Dedekind Zeta Functions and Arithmetical Equivalence: Analytic and Algebraic Approaches

Corso di Laurea Magistrale in Matematica

Dedekind Zeta functions and arithmetical equivalence: analytic and algebraic approaches

Relatore: Prof. Giuseppe Molteni Tesi di laurea di: Francesco Battistoni Matricola 863778

Anno Accademico 2015-2016 Contents

1 Recalls of algebraic number theory 6 1.1 Numberfieldsandringsofintegers ...... 6 1.1.1 Numberfieldsandtheirembeddings ...... 6 1.1.2 Integralityandringsofintegers ...... 8 1.2 Dedekind domains, fractional ideals and class group ...... 9 1.2.1 Prime factorization in Dedekind domains ...... 9 1.2.2 DVR,ramificationandinertiadegrees ...... 11 1.2.3 Fractionalidealsandclassgroup ...... 14 1.3 Somealgebraicinvariants...... 17 1.3.1 Normandtrace...... 17 1.3.2 Thediscriminant ...... 18 1.4 Norm of ideals in K ...... 20 1.4.1 NormofproperidealsO ...... 20 1.4.2 Normoffractionalideals ...... 22 1.5 TheoremsbyMinkowskianDirichlet ...... 22 1.6 Thedifferentideal ...... 23

2 The Riemann Zeta Function 26 2.1 Dirichletseries ...... 26 2.1.1 Convergencetheorems ...... 26 2.1.2 Arithmetical functions and Euler product...... 28 2.2 The Riemann Zeta Function ζ(s) ...... 31 2.3 Thefunctionalequation ...... 32

3 The Dedekind Zeta Functions 35 3.1 Deﬁnitionandﬁrstproperties ...... 35 3.2 The functional equation of ζK and the class number formula . 37 3.2.1 Thefunctionalequation ...... 37 3.2.2 Analyticlemmas ...... 39 3.2.3 Proof of the functional equation for ζ ( ; s)...... 41 K R

1 3.2.4 Computation of the residue in 1: the Class Number Formula ...... 47

4 Arithmetical equivalence: the analytic way 49 4.1 Formulationandfirstresults ...... 49 4.1.1 Theproblem...... 49 4.1.2 Resultonimmersionsanddegrees ...... 50 4.2 The Asymptotic formula for the non trivial zeros of ζK .... 51 4.2.1 Arithmetically equivalent fields have same discriminant 51 4.2.2 The non vanishing of ζK (s)overtwolines ...... 52 4.2.3 Recalls on entire functions and Phragmen-Lindelöf The- orem...... 54 4.2.4 Proofoftheasymptoticformula ...... 55

5 Arithmetical equivalence: the algebraic approach 64 5.1 Decomposition type and Gassmann equivalence ...... 64 5.1.1 Decomposition type of prime numbers in the rings of integers ...... 64 5.1.2 Gassmannequivalence ...... 66 5.1.3 NumberfieldsandGaloisgrouptheory ...... 69 5.1.4 Arithmetically equivalent number fields share same roots ofunity ...... 71 5.2 Arithmetically equivalent fields which are not isomorphic... 73 5.2.1 Cohomologicaltools ...... 73 5.2.2 Anexplicitexample...... 77

6 Appendix 79 6.1 Summationformulas ...... 79 6.2 PropertiesoftheGammafunction...... 80 6.3 Groupactions...... 81 6.4 BasicGaloisTheory ...... 82

2 Introduction

“It’s this simple law, which every writer knows, of taking two opposites and putting them in a room together.” Trey Parker

s The Riemann Zeta function ζ(s) := n n− is a typical example of a mathematical instrument originally related to certain fields of Mathematics and suddenly become fundamental for theP study of other problems. Euler used this function (which for real s is a classic object of study in Analysis) to give a different proof of the existence of infinite prime numbers, while Riemann exploited its meromorphic continuation on the plane C of complex numbers for a deeper study of the distribution of the prime numbers (arriving to the formulation of the Riemann Hypothesis). The results obtained showed how much important can be analytic tools for the investigations of classical Arith- metic and Number Theory.

If one is interested in the study of the number fields K and the rings of s integers , then the Dedekind Zeta functions ζ (s) := N(I)− (where OK K I the sum is made over the ideals of K and N(I) is the norm of the ideal I) play a similar role. The behavior ofO these functions is analoP gue to the Rie- mann Zeta function’s one (in fact, ζ(s) is just the Dedekind Zeta function related to the simplest number field, Q) and their analytic properties can be used to investigate the distribution of the prime ideals in K . The Dedekind Zeta functions are useful also to detect the values of theO many algebraic invariants of K and K , like the discriminant, the class number, the regulator and the number ofO roots of unity contained in the number field. The first aim of this thesis is to give the basic theory of the Dedekind Zeta functions and to proof a bunch of their properties, among which there is the functional equation, which permits to meromorphically extend the Dedekind Zeta function and to find the Class Number formula, an important relation between the invariants of K and . OK

3 At this point one might be tempted to ask: how much the properties of the number field K depend on Dedekind Zeta function ζK (s)? If we are 0 given two number fields K and K which share the same Dedekind Zeta function, do they have something in common in their algebraic structure? Fields of this kind are said to be “arithmetically equivalent” and the problem of “arithmetical equivalence” is the main object of study of the rest of the thesis. It will be shown that two fields with same Dedekind Zeta function have the same number of real and complex embeddings in C (hence same degree over Q), and same discriminant; all these results can be achieved just by exploiting the analytic properties of the functions ζK (s).

The attempt to study further the arithmetical equivalence reveals however that one cannot discover something new just with the help of Analysis, and the way to get out from the impasse is to come back to Algebra. In fact, using concepts typical of Group Theory, it can be shown that two number fields arithmetically equivalent have also the same roots of unity. Moreover it is the theory of Group Cohomology (or at least its basic version) which permits to construct explicit examples of number fields with same Dedekind Zeta functions but which are not isomorphic, showing finally that the arithmetical equivalence is a relation weaker than the isomorphism.

Here is how this work is organized. In Chapter 1 we introduce the algebraic concepts necessary for the definition of Dedekind Zeta functions: number fields, rings of integers, class group, discriminant, embeddings, group of unities, regulator, norm of ideals. Chapter 2 deals with the basic analytic properties of the Dirichlet series (of which the Dedekind Zeta functions are particular examples). Furthermore there is a brief exposition of the Riemann Zeta function and of its properties (like the functional equation and the meromorphic continuation). Dedekind Zeta functions and their analytic properties are the objects of Chapter 3. In this part of the work we focus on the convergence properties and on the proof (the more detailed possible) of the functional equation, necessary to obtain a meromorphic continuation for ζK (s). Then it is com- puted the residue of these functions in 1, which gives the Class Number formula. In Chapter 4 we introduce the problem of arithmetical equivalence and we expose the main results achievable with an analytic approach. It is showed that two number fields with same Dedekind Zeta functions have same number of real embeddings and of complex embeddings in C, and that the degree of the two fields over Q is the same. Later it is proved that two arithmetically equivalent number fields have same discriminant: this follows by the proof of

4 an asymptotic formula for the function counting of the zeros of ζK (s) with real part between 0 and 1. In Chapter 5 it is proved that two number fields with same Dedekind Zeta functions induce the same kind of decomposition of the integer prime numbers in their rings of integers (this is called the splitting type). Later it is introduced the concept of Gassmann equivalence, a group theoretical relation which will be shown to be strictly related to the arithmetical equivalence; this instrument is the fundamental algebraic key to prove that arithmetically equivalent number fields have the same roots of unity and that two Galois extensions of Q have same Dedekind Zeta functions if and only if they are isomorphic. Finally we present a way to construct arithmetically equivalent number fields which are not isomorphic, and an explicit example is given. Chapter 6 is an appendix, which recalls results and concepts on summation formulas, the Gamma function, group actions and basic Galois Theory.

5 Chapter 1

Recalls of algebraic number theory

1.1 Number ﬁelds and rings of integers

1.1.1 Number fields and their embeddings A number field K is a field which is a finite extension of Q, meaning it is a finite dimensional Q-vector space (the dimension is called degree and denoted with [K : Q]).

Remark: Being Q a field of characteristic 0, each of its finite field extensions K is separable, which means that for every α K its minimum polynomial ∈ fα(x) splits without multiple roots when is seen as a polynomial with complex coefficients.

Theorem 1. (Primitive element theorem): Let K be a number ﬁeld. Then there exists an element α K such that there is the ring isomorphism ∈ Q[x] K Q[α] ' ' (fα(x)) where fα(x) is the minimum polynomial of α over Q. Proof. The claim is true for the wider class of separable extensions of a generic ﬁeld L. See Chapter 5 of [Mil] for the proof.

Every number field is a subfield of the field Q of the algebraic numbers (where an algebraic number is a root of a polynomal with rational coefficients)

6 and so it can be embedded in the ﬁeld C of complex numbers, but there is no canonical way to see this embedding.

Theorem 2. (Artin’s Lemma): Let K be a number ﬁeld of degree n. Then there is an isomorphism of C-algebras:

n Ψ: K Q C C . (1.1) ⊗ ' Proof. By the Primitive Element Theorem it follows immediately the isomorphism of C-algebras

Q[x] C[x] C[x] n K Q C Q C C ⊗ ' (f (x)) ⊗ '(1) (f (x)) '(2) x α ' α α α i Yi − where (1) is due to the properties of the tensor product and (2) is a consequence of the Chinese Remainder Theorem and of the separability of the ﬁeld K (which implies that the roots αi’s are all distinct).

Let φ : K K Q C the canonical ring homomorphism which sends x to → ⊗ n x 1 and let πi : C C the canonical projection on the i-th component. Artin’s⊗ lemma implies→ the existance of n distinct ring homomorphisms

σ := π Ψ φ : K C. i i ◦ ◦ → They are injective ring morphisms (obviously, because K is a field) and they are called the embeddings of the number field K in C. Observe that for every i it is σi Q = idQ Q. | → The morphism σi is called a real embedding if σi(K) R, otherwise it is a complex embedding. Every embedding is (by its⊆ construction) related to a root αi of the minimum polynomial defining the number field K, and by Artin’s Lemma it is clear that σi is real/complex if and only if αi is real/complex. Observe then that every complex not real embedding defines another complex embedding by applying complex conjugation to it. If r1 is the number of real embeddings and r2 the number of conjugated couples of complex embeddings, it is obviously n = r1 +2r2 (because the same relation is satisfied by the real and complex roots of the polynomial defining K).

Let us consider two easy examples:

If K = Q(i), then r1 = 0 and r2 = 1, the couple of complex embedding • given by σ(i)= i and σ(i)= i. − 7 If K = Q(√2) then r1 = 2 and r2 = 0, and the two real embeddings • are σ (√2) =√2 and σ (√2) = √2. 1 2 − 1.1.2 Integrality and rings of integers Given a domain R and a ﬁeld K containig R, an element α K is said to be integral over R if there exists a monic polynomial f(T ) ∈ R[T ] such that f(α)=0. ∈ Lemma 1. Let R and K as before, and let α K. The following statements are equivalent: ∈ 1) α is integral over R;

2) The subring R[α] K is finitely generated as R-module; ⊂ 3) There exists a finitely generated faithful R-submodule M K such that αM M (where “faithful” means that the homomorphism⊂ R End(M) which⊂ sends an element x to the linear morphism on M given→ by multiplication with x is injective). Proof. The proof can be found in any set of books or notes of Algebraic Number Theory. For example, see Chapter 2 of [Scho]. Let us define the set S := x K : x is integral over R . The previous lemma can be{ used∈ to show that in fact S} is a subring of the field K, called the subring of integral elements of K over R. We say that the domain R is integrally closed in K if S = R (which means that every integral element of K is contained in R). Lemma 2. S is integrally closed in K.

Proof. Let α K be an integral element over S, so that there exist bn 1,...,b0 ∈n n 1 − ∈ S such that α + bn 1α − + ... + b0 = 0. − By the previous lemma the module M = R[b0,..,bn 1] is finitely generated − over R and the previous relationship implies that M[α] satisfies the hypothesis of the lemma as R-module. So α is integral over R and then α S. ∈ Now, let K be a number field. The set

= α K : p(x) Z[x] monic with p(α)=0 OK { ∈ ∃ ∈ } is called the ring of algebrac integers of K. It is the ring of integral elements of K over Z and it is maybe the most important object of classical

8 algebraic number theory: one typical question is how to compute it given a generic number ﬁeld K. There are two easy classes of examples:

In the trivial case K = Q it is Q = Z because a rational number • satisfies a monic polynomial withO integer coefficients if and only if it is an integer number. Let K be a number field of degree 2: then there exists a squarefree • integer d Z such that K Q(√d). Classical computations show that ∈ '

Z[√d] d =2, 3 mod 4 K = 1+√d O (Z[ 2 ] d =1 mod 4 Remember that, given a domain R, its fraction field is the field of the elements a Frac R := : a R,b R∗ b ∈ ∈ and it is easy to verify that it is then smallest field containino g R. Lemma 3. Let K be a number field. Then Frac = K. OK Proof. The inclusion is obvious by definition of Frac. ⊆ Conversely, let α K: there exist an, . . . , a0 integers, with an = 0, such that n ∈ n 1 6 a α + + a α + a = 0. Multiplying by a − the expression becomes n ··· 1 0 n n n 2 n 1 (a α) + + a − a (a α)+ a − a =0 n ··· n 1 n n 0 so we have anα K , i.e. there exists c K such that anα = c, which implies α = c/a ∈ OFrac . ∈ O n ∈ OK So we can say that K is a domain which is integrally closed in its field of fractions. O The rings of integers are very famous members of an important class of rings called Dedekind domains, which will be the main object of study in the next section.

1.2 Dedekind domains, fractional ideals and class group

1.2.1 Prime factorization in Dedekind domains A Dedekind domain is a ring R which satisﬁes the following four conditions:

9 1) R is a domain;

2) R is a Noetherian ring (i.e. every ideal of R is ﬁnitely generated);

3) Every prime ideal not zero of R is maximal;

4) R is integrally closed in its ﬁeld of fractions.

Why are Dedekind domains so important? Because they allow us to gen- eralize one of the most important facts in mathematics: the fundamental theorem of arithmetic. Greeks already proved that every integer number n Z admits a unique factorization (up to permutation of the factors) as product∈ of prime numbers. Extending this statement is not always possible in generic rings; a famous counterexample is the factorization of 6 in the ring Z[√ 5], being − 6=2 3=(1+√ 5) (1 √ 5) · − · − − but if instead of the elements we work with the ideals then, if we are dealing with Dedekind domains, we can obtain the generalization we were looking for.

Theorem 3. Let R be a Dedekind domain. Then every ideal not zero of R admits a unique factorization (up to permutation of the factors) as a product of prime ideals of R.

Proof. An elementary proof of this fact can be found in Chapter 1, Section 3 of [Jan]. Remark: It is important to observe that, fixed an ideal I not zero, the prime ideals giving its factorization are only in finite number: that is because every prime factor of I contains I and there is only a finite number of prime ideals containing I (this follows almost immediatly from the basic theory of Ar- tinian rings; the reader who doesn’t know it can see Chapter 8 of [AtyMac]).

Let K be a number field. If we could show that its ring of integers is OK a Dedekind domain then our sensation about K as a ring playing in K the same role of Z in Q would be much more justified,O expecially because we could talk about factorization in products of primes (which we remind is one of the most basic and at the same time impressive facts in classical arithmetic). By Lemma 2 we know that K is a domain and is integrally closed in its field of fractions. The propertiesO 2) and 3) are immediate consequences of the two following theorems.

10 Theorem 4. Let R be a noetherian and integrally closed domain, F its ﬁeld of fractions and K a separable extension of F of ﬁnite degree. Let A L be the subring of integral elements of L over R. ⊂

K L

Then A is finitely generated as R-module. Theorem 5. Let R S be a ring inclusion of domains, and suppose that every element of S is⊂ integral over R. If R satisfies condition (3) in p. 8, then S does the same. Proof. The first theorem can be found in Chapter 1 of [Lang1], while the second one is a specific case of a more general statement which can be found in Chapter 5 of [AtyMac]. Remark: In the specific case of the rings of integers, Theorem 4 implies that K is Noetherian, being the ring of integral elements of the Noetherian ring ZO in the number field K. Moreover it is a free abelian group of rank [K : Q] (being a torsion free finitely generated module over the PID Z). Theorem 5 implies condition (3) for K . So we can finally say thatOfor every number field K the ring of integers is a Dedekind domain. OK 1.2.2 DVR, ramification and inertia degrees A discrete valuation ring (or DVR) is a ring R which satisfies the following conditions:

a) R is a local domain (i.e a domain with only one maximal ideal ); M b) R is Noetherian;

c) The only prime ideals of R are (0) and ; M d) R is integrally closed in its ﬁeld of fractions.

Let R be a domain, R a prime ideal. The localization of R in is the ring P ⊂ P a R := : a R,b R . P b ∈ ∈ \P n o 11 If = (0) then R is the usual fraction ﬁeld Frac R. P R Pis a domain contained in Frac R and is a local ring, its maximal ideal P being generated by the elements whose numerator belongs to . If M is an R-module the localization of M in is the R-moduleP P a M := : a M,b R P b ∈ ∈ \P where the fractions are to be intendedn as equivalenceo classes on the set M (R ), with (m, a) (n,b) if there exists an element x such that× x (\Pbm an)=0. ∼ ∈ P Classical· − results of Commutative Algebra show that there is a canonical isomorphism of R-modules

M R R M ⊗ P ' P (see Chapter 3 of [AtyMac]).

