Algebraic Number Theory

Summer Term 2012

Université du Luxembourg

Sara Arias-de-Reyna, Gabor Wiese [email protected] [email protected]

Version of 21st June 2012

1 CONTENTS 2

Contents

1 Motivation 4

2 Linear algebra in ﬁeld extensions 9

3 Rings of integers 15

4 Ideal arithmetic 22

5 Ideals in Dedekind rings 26

6 Geometry of Numbers 32 6.1 Introduction...... 32 6.2 Lattices ...... 35 6.3 Number rings as lattices ...... 40 6.4 Finiteness of the class number ...... 42 6.5 Dirichlet Unit Theorem ...... 47 CONTENTS 3

Preface

These are notes of a one-term course (12 lectures of 90 min each) taught at the University of Lux- embourg in Summer Term 2012. The lecture builds on the lecture Commutative Algebra from the previous term, the lecture notes of which are available on http://maths.pratum.net. The lecture provides an introduction to the most basic classical topics of (global) algebraic number theory:

first cases of Fermat’s Last Theorem, • norms, traces and discriminants of field extensions, • rings of integers, • ideal arithemtic and ideal class groups, • Dedekind rings, • fundamentals of the geometry of numbers, • finiteness of the class number, • Dirichlet’s Unit Theorem. •

In preparing these lectures we used several sources:

Neukirch: Algebraische Zahlentheorie, Springer-Verlag. • Samuel: Algebraic Theory of Numbers. • Bas Edixhoven: Théorie algébrique des nombres (2002), Lecture notes available on Edix- • hoven’s webpage.

Peter Stevenhagen: Number Rings, Lecture notes available on Stevenhagen’s webpage. • Lecture notes of B.H. Matzat: Algebra 1,2 (Universität Heidelberg, 1997/1998). • Lecture notes of lectures on Algebraische Zahlentheorie taught at Universität Duisburg-Essen • in Winter Term 2009/2010.

Luxembourg, June 2012. Sara Arias-de-Reyna, Gabor Wiese 1 MOTIVATION 4

1 Motivation

As a motivation we are going to treat the simples cases of the Fermat equation. Let n N. The n-th ∈ Fermat equation is F (a,b,c)= an + bn cn. n − What are the zeros of this equation in the (positive) integers? n = 1: For any ring R, there is the bijection

(a,b,c) R3 F (a,b,c) = 0 R2, { ∈ | 1 } ↔ given by sending (a,b,c) with F (a,b,c)= a + b c = 0 to (a, b). Its inverse clearly is the map that 1 − sends (a, b) to (a,b,a + b). This clearly describes all solutions. n = 2: A triple (a,b,c) N3 such that F (a,b,c) = a2 + b2 c2 = 0 is called a Pythagorean ∈ 2 − triple. It is called primitive if gcd(a,b,c) = 1 and a is odd (whence b is even). In last term’s course on Commutative Algebra you proved on Sheet 5 (almost) that there is the bijection

(u,v) N2 u>v, gcd(u,v) = 1, 2 uv { ∈ | | } (a,b,c) N3 (a,b,c) primitive Pythagorean triple , ↔ { ∈ | } sending (u,v) to (u2 v2, 2uv,u2 + v2). − We postpone the case n = 3 and continue with: n = 4: Theorem 1.1. There is no (a,b,c) N3 such that a4 + b4 = c4, i.e. F has no solution in positive ∈ >0 4 integers [recall that positive means strictly bigger than 0]. This will immediately follow from the following Proposition. Proposition 1.2. Let (a,b,c) Z3 be such that a4 + b4 = c2. Then abc = 0. ∈ Proof. Since the exponents are all even, we can without loss of generality assume that all a,b,c are non-negative. We assume that the assertion of the proposition is wrong and want to get a contradiction. For that we let c be minimal such that there are a,b > 0 satisfying a4 + b4 = c2. As c is minimal, we have that gcd(a,b,c) = 1; for, if d is the greatest common divisior, then we have a b a4 + b4 c2 c 4 + 4 = = = 2, d d d4 d4 d2 because d2 has to divide c. Now we can reinterpret the equation as (a2, b2,c) being a primitive Pythagorean triple (after possibly exchanging a and b so that a2 is odd). Hence, we may apply the case n = 2. This means that there are u,v N such that u>v, gcd(u,v) = 1 and ∈ a2 = u2 v2, b2 = 2uv, c2 = u2 + v2. − Hence, a2 + v2 = u2, which gives yet another primitive Pythagorean triple, namely (a,v,u) (note that since a is odd, v is even). So, we can again apply n = 2 to obtain r > s such that gcd(r, s) = 1 and a = r2 s2, v = 2rs, u = r2 + s2. − 1 MOTIVATION 5

Plugging in we get: b b2 = 2uv = 4urs, and hence 2 = urs. (1.1) 2 As gcd(u,v) = 1, we also have that gcd(u,rs) = 1 (note: u is odd). As, furthermore, gcd(r, s) = 1, it follows from Equation (1.1) that u, r and s are squares:

u = x2, r = y2, s = z2.

They satisfy: x2 = u = r2 + s2 = y4 + z4. So, we have found a further solution of our equation. But:

c = u2 + v2 = x4 + v2 >x4 x, ≥ contradicting the minimality of c.

In this proof, the gcd played an important role and we used at several places that Z is a unique factorisation domain (UFD), that is, that every non-zero integer is uniquely the product of prime numbers (and 1). − n 3 in C[X]: In order to illustrate one quite obvious (but, failing) attempt at proving that the ≥ Fermat equation has no positive solutions for n 3, we now work for a moment over C[X], where ≥ this strategy actually works. Recall that C[X] is a Euclidean ring, just like Z. Below we will show that 2πi/3 this strategy also works for the Fermat equation F3 over Z because the ring Z[ζ3] with ζ3 = e is a unique factorisation domain and has ‘few’ roots of unity.

Proposition 1.3. Let n 3 and let a,b,c C[X] be such that an + bn = cn. Then a, b and c form ≥ ∈ a trivial solution: they are scalar multiples of one polynomial (a(X) = αf(X), b(X) = βf(X), c(X)= γf(X) for some f(X) C[X] and α,β,γ C). ∈ ∈ Proof. We prove this by obtaining a contradiction. Let us, hence, assume that there are a,b,c C[X] ∈ satisfying an + bn = cn such that

max deg(a), deg(b), deg(c) > 0 and is minimal among all solutions. { } As C[X] is factorial (because it is Euclidean), we can always divide out common divisors. Thus, by the minimality assumption the polynomials a,b,c are pairwise coprime. Also note that at most one of the polynomials can be constant, unless we have a trivial solution. The principal point of this proof is that we can factor the Fermat equation into linear factors because ζ = e2πi/n is an element of C (this, of course, fails over Z, whence in the attempt to use this trick for the original Fermat equation one has to work with Z[ζ], which will not be factorial in general). The factorisation is this one:

n 1 − an = cn bn = (c ζjb). (1.2) − − jY=0 1 MOTIVATION 6

If you have never seen this factorisation, just consider c as a variable and observe that ζjb are n distinct roots of the polynomial cn bn and recall that a polynomial of degree n over an integral domain has − at most n zeros. Recall once more that C[X] is a factorial ring. So it makes sense to ask whether the above factorisation is into pairwise coprime factors. We claim that this is indeed the case. In order to verify this, let j, k 0,...,n 1 be distinct. We have: ∈ { − }

1 j k 1 j j k k b = (c ζ b) (c ζ b) and c = ζ− (c ζ b) ζ− (c ζ b) . ζk ζj − − − ζ j ζ k − − − − − − − Thus, any common divisor of (c ζjb) and (c ζkb) necessarily divides both b and c. As these are − − coprime, the common divisor has to be a constant polynomial, which is the claim. We now look again at Equation (1.2) and use the coprimeness of the factors. It follows that each factor c ζjb has to be an n-th power itself, i.e. there are y C[X] such that − j ∈ yn = c ζjb j − for all j 0,...,n 1 . Of course, the coprimeness of the c ζjb immediately implies that y and ∈ { − } − j y for j = k have no common non-constant divisor. If the degrees of c and b are different, then the k 6 degree of yj is equal to the maximum of the degrees of c and b divided by n for all j. If the degrees are equal, then at most one of the yj can have degree strictly smaller than the degree of b divided by n because this can only happen if the leading coefﬁcient of c equals ζj times the leading coefﬁcient of b. As n 3, we can pick three distinct j, k, ℓ 0,...,n 1 . We do it in such a way that y is ≥ ∈ { − } j non-constant. Now consider the equation

n n j k ℓ n αyj + βyk = α(c + ζ b)+ β(c + ζ b)= c + ζ b = yℓ ,

which we want to solve for 0 = α,β C. Thus, we have to solve 6 ∈ α + β = 1 and αζj + βζk = ζℓ.

A solution obviously is ζℓ ζk α = − and β = 1 α. ζj ζk − − n n In C we can draw n-th roots: α = γ and β = δ . Setting r = γyj, s = δyk and t = yℓ, we obtain

rn + sn = tn,

with polynomials r, s, t C[X]. Let us ﬁrst remark that r is non-constant. The degrees of r, s, t are ∈ less than or equal to the maximum of the degrees of b and c divided by n, hence, the degrees of r, s, t are strictly smaller than the degrees of b and c. As the degree of a has to be at most the maximum of the degrees of b and c, the degrees of r, s, t are strictly smaller than the maximum of the degrees of a,b,c. So, we found another solution with smaller maximum degree. This contradiction proves the proposition.

n=3: We will prove a slightly more general statement. First a lemma. 1 MOTIVATION 7

Lemma 1.4. Let ζ = e2πi/3 = 1 + i √3 C. Consider A := Z[ζ]= a + ζb a, b Z . − 2 2 ∈ { | ∈ } (a) ζ is a root of the irreducible polynomial X2 + X + 1 Z[X]. ∈ (b) The ﬁeld of fractions of A is Q(√ 3). − (c) The norm map N : Q(√ 3) Q, given by a+b√ 3 a2 +3b2 = (a+b√ 3)(a b√ 3) = − → − 7→ − − − (a + b√ 3)(a + b√ 3) is multiplicative and sends any element in A to an element in Z. In − − particular, u A is a unit (i.e. in A ) if and only if N(u) 1, 1 . Moreover, if N(a) is a ∈ × ∈ { − } ± prime number, then a is irreducible.

(d) The unit group A is equal to 1, ζ, ζ2 and is cyclic of order 6. × {± ± ± } (e) The ring A is Euclidean with respect to the norm N and is, hence, by a theorem from last term’s lecture, a unique factorisation domain.

(f) The element λ = 1 ζ is a prime element in A and 3= ζ2λ2. − −

(g) The quotient A/(λ) is isomorphic to F3.

(h) The image of the set A3 = a3 a A under π : A A/(λ4) = A/(9) is equal to { | ∈ } → 0 + (λ4), 1 + (λ4), λ3 + (λ4) . { ± ± } Proof. Exercise on Sheet 1.

Theorem 1.5. Let a,b,c A and u A satisfy ∈ ∈ × a3 + b3 = uc3.

Then abc = 0.

Proof. We essentially rely on the result of Lemma 1.4 that A is a factorial ring. The proof is again by obtaining a contradiction. Let us hence assume that we have a,b,c A and u A satisfying ∈ ∈ × a3 + b3 = uc3 and abc = 0. By dividing out any common factors, we may and do assume that a,b,c 6 are pairwise coprime. Note that consequently at most one of a,b,c can be divisible by λ. Of course, we will use the factorisation

a3 + b3 = (a + b)(a + ζb)(a + ζ2b). (1.3)

We derive our contradiction in several steps and the factorisation is not used before (4). (1) We show λ abc. | Suppose this is not the case. Then a3, b3 and c3 are 1 in A/(λ4) by Lemma 1.4. Consider the ± equation a3 + b3 = uc3 in A/(λ4). The left hand side is thus in (λ4), 2 + (λ4), 2 + (λ4) , the right { − } hand side is u + (λ4). Thus u 2 (mod (λ4)) (as u is a unit, the left hand side cannot be (λ4). ± ≡ ± However, the triangle inequality for the absolute value of C immediately shows that 0 < u 2 3, | ± |≤ which is smaller than λ4 = 9, excluding that λ4 divides u 2 (this conclusion uses, of course, that | | ± the absolute value of any nonzero a A is at least 1; but, that is obvious.). ∈ (2) We may (and do) assume without loss of generality that λ c. | 1 MOTIVATION 8

If λ does not divide c, then by (1) it has to divide a or b (note that it cannot divide both, as it would then divide c as well). We argue similarly as in (1) and take our equation in A/(λ4). The right hand side is again u + (λ4). The left hand side is either 1 + (λ4) or 1+ λ3 + (λ4). Hence, we ± ± ± ± get λ4 divides u + 1 or u + 1+ λ3. But, 0 < u + 1+ λ3 2+ √27 < 8 < 9 = λ4 , ± ± ± | ± ± | ≤ | | so the first possibility is excluded. Because of u + 1 2 < 9 = λ4 , we necessarily have | ± | ≤ | | u = 1, which satisfies u3 = u. So, we have instead consider the equation a3 + (uc)3 = ( b)3 or ± − b3 + (uc)3 = ( a)3. − (3) We show λ2 c. Let r 2 be such that λr c and λr+1 ∤ c. | ≥ | Suppose λ2 ∤ c. We do it again similarly and again reduce the equation modulo λ4. The right hand side is uλ3, but the left hand side is, as in (1), in (λ4), 2 + (λ4) . Hence, again λ4 has to divide ± { ± } uλ3 or uλ3 2, which are both impossible by the same considerations of absolute values as above. ± (4) Replacing b by ζb or by ζ2b, we may (and do) assume λ2 (a + b). | The right hand side of Equation 1.3 is divisible by λ6 (because of (3)). Thus, λ2 divides one of the three factors and making one of the mentioned substitutions we assume it is the first one. (5) We show λ a + ζb, λ a + ζ2b, λ2 ∤ a + ζb and λ2 ∤ ζ2b. | | We only treat a + ζb, the other one works precisely in the same way. Note that ζ 1 (mod (λ)). ≡ Thus, a + ζb a + b 0 (mod (λ)). If λ2 (a + ζb), then because of λ2 (a + b), substracting the ≡ ≡ | | two yields λ2 b(ζ 1) = λb, whence λ b, which is excluded. | − − | (6) We show that the only common prime divisor of any pair of a + b, a + ζb, a + ζ2b is λ (up to multiplying λ by units, of course, i.e. up to associates). This argument is very standard and we only do it for one pair. Suppose that the prime element µ divides a + b and a + ζb. Then µ divides (a + b) (a + ζb)= bλ, whence µ divides b. Moreover, µ − also divides ζ(a + b) (a + ζb)= aλ, whence µ also divides a. As a and b are coprime, we have a − − contradiction. 3 3 (7) We show that there are coprime 0 = a1, b1,c1 A and there is u1 A× such that a1 + b1 = 3 r 6 ∈ ∈ u1c1 and λ ∤ c1. From (5) and (6) we can write using the factoriality of A:

3r 2 3 3 2 3 a + b = ǫ1λ − α , a + ζb = ǫ2λβ , a + ζ b = ǫ3λγ with pairwise coprime α,β,γ and units ǫ , ǫ , ǫ A . Now we compute 1 2 3 ∈ × 2 2 2 3r 3 3 3 2 3 0=(1+ ζ + ζ )(a + b) = (a + b)+ ζ(a + ζb)+ ζ (a + ζ b)= λ ǫ1λ − α + ǫ2ζβ + ǫ3ζ γ .

2 Dividing by λζ ǫ3 we obtain 3 3 3 γ + ǫβ = u1α 4 with units ǫ and u1. The same calculations as in (2) yield (taking the equation modulo (λ )) that ǫ = 1, whence ǫ3 = ǫ. Thus, letting a := γ, b = ǫβ and c = α , we obtain (7). ± 1 1 1 1 (8) End of proof. 2 Repeating steps (1) to (7) with a1, b1,c1 often enough we can achieve that λ ∤ c1, which contra- dicts (3).

The point is that we used everywhere that the rings in which we worked are factorial! This property does not persist (see, e.g. Commutative Algebra, Sheet 5, Exercise 2) and, hence, we need to ﬁnd a substitute. 2 LINEAR ALGEBRA IN FIELD EXTENSIONS 9

2 Linear algebra in ﬁeld extensions

Let L/K be a field extension, i.e. K is a subfield of L. Recall that multiplication in L makes L into a K-vector space. We speak of a finite field extension if [L : K] := dim (L) < . Recall, moreover, K ∞ that an element a L is called algebraic over K if there is a non-zero polynomial m K[X] such ∈ a ∈ that f(a) = 0. If ma is monic (leading coefficient equal to 1) and irreducible, then ma is called the minimal polynomial of a over K. It can be characterised as the unique monic generator of the kernel of the evaluation map f(X) f(a) K[X] 7→ L, −−−−−−−→ which is trivially checked to be a K-algebra homomorphism (i.e. a homomorphism of rings and of K-vector spaces). We now assume that L/K is a finite extension of degree [L : K] = n. Later we will ask it to be separable, too (which is automatic if the characteristic of K (and hence L) is 0). Let a L. Note that ∈ multiplication by a: T : L L, x ax a → 7→ is L-linear and, thus, in particular, K-linear. Once we choose a K-basis of L, we can represent Ta by an n n-matrix with coefficients in K, also denoted T . × a Here is the most simple, non-trivial example. The complex numbers C have the R-basis 1,i and x b { } with respect to this basis, any z C is represented as ( y )= x+iy. Now, take a = ( c )= b+ci C. b c ∈ ∈ We obtain: Ta = c− b , as we can easily check:

b c x bx cy T (z)= az = (b + ci)(x + iy) = (bc cy)+ i(cx + by) and T (z)= ( y )= − . a − a c− b cx+by As an aside: You may have seen this matrix before; namely, writing z = r(cos(ϕ)+i sin(ϕ)), it looks cos(ϕ) sin(ϕ) like r sin(ϕ)− cos(ϕ) , i.e. it is a rotation matrix times a homothety (stretching) factor. We can now do linear algebra with the matrix Ta Matn(K). ∈ Definition 2.1. Let L/K be a finite field extension. Let a L. The trace of a in L/K is defined as ∈ the trace of the matrix T Mat (K) and the norm of a in L/K is defined as the determinant of the a ∈ n matrix T Mat (K): a ∈ n

TrL/K (a) := Tr(Ta) and NormL/K (a) := det(Ta).

Note that trace and norm do not depend on the choice of basis by a standard result from linear algebra.

