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Math 129 - Number Fields

Taught by Barry MazurNotes by Dongryul Kim

Spring 2016

This course was taught by Barry Mazur. We met twice a week on Tues-days and Thursdays from 10:00 to 11:30 in Science Center 507. We used thetextbook Number fields by Daniel M. Marcus. There were 11 students takingthe course. There was an in-class final exam, and the course also required onethirty-minutes-long presentation. The course assistant was Kevin Yang.

Contents

1 January 26, 2016 41.1 Algebraic numbers and integers . . . . . . . . . . . . . . . . . . . 41.2 Quadratic fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 January 28, 2016 72.1 Gauss’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Primitive element theorem . . . . . . . . . . . . . . . . . . . . . . 82.3 Roots of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 February 2, 2016 103.1 The fundamental embedding . . . . . . . . . . . . . . . . . . . . 103.2 Trace and norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Galois size and S-numbers . . . . . . . . . . . . . . . . . . . . . . 12

4 February 4, 2016 134.1 Integral closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Fundamental embedding to K ⊗ R . . . . . . . . . . . . . . . . . 134.3 Weil numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.4 Fermat’s last theroem . . . . . . . . . . . . . . . . . . . . . . . . 15

5 February 9, 2016 165.1 Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6 February 11, 2016 196.1 Reasons for loving the discriminant . . . . . . . . . . . . . . . . . 196.2 S-numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1 Last Update: August 27, 2018

7 February 16, 2016 217.1 Dedekind domain . . . . . . . . . . . . . . . . . . . . . . . . . . . 217.2 Factorization of ideals . . . . . . . . . . . . . . . . . . . . . . . . 22

8 February 18, 2016 258.1 Order of an ideal . . . . . . . . . . . . . . . . . . . . . . . . . . . 258.2 Finite approximation . . . . . . . . . . . . . . . . . . . . . . . . . 268.3 Residue fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268.4 Unique factorization implies principal ideal domain . . . . . . . . 27

9 February 23, 2016 289.1 Quotient of ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 289.2 Ramification indices and residue field degrees . . . . . . . . . . . 299.3 Spectrum of a ring . . . . . . . . . . . . . . . . . . . . . . . . . . 30

10 February 25, 2016 3210.1 Decomposition equation . . . . . . . . . . . . . . . . . . . . . . . 3210.2 Ramification and the discriminant . . . . . . . . . . . . . . . . . 3310.3 The Kronecker-Weber theorem . . . . . . . . . . . . . . . . . . . 34

11 March 1, 2016 3511.1 The decomposition group . . . . . . . . . . . . . . . . . . . . . . 3511.2 The inertia group . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

12 March 3, 2016 3812.1 Ramification in the tower . . . . . . . . . . . . . . . . . . . . . . 3812.2 Ideal class group of Q[ζ23] . . . . . . . . . . . . . . . . . . . . . . 39

13 March 8, 2016 4213.1 Frobenius structure of a Galois group . . . . . . . . . . . . . . . 42

14 March 10, 2016 4414.1 Finiteness of the ideal class group . . . . . . . . . . . . . . . . . . 44

15 March 22, 2016 4615.1 The Minkowski bound . . . . . . . . . . . . . . . . . . . . . . . . 46

16 March 24, 2016 4816.1 Computation of ideal class group . . . . . . . . . . . . . . . . . . 4816.2 Higher ramification groups . . . . . . . . . . . . . . . . . . . . . . 49

17 March 29, 2016 5117.1 Logarithm of the fundamental embedding . . . . . . . . . . . . . 51

18 March 31, 2016 5318.1 Dirichlet unit theorem . . . . . . . . . . . . . . . . . . . . . . . . 5318.2 The different ideal . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2

19 April 5, 2016 5619.1 Counting ideals in ideal classes . . . . . . . . . . . . . . . . . . . 57

20 April 7, 2016 5920.1 Equidistribution of ideals in ideal classes . . . . . . . . . . . . . . 5920.2 The Chebotarev density theorem . . . . . . . . . . . . . . . . . . 60

21 April 12, 2016 6121.1 Distribution of ideals in the class group . . . . . . . . . . . . . . 6121.2 The regulator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

22 April 14, 2016 6422.1 Dirichlet Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6422.2 Minkowski’s thereom . . . . . . . . . . . . . . . . . . . . . . . . . 65

23 April 19, 2016 6623.1 The L-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6623.2 Extending the zeta function . . . . . . . . . . . . . . . . . . . . . 67

24 April 21, 2016 6924.1 Infinite products . . . . . . . . . . . . . . . . . . . . . . . . . . . 6924.2 Density of primes . . . . . . . . . . . . . . . . . . . . . . . . . . . 7024.3 Odlyzko’s bound . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

25 April 26, 2016 7225.1 Polar density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7225.2 A density theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 7325.3 Cyclotomic units . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

A The Chebotarev density theorem 75A.1 The Frobenius element . . . . . . . . . . . . . . . . . . . . . . . . 75A.2 The Chebotarev density theorem . . . . . . . . . . . . . . . . . . 75A.3 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3

Math 129 Notes 4

1 January 26, 2016

There will be half-hour presentations. I have experimented this in Math 124,and it worked well.

1.1 Algebraic numbers and integers

Definition 1.1. An algebraic number is a root of a (monic) polynomialf(X) = Xd + ad−1X

d−1 + · · ·+ a0 for ai ∈ Q. An algebraic integer is a rootof a monic polynomial with integer coefficients.

These have connection with many branches of mathematics. We can look atthe ring A of algebraic integers and talk about SpecA, which we will discuss.This A can be viewed as a lattice in the Euclidean space thus a geometricstructure. For instance, the ring

Z[√−D] = a+ b

√−D : a, b ∈ Z

on the complex plane looks like a rectangular lattice. There are also connectionswith analysis; the zeta function contains information about the nature of primeideals in A.

Proposition 1.2. An algebraic integers that is a rational number is an integer.

Proof. Let mn be an algebraic integer for (m,n) = 1. Then there will be integers

ad−1, ad−2, . . . , a0 such that(mn

)d+ ad−1

(mn

)d−1

+ · · ·+ a0 = 0.

Then multiplying nd to both sides, we get

md + ad−1nmd−1 + · · ·+ a0n

d = 0

and thus n divides m. This means that n = ±1.

Proposition 1.3. If α is an algebraic number then there is an integer N ≥ 1such that Nα is an algebraic number.

Proof. There is are integers ai and N 6= 0 that makes α a root of the polynomial

f(X) = Xd +ad−1

NXd−1 +

ad−2

NXd−2 + · · ·+ a0

N= 0.

Then

NdXd + ad−1Nd−1Xd−1 + ad−2N ·Nd−2Xd−2 + · · ·+Nda0 = 0

and thus Y = Nα is the root of

F (Y ) = Y d + ad−1Yd−1 + ad−2NY

d−2 + · · ·+Nda0 = 0.

Math 129 Notes 5

Proposition 1.4. Let z1, z2 be two complex numbers that are linearly indepen-dent over Q. Suppose that

αz1 = a11z1 + a12z2

αz2 = a21z1 + a22z2

for some rational numbers a11, a12, a21, a22. Then α is an algebraic number.

Proof. Let

A =

(a11 a12

a21 a22

)∈ Mat2(Q).

Then we see that det(α · I2 −A) = 0 and hence α is a root of the characteristicpolynomial

X2 − tr(A)X + det(A).

This completes the proof.

Note that if the aijs were integers, then α should have been an algebraicinteger.

We can generalize it to more variables. In fact, we have the following claim.

Proposition 1.5. Let V ⊂ C be a finite-dimensional Q-vector space, and letM ⊂ C be a finitely generated abelian group. Consider an α ∈ C. If α · V ⊆ V ,then α is an algebraic number. If α ·M ⊆M , then α is an algebraic integer.

Actually we need to use the fundamental theorem of finitely generatedabelian groups. Since M is finitely generated and torsion-free (M is a sub-set of C), it has to be isomorphic to some Zr. Then we can run the exactlysame argument for the previous proposition.

Also note that the converse is also true. Given an algebraic number/integerα ∈ C, we can set V /M to be the vector space/abelian group generated by thepowers of α.

Corollary 1.6. Let K/Q be a field extension of finite degree. Then every ele-ment of K is an algebraic number.

Proof. Just set V = K.

Theorem 1.7. The sum and product of two algebraic numbers are again alge-braic numbers. The sum and product of two algebraic integers are agin algebraicintegers.

Proof. This is because Q[α, β] is a finitely generated vector space over Q, andZ[α, β] is a finitely generated abelian group.

Math 129 Notes 6

1.2 Quadratic fields

For a extension field K over Q, we denote by AK the right of algebraic integersin K. Let us try to describe the structure of AK for a extension field K overQ of degree 2. Consider any α ∈ K. By the quadratic formula, we know thatα = r + s ·

√D for some square-free D ∈ Z. Let α = r − s ·

√D be the Galois

conjugate. Then α is the root of the quadratic polynomial

X2 − 2rX + (r2 −Ds2) = 0.

It follows from this fact that α is an algebraic if and only if both 2r and r2−Ds2

are integers. In fact, we can more explicitly describe the additive group ofalgebraic integers.

Exercise 1.8. Assume that K is a extension field over Q of degree 2. Then thering of algebraic integers AK is either (i) the abelian group generated by 1 and√D or (ii) the abelian group generated by 1 and (1 +

√D)/2.

Math 129 Notes 7

2 January 28, 2016

There is a lemma I would like to begin with:

Lemma 2.1. Let K be a field and let f(X) ∈ K[X] be a polynomial. If f(X)and f ′(X) are relatively prime, then f(X) has no multiple roots.

Let θ ∈ C be an algebraic number. Then there is a unique irreducible monicpolynomial fθ(X) ∈ Q[X] that has θ as a root with smallest degree d. We canwrite

fθ(X) = (X − θ1) · · · (X − θd).

By the above lemma and the fact that f and f ′ has to be relatively prime, wesee that θ1, θ2, . . . , θd are all distinct. We call the set θ1, . . . , θd the full setof Galois conjugates.

If θ and θ′ are Galois conjugates, or in other words, roots of the same irre-ducible polynomial over Q, then we have a natural isomorphism

Q[θ] ∼= Q[θ′]

Q

2.1 Gauss’s lemma

Factorization in Z[X] and Q[X] might be slightly different. Suppose that wehave a polynomial f(X) ∈ Z[X] such that the greatest common divisor of thecoefficients is 1. Assume that we have a factorization

f(X) = g(X) · h(X)

where g(X) and h(X) are in Q[X]. Then we can multiply integers on both sidesand factor out some greatest common divisors and make the equation into

a · f(X) = b · g′(X) · h′(X)

where all f, g, h has the property that the greatest common divisor of the coef-ficients is 1, and a and b are relatively prime. Suppose that there is a prime pthat divides a. Then we can reduce everything modulo p and get

0 = b · g′(X) · h′(X).

This is clearly a contradiction since Fp is a integral domain. So we have a = 1,and likewise, b = 1. This means that the factorization in Q[X] was essentiallya factorization in Z[X].

Math 129 Notes 8

2.2 Primitive element theorem

Theorem 2.2 (Primitive element theorem). Suppose we have a field extensionK/F of finite degree, that has the property that there are only finitely manyintermediate fields. Then there is an element γ ∈ K such that K = F [γ].

Proof. First assume that F is finite. Then K is also finite, and we actuallyknow the structure of any finitely field, and its generated by one element.

Now assume that F is infinite. Let α, β ∈ K be any elements. If we showthat F [α, β] = F [γ] for some γ ∈ K, then we can apply this fact multiple timesto get

F [α1, . . . , αm] = · · · = F [γ]

for some γ. Since K is of finite degree over F , it is generated by finitely manyelements, and this will finish the proof.

So let us show that F [α, β] = F [γ] for some γ. Given a c ∈ F , we defineγc = α + cβ. Then we clearly have F [γc] ⊂ F [α, β] for any c. Since there arefinitely many intermediate subfields, there are two c1 6= c2 such that

F [γc1 ] = F [γc2 ].

This will imply α+ c1β, α+ c2β ∈ F [γc1 ] and thus α, β ∈ F [γc1 ]. Then we have

F [α, β] ⊂ F [γc1 ] ⊂ F [α, β],

and the result follows.

Proposition 2.3. Let K/F be any field extension of finite degree, and assumethat F has characteristic zero. Then there are only finitely many intermediatefield extensions.

To prove this we have to use some Galois theory. We won’t assume knowl-edge, but let me just outline the basic idea.

Proof. Since F has characteristic zero, we see that there is a finite Galois ex-tension L/F where K ⊂ L. Then by the fundamental theorem of Galois theory,the intermediate field extensions corresponds to subgroups of the Galois group,which is finite. So there are only finitely many intermediate field extensions.

2.3 Roots of unity

Take any m ≥ 1. We let ζm = e2πi/m and

µm = 1, ζm, ζ2m, . . . , ζ

m−1m .

Then we easily see that these are the roots of

Xm − 1 =

m∏j=0

(X − ζjm) =∏d|m

( ∏(k,d)=1

(X − ζkd )

).

Math 129 Notes 9

It is quite natural to define

Φd(X) =∏

(k,d)=11≤k≤d−1

(X − ζkd )

and writeXm − 1 =

∏d|m

Φd(X).

Theorem 2.4. Any two primitive mth roots of unity are Galois conjugates.

Proof. We first note that it suffices to show that ζm and ζkm are Galois conju-gates, because the relation is transitive. And we also note that it suffices toshow for prime k = p - m.

Let f(X) be the minimal polynomial ζm and let

Xm − 1 = f(X)g(X).

Assume that ζm and ζpm are not Galois conjugates. Then ζpm cannot be a root off(X) and thus must be a root of g(X). This means that ζm is a root of g(Xm)and thus g(Xm) = f(X) · u(X) for some u(X) ∈ Z[X].

Let us now reduce everything modulo p. We have

g(X)p = g(Xp) = f(X) · u(X)

and thus g and f are not relatively prime in Fp[X]. This means that

Xm − 1 = f(X) · g(X)

has to have a multiple root. But this contradicts the fact that Xm − 1 andits derivative mXm−1 are relatively prime modulo p. Therefore ζm and ζpm areGalois conjugates.

Corollary 2.5. The polynomial Φd(X) is in Z[X] and is irreducible.

Math 129 Notes 10

3 February 2, 2016

3.1 The fundamental embedding

Because C is algebraically closed, given a monic polynomial f(X) ∈ Q[X], wecan always it as

f(X) =

d∏i=1

(X − θi)

where θ1, . . . , θd are algebraic numbers. Because f has a unique factorization

into f(X) =∏dj=1 fj(X) where fj(X) are irreducible. Then we can partition

the θ1, . . . , θd into sets consisting of Galois conjugates. This means that the setof roots of f(X), whether it is irreducible or not, is a union of full sets of Galoisconjugates of algebraic numbers.

We can look at the coefficients of f(X). We have

f(X) = Xd − s1Xd−1 + s2X

d−2 − · · ·

where s1(θ1, . . . , θd) = θ1 + θ2 + · · ·+ θd,

s2(θ1, . . . , θd) =∑i 6=j θiθj ,

...

sd(θ1, . . . , θd) = θ1 · · · θd.

These si are the elementary symmetric polynomials.We have the following proposition.

