George Turcas Number Theory: Lecture Notes Started on ...George Turcas Number Theory: Lecture Notes Started on January 19, 2015 which holds for all q lying above p, so e q = e q=p

George Turcas Number Theory: Lecture Notes Started on January 19, 2015

Lecture Notes Summary

This notes are taken after the course Algebraic Number Theory taught by Dr. Jack Thorne at Part

III of the Mathematical Tripos, Cambridge

This will be a second course in algebraic number theory, with an emphasis on the adele-theoretic

(rather than the classical ideal-theoretic) point of view. We will assume familiarity with the basic notions

of number fields and local fields, which will be reviewed briefly at the beginning of the course.

All the possible errors in the notes are probably mine, so if you find any please notify me at myfirst-

[email protected].

Topics Covered

Page

Extensions of Dedekind domains 1

Extensions of number fields 3

Absolute values and completions 8

Different and discriminant 12

Restricted direct products and the adele ring 18

Cyclotomic fields 24

Dirichlet series and zeta-functions 26

Generalised ideal class group 29

Analytic continuation and functional equation of L(χ, s) 31

Applications of zeta functions 38

Chebotarev density theorem 41

Explicit class number formulae (non-examinable) 47

Extensions of Dedekind domains

We make the following notations:

R is a Dedekind domain and K = Frac(R) is its field of fractions.

Let E/K be a finite separable field extension and let S be the integral closure of R in E.

If q ⊂ S is a non-zero prime ideal, we’ll say that q lies over a prime p ⊂ R if q ∩R = p.

E ⊇ S ⊆ q

⊆ ⊆ ⊆

K ⊇ R ⊇ p

Proposition 1. S is a finitely generated R-module and a Dedekind domain. Also, every nonzero prime ideal

of S lies above a nonzero prime of R, and conversely every non-zero prime of R lies below a non-zero prime

of S.

Extensions of Dedekind domains continued on next page. . . Page 1 of 49


Proof. S is integrally closed by definition. To show that S is Noetherian, we use the trace form trE/K .

Recall that trE/K is a nonzero K-linear map E → K, because E is a separable extension.This allows us to define B(−,−) : E × E → K, a K-bilinear form, by the formula B(x, y) = trE/K(xy).It is easy to see that this K-bilinear form is perfect, since trE/K is nonzero.

Let e1, . . . , en be a K-basis of E. Wlog, we can assume that e1, . . . , en ∈ S (if not, we can just multiplyby elements of R to clear denominators.)

Let f1, . . . , fn be the dual basis of e1, . . . , en with respect to B. Now again, there exists c ∈ R such thatc · fi ∈ S, for all i ∈ {1, . . . , n}.

We claim that S ⊆n⊕i=1

c−1Rei.

To prove the claim, observe that trE/KS ⊂ R. Take z ∈ S and write z =n∑i=1

riei for some ri ∈ K.

Looking at B(z, c · fj) = cṙj = trE/K(z · (c · fj)) ∈ R. The last is true, since z ∈ S and c · fj ∈ S. Now

rj ∈ c−1R and hence S ⊆n⊕i=1

c−1Rei, hence the claim is proved.

Now, since R is Noetherian, we get that S is a finitely generated R-module, hence a finitely generated

R-algebra. Therefore, S is Noetherian, by Hilbert basis theorem. ( 1)

Now let q ⊂ S be a nonzero prime ideal, and let p = q ∩ R. There is an injection R/p ↪→ S/q, thereforep is a prime ideal.

If p is nonzero, choose b ∈ q \ {0} and let bd + a1bd−1 + · · · + ad = 0 be an equation of minimal degreed, with ai ∈ R. Then ad 6= 0 and ad ∈ R ∩ q = p. Then S/q is a finite dimensional R/p- algebra, R/p is afield, therefore S/q is a field, hence q is maximal. Therefore, we proved that S is a Dedekind domain.

Let p ⊂ R be a nonzero prime ideal.We claim that pS 6= S. To prove this, suppose the contrary. Then p−1S = S, hence R = K ∩ S =

K ∩ p−1S ⊇ p−1. But this gives the contradiction p−1 ⊂ R, therefore proving our claim.Since S is a Dedekind domain and pS 6= S, we can factor pS = q1q2 . . . qm, with qi nonzero prime ideals

of S and m > 0.

Now q1 ∩R ⊇ q1 . . . qm ∩R = pS ∩R ⊇ p. Since R is a Dedekind domain and q1 ∩R ⊇ p is an inclusionof nonzero prime ideals, we get that p = q1 ∩R.

If K is a number field, recall that OK is the integral closure of Z in K. But Z is a PID, therefore aDedekind domain and hence OK is a Dedekind domain as well.

Corollary 2. If K is a number field, a ⊆ OK is a nonzero ideal, we can factor a = p1 . . . pn uniquely as aproduct of prime ideals of OK .

Definition If a ⊆ OK is any nonzero ideal, we define its norm N(a) = [OK : a] - the index as additivesubgroup.

The norm is a positive integer, since a ⊇M · OK for some integer N , hence [OK : a] ≤ [OK : M · OK ] ≤M [K:Q], as (OK ,+) ≡ Z[K:Q].

If p ⊂ OK is a nonzero prime ideal, then p ∩ Z = (p), where (p) is some rational prime. This p is calledthe characteristic of the ideal p.

We denote by Fp = OK/p the residue field of p and by fp = [OK : Z/(p)] = [Fp : Fp], the residuedegree of p.

Lemma 3. If p ⊂ OK is a nonzero prime ideal, n ≥ 1 and x ∈ pn \ pn+1, then the map OK/p → pn/pn+1defined by α 7→ αx is an isomorphism.

Proof. The injectivity comes from the fact that the map is nonzero and OK/p is a simple OK module.

Extensions of Dedekind domains continued on next page. . . Page 2 of 49


To prove that the map is also surjective, let y ∈ pn \ pn+1. We know that (x) = pn · a, where a ⊂ OKis some ideal. As x /∈ pn+1, we get that a is coprime to p, hence (a, p) = OK . By the Chinese RemainderTheorem, we know there is β ∈ OK satisfying{

β ≡ y (mod pn+1)β ≡ 0 (mod a)

This implies that (β) ⊂ pn · a, therefore (β) · (x)−1 ⊂ pn · a · (x)−1 = OK . This means that α = βx ∈ OK ,so α 7→ β ≡ y (mod pn+1). Since y was chosen arbitrary, our map is surjective.

Consequence. If p ⊂ OK is prime, n ≥ 1, then N(pn) = (N(p))n. This follows easily from

|OK/pn| =i−1∏i=0

|pi/pi+1| = |OK : p|n.

If a = pe11 . . . penn is a factorization into distinct prime ideals, by Chinese Remainder Theorem

Ok/a ≡n∏i=1

OK/piei .

Let p ∈ Z be a rational prime and factor pOK =∏

p|pOKpep into distinct prime factors. Taking norms,

we get that p[K:Q] =∏

p|pOK(N(p))ep =

∏p|pOK

pfp·ep , where ep is called the absolute ramification index of p

and fp = [Fp : Fp] is the absolute residue degree of p.From the above, we get the equality [K : Q] =

∑p|pOK

ep · fp.

The prime p ∈ Z exibits 3 kinds of behaviour in the number field K, namely

1. Ramification - if ep > 1 for some p|pOK .

2. Residue extension - if fp > 1 for some p|pOK .

3. Splitting - if |{p ⊂ OK : p|pOK}| > 1.

Example K = Q[i], where i2 = −1. We have that [K : Q] = 2, OK = Z[i] which is a PID.2OK = (1 + i)2 and (1 + i) is a prime ideal, so we have ramification.3OK exibits a Residue extension since it is a prime ideal in OK .5OK = (2 + i)(2− i) is factored as a product of two distinct prime ideals, so 5 splits in OK .

Remark More generally, if p is an odd prime, then p splits in Z[i]⇔ p ≡ 1 (mod 4)⇔ ∃x, y ∈ Z such thatp = x2 + y2.

Extensions of number fields

Fix E/K an extension of number fields.

Lemma 4. For any x ∈ OK \ {0}, N(xOK) = |NK/Q(x)|, where the norm in the RHS is the absolute valuenorm on Q. Be aware of the fact that x ∈ OK ⇒ NK/Q(x) ∈ Z.

Extensions of number fields continued on next page. . . Page 3 of 49


Proof. Denote by n = [K : Q] and let α1, . . . , αn be a Z basis of OK . The α′is form a Q basis of K as well.

Write xαj =n∑i=1

γi,jαi for some γi,j ∈ Z.

y OK Zn

xy OK Zn

'

Γ

Γ is the map with matrix Γ = (γi,j)1≤i,j≤n.

The norm N(xOK) = [OK : xOK ] = [Zn : ΓZn]

y K Qn

xy K Qn

'

Γ

Recall that NK/Q(x) = det(y 7→ xy as a Q-linear isomorphism of K) = det(Γ).Our lemma now follows from [Zn : ΓZn] = |det(Γ)|, which follows for example from the theorem on

elementary divisors.

Proposition 5. For any nonzero prime p ⊂ OK and any q|pOE, prime ideal of OE, we define the relativeramification index eq/p to be the positive integer that appears in the factorisation pOE =

∏q|pOE

qeq/p and

the relative residue degree fq/p = [Fq : Fp]. Then, the following relations hold:

1. eq/p · ep = eq and fq/p · fp = fq.

2. For all p ⊂ OK nonzero prime ideal,

∑q|pOE

eq/p · fq/p = [E : K].

Proof. Let p ⊂ OK be a prime and p ∈ Z its residue characteristic. If q ⊂ OE is a prime lying over p, thenwe have

Z/(p) ↪→ OK/p ↪→ OE/q

= = =

Fp ↪→ Fp ↪→ Fq

⇒ [Fq : Fp] = [Fq : Fp] · [Fp : Fp],which is equivalent to fq = fq/p · fp.If p ⊂ OK has residue characteristic p, we can factor

pOE =∏

q|pOE

qeq/p

and also

pOK = pep ·

∏p′|pOK ,p′ 6=p

p′ep′

which gives us that

pOE =

∏q|pOE

qeq/p

ep · [ terms prime to pOE ] = ∏q|pOE

= qeq ,



which holds for all q lying above p, so eq = eq/p · ep and the proof for first part is complete.For the second part, it is enough to prove that N(p)[E:K] = N(p)

∑eq/p·fq/p .

Let y ∈ p \ p2 and, using the Chinese Remainder theorem, we know that we can find x ∈ OK , satisfying{x ≡ y (mod p2)x ≡ 1 (mod p′)

for all primes p′ ⊂ OK , p′ 6= p that have the same residue characteristic p, as p.This implies that xOk = p · a, where N(a) is coprime to p.We can calculate

N(pOE) = N

∏q|pOE

qeq/p

= ∏q|pOE

(N(q))eq/p =

∏q|pOE

(N(p)fq/p

)eq/p= (N(p))

∑q|pOE

eq/p·fq/p.

E

|K 3 x |NE/Q(x)| = |NK/Q ◦NE/K(x)| = |NK/Q(x)|[E:K] by the transitivity of norms.|Q

Therefore, the N(pOE) = the p-part of |NE/Q(x)| = p-part of |NK/Q(x)|[E:K] = (N(p))[E:K]

and we are

done.

Now we assume E/K is Galois and we denote by G = Gal(E/K).

Theorem 6. Let p ⊂ OK be a nonzero prime ideal. Then G acts transitively on the set of primes q ⊂ OElying above p.

Proof. If σ ∈ G, q ⊂ OE is a prime such that q ∩K = p, then σ(q) ∩K = p, therefore σ(q) is also a primelying above p.

