1 A Course In Algebraic Number Theory Robert B. Ash Preface This is a text for a basic course in algebraic number theory, written in accordance with the following objectives. 1. Provide reasonable coverage for a one-semester course. 2. Assume as prerequisite a standard graduate course in algebra, but cover integral ex- tensionsandlocalizationbeforebeginningalgebraicnumbertheory. Forgeneralalgebraic background, see my online text “Abstract Algebra: The Basic Graduate Year”, which can be downloaded from my web site www.math.uiuc.edu/∼ r-ash/ The abstract algebra material is referred to in this text as TBGY. 3. Cover the general theory of factorization of ideals in Dedekind domains, as well as the number field case. 4. Do some detailed calculations illustrating the use of Kummer’s theorem on lifting of prime ideals in extension fields. 5. Give enough details so that the reader can navigate through the intricate proofs of the Dirichlet unit theorem and the Minkowski bounds on element and ideal norms. 6. Cover the factorization of prime ideals in Galois extensions. 7. Coverlocalaswellasglobalfields,includingtheArtin-Whaplesapproximationtheorem and Hensel’s lemma. Especially helpful to me in preparing this work were the beautiful little book by Samuel,“AlgebraicTheoryofNumbers”,Hermann1971,andthetreatmentofcyclotomic fieldsbyJ.Milneinhisonlinetext“AlgebraicNumberTheory”(www.math.lsa.umich.edu/∼ jmilne/) Some other useful references are: Esmonde, J., and Murty, M.R., “Problems in Algebraic Number Theory”, Springer 1999 Fro¨lich, A., and Taylor, M.J., “Algebraic Number Theory”, Cambridge 1991 Janusz, G.J.,“ Algebraic Number Fields”, AMS 1996 Marcus, D.A., “Number Fields”, Springer 1977 Stewart, I., and Tall, D., “Algebraic Number Theory”, Chapman and Hall 1987 (cid:1)ccopyright 2003, by Robert B. Ash. Paper or electronic copies for noncommercial use maybemadefreelywithoutexplicitpermissionoftheauthor. Allotherrightsarereserved. Table of Contents Chapter 1 Introduction 1.1 Integral Extensions 1.2 Localization Chapter 2 Norms, Traces and Discriminants 2.1 Norms and traces 2.2 The Basic Setup For Algebraic Number Theory 2.3 The Discriminant Chapter 3 Dedekind Domains 3.1 The Definition and Some Basic Properties 3.2 Fractional Ideals 3.3 Unique Factorization of Ideals 3.4 Some Arithmetic in Dedekind Domains Chapter 4 Factorization of Prime Ideals in Extensions 4.1 Lifting of Prime Ideals 4.2 Norms of ideals 4.3 A Practical Factorization Theorem Chapter 5 The Ideal Class Group 5.1 Lattices 5.2 A Volume Calculation 5.3 The Canonical Embedding Chapter 6 The Dirichlet Unit Theorem 6.1 Preliminary Results 6.2 Statement and Proof of Dirichlet’s Unit Theorem 6.3 Units in Quadratic Fields 1 2 Chapter 7 Cyclotomic Extensions 7.1 Some Preliminary Calculations 7.2 An Integral Basis of a Cyclotomic Field Chapter 8 Factorization of Prime Ideals in Galois Extensions 8.1 Decomposition and Inertia Groups 8.2 The Frobenius Automorphism 8.3 Applications Chapter 9 Local Fields 9.1 Absolute Values and Discrete Valuations 9.2 Absolute Values on the Rationals 9.3 Artin-Whaples Approximation Theorem 9.4 Completions 9.5 Hensel’s Lemma Chapter 1 Introduction Techniquesofabstractalgebrahavebeenappliedtoproblemsinnumbertheoryforalong time, notably in the effort to prove Fermat’s last theorem. As an introductory example, we will sketch a problem for which an algebraic approach works very well. If p is an odd prime and p ≡ 1 mod 4, we will prove that p is the sum of two squares, that is, p can expressed as x2 +y2 where x and y are integers. Since p−1 is even, it follows that −1 2 is a quadratic residue (that is, a