondents. In the following we will determine all ﬁnitely generated indecomposable k[B]-modules. Another important subgroup of B is the subgroup of diagonal matrices

a−1 0 ∼ T := : a ∈ × −−→= × 0 a Fp Fp a−1 0 0 a 7−→ a . Since the order p−1 of T is prime to p the group ring k[T ] is semisimple. As T is abelian and k is algebraically closed all simple k[T ]-modules are one dimensional. They correspond to the homomorphisms

× i : T −→ k for 0 ≤ i < p − 1 a−1 0 i 0 a 7−→ a Exercise. i. The map

pr B −−→ T a−1 0 a−1 0 c a 7−→ 0 a is a surjective homomorphism of groups with kernel U.

ii. B = TU = UT as sets.

iii. All simple k[B]-modules are one dimensional and correspond to the homomorphisms i ∘ pr for 0 ≤ i ≤ p − 2. (Argue that U acts trivially on any simple k[B]-module since in any nonzero k[B]-module L the subspace {x ∈ L : u+x = x} is nonzero.)

131 Let Si, for 0 ≤ i ≤ p − 2, denote the simple k[B]-module corresponding to i ∘ pr. Lemma 21.8. For any k[B]-module L we have rad(L) = (u+ − 1)L. Proof. According to the above exercise U acts trivially on any simple k[B]- module and hence on L/ rad(L). It follows that (u+ − 1)L ⊆ rad(L). The + i + i Pi i + j binomial formula (u ) − 1 = ((u − 1) + 1) − 1 = j=1 j (u − 1) + P implies that (u − 1)k[U] = u∈U (u − 1)k[U]. Since U is normal in B it follows that (u+ − 1)L is a k[B]-submodule and that the U-action on L/(u+ − 1)L is trivial. Hence k[B] acts on L/(u+ − 1)L through its quotient k[T ], and therefore L/(u+ − 1)L is a semisimple k[B]-module. This implies rad(L) ⊆ (u+ − 1)L. Let L ∕= {0} be any finitely generated k[B]-module. According to Lemma 21.8 we have the sequence of k[B]-submodules F 0(L) := L ⊇ F 1(L) := (u+ − 1)L ⊇ ... ⊇ F i(L) := (u+ − 1)iL ⊇ ... Each subquotient F i(L)/F i+1(L) is a semisimple k[B]-module. It follows m(L) from Prop. 2.1 that there is a smallest m(L) ∈ ℕ such that F (L) = {0} and F i(L) ∕= F i+1(L) for any 0 ≤ i < L(m). Our first example of a finitely generated indecomposable projective k[B]- module is Vp−1 (Prop. 21.7, Lemma 18.5, Lemma 21.3.iii). It follows from Lemma 21.2.ii that i F (Vp−1) = Wp−i, for 0 ≤ i ≤ p, and m(Vp−1) = p. i In particular, the sequence F (Vp−1) is a composition series of the k[B]- module Vp−1 and m(Vp−1) is the length of Vp−1. For any 0 ≤ i ≤ p − 1 and × a ∈ Fp we compute a−1 0 (XiY p−1−i) = (a−1X)i(aY )p−1−i = a−2iXiY p−1−i . 0 a This shows that i i+1 ∼ F (Vp−1)/F (Vp−1) = Wp−i/Wp−i−1 = S2i mod p−1 for 0 ≤ i ≤ p − 1. 1 ∼ In particular, Vp−1/ rad(Vp−1) = Vp−1/F (Vp−1) = S0 is the trivial k[B]- module. Hence Vp−1 is a projective cover of the trivial module. It follows from Prop. 7.4.i that there are exactly p − 1 isomorphism classes {Q0},..., {Qp−2} of finitely generated indecomposable projective k[B]-modules which can be numbered in such a way that 1 ∼ Qj/ rad(Qj) = Qj/F (Qj) = Sj for 0 ≤ j < p − 1.

132 ∼ Lemma 21.9. Qj = Vp−1 ⊗k Sj for any 0 ≤ j < p − 1. Proof. (See section 10 for the tensor product of two modules.) Because of Lemma 6.8 it suﬃces to show that Vp−1 ⊗k Sj is a projective cover of Sj. But Vp−1 ⊗k Sj obviously is ﬁnitely generated. It is indecomposable even as a k[U]-module. Moreover, using Lemma 21.8, we obtain

+ Vp−1 ⊗k Sj/ rad(Vp−1 ⊗k Sj) = (Vp−1 ⊗k Sj)/(u − 1)(Vp−1 ⊗k Sj) + = (Vp−1 ⊗k Sj)/ ((u − 1)Vp−1) ⊗k Sj + = (Vp−1/(u − 1)Vp−1) ⊗k Sj ∼ ∼ = S0 ⊗k Sj = Sj .

Hence, by Remark 6.9, it remains to show that Vp−1 ⊗k Sj is a projective k[B]-module. Let Vp−1 ⊗k Sj

M / N / 0 be an exact “test diagram” of k[B]-modules. Let 0 ≤ j′ < p − 1 such that ′ j ≡ −j mod p − 1. Observing that (Vp−1 ⊗k Sj) ⊗k Sj′ = Vp−1 we deduce the exact diagram

Vp−1 ˜ q q q ⊗idS ′ q j xq q M ⊗k Sj′ / N ⊗k Sj′ / 0 . ⊗id Sj′

Since Vp−1 is projective we ﬁnd a k[B]-module homomorphism ˜ such that the completed diagram commutes. But then := ˜ ⊗ idSj : Vp−1 ⊗k Sj −→ (M ⊗k Sj′ ) ⊗k Sj = M satisﬁes = ∘ .

Using Lemma 21.9 and our knowledge about Vp−1 we deduce the following properties of the indecomposable projective k[B]-modules Qj: ∼ – Qj = k[U] as a k[U]-module. i i+1 ∼ – F (Qj)/F (Qj) = S2i+j mod p−1 for 0 ≤ i ≤ p − 1, in particular, the i F (Qj) form a composition series of Qj. ∼ – Qj/ rad(Qj) = soc(Qj).

133 – The Cartan matrix of k[B] is of size (p − 1) × (p − 1) and has the form ⎛3 0 2 0 2 ...⎞ ⎜0 3 0 2 0 ...⎟ ⎜ ⎟ ⎜2 0 3 0 2 ...⎟ ⎜ ⎟ ⎜0 2 0 3 0 ...⎟ . ⎜ ⎟ ⎜2 0 2 0 3 ...⎟ ⎝. . . . . ⎠ . . . . .

In addition, the modules Qj have the following remarkable property. Definition. Let H be a finite group. A finitely generated k[H]-module is called uniserial if it has a unique composition series. Exercise. In a uniserial module the members of the unique composition series are the only submodules.

Lemma 21.10. Any Qj is a uniserial k[B]-module.

Proof. We show the apparently stronger statement that Qj is uniserial as a ∼ k[U]-module. Since Qj = k[U] it suﬃces to prove that k[U] is uniserial. By our earlier discussion this amounts to showing that k[Z]/Zpk[Z] is uniserial as a module over itself. This is left as an exercise to the reader.

It follows immediately that all

i Qj,i := Qj/F (Qj) for 0 ≤ j < p − 1 and 0 < i ≤ p are indecomposable k[B]-modules. They are pairwise nonisomorphic since any two either have different k-dimension or nonisomorphic factor modules modulo their radical. In this way we obtain p(p − 1) = ∣B∣ isomorphism classes of finitely generated indecomposable k[B]-modules. Remark 21.11. Let H be a finite group and let M be a (finitely generated) ∗ k[H]-module. The k-linear dual M := Homk(M, k) is a (finitely generated) k[H]-module with respect to the H-action defined by H × M ∗ −→ M ∗ (ℎ, l) 7−→ ℎl(x) := l(ℎ−1x) .

Suppose that M is ﬁnitely generated. Then the map N 7−→ N ⊥ := {l ∈ M ∗ : l∣N = 0} is an inclusion reversing bijection between the set of all k[H]-submodules of M and the set of all k[H]-submodules of M ∗ such that N ⊥ = (M/N)∗ and M ∗/N ⊥ = N ∗ .

134 It follows that M is a simple, resp. indecomposable, resp. uniserial, k[H]- module if and only if M ∗ is a simple, resp. indecomposable, resp. uniserial, k[H]-module. Lemma 21.12. Let H be a finite group; if the finitely generated indecomposable projective k[H]-modules are uniserial then all finitely generated indecomposable k[H]-modules are uniserial. Proof. Let M be any finitely generated indecomposable k[H]-module. We consider the family of all pairs of k[H]-submodules L ⊂ N of M such that N/L is nonzero and uniserial. Such pairs exist: Take, for example, any two consecutive submodules in a composition series of M. We fix a pair L ⊂ N in this family for which the length of the factor module N/L is maximal. In a first step we claim that there exists a k[H]-submodule N0 ⊆ N such that N = N0 ⊕ L. Since N/L is uniserial there is a unique k[H]-submodule L ⊆ L1 ⊂ N such that N/L1 is simple. Let : P −→ N/L1 be a projective cover of N/L1. By the projectivity we find a k[H]-module homomorphism : P −→ N such that the diagram

P y yy yy yy |yy pr N / N/L1 is commutative. We deﬁne N0 := im( ). As is surjective we have N0+L1 = pr N. If the composite map P −→ N −−→ N/L were not surjective its image would be contained in L1/L since this is the unique maximal submodule of N/L. This would imply that N0 ⊆ L1 which contradicts the above. The surjectivity of this composite map says that

N0 + L = N.

On the other hand N0, as a factor module of the indecomposable projective module P (cf. Prop. 7.4), is uniserial by assumption. Hence {0} ⊂ N0 is one of the pairs of submodules in the family under consideration. Because of the surjection N0 ↠ N/L and the Jordan-H¨olderProp. 1.2 the length of N0 cannot be smaller than the length of N/L. The maximality of the latter =∼ therefore implies that N0 −−→ N/L is an isomorphism and hence that

N0 ⊕ L = N. The above argument, in particular, says that we may assume our pair of submodules to be of the form {0} ⊂ N. In the second step, we pass to

135 the dual module M ∗. The Remark 21.11 implies that N ⊥ ⊂ M ∗ is a pair of k[H]-submodules of M ∗ such that M ∗/N ⊥ is a nonzero uniserial module of maximal possible length. Hence the argument of the ﬁrst step applied to M ∗ leads to existence of a k[H]-submodule ℳ ⊆ M ∗ such that ℳ ⊕ N ⊥ = M ∗. But with M also M ∗ is indecomposable. We conclude that N ⊥ = {0} and consequently that M = N is uniserial.

Proposition 21.13. Any ﬁnitely generated indecomposable k[B]-module Q is isomorphic to some Qj,i. Proof. It follows from Lemmas 21.10 and 21.12 that Q is uniserial. Let : 1 1 Qj −→ Q/F (Q) be a projective cover of the simple k[B]-module Q/F (Q). We then ﬁnd a commutative diagram of k[B]-module homomorphisms

Qj v vv vv vv vv zvv 1 Q pr / Q/F (Q).

Since is surjective the image of cannot be contained in the unique maximal submodule F 1(Q) of Q. It follows that is surjective and induces ∼ an isomorphism Qj,i = Q for an appropriate i.

Among the p(p − 1) modules Qj,i exactly the p − 1 modules Qj,p = Qj are projective. Hence using the Green correspondence as discussed above we derive the following result.

Proposition 21.14. There are exactly (p − 1)2 isomorphism classes of finitely generated indecomposable nonprojective k[G]-modules; they are the Green correspondents of the k[B]-modules Qj,i for 0 ≤ j < p − 1 and 1 ≤ i < p. ∼ Example. For 0 ≤ n < p we have Vn = Q−n mod p−1,n+1 as k[B]-modules. Next we will compute the Cartan matrix of k[G]. But first we have to establish a useful general fact about indecomposable projective modules. For any finite group H the group ring k[H] is a so called Frobenius algebra in the following way. Using the k-linear form

1 : k[H] −→ k X aℎℎ 7−→ a1 ℎ∈H

136 we introduce the k-bilinear form

k[H] × k[H] −→ k

(x, y) 7−→ 1(xy) .

Remark 21.15. i. The above bilinear form is symmetric, i. e., 1(xy) = 1(yx) for any x, y ∈ k[H]. ii. The map

=∼ ∗ k[H] −−→ k[H] = Homk(k[H], k)

x 7−→ x(y) := 1(xy)

is a k-linear isomorphism. P P Proof. i. If x = ℎ∈H aℎℎ and y = ℎ∈H bℎℎ then we compute X X 1(xy) = aℎbℎ−1 = bℎaℎ−1 = 1(yx) . ℎ∈H ℎ∈H P ii. It suﬃces to show that the map is injective. Let x = ℎ∈H aℎℎ such that x = 0. For any ℎ0 ∈ H we obtain

−1 X −1 X 0 = x(ℎ0 ) = 1 aℎℎℎ0 = 1 aℎℎ0 ℎ = aℎ0 ℎ∈H ℎ∈H and hence x = 0.

Remark 21.16. For any ﬁnitely generated projective k[H]-module P we have:

i. P ∗ is a projective k[H]-module;

ii. P ⊗k M, for any k[H]-module M, is a projective k[H]-module. iii. let M be any k[H]-module and let L ⊆ N ⊆ M be k[H]-submodules ∼ such that N/L = P ; then there exists a k[H]-submodule M0 ⊆ M such that ∼ ∼ ∼ M = P ⊕ M0,M0 ∩ N = L, and M0/M0 ∩ N = M/N .