Lemma 4. (Local property): Let M be an R-module. The following statements are equivalent:

M =0; • M =0 for every prime ideal R; • P P ⊂ M =0 for every maximal ideal R. • M M⊂ Proof. It is clear that the ﬁrst condition implies the second one, which in turn implies the third one. Now suppose M = 0 for every maximal ideal R. If there exists M x M not zero, then the ideal Ann(x) := a R:Max =0 ⊂ is a proper ideal of∈R and so it is contained in a maximal{ ideal∈ . But}x/1 = 0in M , M which means that there exists b R suchM that bx = 0. This implies b Ann(x) , and this is impossible.∈ \M ∈ ⊂M Theorem 6. Let R be a Noetherian domain. Then R is a Dedekind domain if and only if for every prime ideal not zero of R the localization ring R P is a DVR. P

Proof. This is a classical result of Commutative Algebra: there is plenty of books with the proof in (I suggest expecially [AtyMac]). The proof is based on the local property and on two main facts:

A domain R is integrally closed if and only if every localization R is P • integrally closed.

12 R satisﬁes the condition (3) if and only if every localization R satisﬁes P • the condition (c).

Let K be a number field of degree n and K its ring of integers. Let p Z be a prime number. A typical problem inO Number Theory is to determine∈ whether the ideal p K is prime or not, and how it factorizes as a product of prime ideals. O e1 es Let p K = Q1 Qs be the factorization in primes. The primes Qi are said toO be the prime····· ideals lying over p, and each of these contain p . OK The number ei is called the ramification degree of the prime Qi over p and is denoted as e(Q p). i| Being Z K an injective homomorphism which sets K as a free abelian group of⊂ finite O rank, the induced field morphism O

Z/pZ /Q →OK i is a finite degree field extension of the finite field Fp := Z/pZ. The degree fi := [ K /Qi : Z/pZ] is called is called the inertia degree of the prime Q overO p and is denoted with f(Q p). i i| We are going to explicit a very important formula that relates the degrees introduced before.

Theorem 7. Let K be a number ﬁeld of degree n. Let p Z be a prime e1 ∈ number, p = Q Qes its factorization in . Then OK 1 ····· s OK s e f = n. (1.2) i · i i=1 X This implies that for every prime number p there exist at most n = [K : Q] prime ideals of which contain p . OK OK

Proof. We give a sketch of the proof: the idea is to localize the ring K e O in the prime ideals Q , so that pO = Q i O (the other prime factors i KQi i KQi disappear in the factorization because they become invertible). Then one has to show that the F scalar extension of , i.e. the ring p OK Z F , satisﬁes the following ring isomorphisms: OK ⊗ p

s s O s O KQi KQi Z F (O Z F ) . K p KQi p e O ⊗ ' ⊗ ' pOK ' (QiOK ) i i=1 i=1 Qi i=1 Qi Y Y Y 13 The equation above describes an isomorphism of Fp-algebras, so it is an isomorphism of Fp-vectorial spaces and preserves the dimension over the ﬁeld Fp. The term K Z Fp has dimension n = [K : Q] over Fp (because K is a free abelianO group⊗ of such rank), while it can be proved that the iO-th factor in the last product has dimension ei fi over Fp, and this proves the claim. ·

We introduce some notation which will be useful in the next chapters. Let p Z be a prime number. We say that: ∈ p ramiﬁes in K if there exists a prime ideal K and an integer • number e 2O such that e appears in the factorizationP⊂O of p . ≥ P OK p is inert in if p is itself a prime ideal of . • OK OK OK p splits in K if there exist Q1,...,Qn distinct prime ideals of K • such that p O = Q Q . O OK 1 ····· n 1.2.3 Fractional ideals and class group Let R be a domain, K = Frac R.A fractional ideal I is a not zero R- submodule of K such that there exists x K∗ with xI R. 1 ∈ ⊂ Given a fractional ideal I, the set I− := x K∗ : xI R is itself a 1 { ∈ ⊂ } fractional ideal (by its deﬁnition it is yI− R for every y I). ⊂ ∈ If I is an R-submodule of K, I is principal if there exists x K∗ such that I = xR, while I is said to be invertible if there exists∈ a non zero R-submodule J K such that IJ = R. ⊂ Lemma 5. Let I K be a not zero R-submodule. The following implications hold: ⊂

principal invertible ﬁn. generated fractional. →1 →2 →3 If R is Noetherian then the implication 3 can be reversed. 1 Proof. (1) Let I = xR with x R∗; it is clear that, if J = x− R, then IJ = R. ∈ (2) Being I invertible there exists J such that R = IJ and so there exist n i1, . . . , in I and j1,...,jn in I such that 1 = k=1 ikjk. Now, let i I: then∈ ∈ n n P i = i 1= i i j = ij i · · k k k · k Xk=1 Xk=1 and the products ijk are in R, so that I is ﬁnitely generated.

14 (3) Being I finitely generated there exist x1,..,xk in the field K such that I = R xk and there exists z K such that xk = yk/z. Then zI R. · ∈ Suppose⊂ P now that R is Noetherian, and let I be a fractional ideal. There exists x K∗ such that xI R, and in fact the multiplication by x gives an R∈-module isomorphism⊂ between I and xI. But xI is an R-submodule of R, i.e an ideal of R, and so is finitely generated for Noetherianity.

Lemma 6. 1) Let I,J be fractional ideals. Then the product IJ is a fractional ideal;

2) For every fractional ideal I it is I = IR = RI;

1 3) A fractional ideal I is invertible if and only if II− = R.

Proof. (1) If x,y K∗ are such that xI R and yJ R then (xy)IJ = (xI)(yJ) R.∈ ⊂ ⊂ ⊂ (2) This point is trivial.

(3) The claim is true by proving that if there exists a fractional ideal J 1 such that IJ = R, then it’s I− = J. 1 1 1 1 Clearly it is J I− . Furthermore I− = I− R = I− IJ RJ = J. ⊂ ⊂

Remark: The previous lemma shows that the fractional ideals over a ﬁxed domain R are endowed with a monoid structure, where the product is the usual product of R-modules and R is the identity element. The submonoid of the invertible elements becomes a group with the same operation.

Our next goal is to show that if R is a Dedekind domain, like a ring of integers for a number ﬁeld K, then every R-fractional ideal is not only ﬁnitely generated (by Noetherianty) but also invertible, which means that in this case the monoid of fractional ideals is in fact a group.

Lemma 7. (1) Let R be a DVR, I a fractional ideal over R. Then I is principal.

2) Let R be a Noetherian domain, R a prime ideal not zero and I,J fractional ideals over R. Then IP ⊂is a fractional ideal over R and it 1 P 1 P is (IJ) = I J and (I− ) =(I )− . P P P P P

15 Proof. (1) By Noetherianity we have that I is ﬁnitely generated and isomorphic (as R-module) to an ideal of R. But every DVR is a principal ideal domain (see Chapter 7 of [AtyMac]) and so I is principal. (2) Let x K = Frac R such that xI R. For every element i/s I we P have (x/∈ 1) (i/s) R and so the localization⊂ gives us a new fractional∈ P ideal. · ∈ The relation (IJ) = I J is very easy to prove: it follows from the P P P fact that if i I,j J and s,t R then (i/s) (j/t)=(ij/st). 1 ∈ ∈ ∈ \P · Let x I− : for every s R and for every i/t I itis(x/s) (i/t)= ∈ ∈ \P 1 ∈ P 1 · (xi)/(st) R , so we have the inclusion (I− ) (I )− . ∈ P 1 P ⊂ P Now, let x (I )− : by Noetherianity if we have I = i1/1, . . . , in/1 ∈ P P { } where the ij’s are the generators for the fractional ideal I) then x ik/1 R for every k = 1,...,n. This means that for every k there· is an∈ P equation x iksk = rk where sk R and rk R. Thus we have 1 · ∈ \P ∈ 1 xsk I− for every k, and being x =(xsk)/sk (I− ) for every k, we P have∈ the claim. ∈

Remark: In the previous proof the Noetherianity of the domain R is fundamental: it assures that every consideration about the generators of the fractional ideals is well justiﬁed and explainable in a ﬁnite amount of time. There are in fact examples of non Noetherian domains for which the last equality on the fractional ideals is false. Theorem 8. Let R be a Dedekind domain. Then every fractional ideal I over R is invertible.

1 Proof. It is II− = R if and only if (thanks to local property and the previous 1 1 lemma) R = (II− ) = I I− . But by Theorem 6 we know that R is a P P P P DVR and so every fractional idealP over R is invertible. P

Now, suppose K is a number ﬁeld and R = K is its ring of integers. Let I(K) := I K : I is a fractional ideal of O . By the previous theorem { ⊂ OK } (and the fact that K is a Dedekind domain) we get that I(K) is an abelian group with respectO to the multiplication of ideals. Let P (K) I(K) the subgroup of principal fractional ideals. Then the class group⊂ of K is the quotient group I(K) Cl(K) := . P (K)

Theorem 9. Let K be a number ﬁeld, K be its ring of integers. Then K is a principal ideal domain if and only ifOCl(K) 0 . O '{ } 16 Proof. The truth of the claim follows from the fact that every K -fractional ideal is isomorphic (as -module) to a proper -ideal. O OK OK

1.3 Some algebraic invariants

In this section we introduce the main objects for the classification of the rings of integers in number fields Q K. Recall that in Section 1.1.1 we introduced the embeddings σ : K ⊂C. i → 1.3.1 Norm and trace Let K be a number field of degree n over Q, x an element of K and define the function mx : K K which sends y to xy. Itisa Q-linear morphism, and it is an isomorphism→ unless x = 0 (in which case the function is the zero function). Let ω ,...,ω be a Q-basis of K and M the matrix which represents the { 1 n} x linear function mx with respect to this base. We define the norm NmK/Q(x) and the trace TrK/Q(x) of x as the determinant and the trace, respectively, of the matrix Mx.

Remark: It is well known from Linear Algebra that the determinant and the trace of two matrices which represent the same linear morphism with respect to two different bases are equal, so that the concepts introduced before are well defined. Theorem 10. Let K be a number field with degree [K : Q]= n and x K. Then: ∈

(1) NmK/Q(x) and TrK/Q(x) are in Q;

(2) Let σ1,...,σn be the embeddings of K in the field C. Then NmK/Q(x)= n { } n i=1 σi(x) and TrK/Q(x)= i=1 σi(x); (3)Q If x , then its norm andP its trace are in Z; ∈OK (4) The norm is a multiplicative function from K to Q and the trace is a Q-linear morphism from K to Q. Proof. (1) Fixed any Q-basis of K, the matrix representing the linear morphism mx with respect to that base has rational coefficients. Being both the determinant and the trace a polynomial combination of these coefficients, we obtain the claim.

17 (2) The Q-linear morphism m : K K induces a C-linear morphism x → mx 1 : K Q C K Q C which is the C-linear extension of the multiplication⊗ ⊗ by x→. ⊗ Let us ﬁx a Q-base ω ,...,ω of K. Then the set ω 1,...,ω 1 { 1 n} { 1 ⊗ n ⊗ } is a C-base of K Q C and we see that the matrix M related to m ⊗ x x with respect to the Q-base is equal to the matrix Mx 1 related to mx 1 ⊗ ⊗ with respect to the C-base (and so they have same determinant and trace). But by Artin’s lemma we know there is a C-algebras isomorphism

n Ψ: K Q C C ⊗ → k e (σ (k)e,...,σ (k)e) ⊗ → 1 n so that on the C-vectorial space Cn the multiplication by x 1 becomes, via Ψ, the C-linear morphism ⊗ mˆ : Cn Cn x → (e ,...,e ) (e σ (x),...,e σ (x)) 1 n → 1 1 n n and its matrix with respect to the canonical base of Cn is diagonal with the σi(x)’s as diagonal entries: thus the determinant and the trace of this morphism (which are the norm and the trace of x by what we have n n shown) are respectively i=1 σi(x) and i=1 σi(x).

(3) By (2) it follows immediatelyQ that if xP then Nm Q(x) and ∈ OK K/ TrK/Q(x) are algebraic integers. But (1) says they are rational numbers, so they are in Z.

(4) The claim follows by (2): remember that the embeddings σi are Q- algebras morphisms.

1.3.2 The discriminant

Let ω1,...,ωn be a Q-basis of a number ﬁeld K of degree [K : Q] = n. Let {σ ,...,σ } be the embeddings of K in C. { 1 n} The discriminant of the n-ple (ω1,...,ωn) is the number

2 D(ω1,...,ωn) := (det(σi(ωj))i,j=1,..,n) = det(TrK/Q(ωiωj))i,j=1,..,n.

The last equality follows by this simple fact: if M = (σi(ωj))i,j=1,..,n then T M M = (Tr Q(ω ω )) . · K/ i j i,j=1,..,n A straightforward implication is that D(ω1,...,ωn) Q and that if ω1,...,ωn is a Q-base of K made by elements of , then D(∈ω ,...,ω ) Z{. } OK 1 n ∈ 18 Lemma 8. Let K be a number field of degree n over Q. Then there exists a Q-base of K made of elements of . OK Proof. Let ω ,...,ω be a Q-basis of K. It was shown in Lemma 3 that { 1 n} K = Frac K , so that for every i =1,...,n there exist ri,si K such that ω = r /s O. Define s = s s ; then for every i it is sω∈O and the i i i 1 ····· n i ∈ OK sωi’s remain linearly independent over Q. Suppose now ω ,...,ω is a Z-basis of the ring of integers (remind { 1 n} OK that K is a free abelian group of rank n =[K : Q]). This condition implies thatO it is also a Q-basis of K (while it is not always true the converse) and so we can compute the discriminant of this base. In such case we call ∆K/Q := D(ω1,...,ωn) the discriminant of the field K over Q. Theorem 11. (1) Let D : n Z be the function that to each n-ple OK → (ω1,...,ωn) assigns its discriminant D(ω1,...,ωn). Let N be the minimum positive value of the function. Then D(ω1,...,ωn) = N if and only if ω ,...,ω is a base for the free abelian group . { 1 n} OK (2) Let ω ,...,ω and δ ,...,δ be two Q-bases of K made of elements { 1 n} { 1 n} of K . Let M be the matrix that sends the first base in the second one. ThenO

D(δ ,..,δ ) = (det M)2 D(ω ,...,ω ). 1 n · 1 n Proof. This is another typical result of Algebraic Number Theory: for the proof see Chapter 3, Section 5 of [Jar]. Observe that point (2) implies that ∆K/Q is a well deﬁned number.

Theorem 12. Let K be a number ﬁeld, r2 the number of its couples of complex conjugated embeddings and ∆K/Q its discriminant. Then

r2 ∆ Q =( 1) ∆ Q . K/ − | K/ |

Proof. The claim is easily veriﬁed for K = Q (since ∆Q/Q = 1) and for every √ quadratic ﬁeld K = Q( d) (since, if d is a squarefree integer, then ∆Q(√d)/Q is equal to d or 4d depending on d being equal to 1 or 3 modulo 4, so that the sign of the discriminant is the sign of d). 2 Let qK : K Q be the quadratic form x TrK/Q(x ), with associated bilinear form→ which sends (x,y) to → 1 (q (x + y) q (x) q (y)) = Tr Q(xy). 2 K − K − K K/ 19 For every Q-base of K the determinant of the matrix related to the bilinear 2 form is equal to q ∆ Q with q Q∗, so that they share the same sign. · K/ ∈ By R-extension of the scalars it follows that qK Q R : K Q R R which 2 ⊗ ⊗ → sends u R in TrK QR/R(u ) has the same determinant of qK and is a non- ⊗ degenerate∈ quadratic form; thus, if n =[K : Q], the quadratic form has the signature (a, n a). The claim follows by proving that a = r1 + r2. − r1 r2 The R-algebras isomorphism K Q R R C (which can be proved in the same way we proved Artin’s⊗ Lemma)' implies× that the factors of the last product are pairwise orthogonal with respect to qK Q R, and the trace map restricted to each factor is the R-trace of the factor⊗ ring. Then it remains to 2 2 show that TrR/R(x ) has signature (1, 0) and TrC/R(z ) has signature (1, 1) and these are straightforward checks (by taking the R-bases 1 for the real factors and 1, i for the complex ones). { } { }

1.4 Norm of ideals in OK 1.4.1 Norm of proper ideals

Remark: Suppose K = Q, so that K = Z. If I = (n) is a not zero ideal, then the quotient ring Z/I is a finiteO ring with exactly n elements. We are going to show that a similar property holds for every ideal not zero in the ring of integers , for every number field K. OK Lemma 9. Let K be a number field, I K a not zero ideal. Then the quotient ring /I is finite. ⊂ O OK Proof. Let I = e1 es be the factorization of I as product of prime P1 ·····Ps ideals, with i = j if i = j. The Chinese Remainder Theorem induces the ring isomorphismP 6 P 6 K K K O Oe1 O . I ' ×···× es P1 Ps So we are left to prove that for every prime ideal K and for every n N the ring / n is finite and actually is # P/ ⊂On = (# / )n. ∈ OK P OK P OK P We know that K / is a finite field extension of a finite field Z/pZ for a prime number pO(itsP cardinality is the inertia degree f( p)). To prove the claim we work by induction on n (the case P|n = 1 proved above). Suppose the claim is true for every m

Given an ideal I not zero, we deﬁnte the norm of I as N(I) := ⊂ OK # K /I. TheO previous lemma shows that N(IJ)= N(I)N(J) for every couple of not zero ideals I,J K . There is an interesting⊂O consequence of this fact on the structure of fractional ideals over rings of integers.

