# Algebraic Number Theory

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1 Algebraic Number Theory Johan Bosman Notes by Florian Bouyer Copyright (C) Bouyer 011 Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1 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 Contents 1 Introduction and Motivations 11 Motivations 1 Finding Integer Solutions 1 Pell's Equations Fields, Rings and Modules 5 1 Fields 5 Rings and Modules 6 Ring Extensions 7 Norms, Discriminants and Lattices 9 1 Conjugates, Norms and Traces 9 Discriminant 10 Lattices 1 4 Cyclotomic Fields 14 5 Dedekind Domains Euclidean domains 17 5 Dedekind Domain 18 5 Kummer-Dedekind Theorem 4 6 The Geometry of Numbers 6 61 Minkowski's Theorem 6 6 Class Number 7 6 Dirichlet's Unit Theorem 9 1

2 1 Introduction and Motivations Most of the ideas in this section will be made more formal and clearer in later sections 11 Motivations Denition 11 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 Q( ) = { a + b : a, b Q } Q(i) = {a + bi : a, b Q} Q( ) = Q[x]/(x ) 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 K Then Q(α) Kand we will late see that [Q(α) : Q] [K : Q] < So there exists a relation between 1, α,, α n 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 squares? That is p = x + y = (x + iy)(x iy), we guess that an odd prime p is x + y if and only if p mod 4 A square is always 0 or 1 mod 4, so the sum of two squares is either 0, 1 or mod 4 Hence no number that is 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 a bi = a + bi is a ring automorphism We can dene the norm map N : Z[i] Z by α αα, more explicitly, (a + bi) (a + bi)(a bi) = a + b We will later see that N(αβ) = N(α)N(β) Denition 1 Let K be a number eld, a element α K is called a unit if it is invertible That is there exists β K such that αβ = 1 Proposition 1 The units of Z[i] are 1, 1, i, i Proof Let α Z[i] be a unit Then N(α) is a unit in Z, (since there exists β Z[i] such that αβ = 1, hence 1 = N(αβ) = N(α)N(β)) Now let α = a + bi, then N(α) = a + b = ±1 Now 1 is not the sum of two squares hence α {±1, ±i} Denition 14 Let K be a number eld, an element α K is irreducible if α is not a unit, and for all β, γ 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 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 15 If p 1 mod 4 is a prime then there exists x, y Z such that p = x + y = (x + iy)(x iy) = N(x + iy) ( ) Proof First we show that there exists a Z such that p a Since p 1 mod 4 we have p = 1 (see Topics in Number Theory) Let a = p 1!, then a = ( ) ( p 1! p 1 ) (! = 1 p 1 ) ( p 1 ) 1 1 mod p Hence p a + 1 = (a + i)(a i) Is p irreducible in Z[i]? If p were indeed irreducible, then p (a + i) or p (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, β / (Z[i]) and N(p) = p = N(α)N(β) N(α) ±1 N(β) So N(α) = p = N(β) Write α = x + iy, then N(α) = p = x + y

3 1 Finding Integer Solutions Problem 16 Determine all integer solution of x + 1 = y Answer First note x + 1 = (x + i)(x i) = y, 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 δ (x + i) (x i) = i = (1 + i) So if x + i and x i are not coprime, then (1 + i) (x + i), ie,(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 x + 1 mod 8, so x + 1 is even but not divisible by 8, contradicting the fact that is is a cube Hence x + i and x i are coprime in Z[i] So let x + i = ɛπ e1 1 πen n where π i are distinct up to units Now e x i = x + i = ɛπ 1 e 1 π n n So (x + i)(x i) = ɛɛπ e1 1 e πen n π 1 e 1 π n n = y Let y = ɛ q f1 1 qfn n y = ɛ q f1 1 qn fn The q i are some rearrangement of π i, π i up to units Hence we have e i = f j, so x + i =unit times a cube, (Note in Z[i], ±1 = (±1) and ±i = ( i) ) Hence x + i is a cube in Z[i] So let x + i = (a + ib) for some a, b Z Then x + i = a + a bi ab b i = a ab + (a b b )i Solving the imaginary part we have 1 = a b b = b(a b ) So b = ±1 and a b = a 1 = ±1 Now a = is impossible, so we must have a = 0, ie, a = 0 and b = 1 This gives x = a ab = 0 Hence y = 1, x = 0 is the only integer solution to x + 1 = y Theorem 17 (This is False) The equation x + 19 = y has no solutions in Z (Not true as x = 18, y = 17 is a solution since = = 4 = 17 ) Proof of False Theorem This proof is awed as we will explain later on (Try to nd out where it is awed) Consider Z[ 19] = {a + b 19 : a, b Z] Then we dene the conjugation this time to be a + b 19 = a b 19, and similarly we dene a norm function N : Z[ 19] Z by α αα Hence N(a+b 19) = a +19b So we have x + 19 = (x + 19)(x 19) Suppose that these two factors have a common divisor, say δ Then δ 19 Now 19 is irreducible since N( 19) = 19 which is a prime If = αβ with α, β / ( Z[ 19 ), then N(α)N(β) = N() =, so N(α) = which is impossible So is also irreducible Hence we just need to check where x + 19 or 19 x + 19 is possible Suppose 19 x + 19, then x + 19 = 19(a + b 19) = 19b + a 19, so a = 1 and 19 x Hence x mod 19, ie, x + 19 is divisible by 19 but not by 19 so it can't be a cube Suppose x + 19, then x + 19 = a + b 19, which is impossible Hence we have x + 19 and x 19 are coprime, and let x + 19 = ɛπ e1 1 πen x + e 19 = ɛπ 1 e 1 π n n, so (x + 19)(x 19) = ɛɛπ e1 1 e πen n π 1 e 1 π n n then y = ɛ q f1 1 qn fn n Then x 19 = = y If we let y = ɛ q f1 1 qfn n,, so the q i are some rearrangements of π i, π i up to units Hence corresponding e i = f i and so x + 19 =unit times a cube Now units of Z[ 19] = {±1} So x + 19 = (a + b 19) = (a 19ab ) + (a b 19b ) 19 Again comparing 19 coecients we have b(a 19b ) = 1, so b = ±1 and a 19 = ±1 But a = 0 is impossible since 0, and a = 18 = 6 is impossible since 6 is not a square So no solution exists This proof relied on the fact that Z[ 19] is a UFD, which it is not We can see this by considering 4 = 7 = ( )(18 19) Now N(7) = 7 Suppose 7 = αβ with α, β / ( Z[ 19] ) Then N(α)N(β) = 7, so N(α) = 7, but N(a+b 19) = a +19b 7 So 7 is irreducible in Z[ 19] On the other hand N(18+ 19) = 7, and suppose that N(α)N(β) = 7, then without loss of generality N(α) = 7 and N(β) = 7 But we have just seen no elements have N(α) = 7, so is irreducible in Z[ 19] The same argument shows that 18 = 19 is also irreducible in Z[ 19] 1 Pell's Equations Fix d Z >0 with d a for any a Z Then Pell's equation is x dy = 1, with x, y Z Now Z[ d] = {a + b d : a, b Z} This has an automorphism a + b d a b d = a + b d (Note that is just notation, and it does not mean complex conjugation) Again we can dene a function called the norm, N : Z[ d] Z dened by α αα, and explicitly (a + b d) a db Hence Pell's equation comes down to solving N(x + y d) = 1

