Solutions - CFT

Here are some solutions to exercises written by me (Ricardo) and some

fellow students during a course on Class Field Theory. The course took place

in the first semester of 2018 at UFRJ, where we followed the book by Nancy

Childress under the supervision of Professor Aftab Pande.

In this pdf you will find the following exercises of the book:

Chapter 2: 20, 21

Chapter 3: 1, 2, 3, 4, 5

Chapter 4: 1, 2, 3, 6, 7, 8, 9, 11, 12, 14, 15, 20, 21, 22, 25

Chapter 5: 1, 3, 5, 7, 9, 11, 13, 15

Chapter 6: 1, 3

The ones that are not listed on the next page were a contribution of my

colleagues and can be found after page 21.

(All solutions were written by students, so take them at your own discretion)

Class Field Theory - Homework

Book: Class Field Theory, by N. Childress

Student: Ricardo Toso

Chapter 3: 1, 2, 3, 4, 5. (complete)

Chapter 4: 1, 15, 20, 21, 22.

Chapter 5: 1, 3, 5, 7, 9, 11, 15.

Chapter 6: 1, 3.

Exercise 3.1 (p.48): Show that

P+F,m =

{〈α〉 : α ∈ F, α� 0, α

×≡ 1 (mod m)}

={⟨αβ

⟩: α, β ∈ OF prime to m,

α

β� 0, α ≡ β (mod m)

}Proof: To ease notation and make things clear, we name these sets and write them

down thoroughly.

S1 :={〈α〉 : α ∈ F, α� 0, α

×≡ 1 (mod m)}

S2 :={ ⟨α

β

⟩: α, β ∈ OF prime to m,

α

β� 0, α ≡ β (mod m)

}P+F,m =

⟨{〈α〉 : α ∈ OF , α� 0, α ≡ 1 (mod m)

}⟩(Note that the big 〈.〉 outside the bracket denotes the group generated by that set,

while the 〈α〉 inside the same bracket denotes the ideal generated by α.)

We will attain the desired equality via P+F,m ⊂ S2 ⊂ S1 ⊂ P

+F,m.

•[P+F,m ⊂ S2

]By definition of a group generated by a set, every I ∈ P+

F,m can be

written as a finite product

I =<∞∏i

〈γi〉±1 with every γi satisfying γi ∈ OF , γi � 0, γi ≡ 1 (mod m).

Taking α as the product of all γi’s that appear above with a positive exponent, and

β as the product of the ones that appear with a negative exponent, we get:

I = 〈α〉 〈β〉−1 =⟨αβ

⟩where α, β ∈ OF , α, β � 0, α, β ≡ 1 (mod m).

1

But α, β � 0 implies α/β � 0. Furthermore, α, β ≡ 1 (mod m) yields α, β prime to

m and α ≡ β (mod m). Thus I ∈ S2. i.e. We have shown P+F,m ⊂ S2.

•[S2 ⊂ S1

]Given I ∈ S2, by the definition of S2 we can write I =

⟨αβ

⟩for some pair

α, β such that

α, β ∈ OF prime to m,α

β� 0, α ≡ β (mod m).

Since α, β are prime to m, they are invertible as elements of the local rings OF,p’s of

the completions Fp’s (for every prime p|m). Let p′ be the unique maximal ideal of

OF,p (i.e. p′ = pOF,p). In this way, α ≡ β (mod m) implies

α ≡ β (mod p′ ordp(m)),

which (as β is invertible as an element of OF,p) is the same as writing

α

β≡ 1 (mod p′ ordp(m)).

In other wordsα

β

×≡ 1 (mod m).

Therefore, since αβ∈ F and α

β� 0, we have shown

⟨αβ

⟩∈ S1. Hence S2 ⊂ S1.

•[S1 ⊂ P+

F,m

]Given I ∈ S1, by definition we have I =

⟨α⟩

for some

α ∈ F, α� 0, α×≡ 1 (mod m).

Since α×≡ 1 (mod m), we can write α = γ1

γ2with γ1, γ2 ∈ OF and γ1, γ2 ≡ 1 (mod m).

Next, take c ∈ OF such that cγ1 � 0.

Then take m ∈ N such that mc ∈ m.

Then take n ∈ N sufficiently large such that signσ(nmc+ 1) = signσ(c) ∀σ real.

In this way, we have built an element k := nmc + 1 ∈ OF which by construction

satisfies

k ≡ 1 (mod m) and signσ(kγ1) = sign σ(cγ1) > 0 ∀σ real embedding.

Therefore, we have

kγ1 ≡ kγ2 ≡ 1 (mod m) and kγ1 � 0.

2

Furthermore, since α� 0 and kγ1kγ2

= α, we must also have

kγ2 � 0.

In summary, we have found two elements

kγ1, kγ2 ∈ OF such that kγ1, kγ2 � 0, kγ1 ≡ kγ2 ≡ 1 (mod m)

(i.e. they satisfy the conditions for P+F,m) and α = kγ1

kγ2. Therefore

I =⟨α⟩

=⟨kγ1⟩⟨kγ2⟩−1 ∈ P+

F,m.

Hence S1 ⊂ P+F,m.

Exercise 3.2 (p.49): Let F = Q(√m) where m > 1 is a square-free integer. Let

m = OF and let ε be a fundamental unit in OF .

(a) Suppose NF/Q(ε) = −1. Show that the ideal class group of F and the strict

ideal class group of F are isomorphic.

(b) Suppose NF/Q(ε) = 1. Show that the strict ideal class group of F is twice as

large as the ideal class group of F .

Proof: We know from basic algebra that the Galois group of Q(√m)/Q is G = {ι, σ}

where ι is the identity map and σ : (a+ b√m) 7−→ (a− b

√m). Thus

NF/Q(x) =∏σ∈G

σ(x) = ι(x)σ(x)

(a) Here ι(ε)σ(ε) = −1. Therefore either ι(ε) < 0 or σ(ε) < 0 (but not both < 0).

First, let us assume

ι(ε) > 0 and σ(ε) < 0.

In this case, for every 〈α〉 ∈ PF we can check:

• If ι(α) > 0 and σ(α) > 0, then (by definition) we have α� 0. Therefore 〈α〉 ∈ P+F .

• If ι(α) < 0 and σ(α) < 0, then we have −α� 0. Therefore 〈α〉 = 〈−α〉 ∈ P+F .

• If ι(α) > 0 and σ(α) < 0, then we have εα� 0. Therefore 〈α〉 = 〈εα〉 ∈ P+F .

• If ι(α) < 0 and σ(α) > 0, then we have −εα� 0. Therefore 〈α〉 = 〈−εα〉 ∈ P+F .

i.e. we have shown PF ⊂ P+F . Therefore PF = P+

F , and so RF = R+F .

3

Now for the case

ι(ε) < 0 and σ(ε) > 0,

all we have to do is replace ε by −ε and repeat the same argument.

