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Algebraic Number Theory

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Algebraic Number Theory Algebraic Number Theory Andrew Kobin Spring 2016 Contents Contents Contents 0 Introduction 1 0.1 Attempting Fermat's Last Theorem . .1 1 Algebraic Number Fields 4 1.1 Integral Extensions of Rings . .4 1.2 Norm and Trace . .5 1.3 The Discriminant . .6 1.4 Factorization of Ideals . 11 1.5 Ramification . 15 1.6 Cyclotomic Fields and Quadratic Reciprocity . 21 1.7 Lattices . 24 1.8 The Class Group . 26 1.9 The Unit Theorem . 31 2 Local Fields 34 2.1 Discrete Valuation Rings . 34 2.2 The p-adic Numbers . 37 2.3 Absolute Values . 42 2.4 Local Fields . 49 2.5 Henselian Fields . 52 2.6 Ramification Theory . 54 2.7 Extensions of Valuations . 61 2.8 Galois Theory of Valuations . 65 2.9 Higher Ramification Groups . 69 2.10 Discriminant and Different . 75 i 0 Introduction 0 Introduction These notes follow a course on algebraic number theory taught by Dr. Andrew Obus at the University of Virginia in Spring 2016. The main topics covered are: Algebraic number fields (the global case) The ideal class group Structure of the unit group The p-adic numbers (the local case) Hensel's Lemma Ramification theory Further topics, including adeles and ideles The main companion for the course is Neukirch's Algebraic Number Theory. Other great ref- erences include Cassels and Frohlich's Algebraic Number Theory, Janusz's Algebraic Number Fields, Lang's Algebraic Number Theory, Marcus's Number Fields and Weil's Basic Number Theory. 0.1 Attempting Fermat's Last Theorem Algebraic number theory was developed primarily as a set of tools for proving Fermat's Last Theorem. We recall the famous (infamous?) theorem here. Fermat's Last Theorem. The equation xn + yn = zn has no solutions in positive integers for n ≥ 3. In attempting to prove the theorem, we first remark that the n = 4 case is elementary; it's just a matter of parametrizing the Pythagorean triples (x; y; z) that solve x2 + y2 = z2 and noticing that not all three can be perfect squares. With this, we can reduce to the case when n = p, an odd prime. There are two cases: Case 1: x; y; z are all relatively prime to p. Case 2: p divides exactly one of x; y; z. We will show a proof for the first few primes in Case 1; the other case uses similar tech- niques. Let ζ be a primitive pth root of unity (e.g. ζ = e2πi=p) and assume Z[ζ] is a unique factorization domain (UFD). This was the classical approach, but number theorists quickly realized that Z[ζ] is not always a UFD. In fact, it is an open question whether there are an infinite number of primes p for which Z[e2πi=p] is a UFD. In any case, the assumption that Z[ζ] is a UFD holds for p < 23 so we will have proven a number of cases of Fermat's Last Theorem with the following proof. 1 0.1 Attempting Fermat's Last Theorem 0 Introduction Proof. Suppose x; y; z are positive integers satisfying xp + yp = zp. We may assume x; y; z are relatively prime in Z. The equation above may be factored as p Y (x + ζiy) = zp (∗) i=1 For p = 3, the only cubes mod 9 are ±1 and 0 so there are no solutions for (*) where 3 - xyz. So we may assume p ≥ 5. We need the following lemmas: p−1 Y Lemma 0.1.1. p = (1 − ζi). i=1 tp−1 Proof. Consider expanding t−1 in two ways: tp − 1 (t − ζ) · (t − ζp−1) = = tp−1 + ::: + t + 1: t − 1 Then plugging in t = 1 gives the result. Lemma 0.1.2. For any 0 ≤ i < j ≤ p − 1, the elements x + ζiy and x + ζjy are coprime in Z[ζ]. Proof. Suppose that π 2 Z[ζ] is a prime which divides x + ζiy and x + ζjy. Then π divides ζiy(1 − ζj−i). Notice that ζi is a unit and p - y by assumption, but 1 − ζj−i j p. So in particular, π j y and thus π j yp. Since π is a prime, π j y or π j p. Repeating the argument for x shows that π j x or π j p. Since x and y are coprime in Z, we cannot have π j x and π j y simultaneously, so π j p. By assumption we have that π divides xp + yp and therefore