Unknown
If you face any problem to loading this page, click here for PDF file of this chapter. 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
2ax + b
2
≡ 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
χ(a) =
1 ifa(p−1)/2 ≡1 modp
−1 if a(p−1)/2 ≡ −1 modp.
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 ( a
p ) 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 ≡ gr mod p for some r. Therefore a(p−1)/2 ≡ gr(p−1)/2 ≡1 mod p,
so p − 1 divides r(p − 1)/2, hence r/2 is an integer. But then (gr/2)2 = gr ≡ a mod p,
and a is a QR mod p. ♣
compute
ab
p
≡ (ab)(p−1)/2 = a(p−1)/2b(p−1)/2 ≡
a
p
b
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
p
= (−1)(p−1)/2 =
1 ifp ≡1 mod4
−1 if p ≡3 mod4.
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=1
a
p
ζa.
A8. Theorem
τ 2
p = (−1)(p−1)/2 p.
Proof. From the definition of Gauss sum and (A4) we have
τ 2
p =
p−1
a,b=1
ab
p
ζa+b.
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
τ 2
p =
p−1
a=1
p−1
c=1
a2c
p
ζa+ac.
Since
a2
p
= 1, this simplifies to
τ 2
p =
p−1
c=1
p−1
a=1
ζa(1+c)
c
p
.
If 1+c ≡0 mod p then 1, ζ1+c, ζ2(1+c), . . . , ζ(p−1)(1+c) runs through all the roots (zeros)
of Xp − 1 (note that ζp = 1). But the coefficient of Xp−1 is 0, so the sum of the roots is
0. Therefore the sum of ζ1+c, ζ2(1+c), . . . , ζ(p−1)(1+c) is -1. If 1 + c ≡0 mod p, then we
are summing p − 1 ones. Consequently,
τ 2
p = −
p−2
c=1
c
p
+ (p − 1)
−1
p
.
(Note that 1 + c ≡0 mod p ↔ c = p − 1.) We can sum from 1 to p − 1 if we add
−1
p
,
hence
τ 2
p = −
p−1
c=1
c
p
+ p
−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
q
p
= (−1)(p−1)(q−1)/4.
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 Xp − 1 over Fq. Then
τ q
p = (τ 2
p )(q−1)/2 τp
which by (A8) is
(−1)(p−1)/2 p
(q−1)/2
τp.
binomial expansion applied to the definition of τp in (A7),
τ q
p =
p−1
a=1
a
p
ζaq
(Recall that K has characteristic q, so aq = a in K.) Let c ≡ aq mod p and note that
q−1
p
=
q
p
because the product of the two terms is ( 1
p ) = 1. Thus
τ q
p =
p−1
c=1
c
p
q
p
ζc =
q
p
τp.
We now have two expressions (not involving summations) for τ q
p, so
(−1)(p−1)(q−1)/4 p(q−1)/2 =
q
p
.
Since p(q−1)/2 = (p
q ) 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 ( p
q ) to complete
the proof. ♣
A10. The Second Supplementary Law
2
p
= (−1)(p2−1)/8
so
2
p
=
1 ifp ≡ ±1 mod 8
−1 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 = ζ2 + ζ−2 +2. Now ζ2
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
τ p = (τ 2)(p−1)/2 τ = 2(p−1)/2 τ =
2
p
τ =
2
p
(ζ + ζ
−1).
But by definition of τ , τ p = (ζ + ζ−1)p = ζp + ζ−p. Now (again as in C) ζp + 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,
ζp + ζ
−p = (−1)(p2−1)/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 ζ2 + ζ−2 = 0 and
ζp + ζ−p = (−1)(p2−1)/8(ζ + ζ−1). ♣
A11. Example
We determine whether 113 is a QR mod 127:
( 113
127) = (127
113 ) because 113 ≡1 mod 4;
( 127
113) = ( 14
113 ) because ( a
p ) depends only on the residue class of a mod p;
( 14
113) = ( 2
113 )( 7
113) = ( 7
113 ) by (A4) and the fact that 113 ≡1 mod 8;
( 7
113) = (113
7 ) because 113 ≡1 mod 4;
( 113
7 ) = (1
7 ) because 113 ≡1 mod 7
and since ( 1
7 ) = 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
i=1
a
qi
.
It follows directly from the definition that
a
Q
a
Q
=
a
,
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.
A13. Theorem
If Q is an odd positive integer then
−1
Q
= (−1)(Q−1)/2.
Proof. We compute
−1
Q
=
s
j=1
−1
qj
=
s
j=1
(−1)(qj−1)/2 = (−1)
s
j=1(qj−1)/2.
Now if a and b are odd then
ab − 1
2
−
a − 1
2
+ b − 1
2
=
(a − 1)(b − 1)
2
≡0 mod2,
hence
a − 1
2
+ b − 1
2
≡ ab − 1
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
Q
= (−1)(Q2−1)/8.
