1 Number Fields Introduction Algebraic Numbers Algebraic Integers Algebraic Integers Modules over Z...

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1 Contents 1 Number Fields Introduction Algebraic Numbers Algebraic Integers Algebraic Integers Modules over Z Norm, Trace and Discriminant Embeddings in C Discriminant Integral Basis of Number Rings Integral Basis An Integral basis of Q(ζ p ) An algorithm for finding integral basis Factorization of ideals in a number ring Dedekind Domains The Ideal Class Group Unique Factorization of Ideals Ideals in a Dedekind Domain Appendix Decomposition of Prime ideals Splitting of Primes in Extensions Degree of extension, ramification index and inertial degree Normal Extensions, inertia degree and ramification

2 2 CONTENTS 6.4 Ramified primes and discriminant Inertial Degree and order of a prime p (mod q) Finiteness of Class Groups Kummer s Theorem Galois Theory and Prime decomposition Decomposition group and Inertia Group Decomposition field and Inertia field A proof of Quadratic Reciprocity Law Ramified primes revisited Minkowski s Theorem Lattices Minkowski s Theorem Finiteness of Class group Ramification over Z Dirichlet s Unit Theorem Units in the ring of algebraic integers Dirichlet s Unit Theorem

3 Chapter 1 Number Fields 1.1 Introduction Algebraic Number Theory is essentially the study of number fields, which are finite extensions of Q. Such fields can be useful in solving problems which at first appear to involve only rational numbers. Consider for example the following old problem : Example Find all the integer solutions of the equation x 2 + y 2 = z 2, for which x, y, z are pairwise coprime. Solutions. If gcd(x, y) = 1, then it is necessary that z is odd. (If z is even, both x and y are odd. But we know that sums of two odd squares cannot be a square). Next, x 2 + y 2 = (x + iy)(x iy) = z 2. Note that because x, y Z, we know that x + iy, x iy Z[i]. We can prove that Z[i] is a Principal Ideal Domain, and hence, a Unique factorization Domain. This means that every element in Z[i] can be expressed uniquely (up to units) in terms of irreducible elements. We now show that if π is an irreducible element that divides x + iy and x iy then π is a unit. If π x + iy and π x iy, the π 2x. Also, π 2 z 2, implies that π z (Euclid s Lemma). Therefore, π gcd(2x, z). Now, z is odd, and so, 2 z. Furthermore, gcd(x, z) = 1, and hence, gcd(2x, z) = 1. Since π 1, π is a unit. This is a 3

4 4 CHAPTER 1. NUMBER FIELDS contradiction. Hence, x + iy and x iy are relatively prime and therefore, x + iy = αu 2, x iy = βv 2, u, v Z[i] and α, β {±1, ±i}. Observe that if u = m + in, m, n Z and α = 1, then v = m in. Hence, 2x = 2(m 2 n 2 ), or x = m 2 n 2, 2y = 4mn, or y = 2mn. This gives z = m 2 + n 2. The consideration of other α leads to the solutions x = ±2mn, y = ±(m 2 n 2 ), and z = m 2 + n 2. The above example clearly illustrates the need to study number fields other than Q. From the property of Z[i], we are able to determine all the solutions of the Pythagoras equation. Using the fact that Z[ζ 3 ], where ζ 3 = e 2πi/3, is a Unique Factorization Domain, Euler was able to show that there are no non-trivial integer solutions to x 3 + y 3 = z 3. In 1847, the French Mathematician Lamé announced that he had proved the famous Fermat s Last Theorem, which states that there are no nonzero integers x, y, z satisfying x n + y n = z n, n 3. He first reduced the problem to the case when n = p is prime. Then he deduced that x + y, x + ζ p y,, x + ζp p 1 y, with ζ p = e 2πi/p, have no common factors and argued that from x p + y p = z p, one can deduce that x + y = u p 1, x + ζ p y = u p 2,, x + ζ p 1 p = u p p, u i Z[ζ p ]. On this basis, Lamé derived a contradiction. It was immediately pointed out to him by Louiville that Lamé s proof assumed unique factorization in a crucial way and that this unique factorization property may not hold for arbitrary rings. Louiville s fear was later confirmed by Kummer, who showed that Unique factorization property failed for Z[ζ 23 ]. In 1847, Kummer went on to devise his own proof of Fermat s Last Theorem for certain primes p, known as regular primes, surmounting the difficulties of non-uniqueness of factorization by introducing the theory of ideal complex numbers. This marks the beginning of modern algebraic number theory.

5 1.2. ALGEBRAIC NUMBERS Algebraic Numbers We have already seen in Section 1 that numbers arising from finite extensions of Q are useful in solving problems associated with rational numbers or integers. In this section, we begin our study of algebraic numbers. Definition. An algebraic number field is a finite extension of Q. Definition. An algebraic number α is defined as an element which belongs to an algebraic number field. If α is an algebraic number then α lies in an algebraic number field and in particular, Q(α) is a finite extension of Q. Now, a finite extension is algebraic, that is, any element satisfies a polynomial with coefficients in Q. Hence, we have an algebraic number α is a number which satisfies an equation of the form a 0 + a 1 x + + a n x n = 0, a k Q. Conversely, any number satisfying an equation of the above type lies in a finite extension of Q, and is hence, an algebraic number. Example The fields F 1 = Q( 3), F 2 = Q( 2, 7) are all algebraic number fields. Example The field Q(π) is not an algebraic number field. To motivate our next Theorem, let us consider the field Q( 2, 3). By simple calculation, we can show that Q( 2, 3) = Q( 2+ 3). The question is : If α 1, α 2,, α n are algebraic numbers, can we write Q(α 1,, α n ) = Q(θ) for some θ? The answer is yes. Theorem If K is a number field, then K = Q(θ) for some algebraic number θ. Proof. First, note that if K is an algebraic number field, then [K : Q] is finite. Thus, we may pick finitely elements in K such that K = Q(α 1,, α n ). Our aim is to show that K = Q(θ) for some θ. It suffices to show that

6 6 CHAPTER 1. NUMBER FIELDS Q(α, β) = Q(θ). Let p(t) and q(t) be the minimal polynomials of α and β respectively. Suppose and p(t) = (t α)(t α 2 ) (t α n ), q(t) = (t β)(t β 2 ) (t β k ). (The roots of p(t) and q(t) are distinct, why?) Choose c so that α i + cβ j α + cβ, for i = 2, n and j = 2,, k. Define θ = α + cβ. We will show that Q(α, β) = Q(θ). Clearly, Q(α, β) contains θ. Next, α = θ cβ. Then p(θ cβ) = 0 and so, the polynomial r(t) = p(θ ct) has a root at β. Now, r(t) and q(t) has exactly one common root β. For if not, suppose r(β j ) = q(β j ) = 0. Then this implies that θ cβ j is a root of p(t), say α i. This gives θ = α i +cβ j, contradicting our choice of c. Hence, gcd(r(t), q(t)) = t β and t β = q(t)u(t) + r(t)v(t), with u(t), v(t) Q(θ)[t] (r(t) Q(θ)[t]). Hence, β Q(θ). Example The field Q( 2, 3 5) = Q( ). From now on, we can always write K = Q(α) for any number field K and some α K.

