Algebraic Number Theory Johan Bosman Notes by Florian Bouyer Copyright (C) Bouyer 2011. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.3 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license can be found at http://www.gnu.org/licenses/fdl.html Contents 1 Introduction and Motivations 2 1.1 Motivations . 2 1.2 Finding Integer Solutions . 3 1.3 Pell's Equations . 3 2 Fields, Rings and Modules 5 2.1 Fields . 5 2.2 Rings and Modules . 6 2.3 Ring Extensions . 7 3 Norms, Discriminants and Lattices 9 3.1 Conjugates, Norms and Traces . 9 3.2 Discriminant . 10 3.3 Lattices . 12 4 Cyclotomic Fields 14 5 Dedekind Domains 17 5.1 Euclidean domains . 17 5.2 Dedekind Domain . 18 5.3 Kummer-Dedekind Theorem . 24 6 The Geometry of Numbers 26 6.1 Minkowski's Theorem . 26 6.2 Class Number . 27 6.3 Dirichlet's Unit Theorem . 29 1 1 Introduction and Motivations Most of the ideas in this section will be made more formal and clearer in later sections. 1.1 Motivations Denition 1.1. An element α of C is an algebraic number if it is a root of a non-zero polynomial with rational coecients A number eld is a subeld K of C that has nite degree (as a vector space) over Q. We denote the degree by [K : Q]. Example. • Q p p • Q( 2) = a + b 2 : a; b 2 Q • Q(i) = fa + bi : a; b 2 Qg p 3 • Q( 2) = Q[x]=(x3 − 2) Note that every element of a number eld is an algebraic number and every algebraic number is an element of some number eld. The following is a brief explanation of this. Let K be a number eld, α 2 K. Then Q(α) ⊆ Kand we will late see that [Q(α): Q]j[K : Q] < 1. So there exists a relation between 1; α; : : : ; αn for some n. If α is algebraic then there exists a minimal polynomial f for which α is a root. Q(α) =∼ Q[x]=(f) has degree deg(f) over Q. Consider Z[i] ⊂ Q[i], also called the Gaussian integers. A question we may ask, is what prime number p can be written as the sum of 2 squares? That is p = x2 + y2 = (x + iy)(x − iy), we guess that an odd prime p is x2 + y2 if and only if p ≡ 2 mod 4. A square is always 0 or 1 mod 4, so the sum of two squares is either 0; 1 or 2 mod 4. Hence no number that is 3 mod 4 is the sum of two squares. Therefore not all numbers that are 1 mod 4 can be written as the sum of two squares. Notice that there exist complex conjugation in Z[i], that is the map a+bi 7! a−bi = a + bi is a ring automorph- ism. We can dene the norm map N : Z[i] ! Z by α 7! αα, more explicitly, (a + bi) 7! (a + bi)(a − bi) = a2 + b2. We will later see that N(αβ) = N(α)N(β). Denition 1.2. Let K be a number eld, a element α 2 K is called a unit if it is invertible. That is there exists β 2 K such that αβ = 1. Proposition 1.3. The units of Z[i] are 1; −1; i; −i Proof. Let α 2 Z[i] be a unit. Then N(α) is a unit in Z, (since there exists β 2 Z[i] such that αβ = 1, hence 1 = N(αβ) = N(α)N(β)) Now let α = a + bi, then N(α) = a2 + b2 = ±1. Now −1 is not the sum of two squares hence α 2 {±1; ±ig Denition 1.4. Let K be a number eld, an element α 2 K is irreducible if α is not a unit, and for all β; γ 2 Z[i] with α = βγ, we have either β or γ is a unit. Fact. Z[i] is a unique factorization domain, that is every non-zero elements α 2 Z[i] can be written as a product of irreducible elements in a way that is unique up to ordering and multiplication of irreducible elements by units. Theorem 1.5. If p ≡ 1 mod 4 is a prime then there exists x; y 2 Z such that p = x2 + y2 = (x + iy)(x − iy) = N(x + iy) Proof. First we show that there exists such that 2 . Since we have −1 (see Topics a 2 Z pja + 1 p ≡ 1 mod 4 p = 1 in Number Theory). Let p−1 , then 2 p−1 p−1 p−1 p−1 . Hence a = 2 ! a = 2 ! 