CLASS FIELD THEORY AND THE THEORY OF N-FERMAT PRIMES ANDREW KOBIN. A Thesis Submitted to the Graduate Faculty of

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1 CLASS FIELD THEORY AND THE THEORY OF N-FERMAT PRIMES BY ANDREW KOBIN A Thesis Submitted to the Graduate Faculty of WAKE FOREST UNIVERSITY GRADUATE SCHOOL OF ARTS AND SCIENCES in Partial Fulfillment of the Requirements for the Degree of MASTER OF ARTS Mathematics May, 2015 Winston-Salem, North Carolina Approved By: Frank Moore, Ph.D., Advisor Hugh Howards, Ph.D., Chair Jeremy Rouse, Ph.D.

2 Acknowledgments This was probably the hardest page of this thesis to write, as no number of words are sufficient to praise those who have helped and supported me along the way to my Master s degree. First, I would like to thank the members of my committee. Thank you to Dr. Jeremy Rouse for his seemingly infinite wisdom and even greater generosity in sharing his knowledge with me. We had numerous discussions on the finer details of algebraic number theory that helped shape the direction of my research. He is also a fast runner. Thank you to Dr. Hugh Howards for his mentorship and advice going all the way back to 2010 when I was mere freshman at Wake Forest. My identity as a mathematician is in large part due to the dedication of Dr. Howards as an educator. Lastly, thank you to my adviser, Dr. Frank Moore, for his selfless devotion to this project over nearly two years time. I would not be where I am today without his mathematical knowledge, worldly advice and sincere friendship. The Department of Mathematics at Wake Forest has been like a second family to me for some years now. Thank you to everyone here that has helped me to survive and thrive at Wake Forest. To my office mates, Mackenzie, Elliott, Elena and Amelie: thank you for your support and for putting up with my loud music! Finally, I would like to thank my family for their love and for providing me with opportunities in life that have allowed me to succeed. They are my biggest supporters and I love them dearly. ii

3 Table of Contents Acknowledgments Abstract ii v Chapter 1 Algebraic Number Fields Rings of Algebraic Integers Dedekind Domains Ramification of Primes The Decomposition and Inertia Groups Norms of Ideals Discriminant and Different The Class Group The Hilbert Class Field Orders Units in a Number Field Chapter 2 Class Field Theory Valuations and Completions Frobenius Automorphisms and the Artin Map Ray Class Groups L-series and Dirichlet Density The Frobenius Density Theorem The Second Fundamental Inequality The Artin Reciprocity Theorem The Conductor Theorem The Existence and Classification Theorems The Čebotarev Density Theorem Ring Class Fields Chapter 3 Quadratic Forms and n-fermat Primes The Theory of Binary Quadratic Forms The Form Class Group n-fermat Primes iii

4 Bibliography Appendix A Appendix A.1 The Four Squares Theorem A.2 The Snake Lemma A.3 Cyclic Group Cohomology A.4 Helpful Magma Functions Curriculum Vitae iv

5 Abstract Andrew J. Kobin Most problems in number theory are exceedingly simple to state, yet many continue to elude mathematicians even centuries after they were originally posed. Such a question, Given a positive integer n, when can a prime number be written in the form x 2 + ny 2?, was solved by Cox [7], and although the statement is elementary, the solution requires the depth and power of class field theory to understand. In our approach to this question, we will explore a variety of topics, including: algebraic number fields; types of class groups and class fields; two density theorems; the main theorems in class field theory; and the theory of quadratic forms. Our discussion will culminate in Theorem , a full characterization of primes of the form x 2 + ny 2. However, the intrigue doesn t end there. In Chapter 3, we pose the related question: If p is a prime of the form x 2 +ny 2, when is y 2 +nx 2 also prime? This question turns out to be much harder to approach, but we will investigate the symmetric n- Fermat prime question thoroughly. In certain sections (1.8, 2.10 and 3.3) we use the Magma Computational Algebra System to handle large or complicated computations. Many of the basic commands can be found in the Magma handbook, available at usyd.edu.au/magma/handbook/ through the University of Sydney s Computational Algebra Group. v

6 Chapter 1: Algebraic Number Fields In the first chapter we provide a detailed description of the main topics in algebraic number theory: algebraic number fields, rings of integers, the behavior of prime ideals in extensions, norms of ideals, the discriminant and different, the class group, the Hilbert class field, orders and Dirichlet s unit theorem. 1.1 Rings of Algebraic Integers Let Q be an algebraic closure of Q. Then Q is an infinite dimensional Q-vector space and every polynomial f Q[x] splits in Q[x]. An example of such an algebraic closure is Q = {u C f(u) = 0 for some f Q[x]}. Then Q Q C. Note that any two choices of Q are isomorphic. One of the most important elements of a number field we will be working with is: Definition. An element α Q is an algebraic integer if it is a root of some monic polynomial with coefficients in Z. Example is an algebraic integer since it is a root of x 2 2. However, 1 2, π and e are not algebraic integers. We will see in a moment why 1 2 is not algebraic, but the proof for π and e is famously difficult. Note that the set of algebraic integers in Q is precisely the integers Z. In a moment we will generalize this set to fields other than Q. Definition. The minimal polynomial of α Q is the monic polynomial f Q[x] of minimal degree such that f(α) = 0. The minimal polynomial of α is unique, as the following lemma shows. 1

7 Lemma Suppose α Q. Then the minimal polynomial f of α divides any other polynomial h such that h(α) = 0. Proof. Suppose h(α) = 0. Then by the division algorithm, h = fq + r with deg r < deg f. Note that r(α) = h(α) f(α)q(α) = 0 so α is a root of r. But since deg f is minimal among all polynomials of which α is a root, r must be 0. This shows that f divides h. Lemma If α Q is an algebraic integer then the minimal polynomial has coefficients in Z. Proof. Let f Q[x] be the minimal polynomial of α. Since α is an algebraic integer, there is some g Z[x] such that g(α) = 0. By Lemma 1.1.2, g = fh for some monic h Q[x]. Suppose f Z[x]. Then there is some prime p dividing the denominator of at least one of the coefficients of f; let p i be the largest power of p that divides a denominator. Likewise let p j be the largest power of p that divides the denominator of a coefficient of h. Then p i+j g = (p i f)(p j h) and reducing mod p gives 0 on the left, but two nonzero polynomials in F p [x] on the right, a contradiction. Hence f Z[x]. An important characterization of algebraic integers is provided in the following proposition. Proposition α Q is an algebraic integer if and only if Z[α] = is a finitely generated Z-module. { n } c i α i : c i Z, n 0 i=0 Proof. ( = ) Suppose α is integral with minimal polynomial f Z[x], where deg f = k. Then Z[α] is generated by 1, α,..., α k 1. 2