137 Proof. i. By Prop. 6.4 the module P is isomorphic to a direct summand of some free module k[H]m. Hence P ∗ is isomorphic to a direct summand of k[H]∗m. Using Remark 21.15.ii we have the k-linear isomorphism

∼ k[H] −−→= k[H]∗ X X aℎℎ 7−→ aℎℎ−1 . ℎ∈H ℎ∈H The computation

−1 ′−1 ′−1 ℎ′ (ℎ′ℎ)−1 (y) = 1(ℎ ℎ y) = ℎ−1 (ℎ y) = ℎ−1 (y) , for any ℎ, ℎ′ ∈ H and y ∈ k[H], shows that it is, in fact, an isomorphism of k[H]-modules. ii. Again by Prop. 6.4 it suffices to consider the case of a free module P = k[H]m. It further is enough to treat the case m = 1. But in the proof of Prop. 10.4 (applied to the subgroup {1} ⊆ H and W := k, V := M) we have seen that ∼ dimk M k[H] ⊗k M = k[H] holds true. iii. (In case N = M the assertion coincides with the basic characteri- zation of projectivity in Lemma 6.2.) In a first step we consider the other extreme case where L = {0}. Then N is a finitely generated projective submodule of M. Denoting by N −→ M the inclusion map we see, using ∗ ∗ ∗ ∗ ∗ i., that N through : M ↠ N is a projective factor module of M . Lemma 6.2 therefore implies the existence of a k[H]-module homomorphism ∗ ∗ ∗ : N −→ M such that ∘ = idN ∗ . Dualizing again and identifying M in the usual way with a submodule of M ∗∗ we obtain M = N ⊕ ker(∗∣M). In the general case we first use the projectivity of N/L to find a submodule L′ ⊆ N such that N = L′ ⊕ L and L′ =∼ N/L =∼ P . Viewing now L′ as a projective submodule of M we apply the first step to obtain a submodule ′ ′ M0 ⊆ M such that M = L ⊕ M0. Then N = L ⊕ (M0 ∩ N).

Let k][H] = {{P1},..., {Pt}} be the set of isomorphism classes of ﬁnitely generated indecomposable projective k[H]-modules. It follows from Remarks ∗ 21.11 and 21.16.i that {Pi} 7−→ {Pi } induces a permutation of the set k][H]. Remark 21.11 also implies that the k[H]-modules

∼ ∗ ∗ ∗ soc(Pi) = (Pi / rad(Pi ))

138 are simple. Using Prop. 7.4.i we see that

k[[H] = {{P1/ rad(P1)},..., {Pt/ rad(Pt)}} = {{soc(P1)},..., {soc(Pt)}}. ∼ Proposition 21.17. soc(Pi) = Pi/ rad(Pi) for any 1 ≤ i ≤ t. Proof. For any 1 ≤ i ≤ t there is a unique index 1 ≤ i∗ ≤ t such that ∼ ∗ soc(Pi) = Pi∗ / rad(Pi∗ ). We have to show that i = i holds true. For this we ﬁx a decomposition k[H] = N1 ⊕ ... ⊕ Ns into indecomposable (projective) submodules. Deﬁning M Mi := Nj for any 1 ≤ i ≤ t

Nj =∼Pi we obtain a decomposition

k[H] = M1 ⊕ ... ⊕ Mt .

We know from Cor. 7.5 that Mi ∕= {0} (and hence soc(Mi) ∕= {0}) for any 1 ≤ i ≤ t. We also see that

soc(k[H]) = soc(M1) ⊕ ... ⊕ soc(Mt) is the isotypic decomposition of the semisimple k[H]-module soc(k[H]) with soc(Mi) being {soc(Pi)}-isotypic. Let 1 = e1 +...+et with ei ∈ Mi. By Prop. 5.1 the e1, . . . , et are pairwise orthogonal idempotents in k[H] such that Mi = k[H]ei. We claim that

∗ Mi soc(Mj) = {0} whenever i ∕= j .

Let x ∈ soc(Mj) such that Mix ∕= {0}. Since Mix is contained in the {soc(Pj)}-isotypic module soc(Mj) it follows that Mix and a fortiori Mi ⋅x ∼ (through Mi −−→ Mix) have a factor module isomorphic to soc(Pj) = Pj∗ / rad(Pj∗ ). But Mi/ rad(Mi) by construction is {Pi/ rad(Pi)}-isotypic. It follows that necessarily i = j∗. This shows our claim and implies that X ej∗ x = 1 − ei x = x for any x ∈ soc(Mj). i∕=j∗

Suppose that j∗ ∕= j. We then obtain, using Remark 21.15.i, that

1(x) = 1(ej∗ x) = 1(xej∗ ) = 1(xejej∗ ) = 1(0) = 0

139 and therefore that

{0} = 1(k[H]x) = 1(xk[H]) = x(k[H]) for any x ∈ soc(Mj). By Remark 21.15.ii this implies soc(Mj) = {0} which is a contradiction.

Exercise. (Bruhat decomposition) G = B ∪ BwU = B ∪ UwB with w := 0 1 −1 0 .

Lemma 21.18. i. The k[G]-module Vp is uniserial of length 2, and there is an exact sequence of k[G]-modules

0 −→ V1 −→ Vp −→ Vp−2 −→ 0 .

p p p−1 ii. The socle of Vp as a k[U]-module is equal to kY ⊕ k(X − XY ).

Proof. We deﬁne the k-linear map : Vp −→ Vp−2 by

(XiY p−i) := iXi−1Y p−1−i for 0 ≤ i ≤ p.

In order to see that is a k[G]-module homomorphism we have to check that (gv) = g (v) holds true for any g ∈ G and v ∈ Vp. By additivity it suffices to consider the vectors v = XiY p−i for 0 ≤ i ≤ p. Moreover, as a consequence of the Bruhat decomposition it also suffices to consider the group elements g = u+, w and g ∈ T . All these cases are easy one line computations. Obviously is surjective with ker( ) = kXp + kY p. On the other hand we have the injective ring homomorphism : k[X,Y ] −→ k[X,Y ] defined by (X) := Xp, (Y ) := Y p, and ∣k = id. a b For g = c d ∈ G we compute (gX) = (aX + cY ) = aXp + cY p = (aX + cY )p = (gX)p = g(Xp) = g (X) and similarly (gY ) = g (Y ). This shows that is an endomorphism of =∼ k[G]-modules. It obviously restricts to an isomorphism : V1 −−→ ker( ). Hence we have the exact sequence in i. Since V1 and Vp−2 are simple k[G]- modules the length of Vp is 2.

140 Next we prove the assertion ii. The socle of Vp (as a k[U]-module) contains the image under of the socle of V1 and is mapped by into the socle of Vp−2. Using Lemma 21.3.ii we deduce that

p p−2 kY ⊆ soc(Vp) and (soc(Vp)) ⊆ kY .

In particular, soc(Vp) is at most two dimensional. It also follows that

−1 p−2 p p p−1 soc(Vp) ⊆ (kY ) = kY ⊕ kX ⊕ kXY .

We have

u+Y p = Y p, u+Xp = Xp + Y p, u+(XY p−1) = XY p−1 + Y p ,

+ p p−1 p p−1 p p−1 hence u (X −XY ) = X −XY and therefore X −XY ∈ soc(Vp). It remains to show that Vp is uniserial. Suppose that it is not. Then p p ∼ Vp = (kX + kY ) ⊕ N for some k[G]-submodule N = Vp−2. We are going to use the ﬁltration

p p kY = W1 ⊂ W2 ⊂ ... ⊂ Wp ⊂ Wp+1 = Vp with Vp = Wp ⊕ kX introduced before Lemma 21.2. The identity

p p p−1 p kY ⊕ k(X − XY ) = soc(Vp) = kY ⊕ soc(N)

p implies that the socle of N contains a vector of the form u0 + X with p u0 ∈ Wp. Let v = u + aX with u ∈ Wp and a ∈ k be any vector in N. p Then v − a(u0 + X ) = u − au0 ∈ Wp ∩ N. If u − au0 ∕= 0 then this element, by Lemma 21.2.iii, generates, as a k[U]-module, Wi for some 1 ≤ i ≤ p. It p would follow that kY = W1 ⊆ Wi ⊆ N which is a contradiction. Hence p p u = au0, v = a(u0 + X ), and N ⊆ k(u0 + X ). But Vp−2 has k-dimension at least 2, and we have arrived at a contradiction again.

Lemma 21.19. i. For any n ≥ 1 there is an exact sequence of k[G]- modules 0 −→ Vn−1 −→ V1 ⊗k Vn −→ Vn+1 −→ 0 . ∼ ii. For 1 ≤ n ≤ p − 2 we have V1 ⊗k Vn = Vn−1 ⊕ Vn+1. Proof. i. Since G acts on k[X,Y ] by ring automorphisms the multiplication

: k[X,Y ] ⊗k k[X,Y ] −→ k[X,Y ]

f1 ⊗ f2 7−→ f1f2

141 is a homomorphism of k[G]-modules. It restricts to a surjective map

: V1 ⊗k Vn −→ Vn+1 . On the other hand we have the k-linear map

: Vn−1 −→ V1 ⊗k Vn v 7−→ X ⊗ Y v − Y ⊗ Xv .

a b For g = c d ∈ G we compute g (v) = g(X ⊗ Y v − Y ⊗ Xv) = gX ⊗ gY gv − gY ⊗ gXgv = (aX + cY ) ⊗ (bX + dY )gv − (bX + dY ) ⊗ (aX + cY )gv = abX ⊗ Xgv + adX ⊗ Y gv + cbY ⊗ Xgv + cdY ⊗ Y gv − (baX ⊗ Xgv + bcX ⊗ Y gv + daY ⊗ Xgv + dcY ⊗ Y gv) = (ad − bc)X ⊗ Y gv − (ad − bc)Y ⊗ Xgv = X ⊗ Y gv − Y ⊗ Xgv = (gv) .

Hence is a k[G]-module homomorphism. We obviously have im( ) ⊆ ker(). Since V1 ⊗k Vn = X ⊗ Vn ⊕ Y ⊗ Vn it also is clear that is injective. But

dimk Vn−1 = n = 2(n + 1) − (n + 2)

= dimk V1 ⊗k Vn − dimk Vn+1

= dimk ker() . It follows that im( ) = ker() which establishes the exact sequence in i. ii. In the given range of n the modules Vn−1 and Vn+1 are simple and nonisomorphic. As a consequence of i. the assertion ii. therefore is equivalent to V1 ⊗k Vn having a k[G]-submodule which is isomorphic to Vn+1, resp. to the nonvanishing of Homk[G](Vn+1,V1⊗k Vn). For n = p−2 this is clear, again by i., from the projectivity of Vp−1. We now argue by descending induction ∼ with respect to n. Suppose that V1 ⊗k Vn+1 = Vn ⊕ Vn+2 holds true. We ∗ observe that with V1 also V1 is a simple k[G]-module by Remark 21.11. But since V1, up to isomorphism, is the only simple two dimensional k[G]-module ∗ ∼ we must have V1 = V1. We also recall from linear algebra that the map

=∼ ∗ Homk[G](Vn+1,V1 ⊗k Vn) −−→ Homk[G](Vn+1 ⊗k V1 ,Vn)

A 7−→ [v ⊗ l 7−→ (l ⊗ idVn )(A(v))]

142 is bijective. It follows that ∼ ∗ Homk[G](Vn+1,V1 ⊗k Vn) = Homk[G](Vn+1 ⊗k V1 ,Vn) ∼ = Homk[G](Vn+1 ⊗k V1,Vn) ∼ = Homk[G](Vn ⊕ Vn+2,Vn) ∕= {0} .

For the remainder of this section let {P0},..., {Pp−1} be the isomorphism classes of ﬁnitely generated indecomposable projective k[G]-modules numbered in such a way that ∼ Pn/ rad(Pn) = Vn for 0 ≤ n < p (cf. Prop. 7.4). We already know from Prop. 21.7 that ∼ Pp−1 = Vp−1 . ∼ Proposition 21.20. Pp−2 = V1 ⊗k Vp−1.

Proof. With Vp−1 also the k[G]-module V1 ⊗k Vp−1 is projective by Prop. 21.7 and Remark 21.16.ii. Let

V1 ⊗k Vp−1 = L1 ⊕ ... ⊕ Ls be a decomposition into indecomposable projective k[G]-modules Li. By Lemma 21.19.i we have an injective k[G]-module homomorphism : Vp−2 −→ pr V1⊗k Vp−1. Obviously, not all of the composed maps Vp−2 −→ V1⊗k Vp−1 −−→ Li can be equal to the zero map. Since Vp−2 is simple we therefore ﬁnd a 1 ≤ i ≤ s together with an injective k[G]-module homomorphism Vp−2 −→ Li. But according to Prop. 21.17 the only indecomposable projective k[G]- module whose socle is isomorphic to Vp−2 is, up to isomorphism, Pp−2. Hence ∼ Pp−2 = Li, i. e., Pp−2 is isomorphic to a direct summand of V1 ⊗k Vp−1. To prove our assertion it remains to check that Pp−2 and V1 ⊗k Vp−1 have the same k-dimension or rather that dimk Pp−2 ≥ 2p. As already recalled from Prop. 21.17 we have ∼ ∼ Pp−2/ rad(Pp−2) = soc(Pp−2) = Vp−2 .

Since Pp−2 is indecomposable and Vp−2 is simple and nonprojective (cf. Prop. 21.7) we must have soc(Pp−2) ⊆ rad(Pp−2) (cf. Remark 21.22.i below). It follows that dimk Pp−2 ≥ 2 dimk Vp−2 = 2p − 2. On the other hand we know from Lemma 21.6.ii that p divides dimk Pp−2. As p ∕= 2 we conclude that dimk Pp−2 ≥ 2p.