To see that it acts transitively, factor pOE =n∏i=1

qeii . WLOG, it is enough to show that ∃σ ∈ G such that

σ(q1) = q2.

By the Chinese Remainder Theorem, we are allowed to choose α ∈ OE such that αOE = q1 · a, for someideal a ⊂ OE , prime to pOE . We have NE/K(α) =

∏σ∈G

σ(α) ∈ q1 ∩K = p. Hence,∏σ∈G

σ(α)OE ⊂ pOE , i.e.

∏σ∈G

σ(q1)σ(a) ⊂n∏i=1

qeii .

It follows that q2 divides σ(q1)σ(a) for some σ ∈ G. Since σ(a) is prime to pOE , we deduce that q2 = σ(q1)for some σ ∈ G, which is what we wanted to prove.

Remark As a consequence, if p ⊂ OK , then the quantities eq/p and fq/p are independent of the choice ofprime q|pOE . This is because from pOE =

∏qeq/p , we get that pOE = σ(pOE) =

∏σ(q)eq/p , for every

σ ∈ G. Here, of course, the product runs over all the primes q lying above p.By the unique factorization into prime ideas, we get that eσ(q)/p = eq/p, for all σ ∈ G and the rest follows

by transitivity.

Also, by definition we have Fq = OE/q. Now, if σ ∈ G we get



σ : OE/q OE/σ(q)

OK/p

∼

or equivalently

σ : Fq Fσ(q)

Fp

∼

giving [Fq : Fp] = [Fσ(q) : Fp], i. e. fq/p = fσ(q)/p.

Choose q ⊂ OE a nonzero prime ideal lying over p ⊂ OK . We define Dq/p = StabG(q) the decomposi-tion group at q.

An easy argument using the Orbit-Stabiliser theorem and the fact that G acts transitively (i.e. there is

just one orbit) on the prime ideals above p, proves the fact that |Dq/p| = |StabG(q)| = eq/p · fq/p.There is a natural map Dq/p → Gal(Fq/Fp). Recall from a first course in Galois Theory that the group

Gal(Fq/Fp) is cyclic of order fq/p and is generated by the Frobenius automorphism x 7→ xpfp

, i.e. x 7→ x|Fp|.We write Iq/p = ker

(Dq/p → Gal(Fq/Fp)

), and we call this the inertia group.

Proposition 7. The map Dq/p → Gal(Fq/Fp) is surjective.

Proof. Let ED ⊂ E be the fixed field of Dq/p.E

ED

K

G

Dq/p

⇒ E/ED is a Galois extension of group Dq/p. Also qD = q∩ED, so we get a tower

q

qD

p

.

Note that q is the unique prime of OE lying above qD. This is because Gal(E/ED) acts transitively onsuch primes, but Gal(E/ED) = Dq/p fixes q by definition. We observe that

[E : ED] = eq/qD · fq/qD = |Dq/p| = eq/p · fq/p

and because e∗ and f∗ are multiplicative in tower, we get that eq/qD |eq/p and fq/qD |fq/p. Hence, thedivisibility relations are actually equalities. In particular, fq/qD = fq/p implies that fqD/p =

fq/pfq/qD

= 1,

hence the natural inclusion Fp ↪→ FqD is actually an isomorphism.We have the commutative diagram

Dq/qD Gal(Fq/FqD )

Dq/p Gal(Fq/Fp)

'

This means that is enough to show that the map Dq/qD → Gal(Fq/FqD ) is surjective.By the primitive element theorem, there exists x ∈ Fq = OE/q such that Fq = FqD (x) and let x ∈ OE

be a lift of x. Let x1, . . . , xk ∈ OE be the conjugates of x under Dq/qD . The minimal polynomial of x over

ED is given by f(X) =k∏i=1

(X − xi); it is irreducible in ED[X] and Dq/qD acts transitively on its roots.



This imples the result: for τ ∈ Gal(Fq/FqD ), τ(x) is a root of the reduction f(X) ∈ FqD . Hence, thereexists σ ∈ Dq/qD such that σ(x) = τ(x). Because x is a primitive element, this implies that the image of σin Gal(Fq/FqD ) is exactly τ . So, the map Dq/qD → Gal(Fq/FqD ) is surjective, completing the proof.

The proposition shows that there is an exact sequence of groups

1 Iq/p Dq/p Gal(Fq/Fp) 1

eq/p fq/p

order order

We will later show that the group Dq/p can be canonically identified with the Galois group of an extension

of local fields.

We will also see that for all but finitely many primes q of E, the ramification index eq/p = 1, hence the

inertia group is trivial. For such a prime q, the generator x 7→ xpfq/p

therefore lifts canonically to Dq/p. We

call this lifted element Frobq/p ∈ Dq/p ⊂ G the Frobenius element at q. These elements play an esential rolein algebraic number theory.

We now give a result which is extremely useful in practice to calculate the factorization of primes in

extensions of number fields.

Theorem 8 (Kummer-Dedekind). Let E/K be an extension of number fields, and let α ∈ OE be suchthat the index [OE : OK [α]] is finite. Let p be a prime of OK such that Np is prime to this index. Letf(X) ∈ OK [X] be the minimal polynomial of α over K.

Let f(X) (mod p) =k∏i=1

geii be the factorization into distinct irreducibles in Fp[X]. Then, for each

i = 1, . . . , k the ideal qi = (p, gi(α)) ⊂ OE is prime, and pOE =k∏i=1

qeii .

Proof. This is a problem on the first example sheet, so the proof was omitted in lectures.

Example Let E = Q( 3√

2,√−3), the splitting field of X3 − 2 and K = Q( 3

√2), L = Q(

√−3). Then E/Q is

a Galois extension with Galois group S3. We will show later that OK = Z[ 3√

2], while it is very easy to show

directly that OL = Z[α], where α = (1 +√−3)/2 is a root of the polynomial X2 −X + 1.

Let us factor some small primes in OE . We have that 2OK = ( 3√

2)3 = p32, say, while the Kummer-

Dedekind theorem shows us that the ideal 2OL = q2 is already prime (since the polynomial X2 −X + 1 isirreducible in F2[X]). If r2 is a prime of OE above 2, then we have that 3 = ep2/2|er2/2 and 2 = fq2/2|fr2/2and er2/2fr2/2 ≤ [E : Q] = 6. It follows that er2/2 = 3 and fr2/2 = 2 and r2 is the unique prime ideal of OEabove 2. We have Dr2/2 = Gal(E/Q) = S3, and Ir2/2 is cyclic of order 3.

We have X3 − 2 ≡ (X + 1)3 (mod 3), so the Kummer-Dedekind theorem implies that 3OK = p33, wherep3 = (3,

3√

2 + 1). We also have visibly eOL = (√−3)2 = q23, say. Since 2, 3 are coprime, we deduce that

there is aunique r3 of OE above 3, and we have that er3/3 = 6, fr3/3 = 1 and Iq3/3 ≡ S3.X3 − 5 = (X + 2)(X2 + 3X + 4) (mod 5), and these factors are irreducible. It follows that there are two

primes p5 = (5,3√

2 + 2) and p′5 = (5, (3√

2)2 + 3 3√

2 + 4) of OK above 5 and we have fp5/5=1 and fp′5/5 = 2.What happens in E? Let r5 be a prime of OE above p5 and r′5 be a prime of OE above p′5. Writing e = er5/5and f = fr5/5, we have e · f · g = 6, where g denotes the number of primes of OE above 5. We have 2|f andg ≥ 2, so the only posible solution is e = 1, f = 2 and g = 3. Let r′′5 be the extra prime of E above 5. Whichprime of K does it lie above? The extension E/K is quadratic Galois and we have fr5/p5 = 2, fr′5/p′5 = 1. It

follows that the group Gal(E/K) fixes the prime r5, but swaps r′5 and r

′′5 , which both lie above the prime p

′5.

Having seen some examples, we introduce some more terminology. If E/K is an extension of number

fields and p ⊂ OK is a nonzero prime ideal, we say:



1. p is totally ramified in E, if there exists a unique prime q of E above p and eq/p = [E : K], fq/p = 1.

2. p is totally inert in E if the ideal OE is prime, i.e there exists a unique prime q ⊂ OE above p andeq/p = 1, fq/p = [E : K].

3. p splits completely in E/K if for all primes q ⊂ OE above p, we have eq/p = fq/p = 1 and#{q/p} = [E : K].

In the previous example, 3 is totally ramified in E/Q, 2 is totally inert in L/Q and p′5 splits completelyin E/K.

Absolute values and completions

Definition Let K be a field. An absolute value on K is a function | · | : K → R≥0, satisfying the followingproperties:

i) ∀x ∈ K, |x| = 0⇔ x = 0;ii) ∀x, y ∈ K, |xy| = |x| · |y| (i.e it is multiplicative);iii) There exists a > 0 such that ∀x, y ∈ K, |x+ y|a ≤ |x|a + |y|a.

We say that two absolute values | · |1, | · |2 are equivalent if ∃c > 0 such that ∀x ∈ K, |x|2 = |x|c1.

Definition Let K be a field and | · | an absolute value on K. We say that | · | is trivial if ∀x ∈ K \{0}, |x| = 1.We also say that | · | is archimedean, if there exists n ∈ Z such that |n| > 1. Otherwise, we say that thisabsolute value is non-archimedean.

Remark We quickly recall a few results about absolute values on a field K.

1. Two absolute values | · |1, | · |2 on K induce the same topology if and only if they are equivalent.

2. If | · | is an absolute value on K, with respect to which K is complete, then any finite-dimensional Kvector space V is also complete with respect to the natural topology coming from the isomorphism

V ≡ Kn.

3. If |·|1, . . . , |·|n are pairwise inequivalent absolute values on K, then for all � > 0 and for all α1, . . . , αn ∈K, there exits α ∈ K such that for all i ∈ {1, . . . , n} we have that |α− αi| < �.

The complete proofs for the above results can be found in the first course on Local fields or in the

wonderful book A Brief Guide to Algebraic Number Theory by professor Swinnerton-Dyer.

In what follows, suppose K is a number field.

Proposition 9. Any archimedean absolute value | · |a : K → R≥0 is equivalent to one of the form |x|a =|τ(x)|,∀x ∈ K, where τ : K ↪→ C is an embedding of K and | · | is the usual absolute value on C.

If σ, τ : K ↪→ C are embeddings, then the absolute values given by |σ(x)| and |τ(x)| are equivalent if andonly if σ = τ or σ = τ (i.e. the complex conjugate).

Proof. Omitted.

Let α ∈ K be a primitive element over Q, i.e. K = Q(α) and let f(X) ∈ Q[X] be the minimal polynomialof α. Suppose f(X) has r real roots and s pairs of conjugate complex roots in C. Then

{embeddings τ : K ↪→ C} ←→ {roots of f in C}

⊆ ⊆

{embeddings τ : K ↪→ R} ←→ {roots of f in R}

Absolute values and completions continued on next page. . . Page 8 of 49


Corollary 10. The number field K has exactly r + s equivalence classes of archimedean absolute values.

Proposition 11. For any non-Archimedean absolute value | · | on K, there exists 0 < c < 1 such that forall x ∈ K \ {0}, |x| = cm, where m is given by

xOK = pm · a, where p is a prime ideal of OK and a is a fractional ideal of K, prime to p.

The prime ideal p depends only on the absolute value | · | here.

Proof. First, I begin by showing that our non-Archimedean | · | satisfies the ultrametric inequality, i.e.∀α, β ∈ K we have |α+ β| ≤ sup (|α|, |β|).| · | satisfies the ultrametic inequality, if and only if any equivalent absolute value | · |a satisfies it, so

WLOG we can assume that | · | satiesfies the usual triangle inequality, namely |α + β| ≤ |α| + |β| for allα, β ∈ K.