8 = ( ) Suppose α Q and Z[α] is generated by f 1 (α),..., f n (α). Let d M where M = max{deg f i 1 i n}. Then α d = n a i f i (α) i=1 for some choice of a i Z. Hence α is a root of x d n a i f i (x) so it is integral. i=1 Example α = 1 2 is not an algebraic integer since Z [ 1 2] is not finitely generated as a Z-module. Definition. For a given algebraic closure Q of Q, we will denote the set of all algebraic integers in Q by Z. This set inherits the binary operations + and from Q, and an important property is that Z is closed under these operations: Proposition The set Z of all algebraic integers is a ring. Proof. Note that 0 is a root of the zero polynomial, so 0 Z. Then it suffices to prove closure under addition and multiplication. Suppose α, β Z and let m and n be the degrees of their respective minimal polynomial. Then 1, α,..., α m 1 span Z[α] and 1, β,..., β n 1 likewise span Z[β]. So the elements α i β j for 1 i m, 1 j n span Z[α, β], so this Z-module is finitely generated. This implies that the submodules Z[α + β] and Z[αβ] of Z[α, β] are also finitely generated, so it follows by Proposition that α + β and αβ are algebraic integers. The two most important objects of study in algebraic number theory are number fields and their associated rings of integers, which are defined below. 3

9 Definition. A number field is a subfield K Q such that K is a finite dimensional vector space over Q. The dimension of K/Q is called the degree of the field extension, denoted [K : Q]. Definition. The ring of integers of a number field K is O K = K Z = {α K α is an algebraic integer}. Example Q is the unique number field of degree 1, and its ring of integers is the rational integers Z. Example Q(i) is a number field of degree 2. Its ring of integers is Z[i], the Gaussian integers. Example K = Q( 5) has ring of integers O K = Z[(1 + 5)/2]. The reader may recognize this number (1 + 5)/2 as the golden ratio. An object we will study in Section 1.9 is: Definition. An order in O K is any subring O O K such that the quotient O K /O of abelian groups is finite. Example For O Q(i) = Z[i], the subring Z + niz is an order for every nonzero n Z. However, Z Z[i] is not an order since Z does not have finite index in Z[i]. Example For K = Q(α) where α is an algebraic integer, Z[α] is an order in O K but in general Z[α] O K. We study orders in further detail in Section 1.9 and see some important examples where Z[α] O K. Lemma For any number field K, O K Q = Z and QO K = K. Proof. Suppose α O K Q such that α = a b in lowest terms. We may assume b > 0. Since α is integral, Z [ a b ] is finitely generated as a Z-module so b = 1. 4

10 On the other hand, suppose α K with minimal polynomial f(x) Q[x], where deg f = n. For any positive integer d, the minimal polynomial of dα is d n f ( x d). In particular, let d be the least common multiple of the denominators of the coefficients of f. Then d n f ( x d) has integer coeffients, so dα OK. Hence QO K = K. Definition. A lattice in a number field K is a subset L such that QL = K and L is an abelian group of rank [K : Q]. Proposition For any number field K, the ring of integers is a lattice in K. Proof. QO K = K was proven in Lemma , and the second statement can be shown by choosing a basis for K consisting of elements in O K. Corollary O K is a noetherian ring. Proof. By the proposition, O K is finitely generated as a Z-module, so it is clearly finitely generated as a ring. It is well known (cf in [25]) that this implies O K is noetherian. 1.2 Dedekind Domains A nice property of the integers Z is unique factorization: every integer n can be written as a unique product of powers of prime numbers. This unique factorization property fails in general for rings of algebraic integers. However, O K has the special property that every nonzero ideal factors uniquely as a product of prime ideals. Definition. An integral domain R is integrally closed in its field of fractions K if every α K that is a root of a monic polynomial f R[x] is itself in R. Proposition Z is integrally closed and for any number field K, its ring of integers O K is integrally closed. 5

11 Proof. First suppose α Q is integral over Z. Then there is some monic polynomial f(x) in Z[x] such that f(α) = 0. If f(x) = a 0 + a 1 x x n then the a i all lie in O K, where K = Q(a 0,..., a n 1 ). Since O K is finitely generated as a Z-module, so is Z[a 0,..., a n 1 ]. Now f(α) = 0 means that we can write α n as a combination of α i for i < n, with weights c i Z[a 0,..., a n 1 ]. Thus Z[a 0,..., a n 1, α] is also a finitely generated Z-module. But notice that Z[α] is a submodule of Z[a 0,..., a n 1, α], so it too is finitely generated. Hence α is integral over Z, meaning α Z. This proves the first statement, and the second statement now follows easily. Suppose α K is integral over O K. Then since Z is integrally closed, α Z, implying α K Z = O K. This property of O K is important in establishing it as a special type of domain, called a Dedekind domain. Definition. An integral domain R is a Dedekind domain if (1) R is noetherian. (2) R is integrally closed in its field of fractions. (3) Every nonzero prime ideal p R is maximal. Example Z[ 5] is not integrally closed since for example (1+ 5)/2 Q( 5) is integral over Z[ 5] but is not itself an element of Z[ 5]. Therefore Z[ 5] is not a Dedekind domain, but as we shall see in Section 1.9 it is an order of the ring of integers Z[(1 + 5)/2]. Example Any field is (trivially) a Dedekind domain. Example Z is integrally closed and every nonzero prime ideal is maximal, but Z is not noetherian and hence not a Dedekind domain. 6