143 Corollary 21.21. Pp−2 is a uniserial k[G]-module of length 3 such that [Pp−2] = 2[Vp−2] + [V1] in R(k[G]).

Proof. Since we know Vp−2 to be isomorphic to the socle of the indecomposable k[G]-module Pp−2 by Prop. 21.17 it follows from Prop. 21.20 and Lemma 21.19.i that [Pp−2] = [Vp−2] + [Vp] and that Pp−2 is uniserial if Vp is uniserial. It remains to invoke Lemma 21.18.i.

We point out the following general facts which are used repeatedly.

Remark 21.22. Let H be a ﬁnite group, and let M be a ﬁnitely generated projective k[H]-module; we then have:

i. If M is indecomposable but not simple then soc(M) ⊆ rad(M); if moreover, rad(M)/ soc(M) is simple then M is uniserial;

ii. suppose that M has a factor module of the form N1 ⊕ ... ⊕ Nr where the Ni are simple k[H]-modules; let Li ↠ Ni be a projective cover of Ni; then L1 ⊕ ... ⊕ Lr is isomorphic to a direct summand of M. Proof. i. By Prop. 21.17 we have the isomorphic simple modules soc(M) =∼ M/ rad(M). If soc(M) ∕⊆ rad(M) it would follow that the projection map ∼ soc(M) −−→= M/ rad(M) is an isomorphism. Hence M = soc(M) ⊕ rad(M) which is a contradiction. Since M is indecomposable soc(M) is the unique minimal nonzero and rad(M) the unique maximal proper submodule of M. The additional assumption on M therefore guarantees that there are no other nonzero proper submodules. ii. Let M = M1 ⊕ ... ⊕ Ms be a decomposition into indecomposable projective k[H]-modules Mi. Then N1⊕...⊕Nr is a factor module of the semisimple k[H]-module M/ rad(M) = M1/ rad(M1)⊕...⊕Ms/ rad(Ms). The Jordan-H¨olderProp. 1.2 now implies the existence of an injective map : {1, . . . , r} ,→ {1, . . . , s} such that ∼ Ni = M(i)/ rad(M(i)) ∼ and hence Li = M(i) for any 1 ≤ i ≤ r. Proposition 21.23. For p > 3 we have: ∼ i. V1 ⊗k Pp−2 = Pp−3 ⊕ Vp−1 ⊕ Vp−1; ∼ ii. V1 ⊗k Pn = Pn−1 ⊕ Pn+1 for any 2 ≤ n ≤ p − 3;

144 iii. [Pn] = 2[Vn] + [Vp−1−n] + [Vp−3−n] in R(k[G]) for any 1 ≤ n ≤ p − 3; ∼ iv. V1 ⊗k P1 = P0 ⊕ P2 ⊕ Vp−1;

v. P0 is a uniserial k[G]-module of length 3 such that [P0] = 2[V0]+[Vp−3]. ∼ If p = 3 then V1 ⊗k P1 = P0 ⊕ V2 ⊕ V2 ⊕ V2 and [P0] = 3[V0]. Proof. We know from Cor. 21.21 that ∼ ∼ ∼ Pp−2/ rad(Pp−2) = Vp−2, rad(Pp−2)/ soc(Pp−2) = V1, soc(Pp−2) = Vp−2 .

Hence V1 ⊗k Pp−2 has submodules N ⊃ L such that ∼ ∼ V1 ⊗k Pp−2/N = V1 ⊗k Vp−2 = Vp−3 ⊕ Vp−1, ∼ ∼ N/L = V1 ⊗k V1 = V0 ⊕ V2, ∼ ∼ L = V1 ⊗k Vp−2 = Vp−3 ⊕ Vp−1, where the second column of isomorphisms comes from Lemma 21.19.ii. Since Vp−1 is projective we may apply Remark 21.16.iii iteratively to obtain that V1 ⊗k Pp−2 has a factor module which is isomorphic to Vp−3 ⊕ Vp−1 ⊕ Vp−1, resp. V0 ⊕ V2 ⊕ V2 ⊕ V2 if p = 3. Using Remark 21.22.i we then see that the module Pp−3 ⊕ Vp−1 ⊕ Vp−1, resp. P0 ⊕ V2 ⊕ V2 ⊕ V2 if p = 3, is isomorphic to a direct summand of V1 ⊗k Pp−2. It remains to compare k-dimensions. We have dimk V1 ⊗k Pp−2 = 4p. Hence we have to show that dimk Pp−3 ≥ 2p if p > 3, resp. dimk P0 ≥ 3 if p = 3. From Prop. 21.17 and Prop. 21.7 we know (cf. Remark 21.22.i) that

dimk Pp−3 ≥ 2 dimk Vp−3 = 2p − 4 and from Lemma 21.6.ii that p divides dimk Pp−3. Both together obviously imply the asserted inequalities. This establish the assertion i. as well as the ﬁrst half of the case p = 3. We also obtain

[Pp−3] = [V1 ⊗k Pp−2] − 2[Vp−1]

= [Vp−3] + [Vp−1] + [V0] + [V2] + [Vp−3] + [Vp−1] − 2[Vp−1](17)

= 2[Vp−3] + [V2] + [V0] if p > 0, resp. [P0] = [V1 ⊗k P1] − 3[V2] = 3[V0] if p = 3. This is the case n = p − 3 of assertion iii. and the second half of the case p = 3.

145 Next we establish the case n = p−3 of assertion ii. (in particular p ≥ 5). Using (17) and Lemma 21.19.ii we obtain

[V1 ⊗k Pp−3] = 2[V1 ⊗k Vp−3] + [V1 ⊗k V0] + [V1 ⊗k V2]

= 2[Vp−4] + 2[Vp−2] + [V1] + [V1] + [V3]

= 2[Vp−2] + 2[Vp−4] + [V3] + 2[V1] . ∼ On the other hand V1⊗kPp−3 has the factor module V1⊗kVp−3 = Vp−4⊕Vp−2. Hence Remark 21.22.ii says that Pp−4 ⊕ Pp−2 is isomorphic to a direct summand of V1 ⊗k Pp−3. By Cor. 21.21 the summand Pp−2 contributes 2[Vp−2] + [V1] to the above class, whereas the summand Pp−4 contributes at least [Pp−4/ rad(Pp−4)] + [soc(Pp−4)] = 2[Vp−4]. The diﬀerence is [V1] + [V3] which cannot come from other indecomposable projective summands Pm of V1 ⊗k Pp−3 since each of those would contribute another 2[Vm]. It follows that ∼ V1 ⊗k Pp−3 = Pp−4 ⊕ Pp−2 and [Pp−4] = 2[Vp−4] + [V3] + [V1] . This allows us to establish ii. and iii. by descending induction. Suppose that

[Pn] = 2[Vn] + [Vp−1−n] + [Vp−3−n] and

[Pn+1] = 2[Vn+1] + [Vp−2−n] + [Vp−4−n] hold true for some 2 ≤ n ≤ p − 4 (the case n = p − 4 having been settled above). Using Lemma 21.19.ii we obtain on the one hand that

[V1 ⊗k Pn] = 2[V1 ⊗k Vn] + [V1 ⊗k Vp−1−n] + [V1 ⊗k Vp−3−n]

= 2[Vn−1] + 2[Vn+1] + [Vp−2−n] + [Vp−n] + [Vp−4−n] + [Vp−2−n]

= 2[Vn+1] + 2[Vn−1] + [Vp−n] + 2[Vp−2−n] + [Vp−4−n] . ∼ On the other hand V1 ⊗k Pn has the factor module V1 ⊗k Vn = Vn−1 ⊕ Vn+1 and hence, by Remark 21.22.ii, a direct summand isomorphic to Pn−1⊕Pn+1. This summand contributes to the above class at least 2[Vn−1] + 2[Vn+1] + [Vp−2−n]+[Vp−4−n]. Again the diﬀerence [Vp−n]+[Vp−2−n] cannot arise from other indecomposable direct summands of V1 ⊗k Pn. It therefore follows that ∼ V1 ⊗k Pn = Pn−1 ⊕ Pn+1 and [Pn−1] = 2[Vn−1] + [Vp−n] + [Vp−2−n] . It remains to prove the assertions iv. and v. Using iii. for n = 1 and Lemma 21.19.ii we have

[V1 ⊗k P1] = 2[V1 ⊗k V1] + [V1 ⊗k Vp−2] + [V1 ⊗k Vp−4]

= 2[V0] + 2[V2] + [Vp−3] + [Vp−1] + [Vp−5] + [Vp−3]

= [Vp−1] + 2[Vp−3] + [Vp−5] + 2[V2] + 2[V0] .

146 In particular, V1 ⊗k P1 has a subquotient isomorphic to the projective mod- ∼ ule Vp−1. It also has a factor module isomorphic to V1 ⊗k V1 = V0 ⊕ V2. Using Remarks 21.16.iii and 21.22.ii we see ﬁrst that V1 ⊗k P1 has a factor module isomorphic to V0 ⊕ V2 ⊕ Vp−1 and then that it has a direct summand isomorphic to P0 ⊕ P2 ⊕ Vp−1. By iii. for n = 2 the latter contributes to the above class at least 2[V0]+2[V2]+[Vp−3]+[Vp−5]+[Vp−1]. For a third time we argue that the diﬀerence [Vp−3] cannot come from another indecomposable direct summand of V1 ⊗k P1 so that we must have ∼ V1 ⊗k P1 = P0 ⊕ P2 ⊕ Vp−1 and [P0] = 2[V0] + [Vp−3] .

It ﬁnally follows from Remark 21.22.i that P0 is uniserial (for all p ≥ 3).

Remark. i. dimk P0 = dimk Pp−1 = p.

ii. dimk Pn = 2p for 1 ≤ n ≤ p − 2. iii. A closer inspection of the above proof shows that for any of the modules P = P1,...,Pp−3 the subquotient rad(P )/ soc(P ) is semisimple of length 2 so that P is not uniserial.

Exercise. V1 ⊗k P0 = P1. We deduce from Prop. 21.7, Cor. 21.21, and Prop. 21.23 that the Cartan matrix of k[G], which has size p × p, is of the form

⎛3 0 0⎞ ⎝0 3 0⎠ for p = 3 0 0 1 and ⎛ 2 1 0 ⎞ 2 . 1 2 . 1 ⎜ ... ⎟ ⎜ ... ⎟ ⎜ 2 1 . ⎟ ⎜ 3 1 ⎟ ⎜ 1 3 ⎟ for p > 3 . ⎜ . 1 2 ⎟ ⎜ ... ⎟ ⎜ ... ⎟ ⎜ ... ⎟ ⎝ 1 1 2 ⎠ 1 2 0 1

If we reorder the simple modules Vn into the sequence

V0,V(p−1)−2,V2,V(p−1)−4,V4,...,V(p−1)−1,V1,V(p−1)−3,V3,...,Vp−1

147 and correspondingly the Pn then the Cartan matrix becomes block diagonal, with 3 blocks, of the form ⎛ 2 1 ⎞ ⎜ 1 2 1 ⎟ ⎜ 1 2 ⎟ ⎜ ⎟ ⎜ . . . ⎟ ⎜ ...... ⎟ ⎜ ⎟ ⎜ 2 1 ⎟ ⎜ 1 3 0 ⎟ ⎜ ⎟ ⎜ 0 2 1 ⎟ . ⎜ ⎟ ⎜ 1 2 1 ⎟ ⎜ ⎟ ⎜ 1 2 ⎟ ⎜ . . . ⎟ ⎜ ...... ⎟ ⎜ ⎟ ⎜ 2 1 ⎟ ⎝ 1 3 0 ⎠ 0 1 This reﬂects the fact, as we will see later on, that k[G] has exactly three diﬀerent blocks.

22 Green’s indecomposability theorem

We ﬁx a prime number p and an algebraically closed ﬁeld of characteristic p. Before we come to the subject of this section we recall two facts about the matrix algebras Mn×n(k).

Lemma 22.1. Any automorphism T : Mn×n(k) −→ Mn×n(k) of the k- algebra Mn×n(k), for n ≥ 1, is inner, i. e., there exists a matrix T0 ∈ GLn(k) −1 such that T (A) = T0AT0 for any A ∈ Mn×n(k).

Proof. The algebra Mn×n(k) is simple and semisimple. Its, up to isomorphism, unique simple module is kn with the natural action. We now use T to deﬁne a new module structure on kn by n n Mn×n(k) × k −→ k (A, v) 7−→ T (A)v which we denote by T ∗kn. Since any submodule of T ∗kn also is a submodule of kn we obtain that T ∗kn is simple as well. Hence there must exist a module ∼ n = ∗ n isomorphism k −→ T k . This means that we ﬁnd a matrix T0 ∈ GLn(k) such that n T0Av = T (A)T0v for any v ∈ k and A ∈ Mn×n(k).

It follows that T0A = T (A)T0 for any A ∈ Mn×n(k).

148 Lemma 22.2. Let T be an automorphism of the k-algebra Mp×p(k) with the property that

T (ei) = ei+1 for 1 ≤ i ≤ p − 1 and T (ep) = e1 , where ei ∈ Mp×p(k) denotes the diagonal matrix with a 1 for the i-th entry of the diagonal and zeros elsewhere. Then the subalgebra

Q := {A ∈ Mp×p(k): T (A) = A} is local with Q/ Jac(Q) = k.

Proof. According to Lemma 22.1 we ﬁnd a matrix T0 ∈ GLp(k) such that −1 T (A) = T0AT0 for any A ∈ Mp×p(k). In particular

T0ei = ei+1T0 for 1 ≤ i ≤ p − 1 and T0ep = e1T0 .

