The representation ring and the center of the group ring

Transcription

1 The representation ring and the center of the group ring (1) M 1 (12) M (123) M Master s thesis Jos Brakenhoff supervisor Bart de Smit Mathematisch instituut Universiteit Leiden February 28, 2005

2 Contents 1 Introduction 2 2 Comparison of discriminants The representation ring and the center of the group ring Discriminants Divisibility of discriminants Comparison of spectra Spectra The spectrum of the representation ring The spectrum of the center of the group ring Comparison of Q-algebras Q-algebras Finite abelian étale algebras Two categories An equivalence of categories Brauer equivalence Q-algebras, continuation

3 Chapter 1 Introduction Let G be a finite group of order g. For this group we will construct two commutative rings, which we will compare in this thesis. One of these rings, the representation ring, is built up from the representations of G. A representation of the group G is a finite dimensional C-vector space M together with a linear action of G, that is a homomorphism G GL(M). With this action M becomes a C[G]-module. For each representation M of G and an element σ G we look at the trace of the map M M : m σm, which we will denote by Tr M (σ) If σ and τ are conjugate elements of G, then Tr M (σ) = Tr M (τ). Let G/ be the set of conjugacy classes of G. We obtain a map χ M : G/ C x Tr M (σ), with σ x. This map is called the character of M. So we have a map {representations of G} C G/ M χ M. The representation ring R(G) is the subring of C G/ generated by χ M for all representations M. Two representations are isomorphic if they have the same image in C G/. We have two identities one for addition: χ M + χ N = χ M N and the other for multiplication: χ M χ N = χ M N. We can write every C[G]-module in a unique way as a direct sum of simple modules, that is, non-zero modules without proper submodules. The representation ring is a free Z-module, with a basis S consisting of the isomorphism classes of the simple modules, we have R(G) = Z χ S. The number of simple modules is #(G/ ), so R(G) C is isomorphic to C G/. S S The other ring is the center of the group ring Z[G], which we will call Λ(G). This ring is free with basis { σ x σ : x G/ }. For this ring we have an embedding into C S : where Λ(G) C S ( ) #x σ dim C S χ S(x) σ x S S ( ) #x dim C S χ S(σ) is the scalar with which σ x σ acts on S., 2

4 For an abelian group G these two rings are isomorphic. The ring Λ(G) is equal to Z[G] and the representation ring can be identified with Z[Ĝ], where Ĝ is Hom(G, C ), the dual of G. For other groups they are not always isomorphic, but they have some similarities, for example, they are both a free Z-module of rank n = #(G/ ). In this thesis we will compare these two rings on several aspects. We will compare their discriminants, their spectra and the Q-algebra they generate. In the second chapter we compare the discriminants of these rings. The only primes which divide the discriminant of R(G) are the primes which divide the order of G. The same is true for the discriminant of Λ(G). For groups of order less than 512 the quotient (Λ(G)) (R(G)) is in Z. We ask whether this is true for all groups. and shall prove this for groups of order p k and pq, where p and q are primes and k 4. In the third chapter we will give a description of the spectra of R(G) and Λ(G) over A, the subring of C generated by the g-roots of unity. Since all characters have images in A, the spectra of R(G) A and Λ(G) A are easier to compute than the spectra of R(G) and Λ(G). We will get surjective maps Spec(A n ) Spec(R(G) A) Spec(A) and Spec(A n ) Spec(Λ(G) A) Spec(A), such that for all primes p of A not dividing #G there are n points of Spec(R(G) A) respectively Spec(Λ(G) A) which map to p. For R(G) A we will calculate the spectrum and show that it is connected. For Λ(G) A we will give a description of a spectrum between Spec(A n ) and Spec(Λ(G) A); it remains a question whether this spectrum is in fact equal to Spec(Λ(G) A). The spectrum of Λ(G) is also connected. The rings R(G) and Λ(G) are connected by the pairing R(G) ( Λ(G) C χ S, ) σ χ S(σ) dim σ x C S S S, x G/. If we view R(G) and Λ(G) over Q, we see that they are the row span respectively the column span of the matrix ( ) 1 dim C S χ S(σ). S S,[σ] G/ In the last chapter we will generalize this setting and make an equivalence between two categories. For the first category the objects are matrices with entries in C of which both the row and column span over Q are rings. For the other category the objects are two abelian finite étale algebras, that is, finite étale algebras for which the Galois group is abelian, with a pairing. From this we will derive that R(G) Q and Λ(G) Q are abelian finite étale algebras which are Brauer equivalent; see section 4.5 for the definition. Furthermore, we have an action of Γ = Gal (Q ab /Q) on them, which satisfies γm, c = γ M, c = M, γc for all M R(G) Q, c Λ(G) Q and γ Γ, where, is the Q-bilinear pairing R(G) ( Q Λ(G) Q C χ S 1, ) σ 1 χ S(σ) dim σ x C S S S, x G/. 3

5 Chapter 2 Comparison of discriminants In this chapter we will introduce the representation ring R(G) and the center of the group ring Λ(G) of a finite group G. Using the characters, we will give an ring embedding into C n, where n is the number of conjugacy classes of G. With this embedding we can calculate the discriminant of these rings. The question we want to answer, is whether these discriminants divide each other. We will prove for groups of order p k or pq, with p and q prime and k 4 that this indeed the case. 2.1 The representation ring and the center of the group ring First we define the representation ring and center of the group ring and describe their ring structure. Let G be a finite group of order g. Let X = G/ be the set of conjugacy classes of G. Let S be a set of representatives for the isomorphism classes of simple C[G]-modules. For each finitely generated C[G]-module M we have χ M : G C, the character of M, defined by χ M (σ) = Tr(M M : m σm). If σ and τ are conjugate elements of G, then χ M (σ) = χ M (τ). We define χ M (x) = χ M (σ) for x X, where σ is an element of x. Furthermore, if M and N are isomorphic modules, then χ M = χ N. Let R(G) be the Grothendieck group of finitely generated C[G]-modules, that is, the abelian group given by generators the set of isomorphism classes of finitely generated C[G]-modules and relations {[M 2 ] = [M 1 ] + [M 3 ] : 0 M 1 M 2 M 3 0 a short exact sequence}. We will write [M] for the isomorphism class of M. One can prove that [M] = [N] if and only if M and N are isomorphic The group R(G) becomes a ring with the multiplication [M] [N] = [M C N] for M and N finitely generated C[G]-modules [4, sect. 1.5]. As a group R(G) is a free Z-module on {[S] : S S}. Let Λ(G) be the center of the group ring Z[G]. As a group it is a free Z-module on {c x = σ x σ : x X}. Example If G is an abelian group, then X = G and we can take S = {S χ : χ Hom(G, C )}, where S χ is C with G-action gz = χ(g)z for g G and z C. 4

6 Now, R(G) as a group is S S Z[S] and for the multiplication we have [S χ1 ] [S χ2 ] = [S χ1 C S χ2 ] = [S χ1 χ 2 ]. So, We obtain the ring isomorphism R(G) = Z[Hom(G, C )]. Furthermore, Λ(G) = Z[G]. Since G and Hom(G, C ) are isomorphic groups, R(G) and Λ(G) are isomorphic rings. We will see that in general R(G) and Λ(G) are non-isomorphic. To better understand these rings, we are going to give an explicit description of their structure. To do this, we first define the ring homomorphism φ : C[G] Π S S End C (S) σ (s σs) S S, which sends an element of C[G] to all its actions on the simple modules. From representation theory we know that φ is an isomorphism [3, chap. XVIII, sect. 4]. So, the centers of C[G] and Π S S End C (S) are isomorphic. Using the notation Z(R) for the center of the ring R, we have Z(C[G]) = x X c x C and Z(Π S S End C (S)) = Π S S Z(End C (S)) = Π S S I S C, with I S the identity on S. On the center, we can write φ(c x ) = (α S I S ) S, where 1 α S = dim C S Tr(action of c x on S) 1 = χ S (σ) = #x dim C S dim C S χ S(x). σ x The isomorphism of the centers x X c xc Π S S I S C is given by the matrix ( ) #x dim C S χ S(x). (2.1) S S,x X It follows that this matrix is invertible. By restricting the isomorphism of the centers on the left side to Λ(G), we have proved the following lemma. Lemma The map is an injective ring homomorphism. Λ(G) C( S ) #x c x dim C S χ S(x) S S A similar description for R(G) is given by the following lemma. Lemma The map is an injective ring homomorphism.. (2.2) R(G) C X [S] (χ S (x)) x X. (2.3) 5

7 Proof. First note that if M and N are isomorphic modules, then χ M = χ N. So this map is independent of the choice of S. Furthermore, for all finitely generated C[G]-modules M, N, we have χ M + χ N = χ M N and χ M χ N = χ M C N, which can be seen by writing down the corresponding matrices or looking at [4, sect. 2.1, prop, 2]. Finally, since the matrix (2.1) is invertible, the matrix (χ S (x)) S S,x X is invertible, so the C-linear map R(G) C C X [S] (χ S (x)) x X. is a bijection. After restricting the left side to R(G), we obtain an injection. The matrix (χ S (x)) S S,x X is called the character table of G. Example Take G = S 3, the symmetric group on three elements. Then X = {(1), (1 2), (1 2 3)} and S = M 1, M ɛ, M 2, where M 1 and M ɛ are 1-dimensional C-modules, with the following actions G M 1 M 1 (σ, m) m G M ɛ M ɛ (σ, m) ɛ(σ)m where ɛ(σ) is the sign of σ. Let S 3 act on C 3 = v 1 C v 2 C v 3 C by permuting the coordinates. Now, S 3 acts trivially on the vector space (v 1 + v 2 + v 3 )C. The module M 2 is C 3 /(v 1 + v 2 + v 3 )C. Now we can calculate the character table: So χ M1 (σ) = 1 χ Mɛ (σ) = ɛ(σ) ( ) 1 0 χ M2 (1) = Tr = 2 ( 0 1 ) 0 1 χ M2 (1 2) = Tr = ( ) 0 1 χ M2 (1 2 3) = Tr = (χ S (σ)) S S,[σ] X = M 1 M ɛ M 2 We obtain the following ring isomorphism Furthermore, (1) (1 2) (1 2 3) R(S 3 ) = Z(1, 1, 1) Z(1, 1, 1) Z(2, 0, 1) Z 3. ( ) #x dim C S χ S(x) = S S,x X M 1 M ɛ M 2 (1) (1 2) (1 2 3)

8 Which gives the ring isomorphism Λ(S 3 ) = Z Discriminants Z Z Z 3. In this section we calculate the discriminants of R(G) and Λ(G). First the definition of discriminant. Definition Suppose R = n i=1 Z ω i is a ring, then the discriminant (R) is defined as the Z-ideal generated by det(tr R/Z (ω i ω j )) i,j=1...n. In this section we will prove the following two propositions. Proposition Let G be a finite group of order g. The discriminant of R(G) is generated by Q g#x x X #x. Proposition Let G be a finite group of order g. The discriminant of Λ(G) Q is generated by g#x x X #x ( Q. S S dim C S) 2 For the proof of these propositions we first give some lemmas. Lemma Suppose R = n i=1 Z ω i is a ring, then #Hom ring (R, C) n and #Hom ring (R, C) = n if and only if R is a reduced ring, that is, a ring without nilpotent elements. Proof. By extending every morphism of R to R Q, we see that Hom ring (R, C) = Hom ring (R Q, C). The ring R Q = n i=1 Q ω i is artinian and therefore it is a finite product of artinian local rings [1, thm. 8.7]. Write R Q = j R j, where the R j are artinian local rings. Let m j be the maximal ideal of R j. From [3, chap. X, cor. 2.2] we know that m j consists of all the nilpotent elements of R j. So we have Hom ring (R Q, C) = j Hom ring(r j, C) = j Hom ring(r j /m j, C). Therefore #Hom ring (R, C) = j = j #Hom ring (R j /m j, C) # dim Q R j /m j # dim Q R j = n, where the second equality come from the fact that R j /m j is separable over Q [3, chap. V, sect. 4]. Equality holds if and only if m j = 0 for all j, that is, R has no nilpotent elements. Lemma Suppose R = n i=1 Z ω i is a reduced ring. Then (R) is generated by det(fω i ) 2 i,f, where i ranges from 1 to n and f over F = Hom ring(r, C). Proof. Since R is reduced, we have the following ring isomorphism R C C F ω (f(ω)) f. Restricting this morphism to R and taking the trace on both sides, we obtain Tr R/Z (ω) = f f(ω) for all ω R. So we have (Tr R/Z (ω i ω j )) i,j = ( f f(ω iω j )) i,j = (f(ω i )) i,f (f(ω j )) f,j. Therefore (R) is generated by det(f(ω i )) i,f det(f(ω j )) f,j = det(fω i ) 2 i,f. 7

