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)
31
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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:
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
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