Let L/K = C/R and z = x + iy C. Then Tr (z) = 2x = 2 (z) and Norm (z) = ∈ C/R ℜ C/R x2 + y2 = z 2. | | Lemma 2.2. Let L/K be a finite field extension. Let a L. ∈ (a) Tr defines a group homomorphism (L, +) (K, +), i.e. L/K →

TrL/K (a + b) = TrL/K (a)+TrL/K (b). 2 LINEAR ALGEBRA IN FIELD EXTENSIONS 10

(b) Norm deﬁnes a group homomorphism (L , ) (K , ), i.e. L/K × · → × · Norm (a b) = Norm (a) Norm (b). L/K · L/K · L/K

Proof. (a) The trace of a matrix is additive and Ta+b = Ta + Tb because Ta+b(x) = (a + b)x = ax + bx = T (x)+ T (x) for all x L. a b ∈ (b) The determinant of a matrix is multiplicative and Ta b = Ta Tb because Ta b(x) = abx = · ◦ · T T (x) for all x L. a b ∈ Lemma 2.3. Let L/K be a ﬁnite ﬁeld extension. Let a L. ∈ n n 1 (a) Let fa = X + bn 1X − + + b1X + b0 K[X] be the characteristic polynomial of Ta − ··· ∈ n ∈ Matn(K). Then TrL/K (a)= bn 1 and NormL/K (a) = ( 1) b0. − − − d d 1 (b) Let ma = X + cd 1X − + + c1X + c0 K[X] be the minimal polynomial of a over K. − ··· ∈ e Then d =[K(a): K] and with e =[L : K(a)] one has ma(X) = fa(X). Proof. (a) is a general fact from linear algebra that can, for example, be checked on the Jordan normal

form of Ta over an algebraic closure of K, using the fact that trace and determinant are conjugation invariants, that is, do not depend on the choice of basis. f(X) f(a) (b) It is obvious that the evaluation map K[X] 7→ L deﬁnes a ﬁeld isomorphism −−−−−−−→ K[X]/(ma(X)) ∼= K(a), whence the degree of [K(a) : K] equals the degree of m (X) and, moreover, 1,a,a2,...,ad 1 a { − } forms a K-basis of K(a). x ax We now compute the matrix T for the map K(a) 7→ K(a) with respect to the chosen K- a′ −−−→ basis. Very simple checking shows that it is the following matrix:

0 0 0 c0 1 0 ··· 0 −c1 0 1 ··· 0 −c2 Ta′ =  . . .··· −.  ......  0 0 1 cd−1   ··· −  Note that its characteristic polynomial is precisely ma(X). Now let s ,...,s be a K(a)-basis of L. Then a K-basis of L is given by { 1 e} 2 d 1 2 d 1 2 d 1 s , s a, s a ,...,s a − , s , s a, s a ,...,s a − , ... s , s a, s a ,...,s a − . { 1 1 1 1 2 2 2 2 e e e e } K-linear independence is immediately checked and the number of basis elements is OK; this is the way one proves that the ﬁeld degree is multiplicative in towers: [L : K]=[L : K(a)][K(a): K]. With respect to this basis, the matrix Ta is a block matrix consisting of e blocks on the diagonal, each of them equal to Ta′ . This proves (b). We need to use some results from ﬁeld theory. They are gathered in the appendix to this section.

Proposition 2.4. Let L/K be a ﬁnite separable ﬁeld extension, K an algebraic closure of K containing L. Let, furthermore, a L and f the characteristic polynomial of T . Then the following ∈ a a statements hold: 2 LINEAR ALGEBRA IN FIELD EXTENSIONS 11

(a) fa(X)= σ Hom (L,K)(X σ(a)), ∈ K − Q (b) TrL/K (a)= σ Hom (L,K) σ(a), and ∈ K P (c) NormL/K (a)= σ Hom (L,K) σ(a). ∈ K Proof. Let M = K(aQ). We use Equation (2.4) and its notation. By Proposition 2.11 in the appendix, the minimal polynomial of a over K is

m (X) := (X σ (a)). a − i i I Y∈ Let e = #J. We obtain from Lemma 2.3:

f (X)= m (X)e = (X σ (a))e = (X σ (a))e a a − i − i i I i I Y∈ Y∈ = (X σ τ (a)) = (X σ(a)). − i ◦ j − i I j J σ Hom (L,K) Y∈ Y∈ ∈ YK This shows (a). Multiplying out, (b) and (c) are an immediate consequence of the preceding lemma.

Corollary 2.5. Let L/M/K be ﬁnite separable ﬁeld extensions. Then

Tr = Tr Tr and Norm = Norm Norm . L/K M/K ◦ L/M L/K M/K ◦ L/M Proof. We use Equation (2.4) from the appendix and its notation. Then

TrL/K TrM/L(a) = σi TrM/L(a) = σi τj(a) i I i I j J X∈ X∈ X∈ = σ τ (a) = σ τ (a) = Tr (a). i j i ◦ j M/K i I j J i I j J X∈ X∈ X∈ X∈ In the same way, we have

NormL/K NormM/L(a) = σi NormM/L(a) = σi τj(a) i I i I j J Y∈ Y∈ Y∈ = σ τ (a) = σ τ (a) = Norm (a), i j i ◦ j M/K i I j J i I j J Y∈ Y∈ Y∈ Y∈ showing the statement for the norm.

Definition 2.6. Let L/K be a finite separable field extension of degree n = [L : K]. Further, let Hom (L, K) = σ ,...,σ and let α ,...,α L be a K-basis of L. Form the matrix K { 1 n} 1 n ∈ D(α1,...,αn):=(σi(αj))1 i,j n. ≤ ≤ The discriminant of (α1,...,αn) is defined as 2 disc(α1,...,αn) := det D(α1,...,αn) . The trace pairing on L/K is the bilinear pairing

L L K, (x, y) Tr (xy). × → 7→ L/K 2 LINEAR ALGEBRA IN FIELD EXTENSIONS 12

Example 2.7. (a) Let 0, 1 = d Z be a squarefree integer and consider K = Q(√d). Computations 6 ∈ (see Exercise on Sheet 3) show:

1+ √d disc(1, √d) = 4d and disc(1, )= d. 2

(b) Let f(X) = X3 + aX + b Z[X] be an irreducible polynomial and consider K = Q[X]/(f). ∈ Let α C be any root of f, so that we can identify K = Q(α) and 1,α,α2 is a Q-basis of K. ∈ Computations also show disc(1,α,α2)= 4a3 27b2. − − (One can make a brute force computation yielding this result. However, it is easier to identify this discriminant with the discriminant of the polynomial f(X), which is deﬁned by the resultant of f and its formal derivative f ′. This, however, was not treated in last term’s lecture and we do not have time for it here either.)

Proposition 2.8. Let L/K be a ﬁnite separable ﬁeld extension of degree n = [L : K]. Then the following statements hold:

tr (a) Let D := D(α1,...,αn). Then D D is the Gram matrix of the trace pairing with respect to any tr K-basis α1,...,αn. That is, D D = TrL/K (αiαj) 1 i,j n. ≤ ≤ (b) Let α1,...,αn be a K-basis of L. Then

2 tr disc(α1,...,αn) = det(D) = det(D D) = det TrL/K (αiαj) 1 i,j n. ≤ ≤ (c) Let α1,...,αn be a K-basis of L and C = (ci,j)1 i,j n be an n n-matrix with coefﬁcients ≤ ≤ × in K and put βi := Cαi for i = 1,...,n. Then

2 disc(β1,...,βn) = det(C) disc(α1,...,αn).

(d) If L = K(a), then

n 1 2 disc(1,a,...,a − )= σ (a) σ (a) , j − i 1 i

(e) The discriminant disc(α1,...,αn) is non-zero and the trace pairing on L/K is non-degenerate.

Proof. (a) Let σ ,...,σ be the K-homomorphisms L K. Then we have 1 n → n n tr D D = σk(αi)σk(αj) 1 i,j n = σk(αiαj) 1 i,j n = TrL/K (αiαj) 1 i,j n. k=1 ≤ ≤ k=1 ≤ ≤ ≤ ≤ X X tr tr So, the (i, j)-entry of the matrix D D equals Tr(αiαj). Hence, D D is the Gram matrix of the trace pairing with respect to the chosen K-basis. (b) is clear. (c) Exercise on Sheet 3. 2 LINEAR ALGEBRA IN FIELD EXTENSIONS 13

(d) Exercise on Sheet 3. (e) We may always choose some a L such that L = K(a) (this is shown in any standard course ∈ n 1 on Galois theory). From (c) it is obvious that the discriminant disc(1,a,...,a − ) is non-zero and, hence, the trace pairing on L/K is non-degenerate (because by a standard result from linear algebra a bilinear pairing is non-degenerate if and only if its Gram matrix with respect to any basis is invertible). Consequently, disc(α ,...,α ) = 0. 1 n 6 Appendix: Some Galois theory

Let L/K be an algebraic extension of fields (not necessarily finite for the next definition) and K an algebraic closure of K containing L. We pre-suppose here the existence of an algebraic closure, which is not quite easy to prove. However, in the number field case we have C, which we know to be algebraically closed, and in C we can take Q = z C z algebraic over Q , which is an algebraic { ∈ | } closure of Q and also of all number fields. Let f K[X] be a polynomial of degree n. It is called separable if it has n distinct roots in K. It ∈ is very easy to see that f is separable 1 = gcd(f ′, f), ⇔ where f ′ is the formal derivative of f. Otherwise, we say that f is inseparable. If char(K) = 0, then every irreducible polynomial f is separable because gcd(f ′, f) = 1, as the only monic divisor of f of degree < n is 1 and deg(f ) = n 1. Moreover, if K is a finite field ′ − of characteristic p, then every irreducible polynomial f K[X] is also separable. The reason is that ∈ pn the finite field L := K[X]/(f(X)) is a splitting field of the polynomial X X Fp[X], where n − ∈ #L = pn. This implies that f(X) divides Xp X. As the latter polynomial is separable (because n n n − gcd((Xp X) , Xp X) = gcd( 1, Xp X) = 1), also f is separable. A field over which every − ′ − − − irreducible polynomial is separable is called perfect. We have just seen that fields of characteristic 0 and finite fields are perfect. However, not every field is perfect. Consider K = Fp(T ) = Frac(Fp[T ]) and f(X)= Xp T K[X]. The Eisenstein criterion shows that f is irreducible, but, gcd(f , f)= − ∈ ′ gcd(pXp 1, Xp T ) = gcd(0, Xp T )= Xp T = 1, whence f is not separable. In this lecture, − − − − 6 we shall almost entirely be working with number fields, and hence in characteristic 0, so that the phenomenon of inseparability will not occur. Next we explain how irreducible separable polynomials are related to properties of field extensions. We let Hom (L, K) be the set of field homomorphisms (automatically injective!) τ : L K K → such that τ = id , i.e. τ(x) = x for all x K. Such a homomorphism is referred to as a K- |K K ∈ homomorphism. We write [L : K]sep := #HomK (L, K) and call it the separable degree of L/K, for reasons to become clear in a moment. Let now f K[X] be an irreducible polynomial and suppose L = K[X]/(f). We have the ∈ bijection α K f(α) = 0 Hom (L, K), { ∈ | } −→ K given by sending α to the K-homomorphism

σ : K[X]/(f) K, g(X) + (f) g(α). α → 7→ 2 LINEAR ALGEBRA IN FIELD EXTENSIONS 14

Note that it is well-deﬁned because f(α) = 0. The injectivity of the map is clear: α = σα(X +(f)) = σβ(X + (f)) = β. For the surjectivity consider any σ : K[X]/(f) K and put γ = σ(X). As →r σ(f) = 0, we have f(γ) = 0 and it follows that σ = σγ because the X + (f) form a K-generating system of K[X]/(f) on which σ and σγ agree. We have shown for L = K[X]/(f):

[L : K] = # α K f(α) = 0 deg(f)=[L : K]. sep { ∈ | }≤ Now we consider a general algebraic ﬁeld extension L/K again. An element a L is called ∈ separable over K if its minimal polynomial f K[X] is separable. The algebraic ﬁeld extension a ∈ L/K is called separable if every element a L is separable over K. As an immediate consequence ∈ every subextension of a separable extension is separable. The most important technical tool in Galois theory is the following proposition.

Proposition 2.9. Let L/K be an algebraic ﬁeld extension and K an algebraic closure of K containing L. Then any K-homomorphism σ : L K can be extended to a K-homomorphism σ : K K. → → In order to explain the idea behind this proposition, let us take M = L(a) for some a K, ∈ whence M = L[X]/(f) with f the minimal polynomial of a over L, and let us extend σ to M, call it σM . The polynomial f factors into linear factors over K, whence we may choose some α K d i ∈ such that f(α) = 0. Any element of M is of the form i=0 aiX + (f) and we send it via σM to d σ(a )αi in K. Using a Zorn’s lemma argument, one obtains that σ can indeed be extended i=0 i P to K. P Let now L/M/K be algebraic ﬁeld extensions contained inside K and let

Hom (M, K)= σ i I and Hom (L, K)= τ j J . K { i | ∈ } M { j | ∈ } By Proposition 2.9 we may choose σ : K K extending σ for i I. We have i → i ∈ Hom (L, K)= σ τ i I, j J . (2.4) K { i ◦ j | ∈ ∈ }

This is easy to see: ‘ ’ is clear. ‘ ’: Let τ HomK (L, K), then τ M HomK (M, K), whence ⊇ ⊆ 1∈ | ∈ τ = σ for some i I. Now consider σ− τ Hom (L, K), whence there is j J such that |M i ∈ i ◦ ∈ M ∈ τ = σ τ . i ◦ j Moreover, the map I J Hom (L, K), (i, j) σ τ × → K 7→ i ◦ j is a bijection. The surjectivity is precisely the inclusion ‘ ’ shown above. For the injectivity suppose ⊆ σi τj = σk τℓ. Restrict this equality to M and get σi = σk, whence i = k. Having this, multiply ◦ ◦ 1 from the left by σi− and obtain τj = τℓ, whence j = ℓ. As consequence we ﬁnd the multiplicativity of the separable degree in towers of algebraic ﬁeld extensions:

[L : K]sep =[L : K]sep[M : K]sep.

This multiplicivity combined with our calculations for L = K[X]/(f) immediately give for a finite extension L/K: L/K is separable [L : K]=[L : K] , ⇔ sep 3 RINGS OF INTEGERS 15 and the inequality [L : K] [L : K] always holds. ≥ sep One more definition: the set Ksep := x K x is separable over K is called the separable { ∈ | } closure of K in K. It can be seen as the compositum of all finite separable subextensions L/K inside K, whence it clearly is a field.

Proposition 2.10. Let a Ksep such that σ(a)= a for all σ Hom (K, K), then a K. ∈ ∈ K ∈ Proof. If a were not in K, then we let f K[X] be its minimal polynomial and we let α K be a ∈ ∈ root of f. Then we have σ : K(a) K (deﬁned as above) a non-trivial K-homomorphism, which a → we may extend to a non-trivial σ : K K, contradiction. a → This allows us to write down the minimal polynomial of a separable element x Ksep as follows. ∈ Proposition 2.11. Let a Ksep and consider the set ∈ σ ,σ ,...,σ = Hom (K(a), K) { 1 2 n} K

with n =[K(a): K]=[K(a): K]sep. Then the minimal polynomial of a over K is

n f (X) := (X σ (a)). a − i Yi=1 Proof. We extend σ to σ : K K and observe σ(f ) = f (where σ is applied to the coefﬁcients i i → a a of f ) for all K-homomorphisms σ : K K, whence f K[X]. Here we have used that every σ a → a ∈ restricted to K(a) is one of the σi, and, hence, application of σ just permutes the σi in the product. Proposition 2.10 now implies that the coefﬁcients of fa are indeed in K. It remains to see that the polynomial is irreducible. But that is clear for degree reasons. Of course,

a is a zero of fa (one of the σi is the identity on a), fa is monic and its degree is that of [K(a): K].

3 Rings of integers

We recall central deﬁnitions and propositions from last term’s course on commutative algebra.

Deﬁnition 3.1. Let R be a ring and S an extension ring of R (i.e. a ring containing R as a subring). An element a S is called integral over R if there exists a monic polynomial f R[X] such that ∈ ∈ f(a) = 0.

Note that integrality is also a relative notion; an element is integral over some ring. Also note the similarity with algebraic elements; we just added the requirement that the polynomial be monic.

Example 3.2. (a) The elements of Q that are integral over Z are precisely the integers of Z.

(b) √2 R is integral over Z because X2 2 annihilates it. ∈ − (c) 1+√5 R is integral over Z because X2 X 1 annihilates it. 2 ∈ − − 3 RINGS OF INTEGERS 16

1+√ 5 2 3 (d) a := − R is not integral over Z because f = X X + annihilates it. If there were 2 ∈ − 2 a monic polynomial h Z[X] annihilating a, then we would have h = fg with some monic ∈ polynomial g Q[X]. But, now it would follow that both f and g are in Z[X] (see Sheet 4 of last ∈ term’s lecture on Commutative Algebra), which is a contradiction.

(e) Let K be a field and S a ring containing K (e.g. L = S a field) and a L. Then a is integral ∈ over K if and only if a is algebraic over K. Indeed, as K is a field any polynomial with coefficients in K can be made monic by dividing by the leading coefficient. So, if we work over a field, then the new notion of integrality is just the notion of algebraicity from the previous section.

Deﬁnition 3.3. Let S be a ring and R S a subring. ⊆ (a) The set R = a S a is integral over R is called the integral closure of R in S (compare S { ∈ | } with the algebraic closure of R in S – the two notions coincide if R is a ﬁeld). An alternative name is: normalisation of R in S.

(b) S is called an integral ring extension of R if RS = S, i.e. if every element of S is integral over R (compare with algebraic ﬁeld extension – the two notions coincide if R and S are ﬁelds).

(d) An integral domain R is called integrally closed (i.e. without mentioning the ring in which the closure is taken) if R is integrally closed in its fraction ﬁeld.

(e) Let a S for i I (some indexing set). We let R[a i I] (note the square brackets!) be the i ∈ ∈ i | ∈ smallest subring of S containing R and all the a , i I. i ∈ Note that we can see R[a] inside S as the image of the ring homomorphism

d d Φ : R[X] S, c Xi c ai. a → i 7→ i Xi=0 Xi=0 Proposition 3.4. Let R S T be rings. ⊆ ⊆ (a) For a S, the following statements are equivalent: ∈ (i) a is integral over R. (ii) R[a] S is a ﬁnitely generated R-module. ⊆ (b) Let a ,...,a S be elements that are integral over R. Then R[a ,...,a ] S is integral 1 n ∈ 1 n ⊆ over R and it is ﬁnitely generated as an R-module.

(c) Let R S T be rings. Then ‘transitivity of integrality’ holds: ⊆ ⊆ T/R is integral T/S is integral and S/R is integral. ⇔

(d) RS is a subring of S. 3 RINGS OF INTEGERS 17

(e) Any t S that is integral over R lies in R . In other words, R is integrally closed in S ∈ S S S (justifying the name).

Definition 3.5. Recall that a number field K is a finite field extension of Q. The ring of integers of K is the integral closure of Z in K, i.e. Z . An alternative notation is . K OK Example 3.6. Let d = 0, 1 be a squarefree integer. The ring of integers of Q(√d) is 6 (1) Z[√d], if d 2, 3 (mod 4), ≡ (2) Z[ 1+√d ], if d 1 (mod 4). 2 ≡ Proposition 3.7. Every factorial ring (unique factoriation domain) is integrally closed.

Proposition 3.8. Let R be an integral domain, K = Frac(R), L/K a ﬁnite ﬁeld extension and S := RL the integral closure of R in L. Then the following statements hold:

(a) Every a L can be written as a = s with s S and 0 = r R. ∈ r ∈ 6 ∈ (b) L = Frac(S) and S is integrally closed.

(c) If R is integrally closed, then S K = R. ∩ The following proposition was stated but not proved in last term’s lecture.

Proposition 3.9. Let R be an integral domain which is integrally closed (recall: that means integrally closed in K = Frac(R)). Let K be an algebraic closure of K and let a K be separable over K. ∈ Then the following statements are equivalent:

(i) a is integral over R.