9 Lemma For all x, y X we have { χ S (x) g/#x if x = y χ S (y) = 0 if x y. S S where χ S (x) is the complex conjugate of χ S (x). Proof. [4, sect. 2.5, prop. 7] Proof of proposition The representation ring R(G) = S S Z[S] satisfies #Hom ring (R(G), C) = #S, since we have the following distinct ring homomorphisms from lemma R(G) C [S] χ S (x) for all x X. So R(G) is reduced and we can apply lemma Its discriminant is generated by det(χ S (x)) 2 S S,x X. For some number k we have det(χ S (x)) 2 S S,x X = ( 1)k det ( (χ S (x) ) T S S,x X (χ S(x)) S S,y X ). Using lemma 2.2.6, the generator of (R(G)) is det ( (χ S (x) ) T ) S S,x X(χ S (x)) ( ) S S,y X = det χ S (x) χ S (y) S S g #x = det = g #X x X #x. g #x n x,y X where the x i run over X Proof of proposition The center of the group ring Λ(G) = x X Z c x satisfies #Hom ring (Λ(G), C) = #X, since we have the following distinct ring homomorphisms from lemma Λ(G) C c x #x dim C S χ S(x) for all S S. So Λ(G) is reduced and we can apply lemma The discriminant of Λ(G) is ( ) 2 #x (Λ(G)) = det dim C S χ S(x). It follows that (Λ(G)) (R(G)) is generated by ( ) 2 det #x dim C S χ S(σ) S S,x X det(χ S (σ)) 2 S S,[σ] X = S S,x X ( Πx X #x ) 2 Π S S dim C S and (Λ(G)) by ( ) 2 Πx X #x g #X Π S S dim C S x X #x = g#x x X ( #x S S dim C S ) 2. 8

10 Example ( Take G = S 3, then (R(G)) = ) = (36) and (Λ(G)) (R(G)) = ( ) 2 = (9), so (Λ(G)) = (324). We see that Λ(G) and R(G) are not isomorphic as rings, since they have different discriminants. The quotient (Λ(G)) (R(G)) is an integer ideal and we can ask ourselves whether this is always the case. 2.3 Divisibility of discriminants To ease the notation we will use (R) = r for (R) is generated by r. For every g N we consider the following statement Statement For each finite group G of order g, we have Λ(G)/Z R(G)/Z Z. In this section we will prove the following theorems Theorem Let p be a prime. If g = p k with k 4 then statement is true. Theorem Let p, q be primes. If g = pq then statement is true. To show this we will prove the following statement, which is a sufficient condition for statement to be true, for g = p k and g = pq. Statement For all c 1,... c s, d 1,... d t N such that 1. s = t, 2. c i = d 2 j = g, 3. c i g for all i, 4. d j g for all j, 5. #{i c i = 1} g, 6. g #{i c i=1} is not prime, 7. #{i d i = 1} g, we have Q Q i ci j dj Z Theorem For every g N statement implies statement For the proof we need the following lemma. Lemma Let G be a finite group and Z(G) its center. The index [G : Z(G)] is not prime. Proof. If [G : Z(G)] is prime, then G/Z(G) is cyclic. We will prove that if G/Z(G) is cyclic, then G/Z(G) is trivial. Let σ G such that σ generates G/Z(G). Let h G be an element, then we can write h = σ k h for some k N and h Z(G). Now, σh = σσ k h = σ k h σ = hσ, so σ Z(G) and G/Z(G) is trivial. Proof of theorem Let G be a group of order g. Denote by c i the number of elements of the i-th conjugacy class of G and by d j the C-dimension of the j-th simple C[G]-module. Then we have 9

11 1. s = #X = #S = t, 2. c i = x X #x = g and g = dim C C[G] = dim C Π S S End C (S) = S S dim2 C S = d 2 j, 3. c i g, since for the corresponding x X we have #x g, 4. d j g, since for the corresponding S S we have dim C S g [3, chap. XVIII, cor. 4.8], 5. #{i c i = 1} = #Z(G) g, where Z(G) is the center of G, 6. g #{i c i=1} = g #Z(G) is not prime, see lemma 2.3.6, 7. #{i d i = 1} = #G ab g, where G ab is the abelianized G. According to statement 2.3.4, we have Q Q i ci j dj Z. Therefore Λ(G)/Z R(G)/Z = ( ) 2 i c i j d Z. j We shall now prove theorem and by proving statement for g = p k and g = pq. Theorem Let p be a prime. If g = p k with k 4 then statement is true. Proof. Let C m = #{i c i = m} and D m = #{i d i = m} for all m N. Suppose statement is not true for g = p k, then 1. l C p l = l D p l, 2. p k = l C p l pl, 3. p k = l D p l p2l, 4. C 1 = p lc with l c k and l c k 1, 5. D 1 = p l d with l d k, 6. l l C p l < l l D p l. If l c or l d is equal to k, then both of them are, because of equation (1), (2) and (3), then inequality (6) becomes an equality, contradiction. So l c k 2 and l d k 1. Therefore k l c Taking equation (2) modulo p, we get C 1 = 0 mod p, so l c 1. Taking equation (3) modulo p 2, we get D 1 = 0 mod p 2, so l d 2. Therefore k l c When we subtract equation (1) from inequality (6), we get k k 1 2 (l 1) C p l < p lc p l d + (l 1) D p l. (2.4) l=2 The left hand side of 2.4 is non-negative, so the right hand side needs to be greater than 0. For k 4 the right hand side is equal to p lc p l d, so we need l c > l d, so 2 l d < l c k 2 2. This is a contradiction, so there are no solutions for k 4. l=2 10

12 Theorem Let p, q be primes. If g = pq then statement is true. Proof. Let C m = #{i c i = m} and D m = #{i d i = m} for all m N. If p = q, then tells us this theorem is true, so without loss of generality we can assume p < q. Suppose statement is not true for g = pq, then 1. C 1 + C p + C q = D 1 + D p + D q, 2. pq = C 1 + pc p + qc q, 3. pq = D 1 + p 2 D p + q 2 D q, 4. C 1 pq and C 1 p, q, 5. D 1 pq, 6. p Dp q Dq p Cp q Cq, which means D p > C p or D q > C q. If C 1 or D 1 is equal to pq, then both of them are, because of equation (1), (2) and (3) then inequality (6) becomes an equality, so C 1 = 1 and D 1 pq. Since pq < q 2 we have D q = 0, because of equation (3), and therefore D p > C p. Taking equation (3) modulo p, we get D 1 = 0 mod p, so D 1 = p and D p = q 1 p. We are left with the following equations C p + C q = p + q 1 1 p pc p + qc q = pq 1. For which the solution is C p = q 1 p = D p and C q = p 1. We needed D p > C p, so there are no solutions. For g = 12 is statement not true. We can take (c 1... c 6 ) = (1, 1, 1, 3, 3, 3) and (d 1... d 6 ) = (1, 1, 1, 1, 2, 2). Then all the conditions are satisfied, but Q Q i ci j dj = Z. Through exhaustive search, we can prove that this is the only counterexample for g = 12 for statement Statement is also false for g = 18 and for g = 3 5. The following tables give all counterexamples for g = 18 and for g = 3 5, where we use the same notation as in the proof of theorem g = 18 C 1 C 2 C 3 C 6 C 9 D 1 D 2 D g = 3 5 C 1 C 3 C 9 C 27 C 81 D 1 D 3 D A computer program has checked statement for g < 512, so for the above examples, the c i and d j are not the conjugacy class sizes respectively dimensions of simple modules of existing groups. The way to improve this method would be to give more or better conditions for the c i and d j in statement

