FACTORES SOCIOCULTURALES

In this section, we prove the classical finiteness theorems for a number ring R: the Picard group Pic(R) is a finite group, and the unit group R∗ is in many cases finitely generated. These are not properties of arbitrary Dedekind domains, and the proofs rely on the special fact that number rings can be embedded in a natural way as lattices in a finite dimensional real vector space. The key ingredient in the proofs is non-algebraic: it is the theorem of Minkowski on the existence of lattice points in symmetric convex bodies given in 5.1.

Let V be a vector space of finite dimension n over the field R of real numbers, and

h·,·i : V ×V → R a scalar product, i.e. a positive definite bilinear form on V ×V. The scalar product induces a notion of volume on V, which is also known as the Haar measure on V. For a parallelepiped

B =_{r1x1+r2x2+. . .+rnxn : 0≤ri <1} spanned by x1, x2, . . . , xn, the volume is defined by

vol(B) =|det(hxi, xji)ni,j=1|1/2.

This definition shows that the ‘unit cube’ spanned by an orthonormal basis for V has volume 1, and that the image of this cube under a linear map T has volume |det(T)|. If the vectors xi are written with respect to an orthonormal basis for V as xi = (xij)nj=1,

then we have

|det(hxi, xji)ni,j=1|1/2 =|det(M ·Mt)|1/2 =|det(M)| for M = (xij)ni,j=1.

The volume function on parallelepipeds can be uniquely extended to a measure on V. Under the identification V ∼=Rn_{via an orthonormal basis for V, this is the Lebesgue mea-} sure on Rn_{. We usually summarize these properties by saying that V is an n-dimensional} Euclidean space.

A lattice in V is a subgroup of V of the form

L =Z·x1+Z·x2+. . .+Z·xk,

with x1, x2, . . . , xk ∈ V linearly independent. The integer k is the rank of L. It cannot exceed n= dimV, and we say thatL is complete or has maximal rank if it is equal to n. For a complete lattice L_⊂V, the co-volume vol(V /L) of L is defined as the volume of the parallelepiped F spanned by a basis of L. Such a parallelepiped is a fundamental domain for L as every x∈V has a unique representation x =f+l with f ∈F and l ∈L. In fact, vol(V /L) is the volume ofV /L under the induced Haar measure on the factor group V /L.

A subset X _⊂V is said to be symmetric if it satisfies ₋X =_{−x:x_∈X_}=X. 5.1. Minkowski’s theorem. Let L be a complete lattice in an n-dimensional Euclidean space V and X ⊂V a bounded, convex, symmetric subset satisfying

vol(X)>2n·vol(V /L).

ThenX contains a non-zero lattice point. IfX is closed, the same is true under the weaker assumption vol(X)≥2n·vol(V /L).

Proof. By assumption, the set 1₂X = {1

2x : x∈ X} has volume vol(21X) = 2−nvol(X)>

vol(V /L). This implies that the map 1₂X _→V /Lcannot be injective, so there are distinct points x1, x2 ∈ X with 1₂x1 − 1₂x2 = ω ∈ L. As X is symmetric, −x2 is contained in X.

By convexity, we find that the convex combination ω of x1 and −x2 ∈X is in X ∩L.

Under the weaker assumption volume vol(X) ≥ 2nvol(V /L), each of the sets Xk = (1 + 1/k)X withk _∈Z_≥1 contains a non-zero lattice point ωk ∈L. As all ωk are contained in the bounded set 2X, there are only finitely many different possibilities forωk. It follows that there is a lattice element ω ∈ ∩kXk, and for closed X we have ∩kXk =X. Let K be a number field of degreen. ThenK is an n-dimensionalQ-vector space, and by base extension we can map K into the complex vector space

KC=K _⊗Q C∼=

Y σ:K→C

C=Cn

by the canonical map ΦK :x7→(σ(x))σ. Note that ΦK is a ring homomorphism, and that the norm and trace on the free C-algebra KC extend the norm and the trace of the field extension K/Q. The image ΦK[K] of K under the embedding lies in the R-algebra

KR =_{(zσ)σ ∈KC:zσ =zσ}

consisting of the elements of KC invariant under the involution F : (zσ)σ −→ (zσ)σ. Here σ denotes the embedding of K in C that is obtained by composition of σ with complex conjugation.

On KC ∼= Cn, we have the standard hermitian scalar product h ·, · i. It satisfies

hF z1, F z2i = hz1, z2i, so its restriction to KR is a real scalar product that equips KR

with a Euclidean structure. In particular, we have a canonical volume function on KR. It naturally leads us to the following fundamental observation.

5.2. Lemma. LetRbe an order in a number fieldK. ThenΦK[R]is a lattice of co-volume

|∆(R)|1/2 _inKR.