Proposition 3.1. The set θ1, . . . , θd is a finite union of full sets of Galoisconjugate of algebraic integers(resp. algebraic integers) if and only if

si(θ1, . . . , θd) ∈ Q(resp.Z)

for each of i = 1, . . . , d.

Because z 7→ z is an automorphism of C fixing Q, we also have the following.

Proposition 3.2. If θ is an algebraic number, so is θ, its complex conjugate.Moreover, θ is a Galois conjugate of θ.

If we look at a full set of Galois conjugates, then we can always pair upcomplex conjugates. This means that we can write it as

θ1, θ2, . . . , θr, θr+1, θr+1, . . . , θr+s, θr+s.

We have been working on a finite extension field K/Q of degree d. We canthink the set of field homomorphisms Hom(K,C). We first see that any such

Math 129 Notes 11

homomorphism is necessarily an embedding, because the kernel, which is anideal of K, has to be zero. Also, any such homomorphism fixes Q.

K C

Q

σ

There are exactly d such homomorphisms σ; The primitive element theoremtells us that K = Q[θ] for some θ, and the image σ(θ) determines σ. On theother hand, the possible values of σ(θ) are the Galois conjugates of σ. So wehave a correspondence

Hom(K,C) ←→ θ1, . . . , θd.

Suppose that d = r + 2s and the Galois conjugates of θ consists of r realsand s paris of complex conjugates. Then there are r embeddings K → R and shonest complex embeddings K → C and their complex conjugates.

Definition 3.3. We define the fundamental embedding of K into the Eu-clidean space as the ring homomorphism K → Rr × Cs given by

x 7→ (σ1(x), . . . , σr(x), σr+1(x), σr+3(x), . . . , σr+2s−1(x)),

where for each pair of complex conjugate of embeddings, we choose one.

Although we have made a choice, we will later discuss how it can be removed.

3.2 Trace and norm

Definition 3.4. Let K/Q be a finite extension of degree d. For an elementα ∈ K, we define its trace and norm as

Trace(α) = TK/Q(α) =

d∑i=1

σi(α), Norm(α) = NK/Q(α) =

d∏i=1

σi(α).

We note that

T (α+ β) = T (α) + T (β), T (a · α) = a · T (α)

for each a ∈ Q. That is, the map T : K → Q is Q-linear. Using the fundamentalembedding, we can lift this into an R-linear map E → R.

E R

K Q

T

T

This new map T : E → R will be given by

(v1, . . . , vr, vr+1, vr+3, . . . , vr+2s−1) 7→ v1 + v2 + · · ·+ vr+2s.

Math 129 Notes 12

Likewise, using the fact that N is an homomorphism K∗ → Q∗ as multi-plicative groups, we can define N : E∗ → R∗.

E∗ R∗

K∗ Q∗

N

N

3.3 Galois size and S-numbers

Definition 3.5. We define the Galois size of an algebraic number α as

size(α) =d

maxi=1|αi|,

where αi is the set of Galois conjugates.

Theorem 3.6. There are only a finite number of algebraic integers of degree atmost d <∞, and Galois size at most B <∞.

Proof. We look at its minimal polynomial. Because all its roots are bounded,we see that the coefficients are bounded by abusing the triangle inequality.Then there are finitely many such polynomials, and thus there are finitely manyalgebraic numbers.

Corollary 3.7. Any algebraic integer of Galois size at most 1 is a root of unity.

Corollary 3.8. If α is an algebraic integer of Galois size at most 1, then anypower of α will also have Galois size at most 1. Then by the previous theoremαm = αn for some m 6= n, and the result follows.

Definition 3.9. An S-number α is a real algebraic integer greater than 1 suchthat all its Galois conjugates have absolute value less than 1.

I think this is a beautiful introduction to the approximation of algebraicnumbers. For x ∈ R, we define

‖x‖ = distance between x and its nearest integer.

Then we have the following proposition.

Proposition 3.10 (Salem). If α is an S-number, then ‖αn‖ → 0 as n → ∞.In fact,

∑∞n=1‖αn‖2 converges.

Math 129 Notes 13

4 February 4, 2016

4.1 Integral closure

Proposition 4.1. Consider a polynomial equation

f(X) = Xd + ad−1Xd−1 · · ·+ a0

where each ai is an algebraic number(resp. algebraic integer). Then the root off(X) is again an algebraic number(resp. algebraic integer).

Proof. Let f(θ) = 0, and look at the field Q[a0, a1, . . . , ad−1]. Because a0, . . . , ad−1

are algebraic numbers, we see that it is of finite degree over Q. Letting L =Q[a0, . . . , ad−1, θ], we get a finite extension of a finite extension, which will againbe a finite extension field. Multiplication by θ sends L into L, and it followsthat θ is algebraic.

The integral version can be done in the same manner. If we let M =Z[a0, . . . , ad−1, θ], then multiplication by θ sends M to M . This implies that θis an algebraic integer.

Definition 4.2. Let K/Q be an extension of finite degree. Let A0 be a subringof algebraic integers in K. If a number α ∈ K is a root of some

f(X) = Xd + ad−1 + · · ·+ a0

for some ai ∈ A0, then we say that α is integral over A0. The integral closureA is then defined as

A = α ∈ K : α is integral over A0.

If A = A0, we say that A0 is integrally closed.

4.2 Fundamental embedding to K ⊗ RLet us recall the fundamental embedding K → Rr × Cs. We had to make achoice to define the map. We now discuss how to remove this choice.

Definition 4.3. Let U and V be abelian group. Then f : U × V → W , whereW is also an abelian group, is called bilinear if

f(u1 + u2, v) = f(u1, v) + f(u2, v), f(u, v1 + v2) = f(u, v1) + f(u, v2)

for any us in U and vs in V . The tensor product U ⊗ V is defined as theproperty that for any bilinear f : U ×V →W there is a unique homomorphismϕ such that

U ⊗ V

U × V

W

ϕ

F

f

Math 129 Notes 14

commutes. This is given by

U ⊗ V = Z[U × V ]/(ideal generated by R-bilinear relations).

In fact, we can define U ⊗R V for R-modules U and V .Let K/Q be a d-dimensional vector space over Q. Then we see that K⊗ZR.

This is a d-dimensional vector space over R, and also a ring. In other words, itis an R-algebra.

We come back to the fundamental embedding. Instead of looking at Rr×Cs,we embed K into E = K ⊗ R. (It doesn’t matter whether we tensor over Z orQ.) Then we get the following:

K E = K ⊗ R

Q A E = K ⊗ R R

d d

Note that because the ring of integers A in K is an abelian group in E, it formsa lattice. We will later study this beautiful lattice.

Example 4.4. Take K = Q[√D]. Then A is generated by either 1,

√D or

1, (1 +√D)/2. In both cases, we see that A form a nice lattice.

Proposition 4.5. The abelian group A is discrete in Rr × Cs.Proof. Let C be any compact subset of Rr × Cs. Then if α is in C, then allGalois conjugates of α have bounded size. Moreover, the degree is at most d.We have proved last time that this implies that there are only finitely manypossible α. This finishes the proof.

Corollary 4.6. As an abelian group, A is a free abelian group of rank d.

4.3 Weil numbers

Definition 4.7. A Weil number is an algebraic integer α such that for anyGalois conjugate αi of α, its absolute value is

|αi| = pn/2

where p is prime and n ≥ 1.

Let me make a cultural comment about this. Let q = pn be a power of aprime, and consider the equation

y2 = g(x)

over Fq, where g(x) is a monic polynomial with no multiple roots and of degreed+ 2. How many points are there on this curve over Fqν?

Theorem 4.8. The number of solutions over Fqν is

# of solutions/Fqν = q −d∑i=1

ανi ,

where α1, . . . , αd is the set of conjugates of Weil numbers.

Math 129 Notes 15

4.4 Fermat’s last theorem1

Theorem 4.9 (Fermat’s last theorem). Let n ≥ 3 be an integer. There thereexist no nonzero integral solutions of

an + bn = cn.

One easy observation is that proving for n = 4 and n = p suffices. Becausethe case n = 4 was in the homework, we shall look at the case n = p.

Let ω = e2πi/p. We see that every element of Z[ω] can be uniquely repre-sented as

a0 + a1ω + · · ·+ ap−2ωp−2

since Xp−1 +Xp−2 + · · ·+X + 1 is the minimal polynomial of ω.The theorem is hard, and thus we will do as much as we can. Assume first

that Z[ω] is a unique factorization domain. Suppose that there is a solution

xp + yp = zp

where p does not divide any of x, y, z. We can also suppose that x, y, z arerelatively prime in Z.

We can factorize the equation into

zp = xp + yp =

p−1∏i=0

(x+ yωi).

Suppose that x + yωi and x + yωj have a common prime factor. This meansthat both numbers are in some ideal I ⊂ Z[ω] that is not the entire ring. Theny(1− ωj−i) will be in I, and multiplying all the other (1− ωk)s, we see that pyis in I. On the other hand, because x+ yωi divides zp, we see that zp ∈ I. Butbecause py and zp are relatively prime, we get 1 ∈ I, which contradicts the factthat I is not the whole ring. This shows that all x + y, x + yω, . . . , x + yωp−1

are relatively prime, and because they multiply up to a p-th power, we see thatall of them must be a p-th power up to a unit. In other words, we have

x+ yω = uαp

for some u, α ∈ Z[ω], where u is a unit.We now define the class group. Consider the set of all ideals of Z[ω], and we

define an equivalent relation so that A ∼ B for ideals A,B ⊂ Z[ω] if there areprincipal ideals (α) and (β) such that

A · (α) = B · (β).

Proposition 4.10. The equivalence classes form a group under multiplication.The equivalence class consisting only of all principal ideals is the identity.

1This was a presentation by Shyam Narayanan.

Math 129 Notes 16

5 February 9, 2016

We are going to talk about discriminants.

5.1 Discriminant

We have a field K which is of finite degree d over Q. We choose a bunch ofnumbers α1, . . . , αd ∈ K, and embeddings σi : K → C.

Definition 5.1. We define the discriminant as

disc(α1, . . . , αd) =(

det[σi(αj)

])2=(

det[σj(αi)

])2.

Because we square the determinant, we do not have to specify the orderof the numbers and the embeddings. But it depends on the change of basis.Suppose we have α′ =

∑dj=1mijαj for some mij ∈ Q. Then[

σjα′i

]= M ·

[σjαi

]and thus

disc(α′1, . . . , α′d) = (detM)2 disc(α1, . . . , αd).

In particular, we can assume that mij ∈ Z and detM = ±1, or equivalently,the Z-module generated by α1, . . . , αd is same as the Z-module generated byα′1, . . . , α

′d. Then it follows that disc(αi)i = disc(α′i)i.

One thing we notice is that if α1, . . . , αd are linearly dependent over Q,then the the rows (or columns) of [σjαi] are linearly dependent and hencedisc(α1, . . . , αd) = 0. So we may as well assume that α1, . . . , αd is a Q-basis ofK as a Q-vector space.

Definition 5.2. Let A be a Z-module. Then we define its discriminant as

disc(A) = disc(α1, . . . , αd),

where α1, . . . , αd is the basis for A.

Note that we can work over any finite extension field of characteristic zeroinstead of K and Q.

Theorem 5.3. Let T : K → C be the trace. Then

disc(α1, α2, . . . , αd) =∣∣det

[T (αiαj)

]∣∣.Proof. It follows from the fact that[

σjαi][σiαj

]=[∑

σk(αiαj)]

=[T (αiαj)

].

Math 129 Notes 17

We are using the fact this T (x · y) is some kind of a bilinear form. BecauseK is an algebra, multiplication gives a bilinear map, and then T is additivehomomorphism.

Theorem 5.4. Let αii be linearly independent over Q. Then disc(αi)di=1 6= 0.

Proof. Suppose that the determinant det[T (αiαj)] = 0. Then there will berational number aj ∈ Q not all zero, such that

0 =

d∑j=1

ajT (αiαj) =

d∑j=1

T (ajαiαj)

for any i = 1, . . . , d. If we let α =∑dj=1 ajαj , we have

T (α · αi) = 0

for any i = 1, . . . , d.Now we use the fact that α1, . . . , αd form a basis of Q. Then from T (α·αi) =

0, it follows that T (α · x) = 0 for any x ∈ K. But this is clearly not true since

T (α · 1/α) = T (1) =

d∑j=1

σj(1) = d.

Definition 5.5. Let α be an element of K. Then we write

disc(α) = disc(1, α, α2, . . . , αd−1).

We can calculate disc(α) it relatively easily. We have

disc(q, α, α2, . . . , αd−1) = det(σiαj−1)2 = det((σiα)j−1)2 =

∏j<j′

(σjα− σj′α)2

by the Vandermonde identity.

Proposition 5.6. Let α be a primitive element. We have

disc(1, α, . . . , αd−1) = (−1)d(d−1)/2N(f ′(α)),

where f(X) is the monic irreducible polynomial over Q.

Proof. Since f(X) =∏

(X − σiα), we have

f ′(X) =

d∑k=1

∏j 6=k

(X − σjα)

and thusf ′(α) =

∏j 6=1

(α− σjα).

Math 129 Notes 18

Then when we take the norm, it becomes

N(f ′(α)) =

d∏i=1

σi∏j 6=1

(α− σj(α))

=

d∏i=1,j 6=1

(σiα− σiσjα) = (−1)d(d−1)/2∏k<l

(σkα− σlα)2.

The result follows from what we did just before.

Example 5.7. Let m be a square free integer, with m 6= 0,±1. Let d ≥ 2 andconsider

K = Q[ d√m] = Q[X]/(Xd −m).

We calculate disc(A), where

A = d√mZ + ζd

d√mZ + · · ·+ ζd−1

dd√mZ.

Then

disc(A) = disc(1, d√m, ( d√m)2, . . . , ( d

√m)d−1) = ±N(f ′( d

√m))

= ±N(dm(d−1)/d) = ±ddd−1∏k=0

ζkdm(d−1)/d = ±ddmd−1.

In particular, if A = Z[√m] then discA = ±4m, and if A = Z[(1 +

√m)/2]

then discA = ±m.

Math 129 Notes 19

6 February 11, 2016

Let us start by recalling where we are. Let K be a field of degree d over Q.Let α ∈ K be a primitive element. Then 1, α, . . . , αd−1 form a Q be a basis ofK. Let M be the Z-module generated by 1, α, . . . , αd−1. This module will beof rank d.

We can look at the discriminant

disc(M ⊂ K) = disc(1, α, α2, . . . , αd−1) = ±N(f ′(α)).

If you wish, you can think of M as a lattice in K ∼= M ⊗Z Q. There are loadsof corollaries you can make.

Corollary 6.1. Let α = ζp = e2πi/p. Let K = Q[ζp] and M = Z[ζp]. Thendisc(M) = asdfadsf .

Proof. We know that disc(M) = ±N(f ′(ζp)). Since f(X)(X − 1) = Xp − 1, wecan take the derivative and get f ′(ζp) = pζp−1

p . Its norm will be

N(f ′(ζp)) =Np · (Nζp)p−1

N(ζp − 1)=pp−1 · 1

p= pp−2.

Thus it follows that disc(M) = ±pp−2.

Let A be the ring of algebraic integers of K.

Theorem 6.2. A is lattice in Z, and it is a free abelian subgroup of rank d.

