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. 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). If jAj = n then an = 1, hence χ(a)n = 1 for each a 2 A and each character χ 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 (or Hom(A; C∗)). (A) The unit element 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). It would have been possible to develop the theory of characters using the fact that every finite abelian groups is the direct sum of cyclic groups, but we prefer to start from scratch. Let B be a subgroup of A and χ a character on B. By an extension of χ to A 0 0 0 we mean a character χ on A such that χ jB = χ, i.e., χ (b) = χ(b) for b 2 B. 83 Lemma 4.1. Let A be a finite abelian group, B a subgroup of A such that A=B is cyclic, and χ a character on B. Then χ has precisely jAj=jBj extensions to A. Proof. The order of A=B is precisely t := jAj=jBj. Let g 2 A be such that g := gB is a generator of A=B. Then h := gt 2 B. If χ0 is an extension of χ to A, then necessarily χ0(g)t = χ(h). We show that conversely, for each of the t roots ρ of t ρ = χ(h) there is a unique extension χρ of χ to A such that χρ(g) = ρ; this clearly implies our lemma. k Notice that A = fbg : b 2 B; k 2 Zg. The character χρ, if it exists, necessarily k k has to satisfy χρ(bg ) = χ(b)ρ , for b 2 B, k 2 Z. We now define χρ in this way and show that it is well-defined, i.e., independent of the choice of b and k. Indeed, k1 k2 k1−k2 −1 suppose that b1g = b2g , with b1; b2 2 B and k1; k2 2 Z, i.e. g = b1 b2. Then k1−k2 q −1 k1−k2 q g = 1, so q := (k2 − k1)=t 2 Z, hence h = b1 b2. This implies ρ = χ(h) = −1 k2 k1 χ(b1) χ(b2), hence χ(b2)ρ = χ(b1)ρ . This shows that indeed χρ is well-defined. It is easily shown to be a character. Proposition 4.2. Let A be a finite abelian group, B a subgroup of A, and χ a character on B. Then χ has precisely jAj=jBj extensions to A. Proof. We proceed by induction on jAj=jBj. If jAj=jBj = 1 we are done. Assume that jAj=jBj > 1. Choose g 2 A n B and define B0 := Bhgi. Then B0=B is cyclic, so by Lemma 4.1, the character χ has precisely jB0j=jBj extensions to B0. Since jB0j > jBj, we can apply the induction hypothesis and infer that each of these extensions to B0 has precisely jAj=jB0j extensions to A. Thus it follows that χ has precisely jAj=jBj extensions to A. Corollary 4.3. Let A be a finite abelian group. Then jAbj = jAj. Proof. Apply Proposition 4.2 with B = f1g. Corollary 4.4. Let A be a finite abelian group, and g 2 A with g 6= 1. Then there is a character χ on A with χ(g) 6= 1. Proof. Assume g has order r > 1. A character on hgi is uniquely determined by its value in g, so there is precisely one character χ0 on hgi with χ0(g) = 1. By Proposition 4.2, this character has precisely jAj=jhgij = jAj=r extensions to A. Hence there are characters χ on A that do not extend χ0, i.e., for which χ(g) 6= 1. 84 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). Theorem 4.5 (Duality). Let A be a finite abelian group. Then the map a 7! ba defines an isomorphism from A to Ab. b Proof. The map ' : a 7! ba obviously defines a group homomorphism from A to 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 Corollary 4.4 implies that a = 1. So indeed, ' is injective. But then ' is surjective as well, since by Corollary 4.3, jAbj = jAbj = jAj. Hence ' is an isomorphism. Theorem 4.6 (Orthogonality relations for characters). Let A be a finite abelian group. (i) For any two characters χ1; χ2 on A we have X jAj if χ1 = χ2; χ1(a)χ2(a) = 0 if χ1 6= χ2: a2A (ii) For any two elements a; b of A we have X jAj if a = b; χ(a)χ(b) = 0 if a 6= b: χ2Ab Proof. Part (ii) follows by applying part (i) with Ab instead of A, and using The- orem 4.5 and Corollary 4.3. So we prove only (i). Let χ1; χ2 2 Ab and put P −1 P S := a2A χ1(a)χ2(a). Let χ := χ1χ2 = χ1χ2 . Then S = a2A χ(a). Clearly, if (A) (A) χ1 = χ2 then χ = χ0 , hence S = jAj. Let χ1 6= χ2. Then χ 6= χ0 , hence there is g 2 A with χ(g) 6= 1. Further, X χ(g)S = χ(ga) = S; a2A since ga runs through the elements of A. Hence S = 0. This will not be needed later, but for completeness we show that there is also an isomorphism from a finite abelian group A to its character group Ab. But unlike the isomorphism in Theorem 4.5 this is not canonical, since it will depend on a choice of generators for A. 85 Lemma 4.7. Let A be a cyclic group of order n. Then Ab is also a cyclic group of order n. Proof. Let A = hgi. Then A = f1; g; : : : ; gn−1g and gn = 1. A character χ on A is determined by χ(g). Let ρ1 be a primitive n-th root of unity. It is easy to see that (A) n−1 there is a character χ1 on A with χ1(g) = ρ1, that χ0 ; χ1; : : : ; χ1 are distinct, n (A) n and χ1 = χ0 . Further, if χ is any character on A, then χ(g) = 1, which implies that χ is a power of χ1. So Ab = hχ1i is a cyclic group of order n. Lemma 4.8. Let A = A1 × · · · × Ar be the direct product of finite abelian groups A1;:::;Ar. Then Ab is isomorphic to Ac1 × · · · × Acr. Proof. It suffices to prove this for r = 2; then the proof of the lemma can be completed by induction on r. Denote by 1 the unit element of A. Let A = A1 ×A2 = fg1g2 : g1 2 A1; g2 2 A2g where g1g2 = 1 if and only if g1 = g2 = 1. Define a map ' : Ac1 × Ac2 ! Ab :(χ1; χ2) 7! χ1χ2; where χ1χ2(g1g2) := χ1(g1)χ2(g2) for g1 2 A1; g2 2 A2. It is easy to see that ' is a group homomorphism. Substituting g1 = 1, respectively g2 = 1, we see that χ2, χ1 are uniquely determined by χ1χ2. Hence ' is injective. Since Ac1 × Ac2 and Ab have the same cardinality, it follows also that ' is surjective. Proposition 4.9. Every finite abelian group is a direct product of cyclic groups. Proof. See S. Lang, Algebra, Chap.1, x10. Theorem 4.10. Let A be a finite abelian group. Then there exists an isomorphism from A to Ab. Proof. By Proposition 4.9, A is a direct product C1 ×· · ·×Cr of finite cyclic groups. By Lemmas 4.8, 4.7, Ab is isomorphic to Cc1 × · · · × Ccr, where Cbi is a cyclic group of the same order as Ci, for i = 1; : : : ; r. Now the isomorphism from A to Ab can be established by mapping a generator of Ci to one of Cbi, for i = 1; : : : ; r. Remark. The isomorphism constructed above depends on choices for generators of Ci, Cbi, for i = 1; : : : ; r. So it is not canonical. 86 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.