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 |A|.
Let A be a finite abelian group. A character on A is a group homomorphism
χ : A → C∗ (i.e., C \ {0} with multiplication).
If |A| = n then an = 1A , hence χ(a)n = 1 for each a ∈ A and each character on
A. Therefore, a character on A maps A to the roots of unity.
The product χ1 χ2 of two characters χ1 , χ2 on A is defined by (χ1 χ2 )(a) =
χ1 (a)χ2 (a) for a ∈ A. With this product, the characters on A form an abelian
b The unit element
group, the so-called character group of A, which we denote by A.
(A)
b is the trivial character χ0 that maps A to 1. Since any character on A maps
of A
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).
b is also a cyclic group of
Lemma 4.1. Let A be a cyclic group of order n. Then A
order n.
Proof. Let A =< a >= {1, a, a2 , . . . , an−1 } and an = 1. Choose a primitive n-th
root of unity ρ1 . Then χ1 given by χ1 (ak ) = ρk1 for k ∈ Z defines a character on A.
71
Further, χj1 (j = 0, . . . , n − 1) define different characters on A, and χn1 is the trivial
(A)
character χ0 .
b be arbitrary. Then χ(a)n = 1, hence χ(a) = ρj1 for some
Conversely, let χ ∈ A
b =< χ1 > is a cyclic group of order n.
integer j, and so χ = χj1 . It follows that A
Lemma 4.2. Let the abelian group A be a direct product of cyclic groups of orb is also a direct product of cyclic groups of orders
ders n1 , n2 , . . . , nr . Then A
n1 , n2 , . . . , nr .
Proof. By assumption, A is the direct product of cyclic groups < g1 >, . . . , < gr >
of orders n1 , . . . , nr . This means that
A =< g1 > × · · · × < gr >= {g1k1 · · · grkr : ki ∈ Z},
where g1k1 · · · grkr = 1 if and only if ki ≡ 0 (mod ni ) for i = 1, . . . , r.
Clearly, for any roots of unity ρi with ρni i = 1 for i = 1, . . . , r, there is a unique
b with χ(gi ) = ρi for i = 1, . . . , r. Now for i = 1, . . . , r let ρi be a
character χ ∈ A
primitive ni -th root of unity, and define the character χi by
χi (gi ) = ρi ,
χi (gj ) = 1 for j 6= i.
b One easily shows that for any χ ∈ A,
b there are
Clearly, χi has order ni in A.
Q
Qr
(A)
r
k1 , . . . , kr ∈ Z such that χ = i=1 χki i , and moreover, that if i=1 χki i = χ0 for
some ki ∈ Z, then ki ≡ 0 (mod ni ) for i = 1, . . . , r, by evaluating the identities at
g1 , . . . , gr . Hence
b =< χ1 > × · · · × < χr > .
A
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, §10.
Theorem 4.4. Let A be a finite abelian group. Then there exists an isomorphism
b In particular, |A|
b = |A|.
from A to A.
Proof. Obvious from Lemma 4.2 and Proposition 4.3.
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Remark. The isomorphism in Theorem 4.4 is non-canonical, i.e., it ’depends on
b = Cn0 × · · · × Cn0 , where
a choice.’ For we have A = Cn1 × · · · × Cnr and A
r
1
Cni , Cn0 i denote cyclic groups of order ni , for i = 1, . . . , r. We can now establish an
b by mapping a generator of Cn to a generator of Cn0 , for
isomorphism from A to A
i
i
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
b : χ(H) = 1} = {χ ∈ A
b : χ(a) = 1 for a ∈ H}.
H ⊥ := {χ ∈ A
Any homomorphism of abelian groups ϕ : A → B defines a homomorphism between
the character groups in the reverse direction,
b→A
b : χ 7→ χ ◦ ϕ.
ϕ∗ : B
Theorem 4.5. Let ϕ : A → B be a surjective homomorphism of abelian groups.
b→A
b is injective, and
Then ϕ∗ : B
b = (Ker ϕ)⊥ .