Proof. We have

A K E = Rr × Cs.

As a subgroup of E, the group A is discrete.Now next week, Johnnie Han will going to prove that every discrete subgroup

in Rd is a free abelian group of rank at most d. Then it has a have rank exactlyd because for every element of K multiplied by a large integer is an algebraicinteger.

Definition 6.3. We define the discriminant of a number field K as

∆K = discriminant of A.

6.1 Reasons for loving the discriminant

Let K/Q have degree d. We can first find a Q-basis α1, α2, . . . , αd that are alsoalgebraic integers, and then compute the discriminant δ = disc(α1, ,sαd).

Now let α ∈ K be an arbitrary algebraic integer. There will be uniquexi ∈ Q such that α =

∑xiαi. Apply the various embeddings σj , and we get

σjα =∑

xiσjαi.

Math 129 Notes 20

The Cramer’s rule tells us that

xi =det(γi)

det(σj(αi)),

where γi is the matrix obtained bey replacing the ith column of (σj(αi)) byσj(α). When we multiply det(σj(αi)) to both sides, we get

xi · disc(α1, . . . , αd) = det(σj(αi)) det(γi).

Since the left hand side is in Q and the right hand side is an algebraic integer,we have

xi =mi

disc(α1, . . . , αd)

for some integer mi. This enables us to literally compute the basis for the ringof algebraic integers.

Also, when viewed as a lattice, the abelian group A ⊂ Rr × Cs = Rd hasfundamental domain of volume

√∆K .

6.2 S-numbers2

Definition 6.4. Let θ > 1 be a real algebraic integer, and let α1, . . . , αd−1be its Galois conjugates. The number θ is called a S-number if |αi| < 1 for alli. We denote the set of S-numbers by S.

Example 6.5. Every rational number greater than 1 is a trivial S-number.The golden ratio ϕ = (1 +

√5)/2 and the solution ρ to x3 − x− 1 = 0 is also a

S-number, which turns out to be the smallest S-number.

Theorem 6.6. Let ‖x‖ denote the distance from x to its nearest integer. If θis an S-number, then ‖θn‖ → 0 as n→∞.

Proof. Let θ be of degree d, and let α1, . . . , αd−1 be its conjugates. Then since

θn + αn1 + · · ·+ αnd−1

is a rational number, and an algebraic integer, it is an integer. Then

‖θn‖ ≤ |α1|n + · · ·+ |αd−1|n

and because the right hand side goes to zero, ‖θ‖ → 0.

Using a very similar argument, we can in fact also show the following.

Theorem 6.7. Let λ ∈ Q[θ]. Then ‖λθn‖ → 0 as n→∞.

The converse is open in general, but we can prove the following.

Theorem 6.8. Let θ > 1 and λ > 0 be real numbers. If∑‖λθn‖2 <∞, then θ

is an S-number, and moreover λ is an algebraic number in Q[θ].

2This was a presentation by Kat Zhou

Math 129 Notes 21

7 February 16, 2016

7.1 Dedekind domain

Definition 7.1. A ring A is a Dedekind domain if it is an integral domainwith the following three propositions.

1) Every ideal of A is finitely generated.

2) Every nonzero prime ideal is a maximal ideal.

3) A is integrally closed in K, the field of fractions of A.

The main use of Dedekind domains will be illustrated in the following theo-rem.

Theorem 7.2. The ring of (algebraic) integers in a number field (i.e., a fieldof degree d <∞ over Q) is a Dedekind domain.

Proof. We first prove 1). Since A is a free abelian group of rank d, we see thatany ideal I ⊂ A is finitely generated as a Z-module, and thus finitely generatedas a ideal.

We next prove 2). Any ideal in A contains some positive integer m, becausewe can always multiply its inverse times an integer that is in A. Then we have

A A/mA A/I.

Since A/mA ∼= (Z/mZ)d is finite, we see that A/I must also be finite. Nowfor any prime ideal P ⊂ A, we see that A/P is a finite ring, which must be anintegral domain by the definition of a prime ideal. Now exercise 2 on page 82 ofour book states that any finite integral domain is a field. Thus A/P is a fieldand P is a maximal ideal.

We now prove 3). Recall that if α is a root of f(X) with algebraic integercoefficients, then α is also an algebraic integer. Take any element α that satisfiesa monic polynomial equation f(x) = 0 where the coefficients of f are in A.Then α ∈ K, and α is an algebraic integer. Hence α ∈ A and so A is integrallyclosed.

Theorem 7.3. Let k be any field. Then the ring of polynomials k[t] is aDedekind domain, and the ring of power series k[[t]] is also a Dedekind domain.

Dedekind domains is a incredibly important notion, and it is where most ofalgebraic number theory resides.

Let I, J ⊂ A be nonzero ideals. We define their product as

I · J = r∑k=1

αkβk : αk ∈ I, βk ∈ J.

Math 129 Notes 22

This gives a monoid structure on the set of ideals. Indeed, we have the identity(1), as (1) · I = I for any I. A ideal is called a principal ideal if it is of the form(α) for some α. We can define an equivalence relation

I ∼ J ⇔ (α) · I = (β) · J.

Let K be the field of fractions of A, and let M be a finitely generated additivesubgroup of K. If we let m1, . . . ,mr ∈ K be the generates of M , then theremust be a D ∈ K such that Dm1, . . . , Dmr are all in A. We then have abeliansubgroups M ⊂ K and DM ⊂ A. Moreover, if we assume that M is an A-module in K, then DM is also an A-module in A, which is in other words, anideal. That is, we can write

M =1

D· I

where I is an ideal.

7.2 Factorization of ideals

Definition 7.4. A finitely generated A-module in K is called a fractionalideal of A.

Lemma 7.5. Let I ⊂ A be a nonzero ideal and P be a prime ideal. Then P | Iif and only if I ⊂ P .

We will prove this later.

Lemma 7.6. Let I, J ⊂ A be nonzero ideals, and let P be a prime ideal. Then

P | I · J =⇒ P | I or P | J.

Proof. Since P is a maximal ideal, we see that A/P is a field. Under theprojection map

π : A→ A/P,

the images π(I) and π(J) are both ideals, and their product must be zero. Butan ideal of a field is either the entire field of just zero. Thus one of π(I) or π(B)must be zero and thus I ⊂ P or J ⊂ P .

We can also define the inverses of ideals. Let I ⊂ A be a ideal, and let0 6= α ∈ I. Then let

Jα = β ∈ A : β · I ⊂ (α).

Theorem 7.7. The product of I and Jα is I · Jα = (α).

To prove this, we need something from the homework.

Lemma 7.8. Let H ( A ⊂ K be a proper ideal, where A is a Dedekind domainand K is the field of fractions. Then for any γ ∈ K \A, we have γ ·H ⊂ A.

Math 129 Notes 23

Proof of theorem 7.7. We first see that

I · Jα ⊂ (α)

since the definition of Jα makes sure everything lies in (α). In other words, wehave

H =1

α· I · Jα ⊂ A.

Suppose H ( A. Then using the lemma above, we see that there is anγ ∈ K \A such that γH ⊂ A. Then

γ · Jα · I ⊂ (α)

and thus by definition of Jα,γ · Jα ⊂ Jα.

But then, Jα is a Z-module, that is preserved under multiplication by γ, and itfollows that γ is an algebraic integer. This contradicts our definition of γ.

This gives the cancellation property:

Proposition 7.9 (Cancellation). If M , N , I are ideals of A and M · I = N · I,then M = N .

Proof. Consider any Jα such that I · Jα = (α). Then multiplying Jα to eachside we get

αM = M · I · Jα = N · I · Jα = αN

and hence M = N .

Definition 7.10. The ideal class group is the group of equivalence classes ofideals of A (under multiplication). For a number field K, we denote H(K) bythe group of ideal classes of A.

There are wildly open questions about the ideal class groups. For example,let K = Q[

√−D] for D ≥ 1. When does it have trivial ideal class group? It

turns out that the answer is

−D = −1,−2,−3,−7,−11,−19,−43,−67,−163.

Proposition 7.11. Let I and N be ideals. Then I | N if and only if N ⊂ I.

Proof. Clearly, if I | N then N = I · J ⊂ I ·A = I.Suppose now that N ⊆ I and let us construct a J . Let

M =1

α·N · Jα

where α ∈ I and Jα = β ∈ A : β · I ⊂ (α). Then

I ·M =1

α·N · I · Jα = N.

Math 129 Notes 24

Theorem 7.12 (Unique factorization of ideals). Every ideal factors uniquelyas a product of prime ideals.

Proof. We first prove the existence of factorization part. Suppose the contraryand consider a maximal counterexample M . There is a maximal ideal m withM ⊂ m ⊂ A. We first note that since m is itself a prime ideal, M cannot be m.By the divisibility theorem, there has to exist some ideal I such that M = m · I.Then I becomes a ideal bigger than M and hence contradicts our assumptionthat M is maximal.

We now prove the uniqueness, by channeling the old proof of unique factor-ization property of Z. Suppose that we have

P1 · P2 · · · · · Pr = Q1 · · · · ·Qs.

Then P1 divides Q1 · · · · · Qs and we can cross of things. Then inductively wesee that both sides are equal.

Math 129 Notes 25

8 February 18, 2016

Last time we proved the unique factorization of ideals in Dedekind domains.This means that any ideal I can be written as

I =

r∏i=1

P eii

for distinct prime ideals Pi and ei > 0.

8.1 Order of an ideal

Definition 8.1. The order ordIP of an ideal I over a prime ideal P is thelargest exponent e such that P e | I. The order of an element α is defined as

ordP (α) = ordP ((α)).

We can writeI =

∏P

P ordP (I)

in a very nice way. There are infinitely many units, which does not count.The order can be defined even for fractional ideals.

Definition 8.2. Let M = 1αI be a fractional ideal. Then we define

ordP (M) = ordP (I)− ordP (α).

Then likewise we can write

M =∏P

ordP (M).

The fractional ideals form a group, because we can always take the inverse.In this sense, the ord is a homomorphism

ordP : Fract. Ideals Z.

It makes perfect sense to define the greatest common divisor and the leastcommon multiple. For any two ideals I, J ⊆ A, we define

GCD(I, J) = (I, J), LCD(I, J) = I ∩ J.

Then we have

ordP (GCD) = min(ordP (I), ordP (J)), ordP (LCM) = max(ordP (I), ordP (J)).

Math 129 Notes 26

8.2 Finite approximation

Let A be a Dedekind domain. Consider any distinct primes P1, . . . , Pr ande1, . . . , er ≥ 0.

Lemma 8.3. There exists an α ∈ A such that ordPi(α) = ei for i = 1, . . . , r.

Proof. Let us look at

A ⊃r∏i=1

P eii ⊃r∏i=1

P ei+1i .

Because the ideals P ei+1i are pairwise relatively prime, we can use the Chinese

remainder theorem

A A/∏P ei+1i

∼=∏A/P ei+1

i ⊃∏ri=1 P

eii /P

ei+1i

Now choose an r-tuple (u1, r2, . . . , ur) in∏P eii /P

ei+1i so that ui 6= 0. We then

lift it to some α ∈ A. This α will have the desired property.

The consequence of this lemma is that every ideal I ⊆ A is generated bytwo elements. Even better, given any 0 6= α ∈ I, there exists a β such thatI = (α, β). How do we achieve this? Let us first pick any α and let us look at(α). It will be

(α) =

r∏i=1

PEii ·s∏j=1

QFjj

where Ei ≥ ei. Now find a β such that ordPi β = ei and ordQj β = 0. Because

ordP (α, β) = min(ordP α, ordP β),

we see that ordPi = ei, ordQj = 0, and ordR = 0 for any other R. Thus(α, β) = I.

8.3 Residue fields

Let P ⊂ A be a prime ideal. Then A/P must be a field, and because it is finite,it must be isomorphic to k = Fq for some q = pf . Now let us look at P/P 2.Since P and P 2 are both A-modules, it inherits a natural A-module structure.But because the action of P is trivial, we see that P/P 2 is actually a k-vectorspace. More generally, for any e the quotient P e/P e+1 is a nontrivial k-vectorspace.

But what is its dimension? We use the finite approximation property. Con-sider any 0 6= α ∈ P e+1. Then we know that there has to be a β such thatP e = (α, β). But then P e/P e+1 = (β). This shows that

dimk(P e/P e+1) = 1.

Math 129 Notes 27

8.4 Unique factorization implies principal ideal domain

Theorem 8.4. Let A be a Dedekind domain. If A is a unique factorizationdomain, then it is a principal ideal domain.

This is interesting, because any principal ideal domain is a unique factor-ization domain, but not all unique factorization domains are principal ideal do-mains. For instance, the ring C[X,Y ] is not a principal ideal domain, becausethe ideal (X,Y ) is not principal.

Proof. We first note that if A were a UFD, for any α ∈ A, we see that α | δ · εimplies α | δ or α | ε.

Suppose that A is not a PID. Then there exists an ideal I that is not prin-cipal. Then there exists a prime ideal P that is not principal. Since A is aDedekind domain, there is an ideal M such that

P ·M = (α)

is principal; pick the maximal such M .

Math 129 Notes 28

9 February 23, 2016

Let K/Q be a finite degree extension that splits. Let A be the ring of integersin K, which is a Dedekind domain.

9.1 Quotient of ideals

Proposition 9.1. Let 0 6= I ( A be a proper ideal. Then I/I2 is a cyclicA/I-module.

Proof. Since A is a Dedekind domain, for any α ∈ I the ideal I is generated byα and some β ∈ I, i.e., I = (α, β).

Now fix any α ∈ I2. Then we have a β such that I = (α, β). Then I/I2 isgenerated by the image of β.

Corollary 9.2. Let P ⊂ A be prime ideal. Then we have a nice filtration

A = P 0 ⊃ P 1 ⊃ P 2 ⊃ P 3 ⊃ · · · ⊃ Pn ⊃ Pn+1 ⊃ · · · .

We first note that⋂∞n=0 P

n = 0. We call A/P = kP the residue field at P .Because Pn/Pn+1 is a cyclic module, it is a dimension 1 vector space. Thus|Pn/Pn+1| = |kP | = qP .

We note that qP can be the same for distinct P . For instance, (2 + i) and(1 + 2i) in Z[i] have the same qP .

Definition 9.3. We denote ‖I‖ = |A/I|.

Clearly, we have ‖P k‖ = qkP .

Proposition 9.4. Let I, J ⊆ A be ideals. Then ‖I · J‖ = ‖I‖ · ‖J‖.

Proof. Since the Chinese remainder theorem states that

A/∏P

P ordP I =∏P

A/P ordP I ,

we have

‖I‖ = ‖ordP I∏P

‖ =∏P

‖P ordP I‖ =∏P

qordP IP .

Then‖I‖ · ‖J‖ =

∏P

P ordP I+ordP J = ‖I · J‖.

Consider a tower of field extensions

I ·B ⊆ B L

I ⊆ A K

Z Q

n=[L:K]

Math 129 Notes 29

Proposition 9.5. ‖B/IB| = ‖I ·B‖ = ‖I‖n.

Proof. We note that it reduces to the case of I = P . That is, we need onlyprove that ‖P · B‖ = ‖P‖n for prime ideals P . We note that B/PB is aA/P = kP -module. Thus this is equivalent to dimkP B/PB ≤ n.