(ii) The minimal polynomial m K[X] of a over K has coefficients in R. a ∈ Proof. ‘(ii) (i)’: Since by assumption m R[X] is a monic polynomial annihilating a, by defini- ⇒ a ∈ tion a is integral over R. ‘(i) (ii)’: From Proposition 2.11 we know that the minimal polynomial of a over K is ⇒ n m (X)= (X σ (a)), a − i Yi=1 where σ = id,σ ,...,σ = Hom (K(a), K). { 1 2 n} K We assume that a is integral over R, so there is some monic polynomial g R[X] annihilating a. a ∈ It follows that ma divides ga. Consequently, ga(σi(a)) = σi(ga(a)) = σi(0) = 0 for all i = 1,...,n, proving that also σ2(a),σ3(a),...,σn(a) are integral over R. Hence, ma has integral coefficients over R (they are products and sums of the σi(a)). As R is integrally closed in K, the coefficients lie in R.

We now apply norm and trace to integral elements. 3 RINGS OF INTEGERS 18

Lemma 3.10. Let R be an integrally closed integral domain, K its field of fractions, L/K a separable finite field extension and S the integral closure of R in L. Let s S. Then the following statements ∈ hold:

(a) Tr (s) R and Norm (s) R. L/K ∈ L/K ∈ (b) s S Norm (s) R . ∈ × ⇔ L/K ∈ × Proof. (a) directly follows from S K = R. ∩ (b) ‘ ’: Let s, t S such that ts = 1. Then 1 = Norm (1) = Norm (st) = ⇒ ∈ × L/K L/K NormL/K (s)NormL/K (t), exhibiting an inverse of NormL/K (s) in R.

‘ ’: Assume NormL/K (s) R×. Then 1 = rNormL/K (s) = r σ Hom (L,K) σ(s) = ⇐ ∈ ∈ K r id=σ Hom (L,K) σ(s) s = ts, exhibiting an inverse to s in S. 6 ∈ K Q QNext we use the discriminant to show the existence of an integral basis. The discriminant will also be important in the proof of the Noetherian-ness of the ring of integers of a number ﬁeld.

Lemma 3.11. Let R be an integrally closed integral domain, K its field of fractions, L/K a separable finite field extension and S the integral closure of R in L.

(a) For any K-basis α ,...,α of L, there is an element r R 0 such that rα S for all 1 n ∈ \ { } i ∈ i = 1,...,s.

(b) Let α ,...,α S be a K-basis of L and let d = disc(α ,...,α ) be the discriminant of this 1 n ∈ 1 n basis. Then dS Rα + + Rα . ⊆ 1 ··· n si Proof. (a) By Proposition 3.8 (a), we can write αi = with ri R and si S for all i = 1,...,n. ri ∈ ∈ Hence, we may take r = r ... r . 1 · · n (b) Let s = n x α be an element of S with x K for j = 1,...,n. We show ds j=1 j j j ∈ ∈ Rα + + Rα . Note that the elementary properties of the norm yield 1 ··· nP n Tr (α s)= Tr(α α )x S K = R. L/K i i j j ∈ ∩ Xj=1

Tr (α1α1) Tr (α1αn) L/K ··· L/K We can rewrite this in matrix form using M = DtrD = . .. . . Now:  . . .  Tr (αnα1) Tr (αnαn) L/K ··· L/K   n x1 Pj=1 Tr(α1αj )xj . . n M . = . R . xn  n  ∈ Pj=1 Tr(αnαj )xj   Multiplying through with the adjoint matrix M # yields

x1 x1 x1 M #M . = det(M) . = d . Rn. . . . ∈ xn xn xn Thus, dx R for all i = 1,...,n and, consequently, ds Rα + + Rα . i ∈ ∈ 1 ··· n 3 RINGS OF INTEGERS 19

We now need a statement that is very simple and could have been proved in last term’s course on commutative algebra (but, it wasn’t). We give a quick proof.

Theorem 3.12. Let R be a principal ideal domain and M a ﬁnitely generated R-module. Then the following statements hold:

(a) Assume that M is a free R-module of rank m. Then any submodule N of M is ﬁnitely generated and free of rank m. ≤ (b) An element m M is called torsion element if there is 0 = r R such that rm = 0. The set ∈ 6 ∈ M = m M m is a torsion element is an R-submodule of M, the order of which is torsion { ∈ | } ﬁnite.

M = M R ... R . ∼ torsion ⊕ ⊕ ⊕ m times The integer m is called the R-rank of M. | {z }

(e) Let 0 N M Q 0 be a short exact sequence of ﬁnitely generated R-modules. Then → → → → rkR(M)=rkR(N)+rkR(Q). Proof. (a) We give a proof by induction on m. The case m = 0 is clear (the only submodule of the zero-module is the zero-module). Now let m = 1. Then M ∼= R and the submodules of M are the ideals of R under the isomorphism. As R is a principal ideal domain, the rank of the submodules of M is thus equal to 1, unless it is the zero-ideal. Now, suppose we already know the statement for all ranks up to m 1 and we want to prove it for − M of rank m. After an isomorphism, we may suppose M = R ... R. Let π : M = R ... R ⊕ ⊕ ⊕ ⊕ m times be the m-th projection. It sits in the (trivial) exact sequence | {z } 0 R ... R M π R 0. → ⊕ ⊕ → −→ → m 1 times − Let now N M be a submodule and| set {z } ≤ N := N ker π = N R ... R . 1 ∩ ∩ ⊕ ⊕ m 1 times − By induction assumption, N is a free R-module of rank| at most{z n }1. Moreover, π(N) is a submod- 1 − ule of R, hence, by the case m = 1, it is free of rank 0 or 1. We have the exact sequence:

0 N N π π(N) 0. → 1 → −→ → As π(N) is free, it is projective and this sequence splits, yielding

N = N π(N), ∼ 1 ⊕ 3 RINGS OF INTEGERS 20 showing that N is free of rank at most (m 1)+1= m. − (b) is trivial. (c) ‘ ’: Let x ,...,x be a free system of generators of M. Let x = n r x M. If rx = 0 ⇒ 1 n i=1 i i ∈ with R r = 0, then rr = 0 for all i, thus r = 0 for all i, whence x = 0. ∋ 6 i i P ‘ ’: Let x ,...,x be any system of generators of M and let x ,...,x with m n be a ⇐ 1 n 1 m ≤ maximal free subset (possibly after renumbering). If m = n, then M is free, which we want to show. Assume, hence, that m

0 M M M/M 0, → torsion → → torsion → and claim that M/Mtorsion is a free R-module. By (c) it sufﬁces to show that the only torsion element in M/M is 0, which works like this: Let x + M M/M and 0 = r R such that torsion torsion ∈ torsion 6 ∈ r(x+M )= rx+M = 0+M M/M . Then, clearly, rx M , whence torsion torsion torsion ∈ torsion ∈ torsion there is 0 = s R such that s(rx) = (sr)x = 0, yielding x M , as desired. 6 ∈ ∈ torsion As M/Mtorsion is R-free, it is projective and, hence, the above exact sequence splits (see Com- mutative Algebra), yielding the desired assertion. (e) First assume that Q is R-free of rank q. Then the exact sequence splits and one gets M ∼= N Q, making the assertion obvious. If Q = Rq Q , then consider the composite map ⊕ ⊕ torsion π : M ։ Rq Q ։ Rq. We get rk (M) = q + rk (N) with N = ker(π). From the snake ⊕ torsion R R lemma (see exercise) it is obvious that N/N ∼= Qtorsion. From this we want to conclude that rk(N) = rk(N), thene we aree done. We may assume that Ntorsion = 0 (since the torsion part playse no role for the rank), and, hence, Ntorsion = 0, so that N and N are free R-modules. Assume that the rank of N is strictlye smaller than the rank of N. We claim thate there is then some x N which is R-linearly independent of the image of N. For, if no such x ∈ existed,e then there would be 0 = r R such that rN N, hence, rk(N) = rk(rN) rk(e N), which 6 ∈ ⊆ ≤ is impossible. Now, by assumptione there is 0 = r R such that r(x + N)=0+ N, i.e. rx N, 6 ∈ ∈ contradicting the linear independence. e e e

Deﬁnition 3.13. Let R S be an integral ring extension. If S is free as an R-module, then an ⊆ R-basis of S (i.e. a free generating system) exists and is called an integral basis of S over R. 3 RINGS OF INTEGERS 21

We point out that, if S is an integral domain (as it always will be in this lecture), then an R-basis of S is also a K-basis of L = Frac(S) with K = Frac(R). Note that, in general, there is no reason why an integral ring extension S should be free as an R-module. This is, however, the case for the rings of integers, as the following proposition shows.

Proposition 3.14. Let R be a principal ideal domain, K its field of fractions, L/K a finite separable field extension and S the integral closure of R in L. Then every finitely generated S-submodule 0 = M of L is a free R-module of rank [L : K]. In 6 particular, S possesses an integral basis over R.

Proof. As principal ideal domains are unique factorisation domains and, hence, integrally closed, we may apply Lemma 3.11 to obtain a K-basis α ,...,α S of L and we also have dS Rα + + 1 n ∈ ⊆ 1 ··· Rαn =: N with d = disc(α1,...,αn). Note that N is a free R-module of rank n =[L : K]. Let m ,...,m M be a generating system of M L as S-module. As the m are elements 1 k ∈ ⊆ i of L, by Proposition 3.8 (a) there is r R such that rm S for all i = 1, . . . , k, whence rM S. ∈ i ∈ ⊆ Hence, rdM dS N. Consequently, Theorem 3.12 yields that rdM is a free R-module of rank ⊆ ⊆ at most n. Of course, the R-rank of rdM is equal to the R-rank of M. Let 0 = m M. Then 6 ∈ Nm Sm M, showing that n, the R-rank of N (which is equal to the R-rank of Nm) is at most ≤ ≤ the R-rank of M, ﬁnishing the proof.

For the rest of this section we specialise to the case of number ﬁelds.

Deﬁnition 3.15. Let K be a number ﬁeld. A subring of Z is called an order of K if has an O K O integral basis of length [K : Q].

Corollary 3.16. Any order in a number ﬁeld K is a Noetherian integral domain of Krull dimension 1.

Proof. Being a subring of a ﬁeld, is an integral domain. As the ring extension Z is integral O ⊆ O (being contained in the integral extension Z Z ), the Krull dimension of equals the Krull ⊆ K O dimension of Z, which is 1 (see Commutative Algebra). As has an integral basis, we have = O O ∼ Z ... Z. That is Noetherian now follows because Z is Noetherian and ﬁnite direct sums of ⊕ ⊕ O [K:Q] times Noetherian modules are Noetherian (see Commutative Algebra). | {z }

Corollary 3.17. Let K be a number ﬁeld and ZK the ring of integers of K. Then the following statements hold:

(a) ZK is an order of K, also called the maximal order of K.

(b) ZK is a Dedekind ring.

(c) Let 0 ( I E ZK be an ideal. Then I is a free Z-module of rank [K : Q] and the quotient ZK /I is finite (i.e. has finitely many elements; equivalently, the index (ZK : I) is finite).

Proof. (a) It is a trivial consequence of Proposition 3.14 that ZK is a free Z-module of rank [K : Q] because ZK is a ZK -module generated by a single element, namely 1. In particular, ZK has an integral basis and, hence, is an order of K. 4 IDEAL ARITHMETIC 22

(b) From Corollary 3.16 we know that ZK is a Noetherian integral domain of Krull dimension 1. It is also integrally closed (being deﬁned as the integral closure of Z in K), hence, by deﬁnition, a Dedekind ring.

(c) As ZK is Noetherian, the ideal I is finitely generated. Hence, Proposition 3.14 again gives that I is a free Z-module of rank [K : Q]. The quotient of any two free Z-modules of the same rank is finite by Theorem 3.12, proving the final statement.

Definition 3.18. Let K be a number field with ring of integers Z and 0 = a K be a finitely K 6 ⊂ generated ZK -module. The discriminant of a is defined as disc(α1,...,αn) for any Z-basis of the free Z-module a (see Proposition 3.14). (By Proposition 2.8 (c), this definition does not depend on the choice of Z-basis because the basis transformation matrix is invertible with integral entries and thus has determinant 1.) ± The discriminant of K is defined as disc(ZK ).

Proposition 3.19. Let K be a number field and Z its ring of integers. Let 0 = a b K be two K 6 ⊆ ⊂ ZK -modules. Hence, the index (b : a) is finite and satisfies

disc(a) = (b : a)2 disc(b).

Proof. Exercise on Sheet 4.

4 Ideal arithmetic

It is useful, in order to make the set of non-zero ideals of a Dedekind ring into a group with respect to multiplication of ideals, to introduce fractional ideals, which will be needed for the inverses.

Deﬁnition 4.1. Let R be an integral domain and K = Frac(R).

An R-submodule I K is called a fractional ideal of R (or: fractional R-ideal) if • ≤ – I = (0) and 6 – there is x K such that xI R. ∈ × ⊆ Note that x can always be chosen in R 0 . Note also that xI is an ideal of R (in the usual \ { } sense).

A fractional R-ideal I is called an integral ideal if I R. • ⊆ Note that for a subset (0) = I K, one trivially has: 6 ⊂ I E R is an ideal of R in the usual sense I is an integral fractional R-ideal. ⇔

A fractional R-ideal I is called principal if there is x K such that I = Rx. • ∈ × Let I, J be fractional R-ideals. The ideal quotient of I by J is deﬁned as • I : J = (I : J)= x K xJ I . { ∈ | ⊆ } 4 IDEAL ARITHMETIC 23

The inverse ideal of the fractional R-ideal I is deﬁned as • 1 I− := (R : I)= x K xI R . { ∈ | ⊆ } The multiplier ring of the fractional R-ideal I is deﬁned as • r(I):=(I : I)= x K xI I . { ∈ | ⊆ } Example 4.2. The fractional ideals of Z are all of the form I = a Z with a, b Z 0 . Hence, all b ∈ \ { } fractional Z-ideals are principal. a It is clear that b Z is a fractional ideal. Conversely, let I be a fractional ideal such that bI is an a ideal of Z, whence bI = (a)= aZ, so that I = b Z. a c Let I = b Z and J = d Z, then c a ad ad (I : J)= x Q x Z Z = x Q x Z = Z. { ∈ | d ∈ b } { ∈ | ∈ bc } bc In particular, I 1 = b Z and II 1 = Z (because, clearly and 1 II 1). − a − ⊆ ∈ − Lemma 4.3. Let R be an integral domain and K = Frac(R). Let I, J K be fractional R-ideals. ⊂ Then the following sets are fractional R-ideals.

I + J = x + y x I, y J , • { | ∈ ∈ } IJ = n x y n N,x ,...,x I, y ,...,y J , • { i=1 i j | ∈ 1 n ∈ 1 n ∈ } In = I PI ... I, • · · · n times I J,| {z } • ∩ (I : J). • Proof. Exercise.

Lemma 4.4. Let R be an integral domain and H,I,J K fractional R-ideals. Then the following ⊂ properties hold:

(a) IJ I J (assume here that I and J are integral ideals), ⊆ ∩ (b) H + (I + J) = (H + I)+ J = H + I + J,

(d) H(I + J)= HI + HJ.

Proof. Exercise.

Lemma 4.5. Let R be an integral domain and I, J E R be ideals (in the usual sense). If I + J = R, then we call I and J coprime ideals. Suppose now that I and J are coprime. Then the following statements hold: 4 IDEAL ARITHMETIC 24

(a) In and J m are coprime for all n, m N. ∈ (b) I J = IJ. ∩ (c) R/(IJ) = R/I R/J (Chinese Remainder Theorem). ∼ × (d) If IJ = Hn for some n N, then I = (I + H)n, J = (J + H)n and (I + H)(J + H)= H. ∈ In words: If an ideal is an n-th power, then so is each of its coprime factors.

Proof. (a) By assumption 1= i + j for some i I and some j J. Now 1 = 1n+m = (i + j)n+m ∈ ∈ ∈ In + J m. (b) The inclusion ‘ ’ is clear. We now show ‘ ’. Let x I J. Again by assumption 1= i + j ⊇ ⊆ ∈ ∩ for some i I and some j J. Hence, x = x 1 = xi + xj, whence x IJ because xi IJ and ∈ ∈ · ∈ ∈ xj IJ. ∈ (c) That’s just a reminder. It was proved in some of your Algebra lectures. (d) We start with the following computation:

n n n 1 n 2 2 n 1 n (I + H) = I + I − H + I − H + + IH − + H ··· n 1 n 2 n 1 = I(I − + I − H + + H − + J) ··· = IR = I

n n 1 because H = IJ and J and I − are coprime by the Lemma. Deﬁne A = I + H and B = J + H. Then

AB = (I + H)(J + H)= IJ + IH + JH + H2 = Hn + IH + JH + H2 n 1 = H(H − + I + J + H)= HR = H, as required.

Example 4.6. Let us consider the ring R = Z[√ 19]. In this ring, we have the following factorisa- − tions: 182 + 19 = (18 + √ 19)(18 √ 19) = 343 = 73. − − − Let us take the principal ideals I = (18 + √ 19) and J = (18 √ 19), then − − − IJ = (7)3.

The previous lemma now gives:

I = (I + (7))3 = (18 + √ 19, 7)3 and J = (J + (7))3 = (18 √ 19, 7)3. − − − But, one can check, by hand, that the elements 18 + √ 19 and 18 √ 19 are not third powers in R − − − (just take (a + b√ 19)3 = 18 √ 19 and work out that no such a, b Z exist). − − − ∈ In this example we see that ideals behave better than elements. We will extend the phenomenon that we just saw to the unique factorisation of any ideal in a Dedekind ring into a product of prime ideals. 4 IDEAL ARITHMETIC 25

Proposition 4.7. Let R be a Noetherian integral domain, K = Frac(R) and (0) = I K a subset. 6 ⊂ Then the following two statements are equivalent:

(i) I is a fractional R-ideal.

(ii) I is a ﬁnitely generated R-submodule of K (this is the deﬁnition in Neukirch’s book).

Proof. ‘(i) (ii)’: By definition, there is r R 0 such that rI R, hence, rI is an ideal of R ⇒ ∈ \ { } ⊆ in the usual sense. As R is Noetherian, rI is finitely generated, say by a1,...,an. Then I is finitely a1 an generated as R-submodule of K by r ,..., r . a1 an ‘(ii) (i)’: Suppose I is generated as R-submodule of K by ,..., . Then r = r1 ... rn is ⇒ r1 rn · · such that rI R. ⊆ This proposition also shows us how we must think about fractional R-ideals, namely, just as R-linear combinations of a given set of fractions a1 ,..., an (where we may choose a common de- rn rn nominator).

Deﬁnition 4.8. Let R be an integral domain and K = Frac(R). A fractional R-ideal I is called an invertible R-ideal if there is a fractional R-ideal J such that IJ = R.

Note that the term ‘invertible R-ideal’ applies only to fractional R-ideals (which may, of course, be integral).

Lemma 4.9. Let R be an integral domain, K = Frac(R) and I a fractional R-ideal. Then the following statements hold:

(a) II 1 R. − ⊆ (b) I is invertible II 1 = R. ⇔ − 1 (c) Let J be an invertible R-ideal. Then (I : J)= IJ − .