2 ! = 1 ····· 2 · 2 ····· 1 ≡ −1 mod p pja2 + 1 = (a + i)(a − i). Is p irreducible in Z[i]? If p were indeed irreducible, then pj(a + i) or pj(a − i). Not possible since a + i = ∗ p(c + di) = pc + pdi means pd = 1. So p must be reducible in Z[i]. Let p = αβ, α; β2 = (Z[i]) and N(p) = p2 = N(α)N(β) ) N(α) 6= ±1 6= N(β). So N(α) = p = N(β). Write α = x + iy, then N(α) = p = x2 + y2 2 1.2 Finding Integer Solutions Problem 1.6. Determine all integer solution of x2 + 1 = y3 Answer. First note x2 + 1 = (x + i)(x − i) = y3, we'll use this to show that if x + i and x − i are coprime then x + i and x − i are cubes in Z[i]. Suppose that they have a common factor, say δ. Then δj(x + i) − (x − i) = 2i = (1 + i)2. So if x + i and x − i are not coprime, then (1 + i)j(x + i), i.e.,(x + i) = (1 + i)(a + bi) = (a − b) + (a + b)i. Now a + b and a − b are either both even or both odd. We also know that a + b = 1, so they must be both odd, hence x is odd. Now an odd square is always 1 mod 8 . Hence x2 + 1 ≡ 2 mod 8, so x2 + 1 is even but not divisible by 8, contradicting the fact that is is a cube. Hence and are coprime in . So let e1 en where are distinct up to units. Now x + i x − i Z[i] x + i = π1 : : : πn πi f e1 en . So e1 en e1 en 3. Let 0 1 fn 3 x − i = x + i = π1 ::: πn (x + i)(x − i) = π1 : : : πn π1 ::: πn = y y = q1 : : : qn ) y = 3f 03 1 3fn . The are some rearrangement of up to units. Hence we have , so unit times q1 : : : qn qi πi; πi ei = 3fj x + i = a cube, (Note in Z[i], ±1 = (±1)3 and ±i = (∓i)3). Hence x + i is a cube in Z[i]. So let x + i = (a + ib)3 for some a; b 2 Z. Then x + i = a3 + 3a2bi − 3ab2 − b3i = a3 − 3ab2 + (3a2b − b3)i. Solving the imaginary part we have 1 = 3a2b − b3 = b(3a2 − b2). So b = ±1 and 3a2 − b2 = 3a2 − 1 = ±1. Now 3a2 = 2 is impossible, so we must have 3a2 = 0, i.e., a = 0 and b = −1. This gives x = a3 − 3ab2 = 0. Hence y = 1; x = 0 is the only integer solution to x2 + 1 = y3 Theorem 1.7 (This is False). The equation x2 + 19 = y3 has no solutions in Z (Not true as x = 18; y = 17 is a solution since 182 + 19 = 324 + 19 = 343 = 172) Proof of False Theorem. This proof is awed as we will explain later on. (Try to nd out where it is awed) p p p Consider . Then we dene the conjugation this time to be p Z[ −19] = fa + b −19 : a; b 2 Z] p p a + b −19 = , and similarly we dene a norm function by . Hence 2 2. a−b −19 p p N : Z[ −19] ! Z α 7! αα N(a+b −19) = a +19b So we have 2 . x + 19 = (x + −19)(x − −19) p p Suppose that these two factors have a common divisor, say δ. Then δj2 −19. Now −19 is irreducible since p p ∗ N( −19) = 19 which is a prime. If 2 = αβ with α; β2 = [ −19 , then N(α)N(β) = N(2) = 22, so N(α) = 2 Z p p p which is impossible. So 2 is also irreducible. Hence we just need to check where 2jx + −19 or −19jx + −19 is possible. p p p p p p Suppose , then , so and . Hence −19jx + −19 x + −19 = −19(a + b −19) = −19b + a −19 a = 1 19jx p 2 2, i.e., 2 is divisible by but not by 2 so it can't be a cube. Suppose , x + 19 ≡p19 mod 19 p x + 19 19 19 2jx + −19 then , which is impossible. x + −19 = 2a + 2pb −19 p p Hence we have and are coprime, and let e1 en . Then x + −19 x − −19 x + −19 = π1 : : : πn x − 19 = p p p f e1 en , so e1 en e1 en 3. If we let 0 1 fn , x + −19 = π1 ::: πn (x + −19)(x − −19) = π1 : : : πn π1 ::: πn = y y = q1 : : : qn 3f then 3 03 1 3fn , so the are some rearrangements of up to units.