Part II — Number Fields

Based on lectures by I. Grojnowski Notes taken by Dexter Chua Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine.

Part IB Groups, Rings and Modules is essential and Part II Galois Theory is desirable

Definition of algebraic number fields, their integers and units. Norms, bases and discriminants. [3] Ideals, principal and prime ideals, unique factorisation. Norms of ideals. [3] Minkowski’s theorem on convex bodies. Statement of Dirichlet’s unit theorem. Deter- mination of units in quadratic fields. [2] Ideal classes, finiteness of the class group. Calculation of class numbers using statement of the Minkowski bound. [3] Dedekind’s theorem on the factorisation of primes. Application to quadratic fields. [2] Discussion of the cyclotomic field and the Fermat equation or some other topic chosen by the lecturer. [3]

1 Contents II Number Fields

Contents

0 Introduction 3

1 Number fields 4

2 Norm, trace, discriminant, numbers 10

3 Multiplicative structure of ideals 17

4 Norms of ideals 27

5 Structure of prime ideals 32

6 Minkowski bound and finiteness of class group 37

7 Dirichlet’s unit theorem 48

8 L-functions, Dirichlet series* 56

Index 68

2 0 Introduction II Number Fields

0 Introduction

Technically, IID Galois Theory is not a prerequisite of this course. However, many results we have are analogous to what we did in Galois Theory, and we will not refrain from pointing out the correspondence. If you have not learnt Galois Theory, then you can ignore them.

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1 Number fields

The focus of this course is, unsurprisingly, number fields. Before we define what number fields are, we look at some motivating examples. Suppose we wanted to find all numbers of the form x2 + y2, where x, y ∈ Z. For example, if a, b can both be written in this form, does it follow that ab can? In IB Groups, Rings and Modules, we did the clever thing of working with Z[i]. The integers of the form x2 + y2 are exactly the norms of integers in Z[i], where the norm of x + iy is

N(x + iy) = |x + iy|2 = x2 + y2.

Then the previous result is obvious — if a = N(z) and b = N(w), then ab = N(zw). So ab is of the form x2 + y2. Similarly, in the IB Groups, Rings and Modules example√ sheet, we found 2 3 all solutions to the equation x + 2 = y by working in Z[ −2]. This is a√ very general technique — working with these rings, and corresponding fields Q( −d) can tell us a lot about arithmetic we care about. In this chapter, we will begin by writing down some basic definitions and proving elementary properties about number fields. Definition (Field extension). A field extension is an inclusion of fields K ⊆ L. We sometimes write this as L/K.

Definition (Degree of field extension). Let K ⊆ L be fields. Then L is a vector space over K, and the degree of the field extension is

[L : K] = dimK (L).

Definition (Finite extension). A finite field extension is a field extension with finite degree.

Definition (Number field). A number field is a finite field extension over Q. A field is the most boring kind of ring — the only ideals are the trivial one and the whole field itself. Thus, if we want to do something interesting with number fields algebraically, we need to come up with something more interesting. In the case of Q itself, one interesting thing to talk about is the integers Z. It turns out the right generalization to number fields is algebraic integers. Definition (Algebraic integer). Let L be a number field. An algebraic integer is an α ∈ L such that there is some monic f ∈ Z[x] with f(α) = 0. We write OL for the set of algebraic integers in L.

Example. It is a fact that if L = Q(i), then OL = Z[i]. We will prove this in the next chapter after we have the necessary tools. These are in fact the main objects of study in this course. Since we say this is a generalization of Z ⊆ Q, the following had better be true:

Lemma. OQ = Z, i.e. α ∈ Q is an algebraic integer if and only if α ∈ Z.

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Proof. If α ∈ Z, then x − α ∈ Z[x] is a monic polynomial. So α ∈ OQ. On the other hand, let α ∈ Q. Then there is some coprime r, s ∈ Z such that r α = s . If it is an algebraic integer, then there is some n n−1 f(x) = x + an−1x + ··· + a0

n with ai ∈ Z such that f(α) = 0. Substituting in and multiplying by s , we get n n−1 n r + an−1r s + ··· + a0s = 0, | {z } divisible by s So s | rn. But if s =6 1, there is a prime p such that p | s, and hence p | rn. Thus p | r. So p is a common factor of s and r. This is a contradiction. So s = 1, and α is an integer.

How else is this a generalization of Z? We know Z is a ring. So perhaps OL also is.

Theorem. OL is a ring, i.e. if α, β ∈ OL, then so is α ± β and αβ.

Note that in general OL is not a field. For example, Z = OQ is not a field. The proof of this theorem is not as straightforward as the previous one. Recall we have proved a similar theorem in IID Galois Theory before with “algebraic integer” replaced with “algebraic number”, namely that if L/K is a field extension with α, β ∈ L algebraic over K, then so is αβ and α ± β, as well 1 as α if α 6= 0. To prove this, we notice that α ∈ K is algebraic if and only if K[α] is a finite extension — if α is algebraic, with

n f(α) = anα + ··· + a0 = 0, an 6= 0 then K[α] has degree at most n, since αn (and similarly α−1) can be written as a linear combination of 1, α, ··· , αn−1, and thus these generate K[α]. On the other hand, if K[α] is finite, say of degree k, then 1, α, ··· , αk are independent, hence some linear combination of them vanishes, and this gives a polynomial for which α is a root. Moreover, by the same proof, if K0 is any finite extension over K, then any element in K0 is algebraic. Thus, to prove the result, notice that if K[α] is generated by 1, α, ··· , αn and K[β] is generated by 1, β, ··· , βm, then K[α, β] is generated by {αiβj} for 1 ≤ i ≤ n, 1 ≤ j ≤ m. Hence K[α, β] is a finite extension, hence αβ, α ± β ∈ K[α, β] are algebraic. We would like to prove this theorem in an analogous way. We will consider OL as a ring extension of Z. We will formulate the general notion of “being an algebraic integer” in general ring extensions: Definition (Integrality). Let R ⊆ S be rings. We say α ∈ S is integral over R if there is some monic polynomial f ∈ R[x] such that f(α) = 0. We say S is integral over R if all α ∈ S are integral over R. Definition (Finitely-generated). We say S is finitely-generated over R if there n exists elements α1, ··· , αn ∈ S such that the function R → S defined by P (r1, ··· , rn) 7→ riαi is surjective, i.e. every element of S can be written as a R- linear combination of elements α1, ··· , αn. In other words, S is finitely-generated as an R-module.

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This is a refinement of the idea of being algebraic. We allow the use of rings and restrict to monic polynomials. In Galois theory, we showed that finiteness and algebraicity “are the same thing”. We will generalize this to integrality of rings.

Example. In a number field Z ⊆ Q ⊆ L, α ∈ L is an algebraic integer if and only if α is integral over Z by definition, and OL is integral over Z.

Notation. If α1, ··· , αr ∈ S, we write R[α1, ··· , αr] for the subring of S generated by R, α1, ··· , αr. In other words, it is the image of the homomorphism from the polynomial ring R[x1, ··· , xn] → S given by xi 7→ αi. Proposition. (i) Let R ⊆ S be rings. If S = R[s] and s is integral over R, then S is finitely-generated over R.

(ii) If S = R[s1, ··· , sn] with si integral over R, then S is finitely-generated over R. This is the easy direction in identifying integrality with finitely-generated. Proof.

(i) We know S is spanned by 1, s, s2, ··· over R. However, since s is integral, there exists a0, ··· , an ∈ R such that

n n−1 s = a0 + a1s + ··· + an−1s .

So the R-submodule generated by 1, s, ··· , sn−1 is stable under multiplica- tion by s. So it contains sn, sn+1, sn+2, ··· . So it is S.

(ii) Let Si = R[s1, ··· , si]. So Si = Si−1[si]. Since si is integral over R, it is integral over Si−1. By the previous part, Si is finitely-generated over Si−1. To finish, it suffices to show that being finitely-generated is transitive. More precisely, if A ⊆ B ⊆ C are rings, B is finitely generated over A and C is finitely generated over B, then C is finitely generated over A. This is not hard to see, since if x1, ··· , xn generate B over A, and y1, ··· , ym generate C over B, then C is generated by {xiyj}1≤i≤n,1≤j≤m over A. The other direction is harder.

Theorem. If S is finitely-generated over R, then S is integral over R. The idea of the proof is as follows: if s ∈ S, we need to find a monic polynomial which it satisfies. In Galois theory, we have fields and vector spaces, and the proof is easy. We can just consider 1, s, s2, ··· , and linear dependence kicks in and gives us a relation. But even if this worked in our case, there is no way we can make this polynomial monic. Instead, consider the multiplication-by-s map: ms : S → S by γ 7→ sγ. If S were a finite-dimensional vector space over R, then Cayley-Hamilton tells us ms, and thus s, satisfies its characteristic polynomial, which is monic. Even though S is not a finite-dimensional vector space, the proof of Cayley-Hamilton will work.

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Proof. Let α1, ··· , αn generate S as an R-module. wlog take α1 = 1 ∈ S. For any s ∈ S, write X sαi = bijαj

for some bij ∈ R. We write B = (bij). This is the “matrix of multiplication by S”. By construction, we have   α1  .  (sI − B)  .  = 0. (∗) an Now recall for any matrix X, we have adj(X)X = (det X)I, where the i, jth entry of adj(X) is given by the determinant of the matrix obtained by removing the ith row and jth column of X. We now multiply (∗) by adj(sI − B). So we get   α1  .  det(sI − B)  .  = 0 αn

In particular, det(sI −B)α1 = 0. Since we picked α1 = 1, we get det(sI −B) = 0. Hence if f(x) = det(xI − B), then f(x) ∈ R[x], and f(s) = 0. Hence we obtain the following:

Corollary. Let L ⊇ Q be a number field. Then OL is a ring.

Proof. If α, β ∈ OL, then Z[α, β] is finitely-generated by the proposition. But then Z[α, β] is integral over Z, by the previous theorem. So α ± β, αβ ∈ Z[α, β]. Note that it is not necessarily true that if S ⊇ R is an integral extension, then S is finitely-generated over R. For example, if S is the set of all algebraic integers in C, and R = Z, then by definition S is an integral extension of Z, but S is not finitely generated over Z. Thus the following corollary isn’t as trivial as the case with “integral” replaced by “finitely generated”: Corollary. If A ⊆ B ⊆ C be ring extensions such that B over A and C over B are integral extensions. Then C is integral over A. The idea of the proof is that while the extensions might not be finitely gener- ated, only finitely many things are needed to produce the relevant polynomials witnessing integrality. Proof. If c ∈ C, let N X i f(x) = bix ∈ B[x] i=0

be a monic polynomial such that f(c) = 0. Let B0 = A[b0, ··· , bN ] and let C0 = B0[c]. Then B0/A is finitely generated as b0, ··· , bN are integral over A. Also, C0 is finitely-generated over B0, since c is integral over B0. Hence C0 is finitely-generated over A. So c is integral over A. Since c was arbitrary, we know C is integral over A.

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Now how do we recognize algebraic integers? If we want to show something is an algebraic integer, we just have to exhibit a monic polynomial that vanishes on the number. However, if we want to show that something is not an algebraic integer, we have to make sure no monic polynomial kills the number. How can we do so? It turns out to check if something is an algebraic integer, we don’t have to check all monic polynomials. We just have to check one. Recall that if K ⊆ L is a field extensions with α ∈ L, then the minimal polynomial is the monic polynomial pα(x) ∈ K[x] of minimal degree such that pα(α) = 0. Note that we can always make the polynomial monic. It’s just that the coefficients need not lie in Z. Recall that we had the following lemma about minimal polynomials:

Lemma. If f ∈ K[x] with f(α) = 0, then pα | f.

Proof. Write f = pαh + r, with r ∈ K[x] and deg(r) < deg(pα). Then we have

0 = f(α) = p(α)h(α) + r(α) = r(α).

So if r 6= 0, this contradicts the minimality of deg pα.

In particular, this lemma implies pα is unique. One nice application of this result is the following:

Proposition. Let L be a number field. Then α ∈ OL if and only if the minimal polynomial pα(x) ∈ Q[x] for the field extension Q ⊆ L is in fact in Z[x]. This is a nice proposition. This gives us an necessary and sufficient condition for whether something is algebraic. Proof. (⇐) is trivial, since this is just the definition of an algebraic integer. (⇒) Let α ∈ OL and pα ∈ Q[x] be the minimal polynomial of α, and h(x) ∈ Z[x] be a monic polynomial which α satisfies. The idea is to use h to show that the coefficients of pα are algebraic, thus in fact integers. Now there exists a bigger field M ⊇ L such that

pα(x) = (x − α1) ··· (x − αr)

factors in M[x]. But by our lemma, pα | h. So h(αi) = 0 for all αi. So αi ∈ OM is an algebraic integer. But OM is a ring, i.e. sums and products of the αi’s are still algebraic integers. So the coefficients of pα are algebraic integers (in OM ). But they are also in Q. Thus the coefficients must be integers. Alternatively, we can deduce this proposition from the previous lemma plus Gauss’ lemma. Another relation between Z and Q is that Q is the fraction field of Z. This is true for general number fields Lemma. We have α  Frac O = : α, β ∈ O , β 6= 0 = L. L β L

In fact, for any α ∈ L, there is some n ∈ Z such that nα ∈ OL.

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Proof. If α ∈ L, let g(x) ∈ Q[x] be its monic minimal polynomial. Then there exists n ∈ Z non-zero such that ng(x) ∈ Z[x] (pick n to be the least common multiple of the denominators of the coefficients of g(x)). Now the magic is to put  x  h(x) = ndeg(g)g . n Then this is a monic polynomial with integral coefficients — in effect, we have just multiplied the coefficient of xi by ndeg(g)−i! Then h(nα) = 0. So nα is integral.

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2 Norm, trace, discriminant, numbers

Recall that in our motivating example of Z[i], one important tool was the norm of an algebraic integer x + iy, given by N(x + iy) = x2 + y2. This can be generalized to arbitrary number fields, and will prove itself to be a very useful notion to consider. Apart from the norm, we will also consider a number known as the trace, which is also useful. We will also study numbers associated with the number field itself, rather than particular elements of the field, and it turns out they tell us a lot about how the field behaves.

Norm and trace Recall the following definition from IID Galois Theory:

Definition (Norm and trace). Let L/K be a field extension, and α ∈ L. We write mα : L → L for the map ` 7→ α`. Viewing this as a linear map of L vector spaces, we define the norm of α to be

NL/K (α) = det mα,

and the trace to be trL/K (α) = tr mα. The following property is immediate: Proposition. For a field extension L/K and a, b ∈ L, we have N(ab) = N(a)N(b) and tr(a + b) = tr(a) + tr(b).

We can alternatively define the norm and trace as follows:

Proposition. Let pα ∈ K[x] be the minimal polynomial of α. Then the characteristic polynomial of mα is

[L:K(α)] det(xI − mα) = pα

0 Hence if pα(x) splits in some field L ⊇ K(α), say

pα(x) = (x − α1) ··· (x − αr),

then Y X NK(α)/K (α) = αi, trK(α)/K (α) = αi, and hence Y [L:K(α)] X  NL/K (α) = αi , trL/K = [L : K(α)] αi .

This was proved in the IID Galois Theory course, and we will just use it without proving.

Corollary. Let L ⊇ Q be a number field. Then the following are equivalent:

(i) α ∈ OL.

(ii) The minimal polynomial pα is in Z[x]

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(iii) The characteristic polynomial of mα is in Z[x].