This is in the homework due this Friday, so let us prove only for A = Z. LetP = (p). Then B/PB = B/(pB), and since B is a free abelian group of rankn, we see that B/pB is a Fp-vector space of rank n.

Example 9.6. Let A = Z ⊃ (p). If B = Z[i] then

PB =

P · P p ≡ 1 (mod 4)

P 2 p = 2

P p ≡ 3 (mod 4).

Example 9.7. Consider the example:

B = C[√t] C(

√t)

(t− a) ⊆ A = C[t] C(t)

deg.=2

We see that(t− a)B = (

√t−√a)(√t+√a)B.

This ramifies if and only if t = 0.

9.2 Ramification indices and residue field degrees

More generally, we see that for

p ·B ⊆ B L

(p) ⊂ A = Z Q

n

we have

|B/pB| = ‖pB‖ =

r∏i=1

qeiPi ,

where (p)B =∏rii=1 P

eii .

Definition 9.8. We let

f(Pi|p) = (relative residue field degree of Pi over p) = [kPi : Fp]

and e(Pi|p) be the relative ramification index defined by

pB =∏PI

Pe(Pi|p)i .

Math 129 Notes 30

We see that

pn = ‖pB‖ =

r∏i=1

pe(Pi|p)·f(Pi|p).

Therefore we have the following theorem.

Theorem 9.9.[L : Q] =

∑Pi|pB

e(Pi|p)f(Pi|p).

More generally, for any field extensions L/K/Q and its corresponding ring ofintegers B → L and A → K. We can define e(Qj |P ) and f(Qj |P ) for Qj ⊂ Band P ⊂ A by

PB =∏j

Qe(Qj |P )j , f(Qj |P ) = dimkP kQj .

Then likewise we have the following.

Theorem 9.10.[L : K] =

∑Qj |PB

e(Qj |P ) · f(Qj |P ).

Example 9.11. Let K = Q[√D]. From the previous theorem, for any prime p

in Z, we have

p ·A =

r∏i=1

P eii ,

r∑i=1

eifi = 2.

There are three cases:r = 2 : ei = fi = 1 (p) = PP

r = 1 : e1 = 2, f1 = 1 (p) = P 2

r = 1 : e1 = 1, f1 = 2 (p) = P

9.3 Spectrum of a ring

Definition 9.12. For a commutative ring A, we define its spectrum as

SpecA = set of prime ideals of A.

For any ring homomorphism π : A→ B, it induces a map SpecB → SpecAjust by inverse images, because if Q ⊂ B is a prime ideal then π−1(Q) ⊂ A is aprime ideal.

Now let A and B be Dedekind domains.

K L

A B

Math 129 Notes 31

We first note that SpecA is simply the set of maximal ideals of A plus the zeroideal 0, or in other words, SpecA = max SpecA ∪ 0. Likewise, SpecB =max SpecB ∪ 0. Also, clearly 0 ∈ SpecB maps to 0 ∈ SpecZ, so we cansimply look at the map

max SpecB → max SpecA.

This map is surjective, because given any prime ideal P of A, we see that theideal generated by π(P ) is a prime ideal, and then SpecB → SpecA maps thatideal to P .

If the map SpecB → SpecA maps Q 7→ P , then we say that Q lies above Pand P lies below Q. This is something very pictorial. There are many equivalent

SpecC[√t]

SpecC[t]

conditions for Q lying above P . The following five conditions are all equivalentto Q lying above P .

(a) Q | PS (d) Q ∩R = P

(b) Q ⊃ PS (e) Q ∩K = P

(c) Q ⊃ P

Think about this example. Let f(X) ∈ Z[X] be an irreducible monic poly-nomial, and let A = Z[X]/(f(X)). Suppose that this is a Dedekind domain.Let us consider a prime p ∈ Z. Consider f(X) ∈ Fp[X] be the reduction off(X) modulo p. Think about how f(X) factorizes in Fp[X].

Math 129 Notes 32

10 February 25, 2016

We consider the following setting:

I ·B ⊆ B L

I ⊆ A K

Z Q

n=[L:K]

10.1 Decomposition equation

Theorem 10.1. Let I = P be a prime ideal of A, and let

PB =

r∏i=1

Qeii .

Let f(Qi|P ) be the relative field degree. Then

[L : K] =∑i

f(Qi|P )e(Qi|P ).

Note that we have proved this for A = Z. It was done by simply observingthat each sides are the exponent of p in

‖PB‖ = |B/pB| = p[L:K].

But if we prove ‖PB‖ = ‖P‖[L:K], then in general we will get the decompo-sition equation. That is, the following proposition is simply equivalent to thedecomposition equation.

Proposition 10.2.‖PB‖ = ‖P‖[L:K].

We shall prove the proposition using the following statement:

(∗) ‖PB‖ = ‖P‖n for some n ≤ [L : K].

The verification is left as an homework.

Proof. We look at the extension B over A over Z. Consider any prime (p) inZ and we look at the decomposition of pA and let it pA =

∏ri=1 P

eii . Then we

have ‖pA‖ = p[K:Q] and ‖pB‖ = p[K:Q][L:K].Now we have

‖pB‖B =∥∥∥ r∏i=1

P eii B∥∥∥B

=∏‖P eii ‖

niA

Math 129 Notes 33

for some ni ≤ [L : K]. When we look at the exponent of p on both sides, we get

[K : Q][L : K] =∑i

nie(Pi|p)f(Pi|p).

But since [K : Q] =∑i e(Pi|p)f(Pi|p) and ni ≤ [L : K], we see that all

ni = [L : K]. Thus we get the desired result.

Let f(X) be a monic irreducible polynomial in Z[X]. Let A = Z[X]/(f(X))and assume that A is integrally closed. We want to look at how the ideal (p)splits inside A. Let

f(X) ≡r∏i=1

fi(X)ei (mod p)

where each fi(X) is irreducible modulo p. Let Pi = (p, fi(X)) ⊂ A. Then wesee that

A/Pi = Z/pZ[X]/(fi(X))

is a field, and hence each Pi is a prime ideal. We also note that∏P eii = pA.

There are theorems such as the Chebotarev density theorem that describes howprimes split.

10.2 Ramification and the discriminant

Theorem 10.3. If (p) ⊂ Z ramifies in A, then p divides the discriminant ofK.

Proof. Let pA = P eP e22 P e33 · · · where e > 1. Then we can write pA = PI whereI is divisible by all prime ideals of A that divide p. Then I ) pA and thereforewe can pick an element α ∈ I such that α /∈ pA.

Consider α1, . . . , αn ∈ A that generated A as an abelian group. There is arepresentation

α = m1α1 +m2α+ · · ·+mnαn,

and since α /∈ pA, there is a mi that is not divisible by p. Let it be m1 be theone, and let A0 be the free abelian group generated by α, α2, . . . , αn. Then wesee that the index |A/A0| is m1 and thus discA = m2

1 discA0. Since p does notdivide m1, it suffices to that p divides discA.

Now α is in I, and hence α is in any prime ideal of A lying over p. Let L bea field extension of K that is Galois over Q, and let B be the ring of integersin L. Then we see that α is contained in any prime ideal of B, because such anideal will lie over some prime ideal of A lying over p. Let T be any prime idealof B lying over p. Then for any σ ∈ Gal(L/Q), we see that α ∈ σ−1(T ) andthus σ(α) ∈ T . This shows that the discriminant discK is in T and thus is inT ∩ Z = (p).

Math 129 Notes 34

10.3 The Kronecker-Weber theorem

Definition 10.4. A cyclotomic field is any extension of the form Q[ζn] whereζn is a primitive nth root of unity.

We note that Gal(Q[ζn],Q) = (Z/nZ)×.

Theorem 10.5 (Kronecker-Weber theorem). Any abelian extension over Q isa subfield of a cyclotomic extension of Q.

We first look at the case when Q[√a] is an abelian extension. Let

S =∑i

(a

p

)ζip

be the Gauss sum. We have

S2 =∑i,j

(i

p

)(j

p

)ζi+jp =

∑i,j

(ij2

p

)ζj(i+1)p =

∑i,j

(i

p

)ζj(i+1)p =

(−1

p

)p.

This show that√p is in some cyclotomic extension. Therefore any thing that

looks like√a/b is contained in some cyclotomic extension.

Math 129 Notes 35

11 March 1, 2016

11.1 The decomposition group

Let L/K be an Galois extension, and let G = Gal(L/K). Let A,B be the ringsof integers in L and K. Consider a prime ideal P ⊂ A and let Q1, . . . , Qr be allthe prime ideals lying over P . Any automorphism σ ∈ G acts and Qi and givesa prime ideal σQi lying over σP = P . The marvelous fact is the following.

Theorem 11.1. G acts transitively on Q1, . . . , Qr.

Clearly G acts on Q = Q1, . . . , Qr. We can define the decompositiongroup

D = GQi = σ ∈ G : σQi = Qi.

Since G/GQi has a one-to-one correspondence with the orbit of Qi, assumingthe theorem we will have the following corollary.

Corollary 11.2. G/GQi = G/D has a one-to-one correspondence to Q1, . . . , Qr.

Proof of theorem. Suppose that the orbit of Q1 and let Q1, . . . , Qs under theaction of G is smaller than Q, i.e., s < r. Then there is a prime ideal Q′ outsideQ1, ,Qs lying over P .

Now apply the finite approximation theorem and find an element β ∈ b withβ ≡ 1 (mod Qi) for i = 1, . . . , s and β ∈ Q′i. Consider its norm α =

∏σ∈G σβ

in A. Since none of the σβ is in Q1, we have α /∈ Q1. Since P = Q1 ∩ A, wehave α /∈ P . On the other hand, since β ∈ Q′ we have α ∈ Q′. Then A∩Q′ = Pcontains α. Thus we arrive at a contradiction.

Let PB = Qe11 Qe22 · · ·Qerr and [L : K] =

∑ri=1 eifi, where ei = e(Qi|P ) and

fi = f(Qi|P ). Since all prime ideals are Galois conjugates to each other, wehave can let f(Qi|P ) = f and e(Qi|P ). Then simply have

[L : K] = efr.

Now consider GQ = D, and let us look at the tower of extension:

L B Q

LD BD QD

K A P

Now the transitive theorem says that

efr = [L : K] = |G| = |D| · |G/D| = |D| · r

and thus |D| = ef .

Math 129 Notes 37

Example 11.5. Let α = 3√

14 and ζ = ζ3. The splitting field of

X3 − 14 = (X − α)(X − ζα)(X − ζ−1α)

is L = Q(α, ζ), with Galois group isomorphic to S3.Let us take a random prime 13, and see how it splits in L. Since

X3 − 14 ≡ X3 − 1 = (X − 1)(X − ζ)(X − ζ2) (mod 13),

we see that (13) splits into three primes Q1, Q2, Q3. Consider Q = Q1 andlook at the field LD. Then we see that P = (13) splits into PBD = QDR, andPB = Q1Q2Q3.

Math 129 Notes 38

12 March 3, 2016

L B Q kQ

LI BI QI kI

LD BD QD kQD

K A QP P

I C D D/I → Gal(kQ/kP )

This is our setting.

12.1 Ramification in the tower

We know that if we let Q1, . . . , Qr be the set of primes lying over P , thenG = Gal(L/K) acts on the Qi’s transitively. We recall that

G ⊇ GQ = σ : σQ = Q

and IQ is a kernel of the induced map GQ → Gal(kQ/kP ), or more explicitly,

GQ B IQ = σ : σ on B/Q is the identity.

Let us denote Di = GQi and LQi . These are also moved in a natural way byany automorphism of L over K. That is, if σQi = Qiσ , then

σLDi = LDiσ , σLIi = LIiσ , σGQiσ−1 = GQiσ , σIQiσ

−1 = IQiσ .

Proposition 12.1.f(Q|QI) = 1.

Proof. We see that it suffices to prove that any automorphism of B/Q that isthe identity on BI/QI is the identity on B/Q. This is because any finite fieldextension of finite fields is Galois.

B B/Q kQ

Bi BI/QI kQI

Let us consider any α ∈ B and look at the image α under the map B → B/Q.Let

g(X) =∏σ∈I

(X − σα) ∈ B[X]

Math 129 Notes 39

be the characteristic polynomial of α over LI . This has coefficients in LI andthus in BI . If we reduce it modulo QI , then we get

g(X) =∏σ∈I

(X − σ(α)) =∏σ∈I

(X − α).

Therefore the only place α can be sent is α.

Hence we have the exact sequence

0 IQ GQ Gal(kQ/kP ) 0

Example 12.2. Let a > 1 be an integer that is cube-free. Take α = 3√a and

L = Q[ζ3, α] be the splitting field of X3 − a = (X − α)(X − ζα)(X − ζ2α).This is an S3-Galois extension. Let A = Z and B be the ring of integers ofK = Q and L respectively. Consider an prime p not dividing 3a; this is takingan unramified prime.

Since p is unramified, we see that pB =∏ri=1Qi for some primes Qi in

B. Then 6 = f · r and we don’t have many choices. We can exclude the caser = 1 and f = 6, because this would say that kQ/kP is S3, but any Galoisgroup of finite extensions of finite fields is cyclic. So we have the following threepossibilities.

L = Q[ζ3, α]

Q[ζ]

K = Q

Q1 Q2

P1 P2

p

Q1Q2Q3 Q4Q5Q6

P1 P2

p

Q1 Q2 Q3

P

p

p ≡ 1 (mod 3) p ≡ 1 (mod 3) p ≡ −1 (mod 3)a 6≡ x3 (mod p) a ≡ x3 (mod p)

12.2 Ideal class group of Q[ζ23]3

Definition 12.3. Let R be a Dedekind domain. Let G denote the group offractional ideals, and let H be the subgroup of principal fractional ideals. Wedefine

h(R) = G/H

as the ideal class group of R.

We are going to show that the ideal class group of Q[ω], where ω = e2πi/23,is nontrivial. Let θ = (1 +

√−23)/2 and consider the ring of integers of Q[θ],

3This was a presentation by James Hotchkiss.

Math 129 Notes 40

which is Z[θ]. We have the following tower of extensions:

Q[ω] Z[ω]

Q[√−23] Z[θ]

Q Z

Let us take a prime ideal p = (2, θ) ⊂ Z[θ], and consider any q ⊂ Z[ω] lyingover p.

Lemma 12.4. f(q|p) = 11.

Proof. We note that (2) = (2, θ) · (2, θ) in Z[θ], and thus f(p|(2)) = 1. Also,f(q|(2)) is the multiplicative order of 2 modulo 23, and thus is 11. It followsthat f(q|p) = 11.

Lemma 12.5. In Q[ω], we have q = (2, θ) = pZ[ω].

Proof. We have [Q[ω],Q[√−23]] = 11, and since ref = 11 and f = 11, we have

r = e = 1. Thus we have the desired result.

Proposition 12.6. In Z[θ], we have (θ − 2) = p3.

Proof. Recall the ideal norm defined as

NQ[√−23]

Q (I) = Z ∩∏

σ∈Gal(Q[√−23]/Q)

I.

We proved in the homework that this is multiplicative, and N(Q) = pf(Q|(p)),and N((α)) = N(α)Z. Using this, we compute

N((θ − 2)) = (θ − 2)(θ − 2)Z = 8Z.