7 Chapter 2 Algebraic Integers 2.1 Algebraic Integers Historically, Z was discovered before Q. Now that we are given the field Q(α), which is an analogue of Q, we would like to know if there is an analogue of Z in Q(α). The question is, how do we define this analogue? To answer this question, we need the following definition: Definition. An algebraic integer α is a number in C satisfying an equation a 0 + a 1 x + + x n = 0, a k Z. The set of algebraic integers which belong to an algebraic number field K is denoted by O K. It turns out that this is a ring and it is the analogue of Z which we seek above. (This situation O K K is a generalization of the situation Z Q.) Example The set of algebraic integers in Q is Z. Example Show that the set of algebraic integers in Q( 1) is Z[ 1]. 7

8 8 CHAPTER 2. ALGEBRAIC INTEGERS Solutions. Let α = A + ib is an algebraic integer in Q(i). If B = 0, then α = A satisfies x A = 0. Since A is an integer, A Z. If B 0, then α satisfies x 2 2Ax + A 2 + B 2 = 0, with 2A, A 2 + B 2 Z. If A = n/2, n Z. Now, A 2 + B 2 Z, shows that n 2 + 4B 2 = 4m, m Z. 4B 2 is an integer implies that B = k/2 for some k Z. But this says that n 2 + k 2 = 4m, which is possible only when n and k are both even. Hence, A = n/2 = 2n /2 = n and B = k/2 = 2k /2 = k. Hence α = n + k i, n, k Z. Therefore, the ring of integers of Q(i) is Z[i]. In general, we have Example Let m be a squarefree integer. The set of algebraic integers in the quadratic field Q( m) is {a + b m : a, b Z} if m 2 or 3 (mod 4), { } a + b m : a, b Z, a b (mod 2), if m 1 (mod 4). 2 The above examples show that O K is a ring for quadratic field K. We will show that O K is a ring for any number field. 2.2 Modules over Z Definition. Let R be a ring. By an R-module, we mean an abelian group M (written additively), together with a function α : R M M, for which we write α(r, m) = r m (r R, m M), satisfying (a) (r + s)m = rm + sm, (b) r(m + n) = rm + rn, (c) r(sm) = (rs)m, (d) 1m = m,

9 2.2. MODULES OVER Z 9 for all r, s R, m, n M. Note that if R is a field, an R-module is exactly a vector space. An R- module M is said to be finitely generated if there exist m 1, m 2,, m k such that M = Rm 1 + Rm Rm k. When R is a field, this is just a finite dimensional vector space. Recall that an algebraic integer α K, where K is an algebraic number field is an element that satisfies the equation a 0 + a 1 x + + x n = 0, a i Z. We sometimes say that α is integral over Z. Theorem Let α K. The following are equivalent: (1) The element α is integral over Z. (2) The module Z[α] := {r 0 + r 1 α + + r k α k, r j Z} is a finitely generated Z-module. (3) Z[α] is contained in a subring of K, which is a finitely generated Z- module. (4) There is an Z[α]-module M such that M is finitely generated as an Z- module and the only element y of Z[α] such that ym = 0 is y = 0. Proof. (1) = (2). If the monic polynomial in Z[x] satisfied by α has degree n, then Z[α] is generated by 1, α, α 2,, α n 1 over Z. (2) = (3). Take B = Z[α]. (3) = (4). Take M = B. Clearly M is a Z[α]-module. Also, M finitely generated Z-module since B is. Now, 1 B. So, yb = 0 implies that y = 0. (4) = (1). Let m 1,, m k generates M over Z. Let r ij be such that αm i = k r ij m j. j=1

10 10 CHAPTER 2. ALGEBRAIC INTEGERS Let A be the matrix given by A = αi (r ij ). Now, let A be the adjoint of A. Therefore, m 1 m 2 A A. = 0. m k This means that dm = 0, where d = det(a). Note that d Z[α]. By hypothesis, d = 0. Hence det(a) = 0. But this means that α satisfies a monic equation over Z, and so α is integral over Z. Corollary Suppose α, β K, K an algebraic number field, such that α and β are integral over Z. Then α + β and αβ are both integral over Z. In other words, the set O K is a subring of K. Proof. We know that Z[α] and Z[β] are both finitely generated Z-module. Hence, Z[α, β] := (Z[α])[β] is a finitely generated Z-module (generated by α i β j, 0 i n 1, 0 j m 1). Clearly this contains Z[α + β] and Z[αβ]. Therefore by condition (2) of the above Theorem, α + β and αβ are both integral over Z. Knowing now that the set of elements in K integral over Z is a ring, to determine this ring is far from trivial.

11 Chapter 3 Norm, Trace and Discriminant In the previous two Chapters, we defined Algebraic Numbers and Algebraic Integers. We have also shown that the set of all algebraic integers in an algebraic number field K form a subring of K. In order to know more about this ring O K, we need some basic facts from Galois Theory. In this Chapter, we will concentrate on developing some tools which will enable us to continue our investigation of the structure of O K. 3.1 Embeddings in C Let K be a number field of degree n over Q. We have seen that K = Q(θ) for some θ K. Suppose τ : K C is a monomorphism which fixes Q. Then τ is determined by its action on θ. Since f(θ) = 0, f(τ(θ)) = 0. This implies that τ(θ) is a root of f(x) and therefore, τ(θ) = θ i for some i. Furthermore since there are exactly n distinct roots of f(x) we conclude that there exactly n distinct monomorphisms from K to C which fixes Q. We call these the embeddings of K into C. With this, we now define for each α K, n n T K Q (α) = σ j (α), and N K Q (α) = σ j (α), j=1 where σ i are the embeddings of K into C. Example Find all monomorphisms Q( 3 7) C. 11 j=1

12 12 CHAPTER 3. NORM, TRACE AND DISCRIMINANT Norm and Trace are useful, especially in determining O K. We first observe that if α is an algebraic integer, then the minimal polynomial f(x) of α must have integer coefficients. Since the trace gives 1 the coefficient of x n 1 of f(x) and the norm gives the constant term of f(x) up to ±1, we conclude that they are integers. With this observation, we conclude the following: Example Let m be a squarefree integer. The set of algebraic integers in the quadratic field Q( m) is { a + b m 2 {a + b m : a, b Z} if m 2 or 3 (mod 4), } : a, b Z, a b (mod 2), if m 1 (mod 4). In general, it is difficult to determine O K. To have more examples of O K, we need the notion of discriminant. 3.2 Discriminant Once again, suppose K = Q(θ), σ 1,, σ n be the embeddings from K to C and {α 1,, α n } be a basis of K. We define the discriminant of this basis to be [α 1,, α n ] = {det[σ i (α j )]} 2. Note that if we pick another basis {β 1,, β n } for K, then β k = n c ik α i, i=1 for k = 1, 2,, n and det(c ik ) 0. The product formula for determinants, and the fact that σ i fixes Q, shows that [β 1,, β n ] = (det(c ik )) 2 [α 1,, α n ].