also zp in Z, but (p; z) = 1 so the Euclidean algorithm implies that π j 1. Therefore x + ζiy and x + ζjy are relatively prime in Z[ζ]. Now, each factor x + ζiy must be a pth power in Z[ζ], possibly multiplied by a unit. Write x + ζy = utp for u 2 Z[ζ]∗ and t 2 Z[ζ]. Lemma 0.1.3. u=u¯ is a pth root of unity. Proof. It is simple to show that u=u¯ and all of its Galois conjugates have modulus 1 in C; this is then true for all powers of u=u¯ as well. Then the degree of u=u¯ and all of its powers is bounded. Since all of these are algebraic integers, there are only finitely many possible choices for their minimal polynomials. Hence the set f(u=u¯)k : k 2 Ng is finite. This proves u=u¯ is a root of unity in Z[ζ]. In particular, (u=u¯)2p = 1 but we want to show it is a pth root of unity. Suppose (u=u¯)p = −1. Then up = −u¯p. Since u 2 Z[ζ] we may write 2 p−2 u = a0 + a1ζ + a2ζ + ::: + ap−2ζ for unique ai 2 Z; this follows from unique factorization in Z[ζ]. Now p p p p u ≡ a0 + a1 + ::: + ap−2 (mod p) ≡ a0 + a1 + ::: + ap−2 (mod p) by Fermat's Little Theorem: 2 0.1 Attempting Fermat's Last Theorem 0 Introduction In particular, up is conjugate to a real number mod p. Likewise, we can write −u¯ as −u¯ = p−1 2 −(a0 + a1ζ + ::: + ap−2ζ ) so p −u¯ ≡ −a0 − a1 − ::: − ap−2 (mod p): p This implies a0 + a1 + ::: + ap−2 ≡ 0 (mod p) so p j u . However, this is impossible if u is a unit. Therefore (u=u¯)p = 1. Putting these results together, we can now write x + ζy = ζjut¯ p ≡ ζju¯t¯p ≡ ζj(x + ζy¯ ) (mod p): Expanding this out gives us x + ζy − ζjx − ζj−1y ≡ 0 (mod p): (∗∗) ∼ p−1 ∼ p−1 Now Z[ζ]=(p) = Z[x]=(p; x + ::: + x + 1) = Fp[x]=(x + ::: + x + 1). Thus the images p−2 p−1 of 1; x; : : : ; x are Fp-linearly independent in this ring. This implies 1; ζ; : : : ; ζ are Z- linearly independent in Z[ζ]=(p). Since x; y 2 Z, the only possibilities in (**) for j are j = 0; 1; 2; p − 1. If p = 0; 2; p − 1, it is easy to simplify (**) and produce a nontrivial ζ2 term, which is impossible. If j = 1, (**) becomes (x − y)(1 − ζ) ≡ 0 (mod p): Qp−1 i Thus i=2 (1 − ζ ) divides x − y but since x − y 2 Z, it must be that p j (x − y). Rearranging the equation xp + yp = zp to read xp + (−z)p = yp and repeating the argument so far shows that p j (x + z) as well. Thus y ≡ x ≡ −z (mod p). But then 0 = xp + yp − zp ≡ 3xp (mod p) which implies p j x, contradicting the assumption that p 6= 3. Therefore no solutions exist to xp + yp = zp for p > 5 such that p - xyz. This proof fails for general primes p in two places: as we mentioned, not every ring Z[e2πi=p] is a UFD; moreover, there can be many more units than just the roots of unity in Z[e2πi=p]. This motivates the study of ideal class groups { which measure how far from being a PID (and a UFD) a ring of integers is { and unit groups in algebraic number theory. 3 1 Algebraic Number Fields 1 Algebraic Number Fields 1.1 Integral Extensions of Rings Let A ⊆ B be rings. Definition. An element x 2 B is integral over A if it is a root of a monic polynomial with coefficients in A. We say B is integral over A if every element of B is integral over A. Definition. The integral closure of A in B is the set of all x 2 B which are integral over A. If A is equal to its integral closure in B then we say A is integrally closed in B. In particular, if A is a domain and B is the fraction field of A then we simply say that A is integrally closed. Lemma 1.1.1. x 2 B is integral over A if and only if A[x] is a finitely generated A-module. n n−1 n Pn−1 i Proof. ( =) ) If x + an−1x + ::: + a0 for ai 2 A then x 2 M := i=1 Ax which is a finitely generated A-module. By induction, for all m ≥ n, xm 2 M. This implies A[x] = M, so in particular A[x] is finitely generated. ( ) = ) Suppose A[x] is generated by f1(x); : : : ; fn(x) where fi are polynomials in a single n variable over A. Let d ≥ maxfdeg figi=1. Then n d X x = aifi(x) i=1 d Pn for some choice of ai 2 A.
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