12 Proposition O K is a Dedekind domain. Proof. We have shown (Corollary and Proposition 1.2.1) that O K is integrally closed and noetherian, so it suffices to show that prime ideals are maximal. Suppose p is a nonzero prime ideal of O K. Let α p and let f(x) = x n + a n 1 x n a 0 be its minimal polynomial. Then f(α) = 0 so a 0 = α n a n 1 α n 1... a 1 α p. Then a 0 Z p so every element of the quotient O K /p is killed by a 0, which implies O K /p is finite. Since p is prime, O K /p is an integral domain, and every finite integral domain is a field, which proves that p is maximal. Hence O K is a Dedekind domain. The crucial property of Dedekind domains is that their nonzero ideals factor uniquely into prime ideals. In fact, unique factorization holds for a more general class of objects in a Dedekind domain called fractional ideals. Definition. Let R be a Dedekind domain and K be its field of fractions. A fractional ideal of R is a nonzero R-submodule of K that is finitely generated as an R-module. Note that since fractional ideals are finitely generated, we can clear denominators of a generating set to realize every fractional ideal in the form ai = {ab b I} where a K and I is an integral ideal of the ring R. Example Z is a fractional ideal of Z. 2 Lemma Let R be a Dedekind domain. For every nonzero ideal I R, there exist prime ideals p 1,..., p n such that p 1 p n I. 7

13 Proof. Let S be the set of nonzero ideals in R that do not satisfy the conclusion of the lemma. The idea here is to use the fact that R is noetherian to show that S must be empty. Supposing to the contrary that S is not empty, the noetherian property allows us to choose a maximal element I S. If I were prime, it would trivially contain a product of primes so we know this is not the case. Then there exist a, b R I such that ab I. Let J 1 = I + (a) and J 2 = I + (b). Then neither J 1 nor J 2 is in S since I is maximal, so each contains the product of primes, say p 1 p r J 1 and q 1 q s J 2. Then p 1 p r q 1 q s J 1 J 2 = I 2 + I(b) + (a)i + (ab) I. We have shown I to contain a product of primes, producing the necessary contradiction to show that S is empty. Hence every nonzero ideal of R contains a product of primes. The critical property of fractional ideals is proven next. Theorem The set of fractional ideals of a Dedekind domain R forms an abelian group under ideal multiplication, with identity R. Proof. The product of two fractional ideals is again finitely generated, hence a fractional ideal. Also, for any nonzero ideal I, IR = R so it suffices to show the existence of inverses. First we prove that if p K is prime, it has an inverse. Let I = {a K ap R}; we will show this is an inverse of p. Fix a nonzero b p. Since I is an R-module, bi is an ideal in R. And since R I we have p Ip R, but p is maximal (R is a Dedekind domain) so either p = Ip or Ip = R. If Ip = R then I is an inverse of p and we re done. Instead suppose Ip = p. By Lemma we can choose a minimal product of prime ideals p 1 p 2 p m (b) p. If no p i is contained in p then for each i there is some a i p i with a i p, but a i p 8

14 which contradicts that p is a prime ideal. Thus there is some p i p. However, every prime is maximal so p i = p. Since m was minimal, p 2 p 3 p m (b) and so there is some c (b) that lies in p 2 p 3 p m. Then p(c) (b) so we have d := c b I. However d R since if it were, it would lie in (b). But note that d preserves p as an R-module that is, dp p since d = c b so d must be in R, a contradiction. Hence Ip = R, so every prime ideal has an inverse in R. Now we turn to fractional ideals. Every fractional ideal is of the form ai for some a K and I an ideal of R. Since the prime ideals are maximal in R, I p for some prime p. Multiplying both sides of this containment by p 1, we have I p 1 I p 1 p = R. By the same argument as above, p 1 I = R so every fractional ideal has an inverse. In the next two theorems we show that unique factorization of ideals holds in any Dedekind domain. Theorem Every nonzero ideal I in a Dedekind domain R can be written as a unique (up to order) product of prime ideals. Proof. Suppose I is maximal among those ideals that cannot be factored into primes. Every ideal is contained in a maximal ideal so I p for some maximal p which is also prime. If Ip 1 = I then p 1 = R by group properties, but this is impossible. However, R p 1 which implies I Ip 1. By maximality of I, Ip 1 = p 1 p n for prime ideals p i. Then I = p 1 p n p, which shows I can in fact be written as a product of primes, contradicting our initial assumption. Hence every ideal has a prime factorization. To prove uniqueness, suppose p 1 p n = q 1 q m. If no q i is contained in p 1 then for each i there is some a i q i p 1. But then a 1 a m q 1 q m = p 1 p n p 1 9

15 which contradicts primality of p 1. Thus p 1 = q i for some i, and this argument can be repeated for each p j to show that p j = q i for some i. Thus the factorization is unique up to order. Theorem If I is a fractional ideal of R then there exist prime ideals p 1,..., p n and q 1,..., q m so that I = (p 1 p n )(q 1 q m ) 1 and this factorization is unique up to order. Proof. We can clear denominators to write ai = J for some a R and J an integral ideal of R. Apply unique factorization to J and (a) and the result follows from Theorem Example Let K = Q( 6) with ring of integers O K = Z[ 6]. If ab = 6 with neither a unit, then Norm(a)Norm(b) = 6 (see Section 1.5). Without loss of generality let Norm(a) = 2 and Norm(b) = 3. If a = x + y 6 then Norm(a) = x 2 + 6y 2 = 2 which has no solutions in Z. This shows that 6 is irreducible, and even if a or b were a unit, the other would equal 6 so 6 would be irreducible anyways. So 6 cannot be written as a product of irreducibles in O K. However, (6) factors into prime ideals as Theorem suggests: (6) = (2, 2 + 6) 2 (3, 3 + 6) 2. This is not trivial to calculate, but we will develop the techniques required to determine such a factorization in subsequent sections. A special case of a Dedekind domain is: Definition. An integral domain R is a discrete valuation ring if it is noetherian, integrally closed and contains exactly one nonzero prime ideal. 10