Proof. As in (A13),
2
Q
=
s
j=1
2
qj
=
s
j=1
(−1)(q2
j
−1)/8.
But if a and b are odd then
a2b2 − 1
8
−
a2 − 1
8
+ b2 − 1
8
=
(a2 − 1)(b2 − 1)
8
≡0 mod8.
In fact a2−1
8 = (a−1)(a+1)
8 is an integer and b2 −1 ≡ 0 mod 8. (Just plug in b = 1, 3, 5, 7.)
Thus
a2 − 1
8
+ b2 − 1
8
≡ a2b2 − 1
8
mod 8
and we can apply this repeatedly to get the desired result. ♣
A15. Theorem
If P and Q are odd, relatively prime positive integers, then
P
Q
Q
P
= (−1)(P−1)(Q−1)/4.
Proof. Let the prime factorizations of P and Q be P =
r
i=1 pi and Q =
s
j=1 qj . Then
P
Q
Q
P
=
i,j
pi
qj
qj
pi
= (−1)
i,j (pi−1)(qj−1)/4.
But as in (A13),
r
i=1
(pi − 1)/2 ≡ (P − 1)/2 mod2,
s
j=1
(qj − 1)/2 ≡ (Q − 1)/2 mod2.
Therefore
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) = (2
3 )( 2
5) = (−1)(−1) = 1, but 2 is a QNR mod 15.
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 Cc = 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
|a + b|n =
n
j=0
n
j
ajbn−j
≤ 2(n +1) max
j
n
j
|a|j |b|n−j
.
But |m| = |1 + 1 + · · · + 1| ≤ 2m for m ≥ 1, so
|a + b|n ≤ 4(n +1) max
j
n
j
|a|j |b|n−j
.
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. ♣
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
basis for L over K. If a =
n
i=1 cibi, then the max norm on L is defined by
|a|0 = max
1≤i≤n
|ci|.
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,Δ > 0
such that for all a ∈ S we have
o < δ ≤ |a| ≤ Δ.
If 0 = a ∈ L we can find c ∈ K such that |a|0 = |c|. (We have a =
n
i=1 cibi, 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
≤ Δ.
Now
|a/c|
|a/c|0
=
|a|
|a|0
because |a/c|0 = |a|0/|c|. Therefore
0 < δ ≤
|a|
|a|0
≤ Δ.
( Now suppose |a| ≤ 1, so |a|0 ≤ |a|/δ ≤ δ−1. Thus
|1 + a| ≤ Δ|1 + a|0
≤ Δ[|1|0 + |a|0] (1)
≤ Δ[|1|0 + δ
−1] = C
where step (1) follows because | |0 is a norm. ♣
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(ωiωj )), where T stands
for trace. Define
D−1 = {x ∈ K : T(xOK) ⊆ Z}.
C2. Theorem
D−1 is a fractional ideal with Z-basis ω∗
1, . . . ,ω∗
n, the dual basis of ω1, . . . , ωn referred to
the vector space K over Q. [The dual basis is determined by T(ωiω∗
j) = 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.
By (2.2.2), the trace of ω∗
i ωj is an integer for all j. Thus Zω∗
1 +· · ·+Zω∗
n
⊆ D−1. We
must prove the reverse inclusion. Let x ∈ D−1, so x =
n
i=1 aiω∗
i , ai ∈ Q. Then
T(xωj) = T(
n
i=1
aiω
∗
i ωj) = aj .
But x ∈ D−1 implies that T(xωj) ∈ Z, so aj ∈ Z and
D−1 =
n
i=1
Zω
∗
i . ♣
C3. Remarks
Since K is the fraction field of OK, for each i there exists ai ∈ OK such that aiω∗
i
∈ OK.
By (2.2.8), we can take each ai to be an integer. If m =
n
i=1 ai, then mD−1 ⊆ OK,
which gives another proof that D−1 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.
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
mω
∗
i =
n
j=1
aijωj
ωi =
n
j=1
bijω
∗
j
so there is a matrix equation (bij) = (aij/m)−1. Now
T(ωiωj) =
n
k=1
bikT(ω
∗
kωj) =
n
k=1
bikδkj = bij
so
det(bij) = dK.
By (C2), mD−1 is an ideal of OK with Z-basis mω∗
i , i = 1. . . , n, so by (4.2.5),
dK/Q(mω
∗
1, . . . ,mω
∗
n) = N(mD−1)2dK
and by (2.3.2), the left side of this equation is (det aij)2dK. Thus
| det(aij)| = N(mD−1) = mnN(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], ω= −1
2
+
1
2
√
−3
and since OK is a PID, D = (a + bω) for some a, b ∈ Z. Taking norms, we get
3 =
a − b
2
2
+ 3
b
2
2
, (2a − b)2 + 3b2 = 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.
If you face any problem to loading this page, click here for PDF file of this chapter.
You Might Also Like
Posted in: Linear Algebra