Also, N(p) = 2Z. But since the only primes lying over 2 is p and p, we see thateither (θ − 2) is either p3, p2p, pp2, or p3. Because (θ − 2) is in p but does notcontain pp = (2), we see that (θ − 2) = p3.

Lemma 12.7. f(q|p) = 1.

Proof. We see that f(q|(2)) is the order of 2 in (Z/23Z)×, which is 11. Thusf(p|(2)) is 1 or 11, but we can easily check that it cannot be 11. Thereforef(p|(2) is 1, and f(q|p) = 11.

Lemma 12.8. If K ⊂ L then there is an homomorphism ϕ : H(L) → H(K)given by

[I] 7→ [NLK(I)].

Math 129 Notes 41

Proof. We check that if I ∼ J then there exists α, β with αIβJ and thenN(α)N(I) = N(β)N(J).

Now let us look at NLK(q) = pf(q|p). It follows that ‖p‖ divides ‖q‖ · f(q|p).

Thus 3 divides 11 times the order of q in the ideal class group, and it followsthat H(L) is nontrivial.

Math 129 Notes 42

13 March 8, 2016

Suppose that L/K is Galois. We look at an intermediate extension L/K ′. Thenwe see that it is also Galois and from the definition we directly get

Lemma 13.1.

IQ(L/K ′) = IQ(L/K) ∩Gal(L/K ′),

GQ(L/K ′) = GQ(L/K) ∩Gal(L/K) ⊂ Gal(L/K).

Conversely, suppose that L′/K is a Galois sub extension so that L ⊃ L′ ⊃ K.We have a surjective map G(L/K) G(L′/K).

Lemma 13.2.G(L/K) G(L′/K)

GQ(L/K) GQ′(L′/K)

IQ(L/K) IQ′(L′/K)

Proof. We see that the map G(L/K)→ G(L′/K) indeed maps things into whatwe want. Now we need to show that the induced maps are surjective.

For the surjectivity of the map GQ(L/K) GQ′(L′/K), we look at the

action of Gal(L/L′) on the set of prime ideals Q1, . . . , Qr lying over Q′. Nowwe find a σ such that σ|L′ = σ.

The surjectivity of the map IQ(L/K) IQ′(L′/K) is left as an exercise.

13.1 Frobenius structure of a Galois group

Let P be a prime ideal in K and Q1, . . . , Qr be the prime ideals in L lying overK. We see that G acts transitively on Q1, . . . , Qr, and hence GQj and GQiare conjugates, and likewise IQj and IQi are conjugates. Thus we can simplyintroduce a notation GP and IP , which are defined up to conjugacy. ThenQP /IP = Gal(kQ/kP ) for any Q lying over P . Because only finitely many primeideals P are ramified in L/K, for any unramified P , we get QP ∼= Gal(kQ/kP ).

We know that for any finite field extensions Fqm/Fq, its Galois group iscyclic of order ω with the generator ϕq : x 7→ xq, which is called the relativeFrobenius map.

Now we have

Gal(L/K) ⊃ GP ∼= Gal(kQ/kP ) 3 ϕPfor any unramified P . Now this gives a canonical map

P 7→ ϕP conj. ∈ ConjClassGal(L/K).

This is called the Frobenius structure for the Galois group.The Frobenius structure gives a complete description of how the prime P

splits, because Gal(L/LD) = 〈ϕp〉 ⊂ Gal(L/K). For example, r = n can happenif and only if ϕP = 1 in Gal(L/K).

Math 129 Notes 43

Example 13.3. Let us look at L/K where L = Q(ζm) and K = Q. In this casethe Galois group is canonically G ∼= (Z/mZ)∗. For a prime p, the conjugacyclass ϕp is simply p modulo m.

The Chebotarev density theorem states that for any conjugacy class there areinfinitely many primes. From the previous example, we see that the Cheboratevdensity theorem implies the Dirichlet’s theorem on arithmetic progressions.

Math 129 Notes 44

14 March 10, 2016

Let L/K be a Galois extension and G = Gal(L/K). For any unramified primeP , there is a map

P 7→ φP ∈ ConjClass(Gal(L/K)),

and this φP completely determines the splitting behavior of P in L/K. Forinstance, what does φP = id mean? Since f is the order of φP in Gal(kQ/kP ),we see that φP = id is equivalent to P splitting completely in K/L.

Let us look at cyclotomic fields. Consider L = Q(ζn) over K = Q. Thenthe Frobenius element φ for p is the automorphism mapping ζn 7→ ζpn. Then psplitting completely in K is equivalent to p = 1 + mn for some m. Likewise,p being inert (remaining prime in L) is equivalent to p being a generator of(Z/nZ)×.

For fun, let us also look at when p totally ramifies in Q(ζn). If n = pa ·m, thefield Q(ζn) contains Q(ζm). Since p does not divide m, we see that p unramifiesQ(ζm). (We can simply compute the discriminant to see this.) Then the degreeϕ(m) = [Q(ζm) : Q] must be 1, and we conclude that m = 1 or 2. In fact, it isa nice exercise to check that p totally ramifies if and only if m = 1 or 2.

14.1 Finiteness of the ideal class group

Let A be the ring of integers of a field K over Q. Recall that ‖I‖ = |A/I|, andthat

‖(α)‖ = |A/αA| = |NK/Q(α)|.Also, we note that for a fixed B < +∞, there are only finitely many idealsI ⊂ A with ‖I‖ ≤ B.

Let us fix A ⊆ K.

Lemma 14.1. There exists a λ < +∞ such that for any nonzero ideal I, thereexists a α ∈ I such that |NK/Q(α)| ≤ λ‖I‖.

Proof. Let α1, . . . , αn be a Z-module basis of A. We are going to make a box ∑0≤hi≤m

hiαi

for an integer mn ≤ ‖I‖ < (m+1)n. Then the box maps to A/I, and the numberof elements in the box is greater than A/I. Now Dirichlet’s box principle saysthat there are two elements γ, δ that maps to the same element. That is,

0 6= γ − δ =∑

ciαi ∈ I

with |ci| ≤ m. I can now estimate the norm. From

NK/Qα =

n∏j=1

σj

(∑i

ciαi

)=

n∏j=1

∑i

ciσjαi

Math 129 Notes 45

it follows that

|NK/Qα| =∏j

∣∣∣∑i

ciσjαi =∏j

∑i

|ci||σjαi| ≤ mn∏j

∑i

|σjαi| ≤ ‖I‖λ

for λ =∏j

∑i|σjαi| not dependent on I.

Corollary 14.2. The ideal class group H(K) is finite.

Proof. For any class of ideals in H(K), let us pick an ideal I in it. By the lemmaabove, we have an ideal α ∈ I with small |NK/Q(α)|/‖I‖. If we let (α) = I · J ,then we have

|NK/Qα| = ‖I · J‖ = ‖I‖‖J‖ ≤ λ‖I‖.

Then ‖J‖ ≤ λ. This shows that there are only finitely many possibilities forthe inverse element of I in the ideal class group. Thus the ideal class group isfinite.

Example 14.3. Let us compute the ideal class group of K = Q(√d) for d =

2, 3,−5. In each case, we have λ = 5.8 . . . , 7.5 . . . , 10.7 . . . . Thus we can onlylook at ideals ‖I‖ with ‖I‖ ≤ 5, 7, 10. For such ideals, I, we have that

NK/QI = ‖I‖Z.

Thus we need only look at the prime factorization of (p) in L for p ≤ 5, 7, 10.For d = 2, we have

(2) = (√

2)2 (5) = (5)

(3) = (3)

and because all are principal, we see that H(K) is trivial, i.e., K = Q(√

2) is aprincipal ideal domain.

Likewise

(2) = (−1 +√

3)(1 +√

3) (5) = (5)

(3) = (√

3)2 (7) = (7)

and hence again H(K) is trivial and Q(√

3) is a principal ideal domain.In the d = −5 case, we have two ideals

I2 = (2, 1 +√

5), I3 = (3, 1 +√

5)

each with ‖I2‖ = 2 and ‖I3‖ = 3. With a little more work, we see that theseare inverses in the ideal class group and thus |H(K)| = 2.

Math 129 Notes 46

15 March 22, 2016

15.1 The Minkowski bound

Let K/Q be a finite extension of degree n. Recall that there are exactly nembeddings of K into C, with r of them real and 2s of them literally complex.Let σ1, . . . , σr : K → R be the real embeddings and let τ1, τ1, . . . , τs, τs : K → Cbe the purely complex embeddings. Then we have a fundamental embedding

K K ⊗Q R ∼= Rr × Cs ∼= Rn

than sends

α 7→ (σ1(α), . . . , σr(α), τ1(α), . . . , τs(α))

= (σ1(α), . . . , σr(α),<(τ1(α)),=(τ1(α)), . . . ,<(τs(α)),=(τs(α))).

Now with a suitable change of basis, we can change <(τi(α)) and =(τi(α)) intoτi(α) and τi(α). The determinant of this coordinate change matrix will be (2i)s.

Let A = α1Z + · · · + anZ. Then we see that the square of the determinantof the matrixσ1(α) · · · σr(α) τ1(α) τ1(α) · · · τs(α) τs(α)

.... . .

......

.... . .

......

σ1(α) · · · σr(α) τ1(α) τ1(α) · · · τs(α) τs(α)

is the discriminant of K.

Consider the image Λ of A under the fundamental embedding. This is alattice in Rn, and we can look at vol(Rn/Λ). This is defined as the volume ofthe fundamental parallelotope for Λ

F = t1α1 + · · ·+ tnαn : 0 ≤ ti < 1.

Then Rn is the disjoint union Rn = qλ∈Λ(F + λ). Because vol(Rn/Λ) =|Λ0/Λ| vol(Rn/Λ0), we immediately have the following theorem.

Theorem 15.1.vol(Rn/A) = 2−s

√|discA|.

Corollary 15.2.vol(Rn/I) = 2−s

√|discA| · ‖I‖.

This corollary can help us improve λ. Let us define a funny norm on Rn as

N(x1, . . . , xr, yr+1, . . . , yn) = x1 · · ·xr(y2r+1 + y2

r+2) · · · (y2n−1 + y2

n).

This is made so that if x is the image of α under the fundamental embedding,then N(x) = NK/Q(α).

Let us consider the ball

x ∈ Rn : |N(x)| ≤ t

with respect the the norm. We apply Minkowski’s theorem.

Math 129 Notes 47

Theorem 15.3 (Minkowski). Let Ω ⊂ Rn be a compact, convex, and centrallysymmetric set with positive volume. Let Λ be any lattice in Rn. If vol(Ω) ≥2n · vol(Rn/Λ) then Ω contains a nonzero lattice point in Λ.

Because we want to make sure something is in the ball, we want to find aconvex, symmetric Ω ⊂ x : |N(x)| ≤ 1 to have as large volume as possible.This is done by looking at the arithmetic norm

AN(x) =1

n

( s∑i=1

|xi|+ 2√y2r+1 + y2

r+2 + · · ·+ 2√y2n−1 + y2

n

).

Clearly, x : |N(x)| ≤ 1 ⊃ x : |AN(x)| ≤ 1, and Ω(1) = x : |AN(x)| ≤ 1 isconvex and centrally symmetric. Moreover, we can calculate its volume

vol(Ω(1)) =nn

n!2r(π

2

)s.

Because tΩ(1) = x : |AN(x)| ≤ t ⊂ x : |N(x)| ≤ tn, we can apply theMinkowski theorem to tΩ(1) and get the following corollaries.

Corollary 15.4. I contains a point x with

|N(x)| ≤ n!

nn

( 8

π

)svol(Rn/I).

Corollary 15.5. An ideal I contains a nonzero element α with

|NK/Q(α)| ≤ n!

nn

( 4

π

)s√|discA| · ‖I‖.

Corollary 15.6. Every ideal class contains an ideal J with

‖J‖ ≤ n!

nn

( 4

π

)s√|discA|.

Corollary 15.7.

|discA| ≥ nn

n!

(π4

)s.

Now the right hand side is strictly greater than 1 when n > 1.

Corollary 15.8. Every nontrivial extension K/Q has discriminant greater than1, i.e., has some prime that is ramified.

Math 129 Notes 48

16 March 24, 2016

Let K/Q be an extension of degree n. Let r and 2s be the number of real andcomplex embeddings. The fundamental embedding is the map K → K ⊗Q R ∼=Rr × Cs.

We used the “Existence of an element lemma” to get the Minkowski bound.

Lemma 16.1. Let Λ ⊂ Rn be a lattice. There exists an 0 6= x ∈ Λ such that|N(x)| ≤ 2n

V vol(R/Λ), where N is the funny norm.

Theorem 16.2. Every ideal class is represented by an ideal I with

‖I‖ ≤ n!

nn

( 4

π

)s√|disc(K)|.

16.1 Computation of ideal class group

Let us denote ∆ = disc(K).

Corollary 16.3. If

|∆| < 4n2n

(n!)2

(π4

)2s

,

then the ring of integers has trivial ideal class group, i.e., its class number is 1.In particular, this is true for K = Q(

√m) for m = 2, 3, 5, 13;−1,−2,−3,−7.

There is a big difference between the real quadratic case and the imaginaryquadratic case.

Theorem 16.4. The only imaginary quadratic fields of class number 1 is Q(√m)

for m = −1,−2,−3,−7,−11,−19,−43,−67,−163.

Conjecture 16.5. About 3/4 of the real quadric fields are of class number 1.

1. K = Q(√

7)

The discriminant is ∆ = 28 and we only have to check ideals of ‖I‖ ≤12

√∆ = 2.6 . . . . Then becuase (3 +

√7)(3 −

√7) = (2), we see that the

class number is 1.

2. K = Q(√−30)

The discrimant is ∆ = −120. Then because 12

√120 = 5.47 . . . , we have

to check for primes 2, 3, 5. Because they all divide 30, they all ramify into

(2) = P 22 , (3) = P 2

3 , (5) = P 25 .

They are all nontrivial, and also (√−30) = P2P3P5. This implies that the

ideal class group is isomorphic to C2 × C2.

Math 129 Notes 49

16.2 Higher ramification groups4

Let L/K be a Galois extension, and let S,R be the ring of integers in L and R.Consider a prime ideal p ⊂ R and a prime q ⊂ S lying over p. We define thehigher ramification group as

Vm = σ ∈ G : σ(α) ≡ α (mod qm+1).

We see that any σ ∈ G = Gal(L/K) induces an automorphism σ of S/qm+1

over R/qm+1 and hence we have a map

ι : G→ Aut((S/qm+1)/(R/qm+1)).

Lemma 16.6. The kernel of ι is ker ι = Vm.

We have a tower of normal subgroups of D

D ⊃ V0 = I ⊃ V1 ⊃ V2 ⊃ · · · .

We also have the following lemma, because the intersection of all of them istrivial.

Lemma 16.7. For all sufficiently large m, Vm = 1.

Theorem 16.8. Choose an π ∈ q−q2, and let σ ∈ I. Then σ ∈ Vm if and onlyif σ(π) ≡ π (mod qm+1).

Proof. We will first prove a slightly weaker statement.

Proposition 16.9. Choose an π ∈ q− q2, and let σ ∈ Vm−1. Then σ ∈ Vm ifand only if σ(π) ≡ π (mod qm+1).