16 In Section 2.1, we will see where the name discrete valuation ring comes from, as well as study some of the properties of a DVR as they relate to absolute values on a field. We proved in Theorem that the set of fractional ideals of a Dedekind domain forms a group under ideal multiplication, but there is an even stronger characterization. Theorem Let R be an integral domain. Then the following are equivalent: (1) R is a Dedekind domain. (2) For every prime ideal p R, the local ring R p is a discrete valuation ring. (3) The fractional ideals of R form a group. (4) For every fractional ideal I R there is an ideal J R such that IJ = R. Proof. See VIII.6.10 in [14]. In the case of O K, there are some important groups that arise from fractional ideals, the most important being the class group. Definition. Let I K denote the group of fractional ideals of O K and let P K denote the subgroup of all principal fractional ideals of O K : P K = {αo K α K }. Then the quotient I K /P K is called the ideal class group of K, denoted C(O K ). In Section 1.7 we explore this group fully. A major result we will prove is Theorem. C(O K ) is a finite group. 11

17 = 1.3 Ramification of Primes Let K be a number field and suppose L/K is any finite extension. If p is a prime ideal of O K then po L is an ideal of O L and hence has prime factorization po L = P e 1 1 P eg g where P i are the distinct prime ideals of O L containing p. We will sometimes say a prime P i lies over p, P i contains p or P i divides po L. Definition. For each P i, the integer e i is called the ramification index of p in P i. If any of these are greater than 1, p is said to ramify in L. Definition. Each ideal P i lying over p gives a residue field extension O L /P i O K /p. The degree of this extension, denoted f i, is called the inertial degree of p in P i. Definition. A prime p is said to split completely in L if e i = f i = 1 for all P i in the prime factorization of po L. If in addition po L is itself a prime ideal, i.e. g = 1, we say p is inert. The set of all prime ideals of a ring R together with (0) is called the spectrum of R, denoted Spec(R). We will also occasionally use Spec(p) to denote the set of primes P O L lying over p O K. Example In Z[i], (2) = (1 + i) 2 so (2) ramifies with e 1 = 2. By contrast, (3) is inert in Q(i) with residue field Z[i]/(3) = F 9, and (5) = (2 + i)(2 i) is unramified. The next lemma characterizes the primes of O L which divide po L. Lemma A prime ideal P O L divides po L if and only if p = P K. Proof. ( = ) Clearly p P K O K. Since p is maximal, this implies p = P K. ( ) If p P then we have seen that po L P and this implies that P occurs in the prime factorization of po L. 12

18 There is an important relation between the ramification indices, inertial degrees and number of primes in Spec(p) that is described in the next theorem, known as the ef g theorem. Theorem Let m = [L : K] and let P 1,..., P g be the prime O L -ideals containing p O K. Then g e i f i = m. i=1 Furthermore, if L/K is Galois, then all the ramification indices are equal to e = e 1 and all the inertial degrees are equal to f = f 1, so efg = m. Proof. The first statement is proven by showing both sides are equal to [O L /po L : O K /p]. By the Chinese remainder theorem, O L po L = O L P e i i OL =. For each i = 1,..., g, f i is the degree of the extension O L /P i O K /p, and for each r i, P r i i /Pr i+1 i is an O L /P i -module. Since there is no ideal between P r i i and P r i+1 i (O L ) Pi is a DVR this module has dimension 1 as an O L /P i -vector space, and hence dimension f i as an O K /p-vector space. Therefore each quotient in the chain P e i i O L P i P 2 i P e i i has dimension f i over O K /p. Thus [O L /P e i i side equals [O L /po L : O K /p]. : O K /p] = e i f i. This shows that the left For the other equality, we first prove it when O L is a free O K -module (e.g. when O K is a PID). On one hand, O n K = O L induces an isomorphism K n L which shows that n = m. On the other hand, O n K = O L also induces an isomorphism (O K /p) n O L /po L which shows that m = n = [O L /po L : O K /p]. In the general 13

19 case, localize O K at p to obtain a DVR O K = (O K) p. Since a DVR is always a PID, O L = (O L) p satisfies po L = (P i O L) e i so [O L /po L : O K /po K ] = m. This completes the first part of the proof. Now assume L is Galois over K. Take σ G = Gal(L/K). Then if P O L is a prime ideal, so is σ(p). Moreover, if P contains p then by Lemma so must σ(p). Clearly e(σ(p) p) = e(p p) and f(σ(p) p) = f(p p). To complete the proof, we will show that G acts transitively on Spec(p), the set of prime ideals of O L lying over p. Suppose P and Q both contain p but are not Galois conjugates. By the Chinese remainder theorem we can find an element β Q that does not lie in σ(p) for any σ G. Define b = N(β), where N denotes the norm (see Section 1.5). Then b O K and since β Q, b Q as well. Thus b O K Q = p. On the other hand, β σ 1 (P) for any σ G so σ(β) P. However, N(σ(β)) = N(β) = b p so we have p P which contradicts primality of p. Hence Gal(L/K) acts transitively on the primes containing p and the result follows by the preceding paragraph since e and f are invariant under Gal(L/K). As we saw in Example , it is hardly easy to determine the factorization of ideals in a number field. The next theorem will be of immense importance going forward, as it allows us to describe the splitting behavior of a prime p O K as we pass to an extension L/K. 14