Proof. Choose a Z-basis _{x1, x2, . . . , xn} for R. Then ΦK[R] is spanned by the vectors (σxi)σ ∈ KR, and using the matrix X = (σi(xj))ni,j=1 from the proof of 4.6, we see that

the co-volume of ΦK[R] equals

|det(h(σxi)σ,(σxj)σi)ni,j=1|1/2 =|det(Xt·X)|1/2 =|∆(R)|1/2. If I _⊂ R is a non-zero ideal of norm N(I) = [R : I] _∈ Z, then 5.2 implies that ΦK[I] is a lattice of co-volume N(I) · |∆(R)|1/2 _{in KR. To this lattice in KR we will apply}

Minkowski’s theorem 5.1, which states that every sufficiently large symmetric box in KR

contains a non-zero element of ΦK[I].

In order to compute volumes in KR, we have a closer look at its Euclidean structure. Denote the real embeddings of K in C by σ1, σ2, . . . , σr and the pairs of complex embed- dings of K by σr+1, σr+1, σr+2, σr+2, . . . , σr+s, σr+s. Then we have r+ 2s =n= [K :Q], and an isomorphism of R-algebras

KR _−→Rr_×Cs (5.3)

(zσ)σ 7−→(zσi) r+s i=1.

The inner product on KR is taken componentwise, with the understanding that at a complex component, the inner product of elements z1 =x1+iy1 and z2 = x2+iy2 ∈ C

equals h z1 z1 , z2 z2 i=z1z2+z1z2 = 2 Re(z1z2) = 2(x1x2+y1y2).

Note that this differs by a factor 2 from the inner product under the ‘standard identifica- tion’ C = R2 of C as the ‘complex plane’. For this reason, volumes in KR are 2s times larger than they are in Rr_×Cs _{with the ‘standard’ Euclidean structure.}

5.4. Theorem. LetRbe an order in a number fieldKwithspairs of complex embeddings. Then every ideal class in the Picard group Pic(R) contains an integral ideal of norm at most 2_πs|∆(R)|1/2_{, and Pic(R) is a finite abelian group.}

Proof. Let I _⊂ R be an ideal and Xt ⊂ KR the closed box consisting of elements (zσ)σ with_|zσ| ≤tfor allσ. ThenXt ⊂Rr×Csis a product ofr real intervals [−t, t] andsdisks in Cof radius t, so its canonical volume in KR is 2s·(2t)r(πt2)s= 2r+sπstn. Minkowski’s theorem 5.1 implies that Xt contains a non-zero element ΦK(x) ∈ ΦK[I]∩Xt, provided that we have 2r+s_πs_tn_≥2n_N(I)|∆(R)|1/2_{. We therefore choose t∈R}

>0 to satisfy

tn = _π2sN(I)_{· |}∆(R)_|1/2.

The norm of the element x ∈ I obtained satisfies |NK/Q(x)| =

Q

σ|σ(x)| ≤ tn since all

|σ(x)_| are bounded by t, so we conclude that every integral R-ideal I contains a non-zero element x of norm _|N_K/Q(x)| ≤ _π2

s

N(I)_{· |}∆(R)_|1/2_.

If I is invertible, the element x∈ I we just found gives rise to an integral ideal xI−1

in the ideal class [I−1]_∈ Pic(R) of norm at most _π2s_|∆(R)_|1/2. As I was arbitrary, this implies that every ideal class in Pic(R) contains an integral ideal satisfying this norm bound. There are only finitely many ideals in R having norm below a given bound, so we

find that Pic(R) is finite.

For R=_OK the ring of integers of a number fieldK, the Picard group Pic(R) is the class group

Cl(K) =Cl(OK)

of K, and its order is known as the class number hK of K. It is a fundamental invariant of the number field K.

Exercise 1. Show that Pic(R) is finite for every Dedekind domainR⊂K. [Hint: use exercise 3.28.] As the primary ideals occurring in the factorization of an invertible integral ideal I ⊂ R

have norm at most N(I), it follows from 5.4 that for an order R, the Picard group is generated by primary ideals of norm at most _π2s_|∆(R)_|1/2_{. For the ring of integers of}

K, we find that Cl(OK) is generated by prime ideals of norm at most _π2s|∆K|1/2. This enables us to find an explicit set of generators of the class group by factoring the rational primes up to this bound.

5.5. Examples. For the Dedekind domain R=Z[√₋5] we have ∆(R) =₋20, so Pic(R) is generated by the ideal classes of the primes of norm at most _π2√20 < 3. The unique

prime p = (2,1 +√−5) of norm 2 in R is non-principal, but its square is generated by 2. We find Pic(R)∼=Z/2Z, so Q(√₋5) is a field of class number 2.