This in particular implies NL/Q(α) ∈ Z and trL/Q(α) ∈ Z. Proof. The equivalence between the first two was already proven. For the equivalence between (ii) and (iii), if mα ∈ Z[x], then α ∈ OL since it vanishes on a monic polynomial in Z[x]. On the other hand, if pα ∈ Z[x], then so is the N characteristic polynomial, since it is just pα . The final implication comes from the fact that the norm and trace are just coefficients of the characteristic polynomial. It would be nice if the last implication is an if and only if. This is in general not true, but it occurs, obviously, when the characteristic polynomial is quadratic, since the norm and trace would be the only coefficients. √ 2 Example. Let L = K( d) = K[z]/√(z − d), where d is not a square in K. As a vector space over K, we can take 1, d as our basis. So every α can be written as √ α = x + y d. Hence the matrix of multiplication by α is

x dy m = . α y x

So the trace and norm are given by √ √ √ trL/K (x + y d) = 2x = (x + y d) + (x − y d) √ √ √ 2 2 NL/K (x + y d) = x − dy = (x + y d)(x − y d)

We can also√ obtain this by consider the roots of the minimal√ polynomial of α = x + y d, namely (α − x)2 − y2d = 0, which has roots x ± y d. √ In particular, if L = Q( d), with d < 0, then the norm of an element is just the norm of it as a complex number. Now that we have computed the general trace and norm, we can use the proposition to find out what the algebraic integers are. It turns out the result is (slightly) unexpected: √ Lemma. Let L = Q( d), where d ∈ Z is not 0, 1 and is square-free. Then ( √ Z[ d] d ≡ 2 or 3 (mod 4) √ OL = h 1 i Z 2 (1 + d) d ≡ 1 (mod 4) √ 2 2 Proof. We know x + y λ ∈ OL if and only if 2x, x − dy ∈ Z by the previous 2 r example. These imply 4dy ∈ Z. So if y = s with r, s coprime, r, s ∈ Z, then we must have s2 | 4d. But d is square-free. So s = 1 or 2. So u v x = , y = 2 2

for some u, v ∈ Z. Then we know u2 − dv2 ∈ 4Z, i.e. u2 ≡ dv2 (mod 4). But we know the squares mod 4 are always 0 and 1. So if d 6≡ 1 (mod 4), then u2 ≡ dv2

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2 2 (mod 4) imply that√ u = v = 0 (mod 4), and hence u, v are even. So x, y ∈ Z, giving OL = Z[ d]. On the other hand,√ if d ≡ 1 (mod 4), then u, v have the same√ parity mod 2, i.e. we can write x + y d as a -combination of 1 and 1 (1 + d). Z 2 √ 1 As a sanity check, we find that the minimal polynomial of 2 (1 + d) is 2 1 x − x + 4 (1 − d) which is in Z if and only if d ≡ 1 (mod 4).

Field embeddings Recall the following theorem from IID Galois Theory: Theorem (Primitive element theorem). Let K ⊆ L be a separable field extension. Then there exists an α ∈ L such that K(α) = L. √ √ √ √ For example, Q( 2, 3) = Q( 2 + 3). Since Q has characteristic zero, it follows that all number fields are separable extensions. So any number field L/Q is of the form L = Q(α). This makes it much easier to study number fields, as the only extra “stuff” we have on top of Q. One particular thing we can do√ is to look at the number of ways we can embed√L,→ C. For example, for Q( √−1), there are two such embeddings — one sends −1 to i and the other sends −1 to −i.

Lemma. The degree [L : Q] = n of a number field is the number of field embeddings L,→ C.

Proof. Let α be a primitive element, and pα(x) ∈ Q[x] its minimal polynomial. 2 n−1 Then by we have deg pα = [L : Q] = n, as 1, α, α , ··· , α is a basis. Moreover,

Q[x] =∼ Q(α) = L. (pα)

Since L/Q is separable, we know pα has n distinct roots in C. Write

pα(x) = (x − α1) ··· (x − αn).

Now an embedding Q[x]/(pα) ,→ C is uniquely determined by the image of x, and x must be sent to one of the roots of pα. So for each i, the map x 7→ αi gives us a field embedding, and these are all. So there are n of them. Using these field embeddings, we can come up with the following alternative formula for the norm and trace.

Corollary. Let L/Q be a number field. If σ1, ··· , σn : L → C are the different field embeddings and β ∈ L, then X Y trL/Q(β) = σi(β),NL/Q(β) = σi(β). i

We call σ1(β), ··· , σn(β) the conjugates of β in C. Proof is in the Galois theory course. Using this characterization, we have the following very concrete test for when something is a unit.

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Lemma. Let x ∈ OL. Then x is a unit if and only if NL/Q(x) = ±1. × −1 Notation. Write OL = {x ∈ OL : x ∈ OL}, the units in OL. × Proof. (⇒) We know N(ab) = N(a)N(b). So if x ∈ OL , then there is some y ∈ OL such that xy = 1. So N(x)N(y) = 1. So N(x) is a unit in Z, i.e. ±1. (⇐) Let σ1, ··· , σn : L → C be the n embeddings of L in C. For notational convenience, We suppose that L is already subfield of C, and σ1 is the inclusion map. Then for each x ∈ OL, we have

N(x) = xσ2(x) ··· σn(x).

−1 −1 Now if N(x) = ±1, then x = ±σ2(x) ··· σn(x). So we have x ∈ OL, since this is a product of algebraic integers. So x is a unit in OL.

Corollary. If x ∈ OL is such that N(x) is prime, then x is irreducible. Proof. If x = ab, then N(a)N(b) = N(x). Since N(x) is prime, either N(a) = ±1 or N(b) = ±1. So a or b is a unit. We can consider a more refined notion than just the number of field embed- dings.

Definition (r and s). We write r for the number of field embeddings L,→ R, and s the number of pairs of non-real field embeddings L,→ C. Then n = r + 2s.

Alternatively, r is the number of real roots of pα, and s is the number of pairs of complex conjugate roots. The distinction between real embeddings and complex embeddings will be important in the second half of the course.

Discriminant The final invariant we will look at in this chapter is the discriminant. It is based on the following observation: Proposition. Let L/K be a separable extension. Then a K-bilinear form L × L → K defined by (x, y) 7→ trL/K (xy) is non-degenerate. Equivalent, if α1, ··· , αn are a K-basis for L, the Gram matrix (tr(αiαj))i,j=1,··· ,n has non-zero determinant.

Recall from Galois theory that if L/K is not separable, then trL/K = 0, and it is very very degenerate. Also, note that if K is of characteristic 0, then there is a quick and dirty proof of this fact — the trace map is non-degenerate, because −1 for any x ∈ K, we have trL/K (x · x ) = n =6 0. This is really the only case we care about, but in the proof of the general result, we will also find a useful formula for the discriminant when the basis is 1, θ, θ2, . . . , θn−1. We will use the following important notation: Notation. ∆(α1, ··· , αn) = det(trL/K (αiαj)).

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Proof. Let σ1, ··· , σn : L → K¯ be the n distinct K-linear field embeddings L,→ K¯ . Put   σ1(α1) ··· σ1(αn)  . .. .  S = (σi(αj))i,j=1,··· ,n =  . . .  σn(α1) ··· σn(αn). Then n ! T X S S = σk(αi)σk(αj) . k=1 i,j=1,···n

We know σk is a field homomorphism. So n n X X σk(αi)σk(αj) = σk(αiαj) = trL/K (αiαj). k=1 k=1 So T S S = (tr(αiαj))i,j=1,··· ,n. So we have T 2 ∆(α1, ··· , αn) = det(S S) = det(S) . Now we use the theorem of primitive elements to write L = K(θ) such that 1, θ, ··· , θn−1 is a basis for L over K, with [L : K] = n. Now S is just

 n−1  1 σ1(θ) ··· σ1(θ) . . .. .  S = . . . .  . n−1 1 σn(θ) ··· σn(θ) This is a Vandermonde matrix, and so

n−1 2 Y 2 ∆(1, θ, ··· , θ ) = (det S) = (σi(θ) − σj(θ)) . i

Since the field extension is separable, and hence σi 6= σj for all i, j, this implies σi(θ) 6= σj(θ), since θ generates the field. So the product above is non-zero. So we have this nice canonical bilinear map. However, this determinant is 0 0 not canonical. Recall that if α1, ··· , αn is a basis for L/K, and α1, ··· , αn is another basis, then 0 X αi = aijαj

for some A = (aij) ∈ GLn(K). So 0 0 2 ∆(α1, ··· , αn) = (det A) ∆(α1, ··· , αn). However, for number fields, we shall see that we can pick a “canonical” basis, and get a canonical value for ∆. We will call this the discriminant.

Definition (Integral basis). Let L/Q be a number field. Then a basis α1, ··· , αn of L is an integral basis if

( n ) n X M OL = miαi : mi ∈ Z = Zαi. i=1 1

In other words, it is simultaneously a basis for L over Q and OL over Z.

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Note that integral bases are not unique, just as with usual bases. Given one basis, you can get any other by acting by GLn(Z). √ ∼ Example. Consider Q(√ d) with d square-free, d 6= 0, 1. If d = 1 (mod 4), we’ve seen that 1, 1 (1 + λ) is an integral basis. Otherwise, if d =∼ 2, 3 (mod 4), √ 2 then 1, d is an integral basis. The important theorem is that an integral basis always exists.

Theorem. Let Q/L be a number field. Then there exists an integral basis for ∼ n OL. In particular, OL = Z with n = [L : Q].

Proof. Let α1, ··· , αn be any basis of L over Q. We have proved that there is some ni ∈ Z such that niαi ∈ OL. So wlog α1, ··· , αn ∈ OL, and are an basis of L over Q. Since αi are integral, so are αiαj, and so all these have integer trace, as we have previously shown. Hence ∆(α1, ··· , αn), being the determinant of a matrix with integer entries, is an integer. Now choose a Q-basis α1, ··· , αn ∈ OL such that ∆(α1, ··· , αn) ∈ Z \{0} has minimal absolute value. We will show that these are an integral basis. Let x ∈ OL, and write X x = λiαi

for some λi ∈ Q. These λi are necessarily unique since α1, ··· , αn is a basis. Suppose some λi 6∈ Z. wlog say λ1 6∈ Z. We write

λ1 = n1 + ε1,

for n1 ∈ Z and 0 < ε1 < 1. We put

0 α1 = x − n1α1 = ε1α1 + λ2α2 + ··· + λnαn ∈ OL.

0 So α1, α2, ··· , αn is still a basis for L/Q, and are still in OL. But then

0 2 ∆(α1, ··· , αn) = ε1 · ∆(α1, ··· , αn) < ∆(α1, ··· , αn).

This contradicts minimality. So we must have λi ∈ Z for all Z. So this is a basis for OL.

0 0 Now if α1, ··· , αn is another integral basis of L over Q, then there is some 0 g ∈ GLn(Z) such that gαi = αi. Since det(g) is invertible in Z, it must be 1 or −1, and hence

0 0 2 det ∆(α1, ··· , αn) = det(g) ∆(α1, ··· , αn) = ∆(α1, ··· , αn)

and is independent of the choice of integral basis.

Definition (Discriminant). The discriminant DL of a number field L is defined as DL = ∆(α1, ··· , αn)

for any integral basis α1, ··· , αn.

15 2 Norm, trace, discriminant, numbers II Number Fields

√ ∼ Example. Let L = Q( d), where d 6=√ 0, 1 and d is square-free. If d = 2, 3 (mod 4), then it has an integral basis 1, d. So

√ 2 1 d  D = det √ = 4d. L 1 − d

Otherwise, if d =∼ 1 (mod 4), then

√ 2 1 1 (1 + d) 2 √ DL = det 1 = d. 1 2 (1 − d) Recall that we have seen the word discriminant before, and let’s make sure these concepts are more-or-less consistent. Recall that the discriminant of a Q polynomial f(x) = (x − αi) is defined as

Y 2 n(n−1)/2 Y disc(f) = (αi − αj) = (−1) (αi − αj). i

If pθ(x) ∈ K[x] is the minimal polynomial of θ (where L = K[θ]), then the roots of pθ are σi(θ). Hence we get

Y 2 disc(pθ) = (σi(θ) − σj(θ)) . i

In other words, n−1 disc(pθ) = ∆(1, θ, ··· , θ ). So this makes sense.

16 3 Multiplicative structure of ideals II Number Fields

3 Multiplicative structure of ideals

Again, let L/Q be a number field. It turns out that in general, the integral ring OL is not too well-behaved as a ring. In particular, it fails to be a UFD in general. √ √ Example. Let L = Q( 5). Then OL = Z[ −5]. Then we find √ √ 3 · 7 = (1 + 2 −5)(1 − 2 −5). √ These have norms 9, 49,√21, 21. So none of 3, 7, 1 + 2 5 are associates. Moreover, 3, 7, 1 ± 2 −5 are all irreducibles. The proof is just a straightfor- ward check on the norms. For example, to show that 3 is irreducible, if 3 = αβ, then 9 = N(3) = N(α)N(β). Since none of the terms on the right are ±1, we must have N(α) = ±3. But there are no solutions to

x2 + 5y2 = ±3 √ where x, y are integers. So there is no α = x + y −5 such that N(α) = ±3. So unique factorization fails. Note that it is still possible to factor any element into irreducibles, just not uniquely — we induct on |N(α)|. If |N(α)| = 1, then α is a unit. Otherwise, α is either irreducible, or α = βγ. Since N(β)N(γ) = N(α), and none of them are ±1, we must have |N(β)|, |N(γ)| < |N(α)|. So done by induction. To fix the lack of unique factorization, we instead look at ideals in OL. This has a natural multiplicative structure — the product of two ideals a, b is generated by products ab, with a ∈ a, b ∈ b. The big theorem is that every ideal can be written uniquely as a product of prime ideals.

Definition (Ideal multiplication). Let a, b C OL be ideals. Then we define the product ab as   X  ab = αiβj : αi ∈ a, βj ∈ b .  i,j  We write a | b if there is some ideal c such that ac = b, and say a divides b.

The proof of unique factorization is the same as the proof that Z is a UFD. Usually, when we want to prove factorization is unique, we write an object as

a = x1x2 ··· xm = y1y2 ··· yn.

We then use primality to argue that x1 must be equal to some of the yi, and then cancel them from both sides. We can usually do this because we are working with an integral domain. However, we don’t have this luxury when working with ideals. Thus, what we are going to do is to find inverses for our ideals. Of course, −1 −1 given any ideal a, there is no ideal a such that aa = OL, as for any b, we know ab is contained in a. Thus we are going to consider more general objects known as fractional ideals, and then this will allow us to prove unique factorization.

17 3 Multiplicative structure of ideals II Number Fields

Even better, we will show that a | b is equivalent to b ⊆ a. This is a very useful result, since it is often very easy to show that b ⊆ a, but it is usually very hard to actually find the quotient a−1b. We first look at some examples of multiplication and factorization of ideals to get a feel of what these things look like. Example. We have

hx1, ··· , xnihy1, ··· , ymi = hxiyj : 1 ≤ i ≤ n, 1 ≤ j ≤ mi.

In particular, hxihyi = hxyi. It is also an easy exercise to check (ab)c = a(bc). √ Example. In Z[ −5], we claim that we have √ √ h3i = h3, 1 + −5ih3, 1 − −5i.

So h3i is not irreducible. Indeed, we can compute √ √ √ √ h3, 1 + −5ih3, 1 − −5i = h9, 3(1 + 2 −5), 3(1 − 2 −5), 21i.

But we know gcd(9, 21) = 3. So h9, 21i = h3i by Euclid’s algorithm. So this is in fact equal to h3i.

Notice that when we worked with elements, the number 3 was√ irreducible,√ as there is no element of norm 3. Thus, scenarios such as 2·3 = (1+ −5)(1− −5) could appear and mess up unique factorization. By passing on to ideals, we can further factorize h3i into a product of smaller ideals. Of course, these cannot be principal ideals, or else we would have obtained a factorization of 3 itself. So we can think of these ideals as “generalized elements” that allow us to further break elements down. Indeed, given any element in α ∈ OL, we obtain an ideal hαi corresponding to α. This map is not injective — if two elements differ by a unit, i.e. they are associates, then they would give us the same ideal. However, this is fine, as we usually think of associates as being “the same”. We recall the following definition: Definition (Prime ideal). Let R be a ring. An ideal p ⊆ R is prime if R/p is an integral domain. Alternatively, for all x, y ∈ R, xy ∈ p implies x ∈ p or y ∈ p. In this course, we take the convention that a prime ideal is non-zero. This is not standard, but it saves us from saying “non-zero” all the time.

It turns out that the ring of integers OL is a very special kind of rings, known as Dedekind domains:

Definition (Dedekind domain). A ring R is a Dedekind domain if (i) R is an integral domain. (ii) R is a Noetherian ring.

18 3 Multiplicative structure of ideals II Number Fields

(iii) R is integrally closed in Frac R, i.e. if x ∈ Frac R is integral over R, then x ∈ R. (iv) Every proper prime ideal is maximal.

This is a rather specific list of properties OL happens to satisfy, and it turns out most interesting properties of OL can be extended to arbitrary Dedekind domains. However, we will not do the general theory, and just study number fields in particular. The important result is, of course:

Proposition. Let L/Q be a number field, and OL be its ring of integers. Then OL is a Dedekind domain. The first three parts of the definition are just bookkeeping and not too interesting. The last one is what we really want. This says that OL is “one dimensional”, if you know enough algebraic geometry. Proof of (i) to (iii).

(i) Obvious, since OL ⊆ L.

n (ii) We showed that as an abelian group, OL = Z . So if a ≤ OL is an ideal, then a ≤ Zn as a subgroup. So it is finitely generated as an abelian group, and hence finitely generated as an ideal.

(iii) Note that Frac OL = L. If x ∈ L is integral over OL, as OL is integral over Z, x is also integral over Z. So x ∈ OL, by definition of OL. To prove the last part, we need the following lemma, which is also very important on its own right.

Lemma. Let a C OL be a non-zero ideal. Then a ∩ Z 6= {0} and OL/a is finite. Proof. Let α ∈ a and α 6= 0. Let

m m−1 pα = x + am−1x + ··· + a0

be its minimal polynomial. Then pα ∈ Z[x]. We know a0 =6 0 as pα is irreducible. Since pα(α) = 0, we know

m−1 m−2 a0 = −α(α + am−1α + ··· + a2α + a1).

We know α ∈ a by assumption, and the mess in the brackets is in OL. So the whole thing is in a. But a0 ∈ Z. So a0 ∈ Z ∩ a. Thus, we know ha0i ⊆ a. Thus we get a surjection O O L → L . ha0i a

Hence it suffices to show that OL/ha0i is finite. But for every d ∈ Z, we know O n  n L = Z = Z , hdi dZn dZ which is finite.

19 3 Multiplicative structure of ideals II Number Fields

Finally, recall that a finite integral domain must be a field — let x ∈ R with x 6= 0. Then mx : y 7→ xy is injective, as R is an integral domain. So it is a bijection, as R is finite. So there is some y ∈ R such that xy = 1. This allows us to prove the last part

Proof of (iv). Let p be a prime ideal. Then OL/p is an integral domain. Since the lemma says OL/p is finite, we know OL/p is a field. So p is maximal. We now continue on to prove a few more technical results.

Lemma. Let p be a prime ideal in a ring R. Then for a, b C R ideals, then ab ⊆ p implies a ⊆ p or b ⊆ p. Once we’ve shown that inclusion of ideals is equivalent to divisibility, this in effect says “prime ideals are primes”.

Proof. If not, then there is some a ∈ a \ p and b ∈ b \ p. Then ab ∈ ab ⊆ p. But then a ∈ p or b ∈ p. Contradiction. Eventually, we will prove that every ideal is a product of prime ideals. However, we cannot prove that just yet. Instead, we will prove the following “weaker” version of that result:

Lemma. Let 0 6= a C OL a non-zero ideal. Then there is a subset of a that is a product of prime ideals. The proof is some unenlightening abstract nonsense.

Proof. We are going to use the fact that OL is Noetherian. If this does not hold, then there must exist a maximal ideal a not containing a product of prime ideals (by which we mean any ideal greater than a contains a product of prime ideals, not that a is itself a maximal ideal). In particular, a is not prime. So there are some x, y ∈ OL such that x, y 6∈ a but xy ∈ a. Consider a + hxi. This is an ideal, strictly bigger than a. So there exists prime ideals p1, ··· , pr such that p1 ··· pr ⊆ a + hxi, by definition. Similarly, there exists q1, ··· qs such that q1 ··· qs ⊆ a + hyi. But then

p1 ··· prq1 ··· qs ⊆ (a + hxi)(a + hyi) ⊆ a + hxyi = a

So a contains a product of prime ideals. Contradiction.

Recall that for integers, we can multiply, but not divide. To make life easier, we would like to formally add inverses to the elements. If we do so, we obtain 1 things like 3 , and obtain the rationals. Now we have ideals. What can we do? We can formally add some inverse and impose some nonsense rules to make sure it is consistent, but it is helpful to actually construct something explicitly that acts as an inverse. We can then understand what significance these inverses have in terms of the rings. Proposition.

(i) Let 0 6= a C OL be an ideal. If x ∈ L has xa ⊆ a, then x ∈ OL.

20 3 Multiplicative structure of ideals II Number Fields

(ii) Let 0 6= a C OL be a proper ideal. Then

{y ∈ L : ya ≤ OL}

contains elements that are not in OL. In other words, {y ∈ L : ya ≤ O } L 6= 0. OL

We will see that the object {y ∈ L : ya ≤ OL} is in some sense an inverse to a. Before we prove this, it is helpful to see what this means in a concrete setting.

a Example. Consider OL = Z and a = 3Z. Then the first part says if b ·3Z ⊆ 3Z, a then b ∈ Z. The second says na a o : · 3 ∈ b b Z 1 contains something not in Z, say 3 . These are both “obviously true”. Proof.

(i) Let a ⊆ OL. Then since OL is Noetherian, we know a is finitely generated, say by α1, ··· , αm. We consider the multiplication-by-x map mx : a → a, i.e. write X xαi = aijαj,

where A = (aij) is a matrix in OL. So we know   α1  .  (xI − A)  .  = 0. αn

By multiplying by the adjugate matrix, this implies det(xI − A) = 0. So x satisfies a monic polynomial with coefficients in OL, i.e. x is integral over OL. Since OL is integrally closed, x ∈ OL. (ii) It is clear that if the result is true for a, then it is true for all a0 ⊆ a. So it is enough to prove this for a = p, a maximal, and in particular prime, ideal. Let α ∈ p be non-zero. By the previous lemma, there exists prime ideals p1, ··· , pr such that p1 ··· pr ⊆ hαi. We also have that hαi ⊆ p by definition. Assume r is minimal with this property. Since p is prime, there is some i such that pi ⊆ p. wlog, we may as well assume i = 1, i.e. p1 ⊆ p. But p1 is a prime ideal, and hence maximal. So p1 = p.

Also, since r is minimal, we know p2 ··· pr 6⊆ hai.

Pick β ∈ p2 ··· pr \ hai. Then

βp = βp1 ⊆ p1p2 ··· pr ⊆ hαi.

β β Dividing by α, we get α p ⊆ OL. But β 6∈ hαi. So we know α 6∈ OL. So done.

21 3 Multiplicative structure of ideals II Number Fields

What is this {x ∈ L : xa ≤ OL}? This is not an ideal, but it almost is. The only way in which it fails to be an ideal is that it is not contained inside OL. By this we mean it is closed under addition and multiplication by elements in OL. So it is an OL module, which is finitely generated (we will see this in a second), and a subset of L. We call this a “fractional ideal”.

Definition (Fractional ideal). A fractional ideal of OL is a subset of L that is also an OL module and is finitely generated.

Definition (Integral/honest ideal). If we want to emphasize that a C OL is an ideal, we say it is an integral or honest ideal. But we never use “ideal” to mean fractional ideal.

Note that the definition of fractional ideal makes sense only because OL is Noetherian. Otherwise, the non-finitely-generated honest ideals would not qualify as fractional ideals, which is bad. Rather, in the general case, the following characterization is more helpful:

Lemma. An OL module q ⊆ L is a fractional ideal if and only if there is some × c ∈ L such that cq is an ideal in OL. Moreover, we can pick c such that c ∈ Z. 1 In other words, each fractional ideal is of the form c a for some honest ideal a and integer c. Proof. (⇐) We have to prove that q is finitely generated. If q ⊆ L×, c ∈ L non-zero, ∼ then cq = q as an OL module. Since OL is Noetherian, every ideal is finitely-generated. So cq, and hence q is finitely generated.

yi (⇒) Suppose x1, ··· , xn generate q as an OL-module. Write xi = , with ni yi ∈ OL and ni ∈ Z, ni 6= 0, which we have previously shown is possible.

We let c = lcm(n1, ··· , nk). Then cq ⊆ OL, and is an OL-submodule of OL, i.e. an ideal.

Corollary. Let q be a fractional ideal. Then as an abelian group, q =∼ Zn, where n = [L : Q]. × ∼ Proof. There is some c ∈ L such that cqCOL as an ideal, and cq = q as abelian groups. So it suffices to show that any non-zero ideal q ≤ OL is isomorphic to n ∼ n ∼ m Z . Since q ≤ OL = Z as abelian groups, we know q = Z for some m. But n ∼ also there is some a0 ∈ Z ∩ q, and Z = ha0i ≤ q. So we must have n = m, and q =∼ Zn.

Corollary. Let a ≤ OL be a proper ideal. Then {x ∈ L : xa ≤ OL} is a fractional ideal.

Proof. Pick a ∈ a. Then a ·{x ∈ L : xa ≤ OL} ⊆ OL and is an ideal in OL. Finally, we can state the proposition we want to prove, after all that nonsense work. Definition (Invertible fractional ideal). A fractional ideal q is invertible if there exists a fractional ideal r such that qr = OL = h1i.

22 3 Multiplicative structure of ideals II Number Fields

Notice we can multiply fractional ideals using the same definition as for integral ideals. Proposition. Every non-zero fractional ideal is invertible. The inverse of q is

{x ∈ L : xq ⊆ OL}.

This is good. 1 1 Note that if q = n a and r = m b, and a, b C OL are integral ideals, then 1 qr = ab = O mn L if and only if ab = hmni. So the proposition is equivalent to the statement that for every a C OL, there exists an ideal b C OL such that ab is principal.

Proof. Note that for any n ∈ OL non-zero, we know q is invertible if and only if nq is invertible. So if the proposition is false, there is an integral ideal a C OL which is not invertible. Moreover, as OL is Noetherian, we can assume a is 0 0 maximal with this property, i.e. if a < a < OL, then a is invertible. Let b = {x ∈ L : xa ⊆ OL}, a fractional ideal. We clearly have OL ⊆ b, and by our previous proposition, we know this inclusion is strict. As OL ⊆ b, we know a ⊆ ab. Again, this inclusion is strict — if ab = a, then for all x ∈ b, we have xa ⊆ a, and we have shown that this implies x ∈ OL, but we cannot have b ⊆ OL. So a ( ab. By assumption, we also have ab ⊆ OL, and since a is not invertible, this is strict. But then by definition of a, we know ab is invertible, which implies a is invertible (if c is an inverse of ab, then bc is an inverse of a). This is a contradiction. So all fractional ideals must be invertible. Finally, we have to show that the formula for the inverse holds. We write

c = {x ∈ L : xq ⊆ OL}.

Then by definition, we know q−1 ⊆ c. So

−1 OL = qq ⊆ qc ⊆ OL.

−1 Hence we must have qc = OL, i.e. c = q . We’re now done with the annoying commutative algebra, and can finally prove something interesting.

Corollary. Let a, b, c C OL be ideals, c 6= 0. Then (i) b ⊆ a if and only if bc ⊆ ac (ii) a | b if and only if ac | bc (iii) a | b if and only if b ⊆ a. Proof. (i)( ⇒) is clear, and (⇐) is obtained by multiplying with c−1. (ii)( ⇒) is clear, and (⇐) is obtained by multiplying with c−1.

23 3 Multiplicative structure of ideals II Number Fields

(iii) (⇒) is clear. For the other direction, we notice that the result is easy if a = hαi is principal. Indeed, if b = hβ1, ··· , βri, then b ⊆ hαi means there 0 0 0 are some βa, ··· , βr ∈ OL such that βi = βiα. But this says

0 0 hβa, ··· , βri = hβ1, ··· , βrihαi, So hαi | b. In general, suppose we have b ⊆ a. By the proposition, there exists an ideal c C OL such that ac = hαi is principal with α ∈ OL, α 6= 0. Then – b ⊆ a if and only if bc ⊆ hαi by (i); and – a | b if and only if hαi | bc by (ii). So the result follows. Finally, we can prove the unique factorization of prime ideals:

Theorem. Let a C OL be an ideal, a 6= 0. Then a can be written uniquely as a product of prime ideals. Proof. To show existence, if a is prime, then there is nothing to do. Otherwise, if a is not prime, then it is not maximal. So there is some b ) a with b C OL. Hence b | a, i.e. there is some c C OL with a = bc, and c ⊇ a. We can continue factoring this way, and it must stop eventually, or else we have an infinite chain of strictly ascending ideals. We prove uniqueness the usual way. We have shown p | ab implies p | a or p | b. So if p1 ··· pr = q1 ··· qs, with pi, qj prime, then we know p1 | q1 ··· qs, which implies p1 | qi for some i, and wlog i = 1. So q1 ⊆ p1. But q1 is prime and hence maximal. So p1 = q1. −1 Multiply the equation p1 ··· pr = q1 ··· qs by p1 , and we get p2 ··· pr = q2 ··· qs. Repeat, and we get r = s and pi = qi for all i (after renumbering). Corollary. The non-zero fractional ideals form a group under multiplication. We denote this IL. This is a free abelian group generated by the prime ideals, a1 qr i.e. any fractional ideal q can be written uniquely as p1 ··· pr , with pi distinct prime ideals and ai ∈ Z. Moreover, if q is an integral ideal, i.e. q C OL, then ai, ··· , ar ≥ 0. Proof. We already have unique factorization of honest ideals. Now take any −1 fractional ideal, and write it as q = ab , with a, b ∈ OL (e.g. take b = hni for some n), and the result follows. Unimportant side note: we have shown that there are two ways we can partially order the ideals of OL — by inclusion and by division. We have shown that these two orders are actually the same. Thus, it follows that the “least common multiple” of two ideals a, b is their intersection a ∩ b, and the “greatest common divisor” of two ideals is their sum

a + b = {a + b : a ∈ a, b ∈ b}. √ Example. Again let [L : Q] = 2, i.e. L = Q( d) with d 6= 0, 1 and square-free. While we proved that every ideal can be factorized into prime ideals, we have completely no idea what prime ideals look like. We just used their very abstract

24 3 Multiplicative structure of ideals II Number Fields

properties like being prime and maximal. So we would like to play with some actual ideals. Recall we had the example √ √ h3, 1 + 2 −5ih3, 1 − 2 −5i = h3i.

