Chapter 4 Characters and Gauss sums 4.1 Characters on finite abelian groups In what follows, abelian groups are multiplicatively written, and the unit element of an abelian group A is denoted by 1 or 1A. We denote the order (number of elements) of A by jAj. Let A be a finite abelian group. A character on A is a group homomorphism χ : A ! C∗ (i.e., C n f0g with multiplication). n n If jAj = n then a = 1A, hence χ(a) = 1 for each a 2 A and each character on A. Therefore, a character on A maps A to the roots of unity. The product χ1χ2 of two characters χ1; χ2 on A is defined by (χ1χ2)(a) = χ1(a)χ2(a) for a 2 A. With this product, the characters on A form an abelian group, the so-called character group of A, which we denote by Ab. The unit element (A) of Ab is the trivial character χ0 that maps A to 1. Since any character on A maps A to the roots of unity, the inverse χ−1 : a 7! χ(a)−1 of a character χ is equal to its complex conjugate χ : a 7! χ(a). Lemma 4.1. Let A be a cyclic group of order n. Then Ab is also a cyclic group of order n. Proof. Let A =< a >= f1; a; a2; : : : ; an−1g and an = 1. Choose a primitive n-th k k root of unity ρ1. Then χ1 given by χ1(a ) = ρ1 for k 2 Z defines a character on A. 71 j n Further, χ1 (j = 0; : : : ; n − 1) define different characters on A, and χ1 is the trivial (A) character χ0 . n j Conversely, let χ 2 Ab be arbitrary. Then χ(a) = 1, hence χ(a) = ρ1 for some j integer j, and so χ = χ1. It follows that Ab =< χ1 > is a cyclic group of order n. Lemma 4.2. Let the abelian group A be a direct product of cyclic groups of or- ders n1; n2; : : : ; nr. Then Ab is also a direct product of cyclic groups of orders n1; n2; : : : ; nr. Proof. By assumption, A is the direct product of cyclic groups < g1 >; : : : ; < gr > of orders n1; : : : ; nr. This means that k1 kr A =< g1 > × · · · × < gr >= fg1 ··· gr : ki 2 Zg; k1 kr where g1 ··· gr = 1 if and only if ki ≡ 0 (mod ni) for i = 1; : : : ; r. ni Clearly, for any roots of unity ρi with ρi = 1 for i = 1; : : : ; r, there is a unique character χ 2 Ab with χ(gi) = ρi for i = 1; : : : ; r. Now for i = 1; : : : ; r let ρi be a primitive ni-th root of unity, and define the character χi by χi(gi) = ρi; χi(gj) = 1 for j 6= i: Clearly, χi has order ni in Ab. One easily shows that for any χ 2 Ab, there are Qr ki Qr ki (A) k1; : : : ; kr 2 Z such that χ = i=1 χi , and moreover, that if i=1 χi = χ0 for some ki 2 Z, then ki ≡ 0 (mod ni) for i = 1; : : : ; r, by evaluating the identities at g1; : : : ; gr. Hence Ab =< χ1 > × · · · × < χr > : This proves our Lemma. Proposition 4.3. Every finite abelian group is isomorphic to a direct product of cyclic groups. Proof. See S. Lang, Algebra, Chap.1, x10. Theorem 4.4. Let A be a finite abelian group. Then there exists an isomorphism from A to Ab. In particular, jAbj = jAj. Proof. Obvious from Lemma 4.2 and Proposition 4.3. 72 Remark. The isomorphism in Theorem 4.4 is non-canonical, i.e., it 'depends on 0 0 a choice.' For we have A = Cn1 × · · · × Cnr and Ab = Cn1 × · · · × Cnr , where C ;C0 denote cyclic groups of order n , for i = 1; : : : ; r. We can now establish an ni ni i isomorphism from A to A by mapping a generator of C to a generator of C0 , for b ni ni i = 1; : : : ; r. Of course, this isomorphism depends on the choices of the generators. Let A be a finite abelian group. For a subgroup H of A, we define H? := fχ 2 Ab : χ(H) = 1g = fχ 2 Ab : χ(a) = 1 for a 2 Hg: Any homomorphism of abelian groups ' : A ! B defines a homomorphism between the character groups in the reverse direction, '∗ : Bb ! Ab : χ 7! χ ◦ ': Theorem 4.5. Let ' : A ! B be a surjective homomorphism of abelian groups. Then '∗ : Bb ! Ab is injective, and '∗(Bb) = (Ker ')?: In particular, if ' is an isomorphism from A to B, then '∗ is an isomorphism from Bb to Ab. ∗ ∗ (A) Proof. We prove that ' is injective. Let χ 2 Ker ' , i.e., χ ◦ ' = χ0 . Let b 2 B (B) and choose a 2 A with '(a) = b. Then χ(b) = χ('(a)) = 1. Hence χ = χ0 . As for the image, let H := Ker '. If χ = '∗χ0 for some χ0 2 Bb, then for a 2 H we have χ(a) = χ0('(a)) = 1; ? ? hence χ 2 H . Conversely, if χ 2 H , then χ(a1) = χ(a2) for any two a1; a2 with 0 '(a1) = '(a2). Hence χ : b 7! χ(a) for any a 2 A with '(a) = b gives a well-defined character on B with χ = '∗'0. So χ 2 '∗(Bb). It follows that '∗(Bb) = H. Corollary 4.6. Let A be a finite abelian group, and a 2 A with a 6= 1. Then there is χ 2 Ab with χ(a) 6= 1. Proof. Let B := A= < a > and let ' : A ! B be the canonical homomorphism. Then by the previous Theorem, '∗ establishes an isomorphism from Bb to < a >?= fχ 2 Ab : χ(a) = 1g. Hence j < a >? j = jBbj = jBj = jAj=j < a > j < jAj: This shows that there is χ 2 Ab with χ(a) 6= 1. 73 For a finite abelian group A, let Ab denote the character group of Ab. Each element a 2 A gives rise to a character ba on Ab, given by ba(χ) := χ(a). b Theorem 4.7 (Duality). a 7! ba defines a canonical isomorphism from A to Ab. Proof. The map ' : a 7! ba obviously defines a group homomorphism from A to b Ab. We show that it is injective. Let a 2 Ker('); then ba(χ) = 1 for all χ 2 Ab, i.e., χ(a) = 1 for all χ 2 Ab, which by the previous Corollary implies that a = 1. So indeed, ' is injective. But then ' is surjective as well, since by Theorem 4.4, jAbj = jAbj = jAj. Hence ' is an isomorphism. Theorem 4.8 (Orthogonality relations). Let A be a finite abelian group. (i) For χ1; χ2 2 Ab we have X jAj if χ1 = χ2; χ1(a)χ2(a) = 0 if χ1 6= χ2: a2A (ii) For a; b 2 A we have X jAj if a = b; χ(a)χ(b) = 0 if a 6= b: χ2Ab −1 −1 Proof. (i) Let χ = χ1χ2 . Then χ1(a)χ2(a) = χ1(a)χ2(a) = χ(a), hence the left-hand side is equal to X S := χ(a): a2A Clearly, if χ1 = χ2 then χ(a) = 1 for all a 2 A, hence S = jAj. Suppose that χ1 6= χ2. Then χ is not the trivial character, hence there is c 2 A with χ(c) 6= 1. Now X X χ(c)S = χ(ac) = χ(a) = S; a2A a2A hence S = 0. (ii) Apply (i) with Ab instead of A. Then by Theorem 4.7 we get for a; b 2 A, ( X X jAbj = jAj if a = b; χ(a)χ(b) = a(χ)b(χ) = b 0 if a 6= b: χ2Ab χ2Ab 74 4.2 Dirichlet characters Let q 2 Z>2. Denote the residue class of a mod q by a. Recall that the prime residue classes mod q,(Z=qZ)∗ = fa : gcd(a; q) = 1g form a group of order '(q) ∗ under multiplication of residue classes. We can lift any character χe on (Z=qZ) to a map χ : Z ! C by setting χ(a) if gcd(a; q) = 1; χ(a) := e 0 if gcd(a; q) > 1: Notice that χ has the following properties: (i) χ(1) = 1; (ii) χ(ab) = χ(a)χ(b) for a; b 2 Z; (iii) χ(a) = χ(b) if a ≡ b (mod q); (iv) χ(a) = 0 if gcd(a; q) > 1. Any map χ : Z ! C with properties (i){(iv) is called a (Dirichlet) character modulo ∗ q. Conversely, from a character mod q χ one easily obtains a character χe on (Z=qZ) by setting χe(a) := χ(a) for a 2 Z with gcd(a; q) = 1. Let G(q) be the set of characters modulo q. We define the product χ1χ2 of χ1; χ2 2 G(q) by (χ1χ2)(a) = χ1(a)χ2(a) for a 2 Z. With this operation, G(q) becomes a group, with unit element the principal character modulo q given by 1 if gcd(a; q) = 1; χ(q)(a) = 0 0 if gcd(a; q) > 1: The inverse of χ 2 G(q) is its complex conjugate χ : a 7! χ(a): It is clear, that this makes G(q) into a group that is isomorphic to the character group of (Z=qZ)∗. One of the advantages of viewing characters as maps from Z to C is that this allows to multiply characters of different moduli: if χ1 is a character mod q1 and χ2 a character mod q2, their product χ1χ2 is a character mod lcm(q1; q2).