square) mod p. To see this, pair each of the numbers 2,3,... ,p−2 with its inverse mod p, and pair 1 with p−1≡−1 mod p. The product of the numbers 1 through p−1 is, mod p, p−1 p−1 1×2×···× ×−1×−2···×− 2 2 and therefore p−1 [( )!]2 ≡−1 mod p. 2 If −1≡x2 mod p, then p divides x2+1. Now we enter the ring Z[i] of Gaussian integers and factor x2+1 as (x+i)(x−i). Since p can divide neither factor, it follows that p is not prime in Z[i]. Since the Gaussian integers form a unique factorization domain, p is not irreducible, and we can write p=αβ where neither α nor β is a unit. Define the norm of γ =a+bi as N(γ)=a2+b2. Then N(γ)=1 iff γ is 1,-1,i or −i, equivalently, iff γ is a unit. Thus p2 =N(p)=N(α)N(β) with N(α)>1 and N(β)>1, so N(α)=N(β)=p. If α=x+iy, then p=x2+y2. Conversely, if p is an odd prime and p=x2+y2, then p is congruent to 1 mod 4. [If x is even, then x2 ≡0 mod 4, and if x is odd, then x2 ≡1 mod 4. We cannot have x and y both even or both odd, since p is odd.] Itisnaturaltoconjecturethatwecanidentifythoseprimesthatc√anberepresentedas x2+|m|y2, where m is a negative integer, by working in the ring Z[ m]. But the above argument depends critically on unique factorization, which does not hold in general. A 1 2 CHAPTER 1. INTRODUCTION √ √ √ standard example is 2×3=(1+ −5)(1− −5) in Z[ −5]. Difficulties of this sort led Kummer to invent “ideal numbers”, which became ideals at the hands of Dedekind. We willseethatalthougharingofalgebraicintegersneednotbeaUFD,uniquefactorization of ideals will always hold. 1.1 Integral Extensions If E/F is a field extension and α ∈ E, then α is algebraic over F iff α is a root of a nonconstant polynomial with coefficients in F. We can assume if we like that the polynomial is monic, and this turns out to be crucial in generalizing the idea to ring extensions. 1.1.1 Definitions and Comments All rings are assumed commutative. Let A be a subring of the ring R, and let x∈R. We say that x is integral over A if x is a root of a monic polynomial f with coefficients in A. The equation f(X)=0 is called an equation of integral dependence for x over A. If x is a real or complex numbe√r that is integral over Z, then x is called an algebraic integer. Thus for every integer d, d is an algebraic integer, as is any nth root of unity. (The monic polynomials are, respectively, X2−d and Xn−1.) The next results gives several conditions equivalent to integrality. 1.1.2 Theorem Let A be a subring of R, and let x∈R. The following conditions are equivalent: (i) The element x is integral over A; (ii) The A-module A[x] is finitely generated; (iii) The element x belongs to a subring B of R such that A ⊆ B and B is a finitely generated A-module; (iv)ThereisasubringB ofRsuchthatB isafinitelygeneratedA-moduleandxstabilizes B, that is, xB ⊆B. (If R is a field, the assumption that B is a subring can be dropped, as long as B (cid:9)=0); (v) There is a faithful A[x]-module B that is finitely generated as an A-module. (Recall that a faithful module is one whose annihilator is 0.) Proof. (i)implies (ii): If x is a root of a monic polynomial of degree n over A, then xn and all higher powers of x can be expressed as linear combinations of lower powers of x. Thus 1,x,x2,... ,xn−1 generate A[x] over A. (ii) implies (iii): Take B =A[x]. (iii) implies (i)(cid:2): If β1,... ,βn generate B over A, then xβi is a linear combination of the β , say xβ = n c β . Thus if β is a column vector whose components are the β , I j i j=1 ij j i is an n by n identity matrix, and C =[c ], then ij (xI−C)β =0, 1.1. INTEGRAL EXTENSIONS 3 and if we premultiply by the adjoint matrix of xI−C (as in Cramer’s rule), we