This is an example where we multiply two√ ideals together to get a principal√ ideal, and the key to this working is that 1 + 2 −5 is conjugate to 1 − 2 −5. We will use this idea to prove the previous result for number fields of this form. Let a C OL be a non-zero ideal. We want to find some b C OL such that ab is principal. ∼ 2 We know OL = Z , and a ≤ OL as a subgroup. Moreover, we have shown that we must have a =∼ Z2 as abelian groups. So it is generated by 2 elements as 2 subgroups of Z . Since Z is a subring of OL, we know a must be generated by at most 2 elements as OL-modules, i.e. as ideals of OL. If it is generated by one element, then it is already principal. Otherwise, suppose a = hα, βi for some α, β ∈ OL. √ Further, we claim√ that we can pick α, β such that β ∈ Z. We write α = a+b d and β = a0 + b0 d. Then let ` = gcd(b, b0) = mb + m0b0, with m, m0 ∈ Z (by Euclid’s algorithm). We set −b0 β0 = (mα + m0β) · + β ` √ −b0 √ = (ma + m0a0 + ` d) + a0 + b0 d ` −b0 = (ma + m0a0) + a0 ∈ ` Z

b0 0 using the fact that that − ` ∈ Z. Then hα, β i = hα, βi. So suppose a = hb, αi, with b ∈ Z and α ∈ OL. We now claim hb, αihb, α¯i √ √ is principal (where α = x + y d, α¯ = x − y d). In particular, if a C OL, then aa¯ is principal, so the proposition is proved by hand. To show this, we can manually check

hb, αihb, α¯i = hb2, bα, bα,¯ αα¯i = hb2, bα, b tr(α),N(α)i,

using the fact that tr(α) = α + α¯ and N(α) = αα¯. Now note that b2, b tr(α) and N(α) are all integers. So we can take the gcd c = gcd(b2, b tr(α),N(α)). Then this ideal is equal to

= hc, bαi.

Finally, we claim that bα ∈ hci. b b Write bα = cx, with x ∈ L. Then tr x = c tr α ∈ Z since c ∈ Z by definition, and bα b2N(α) b2 N(α) N(x) = N = = ∈ . c c2 c c Z

So x ∈ OL. So c | bα in OL. So hc, bαi = hci.

25 3 Multiplicative structure of ideals II Number Fields

Finally, after all these results, we can get to the important definition of the course.

Definition (Class group). . The class group or ideal class group of a number field L is clL = IL/PL, where IL is the group of fractional ideals, and PL is the subgroup of principal fractional ideals.

If a ∈ IL, we write [a] for its equivalence class in clL. So [a] = [b] if and only if there is some γ ∈ L× such that γa = b. The significance is that clL measures the failure of unique factorization: Theorem. The following are equivalent:

(i) OL is a principal ideal domain

(ii) OL is a unique factorization domain

(iii) clL is trivial. Proof. (i) and (iii) are equivalent by definition, while (i) implies (ii) is well-known from IB Groups, Rings and Modules. So the real content is (ii) to (i), which is specific to Dedekind domains. If p C OL is prime, and x ∈ p \{0}, we factor x = α1 ··· αk such that αi is irreducible in OL. As p is prime, there is some αi ∈ p. But then hαii ⊆ p, and hαii is prime as OL is a UFD. So we must have hαii = p as prime ideals are maximal. So p is principal. In the next few chapters, we will come up with methods to explicitly compute the class group of any number field.

26 4 Norms of ideals II Number Fields

4 Norms of ideals

In the previous chapter, we defined the class group, and we know it is generated by prime ideals of OL. So we now want to figure out what the prime ideals are. In the case of finding irreducible elements, one very handy tool was the norm — we know an element of OL is a unit iff it has norm 1. So if x ∈ OL is not irreducible, then there must be some element whose norm strictly divides N(x). Similarly, we would want to come up with the notion of the norm of an ideal, which turns out to be an incredibly useful notion.

Definition (Norm of ideal). Let a C OL be an ideal. We define

|N(a)| = |OL/a| ∈ N.

Recall that we’ve already proved that |OL/a| is finite. So this definition makes sense. It is also clear that N(a) = 1 iff a = OL (i.e. a is a “unit”). ∼ n ∼ n Example. Let d ∈ Z. Then since OL = Z , we have dOL = dZ . So we have

n n n n N(hdi) = |Z /(dZ) | = |Z/dZ| = d . We start with a simple observation:

Proposition. For any ideal a, we have N(a) ∈ a ∩ Z.

Proof. It suffices to show that N(a) ∈ a. Viewing OL/a as an additive group, the order of 1 is a factor of N(a). So N(a) = N(a) · 1 = 0 ∈ OL/a. Hence N(a) ∈ a. The most important property of the norm is the following:

Proposition. Let a, b C OL be ideals. Then N(ab) = N(a)N(b). We will provide two proofs of the result. Proof. By the factorization into prime ideals, it suffices to prove this for b = p prime, i.e. N(ap) = N(a)N(p). In other words, we need to show that

OL OL . OL = . a ap p By the third isomorphism theorem, we already know that O O   a  L =∼ L  . a ap ap ∼ So it suffices to show that OL/p = a/ap as abelian groups. In the case of the integers, we could have, say, p = 7Z, a = 12Z. We would then simply define Z 12Z 7Z 7 · 12Z x 12x

27 4 Norms of ideals II Number Fields

However, in general, we do not know that a is principal, but it turns out it doesn’t really matter. We can just pick an arbitrary element to multiply with. By unique factorization, we know a 6= ap. So we can find some α ∈ a \ ap. We now claim that the homomorphism of abelian groups

OL a p ap x + p αx + ap

is an isomorphism. We first check this is well-defined — if p ∈ p, then αp ∈ ap since α ∈ a. So the image of x+p and (x+p)+p are equal. So this is well-defined. To prove our claim, we have to show injectivity and surjectivity. To show injectivity, since hαi ⊆ a, we have a | hai, i.e. there is an ideal c ⊆ OL such that ac = hαi. If x ∈ OL is in the kernel of the map, then αx ∈ ap. So xac ⊆ ap.

So xc ⊆ p. As p is prime, either c ⊆ p or x ∈ p. But c ⊆ p implies hαi = ac ⊆ ap, contradicting the choice of α. So we must have x ∈ p, and the map is injective. To show this is surjective, we notice that surjectivity means hαi/ap = a/ap, or equivalently ap + hαi = a. Using our knowledge of fractional ideals, this is equivalent to saying (ap + −1 hαi)a = OL. But we know ap < ap + hαi ⊆ a.

We now multiply by a−1 to obtain

−1 p < (ap + hαi)a = p + c ⊆ OL. Since p is a prime, and hence maximal ideal, the last inclusion must be an equality. So ap + hαi = a, and we are done. Now we provide the sketch of a proof that makes sense. The details are left as an exercise in the second example sheet.

a1 ar a1 ar Proof. It is enough to show that N(p1 ··· pr ) = N(p1) ··· N(pr) by unique factorization. By the Chinese remainder theorem, we have

OL ∼ OL OL a1 ar = a1 × · · · × ar p1 ··· pr p1 pr where p1, ··· , pr are distinct prime ideals. Next, we show by hand that

r−1 r OL OL p p OL = × × · · · × = , pr p p2 pr p

k k+1 by showing that p /p is a 1-dimensional vector space over the field OL/p. Then the result follows.

28 4 Norms of ideals II Number Fields

This is actually the same proof, but written in a much saner form. This is better because we are combining a general statement (the Chinese remainder theorem), with a special property of the integral rings. In the first proof, what we really did was simultaneously proving two parts using algebraic magic. We’ve taken an obvious invariant of an ideal, the size, and found it is multiplicative. How does this relate to the other invariants? Recall that

2 ∆(α1, ··· , αn) = det(trL/Q(αiαj)) = det(σi(αj)) .

Proposition. Let a C OL be an ideal, n = [L : Q]. Then

(i) There exists α1, ··· , αn ∈ a such that

n nX o M a = riαi : ri ∈ Z = αiZ, 1

and α1, ··· , αn are a basis of L over Q. In particular, a is a free Z-module of n generators.

(ii) For any such α1, ··· , αn,

2 ∆(α1, ··· , αn) = N(a) DL.

The prove this, we recall the following lemma from IB Groups, Rings and Modules:

Lemma. Let M be a Z-module (i.e. abelian group), and suppose M ≤ Zn. Then M =∼ Zr for some 0 ≤ r ≤ n. Moreover, if r = n, then we can choose a basis v1, ··· , vn of M such that the change of basis matrix A = (aij) ∈ Mn×n(Z) is upper triangular, where X vj = aijei,

n where e1, ··· , en is the standard basis of Z . In particular, n |Z /M| = |a11a12 ··· ann| = | det A|.

Proof of proposition. Let d ∈ a ∩ Z, say d = N(α). Then dOL ⊆ a ⊆ OL. As 0 0 abelian groups, after picking an integral basis α1, ··· , αn of OL, we have

n n n Z =∼ dZ ≤ a ≤ Z . ∼ n So a = Z . Then the lemma gives us a basis α1, ··· , αn of a as a Z-module. As 0 a Q-module, since the αi are obtained from linear combinations of αi, by basic linear algebra, α1, ··· , αn is also a basis of L over Q. Moreover, we know that we have

2 0 0 ∆(α1, ··· , αn) = det(A) ∆(α1, ··· , αn).

2 2 0 0 Since det(A) = |OL/a| = N(a) and DL = ∆(α1, ··· , αn) by definition, the second part follows.

29 4 Norms of ideals II Number Fields

This result is very useful for the following reason:

Corollary. Suppose a C OL has basis α1, ··· , αn, and ∆(α1, ··· , αn) is square- free. Then a = OL (and DL is square-free). This is a nice trick, since it allows us to determine immediately whether a particular basis is an integral basis. Proof. Immediate, since this forces N(a)2 = 1.

Note that nothing above required a to be an actual ideal. It merely had n to be a subgroup of OL that is isomorphic to Z , since the quotient OL/a is well-defined as long as a is a subgroup. With this, we can have the following useful result:

Example. Let α be an algebraic integer and L = Q(α). Let n = [Q(α): Q]. Then a = Z[α] C OL. We have

2 n−1 disc(pα) = ∆(1, α, α , ··· , α ) = discriminant of minimal polynomial of α.

Thus if disc(pα) is square-free, then Z[α] = OL. 2 Even if disc(pα) is not square-free, it still says something: let d | disc(pα) 2 be such that disc(pα)/d is square-free. Then |N(Z[α])| divides d. Let x ∈ OL. Then the order of x + Z[α] ∈ OL/Z[α] divides N(Z[α]), hence 1 d. So d · x ∈ Z[α]. So x ∈ d Z[α]. Hence we have 1 [α] ⊆ O ⊆ [α]. Z L dZ √ √ For example, if α = a for some square-free a, then disc( a) is the discriminant of x2 − a, which is 4a. So the d above is 2, and we have 1 Z[α] ⊆ O √ ⊆ Z[α], Q( d) 2 as we have previously seen. We shall prove one more lemma, and start factoring things. Recall that we had two different notions of norm. Given α ∈ OL, we can take the norm N(hαi),

or NL/Q(α). It would be great if they are related, like if they are equal. However, that cannot possibly be true, since N(hαi) is always positive, but NL/Q(α) can be negative. So we take the absolute value.

Lemma. If α ∈ OL, then

N(hαi) = |NL/Q(α)|.

Proof. Let α1, ··· , αn be an integral basis of OL. Then αα1, .., ααn is an integral basis of hαi. So by the previous lemma,

2 ∆(αα1, ··· , ααn) = N(hαi) DL.

30 4 Norms of ideals II Number Fields

But

2 ∆(αα1, ··· , ααn) = det(σi(ααj)ij) 2 = det(σi(α)σi(αj)) n !2 Y = σi(α) ∆(α1, ··· , αn) i=1 2 = NL/Q(α) DL. So 2 2 NL/Q(α) = N(hαi) . But N(hαi) is positive. So the result follows.

31 5 Structure of prime ideals II Number Fields

5 Structure of prime ideals

We can now move on to find all prime ideals of OL. We know that every ideal factors as a product of prime ideals, but we don’t know what the prime ideals are. The only obvious way we’ve had to obtain prime ideals is to take a usual prime, take its principal ideal and factor it in OL, and get the resultant prime ideals. It turns out this gives us all prime ideals.

Lemma. Let p C OL be a prime ideal. Then there exists a unique p ∈ Z, p prime, with p | hpi. Moreover, N(p) = pf for some 1 ≤ f ≤ n. This is not really too exciting, as soon as we realize that p | hpi is the same as saying hpi ⊆ p, and we already know p ∩ Z is non-empty. Proof. Well p ∩ Z is an ideal in Z, and hence principal. So p ∩ Z = pZ for some p ∈ Z. We now claim p is a prime integer. If p = ab with ab ∈ Z. Then since p ∈ p, either a ∈ p or b ∈ p. So a ∈ p ∩ Z = pZ or b ∈ p ∩ Z = pZ. So p | a or p | b. Since hpi ⊆ p, we know hpi = pa for some ideal a by factorization. Taking norms, we get pn = N(hpi) = N(p)N(a). So the result follows. This is all good. So all we have to do is to figure out how principal ideals hpi factor into prime ideals. We write e1 em hpi = p1 ··· pm

fi for some distinct prime ideals pi, with N(pi) = p for some positive integers ei. Taking norms, we get Y pn = pfiei . So X n = eifi. We start by giving some names to the possible scenarios.

e1 em Definition (Ramification indices). Let hpi = p1 ··· pm be the factorization into prime ideals. Then e1, ··· , em are the ramification indices.

Definition (Ramified prime). We say p is ramified if some ei > 1.

Definition (Inert prime). We say p is inert if m = 1 and em = 1, i.e. hpi remains prime.

Definition (Splitting prime). We say p splits completely if e1 = ··· = em = 1 = f1 = ··· = fm. So m = n. Note that this does not exhaust all possibilities. The importance of these terms, especially ramification, will become clear later. So how do we actually compute pi and ei? In other words, how can we factor the ideal hpi C OL into prime ideals? The answer is given very concretely by Dedekind’s criterion.

32 5 Structure of prime ideals II Number Fields

Theorem (Dedekind’s criterion). Let α ∈ OL and g(x) ∈ Z[x] be its minimal polynomial. Suppose Z[α] ⊆ OL has finite index, coprime to p (i.e. p - |OL/Z[α]|). We write g¯(x) = g(x) (mod p),

sog ¯(x) ∈ Fp[x]. We factor

e1 em g¯(x) = ϕ1 ··· ϕm

into distinct irreducibles in Fp[x]. We define the ideal

pi = hp, ϕ˜i(α)i C OL,

generated by p and ϕ˜i, where ϕ˜i is any polynomial in Z[x] such that ϕ˜i mod p = 0 0 ϕi. Notice that if ϕ˜ is another such polynomial, then p | (ϕ˜i −ϕi), so hp, ϕ˜ (α)i = hp, ϕ˜(α)i. Then the pi are prime, and

e1 em hpi = p1 ··· pm .

deg ϕi Moreover, fi = deg ϕi, so N(a) = p .

If we are lucky, we might just find an α such that Z[α] = OL. If not, we can find something close, and as long as p is not involved, we are fine. After finding α, we get its minimal polynomial, factor it, and immediately get the prime factorization of hpi. √ Example.√ Consider L = Q( −11). We want to factor h5i in OL. We consider Z[ −11] ⊆ OL. This has index 2, and (hopefully) 5 - 2. So this is good enough. The minimal polynomial is x2 + 11. Taking mod 5, this reduces to x2 − 4 = (x − 2)(x + 2). So Dedekind says √ √ h5i = h5, −11 + 2ih5, −11 − 2i. √ In general, consider√ L = Q( d), d 6= 0, 1 and square-free, and p an odd prime. Then Z[ d] ⊆ OL has index 1 or 2, both of which are coprime to p. So Dedekind says factor x2 − d mod p. What are the possibilities?

 d  (i) There are two distinct roots mod p, i.e. d is a square mod p, i.e. p = 1. Then x2 − d = (x + r)(x − r) (mod p) for some r. So Dedekind says

hpi = p1p2,

where √ √ p1 = hp, d − ri, p2 = hp, d + ri,

and N(p1) = N(p2) = p. So p splits.

