Kummer Theory and Reciprocity Laws
Peter Stevenhagen
Abstract. Insert abstract here.
1. Introduction
How can we find abelian extensions of a number field? For any such field K, we
have the cyclotomic extension K ⊂ K(ζm ); the Galois group will be abelian and a
∗
subgroup of (Z/mZ)
. We might also adjoin a square root, but should we adjoin
√
n
a general root a, we no longer even have a Galois extension, unless ζn ∈ K
already.
But in fact, for K = Q, all quadratic extensions of Q are already cyclotomic:
for p 6= 2, there is only one cyclotomic extension which is ramified only at p
∗ ∼
and only tamely ramified, namely, Q(ζp ). This field has Galois group (Z/pZ)
√ ∗=
Z/(p−1)Z, so it has a unique quadratic subfield, which is easily seen to be Q( p ),
where p∗ = (−1)(p−1)/2
it is the unique quadratic field ramified only at p. In
√ p: √
√
addition, we have Q( −1, 2) = Q(ζ8 ), so we find Q( a) ⊂ Q(ζ4|a| ).
Already this discussion of quadratic subextensions of cyclotomic extensions
gives us the classical quadratic reciprocity law. Given distinct odd primes p, q,
then there is a Legendre symbol (p/q) which is 1 or −1, respectively, depending on
whether or not p is a square modulo q or not, respectively. One has the quadratic
reciprocity law where
 q


,
p or q ≡ 1 (mod 4),

p
p =
q

q

, else.
−
p
We can write this more compactly as
p
q
= (−1)(p−1)(q−1)/4 .
q
p
To prove this, consider first the case q ≡ 1 (mod 4). We consider the cyclotomic extension Q(ζq )/Q, with Galois group (Z/qZ)∗ . This extension contains the
√
quadratic extension Q( q). Let σ ∈ (Z/qZ)∗ be the Frobenius at p, and note that
√ √
(p/q) = 1 if and only if σp ( q)/ q = 1. By the property of the Frobenius, we
2
Peter Stevenhagen
have
√ √
√
σp ( q)/ q ≡ ( q)p−1
(mod p),
so this is equal to 1 if and only if q (p−1)/2 ≡ 1 (mod p), which holds if and only
if (q/p) = 1 by Euler’s criterion. This proves quadratic reciprocity. The case when
√
q ≡ 3 (mod 4) is similar, except now we work in Q( −q), and we end up with an
extra factor (−1)(p−1)/2 .
Seen one way, then, the quadratic reciprocity law is none other than the
statement that the quadratic extensions are all contained in cyclotomic extensions,
over which we have control.
2. Kummer Theory
We would like to generalize the quadratic reciprocity law; to do so, we need to
construct abelian extensions of number fields K.
√
Throughout we assume that ζn ∈ K. Then L = K( n a) (the splitting field of
X n − a) is abelian over K, for we have a map
Gal(L/K) ,→ hζn i
√
σ( n a)
σ 7→ √
.
n
a
√
Note that this is independent of the choice of n a (they differ by a root of unity,
which since they are in K, drops out), and by a little work we see that in fact this
is a group homomorphism.
Kummer theory gives a certain converse to this statement: If L/K is cyclic of
order n, and ζn ∈ K, then there exists an α ∈ L such that L = K(α) and αn ∈ K.
This was basically discovered by Lagrange: If G = hσi, then for any x write down
the Lagrange resultant
α = x + ζn−1 σ(x) + ζn−2 σ 2 (x) + · · · + ζ 1−n σ n−1 (x).
We see that σ(α) = ζn α, and if α 6= 0, then such an α will generate the extension.
We can always find such an α 6= 0; this is a result from Galois theory (known as
‘Artin-Dedekind’ or linear independence of characters).