Now, for all α, β ∈ K and N ≥ 1, we have

|(α+ β)N | ≤N∑j=0

∣∣∣∣(Nj)αjβN−J

∣∣∣∣ ≤ N∑j=0

|αjβN−j | ≤ (N + 1) sup (|α|, |β|)N .

For the second inequality above, I used the non-Archimedean property, i.e. |m| ≤ 1 for all m ∈ Z.Letting N →∞ after taking N -th roots in the above inequality, we get exactly

|α+ β| ≤ sup (|α|, |β|)

Hence, our norm satisfies the ultra-metric inequality.

Observe that for all α, β ∈ K such that |β| < |α|, we have that |α+ β| = |α|.Let α ∈ OK . By definition, α satiesfies an integral equaltion αm + a1αm−1 + · · ·+ am = 0, where ai ∈ Z,

for all i ∈ {1, . . . ,m}. If |α| > 1, then 0 = |αm + a1αm−1 + · · ·+ am| = |α|m. This is because |ai| ≤ 1, for allai’s ( non-Archimedean property). Obviously, we reached a contradiction, therefore proving that |α| ≤ 1for all α ∈ OK .

Since | · | is non-trivial, we can find x ∈ OK such that |x| < 1. Therefore, let p = {y ∈ OK : |x| < 1}. Itis easy to see that p is a non-zero prime ideal of OK .

Choose π ∈ p \ p2. Now, we intend to show that |x| = |π|m, where m is defined by x ∈ pm \ pm+1.To do that, factor (x) = pm · a, where a ⊂ OK is prime to p. Notice that (π) = p · b, where b ⊂ OK is

prime to π.

Observe that (π)m

(x) =pm·bmpm·a = b

m · a−1 is a fractional ideal of K, prime to p.By the Chinese Remainder Theorem, we can find y ∈ OK such that y ≡ 1 (mod p) and y ≡ 0 (mod a).

Now, the ideal generated by(y·πmx

)= bm · a−1 · (y) ⊂ OK .

As y ∈ a, the element y·πm

x ∈ OK \ p, therefore∣∣∣y·πmx ∣∣∣ = 1. y was chosen to be in OK \ p, so |y| = 1,

giving us that |x| = |π|m. Finally, our proof is completed if we just choose c = |π|.

Definition A place of a number field K is an equivalence class of non-trivial absolute values on K. We

call a place infinite if it consists of archimedean absolute values, respectively finite if it consists of non-

archimedean absolute values.

Remark By Corollary 10, there are r + s infinite places of a number field K. Also, the finite places are

in bijection with the non-zero prime ideals p ⊂ OK . We will generally use letters v, w to denote places of Kand | · |v, | · |w their representative absolute values.



It should be clear that if v is a place of K, the topology induced by | · |v on K depends only on theequivalence class v. In fact, this determines the place v.

Denote by Kv the completion of K with respect to this topology.

We can represent Kv as{Cauchy sequences wrt to |·|v}{Null sequences wrt to |·|v} . But what does Kv really look like?

1. If v is an infinite place corresponding to τ : K ↪→ R, then τ induces a canonical isomorphism Kv ∼−→ R.

2. If v is an infinite place corresponding to a pair of complex conjugates τ, τ : K ↪→ C, then τ induces anisomorphism Kv ∼−→ C.

3. If v is a finite place corresponding to a non-zero prime ideal p ⊂ OK , let OKv = {x ∈ Kv : |x|v ≤ 1}and pv = {x ∈ Kv : |x|v < 1}.

There is an identification,

Kv =

∞∑

j>>−∞bjπ

j |bj ∈ {a1, . . . , aN}

,OKv = lim←−

i

OK/pi =

∞∑j=0

bjπj |bj ∈ {a1, . . . , aN}

,and pv = πOKv , where π ∈ p \ p2 and a1, . . . , aN are just representatives for the residue classes of OKmodulo p.

OKv is a local ring, with unique non-zero prime ideal pv, that lies above p.If we are give a = pa ·

s∏i=1

paii , where pi 6= p, for all i ∈ {1, . . . , s} then aOKv = pav .

Remark A consequence of this is the fact that to know the factorization of the ideal a it equivalent to know

the factorization of aOKv for every finite place v of K.

Let E/K be a finite extension of number fields, v a finite place of K corresponding to the ideal p ⊂ OK .Let q1, . . . , qg ⊂ OE be the primes of OE lying above p and denote by w1, . . . , wg their corresponding finiteplaces of E.

Observe that the restrictions | · |wi to K are equivalent to | · |v, since the primes qi lie all above p. Hence,the field embedding K ↪→ E extends to an embedding of completions

Kv ↪→ Ewi

⊆ ⊆

OKv ↪→ OEwi

for each i ∈ {1, . . . , g}

Proposition 12. The degree of the extension Ewi/Kv is equal to eqi/p · fqi/p.

Proof. As OKv is a PID, OEwi is a free OKv module. Then,

[Ewi : Kv] = dimKv Ewi = rankOKvOEwi = dimOKv/pv

(OEwipvOEwi

).

Notice that OKv/pv ≡ OK/p = Fp.Hence, looking at

OEwipvOEwi

as a Fp vector space we have∣∣∣∣∣ OEwipvOEwi∣∣∣∣∣ = |Fp|[Ewi :Kv ].

On another hand,



OEwipvOEwi

= lim←−j

OEpOE + qji

,

but for j sufficiently large, namely for j > eqi/p , the ideal pOE+qji =

g∏t=1

qeqt/pt +q

ji = q

eqi/pi . Subsituting

this in the limit above, one gets thatOEwipvOEwi

=OEqeqi/pi

.

Now, the quotient∣∣∣∣∣ OEwipvOEwi∣∣∣∣∣ =

∣∣∣∣∣ OEqeqi/pi∣∣∣∣∣ = N (qeqi/pi ) = (N(p))eqi/p∗fqi/p = |Fp|eqi/p·fqi/p

sothe proof is complete.

Proposition 13. Using the same notations, let α ∈ E be a primitive element, i.e E = K(α) and f(X) ∈K[X] be the minimal polynomial of α. Factor f(X) =

s∏i=1

fi(X) as a product of irreducible polynomials in

the completion Kv[X]. Then s = g and, after relabeling, fi(X) is the minimal polynomial of α ∈ Ewi overKv. Moreover, there is an isomorphism

E ⊗K Kv ∼−→g∏i=1

Ewi .

Proof. Observe that for each i ∈ {1, . . . , g} we have Ewi = Kv(α). The reasons for this are the following.Kv(α) = Kv · E, as subfields of Ewi . Kv(α) is a finite dimensional Kv vector space, therefore Kv(α) iscomplete in the natural topology. It is dense and complete inside Ewi , therefore Kv(α) = Ewi .

There is a natural continuous map ρ : E⊗KKv −→g∏i=1

Ewi , which is given by the product of the natural

maps E −→g∏i=1

Ewi and Kv −→g∏i=1

Ewi .

We have an isomorphsim

E ⊗K Kv ≡ Kv[X]/(f(X)) ≡s∏i=1

Kv[X]/(fi(X))

By the definition of the tensor product, dimKv E ⊗K Kv = dimkE = [E : K].

We saw in the previous problem that dimKvg∏i=1

Ewi =g∑i=1

dimKv Ewi =g∑i=1

eqi/p · fqi/p = [E : K].

Since the dimensions of the codomain and domain of ρ are the same, it is enough to prove that ρ is

injective.

For this, take an element α ∈ E ⊗K Kv be an element in the kernel of ρ. Since E is dense in E ⊗K Kv,we can find a sequence (αj)j≥1 of elements in E, such that αj → α in E ⊗K Kv. Since ρ is continous,

rho(αj)→ 0 ing∏i=1

Ewi , in other words αj → 0 in each Ewi . But this means that |αj |wi → 0 as j →∞.

This tells us that the fractional ideal αjOE is divisible by higher and higher powers of the primes qi, forevery i, as j → ∞. Replacing the sequence (αj)j≥1 by a subsequence, we can assume that the fractionalideal αjOE is divisible by pjOE , for every j ≥ 1.

Now, if we let π ∈ p \ p2, each αj = πjβj where βj ∈ E is integral at all the primes q1, . . . , qg, i.e.βj ∈ OEwi for all i ∈ {1, . . . , g}. But this implies that αj → 0 in our topology on E, hence α = 0.

Thus, ρ is an isomorphism from



E ⊗K Kv ≡s∏i=1

Kv[X]/(fi(X)) ∼−→g∏i=1

Ewi .

This implies that s = g and that ρ is a product of isomorphism Kv[X]/(fi(X)) ∼−→ Ewi , at least afterrelabeling the fi(X).

Now, suppose E/K is a Galois extension of number fields, v is a finite place of K that corresponds to

the prime ideal p ⊂ OK , and w corresponding to q are the finite place and its corresponding prime ideal ofE lying above v and p. Denote again by G = Gal(E/K).

The decomposition group Dq/p = StabG(q) leaves | · |w invariant, so by passage to completion we get

Dq/p ↪→ AutKv (Ew),

which is injective, because E ↪→ Ew.

Proposition 14. The extension Ew/Kv is Galois and there is an induced isomorphism Dq/p ∼−→ Gal(Ew/Kv).

Proof. Let D ⊂ AutKv (Ew) be the image of Dq/p. By Galois Theory, Ew/EDw is Galois of Galois group D.We have the tower of extensions Ew/E

Dw /Kv and [Ew : Kv] = eq/p · fq/d. But, since the eq/p · fq/d =

|Dq/p = |D| = [Ew : EDw ], we have that EDw = Kv.

Example Let f(X) = X5 + 5X + 5 ∈ Z[X]. Let E be the splitting field of f over Q and K = Q(α) whereα is a root of f in E. f is Eisenstein at 5, so if v5 is a place of K above 5, then Kv5/Q5 is totally ramifiedof degree 5.

If w5 is a place of E above 5, then 5 = ev5/5 divides |Dw5/5|, hence divides the order of G, whereG = Gal(E/Q) ⊂ S5. By Cauchy’s theorem for groups, G must contain a 5-cycle.

In F2[X], f(X) factors into irreducibles as f(X) = (X2 + X + 1)(X3 + X2 + 1). If w2 is a place of Eabove 2, then Frobw2/2 ∈ Dw2/2 ⊂ G ⊂ S5. The Frobenius has cycle type (2)(3) so G = S5.

For each number field K, denote by MK the set of places of K, i.e. the set of equivalence classes of

non-trivial absolute values of K. For each place v, fix | · |v ti be a distinguished (representative) absolutevalue for this place. In what follows, we will always refer to this distinguished representative.

If v is an infinite place corresponding to τ : K ↪→ R, then we take |x|v = |τ(x)|, for all x ∈ K. If v is aninfinite place corresponding to τ, τ : K ↪→ C, then we take |x|v = |τ(x)|2 ( the square of the ”usual” absolutevalue on C).

On the other hand, if v is a finite place corresponding to a non-zero prime ideal p ⊂ OK , then we take| · |v to be the absolute value |π|v = (N(p))−1, where π ∈ p \ p2.

Lemma 15 (Product formula). For all x ∈ K \ {0}, we have the equality∏

v∈MK|x|v = 1.

Proof. Exercise.

Different and discriminant

If E/K is an extension of number fields, then we are going to definte the different ideal DE/K , an ideal of

OE which ”measures the ramification” of E/K.We begin with the abstract setting, where R is a Dedekind domain, K = Frac(R), E/K is just a finite

separable extension and S is the integral closure of R in E.