13 Chapter 3 Comparison of spectra In this chapter we are going to calculate the spectra of R(G) and Λ(G). We view the rings over the subring of C generated by the g-th roots of unity, with g the order of G. We will this ring A. Since all characters of representations of G have images in A, the spectra of R(G) A and Λ(G) A are easier to compute than the spectra of R(G) and Λ(G). After some general notions about spectra we will calculate the spectrum of R(G) A. For the spectrum of Λ(G) A we will give an approximation. 3.1 Spectra First some general theory about spectra. Definition Let R be a commutative ring. The spectrum of R, denoted by Spec(R), is the topological space consisting of all prime ideals of R, with topology defined by the closed sets C(I) = {p prime : p I}, for each ideal I of R. This topology is called the Zariski topology. Proposition If φ : R 1 R 2 is a ring homomorphism, then we have an induced continuous map φ : Spec(R 2 ) Spec(R 1 ) p φ 1 (p). Proof. We need to prove that φ 1 (p) is a prime ideal of R 1. The map φ : R 1 R 2 R 2 /p gives to following injection into the domain R 2 /p R 1 / ker(φ ) R 2 /p. So R 1 / ker(φ ) is also a domain and ker(φ ) = φ 1 (p) is a prime ideal. Furthermore, to see that φ is continuous, let V f = {p Spec(R 1 ) : f p} for every element f R 1. These sets are open in Spec(R 1 ), since Vf c = C(fR), where V c is the complement of the set V. They also form a basis for the topology of Spec(R 1 ), since C(I) c = f I V f for every ideal I. Let W g = {p Spec(R 2 ) : g p} for every element g R 2. Now we have φ 1 (V f ) = {p Spec(R 2 ) : φ (p) V f } 12

14 = {p : f φ 1 (p)} = {p : φ 1 φf φ 1 (p)} = {p : φf (p)} = W φ(f). So, φ is continuous. Proposition If R 1 and R 2 are commutative rings, then Spec(R 1 R 2 ) = Spec(R 1 ) Spec(R 2 ). Proof. If p 1 is a prime ideal of R 1, then (R 1 R 2 )/(p 1 R 2 ) = R 1 /p 1 is a domain. So (p 1 R 2 ) is a prime ideal of (R 1 R 2 ). In the same way, if p 2 is a prime ideal of R 2, then R 1 (p 2 ) is a prime ideal of (R 1 R 2 ). If p is a prime ideal of R 1 R 2, then (R 1 R 2 )/p is a domain. In this domain we have (1, 0) (0, 1) = (0, 0), so (1, 0) = (0, 0) or (0, 1) = (0, 0). If (1, 0) = (0, 0), then R 1 0 p, so p = R 1 p 2, where p 2 is a prime ideal of R 2. Since (R 1 R 2 )/(R 1 p 2 ) is a domain, p 2 is a prime ideal of R 2. In the same way if (0, 1) = (0, 0), then p = p 1 R 2, where p 1 is a prime ideal of R 1. Definition Let R be a commutative ring. Let p 0 p 1... p k be a chain of prime ideals of R. We call k the length of such a chain. Define the dimension of R to be the maximal length of all such chains. Now, let A be an order in a number field and let B be a ring such that we have a finite set V and injective A-algebra morphisms A B A V, such that the index [A V : B] is finite. Since each non-zero prime ideal of A is maximal, the dimension of A is 1. We can think of Spec(A) as a line. Furthermore, by proposition 3.1.3, we have Spec(A V ) = V Spec(A), so we can think of Spec(A V ) as #V lines. We want to determine Spec(B). We have ring homomorphisms A B A V, so according to proposition 3.1.2, we have continuous maps Spec(A V ) Spec(B) Spec(A). Let π be the map Spec(A) V Spec(B). Proposition The map Spec(A V ) Spec(B) is surjective. Proof. Examine the extension A A V a (a, a,..., a). Let α = (a 1, a 2,..., a n ) A V and f = Π i (X a i ) A[X], then f(α) = 0. So A A V is integral. So A V is also integral over B. According to the going-uptheorem [1, thm. 5.10] Spec(A V ) Spec(B) is surjective. We now know that Spec(B) is a quotient set of Spec(A V ). If two elements (v 1, p 1 ), (v 2, p 2 ) Spec(A V ) are in the same equivalence class, then p 1 = p 2. The following proposition tells us for which primes p the equivalence class (v, p) consists of one point, for all v V. Proposition Let p be a non-zero prime ideal of A and p the characteristic of A/p. Suppose p [A V : B] = t, then B is totally split at p, which means that the equivalence class (v, p) consists of one point, for all v V. Proof. We have ta V B A V and since localisation is exact, we have ta V p B p A V p. Now, since p t, we have t (A V p ), so B p = A V p. Therefore we have the following ring isomorphisms B p (A p /p p ) = A V p (A p /p p ). 13

15 So, we have B/p = B p /p p = B p (A p /p p ) = A V p (A p /p p ) = (A p /p p ) V = (A/p) V. Therefore is the number of primes of B which map to the prime p of A equal to #V. The following proposition gives us a way of computing Spec(B) in case we have an explicit description. Proposition Suppose B can be written as B = #V i=1 c i A, with c i = (c vi ) v A V. The points (v 1, p), (v 2, p) Spec(A V ) have the same image in Spec(B) if and only if we have c v1i c v2i mod p for all i. Proof. Suppose c v1i c v2i mod p for all i. Let b π(v 1, p) be an element, we can write b = i a ic i = ( i a ic vi ) v with a i A and i a ic v1i p. Since i a ic v1i i a ic v2i mod p, we have i a ic v2i p. Therefore b π(v 2, p) and π(v 1, p) = π(v 2, p). On the other hand, if π(v 1, p) = π(v 2, p), then c i c v1i 1 = (c vi c v1i) v π(v 1, p) = π(v 2, p) for all i, so c v2i c v1i p for all i. Example Let G be S 3, the symmetric group on three elements. From example we have the following ring isomorphism Λ(G) = Z Z Z Z 3. The points (M 1, p) and (M ɛ, p) are in the same equivalence class if and only if (1, 3, 2) = (1, 3, 2) in (Z/pZ) 3, that is, when p is 2 or 3. The points (M 1, p) and (M 2, p) are in the same equivalence class if and only if (1, 3, 2) = (1, 0, 1) in (Z/pZ) 3, that is, when p is 3. The points (M ɛ, p) and (M 2, p) are in the same equivalence class if and only if (1, 3, 2) = (1, 0, 1) in (Z/pZ) 3, that is, when p is The spectrum of the representation ring Let G be a finite group of order g and A the subring of C generated by the g-th roots of unity. First we calculate the spectrum of R(G) A. We are going to embed R(G) A in A X, for this we use the following lemma. Lemma Let M be a representation of G and σ G, then χ M (σ) A. Proof. We have χ M (σ) = Tr(M M : m σm), which is the sum of the eigenvalues counted with their multiplicity. Since (M M : m σm) has order a divisor of g, all its eigenvalues have order a divisor of g. So all eigenvalues are a g-th root of unity. So, we can embed R(G) A in A X by the injective A-algebra morphisms A R(G) A A X a 1 a [M] 1 (χ M (x)) x X. 14

