1
Appendices
We collect some results that might be covered in a first course in algebraic number theory.
A. Quadratic Reciprocity Via Gauss Sums
A1. Introduction
In this appendix, p is an odd prime unless otherwise specified. A quadratic equation
modulo p looks like ax2 + bx + c = 0 in Fp . Multiplying by 4a, we have
2
2ax + b
≡ b2 − 4ac mod p
Thus in studying quadratic equations mod p, it suffices to consider equations of the form
x2 ≡ a mod p.
If p|a we have the uninteresting equation x2 ≡ 0, hence x ≡ 0, mod p. Thus assume that
p does not divide a.
A2. Definition
The Legendre symbol
χ(a) =
a
p
is given by
1
if a(p−1)/2 ≡ 1 mod p
χ(a) =
−1 if a(p−1)/2 ≡ −1 mod p.
If b = a(p−1)/2 then b2 = ap−1 ≡ 1 mod p, so b ≡ ±1 mod p and χ is well-defined. Thus
χ(a) ≡ a(p−1)/2
mod p.
A3. Theorem
The Legendre symbol ( ap ) is 1 if and only if a is a quadratic residue (from now on abbreviated QR) mod p.
Proof. If a ≡ x2 mod p then a(p−1)/2 ≡ xp−1 ≡ 1 mod p. (Note that if p divides x then
p divides a, a contradiction.) Conversely, suppose a(p−1)/2 ≡ 1 mod p. If g is a primitive
root mod p, then a ≡ g r mod p for some r. Therefore a(p−1)/2 ≡ g r(p−1)/2 ≡ 1 mod p,
so p − 1 divides r(p − 1)/2, hence r/2 is an integer. But then (g r/2 )2 = g r ≡ a mod p,
and a is a QR mod p. ♣
2
A4. Theorem
The mapping a → χ(a) is a homomorphism from Fp to {±1}.
Proof. We compute
ab
p
≡ (ab)(p−1)/2 = a(p−1)/2 b(p−1)/2 ≡
b
a
p
p
so χ(ab) = χ(a)χ(b). ♣
A5. Theorem
If g is a primitive root mod p, then χ(g) = −1, so g is a quadratic nonresidue (from now
on abbreviated QNR) mod p. Consequently, exactly half of the integers 1,2,...,p − 1 are
QR’s and half are QNR’s.
Proof. If h2 ≡ g mod p then h is a primitive root with twice the period of g, which is
impossible. Thus by (A4), χ(ag) = −χ(a), so a → ag gives a bijection between QR’s and
QNR’s. ♣
A6. The First Supplementary Law
From the definition (A2) and the fact that (p − 1)/2 is even if p ≡ 1 mod 4 and odd if
p ≡ 3 mod 4, we have
1
if p ≡ 1 mod 4
−1
(p−1)/2
= (−1)
=
p
−1 if p ≡ 3 mod 4.
A7. Definition
Let K be a field of characteristic = p such that K contains the p-th roots of unity. Let
ζ ∈ K be a primitive p-th root of unity. Define the Gauss sum by
τp =
p−1 a
a=1
p
ζ a.
A8. Theorem
τp2 = (−1)(p−1)/2 p.
Proof. From the definition of Gauss sum and (A4) we have
τp2 =
p−1 ab a+b
.
ζ
p
a,b=1
3
For each a, we can sum over all c such that b ≡ ac mod p. (As c ranges over 1, . . . , p − 1,
ac also takes all values 1, . . . , p − 1.) Thus
τp2
Since
=
a=1 c=1
a2
p
p−1 p−1 2 a c
p
ζ a+ac .