Lemma 10. Let K be a number ﬁeld of degree n over Q, and let I be a fractional ideal over . Then I is a free abelian group of rank n. OK

Proof. We can suppose I is an ideal of K (thanks to the results proven in Section 1.2.3). Being a free group ofO rank n and I an ideal, and thus a OK subgroup of K , the general theory of modules over PID’s (like Z) implies that I is free,O of rank m n. To see that its rank is equal≤ to n, consider the exact sequence of abelian groups

0 I /I 0. → →OK →OK → By tensorization with Q we obtain the exact sequence of Q-vector spaces

0 I Z Q Z Q /I Z Q 0. → ⊗ →OK ⊗ →OK ⊗ → m n Now I Z Q Q , K Z Q Q and (because the quotient ring is ﬁnite ⊗ ' O ⊗ ' m n for the previous theorem) K /I Z Q 0, so that Q Q and therefore m = n. O ⊗ ' '

Theorem 13. Let K be a number ﬁeld, (x) be a non zero principal ⊂ OK ideal in the ring of integers . Then N((x)) = Nm Q(x) . OK | K/ | Proof. Let M : be the map of multiplication with α. Then the α OK → OK image α K is an ideal of K , so a free abelian group of rank n, and the only thingO to show is that #O /α = det M which we know is equal to OK OK | α| Nm Q(α) . For this see Chapter 7 of [Jar]. | K/ |

21 1.4.2 Norm of fractional ideals

Let I be a K -fractional ideal. It is a consequence of the results in Section 1.2.3 that theO fractional ideals admit a unique factorization in product of powers (positive or negative) of prime ideals. Then, by the complete multiplicativity of the norm of ideals showed before, we can immediately deﬁne the norm of a fractional ideal I: if I = a1 P1 · as with a Z for every i, then ····Ps i ∈ N(I) := N( )a1 ( )as P1 ····· Ps where on the prime ideals the norms are the one deﬁned before; thus the norm becomes a homomorphism

N : I(K) Q∗ . → >0 1.5 Theorems by Minkowski an Dirichlet

The aim of this section is to present two classical and fundamental results in Algebraic Number Theory: Minkowski’s Convex Body Theorem (which implies that for every number field K the class group Cl(K) is finite) and Dirichlet’s Unit Theorem (which provides a canonical structural isomorphism for the group of integer units K∗ ). The proofs of these theoremsO are rather technical but the reader interested can found them in Chapter 4 of [FröTay]. Theorem 14. (Minkowski): Let K be a number field of degree n = [K : Q]= r1 +2r2. The following statements hold:

If I K is a non zero ideal, then there exists x I not zero such • that ⊂ O ∈ r2 n! 4 1/2 0 < Nm Q(x) ∆ Q N(I); | K/ |≤ nn π | K/ | For every class Cl(K) there exists I such that I K e • r2R ∈ ∈ R ⊆ O n! 4 1/2 N(I) ∆ Q ; ≤ nn π | K/ | The class group Cl(F ) is finite for every number field F . • Theorem 15. (Dirichlet): Let K be a number field of degree n =[K : Q], with r1 real embeddings and r2 couples of complex not real embeddings. The group morphism

r1+r2 ψ : ∗ R OK → x (log σ (x) ,..., log σ (x) , 2log σ (x) ,..., 2log σ (x) ) → | 1 | | r1 | | r1+1 | | r1+r2 |

22 induces the group isomorphsim

r1+r2 1 ∗ Z − µ OK ' × K where µK is the ﬁnite group of the roots of unity contained in K.

Let ε1,...,εr1+r2 1 be the Z-linear independent elements of K∗ whose im- − r1+r2 1 O ages realize the Z-base of the lattice Z − in the isomorphism of Dirichlet’s theorem. The ε ’s are called fundamental units of . i OK Let h be any index in 1,...,r + r 1 . The regulator of K is deﬁned as { 1 2 − }

RK := det(log σj(εi) )j=h . (1.3) | | | 6 | It can be proved that it does not depend on the index h chosen.

Lemma 11. Consider the lattice ψ( ∗ ) contained in the hyperplane H := OK (x ,...,x ): x =0 Rr1+r2 and let E be the fundamental domain { 1 r1+r2 i i }⊆ (which means the set of representatives of the quotient group H/ψ( K∗ ) which contains 0). P O Then r2 vol(E)=2− R . · K Proof. See Chapter 8 of [Scho].

1.6 The diﬀerent ideal

Let K be a number ﬁeld of degree [K : Q]= n, and let L K be an additive subgroup. ⊆ We deﬁne its complement with respect to the trace as the set

0 L := x K : TrK/Q(xL) Z . 0 { ∈ ⊂ } It is easy to show that L is an additive subgroup. If M is another additive 0 0 subgroup of K such that L M then M L . 0 If L = , then L L , but⊂ usually they⊂ are not equal. OK ⊂

Lemma 12. Let ω1,...,ωn be a Q-basis of the number ﬁeld K, and L = { 0 }0 0 0 0 Zω1 Zωn. Then L = Zω1 Zωn where ω1,...,ωn is the dual base ⊕···⊕ ⊕···⊕ { } 0 with respect to the trace (i.e. they are the elements such that TrK/Q(ωiωj)= δij with δij the Kronecker symbol).

23 Proof. By the deﬁnition of dual basis and the Q-linearity of the trace mor- 0 0 0 phism we immediately see that Zω1 Zωn L . ⊕···⊕ 0 ⊂ To prove the opposite inclusion, suppose α L and let ai = TrK/Q(αωi) Z ∈ 0 ∈ for every i = 1,...,n. We want to show that α = aiω . To prove this, 0 i i note that TrK/Q[(α i aiωi)ωj] = 0 for every j = 1,...,n, which means that this linear operator− is the zero function (being linearP and equal to 0 on P a basis of K). But if the operator TrK/Q(x ): K Q is the zero function ·− → 1 then it must be x = 0 (otherwise [K : Q] = TrK/Q(x x− ) = 0, and this is impossible). ·

As an example, suppose K = Q(√2) and L = K = Z[√2] = Z Z√2. Then 0 the previous lemma implies that L = Z 1 ZO1 which strictly⊕ contains L. 2 ⊕ 2√2 0 Lemma 13. Let I be a K -fractional ideal. Then its complement I is a -fractional ideal. O OK 0 0 Proof. Clearly I K and is a K -submodule of K: if x I and a K , then ⊂ O ∈ ∈O axI = xaI xI Z 0 ⊂ ⊂ so that ax I . Now, there∈ exist ω ,...,ω K such that I = Zω Zω : being 1 n ∈ 1 ⊕···⊕ n the ωi’s independent over Z they are independent over Q (this follows by tensorization with Q: remember that K Z Q K). So by the previous 0 0 0 O ⊗ '0 lemma one has I = Zω1 Zωn where the ωi’s form the dual base with respect to the trace. ⊕···⊕ But these new elements are in K and independent over Q, so there exists 0 x K such that xωi K for every i =1,...,n, and therefore the theorem is∈O proved. ∈O

We deﬁne the diﬀerent ideal of K over Q as the fractional ideal

0 1 K /Z := ( K )− . DO O Theorem 16. Let I be a -fractional ideal. Then OK 0 0 1 1 I = K I− =( K /Z I)− . O · DO · 0 0 Proof. Let us prove the inclusion by proving that I I K (which is equivalent); one has ⊆ · ⊆ O

0 0 Tr Q((I I) ) = Tr Q(I I) Z. K/ · ·OK K/ · ⊆

24 Now we prove the opposite inclusion: we have

0 1 0 Tr Q(( I− ) I) = Tr Q( ) Z. K/ OK · · K/ OK ·OK ⊆

Remark: This factorization of the complement of a fractional ideals in terms of the diﬀerent will be very useful in the next chapters. Moreover, we remember that in the previous chapter we extended the concept of norm even to fractional ideals. We exploit this to state the following lemma (very important for our purposes):

Lemma 14. Let /Z be the diﬀerent ideal of K over Q. Then DOK

N( /Z)= ∆K/Q =: dK . DOK | | Proof. See Chapter 3 of [Lang1].

Remark: For every fractional ideal I I(K) deﬁne the quantity dI := 2 0 ∈ N(I) dK . Then dI = 1/dI (it is a trivial computation by the previous lemma)· and it will be used in the fundamental theorem of Chapter 3. Furthermore, if η1,...,ηn is a Z-base of the fractional ideal, then the properties of the discriminant{ showed} in Section 1.3 imply that d = D(η ,...,η ) . I | 1 n |

25 Chapter 2

The Riemann Zeta Function

2.1 Dirichlet series

2.1.1 Convergence theorems A Dirichlet series is a function of complex variable of the form + ∞ f(n) D (s) := f ns n=1 X where f : N C is an arithmetical function (i.e. a sequence of complex numbers). → Dirichlet series are typical objects of study in Analytic Number Theory: their analytic properties (such as convergence) are strictly related to the arithmetical nature of the sequence f(n).

Lemma 15. Let Df (s) be a Dirichlet series. Suppose it converges for a complex number s = σ + it . Then D (s) converges for every s C such 0 0 0 f ∈ that Re s>σ0 and the convergence is uniform in the sectors Sl := s : Im(s s ) l Re(s s ) , where l> 0. { | − 0 |≤ · − 0 } Proof. Let s C such that Re s>σ0. The series Df (s) converges if and only if for every∈ ε> 0 there exists k 0 such that, for every N,M integers greater than k with N>M, one has≥ N f(n) < ε. ns n=M+1 X Deﬁne the tail sum function (for x 1) ≥ + ∞ f(n) R(x) := . ns0 n= x Xb c 26 Then by partial summation formula (see Appendix) we get

N f(n) N 1 f(n) = ns ns so ns0 n=M+1 n=M+1 − X X R(M) R(N) N R(x) = +(s s0) dx M s so − N s so − xs so+1 − − ZM − Take M such that R(x) <ε for every x M: then | | ≥

N f(n) R(M) R(N) N R(x) | | + | | + s s0 | | dx ns ≤ M Re(s s0) N Re(s s0) | − | xRe(s s0)+1 n=M+1 − − M − X Z N 1 ε 2+ s s0 dx ≤ | − | xRe(s s0)+1 ZM − s s Ims s ε 2+ | − 0| ε 3+ | − 0| . ≤ Re(s s0) ≤ Re(s s0) − − These inequalities prove the convergence of the series, and if s Sl then the last term is less than ε(3 + l), and so the convergence is uniform.∈

The convergence of the series in one point s0 implies the convergence in the half-plane to the right of the point (in general nothing can be said about the convergence over the line Re s = Re s0). The situation gets better when we deal with absolute convergence.

Lemma 16. Let Df be a Dirichlet series which converges absolutely in a point s . Then the series converges absolutely in the half-plane Re s Re s . 0 ≥ 0 Proof. The proof is an easy corollary of the previous lemma: the only new thing is the convergence on the vertical line containing s0 (but this follows because ns0 = nRe s0 ). | |

Theorem 17. (Morera): Let f : Ω C be a continuous function on a connected open set Ω. Suppose that for→ every closed simple curve γ Ω ⊂ f(z)dz =0. Zγ Then f is an holomorphic function.

Proof. See Chapter 4 of [Ahl].

27 Remark: The Dirichlet series Df (s) is an holomorphic function on every half-plane in which it converges. In fact it is continuous, because it is a sum of continuous functions uniformly convergent on compact subsets, and for every simple closed curve γ contained in the half-plane one has

+ + ∞ f(n) ∞ f(n) dz = dz =0 nz nz γ n=1 n=1 γ Z X X Z and this assures the holomorphicity of Df (by Morera’s Theorem).

Lemma 17. Let f,g be arithmetical functions. Suppose both Df (s) and Dg(s) converge in a half-plane Re s> Re s0 and that in this domain we have Df (s)= Dg(s). Then f(n)= g(n) for every n N. ∈ Proof. The proof is inductive: we exploit the uniform convergence of the Dirichlet series in the sectors Sl. Take s a real number with s> Re s0. Then

+ + ∞ f(n) ∞ g(n) D (s)= f(1) + = g(1) + = D (s). f ns ns g n=2 n=2 X X Let us send s to + in both terms. By the uniform convergence of the ∞ Dirichlet series on the sectors Sl we can exchange the sums with the limit and obtain

+ + ∞ f(n) ∞ g(n) f(1) = f(1) + lim = g(1) + lim = g(1). s + ns s + ns n=2 → ∞ n=2 → ∞ X X Now let j 2 and suppose f(k)= g(k) for every k j 1. Then D (s)= ≥ ≤ − f Dg(s) implies

+ + ∞ f(n) ∞ g(n) = . ns ns n=j n=j X X The claim now follows multiplying each side of the equality with js and repeating the previous passages.

2.1.2 Arithmetical functions and Euler product Let f : N C be an arithmetical function. The function→ f is said to be multiplicative if f(m n) = f(m) f(n) for · · 28 every pair of coprime integers m,n. The function is completely multiplicative if f(m n)= f(m)f(n) for every m,n N. · An example∈ of completely multiplicative function is

1(n):=1 n N. ∀ ∈ Now, for every n N, deﬁne ω(n) as the arithmetical function ∈ ω(n):=# distinct prime divisors of n { } The M¨obius function is the arithmetical function

( 1)ω(n) if n is squarefree µ(n) := − . (0 otherwise

It is a multiplicative function and, together with the function 1, satisﬁes an important relation:

Theorem 18. (Inversion Theorem):

1) For every n N it holds ∈ 1 n =1 µ(d)= ; 0 n =1 d n ( X| 6 2) If f and g are arithmetical functions, then

f(n)= g(d) (2.1) d n X| if and only if n g(n)= f(d)µ . (2.2) d d n X| Proof. 1) For n = 1 the claim is trivial. The M¨obius function is multiplicative, and so its values for n 2 are uniquely determined by the values for the numbers pk, where ≥p is prime, k 1. But: ≥ k µ(pn)=1 1=0. − n=0 X

29 2) The relation (2.1) and the point 1 imply that n n n f(d)µ = g(e) µ = g(e) µ = g(n) d d d d n d n e d e n e d n X| X| X| X| X| | and in the same way one proves the opposite implication.

We are interested in the Dirichlet series associated to multiplicative functions: they are very interesting because they can be reformulated as an inﬁnite product on the half-plane of convergence.

Theorem 19. Let f be a multiplicative arithmetical function, Df (s) its associated Dirichlet series. Suppose that Df (s) converges absolutely in a half- plane Re s>σ0. Then in this half-plane we have the equality

+ + ∞ f(n) ∞ f(pk) = 1+ . ns pks n=1 p X Y Xk=1 The term in the right side is called Euler product.

Proof. First, we show that the the Euler product converges absolutely on the half-plane Re s>σ0. Let N > 0. Thanks to the multiplicativity of f we have

+ + ∞ f(pk) f(n) ∞ f(n) 1+ = | | < + pks ns ≤ nRes ∞ p N k=1 n S n=1 ≤ ∈ N Y X X X where SN is the set of natural numbers with prime factors are less or equal than N. So our product absolutely converges on the half-plane. To prove the claim one just has to note the validity of the equality

+ ∞ f(pk) f(n) 1+ = . pks ns p N k=1 n S Y≤ X X∈ N Both the expressions converge for N + , the ﬁrst one to the Euler → ∞ product and the second one to Df (s). Remark: If f is a completely multiplicative function, then f(pk) = f(p)k and (due to the properties of geometric series) in the half-plane of convergence we have the stronger equality

30 + ∞ f(n) f(p) 1 = 1 − . (2.3) ns − ps n=1 p X Y It is important to observe that all the results in this section are consequence of the unique factorization of the natural numbers in product of prime numbers.

2.2 The Riemann Zeta Function ζ(s)

Originally studied (among the others) by Euler, the Zeta function is now associated to Riemann because of his fundamental paper of 1859 [Rie]. It is deﬁned as

+ ∞ 1 ζ(s) := (2.4) ns n=1 X that is the Dirichlet series related to the arithmetical function 1 (which is obviously a completely multiplicative arithmetical function). It converges absolutely in the half-plane Re s> 1 and there (by the results of the previous section) we have the Euler product equality:

s 1 ζ(s)= (1 p− )− . (2.5) − p Y Therefore we can say that the Riemann Zeta function is holomorphic and not zero for Re s> 1.

Remark: Equation (2.4) was used by Euler, together with the divergence of the harmonic series, to give a diﬀerent proof of the existence of inﬁnite prime numbers.

The breakthrough work of Riemann consisted in an explicit way to provide an analytic continuation for the Zeta function over the whole complex plane, with the only exception of the point 1, which is a simple pole (as we shall see). The ﬁrst step is to show how to extend ζ(s) for 0 < Re s 1, by means of the Gamma Function Γ(s) (remember that it is a meromorphic≤ function, with no zeros on the half-plane Re s > 0 and with simple poles on the negative integers: see the Appendix). Let s be with Re s> 1. Then:

31 + s + 1 ∞ x x 1 ∞ nx s 1 − − − − Γ(s) s = e x dx = e x dx n 0 n 0 Z Z and because of the absolute convergence of the Riemann Zeta function we can exchange the sum and the integral to obtain

+ + ∞ + s 1 ∞ nx s 1 ∞ x − Γ(s)ζ(s)= e− x − dx = dx ex 1 Z0 n=1 Z0 − 1 X 1 + s 1 1 1 s 1 s 2 ∞ x − − − = x x dx + x dx + x dx. 0 e 1 − x 0 1 e 1 Z − Z Z − The ﬁrst and the third integral in the equation above deﬁne holomorphic 1 functions for Re s> 0, while the second one is equal to (s 1)− (remember that we are still assuming Re s> 1); so we have −

1 + s 1 1 1 1 s 1 1 ∞ x − − ζ(s)= x x dx + + x dx Γ(s) 0 e 1 − x s 1 1 e 1 Z − − Z − and thus we have deﬁned an analytic continuation for ζ(s) over the half-plane Re s> 0.

2.3 The functional equation

By now we extended the domain for the Zeta function only in the positive half-plane and all the results discovered could have been accomplished with other instruments: for example starting from (2.4) and applying Euler- McLaurin summation formula (6.3) (see Appendix) one gets the analytic continuation on the positive half-plane (written in another way) and the po- lar nature of the point s = 1. In fact it is possible, by much more precise estimates on the error term of the Euler-McLaurin formula, to obtain inductively the analytic continuation for ζ(s) over all the complex plane, but it is to say that this method is not very satisfying, because it does not allow to obtain the values for ζ(s) for Re s 0 in a way which is not tricky. The best≤ (and the more enlightening) method to reach our goal is to prove the so called functional equation which will permit us to see in clear way the holomorphic continuation of ζ(s) in the region Re s 0 (together with an easy formula for the computation in this region). ≤ The complete details of the following proof can be found in [Iwa].