13 3.2. DISCRIMINANT 13 Theorem The discriminant of any basis in K = Q(θ) is rational and nonzero. Proof. Let {1, θ, θ 2,, θ n 1 } be a basis for K. If the conjugates of θ are θ 1,, θ n then [1, θ,, θ n 1 ] = (det θ j i )2. A determinant of the form D = det(t j i ) is called a Vandermonde determinant, and has value D = (t i t j ). Hence, 1 i<j n = [1, θ,, θ n 1 ] = ( (θ i θ j )) 2. Now, D 2 is symmetric in t i, and so is rational. Since the θ i are distinct, 0. Now, by the remarks before the theorem, we find that the discriminant of any basis is rational and non-zero. Example Let K = Q( 2). Let {1, 2} be the basis. Then (σ i (α j )) = ( ) Hence [1, 2] = 8. Example Let K = Q(ζ p ), where ζ p = e 2πi/p. Choose the basis {1, ζ p,, ζp p 1 }. Let p 1 Φ p (x) = (x ζp), j j=1 be the cyclotomic polynomial of degree p. (Warning, the word degree does not refer to the degree of the polynomial!) Now, differentiate Φ p (x) with

14 14 CHAPTER 3. NORM, TRACE AND DISCRIMINANT respect to x to obtain Therefore, p 1 p 1 Φ p(x) = (t ζp). i j=1 i=1 i j p 1 Φ p(ζp) j = (ζp j ζp). i i=1 i j Multiplying all these equations for j = 1,, n, we obtain p 1 p 1 Φ p(ζp) j = (ζp j ζp). i j=1 i,j=1 The left hand side is the N K Q (Φ p(ζ p )) and each factor (ζp j ζp) i on the right hand side appears twice, once as (ζp j ζp), i and once as (ζp i ζp). j Therefore, the right hand is (ζp i ζp) j 2 = ( 1) (p 1)(p 2)/2 [1, ζ p,, ζp p 1 ]. 1 i<j p 1 Hence, Now, Hence, Therefore, = ( 1) (p 1)(p 2)/2 N K Q (Φ p(ζ p )). N(Φ p(ζ p )) = N Φ p (x) = xp 1 x 1. ( pζ p 1 ) p = p p 2. (ζ p 1) = ( 1) (p 1)/2 p p 2.

15 Chapter 4 Integral Basis of Number Rings 4.1 Integral Basis We have shown that K is a number field, then O K is a ring. We show next that O K is a finitely generated Z-module with rank [K : Q] = n. First, from Chapter 2, exercise 3, we know that K = Q(θ) where θ is an algebraic integer. This means that we can always choose {α 1,, α n } (e.g. the set {1, θ,, θ n 1 }) to be a Q-basis for K, where α i O K. The set A = {m 1 α m n α n }, n = [K : Q], is a Z-module of rank n. Furthermore, A O K. This shows that the Z-rank of O K is greater than n. Next, let {α 1,, α n } be a basis for K over Q consisting of entirely algebraic integers. Suppose α O K. Then α = x 1 α x n α n, with x i Q. Applying σ j to the above equation for j = 1,, n, we obtain a set of n equations with n variables. Solving the equations for x i, we conclude that x i = n i δ, where δ = det(σ j (α i )). Write x i = m i d, 15

16 16 CHAPTER 4. INTEGRAL BASIS OF NUMBER RINGS where δ 2 = d, d being the discriminant with respect to the basis {α 1,, α n }. Since d is rational (Theorem 3.2.1), we conclude that x i d Z. Therefore, α is of the form α = m 1α 1 + m n α n. d Therefore O K 1 d A. Since the rank of 1 A is n, we conclude that d Theorem The ring O K is a Z-module of rank n. The above Theorem shows that O K has a basis over Z. We call the basis an integral basis of K (or O K ). For an arbitrary K, finding its integral basis is very challenging. We will first illustrate by finding the integral basis for the cyclotomic field Q(ζ p ), where p is a prime. 4.2 An Integral basis of Q(ζ p ) The purpose of this section is to find the ring of integers of K = Q(ζ p ). Before we continue, let us take a closer look at Example Note that in the case m 1 (mod 4), the denominator of an algebraic integer can be some number other than 1. In that case, the denominator is 2. This is rather counter-intuitive since our notion of integers refer to a rational number with trivial denominator (i.e. 1 or 1). With this example in mind, we cannot conclude that the ring of integers O K is of the form Z[ζ p ] since some algebraic integers in O K may have non-trivial denominators. We will show in the next Theorem that this is impossible. Theorem The ring of integers of Q(ζ p ) is Z[ζ p ]. Proof. Step 1. Let ζ = ζ p. We first check that T K Q (ζ) = 1 and N K Q (ζ) = 1. We will now compute T K Q (1 ζ) and N K Q (1 ζ). In the equation x p 1 + x p , replace x by 1 y. Then g(x) := (1 y) p 1 + (1 y) p = y p 1 py p p.

17 4.2. AN INTEGRAL BASIS OF Q(ζ P ) 17 Note that 1 ζ is a root of g(x), which is irreducible. Hence, N K Q (1 ζ) = p and T K Q (1 ζ) = p. Since 1 ζ j, j = 1,, p 1 are also roots of g(x), we conclude that N K Q (1 ζ) = (1 ζ)(1 ζ 2 ) (1 ζ p 1 ) = p. (4.2.1) Step 2. Let O be the ring of integers in Q(ζ). Then O contains ζ and its powers. Our next aim is to show that (1 ζ)o Z = pz. (4.2.2) (Recall that ur, with u R, is the principal ideal generated by u.) We know that p (1 ζ)o by (4.2.1). Thus pz (1 ζ)o Z. Since pz is a maximal ideal of Z, (1 ζ)o Z = pz or Z. If the latter is true, then 1 (1 ζ)o and hence (1 ζ) is a unit in O, or (1 ζ)w = 1, w O. Let σ j Gal(K Q) and applying this to the above yields (1 ζ j )σ j (w) = 1. This implies that (1 ζ j ) are also units, j = 1,, p 1. But this contradicts (4.2.1) as it implies that p is a unit. Hence, (1 ζ)o Z = pz. Step 3. Next, we prove that for any y O, T K Q (y(1 ζ)) pz. p 1 p 1 T K Q (y(1 ζ)) = σ j (y(1 ζ)) = σ j (y)σ j (1 ζ) j=1 j=1 p 1 = σ j (y)(1 ζ j ) j=1 p 1 = (1 ζ) σ j (y)(1 + ζ + + ζ j 1 ) (1 ζ)o. j=1