Proof. Suppose that σ(π) ≡ π (mod qm+1). For any α ∈ πS, we have α = πβfor some β. Then

σ(πβ) = σ(π)σ(β) ≡ πβ (mod qm+1)

since σ ∈ Vm−1.Next we show σ(α) ≡ α (mod qm+1) for any α ∈ q. Because π ∈ q− q2, the

principal ideal (π) factorizes as

(π) = q∏i

qeii .

If we pick some β ∈∏

qeii then αβ ∈ πS while β /∈ q. Since

βσ(α) ≡ σ(β)σ(α) = σ(βα) ≡ βα (mod qm+1),

we can cancel out β and get the desired result.Lastly, we look at an arbitrary α ∈ S. This follows from the fact that it is

true for α ∈ q.5

4This was a presentation by Vikram Sundar.5I did not understand the argument.

Math 129 Notes 50

Now returning to the original problem, assume that σ ∈ I and σ(π) ≡π (mod qm+1). Suppose that σ ∈ Vi − Vi+1. Then because of the previousproposition, we see that i ≥ m. Hence σ ∈ Vm.

We now look at the structure of the ramification groups.

Theorem 16.10. For m ≥ 2, there is a canonical injection Vm−1/Vm → S/q.

Proof. We start with a lemma:

Lemma 16.11. Fix π ∈ q − q2. For each σ ∈ Vm−1, there exists an ασ ∈ Ssuch that

σ(π) ≡ π + ασπm (mod qm+1).

Moreover, ασ is unique modulo q.

Proof. Note that πm = qm − qm+1. Then σ(π) ≡ π (mod qm) by definition.Let (πm) = qm

∏qeii . By the Chinese remainder theorem, there exists a unique

x up to modulo q(πm) such that

x ≡ σ(π)− π (mod qm+1), x ≡ 0 (mod∏

peii ).

Letting x = απm, we get an α up to modulo q. It is clear that the congruenceabove is equivalent to σ(π) ≡ π + απm (mod qm+1).

This lemma gives a map Vm−1 → S/q. If fact, this is a homomorphismbecause

σ(τ(π)) ≡ σ(π + ατπm) ≡ σ(π) + σ(ατ )πm ≡ π + (ασ + ατ )πm (mod qm+1).

Also, the kernel is Vm by definition. This induces an injective map Vm−1/Vm →S/q.

Corollary 16.12. Let p = char(S/q). The quotient Vm−1/Vm is an abeliangroup of order a power of p.

Similarly, we have the following.

Theorem 16.13. There is a canonical injection I/V1 → (S/q)×.

Corollary 16.14. The quotient I/V1 is a cyclic group of order dividing |S/q|−1.

Math 129 Notes 51

17 March 29, 2016

We have the fundamental embedding K → E ∼= Rr × Cs as usual.

Lemma 17.1 (Pliability lemma). Let C1, . . . , Cr+s > 0 be real numbers andassume that

∏Ci ≥ (2/π)s

√|disc(A)|. Then there exists an element α with

|NK/Q(α)| < (2/π)s√|disc(A)|, and

|xi| ≤ Ci (i = 1, . . . , r), y2r+1 + y2

r+2 ≤ Cr+1, . . . , y22n−1 + y2

2n ≤ Cr+s.

Proof. The set |xi| ≤ Ci, y2r+2i−1 + y2

r+2i ≤ Cr+i is convex and centrally sym-metric. We can apply Minkowski’s theorem.

17.1 Logarithm of the fundamental embedding

The fundamental embedding maps A ⊂ K → Rr × Cs ∼= Rn. This map ishomomorphism between rings, and this induces a homomorphism of groupsA∗ ⊂ K∗ → (R∗)r × (C∗)s. I am going to write the group of units A∗ = U and(R∗)r × (C∗)s. Now we can map this to

A∗ ⊂ K∗ (R∗)r × (C∗)s Rr+s“log”

by sending

(x1, . . . , xr,yr+1, yr+2, . . . , yn−1, yn)

7→ (log|x1|, . . . , log|xr|, log(y2r+1 + y2

r+2), . . . , log(y22n−1 + y2

2n))

in the logarithm map. I am happy to call the composition also as log.

Proposition 17.2. If for α ∈ K∗ and log(α) = (a1, . . . , ar+s), then

r+s∑i=1

ai = log|N(α)|.

Proof. This is straightforward given the definition of the logarithm map.

Corollary 17.3. The map log maps U to H ⊂ Rr+s, where H = (a1, . . . , ar+s) :∑ai = 0.

The first question is, what is the kernel of U → H? It consists of roots ofunity, and hence is a finite cyclic group of roots of unity.

0

finitecyclicgroup

U H

Moreover, it has discrete image. (Think about this.)

Math 129 Notes 52

Theorem 17.4 (Dirichlet unit theorem). The image of U is a full lattice inH ∼= Rr+s−1.

Corollary 17.5. U is an abelian group isomorphic to the product of a finitecyclic group and Zr+s−1.

I will give the proof next time, and in the remaining minute, let us work outan example.

Example 17.6. Consider the case r + s − 1 = 1. If (r, s) = (2, 0), it is a realquadratic field. If (r, s) = (1, 1), it is a cubic filed with unique real embedding.In these cases, U is Z times a finite cyclic group.

Math 129 Notes 53

18 March 31, 2016

Recall the pliability lemma:

Lemma 18.1 (Pliability lemma). For any positive number C1, . . . , Cr+s with∏Ci ≥ (2/π)s

√|∆| = B, there is an α ∈ A such that |NK/Q(α)| ≤ B and

|σi(α)| ≤ Ci and |τj(α)|2 ≤ Cr+j.

The logarithm map is the group homomorphism defined as

U = A∗ K∗ (R∗)r × (C∗)s Rr+s.log

We noted that the image lies in the hyperplane H = ∑ai = 0, and the kernel

is the roots of unity.

18.1 Dirichlet unit theorem

Theorem 18.2 (Dirichlet unit theorem). The image of U under log, which isisomorphic to U/ ker log, is a lattice of rank r + s− 1.

Lemma 18.3. Let k be an integer with 1 ≤ k ≤ r + s and α ∈ A be a nonzerointeger. Let logα = (a1, . . . , ar+s). Then there exists a nonzero β ∈ A withlog β = (b1, . . . , br+s) such that |NK/Q(β)| ≤ (2/π)s

√|∆| and bi < ai for i 6= k.

The point is that we know nothing about bk at all!

Proof. Take Ci smaller than eai for i ≤ k and take any Ck so that∏Ci =

(2/π)s√|∆|. Then we get what we want from the pliability lemma.

Proof of the Dirichlet unit theorem. Take any nonzero α1 ∈ A. Using thelemma, we can inductively define a sequence of nonzero integers α1, α2, . . .

with logαi = (ai1 , . . . , a

ir+s) such that |NK/Q(αm)| ≤ B and

a1i > a

2i > a

3i > · · ·

for all i 6= k.Note that ‖(αm)‖ = |NK/Qαm| ≤ B. The sequence of ideals generated

by αm has bounded norm, and therefore there must be m < n such that(αm) = (αn). Then αm = ukα

nfor some uk, and this uk must be aunit. If we let log uk = (v1, . . . , vr+s), then by definition vi < 0 for i 6= k. Itautomatically follows that vk > 0 since they must add up to zero.

We claim that uk is linearly independent in the logarithm space (or equiva-lently, under multiplication). Consider the matrix

u1

...ur+s

v1,1 · · · v1,r+s

.... . .

...vr+s,1 · · · vr+s,r+s

Math 129 Notes 54

We want to prove that this matrix has rank r+ s− 1. It has a special property:all diagonal entires are positive and the off-diagonal entries are negative. Alsothe sum of each row is zero. This gives a linear relation above columns

∑ci = 0.

Suppose that there exists another linear relation∑tici = 0. We can suppose

that tk = 1 is the maximum and then 1 = tk ≥ ti for all other i. Let us look atthe kth entry. Then

0 =∑

tivk,i = vk,k +∑i6=k

tivk,i ≤ vk,k +∑i 6=k

vk,i = 0

where equality can hold only if all t1 = t2 = · · · = tr+s = 1. Therefore the rankis r + s− 1.

Example 18.4. Let K = Q[√D] with 0 < D 6≡ 1 (mod 4) squarefree. Then

because s = 0 and r = 2, we see that the set of units is ±1 × uZ for someu > 1.

18.2 The different ideal6

For a number field K, we denote by OK its ring of integers.

Example 18.5. Let α3 − α − 1 = 0 and consider the field K = Q(α). Thendisc(Z[α]) = 23. and so OK = Z[α] and discK = 23. We know that 23 ramifiesand because x3 − x− 1 = (x− 3)(x− 10)2 (mod 23), it splits into pq2. But wedon’t know which one ramifies.

Definition 18.6. Let L be a lattice in K. We define the dual lattice as

L∨ = α ∈ K : TK/Q(αL) ⊂ Z.

For a lattice L =⊕

Zei, we have L∨ =⊕

Ze∨i , where e∨i is the dual basisof ei with respect to the trace.

Proposition 18.7. For any lattice L, we have the following.

(1) L∨∨ = L

(2) L1 ⊂ L2 if and only if L∨1 ⊃ L∨2

(3) (L1 + L2)∨ = L∨1 ∩ L∨2

(4) (L1 ∩ L2)∨ = L∨1 + L∨2

(5) (αL)∨ = α−1L∨

Proposition 18.8. For a fractional ideal a in K, its dual a∨ is also a fractionalideal. Moreover, a∨ = a−1O∨K .

Proposition 18.9. OK is the largest fractional ideal in K whose elements allhave trace in Z.

6This was a presentation by Amanda Glazer

Math 129 Notes 55

Definition 18.10. We define the different ideal of K as

DK = (O∨K)−1 = x ∈ K : xO∨K ⊂ OK.

Example 18.11. Let K = Q(i). ThenOK = Z[i] and we see that Z[i]∨ = 12Z[i].

Then D = 2Z[i].

Theorem 18.12. The norm of the different ideal is ‖DK‖ = |disc(K)|.

Lemma 18.13. For any nonzero ideal a in OK , a | DK if and only if T (a−1) ⊆Z.

Theorem 18.14 (Dedekind). The prime ideal factors of DK are the primes inK that ramify over Q. More precisely, for any prime ideal p in OK lying overa prime number p with ramification index e = e(p | p), the exact power of p inDK is pe−1 if p - e and pe if p | e.

Example 18.15. Let us go back to the example α3 − α − 1 = 0. Because wehave 23 = (pq2) and N(DK) = |disc(K)| = 23, we have DK = q. Hence we cansimply compute DK to get q.

Math 129 Notes 56

19 April 5, 2016

Recall Minkowski’s bound: every ideal class has a representative ideal I with

‖I‖ ≤( 4

π

)r n!

nn

√|∆|.

A corollary is that every nontrivial field extension has a prime that ramifies.There is a function field analogue of the theory. Consider C[t] ⊂ C(t) be

the ring of integers and let K be a finite field extension of C(t) with A its ringof integers. The splitting of primes can be described geometry. Suppose thatA = C[t, y]/(y2−g(t)) where g(t) = t(t−1)(t−2). Each prime in SpecC[t] = C(we ignore the zero ideal) either splits into two discint primes, ramifies, or hasresidue degree 2. But there is no such thing as residue degree 2, because C isalgebaically closed.

SpecC[t]0 1 2

Now there is a geometric reason why there always has to be a ramification.Becasue C has trivial fundamental group, there cannot be a covering space withno ramification. Likewise in the Q case, what we are saying when we statethat every extension has a ramification is that SpecZ has trivial “fundamentalgroup.” But in this case, things are more subtle.

Example 19.1. Consider α be the root ofX5−X+1 and letK = Q[α]. BecausediscZ[α] = 5356 is squarefree, it is the ring of integers. The Minkowski boundtells us that to find H(K) we need only look at primes P with ‖P‖ < 4. Butthere are no prime ideals with norm 2 or 3, because the polynomial X5−X + 1has no root in F2 and F3. Therefore we conclude that A is an principal idealdomain.

Now let us look at the Galois closure L of K.

L

Q[√

19 · 151] K

Q

A5

5

It can be checked that L over Q[√

19 · 151] is unramified.

Math 129 Notes 57

Example 19.2. Let ζp be the pth root of unity. We have

L = Q(ζp)

L+ = Q(ζp)+ = Q(ζp + ζ−1

p )

Q

2

±1

p−1

p−12

This L+ is totally real, i.e., has only real embeddings. By the Dirichlet unittheorem, the unit group U(L+) has rank (p− 1)/2 + 0− 1 = (p− 3)/2 = ρ andL also has rank 0 + (p− 1)/2− 1 = ρ. Furthermore, because the roots of unitycontained in both fields are ±1 and µp, we have

U(L+) ∼= ±1 × Zρ ( µp × Zρ ∼= U(L).

For instance, the units of Z[ζ5] are just a root of unity times a unit of Z[(1 +√5)/2].

19.1 Counting ideals in ideal classes

Definei(t) =

∣∣ideals I in A with ‖I‖ ≤ t∣∣

and for an ideal class C ∈ H(K), define

iC(t) =∣∣ideals I in A with ‖I‖ ≤ t representing C

∣∣.Clearly we have

i(t) =∑

C∈H(K)

iC(t).

Theorem 19.3 (Equidistribution theorem).

iC(t)

t= κK +O(t−n

−1

),

where κK only depends on K and not on C.

If K is real and quadratic, then

κ =2 log u√|∆|

.

Also, we have a corollary i(t) = hκt+O(t1−n−1

).

Math 129 Notes 58

First fix C ∈ H(K) and an ideal J representing the inverse class C−1 ∈H(K). Then for any I ∈ C, the product IJ is principal and hence there is anumber αI for which

I · J = (αI).

This αI is well-defined only up to multiplication by U = A∗. The coset U · αIis really well-defined. Since

‖I · J‖ = ‖I‖ · ‖J‖ = ‖(αI)‖ = |NK/QαI |,

there is a correspondenceideals I ∈ Cwith ‖I‖ ≤ t

←→

UαI ⊆ J with

|NK/Q(αI)| ≤ t · ‖J‖

.

Let us look at the easy case K = Q(√−D). Then ∆ = −D,−4D and

w = |U | = 2, 4, 6. Also, ‖(α)‖ = |α|2. Because U is finite, we have a formula

iC(t) =|α ∈ J : |α|2 ≤ t‖J‖

w.

This is couting the elements of the lattice J ⊂ C that is contained in the circleof radius

√t‖J‖.

Math 129 Notes 59

20 April 7, 2016

Let K be a extension of Q and let H(K) be its ideal class group. Fix a classC ∈ H(K), and define

iC(t) = |I representing C : ‖I‖ ≤ t|.

20.1 Equidistribution of ideals in ideal classes

Theorem 20.1.iC(t) = κt+O(t1−

1n ).

Fix an ideal J representing C−1. Then for any I, there exists an αI suchthat IJ = (αI). Then

|NK/Q| = ‖(αI)‖ = ‖I‖‖J‖.

That is, there is a bijection

I : I ∈ C, ‖I‖ ≤ t ←→ UαI ⊂ J : |NK/QαI | ≤ t‖J‖

where U is the unit group. If we decompose U as U = W × V where W is afinite cyclic group and V is a free abelian group of rank r + s− 1, then we caneven write

|I : I ∈ C, ‖I‖ ≤ t| = |UαI ⊂ J : |NK/QαI | ≤ t‖J‖|

=|V αI ⊂ J : |NK/QαI | ≤ t‖J‖|

|W |.

