René Schoof, Number Theory Rome, Spring 2003

1. Commutative algebra. In this first section we recall some basic facts from commutative algebra. All rings are supposed to be commutative and have a unit element 1.

Proposition 1.1. Let R be a ring.

(i) If R is not the zero ring, it admits a maximal ideal. (ii) An element x ∈ R is a unit if and only if it is not contained in any maximal ideal

of R.

Proof. (i) Consider the set Ω of all ideals I ⊂ R that are not equal to R itself. Then Ω is partially ordered by inclusion. Every chain {I_{α} : α ∈ A} has the upper bound ∪_{α∈A}I_{α}. It is also contained in Ω since it does not contain 1. By Zorn’s Lemma there is therefore an element 𝔪 ∈ Ω that is maximal for the inclusion relation. This means precisely that it is a maximal ideal of R. (ii) If the ring R/(x) is the zero-ring, then 1 ∈ (x) and x is a unit. If not, then part (i) implies that it possesses a maximal ideal. This ideal is of the form 𝔪/(x) for some maximal R-ideal 𝔪 containing x.

Proposition 1.2. Let R be a ring. Then the nilradical

Nil(R) = {z ∈ R : z^{n} = 0 for some n ≥ 1}

of R is equal to the intersection of all prime ideals of R.

Proof. It is clear that Nil(R) ⊂ 𝔭 for every prime ideal 𝔭 ⊂ R. Conversely, let z ∉ Nil(R) and consider

Ω = {I ⊂ R : z^{n} ∉ I for any n ≥ 1}.

Since {0} ∈ Ω, this is a non-empty set, partially ordered by inclusion. Since every chain has an upper bound in Ω, Zorn’s Lemma applies and there is an ideal I ⊂ Ω that is maximal with respect to the inclusion. We claim that I is prime. Indeed, if not, then let x, y ∉ I while xy ∈ I. Since I + (x) and I + (y) are strictly larger than I, they contain each a power of z. But then so does I since it contains (I + (x))(I + (y)). Contradiction. Therefore I is prime. Since z ∉ I, the proof of the proposition is complete.

Proposition 1.3. (Chinese Remainder Theorem). Let R be a ring and let I, J ⊂ R be two ideals for which I + J = R. Then the natural map

R/IJ ⟶ R/I × R/J

given by (x (mod IJ)) ↦ (x (mod I), x (mod J)) is a well-defined isomorphism of Ralgebras.

Proof. The map is a well-defined R-algebra homomorphism. Let λ ∈ I and μ ∈ J such that λ + μ = 1. Let x, y ∈ R. Since the element z = μx + λy ∈ R has the property that z ≡ μx = (1 − λ)x ≡ x (mod I) and z ≡ λy = (1 − μ)y ≡ y (mod J), the image of (z (mod IJ)) is (x (mod I), x (mod J)). It follows that the map is surjective.

The kernel of the map R ⟶ R/I × R/J given by x ↦ (x (mod I), y (mod J)) is equal to I ∩ J. We have that IJ ⊂ I ∩ J. To see that equality holds, let x ∈ I ∩ J. Then x = x(λ + μ) is also contained in IJ. This proves the proposition.

Definitions. Let R be a ring. A module M is said to be generated by a subset S ⊂ M, if any m ∈ M is of the form λ_{1}m_{1} + … + λ_{t}m_{t} for some elements m_{1}, …, m_{t} ∈ S and λ_{1}, …, λ_{t} ∈ R. We write M = ∑_{m∈S} mR. If S can be taken finite, then M is said to be finitely generated. If M can be generated by one element m, it is denoted by mR. In particular, the principal ideal generated by x ∈ R is denoted by xR.

An R-module is called free, if it is of the form ⊕_{s∈Σ}R for some index set Σ. A finitely generated free module is of the form R^{n} for some n ≥ 0. A projective R-module is a direct summand of a free module. Free modules are themselves projective. Since a free module P has the property that every exact sequence of the form

0 ⟶ N ⟶ M ⟶ P ⟶ 0

is split, the same is true when P is merely projective.

An element m in an R-module M is called torsion if there is a non-zero λ ∈ R for which λm = 0. An R-module all whose elements are torsion is called a torsion module. An R-module none of whose non-zero elements are torsion is called torsion-free. Since free modules are torsion-free, so are projective modules.

Definition. Let R be a domain with quotient field F. The rank of an R-module M is defined as rank(M) = dim_{F} (M ⊗_{R} F). If M is finitely generated, we have that rank(M) = dim_{F} Hom_{R}(M, F). Ranks are additive. If N is a submodule or quotient module of M, then rank(N) ≤ rank(M).

Definition. A Noetherian ring is a ring all of whose ideals are finitely generated.

Proposition 1.4. Let R be a ring. The following are equivalent.

(i) R is Noetherian. (ii) Every chain of R-ideals I_{1} ⊂ I_{2} ⊂ … stabilizes. (iii) Every non-empty of ideals Ω possesses an element that is maximal for the inclusion

ordering.

Proof. Let I_{1} ⊂ I_{2} ⊂ … be a chain of ideals of R. The union is finitely generated and a finite set of generators is contained in I_{n} for some n ≥ 1. Clearly I_{k} = I_{n} for all k ≥ n so that the chain stabilizes. This shows that (i) implies (ii). Let Ω be a non-empty collection of R-ideals. If it does not contain a maximal element, then we can choose a infinite chain

I_{1} ⊂_{≠} I_{2} ⊂_{≠} I_{3} …,

contradicting (ii). Therefore (ii) implies (iii). Finally, let I be an R-ideal and consider the set of ideals {J ⊂ I : J is finitely generated}. If (iii) holds, there is a maximal element J_{0} ∈ Ω. Then J_{0} = I so that I itself is finitely generated. Indeed, if not, then pick x ∈ I − J_{0} and consider the ideal generated by J_{0} and x. It is an element of Ω that is strictly larger than J_{0}. Contradiction. Therefore (i) holds.

This proves the proposition.

Proposition 1.5. Let R be a Noetherian ring. Then

(i) For any ideal I ⊂ R, the quotient ring R/I is Noetherian. (ii) Let M be a finitely generated R-module. Then any submodule N ⊂ M is also finitely

generated.

Proof. (i) Indeed, any ideal of R/I is of the form J/I where J ⊂ R is an ideal containing I. Since R is Noetherian, J is finitely generated. The same generators generate the R/I ideal J modulo I. (ii) We proceed with induction with respect to the number of generators of M. If M is generated by one element, then M ≅ R/I for some ideal I and we are done by (i). If M is generated by m_{1}, …, m_{t} for some t ≥ 2, then we have the following commutative diagram with exact rows.

0 ⟶ N ∩ Rm_{1} ⟶ N ⟶ N/(N ∩ Rm_{1}) ⟶ 0
↓^{⊂}  ↓^{⊂}  ↓^{⊂}
0 ⟶  Rm_{1}  ⟶ M ⟶  M/Rm_{1}  ⟶ 0

Since R is Noetherian, the module N ∩Rm_{1} admits finitely many generators. By induction, the submodule N/(N ∩ Rm_{1}) of M/Rm_{1} is also finitely generated. The set of generators of N ∩Rm_{1} together with any lift of a finite set of generators of N/(N ∩Rm_{1}), generate N. This proves the proposition.

Proposition 1.6. (Hilbert Basissatz) Let R be a Noetherian ring. Then the polynomial ring R[X] is also Noetherian.

Proof. Let I ⊂ R[X] be an ideal. For n ≥ 1 consider the R-ideals

J_{n} = {leading coefficients of f ∈ I with deg f ≤ n}

We have that

J_{1} ⊂ J_{2} ⊂ J_{3} ⊂ …

Since R is Noetherian, this sequence stabilizes at J_{n_{0}}, say. Consider the R-module M = {f ∈ I : deg f ≤ n_{0}}. Since M is a submodule of the finitely generated free R-module of all polynomials of degree n_{0}, Prop. 1.5 implies that it is itself finitely generated over R. Let f_{1}, …, f_{t} be generators. Note that their leading coefficients a_{1}, …, a_{t} generate J_{n_{0}}.

We claim that f_{1}, …, f_{t} generate the R[X]-ideal I. Indeed, let φ(X) ∈ I. If deg φ ≤ n_{0}, then φ ∈ M and φ is even an R-linear (rather than R[X]-linear) combination of f_{1}, …, f_{t}. If deg φ = n > n_{0}, then the leading coefficient a of φ is contained in J_{n} and hence in J_{n_{0}}. This means that a = λ_{1}a_{1} + … + λ_{t}a_{t} for certain λ_{1}, …, λ_{t} ∈ R. Consider now the polynomial ψ(X) = φ(X) − ∑_{i=1}^{t} λ_{i}X^{n−deg f_{i}}f_{i}(X). Then ψ ∈ I and its degree is smaller than n because its n-th degree coefficient is equal to a − ∑_{i=1}^{t} λ_{i}a_{i} = 0.

The proof of the proposition is now completed with respect to induction of the degree of φ.

Corollary 1.7. For any n ≥ 0 and any ideal I of Z[X_{1}, …, X_{n}] the ring Z[X_{1}, …, X_{n}]/I is Noetherian. For any field K, any n ≥ 0 and and any ideal J ⊂ K[X_{1}, …, X_{n}], the ring K[X_{1}, …, X_{n}]/J is Noetherian.

Proof. This follows from Proposition 1.5 and Theorem 1.6.

Definition. Let R be a ring and let M be an R-module. Then M is called faithful if the natural homomorphism R ⟶ End(M) that maps x ∈ R to the multiplication by x map, is injective.

Definition. Let R ⊂ S be an inclusion of rings. An element x ∈ S is called integral over R if there exists a monic polynomial f ∈ R[X] with f(x) = 0.

Proposition 1.8. Let R ⊂ S be an inclusion of rings and let x ∈ S. The following are equivalent.

(i) x is integral over R. (ii) The subring R[x] of S is finitely generated as an R-module. (iii) There exists a finitely generated faithful R-submodule M ⊂ S for which xM ⊂ M.

Proof. If x is zero of a monic polynomial in R[X] of degree n, the ring R[x] is generated as an R-module by the elements 1, x, …, x^{n−1}. This shows that (i) implies (ii). Since R[x] contains 1, it is a faithful R-module. Therefore (ii) implies (iii). To show that (iii) implies (i), let e_{1}, e_{2}, …, e_{t} denote generators of M as an R-module. Since xM = {xm : m ∈ M} is contained in M, there exist for every i = 1, 2, …, t coefficients λ_{i1}, λ_{i2}, …, λ_{it} ∈ R such that

xe_{i} = ∑_{j=1}^{t} λ_{ij}e_{j}.

It follows that det(λ_{ij} − xδ_{ij})e_{i} = 0 for all i. Here δ_{ij} denotes the Kronecker δ-function: δ_{ij} = 1 when i = j and is 0 otherwise. As a consequence det(λ_{ij} − xδ_{ij}) kills every element of M. Since M is faithful it follows that det(λ_{ij} − xδ_{ij}) is zero and hence that the monic polynomial det(λ_{ij} − Xδ_{ij}) ∈ R[X] has x as a zero. This proves the proposition.

Corollary 1.9. Let R ⊂ S be an inclusion of rings. The elements in S that are integral over R form a subring of S that contains R.

Proof. Since any a ∈ R is zero of the polynomial X −a ∈ R[X], the last statement is clear. We need to show that for any two integral elements x, y ∈ S, both sum and product are integral as well. The subrings R[x] and R[y] are generated as R-modules by 1, x, …, x^{n−1} and 1, y, …, y^{m−1} respectively. It follows that the subring R[x, y] is generated as an Rmodule by the monomials x^{i}y^{j} with 0 ≤ i ≤ n − 1 and 0 ≤ j ≤ m − 1. Since x + y and xy are contained in R[x, y], the result now follows from the previous proposition.

Definition. A domain R is called integrally closed or normal if every element in its quotient field that is integral over R, is already contained in R.

Proposition 1.10. Principal ideal domains are integrally closed.

Proof. Any element x in its quotient field is of the form x = u/v with coprime u, v ∈ R, i.e. with u, v such that uR + vR = R. Suppose f(X) = X^{n} + … + a_{1}X + a_{0} ∈ R[X] has x as a zero. Then

u^{n} + … + a_{1}uv^{n−1} + a_{0}v^{n} = 0.

If v is not a unit, then its is contained in some maximal ideal 𝔪 = πR of R. It follows that v^{n} and hence v are in πR. This contradicts that fact that uR + vR = R. therefore v is a unit and x is integral, as required.

Definition. The Krull dimension dim R of a domain R is defined as the supremum of the n ≥ 0 for which there exists a chain of prime ideals

(0) = 𝔭_{0} ⊂_{≠} 𝔭_{1} ⊂_{≠} … ⊂_{≠} 𝔭_{n} ⊂_{≠} R

of R.

Proposition 1.11. Let R be a domain. Then

(i) dim R = 0 if and only if R is a field. (ii) dim R ≤ 1 if R is a principal ideal domain.

Proof. If R is a field, the only prime ideal is its zero ideal. Conversely, a domain that does not contain any non-zero prime ideals is a field. This follows from Prop. 1.1. This proves (i). To prove (ii), let R be a principal ideal domain and let 𝔭 = aR be a non-zero prime ideal of R. We want to show it is a maximal ideal. Let 𝔪 = bR be a maximal ideal containing a. Then a = λb for some λ ∈ R. If λ ∈ 𝔭, we have that λ = μa for some μ ∈ R and hence a = λb = μab and hence μb = 1. However, this is impossible, because b is not a unit. Therefore λ ∉ 𝔭. Since a = λb ∈ 𝔭, this implies that b ∈ 𝔭 and hence 𝔪 = 𝔭 as required.

Exercises

1.1 Let R be a ring. Show that for every R-module M there exists a free R-module F and a surjective R-homomorphism F → → M. Does there always exist a free R-module F and an injective R-homomorphism M ↪ F?

1.2 If possible, give an example of an exact sequence of finite abelian groups

0 ⟶ N ⟶ (Z/2Z) × (Z/8Z) ⟶ M ⟶ 0.

with

(i) M ≅ (Z/2Z) × (Z/2Z) and N ≅ Z/4Z; (ii) M ≅ Z/4Z and N ≅ (Z/2Z) × (Z/2Z); (iii) M ≅ (Z/2Z) × (Z/2Z) and N ≅ (Z/2Z) × (Z/2Z); (iv) M ≅ Z/4Z and N ≅ Z/4Z.

1.3 Show that the dimension of the polynomial ring R[X_{1}, …, X_{n}] is at least n. Show that the dimension of the polynomial ring Z[X_{1}, …, X_{n}] is at least n + 1.

1.4 Let R be a ring. Show that Z, Q, Q/Z and ⊕_{n≥1}Z/nZ are faithful Z-modules. Give an example of a Z-module that is not faithful.

1.5 Let R be a domain with quotient field F. The integral closure of R is the subring R′ of F that consists of all x ∈ F that are integral over R. Show that R′ is integrally closed.

1.6 Let n ∈ Z_{>0}; consider the ring R = Z[X]/(X^{2}, nX).

(i) Determine the unit group R^{∗}. (ii) Let a, b ∈ Z. Show that the R-ideal aXR is contained in bXR if and only if gcd(b, n)

divides gcd(a, n). (iii) Find n ∈ Z_{>0} and a, b ∈ Z for which the ideals aXR and bXR of R are equal, while

there does not exist a unit u ∈ R^{∗} with uaX = bX.

1.7 Let R be a ring and let M, N be R-modules.

(i) Show that the abelian group Hom_{R}(M, N) of R-homomorphisms has the structure of an R-module given by (λf)(m) = λf(m) (for λ ∈ R, m ∈ M and f ∈ Hom_{R}(M, N)).

(ii) Let f : M ⟶ N be an R-homomorphism. and let P be a third R-module. Show that the map f_{P} : Hom_{R}(P, M) ⟶ Hom_{R}(P, N) given by f_{P} (φ) = f · φ is a group homorphism.

(iii) Suppose that

0 ⟶ L ⟶^{f} M ⟶^{g} N ⟶ 0.

is an exact sequence of R-modules. Show that the induced sequence

0 ⟶ Hom_{R}(P, L) ⟶^{f_{P}} Hom_{R}(P, M) ⟶^{g_{P}} Hom_{R}(P, N) ⟶ 0

is left-exact (i.e., exact except perhaps at Hom_{R}(P, N)). (iv) Show that the sequence of part (iii) is exact if P is projective.

(v) Let R = Z and P = Z/2Z. Give an example of a short exact sequence 0 → L → M → N → 0 of abelian groups, for which the sequence

0 ⟶ Hom_{R}(Z/2Z, L) ⟶^{f_{Z/2Z}} Hom_{R}(Z/2Z, M) ⟶^{g_{Z/2Z}} Hom_{R}(Z/2Z, N) ⟶ 0

is not exact.

2. Dedekind rings. Rings of integers of algebraic number fields are Noetherian integrally closed domains of Krull dimension 1. Rings having these properties are called Dedekind rings. In this section we discuss their basic properties.

Definition. A ring R is called a Dedekind ring if it is a Noetherian, integrally closed domain of Krull dimension ≤ 1.

Proposition 2.1. Prinicipal ideal domains are Dedekind domain.

Proof. Principal ideal domains are clearly Noetherian. It follows therefore from Propositions 1.10 and 1.11 that they are Dedekind rings.

Theorem 2.2. Let R be a Dedekind domain. For every two non-zero R-ideals I ⊂ J, there exists a unique R-ideal J′ so that JJ′ = I.

Proof. The unicity of J′ follows from the existence: pick a non-zero x ∈ J. Then there is an ideal J′′ for which JJ′′ = xR. If there were two ideals J′_{1} and J′_{2} with I = JJ′_{1} = JJ′_{2}, then multiplying by J′′ gives that xJ′_{1} = xJ′_{2} and hence J′_{1} = J′_{2}. Claim. Every non-zero R-ideal contains a product of non-zero prime ideals. Proof. Suppose not. Let I be a maximal element in the non-empty partially ordered set Ω of ideals for which the statement is false. Then I is certainly not itself a prime ideal. Therefore there are x, y not in I with xy ∈ I. The strictly larger ideals I + xR and I + yR each contain a product of non-zero prime ideals. Since I contains the product of I + xR and I + yR, so does I. Contradiction.

Suppose I ⊂ J are R-ideals for which the statement of the theorem does not hold. Since R is Noetherian, we may by Prop, 1.4 (iii) assume that J is maximal with respect to this property, i.e. that for every strictly larger J and any ideal I ⊂ J, the statement of the theorem is true. Let y be non-zero element of J. By the claim, the principal ideal yR contains a product of non-zero prime ideals

∏_{i=1}^{s} 𝔭_{i} ⊂ yR,

which we assume has a minimal number of factors s ≥ 1. The ideal yR is contained in a maximal ideal 𝔪. Then 𝔭_{1} ⊂ 𝔪, say. Since R has dimension ≤ 1, it follows that 𝔪 = 𝔭_{1}. Since the number of factors in the product is minimal, we have that

∏_{i=2}^{s} 𝔭_{i} ⊄ yR.

Let x ∈ ∏_{i=2}^{s} 𝔭_{i} − yR. Note that this means that the fraction \frac{x}{y} is not contained in R. However, \frac{x}{y}J ⊂ R because xJ ⊂ x𝔭_{1} ⊂ ∏_{i=1}^{s} ⊂ yR. This implies that the R-ideal J + \frac{x}{y}J strictly contains J. Indeed, if not we would have that \frac{x}{y}J ⊂ J so that \frac{x}{y} would be integral over R and hence be contained in R.

Since

I ⊂ J ⊂_{≠} J + \frac{x}{y}J,

there exists by the maximality of J an ideal J′ so that I = J′(J + \frac{x}{y}J). Consider now J′′ = J′ + \frac{x}{y}J′. Then J′′J = (J′ + \frac{x}{y}J′)J = I. In particular, J′′J ⊂ J so that every x ∈ J′′ is integral and hence J′′ ⊂ R. This proves the theorem.

Corollary 2.3. Let R be a Dedekind domain. Then every non-zero ideal is a product of prime ideals. Up to permutation of the prime ideals, this factorization of I is unique.

Proof. Let I be a non-zero ideal of R. If I ≠ R, it is contained in some maximal ideal 𝔭. By the Theorem, there is an R-ideal I_{1} so that I = 𝔭I_{1}. The ideal I_{1} strictly contains I. Repeat this with I_{1} instead of I. Since R is Noetherian, this process must stop at some ideal I_{n}. Then I_{n} = R. Indeed, if we had that I_{n} = 𝔭I_{n+1} = 𝔭I_{n} for some prime 𝔭, then 𝔭 = R by the unicity statement of Theorem 2.2. This shows the existence of a prime factorization.

If we have two prime factorizations, then any prime 𝔭 occurring in one factorization must occur also in the other. Pick 0 ≠ x ∈ 𝔭 and let J ⊂ R be the ideal for which J𝔭 = xR. After multiplying the factorizations by J, we can divide by x and are left with two factorizations with fewer factors. This process eventually stops, at which point we conclude that the factorizations were equal as required.

Definition. Let R be a Dedekind domain with quotient field F. For every non-zero Rideal J and every non-zero prime ideal of R we put ord_{𝔭}(J) = a_{𝔭}, where the a_{𝔭} ∈ Z_{≥0} is the exponent that occurs in the prime factorization of the ideal J:

J = ∏_{𝔮} 𝔮^{a_{𝔮}}.

A fractional R-ideal is a subset I of F for which there exist x ∈ O_{F} so that xI = {xy : y ∈ I} is a non-zero ideal of R. For any non-zero prime ideal 𝔭 of R we put

ord_{𝔭}(I) = ord_{𝔭}(xI) − ord_{𝔭}(xR).

The product of two fractional ideals I and J is the submodule of F generated by elements of the form xy with x ∈ I and y ∈ J. By Corollary 2.3, the fractional ideals form a group isomorphic to ⊕_{𝔭}Z. For any x ∈ F^{∗}, the principal fractional ideal generated by x is the set xR = {xy : y ∈ R}. For x ∈ F^{∗} and any non-zero prime ideal 𝔭 of R we put

ord_{𝔭}(x) = ord_{𝔭}(xR).

This is a homomorphism from F^{∗} to Z. An element x ∈ F^{∗} is contained in R if and only if ord_{𝔭}(x) ≥ 0 for all prime ideals 𝔭. It is contained in the unit group R^{∗} if and only if ord_{𝔭}(x) = 0 for all prime ideals 𝔭. Putting all these homomorphisms together, we obtain therefore an exact sequence

0 ⟶ R^{∗} ⟶ F^{∗} ⟶ ⊕_{𝔭} Z ⟶ Cl(R) ⟶ 0,

which we use to define the class group Cl(R) of R.

Two ideals I, J ⊂ R have the same image in Cl(R) if and only if there are non-zero x, y ∈ R so that xI = yJ. Since any R-homomorphism I ⟶ J is given by multiplication by an element in F^{∗}, this happens if and only if I and J are isomorphic R-modules. Therefore the map

{non-zero R-ideals up to R-isomorphism} ⟶^{≅} Cl(R)

that send an ideal to its class, is a bijection.

Proposition 2.4. Let R be a Dedekind ring. Then the following are equivalent.

(i) The class group Cl(R) is trivial. (ii) R is a principal ideal domain. (iii) R is a unique factorization domain.

Proof. For any ring (ii) implies (iii). To show that (i) implies (ii), let I ⊂ R be a non-zero ideal. Let I = ∏_{𝔭} 𝔭^{a_{𝔭}} be its factorization into a product of non-zero prime ideals of R. Since Cl(R) is trivial, there is an element x ∈ F^{∗} for which ord_{𝔭}(x) = a_{𝔭}. The implies that x ∈ R and that I = xR so that I is principal. Finally, to show that (iii) implies (i), it suffices to show that every non-zero prime of R is principal. This implies that every vector (a_{𝔭}) ∈ ⊕_{𝔭}Z is the image of a suitable element in F^{∗}. Let therefore 𝔭 be a non-zero prime and let 0 ≠ x ∈ 𝔭. Since x is a product of irreducible elements, it follows that 𝔭 contains an irreducible element π. But then πR ⊂ 𝔭 are two non-zero prime ideals of R. Since R has dimension 1, they are therefore equal.

This proves the proposition.

Lemma 2.5. Let R be a Dedekind ring and let I ⊂ R be a non-zero ideal. Then R/I is a principal ideal ring. A local Dedekind ring is a discrete valuation domain.