32 We deﬁne the complex function

s/2 Λ(s) := π− Γ(s/2)ζ(s). The results of the previous section imply that Λ(s) is a well deﬁned and meromorphic function on Re s> 0, with only a simple pole located in 1 and coming from ζ(s). Assume now Re s> 1: it is an easy computation to see that

+ s/2 s ∞ πn2x s/2 1 π− Γ(s/2)n− = e− x − dx. Z0 The absolute convergence of the Zeta function for Re s> 1 implies that

+ s/2 ∞ s/2 1 π− Γ(s/2)ζ(s)= ω(x)x − dx (2.6) Z0 where it has been introduced the function

+ ∞ πn2x ω(x) := e− . n=1 X This is strictly related to the theta function

πn2x θ(x) := e− . n Z X∈ They are both well deﬁned and smooth functions on (0, + ) and they satisfy the relation θ(x)=2ω(x) + 1. Moreover the Poisson summation∞ formula (which we recall in Section 3.2.2) applied to the theta function provides the functional equation

1 1 θ(x)= x− 2 θ(x− ) · and the immediate counterpart for the ω(x)

1 1 1 1 ω(x− )= x 2 ω(x)+ x 2 1 . (2.7) 2 − Coming back to equation (2.6) and applying (2.7) we obtain

33 1 + s/2 s/2 1 ∞ s/2 1 π− Γ(s/2)ζ(s)= ω(x)x − dx + ω(x)x − dx = 0 1 + Z Z + ∞ 1/2 1 s/2 s/2 1 ∞ s/2 1 = x ω(x)+ (x 1) x− − dx + ω(x)x − dx = 1 2 − 1 Z + Z ∞ 1−s 1 = ω(x)(xs/2 + x 2 )dx + . s(s 1) Z1 − Thus, thanks to the convergence of the integral (due to the fact that ω(x)= πx O(e− ) for x> 0) we see that:

Λ(s) is a meromorphic function over C with singularities only in 0 and • 1, which are simple poles;

Λ(s) satisﬁes the functional equation Λ(1 s) = Λ(s) which can be • expressed as −

s/2 (1 s)/2 1 s π− Γ(s/2)ζ(s)= π− − Γ − ζ(1 s). 2 − This is the famous functional equation of the Riemann Zeta function and by exploiting Gamma function’s properties (see the Ap- pendix) we can state it in the more explicit equation:

s s 1 πs ζ(s)=2 π − Γ(1 s)sin ζ(1 s). (2.8) − 2 − It is clear from this formula that the Zeta function admits the desired analytic continuation in the negative half-plane Re s 0, and that the points where ζ(s) is zero in this region rise just from≤ the contribution of the sine term: they are the negative even integers and because of the easy way to detect them they are called trivial zeros.

Remark: Thanks to the independent works of Hadamard [Had] and of de la ValléèPoussin [Val] (which assert that ζ(1 + it) = 0 for every real t = 0) and to (2.8) it follows immediately that ζ(it) = 0 for6 such values of t. 6 But (2.8) implies also that ζ(0) is finite and6 not zero: that is because for 1 s 0 it is sin(πs/2) πs/2 and ζ(1 s) s− (so that the computations show→ that ζ(0) = 1/∼2). − ∼− −

34 Chapter 3

The Dedekind Zeta Functions

3.1 Deﬁnition and ﬁrst properties

Let Q K be a number field, K the ring of integers of the field K. We define the⊆ Dedekind Zeta functionO related to K as 1 ζ (s) := K N(I)s I X⊂OK where the sum is made over all the non zero ideals of the ring K and N(I) is the norm of the ideal I. Our goal is to study this series definedO as a function of complex variable.

Theorem 20. Let K be a number ﬁeld, ζK its related Dedekind Zeta function. Then ζK (s) converges absolutely in the half-plane Re s> 1 and there deﬁnes a holomorphic function. Moreover in this half-plane there is the equality

1 s − ζ (s)= 1 N( )− (3.1) K − P P⊂OYK where the product ranges over the non zero prime ideals of . OK

Remark: From now on the words “ideal” and “prime ideal” will mean respectively “not zero ideal” and “not zero prime ideal”. Proof. We start studying the Euler product and showing that it converges absolutely in the half-plane Re s > 1. Remember that N( ) 2 for every P ≥ prime ideal of K . First of all,P thanksO to the triangular inequality one gets 1 s − 1 1 1 N( )− = 1+ 1+ . − P N( )s 1 ≤ N( )Re s 1 P⊂OK P⊂OK P − P⊂OK P − Y Y Y

35 Let us show that the last product converges absolutely. Let x = Re s. Our claim is true if and only if the series 1 N( )x 1 P⊂OXK P − converges absolutely in the same region. Now, we know a strong connection between the natural prime numbers and the prime ideals of K : for every prime number p there exist at least one and at most [K : Q] primeO ideals fp P lying over p, so that N( )= p for some natural number fp. Therefore we have P

1 2 1 2 [K : Q] < + . N( )x 1 ≤ N( )x ≤ · px ∞ p P⊂OXK P − P⊂OXK P X

Now let us show that the usual writing of ζK converges as a series. Let Sm be the set of ideals of K which have prime factors with norm less or equal than m. We have O

1 1 1 1 1+ 1+ = 1 − . N(I)x ≤ N(I)x − N( )x N(I) m Sm N( ) m X≤ X YP ≤ P

Letting m goes to + we get the absolute convergence of ζK (s) in Re s> 1 by the absolute convergence∞ in this half-plane of the Euler product. Finally, the series expression and the product expression of ζK are equal because of the equality 1 1 1+ = 1 . N(I)x − N( )x Sm Sm X P∈Y P Both expressions converge absolutely, the ﬁrst one to ζK (s) and the second one to the Euler product.

Remark: If K = Q, being OQ = Z and N(nZ) = n, it follows that ζK is just the Riemann Zeta function ζ(s), and we know that its expression as Dirichlet series converges absolutely in the half-plane Re s> 1; moreover in this region one has the equality

1 s − ζ(s)= 1 p− − p Y 36 where the product is made over the positive prime numbers.

Remark: It is possible to write ζK as a usual Dirichlet series ranging over the natural numbers, simply deﬁning the function m(n) := # I K N(I) = n so that { ⊂ O | } + ∞ m(n) ζ (s)= . (3.2) K ns n=1 X Note that, even if the norm function is completely multiplicative on the ideals, the function m(n) is not completely multiplicative: for example if K = Q[i] there are no ideals of norm equal to 3 but the ideal 3 OQ[i] has norm equal to 9. Nonetheless, it will be shown in Chapter 4 that·m(n) is a multiplicative function, and so the series (3.2) can be expressed as a Euler product ranging over the natural prime numbers.

3.2 The functional equation of ζK and the class number formula

3.2.1 The functional equation The goal of this section is to obtain a meromorphic analytic continuation for ζK just as we have done for the Riemann Zeta function: in order to do this we have to show the following theorem.

Theorem 21. Let K be a number ﬁeld of degree n over Q, with r1 real embeddings and r2 couples of complex conjugated embeddings, and let dK := ∆K/Q . |Deﬁne| the function

r1 r2 s/2 s/2 s s Λ (s) := d (π)− Γ (2π)− Γ(s) ζ (s). K K · 2 K h i h i Then ΛK (s) can be analytically extended to a meromorphic function, holomorphic over the whole complex plane except for two simple poles, in 0 and 1. Furthermore, for s =0, 1 one has the functional equation 6 Λ (s)=Λ (1 s). (3.3) K K − The analytic continuation allows us to deﬁne the Dedekind Zeta Function over the half-plane Re s > 0 (just as it has been done for the Riemann zeta

37 function) and to detect the simple pole in 1, while it is the functional equation of ΛK which permits to deﬁne ζK over the half-plane Re s 0. In fact, the application of the properties of Gamma function≤s to (3.3) give

1 s Γ − r1 r2 1/2 s s 1/2 2 s 1/2 Γ(1 s) ζK (s)= dK − π − (2π) − − ζK (1 s) Γ s Γ(s) − h 2 i h i r1 r2 1/2 s s s 1 πs 2s 2s 2 2 = d − 2 π − Γ(1 s)sin 2 π − Γ(1 s) sin(πs) ζ (1 s) K − 2 − K − r1 1/2 sh s s 1 n πsi h r2 i = d − [2 π − Γ(1 s)] sin sin(πs) ζ (1 s). K − 2 K − This formulation provides an explicit functional equation for ζK (s) and we immediately see that:

In the half-plane Re s < 0 the function ζ (s) has zeros of multiplicity • K r2 in the negative odd integers and zeros of multiplicity r1 + r2 in the negative even integers. These are called the trivial zeros of ζK (s) and there are no other zeros in the negative half-plane.

The function ζK (s) is holomorphic in 0 and ζK (0) = 0 (with the multi- • plicity of the zero equal to r +r 1) unless K = Q or K is a quadratic 1 2 − imaginary ﬁeld (because in these cases r1 +r2 = 1 and so the sine terms contribute with a simple zero in 0 which is annihilated by the simple pole of ζK (s) in 1). How to prove the theorem? The idea could be that of using a generalization of the Theta function seen in the previous chapter. However it is not advis- able to go straightforward this way because in this setting fundamental roles are played by arithmetical factors which are not remarkable for the Riemann Zeta function (as the discriminant or the diﬀerent ideal). So it is easier to start focusing not on ζK but on a sum with less terms, depending on the elements of the class group.

In fact, let be a class of the Class Group of K . We deﬁne the partial DedekindR Zeta function as O

1 ζ ( ; s) := . K R N(I)s I X∈R The sum is made over the ideals of K contained in the class and so for Re s > 1 it converges (being estimatedO in module by the convergentR series

38 Re s I N(I)− ). P Theorem 22. Let be a class in the Class Group of K. Deﬁne the function R r1 r2 s/2 s/2 s s Λ ( ; s)= d (π)− Γ (2π)− Γ(s) ζ ( ; s). K R K · 2 K R h i h i Then ΛK ( ; s) is a meromorphic function, analytic over the whole complex plane exceptR for two simple poles, in 0 and 1. Furthermore for s =0, 1 one has the functional equation 6 0 ΛK ( ; s)=ΛK ( ;1 s) (3.4) 0 R R − where is the class in Cl(K) containing the complements (with respect to the trace)R of the members of . R

Theorem 21 follows at once by Theorem 22 because ΛK (s) = ΛK ( ; s) 0 and the assignment is a bijection in the class group. R R Before starting withR→R the proof we need a couple of lemmas. P

3.2.2 Analytic lemmas A function f : Rn C is a Schwartz function if f is inﬁnitely derivable and for every couple→α, β of multi-indexes it holds

α β sup x ∂x f(x) < + . x Rn | | | | ∞ ∈ The set (Rn) := f : Rn C: f is a Schwartz function is called the SchwartzS space and{ is a C-vectorial→ space. } Let f : Rn C; the Fourier transform of f is the function →

2πi η,x fˆ(η) := f(x)e− h idx. n ZR For every f (Rn) the Fourier transform fˆ is a well-deﬁned function and fˆ (Rn). ∈ S ∈S πα x 2 Lemma 18. Let f(x)= e− | | , where α> 0. Then

1 π x 2 fˆ(x)= e− α | | . αN/2 Proof. This is a classical result of harmonic analysis: see Chapter 1 of [Pel].

39 Now, let Q(x) := i,j n aijxi xj be a real and positive defined quadratic form. Given α> 0, we want≤ to compute· the Fourier transform of the function P παQ(x) f(x) := e− . Remember that Q(x)= Ax, x where A is the matrix whose ij-th coefficient h i is defined as (aij + aji)/2, and that all the eigenvalues of A are positive. Lemma 19. If f(x) is the function above, then

0 1 1 π Q (x) fˆ(x)= e− α αn/2 det A 1/2 0 | | 1 where Q (x)= A− x,x . h i Proof. Let us recall that, being the matrix A symmetric and with positive eigenvalues, there exists a symmetric matrix B such that A = B2 (see [Lang2]). Hence

παQ(y) 2πi y,x πα Ay,y 2πi y,x fˆ(x)= e− e− h idy = e− h ie− h idy. n n ZR ZR The assumptions above imply that Ay, y = B2y,y = By,By ,and the Diﬀeomorphisms theorem implies thath (withi theh changei of varh iablesiz = Bx)

πα By,By 2πi y,x πα z,z 2πi B−1z,x dz fˆ(x)= e− h ie− h idy = e− h ie− h i n n | det B ZR ZR | | πα z,z 2πi B−1z,x dz 1 πα z,z 2πi z,(B−1)T x = e− h ie− h i = e− h ie− h idz. n det B det B n ZR | | | | ZR πα x 2 The last term is the Fourier transform of the Gaussian function e− | | eval- 1 T 1 1/2 uated in the point (B− ) x = B− x and divided by det B = det A , so that | | | |

1 1 π B−1x,B−1x fˆ(x)= e− α h i αn/2 det A 1/2 1 1 1 | | and being B− x,B− x = A− x,x the truth of the claim is proved. h i h i We conclude this section with another technical result. Theorem 23. (Poisson Summation Formula): Let f (Rn), fˆ its Fourier transform. Then it holds ∈ S

f(x)= fˆ(x). x Zn x Zn X∈ X∈ Proof. The proof can be found in Chapter 6 of [Kat].

40 3.2.3 Proof of the functional equation for ζK( ; s) R 1 Let be a class of Cl(K) and let I be an ideal in the inverse class − : the mapR R

principal ideals divisible by I R → { } J IJ =(η) → induces a bijection between ideals in and elements η R(I), where R(I) denotes the set of elements of the idealR I up to multiplication∈ by a unit of . Thus we can write OK

s 1 s 1 ζK ( ; s)= N(I) s = N(I) s . R N((η)) NmK/Q(η) η R(I) η R(I) ∈X ∈X | |

We will prove the existence of a functional equation for ζK (s), as in the Riemann Zeta function case, starting by this series and then collecting each expression for every class of the class Group (remember that it’s a ﬁnite group). R

2 Let σv be a real embedding of K in C. Recall the quantity dI := N(I) dK introduced at the end of Chapter 1. Then one has the following relation:·

1/2n s 1/2 1/2n 1/n s Γ(s/2) dI s − (π dK N(I) ) s = 1/2 Γ · σv(η) π σv(η) · 2 | | | | + 1/2n s ∞ x s/2 dI dx − = e x 1/2 0 · π σv(η) x Z + | | ∞ 1/n 2 s/2 dy = exp( πdI− σv(η) y) y . 0 − | | · y Z (3.5) where in the last equality it has been used the change of variables

1/2n 2 1/n dI dI y := x 1/2 = x 2 . · π σv(η) · π σv(η) | | | | Now let σv be a complex embedding (in this case the index v counts at the same time σ and its conjugate). By means of completely similar computations we obtain the expression

41 1/n s 1 1 1/n 2/n s Γ(s) dI − − (2 π dK N(I) ) 2s = 2 Γ(s) · σv(η) 2π σv(η) · | | | | + 1/n s ∞ x s dI dx − = e x 2 0 · 2π σv(η) x Z + | | ∞ 1/n 2 s/2 dy = exp( πdI− 2 σv(η) y) y . 0 − | | · y Z (3.6) and in the last equality the change of variables is

1/n dI y := x 2 . · 2π σv(η) | | Multiplying (3.5) r1 times (one for each real embedding) and (3.6) r2 times (one for each couple of conjugated complex embeddings) gives

r1 r2 s s/2 s/2 s s N(I) d (π)− Γ (2π)− Γ(s) K · 2 N((η))s h+ + i h i ∞ ∞ 1/n 2 s/2 dy = exp πd− N σ (η) y y ··· − I v| v | · v · || || y 0 0 v Z Z X where the integrals are made (r1 + r2) times; it is

1 σv real embedding Nv = (2 σv complex embedding and y s/2 = yNv , dy/y = d /y . || || v v v yv v We want to exchangeQ these integralsQ with the sum made over the elements η R(I). This is possible because for Re s > 1 the sum in the integrals is a sum∈ of positive terms and so we obtain a new expression for the function Λ ( ; s): K R

r1 r2 s/2 s/2 s s Λ ( ; s)= d (π)− Γ (2π)− Γ(s) ζ ( ; s) K R K · 2 K R + h + i h i ∞ ∞ 1/n 2 s/2 dy = exp πd− Nv σv(η) yv y . ··· − A | | · || || y 0 0 η R(I) v Z Z ∈X X 42 + + Let us denote G := R R , where the product is taken (r1 +r2) times; thus we can write ×···×

s/2 Λ ( ; s)= f(y) y d∗y K R · || || ZG where the measure is d∗y = dy/y,and

1/n 2 f(y)= exp πd− N σ (η) y . − I v| v | · v η R(I) v ∈X X

Now, recalling Dirichlet’s unit theorem, if U = K∗ is the group of units in , let V be the image of U of via the map O OK

K∗ G → u ( σ (u) ) . → | v | v

The kernel of this group homomorphism is the torsion subgroup µK made by the roots of unity and the image V (which is a discrete subgroup of Rr1+r2 ) is contained in the subgroup

G := y G: y =1 . 0 { ∈ || || } Furthemore the quotient group G0/V , which is in bijection with the fundamental domain of the lattice V , is compact (it is a consequence of Dirichlet’s theorem) and, by the fact that any element of G can be written in a unique way as a product t1/nc with t R+ and c G , we have the decomposition ∈ ∈ 0 G = R+ G × 0 and this induces a different way to write the integrals above. In fact, by changing variables (posing y = t) we have || || + ∞ 1/n s/2 dt Λ ( ; s)= f(t c)t d∗c K R t Z0 ZG0 where d∗c has now to be intended as the restriction of the previous measure on the subgroup G0. The very definition of f shows that this function is invariant under the action of the group V and so (remembering what is the definition of the Haar 0 measure for the quotient set E = G /V and that c =(cv)v) we have

43 ΛK ( ; s) R+ ∞ 1/n 2 2 1/n s/2 dt = exp πd− N σ (η) σ (x) t c t d∗c − I v| v | | v | · v · t 0 E x V η R(I) v Z Z X∈ ∈X X + ∞ 1 1/n 2 1/n s/2 dt = exp πdI− Nv σv(ηu) t cv t d∗c wK − | | · · t 0 E u U η R(I) v Z Z X∈ ∈X X + ∞ 1 1/n s/2 dt = [Θ(I; t c) 1]t d∗c w − t Z0 ZE K 1/n where wK is the number of roots of unity in K and Θ(I; t c) is the Hecke theta function O

1/n 1/n 2 1/n Θ(I; t c) := exp πd− N σ (η) t c . − I v| v | · v η I v X∈ X Just as for the Theta function of Chapter 2, it can be proved that there exists D> 0 such that, for t going to + , one has ∞ 1/n πDt1/n Θ(I; t c) 1= O(e− ). (3.7) − and similarly to the classical case the Hecke Theta function admits a very useful functional equation. Lemma 20. Let Θ(I; t1/nc) as above: then it holds

1/n 1 0 1/n 1 Θ(I; t c)= Θ(I ; t− c− ) √t 0 1 where I = (A K Q)− is the complement, with respect to the trace, of I 1 ·D | and c− denotes the point in G0 which has as coordinates the inverses of the coordinates of c.