(b) Here ι(ε)σ(ε) = 1 and we begin by showing that 〈√m〉 6∈ P+

F . Indeed, lets

assume (ad absurdum) that 〈√m〉 ∈ P+

F . Then, by definition there exists c ∈ OFsuch that c � 0 and 〈c〉 = 〈

√m〉. Hence c = u

√m for some u ∈ O×F . But since

O×F = (±1)εZ, we must have ι(u)σ(u) = 1. Therefore

ι(c)σ(c) = ι(u√m)σ(u

√m) = ι(

√m)σ(

√m) = −m,

which implies c 6� 0 (absurd). In this way we have shown 〈√m〉 6∈ P+

F , which yields

[PF : P+F ] > 1.

Next we have to show that 〈√m〉 is the only class in PF/P+

F other than 〈1〉 (and

therefore the quotient has order 2).

Given 〈α〉 ∈ PF − P+F , since 〈α〉 = 〈−α〉 (changing the sign of α if necessary) we

may assume that

ι(α) > 0 and σ(α) < 0.

Hence

ι(α√m) = ι(α)

√m > 0 and σ(α

√m) = σ(α)(−

√m) > 0

i.e. α√m � 0. Therefore 〈α

√m〉 ∈ P+

F , and so 〈α√m〉 = 〈1〉 in PF/P+

F . In other

words, we have shown that 〈α〉 = 〈√m〉−1

in PF/P+F . As 〈α〉 ∈ PF − P+

F was

taken generically this implies that there is only one nontrivial class, namely the class

〈√m〉 = 〈

√m〉−1

. Therefore PF/P+F =

{〈1〉, 〈

√m〉}

. Thus [PF : P+F ] = 2, which

yields

[IF : P+F ] = 2 [IF : PF ].

Exercise 3.3 (p.51): Show that the map ψ : F (m) −→ {±1}r1(OF/m)×

given by

α 7−→(signσ1(α), . . . , signσr1(α), α + m

)(which appears during the proof of Ch.3 Proposition 2.1 ) is surjective.

Proof: Given V ∈ {±1}r1(OF/m)×

, say V = (s1, ..., sr1 , c + m), we want to find

α ∈ F (m) such that ψ(α) = V .

4

Decomposing m = pe11 ... penn we have the isomorphism

OFm∼=OFpe

1

1

× ...× OFpenn

which guarantees the class c+ m is completely determined by the classes c+ peii .

Now, recalling that the p-adic absolute values are given by |x|pj = ( 1Npj

)ordpj (x),

we take

ε := Inf

{1

2,

1

Np1e1, ... ,

1

Npnen

}.

By the Approximation Theorem (Ch.3 Theorem 1.1), we know there exists an α ∈ Fsuch that

|α− si|σi < ε and |α− c|pj < ε ∀i, j.

In this way:

• ∀i we have |α− si|σi < 1/2, which is the same as |σi(α)− si| < 1/2, that implies

signσi(α) = si.

• ∀j we have |α − c|pj < 1Npj

ej , which is the same as ordpj(α − c) > ej, which is the

same as (α− c) ∈ pejj , and so α ≡ c (mod p

ejj ). Thus

α ≡ c (mod m).

Therefore we have shown

ψ(α) =(signσ1(α), . . . , signσr1(α), α + m

)=(s1, ..., sn, c+ m

)= V

Exercise 3.4 (p.53): Let F = Q(√

5), m = OF . Find the fundamental units for F ,

and determine R+F,m (up to isomorphism of groups).

Proof: Recall that for F = Q(√

5) we have OF = Z[1+√5

2], and let θ := 1+

√5

2.

As(

1+√5

2

)(−1+

√5

2

)= 1, we see that θ is an unit. Now to check if it is funda-

mental, note that every element in OF can be written as (a + b√

5)/2 with a, b ∈ Z,

therefore one of the fundamental units can be written as (a + b√

5)/2 with a, b ∈ N.

On the other hand, a straightforward computation shows that for an element

an + bn√

5

2:=

(a+ b

√5

2

)n

with a, b ∈ N− {0}

5

we always have bn ≥ b (i.e. the b part never decreases when we take powers). So, as

θ is represented by a = 1 b = 1 there can be no ε = (a′ + b′√

5)/2 with a, b ∈ N such

that εn = θ. Therefore θ is a fundamental unit.

Now we can show that R+F,m =

{ ¯〈1〉}

. Here we could use the formula from Ch.3

Proposition 2.1, but since m = OF , that will only give us as much information as

Exercise 3.2. Indeed, applying Exercise 3.2 to

NF/Q(θ) = ι(θ)σ(θ) = −1

yields R+F,m∼= RF,m = CF . Therefore all we have to do is compute CF . To do so, we

use Minkowski’s bound [FT, Ch.IV Theorem 35] which states that every ideal class

in CF contains an ideal a such that

Na ≤(

4

π

)r2 n!

nn

√|dF |

where r2 is the number of pairs of complex embeddings, while n and dF are the degree

and the discriminant of F/Q. So for F = Q(√

5) we get that every ideal class contains

an ideal a such that

Na ≤(

4

π

)02!

22

√|5| =

1

2

√5 ≈ 1.12

i.e. Na = 1, i.e. a = OF . Thus every class in CF contains OF and therefore are all

the same. Hence CF ={ ¯〈1〉

}.

Exercise 3.5 (p.54): Let F be a number field and let n, m be (not necessarily dis-

tinct) ideals of OF . Let P+F,n < H1 < IF (n), and P+

F,m < H2 < IF (m). If H1 6= H2, is

it possible for them to have the same class field over F?

Proof: Yes, it is possible. Take for example F = Q, n = Z and m = 2Z. Let

H1 := PF,n = PQ = IQ ={〈α〉 : α ∈ Q×

}H2 := P+

F,m ={〈α〉 : α ∈ Q, α� 0, α

×≡ 1 (mod 2Z)}.

In this way we have 〈2〉 ∈ H1 while 〈2〉 6∈ H2. Thus H1 6= H2. Nonetheless:

• H1 contains every prime ideal of Q, therefore its class field is Q itself.

• H2 is the group from Ch.3 Example 3, therefore its class field is Q(ζ2) = Q.

6

Exercise 4.1 (p.64): For any such absolute value ‖ · ‖ (described in the book), show

that there exists a positive real number λ such that the absolute value ‖ · ‖λ satis-

fies the triangle inequality (i.e., is an absolute value in the stricter sense of Chapter 1).

Proof: Given a pair a, b ∈ F× we assume without loss that ‖b‖≥‖a‖ and set x = a/b.

In this way ‖x‖≤ 1, and so have ‖1 + x‖≤ c. Multiplying both sides by ‖b‖ yields

‖b+ a‖≤ c ‖b‖.

Therefore, if c=1 then we already have the triangle inequality (since ‖b‖≤ ‖b‖+‖a‖).Thus we may assume c > 1 and take λ := logc 2. So raising to the power of λ we get

‖b+ a‖λ≤ 2 ‖b‖λ

Hence ‖ · ‖λ satisfies the weak maximum inequality

‖b+ a‖λ≤ 2 max{‖a‖λ, ‖b‖λ},

which implies the triangle inequality.

(Neither [FT] nor [Ne] mention the weak maximum inequality and this implica-

tion, therefore we provide a proof for the above claim. The following proof is takes

from J. Voloch’s notes on number theory.)