4 ( Now recall that α Z[, d]) then there exists β such that αβ = 1 So N(α)N(β) = 1, so N(α) = ±1 On the other hand if N(α) = ±1, then αα = ±1, so ±α = α 1, hence α is a unit Example d = Then x y = 1 y + 1 = x y = 0 y + 1 = 1 This is ok, it leads to (1, 0) which correspond to 1 Z[ ] y = 1 y + 1 = 4 This is ok, it leads to (, 1) which gives + Z[ ] y = y + 1 = 1 y = y + 1 = 8 y = 4 y + 1 = 49 This is ok, it leads to (7,4) which gives Z[ ] Note that if ɛ is a unit in Z[ d], then ±ɛ n is a unit for all n Z (For example (+ ) = + + = 7+4 If x, y is a solution, then of course ( x, y) is a solution as well Hence there are innitely many solutions Theorem 18 Let d Z >0 with d a Then there exists ɛ d Z[ d], ɛ d ±1 such that every unit can be written as ±ɛ n d, n Z Such an ɛ d is called a Fundamental Unit of Z[ d] If ɛ d is a fundamental unit, then so is ±ɛ 1 d Proof This is a consequence of Dirichlet's Unit Theorem, which we will prove at the end of the course Example We will show that ɛ = + Z[ ] Let x 1 + y 1 Z[ d] be a fundamental unit Without any lost of generality we can assume that x1 0 Now (x 1 + y 1 ) 1 x = 1 y 1 (x 1+y 1 )(x1 y 1 ) = ±(x 1 y 1 ) So without loss of generality we can also assume y1 0 Put x n + y n = (x1 + y 1 ) = x n 1 + nx n 1 1 y + So x n = x n 1 + x n 1 and y n = nx n 1 1 y 1 If x 1 = 0 then y 1 = ±1 which is not possible Similarly if y 1 = 0 then x 1 = 1 x 1 = ±1 and ɛ = ±1 which is impossible by denition So x 1 1, y 1 1 For n : x n x n 1 x 1 and y n = nx n 1 1 y 1 > ny 1 > y 1 Conclusion: A solution (x, y) of x y = ±1 with y 1 minimal is a Fundamental unit for Z[ ] Hence + is a fundamental unit for Z[ ], so all solution for x + y = ±1 are obtained by (x, y) = (±x n, ±y n ) where x n + y n = ( + ) n 4

5 Fields, Rings and Modules 1 Fields Denition 1 If K is a eld then by a eld extension of K, we mean a eld L that contains K We will denote this by L/K If L/K is a eld extension, then multiplication of K on L denes a K-vector space structure on L The degree [L : K] of L/K is the dimension dim K (L) Example [K : K] = 1 [C : R] = [R : Q] = (uncountably innite) The Tower Law If L/K and M/K are elds extensions with L M, then [M : K] = [M : L][L : K] Proof Let {x α : α I} be a basis for L/K and let {y β : β : J} be a basis for M/L Dene z αβ = x α y β M We claim that {z αβ } is a basis for M/K We show that they are linearly independent If α,β a αβz αβ = 0 with nitely many a αβ K non-zero Then β ( α a αβx α )y β = 0, since the y β are linearly independent over L we have α a αβx α = 0 for all β Since the x α are linearly independent over K we have a αβ = 0 for all α, β We show spanning If z M, then z = λ β y β for λ β L For each λ β = a αβ x α So x = β ( α a αβx α )y β = α,β a αβx α y β = a αβ x αβ So {z αβ } is a basis for M over K, so [M : K] = [M : L][L : K] Corollary If K L M are elds with [M : K] < then [L : K] [M : K] Denition L/K is called nite if [L : K] < If K is a eld and x is an indeterminate variable, then K(x) denotes the eld of rational functions in x with coecients in K That is { } f(x) K(x) = : f, g K[x], g 0 g(x) If L/K is a eld extension, α L Then K(α) is the subeld of L generated by K and α { } f(α) K(α) = : f, g K[x], g(α) 0 = M g(α) K M L,α M Let L/K be a eld extension, α L We say that α is algebraic over K if there exists a non-zero polynomial f K[x] with f(α) = 0 Theorem Let L/K be a eld extension and α L Then α is algebraic over K if and only if K(α)/K is a nite extension Proof ) Let n = [K(α) : K] and consider 1, α,, α n K(α) Notice that there are n + 1 of them, so they must be linearly dependent since the dimension of the vector space is n So there exists a i K such that a 0 + a 1 α + + a n α n = 0 with a i not all zero Hence by denition α is algebraic ) Assume that there exists f 0 K[x] such that f(α) = 0, and assume that f has minimal degree n We claim that f K[x] is irreducible Suppose that f = gh, with g, h non-constant Then 0 = f(α) = g(α)h(α), so without loss of generality g(α) = 0, but deg(g) < deg(f) This is a contradiction Let f = a n x n + + a 0 with a n 0 Then f(α) = 0 a n α n + + a 0 = 0 α n = 1 a n (a n 1 α n a 0 ) So we can reduce any polynomial expression in α of degree n to one of degree { n 1 } Hence K(α) = b0+ +b n 1α n 1 c 0+ +c n 1α : b n 1 i, c i K Pick b(α) c(α) K(α), now deg(c) n 1 < deg f and c(α) 0 Hence gcd(c, f) = 1, so there exists λ, µ K[x] with λ(x)c(x) + µ(x)f(x) = 1 In particular 1 = λ(α)c(α) + µ(α)f(α) = λ(α)c(α), hence λ(α) = 1 c(α) K[α] Any elements of K(α) is a polynomial in α of degree n 1 So if α is algebraic over K, we have just shown that K(α) = K[α] and 1, α,, α n 1 is a basis for K[α]/K, hence [K(α) : K] = n 5