(d) If 0 = i I such that i 1 I 1, then I = (i). 6 ∈ − ∈ − Proof. (a) holds by definition. (b) ‘ ’: Let J be a fractional R-ideal such that IJ = R (exists by definition of I being invertible). ⇒ 1 1 1 1 Then, on the one hand, by the definition of I− we have J I− . On the other hand, I− = I− IJ 1 ⊆ ⊆ RJ = J, showing J = I− . ‘ ’: is trivial. ⇐ (c) We show both inclusions of x K xJ I = IJ 1. { ∈ | ⊆ } − ‘ ’: Let x K such that xJ I. This implies x xR = xJJ 1 IJ 1. ⊆ ∈ ⊆ ∈ − ⊆ − ‘ ’: We have (IJ 1)J = I(JJ 1)= I I, whence IJ 1 (I : J). ⊇ − − ⊆ − ⊆ (d) We have I = i(i 1I) iI 1I iR = (i) I. − ⊆ − ⊆ ⊆ We include the next lemma to avoid writing down the Noetherian hypothesis in the next corollary and the subsequent definition.

Lemma 4.10. Let R be an integral domain with K = Frac(R). Then any invertible R-ideal is ﬁnitely generated. 5 IDEALS IN DEDEKIND RINGS 26

Proof. Let IJ = R. In particular, 1 is in IJ, whence there are i I and j J for k = 1,...,n k ∈ k ∈ (some n N) such that 1= n i j . Let x I. Then ∈ k=1 k k ∈ P n n x = x 1= (xj )i Ri , · k k ∈ k Xk=1 Xk=1 n hence, I = k=1 Rik. Corollary 4.11.P Let R be an integral domain. The set (R) of invertible fractional R-ideals forms I an abelian group with respect to multiplication of ideals, with R being the neutral element, and the inverse of I (R) being I 1. ∈ I − The set (R) := xR x K of principal fractional R-ideals forms a subgroup of (R). P { | ∈ ×} I Proof. This just summarises what we have seen. That (R) is a subgroup is clear. P Deﬁnition 4.12. Let R be an integral domain. One calls (R) the group of invertible R-ideals and I (R) the subgroup of principal invertible R-ideals. P The quotient group Pic(R) := (R)/ (R) is called the Picard group of R. I P If K is a number ﬁeld and ZK its ring of integers, one also writes CL(K) := Pic(ZK ), and calls it the ideal class group of K.

Corollary 4.13. Let R be an integral domain and K = Frac(R). Then we have the exact sequence of abelian groups f proj 1 R× K× (R) Pic(R) 1, → → −→I −−→ → where f(x) is the principal fractional R-ideal xR.

Proof. The exactness is trivially checked. Note, in particular, that xR = R (the neutral element in the group) if and only if x R . ∈ × Corollary 4.14. Let R be a principal ideal domain. Then Pic(R) = R (the group with one ele- { } ment).

Proof. This is the case by deﬁnition: that every ideal is principal implies that every fractional ideal is principal, i.e. (R)= (R), whence their quotient is the group with one element. I P Example 4.15. The groups CL(Q) = Pic(Z) and Pic(K[X]) (for K a ﬁeld) are trivial.

5 Ideals in Dedekind rings

We will now giving a ‘local characterisation’ of invertible ideals. Recall that, if R is a ring and p 1 is a prime ideal, we deﬁned the localisation of R at p as Rp := S− R, where the multiplicatively closed subset S R is given as S = R p (the multiplicative closedness being precisely the property ⊆ \ 1 of p being a prime ideal). For any R-module, we deﬁned its localisation at p as Mp = S− M. 1 Consequently, if I is a fractional R-ideal, then Ip K (note that S K = K and thus the embedding ⊆ − = I ֒ K gives rise to an embedding Ip ֒ K). If I E R is an ideal in the usual sense, then Ip →1 1 → S I S R = Rp K. See the lecture on Commutative Algebra for more details on localisation. − ⊆ − ⊆ 5 IDEALS IN DEDEKIND RINGS 27

r i Very concretely, we have Rp = s K r R, s S and Ip = s K i I, s S . 1 1 { ∈ | ∈ ∈ } { ∈ | ∈ ∈ } Moreover, we have (Ip)− = (I− )p. We ﬁrst prove that the invertibility of an ideal is a local property.

Theorem 5.1. Let R be an integral domain and I a fractional R-ideal. Then the following statements are equivalent:

(i) I is invertible.

(ii) I is ﬁnitely generated as R-submodule of K (this assumption is unnecessary if R is Noethe- • rian by Proposition 4.7) and

Im is a principal fractional Rm-ideal for all maximal ideals m ¡ R. • Proof. ‘ ’: Let I be invertible. Then Lemma 4.10 implies that I is finitely generated. Since II 1 = ⇒ − R, there are i I and j I 1 for k = 1,...,n and for some n N such that 1 = n i j . k ∈ k ∈ − ∈ k=1 k k Let m be any maximal ideal. There is some index k such that i j m, as otherwise 1 m. Hence, k k P 1 jk 1 6∈ ∈ ikjk =: s R m, so that ik− = s Im− . Lemma 4.9 (d) implies Im = iRm. ∈ \ ∈ 1 ‘ ’: Let us assume the contrary, i.e. II− ( R. Then there is a maximal ideal m ¡ R such that 1⇐ II m. By assumption we have Im = xRm with some x I (after clearing denominators). The − ⊆ ∈ finite generation of I implies I = (i ,...,i ) for some n N. For each k = 1,...,n we find r R 1 n ∈ k ∈ and we find s R m such that ∈ \ r i = x k (same denominator without loss of generality). k s Hence, we have R r = si x 1 for all k = 1,...,n. Thus, we have sx 1I R, whence ∋ k k − − ⊆ sx 1 I 1. We conclude s xI 1 II 1 m, which is a contradiction because s is not − ∈ − ∈ − ⊆ − ⊆ in m.

The property (ii) is called: ‘I is locally free of rank 1’. In Algebraic Geometry one usually takes this property as the deﬁning property of invertibility: one deﬁnes invertible sheaves as those sheaves that are locally free of rank 1.

Example 5.2. We continue Example 4.6. Hence, R = Z[√ 19] and we consider the ideal I := − (18 + √ 19, 7) = (7, 3 √ 19). − − − We ﬁrst show that I is maximal. That we do as follows. Consider the ring homomorphism

X 3 α : Z[X] 7→ F . −−−→ 7 Its kernel clearly is (7, X 3). Moreover, consider the natural projection − X √ 19 π : Z[X] ։ Z[X]/(X2 + 19) ∼ 7→ − Z[√ 19]. −−−−−−−−→ − Also consider the surjection

φ : Z[√ 19] F , a + b√ 19 a + b3. − → 7 − 7→ 5 IDEALS IN DEDEKIND RINGS 28

We note that α = φ π, from which we conclude that the kernel of is the image under π of (7, X 3), ◦ − hence, is equal to (7, √ 19 3) = I as claimed. Hence, I is maximal because the quotient R/I = F − − 7 is a ﬁeld. Next, we compute the localisation of I at a maximal ideal m ¡ Z[√ 19]. − First case: m = I. Then there is x I m, so that Im = Rm because Im contains a unit of Rm. 6 ∈ \ Second case: m = I. Then we claim that Im = 7Rm. For this, we have to show that 3 √ 19 − − ∈ 7Rm. We have: 3+ √ 19 7= − (3 √ 19). 4 − − Note that 4 I and 3+ √ 19 I (to see the former, observe that in the contrary case 2 4 7 = 6∈ − 6∈ · − 1 I; to see the latter observe that in the contrary case 7 (3 + √ 19) (3 √ 19) = 1 I). ∈ 3+√ 19 − − − − − ∈ Hence, 4− is a unit in Rm, proving the claim. Lemma 5.3. Let R be a Noetherian integral domain with ﬁeld of fractions K. For every ideal 0 = 6 I E R, there is n N and there are non-zero prime ideals p ,..., p such that ∈ 1 n p p ... p I. 1 · 2 · · n ⊆ Proof. Consider the set

:= 0 = I E R the assertion is wrong for I . M { 6 | } We want to show = . So, let us assume = . We want to apply Zorn’s lemma to obtain a M ∅ M 6 ∅ maximal element J in , i.e. an element J such that for all ideals J ( I we have I . M ∈M 6∈ M Note that has a partial ordering given by . For Zorn’s Lemma we have to check that every M ⊆ ascending chain I I ... 1 ⊆ 2 ⊆ with I for i = 1, 2,... has an upper bound, that is, an element I containing all the I . i ∈ M ∈ M i That is the case since R is Noetherian and, thus, the ideal chain becomes stationary. So, let J ∈ M be such a maximal element. We distinguish two cases. First case: J is a prime ideal. Then J J implies J , contradiction. Hence, we are in the ⊆ 6∈ M Second case: J is not a prime ideal. Consequently, there are two elements x, y R such that ∈ xy J but x, y J. This allows us to consider the ideals ∈ 6∈

J1 := (J, x) ) J and J2 := (J, y) ) J.

Due to the maximality of J , we have that J and J are not in . Consequently, there are ∈ M 1 2 M p1,..., pn and q1,..., qm non-zero prime ideals of R such that

p ... p J and q ... q J . 1 · · n ⊆ 1 1 · · m ⊆ 2 This implies p ... p q ... q J J = (J, x)(J, y) J, 1 · · n · 1 · · m ⊆ 1 2 ⊆ which is also a contradiction. Hence, = . M ∅ 5 IDEALS IN DEDEKIND RINGS 29

Corollary 5.4. Let R be a local Noetherian integral domain of Krull dimension 1. Then every nonzero ideal I E R contains a power of the maximal ideal p.

Proof. Since R is a local Noetherian integral domain of Krull dimension 1, its only non-zero prime ideal is p. Hence, the assertion follows directly from Lemma 5.3.

Corollary 5.5. Let R be a Noetherian integral domain of Krull dimension 1. Then every non-zero ideal I ¡ R with I = R is contained in only ﬁnitely many maximal ideals of R. More precisely, if 6 p ... p I, then I is not contained in any maximal ideal different from p ,..., p . 1 · · n ⊆ 1 n Proof. By Lemma 5.3 there are non-zero prime ideals p ,..., p such that p ... p I. Let now 1 n 1 · · n ⊆ m be a maximal ideal of R containing I. We want to show that m is equal to one of the pi, which proves the assertions. Assume, hence, that m is none of the pi. As the Krull dimension is 1, none of the pi can be contained in m. Consequently, for each i = 1,...,n the ideal pi is coprime to m. There are thus x p and y m such that 1= x + y . We conclude i ∈ i i ∈ i i m p ... p x ... x = (1 y ) ... (1 y ) 1+ m, ⊇ 1 · · n ∋ 1 · · n − 1 · · − n ∈ which is the desired contradiction.

Lemma 5.6. Let R be an integral domain and I a fractional R-ideal. Then I = Im m¡R maximal ⊂ K. T Proof. We show both inclusions. ‘ ’: is trivial because I Im for all prime ideals (and, hence, in particular, all maximal ideals) m, ⊆ ⊆ as K is an integral domain. ‘ ’: Let x Im and consider the ideal J := r R rx I E R. We want ⊇ ∈ m¡R maximal { ∈ | ∈ } to show J = R because then x I. If J = R, then J is contained in some maximal ideal m ¡ R. T ∈ 6 Write x = a with a I and s R m. Because sx = a I, it follows s J m, which is a s ∈ ∈ \ ∈ ∈ ⊆ contradiction.

Theorem 5.7. Let R be a Noetherian integral domain of Krull dimension 1. Then the map

Φ: (R) (Rp), I (...,Ip,...), I → P 7→ 0=p¡R prime ideal 6 M is an isomorphism of abelian groups.

The meaning of this theorem is that any non-zero invertible ideal I ¡ R is uniquely determined by all its localisations Ip (at the non-zero prime ideals of R).

Proof. There are four things to show.

Φ is well-defined. First recall that Theorem 5.1 shows that Ip is a principal ideal. Second, recall • that an element of a direct sum only has finitely many components different from the identity; the identity of (Rp) is (1) = Rp. P We first show that the statement is correct for an integral ideal 0 = I E R. Suppose thus that 6 Ip ( Rp. Then I p (all elements of R p are units in Rp). Corollary 5.5 implies that there ⊆ \ 5 IDEALS IN DEDEKIND RINGS 30

are only ﬁnitely many such p. Now let us suppose that I is a fractional R-ideal. Then there is some r R 0 such that 0 = rI E R is an integral ideal. Thus, we may (and do) apply ∈ \ { } 6 the previous reasoning to the integral ideals rI and (r) = rR, and we obtain that for all prime ideals but possibly ﬁnitely many (rI)p = Rp and (r)p = rRp = Rp. For any such p we hence have Rp = (rI)p = (rRp) Ip = Rp Ip = Ip, proving the assertion. · · Φ is a group homomorphism. This is a property of localisations (already used in the previous • item): Let S = R p. Then (S 1I )(S 1I )= S 1(I I ), i.e. Φ(I I ) = Φ(I )Φ(I ). \ − 1 − 2 − 1 2 1 2 1 2

Φ is injective. Suppose Ip = Rp for all non-zero prime ideals p of R. Then we have •

I = Ip = Rp = R 0=p¡R prime ideal 0=p¡R prime ideal 6 \ 6 \ by Lemma 5.6.

Φ is surjective. As Φ is a group homomorphism, it sufﬁces to construct an invertible ideal • J (R) such that, for given maximal ideal m ¡ R and given principal ideal a ¡ Rm, we have ∈ I Jp = Rp for all nonzero prime ideals p = m and Jm = a. 6 We set J := R a. We ﬁrst claim Jm = a. ∩ ‘ ’: Let r R a, that means r a, whence r a for all s S = R m. ⊆ ∈ ∩ 1 ∈ s ∈ ∈ \ a a a a ‘ ’: Let a with a R and s S = R m. Then s = a R, whence Jm, ⊇ s ∈ ∈ ∈ \ s 1 ∈ ∩ s ∈ proving the claim. n By Corollary 5.4, there is n N such that (mRm) Jm. Recall that mRm is the maximal ∈ n n ⊆ n n ideal of Rm. It is clear that we have m (mRm) R. Consequently, m R (mRm) ⊆ ∩ ⊆ ∩ ⊆ Jm R = J. By Corollary 5.5 we have that J p for all maximal ideals p = m, whence ∩ 6⊆ 6 Jp = Rp. This concludes the proof.

We are now going to apply the above to Dedekind rings. For this, we recall the following characterisation from the lecture on Commutative Algebra.

Proposition 5.8. Let R be a Noetherian integral domain of Krull dimension 1. Then the following assertions are equivalent:

(i) R is a Dedekind ring.

(ii) R is integrally closed.

(iii) Rm is integrally closed for all maximal ideals m ¡ R.

(iv) Rm is regular for all maximal ideals m ¡ R.

(v) Rm is a principal ideal domain for all maximal ideals m ¡ R.

We will mostly be interested in (iv). Hence, it is useful to quickly recall the deﬁnition of a regular local ring and the main property of such rings in our case of Krull dimension 1. 5 IDEALS IN DEDEKIND RINGS 31

2 Deﬁnition 5.9. A Noetherian local ring with maximal ideal m is called regular if dimR/m(m/m ) equals the Krull dimension of R.

Proposition 5.10. Let R be a regular local ring of Krull dimension 1. Then there is x R such that ∈ all non-zero ideals are of the form (xn) for some n N. ∈ Such a ring is called a discrete valuation ring.

Corollary 5.11. Let R be a regular local ring of Krull dimension 1 and let p be its maximal ideal. Then there is x R such that all fractional ideals of R are of the form (x)n = pn for some n Z. ∈ ∈ Moreover, the map Z (R), n pn → I 7→ is an isomorphism of abelian groups.

Proof. By Proposition 5.10, the unique maximal ideal p is equal to (x), and, hence, all integral ideals of R are of the form pn for some n N. It is clear that (xn) = pn is invertible with inverse 1 n n ∈ (( x ) ) = (x)− . The ﬁnal statement is an immediate consequence. Deﬁnition 5.12. Let R be a Dedekind ring and I be an invertible R-ideal. For a maximal ideal p¡R, n by Proposition 5.10, there is a unique integer n 0 such that Ip = (pRp) . We write ordp(I) := n. ≥ Now we can prove unique ideal factorisation.

Theorem 5.13. Let R be a Dedekind ring. The map

Φ: (R) Z, I (..., ordp(I),...) I → 7→ 0=p¡R prime ideal 6 M is an isomorphism of abelian groups. Every I (R) can be uniquely written as ∈ I I = pordp(I) 0=p¡R prime ideal 6 Y (note that the product is ﬁnite).

Proof. The first statement follows from composing the isomorphism of Theorem 5.7 (which also implies the finiteness of the product) with the isomorphism (Rp) Z, given by ordp (the inverse P → to the isomorphism from Corollary 5.11). It suffices to show the final claim for invertible integral ideals because we can write any fractional R-ideal as a quotient of two integral ones: rI E R for some r R 0 , whence I = (rI) (r) 1. ∈ \ { } · − To see the final claim, for I E R we compute

ordp(I) I = Ip = (Ip R)= ((pRp) R) ∩ ∩ 0=p¡R prime ideal 0=p¡R prime ideal 0=p¡R prime ideal 6 \ 6 \ 6 \ = pordp(I) = pordp(I), 0=p¡R prime ideal 0=p¡R prime ideal 6 \ 6 Y where we used Lemma 5.6 and the pairwise coprimeness of the maximal ideals, so that the intersection n n becomes a product. We also used (pRp) R = p (see Remark 5.15). ∩ 6 GEOMETRY OF NUMBERS 32

Remark 5.14. Theorem 5.13 is a generalisation of unique factorisation in a principal ideal domain.

Remark 5.15. Let I be an invertible integral R-ideal of a Noetherian integral domain of Krull dimension 1. For a maximal ideal p ¡ R we deﬁne the p-primary component of I as I := Ip R. The (p) ∩ calculations made in the proof of Theorem 5.7 show that the localisation at a maximal ideal m is the following one:

Ip if p = m, (I(p))m = (Rm if p = m. 6 Moreover, the primary components behave ‘multiplicatively’:

(IJ)(p) = I(p)J(p) for any invertible integral R-ideals I and J. This is easy to see by working locally at all maximal ideals p (which sufﬁces by Theorem 5.7): the ideals on both sides have the same local components at all maximal ideals m. The multiplicativity implies, in particular, that

n n (pRp) R = p ∩ for an invertible maximal ideal p, which we used in the proof of Theorem 5.13, because pRp R = p ∩ (this equality can either be checked locally or directly, like this: if x = r with x p, r R and s 1 ∈ ∈ s R p, then x = rs p, whence r p by the prime ideal property of p). ∈ \ ∈ ∈ We ﬁnish with one corollary that we should have stated immediately after Proposition 5.8.

Corollary 5.16. Let R be a Dedekind ring. Then any fractional R-ideal is invertible.

Proof. By Proposition 5.8 we know that Rm is a principal ideal domain for all maximal ideals m ¡ R. Hence, given any fractional R-ideal I, we have that Im is principal for all m, which by Theorem 5.7) implies that I is invertible.