get [det(xI−C)]Iβ =0 hencedet(xI−C)b=0foreveryb∈B. SinceB isaring, wemaysetb=1andconclude that x is a root of the monic polynomial det(XI−C) in A[X]. If we replace (iii) by (iv), the same proofs work. If R is a field, then in (iv)⇒(i), x is an eigenvalue of C, so det(xI−C)=0. If we replace (iii) by (v), the proofs go through as before. [Since B is an A[x]-module, in (v)⇒(i) we have xβ ∈ B. When we obtain [det(xI −C)]b = 0 for every b ∈ B, the i hypothesis that B is faithful yields det(xI−C)=0.] ♣ Wearegoingtoproveatransitivitypropertyforintegralextensions,andthefollowing result will be helpful. 1.1.3 Lemma Let A be a subring of R, with x ,... ,x ∈ R. If x is integral over A, x is integral 1 n 1 2 over A[x1],..., and xn is integral over A[x1,... ,xn−1], then A[x1,... ,xn] is a finitely generated A-module. Proof. Then=1casefollowsfrom(1.1.2),condition(ii). Goingfromn−1tonamounts to proving that if A,B and C are rings, with C a finitely generated B-module and B a finitely generated A-module, then C is a finitely generated A-module. This follows by a brief computation: (cid:3)r (cid:3)s (cid:3)r (cid:3)s C = By , B = Ax , so C = Ay x . ♣ j k j k j=1 k=1 j=1k=1 1.1.4 Transitivity of Integral Extensions Let A, B and C be subrings of R. If C is integral over B, that is, every element of C is integral over B, and B is integral over A, then C is integral over A. Proof. Let x ∈ C, with xn+bn−1xn−1+···+b1x+b0 = 0, bi ∈ B. Then x is integral over A[b0,... ,bn−1]. Each bi is integral over A, hence over A[b0,... ,bi−1]. By (1.1.3), A[b0,... ,bn−1,x]isafinitelygeneratedA-module. Itfollowsfromcondition(iii)of(1.1.2) that x is integral over A. ♣ 1.1.5 Definitions and Comments If A is a subring of R, the integral closure of A in R is the set A of elements of R that c areintegraloverA. NotethatA⊆A becauseeacha∈AisarootofX−a. Wesaythat c A is integrally closed in R if A =A. If we simply say that A is integrally closed without c reference to R, we assume that A is an integral domain with fraction field K, and A is integrally closed in K. If x and y are integral over A, then just as in the proof of (1.1.4), it follows from (1.1.3) that A[x,y] is a finitely generated A-module. Since x+y,x−y and xy belong to 4 CHAPTER 1. INTRODUCTION thismodule,theyareintegraloverAby(1.1.2),condition(iii). Theimportantconclusion is that A is a subring of R containing A. c If we take the integral closure of the integral closure, we get nothing new. 1.1.6 Proposition The integral closure A of A in R is integrally closed in R. c Proof. By definition, A is integral over A. If x is integral over A , then as in the proof c c of (1.1.4), x is integral over A, and therefore x∈A . ♣ c We can identify a large class of integrally closed rings. 1.1.7 Proposition If A is a UFD, then A is integrally closed. Proof. If x belongs to the fraction field K, then we can write x = a/b where a,b ∈ A, withaandbrelativelyprime. IfxisintegraloverA,thenthereisanequationoftheform (a/b)n+an−1(a/b)n−1+···+a1(a/b)+a0 =0 with all a belonging to A. Multiplying by bn, we have an+bc = 0, with c ∈ A. Thus b i dividesan,whichcannothappenforrelativelyprimeaandbunlessbhasnoprimefactors at all, in other words, b is a unit. But then x=ab−1 ∈A. ♣ Problems For Section 1.1 Let A be a subring of the integral domain B, with B integral over A. In Problems 1-3, we are going to show that A is a field if and only if B is a field. 