20 Theorem Let L/K be Galois, where L = K(α) for some α O L. Let f(x) O K [x] be the minimal polynomial of α over K. Suppose p is a prime ideal of O K and f(x) is separable mod p. Then (1) p is unramified in L. (2) If f(x) f 1 (x) f g (x) mod p for distinct f i (x) which are irreducible mod p, then P i = po L + f i (α)o L is a prime ideal of O L, and the prime factorization of po L is po L = P 1 P g. Furthermore, deg f i = f(p i p) for all i, and since L/K is Galois, these are all the same. (3) p splits completely in L f(x) 0 mod p has a solution in O K. Proof. (1) and (3) will follow during the course of proving (2). To prove (2), observe that since f(x) is separable mod p, f(x) f 1 (x) f g (x) mod p for distinct, irreducible (mod p) polynomials f i (x). If P O L is a prime lying over p, then f i (α) P for some i; we may relabel the f i so that f 1 (α) P. Then by Galois theory, [O L /P : O K /p] [L : K] = deg f. Now for any σ Gal(L/K) such that σ(p) = P, f 1 (σ(α)) P and f 1 (x) is separable by hypothesis, so deg f 1 ef, where e and f are the ramification index and inertial degree, respectively, of p in P. This shows that e = 1 and f = deg f 1, so (1) is proved. Now let po L = P 1 P g be the prime factorization of po L into prime ideals of O L. Theorem implies that deg f i = f for all i, so it remains to prove that each P i is generated by p and f i (α). On one hand, po L + f i (α)o L is contained in P i since f i (α) P i (reindexing if necessary). On the other hand, (po L + f i (α)o L ) po L. 15

21 Each ideal on the left is contained in a prime ideal in the factorization of po L, and this must be P i for each i. This completes the proof of (2). We will develop further techniques for deciding when a prime ramifies/splits/stays inert in Section 1.6. For the moment, we do not even know if there are an infinite number of primes splitting in an extension L/K; this question will finally be given an answer in Section Example In this example we provide a full characterization of the splitting behavior of primes in quadratic extensions. Suppose K = Q( n) where n is a squarefree integer. Then K/Q is Galois, so for each prime p Z we have 2 = efg by Theorem There are exactly three possibilities for e, f and g: e = 2 and f, g = 1. In this case p ramifies in O K so po K = P 2 for some prime ideal P. It turns out that there are only finitely many such primes since by (3) of the previous theorem, p ramifies in K if and only if x 2 + n 0 (mod p) has a multiple root. This ties in with the idea that the discriminant of a polynomial determines its number of roots in Section 1.6, we will see that the connection between ramification and discriminants runs even deeper. f = 2 and e, g = 1. In this case p is inert, so po K is prime. It turns out that this happens half the time (minus the finitely many cases when a prime ramifies). g = 2 and e, f = 1. Here p splits completely in O K, so po K = P 1 P 2 for prime ideals P 1 P 2. This happens the other half of the time. Definition. For a quadratic field K = Q( n), the discriminant of K is d K = { n if n 1 (mod 4) 4n otherwise. 16

22 For any integer q we also define the Kronecker symbol by ( q 0 if q 0 (mod 4) = 1 if q 1 (mod 8) 2) 1 if q 5 (mod 8). As a consequence of the above characterization of primes in O K, where K = Q( n), we have the following characterization of the splitting of primes in a quadratic extension. Proposition A prime p ramifies in K = Q( n) if and only if p d K, and p ( ) d splits completely in K if and only if Kp = 1. The first statement follows from the general case in Sections 1.6 and the second ( ) ( ) is a consequence of (3) of Theorem 1.3.4, since = = 1 if and only if 4n p n p x 2 + n 0 (mod p) for some integer x. For now, let s take a look at a familiar example. Example Let K = Q(i) and recall that the Gaussian integers Z[i] are the ring of integers for K. In this example we will describe the splitting behavior of primes in Z[i]. From the last few results, we claim that for an odd prime integer p (excluding p = 2) the following are equivalent: (i) p 1 (mod 4). (ii) (p) splits completely in Z[i]. (iii) p = x 2 + y 2 for some integers x, y. Proof. To prove our claim, note that Z[i] is the ring of integers for K = Q(i) so we may take the α in Theorem to be i, which has minimal polynomial x over Q. Thus (p) splits completely in Z[i] if and only if x splits mod p. This in turn 17

23 happens if and only if F p contains a fourth root of unity, i.e. F p contains an element of order 4. Since F p has order p 1, this means 4 p 1 and so (i) (ii) is proven. Next suppose (p) splits in Z[i]; let (p) = p 1 p 2 for prime ideals p 1, p 2 Z[i]. In Example 1.7.2, we will prove that the ring of Gaussian integers Z[i] is a PID. Using this fact, we know p 1 = (x + yi) for integers x and y, but then p 2 must be (x yi). Therefore p = x 2 + y 2 up to multiplication by a unit in Z[i]. However the only units are ±1, ±i so clearly p must just be x 2 + y 2. Conversely, if p = x 2 + y 2 then p = (x + yi)(x yi) in Z[i]. Note that this solves Fermat s theorem characterizing primes of the form x 2 + y 2. It will be a continuing theme in these notes to fully characterize primes of the form x 2 + ny 2 for all integers n. 1.4 The Decomposition and Inertia Groups In this section we describe two important subgroups of Gal(L/K) for a Galois extension L/K of number fields. Definition. For a Galois extension L/K and a prime ideal P O L lying over p O K, the decomposition group of P is D P = {σ Gal(L/K) σ(p) = P} and the inertia group of P is I P = {σ Gal(L/K) σ(α) α mod P for all α O L }. Let k = O K /p and l = O L /P denote the respective residue fields of p and P. We will prove that there is an exact sequence 1 I P D P Gal(l/k) 1. 18

24 Recall from the proof of Theorem that G = Gal(L/K) acts transitively on Spec(p). Then we can interpret D P as the stabilizer of P under this action. The Orbit-Stabilizer Theorem tells us that [G : D P ] = g, where g is the number of distinct primes in the factorization of po L. Hence D p = ef. Lemma For a fixed prime ideal p O K, the decomposition groups D P of the prime ideals lying over p are conjugate subgroups of Gal(L/K). Proof. This is a more general fact about the stabilizers of a transitive group action. Note that for σ, τ Gal(L/K), τ 1 στ D P τ 1 στp = P στp = τp σ D τp which implies that σ D P τστ 1 D P. Hence τd P τ 1 = D τp. The decomposition group is useful because we can view an extension L/K as a tower of extensions so that we understand the splitting of primes better in each step of the tower. Proposition Let L/K be a Galois extension and fix a prime p O K. Let D = D P be the decomposition group for a particular prime P lying over p. Then the fixed field L D = {α L σ(α) = α for all σ D} is the smallest subfield E of L such that g = 1 for P O E. Proof. First suppose E = L D. By Galois theory, Gal(L/E) = D and as in the last section, D acts transitively on the set of primes of O L lying over P E := P O E. One of these primes is P itself, and D fixes P by definition, so this must be the only prime lying over P E, i.e. g = 1. On the other hand, if g = 1 for P E then Gal(L/E) fixes P: it s the only prime over P E. So Gal(L/E) D and by Galois correspondence, L D E. 19