6 Geometry of Numbers

6.1 Introduction

Up to this point, we have been studying Dedekind domains in quite some generality. In this last part of the series of lectures, we will focus on the case of rings of integers of number ﬁelds. Recall (cf. Corollary 4.13) that, for any integral domain R, we have the following exact sequence

/ / f / proj / / 1 / R / K / (R) / Pic(R) / 1 × × I where:

K is the ﬁeld of fractions of R. • (R) is the group of invertible ideals of R. •I 6 GEOMETRY OF NUMBERS 33

Pic(R) is the Picard group of R, that is to say, the quotient of (R) modulo the group (R) of • I P principal fractionals ideals of R.

f : K (R) maps an element x K to the principal fractional ideal xR. • × → I ∈ proj : (R) (R)/ (R) = Pic(R) is the projection. • I → I P We want to study this exact sequence in the particular case where R = ZK is the ring of integers of a number ﬁeld K. Since ZK is a Dedekind domain, all nonzero fractional ideals are invertible (see Corollary 5.16). Hence (Z ) is the set of all nonzero fractional ideals. Recall also that we denote I K Pic(ZK ) = CL(K) and call it the class group of K. The exact sequence boils down to:

/ / f / proj / / 1 Z× K (ZK ) CL(K) 1 (6.5) K × I The group CL(K) measures the failure of ZK to be a principal ideal domain. More precisely, if CL(K) has just one element, then the map f : K (R) is surjective, so that each nonzero × → I fractional ideal can be expressed as xR for some x K . In other words, every fractional ideal is ∈ × principal. On the other hand, the greater CL(K) is, the further is f from being surjective, meaning there will be “many” fractional ideals which are not principal.

One of the fundamental results that we will prove is that CL(K) is ﬁnite (hence, although ZK is not a principal ideal domain, it is also “not too far” from it). Another important result will be that ZK× is ﬁnitely generated. As a motivation to study ZK× , consider the following example. Example 6.1. Let d be a rational integer which is not a square. Consider the equation x2 = dy2 + 1.

Question: Find all the solutions (x, y) Z Z of x2 = dy2 + 1. ∈ × This equation is called Pell’s equation, and was already considered by Archimedes (287? BC– 212?BC). Actually, Exercise Sheet 8 is devoted to the Problem of the Cattle of the Sun, that Archimedes proposes in a letter to Eratóstenes of Cirene. If d 0, then we can rewrite the equation as x2 +( d)y2 = 1, and it only has the trivial solutions ≤ − ( 1, 0) for d = 1 and ( 1, 0), (0, 1) for d = 1. But if d > 0, it is not obvious whether this ± 6 − ± ± − equation has a solution (different from ( 1, 0)) or not, much less to ﬁnd all solutions of the equation. ± Actually, without making use of any machinery at all, we can prove that for d> 0 Pell’s equation always admits a nontrivial solution. We need the following auxiliary lemma.

Lemma 6.2. Let d be a positive rational integer which is not a square. There exist inﬁnitely many pairs of integers (x, y) such that 0 < x2 dy2 < 1 + 2√d. | − | Proof. First let us see that there exists a pair of positive integers (x, y) with 0 < x2 dy2 < 1+2√d, | − | later we will see there are inﬁnitely many. Let m> 1 be a positive integer. For each i 1,...,m , ∈ { } let x Z be such that 0 x i√d< 1 (that is to say, we approximate √d by a quotient of integers, i ∈ ≤ i − where the denominator is i). This can always be done: namely, consider the fractional part of √d, that is to say, √d := √d √d . This lies in the interval [0, 1). If we cut out the interval in i equal { } −⌊ ⌋ subintervals, namely 1 1 2 i 1 [0, 1) = 0, , − , 1 , i ∪ i i ∪···∪ i 6 GEOMETRY OF NUMBERS 34

j 1 j j 1 there is a unique j 1,...,i such that √d [ − , ). Then 0 √d < , and ∈ { } { } ∈ i i ≤ i − { } i therefore 0 j i √d < 1. To approximate √d we just sum and substract i √d , and we get ≤ − { } ⌊ ⌋ 0 (i √d + j) i√d< 1. We can take x = i √d + j. ≤ ⌊ ⌋ − i ⌊ ⌋ Now divide the interval 1 1 2 m 2 [0, 1) = 0, , − , 1 . m 1 ∪ m 1 m 1 ∪···∪ m 1 − − − − There are m 1 intervals, but m pairs (x ,i). Hence (by Dirichlet’s Pidgeonhole Principle), there is − i one interval which contains both x i√d and x j√d with i = j. Assume x i√d x j√d i − j − 6 i − ≥ j − (otherwise swap i and j). Call x = x x , y = i j. Hence i − j − 1 x y√d = (x x ) (i j)√d = (x i√d) (x j√d) , − i − j − − i − − j − ≤ m 1 − thus 1 0 x y√d . ≤ − ≤ m 1 − √ 1 1 Since 1 i, j m, we have 0 < y < m, hence x y d m 1 y . Now we can bound ≤ ≤ | | − ≤ − ≤ | | 0 x2 dy2 = (x + y√d)(x y√d) = (x y√d + 2y√d) (x y√d) ≤| − | | − | | − | − y y = (x y√d)2 + 2 y √d(x y√d) 1 + 2 | | √d 1 + 2| |√d =1+2√d. − | | − ≤ m 1 ≤ y − | | Moreover we know that, since d is not a square, x2 dy2 = 0, and x2 y2d =1+2√d. − 6 | − | 6 Suppose now that the set A = (x, y) Z Z such that 0 < x2 dy2 < 1 + 2√d is { ∈ × | − | } ﬁnite. Then choosing an m N such that 1 is smaller than x y√d for all (x, y) A, the ∈ m 1 − ∈ previous construction provides us with a pair (x− , y ) A satisfying x y √d < 1 , which is a ′ ′ ∈ ′ − ′ m 1 contradiction. −

Proposition 6.3. Let d be a positive rational integer which is not a square. There exists pair of rational integers (x, y) with y = 0 such that x2 dy2 = 1. 6 − Proof. Since the number of integers in ( 1 2√d, 1 + 2√d) 0 is finite, by Lemma 6.2 there − − \ { } exists one k in this set such that there are infinitely many pairs (x, y) with x2 dy2 = k. By − definition k = 0. Moreover, since there are only finitely many residue clases in Z/kZ, we can assume 6 that there are α,β Z/kZ such that there are infinitely many pairs (x, y) with x2 dy2 = k and ∈ − x α (mod k), y β (mod k). Take two such pairs, (x , y ) and (x , y ). Consider the product ≡ ≡ 1 1 2 2 (x y √d)(x + y √d) = (x x y y d) + (x y x y )√d. 1 − 1 2 2 1 2 − 1 2 1 2 − 2 1 Note that k divides both x (x x )+k+dy (y y )= x (x x )+(x2 dy2)+dy (y y )= 1 2 − 1 1 1 − 2 1 2 − 1 1 − 1 1 1 − 2 x x dy y and (x x )y (y y )x = x y x y . Hence we can write 1 2 − 1 2 1 − 2 2 − 1 − 2 2 1 2 − 2 1 (x y √d)(x + y √d)= k(t + u√d) 1 − 1 2 2 for some integers t, u. Moreover note that

(x + y √d)(x y √d)= k(t u√d), 1 1 2 − 2 − 6 GEOMETRY OF NUMBERS 35 thus k2 = (x2 y2d)(x2 y2d)= k2(t2 u2d), 1 − 1 2 − 2 − so that dividing by k (which is nonzero), we get t2 u2d = 1. − This reasoning is valid for all (x , y ) and (x , y ) satisfying y2 dx2 = k and x α (mod k), 1 1 2 2 i − i i ≡ y β (mod k) for i = 1, 2. It remains to see that we can choose (x , y ) and (x , y ) so that the i ≡ 1 1 2 2 corresponding u is nonzero. Note that, if u = 0, then t = 1, so ± (x y √d)(x + y √d)= k(t + u√d)= k 1 − 1 2 2 ± On the other hand we have

(x y √d)(x + y √d)= x2 y2d = k. 1 − 1 1 1 1 − 1 Therefore we get x + y √d = (x + y √d) 1 1 ± 2 2 Fix one pair (x1, y1). Since we can choose (x2, y2) from an inﬁnity of pairs, we can assume, without loss of generality, that x + y √d = (x + y √d) (just take x = x , y = y ), and 2 2 6 ± 1 1 2 6 ± 1 2 6 ± 1 hence the solution (t,u) that we obtain satisﬁes u = 0. 6 Remark 6.4. Let d be a positive rational integer which is not a square. Consider the ring of integers Z of K = Q(√d). Recall that Z is Z[√d] if d 2, 3 (mod 4), Z[ 1+√d ] if d 1 (mod 4) (see K K ≡ 2 ≡ Example 3.6).

In the ﬁrst case, the elements of (ZK )× are precisely the set of elements x + √dy such that x2 dy2 = 1. In the second case the elements of (Z ) are those elements x + y 1+√d such that − ± K × 2 y 2 y 2 x + d = 1, 2 − 2 ± that is to say, (2x + y)2 dy2 = 4, − ± In both cases the knowledge of the group of unities of quadratic ﬁelds completely determines the set of solutions of the Pell equation.

The tool that we will use to study the exact sequence (6.5) is called Geometry of Numbers. This consists of viewing rings of integers as special subsets of Rn (namely lattices), and using some analytic tools (computing volumes) to obtain results concerning ZK .

6.2 Lattices

In this section we work with (Rn), endowed with the following structures:

A R-vector space structure (Rn, +, ), where + and are deﬁned componentwise. • · · A Z-module structure (Rn, +), obtained from the vector structure above by forgetting the scalar • multiplication. 6 GEOMETRY OF NUMBERS 36

A normed vector space structure (Rn, +, , ), where the R-vector space structure is the one • · k · k2 above and the norm is deﬁned as

: Rn R k · k2 → (a ,...,a ) = a 2 + + a 2. k 1 n k2 | 1| ··· | n| p We will denote by e ,...,e the canonical basis of Rn as R-vector space, so that n a e = { 1 n} i+1 i i (a ,...,a ). 1 n P Given a vector v Rn and a positive real number r, we denote by B(v; r) := w Rn : ∈ { ∈ w v < r the open ball of radius r centered at v and B(v; r) := w Rn : w v r k − k2 } { ∈ k − k2 ≤ } the closed ball of radius r centered at v. The set of all balls B(v; r) : v Rn,r 0 is a basis for { ∈ ≥ } the topology in Rn. We say that a set A Rn is bounded if it is contained in some ball centered at ⊂ 0 Rn. Recall that a set is compact if and only if it is closed and bounded (Theorem of Heine-Borel). ∈ We will usually work with subgroups of (R, +) which are not subvector spaces. For instance, n n Z is one such subgroup. Given v1,...,vr R , we will denote by v1,...,vr Z the Z-module n ∈ h i generated by v ,...,v inside R . Note that v ,...,v Z is a countable subset, while the vector 1 r h 1 ri space generated by v ,...,v has cardinality R . On the other hand, whenever we talk about linear 1 r | | dependence of elements of Rn, we will always be considering Rn with the structure of R-vector space. For x R, we denote by x the integer part of x, that is, the maximum m Z such that m x. ∈ ⌊ ⌋ ∈ ≤ Deﬁnition 6.5. A half-open parallelotope (resp. closed parallelotope) is a subset of Rn of the form

m P := v Rn : v = a v with 0 a < 1 for all i , { ∈ i i ≤ i } i=1 X m n resp. P := v R : v = aivi with 0 ai 1 for all i { ∈ ≤ ≤ }! Xi=1 where v ,...,v Rn are linearly independent. 1 m ∈ Remark 6.6. Note that closed parallelotopes are compact sets.

The point of this section is to compute volumes of parallelotopes in Rn. We mean by this the Lebesgue measure of the parallelotope. We will denote by µ the Lebesgue measure on Rn. We will not recall here its deﬁnition, but just one very important property: it is invariant under translation; that is, for all measurable sets A and all vectors v Rn, the set A + v := w + v : w A is measurable and we have ∈ { ∈ } µ(A)= µ(A + v).

Moreover the measure is normalized so that the measure of the standard cube n λ e : 0 λ { i=1 i i ≤ i ≤ 1 is equal to 1. } P The following lemma can be proven in an elementary calculus course.

n Lemma 6.7. Let P be the parallelotope deﬁned by n linearly independent vectors v1,...,vn R , n ∈ where each vi = j=1 aijej. Then µ(P )= det((aij)1 i,j n) . | ≤ ≤ | P 6 GEOMETRY OF NUMBERS 37

Deﬁnition 6.8. A subgroup H Rn is called discrete if, for any compact subset K Rn, H K is ⊂ ⊂ ∩ a ﬁnite set.

Remark 6.9. Since a subset of Rn is compact if and only if it is closed and bounded, then a subgroup H Rn is discrete if and only if for every r> 0, H B(0; r) is finite. ⊂ ∩ n Example 6.10. Let v ,...,v R be m linearly independent vectors. Then v ,...,v Z • 1 m ∈ h 1 mi is a discrete subgroup. Indeed, given any r > 0, we can show that v ,...,v Z B(0; r) is h 1 mi ∩ finite as follows: n First of all, complete v1,...,vm to a basis v1,...,vn of R . It suffices to show that the intersection v ,...,v Z B(0; r) is finite. h 1 ni ∩ Consider the linear map f : Rn Rn → v e for all i = 1,...n. i 7→ i n n Thus f( i=1 λivi)= i=1 λiei. Linear mapsP between finiteP dimensional finite-dimensional R-vector spaces are bounded oper- ators; that is to say there exists a constant C such that, for all v Rn, ∈ f(v) C v k k2 ≤ · k k2 Indeed, we have that n n n f( a e ) a f(e ) max a : 1 i n f(e ) k i i k2 ≤ | i| · k i k2 ≤ {| i| ≤ ≤ }· k i k2 Xi=1 Xi=1 Xi=1 Taking C = n f(e ) , it suffices to observe that i=1 k i k2 P n n max a : 1 i n a 2 = a e . {| i| ≤ ≤ }≤ v | i| k i ik2 ui=1 i=1 uX X t Therefore, we have that n n λ e C λ v . (6.6) k i ik2 ≤ · k i ik2 Xi=1 Xi=1 n r Assume now that v = i=1 λivi with some λi0 > C . Then Equation 6.6 implies that n P 1 1 v λ e λ > r, k k2 ≥ C k i ik2 ≥ C | i0 | Xi=1 hence v B(0; r). Thus 6∈ n r v1,...,vn Z B(0; r) λivi : λi for all i , h i ∩ ⊂ ( ≤ C ) Xi=1 which is a finite set. 6 GEOMETRY OF NUMBERS 38

n n Let v R be a nonzero vector. Then v, √2v Z is not a discrete subgroup of R (see Sheet 9). • ∈ h i The ﬁrst proposition is a characterisation of discrete subgroups of Rn.

Proposition 6.11. Let H be a discrete subgroup of Rn. Then H is generated as a Z-module by m linearly independent vectors for some m n. ≤ Proof. We can assume without loss of generality that H = 0 . Let 6 { } m := max r : there exist v ,...,v H linearly independent in Rn . (6.7) { 1 r ∈ } Since the numbers r appearing in (6.7) are bounded by n, we have that m is a finite number between 0 and n. Since H = 0, we have that m 1. Now let u ,...,u H be m vectors which are linearly 6 ≥ 1 m ∈ independent in Rn. Fix any v H nonzero. Then the set u ,...,u ,v is linearly dependent by ∈ { 1 m } the maximality of m, so there exist λ ,...,λ R such that v = m λ u . For each j N, we 1 m ∈ i=1 i i ∈ consider m P v := (jλ jλ )u = jv jλ u H. j i −⌊ i⌋ i − ⌊ i⌋ i ∈ Xi=1 X On the other hand, v w Rn : w = m a u with 0 a 1 =: P , and the set P is compact j ∈ { ∈ i=1 i i ≤ i ≤ } (see Remark 6.6) so v belongs to the finite set H P . This implies already that H is a Z-module of j P ∩ finite type (more precisely, we have proven that every v in H can be written as v + m λ u , so 1 i=1⌊ i⌋ i H is generated as a Z-module by the finite set = (H P ) u ,...,u ). G ∩ ∪ { 1 m} P Since the set v : j N is finite, there must exist j, k different natural numbers such that { j ∈ } v = v , that is m (jλ jλ )u = m (kλ kλ )u . Since the u ’s are linearly independent, j k i=1 i−⌊ i⌋ i i=1 i−⌊ i⌋ i i we get that for all i, (j k)λ = jλ kλ . In particular, for all i, λ Q. Since this is valid for P − i ⌊ i⌋−⌊P i⌋ i ∈ all v H, we get that H is a finitely generated Z-module contained in the Q-vector space generated ∈ by u ,...,u . Pick a finite number of generators of H as Z-module (for example ), write each of 1 m G them as r λ u for λ Q and pick a common denominator d for all the coefficients λ ’s of all the i=1 i i i ∈ i generators. Then we have dH u ,...,u Z. We now apply Theorem 3.12 to conclude that dH P ⊂ h 1 mi is a free Z-module of rank smaller than or equal to m. Since we know that dH contains the free Z- module generated by du ,...,du , the rank must be precisely m. Let u ,...,u dH be such that 1 m 1′ m′ ∈ u ,...,u Z = dH. Since dH contains the m linearly independent vectors du ,...,du , it follows h 1′ m′ i 1 m that u1′ ,...,um′ must span a R-space of dimension m, hence they are linearly independent over R. 1 1 1 1 Finally, u ,..., u H are linearly independent vectors such that u ,..., u Z = H. d 1′ d m′ ∈ h d 1′ d m′ i From all the discrete subgroups, we will be interested in those that are generated by n linearly independent vectors.

Deﬁnition 6.12. A lattice in Rn is a discrete subgroup H Rn of rank n as a Z-module. • ⊂ Equivalently, a lattice in Rn is a Z-module generated by n linearly independent vectors.

A basis of a lattice H Rn will be basis of H as a Z-module. • ⊂ Note that a basis of a lattice H consists of n linearly independent vectors of Rn, so in particular is a basis of Rn as R-vector space. 6 GEOMETRY OF NUMBERS 39

Deﬁnition 6.13. Let H Rn be a lattice, and = u ,...,u a basis of H. We will say that the ⊂ U { 1 n} (half-open) parallelotope P determined by is a fundamental domain for H. U U Remark 6.14. One lattice has different fundamental domains; in other words, fundamental domains are not unique.

Lemma 6.15. Let H Rn be a lattice, P , P fundamental domains for H. Then µ(P )= µ(P ). ⊂ ′ ′

Proof. Let = u1,...,un (resp. ′ = u1′ ,...,un′ ) be a basis of H deﬁning P (resp. P ′) and B { } B n { } n n let e1,...,en the canonical basis of R . Write ui′ = j=1 aijuj with aij Z, ui = j=1 bijej, { n } ∈ ui′ = j=1 cijej with bij,cij R and set A = (aij)1 i,j n, B = (bij)1 i,j n, C = (cij)1 i,j n. ∈ ≤P ≤ ≤ ≤ P ≤ ≤ We have C = AB. Since both and ′ are Z-bases of H, we have det((aij)1 i,j n)= 1. And by P B B ≤ ≤ ± Lemma 6.7 µ(P )= det(B) = det(B) det(A) = det(C) = µ(P ′). | | | |·| | | |

Deﬁnition 6.16. Let H Rn be a lattice. We deﬁne the volume of H as ⊂ v(H) := µ(P ),

for some fundamental domain P of H.

