Algebraic Number Theory

Home , Algebraic element, Algebraic extension, Cyclotomic field, Different ideal, Field extension

Lecture Notes in Algebraic Number Theory

Lectures by Dr. Sheng-Chi Liu

Throughout these notes, signiﬁes end proof, and N signiﬁes end of example.

Table of Contents

Table of Contents i

Lecture 1 Review 1 1.1 Field Extensions ...... 1

Lecture 2 Ring of Integers 4 2.1 Understanding Algebraic Integers ...... 4 2.2 The Cyclotomic Fields ...... 6 2.3 Embeddings in C ...... 7 Lecture 3 Traces, Norms, and Discriminants 8 3.1 Traces and Norms ...... 8 3.2 Relative Trace and Norm ...... 10 3.3 Discriminants ...... 11

Lecture 4 The Additive Structure of the Ring of Integers 12 4.1 More on Discriminants ...... 12 4.2 The Additive Structure of the Ring of Integers ...... 14

Lecture 5 Integral Bases 16 5.1 Integral Bases ...... 16 5.2 Composite Field ...... 18

Lecture 6 Composition Fields 19 6.1 Cyclotomic Fields again ...... 19 6.2 Prime Decomposition in Rings of Integers ...... 21

Lecture 7 Ideal Factorisation 22 7.1 Unique Factorisation of Ideals ...... 22

Lecture 8 Ramiﬁcation 26 8.1 Ramiﬁcation Index ...... 26 Notes by Jakob Streipel. Last updated June 13, 2021.

i TABLE OF CONTENTS ii

Lecture 9 Ramiﬁcation Index continued 30 9.1 Proof, continued ...... 30

Lecture 10 Ramiﬁed Primes 32 10.1 When Do Primes Split ...... 32

Lecture 11 The Ideal Class Group 35 11.1 When are Primes Ramiﬁed in Cyclotomic Fields ...... 35 11.2 The Ideal Class Group and Unit Group ...... 36

Lecture 12 Minkowski’s Theorem 38 12.1 Using Geometry to Improve λ ...... 38

Lecture 13 Toward Minkowski’s Theorem 42 13.1 Proving Minkowski’s Theorem ...... 42 13.2 The Dirichlet Unit Theorem ...... 45

Lecture 14 Dirichlet Unit Theorem 46 14.1 Dirichlet Unit Theorem ...... 46 14.2 Fermat’s Last Theorem ...... 48

Lecture 15 Kummer’s Theorem 49 15.1 Using the Dirichlet Unit Theorem ...... 49 15.2 Kummer’s Theorem ...... 49

Lecture 16 Kummer’s Theorem, continued 51 16.1 Proof continued ...... 51

Lecture 17 Local Fields 53 17.1 Valuations ...... 53

Lecture 18 Ostrowski’s Theorem 56 18.1 Ostrowski’s Theorem ...... 56 18.2 Completions ...... 57

Lecture 19 Ostrowski’s Theorem continued 58 19.1 Proof of Ostrowki’s Theorem ...... 58

Lecture 20 Local Fields 62 20.1 Local Fields ...... 62 20.2 p-adic Numbers ...... 63

Lecture 21 Hensel’s Lemma 65 21.1 Generalisation of p-adic Numbers ...... 65 21.2 Topology of Local Fields ...... 66 21.3 Hensel’s Lemma ...... 67

Lecture 22 Hensel’s Lemma revisited 69 22.1 Second Form of Hensel’s Lemma ...... 69 22.2 Extension of Valuations ...... 71

Lecture 23 Krasner’s Lemma 72 TABLE OF CONTENTS iii

23.1 Proof continued ...... 72 23.2 Krasner’s Lemma ...... 74

Lecture 24 Eisenstein Extensions 76 24.1 Proof continued ...... 76 24.2 Eisenstein Extension ...... 77

Lecture 25 Totally Ramiﬁed Extensions 79 25.1 Proof of Lemma ...... 79 25.2 Unramiﬁed Extensions ...... 81

Lecture 26 Unramiﬁed Extensions 81 26.1 Classifying Unramiﬁed Extensions ...... 81

Lecture 27 Unramiﬁed Extensions 84 27.1 Tamely and Wildly Ramiﬁed Extensions ...... 84 27.2 Decomposition Group and Intertia Group ...... 87

Lecture 28 Decomposition and Inertia Group 88 28.1 Decomposition and Inertia ...... 88

Lecture 29 More on Ramiﬁcation of Extensions 91 29.1 Splitting of Primes ...... 91

Lecture 30 The Diﬀerent Ideal 94 30.1 Duals ...... 94

Index 97 REVIEW 1

Lecture 1 Review

Algebraic number theory is fundamentally the study of finite extensions of the rational numbers Q, called number fields. To this end we will first review some of the theory of field extensions.

1.1 Field Extensions Definition 1.1.1 (Finite field extension). A finite field extension of a field K is a field L ⊃ K such that dimK L is finite (as a vector space over K). The degree of the extension L/K is defined to be deg(L/K) := dimK L. Fact. If we have a tower of finite field extensions, say K ⊂ L ⊂ F , then deg(F/K) = deg(F/L) · deg(L/K). The proof of this boils down to considering a basis of either extension and expressing one with the help of the other, and the amount of elements in the basis between F and K will be the product of the number of basis elements in either intermediate extensions.

Definition 1.1.2 (Algebraic element). Let L/K be a field extension, and let α ∈ L. Then α is called algebraic over K if it satisfies f(α) = 0 for some f ∈ K[x]. Fact. Any element of a finite extension of K is algebraic over K.

To see this, consider an extension L ⊃ K of degree n and let α ∈ L. Then 1, α, α2, . . . , αn are linearly dependent in K, since there are more than n of them, and therefore there is a nontrivial linear combination over K of them that adds to 0, which in turn gives us a polynomial in K[x] with α as a root. Fact. An algebraic element α ∈ L has a minimal polynomial over K, i.e. there exists a unique pα(x) ∈ K[x] such that if f(x) ∈ K[x] with f(α) = 0, then pα(x) | f(x), with the uniqueness being up to multiplication by an element in K. Equivalently, therefore, there is a unique monic polynomial of least degree with α as its root.

Fact. For an algebraic element α ∈ L with minimal polynomial pα(x), we have the isomorphism K[x] K(α) =∼ , hpα(x)i with the isomorphism being the map that evaluates x in α. Moreover

deg(K(α)/K) = deg(pα(x)) = n and 1, α, α2, . . . , αn−1 is a bases for K(α) over K. Thus K(α) = K[α], and in the latter αn and higher degrees can be reduced to lower degrees.

Definition 1.1.3 (Number field). An (algebraic) number field K (⊂ C) is a finite extension of Q. Date: January 9th, 2018. 1.1 Field Extensions 2

Theorem 1.1.4 (Primitive element theorem). Let K be a number field. Then K = Q(α) = Q[α] for some α ∈ K. Two famous examples of number fields are the following: √ Example 1.1.5 (Quadratic fields). Consider K = Q( D), where D is a square-√ free integer (otherwise, simply pull out the square part).√ If D > 0, then Q( D) is called a real quadratic field, and if D < 0 then Q( D) is an imaginary quadratic field. √ These extensions are of degree 2, since the minimal polynomial of D is x2 − D, and moreover they are Galois extensions since they are both separable and normal. N

Example 1.1.6 (Cyclotomic ﬁelds). Consider K = Q(ξm), where ξm is a prim- m n itive mth root of unity, meaning that ξm = 1 and ξm 6= 1 for 0 < n < m. 2πi/m For instance, and by convention what we’ll always use, ξm = e . The other k primitive mth roots of unity are ξm where gcd(m, k) = 1, so deg(K/Q) = ϕ(m), where by ϕ we mean Euler’s totient function. Note that if m = 1, ξ1 = 1 and if m = 2 we have ξ2 = −1, so K = Q in both cases, making neither of these interesting. Hence when discussing cyclotomic 2πi/3 ﬁelds we are interested in m ≥ 3, since when m = 3 we get ξ3 = e yielding K = Q(ξ3) which is an extension of degree ϕ(3) = 2. 4 2 2 Now if m = 6, then ξ6 = −ξ6 = −(ξ6 ) , by straight forward computation, 2 2 meaning that Q(ξ6 ) = Q(ξ6), but note that ξ6 = ξ3, so in fact Q(ξ3) = Q(ξ6). N This last idea generalises:

Proposition 1.1.7. For odd m, Q(ξm) = Q(ξ2m). 2πi/m 2πi/(2m) 2 Proof. Take ξm = e and likewise ξ2m = e . Then clearly ξ2m = ξm, 2 so Q(ξm) = Q(ξ2m) ⊂ Q(ξ2m). m πi m+1 On the other hand, ξ2m = e = −1, meaning that ξ2m = −ξ2m. Since m+1 2 m is odd, m + 1 is even, and therefore ξ2m = −ξ2m ∈ Q(ξ2m), implying that 2 Q(ξ2m) ⊂ Q(ξ2m) = Q(ξm).

Remark 1.1.8. We will show that for even m, all Q(ξm) are distinct. Note also ∼ × that Gal(Q(ξm)/Q) = (Z/mZ) . Deﬁnition 1.1.9 (Algebraic integer). An element α ∈ C is called an algebraic integer if it is a root of a monic polynomial f(x) ∈ Z[x]. √ Example 1.1.10. Many elements are algebraic integers. For instance, 2 is, 2 since its minimal polynomial is x −2, which√ is monic. Similarly 5 is an algebraic integer, since it is the root of x − 5, and D is the root of x2 − D. m The primitive roots of unity are too, since ξm is a root of f(x) = x − 1. N This notion is equivalent with the notion from commutative algebra of α being integral over Z. Note also that since there are only countably many polynomials over Z, the fundamental theorem of algebra implies that there are only countably many algebraic integers.

Theorem 1.1.11. The following are equivalent for α ∈ C: 1.1 Field Extensions 3

(i) α is an algebraic integer.

(ii) Z[α] is a finitely generated Z-module (i.e. Abelian group). (iii) There exists a subring of C containing α that is finitely generated. (iv) There exists a finitely generated nonzero Z-module M ⊂ C such that αM ⊂ M. Proof. That (i) implies (ii) is clear: since α is an algebraic integer we have an equation n n−1 α + cn−1α + ... + c1α + c0 = 0 n for some ci ∈ Z. Rearranging this for α we have

n n−1 α = −(cn−1α + ... + c1α + c0) meaning that Z[α] is ﬁnitely generated by 1, α, α2, . . . , αn−1. (ii) trivially implies (iii) by just taking Z[α] itself, and in the same way (iii) implies (iv) using the subring from (iii) itself. Finally (iv) implies (i) by letting x1, x2, . . . , xn be generators of M, and since αM ⊂ M, we have

αxi = ai1x1 + ai2x2 + ... + ainxn for all i = 1, 2, . . . , n, with aij ∈ Z. As matrix equations we then have       αx1 a11 a12 ··· a1n x1 αx2  a21 a22 ··· a2n  x2    =     ,  .   . . .. .   .   .   . . . .   .  αxn an1 an2 ··· ann xn which rearranged and calling the coeﬃcient matrix A and the xi-vector x, that (αIn − A)x = 0, where In is the identity matrix. Now since x1, x2, . . . , xn generate a nontrivial Z-module, x 6= 0, so (αIn − A) has linearly dependent columns, making its determinant 0, so

n 0 = det(αIn − A) = α + lower order terms in α.

Corollary 1.1.12. Let A be the set of algebraic integers in C. Then A is a ring. Proof. We need to show that α + β and αβ are both in A if α and β are in A. This follows almost trivially from the above theorem, however: we have that Z[α] and Z[β] are ﬁnitely generated Z-modules, and therefore Z[α, β] is ﬁnitely generated as a Z-module as well. This latter ring contains α + β and αβ, and therefore they are algebraic integers.

Deﬁnition 1.1.13. Let K be a number ﬁeld. The ring of integers OK of K is the set of algebraic integers in K, i.e. OK = A ∩ K. By the above corollary, it is a subring of K. Note also that in the language of commutative algebra, this makes OK the integral closure of Z in K, meaning in particular that OK is integrally closed. RING OF INTEGERS 4

Theorem 1.1.14. Let K be a number ﬁeld, and let α ∈ K. Then α can be written as α = β/d, where β ∈ OK and d ∈ Z. Hence K is the quotient ﬁeld of OK .

Proof. Since α ∈ K is algebraic over Q we have

n n−1 α + cn−1α + ... + c1α + c0 = 0 with ci ∈ Q. We can write all of the fractions ci with common denominator, say ci = bi/d, with bi, d ∈ Z, and d 6= 0. Hence b b b αn + n−1 αn−1 + ... + i α + 0 = 0. d d d Multiplying this by dn we get

n n−1 n−2 n−1 (dα) + bn−1(dα) + ... + b1d (dα) + b0d = 0, which we can interpret as a monic polynomial in dα, meaning that dα = β ∈ OK , and so α = β/d.

Lecture 2 Ring of Integers

2.1 Understanding Algebraic Integers

Recall that α is an algebraic integer, or integral over Z, if f(α) for some monic f(x) ∈ Z[x]. The question with which we will concern√ ourselves for the foreseeable future is this: what is OK if K is, say, Q, Q( D), or Q(ξm). The answers, respectively, are Z, depends on whether D ≡ 1 (mod 4), and Z[ξm]. We will spend some time proving these claims.

Theorem 2.1.1. Let α be an algebraic integer. Let f(x) ∈ Z[x] be a monic polynomial of least degree such that f(α) = 0. (Note that this implies that f is irreducible over Z.) Then f(x) is irreducible over Q. We will prove this in a moment, but first note the following straight-forward corollary: Corollary 2.1.2. The number α is an algebraic integer if and only if the monic minimal polynomial of α over Q has coefficients in Z. Proof. The latter implying the former is true by definition, and the former implying the latter is the above theorem. We will use the following result to prove the theorem:

Lemma 2.1.3 (Gauss Lemma). Let f(x) ∈ Z[x] be a monic polynomial. Sup- pose f(x) = g(x)h(x) with g(x), h(x) ∈ Q[x] monic. Then g(x), h(x) ∈ Z[x]. More generally, we can replace Z by any unique factorisation domain and Q by its ﬁeld of fractions. Date: January 11th, 2018. 2.1 Understanding Algebraic Integers 5

Proof. Let m and n be the smallest positive integers such that mg(x) and nh(x) have coefficients in Z. Hence the coefficients of mg(x) have no common factors, and the same is true for the coefficients of nh(x). We claim that m = n = 1. Suppose not, i.e. mn > 1. Take a prime p such that p | mn, and consider mnf(x) = (mg(x))(nh(x)). Reducing this modulo p we get 0 = mg(x)·nh(x) ∈ Z/pZ[x]. Now Z/pZ is a field, which in particular makes it an integral domain, so the polynomial ring is an integral domain too. Hence mg(x) = 0 or nh(x) = 0. Therefore p divides all coefficients of mg(x), or p divides all coefficients of nh(x), which is a contradiction since we chose m and n minimal in order to make sure the coefficients had no common factors. Using this we are equipped to prove the theorem:

Proof of 2.1.1. Suppose f(x) is reducible over Q, i.e. f(x) = g(x)h(x), where g(x), h(x) ∈ Q[x]. Since f(x) is monic and since Q is a ﬁeld, we can assume without loss of generality that g and h are monic too. Gauss Lemma therefore implies that g(x), h(x) ∈ Z[x]. Moreover deg g, deg h < deg f, but since 0 = f(α) = g(α)h(α) implies that g(α) = 0 or h(α) = 0, we have a contradiction, for f was chosen to be of minimal degree with α as a zero.

Corollary 2.1.4. The only algebraic integers in Q is Z. Proof. The minimal monic polynomial of α ∈ Q over Q is f(x) = x − α, naturally. Now α is an algebraic integer if and only if f(x) ∈ Z[x], if and only if α ∈ Z. We are now ready to ﬁnd the ring of integers for quadratic integers. √ Corollary 2.1.5. Consider K = Q( D), with D being square-free. Then ( √ √ [ D] = { a + b D | a, b ∈ }, if D 6≡ 1 (mod 4), Z √ √ Z OK = 1+ D 1+ D Z[ 2 ] = { a + b 2 | a, b ∈ Z }, if D ≡ 1 (mod 4). Proof. Let ( √ √ [ D] = { a + b D | a, b ∈ }, if D 6≡ 1 (mod 4), Z √ √ Z R = 1+ D 1+ D Z[ 2 ] = { a + b 2 | a, b ∈ Z }, if D ≡ 1 (mod 4). √ 2 First we√ claim that OK ⊃ R. The minimal√ polynomial of D is f(x) = x −D ∈ Z[x], so D ∈ OK , implying that Z[ D] ⊂ OK . (Regardless of whether D is congruent to 1 modulo 4 or not, as it happens). If D ≡ 1 (mod 4), then √ √ 1 + D 1 − D D − 1 f(x) = x − x − = x2 − x + ∈ [x] 2 2 4 Z √ √ 1− D since D−1 is a multiple of 4, implying that (1+ D)/2 ∈ OK , and so Z[ 2 ] ⊂ OK . √ Secondly we show that R ⊃ OK . Suppose α = s+t D ∈ OK , with s, t ∈ Q. 2.2 The Cyclotomic Fields 6

If t = 0, α = s ∈ Q ∩ OK = Z implying, by the previous theorem, that α ∈ Z ⊂ R. Hence assume t 6= 0. Then α ∈ OK if and only if the minimal polynomial of α over Q, namely √ √ 2 2 2 pα(x) = (x − (s + t D))(x − (s − t D)) = x + 2sx + s − t D is in Z[x]. This in turn is equivalent with 2s ∈ Z and s2 − t2D ∈ Z. Hence set s0 = 2s, or s = s0/2. We have two possibilities: s0 is either even or odd. 0 2 Suppose s is even. Then√s ∈ Z, so t D ∈ Z which, since D is square-free, means t ∈ Z. Thus α = s + t D ∈ R since s, t ∈ Z, so OK ⊂ R. Next suppose s0 is odd. Let n = s2 − t2D ∈ Z, which we rewrite as n = (s0/2)2 − t2D, or 4n = (s0)2 − (2t)2D meaning that (2t)2D ∈ Z which, again by D being square-free, means 2t ∈ Z. Set t0 = 2t ∈ Z. Then 4n = (s0)2 − (t0)2D. Reducing this modulo 4 we get 0 ≡ 1 − (t0)2D (mod 4) since s0 is odd, so (t0)2D ≡ 1 (mod 4). Now (t0)2 is either 0 or 1 modulo 4, but we can’t have it being 0 since the product is 1, so (t0)2 ≡ 1 (mod 4), and D ≡ 1 (mod 4). Hence t0 is odd, and we learned along the way that this second case, with s0 being odd, never happens if D 6≡ 1 (mod 4). Now D ≡ 1 (mod 4) means √ √ s0 t0 √ s0 − t0 1 + D α = s + t D = + D = +t0 ∈ R 2 2 2 2 | {z } ∈Z since s0 and t0 are both odd.

2.2 The Cyclotomic Fields

We learned last time that it suffices to consider K = Q(ξm) with m even, since Q(ξm) = Q(ξ2m) if m is odd. There are two things we aspire to show: first, that Q(ξm) are all distinct for even m, and secondly that OK = Z[ξm]. We’ll recall some basic properties we’ll need. Let ω = e2πi/m. 1.A (Galois) conjugate of ω is a root of a minimal polynomial of ω over Q. Note that this has coefficients in Z since ω is an algebraic integer. 2. Every conjugate of ω is a primitive mth root of unity, i.e. a root of xm − 1 but not a root of xn − 1 for n < m. 3. The conjugates are ωk for 1 ≤ k ≤ m, gcd(k, m) = 1.

Let ξm be a primitive mth root of unity and let K = Q(ξm). Then

4. deg(K/Q) = ϕ(m), 5. Gal(K/Q) =∼ (Z/mZ)×. 2.3 Embeddings in C 7

Proposition 2.2.1. If m is even, then the only roots of unity in Q(ξm) are n m mth roots of unity. In other words, θ ∈ Q(ξm) with θ = 1 implies θ = 1. In addition, if m is oﬀ, then the only roots of unity in Q(ξm) are (2m)th roots of unity.

Proof. The second result follows immediately from the ﬁrst since Q(ξm) = Q(ξ2m) for m odd. Hence assume m is even. Let θ ∈ Q(ξm) be a primitive kth root of unity. Then Q(ξm) contains a primitive rth root of unity where r = lcm(k, m), since 2πi/k 2πi/m if θ = e and ξm = e and gcd(k, m) = d = ak + bm, we have

b a bm+ak d b a 2πi( k + m ) 2πi mk 2πi mk 2πi/ lcm(m,k) 2πi/r ξr = θ ξm = e = e = e = e = e .

Now ϕ(r) | ϕ(m). However m | r, meaning that m = r so k | m.

Corollary 2.2.2. Let m be even. Then the mth cyclotomic ﬁelds Q(ξm) are all distinct.

In order to prove OK = Z[ξm], we ﬁrst need to discuss embeddings in C.

2.3 Embeddings in C Let K = Q(α) be a number ﬁeld of degree n, meaning that α has n conjugates (since its minimal polynomial is of degree n). Each conjugate of β determines a unique embedding of K in C, namely σ : K → C, α 7→ β. Hence there are exactly n embeddings of K in C. √ √ Example 2.3.1. Let√ K = Q( D√) with D square-free. Then D has two conjugates,√ namely√ D itself and − D. So we have two embeddings,√ σ1 : K√ → C by D 7→ D, namely the identity map, and σ2 : K 7→ C by D 7→ − D. Since both of the embeddings are contained in K, this is a Galois extension. N √ √ 3 3 Example 2.3.2. Let K = Q( 2). Since the minimal polynomial of 2 over Q 3 is x − 2, there are three conjugates,√ √ and we√ can see them all by factoring√ this 3 3 3 2 polynomial over C. We get 2, 2ω, and 2ω , where√ ω =√ (−1√ + −√3)/2. 3 3 3 3 Hence√ we have√ three embeddings, σ1, σ2, σ3 : K → C by 2 7→ 2, 2 7→ 2ω, 3 3 2 and 2 7→ 2ω respectively. √ √ 3 3 2 This extension is not Galois since 2ω and 2ω are not in K. N More generally, if K m L n

Q meaning that K and L are number ﬁelds, then every embedding σ : L → C can be extended to m embeddings of K in C. In particular K has m embeddings in C that ﬁx L pointwise. TRACES, NORMS, AND DISCRIMINANTS 8

Lecture 3 Traces, Norms, and Discriminants

3.1 Traces and Norms

Let K be a number ﬁeld of degree n, and let σ1, σ2, . . . , σn be the embeddings of K in C.

Deﬁnition 3.1.1. If σi(K) ⊂ R, then σi is called a real embedding. Otherwise σi is called a complex embedding. Fact. The complex embeddings come in pairs. Since σ(x) := σ(x) is also a complex embedding of K in C.

Letting r1 denote the number of real embeddings and r2 denote the number of conjugate pairs of complex embeddings, we of course have n = r1 + 2r2. Deﬁnition 3.1.2 (Trace and norm). For any α ∈ K, we deﬁne the trace of α to be

TrK/Q(α) = σ1(α) + σ2(α) + ... + σn(α) and the norm of α to be

NK/Q(α) = σ1(α)σ2(α) · ... · σn(α). When it is obvious from context what ﬁeld K we are working over we will omit the subscript K/Q.

Exercise 3.1.3. Consider the Q-linear map Lα : K → K deﬁned by Lα(x) = αx. Then TrK/Q(α) = tr(Lα) and NK/Q(α) = det(Lα). The trace and norm behave largely as expected:

1. Tr(α + β) = Tr(α) + Tr(β),

2. If r ∈ Q, then Tr(rα) = r Tr(α) and Tr(r) = nr, 3.N( αβ) = N(α) N(β), and

4. If r ∈ Q, then N(rα) = rn N(α) and N(r) = rn. Theorem 3.1.4. Let K be a number ﬁeld of degree n and let α ∈ K. Then n Tr (α) = Tr (α) K/Q d Q(α)/Q and n/d NK/Q(α) = NQ(α)/Q(α) where d = [Q(α): Q], meaning that d | n.

Proof. There exist d embeddings of Q(α) in C, say σ1, σ2, . . . , σd. Each σi extends to exactly n/d embeddings of K in C, say σi,1, σi,2, . . . , σi,n/d. Then

{ σi,j } 1≤i≤d 1≤j≤n/d

Date: January 16th, 2018. 3.1 Traces and Norms 9 gives all n embeddings of K in C. Hence X X n n Tr (α) = σ (α) = σ (α) = Tr (α) K/Q i,j d i d Q(α)/Q i,j i since for α ∈ K we have σi,j(α) = σi(α) for all j. Similarly

n/d Y Y n/d NK/Q(α) = σi,j(α) = (σi(α)) = NQ(α)/Q(α) . i,j i

Corollary 3.1.5. For α ∈ K we have

TrK/Q(α), NK/Q(α) ∈ Q and if α ∈ OK we have

TrK/Q(α), NK/Q(α) ∈ Z.

d d−1 Proof. Let pα(x) = x + ad−1x + ... + a1x + a0 be the monic minimal polynomial of α over Q. Then we know moreover that

pα(x) = (x − σ1(α))(x − σ2(α)) · ... · (x − σd(α)) since σi(α) are the Galois conjugates of α. Hence, multiplying out,

TrQ(α)/Q(α) = −ad−1 ∈ Q and

NQ(α)/Q(α) = ±a0 ∈ Q (with the plus or minus depending on the degree d). By Theorem 3.1.4 we therefore moreover have TrK/Q(α) and NK/Q(α) in Q, since they are rationals times integers. If α ∈ OK , then ai ∈ Z by deﬁnition, whence the above become the products of integers, hence integers. Note that

TrK/Q : K → (Q, +) is an additive homomorphism and that

∗ NK/Q : K → (Q, ·) is a multiplicative homomorphism, whereby K∗ we mean K \{ 0 }. √ √ Example 3.1.6. Let K = Q( D) with D square-free, and let α = a + b D with a, b ∈ Q. Then √ √ Tr(α) = (a + b D) + (a − b D) = 2a and √ √ 2 2 N(α) = (a + b D)(a − b D) = a − b D. N 3.2 Relative Trace and Norm 10

These constructs are not just idle curiosities. Indeed, computing norms can oﬀer insight into whether something is a unit. −1 Let α ∈ OK be a unit. Then N(α) = ±1. Seeing this is easy: α, α ∈ OK implies N(αα−1) = N(1) = 1, meaning that N(α) N(α−1) = 1, so N(α) | 1. Another observation: let α ∈ OOK so that N(α) ∈ Z is a prime. Then α is irreducible, since α = β1β2 is equivalent with N(α) = N(β1) N(β2), so if the left-hand side is a prime, one of the elements in the right-hand side must be a unit. √ Example 3.1.7. This tells us that, for instance, Z[ −2]× = { 1, −1 } since 2 2 N(α) = a + 2b = 1, forcing b =√ 0, and hence a = ±1.√ √ It also tells us that, say, 9 + 10 is irreducible in Z[ 10] since N(9 + 10) = 71, which is prime. N

3.2 Relative Trace and Norm

We don’t necessarily have to consider traces and norms with respect to Q. Definition 3.2.1 (Relative trace and norm). Let K and L be two number fields such that K ⊂ L and [L : K] = n. Let σ1, σ2, . . . , σn be the n embeddings of L in C that fix K. For all α ∈ L we define the relative trace

TrL/K (α) = σ1(α) + σ2(α) + ... + σn(α) and relative norm by

NL/K (α)σ1(α)σ2(α) · ... · σn(α).

As before, these are homomorphisms, and if α ∈ OL, then TrL/K (α) ∈ OK . Theorem 3.2.2 (Transitivity). Let K, L, and M be number ﬁelds with K ⊂ L ⊂ M. Then for α ∈ M we have

TrL/K (TrM/L(α)) = TrM/K (α) along with NL/K (NM/L(α)) = NM/K (α).

Proof. Let [M : L] = m and [L : K] = n, and let τ1, τ2, . . . , τm be the embeddings of M in C which ﬁx L, and σ1, σ2, . . . , σn be the embeddings of L in C which ﬁx K. Take N to be a normal extension of Q such that N ⊃ M. The extension being normal means that every embedding can be extended to it, whereby we letτ ¯i be the extension of τi to N andσ ¯j the extension of σj to N. In other words they are automorphisms from N to N. Then we have X X X X X TrL/K (TrM/K (α)) = TrL/K τi(α) = σj(τi(α)) = σ¯j(¯τi(α)) i j i j i

(where we brought the inner sum outside since these are homomorphisms). We now need to show that this is the same as TrM/K (α). 3.3 Discriminants 11

Now we know by degree considerations that there are nm embeddings of M in C that ﬁx K, so if we can show that

{ σ¯jτ¯i }1≤i≤m 1≤j≤n are all distinct, then they must be the mn embeddings of M in C that ﬁx K.

Suppose therefore thatσ ¯j1 τ¯i1 =σ ¯j2 τ¯i2 on M. For x ∈ L,

σ¯j1 (¯τi1 (x)) =σ ¯j1 (τi1 (x)) =σ ¯j1 (x) and

σ¯j2 (¯τi2 (x)) =σ ¯j2 (τi2 (x)) =σ ¯j2 (x) since in L,τ ¯i(x) = τi(x) = x. Moreover these two remain equal, soσ ¯j1 (x) =

σ¯j2 (x) for all x ∈ L. By the same argument again therefore this means σj1 (x) =

σj2 (x), so j1 = j2, whenceτ ¯i1 =τ ¯i2 on M, so i1 = i2. The argument for the norm is almost identical, but with products in place of sums.

3.3 Discriminants Deﬁnition 3.3.1 (Discriminant). Let K be a number ﬁeld of degree n and let σ1, σ2, . . . , σn be the n embeddings of K in C. For any n elements

α1, α2, . . . , αn ∈ K, we define the discriminant of α1, α2, . . . , αn by 2 disc(α1, α2, . . . , αn) = det[σi(αj)]1≤i≤n . 1≤j≤n Note that because of the square the discriminant is independent of the order of embeddings or elements, since permuting these simply changes the sign, but squaring nullifies this. Theorem 3.3.2. With the setup as in the definition above, we have

disc(α1, α2, . . . , αn) = det([TrK/Q(αiαj)]). Proof. Note that

det([TrK/Q(αiαj)]) = [σj(αi)][σi(αj)] where the two matrices in the right-hand side are each others transposes. Taking determinants now we get the expression sought.

Corollary 3.3.3. Again with the setup as above, we have disc(α1, α2, . . . , αn) ∈ Q and if αi ∈ OK , then disc(α1, α2, . . . , αn) ∈ Z. THE ADDITIVE STRUCTURE OF THE RING OF INTEGERS 12

Lecture 4 The Additive Structure of the Ring of Integers

4.1 More on Discriminants

Theorem 4.1.1. With the same setup as above, disc(α1, α2, . . . , αn) = 0 if and only if α1, α2, . . . , αn are linearly dependent over Q.

Proof. We start by assuming α1, α2, . . . , αn are linearly dependent over Q. This means that [σi(αj)] has linearly dependent column vectors over Q, whence its determinant is 0, and so the discriminant is 0 too. For the converse, assume disc(α1, α2, . . . , αn) = 0. This means that the matrix [Tr(αiαj)] has linearly dependent row vectors. Let Ri be the ith row vector of this matrix. Then

a1R1 + a2R2 + ... + anRn = 0 (4.1.1) for some ai ∈ Q with not all ai = 0. Let

α = a1α1 + a2α2 + ... + anαn.

The jth coordinate of (4.1.1) is

a1 Tr(α1αj) + a2 Tr(α2αj) + ... + an Tr(αnαj) = 0 which by linearity of trace we can rewrite as

Tr((a1α1 + a2α2 + ... + anαn)αj) = 0 and so in all Tr(ααj) = 0 for all j. Now suppose α1, α2, . . . , αn are linearly independent over Q, meaning that they form a basis for K over Q and α 6= 0 implies αα1, αα2, . . . , ααn is also a basis for K over Q. But the last equation above then means that Tr is 0 on a basis, whence Tr(x) = 0 for all x ∈ K. This is a contradiction, since we know Tr(1) = n 6= 0.

We get a special case of the discriminant if we select α1, α2, . . . , αn to be powers of α:

Proposition 4.1.2. Let K = Q(α) with [K : Q] = n. Then 1, α, α2, . . . , αn−1 is a basis for K over Q, and

2 n−1 Y 2 disc(α) := disc(1, α, α , . . . , α ) = (αi − αj) 1≤i

Proof. We have j 2 j 2 disc(α) = det([σi(α )]) = det([αi ]) which is a Vandermode matrix, so this is equal to

2 Y 2 Y n(n−1)/2 Y Y (αi − αj) = (αi − αj) = (−1) (αi − αj) 1≤i

2πi/p Example 4.1.3. Let K = Q(ξp) with ξp = e and p a prime. Say p ≥ 3, since otherwise K = Q. Then [K : Q] = ϕ(p) = p − 1, whence we consider

p−2 disc(ξp) = disc(1, ξp, . . . , ξp ).

The monic minimal polynomial of ξp over Q is f(x) = xp−1 + xp−2 + ... + x + 1

0 0 and we would like to compute f (ξp) and in particular NK/Q(f (ξp)). Rather than doing this directly, note that xp − 1 = (x − 1)f(x), so by the product rule

pxp−1 = f(x) + (x − 1)f 0(x) whereby if we plug in x = ξp we have

p−1 0 pξp = (ξ − 1)f (ξp) which we rearrange to p−1 0 pξp f (ξp) = . ξp − 1 Taking norms and recalling that it is a multiplicative homomorphism, we get

p−1 0 N(p) N(ξp) N(f (ξp)) = . N(ξp − 1)

We evaluate the factors of the right-hand side one at a time. First, N(p) = pp−1. Next N(ξp) by deﬁnition is the product of then conjugates of ξp, so

2 p−1 p(p−1)/2 p (p−1)/2 N(ξp) = ξpξp · ... · ξp = ξp = (ξp ) = 1.

Finally

p−1 p−2 2 p−1 f(x) = x + x + ... + x + 1 = (x − ξp)(x − ξp) · ... · (x − ξp ) whereby

2 p−1 f(1) = p = (1 − ξp)(1 − ξp) · ... · (1 − ξp ) = N(1 − ξp) = N(ξp − 1). 4.2 The Additive Structure of the Ring of Integers 14

Hence pp−1 · 1 N(f 0(ξ )) = = pp−2, p p whereby p(p−1)/2 p−2 disc(ξp) = (−1) p , p−2 meaning that disc(ξp) | p . N Something more general is true: 2πi/m ϕ(m) Example 4.1.4. Let K = Q(ξm), with ξm = e . Then disc(ξm) | m . m Let f(x) be the monic minimal polynomial of ξm over Q again. Then x − 1 = f(x)g(x), wherein f(x), g(x) ∈ Z[x] by Gauss lemma. Therefore again by the product rule mxm−1 = f 0(x)g(x) + f(x)g0(x) m−1 0 whence mξm = f (ξm)g(ξm), since ξm is a root of f(x) itself. Multiplying both sides by ξm we get 0 m = ξmf (ξm)g(ξm) which if we take norms becomes

ϕ(m) 0 m = N(f (ξm)) N(g(ξm)ξm) wherein g(ξm)ξm ∈ OK so the second norm is in Z. The same is true for the ϕ(m) ﬁrst one, in both cases since ξm is an algebraic integer. Hence disc(ξm) | m , since note that the ﬁrst norm is disc(ξm) save for maybe a sign. N

4.2 The Additive Structure of the Ring of Integers As per usual, let K be a number ﬁeld of degree n. The goal now is to show that OK is a free abelian group of rank n (or in other words a free Z-module of rank n). First recall some facts about such matters: Fact. A free abelian group of ﬁnite rank n is isomorphic to

n Z ⊕ Z ⊕ ... ⊕ Z = Z . Fact. Every subgroup of a free abelian group of rank n is also a free abelian group of rank m ≤ n. For an example that doesn’t decrease the rank, consider for instance replacing one of the Z’s above with 2Z.

Corollary 4.2.1. Let G1 ⊂ G2 ⊂ G3 be groups. If G1 and G3 are free abelian groups of rank n, then so is G2.

Hence our strategy is to show that OK is between two free abelian groups of rank n.

Remark 4.2.2. We know that K has a basis α1, α2, . . . , αn over Q. In fact, we can arrange for this basis to be elements from OK , since any αi = βi/mi with βi ∈ OK and mi ∈ Z, so scaling to get rid of the denominator yields a basis in OK . Hence K = Q(α), and we can assume α ∈ OK , and the basis is 1, α, . . . , αn−1. 4.2 The Additive Structure of the Ring of Integers 15

Notice now that K ⊃ OK )Z[α]. The latter is a free abelian group of rank n though, so half of our inclusion is trivially true. Letting α1, α2, . . . , αn be a basis contained in OK , consider

G = Zα1 ⊕ Zα2 ⊕ ... ⊕ Zαn = { m1α1 + m2α2 + ... + mnαn | mi ∈ Q }.

This is a free abelian group of rank n on α1, α2, . . . , αn. We claim, and soon prove, that 1 α α α G ⊂ O ⊂ G = 1 ⊕ 2 ⊕ ... ⊕ n , K d Z d Z d Z d where d = disc(α1, α2, . . . , αn).

1 Theorem 4.2.3. Let d = disc(α1, α2, . . . , αn). Then OK ⊂ d G. In fact, every α ∈ OK can be written as m α + m α + ... + m α α = 1 1 2 2 n n d

2 with mi ∈ Z and d | mi .

Before we prove this, note two things: d 6= 0 since αi is a basis, and d ∈ Z since αi ∈ OK . Note also that this being true, we have the immediate corollary we want:

Corollary 4.2.4. Let K be a number ﬁeld of degree n. Then OK is a free abelian group of rank n.

Proof of theorem. Let α ∈ OK , whereby α = x1α1 + x2α2 + ... + xnαn for some xi ∈ Q with αi forming a basis. Moreover let σ1, σ2, . . . , σn be the embeddings of K in C. Then we have the following system of equations  σ1(α1)x1 + σ1(α2)x2 + ... + σ1(αn)xn = σ1(α)  σ2(α1)x1 + σ2(α2)x2 + ... + σ2(αn)xn = σ2(α) .  .  σn(α1)x1 + σn(α2)x2 + ... + σn(αn)xn = σn(α) which we rewrite as     x1 σ1(α) x2  σ2(α) σ (α )   =   . i j  .   .   .   .  xn σn(α) 2 If we solve for xi using Cramer’s rule, letting δ = det([σi(αj)]), whence δ = d ∈ Z, we have xi = γi/δ, where σi is the determinant of the original matrix [σi(αj)] except with the ith column replaced by the right-hand side of the system. 2 Hence dxi = δ γi/δ = δγi. INTEGRAL BASES 16

Lecture 5 Integral Bases

5.1 Integral Bases We start by ﬁnishing the proof from last time.

Proof continued. We have dxi = δγi, where since d ∈ Z and xi ∈ Q the left- hand side is in Q, and the right-hand side is an algebraic integer if α ∈ OK , so indeed it is an integer. Then dxi = mi ∈ Z, and we write xi = mi/d, whence m α + m α + ... + m α α = 1 1 2 2 n n . d Next we have m2 d2x2 γ2 i = i = dx2 = δ2 i = γ2, d d i δ2 i 2 which is an algebraic integer in Q, so it’s an integer. Hence d | mi . Deﬁnition 5.1.1 (Integral basis). Let K be a number ﬁeld of dimension n. Since OK has a basis over Z (since it’s a free abelian group) there exists β1, β2, . . . , βn ∈ OK such that every α ∈ OK can be written as

α = m1β1 + m2β2 + ... + mnβn, with mi ∈ Z. Such a basis { β1, β2, . . . , βn } is called an integral basis for OK (or K). Note that if we take mi ∈ Q, the above will span K. √ Example√ 5.1.2. Let K = Q( D),√ with D being square-free. Then OK is Z[ D] if D 6≡ 1 (mod 4) and Z[(1 + D)√/2] if D ≡ 1 (mod 4). √ Hence OK has the integral basis { 1, D } in the former case and { 1, (1 + D)/2 } in the latter. N 2πi/pr r Theorem 5.1.3. Let K = Q(ξpr ), with ξpr = e and p ≥ 3 for p prime. Then OK = Z[ξpr ]. We will need two lemmas to prove this.

Lemma 5.1.4. With ξ = ξpr as above, Z[1−ξ] = Z[ξ], and disc(1−ξ) = disc(ξ). Proof. The ﬁrst statement is clear: Z[ξ] ⊂ Z[1 − ξ] since 1 − ξ ∈ Z[ξ], and since 1 − (1 − ξ) = ξ the opposite inclusion is true as well. For the second statement we have

Y 2 Y 2 disc(ξ) = (αi − αj) = ((1 − αi) − (1 − αj)) = disc(1 − ξ) 1≤i

Lemma 5.1.5. With the same setup, Y (1 − ξk) = p 1≤k≤pr p-k meaning that N(1 − ξ) = p. Proof. The monic minimal polynomial of ξ over Q is r Y xp − 1 tp − 1 f(x) = (x − ξk) = = = tp−1 + tp−2 + ... + t + 1 xpr−1 − 1 t − 1 1≤k≤pr p-k

r−1 with t = xp . Hence taking x = 1 gives t = 1 i.e. Y f(1) = p = (1 − ξk). 1≤k≤pr p-k We are now ready to prove the theorem:

2 n−1 Proof of theorem. Taking, again, ξ = ξpr , we have { 1, ξ, ξ , . . . , ξ } ⊂ OK with n = ϕ(pr) is a basis for K over Q. So therefore is 2 n−1 { 1, (1 − ξ), (1 − ξ) ,..., (1 − ξ) } ⊂ OK .

Hence we can write α ∈ OK as m + m (1 − ξ) + ... + m (1 − ξ)n−1 α = 1 2 n , d where d = disc(1 − ξ) = disc(ξ) by the first lemma. We now claim that OK = Z[1 − ξ], which by the first lemma is the same as Z[ξ]. Suppose OK is strictly larger than Z[1 − ξ], i.e. there exists some α ∈ OK r ϕ(pr ) r such that d - mi for some i. Note that d | (p ) , so d = p for some r. There must exist some element in OK of the form t (1 − ξ)i−1 + ... + t (1 − ξ)n−1 β = i n p with t - ti. r Hint. Let mj1 , ··· , mjk be all mis that are not divisible by d = p . Write rjs mjs = p tjs with (tjs , p) = 1 and rjs < r. Take i to be the smallest index such r−ri−1 that ri = min{ rjs }. Consider p α subtracting its first (i − 1) terms. r Note that 1 − ξ | 1 − ξk for all k, so (1 − ξ)ϕ(p ) = (1 − ξ)n | p, meaning that p/(1 − ξ)i ∈ Z[ξ] for 1 ≤ i ≤ n. Therefore p ti β = + t + t (1 − ξ) + ... + t (1 − ξ)n−i−1, (1 − ξ)i 1 − ξ i+1 i+2 n where the sum of the latter terms are in Z[1 − ξ] ⊂ OK . This implies that ti/(1 − ξ) ∈ OK , whence t N i ∈ K/Q 1 − ξ Z n n so ti / N(1 − ξ) = ti /p ∈ Z, meaning that p | ti, which is a contradiction. Hence OK = Z[1 − ξ] = Z[ξ]. 5.2 Composite Field 18

Theorem 5.1.6. Let { α1, α2, . . . , αn } and { β1, β2, . . . , βn } be two integral bases for OK . Then disc(α1, α2, . . . , αn) = disc(β1, β2, . . . , βn).

Proof. Express the αi in terms of βj’s, say     α1 β1 α2  β2    = M    .   .   .   .  αn βn where M is an n × n matrix with elements in Z. Hence 2 2 2 disc(α1, α2, . . . , αn) = det([σi(αj)]) = det(M[σi(βj)]) = det(M) disc(β1, β2, . . . , βn) whence disc(β1, β2, . . . , βn) | disc(α1, α2, . . . , αn) and by reversing the argument we have the opposite divisibility as well. Now since they both have the same sign, since det(M)2 > 0, they must be equal.

Definition 5.1.7. Let K be a number field, and let { α1, α2, . . . , αn } be an integral basis for OK . Define the discriminant of K (or OK ) by

disc(K) = disc(OK ) = disc(α1, α2, . . . , αn) which by the above theorem is independent of the choice of integral basis.

Exercise 5.1.8. Consider { α1, α2, . . . , αn } ⊂ OK . Then { α1, α2, . . . , αn } is an integral basis for OK if and only of disc(α1, α2, . . . , αn) = disc(K).

5.2 Composite Field Definition 5.2.1 (Composite Field). Let K be a number field of degree m and let L be a number field of degree n. Then

n r o X KL = αiβi αi ∈ K, βi ∈ L i=1 is a subﬁeld of C called the composition of K and L. We have the following inclusions:

K L

K ∩ L

Q A few facts follow immediately from this deﬁnition: COMPOSITION FIELDS 19

• [KL : K] ≤ [L : L ∩ K],

• [KL : Q] ≤ [K : Q] · [L : Q], • Deﬁne n r o X OK ·OL = αiβi αi ∈ OK , βi ∈ OL i=1

then OK ·OL ⊂ OKL.

Lecture 6 Composition Fields

6.1 Cyclotomic Fields again We are soon ready to prove what has been our goal for quite some time now, namely that if K = Q(ξm), then OK = Z[ξm]. We know that this is true for m = pr, so what about general m? The strategy is to prime factorise m, because r s if, say, m = p q with p and q distinct primes, and L = Q(ξpr ) and M = Q(ξqs ), we have K = Q(ξm)

L = Q(ξpr ) M = Q(ξqs )

Q and we will show that K = LM and moreover OK = Z[ξpr ]Z[ξqs ] = Z[ξm]. Proposition 6.1.1. Let K and L be two number ﬁelds with [K : Q] = m and [L : Q] = n. Assume [KL : Q] = mn, i.e. K ∩ L = Q, and let d = gcd(disc(K), disc(L)). Then 1 O ⊂ O ·O . KL d K L An immediate result that follows is

Corollary 6.1.2. Under the same assumptions, if d = 1, we have OK ·OL = n m OKL and disc(KL) = disc(K) disc(L) . To prove the proposition, we ﬁrst establish the following lemma:

Lemma 6.1.3. Assume [KL : Q] = mn. Let σ be an embedding of K in C and let τ be an embedding of L in C. Then there exists an embedding of KL in C which restricted to K is σ and restricted to L is τ.

Proof. We know that σ has n distinct extensions to embeddings of KL in C, sayσ ˜1,..., σ˜n, and moreoverσ ˜i|L 6=σ ˜j|L for i 6= j since otherwiseσ ˜i =σ ˜j on KL, but they’re all distinct. Henceσ ˜i|L give n distinct embeddings of L in C, but there are only n distinct embeddings of L in C since [L : Q] = n, so we must haveσ ˜i|L = τ for some i. Date: January 25th, 2018. 6.1 Cyclotomic Fields again 20

Proof of proposition. Let { α1, α2, . . . , αm } be an integral basis for OK and let { β1, β2, . . . , βn } be an integral basis for OL. Then { αiβj } for 1 ≤ i ≤ m and 1 ≤ j ≤ n is a basis for KL over Q. It is also a basis for OK ·OL over Z, and { αiβj } ⊂ OKL. By Theorem 4.2.3, for γ ∈ OKL we can write

X aij γ = α β M i j i,j where M, aij ∈ Z and we have reduced in such a way that

gcd(M, a11, . . . , amn) = 1, and we know that M | disc(αiβj). 1 Our goal is to show that M | d since then OKL ⊂ d OK OL. Let σ1, σ2, . . . , σn be the embeddings of K in C. For each σi there is an extensionσ ˜i to KL such thatσ ˜i|L = IdL by the lemma. Apply this to γ above, whence X aij X σ˜ (γ) = σ (α )β = σ (α )x k M k i j k i i i,j j where X aij x = β . i M j j Setting this up as a matrix equation we have     x1 σ˜1(γ)  x2   σ˜2(γ)  σ (α )   =   k i  .   .   .   .  xm σ˜m(γ)

2 which we solve using Cramer’s rule by taking δ = det([σk(αi)]), so δ = disc(K) and letting γi be the determinant of [σk(αi)] with the ith column replaced by the right-hand side above. Then xi = γi/δ. Note that δ, γi ∈ A, the algebraic integrals in C, and so γ disc(K)x = δ2 i = δx ∈ A, i δ i but the left-hand side is

X disc Kaij β ∈ L ∩ A = O M j L j since βj form an integral basis for OL. Hence disc(K)a ij ∈ , M Z so M | disc(K) since by construction M and aij are all coprime. By the exact same argument but replacing K with L we have M | disc(L), so M | gcd(disc(K), disc(L)). If d = 1, then M = ±1, so { αiβj } is an integral basis for OKL, and then by computation n m disc(KL) = disc(αiβj) = disc(K) disc(L) . 6.2 Prime Decomposition in Rings of Integers 21

2πi/m Corollary 6.1.4. Let K = Q(ξm) with ξm = e . Then OK = Z[ξm].

r1 r2 Proof. It suﬃces to consider m = p1 p2 with p1 and p2 distinct primes. Let r1 r2 m1 = p1 and m2 = p2 , and let K1 = Q(ξm1 ) and K2 = Q(ξm2 ). Then m1 OK1 = Z[ξm1 ] and OK2 = Z[ξm2 ]. Note that gcd(m1, m2) = 1 and ξm = ξm2 and ξm2 = ξ , and so ξ = ξs ξt where sm + tm = 1, s, t ∈ . m m1 m m1 m2 1 2 Z ϕ(mi) Since disc(Ki) | mi , we have gcd(disc(K1), disc(K2)) = 1, and hence by the corollary above OK = OK1 ·OK2 = Z[ξm1 ]Z[ξm2 ] = Z[ξm].

Remark 6.1.5. With K = Q(α) ⊂ OK , we do not in general have OK = Z[α].

6.2 Prime Decomposition in Rings of Integers Before proceeding we need some prerequisites from commutative algebra. Definition 6.2.1 (Dedekind domain). A Dedekind domain D is a Noethe- rian, integrally closed domain of dimension 1. Definition 6.2.2 (Noetherian ring). A Noetherian ring R is a ring that satisfies the ascending chain condition, meaning that a chain of ideals

I1 ⊂ I2 ⊂ I3 ⊂ ... is stationary, i.e. eventually Ik = Ik+1 = .... This is in turn equivalent with every ideal I ⊂ R being finitely generated as an R-module. Definition 6.2.3 (Integrally closed). Let D be an integral domain and let K be its quotient field. Then D is integrally closed if α ∈ K and f(α) = 0 for some monic f(x) ∈ D[x] implies α ∈ D. Definition 6.2.4 (Dimension 1). A Noetherian domain R is of dimension 1 if every nonzero prime ideal is maximal.

Theorem 6.2.5. The ring OK is a Dedekind domain. ∼ n Proof. To see that OK is Noetherian, take I ⊂ OK = Z , with this isomorphism being as Z-modules. Then I is finitely generated as a Z-module, and since Z ⊂ OK , it is also finitely generated as an OK -module. By definition of OK , it is the integral closure of Z in K and hence it is integrally closed. Finally to see that OK is of dimension 1, take P ⊂ OK a nonzero prime ideal. We want to show that P is maximal, which is equivalent to OK /P being a field. Since P is prime by assumption, we know first of all that OK /P is an integral domain. It suffices to show that OK /P is finite since every finite integral domain is a field. Let α 6= 0 in P and let Z 3 m = NK/Q(α) = αβ, where by β we mean the product of the remaining Galois conjugates of α. Then m β = ∈ K ∩ A = O α K whence m = αβ ∈ P , since α ∈ P and β ∈ OK . Hence (m) ⊂ P ⊂ OK , and so O O n K ≤ K = Z < ∞ P (m) nZn by the aforementioned isomorphism. IDEAL FACTORISATION 22

Deﬁnition 6.2.6. Let D be a Dedekind domain and K its ﬁeld of fractions. Let I,J ⊂ D be ideals. We say that I divides J, denoted I | J if I ⊃ J. We also call gcd(I,J) = I + J, lcm(I,J) = I ∩ J, and n o X IJ = aibi ai ∈ I, bi ∈ J i and then I ∩ J = IJ if gcd(I,J) = 1 = (1) = D, i.e. I and J are coprime.

Lecture 7 Ideal Factorisation

7.1 Unique Factorisation of Ideals In the following discussion, D is a Dedekind domain and K is its ﬁeld of fractions. Lemma 7.1.1. For every ideal I ⊂ D, I 6= 0, there exists nonzero prime ideals P1,P2,...,Pr such that P1 · P2 · ... · Pr ⊂ I,

P1,P2,...,Pr not necessarily distinct. Note how this corresponds to how every integer has at least one prime factor. Proof. Let S be the set of proper ideals of D for which the assumption in the lemma is false. Assume S 6= ∅—we clearly wish for this to lead to a contradiction. Since D is Noetherian, S has a maximal element, say M. Then M is not a prime ideal, for otherwise it is its own factor P1 above. Hence there exists some b1, b2 ∈ D \ M such that b1b2 ∈ M. Now consider I1 = (b1) + M ) M and I2 = (b2) + M ) M. Since M is maximal in S, we must have I1,I2 6∈ S. Therefore they do contain products of prime ideals, so there exist P1 · P2 · ... · Pr ⊂ I1 and Q1 · Q2 · ... · Qs ⊂ I2, where Pi and Qj are prime ideals. Then P1P2 ··· PrQ1Q2 ··· Qs ⊂ I1I2 ⊂ M, since b1b2 ∈ M, which is a contradiciton. Hence S = ∅. Lemma 7.1.2. Let P be a nonzero prime ideal in D and deﬁne

P −1 = { x ∈ K | xP ⊂ D }.

Then IP −1 6= I for every nonzero ideal I ⊂ D. Note that n r o −1 X −1 IP = aixi ai ∈ I, xi ∈ P i=1 so, for example, 1 P −1 = (p )−1 = Z pZ in Z. Date: January 30th, 2018. 7.1 Unique Factorisation of Ideals 23

Proof. We claim ﬁrst that P −1 6⊂ D. Let 0 6= a ∈ P with (a) ( P . Take

P1P2 ··· Pr ⊂ (a) ( P with r as small as possible, which is of course possible by the above lemma. Note r ≥ 2. Then if Pi 6⊂ P for every i = 1, 2, . . . , r, then there exist αi ∈ Pi, hence αi ∈ P for every i = 1, 2, . . . , r, such that

α1α2 ··· αr 6∈ P.

But since P is a prime ideal, this is a contradiction. Hence Pi ⊂ P for some i. But in a Dedekind domain every nonzero prime ideal is maximal, so Pi = P . Without loss of generality, let us assume P1 = P (if not, just rearrange and relabel). Now P2P3 ··· Pr 6⊂ (a) since r was chosen as small as possible. Take b ∈ P2P3 ··· Pr, b 6∈ (a). Then b P ⊂ D, a which means by deﬁnition that b/a ∈ P −1, but b/a 6∈ D since b 6∈ (a). Now let I 6= 0, I ⊂ D, and let α1, α2, . . . , αn be the generators of I as a D-module (which must exist since D is Noetherian). Suppose IP −1 = I. For any x ∈ P −1, we have

n X xαi = aijαj j=1 with aij ∈ D. Let A = [aij] and so     α1 α1 α2  α2  A   = x    .   .   .   .  αn αn meaning that   α1 α2  (xI − A)   = 0  .   .  αn and so det(xI − A) = 0, and expanding this determinant we get some monic polynomial xn + ... ∈ D[x], whence x is integral over D. Since D is Dedekind it is also integrally closed, meaning that x ∈ D, whereby P −1 ⊂ D. That’s a contradiction, which means that IP −1 6= I.

Theorem 7.1.3. Every nonzero proper ideal I ( D had a unique factorisation

e1 e2 er I = P1 P2 ··· Pr with ei > 0 and Pi distinct primes. 7.1 Unique Factorisation of Ideals 24

Proof. We start with the existence proof. Let S be the set of nonzero proper ideals in D such that they do not have a prime factorisation. Assume, as before, that S 6= ∅. Then since D is Noetherian there must be a maximal element in S, call it M. Then we must have M ⊂ P ⊂ D for some maximal (and prime) ideal P . By deﬁnition, D ⊂ P −1, and by the lemma

−1 −1 M ( MP ⊂ PP = D. The last equality comes from P ⊂ PP −1 ⊂ D, and since P is maximal we must have P = PP −1 or PP −1 = D, and by the lemma it can’t be the former. Since M ( P , and MP −1 6= D, for otherwise MP −1 = D which implies P = DP = MP −1P = M, which is a contradiction. Thus M ( MP −1 ( D, and since M is maximal in S this means that MP −1 6∈ S, so it by assumption has a unique prime factorisation, say

−1 e1 e2 er MP = P1 P2 ··· Pr , where Pi are prime ideals and ei > 0. But if we just multiply by P we get

e1 e2 er M = PP1 P2 ··· Pr which is a contradiction since M ∈ S, and so therefore S = ∅. Now let us consider uniqueness. Suppose

e1 e2 er a1 a2 a2 I = P1 P2 ··· Pr = Q1 Q2 ··· Qs . This means that a1 a2 a2 Q1 Q2 ··· Qs ⊂ P1 implying that Qi = Pi for some i; let’s say Q1 = P1. Then

−1 e1−1 e2 er a1−1 a2 a2 P1 I = P1 P2 ··· Pr = Q1 Q2 ··· Qs .

Repeat this, yielding eventually Pi = Qi, ei = ai, and r = s. Note that the uniqueness proof is essentially the same as the standard proof of the same for integers. Note also that, as with integers, it is often very hard to actually factor ideals, even though we know we can. Definition 7.1.4 (Fractional ideal). Let D be a Dedekind domain and K its field of fractions. A fractional ideal of K is a finitely generated D-submodule of K. Example 7.1.5. Any ideal I ⊂ D is a fractional ideal. In particular, these are called integral ideals of K. N

Remark 7.1.6. Consider I = (α1, α2, . . . , αk) as a D-module, with αi ∈ K. Then there exist di ∈ D such that diαi ∈ D, and so there exists a common denominator d ∈ D such that dI is an integral ideal. Lemma 7.1.7. For any ideal I ⊂ D, I 6= 0, I−1 = { x ∈ K | xI ⊂ D } is a fractional ideal. 7.1 Unique Factorisation of Ideals 25

Proof. Let x 6= 0 in I. Then xI−1 ⊂ D by the deﬁnition of I−1. Since D is Noetherian, we moreover know that xI−1 is ﬁnitely generated as a D-module, −1 say by xα1, xα2, . . . , xαn. Then for any y ∈ I we have

n X xy = aixαi i=1 where ai ∈ D. Since D is an integral domain, we have cancellation laws, so

n X y = aiαi, i=1

−1 meaning that I is generated by α1, α2, . . . , αn as a D-module, whence it’s a fractional ideal.

Theorem 7.1.8. Let IK denote the set of all nonzero fractional ideals of K. Then IK is an abelian group under multiplication with identity D = (1) and the −1 inverse of I ⊂ IK is I = { x ∈ K | xI ⊂ D }.

Proof. First we prove that it is closed under multiplication. For any I,J ∈ IK , if I is generated by α1, α2, . . . , αn and J by β1, β2, . . . , βm, then IJ is generated by αiβj. Hence IJ ∈ IK . Secondly, let us show that D = (1) is the identity, i.e. that DI = I for any I ⊂ IK . But this is immediately true since I is a D-module. Finally let us verify that the inverse does what it should. Let P be a prime ideal in D. Then P ( PP −1 ⊂ D, with the previous lemma guarantueeing P ( PP −1. Now since P is maximal, we must have PP −1 = D. For I ⊂ D, use the prime factorisation to conclude II−1 = D. Now for general I ∈ IK , there exists some d ∈ D such that dI ⊂ D, so (dI)(dI)−1 = D.

Remark 7.1.9. Every nonzero fractional ideal I has a unique factorisation

e1 e2 ek I = P1 P2 ··· Pk where, as compared to I ⊂ D, ei ∈ Z \{ 0 }, i.e. we allow negative powers. In other words IK is a free abelian group on the set of nonzero prime ideals of D. Definition 7.1.10 (Principal fractional ideal). The principal fractional ideals of K are the D-modules in K of the form xD = (x), with x ∈ K. Let PK denote the set of nonzero principal fractional ideals in K. Then PK is a subgroup of IK . Definition 7.1.11 (Class group, class number). The class group of K is Cl(K) = IK /PK . Moreover h(K) = # Cl(K) is called the class number of K. Remark 7.1.12. The class group Cl(K) measures the failure of unique factorisation in D, since if Cl(K) is trivial, i.e. h(K) = 1, then D is a principal ideal domain, which for a Dedekind domain means its a unique factorisation domain. We will show later that if K is a number field, then h(K) < ∞. RAMIFICATION 26

Example 7.1.13. Note that prime ideals in Z are√ not, when lifted, necessarily prime ideal in number ﬁelds. Suppose, say K = Q( −5), and consider √ √ 2 K ⊃ OK = Z[ −5] ⊃ 2OK = (2, 1 + −5)

Q ⊃ Z ⊃ 2Z.

Here 2Z√is a prime ideal in Z, but 2OK is not a prime ideal in OK . However, (2, 1 + −5) is. N

Lecture 8 Ramiﬁcation

8.1 Ramiﬁcation Index

Given a number ﬁeld K, with a prime ideal pZ in Z ⊂ Q, it in general won’t be the case that pOK is a prime. However, by the discussion last time, it must be the case that e1 e2 er pOK = q1 q2 ··· qr where qi ⊂ OK are prime ideals.

Definition 8.1.1 (Ramification). Let L ⊃ K be number fields. Let P ⊂ OK be a prime ideal.

e1 e2 er (i) If P OL = q1 q2 ··· qr , then ei is called the ramiﬁcation index of P at qi. We denote this by e(qi/P ).

(ii) The prime P is said to be ramiﬁed in L if some ei > 1.

Proposition 8.1.2. Let L ⊃ K be number ﬁelds. Let P ⊂ OK be a prime in OK , and q ⊂ OL be a prime in OL. The following are equivalent:

(i) q | P OL, i.e. P OL ⊂ q; (ii) q ⊃ P ;

(iii) q ∩ OK = P ; and (iv) q ∩ K = P .

Remark 8.1.3. We say that q lies over P or P lies under q in the above proposition. Proof. (i) and (ii) are trivially equivalent by definition. Likewise (iii) and (iv) are equivalent since q ⊂ OL and OL ∩ K = OK . (iii) implies (ii) again trivially, and finally (ii) implies (iii) by noting that P ⊂ q ∩ OK ( OK , because it q ∩ OK = OK = (1), then 1 ∈ q meaning that q = OL, a contradiction. So P in the left-hand side is nonzero, making it maximal, so P = q ∩ OK . Theorem 8.1.4. Let L ⊃ K be number fields.

Date: February 1st, 2018. 8.1 Ramiﬁcation Index 27

(i) Every prime q ⊂ OL lies over a unique prime P ⊂ OK .

(ii) Every prime P ⊂ OK lies under at least one prime q ⊂ OL.

Proof. (i) follows from q ∩ OK being a prime in OK . Similarly, (ii) follows from P OL 6= OL, hence has a prime factorisation. −1 −1 Suppose P OL = OL. Then 1 ∈ P OL. Consider P ⊂ K, P 6⊂ OK . −1 Then there exists some e ∈ K \OK , r ∈ P , i.e. rP ∈ OK . Then rP OL ⊂ OK OL = OL, but 1 ∈ P OL, meaning that r ∈ OL ∩K = OK , a contradiction. In light of this, consider the following setting:

L ⊃ OL ⊃ q

K ⊃ OK ⊃ P.

Consider the ring homomorphism O ϕ: O → L , K q where since q is a nonzero prime ideal it is also maximal, whence OL/q is a ﬁeld. The kernel of this map is ker ϕ = OK ∩ q = P , so this induces a homomorphism O O ϕ¯: K ,→ L P q where both of the above quotients are ﬁelds. Hence OL q

OK P is a ﬁeld extension. We call OL/q the residue ﬁeld associated with q. The dimension of OL/q over OK /P as a fector space is called the inertial degree of q over P , denoted f(q/P ). Example 8.1.5. Consider

2 K = Q(i) ⊃ Z[i] = OK ⊃ 2OK = (1 − i)

Q ⊃ Z ⊃ (2), where we then consider Z[i] (1−i)

Z . 2Z 8.1 Ramiﬁcation Index 28

2 Then we have 2OK = (1 − i) , so e((1 − i)/2) = 2 and

[i] Z = 2 (1 − i) and |Z/2Z| = 2, so f((1 − i)/2) = 1. N Exercise 8.1.6. As expected, if we let M ⊃ L ⊃ K be number ﬁelds, and let R ⊃ q ⊃ P be primes in OM ⊃ OL ⊃ OK respectively, such that

M ⊃ OM ⊃ R

L ⊃ OL ⊃ q

K ⊃ OK ⊃ P, we have e(R/P ) = e(R/q)e(q/P ) and f(R/P ) = f(R/q)f(q/P ). Theorem 8.1.7. Let

e1 e2 er L ⊃ OL ⊃ P OL = q1 q2 ··· qr n

K ⊃ OK ⊃ P and hence OL/qi

OK /P with fi = f(qi/P ) and ei = e(qi/P ). Then

r X eifi = n. i=1 Before we prove this, let us deﬁne a new notion and state a proposition which we’ll prove alongside the theorem.

Definition 8.1.8 (Norm of ideal). Let K be a number field and let I ⊂ OK be an ideal. The norm of I is O N(I) = N (I) = K . K I Proposition 8.1.9. Let L ⊃ K be number fields with [L : K] = n.

(i) For ideals I,J ⊂ OK , N(IJ) = N(I) N(J). 8.1 Ramiﬁcation Index 29

n (ii) Let I ⊂ OK . Then NL(IOL) = NK (I) .

(iii) Let α 6= 0, α ∈ OK . For the principal ideal (α) we have NK ((α)) =

|NK/Q(α)|, where the ﬁrst norm is that of an ideal, and the second is that of an element. Proof. We start by proving (i) from the proposition. First assume I and J are coprime, i.e. I +J = OK , or equivalently IJ = I ∩J. By the Chinese remainder theorem we have that OK /IJ = OK /I∩J is isomorphic to OK /I×OK /J. Hence O O O K = K · K , IJ I J whereby N(IJ) = N(I) N(J). m Secondly, assume I = P for P ⊂ OK prime, i.e. I is a prime power. Then the goal is to show that N(I) = N(P )m, and

O N(I) = K P m by deﬁnition. Now note that

2 3 m OK ⊃ P ⊃ P ⊃ P ⊃ ... ⊃ P , where the norm between each is N(P ). Hence the claim is true if

O P k N(P ) = K = . P P k+1 Fix any α ∈ P k \ P k+1, i.e. nonzero in the latter quotient. We have an isomorphism OK /P → αOK /αP which is induced by the natural projection OK → αOK /αP. Now consider k k+1 the homomorphism ϕ: αOK → P /P , which is the natural inclusion since k αOK ⊂ P . k+1 Then ker ϕ = αOK ∩ P = αP , and

αO + P k+1 P k Im ϕ = K = P k+1 P k+1

k+1 k+1 k since P ( αOK + P = P . k k+1 Therefore ϕ induces an isomorphism between αOK /(αP ) and P /P , whence O αO P k K = K = . P αP P k+1 Hence N is multiplicative for coprime ideals and prime powers, so for two arbi- trary ideals, ﬁrst factor them, then apply these two. Next let us tackle the theorem for the special case K = Q. Then P = pZ, with p a prime, and

e1 e2 er NL(pOL) = N(q1) N(q2) ··· N(qr) RAMIFICATION INDEX CONTINUED 30

fi by the above. Hence N(qi) = p , and

r Pr Y fiei fiei NL(pOL) = p = p i=1 . i=1 Moreover O n N (pO ) = L = Z = pn, L L n pOL pZ whereby r X n = eifi, i=1 so the theorem holds for K = Q.

Lecture 9 Ramiﬁcation Index continued

9.1 Proof, continued

We showed last time that the theorem we are working on is true for K = Q. Now let us use what we know so far to tackle the second part of the proposition, but first a lemma: Lemma 9.1.1. Let D be a Dedekind domain with the field of fractions K. Let B ⊂ A ( D be ideals. Then there exists r ∈ K such that rB ⊂ D but rB 6⊂ A. Proof. We know from last time that BB−1 = (1) = D, where B−1 is a fractional ideal. Hence there exists some α ∈ D such that αB−1 ⊂ D, and hence (αB−1)B = αD. Now since A ( D we have aA ( αD, and we select β ∈ αB−1 such that βB ⊂ αD but βB 6⊂ αA. Take r = β/α. Then rB = β/αB ⊂ D but rB 6⊂ A. Proof of Proposition (ii). Since N is multiplicative, it suffices to consider I = P , a prime ideal. Then

OL/P OL

OK /P where OL/P OL is a vector space over OK /P since the latter is a ﬁeld. Our goal is to prove that the dimension of this vector space is n. First let us establish that the dimension is bounded above by n, i.e.

[OL/P OL : OK /P ] ≤ n.

Let α1, α2, . . . , αn+1 ∈ OL. We then want to show that the corresponding elements in OL/P OL are linearly dependent over OK /P . Note ﬁrst of all that the elements are linearly dependent over K, since [L : K] = n, and hence they are also linearly dependent over OK (to see this, ﬁnd a common denominator of the linear sum in K). Hence

β1α1 + β2α2 + ... + βn+1αn+1 = 0 Date: February 6th, 2018. 9.1 Proof, continued 31

for some βi ∈ OK . It is then our goal to show that not all βi ≡ 0 (mod P ). Applying the lemma with A = P and B = (β1, β2, . . . , βn+1), there exists some r ∈ K such that rB ⊂ OK , i.e. rβi ∈ OK for all i, but at the same time rB 6⊂ P , i.e. not all rβi ≡ 0 (mod P )¿ Hence

(rβ1)α1 + (rβ2)α2 + ... + (rβn+1)αn+1 = 0 and we are done. Using this we can prove that in fact [OL/P OL : OK /P ] = n. Consider P ∩ Z = pZ, with p a prime number. Then

e1 e2 er pOK = P1 P2 ··· Pr , and one of them must be equal to P , say P1 = P . Let fi = f(Pi/p), whence r Y ei pOL = pOK ·OL = Pi OL i=1 whence by the ﬁrst part of the proposition r Y ei NL(pOL) = NL(PiOL) . i=1

Set ni = [OL/PiOLLOK /Pi]. Then ni ≤ n for all i by the first claim above, ni fi and NL(PiOL) = NK (Pi) , and moreover NK (Pi) = p . Hence r r Y niei Y finiei NL(pOL) = NK (Pi) = p . i=1 i=1 But we also have O mn N (pO ) = L = Z = pmn, L L mn pOL pZ where m = [K : Q], whereby r X finiei = nm, i=1 and since the theorem is true for the base field being Q we have r X eifi = m, i=1 so, since ni ≤ n, we get ni = n for all i. We are finally ready to prove the general case of the theorem.

e1 e2 er Proof of Theorem. Consider the factorisation P OL = q1 q2 ··· qr in L. By n (ii) in the proposition, NL(P OL) = NK (¶) , where n = [L : K]. At the same time r r Pr Y ei Y fiei fiei NL(P OL) = NL(qi) = NK (P ) = NK (P ) i=1 , i=1 i=1 so r X eifi = n. i=1 RAMIFIED PRIMES 32

We end the lecture by proving the last part of the proposition.

Proof of Proposition (iii). Let M be a normal extension of Q containing K, say

M n K m

Q and let σ : K,→ C extend toσ ¯ : M,→ C (actuallyσ ¯ : M,→ M since M is normal). Moreoverσ ¯(OM ) = OM , hence

NM (αOM ) = N(σ(α)OM ) for α ∈ K. Let a = NK/Q(α) = σ1(α)σ2(α) ··· σm(α) ∈ Z. Then

NM (aOM ) = NM (σ1(α)OM )NM (σ2(α)OM ) ··· NM (σm(α)OM ) and since the extension is Galois all the embeddings just permute themselves, so this is equal to m mn NM (αOM ) = NK (αOK ) . mn mn Now since a ∈ Z we have NM (aOM ) = NQ(aZ) = |a| , whence NK (αOK ) = |a| = |NK/Q(α)|.

Lecture 10 Ramiﬁed Primes

10.1 When Do Primes Split Proposition 10.1.1. Let L ⊃ K be a Galois extension, and let G = Gal(L/K). Then G acts transitively on the set of prime ideals q of OL lying over P ⊂ OK .

e1 e2 er The idea, then, is that P OL = q1 q2 ··· qr and if σ ∈ Gal(L/K), then σ : L → L and σ(OL) = OL. The proposition says that for each pair qi and qj there exists some σ ∈ Gal(L/K) such that σ(qi) = qj.

0 Proof. Suppose there exists q and q being prime ideals in OL lying above P such that q 6= σ(q0) for all σ ∈ G. Then by the Chinese remainder theorem there exists some x ∈ OL such that ( x ≡ 0 (mod q) x ≡ 1 (mod σ(q0)) for all σ ∈ G. Now Y NL/K (x) = σ(x) ∈ q ∩ K = P, σ∈G meaning that x 6∈ σ(q0) for all σ ∈ G, so σ−1(x) 6∈ q0 for every σ ∈ G.

Date: February 8th, 2018. 10.1 When Do Primes Split 33

Therefore σ(x) 6∈ q0 for every σ ∈ G, so

Y 0 NL/K (x) = σ(x) 6∈ q σ∈G since q0 is prime, but q0 ∩ K = P , so this is a contradiction.

Corollary 10.1.2. Suppose L ⊃ K is a Galois extension, and let P ⊂ OK be a nonzero prime ideal. Let q1, q2 ⊂ OL be two prime ideals lying over P . Then e(q1/P ) = e(q2/P ) and f(q1/P ) = f(q2/P ), and hence [L : K] = ref where r is the number of primes q ⊂ OL lying above P . The fundamental question we wish to ask and answer is this: Let K be a number field. For which primes p ∈ Z is p ramified in K, i.e. e(q/p) > 1 for some q ∈ OK ? The answer is very simple to state, but not nearly as simple to prove: p is ramified in K if and only if p | disc(K). We will prove the forward direction today, and the converse much, much later.

Theorem 10.1.3. Let p be a prime in Z. Suppose p is ramiﬁed in a number ﬁeld K. Then p | disc(K).

Proof. Let q ∈ OK lie above p with e(q/p) > 1. We have

e e1 e2 er e−1 e1 e2 er pOK = q q1 q2 ··· qr = q(q q1 q2 ··· qr ). | {z } = I

Then pOK = qI with pOK ( I and I is divisible by all primes in OK lying above p by construction. Let L be a Galois extension of Q such that K ⊂ L, and let σ1, σ2, . . . , σn be the embeddings of K in C. Each of them extends to L, call itσ ¯i. Let α1, α2, . . . , αn be any integral basis for OK , and let d = disc(K) = disc(α1, α2, . . . , αn). Since pOK ( I, we can ﬁnd α ∈ I \ pOK . Write

α = m1α1 + m2α2 + ... + mnαn with mi ∈ Z, not all mi ≡ 0 (mod p) since α 6∈ pOK . Without loss of generality, say m1 6≡ 0 (mod p), i.e. p - m1. Then

2 disc(α, α2, α3, . . . , αn) = disc(m1α1, α2, . . . , αn) = m1 disc(α1, α2, . . . , αn).

It then suﬃces to show p | disc(α, α2, . . . , αn) since p - m1. Now ﬁx any prime P ⊂ OL lying above p. Then α ∈ I means that α ∈ qi ⊂ OK lying above p for every i, and so α ∈ Pi ⊂ OL lie above p for every i too. Hence α ∈ P1 ∩ P2 ∩ ... ∩ Ps whereby σ(α) ∈ P1 ∩ P2 ∩ ... ∩ Ps for every σ ∈ Gal(L/Q). Fix P = P1. Then σ(α) ∈ P for all σ ∈ Gal(L/Q), and in particular σi(α) ∈ P for σi : K,→ C. 10.1 When Do Primes Split 34

Hence

2 m1 disc(α1, α2, . . . , αn) = disc(α, α2, . . . , αn) ∈ P ∩ Z = pZ where the left-hand side is in Z, so p | disc(α, α2, . . . , αn).

Corollary 10.1.4. Let K = Q(α) with α ∈ OK . Let f(x) be the monic minimal 0 polynomial of α over Q. Suppose p ∈ Z is a prime such that p - NK/Q(f (α)). Then p is unramiﬁed in K.

Proof. We have K = Q(α) ⊃ OK ⊃ Z[α], and so 2 disc(α) = [OK : Z[α]] disc(K),

0 0 where the left-hand side is ± NK/Q(f (α)). Hence p - NK/Q(f (α)) implies p - disc(K), and so by the contrapositive of the theorem p is unramiﬁed.

Corollary 10.1.5. There are only finitely many primes in Z that are ramified in a number field K. Corollary 10.1.6. Let L ⊃ K be number fields. Then there are only finitely many prime ideals that are unramified in OL. √ Example 10.1.7. Let K = Q( D), with D being square-free. We know that ( √ [ D], if D 6≡ 1 (mod 4), Z √ OK = 1+ D Z[ 2 ] if D ≡ 1 (mod 4), along with ( 4D, if D 6≡ 1 (mod 4), disc(K) = D if D ≡ 1 (mod 4).

Let p ∈ Z be prime. Then

 2 P if p | disc(K),  disc(K) pOK = P if p - disc(K), p = −1,  disc(K) P1P2 if p - disc(K), p = 1.

We call the second case inert and the third case split. N 2πi/m Example 10.1.8. Consider the cyclotomic ﬁeld K = Q(ξm) where ξm = e , and m ≥ 3. We know that OK = Z[ξm]. First suppose m = pr, with p ∈ Z a prime. Then we know that disc(K) | mϕ(m) = pn for some power n, and so disc(K) = pv for some power v. Hence p is the only prime ramiﬁed in K. r1 r2 Next suppose m = p1 p2 , with p1 6= p2 primes in Z. Then as we know

L1 = (ξ r1 ) L2 = (ξ r2 ) Q p1 Q p2

n1 n2

Q THE IDEAL CLASS GROUP 35

n2 n1 n1 n2 and disc(K) = disc(L1) disc(L2) = p2 p1 so p is ramiﬁed in K if and only r1 r2 if p | m where m = p1 p2 for some powers r1 and r2. Inductively, for any m ≥ 3, p is ramiﬁed in Q(ξm) if and only if p | m. N

Note that if K = Q(ξm) we also have that K is a Galois extension of Q, and so e pOK = (q1q2 ··· qr) and f = f(qi/p) for all i, and so ref = ϕ(m). Now let us see if we can determine r, e, and f. The first case to consider is p - m. Then p is unramified in K, so e = 1, and we claim that f is the multiplicative order of p modulo m. The way to think about this is that if gcd(p, m) = 1, then p ∈ (Z/mZ)×, whereby pf ≡ (mod m), so r = ϕ(m)/f. So assume first that m = qk for q ∈ Z a prime such that q 6= p. Then if Q lies above P in OK , then OK /Q = Fpf

Z/pZ = Fp

p and Gal(Fpf /Fp) = (τ) is cyclic, and in particular τ(x) = x is the Frobenius automorphism, and naturally |(τ)| = f. ∼ × a On the other hand, Gal(K/Q) = (Z/mZ) by the isomorphism σ1(x) = x p corresponding to a (mod m), and in particular σp(x) = x corresponding to p (mod m).

Lecture 11 The Ideal Class Group

11.1 When are Primes Ramiﬁed in Cyclotomic Fields

Let K = Q(ξm) and let p ∈ Z be a prime. Then p is ramified in K if and only if p | m. e We know that K is a Galois extension of Q, so pOK = (q1q2 ··· qr) , and f = f(q1/p) = ... = f(qr/p). We therefore want to determine r, e, and f, knowing that ref = n = ϕ(m). Let us first consider the case p - m, so that p is unramified, meaning that e = 1, and so pOK = q1q2 ··· qr. We claim that f is the multiplicative order of p modulo m. × This makes sense since p - m means that (p, m) = 1, and thereby p ∈ Zm, and we claim that f is the smallest positive integer such that pd ≡ 1 (mod m). k Assume m = q for q ∈ Z a prime, but p 6= q. Let ξ = ξqk . We then have ∼ OK /q1 = Fpf

f ∼ Z/pZ = Fp

p and Gal(Fpd /Fp) = (τ), where τ : Fpf → Fpf , τ(x) = x is the Frobenius automorphism, and of course |(τ)| = f. We also have Gal(K/Z) =∼ (Z/mZ)×, a p by σa(ξ) = ξ ↔ a (mod m), and so in particular σp(ξ) = ξ ↔ p (mod m).

Date: February 11th, 2018. 11.2 The Ideal Class Group and Unit Group 36

The order of p modulo m is equal to the order of σp in Gal(K/Q). Hence we want to show that the order of σp in Gal(K/Q) is in turn equal to the order of τ in Gal(Fpf /Fp). a a This is the same as saying σp = Id if and only if τ = Id, for every a ∈ N. a a pa First σp = Id is equivalent with σp (ξ) = ξ if and only if ξ = ξ which is the case if and only if pa ≡ 1 (mod m) since ξ is a primitive mth root of unity. a a pa Secondly, τ = Id if and only if τ (ξ) ≡ ξ (mod q1), if and only if ξ ≡ ξ a (mod q1). We claim that this, in turn, is equivalent with p ≡ 1 (mod m). The reverse direction is trivial since ξ is an mth root of unity. pa The forward direction is a little trickier. Assume ξ ≡ ξ (mod q1), and a write pa ≡ b (mod m) for 0 ≤ b < m, meaning that ξp = ξb. Then b 6= 0 since otherwise ξ = 1 (and indeed p | m if it’s the case). × × m pa−1 We have ξ ∈ OK = Z[ξ] since N(ξ) = 1 (and ξ = 1). So ξ ≡ 1 b−1 (mod q1), implying that ξ ≡ 1 (mod q1). Then (1 − ξ)(1 − ξ2) ··· (1 − ξm−1) = m since xm−1 + xm−2 + ... + x + 1 = (x − ξ)(x − ξ2) ··· (x − ξm−1) so we plug in x = 1. b−1 Now if b > 1, 1−ξ appears somewhere in the above, whence m ∈ q1 ∩Z = pZ, but this is a contradiction since p - m. Therefore b = 1. So the order of σp is the same as the order of τ. This same argument works for any (m, p) = 1¡ so for p - m, with e = 1, we know f, and therefore we can ﬁnd r. Now consider instead the case p | m, whereby p is ramiﬁed in K, so

e pOK = (q1q2 ··· qr) k ϕ(m) with e > 1. Assume m = p . We claim pOK = (1 − ξ) . We have N(1 − ξ) = p, and moreover

Y k Y 1−ξk ϕ(m) N(1 − ξ) = (1 − ξ ) = 1−ξ × (1 − ξ) = u(1 − ξ) , ∈OK (k,m)=1 (k,m)=1 1≤k

ϕ(m) ϕ(m) so pOK = u(1−ξ) OK = (1−ξ) , so ref = ϕ(m) and e ≥ ϕ(m) (because (1−ξ) could in principle split), but since the we have ref = ϕ(m) we must have e = ϕ(m) and so r = f = 1, meaning that (1 − ξ) is a prime ideal in OK . The case we haven’t dealt with is the possibility that m = pk · n, where (n, p) = 1.

11.2 The Ideal Class Group and Unit Group

Let K be a number ﬁeld and let IK denote the set of fractional ideals in K and PK denote the set of principal fractional ideals in K. The class group is Cl(K) = IK /PK , and the class number is h(K) = # Cl(K). The goal of this discussion is to show that h(K) < ∞. Theorem 11.2.1. Let K be a number ﬁeld. Then there exists λ > 0 (depending only on K) such that for every nonzero I ⊂ OK there exists a nonzero α ∈ I with

|NK/Q(α)| ≤ λ NK (I). 11.2 The Ideal Class Group and Unit Group 37

Proof. Let n = [K : Q]. Let α1, α2, . . . , αn be an integral basis for OK and let σ1, σ2, . . . , σn be the embeddings of K in C. Set n n Y X λ = |σi(αj)| i=1 j=1 depending on K. n For any nonzero ideal I ⊂ OK , let m ∈ N such that m ≤ NK (I) = |OK /I| < (m + 1)n. n Consider the following (m + 1) elements in OK deﬁend by n X mjαj j=1 with 0 ≤ mj ≤ m. By construction this is a larger set of elements than NK (I), so two must be equal modulo I. Take the diﬀerence of those two elements, call it n X rjαj ∈ I j=1 with |rj| ≤ m, and call this element our α. Then we have n n n n Y X Y X |N (α)| = r σ (α ) ≤ mn |σ (α )| ≤ λ N (I) K/Q j i j i j K i=1 j=1 i=1 j=1 since mn ≤ NK (I) by construction.

Corollary 11.2.2. Every ideal class of K contains an integral ideal J ⊂ OK with NK (J) ≤ λ, with λ as above. −1 Proof. Given an ideal class C ∈ Cl(K) = IK /PK , we know that C exists since the class group is indeed a group. Fix any integral ideal I ∈ C−1. By the previous theorem there exists some nonzero α ∈ I such that

|NK/Q(α)| ≤ λ NK (I).

Note that (α) ⊂ I ⊂ OK , meaning that (α) = IJ for some J ⊂ OK . Consider this equation modulo PK , i.e. in IK /PK , we get IJ ≡ 1 (mod PK ) since (α) is principal. Now I is a representative for the coset C−1 and J one for C, so J ∈ C. Then

|NK/Q(α)| = N((α)) = N(I) N(J) ≤ λ N(I) and so N(J) ≤ λ. Corollary 11.2.3. The class number h(K) is finite. Proof. There are only finitely many primes p ∈ Z such that |p| < λ, and above each of those there is a finite number of ideals J, whose norms are greater than the size of the prime they lie above, and so there are only finitely many ideals in OK with norm less than λ. Hence by the above corollary there are only finitely many ideal classes, since each ideal class contains an integral ideal with norm less than or equal to λ. MINKOWSKI’S THEOREM 38

√ √ Example√ 11.2.4. Let K = Q( 2), so OK = Z[ 2], with the integral basis { 1, 2 }. Then √ √ √ λ = (1 + 2)(1 + |− 2|) = (1 + 2)2 ≈ 5.8 so the only candidate√ primes we need consider are 2, 3, and 5. 2 Now 2OK = ( 2) factors, but 3OK and 5OK are prime ideals. For the 2 latter two we must therefore have e = 1 and f = 2, and N(3OK ) = 3 = 9 along 2 with N(5OK ) = 5 = 25, both of which exceed λ. √ √ Hence the ideals in OK with norms less than λ are OK = (1), 2OK = ( 2), and 2OK = (2), all of which are principal, so OK is a principal ideal domain. N

Lecture 12 Minkowski’s Theorem

12.1 Using Geometry to Improve λ We showed last time that given a number ﬁeld K, there exists some λ (depending on K) such that for every nonzero ideal I ⊂ OK there is some nonzero α ∈ I for which

|NK/Q(α)| ≤ λ NK (I). Last time we showed that

n n Y X λ = |σi(αj)| i=1 j=1 works, where { αi } is an integral basis for OK and σj : K,→ C are the embeddings of K into C. The choice of this λ, as we saw at the end last time, decides how many primes we need to study in order to tell if OK is a principal ideal domain or not. It is therefore in our interest to sharpen λ, making it smaller.

Deﬁnition 12.1.1 (Lattice). A lattice Λ in V = Rn is an additive subgroup of V of the form Λ = Zv1 + Zv2 + ... + Zvn with v1, v2, . . . , vn linearly independent over R. Example 12.1.2. The standard lattice that comes to mind is the integer lattice, 2 2 i.e. in V = R we have Λ = Z = Ze1 + Ze2, where e1 = (1, 0) and e2 = (0, 1). N

Let K be a number ﬁeld of degree n, and let σ1, σ2, . . . , σr be the real embeddings of K, with τ1, τ1, . . . , τ2, τs being the complex embeddings of K Consider the following function ϕ: K → Rn deﬁned by

ϕ(α) = (σ1(α), σ2(α), . . . , σr(α), Re τ1(α), Im τ1(α),..., Re τs(α), Im τs(α)).

This is an additive homomorphism with ker ϕ = { 0 }, and hence ϕ is an embedding.

Date: February 15th, 2018. 12.1 Using Geometry to Improve λ 39

Proposition 12.1.3. Let K be a number ﬁeld with ring of integers OK . Then n Λ = ΛOK = ϕ(OK ) is a lattice in R .

Proof. Let α1, α2, . . . , αn be an integral basis for OK , and set vi = ϕ(αi) for i = 1, 2, . . . , n. Then Λ = Zv1 + Zv2 + ... + Zvn, since OK = Zα1 + Zα2 + ... + Zαn. We need to verify that vi are linearly independent over R, so let us compute the determinant of the matrix made up of them as rows. Let   v1 v2  δ = det    .   .  vn which expanded is

σ1(α1) ··· σr(α1) Re τ1(α1) Im τ1(α1) ··· Re τs(α1) Im τs(α1)

. . . . .

σ1(αn) ··· σr(αn) Re τ1(αn) Im τ1(αn) ··· Re τs(αn) Im τs(αn) If we multiply the ﬁrst Im column by i, and then add the Im-column to the Re-column, we get

σ1(α1) ··· σr(α1) τ1(α1) i Im τ1(α1) ··· Re τs(α1) Im τs(α1) 1 . . . . . i σ1(αn) ··· σr(αn) τ1(αn) i Im τ1(αn) ··· Re τs(αn) Im τs(αn)

Next add −2 times the τ1-column to the i Im-column, whence we have

σ1(α1) ··· σr(α1) τ1(α1) τ1(α1) ··· Re τs(α1) Im τs(α1) 1 . . . . . −2i σ1(αn) ··· σr(αn) τ1(αn) τ1(αn) ··· Re τs(αn) Im τs(αn) Repeat this for tall the columns corresponding to complex embeddings to get

σ1(α1) ··· σr(α1) τ1(α1) τ1(α1) ··· τs(α1) τs(α1) 1 . . . s . . (−2i) σ1(αn) ··· σr(αn) τ1(αn) τ1(αn) ··· τs(αn) τs(αn) But the determinant there is the discriminant of K, so 1 |δ|2 = |disc(K)|= 6 0 22s and so δ 6= 0, meaning that vi are indeed linearly independent over R. Deﬁnition 12.1.4 (Fundamental parallelotope). A fundamental parallelotope for Λ is the set n n o X D = aivi 0 ≤ ai < 1 . i=1 We denote D = Rn/Λ. 12.1 Using Geometry to Improve λ 40

Example 12.1.5. With V = R2 and Λ = Z2, the fundamental parallelotope is the unit square between (0, 0), (1, 0), (1, 1), and (0, 1), with the right and top edges excluded. N Note that G V = (λ + D) λ∈Λ meaning that shifted D tile the space V . Proposition 12.1.6. We have   v1 v2  Vol( n/Λ) = det   R  .   .  vn

n and Vol(R /Λ) is independent of the choice of vi. These are elementary linear algebra considerations. The ﬁrst one can in fact be taken as the deﬁnition of the determinant, and the second one follows from there being a transformation between any two integral bases whose determinant is ±1.

n Theorem 12.1.7. The map ϕ: K,→ R sends OK to a lattice Λ = ϕ(OK ) p with Vol(Rn/Λ) = 2−s |disc(K)|. Moreover

ϕ(K) = Qv1 + Qv2 + ... + Qvn is dense in Rn. n Let V = R and assume Λ1 ⊂ Λ2 ⊂ V are lattices. Then Λ2/Λ1 is a ﬁnite group, and n n Vol(R /Λ1) = Vol(R /Λ2) · |Λ2/Λ1|. 2 Example 12.1.8. Consider V = R and Λ2 = Z × Z ⊃ Λ1 = 2Z × Z, i.e. in the second one we only have even integers in our x-coordinates. Then the volume of the ﬁrst fundamental parallelotope is 1·1 = 1, and the second one is 2·1 = 2, and the factor between them of course is 2. N We are now almost at the point where this excursion in geometry ties into our number theory:

n Example 12.1.9. Let ϕ: K,→ R be as before, and let Λ = ϕ(OK ) be the same lattice. Let I ⊂ OK be an ideal. Then ΛI = ϕ(I) ⊂ Λ is a sublattice, and so

n n n Vol(R /ΛI ) = Vol(R /Λ) · |OK /I| = Vol(R /Λ) NK (I) 1 = p|disc(K)| N (I). 2s K N Let us now define a special norm on Rn depending on K, a number field of degree n with r real embeddings and 2s complex embeddings. For x = n (x1, . . . , xr, xr+1, . . . , rn) ∈ R we define 2 2 2 2 2 2 N(x) = x1x2 ··· xr(xr+1 + xr+2)(xr+3 + xr+4) ··· (xn−1 + xn) 12.1 Using Geometry to Improve λ 41

n which is of note since if ϕ: K,→ R by α 7→ ϕ(α) = x, then NK/Q(α) = N(x) since r s Y Y NK/Q(α) = σi(α) τj(α)τj(α). i=1 j=1 Note that whilst we defined this in terms of a number field K, we could equally well forget about the number field and define such a norm on Rn for any partition n = r + 2s.

Theorem 12.1.10 (Minkowski). Let n = r + 2s, r, s ∈ Z. For any lattice Λ ⊂ Rn there exists a nonzero x ∈ Λ such that n! 8 s |N(x)| ≤ Vol( n/Λ). nn π R Notice the complete lack of number ﬁelds above. It works for any partition n = r + 2s; it’s pure analysis.

Theorem 12.1.11. For every nonzero ideal I ⊂ OK , there exists a nonzero α ∈ I such that n! 4 s |N (α)| ≤ p|disc(K)| N (I). K/Q nn π K

Proof. Simply apply Minkowski’s theorem to ΛI = ϕ(I). In other words we can use n! 4 s λ = p|disc(K)| nn π in the type of considerations made last time:

Corollary 12.1.12. Every ideal class of K contains an integral ideal J ⊂ OK with n! 4 s N (J) ≤ p|disc(K)|. K nn π Remark 12.1.13. The quantity n! 4 s nn π is called the Minkowski constant and notably decays very, very quickly, meaning that for number ﬁelds of ﬁxed discriminant, as n → ∞ almost all of them are principal ideal domains.

Example 12.1.14. Let K = Q(ξ5) with OK = Z[ξ5] and [K : Q] = ϕ(5) = 4 = n. We have disc(K) = 55−2 = 125 since s = 2, and so 4! 4 2√ N (J) ≤ 125 < 2 K 44 π and therefore NK (J) = 1, so J = OK , meaning that every ideal class contains OK , and therefore OK is a principal ideal domain. N Let us repeat the computation from last time as well, to show how much better this λ is. TOWARD MINKOWSKI’S THEOREM 42

√ Example 12.1.15. Let K = Q( 2), so s = 0 and n = 2, and disc(K) = 4 · 2 = 8. Then 2! 4 0√ √ N (J) ≤ 8 = 2, K 22 π so NK (J) = 1, and OK is a principal ideal domain. Notice how this time we didn’t have to try to factor a single prime ideal in OK . N Corollary 12.1.16. For K 6= Q, i.e. [K : Q] > 1, we have |disc(K)| > 1.

Proof. Simyply use the last corollary. Since NK (J) ≥ 1, we have n! 4 s 1 ≤ p|disc(K)| nn π which rearranged becomes nn π s p|disc(K)| ≥ > 1 n! 4 for every n > 1. It remains to prove Minkowski’s theorem. This will take some work, but ﬁrst:

Lemma 12.1.17. Let Λ ⊂ Rn be a lattice. Let E be a convex, measurable, symmetric subset of Rn. If Vol(E) > 2n Vol(Rn/Λ), then E contains a nonzero element of Λ. Moreover if E is also compact, then the assumption can be weakened to Vol(E) ≥ 2n Vol(Rn/Λ). Before we prove this, let us ﬁrst recall what the various properties above mean. We call a set convex if for every x, y ∈ E we also have ax + (1 − a)y ∈ E for every 0 ≤ a ≤ 1, i.e. the line segment between any two points in the set is also contained in the set. By measurable we mean a Lebesgue measurable set, so it is in the σ-algebra generated by the opens sets in Rn. When we say symmetric we mean that if x ∈ E then −x ∈ E as well.

Lecture 13 Toward Minkowski’s Theorem

13.1 Proving Minkowski’s Theorem We’ll start by proving the lemma stated at the end of last lecture.

Proof. Let F = Rn/Λ be the fundamental parallelotope for Λ. Then

n G R = x + F x∈Λ meaning that 1 G 1 E = E ∩ (x + F ) . 2 2 x∈Λ

Date: February 20th, 2018. 13.1 Proving Minkowski’s Theorem 43

By assumption, 1 X X Vol(F ) < Vol(E) = Vol(1/2E) = Vol(1/2E∩(x+F )) = Vol((1/2E−x)∩F ) 2n x∈Λ x∈Λ since the Lebesgue measure is translation invariant. Hence (1/2E −x)∩(1/2E − y) 6= ∅ for some x, y ∈ Λ, x 6= y, since otherwise the volume would be less than or equal to Vol(F ), which is a contradiction. So 1/2e1 − x = 1/2e2 − y for some e1, e2 ∈ E, meaning that 1 1 1 x − y = e − e = (e + (−e )) ∈ E 2 1 2 2 2 1 2 since E is symmetric, so −e2 ∈ E, and E is convex, with the last step above being a convex combination of e1 and −e2. Moreover x − y ∈ Λ since Λ is a group, whence x − y 6= 0 is in Λ ∩ E. Suppose now that E is compact and Vol(E) ≥ 2n Vol(Rn/Λ). For m = 1, 2,..., we have

n n n Vol((1 + 1/m)E) = (1 + 1/m) Vol(E) > 2 Vol(R /Λ), so by the above there exists some nonzero xm ∈ Λ ∩ (1 + 1/m)E. ∞ Hence { xm }m=1 ⊂ Λ ∩ (2E), where 2E is compact (it’s closed and bounded n in R ), and Λ ∩ (2E) is ﬁnite since Λ is discrete. Hence { xm } takes ﬁnitely many values, meaning that there exists some subsequence { xk } with xk = x for all k. Therefore x = xk ∈ (1 + 1/k)E for all k, so when in the limit (1 + 1/k)E goes to E = E since E is compact. Hence x ∈ E ∩ Λ.

Corollary 13.1.1. Let A be a compact, convex, symmetric subset of Rn with Vol(A) > 0 and |N(a)| ≤ 1 for all a ∈ A, where by N we mean the special norm on Rn deﬁned by n = r + 2s as before. Then for every lattive Λ ⊂ Rn there exists some nonzero x ∈ Λ with 2n |N(x)| ≤ Vol( n/Λ). Vol(A) R Proof. Let E = tA where 2n tn = Vol( n/Λ). Vol(A) R Then n n n Vol(E) = Vol(tA) = t Vol(A) = 2 Vol(R /Λ). By the lemma, there exists a nonzero x ∈ Λ ∩ E such that x = ta for some a ∈ A. Then 2n |N(x)| = |N(ta)| = |tn N(a)| ≤ Vol( n/Λ). Vol(A) R

Theorem 13.1.2 (Minkowski). For every lattice Λ ⊂ Rn there exists a nonzero x ∈ Λ such that n! 8 s |N(x)| ≤ Vol( n/Λ) nn π R where n = r + 2s. 13.1 Proving Minkowski’s Theorem 44

Proof. Deﬁne the set A by q q 2 2 2 2 |x1| + |x2| + ... + |xr| + 2( xr+1 + xr+2 + ... + xn−1 + xn) ≤ n, where n = r + 2s. Then A is compact (it’s certainly closed and bounded), it is convex (it suﬃces to study the middle points between any two points), and it’s trivially symmetric. We also have Vol(A) > 0 since we at the very least contain, say, some small intervals in each coordinate. Moreover for a ∈ A we ahve |N(a)| ≤ 1 since by the AM-GM inequality q q 2 2 2 2 |x1| + |x2| + ... + |xr| + 2( xr+1 + xr+2 + ... + xn−1 + xn) 1 ≥ n 2 2 2 2 1/n 1/n ≥ (|x1||x2| · · · |xr|(xr+1 + xr+2) ··· (xn−1 + xn)) = N(a) .

We now claim nn π s Vol(A) = 2r . n! 2 Then the theorem follows from the corollary above. r+2s Let Vr,s(t) be the volume of the subset of R deﬁned by q q 2 2 2 2 |x1| + |x2| + ... + |xr| + 2( xr+1 + xr+2 + ... + xn−1 + xn) ≤ t.

r+2s Then Vol(A) = Vr,s(n) and by scaling Vr,s(t) = t Vr,s(1). We will show that 1 π s V (1) = 2r . r,2 (r + 2s)! 2

Now if r > 0, we have could move |xr| to the right-hand side and then bound by 1 − t, so

Z 1 Z 1 r−1+2s Vr,s(1) = 2 Vr−1,s(1 − t) dt = 2 (1 − t) Vr−1,s(1) dt 0 0 2 = V (1), r + 2s r−1,s which by induction becomes 2r V (1) = V (1). r,s (r + 2s)(r + 2s − 1) ··· (2s + 1) 0,s

Note that if s = 0 we deﬁne V0,0(1) = 1 (since V1,0(1) = 2 = 2V0,0(1)), and we are done. Now if s > 0, V0,s(1) is given by q q 2 2 2 2 2( xr+1 + xr+2 + ... + xn−1 + xn) ≤ 1, which if we move one of the square roots over yields ZZ p 2 2 V0,s(1) = V0,s−1(1 − 2 x + y ) dx dy R 13.2 The Dirichlet Unit Theorem 45 where R = { x2 + y2 ≤ 1/4 } is the circle of radius 1/2. We therefore switch to polar coordinates, x = ρ cos(θ) and y = ρ sin(θ), with dx dy = ρ dρ dθ, whence our integral becomes

Z 2π Z 1/2 V0,s(1) = V0,s−1(1 − 2ρ)ρ dρ dθ 0 0 Z 2π Z 1/2 2(s−1) = (1 − 2ρ) V0,s−1(1)ρ dρ dθ 0 0 π 1 = V (1), 2 (2s)(2s − 1) 0,s−1 so by induction π s 1 V (1) = V (1) 0,s 2 (2s)(2s − 1) ··· 2 · 1 0,0 whence 1 π s V (1) = 2r . r,s (r + 2s)! 2

13.2 The Dirichlet Unit Theorem

× Let K be a number ﬁeld and OK its ring of integers. Let OK be the set of units in OK . Next let σ1, σ2, . . . , σr be the real embeddins of K and τ1, τ2, . . . , τs be the distinct nonconjugate compled embeddings of K, with n = r + 2s. Deﬁne a logarithm map log: K× → Rr+2 by log(α) = log|σ1(α)|, log|σ2(α)|,..., log|σr(α)|, 2 log|τ1(α)|,..., 2 log|τs(α)| .

It is easy to check log(αβ) = log(α)+log(β), making it a multiplicative-additive group homomorphism. × r+s × Therefore log(OK ) is an additive subgroup of R . For α ∈ OK , we have

2 2 1 = |NK/Q(α)| = |σ1(α)σ2(α) ··· σr(α)|τ1(α)| · · · |τs(α)| |, means that

0 = log|σ1(α)| + ··· + log|σr(α)| + 2 log|τ1(α)| + ... + 2 log|τs(α)| which means that

× r+s log(OK ) ⊂ H = { x ∈ R | x1 + x2 + ... + xr+s = 0 }, a hyperplane with dimR H = r + s − 1.

Proposition 13.2.1. Consider log| × , by which we mean the logarithm re- OK × stricted to OK . Then (i) ker(log) = µ(K) is the group of roots of unity in K, which is ﬁnite.

(ii) Im(log) is a lattice in H. Hence Im(log) =∼ Zr+s−1. DIRICHLET UNIT THEOREM 46

× ∼ r+s−1 Corollary 13.2.2 (Dirichlet unit theorem). We have OK = µ(K) × Z . To prove these, we ﬁrst need the following lemma.

Lemma 13.2.3. The number of elements on OK with bounded embeddings is ﬁnite, i.e.

#{ α ∈ OK | |σi(α)| < M, |τj(α)| < M, i = 1, 2, . . . , r, j = 1, 2, . . . , s } is ﬁnite for every M > 0.

Proof. Note that α ∈ OK is a root of a monic polynomial in Z[x] of degree less than or equal to n, with coeﬃcients bounded by

n max M i 1≤i≤n i by the binomial theorem. There exists only ﬁnitely many such polynomials, and therefore only ﬁnitely many such roots.

Lecture 14 Dirichlet Unit Theorem

14.1 Dirichlet Unit Theorem We start by proving the proposition stated last time. Proof. The ﬁrst part, that µ(K) ⊂ ker(log), is trivial, since roots of unity have absolute value 1, so the constituent logarithms all become 0. For the opposite inclusion, ker(log) ⊂ µ(K), consider

× ker(log) = { α ∈ OK | |σi(α)| = 1, |τj(α)| = 1, i = 1, 2, . . . , r, j = 1, 2, . . . , s }. By the lemma this is ﬁnite, i.e. # ker(log) < ∞. So if we pick some α ∈ ker(log), the set { α, α2, α3,... } must be ﬁnite—all of these elements are in the kernel, since log is a homomorphism. Hence αm = αh for some m > h, so αm−h = 1, whence α is a root of unity, meaning that α ∈ µ(K). × For the second part we need to show that log(OK ) is discrete, and that × log(OK ) spans H over R. For any bounded set in H, it is contained in A = [−M,M]r+s ∩ H for some −1 M. Now the pre-image log (A) in OK is contained in

M M/2 { α ∈ OK | |σi(α)| < e , |τj(α)| < e , i = 1, 2, . . . , r, j = 1, 2, . . . , s }

× × which by the lemma is ﬁnite. Hence log(OK ) ∩ A is a ﬁnite set, and so log(OK ) is discrete. To prove the second part we need two lemmata.

Lemma 14.1.1. Fix any k, 1 ≤ k ≤ r + s. For each α ∈ OK , there exists some nonzero β ∈ OK depending on α such that 2 s |N (β)| ≤ p|disc(K)| K/Q π Date: February 22nd, 2018. 14.1 Dirichlet Unit Theorem 47 independent of α, and if

log(α) = (a1, a2, . . . , ar+s) and log(β) = (b1, b2, . . . , br+1) then bi < ai for all i 6= k.

2 Proof. Take E as the subset deﬁned by |xi| ≤ ci for i = 1, 2, . . . , r, and xr+1 + 2 2 2 ai xr+2 ≤ cr+1, . . . , xn−1 + xn ≤ cr+s, where ci are chosen to satisfy 0 < ci < e for all i 6= k, and moreover 2 s c c ··· c = p|disc(K)|, 1 2 r+s π which then gives us ck. Then r s n n Vol(E) = 2 π c1c2 ··· cr+s = 2 Vol(R /ΛOK ), n −sp where we know Vol(R /ΛOK ) = 2 |disc(K)|. This set E is convex, compact, symmetric, and Lebesgue measurable, so we can apply Minkowski’s theorem, saying that there exists a nonzero x ∈ E ∩ΛOK . n Now take β ∈ OK such that ϕ(β) = x, where ϕ: OK ,→ R ∩ E simply sends β a vector with each component being an embedding of β. Then 2 s |N (β)| = |N(x)| ≤ p|disc(K)| K/Q π and β satisﬁes pur requirements by the choice of ci.

× Lemma 14.1.2. Fix any k, 1 ≤ k ≤ r + s. Then there exists a unit u ∈ OK such that if log(u) = (y1, y2, . . . , yr+s), then yi < 0 for all i 6= k and yk > 0 (they couldn’t all be negative, since their sum must be 0).

Proof. Take any α1 6= 0 in OK and apply the last lemma repeatedly to construct α2, then α3, and so forth. Thus for each i 6= k the ith coordinate of log(αj+1) is less than the ith coordinate of log(αj). Now since 2 s N ((α )) = |N (α )| ≤ p|disc(K)| K j K/Q j π if a uniform bound, and since there exists only ﬁnitely many distinct ideals of bounded norm (since they all lie above a ﬁnite number of primes less than some bound), we must have (αj) = (αh) for some j < h, and so αj = uαj for some × unit u ∈ OK . Hence log(αh) = log(uαj) = log(u) + log(αj).

Call the ith coordinate of the left-hand side ah,i and the ith coordinates of the two addends in the right-hand side yi and aj,i respectively. Then for j < h we have ah,i = yi + aj,i for i 6= k, and so ah,i < aj,i meaning that yi < 0 for every i 6= k. 14.2 Fermat’s Last Theorem 48

With this in hand we apply the second lemma for each 1 ≤ k ≤ r + s − 1 to × obtain u1, u2, . . . , ur+s−1 ∈ OK , where log(uj) has positive jth coordinate and all other coordinates negative. Set vi = log(ui) = (yi,1, yi,2, . . . , yi,r+s−1), so vi,j < 0 if i 6= j and yi,i > 0. We claim that the set { v1, v2, . . . , vr+s−1 } spans H over R. In other words we want to show that it is a linearly independent set, since the dimension of H over R is exactly r + s − 1. Suppose t1w1 + t2w2 + ... + tr+s−1wr+s−1 = 0 for ti ∈ R, not all ti = 0, where wi are the columns of the matrix whose rows are vi. We may assume tk = 1 for some k, and |ti| ≤ 1 for i 6= k (simply ﬁnd the largest coeﬃcient and divide both sides by it). Then the kth coordinate is

r+s−1 X X 0 = tiyi,k = yk,k + tiyi,k i=1 i6=k where yk,k > 0 and yi,k < 0. Note that at least one tj 6= 0 for some j 6= k since otherwise we’d have 0 = yk,k > 0, which is impossible. Hence X 0 > yk,k + yi,k = 0 i6=k since log(vi) ∈ H. Hence vi are linearly independent, and there are as many of them as there are dimensions in H, so they span H.

14.2 Fermat’s Last Theorem Theorem 14.2.1 (Fermat’s last theorem). Let n ≥ 3 be an integer. Then xn + yn = zn has no nontrivial solutions in Z (nontrivial meaning xyz 6= 0). There is a direct reduction to be made by considering primes p | n, since if n = pm, then (xm)p + (ym)p = (zm)p. Hence if xn + yn = zn has a nontrivial solution, then xp + yp = zp also has a nontrivial solution. Hence by contrapositive for any oﬀ prime p, xp +yp = zp having no nontrivial solutions implies that Fermat’s last theorem is true. Now assume gcd(x, y, z) = 1 (otherwise factor the gcd out). We have two cases: First, p - xyz, and secondly p divides exactly one of x, y, and z. For the second case, if p | x, then yp ≡ zp (mod p), and so y ≡ z (mod p), so if p | x and p | y, then gcd(x, y, z) > 1. The ﬁrst case is provided a partial answer by Kummer’s theorem, namely with p - h(K), for K = Q(ζp). KUMMER’S THEOREM 49

Lecture 15 Kummer’s Theorem

15.1 Using the Dirichlet Unit Theorem

× ∼ r+s−1 The Dirichlet unit theorem says that OK = µ(K) × Z , which in turn is isomorphic to Z/ωZ × Zr+s−1, where |µ(K)| = ω. We call Z/ωZ the torsion part and Zr+s−1 the free part. × n1 n2 nr+s−1 For any u ∈ OK , we therefore have u = ξε1 ε2 ··· εr+s−1 , where ξ ∈ µ(K) and ni, with ε1, ε2, . . . , εr+s−1 being free generators. Then { ε1, ε2, . . . , εr+s−1 } is called a fundamental system of units. √ Example 15.1.1. Let K = Q( D) with D being square free. We have two cases: × 2 First, D < 0. We know that OK is { ±1, ±i } if D = −1, it’s { ±1, ±ξ3, ±ξ3 } is D = −3, and it’s { ±1 } otherwise. × ∼ We have n = 2, r = 0, and s = 1, so r+s−1 = 0, i.e. rank 0, so OK = µ(K). Next consider D > 0. Then n = 2, r = 2, and s = 0, so r + s − 1 = 1. We have µ(K) = { ±1 }, and so O× ∼ { ±1 } × . K = Z √ Let ε be a fundamental unit of K with ε > 1. Then ε = a + b D if D 6≡ 1 (mod 4) (and similar but with half integers for D ≡ 1 (mod 4)). Then N(ε) = a2 −b2D = ±1, and ε is a fundamental solution of this equation since all units u = ±εm for some m ∈ Z. This equation is known as Pell’s equation. N

15.2 Kummer’s Theorem

Theorem 15.2.1 (Kummer’s theorem). Suppose p > 2 is prime. If p - h(K), p p p with K = Q(ξp), then there are no nontrivial solutions to x + y = z with p - xyz. Remark 15.2.2. A prime p satisfying the above is called a regular prime, so this theorem says that Fermat’s last theorem is true for all regular primes.

Before we proceed we recall some properties of the cyclotomic ﬁeld K = Q(ξp).

1. OK = Z[ξp], and (1 − ξp)OK is a prime ideal. n p−1 2. (1 − ξp) = (1 − ξp) = pOK , so p is totally ramiﬁed—this is what implies that (1 − ξp)O)K is prime.

3.N K/Q(1 − ξp) = ±p. ∼ 4. Z/pZ = OK /(1 − ξp) since f = 1. s 5. µ(K) = { ±ξp }0≤s≤p−1. In order to prove Kummer’s theorem, we ﬁrst need the following lemma.

Lemma 15.2.3. (i) For each α ∈ OK , there exists some a ∈ Z such that p p p α ≡ a (mod (1 − ξp) ).

× r (ii) Every unit ε ∈ OK is of the form ε = u · ξp with u ∈ R and r ∈ Z. Date: February 27th, 2018. 15.2 Kummer’s Theorem 50

Proof. (i) By the fourth property above, there exists some a ∈ Z such that α = a + (1 − ξp)β for β ∈ OK , so

p−1 X p αp = ap + an((1 − ξ )β)p−n + (1 − ξ )pβp. n p p n=1

p Since p | n for 1 ≤ n ≤ p, and each such p has p − 1 factors of (1 − ξp) by the second property, and the last coming from (1 − ξp), we have that the sum in the p middle vanishes modulo (1 − ξp) , as, of course, does the last term. p p p Hence α ≡ a (mod (1 − ξp) ). × (ii) Let ε ∈ OK ⊂ Z[ξp]. Thus ε = f(ξp) for some f(x) ∈ Z[x]. n For each i ≤ n ≤ p − 1, let εn = f(ξp ) ∈ OK , calling ε = ε1. n Since ξp is a Galois conjugate of ξp, this εn is a Galois conjugate of ε, since × embeddings are homomorphisms, and so εn ∈ OK . n −n Note that the complex conjugate εn is precisely εn = εp−n since ξp = ξp = p−n ξp . × Therefore εn/εp−n ∈ OK , and |εn/εp−n| = 1, the Galois conjugates of which are εm/εp−m. Hence the roots of the minimal polynomial of εn/εp−n over Q all have absolute value 1, and therefore they must be pth roots of unity. s So ε/εp−1 = ±ξp for some s. Since p is an odd prime, gcd(p, 2) = 1, and so we can ﬁnd some r such that s ≡ 2r (mod p), whence we can write this in turn 2r as ±ξp . Now by the fourth property there exists some a such that ε ≡ a (mod (1 − ξp)), and taking complex conjugates we get ε = εp−1 ≡ a (mod (1 − ξp)). Strictly speaking we should conjugate the ideal in the modulus too, but we have shown before that we get the same ideal. Hence ε ≡ ε (mod (1 − ξp)), and we can write

2r εp−1 ≡ ε ≡ ±ξp εp−1 ≡ ±εp−1 (mod (1 − ξp))

2r since ξp ≡ 1 (mod (1 − ξp)). Let us now decide whether ± should be positive or negative. Suppose εp−1 ≡ −εp−1 (mod (1 − ξp)). This implies 2εp−1 ≡ 0 (mod (1 − ξp)). p−1 Taking norms we get N(1 − ξp) | N(2εp−1), so ±p | ±2 , which is a contadiction. 2r So we must have ε/εp−1 = ξp . −r r −r −r Hence εξp = εp−1ξp = εξp , so εξp is its own complex conjugate, ergo −r r real, so εξp = u ∈ R and hence ε = uξp. Proof of Kummer’s theorem. We should assume x 6≡ y (mod p). To see why, suppose x ≡ y ≡ −z (mod p). Then xp + yp = zp implies −z + (−z) ≡ z (mod p), so 3z ≡ 0 (mod p). But p - xyz, so p | 3, and this is a contradiction for p 6= 3. Secondly, suppose x ≡ y 6≡ −z (mod p). Then xp + yp = zp, which rearranged is xp +(−z)p = (−y)p, so if we rename −z = Y and −y = Z, then x 6≡ Y (mod p). So we may indeed assume x 6≡ y (mod p). We also assume x, y, and z are pairwise coprime, which we have discussed before. KUMMER’S THEOREM, CONTINUED 51

p p p We go about factorising x + y = z over K = Q(ξp), getting

p−1 Y (x + ξiy) = zp, i=0 where we’ve called ξ = ξp. Then as ideals

p−1 Y i p (x + ξ y)OK = (zOK ) . i=0 We’ll prove the theorem in ﬁve steps. i Step 1: The ideals (x + ξ y)OK are pairwise coprime. j i To see this, note that the gcd of (x + ξ y)OK and (x + ξ y)OK must contain (x + xijy) − (x + ξiy) = ξjy(1 − ξi−j), supposing i > j. Hence a common prime factor must necessarily divide ξjy(1 − ξi−j), which means it must divide (y) or (1 − ξi−j), since ξj is a unit. For the latter we have (1 − ξi−j) = (1 − ξ) since

1 − ξi−j ∈ O× . 1 − ξ K

Now if it divides (y), then it must also divide (z) in OK , but this is a contradiciton since gcd(y, z) = 1. On the other hand, if it divides (1 − ξ), which is a prime ideal, then it must be equal to (1 − ξ) by maximality of nonzero prime ideals in Dedekind domains. Hence it implies (1 − ξ) | zOK , taking norms yields N(1 − ξ) | N(z), or in other words ±p | zp, implying p | z, which is a contradiction since p - xyz.

Lecture 16 Kummer’s Theorem, continued

16.1 Proof continued Proof of Kummer’s theorem, continued. Step 2: We show that

i p (xξy) = Ii where Ii is a principal ideal. i p We know from Step 1 that (x + ξ y) = Ii for some ideal Ii ⊂ OK , by prime factorisation of ideals. Now since p - h(K), then there exist a, b ∈ Z such that ap + bh(K) = 1. In Cl(K), we have

ap+bh(K) a p a i a [Ii] = [Ii] = [Ii] p = [Ii ] = [(x + ξ y)] = 1 since h(K) is the order of the class group, and since (x+ξiy) is clearly principal. Hence Ii is principal, because it is in the same ideal class as a principal ideal. i p p Therefore (x + ξ y) = (αi) = (αi ) for some αi ∈ OK . For i = 1, we have p × x + ξy = εα , with α = α1 and ε ∈ OK . Date: March 1st, 2018. 16.1 Proof continued 52

By the lemma, ε = u · ξr with u ∈ R and r ∈ Z, and αp ≡ ap ≡ d (mod (1 − ξ)p). Hence x + ξy ≡ uξrd (mod (1 − ξ)p). Step 3: Let us show that r 6≡ 0, 1 (mod p), and if p > 3, then 2r 6≡ 1 (mod p). −1 p p To show this, note that (1 − ξ ) = (1 − ξ) as ideals in OK since 1 − ξ−1 ∈ O× , 1 − ξ K so they diﬀer by a unit. Next ξ−r(x + ξy) ≡ ud (mod (1 − ξ)p), and taking complex conjugates we have ξ−r(x + ξy) ≡ ξr(x + ξ−1y) (mod (1 − ξ)p), where again we should strictly speaking conjugate the ideal, but by the above they’re the same. Expanding we thus have ξ−rx + ξ1−ry ≡ ξrx + ξr−1y (mod (1 − ξ)p). Therefore if r ≡ 0 (mod p), then ξr ≡ 1, in which case ξy ≡ ξ−1y (mod (1−ξ)p), so y(ξ − ξ−1) ≡ 0 (mod (10ξ)p), but 1 − ξ2 ξ − ξ−1 = −ξ−1(1 − ξ2) = −ξ−1 (1 − ξ) 1 − xi

| {z× } ∈OK so (ξ − ξ−1) = (1 − ξ) as ideals, so y ≡ 0 (mod (1 − ξ)p−1), whence (1 − ξ)p−1 = p−1 p−1 pOK | (y). Taking norms p | y , so p | y, a contradiction. Hence r 6≡ 0 (mod p). If r ≡ 1 (mod p), then ξr ≡ r, and we get x(ξ−1 − ξ) ≡ 0 (mod (1 − ξ)p), so by identical argument p | x, a contradiction. Finally let p > 3. Then if 2r ≡ 1 (mod p), ξ2r ≡ ξ, whence ξ2r−1 = 1. Multiplying the equation we’ve been working with by ξr, then, yields x + ξy = ξ2rx + ξ2r−1y ≡ ξx + y (mod (1 − ξ)p). Hence (x − y) + (y − x)ξ = (x − y)(1 − ξ) ≡ 0 (mod (10ξ)p). p Since (1 − ξ) = (1 − ξ)pOK , this if the same as x − y ≡ 0 (mod p), so x ≡ y (mod p), a contradiction. Step 4: The theorem is true for p > 3. By Step 3, for p > 3, all of ξ−r, ξ1−r, ξr, and ξr−1 are all distinct. Hence they are linearly independent over Z. Therefore xξ−r + yξ1−r − xξr − yξr−1 ≡ 0 (mod (1 − ξ)p), factoring (1 − ξ) we have x ≡ 0 (mod p), a contradiction. Step 5: The theorem is true for all odd p. It only remains to consider p = 3. We must have x3 ≡ ±1 (mod 9) (just consider x = 0, 1, 2), but x = 0 is not an option since xyz 6= 0. Hence x3 + y3 ≡ (±1) + (±1) (mod 9), so it’s ±2 or 0, but z3 can’t be either of those by the same reasoning. LOCAL FIELDS 53

Lecture 17 Local Fields

17.1 Valuations Deﬁnition 17.1.1 (Valuation). Let K be a ﬁeld. A valuation on K is a function |·|: K → R≥0 such that (i) |x| ≥ 0 and |x| = 0 if and only if x = 0,

(ii) |xy| = |x||y|, and (iii) |x + y| ≤ |x| + |y|, i.e. the triangle inequality. Clearly |1| = |−1| = 1 (take x = y = ±1 in (ii)). