Unknown
If you face any problem to loading this page, click here for PDF file of this chapter. 8.1 Decomposition and Inertia Groups
We return to the general AKLB setup: A is a Dedekind domain with fraction field K, L is
a finite separable extension of K, and B is the integral closure of A in L.But now we add
the condition that the extension L/K is normal, hence Galois.W e will see shortly that
the Galois assumption imposes a severe constraint on the numbers ei and fi in the ram-rel
identity (4.1.6). Throughout this chapter, G will denote the Galois group Gal(L/K).
8.1.1 Proposition
If σ ∈ G, then σ(B) = B.If Q is a prime ideal of B, then so is σ(Q).Moreo ver, if Q
lies above the nonzero prime ideal P of A, then so does σ(Q).Th us G acts on the set of
prime ideals lying above P.
Proof.If x ∈ B, then σ(x) ∈ B (apply σ to an equation of integral dependence).Th us
σ(B) ⊆ B.But σ−1(B) is also contained in B, hence B = σσ− 1(B) ⊆ σ(B).If PB =
Qei
i , then apply σ to get PB =
σ(Qi)ei.The σ(Qi) must be prime ideals because σ
preserves all algebraic relations.Note also that σ is a K-automorphism, hence fixes every
element of A (and of P).Therefore Q ∩ A = P ⇒ σ(Q) ∩ A = P. ♣
We now show that the action of G is transitive.
8.1.2 Theorem
Let Q and Q1 be prime ideals lying above P.Then for some σ ∈ G we have σ(Q) = Q1.
Proof.If the assertion is false, then for each σ, the ideals Q1 and σ(Q) are maximal and
distinct, so Q1 ⊆ σ(Q).By the prime avoidance lemma (Section 3.1, exercises), there is
an element x ∈ Q1 belonging to none of the σ(Q).Computing the norm of x relative to
L/K, we have N(x) =
σ∈G σ(x) by (2.1.6). But one of the σ’s is the identity, Q1 is an
ideal, and [by (8.1.1)] σ(x) ∈ B for all σ.Consequen tly, N(x) ∈ Q1.But N(x) ∈ A by (2.2.2), so N(x) ∈ Q1 ∩ A = P = Q ∩ A.Th us N(x) belongs to the prime ideal Q, and
therefore some σ−1(x) belongs to Q as well.This gives x ∈ σ(Q), a contradiction. ♣
8.1.3Corollary
In the factorization PB =
g
i=1 Pei
i of the nonzero prime ideal P, the ramification indices
ei are the same for all i, as are the relative degrees fi.Th us the ram-rel identity simplifies
to efg = n, where n = [L : K] = |G|.
Proof. This follows from (8.1.2), along with the observation that an automorphism σ
preserves all algebraic relations. ♣
Since we have a group G acting on the prime factors of PB, it is natural to consider
the stabilizer subgroup of each prime factor Q.
8.1.4 Definitions and Comments
We say that the prime ideals σ(Q), σ ∈ G, are the conjugates of Q. Thus (8.1.2) says that
all prime factors of PB are conjugate.The decomposition group of Q is the subgroup D
of G consisting of those σ ∈ G such that σ(Q) = Q.(This does not mean that σ fixes
every element of Q.) By the orbit-stabilizer theorem, the size of the orbit of Q is the
index of the stabilizer subgroup D.Since there is only one orbit, of size g,
g = [G : D] = |G|/|D|, hence |D| = n/g = efg/g = ef,
independent of Q.Note also that distinct conjugates of Q determine distinct cosets of D.
For if σ1D = σ2D, then σ
−1
2 σ1 ∈ D, so σ1(Q) = σ2(Q).
There is a particular subgroup of D that will be of interest. By (8.1.1), σ(B) = B for
every σ ∈ G.If σ ∈ D, then σ(Q) = Q.It follows that σ induces an automorphism σ of
B/Q.(Note that x ≡ y mod Q iff σx ≡ σy mod Q.) Since σ is a K-automorphism, σ
is an A/P-automorphism.The mapping σ → σ is a group homomorphism from D to the
group of A/P-automorphisms of B/Q.
8.1.5 Definition
The kernel I of the above homomorphism, that is, the set of all σ ∈ D such that σ is
trivial, is called the inertia group of Q.
8.1.6 Remarks
The inertia group is a normal subgroup of the decomposition group, as it is the kernel of
a homomorphism.It is given explicitly by
I = {σ ∈ D : σ(x) + Q = x + Q ∀x ∈ B} = {σ ∈ D : σ(x) − x ∈ Q ∀x ∈ B}.