16 So we have continuous maps Spec(A X ) = X Spec(A) π Spec(R(G) A) Spec(A) and Spec(R(G) A) is a quotient space of Spec(A X ). We want to know which equivalence classes of Spec(R(G) A) consist of more than one point. First a lemma which restricts the primes we need to look at. Lemma If a prime p divides [A X : R(G) A], then it divides g = #G. Proof. For this proof we use the generalized notion of discriminant from a book by Serre, which defines the discriminant and index for lattices over a Dedekind domain [5, chap. III, sect. 2]. We will denote the discriminant of a lattice L over the Dedekind domain A as A (L) and the index of lattice L and L as [L : L ] A. From [5, chap. III, sect. 2, prop. 5] we have the following formula for lattices L L over A A (L ) = A (L)[L : L ] 2 A. (3.1) Using this formula for L = A X and L = R(G) A and proposition 2.2.2, we obtain [A X : R(G) A] 2 = [A X : R(G) A] 2 A = ( ) A(R(G) A) g #X A (A X = ) x X #x. So, according to proposition 3.1.6, if a prime p of A does not divide the order of G, then the equivalence classes of Spec(R(G) A) above p consist of one element. Next, we will calculate Spec(R(G) A) in the same way as [4, sect. 11.4]. Lemma Let p be a prime number and G a finite group, then each x G can be written in a unique way as x = x u x r where x u is a p-unipotent element, that is, it has order a power of p and x r is a p-regular element, that is, it has order prime to p. Proof. To see that there is a pair x u and x r, decompose the cyclic subgroup generated by x as a direct product H 1 H 2 of two subgroups, where the order of H 1 is a power of p and the order of H 2 is prime to p. To see this is the only way, suppose x = x u x r, with x u a p-unipotent element and x r a p-regular element. Let H 1 be the subgroup generated by x and let H 2 be the subgroup generated by x u and x r. Both H 1 and H 2 are cyclic of order ord(x). Since x H 2, we have H 1 = H 2, so x u and x r are powers of x. The element x u (respectively x r ) is called the p-component (respectively the p -component) of x. Note that x u and x r commute. Lemma Let p be a prime of A with char(a/p) = p, let χ be the image of an element of R(G) A in A X, let x G, and let x r be the p -component of x. Then χ(x) χ(x r ) mod p. Proof. The character χ is also the character of an element of R(H) A for every subgroup H of G. We will prove the lemma using the subgroup generated by x, which we will call H. Now χ = χ H = i a iχ i, with a i A and χ i running over the distinct characters of degree 1 of H. If q is a sufficiently large power of the norm of p, we have x q = x q r and thus χ i(x) q = χ i (x r ) q for all i. Therefore χ(x) q = ( i a iχ i (x)) q i aq i χ i(x) q = i aq i χ i(x r ) q ( i a iχ i (x r )) q = χ(x r ) q mod p, hence χ(x) = χ(x r ) mod p, since a q a mod p for all a A. 15

17 Lemma Let x be a p -element of G, that is, an element of order coprime to p. Then there is an element M R(G) A for which the character has the following properties: χ(x) 0 mod p χ(s) = 0 for each p -element of G which is not conjugate to x. Proof. [4, sect. 10.3, lemma 8] Theorem Let p be a prime ideal of A and p the characteristic of A/p, furthermore let c 1 and c 2 be conjugacy classes of G. Let c 1 (respectively c 2 ) be the class consisting of the p -components of the elements of c 1 (respectively c 2 ). Let π be that map X Spec(A) Spec(R(G) A) defined previously. Then we have π(c 1, p) = π(c 2, p) if and only if c 1 = c 2. Proof. According to proposition 3.1.7, the two primes π(c 1, p) and π(c 2, p) are the same if and only if for all simple C[G]-modules S we have χ S (c 1 ) = χ S (c 2 ) mod p. If c 1 = c 2, lemma shows that for every C[G]-module M we have Tr M (c 1 ) = Tr M (c 1) = Tr M (c 2) = Tr M (c 2 ) mod p, hence π(c 1, p) = π(c 2, p). If c 1 c 2, then lemma gives an element M R(G) A, such that its character χ satisfies χ(c 1) 0 mod p χ(c 2 ) = 0, which implies there is a simple module S for which χ S (c 1 ) χ S (c 2 ) mod p, hence π(c 1, p) π(c 2, p). Example Let G be S 3, the symmetric group on three elements, then #G = 6, so according to lemma it suffices to look at primes of residue-characteristic 2 or 3. In the following table are the p -components for p equal 2 or 3 for all conjugacy classes of G. conjugacy class (1) (1 2) (1 2 3) p = 2 (1) (1) (1 2 3) p = 3 (1) (1 2) (1) So, for all p of A of residue-characteristic 2, we have π((1), p) = π((1 2), p) and for all p of A of residue-characteristic 3, we have π((1), p) = π((1 2 3), p). We could also have calculated this spectrum using proposition Since all the characters of S 3 have image in Z 3, we would have gotten the same result for R(S 3 ). So the spectrum of R(S 3 ) looks like 16

18 (1) (12) (123) 2 3 The spectrum of R(S 3 A) looks the same, with the exception that there are more primes of residue-characteristic 2 or 3. The spectrum we obtained in the previous example is connected. The following theorem tells us this is the case for all finite groups. Theorem The spectrum Spec(R(G) A) is connected in the Zariski topology. Proof. Let x be an element of G and let p k1 1 pk pk l l be the prime decomposition of the order of x. The element x can be written as x = x u x r, where x u has order a power of p and x r has order prime to p. Using this for every prime, we get x = x p1 x p2... x pl, where x pi has order p ki i. Since x and x p2... x pl have the same p -components, theorem tells us that (x, p 1 ) and (x p2... x pl, p 1 ) are in the same equivalence class of R(G) A. Furthermore π({y} Spec(A)) is connected for all y X, since it is isomorphic to Spec(A). So, continuing in the same way, we obtain that π(x, Spec(A)) is connected to π(1, Spec(A)). So Spec(R(G) A) is connected. Corollary The spectrum Spec(R(G)) is connected in the Zariski topology. Proof. From the ring homomorphism R(G) R(G) A we obtain a surjective continuous map Spec(R(G) A) Spec(R(G)). The spectrum Spec(R(G)) is the image of a connected space under a continuous map and is therefore connected. 3.3 The spectrum of the center of the group ring For the spectrum of Λ(G) A we will give a criterion for when an equivalence class certainly consists of more than one element. We want to embed Λ(G) A in A S, for that we will use the following lemma. Lemma Let x X be a conjugacy class of G, then #x dim C S χ S(x) A. Proof. From lemma we know that χ S (x) A. Furthermore, the characteristic polynomial of the matrix of the map Λ(G) Λ(G) c c x c. #x is monic and c x is a zero of it and therefore is c x integral over Z. Since dim C S χ S(x) is #x the scalar by which c x acts on S, we have dim C S χ S(x) integral over Z and therefore it is an element of A. 17