Definition. The ideals of R/I are of the form J/I where J ⊂ R is an ideal containing I. Every such J is a product of prime ideals 𝔭. It suffices to show that each such prime is of the form 𝔭 = I + xR for some x ∈ R. By the Chinese Remainder Theorem, there exists an element x ∈ 𝔭 − 𝔭^{2} which is congruent to 1 modulo the remaining primes that occur in the factorization of J. Then I + xR ⊂ 𝔭. By Theorem 2.2 there exists therefore an ideal J′ ⊂ R with J′𝔭 = I + xR. Any prime occuring in the prime factorization of J′ occurs in the prime factorization of I and of xR. Therefore only 𝔭 can occur. However, since I + xRnot ⊂ 𝔭^{2}, the prime 𝔭 does not occur either. Therefore J′ is the unit ideal and 𝔭 = I + xR as required.

If R is a local Dedekind ring with maximal ideal 𝔪, then R/𝔪^{2} is a principal ideal ring. Let π ∈ R be a generator of 𝔪/𝔪^{2}. Then πR = 𝔪 is the unique factorization into prime ideals of the ideal πR. It follows that R is a discrete valuation ring, the valuation F^{∗} ⟶ Z being given by x ↦ ord_{𝔪}(x).

This proves the lemma.

Proposition 2.6. Let R be a Dedekind ring and let I, J ⊂ R be two non-zero ideals. Then

I ⊕ J ≅ R ⊕ IJ, as R-modules.

Proof. If I + J = R, we have that IJ = I ∩ J and the result follows from the following split-exact sequence

0 ⟶ IJ ⟶ I ⊕ J ⟶ R ⟶ 0.

To get the statement in general, it suffices to show that for any two R-ideals I, J there exists an ideal J′ ⊂ R in the same ideal class as J that is coprime to I. Pick a non-zero element x ∈ J. By Theorem 2.2 there exists an ideal J_{1} ⊂ R with JJ_{1} = xR. Then IJ_{1} ⊂ J_{1}, so by Prop. 2.5 there exists y ∈ J_{1} so that IJ_{1} + yR = J_{1}. Multiplying this relation by the ideal J we get

IJ_{1}J + yJ = J_{1}J.

since yJ ⊂ J_{1}J = xR, we have that yJ = xJ′ for some ideal J′ ⊂ R. In particular, J and J′ are in the same ideal class of R. Replacing yJ by xJ′ and JJ_{1} by xR in the formula leads to the equality xI + xJ′ = xR, which after division by x gives the required result.

Corollary 2.7. Every ideal of a Dedekind ring R is projective.

Proof. Let I ⊂ R be an ideal. If I = 0, everything is clear. If not, pick 0 ≠ x ∈ I. By Theorem 2.2 there is an ideal J ⊂ R with IJ = xR. The proposition implies that

I ⊕ J ≅ R ⊕ xR ≅ R^{2},

showing that I is projective as required.

Lemma 2.8. Let R be a Dedekind ring. Let M be a torsion-free R-module. Then rank(M) = 1 if and only if M is isomorphic to a non-zero R-ideal.

Proof. For any non-zero ideal I ⊂ R we have that Hom_{R}(I, F) ≅ F, the isomorphism being given by f ↦ f(x)/x where x ∈ I is any non-zero element. On the other hand, if M is a torsion free R-module for which Hom_{R}(M, F) ≅ F is 1-dimensional, then we can find an R-linear f : M ⟶ F whose image is a non-zero ideal J in R. Since J is projective, the sequence 0 → ker f → M → J → 0 splits. This implies that ker f has rank 0. Since M is torsion-free, it follows that ker f is zero and hence that M ≅ J as required.

Theorem 2.9. Let R be a Dedekind domain. Suppose that M is a direct product of R-ideals and that N ⊂ M is a submodule. Then

(i) the submodule N is also isomorphic to a product of R-ideals. (ii) There exist n ≥ 0 and elements e_{i} ∈ M and R-ideals J_{i} ⊂ I_{i} for i = 1, …, n so that

\begin{aligned}M & = I_{1}e_{1} ⊕ ⋯ ⊕ I_{n}e_{n}, \\ N & = J_{1}e_{1} ⊕ ⋯ ⊕ J_{n}e_{n}. \end{aligned}

Proof. If N ≠ 0, then there is a non-torsion element x ∈ N and hence an isomorphism xR ⟶ R. This isomorphism extends to a morphism of N and even of M to R. We denote it by φ. The images φ(M) and φ(N) are non-zero ideals I and J of R. Let K ⊂ M denote the kernel of φ. We have the following commutative diagram with split-exact rows.

0 ⟶ K ∩ N ⟶ N ⟶^{φ} J ⟶ 0
↓^{⊂}  ↓^{⊂}  ↓^{⊂}
0 ⟶  K  ⟶ M ⟶^{φ} I ⟶ 0

(i) We proceed with induction with respect to the rank of N. If the rank of N is 0, then N, being torsion free, is itself 0 and there is nothing to prove. If rank(N) ≥ 1, then the above applies. Since the rank of J is 1, the rank of K ∩ N is strictly smaller than rank(N) and we are done, since the sequence splits. (ii) This part is proved with induction with respect to the rank of M. If rank(M) = 1, then M is isomorphic to an ideal of R and everything is clear. If rank(M) > 1, then the above applies. By part (i) the module K is isomorphic to a product of R-ideals. Since rank(J) = 1 the result now follows by induction.

Corollary 2.10. Let R be a Dedekind ring. Then any finitely generated R-module is a direct product of a torsion module T and a projective module P. Moreover, there are R-ideals I_{1}, …, I_{s} so that

T ≅ (R/I_{1}) × ⋯ (R/I_{s}),

and for some r ≥ 0 and some ideal I ⊂ R we have that

P ≅ R^{r} × I

Proof. Let M be a finitely generated R-module and choose a surjective R-morphism R^{n} ⟶ M. Let N be the kernel. Then M ≅ R^{n}/N and it follows from Theorem 2.9 that M is isomorphic to I_{1}/J_{1} ⊕ ⋯ ⊕ I_{s}/J_{s} for certain ideals J_{i} ⊂ I_{i} (for i = 1, …, s). The summands that have J_{i} = 0 is isomorphic to a sum of ideals of R. By Prop. 2.6, this sum is either zero or is isomorphic to R^{r} ⊕ I for some r ≥ 0 and some ideal I. For the summands for which J_{i} ≠ 0 part, we have by Prop. 2.5 that I_{i} = J_{i} + xR for some x ∈ R. This implies that I_{i}/J_{i} ≅ (J_{i} + xR)/J_{i} ≅ xR/J where J = xR ∩ J_{i}.

Corollary 2.11. Let R be a Dedekind ring. Then any finitely generated R-module is projective if and only if it is torsion-free.

Corollary 2.12. Let R be a principal ideal domain and let M be a finitely generated R-module. Then

(i) M is free if and only if it is projective and if and only if it is torsion-free. (ii) There exist elements a_{1}, a_{2}, …, a_{t} ∈ R so that M is isomorphic to the R-module

R/a_{1}R × … × R/a_{t}R.

Proof. If M is free, it is projective and hence torsion-free. Conversely, if M is torsion-free it is projective. By Prop. 2.4 it is therefore free. This proves (i). Part (ii) is a consequence of the fact that all R-ideals are principal.

Proposition 2.13. (Gauss’ Lemma) Let R be a Dedekind ring with quotient field F. Suppose that f ∈ R[X] is a product of two monic polynomials g, h ∈ F [X]. Then g and h are contained in R[X].

Proof. Suppose that f = g · h with monic polynomials g, h ∈ F [X] and let 𝔭 be a prime ideal. We claim that all coefficients of g and h are integral at 𝔭. Indeed, suppose not. Let i ≥ 0 be the smallest power so that 𝔭^{i}g ⊂ R[X] and let j ≥ 0 be the smallest power so that 𝔭^{i}h ⊂ R[X]. Let a ∈ 𝔭 − 𝔭^{2} and consider the relation

a^{i+j}f = (a^{i}g) · (a^{j}h).

By assumption i + j ≥ 1. This implies that the left hand side is zero in the ring (R/𝔭)[X]. On the other hand, the polynomial a^{i}g ≢ 0 (mod 𝔭). Indeed, if i = 0 this follows from the fact that g is monic and if i > 0 there is by definition of i a coefficient b of g for which ord_{𝔭}(b) = −i. This implies that the coefficient a^{i}b of a^{i}g has its ord_{𝔭} equal to zero and is therefore not zero modulo 𝔭. Similarly, a^{j}h ≢ 0 (mod 𝔭). This contradicts the fact that (R/𝔭)[X] is a domain. This proves the proposition.

Exercises.

2.1 Let R = Z, let M = Z^{2} and N = ( \begin{matrix}2 \\ 5 \end{matrix} )Z + ( \begin{matrix}7 \\ 2 \end{matrix} )Z. Find a basis for M as in Theorem 2.8.

Do the same for N′ = ( \begin{matrix}4 \\ 8 \end{matrix} )Z + ( \begin{matrix}6 \\ 12 \end{matrix} )Z.

2.2 (Jordan Normal Form). Let A be an n × n-matrix with complex coefficients. Prove that there exists an invertible matrix B so that BAB^{−1} has a ‘block form’ with each block of the shape

( \begin{matrix}λ & 1 & 0 & ⋯ & 0 & 0 \\ 0 & λ & 1 & ⋯ & 0 & 0 \\ 0 & 0 & λ & ⋯ & 0 & 0 \\ ⋮ & & & ... & & ⋮ \\ 0 & 0 & & ⋯ & λ & 1 \\ 0 & 0 & & ⋯ & 0 & λ \end{matrix} )

for some λ ∈ C. (Sugg. Take R = C[X] and provide C^{n} with the structure of a C[X]-module by defining X · v = Av for every v ∈ C^{n}. Then apply Cor. 2.12 and the Chinese Remainder Theorem).

2.3 Let R be a Dedekind ring and let I_{1}, …, I_{r}, J_{1}, …, J_{s} be non-zero R-ideals. Show that the R-modules I_{1} × … × I_{r} and J_{1} × … × J_{s} are isomorphic if and only if r = s and the ideal classes of the products I_{1} · . . . · I_{r} and J_{1} · . . . · J_{s} are equal.

2.4 Let d ∈ Z be squarefree. We define the Norm of an element x ∈ Q(\sqrt{d}) by N(x) = a^{2} − db^{2}. Here a, b ∈ Q are taken so that x = a + b\sqrt{d}. Show that the norm map is multiplicative: N(xy) = N(x)N(y) for all x, y ∈ Q(\sqrt{d}.

2.5 Let R = Z[\sqrt{−6}].

(i) Show that 10 = 2·5 and 10 = (2+\sqrt{−6})(2−\sqrt{−6}) are two factorizations of 10 ∈ Z[\sqrt{−6}] that are essentially distinct in the sense that the factorizations cannot be transformed into one another by multiplying the factors by units.

(ii) Find the factorizations of the principal ideals (2), (5), (2 + \sqrt{−6}) and (2 − \sqrt{−6}) into a

product of prime ideals of R = Z[\sqrt{−6}].

2.6 Let R be a domain and let N : R − {0} ⟶ R_{>0} be a multiplicative function. The ring R is called Euclidean with respect to the map N if for every x, y ∈ R with y ≠ 0, there exist q, r ∈ R with y = qx + r and either r = 0 or N(r) < N(y).

(i) Show that a Euclidean domain is a PID.

(ii) Show that the ring Z is Euclidean with respect to the map N(n) = |n|.

(iii) Let K be a field. Show that the polynomial ring K[X] is Euclidean with respect to the

map N(f) = e^{deg f}. (iv) Show that the ring of integers Z[i] of Gauss is Euclidean with respect to the map

N(a + bi) = a^{2} + b^{2} (where a, b ∈ Z).

2.7. Let R be a domain with quotient field F. Suppose that R is equipped with a multiplicative function N : R − {0} ⟶ R_{>0}. Show that N can be extended to a a homomorphism F^{∗} ⟶ R_{>0}. Show that R is Euclidean if and only if for every x ∈ F, either x ∈ R or there exists an element x ∈ R with N(x − y) < 1.

3. Finite free algebras. In this section we discuss finite free algebras over a base ring R.

Definition. Let R be a ring. A finite free R-algebra is an R-algebra that is finitely generated and free as an R-module.

Examples of finite free algebras are R-algebras of the form R[X]/(φ(X)) where φ(X) ∈ R[X] is a monic polynomial. If R is a field, any finite extension of R is automatically a finite free R-algebra.

Definition. Let R be a ring and let A be a finite free R-algebra. For any a ∈ A, multplication by a is an R-linear map. With respect to an R-basis of A, it can be described by a the square matrix M with entries in R. We define the norm of a by N(a) = det(A) and the trace of a by Tr(a) = Trace(A). The characteristic polynomial f_{char}^{a}(X) of a is defined as the characteristic polynomial det(X · id − M) of M.

Definition. The discriminant of a finite free R-algebra is defined by

Δ(A/R) = det(Tr(ω_{i}ω_{j}))

where ω_{1}, …, ω_{n} is any R-basis for A. Changing the basis, changes Δ(A/R) by the square of the determinant of the base change matrix M. Therefore Δ(A/R) is an element of R that is well defined up to multiplication by squares of units of R.

Proposition 3.1. Let R be a ring and φ(X) ∈ R[X] be a monic polynomial. Then the discriminant of A = R[X]/(φ(X)) is equal to the discriminant of φ. In other words, if λ_{1}, …, λ_{n} denote the zeroes of φ, then Δ(A/R) = ∏_{1≤i<j≤n}(λ_{i} − λ_{j})^{2}.

Proof. We use the R-basis ω_{1}, …, ω_{n} = 1, X, …, X^{n−1} of A. The multiplication by X matrix A has an almost diagonal form and one easily computes that its characteristic polynomial is equal to φ. Counting multiplicities, let λ_{1}, …, λ_{n} be the eigenvalues of A. Then Tr(A^{k}) = λ_{1}^{k} + … + λ_{n}^{k} and

( \begin{matrix}Tr(ω_{1}^{2}) & ⋯ & Tr(ω_{n}ω_{1}) \\ Tr(ω_{1}ω_{2}) & ⋯ & Tr(ω_{n}ω_{2}) \\ ⋮ & & ⋮ \\ Tr(ω_{1}ω_{n}) & ⋯ & Tr(ω_{n}^{2}) \end{matrix} ) = M^{t}M, where M = ( \begin{matrix}1 & ⋯ & 1 \\ λ_{1} & ⋯ & λ_{n} \\ ⋮ & & ⋮ \\ λ_{1}^{n−1} & ⋯ & λ_{n}^{n−1} \end{matrix} )

The determinant of the Vandermonde matrix M is equal to ∏_{1≤i<j≤n}(λ_{i} − λ_{j}) and the result follows.

Definition. For any R-algebra A, its module of Kähler differentials Ω_{A/R}^{1} is the A-module generated by symbols da for a ∈ A modulo the A-submodule generated by the relations dλ for λ ∈ R and d(a + b) − da − db and d(ab) − adb − bda for a, b ∈ A

Example. Let R be a ring and φ(X) ∈ R[X] be a monic polynomial and let A = R[X]/(φ(X)). Then the module of Kähler differentials Ω_{A/R}^{1} is given by AdX/φ′(X)dX. It is isomorphic to the A-mdoule R[X]/(φ(X), φ′(X)).

Definition. Let F be a field. A polynomial f ∈ F [X] is called separable if it has no double zeroes in \overline{F} or, equivalently, if gcd(f, f′) = 1. The field F is called perfect if every irreducible polynomial f ∈ F [X] is separable.

Proposition 3.2. (Theorem of the primitive element). Let F be a perfect field. Then every finite field extension K of F is of the form F (γ) for some so-called primitive element γ ∈ K.

Proof. If F is finite, we let γ denote a generator of the multiplicative group L^{∗}. Then we have that L = F (γ). If F is infinite, we proceed by induction with respect to [L : F]. It suffices to show that any field of the form F (α, β) with α, β ∈ F is also of the form F (γ) for some γ ∈ L. Let f, g ∈ F [X] be the minimum polynomials of α and β respectively. Since F is separable, both f and g have distinct zeroes. Let λ ∈ F^{∗} be distinct from all numbers

\frac{α − α′}{β − β′}

where α′ ≠ α denotes a zero of f and β′ ≠ β a zero of g. Consider γ = α+λβ. It is distinct from α′ + λβ′ for all choices of α′ and β′. This implies that the polynomials f(γ − λT) and g(T) have only the zero β in common. Since both polynomials are contained in the ring F (γ)[T], this implies that β ∈ F (γ). It follows that α ∈ F (γ) as well and the proposition follows.

Theorem 3.3. Let F be a field and let A be a finite F -algebra. Then A is isomorphic to a product of local F -algebras with nilpotent maximal ideals.

Proof. Suppose that 𝔪_{1}, …, 𝔪_{t} are distinct maximal ideals of A. Then, if 𝔪 is yet another maximal ideal, we have that 𝔪_{1}∩…∩𝔪_{t} ⊄ 𝔪 because the product 𝔪_{1} ⋯ 𝔪_{t} is not contained in 𝔪. It follows that 𝔪_{1} ∩ … ∩ 𝔪_{t} ∩ 𝔪 is strictly smaller than 𝔪_{1} ∩ … ∩ 𝔪_{t}. Since the F -dimension of A is finite, this process must eventually stop. This shows that A admits only finitely many maximal ideals 𝔪_{1}, …, 𝔪_{t} say. Since all prime ideals of A are maximal, Lemma 3.1. implies that the product 𝔪_{1}, ⋯, 𝔪_{t} is nilpotent. Let n ≥ 1 be an exponent for which (𝔪_{1} ⋯ 𝔪_{t})^{n} = 0. Since 𝔪_{1} +𝔪_{2} ⋯ 𝔪_{t} = A, we also have that 𝔪_{1}^{n} +(𝔪_{2} ⋯ 𝔪_{t})^{n} = A. The Chinese Remainder Theorem provides us then with an isomorphism of A-algebras

A ≅ A/𝔪_{1}^{n} × A/(𝔪_{2} ⋯ 𝔪_{t})^{n}.

The F -algebra A/𝔪_{1}^{n} is local and its maximal ideal 𝔪_{1}/𝔪_{1}^{n} is nilpotent. The result now follows by induction with respect to the F -dimension of A.

Definition. An étale R-algebra is an algebra that satisfies any of the conditions of the following theorem.

Theorem 3.4. Let F be a perfect field and let A be a finite F -algebra. Then the following are equivalent.

(i) The discriminant Δ(A/F) is not zero; (ii) The trace map A ⟶ F is non-degenerate; (iii) A is isomorphic to a product of finite field extensions of F; (iv) The module of Kähler differentials Ω_{A/F}^{1} vanishes;

(v) A is reduced, i.e., its nilradical is zero.

Proof. We write A as a product of local F -algebras A_{i} with nilpotent maximal ideals 𝔪_{i}. We show: (iii) → (ii) → (i) →(iii) and then (iii) → (iv) → (v) →(iii).

To show that (iii) implies (ii), it suffices to show that the trace map A_{i} ⟶ F is non-degenerate for each of the local factors A_{i}. Since by assumption, each A_{i} is a field, it suffices to show that for a finite field extension F ⊂ L, the trace map L ⟶ F is not identically zero. Let x ∈ L and let f ∈ F [X] denote its characteristic polynomial. Let n = [L : F]. The zeroes λ_{1} …, λ_{n} of f(T) are the eigenvalues, with multiplicities, of the matrix A that corresponds to the of the multiplication by a map. We have the following identity in the power series ring F [[T]]:

\frac{T^{n−1}f′(1/T)}{T^{n}f(1/T)} = ∑_{i=1}^{n} \frac{1}{1 − λ_{i}T} = ∑_{k=1}^{∞} Tr(A^{k})T^{k}.

Suppose that the trace map is zero. Then the right hand side is zero. Since T^{n}f(1/T) is a unit in the ring F [[T]], it follows that f′(T) = 0. When we take for x ∈ L a primitive element, the polynomial f is equal to the minimum polynomial of x. Since F is perfect, this contradicts the fact that f′ = 0..

To show that (ii) implies (i), we let ω_{1}, …, ω_{n} denote an F -basis of A. If the discriminant of A is zero, there is a non-trivial F -linear relation between the columns of the matrix (Tr(ω_{i}ω_{j}))_{ij}. This implies that there is an F -linear combination a = λ_{1}ω_{1} + … + λ_{n}ω_{n} of the ω_{i} that is not zero, but has the property that Tr(ab) = 0 for all b ∈ A as required.

To see that (i) implies (iii), we assume that A is not a product of fields. Then there is a non-zero nilpotent element a ∈ A, which we can take as an element in an F -basis of A. Since mutiples of a are also nilpotent and nilpotent elements have trace zero, this leads to a zero column in the matrix (Tr(ω_{i}ω_{j}))_{ij} and hence Δ(A/F) = 0.

To show that (iii) implies (iv) it suffices to prove that Ω_{L/F}^{1} = 0 when F ⊂ L is a finite field extension. Since F is perfect, L ≅ F [X]/(φ(X) for some irreducible polynomial φ(X). The Kähler differentials are then equal to K[X]/(φ, φ′) which vanishes, because φ has no double zeroes.

To show that (iv) implies (v), we assume that A is not reduced and show that Ω_{A/F}^{1} does not vanish. Since for any surjective F -algebra homomorphism A → → B the natural map Ω_{A/F}^{1} ⟶ Ω_{B/F}^{1} is also surjective, it suffices to construct a suitable quotient algebra B. We first project A on any of its local factors A_{i} that contains non-zero nilpotent elements. Then we take the quotient by 𝔪_{i}^{2}. The maximal ideal 𝔪 of the resulting local F -algebra B satisfies 𝔪^{2} = 0. Let L = A/𝔪 be the residue field of A′. Since F is perfect, we have that L = K(x) for some primitive element x ∈ L. Let f ∈ K[X] denote the minimum polynomial of x. Let a ∈ B be some lift of x to B. Then f(a) ≡ 0 (mod 𝔪) and f′(a) ≢ (mod 𝔪). Let a′ = a − f(a)/f′(a). Since 𝔪^{2} = 0, we have that f(a)^{2} = 0 and hence that

f(a′) = f (a − \frac{f(a)}{f′(a)}) = f(a) − \frac{f(a)}{f′(a)}f′(a) = 0.

It follows that the natural map from B to its residue field L admits a section and hence that B is isomorphic to the F -algebra L[X_{1}, …, X_{m}]/J for a certain ideal J satisfying (X_{1}, …, X_{m})^{2} ⊂ J ⊂ (X_{1}, …, X_{m}). It follows that

Ω_{B/F}^{1} = ⊕_{i=1}^{n} BdX_{i}/⟨dg : g ∈ J⟩ ≅ ⊕_{i=1}^{n} LdX_{i}/⟨∑_{i=1}^{n} \frac{∂g}{∂X_{i}} (0)dX_{i} : g ∈ J⟩ ≅ (X_{1}, …, X_{n})/J.

Since J is strictly contained in the ideal (X_{1}, …, X_{n}), this does not vanish and we are done.

Finally, the fact that that (v) implies (iii) follows at once from the fact that A is isomorphic to a product of local F -algebras with nilpotent maximal ideals.

Exercises.

3.1 Let α = ζ_{5} + ζ_{5}^{−1} ∈ Q(ζ_{5}) where ζ_{5} denotes a primitive 5th root of unity. Calculate the

characteristic polynomial of α ∈ Q(ζ_{5}). 3.2 Let F be a number field of degree n and let x ∈ F. Show that for q ∈ Q ⊂ F one has that

\begin{aligned}Tr(qx) & = qTr(x), \\ Tr(q) & = nq, \\ N(q) & = q^{n}. \end{aligned}

Show that the map Tr : F ⟶ Q is surjective. Show that the norm N : F^{∗} ⟶ Q^{∗} is, in general, not surjective.

3.3 Let α be a zero of the polynomial X^{3} − X + 1. Put A = Z[α]. Show that Δ(A/Z) = −23. 3.4 Let 1 ≠ d ∈ Z be squarefree and put F = Q(\sqrt{d}). Show that

O_{F} = \{ \begin{matrix}Z[\sqrt{d}], & if d ≡ 2, 3 (mod 4); \\ Z[ \frac{1+\sqrt{d}}{2} ], & if d ≡ 1 (mod 4). \end{matrix}

3.5 Let F be a number field of degree n and let α ∈ F. Show that for q ∈ Q one has that

N(q − α) = f_{char}^{α}(q). Show that for q, r ∈ Q one has that N(q − rα) = r^{n}f_{char}^{α}(q/r). 3.6 Let A be a finite algebra over a field F. Suppose that A is a field. Show that for every a ∈ A,

the characteristic polynomial f_{char}^{a} ∈ F [X] is a power of the minimal polynomial of a. 3.7 Let A and B be finite free R-algebras.