Claim: Assume | · | : F → [0,∞) is a function that satisfies

|x| = 0 iff x = 0,

|ab| = |a||b| ∀a, b ∈ F,

|a+ b| ≤ 2 max{|a|, |b|} ∀a, b ∈ F.

Then | · | satisfies

|a+ b| ≤ |a|+ |b| ∀a, b ∈ F.

Proof: Applying induction to the weak maximum inequality yields

|x1 + x2 + · · · + x2n| ≤ 2n max{|x1|, |x2|, . . . , |x2n|} ∀x1, x2,..., x2n ∈ F.

Given a positive integer N , take n such that 2n−1 ≤ N ≤ 2n. In this way, for any set

x1, x2,..., xN ∈ F , by letting xN+1, xN+2, ..., x2n = 0 we see

7

|x1 + x2 + · · · + xN | ≤ 2n max{|x1|, |x2|, . . . , |xN |}

≤ 2N max{|x1|, |x2|, . . . , |xN |} ∀x1, x2,..., xN ∈ F.

In particular, taking all xi = 1 gives

|N | ≤ 2N.

Now, for any a, b ∈ F and any positive integer L we can compute:

|a+ b|L = |(a+ b)L|

=∣∣∣ L∑l=0

(L

l

)albL−l

∣∣∣≤ 2(L+ 1) max

0≤l≤L

{∣∣∣(Ll

)albL−l

∣∣∣}≤ 4(L+ 1) max

0≤l≤L

{(Ll

)|a|l|b|L−l

}= 4(L+ 1)( |a|+ |b| )L

Therefore, raising to the power of 1L

we have

|a+ b| ≤(4(L+ 1)

)1/L(|a|+ |b|),

and so, letting L→∞ yields

|a+ b| ≤ |a|+ |b|.

Exercise 4.15 (p.71): Show that F×, (viewed as a subset of JF , so identified with

ι(F×)), is a discrete subgroup of JF . It is called the subgroup of pricipal ideles and

CF = JF/F× is called the group of ideles classes.

Proof: To prove that it is discrete, given a generic k ∈ F× we have to find an open

set Ak ⊆ JF that contains ι(k) = (..., k, k, k, ...) and no other point of ι(F×). Take

r = 12

and denote by Bvr (k) the ball of radius r and center k in F×v . We claim that

the set

Ak :=∏v <∞

ordv(k)=0

Uv∏v <∞

ordv(k)6=0

Bvr (k)

∏v ∈∞

Bvr (k)

8

has such property.

To see that, first we note that because ordv(k) = 0 implies k ∈ Uv, we have

(..., k, k, k, ...) ∈ Ak. Next, suppose we have another element α ∈ F× such that

(..., α, α, α, ...) ∈ Ak. In this case:

• |α− k|v < r for all infinite v (because for such v, α ∈ Bvr (k))

• |α− k|v < r for all finite v such that ordv(k) 6= 0 (because for such v, α ∈ Bvr (k))

• |α−k|v ≤ 1 for all finite v such that ordv(k) = 0 (because for such v, α and k ∈ Uv)

Therefore, since r = 12, we get ∏

v∈VF

|α− k|v < 1

which would contradict the product formula if we had α − k 6= 0. Hence we must

have α = k.

Exercise 4.20 (p.83): Show that this action (described in the book) is consistent

with the usual action of G on K×, were we view K× ⊆ JK as before (i.e. via the

diagonal embedding ι :K× ↪→ JK).

Proof: For any place w ∈ VK let ιw :K× ↪→ K×w be the natural embedding. So, the

way how G was defined to act on VK (while permuting the Kw’s) translates to

σ(ισ−1w(k)

)= ιw(σ(k)).

Meanwhile, the action of G on JK was defined by

σ : a = ( . . . , aw , . . . ) 7→ ( . . . , σ(aσ−1w) , . . . ).

Thus, for the diagonal embedding ι :K× ↪→ JK we can compute

σ(ι(k)

)= ( . . . , σ

(ι(k)σ−1w

), . . . )

= ( . . . , σ(ισ−1w(k)

), . . . )

= ( . . . , ιw(σ(k)) , . . . )

= ι(σ(k)

)i.e. the action of G that we defined on JK is consistent with the action of G on K.

9

Exercise 4.21 (p.87): If A and B are G-modules that have Herbrand quotients,

show that

QG(A×B) = QG(A)QG(B).

Proof: From basic algebra we have

A×B0×B

∼= A.

Therefore, Ch 4 Proposition 4.2 yields

QG(A×B)

QG(0×B)= QG(A).

Hence the desired result follows from 0×B ∼= B.

Exercise 4.22 (p.91): Let k2/k1 be an extension of local fields above Qp. Show that,

with respect to the p-adic topology, Nk2/k1 : k2 → k1 is continuous.

Proof: Recall that on a topological group, the product of continuous functions is

also continuous. So, as

Nk2/k1(α) =∏σ∈G

σ(α),

all we have to do is prove that every σ ∈ G is continuous.

To see that, it is enough to show that the pre-image of every open ball is open.

But the open balls of k2 are of the form

Br(α) = α + per

for some er (which depends on r) and where p is the unique maximal ideal of Ok2 .Hence for any open ball and σ we can simply compute

σ−1(Br(α)) = σ−1(α + per)

= σ−1(α) + σ−1(per)

= σ−1(α) + per

= Br(σ−1(α)),

which is open. Therefore σ is continuous.

10

Exercise 5.1 (p.106): Show that E+F,mE+F,n = E+F,(m,n).

Proof: We have already seen that

E+F,m =∏

v imaginary

C×∏v real

R×+∏pv6 |m

Uv∏pv |m

(1 + pordv(m)v ).

Now, on the one hand, for any pair N > n > 0 and any finite v ∈ VF , we always have

(1 + pNv ) ⊂ (1 + pnv ). Thus (1 + pNv )(1 + pnv ) = (1 + pnv ). Therefore

(1 + pordv(m)v )(1 + pordv(n)v ) = (1 + pmin{ordv(m),ordv(n)}

v ) = (1 + pordv(gcd(m,n))v ).

On the other hand, trivially

Uv (1 + pordv(m)v ) = Uv.

So, putting things together

E+F,mE+F,n =

∏v imaginary

C×C×∏v real

R×+R×+∏pv6 |mand/or

pv6 | n

UvUv or Uv(1 + pordv(m or n)v )

∏pv |mandpv |n

(1 + pordv(m)v )(1 + pordv(n)v )

=∏

v imaginary

C×∏v real

R×+∏pv6 |mand/or

pv6 | n

Uv∏pv |mandpv |n

(1 + pordv(gcd(m,n))v )

=∏

v imaginary

C×∏v real

R×+∏

pv6 | gcd(m,n)

Uv∏

pv | gcd(m,n)

(1 + pordv(gcd(m,n))v )

= E+F, gcd(m,n)

Exercise 5.3 (p.106): Suppose F ⊆ E ⊆ K are number fields and K/F is abelian.