6 Theorem 4 Let L/K be a eld extension, then the set M of all α L that are algebraic over K is a subeld of L containing K Proof First K M, as α K is a root of x α K[x] So take α, β M, we need to show that α β M and α β M if β 0 Consider the subeld K(α, β) L Now [K(α)(β) : K] = [K(α, β) : K(α)][K(α) : K] We have [K(α)(β) : K(α)] [K(β) : K] since the rst one is the degree of the minimal polynomial of β over K(α), and β is algebraic, so there is f K[x] K[α] such that f(β) = 0 Now α β K(α)(β) and if β 0, α β K(α)(β) This implies that K(α β) K(α, β) [K(α β) : ( ) ( ) α α K] [K(α, β) : K] < and K β K(α, β) [K β : K] [K(α, β) : K] < Hence α β and α β are algebraic over K Corollary 5 The set of algebraic number is a eld We denote this with Q For any subeld K C, we let K denote the algebraic closure of K in C, ie, the set of α C that are algebraic over K For example R = C = R(i) We also conclude that Q = K number field K Also [Q : Q] = so Q itself is not a number eld Rings and Modules In this course we use the following convention for rings Every ring R is assumed to be commutative and has 1 We also allow 1 to be 0, in which case R = 0 = {0} A ring homomorphism φ : R S is assumed to send 1 R to 1 S A subring R of a ring S is assumed to satisfy 1 R = 1 S Example Let R 1 and R be two non-zero rings Then we have a ring R = R 1 R with 1 R = (1 R1, 1 R ) Note that R 1 = R 1 {0} R is a ring, but 1 R 1 = (1, 0) 1 R so R 1 is not a subring of R Finally φ : R 1 R dened by r (r, 0) is not a ring homomorphism Denition 6 Let R be a ring then a module over R is an abelian group M with scalar multiplication by R, satisfying 1 m = m (r + s)m = rm + sm r(m + n) = rm + rn (rs)m = r(sm) For all r, s R, m, n M An homomorphism of R-modules is a homomorphism of abelian group that satises φ(rm) = rφ(m) for all r R, m M Example If R is a eld, then modules are the same as vector spaces Any ideal I of R is an R-module Any quotient R/I is an R-module If R S are both rings, then S is an R-module Let R = Z Then any abelian group is a Z-module Denition 7 A module is free of rank n if it is isomorphism to R n Theorem 8 If R 0, the rank of a free module over R is uniquely determined, ie, R m = R n m = n Proof This is not proven in this module Denition 9 If R is a ring then an R-module M is nite if it can be generated by nitely many elements Example R = Z, M = Z[i] is nite with generators 1 and i R = Z[i], M = Z[i] This is also nite with generators 1 and i, but it is not free R = Z, M = Z [ ] { 1 = n : x Z, m 0 } Q This is not nite as any nite set has a maximum power of m occurring in the denominator 6

7 Ring Extensions Denition 10 Let R be a ring, then a ring extension of R is a ring S that has R as a subring A ring extension R S is nite if S is nite as an R-module Let R S be a ring extension, s S Then s is said to be integral over R if there exists a monic polynomial f = x n + a n 1 x n a 0 R[x] with f(s) = 0 Theorem 11 Let R S be a ring extension, s S Then the following are equivalent: 1 s is integral over R R[s] is a nite extension of R There exists a ring S such that R S S, S is nite over R and s S Proof Not proven in this modules Some of these are obvious (See Commutative Algebra Theorem 4) Theorem 1 If R S is a ring extension, then the set S of s S that are integral over R is a ring extension of R inside S Proof Note that R S since r R is a root of x r R[x] Given s 1, s S we want to prove that s 1 s, s 1 s S We have R R[s 1 ] R[s 1, s ] S, now the rst ring extension is nite since s 1 is integral over R We also have s is integral over R so in particular it is integral over R[s 1 ] Take the generators for R[s 1 ] as an R-module: } 1,, s n 1 and take the generators for R[s 1, s ] as an R[s 1 ]-module: 1,, s m Then {s i 1s j : 1 j m, 1 i n is a set of generators for R[s 1, s ] as an R-module Hence we conclude that R[s 1, s ] is a nite extension of R Now s 1 s, s 1 s R[s 1, s ] So if we apply the previous theorem, we have s 1 s, s 1 s are integral over R Denition 1 Let R S be an extension of rings, then the ring of R integral elements of S is called the integral closure of R in S Given an extension of rings R S then we say that R is integrally closed in S if the integral closure of R in S is R itself Theorem 14 Let R S be a ring extension and let R S be the integral closure of R in S Then R is integrally closed in S Proof Take s S integral over R We want to show that s is integral over R Take f = x n +a n 1 x n 1 + +a 0 R [x] with f(s) = 0 Consider a subring of R R[a 0, a 1,, a n 1 ] R Now R R[a 0 ] R[a 0, a 1 ] R[a 0,, a n 1 ] Now f R[a 0,, a n 1 ][x] So s is integral over R[a 0,, a n 1 ], hence R[a 0,, a n 1 ][s] is nite over R[a 0,, a n 1 ] and hence nite over R So by Theorem 1, we have that s is integral over R Denition 15 An element α C is an algebraic integer if it is integral over Z The ring of algebraic integers is denoted by Z If K is a number eld, then the ring of integers in K is denoted O K = Z K = integral closure of Z in K Example Let K = Q Take p/q Q integral over Z (assume that gcd(p, q) = 1), then there exists f(x) Z[x] such that f(p/q) = 0 So x p/q is a factor of f in Q[x], but Gauss' Lemma states if f Z[x] is monic and f = g h with g, h Q[x] then g, h Z[x] So x p/q Z[x], that is p/q Z So O Q = Z Consider K = Q( d), with d 1 and d is square free Consider α K, α = a + b d, a, b Q and suppose that α is an algebraic integer Assume that deg(α) =, that is the minimum monic polynomial f of α in Q[x] has degree Then by Gauss, we know f Z[z], furthermore f = (x (a + b d))(x (a b d)) = x ax + a db So we want a Z and a db Z So a Z a = a with a Z Then a b d = is square-free) d(b) Z b Z and (a ) d(b ) ( a If a is even, then a Z, so b is even and thus b Z If a is odd, then (a ) 1 mod 4, so b is odd as well and d 1 mod 4 We have just proven the following: ) b d = (a ) d(b) 4Z So (using the fact that d mod 4 So we conclude: 7

