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Math 6370: Algebraic Number Theory

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Math 6370: Algebraic Number Theory Math 6370: Algebraic Number Theory Taught by David Zywina Notes by David Mehrle [email protected] Cornell University Spring 2018 Last updated May 13, 2018. The latest version is online here. Contents 1 Introduction................................. 5 1.1 Administrivia............................. 8 2 Algebraic integers.............................. 9 2.1 Trace and Norm............................. 11 2.2 Complex embeddings ........................ 13 2.3 OK as an abelian group........................ 16 2.4 Discriminants............................. 18 2.5 Example: Cyclotomic Integers.................... 25 3 The ideal class group............................ 28 3.1 Unique Factorization......................... 29 3.2 Fractional Ideals............................ 34 3.3 Ramification.............................. 40 3.4 Dedekind Domains.......................... 44 3.5 Discrete Valuation Rings....................... 46 3.6 Extensions of number fields. .................... 48 4 Geometry of Numbers (Minkowski Theory)............... 50 4.1 Algebraic integers as a lattice .................... 53 4.2 Minkowski’s Theorem........................ 54 4.3 The class group is finite........................ 57 4.4 Bounds on the Discriminant...................... 61 4.5 Bounds on the Discriminant..................... 66 5 The units of OK ............................... 68 5.1 Dirichlet Unit Theorem and Examples............... 69 5.2 Pell’s Equation.............................. 71 5.3 Proof of the Dirichlet Unit Theorem ................ 73 6 Galois extensions of number fields.................... 77 6.1 Galois Groups............................. 78 6.2 Splitting in Galois Number Fields................... 81 6.3 Frobenius elements and quadratic reciprocity........... 84 6.4 Proof of Quadratic Reciprocity ................... 87 6.5 Chebotarev Density Theorem.................... 90 7 Dedekind Zeta Functions.......................... 94 7.1 Counting Ideals............................ 96 7.2 Dirichlet Series and L-functions................... 99 1 8 Local Fields ................................. 105 8.1 p-adic fields .............................. 105 8.2 How do you write down elements of Kp or Op? . 109 8.3 Extensions of Qp . 110 8.4 Global and Local class field theory . 113 2 Contents by Lecture Lecture 01 on 25 January 2018......................... 5 Lecture 02 on 30 January 2018......................... 8 Lecture 03 on 1 February 2018......................... 12 Lecture 04 on 8 February 2018......................... 16 Lecture 05 on 13 February 2018 ........................ 20 Lecture 06 on 15 February 2018 ........................ 24 Lecture 07 on 22 February 2018 ........................ 29 Lecture 08 on 27 February 2018 ........................ 32 Lecture 09 on 1 March 2018 .......................... 36 Lecture 10 on 6 March 2018 .......................... 40 Lecture 11 on 8 March 2018 .......................... 44 Lecture 12 on 13 March 2018.......................... 47 Lecture 13 on 15 March 2018.......................... 54 Lecture 14 on 20 March 2018.......................... 58 Lecture 15 on 22 March 2018.......................... 62 Lecture 14 on 20 March 2018.......................... 62 Lecture 16 on 27 March 2018.......................... 69 Lecture 17 on 29 March 2018.......................... 73 Lecture 18 on 10 April 2018 .......................... 77 Lecture 19 on 12 April 2018 .......................... 82 Lecture 20 on 17 April 2018 .......................... 86 Lecture 21 on 19 April 2018 .......................... 90 Lecture 22 on 24 April 2018 .......................... 93 3 Lecture 23 on 26 April 2018 .......................... 99 Lecture 24 on 03 May 2018........................... 105 Lecture 25 on 08 May 2018........................... 110 4 Lecture 01: Introduction 25 January 2018 1 Introduction The main objects of algebraic number theory are number fields. Definition 1.1. A number field is an extension field of Q of finite degree, i.e. K ⊇ Q with [K: Q] = dimQ K < . p p p Example 1.2. Q, Q( 2), Q( -3), Q( 3 5). 