2  d  (ii) x − d is irreducible, i.e. d is not a square mod p, i.e. p = −1. Then Dedekind says hpi = p is prime in OL. So p is inert.

33 5 Structure of prime ideals II Number Fields

2  d  (iii) x − d has a repeated root mod p, i.e. p | d, or alternatively p = 0. Then by Dedekind, we know hpi = p2,

where √ p = hp, di. So p ramifies. So in fact, we see that the Legendre symbol encodes the ramification behaviour of the primes. √ What about the case where p = 2, for L = Q( d)? How do we factor h2i? √ Lemma. In L = Q( d), (i) 2 splits in L if and only if d ≡ 1 (mod 8); (ii) 2 is inert in L if and only if d ≡ 5 (mod 8);

(iii) 2 ramifies in L if d ≡ 2, 3 (mod 4). Proof. √ 1 – If d ≡ 1 (mod 4), then then OL = Z[α], where α = 2 (1 + d). This has minimal polynomial 1 x2 − x + (1 − d). 4 We reduce this mod 2.

◦ If d ≡ 1 (mod 8), we get x(x + 1). So 2 splits. ◦ If d ≡ 5 (mod 8), then we get x2 + x + 1, which is irreducible. So h2i is prime, hence 2 is inert. √ 2 – If d ≡ 2, 3 (mod 4), then OL = Z[ d], and x −d is the minimal polynomial. Taking mod 2, we get x2 or x2 +1 = (x+1)2. In both cases, 2 ramifies. √ Note how important p - |OL/Z[α]| is. If we used Z[ d] when d ≡ 1 (mod 4), we would have gotten the wrong answer. Recall ( 4d d ≡ 2, 3 (mod 4) DL = d d ≡ 1 (mod 4)

The above computations show that p | DL if and only if p ramifies in L. This happens to be true in general. This starts to hint how important these invariants like DL are. Now we get to prove Dedekind’s theorem. Proof of Dedekind’s criterion. The key claim is that Claim. We have OL Fp[x] =∼ . pi hϕii

34 5 Structure of prime ideals II Number Fields

Fp[x] Suppose this is true. Then since ϕi is irreducible, we know is a field. hϕii So pi is maximal, hence prime. Next notice that

e1 ei ei p1 = hp, ϕ˜i(α)i ⊆ hp, ϕ˜i(α) i. So we have

e1 em e1 em p1 ··· pm ⊆ hp, ϕ˜1(α) ··· ϕ˜m(α) i = hp, g(α)i = hpi, using the fact that g(α) = 0. fi So to prove equality, we notice that if we put fi = deg ϕi, then N(pi) = p , and P e1 em e1 em eifi deg g N(p1 ··· pm ) = N(p1) ··· N(pm) = p = p . n Since N(hpi) = p , it suffices to show that deg g = n. Since Z[α] ⊆ OL has finite index, we know Z[α] =∼ Zn. So 1, α, ··· , αn−1 are independent over Z, hence Q. So deg g = [Q(α): Q] = n = [L : Q], and we are done. So it remains to prove that

OL Z[α] Fp[x] =∼ =∼ . pi pi ∩ Z[α] hϕii The second isomorphism is clear, since

Z[α] Z[x] Fp[x] Fp[x] Fp[x] =∼ =∼ = = . hp, ϕ˜i(α)i hp, ϕ˜i(x), g(x)i hϕ˜i(x), g(x)i hϕi(x), g¯(x)i hϕii To prove the first isomorphism, it suffices to show that the following map is an isomorphism: [α] O Z → L (∗) pZ[α] pOL x + pZ[α] 7→ x + pOL

If this is true, then quotienting further byϕ ˜i gives the desired isomorphism. To prove the claim, we consider a slightly different map. We notice p - |OL/Z[α]| means the “multiplication by p” map

O p O L L (†) Z[α] Z[α]

is injective. But OL/Z[α] is a finite abelian group. So the map is an isomorphism. By injectivity of (†), we have Z[α] ∩ pOL = pZ[α]. By surjectivity, we have Z[α] + pOL = OL. It thus follows that (∗) is injective and surjective respectively. So it is an isomorphism. We have basically applied the snake lemma to the diagram

OL 0 [α] OL 0 Z Z[α] p p p

OL 0 [α] OL 0 Z Z[α]

35 5 Structure of prime ideals II Number Fields

Corollary. If p is prime and p < n = [L : Q], and Z[α] ⊆ OL has finite index coprime to p, then p does not split completely in OL. Proof. By Dedekind’s theorem, if g(x) is the minimal polynomial of α, then the factorization of g¯(x) = g(x) mod p determines the factorization of hpi into prime ideals. In particular, p splits completely if and only if g¯ factors into distinct linear factors, i.e. g¯(x) = (x − α1) ··· (x − αn), where αi ∈ Fp and αi are distinct. But if p < n, then there aren’t n distinct elements of Fp! Example. Let L = Q(α), where α has minimal polynomial x3 − x2 − 2x − 8. This is the case where n = 3 > 2 = p. On example sheet 2, you will see that 2 splits completely, i.e. OL/2OL = F2 × F2 × F2. But then this corollary shows that for all β ∈ OL, Z[β] ⊆ OL has even index, i.e. there does not exist an β ∈ OL with |OL/Z[β]| odd. As we previously alluded to, the following is true:

Theorem. p | DL if and only if p ramifies in OL. We will not prove this.

36 6 Minkowski bound and finiteness of class group II Number Fields

6 Minkowski bound and finiteness of class group

Dedekind’s criterion allowed us to find all prime factors of hpi, but if we want to figure out if, say, the class group of a number field is trivial, or even finite, we still have no idea how to do so, because we cannot go and check every single prime p and see what happens. What we are now going to do is the following — we are going to use purely geometric arguments to reason about ideals, and figure that each element of the class group clL = IL/PL has a representative whose norm is bounded by some number cL, which we will find rather explicitly. After finding the cL, to understand the class group, we just need to factor all prime numbers less than cL and see what they look like. We are first going to do the case of quadratic extensions explicitly, since 2-dimensional pictures are easier to draw. We will then do the full general case afterwards.

Quadratic extensions √ Consider again the case L = Q( d), where d < 0. Then OL = Z[α], where (√ d d ≡ 2, 3 (mod 4) √ α = 1 2 (1 + d) d ≡ 1 (mod 4)

We can embed this as a subfield L ⊆ C. We can then plot the points on the complex plane. For example, if d ≡ 2, 3 (mod 4), then the points look like this:

√ 2 d

√ √ d 1 + d

0 1

√ Then an ideal of OL, say a = h2, di, would then be the sub-lattice given by the blue crosses. × × ×

× × ×

× × ×

× × ×

× × ×

37 6 Minkowski bound and finiteness of class group II Number Fields

2 We always get this picture, since any ideal of OL is isomorphic to Z as an abelian group. If we are in the case where d ≡ 1 (mod 4), then the lattice is hexagonal:

√ d

√ 1 2 (1 + d)

0 1

The key result is the following purely geometric lemma:

2 Lemma (Minskowski’s lemma). Let Λ = Zv1 + Zv2 ⊆ R be a lattice, with 2 v1, v2 linearly independent in R (i.e. Rv1 +Rv2 = R ). We write vi = aie1 +bie2. Then let   a1 a2 A(Λ) = area of fundamental parallelogram = det , b1 b2 where the fundamental parallelogram is the following:

v1 + v2

v1

v2

Then a closed disc S around 0 contains a non-zero point of Λ if

area(S) ≥ 4A(Λ).

In particular, there exists an α ∈ Λ with α 6= 0, such that 4A(Λ) 0 < |α|2 ≤ . π This is just an easy piece of geometry. What is remarkable is that the radius of the disc needed depends only on the area of the fundamental parallelogram, and not its shape. Proof. We will prove a general result in any dimensions later.

2 We now apply this to ideals a ≤ OL, regarded as a subset of C = R via some embedding L,→ C. The following proposition gives us the areas of the relevant lattices: Proposition. √ (i) If α = a + b λ, then as a complex number, √ √ |α|2 = (a + b λ)(a − b λ) = N(α).

38 6 Minkowski bound and finiteness of class group II Number Fields

(ii) For OL, we have 1 A(O ) = p|D |. L 2 L (iii) In general, we have 1 A(a) = p|∆(α , α )|, 2 1 2

where α1, α2 are the integral basis of a. (iv) We have A(a) = N(a)A(OL).

Proof. (i) This is clear.

(ii) We know OL has basis 1, α, where again (√ d d ≡ 2, 3 (mod 4) √ α = 1 . 2 (1 + d) d ≡ 1 (mod 4) So we can just look at the picture of the lattice, and compute to get ( p|d| d ≡ 2, 3 (mod 4) 1 A(O ) = = p|D |. L 1 p 2 L 2 |d| d ≡ 1 (mod 4)

(iii) If α1, α2 are the integral basis of a, then the lattice of a is in fact spanned 0 0 by the vectors α1 = a + bi, α2 = a + b i. This has area

 a b  A(a) = det , a0 b0

whereas we have  2 α1 α¯1 ∆(α1, α2) = det α2 α¯2 2 = (α1α¯2 − α2α¯1) 2 = Im(2α1α¯2) = 4(a0b − ab0)2 = 4A(a)2.

(iv) This follows from (ii) and (iii), as

2 ∆(α1, ··· , αn) = N(a) DL

in general. Now what does Minkowski’s lemma tell us? We know there is an α ∈ a such that 4A(a) N(α) ≤ = N(a)c , π L

39 6 Minkowski bound and finiteness of class group II Number Fields where 2p|D | c = L . L π But α ∈ a implies hαi ⊆ a, which implies hαi = ab for some ideal b. So |N(α)| = N(hαi) = N(a)N(b). So this implies 2p|D | N(b) ≤ c = L . L π Recall that the class group is clL = IL/PL, the fractional ideals quotiented by principal ideals, and we write [a] for the class of a in clL. Then if ab = hαi, then we have [b] = [a−1]

in clL. So we have just shown,

Proposition (Minkowski bound). For all [a] ∈ clL, there is a representative b of [a] (i.e. an ideal b ≤ OL such that [b] = [a]) such that 2p|D | N(b) ≤ c = L . L π −1 −1 Proof. Find the b such that [b] = [(a ) ] and N(b) ≤ cL. Combining this with the following easy lemma, we know that the class group is finite!

Lemma. For every n ∈ Z, there are only finitely many ideals a ≤ OL with N(a) = m.

Proof. If N(a) = m, then by definition |OL/a| = m. So m ∈ a by Lagrange’s theorem. So hmi ⊆ a, i.e. a | hmi. Hence a is a factor of hmi. By unique factorization of prime ideals, there are only finitely many such ideals. Another proof is as follows:

n Proof. Each ideal bijects with an ideal in OL/mOL = (Z/m) . So there are only finitely many. Thus, we have proved

Theorem. The class group clL is a finite group, and the divisors of ideals of the form hpi for p ∈ Z, p a prime, and 0 < p < cL, collectively generate clL. Proof. p (i) Each element is represented by an ideal of norm less than 2 |DL|/π, and there are only finitely many ideals of each norm.

(ii) Given any element of clL, we pick a representative a such that N(a) < cL. We factorize e1 er a = p1 ··· pr . Then N(pi) ≤ N(a) < cL.

Suppose pi | hpi. Then N(p) is a power of p, and is thus at least p. So p < cL.

40 6 Minkowski bound and finiteness of class group II Number Fields

We now try to work with some explicit examples, utilizing Dedekind’s criterion and the Minkowski bound. √ Example. Consider d = −7. So Q( −7) = L, and DL = −7. Then we have √ 2 7 1 < c = < 2. L π

So clL = {1}, since there are no primes p < cL. So OL is a UFD. Similarly, if d = −1, −2, −3, then OL is a UFD.

Example. Let d = −5. Then DL = −20. We have √ 4 5 2 < c = < 3. L π

So clL is generated by primes dividing h2i. Recall that Dirichlet’s theorem implies √ h2i = h2, 1 + −5i2 = p2. √ Also,√p = h2, 1 + −5i is not principal. If it were, then p = hβi, with β = x + y −5, and N(β) = 2. But there are no solutions in Z of x2 + 5y2 = 2. So clL = hpi = Z/2.

Example. Consider d = −17 ≡ 3 (mod 4). So cL ≈ 5.3. So clL is generated by primes dividing by h2i, h3i, h5i. We factor

x2 + 17 ≡ x2 + 1 ≡ (x + 1)2 (mod 2).

So √ h2i = p2 = h2, 1 + di2. Doing this mod 3, we have

x2 + 17 ≡ x2 − 1 ≡ (x − 1)(x + 1) (mod 3).

So we have √ √ h3i = q¯q = h3, 1 + dih3, 1 − di. Finally, mod 5, we have

x2 + 17 ≡ x2 + 2 (mod 5).

So 5 is inert, and [h5i] = 1 in clL. So

clL = h[p], [q]i,

and we need to compute what this is. We can just compute powers q2, q3, ··· , pq, pq2, ··· , and see what happens. But a faster way is to look for principal ideals with small norms that are multiples of 2 and 3. For example, √ N(h1 + di) = 18 = 2 · 32.

41 6 Minkowski bound and finiteness of class group II Number Fields

But we have √ 1 + d ∈ p, q. √ √ So p, q | h1 + di. Thus we know pq | h1 + di. We have N(pq) = 2 · 3 = 6. So there is another factor of 3 to account for. In fact, we have √ h1 + di = pq2, which we can show by either thinking hard or expanding it out. So we must have

[p] = [q]−2

−2 in clL. So we have clL = √h[q]i. Also, [q] = [p] 6= 1 in clL, as if it did, then p is principal, i.e. p = hx + y di, but 2 = N(p) = x2 + 7y2 has no solution in the integers. Also, we know [p]2 = [1]. So we know

clL = Z/4Z. In fact, we have √ Theorem. Let L = Q( d) with d < 0. Then OL is a UFD if −d ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163}.

Moreover, this is actually an “if and only if”.

The first part is a straightforward generalization of what we have been doing, but the proof that no other d’s work is hard.

General case Now we want to extend these ideas to higher dimensions. We are really just doing the same thing, but we need a bit harder geometry and proper definitions.

Definition (Discrete subset). A subset X ⊆ Rn is discrete if for every x ∈ X, there is some ε > 0 such that Bε(x) ∩ X = {x}. This is true if and only if for every compact K ⊆ Rn, K ∩ X is finite. We have the following very useful characterization of discrete subgroups of Rn: Proposition. Suppose Λ ⊆ Rn is a subgroup. Then Λ is a discrete subgroup of (Rn, +) if and only if ( m ) X Λ = nixi : ni ∈ Z 1

for some x1, ··· , xm linearly independent over R. √ √ Note that linear independence is important. For example, Z 2 + Z 3 ⊆√R is not discrete. On the other hand, if Λ = a C OL is an ideal, where L = Q( d) and d < 0, then this is discrete.