8 { Z[ d] d 1 mod 4 Theorem 16 Let d Z, with d 1 and square free Then O Q( d) = [ ] Z d 1 mod 4 Theorem 17 Let R be a UFD Then R is integrally closed in its fraction eld (the converse does not hold) Proof Take s = r1 r integral over R, and assume that r 1, r are coprime (well dened since R is a UFD), we have to show that r R If r / R, then let π R be any factor of r Now s is integral, so there exists a i and n such that s n + a n 1 s n a 0 = 0 Multiplying through by r n we have r1 n + a n 1 r1 n 1 r + + a 0 r n = 0 Now since r 0 mod π, if we take mod both side we have r1 n 0 mod π Hence π r1 n π r 1 This is a contradiction The converse of this theorem is not true, as an example O Q( 5) = Z[ 5] is integrally closed but not a UFD since 6 = = (1 + 5)(1 5) 1+ d 8

9 Norms, Discriminants and Lattices 1 Conjugates, Norms and Traces The Theorem of Primitive Elements Any number eld K can be generated by a single elements θ K That is K = Q(θ) Proof See any courses in Galois Theory Consider a number eld K = Q(θ) This θ has a monic minimal polynomial, say f θ Q[x] We can factor f θ over C, say f θ = (x θ 1 )(x θ ) (x θ n ), where θ 1 = θ and all the θ i are distinct For each i we have a eld embedding, which we denote σ i : K C dened by θ θ i These are all possible embedding of K C Example K = Q[ d], then f θ = x d = (x d)(x + d) So we have σ 1 = id and σ = a + b d a b d K = Q[ ], then f θ = x = (x )(x ζ )(x ζ ) where ζ = e πi a third root of unity So we have: σ 1 : (ie, the identity map), σ : ζ σ : ζ Denition 1 Let K be a number eld and σ 1,, σ n all the embeddings K C Let α K Then the elements σ i (α) are called the conjugates of α Theorem Let K be a number eld, n = [K : Q] Take α K, consider the multiplication by α as a linear map from the Q-vector space K to itself That is α : K K is dened by β αβ Then the characteristic polynomial of this map is equal to P α (x) = n i=1 (x σ i(α)) Proof Let K = Q(θ) and consider the basis: 1, θ, θ,, θ n 1 Let M α be the matrix that describes the linear map α relative to this basis First consider α = θ Let f θ = x n + a n 1 x n a 0 Then we have a a 1 M θ = a a n 1 We now calculated the characteristic polynomial of M θ : x 0 0 a 0 1 x 0 a 1 det(x I n M θ ) = det a = a k x k x + a n 1 Hence the characteristic polynomial of M θ = f θ = n i=1 (x σ i(θ)) as required Hence we know from Linear Algebra that there exists an invertible matrix A such that: σ 1 (θ) σ (θ) 0 M θ = A 0 0 σ n (θ) A 1 9

10 Now note that M α±β = M α ±M β and M αβ = M α M β (basic linear algebra) So if we have a polynomial g Q[x], then M gα = g(m α ) Now we can write any α K as g(θ) for some g Q[X] Hence we have g(σ 1 (θ)) g(σ (θ)) 0 M α = g(m θ ) = A 0 0 g(σ n (θ)) σ 1 (g(θ)) σ (g(θ)) 0 = A 0 0 σ n (g(θ)) σ 1 (α) σ (α) 0 = A A σ n (α) Hence, the characteristic polynomial of M α is n i=1 (x σ i(α)) as required Corollary For α K, the coecients of n i=1 (x σ i(α)) are in Q A 1 A 1 Denition 4 Let K be a number eld, α K We dene the norm of α as N(α) = N K/Q (α) = n i=1 σ i(α) Q Corollary 5 N(α) = det( α) = det(m α ) We can see that the norm is a multiplicative function, ie, N(αβ) = N(α)N(β) Denition 6 Let K be a number eld and α K We dene the trace of α as Tr(α) = Tr K/Q (α) = n i=1 σ i(α) Q Corollary 7 Tr(α) = Tr( α) = Tr(M α ) We can see that the trace is an additive function, ie, Tr(α + β) = Tr(α) + Tr(β) Example Let K = Q( d) Then we have: Tr(a + b d) = (a + b d) + (a b d) = a N(a + b d) = (a + b d)(a b d) = a db Let K = Q( ) and recall that x = (x )(x ζ )(x ζ Then we have: Tr(a + b + c 4) = a + b (1 + ζ + ζ ) + c 4(1 + ζ + ζ ) = a ) where ζ = e πi a third root of unity N(a + b + c 4) = (a + b + c 4)(a + bζ + cζ 4)(a + bζ + cζ 4) = a + b + 4c + 6abc Discriminant Denition 8 Let K be a number eld and α 1,, α n be a basis for K Let σ 1,, σ n : K C be all the embeddings The discriminant of (α 1,, α n ) is dened as We denote this by (α 1,, α n ) or by disc(α 1,, α n ) σ 1 (α 1 ) σ 1 (α ) σ 1 (α n ) det σ (α 1 ) σ (α ) σ (α n ) σ n (α 1 ) σ n (α ) σ n (α n ) 10

11 Theorem 9 We have Tr(α 1 α 1 ) Tr(α 1 α ) Tr(α 1 α n ) Tr(α α 1 ) Tr(α α ) Tr(α α n ) (α 1,, α n ) = det Tr(α n α 1 ) Tr(α n α ) Tr(α n α n ) Proof Let M = (σ i (α j )) ij Then we have (α 1,, α n ) = det(m) = det(m ) = det(m T M) But note that the entries of M T M at (i, j) is n k=1 σ k(α i ) σ k (α j ) = n k=1 σ k(α i α j ) = Tr(α i α j ) Corollary 10 We have (α 1,, α n ) Q Theorem 11 We have (α 1,, α n ) 0 Tr(α 1 α 1 ) Proof Suppose that (α 1,, α n ) = 0 Then there exists not all zero c 1,, c n Q with c Tr(α n α 1 ) Tr(α n α 1 ) Tr(α 1 cj α j ) c n = 0 Hence = 0 Put α = c j α j, we have just shown that Tr(α i α) = 0 i Tr(α n α n ) Tr(α n cj α j ) But we have that α i forms a basis for K over Q, hence Tr(βα) = 0 β K We have α 0, so let β = α 1, then Tr(βα) = Tr(1) = n = [K : Q] which is a contradiction Denition 1 The map K K Q dened by (α, β) Tr(αβ) is know as the trace pairing on K It is bilinear 1 x 1 x 1 x n 1 1 Let K = Q(θ), this has basis 1,, θ n 1 In general det is called a Vandemonde 1 x n x n x n 1 n determinant and it is equal to 1 i<j n (x j x i ) (See Linear Algebra or Algebra I for a proof by induction) So in our case, (1, θ,, θ n 1 ) = 1 i<j n (σ i(θ) σ j (θ)) Also note that (f θ ) := (1, θ,, θ n 1 ) (Generally, if f = (x α 1 ) (x α n ) then (f) := 1 i<j n (α i α j ), check with the denition of a discriminant of a quadratic) Example Let K = Q( d) Consider the basis 1, d We calculate the discriminant in two ways: (1, ( ) 1 d d) = det 1 = ( d) = 4d d (1, ( ) ( ) Tr(1) Tr( d) d) = det Tr( 0 = det = 4d d) Tr(d) 0 d Now consider the basis 1, 1+ d Then (1, 1+ d ) = ( d) = d Let K = Q( d), with basis 1, d, d Then we have (1, d, Tr(1) Tr( d) Tr( d ) d ) = det Tr( d) Tr( d ) Tr(d) Tr( d ) Tr(d) Tr( d) = det d 0 d 0 = 7d 11