How are f(K/F ) and f(E/F ) related?

Proof: Recall that for such tower of extensions we have [Ne, Ch.1 Cor.2.7 ]:

NK/F k = NE/F (NK/E k) ∀ k ∈ K,

which (by the definition of the idelic norm) translates to

NK/F a = NE/F (NK/E a) ∀ a ∈ JK .

11

Thus NK/FJK ⊂ NE/FJE, and therefore

E+F,m ⊂ F×NK/FJK implies E+F,m ⊂ F×NE/FJE.

Hence, by the definition of f, we must have

f(E/F )∣∣ f(K/F ).

Exercise 5.5 (p.111): Let K = Q(√

5, i). Then Gal(K/Q) = {1, σ, τ, στ}, where

σ is complex conjugation, while τ fixes i and sends√

5 7→ −√

5. Suppose pZ is an

unramified prime in K/Q.

(a) Compute(

pZK/Q

), i.e., give conditions (in terms of congruences) on p that de-

termine whether the Artin symbol is 1, σ, τ or στ . (HINT: If you can find some

cyclotomic field that contains K, then Ch.1 Example 1 may be of use.)

(b) Give necessary and sufficient conditions (in terms of congruences) for the prime

pZ to split completely in K/Q. Compare your answer with part a and Theorem 1.1.8.

(c) Suppose pZ is inert in Q(i)/Q. What can you say about(

pZ[i]K/Q(i)

)?

(d) Suppose pZ splits in Q(i)/Q, say pZ[i] = pp′. What can you say about(

pK/Q(i)

)?

Proof: (a) As hinted, we shall make use of the discussion from Ch.1 Example 1.

That is: if m divisible by 4 then the primes that ramify in Q(ζm)/Q are exactly the

divisors of m. And in that case, for every p unramified the Artin automorphism is

simply the pth power map. i.e.,(

pZQ(ζm)/Q

)is given by ζm 7−→ ζpm.

Now for our case, since

i = ζ4 = ζ520 and√

5 = ζ5 − ζ25 − ζ35 + ζ45 = ζ420 − ζ820 − ζ1220 + ζ1620

we see that K ⊂ Q(ζ20). Thus, for every prime p unramified in Q(ζ20)/Q (i.e., for

every p 6= 2 and p 6= 5) we can apply Corollary 1.2 followed by the above discussion

to get

σp :=

(pZK/Q

)=

(pZ

Q(ζm)/Q

) ∣∣∣K

= (ζ20 7→ ζp20).

12

This way, we can compute σp(i) = σp(ζ420) = (ζ420)

p = i p, and therefore

p ≡ 1 (mod 4) =⇒ σp(i) = i

p ≡ 3 (mod 4) =⇒ σp(i) = −i.

Analogously, we can compute σp(√

5) = σp(ζ420− ζ820− ζ1220 + ζ1620 ) = ζ4p20 − ζ

8p20 − ζ

12p20 +

ζ16p20 = ζp5 − ζ2p5 − ζ

3p5 + ζ4p5 , and therefore

p ≡ 1 (mod 5) =⇒ σp(√

5) = ζ5 − ζ25 − ζ35 + ζ45 =√

5

p ≡ 2 (mod 5) =⇒ σp(√

5) = ζ25 − ζ45 − ζ5 + ζ35 = −√

5

p ≡ 3 (mod 5) =⇒ σp(√

5) = ζ35 − ζ5 − ζ45 + ζ25 = −√

5

p ≡ 4 (mod 5) =⇒ σp(√

5) = ζ45 − ζ35 − ζ25 + ζ5 =√

5.

Now, putting all this information together modulo 20 (while ignoring the primes 2

and 5 because they ramify), we get

p ≡ 1 ⇒ p ≡ 1 (mod 4), p ≡ 1 (mod 5)⇒ σp(i) = i, σp(√

5) =√

5 ⇒ σp = 1

p ≡ 3 ⇒ p ≡ 3 (mod 4), p ≡ 3 (mod 5)⇒ σp(i) = −i, σp(√

5) = −√

5 ⇒ σp = στ

p ≡ 7 ⇒ p ≡ 3 (mod 4), p ≡ 2 (mod 5)⇒ σp(i) = −i, σp(√

5) = −√

5 ⇒ σp = στ

p ≡ 9 ⇒ p ≡ 1 (mod 4), p ≡ 4 (mod 5)⇒ σp(i) = i, σp(√

5) =√

5 ⇒ σp = 1

p ≡ 11 ⇒ p ≡ 3 (mod 4), p ≡ 1 (mod 5)⇒ σp(i) = −i, σp(√

5) =√

5 ⇒ σp = σ

p ≡ 13 ⇒ p ≡ 1 (mod 4), p ≡ 3 (mod 5)⇒ σp(i) = i, σp(√

5) = −√

5 ⇒ σp = τ

p ≡ 17 ⇒ p ≡ 1 (mod 4), p ≡ 2 (mod 5)⇒ σp(i) = i, σp(√

5) = −√

5 ⇒ σp = τ

p ≡ 19 ⇒ p ≡ 3 (mod 4), p ≡ 4 (mod 5)⇒ σp(i) = −i, σp(√

5) =√

5 ⇒ σp = σ.

(b) On the one hand, recall that a prime splits completely if and only if its

decomposition group is trivial. And on the other hand,(

pZK/Q

)= σp generates such

decomposition group. Thus, using item (a) we can summarize the whole thing as

pZ splits completely ⇐⇒ its decomposition group is trivial

⇐⇒(

pZK/Q

)= Id

⇐⇒ p ≡ 1 or 9 (mod 20)

13

typo, should be the 5th power

(c) Applying Corollary 1.3 to the tower K ⊃ Q(i) ⊃ Q we get(pZ[i]

K/Q(i)

)=

(NQ(i)/Q pZ[i]

K/Q

)=

(pZK/Q

)f(pZ[i]/pZ).

Now, since pZ is inert, f(pZ[i]/pZ) = [Q(i) : Q] = 2. However, every element of

Gal(K/Q) has order 2. Thus we conclude(pZ[i]

K/Q(i)

)=

(pZK/Q

)2

= Id

(d) The fundamental equation grants [Q(i) : Q] = efg. But here [Q(i) : Q] = 2,

thus pZ splitting implies g = 2 and ef = 1. Therefore, applying Corollary 1.3 to the

tower K ⊃ Q(i) ⊃ Q yields(p

K/Q(i)

)=

(NQ(i)/Q p

K/Q

)=

(pZK/Q

)f(p/pZ)=

(pZK/Q

),

and the later has already been calculated in (a).

Exercise 5.7 (p.123): Let K/F be an abelian extension of number fields. Show that

if m is any admissible ideal for K/F then

IF (m)

P+F,mNK/F (m)

∼= Gal(K/F )

and in particular, the conductor f(K/F ) satisfies Artin Reciprocity.

Proof: Since m is admissible, it is divisible by every prime that ramifies in K/F .