This forces T0 to be of the form ⎛ ⎞ 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ tp ⎜ . ⎟ ⎜ t1 0 . ⎟ ⎜ ⎟ ⎜ . .. . ⎟ with t , . . . , t ∈ k×, ⎜ . t2 . . ⎟ 1 p ⎜ ⎟ ⎜ . .. . ⎟ ⎝ . . 0 . ⎠ 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ tp−1 0 which is conjugate to

⎛ 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ tp ⎞ ⎜ . ⎟ ⎜ 1 0 . ⎟ ⎜ ⎟ ⎜ . .. . ⎟ p ⎜ . 1 . . ⎟ where t = t1 ⋅ ... ⋅ tp . ⎜ ⎟ ⎜ . .. . ⎟ ⎝ . . 0 . ⎠ 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 1 0

The minimal polynomial of this latter matrix is Xp −tp = (X −t)p. It follows that the Jordan normal form of T0 is ⎛ t 0 ⎞ ⎜ 1 t ⎟ ⎜ ⎟ ⎜ .. .. ⎟ ⎜ . . ⎟ , ⎜ ⎟ ⎜ .. ⎟ ⎝ . t ⎠ 0 1 t

149 and hence that the algebra Q is isomorphic to the algebra

t ! t ! 1 t 1 t Q := A ∈ Mp×p(k): ⋅ ⋅ A = A ⋅ ⋅ . e ⋅ t ⋅ t 1 t 1 t We leave it to the reader to check that: a 0 0 ! ∗ a – Qe is the subalgebra of all matrices of the form ⋅ . a 0 ∗ ∗ a

0 0 ! ∗ 0 – I := ⋅ is a nilpotent two-sided ideal in Qe. 0 ∗ ∗ 0

– Q/Ie = k. In particular, Qe is local with I = Jac(Qe) (cf. Prop. 2.1.v and Prop. 4.1). We now let R be a noetherian, complete, local commutative ring such that R/ Jac(R) = k (e. g., R = k or R a (0, p)-ring for k), and we let G be a finite group. We fix a normal subgroup N ⊆ G as well as a finitely generated indecomposable R[N]-module L. In this situation Mackey’s Prop. 19.4 simplifies as follows. Let {g1, . . . , gm} ⊆ G be a set of representatives for the cosets in N∖G/N = G/N. We identify L with the R[N]-submodule G 1 ⊗ L ⊆ R[G) ⊗R[N] L = IndN (L). Then G ∼ −1 ∗ (18) IndN (L) = g1L ⊕ ... ⊕ gmL and giL = (gi ) L as R[N]-modules (cf. Remark 12.2.i). With L any giL is an indecomposable R[N]-module (cf. Remark 12.2). Hence (18) is, in fact, a decomposition of G IndN (L) into indecomposable R[N]-modules. In slight generalization of the discussion at the beginning of section 12 we have that

∗ ∼ IG(L) := {g ∈ G : g L = L as R[N]-modules} is a subgroup of G which contains N.

G Lemma 22.3. If IG(L) = N then the R[G]-module IndN (L) is indecomposable.

G Proof. Suppose that the R[G]-module IndN (L) = M1 ⊕ M2 is the direct sum of two submodules. By the Krull-Remak-Schmidt Thm. 4.7 one of this summands, say M1, has a direct summand M0, as an R[N]-module, which is ∼ −1 ∗ ∼ −1 ∗ ∼ isomorphic to L. Then giL = (gi ) L = (gi ) M0 = giM0 is (isomorphic to)

150 a direct summand of giM1 = M1. Using again Thm. 4.7 we see that in any decomposition of M1 into indecomposable R[N]-modules all the giL must occur up to isomorphism. On the other hand our assumption means that the R[N]-modules giL are pairwise nonisomorphic, so that any giL occurs G in IndN (L) with multiplicity one. This shows that necessarily M2 = {0}.

Remark 22.4. EndR[N](L) is a local ring with

EndR[N](L)/ Jac(EndR[N](L)) = k .

Proof. The ﬁrst half of the assertion is a consequence of Prop. 3.6 and Prop. 4.5. From Lemma 3.5.ii/iii we know that EndR[N](L) is ﬁnitely generated as an R-module and that

Jac(R) EndR[N](L) ⊆ Jac(EndR[N](L)) .

It follows that the skew ﬁeld D := EndR[N](L)/ Jac(EndR[N](L)) is a ﬁnite dimensional vector space over R/ Jac(R) = k. Since k is algebraically closed we must have D = k.

Lemma 22.5. Suppose that [G : N] = p and that IG(L) = G; then the ring G EG := EndR[G](IndN (L)) is local with EG/ Jac(EG) = k. Proof. We have the inclusions of rings

G G EG = EndR[G](IndN (L)) ⊆ EN := EndR[N](IndN (L)) G ⊆ E := EndR(IndN (L)) .

Fixing an element ℎ ∈ G such that ℎN generates the cyclic group G/N G we note that the action of ℎ on IndN (L) is an R-linear automorphism and therefore deﬁnes a unit ∈ E×. We obviously have

EG = { ∈ EN : = } .

We claim that the ring automorphism

Θ: E −→ E 7−→ −1 satisﬁes Θ(EN ) = EN and therefore restricts to a ring automorphism Θ : EN −→ EN . Let ∈ EN , which means that ∈ E satisﬁes (gx) = g (x)

151 G for any g ∈ N and x ∈ IndN (L). Then −1(gx) = ℎ (ℎ−1gx) = ℎ (ℎ−1gℎℎ−1x) = ℎℎ−1gℎ (ℎ−1x) = gℎ (ℎ−1x) = g( −1)(x)

G −1 −1 for any g ∈ N and x ∈ IndN (L). Hence ∈ EN , and similarly ∈ EN . This proves our claim, and we deduce that

EG = { ∈ EN : Θ( ) = } . An immediate consequence of this identity is the fact that

× × EG = EG ∩ EN . Using Prop. 2.1.iv it follows that

× EG ∩ Jac(EN ) = { ∈ EG : 1 + EN ⊆ EN } × ⊆ { ∈ EG : 1 + EG ⊆ EN ∩ EG} × = { ∈ EG : 1 + EG ⊆ EG } = Jac(EG) .

p−1 In order to compute EN we note that {1, ℎ, . . . , ℎ } is a set of representatives for the cosets in G/N. The decomposition (18) becomes

G p−1 IndN (L) = L ⊕ ℎL ⊕ ... ⊕ ℎ L. ∼ −1 ∗ ∼ Furthermore, by our assumption that IG(L) = G we have ℎL = (ℎ ) L = L and hence G ∼ IndN (L) = L ⊕ ... ⊕ L as R[N]-modules with p summands L on the right hand side. We ﬁx such an isomorphism. It induces a ring isomorphism

=∼ 0 0 EN −−→ Mp×p(EN ) with EN := EndR[N](L) 11 ⋅⋅⋅ 1p ! . . 7−→ . . p1 ⋅⋅⋅ pp

G where, if x ∈ IndN (L) corresponds to (x1, . . . , xp) ∈ L ⊕ ... ⊕ L, then (x) corresponds to p p X X 1j(xj),..., pj(xj) . j=1 j=1

152 The automorphism Θ of EN then corresponds to an automorphism, which 0 we denote by T , of Mp×p(EN ). On the other hand, using Lemma 2.5 and Remark 22.4, we have

=∼ 0 0 EN / Jac(EN ) −−→ Mp×p(EN / Jac(EN )) = Mp×p(k) . As any ring automorphism, T respects the Jacobson radical and therefore induces a ring automorphism T of Mp×p(k). Introducing the subring

Q := {A ∈ Mp×p(k): T (A) = A} we obtain the commutative diagram of injective ring homomorphisms EG/EG ∩ Jac(EN ) / Q _ _

=∼ EN / Jac(EN ) / Mp×p(k) .

We remark that Jac(R)EN ⊆ Jac(EN ) by Lemma 3.5.ii/iii. It follows that the above diagram in fact is a diagram of R/ Jac(R) = k-algebras. Next we observe that the automorphism Θ has the following property. Let i ∈ EN , for 1 ≤ i ≤ p, be the endomorphism such that ( j id if j = i − 1, i∣ℎ L = 0 otherwise.

Since (ℎjL) = ℎj+1L for 0 ≤ j < p − 1 and (ℎp−1L) = L we obtain

−1 −1 i = i+1 for 1 ≤ i < p and p = 1 and hence Θ( i) = i+1 for 1 ≤ i < p and Θ( p) = 1 . 0 The matrix in Mp×p(EN ) corresponding to i is the diagonal matrix with idL for the i-th entry of the diagonal and zeros elsewhere. It immediately follows that the k-algebra automorphism T of Mp×p(k) satisﬁes the assumption of Lemma 22.2. We conclude that the k-algebra Q is local with Q/ Jac(Q) = k. Let EG ∩Jac(EN ) ⊆ J ⊆ EG denote the preimage of Jac(Q). Since Jac(Q) is m nilpotent by Prop. 2.1.vi we have J ⊆ EG ∩ Jac(EN ) ⊆ Jac(EG) for some m ≥ 1. Using Prop. 2.1.v it follows that J ⊆ Jac(EG). The existence of the injective k-algebra homomorphism EG/J ,→ Q/ Jac(Q) = k then shows that EG/ Jac(EG) = k must hold true. Prop. 4.1 ﬁnally implies that EG is a local ring.

153 Theorem 22.6. (Green) Let N ⊆ G be a normal subgroup such that the index [G : N] is a power of p; for any ﬁnitely generated indecomposable G R[N]-module L the R[G]-module IndN (L) is indecomposable. Proof. Since G/N is a p-group we ﬁnd a sequence of normal subgroups

N = N0 ⊂ N1 ⊂ ... ⊂ Nl = G in G such that [Ni : Ni−1] = p for any 1 ≤ i ≤ l. By induction we therefore may assume that [G : N] = p. If IG(L) = N the assertion follows from Lemma 22.3. Suppose therefore that IG(L) = G. Then the ring EG := G EndR[G](IndN (L)) is local with EG/ Jac(EG) = k by Lemma 22.5. It also is complete by Prop. 3.6.iii. Hence Prop. 5.11 says that 1 is the only idempotent G in EG. On the other hand, the projection of IndN (L) onto any nonzero direct summand as an R[G]-module is obviously an idempotent in EG. It follows G that IndN (L) must be indecomposable.

154 Chapter V Blocks

Throughout this chapter we fix a prime number p, an algebraically closed field k of characteristic p, and a finite group G. Let E := E(G) := {e1, . . . , er} be the set of all primitive idempotents in the center Z(k[G]) of the group ring k[G]. We know from Prop. 5.5.iii/iv that the ei are pairwise orthogonal and satisfy e1 + ... + er = 1. We recall that the ei-block of k[G] consists of all k[G]-modules M such that eiM = M. An arbitrary k[G]-module M decomposes uniquely and naturally into submodules

M = e1M ⊕ ... ⊕ erM where eiM belongs to the ei-block. In particular, we have: – If a module M belongs to a block then any submodule and any factor module of M belongs to the same block.

– Any indecomposable module belongs to a unique block.

By Prop. 5.3 the block decomposition

k[G] = k[G]e1 ⊕ ... ⊕ k[G]er of k[G] is a decomposition into two-sided ideals. By Cor. 5.4 it is the ﬁnest such decomposition in the sense that no k[G]ei can be written as the direct sum of two nonzero two-sided ideals.

23 Blocks and simple modules

We deﬁne an equivalence relation on the set kg[G] as follows. For {P }, {Q} ∈ kg[G] we let {P } ∼ {Q} if there exists a sequence {P0},..., {Ps} in kg[G] such ∼ ∼ that P0 = P , Ps = Q, and

Homk[G](Pi−1,Pi) ∕= {0} or Homk[G](Pi,Pi−1) ∕= {0} for any 1 ≤ i ≤ s. We immediately observe that, if {P } and {Q} lie in diﬀerent equivalence classes, then

Homk[G](P,Q) = Homk[G](Q, P ) = {0} .

155 Let k[G] = Q1 ⊕ ... ⊕ Qm be a decomposition into indecomposable (projective) submodules. For any equivalence class C ⊆ kg[G] we deﬁne M PC := Qi ,

{Qi}∈C and we obtain the decomposition M k[G] = PC . C⊆kg[G]

Lemma 23.1. For any equivalence class C ⊆ kg[G] there exists a unique eC ∈ E such that PC = k[G]eC; the map

∼ set of equivalence classes in kg[G] −−→ E

C 7−→ eC is bijective.

Proof. First let {P }, {Q} ∈ kg[G] such that there exists a nonzero k[G]- module homomorphism f : P −→ Q. Suppose that P and Q belong to the e- and e′-block, respectively. Then P/ ker(f) belongs to the e-block and im(f) to the e′-block. Since P/ ker(f) =∼ im(f) ∕= {0} we must have e = e′. This easily implies that for any equivalence class C there exists a unique eC ∈ E such that PC belongs to the eC-block. Secondly, as already pointed out we have Homk[G](PC,PC′ ) = {0} for any two equivalence classes C ∕= C′. This implies that any k[G]-module endomorphism f : k[G] −→ k[G] satisﬁes f(PC) ⊆ PC for any C. Since multiplication from the right by any element in k[G] is such an endomorphism it follows that each PC is a two-sided ideal of k[G]. Hence, for any e ∈ E, M k[G]e = ePC C⊆kg[G] is a decomposition into two-sided ideals. It follows that there is a unique equivalence class C(e) such that ePC(e) ∕= {0}; then, in fact, k[G]e = ePC(e). Since PC(e) belongs to the eC(e)-block we must have e = eC(e) and k[G]eC(e) = PC(e). On the other hand, given any equivalence class C we have PC ∕= {0} by Cor. 7.5. Hence PC = eCPC implies C(eC) = C.