18 18 CHAPTER 4. INTEGRAL BASIS OF NUMBER RINGS Now, if u O is an algebraic integer in Q, then u satisfies x a = 0 for a Z. Hence, O Q = Z. Since T K Q (y(1 ζ)) O Q, T K Q (y(1 ζ)) Z. Therefore, Step 4. Now, take any u O. Then Hence, T K Q (y(1 ζ)) Z (1 ζ)o = pz. u = a 0 + a 1 ζ + + a p 2 ζ p 2. u(1 ζ) = a 0 (1 ζ) + + a p 2 (ζ p 2 ζ p 1 ). Taking Trace, we conclude that By Previous step, we find that Hence a 0 Z. Next, write T K Q (u(1 ζ)) = a 0 T K Q (1 ζ). a 0 p pz. (u a 0 )ζ 1 = a 1 + a 2 ζ + + a p 2 ζ p 3. Since, ζ 1 O, we may apply the same argument successively and conclude that a j Z. This completes the proof. 4.3 An algorithm for finding integral basis Theorem is non-trivial. This shows that finding the ring of integers of an algebraic number field is not an easy task. In this section, we will show using several examples, how we can use Norm, Trace and Discriminant to find integral basis of certain number fields. Let us now quote some facts which will enable us to construct integral basis for certain algebraic number fields. Our examples will show that it is rare that {1, θ, θ 2,, θ n } forms an integral basis, although this is usually a basis for Q(θ).

19 4.3. AN ALGORITHM FOR FINDING INTEGRAL BASIS 19 Let O K µ 1 Z µ n Z. Let G is a subgroup of O K of rank n. Then it can be shown by induction that G d 1 µ 1 Z d n µ n Z, d i > 0. Note that O K : G = d 1 d 2 d n. Let be the discriminant of the basis {µ 1,, µ n } and define Then G := [d 1 µ 1,, d n µ n ]. G = (d 1 d n ) 2. Therefore the index O K : G 2 divides G. Now, suppose O K : G > 1. Then there exist a prime p O K : G. By Cayley s Theorem, there exist an element u + G O K /G such that pu + G = G, or pu G. This means that u = g/p for some g G. We may now use this observation to determine algebraic integers of a number field. First, we take an initial guess for the basis of O K, say, {α 1,, α n }. We compute the discriminant D = [α 1,, α n ] and since p 2, if we can show that there are no algebraic integers of the type g/p, with g = a 1 α a n α n, a i Z, then we conclude that {α 1,, α n } is a basis for O K. Otherwise we replace one of the α j s by the new integer g/p and repeat the procedure. Example Find the ring of integers of Q( 3 5). Our initial guess is {1, 3 5, }. The discriminant is and so the possible candidates for p are p = 3 or 5. Let θ = 3 5. (a) Can α = 1 3 (λ 1 + λ 2 θ + λ 3 θ 2 )? (b) Can α = 1 5 (λ 1 + λ 2 θ + λ 3 θ 2 )? Example Find the ring of integers of Q( 3 10).

20 20 CHAPTER 4. INTEGRAL BASIS OF NUMBER RINGS

21 Chapter 5 Factorization of ideals in a number ring 5.1 Dedekind Domains We have seen that not all number rings are Unique Factorization Domain. We will show however that all ideals in a Number Ring factor uniquely into prime ideals. This can be regarded as a generalization in Z, where the ideals are just the principal ideals (n) and the prime ideals are (p), with p being prime. We will first show that Number Rings are Dedekind Domains and that ideals in Dedekind Domains factor uniquely into prime ideals. Definition. A Dedekind Domain is an integral domain R which satisfies 1. Every ideal is finitely generated 2. Every nonzero prime ideal is a maximal ideal 3. R is integrally closed in its field of fractions K = {α/β : α, β R, β 0}. The first condition is equivalent to saying that every increasing sequence of ideals is eventually constant. It is also equivalent to the statement that every non-empty set S of ideals has a (not necessarily unique) maximal member. A ring satisfying condition 1 or its equivalence is called a Noetherian ring. 21

22 22 CHAPTER 5. FACTORIZATION OF IDEALS IN A NUMBER RING Condition 3 says simply that if α/β K is a root of certain monic polynomial with coefficients in R, then α/β R. Theorem Every number ring is a Dedekind Domain. Proof. Condition 1. Let K be a number field and O K be its number ring. We know that every number ring is a free abelian group of rank n = [K : Q] (Theorem 4.1.1). Let I be an ideal of O K. Then I is also free of finite rank, I = Zα Zα m O K α O K α m. Therefore, I is a finitely generated O K -ideal. Condition 2. Let p be a prime ideal. We wish to show that p is a maximal ideal. We note that it suffices to show that O K /p is finite. (Note that every finite integral domain is a field. Exercise.) More generally, let I be any nonzero ideal in a number ring. We will show that O K /p is finite. Let α I and let m = N K Q (α). Note that m = α β for some β O K. Since I is an ideal, m I. Now, O K /(m)o K = m n, and (m) I implies that O K /I O K /(m)o K. This implies that O K /I is a finite set. Condition 3. In Corollary 2.2.2, we see that if α and β are algebraic integers, then Z[α, β] is a finitely generated Z-module. By induction Z[α 1,, α m ] is a finitely generated Z-module. Let u/v K be integral over O K, i.e., u/v satisfies a monic polynomial equation with coefficients in O K. This means that u/v Z[α 1,, α m, u/v] where α i are the coefficients appearing in the polynomial satisfied by u/v. Note that B[u/v] is a finitely generated B-module where B = Z[α 1,, α m ]. But B is a finitely generated Z-module. Therefore, Z[α 1,, α m, u/v]

23 5.2. THE IDEAL CLASS GROUP 23 is a finitely generated Z-module. By Theorem 2.2.1, u/v is an algebraic integer. Therefore O K is integrally closed in its field of fractions. With the above definition, we see that Z[ 1] is a Dedekind Domain but Z[ 3] is not a Dedekind Domain. (Z[ 3] is not Dedekind because the number 1+ 3 is integral over Z, and therefore over O 2 K, with K = Q( 3) but not in O K. 5.2 The Ideal Class Group We begin with an important fact. Theorem Let I be an ideal in a Dedekind Domain R. Then there is an ideal J such that IJ is principal. Before we prove the Theorem, we need two lemmas. Lemma In a Dedekind Domain, every ideal contains a product of prime ideals. Proof. Suppose not. Then the set of ideals which do not contain such products is nonempty, and consequently has a maximal member M by condition (1 ). The ideal M is not a prime since it does not contain a product of primes, so, there exist r, s, R M such that rs M. The ideals M + (r) and M + (s) are strictly bigger than M and so they must contain product of primes. But this means that (M + (r))(m + (s)) contains a product of primes. Since (M + (r))(m + (s)) M, we conclude that M contains a product of prime ideals, which is a contradiction. Lemma Let a be a proper ideal in a Dedekind Domain R with field of fractions K. Then there is an element γ K R such that γa R. Proof. Fix any nonzero element a a. By Lemma 5.2.2, the principal ideal (a) contains a product of primes. Fix primes p 1,, p r such that p 1 p 2 p r (a), with minimal r.