Therefore it satisfies (i) and (iii) from Theorem 2.1. Now, although we may not have

(ii), if we can somehow show that P+F,m ⊂ ker(A), then we can proceed anyway in

accordance with the comments in the statement of Theorem 2.1, thus getting the

desired isomorphism. i.e., all we have to do is prove P+F,m ⊂ ker(A).

To do that, let E/F be a cyclic subextension (i.e., K ⊃ E). Since m is admissible

for K/F , it is in particular divisible by every prime that ramifies in E/F and

E+F,m ⊂ NK/FEK = NE/F (NK/EEK) ⊂ NE/FJE.

14

Therefore we can apply Proposition 2.2 to the extension E/F to get

ker(AE/F ) ⊂ P+F,mNE/F (m).

On the other hand, the Cyclic Norm Equality assures

[IF (m) : ker(AE/F )] = #Gal(E/F ) = [E : F ] = [IF (m) : P+F,mNE/F (m)].

Thus

ker(AE/F ) = P+F,mNE/F (m),

and so P+F,m ⊂ ker(AE/F ). In other words, we have shown that

P+F,m ⊂ ker(AE/F ) ∀ E/F cyclic subextension.

Therefore, for every a ∈ P+F,m we have(

a

E/F

)= 1 ∀ E/F cyclic subextension.

However, we have already seen (during the last three paragraphs of the proof of

Theorem 2.1 ) that this implies(

aK/F

)= 1. Hence

P+F,m ⊂ ker(AK/F ) = ker(A).

Exercise 5.9 (p.125): Let K/F be an abelian extension of number fields with

Gal(K/F ) = G. Let σ ∈ G and define

Sσ = {primes p of OF : p is unramified in K/F and

(p

K/F

)= σ}.

Show that δF (Sσ) = 1[K:F ]

.

Proof: Here we have to assume the claims stated in the book (right before this ex-

ercise), so that we are under sufficient conditions to use Ch.3 Proposition 2.3.

Let m ∈ IF be an ideal divisible by all the ramified primes and no others, and

suppose that m satisfies Artin Reciprocity. Since the Artin map is surjective, there

exists b ∈ IF (m) such that(

bK/F

)= σ−1. Furthermore, if a prime p of OF is such

that pb ∈ IF (m), then by the unique factorization we must have p ∈ IF (m), and so

p is unramified.

15

Therefore, for every prime p of OF we can argue

p ∈ Sσ ⇐⇒ p is unramified in K/F and

(p

K/F

)= σ

⇐⇒ p is unramified in K/F and

(pb

K/F

)= 1

⇐⇒ pb ∈ ker(A) = P+F,mNK/F (m) ⊂ IF (m)

⇐⇒ p ∈ b−1P+F,mNK/F (m)

i.e., we have proven

Sσ = {primes p of OF : p ∈ b−1P+F,mNK/F (m)}.

Thus, applying Ch.3 Proposition 2.3 (with a = b−1 and H = P+F,mNK/F (m)), we get

δF (Sσ) =1

[IF (m) : P+F,mNK/F (m)]

.

On the other hand, we have already seen in Corollary 2.9 that TK/F = SK/F . There-

fore we also have

1

[IF (m) : P+F,mNK/F (m)]

= δF (TK/F ) = δF (SK/F ) =1

[K : F ]

(where the first equality is again due to Ch.3 Proposition 2.3 (this time with a = 1),

while the last one is Ch.1 Theorem 5.1 ).

Exercise 5.11 (p.125): Let F be a number field, and let f(X) ∈ F [X] be a poly-

nomial. Suppose f(X) splits into linear factors modulo p for all but finitely many

prime ideals p of OF . Use the Chebotarev Density Theorem, applied to the splitting

field of f(X), to show that f(X) splits in F [X].

Proof: Let us assume (ad absurdum) that f(X) is irreducible in F [X], and let K

be its splitting field. So G := Gal(K/F ) permutes transitively all the roots of f(X).

Thus, by Jordan’s Theorem, there exists an element σ0 ∈ G that does not fix any of

the roots of f(X).

Now on the one hand, by the Cheboratev Density Theorem, there are infinitely

many primes P of K unramified in K/F with(

PK/F

)∈ [σ0]G. And since G also

16

permutes transitively all the prime ideals P’s above a prime p of F , we get that:

• There are infinitely many primes p of F with a P above that satisfies(

PK/F

)= σ0.

On the other hand, by hypothesis:

• f(X) splits into linear factors modulo p for all but finitely many primes p of F .

Since the first set is infinite and the second one excludes only finitely many primes,

we know we can find a prime p0 of F that is in both. i.e., There exists a pair of primes

P0|p0 of K/F such that f(X) splits into linear factors modulo p0 and(

P0

K/F

)= σ0.

Which translates to:

f(X) has a root (modulo p0) and Gal

(OKP0

/OFp0

)= 〈σ0〉

∣∣OKP0

.

Therefore σ0 has to fix such root. But this contradicts our choice of σ0 (taken not

fixing any root). Hence f(X) cannot be irreducible.

Exercise 5.15 (p.127): Let F ⊂ L ⊂ K, F ⊂ E ⊂ K be number fields and suppose

K/F is abelian. If a ∈ JE, do we have(a

K/E

) ∣∣∣L

=

(NE/F (a)

L/F

)as we did with the classical Artin maps on fractional ideals?

Proof: Yes. Let m be an ideal admissible for K/F which is big enough so that

mOE is admissible for K/E (this is possible by the same argument used to guarantee

the existence of admissible ideals, see the discussion right after Exercise 5.3 ). In

particular, m is also admissible for the lower subextension L/F . Thus the Artin

reciprocity works for both cases

IE(mOE) −→ Gal(K/E) and IF (m) −→ Gal(L/F ).

Therefore, the idelic Artin map for both cases can be defined using these two ideals

and the map 〈 · 〉 (introduced a few paragraphs before this exercise). On the other

hand, a straightforward verification shows that

b ∈ J+E,mOE

=⇒ NE/F (b) ∈ J+F,m and NE/F 〈b〉 =

⟨NE/F (b)

⟩.

(where the second N above is the ideal norm, while the others are the idelic norm)

17

Therefore we can compute(a

K/E

) ∣∣∣L

=

(〈αa〉K/E

) ∣∣∣L

(for an α such that αa ∈ J+E,mOE

)

=

(NE/F 〈αa〉L/F

)(by the Consistency Property)

=

(⟨NE/F (α)NE/F (a)

⟩L/F

)(by the above comments)

=

(⟨β NE/F (a)

⟩L/F

)(taking β := NE/F (α))

=

(NE/F (a)

L/F

)(because by the above comments, αa ∈ J+

E,mOE

=⇒ β NE/F (a) = NE/F (αa) ∈ J+F,m )

Exercise 6.1 (p.141): What are JF,S∞ and FS∞?

Proof: Recall that, for every place v ∈ S∞ by definition Uv = F×v . Therefore

JF,S∞ =∏v∈S∞

F×v ×∏v/∈S∞

Uv =∏v∈S∞

Uv ×∏v/∈S∞

Uv =∏v∈VF

Uv = EF .

And so

FS∞ = JF,S∞ ∩ F× = EF ∩ F× = O×F .