42 6 Minkowski bound and finiteness of class group II Number Fields

Proof. Suppose Λ is generated by x1, ··· , xm. By linear independence, there is some g ∈ GLn(R) such that gxi = ei for all 1 ≤ i ≤ m, where e1, ··· , en is the standard basis. We know acting by g preserves discreteness, since it is a m m n−m 1 homeomorphism, and gΛ = Z ⊆ R × R is clearly discrete (take ε = 2 ). So this direction follows. For the other direction, suppose Λ is discrete. We pick y1, ··· , ym ∈ Λ which are linearly independent over R, with m maximal (so m ≤ n). Then by maximality, we know

( m ) ( m ) X X λiyi : λi ∈ R = λizi : λi ∈ R, zi ∈ Λ , i=1 1 and this is the smallest vector subspace of Rn containing Λ. We now let ( m ) X ∼ m X = λiyi : λi ∈ [0, 1] = [0, 1] . i=1 This is closed and bounded, and hence compact. So X ∩ Λ is finite. Also, we know M m Zyi = Z ⊆ Λ,

and if γ is any element of Λ, we can write it as γ = γ0 + γ1, where γ0 ∈ X and γ ∈ m. So 1 Z Λ ≤ |X ∩ Λ| < ∞. Zm m m 1 m So let d = |Λ/Z |. Then dΛ ⊆ Z , i.e. Λ ⊆ d Z . So 1 m ⊆ Λ ⊆ m. Z dZ 1 m So Λ is a free abelian group of rank m. So there exists x1, ··· , xm ∈ d Z which is an integral basis of Λ and are linearly independent over R. Definition (Lattice). If rank Λ = n = dim Rn, then Λ is a lattice in Rn. Definition (Covolume and fundamental domain). Let Λ ⊆ Rn be a lattice, and x1, ··· , xn be a basis of Λ, then let ( n ) X P = λixi : λi ∈ [0, 1] , i=1 and define the covolume of Λ to be covol(Λ) = vol(P ) = | det A|, P where A is the matrix such that xi = aijej. We say P is a fundamental domain for the action of Λ on Rn, i.e. n [ R = (γ + P ), γ∈Λ and (γ + P ) ∩ (µ + P ) ⊆ ∂(γ + P ). In particular, the intersection has zero volume.

43 6 Minkowski bound and finiteness of class group II Number Fields

This is called the covolume since if we consider the space Rn/Λ, which is an n-dimensional torus, then this has volume covol(Λ). 0 0 Observe now that if x1, ··· , xn is a different basis of Λ, then the transition 0 P matrix xi = bijxj has B ∈ GLn(Z). So we have det B = ±1, and covol(Λ) is independent of the basis choice. With these notations, we can now state Minkowski’s theorem.

Theorem (Minkowski’s theorem). Let Λ ⊆ Rn be a lattice, and P be a funda- mental domain. We let S ⊆ Rn be a measurable set, i.e. one for which vol(S) is defined. (i) Suppose vol(S) > covol(Λ). Then there exists distinct x, y ∈ S such that x − y ∈ Λ. (ii) Suppose 0 ∈ S, and S is symmetric around 0, i.e. s ∈ S if and only if −s ∈ S, and S is convex, i.e. for all x, y ∈ S and λ ∈ [0, 1], then

λx + (1 − λ)y ∈ S.

Then suppose either (a) vol(S) > 2n covol(Λ); or (b) vol(S) ≥ 2n covol(Λ) and S is closed. Then S contains a γ ∈ Λ with γ 6= 0. Note that for n = 2, this is what we used for quadratic fields. By considering Λ = Zn ⊆ Rn and S = [−1, 1]n, we know the bounds are sharp.

Proof.

(i) Suppose vol(S) > covol(Λ) = vol(P ). Since P ⊆ Rn is a fundamental domain, we have   n X X vol(S) = vol(S ∩ R ) = vol S ∩ (P + γ) = vol(S ∩ (P + γ)). γ∈Λ γ∈Λ

Also, we know

vol(S ∩ (P + γ)) = vol((S − γ) ∩ P ),

as volume is translation invariant. We now claim the sets (S − γ) ∩ P for γ ∈ Λ are not pairwise disjoint. If they were, then X X vol(P ) ≥ vol((S − γ) ∩ P ) = vol(S ∩ (P + γ)) = vol(S), γ∈Λ γ∈Λ

contradicting our assumption. Then in particular, there are some distinct γ and µ such that (S − γ) and (S − µ) are not disjoint. In other words, there are x, y ∈ S such that x − γ = y − µ, i.e. x − y = γ − µ ∈ Λ 6= 0.

44 6 Minkowski bound and finiteness of class group II Number Fields

(ii) We now let 1 1  S0 = S = s : s ∈ S . 2 2 So we have vol(S0) = 2−n vol(S) > covol(Λ), by assumption. (a) So there exists some distinct y, z ∈ S0 such that y − z ∈ Λ \{0}. We now write 1 y − z = (2y + (−2z)), 2 Since 2z ∈ S implies −2z ∈ S by symmetry around 0, so we know y − z ∈ S by convexity. 1
(b) We apply the previous part to Sm = 1 + m S for all m ∈ N, m > 0. So we get a non-zero γm ∈ Sm ∩ Λ.

By convexity, we know Sm ⊆ S1 = 2S for all m. So γ1, γ2, · · · ∈ S1 ∩Λ. But S1 is compact set. So S1 ∩ Λ is finite. So there exists γ such that γm is γ infinitely often. So \ γ ∈ Sm = S. m≥0

So γ ∈ S. We are now going to use this to mimic our previous proof that the class group of an imaginary quadratic field is finite. To begin with, we need to produce lattices from ideals of OL. Let L be a number field, and [L : Q] = n. We let σ1, ··· , σr : L → R be the real embeddings, and σr+1, ··· , σr+s, σ¯r+1, ··· , σ¯r+s : L → C be the complex embeddings (note that which embedding is σr+i and which isσ ¯r+i is an arbitrary choice). Then this defines an embedding

r s ∼ r 2s r+2s n σ = (σ1, σ2, ··· , σr, σr+1, ··· , σr+s): L,→ R × C = R × R = R = R ,

under the isomorphism C → R2 by x + iy 7→ (x, y). Just as we did for quadratic fields, we can relate the norm of ideals to their covolume.

Lemma.

n −s 1 (i) σ(OL) is a lattice in R of covolume 2 |DL| 2 .

(ii) More generally, if aCOL is an ideal, then σ(a) is a lattice and the covolume

−s 1 covol(σ(a)) = 2 |DL| 2 N(a).

Proof. Obviously (ii) implies (i). So we just prove (ii). Recall that a has an integral basis γ1, ··· , γn. Then a is the integer span of the vectors

(σ1(γi), σ2(γi), ··· , σr+s(γi))

45 6 Minkowski bound and finiteness of class group II Number Fields

for i = 1, ··· , n, and they are independent as we will soon see when we compute the determinant. So it is a lattice. We also know that

2 2 ∆(γ1, ··· , γn) = det(σi(γj)) = N(a) DL, where the σi run over all σ1, ··· , σr, σr+1, ··· , σr+s, σ¯r+1, ··· σ¯r+s. So we know 1 | det(σi(γj))| = N(a)|DL| 2 .

So what we have to do is to relate det(σi(γj)) to the covolume of σ(a). But these two expressions are very similar. In the σi(γj) matrix, we have columns that look like

σr+i(γj)σ ¯r+i(γj) = z z¯ .

On the other hand, the matrix of σ(γ) has corresponding entries

1 1 1  z Re(z) Im(z)
= 1 (z +z ¯) i (¯z − z)
= 2 2 2 i −i z¯

1 1 1  We call the last matrix A = . We can compute the determinant as 2 i −i   1 1 1 1 | det A| = det = . 2 i −i 2

Hence the change of basis matrix from (σi(γj)) to σ(γ) is s diagonal copies of A, so has determinant 2−s. So this proves the lemma.

Proposition. Let a C OL be an ideal. Then there exists an α ∈ a with α 6= 0 such that |N(α)| ≤ cLN(a), where  s 4 n! 1 c = |D | 2 . L π nn L This is the Minkowski bound. Proof. Let

n r s X X o Br,s(t) = (y1, ··· , yr, z1, ··· , zs) ∈ R × C : |yi| + 2 |zi| ≤ t .

This (i) is closed and bounded; (ii) is measurable (it is defined by polynomial inequalities); (iii) has volume π s tn vol(B (t)) = 2r ; r,s 2 n! (iv) is convex and symmetric about 0.

46 6 Minkowski bound and finiteness of class group II Number Fields

Only (iii) requires proof, and it is on the second example sheet, i.e. we are not doing it here. It is just doing the integral. We now choose t so that

n vol Br,s(t) = 2 covol(σ(a)).

Explicitly, we let  4 s tn = n!|D |1/2N(a). π L Then by Minkowski’s lemma, there is some α ∈ a non-zero such that σ(α) ∈ Br,s(t). We write σ(α) = (y1, ··· , yr, z1, ··· , zs). Then we observe

Y Y 2 N(α) = y1 ··· yrz1z¯1z2z¯2 ··· zsz¯s = yi |zj| .

By the AM-GM inequality, we know

1 X X  t |N(α)|1/n ≤ y + 2 |z | ≤ , n i j n

as we know σ(a) ∈ Br,s(t). So we get tn |N(α)| ≤ = c N(a). nn L

Corollary. Every [a] ∈ clL has a representative a ∈ OL with N(a) ≤ cL.

Theorem (Dirichlet). The class group clL is finite, and is generated by prime ideals of norm ≤ cL. Proof. Just as the case for imaginary quadratic fields.

47 7 Dirichlet’s unit theorem II Number Fields

7 Dirichlet’s unit theorem

We have previously characterized the units on OL as the elements with unit norm, i.e. α ∈ OL is a unit if and only if |N(α)| = 1. However, this doesn’t tell us much about how many units there are, and how they are distributed. The answer to this question is given by Dirichlet’s unit theorem. Theorem (Dirichlet unit theorem). We have the isomorphism

× ∼ r+s−1 OL = µL × Z , where N µL = {α ∈ L : α = 1 for some N > 0} is the group of roots of unity in L, and is a finite cyclic group.

Just as in the finiteness of the class group, we do it for an example first, or else it will be utterly incomprehensible. √ We do the example of real quadratic fields, L = Q( d), where d > 1 is square-free. So r = 2, s = 0, and L ⊆ R implies µL = {±1}. So × ∼ OL = {±1} × Z. Also, we know that √ √ √ N(x + y d) = (x + y d)(x − y d) = x2 − dy2.

So Dirichlet’s theorem is saying that there are infinitely many solutions of x2 − dy2 = ±1, and are all (plus or minus) the powers of one single element. √ Theorem (Pell’s equation). There are infinitely many x + y d ∈ OL such that x2 − dy2 = ±1. You might have seen this in IIC Number Theory, where we proved it directly by continued fractions. We will provide a totally unconstructive proof here, since this is more easily generalized to arbitrary number fields. This is actually just half of Dirichlet’s theorem. The remaining part is to show that they are all powers of the same element.

2 Proof. Recall that σ : OL → R sends √ √ √ α = x + y d 7→ (σ1(α), σ2(α)) = (x + y d, x − y d). √ (in the domain, d is a formal symbol, while in the codomain, it is a real number, namely the positive square root of d) Also, we know 1 covol(σ(OL)) = |DL| 2 .

48 7 Dirichlet’s unit theorem II Number Fields

Z[X]

(1, 1) N(α) = 1

Consider  |D |1/2  s = (y , y ) ∈ 2 : |y | ≤ t, |y | ≤ L . t 1 2 R 1 2 t So 1 n vol(st) = 4|DL| 2 = 2 covol(OL),

as n = [L : Q] = 2. Now Minkowski implies there is an α ∈ OL non-zero such that σ(α) ∈ st. Also, if we write

σ(α) = (y1, y2), then N(α) = y1y2. So such an α will satisfy

1/2 1 ≤ |N(α)| ≤ |DL| . This is not quite what we want, since we need |N(α)| = 1 exactly. Nevertheless, this is a good start. So let’s try to find infinitely such elements. First notice that no points on the lattice (apart√ from the√ origin) hits the x or y axis, since any such point must satisfy x ± y d = 0, but d is not rational. Also, st is compact. So st ∩ σ(OL) contains finitely many points. So we can find a t2 such that for each (y1, y2) ∈ st ∩ OL, we have |y1| > t2. In particular,

st2 does not contain any point in st ∩ σ(OL). So we get a new set of points 1/2 α ∈ st2 ∩ OL such that 1 ≤ |N(α)| ≤ |DL| .

s2

s1

49 7 Dirichlet’s unit theorem II Number Fields

We can do the same thing for st2 and get a new t3. In general, given t1 > ··· > tn, pick tn+1 be such that

( n ) [ 0 < tn+1 < min |y1| :(y1, y2) ∈ sti ∩ σ(OL) , i=1

and the minimum is finite since st is compact and hence contains finitely many lattice points on σ(OL).

Then we get an infinite sequence of ti such that sti ∩ σ(OL) are disjoint for different i. Since each must contain at least one point, we have got infinitely 1/2 many points in OL satisfying 1 ≤ |N(α)| ≤ |DL| . 1/2 Since there are only finitely many integers between 1 and |DL| , we can apply the pigeonhole principle, and get that there is some integer satisfying 1/2 1 ≤ |m| ≤ |DL| such that there exists infinitely many α ∈ OL with N(α) = m. This is not quite good enough. We consider

∼ [L: ] OL/mOL = (Z/mZ) Q ,

another finite set. We notice that each α ∈ OL must fall into one of finitely many the cosets of mOL in OL. In particular, each α such that N(α) = m must belong to one of these cosets. So again by the pigeonhole principle, there exists a β ∈ OL with N(β) = m, and infinitely many α ∈ OL with N(α) = m and α = β (mod mOL). Now of course α and β are not necessarily units, if m 6= 1. However, we will show that α/β is. The hard part is of course showing that it is in OL itself, because it is clear that α/β has norm 1 (alternatively, by symmetry, β/α is in OL, so an inverse exists). Hence all it remains is to prove the general fact that if

α = β + mγ, where α, β, γ ∈ OL and N(α) = N(β) = m, then α/β ∈ OL. To show this, we just have to compute

α m N(β) = 1 + γ = 1 + γ = 1 + βγ¯ ∈ O , β β β L

since N(β) = ββ¯. So done. We have thus constructed infinitely many units. We now prove the remaining part √ Theorem (Dirchlet’s unit theorem for real quadratic fields). Let L = Q( d). × Then there is some ε0 ∈ OL such that

× n OL = {±ε0 : n ∈ Z}.

We call such an ε0 a fundamental unit (which is not unique). So

× ∼ OL = {±1} × Z.

50 7 Dirichlet’s unit theorem II Number Fields

Proof. We have just proved the really powerful theorem that there are infinitely many ε with N(ε) = 1. We are not going to need the full theorem. All we need is that there are three — in particular, something that is not ±1. × We pick some ε ∈ OL with ε 6= ±1. This exists by what we just proved. Then we know |σ1(ε)|= 6 1, −1 as |σ1(ε)| = 1 if and only if ε = ±1. Replacing by ε if necessary, we wlog E = |σ1(ε)| > 1. Now consider

{α ∈ OL : N(α) = ±1, 1 ≤ |σ1(α)| ≤ E}.

This is again finite, since it is specified by a compact subset of the OL-lattice. So we pick ε0 in this set with ε0 6= ±1 and |σ1(ε0)| minimal (> 1). Replacing ε0 by −ε0 if necessary, we can assume σ1(ε) > 1. × N Finally, we claim that if ε ∈ OL and σ1(ε) > 0, then ε = ε0 for some N ∈ Z. This is obvious if we have addition instead of multiplication. So we take logs. Suppose log ε = N + γ, log ε0 where N ∈ Z and 0 ≤ γ < 1. Then we know

−N γ × εε0 = ε0 ∈ OL ,

γ γ but |ε0 | = |ε0| < |ε0|, as |ε0| > 1. By our choice of ε0, we must have γ = 0. So done.