Strangely, the natural way to introduce the different ideal si by introducing its inverse first.

Different and discriminant continued on next page. . . Page 12 of 49


Definition The inverse different D−1S/R = {x ∈ E|trE/K(xS) ⊂ R}. As one can guess, the different ideal is

DS/R =(D−1S/R

)−1.

Let us first see that DS/R is a well-defined, non-zero ideal of S. For this, let e1, . . . , en be a K-basis of

E such that S ⊇n⊕i=1

R · ei and let f1, . . . , fn be a dual basis with respect to e1, . . . , en and the bilinear form

B(x, y) = trE/K(xy), which is always non-degenerate, because E/K is separable.

By the choice of our basis e1, . . . , en, we can observe that the inverse different

D−1S/R ⊆ {x ∈ E|trE/K(x · ei) ∈ R,∀i ∈ {1, . . . , n}} =n⊕i=1

R · fi.

It follows from the above that D−1S/R is a finitely generated R-module, hence a finitely generated S−module. We have that S ⊆ D−1S/R, since trE/K(S) ⊆ R. We just proved that the inverse different is afractional ideal of E which contains S, hence taking inverses we get that DS/R ⊆ S is an ideal of S.

Proposition 16. Let α ∈ S such that S = R[α] and let f(X) ∈ R[X] be the minimal polynomial of α overR. Then, DS/R = f

′(α) · S, where f ′ is just the polynomial derivative of f .

Proof. Let us factor f(X) = (X − α)(bn−1Xn−1 + · · ·+ b0) in S[X]. We have S =n−1⊕i=0

R · αi.

Claim. The dual K- basis of 1, α, . . . , αn−1 with respect to the trace-pairing form B defined before is{b0

f ′(α), . . . ,

bn−1f ′(α)

}.

Proof of claim. Let Egal/E be a Galois closure of E and let τ1, . . . , τn : E ↪→ Egal be the distinctK-embeddings of E. Notice that there must be n = [E : K] of them using Galois Theory. Let us denote by

α1 = τ1(α), . . . , αn = τn(α).

We can use the formula

n∑k=1

f(X)

(X − αk)·

αjkf ′(αk)

= Xj ,

for all j ∈ {0, 1, . . . , n − 1}. To convince yourself that this is true, notice that both of the sides arepolynomials of degree less than n, which have the same values at X = α1, . . . , αn.

Therefore,

Xj =

n∑k=1

f(X)

(X − αk)·

αjkf ′(αk)

=

n∑k=1

τk

(f(X)

X − α· α

j

f ′(α)

)=

n∑k=1

τk

((bn−1X

n−1 + · · ·+ b0) ·αj

f ′(α)

)=

=

n−1∑i=0

trE/K

(biα

j

f ′(α)

)Xi.

Comparing powers of X on each side, we get that trE/K

(biα

j

f ′(α)

)= δi,j , which ends the proof for our

claim.

To deduce the proposition, use the fact that bn−1 = 1 and bi − αbi+1 = ai+1, hence αi−1 = bn−1 +∑1≤j


D−1S/R =

n−1⊕i=0

R · bif ′(α)

=1

f ′(α)

n−1⊕i=0

R · bi = f ′(α)−1S.

Inverting, we get that DS/R = f′(α)S ⊂ S.

In what follows, we are goint to discuss a transitivity property of the different. Let us take another

finite and separable extension L/E, with T , being the integral closure of S in L, which is the same as

the integral closure of R in L, by the transitivity of integral closure. Just to recall, L/E/K with T/S/R,

where S, T are the integral closures of R in E and L respectively.

Lemma 17. In the hypothesis described above, we have DT/R = DS/R ·DT/S regarded as ideals of T .

Proof. For any fractional ideal a ⊆ E, we have a ⊆ D−1S/R if and only if trE/K(a) ⊆ R. The if part of theabove holds by definition, because trE/KD

−1S/R ⊆ R and the only if part is because x ∈ a implies xS ⊆ a, so

trE/K(xS) ⊆ R which means that x ∈ D−1S/R.As a consequence, D−1S/R is the largest fractional ideal I of E such that trE/K(I) ⊆ R.Now, let b be a fractional ideal of L. We have that

b ⊆ D−1T/S ⇔ trL/E(b) ⊆ S ⇔ D−1S/RtrL/E(b) ⊆ D

−1S/R

We got the last inclusion by multiplying both sides with D−1S/R, which is a fractional ideal of E. The

chain of equivalences can continue with

D−1S/RtrL/E(b) ⊆ D−1S/R ⇔ trE/K

(D−1S/RtrL/E(b)

)⊆ R⇔ trL/K

(D−1S/R · b

)⊆ R,

which is equivalent to D−1S/R · b ⊆ D−1T/R. Since b was chosen arbitrary, we proved in fact that D

−1T/S =

DS/R ·D−1T/R, i.e. DT/R = DS/R ·DT/S .

Now suppose E/K is an extension of number fields. We will write DE/K instead of DOE/OK ⊆ OE . IfL/E is another number field extension, then DL/K = DL/E ·DE/K as ideals of OL, by the lemma we justproved.

Remark An important observation is the fact that DE/K can be calculated locally as we can see from the

following lemma.

Lemma 18. Let w be a finite place of E that lies above the place v of K. Then, DE/KOEw = DOEw/OKv .

Proof. Let w = w1, . . . , wg be the finite places of E that lie above v. If α ∈ E, then

trE/K(α) =

g∑i=1

trEwi/Kv (α).

Notice that in the equality above LHS ∈ K, RHS ∈ Kv and we have K ⊆ Kv. The equality holds,because trE/K(α) = trK (x · α : E → E) and E ⊗K Kv =

g∏i=1

Ewi .

We are first going to show that D−1E/K · OEw ⊆ D−1OEw/OKv

.

Let x ∈ D−1E/K ·OEw and y ∈ OEw , in order for x to be in the inverse of the different ideal that we desire,we have to check that trEw/Kv (xy) ∈ OKv .

For all � > 0 there exists z ∈ OE such that



|z − y|w < �and

|z|wi < �,∀i ∈ {2, . . . , g}

where we just used the Chinese Remainder Theorem.

Now we have trE/K(xz) ∈ OK ⊆ OKv . This means that

1 ≥ |trE/K(xz)|v =

∣∣∣∣∣g∑i=1

trEwi/Kv (xz)

∣∣∣∣∣ =∣∣∣∣∣trEw/Kv (xy) + trEw/Kv (x(z − y)) +

g∑i=2

trEwi/Kv (xz)

∣∣∣∣∣v

.

The local trace maps are continuous, so for � > 0 sufficiently small, we are going to get that |trEw/Kvx(z−y)|v ≤ 1 and |trEwi/Kv(xz)|v ≤ 1 for all i ∈ 2, . . . , g and by the use of ultrametric inequality it follows that|trEw/Kv (xy)|v ≤ 1. But this just means that trEw/Kv ∈ OKv . y was chosen arbitrary, so by previousremarks we have that x ∈ D−1OEw/OKv so we managed to prove one inclusion.

Now we are showing the reverse, namely D−1OEw/OKv⊆ D−1E/K · OEw . Using the Chinese Remainder

Theorem and the fact that OEw is a PID, we can choose x ∈ E such that (x)OEw = D−1OEw/OKv and|x|w′ ≤ 1 for all the remaining finite places w′ 6= w of E above v. It is enough to show that x ∈ D−1E/K or,equivalently that trE/K(xOE) ⊆ OK .

Let y ∈ OE be arbitrary. It is enough to show that for every finite place v′ of K, |trE/K(xy)|v′ ≤ 1,because we can characterise OK = {α ∈ K : |α|v′ ≤ 1 for all finite places v′ ∈MK}.

If v′ 6= v, then

|trE/K(xy)|v′ = |∑

w′|v′ and w′∈ME

trEw′/Kv′ (xy)|v′ .

By the way we chose it, x ∈ OE′w for every w′ 6= w, which implies that |trE/K(xy)|v′ ≤ 1, using ultrametric

inequality in the relation above.

If v′ = v, then

|trE/K(xy)|v =

∣∣∣∣∣g∑i=1

trEwi/Kv (xy)

∣∣∣∣∣v

=

∣∣∣∣∣trEw/Kv (xy) +g∑i=2

trEwi/Kv (xy)

∣∣∣∣∣v

.

But trEwi/Kv (xy) ∈ OKv for all i ∈ {2, . . . , g}, hence using the ultrametric inequality, we get that|trE/K(xy)|v ≤ sup

(1, |trE/K(xy)|v

)= 1, because x ∈ D−1OEw/OKv by hypothesis.

Theorem 19. Let w be a finite place of E, corresponding to a prime ideal q ⊆ OE, that lies above the finiteplacev of K which corresponds to the prime p ⊆ OK . If we denote by e := eq/p, the ramification index, wehave that qe−1 divides DE/K . In addition, q

e divides DE/K if and only if e ∈ q ∩ Z, i.e. e is divisible bythe residue characteristic of q.

Remark In particular, q - DE/K if and only if e = 1, i.e. q is unramified over p.

Proof. First assume that e = 1, i.e. Ew/Kv is unramified. Choose a primitive element α ∈ Fq of theextension Fq/Fp and let α ∈ OEw be any lift of α. Let f(X) ∈ OEw [X] be the minimal polynomial of α overKv. From the theory of p− adic extensions it follows that OEw = OKv [α] and f(X) = f(x) (mod p) is theminimal polynomial of α over Fp.

We know that DOEw/OKv = f′(α)OEw . But f ′(α) (mod qw) = f

′(α) 6= 0 in Fq, as f(X) has distinct

roots. Thus, f ′(α)OEw = OEw , so DE/KOEw = OEw , therefore q - DE/K .



In general, there exists an intermediate field Ew/L/Kv such that L/Kv is unramified and Ew/L is

totally ramified. If π ∈ OEw is a generator of qw ⊆ OEw then OEw = OL[π] and the minimal polynomialg(X) ∈ OL[X] of π over L is Eisenstein.

I we denote by qL ⊆ OL the unique maximal ideal,

g(X) = Xe + a1Xe−1 + · · ·+ ae, where ai ∈ qL,∀i ∈ 1, . . . , e and also ae ∈ qL \ q2L.

NowDOEw/OL = g

′(π)OEwg′(π) (mod πe) = e · πe−1 (mod πe)

}⇒ DOEw/OL is divisible by π

e−1

and DOEw/OL is divisible by πe if and only if e ∈ qw. We then use transitivity of the different to get that

DOEw/OKv = DL/OKv ·DOEw/OL . Now if we look at the particular case e = 1 we proved at the begining,we see that DL/OKv = OL, hence we get the result.

Remark A trivial observation would be that if q ⊂ OE is a non-zero prime ideal, then q|DE/K ⇔ q isramified over K.

Definition Let K be a number field, n = [K : Q] and α1, . . . , αn ∈ K is a Q-basis of K. If M =n⊕i=1

Z[αi]

then we define the discriminant of M to be disc(M) = det2[A(α1, . . . , αn)], where A(α1, . . . , αn) is the n×nmatrix with entries τi(αj). Here τ1, . . . , τn : K ↪→ C are the n distinct embeddings of K into C. In theparticular case M = OK , we write dK = disc(OK).

Proposition 20. 1.) disc(M) only depends on M and not on the choice of its Z− basis α1, . . . , αn. Wealways have disc(M) ∈ Q and dK ∈ Z.