(i) Show that A × B is also a finite free R-algebra. (ii) Show that Δ((A × B)/R) = Δ(A/R)Δ(B/R).

(iii) Show that the characteristic polynomial of (a, b) ∈ A × B is equal to the product of the characteristic polynomials of a ∈ A and b ∈ B.

3.8 Prove that Disc(T^{n} − a) = n^{n}a^{n−1}. Compute Disc(T^{2} + bT + c) and Disc(T^{3} + bT + c).

3.9 (Newton’s formulas) Let K be a field and let α_{1}, α_{2}, …, α_{n} ∈ K. We define the symmetric functions s_{k} of the α_{i} by

∏_{i=1}^{n}(T − α_{i}) = T^{n} − s_{1}T^{n−1} + s_{2}T^{n−2} + … + (−1)^{n}s_{n}.

We extend the definition by putting s_{k} = 0 whenever k > n. We define the power sums p_{k} by

p_{k} = ∑_{i=1}^{n} α_{i}^{k} for k ≥ 0.

Show that for every k ≥ 1 one has that

(−1)^{k}ks_{k} = p_{k} − p_{k−1}s_{1} + p_{k−2}s_{2} − p_{k−3}s_{3} + ….

In particular

\begin{aligned}s_{1} & = p_{1} \\ −2s_{2} & = p_{2} − p_{1}s_{1} \\ 3s_{3} & = p_{3} − p_{2}s_{1} + p_{1}s_{2} \\ −4s_{4} & = p_{4} − p_{3}s_{1} + p_{2}s_{2} − p_{1}s_{3} \\ 5s_{5} & = … \end{aligned}

(Hint: Take the logarithmic derivative of ∏_{i=1}^{n}(1 − α_{i}T).)

3.10 Let f(X) ∈ Z[X] be an irreducible polynomial of degree n and let F = Q(α). Show that

Δ(Z[α]/Z) = det((p_{i+j−2})_{1≤i,j≤n}).

Here p_{k} denotes the power sum α_{1}^{k} + … + α_{n}^{k} of the zeroes α = α_{1}, α_{2}, …, α_{n} are the zeroes of f(X).

3.10 Show that the polynomial T^{5} + T^{3} − 2T + 1 ∈ Z[T] is irreducible. Compute its discriminant.

4. Lattices. In this section we discuss the basic properties of lattices.

Definition. A (Euclidean) lattice is a free abelian group L of finite rank together with a scalar product ⟨−, −⟩ on the vector space R ⊗_{Z} L. Two lattices L and L′ are called isometric if there is a Z-linear bijection A : L ⟶ L′ compatible with the scalar products: ⟨A(v), A(w)⟩′ = ⟨v, w⟩ for all v, w ∈ L.

Here all scalar products are supposed to be positive definite. Since for every scalar product the underlying vector space admits an orthonormal basis, every lattice is isomorphic to a lattice of the form

L = Z ( a_{11} ; ⋮ ; a_{1n} ) + … + Z ( a_{n1} ; ⋮ ; a_{nn} )

equipped with the usual scalar product on R^{n}. It follows that the lattices of rank n are parametrized by the points of the symmetric space GL_{2}(R)/O_{n}(R). For example, for n = 1 this is R^{∗}/{±1} ≅ R_{>0}^{∗} with t ∈ R_{>0}^{∗} corresponding to the lattice tZ ⊂ R.

Definition. The covolume covol(L) of a lattice L is the volume of V/L. Here V = L⊗_{Z} R. Alternatively, it is the volume of a fundamental domain: if L = ⊕_{i=1}^{n}Ze_{i}, then covol(L) = vol (⊕_{i=1}^{n}[0, 1]e_{i}).

The covolume of a lattice L ⊂ R^{n} of the form

L = Z ( a_{11} ; ⋮ ; a_{1n} ) + … + Z ( a_{n1} ; ⋮ ; a_{nn} )

is equal to the absolute value of det(a_{ij}). Here R^{n} is equipped with its usual scalar product.

Proposition 4.1. Let V be a real vector space equipped with scalar product and let L ⊂ V be an additive subgroup. The following are equivalent:

(i) L is a lattice; (ii) L is discrete and cocompact; (iii) L contains a basis of V and L ⊂ B is finite for every bounded subset B ⊂ V.

Proof. Let e_{1}, …, e_{n} be a Z-basis of L. Since for every x ∈ L, the intersection of L and the open set ⊕_{i=1}^{n}(−1/2, 1/2)e_{i} is equal to {x}, the group L is discrete. The canonical map ⊕_{i=1}^{n}[0, 1]e_{i} → → V/L is continuous and surjective. This shows that V/L is compact. This shows that (i) implies (ii). To see that (ii) implies (iii), let W be the subspace of V generated by L. Then there is a continuous surjective map V/L → → V/W. This shows that the vector space V/W is compact. Therefore it is zero and L contains a basis. If for some bounded set B, the intersection B ∩ L were infinite, then L would not be discrete.

Finally we show that (iii) implies (i). Let e_{1}, …, e_{n} ∈ L be a basis of V. The intersection of the L with the bounded set B = ⊕_{i=1}^{n}[0, 1]e_{i} is finite. Then L is a finite union:

L = ∪_{x∈B∩L} (x + ⊕_{i=1}^{n}Ze_{i})

It follows that the index m = [L : ⊕_{i=1}Ze_{i}] is finite. This implies that mL ⊂ ⊕_{i=1}Ze_{i}. Therefore L, being a subgroup of finite index of a free group of rank n, is itself free of rank n as required. This follows from Cor. 2.12 applied to the principal ideal domain Z.

Theorem 4.2. (Minkowski’s Convex Body Theorem) Let L be a lattice and let B ⊂ V = L ⊗_{Z} R be a bounded symmetric convex set containing 0. If the volume of B exceeds 2^{n}covol(L), then there is a non-zero vector in B ∩ L.

Proof. Consider the combined map

B ↪ V → → V/2L.

It preserves distances. Since vol(B) is strictly larger than vol(V/2L) = 2^{n}covol(L), the map cannot be injective. There are therefore two distinct elements b_{1}, b_{2} ∈ B that map to the same element in V/2L. In other words, x = \frac{1}{2}(b_{1} + (−b_{2})) ∈ L. Since b_{1} ≠ b_{2}, the vector x is not zero. Since B is symmetric, the vector −b_{2} is contained in B and since B is convex, it contains x. Therefore x is a non-zero vector contained in B ∩ L. This proves the theorem.

Definition. Let L be a lattice and put V = L ⊗_{Z} R. The dual or Z-dual L^{∨} of L is given by

L^{∨} = {v ∈ V : ⟨v, w⟩ ∈ Z for all w ∈ L}.

When V = R^{n} and L is generated by the columns of an invertible matrix (a_{ij}) as above, then the functionals f_{i} : L ⟶ Z defined by f_{i}(w) = ⟨b_{i}, w⟩ where b_{i} denotes the i-th row of the inverse of the matrix(a_{ij}) are clearly a Z-basis for Hom(L, Z). This shows that L^{∨} is the lattice generated by the columns of the matrix ^{t}(a_{ij})^{−1} in R^{n}. The covolume of L^{∨} is equal to the absolute value of the determinant of the matrix ^{t}(a_{ij})^{−1}, which in turn is equal to covol(L)^{−1}.

Theorem 4.3. (Poisson summation formula) Let L be a lattice and let L^{∨} denote its Z-dual. Then

∑_{x∈L} e^{−π||x||^{2}} = \frac{1}{covol(L)} ∑_{x∈L^{∨}} e^{−π||x||^{2}}.

In order to prove this theorem, we consider the Schwartz space of rapidly decreasing functions.

S = {f : R^{n} ⟶ C : for every polynomial g ∈ C[X_{1}, …, X_{n}] and every (higher) partial derivative ∂f of f, the function g · ∂f is bounded}

The Fourier transform f̂ of a function f ∈ S is defined by

f̂(x) = ∫_{R^{n}} f(t)e^{−2πi⟨x,t⟩}dt.

Here x = (x_{1}, …, x_{n}) and t = (t_{1}, …, t_{n}). Similarly ‘dt’ indicates dt_{1} ⋯ dt_{n}.

Lemma 4.4. If f ∈ S then also f̂ ∈ S.

Proof. Let f ∈ S. Then it satisfies

f(x_{1}, …, x_{n}) ≤ \frac{C}{(1 + x_{1}^{2}) ⋯ (1 + x_{n}^{2})}

for some constant C that depends on f. Therefore |f̂| ≤ ∫_{R^{n}} |f|dx_{1} … dx_{n} is bounded. We need to show that not only f̂, but that any function g · ∂f̂ with g ∈ C[X_{1}, …, X_{n}] and any (high) partial derivative of f is bounded.

We have that

\frac{∂f̂}{∂x_{1}} = ∫_{R^{n}} f(t)e^{−2πix·t}dt = ∫_{R^{n}}(−2πit_{1})f(t)e^{−2πix·t}dt.

Since the function (−2πit_{1})f(t) is contained in S, the integral is bounded. It follows inductively that all higher derivatives of f̂ are contained in S. In a similar way, integrating by parts gives that

∫_{R^{n}} \frac{∂f}{∂t_{1}} e^{−2πix·t}dt = 0 − ∫_{R^{n}} f · (−2πix_{1})e^{−2πix·t}dt = 2πix_{1}f.

This shows that x_{1}f̂ ∈ S. Similarly and inductively, g · f̂ ∈ S for every polynomial g ∈ C[X_{1}, …, X_{n}].

This proves the lemma.

Lemma 4.5. The function R^{n} ⟶ C given by x ↦ e^{−π||x||^{2}} has a Fourier transform equal to itself.

Proof. We want to show that

∫_{R^{n}} e^{−2πi⟨x,t⟩}e^{−π||t||^{2}}dt = e^{−π||x||^{2}}.

Proof. Since both sides can be written as products of expressions that depend only on one variable, it suffices to deal with the case of one variable: we need to show that

∫_{−∞}^{∞} e^{−2πixt−πt^{2}}dt = e^{−πx^{2}}.

We integrate the function e^{−πz^{2}} of a complex variable z over the contour in C given by −A → A → A + ix → −A + ix → −A and let A tend to infinity. The integrals over the vertical segments tend to zero. The integral from −A to A tends to ∫_{−∞}^{∞} e^{−πt^{2}}dt = 1. Therefore the integral from A + ix to −A + ix tends to

∫_{−∞}^{∞} e^{−(t+ix)^{2}}dt = 1.

This proves the lemma.

Theorem 4.6. Let f be a function in the Schwartz space S of rapidly decreasing functions R^{n} ⟶ C. Then

∑_{x∈Z^{n}} f(x) = ∑_{x∈Z^{n}} f̂(x).

Proof. Consider the function

g(x) = ∑_{m∈Z^{n}} f(m + x).

Since the sum converges absolutely and uniformly, this is a well-defined function on the torus T^{n} = R^{n}/Z^{n}. For k ∈ Z^{n} we define its Fourier coefficient c_{k} by

c_{k} = ∫_{T^{n}} g(x)e^{−2πikx}dx.

Since for every d > 0 there is a constant C(d, g) > 0 for which |c_{k}| ≤ C(g, d)/||k||^{d}, we have that

| ∑_{||k||>N} c_{k}e^{2πikx}| ≤ C(d, g) ∑_{||k||>N} \frac{1}{||k||^{d}} = O(\frac{1}{N^{d−1}} ),

and we conclude that the Fourier series ∑_{k∈Z^{n}} c_{k}e^{2πikx} converges uniformly and hence pointwise to g(x). In particular,

∑_{m∈Z^{n}} f(m) = g(0) = ∑_{k∈Z^{n}} c_{k}.

Moreover,

c_{k} = ∫_{T^{n}} ∑_{m∈Z^{n}} f(m + x)e^{−2πi⟨k,x⟩}dx,

= ∫_{T^{n}} ∑_{m∈Z^{n}} f(m + x)e^{−2πi⟨k,x+m⟩}dx,

= ∫_{R^{n}} f(x)e^{−2πi⟨k,x⟩}dx,

= f̂(k).

This proves the theorem.

Proof of Theorem 4.3. We may assume that V = L ⊗_{Z} R is just R^{n} with its usual scalar product and that L = A(Z) for some matrix A ∈ GL_{n}(R). In this way the Z-dual lattice L^{∨} is equal to ^{t}A^{−1}(Z). The left hand side of the Poisson summation formula is equal to

∑_{x∈L} e^{−π||x||^{2}} = ∑_{k∈Z^{n}} e^{−π||A(x)||^{2}}.

The function f(x) = e^{−π||A(x)||^{2}} is in the Schwartz space S and by Lemma 4.5 its Fourier transform is equal to

f̂(t) = ∫_{∞}^{∞} e^{−2πi⟨x,t⟩}e^{−π||A(x)||^{2}}dx,

= \frac{1}{|det(A)|} ∫_{∞}^{∞} e^{−2πi⟨A^{−1}(x),t⟩}e^{−π||x||^{2}}dx,

= \frac{1}{|det(A)|} ∫_{∞}^{∞} e^{−2πi⟨x,^{t}A^{−1}(t)⟩}e^{−π||x||^{2}}dx,

= \frac{1}{|det(A)|}e^{−π||^{t}A^{−1}(x)||^{2}}.

It follows that

∑_{k∈Z^{n}} e^{−π||A(x)||^{2}} = \frac{1}{|det(A)|} ∑_{x∈L^{∨}} e^{−π||x||^{2}}

and the result follows.

Exercises.

4.1 Let L′ ⊂ L be two lattices in R^{n}. Show that covol(L) = [L : L′]covol(L). 4.2 Let

L = \{( \begin{matrix}x \\ y \\ z \end{matrix} ) ∈ Z^{3} : x + y + z ≡ 0 (mod 7)\}.

Show that L ⊂ R^{3} is a lattice and compute its covolume. 4.3 Let L ⊂ R^{n} be a lattice. Let A be an invertible n × n-matrix. Show that A(L) is a

lattice. Show that covol(A(L)) = |det(A)|covol(L). Let m ∈ R_{>0}; show that covol(mL) = m^{n}covol(L).

4.4 Identify the quaternions H = {a + bi + cj + dk : a, b, c, d ∈ R} with R^{4} via a + bi + cj + dk ↔ (a, b, c, d). What is the covolume of the ring of Hurwitz integers

Z[i, j, k, \frac{1 + i + j + k}{2} ]

in H ≅ R^{4}? 4.5 Let F be a number field. Suppose R ⊂ F is a subring with the property that its image in

F ⊗ R is a lattice. Show that R ⊂ O_{F}. 4.6 (Euclidean complex quadratic rings.) Let F be an imaginary quadratic number field. We

embed F in C.

(i) Show that O_{F} is Euclidean for the norm if and only if the disks with radius 1 and centers in O_{F} cover C.

(ii) Show that O_{F} is Euclidean for the norm if and only if Δ_{F} = −3, −4, −7, −8 or −11. 4.7 Show that the symmetric space is GL_{2}(R)/O_{2}(R) homeomorphic to R_{>0}^{∗} ×SL_{2}(R)/SO_{2}(R).

Show that the map SL_{2}(R)/SO_{2}(R) ⟶ H = {z ∈ C : Im(z) > 0} given by

( \begin{matrix}a & b \\ c & d \end{matrix} ) ↦ \frac{ai + b}{ci + d}

is a homeomorphism.

5. Number fields. In this section we ‘generalize’ the three rings Z ⊂ Q ⊂ R to an arbitrary number field F.

Definition. Let F be a number field of degree n over Q. The algebra F_{R} is defined by

F_{R} = F ⊗_{Q} R.

In more explicit terms, writing F = Q(α) with f(X) ∈ Q[X] the minimum polynomial of α, we have that F = Q[X]/(f(X)) and hence that F_{R} = R[X]/(f(X)).

Since irreducible polynomials in Q[X] don’t have double zeroes, the Chinese Remainder theorem implies that the R-algebra is isomorphic to a product of copies of R and C. In these terms the natural map F ⟶ F_{R} can be described as follows. Since a field homomorphism σ : F ⟶ C is entirely determined by the zero z in C of f(X) for which σ(α) = z, there are exactly n distinct embeddings

σ : F ⟶ C.

Since f(X) ∈ R[X], the ring homomorphism \overline{σ} given by \overline{σ}(α) = \overline{σ(α)} is an embedding, whenever σ is. If z = σ(α) ∈ R, we have that \overline{σ} = σ. The σ corresponding to z ∉ R come in complex conjugate pairs.

Definition. An infinite prime of a number field F is an embedding σ : F ↪ C considered up to complex conjugation. An infinite prime σ is called real if σ(F) ⊂ R and complex if σ(F) ⊄ R. The number of real and complex infinite primes is denoted by r_{1} and r_{2} respectively.

We have that r_{1} + 2r_{2} = n = [F : Q]. The R-algebra morphism

F_{R} = R[X]/(f(X)) ⟶^{≅} ∏_{σ} F_{σ}

given by mapping X ↦ (σ(α)) is an isomorphism. Here F_{σ} denotes R or C depending on whether σ is real or complex. The product runs over the infinite primes σ : F ↪ C.

Lemma 5.1. Let F be a number field and let x ∈ F. Then

f_{char}^{x}(T) = ∏_{all σ}(T − σ(x))

and hence N(x) = ∏_{all σ} σ(x) and Tr(x) = ∑_{all σ} σ(x). Here “all σ” means that the sum is extended over all embeddings F ↪ C, not merely up to complex conjugation. In terms of infinite primes σ one has that N(x) = ∏_{σ} σ(x)^{deg(σ)} and that Tr(x) = ∑_{σ} deg(σ)σ(x)

Proof. The characteristic polynomial of x viewed as element of the Q-algebra F is the same as the one of x viewed as element of the R-algebra F_{R}. We compute the characteristic polynomial by writing F_{R} as a product of copies of R and C. In this way the element x ∈ F is identified with the vector (σ(x)) where σ runs over the embeddings σ : F ↪ C up to complex conjugation. Since the characteristic polynomial of x is equal to the product of the characteristic polynomials of the various σ(x), we may proceed coordinate by coordinate. For the real coordinates the characteristic polynomial is simply T − σ(x). For complex coordinates, we use the R-basis {1, i} of C and it is the characteristic polynomial of the 2-by-2 matrix

( \begin{matrix}Re σ(x) & −Im σ(x) \\ Im σ(x) & Re σ(x) \end{matrix} ) ,

which is (T − σ(x))(T − \overline{σ(x)}). Taking the product now gives the result.

Definition. Any étale R-algebra A admits a canonical involution and a canonical scalar product as follows. Writing A as a product of copies of R and C, the involution is given by complex conjugation on each coordinate. This involution is functorial. It is a functor on the category of étale R-algebras. We denote it by a ↦ \overline{a} and in terms of it we define a scalar product by

⟨a, b⟩ = Tr(a\overline{b}), for a, b ∈ A.

Definition. Let F be a number field. Its ring of integers O_{F} is the integral closure of Z in F:

O_{F} = {x ∈ F : x is integral over Z}.

By Cor. 1.9 the ring O_{F} is a subring of F. We view it as a subring of the algebra F_{R} through the natural map F ⟶ F_{R}.

Definition. Let F be a number field. The discriminant Δ_{F} of F is the discriminant Δ(O_{F}) of the Z-algebra O_{F}.

Lemma 5.2. Let F be a number field. Let x ∈ F. The following are equaivalent.

(i) the element x is contained in O_{F}; (ii) the minimum polynomial of x is in Z[X]; (iii) the characteristic polynomial of x is in Z[X].

Proof. Gauss’s Lemma (Prop. 2.13 with R = Z) shows that (i) implies (ii). Part (iii) follows from (ii) because f_{char}^{x} is a power of f_{min}^{x} by Exer. 3.2. Finally, it is trivial that (iii) implies (i).

Proposition 5.3. The ring O_{F} is a lattice in F_{R}. Its covolume is equal to \sqrt{|Δ_{F} |}.

Proof. Suppose ω_{1}, …, ω_{n} is a basis for F as a Q-vector space. Then it is also a basis for F_{R} as a R-vector space. Multiplying the ω_{i} by a large integer M does not change these properties. However, we can choose M in such a way that all Mω_{i} are integral. This shows that O_{F} contains an R-basis for F_{R}.

Let μ > 0 and consider the bounded subset B = {x ∈ F_{R} : ||x|| < μ} of F_{R}. Suppose that x is contained in O_{F} ∩B. Then Tr(x\overline{x}) < μ^{2} and hence σ(x) < μ for every embedding σ : F ↪ C. It follows then from Lemma 5.1 that the coefficients of the characteristic polynomial of x are bounded by \binom{2n}{n}μ^{n}. By Lemma 5.2 all these coefficients are in Z. Therefore there are only finitely many possibilities for f_{char}^{x} and hence for x. It follows that O_{F} ∩ B is finite.

The criterion of Prop. 4.1 implies then that O_{F} is a lattice in F_{R}. Its covolume is the absolute value of the discriminant of the matrix

Corollary 5.4. Let F be a number field of degree n. Then there are elements ω_{1}, …, ω_{n} ∈ F so that

\begin{aligned}O_{F} & = Zω_{1} ⊕ ⋯ ⊕ Zω_{n}, \\ ⊂ & ↓ \\ F & = Qω_{1} ⊕ ⋯ ⊕ Qω_{n}, \\ ⊂ & ↓ \\ F_{R} & = Rω_{1} ⊕ ⋯ ⊕ Rω_{n}. \end{aligned}

Definition. Let F be a number field and let I ⊂ O_{F} be a non-zero ideal. The norm N(I) of I is defined as N(I) = [O_{F} : I].

The following proposition shows that N(I) is a well defined natural number.

Proposition 5.5. Let F be a number field. Then

(i) for any non-zero ideal I ⊂ O_{F} the index of I in O_{F} is finite; (ii) any non-zero ideal I ⊂ O_{F} is a lattice in F_{R}. Its covolume is given by

covol(I) = N(I)|Δ_{F} |^{1/2}.

(iii) There are only finitely many O_{F} -ideals I ⊂ O_{F} of bounded norm.

Proof. (i) Let 0 ≠ x ∈ I. Then the constant term a_{0} of its minimum polynomial f is not zero. It is contained in I because a_{0} = a_{0} − f(x) is a multiple of x. Therefore a_{0}O_{F} ⊂ I. Since O_{F} /a_{0}O_{F} ≅ Z/(a_{0}) × … × Z/(a_{0}) is finite, so is O_{F} /I. (ii) Since I ⊂ O_{F}, it is a discrete subgroup of F_{R}. Since I contains a group of the form a_{0}O_{F}, it contains an F -basis. It follows that I ⊂ F_{R} is a lattice. By Exercise 4.2 its covolume is equal to covolume(O_{F}) times [O_{F} : I]. The result now follows from Prop. 5.3. (iii) Any ideal I ⊂ O_{F} of index m contains mO_{F}. Since O_{F} /mO_{F} is a finite group, there are only finitely many ideals of index m.

This proves the proposition

Theorem 5.6. Let F be a number field. Then the ring of integers O_{F} is a Dedekind ring.

Proof. Since Z is Noetherian and since O_{F} is a free abelian group of rank n, any ideal I ⊂ O_{F} is finitely generated group. So it is certainly also finitely generted as an O_{F} - module. Therefore O_{F} is Noetherian. Any non-zero prime ideal of O_{F} has finite index. Therefore it is maximal. This shows that the Krull dimension of O_{F} is 1. Finally, O_{F} is the integral closure of Z in F. By Exerc.1.5 it is therefore a normal ring.

This proves the theorem.

Proposition 5.7. Let F be a number field.

(i) For any non-zero x ∈ O_{F} we have that |N(x)| = N((x)); (ii) N(IJ) = N(I)N(J) for any two non-zero ideals I, J ⊂ O_{F}.

Proof. (i) We prove that |N(x)|N(I) = N(xI) for any non-zero O_{F} -ideal I. Let ω_{1}, …, ω_{n} be a Z-basis for O_{F}, let ∑_{j} b_{ij}ω_{j} (for i = 1, …, n) be a Z-basis for I and let A be the matrix of the multiplication my x map with respect to the basis ω_{1}, …, ω_{n}. Then N(I) = |det(A)| and N(xI) = |det(AB)|. Since N(x) = det(A), the result follows. (ii) We use the fact that O_{F} is a Dedekind ring. By the proof of Prop. 2.6, there are non-zero x, y ∈ O_{F} and there is an ideal J′ ⊂ O_{F} that is coprime to I so that xJ = yJ′. By the Chinese Remainder Theorem we have that O_{F} /IJ′ ≅ (O_{F} /I) × (O_{F} /J′) so that N(IJ′) = N(I)N(J′). Therefore

\begin{aligned}|N(x)|N(IJ) & = N(xIJ) = N(yIJ′) = |N(y)|N(I)N(J′) = \\ & = N(I)N(yJ′) = N(I)N(xJ) = |N(x)|N(I)N(J), \end{aligned}

as required.