Lemma 6.17. Let H Rn be a lattice and P be a fundamental domain. ⊂ For each v Rn there exists a unique u P such that v u H. • ∈ ∈ − ∈ n R is the disjoint union of the family P + u u H . • { } ∈ Proof. See Sheet 9.

Now we will state the fundamental result of this section. The idea is the following: given a lattice H, if a measurable set S Rn is big enough (with respecto to µ), no matter what it looks like, it ⊂ must contain two elements which are “equivalent modulo H”, that is to say, two different elements v ,v S with v v H. 1 2 ∈ 1 − 2 ∈ Theorem 6.18 (Minkowsky). Let H Rn be a lattice and S Rn be a measurable subset of Rn ⊂ ⊂ satisfying µ(S) >v(H). Then there exist v ,v S different elements with v v H. 1 2 ∈ 1 − 2 ∈ n Proof. Sine P is a fundamental domain for H, Lemma 6.17 implies that R = u H (P + u). Intersecting both sides with S yields ∈ F S = (S (P + u)). ∩ u H G∈ Recall that H is countable. Therefore by the countable additivity of µ, we get

µ(S)= µ(S (P + u)). ∩ u H X∈ 6 GEOMETRY OF NUMBERS 40

Since µ is invariant by translation, we get that, for all u H, µ(S (P + u)) = µ((S u) P ). ∈ ∩ − ∩ Now if the family of sets (S u) P u H were disjoint, we would get, using the countable additivity { − ∩ } ∈ of µ again, that u H µ((S u) P )= µ( u H (S u) P )) µ(P ). Hence ∈ − ∩ ∈ − ∩ ≤ µ(S)= P µ(S (P + u)) = µ((FS u) P )= µ( ((S u) P ))) µ(P ) ∩ − ∩ − ∩ ≤ u H u H u H X∈ X∈ G∈ contradicting that µ(S) > v(H). Thus the family (S u) P u H is not disjoint, that is to { − ∩ } ∈ say, there exist u ,u H, u = u , with ((S u ) P ) ((S u ) P ) = . Let w 1 2 ∈ 1 6 2 − 1 ∩ ∩ − 2 ∩ 6 ∅ ∈ (S u ) P ) ((S u ) P ). Then w = v u = v u for some v ,v S. And − 1 ∩ ∩ − 2 ∩ 1 − 1 2 − 2 1 2 ∈ v v = u u H is nonzero. 1 − 2 1 − 2 ∈ We will use a particular case of this theorem, when S has some special properties. Deﬁnition 6.19. Let S Rn. ⊂ S is centrally symmetric if, for all v S, v S. • ∈ − ∈ S is convex if, for all v ,v S, for all λ [0, 1], λv + (1 λ)v S. • 1 2 ∈ ∈ 1 − 2 ∈ Corollary 6.20. Let H Rn be a lattice and S Rn be a centrally symmetric, convex, measurable ⊂ ⊂ set such that µ(S) > 2nv(H). Then S (H 0 ) = . ∩ \ { } 6 ∅ Proof. Let S = 1 S := 1 v : v S . Note that µ(S ) = 1 µ(S) > v(H). Hence we can apply ′ 2 { 2 ∈ } ′ 2n Theorem 6.18 to S and conclude that there are elements v ,v S with v v H 0 . Note ′ 1 2 ∈ ′ 1 − 2 ∈ \ { } furthermore that v ,v S implies that 2v , 2v S, and since S is centrally symmetric, also 1 2 ∈ ′ 1 2 ∈ 2v S. The convexity of S now implies that v v = 1 (2v )+ 1 1 ( 2v ) S. Hence − 2 ∈ 1 − 2 2 1 − 2 − 2 ∈ v v S (H 0 ). 1 − 2 ∈ ∩ \ { } 6.3 Number rings as lattices

Let C be the field of complex numbers. Inside C we have the subfield of rational numbers Q, which can be characterised as the smallest subfield of C (or, in other words, the prime field of C, that is to say, the intersection of all subfields of C). We also have the subfield of C defined as Q := z C : { ∈ z is algebraic over Q . Q is an algebraically closed field, and clearly it is the smallest subfield of C } containing Q which is algebraically closed, hence an algebraic closure of Q. Let K/Q be a number field of degree n and let K be an algebraic closure. Since K is algebraic over Q, K is also an algebraic closure of Q and hence isomorphic to Q. Fixing one such isomorphism, we can identify K with Q and K with a subfield of Q C. ⊂ Since char(K) = 0, K is separable, and therefore (see the Appendix to section 2) there exist n different ring homomorphism (necessarily injective) from K to Q fixing Q. Since the image of any ring homomorphism σ : K C must be contained in Q, we have that there are exactly n different → .ring homomorphisms σ : K ֒ C fixing Q → Let α : C C be the complex conjugation. Then, for all σ HomQ(K, C), we have that → ∈ α σ HomQ(K, C), and α σ = σ if and only if σ(K) R. Call r the number of ring ◦ ∈ ◦ ⊂ 1 homomorphisms σ : K C such that α σ = σ. The remaining homomorphisms can be collected in → ◦ pairs σ, α σ , so there is an even number of them. Let us call 2r this number, so that n = r +2r . { ◦ } 2 1 2 Let us enumerate the n homomorphisms in Hom(K, C) in the following way: 6 GEOMETRY OF NUMBERS 41

Let σ ,...,σ be the r homomorphisms with image contained in R. • 1 r1 1 Let us enumerate the r pairs σ, α σ and, for each pair, choose one of the two homomor- • 2 { ◦ } phisms. The chosen homomorphism of the i-th pair (1 i r ) will be σ , the other one ≤ ≤ 2 r1+i will be σr1+r2+i.

Now we can deﬁne a ring homomorphism

Φ : K Rr1 Cr2 0 → × x (σ (x),...,σ (x),σ (x),...,σ (x)) 7→ 1 r1 r1+1 r1+r2 Definition 6.21. For z = x + iy C, denote by Rez := x the real part of z and Imz := y the ∈ imaginary part of z. The map C R R defined as z (Rez, Imz) is an isomorphism of R-vector → × 7→ spaces. Define the map

Φ: K Rr1 R2r2 → × x (σ (x),...,σ (x), Reσ (x), Imσ (x) ..., Reσ (x), Imσ (x)). 7→ 1 r1 r1+1 r1+1 r1+r2 r1+r2 Remark 6.22. The map Φ above is injective (because each σ is injective), and a group ho- • i momorphism (of the additive groups (K, +) and (Rn, +)). Moreover, both K and Rn have a Q-vector space structure, and Φ preserves it.

Φ provides us with a way to see number ﬁelds inside n-dimensional R-vector spaces. We are • interested in subgroups of K that give rise to lattices in Rn.

Proposition 6.23. Let M K be a free Z-module of rank n, say with basis x ,...,x . Then ⊂ { 1 n} Φ(M) is a lattice in Rn. • r2 Let A = (σi(xj))1 i,j n. Then v(Φ(M)) = 2− det A . • ≤ ≤ | | Remark 6.24. With the notations above, the discriminant of the tuple (x ,...,x ) Kn is deﬁned as 1 n ∈ the square of det A. Moreover (see Proposition 2.8-(e)) the discriminant of (x1,...,xn) is nonzero.

Proof. Φ : K Rn is an injective morphism from (K, +) to (Rn, +), hence it carries free Z- → modules into free Z-modules, and transforms Z-bases into Z-bases. Therefore Φ(M) is a Z-module n of rank n in R with basis Φ(x1),..., Φ(xn). To prove that it is a lattice, we need to see that the n vectors Φ(x1),..., Φ(xn) are linearly independent over R. The coordinates of Φ(xi) are

(σ1(xi),...,σr1+1(xi), Reσr1+1(xi), Imσr1+1(xi) ..., Reσr1+r2 (xi), Imσr1+r2 (x))

Let B be the matrix with i-th row as above, for all i 1,...,n . We will prove that det B = 0, thus ∈ { } 6 showing that the vectors Φ(x1),..., Φ(xn) are linearly independent over R. z For j = 1,...,r2, call j the column vector with entries (σr1+j(xi))i=1,...,n, and denote the column vector whose entries are the complex conjugates of the entries of zj by zj. Then we have that

. . . z z z z . B = . z z . = . j + j j j . . Re j Im j . . 2 2−i . 6 GEOMETRY OF NUMBERS 42

Hence

. z z . . z z . det B = det . j j . + det . j j . . 2 2i . . 2 −2i . . z z . . z z . + det . j j . + det . j j . . 2 2i . . 2 −2i .

1 . . 1 . . 1 . . = − det . z z . + det . z z . = − det . z z . . 4i . j j . 4i . j j . 2i . j j . Repeating this process for all i= 1,...,r2, we get 1 r2 det B = − det A′, 2i where A′ is the matrix with i-th row given by

(σ (x ),...,σ (x ),σ (x ), α σ (x ) ...,σ (x ), α σ (x )). 1 i r1+1 i r1+1 i ◦ r1+1 i r1+r2 i ◦ r1+r2 i Since the columns of A and A coincide up to a permutation, we have det A = det A = 0. ′ | ′| | | 6 This proves that Φ(M) is a lattice. Moreover v(Φ(M)) = det B = 2 r2 det A . | | − | | Definition 6.25. Let K be a number field. Let a Z be a nonzero integral ideal. We define the norm of a as N(a)=[Z : a]. ⊂ K K Let I K be a nonzero fractional ideal. We define the norm of I as N(I) = N(xI)/N (x), ⊂ K/Q where x Z is some element different from zero such that xI is an integral ideal. ∈ K Remark 6.26. Let K be a number field. Then N : (Z ) Q× is a group homomorphism (See I K → Sheet 9).

Corollary 6.27. Let K/Q be a number ﬁeld of degre n = r1 + 2r2 and a an integral ideal of ZK . n Then we have that Φ(ZK ), Φ(a) are lattices of R and

r2 r2 v(Φ(Z )) = 2− disc(Z ) , v(Φ(a)) = 2− disc(Z ) N(a). K | K | | K | p p Proof. Since ZK is an order of K (see Corollary 3.17-(a)), it is a free Z-modules of rank n. By Corollary 3.17-(c), a is also a free Z-module of rank n. The formula for the volume of Φ(ZK ) follows directly from the deﬁnition of disc(ZK ); the formula for the volume of Φ(a) follows from Proposition 3.19.

6.4 Finiteness of the class number

Let K be a number ﬁeld of degree n. As in the previous section, we denote by r1 the number of .embeddings of K ֒ R and r = (n r )/2 → 2 − 1 Proposition 6.28. Let a Z be a nonzero integral ideal. There exists a a different from zero ⊂ K ∈ such that 2 r2 N (a) disc(Z ) N(a). | K/Q |≤ π | K | p 6 GEOMETRY OF NUMBERS 43

Proof. We will apply Corollary 6.20 in Rn. First we deﬁne the measurable set S as follows: Let A ,...,A and B ,...,B be some positive real numbers. Consider the set S Rn deﬁned by 1 r1 1 r2 ⊂

S = (x ,...,x , y , y′ ,...,y , y′ ): { 1 r1 1 1 r2 r2 x A for all i = 1,...,r , y2 + y 2 B for all j = 1,...,r . (6.8) | i|≤ i 1 j j′ ≤ j 2} q The set S is centrally symmetric (clear) and convex: if we have (x1,...,xr1 , y1, y′1,...,yr2 , y′r2 ) and (˜x ,..., x˜ , y˜ , y˜ ,..., y˜ , y˜ ) in S, then for any λ (0, 1), 1 r1 1 1′ r2 r′ 2 ∈ λx + (1 λ)˜x λ x + 1 λ x˜ A , | i − i|≤| |·| i| | − |·| i|≤ i and

(λy + (1 λ)˜y )2 + (λy + (1 λ)˜y )2 j − j j′ − j′ ≤ q (λy )2 + (λy )2 + ((1 λ)˜y )2 + ((1 λ)˜y )2 j j′ − j − j′ ≤ q q λ y2 + y 2 + 1 λ y˜2 + (˜y )2 B . | |· j j′ | − |· j j′ ≤ j q q Its Lebesgue measure can be computed as

r1 r2 r1 r2 µ(S)= (2A ) (πB2) = 2r1 πr2 A B2. i · j i j Yi=1 jY=1 Yi=1 jY=1 On the other hand, we can embed K ֒ Rn via the map Φ from Definition 6.21. H = Φ(a) is a → lattice of volume v(H) = 2 r2 disc(Z ) N(a) (Corollary 6.27). − | K | Let ε> 0. Choose A ,...,A , B ,...,B positive integers such that 1 pr1 1 r2 r1 r2 2 r2 A B2 = disc(Z ) N(a)+ ε, i j π | K | i=1 j=1 Y Y p and call Sε the set defined by (6.8). n Then it holds that v(H) > 2 µ(Sε), so we can apply Corollary 6.20 and conclude that there exists some nonzero v S H. Let a a such that Φ(a) = v. The fact that Φ(a) S means that, for ∈ ε ∩ ∈ ∈ ε all i = 1,...,r , σ (a) A , and for all j = 1,...,r , (Reσ (a))2 + (Imσ (a))2 B . 1 | i | ≤ i 2 r1+j r1+j ≤ j Therefore p r1 r2 r1 r2 2 r2 N (a) = σ (a) σ (a) 2 A B2 = disc(Z ) N(a)+ ε | K/Q | | i |· | j | ≤ i j π | K | i=1 j=1 i=1 j=1 Y Y Y Y p Now for all ε there exists an a a such that N (a) satisfies the inequality above. But ∈ | K/Q | this norm is an integer, so taking ε small enough, we will get an a a such that NK/Q(a) r2 ∈ | | ≤ 2 disc(Z ) N(a). π | K | Proposition p 6.29. Let a Z a nonzero integral ideal. There exists a a different from zero such ⊂ K ∈ that 4 r2 n! N (a) disc(Z ) N(a). | K/Q |≤ π nn | K | p 6 GEOMETRY OF NUMBERS 44

Proof. See Sheet 10.

Proposition 6.28 (or Proposition 6.29) will be a key ingredient in the proof of the following result.

Theorem 6.30 (Dirichlet). Let K be a number ﬁeld. The class group CL(K) = (Z )/ (Z ) is I K P K ﬁnite.

Before proceeding to the proof, let us establish a technical lemma.

Lemma 6.31. Let K be a number ﬁeld, and C CL(K) be a class of ideals. Then there exists a ∈ nonzero integral ideal a of ZK which belongs to C and satisﬁes 2 r2 N(a) disc(Z ) . ≤ π | K | p Proof. Let I be a nonzero fractional ideal in C. Then I 1 = a Z : aI Z is also a nonzero − { ∈ K ⊂ K } fractional ideal. Therefore there exists x K such that b = xI 1 is a nonzero integral ideal. We can ∈ − apply Proposition 6.28 to the ideal b; there exists b b such that ∈ r2 r2 2 2 1 N (b) disc(Z ) N(b)= disc(Z ) N (x) N(I)− . | K/Q |≤ π | K | π | K || K/Q | p p b Therefore the ideal a = x I belongs to the class C and furthermore

r2 NK/Q(b) 2 N(a)= | | N(I) disc(Z ) . N (x) ≤ π | K | K/Q | | p

Proof of Theorem 6.30. Since every class C CL(K) contains a nonzero integral ideal of norm r2 ∈ smaller than 2 disc(Z ) (because of Lemma 6.31), it suffices to prove that, for any M N, π | K | ∈ there are only finitely many integral ideals of norm smaller than M. First of all, note that it suffices p to see that there are only finitely many prime integral ideals of norm smaller than M; indeed if r ei r ei a = i=1 pi is a factorisation of a into a product of prime ideals, then N(a) = i=1 N(pi) , so if N(a) is smaller than M, the only prime ideals that can occur in the factorisation of a are those Q Q with norm smaller than M, and the exponents ei that can occur must also be smaller than M. Assume now that p is a prime integral ideal of norm smaller than M, say m. Then 1 Z /p ∈ K satisfies that m 1 = 0 Z /p, thus m p. But we know that that there are only a finite number of · ∈ K ∈ maximal ideals of ZK containing a given ideal I (Corollary 5.5). In particular, for I = (m), we get that there are only finitely many prime ideals p of ZK of norm m.

Remark 6.32. Let K be a number ﬁeld. Then CL(K) is generated by the classes of the prime • r2 ideals p (Z ) such that N(p) 2 disc(Z ) . This allows one to compute explicitly ∈ I K ≤ π | K | the class group of a given number ﬁeld, provided one knows how to compute the prime ideals p of given norm.

The same proof, but using the better bound of Proposition 6.29, shows that CL(K) is generated • r2 by the classes of the prime ideals p (Z ) such that N(p) 4 n! disc(Z ) . ∈ I K ≤ π nn | K | p 6 GEOMETRY OF NUMBERS 45

Remark 6.33. Let E/K be an extension of number ﬁelds, and let p Z be a nonzero prime • ⊂ K ideal. The ideal pZE generated by the elements of p inside ZE need not be prime anymore, but, since ZE is a Dedekind domain, it will factor in a unique way as a product of primes

r ei pZE = Pi . Yi=1 The ideals Pi are the prime ideals of ZE containing pZK (Corollary 5.5). We will say that P1,..., Pr are the prime ideals of ZE lying above p.

More generally, if A is a Dedekind ring, p a nonzero prime ideal of A and B A a Noetherian • ⊃ integral domain of Krull dimension 1, we will say that the prime ideals of B lying above p are the prime ideals P of B such that pB P. By Corollary 5.5 we know there are only ﬁnitely ⊂ many prime ideals of B lying above a nonzero prime ideal p of A.

Proposition 6.34. Let A be a Dedekind ring, K its fraction field, E/K a finite separable extension of K and B E the integral closure of A in E. Assume there exists α B such that B = A[α], and ⊂ ∈ call f(x) the minimal polynomial of α over K. Let p be a nonzero prime ideal of A, let k = A/p be the residue field, let f(X) k[X] be the ∈ reduction of f(X) mod p, and let r

f(X)= qi(X) Yi=1 be a factorisation of f(X) into monic irreducible polynomials in k[X]. For each i = 1,...,r, choose q (X) A[X] reducing to q (x) mod p. Then the prime ideals in B above p are given by i ∈ i

Pi := pB + qi(α)B, i = 1,...,r.

Proof. Let i 1,...,r , and fix a root β k of q (X). Consider the ring homomorphism ∈ { } ∈ i φ : B = A[α] k[β] → α β 7→ a A a k = (A/p). ∈ 7→ ∈ Let P = ker φ. Since p is a prime ideal of A, k is a field and k[β] k is an integral domain. Thus ⊂ .B/P ֒ k is an integral domain, and P is a prime ideal. We will now show that P = pB + q (α)B → i Clearly φ(a) = 0 for all a p and φ(q (α)) = q (β) = 0, hence we have the inclusion. ⊇ ∈ i i Let b P, say b = g(α) for some g(X) A[X]. Then 0 = φ(b) = φ(g(α)) = g(φ(α)) = ⊆ ∈ ∈ g(β), where g(X) k[X] is the reduction of g(X) modulo p. Thus g(X) is divisible by the ∈ minimal polynomial of β over k, that is q (X), say g(X)= q (X)h(X). Taking h(X) A[X] i i ∈ reducing to h(X), we have that g(X) q (X)h(X) A[X] has coefficients in p, and therefore − i ∈ g(α) q (α)B + pB. This proves the other inclusion. ∈ i 6 GEOMETRY OF NUMBERS 46

This proves that the r primes Pi are primes of B above p. Reciprocally, let P be a prime over p, and consider the projection φ : B B/P. We know that B/P is a field, and we have a natural → inclusion k = A/p B/P. Since f(α) = 0, then f(α) = 0, therefore α is a root of some of → the qi(X), the projection φ is the composition of one of the projections φi considered above with an isomorphism B/P B/P fixing k, say τ φ , and P = ker(τ φ ) = ker φ = P . i ≃ ◦ i ◦ i i i Remark 6.35. Let K be a number field, p Z a nonzero prime. Then the prime ideals of Z above ∈ K (p) are those whose norm is a power of p.