= 1, this simplifies to
τp2 =
p−1 p−1
c=1
a=1
ζ a(1+c)
c
.
p
If 1 + c ≡ 0 mod p then 1, ζ 1+c , ζ 2(1+c) , . . . , ζ (p−1)(1+c) runs through all the roots (zeros)
of X p − 1 (note that ζ p = 1). But the coefficient of X p−1 is 0, so the sum of the roots is
0. Therefore the sum of ζ 1+c , ζ 2(1+c) , . . . , ζ (p−1)(1+c) is -1. If 1 + c ≡ 0 mod p, then we
are summing p − 1 ones. Consequently,
τp2 = −
p−2 c
c=1
p
−1
+ (p − 1)
.
p
(Note that 1 + c ≡ 0 mod p ↔ c = p − 1.) We can sum from 1 to p − 1 if we add
hence
p−1 c
−1
2
τp = −
+p
.
p
p
c=1
−1 p
,
The sum is 0 by (A5), and the result follows from (A6). ♣
A9. The Law of Quadratic Reciprocity
Let p and q be odd primes, with p = q. Then
p
q
= (−1)(p−1)(q−1)/4 .
q
p
Thus if either p or q is congruent to 1 mod 4, then p is a QR mod q if and only if q is a
QR mod p; and if both p and q are congruent to 3 mod 4, then p is a QR mod q if and
only if q is a QNR mod p.
Proof. Let K have characteristic q and contain the p-th roots of unity. For example, take
K to be the splitting field of X p − 1 over Fq . Then
τpq = (τp2 )(q−1)/2 τp
which by (A8) is
(q−1)/2
(−1)(p−1)/2 p
τp .
4
Thus
τpq = (−1)(p−1)(q−1)/4 p(q−1)/2 τp .
But by the binomial expansion applied to the definition of τp in (A7),
τpq =
p−1 a
a=1
p
ζ aq
(Recall that K has characteristic q, so aq = a in K.) Let c ≡ aq mod p and note that
−1 q
q
=
p
p
because the product of the two terms is ( p1 ) = 1. Thus
τpq
=
p−1 c
q
c=1
p
q
ζ =
τp .
p
p
c
We now have two expressions (not involving summations) for τpq , so
q
(−1)(p−1)(q−1)/4 p(q−1)/2 =
.
p
Since p(q−1)/2 = ( pq ) by (A2), the above equation holds not only in K but in Fq , hence
can be written as a congruence mod q. Finally, multiply both sides by ( pq ) to complete
the proof. ♣
A10. The Second Supplementary Law
2
2
= (−1)(p −1)/8
p
so
1
2
=
p
−1
if p ≡ ±1 mod 8
if p ≡ ±3 mod 8.
Thus if p ≡ 1 or 7 mod 8, then 2 is a QR mod p, and if p ≡ 3 or 5 mod 8, then 2 is a
QNR mod p.
Proof. Let K be a field of characteristic p containing the 8th roots of unity, and let ζ
be a primitive 8th root of unity. Define τ = ζ + ζ −1 . Then τ 2 = ζ 2 + ζ −2 + 2. Now ζ 2
and ζ −2 are distinct 4th roots of unity, not ±1, so they must be negatives of each other
(analogous to i and −i in C). Therefore τ 2 = 2. Modulo p we have
2
2
τ p = (τ 2 )(p−1)/2 τ = 2(p−1)/2 τ =
τ=
(ζ + ζ −1 ).
p
p
5
But by definition of τ , τ p = (ζ + ζ −1 )p = ζ p + ζ −p . Now (again as in C) ζ p + ζ −p and
ζ + ζ −1 will coincide if p ≡ ±1 mod 8, and will be negatives of each other if p ≡ ±3 mod
8. In other words,
ζ p + ζ −p = (−1)(p
2
−1)/8
(ζ + ζ −1 ).
Equating the two expressions for τ p , we get the desired result. We can justify the appeal
to the complex plane by requiring that K satisfy the constraints ζ 2 + ζ −2 = 0 and
2
ζ p + ζ −p = (−1)(p −1)/8 (ζ + ζ −1 ). ♣
A11. Example
We determine whether 113 is a QR mod 127:
127
( 113
127 ) = ( 113 ) because 113 ≡ 1 mod 4;
14
a
( 127
113 ) = ( 113 ) because ( p ) depends only on the residue class of a mod p;
14
2
7
7
( 113
) = ( 113
)( 113
) = ( 113
) by (A4) and the fact that 113 ≡ 1 mod 8;
7
) = ( 113
( 113
7 ) because 113 ≡ 1 mod 4;
1
( 113
7 ) = ( 7 ) because 113 ≡ 1 mod 7
and since ( 17 ) = 1 by inspection, 113 is a QR mod 127. Explicitly,
113 + 13(127) = 1764 = 422 .