ϕ∗ (B)
In particular, if ϕ is an isomorphism from A to B, then ϕ∗ is an isomorphism from
b to A.
b
B
(A)
Proof. We prove that ϕ∗ is injective. Let χ ∈ Ker ϕ∗ , i.e., χ ◦ ϕ = χ0 . Let b ∈ B
(B)
and choose a ∈ A with ϕ(a) = b. Then χ(b) = χ(ϕ(a)) = 1. Hence χ = χ0 .
b then for a ∈ H
As for the image, let H := Ker ϕ. If χ = ϕ∗ χ0 for some χ0 ∈ B,
we have
χ(a) = χ0 (ϕ(a)) = 1,
hence χ ∈ H ⊥ . Conversely, if χ ∈ H ⊥ , then χ(a1 ) = χ(a2 ) for any two a1 , a2 with
ϕ(a1 ) = ϕ(a2 ). Hence χ0 : b 7→ χ(a) for any a ∈ A with ϕ(a) = b gives a well-defined
b It follows that ϕ∗ (B)
b = H.
character on B with χ = ϕ∗ ϕ0 . So χ ∈ ϕ∗ (B).
Corollary 4.6. Let A be a finite abelian group, and a ∈ A with a 6= 1. Then there
b with χ(a) 6= 1.
is χ ∈ A
Proof. Let B := A/ < a > and let ϕ : A → B be the canonical homomorphism.
b to < a >⊥ =
Then by the previous Theorem, ϕ∗ establishes an isomorphism from B
b : χ(a) = 1}. Hence
{χ ∈ A
b = |B| = |A|/| < a > | < |A|.
| < a >⊥ | = |B|
b with χ(a) 6= 1.
This shows that there is χ ∈ A
73
b
b denote the character group of A.
b Each element
For a finite abelian group A, let A
b given by b
a ∈ A gives rise to a character b
a on A,
a(χ) := χ(a).
b
b
Theorem 4.7 (Duality). a 7→ b
a defines a canonical isomorphism from A to A.
Proof. The map ϕ : a 7→ b
a obviously defines a group homomorphism from A to
b
b We show that it is injective. Let a ∈ Ker(ϕ); then b
b
A.
a(χ) = 1 for all χ ∈ A,
b which by the previous Corollary implies that a = 1.
i.e., χ(a) = 1 for all χ ∈ A,
So indeed, ϕ is injective. But then ϕ is surjective as well, since by Theorem 4.4,
b
b = |A|
b = |A|. Hence ϕ is an isomorphism.
|A|
Theorem 4.8 (Orthogonality relations). Let A be a finite abelian group.
b we have
(i) For χ1 , χ2 ∈ A
X
|A| if χ1 = χ2 ;
χ1 (a)χ2 (a) =
0 if χ1 6= χ2 .
a∈A
(ii) For a, b ∈ A we have
X
χ(a)χ(b) =
|A| if a = b;
0 if a 6= b.
b
χ∈A
−1
Proof. (i) Let χ = χ1 χ−1
= χ(a), hence the
2 . Then χ1 (a)χ2 (a) = χ1 (a)χ2 (a)
left-hand side is equal to
X
S :=
χ(a).
a∈A
Clearly, if χ1 = χ2 then χ(a) = 1 for all a ∈ A, hence S = |A|. Suppose that
χ1 6= χ2 . Then χ is not the trivial character, hence there is c ∈ A with χ(c) 6= 1.
Now
X
X
χ(c)S =
χ(ac) =
χ(a) = S,
a∈A
a∈A
hence S = 0.
b instead of A. Then by Theorem 4.7 we get for a, b ∈ A,
(ii) Apply (i) with A
(
X
X
b = |A| if a = b,
|A|
χ(a)χ(b) =
b
a(χ)bb(χ) =
0
if a 6= b.
b
b
χ∈A
χ∈A
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4.2
Dirichlet characters
Let q ∈ Z>2 . Denote the residue class of a mod q by a. Recall that the prime
residue classes mod q, (Z/qZ)∗ = {a : gcd(a, q) = 1} form a group of order ϕ(q)
under multiplication of residue classes. We can lift any character χ
e on (Z/qZ)∗ to a
map χ : Z → C by setting
χ
e(a) if gcd(a, q) = 1;
χ(a) :=
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 ∈ 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 χ
e on (Z/qZ)∗
by setting χ
e(a) := χ(a) for a ∈ Z with gcd(a, q) = 1.
Let G(q) be the set of characters modulo q. We define the product χ1 χ2 of
χ1 , χ2 ∈ G(q) by (χ1 χ2 )(a) = χ1 (a)χ2 (a) for a ∈ 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;
(q)
χ0 (a) =
0 if gcd(a, q) > 1.