2.) If M ′ ⊆M is a finite index subgroup, then disc(M ′) = [M : M ′] · disc(M).3.) If K = Q(α) and α ∈ OK , f(X) ∈ Z[X] is the minimal polynomial of α, then

disc(Z[α]) = (−1)n(n−1)

2 NK/Q (f′(α)) .

Proof. 1.) Let α1, . . . , αn be a Z− basis of M . If we permute the embeddings τ1, . . . , τn : K ↪→ C then webasically permute the rows of A(α1, . . . , αn), which does not change the square of its determinant.

If β1, . . . , βn is another choice of Z- basis for M . Write βj =n∑i=1

bijαi, for all j ∈ {1, . . . , n}. The matrix

B := (bij)1≤i,j≤n has determinant ±1 because it is integral and invertible, hencedet2[A(β1, . . . , βn)] = det

2[A(α1, . . . , αn)] · det2(B) = det2[A(α1, . . . , αn)]. This shows that disc(M)depends only on M .

LetKgal be the compositum of τ1(K), . . . , τn(K) as subfields of C. ThenKgal/Q is Galois andA(α1, . . . , αn)has entries inKgal. Acting onA(α1, . . . , αn) by an element ofGal(K

gal,Q) permutes the rows ofA(α1, . . . , αn)so it fixes the square of its determinant. Therefore disc(M) ∈ Fix(Gal(Kgal/Q)) = Q.

If M = OK then A(α1, . . . , αn) has entries in OKgal , hence disc(M) ∈ Q ∩ OKgal = Z, so the first partof the proposition is proved.

2.) Let α1, . . . , αn be a Z− basis of M and α′1, . . . , α′n a Z- basis of M ′.Write α′j =

n∑i=1

bijαi for all j ∈ {1, . . . , n}. Denote B := (bij)1≤i,j≤n.

Since the values of disc(M ′) and disc(M) are invertible under invertible integral transformations, we

may use the Smith normal form to suppose that B is a scalar matrix, with integral entries. It is easy to see

that those integral on the diagonal of B multiply together to give [M : M ′].

Hence det2[A(α′1, . . . , α′n)] = det

2[A(α1, . . . , αn)] · det2(B), which is equivalent to disc(M ′) = disc(M) ·[M : M ′]2.

3.) disc(Z[α]) = det2[A(1, α, . . . , αn−1]. Denote by αi := τi(α).



Then we have a Vandermonde determinant,

det[A(1, α, . . . , αn−1)]2 =∏i


Restricted direct products and the adele ring

Let I be a set and S ⊆ I a finite subset. Suppose we are give the following data:

1. for all i ∈ I, a topological group Gi;

2. for all i ∈ I \ S, an open subgroup Hi ⊆ Gi.

In this situation, we define the restricted direct product

∏i∈I

′Gi =

{(gi)i∈I ∈

∏i∈I

Gi∣∣gi ∈ Hi for all but finitely many i ∈ I \ S} .

We topologize∏i∈I

′Gi as follows. For every finite set T ⊆ I such that S ⊆ T , define

G(T ) =∏i∈T

Gi ×∏i∈I\T

Hi,

endowed with the product topology.

Then we have∏i∈I

′Gi =⋃

T⊇S,T−finiteG(T ) as sets, and we declare U ⊆

∏i∈I

′Gi to be open, exactly when

U ∩G(T ) is open for all T as above.

Remark 1.∏i∈I

′Gi depends on the choice of Hi, but we ommit these from the notation.

2. If Gi are all locally compact groups and Hi ⊆ Gi are open compact subgroups then∏i∈I

′Gi is still

compact (satisfy Pontryagin duality, etc).

In what follows, let K be a number field and let S∞ ⊂MK be the set of infinite places of K.

Definition The adele ring of K is AK =∏

v∈MK

′Kv, taken with respect to the open subgroups OKv ⊆ Kv

for all v ∈MK \ S∞.

Remark OKv = {x ∈ Kv∣∣|x|v ≤ 1} is an open compact subring, hence the adele ring AK is compact.

If T ⊆MK is a finite set containing S∞ then AK(T ) =∏

v∈MK\TOKv ×

∏v∈T

Kv. Obviously, AK(T ) ⊂ AK

is an open subset and we have AK =⋃

T⊇S∞,T−finiteAK(T ).

There is a natural embedding, induced by the diagonal map, K ↪→ AK , as for all x ∈ K we have |x|v ≤ 1for all but finitely many places v ∈ MK . This way we can identify K with a subset of AK and we call theelements of AK that are in the image of this map the principal adeles. We will write a general elementx ∈ AK as x = (xv)v ∈MK .

Theorem 22. The subspace topology on K ⊂ AK is the discrete topology and AK/K is compact.

Proof. (AK ,+) is a topological group, hence the translations are homeomorphisms. To show that K has thediscrete topology, it is enough to show that there is a neighbourhood of 0 in AK which contains no otherpoint of K.

Let us write

U = {(xv)v ∈ AK(S∞)|∀v ∈ S∞, |xv|v < 1} .

This is an open set of AK(S∞) =∏

v−finiteOKv ×

∏v−infinite

Kv, hence an open set of AK .

Restricted direct products and the adele ring continued on next page. . . Page 18 of 49


If x ∈ K ∩ U , then |x|v ≤ 1 for all finite places of K. Therefore, x ∈ OKv for all finite places of K andhence NK/Q(x) = m ∈ Z.

We have that |m|∞ =∏

v∈S∞|x|v < 1 ⇒ m = 0 ⇒ x = 0. Hence K ∩ U = {0}. Therefore, the subspace

topology on K ⊂ AK is the discrete topology.To prove that AK/K is compact, it is enough to find a compact subset W ⊆ AK such that the composite

W → AK → AK/K is surjective.Claim: AK = K + AK(S∞). In other words, given any adele (xv)v, there exists an element x ∈ K such

that each xv − x is on OKv .To see that, choose x = (xv)v∈MK ∈ AK and let m ∈ Z be such that |mxv|v ≤ 1 for all finite places

v ∈MK .Let � > 0. By the Chinese Remainder Theorem, we can find an element α ∈ OK such that |mxv−α|v < �

for all the finite places v of K for which |m|v < 1. Then, we observe that |x− αm |v = |m|−1v · |mx− α|v for

all v ∈MK .If v ∈ MK \ S∞ such that |m|v = 1, then since mx and α are in OKv , we have that mx − α is in OKv

and hence |mx− α|v = |x− αm |v ≤ 1.If v ∈MK \ S∞ such that |m|v < 1, then |x− αm |v < |m|

−1v · �.

Therefore, for � sufficiently small, we get an α ∈ OK and m ∈ Z such that |x − αm |v ≤ 1 for all finiteplaces v of K. Therefore, x ∈ αm + AK(S∞), hence this is the end of the proof for our claim.

As a consequence, by the second isomorphism theorem AK/K ≡ AK(S∞)K∩AK(S∞) ≡ AK(S∞)/OK .Now observe that

K ⊗Q R ≡∏v∈S∞

Kv ≡ Rr1 × Cr2 ,

where r1 is the number of real embeddings of K and r2 is the number of pairs of complex embeddings.

The above is a R vector space of dimension r1 + 2r2 = [K : Q]. It has a free abelian subgroup OK of rank[K : Q] which spans over R.

As a consequence, the quotient (K ⊗Q R) /OK is homeomorphic to a product of [K : Q] circles. Inparticular, we can find a compact subset W∞ ⊆ K ⊗Q R such that the composite

W∞ → K ⊗Q R→ (K ⊗Q R) /OK

is surjective.

Now, take W =∏

v∈MK\S∞OKv ×W∞ ⊆ AK(S∞). This is compact, since it is a product of compact sets.

Moreover, the composite

W → AK(S∞)→ AK(S∞)/OK ≡ AK/K

is surjective.

Lemma 23 (The adelic Minkowski lemma). Let x = (xv) ∈ A×K be a unit in the adele ring, i.e. |xv|v = 1for almost all places v. There exists C = CK > 0 such that if

∏v∈MK

|xv|v > C then there exists a y ∈ K×

such that |y|v ≤ |xv|v for all v ∈MK .

Proof. If x = (xv) ∈ A×K , let L(x) = {y ∈ K×∣∣|y|v ≤ |xv|v,∀v ∈ MK}. Observe that since by the product

formula, we have∏

v∈MK|xv|v =

∏v∈MK

|α · xv|v for all α ∈ K×, there is a bijection L(x) ∼−→ L(αx) sending

y 7→ αy for all α ∈ K×.Hence, we are free to replace x by α · x for any α ∈ K×.Let α1, . . . , αn be a Z- basis of OK . For each v ∈ S∞, let Cv := sup

i=1,...,n|n · αi|v.



We will prove the lemma for C := 2r1+r2∏v∈S∞

Cv , where r1 is the number of real embeddings and r2 is

the number of pairs of complex embeddings of the number field K.

Let x = (xv)v ∈ A×K be as in the lemma, with∏

v∈MK|xv|v > C. By the weak approximation lemma

(Remark 3, on page 8), we find α ∈ K× such that ∀v ∈ S∞,

Cv < |αxv|v < 2 · Cv.

We can also find an integer a ≥ 1 such that

|a · xv|v ≤ 1 for all v ∈MK \ S∞.

Let y := a · α · x and since a · α ∈ K×, we saw that L(x) ≡ L(y).Now

aCv < |yv|v < 2aCv for all real places v,

a2Cv < |yv|v < 2a2Cv for all complex places v

and

|yv|v ≤ 1 for all finite places v of K.

Let M =

{n∑i=0

aiαi|ai ∈ {0, 1, . . . , a}}⊆ OK be a subgroup of cardinality (a+ 1)n.

A consequence of the construction of y = (yv)v∈MK is that if (a + 1)n >

∣∣∣∣∣ ∏v∈MK\S∞OKv/yvOKv∣∣∣∣∣ (†),

then we can find distinct elements β, γ ∈M such that they have the same image in∏

v∈MK\S∞OKv/yvOKv .

Notice that since |yv|v ≤ 1 for all finite places v, the quotient OKv/yvOKv makes sense. If |yv|v = 1, as it isthe case for almost all v ∈MK \ S∞ we get that the quotient OKv/yvOKv is trivial. So, in fact the product∏v∈MK\S∞

OKv/yvOKv is a finite product.

If (†) hols, let δ := β − γ ∈ OK \ {0}, so δ =n∑i=1

biαi, where bi ∈ {−a, . . . , a}. If v is finite, then

δ ∈ yvOKv ⇒ |δ|v ≤ |yv|v. If v is a real place, then |δ|v ≤ n · a · supi=1,...,n

|αi|v = a · Cv < |yv|v. Similarly, if v

is a complex place, then |δ|v ≤ n2 · a2 · supi=1,...,n

|αi|v = a2 · Cv < |yv|v. Therefore, δ ∈ L(y) ≡ L(x).

It remains to show that if∏

v∈MK|xv|v > C, then the inequality (†) holds. By definition,∣∣∣∣∣∣∏

v−finite

OKv/yvOKv

∣∣∣∣∣∣ =∏

v−finite

|yv|−1v .

So (†) holds if an∏

v−finite|yv|v > 1. Now observe that an

∏v−finite

|yv|v =∏

v−infinite|a|v ·

∏v−finite

|a·α·xv|v =∏v−finite

|αxv|v, by the product formula.∏v−finite

|α · xv|v > 1⇔∏

v∈MK

|α · xv|v >∏

v−infinite

|α · xv|v ⇔

by the product formula,



∏v∈MK

|xv|v >∏

v−infinite

|αxv|v.(††)

If we use now that |αxv|v < 2 ·Cv for every infinite place v, we get that LHS < 2r1+r2 ·Cv = C. But weknow that

∏v∈MK

|xv|v > C so, (††) and hence (†) are indeed true.