19 So, we can embed Λ(G) A in A S with the A-algebra homomorphisms So we have continuous maps A Λ(G) A A S a 1 a ( ) #x c x 1 dim C S χ S(x). S S Spec(A S ) = S Spec(A) π Spec(Λ(G) A) Spec(A). We want to know for which elements of Spec(A S ) we have π(m 1, p) = π(m 2, p). First a lemma which restricts the primes we need to look at. Lemma If a prime p divides [A X : Λ(G) A], then it divides g = #G. Proof. Using formula 3.1 for L = A X and L = Λ(G) A, and proposition we obtain [A X : Λ(G) A] 2 = A(Λ(G) A) A (A X = g#x x X #x ) ( S S dim C S ) 2. So, according to proposition 3.1.6, if a prime p doesn t divide the order of G, then the equivalence classes of Spec(Λ(G) A) above p consist of one element. Theorem Let M and N be two A[G]-modules, such that M A C and N A C are simple C[G]-modules and let p be a non-zero prime of A. Define M = M A A/pA and N = N A A/pA. If M and N have a common non-trivial A/pA-subquotient then π(m, p) = π(n, p). Proof. Each element c Λ(G) A acts as a scalar of A on M and N, therefore c will act as a scalar of A/pA on M and N, say c M and c N respectively. According to proposition 3.1.7, the two primes π(m, p) and π(n, p) are the same if for all c Λ(G) A we have c M = c N. If M and N have a non-trivial common subquotient then each c acts as a scalar on that subquotient, say c S. We get c M = c S = c N. Note: if M and N have a non-trivial common subquotient, then they certainly have a common simple subquotient, so it suffices to look at simple subquotients of M and N. It is proven in [4, section 15.2] that for each simple C[G]-module M C we can find an A[G]-module M A, such that M A C = M C and that its simple subquotients do not depend on the choice of M A. So it is sufficient to construct one A[G]-module for each simple C[G]-module and compare only those modules. Example Again, let G be S 3, then #G = 6, so according to lemma it suffices to look at primes of residue-characteristic 2 or 3. Recall from section 2.1 the three simple modules M 1, M ɛ and M 2. The A[G]-modules we will use are the module generated by 1 for M 1 and M ɛ. For M 2 we will use the A[G]-module generated by v 1 and v 2. For primes of residue-characteristic 2 the modules M 1 and M ɛ are equal, since (1, 1, 1) = (1, 1, 1) in (Z/2Z) 3. So they certainly have a common non-trivial subquotient. The module M 2 does not have a common non-trivial subquotient with M 1 or M ɛ, since if it would, then there would be a submodule of dimension 1 for which 18

20 (1 2 3) acts trivially. There is no such submodule, since M 2 consists of four elements, 0,v 1, v 2 and v 1 + v 2, and (1 2 3) acts as a cyclic permutation of v 1, v 2 and v 1 + v 2. For primes of residue-characteristic 3, let N be the submodule of M2 spanned by v 1 +2v 2, then G acts as the sign on N, since (1 2) v 1 +2v 2 = v 2 +2v 1 = (v 1 +2v 2 ) and (1 2 3) v 1 + 2v 2 = v 2 + 2v 3 = v 2 + 2( v 1 v 2 ) = v 1 + 2v 2. So M 2 and M ɛ have a common non-trivial subquotient. Furthermore, G acts on M 2 /N trivially, since we have (1 2) v 1 N = v 2 N = (v 2 + v 1 + 2v 2 )N = v 1 N and (1 2 3) v 1 N = v 2 N = v 1 N. So M 2 and M 1 also have a common non-trivial subquotient. Note that M1 and M ɛ do not have a common non-trivial subquotient, since (1, 1, 1) (1, 1, 1) in (Z/3Z) 3. Still, for primes of residue-characteristic 3, we have π(m 1 p) = π(m 2, p) = π(m ɛ, p). Apparently, the relation M and N have a common non-trivial subquotient is not transitive, so we need to take the transitive closure to get an quotient space of Spec(A S ). This space is an approximation of Spec(Λ(G) A). Let us call the spectrum we just calculated Spec(B ). From theorem we know we have surjective continuous maps Spec(A S ) Spec(B ) Spec(Λ(S 3 ) A). In fact we have Spec(B ) = Spec(Λ(S 3 ) A), since we could also have calculated the spectrum of Λ(S 3 ) A using proposition Since we know from example that Λ(S 3 ) A has image in Z 3, we would have gotten the same result for Spec(Λ(S 3 )), which we calculated in example Both Spec(Λ(S 3 )) and Spec(B ) look like M 1 M M There are more groups for which theorem gives not only a necessary, but also sufficient condition, but it is not known to the author whether this is true for all groups. The example tells us that Spec(Λ(S 3 )) is connected. This is the case for all groups as we shall see from theorem First some lemmas we need to prove this theorem. Definition A ring R is local if is has a unique maximal left ideal and a unique maximal right ideal and these two ideals coincide [6, thm ]. Lemma Let H be a group of order p k with p a prime. The ring F p [H] is a local ring. Proof. Let m be a maximal left ideal of F p [H]. Let I be the ideal generated by 19

21 {h 1 : h H}; it is the kernel of the map F p [H] F p a h h h h a h, so I is a maximal ideal. Let m be a maximal left ideal of F p [H]. Now, M = F p [H]/m is a simple left F p [H]-module. From [3, chap. I, thm 6.5] we know that the center of H contains a non-trivial element c. Since c pk = 1, for some k, we have (c 1) pk = 0 in F p [H]. So the left module automorphism φ : M M m (c 1)m is not surjective. Therefore is the image of φ equal to 0. So c acts trivially on M and M is a simple F p [H/ c ]-module, where c is the subgroup of H generated by c. By induction to the order of H, we get that M is a simple F p -module and I = h 1 : h H m. Since I is maximal, we have I = m. In the same way we prove that I is the unique maximal right ideal and therefore is F p [H] a local ring. Lemma Let G be a finite group and let P be a finitely generated projective Z[G]-module, then #G divides the Z-rank of P. Proof. Let p be a prime dividing the order of G. Let H be the Sylow-p-group of G, then P is also a Z[H]-module. The module P F p is a projective F p [H]-module. From the above lemma we know that F p [H] is a local ring, so P is a free module [6, th ] and the rank of P is a multiple of #H. Since this is true for all primes p we have #G dividing the Z-rank of P. Theorem Let G be a finite group, then Spec(Λ(G)) is connected in the Zariski topology. Proof. Suppose Spec(Λ(G)) is not connected, then we can write Λ(G) = L 1 L 2, with L 1 and L 2 proper quotient rings of Λ(G). Let e be the unit of L 1, then we can write Λ(G) = e Λ(G) (1 e)λ(g), with e not 0 or 1. The module e Z[G] is a finitely generated projective Z[G]-module of rank which does not divide #G. This is a contradiction with the previous lemma. 20