1 Theorem 1.3 (Primitive Element). For any number field K, K = Q(α) for some α. In number theory, we study the integers Z ⊆ Q. The integers are nice because we can factor any element into a product of primes. When we work over other number fields, we want something analogous: what do the integers look like in a number field? Let K be a finite field extension of Q of degree n. Take any α 2 K. Define a homomorphism of Q-algebras Q[x] K x α Since Q[x] is a principal ideal domain (PID), the kernel of this map is gener- ated by a single element, say hpα(x)i. The polynomial pα(x) is the minimal polynomial of α, a monic polynomial in Q[x] of minimal degree with root α. Q[x] We write the image of in K as Q[α] to denote the Q-algebra hpα(x)i generated by α; it is a field. We often write Q(α) = Q[α]. Definition 1.4 (Working definition). We say that α 2 K is an algebraic integer if pα(x) has coefficients in Z. Definition 1.5. Let OK be the set of algebraic integers in K. This is the ring of algebraic integers of K. We will prove this later that OK is a ring. This fact isn’t obvious. Example 1.6. If K = Q, and α 2 Q, then pα(x) = (x - α). The polynomial pα(x) has integer coefficients if and only if α 2 Z. Therefore, OQ = Z p p Example 1.7. Let K = Q( -2) ⊆ C. Elements of this field are a + b -2 for a, b rational. p Let α = a + b -2. If b = 0, then α 2 Q and pα(x) = (x - a). So α 2 OK in this case precisely when a 2 Z. If b 6= 0, then p p 2 2 2 pα(x) = (x -(a + b -2))(x -(a - b -2)) = x - 2ax + (a + 2b ). 5 Lecture 01: Introduction 25 January 2018 2 2 In this case, α 2 OK if and only if 2a 2 Z and a + 2b 2 Z. Exercise: this is equivalent to a, b 2 Z. p p Therefore, OK = Z[ -2] = a + b -2 j a, b 2 Z . p p If we had replaced -2 by -3 in the previous example, the algebraic inte- gersp would be slightly larger than the na¨ıve thing we might expect. Moreover, Z[ -2], like Z, is a unique factorization domain (UFD), but this isn’t always the case. Number fields are useful to solve Diophantine equations, as the next example shows. Example 1.8. Find all solutions of y2 = x3 - 2 with x, y 2 Z. Right away, we can see a few solutions for small numbers: (3, ±5), but there aren’t other obvious ones. Are these the only ones? p To solve this, work in the larger ring Z[ -2]. We may factor this equation p p x3 = y2 + 2 = (y + -2)(y - -2). p p x Z[ -2] x = u πe1 ··· πer u 2 Z[ -2]× = f±1g We mayp factor in as 1 1 r with i , πi 2 Z[ -2] irreducible, and ei ≥ 0. Assume moreover that πi 6= ±πj for i 6= j. When we substitute this expression into the equation above, p p 3 3e1 3er 3 ui π1 ··· πr = x = (y + -2)(y - -2). p p p Claim 1.9. y + -2 and y - -2 are relatively prime in Z[ -2], i.e. the only common divisors are units. Proof.pTo prove thisp claim, takep an irreduciblep π dividing both. Then π divides 3 (y + -2)-(p y - -2) = 2 -2 = -(p -2) . We may assumep by uniquenessp thatp π = -2. Thenp π divides y + -2 implies that y + -2 = -2(a + b -2) = -2b + a -2 for some a, b 2 Z. Hence, y = -2b, so x3 = 2 + y2 ≡ 2 (mod 4). This is impossible in the integers. So we have demonstrated the claim. 1 ≤ i ≤ r π3ei Nowp we mayp use the claim in the example. For each , i divides y + -2 or y - -2. Therefore, p 3ei y + -2 = u πi i2I Y p p × for somep I ⊆ f1, ::: , rg and u 2 Z[ -2] = f±1g. Therefore, y + -2 is a cube in Z[ -2]. p p y + -2 = (a + b -2)3 for some a, b 2 Z. Expand: p p y + -2 = (a3 - 6ab2) + (3a2b - 2b3) -2 6 Lecture 01: Introduction 25 January 2018 Returning to Z, we find that y = a3 - 6ab2 for integers a, b such that 3a2b - 2b3 = 1. For this second equation, we can factor out b to learn that b(3a2 - 2b2) = 1, but b is an integer, so b = ±1. This reduces the three equations and four unknowns to three equations with three unknowns and two cases. If b = +1, then we have a system of equations y = a3 - 6a 83a2 - 2 = 1 > <>y2 = x3 - 2 > We may solve these first two equations:> to discover that y = ±5, and we can solve for x to see that there are solutions (3, ±5). If b = -1, then 3a2(-1)- 2(-1)3 = 1, or equivalently 3a2 = 1. This has no solutions in the integers. So the only solutions to y2 = x3 - 2 in the integers is (3, ±5). The point of this example is that, even when working over the integers, we are obliged to consider larger rings. 3 3 Example 1.10 (Non-example). One could try the samep thing for y = x - 61. Solve for x3, and then factor the right hand side using -61. p p x3 = y2 + 61 = (y - -61)(y + -61). p Assuming that Z[ -61] is a UFD, one can show that there are no solutions for x, y 2 Z. p However, 82 = 53 - 61; Z[ -61] is not a UFD. Indeed, p p 53 = (8 + -61)(8 - -61) p p and 5, 8 + -61 and 8 - -61 are all three irreducible. How do we cope with the loss of unique factorization? Here’s an idea due to Kummer from around 1846: if unique factorization fails, its because we don’t have enough numbers yet; we should be able to add some “idealp numbers”p and 3 recover unique factorization.p For example, wherep 5 = (8 + -61)(8 - -61), we may have 5 = pq, 8 + -61 = p3 and 8 - -61 = q3.
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