25 This shows that p does not split when moving from L D to L: it either ramifies or stays inert. Let E = L D and denote P O E by P E, e = e(p p), f = f(p p) and g = g(p p). To piece together more of the puzzle, we have the following. Proposition Given K, L, E, p, P and P E as above, e(p E p) = f(p E p) = 1, g = [E : K], e = e(p P E ) and f = f(p P E ). Proof. As mentioned in the remarks preceding Lemma 1.4.1, the Orbit-Stabilizer Theorem implies that g = [Gal(L/K) : D]. Then by Galois theory, [Gal(L/K) : D] = [E : K], so this equals g. The previous proposition gives us g(p P E ) = 1, and by Theorem we have e(p P E )f(p P E ) = [L : E] = [L : K] [E : K] = efg [E : K] = ef. Now e(p P E ) e and f(p P E ) f, so we must have that e(p P E ) = e and f(p P E ) = f. It follows easily that e(p E p) and f(p p) are 1. Fix a prime P O L lying over p O K. Observe that each σ D P acts on the finite field l = O L /P and fixes k = O K /p so we obtain a group homomorphism ϕ : D P Gal(l/k). The next two results establish that ϕ is surjective, which we will use to prove exactness of the sequence described at the start of this section. Lemma The residue field extension l/k is Galois. Proof. First we show that l/k is normal. To do this, take any α l and let f(x) be 20

26 its minimal polynomial over k. Let α O L be a lift of α. Then f(x) = σ D P (x σ(α)) O K [x] splits completely over O K and has α as a root, when taken mod p. Thus l/k is normal. Furthermore, l/k will be Galois whenever it is separable, but since O K /p is a finite field, it is perfect and therefore any finite extension is separable [10]. Proposition ϕ is surjective. Proof. By the lemma, l/k is Galois. We will show that ϕ(d P ) acts transitively on the conjugates of α over k. By the Chinese remainder theorem, one may choose α O L such that α { α mod P 0 mod P for any other P lying over p. Then for any σ G D P we have α 0 mod σ 1 P and hence σ(α) 0 mod P. This implies that f(x) = σ D P = ( x σ(α) ) x σ D P σ D P (x ϕ(σ)( α)) σ D P x which lies in k[x]. Notice that the first product lies in k[x], so it is divisible by the minimal polynomial of α over k. So given any conjugate α of α, (x α ) divides the first product above and thus α must equal ϕ(σ)( α) for some σ D P. Hence the action of ϕ(d P ) on the conjugates of α is transitive and it follows that ϕ is surjective since the image has at least [l : k] = Gal(l/k) elements. Next we relate the inertia group I P to the map ϕ, and use it to prove that the original sequence we defined is exact. 21

27 Proposition The inertia group I P is the kernel of ϕ : D P Gal(l, k). Proof. By definition ker ϕ = {σ D P σ(α) α mod P for all α O L } so it suffices to show that if σ D P then there exists an α O L such that σ(α) α mod P. If σ D P then of course σ 1 D P so σ 1 (P) P. Since both σ 1 (P) and P are maximal ideals, there exists some α P with α σ 1 (P), which implies σ(α) P. Thus σ(α) α mod P and it follows that I P = ker ϕ. We now summarize our findings. Corollary If L/K is a Galois extension, the sequence 1 I P D P Gal(l/k) 1 is exact. Moreover, I P = e and D P = ef, where e and f are the ramification index and inertial degree, respectively, for L/K. Notice that the inertia group is a very useful measure of how a prime p ramifies in a Galois extension L. This is a common theme in algebraic number theory: the behavior of primes in an extension is often encoded in the automorphisms of the field itself. 1.5 Norms of Ideals In this section we define the norm of an ideal. As in previous sections, all of these definitions and results generalize to any Dedekind domain A with integral closure B see [19] for the general cases. In our context, we will replace A with O K and B with O L, which have fields of fractions K and L, respectively. 22

28 Let I K and I L denote the groups of fractional ideals of O K and O L, respectively. We want to define a group homomorphism N : I L I K. Since I L is the free abelian group on the set of prime ideals in O L, we only have to define N for p prime. Let p be a prime ideal of O L and factor po L = P e i i for P i prime. Suppose p = (π) is principal. Then we should have N (po L ) = N (πo L ) = N (π)o K = (π) m = p m where m = [L : K]. We also want N to be a homomorphism, so we must have N (po L ) = N ( ) P e i i = N (P i ) e i. Recall that m = e i f i, so the correct definition for N is Definition. For a prime P O L lying over p O K, the norm of P is defined to be N (P) = p f where f = [O L /P : O K /p]. To distinguish this norm from a similar norm to be defined shortly, we will sometimes refer to N as the ideal norm. If the norm is taken with respective to an extension L/K, we write N L/K but when the context is clear we will often drop the decoration. Remark. By the properties of inertial degree f, it is easy to see that for a tower M L K, N L/K (N M/L (a)) = N M/K (a). Next we check that the properties discussed above hold for the norm we have defined. 23