Proof. Let η1,..,ηn be a Z-basis of the fractional ideal I. We can rewrite our function{ as a sum} over the lattice of integer points z =(z ,..,z ) Zn: 1 n ∈

2 1/n 1/n 1/n Θ(I; t c)= exp πd− N z σ (η ) t c − I v i v i · v z Zn v i X∈ X X 1/n 1/n = exp πd− t σ (η )σ (η )c z z − I v i v j v i j z Zn i,j v ∈ X X Xn 1/n 1/n = exp πd− t σ (η )σ (η )c z z . − I k i k j k i j z Zn i,j k=1 X∈ X X

44 Observe that in the last equation we have changed the indexes: the counting is now made from 1 to n, because for every complex not real embedding σv we add its complex conjugated σv. If σi,σj are conjugated we impose xi = xj. The exponentials are then evaluated on the quadratic form related to the matrix M =(bij)1 i,j n where ≤ ≤ 1 n b = c (σ (η )σ (η )+ σ (η )σ (η )). ij 2 k · k i k j k j k i Xk=1 Now, consider the matrix N =(lij)1 i,j n where ≤ ≤ l = √c σ (η ). ij i · i j It can be veriﬁed that N multiplied with its conjugated transposed matrix gives M as a result, so that

det M = det N 2 = c c c2 2 D(η ,...,η ) =1 D(η ,...,η ) . | | | | 1 ··· r1 · r1+1 ···r1+r2 | 1 n | ·| 1 n | 0 0 0 Let η1,...,ηn be the basis of I , i.e. the dual basis with respect to the trace{ of the Z-basis} of I: the inverse matrix of N has i, j-th coeﬃcient equal 1/2 0 to ci− σi(ηj): so applying the analytic lemmas of the previous section and 0 1 remembering that dI = dI− we obtain

1/2 1/n dI 0 1/n 1 Θ(I; t c)= Θ(I ; t− c− ) √t D(η ,...,η ) 1/2 ·| 1 n | and the truth of the claim follows remembering that d = D(η ,...,η ) . I | 1 n |

Let us split the ﬁrst integral in the writing of Λ ( ; s): we have K R

+ ∞ 1 1/n s/2 dt ΛK ( ; s)= [Θ(I; t c) 1]t d∗c R 1 E wK − t Z 1 Z 1 1/n s/2 dt + Θ(I; t c)t d∗c 0 E wK t Z 1 Z 1 s/2 dt t d∗c . (3.8) − w t Z0 ZE K Before completing the prove of the functional equation, we need a last technical lemma.

45 Lemma 21. The previous hypotheses imply

r1 1 d∗c =2 − R · K ZE where RK is the regulator, as deﬁned in (1.3). Proof. There is an isomorphism of groups

r1+r2 1 R − G → 0

a1 ar1+r2−1 (a1, . . . , ar1+r2 1) [σ1(ε1)] [σr1+r2 1(εr1+r2 1)] − → ··· − − where the εj’s are the fundamental units of K∗ . The Jacobian of this map is then equal to O r1+r2 1 r2 r1 1 2 − 2− R =2 − R · K · K and the measure of E with respect to d∗c is nothing but the volume of the r1+r2 1 unit cube in R − multiplied by this Jacobian (see Chapter 13, Section 3 of [Lang1] for further details).

Let us ﬁnally prove the functional equation of ΛK ( ; s). Consider the right side of the equation (3.8). The ﬁrstR term, by the convergence properties of the Hecke Theta function shown in (3.7), is an absolutely convergent integral for every complex number s, and thus is an entire function in the variable s. By lemma 21 the third term is trivially seen to be

r1 2 RK − s w · K and so we are left with the study of the second term: but lemma 20 on Hecke Theta functions implies

1 1 1 1/n s/2 dt 1 0 1/n 1 s/2 dt Θ(I; t c)t d∗c = Θ(I ; t− c− )t d∗c . 0 E wK t 0 E wK √t t Z Z Z Z (3.9) 1 Now, observe that the morphsim c c− ﬁx the set E and does not change the measure; then with the change→ of variables 1/t = ρ the integral above becomes

46 + ∞ 1 0 1/n s/2 dt √t Θ(I ; t c)t− d∗c 1 E wK · t Z +Z + ∞ 1 0 1/n (1 s)/2 dt ∞ 1 (1 s)/2 dt = [Θ(I ; t c) 1]t − d∗c + t − d∗c w − t w t Z1 ZE K Z1 ZE K + r1 ∞ 1 0 1/n (1 s)/2 dt 2 RK = [Θ(I ; t c) 1]t − d∗c . w − t − (1 s)w Z1 ZE K − K So we ﬁnally obtain the meromorphic continuation for we were looking for:

+ ∞ 1 1/n s/2 dt Λ ( ; s)= [Θ(I; t c) 1]t d∗c K R w − t Z1 ZE K + r1 ∞ 1 0 1/n (1 s)/2 dt 2 RK + [Θ(I ; t c) 1]t − d∗c + . w − t w s(s 1) Z1 ZE K K − The equation is valid for Re s> 1, but the right term is actually a meromorphic function, with simple poles in 0 and 1. Moreover this formulation immediately holds the functional equation (3.4), and so we are done with the proof.

3.2.4 Computation of the residue in 1: the Class Num- ber Formula

In the previous section we showed that ζK is analytic over the whole complex plane except for the point s = 1, which is a pole of simple order. This fact is important not only for analytical reasons but for algebraic ones too: the residue in the point is related to many algebraic invariants of the number ﬁeld K and of its ring of integers . OK Theorem 24. (Class number formula): Let ζK be the Dedekind Zeta function related to the number ﬁeld K. Then

r1+r2 r2 2 π RK hK lim (s 1)ζK (s)= (3.10) s 1 − 1/2 → wK dK where: r is the number of real embeddings of K; • 1 r is the number of couples of complex conjugated embeddings of K; • 2 R is the regulator of K; • K 47 h is the order of the class group Cl(K); • K w is the number of roots of unity in K; • K

d is the absolute value of the discriminant ∆ Q. • K K/ Proof. The theorem follows almost immediately from the considerations ex- posed in the previous section. In fact, for every partial Dedekind zeta function we just have to compute the limit

r1 r2 s/2 s/2 s − s − lim(s 1)ζK ( ; s) = lim(s 1)ΛK ( ; s)d− (π)− Γ (2π)− Γ(s) . s 1 − R s 1 − R K 2 → → r1 The residue of ΛK ( ; s) in 1 is equal to (2 RK )/wK and the remaining R r2 product is equal to (2π) /√dK when is evaluated in 1. The claim follows by adding over all the class of the class group Cl(K).

48 Chapter 4

Arithmetical equivalence: the analytic way

4.1 Formulation and ﬁrst results

4.1.1 The problem

0 0 Suppose we are given two number fields K and K . Clearly if K and K are isomorphic fields the Dedekind Zeta functions related to the two fields are equal. 0 This is obvious because an isomorphism between K and K induces an iso- 0 morphism between the rings of integers K and OK , and so there is a bijection between the prime ideals of theO two rings which preserves inertia and ramification degrees. These remarks lead to the equality of the Euler products 1 1 1 s = 1 0 s − N( ) 0 − N( ) K P O 0 P P⊂OY P Y⊂ K

0 deﬁning ζK and ζK , which implies the equality of meromorphic continuation for the two functions.

In this chapter we are interested in the inverse problem: suppose that we 0 know that K and K are arithmetically equivalent number fields, i.e. they have equal zeta functions. What can be said then about the two fields? Are they isomorphic? Do they share some algebraic property, like the same value for their algebraic invariants? Briefly, how do the analytic behaviour of the Dedekind Zeta functions deter- mines the algebraic behaviour of the number field related?

49 The equality of the two Dedekind Zeta functions implies the equality of the residues in the point 1, which means that the Class Number Formulas for the two Zeta functions are equal:

0 0 0 r1+r2 r2 r1+r2 r2 0 0 2 π RK hK 2 π RK hK 1/2 = 1/2 . (4.1) w d w 0 d 0 K K K K This a first relation between the algebraic invariants of the two fields: the main purpose of Chapters 4 and 5 of this work is to show that for two arithmetically equivalent number fields this is just the beginning. In fact, if 0 ζK = ζK then:

1) the ﬁelds have the same number of couples of complex, not real, conjugated embeddings;

2) the ﬁelds have the same number of real embeddings and the same degree over Q;

3) the ﬁelds have the same absolute value for the discriminant, and actually they have the same discriminant;

4) almost all the prime numbers of Z decompose in the same way when embedded in the rings of integers of the two ﬁelds;

5) the two ﬁelds share the same roots of unity.

4.1.2 Result on immersions and degrees In the previous chapter we showed that for every complex number s with Re s 0 the Dedekind Zeta function of the number ﬁeld K satisﬁes the functional≤ equation

r1 1/2 s s s 1 n πs r2 ζ (s)= d − [2 π − Γ(1 s)] sin sin(πs) ζ (1 s) (4.2) K K − 2 K − where n = [K : Q] is the degree of the ﬁeld K, r1 is its number of real embeddings , r2 is its number of couples of complex, not real, conjugated embeddings (so that n = r1 +2r2) and dK is the absolute value of the discriminant ∆K/Q.

0 Lemma 22. Let K and K be arithmetically equivalent number ﬁelds. Then:

50 0 K and K have same number r2 of couples of complex, not real, conju- • gated embeddings;

0 K and K have same number r of real embeddings; • 1 0 K and K have same degree n over Q. •

0 Proof. Suppose that ζK (s)= ζK (s). Then for Res 0 the functional equation (4.2) holds for both the functions and ≤

r1 1/2 s s s 1 n πs r2 d − [2 π − Γ(1 s)] sin sin(πs) ζ (1 s) K − 2 K − 0 0 0 πs r1 1/2 s s s 1 n r2 = d 0 − [2 π − Γ(1 s)] sin sin(πs) ζ 0 (1 s). K − 2 K − The equality of the two Zeta functions implies that they have the same trivial zeros with the same multiplicity in the half-plane Res 0. The functional ≤ equation shows that the negative odd integers are zeros of multiplicity r2 for 0 0 0 ζK and of multiplicity r2 for ζK , and this implies r2 = r2; thus the ﬁrst claim is proved. The truth of the second claim follows in the same way, by considering the multiplicity of the the negative even integers as zeros for the two functions. Being n = r1 +2r2 the third point of the theorem is immediately seen to be true.

4.2 The Asymptotic formula for the non triv-

ial zeros of ζK 4.2.1 Arithmetically equivalent ﬁelds have same discriminant

0 We have just seen that if K and K are arithmetically equivalent then (with the exception of the absolute value of the discriminant) the arithmetic factors involved in the functional equation (4.2) are the same for both the ﬁelds. The equality of the Class Number Formulas (4.1) for the two ﬁelds is then reduced to the equality

0 0 RK hK RK hK 1/2 = 1/2 . w d w 0 d 0 K K K K In this section it will be proved that the ﬁelds have also same absolute value of the discriminant (and by Theorem 12 they have same discriminant). Usually

51 this result is proved by considerations on the functional equation satisfied by the factors of the Euler product defining ζK . In this work the result will be proved in a different way, with a generalization of the classical asymptotic formula for the computation of the number of zeros for the Riemann Zeta function in the strip 0 < Re s< 1.

Theorem 25. (Asymptotic formula): Let K be a number ﬁeld and ζK its related Dedekind Zeta function. Let N(T ) be the function which counts (with multiplicity) the number of zeros of ζK (s) such that 0 < Re s< 1 and Im s

4.2.2 The non vanishing of ζK(s) over two lines

From the Euler product expression (3.1) of ζK (s) it is easily seen that the Dedekind Zeta function does not vanish over the half-plane Re s > 1. The functional equation (4.2) shows the existence of the trivial zeros in the non positive integers (including 0 for every ﬁeld which is neither Q nor the quadratic imaginary ﬁelds). We are interested in an approximated computation of the non trivial zeros of ζK (s) with real part between 0 and 1. But what can be said for the points 1+ it and it with t R, t = 0? Are they zeros for ζ (s)? ∈ 6 K Lemma 23. For every real σ > 1, for every real t =0 it holds 6 ζ (σ)3 ζ (σ + it) 4 ζ (σ +2it) 2 1. (4.4) K ·| K | ·| K | ≥ Theorem 26. ζ (1 + it) =0 and ζ (it) =0 for every t =0. K 6 K 6 6 Proof. (Theorem): Suppose there exists t0 = 0 such that ζK (1 + it0) = 0. Then there exists 6 n + a C∗ such that ζK (σ + it0) a(σ 1) for σ 1 , with n N, n 1. ∈ ∼ − 1 → + ∈ ≥ Furthermore we know that ζ (σ) c(σ 1)− for σ 1 , where c is the K ∼ − → 52 residue arising from the Class Number Formula (3.10). Being t = 0, the Dedekind Zeta function is holomorphic in the point 1+2it 0 6 0 and there exists a neighborhood of the point where ζK (s) is bounded in its modulus by a number B > 0. Therefore, given the Lemma, for σ 1+, one has → 3 4 2 4n 3 1 ζ (σ) ζ (σ + it ) ζ (σ +2it ) (σ 1) − ≤ K ·| K 0 | ·| K 0 | − and being 4n 3 > 0 the right term in the equation goes to 0, but this is contradictory.− We conclude that ζK (s) does not vanish on the points 1+ it with t = 0 and the claim for the points it follows immediately by the non vanishing6 of the factors in the functional equation (4.2) over the line Re s = 0. Proof. (Lemma): Every factor in the left term of (4.4) is a positive real number: the claim is equivalent to show that

Log(ζ (σ)3 ζ (σ + it ) 4 ζ (σ +2it ) 2) 0 (4.5) K | K 0 | | K 0 | ≥ where Log is the complex logarithm taken with the principal cut. σ it Observe that ζK (σ + it)= ζK (σ it) (because N(I) − is the conjugated of N(I)σ+it and the conjugation is− a continuous function, so that it commutes with the series deﬁning ζK (s)). Furthermore it is known that there is the power series equality

+ ∞ zm Log(1 z)= − − m m=1 X when z < 1. Mixing these facts together with the Euler product expression (3.1),| the| left term in (4.5) becomes 3Log(ζ (σ)) + 2Log(ζ (σ + it)) + 2Log(ζ (σ it)) K K K − + Log(ζ (σ +2it)) + Log(ζ (σ 2it)) K K − 1 1 1 = 3Log 1 σ 2Log 1 σ+it 2Log 1 σ it − − N( ) − − N( ) − − N( ) − P⊂OXK h P P P 1 1 Log 1 σ+2it Log 1 σ 2it − − N( ) − − N( ) − + P P i ∞ 1 imt imt 2imt 2imt = (3+2N( )− +2N( ) + N( )− +2N( ) ) mN( )mσ P P P P m=1 P⊂OXK X P + ∞ 1 imt imt 2 = (1 + N( ) + N( )− ) 0 mN( )mσ P P ≥ m=1 P⊂OXK X P 53 because the term in the parentheses is real.

4.2.3 Recalls on entire functions and Phragmen-Lindelöf Theorem A function f : C C is said to be entire if it is holomorphic over the whole complex plane. Typical→ examples of entire functions are the polynomials or the exponential es. An entire function can have no zeros (like es), finitely many (like the plynomials) or an infinite countable number (like sin s). An entire function has finite order ρ if for s + and ε> 0 one has | | → ∞ ∀ s ρ+ε f(s)= O(e| | ). Theorem 27. (Hadamard): Let f be an entire function of finite order ρ, and let z be the sequence of its zeros =0. Then there is the factorization n 6 s f(s)= eg(s)zm 1 eP (s/zn) (4.6) − zn where m is the order of f in 0, g isY a polynomial of degree ρ and P (z)= z + z2/2+ + zk/(k 1) and k is the smallest integer > ρ≤. ··· − Proof. This result is proved in Chapter 13.3 of [Lang3]. Next, in order to prove the asymptotical formula, one has to recall the Phragmen-Lindelöf Theorem which permits to obtain a bound for a holomorphic function on a strip by the bound on the sides of the strip. Theorem 28. (Phragmen-Lindelöf): Let f be a holomorphic function on a strip Re s [σ ,σ ]. Suppose there exists A > 0 such that f A on ∈ 1 2 | | ≤ the lines Re s = σ1 and Re s = σ2, and that there exists ρ > 0 such that f(s)= O(exp( s ρ)) for s + . Then f A in| | the strip.| | → ∞ | |≤ Theorem 29. (Phragmen-Lindelöf Corollary): Suppose f(s)= O(exp( s ρ)) | | in a strip Re s [σ1,σ2] and that there exists a positive integer M such that ∈ M M f(σ1 + it)= O( t ) and f(σ2 + it)= O( t ) for t + Then in the strip| | we have f(s)= O( s M )| for| s | |+ → . ∞ | | | | → ∞ Theorem 30. (Second Convexity Theorem): Suppose the hypotheses of the previous theorem hold, and for every σ [σ1,σ2] let ψ(σ) be the least number 0 such that ∈ ≥ f(σ + it) t ψ(σ)+ε | | for every ε> 0. Then ψ(σ) is a continuous convex function on [σ1,σ2]. Proof. See Chapter 12.6 of [Lang3].