Definition The idele groups is IK :=∏

v∈MK

′K×v , where the restricted direct product is taken with respect

tot the open compact subgroups O×Kv ⊆ K×v for all v ∈MK \ S∞.

This is indeed a group, consisting of elements of the form α = (αv)v∈MK ∈ AK such that αv 6= 0 for allv and αv is a unit for all but finitely many v. In particular, this imples |αv|v = 1 for all but finitely many v.

Although IK is a subset of AK , we must not give it the supspace topology. This is, because the mapx 7→ x−1 from IK → IK is not continuous in that topology. For example, take K = Q and let (α(p)) ∈ IQbe the sequence of ideles with components α

(p)p = p and all other components 1. In the adelic toplogy (i.e.

subspace topology), α(p) → 1 as p→∞ but (α(p))−1 does not tend to a limit.We give to IK the topology induced by the restricted direct product, i.e. the topology induced by

regardint IK as a group of operators on the additive group AK : that is, a base for the open sets in IK isgiven by

∏v∈MK

Uv where each Uv is open in K×v and Uv = O×Kv for all but finitely many v.

This topology is stricly finer than the subspace topology, so the inclusion IK ↪→ AK and the multiplicationmap IK × AK → AK are continuous, and with it IK is a locally compact topological group.

Remark One can check that the embedding IK = A×K ↪→ AK × AK given by x 7→ (x, x−1) does give thecorrect topology on IK that we described above.

There is a map || · ||K : IK → R>0 given by x = (xv)v∈MK 7→∏

v∈MK|xv|v. Since only finitely many terms

in the product are not 1, there is no issue with the convergence.

Lemma 24. 1.) || · ||K is a continuous, surjective group homomorphism.2.) If we let I1K = ker || · ||K , a closed subgroup of IK , then the map I1K ↪→ AK is a homeomorphism onto

a closed subset of AK .

Proof. 1.) It suffices to show that ||·||K is a continous surjective homomorphism after restriction to subgroupsIK(T ) =

∏v∈T

K×v ×∏

v∈MK\TO×Kv for finite sets S∞ ⊆ T ⊆MK .

If x = (xv)v∈MK ∈ IK(T ), then ||x||K =∏v∈T|xv|v, so this is clearly a continous homomorphism. It is

surjective since the maps

| · |v : Kv −→ R>0 are surjective for all v ∈ S∞.

2) At first, let us observe that if v ∈MK \ S∞ and x ∈ Kv satisfies |x|v < 1, then

|x|v ≤ (N(p))−1 ≤1

2.

For any x = (xv)v∈MK ∈ AK there are only finitely many v ∈MK such that |xv|v > 1, hence the product∏v∈MK

|xv|v converges. If there are infinitely many v ∈MK such that |xv|v < 1 then it converges to 0 and if

there are finitely many x ∈MK such that |xv| < 1 then the product converges to a non-zero value providedxv 6= 0 for all v, as it happens in the case x ∈ IK .

In particular, if x ∈ AK satisfies∏

v∈MK|xv|v = 1 then we get automatically that x ∈ IK , and actually

x ∈ I1K . Let us show that I1K is closed in AK by showing that AK \ I1K is open.



Let x ∈ AK \ I1K and we will construct an open neighbourhood W of x in AK \ I1K . Since x 6∈ I1K , i.e.∏v∈MK

|xv|v 6= 1 we can split the rest of the proof in two cases.

Case 1. If∏

v∈MK|xv|v < 1, let S ⊇ S∞ be a finite set of places of K containing all places v such that

|xv|v > 1 and such that∏v∈S|xv|v < 1. For � > 0, define an open neighbourhood W� of x in AK \ I1K by

W� :=

{y = (yv)v ∈ AK

∣∣∣∣ |xv − yv|v < �,∀v ∈ S|yv|v ≤ 1,∀v 6∈ S}.

Now, if we take a point y ∈W�, the product∏v∈MK

|yv|v ≤∏v∈S|yv|v ≤

∏v∈S

(|xv|v + �) ,

and since S is finite, we can choose � sufficiently small so that∏v∈S

(|xv|v + �) < 1. Therefore, W := W� ⊆

AK \ I1K is an open neighbourhood of x.Case 2. If

∏v∈MK

|xv| > 1, denote∏v∈MK |xv|v = C > 1. Let S ⊇ S∞ be a finite set of places of K

containing all the places such that |xv|v 6= 1 and all the places v such that the corresponding prime ideal pof v has norm N(p) ≤ 2C.

The construction of S implies in particular that if v 6∈ S, then the corresponding ideal p has normN(p) > 2C. (Why there are only finitely many?)

For � > 0 define again

W� :=

{y = (yv)v ∈ AK

∣∣∣∣ |xv − yv|v < �,∀v ∈ S|yv|v ≤ 1,∀v 6∈ S}.

Since∏

v∈MK|xv|v =

∏v∈S|xv|v = C > 1, we can choose � sufficiently small such that for all y ∈W�,

1 <∏v∈S

(|xv|v − �) ≤∏v∈S|yv|v ≤

∏v∈S

(|xv|v + �) < 2C.

Then, W := W� is an open neighbourhood of x ∈ AK and we are just about to prove that W ⊆ AK \ I1K .To complete the proof, let y ∈W and consider two subcases.

Case 2.a. If |yv|v = 1 for all v 6∈ S, then the product∏

v∈MK|yv|v =

∏v∈S|yv|v > 1 and hence y 6∈ AK \ I1K .

Case 2.b. If there exists v0 6∈ S such that |yv0 |v0 < 1 then, by the observation we made at the beginingand by the construction of S, we get that |yv0 |v0 ≤ 1N(p0) <

12C , where p0 is the prime corresponding to the

finite place v0. Then,

∏v∈MK

|yv|v ≤

(∏v∈S|yv|v

)· |yv0 |v0 <

2C

2C= 1,

hence y ∈ AK \ I1K .This shows that I1K is a closed subset of AK .It remains to check that the subspace topology on I1K ⊆ AK agrees with the natural topology on I1K .AK is a topological ring, therefore multiplication by elements of A×K acts on AK by homeomorphisms.

Knowing this, it is enough to show that for all open neighbourhoods U of 1 ∈ I1K , there exists an open subsetW ⊆ AK such that 1 ∈W and W ∩ I1K ⊆ U .

We are free to shring U , so we can assume it is of the form

U =

{y = (yv)v ∈ I1K

∣∣∣∣ |yv − 1|v < �,∀v ∈ S|yv| = 1,∀v 6∈ S}



for some � > 0 and some finite set S ⊇ S∞.After shrinking � > 0 we can assume that

∏v∈S|yv|v < 2 for all y ∈ U . Take now

W :=

{x = (xv)v ∈ AK

∣∣∣∣ |xv − 1|v < �,∀v ∈ S|xv|v ≤ 1,∀v 6∈ S}

Claim: W ∩ I1K = U . Indeed, the inclusion U ⊆ W ∩ I1K can be seen from the construction of the sets.For the other inclusion, let y ∈W ∩ I1K . Then, either we have |yv|v = 1 for all v 6∈ S, in which case y ∈ U orthere exits v0 6∈ S such that |yv0 |v0 < 1. But, by the observation at the begining, the later case is impossiblesince it would imply that |yv0 |v0 ≤ 12 which implies again that

∏v∈MK

|yv| ≤

(∏v∈S|yv|v

)· |yv0 |v0 < 2 ·

1

2= 1,

which contradicts y ∈ I1K .Hence the natural topology on I1K is the subspace topology coming from AK and using what we proved

so far, we can tell that I1K ↪→ AK maps homeomorphically into a closed subset of AK .We showed previously that AK/K is compact. Since by the product rule, x ∈ K× imples || · ||K = 1, we

have that K× ↪→ I1KThe idele norm, gives a surjection

IK/K× � R>0.

But we will see that the ”idele class” group IK/K×, endowed with the quotient topology is non-compact,but I1K/K× is compact.

Remark Observe that since the topology induced on K× as a subset of IK is finer than the one induced onit as a subset of AK , since the latter is already the discrete topology (Theorem 22.), K× is also a discretesubgroup of IK .

The point is that if we restrict to I1K we no longer have the nuisance of having two distinct inducedtoplogy, because the ones induced from AK and from IK coincide.

If W ⊆ AK is a compact subset, since W ∩ I1K ⊆ AK is closed we get that W ∩ I1K is compact in the AKtopology and therefore also in the I1K topology, since they coincide.

Theorem 25. I1K/K× is a compact topological group.

Proof. It is enough to find a compact subset W ⊆ AK such that the map W ∩ I1K → I1K/K× is surjective.By the adelic Minkowski lemma (Lemma 23), there exists C > 0 which depends only on K, such that

for any x = (xv) ∈ IK with idelic norm ||x||K > C, there exists α ∈ K× such that |α|v ≤ |xv|v for all placesv ∈MK .

Choose any x ∈ IK such that ||x||K > C and define W ={y = (yv)v ∈ AK

∣∣|yv| ≤ |xv|v,∀v ∈MK}. Thisis a closed and hence compact subset of AK .

Let θ ∈ I1K be any element. Then ||x · θ−1||K = ||x||K > C and hence, by adelic Minkowski lemma, thereexists α ∈ K× such that |α|v < |xv · θ−1v |v, for all v ∈MK .

This is equivalent to |α · θv| ≤ |xv|v, for all v ∈ MK , which means that α · θ ∈ W , in particularα · θ ∈W ∩ I1K . This proves that θ ∈ K× ·W ∩ I1K .

Since θ ∈ I1K was chosen arbitrary, the map W ∩ I1K → I1K/K× is surjective. Therefore, I1K/K× iscompact.

Theorem 26. 1.) The ideal class group Cl(OK) is finite.2.) The unit group O×K is a finitely generated abelian group with free part of rank r1 + r2 − 1.



Proof. 1.) Let I be the group of non-zero fractional ideals and P ⊆ I be the subgroup of principalfractional ideals of OK .

Endow I ,P and Cl(OK) = I /P with the discrete topology.We now define a homomorphism

IK → I(xv)v 7→ a

where a is the unique fractional ideal such that aOKv = xvOKv as fractional ideals of Kv for everyv ∈MK \ S∞.

This homomorphism is surjective and continuous. Its kernel contains the open subgroup

IK(S∞) =∏

v−finite

O×Kv ×∏

v−infinite

K×v .

As K× →P, passing to the quotient gives a continuous surjection IK/K× → I /P = Cl(OK).The map above is still surjective when restricted to I1K/K×.To see this, let K×∞ :=

∏v∈S∞

K×v . Then K×∞ ⊆ ker(IK → I ) and IK = K×∞ · I1K .

So, we found a continuous surjection I1K/K× → Cl(OK). But Cl(OK) was endowed with the discretetopology, hence it is discrete. I1K/K× is compact, therefore Cl(OK) is also compact. Hence Cl(OK) is finite.

2.) Define I1K(S∞) := I1K ∩ IK(S∞) = {x ∈ I1K∣∣|xv|v = 1,∀v 6∈ S∞}, which is an open subgroup of I1K .

Then O×K = K× ∩ I1K(S∞) is a discrete subgroup of I1K(S∞) and I1K(S∞)/O×K = I1K(S∞)/(K× ∩ I1K(S∞)) is

open in I1K/K×. Since it is a subgroup, it is also closed because its complement is just a union of distinctcosets, which are open. So, I1K(S∞)/O

×K is compact.