1. Assume that B is a field, and let a be a nonzero element of A. Then since a−1 ∈ B, there is an equation of the form (a−1)n+cn−1(a−1)n−1+···+c1a−1+c0 =0 with all c belonging to A. Show that a−1 ∈A, proving that A is a field. i 2. Now assume that A is a field, and let b be a nonzero element of B. By condition (ii) of (1.1.2), A[b] is a finite-dimensional vector space over A. Let f be the A-linear transformation on this vector space given by multiplication by b, in other words, f(z) = bz, z ∈A[b]. Show that f is injective. 3. Show that f is surjective as well, and conclude that B is a field. In Problems 4-6, let A be a subring of B, with B integral over A. Let Q be a prime ideal of B and let P =Q∩A. 4. ShowthatP isaprimeidealofA,andthatA/P canberegardedasasubringofB/Q. 5. Show that B/Q is integral over A/P. 6. Show that P is a maximal ideal of A if and only if Q is a maximal ideal of B. 1.2. LOCALIZATION 5 1.2 Localization Let S be a subset of the ring R, and assume that S is multiplicative, in other words, 0∈/ S, 1∈S, and if a and b belong to S, so does ab. In the case of interest to us, S will be the complement of a prime ideal. We would like to divide elements of R by elements of S to form the localized ring S−1R, also called the ring of fractions of R by S. There is no difficulty when R is an integral domain, because in this case all division takes place in the fraction field of R. Although we will not need the general construction for arbitrary rings R, we will give a sketch. For full details, see TBGY, Section 2.8. 1.2.1 Construction of the Localized Ring If S is a multiplicative subset of the ring R, we define an equivalence relation on R×S by (a,b)∼(c,d) iff for some s∈S we have s(ad−bc)=0. If a∈R and b∈S, we define the fraction a/b as the equivalence class of (a,b). We make the set of fractions into a ring in a natural way. The sum of a/b and c/d is defined as (ad+bc)/bd, and the product of a/bandc/disdefinedasac/bd. Theadditiveidentityis0/1,whichcoincideswith0/sfor every s∈S. The additive inverse of a/b is −(a/b)=(−a)/b. The multiplicative identity is 1/1, which coincides with s/s for every s∈S. To summarize: S−1R isaring. IfR isanintegraldomain, soisS−1R. IfR isanintegraldomainand S =R\{0}, then S−1R is a field, the fraction field of R. There is a natural ring homomorphism h : R → S−1R given by h(a) = a/1. If S has no zero-divisors, then h is a monomorphism, so R can be embedded in S−1R. In particular, a ring R can be embedded in its full ring of fractions S−1R, where S consists of all non-divisors of 0 in R. An integral domain can be embedded in its fraction field. Our goal is to study the relation between prime ideals ofR and prime ideals of S−1R. 1.2.2 Lemma If X is any subset of R, define S−1X = {x/s : x ∈ X,s ∈ S}. If I is an ideal of R, then S−1I is an ideal of S−1R. If J is another ideal of R, then (i) S−1(I+J)=S−1I+S−1J; (ii) S−1(IJ)=(S−1I)(S−1J); (iii) S−1(I∩J)=(S−1I)∩(S−1J); (iv) S−1I is a proper ideal iff S∩I =∅. Proof. The definitions of addition and multiplication in S−1R imply that S−1R is an ideal, and that in (i), (ii) and (iii), the left side is contained in the right side. The reverse inclusions in (i) and (ii) follow from a b at+bs ab ab + = , = . s t st s t st To prove (iii), let a/s = b/t, where a ∈ I, b ∈ J, s,t ∈ S. There exists u ∈ S such that u(at−bs)=0. Then a/s=uat/ust=ubs/ust∈S−1(I∩J). Finally, if s ∈ S ∩ I, then 1/1 = s/s ∈ S−1I, so S−1I = S−1R. Conversely, if S−1I = S−1R, then 1/1 = a/s for some a ∈ I, s ∈ S. There exists t ∈ S such that t(s−a)=0, so at=st∈S∩I. ♣ 6 CHAPTER 1. INTRODUCTION Ideals in S−1R must be of a special form. 1.2.3 Lemma Let h be the natural homomorphism from R to S−1R [see (1.2.1)]. If J is an ideal of S−1R and I =h−1(J), then I is an ideal of R and S−1I =J. Proof. I is an ideal by the basic properties of preimages of sets. Let a/s ∈ S−1I, with a∈I ands∈S. Thena/1=h(a)∈J, soa/s=(a/1)(1/s)∈J. Conversely, leta/s∈J, with a∈R,s∈S. Then h(a)=a/1=(a/s)(s/1)∈J, so a∈I and a/s∈S−1I. ♣ Prime ideals yield sharper results. 