29 Proposition Let L/K, O K and O L be as above. (a) For any nonzero ideal a O K, N (ao L ) = a m where m = [L : K]. (b) If L/K is Galois and P O L is any nonzero prime ideal with p = P O K and po L = (P 1 P g ) e, then N (P) = (P 1 P g ) ef = σ(p). σ Gal(L/K) (c) For any nonzero element β O L, N(β)O K = N (βo L ), where N denotes the regular field norm. Proof. (a) It suffices to prove this for prime ideals, for which we have N (po L ) = N ( ) P e i i = p e i f i = p m using Theorem (b) Since N (P i ) = p f for any prime P i in the prime factorization of po L, the left equality is clear. Recall that G = Gal(L/K) acts transitively on the set Spec(p) = {P 1,..., P g }. Then by the Orbit-Stabilizer Theorem, each P i occurs Gal(L/K) Spec(p) = m g = ef times in the collection {σ(p) σ G}, which implies the right equality. (c) First suppose L/K is Galois. Denote βo L by b. The map I K I L given by a ao L is injective since I K and I L are free on nonzero prime ideals, so it suffices to show that N(β)O L = N (b). But by (b), N (b) = σ(b) = L ) = σ G σ G(σ(β)O 24 ( ) σ(β) O L = N(β)O L. σ G

30 In the general case, let E be a finite Galois extension of K containing L, with d = [E : L] and O E the integral closure of O L in E. Then we have N L/K (βo L ) d = N E/K (βo E ) by the remark = N E/K (β)o K by the Galois case = N L/K (β) d O K. Lastly since I K is torsion-free, the above implies that N L/K (βo L ) = N L/K (β)o K for all nonzero β O L. For a Galois extension K/Q, we define a different norm taking ideals of O K to integers. We will see that the definition below coincides with the ideal norm. Definition. Let a O K be a nonzero ideal. The numerical norm of a is its index in the lattice of integers: N(a) = [O K : a]. In order to justify this definition, we need to check that [O K : a] is always finite. Proposition Every nonzero ideal a in O K has finite index in the lattice O K. Proof. Let a be a nonzero O K -ideal. The proof of Proposition shows that a contains a nonzero integer m (a 0 from that proof). So consider ϕ : O K /mo K O K /a, which is clearly surjective. By Proposition , O K is a free Z-module of rank n = [K : Q]. This means that Since ϕ is surjective, it follows that O K /a m n <. O K mo K = Z n mz n is a finite quotient of order mn. Notice that the ideal norm is defined for any extension L/K and outputs an ideal of O K. On the other hand, the numerical norm is defined on K/Q and outputs an integer in Z. The connection between the two norms is described in the next proposition. 25

31 Proposition Let K be any number field. (a) For any ideal a O K, N K/Q (a) = (N(a)) and therefore N(ab) = N(a)N(b). (b) For any fractional ideals b a of O K, [a : b] = N(a 1 b). Proof. (a) Write a = p e i i and let f i = f(p i p i ) where (p i ) = Z p i. Then N (p i ) = (p i ) f i. By the Chinese remainder theorem, O K /a = O K /p e i i and thus [O K : a] = [O K : p e i i ]. We previously proved that [O K : p e i i ] = pe if i i, thus [O K : a] = (p e if i i ) = N K/Q (a). When we identify the set of nonzero ideals of Z with the set of positive integer generators, N and N are seen to coincide, and multiplicativity of N follows from the same property of the ideal norm. (b) We can multiply by some integer d to make a and b integral ideals. Then part (a) gives us [a : b] = [da : db] = [O K : db] [O K : da] = N(db) N(da) = N(a 1 b). 1.6 Discriminant and Different One may recall the definition of discriminant from field theory. Here we present it in the context of extensions of number fields. Definition. For a number field extension L/K with rings of integers O L O K, suppose O L has a basis {β 1,..., β m } over O K. Then the discriminant of O L is D(O L ) = D(β 1,..., β m ) = det(t r L/K (β i β j )) 26

32 where T r denotes the trace. In this section we use the discriminant to characterize which primes ramify in Galois extensions of a number field. Definition. Let D = D(O L ) as above. The discriminant ideal of O L, denoted (L/K) or simply, is the ideal of O K generated by D. We will prove Theorem. The primes which ramify in O L are those that divide. First, we establish some properties of the discriminant. The details can be found in [19]. Definition. Let L = K(β) for some β L and let f be the minimal polynomial of β over K, setting deg f = m. Then the discriminant of f is defined to be D(f) := D(1, β, β 2,..., β m 1 ) = ( 1) m(m 1)/2 N L/K (f (β)). Proposition D(f) = 0 if and only if f has a repeated root, i.e. is not separable. Lemma Let O L have basis {β 1,..., β m } over O K. Then for any O K -ideal a, { β 1,..., β m } is a basis for O L /ao L over O K /a and the discriminant satisfies D( β 1,..., β m ) D(β 1,..., β m ) mod a, where the discriminant on the left is taken with respect to O L /ao L (as a module over O K /a) and on the right with respect to O L over O K. Proof. See 3.36 in [19]. We are now ready to prove the main result. 27

33 Theorem A prime p O K ramifies in O L if and only if p (L/K). Proof. By definition = (L/K) is the ideal generated by D = D(O L ). Thus p if and only if D p, which in turn happens if and only if D( β 1,..., β m ) = det(t r( β i βj )) = 0 in O K /p by Lemma Let p have factorization po L = P e 1 1 P eg g. By the Chinese remainder theorem, O L /po L = OL /P e 1 1 O L /P eg g. First suppose p is not ramified in L. Then each e i = 1 and O L /P i is a separable extension of O K /p. Let t i denote the trace map O L /P i O K /p. Select a basis {u i } for O L /po L such that {u 1,..., u k } is a basis for O L /P 1, {u k+1,..., u k+l } is a basis for O L /P 2, etc. Then for each ȳ O L /po L, ȳ = y y g with y i O L /P i. Each multiplication map r i : x xy i takes O L /P i to itself, and if r i has standard matrix A i then the matrix for r : x xȳ decomposes into the block matrix A = A 1 A A g Then T r(ȳ) = t 1 (y 1 ) t g (y g ). More importantly, the discriminant matrix has block form 1 2 B =... g where i is the discriminant of the chosen basis of O L /P i over O K /p. But O L /P i 28