54 4.2.4 Proof of the asymptotic formula Let us recall the function

r1 r2 s/2 s/2 s s Λ (s) := d (π)− Γ (2π)− Γ(s) ζ (s) K K · 2 K defined in the Section 3.2.1.h i h i It is a meromorphic function with only two singular points, 0 and 1, which are simple poles. In the previous chapter we proved the functional equation Λ (s)=Λ (1 s) (valid for s different from 0 and 1), and the equation (4.2) K K − for the Dedekind Zeta function guarantees that ΛK (s) does not vanish on the negative integers (because the trivial zeros of ζK (s) delete the singularities arising from the Gamma factors). Consider now the function (s) := s(s 1)Λ (s). The previous discussion LK − K implies that K (s) is an entire function which satisfies the functional equation (s) =L (1 s) and its only possible zeros must have real part in LK LK − the interval (0, 1); by definition of ΛK (s), if these zeros exist they are indeed the non trivial zeros of ζK (s) we are looking for.

Step 1: Existence of inﬁnite zeros for K (s). This claim will be a straightforward consequenceL of the following:

Theorem 31. The function K (s) is entire of order 1, and it has a countable infinite number of zeros, eachL one with Re s (0, 1). ∈ Proof. Let us begin showing that each entire function which is a factor in the product defining (s) is a function of finite order ρ 1. We assume LK ≤ Re s 1/2 because of the symmetry due to the functional equation K (1 s)= ≥ (s): L − LK s is a polynomial, and so is an entire function of order 0. • s/2 s/2 s dK , π− and (2π)− are nothing but exponential functions of order • 1.

The Stirling formula (6.4) introduced in the Appendix implies that • every Gamma function in the factorization is asymptotic to exp(A s Log s) for some A> 0 depending on the factor. ·

It remains to estimate the order of the entire function • f(s)=(s 1)ζ (s). − K

55 The idea is to show that it is polynomially bounded on C by using the properties of Dedekind Zeta function; in order to do this we recall the function of real variable ψ(σ) := inf ρ> 0: (σ+it 1)ζ (σ+it) t ρ+ε as t + , ε> 0 { | − K || | | | → ∞ ∀ } introduced in Theorem 30.

Let us ﬁx ε> 0. If σ 1+ ε then ≥ 1 1 ζ (σ + it) = < + | K | N(I)σ+it ≤ N(I)σ ∞ I I ⊂OK ⊂OK X X and so f(σ + it) t and ψ(σ)=1. If σ 1 (1 + ε|) =| ε, thanks to the functional equation (3.3) one has ≤ − −

r1 r2 1 s Γ − Γ 1 s 1/2 s 1 2 (s 2 )r1 (2s 1)r2 − (s 1)ζK (s)=(s 1)d − π − (2π) − ζK (1 s). K s r1 r2 − − Γ( 2 ) Γ(s) − We assume without loss of generality that t + (because by Schwarz principle it is ζ (σ it)= ζ (σ + it)). → ∞ K − K On the line Re s = ε the exponential terms and ζK (1 s) are bounded, so one has to investigate− the behavior of the Gamma− factors. Stirling’s formula (6.4) implies that

(σ+it 1/2) Log(σ+it) (σ+it)+o(1) πt/2 σ 1/2 Γ(σ + it)= e − − C(σ)e− t − ∼ for t + (where C(σ) C∗) and so → ∞ ∈

r1 r2 1 s Γ −2 Γ 1 s r1(1/2 σ) 2r2(1/2 σ) − D(σ)t − t − s r1 r2 Γ( 2 ) Γ(s) ∼ with D(σ) C∗. This implies ∈ 1 ψ(σ)=[K : Q] σ + ψ(1 σ). 2 − − If one proves that f(s) satisfies the hypotheses required by the Phragmen- Lindelöf principle (i.e. that f(s) exp( s ρ) in the strip Re s (0, 1) for some finite ρ) we obtain the polynomial | | bound in the strip. But∈ this follows by the expression for ζK (s) we obtained in Chapter 3 (when it was needed to extend meromorphically the function ΛK (s)).

56 Remark: The second convexity theorem says that ψ(σ) is a continuous function: thus one obtains that for the function f(s)=(s 1)ζK (s), it is ψ(1) = 1 and ψ(0) = [K : Q]/2. −

Mixing the results above we can conclude that (s) is an entire function of LK order 1. Observe that if K (s) had only a ﬁnite number of zeros z1,...,zk , then by Hadamard’s FactorizationL Theorem it would be { }

k s (s)= eAs+B 1 es/zn LK − z n=1 n Y Cs ad so K (s) e for some C > 0. But this is not acceptable due to the presenceL of the Gamma factors, because for real s going to + they contribute as exp(D s Log s) for some D> 0. ∞ · Let zn n be the sequence of zeros of K (s) (i.e. the non trivial zeros of the Dedekind{ } Zeta function). L The Hadamard factorization of (s) is equal to LK + ∞ s (s)= eAs+B 1 es/zn . (4.7) LK − z n=1 n Y Observe that some zn’s could be equal if there exist zeros with multiplicity greater than 1. Step 2: Principle of the argument We know that if f is a holomorphic function on a open connected region D C and γ D is a closed simple curve, then Jordan’s Theorem guarantees⊆ that we can⊂ separate D in two regions I and E (interior and exterior). The argument principle says that

0 1 f (z) # s I : f(s)=0 = dz { ∈ } 2πi f(z) Zγ and the zeros are counted with their multiplicity. Let T > 0 such that ζ (σ + iT ) = 0. Deﬁne K 6 N(T ):=# s C: Re s (0, 1) , Ims < T, (s)=0 { ∈ ∈ | | LK } where the counting is made with multiplicity. Let R be the rectangle with vertexes the points 3/2+ iT , 1/2+ iT , 1/2 iT ,3/2 iT . The argument principle then implies that − − − −

0 (z) 2πiN(T )= LK dz. (4.8) (z) Z∂R LK 57 0 Being K (1 z) = K (z), the derivative satisﬁes the relation (1 z) = 0 K (zL) and− so L L − −LK 0 0 (1 z) (z) LK − = LK . (1 z) − (z) LK − LK Let γK be the line in ∂R which connects the points 3/2 and 1/2+ iT . The 0 symmetry of K (z)/ K (z), together with the Schwarz reﬂection principle, permits to reduceL theL computation to the line integral of this logarithmic derivative made over γK , and turning back to (4.8) one has

0 π (z) N(T )=Im LK dz . 2 (z) ZγK LK 0 0 0 By the fact that (fg) /(fg)= f /f + g /g one has

0 0 K (s) 1 1 1 r1 Γ s L = + + log dK log π r2 log(2π)+ r1 K (s) s s 1 2 − 2 − Γ 2 L 0 − 0 Γ ζK + r2 (s)+ (s) Γ ζK and therefore one can study the function N(T ) by applying separately the argument principle to every factor of (s). LK The term 1/s is such that •

dz π 2T 1 = Log(1/2+ iT ) Log(3/2) = i + Log + O . γ z − 2 3 T Z K The computation for 1/(s 1) is completely similar and one sees that both terms contribute in the− computation of N(T ) with a quantity

π 1 + O . 2 T For the term log d /2 one gets • K log d 1 K dz = (iT 1)log d 2 2 − K ZγK and so the contribution of this term is equal to T log d + O(1). 2 K 58 In the same way one gets for r1 log π/2 and r2 log(2π) the contribu- tions − − T T r log π + O(1) and 2r log(2π)+ O(1). − 1 2 − 2 2 The Gamma factors can be separated in two cases, remembering that 0 0 • (Γ /Γ)(s) = (LogΓ) (s) and applying Stirling’s formula (6.4). For every factor related to a real embedding of K in C the line integral of the logarithmic derivative over γK is equal to 1 T 1 1 T 1 T + i Log + i i + R (T ) 4 2 − 2 4 2 − 4 − 2 1 where R1(T ) is real, so that the contribution required is equal to T T T log + O(1). 2 2 − 2 Analogues computations for every Gamma factor related to the complex embeddings of K in C give rise to the contribution

T log T T + O(1). − and so the total contribution of the Gamma factors is equal to T T T T T T r log +2r log + O(1). 1 2 2 − 2 2 2 2 − 2

Assume that the contribution of ζK (T ) in the argument principle is SK (T ); then gathering all the terms studied before one gets

π π π T T T N(T )= + + log d r log π 2r log(2π) 2 2 2 2 K − 1 2 − 2 2 T T T T T T + r log +2r log + S (T )+ O(1) 1 2 2 − 2 2 2 2 − 2 K T T T T = π + log d +[K : Q] log + S (T )+ O(1) 2 K 2 2π − 2 K T T T = π + log d +[K : Q] log + S (T )+ O(1). 2 K 2 2πe K

The asymptotic formula (4.3) is proved once one is able to show that SK (T )= O(log T ). This is not as simple as the previous computations and one has to recall some properties of the non-trivial zeros of ζK (s).

59 Step 3: Estimate of SK (T ). We want to show that the imaginary part of the integral

0 ζ (z) K dz ζ (z) ZγK K is bounded by C log T for some C > 0. We divide the estimate in two parts: the one for the integral on the vertical line γ1 connecting 3/2 to 3/2 + iT and the one for the integral over the horizontal line γ2 connecting 3/2+ iT to 1/2+ iT .

0 0 Remember that for Re s> 1 it is (ζK (s)/ζK (s)) = (Log ζK ) (s) and so, • exploiting the Euler product (3.1) one gets

0 3/2+iT ζK (z) 3 3 dz = Log ζK (z) = Log ζK + iT Log ζK γ1 ζK (z) 3/2 2 − 2 Z 1 1 = Log 1 Log 1 − − N( )3/2+iT − − N( )3/2 P⊂OXK h P P i + ∞ 1 1 1 = m N( )m(3/2+iT ) − N( )m(3/2) m=1 P⊂OXK X P P and the last term of the equation is uniformly bounded in T by the convergent series

+ ∞ 1 2 =2 Log ζ (3/2). mN( )m(3/2) · K m=1 P⊂OXK X P So the contribution along the vertical line γ1 has been easily estimated as O(1).

We are left with the study of the imaginary part of the integral •

0 ζ (z) K dz. ζ (z) Zγ2 K We have to recall the Hadamard factorization (4.7) of the entire function (s); assuming Re s> 1, it is LK

+ ∞ s s Log (s)= As + Log 1 + LK − z z n=1 n n X h i 60 and thus the logarithmic derivative of (s) is LK

0 + ∞ 1 1 LK (s)= A + + . K zn s zn L n=1 − X 0 Recalling the diﬀerent expression of ( K / K )(s) previously used we obtain the identity L L

0 + ζ ∞ 1 1 1 1 1 r K (s)= A + + log d + 1 log π+ ζ z s z − s − s 1 − 2 K 2 K n=1 n − n − X 0 0 Γ s Γ + r log(2π) r r (s) 2 −− 1 Γ 2 − 2 Γ which is valid for s = σ+iT with σ [1/2, 3/2] (this causes no problem because of the assumption that the∈ Zeta function has no zeros with imaginary component equal to T ). The idea is to estimate the integrals over γ2 of every term in the right side of the equation. Clearly the integral of the term A is of no interest in this discussion, because 1 3 Adz = A + iT iT = A = O(1) γ2 2 − 2 − − Z and the same result is obtained by integrating the remaining constant terms (the ones which do not depend on s or T ). The integral of the term 1/s is well bounded being

dz 1 1 = Log 1+ = O γ2 z 1/2+ iT T Z and one gets the same bound for 1/(s 1). For the Gamma factors related to the− real embeddings, by means of the Stirling formula one has

0 1 T Γ z Γ( 4 + i 2 ) 1/4 3/4 dz = Log 3 T log(T − ) = O(log T ) γ2 Γ 2 Γ( + i ) | | Z 4 2 while for the Gamma factors related to the complex embeddings the recursive formula of Γ(s) implies:

61 0 1 Γ Γ( 2 + iT ) 1 (z)dz = Log = Log 1 = O(log T ) γ2 Γ Γ(3/2+ iT ) + i Z 2T So we are ﬁnally left with the estimate of

+ ∞ 1 1 + dz. z z z γ2 n=1 n n Z X − Lemma 24. There exists F > 0 such that for every s = σ + iT with σ [1, 3/2] and T 2 it holds ∈ ≥ 0 + ζ ∞ 1 1 Re K (s)

Lemma 25. 1) Let zn = αn + iβn the non trivial zeros of ζK . Then for a sufficiently large T (with T > β ) one has | 1| 1 = O(log T ); 1+(T β )2 n n X − 2 2) The sum n T βn − made over the zeros zn such that T βn 1 is O(log T );| − | | − |≥ P 3) The number of non trivial zeros of ζK such that T βn < 1 is O(log T ); | − | 4) If s = σ + iT with σ [ 1/2, 3/2] and T sufficiently large, then ∈ − 0 ζ (s) 1 K = 0 + O(log T ) ζ (s) s z K n n X − where the first sum is made over the zeros such that T β < 1. | − n| Proof. 1) From the previous lemma, if s = 3/2+ iT , the series with positive terms Re(1/(s z )) is O(log T ). But n − n + + 3 + 1 ∞ 1 P ∞ αn ∞ Re = 2 − 2 s z ( 3 α )2 +(T β )2 ≥ ( 9 +(T β ))2 n=1 n n=1 2 n n n=1 4 n X − X − − X − 62 and so the claim is proved. 2) and 3) are immediate consequences of 1). 4) If s = σ + iT with σ [ 1/2, 3/2], then ∈ − 0 0 + ζ ζ ∞ 1 1 K (s) K (3/2+ iT )= + O(log T ). ζ − ζ s z − 3/2+ iT z K K n=1 n n X − − If T β 1 then | − n|≥

1 1 3/2 σ 3/2+1/2 = − 2 s zn − 3/2+ iT zn s zn 3/2+ iT zn ≤ s zn − − | − || − | | − | and so the sum over this kind of zeros is O(log T ) by 2). If T β < 1 then 3/2+ iT z 1 and so by 3) we obtain 4). | − n| | − n|≥ We are now able to get the estimate we want (and the asymptotic formula (4.3) as a consequence):

0 ζ 1 Im K (z)dz = 0 Im dz + O(log T ) dz . γ2 ζK γ2 z zn γ2 | | Z X Z − Z The second term is again O(log T ) (being estimated in module by C log T 1/2+ iT 3/2 iT where C > 0) and the ﬁrst one is equal to ·| − − |

1/2 0 T βn 0 1 2 − 2 dx 2 = O(log T ). 3/2 (x αn) +(T βn) (T βn) X Z − − X − This implies that SK (T ) = O(log T ) and so the asymptotic formula is proved.

63 Chapter 5

Arithmetical equivalence: the algebraic approach

5.1 Decomposition type and Gassmann equivalence

5.1.1 Decomposition type of prime numbers in the rings of integers Thanks to the result obtained in the previous chapter, the equality for the 0 Class Number Formulas of the ﬁelds K and K is reduced to the even simpler equality

R h R 0 h 0 K K = K K . 0 wK wK The ﬁrst purpose of this chapter is to show that two arithmetically equivalent number ﬁelds contain the same number of roots of unity. In order to do this one can not rely on only analytic tools but also on algebraic ones, typical of Group Theory or Cohomology Theory.

Let K be a number ﬁeld. Remember that for Re s > 1 the Dedekind Zeta function related to K admits an expression as Dirichlet series:

+ ∞ m(n) ζ (s)= K ns n=1 X where m(n):=# I K : N(I)= n . { ⊂O } 0 Lemma 17 in Chapter 2 implies that, if ζK = ζK , then the two functions

64 0 m(n) and m (n) are equal for every n N: two number ﬁelds are arith- ∈ metically equivalent if and only if the number of K -ideals of ﬁxed norm n is the same for every n. O

e1 er Let p Z be a prime number, p K = 1 r its decomposition as a product∈ of prime ideals in , withO =P ···Pif i = j. Suppose the indexes OK Pi 6 Pj 6 are ordered such that, if fi = f( i p) is the inertia degree of the prime i over p, it is f f . P | P 1 ≤···≤ r We deﬁne the decomposition type of p as the r-ple f =(f1,...,fr).

If p Z is a prime number and f N we deﬁne the set ∈ ∈ (f) := prime : Z = pZ and f( p)= f Bp {P ⊂ OK P∩ P| } and we denote its cardinality as bp(f).