1.2.4 Lemma IfI isanyidealofR,thenI ⊆h−1(S−1I). TherewillbeequalityifI isprimeanddisjoint from S. Proof. If a∈I, then h(a)=a/1∈S−1I. Thus assume that I is prime and disjoint from S, and let a∈h−1(S−1I). Then h(a)=a/1∈S−1I, so a/1=b/s for some b∈I,s∈S. There exists t ∈ S such that t(as−b) = 0. Thus ast = bt ∈ I, with st ∈/ I because S∩I =∅. Since I is prime, we have a∈I. ♣ 1.2.5 Lemma If I is a prime ideal of R disjoint from S, then S−1I is a prime ideal of S−1R. Proof. By part (iv) of (1.2.2), S−1I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S−1I, with a,b ∈ R,s,t ∈ S. Then ab/st = c/u for some c ∈ I,u ∈ S. There exists v ∈ S such that v(abu−cst) = 0. Thus abuv = cstv ∈ I, and uv ∈/ I because S∩I = ∅. Since I is prime, ab∈I, hence a∈I or b∈I. Therefore either a/s or b/t belongs to S−1I. ♣ The sequence of lemmas can be assembled to give a precise conclusion. 1.2.6 Theorem There is a one-to-one correspondence between prime ideals P of R that are disjoint from S and prime ideals Q of S−1R, given by P →S−1P and Q→h−1(Q). Proof. By (1.2.3), S−1(h−1(Q))=Q, and by (1.2.4), h−1(S−1P)=P. By (1.2.5), S−1P is a prime ideal, and h−1(Q) is a prime ideal by the basic properties of preimages of sets. If h−1(Q) meets S, then by (1.2.2) part (iv), Q=S−1(h−1(Q))=S−1R, a contradiction. Thus the maps P → S−1P and Q → h−1(Q) are inverses of each other, and the result follows. ♣ 1.2. LOCALIZATION 7 1.2.7 Definitions and Comments If P is a prime ideal of R, then S = R\P is a multiplicative set. In this case, we write R for S−1R, and call it the localization of R at P. We are going to show that R is P P a local ring, that is, a ring with a unique maximal ideal. First, we give some conditions equivalent to the definition of a local ring. 1.2.8 Proposition For a ring R, the following conditions are equivalent. (i) R is a local ring; (ii) There is a proper ideal I of R that contains all nonunits of R; (iii) The set of nonunits of R is an ideal. Proof. (i) implies (ii): If a is a nonunit, then (a) is a proper ideal, hence is contained in the unique maximal ideal I. (ii) implies (iii): If a and b are nonunits, so are a+b and ra. If not, then I contains a unit, so I =R, contradicting the hypothesis. (iii)implies(i): IfI istheidealofnonunits,thenI ismaximal,becauseanylargeridealJ would have to contain a unit, so J =R. If H is any proper ideal, then H cannot contain a unit, so H ⊆I. Therefore I is the unique maximal ideal. ♣ 1.2.9 Theorem R is a local ring. P Proof. Let Q be a maximal ideal of R . Then Q is prime, so by (1.2.6), Q = S−1I P for some prime ideal I of R that is disjoint from S = R\P. In other words, I ⊆ P. Consequently, Q = S−1I ⊆ S−1P. If S−1P = R = S−1R, then by (1.2.2) part (iv), P P is not disjoint from S = R\P, which is impossible. Therefore S−1P is a proper ideal containing every maximal ideal, so it must be the unique maximal ideal. ♣ 1.2.10 Remark It is convenient to write the ideal S−1I as IR . There is no ambiguity, because the P product of an element of I and an arbitrary element of R belongs to I. 1.2.11 Localization of Modules If M is an R-module and S a multiplicative subset of R, we can essentially repeat the construction of (1.2.1) to form the localization of M by S, and thereby divide elements of M by elements of S. If x,y ∈M and s,t∈S, we call (x,s) and (y,t) equivalent if for some u ∈ S, we have u(tx−sy) = 0. The equivalence class of (x,s) is denoted by x/s, and addition is defined by x y tx+sy + = . s t st