The inverse of χ ∈ 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 χ1 χ2 is a character mod lcm(q1 , q2 ).
We can easily translate the orthogonality relations for characters of (Z/qZ)∗ into
orthogonality relations for Dirichlet characters modulo q. Recall that a complete
75
residue system modulo q is a set, consisting of precisely one integer from every
residue class modulo q, e.g., {3, 5, 11, 22, 104} is a complete residue system modulo
5.
Theorem 4.9. Let q ∈ Z>2 , and let Sq be a complete residue system modulo q.
(i) Let χ1 , χ2 ∈ G(q). Then
X
χ1 (a)χ2 (a) =
a∈Sq
ϕ(q) if χ1 = χ2 ;
0
if χ1 6= χ2 .
(ii) Let a, b ∈ Z. Then

 ϕ(q) if gcd(ab, q) = 1, a ≡ b (mod q);
χ(a)χ(b) =
0
if gcd(ab, q) = 1, a ≡
6 b (mod q);

χ∈G(q)
0
if gcd(ab, q) > 1.
X
Proof. Exercise.
Let χ be a character mod q and d a positive divisor of q.
We say that q is induced by a character χ0 mod d if χ(a) = χ0 (a) for every a ∈ Z
(q)
with gcd(a, q) = 1. Equivalently stated, χ is induced by χ0 if χ = χ0 χ0 . Notice
that if gcd(a, d) = 1 and gcd(a, q) > 1, then χ0 (a) 6= 0 but χ(a) = 0.
The character χ is called primitive if it is not induced by a character mod d for
any divisor d < q of q.
The conductor fχ of a χ is the gcd of all divisors d of q such that χ is induced
(1)
by a character modulo d. If we allow the character mod 1, given by χ0 (a) = 1 for
a ∈ Z, we see that for q > 2 the principal character mod q has conductor 1.
The next lemma gives a method to verify if a character χ is induced by a character
mod d.
Theorem 4.10. Let χ be a character mod q, and d a divisor of q. Then the following assertions are equivalent:
(i) χ(a) = 1 for all a ∈ Z with a ≡ 1 (mod d) and gcd(a, q) = 1;
(ii) χ is induced by a character mod d.
Moreover, if (ii) holds, the character mod d by which χ is induced is uniquely determined.
76
Proof. We first observe that if a is an integer coprime with d then there is an integer
b with b ≡ a (mod d) such that b is coprime with q. Indeed, write q = q1 q2 , where
q1 is composed of the primes occurring in the factorization of d, and where q2 is
composed of primes not dividing d. By the Chinese Remainder Theorem, there is
b ∈ Z with
b ≡ a (mod d), b ≡ 1 (mod q2 ).
This integer b is coprime with d, hence with q1 , and also coprime with q2 , so it is
coprime with q.
The remaining part of the proof is a simple application of Theorem 4.5. Our
above observation implies that
ϕd : (Z/qZ)∗ → (Z/dZ)∗ : a(mod q) 7→ a(mod d)
is a surjective homomorphism. The kernel of ϕd is
Hd := {a(mod q) ∈ (Z/qZ)∗ : a ≡ 1 (mod d)}.
e
Now χ is induced by a character χ0 mod d if and only if χ
e = χe0 ◦ ϕd = ϕ∗d χe0 , where χ
∗
0
is the character on (Z/qZ) corresponding to χ, and χe is the character on (Z/dZ)∗
corresponding to χ0 . By Theorem 4.5, such a character χe0 exists, and if it does is
unique, if and only if χ
e(Hd ) = 1. But this is precisely the equivalence (i)⇐⇒(ii) in
another language.
Theorem 4.11. Let χ be a character mod q. Suppose χ is induced by a character
mod d1 and by a character mod d2 . Then χ is induced by a character mod gcd(d1 , d2 ).
We need the following lemma.
Lemma 4.12. Let d = gcd(d1 , d2 ). Let a be an integer with gcd(a, q) = 1 and
a ≡ 1 (mod d). Then there are integers u, v such that a = 1 + ud1 + vd2 and
gcd(1 + ud1 , q) = 1.