A12. The Jacobi Symbol
Let Q be an odd positive integer with prime factorization Q = q1 · · · qs . The Jacobi symbol
is defined by
a
Q
s a
=
i=1
It follows directly from the definition that
a
a
a
=
,
Q
Q
QQ
qi
a
Q
.
a
Q
=
aa
.
Q
Also,
a≡a
mod Q ⇒
a
Q
=
a
Q
because if a ≡ a mod Q then a ≡ a mod qi for all i = 1, . . . , s.
Quadratic reciprocity and the two supplementary laws can be extended to the Jacobi
symbol, if we are careful.
6
A13. Theorem
If Q is an odd positive integer then
−1
= (−1)(Q−1)/2 .
Q
Proof. We compute
−1
Q
=
s −1
qj
j=1
=
s
s
(−1)(qj −1)/2 = (−1)
j=1 (qj −1)/2
.
j=1
Now if a and b are odd then
ab − 1
a−1 b−1
(a − 1)(b − 1)
−
+
=
≡0
2
2
2
2
mod 2,
hence
a−1 b−1
ab − 1
+
≡
2
2
2
mod 2.
We can apply this result repeatedly (starting with a = q1 , b = q2 ) to get the desired
formula. ♣
A14. Theorem
If Q is an odd positive integer, then
2
2
= (−1)(Q −1)/8 .
Q
Proof. As in (A13),
2
Q
=
s s
2
2
=
(−1)(qj −1)/8 .
q
j
j=1
j=1
But if a and b are odd then
2
a2 b2 − 1
a − 1 b2 − 1
(a2 − 1)(b2 − 1)
−
+
=
≡0
8
8
8
8
In fact
Thus
a2 −1
8
=
(a−1)(a+1)
8
mod 8.
is an integer and b2 − 1 ≡ 0 mod 8. (Just plug in b = 1, 3, 5, 7.)
a2 − 1 b2 − 1
a2 b2 − 1
+
≡
8
8
8
mod 8
and we can apply this repeatedly to get the desired result. ♣
7
A15. Theorem
If P and Q are odd, relatively prime positive integers, then
P
Q
= (−1)(P −1)(Q−1)/4 .
Q
P
r
s
Proof. Let the prime factorizations of P and Q be P = i=1 pi and Q = j=1 qj . Then
P
Q
Q
P
=
pi qj qj
i,j
pi
= (−1)
i,j (pi −1)(qj −1)/4
.
But as in (A13),
r
(pi − 1)/2 ≡ (P − 1)/2
mod 2,
i=1
Therefore
s
(qj − 1)/2 ≡ (Q − 1)/2 mod 2.
j=1
P
Q
Q
P
= (−1)[(P −1)/2][(Q−1)/2]
as desired. ♣
A16. Remarks
Not every property of the Legendre symbol extends to the Jacobi symbol. for example,
2
( 15
) = ( 23 )( 25 ) = (−1)(−1) = 1, but 2 is a QNR mod 15.
8
B. Extension of Absolute Values
B1. Theorem
Let L/K be a finite extension of fields, with n = [L : K]. If | | is an absolute value on K
and K is locally compact (hence complete) in the topology induced by | |, then there is
exactly one extension of | | to an absolute value on L, namely
|a| = |NL/K (a)|1/n .
We will need to do some preliminary work.
B2. Lemma
Suppose we are trying to prove that | | is an absolute value on L. Assume that | | satisfies
the first two requirements in the definition of absolute value in (9.1.1). If we find a real
number C > 0 such that for all a ∈ L,
|a| ≤ 1 ⇒ |1 + a| ≤ C.