Exercise 6.2 (p.142): The group of S-idele classes of F is

CF,S =JF,SFS

.

We have JF,S ⊂ JF and FS ⊂ F×. So there is a natural monomorphism CF,S ↪→ CF .

Show that there is a topological and algebraic isomorphism

JFF×JF,S

∼=CFCF,S

.

18

Proof: Algebraically we have

CFCF,S

=JF/F

×

JF,S/FS=

JF/F×

JF,S/(JF,S ∩ F×)(by using just the definitions)

∼=JF/F

×

F×JF,S/F×(by the natural identification of the denominators

as subgroups of JF/F×, via the second isomor-

phism theorem).

∼=JF

F×JF,S(by the third isomorphism theorem).

Now, from the theory of topological groups, we know that the third isomorphism

theorem still holds for topological groups (i.e., the group isomorphism it provides is

also a homeomorphism). Therefore the last “∼= ” above is also topological. On the

other hand, although the second isomorphism theorem is not true topologically, we

actually don’t need it for first “∼=” above, as that step follows from the definition of

the quotient JF /F×

JF,S/(JF,S∩F×).

Indeed, to clarify things up, the quotient JF /F×

JF,S/(JF,S∩F×)is defined through the

natural identification of JF,S/(JF,S ∩ F×) with F×JF,S/F× as a subgroup of JF/F

×

(because in reality JF,S/(JF,S ∩ F×) is not even a subset of JF/F×). Therefore, when

viewing JF,S/(JF,S ∩ F×) as a topological subgroup of JF/F× we are actually regard-

ing it with the topology that it inherits from the identification with F×JF,S/F× (in

fact, any other topology would fail to make it a topological subspace). Hence the “∼=”

in question is trivially an isomorphism of topological groups (as we are considering

both denominators as being the same subgroup and with the same topology).

Exercise 6.3 (p.142): Show that FS ∼= O×F × γ(FS).

Proof: Using the exact sequence that was mentioned a few lines before this exercise

and the first isomorphism theorem, we get

γ(FS) ∼=FSUF

.

On the other hand, by definition O×F = UF . Thus our problem is the same as showing

FS ∼= UF ×FSUF

,

19

which is straightforward commutative algebra. Indeed, since FSUF

is a free Z-module

of finite rank, it is projective. Therefore, by the lifting property the short exact se-

quence given by the canonical projection is split, which translates exactly into the

above isomorphism.

(With that being said, a few days ago before the class we were trying to figure out

this isomorphism explicitly and got somewhat confused. Thus, to clear things up, it

might be useful to see an explicit proof. In what follows, we shall state it generically

and prove it explicitly.)

Claim: Let R be a ring (commutative with unit). Let M be an R-module. Let N

be an R-submodule of M such that MN

is free of finite rank. Then

M ∼= N × M

N

Proof: Let π : M → MN

be the canonical projection. Since MN

is free of rank r <∞,

we can take a basis {b1, ... , br} for it. Via π we can lift such basis to a linearly inde-

pendent set {b1, ... , br} ⊂M . Let M ⊂M be the submodule spanned by {b1, ... , br}.With this setup we have a trivial isomorphism

π|M : M = Span{b1, ... , br} −→ Span{b1, ... , br} =M

N,

with inverse

π :M

N= Span{b1, ... , br} −→ Span{b1, ... , br} = M.

(Note that by construction π ◦ π = IdMN

and π ◦ π|M = IdM , but π ◦ π 6= IdM .)

We assert that the function defined by

φ : M −−−−−−→ N × M

N

m 7−→(m− π ◦ π(m), π(m)

)is the desired isomorphism.

To see that, all we have to do is verify: (i) it is well defined, (ii) it is a homomor-

phism, (iii) it is injective, (iv) it is surjective. These are all straightforward, but to

avoid confusion we prove them.

(i) Here one has to show that ∀m ∈M we have (m − π ◦ π(m)) ∈ N. And indeed,

π(m − π ◦ π(m)

)= π(m) − π ◦ π ◦ π(m) = π(m) − π(m) = 0, which implies

(m− π ◦ π(m)) ∈ kerπ = N .

20

(ii) This is immediate from the fact that all the functions in the definition of φ are

homomorphisms.

(iii) If φ(m) = φ(m′), then from the second coordinate we get π(m) = π(m′). So, in

particular π ◦ π(m) = π ◦ π(m′). On the other hand, using the first coordinate we

get m − π ◦ π(m) = m′ − π ◦ π(m′). Thus, canceling on both sides yields m = m′.

Therefore φ is injective.

(iv) Given (n,m) ∈ N × MN

, take m ∈ M such that π(m) = m, and recall that in

this case π ◦ π(m) = m (because m ∈ M). Then we can simply compute φ(n+m) =(n+m−π◦π(n+m), π(n+m)

)=(n+m−π◦π(m), π(m)

)=(n+m−m, m

)= (n,m).

Therefore φ is surjective.

[FT] Frohlich, A. and Taylor, M., Algebraic Number Theory, Cambridge Univer-

sity Press, 1991.

[Ne] Neukirch, J., Algebraic Number Theory, Volume 322 of Grundlehren der

methematischen Wissenschaften, Srpinger-Verlag, Berlin, 1999 (Transla-

tion from the 1992 German original).

21

Class Field Theory - Homework: Chapter 4Book: Class Field Theory by N. Childress

Student: Andre Lelis

Exercise 4.2 Let F = Q(√2) and K = Q( 4

√2,√3). Find all infinite places

of F and K, groupings the places of K according to which places of F theyextend.

Solution:Q(√2) has two different embeddings in R.

σ1(√2) =

√2 and σ2(

√2) = −

√2.

K has more embeddings:σ1,1 = Idσ1,2 : σ( 4

√2) = 4

√2 and σ(

√3) = −

√3

σ1,3 : σ( 4√2) = − 4

√2 and σ(

√3) =

√3

σ1,4 : σ( 4√2) = − 4

√2 and σ1,4(

√3) = −

√3.

Now look at σ2. In this embedding, we have that σ2(√2) = −

√2. But we

know that 4√2 4√2 =√2. Then this embedding will extend to embedding in C.

σ2,1 : σ2,1(4√2) = i 4

√2 and σ2,1(

√3) =

√3

σ2,2 : σ2,2(4√2) = i 4

√2 and σ2,2(

√3) = −

√3.

σ2,3 : σ2,3(4√2) = −i 4

√2 and σ2,3(

√3) =

√3.

σ2,4 : σ2,4(4√2) = −i 4

√2 and σ2,4(

√3) = −

√3.

Exercise 4.3 Let F = Q(i) and x = 2− i. Compute ‖x‖v for all places v ∈ VFand verify that the prdouct formula holds for x.

Solution: We know that x = 2 − i is prime in OQ(i) because the norm is 5.Actually, 5 splits into (2− i)(2 + i).

We have σ1 : σ(i) = i and σ2 : σ(i) = −i, in this case we have ‖x‖σ1=

‖x‖σ2 = 5.We also have that ‖x‖2−i = N(2− i)−ord2−i(x) = 25−1

Then the product rule holds.Exercise 4.4

Exercise 4.6 Show that if H is a subgroup of the topological group G, thenthe closure of H is also subgroup of G.