Proof. Write q = q1 q2 , where q1 is composed of primes dividing d2 , and q2 is composed of primes not dividing d2 . Then gcd(q2 d1 , d2 ) = d. There are integers t, x, y
such that a = 1 + td and d = xq2 d1 + yd2 . This implies
a = 1 + txq2 d1 + tyd2 = 1 + ud1 + vd2 with u = txq2 , v = ty.
77
It is clear that 1 + ud1 is coprime with q2 . Further, it is coprime with d2 , since
1 + ud1 = a − vd2 ,
and since a is coprime with q, hence with d2 . So 1 + ud1 is coprime with q1 . It
follows that 1 + ud1 is coprime with q1 q2 = q.
Proof of Theorem 4.11. Let a be any integer with gcd(a, q) = 1 and a ≡ 1 (mod d).
Choose integers u, v such that a = 1 + ud1 + vd2 and gcd(1 + ud1 , q) = 1. Using
subsequently that χ is induced by characters mod d2 , d1 , it follows that
χ(a) = χ(1 + ud1 ) = χ(1) = 1.
Now by Theorem 4.10, χ is induced by a character mod d.
Corollary 4.13. Let χ be a character mod q. Then there is a unique character mod
fχ that induces χ.
Proof. Obvious.
The orthogonality relations in Theorem 4.9 will play a crucial role in the proof
of the Prime Number Theorem for arithmetic progressions. Theorem 4.9 is in fact
the special case A = (Z/qZ)∗ of Theorem 4.8, and in the proof of this last theorem,
we used a rather involved result from algebra, which we stated without proof, that
is, that every finite abelian group is a direct product of cyclic groups.
For the orthogonality relations in Theorem 4.9 we need only that (Z/qZ)∗ is a
direct product of cyclic groups. This is an easy consequence of the result below.
We denote by a the residue class a (mod q), if from the context it is clear which
integer q we are considering.
Theorem 4.14. (i) Let q = pk11 · · · pkt t , where the pi are distinct primes and the ki
positive integers. Then
(Z/qZ)∗ ∼
= (Z/pk11 Z)∗ × · · · × (Z/pkt t Z)∗ .
(ii) Let p be a prime > 3. Then (Z/pk Z)∗ is cyclic of order pk−1 (p − 1).
(iii) (Z/4Z)∗ is cyclic of order 2.
Further, if k > 3 then (Z/2k Z)∗ =< −1 > × < 5 > is the direct product of a cyclic
group of order 2 and a cyclic group of order 2k−2 .
78
We skip the proof of (i) and the proof of the case k = 1 of (ii), which belong to
a basic algebra course. For the proof of the remaining parts, we need a lemma.
For a prime number p, and for a ∈ Z \ {0}, we denote by ordp (a) the largest
integer k such that pk divides a.
Lemma 4.15. Let p be a prime number and a an integer such that ordp (a − 1) > 1
if p > 3 and ordp (a − 1) > 2 if p = 2. Then
k
ordp (ap − 1) = ordp (a − 1) + k.
Proof. We prove the assertion only for k = 1; then the general statement follows
easily by induction on k. Our assumption on a implies that a = 1 + pt b, where t > 1
if p > 3, t > 2 if p = 2, and where b is an integer not divisible by p. Now by the
binomial formula,
k X
p
a −1=
(pt b)j = pt+1 bj + pt+2 (· · · ).
j
j=1
p
Here we have used that all binomial coefficients pj are divisible by p except the
last. But the last term (pt b)p is divisible by ppt , and the exponent pt is larger than
t + 1 (the assumption t > 2 for p = 2 is needed to ensure this). This shows that
ordp (ap − 1) = t + 1.
Proof of Theorem 4.14. (ii) We take for granted that (Z/pZ)∗ is cyclic of order p−1,
and assume that k > 2. We construct a generator for (Z/pk Z)∗ . Let g be an integer
such that g (mod p) is a generator of (Z/pZ)∗ . We show that we can choose g such
that ordp (g p−1 − 1) = 1. Indeed, assume that ordp (g p−1 − 1) > 2 and take g + p.
Then
p−1 X
p − 1 p−1−j j
p−1
(g + p)
−1 =
g
p = (p − 1)g p−2 p + p2 (· · · )
j
j=1
= = −g p−2 p + p2 (· · · )
hence ordp ((g + p)p−1 − 1) = 1. So, replacing g by g + p if need be, we get an integer
g such that g (mod p) generates (Z/pZ)∗ and ordp (g p−1 − 1) = 1.