Let us look at the special case K = Q(−√D) where D is squarefree and is

positive. Our problem is to count α ∈ J such that |α|2|t‖J‖. That is, we needto compute

iC(t) = |α ∈ J : |α|2 ≤ t‖J‖|.

In a more general context, for any lattice Λ ⊂ C let us count the latticepoint in the circle of radius ρ. If we denote the fundamental parallelogram by

F = t1λ1 + t2λ2 : 0 ≤ ti < 1,

where λ1, λ2 is the basis, we can tile the whole plane by translates of F as∐λ∈Λ(F + λ). Let

n(ρ) = #lattice-points in this circle,

n−(ρ) = #translates F + λ contained in this circle.

n+(ρ) = #translates F + λ that meet the circle.

Clearly n−(ρ) ≤ n(ρ) ≤ n+(ρ) and n−(ρ) vol(F ) ≤ πρ2 ≤ n+(ρ) vol(F ). More-over, n+(ρ)− n−(ρ) is the number of F + λ that intersect the boundary of the

Math 129 Notes 60

circle, and thus is contained in some “shell” around the boundary. If we denoteby δ the longest diagonal of F , we see that

(n+(ρ)− n−(ρ)) vol(F ) ≤ π(ρ+ δ)2 − π(ρ− δ)2.

It follows that

|n(ρ) vol(F )− πρ2| ≤ Cρ = O(ρ), n(ρ) =πρ2

vol(F )+O(ρ).

In out case, ρ =√t‖J‖ and vol(F ) = 1

2

√∆‖J‖. When we plug this in, we get

iC(t) =2π

|W |√|∆|

t+O(t1/2).

That is, κ = 2π/|W |√|∆| is independent of C. We have not used anything

about the ideal class group but somehow still have derived this result!Let us do some proof analysis. How much information about the geometry

of the circle do we need? If we have any region (closure of an open set) B anda lattice Λ, then

|Λ ∩ tB| ≤ vol(tB)

vol(Rn/Λ)+ ε(t)

by the same shell technique.

20.2 The Chebotarev density theorem

This was my presentation. See appendix.

Math 129 Notes 61

21 April 12, 2016

Let A ⊂ K be the ring of integers, and let C be an ideal class in the ideal classgroup H(K). Pick J be an ideal in the ideal class C−1. Then for all I thereis an αI such that IJ = αIU , where U is the unit group. The Dirichlet unittheorem states that U = W ×V where V is the free abelian group generated byu1, . . . , ur+s−1 and W is a finite cyclic group with w = |W |. We then knowthat

iC(t) = |I ∈ C : ‖I‖ ≤ t|= |(αI) : αI ∈ J and ‖(αI)‖ ≤ t‖J‖|= |αU : α ∈ J and |N(α)| ≤ t‖J‖|

=1

w|αV : α ∈ J and |N(α)| ≤ t‖J‖|.

21.1 Distribution of ideals in the class group

We also have the following diagram.

V (R∗)r × (C∗)s

0 H Rr+s R 0

log log

So far, everything is pretty much canonical, but we will now make a choiceand construct a splitting of the exact sequence on the bottom row. Let

Rr+s = H × R.

Since translates of the fundamental parallelotope tile H, we see that

Rr+s =∐λ∈ΛV

(FV × R + λ).

Define B = log−1(FV × R). Then it follows that

(R∗)r × (C∗)s = qu∈V uB.

Now this relevant to our discussion on iC(t). If we denote B(t) = x ∈ B :|N(x)| ≤ T, we have

iC(t) =1

w|αV : α ∈ J, |N(α)| ≤ T = t‖J‖|

=1

w|α ∈ B : α ∈ J, |N(α)| ≤ T|

=1

w|B(T ) ∩ J |.

Math 129 Notes 62

Here, B(T ) = t1/nB(1).Let us look at the case of the real quadratic field K. In this case, r = 2 and

s = 0, so the logarithm map is like log : (R∗)2 → R2 given by

(u, u′) 7→ (log|u|, log|u′|).

Then we have the following pictures.

B

H

FV × R

(R∗)2 R2

We have the general lemma:

Lemma 21.1. If Λ ⊂ Rn is a lattice and D is a “nice” region, then

|Λ ∩ tD| = vol(D)

vol(Rn/Λ)tn +O(tn−1).

Applying to our case, we get

|J ∩B(T )| = vol(B(‖J‖))vol(Rn/J)

t+O(t1−n−1

) =vol(B(1))

vol(Rn/A)t+O(t1−n

−1

).

Because we know that vol(R/A) = 2−s√|∆|, we finally get

iC(t) = 2svol(B(1))√|∆|w

t+O(t1−n−1

).

In the real quadratic case, the volume of the curved triangle can be computedeasily.

21.2 The regulator

We define the regulator as a generalization of log u. Let u1, . . . , ur+s−1 be thebasis for V . We are going to look at the matrix with rows(

log|σ1(uk)| · · · log|σr(uk)| 2 log|τ1(uk)| · · · 2 log|τs(uk)|).

This is an m× (m− 1) matrix, but all the row-sums are 0.We are now going to build a determinant out of this. Let that matrix be M ,

and let us augment this by adding a last row a = (a1, . . . , am) and make it intoM = M(a). Because all the row-sums are 0, if

∑ai = 0 then detM(a) = 0.

That is, M as a linear functional is a constant times∑ai.

Math 129 Notes 63

Definition 21.2. We define the regulator of K as

Reg(K) = R(K) = detM(a)

for any a with∑ai = 1.

Math 129 Notes 64

22 April 14, 2016

Let K be a number field with degree [K : Q] = d. For any class C, we have

iC(t) = |I ∈ C : ‖I‖ ≤ t| = vol(B(‖J‖))vol(Rd/J)w

t+O(t1−1/d)

=2s vol(B(1))√

|∆|wt+O(t1−1/d),

where B(T ) = x ∈ B : |Nx| ≤ T. We have defined something called aregulator, and in fact, there is the following proposition.

Proposition 22.1. vol(B(1)) = 2r · πs · Reg(K).

We shall not prove this, but you should read the proof in Marcus.There is a relation between the regulator and the class number.

Theorem 22.2 (Brouer-Siegel theorem). Let K1,K2, . . . be a sequence of Ga-lois extensions of Q. Let ni = [Ki : Q], hi = |H(Ki)|, Ri = Reg(Ki),∆i = |disc(Ki)|. If ni/ log|∆i| → 0 as i→∞, then

limi→∞

log(hiRi)

log√|∆i|

= 1.

That is, the product hR is “going to be” more regular than h or R separately.As a related topic, there is the Gauss class number problem. There is a

beautiful expository paper written by D. Goldfeld, Gauss’ Class Number Prob-lem and Imaginary Quadratic Fields, BAMS (1985). It will be totally accessibleto you, and it gives the effective estimate h(D) > c(log|∆|)1−ε for every ε > 0and some c = cε.

22.1 Dirichlet Series

Let us now move on to the next topic. Consider the series

∞∑n=1

anns

for a complex s = x+iy. We have to check whether the series converge absolutelyand uniformly on some compact set.

Lemma 22.3. Let An = a1 + · · · + an. If An = O(nr) for some r ≥ 0, thenD(s) converges for <(s) = x > r.

Proof. We have

M∑n=m

anns

=

M∑n=m

Anns−

M∑n=m

An−1

ns

=AMMs− Am−1

ms+

M−1∑n=m

An

( 1

ns− 1

(n+ 1)s

).

Math 129 Notes 65

Because∣∣∣( 1

ns− 1

(n+ 1)s

)∣∣∣ =∣∣∣ ∫ n+1

t=n

sdt

ts+1

∣∣∣ ≤ |s|∣∣∣ ∫ n+1

n

At

tx+1

∣∣∣ ≤ |s|n−(x+1),

we have the estimate

M∑n=m

ann3

=AMMs− AM−1

ms+

M−1∑n=m

An

( 1

ns− 1

(n+ 1)s

)≤ B

(Mr

Mx+mr

mx+ |s|

M∑n=m

nr−x−1

).

It follows that the sum converges absolutely.

We can now define the functinos we are really intereseted in.

Definition 22.4. Let K/Q be a finite extension. We define the Dedekindzeta function as

ζK(s) =

∞∑n=1

jnns

where jn = I ⊂ A : ‖I‖ = n.

22.2 Minkowski’s theorem7

Minkowski’s theorem was proved by 1889 and was used to prove Lagrange’sSquare theorem. The statement is as follows:

Theorem 22.5. Let Ω ⊂ RN be a convex, centrally symmetric body with volumevol(Ω) > 2N . Then Ω contains a non-zero lattice point.

Proof. We can write the volume condition as vol(12Ω) > 1. Then there exist

two points P,Q ∈ 12Ω whose difference has integer coordinates. It follows that

P −Q = 12 (2P + (−2Q)) is in Ω.

7This was a presentation by Salash Nabaala.

Math 129 Notes 66

23 April 19, 2016

Recall that we defined the Dirichlet series as a series of the form

D(s) =

∞∑n=1

anns

for s ∈ C, and we have shown the following lemma.

Lemma 23.1. If a1 + · · · + an = O(nr), then D(s) converges uniformly in acompact subset of the half-plane <(s) > r.

We have also defined the Dedekind zeta function for a given number field Kas

ζK(s) =

∞∑n=1

jnns

where jn = |I ⊂ A : ‖I‖ = n| is the number of ideals with norm exactly n.We already have an estimate of the number of ideals with norm at most t:

i(t) =∑k≤t

jk = hK · κ · t+O(t1−1/n).

Then An = O(n) and hence we have the following.

Proposition 23.2. The Dedekind zeta function ζK(s) converges (absolutely,uniformly in a compact subset of) in the half-plane <(s) > 1.

23.1 The L-function

We can also define the partial Dedekind zeta function. Fix a C ∈ H(K) anddefine jn,C = |I ∈ C : ‖I‖ = n|. Let

ζK,C(s) =

∞∑n=1

jn,Cns

.

Then by the equidistribution theorem, we again have the same result for ζK,C(s).Now consider a character, i.e., a group homomorphism

χ : H(K)→ C∗.

We are going to “weight” the partial Dedekind zeta functions by this characterand define the L-function

L(K,χ; s) =∑

C∈H(K)

χ(c)ζK,C(s) =

∞∑n=1

∑C∈H(K) χ(C)jn,C

ns.

Math 129 Notes 67

Let us check where this converges. By the equidistribution theorem,

a1 + · · ·+an =∑k≤n

jk,χ =∑

C∈H(K)

χ(C)iC(n) =

( ∑C∈H(K)

χ(C)

)κt+O(t1−1/n).

But the sum of the character is 0 unless χ is the trivial character 1.8 So if χ istrivial, we retrieve

L(K,1; s) = ζK(s)

and if χ is nontrivial, then L(K,χ; s) converge for <(s) > 1−1/n. In particular,it converges in a neighborhood of s = 1.

As a general remark, the Riemann zeta function is the case K = Q andis simply ζ(s) = ζQ(s) =

∑∞n=1 1/ns. This actually extends to be a holomorphic

function on C \ 1 and is meromorphic at s = 1 with a simple pole. That is,(s− 1)ζ(s) is holomorphic everywhere.

23.2 Extending the zeta function

Going back to our discussion on the Dirichlet zeta function, we want to extendthe range of convergence of the Dirichlet zeta function. We first look at theRiemann zeta function ζ(s). Let us consider the analytic function

εa(s) = 1− a1−s = 1− e(log a)(1−s).

The zeros are the number of the form sk = 1 + 2πik/ log a, where k ∈ Z. Notethat if p and q are distinct primes, then εp(s) and εq(s) have exactly one commonzero, namely s = 1. We are going to look at the “p-deprived” Riemann zetafunction

ζ(p)(s) = (1− p1−s)ζ(s) = (1− p1−s)

∞∑n=1

1

ns=

∞∑n=1

cp,nns

,

where cp,n = 1 if p - n and cp,n = 1 − p. The interesting fact is that inthis case, a1 + · · · + an = O(n0)! It follows that (1 − p1−s)ζ(s) is convergenton (uniformly on a compact subset of) the domain <(s) > 0. That is, ζ(s)is at least meromorphic for <(s) > 0 and has poles possibly at the zeros ofεp(s) = 1− p1−s. The only common pole is placed at s = 1, and hence we havethe following proposition.

Proposition 23.3. The Riemann zeta function ζ(s) extends to a meromorphicfunction on the half-plane <(s) > 0 and as a simple pole at s = 1.

Using this fact, we can extend the Dedekind zeta function to a larger domain.

Theorem 23.4. The Dedekind zeta function ζK(s) extends to the half-plane<(s) > 1− 1/n and has a simple pole at s = 1.

8Let G be any finite group and let χ : G→ C∗ be a character. If χ is nontrivial, i.e., thereexists a g0 with χ(g0) 6= 1, then

∑χ(g) =

∑χ(g0g) = χ(g0)

∑χ(g) and hence the sum must

be zero.

Math 129 Notes 68

Proof. Let us consider a single ideal class C. We have

ζK,C(s) =

∞∑n=1

jn,Cns

=

∂∑n=1

jn,Cκ

+ κ

∞∑n=1

1

ns.

The first sum converges on <(s) > 1 − 1/[K : Q], because j1,C + · · · + jn,C =O(n1−1/[K:Q]). The second some, as we have shown above, can be extendedto <(s) > 0 with a pole at s = 1. Therefore, ζK,C(s) can be extended to ameromorphic function on <(s) > 1 − 1/[K : Q] with a simple pole at s = 1.Then we can add all these up to see that ζK(s) can be extended to <(s) >1− 1/[K : Q].

Math 129 Notes 69

24 April 21, 2016

If you remember, we used 1− a1−s = fa(s) as a tool to extend analyticity. Weestimate

fa(s) = 1− exp((log a)(1− s)) = 1− 1− (log a)(1− s)(1 +O(1− s))= (1− s) · (c+O(1− s)).

Then fp(s)ζ(s) is defined analytically near s = 1. It follows that ζ(s) · (s − 1)is analytic near s = 1 and is nonzero at s = 1. Using this we showed that ζK,Cextends to a meromorphic function at <(s) > 1− 1/[K : Q] with a simple poleat s = 1. We simply collected a κ from all the terms and made

∞∑n=1

jn,Cns

=∞∑n=1

jn,C − κns

+ κζ(s).

Then the partial sums are O(n1−1/[K:Q]).

24.1 Infinite products

We want to change the zeta functions into infinite products like

m∏i=1

(1− ai)−1 =∑

(r1,...,rj)

m∏i=1

arii .

But we need to make sure the infinite product as m goes to ∞ makes sense.To achieve this, assume that |ai| < 1 and

∑∞i=1|ai| < ∞. We want to first

make sure that∏∞i=1(1 − ai)−1 makes sense. In the case in which all ai are

positive reals, we can use the fact that

log(1− x)

x→ 1

as x → 0. Then the fact that∑|ai| converges implies that the product con-

verges. In the complex case, there is an issue of log z being defined only up toa integer multiple of 2πi, but we can simply choose the one closest to 0 and dothe same thing. Once we have that the sum of the logs converges, then we canexponentiate to get what we want.