156 Remark 23.2. For any {P }, {Q} ∈ kg[G] we have Homk[G](P,Q) ∕= {0} if and only if the simple module P/ rad(P ) is isomorphic to a subquotient in some composition series of Q.

Proof. First we suppose that there exists a nonzero k[G]-module homomorphism f : P −→ Q. The kernel of f then must be contained in the unique ∼ maximal submodule rad(P ) of P . Hence P/ rad(P ) −−→= f(P )/f(rad(P )). f Vice versa, let us suppose that there are submodules L ⊂ N ⊆ Q such that N/L =∼ P/ rad(P ). We then have a surjective k[G]-module homomorphism P −→ N/L and therefore, by the projectivity of P , a commutative diagram of k[G]-module homomorphisms

P x f xx xx xx |xx N pr / / N/L , where f cannot be the zero map. It follows that Homk[G](P,Q) contains the f ⊆ nonzero composite P −→ N −−→ Q.

Proposition 23.3. Let C ⊆ kg[G] be an equivalence class; for any simple k[G]-module V the following conditions are equivalent:

i. V belongs to the eC-block; ii. V is isomorphic to P/ rad(P ) for some {P } ∈ C;

iii. V is isomorphic to a subquotient of Q for some {Q} ∈ C.

Proof. i. =⇒ ii. Let V belong to the eC-block. Picking a nonzero vector v ∈ V we obtain the surjective k[G]-module homomorphism

k[G] −→ V x 7−→ xv .

It restricts to a surjective k[G]-module homomorphism k[G]eC −→ eCV = V . Lemma 23.1 says that k[G]eC = PC is a direct sum of modules P such that {P } ∈ C. Hence V must be isomorphic to a factor module of one of these P . ii. =⇒ iii. This is trivial. iii. =⇒ i. By assumption V belongs to the same block as some Q with {Q} ∈ C. But Lemma 23.1 says that Q, and hence V , belongs to the eC-block.

157 Corollary 23.4. If the simple k[G]-module V is projective then it is, up to isomorphism, the only simple module in its block.

Proof. We consider any {Q} ∈ kg[G] diﬀerent from {V }. By Remark 21.16.iii the projective module V cannot be isomorphic to a subquotient of Q. Hence Homk[G](Q, V ) = Homk[G](V,Q) = {0}. It follows that the equivalence class of {V } consists of {V } alone.

Let us consider again the example of the group G = SL2(Fp) for p > 2. The simple k[G]-modules V0,V1,...,Vp−1, as constructed in Prop. 21.4, are distinguished by their k-dimension dimk Vn = n + 1. Let Pn −→ Vn be a projective cover. In Prop. 21.7, Cor. 21.21, and Prop. 21.23.iii/v we have determined the simple subquotients of each Pn. By using Remark 23.2 in order to translate the existence of certain subquotients into the existence of certain nonzero k[G]-module homomorphisms between the Pn we deduce the following facts:

– All simple subquotients of Pn for n even, resp. odd, are odd, resp. even, dimensional. This implies that Homk[G](Pn,Pm) = {0} whenever n and m have diﬀerent parity.

– Vp−1 is projective. – There exist nonzero k[G]-module homomorphisms:

Pp−3 / P0

Pp−5 / P2

. / P4

P2 / .

Pp−3

158 and Pp−4 / P1

Pp−6 / P3

. / P5

P1 / .

Pp−2

Together they imply that the equivalence classes in kg[G] are {P0}, {P2},..., {Pp−3} , {P1}, {P3},..., {Pp−2} , and {{Vp−1}}.

We conclude that SL2(Fp) has three blocks. More precisely, we have

E = {eeven, eodd, eproj} such that a simple module V lies in the

eeven-block ⇐⇒ dimk V is even,

eodd-block ⇐⇒ dimk V is odd and ∕= p, ∼ eproj-block ⇐⇒ V = Vp−1 .

24 Central characters

We have the decomposition Y Z(k[G]) = Ze with Ze := Z(k[G])e e∈E of the center as a direct product of the rings Ze (with unit element e). Since E is the set of all primitive idempotents in Z(k[G]) the rings Ze do not contain any other idempotent besides their unit elements. Therefore, by

159 Prop. 5.11, each Ze is a local ring. Moreover, the skew fields Ze/ Jac(Ze) are finite dimensional over the algebraically closed field k. Hence the maps

e : k −→ Ze/ Jac(Ze)

a 7−→ ae + Jac(Ze) are isomorphisms. This allows us to introduce the k-algebra homomorphisms

−1 pr pr e e : Z(k[G]) −−→ Ze −−→ Ze/ Jac(Ze) −−−→ k , which are called the central characters of k[G].

Exercise. The e correspond bijectively to the simple Z(k[G])-modules which all are one dimensional.

Proposition 24.1. Let e ∈ E, and let V be a simple k[G]-module; then V belongs to the e-block if and only if zv = e(z)v for any z ∈ Z(k[G]) and v ∈ V .

Proof. The equation ev = e(e)v = 1v = v immediately implies eV = V and therefore that V belongs to the e-block. Since k is algebraically closed Schur’s lemma implies that Endk[G](V ) = k idV . This means that the homomorphism Z(k[G]) −→ Endk[G](V ) induced by the action of Z(k[G]) on V can be viewed as a k-algebra homomorphism : Z(k[G]) −→ k such that zv = (z)v for any z ∈ Z(k[G]) and v ∈ V . The Q Jacobson radical Jac(Z(k[G])) = e Jac(Ze), of course, lies in the kernel of . Suppose that V belongs to the e-block. Then ev = v and e′v = 0 for e ∕= e′ ∈ E and any v ∈ V . It follows that (e) = 1 and (e′) = 0 for any ′ e ∕= e, and hence that = e. Let O(G) denote the set of conjugacy classes of G. For any conjugacy class O ⊆ G we deﬁne the element X Oˆ := g ∈ k[G] . g∈O

We recall that {Oˆ : O ∈ O(G)} is a k-basis of the center Z(k[G]). The multiplication in Z(k[G]) is determined by the equations X Oˆ1Oˆ2 = (O1, O2; O)Oˆ with (O1, O2; O) ∈ k. O∈O(G)

(Of course, the (O1, O2; O) lie in the prime ﬁeld Fp.)

160 25 Defect groups

The group G × G acts on k[G] by

(G × G) × k[G] −→ k[G] ((g, ℎ), x) 7−→ gxℎ−1 .

In this way k[G] becomes a k[G × G]-module. The two-sided ideals in the ring k[G] coincide with the k[G × G]-submodules of k[G]. We see that the block decomposition M k[G] = k[G]e e∈E is a decomposition of the k[G × G]-module k[G] into indecomposable submodules. For any e ∈ E we therefore may consider the set V0(k[G]e) of vertices of the indecomposable k[G × G]-module k[G]e. Of course, these vertices are subgroups of G×G. But we have the following result. To formulate it we need the “diagonal” group homomorphism

: G −→ G × G g 7−→ (g, g) .

Proposition 25.1. The k[G×G]-module k[G] is relatively k[(G)]-projective. Proof. Because of (gxℎ−1, 1)(G) = (gx, ℎ)(G) the map

G −−→∼ (G × G)/(G) x 7−→ (x, 1)(G) is an isomorphism of G×G-sets. By Lemma 11.6.i it induces an isomorphism

=∼ G×G k[G] −−→ Ind(G) (k) of k[G × G]-modules. The assertion therefore follows from Prop. 18.6.

Corollary 25.2. For any e ∈ E the indecomposable k[G × G]-module k[G]e has a vertex of the form (H) for some subgroup H ⊆ G; if H′ ⊆ G is another subgroup such that (H′) is a vertex of k[G]e then H and H′ are conjugate in G. Proof. By Prop. 25.1 and Lemma 18.2 the k[G × G]-module k[G]e, being a direct summand of k[G], is relatively k[(G)]-projective. This proves the ﬁrst half of the assertion. By Prop. 19.5 there exists an element (g, ℎ) ∈ G × G such that (H′) = (g, ℎ)(H)(g, ℎ)−1. It follows that H′ = gHg−1.

161 Deﬁnition. Let e ∈ E; the subgroups D ⊆ G such that (D) ∈ V0(k[G]e) are called the defect groups of the e-block.

Defect groups exist and are p-subgroups of G by Lemma 19.3. The defect groups of a single block form a conjugacy class of subgroups of G.

Lemma 25.3. Let e ∈ E, and let D be a defect group of the e-block; then k[G]e as a k[(G)]-module is relatively k[(D)]-projective.

Proof. Let k[G]e = L1 ⊕ ... ⊕ Ls be a decomposition of the indecomposable k[G × G]-module k[G]e into indecomposable k[(G)]-modules. By Lemma 18.2 it suﬃces to show that (D) ∈ V(Li) for any 1 ≤ i ≤ s. According to Lemma 20.1.i we ﬁnd elements (gi, ℎi) ∈ G × G such that

−1 (gi, ℎi)(D)(gi, ℎi) ∩ (G) ∈ V(Li) .

−1 But the equation (gi, ℎi)(d, d)(gi, ℎi) = (g, g) with d ∈ D and g ∈ G −1 −1 implies gidgi = ℎidℎi and therefore

−1 −1 (gi, ℎi)(d)(gi, ℎi) = (gi)(d)(gi) .

It follows that

−1 −1 (gi, ℎi)(D)(gi, ℎi) ∩ (G) ⊆ (gi)(D)(gi) and hence that (D) ∈ V(Li) (cf. Ex. 19.2.ii and Lemma 19.1). Proposition 25.4. Let e ∈ E, and let D be a defect group of the e-block; any k[G]-module M belonging to the e-block is relatively k[D]-projective.

Proof. We denote by (k[G]e)ad the k-vector space k[G]e viewed as a k[G]- ∼ module through the group isomorphism G −−→= (G). This means that G acts on (k[G]e)ad by

G × (k[G]e)ad −→ (k[G]e)ad (g, x) 7−→ gxg−1 .

As a consequence of Lemma 25.3 the module (k[G]e)ad is relatively k[D]- projective. We therefore ﬁnd, by Prop. 18.6, a k[D]-module L such that

162 ad G (k[G]e) is isomorphic to a direct summand of IndD(L). On the other hand we consider the k-linear maps

ad M −→ (k[G]e) ⊗k M −→ M v 7−→ e ⊗ v x ⊗ v 7−→ xv .

Because of

(gv) = e ⊗ gv = geg−1 ⊗ gv = g(e ⊗ v) = g (v) and

(g(x ⊗ v)) = (gxg−1 ⊗ gv) = gxg−1gv = gxv = g (x ⊗ v) both maps are k[G]-module homomorphisms. The composite map satisﬁes (v) = ev = v and hence is the identity map. It follows that M is isomor- ad phic to a direct summand of (k[G]e) ⊗k M. Together we obtain that M G is isomorphic to a direct summand of IndD(L) ⊗k M. But, as we have used before in the proof of Prop. 10.4, there are isomorphisms of k[G]-modules

G ∼ ∼ ∼ G IndD(L)⊗k M = (k[G]⊗k[D] L)⊗k M = k[G]⊗k[D] (L⊗k M) = IndD(L⊗k M).

Therefore, Prop. 18.6 implies that M is relatively k[D]-projective.

The last result says that a defect group of an e-block contains a vertex of any ﬁnitely generated indecomposable module in this block. Later on (Prop. 26.7) we will see that the defect group occurs among these vertices. Hence defect groups can be characterized as being the largest such vertices. At this point we start again from a diﬀerent end. For any x ∈ G we let CG(x) := {g ∈ G : gx = xg} denote the centralizer of x. Let O ⊆ G be a conjugacy class.

Deﬁnition. The p-Sylow subgroups of the centralizers CG(x) for x ∈ O are called the defect groups of the conjugacy class O.

For any p-subgroup P ⊆ G we deﬁne X IP := {kOˆ : P contains a defect group of O} ⊆ Z(k[G]) .

Exercise 25.5. i. Any two defect groups of O are conjugate in G.

ii. If P is a p-Sylow subgroup then IP = Z(k[G]).

163 ′ iii. If P ⊆ P then IP ⊆ IP ′ .

iv. IP only depends on the conjugacy class of P .

Lemma 25.6. Let O1, O2, and O in O(G) be such that (O1, O2; O) ∕= 0; if the p-subgroup P ⊆ G centralizes an element of O then it also centralizes elements of O1 and O2. Proof. Suppose that P centralizes x ∈ O. We deﬁne

X := {(y1, y2) ∈ O1 × O2 : y1y2 = x} as a P -set through P × X −→ X −1 −1 (g, (y1, y2)) 7−→ (gy1g , gy2g ) . One checks that ∣X∣ = (O1, O2; O) ∕= 0 in k .

On the other hand let X = X1 ∪ ... ∪ Xm be the decomposition of X into its P -orbits. Since P is a p-group the ∣Xi∣ all are powers of p. Hence either ∣Xi∣ = 0 in k or ∣Xi∣ = 1. We see that we must have at least one P -orbit Y = {(y1, y2)} ⊆ X consisting of one point. This means that P centralizes y1 ∈ O1 and y2 ∈ O2.