Theorem 32. Let K be a number ﬁeld. Then ζK is completely determined by the numbers bp(f), for every f N and for every prime number p Z, and the converse it also true. ∈ ∈

Proof. We begin by showing that the arithmetical function m(n) deﬁning the Dirichlet series expression for ζK is multiplicative (so that its values are uniquely determined by the evaluations on the powers of prime numbers). In fact, remembering that for every prime number p Z there exists only ∈ a ﬁnite number of prime ideals of K such that lies over p, and their norm is equal to pf(p) (where f(pP) is theO inertia degreeP of over p) the Euler product (3.1) implies P

+ + ∞ m(n) 1 1 1 ∞ 1 = = 1 − = ns N(I)s − N( )s N( )ks n=1 I k=0 X X⊂OK P⊂OYK P P⊂OYK X P + + ∞ 1 ∞ m(pk) = f( p)ks = ks p P| p p p k=0 p k=0 Y YP| X Y X (the last equality is true being m(pk) the function which counts the number of terms 1/pks appearing in the last product). Let us compute m(pf ), where p is a prime number and f N. Thanks to the unique factorization of ideals as product of prime ideals∈ in we have OK

65 f f m(p )=# I K : N(I)= p { ⊂O } r =# I : I = , with prime ideals, f( p)= f { ⊂OK P1 ···Pr Pj Pj| } j=1 X s =# I : I = I I , I = , with B (f ), r f = f { ⊂OK 1 ··· s i Pi1 ···Piri Pij ∈ p i i i } i=1 X = # I : I = I I , I = , with B (f ) { 1 ··· s i Pi1 ···Piri Pij ∈ p i } r1f1+ +rsfs=f ··· 1 f1X<

5.1.2 Gassmann equivalence In this section we give some group theoretic definitions which will be very useful in the study of arithmetical equivalence: it seems that at this stage a stronger analytic approach is no more possible. The work of this section is based on Robert Perlis’ article [Per]. From now on, if X is a set, we denote its cardinality #X with X . | | Let G be a finite group, H G a subgroup and C = c G a cyclic subgroup. We say that g h ⊂g in G if there exist h H,kh i ⊂N such that 1 2 ∈ ∈ 66 k g2 = hg1c . This is an equivalence relation and an equivalence class HτC is called a double coset of H and C. Let τ1,...,τn be the representatives of these double cosets, so that G = n i=1 HτiC. We define the coset type of (G,H,C) as the n-uple A = (f1,...,fn) of numbers such that HτiC = H fi and fi fi+1. The fol- ` | | | |· ≤ lowing Remark a) shows that each fi is actually an integer.

0 Now let G be a ﬁnite group and H,H be two subgroups. We say that 0 H and H are Gassmann equivalent if for every conjugacy class xG of G 0 it is H xG = H xG . | ∩ | | ∩ | Remark: a) Let Cg be the image of C under the conjugation by g. Then the cardinality of a double coset is equal to

g 1 g H C HgC = HgCg− = HC = | || | (5.1) | | | | | | H Cg | ∩ | where the ﬁrst equality holds because, if G is a group, then the right multiplication by an element g G is a bijection of G with itself. ∈ 0 0 b) If H and H are Gassmann equivalent, then H = H ; in fact, if | | | | x1,..,xk are the distinct representatives of the conjugacy classes of G, k G k G 0 k 0 G from G = i=1 xi one has H = i=1 H xi and H = i=1 H xi , so that the Gassmann equivalence implies:∩ ∩ ` ` `

k k 0 0 H = H xG = H xG = H ; | | | ∩ i | | ∩ i | | | i=1 i=1 X X c) Given x G, the theorems on the group actions (see the Appendix) imply that∈

1 G g G: gxg− H = x H C (x) |{ ∈ ∈ }| | ∩ |·| G | where C (x)= y G: yx = xy is the centralizer of x in G. G { ∈ } The following lemma is the fundamental tool for the group theoretic charac- terization of arithmetical equivalence.

0 Lemma 26. Let H and H be two subgroups of a ﬁnite group G. Then H and 0 H are Gassmann equivalent if and only if for every cyclic subgroup C G 0 the coset types of (G,H,C) and (G,H ,C) coincide. ⊂

67 Proof. We begin by observing that the condition on the coset types implies 0 that H = H : if C = 1 then the double cosets of H and C are simply | | | | { } the usual cosets of H and so the numbers fi deﬁning the coset type are all 0 0 equal to the index [G : H]. If fi = f for every i then [G : H]=[G : H ] and 0 i so H = H . In| order| to| prove| the claim, given C = c G a cyclic subgroup, we deﬁne the arithmetical function h i⊂

l (n)= g G: HgC = H n C |{ ∈ | | | |· }| = double cosets HgC : HgC = H n H n |{ | | | |· }| · | |· where the last equality holds because if g contributes to lC (n) and HgC = 0 0 Hg C, then also g contributes. We deﬁne now the arithmetical function

k (n):= l (d)= g G: HgC divides H n C C |{ ∈ | | | |· }| d n X| = g G: Cg / H Cg divides n (1) |{ ∈ | | | ∩ | }| = g G: cn g H Cg (2) |{ ∈ h i ⊂ ∩ }| n 1 n G n = g G: gc g− H = (c ) H C (c ) |{ ∈ ∈ }| (3) | ∩ |·| G | where equality (1) is true because of Remark a) and (5.1); • equality (2) follows by the fact that the order of the quotient group • g g 1 n n 1 g C /H C divides n and so (gcg− ) = gc g− H C ; ∩ ∈ ∩ equality (3) is a consequence of Remark c). • By construction we observe that, for every cyclic subgroup C G, the func- ⊂ tion lC is completely defined by the coset type (G,H,C) while the equalities above show that kC is uniquely determined by the intersections of the conjucacy classes with H. Furthermore the Möbius inversion theorem (which is Theorem 18) implies that n k (n)= µ l (d) C d C d n X| and so the two functions are mutually determined. 0 Now suppose H and H satisfy the condition on coset types: then for every 0 cyclic subgroup C G one gets l = l (this is due to the equality H = ⊂ C C | | 68 0 0 H ) and so kC = kC , and this implies that the two groups are Gassmann |equivalent.| 0 In the same way the Gassmann equivalence implies kC = k and so (by 0 C Möbius inversion) lC = lC , i.e. the condition on coset types.

5.1.3 Number fields and Galois group theory Let Q N be a finite Galois fields extension, G = Gal(N Q) the Galois group of⊂ this extension. Galois’ fundamental theorem (see Ap| pendix) gives a bijection between fields extensions Q K N and subgroups of G that inverts the inclusions. ⊂ ⊂

Lemma 27. Let Q N be a Galois extension. Then every prime number p has the factorization⊂ p =( )e ON P1 ···Pr and every prime of the factorization has the same inertia degree. Moreover, the Galois group G = Gal(N Q) acts transitively on the ’s. | Pj Proof. See Chapter 1 of [Lang1].

If G = Gal(N Q) and N is a prime ideal, the decomposition group of is the subgroup| ofP⊂OG P G := σ G: σ = P { ∈ P P} while the inertia subgroup is the subgroup of G P

I := σ G : σ = id / P { ∈ P ON P } where σ : / / is the map on the residue ﬁelds induced by σ. ON P→ON P Lemma 28. Let Q N be a Galois extension, let p be a prime number and ⊂ a prime ideal in N which lies above p, with e( p)= e and f( p)= f. PThen G = ef andO I = e and there is the shortP| exact sequenceP| | P | | P |

0 I G Gal( N / Z/pZ). → P → P → O P| If is unramiﬁed (i.e. e = 1) then G Gal( N / Z/pZ) and so it is a P cyclicP group. ' O P|

Proof. See the reference indicated in the previous lemma. We are going to prove the fundamental theorem which connects the arithmetical equivalence to the group theoretic issues introduced above.

69 Theorem 33. (Hasse): Let K be a number field and and let N be a finite Galois extension of Q such that Q K N, with Galois groups G = Gal(N Q) and H = Gal(N K). ⊂ ⊂ Let p be| a prime number which| is not ramified in and let a prime of ON P N which lies over p. Let G be the decomposition group of . P 0 OLet A be the coset type of (G,H,G ) and let A be the decompositionP type of 0 P p in K. Then A = A .

Proof. Let p K = q1 qg be the factorization of p in K , with the primes O 0 ··· 0 0 O ordered such that A =(f1,...,fg) is the decomposition type of p. Let us fix a q . The group G acts transitively on the primes of lying over i ON p, and so there exists τi G such that τi lies over qi. Being G = (by the∈ definition of decompositionP group) we have that P P P also τiG lies over qi. P The multiplicativityP of the ramification index implies that the factorization of q in is non ramified and so the group H acts transitively on the primes i ON of this factorization. Therefore the set HτiG lies over qi. P P Let σ G such that σ lies over qi. Then by the transitivity of the action of ∈ P 1 1 H there exists ρ H such that σ = ρτi : this implies that τi− ρ− σ G ∈ P P ∈ P and therefore that σ HτiG . ∈ P By the lines above we have that HτiG HτjG = if i = j and G = g P ∩ P ∅ 6 i=1 HτiG , providing the double cosets of H and G . P P 0 Now, let f = f( p) and hi = f(τi qi). Then f = hi fi . ` P| P|1 · Remembering that Gτi = τiG τi− (by the transitivity of the action of G), P P if A =(f1,..,fg) is the coset type of H and G , we finally have: P

1 H fi = HτiG = HτiG τi− | |· | P | | P | Gτ iP = HGτi = H | | | P | | |· H Gτ | ∩ iP | f 0 = H = H fi | |· hi | |· and so the claim is proved. We are now ready to connect the analytic point of view on the arithmetical equivalence (given by the Dedekind Zeta functions) to the algebraic one.

0 Theorem 34. Let K and K be two number ﬁelds, and let N be a Galois 0 extension over Q which contains both K and K and with Galois group G = Gal(N Q). The following| statements are equivalent:

70 0 1) ζK (s)= ζK (s); 2) The decomposition type of every prime number is the same in K and 0 K ;

0 0 3) The subgroups H = Gal(K Q) and H = Gal(K Q) are Gassmann equivalent. | | Proof. The equivalence between 1) and 2) has been proved in Theorem 32. To prove the equivalence between 2) and 3) we have to notice that, given a Galois extension Q N, it can be proved (by means of Cebotarev’s density theorem) that every⊂ cyclic subgroup of Gal(N Q) is a decomposition group for some prime of N (see Chapter 1 of [Kli]| for the details). Thus the equivalenceP between 2) and 3) is a direct consequence of Lemma 26 (which connects the Gassmann equivalence to the conditions on the coset types) and of Hasse’s Theorem (which connects the coset type with respect to the decomposition groups to the decomposition type of the prime ideals).

Remark: The equality of the signature (r1,r2) and for the discriminant of two arithmetically equivalent number ﬁelds can be shown as a direct consequence of the Gassmann equivalence theory; nonetheless this approach has the problem that does not focus on the analytic properties of the Dedekind Zeta function, and one can not discover the asymptotic formula for counting the non trivial zeros in this way.

5.1.4 Arithmetically equivalent number ﬁelds share same roots of unity

0 In this section it will be proved that, if two number ﬁelds K and K are arithmetically equivalent, then they have the same group µK of roots of unity (and by Dirichlet’s Theorem and the results in Chapter 4 there is a

0 group isomorphism K∗ K∗ ). We recall two usefulO deﬁnitions:'O

The Galois kernel of a number ﬁeld K is the greatest Galois extension • Q L contained in K. ⊆ Given a ﬁnite group G and a subgroup H, the smallest normal subgroup • of G containing H is

G σ H := σ GH ∪ ∈ σ 1 where H = σhσ− : h H . { ∈ } 71 0 Theorem 35. Let K and K be two arithmetically equivalent number ﬁelds, 0 Q N a Galois extension that contains both K and K . 0 Then⊆ K and K have the same Galois kernel and the same group of roots of unity.

0 0 Proof. Let G := Gal(N Q), H := Gal(N K) and H := Gal(N K ). Galois 0 theory asserts that the Galois| kernels of K| and K are the fixed| fields of the 0 groups HG and H G, and so it suffices to show that these two groups are in fact the same. 0 0 Being K and K arithmetically equivalent, the groups H and H are Gassmann equivalent. Let h H: being h1G = h H we have ∈ ∈ 0 hG H = hG H 1 | ∩ | | ∩ |≥ 0 and thus there exists σ G such that hσ H . The reasoning works for 0 0 0 every h H, so that HG∈ H G, and it can∈ be reversed for every h H , 0 so that H∈G = H G. ⊆ ∈ Now, let µK = roots of unity in K . The field extension Q Q(µK ) is 0 Galois and contained{ in the Galois kernel} of K. But K and K⊆ have the 0 same Galois kernel, so that µK K . ⊂ 0 By taking the opposite inclusions with the set µK we finally prove the claim on the roots of unity. Remark: With the notations introduced in the previous theorems, it can 0 be shown that K and K are arithmetically equivalent number fields if and 0 only if HG = H G (for the proof see Chapter 2 of [Kli]).

We conclude this section observing that Theorem 35 and the previous remark imply two important facts:

If Q L is a Galois ﬁeld extension, the Galois sub-extensions Q K • of L ⊆correspond uniquely to the normal subgroups N of the group⊆ G = Gal(L Q). But for a normal| subgroup one has N G = N and this implies that two Galois extensions of Q are arithmetically equivalent if and only if they are isomorphic.

0 Theorem 35 shows that ωK = ωK for two arithmetically equivalent • number ﬁelds, and so the Class Number equality (4.1) can be reduced to h R = h 0 R 0 . (5.2) K · K K · K

72 5.2 Arithmetically equivalent ﬁelds which are not isomorphic

5.2.1 Cohomological tools

0 Are there two number ﬁelds K and K arithmetically equivalent but which are not isomorphic? The results of the previous section tell us that ﬁelds of this kind must not be Galois extensions of Q. Moreover, Perlis in his article [Per] proved the following:

0 Theorem 36. Let K and K two number ﬁelds of degree 6. Then K and 0 K are arithmetically equivalent if and only if they are isomorphic.≤

Nonetheless, the answer to the previous question is yes: this kind of ﬁelds do exists, and it is the aim of this section to provide a way to construct some example of arithmetically equivalent and not isomorphic number ﬁelds.

Remark: It is possible to construct arithmetically equivalent number fields which not only are not isomorphic but also have different class number (Perlis and De Smith provided some example in [DesPer]). This fact imply that the simplified Class Number equality (5.2) cannot be reduced further.

Let us begin with some deﬁnitions. Given a group H and an abelian group A, we say that A is an H-module if there is a group morphism H Aut(A) → h φ → h h and for every a A we write φh(a) as a . Observe that this deﬁnition induces an action of H ∈on A. Given an H-module A, a cocycle is a function χ : H A such that → χ(gh)= χ(g)h χ(h) · for every g,h H. The set of cocycles is denoted as Z1(H,A) and is an abelian group.∈ A function ψ : H A is said to be a coboundary if there exists a A such that → ∈

h 1 ψ(h)= a a− . · 73 Every coboundary is a cocycle and the set B1(H,A) of cocycles is an abelian subgroup of Z1(H,A). The fundamental tool in this discussion will be the quotient group

Z1(H,A) H1(H,A) := . B1(H,A)

Remark: The quotient group deﬁned above rises in a natural way as the ﬁrst group of cohomology of the H-module A, and so one can study its properties by means of the derived functor of HomH (Z, ) (where the structure of Z as an H-module is given by the identity action− of H on Z). However, in this setting we are not interested in a so generic view: more details on this kind of treatment can be found in Chapter 9 of [Rot].

Suppose now G is a ﬁnite group, H G a subgroup and A G an abelian normal subgroup. The group G is the⊂ semidirect product⊂ of H and A if there exists a short exact sequence of groups

1 Ai Gk H 1 and a morphism σ : H G, called section, such that k σ = idH . In this case there is an→ action of H on A: for every h H◦and a A we put h 1 ∈ ∈1 a := g− ag where g is a generic element of the counterimage k− (h). This action is well deﬁned and endows A of a structure of H-module. Given two sections σ and τ we say that σ τ if there exists an a A ∼ 1 ∈ such that for every h H one has σ(h) = aτ(h)a− . Let M be the set of equivalence classes of the∈ sections. We can deﬁne an action of the group H1(H,A) on M:

H1(H,A) M M × → ([χ], [σ]) [χ σ] → · where (χ σ)(h)= χ(h)σ(h). · Lemma 29. The action written above is well deﬁned, free and transitive.

Proof. Well deﬁned: If two sections σ and τ are equivalent, then there• exists a A such that ∈ 1 1 (χ σ)(h)= χ(h)a− τ(h)a = a− χ(h)τ(h)a · where the last equality holds because every cocycle χ has values in the abelian group A, and so χ σ χ τ. · ∼ · 74 1 h 1 If ψ [χ] there exists a A such that χ(h)ψ(h)− = a a− = ∈ 1 1 ∈ σ(h)aσ(h)− a− for every h H and for every σ M. Then ∈ ∈

(χ σ)(h)a = χ(h)σ(h)= ψ(h)aσ(h)= aψ(h)σ(h)= a(ψ σ)(h) · · and so [χ σ]=[ψ σ]. · · The action is free: infactlet[χ] H1(H,A) such that [σ]=[χ σ] for • every [σ] M. Then there exists a∈ A such that a(χ σ)(h)= χ· (h)a, and so ∈ ∈ · 1 1 1 h χ(h)= a− σ(h)aσ(h)− = a− a i.e. the function χ is a coboundary and so [χ]=1.

The action is transitive: let [σ], [τ] M be two classes of sections. • ∈ 1 Let us take the function χ : H A such that χ(h)= τ(h)σ(h)− . → 1 This function is well deﬁned because k(χ(h)) = k(τ(h)σ(h)− )= k(τ(h)) 1 · k(σ(h)− )=1 1=1. It is enough to· prove that χ is a cocycle, because then [τ]=[χ σ]. In fact, one has ·

1 χ(hg)= τ(hg)σ− (hg) g g 1 = τ(h) τ(g)(σ(h) σ(g))− g 1 g 1 = τ(h) τ(g)σ(g)− (σ(h) )− 1 1 1 1 = τ(g)τ(h)τ(g)− τ(g)σ(g)− (σ(g)σ(h)σ(g)− )− 1 1 = τ(g)τ(h)σ(h)− σ(g)− 1 g 1 =(τ(h)σ(h)− ) τ(g)σ(g)− = χ(h)gχ(g).

For every h H, the H-module A is also a h -module (the action is just the restriction of∈ the action of H to the subgrouph i h ). Consider the function h i Z1(H,A) Z1( h ,A) → h i

χ χ h . → |h i It is a well deﬁned function (the restriction of a cocycle remains a cocycle) which sends coboundaries in coboundaries, and so it induces a group morphism H1(H,A) H1( h ,A) → h i 75 [χ] [χ]h := [χ h ]. → |h i Collecting these morphisms for every h H we get the group morphism ∈ ρ : H1(H,A) H1( h ,A) A → h i h H Y∈ χ ([χ] ) . → h h Theorem 37. Let G, H and A as before. If the morphism ρA is not injective, then there exist two sections σ ,σ : H G such that 1 2 → 1) [σ ] =[σ ]; 1 6 2 2) [σ ] =[σ ] for every h H; 1 h 2 h ∈ 3) The subgroups σ1(H) and σ2(H) are Gassmann equivalent but not conjugate.

Proof. Let [χ] ker ρA be a non trivial class in the kernel. Let [σ1] M and [σ ]=[χ σ ]. ∈ ∈ 2 · 1 1) In the previous lemma we proved that the action of H1(H,A) on M is free: then, being [χ] = [1], we have [σ ] =[σ ]. 6 1 6 2 2) The restriction [χ ] to every h H is the trivial class because [χ] h ∈ ∈ ker ρA. But the morphism ρA commutes with the action, and so [σ ] =[χ σ ] =[σ ] . 2 h h · 1 h 1 h G 3) Let H1 = σ1(H) and H2 = σ2(H). If x is a conjucacy class of G such G G that x H1 = x H2 = , then the two sets have clearly the same cardinality.∩ ∩ ∅ G Suppose that there exists h H such that σ1(h) σ1(H) x . Then it holds ∈ ∈ ∩ (σ k)(σ (h)) = σ (k σ )(h)= σ id(h)= σ (h) H . 2 ◦ 1 2 ◦ ◦ 1 2 ◦ 2 ∈ 2

The previous point of the theorem implies that for every h H there 1 ∈ G exists a A such that σ2(h) = aσ1(h)a− , but being σ1(h) x and ∈ G ∈ a A subgroup of G then σ2(h) x . ∈ ∈ G Thus the function σ2 k is a well deﬁned function from σ1(H) x to G ◦ ∩ G σ2(H) x with inverse function equal to σ1 k, and so H1 x = ∩G ◦ | ∩ | H2 x . |The∩ two| groups are Gassmann equivalent but not conjugate (because the two sections σ1 and σ2 are not equivalent).

76 5.2.2 An explicit example Let us now give an example of arithmetically equivalent number fields which are not isomorphic. Let θ := √8 15 and K = Q(θ), which is a number field of degree 8 over − Q. Let ζ8 be the primitive eighth root of unity and L = Q(ζ8) = Q(i, √2), cyclotomic field which is a Galois extension of Q with degree 4. Let N = K L be the smallest field containg both K and L: the extension Q N is Galois· because N is the splitting field of the polynomial x8 + 15. ⊂ N

K L

Q Let G = Gal(N Q),H = Gal(L Q) and A = Gal(N K). | | | The group A is isomorphic to the cyclic group µ8 of the eighth roots of unity, where to every morphism a A is uniquely associated the root a(θ)/θ (the isomorphism works because ∈N L[X]/(X8 + 15)). Being L K = Q there is the group' isomorphism Gal(N K) H, given by the restriction∩ to L of the elements σ of Gal(N K) (this| is an' isomorphism because the condition on the intersection assures| that every element in A is uniquely determined up to a unity in L). Moreover H (Z/2Z) (Z/2Z) and an explicit presentation of this group is ' ×

H id, h ,h ,h '{ 1 2 3} 2j+1 where hj(ζ8)= ζ8 . Let us now ﬁnd a non trivial element in the kernel of ρA. Consider the map

χ : H A → σ(√2) σ . → √2 This is a cocycle which is not a coboundary, while its restrictions are. In

77 1 fact, being √2= ζ8 + ζ8− one gets

χ(id)=1 3 3 ζ8 + ζ8− χ(h1)= χ(h2)= = 1 √2 − 1 ζ8 + ζ8− χ(h3)= =1. √2 Finally, let σ : H Gal(N K) be a section. By definition the fixed field of σ(H) is equal to K→= Q(θ).| Let us now consider the section τ := χ σ: we know by the previous discussion that it is not equivalent to σ, and· so τ(H) is not Gassmann equivalent to σ(H); our goal is to determine the fixed field of τ(H), which will be arithmetically equivalent but not isomorphic to K. Since for every non trivial h H it is h(ζ )= ζj for some odd j, we have ∈ 8 8

τ(h)(√2θ)= χ(h) σ(h)(√2) σ(h)(θ) · · j j ζ8 + ζ8− j j = 1 (ζ8 + ζ8− )θ ζ8 + ζ8− · j j 2 (ζ + ζ− ) = 8 8 θ √2 2 = θ = √2θ √2 and so the ﬁxed ﬁeld of τ(H) is equal to Q(√2θ) = Q(√8 240). This is arithmetically equivalent to Q(√8 15) but they are not isomorphic− by construction. −

0 Remark: The number ﬁelds K = Q(√8 15) and K = Q(√8 240) are arithmetically equivalent, and so they have same− degree, same num−ber of real and complex embeddings, same discriminant and same number of roots of unity. However, they have diﬀerent class numbers: Perlis and De Smit proved in 0 [DesPer] that hK =2hK . This implies that there is no way to simplify the reduced Class Number equality (5.2).

78 Chapter 6

Appendix

6.1 Summation formulas

In this section we provide two useful formulas that compute the values of ﬁnite sums of complex numbers.

Theorem 38. Let an n and bn n be two sequences of complex numbers. Deﬁne the function { } { }

n x an x 1 A(x) := ≤ ≥ 0 x< 1. (P Then for every integer N 1 it holds: ≥ N N 1 − a b = A(N) b A(n) (b b ). (6.1) n · n · N − · n+1 − n n=1 n=1 X X Theorem 39. Let an n be a sequence of complex numbers and f : [1,N] C be a C1 function{ (where} N 1 is integer). Then → ≥ N N 0 a f(n)= A(N) f(N) A(x) f (x)dx (6.2) n · · − · n=1 1 X Z Theorem 40. (Euler-McLaurin): Let f : [1,N] C be a C1-function. Then it holds: →

N N N 1 0 1 f(n)= f(x)dx+ (f(1)+f(N))+ f (x) x [x] dx (6.3) 2 · − − 2 n=1 1 1 X Z Z where [x] is the integer part of x. All these formulas are proved in [IwaKow].

79 6.2 Properties of the Gamma function

The Gamma function is a complex function deﬁned, for Re s> 0, as

+ ∞ x s dx Γ(s) := e− x . x Z0 In this half-plane the integral deﬁning Γ(s) is absolutely convergent and therefore (by Lebesgue’s Dominated Convergence Theorem) Γ(s) is holomorphic in such a region. Applying integration by parts and changing the variables it is easy to show that

Γ(s +1) = s Γ(s) Re s> 0 · ∀ Γ(n)=(n 1)! n N, n 1 − ∀ ∈ ≥ Γ(1/2) = √π The Gamma function can be uniquely extended to a meromorphic function with simple poles in the non positive integers: we have the analytic extension

+ e γs ∞ s 1 Γ(s)= − 1+ − ens s n n=1 n Y where γ := limn + ( 1/k log n) is the Euler-Mascheroni number. → ∞ k=1 By means of this writing (and remembering− the expression of the sin function as an inﬁnite product)P one can prove the following fundamental identities of Gamma function: π Γ(s)Γ(1 s)= − sin(πs)

1 1 2s Γ(s)Γ s + =2 − √π Γ(2s) 2 · Moreover, for Arg s = π and s + one can get the Stirling asymptotic formula for the6 − Gamma| | →function∞ : 1 1 1 Log Γ(s)= s Log s s + log(2π)+ O (6.4) − 2 − 2 s All the proofs for the statements above can be found in [Art] and [Jam].

80 6.3 Group actions

Let Ω be a set and G be a group. We say that there is an action of G on Ω (or that G acts on Ω) if there is a function

G Ω Ω × → (g,x) g x → · such that, For every g,h G and for every x Ω, the following two properties hold: ∈ ∈

a) (gh) x = g (h x); · · · b) 1 x = x. G · For every x Ω the set G x := g x: g G is called the orbit of G; two orbits are either∈ the same· or are{ disjoint.· ∈ } For every x Ω the group Gx := g G: g x = x is called the stabilizer of x. It can∈ be proved that, if G {is a∈ ﬁnite group,· } then

#G = (#G x)#G . · x An example of group action is the conjugation: if G is a group, the function

G Ω Ω × → 1 (g,h) g h := g− hg → · provides an action of the group G on itself, with orbits the congjucacy classes G 1 h := g− hg : g G . An action{ is said∈ to be} free if the only element g G such that g x = x for ∈ · every x Ω is 1G. An action∈ is said to be transitive if for every x,y Ω there exists g G such that y = g x (i.e. there is only one orbit). ∈ ∈ · Lemma 30. Suppose that a group G acts transitively on a set Ω. Then the stabilizers Gx are conjugated subgroups. Further details on these topics can be found in any good book on Group Theory (see for example [Rob]).

81 6.4 Basic Galois Theory

Let K F be a field extension of finite degree (we are not interested on the characteristic⊆ of the fields in this setting). The extension is separable if the minimum polynomial of every element of F is separable (meaning that its irreducible factors have no multiple roots in the algebraic closure of F ). The extension is said to be normal if F is the splitting field of the minimum polynomials of every element of F .

Let G be a subgroup of the group AutF of the field automorphisms of F . The set F G := x F : σx = x, σ G { ∈ ∀ ∈ } is a subfield of F denoted as the fixed field of G. Lemma 31. (Artin): Let G be a finite group of automorphisms for a field F , and let K = F G. Then [F : K] G . ≤| | Let K F be a finite degree field extension. The group of K-automorphisms of F is⊆ the group of field automorphisms of F

Aut(F K) := σ : F F : σ(x)= x x K . | { → ∀ ∈ } The previous lemma implies that (in this setting) this group is ﬁnite.

Aut(F K) A finite degree field extension K F is said to be Galois if K = F | , and in this case the group of K-automorphisms⊆ of F is denoted as Gal(F K). | Theorem 41. Let K F be a field extension. Then the following statements are equivalent: ⊆ 1) F is the splitting field of a separable polynomial f K[T ]; ∈ 2) K = F G for some finite group G of automorphisms of F ;

3) F is normal, separable and of ﬁnite degree over K;

4) F is Galois over K. Theorem 42. (Fundamental Theorem of Galois Theory): Let K F be a Galois extension of ﬁnite degree, with G = Gal(F K). The maps⊆ H F H and M Gal(E/M) are inverse bijections between| the sets: → → subgroups of G and intermediate ﬁelds K M F . { } { ⊆ ⊆ } Furthermore:

82 a) The correspondence is inclusion-reversing (meaning H H if and 1 ⊂ 2 only if EH2 EH1 ); ⊂ H1 H2 b) [H2 : H1]=[E : E ] (the ﬁrst one being a group index and the second one the degree of the extension);

c) H is a normal subgroup of G if and only if the ﬁeld extension K F H is Galois. ⊆

All the details for the proofs are in Chapter 3 of [Mil].

83 Bibliography

[Ahl] Ahlfors L., Complex Analysis, Third Edition, McGraw-Hill Science (1979)

[Art] Artin E. The Gamma Function, Athena Series, Selected Topics in Mathematics, Holt, Rinehart and Winston (1964).

[AtyMac] Atiyah M. F., MacDonald I. G., Introduction to Commutative Al- gebra, Addison-Wesley Series in Mathematics (1994).

[Dav] Davenport H., Multiplicative Number Theory, Springer New York (1980).

[DesPer] De Smit B., Perlis R., Zeta Functions do Not Determine Class Numbers, Bulletin of the American Mathematical Society, Volume 31, Number 2, pg. 213-215 (October 1994).

[Fr¨oTay] Fr¨ohlich A., Taylor M.J., Algebraic Number Theory, Cambridge University Press, Cambridge Studies in Advanced Mathematics (1992).

[Had] Hadamard J., Sur la distribution des z`eros de la function ζ(s) et ses cons`equences arithmetiques, Oeuvres de Jacques Hadamard, C.N.R.S., Paris, vol 1, pg. 189–210 (1968).

[Iwa] Iwaniec H., Lectures On The Riemann Zeta Function, American Math- ematical Society, University Lectures (2014).

[IwaKow] Iwaniec H., Kowalski E., Analytic Number Theory, Colloquium Pubblications, Vol. 53, American Mathematical Society (2004).

[Jam] Jameson G.J.O., A simple proof of Stirling’s Formula for the Gamma Function, Math. Gazette 99 (2015), pg. 68–74

[Jan] Janusz G.J., Algebraic Number Fields, Academic Press, Pure and Ap- plied Mathematics (1973).

84 [Jar] Jarvis F., Algebraic Number Theory, Springer International Publishing, Springer Undergraduate Mathematics Series (1994). [Kat] Katznelson Y., An Introduction to Harmonic Analysis, Cambridge University Press (1968). [Kli] Klingen N., Arithmetical similarities: Prime Decomposition and Finite Group Theory, Oxford University Press (1998). [Lang1] Lang S., Algebraic Number Theory, Springer, Graduate Text in Mathematics (1994). [Lang2] Lang S., Linear Algebra, Third Edition, Springer, Undergraduate Text in Mathematics (1987). [Lang3] Lang S., Complex Analysis, Springer, Graduate Text in Mathematics (2003). [Titch] Titchmarsh E.C., Heath-Brown D.R., The Theory of The Riemann Zeta Function, Oxford University Press, Oxford Science Publications (1986). [Mil] Milne J.S., Fields and Galois Theory, notes available from www. jmilne.org/math [Pel] Peloso M., Notes for the course “Harmonic Analysis”, available on http://users.mat.unimi.it/users/peloso/

0 [Per] Perlis R., On the Equation ζK (s)= ζK (s), Journal of Number Theory, 9, (1977). [Rie] Riemann B., Ueber die anzhal der Primezahlen unter einer gegebenen Grösse, Monatsberichte der Berliner Akademie (1859). [Rot] Rotman J.J., An Introduction to Homological Algebra, 2nd edition, Springer (1994). [Rob] Robinson D.J.S., A Course in The Theory of Groups, Second Edition, Springer-Verlag, Graduate Texts in Mathematics (1995). [Scho] Schoof R., Algebraic Number Theory, notes available from www.mat. uniroma2.it/~eal/tn2010.html (2003). [Val] de la ValléèPoussin, C.J., Recherches analytiques sur la théorie des nombres (premiére partie), Ann. Soc. Sci. Bruxelles [I], 20, pg. 183-256 (1896).

85 Ringraziamenti

Scrivere questa tesi èstato un lavoro allo stesso tempo istruttivo ed entusi- asmante, il culmine di 5 intensi anni universitari. Desidero innanzitutto ringraziare il Professor Giuseppe Molteni per avere accettato di supervisionare il mio lavoro, per la sua disponibilitànel chiarire ogni mio dubbio, per l’interesse e la curiositàmostrati in ogni fase della stesura e per la fiducia accordatami, oltre che per tutto il sostegno fornito alla mia candidatura al Dottorato di Ricerca.

L’impostazione di questa tesi (specialmente del primo capitolo) deve molto anche ai professori Fabrizio Andreatta e Marco Seveso, che con le loro lezioni di Algebra Commutativa e Teoria dei Numeri mi hanno insegnato, fra le altre cose, cosa signiﬁchi “alzare l’asticella” in Matematica. Ringrazio poi i professori Luca Barbieri Viale e Mariagrazia Bianchi per avermi scelto come tutor per il corso di Algebra 1, dandomi la possibilit`adi capire come “abbassare l’asticella” per introdurre i profani al mondo delle strutture algebriche. Un grazie anche agli studenti di Algebra 1 che hanno accettato di spendere il loro tempo seguendo le mie lezioni; spero di essere stato per voi un valido aiuto.

Ringrazio di cuore Edo, Emi, Andre, Filo e Davide, che come me hanno lottato con la tesi in questi ultimi mesi da studente; un grazie anche a Ema, Riccardo, Alberto e Iacopo per avere ascoltato pazientemente i progressi del mio lavoro. Ringrazio inoltre le persone incontrate in questi anni vissuti fra i banchi e il cortile di Via Saldini 50: con voi ho condiviso studi e domande in biblioteca, seminari, ripassi alla lavagna, partite a calcetto, Fifa, briscola e Bang, aper- itivi e pizze gozze, camminate, stanchezze, treni in ritardo e gioie. Vi sar`o sempre grato per tutto questo.

Ringrazio Walter Macchi per avermi fatto scoprire cosa sia veramente la Matematica, per essere stato il mio tutore durante il Tirocinio Didattico

86 e per avermi insegnato che, parafrasando Jos`eMourinho, “chi sa solo di Matematica non sa niente di Matematica”.

Non sarei qui senza tutto quello che ho vissuto (sia durante gli anni del liceo che dopo) con la mitica sezione A del Galilei di Caravaggio: ringrazio perci`oi miei ex compagni Alessandro, Pippo, Giulia, Fiﬁ, Nico, Alessia, Bea, Gianluca, Fra, Oldo, Robu e Teo.

Ulteriori ringraziamenti vanno anche a:

Lorenzo, Andrea e Fabiola, amici e compagni musicisti nei Gulliver. • Gabri, che da piùdi un decennio èper me un punto fermo. • I ragazzi del calcetto di Caravaggio, per avermi dato quasi ogni setti- • mana la possibilitàdi non pensare alla Matematica.

Giachi, Tia, Fil, Ari, Dario e Sof per avermi sempre supportato, chi da • vicino e chi da lontano.

Giulia, per la splendida cambusa a Parma e le passeggiate a Milano. • Gianandrea, che èla roccia con cui mi posso confidare. • I miei cugini Linda, Luca, Alberto e Lorenzo, mio nonno Gino e tutti • i miei parenti sparsi fra la bergamasca e Roma, per l’affetto che mi hanno donato.

Infine ringrazio mio padre Giuseppe, mia madre Katia, mia sorella Maria e mio fratello Paolo per il loro amore e per essermi sempre stati vicini in questi 5 anni, soprattutto nei momenti piùdifficili.