The ring R=Z[√₋19] is not a Dedekind domain. By 3.11 the prime 3 is inert in R, so the only primary R-ideals of norm at most _π2√4·19<6 are the non-invertible prime

p₂ = (2,1 +√−19), the principal ideal (2) and the prime ideals p₅ = (5,1 +√−19) and

q5 = (5,1−√−19) lying over 5. Asp5q5 = (5) is principal, we find that Pic(R) is generated

byp5. Asp5 is non-principal with cubep35 = (7+2 √

−19), the Picard group Pic(R) is cyclic of order 3.

Let _O = Z[(1 +√₋19)/2] be the integral closure of Z[√₋19]. Then 2 is inert in _O, so there are no primes in _O of norm smaller than 2_π√19<3. It follows that no primes are needed to generate the class group, so Pic(O) is trivial and Q(√−19) has class number 1. The constant 2_πsin 5.4 can be improved if we can replaceXt by a box of the same volume containing elements of smaller norm.

xy=±1 x y xy=_±1₂ x y

A look at the real quadratic case above suggests that we should take Xt ⊂KR of the form

(5.6) Xt ={(zσ)σ ∈KR :

X σ

|zσ| ≤t}. Computing the volume of this box is an exercise in integration.

5.7. Lemma. The box Xt ⊂KR in 5.6 has canonical volume vol(Xt) = 2rπs t n n!.

Proof. Inside Rr_×Cs, our box takes the shape

Xt ={((xi)i,(zj)j)∈Rr×Cs : X

i|xi|+ 2 X

j|zj| ≤t}, and we need to show that this box has ‘standard volume’Vr,s(t) = 2r π₂s t

n

n!. As the volume

function is a real valued function depending on two integral parameters r and s, we can apply induction with respect to each of these parameters.

We note first that V1,0(t) = vol([−t, t]) = 2t and V0,1(t) = vol({|z| ≤ t/2}) = πt2/4

have the required value. For Vr+1,s(t) we use the induction hypothesis to obtain

Vr+1,s(t) = Z t x=−t Vr,s(t− |x|)dx= 2r π 2 s ·2 Z t x=0 (t−x)n n! dx= 2 r+1 π 2 s tn+1 (n+ 1)!. For Vr,s+1(t) the method is similar, but we have to integrate with respect to a complex

variable z =x+iy =ρeiθ to obtain its value Z |z|<t/2 Vr,s(t−2|z|)dxdy = 2r π 2 s ·2π Z t/2 ρ=0 (t−2ρ)n n! ρdρ= 2 r π 2 s+1 tn+2 (n+ 2)!.

This establishes the result.

5.8. Theorem. Let R be an order in a number field K of degree n with s pairs of complex embeddings. Then every ideal class of Pic(R) contains an integral ideal of norm not exceeding the Minkowski constant

MR= 4 π s n! nn · |∆(R)| 1/2 of the order R.

Proof. As in the proof of 5.4, we need to bound the absolute value of the norm of an element x= (zσ)σ ∈Xt, where t is chosen such that vol(Xt) = 2nN(I)|∆K|1/2. Using the value vol(Xt) = 2rπs t

n

n! from 5.7 and the fact that the geometric mean (

Q

σ|zσ|)1/n does not exceed the arithmetic mean _n1 P_σ|zσ|, we obtain

|N(x)_|=Y σ |zσ| ≤( 1 n X σ |zσ|)n ≤ tn nn = 4 π sn! nn · |∆(R)| 1/2_.

The result follows as in 5.4.

5.9. Corollary. The class group of a number field K is generated by the classes of the prime ideals of norm at most MK = _π4s n_nn!|∆K|1/2. A second corollary of 5.8 is the existence of a lower bound for discriminants of orders in terms of their rank.

5.10. Corollary. Let R be an order of rank n over Z. Then we have

|∆(R)| ≥ π 4 2s n2n (n!)2 ≥bn def = π 4 n n2n (n!)2.

One has bn ≥πn/4 for all n, and limn→∞b1n/n =πe2/4≈5.803.

Proof. As every integral R-ideal has integral norm, we must have MR ≥ 1 in 5.8. This yields the inequality. From the identity bn+1/bn = (π/4)·(1 + _n1)2n ≥ π we obtain both the lower bound for bn and the limit behavior limn→∞b1n/n = limn→∞ bn+1_bn =πe2/4.

Better lower bounds for discriminants have been derived around 1975 by complex analytic methods [25]. Our bounds already imply the famous result of Minkowski that there are no number fields K 6=Q of discriminant 1.