Proposition 25.7. Let P1,P2 ⊆ G be p-subgroups; then X −1 IP1 IP2 ⊆ IP1∩gP2g . g∈G

Proof. For i = 1, 2 let Oi ∈ O(G) such that Pi contains a defect subgroup Di of Oi. We have to show that X X ˆ ˆ ˆ −1 O1O2 = (O1, O2; O)O ∈ IP1∩gP2g . O∈O(G) g∈G

Obviously we only need to consider any O ∈ O(G) such that (O1, O2; O) ∕= 0. We pick a defect subgroup D of O. By deﬁnition D centralizes an element of O. Lemma 25.6 says that D also centralizes elements of O1 and O2. We therefore ﬁnd elements gi ∈ G, for i = 1, 2, such that −1 −1 −1 −1 D ⊆ g1D1g1 ∩ g2D2g2 ⊆ g1P1g1 ∩ g2P2g2 . −1 Setting g := g1 g2 and using Ex. 25.5.iv it follows that ˆ kO ⊆ I −1 −1 = IP ∩gP g−1 . g1P1g1 ∩g2P2g2 1 2

164 Corollary 25.8. IP , for any p-subgroup P ⊆ G, is an ideal in Z(k[G]). Proof. Let P ′ ⊆ G be a ﬁxed p-Sylow subgroup. Using Ex. 25.5.ii/iii we deduce from Prop. 25.7 that X IP Z(k[G]) = IP IP ′ ⊆ IP ∩gP ′g−1 ⊆ IP . g∈G

Remark 25.9. Let e ∈ E, and let P be a set of p-subgroups of G; if e ∈ P P ∈P IP then IQe = Ze for some Q ∈ P. Proof. We have X e = ee ∈ IP e ⊆ Ze . P ∈P

As noted at the beginning of section 24 the ring Ze is local. By Cor. 25.8 each IP e is an ideal in Ze. Since e∈ / Jac(Ze) we must have IQe ∕⊆ Jac(Ze) for at least one Q ∈ P. But IQe then contains a unit so that necessarily IQe = Ze. For any subgroup H ⊆ G we have in k[G] the subring

K[G]ad(H) := {x ∈ k[G]: ℎx = xℎ for any ℎ ∈ H} .

We note that k[G]ad(G) = Z(k[G]). Moreover, there is the k-linear “trace map”

ad(H) trH : k[G] −→ Z(k[G]) X x 7−→ gxg−1 . g∈G/H

It satisﬁes trH (yx) = trH (xy) = y trH (x) for any x ∈ k[G]ad(H) and y ∈ Z(k[G]).

Lemma 25.10. IP ⊆ im(trP ) for any p-subgroup P ⊆ G.

Proof. It suﬃces to show that Oˆ lies in the image of trP for any conjugacy class O ∈ O(G) such that Q := P ∩ CG(x) is a p-Sylow subgroup of CG(x)

165 for some x ∈ O. Since [CG(x): Q] is prime to p and hence invertible in k we may deﬁne the element

1 X y := ℎxℎ−1 ∈ k[G]ad(P ) . [CG(x): Q] ℎ∈P/Q

We compute

X 1 X −1 −1 1 X −1 trP (y) = g ℎxℎ g = gxg [CG(x): Q] [CG(x): Q] g∈G/P ℎ∈P/Q g∈G/Q X = gxg−1 = Oˆ .

g∈G/CG(x)

Proposition 25.11. Let P ⊆ G be a p-subgroup, and let e ∈ IP be an idempotent; for any k[G]-module M the k[G]-submodule eM is relatively k[P ]-projective.

Proof. According to Lemma 25.10 there exists an element y ∈ k[G]ad(P ) such that e = trP (y). Since y commutes with the elements in P the map

: eM −→ eM v 7−→ yv is a k[P ]-module homomorphism. Let {g1, . . . , gm} be a set of representatives Pm −1 for the cosets in G/P . Then i=1 giygi = e, which translates into the identity m X −1 gi gi = ideM . i=1 Our assertion therefore follows from Lemma 18.7.

Corollary 25.12. Let e ∈ E, and let P ⊆ G be a p-subgroup; if e ∈ IP then P contains a defect subgroup of the e-block.

Proof. By assumption we have

s X e = aiOˆi i=1

166 with ai ∈ k and such that O1,..., Os are all conjugacy classes such that P ∩ CG(xi), for some xi ∈ Oi, is a p-Sylow subgroup of CG(xi). We deﬁne a central idempotent " in k[G × G] as follows. Obviously

O(G × G) = O(G) × O(G) .

For O ∈ O(G) we let O−1 := {g−1 : g ∈ O} ∈ O(G). We put

s X −1 ∧ " := aiaj(Oi, Oj ) ∈ Z(k[G × G]) . i,j=1

With e also " is nonzero. To compute "2 it is more convenient to write P e = g∈G agg. We note that

– ag = 0 if g∈ / O1 ∪ ... ∪ Os, and – that e2 = e implies P a a = a . g1g2=g g1 g2 g P −1 Using the former we have " = g,ℎ∈G agaℎ(g, ℎ ). The computation

2 X −1 X −1 " = ag1 aℎ1 (g1, ℎ1 ) ag2 aℎ2 (g2, ℎ2 ) g1,ℎ1∈G g2,ℎ2∈G X X −1 = ag1 ag2 aℎ1 aℎ2 (g, ℎ ) g,ℎ∈G g1g2=g,ℎ1ℎ2=ℎ X X X −1 = ag1 ag2 aℎ1 aℎ2 (g, ℎ ) g,ℎ∈G g1g2=g ℎ1ℎ2=ℎ X −1 = agaℎ(g, ℎ ) g,ℎ∈G = " now shows that " indeed is an idempotent. For any pair (i, j) the group

−1 −1 (P × P ) ∩ CG×G((xi, xj )) = (P ∩ CG(xi)) × (P ∩ CG(xj ))

= (P ∩ CG(xi)) × (P ∩ CG(xj))

−1 is a p-Sylow subgroup of CG×G((xi, xj )) = CG(xi)×CG(xj). It follows that

" ∈ IP ×P ⊆ Z(k[G × G]) .

Hence we may apply Prop. 25.11 to P × P ⊆ G × G and " and obtain that the k[G × G]-module "k[G] is relatively k[P × P ]-projective. But applying

167 " = P a a P P (x, y−1) to any v ∈ k[G] gives i,j i j x∈Oi y∈Oj X X X X "v = aiaj xvy = aiajOˆivOˆj

i,j x∈Oi y∈Oj i,j X X = aiOˆi v ajOˆj i j = eve and shows that

"k[G] = ek[G]e = k[G]e .

It follows that P × P contains a subgroup of the form (g, ℎ)(D)(g, ℎ)−1 with g, ℎ ∈ G and D a defect subgroup of the e-block. We deduce that P contains the defect subgroup gDg−1.

Later on (Thm. 27.8.i) we will establish the converse of Cor. 25.12. Hence the defect groups of the e-block can also be characterized as being the smallest p-subgroups D ⊆ G such that e ∈ ID.

26 The Brauer correspondence

Let e ∈ E, let D be a defect group of the e-block, and put N := NG(D). Obviously N × N contains the normalizer NG×G((D)). Hence, by Green’s Thm. 20.6, the indecomposable k[G × G]-module k[G]e has a Green correspondent which is the, up to isomorphism, unique indecomposable direct summand with vertex (D) of k[G]e as a k[N × N]-module. Can we identify this Green correspondent? We ﬁrst establish the following auxiliary but general result.

Lemma 26.1. Let P be any ﬁnite p-group and Q ⊆ P be any subgroup; the P k[P ]-module IndQ(k) is indecomposable with vertex Q. Proof. Using the second Frobenius reciprocity in section 10 we obtain

P Homk[P ](k, IndQ(k)) = Homk[Q](k, k) = k id .

Since the trivial module k is the only simple k[P ]-module by Prop. 9.7 this P implies that the socle of the k[P ]-module IndQ(k) is one dimensional. But by the Jordan-H¨older Prop. 1.2 any nonzero k[P ]-module has a nonzero socle. It P follows that IndQ(k) must be indecomposable. In particular, by Prop. 18.6,

168 P P IndQ(k) has a vertex V contained in Q. So, again by Prop. 18.6, IndQ(k) is isomorphic to a direct summand of

m P P ∼ M P V −1 ∗ IndV (IndQ(k)) = IndV (Ind −1 ((gi ) k)) V ∩giQgi i=1 m M P = Ind −1 (k) . V ∩giQgi i=1

Here {g1, . . . , gm} ⊆ G is a set of representatives for the double cosets in V ∖G/Q, and the decomposition comes from Mackey’s Prop. 19.4. By what we have shown already each summand is indecomposable. Hence the Krull- P ∼ P Remak-Schmidt Thm. 4.7 implies that IndQ(k) = Ind −1 (k) for some V ∩giQgi 1 ≤ i ≤ m. Comparing k-dimensions we deduce that ∣Q∣ ≤ ∣V ∣ and consequently that Q = V .

Remark 26.2. In passing we observe that any p-subgroup Q ⊆ G occurs as a vertex of some ﬁnitely generated indecomposable k[G]-module. Let P ⊆ G be a p-Sylow subgroup containing Q. Lemma 26.1 says that the indecompos- P able k[P ]-module IndQ(k) has the vertex Q. It then follows from Lemma 20.4 G that IndQ(k) has an indecomposable direct summand with vertex Q. Since G the trivial k[G]-module k is a direct summand of IndP (k) (as a p ∤ [G : P ]) we, in particular, see that the p-Sylow subgroups of G are the vertices of k. We further deduce, using Prop. 25.4, that the p-Sylow subgroups of G also are the defect groups of the block to which the trivial module k belongs. We also need a few technical facts about the k[G×G]-module k[G] when we view it as a k[H × H]-module for some subgroup H ⊆ G. For any y ∈ G the double coset HyH ⊆ G is an H × H-orbit in G for the action

(H × H) × G −→ G −1 ((ℎ1, ℎ2), g) 7−→ ℎ1gℎ2 , and so k[HyH] is a k[H × H]-submodule of k[G]. We remind the reader that the centralizer of a subgroup Q ⊆ G is the subgroup CG(Q) = {g ∈ G : gℎ = ℎg for any ℎ ∈ G}. ∼ H×H Lemma 26.3. i. k[HyH] = Ind(1,y)−1(H∩yHy−1)(1,y)(k) as k[H × H]- modules.

ii. If H is a p-group then k[HyH] is an indecomposable k[H × H]-module with vertex (1, y)−1(H ∩ yHy−1)(1, y).

169 iii. Suppose that y∈ / H and that CG(Q) ⊆ H for some p-subgroup Q ⊆ H; then no indecomposable direct summand of the k[H × H]-module k[HyH] has a vertex containing (Q).

Proof. i. We observe that

−1 {(ℎ1, ℎ2) ∈ H × H :(ℎ1, ℎ2)y = y} = {(ℎ1, ℎ2) ∈ H × H : ℎ1yℎ2 = y} −1 = {(ℎ1, ℎ2) ∈ H × H : ℎ2 = y ℎ1y} = {(ℎ, y−1ℎy): ℎ ∈ H, y−1ℎy ∈ H} = {(1, y)−1(ℎ, ℎ)(1, y): ℎ ∈ H ∩ yHy−1} = (1, y)−1(H ∩ yHy−1)(1, y) .

Hence, by Remark 11.2, the map

H × H/(1, y)−1(H ∩ yHy−1)(1, y) −−→≃ HyH −1 (ℎ1, ℎ2) ... 7−→ ℎ1yℎ2 is an isomorphism of H × H-sets. The assertion now follows from Lemma 11.6.i. ii. This follows from i. and Lemma 26.1. iii. By Prop. 18.6 the assertion i. implies that each indecomposable summand in question has a vertex contained in (1, y)−1(H ∩ yHy−1)(1, y). Suppose therefore that this latter group contains an H × H-conjugate of (Q). We then ﬁnd elements ℎ1, ℎ2 ∈ H such that

−1 −1 −1 −1 (ℎ1, ℎ2)(Q)(ℎ1, ℎ2) ⊆ (1, y) (H ∩ yHy )(1, y) ⊆ (1, y) (G)(1, y) , hence −1 (ℎ1, yℎ2)(Q)(ℎ1, yℎ2) ⊆ (G) , and therefore

−1 −1 ℎ1gℎ1 = yℎ2g(yℎ2) for any g ∈ Q.

−1 It follows that ℎ1 yℎ2 ∈ CG(Q) ⊆ H. But this implies y ∈ H which is a contradiction.

For any p-subgroup D ⊆ G we put

ED(G) := {e ∈ E : D is a defect subgroup of the e-block} .

170 Theorem 26.4. (Brauer’s ﬁrst main theorem) For any p-subgroup D ⊆ G and any subgroup H ⊆ G such that NG(D) ⊆ H we have a bijection

∼ B : ED(G) −−→ ED(H) such that the k[H × H]-module k[H]B(e) is the Green correspondent of the k[G × G]-module k[G]e.

Proof. First of all we point out that, for any two idempotents e ∕= e′ in E(G), the k[G]-modules k[G]e and k[G]e′ belong to two diﬀerent blocks and therefore cannot be isomorphic. A fortiori they cannot be isomorphic as k[G × G]-modules. A corresponding statement holds, of course, for the k[H × H]-modules k[H]f with f ∈ E(H). Let e ∈ ED(G). Since H × H contains NG×G((D)) the Green correspondent Γ(k[G]e) of the indecomposable k[G × G]-module k[G]e exists, is an indecomposable direct summand of k[G]e as a k[H ×H]-module, and has the vertex (D) (see Thm. 20.6). But we have the decomposition M (19) k[G] = k[HyH] y∈H∖G/H as a k[H × H]-module. The Krull-Remak-Schmidt Thm. 4.7 implies that Γ(k[G]e) is isomorphic to a direct summand of k[HyH] for some y ∈ G. By Lemma 26.3.iii we must have y ∈ H. Hence Γ(k[G]e) is isomorphic to a direct summand of the k[H × H]-module k[H]. It follows that

Γ(k[G]e) =∼ k[H]f for some f ∈ ED(H). Our initial observation applied to H says that B(e) := f is uniquely determined. ′ ′ For e ∕= e in ED(G) the k[G × G]-modules k[G]e and k[G]e are nonisomorphic. By the injectivity of the Green correspondence the k[H × H]- modules Γ(k[G]e) and Γ(k[G]e′) are nonisomorphic as well. Hence B(e) ∕= B(e′). This shows that B is injective. For the surjectivity of B let f ∈ ED(H). The decomposition (19) shows that k[H]f, as a k[H×H]-module, is isomorphic to a direct summand of k[G] and hence, by the Krull-Remak-Schmidt Thm. 4.7, to a direct summand of k[G]e for some e ∈ E(G). It follows from Prop. 20.9 that necessarily ∼ k[H]f = Γ(k[G]e) and e ∈ ED(G). In particular, f = B(e). The bijection B in Thm. 26.4 is a particular case of the Brauer correspondence whose existence comes from the following result.