We show that g := g (mod pk ) generates (Z/pk Z)∗ . Let n be he order of g in
(Z/pk Z)∗ ; that is, n is the smallest positive integer with g n ≡ 1 (mod pk ). On the
79
one hand, g n ≡ 1 (mod p), hence p − 1 divides n. On the other hand, n divides the
order of (Z/pk Z)∗ , that is, pk−1 (p − 1). So n = ps (p − 1) with s 6 k − 1. By Lemma
4.15 we have
ordp (g n − 1) = ordp (g p−1 − 1) + s = s + 1.
This has to be equal to k, so s = k − 1. Hence n = pk−1 (p − 1) is equal to the order
of (Z/pk Z)∗ . It follows that (Z/pk Z)∗ =< g >.
(iii) Assume that k > 3. Define the subgroup of index 2,
H := {a ∈ (Z/2k Z)∗ : a ≡ 1 (mod 4)}.
Then
(Z/2k Z)∗ = H ∪ (−H) = {(−1)k a : k ∈ {0, 1}, a ∈ H}
and (−1)k a = 1 if and only if k = 0 and a = 1. Hence (Z/2k Z)∗ =< −1 > ×H.
Similarly as above, one shows that H is cyclic of order 2k−2 , and that H =< 5 >.
4.3
Gauss sums
Let q ∈ Z>2 . For a character χ mod q and for b ∈ Z, we define the Gauss sum
τ (b, χ) :=
X
χ(x)e2πibx/q ,
x∈Sq
where Sq is a full system of representatives modulo q. This does not depend on the
choice of Sq . Gauss sums occur for instance in functional equations for L-functions
P
−s
(later).
L(s, χ) = ∞
n=1 χ(n)n
We prove some basic properties for Gauss sums.
Theorem 4.16. Let q ∈ Z>2 and let χ be a character mod q. Further, let b ∈ Z.
(i) If gcd(b, q) = 1, then τ (b, χ) = χ(b) · τ (1, χ).
(ii) If gcd(b, q) > 1 and χ is primitive, then τ (b, χ) = χ(b) · τ (1, χ) = 0.
Proof. (i) Suppose gcd(b, q) = 1. If x runs through a complete residue system Sq
mod q, then bx runs to another complete residue system Sq0 mod q. Write y = bx.
80
Then χ(y) = χ(b)χ(x), hence χ(x) = χ(b)χ(y). Therefore,
τ (b, χ) =
X
χ(x)e2πibx/q =
X
χ(b)χ(y)e2πiy/q
y∈Sq0
x∈Sq
= χ(b)τ (1, χ).
(ii) We use the following observation: if q1 is any divisor of q with 1 6 q1 6 q,
then there is c ∈ Z such that c ≡ 1 (mod q1 ), gcd(c, q) = 1, and χ(c) 6= 1. Indeed,
this is obvious if q1 = q. If q1 < q, then Theorem 4.10 implies that if there is no
such integer c then χ is induced by a character mod q1 , contrary to our assumption
that χ is primitive.
Now let d := gcd(b, q), put b1 := b/d, q1 := q/d, and choose c according to the
observation. Then
X
χ(cx)e2πibx/q .
χ(c)τ (b, χ) =
x∈Sq
If x runs through a complete residue system Sq mod q, then y := cx runs through
another complete residue system Sq0 mod q. Further, since c ≡ 1 (mod q1 ) we have
e2πixb/q = e2πixb1 /q1 = e2πicxb1 /q1 = e2πiyb/q .
Hence
χ(c)τ (b, χ) =
X
χ(y)e2πiby/q = τ (b, χ).
y∈Sq
Since χ(c) 6= 1 this implies that τ (b, χ) = 0.
Theorem 4.17. Let q ∈ Z>2 and let χ be a primitive character mod q. Then
|τ (1, χ)| =
81
√
q.
Proof. We have by Theorem 4.16,
2
|τ (1, χ)|
= τ (1, χ) · τ (1, χ) =
q−1
X
χ(x)e−2πix/q τ (1, χ)
x=0
=
q−1
X
e−2πix/q τ (x, χ) =
q−1
q−1
X
X
=
χ(y)
!