24 24 CHAPTER 5. FACTORIZATION OF IDEALS IN A NUMBER RING Next, every proper ideal is contained in a maximal ideal, which is necessarily a prime. Hence a p for some prime ideal p. Therefore, p 1 p 2 p r p. Now we claim that p must contain one of p i. If not, then there exist a j p j such that a 1 a 2 a r p but none of a j p, which contradicts the fact that p is a prime ideal. Without loss of generality suppose p 1 p. By condition (2) for Dedekind Domain, we must have p 1 = p. Now, recall that (a) cannot contain a product of fewer than r primes. In particular, there exist an element b p 2 p r and b (a). Now take γ = b/a. Note that γ is not in R and γa γp 1 1 a p 1p 2 p r 1 (a) R. a Proof of Theorem Let α be any nonzero element of I and let J := {β R : βi (α)}. Now check that J is an ideal of R. Clearly IJ (α). Next, let a = 1 α IJ. The set a is contained in R and it is an ideal (Check). If a = R, then IJ = (α) and we are done. Otherwise a is a proper ideal and we apply Lemma to conclude that γa R, for some γ K R. Since R is integrally closed, we will show that γ satisfies a monic polynomial over R and conclude that γ R, which leads to a contradiction. Observe that a contains J since α I (j J implies that j = 1 αj a) α and thus, γj γa R. Now, γa R, which means that γij (α). This implies that (γj)i (α). Since γj R, this gives γj J by definition of J. Now choose a set of generators m 1,, m k for J. The above inclusion then gives m 1 m 2 m k m 1 m 2 γ. = M., for some matrix M. By computing the determinant of γi M we arrive at a monic polynomial satisfied by γ. m k

25 5.3. UNIQUE FACTORIZATION OF IDEALS 25 Definition. Let I and J be two ideals of a number ring O K. We say that I J (I is related to J) if ai = bj for some a, b O K. It turns out that this is an equivalence relation on the set of ideals of O K. Now, let C K := {[I 1 ],, [I h ] } denote the set of equivalent classes under. We recall that given two ideals I and J in a ring R, IJ is defined as the set of elements in R which are linear combinations of ij with i I and j J. It is an ideal and so, we may define [I] [J] = [IJ]. We will now show that (C K, ) forms an abelian group. We first show that if p = (a) is a principal ideal, then [p] [I] = [I]. This is clear since ai = ((a)i), which means that [I] [(a)i]. The associativity is clear since [I] [JK] = [IJK] while [IJ] [K] = [IJK]. The most difficult part is perhaps the existence of an inverse for any equivalent class [I]. But this amounts to seeking J so that [IJ] = [p], the class containing the principal ideals. The existence of such a J is guaranteed by Theorem Hence, Theorem The set C K with binary operation is an abelian group. It will turn out that C K is a finite group. But we will defer this to a later chapter. Example The group C K is the trivial group if K = Q( 1) since all ideals are principal. Conversely, if we can show that C K = 1 for a chosen K, then we may conclude that O K is a Principal Ideal Domain. It turns out that when K is imaginary quadratic, there are only nine fields for which O K is a Principal Ideal Domain. These are Q( m) for m = 1, 2, 3, 7, 11, 19, 43, 67, Unique Factorization of Ideals Definition. Given two nonzero ideals I, J of R, we say that I J if there exist an ideal K in R such that J = IK. Lemma If a and b are ideals in a Dedekind Domain R, then a b if and only if b a.

26 26 CHAPTER 5. FACTORIZATION OF IDEALS IN A NUMBER RING Proof. Clearly, if a b, then b = ac for some ideal c and so b a. Conversely, suppose b a, then there exist an ideal J such that Ja = (α) is principal by Theorem and Ja contains Jb. Write c = 1 Jb. The set c is an ideal in α R and ca = b. Hence a b. Theorem Every proper ideal in a Dedekind Domain R is uniquely representable as a product of prime ideals. Proof. We first show that every proper ideal is a product of prime ideals. Suppose not. Then the set of proper ideals not representable as a product of primes is nonempty and has a maximal member m. Now, m is contained in some maximal ideal p (which is also prime). By Lemma 5.3.1, we conclude that m = pi for some ideal I. Now, m I for otherwise, we can find n so that mn = mi = (α), which implies that R = p, a contradiction. Since I contains m it is a product of primes and m = pi shows that m is a product of primes, a contradiction. It remains to show that the product representation is unique. Suppose p 1 p 2 p r = q 1 q 2 q s, where p i and q j are primes, not necessarily distinct. Then q 1 q s p 1. This implies that q j p 1 for some j. (If not, pick a j p 1 q j and arrive at a contradiction.) By renaming the indices, we may assume that q 1 p 1. But q 1 is a maximal ideal implies that q 1 = p 1. By Theorem we can find P so that pp 1 = (α) and so multiplying both sides of the representations by P, we find that p 2 p r = q 2 q s. Continuing, we find that q i = P i for all i and we are done. Example We have seen that Unique Factorization of elements does not hold for Z[ 5]. For example, 2 3 = (1 + 5)(1 5). However, (2) = (2, 1 + 5) 2, (3) = (3, 1 + 5)(3, 1 5) as ideals. Furthermore, (1 + 5) = (2, 1 + 5)(3, 1 + 5) and (1 5) = (2, 1 + 5)(3, 1 5). And so, (6) does indeed factor into unique product of prime ideals. 5.4 Ideals in a Dedekind Domain In this section, we will show that every proper ideal in a Dedekind Domain is generated by at most two elements. More precisely, if I is a nonzero ideal then there exist a, b I such that I = ao K + bo K.