171 Lemma 26.5. Let f ∈ E(H) such that CG(D) ⊆ H for one (or equivalently any) defect group D of the f-block; we then have:

i. There is a unique (f) := e ∈ E(G) such that k[H]f is isomorphic, as a k[H × H]-module, to a direct summand of k[G]e;

ii. any defect group of the f-block is contained in a defect group of the e-block.

Proof. i. The argument for the existence of e was already given at the end of the proof of Thm. 26.4. Suppose that k[H]f also is isomorphic to a direct summand of k[G]e′ for a second idempotent e ∕= e′ ∈ E(G). Then k[H]f ⊕ L k[H]f is isomorphic to a direct summand of k[G] = e˜∈E(G) k[G]˜e. Since k[H]f, but not k[H]f ⊕ k[H]f is isomorphic to a direct summand of k[H] the decomposition (19) shows that k[H]f must be isomorphic to a direct summand of k[HyH] for some y ∈ G ∖ H. But this is impossible according to Lemma 26.3.iii (applied with Q = D). ii. Let D˜ be a defect group of the e-block. By Lemma 20.1.i applied to ∼ k[G]e = k[H]f ⊕ ... and H × H ⊆ G × G we ﬁnd a = (g1, g2) ∈ G × G ˜ −1 ˜ −1 such that (D) ⊆ (D) . Then D ⊆ g1Dg1 . For any p-subgroup D ⊆ G we put

E≥D(G) := {e ∈ E : D is contained in a defect group of the e-block} .

We fix a subgroup H ⊆ G and a p-subgroup D ⊆ H, and we suppose that ′ ′ CG(D) ⊆ H. Any other p-subgroup D ⊆ D ⊆ H then satisfies CG(D ) ⊆ CG(D) ⊆ H as well. Therefore, according to Lemma 26.5, we have the well defined map : E≥D(H) −→ E≥D(G) . Theorem 26.6. Let f ∈ E(H), and suppose that there is a finitely generated indecomposable k[H]-module N belonging to the f-block which has a vertex V that satisfies CG(V ) ⊆ H; we then have:

i. CG(D) ⊆ H for any defect group D of the f-block; in particular, (f) is deﬁned;

ii. if a ﬁnitely generated indecomposable k[G]-module M has a direct summand, as a k[H]-module, isomorphic to N then M belongs to the (f)- block.

172 Proof. By Prop. 25.4 we may assume that V ⊆ D. Then CG(D) ⊆ CG(V ) ⊆ H which proves i. Let e ∈ E(G) such that M belongs to the e-block. Suppose that e ∕= (f). We deduce from (19), by multiplying by f, the decomposition M fk[G] = k[H]f ⊕ fX with X := k[HyH] . y∈H∖G/H,y∈ /H

Since f commutes with the elements in H this is a decomposition of k[H×H]- modules. Similarly we have the decomposition

fk[G] = fk[G]e ⊕ fk[G](1 − e) as a k[H × H]-module. So fk[G]e is a direct summand of k[H]f ⊕ fX. By assumption the indecomposable k[H × H]-module k[H]f is not isomorphic to a direct summand of k[G]e = fk[G]e ⊕ (1 − f)k[G]e. Hence it is not isomorphic to a direct summand of fk[G]e. It therefore follows from the Krull-Remak-Schmidt Thm. 4.7 that fk[G]e must be isomorphic to a direct summand of fX and hence of X = fX ⊕ (1 − f)X as a k[H × H]-module. Using Lemma 26.3.iii we conclude that no indecomposable direct summand Y of the k[H ×H]-module fk[G]e has a vertex containing (V ). We consider any indecomposable direct summand Z of Y as a k[(H)]-module. Lemma 20.1.i applied to Y = Z ⊕ ... and (H) ⊆ H × H implies that any vertex of Z is contained in some vertex of Y (as a k[H × H]-module). It follows that no vertex of Z contains (V ). Using the notation introduced in the proof of Prop. 25.4 we conclude that no indecomposable direct summand of the k[H]-module (fk[G]e)ad has a vertex containing V . If we analyze the arguments in the proof of Prop. 25.4 then, in the present context, they give the following facts:

ad – fM = efM is isomorphic to a direct summand of (fk[G]e) ⊗k fM. – If Z is an indecomposable direct summand of (fk[G]e)ad with vertex U then Z ⊗k fM is relatively k[U]-projective. This implies that the ad vertices of the indecomposable direct summands of (fk[G]e) ⊗k fM are contained in vertices of the indecomposable direct summands of (fk[G]e)ad.

We deduce that the indecomposable direct summands of fM have no vertices containing V . But N = fN is a direct summand of M and hence of fM and it does have vertex V . This is a contradiction. We therefore must have e = (f).

173 Proposition 26.7. Let e ∈ E(G) and let D be a defect group of the e- block; then there exists a ﬁnitely generated indecomposable k[G]-module M belonging to the e-block which has the vertex D.

Proof. Let H = NG(D) and let X be any simple k[H]-module in the B(e)- block of k[H]. We claim that X has the vertex D. But D is a normal p- subgroup of H. Hence as a k[D]-module X is a direct sum

X = k ⊕ ... ⊕ k of trivial k[D]-modules k (cf. Prop. 9.7 and Thm. 12.3.i). On the other hand the B(e)-block has the defect group D so that, by Prop. 25.4, X is relatively k[D]-projective. We therefore may apply Lemma 20.1 to D ⊆ H and the above decomposition of X and obtain that any vertex of the trivial k[D]- module also is a vertex of the k[H]-module X. The former, by Lemma 26.1, are equal to D. Hence X has the vertex D. Green’s Thm. 20.6 now tells us that X is the Green correspondent of some ﬁnitely generated indecomposable k[G]-module M with vertex D. Moreover, Thm. 26.6 implies that M belongs to the (B(e)) = e-block.

Proposition 26.8. For any e ∈ E(G) the following conditions are equivalent:

i. The e-block has the defect group {1};

ii. any k[G]-module belonging to the e-block is semisimple;

iii. the k-algebra k[G]e is semisimple; ∼ iv. k[G]e = Mn×n(k) as k-algebras for some n ≥ 1; v. there is a simple k[G]-module X belonging to the e-block which is projective.

Proof. i. =⇒ ii. By Prop. 25.4 any module M belonging to the e-block is relatively k[{1}]-projective and hence projective. In particular, M/N is projective for any submodule N ⊆ M; it follows that M =∼ N ⊕ M/N. This implies that M is semisimple (cf. Prop. 1.4). ii. =⇒ iii. k[G]e as a k[G]-module belongs to the e-block and hence is semisimple. iii. =⇒ iv. Since e is primitive in Z(k[G]) the semisimple k-algebra k[G]e must, in fact, be simple (cf. Cor. 5.4). Since k is algebraically closed any ﬁnite dimensional simple k-algebra is a matrix algebra.

174 iv. =⇒ v. The simple module of a matrix algebra is projective. v. =⇒ i. We have seen in Cor. 23.4 that X is, up to isomorphism, the only simple module belonging to the e-block. This implies that any ﬁnitely generated indecomposable module belonging to the e-block is isomorphic to X and hence has the vertex {1}. Therefore the e-block has the defect group {1} by Prop. 26.7.

27 Brauer homomorphisms

Let D ⊆ G be a p-subgroup. There is the obvious k-linear map

s˜ : k[G] −→ k[CG(D)] ⊆ k[NG(D)] X X agg 7−→ agg .

g∈G g∈CG(D)

We observe that the centralizer CG(D) is a normal subgroup of the normalizer NG(D). This implies that the intersection O∩CG(D), for any conjugacy class O ∈ O(G), on the one hand is contained in CG(D), of course, but on the other hand is a union of full conjugacy classes in O(NG(D)). We conclude thats ˜ restricts to a k-linear map

s : Z(k[G]) −→ Z(k[NG(D)]) X Oˆ 7−→ oˆ .

o∈O(NG(D)), o⊆O∩CG(D)

Lemma 27.1. s is a homomorphism of k-algebras.

Proof. Clearly s respects the unit element. It remains to show that s is multiplicative. Let O1, O2 ∈ O(G). We have X Oˆ1Oˆ2 = ∣Bg∣g with Bg := {(g1, g2) ∈ O1 × O2 : g1g2 = g} g∈G and hence X s(Oˆ1Oˆ2) = ∣Bg∣g .

g∈CG(D) On the other hand X X X s(Oˆ1)s(Oˆ2) = g1 g2 = ∣Cg∣g

g1∈O1∩CG(D) g2∈O2∩CG(D) g∈CG(D)

175 with

Cg := {(g1, g2) ∈ O1 × O2 : g1, g2 ∈ CG(D) and g1g2 = g} .

Obviously Cg ⊆ Bg. We will show that, for g ∈ CG(D), the integer ∣Bg ∖ Cg∣ is divisible by p and hence is equal to zero in k. Since D centralizes g we may view Bg as a D-set via the action

D × Bg −→ Bg −1 −1 (ℎ, (g1, g2)) 7−→ (ℎg1ℎ , ℎg2ℎ ) .

The subset Cg consists of exactly all the ﬁxed points of this action. The complement Bg ∖ Cg therefore is the union of all D-orbits with more than one element. But since D is a p-group any D-orbit has a power of p many elements.

Literally the same reasoning works for any subgroup CG(D) ⊆ H ⊆ NG(D) and shows that we may view s also as a homomorphism of k-algebras

sD,H := s : Z(k[G]) −→ Z(k[H]) .

It is called the Brauer homomorphism (of G with respect to D and H). For any e ∈ E(G) its image s(e) either is equal to zero or is an idempotent. We want to establish a criterion for the nonvanishing of s(e).

Lemma 27.2. For any O ∈ O(G) we have:

i. O ∩ CG(D) ∕= ∅ if and only if D is contained in a defect group of O;

ii. if D is a defect group of O, then O ∩CG(D) is a single conjugacy class in NG(D);

iii. let o ∈ O(NG(D)) such that o ⊆ O; if D is a defect group of o then D also is a defect group of O and o = O ∩ CG(D).

Proof. i. First let x ∈ O ∩ CG(D). Then D ⊆ CG(X), and D is contained in a p-Sylow subgroup of CG(x). Conversely, let D be contained in a p- Sylow subgroup of CG(x) for some x ∈ O. Then D centralizes x and hence x ∈ O ∩ CG(D). ii. We assume that D is a p-Sylow subgroup of CG(x) for some x ∈ O. Then x ∈ O ∩ CG(D). Let y ∈ O ∩ CG(D) be any other point and let g ∈ G such that x = gyg−1. Since D centralizes y the conjugate group gDg−1 centralizes x. It follows that D as well as gDg−1 are p-Sylow subgroups of

176 −1 −1 CG(x). Hence we ﬁnd an ℎ ∈ CG(x) such that ℎDℎ = gDg . We see that −1 −1 −1 −1 −1 −1 −1 ℎ g ∈ NG(D) and (ℎ g)y(ℎ g) = ℎ gyg ℎ = ℎ xℎ = x.

iii. We now assume that D is a p-Sylow subgroup of CNG(D)(x) for some x ∈ o. Then D centralizes x and hence x ∈ O ∩ CG(D). It follows from i. that D is contained in a defect group of O and, more precisely, in a p-Sylow subgroup P of CG(x). If we show that D = P then D is a defect group of O and ii. implies that o = O ∩ CG(D). Let us therefore assume that D ⫋ P . Then, as in any p-group, also D ⫋ NP (D). We pick an element ℎ ∈ NP (D) ∖ D and let Q denote the p-subgroup generated by D and ℎ. We then have D ⫋ Q ⊆ NP (D) ⊆ P ⊆ CG(x) , from which we deduce that D ⫋ Q ⊆ CNG (x). But D was a p-Sylow subgroup of CNG (x). This is a contradiction.

Lemma 27.3. Let e ∈ E(G) with corresponding central character e; if D is minimal with respect to e ∈ ID then we have: i. s(e) ∕= 0;

ii. e(IQ) = {0} for any proper subgroup Q ⫋ D;

iii. if e(Oˆ) ∕= 0 for some O ∈ O(G) then D is contained in a defect group of O. Proof. i. Let S ⊆ O(G) denote the subset of all conjugacy classes O such that D contains a defect group DO of O. By assumption we have X e = aOOˆ with aO ∈ k. O∈S Suppose that X X X s(e) = aOs(Oˆ) = aO x = 0 .

O∈S O∈S x∈O∩CG(D)

It follows that, for any O ∈ S, we have aO = 0 or O ∩ CG(D) = ∅. The latter, by Lemma 27.2.i, means that D is not contained in a defect group of O. We therefore obtain that X ˆ X e = aOO ∈ IDO . O∈S,DO⫋D O∈S,DO⫋D

Remark 25.9 then implies that e ∈ IDO for some O ∈ S such that DO ⫋ D. This contradicts the minimality of D.