Let H ⊂ RS∞ denote the hyperplane such that all coordinates sum to 0. We define L : I1K(S∞)→ H bythe formula x 7→ (log |xv|v)v∈S∞ . It is easy to see that L is both continuous and surjective, since it factorsthrough the projection to the infinite components. It is also easy to see that L is proper (i.e. the pre-image

of a compact set is compact). This implies that ker(L)∩O×K is finite (since it is both compact and discrete)and L(O×K) is discrete (since its intersection with any compact set is finite and H is locally compact).

This implies the theorem: L(O×K) ⊂ H is discrete, hence a free Z-module of finite rank. The quotientH/L(O×K) is compact, being the image under L of I1K(S∞)/O

×K . This is only possible if L(O

×K) has rank

equal to the dimRH = r1 + r2 − 1. This concludes the proof.

Cyclotomic fields

Definition If m ≥ 1, the m-th cyclotomic field is defined as the splitting field of Xm − 1. We write it asQ(ζm), where ζm is any choice of primitive m−th root of unity.

We will study K := Q(ζm) when m = pr, for p an odd prime. Define Φ(x) = xpr−1

xpr−1−1, to be the

cyclotomic polynomial. Its roots are exactly the primitive pr−th roots of unity. From now on, we will justwrite ζ := ζpr .

Evaluating Φ at x = 1 we get that∏a∈(Z/prZ)×

(1− ζa) = Φ(1) = p, in Q(ζ)

If a, b ∈ (Z/prZ)× satisfy ab = 1 then{(1− ζa) = (1− ζ)(1 + ζ + · · ·+ ζa−1)(1− ζ) = (1− ζab) = (1− ζa)(1 + ζa + · · ·+ ζa(b−1))

Cyclotomic fields continued on next page. . . Page 24 of 49


therefore, as ideals{(1− ζ)OK

∣∣(1− ζa)OK and(1− ζa)OK

∣∣(1− ζ)OK ⇒ (1− ζ)OK = (1− ζa)OK for any a ∈ (Z/prZ)× ..

Now, using the identity∏

a∈(Z/prZ)×(1− ζa) = p we deduce that as ideals,

(1− ζ)φ(pr)OK = pOK ⇔ (1− ζ)p

r−1(p−1)OK = pOK ,

where φ above is the Euler-totient function. This implies that for any prime ideal p ⊆ OK that lies abovethe prime p, we have that ep ≥ pr−1(p− 1).

But it is known that Φ(x) is irreducible and K = Q(ζ) is obtained by adjoining a single root of Φ(x).Hence pr−1(p− 1) = deg Φ(x) = [K : Q] ≥ ep ≥ pr−1(p− 1) = deg Φ(x). Thus equality must hold above, soep = [K : Q], i.e. p is totally ramified and (1− ζ)OK is the unique prime ideal of OK that lies above p.

From Part II Galois Theory, we know that the map

Gal(K/Q)→ (Z/prZ)×

σ 7→ the unique a such that σ(ζ) = ζa

is an injective group homomorphism and since [K : Q] = # (Z/prZ)×, the above map is an isomorphism.

Proposition 27. We have that OK = Z[ζ], DK/Q = (1 − ζ)N and the discriminant |dK | = pN , whereN := pr−1 (r(p− 1)− 1).

Proof. We first calculate disc (Z[ζ]) = ±NmK/Q (Φ′(ζ)). Let u(X) = Xpr − 1 and v(X) = Xpr−1 − 1, so

Φ(X) = u(X)v(X) .

Using this notation, disc (Z[ζ]) = ±NmK/Q(u′(ζ)v(ζ)

). Since u′(ζ) = prζp

r−1, by the multiplicity of the

norm

NmK/Q (u′(ζ)) = NmK/Q (p

r) ·(NmK/Qζ

)pr−1= ±pr·[K:Q] = ±pr·p

r−1(p−1).

Now, v(ζ) = ζpr−1 − 1 = ζp − 1, where ζp is a primitve p-th root of unity.

We have a towerK = Q(ζ)

Q(ζp)

Q

pr−1

p−1

And therefore

NmK/Q (v(ζ)) = NmK/Q (ζp − 1) = NmQ(ζp)/Q (ζp − 1)pr−1

= ±ppr−1

,

because NmQ(ζp)/Q (ζp − 1) = ±p., thus

disc (Z[ζ]) = ±ppr−1(r(p−1)−1) = ±pN .

This shows that [OK : Z[ζ]] is a power of p. To show that this index is 1, it is enough to prove tat it iscoprime to p.

For this, let π = 1−ζ. Clearly Z[ζ] = Z[π] and observe that 1, π, . . . , πM , whereM = [K : Q] = (p−1)pr−1form a Q-basis for our number field K = Q(ζ).

Cyclotomic fields continued on next page. . . Page 25 of 49


It is easy to see that the natural map Z[π]→ OK/pOK has kernel pZ[π]. This means that pOK ∩Z[π] =pZ[π], so the quotient group OK/Z[π] has no elements of order p, therefore the index [OK : Z[π]] is coprimeto p and hence OK = Z[π] = Z[ζ].

Suppose that q is a prime of OK above a rational prime q 6= p. Then q - dK ⇒ eq/q = 1.This implies that the decomposition group Dq/q ⊆ Gal(K/Q) is cyclic, generate by the Frobq/q which is

characterised uniquely by its action on Fq, namely x 7→ xqfq/q

Lemma 28. 1.) Frobq = Frobq/q is independent on the choice of q ⊆ OK above q.2.) Frobq ∈ Gal(K/Q) ' (Z/prZ)× corresponds to the element q (mod pr).

Proof. 1.)This result holds for any Galois extension with abelian Galois group, since Frobσ(q)/q = σFrobq/qσ−1

for all σ ∈ Gal(K/Q). Now since Gal(K/Q) acts transitively on the set of primes q above q, the result follows.2.) We have an injection O×K [pr] ↪→ F×q .By definition, this is compatible with the action of Frobq/q on the left-hand side and the map x 7→ xq on

the right hand side. It follows that Frobq/q agrees with multiplication by q on O×K [pr], which is equivalentto saying that Frobq/q = q in Gal(K/Q).

Let p be an odd prime so that (Z/pZ)× is cyclic with (p−1) elements. It has a unique index two subgroup,formed by the quadratic residues. This shows that the field K = Q (ζp) has a unique quadratic subfield L.We know L is ramified only at p, since K is so, therefore we must have L = Q (

√p∗), where p∗ =

(−1p

)p,

by the calculation of discriminants of quadratic fields.

We will show that(p∗

q

)=(qp

)by looking at how q splits in Q (

√p∗).

By Kummer-Dedekind to say that q splits in Q (√p∗) if and only if p∗ is a quadratic residue modulo q.

On the other hand q splits in Q (√p∗) if and only if Dq/q ⊆ Gal

(Q(ζp)

/Q (√p∗)), if and only if q is a

quadratic residue modulo p.

Dirichlet series and zeta-functions

Definition∞∑n=1

ann−s, where an ∈ C and s ∈ C is a complex variable is called a Dirichlet series.

Following Riemann, we write s = σ+i·t, for σ, t ∈ R, thus the region ”σ > 0” is the set {s ∈ C|Re(s) > 0}.

Lemma 29. Let f(s) =∑nann

−s be a Dirichlet series.

1.) If the series converges absolutely for s0 = σ0 +i ·t0 ∈ C, then it converges absolutely for all s = σ+i ·tsuch that σ > σ0.

2.) If the series converges for some s0 = σ0 + i · t0 ∈ C, then it converges for all s = σ+ i · t with σ > σ0.

Summation by parts identity: Given a1, . . . , aN , b1, . . . , bN ∈ C, the following identity holdsN∑n=1

anbn = ANbN +N−1∑n=1

An(bn − bn+1), where An =n∑i=1

ai.

Proof. 1.) n−s = exp(−s · log n)⇒ |n−s|n−σ. If σ > σ0, then

∞∑n=1

|ann−s| =∞∑n=1

|an|n−σ <∞∑n=1

|an|n−σ0 =∞∑n=1

|ann−s0 |,

which converges by hypothesis.

Dirichlet series and zeta-functions continued on next page. . . Page 26 of 49


2.) Fix M > 0. For N > M , define PN (s) =N∑

n=M

an · n−s. We will show that for s = σ + i · t with

σ > σ0, the partial sum of∞∑n=1

an · n−s is Cauchy by estimating PN (s).

Let us write PN (s) =N∑

n=M

[an · n−s0 ] · [ns0−s] and apply the summation by parts identity.

PN (s) = PN (s0) ·Ns0−s +N−1∑n=M

Pn(s0)(ns0−s − (n+ 1)s0−s

).

By assumption, the series converges for s0, so the terms Pn(s0) above are bounded. We also have that

1

ns−s0− 1

(n+ 1)s−s0= (s− s0)

∫ n+1x=n

xs0−s−1dx.

All of the above, together with the triangle inequality gives that

|PN (s)| ≤ |PN (s0)| ·Nσ0−σ +N∑

n=M

|Pn(s0)||s− s0|nσ0−σ−1.

Since σ > σ0,∞∑n=1

nσ0−σ−1 is convergent, so which gives that limM→∞

|PN (s)| = 0, i.e. the partial sum of∞∑n=1

ann−s is Cauchy.

This proof also shows that if a Dirichlet series∞∑n=1

ann−s converges at s0, then it converges uniformly in

compact subsets in the region σ > σ0, as |s− s0| is bounded in compact sets.We define the{

σa = inf{σ ∈ R|f(σ) converges absolutely} − abscisa of absolute convergenceσc = inf{σ ∈ R|f(σ) converges} − abscisa of convergence

Lemma 29 says that f(s) converges absoltely in the region σ > σa and f(s) converges in the region

σ > σc.

Uniform convergence in compact subsets implies that f(s) converges to a holomorphic function in the

region σ > σc.

Remark If σa, σc are both finite, then σc ∈ [σa − 1, σa].σc ≤ σa is clear by definition. For the second inequality, note that if

∑ann

−σ converges, the terms

|ann−σ| are bounded. The absolute convergence of∑ann

−(σ+1+�) =∑ann

−σn−(1+�) follows from the

convergence of∑n−(1+�) and the product rule for series.

Lemma 30. Let f(s) =∞∑n=1

ann−s be a Dirichlet series with an ∈ R≥0, non-negative for all n ∈ N. Suppose

f(s) converges in σ > σ0 and admits an analytic continuation to a disc D(σ0, �), centered at σ0 of radius �.

Then f(s) converges for some s = σ + i · t with σ < σ0.

Proof. Let 0 < δ < �/2. Let c = σ0 + δ. We can write

f(s) =

∞∑n=1

ann−s =

∞∑n=1

an · n−c · nc−s =∞∑n=1

an · n−c · exp[(c− s) log n]

COMPLETE THIS PROOF LATER

Dirichlet series and zeta-functions continued on next page. . . Page 27 of 49


Examples of Dirichlet series 1.) L-series associated to a Dirichlet character. Fix N ≥ 1, and a characterχ : (Z/NZ)× → C×. We extend χ to χ : Z→ C by setting

χ(a) =

{χ(a mod N), if (a,N) = 1

0, otherwise

We associate to χ the Dirichlet series L(χ, s) =∞∑n=1

χ(n)n−s. By comparison to the Riemann-zeta

function, L(χ, s) is absolutely convergent in the region σ > 1.

If N = 1, then L(χ, s) = ζ(s) is the Riemann-zeta function.

2.) If K is a number field, then

ζK(s) =∑a6=0

N(a)−s =

∞∑n=1

ann−s,

where the first sum is over all non-zero ideals of OK and an = #{a ⊆ OK |N(a) = n}.

Remark If K = Q, then ζK(s) = ζ(s) is again the Riemann zeta function. In general, ζK is called aDedekind zeta function. We will show that both L(χ, s) - introduced in example 1.) and ζK(s) are special

cases of L-series associated to the generalised ideal class characters.