Solution: Let g, h ∈ H . We know that the product is continuous µ, thenlet U be an open neighborhood of gh, so the inverse image of U by productfunction is open. On the other hand, as g, h ∈ H we have that there is h′ ∈V1 ⊂ H ∩ µ−1(u) and g ∈ V1. With the same argument we can prove that thereis g′ and V2, with V1 × V2 ⊂ µ−1(U). Thus h′g′ ∈ U , then gh ∈ H .

Same idea holds for inverse.

Exercise 4.7 LetG be a topological group with identify 1, and suppose thereis some compact neigborhood A of 1 in G. Show that G is locally compact.

Solution: Let g ∈ G. Then g ∈ gA. But gA is homeomorphic to A, then gAis a compact neigborhood of g. Thus G is locally compact.

1

Exercise 4.8 Let H be a subgroup of the topological group G.(a) Show that if H is open in G, then H is closed.

(b) Show that if H is closed with finite index, then H is also open.

(c) Show that if G is compact and H is open, then [G : H] is finite.

Solution: (a) We can write F = ∪x/∈HxH . We know that each xH is opensince xH is homeomorphic to H . Thus H = G \ F is closed.

(b) Again we can write F = ∪x/∈HxH , but since the index is finite, the F =∪ni=1xiH . Thus F is closed and H = G \ F is open.

(c) ∪x/∈HxH ∪H is an open cover of G. But G is compact, then ∪x/∈HxH =

∪ji=1xiH for some xi.Exercise 4.9 Let G be a topological group with identify 1. Prove or disproveand salvage:

(a) IfH is a subgroup ofG that contains a neigborhood of 1, thenH is open.(b) If A is an open neighborhood of 1 in G, then A is a subgroup of G.Solutions by Manuel:(a) Let V be an open neighborhood of 1 such that V ⊂ H . If g ∈ H , then

g ∈ gV and gV ⊂ H , because group is closed by product. Then ∪g∈HgV ⊂ H .On the other hand, if g ∈ H then g ∈ gV , thus g ∈ ∪g∈HgV .

So, H = ∪g∈HgV which is open.(b) R∗+ is a multiplicative group, but ( 12 ,∞) isn’t a subgroup.

2

1

Aluno: Manuel Saavedra

Exercise 4.11a) For a P JF and A open in EF , aA is open in JF .

p P aA

a�1p P A

a�1p P¹vRJUv

¹vPJ

Wv � A

J is finite, Wv is open in Uv for v P J . Since a P JF , there is I finite such thatav P Uv for v R I.

a�1p P¹

vRpJYIq

Uv

¹vPpIzJ q

Uv

¹vPJ

Wv � A

then

p P¹

vRpJYIq

Uv

¹vPpIzJ q

avUv

¹vPJ

avWv � aA

but IzJ and J are finite, avUv and avWv are open in F�v . Then aA is open in JF .

b) taA : a P JF , and A is an open subset of EF u is a basis de open sets for JF

• For x P JF , x P xEF .• Let p P aA X bB, them a�1p P A and b�1p P B. There are J, I set finite such

that

a�1p P¹vRI

Uv

¹vPI

Wv

b�1p P¹vRJ

Uv

¹vPJ

Zv

For every v there is ζv P Uv, with av � ζvbv, them avUv � bvUv. Then

p P¹

vRpIYJq

avUv

¹vPJzI

bvZv

¹vPIzJ

avWv

¹vPpIXJq

avpWv X ζ�1v Zvq

let the open set in EF

D �¹

vRpIYJq

Uv

¹vPJzI

Zv

¹vPIzJ

Wv

¹vPpIXJq

pWv X ζ�1v Zvq

2

and let c � pcvqv, with

cv � av, v P pI Y JqC Y I

cv � bv, v P pJzIq

then p P cD � aAX bB.• Let x P U open in JF ,there is I finite such that

x P¹vRI

Uv

¹vPI

Wr � U

since that EF is subgroup de JF them there is p P JF with x P pEF . For v P I,xv � pvUv and xv P Wv, them

xv P pv

�Uv X p�1

v Wv

�� pvZv � Wv

note that Zv � Uv is open. Let

A �¹vRI

Uv

¹vPI

Zvpis open in EF q

them

x P aA � U

with a � pavqv, av � 1 if v R I and av � pv if v P I.

Exercise 4.14Let contentpaq �

±vPVF

}av}v

For a, b P JF there are I and J subsets finite of VF such that

v R I ô av R Uv

v R J ô bv R Uv

then

contentpabq � contentppavbvqvq �¹

vPpIYJq

}avbv}

�¹vPI

}av}v

¹vPJ

}bv}v �¹

v

}a}¹

v

}b}

� contentpaqcontentpbq

Let a P JF and r � contentpaq ¡ 0.

contentpaq P

�r

2 ,3r2

� R�

�

note que a P aEF (open in JF ) and contentpaEF q � r P pr{2, 3r{2q. Then content isa continuous homomorphism.

3

Exercise 4.12Let x P JF , x�1 P W open

JF ÝÑ JF

x ÞÝÑ x�1 � W

there is a P JF and A � EF open such that x�1 � aA � W . Let B � A�1 open inEF and b � a�1, then

x P bB ÞÝÑ x�1 P pbBq�1 � aA � W

Let a, b P JF , ab P D open

JF � JF ÝÑ JF

pa, bq ÞÝÑ ab P D

there is an open

ab P¹vRI

Uv

¹vPI

Wv � D

then

ab P¹vRI

Uv

¹vPI

avbv

�Uv X pavbvq

�1Wv

�� D

For v P I, Zv � Uv X pavbvWvq(open), note que 1 P Zv, there is Yv open withYvYv � Zv. Then

a P¹vRI

Uv

¹vPI

pavYvq � H pis open inJF q

b P¹vRI

Uv

¹vPI

pbvYvq � G pis open inJF q

with

ab P HG �¹vRI

Uv

¹vPI

Wv � D

Student: Claudio da Silva Velasque

Exercise 2.20:Define S = {p prime of Z; I | pOF , I ∈ SK|F }. Let p ∈ S, I ∈ SK|F ; I | pOF . Since

f(K |P ) = f(K | I).f(I |P )e(K |P ) = e(K | I). e(I |P )

We have that S ∩ SF = SK . The function s 7→∑

p∈s,f>1=e

f−1. p−f.s is analytic in R e s > 2−1, then

δF (SK|F ) = lims→1+

∑I∈SK|F

NI−s

log(

1s−1

) = lims→1+

∑p∈S

[F : Q]

f.e.p−f.s

log(

1s−1

)