12 Lattices Denition 1 Let K be a number eld A lattice Λ in K is a subgroup generated by Q-linearly independent elements of K That is Λ = {n 1 α n r α r n i Z} where α i are linearly independent over Q We always have r [K : Q] The number r is called the rank of the lattice, this is sometimes denoted rk(λ) Example Z[i] is a lattice in Q(i) Theorem 14 Any nitely generated subgroup of a number eld K is a latice Proof Let Λ be a nitely generated subgroup of K By the Fundamental Theorem of Finitely Generated Abelian Group, we have Λ = T Z r, where T is the torsion As K is a Q-vector space, we have T = 0, so Λ = Z r Let φ : Z r Λ be an isomorphism Claim: α i = φ(e i ) is a basis (ie, linearly independent generating set) for Λ, where e i is the standard basis for Z r Now φ(c 1,, c r ) = n i=1 c iα i Since φ is surjective, all elements of Λ are reached If c i α i = 0 for c i Q multiply c i by the common denominator, then without loss of generality, we can assume c i Z But we know that φ is injective, so for all i, c i = 0 Denition 15 A lattice of K is said to be full rank if its rank r = [K : Q] Theorem 16 Let Λ K be a full rank lattice Then (α 1,, α r ) is the same for every basis α 1,, α r of Λ Proof Suppose (α i ) i and (β i ) i are basis for Λ Then each β i can be written as a linear combination of α j with coecients in Z, ie β 1 = A β r α 1 α r with A an r r matrix with coecients in Z Similarly α 1 α r = B Tr(α 1 α 1 ) Tr(α 1 α r ) Hence we have AB = I r, so A GL r (Z), so det(a) = ±1 Put S = Then Tr(α r α 1 ) Tr(α r α r ) Tr(β 1 β 1 ) Tr(β 1 β r ) = A T SA (Base change for matrices describing symmetric bilinear forms, see Algebra Tr(β r β 1 ) Tr(β r β r ) I) So we have (β 1,, β r ) = det(a T SA) = det(a ) det(s) = det(s) = (α 1,, α r ) Denition 17 Let Λ K be a full rank lattice, then we dene (Λ) to be the discriminant of any basis of Λ Theorem 18 Let K be a number eld and Λ K be a full rank lattice with Λ O K Then (Λ) Z Proof We have (Λ) = det((tr(α i α j ) ij ) with α i O K If α O K, then Tr(α) = n i=1 σ i(α) Z Q = Z Hence (Λ) Z Theorem 19 Let K be a number eld and Λ Λ be two full rank lattices Then the index (Λ : Λ) is nite and (Λ) = (Λ : Λ) (Λ ) Proof All the elements of Λ can be written as an integral linear combination of some chosen basis of Λ So there exists A M n (Z) with Λ = AΛ Consider Λ /Λ = Z n /AZ n, this is a nitely generated abelian group so by FTFGAG Λ /Λ = Z/d 1 Z Z/d m Z Z r with d 1 d d m So (by Smith Normal Form from Algebra I) there d exists B, B GL n (Z) with BAB 0 d = As we have rk(λ ) = rk(λ), we have that r = 0, and 0 d n thus det(a) = d 1 d m = Z n /AZ n = (Λ : Λ) Furthermore (Λ) = (AΛ ) = (det A) (Λ ) Theorem 0 Let K be a number eld with n = [K : Q] Then there exists a basis ω 1,, ω n of K/Q such that O K = Zω Zω n = { a i ω i a i Z} (That is O K is a full rank lattice in K) β 1 β r 1

13 Proof We consider all Λ O K that are full rank lattices in K The rst question is: do such Λ exists? Write K = Q(θ), θ K and f θ = x n + a n 1 x n a 0 with a i Q Now let d be a common denominator of the a i, then dθ O K Also note that Q(θ) = Q(dθ), so without loss of generality we can assume θ O K Then Z[θ] O K, furthermore 1, θ,, θ n 1 are linearly independent over Z, hence Z[θ] is a full rank lattice Of all such Λ, we have that (Λ) Z (by Theorem 18) So consider Λ with (Λ) minimal Claim: Λ = O K Suppose Λ O K We do have Λ O K, so take α O K \ Λ Then Λ := Λ + Zα is nitely generated as an abelian group of K and thus Λ is a lattice of full rank Also Λ O K But we have (Λ) = (Λ : Λ) (Λ), and since Λ Λ, we nd (Λ) > (Λ ), which is a contradiction Denition 1 The discriminant of a number eld K/Q is dened as (K/Q) = (O K ) Example Let K = Q( { 4d d 1 mod 4 d) with d 1 and square free Then (K/Q) = (O K ) = d d 1 mod 4 Note that if Λ O K is a full rank sublattice, then (Λ) = (O K : Λ) (O K ) by Theorem 19 Corollary If Λ O K and (Λ) is square free then Λ = O K 1