27 5.5. APPENDIX 27 First, we need a definition. We say that two ideals I, J are coprime if I + J = O K. Note that if p is a prime ideal and m is any nonzero ideal not contained in p then p + m = O K (since this is larger than the maximal ideal p.) Now, 1 p + m implies that 1 = p + m, and so, 1 = (p + m) n. This implies that 1 p n + m for any integer n. Hence, we also have p n + m = O K. Next, suppose m and n are ideals which are coprime. It is clear that mn m n. Conversely, if x m n then x = a m and x = b n. Since 1 m + n, 1 = a + b for some a m and b n. Therefore, x = xa + xb = ba + ab mn. Now, let I be a proper ideal of O K. Then I = p α i i. Pick a I. Then the ideal (a) I and so, (a) = I J, for some ideal J. Write (a) = i pδ i i j qβ j j, where p i I, δ i α i, and q j I. Now, the Chinese Remainder Theorem (see Appendix) states that given prime ideals m i, O K / m k i i = O K / m k i i i O K /m k i i. This means that we can find b O K, such that b b i (mod p α i+1 i ) and where This means that and that b 1(mod q j ), b i p α i i p α i+1 i. (b) = i p α i i k (a) + (b) = I. n γ k k, 5.5 Appendix We will prove the Chinese Remainder Theorem in the following form : O K /(I J) = O K /I O K /J,

28 28 CHAPTER 5. FACTORIZATION OF IDEALS IN A NUMBER RING if I + J = O K. The general case follows by induction. Consider the natural map f : O K O K /I O K /J. The kernel of this map is clearly I J = IJ. To show the isomorphism, we need to show that the map is surjective. Since I + J = O K, we know that a + b = 1 for some a I and b J. Now, given r 1 + I, and r 2 + J, set r = r 1 b + r 2 a. Therefore r is mapped to (r + I, r + J) = (r 1 (1 a) + I, r 2 (1 b) + J) = (r 1 + I, r 2 + J).

29 Chapter 6 Decomposition of Prime ideals 6.1 Splitting of Primes in Extensions For an arbitrary field extension K, denote S K as the set of prime ideals in O K. Suppose L is a finite extension of K. We want to relate S L and S K, i.e., to understand the relation between prime ideals in O K and that in O L. Before we continue, consider the following example. Suppose L = Q( 1) and K = Q. Then the primes in Z are of the form (p), with p being prime. Now, pz is a prime ideal in Z but the ideal po L may not be a prime ideal in O L. For example, 5Z = (2 + i)o L (2 i)o L. Note that (2 + i)o L is a prime ideal in O L. So there is a relation between primes in O L and primes in O K. In the above case, we say that 5Z splits in O L. As another example, we consider 3Z. This ideal does not split in O L since there is no solution to x 2 + y 2 = 3. Hence 3O L remains a prime ideal in O L. We say that the prime 3Z is inert in O L. Finally, consider the example 2Z. In this case, 2O L = ((1 + i)o L ) 2. This is different from the case of 5Z where the two prime ideals are distinct. If a prime ideal splits into primes which are not distinct, we say that the prime ideal ramifies in O L. In this case, 2Z ramifies in O L. We now prove a simple result, which relates elements in S K and elements in S L. Theorem Let L be an extension of K. Let p be a prime ideal of O K and q a prime ideal of O L. Then the following conditions are equivalent: 29

30 30 CHAPTER 6. DECOMPOSITION OF PRIME IDEALS (1) q po L (2) po L q (3) p q (4) q O K = p (5) q K = p. Proof. The equivalence of (1) and (2) follows from Lemma The equivalence of (2) and (3) is trivial since q is an ideal in O L. Clearly, (4) implies (3) and that (4) is equivalent to (5). It remains to prove (3) implies (4). Since p q O K and p is a maximal ideal, q O K = O K or p. If q O K = O K then 1 q and q = O L, a contradiction. Hence the result. When any one of the conditions in the above Theorem holds, we say that q lies over p or p lies under q. For example, (2 + i)z[i] lies over 5Z. Theorem Every prime ideal q of O L lies over a unique prime ideal p of O K ; every prime p of O K lies under at least one prime ideal q of O L. Proof. The first part is equivalent to showing that a = q O K is a prime ideal in O K. Since 1 q, a is not O K. Let a, b O K and ab a. If a a then a q. Since q is a prime ideal, b q. Hence, b a and a is a prime ideal. Also, a is a nonzero ideal; this can be seen by taking a nonzero element from q and observe that its norm is in O K, which is nonzero. Next, let p be a prime ideal in O K. If po L O L, then po L is a product of prime ideals in O L and these prime ideals satisfy Condition (1) of Theorem So, it suffices to show that po L O L. By Lemma 5.2.3, there exist a γ K O K such that γp O K. Then γpo L O L. If 1 po L, then γ O L and γ is an algebraic integer, contradiction to our choice of γ. As we have noted, the primes lying over a given p are the ones that divide po L. Suppose po L = q e 1 1 q er r, where q i are prime ideals in O L, we call the numbers e i := e i (q i p) the ramification index of q i over p. Example We have seen that if K = Q and L = Q(ζ p ), the cyclotomic field, then (p) = (1 ζ)(1 ζ 2 ) (1 ζ p 1 ).

31 6.2. DEGREE OF EXTENSION, RAMIFICATION INDEX AND INERTIAL DEGREE31 Note that (1 ζ j )/(1 ζ) is always a unit (Check) and hence, (1 ζ j )O L = (1 ζ)o L. Hence, (p)o L = ((1 ζ)o L ) p 1 and the ramification index (we will see later that (1 ζ p )O L is a prime ideal) is e((1 ζ) (p)) = p 1. Next, consider the prime p in O K. Since p is also a maximal ideal of O K, we find that O K /p is a field. Similarly, if q lies over p then O L /q is also a field. Now, consider the homomorphism ϕ : O K O L /q, defined by a a + q. The kernel of ϕ is q O K = p. Hence, O K /p may be viewed as a subfield of O L /q. The fields O K /p and O L /q are called the residue fields associated with p and q respectively. We know that these are finite fields (we have shown this when we proved that a prime ideal is a maximal ideal). Hence O L /q is an extension of finite degree over O K /p. Let f = f(q p) be the degree of the extension. This is known as the inertial degree of q over p. Example Let L = Q( 1) and K = Q. Let p = (3) and q = 3O L. We have f(3o L 3O K ) = Degree of extension, ramification index and inertial degree In this section, we will prove a very interesting result that relates the degree of extension, ramification index and the inertial degree. Theorem Let n = [L : K]. Let q 1, q 2,, q r be the primes of O L lying over p of O K. Denote the corresponding ramification indices and inertial degrees by e i, and f i, 1 i r. Then n = r e i f i. i=1 We will prove the above theorem as we establish the following result:

32 32 CHAPTER 6. DECOMPOSITION OF PRIME IDEALS Theorem Let O L, O K, K, L be as in the above Theorem, n = [L : K]. Let I be an ideal of a ring R and denote N (I) := R/I. (a) For ideals I and J in O K, (b) Let I be an ideal in O K. Then N (IJ) = N (I)N (J). N (IO L ) = N (IO K ) n. (c) Let α O K, α 0. For the principal ideal (α), N ((α)o K ) = N K Q (α). The expression N ( ) is called the norm of an ideal and (c) establishes the relationship between this ideal norm and the usual norm of an element over Q. Proof of Theorem (a). We first prove this for I and J relatively prime and then show that N (p m ) = N (p) m for all prime ideals p. This will imply that N (p m 1 1 p mr r ) = N (p m 1 1 ) N (p mr r ) = N (p 1 ) m1 N (p r ) mr. Factoring I and J into prime ideals and applying the above result yields (a). So, assume I and J are relatively prime. Then I+J = O K and IJ = I J. By the Chinese Remainder Theorem, O K /IJ O K /I O K /J. Hence the result. Next consider N (p m ). We have a chain of ideals p m p 2 p O K. Note that N (p m ) = O K /p p/p 2 p m 1 /p m. We now show that O K /p = p k /p k+1, for 1 k m 1. To show this, we consider first the isomorphism O K /p αo K /αp, where α p k p k+1. Now, αo K p k induces the homomorphism ϕ : αo K p k /p k+1,

33 6.2. DEGREE OF EXTENSION, RAMIFICATION INDEX AND INERTIAL DEGREE33 whose kernel is αo K p k+1. Hence the image of ϕ is isomorphic to αo K /(αo K p k+1 ) (αo K + p k+1 )/p k+1, by second isomorphism theorem. Now the following calculations conclude our result : Let (α) = p k m, where m is relatively prime to p. Then (α)o K + p k+1 = p k (m + p) = p k. Also, (α)o K p k+1 = p k (m p) = p k mp = (α)p. Proof of Theorem 6.2.1, special case. We will prove the Theorem for K = Q. In this case, p = pz for some prime p Z. We have Hence On the other hand, we know that hence the result. po L = q e i i. N (po L ) = N (q) e i = p f ie i. N (po L ) = p n, Proof of Theorem (b). it suffices to prove the result for prime ideal p. Notice that O L /po L is a vector space over O K /p. We claim that the dimension of this vector space is n. First, we show that the dimension is at most n. It will be sufficient to prove that any n + 1 elements are linearly dependent. Fix α 1,, α n+1 + po L O L /po L. We must show that they are linearly dependent over O K /p. Now, we know that α 1,, α n+1 are linearly dependent over K, hence over O K. Hence, there exist β 1,, β n+1 O K such that n+1 β j α j = 0, (6.2.1) j=1 with at least one β i 0. Our task is to show that the corresponding elements of β i in O K /p are not all zeros. Consider the principal ideals B i = (β i )O K.

34 34 CHAPTER 6. DECOMPOSITION OF PRIME IDEALS We let m = B i. First, if m = O K, this means that there is at least one β i that is not in p which when modulo p is nonzero. Now suppose m is a proper ideal contained in p. Let B be such that Bm = (β) (β)p. This means that there is an element α B such that αm βp. Let γ = α/β K. This element has the property that γm O K and γm p. Multiplying γ into (6.2.1), we conclude that βi γ α i = 0 hold. Now the coefficients in this equation when modulo p do not vanish for otherwise, γm po L. Hence, we conclude that α i, 1 i n + 1 are linearly dependent over O K /p and the dimension of O L /po L over O K /po K is at most n. To establish equality, we first consider p Z = pz. Let p i be all the primes of O K lying over p. We know that O L /po L is a vector space of dimension n i n over O K /p. Set e i = e(p i p) and f i = f(p i p). Then by special case of Theorem 6.2.1, we conclude that m = e i f i, m = [K : Q]. Now, po K = p e i i, and hence pol = (piol) ei. Hence, by Theorem (a), N (po L ) = N (p i O L ) e i = N (p i O K ) n ie i = (p f i ) n ie i. On the other hand, N (po L ) = N (po K ) n = N (pz) nm = p nm. Hence, mn = e i f i n i. Since n i n and e i f i = m, we conclude that n i = n. Therefore, N (po L ) = N (po K ) n. Completion of the proof of Theorem We have po L = q e i i. Hence, N (po L ) = N (q i O L ) e i = N (po K ) f ie i, by Theorem (a). On the other hand Theorem (b) shows that N (po L ) = N (po K ) n.

35 6.2. DEGREE OF EXTENSION, RAMIFICATION INDEX AND INERTIAL DEGREE35 Hence the result. Proof of Theorem (c). Extend K to a normal extension M of Q. For each embedding σ : K C, we have N (σ(α)o M ) = N (αo M ). This is true since O M /αo M O M /σ(α)o M. Now set N = N K Q (α). Then by Theorem (a), we have N (NO M ) = N (σ(α)o M ) = N (αo M ) n. Using Theorem (b), we conclude that N (NO M ) = N mn, where m = [M : K]. Again Theorem (b) shows that N (αo M ) = N (αo K ) m. Therefore, N = N (αo K ). Application. As an application of these results, we now show that (1 ζ p ) in Z[ζ p ] (ζ p = e 2πi/p ) is a prime in two ways. We have seen that (1 ζ p ) p 1 = pz[ζ p ]. Since p 1 is the degree of Q(ζ p ) over Q, we conclude that there is no more splitting occurring in Z[ζ p ] for p. Hence (1 ζ p ) is a prime ideal. We can also conclude the fact that (1 ζ p ) is prime by computing the norm of the ideal. Note that by Theorem we conclude that Hence, and we conclude that N ((1 ζ p )) p 1 = N ((1 ζ p ) p 1 ) = N ((p)) = p p 1. N ((1 ζ p )) = p, N ((1 ζ p )) = p. This shows that (1 ζ p ) is prime by Theorem (a).