Exercises.

5.1 Let F be a number field.

(i) Let α ∈ F. Show that there exist an integer 0 ≠ m ∈ Z such that mα ∈ O_{F}.

(ii) Show that for every number field F there exists an integral element α ∈ O_{F} such that

F = Q(α). (iii) Show that the field of fractions of O_{F} is F. (iv) Let F ⊂ K be an extension of number fields. Show that O_{K} ∩ F = O_{F}.

5.2 Let F be a number field. Show that every ideal I ≠ 0 of O_{F} contains a non-zero integer

m ∈ Z. 5.3 Let F be a number field and let α ∈ O_{F}. Show that N(α) = ±1 if and only if α is a unit of

the ring O_{F}. 5.4 Let F be a number field. Let r_{1} and r_{2} denote the number of real and complex infinite primes

respectively. Show that the sign of Δ_{F} is (−1)^{r_{2}}. 5.5 Let A be an étale R-algebra. Show that Tr(a\overline{b}) defines a scalar product on A: show that it

is symmetric, bilinear and positive definite. 5.6 Let F be a number field. Show that \overline{L} = \{\overline{x} : x ∈ L\} ⊂ F_{R} is a lattice if and only L is.

Moreover, L and \overline{L} have the same covolume. 5.7 Let F be a number field.

(i) Show that ||x|| = ||\overline{x}|| for all x ∈ F_{R}.

(ii) For x = (x_{σ}) ∈ F_{R} we let |x| ∈ F_{R} the element whose σ-th component is equal to |x_{σ}|.

Show that ||x|| = |||x|||. (iii) Let φ : F_{R} ⟶ F_{R} be F_{R}-linear. Show that φ is given by multiplication by some

element x ∈ F_{R}. (iv) Let x ∈ F_{R}. Show that the multiplication by x map preserves the canonical scalar

product if and only if |x| = 1.

5.8*(Stickelberger 1923) Let F be a number field of degree n. Let {ω_{1}, ω_{2}, …, ω_{n}} be a Z-basis for the ring of integers of F. Let σ_{i} : F ↪ C denote the n distinct embeddings of F into C.

By A_{n} we denote the normal subgroup of even permutations of the symmetric group S_{n}. We define Δ^{+} = ∑_{τ∈A_{n}} ∏_{i=1}^{n} ϕ_{i}(ω_{τ(i)}) and Δ^{−} = ∑_{τ∈S_{n}−A_{n}} ∏_{i=1}^{n} ϕ_{i}(ω_{τ(i)}). Prove, using Galois theory, that Δ^{+}+Δ^{−} e Δ^{+}Δ^{−} are in Z. Conclude that Δ_{F} = (Δ^{+}+Δ^{−})^{2}−4Δ^{+}Δ^{−} ≡ 0 or 1 (mod 4).

6. Rings of integers. In this section we explain how to compute the ring of integers O_{F}. Once we have a presentation of O_{F} as a Z-algebra, we show how to compute explicit generators for the prime ideals of O_{F}.

Suppose that F is a number field F and that it is given as Q(α) for some element α ∈ F. Suppose in addition that α ∈ O_{F}. This is not a real restriction since Mα is integral for an appropriate choice of a non-zero integer M. Since Z[α] contains a Q-basis for F and is contained in O_{F}, it is a lattice and has the same rank as O_{F}. Therefore the index [O_{F} : Z[α]] is finite.

We consider, in general, a subring R ⊂ O_{F} of finite index [O_{F} : R]. In the applications, R is explicitly given and we want to either decide that R = O_{F} or, if it isn’t, construct generators for O_{F}. This is done inductively. The ring R is called maximal at a prime number p, if p does not divide [O_{F} : R]. We describe an algorithm that for each prime number p either proves that R is maximal at p or exhibits a strictly larger ring R′ ⊂ O_{F} with [R′ : R] a power of p. Applying this algorithm inductively, we can compute generators for O_{F}.

We first make an easy observation.

Lemma 6.1. Let F be a number field and let R be a subring of O_{F} of finite index. If R is not maximal at a prime p, then p^{2} divides Δ(R/Z). In particular, if Δ(R/Z) is square-free, we have that R = O_{F}.

Proof. This follows from the fact that

Δ(R/Z) = [O_{F} : R]^{2}Δ(O_{F} /Z).

This proves the lemma.

Proposition 6.2. Let F be a number field, let p be a prime and let R be a subring of O_{F} of finite index. Let N = {x ∈ R : x (mod pR) is contained in Nil(R/pR)}. Put R′ = {x ∈ O_{F} : xN ⊂ N} and R′′ = {x ∈ O_{F} : px ∈ R}. Then

R′ = R′′.

Proof. Since p ∈ N, every x ∈ R′ has the property that px ∈ N ⊂ R. So we have that R ⊂ R′ ⊂ R′′. To show that R′′ ⊂ R′, let x ∈ R′′ and let y ∈ N. Then y^{m} ≡ 0 (mod pR). It follows that xy^{m} ∈ R and hence that y^{mk}x^{k} = y^{m}(xy^{m})^{k} ∈ pR for all k ≥ 1. Since x is integral, it satisfies a relation of the form x^{n} + … + a_{1}x = a_{0} = 0 with a_{i} ∈ Z ⊂ R. This implies that

y^{mn}x^{k} ∈ pR, for all k ≥ 1.

Indeed y^{mn}x^{k} ∈ pR for all k ≤ n − 1 and the integrality relation then gives the same for all k ≥ 1.

In particular, (xy)^{mn} ∈ pR showing that xy ∈ N. Since y was chosen arbitrarily, this implies that x ∈ R′ as required.

Corollary 6.3. In the notation of Proposition 6.2, we have that R = R′ if and only if the homomorphism

m : R/pR ⟶ End(N/pN)

given by mapping x ∈ R to the multiplication by x map, is injective. Moreover, if the map is not injective, then for any x ∈ R for which x (mod pR) is contained in ker h, we have that

R ⊂_{≠} R[\frac{x}{p}] ⊂ O_{F}

and the index [R[\frac{x}{p}] : R] is a power of p.

Proof. Suppose that m is injective and let x ∈ R′. Then xN ⊂ N and hence xpN ⊂ pN. Since x ∈ R′ ⊂ R′′, the element xp is contained in R. It follows that xp is in the kernel of m. Since m is injective, we have that px = px′ for some x′ ∈ R. It follows that x = x′ ∈ R showing that R = R′.

Conversely, suppose that R = R′ and let x ∈ R with m_{x} the zero-map. Then xN ⊂ pN. This implies that \frac{x}{p}N ⊂ N. Since N is finitely generated, this implies that \frac{x}{p} ∈ O_{F}. By definition, it is contained in R′ = R. In other words, x = px′ for some x′ ∈ R. This shows that x (mod pR) is zero and that m is injective.

The second part of this proof also shows that \frac{x}{p} ∈ O_{F} for every x ∈ R in the kernel of m. Since p\frac{x}{p} = x ∈ R, the index is a power of p as required.

The algorithm mentioned above proceeds as follows. Let F = Q(α) be a number field with α ∈ O_{F}. First we apply Lemma 6.1 and compute the discriminant of R and make a list of the primes p for which p^{2} divides Δ(R/Z). For each prime p in the list we compute the nilradical of the finite ring R/pR and check whether the map m of Cor. 6.3 is injective. This involves only linear algebra over F_{p}. When m is injective, we know that p does not divide [O_{F} : R] and we are done with p. If on the other hand m is injective, then we pick x ∈ R with x ≢ 0 (mod p) but m(x) = 0 and replace R by the strictly larger ring R[\frac{x}{p}]. If the discriminant of this new ring is not divisible by p^{2} we are once again done with p. If not we repeat this procedure.

Since each time the ring R ‘becomes’ strictly larger, the algorithm terminates. At the end we have that R = O_{F}.

Corollary 6.4. Let F = Q(α) be a number field and let p be a prime. Suppose that α ∈ O_{F} and that the minimum polynomial of α is an Eisenstein polynomial with respect to p. Then p does not divide the index [O_{F} : Z[α]]

Proof. Put R = Z[α]. Since the minimum polynomial f(X) of α is Eisenstein, the ring R/pR is isomorphic to (Z/pZ)[X]/(X^{n}). Here n = [F : Q]. The ideal N of Prop. 6.2 is therefore equal to (α, p) ⊂ R. We check that the map m of Cor. 6.3 is injective. Let x ∈ R be in the kernel of m. Then x = g(α) for some polynomial g(X) ∈ Z[X] of degree at most n − 1 and g(α)N ⊂ pN. This implies that

g(X)X = p^{2}h_{1}(X) + pXh_{2}(X) + f(X)h_{3}(X)

for certain polynomials h_{1}, h_{2}, h_{3} ∈ Z[X]. Putting X = 0, we see that f(0)h_{3}(0) ≡ 0 (mod p^{2}). Since f is Eisenstein, this implies that p divides h_{3}(0). Now take the equation modulo p. This leads to the congruence g(X)X ≡ X^{n}h_{3}(X) (mod p). The degree of the polymonial on the left is at most n while the degree of the polynomial on the right is at least n + 1. This shows that g(X) ≡ 0 (mod p) and hence that x ∈ pR and that m is injective.

The result now follows from Cor. 6.3.

Example. Consider F = Q(\sqrt[3]{17}) and put R = Z[\sqrt[3]{17}]. Then R ⊂ O_{F} has finite index. A Z-basis for R is given by \{1, \sqrt[3]{17}, (\sqrt[3]{17})^{2}]\} and the discriminant of R is equal to the determinant of the matrix of traces

( \begin{matrix}3 & 0 & 0 \\ 0 & 0 & 3 · 17 \\ 0 & 3 · 17 & 0 \end{matrix} ) = −3^{3} · 17^{2}.

Therefore R is maximal at all primes p except possibly p = 3 and 17. Since the minimum polynomial X^{3} −17 of \sqrt[3]{17} is Eisenstein with respect to the prime 17, Cor. 6.4 implies that R is maximal at 17. In order to study the prime p = 3, we compute the nilradical of the ring R/3R ≅ Z[X]/(X^{3} − 17). Since X^{3} − 17 ≡ (X + 1)^{3} (mod 3), the nilradical of R/3R is generated by X + 1 and the R-ideal N of Cor. 6.3 is equal to (3, α) where α = \sqrt[3]{17} + 1. Note that α^{3} − 3α^{2} + 3α − 18 = (α − 1)^{3} − 17 = 0

A Z-basis for R is given by {1, α, α^{2}} and a Z-basis for N by {3, α, α^{2}}. Next we study kernel of the homomorphism

m : R/3R ⟶ End(N/3N)

that maps x ∈ R to the multiplication by x map. Suppose that a, b, c ∈ Z and x = a + bα + cα^{2} ∈ R. We multiply the three F_{3}-basis vectors of N/3N by x. We have that x · 3 = 3a + 3bα + 3cα^{2} ≡ a · 3 (mod 3N). Since α^{3} ≡ 0 (mod 3N), we have that x·α = a·α+b·α^{2} and x·α^{2} = a·α^{2}. Therefore the matrix that describes the multplication by x map with respect to the basis {3, α, α^{2}} is given by

( \begin{matrix}a & 0 & 0 \\ 0 & a & 0 \\ b & a & 0 \end{matrix} ) .

It follows that ker(m) is a 1-dimensional F_{3}-vector space generated by α^{2}. By Corollary 6.3, the ring

R′ = Z[\sqrt[3]{17}, β],  with  β = \frac{α^{2}}{3} = \frac{(1 + \sqrt[3]{17})^{2}}{3} ,

is a subring of O_{F} strictly containing R. Since Δ(R/Z) = 3^{3} · 17^{2}, the index satisfies [R′ : R] = 3 and the discriminant of R′ is equal to 3 · 17^{2}. It follows that O_{F} = R′.

In the remainder of this section we explain how one can explicitly compute the prime ideals of the ring of integers O_{F} of a number field F.

By Proposition 5.5 any non-zero prime ideal contains an integer and hence a prime number p. Therefore, fixing a prime number p, the prime ideals containing p correspond exactly the prime ideals of the quotient ring O_{F} /pO_{F}. More precisely, the prime ideals 𝔭 dividing p are of the form

𝔭 = {x ∈ O_{F} : x (mod p) ∈ 𝔓}, for some prime ideal 𝔓 of R/pR.

The ring R/pR is a finite F_{p}-algebra and therefore it is a product of local F_{p}-algebras R/pR ≅ ∏_{i} A_{i}, where A_{i} is local with nilpotent maximal ideal. This is a computation with a finite F_{p}-algebra that is in practice not very difficult.

It is easy to make this explicit in terms of the prime ideals 𝔭 of O_{F} that divide p. Indeed, since O_{F} is a Dedekind ring, the O_{F} -ideal (p) admits a factorization as a product of prime ideals (p) = ∏_{𝔭} 𝔭^{e_{𝔭}}. Here the product runs over distinct prime ideals 𝔭. By the Chinese Remainder Theorem we have then that

R/pR ≅ ∏_{𝔭|p} O_{F} /𝔭^{e_{𝔭}}.

Each of the factors is a local F_{p}-algebra with maximal nilpotent ideal 𝔭/𝔭^{e_{𝔭}}.

Before we give some explicit examples, we introduce some terminology. For a prime ideal 𝔭 oif O_{F}, the exponent e_{𝔭} is called the inertia index of 𝔭. Prime ideals 𝔭 for which e_{𝔭} exceeds 1 are said to be ramified. If 𝔭 contains the prime number p, then the residue field O_{F} /𝔭 is a finite field extension of F_{p}. We denote by f_{𝔭} the degree of this extension. We have that N(𝔭) = p^{f_{𝔭}}. A prime number p is said to be inert if the O_{F} -ideal (p) is a prime ideal of O_{F}. It is said to be ramified if some prime ideal divising it is ramified. A prime number is said to be totally split if (p) = ∏_{𝔭} 𝔭 and f_{𝔭} = 1 for every 𝔭.

Proposition 6.5. Let F be a number field and let p be a prime number. Then

(i) Denoting by 𝔭 the prime divisors of the O_{F} -ideal (p), we have that

∑_{𝔭|p} f_{𝔭}e_{𝔭} = n = [F : Q].

(ii) The prime p is ramified if and only if it divides Δ(O_{F} /Z). In particular, there are only finitely many ramified primes.

Proof. To prove (i), just take the norm of the relation (p) = ∏_{𝔭} 𝔭^{e_{𝔭}}. To prove part (ii), note that the discriminant of the F_{p}-algebra O_{F} /(p) is just Δ(O_{F} /Z) (mod p). Therefore p divides the discriminant of O_{F} if and only the F_{p}-algebra is not a product of fields. This means that one of the local factors is of the form O_{F} /𝔭^{e_{𝔭}} with e_{𝔭} stricly larger than 1. In other words, if and only if p is ramified.

Example. Take F = Q(\sqrt[3]{17}) and p ≠ 3. Since the index of the ring R = Z[ ^{3} \sqrt{1}7] inside O_{F} is 3, which is prime to p, the natural map R/pR ⟶ O_{F} /(p) is an isomorphism. It is convenient to work with R rather than O_{F}. For instance R/2R ≅ F_{2}[X]/(X^{3} − 17) ≅ F_{2}[X]/(X + 1) × F_{2}[X]/(X^{2} + X + 1) since X^{3} − 17 ≡ (X + 1)(X^{2} + X + 1) (mod 2).

Therefore (2) = 𝔭_{2}𝔭_{4} with 𝔭_{2} = (2, 1 + \sqrt[3]{17}) and 𝔭_{4} = (2, 1 + \sqrt[3]{17} + \sqrt[3]{17}^{2}). We have that f_{𝔭_{2}} = 1 and f_{𝔭_{4}} = 2. Since R/2R is a product of fields, the inertia indices e_{𝔭_{2}} and e_{𝔭_{4}} are equal to 1. Indeed, we have that f_{𝔭_{2}}e_{𝔭_{2}} + f_{𝔭_{4}}e_{𝔭_{4}} = 1 + 2 = [F : Q].

For the prime 3 we do not use the ring R and turn to the ring O_{F} instead. We have shown above that O_{F} = Z[α, β] with α = \sqrt[3]{17} + 1 and β = (\sqrt[3]{17} + 1)^{2}/3. In order to write a presentation of O_{F} as a Z-algebra, we compute all possible products between the two generators:

\begin{aligned}α^{2} & = 3β, \\ αβ & = \frac{α^{3}}{3} = \frac{3α^{2} − 3α + 18}{3} = 3β − α + 6 \\ β^{2} & = \frac{α^{4}}{9} = \frac{3βα − α^{2} + 6α}{3} = βα − β + 2α = 2β + α + 6. \end{aligned}

This means that

O_{F} ≅ Z[X, Y]/(X^{2} − 3Y, XY − 3Y + X − 6, Y^{2} − 2Y − X + 6)

and hence

O_{F} /(3) ≅ F_{3}[X, Y]/(X^{2}, XY + X, Y^{2} + Y − X),
≅ F_{3}[Y]/((Y^{2} + Y)^{2}, (Y^{2} + Y)(Y + 1)),
≅ F_{3}[Y]/((Y^{2} + Y)(Y + 1)),
≅ F_{3}[Y]/(Y) × F_{3}[Y]/((Y + 1)^{2}).

It follows that (3) = 𝔭_{3}^{2}𝔭′_{3} with 𝔭_{3} = (β + 1, 3) and 𝔭′_{3} = (β, 3).

Exercises.

6.1 (Kummer’s Lemma) Suppose f ∈ Z[T] is an irreducible polynomial. Let α denote a zero of f and let F = Q(α). Let p be a prime number not dividing the index [O_{F} : Z[α]]. Suppose the polynomial f factors in F_{p}[T] as

f(T) = h_{1}(T)^{e_{1}} · . . . · h_{g}(T)^{e_{g}}

where the polynomials h_{1}, …, h_{g} are the distinct irreducible factors of f modulo p. Show that the prime factorization of the ideal (p) in O_{F} is given by

pO_{F} = 𝔭_{1}^{e_{1}} · . . . · 𝔭_{g}^{e_{g}} ,

where the 𝔭_{i} = (h_{i}(α), p) are distinct prime ideals with N(𝔭_{i}) = p^{deg(h_{i})}. 6.2 Let F = Q(α) where α be a zero of the polynomial T^{3} −T −1. Show that the ring of integers

of F is Z[α]. Find the factorizations in Z[α] of the primes less than 10. 6.3 Let d be a squarefree integer and let F = Q(\sqrt{d}) be a quadratic field. Show that for odd

primes p one has that p splits (is inert, ramifies) in F over Q if and only if d is a square (non-square, zero) modulo p.

6.4 Let ζ_{5} denote a primitive 5th root of unity. Determine the decomposition into prime factors in Q(ζ_{5}) of the primes less than 14.

6.5 Show that the ring of integers of F = Q(\sqrt[3]{20}) is equal to Z[\sqrt[3]{20}, \sqrt[3]{50}]. Show there is no α ∈ O_{F} such that O_{F} = Z[α].

6.6 Show that the following three polynomials have the same discriminant:

T^{3} − 18T − 6,
T^{3} − 36T − 78,
T^{3} − 54T − 150.

Let α, β and γ denote zeroes of the respective polynomials. Show that the fields Q(α), Q(β) and Q(γ) have the same discriminants, but are not isomorphic. (Hint: the splitting behavior of the primes is not the same.)

6.7*(Samuel) Let f(T) = T^{3} + T^{2} − 2T + 8 ∈ Z[T]. Show that f is irreducible.

(i) Show that Disc(f) = −4 · 503. Show that the ring of integers of F = Q(α) admits 1, α, β = (α^{2} − α)/2 as a Z-basis.

(ii) Show that O_{F} has precisely three distinct ideals of index 2. Conclude that 2 splits

completely in F over Q. (iii) Show that there is no α ∈ F such that O_{F} = Z[α]. Show that for every α ∈ O_{F} − Z,

the prime 2 divides the index [O_{F} : Z[α]].

6.8*(Dedekind’s Criterion.) Suppose α is an algebraic integer with minimum polynomial over f(T) ∈ Z[T]. Let F = Q(α). For p be a prime number, let f_{1}, …, f_{g} ∈ Z[T] and e_{1}, …, e_{g} ∈ Z_{≥1} such that f = f_{1}^{e_{1}} ·. . .·f_{g}^{e_{g}} is the decomposition of f into distinct irreducible polynomials f_{i} modulo p. Show that

p divides the index [O_{F} : Z[α]]

if and only if there is an index j such that

f_{j} divides (\frac{f(T) − ∏_{j} f_{j}(T)^{e_{j}}}{p} ) in F_{p}[T]  and  e_{j} ≥ 2.

7. Arakelov divisors. In this section we introduce the Arakelov class group. Let F be a number field. To every prime ideal 𝔭 of its ring of integers O_{F} we associate a 𝔭-adic absolute value on F by putting

|x|_{𝔭} = e^{−ord_{𝔭}x}, for x ∈ F^{∗}

and |0|_{𝔭} = 0. We have the following properties:

\begin{aligned}|xy|_{𝔭} & = |x|_{𝔭}|y|_{𝔭}, \\ |x + y|_{𝔭} & ≤ max(|x|_{𝔭}, |y|_{𝔭}) ≤ |x|_{𝔭} + |y|_{𝔭} \end{aligned}

for x, y ∈ F. The second property is called the Triangle Inequality. In this way, F acquires the structure of a metric toplogical space. The two properties are shared by the usual absolute values on R or C. Inverting the process, we now define for each infinite prime the homomorphism ord_{σ} : F^{∗} ⟶ R given by

ord_{σ}(x) = − log |σ(x)|.

Definition. Let F be a number field. The Arakelov divisor group or divisor group Div_{F} of F is defined as

Div_{F} = ⊕_{𝔭} Z × ⊕_{σ} R.

Here the first sum runs over the non-zero prime ideals of the ring of integers O_{F} of F and the second sum runs over the infinite primes σ : F ⟶ C. Elements of Div_{F} are called divisors or Arakelov divisors. We write elements D ∈ Div_{F} as finite formal sums

D = ∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ

with x_{σ} ∈ R and n_{𝔭} ∈ Z. All but finitely many of the n_{𝔭} are zero. The support of D is the set of primes 𝔭 and σ with non-zero coefficients.

Consider the homomorphism

d : F^{∗} ⟶ Div_{F}

that maps x ∈ F^{∗} to the principal divisor (x) = ∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ given by n_{𝔭} = ord_{𝔭}(x) for primes 𝔭 of O_{F} and x_{σ} = ord_{σ}(x) for infinite primes σ.

Proposition 7.1. Let F be a number field. The kernel of the map d : F^{∗} ⟶ Div_{F} given by x ↦ (x) is a finite group equal to the group of roots of unity μ_{F} of F.

Proof. Let x ∈ μ_{F}. Then x^{m} = 1 for some integer m ≥ 1. Then x is a unit so that ord_{𝔭}x = 0 for all primes 𝔭. In addition, |σ(x)|^{m} = 1 and hence |σ(x)| = 1 so that ord_{σ}(x) = 0 for every σ. This shows that μ_{F} is contained in the kernel of d. Now we show that ker(d) is finite. Any x ∈ ker(d) satisfies ord_{𝔭}(x) = 0 for all primes 𝔭 and |σ(x)| = 1 for every σ. This implies that x is in O_{F} and that x ∈ O_{F} ⊂ F_{R} is contained in the bounded set of vectors (v_{σ}) ∈ F_{R} = ∏_{σ} F_{σ} satsifying |v_{σ}| ≤ 1. Since O_{F} is a lattice, there are only finitely many possibilities for x and hence the kernel of d is a finite group. It follows that there is an integer m ≥ 1 for which x^{m} = 1 for all x in ker d, showing that the kernel of d is contained in μ_{F} as required.

We identify the real vector space ⊕_{σ}R with the subalgebra ∏_{σ} R of F_{R} = ∏_{σ} F_{σ} and equip it with the canonical scalar product of F_{R}. It induces the usual topological group structure on ⊕_{σ}R. Providing the group ⊕_{𝔭}Z with the discrete topology, the group of Arakelov divisors Div_{F} acquires the structure of a locally compact Hausdorff topological group. It is a countable union of copies of the vector space ⊕_{σ}R. Since we have a scalar product on ⊕_{σ}R, the group Div_{F} is actually a Riemannian manifold.

Proposition 7.2. Let F be a number field. Then the image of the map d : F^{∗} ⟶ Div_{F} is a closed discrete subset of Div_{F}.