22 Chapter 4 Comparison of Q-algebras In this final chapter we view our rings over Q. We will give an equivalence between two categories. The first will generalize the idea of a character table, the second one will consist of a pairing between two abelian finite étale algebras. From this equivalence we will see that R(G) Q and Λ(G) Q are abelian finite étale Q- algebras which are Brauer equivalent. In this chapter we will use several notions from category theory. For definitions, see [2, chap. 2]. 4.1 Q-algebras We are going to examine the rings R(G) Q and Λ(G) Q. By tensoring the homomorphism 2.3 with Q, we obtain the following Q-algebra isomorphism R(G) Q Q-span rows (Tr S (σ)) S,σ C X. Since we are taking the Q-span of the rows, we can multiply a row with a number from Q, without changing the algebra, so we may replace (Tr S (σ)) S,σ by ) to obtain ( TrS(σ) dim C S S,σ R(G) Q Q-span rows ( ) TrS (σ) C X. dim C S S,σ In the same way, by tensoring the homomorphism 2.2 with Q, we obtain the Q-algebra isomorphism ( Λ(G) Q Q-span columns Tr S (σ) #[σ] ) C S. dim S S,σ ( ) ( ) We can replace Tr S (σ) #[σ] dim S by TrS(σ) dim S,σ C S to obtain S,σ ( ) Λ(G) Q TrS (σ) Q-span columns C S. dim S ( ) So the C-valued matrix TrS(σ) dim S has the property that both the Q-span of S,σ the rows and the Q-span of the columns is a ring. We are going the study this kind of matrices and see what we can tell about the Q-spans of rows and columns. S,σ 21

23 4.2 Finite abelian étale algebras First we need some new terminology and theory about abelian finite étale algebras. Let K be a field, K an algebraic closure of K. Let K sep be the maximal separable extension of K within K and K ab K sep the maximal abelian extension of K within K sep. Let Γ and Γ ab be the Galois groups of K sep /K and K ab /K respectively. A finite étale K-algebra is a finite product i E i where the E i are finite separable field extensions of K. An abelian finite étale K-algebra is a finite étale K-algebra where the field extensions are abelian over K. Lemma Let E be a abelian finite étale K-algebra. 1. There is a unique Γ ab -action on the set E, such that every K-algebra homomorphism E K ab is Γ ab -equivariant. 2. For this Γ ab -action the map E E e γe is a K-algebra homomorphism for all γ Γ ab. Proof. Let E i be a finite abelian extension of K. Let σ 1 : E i K ab be a K-algebra homomorphism. 1. The only action of Γ ab on E i which satisfies the requirements is Γ ab E i E i (γ, e) σ1 1 1e. We want to prove that this action is independent of the choice of σ 1. Let σ 2 : E i K ab be another K-algebra homomorphism. There is a γ Γ ab such that σ 2 = γσ 1. We now have Γ ab E i E i (γ, e) σ2 1 2e = σ1 1 γ γσ 1 e = σ1 1 1 e = σ1 1 1e. So the action on E i is independent of the choice of σ. Since every K-algebra homomorphism E K ab is composed of a projection E E i and a K-algebra homomorphism E i K ab, the only action of Γ ab on E which satisfies the requirements is the componentwise action on the E i. 2. A projection E E i and the map are K-algebra homomorphisms. Γ ab E i E i (γ, e) σ1 1 1e 22

24 4.3 Two categories In this section we define two categories. Let L be a field, such that K ab L. Define the category C in the following way. The objects of C are triples (S, T, A) with S and T finite sets and A = [a st ] s S,t T Map(S T, L) an invertible matrix such that s K(s-th row of A) LT and t K(t-th column of A) LS are subrings. A morphism (S, T, A) (S, T, A ) consists of a map φ S : S S and a map φ T : T T such that a φs(s )t = a s φ T (t) for all s S, t T. It usually easier to think of a C-morphism as a diagram. A S T L φ S φ T id L A S T L Example We take K = Q. Let a and b be non-zero natural numbers. Take for S and T the set {1, 2, 3, 4, 5, 6, 7} and a a a a a a a a A = b b b b b b b b ab ab ab ab ab ab ab ab Then (S, T, A) is an element of C. To see this we need to show: 1. the element (1, 1, 1, 1, 1, 1, 1) is in the row space. 2. the element (1, 1, 1, 1, 1, 1, 1) is in the column space. 3. if we multiply two rows we get a Q-linear combination of the rows. 4. if we multiply two columns we get a Q-linear combination of the columns. 5. the matrix A is invertible. Let r i be the i-th row of A and c j the j-th column. 1. We have: (1, 1, 1, 1, 1, 1, 1) = r (r 2 + r 3 + r 4 + r 5 + r 6 + r 7 ). 2. We have: (1, 1, 1, 1, 1, 1, 1) = 1 4 (c 1 + c 2 + c 3 + c 4 ) + c 5 + c 6 + c There are several types of rows. Below is for every combination of types one example. r 1 r 1 = r 1, r 1 r 2 = 1 2 (r 2 r 3 ), 23

25 r 2 r 2 = ar (r 2 + r 3 ), r 2 r 6 = ar (r 4 + r 5 ). 4. There are several types of columns. Below is for every combination of types one example. c 1 c 2 = 1 4 (c 1 + c 2 + c 3 + c 4 ) + ac 5 bc 6 abc 7, c 1 c 5 = 1 4 (c 1 + c 2 + c 3 + c 4 ), c 5 c 5 = c 5, c 5 c 6 = The determinant of A is 128ab. Now, define the category D. The objects of D are triples (E, F,, ) where E and F are abelian finite étale K-algebras, and, is a non-degenerate K-bilinear pairing E F, K ab, which satisfies γe, f = e, γf = γ e, f for all e E, f F and γ Γ ab. A morphism (E, F,, ) (E, F,, ) consists of K-algebra homomorphisms φ E : E E and φ F : F F such that φ E (e ), f = e, φ F (f) for all e E and f F. It usually easier to think of a D-morphism as a diagram., E F K ab φ E φ F id K ab, E F K ab Example We take K = Q. Let a and b be two non-square integers. Take and The pairing will be defined by setting E = Q Q( a) Q( b) Q( ab) F = Q( a, b) Q 3. e = (t, u + v a, w + x b, y + z ab) E and and taking f = (t + v a + x b + z ab, u, w, y ) F, : E F Q ab (e, f) tt + uu + vv a + ww + xx b + yy + zz ab. 24