So taking m to ∞, and using unique factorization, we can say the following.

Theorem 24.1. If <(s) > 1, then

ζK(s) =∏P

(1− 1

‖P‖s)−1

.

If we can express ζK(s) as the infinite product like the above, then how canζK,C(s) be expressed? If we simply take the product over primes in C, then∏

P∈C

(1− 1

‖P‖s)−1

=∑

I divisible only by P∈C

1

‖I‖s,

Math 129 Notes 70

which is not really satisfactory. Instead, if we look at the L-function,

L(K,χ, s) =∏P

(1− χ(P )

‖P‖s)−1

because χ(∏Pi) =

∏χ(Pi).

24.2 Density of primes

Generally, let Π be the set of all primes, and let P be any subset of Π. Our goalis to measure how “large” is P compared to Π.

1. Natural density:We define

δnatural = limX→∞

|P ∈ P : ‖P‖ ≤ X||P ∈ Π : ‖P‖ ≤ X|

.

This is the only density that really speaks to you. But this is veryhard to show that the natural density exists. So we sometimes use other“smoother” densities.

2. Polar density:We first define a zeta function

ζK,P(s) =∏P∈P

(1− 1

‖P‖s)−1

.

We say that P has polar densitym/n if ζK,P(s)n extends to a meromorphicfunction about s = 1 and has a pole of order m at s = 1.

24.3 Odlyzko’s bound

Let r1 and r2 be r and s in the usual class notation. (We need to reserve s forthe complex variable.) The Minkowski bound says that

log|Dk| ≥ (2− o(1))n− 2(log(4/π))r2.

We’ve only extended the zeta function to <(s) > 1−ε, but we have the functionalequation

ΛK(s) =

(√|DK |

2r2πn/2

)sΓ(s

2

)r1Γ(s)r2ζK(s)

=

(√|DK |

2r2πn/2

)1−s

Γ(1− s

2

)Γ(1− s)r2ζK(1− s).

If we define Λ∗K(s) = s(s− 1)ΛK(s), there is a Hadamard product formula

Λ∗K(s) = ea+bs∏ρ

(1− s

ρ

)es/ρ,

Math 129 Notes 71

where we take the product over the nontrivial zeros ρ.Now when you take the log of the Hadamard product formula and take the

derivative and do some stuff, you get

log|DK | = r1

(log π − Γ′

Γ

(s2

))+ 2r2(log 2π − Γ′

Γ(s))

− 2

s− 1− 2

s− 2

ζ ′KζK

(s) + 2∑ρ

< 1

(s− ρ).

for s ∈ R and is greater than 1. (This formula is due to Artin.)We need a lemma:

Lemma 24.2. The inequality

∂m

∂sm

(− ζ ′KζK

)(s) > 0

holds if and only if m is even.

If we take the mth derivative with respect to s, then we get

δm log|DK | > (−1)m[r1∂m

∂sm

(log π − Γ′

Γ(x/2) + r2

∂m

∂ms

(log 2π − Γ′

Γ(s)

]+m!,

where δm is 1 if m = 0 and 0 otherwise. Plugging some appropriate values in,we get

|Dk|1/n > (22.38)r1/n(11.9)2r2/n.

If you do some more work, you get

|Dk|1/n > (215)r1/n(44.7)2r2/n,

and this is Odlyzko’s bound.

Math 129 Notes 72

25 April 26, 2016

We have the curious lemma that says if∑|ai| <∞ and |ai| < 1 then

∏(1−ai)−1

converges. Using this, we established the infinite product structure

ζK(s) =∏P

(1− 1

‖P‖s)−1

=∑I

1

‖I‖s

of the zeta function.Let S be a finite set of primes of K. Let us define

ζSK =

∏P /∈S

(1− 1

‖P‖s)−1

.

Then this can be written as a product of a harmless function and ζSK as

ζSK (s) =

∏P∈S

(1− 1

‖P‖s)ζSK (s).

Then because ζK is meromorphic on <(s) > 1−1/n with a simple pole at s = 1,

the new ζSK also has the same behavior.

25.1 Polar density

Let P be a set of primes of K. We are going the define a notion of some densityof P in the set of all primes of K. Let us define

ζK,P(s) =∏P∈P

(1− 1

‖P‖s)−1

.

We would like the have an “analytic” extension to a neighborhood of s = 1.

Definition 25.1. If (ζK,P(s))n can be extended to a neighborhood of s = 1and has a pole of order m at s = 1, then we say

δpolar(P) =m

n.

This is the same as saying that (s−1)mf(s)n can be extended to an analyticfunction around 1 with nonzero at s = 1. The density is well-defined, because(s − 1)mλf(s)nλ = ((s − 1)mf(s)n)λ. Also, with the same type of argument,you can check that if P1 and P2 are disjoint, then

δpolar(P1 ∪ P2) = δpolar(P1) + δpolar(P2),

assuming that two of them exist.We also note that (this is in one of the problem set) if P has a polar density,

then the Dirichlet density

δDir(P) = lims→1+

∑P∈P‖P‖−s∑P ‖P‖−s

exists and is equal to the polar density.

Math 129 Notes 73

25.2 A density theorem

Theorem 25.2. Let L/K be a Galois extension, and let P the set of primesof K that split completely in L. Then P has a polar density, and it is equal to1/[L : K].

Corollary 25.3. Any L/K has the property that infinitely many primes of Ksplit completely in L.

Proof. Let M be the Galois closure of L/K. Then a prime P of K splitscompletely in F if and only if it splits completely in L. The corollary follows.

Example 25.4. Any cubic (non Galois) extension of Q has one sixth of the setof rational primes completely splitting in it.

Because there are only finitely many primes that ramify and we don’t wantto see them, we let

ζ ′K,P(s) =∏P∈P,

P unramified

(1− 1

‖P‖s)−1

.

Let R be any set of primes of L that contain all primes that split completelyover Q. Then

ζ ′L,R(s) ·D(s) = ζ ′L(s)

where

D(s) =∏

P /∈R,unramified

(1− 1

‖P‖s)−1

.

Because all such P will have norm ‖P‖ = pe for e > 1, we see that D(s) can beextended to an analytic function around s = 1.

Now let Q be the set of primes of L that have split completely in L/K andlet P be the set of primes of K that split completely in L. Then there is a[L : K]-to-1 surjection

Q → P,

and if Q lies over P then ‖Q‖ = ‖P‖. It follows that

ζL,Q(s) = ζK,P(s)[L:K].

This is true for <s > 1, so if we extend it to a neighborhood of s = 1 they willalso agree. Hence we get the theorem.

25.3 Cyclotomic units9

Let ζn be a nth root of unity, and let K∗ denote the units of A ∩ K and K+

denote the maximal real subfield.

9This was a presentation by David Mende.

Math 129 Notes 74

Definition 25.5. Let Vn be the multiplicative group generated by

±ζn ∪ 1− ζan : 1 ≤ a < n.

Then define Cn = Vn ∩Q[ζn]∗ and in general, for any K ⊂ Q[ζn], define CK =Cn ∩K∗.

Our goal is to show that [Q(ζn)∗ : Cn] is finite. We first do it for n = pm.

Proposition 25.6. The set −1, ξa generate C+Q(ζn), where

ξa = ζ(1−a)/2 1− ζa

1− ζ= ± sin(πa/pm)

sin(π/pm)

for gcd(a, p) = 1 and 1 < a < pm/2.

Sketch of proof. A general element of Vpm will have the form

ξ = ±ζd∏

1≤a<pm/2gcd(a,p)=1

(1− ζa)ca .

Using some facts, we see that∑ca = 0. Then

ξ = ±ζe∏(1− ζa

1− ζ

)ca= ±ζe

∏ξcaa .

Because ξ ∈ C+Q(ζn) must be real, we see that ξ is generated by ξas and −1.

To show that the set ξa actually generate CQ(ζ)+ , we compute the regulatorof ξa. It is

Reg(ξa) = ±det(log|ξ+a |) =

∏ ∑1≤a<pm/2

χ(a) log|1− ζa|

=∏−τ(χ)L(1, χ)/2 6= 0.

Math 129 Notes 75

A The Chebotarev density theorem

A.1 The Frobenius element

Let L/K be a Galois extension, and let P be a prime lying over p. Recall thatthe decomposition and interia groups associated to P are defined as thus:

D(P|p) = σ ∈ Gal(L/K) : σP = P,I(P|p) = σ ∈ Gal(L/K) : σα ≡ α (mod P) for all α ∈ OL.

We have a tower of extensions.

L OL P

LI OLI PI

LD OLD PD

K OK p

e total ram., no res.

f no ram., total res.

r no ram., no res.

If we assume that p does not ramify, then in that case I is trivial and the map

D(P|p) Gal(kP/kp)

is an isomophism. The Galois group Gal(kP/kp) is generated by the Frobeniusmap α 7→ α‖p‖. Using the isomorphism, we can canonically define the Frobeniuselement φ(P|p) ∈ D(P|p) ⊆ Gal(L/K) that corresponds to the Frobenius mapon the residue fields. Since φ(σP|p) = σφ(P|p)σ−1 for any σ ∈ Gal(L/K), anunramified prime p canonically gives a conjugacy class

φ(P|p)conj. ∈ ConjClass(Gal(L/K))

of the Galois group. We shall denote this conjugacy class by Frobp.

A.2 The Chebotarev density theorem

The Chebotarive density theorem describes the distribution of primes with agiven Frobineous element. Chebotarev proved the theroem in 1922, and theideas used in the proof inspired Artin to prove his reciprocity theorem.

Before giving the statement of the theorem, we need a notion for measuringhow many primes there are.

Definition A.1. Let M be a set of prime ideals of K. We define the Dirichletdensity of M as

d(M) = lims→1+

∑p∈M‖p‖−s∑p‖p‖−s

if the limit exists. Likewise, we define the natural density of M as

δ(M) = limx→∞

#p ∈M : ‖p‖ ≤ x#p : ‖p‖ ≤ x

.

Math 129 Notes 76

If the natural density exists, then the Dirichlet density also exists and theyare the same. However, the existence of the Dirichlet density does not implythe existence of the natural density.

Theorem A.2 (Chebotarev density theorem). Let L/K be a finite Galois ex-tension between number fields and let C be a conjugacy class of Gal(L/K).Then

d(p : Frobp = C) =|C|

[L : K].

In fact, the natural density of the set also exists and is equal to |C|/[L : K].

A.3 Consequences

There are direct corollaries to the theorem concerning the densities of certainclasses of primes. Historically, the Dirichlet density theorem and the Frobeniusdensity theorem was generalized by Chebotarev, but it is still worthwhile to seehow Chebotarev’s theorem implies the other ones.

Theorem A.3 (Dirichlet density theorem). Let m ≥ 2 be an integer and abe relatively prime to m. Then the primes of the form mk + a have Dirichletdensity 1/ϕ(m).

Proof. Consider the field L = Q[ζm]. Because this is the splitting field of Xm−1,it is Galois, and its Galois group can be identified with (Z/mZ)× through themap (Z/mZ)× → Gal(L/Q) given by

a 7→ (ζm 7→ ζam).

Consider a rational prime p ∈ Z that is unramified. Because Gal(L/Q) isabelian, all conjugacy classes have size 1. It follows that if Frobp = φ then

φ(α) ≡ αp (mod P)

for any α ∈ OL and all primes P lying over p. Because pOL is the product ofall such P, we further have

φ(α) ≡ αp (mod pOL).

We note the only automorphism given by ζm 7→ ζpm onlysatisfies that criterion.Therefore Frobp = p ⊂ (Z/mZ)×.

Applying the Chebotarev density theorem, we immediately get that the setof primes that are a modulo m has density 1/ϕ(m), because [L : Q] = ϕ(m).

Consider a monic polynomial f with integer coefficients that does not havea double root. For each p, the reduced polynomial f ∈ Z/pZ[X] might or mightnot be reducible. If the factorization has polynomials of degree n1, . . . , nt(wheren1 + · · ·+ nt = n), then say that p has decompoisition type (n1, . . . , nt).

Let L be the splitting field of f over Q. Then L is Galois over Q andcontains all the roots α1, . . . , αn, where deg f = n. Each element σ ∈ Gal(L/Q)

Math 129 Notes 77

permutes these roots and thus divides them into disjoint cycles. We say thatσ ∈ Gal(L/Q) has cyclic pattern (n1, . . . , nt) if the action of σ on α1, . . . , αnis composed of cyclics of size n1, . . . , nt.

Theorem A.4 (Frobenius density theorem). The density of the set of primesp for which f has a given decomposition type (n1, . . . , nt) exists, and is equal to1/[L : Q] times the number of σ ∈ Gal(L/Q) with cyclic pattern (n1, . . . , nt).

Proof. Clearly the set

S(n1,...,nt) = σ ∈ Gal(L/Q) : σ has cyclic pattern (n1, . . . , nt)

is stable under conjugation and hence is a disjoint union of conjugacy classes.Consider a prime p with decomposition type (n1, . . . , nt). Because we are

concerned with the density of primes, we may further assume that p does notdivide the discriminant of f . It suffices to prove that the Frobenius elementFrobp consists of automorphisms of cyclic pattern (n1, . . . , nt). Then we wouldbe able use the Chebotarev density theorem to conclude that the density ofprimes with decomposition type (n1, . . . , nt) is |S(n1,...,nt)|/[L : Q].

Let us look at one prime P lying over p. Let r1, . . . , rd be the roots of f sothat

f(X) = (X − r1)(X − r2) · · · (X − rd).

When we reduce it modulo P, we get

f(X) = (X − r1)(X − r2) · · · (X − rd) ∈ (OL/P)[X]

where ri is the image of ri under the projection map OL → OL/P. It fol-lows from the assumption that p does not divide disc f that all ri are distinct.(If ri = rj then ri − rj ∈ P and hence disc f ∈ P.) Because the Galoisgroup Gal((OL/P)/(Z/pZ)) is generated by σ, a polynomial in (OL/P)[X] isin (Z/pZ)[X] if and only if it is invariant under σ. If σ has a cyclic structure

σ : r1 7→ r2 7→ · · · 7→ rn 7→ r1,

then

σ

( n∏i=1

(X − ri))

=

n∏i=1

(X − ri+1) =

n∏i=1

(X − ri)

and hence (X − r1) · · · (X − rn) is fixed under σ. However, every proper divisorof (X − r1) · · · (X − rn) that is not 1 is not fixed by σ. Therefore

(X − r1) · · · (X − rn) ∈ (Z/pZ)[X]

is irreducible. It follows that f(X) factors modulo p exactly accoriding to thecyclic structure of σ.

Index

L-function, 66S-number, 12, 20

algebraic integer, 4algebraic number, 4

cyclotomic polynomial, 9

Dedekind domain, 21Dedekind zeta function, 65density of primes, 70different ideal, 55Dirichlet unit theorem, 53discriminant, 16

of a number field, 19dual lattice, 54

fractional ideal, 22Frobenius structure, 42fundamental embedding, 11, 14

Galois conjugate, 7Galois size, 12

ideal class group, 23inertia group, 36integral closure, 13

norm, 11

order, 25

regulator, 63residue field, 28Riemann zeta function, 67

spectrum of a ring, 30

tensor product, 13trace, 11

Weil number, 14

78

math 129 - number fields · 2018-12-12 · math 129 - number fields taught by barry mazur notes by...

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