χ(y)e2πixy/q
y=0
!
χ(y)e2πix(y−1)/q
Pq−1
x=0
q−1
X
!
2πix(y−1)/q
e
=
q−1
X
χ(y)S(y), say.
y=0
x=0
y=0
If y = 1, then S(y) =
q−1
X
y=0
x=0
q−1
X
e−2πix/q
x=0
x=0
=
q−1
X
1 = q, while if y 6= 1, then
S(y) =
e2πi(y−1) − 1
= 0.
e2πi(y−1)/q − 1
Hence |τ (1, χ)|2 = χ(1)q = q.
√
Remark. Theorem 4.17 implies that εχ := τ (1, χ)/ q lies on the unit circle. Gauss
gave a simple expression for εχ in the case that εχ is a primitive real character mod
q, i.e., χ(a) ∈ {±1} if gcd(a, q) = 1. There is no general method known to compute
εχ for non-real characters χ modulo large values of q.
4.4
Quadratic reciprocity
For the interested reader, we give a proof of Gauss’ Quadratic Reciprocity Theorem
using Gauss sums. This section requires a little bit more algebraic background.
Let p > 2 be a prime number. For a ∈ Z, we define the Legendre symbol
1 if x2 ≡ a (mod p) is solvable in x ∈ Z and p - a;
:=
−1 if x2 ≡ a (mod p) is not solvable in x ∈ Z;

p
0 if p|a.
a


82
Lemma 4.18.
·
p
is a character mod p, and
a
p
≡ a(p−1)/2 (mod p) for a ∈ Z.
Proof. The group (Z/pZ)∗ is cyclic of order p − 1. Let g(mod p) be a generator
of this group, and take a ∈ Z with gcd(a, p) = 1. Then there is t ∈ Z such that
2
a ≡ g t (mod p).
is solvable
in x ∈ Z if and only if t is
Now clearly, x ≡ a (mod p) a
·
t
even. Hence p = (−1) . This implies that p is a character mod p.
Note that (g (p−1)/2 )2 ≡ 1 (mod p), hence g (p−1)/2 ≡ ±1 (mod p). But g (p−1)/2 6≡
1 (mod p) since g (mod p) is a generator of (Z/pZ)∗ . Hence g (p−1)/2 ≡ −1 (mod p).
It follows that
a
(p−1)/2
t
a
≡ (−1) ≡
(mod p).
p
Gauss’ Quadratic Reciprocity Theorem is as follows:
Theorem 4.19. Let p, q be distinct primes > 2. Then
p q −1 if p ≡ q ≡ 3 (mod 4),
(p−1)(q−1)/4
= (−1)
=
1 otherwise.
q p
We first make some preparations and then prove some lemmas.
Let Q[X] denote the ring of polynomials with coefficients in Q. A number α ∈ C
is called algebraic if there is a non-zero polynomial f ∈ Q[X] such that f (α) = 0.
Among all non-zero polynomials from Q[X] having α as a zero, we choose one of
minimal degree. By multiplying such a polynomial with a suitable constant, we
obtain one which is monic, i.e., of which the coefficient of the highest power of X
is 1. There is only one monic polynomial in Q[X] of minimal degree having α as
a zero, for if there were two, their difference would give a non-zero polynomial in
Q[X] of smaller degree having α as a zero. This unique monic polynomial in Q[X]
of minimal degree having α as a zero is called the minimal polynomial of α, denoted
by fα .
We observe that fα must be irreducible in Q[X], that is, not a product of two
non-constant polynomials from Q[X]. For otherwise, α would be a zero of one of
these polynomials, which has degree smaller than that of fα .
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Let q be a prime number > 2. We write ζq := e2πi/q . Define
( r
)
X
Rq := Z[ζq ] =
ai ζqi : ai ∈ Z, r > 0 .
i=0
This set is closed under addition and multiplication, hence it is a ring.
Lemma 4.20. Rq ∩ Q = Z.