Now we get to prove the Dirichlet unit theorem in its full glory. Theorem (Dirichlet unit theorem). We have the isomorphism

× ∼ r+s−1 OL = µL × Z , where N µL = {α ∈ L : α = 1 for some N > 0} is the group of roots of unity in L, and is a finite cyclic group.

Proof. We do the proof in the opposite order. We throw in the logarithm at the very beginning. We define × r+s ` : OL → R by

x 7→ (log |σ1(x)|, ··· , log |σr(x)|, 2 log |σr+1(x)|, ··· , 2 log |σr+s(x)|).

Note that |σr+i(x)| = |σr+`(x)|. So this is independent of the choice of one of σr+i, σ¯r+i. r+s Claim. We now claim that im ` is a discrete group in R and ker ` = µL is a finite cyclic group.

51 7 Dirichlet’s unit theorem II Number Fields

We note that log |ab| = log |a| + log |b|. So this is a group homomorphism, and the image is a subgroup. To prove the first part, it suffices to show that im ` ∩ [−A, A]r+s is finite for all A > 0. We notice ` factors as

× σ r s j r+s OL OL R × C R . where σ maps α 7→ (σ1(α), ··· , σr+s(α)), and

j :(y1, ··· , yr, z1, ··· , zs) 7→ (log |y1|, ··· , log |yr|, 2 log |z1|, ··· , 2 log |z2|).

We see

−1 r+s −A A −A A j ([−A, A] ) = {(yi, zj): e ≤ |yi| ≤ e , e ≤ 2|zj| ≤ e }

is a compact set, and σ(OL) is a lattice, in particular discrete. So σ(OL) ∩ j−1([−A, A]r+s) is finite. This also shows the kernel is finite, since the kernel is the inverse image of a compact set. Now as ker ` is finite, all elements are of finite order. So ker ` ⊆ µL. Con- versely, it is clear that µL ⊆ ker `. So it remains to show that µL is cyclic. Since L embeds in C, we know µL is contained in the roots of unity in C. Since µL is finite, we know L is generated by a root of unity with the smallest argument (from, say, IA Groups). Claim. We claim that

n X o ∼ r+s−1 im ` ⊆ (y1, ··· , yr+s): yi = 0 = R .

× To show this, note that if α ∈ OL , then

n s Y Y N(α) = σi(α) σr+`(α)¯σr+` = ±1. i=1 `=1 Taking the log of the absolute values, we get X X 0 = log |σi(α)| + 2 log |σr+i(α)|.

So we know im ` ⊆ Rr+s−1 as a discrete subgroup. So it is isomorphic to Za for some a ≤ r + s − 1. Then what we want to show is that im ` ⊆ Rr+s−1 is a lattice, i.e. it is congruent to Zr+s−1. Note that so far what we have done is the second part of what we did for the real quadratic fields. We took the logarithm to show that these form a discrete subgroup. Next, we want to find r + s − 1 independent elements to show it is a lattice.

Claim. Fix a k such that 1 ≤ k ≤ r + s and α ∈ OL with α 6= 0. Then there exists a β ∈ OL such that

 2 s |N(β)| ≤ |D |1/2, π L

52 7 Dirichlet’s unit theorem II Number Fields

and moreover if we write

`(α) = (a1, ··· , ar+s)

`(β) = (b1, ··· , br+s),

then we have bi < ai for all i 6= k. We can apply Minkowski to the region

r s S = {(y1, ··· , yr, z1, ··· , zs) ∈ R × C : |yi| ≤ ci, |zj| ≤ cr+j}

(we will decide what values of ci to take later). Then this has volume

r s vol(S) = 2 π c1 ··· cr+s.

ai We notice S is convex and symmetric around 0. So if we choose 0 < ci < e for i 6= k, and choose  s 2 1/2 1 ck = |DL| . π c1 ··· cˆk ··· cr+s

Then Minkowski gives β ∈ σ(OL) ∩ S, satisfying the two conditions above. × Claim. For any k = 1, ··· , r + s, there is a unit uk ∈ O such that if `(uk) = L P (y1, ··· , yr+s, then yi < 0 for all i 6= k (and hence yk > 0 since yi = 0). This is just as in the proof for the real quadratic case. We can repeatedly apply the previous claim to get a sequence α1, α2, · · · ∈ OL such that N(αt) is bounded for all t, and for all i 6= k, the ith coordinate of `(α1), `(α2), ··· is strictly decreasing. But then as with real quadratic fields, the pigeonhole principle implies we can find t, t0 such that

N(αt) = N(αt0 ) = m,

say, and αt ≡ αt0 (mod mOL),

i.e. αt = αt0 in OL/mOL. Hence for each k, we get a unit uk = αt/αt0 such that

0 `(uk) = `(αt) − `(αt) = (y1, ··· , yr+s) P has yi < 0 if i 6= k (and hence yk > 0, since yi = 0). We need a final trick to show the following: r+s−1 Claim. The units u1, ··· , ur+s−1 are linearly independent in R . Hence × the rank of `(OL ) = r + s − 1, and Dirichlet’s theorem is proved.

We let A be the (r + s) × (r + s) matrix whose jth row is `(uj), and apply the following lemma:

Claim. Let A ∈ Matm(R) be such that aii > 0 for all i and aij < 0 for all i 6= j, P and j aij ≥ 0 for each i. Then rank(A) ≥ m − 1.

53 7 Dirichlet’s unit theorem II Number Fields

To show this, we let vi be the ith column of A. We show that v1, ··· , vm−1 are linearly independent. If not, there exists a sequence ti ∈ R such that

m−1 X tivi = 0, (∗) i=1 with not all of the ti non-zero. We choose k so that |tk| is maximal among the t1, ··· , tm−1’s. We divide the whole equation by tk. So we can wlog assume tk = 1, ti ≤ 1 for all i. Now consider the kth row of (∗). We get

m−1 m−1 X X 0 = tiaki ≥ aki, i=1 i=1

as a < 0 and t ≤ 1 implies at ≥ a. Moreover, we know ami > 0 strictly. So we get m X 0 > aki ≥ 0. i=1 This is a contradiction. So done. You should not expect this to be examinable. We make a quick definition that we will need later. Definition (Regulator). The regulator of a number field L is

× r+s−1 RL = covol(`(OL ) ⊆ R ).

× More concretely, we pick fundamental units ε1, ··· , εr+s−1 ∈ OL so that

× n1 nr+s−1 OL = µL × {ε1 ··· εr+s−1 : ni ∈ Z}.
We take any (r + s − 1)(r + s − 1) subminor of the matrix `(ε1) ··· `(εr+s) . Their determinants all have the same absolute value, and

| det(subminor)| = RL.

This is a definition we will need later. √ We quickly look at some examples with quadratic fields. Consider L = Q( d), where d 6= 0, 1 square-free.

× Example. If d < 0, then r = 0 and s = 1. So r + s − 1 = 0. So OL = µL is a finite group. So RL = 1. Lemma.

(i) If d = −1, then Z[i]× = {±1, ±i} = Z/4Z. √ 1 6 × (ii) If d = −3, then let ω = 2 (1 + d), and we have ω = 1. So Z[ω] = {1, ω, ··· , ω5} =∼ Z/6Z. × (iii) For any other d < 0, we have OL = {±1}.

54 7 Dirichlet’s unit theorem II Number Fields

Proof. This is just a direct check. If d ≡ 2, 3 (mod 4), then by looking at the solution of x2 − dy2 = ±1 in the integers, we get (i) and (iii). y
2 d 2 If d ≡ 1 (mod 4), then by looking at the solutions to x + 2 − 4 y = ±1 in the integers, we get (ii) and (iii).

Now if d > 0, then RL = | log |ε||, where ε is a fundamental unit. So how do we find a fundamental unit? In general, there is no good algorithm for finding the fundamental unit of a fundamental field. The best algorithm takes exponential time. We do have a good algorithm for quadratic fields using continued fractions, but we are not allowed to use that. Instead, we could just guess a solution — we find a unit by guessing, and then show there is no smaller one by direct check. √ √ Example. Consider the field Q( 2). We can try ε = 1 + 2. We have N(ε) = 1 − 2 = −1. So this√ is a unit. We claim this is fundamental. If not, there there exists u = a + b 2, where a, b ∈ Z and 1 < u < ε (as real numbers). Then we have √ u¯ = a − b 2 has uu¯ = ±1. Since u > 1, we know |u¯| < 1. Then we must have u ± u¯ > 0. So we need a, b > 0. We know can only be finitely many possibilities for √ √ 1 < a + b 2 < 1 + 2, where a, b are positive integers. But there actually are none. So done. √ √ Example. Consider Q( 11). We guess ε = 10 − 3 11 is a unit. We can compute N(ε) = 100 − 99 = 1. Note that ε < 1 and ε−1 > 1. Suppose this is not fundamental. Then we have some u such that √ √ 1 < u = a + b 11 < 10 + 3 11 = ε−1 < 20. (∗)

We can check all the cases, but there is a faster way. We must have N(u) = ±1. If N(u) = −1, then a2 − 11b2 = −1. But −1 is not a square mod 11. So there we must have N(u) = 1. Then u−1 = u¯. We get 0 < ε < u−1 = u¯ < 1 also. So √ −1 < −a + b 11 < 0. Adding this to (∗), we get √ √ √ 0 < 2b 11 < 10 + 3 11 < 7 11.

So b = 1, 2 or 3, but 11b2 + 1 is not a square for each of these. So done.

55 8 L-functions, Dirichlet series* II Number Fields

8 L-functions, Dirichlet series*

This section is non-examinable. We start by proving the exciting fact that there are infinitely many primes. Theorem (Euclid). There are infinitely many primes.

Proof. Consider the function

−1 Y  1 Y  1 1  X 1 1 − = 1 + + + ··· = . p p p2 n p primes p prime n>0

e1 er This is since every n = p1 ··· pr factors uniquely as a product of primes, and each such product appears exactly once in this. If there were finitely many −1 P 1  1  primes, as pn converges to 1 − p , the sum

X 1 Y  1 = 1 − n p n≥1 p prime

must be finite. But the harmonic series diverges. This is a contradiction.

We all knew that. What we now want to prove is something more interesting.

Theorem (Dirichlet’s theorem). Let a, q ∈ Z be coprime. Then there exists infinitely many primes in the sequence

a, a + q, a + 2q, ··· ,

i.e. there are infinitely many primes in any such arithmetic progression. We want to imitate the Euler proof, but then that would amount to showing that −1 Y  1 1 − p p≡a mod q p prime is divergent, and there is no nice expression for this. So it will be a lot more work. To begin with, we define the Riemann zeta function.

Definition (Riemann zeta function). The Riemann zeta function is defined as X ζ(s) = n−s n≥1

for s ∈ C. There are some properties we will show (or assert): Proposition. (i) The Riemann zeta function ζ(s) converges for Re(s) > 1.

56 8 L-functions, Dirichlet series* II Number Fields

(ii) The function 1 ζ(s) − s − 1 extends to a holomorphic function when Re(s) > 0. In other words, ζ(s) extends to a meromorphic function on Re(s) > 0 with a simple pole at 1 with residue 1. (iii) We have the expression

−1 Y  1  ζ(s) = 1 − ps p prime

for Re(s) > 1, and the product is absolutely convergent. This is the Euler product. The first part follows from the following general fact about Dirichlet series.

P −s Definition (Dirichlet series). A Dirichlet series is a series of the form ann , where a1, a2, · · · ∈ C. Lemma. If there is a real number r ∈ R such that

r a1 + ··· + aN = O(N ),

then X −s ann converges for Re(s) > r, and is a holomorphic function there. Then (i) is immediate by picking r = 1, since in the Riemann zeta function, a1 = a2 = ··· = 1. Recall that xs = es log x has

|xs| = |xRe(s)|

if x ∈ R, x > 0. Proof. This is just IA Analysis. Suppose Re(s) > r. Then we can write

N X −s −s −s −s −s ann = a1(1 − 2 ) + (a1 + a2)(2 − 3 ) + ··· n=1 −s −s + (a1 + ··· + aN−1)((N − 1) − N ) + RN , where a + ··· + a R = 1 N . N N s This is getting annoying, so let’s write

T (N) = a1 + ··· + aN .

We know

T (N) T (N) 1 = → 0 N s N r N Re(s)−r

57 8 L-functions, Dirichlet series* II Number Fields

as N → ∞, by assumption. Thus we have

X −s X −s −s ann = T (n)(n − (n + 1) ) n≥1 n≥1 if Re(s) > r. But again by assumption, T (n) ≤ B · nr for some constant B and all n. So it is enough to show that X nr(n−s − (n + 1)−s) n converges. But Z n+1 −s −s s n − (n + 1) = s+1 dx, n x and if x ∈ [n, n + 1], then nr ≤ xr. So we have Z n+1 Z n+1 r −s −s r s dx n (n − (n + 1) ) ≤ x s+1 dx = s s+1−r . n x n x It thus suffices to show that Z n dx s+1−r 1 x s converges, which it does (to s−r ). We omit the proof of (ii). The idea is to write

∞ 1 X Z n+1 dx = , s − 1 xs n=1 n P and show that φn is uniformly convergent when Re(s) > 0, where Z n+1 −s dx φn = n − s . n x

For (iii), consider the first r primes p1, ··· , pr, and r Y −s −1 X −s (1 − pi ) = n , i=1 where the sum is over the positive integers n whose prime divisors are among p1, ··· , pr. Notice that 1, ··· , r are certainly in the set. So

r Y −s −1 X −s X − Re(s) ζ(s) − (1 − pi ) ≤ |n | = n . i=1 n≥r n≥r P − Re(s) But n≥r n → 0 as r → ∞, proving the result, if we also show that it converges absolutely. We omit this proof, but it follows from the fact that X X p−s ≤ n−s. p prime n Q and the latter converges absolutely, plus the fact that (1 − an) converges if P and only if an converges, by IA Analysis I. This is good, but not what we want. Let’s mimic this definition for an arbitrary number field!

58 8 L-functions, Dirichlet series* II Number Fields

Definition (Zeta function). Let L ⊇ Q be a number field, and [L : Q] = n. We define the zeta function of L by

X −s ζL(s) = N(a) . aCOL

It is clear that if L = Q and OL = Z, then this is just the Riemann zeta function. Theorem.

(i) ζL(s) converges to a holomorphic function if Re(s) > 1.

(ii) Analytic class number formula: ζL(s) is a meromorphic function if Re(s) > 1 1 − n and has a simple pole at s = 1 with residue | cl |2r(2π)sR L L , 1/2 |DL| |µL|

where clL is the class group, r and s are the number of real and complex embeddings, you know what π is, RL is the regulator, DL is the discriminant and µL is the roots of unity in L. (iii) Y −s −1 ζL(s) = (1 − N(p) ) . pCOL prime ideal This is again known as the Euler product. We will not prove this, but the proof does not actually require any new ideas. Note that X Y N(a)−s = (1 − N(p)−s)−1

aCOL pCOL,p prime holds “formally”, as in the terms match up when you expand, as an immediate consequence of the unique factorization of ideals into a product of prime ideals. The issue is to study convergence of P N(a)−s, and this comes down to estimating the number of ideals of fixed norm geometrically, and that is where all the factors in the pole come it. √ Example. We try to compute ζL(s), where L = Q( d). This has discriminant D, which may be d or 4d. We first look at the prime ideals. If p is a prime ideal in OL, then p | hpi for a unique p. So let’s enumerate the factors of ηL controlled by p ∈ Z. 2 Now if p | |DL|, then hpi = p ramifies, and N(p) = p. So this contributes a factor of (1 − p−s)−1. Now if p remains prime, then we have N(hpi) = p2. So we get a factor of

(1 − p−2s)−1 = (1 − p−s)−1(1 + p−s)−1.