Proof. For any divisor D = ∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ and any ε > 0, the set

U(D) = {∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ : |y_{σ} − x_{σ}| < ε}

is an open neighborhood of D in Div_{F}. Moreover, such sets form a basis for the topology on Div_{F}. It suffices to show that these kind of sets contain only finitely many principal divisors. Suppose there is at least one such divisor (x) in U(D) and let (y) be a second one. Then u = y/x has the property that ord_{𝔭}(u) = n_{𝔭} − n_{𝔭} = 0 for all primes 𝔭. Therefore u is contained in O_{F}^{∗}. From | − log |σ(x)| − x_{σ}| < ε and | − log |σ(y)| − x_{σ}| < ε we deduce that | log |σ(u)|| < 2ε for every σ. It follows that the collection of such units u is contained in a bounded subset of F_{R}. Therefore there are only finitely many and the proposition follows.

Definition. Let F be a number field. By Pic_{F} we denote the quotient of the group Div_{F} by the image of F^{∗} under the map d. In other words, Pic_{F} is the group of Arakelov divisors modulo principal Arakelov divisors.

Since the image of F^{∗} is closed in Div_{F}, the group Pic_{F} has a natural induced structure of a topological group. It is Hausdorff and locally compact. In addition, Pic_{F} inherits the Riemannian structure from Div_{F}.

Definition. Let F be a number field. The degree of a non-zero prime ideal is defined by

deg(𝔭) = log N(𝔭).

The degree of an infinite prime σ is defined to be 1 or 2 depending on whether σ is real or complex. We extend the degree linearly to the entire divisor group and in this way we obtain a continuous homomorphism

deg : Div_{F} ⟶ R.

The degree map is surjective. Its kernel is denoted by Div_{F}^{0} .

Proposition 7.3. (Product Formula) Principal Arakelov divisors have degree zero.

Proof. Let x ∈ F^{∗}. By Prop. 5.7 we have that

N((x)) = |N(x)|.

Let xO_{F} = ∏_{𝔭} 𝔭^{n_{𝔭}} be the prime decomposition of the principal ideal generated by x. It follows from the multiplicativity of the norm map and Lemma 5.1 that

∏_{𝔭} N(𝔭)^{n_{𝔭}} = ∏_{σ} |σ(x)|^{deg σ}.

Taking logarithms implies the result.

Proposition 7.3 says that the image of F^{∗} under the map d is contained in the subgroup Div_{F}^{0} . This leads to the following definition.

Definition. Let F be a number field. The Picard-Arakelov class group or Arakelov class group Pic_{F}^{0} is the group Div_{F}^{0} modulo the image of F^{∗} under the map d.

There is a natural exact sequence

0 ⟶ Pic_{F}^{0} ⟶ Pic_{F} ⟶^{deg} R ⟶ 0.

The Arakelov class group inherits its structure of a topological group and Riemannian manifold from the group Pic_{F}. In the rest of this section we relate it to the class group and the unit group of the Dedekind ring O_{F}. The image of the principal Arakelov divisors under the projection map from Div_{F} to the group of fractional ideals ⊕_{𝔭}Z is precisely the group Pr_{F} of principal fractional ideals. This leads to the following commutative diagram with exact rows and columns.

0  0  0
↓  ↓  ↓
0 ⟶ O_{F}^{∗} /μ_{F} ⟶ F^{∗}/μ_{F} ⟶ Pr_{F} ⟶ 0
↓  ↓  ↓
0 ⟶ ⊕_{σ} R ⟶ Div_{F} ⟶ ⊕_{𝔭} Z ⟶ 0
↓  ↓  ↓
0 ⟶  T  ⟶ Pic_{F} ⟶ Cl(O_{F}) ⟶ 0
↓  ↓  ↓
0  0  0

Here T is defined as the cokernel of the homomorphism O_{F}^{∗} ⟶ ⊕_{σ}R. Recall that the latter homomorphism maps a unit u to the vector(−log|σ(u)|). Note that the quotient topology on Cl(O_{F}) is discrete. All homomorphisms in the diagram all continuous. This is either trivially so or it follows from the universal property of the quotient topology.

Now we take degree zero parts. First in the central column. Since F has at least one infinite prime, the projection map Div_{F}^{0} ⟶ ⊕_{𝔭}Z is still surjective. Its kernel is given by

(⊕_{σ}R)^{0} = {(x_{σ}) ∈ ⊕_{σ}R : ∑_{σ} deg(σ)x_{σ} = 0}.

Denoting the cokernel of the homomorphism O_{F}^{∗} /μ_{F} ⟶ (⊕_{σ}R)^{0} by T^{0} we obtain the following commutative diagram with exact rows and columns and continuous homomorphisms. The bottom row shows that Pic_{F}^{0} is an extension of the discrete ideal class group Cl(O_{F}) by the connected group T^{0}.

0  0  0
↓  ↓  ↓
0 ⟶ O_{F}^{∗} /μ_{F} ⟶ F^{∗}/μ_{F} ⟶ Pr_{F} ⟶ 0
↓  ↓  ↓
0 ⟶ (⊕_{σ} R)^{0} ⟶ Div_{F}^{0} ⟶ ⊕_{𝔭} Z ⟶ 0
↓  ↓  ↓
0 ⟶  T^{0}  ⟶ Pic_{F}^{0} ⟶ Cl(O_{F}) ⟶ 0
↓  ↓  ↓
0  0  0

For F = Q there is only one infinite prime and therefore T^{0} is trivial. Since Z is a unique factorization domain, the class group is trivial as well. It follows that Pic_{Q} is trivial and that

Pic_{Q} ⟶_{≅}^{deg} R

is an isomorphism. See sections 10 and 13 for non-trivial examples of Arakelov-Picard class groups.

Definition. Let F be a number field. We define a norm on the connected component T of Pic_{F} by putting

||x||_{Pic} = min_{ε∈O_{F}^{∗}} ||εx||.

Here ||εx|| denotes the length of εx ∈ F_{R} with respect to the canonical scalar product.

The norm on Pic_{F} satisfies the triangle inequality and induces the usual metric structure on Pic_{F}.

Exercises.

7.1 Let G be a topological group with neutral element 1.

(i) Show that for every open neighborhood U of 1 ∈ G, there is an open neighborhood V of 1 for with V^{2} ⊂ U. Here V^{2} = {vw : v, w ∈ V }. (ii) Show that G is Hausdorff if and only if 1 is closed.

7.2 Let G be a topological group.

(i) Show that every open subgroup of G is also closed.

(ii) Let H ⊂ G be a normal subgroup. Show that G/H, equipped with the quotient

topology, is Hausdorff if and only if H is closed in G. 7.3 Let F = Q(\sqrt[3]{−2}). Let x = 1 − \sqrt[3]{−2}. Compute the coordinates of the principal

Arakelov divisor (x). Check that deg((x)) = 0. 7.4 Show that the Pic_{F}^{0} = 0 for F = Q(i). 7.5 Let F be a number field and let x, y ∈ T = (⊕_{σ}R) /im(O_{F}^{∗}). Show that ||x + y||_{Pic} ≤

||x||_{Pic} + ||y||_{Pic}.

8. Ideal lattices. In this section we interpret Arakelov divisors as ideal lattices. We show that the Arakelov class group classifies ideal lattices up to isometry. We first introduce the Hermitian line bundle associated to an Arakelov divisor.

Let F be a number field and let D = ∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ be an Arakelov divisor. We put I = ∏_{𝔭} 𝔭^{−n_{𝔭}} and call it the ideal associated to D. Then we let u denote the vector (u_{σ}) ∈ ∏_{σ} R_{>0}^{∗} for which u_{σ} = exp(−x_{σ}) for each σ. Since the group ∏_{σ} R_{>0}^{∗} is contained in F_{R}^{∗} ≅ ∏_{σ} F_{σ}^{∗}, we may view u as an element of F_{R}^{∗}. It is totally positive in the sense that each coordinate u_{σ} is contained in the subgroup R_{>0}^{∗} of F_{σ}^{∗}.

Notation. Let F be a number field and letD = ∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ be an Arakelov divisor. Then the Hermitian line bundle associated to D is the pair (I, u) where I is a fractional ideal and u is a totally positive unit in F_{R}^{∗} as explained above.

For any Arakelov divisor D = (I, u) we have that N(D)^{−1} = N(u)N(I). To D we associate the O_{F} -submodule

uI = {ux : x ∈ I} ⊂ F_{R}.

Since I is a lattice in F_{R}, so is uI. We have by Prop. 5.5 that

covol(uI) = N(u)N(I)\sqrt{|Δ_{F} |} = e^{−deg(D)}\sqrt{|Δ_{F} |}.

Definition. Let F be a number field. An ideal lattice associated to F is a projective O_{F} -module M of rank 1 together with a scalar product on the F_{R}-module L ⊗_{Z} R that satisfies ⟨λx, y⟩ = ⟨x, \overline{λ}y⟩ for all x, y ∈ L ⊗_{Z} R and all λ ∈ F_{R}. Two ideal lattices L and L′ are said to be isometric if there is an isomorphism f : L ⟶ L′ of O_{F} -modules that is compatible with the scalar products on L ⊗_{Z} R and L′ ⊗_{Z} R.

Proposition 8.1. Let F be a number field. Then two Arakelov divisors D = (I, u) and D′ = (I′, u′) have the same classes in Pic_{F} if and only if the associated ideal lattices uI and u′I′ are isometric.

Proof. If D + (f) = D′ for some f ∈ F^{∗}, then we have that I = fI′ and u|f| = u′. This means that the map F_{R} ⟶ F_{R} given by multiplication by u^{−1}f^{−1}u′ is O_{F} -linear and maps I to I′. Since |u^{−1}f^{−1}u′| = 1 it also preserves the canonical scalar product on F_{R}. This is the content of by Exer. 5.7. Conversely, suppose that φ : uI ⟶ u′I′ is O_{F} -linear and compatible with the scalar products on uI ⊗_{Z} R = u′I′ ⊗_{Z} R = F_{R}. Then the induced map F_{R} ⟶ F_{R} is F_{R}-linear and compatible with the canonical scalar product. By Exer. 5.7 it is therefore given by multiplication my some element x ∈ F_{R}^{∗} with |x| = 1. It follows that the element f = u^{−1}x^{−1}u′ is in F^{∗} and satisfies |u^{−1}f^{−1}u′| = u^{−1}|f|^{−1}u′ = 1. Since fI′ = I, we see that D + (f) = D′. This proves the proposition.

Proposition 8.2. Let F be a number field. The map that associates to an Arakelov divisor D = (I, u) the ideal lattice uI induces a bijection

Pic_{F} ⟶^{≅} {ideal lattices up to isometry}.

Proof. By the previous proposition the map that associates the lattice uI to an Arakelov divisor D = (I, u) is well defined and injective. We need to show it is surjective. Let L be an ideal lattice. Then there is an isomorphism of O_{F} -modules L ≅ I for some fractional ideal I. By means of this isomorphism we identify L ⊗_{Z} R with I ⊗_{Z} R = F_{R}. In this way, the scalar product on L ⊗_{Z} R leads to a scalar product ⟨−, −⟩_{L} on F_{R} that satisfies ⟨λx, y⟩_{L} = ⟨x, \overline{λ}y⟩_{L} for x, y, λ ∈ F_{R}. For every infinite prime σ, let e_{σ} ∈ F_{R} denote the idempotent that has all its coordinates in F_{R} ≅ ∏_{σ} F_{σ} equal to zero, except the σ-th one, which is 1. Put u = ∑_{σ}⟨e_{σ}, e_{σ}⟩_{L}^{1/2}e_{σ} ∈ F_{R}^{∗}. Since the idempotents e_{σ} are orthogonal, we have for every x, y ∈ F_{R} that

⟨ux, uy⟩ = ∑_{σ} u_{σ}^{2}x_{σ}\overline{y}_{σ} = ∑_{σ} x_{σ}\overline{y}_{σ}⟨e_{σ}, e_{σ}⟩_{L} = ⟨x, y⟩_{L}.

Therefore the isomorphism L ⟶ I above induces an isomorphsm of ideal lattices (I, u) ≅ L as required.

The following result says that ideal lattices are ‘beautiful’ in the sense that they admit bases that are not very skew. In other words, ideal lattices do not contain any very short vectors.

Proposition 8.3. Let F be a number field of degree n and let D = (I, u) be an Arakelov divisor. Then for every non-zero y in the associated ideal lattice uI we have that

||y||^{2} ≥ ne^{− \frac{2}{n} deg(D)}.

In particular, when deg(D) = 0, the shortest non-zero vectors of the lattice uI have length at least \sqrt{n}.

Proof. Let 0 ≠ y ∈ uI. By the Arithmetic-Geometric Mean inequality we have that

||y||^{2} = ∑_{σ} deg(σ)|y_{σ}|^{2} ≥ n (∏_{σ} |y_{σ}|^{2deg(σ)})^{1/n} = n|N(y)|^{2/n}.

Since y = ux for some non-zero x ∈ I and since |N(x)| ≥ N(I) we have therefore that

||y||^{2} ≥ n|N(u)N(x)|^{2/n} ≥ n|N(u)N(I)|^{2/n} = ne^{− \frac{2}{n} deg(D)}.

The last inequality follows from the fact that e^{−deg(D)} = N(D)^{−1} = N(U)N(I). This proves the proposition.

Exercises.

8.1 Let D = (I, u) and D′ = (I′, u′) be two Arakelov divisors of a number field F. Check that D + D′ = (II′, uu′). Check that the neutral element of Div_{F} is the pair (O_{F} , 1). Show that the degree of D = (I, u) is equal to −log(N(u)N(I)).

8.2 Let F be a number field of degree n. If the lattice associated to an Arakelov divisor D = (I, u)

of degree 0 contains a vector ux of length \sqrt{n}, then D is the principal divisor generated by x. 8.3 Show that under the isomorphism deg : Pic_{Q} ≅ R, the Arakelov divisor corresponding to

x ∈ R is x·σ where σ : Q ↪ R is the inclusion map. Show that the ideal lattice corresponding to x ∈ R is given by e^{−x}Z.

8.4 Let F = Q(\sqrt{−5}). Let I be the O_{F} -ideal generated by 2 and 1 + \sqrt{−5}. Show that D = (I, 1/\sqrt{2}) is a Hermitian line bundle of degree 0. Draw pictures of the ideal lattices associated to (O_{F} , 1) and D.

9. Minkowski and Dirichlet.

In this section we show that the Arakelov divisor class group Pic_{F}^{0} of a number field F is a compact topological group. The proof is constructive in the sense that we cover Pic_{F}^{0} with an explicit finite set of closed simplices. As a consequence we obtain Minkowski’s Theorem that the ideal class group Cl(O_{F}) of the ring of integers O_{F} of a number field F is finite and Dirichlet’s Unit Theorem that the unit group O_{F}^{∗} is finitely generated.

Proposition 9.1. Let F be a number field. For any divisor D = (I, u) in Div_{F}^{0} there exists an element x ∈ I, so that the O_{F} -ideal J = xI^{−1} satisfies

N(J) ≤ (\frac{2}{π} )^{r_{2}} \sqrt{|Δ_{F} |}

and v = u|x| ∈ F_{R}^{∗} satisfies

log |v_{σ}| = log |u_{σ}σ(x)| ≤ \frac{1}{n} log(\sqrt{|Δ_{F} |}) − \frac{r_{2}}{n} log(\frac{π}{2} ), for all infinite primes σ.

Proof. Let D = (I, u) be a divisor of degree 0. The corresponding ideal lattice uI ⊂ F_{R} has covolume \sqrt{|Δ_{F} |}. The volume of the bounded symmetric and convex ‘box’

B(R) = {(x_{σ}) ∈ F_{R} : |x_{σ}| ≤ R}

is equal to 2^{r_{1}}(2π)^{r_{2}}R^{n}. Here r_{1} and r_{2} denote the number of real and complex primes of F respectively. Let ε ≥ 0 and let R > 0 be determined by 2^{r_{1}}(2π)^{r_{2}}R^{n} = 2^{n}\sqrt{|Δ_{F} |} + ε. By Minkowski’s Convex Body Theorem Thm. 4.2, there is a non-zero element in uI ∩B(R) when ε > 0. Since uI ∩ B(R) is finite, it follows then that there exists also a non-zero x ∈ uI ∩ B(R) when ε = 0.

In other words, there is a non-zero x ∈ I for which |u_{σ}σ(x)| ≤ R for all σ and hence

|N(x)N(u)| = ∏_{σ} |u_{σ}σ(x)|^{deg(σ)} ≤ R^{n} = \frac{2^{n}\sqrt{|Δ_{F} |}}{2^{r_{1}}(2π)^{r_{2}}}.

We put J = xI^{−1}. Since deg(D) = 0 we have that |N(u)| = N(I)^{−1} and we find that

N(J) = |N(x)|/N(I) ≤ (\frac{2}{π} )^{r_{2}} \sqrt{|Δ_{F} |}.

Moreover

log |u_{σ}σ(x)| ≤ \frac{1}{n} log ((\frac{2}{π})^{r_{2}} \sqrt{|Δ_{F} |}) , for all σ,

as required.

Theorem 9.2. Let F be a number field. Then the group Pic_{F}^{0} is compact.

Proof. By Prop. 9.1 there exists for every divisor D = (I, u) of degree 0 a principal divisor (x) and a divisor of the form (J^{−1}, v) with J ⊂ O_{F} and v ∈ F_{R}^{∗} totally positive satisfying

N(J) ≤ (\frac{2}{π} )^{r_{2}} \sqrt{|Δ_{F} |},
log(v_{σ}) ≤ \frac{1}{n} log(\sqrt{|Δ_{F} |}) − \frac{r_{2}}{n} log(\frac{π}{2} ) for all σ,

so that

D + (x) = (J^{−1}, v).

In other words, Pic_{F}^{0} is the continuous image of the subset S of Arakelov divisors of degree zero that are of the form (J^{−1}, v) with J and v as above.

Since deg(J^{−1}, v) = 0, we have that ∑_{σ} deg(σ) log(v_{σ}) = 0. Therefore the elements v ∈ F_{R}^{∗} satisfying the conditions above form a compact simplex. Indeed, the inequality of Exercise 9.1 implies that

||v||_{Pic}^{2} ≤ ∑_{σ} deg(σ) log^{2} |u_{σ}σ(x)| ≤ \frac{n(n − 1)}{n^{2}} log^{2} \sqrt{|Δ_{F} |}

and hence ||v||_{Pic} ≤ \frac{1}{2}log|Δ_{F} |.

By Prop. 5.7 there are only finitely many ideals J ⊂ O_{F} of bounded norm. This implies the set S is a compact subset of Div_{F}. It follows that Pic_{F}^{0} is itself compact.

Corollary 9.3. Let F be a number field. Then

(i) (Minkowski) the ideal class group Cl(O_{F}) is finite; (ii) (Dirichlet) the image of O_{F}^{∗} in the Euclidean space (⊕_{σ}R)^{0} is a lattice. In particular,

O_{F} is a finitely generated abelian group isomorphic to Z^{r_{1}+r_{2}−1} × μ_{F}.

Proof. By section 7 there is an exact sequence

0 ⟶ T^{0} ⟶ Pic_{F}^{0} ⟶ Cl_{F} ⟶ 0.

with continuous homomorphisms. Since Pic_{F}^{0} is compact, so is Cl_{F}. On the other hand Cl_{F} has the discrete topology. It follows that Cl_{F} is finite. This proves (i). The T^{0} is of the form (⊕_{σ}R)^{0} modulo the discrete subgroup formed by the image of the unit group. Since T^{0} is a closed subgroup of Pic_{F}^{0} , it is compact. It follows that the image of the unit group is a lattice inside the real vector space (⊕_{σ}R)^{0} of dimension r_{1} + r_{2} − 1. Since the kernel of the map O_{F}^{∗} ⟶ (⊕_{σ}R)^{0} given by ε ↦ (log |σ(ε)|)_{σ} consists precisely of the subgroup of roots of unity, part (ii) follows.

Proposition 9.4. Let F be a number field of degree n with r_{1} real and r_{2} complex infinite primes. Then the natural volume of the Arakelov class group is equal to

2^{\frac{r_{2}}{2}} \frac{h}{\sqrt{n}}|det ( \begin{matrix}1 & log |σ_{1}ε_{1}| & … & log |σ_{1}ε_{r−1}| \\ 1 & log |σ_{2}ε_{1}| & … & log |σ_{2}ε_{r−1}| \\ ⋮ & ⋮ & & ⋮ \\ 1 & log |σ_{r}ε_{1}| & … & log |σ_{r}ε_{r−1}| \end{matrix} ) |.

Here r = r_{1} + r_{2} denotes the number of infinite primes σ_{i}. By h = #Cl(O_{F}) we denote the class number of F and by ε_{1}, …, ε_{r−1} a set of generators of the unit group O_{F}^{∗} modulo torsion. The matrix is therefore an r by r-matrix.

Proof. The vector 1 ∈ ⊕_{σ}R is orthogonal to (⊕_{σ}R)^{0} and has length \sqrt{n}. Therefore the volume of the torus T^{0} is equal to the determinant above times the volume of the unit block \{(v_{σ}) ∈ F_{R} : |v_{σ}| ≤ 1 for all σ\} divided by \sqrt{n}. The unit block has volume 2^{r_{2}/2}. The proposition now follows from the fact that the volume of Pic_{F}^{0} is equal to the volume T^{0} times the class number.

It follows from Theorem 9.1 that for a number field F, every ideal class in Cl(O_{F}) contains an integral ideal of norm at most (\frac{2}{π} )^{r_{2}} \sqrt{|Δ_{F} |}. This result can be improved somewhat by choosing another kind of ‘box’ in F_{R}.

Proposition 9.5. Let F be a number field of degree n with r_{1} real and r_{2} complex infinite primes. Then every ideal class contains an ideal I ⊂ O_{F} with

N(I) ≤ \frac{n!}{n^{n}} (\frac{4}{π} )^{r_{2}} \sqrt{|Δ_{F} |}.

Proof. Let R > 0 and consider the set

B′(R) = {(x_{σ}) ∈ F_{R} : Tr|x_{σ}| ≤ R}.

This a bounded symmetric convex set and by Exercise 9.3, its volume is equal to

vol(B′(R)) = 2^{r_{1}}π^{r_{2}} \frac{R^{n}}{n!} .

Let I′ be an ideal in the inverse of a given ideal class in Cl(O_{F}). The covolume of the lattice I′ ⊂ F ⊂ F_{R} is N(I′)\sqrt{|Δ_{F} |}. Let ε ≥ 0 and let R be such that

2^{r_{1}}π^{r_{2}} \frac{R^{n}}{n!} = 2^{n}N(I′)\sqrt{|Δ_{F} |} + ε.

If ε > 0 there is by Theorem 4.2 a non-zero vector x ∈ I′ ∩ B′(R). This implies that the ideal I = xI′^{−1} is contained in O_{F}. It is contained in the ideal class that was given above. By an argument similar to the one used in the proof of Thm. 9.1, the same is true with ε = 0. By the Arithmetic-Geometric Mean inequality of Exer. 9.2 we have that

|N(x)| = ∏_{σ} |σ(x)|^{deg(σ)} ≤ (\frac{R}{n})^{n}

and hence

N(I) = |N(x)|/N(I′) ≤ (\frac{R}{n})^{n} \frac{1}{N(I′)} = \frac{(2^{n}N(I′)\sqrt{|Δ_{F} |})n!}{n^{n}2^{r_{1}}π^{r_{2}}N(I′)}.

This proves the proposition.

Corollary 9.6. Let F be a number field of degree n with r_{1} real and r_{2} complex primes. Then (i)

|Δ_{F} | ≥ (\frac{n^{n}}{n!} (\frac{π}{4} )^{r_{2}})^{2} ;

(ii) |Δ_{F} | ≥ \frac{π^{n}}{4} . In particular, |Δ_{F} | > 1 whenever F ≠ Q.

Proof. Part (i) follows by applying the proposition to the trivial class. To prove (ii) one verifies inductively that n^{n} ≥ 2^{n−1}n! for all n ≥ 1. Part (i) implies then that

|Δ_{F} | ≥ (\frac{n^{n}}{n!} )^{2} (\frac{π}{4} )^{2r_{2}} ≥ (2^{n−1})^{2} (\frac{π}{4} )^{n} = \frac{π^{n}}{4} .

This proves the corollary.

In the rest of this section we consider the covering of Pic_{F}^{0} of Theorem 9.2 in some more detail. Let F be a number field of degree n. For every ideal J ⊂ O_{F}, we let Σ_{J} denote the set of divisors given by

Σ_{J} = \{(J^{−1}, v) ∈ Div_{F}^{0} : log(v_{σ}) ≤ \frac{1}{n} log(\sqrt{|Δ_{F} |}) − \frac{r_{2}}{n} log(\frac{π}{2} ) for all σ\}.