5.11. Corollary. Let K 6= Q be a number field. Then there exists a prime p that is ramified in K.

Proof. For K ₆=Q we have n_≥2 and_|∆K| ≥π2/4>2, so there is a prime dividing ∆K.

This prime ramifies in K/Q by 4.14.

It follows from 5.11 and 4.5 that the discriminant of any order R ₆= Z in a number field satisfies |∆(R)| > 1. There are other restrictions on the values of discriminants, such as the Stickelberger criterion ∆(R)_≡0,1 mod 4 from exercise 4.13. IfR is quadratic over Z, it is uniquely determined by its discriminant. This is not true for orders of higher rank (exercise 18), but the geometry of numbers yields the following finiteness result for orders ‘with bounded ramification’.

5.12. Hermite’s theorem. LetD_∈Zbe an integer. Then there are, up to isomorphism, only finitely many orders R of discriminant∆(R) =D.

Proof. As the ring of integers _OK of a number fieldK contains only finitely many orders of bounded index, it suffices in view of 4.10 to show that there are only finitely many isomorphism classes of number fields of discriminant ∆K =D. The degree n of such K is bounded by 5.10, so we may assume that both D and n are fixed.

SupposeK is a field of degreenwith absolute discriminantD. Then we construct a box

X ⊂KR that is bounded in terms of D and contains an element Φ(x) with K = Q(x) as follows. IfK has a real embeddingτ :K _→R we letX consist of the elements (zσ)σ ∈KR satisfying

|zτ| ≤C and |zσ| ≤1 if σ 6=τ, where C ∈ R>0 is chosen to have vol(X) = 2n

p

|D|. If K has only complex embeddings and τ :K →Cis one of them, we define X as consisting of the elements with

zτ, zτ ∈[−1,1] + [−C, C]i⊂C and |zσ| ≤1 if σ 6=τ, τ , with C _∈ R>0 such that again vol(X) = 2n

p

|D_|. In both cases, X is symmetric and convex and C depends on D but not onK.

By 5.1 and 5.2, there exists a non-zero element x ∈ OK such that Φ(x) = (σ(x))σ is contained in X. We will show thatK =Q(x) for thisx. As the embeddingsσ :K _→Care the [K :Q(x)] extensions toK of each of the [Q(x) :Q] embeddingsQ(x)_→C, it suffices to show that for our chosenτ, we have σ(x)6=τ(x) wheneverσ 6=τ. In the case thatτ is a real embedding, we cannot have τ(x) =σ(x) for any σ =₆ τ as this would imply _|σ(x)_|<1 for all embeddings σ : K _→ C and therefore _|N_K/Q(x)| = Q_σ|σ(x)|< 1. This obviously

cannot happen for the non-zero integer NK/Q(x) ∈ Z. In the case that τ is a complex

embedding, the same argument shows that τ(x) ₆= σ(x) if σ ₆= τ, τ. We also cannot have

τ(x) =τ(x), since this would imply that τ(x) is a real element in (₋1,1) + (₋C, C)i, which leads to the same impossible estimate |τ(x)|<1.

We conclude that K is generated overQ by an elementx∈ OK with Φ(x)∈X. This means that the zeroes σ(x)_∈C of the irreducible polynomialfx

Q ∈Z[X] have an absolute

value that can be bounded in terms of D. As the degree n of fx

bound the coefficients of fQx in Z. It follows that, given D, there are only finitely many

possibilities for f_Qx, and therefore the number of possible K _⊂Q is also finite. Given r, s ∈ Z≥0, it is a challenging problem to count up to isomorphism the number

Sr,s(X) of number fields K having r real and 2s complex embeddings and satisfying

|∆K|< X. One can do this exactly for given X, or asymptotically for X → ∞. Her- mite’s theorem, though in principle constructive, does not provide a direct answer to these questions. Only recently one has been able to obtain answers in cases withn=r+ 2s >3.

Exercise 2. Find asymptotic expressions forS2,0(X) andS0,1(X).

As a final application of Minkowski’s theorem, we show that the unit group of an order R

in a number field with r real and 2s complex embeddings isfinitely generated of free rank

r+s₋1. The classical formulation of this result is as follows.

5.13. Dirichlet unit theorem. Let R be an order admitting r real and 2s complex embeddings, and write µR for the group of roots of unity in R. Then µR is finite, and

R/µR is a free abelian group of rank r+s−1.

More explicitly but less canonically, 5.13 states that there exists a finite set η1, η2, . . .,

ηr+s−1 of fundamental units such that we have

R∗ =µR× hη1i × hη2i ×. . .× hηr+s−1i.

Such a system of fundamental units, which forms a Z-basis for R∗/µR, is only unique up

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