If p splits completely, then hpi = p1p2. So N(pi) = p,

59 8 L-functions, Dirichlet series* II Number Fields

and so we get a factor of

(1 − p−s)−1(1 − p−s)−1.

So we find that ζL(s) = ζ(s)L(χD, s), where we define Definition (L-function). We define the L-function by Y L(χ, s) = (1 − χ(p)p−s)−1. p prime

In our case, χ is given by  0 p | D ( D   p p is odd χD(p) = −1 p remains prime = . depends on d mod 8 p = 2 1 p splits √ Example. If L = Q( −1), then we know

−4 −1 p−1 = = (−1) 2 if p 6= 2, p p

and χD(2) = 0 as 2 ramifies. We then have

Y p−1 −s −1 1 1 1 L(χ , s) = (1 − (−1) 2 p ) = 1 − + − + ··· . D 3s 5s 7s p>2 prime

Note that χD was defined for primes only, but we can extend it to a function χD : Z → C by imposing

χD(nm) = χD(n)χD(m),

i.e. we define e1 er e1 er χD(p1 ··· pr ) = χD(p1) ··· χD(pr) . √ Example. Let L = Q( −1). Then

( m−1 (−1) 2 m odd χ−4(m) = 0 m even.

It is an exercise to show that this is really the extension, i.e.

χ−4(mn) = χ−4(m)χ−4(n).

Notice that this has the property that

χ−4(m − 4) = χ−4(m).

We give these some special names

60 8 L-functions, Dirichlet series* II Number Fields

Definition (Dirichlet character). A function χ : Z → C is a Dirichlet character of modulus D if there exists a group homomorphism

 × Z × w : → C DZ such that ( w(m mod D) gcd(m, D) = 1 χ(m) = . 0 otherwise We say χ is non-trivial if ω is non-trivial.

Example. χ−4 is a Dirichlet character of modulus 4. Note that χ(mn) = χ(m)χ(n) for such Dirichlet characters, and so

Y X χ(n) L(χ, s) = (1 − χ(p)p−s)−1 = ns p prime n≥1

for such χ. √ Proposition. χD, as defined for L = Q( d) is a Dirichlet character of modulus D. Note that this is a very special Dirichlet character, as it only takes values 0, ±1. We call this a quadratic Dirichlet character. Proof. We must show that

χD(p + Da) = χD(p)

for all p, a. (i) If d ≡ 3 (mod 4), then D = 4d. Then

χD(2) = 0,

as (2) ramifies. So χD(even) = 0. For p > 2, we have

D  d p p−1 χ (p) = = = (−1) 2 D p p d

d−1 as 2 ≡ 1 (mod 2), by quadratic reciprocity. So   p + Da p−1 4da/2 χ (p + Da) = (−1) 2 (−1) = χ (p). D d D

(ii) If d ≡ 1, 2 (mod 4), see example sheet.

Lemma. Let χ be any non-trivial Dirichlet character. Then L(χ, s) is holomor- phic for Re(s) > 0.

61 8 L-functions, Dirichlet series* II Number Fields

Proof. By our lemma on convergence of Dirichlet series, we have to show that

N X χ(i) = O(1), i=1 i.e. it is bounded. Recall from Representation Theory that distinct irreducible characters of a finite group G are orthogonal, i.e. ( 1 X 1 χ1 = χ2 χ1(g)χ2(g) = . |G| 0 otherwise g∈G

× We apply this to G = (Z/DZ) , where χ1 is trivial and χ2 = χ. So orthogonality gives X X χ(i) = χ(i) = 0, × aD

Corollary. For quadratic characters χD, we have

L(χD, 1) 6= 0.

For example, if D < 0, then

2π| cl √ | L(χ , 1) = Q( d) . D |D|1/2|µ √ | Q( d) Proof. We have shown that

√ ζ (s) = ζ (s)L(χD, s). Q( d) Q

√ Note that ζ (s) and ζ (s) have simple poles at s = 1, while L(χD, s) is Q( d) Q holomorphic at s = 1. Since the residue of ζ (s) at s = 1 is 1, while the residue of ζ √ at s = 1 Q Q( D) is non-zero by the analytic class number formula. So L(χD, 1) is non-zero, and given by the analytic class number formula. √ Example. If L = Q( −1), then 1 1 1 2π · 1 π 1 − + − + ··· = = . 3 5 7 2 · 4 4 In general, for any field whose class number we know, we can get a series expansion for π. And it converges incredibly slow. Note that this corollary required two things — the analytic input for the analytic class number formula, and quadratic reciprocity (to show that χD is a Dirichlet character). More ambitiously, we now compute the zeta function of a cyclotomic field, L = Q(ωq), where ωq is the primitive qth root of unity and q ∈ N. We need to know the following facts about cyclotomic extensions: Proposition.

62 8 L-functions, Dirichlet series* II Number Fields

(i) We have [L : Q] = ϕ(q), where × ϕ(q) = |(Z/qZ) |.

(ii) L ⊇ Q is a Galois extension, with × Gal(L/Q) = (Z/qZ) ,

× r where if r ∈ (Z/qZ) , then r acts on Q(wQ) by sending ωq 7→ ωq . This is what plays the role of quadratic reciprocity for cyclotomic fields. (iii) The ring of integers is

OL = Z[ωq] = Z[x]/Φq(x), where xq − 1 Φq(x) = Q d|q,d6=q Φd(x) is the qth cyclotomic polynomial.

(iv) Let p be a prime. Then p ramifies in OL if and only if p | DL, if and only if p | q. So while D might be messy, the prime factors of D are the prime factors of q.

(v) Let p be a prime and p - q. Then hpi factors as a product of ϕ(q)/f distinct prime ideals, each of norm pf , where f is the order of p in (Z/qZ)×. Proof. (i) In the Galois theory course. (ii) In the Galois theory course. (iii) In the example sheet. (iv) In the example sheet. (v) Requires proof, but is easy Galois theory, and is omitted. Example. Let q = 8. Then x8 − 1 x8 − 1 Φ = = = x4 + 1. 8 (x + 1)(x − 1)(x2 + 1) x4 − 1 So given a prime p (that is not 2), we need to understand

Fp[x] OL/p = , Φ8

i.e. how Φ8 factors factors mod p (Dedekind’s criterion). We have

× (Z/8) = {1, 3, 5, 7} = {1, 3, −3, −1} = Z/2 × Z/2. Then (v) says if p = 17, then x4 factors into 4 linear factors, which it does. If p = 3, then (v) says x4 factors into 2 quadratic factors. Indeed, we have

(x2 − x − 1)(x2 + x − 1) = (x2 − 1)2 − x2 = x4 + 1.

63 8 L-functions, Dirichlet series* II Number Fields

Given all of these, let’s compute the zeta function! Recall that Y ζ (s) = (1 − N(p)−s)−1. Q(ωq ) p

We consider the prime ideals p dividing hpi, where p is a fixed integer prime number. If p - q, then (v) says this contributes a factor of (1 − p−fs)−ϕ(q)/f ,

to the zeta function, where f is the order of p in (Z/qZ)×. We observe that this thing factors, since Y 1 − tf = (1 − γt),

γ∈µf with f µf = {γ ∈ C : γ = 1}, and we can put t = p−s. We let × × ω1, ··· , ωϕ(q) :(Z/qZ) → C be the distinct irreducible (one-dimensional) representations of (Z/qZ)×, with × ω1 being the trivial representation, i.e. ω1(a) = 1 for all a ∈ (Z/qZ) . The claim is that ω1(p), ··· , ωϕ(q)(p) are fth roots of 1, each repeated ϕ(q)/f times. We either say this is obvious, or we can use some representation theory. We know p generates a cyclic subgroup hpi of (Z/qZ)× of order f, by definition of f. So this is equivalent to saying the restrictions of ω1, ··· , ωϕ(q) to p are the f distinct irreducible characters of hpi =∼ Z/f, each repeated ϕ(q)/f times. Equivalently, note that

× × (Z/qZ) (Z/qZ) × Reshpi (ω1 ⊕ · · · ⊕ ωϕ(q)) = Reshpi (regular representation of (Z/qZ) ).

So this claims that

× ϕ(q) Res(Z/qZ) (regular rep. of ( /q )×) = (regular rep. of /f). hpi Z Z f Z But this is true for any group, since

G ResH CG = |G/H|CH,

as the character of both sides is |G|δe. So we have

ϕ(q) −fs −ϕ(q)/f Y −s −1 (1 − p ) = (1 − ωi(p)p ) . i=1 So we let ( wi(n mod q) gcd(n, q) = 1 χi(n) = 0 otherwise be the corresponding Dirichlet characters. So we have just shown that

64 8 L-functions, Dirichlet series* II Number Fields

Proposition. We have

ϕ(q) ϕ(q) Y Y ζ (s) = L(χ , s) · (corr. factor) = ζ (s) L(χ , s) · (corr. factor) Q(ωq ) i Q i i=1 i=2 where the correction factor is a finite product coming from the primes that divide q. By defining the L functions in a slightly more clever way, we can hide the correction factors into the L(χ, s), and then the ζ function is just the product of these L-functions.

Proof. Our analysis covered all primes p - q, and the correction factor is just to include the terms with p | q. The second part is just saying that

Y −s −1 ζQ(s) = L(χ1, s) (1 − p ) . p|q

This allows us to improve our result on the non-vanishing of L(χ, 1) to all Dirichlet characters, and not just quadratic Dirichlet characters. Corollary. If χ is any non-trivial Dirichlet character, then L(χ, 1) 6= 0. Proof. By definition, Dirichlet characters come from representations of some (Z/qZ)×, so they appear in the formula of the ζ function of some cyclotomic extension. Consider the formula

ϕ(q) Y ζ (s) = ζ (s) L(χ , s) · (corr. factor) Q(ωq ) Q i i=2

at s = 1. We know that the L(χi, s) are all holomorphic at s = 1. Moreover, both ζ and ζ have a simple pole at 0. Since the correction terms are finite, Q(ωq ) Q it must be the case that all L(χi, s) are non-zero.

Theorem (Dirichlet, 1839). Let a, q ∈ N be coprime, i.e. gcd(a, q) = 1. Then there are infinitely many primes in the arithmetic progression

a, a + q, a + 2q, a + 3q, ··· .

Proof. As before, let

× × ω1, ··· , ωϕ(q) :(Z/qZ) → C be the irreducible characters, and let

χ1, ··· , χϕ(q) : Z → C

be the corresponding Dirichlet character, with ω1 the trivial one. Recall the orthogonality of columns of the character table, which says that if gcd(p, q) = 1, then ( 1 X 1 a ≡ p (mod q) ωi(a)ωi(p) = . ϕ(q) 0 otherwise i

65 8 L-functions, Dirichlet series* II Number Fields

Hence we know ( 1 X 1 a ≡ p (mod q) χi(a)χi(p) = , ϕ(q) 0 otherwise i

even if gcd(p, q) 6= 1, as then χi(p) = 0. So X 1 X X p−s = χ (a) χ (p)p−s. (‡) ϕ(q) i i p≡a mod q i all primes p p prime We want to show this has a pole at s = 1, as in Euclid’s proof. P −s To do so, we show that p χi(p)p is “essentially” log L(χi, s), up to some bounded terms. We Taylor expand

X X χ(p)n X χ(pn) log L(χ, s) = − log(1 − χ(p)p−s) = = . npns npns n≥1 n≥1 p prime p prime What we care about is the n = 1 term. So we claim that X χ(pn) npns n≥2,p prime converges at s = 1. This follows from the geometric sum

n X X χ(p ) X X −ns X 1 X 1 ≤ p = ≤ < ∞. npns ps(ps − 1) ns(ns − 1) p n≥2 p n≥2 p prime n≥2 Hence we know

X −s log L(χ, s) = χi(p)p + bounded stuff p near s = 1. So at s = 1, we have 1 X (‡) ∼ χ (a) log L(χ , s). ϕ(q) i i i and we have to show that the right hand side has a pole at s = 1. We know that for i =6 1, i.e. χi non-trivial, L(χi, s) is holomorphic and non-zero at s = 1. So we just have to show that log L(χ1, s) has a pole. Note

that L(χ1, s) is essentially ζQ(s). Precisely, we have

Y −s L(χ1, s) = ζQ(s) (1 − p ). p|q

Moreover, we already know that ζQ(s) blows up at s = 1. We have 1 ζ (s) = + holomorphic function Q s − 1 1 = (1 + (s − 1)(holomorphic function)). s − 1

66 8 L-functions, Dirichlet series* II Number Fields

So we know  1  log L(χ , s) ∼ log ζ (s) ∼ log , 1 Q s − 1 and this does blow up at s = 1.

So far, we have been working with abelian extensions over Q, i.e. extensions L/Q whose Galois group is abelian. By the Kronecker–Weber theorem, every abelian extension of Q is contained within some cyclotomic extension. So in some sense, we have considered the “most general” abelian extension. Can we move on to consider more complicated number fields? In general, suppose L/Q is Galois, and G = Gal(L/Q). We can still make sense of the ζ functions, and it turns out it always factors as

Y dim ρ ζL(s) = L(ρ, s) , ρ where ρ ranges over all the irreducible representations of G, and L(ρ, s) is the Artin L-function. It takes some effort to define the Artin L-function, and we

shall not do so here. However, it is worth noting that L(1, s) is just ζQ(s), and for ρ 6= 1, we still have a factorization of the form Y L(ρ, s) = Lp(ρ, s). p prime

This Lp(ρ, s) is known as the Euler factor. One can show that L(ρ, s) is always a meromorphic function of s, and is conjectured to be holomorphic for all s (if ρ 6= 1, of course). If ρ is one-dimensional, then L(ρ, s) is a Dirichlet series L(χ, s) for some χ. Recall that to establish this fact for quadratic fields, we had to use quadratic reciprocity. In general given a ρ, finding χ is a higher version of “quadratic reciprocity”. This area is known as class field theory. If dim ρ > 1, then this is “non-abelian class field theory”, known as Langlands programme.

67 Index II Number Fields

Index

DL, 15 fractional, 22 IL, 24 honest, 22 L-function, 60 integral, 22 clL, 26 invertible, 22 ζ, 56 multiplication, 17 ζL, 59 norm, 27 pα, 8 prime, 18 r, 13 sum, 24 s, 13 unique factorization, 24 ideal class group, 26 addition of ideals, 24 inert prime, 32 algebraic integer, 4 integral, 5 analytic class number formula, 59 integral basis, 14 Artin L-function, 67 integral ideal, 22 associates, 18 invertible fractional ideal, 22

class field theory, 67 Kronecker–Weber theorem, 67 class group, 26 finiteness, 47 Langlands programme, 67 conjugate, 12 lattice, 43 covolume, 43 minimal polynomial, 8 Dedekind domain, 18 Minkowski bound, 40, 46 Dedekind’s criterion, 33 Minkowski’s lemma, 38 degree, 4 Minkowski’s theorem, 44 Dirichlet character, 61 multiplication of ideals, 17 Dirichlet series, 57 Dirichlet unit theorem, 48 norm, 10 Dirichlet’s theorem, 56 of ideal, 27 Dirichlet’s theorem on primes in number field, 4 AP, 65 Pell’s equation, 48 Dirichlet’s unit theorem, 50, 51 prime ideal, 18 discrete subset, 42 primitive element theorem, 12 discriminant, 15 divides, 17 quadratic Dirichlet character, 61

Euler factor, 67 ramification index, 32 ramified prime, 32 field extension, 4 regulator, 54 finite extension, 4 Riemann zeta function, 56 finitely-generated, 5 finiteness of class group, 47 splitting prime, 32 fractional ideal, 22 sum of ideals, 24 invertible, 22 fundamental domain, 43 trace, 10

honest ideal, 22 unique factorization of ideals, 24

ideal zeta function, 59

68