Unique multiplication of integers and multiplicative property of χ, i.e. χ(mn) = χ(m)χ(n) for all

m,n ≥ 1, implies that formally

L(χ, s) =∏

p−prime

(1− χ(p)p−s

)−1=

∏p−prime

(1 + χ(p)p−s + χ(p)2p−2s + . . .

)In the region σ > 1, both of the products above converge and they are equal. Recall that an infinite

product∏n≥1

(1 + bn) with bn 6= 1 for all n, is said to converge to a non-zero value, when the limit limN→∞

N∏n=1

exists. It converges absolutely if the product∏n≥1

(1 + |bn|) converges.

It is a fact that absolute convergence of such a product is equivalent to the absolute convergence of the

series∑n≥1 |bn|. From the theory of absolute convergence series it follows that, when the product converges

absolutely, changing the order of bn’s does not change the value of the product.

We now apply this to the Dedekind zeta-function ζK , where K is a number field.

ζK(s) =∑a6=0

N(a)−s.

We can rewrite this as an Euler product over the primes p ⊆ OK .

ζK(s) =∏p

(1−N(p)−s

)−1=∏p

(1 +N(p)−s +N(p)−2s + . . .

).

This identity holds formally because of unique factorization. In fact, both of ζK(s) and∏

p (1−N(p)−s)−1

converge absolutely in the region σ > 1 and they give the same value. To see this, notice that the product∏p(1−N(p)−σ) converges absolutely in σ > 1, since the series

∑pN(p)

−σ ≤ [K : Q]∑p p−σ 1.

For any m > 1, we have∑a⊆OK ,N(a)


Letting m → ∞, we see that σK(s) converges absolutely for σ > 1 and σK(s) =∏

p (1−N(p)−s)−1

in

this region.

Generalised ideal class group

Let K be a number field.

Definition A divisor (or cycle, in Serge Lang’s notation) of K is a formal product C = C0 · C∞, where C0is a non-zero ideal of OK and C∞ is a (posibly empty) set of real infinite places of K.

If v ∈MK , we define

mv(C) =

power of pv dividing C0OKv , if v is finite1, if v ∈ C∞, 0 if v 6∈ C∞, if v is real0, if v is complex

If C is a divisor of K, we introduce the following notation KC = {α ∈ K×∣∣α ∈ O×Kv and α ≡

1 in(OKv/p

mv(C)v

)×for all v is finite with mv(C) > 0 and τv(α) > 0, for the corresponding embedding τv :

K ↪→ R, for all real places v such that mv(C) > 0}It is clear that KC defined above is a subgroup of K

×. We also observe (we actually include this

redundantly in the definition) that the elements of KC are necesarily v− units if v is a finite place dividingC (i.e. mv(C) > 0). As a matter of notation, in general we write X(C) to denote the subset of X consisting

of those elements prime to C and by XC to denote the subset of X consisting of those elements satisfying

the relations given in the definition of the set KC.

K(C) = {α ∈ K×∣∣αOK is prime to C0}.

We denote by I the group of fractional ideals of OK and by P the group of principal fractional ideals.Denote also by I(C) = {a ∈ I

∣∣a is prime to C0} and by P(C) = {a ∈ P |a is prime to C0} = P ∩ I(C) ={αOK

∣∣α ∈ K(C)}.Definition The generalised ideal class group of level C is the quotient HC = I(C)/PC, where PC ={αOK

∣∣α ∈ KC}. If C = 1, it is agreed by convention that I(1) = I and P1 = P.Clearly PC is a normal subgroup of P(C) and there is a natural injection

I(C)/P(C) ↪→ I/P = Cl(OK),

hence a natural map

HC � I(C)/P(C) ↪→ Cl(OK),

where the first arrow is a surjection and the second one is an injection.

Proposition 31. 1.) I(C)/P(C) ↪→ I/P = Cl(OK) is also surjective. Of course, this would imply that thecomposition HC � Cl(OK) is a surjection.

2.) There are short exact sequences of finite abelian groups

0→ P(C)/PC → HC → Cl(OK)→ 0

and

0→ U/UC →∏

v-∞,mv(C)>0

(OKv/pmv(C)v

)××∏v∈C∞

{±1} → P(C)/PC → 0

Generalised ideal class group continued on next page. . . Page 29 of 49


where U = O×K and UC = U ∩KC. In particular, if hK = #Cl(OK) and hC = #HC, then

hC =hKφ(C)

[U : UC],

where φ(C) = #

[ ∏v-∞,mv(C)>0

(OKv/p

mv(C)v

)××

∏v∈C∞

{±1}

].

Proof. Recall the ”weak approximation theorem”, which tells us that for any finite set S ⊂ MK ,for any� > 0, for any (αv)v∈S ∈

∏v∈S Kv, there exists α ∈ K such that for all v ∈ S, |α− αv|v < �.

1.) Now the first assertion follows easily, because I(C)→ I/P is surjective, means that for all a ∈ I, thereexits α ∈ K× such that αa is prime to C0. This follows immediately either from the ”weak approximationlemma” or the chinese remainder theorem.

2.) There is a commutative diagram with exact rows

0 U K(C) P(C) 0

0 UC KC PC 0

(1)

The second row is exact by definition and the columns are just natural inclusions.

By 1.) there is also an exact sequence

0 P(C)/PC I(C)/PC ≡ HC I(C)/P (C) ≡ Cl(OK) 0 .

To get the other exact sequence, we apply Snake Lemma to (1) to obtain another exact sequence

0 U/UC K(C)/KC P(C)/PC 0 (2)

To finish the proof, we must show that K(C)/KC ≡∏

v-∞,mv(C)>0

(OKv/p

mv(C)v

)××

∏v∈C∞

{±1}.

There is a map

K(C)/KC →∏

v-∞,mv(C)>0

(OKv/pmv(C)v

)××∏v∈C∞

K×v /(K×v )

2

which is injective by definition. It is also surjective, by the weak approximation theorem. If v ∈ C∞,then K×v ≡ R×, so K×v /(K×v )2 ≡ R×/R>0 ≡ {±1}.

Example A typical example is K = Q. Each prime ideal is represented by a prime number p, and we let∞ denote the real absolute value. Let N be an integer > 1, representing an ideal (m) and let C = (N) · ∞.Then U = Z× = {±1}, UC = 1 and Cl(Z) = 0.

The exact sequence (2) yields

0 {±1} (Z/NZ)× × {±1} HC 0 ,

so HC ≡ (Z/NZ)×.Another example would be if we choose K - a real quadratic extension with trivial class group. Let

C = C∞ = {∞1,∞2} consists of the two finite places of K. Then, the proposition before yields an exactsequence

0O×K

O×K∩KC{±1} × {±1} HC 0

Generalised ideal class group continued on next page. . . Page 30 of 49


By Dirichlet’s unit theorem, as the only roots of unity in K are ±1, we know that O×K = {±�Z}, where� ∈ O×K is a fundamental unit.

Now, if τ1(�) and τ2(�) have opposite sign, i.e. NK/Q(�) = −1, then O×K ∩KC = {�2Z} and hence HC = 0is trivial.

On the other hand, if τ1(�) and τ2(�) have the same sign, i.e NK/Q(�) = 1, then O×K ∩ KC = {eZ}, soHC ' Z/2Z.

Both of the above possibilities can occur. For example, if K = Q(√

2) the fundamental unit � = 1 +√

2

has norm NK/Q = −1 and if K = Q(√

3), the fundamental unit � = 2 +√

3 has norm NK/Q = 1.

Definition A character χ : HC → C× is called a generalized ideal class character of level C. Given such acharacter χ, we define

L(χ, s) =∏p-C0

(1− χ([p])N(p)−s

)−1,

where [p] denotes the class of a prime ideal p ∈ I(C). It is clear that the infinite Euler product convergesabsolutely and uniformly in the same manner as the product for the Dedekind zeta function ζK(s) does.

If K = Q, C = (N) ×∞, χ : HC ' (Z/NZ)× → C×. In this case, the new definition of L(χ, s) agreeswith the previous one for Dirichlet characters.

If K is a general number field and C = 1 is trivial, if we choose χ to be the trivial character of HC =

Cl(OK) then L(χ, s) = ζK(s) is the Dedekind-zeta function defined before.

Definition If K is a number field, C and T are divisors of K, we write T|C if T0|C0 and T∞ ⊆ C∞.In this case, we have I(C) ⊆ I(T) and PC ⊂ PT, so there is a natural map HC � HT, which is surjective

by the weak approximation theorem.

Definition We say that a character χ : HC → C× is primitive if it does not arrise by inflation from acharacter χ′ : HT → C×, for any T|C, T 6= C.

Analytic continuation and functional equation of L(χ, s)

Let K be a number field, C a divisor of K and χ : HC → C× a primitive character. Γ(s) is the usual gammafunction, ΓR(s) = π

−s/2Γ(s/2) and ΓC(s) = (2π)−sΓ(s).

Theorem 32. The function ζ(χ, s), defined by

ζ(χ, s) := (|dK |N(C0))s/2 ·

∏v real,v 6∈C∞

ΓR(s) ·∏

v real,v∈C∞

ΓR(s+ 1) ·∏

v complex

ΓC(s)

· L(χ, s),admits a meromorphic continuation to the whole complex plane. Notice that, by definition ζ(χ, s) is

analytic in the region σ > 1.

If χ 6= 1, then ζ(χ, s) is analytic everywhere. If χ = 1, then ζ(χ, s) has simple poles at s = 0 and s = 1and no other poles. In any case, there exists w(χ) ∈ C of absolute value 1 such that

ζ(χ, s) = w(χ)ζ(χ−1, 1− s).

We are going to give Hecke’s proof for the case of a trivial divisor C = 1. There is also an adelic proof,

called Tate’s thesis, which is much more transparent in the case where C is non-trivial.

Analytic continuation and functional equation of L(χ, s) continued on next page. . . Page 31 of 49


So, as previously noted, we assume from now on that C = 1, so HC = Cl(OK). Therefore,

ζ(χ, s) := |dK |s/2ΓR(s)r1ΓC(s)r2 · L(χ, s),

and L(χ, s) coincides with the Riemann-zeta function ζK(s), as we previously remarked.

Definition For each class R ∈ Cl(OK), we define a ”partial zeta function”

ζ(R, s) =∑a∈R

1

N(a)s.

This is analytic in the region σ > 1 and notice that

ζK(s) =∑

R∈Cl(OK)

ζ(R, s).

Notice that if we write ζ(R, s) =∑ an

ns , then the partial sum An = a1 + · · ·+ an is equal to the numberof ideals a in R such that N(a) 6= n.

Theorem 33. For our number field K such that N = [K : Q], let r1 be the number of real embeddings andr2 the number of pairs of complex embeddings. If we denote by A := 2

−r2 |dK |1/2π−N/2 and by

ζmer(R, s) := AsΓ(s/2)r1 · Γ(s)r2ζ(R, s),

then ζmer(R, s) has a meromorphic continuation to the whole complex plane, with simple poles at s = 0and s = 1 and no other poles. ζmer also satisfies the functional equation

ζmer(R, s) = ζmer(R′, 1− s

),

where R′ := R−1[DK/Q]−1, DK/Q being the different ideal.

Why does this imply the main teorem? We can write

L(χ, s) =∏p

(1− χ([p]N(p)−s)−1 =∑

a⊆OK ,a6=0

χ([a])N(a)−

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George Turcas Number Theory: Lecture Notes Started on ...George Turcas Number Theory: Lecture Notes Started on January 19, 2015 which holds for all q lying above p, so e q = e q=p