Corollary 6.36. Let K be a number ﬁeld, and assume that there exists α Z such that Z[α]= Z . ∈ K K Call f(X) Z[X] be the minimal polynomial of α over Q. Let p be a prime, let f(X) F [X] be ∈ ∈ p the reduction of f(X) mod p, and let

f(X)= qi(X) Yi=1 be a factorisation of f(X) into monic irreducible polynomials in Fp[X]. For each i = 1,...,r, choose q (X) Z[X] reducing to q (x) mod p. Then the prime ideals of Z of norm equal to a i ∈ i K power of p are given by

Pi := (p, qi(α))ZK , i = 1,...,r.

Example 6.37. Let K = Q(√7). Then ZK = Z[√7], and disc(ZK ) = 4 7. Since K R, • r2 · ⊂ r = 0 and n = r = 2. The quantity C = 4 n! disc(Z ) satisﬁes 2

– Prime ideals of norm a power of 2: We apply Corollary 6.36: α = √7 satisﬁes ZK = Z[√7]. The minimal polynomial of α over Q is f(x)= x2 7. Now x2 7 x2+1=(x+ 2 − − ≡ 1) (mod 2), hence the only prime ideal of ZK above (2) is p = (2, 1+√7) = (3+√7).

Therefore CL(K) is generated by the classes of principal ideals. Thus CL(K)= 1 . { }

Let K = Q(√ 5). Then ZK = Z[√ 5], and disc(ZK ) = 20. Now K R, and therefore • − − r2 − 6⊆ n = r = 2, r = 0. The quantity C = 4 n! disc(Z ) satisﬁes 2

6.5 Dirichlet Unit Theorem

The aim of this section is to prove the following result:

Theorem 6.38 (Dirichlet). Let K be a number ﬁeld of degree n = r1 + 2r2. Then there is a group isomorphism r1+r2 1 Z× µ Z − , K ≃ K × where µK is the (ﬁnite) subgroup of ZK× consisting of roots of unity.

Remark 6.39. Note that, in both ZK× and µK the group structure is written multiplicatively, whereas r1+r2 1 in Z − the group structure is written additively.

The proof of this theorem will be given gradually through a series of steps (Lemmas 6.41 6.42, 6.45, 6.46, 6.47 and Corollaries 6.43, 6.44). Consider the following map

/ / K / Rr1 Cr2 / Rr1+r2 ×

a / Φ (a) = (σ (a),...,σ (a)) / ( σ (a) ,..., σ (a) ), 0 1 r1+r2 | 1 | | r1+r2 | where Φ is the map considered before Definition 6.21 and, in the second map, : R R is the 0 |·| → usual absolute value, and : C R is the norm given by x + iy = x2 + y2 for all x, y R. |·| → | | ∈ p Definition 6.40. Let K be a number field of degree n = r1+2r2. We define the logarithmic embedding as the group morphism

r1+r2 Φ : K× R log → a (log σ (a) ,..., log σ (a) ). 7→ | 1 | | r1+r2 |

Recall that, if K is a number ﬁeld and a Z , then a Z× if and only if N (a) = 1 (cf. ∈ K ∈ K K/Q ± Lemma 3.10).

Lemma 6.41. Let K be a number ﬁeld of degree n = r + 2r and B Rr1+r2 a compact set. 1 2 ⊂ Consider the set B′ := a Z× : Φ (a) B . { ∈ K log ∈ } Then there exists an M > 1 such that, for all a B and all i = 1,...,r + r , ∈ ′ 1 2 1 < σ (a) < M. M | i | Proof. Since B is bounded, there exists an N such that, for all y = (y ,...,y ) B, y < N 1 r1+r2 ∈ | i| for all i = 1,...,r + r . If a B , then Φ (a) B, and therefore log σ (a) N for all 1 2 ∈ ′ log ∈ | | i || ≤ i = 1,...,r1 + r2. Hence

N N e− < σ (a) < e for all i = 1,...,r + r . | i | 1 2 Take M = eN . 6 GEOMETRY OF NUMBERS 48

Lemma 6.42. Let K be a number ﬁeld of degree n = r1 + 2r2 and B, B′ as in Lemma 6.41. Then B′ is ﬁnite.

Proof. By Lemma 6.41, there exists M > 1 such that, for all i = 1,...,r + r , σ (a)

Corollary 6.43. Φlog(ZK× ) is a discrete subgroup, hence a free Z-module of rank less than or equal to r1 + r2.

Proof. This follows from Proposition 6.11.

Corollary 6.44. The kernel of Φlog Z× is a ﬁnite group, consisting of the roots of unity contained in | K ZK .

r1+r2 Proof. Take any compact B of R containing 0. Then ker(Φlog Z× ) B′, hence it is finite. If | K ⊂ s a Z× belongs to a finite subgroup, it must have finite order, so there exists s N with a = 1. In ∈ K ∈ other words, a is a root of unity. Reciprocally, if a Z is a root of unity, then it satisfies that, for some s N, as = 1. Therefore, ∈ K ∈ for all i = 1,...,r + r , σ (a)s = 1, thus log σ (x) =log1=0, and Φ (a) = 0. 1 2 i | i | log Lemma 6.45. Let K be a number field. Then

Z× µ Φ (Z× ) K ≃ K × log K Proof. We have the exact sequence of groups

× 1 ker(Φlog Z ) ZK× Φlog(ZK× ) 0. → | K → → →

× By Corollary 6.44 we know that ker(Φlog Z )= µK , and by Corollary 6.43 we know that Φlog(ZK× ) | K is a free Z-module, hence the exact sequence splits.

Lemma 6.46. Let K be a number ﬁeld of degree n = r1 + 2r2. The rank of Φlog(ZK× ) is less than or equal to r + r 1. 1 2 − 6 GEOMETRY OF NUMBERS 49

Proof. Let a Z× . Then the norm of a is 1, thus ∈ K ± r1 r1+r2 1= N (a)= σ (a) σ (a)(α σ )(a) ± K/Q i · i ◦ i Yi=1 i=Yr1+1 where α : C C denotes the complex conjugation. Applying log to both sides, we get → |·| r1 r1+r2 0= log σ (a) + 2 log σ (a) . | i | | i | Xi=1 i=Xr1+1 r1+r2 r1 Therefore Φlog(a) belongs to the subspace W := (y1,...,yr1+r2 ) R : i=1 y + r1+r2 { ∈ 2 y = 0 . Therefore Φ (a) must have rank smaller than or equal to dimR W = r + i=r1+1 i } log P 1 r 1. 2P−

Up to this point, we have proven that ZK× is not very big, that is, it is ﬁnitely generated, and we even have a bound for the number of generators of the free part. That was the easy part. Note that, up to now, we have not used Minkowsky’s Theorem 6.18 or its corollary. The hard part is to show that, indeed, the torsion-free part of the group Z× has r + r 1 free generators; and for this we will need K 1 2 − Corollary 6.20.

Lemma 6.47. Let K be a number field of degree n = r1 + 2r2. The rank of Φlog(ZK× ) is equal to r + r 1. 1 2 − Proof. We already know one inequality by Lemma 6.46. To show the other inequality, we will prove that Φ (Z× ) cannot be contained in any proper vector subspace of W := (y ,...,y ) log K { 1 r1+r2 ∈ Rr1+r2 : r1 y + 2 r1+r2 y = 0 . i=1 i=r1+1 i } Assume then that there exists W Rr1+r2 a proper subvector space of W containing Φ (Z ). P P 0 log K× r1+r2 1 ⊂ The projection W R − given by (y1,...,yr1+r2 ) (y1,...,yr1+r2 1) is an isomorphism of − → 7→ r1+r2 1 R-vector spaces. Via this projection, W0 corresponds to a subvector space of R − . In particular, r1+r2 1 r1+r2 1 there exists a vector (c1,...,cr1+r2 1) R − such that, for all w W0, i=1 − ciwi = 0. − ∈ ∈ We will find an u Z× such that ∈ K P r1+r2 1 − c log σ (u) = 0. i | i | 6 Xi=1 Let us fix some constant 2 r2 M > disc(Z ) . π | K | p A The main step in the proof of this lemma is to show that, for any tuple = (A1,...,Ar1+r2 1) r1+r2 1 − ∈ R − of positive real numbers, there exists an a Z such that N (a) M and >0 ∈ K | K/Q |≤ r1+r2 1 r1+r2 1 r1+r2 1 − − − ci log σi(a) ci log Ai ci log M. (6.9) | |− ≤ | | Xi=1 Xi=1 Xi=1 A We proceed as follows: given = (A1,...,Ar1+r2 1), set − M A := . r1+r2 r1 r2 1 2 s i=1 2Ai j=−r1+1 Aj Q Q 6 GEOMETRY OF NUMBERS 50

Then, like in the proof of Proposition 6.28, we consider the set S Rr1+2r2 deﬁned by ⊂

S = (x ,...,x , y , y′ ,...,y , y′ ): { 1 r1 1 1 r2 r2 x A for all i = 1,...,r , y2 + y 2 A for all j = r + 1,...,r + r . | i|≤ i 1 j j′ ≤ j 1 1 2} q We already saw in the proof of Proposition 6.28 that S is a centrally symmetric and convex set of Lebesgue measure

r1 r2 r1 r1+r2 µ(S)= (2A ) (πB2) = 2r1 πr2 A A2 = 2r1 πr2 M > 2r1+r2 v(Φ(Z )). i · j i j K Yi=1 jY=1 Yi=1 j=Yr1+1 Therefore by Corollary 6.20 there exists aA Z such that Φ(aA) S. That means that ∈ K ∈ σ (aA) A for all i = 1,...,r + r | i |≤ i 1 2 Now we will play around with these inequalities. First note that

n r1 r1+r2 r1 r1+r2 2 2 NK/Q(aA) = σi(aA) = σi(aA) σi(aA) Ai Ai = M. (6.10) | | 1 | | | | | | ≤ Yi= Yi=1 i=Yr1+1 Yi=1 i=Yr1+1 To complete the main step, we need to check that Equation (6.9) holds for a = aA. On the one hand, since aA Z , its norm satisﬁes N (aA) 1, and on the other hand, ∈ K | K/Q | ≥ since aA S, we have that ∈ 1 1 − − 1 σ (aA) = N (aA) σ (aA) 1 σ (aA) A M − | i | | K/Q |·  | j | ≥ ·  | j | ≥ i j=i j=i Y6 Y6     Therefore we have, for all i = 1,...,n,

1 A M − σ (aA) A i ≤| i |≤ i

We now take logarithms in this equation (recall that all Ai are positive numbers)

log A log M log σ (aA) log A i − ≤ | i |≤ i Multiplying by 1 and summing log A we obtain that, for all i = 1,...,n, − i 0 log A log σ (a) log M. ≤ i − | i |≤ r1+r2 1 r1+r2 1 Now we can estimate the difference between − c log σ (aA) and − c log A as i=1 i | i | i=1 i i follows: P P

6 GEOMETRY OF NUMBERS 51

This completes the main step. r1+r2 1 A Let M1 > i=1 − ci log M. Now we will apply the main step to the following tuples : | (|m) (m) r1+r2 1 (m) For each m N, choose A1 ,...,Ar1+r2 1 > 0 such that i=1 − ci log Ai = 2mM1, and ∈ P − set A(m) := (A(m),...,A(m) ). Then (by the main step) there exists a Z satisfying that 1 r1+r2 1 P m ∈ K N (a ) M and Equation− (6.9), that is to say, | K/Q m |≤

r1+r2 1 − ci log σi(am) 2M1m M1. | |− ≤ i=1 X

Therefore we have that

r1+r2 1 − c log σ (a ) ((2m 1)M , (2m + 1)M ). i | i m | ∈ − 1 1 Xi=1 r1+r2 1 This implies that the sequence of numbers i=1 − ci log σi(am) m N is strictly increasing. { | |} ∈ But, on the other hand, the principal ideals a Z have all norm bounded by M, and we know P m K that there are only a ﬁnite number of integral ideals with bounded norm (see the proof of Theorem 6.30). Therefore there exist m = m such that a Z = a Z . Hence there is a unit u Z× 1 6 2 m1 K m2 K ∈ K such that am1 = uam2 , and

r1+r2 1 r1+r2 1 − − c log σ (a ) = c log σ (ua ) = i | i m1 | i | i m2 | Xi=1 Xi=1 r1+r2 1 r1+r2 1 − − c log σ (u) + c log σ (a ) , i | i | i | i m2 | Xi=1 Xi=1 thus

r1+r2 1 r1+r2 1 r1+r2 1 − − − c log σ (u) = c log σ (a ) c log σ (a ) = 0. i | i | i | i m1 |− i | i m2 | 6 Xi=1 Xi=1 Xi=1 This shows that u W , and concludes the proof of Theorem 6.38 6∈ 0

Definition 6.48. Let K be a number field of degree n = r1 + 2r2. r1+r2 1 We will say that a tuple (ξ1,...,ξr1+r2 1) (Z× ) − is a fundamental system of units if, for − ∈ K all u Z× there exist a root of unity µ ZK and n1,...,nr1+r2 1 Z such that ∈ K ∈ − ∈ n1 nr1+r2−1 u = µ ξ1 ξr1+r2 1 . · ····· − To finish this section we will see how Dirichlet Unit Theorem applies to the case of real quadratic fields, allowing a complete description of the solutions of the Pell equation considered in Example 6.1. Let d Z be a squarefree, positive number, and let K = Q(√d). For the rest of the section, fix ∈ .an embedding K ֒ R. We have that n := [K : Q] = 2, and, since K R, r = 0 and r = 2 → ⊂ 2 1 Therefore r + r 1 = 1, and from Dirichlet Unit Theorem we obtain: 1 2 − 6 GEOMETRY OF NUMBERS 52

Corollary 6.49. Let K be a real quadratic field. Then Z× µ Z. K ≃ K × 2πir Note that the only roots of unity in R are 1 (since the m-th roots of unity in C are e m , r = ± 1,...,m, and of these only 1 are real). In particular, since K R, the only roots of unity of K are ± ⊂ 1. Hence ± Z× 1 Z. K ≃ {± }× 1 1 For each z ZK× , we have that z,z− , z− also belong to ZK× . Assume that z > 0 (otherwise, ∈ 1 − − 1 interchange z and z). Then z− > 0, z, z− < 0. Moreover, if z = 1, one of the two numbers 1 − − − 6 1 z,z− must be greater than 1, the other smaller than 1. Interchanging z and z− if necessary, we can assume z > 1. Then 1 1 z > 1 >z− > 0 > z− > 1 > z. − − − If we consider only the units which are 0 , then these form a group isomorphic to Z, say ZK,>× 0. 1 ≥ There are two elements z,z Z× that generate the group (those corresponding to 1 Z). − ∈ K,>0 ± ∈ The neutral element in Z, which is 0, corresponds to the neutral element of ZK,>× 0, which is 1, so z = 1, and therefore one of the two numbers z,z 1 R is greater than 1, and the other smaller than 6 − ∈ 1. Denote by ZK,>1 the units that are > 1. We call the fundamental unit of ZK the generator of ZK,>× 0 that belongs to ZK,>1 (note that this terminology differs slightly from Definition 6.48, and note also that it depends on our choice of embedding K R). Thus in order to find all units of Z , it is enough ⊂ K to find the fundamental unit z = a + b √d Z× ; then 1 1 1 ∈ K,>1 m Z× = (a + b √d) : m Z K {± 1 1 ∈ } m Z× = (a + b √d) : m Z K,>0 { 1 1 ∈ } m Z× = (a + b √d) : m N K,>1 { 1 1 ∈ } Note that, since N (z ) = (a + b √d)(a b √d)= 1, K/Q 1 1 1 1 − 1 ± 1 1 1 1 either z− = a b √d (and z− = a +b √d), or z− = a +b √d (and z− = a b √d). 1 1 − 1 − 1 − 1 1 1 − 1 1 − 1 1 − 1 We have

1 1 z ,z− , z , z− = a + b √d,a b √d, a + b √d, a b √d . { 1 1 − 1 − 1 } { 1 1 1 − 1 − 1 1 − 1 − 1 } Of these four numbers the biggest is a + b √d. Therefore we conclude that a , b 0, and the | 1| | 1| 1 1 ≥ equation 1= a2 b2d, together with the fact that z = 0, implies that b > 0. ± 1 − 1 1 6 1 Call zm = am + bm√d, then

am+1 := ama1 + dbmb1

(bm+1 := amb1 + a1bm

2 Note that the sequence bm m N is increasing. Hence b1 := min b N : a N such that a { } ∈ { ∈ ∃ ∈ − db2 = 1 . In this way one can explicitly ﬁnd the fundamental unit z . ± } 1 We now distinguish two cases:

d 2, 3 (mod 4). Then Z = Z[√d]. • ≡ K There are two possibilities: 6 GEOMETRY OF NUMBERS 53

– If N (z ) = 1, then the equation a2 db2 = 1 does not have solutions in Z, and K/Q 1 − − 2 2 a + b√d Z× a db = 1. ∈ K ⇔ − – If N (z )= 1, then K/Q 1 − 2 2 a + b√d Z× a db = 1. ∈ K ⇔ − ± 2 2 and the subgroup 1,z Z× corresponds to the solutions of a db = 1. h− 2i⊂ K − d 1 (mod 4). Then Z = Z[ 1+√d ]. • ≡ K 2 We can write ZK as

1+ √d 1 1 a + b : a, b Z = x + y√d : x, y Z and y x 0 (mod 2) 2 ∈ 2 2 ∈ − ≡ ( )

1 2 2 If we have (x + y√d) Z× , then it holds that x dy = 4. 2 ∈ K − ± 1 m Let z = (x + y √d) be the fundamental unit of Z . Then Z× = z : m N . Call 1 2 1 1 K K,>1 { 1 ∈ } 1 √ m zm = 2 (xm + ym d) := z1 .

– If N (z ) = 4, then the equation x2 dy2 = 4 does not have solutions in Z, and K/Q 1 − −

1 2 2 (x + y√d) Z× x dy = 4. 2 ∈ K ⇔ − – If N (z )= 4, then K/Q 1 −

1 2 2 (x + y√d) Z× x dy = 4. 2 ∈ K ⇔ − ± 2 2 and the subgroup 1,z Z× corresponds to the solutions of x dy = 4. h− 2i⊂ K − But we are interested in the solutions of x2 dy2 = 1. There are two possibilities: − ± – If x and y are both even, then calling x = 1 x and y = 1 y , we have that (x )2 1 1 1′ 2 1 1′ 2 1 1′ − d(y )2 = 1, and all solutions of x2 dy2 = 1 are obtained as 1′ ± − ± x := xm m′ ± 2 ym (y′ := m ± 2 Taking the sign into account, we obtain: If N (z ) = 4, then ∗ K/Q 1 2 2 (x′) d(y′) = 1 x′ + y′√d 1,z Z× . − ⇔ ∈ h− 1i⊂ K If N (z )= 4, then ∗ K/Q 1 − 2 2 (x′) d(y′) = 1 x′ + y′√d 1,z Z× . − ⇔ ∈ h− 2i⊂ K 6 GEOMETRY OF NUMBERS 54

– If x1 and y1 are odd, then

1 1 x2 + y2d 2x y 1 x2 + y2d z = (x + y √d)2 = ( 1 1 + 1 1 √d)= ( 1 1 + x y √d) 2 22 1 1 2 2 2 2 2 1 1

2 2 Note that, since d 1 (mod 4), x1 + y1d is divisible once and only once by 2, hence 2 2 ≡ x1+y1 d x2 = 2 and y2 = x1y1 are both odd.

