Algebraic Number Theory

Transcription

1 Algebraic Number Theory Andrew Kobin Spring 2016

2 Contents Contents Contents 0 Introduction Attempting Fermat s Last Theorem Algebraic Number Fields Integral Extensions of Rings Norm and Trace The Discriminant Factorization of Ideals Ramification Cyclotomic Fields and Quadratic Reciprocity Lattices The Class Group The Unit Theorem Local Fields Discrete Valuation Rings The p-adic Numbers Absolute Values Local Fields Henselian Fields Ramification Theory Extensions of Valuations Galois Theory of Valuations Higher Ramification Groups Discriminant and Different i

3 0 Introduction 0 Introduction These notes follow a course on algebraic number theory taught by Dr. Andrew Obus at the University of Virginia in Spring The main topics covered are: ˆ Algebraic number fields (the global case) ˆ The ideal class group ˆ Structure of the unit group ˆ The p-adic numbers (the local case) ˆ Hensel s Lemma ˆ Ramification theory ˆ Further topics, including adeles and ideles The main companion for the course is Neukirch s Algebraic Number Theory. Other great references include Cassels and Frohlich s Algebraic Number Theory, Janusz s Algebraic Number Fields, Lang s Algebraic Number Theory, Marcus s Number Fields and Weil s Basic Number Theory. 0.1 Attempting Fermat s Last Theorem Algebraic number theory was developed primarily as a set of tools for proving Fermat s Last Theorem. We recall the famous (infamous?) theorem here. Fermat s Last Theorem. The equation x n + y n = z n has no solutions in positive integers for n 3. In attempting to prove the theorem, we first remark that the n = 4 case is elementary; it s just a matter of parametrizing the Pythagorean triples (x, y, z) that solve x 2 + y 2 = z 2 and noticing that not all three can be perfect squares. With this, we can reduce to the case when n = p, an odd prime. There are two cases: ˆ Case 1: x, y, z are all relatively prime to p. ˆ Case 2: p divides exactly one of x, y, z. We will show a proof for the first few primes in Case 1; the other case uses similar techniques. Let ζ be a primitive pth root of unity (e.g. ζ = e 2πi/p ) and assume Z[ζ] is a unique factorization domain (UFD). This was the classical approach, but number theorists quickly realized that Z[ζ] is not always a UFD. In fact, it is an open question whether there are an infinite number of primes p for which Z[e 2πi/p ] is a UFD. In any case, the assumption that Z[ζ] is a UFD holds for p < 23 so we will have proven a number of cases of Fermat s Last Theorem with the following proof. 1

4 0.1 Attempting Fermat s Last Theorem 0 Introduction Proof. Suppose x, y, z are positive integers satisfying x p + y p = z p. We may assume x, y, z are relatively prime in Z. The equation above may be factored as p (x + ζ i y) = z p ( ) i=1 For p = 3, the only cubes mod 9 are ±1 and 0 so there are no solutions for (*) where 3 xyz. So we may assume p 5. We need the following lemmas: p 1 Lemma p = (1 ζ i ). i=1 Proof. Consider expanding tp 1 t 1 in two ways: Then plugging in t = 1 gives the result. (t ζ) (t ζ p 1 ) = tp 1 t 1 = tp t + 1. Lemma For any 0 i < j p 1, the elements x + ζ i y and x + ζ j y are coprime in Z[ζ]. Proof. Suppose that π Z[ζ] is a prime which divides x + ζ i y and x + ζ j y. Then π divides ζ i y(1 ζ j i ). Notice that ζ i is a unit and p y by assumption, but 1 ζ j i p. So in particular, π y and thus π yp. Since π is a prime, π y or π p. Repeating the argument for x shows that π x or π p. Since x and y are coprime in Z, we cannot have π x and π y simultaneously, so π p. By assumption we have that π divides x p + y p and therefore also z p in Z, but (p, z) = 1 so the Euclidean algorithm implies that π 1. Therefore x + ζ i y and x + ζ j y are relatively prime in Z[ζ]. Now, each factor x + ζ i y must be a pth power in Z[ζ], possibly multiplied by a unit. Write x + ζy = ut p for u Z[ζ] and t Z[ζ]. Lemma u/ū is a pth root of unity. Proof. It is simple to show that u/ū and all of its Galois conjugates have modulus 1 in C; this is then true for all powers of u/ū as well. Then the degree of u/ū and all of its powers is bounded. Since all of these are algebraic integers, there are only finitely many possible choices for their minimal polynomials. Hence the set {(u/ū) k : k N} is finite. This proves u/ū is a root of unity in Z[ζ]. In particular, (u/ū) 2p = 1 but we want to show it is a pth root of unity. Suppose (u/ū) p = 1. Then u p = ū p. Since u Z[ζ] we may write u = a 0 + a 1 ζ + a 2 ζ a p 2 ζ p 2 for unique a i Z; this follows from unique factorization in Z[ζ]. Now u p a p 0 + a p a p p 2 (mod p) a 0 + a a p 2 (mod p) by Fermat s Little Theorem. 2

5 0.1 Attempting Fermat s Last Theorem 0 Introduction In particular, u p is conjugate to a real number mod p. Likewise, we can write ū as ū = (a 0 + a 1 ζ p a p 2 ζ 2 ) so ū p a 0 a 1... a p 2 (mod p). This implies a 0 + a a p 2 0 (mod p) so p u p. However, this is impossible if u is a unit. Therefore (u/ū) p = 1. Putting these results together, we can now write x + ζy = ζ j ūt p ζ j ū t p ζ j (x + ζy) (mod p). Expanding this out gives us x + ζy ζ j x ζ j 1 y 0 (mod p). ( ) Now Z[ζ]/(p) = Z[x]/(p, x p x + 1) = F p [x]/(x p x + 1). Thus the images of 1, x,..., x p 2 are F p -linearly independent in this ring. This implies 1, ζ,..., ζ p 1 are Z- linearly independent in Z[ζ]/(p). Since x, y Z, the only possibilities in (**) for j are j = 0, 1, 2, p 1. If p = 0, 2, p 1, it is easy to simplify (**) and produce a nontrivial ζ 2 term, which is impossible. If j = 1, (**) becomes (x y)(1 ζ) 0 (mod p). Thus p 1 i=2 (1 ζi ) divides x y but since x y Z, it must be that p (x y). Rearranging the equation x p + y p = z p to read x p + ( z) p = y p and repeating the argument so far shows that p (x + z) as well. Thus y x z (mod p). But then 0 = x p + y p z p 3x p (mod p) which implies p x, contradicting the assumption that p 3. Therefore no solutions exist to x p + y p = z p for p > 5 such that p xyz. This proof fails for general primes p in two places: as we mentioned, not every ring Z[e 2πi/p ] is a UFD; moreover, there can be many more units than just the roots of unity in Z[e 2πi/p ]. This motivates the study of ideal class groups which measure how far from being a PID (and a UFD) a ring of integers is and unit groups in algebraic number theory. 3

6 1 Algebraic Number Fields 1 Algebraic Number Fields 1.1 Integral Extensions of Rings Let A B be rings. Definition. An element x B is integral over A if it is a root of a monic polynomial with coefficients in A. We say B is integral over A if every element of B is integral over A. Definition. The integral closure of A in B is the set of all x B which are integral over A. If A is equal to its integral closure in B then we say A is integrally closed in B. In particular, if A is a domain and B is the fraction field of A then we simply say that A is integrally closed. Lemma x B is integral over A if and only if A[x] is a finitely generated A-module. Proof. ( = ) If x n + a n 1 x n a 0 for a i A then x n M := n 1 i=1 Axi which is a finitely generated A-module. By induction, for all m n, x m M. This implies A[x] = M, so in particular A[x] is finitely generated. ( ) Suppose A[x] is generated by f 1 (x),..., f n (x) where f i are polynomials in a single variable over A. Let d max{deg f i } n i=1. Then = x d = n a i f i (x) i=1 for some choice of a i A. This shows that x is a root of the polynomial t d n i=1 a if i (t) so x is integral over A. Theorem The integral closure of A in B is a ring. Proof. It suffices to prove that the integral closure Ā is closed under the addition and multiplication of B. If x, y Ā, Lemma shows A[x, y] is finitely generated. This implies that the submodules A[x + y] and A[xy] are also finitely generated, so x + y, xy Ā. Hence Ā is a ring. Let A B be a subring. We will make use of the following facts about integral extensions of rings: ˆ Every UFD is integrally closed. ˆ If A is a domain, B is finite over A if and only if B is integral over A and B is finitely generated as an A-module. ˆ Suppose C B A are all rings. If C is integral over B and B is integral over A then C is integral over A. ˆ If B is integral over A then S 1 B is integral over S 1 A for any multiplicatively closed subset S A. 4

7 1.2 Norm and Trace 1 Algebraic Number Fields The two most important objects in global algebraic number theory are defined next. Definition. K is a number field if K is a finite field extension of Q. Definition. For a number field K Q, the integral closure of Z in K is called the ring of integers of K, written O K. Examples. 1 The ring of integers of Q is Z. 2 For K = Q( [ 1+ 3), the ring of integers is O K = Z ] For a prime p, the cyclotomic field K = Q(ζ p ) = Q(e 2πi/p ) has ring of integers O K = Z[ζ p ]. It turns out that O K is always a free Z-module of rank [K : Q]. Thus we can think of O K as a lattice embedded in the vector space K. 1.2 Norm and Trace Two important maps for understanding number fields are introduced in this section. Let L/K be a finite field extension and fix x L. Definition. The norm of x is the element N L/K (x) = det T x K, where T x : L L is the K-linear map T x (l) = xl. Definition. The trace of x is Tr L/K (x) = tr T x, where tr denotes the trace. Note that the norm and trace are defined for any finite extension L/K, not just number fields. We will often drop the subscript and write N(x) and Tr(x) when the extension is understood. Lemma The norm map N L/K : L K is a homomorphism of multiplicative groups, and the trace map Tr L/K : (L, +) (K, +) is a homomorphism of abelian groups. Theorem Suppose L/K is a finite, separable extension of fields. Let σ 1,..., σ n be the distinct embeddings L K where K is the algebraic closure of K. Then for all x L, N L/K (x) = n σ i (x) and Tr L/K (x) = i=1 n σ i (x). Proof. Assume σ i (x) σ j (x) when i j. A basis of L/K is 1, x,..., x n 1 and the matrix for T x in this basis is a a a a n 1 5 i=1

8 1.3 The Discriminant 1 Algebraic Number Fields where f(x) = a 0 + a 1 x a n x n is the minimal polynomial of x over K. In this case f is also the characteristic polynomial of x, so by linear algebra, Tr(x) is equal to the sum of the roots of f and N(x) is equal to the product of the roots of f. This implies the result. Example Let K = Q( d) for d a squarefree integer (this means d = ±p 1 p 2 p r in its prime factorization). Then an element x = a + b d Q( d) has norm N(x) = a 2 b 2 d and trace Tr(x) = 2a. 1.3 The Discriminant In this section let L/K be a finite, separable extension of fields and let {α 1,..., α n } be a K-basis of L, so that [L : K] = n. Also denote by σ 1,..., σ n : L K the n distinct K-embeddings of L into the algebraic closure of K. Definition. The discriminant of the basis {α 1,..., α n } is d L/K (α 1,..., α n ) = [det(σ i (α j ))] 2. Proposition Let A = [Tr L/K (α i α j )]. Then d L/K (α 1,..., α n ) = det A. In particular, d L/K (α 1,..., α n ) lies in K. Proof. By Theorem 1.2.2, Tr L/K (α i α j ) = n k=1 σ k(α i )σ k (α j ). Thus A = BC, where B = (σ k (α i )) T and C = (σ k (α j )). Taking the determinant gives us det A = (det B)(det C) = (det C) 2 = d L/K (α 1,..., α n ). One case of interest is when L = K(α) is a simple extension and {1, α, α 2,..., α n 1 } is a basis for L as a K-vector space. Then the discriminant of α is defined to be d L/K (α) := d L/K (1, α, α 2,..., α n 1 ). Lemma For any algebraic element α over K, d L/K (α) equals the discriminant of the minimal polynomial of α. Proof. Set L = K(α) and let α i = σ i (α) for each embedding σ i : L K. Then 1 α 1 α1 n 1 1 α 2 α n 1 2 d L/K (α) = det α n αn n 1 This is a Vandermonde determinant, which evaluates to d L/K (α) = (α i α j ) = 1 i,j n i j 1 i<j n (α i α j ) 2. Since K(α)/K is separable, d L/K (α) 0. In fact, the product formula above is precisely the discriminant of f, the minimal polynomial of α over K. 6

9 1.3 The Discriminant 1 Algebraic Number Fields Proposition For any K-basis {α 1,..., α n } of L, d L/K (α 1,..., α n ) 0. Proof. Since L/K is finite and separable, L = K(θ) for some θ L by the primitive element theorem. Then by Lemma 1.3.2, d L/K (1, θ,..., θ n 1 ) 0. Let A GL n (K) be the change of basis matrix from {α 1,..., α n } to {1, θ,..., θ n 1 }. Then for each 1 i, j n, det(σ i (α j )) = (det A)(det(σ i (θ j 1 )). Both determinants on the right are nonzero, so det(σ i (α j )) 0 which implies finally that d L/K (α 1,..., α n ) 0 by the definition of disciminant. The proof of Proposition gives us the following useful formula: If A, B are two K-bases for L with change of basis matrix A, then d L/K (A) = (det A) 2 d L/K (B). Example Take our favourite example, K = Q( d) over Q, where d is a squarefree integer. Then {1, d} is a basis for L, and its discriminant is d K/Q (1, d) = ( ( )) 2 1 d det 1 = ( 2 d) 2 = 4d. d This matches the fact that the discriminant of x 2 d is 4d. Suppose A K is integrally closed with fraction field K. Let B be the integral closure of A in L. Observe that if x B then all conjugates of x in K are integral over K. Thus N L/K (x) A and Tr L/K (x) A since A is integrally closed. Lemma If x B then N L/K (x) A. Proof. By Lemma 1.2.1, N L/K is a homomorphism of groups. Lemma Suppose α 1,..., α n B form a K-basis of L. Let d = d L/K (α 1,..., α n ). Then db Aα Aα n. Proof. Let a 1,..., a n K such that α := n i=1 a iα i B. Then (a 1,..., a n ) is a solution to the system of linear equations Tr L/K (α i α) = n Tr L/K (α i α j )x j, 1 i n. j=1 The matrix corresponding to this system has determinant d by Proposition Thus each a j can be written as 1 times an A-linear combination of Tr(α d iα). Since α i, α B, Tr(α i α) A so dα j A for each j. Thus dα = n da j α j Aα Aα n. j=1 Since α B was arbitrary, we have shown that db Aα Aα n. 7

10 1.3 The Discriminant 1 Algebraic Number Fields Proposition If A is a PID and M L is a finitely generated B-module, then M is free of rank n = [L : K] as an A-module. In particular, B is free of rank n as an A-module. Proof. Let {α 1,..., α n } B be a basis for L/K. We know the rank of B, which is welldefined over a PID, is at most n. On the other hand, since the α i are linearly independent, the rank of B is at least n. Thus the rank of B equals n. Now suppose M is finitely generated as a B-module, say by elements µ 1,..., µ r. Then there exists an a A such that aµ i B for each i (by a homework problem). Thus daµ i M 0 := Aα Aα n by Lemma So dam M 0. By the structure theory of modules over a PID, since M 0 is free, dam is also free, so M is free of rank at most n. On the other hand, rank M rank B = n by assumption so rank M = n. The second statement follows from taking M = B. Definition. In the situation above, an A-basis for B is called an integral basis. The most important case of Proposition is when K = Q, A = Z and B L is the integral closure of A in some number field L. Proposition Let {α 1,..., α n } and {β 1,..., β n } be two integral bases for B/Z. Then d L/Q (α 1,..., α n ) = d L/Q (β 1,..., β n ). Proof. Let d = d L/Q (α 1,..., α n ) and d = d L/Q (β 1,..., β n ). Then d = (det M) 2 d for some M GL n (Z). Thus det M = ±1 so d = d. This allows us to define: Definition. The discriminant of a number field K/Q is for any Z-basis {α 1,..., α n } of O K. d K := d K/Q (α 1,..., α n ) Example The quadratic field K = Q( 2) has integral basis {1, 2}. Then d K = 8. In general, for a quadratic extension K = Q( d), the discriminant is given by { 4d, d 2, 3 (mod 4) d K = d, d 1 (mod 4) Example Let ζ be a primitive p r th root of unity and let K = Q(ζ). We know that [K : Q] = ϕ(p r ) = p r 1 (p 1). Set n = ϕ(p r ). We will show that O K = Z[ζ] for every prime power p r. First, we compute the discriminant d K = d K/Q (ζ). To do so, we need the following lemma. Lemma If f is the minimal polynomial of α and K = Q(α), d K/Q (α) = ±N K/Q (f (α)). 8

11 1.3 The Discriminant 1 Algebraic Number Fields Proof. Let σ 1,..., σ n be the Q-embeddings of K into Q. We may assume σ 1 = id : K Q. Then by Lemma 1.3.2, d K/Q (α) = (σ i (α) σ j (α)) 2. 1 i<j n By the product rule for differentiation, we have f (α) = n (α σ i (α)). i=2 Finally, compute the norm of f (α): N K/Q (f (α)) = ( n ) σ j (α σ i (α)) 1 j n = 1 i j n i=2 (σ i (α) σ j (α)) = ± (σ i (α) σ j (α)) 2 = d K/Q (α). 1 i<j n (Note: in fact, Lemma holds for any finite separable extension L/K.) Now, to compute d K/Q (ζ) in our example, let f(x) be the minimal polynomial of ζ over Q. We may write this in two ways: f(x) = xpr 1 x pr 1 1 or (x pr 1 1)f(x) = x pr 1. Differentiating the second expression gives us f (x)(x pr 1 1) + f(x)(p r 1 x pr 1 1 ) = p r x pr 1. Then plugging in ζ and solving for f (ζ) produces Take the norm of this expression: f (ζ) = pr ζ r 1 ζ pr 1 1. N(f (ζ)) = ± N(pr ) N(ζ pr 1 1) = ±pa for some a Z. Thus by Lemma , d K/Q (ζ) = ±p a for some a Z, where a ϕ(p r )p r. It turns out that it s easier to work with 1 ζ in this example. In general this creates no obstacles, since d K/Q (1 ζ) = d K/Q (ζ). In our case, we observe that d K/Q (1 ζ) = 1 i<j n (1 σ i (ζ) (1 σ j (ζ))) 2 = 1 i<j n (σ i (ζ) σ j (ζ)) 2 = d K/Q (ζ). Thus d K/Q (1 ζ) = ±p a. To proceed, we need the following generalization of Lemma

12 1.3 The Discriminant 1 Algebraic Number Fields Lemma Proof. Consider (1 ζ k ) = p. p k 1 k p r Plugging in x = 1 gives the result. f(x) = xpr 1 x pr 1 1 = 1 + xpr 1 + x 2pr x (p 1)pr 1. Observe that for any two k 1, k 2 N not divisible by p, 1 ζ k 1 1 ζ k 2 Z[ζ]. Then by symmetry, 1 ζk 1 is a unit in Z[ζ] for all such k 1 ζ k 2 1, k 2. We will now show O K = Z[1 ζ]. Consider the basis {1, 1 ζ, (1 ζ) 2,..., (1 ζ) n 1 } for K/Q. If x O K, we can write x in the following manner by Lemma 1.3.6: x = n 1 i=0 b i p a (1 ζ)i for b i Z, using the fact that d K/Q (1 ζ) = ±p a. If b i Z for each i, then we re done. If not, multiply p a by some p c b so that all i p c 1Z but not all of them lie in Z. Note that p a p pc x O K, so we may replace x with p c x and write x = n 1 i=0 b i p (1 ζ)i, b i Z. Suppose x Z[1 ζ]. Subtracting off the terms where p b i if necessary, we may assume b i = 0 whenever p b i. Let j be the smallest index with p b j. Then p x = n 1 i=j b i p (1 ζ)j, p b j. The element lies in Z[1 ζ] since j + 1 n and (1 ζ) n p by Lemma (1 ζ) j+1 Therefore we may multiply the expression for x by to obtain Note that N ( ) bj 1 ζ Z but this contradicts the fact that O K = Z[1 ζ] = Z[ζ]. x = = bn j = bn j N(1 ζ) p b j 1 ζ b j 1 ζ p (1 ζ) j+1 + (terms in Z[1 ζ]). ( ) is not divisible by p, Thus N bj 1 ζ O K. Hence x Z[1 ζ] which finally proves the claim that 10

13 1.4 Factorization of Ideals 1 Algebraic Number Fields The following theorem allows us to generalize Example to all Q(ζ) where ζ is a primitive nth root of unity. Theorem Let A be an integrally closed integral domain with field of fractions K and suppose L/K and M/K are finite separable extensions with ω 1,..., ω n an integral basis for L with respect to A and α 1,..., α m an integral basis for M with respect to A. Further suppose d L/K (ω 1,..., ω n ) and d M/K (α 1,..., α m ) are relatively prime in A. Then {ω i α j } is an integral basis for the compositum LM over A and d LM/K (ω i α j ) = d L/K (ω i ) m d M/K (α j ) n. Corollary If ζ m is an mth root of unity then O Q(ζm) = Z[ζ m ]. Proof. Factor m = p a 1 1 p ar r. Then Q(ζ m ) = Q(ζ a ) Q(ζ p 1 p ar ). Moreover, for distinct 1 r primes p q, d Q(ζp)/Q and d Q(ζq)/Q are relatively prime. Therefore by Theorem , the ring of integers of Q(ζ m ) is O Q(ζm) = Z[ζ a,..., ζ p 1 p ar ] = Z[ζ 1 r m]. 1.4 Factorization of Ideals Let K be a number field. We have seen that unique factorization may fail in O K, as we recall in the example below. Example The quadratic field K = Q( 5) has ring of integers O K = Z[ 5]. In this ring, 6 has two different factorizations: 6 = 2 3 = (1 + 5)(1 5). Therefore unique factorization fails in Z[ 5]. To see that these are the only two factorizations of 6, observe that N(1 + 5) = N(1 5) = 6, but there are no solutions in integers to the equation N(a + b 5) = a 2 + 5b 2 = 2, 3. It is our goal in this section to in some fashion repair the failure of unique factorization in O K, and in an arbitrary Dedekind domain A (to be defined below). Then we will further study the problem of determining all factorizations of an element in an integral extension. For the unique factorization problem, it would be nice (even ideal ) if there were some objects p 1, p 2, p 3, p 4 Z[ 5] such that 2 = p 1 p = p 1 p 3 3 = p 3 p = p 2 p 4. In fact, the exact objects we are looking for are prime ideals in O K. In order to describe a unique factorization into prime ideals, recall that for ideals I, J A, their ideal product is { n } IJ = x i y i : x i I, y i J. i=1 11

14 1.4 Factorization of Ideals 1 Algebraic Number Fields Definition. An integral domain A is a Dedekind domain if (1) A is integrally closed. (2) A is Noetherian. (3) All nonzero prime ideals of A are maximal. The main theorem we will prove is: Theorem If A is a Dedekind domain, then every nonzero ideal I A has a factorization I = n i=1 pa i i for distinct prime ideals p i A which are unique up to ordering. Theorem For every number field K, O K is a Dedekind domain. Proof. (1) O K is integrally closed since by definition it is the integral closure of Z in K. (2) We have seen (Prop ) that O K lies inside a finitely generated Z-module. By commutative algebra, this is sufficient to conclude that O K is Noetherian. (3) The property of nonzero prime ideals being maximal is alternatively known as Krull dimension 1. It is known that finite integral extensions which are integrally closed preserve Krull dimension, e.g. by the going up theorem. Since Z is Dedekind, integer unique factorization can be captured by Theorem by associating a prime integer p Z with the principal prime ideal it generates: (p) Z. For example, in Z[ 5] we have (6) = (2, 1 + 5)(2, 1 5)(3, 1 + 5)(3, 1 5). Lemma If A is Dedekind, every nonzero ideal I A contains a finite product of prime ideals. Proof. Let M be the set of nonzero ideals of A not divisible by a finite product of primes. Since A is Noetherian, there exists a maximal element a M. Then a must not be prime, so there exist elements b 1, b 2 A a such that b 1 b 2 a. Consider the ideals a + (b 1 ) and a + (b 2 ). Since a is maximal in M, each of these contains a finite product of prime ideals. Then (a + (b 1 ))(a + (b 2 )) a contains a product of primes, a contradiction. Hence M is empty. The classic proof of unique factorization of integers relies on being able to cancel out primes (by dividing), so to mimic this in our proof of Theorem 1.4.2, we define an analogy of inverses for ideals. Definition. If J A is an ideal, the fractional ideal generated by J is the A-module J 1 := {x K xj A}. Lemma For every ideal J A, J 1 is an A-submodule of K. Notice that J 1 A so for any proper ideal J A, J 1 is not an ideal of A. 12

15 1.4 Factorization of Ideals 1 Algebraic Number Fields Lemma If p A is a prime ideal, p 1 A. Proof. Let x p. Then By Lemma 1.4.4, (x) p 1 p r for prime ideals p i A. Assume r is minimal among such products of primes contained in (x). We claim that p = p i for some 1 i r. If not, there exists an a i p i p for each i, by maximality of prime ideals. Then a 1 a r p 1 p r (x) p, a contradiction. Thus p = p i for some i. Assume p = p 1. Then by minimality of r, we know (x) p 2 p r. Let b p 2 p r (x). Then x 1 b A, but x 1 bp A. So x 1 b p 1 A. Lemma If a A is an ideal and p A is a prime ideal, then ap 1 a. Proof. Certainly ap 1 a since p 1 A. Suppose ap 1 = a and let x p 1. Then xa a, so in particular, left multiplication by x is an element of the A-algebra End A (a). Since A is Noetherian, End A (a) is finitely generated. Clearly End A (a) is also a faithful A-module, so by a well-known characterization of integrality (cf. Atiyah-Macdonald), x is integral over A. Then since A is integrally closed, x A. We have shown p 1 = A, but this contradicts Lemma Therefore ap 1 a. Corollary For any prime ideal p A, pp 1 = A. Proof. By Lemma 1.4.7, we have p pp 1 A, but primes are maximal in a Dedekind domain, so pp 1 = A. Corollary For an ideal a A and a prime ideal p a, ap 1 A. Proof. If ap 1 = A then p = app 1 = a, a contradiction. We are now prepared to prove the unique factorization theorem for nonzero ideals in a Dedekind domain. Proof. (of Theorem 1.4.2) Let M be the collection of nonzero, non-unital ideals in A that do not have a factorization into prime ideals. Since A is Noetherian, M has a maximal element a. As before, a cannot be prime so it is contained in a prime ideal p. By Lemma 1.4.7, ap 1 a so ap 1 M. On the other hand, by Corollary we have ap 1 (1) so ap 1 = p 1 p r. Multiplying by p gives us a = app 1 = pp 1 p r which shows a has a prime factorization. Thus M must be empty. This proves the existence part of the theorem. For uniqueness, suppose a = p 1 p r = q 1 q s for prime ideals p i, q j A. Then by the proof of Lemma 1.4.6, p 1 j q j implies p 1 = q j for some 1 j s. Multiplying both sides by p1 1 cancels out terms, yielding shorter prime factorizations of a which are equal by induction. The base case of this induction is easy: if a is prime then it only has the trivial factorization a = a. This finishes the proof of unique factorization of ideals in a Dedekind domain. Remark. For an ideal a A and a prime ideal p A, we will use the expressions p a ( p contains a ) and p a ( p divides a ) to mean the same thing: p appears in the prime factorization of a. If I, J A are ideals, we will write (I, J) = 1 if I + J = (1), that is, if I and J are relatively prime in A. 13

16 1.4 Factorization of Ideals 1 Algebraic Number Fields Definition. A fractional ideal of A is any finitely generated A-submodule of K. Example For any ideal J A, J 1 is a fractional ideal. Proposition The nonzero fractional ideals of A form a group under ideal multiplication, with identity (1). Proof. By Theorem 1.4.2, fractional ideals of the form p a i i, with a i Z and p i A prime, form a group which is isomorphic to a direct sum of copies of Z. Let M be any fractional ideal. Then M is finitely generated, so there exists an element x K such that xm A is an ideal. Since (x) and xm have prime factorizations, so does M = (x) 1 xm. Hence all fractional ideals form a group under multiplication. Corollary shows that (1) is the identity element in this group. Proposition If I A is an ideal then II 1 = (1). Proof. Suppose I = p 1 p r is the prime factorization of I. Then J = p 1 1 p 1 r is a fractional ideal, and by Corollary 1.4.8, IJ = (1). It remains to show J = I 1. First, since IJ = (1) we have J I 1. If x I 1 then xi A so xip 1 1 p 1 r p 1 1 p 1 r = J. Thus Ax J so x J. This proves J = I 1 as required. Corollary If A is a Dedekind domain and a unique factorization domain, then A is also a PID. Definition. For a Dedekind domain A, let J A denote the group of fractional ideals of A. The ideal class group of A is defined as the quotient group C A = J A /P A where P A is the subgroup of J A consisting of the principal fractional ideals of A. Clearly C A = 1 if and only if A is a PID (and therefore a UFD), so the ideal class group is a direct measure of the failure of unique factorization in A. Moreover, the ideal class group corresponds to an exact sequence of groups 1 A K J A C A 1. We will study this further when we characterize the unit group K in Section 1.9. Lemma Every class in C A can be represented by an ideal I A. Example The ring A = C[x, y]/(y 2 x 3 x) is a Dedekind domain. It turns out that the ideal class group C A has cardinality equal to C, so this example shows that ideal class groups can be particularly bad. In particular, unique factorization fails in A: x 3 x = y 2 = x(x 1)(x + 1). One of the most important results in algebraic number theory is the following theorem, which we will prove in Section

17 1.5 Ramification 1 Algebraic Number Fields Theorem. For a number field K, the class group C K := C OK is finite. Example Let K = Q( 5) and recall that 6 = 2 3 = (1 + 5)(1 5). What do 2 and 3 split into as ideals in O K = Z[ 5]? It turns out that 2O K = (2, 1 + 5)(2, 1 5) and 3O K = (3, 1 + 5)(3, 1 5). The underlying principle governing this splitting behavior is the fact that the minimal polynomial x of 5 splits differently mod 2 and 3: x (x + 1) 2 (mod 2) and x (x + 1)(x 1) (mod 3). 1.5 Ramification In this section let L/K be a finite separable field extension, let O K be a Dedekind domain with field of fractions K and let O L be the integral closure of O K in L. Put n = [L : K]. Lemma O L is a Dedekind domain. Proof. This is the same proof as for Theorem Lemma If p O K is a prime ideal then po L O L. Proof. Take x p 1 O K, which exists by Lemma Then xp O K so xpo L O L. If po L = O L then we have xpo L = xo L O L, a contradiction. Therefore po L O L. Now fix a nonzero prime ideal p O K. By Theorem 1.4.2, p considered as an ideal of O L has a unique factorization po L = P e 1 1 P eg g where the P i O L are distinct primes and each e i > 1. Note that for each i, O L /P i is a finite dimensional O K /p-vector space. (This follows from the fact that P i O K = p.) We say the P i are the primes of O L lying over p. By unique factorization, these are the only primes lying over p. Definition. For a prime P i in the factorization of po L, the index f i = [O L /P i : O K /p] is called the inertial degree of P i (over p) and the exponent e i is called the ramification index of P i (over p). We say the prime p is totally split if e i = f i = 1 for all 1 i g; p is totally ramified if g = 1 and f 1 = 1; and p is inert if g = 1 and e 1 = 1. Definition. If any e i > 1 or (O L /P i )/(O K /p) is inseparable, we say the prime p is ramified (in O L ). Otherwise p is unramified. Theorem For any prime p O K with prime factorization po L = g g i=1 e if i = n = [L : K]. i=1 Pe i i, we have 15

18 1.5 Ramification 1 Algebraic Number Fields Proof. By the Chinese remainder theorem, we can write g g O L /po L = O L / = i=1 P e i i i=1 O L /P e i i. To prove the theorem, we show that [O L /po L : O K /p] = n and [O L /P e i i : O K /p] = e i f i for each 1 i g. For the first equality, take {ω 1,..., ω m } to be a basis for O L /po L as an O K /p-vector space. Lift these elements to ω 1,..., ω m O L. Suppose a 1 ω a m ω m = 0 for coefficients a i O K. Let a = (a 1,..., a m ) O K and let x a 1 ap; such an element exists by Lemma Then xa i O K for all i, but xa i p for some i. Replacing a i with xa i and reducing mod p gives us a linear dependence, contradicting the assumption that ω 1,..., ω m are a basis of O L /po L. Hence ω 1,..., ω m must be linearly independent in O K. To show they span O K, let M = ω 1 O K ω m O K O L. Since the ω i generated O L /po L, we get M + po L = O L. In other words, p(o L /M) = O L /M. By Nakayama s Lemma, this means O L /M is killed by some x 1 + p. In particular, such an x is necessarily nonzero so xo L M. Thus ω 1 xo K ωm O x K O L. This implies that ω 1 x K+...+ ωm K = x L so the ω i span O L as an O K -vector space. This of course is only possible if m = n, so we have the first equality. Now consider the sequence O L /P e i i P i /P e i i P 2 i /P e i i P e i 1 i /P e i i 0. Taking each quotient in the chain yields something of the form P ν i /P ν+1 i, and by unique factorization, each of these quotients is nontrivial. Thus we can choose x P ν i P ν+1 i. Consider the map ϕ : O L P ν i /P ν+1 i α xα. Certainly ker ϕ = P i since P i ker ϕ and primes are maximal in O L. Also, ϕ is surjective since P ν+1 i (x)+p ν+1 i P ν i which implies (x)+p ν+1 i = P ν i. Therefore O L /P i = P ν i /P ν+1 i as O K /p-vector spaces. Adding these up gives us dim OK /p O L /P e i i = e i j=1 dim OK /p O L /P i = e i f i. This proves both claims, and this is of course enough to conclude that n = g i=1 e if i. Let θ O L be a primitive element of L/K, that is, L = K(θ). It is not always guaranteed that O K [θ] = O L. However, we have a way of measuring how far off from the whole ring O L the submodule O K [θ] really is. Definition. The conductor of the extension L/K is the ideal where L = K(θ). f := {α O K αo K O K [θ]} O K 16

19 1.5 Ramification 1 Algebraic Number Fields Example If O K [θ] = O L, then f = (1). Example For K = Q and L = Q( 3), the conductor is f = (2, 1 + 3). Note that f is always nonzero. Theorem Let L/K be a finite separable extension with L = K(θ). Suppose p O K is prime and po L + f = (1), where f is the conductor of the extension L/K. Let ϕ(x) be the minimal polynomial of θ over K. If ϕ(x) factors completely in (O K /p)[x] as ϕ(x) = ϕ 1 (x) e1 ϕ g (x) eg mod p with deg ϕ i = f i, then the factorization of p in O L is po L = g i=1 Pe i i where for each i, P i is a prime ideal with ramification index e i and inertia degree f i, given explicitly by P i = ϕ i (θ)o L + po L for any lift ϕ i (x) of ϕ i (x) in O K [x]. Proof. Set O = O K [θ]. We will prove the following isomorphisms: O L /po L = O /po = (OK /p)[x]/ϕ(x) where again ϕ(x) is the minimal polynomial of θ over K. Clearly O O L so we have a map O /po O L /po L. By assumption, po L + f = O L but f O L so we have po L + O = O L. Hence the map is surjective. On the other hand, po po L O and po L O = (po L O )(po L + f) po + fpo L po. This proves injectivity, so the first isomorphism is proven. The second isomorphism is immediate from the fact that O /po = OK [x]/(ϕ(x), p) = (O K /p)[x]/ϕ(x). Now by the Chinese remainder theorem, we may write O L /po L L = (O K /p)[x] = g (O K /p)[x]/ϕ(x) e i. i=1 The prime ideals on the right are just the ideals (ϕ i (x)). Set R = (O K /p)[x]/ϕ(x) and notice that [R/(ϕ i (x)) : O K /p] = f i = deg ϕ i and g i=1 ϕ i(x) e i = 0. The primes in O L /po L corresponding under the above isomorphism to the ϕ i (θ) are P i := ϕ i (θ)o L + po L. Notice that g i=1 Pe i i po L, but since dim OK /p O L /po L = g e i f i = dim OK /p O L / i=1 we have g i=1 Pe i i = po L. This proves the theorem. g i=1 P e i i, 17

20 1.5 Ramification 1 Algebraic Number Fields Example Let O K = Z[i] be the Gaussian integers. Here the conductor is f = (1). Consider how x splits mod 13: x (x 5)(x + 5) (mod 13). Then by Theorem 1.5.6, the ideal (13) splits in Z[i] in the following way: 13Z[i] = (13, 5 + i)(13, 5 + i) = (3 + 2i)(3 2i). For a prime p O K and a prime P O L lying over p, write k p = O K /p and l P = O L /P. Proposition If p O K is a nonzero prime such that p + f = (1) in O L, then p is unramified if and only if p does not divide the principal ideal (d L/K (θ)) generated by the discriminant of L/K. Proof. We know d L/K (θ) = i<j (θ i θ j ) 2 where θ i are all the K-embeddings of θ in K. Since p + f = (1), p splits in O L based on how ϕ splits mod p, where ϕ is the minimal polynomial of θ over K. Explicitly, Theorem tells us that p is unramified if and only if there are no repeated factors in the factorization of ϕ mod p and ϕ mod p is separable. This is equivalent to all the roots of ϕ mod p having multiplicity 1 in k p, which in turn is equivalent to θ i θ j mod p for all i j. Now consider θ i θ j for all i j i j(θ i θ j ) is relatively prime to p in O K p is relatively prime to each θ i θ j in O M for some normal closure M of L/K p (d L/K (θ)) in O K. Hence p is unramified precisely when p (d L/K (θ)). We now discuss Hilbert s program for ramification theory. Assume that L/K is Galois and let G = Gal(L/K). Note that σ(o L ) = O L for all σ G. If p O K is a prime and po L = g i=1 Pe i i, then each σ G acts on the primes lying over p: σ(p i) = P j for some 1 j g. The key observation is that this action is transitive. Proposition For any prime p O K, G = Gal(L/K) acts transitively on the set of primes of O L lying over p. Proof. Suppose not. Then there is some pair of primes P i, P j lying over p such that σp j P i for all σ G. By the Chinese remainder theorem, pick x P j such that x 1 (mod σp i ) for all σ G. Then N L/K (x) P j O K = p. On the other hand, N L/K (x) = σ G σ(x) but σ(x) P i for any σ, so N L/K (x) p. This is impossible, so there is some σ G such that σp j = P i. Corollary When L/K is Galois, for any prime p O K, all ramification indices e i and all inertia degrees f i for primes over p are equal, and therefore [L : K] = efg, where e = e i and f = f i for any prime P i p. 18

21 1.5 Ramification 1 Algebraic Number Fields Proof. An ideal P ν i divides po L if and only if σp ν i divides po L for all σ G, which by Proposition is equivalent to P ν j dividing po L for all 1 j g. Therefore the ramification indices are all equal; let e denote any one of them. Now given 1 i, j g, suppose σ G is a permutation taking P j to P i, that is, P i = σp j. Then σ determines an isomorphism O L /P j O L /P i. Therefore f i = f j. Let f denote any of the inertial degrees. Then finally, by Theorem we have [L : K] = g i=1 ef = efg. Fix a prime P O L lying over p. Definition. The subgroup D P = {σ G σ(p) = P} of G is called the decomposition group of P. Clearly by the orbit-stabilizer theorem, D P = ef where e and f are the ramification index and inertia degree of p, respectively. By Galois theory, there is a field extension Z P /K corresponding to the subgroup D P G, which is explicitly the fixed field Z P = L D P. Definition. For a prime P p, the field Z P is called the decomposition field of P. D P L Z P G K Lemma If σp = P for two primes P, P lying over p, then D P some σ G. = σd P σ 1 for 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 ramification index and inertia degree are transitive in any tower of Galois field extensions: Lemma For a Galois tower of number fields M L K and a prime p O K, let Q O M be a prime lying over p and set P = Q L. Then e(q p) = e(q P)e(P p) and f(q p) = f(q P)f(P p). Proof. Clearly P is a prime lying over p in O L, so e(p p) and f(p p) are defined. Then e(q p) = e(q P)e(P p) is immediate by unique factorization in Dedekind domains, and f(q p) = f(q P)f(P p) follows from Corollary and the fact that [M : K] = [M : L][L : K]. The decomposition field is characterized by the following proposition. 19

22 1.5 Ramification 1 Algebraic Number Fields Proposition Let a = P Z P be a prime below P in Z P. Then (1) P is the only prime in O L lying over a. (2) If e = e(p p) and f = f(p p) then e(p a) = e, f(p a) = f and e(a p) = 1 = f(a p). L P e f Z P a 1 1 K p Proof. (1) For all σ D P = Gal(L/Z P ), σp = P. By Proposition 1.5.9, D P acts transitively on the primes over a, so P must be the unique one. (2) Since D P = ef, e(p a)f(p a) = ef but by Lemma , e(p a) divides e and f(p a) divides f. Therefore e(p a) = e and f(p a) = f, and the others are 1 by Lemma Remark. Every σ D P induces an automorphism ϕ σ : O L /P O L /P which fixes k p = O K /p l P = O L /P. Thus we get a map ϕ : D P Aut(l P /k p ) σ ϕ σ. Proposition ϕ : D P Aut(l P /k p ) is surjective and l P /k p is a normal extension. Proof. By Proposition , k a = k p for any prime ideal a in the ring of integers of the decomposition field, so we can replace K with Z = Z P and G with D P. Thus P is the only prime lying over p. Take θ l P and let θ O L be any lift, with minimal polynomials ḡ(x) k p [x] and f(x) K[x], respectively. Certainly f( θ) = 0 mod p so ḡ f in kp [x]. Since L/K is normal (it is a Galois extension), f splits over L. This implies f splits over l P, so ḡ splits as well. This proves l P /k p is a normal extension. Now choose θ generating the separable closure of k p in l P. Let σ Aut(l P /k p ). Then σ θ is a root of ḡ and thus of f. Since f splits in L, there exists a root α L of f such that ᾱ = σ θ in l P. Choose σ G = D P such that σθ = α, which is possible since L/K is normal. Then ϕ(σ) = σ because θ generates the separable closure of k p in l P. This proves ϕ is surjective. Definition. The kernel I P = ker ϕ D P is called the inertia group of P. Explicitly, I P = {σ G σ(x) x mod P for all x O L }. Definition. The fixed field T P = L I P is called the inertia field of P (over p). 20

23 1.6 Cyclotomic Fields and Quadratic Reciprocity 1 Algebraic Number Fields By Proposition , we have an exact sequence 1 I P D P Gal(l P /k p ) 1. Proposition Let b = P T P and a = P Z P = b Z P be prime ideals in the inertia and decomposition fields, respectively. Set e = e(p p) and f = f(p p). Then e(p b) = e, f(b a) = f and e(b a) = f(p b) = 1. L T P Z P K P e 1 b 1 f a 1 1 p Proof. Let Z = Z P and T = T P. In light of Proposition , it s enough to show l P = O T /b and D P /I P = f. By the exact sequence 1 I P D P Gal(l P /k p ) 1, D P /I P = Gal(lP /k p ) so if l P = O T /b, then D P /I P = Gal((O T /b)/k p ) = Gal((O T /b)/(o Z /a)) = f. Therefore it suffices to prove the former statement, that is, l P = O T /b. The decomposition/inertia group exact sequence for the extension L/T is which implies l P = O T /b as claimed. 1 I P I P Gal(l P /(O T /b)) Cyclotomic Fields and Quadratic Reciprocity Recall that when ζ m is a primitive mth root of unity and K = Q(ζ m ), the ring of integers of this cyclotomic number field is O K = Z[ζ m ]. This was proven in Corollary In this section, we further elaborate on the properties of Q(ζ m ) and Z[ζ m ] and use algebraic number theory to prove Gauss s celebrated quadratic reciprocity law. Recall the following definition from elementary number theory. Definition. If p is an odd prime and a Z, the Legendre symbol of a mod p is ( ) a 1 p a and a is a square (mod p) = 1 a is not a square (mod p) p 0 p a. 21

24 1.6 Cyclotomic Fields and Quadratic Reciprocity 1 Algebraic Number Fields Proposition Suppose a, p Z with p prime and (2a, p) = 1. Then only if the prime ideal (p) splits completely in O Q( a). ( a p ) = 1 if and Proof. The conductor f a divides 2, so (p) splitting in O Q( a) is equivalent to (p) splitting in Z[ a]. This, in turn, is equivalent to x 2 a splitting mod p, by Theorem 1.5.6, i.e. a x 2 (mod p) for some x Z. Hence (p) splits in O Q( a) if and only if a is a square mod p. ( ) Remark. Since F a p is cyclic, it s easy to show that a (p 1)/2 (mod p) for any a Z p this is called Euler s criterion. In particular, for a = 1 we have ( ) { 1 1, p 1 (mod 4) = p 1, p 1 (mod 4). Most elementary proofs of quadratic reciprocity exploit this characterization of the Legendre symbol in some fashion. Here we prove the reciprocity law by considering the factorization of (p) in the ring Z[ζ q ]. Theorem (Quadratic Reciprocity). Let p and q be distinct, odd primes. Then ( ) ( ) p q = ( 1) (p 1)(q 1)/4. q p ( ) ( ) First, set q = ( 1) (q 1)/2 q q so that q = ( 1) (p 1)(q 1)/4 by Euler s criterion. The p p ( ) ) statement we must then prove is that =. p q ( q p Example The beauty of the quadratic reciprocity law is that it allows for fast computations of the Legendre symbol. For example, is 15 a square mod 37? Rather than trying to compute all squares mod 37, or trying to factor x 2 37 in F 37, we can use reciprocity. Since 37 1 (mod 4), we have: ( ) = ( ) ( ) So 15 is not a square mod 37. = ( 37 5 ) ( ) 37 = 3 ( 2 5 ) ( ) 1 = ( 1)(1) = 1. 3 Lemma Suppose n 2 is an integer with prime factorization n = p ν(p), where the product is over all primes p and ν(p) 0 for all p. For each prime p, let f p be the multiplicative order of p mod n/p ν(p). Then in R = Z[ζ n ] we have pr = (p 1 p r ) ϕ(pν(p)) for distinct prime ideals p 1,..., p r R such that f(p i p) = f p for each 1 i r. Proof. Fix a prime p and set m = n/p ν(p) so that n = p ν(p) m. Consider the number field K = Q(ζ n ). We know the conductor of ζ n in O K is f = 1. Let γ n be the nth cyclotomic polynomial and let {α i } be the primitive p ν(p) th roots of unity and {β j } be the primitive mth roots of unity. Then by the Chinese remainder theorem, (Z/nZ) = (Z/p ν(p) Z) (Z/mZ) 22

25 1.6 Cyclotomic Fields and Quadratic Reciprocity 1 Algebraic Number Fields so we can write γ n (x) = i,j (x α i β j ). Note that all the α i are 1 in any field of characteristic p. Thus, modulo p, γ n (x) j (x β j ) ϕ(pν(p)) = γ m (x) ϕ(pν(p)). This allows us to reduce to the case when m = n, that is, the case when p ν(p) = 1. Let γ m (x) denote the factorization of γ m (x) mod p. Since x m 1 is separable over F p (m is relatively prime to p) and γ m (x) x m 1, we have that γ m (x) is also separable over F p. The smallest extension of F p containing a primitive mth root of unity (and thus all of them) is F p fp. Thus γ m splits over F p fp and each irreducible factor of γ m over F p is the minimal polynomial of some primitive mth root of unity, each of which having degree f p. This implies γ m is a product of degree f p irreducible polynomials over F p. By Theorem 1.5.6, we have pr = (p 1 p r ) ϕ(pν(p)). Remark. In general, Theorem implies that r = ϕ(n) ϕ(p ν(p) )f p. Corollary An odd prime integer p is ramified in Q(ζ n )/Q if and only if p n, and p = 2 is ramified if and only if 4 n. Lemma If q is an odd prime integer, then ( 1) (q 1)/2 q Q(ζ q ). Proof. Set τ = Then τ Q(ζ q ) and τ 2 = ( 1) (q 1)/2 q. a (Z/qZ) ( ) a ζq a. q We are now able to prove quadratic reciprocity (Theorem 1.6.2). Proof. Let p and q be distinct odd primes and set q = ( 1) (q 1)/2 q. Consider the tower of number fields Q(ζ q ) Q( q ) Q, with Galois groups as shown: Z/(q 1)Z Q(ζ q ) Z/((q 1)/2)Z Q( q ) Z/2Z Q 23

26 1.7 Lattices 1 Algebraic Number Fields Then we determine the reciprocity law for ( ) q as follows: p ( ) q = 1 (p) splits in Q( q p ) by Proposition Q( q ) Z P, the decomposition field for any prime P over (p) there exist an even number of primes in Z[ζ q ] lying over (p) q 1 f p is even, where f p is the multiplicative order of p mod q f p divides q 1 2 p q (mod q) ( ) p = 1. q Thus quadratic reciprocity is proven. Corollary If q is an odd prime, then ( ) 2 = q { 1, q 1, 7 (mod 8) 1, q 3, 5 (mod 8). Proof. Set q = ( 1) q 1 2 q, so that q 1 (mod 4). Then ( ) [ ] 2 = 1 (2) splits in O 1 + q q Q( q ) = Z 2 f(x) = x 2 x + 1 q splits mod 2 4 q 1 (mod 8) q = 1, 7 (mod 8). 1.7 Lattices One perspective on rings of algebraic integers is to view them as lattices. For example, Z[i] is very clearly a lattice in C spanned by the vectors 1 and i. We will show that any ring of integers O K in a number field K/Q is a lattice in some R n. This is the beginning of Minkowski s so-called theory of geometry of numbers. Definition. A Z-module Γ R n is a lattice of rank m if Γ = Zv 1 + ldots + Zv m for R-linearly independent vectors v 1,..., v m. If m = n then we say Γ is a complete lattice, or has full rank in R n. 24

27 1.7 Lattices 1 Algebraic Number Fields Definition. For a lattice Γ R n, the set Φ = {x 1 v x m v m 0 x i < 1} is called the fundamental domain of Γ, also sometimes called the fundamental parallelopiped. Observe that Γ is a complete lattice in R n if and only if Γ + Φ = R n. Definition. A subgroup W R n is said to be discrete if every point in W is open in the topology on R n, that is, if every point x W has a neighborhood U in R n such that U W = {x}. Proposition If Γ R n is a subgroup, then Γ is discrete if and only if Γ is a lattice. Proposition If Γ R n is a lattice, then Γ is complete if and only if there exists a bounded set M such that Γ + M = R n. Proof. ( = ) When Γ is complete, M = Φ works. ( ) If Γ is not complete, let V R n be the R-span of Γ. Then V lies in some hyperplane H in R n. Choose d > 0. Then for any bounded set of diameter diam(m) < d, all points further than d from H do not lie in Γ + M H + M. Hence Γ + M R n. = Definition. If Γ = Zv Zv n is a complete lattice in R n, we define the volume of Γ to be the volume of the parallelopiped spanned by v 1,..., v n : vol(γ) := vol(φ) = det A where A = v 1 v n. Note that since det(a T A) = (det A) 2, we can write the volume formula as vol(γ) = det(v i v j ). Definition. A region Ω R n is centrally-symmetric if x Ω implies x Ω. Minkowski s theorem is the key result in the geometry of numbers which allows us to describe lattices like O K and U K, the ring of integers and unit group, respectively, in a number field K/Q. Theorem (Minkowski). If Γ is a complete lattice in R n and X is a centrally-symmetric, convex region of R n such that vol(x) > 2 n vol(γ), then X contains a nonzero point of Γ. Proof. By a linear change of variables, we may assume Γ = Z n. Then vol(γ) = det(i) = 1. Suppose X is as described, with vol(x) > 2 n. Then vol( ( 1 2) > 1. We claim that there exist 25