34 is separable over O K /p if and only if det i 0. Hence if we view the B above as a change of basis matrix, we have D( β 1,..., β g ) = (det B) 2 (det 1 )(det 2 ) (det g ) 0. By the initial comments, this shows that when p is unramified, p. On the other hand, if some e i > 1 then O L /P i is not separable over O K /p. We may reindex the primes lying over p so that e 1 > 1. Choose a basis {v i } for O L /P e 1 1 such that v 1 P 1 /P e 1 1. In the quotient, v e 1 1 = 0 so the multiplication map r v1 : x xv 1 from above (now defined for the v i ) is trivial. Moreover, (v 1 v j ) e 1 = 0 so the characteristic polynomial for the map r v1 v j only has roots for its eigenvalues. Thus t i (v 1 v j ) = T r(r v1 v j ) = 0 so the discriminant matrix for O L /P 1 over O K /p has a row of zeros. Hence det 1 = 0 which implies D( β 1,..., β m ) = 0. By the preliminary comments, this shows that p divides. This tells us when a prime in O K ramifies in L, but the discriminant misses some critical information: Which primes in O L lying over p ramify? That is, which primes P in the factorization of po L have ramification index greater than 1? How do we determine the multiplicity of a prime dividing the discriminant? The rest of this section follows K. Conrad s paper The Different Ideal, which outlines the techniques required to answer these questions. To motivate the problem, consider the following example. Example Let K = Q(α) where α is a root of f(x) = x 3 x 1. This polynomial has discriminant 23 so 23 is the only integer prime which ramifies in 29

35 O K. Since [K : Q] = 3 and x 3 x 1 (x 3)(x 10) 2 mod 23, the factorization we obtain from Theorem is 23O K = pq 2 where p q and both are prime. In general, how do we know that q ramifies but p doesn t? Definition. For a lattice L in a number field K, its dual lattice is L = {α K T r K/Q (αl) Z}. Proposition For a lattice L with Z-basis {e 1,..., e n }, the dual lattice may be written L = n i=1 Ze i where {e i } is the dual basis of {e i } relative to the trace product on K/Q. In particular, L is a lattice. Proof. See 3.4 in [6]. We will see in the next section just how useful lattices can be in algebraic number theory, but for now we will focus on the lattice O K and its fractional ideals. Consider the dual lattice O K. First, we should recognize that O K is not just the elements of K with trace in Z actually it s smaller. But since algebraic integers have integral trace, we see that O K O K. Proposition For any fractional ideal a in K, a is a fractional ideal and a = a 1 O K. Moreover, O K is the largest fractional ideal of K whose elements all have integral trace. Proof. By definition, a = {α K T r K/Q (αa) Z}. First, since any dual lattice is a lattice by the previous proposition, we know a is a finitely generated Z-module. 30

36 Take α a and x O K. Then for any β a, T r K/Q ((xα)β) = T r K/Q (α(xβ)) Z since xβ a and α a. Thus xα a so a is a fractional O K -ideal. Next we check the formula for a. Take α a again. Then for any β a, T r(αβo K ) Z since βo K a. Thus αa O K which implies α a 1 O K. This shows a a 1 O K and the reverse containment is similarly shown. For the last statement, note that any fractional O K -ideal satisfies a = ao K. Therefore T r(a) Z T r(ao K ) Z a O K. Since O K is a fractional ideal containing O K, its inverse is an integral ideal contained in O K, from which we define: Definition. The different of K is the ideal D K = (O K) 1 = {x K xo K O K }. Example For K = Q(i) with O K = Z[i], T r(a + bi) Z precisely when 2a Z, so we see that Z[i] = 1 Z[i]. Thus the different of K is 2Z[i]. This can be 2 verified with the next proposition. Proposition If O K = Z[α] then D K = (f (α)) where f(x) is the minimal polynomial of α over Q. Proof. See 4.3 in [6]. 31

37 Example For a quadratic field K = Q( n) where n is squarefree, D K = { (2 n) if n 1 (mod 4) ( n) if n 1 (mod 4). The different is related to the field discriminant d K by the following. Theorem For a number field K of discriminant d K, N K/Q (D K ) = d K. Proof. Let {β 1,..., β n } be a Z-basis for O K so that we have O K = n Zβ i. i=1 n Then D 1 K = O K = have i=1 Zβ i by Proposition Using the definition of norm, we N (D K ) = [O K : D K ] = [D 1 K : O K] = [O K : O K ]. We can calculate [O K : O K] by finding det A where A is the matrix expressing the basis {β 1,..., β n } in terms of the dual basis {β 1,..., β n }. Since {β i } is a dual basis of {β i } it follows that A = (T r K/Q (β i β j )) and by definition det A = D(O K ) = d K. The result follows. Lemma For any nonzero ideal a O K, a D K if and only if T r(a 1 ) Z. Proof. This may be stated as a D K = (O K ) 1 which in turn is equivalent to a 1 O K. By Proposition 1.6.6, this is equivalent to T r(a 1 ) Z. Dedekind proved the following characterization of ramified primes in terms of the different ideal. The proof can be found in [6]. 32

38 Theorem (Dedekind). The prime factors of D K are exactly the primes in K that ramify over Q. In particular, for any prime ideal p O K lying over a prime p Z with ramification index e, the multiplicity of p in D K is e 1 when e 0 (mod p), and at least e when p e. Corollary The primes in Z that ramify in K are precisely the prime divisors of d K. Proof. Use the fact that d K = N (D K ) and Theorem Note that this also proves the rest of Proposition which characterized ramified primes in quadratic extensions. The theorem we just proved showed the true power of the different: while the discriminant also tells us if a prime ramifies in an extension, it does not tell us anything about the ramification indices of the primes in the larger field. This information is conveyed by the different, but if we know the full factorization of po K, we can relate this multiplicity to d K : Corollary Suppose po K = p e 1 1 p eg g with inertial degrees denoted by f i. Then the multiplicity of p in d K is at least (e 1 1)f (e g 1)f g = n (f f g ). Furthermore, if p e i for all i then this is the exact multiplicity of p in d K. It turns out that the multiplicity of a prime p D K is bounded by e 1 ord p (D K ) e 1 + e ord p (e). The left is Theorem and the right was proven by Hensel (see [6]). 33