1 1 z = (x + y √d)3 = (x3 + 3x y2d + (3x2y + y3d)√d)= 3 23 1 1 8 1 1 1 1 1 1 1 (x (x2 + 3y2d)+ y (3x2 + y2d)√d) 8 1 1 1 1 1 1

Now both x2 +3y2d = ( 4+y2d)+3y2d = 4( 1+y d) and 3x2 +y2d = 3x2 +( 4+ 1 1 ± 1 1 ± 1 1 1 1 ± x2) = 4(x2 1) are divisible by 8, hence x , y are both even, and x = x3 and y = y3 1 1 ± 3 3 3′ 2 3′ 2 is a solution of x2 dy2 = 1. In this case, the solutions of x2 dy2 = 1 are given by − ± − ±

x := x3m m′ ± 2 y3m (y′ := m ± 2 Taking the sign into account, we obtain: If N (z ) = 4, then ∗ K/Q 1 2 2 (x′) d(y′) = 1 x′ + y′√d 1,z Z× . − ⇔ ∈ h− 3i⊂ K If N (z )= 4, then ∗ K/Q 1 − 2 2 (x′) d(y′) = 1 x′ + y′√d 1,z Z× . − ⇔ ∈ h− 6i⊂ K Remark 6.50. The smallest solution to the Problem of the Cattle of the Sun (see Example 6.1 and Sheet 8) has 206545 digits (in base ten). Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 1 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 20/02/2012

3 1. Let ζ = e2πi/3 = 1 + i √ C. Consider A := Z[ζ] = a + ζb a, b Z . Show the following − 2 2 ∈ { | ∈ } statements:

(a) ζ is a root of the irreducible polynomial X2 + X + 1 Z[X]. ∈ (b) The ﬁeld of fractions of A is Q(√ 3) = a + b√ 3 a, b Q C. − { − | ∈ }⊂ (c) The norm map N : Q(√ 3) Q, given by − → 2 2 a + b√ 3 a + 3b = (a + b√ 3)(a b√ 3) = (a + b√ 3)(a + b√ 3) − 7→ − − − − − is multiplicative and sends any element in A to an element in Z. In particular, u A is a unit ∈ (i.e. in A ) if and only if N(u) 1, 1 . Moreover, if N(a) is a prime number, then a is × ∈ { − } ± irreducible. (d) The unit group A is equal to 1, ζ, ζ2 and is cyclic of order 6. × {± ± ± } (e) The ring A is Euclidean with respect to the norm N and is, hence, by a theorem from last term’s lecture, a unique factorisation domain. Hint: Consider the lattice in C spanned by 1 and ζ. Compute (or bound from above) the maximum distance between any point in C and the closest lattice point. Use this to show that a division with remainder exists. (f) The element λ = 1 ζ is a prime element in A and 3= ζ2λ2. − − (g) The quotient A/(λ) is isomorphic to F3. (h) The image of the set A3 = a3 a A under π : A A/(λ4) = A/(9) is equal to { | ∈ } → 0 + (λ4), 1 + (λ4), λ3 + (λ4) . { ± ± } 2. Show that A (from the previous exercise) is the ring of integers of Q(√ 3). − We recommend reading Simon Singh’s novel (not a textbook!) on Fermat’s Last Theorem in order to know how the story continues after the cases n = 2, 3, 4 treated in the lecture. Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 2 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 27/02/2012

1. (a) Show that there exist infinitely many prime numbers p 1 mod 3. ≡− Hint: Imitate Euclid’s proof of the infinitude of the number of primes. (You don’t need any commutative algebra here.) (b) Let a,n N with n 2 such that an 1 is a prime number. Show that a = 2 and n is a prime ∈ ≥ − number. Such primes are called Mersenne primes. 2. (a) Let 0, 1 = d Z be a squarefree integer and let K = Q(√d). It is a quadratic field extension 6 ∈ of Q. For a general element x = a + b√d with a, b Q compute Tr (x) and Norm (x). ∈ K/Q K/Q (b) Let α = √3 2 and let K = Q(α). It is a cubic field extension of Q. For a general element x = a + bα + cα2 with a,b,c Q compute Tr (x) and Norm (x). ∈ K/Q K/Q Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 3 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 05/03/2012

1. (a) Let 0, 1 = d Z be a squarefree integer and consider K = Q(√d). Show: 6 ∈ 1+ √d disc(1, √d) = 4d and disc(1, )= d. 2

(b) Let α = √3 2 and let K = Q(α). It is a cubic ﬁeld extension of Q with Q-basis 1,α,α2. Compute disc(1,α,α2).

2. (a) Let L/K a finite separable field extension, α1,...,αn a K-basis of L and C = (ci,j)1≤i,j≤n an n n-matrix with coefficients in K. We view C as a K-linear map L L via the fixed choice of × → basis, and put βi := C(αi) for i = 1,...,n. 2 Then disc(β1,...,βn) = det(C) disc(α1,...,αn).

(b) Let L/K be a ﬁnite separable ﬁeld extension of degree n =[L : K] and denote by σ1,...,σn the K-homomorphisms L K. Assume L = K(a) for some a L. Show: → ∈ n−1 2 disc(1,a,...,a )= Y σj(a) σi(a) . − 1≤i

Hint: One obtains a Vandermonde determinant. Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 4 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 12/03/2012

1. Find an integral ring extension Z ⊆ S such that S is not free as Z-module.

2. Snake lemma. Let R be a ring, let Mi,Ni for i = 1, 2, 3 be R-modules, and let φi : Mi → Ni be R-module homomorphisms such that the diagram

/ / / / 0 / M1 / M2 / M3 / 0

φ1 φ2 φ3 / / / / 0 / N1 / N2 / N3 / 0

is commutative and has exact rows. Show that there is an exact sequence

δ 0 → ker(φ1) → ker(φ2) → ker(φ3) −→ coker(φ1) → coker(φ2) → coker(φ3) → 0.

(The cokernel of a homomorphism α : M → N is deﬁned as N/ im(α).)

3. Let K be a number field and ZK its ring of integers. Let 0 =6 a ⊆ b ⊂ K be two ZK -modules. Show that the index (b : a) is finite and satisfies

2 disc(a) = (b : a) disc(b). Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 5 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 19/03/2012

1. Let K be a number ﬁeld and {0}= 6 O ⊆ ZK be a subring. Show that the following statements are equivalent:

(i) O is an order of K. (ii) Frac(O)= K.

2. Let R be an integral domain and K = Frac(R). Let I, J ⊂ K be fractional R-ideals. Show that the following sets are fractional R-ideals of R.

(a) I + J = {x + y | x ∈ I, y ∈ J}, n N (b) IJ = {Pi=1 xiyj | n ∈ ,x1,...,xn ∈ I, y1,...,yn ∈ J}, (c) In = I · I · ... · I, | n times{z } (d) I ∩ J, (e) (I : J).

3. Let R be an integral domain and H,I,J ⊂ K fractional R-ideals. Show that the following properties hold:

(a) IJ ⊆ I ∩ J (assume here that I and J are integral ideals), (b) H + (I + J) = (H + I)+ J = H + I + J, (c) H(IJ) = (HI)J, (d) H(I + J)= HI + HJ. Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 6 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 26/03/2012

1. Let R be a ring. Show that R is a principal ideal domain if and only if R is a Dedekind ring with Pic(R) = 0. Hint: It sufﬁces to combine propositions and theorems from the lecture. 2. Consider the ring R = Z[√ 61]. Show that (2, 3+ √ 61) and (5, 3+ √ 61) are invertible ideals − − − in R and determine their order in Pic(R). 3. Consider the ring R = Z[√ 19]. Use for this exercise that Pic(R) is a ﬁnite group of order 3. − Determine all integral solutions of the equation x2 +19 = y5. Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 7 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 16/04/2012

1. Let R be a Noetherian integral domain of Krull dimension 1 and (0) 6= I ¢ R be an ideal. For a

maximal ideal m of R, let I(m) := Im ∩ R be the m-primary part of I.

Show that I has a primary decomposition, i.e. I = Tm⊇I I(m). 2. Let K be a ﬁeld. A subring R ⊆ K is called a valuation ring of K if for each x ∈ K× we have x ∈ R or x−1 ∈ R.

(a) Show that every valuation ring of K is a local ring. (b) Show that every valuation ring of K is integrally closed.

3. (a) A local integral domain R is called a discrete valuation ring if there is π ∈ R such that all non-zero ideals of R are of the form (πn) for some n ∈ N. Let R be a discrete valuation ring and K its field of fractions. Denote by ord(r) for r ∈ R \ {0} the maximum integer n such that r ∈ (πn). r × For x = s ∈ K (with r ∈ R and s ∈ R \ {0}), let ord(x) := ord(r) − ord(s). Finally, let ord(0) := ∞. Show that the map v : K → Z, x 7→ ord(x) (1) satisfies v(1) = 0 v(xy)= v(x)+ v(y) for all x, y ∈ K. (2) v(x + y)) ≥ min(v(x),v(y)) for all x, y ∈ K. The map v is called a discrete valuation. (b) Let K be a field together with a discrete valuation v as in (1) satisfying the three statements in (2). Show that Rv := {x ∈ K | v(x) ≥ 0} is a discrete valuation ring. What is its maximal ideal? (c) Show that every discrete valuation ring is a valuation ring. Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 8 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 23/04/2012

1. First exercise. The problem of the cattle of the Sun From Archimedes to Eratóstenes of Cirene:

“If thou art diligent and wise, O stranger, compute the number of cattle of the Sun, who once upon a time grazed on the fields of the Thrinacian isle of Sicily, divided into four herds of different colours, one milk white, another a glossy black, a third yellow and the last dappled. In each herd were bulls, mighty in number according to these proportions: Understand, stranger, that the white bulls were equal to a half and a third of the black together with the whole of the yellow, while the black were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow. Observe further that the remaining bulls, the dappled, were equal to a sixth part of the white and a seventh, together with all of the yellow. These were the proportions of the cows: The white cows were precisely equal to the third part and a fourth of the whole herd of the black; while the black cows were equal to the fourth part once more and with it a fifth part of the dappled, when all, including the bulls, went to pasture together. Now the dappled cows were equal in number to a fifth part and a sixth of the yellow herd. Finally the yellow cows were in number equal to a sixth part and a seventh of the white herd. If thou canst accurately tell, O stranger, the number of cattle of the Sun, giving separately the number of well-fed bulls and again the number of females according to each colour, thou wouldst not be called unskilled or ignorant of numbers, but not yet shalt thou be numbered among the wise. But come, understand also all these conditions regarding the cattle of the Sun. When the white bulls mingled their number with the black, they stood firm, equal in depth and breadth, and the plains of Thrinacia, stretching far in all ways, were filled with their multitude. Again, when the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular figure, there being no bulls of other colours in their midst nor none of them lacking. If thou art able, O stranger, to find out all these things and gather them together in your mind, giving all the relations, thou shalt depart crowned with glory and knowing that thou hast been adjudged.”

(a) Let Wb, Bb,Yb,Db (resp. Wc, Bc,Dc,Yc) the number of white, black, yellow and dappled bulls (resp. cows). Write out the seven equations indicated in the ﬁrst part of the problem that relate these quantities. (b) Check (using a computer!) that the solutions of the system of the previous paragraph in terms of Yb is given by 543694 Bc := Yb 178 461043 Bb := Yb 99 2402120 Wc := Yb 742 1383129 Wb := Yb 297 604357 Yc := Yb 1580 461043 Db := Yb 891 106540 Dc := Yb 125739 (c) Observe that the system has more than one solution (one for each value of Yb). On the other hand, not every solution of the system is a solution of the problem, since the number of bulls and the number of cows must be integers! Write out an inﬁnite family of integer solutions of the problem, depending on a parameter t that takes values in Z.

If you have done the exercise so far, you are not unskilled or ignorant of numbers. But you have not yet proved that you are wise!

(d) The two conditions in the second paragraph of the problem involve polynomial equations of degree 2. Write out what these extra conditions look like for a generic member of your inﬁnite family. (Hint: Triangular numbers are those of the form x(x + 1)/2) (e) Substituting one equation into the other one, merge your two equations into one equation of the form Au2 = Bv(v + 1) for some A, B ∈ Z and some variables u and v. Using the equality 1 2 1 2 2 v(v + 1) = (v + 2 ) − 4 you can rewrite your equation as A(2u) = B((2v + 1) − 1). (f) Making an adequate change of variables, rewrite your equation as x2 = dy2 + 1 for some integer d, in such a way that integer solutions (x, y) ∈ Z allow you to ﬁnd an integer number of bulls satisfying the conditions.

We have not yet solved the problem! But we have transformed it into an equation whose integer solutions will be studied in the rest of the semester. Hopefully at the end of it we will be counted among the wise...

The main difﬁculty of the problem, as you have seen, is that we do not want any solution, but only those that are natural numbers. This kind of question goes back to Diophantus of Alexandria, who wrote a book called Arithmetica, which consists of a list of problems of the kind: ﬁnd an integer solution to the following (set of) algebraic equations. The solutions that he gives are highly subtle and clever. It is in the margin of his copy of the Arithmetica where Fermat wrote his famous Last Theorem (and many other theorems). If you cannot wait for the end of the semester to learn how to solve the problem of the cattle of the sun, you can check the references

• Lenstra, Hendrik W., Jr. “Solving the Pell equation. Algorithmic number theory: lattices, number ﬁelds, curves and cryptography”, 1–23, Math. Sci. Res. Inst. Publ., 44, Cambridge Univ. Press, Cambridge, 2008.

• http://www.math.nyu.edu/ ˜ crorres/Archimedes/Cattle/Statement.html Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 9 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 30/04/2012

n n 1. Let v R be a nonzero vector. Prove that v, √2v Z is not a discrete subgroup of R . ∈ h i Hint: √2 can be approximated by rational numbers with as much precision as you like. 2. Let H Rn be a lattice and P be a fundamental domain. ⊂ (a) Prove that for each v Rn there exists a unique u P such that v u H. ∈ ∈ − ∈ n (b) Prove that R is the disjoint union of the family P + u ∈ . { }u H 3. Let K be a number field, a Z a nonzero integral ideal. We define the norm of a as N(a)=[Z : a]. ⊂ K K (a) Let x Z different from 0. Prove that the norm of the principal ideal (x) equals N (x). ∈ K K/Q Hint: Consider a Z-basis of Z , say y1,...,y , such that there exist λ1,...,λ Z with K { n} n ∈ λ1y1,...,λ y a Z-basis of the Z-submodule xZ Z . Relate the map T : Z Z { n n} K ⊂ K x K → K defined by T (z)= xz with the map f : Z Z defined by y λ y for i = 1,...,n. x K → K i 7→ i i (b) Prove that if m is a maximal ideal of Z , then N(a m)= N(a) N(m). K · · Hint: Call k = Z /m; show that a/(a m) is a k-vector space of dimension one, and hence K · isomorphic to ZK /m as k-vector spaces. (c) Let b Z be another nonzero integral ideal. Prove that N(a b)= N(a) N(b). ⊂ K · · (d) Let I K be a nonzer fractional ideal. We define the norm of I as N(I)= N(xI)/ N (x) , ⊂ | K/Q | where x Z is some element such that xI is a nonzero integral ideal. Show that the norm of a ∈ K fractional ideal is well-defined. (e) Show that N : (Z ) Q× is a group homomorphism. I K → Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 10 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 07/05/2012

1. Let K be a number ﬁeld and let a ⊂ ZK be a nonzero integral ideal. Prove that there exists a ∈ a different from zero such that 4 r2 n! |N (a)|≤ |disc(Z )|N(a). K/Q π nn K p Hints: Use (without proving them) the following results:

• Let r1,r2 ∈ N, n = r1 + 2r2. For each t ∈ R, deﬁne the set

r1 r2 ′ ′ n 2 ′2 St := {(x1,...,xr1 , y1, y 1,...,yr2 , y r2 ) ∈ R : |xi| + 2 yj + y j ≤ t}. Xi=1 Xj=1 q

Then r2 n r1 π t µ(St) = 2 2 n! ′ ′ n • (Arithmetic Mean-Geometric Mean inequality): For all (x1,...,xr1 , y1, y 1,...,yr2 , y r2 ) ∈ R , it holds that

1 n r1 r2 r1 r2 2 ′2 1 2 ′2  |xi|· (y + y ) ≤  |xi| + 2 y + y  j j n j j Yi=1 jY=1 Xi=1 Xj=1 q    

2. Let K be a number ﬁeld different from Q. Prove that disc(ZK ) > 1. In particular, there exists a rational prime p such that p|disc(ZK ). Hints:

• Use Exercise 1. • Use that π < 4 and π2 > 8. • You may want to prove, as an auxiliary lemma, that the function f : N → R deﬁned by π n nn f(n) := 4 n! is strictly increasing. Exercises in Algebraic Number Theory Summer Term 2012

Université du Luxembourg Sheet 11 Prof. Dr. Gabor Wiese Dr. Sara Arias-de-Reyna 14/05/2012

1. Let K = Q(√ 5). Prove that CL(K) has order 2. − 2. Let K = Q(√ 19). Prove that CL(K)= 1 . − { } 3. Let K be a number ﬁeld, α ZK and f(X) Z[X] the minimal polynomial of α, and let A = Z[α]. ∈ ∈ Let p Z be a prime number, let f(X) Fp[X] be the reduction of f(X) (mod p), and let ∈ ∈ r

f(X)= qi(X) Yi=1

be a factorisation of f(X) into irreducible polynomials in Fp[X] with leading coefﬁcient 1. For each i = 1,...,r, choose qi(X) Z[X] reducing to q (x) (mod p). Then the prime ideals in A above (p) ∈ i are given by pi := (p, qi(α))A, i = 1,...,r.

4. In this exercise we will complete the study of the integral solutions of x2 +19 = y5 that we started in Exercise 3 of Sheet 6.

(a) Show that the map Φ : Q(√ 19) R2 (Deﬁnition 7.1 of the Lecture notes) maps Z[√ 19] − → − into a lattice H = Φ(Z[√ 19]) of volume v(H)= √19. − (b) Knowing that Pic(Z[√ 19]) is generated by the classes of invertible prime integral ideals of − norm less than or equal to 2 √4 19 < 6 π · (which, if you like, you can check by following the proofs of Proposition 8.1 and Lemma 8.4 of the Lecture notes, and adapt them to this case), and knowing that the ideal (2, 1+ √ 19) is not − invertible, prove that Pic(Z[√ 19]) is a group of order 3. − Hints: Use Exercise 2 above. • To prove that the class of a nonprincipal ideal I has order 3, it sufﬁces to prove that I3 is • principal (because then I2 cannot be principal).