14 4 Cyclotomic Fields Denition 41 Let n be a positive integer Then the n-cyclotomic eld is Q(ζ n ) where ζ n = e πi n For simplicity we are going to assume that n = p r with p being a prime Theorem 4 The minimal polynomial of ζ p r is Φ p r = p r k=1,p k (x ζ k p r) Proof Firs note that Φ p r(ζ p r) = 0 In general, n k=1 (x ζk n) = x n 1 We see this by noticing that every zero of the LHS is a zero of the RHS, the degree of both sides are the same and they both have the same leading coecients Consider and notice that ζ p p r Φ p r = p r k=1,p k = ζ pr 1 This means we can rewrite Φ p r = p r p r (x ζp k r) = k=1 (x ζk p r) p r 1 k=1 k=1 (x ζk p r) p r 1 k=1 (x ζk p ) = xp 1 x pr 1 1 = x(p 1)p r 1 r (x ζpk p r ) r 1 + x (p )pr Hence we have Φ p r Z[x] We nally show that Φ p r is irreducible Suppose that Φ p r = fg with f, g Z[x], f, g are both monic and non constant Consider this mod p, we have Φ p r = xpr 1 x pr 1 1 r (x 1)p (x 1) (p 1)(p r 1 ) (x 1) pr 1 mod p (using Fermat's Little Theorem) Let f, g denoted the reduction of f, g mod p, hence we have fg = (x 1) (p 1)pr 1 mod p Now F p is a UFD, so we have f = (x 1) m and g = (x 1) k such that m+k = (p 1)p r 1 Hence we have f = (x 1) m +pf and g = (x 1) k +pg for some F, G Z[x], that is, fg = (x 1) m+k +p(x 1) k F +p(x 1) m G+p F G Now consider x = 1, we get f(1)g(1) = p F (1)G(1) on one hand and Φ p r(1) = 1 (p 1)pr = p on the other hand But p p, so we have a contradiction and Φ p r is irreducible Note that Z[ζ p r] O Q(ζp r ) Problem What is (Z[ζ p r])? Let us denote ζ p r by ζ By denition we have (Z[ζ]) = Let us x k, we want to compute p r F k = p r p r k=1,p k m=1,p m,m k (ζ k ζ m ) m=1,p m,m k (ζk ζ m ) We do this by considering p r m=1,p m,m k (x ζ m ) = Φ p r(x) x ζ k = x p r 1 (x pr 1 1)(x ζ k ) Now F k (ζ k ) = 0 0, so we need to use l'hospital's rule We calculate Φ p r(x) = pr x pr 1 (x pr 1 1) p r 1 x pr 1 1 (x pr 1) (x pr 1 1) Now the roots of x pr 1 1 are powers of ζ p r 1 = ζ p, so ζ k is not a root of (x pr 1 1) Hence F k (ζ k ) = Φ p r(ζk ) = pr ζ k(pr 1) ζ kpr

15 Hence Φ p r(ζ p k) = r, so we have ζ kpr 1 1 Hence we nally compute p r k=1,p k (Z[ζ]) = (x ζ kpr 1 ) = p r k=1,p k p r k=1,p k p r r p r 1 ) ζ kpr 1 1 = pr(p ζ kp r 1 1 (x ζ k p ) = ( p 1 (x ζp k ) k=1 ) p r 1 = (Φ p (x)) pr 1 Plucking in x = 1, we get Φ p (x) pr 1 = p pr 1 Hence we conclude (Z[ζ]) = p rpr rp r 1 p r 1 = p pr 1 (rp r 1) Now it is not important to remember what exactly it is, the key idea is that it is a power of p, the exact exponent does not matter In particular if r = 1 we get (Z[ζ p ]) = p p Theorem 4 For any n we have O Q(ζn) = Z[ζ n ] Proof We will only prove this for n = p, with p prime We already know that Z[ζ p ] O Q(ζp) We also know that p p = (Z[ζ p ]) = ( O Q(ζp) : Z[ζ p ] ) (OQ(ζp)) (by Theorem 19) Suppose that Z[ζ p ] O Q(ζp) then (O Q(ζp) : Z[ζ p ]) = p, where is an unknown exponent Then O Q(ζp)/Z[ζ p ] is an abelian group of order divisible by p Hence there exists α O Q(ζp)/Z[ζ p ] with order p, ie, there exists α O Q(ζp) with pα Z[ζ p ] We want to show that for any α O Q(ζp) such that pα Z[ζ p ] then we already have α Z[ζ p ] Note that Z[ζ p ] = Z[1 ζ p ] Now N(1 ζ p ) = p 1 i=1 σ i(1 ζ p ) = p 1 i=1 (1 ζi p) = Φ p (1) = p Hence we have that p factors as p 1 i=1 (1 ζi p) Now for all i, we have N(1 ζp) i = p 1 j=1 (1 σ j(ζp)) i = p 1 j=1 (1 ζij p ) = N(1 ζ p ) = p, ( 1 ζ i ) p hence in particular we have N 1 ζ p = 1, so 1 ζi p 1 ζ p is a unit for all i Putting all of this together we have p = (1 ζ i p ) (1 ζ p) p 1 (1 ζ p ) p 1 = unit (1 ζ p ) p 1 We can write pα as a 0 + a 1 (1 ζ p ) + + a p (1 ζ p ) p ( ) with a i Z We want to show that p a i for all i For a Z we have p a if and only if (1 ζ p ) a in O Q(ζp) One direction follows from the fact that 1 ζ p p For the other implication, suppose (1 ζ p ) a, then N(1 ζ p ) N(a) p a p 1, hence p a (Note for any number eld and a Q, we have N(a) = a [K:Q] ) We have now the tools to do a prove by induction to show that a n is divisible by p Let n = 0 and consider ( ) module 1 ζ p We have pα 0 mod (1 ζ p ), also for i 1 we have a i (1 ζ p ) 0 mod (1 ζ p ) Hence we nd that a 0 0 mod (1 ζ p ), so (1 ζ p ) a 0 and hence p a 0 Now suppose that p a 0, a 1,, a n 1 and that n p We have that pα is divisible by (1 ζ p ) n+1, but so is a 0, (1 ζ p )a 1,, (1 ζ p ) n 1 a n 1 and a i (1 ζ p ) i for i > n Hence we have (1 ζ p ) n a i 0 mod (1 ζ p ) n+1 Hence there exists β O Q(ζp) with β(1 ζ p ) n+1 = (1 ζ p ) n a n β(1 ζ p ) = a n, so we have (1 ζ p ) a n Hence we have shown by induction that p a i i Hence pα pz[ζ p ] α Z[ζ p ] So to recap, we have shown if Z[ζ p ] O Q(ζp), then we must have α O Q(ζp) \ Z[ζ p ] such that pα Z[ζ p ] But we also shown that if α O Q(ζp) with pα Z[ζ p ] then α Z[ζ p ], hence we have a contradiction Example (Of the proof in action) What is O Q( )? We know that Z[ ] O Q( ), we also know that (Z[ ]) = 7( ) = = (O Q( ) : Z[ ]) (O Q( ) ) Hence if Z[ ] O Q( ), then either divides the index or divides the index Suppose that divides the index Then there exists α O Q( ) \ Z[ ] with α Z[ ] Note that in O Q( ) we have = For a Z we have a if and only if a in O Q( ) Let α = a 0 + a 1 + a 4 Consider this modulo, we have 0 a 0 mod Hence a 0 Now considering this modulo 4, we have 0 a 1 mod 4, again implying that a 1, hence a 1 So nally considering this modulo, we see that a Hence α Z[ ], ie, α Z[ ] So does not divide the index Now suppose that divides the index We claim that = (1+ ) unit Now (1+ ) = = ( ) Now N(1 + ) = =, so N((1 + ) ) = = N() and hence ( ) is a unit, proving our claim Hence we have that for α Z, α if and only if (1 + ) α in O Q( ) So consider α O Q( ) \ Z[ ] such that α Z[ ] and write α = a 0 + a 1 (1 + ) + a (1 + ) (by changing the basis of 15