Proof. We use without proof, that the minimal polynomial of ζq is (X q −1)/(X−1) =
P
j
X q−1 + · · · + X + 1. Hence ζqq−1 = − q−2
j=0 ζq . By repeatedly substituting this into
Pr
P
j
an expression i=0 ai ζqi with ai ∈ Z, we eventually get an expression q−2
j=0 bj ζq with
bj ∈ Z for all j. Hence all elements of Rq can be expressed in this form. Now if
α ∈ Rq ∩ Q, we get
q−2
X
α=
bj ζqj
j=0
with α ∈ Q and bj ∈ Z for all j. This implies that ζq is a zero of the polynomial
bq−2 X q−2 + · · · + b0 − α. Since the minimal polynomial of ζq has degree q − 1, this
is possible only if b0 = α and b1 = · · · = bq−2 = 0. Hence α ∈ Z.
Given α, β ∈ Rq and n ∈ Z>0 , we write α ≡ β (mod n) in Rq if (α − β)/n ∈ Rq .
Further, we write α ≡ β (mod n) in Z if (α − β)/n ∈ Z. By the Lemma we just
proved, for α, β ∈ Z we have that α ≡ β (mod n) in Rq if and only if α ≡ β (mod n)
in Z.
Lemma 4.21. Let p be any prime number. Then for α1 , . . . , αr ∈ Rq we have
(α1 + · · · + αr )p ≡ α1p + · · · + αrp (mod p) in Rq .
Proof. By the multinomial theorem,
(α1 + · · · + αr )p =
X
i1 +···+ir
p!
α1i1 · · · αrir .
i ! · · · ir !
=p 1
All multinomial coefficients are divisible by p, except those where one index ij = p
and the others are 0.
For notational convenience we write χq for q· . We now invoke Gauss sums.
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Lemma 4.22. τ (1, χq )2 = (−1)(q−1)/2 q.
Proof. The proof is almost the same as that of Theorem 4.17. The character χq is
(q)
primitive, since q is a prime and χq 6= χ0 . Using Theorem 4.16 we have, since χq
is a real character,
τ (1, χq )2 =
q−1
X
χq (x)e2πix/q τ (1, χq ) =
q−1
X
e2πix/q τ (x, χq ).
x=0
x=0
The same reasoning as in the proof of Theorem 4.17, but with e2πix/q instead of
e−2πix/q gives
! q−1
q−1
q−1
X
X
X
χ(y)S 0 (y),
χ(y)
e2πix(y+1)/q =
τ (1, χq )2 =
y=0
y=0
x=0
where S 0 (y) = q if y = −1, and S 0 (y) = 0 otherwise. Hence
τ (1, χq )2 = χq (−1) · q = (−1)(q−1)/2 q.
Proof of Theorem 4.19. We work in the ring Rq . Notice that by Lemma 4.21 and
Theorem 4.16,
q−1
X
p
τ (1, χq ) ≡
χq (x)p ζqpx
x=0
≡
q−1
X
χq (x)ζqpx
≡ τ (p, χq ) ≡
x=0
p
q
τ (1, χq ) (mod p) in Rq .
On multiplying with τ (1, χq ) and applying Lemma 4.22, we obtain
p
(mod p) in Rq .
τ (1, χq )p+1 ≡ (−1)(q−1)/2 q ·
q
On the other hand, by Lemmas 4.22, 4.20,
τ (1, χq )p+1 = (−1)(q−1)(p+1)/4 q (p+1)/2 = (−1)(q−1)/2 q · (−1)(q−1)(p−1)/4 q (p−1)/2
(q−1)/2
(p−1)(q−1)/4 q
≡ (−1)
q · (−1)
(mod p) in Rq .
p
As a consequence,
(q−1)/2
(−1)
q·
p
q
(q−1)/2
≡ (−1)
(q−1)(p−1)/4
q · (−1)
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q p
(mod p) in Z.
Since q is coprime with p, this gives
p
q ≡ (−1)(p−1)(q−1)/4
(mod p) in Z.
q
p
Since integers equal to ±1 can be congruent modulo p only if they are equal, this
implies Theorem 4.19.
We finish with the following result.
Theorem 4.23. Let p be a prime > 2. Then
2 1 if p ≡ ±1 (mod 8),
=
−1 if p ≡ ±3 (mod 8).
p
Proof. Exercise.
You have to follow the proof given above, but instead of Rq , χq , you have to use the
ring R8 = Z[ζ8 ] where ζ8 = e2πi/8 , and the character χ8 mod 8, given by


1 if a ≡ ±1 (mod 8),
χ8 (a) =
−1 if a ≡ ±3 (mod 8),

0 if a ≡ 0 (mod 2).
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