177 ii. By Remark 25.9 it follows from e ∈ ID that IDe = Ze is a local ring. Then IQe, for any subgroup Q ⊆ D, is an ideal in this local ring. But if Q ∕= D then the minimality property of D says that e∈ / IQe ⊆ IQ. It follows that IQe ⊆ Jac(Ze) and hence that e(IQ) = e(IQe) = {0}. iii. Let DO ⊆ G be a defect group of O. Using Prop. 25.7 we have X −1 eIDO ⊆ IDIDO ⊆ ID∩gDOg . g∈G

ˆ Because of O ∈ IDO the central character e cannot vanish on the right hand −1 sum. Hence, by ii., there must exist a g ∈ G such that D ⊆ gDOg . Lemma 27.4. Let e ∈ E(G) and O ∈ O(G); if the p-subgroup P is normal in G then we have:

i. If e ∈ IP then (P ) is a vertex of the k[G × G]-module k[G]e;

ii. if e ∈ IQ for some p-subgroup Q ⊆ G then P ⊆ Q;

iii. if O ∩ CG(P ) = ∅ then Oˆ ∈ Jac(Z(k[G])). Proof. i. According to the proof of Cor. 25.12 the indecomposable k[G × G]- module k[G]e is relatively k[P ×P ]-projective. Lemma 20.1 then implies that k[G]e as a k[G × G]-module and some indecomposable direct summand of k[G]e as a k[P × P ]-module have a common vertex. But by Lemma 26.3.ii the summands in the decomposition M k[G] = k[P yP ] y∈P ∖G/P are indecomposable k[P × P ]-modules having the vertex

(1, y)−1(P ∩ yP y−1)(1, y) .

It follows that k[G]e has a vertex of the form (P ∩ yP y−1) for some y ∈ G. Since P is assumed to be normal in G the latter group is equal to (P ). ii. We know from Prop. 25.11 that any k[G]-moudule M belonging to the e-block has a vertex contained in Q. Let X be a simple k[G]-module belonging to the e-block (cf. Prop. 7.4.i and Cor. 7.5 for the existence). By our assumption that P is a normal p-subgroup of G we may argue similarly as in the ﬁrst half of the proof of Prop. 26.7 to see that P is contained in any vertex of X.

178 iii. Let f , for f ∈ E(G), be any central character, and let the p-subgroup

Df ⊆ G be minimal with respect to f ∈ IDf . From ii. we know that P ⊆ Df . On the other hand P cannot be contained in a defect group of O by Lemma 27.2.i. Hence Df is not contained in a defect group of O, which implies that f (Oˆ) = 0 by Lemma 27.3.iii. We see that Oˆ lies in the kernel of every central character, and we deduce that Oˆ ∈ Jac(Z(k[G])).

In the following we need to consider the ideals IP for the same p-subgroup P in various rings Z(k[H]) for P ⊆ H ⊆ G. We therefore write IP (k[H]) in order to refer to the ideal IP in the ring Z(k[H]). We ﬁx a subgroup

DCG(D) ⊆ H ⊆ NG(D) and consider the corresponding Brauer homomorphism

sD,H : Z(k[G]) −→ Z(k[H]) .

Let e ∈ E(G) be such that sD,H (e) ∕= 0 (for example, by Lemma 27.3.i, if D is minimal with respect to e ∈ ID(k[G])). We may decompose

sD,H (e) = e1 + ... + er uniquely into a sum of primitive idempotents ei ∈ E(H) (cf. Prop. 5.5).

Lemma 27.5. If e ∈ ID(k[G]) then e1, . . . , er ∈ ID(k[H]). Proof. Again let S ⊆ O(G) denote the subset of all conjugacy classes O such that D contains a defect group of O. By assumption e is a linear combination of the Oˆ for O ∈ S. Hence sD,H (e) is a linear combination ofo ˆ for o ∈ O(H) such that o ⊆ O for some O ∈ S. Any defect group of such an o is contained in a defect group of O and therefore in gDg−1 for some g ∈ G (cf. Ex. 25.5.i). It follows that X e1 + ... + er = sD,H (e) ∈ IgDg−1∩H (k[H]) g∈G and consequently, by Cor. 25.8, that X ei = ei(e1 + ... + er) ∈ IgDg−1∩H (k[H)] for any 1 ≤ i ≤ r. g∈G

Remark 25.9 now implies that for any 1 ≤ i ≤ r there is a gi ∈ G such that

ei ∈ I −1 (k[H]) . giDgi ∩H Since D is normal in H we may apply Lemma 27.4.ii and obtain that D ⊆ −1 −1 giDgi ∩ H and hence, in fact, that D = giDgi ∩ H.

179 We observe that r M M k[H]sD,H (e) = k[H]f(e1 + ... + er) = k[H]ei . f∈E(H) i=1

Since D is normal in H it follows from Lemma 27.5 and Lemma 27.4.i that (D) is a vertex of the indecomposable k[H × H]-module k[H]ei for any 1 ≤ i ≤ r. On the other hand, G as an H ×H-set decomposes into G = H ∪(G∖H). Correspondingly, k[G] as a k[H ×H]-module decomposes into k[G] = k[H]⊕ k[G ∖ H]. We let H : k[G] −→ k[H] denote the associated projection map, which is a k[H × H]-module homomorphism.

Lemma 27.6. H (e) ∈ sD,H (e) + Jac(Z(k[H])). P ˆ Proof. Let e = O∈O(G) aOO with aO ∈ k. Then X X H (e) = aO x O∈O(G) x∈O∩H whereas s (e) = P a P x. It follows that (e) − D,H O∈O(G) O x∈O∩CG(D) H sD,H (e) is of the form X booˆ with bo ∈ k.

o∈O(H), o∩CG(D)=∅

Lemma 27.4.iii (applied to D ⊆ H) shows that this sum lies in Jac(Z(k[H])).

Proposition 27.7. The k[H × H]-module k[H]sD,H (e) is isomorphic to a direct summand of k[G]e.

Proof. We have the k[H × H]-module homomorphism

: k[H]sD,H (e) −→ k[G]e ⊗k[H] k[H]sD,H (e) v 7−→ e ⊗ v .

By Lemma 27.6 the composite := (H ⊗id)∘ ∈ Endk[H×H](k[H]sD,H (e)) satisﬁes

(v) = (H ⊗ id) ∘ (v) = H (e)v = (sD,H (e) + z)v

180 for any v ∈ k[H]sD,H (e) and some element z ∈ Jac(Z(k[H])). We note that (ei + z)ei = ei + zei, for any 1 ≤ i ≤ r, is invertible in the local ring Z(k[H])ei. Therefore the element

r Y −1 Y × y := (ei + zei) × 0 ∈ Z(k[H]) sD,H (e)

i=1 f∈E(H), f∕=ei is well deﬁned and satisﬁes

(sD,H (e) + z)y = sD,H (e) .

Then

(sD,H (e) + z)k[H]sD,H (e) = (sD,H (e) + z)yk[H]sD,H (e) = k[H]sD,H (e) which implies that the map is surjective. Since k[H]sD,H (e) is ﬁnite dimensional over k the map must be bijective. It follows that k[H]sD,H (e) is isomorphic to a direct summand of k[G]e ⊗k[H] k[H]sD,H (e). Furthermore the latter is a direct summand of

k[G]e = k[G]e ⊗k[H] k[H] = k[G]e ⊗k[H] k[H]sD,H (e) ⊕ k[G]e ⊗k[H] k[H](1 − sD,H (e)) .

Theorem 27.8. For any e ∈ E(G) and any defect group D0 of the e-block we have:

i. e ∈ ID0 (k[G]);

ii. if H = NG(D0) then sD0,H (e) ∈ E(H) and B(e) = sD0,H (e). Proof. With the notations from before Lemma 27.5, where D ⊆ G is a p- subgroup which is minimal with respect to e ∈ ID(k[G]), we know from Lemma 27.5 that e1, . . . , er ∈ ID(k[H]). Applying Lemma 27.4.i to the normal p-subgroup D of H we obtain that (D) is a vertex of the indecomposable k[H × H]-modules k[H]ei. But by Prop. 27.7 these k[H]ei are isomorphic to direct summands of k[G]e. Hence it follows from Lemma 20.1 −1 −1 that D ⊆ gD0g for some g ∈ G. We see that e ∈ ID ⊆ IgD0g = ID0 which proves i. In fact Cor. 25.12 says that D has to contain some defect subgroup −1 of the e-block. We conclude that D = gD0g and therefore that D0 is

181 minimal with respect to e ∈ ID0 (k[G]) as well. This means that everywhere in the above reasoning we may replace D by D0 (so that, in particular, D0CG(D0) ⊆ H ⊆ NG(D0)). We obtain that D0 also is a defect group of the ei-blocks for 1 ≤ i ≤ r. Let H = NG(D0) so that NG×G((D0)) ⊆ H ×H. By the Green correspondence in Thm. 20.6 the indecomposable k[G×G]-module k[G]e with vertex D0 has, up to isomorphism, a unique direct summand, as a k[H × H]-module, with vertex D0. It follows that r = 1, which means that sD0,H (e) = e1 ∈ E(H), and that this summand is isomorphic to k[H]e1, which means that B(e) = e1. This proves ii.

Proposition 27.9. For any e ∈ E(G) and any defect group D0 of the e- block we have:

i. If the p-Sylow subgroup Q ⊆ G contains D0 then there exists a g ∈ G −1 such that D0 = Q ∩ gQg ;

ii. any normal p-subgroup P ⊆ G is contained in D0. Proof. i. By assumption and Ex. 19.2.iii the indecomposable k[G × G]- module k[G]e is relatively k[Q×Q]-projective. Lemma 20.2 therefore implies that k[G]e as a k[Q × Q]-module has an indecomposable direct summand X with vertex (D0). Applying Lemma 26.3.ii to the decomposition M k[G] = k[QyQ] y∈Q∖G/Q we see that X also has a vertex of the form (1, y)−1(Q ∩ yQy−1)(1, y) for some y ∈ G. It follows that −1 −1 −1 (D0) = (q1, q2)(1, y) (Q ∩ yQy )(1, y)(q1, q2) −1 for some (q1, q2) ∈ Q×Q. We obtain on the one hand that ∣D0∣ = ∣Q∩yQy ∣ and on the other hand that −1 −1 −1 −1 −1 q1 dq1 = yq2 d(yq2 ) ∈ Q ∩ yQy for any d ∈ D0. −1 −1 The element g := q1yq2 then lies in CG(D0). Hence D0 ⊆ Q ∩ gQg . Moreover −1 −1 −1 −1 ∣Q ∩ gQg ∣ = ∣Q ∩ q1yq2 Qq2y q1 ∣ −1 −1 −1 −1 = ∣q1 Qq1 ∩ yq2 Qq2y ∣ = ∣Q ∩ yQy ∣ = ∣D0∣ . ii. Since P is contained in any p-Sylow subgroup of G it follows from i. that P ⊆ D0.

182 Theorem 27.10. Let D ⊆ G be a p-subgroup and ﬁx a subgroup DCG(D) ⊆ H ⊆ NG(D). We then have:

i. E≥D(H) = E(H);

ii. the Brauer correspondence : E(H) −→ E≥D(G) is characterized in terms of central characters by

(f) = f ∘ sD,H for any f ∈ E(H);

iii. if e ∈ E(G) such that sD,H (e) ∕= 0 then e ∈ im( ) and X sD,H (e) = f ; f∈ −1(e)

iv. ED(G) ⊆ im( ). Proof. i. Since D is normal in H this is immediate from Prop. 27.9.ii. ii. For any f ∈ E(H) the composed k-algebra homomorphism f ∘ sD,H must be a central character of k[G], i. e., f ∘ sD,H = e for some e ∈ E(G). Since 1 = e(e) = f (sD,H (e)) we have sD,H (e) ∕= 0. Let sD,H (e) = e1 + ... + er with ei ∈ E(H). Then 1 = f (e1) + ... + f (er) and hence f = ej for some 1 ≤ j ≤ r. By Prop. 27.7 the indecomposable k[H × H]-module k[H]f is isomorphic to a direct summand of k[G]e. It follows that e = (f). iii. Let sD,H (e) = e1 + ... + er with ei ∈ E(H). Again it follows from Prop. 27.7 that e = (e1) = ... = (er). On the other hand, if e = (f) for some f ∈ E(H) then we have seen in the proof of ii. that necessarily f = ej for some 1 ≤ j ≤ r. iv. If e ∈ ED(G) then D is minimal with respect to e ∈ ID(k[G]) by Cor. 25.12 and Thm. 27.8.i. Hence sD,H (e) ∕= 0 by Lemma 27.3.i.

Proposition 27.11. For any e ∈ E(G) and any defect group D0 of the e-block we have:

i. If e(Oˆ) ∕= 0 for some O ∈ O(G) then D0 is contained in a defect group of O;

ii. there is an O ∈ O(G) such that e(Oˆ) ∕= 0 and D0 is a defect group of O.

Proof. i. By Cor. 25.12 and Thm. 27.8.i the defect group D0 is minimal with respect to e ∈ ID0 (k[G]). The assertion therefore follows from Lemma 27.3.iii.

183 ii. Since e ∈ ID0 (k[G]) we have X e = aOOˆ with aO ∈ k O∈S where S ⊆ O(G) is the subset of all O such that D0 contains a defect group P ˆ of O. Then 1 = e(e) = O∈S aO e(O). Hence there must exist an O ∈ S such that e(Oˆ) ∕= 0. This O has a defect group contained in D0 but by i. also one which contains D0. It follows that D0 is a defect group of O.

184 References

[TdA] Schneider P.: Die Theorie des Anstiegs. Course at M¨unster, 2006/2007.

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