We now introduce an intermediate field and ring into the basic AKLB setup, as follows.
L B
KD AD
K A
Take KD to be the fixed field of D, and let AD = B ∩KD be the integral closure of A in
KD.Let PD be the prime ideal Q ∩ AD.Note that Q is the only prime factor of PDB.
This is because all primes in the factorization are conjugate, and σ(Q) = Q for all σ ∈ D,
by definition of D.
8.1.7 Lemma
Let PDB = Qe
and f = [B/Q : AD/PD].Then e = e and f = f.Moreo ver,
A/P
∼=
AD/PD.
Proof. First, observe that by the ram-rel identity [see (8.1.3)], e f = [L : KD], which is
|D| by the fundamental theorem of Galois theory.But |D| = ef by (8.1.4), so e f = ef.
Now as in (4.1.3)-(4.1.5), A/P ⊆ AD/PD ⊆ B/Q, so f ≤ f.Also, PAD ⊆ PD, so PD
divides PAD, hence PDB divides PADB = PB.Consequen tly, e ≤ e, and this forces
e = e and f = f.Th us the dimension of B/Q over AD/PD is the same as the dimension
of B/Q over A/P.Since A/P can be regarded as a subfield of AD/PD, the proof is
complete. ♣
8.1.8 Theorem
Assume (B/Q)/(A/P) separable.The homomorphism σ → σ of D to Gal[(B/Q)/(A/P)]
introduced in (8.1.4) is surjective with kernel I.Therefore Gal[(B/Q)/(A/P)] ∼=
D/I.
Proof.Let x be a primitive element of B/Q over A/P.Let x ∈ B be a representative
of x.Let h(X) = Xr + ar−1Xr−1 + · + a0 be the minimal polynomial of x over KD; the
coefficients ai belong to AD by (2.2.2). The roots of h are all of the form σ(x), σ ∈ D.
(We are working in the extension L/KD, with Galois group D.) By (8.1.7), if we reduce
the coefficients of h mod PD, the resulting polynomial h(X) has coefficients in A/P.The
roots of h are of the form σ(x), σ ∈ D (because x is a primitive element).Since σ ∈ D
means that σ(Q) = Q, all conjugates of x over A/P lie in B/Q.By the basic theory of
splitting fields, B/Q is a Galois extension of A/P.
To summarize, since every conjugate of x over A/P is of the form σ(x), every A/Pautomorphism
of B/Q (necessarily determined by its action on x), is of the form σ where
σ ∈ D.Since σ is trivial iff σ ∈ I, it follows that the map σ → σ is surjective and has
kernel I. ♣
8.1.9 Corollary
The order of I is e.Th us the prime ideal P does not ramify if and only if the inertia
group of every prime ideal Q lying over P is trivial.
Proof.By definition of relative degree, the order of Gal[(B/Q)/(A/P)] is f. By (8.1.4),
the order of D is ef. Thus by (8.1.8), the order of I must be e. ♣
Problems For Section 8.1
1.Let D(Q) be the decomposition group of the prime ideal Q.It follows from the
definition of stabilizer subgroup that D(σ(Q)) = σD(Q)σ−1 for every σ ∈ G.Sho w that
the inertia subgroup also behaves in this manner, that is, I(σ(Q)) = σI(Q)σ−1.
2.If L/K is an abelian extension (the Galois group G = Gal(L/K) is abelian), show that
the groups D(σ(Q)), σ ∈ G, are all equal, as are the I(σ(Q)), σ ∈ G.Sho w also that the
groups depend only on the prime ideal P of A.
8.2 The Frobenius Automorphism
In the basic AKLB setup, with L/K a Galois extension, we now assume that K and L
are number fields.
8.2.1 Definitions and Comments
Let P be a prime ideal of A that does not ramify in B, and let Q be a prime lying over P.
By (8.1.9), the inertia group I(Q) is trivial, so by (8.1.8), Gal[(B/Q)/(A/P)] is isomorphic
to the decomposition group D(Q).But B/Q is a finite extension of the finite field A/P [see
(4.1.3)], so the Galois group is cyclic. Moreover, there is a canonical generator given by
x+Q → xq+Q, x ∈ B, where q = |A/P|.Th us we have identified a distinguished element
σ ∈ D(Q), called the Frobenius automorphism, or simply the Frobenius, of Q, relative to
the extension L/K.The Frobenius automorphism is determined by the requirement that
for every x ∈ B,
σ(x) ≡ xq mod Q.
We use the notation
L/K
Q
for the Frobenius automorphism.The behavior of the Frobenius
under conjugation is similar to the behavior of the decomposition group as a whole
(see the exercises in Section 8.1).
8.2.2 Proposition
If τ ∈ G, then
L/K
τ(Q)
= τ
L/K
Q
τ−1.
Proof.If x ∈ B, then
L/K
Q
τ−1x ≡ (τ−1x)q = τ−1xq mod Q.Apply τ to both sides to
conclude that τ
L/K
Q
τ−1 satisfies the defining equation for
L/K
τ(Q)
.Since the Frobenius
is determined by its defining equation, the result follows. ♣
8.2.3Corollary
If L/K is abelian, then
L/K
Q
depends only on P, and we write the Frobenius automorphism
as
L/K
P
, and sometimes call it the Artin symbol.
Proof. By (8.2.2), the Frobenius is the same for all conjugate ideals τ (Q), τ ∈ G, hence
by (8.1.2), for all prime ideals lying over P. ♣
8.2.4 Intermediate Fields
We now introduce an intermediate field between K and L, call it F.W e can then lift P
to the ring of algebraic integers in F, namely B ∩ F.A prime ideal lying over P has the
form Q∩F, where Q is a prime ideal of PB.W e will compare decomposition groups with
respect to the fields L and F, with the aid of the identity
[B/Q : A/P] = [B/Q : (B ∩ F)/(Q ∩ F)][(B ∩ F)/(Q ∩ F) : A/P].
The term on the left is the order of the decomposition group of Q over P, denoted by
D(Q, P).(W e are assuming that P does not ramify, so e = 1.) The first term on the
right is the order of the decomposition group of Q over Q ∩ F.The second term on the
right is the relative degree of Q ∩ F over P, call it f.Th us
|D(Q,Q ∩ F)| = |D(Q, P)|/f
Since D = D(Q, P) is cyclic and is generated by the Frobenius automorphism σ, the
unique subgroup of D with order |D|/f is generated by σf.Note that D(Q,Q ∩ F) is
a subgroup of D(Q, P), because Gal(L/F) is a subgroup of Gal(L/K).It is natural to
expect that the Frobenius automorphism of Q, relative to the extension L/F, is σf .
8.2.5 Proposition
L/F
Q
=
L/K
Q
f
.
Proof.Let σ =
L/K
Q
.Then σ ∈ D, so σ(Q) = Q; also σ(x) ≡ xq mod Q, x ∈ B, where
q = |A/P|.Th us σf (Q) = Q and σf (x) ≡ xqf .Since qf is the cardinality of the field
(B ∩ F)/(Q ∩ F), the result follows. ♣
8.2.6 Proposition
If the extension F/K is Galois, then the restriction of σ =
L/K
Q
to F is
F/K
Q∩F
.
Proof.Let σ1 be the restriction of σ to F.Since σ(Q) = Q, it follows that σ1(Q ∩ F) =
Q ∩ F.(Note that F/K is normal, so σ1 is an automorphism of F. ) Thus σ1 belongs
to D(Q ∩ F,P).Since σ(x) ≡ xq mod Q, we have σ1(x) ≡ xq mod (Q ∩ F), where
q = |A/P|.Consequen tly, σ1 =
F/K
Q∩F
. ♣
8.2.7 Definitions and Comments
We may view the lifting from the base field K to the extension field L as occurring in
three distinct steps.Let FD be the decomposition field of the extension, that is, the fixed
field of the decomposition group D, and let FI be the inertia field, the fixed field of the
inertia group I.W e have the following diagram:
L
e=|I|
FI
f=|D|/e
FD
g=n/ef
K
All ramification takes place at the top (call it level 3), and all splitting at the bottom
(level 1).There is inertia in the middle (level 2).Alternativ ely, the results can be
expressed in tabular form:
e f g
Level 1 1 1 g
2 1 f 1
3 e 1 1
As we move up the diagram, we multiply the ramification indices and relative degrees.
This is often expressed by saying that e and f are multiplicative in towers.The basic
point is that if Q = Qe1
1
· · · and Q1 = Qe2
2
· · · , then Q = Qe1e2
2
· · · .The multiplicativity
of f follows because f is a vector space dimension.
8.3Applications
8.3.1 Cyclotomic Fields
Let ζ be a primitive mth root of unity, and let L = Q(ζ) be the corresponding cyclotomic
field.(W e are in the AKLB setup with A = Z and K = Q.) Assume that p is a rational
prime that does not divide m. Then by (7.2.5) and the exercises for Section 4.2, p is
unramified.Th us (p) factors in B as Q1 · · ·Qg, where the Qi are distinct prime ideals.