Then | | satisfies the triangle inequality (|a + b| ≤ |a| + |b|).
Proof. If |a1 | ≥ |a2 |, then a = a2 /a1 satisfies |a| ≤ 1. We can take C = 2 without loss of
generality (because we can replace C by C c = 2). Thus
|a1 + a2 | ≤ 2a1 = 2 max{|a1 |, |a2 |}
so by induction,
|a1 + · · · + a2r | ≤ 2r max |aj |.
If n is any positive integer, choose r so that 2r−1 ≤ n ≤ 2r . Then
|a1 + · · · + an | ≤ 2r max |aj | ≤ 2n max |aj |.
(Note that 2r−1 ≤ n ⇒ 2r ≤ 2n. Also, we can essentially regard n as 2r by introducing
zeros.) Now
n j n−j
n n j n−j n
|a + b| = a b
.
j |a| |b|
≤ 2(n + 1) max
j
j
j=0
But |m| = |1 + 1 + · · · + 1| ≤ 2m for m ≥ 1, so
n
j
n−j
|a + b| ≤ 4(n + 1) max
|a| |b|
.
j
j
n
The expression in braces is a single term in a binomial expansion, hence
|a + b|n ≤ 4(n + 1)(|a| + |b|)n .
Taking n-th roots, we have
|a + b| ≤ [4(n + 1)]1/n (|a| + |b|).
The right hand side approaches (|a| + |b|) as n → ∞ (take logarithms), and the result
follows. ♣
9
B3. Uniqueness
Since L is a finite-dimensional vector space over K, any two extensions to L are equivalent
as norms, and therefore induce the same topology. Thus (see Section 9.1, Problem 3) for
some c > 0 we have | |1 = (| |2 )c . But |a|1 = |a|2 for every a ∈ K, so c must be 1.
B4. Proof of Theorem B1
By (B2), it suffices to find
C > 0 such that |a| ≤ 1 ⇒ |1 + a| ≤ C. Let b1 , . . . , bn be a
n
basis for L over K. If a = i=1 ci bi , then the max norm on L is defined by
|a|0 = max |ci |.
1≤i≤n
The topology induced by | |0 is the product topology determined by n copies of K. With
respect to this topology, NL/K is continuous (it is a polynomial). Thus a → |a| is a
composition of continuous functions, hence continuous. Consequently, | | is a nonzero
continuous function on the compact set S = {a ∈ L : |a|0 = 1}. So there exist δ, ∆ > 0
such that for all a ∈ S we have
o < δ ≤ |a| ≤ ∆.
n
If 0 = a ∈ L we can find c ∈ K such that |a|0 = |c|. (We have a = i=1 ci bi , and if |ci | is
the maximum of the |cj |, 1 ≤ j ≤ n, take c = ci .) Then |a/c|0 = 1, so a/c ∈ S and
0 < δ ≤ |a/c| =
|a/c|
≤ ∆.
|a/c|0
Now
|a/c|
|a|
=
|a/c|0
|a|0
because |a/c|0 = |a|0 /|c|. Therefore
0<δ≤
|a|
≤ ∆.
|a|0
( Now suppose |a| ≤ 1, so |a|0 ≤ |a|/δ ≤ δ −1 . Thus
|1 + a| ≤ ∆|1 + a|0
≤ ∆[|1|0 + |a|0 ]
≤ ∆[|1|0 + δ
where step (1) follows because | |0 is a norm. ♣
−1
]=C
(1)
10
C. The Different
C1. Definition
Let OK be the ring of algebraic integers in the number field K. Let ω1 , . . . , ωn be an
integral basis for OK , so that the field discriminant dK is det T (ωi ωj )), where T stands
for trace. Define
D−1 = {x ∈ K : T (xOK ) ⊆ Z}.