If N(J) ≤ (\frac{π}{2} )^{r_{2}} \sqrt{|Δ_{F} |}, the set Σ_{J} is a simplex. To see this, notice that in that case Σ_{J} contains the divisor (J^{−1}, N(J)^{1/n}) and that the elements in Σ_{J} are sums of (J^{−1}, N(J)^{1/n}) and an element in the following subset of (⊕_{σ}R)^{0} ⊂ Div_{F}^{0} :

\{(w_{σ}) ∈ (⊕_{σ} R)^{0} : w_{σ} ≤ \frac{1}{n} log(\sqrt{|Δ_{F} |}/N(J)) − \frac{r_{2}}{n} log(\frac{π}{2} )\}.

In order to compare the size of the images of the various Σ_{J} in Pic_{F}^{0} , we fix a connected component, i.e. a coset of the torus T^{0} = (⊕_{σ}R)^{0} /im(O_{F}^{∗}), and choose an ideal I ⊂ O_{F} in the ideal class that corresponds to it. Then the divisor (I^{−1}, N(I)^{1/n}) lies on the component. Let J ⊂ O_{F} in the same ideal class as I. We have that I = βJ for some β ∈ F^{∗}. The divisors of degree zero of the form (J^{−1}, v) lie on the component (I^{−1}, N(I)^{1/n}) + T^{0} and we have that

\begin{aligned}(J^{−1}, v) & = (βI^{−1}, v) ∼ (I^{−1}, |β|v), \\ & ∼ (I^{−1}, N(I)^{1/n}) + (O_{F} , \frac{|β|}{N(β)^{1/n}}) + (O_{F} , \frac{v}{N(J)^{1/n}}). \end{aligned}

Suppose now that N(J) < (\frac{π}{2} )^{r_{2}} \sqrt{|Δ_{F} |} and consider the image of the simplex Σ_{J} in the component (I^{−1}, N(I)^{1/n}) + T^{0}. The elemenst of Σ_{J} are sums of (I^{−1}, N(I)^{1/n}) and elements of the set

\{(log(\frac{|σ(β)|}{N(β)^{1/n}})) + (w_{σ})\} ⊂ (⊕_{σ} R)^{0}

with w_{σ} ≤ \frac{1}{n} log(^{\frac{\sqrt{|Δ_{F} |}}{N(J)}}) − \frac{r_{2}}{n} log(\frac{π}{2} ) and ∑_{σ} deg(σ)w_{σ} = 0. The size of the simplex gets smaller as the norm N(J) of J becomes larger.

Exercises.

9.1 Let x_{i} ∈ R for 1 ≤ i ≤ n. Suppose that ∑_{i} x_{i} ≥ 0 and that x_{i} ≤ a for some a ∈ R. Then

∑_{i} x_{i}^{2} ≤ n(n − 1)a^{2}. 9.2 (Arithmetic-Geometric Mean Inequality) Let x_{i} ∈ R_{≥0} for 1 ≤ i ≤ n Then

(x_{1} ⋯ x_{n})^{1/n} ≤ \frac{x_{1} + … + x_{n}}{n} .

(Hint: Put μ = \frac{x_{1}+...+x_{n}}{n} and show that e^{x_{i}/μ−1} ≥ \frac{x_{i}}{μ} for each i with equality holding if and

only if x_{i} = μ.

9.3 Let r_{1}, r_{2} ∈ Z_{≥0} and put r_{1} + r_{2} = n. Let R ≥ 0 and put

W (r_{1}, r_{2}, R) = \{(x_{1}, …, x_{r_{1}}, z_{1}, …, z_{r_{2}}) ∈ R^{r_{1}} × C^{r_{2}} : ∑_{i=1}^{r_{1}} |x_{i}| + \sqrt{2} ∑_{j=1}^{r_{2}} |z_{j}| ≤ R\}

(i) Show that the volume of the subset B′(R) of F_{R} is equal to the volume of W (r_{1}, r_{2}, R) (with respect to the usual metrics on R and C.

(ii) Show that vol(W (1, 0, R)) = 2R and that vol(W (0, 1, R)) = πR^{2}.

(iii) Show that

vol(W (r_{1}, r_{2}, R)) = 2^{r_{1}}π^{r_{2}} \frac{R^{n}}{n!} .

(Hint: proceed by induction.)

10. Explicit computation. In this section we present two explicit examples. We compute the Arakelov class groups of two quadratic fields. Example 10.1. F = Q(\sqrt{−26}). The field F is a complex quadratic field. This means that it cannot be embedded into R and admits, up to conjugation, only one embedding into C. In other words, F has r_{1} = 0 real infinite primes and r_{2} = 1 infinite complex prime. It follows that the vector space (∏_{σ} R)^{0} and hence the torus T^{0} are both trivial. As a consequence the natural map

Pic_{F}^{0} ⟶^{≅} Cl(O_{F})

that maps Arakelov divisors D = (I, u) to the class of the ideal I, is an isomorphism. Since −26 ≢ 1 (mod 4), Exer. 3.4 implies that the ring of integers of F is given by O_{F} = Z[\sqrt{−26}]. The minimum polynomial of α = \sqrt{−26} is equal to f(T) = T^{2} + 26. The unit group O_{F}^{∗} is finite. Any unit ε = x + yα ∈ O_{F}^{∗} has norm x^{2} + 26y^{2} equal to 1. In order to compute them we solve the equation

x^{2} + 26y^{2} = 1, for x, y ∈ Z.

The only solutions are x = ±1 and y = 0. It follows that O_{F}^{∗} = {±1}. The discriminant Δ_{F} is defined as the discriminant of the free Z-algebra O_{F} over Z. By Prop. 3.1 is equal to the discriminant of f(T), i.e., we have that Δ_{F} = −4 · 26 = −104. Alternatively, one may compute Δ_{F} using its definition. Since {1, α} is a Z-basis for O_{F} and since Tr(α) = 0, we have that

Δ_{F} = det ( \begin{matrix}Tr(1) & Tr(α) \\ Tr(α) & Tr(α^{2}) \end{matrix} ) = det ( \begin{matrix}2 & 0 \\ 0 & −52 \end{matrix} ) = −104.

By Prop. 6.5 the only ramified primes are 2 and 13. By Prop. 9.5, the ideal class group Cl(O_{F}) is generated by the primes of norm less than \frac{2}{π} \sqrt{104} = 6.49 . . .. We now compute explicit generators for these prime ideals and find relations among them by factoring suitable principal ideals into products of prime ideals. This is conveniently done simultaneously as follows.

We first compute f(k) = k^{2} + 26 for several small integers k and factor these numbers into a product of prime numbers. The result is given below.

Table 10.2.

k f(k) = k^{2} + 26 (α − k)
(i) 0 26 = 2 · 13  𝔭_{2}𝔭_{13}
(ii) 1 27 = 3^{3}  𝔭_{3}^{3}
(iii) 2 30 = 2 · 3 · 5  𝔭_{2}𝔭′_{3}𝔭_{5}
(iv) 3 35 = 5 · 7  𝔭′_{5}𝔭_{7}

Since 2 is ramified there is only one prime 𝔭_{2} that divides 2. We have that 𝔭_{2} = (2, α) and 𝔭_{2}^{2} = (2). It follows from the table that the polynomial f(T) has zeroes modulo the primes 3 and 5. This implies that these primes split in F. We have that (3) = 𝔭_{3}𝔭′_{3} with 𝔭_{3} = (α − 1, 3) and 𝔭′_{3} = (α + 1, 3). This follows from the fact that the maximal ideals of the ring O_{F} /(3) ≅ F_{3}[X]/(X^{2} − 1) ≅ F_{3}[X]/(X − 1) × F_{3}[X]/(X + 1) are generated by X − 1 and X + 1 respectively. The ideals 𝔭_{3} and 𝔭′_{3} are then the kernels of the maps O_{F} → → F_{3}[X]/(X − 1) and O_{F} → → F_{3}[X]/(X + 1) respectively. Similarly, (5) = 𝔭_{5}𝔭′_{5} with 𝔭_{5} = (α − 2, 5) and 𝔭′_{5} = (α + 2, 5).

The class group Cl_{F} is therefore generated by the classes of the ideals 𝔭′_{2}, 𝔭_{3}, 𝔭′_{3}, 𝔭_{5} and 𝔭′_{5}. In order to determine the structure of Cl_{F} factor certain principal ideals into products of prime ideals. The factorizations (2) = 𝔭_{2}^{2} and the ones of (p) for p = 3, 5 and 7 give rise to the relation 𝔭_{2}^{2} ∼ 1 and the relations 𝔭′_{3} ∼ 𝔭_{3}^{−1} and 𝔭′_{5} ∼ 𝔭_{5}^{−1} respectively. Here “∼” denotes equality up to principal ideals and ‘1’ denotes the trivial class. It follows at once that Cl_{F} is already generated by the classes of 𝔭_{2}, 𝔭_{3} and 𝔭_{5}. We use Table 10.2 to obtain more relations. Entry (iii) of the table implies that 𝔭_{5} ∼ 𝔭_{2}^{−1}𝔭′_{3}^{−1} ∼ 𝔭_{2}𝔭_{3}. Therefore 𝔭_{5} is contained in the subgroup generated by 𝔭_{2} and 𝔭_{3} and we don’t need it as a generator. Entry (ii) of the table implies that 𝔭_{3}^{3} ∼ 1. It follows from entry (ii) and the fact that 𝔭_{2}^{2} ∼ 1 that Cl_{F} is a quotient of the free group generated by 𝔭_{2} and 𝔭_{3} modulo the relations 𝔭_{2}^{2} ∼ 1 and 𝔭_{3}^{3} ∼ 1. Therefore Cl_{F} is a quotient of the group Z/6Z.

When one does more computations like this, one keeps finding relations that are implied by the ones already found. For instance, since f(7) = 49 + 26 = 75 = 3 · 5^{2}, the principal ideal (α − 7) factors as 𝔭_{5}^{2}𝔭_{3}. It follows that 𝔭_{5}^{2} ∼ 𝔭_{3}^{−1}. But this already follows from entry (iii). Indeed, from 𝔭_{5} ∼ 𝔭_{3}𝔭_{2}^{−1} we deduce that 𝔭_{5}^{2} ∼ 𝔭_{3}^{2}𝔭_{2}^{−2} ∼ 𝔭_{3}^{−1}. Instead, we are going to try to prove that Cl_{F} ≅ Z/6Z. It suffices to show that 𝔭_{2} and 𝔭_{3} are not principal. Suppose that 𝔭_{2} is principal. Then 𝔭_{2} = (β) for some β ∈ O_{F}. Since O_{F} = Z[\sqrt{−26}], we have that β = x + y\sqrt{−26} for some x, y ∈ Z. Taking norms we find that

2 = x^{2} + 26y^{2}.

It is easy to see that this equation has no solutions x, y ∈ Z. Therefore β does not exist and 𝔭_{2} is not principal. One shows in a similar way that 𝔭_{3} is not principal. In this case it is the Diophantine equation x^{2} + 26y^{2} = 3 that does not admit any solutions x, y ∈ Z with y ≠ 0. Finally we notice that the six ideal classes are the classes of O_{F}, 𝔭_{3}, 𝔭_{3}], 𝔭_{2}, 𝔭_{2}𝔭_{3} and 𝔭_{2}𝔭′_{3}. The two ideals 𝔭_{5} and 𝔭′_{5} are contained in the classes of 𝔭_{2}𝔭_{3} and 𝔭_{2}𝔭′_{3} respectively.

Example 10.4. F = Q(\sqrt{105}). The field F is an real quadratic field. It admits two embeddings σ and σ′ into R. This implies that r_{1} = 2 and r_{2} = 0. The Arakelov class group fits therefore in an exact sequence

0 ⟶ (R × R)^{0} /im(O_{F}^{∗}) ⟶ Pic_{F}^{0} ⟶ Cl_{F} ⟶ 0.

Here (R × R)^{0} is the one dimensional vector space {(x, −x) : x ∈ R} and by im(O_{F}^{∗}) we denote the discrete subgroup of elements of the form (− log |σ(u)|, − log |σ′(u)|) where u ∈ O_{F}^{∗}. As a Riemannian manifold, Pic_{F}^{0} is a finite union of circles having the same circumference.

Since 105 ≡ 1 (mod 4), the ring of integers of F is equal to Z[α] with α = \frac{1+\sqrt{105}}{2}. The minimum polynomial of α is given by f(T) = T^{2} −T −26 and the discriminant of F is equal to the discriminant of f(T) which in turn is equal to 105. We observe that the only ramified primes are 3, 5 and 7. Let 𝔭_{3} = (3, α+1), 𝔭_{5} = (5, α+2) and 𝔭_{7} = (7, α+3) denote the unique primes of norm 3,5 and 7 respectively. We have that 𝔭_{3}^{2} = (3), 𝔭_{5}^{2} = (5) and 𝔭_{7}^{2} = (7). Since f(T) ≡ T (T + 1) (mod 2), the ring O_{F} /(2) is isomorphic to a a product of two finite fields of order 2. Therefore 2 = 𝔭_{2}𝔭′_{2} where 𝔭_{2} = (2, α) and 𝔭′_{2} = (2, α + 1) are two distinct prime ideals of norm 2. By Theorem 9.3, the group Pic_{F}^{0} can be covered by open intervals of the form (J^{−1}, u) where J ⊂ O_{F} is an ideal of norm N(J) at most \sqrt{105} = 10.2469 … and log |u_{σ}|, log |u_{σ}| ≤ \frac{1}{2} log \sqrt{105} = 1.16 . . .. In particular, the ideal class group Cl_{F} is generated by the prime ideals in the set P = {𝔭_{2}, 𝔭′_{2}, 𝔭_{3}, 𝔭_{5}, 𝔭_{7}}. This implies that there is an isomorphism

Pic_{F}^{0}  ≅ ( ⊕_{𝔭∈P} Z × R × R)^{0} /Princ(P)

where Princ(P) denotes the subgroup of principal divisors that are supported in the set P and the two infinite primes. Next we search for elements in Pr(P) by factoring principal ideals of O_{F}. As in the previous example, we consider elements of the form α − k with k ∈ Z, because by Exer. 3.5 their norms are equal to k^{2} − k − 26 and are easy to compute. We take k close to the zero \frac{1+\sqrt{105}}{2} = 5.6234 … of f(T) = T^{2} − T − 26.

Table 10.5.

k f(k) = k^{2} − k − 26 (α − k)
(i) 3 −20 = −2^{2} · 5  𝔭′_{2}^{2}𝔭_{5}
(ii) 4 −14 = −2 · 7  𝔭_{2}𝔭_{7}
(iii) 5 −6 = −2 · 3  𝔭′_{2}𝔭_{3}
(iv) 6 4 = 2^{2} 𝔭_{2}^{2}
(v) 7 16 = 2^{4}  𝔭′_{2}^{4}
(vi) 8 30 = 2 · 3 · 5  𝔭_{2}𝔭_{3}𝔭_{5}

Table 10.6.

x  𝔭_{2} 𝔭′_{2} 𝔭_{3} 𝔭_{5} 𝔭_{7}  σ  σ′
2  1 1 0 0 0  − log 2  − log 2
3  0 0 2 0 0  − log 3  − log 3
5  0 0 0 2 0  − log 5  − log 5
7  0 0 0 0 2  − log 7  − log 7
(i) α − 3 0 2 0 1 0 − log |\frac{−5+\sqrt{105}}{2}| − log |\frac{−5−\sqrt{105}}{2}|
(ii) α − 4 1 0 0 0 1 − log |\frac{−7+\sqrt{105}}{2}| − log |\frac{−7−\sqrt{105}}{2}|
(iii) α − 5 0 1 1 0 0 − log |\frac{−9+\sqrt{105}}{2}| − log |\frac{−9−\sqrt{105}}{2}|
(iv) α − 6 2 0 0 0 0 − log |\frac{−11+\sqrt{105}}{2}| − log |\frac{−11−\sqrt{105}}{2}|
(v) α − 7 0 4 0 0 0 − log |\frac{−13+\sqrt{105}}{2}| − log |\frac{−13−\sqrt{105}}{2}|
(vi) α − 8 1 0 1 1 0 − log |\frac{−15+\sqrt{105}}{2}| − log |\frac{−15−\sqrt{105}}{2}|

In Table 10.6 we list the ten principal Arakelov divisors that correspond to the six entries of Table 10.5 plus the four ‘trivial’ factorizations (2) = 𝔭_{2}𝔭′_{2} and (p) = 𝔭_{p}^{2} for p = 3, 5 and 7. They happen all to be supported in the set Princ(P) mentioned above. The row corresponding to an element x contains the coefficients n_{𝔭} and x_{σ} of the principal divisor (x).

These principal divisors lead to relations between the generators of Pic_{F}^{0} . In this way we can eliminate some of the generators. More precisely, the principal divisors (ii), (i) and (iii) express 𝔭_{7}, 𝔭_{5} and 𝔭_{3} in terms of 𝔭_{2}, 𝔭′_{2} and the infinite primes in Pic_{F}^{0} . Similarly, the principal divisor (2) expresses 𝔭′_{2} in terms of 𝔭_{2} and the infinite primes. We apply the Gaussian elimination method. First eliminate 𝔭_{7}: subtract relation (ii) twice from the fourth row and then omit it. Eliminate 𝔭_{5}: subtract relation (i) once from relation (vi) and twice from the third row. Then omit it. In a similar way one eliminates the primes 𝔭_{3} and 𝔭′_{2}. Of course, in the x column one multiplies and divides rather than adds and subtracts.

This leads to the following list of six principal divisors supported in 𝔭_{2}, σ and σ′. The numbers in the last column are merely approximations.

Table 10.7.

\begin{matrix}x & 𝔭_{2} & σ & σ′ \\ 12/(α − 5)^{2} & 2 & −3.4298 & 2.0435 \\ 80/(α − 3)^{2} & 4 & −2.4530 & −0.3196 \\ 7/(α − 4)^{2} & −2 & −0.9768 & 2.3631 \\ α − 6 & 2 & 0.9768 & −2.3631 \\ (α − 7)/16 & −4 & 2.4530 & 0.3196 \\ 8(α − 8)/(α − 3)(α − 5) & 4 & −2.4530 & −0.3196 \end{matrix}

There is an isomorphism

Pic_{F}^{0}  ≅ (Z × R × R)^{0} /Pr(P′)

where the factor Z corresponds to the prime 𝔭_{2} and Pr(P′) denotes the group of principal divisors that are supported in 𝔭_{2}, σ and σ′. In particular, the class group Cl_{F} is generated by 𝔭_{2}. In the 𝔭_{2}-column all entries are even. This means that we cannot eliminate 𝔭_{2}. If we were to search for additional relations, we would continue to find even entries in the 𝔭_{2}-column.

Table 10.8.

\begin{matrix}x & σ & σ′ \\ 12/(α − 5)^{2}(α − 6) & −4.4066 & 4.4066 \\ 80/(α − 3)^{2}(α − 6)^{2} & −4.4066 & 4.4066 \\ 7(α − 6)/(α − 4)^{2} & 0 & 0 \\ (α − 7)(α − 6)^{2}/16 & 4.4066 & −4.4066 \\ 8(α − 8)/(α − 6)^{2}(α − 3)(α − 5) & −4.4066 & 4.4066 \end{matrix}

We proceed as follows. The fourth row of Table 10.6 says that 𝔭_{2}^{2} = (α − 6). Therefore the ideal class group Cl_{F} has order at most 2. Before showing that Cl_{F} ≅ Z/2Z, we perform Gaussian elimination with the fourth row and obtain Table 10.8: a list of Arakelov divisors (x) that are supported in the infinite primes. All elements x in this table are units of O_{F}. The element in the third row is equal to ±1 because both coefficients at the infinite primes vanish. Looking at the other entries in the last two columns, one sees that the corresponding units are equal to ±ε^{±1} for some unit ε. Using the fact that α^{2} −α−12 = 0, we simplify the expression for the unit in the fourth row.

ε = \frac{(α − 7)(α − 6)^{2}}{16} = \frac{128α − 720}{16} = 8α − 45 = −41 + 4\sqrt{105}.

If we were to search for any additional units, we would continue to find the same unit ε or powers of it. Therefore we suspect that the unit group O_{F}^{∗} is actually generated by ε and −1.

Class group computation. Before determining the unit group, we come back to the issue of whether the class group is trivial or isomorphic to Z/2Z. We suspect that Cl_{F} is not trivial. Suppose the contrary is true. Then 𝔭_{2} = (β) for some β ∈ O_{F}^{∗}. The Diophantine equation that expresses the fact that β has norm 2 is

x^{2} − 26y^{2} = ±2, for x, y ∈ Z,

is not so easy to solve as in the previous example. We proceed in a different way. Since 𝔭_{2}^{2} = (α − 6), we would have that β^{2} is equal to α − 6 times a unit u ∈ O_{F}^{∗}. We suspect that O_{F}^{∗} is generated by ε = −41 + 4\sqrt{105} and −1, so that we would have that

(α − 6)(−1)^{l}ε^{m} is a square for some l, m ∈ Z.  (∗)

The multiplicative group U generated by α − 6, −1 and ε modulo squares is a vector space over F_{2}. We now show that its dimension is equal to 3. This implies first of all that −1 and ε generate O_{F}^{∗} modulo squares. But then the hypothetical unit u above is actually equal to (−1)^{l}ε^{m} times a square. This means that we the relation (*) actually would hold. However, if U has dimension 3 this is impossible.

To show that U has dimension 3, we construct a homomorphism to a suitable explicit vector space V over F_{2} and compute the image. We put

V = (R^{∗}/R^{∗}^{2}) × (R^{∗}/R^{∗}^{2}) × (O_{F} /𝔭_{3})^{∗} × (O_{F} /𝔭_{5})^{∗}/(O_{F} /𝔭_{5})^{∗}^{2}.

This is a vector space of dimension 4 over F_{2}. The homomorphism U ⟶ V is given by u ↦ (σ(u), σ′(u), u (mod 𝔭_{3}), u (mod 𝔭_{5})). It so happens that all three elements α − 6, −1 and ε have the property that their images under σ and σ′ are all negative. The element −1 is a square modulo 𝔭_{5} bot not modulo 𝔭_{3}. The unit ε = −41 + 4\sqrt{105} is congruent to 1 (mod 3) and congruent to −1 (mod 5). Therefore it is a square modulo both primes. Finally α − 6 is not a square modulo any of them. Identifying V with F_{2}^{4} in the obvious way, the image of U is therefore spanned by the rows of the matrix

( \begin{matrix}1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{matrix} ) .

Since the rank of this matrix is 3 over F_{2}, the dimension of U is equal to 3. We conclude that Cl_{F} has order 2 as required.

Unit group computation. Next we show that −1 and ε generate O_{F}^{∗}. Since we already know that this is so modulo squares, we must have that

ε = ±η^{k}, for some integer k ≥ 3

if −1 and ε do not generate O_{F}^{∗}. This implies that

\begin{aligned}|σ(η)| & ≤ |σ(ε)|^{1/3} = 4.3442 …, \\ |σ′(η)| & ≤ |σ′(ε)|^{1/3} = 0.2302 …, \end{aligned}

Writing η = x + yα for some x, y ∈ Z we have that ||η||^{2} = (x + yσ(α))^{2} + (x + yσ′(α))^{2} = 2x^{2} + 2xy + 53y^{2}. It follows that

2x^{2} + 2xy + 53y^{2} = ||η||^{2} ≤ (4.3442 …)^{2} + (0.2302 …)^{2} = 18.9256 …

However it follows easily by completing the square of the quadratic form that necessarily y = 0, so that η ∈ Z which is absurd. We conclude that O_{F}^{∗} is generated by −1 and ε.

The group Pic_{F}^{0} is therefore isomorphic to

(Z × R × R)^{0} /⟨v_{1}, v_{2}⟩

where

v_{1} = (2, − ln |(−11 + \sqrt{105})/2|, − ln |(−11 − \sqrt{105})/2|),
v_{2} = (0, − ln | − 41 + 4\sqrt{105}|, − ln | − 41 − 4\sqrt{105}|).

Its volume is equal to 2\sqrt{2} · ln |41 + 4\sqrt{105}| = 12.4636 . . .. Finally we project the group (Z × R × R)^{0} on its first two coordinates. This simplifies the presentation:

Pic_{F}^{0} ≅ (Z × R) /⟨w_{1}, w_{2}⟩,

with

\begin{aligned}w_{1} & = (2, − ln |(−11 + \sqrt{105})/2|), \\ w_{2} & = (0, − ln | − 41 + 4\sqrt{105}|). \end{aligned}

The topological group Pic_{F}^{0} consists of two circles Putting R = log |41 + 4\sqrt{105}|, these are the connected component of identity {(0, y) : y ∈ R/RZ} and its non-trivial coset {(1, y) : y ∈ R/RZ}.