16 Z[ ] to Z[1 + ]) Then if we consider the equation modulo successive powers of (1 + ), we nd that each a i is divisible by (1 + ) and thus by Again this leads to a contradiction Hence we have that Z[ ] = O Q( ) 16

17 5 Dedekind Domains 51 Euclidean domains Denition 51 Let R be a domain (that is 0 1 and there are no non-trivial solutions to ab = 0) An Euclidean function on R is a function φ : R \ {0} Z 0 such that for all a, b R with b 0, there exists q, r R with a = qb + r and either r = 0 or φ(r) < φ(b) Example R = Z, and φ(n) = n R = k[x] where k is any eld and φ(f(x)) = deg(f) R = Z[i] and φ(α) = N(α) Denition 5 A domain on which there is an Euclidean function is called an Euclidean domain Theorem 5 If R is an Euclidean domain then R is a principal ideal domain (PID), ie, every ideal of R can be generated by one element Proof Let I 0 be a non-zero ideal of R Take 0 b I to be an element for which φ(b) is minimal We claim that I = (b) Let a I \ {0} be another element Then there exists q R with a qb either 0 or φ(a qb) < φ(b) As b is an element with φ(b) minimal, we have that a qb is 0, hence a = qb, ie, a (b) Lemma 54 If R is a PID and π R an irreducible element, then for a, b R we have π ab π a or π b Proof Suppose that π a, we want to show that π b Consider the ideal I = (π, a) Let δ R be a generator for I, ie, (π, a) = (δ) There exists x, y R with xπ + ya = δ Also π (δ) so δ π This means that either δ 1 or δ π But the case δ π can not occur since π a but δ a So without loss of generality, assume that δ = 1 Thus xπ + ya = 1, hence xπb + yab = b, but since π ab, we have π b Theorem 55 A PID is a UFD Proof Take a R \ {0}, such that a is not a unit Assume that a = ɛπ 1 π n = ɛ π 1 π m are two distinct factorisation of a into irreducible Without loss of generality we may assume that n is minimal amongst all elements a with non-unique factorisation We have π 1 π 1 π m so by the lemma π 1 π i for some i Without loss of generality we can assume that i = 1, so π 1 π 1 but both are irreducible, hence π 1 π 1 Without loss of generality we can assume that π 1 = π 1 But then π π n = ɛπ π m and π π n has n 1 irreducible factors, so by minimality of n, this factorisation into irreducible is unique ] We show that O Q( ) = Z is Euclidean We claim that the Euclidean function is the Norm N(a + [ 1+ b 1+ ) = (a + b 1+ )(a + b 1 and this ts we the previous line as N(a + b 1+ ) = N(a + b + b ) = a + ab + b (Note that we had over Q( ) N(c + d ) = c + d ) ) = (a + b ) + b 4 = a + ab + b ) Suppose we are given α = a + b 1+ and β = c + d 1+ with β 0 Then α β = a + b 1+ = c + d (a + b )(c d 1+ ) = e + f 1 + N(β) Q[ 1 + ] (so note e, f Q) Then pick g, h Z such that g e, h f 1 and set q = g + h 1 + r = α βq 17

18 Then we have α = βq + r and furthermore if r 0 N(r) = N(α β(g + h 1 + )) = N(β(e + f 1 + g h 1 + )) = N(β)N((e g) + (f h) 1 + ) = N(β)[(e g) + (e g)(f h) + (f h) ] 4 N(β) < N(β) Similar arguments works for O Q( d) with d { 1,,, 7, 11} (you might need to change the bound) Theorem 56 If d < 11 then O Q( d) is not a Euclidean domain (but for d { 19, 4, 67, 16} it is a PID) Proof Assume that φ : R Z 0 is Euclidean, where R = O Q( d) Now R = {±1} Take an element b R\{0, ±1} with φ(b) as small as possible For all a R there exists q, r R with r = a qb and φ(r) < φ(b) or r = 0 Now since φ(b) is as small as possible, we have that r {0, 1, 1}, for all a R We also have that a r mod b, hence R/(b) has at most elements On the other hand the number of elements of (R/(b)) = (R : (b)) ((b)) = (R : (b)) (R) (by Theorem { d d 1 mod 4 19 since (b) R) Let R = Z + Zθ where θ = 1+ Then we have (b) = Zb + Zbθ Now d d 1 mod 4 ( ) b θb ((b)) = det = (bbθ bbθ) b θb = (bb) (θ θ) = N(b) (R) Hence we have (R : (b)) = N(b), that is (R : (b)) = N(b) (since the norm is positive) So if we show that b R \ {0, ±1} we have N(b) > then R/(b) has more than three elements, contradicting the rst paragraph Now we always have N(a + b d) = a + d b Suppose d 1 mod 4, then for a + b d to be in R we need a, b Z Suppose that a + d b then a 1 and d > 11, so b = 0, but a + b d {0, ±1} If d 1 mod 4 we can also have a = a, b = b where a, b Z and a b mod Then N(a + b ( ) d) = N a +b d = 1 (a + d b ) Suppose N(a + b d) then a + d b 1 But d 1, so again b = 0 and a 1 so a Hence a {, 0, }, implying a + b d {0, ±1} Conjecture Let K be a number eld that is not Q( d) for some d < 0 then if O K is a UFD, then it is Euclidean Remark In general φ = N does not work, then φ is very dicult to nd 5 Dedekind Domain Denition 57 A prime ideal is an ideal P R satisfying P R and a, b R with ab P then either a P or b P Fact P R is prime if and only if R/I is a domain Denition 58 A maximal ideal is an ideal M R satisfying M R and there are no ideals I R with M I R Fact M R is a maximal ideal if and only if R/M is a eld Every proper ideal I R is contained in a maximal ideal Corollaries) (See commutative Algebra Theorem 14 and its Example Let R = Z Then its prime ideals are (0) and (p) where p is prime Its maximal ideals are (p) (as Z/(p) = F p is a eld) Denition 59 A ring R is Noetherian if one and thus both of the following equivalent conditions holds 1 Every ideal of R is nitely generated 18