More generally, Kummer theory tells us that if L/K is Galois with group
G which is abelian of√exponent n (meaning that for all σ ∈ G, σ n = idL ), and
ζn ∈ K, then L = K( n W ) for some subgroup (K ∗ )n ⊂ W ⊂ K ∗ . Then Kummer
theory tells us that (within a fixed algebraic closure) there is a bijection
{W : (K ∗ )n ⊂ W ⊂ K ∗ } o
W
(L∗ )n ∩ K ∗ o
/ {L ⊃ K abelian, exponent n (inside K)}
√
/ K( n W )
L.
Kummer Theory and Reciprocity Laws
3
In this case, if W ↔ L, then we have a perfect pairing
Gal(L/K) × W/(K ∗ )n → hζn i
√
σ( n a)
(σ, a) 7→ √
;
n
a
that is to say,
Gal(L/K) ∼
= Hom(W/(K ∗ )n , hζn i).
If L/K is a cyclic extension of degree n but ζn 6∈ K, it can be very hard to
describe. However, when one adjoins ζn to K, by √
Kummer theory, the extension
L(ζn ) is now Kummer over K(ζn ), so L = K(ζn , n a) for some a ∈ K(ζn ). Note
that this extension is not abelian, but it is abelian in two steps, which is usually
good enough.
√
L(ζn ) = K(ζn , n a)
p
pp
ppp
p
p
pp
ppp
K(ζn )
L
o
o
o
oo
ooo
o
o
oo
K
3. Norm Residue Symbol
We now look at the local situation. Recall the statement of local class field theory:
if F = Kp is a local field, and E ⊃ F is a finite abelian extension, then
∼
→ Gal(E/F )
F ∗ /NE/F E ∗ −
and we have a bijection between open subgroups of F ∗ and finite abelian extensions
of F .
√ By Kummer theory, if ζn ∈ F , then E ⊃ F of exponent n arises as E =
F ( n W ) where (F ∗ )n ⊂ W ⊂ F ∗ . But by class field theory, E ⊃ F of exponent n
arises as (F ∗ )n ⊂ NE/F E ∗ ⊂ F ∗ . These two ways of viewing extensions are dual
to each other.
√
The maximal exponent n extension by Kummer theory is E = F ( n F ∗ ). This
is a finite extension, for in the local case (F ∗ )n has finite index in F ∗ . Since the
bijection in class field theory is inclusion-reversing, N E ∗ = (F ∗ )n . By the perfect
pairing, we have
F ∗ /(F ∗ )n × F ∗ /(F ∗ )n → hζn i
√
σα ( n β)
√
(α, β) 7→
n
β
This is called the nth power norm residue symbol, and we denote it F , (−, −)n,F .
It has two useful properties:
4
Peter Stevenhagen
√
• From local class field theory, (α, β)n,F = 1 if and only if α ∈ N (F ( n β)∗ ).
∼
• Since A∗F /NE/F A∗E −
→ I, the inertia group, if the extension E/F is unramified and α ∈ A∗F , then (α, β)n,F = 1.
Example 3.1. Take the case F = Qp , n = 2, p 6= 2. Then Q∗p = hpi × Z∗p , so
(Q∗p )2 = hp2 i × (Z∗p )2 , and so (Q∗p )2 has index 4 in Q∗p , where
Q∗p /(Q∗p )2 ∼
= hpi × hai ∼
= Z/2Z × Z/2Z,
for a ∈ Z with (a/p) = −1.
√ √
We have (x, y)2,Qp ) = σx ( y)/ y. In the unramified case, y = a, we see
that (a, a) = 1, and since πF 7→ FrobE/F , we see (p, a) = −1. In the ramified case,
y = p the norms must generate a subgroup of index 2, so (a, p) = −1; and finally we
√
√
wish to know if p ∈ N (Qp ( p)∗ ): but (p, p)(−1, p) = (−p, p) = 1 (N ( p) = −p),
so (p, p) = (−1, p) = (−1)(p−1)/2 . We summarize the values of (x, y)2,Qp in the
following table:
x\y a
−p
a
1
−1
p
−1 (−1)(p−1)/2
The symbol has other important properties:
• (−α, α) = 1 for all α ∈ F ∗ ;
• (1 − α, α) = 1 for α ∈ F ∗ \ {1};
• (α, β) = (β, α)−1 .
The latter property, for example, follows from
(−αβ, αβ) = (−α, α)(β, α)(α, β)(−β, β) = (β, α)(α, β) = 1.
Note that we also have this pairing for archimedean F . For F = C this is
trivial, for F = R we have the Artin isomorphism
Gal(C/R)
R∗ /N C∗ −−−−−−→
and a corresponding pairing
R∗ /R∗2 × R∗ /R∗2 → h−1i
with (−1, −1)∞ = −1.
Now, for a number field K such that ζn ∈ K, for any prime p of K, there is
a norm residue symbol
(−, −)n,p : Kp∗ /(Kp∗ )n × Kp∗ /(Kp∗ )n → hζn i.
We then have:
Proposition 3.2 (Product formula). For α, β ∈ K ∗ ,
Y
(α, β)n,p = 1.
p≤∞
Kummer Theory and Reciprocity Laws
5
Example 3.3. To complete the picture, we compute the quadratic symbols for Q∞ =
R and Q2 . In the first case, R∗ /(R∗ )2 = h−1i, and (−1, −1)2,∞ = −1. For p = 2,
Q∗2 = h2i × Z∗2 , and Q∗2 /(Q∗2 )2 = h22 i × (Z∗2 )2 , now (Z∗2 )2 = 1 + 8Z2 , so we see
Q∗2 /(Q∗2 )2 ∼
= h2i × h−1i × h5i.
We summarize the values of (x, y)2,2 in the following table:
x\y
2
−1
5
2 −1 5
1
1 −1
1 −1 1
−1 1
1
Proof of the product formula. Consider the following diagram:
Y0
p≤∞
Kp∗
/
M
p≤∞
p
Gal(Kp ( n β)/Kp )
n
/ Gal(K( √
β)/K)
J
The left vertical map is the definition of the idèles. The top map is α 7→ ((α, β)n,p )p ,
which one sees is trivial for almost all p. Q
The horizontal bottom arrow is the Artin
map. The right vertical map is (σp )p 7→ p σp .
The statement is then that for α ∈ K ∗ ⊂ J, √
the global Artin map (the
bottom arrow) factors as J → J/K ∗ = CK → Gal(K( n β)/K); this is a nontrivial
statement from class field theory.
Example 3.4. For the case n = 2, K = Q, p, q distinct odd primes, writing
(p, q)2,` = (p, q)` , we have
Y
(p, q)` = (p, q)2 (p, q)p (p, q)q (p, q)∞ = 1.
`≤∞
Since p, q > 0, (p, q)∞ = 1. We see from the computation in the above table that
(p, q)2 = 1 unless p, q ≡ 3 (mod 4), i.e. (p, q)2 = (−1)(p−1)(q−1)/4 . Also, we see
√ p
q
q
(p, q)p = √ = q (p−1)/2 ≡
(mod p),
q
p
and similarly (q, p)q ≡ pq (mod q). Altogether,
(−1)(p−1)(q−1)/4
q
p
=1
p
q
which is the statement of quadratic reciprocity.
6
Peter Stevenhagen
4. Higher Reciprocity Laws
Are there higher reciprocity laws? In the nineteenth century, such a question was
asked. But what should this mean? We begin with the cubic reciprocity law: for
which p is 2 a cube modulo p?
For F∗p ⊃ (F∗p )3 , if p ≡ 2 (mod 3) we see that already F∗p = (F∗p )3 , and this
question is not interesting, for 2 is always√a cube. We assume therefore that p ≡ 1
(mod 3). Then p splits as p = ππ in Q( −3). In this case, 2 is a cube modulo
p if and only if X 3 − 2 splits completely
modulo p, which happens if and only if
√
p splits completely in L = Q(ζ3 , 3 2). This is a Galois extension of degree 6 with
Galois group S3 .
√
L = Q(ζ3 , 3 2)
NNN
oo
NNN
ooo
o
NNN
o
o
o
o
N
oo
√
K = Q(ζ3 )
Q( 3 2)
PPP
pp
PPP
PPP
ppp
p
p
PPP
p
PP pppp
Q
√
3
But now the extension L/K = Q(ζn , 2)/Q(ζ3 ) is Kummer, so we can get a handle
on it. In fact, one can show that this is the ray class field of Q(ζ3 ) of conductor 6,
and it follows from this that p splits completely in L if and only if p = x2 + 27y 2 .
For an nth power reciprocity law, we assume that ζn ∈ K. What is the
analogue of the Legendre symbol for nth powers? Given a prime p - n∞, we have
the unit group (O/p)∗ , which has order divisible by n since hζn i ,→ (O/p)∗ (the
discriminant of the polynomial X n − 1 is prime to p). Therefore we define:
−
: (O/p)∗ → hζn i
p n
α
≡ α(N p−1)/n (mod p).
p n
For the primes in
S(α) = {p : p | n∞ or ordp (α) 6= 0}
we do not define the symbol. We extend the symbol to any fractional OK -ideal b
of K by multiplicativity,
α
Y α ordp (b)
=
.
b n
p
p6∈S(α)
This symbol is then multiplicative in b on the group I(α) generated by p 6∈ S(α).
For any element β ∈ K ∗ , we also define
α
α
=
.
β n
(β) n
Kummer Theory and Reciprocity Laws
7
Lemma 4.1. For α, β ∈ K ∗ , with ζn ∈ K, we have
−1
Y
β
α
=
(α, β)p .
β n α n
p∈S(α)∩S(β)
In particular, if α and β are coprime, then
−1
Y
α
β
=
(α, β)p .
β n α n
p|n∞
To prove this, we use the following fact:
Lemma 4.2. If b ∈ I(α), then
α
b
=
n
√
σb ( n α)
√
,
n
α
√
n
where σb ∈ Gal(K( α)/K) is the Artin symbol of b.
√
Proof. Suppose b = p ∈ I(α). Now α is a unit at p, so Kp ( n α) is an unramified
extension of Kp . The Artin symbol of p in this case is
√
( n α)N p
√
σp α =
= α(N p−1)/n (mod p)
n
α
which is exactly the definition of the power residue symbol.
Proof of power reciprocity. By the lemma,
Y
α
=
(β, α)n,p .
β n
p6∈S(α)
Similarly,
Y
Y
β
=
(α, β)n,p =
(α, β)−1
n,f rakp
α n
p6∈S(β)
p∈S(β)
by the product formula. For p 6∈ S(α) ∪ S(β), we clearly have (α, β)n,p = 1. For
the quotient
−1
α
β
β n α n
we therefore obtain the quantity
Y
p∈S(β)\S(α)
(β, α)n,p
Y
(α, β)n,p .
p∈S(β)
Using (β, α)(α, β)−1 = 1 at the primes p ∈ S(β) \ S(α), the result follows.
8
Peter Stevenhagen
The expression for power reciprocity may look complicated, but F ∗ /(F ∗ )n is
a finite abelian group, so the norm residue symbol relies on only a finite amount of
information. For example, we have the supplementary law for a odd and positive:

1,
2
a ≡ ±1 (mod 8)
2
= (−1)(a −1)/8 =
−1, a ≡ ±3 (mod 8).
a
We may now prove cubic reciprocity, a result originally due to Eisenstein.
Suppose that α, β ∈ Z[ζ3 ] are coprime and coprime to 3. Since the ideal (3) ramifies
in Q(ζ3 ) as (3) = (1 − ζ3 )2 , we have a map
Z[ζ3 ] → Z[ζ3 ]/3Z[ζ3 ] = Z[ζ3 ]/(1 − ζ3 )2 Z.
This latter group is of order 6, so the map takes hζ6 i = h−ζ3 i injectively into the
quotient. Replacing α by ζ6k α we may assume α ≡ 1 (mod 3), and similarly with
β. In this case, we actually have
α
β
=
.
β 3
α 3
This follows from the power reciprocity law:
−1
β
α
= (α, β)3,(1−ζ3 ) ;
β 3 α 3
as this field is totally complex, there are no infinite primes. Evaluating this entry,
we see that it is equal to 1.
Example 4.1. As another example, we can take quintic reciprocity. Take α, β ∈
Z[ζ5 ] coprime to each other and coprime to 5. We have (5) = (1 − ζ5 )4 , and
−1
β
α
= (α, β)5,(1−ζ5 ) .
β
α
Multiplying α, β by units, for which we have
Z[ζ5 ]∗ = h−ζ5 i × h(1 +
√
5)/2i,
3
we can obtain α, β ≡ 1 (mod (1 − ζ5 ) ). Then (α, β)5,1−ζ5 = 1, and
β
α
=
.
β 5
α 5
Exercises
Exercise 3.1. For odd primes p and q, we showed how to deduce the quadratic
reciprocity law
1
1
p
q
= (−1)p− 2 q− 2
q
p
Kummer Theory and Reciprocity Laws
9
r −1
q) ⊂ Q(ζq ) and properties of the Artin symbol. Give
from the inclusion Q(
q
a similar proof for the supplementary law


1
p ≡ ±1 mod 8,
2
=
−1 p ≡ ±3 mod 8,
p
where p denotes an odd prime.
√
field generated by
In the next two exercises, let K√= Q( −3) be the number
√
3
the cube root of unity
√ ζ3 = (−1 + −3)/2, and L = K( 2) the extension of K
generated by a root 3 2 of X 3 − 2. The unique primes of K lying over 2 and 3 are
denoted by 2 and t, respectively.
Exercise 3.2.
(a) Prove that K ⊂ L is√ cyclic
√ of degree 3, and that : Gal(L/K) → hζ3 i
defined by (σ) = σ( 3 2)/ 3 2 is a group isomorphism.
(b) Show that the conductor F(L/K) divides 2t4 .
(c) Let p be a finite prime of K not dividing 2t, and let N p be the cardinality
of its residue class field. Prove that ((p, L/K)) is the unique element of
hζ3 i that is congruent to 2(N p−1)/3 modulo p.
(d) Show that L is the ray class field of K with modulus 6 (= 2 · t2 ).
Exercise 3.3. Let p be a prime number, and let m be the number of distinct roots
of X 3 − 2 in Fp . Prove the following:
(a)
(b)
(c)
(d)
m = 0 if and only if p ≡ 1 mod 3 and p is not represented by X 2 + 27Y 2 .
m = 1 if and only if p ≡
6 1 mod 3.
m 6= 2.
m = 3 if and only if p is represented by X 2 + 27Y 2 .
Exercise 3.4. Let K be a number field, and r the number of real primes of K.
Denote by Cl the class group of K, and by Cl∗ the strict
(or narrow ) class group
Q
of K; i.e., the ray class group modulo the cycle f = p real p. The 2-rank rk2 A of
a multiplicatively written abelian group A is defined to be the dimension of the
F2 -vector space A/A2 .
(a) Prove that rk2 Cl ≤ rk2√Cl∗ ≤ rk2 Cl +r − 1 if r > 0.
(b) Let H = {x ∈ K ∗ : K( x) is unramified over K at all finite primes}, and
let H + be the set of elements of H that are positive at all real primes of
K. Prove that rk2 Cl∗ equals the dimension of the F2 -vector space H/K ∗2 ,
and that rk2 Cl equals the dimension of the F2 -vector space H + /K ∗2 .
Exercise 3.5. Let the notation be as in the previous exercise.
(a) Let a, b ∈ H. Prove that the norm Q
residue symbol (a, b)2,Q equals 1 for
every finite prime Q of K, and that p real (a, b)2,p = 1.
(b) Prove that rk2 Cl∗ ≤ rk2 Cl +br/2c.
10
Peter Stevenhagen
Mathematisch Instituut,
Universiteit Leiden,
Postbus 9512,
2300 RA Leiden,
The Netherlands
E-mail address: [email protected]