26 The triple (E, F,, ) is an element of D. To see this we need to show that γe, f = e, γf = γ e, f for all γ Γ ab. The only elements of Γ ab we need to consider are γ 1 : γ 2 : γ 3 : ( a a, b b), ( a a, b b) and ( a a, b b). We have γ 1 e, f = e, γ 1 f = γ 1 e, f = tt + uu + vv a + ww xx b + yy zz ab, γ 2 e, f = e, γ 2 f = γ 2 e, f = tt + uu vv a + ww + xx b + yy zz ab, γ 3 e, f = e, γ 3 f = γ 3 e, f = tt + uu vv a + ww xx b + yy + zz ab. 4.4 An equivalence of categories We are going to give an equivalence between the categories C and D, defined in the previous section. First we will construct a functor C D. Theorem will later tell us this functor is an equivalence. Lemma Let (S, T, A) be an object of C, define E = s K(s-th row of A) and F = t K(t-th column of A) then (E, F,, ), with, defined through E F, K ab (s-th row of A, t-th column of A) a st for all s S, t T is an element of D. Proof. First observe that the rows of A generate a ring of finite dimension over K, therefore, all of the elements of A are algebraic. Write e s = s-th row of A for s S, and f t = t-th column of A, for t T. Write X = Hom K-alg(E, L) and Y = Hom K-alg(F, L). Let t T and let π t : L T L be the projection on the t-th coordinate. Define a K-algebra morphism x t : E L, such that the following diagram of K-algebra morphisms commutes. E x t L T π t K L We have x t (e s ) = a st for all s S and t T. All the x t are elements of X and since no two columns of A are the same, all these maps are different. We obtain #X n = dim K E. Since E is Artinian, we obtain from [3, chap. X, thm. 7.7] that E is the direct product of local, Artinian rings, E = i E i. Let m be the maximal ideal of E 1. The sequence m m 2 m 3... has a finite number of ideals, therefore we get from Nakayama s lemma [3, chap. X, lemma 4.1] that m k = 0 for some k. We get m is nilpotent. Since L T has no nilpotent elements, E 1 has no nilpotent element. So m = 0 and E 1 is a field. We obtain E is a finite product of finite field extensions of K. 25

27 We now have dim K E #X = #Hom K-alg(E, L) = i #Hom K-alg(E i, L) ( ) i i #Hom K (E i, K) dim K E i = dim K E. So we obtain equality at (*). Therefore all of the E i are separable [3, chap. V, sect. 4]. We now have a st K sep for all s S, t T. Observe that X = {x t : t T }. We can let Γ act on X in the following way. Γ X X (γ, x) γx. For all t T we have f t = (a st ) s S = (x t (e s )) s S. Let K[X] be the permutation module of X, in other words the K[Γ]-module with basis X. We have a K-module isomorphism i : K[X] F (K sep ) S x (x(e s )) s S and since i(γx) = ((γx)(e s )) s = (γ(x(e s ))) s = γix for all γ Γ, we see that i is a K[Γ]-module isomorphism. In the same way we can define a K[Γ]-module isomorphism K[Y ] E (K sep ) T. Now we have γ e s, f t = γ(x t (e s )) = (γx t )(e s ) = e s, γf t and in the same way γ e s, f t = γe s, f t for all γ Γ, s S and t T. Since {e s : s S} and {f t : t T } are bases for E and F, we can extend their this property to E and F. So, we have e, γf = γ e, f = γe, f for all γ Γ, e E and f F. Now, we have γ 1 γ 2 e, f = γ 2 γ 1 e, f = γ 1 e, γ 2 f = γ 1 e, γ 2 f = γ 2 γ 1 e, f for all γ 1, γ 2 Γ, e E, f F. The action of Γ factors via its abelian quotient Γ ab. Therefore we can factor all our Γ-actions through Γ ab. The algebras E and F are abelian finite étale algebras. and our constructed Γ ab -action on E satisfies x(γe) = γ(x(e)) for all γ Γ ab, e E and x X. So this action is the unique Γ-action on E from lemma The same holds for the Γ ab -action on F. Given a morphism (S, T, A) (S, T, A ) of C, let (E, F,, ) and (E, F,, ) be the objects of D acquired via the process in lemma From the map φ S : S S we define an induced map φ S : ( K ab) S ( K ab) S (a s ) s S (a ) φs(s s S. For all columns f t of A we have φ S (f t) = φ S ((ast) s) = (aφ S (s )t) s = (a s φ T (t)) s, which is the φ T (t)-th column of A, so we can restrict φ S to F. We have a map φ F : F F. In the same way, from the map φ T : T T we can construct a map φ E : E E 26

28 Lemma The maps φ E and φ F are a morphism (E, F,, ) (E, F,, ) of D. Proof. First we shall prove that φ F is a K-algebra homomorphism. It is a ring homomorphism, since it is a restriction of φ S. So it suffices to show that for all y Hom K-alg(F, K ab ) we have φ F y Hom K-alg(F, K ab ). From the proof of lemma we know that every y Hom K-alg(F, K ab ) is the restriction of a projection on one of the coordinates of ( K ab) S So we have the following commutative diagram F ( K ab ) S φ F φ S F ( ) K ab S y s π s K ab and since the map φ S π s is the projection on the φ S(s )-th coordinate, the map φ F y s is an element of Hom K-alg(F, K ab ). In the same way, we prove that φ E is a K-algebra morphism. It remains to show that φ E (e ), f = e, φ F (f) for all e E and f F. Let e s E be the s -th row of A and f t F the t-th columns of A. We have seen that φ F (f t ) is the φ T (t)-th column of A and in the same way is φ E (e s ) the φ S (s )-the row of A. So we have φ E (e s ), f t = a φs(s )t = a s φ T (t) = e s, φ F (f t ) and since {e s : s S } are a basis of E and {f t : t T } are a basis of F, the identity φ E (e ), f = e, φ F (f) is true for all e E and f F Define the functor ψ : C D as follows: on objects it applies the process in lemma 4.4.1, on morphisms it applies the process in lemma Now we are going to construct a functor D C. This functor will become the inverse of ψ. We first show that E and F have a natural basis. The Gram-matrix, with respect to these bases is the matrix of the associated element of C. Let (E, F,, ) be an object of D. We can give X = Hom K-alg(E, K ab ) a Γ ab -action by Γ ab X X (γ, x) γ x. From the pairing, we can define an isomorphism F Hom K[Γ ab ](E, K ab ) f (e e, f ). 27