Covering the Arakelov class group. We now verify that Pic_{F}^{0} is covered by simplices of the form (J^{−1}, v) where J ⊂ O_{F} is an ideal of norm at most \sqrt{105} = 10.2 … and v has the property that log |v_{σ}| and log |v_{σ′}| are at most \frac{1}{4} log(105) = 1.16 . . .. The simplices are 1-dimensional intervals in this case. We show that the intervals associated to the ideals J of norm at most 7 already suffice to cover Pic_{F}^{0} . There are ten such ideals.

It follows from the factorizations in Table 10.4 that for the ideals J = O_{F}, (2), 𝔭_{2}^{2} and 𝔭′_{2}^{2}, 𝔭_{5}, 𝔭_{2}𝔭_{3} and 𝔭′_{2}𝔭_{3} the simplex (J^{−1}, v) is contained in the connected component of the identity of Pic_{F}^{0} . On this component we choose, in the notation of section 9, I = O_{F}. The elements β for which I = β_{J} are listed in the second column of Table 10.8. It is convenient to project the intervals on the first coordinate of T^{0} = (⊕_{σ}R)^{0}. The third column contains the center log |\frac{β}{N(β)^{1/n}} of the intervals and the fourth column contains the

endpoints log |\frac{β}{N(β)^{1/n}} | ± \frac{1}{n} log(\frac{\sqrt{Δ}}{N(J)}).

Table 10.9.

J  β^{−1}  center  simplex
O_{F}  1  0 (−1.163, 1.163)
𝔭_{2}^{2}  \frac{11−\sqrt{105}}{2} log |\frac{4}{11−\sqrt{105}}| = 1.670  (1.199, 2.140)
𝔭′_{2}^{2}  \frac{11+\sqrt{105}}{2} log |\frac{4}{11+\sqrt{105}}| = −1.670 (−2.140, −1.199)
(2) 2 0 (−0.470, 0.470)
𝔭_{5} 10 − \sqrt{105} log |\frac{\sqrt{5}}{10−\sqrt{105}}| = 2.203  (1.845, 2.562)
𝔭_{2}𝔭_{3}  \frac{9+\sqrt{105}}{2} log |\frac{2\sqrt{6}}{9+\sqrt{105}}| = −1.368 (−1.636, −1.101)
𝔭′_{2}𝔭_{3}  \frac{9−\sqrt{105}}{2} log |\frac{2\sqrt{6}}{9−\sqrt{105}}| = 1.368  (1.101, 1.636)

On the other component we choose I = 𝔭_{3} and the divisor (𝔭_{3}^{−1}, \sqrt{3}) as our point of reference. Recall that both cosets are circles with circumference equal to R = 41+4\sqrt{105} = 4.4065 … The reader easily checks from the data in the tables that both circles are covered by the intervals listed in the fourth column.

Table 10.10.

J  β^{−1}  center  simplex
𝔭_{3}  1  0 (−1.228, 1.228)
𝔭_{2}  \frac{9+\sqrt{105}}{6}  log |\frac{2\sqrt{6}}{9+\sqrt{105}}| = −1.368 (−2.185, −0.551)
𝔭′_{2}  \frac{9−\sqrt{105}}{6}  log |\frac{2\sqrt{6}}{9−\sqrt{105}}| = 1.368  (0.551, 2.185)
𝔭_{7}  \frac{21−2\sqrt{105}}{6} log |\frac{\sqrt{21}}{21−2\sqrt{105}}| = 2.203  (2.012, 2.394)

Exercises.

10.1 Compute the ideal class group Cl(O_{F}) and the unit group O_{F}^{∗} for the number fields F =

Q(\sqrt{d}) for d = −47, −71, −163 and −147. 10.2 Compute the ideal class group Cl(O_{F})and the unit group O_{F}^{∗} for the number fields F =

Q(\sqrt{d}) for d = 19, 145, 46 and 200.

11. The Riemann-Roch Theorem. The main result of this section is Theorem 11.4. At the end of the section we explain why it should be viewed as the arithmetic analogue of the Riemann-Roch theorem.

Definition. Let F be a number field. Let D = (I, u) be an Arakelov divisor. Then we put

h^{0}(D) = log ( ∑_{y∈uI} e^{−π||y||^{2}} ) .

We have that h^{0}(D) ∈ R_{>0}. Since h^{0}(D) only depends on the lattice uI ⊂ F_{R} associated to D. By Prop. 8.1 lattices associated to equivalent divisors are isometric. Therefore the map D ↦ h^{0}(D) induces a well defined function on the quotient group Pic_{F}. We

Lemma 11.1. Let F be a number field of degree n and let D = (I, u) be an Arakelov divisor. Let λ denote the length of the shortest non-zero vector in the ideal lattice uI. Then

∑_{{y∈uI \atop y≠0}} e^{−π||y||^{2}} ≤ π ∫_{λ^{2}}^{∞}(1 + \frac{2\sqrt{t}}{λ})^{n}e^{−πt}dt.

Proof. We have that

∑_{{y∈uI \atop y≠0}} e^{−π||y||^{2}} = ∑_{{y∈uI \atop y≠0}} ∫_{||y||^{2}}^{∞} πe^{−πt}dt ≤ π ∫_{λ^{2}}^{∞} #B_{t}e^{−πt}dt,

where B_{t} = {y ∈ uI : ||y||^{2} ≤ t}. The balls with centers in y ∈ B_{t} and radius λ/2 are disjoint. Their union is contained in a ball with center 0 and radius \sqrt{t} + λ/2. Therefore

(\frac{λ}{2})^{n}#B_{t} ≤ (\sqrt{t} + λ/2)^{n}

and hence #B_{t} ≤ (1 + \frac{2\sqrt{t}}{λ})^{n}. This implies the lemma.

Proposition 11.2. Let F be a number field of degree n and let D be an Arakelov divisor. Then (i)

h^{0}(D) ≤ deg(D) + 1, when deg(D) ≥ 0;

(ii)

h^{0}(D) ≤ e^{h^{0}(D)} − 1 ≤ 2 · 3^{n} · exp (−πne^{− \frac{2}{n} deg(D)}) , when deg(D) ≤ 0.

Proof. (i) We use the lemma and the fact that 1 + \frac{2\sqrt{t}}{λ} ≤ \frac{3\sqrt{t}}{λ} for t ≥ λ^{2}. This leads to

\begin{aligned}∑_{{y∈uI \atop y≠0}} e^{−π||y||^{2}} & ≤ π ∫_{λ^{2}}^{∞} (3\frac{\sqrt{t}}{λ})^{n} e^{−πt}dt = ∫ ∞_{\frac{λ^{2}}{π}} (3\frac{\sqrt{x}}{\sqrt{π}λ})^{n} e^{−x}dx, \\ & ≤ \frac{3^{n}}{π^{n/2}λ^{n}} ∫_{0}^{∞} x^{n/2}e^{−x}dx = \frac{3^{n}Γ(\frac{n}{2} − 1)}{π^{n/2}}λ^{−n}. \end{aligned}

Here Γ denotes the Gamma function. See Exer 11.3. Since λ satisfies λ ≥ \sqrt{n}N(D)^{−1/n} we find that

∑_{{y∈uI \atop y≠0}} e^{−π||y||^{2}} ≤ \frac{3^{n}Γ(\frac{n}{2} − 1)}{π^{n/2}n^{n/2}}N(D) ≤ \frac{3}{2}N(D).

Since deg(D) ≥ 0, this implies that

h(D) ≤ log(1 + \frac{3}{2}N(D)) ≤ log(N(D)) + 1 = deg(D) + 1,

as required. (ii) The first inequality is clear. For the second we use the estimate of the lemma together with the the rough estimate 1 + 2(t/λ^{2})^{1/2} ≤ 3^{t/λ^{2}} for t ≥ λ^{2}. This leads to

∑_{{y∈uI \atop y≠0}} e^{−π||y||^{2}} ≤ π ∫_{λ^{2}}^{∞} 3^{nt/λ^{2}}e^{−πt}dt = \frac{πe^{n log(3)−πλ^{2}}}{π − n log(3)/λ^{2}} .

Since deg(D) ≤ 0 we have that λ^{2} ≥ n and hence

∑_{{y∈uI \atop y≠0}} e^{−π||y||^{2}} ≤ \frac{π3^{n}}{π − log(3)}e^{−πλ^{2}}.

The result now follows from the estimate for λ provided by the lemma.

By section 4, the Z-dual of the lattice uI associated to an Arakelov divisor D = (I, u) is the lattice

\{y ∈ F_{R} : Tr(y\overline{z}) ∈ Z for all z ∈ I\} ⊂ F_{R}.

It is in general not an ideal lattice because multiplication by O_{F} does not map \overline{I} to itself. However the conjugate of the Z-dual is. It is given by

\overline{uI}^{∨} = \{y ∈ F_{R} : Tr(yz) ∈ Z for all z ∈ I\}.

Definition. Let F be a number field. The canonical divisor class κ of F is the divisor class corresponding to the ideal lattice \overline{O_{F}}^{∨}.

The ideal lattice

κ = {y ∈ F_{R} : Tr(yz) ∈ Z for all z ∈ O_{F} }

contains O_{F} and is contained in \frac{1}{|Δ_{F} |}O_{F}. The second inclusion follows by choosing a Zbasis ω_{1}, …, ω_{n} for O_{F} and writing z = λ_{1}ω_{1} + … + λ_{n}ω_{n} with λ_{i} ∈ R. The conditions Tr(yz) ∈ Z for z = ω_{i} lead to a linear system with coefficient matrix (Tr(ω_{i}ω_{j})). By Cramer’s rule, the solutions λ_{i} are rational numbers with denominators dividing Δ_{F} = det(Tr(ω_{i}ω_{j})).

Therefore κ is the ideal lattice associated to the divisor (∂^{−1}, 1) where ∂ ⊂ O_{F} is the different of F. It is the inverse of the fractional ideal {y ∈ F : Tr(yz) ∈ Z for all z ∈ O_{F} }. We often call (∂^{−1}, 1) itself the canonical divisor and denote it by κ.

Proposition 11.3. Let F be a number field with canonical divisor κ. Then

(i) deg(κ) = log |Δ_{F} | and N(∂) = |Δ|. (ii) Let D = (I, u) be an Arakelov divisor. Then the Arakelov divisor class corresponding

to the ideal lattice \overline{uI}^{∨} is κ − D. (iii) There is an isomorphism of O_{F} -modules Ω_{O_{F} /Z}^{1} ≅ O_{F} /∂.

Proof. The covolume of κ is equal to 1/covol(O_{F}) = |Δ_{F} |^{−1/2}. Since covol(κ) = e^{−deg(κ)}\sqrt{|Δ_{F} |}, part (i) follows. The ideal lattice associated to the divisor κ − D is given by u^{−1}∂^{−1}I^{−1}. It is contained in \overline{uI}^{∨} because Tr(u^{−1}xyuz) ∈ Z for all x ∈ ∂^{−1}, y ∈ I^{−1} and z ∈ I. On the other hand, by Prop 5.5 the covolume of the lattice associated to κ − D is e^{−deg(κ−D)}\sqrt{|Δ_{F} |} which by part (i) is equal to 1/covol(D). This proves the proposition.

Theorem 11.4. (Riemann-Roch) Let F be a number field. Then for every Arakelov divisor D we have that

h^{0}(D) − h^{0}(κ − D) = deg(D) − \frac{1}{2} log |Δ_{F} |.

Proof. Since the ideal lattice associated to κ−D is the conjugate of the dual of the lattice associated to D = (I, U), an application of the Poisson summation formula of Thm. 4.3 to the lattice uI gives us that

∑_{y∈uI} e^{−π||y||^{2}} = \frac{1}{covol(D)} ∑_{y∈u^{−1}∂^{−1}I^{−1}} e^{−π||y||^{2}}.

Taking logarithms gives the required result.

In the remainder of this section we explain why Theorem 11.4is analogous to the Riemann-Roch Theorem for algebraic curves. First we introduce the analogue of the geometric notion of an effective divisor D on a curve. The corresponding arithmetic notion eff(D) is a real number between 0 and 1. It varies continuously in the coefficients of the infinite primes in the support of D. If eff(D) is 0 or close to zero, D should be thought of as being not effective or ‘close to being not effective’. On the other hand, if eff(D) is close to 1, the divisor D is close to being effective.

Definition. Let F be a number field. The effectivity eff(D) of an Arakelov divisor D = (I, u) is defined as

eff(D) = \{ \begin{matrix}e^{−π||u||^{2}}; & if O_{F} ⊂ I, \\ 0; & otherwise. \end{matrix}

If D = ∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ, the condition that O_{F} ⊂ I = ∏_{𝔭} 𝔭^{n_{𝔭}} means precisely that all coefficients n_{𝔭} are non-negative.

We have that

e^{−π||u||^{2}} = exp (−π ∑_{σ} deg(σ)e^{−2x_{σ}})

and this shows that the effectivity function behaves in a way that is analogous to the geometric function. Indeed, if D = ∑_{𝔭} n_{𝔭}𝔭 + ∑_{σ} x_{σ}σ is a divisor and we have that n_{𝔭} < 0 for some prime 𝔭, then eff(D) = 0. If all n_{𝔭} are non-negative, but at least one coeffient x_{σ} is negative, the exponential e^{−2x_{σ}} is large and so is the sum ∑_{σ} deg(σ)e^{−2x_{σ}}. Therefore eff(D) is a small positive number. The more negative x_{σ} becomes, the closer it gets to 0. Conversely, if n_{𝔭} ≥ 0 for all 𝔭 and the coefficients x_{σ} are positive, than the exponents e^{−2x_{σ}} are all small, so that eff(D) is close to 1. The larger the coefficients x_{σ} become, the closer the effectivity eff(D) is to 1.

Definition. Let F be a number field. Let D = (I, u) be an Arakelov divisor. Then we put

H^{0}(D) = {x ∈ F^{∗} : eff(D + (x)) > 0} ∪ {0}.

We have that eff(D + (x)) > 0 if and only if O_{F} ⊂ I. Since this is precisely the case when x ∈ I, the set H^{0}(D) is equal to the ideal I. We view H^{0}(D) as being analogous to the space of sections of the line bundle associated to the divisor on an algebraic curve. In order to measure the size of H^{0}(D) in some way, we weight its elements x by the effectivity eff(D + (x)) of the divisor D + (x) to which they give rise. Since D + (x) = (x^{−1}I, |x|u) we have that

eff(D + (x)) = e^{−π||xu||^{2}}.

The logarithm of the sum of all these ‘effectivities’ should be viewed as some kind of dimension of H^{0}(D). Adding 1 for the contribution of 0 ∈ I we recover the definition of h^{0}(D) given above.

h^{0}(D) = log (1 + ∑_{{x∈I \atop x≠0}} e^{−π||xu||^{2}} ) = log ( ∑_{y∈uI} e^{−π||y||^{2}} ) .

The canonical divisor class κ defined above is the analogue of the canonical divisor κ_{X} of an algebraic curve X. We briefly explain why. In the geometric situation, the degree of the canonical divisor is equal to 2g −2, where g is the genus of X. The functional equation of the zeta function Z_{X}(s) and algebraic curve X over F_{q} can be expressed by saying that the function

q^{(g−1)s}Z_{X}(s)

is invariant under the substitution s ↔ 1 − s. In Chapter 12 it is shown that the zeta functionZ_{F} (s) of a number field F has the property that |Δ_{F} |^{s/2}Z_{F} (s) is invariant under the substitution s ↔ 1 − s. Therefore the analogue of g − 1 is \frac{1}{2} log |Δ_{F} |. This means that for F = Q, the degree of κ must be zero. In section 7 we saw that the degree map induces an isomorphism Pic_{Q}^{0} ≅ R. Therefore we put κ = 0 for the number field Q. The Riemann-Hurwitz formula says that for a finite cover π : Y ⟶ X of curves we have that κ_{Y} = π^{∗}κ_{X} + δ_{Y/X}, where δ_{Y/X} denotes the different. The arithmetic analogue of this formula for π : Spec(O_{F}) ⟶ Z leads to our definition of the canonical divisor.

Exercises.

11.1 Let D = (I, u) be an Arakelov divisor associated to a number field F. Show that (\overline{uI})^{∨} = \overline{(uI^{∨})}. Show that \overline{uI}^{∨} is an O_{F} -submodule of F_{R}.

12. Zeta functions. In this section we discuss the zeta functions associated to number fields. We derive their functional equations and determine their residues in s = 1. Definition. The Riemann zeta function ζ(s) is defined as

ζ(s) = ∑_{n=1}^{∞} \frac{1}{n^{s}} , for s ∈ C with Re(s) > 1.

By comparing ζ(s) to the integral ∫_{1}^{∞} \frac{dx}{x^{s}} it follows easily that ζ(s) converges absolutely and uniformly on compact subsets of {s ∈ C : Re(s) > 1}.

Proposition 12.1. We have that

ζ(s) = ∏_{p prime} \frac{1}{1 − p^{−s}} , for s ∈ C with Re(s) > 1.

More precisely, the product converges absolutely and its limit is ζ(s).

Proof. Let s ∈ C with Re(s) > 1. We take the logarithm of the absolute value of the product. Since |1 − p^{−s}| = 1 − p^{−Re(s)}|, this gives

∑_{p prime} log (1 − \frac{1}{p^{Re(s)}} ) ≤ ∑_{p prime} (\frac{1}{p^{Re(s)}} + \frac{1}{2p^{2Re(s)}} + …)

≤ ∑_{p prime} \frac{1}{p^{Re(s)} − 1} ≤ ∑_{n=2}^{∞} \frac{1}{n^{Re(s)} − 1}.

Comparing the latter sum to the integral ∫_{2}^{∞} \frac{dx}{x^{Re(s)}−1} shows that the product converges absolutely when Re(s) > 1. To show that the limit is ζ(s), we consider the partial products. Since every integer n > 0 is a product of prime numbers in a unique way, for every n < M there is exactly one term n^{−s} in

∏_{p<M} \frac{1}{1 − p^{−s}} = ∏_{p<M} (1 + \frac{1}{p^{s}} + \frac{1}{p^{2s}} + …) = ∑_{n is product of primes p < M} \frac{1}{n^{s}} .

Therefore

| ∏_{p<M} \frac{1}{1 − p^{−s}} − ζ(s)| ≤ ∑_{n≥M} \frac{1}{n^{Re(s)}} ,

which tends to zero as M → ∞, because the sum converges absolutely. This proves the proposition. Definition. The Dedekind zeta function ζ_{F} (s) is defined as

ζ_{F} (s) = ∑_{0≠I⊂O_{F}} \frac{1}{N(I)^{s}} , for s ∈ C with Re(s) > 1.

Here I runs over the non-zero ideals of O_{F}.

The convergence of ζ_{F} (s) is guaranteed by the next proposition.

Proposition 12.2. Let F be a number field. For every s ∈ C with Re(s) > 1 we have that

ζ_{F} (s) = ∑_{0≠I⊂O_{F}} \frac{1}{N(I)^{s}} = ∏_{0≠𝔭⊂O_{F}} \frac{1}{1 − N(𝔭)^{−s}} .

(the product runs over the non-zero prime ideals of O_{F}). Both product and sum converge absolutely and uniformly on compact subsets of {s ∈ C : Re(s) > 1} and the limits are equal.

Proof. By Prop. 6.5 we have for every prime number p that

(p) = ∏_{𝔭|p} 𝔭^{e_{𝔭}}

and that ∑_{𝔭|p} e_{𝔭}f_{𝔭} = n = [F : Q]. Here f_{𝔭} is the degree of the finite field O_{F} /𝔭 over F_{p}. In other words, N(𝔭) = p^{f_{𝔭}}. Since e_{𝔭} and f_{𝔭} are at least 1, we see that N(𝔭) ≥ p and that (p) admits at most n distinct prime divisors 𝔭. Therefore

log | ∏_{0≠𝔭⊂O_{F}} \frac{1}{1 − N(𝔭)^{−s}} | = ∑_{0≠𝔭⊂O_{F}} − log(1 − N(𝔭)^{−Re(s)}),

= ∑_{0≠𝔭⊂O_{F}} (\frac{1}{N(𝔭)^{Re(s)}} + \frac{1}{2N(𝔭)^{2Re(s)}} + …) ,

≤ ∑_{0≠𝔭⊂O_{F}} \frac{1}{N(𝔭)^{Re(s)} − 1} = ∑_{p prime} ∑_{𝔭|p} \frac{1}{N(𝔭)^{Re(s)} − 1},

≤ ∑_{p prime} \frac{n}{p^{Re(s)} − 1} ≤ ∑_{m≥2} \frac{n}{m^{Re(s)} − 1}.

Comparing with the integral ∫_{2}^{∞} \frac{dx}{x^{Re(s)}−1}, it follows that the product converges absolutely when Re(s) > 1. By Thm. 2.3 every non-zero ideal of the Dedekind ring O_{F} is a product of prime ideals in a unique way. It follows that

∑_{N(I)<M} N(I)^{−Re(s)} ≤ ∏_{0≠𝔭⊂O_{F}} \frac{1}{1 − N(𝔭)^{−Re(s)}}

Therefore the series converges absolutely when Re(s) > 1. The infinite product converges to the same limit because

| ∑_{0≠I⊂O_{F}} \frac{1}{N(I)^{s}} − ∏_{N(𝔭)<M} \frac{1}{1 − N(𝔭)^{−s}} | ≤ ∑_{N(I)≥M} \frac{1}{N(I)^{Re(s)}}

which tends to zero as M → ∞. This proves the Proposition.

For an irreducible projective smooth algebraic curve X over a finite field F_{q}, the zeta function Z_{X}(s) is defined as

Z_{X}(s) = ∑_{D≥0} N(D)^{−s}, for s ∈ C with Re(s) > 1.

Here N(D) = e^{−deg(D)} and the sum runs over the effective divisors D of X. This suggests to involve the effectivity of section 11 and leads to the following definition.

Definition. Let F be a number field. The zeta function Z_{F} (s) associated to F is the function of a complex variable defined by

Z_{F} (s) = ∫_{Div_{F}} N(D)^{−s}eff(D)dD, for Re(s) > 1.

To see that Z_{F} (s) converges for Re(s) > 1, we write D = (I, u) and observe that N(D)^{−1} = NuN(I). Since the integrand is non- zero only when O_{F} ⊂ I, putting J = I^{−1} this leads to

Z_{F} (s) = ∑_{0≠J⊂O_{F}} N(J)^{−s} ∫_{∏_{σ} R_{>0}^{∗}} N(u)^{s}e^{−π||u||^{2}}dμ

Here dμ denotes the measure induced by the natural metric on ∏_{σ} R ≅ ∏_{σ} R_{>0}^{∗}. Since N(u) = ∏_{σ} u_{σ}^{deg σ} and ||u||^{2} = ∑_{σ} deg(σ)u_{σ}^{2}, the integral over ∏_{σ} R_{>0}^{∗} is a product of integrals of the form

\sqrt{deg(σ)} ∫_{0}^{∞} t^{deg(σ)s}e^{−πdeg(σ)t^{2}} \frac{dt}{t}.

This is equal to

\frac{1}{2}π^{− \frac{s}{2}} Γ(\frac{s}{2}) or \frac{1}{\sqrt{2}}(2π)^{−s}Γ(s)

depending on whether σ is real or complex. Since the sum ∑_{0≠J⊂O_{F}} N(J)^{−s} converges to the Dedekind zeta function ζ_{F} (s) when Re(s) > 1 and since the integral defining the Gamma function converges on {s ∈ C : Re(s) > 0}, the integral that defines Z_{F} (s) converges for s ∈ C with Re(s) > 1.

Theorem 12.3. Let F be a number field of degree n and discriminant Δ_{F}. The zeta function Z_{F} (s) admits a meromorphic continuation to C. It satisfies the functional equation: the function

|Δ_{F} |^{\frac{s}{2}} Z_{F} (s)

is invariant under the substitution s ↔ 1 − s. The only poles of Z_{F} (s) are at s = 0 and s = 1. They are of order 1. We have that

Res_{s=1} Z_{F} (s) = \frac{vol(Pic_{F}^{0} )}{w_{F} \sqrt{n}\sqrt{|Δ_{F} |}}  and  Res_{s=0} Z_{F} (s) = \frac{vol(Pic_{F}^{0} )}{w_{F} \sqrt{n}}.

Proof. We compute Z_{F} (s) by first integrating the function N(D)^{−s}eff(D)dD over each coset of the subgroup of principal divisors of Div_{F} and then integrating over Pic_{F}. We write the integral over the cosets of the discrete subgroup of principal divisors as sums rather than integrals. Since N(D) only depends on the class of D in Pic_{F}, this leads to

Z_{F} (s) = ∫_{Pic_{F}} (∑_{(x)} eff(D + (x))) N(D)^{−s}dD,

= ∫_{Pic_{F}} (\frac{1}{w_{F}} ∑_{x∈F^{∗}} eff(D + (x))) N(D)^{−s}dD.