Moreover, the relative degree f is the same for all Qi, because the extension L/Q is
Galois.In order to say more about f, we find the Frobenius automorphism σ explicitly.
The defining equation is σ(x) ≡ xp mod Qi for all i, and consequently
σ(ζ) = ζp.
(The idea is that the roots of unity remain distinct when reduced mod Qi, because the
polynomial Xn − 1 is separable over Fp.)
Now the order of σ is the size of the decomposition group D, which is f.Th us f is
the smallest positive integer such that σf (ζ) = ζ.Since ζ is a primitive mth root of unity,
we conclude that
f is the smallest positive integer such that pf ≡1 mod m.
Once we know f, we can find the number of prime factors g = n/f, where n = ϕ(m).
(We already know that e = 1 because p is unramified.)
When p divides m, the analysis is more complicated, and we will only state the result.
Say m = pam1, where p does not divide m1.Then f is the smallest positive integer such
that pf ≡1 mod m1.The factorization is (p) = (Q1 · · ·Qg)e, with e = ϕ(pa).The Qi are
distinct prime ideals, each with relative degree f.The number of distinct prime factors
is g = ϕ(m1)/f.
We will now give a proof of Gauss’ law of quadratic reciprocity.
8.3.2 Proposition
Let q be an odd prime, and let L = Q(ζq) be the cyclotomic field generated by a primitive
qth root of unity.Then L has a unique quadratic subfield F.Explicitly , if q ≡1 mod 4,
then the quadratic subfield is Q(
√
q), and if q ≡ 3 mod 4, it is Q(
√
−q).More compactly,
F = Q(
√
q∗), where q∗ = (−1)q−1)/2q.
Proof.The Galois group of the extension is cyclic of even order q −1, hence has a unique
subgroup of index 2.Therefore L has a unique quadratic subfield. By (7.1.7) and the
exercises to Section 7.1, the field discriminant is d = (−1)(q−1)/2qq−2 ∈ Q.But
√
d /∈ Q,
because d has an odd number of factors of q.If q ≡ 1 mod 4, then the sign of d is
positive and Q(
√
d) = Q(
√
q).Similarly , if q ≡ 3 mod 4, then the sign of d is negative
and Q(
√
d) = Q(
√
−q).[Note that the roots of the cyclotomic polynomial belong to L,
hence so does
√
d; see (2.3.5).] ♣
8.3.3 Remarks
Let σp be the Frobenius automorphism
F/Q
p
, where F is the unique quadratic subfield
of L, and p is an odd prime unequal to q. By (4.3.2), case (a1), if q∗ is a quadratic residue
mod p, then p splits, so g = 2 and therefore f = 1.Th us the decomposition group D
is trivial, and since σp generates D, σp is the identity.If q∗ is not a quadratic residue
mod p, then by (4.3.2), case (a2), p is inert, so g = 1, f = 2, and σp is nontrivial.Since
the Galois group of F/Q has only two elements, it may be identified with {1,−1} under
multiplication, and we may write (using the standard Legendre symbol) σp = (q
∗
p ).On
the other hand, σp is the restriction of σ =
L/Q
p
to F, by (8.2.6). Thus σp is the identity
on F iff σ belongs to H, the unique subgroup of Gal(L/Q) of index 2.This will happen iff
σ is a square.No w the Frobenius may be viewed as a lifting of the map x → xp mod q.
[As in (8.3.1), σ(ζq) = ζp
q. ] Thus σ will belong to H iff p is a quadratic residue mod q.In
other words, σp = (p
q ).
8.3.4 Quadratic Reciprocity
If p and q are distinct odd primes, then
p
q
= (−1)(p−1)(q−1)/4
q
p
.
Proof. By (8.3.3),
p
q
=
q∗
p
=
(−1)(q−1)/2
p
q
p
=
−1
p
(q−1)/2
q
p
.
But by elementary number theory, or by the discussion in the introduction to Chapter 1,
−1
p
= (−1)(p−1)/2,
and the result follows. ♣
8.3.5 Remark
Let L = Q(ζ), where ζ is a primitive pth root of unity, p prime.As usual, B is the ring
of algebraic integers of L.In this case, we can factor (p) in B explicitly. By (7.1.3) and
(7.1.5),
(p) = (1 − ζ)p−1.
Thus the ramification index e = p−1 coincides with the degree of the extension.W e say
that p is totally ramified.
If you face any problem to loading this page, click here for PDF file of this chapter.
Posted in: Linear Algebra