36 36 CHAPTER 6. DECOMPOSITION OF PRIME IDEALS In the following examples, we give the decompositions of certain ideals into primes in certain extension of Q. We will give in the final section a method to determine such decompositions. Meanwhile we will just state the results as illustrations. Example Let K = Q(2 1/3 ). Note that 2O K = (2 1/3 ) 3, and so, f((α) 2) = 1. In this same field, 5O K = [5, α + 2][5, α 2 + 3α 1] = pq. Note that in this case, f(p 5) = 1 and f(q 5) = 2. Example Let α satisfies the equation α 3 α 1 = 0. Suppose K = Q(α). Then 23O K = [23, α 10] 2 [23, α 3] = p 2 q. Note that the ramification indices of p and q are different. In Example and Example 6.2.2, we see that ramification index and inertia degrees may differ. However, this only happens in non-normal extension. We will show in the next section that inertia degree and ramification indices are the same for primes in L lying over p K if L is normal over K. 6.3 Normal Extensions, inertia degree and ramification Let L be a normal extension of K and p be a prime ideal in O K. Suppose q lies over p then σ(q) is also a prime ideal lying over p for σ Gal(L K). For if p q then σ(p) = p σ(q). Also, if ab σ(q), then σ 1 (a)σ 1 (b) q. If a σ(q) then σ 1 (a) q, and hence, σ 1 (b) q, since q is a prime ideal. This implies that b σ(q) and hence, σ(q) is also a prime ideal. We now show that Gal(L K) permutes transitively the primes lying over p. Theorem Let q and q be two prime ideals lying over p. Then there exists a σ G = Gal(L K) such that σ(q) = q. Proof. Suppose not. Then σ(q) q for all σ G. By Chinese Remainder Theorem, there is a solution to the system of congruences x 0 (mod q ),

37 6.4. RAMIFIED PRIMES AND DISCRIMINANT 37 x 1 (mod σ(q)) for all σ G. Letting α O L be such a solution we have N L K (α) O K q = p, since one of the factors of N L K (α) is α q. On the other hand, we have α σ(q) for each σ G, hence, σ 1 (α) q. We can express N L K (α) as product of σ 1 (α) and since none of these are in q, it follows that N L K (α) q. But we have already seen that N L K (α) p q. Corollary If L is normal over K and q and q are prime ideals over p, then e(q p) = e(q p) and f(q p) = f(q p). The first statement follows from the unique factorization of ideals and the transitive action of G. The second follows from the isomorphism O L /q O L /σ(q) via x + q σ(x) + σ(q), where σ(q) = q. The Corollary shows that po L = (q 1 q g ) e, and each q i has the same inertial degree f over p. Hence, efg = [L : K]. 6.4 Ramified primes and discriminant Definition.Let K, L, O K, O L be as usual; a prime p of O K is said to be ramified in O L if and only if e(q p) > 1 for some prime q of O L lying over p. We have seen that p ramifies in O Q(ζp) and that p ramifies in O Q( ±p). Note that these primes divide the discriminant of the field, i.e., the discriminant of a integral basis of L over Z. We prove in this section the following general statement that

38 38 CHAPTER 6. DECOMPOSITION OF PRIME IDEALS Theorem If p Z ramifies in O L, then p (α 1,, α n ) where {α 1,, α n } is an integral basis of L. (We shall write (O L ) to denote (α 1,, α n ). The converse of the above statement is also true. Indeed if p (O L ), then p ramifies in L. Proof. Let p be a prime of O L lying over p such that e(p p) > 1. Then po L = pi, with I divisible by all primes lying above p in O L. Let σ 1,, σ n be all the embeddings from K to C. As usual, extend σ i to automorphisms of some extension M of L which is normal over Q. Let α 1,, α n be any integral basis for O L. Choose α I po L. In particular, α is in all primes lying above p but not in p. If we write α = m 1 α m n α n then p m 1, (after renaming). Set d = (O L ). Then using the fact that (α 1 + l 2 α l n α n, α 2,, α n ) = (α 1,, α n ), we deduce that (m 1 α 1,, α n ) = (α, α 2,, α n ) and hence, (α, α 2,, α n ) = m 2 1d. Since p m 1, it suffices to show that p (α, α 2,, α n ). Recall α is in every prime ideal in O L lying over p. Let q be a prime ideal in O M lying over p. Then σ(q) lies over p and therefore lies over some prime ideals containing p in O L. In particular, α σ 1 (q) for all σ Gal(M Q). σ(α) q. So, q contains (α, α 2,, α n ). Since the discriminant is in Z, this shows that (α, α 2,, α n ) q Z = pz and we are done. Corollary Only finitely many primes of Z are ramified in a number ring O L.

39 6.5. INERTIAL DEGREE AND ORDER OF A PRIME P (MOD Q) Inertial Degree and order of a prime p (mod q) In this section we prove an interesting fact relating Inertial Degree and order of a prime p (mod q) (p q, q odd). (Here, we say that m is the order of p (mod q) if m is the smallest integer satisfying p m 1 (mod q).) Let p and q be odd primes. Let L = Q(ζ q ). We know that the only primes that divide the discriminant of the ring O L is q. Suppose p q, then we conclude that p is unramified in O L (or simply L). Theorem Let p be a prime lying above p in O L. Then the order of p (mod q) is f(p p). Proof. Let p lies over p and consider Z[ζ q ]/p. This is a field extension of Z p of degree f. The group G = Gal(Z[ζ q ]/p Z p ) is cyclic of order f. Consider the map σ : Z[ζ q ]/p Z[ζ q ]/p, given by x + p x p + p. Note that since Z[ζ q ]/p is finite and σ is injective (because p is a prime ideal), we conclude that σ G. We claim that σ generates G. Suppose not. Then σ f = 1 for some f f. In particular, σ f (x + p) = (x + p) pf = x + p. So each element x + p Z[ζ q ]/p satisfies the equation X pf X = 0. Since we are in a field, there are at most p f elements satisfying this equation, and so Z[ζ q ]/p p f < p f, by assumption that f < f. Hence a contradiction. Therefore, σ has order exactly f. Next, ζ q + p generates the field Z[ζ q ]/p. So, the generator σ of the Galois group must satisfy σ f (ζ q + p) ζ q + p, f < f, for otherwise, σ f = 1. Hence, f is the minimal number for which (ζ q +p) pf = (ζ q + p). Note that this means that ζ pf q ζ q (mod p).

40 40 CHAPTER 6. DECOMPOSITION OF PRIME IDEALS We want to conclude that ζq pf We first note that If ζ pf 1 q = ζ q and hence that p f 1 (mod q). q 1 q = (1 ζq j ). (6.5.1) j=1 1 0 (mod p) for with ζ pf 1 q q 0 1, then (mod p). By (6.5.1), we conclude that p q, which is a contradiction. Example Let p = 11 and q = 7. To find the minimal f for which 11 f 1 (mod 23), we consider the splitting of p in Z[ζ 7 ], namely, that pz[ζ 7 ] = p 1 p 2. This implies that f = 3. A quick check shows that indeed (mod 7). 6.6 Finiteness of Class Groups In this section we use Theorem to prove the finiteness of Class group C K. First we establish the following Theorem: Theorem Let K be a number field. There is a positive real number λ (depending on K) such that every nonzero ideal I of O K contains a nonzero element α with N K Q (α) λn (I). Proof. Fix an integral basis α 1,, α n for O K and let σ 1,, σ n denote the embeddings of K in C. We claim that λ can be taken to be n n σ i α j. i=1 j=1 For any ideal I, let m be the unique positive integer satisfying m n N (I) < (m + 1) n, and consider the (m + 1) n members of O K n m j α j, m j Z, 0 m j m. j=1