=

lims→1+

∑p∈SK

p−s +∑

f>1=e

f−1p−fs +∑e>1

(fepfs

)−1log(

1s−1

) [F : Q] = [F : Q] . lim

s→1+

∑p∈Sk

p−s

log(

1s−1

)= [F : Q].δ(SK) = [F : Q].[K : Q]−1 = [K : F ]−1

Exercise 2.21: Suppose F | Q Galois(a) δF (P ∈ Spec OF ; f(P | P ∩ Z) = 1) =

= lims→1+

∑f=1

NP−s

log(

1s−1

) = lims→1+

∑f=1

[F : Q]e−1.p−s

log(

1s−1

) =

= [F : Q]. lims→1+

∑f=e=1

p−s +∑

f=1<e

e−1.p−s

log(

1s−1

) = [F : Q].δ(SF ) = 1

(b) (⇐)Obvius(⇒)False . Counterexample:Let A = {P ∈ Spec OF ; f(P | P ∩ Z) > 1}

B = {P ∈ Spec OF ; e(P | P ∩ Z) > 1}

δ(p ∈ Z ; f(F | p) = 1) = lims→1+

∑f=1

p−s

log( 1s−1 )

= lims→1+

∑f=e=1

p−s +∑

f=1<e

p−s

log( 1s−1 )

= lims→1+

∑f=e=1

p−s

log( 1s−1 )

=

= δ(SF ) =1

[F : Q]⇒ 1 = δ(p prime of Z) = δ(p ∈ Z ; f(F | p) > 1) + [F : Q]−1

⇒ δ(p ∈ Z ; f(F | p) > 1) = 1 − [F : Q]−1 > 0⇒ # {p ∈ Z ; f(F | p) > 1} = +∞The function ϕ : A→ {p ∈ Z ; f(F | p) > 1} is surjective , then # A = +∞.

P 7→ P ∩ Z# B <∞⇒ # B rA <∞⇒ δF (B rA) = 0

The function s 7→∑

f>1=e

f−1. p−f.s is analytic in R e s > 2−1 , then

δF (ArB) = lims→1+

∑f>1=e

[F : Q].f−1.p−fs

log( 1s−1 )

= 0.

Therefore A ≈ B and # ArB = +∞.

1

Student: Claudio da Silva Veasque

Exercise 4.1Let B(‖ · ‖) = {C ∈ (1,∞); ‖ 1 + x ‖6 C, whenever ‖ x ‖6 1}.

Step 1 : If C ∈ B(‖ · ‖), then ‖ a+ b ‖6 C.max {‖ a ‖, ‖ b ‖}.Proof: Suppose ‖ a ‖= max{‖ a ‖, ‖ b ‖}. Then ‖ a+ b ‖=‖ a.(1 + b

a) ‖=‖ a ‖ . ‖ 1 + b

a‖6 C. ‖ a ‖

Step 2 : If C ∈ B(‖ · ‖) and λ = log 2log C

, then 2 ∈ B(‖ · ‖λ).Proof : ‖ x ‖λ6 1⇒‖ x ‖= (‖ x ‖λ) 1

λ 6 1⇒‖ 1 + x ‖6 C ⇒‖ 1 + x ‖λ6 Cλ = 2

Step 3 : If 2 ∈ B(‖ · ‖) , then ‖n∑i=1

ai ‖6 2n ·max ‖ ai ‖

Proof : By step 1 , ‖ a1 + a2 ‖6 2.max{‖ a1 ‖, ‖ a2 ‖}By induction ,‖ a1 + · · · + a2r ‖6 2r.max {‖ a1 ‖, · · · , ‖ a2r ‖}. n > 1 ⇒ ∃ ! r > 0 ; 2r 6 n <2r+1 ⇒‖ a1 + · · ·+ an ‖=‖ a1 + · · ·+ an + 0 + · · ·+ 0 ‖ (2r+1terms)

6 2r+1.max ‖ ai ‖6 2. n.max ‖ ai ‖

Step 4 : If 2 ∈ B(‖ · ‖) ,then ‖ a+ b ‖6‖ a ‖ + ‖ b ‖ .

Proof : ‖ a+ b ‖n = ‖n∑i=0

(ni) an−i. bi ‖

6 2(n+ 1)max ‖ (ni)an−i. bi ‖

6 4(n+ 1)

n∑i=0

(ni) ‖ a ‖n−i . ‖ b ‖i

= 4(n+ 1).(‖ a ‖ + ‖ b ‖)n⇒‖ a+ b ‖6 (‖ a ‖ + ‖ b ‖). n

√4(n+ 1) −−−−→n→∞ ‖ a ‖ + ‖ b ‖ .

Reference: Weiss , Algebraic Number Theory.

1

Student: Claudio da Silva Velasque

Exercise 4.25:

Let ϕ1 : JF = FX .J+F,m � J+

F,m/F+m

α.a 7→ aϕ2 : J+

F,m/F+m −→ IF (m)/P+

F,m

a 7→ < a >α.a, β.a ∈ J+

F,m ⇒ α.a(β.a)−1 = α.β−1 ∈ F x ∩ J+F,m = F+

m ⇒ ϕ1 is well-defined .

a, b ∈ J+F,m ; a ≡ b mod F+

m ⇒ a.b−1 = ε ∈ F+m ⇒ < a > . < b >−1=< ε >∈< F+

m >= P+F,m ⇒< a >≡<

b > mod P+F,m ⇒ ϕ2 is well-defined .

Let a ∈ IF (m)/P+F,m. Since < J+

F,m >= IF (m), we have that ∃ α ∈ J+F,m;< α >= a, then ϕ2(α) = < α > =

a, hence ϕ2(J+F,m/F

+m) = R+

F,m. Therefore ϕ = ϕ2 ◦ ϕ1 is well-defined and surjective .

α.a ∈kerϕ⇒< a >∈ P+F,m ⇒

{a ∈ F+

m , if a ∈ F+

a ∈ ε+F,m, otherwise⇒ ker ϕ = F+

m .ε+F,m ⇒ ϕ : JF

Fx.ε+F,m

∼=−→ IF (m)

P+F,m

1

Student: Claudio da Silva Velasque

Exercise 5.13: Let ϕ : JF � IF (m)/P+F,m,such that ϕ(a) =< αa > P+

F,m. By exercise 4.5 , ker ϕ = F×E+F,m.Then ϕ : JF /F

×E+F,m∼=−→ IF (m)/P+

F,m.

Let π : IF (m)

P+F,m

�IF (m)/P+

F,m

P+F,mNK/F (m)/P+

F,m

∼= IF (m)

P+F,mNK/F (m)

be the canonical surjection. By proposition 5.6 ,ϕ :

F×NK/F JK

F×E+F,m

∼=−→ P+F,mNK/F (m)

P+F,m

. Then ker π ◦ ϕ =F×NK/F JK

F×E+F,m

.

Hence π ◦ ϕ : JF /F×NK/FJK

∼=−→ IF (m)/P+F,mNK/F (m) because JF

F×NK/F JK∼=

JF /F×E+F,m

F×NK/F JK/F×E+F,m

.

aF×NK/FJK 7→< αa > P+F,mNK/F (m)

Therefore JF � JF /F×NK/FJK

π◦ϕ−→ IF (m)/P+F,mNK/F (m)

A−→ Gal(K/F )

a 7→ aF×NK/FJK 7→ < αa > P+F,mNK/F (m) 7→

(<αa>K/F

)

1