19 Every ascending chains of ideals I 0 I 1 is stationary, ie, there exists r > 0 such that I i = I j for all i, j > r Denition 510 Let R be a domain Then R is a Dedekind Domain if: 1 R is Noetherian R is integrally closed in its eld of fractions Every non-zero prime ideal is a maximal ideal Example Every eld is a Dedekind domain (the only ideals are: (0), (1)) Lemma 511 Every nite domain is a eld Proof Let R be a nite domain Take 0 a R, we need to show there exists x R with ax = 1 Consider the map R a R dened by x ax We note that a is injective, if ab = ac then a(b c) = 0, hence b c = 0 since R is a domain As R is nite, a is also surjective Hence there exists x with ax = 1 Theorem 51 If K is a number eld, then O K is a Dedekind domain Proof Let I O K be an ideal If I = (0) then it is nitely generated, so assume I is non-zero Hence there exists 0 a I, so ao K is a full rank lattice in O K We have ao K I O K, so I is a full rank lattice as well It has [K : Q] < generators as a free abelian group and the same elements generates it as an ideal So O K is Noetherian We know that O K = Z K Furthermore the integral closure of a ring R in an extension S is in fact integrally closed in S So O K is integrally closed in K Let P O K be a non-zero prime ideal P is a full rank lattice so (O K : P ) < Hence O K /P is a nite domain So by the above lemma, O K /P is a eld and hence P is maximal Denition 51 Let R be a domain Then a fractional ideal I of R is a R-submodule of the elds of fractions of R, such that there exists 0 a R with ai R Example Let us work out the fractional ideals of Z The ideals of Z are (n) with n Z So fractional ideals are I Q such that a Z with ai = (n) for some n Z That is I = n a Z Q Note that Q is not a fractional ideal, as elements of Q have arbitrary large denominators If R is a ring, I, J R are ideals, then IJ is the ideal generated by {ij : i I, j J} If R is a domain, I, J fractional ideals of R and K the eld of fraction of R, then IJ is a K-submodule generated by {ij : i I, j J} It is a fractional ideal as abij R (where a, b are such that ai, bj R) Example Let R = Z and consider I = (a), J = (b) with a, b Q Then IJ = (ab) Denition 514 Let R be a domain, K its eld of fraction, I K a fractional ideal Then I is called invertible if there exists a fractional ideal J K such that IJ = R = (1) Example Every non-zero fractional ideal of Z is invertible Every principal non-zero fractional ideal (a) of R is invertible, consider (a)(a 1 ) = (1) Theorem 515 The invertible ideals of a domain R forms a group with respect to fractional ideal multiplication, with unit element R = (1) and inverse I 1 = {a K ai R} (K is the eld of fractions of R) Proof Let I K be invertible, then there exists J with IJ = R We want to show: if a J then ai R and if ai R then a J The rst one follows directly Consider aij = ar and aij J, so ar J means a J Hence J = I 1 If I 1, I, I are fractional ideals then I 1 (I I ) = (I 1 I )I Finally we show that if I, J are invertible then so is IJ 1 We claim (IJ 1 ) 1 = JI 1 To see this consider (IJ 1 )(JI 1 ) = IRI 1 = II 1 = R Theorem 516 Let R be a domain Then the following conditions on R are equivalent 1 R is Dedekind Every non-zero fractional ideals of R is invertible 19

20 Every non-zero ideals of R is the product of prime ideals 4 Every non-zero ideal of R is the product of prime ideals uniquely We will prove this after some examples Example O Q( 5) = Z[ 5] is not a UFD, we have 6 = = (1 + 5)(1 5) But since Z[ 5] is Dedekind (by Theorem 51), we can write (6) as the product of prime ideal uniquely In fact (6) = () () = (1+ 5)(1 5) = (, 1+ 5)(, 1 5)(, 1+ 5)(, 1 5) We check that (, 1+ 5) is prime Now Z[ 5]/(, 1+ 5) = Z[x]/(x +5,, 1+x) Now (, x+1, x +5) = (, x+1, x +5 x(x+1) = (, x+1, x+5) = (, x + 1) Hence Z[ 5]/(, 1 + 5) = Z[x]/(, x + 1) = F [x]/(x + 1) = F, which is a eld Thus (, 1 + 5) is maximal Denition 517 If R is a domain and K its eld of fraction Let I be a non-zero fractional ideal then R : I = {a K : ai R} Note that from Theorem 515, we see that I is invertible if and only if (R : I) I = R Example 518 R = Z[ ] is not Dedekind (As it is not algebraically closed) We show that the ideal I = (, 1 + ) is not invertible R : I = {a + b Q( ) : (a + b ) Z[ ], (1 + )(a + b ) Z[ ]) From the rst condition, we can rewrite a = a, b = b with a, b Z So consider the second condition So a b (1 + )( a + b a ) = + a + b b mod, ie, Z[ ] : (, 1 + { a + b ) = } : a, b Z, a b mod = Z [ ] 1 + Now [ ] 1 + Z (, 1 + ) = = ( 1, 1 + ) (, 1 + ) (, 1 +, 1 + (1 + ) = (, 1 +, 1) = (, 1 + ) = I R ) Hence I is not invertible We now show that I = () can not be written as the product of prime ideals Suppose I = P 1 P P n, then I P i for all i Now {ideals of R containing I} {ideals of R/I} The bijection is dened by J J/I R/I and {x : x J} J In our case R/I = Z[ ]/() = Z[x] /(x +,) = F[x] /(x +1) = F /(x+1) = F[x] /(x) = {a + bɛ : ab F, ɛ = 0} The ideals in R/I are (0), (1) = (1 + ɛ) and (ɛ) Which of these ideal is prime? (1) is never prime, and (0) is not prime as it is not a domain So (ɛ) is the only maximal ideal and hence must be the only prime R/I has Clearly () (, 1 + ), which we saw maximal and so must be the only prime ideal which contains () 0

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