C2. Theorem
D−1 is a fractional ideal with Z-basis ω1∗ , . . . , ωn∗ , the dual basis of ω1 , . . . , ωn referred to
the vector space K over Q. [The dual basis is determined by T (ωi ωj∗ ) = δij , see (2.2.9).]
Proof. In view of (3.2.5), if we can show that D−1 is an OK -module, it will follow that
D−1 is a fractional ideal. We have
x ∈ OK , y ∈ D−1 ⇒ T (xyOK ) ⊆ T (yOK ) ⊆ Z
so xy ∈ D−1 .
∗
∗
−1
By (2.2.2), the trace of ωi∗ ωj is an integer for all j. Thus
n Zω1∗+ · · · + Zωn ⊆ D . We
−1
must prove the reverse inclusion. Let x ∈ D , so x = i=1 ai ωi , ai ∈ Q. Then
n
T (xωj ) = T (
ai ωi∗ ωj ) = aj .
i=1
But x ∈ D−1 implies that T (xωj ) ∈ Z, so aj ∈ Z and
D−1 =
n
Zωi∗ . ♣
i=1
C3. Remarks
Since K is the fraction field of OK , for each i there exists ai
∈ OK such that ai ωi∗ ∈ OK .
n
By (2.2.8), we can take each ai to be an integer. If m = i=1 ai , then mD−1 ⊆ OK ,
−1
which gives another proof that D is a fractional ideal.
C4. Definition and Discussion
The different of K, denoted by D, is the fractional ideal that is inverse to D−1 ; D−1
is called the co-different. In fact, D is an integral ideal of OK . We have 1 ∈ D−1 by
definition of D−1 and (2.2.2). Thus
D = D1 ⊆ DD−1 = OK .
The different can be defined in the general AKLB setup if A is integrally closed, so (2.2.2)
applies.
11
C5. Theorem
The norm of D is N (D) = |dK |.
Proof. Let m be a positive integer such that mD−1 ⊆ OK [(see (C3)]. We have
mωi∗ =
ωi =
n
j=1
n
aij ωj
bij ωj∗
j=1
so there is a matrix equation (bij ) = (aij /m)−1 . Now
T (ωi ωj ) =
n
bik T (ωk∗ ωj ) =
k=1
n
bik δkj = bij
k=1
so
det(bij ) = dK .
By (C2), mD−1 is an ideal of OK with Z-basis mωi∗ , i = 1 . . . , n, so by (4.2.5),
dK/Q (mω1∗ , . . . , mωn∗ ) = N (mD−1 )2 dK
and by (2.3.2), the left side of this equation is (det aij )2 dK . Thus
| det(aij )| = N (mD−1 ) = mn N (D−1 )
where the last step follows because |B/I| = |B/mI|/|I/mI|. Now DD−1 = OK implies
that N (D−1 ) = N (D)−1 , so
|dK | = | det(bij )| = | det(aij /m)|−1 = [N (D−1 )]−1 = N (D). ♣
C6. Some Computations
√
We calculate
√ the different of K = Q( −2). By (C5), N (DK ) = |dK | = 4 × 2 = 8.√Now
OK = Z[ −2] is a principal ideal domain (in fact a Euclidean domain) so D = (a+b −2)
for some a, b ∈ Z. Taking norms, we have 8 = a2 + 2b2 , and the only integer solution is
a = 0, b = ±2. Thus
√
√
D = (2 −2) = (−2 −2).
√
We calculate the different of K = Q( −3). By (C5), N (DK ) = |dK | = 3. Now
OK = Z[ω],
1 1√
ω=− +
−3
2 2
and since OK is a PID, D = (a + bω) for some a, b ∈ Z. Taking norms, we get
2
2
b
b
3= a−
+3
, (2a − b)2 + 3b2 = 12.
2
2
12
There are 6 integer solutions:
2 + ω,
−1 + ω,
−2 − ω,
1 − ω,
1 + 2ω,
−1 − 2ω
but all of these elements are associates, so they generate the same principal ideal. Thus
D = (2 + ω).
It can be shown that a prime p ramifies in the number field K if and only if p divides
the different of K.