Here w_{F} denotes the order of the group μ_{F} of roots of unity. The second equality follows from the fact that the kernel of the map x ↦ (x) from F^{∗} to Div_{F} is equal to μ_{F}. In terms of the function h^{0}(D) introduced in section 11, we have therefore that

Z_{F} (s) = ∫_{Pic_{F}} (\frac{e^{h^{0}(D)} − 1}{w_{F}}) N(D)^{−s}dD, for s ∈ C with Re(s) > 1.

Next we split the integral into two parts: the first integrates over the divisor classes D of deg(D) < \frac{1}{2}deg(κ) or equivalently with N(D) < N(κ)^{1/2} = \sqrt{|Δ_{F} |}. The second integral involves the divisor classes D with N(D) ≥ \sqrt{|Δ_{F} |}. In the second integral we make a change of variables: we replace D by κ − D. Since N(κ − D) ≤ N(κ)^{1/2} whenever N(D) ≥ N(κ)^{1/2}, this gives us the following expression for Z_{F} (s) when Re(s) > 1.

w_{F} Z_{F} (s) = ∫_{{D∈Pic_{F} \atop N(D)<N(κ)^{1/2}}}(e^{h^{0}(D)} − 1) N(D)^{−s}dD + ∫_{{D∈Pic_{F} \atop N(D)<N(κ)^{1/2}}}(e^{h^{0}(κ−D)} − 1) (\frac{N(κ)}{N(D)})^{−s} dD.

The key step is an application of the Riemann-Roch Theorem: in the second integral we use the equality

e^{h^{0}(κ−D)} = e^{h^{0}(D)}N(κ)^{1/2}/N(D).

The second integral becomes equal to

∫_{{D∈Pic_{F} \atop N(D)<N(κ)^{1/2}}} (e^{h^{0}(D)} − 1) \frac{N(κ)^{1/2}}{N(D)} (\frac{N(κ)}{N(D)})^{−s}dD + ∫_{{D∈Pic_{F} \atop N(D)<N(κ)^{1/2}}} (\frac{N(κ)^{1/2}}{N(D)} − 1) (\frac{N(κ)}{N(D)})^{−s}dD.

This leads to the following expression for the zeta function.

\begin{aligned}w_{F} N(κ)^{s/2}Z_{F} (s) = & ∫_{{D∈Pic_{F} \atop N(D)<N(κ)^{1/2}}} (e^{h^{0}(D)} − 1) ((\frac{N(κ)^{1/2}}{N(D)} )^{s} + (\frac{N(κ)^{1/2}}{N(D)} )^{1−s}) dD \\ & + ∫_{{D∈Pic_{F} \atop N(D)<N(κ)^{1/2}}}(\frac{N(κ)^{1/2}}{N(D)} − 1) (\frac{N(κ)^{1/2}}{N(D)} )^{−s} dD \end{aligned}

We evaluate the second integral. Since the integrand does depend on the norm N(D) rather than D itself, we use the exact sequence

0 ⟶ Pic_{F}^{0} ⟶ Pic_{F} ⟶^{deg} R ⟶ 0

Therefore the second integral is equal to the volume of Pic_{F}^{0} times an integral over R. Note however that the degree map is not compatible with the metrics on Pic_{F} and R. Indeed, the element 1 ∈ ⊕_{σ}R ⊂ F_{R} has length \sqrt{n}, but its degree n ∈ R has length n. Therefore the integral is smaller by a factor \sqrt{n} when we integrate with respect to the usual Haar measure on R. Switching to the multiplicative variable t = e^{x} = e^{deg(D)} we have that dx = \frac{dt}{t}. The second integral becomes

vol(Pic_{F}^{0} )\frac{1}{\sqrt{n}} (N(κ)^{(1−s)/2} ∫_{−∞}^{√\overline{N(κ)}} t^{s−1} \frac{dt}{t} − N(κ)^{−s/2} ∫_{−∞}^{√\overline{N(κ)}} t^{s} \frac{dt}{t})

which is equal to vol(Pic_{F}^{0} )/s(s − 1)\sqrt{n}. We find that

w_{F} N(κ)^{s/2}Z_{F} (s) = \frac{vol(Pic_{F}^{0} )}{s(s − 1)\sqrt{n}} + ∫_{{D∈Pic_{F} \atop N(D)<N(κ)^{1/2}}} (e^{h^{0}(D)} − 1)((\frac{N(κ)^{1/2}}{N(D)} )^{s}+ (\frac{N(κ)^{1/2}}{N(D)} )^{1−s})dD.

We claim that the integral expression on the right converges absolutely and uniformly in compact subsets of C. This means that we have found a meromorphic continuation for Z_{F} (s). The symmetry under s ↔ 1 − s is evident and since the volume of Pic_{F}^{0} is not zero, there are two simple poles at s = 0 and s = 1 with the required residues.

The convergence follows from the estimate

h^{0}(D) ≤ 2 · 3^{n} · exp (−πne^{− \frac{2}{n} deg(D)})

of Propososition 11.2. It shows that the absolute value of the integral is at most

2 · 3^{n} \frac{vol(Pic_{F}^{0} )}{\sqrt{n}} ∫_{0}^{√\overline{N(κ)}} e^{−πnt^{− \frac{2}{n}}} ((\frac{N(κ)^{1/2}}{t} )^{s}+ (\frac{N(κ)^{1/2}}{t} )^{1−s}) \frac{dt}{t}.

It suffices to show that the integral

∫_{0}^{√\overline{N(κ)}} e^{−πnt^{− \frac{2}{n}}} t^{s} \frac{dt}{t}

has the required convergence properties. This becomes clear when we make the change of variable y = t^{−2/n}. The integral becomes

∫ ∞_{\frac{1}{^{n} \sqrt{N(κ)}}} e^{−πny}y^{ns/2} \frac{dy}{y}

and is easily seen to converge absolutely and uniformly in compact subsets of C. This proves the theorem.

Since the product expression for the Dedekind zeta function

ζ_{F} (s) = ∏_{0≠𝔭⊂O_{F}} \frac{1}{1 − N(𝔭)^{−s}}

converges for s ∈ C with Re(s) > 1, it follows from the functional equation that Z_{F} (s) has all its zeroes in the critical strip {s ∈ C : 0 ≤ Re(s) ≤ 1}. Conjecturally all zeroes are in the center of the critical strip:

Generalized Riemann Hypothesis. Let F be a number field. Then every zero of Z_{F} (s) has real part equal to \frac{1}{2}.

The zeroes of Z_{F} (s) form a discrete subset of C. For F = Q, the first few zeroes are

\frac{1}{2} ± 14.134725 … i,
\frac{1}{2} ± 21.022040 … i.
\frac{1}{2} ± 25.010856 … i.
\frac{1}{2} ± 30.424878 … i.
⋮

The Riemann Hypothesis was formulated by G.B. Riemann for the Riemann zeta function ζ(s) = ζ_{Q}(s) in his 1859 paper on the distribution of prime numbers. It appears in D. Hilbert’s famous list of problems presented at the international congress in 1900 in Paris. More recently, the Clay Institute (www.claymath.org) included the conjecture in its list of seven major unsolved problems. The institute pays $1,000,000 for a correct proof of the Riemann Hypothesis.

Recently it has been verified by computer that the first 10^{11} zeroes are on the critical

line (www.zetagrid.net/zeta/math/zeta.result.100billion.zeros.html). The com-

putation implies that all zeroes s with |Im(s)| < 29, 538, 618, 432.236 … have real part equal to \frac{1}{2}.

For curves over finite fields the analogue of the Riemann Hypothesis has been proved by A. Weil in 1940–1948.

Exercises.

12.1 Let Γ(s) = ∫_{0}^{∞} t^{s}e^{−t} \frac{dt}{t} denote the Gamma function. (i) Show that the integral converges for s ∈ C with Re(s) > 0. (ii) Show that Γ(1) = 1 and that Γ( \frac{1}{2} ) = \sqrt{π}. (iii) Show that Γ(s + 1) = sΓ(s) for s ∈ C with Re(s) > 0. Deduce that Γ(n + 1) = n! for

every integer n ≥ 1. (iv) For any s ∈ C that is not a negative integer we define

Γ(s) = \frac{Γ(s + k + 1)}{s(s + 1) ⋯ (s + k)} ,

where k is any integer for which Re(s) > −k. Show that this defines a meromorphic continuation of Γ(s) to C. Show that Γ(s) only admits poles at k = 0, −1, −2, . . .. Show that the poles are simple and determine the residues.

(v) Show that Γ(s)Γ(1 − s) = π/sin(πs). (Hint: it suffices to show this for s ∈ C with 0 < Re(s) < 1.)

12.2

13. An explicit example.

Consider the following (randomly selected, Trento, december 1990) polynomial

f(T) = T^{4} − 2T^{2} + 3T − 7 ∈ Z[T].

This polynomial is irreducible modulo 2. This follows from the fact that it is an Artin-Schreier polynomial, but it can also, easily, be checked directly. We study the number field F = Q(α), where α is a zero of f(T). In order

to evaluate the discriminant of f(T), we compute the sums p_{i} of the ith powers of its roots in C using Newton’s relations (Exer.3.9):

\begin{aligned}p_{1} & = 0 \\ p_{2} & = −2s_{2} + p_{1}s_{1} = −2 · 2 + 0 = 4 \\ p_{3} & = 3s_{3} + p_{2}s_{1} − p_{1}s_{2} = 3 · (−3) + 0 + 0 = −9 \\ p_{4} & = 2p_{2} − 3p_{1} + 7p_{0} = 2 · 4 − 0 + 7 · 4 = 36 \\ p_{5} & = 2p_{3} − 3p_{2} + 7p_{1} = 2 · (−9) − 3 · 4 + 0 = −30 \\ p_{6} & = 2p_{4} − 3p_{3} + 7p_{2} = 2 · 36 − 3 · (−9) + 7 · 4 = 127 \end{aligned}

We have that

Disc(f(X)) = det ( \begin{matrix}4 & 0 & 4 & −9 \\ 0 & 4 & −9 & 36 \\ 4 & −9 & 36 & −30 \\ −9 & 36 & −30 & 127 \end{matrix} ) = −98443

which is a prime number. We conclude from Prop.4.8 that Δ_{F} = −98443 and that O_{F} = Z[α]. From the fact that the complex zeroes of f(X) are approximately equal to

\begin{aligned}α_{1} & = −2.195251731 …, \\ α_{2} & = 1.655743097 …, \\ α_{3}, \overline{α}_{3} & = 0.269754317 … ± 1.361277001 … i. \end{aligned}

we deduce that r_{1} = 2 and that r_{2} = 1.

Next we substitute all integers n with −18 ≤ n ≤ 18 in f(T) and factor the result into a product of prime numbers:

Table 13.1.^{\begin{matrix}n & f(n) = N(n − α) \\ 0 & −7 \\ 1 & −5 \\ 2 & 7 \\ 3 & 5 · 13 \\ 4 & 229 \\ 5 & 11 · 53 \\ 6 & 5 · 13 · 19 \\ 7 & 7 · 331 \\ 8 & 5 · 797 \\ 9 & 7^{2} · 131 \\ 10 & 11 · 19 · 47 \\ 11 & 5^{2} · 577 \\ 12 & 20477 \\ 13 & 5 · 5651 \\ 14 & 7 · 5437 \\ 15 & 149 · 337 \end{matrix} \begin{matrix}n & f(n) = N(n − α) \\ 0 & −7 \\ −1 & −11 \\ −2 & −5 \\ −3 & 47 \\ −4 & 5 · 41 \\ −5 & 7 · 79 \\ −6 & 11 · 109 \\ −7 & 5^{2} · 7 · 13 \\ −8 & 31 · 127 \\ −9 & 5 · 19 · 67 \\ −10 & 13 · 751 \\ −11 & 83 · 173 \\ −12 & 5 · 7 · 11 · 53 \\ −13 & 19 · 1483 \\ −14 & 5^{2} · 7^{2} · 31 \\ −15 & 50123 \end{matrix}}

By Minkowski’s Theorem (Prop. 9.5), the class group Cl(O_{F}) is generated by ideal I ⊂ O_{F} of norm at most

\frac{4!}{4^{4}} \frac{4}{π} \sqrt{98443} = 37.45189 ….

In order to calculate the class group, we determine the primes of small norm first.

We see in Table 13.1 that the polynomial f(T) has no zeroes modulo p for the primes p = 2, 3, 17, 23 and 29. We leave the verification that f(T) has no zeroes modulo 37 either, to the reader. By Kummer’s Lemma (Exer. 6.1) we conclude that there are no prime ideals of norm p for these primes p. It is easily checked that f(T) is irreducible modulo 2 and 3 and that f(T) ≡ (T − 1)(T + 2)(T^{2} − T + 1) (mod 5). The polynomial T^{2} − T + 1 is irreducible mod 5.

This gives us the following list of all prime ideals of norm less than 37.45 …: the ideals (2) and (3) are prime and (5) = 𝔭_{5}𝔭′_{5}𝔭_{25}, where 𝔭_{5} and 𝔭′_{5} have norm 5 and 𝔭_{25} is a prime of norm 25. The other primes 𝔭_{p} and 𝔭′_{p} of norm less 37.45 … have prime norm p. They are listed in Table 13.2 and are easily computed from Table 13.1.

Table 13.2.

\begin{matrix}𝔭_{5} = (5, α − 1) & 𝔭′_{5} = (5, α + 2) \\ 𝔭_{7} = (7, α) & 𝔭′_{7} = (7, α − 2) \\ 𝔭_{11} = (11, α + 1) & 𝔭′_{11} = (11, α − 5) \\ 𝔭_{13} = (13, α − 3) & 𝔭′_{13} = (13, α − 6) \\ 𝔭_{19} = (19, α − 6) & 𝔭′_{19} = (19, α + 9) \\ 𝔭_{31} = (31, α + 8) & 𝔭′_{31} = (31, α + 14) \end{matrix}

The class group is generated by the classes of these primes and the class of 𝔭_{25}. There exist, however, many relations between these classes. In the following table we list the factorizations of some numbers of the form q − pα, where p, q ∈ Z. We have chosen numbers of this form because the norms N(q − pα) = p^{4}f(q/p) can be computed so easily (Exer. 3.5). The factorizations into prime ideals of the principal ideals (q − pα) give rise to relations in the class group. For instance N(1 − 4α) = −2015 = −5 · 13 · 31 and (1 − 4α) = 𝔭_{5}𝔭_{13}𝔭_{31}. This shows that the ideal class of 𝔭_{5}𝔭_{13}𝔭_{31} is trivial. Therefore the class of 𝔭_{31} can be expressed in terms of classes of prime ideals of smaller norm:

𝔭_{31} ∼ 𝔭_{5}^{−1}𝔭_{13}^{−1}.

We conclude that the ideal 𝔭_{31} is not needed to generate the ideal class group. In a similar way one deduces from Table 13.3 below that the ideal classes of the primes of norm 31,19,13 and 11, can all be expressed in terms of ideal classes of primes of smaller norm.

Table 13.3.

β  N(β)  (β)
(i)  4α + 1  −5 · 31 · 13  𝔭_{5}𝔭_{13}𝔭_{31}
(ii)  3α − 2  −31  𝔭′_{31}
(iii)  α − 6  5 · 13 · 19  𝔭_{5}𝔭′_{13}𝔭_{19}
(iv)  2α − 1  −5 · 19  𝔭′_{5}𝔭′_{19}
(v)  α + 7  5^{2} · 7 · 13  𝔭′_{5}^{2}𝔭′_{7}𝔭′_{13}
(vi)  3α − 5  13  𝔭′_{13}
(vii)  α − 3  −5 · 13  𝔭′_{5}𝔭_{13}
(viii)  α + 1  −11  𝔭_{11}
(ix)  3α − 4  5^{2} · 11  𝔭′_{5}^{2}𝔭′_{11}

We conclude that Cl(O_{F}) is generated by the primes 𝔭_{5}, 𝔭′_{5}, 𝔭_{7}, 𝔭′_{7} and 𝔭_{25}. One does not need entry (vi) to conclude this, but this entry will be useful later. The primes of norm 5 and 7 are all principal. This follows form the first few lines of Table 13.1. Finally, since 𝔭_{5}𝔭′_{5}𝔭_{25} = (5), one concludes that 𝔭_{25} is principal. We have proved that the class group of Q(α) is trivial.

By Dirichlet’s Unit Theorem, the unit group has rank r_{1} + r_{2} − 1 = 2 + 1 − 1 = 2. The group of roots of unity is just {±1}. In all our calculations, we have not encountered a single unit yet! To find units, it is convenient to calculate the norms of some elements of the form a + bα + cα^{2} with a, b, c ∈ Z. This can be done as follows. We use the accurate approximations of the roots α_{1}, α_{2}, α_{3}, \overline{α_{3}} of f in C. By Lemma 5.1 one has that

N(a + bα + cα^{2}) = (a + bα_{1} + cα_{1}^{2}) (a + bα_{2} + cα_{2}^{2}) |a + bα_{3} + cα_{3}^{2}|^{2} .

Calculating norms of some small elements of the form a + bα + cα^{2} one soon finds that N(1 + α − α^{2}) = 5. This shows that the ideals 1 + α − α^{2} and 𝔭′_{5} are equal. In Table 13.1, we read that 𝔭′_{5} = (α + 2). We conclude that

ε_{1} = \frac{1 + α − α^{2}}{α + 2} = α^{3} − 2α^{2} + 3α − 4

is a unit. Similarly one finds that N(2−2α+α^{2}) = 65. One easily checks that (2−2α+α^{2}) = 𝔭′_{5}𝔭′_{13}. In Table III(vi) we see that 𝔭′_{13} = (3α − 5). We conclude that the principal ideals (2 − 2α + α^{2}) and ((α + 2)(3α − 5)) are equal. This implies that

ε_{2} = \frac{2 − 2α + α^{2}}{(3α − 5)(α + 2)} = α^{3} + α^{2} + α + 3

is a unit.

Rather then proving that the units ε_{1}, ε_{2} and −1 generate the unit group, we provide merely evidence that these units generate the whole group. For this we use the main result of the section 12. We use the ζ-function of the field F. Theorem 12.3 gives us an expression for the residue of the zeta function Z_{F} (s) associated to F at s = 1. Since the Riemann ζ-function ζ_{Q}(s) has a residue equal to 1 at s = 1 and the since the Gamma factors \frac{1}{2}π^{s/2}Γ(\frac{s}{2}) and \frac{1}{\sqrt{2}}(2π)^{s}Γ(s) have values in s = 1 equal to \frac{1}{2} and \frac{1}{2\sqrt{2}} respectively, one can express the content of Theorem 12.3 for a number field of degree n as follows

lim_{s→1} \frac{ζ_{F} (s)}{ζ_{Q}(s)} = \frac{2^{r_{1}}(2π\sqrt{2})^{r_{2}}}{w_{F} \sqrt{n}\sqrt{|Δ_{F} |}}vol(Pic_{F}^{0} ).

Using the Euler product formula for the ζ-functions and ignoring problems of convergence this gives rise to

∏_{p} \begin{matrix}∏ & 𝔭|p & (1 − \frac{1}{N(𝔭)}) & −1 \\ (1 − \frac{1}{p}) & −1 \end{matrix} = \frac{2^{r_{1}}(2π\sqrt{2})^{r_{2}}}{w_{F} \sqrt{n}\sqrt{|Δ_{F} |}}vol(Pic_{F}^{0} ).

We can compute the right hand side: r_{1} = 2, r_{2} = 1, w_{F} = 2 and Δ = −98443. By the calculation above we have that h_{F} = 1. If we assume that the units ε_{1}, ε_{2} are fundamental, we can compute the regulator using the two real embeddings ϕ_{1}, ϕ_{2} : F ↪ R given by α ↦ α_{1} and α ↦ α_{2} respectively. Denoting the three infinite primes by σ_{1}, σ_{2} and σ_{3} this gives

vol(Pic_{F}^{0} ) = 2^{\frac{r_{2}}{2}} \frac{h}{\sqrt{n}}|det ( \begin{matrix}1 & log |σ_{1}(ε_{1})| & log |σ_{1}(ε_{2})| \\ 1 & log |σ_{2}(ε_{1})| & log |σ_{2}(ε_{2})| \\ 1 & log |σ_{3}(ε_{1})| & log |σ_{3}(ε_{2})| \end{matrix} ) |.

Substituting approximations to the σ_{j}(ε_{i}) this gives

vol(Pic_{F}^{0} ) = \frac{1}{\sqrt{2}}|det ( \begin{matrix}1 & 3.427619209 & 1.600462837 \\ 1 & −3.752710586 & 2.479594524 \\ 1 & 0.1625456885 & −2.0400286805 \end{matrix} ) | ≈ 20.513421788

So, assuming that the units ε_{1}, ε_{2} are fundamental we find that the right hand side of the equation is equal to

\frac{4 · 2π\sqrt{2}}{2\sqrt{4}\sqrt{98443}} · 20.513421788 ≈ 0.5809524077.

If the units would not be fundamental, the volume of Pic_{F}^{0} would be k times as small, for some positive integer k. This would imply that the value 0.5809524077 would be replaced by 0.2904762039 (when k = 2) or 0.1936508026 (when k = 3) or … etc.

It can be shown that the Euler product on the left hand side converges. It converges only slowly and not absolutely: the contributions of the various prime ideals 𝔭 must be multiplied in increasing order of N(𝔭). We explicitly determine the factors in the Euler product on the left hand side. For a given prime p, the factor is

(1 − \frac{1}{p}) ∏_{𝔭|p} (1 − \frac{1}{N(𝔭)})^{−1} .

To determine it, we must find the way the prime p splits in the extension F over Q. Apart from the ramified prime 98443, there are five possibilities. Using Kummer’s Lemma they can be distinguished by the factorization of f(T) ∈ F_{p}[T]:

pO_{F} = \{ \begin{matrix}(i) 𝔭_{p}𝔭′_{p}𝔭′′_{p}𝔭′′′_{p}, & if f(T) has 4 zeroes mod p, \\ (ii) 𝔭_{p}𝔭′_{p}𝔭_{p^{2}}, & if f(T) has exactly 2 zeroes mod p, \\ (iii) 𝔭_{p}𝔭_{p^{3}}, & if f(T) has only one zero mod p, \\ (iv) 𝔭_{p^{2}}𝔭′_{p^{2}} & if f(T) has two irreducible quadratic factors mod p, \\ (v) (p), & if f(T) is irreducible mod p. \end{matrix}

here 𝔭_{p}, 𝔭_{p^{2}}, etc. denote primes of norm p, p^{2} etc. We find that the product on the left hand side is equal to

∏_{p} F (p)^{−1}

where

F (p) = (1 − \frac{1}{p})^{3}  in case (i),

= (1 − \frac{1}{p}) (1 − \frac{1}{p^{2}} ) in case (ii),

= (1 − \frac{1}{p^{3}} ) in case (iii),

= (1 + \frac{1}{p}) (1 − \frac{1}{p^{2}} ) in case (iv),

= (1 + \frac{1}{p} + \frac{1}{p^{2}} + \frac{1}{p^{3}} ) in case (v).

We approximate the left hand side product by evaluating the contributions of the primes less than a certain moderately large number. A short computer program enables one to compute this product with some precision. It suffices to count the zeroes of f(T) modulo p. This done by computing gcd(T^{p} − T, f(T)) in the ring F_{p}[T]. Most of the work is the calculation of T^{p} in the ring F_{p}[T]/(f(T)). To distinguish between cases (iv) and (v) one observes that in case (iv), the discriminant of f(T) is a square modulo p, while in case (v) it isn’t.

Using the primes less than 1657 one finds 0.5815983 for the value of the Euler product. This is close to the number 0.5809524077 that we found above. In view of the slow convergence of the Euler product, the error is not unusually large. It is rather unlikely that the final value will be two times, three times or even more times as small. This indicates, but does not prove, that the units ε_{1} and ε_{2} do indeed generate the unit group O_{F}^{∗} modulo torsion. To prove that they do, one should employ different techniques, related to methods for searching short vectors in lattices.

Exercises.

13.1 Let F = Q(\sqrt[5]{2}). Compute ring of integers O_{F}, ideal class group Cl(O_{F}) and unit group O_{F}^{∗}. 13.2 Let f(T) = T^{3} + T^{2} + 5T − 16. Show that f is irreducible and let F = Q(α) where α is a

zero of f(T). Compute ring of integers O_{F}, ideal class group Cl(O_{F}) and unit group O_{F}^{∗}.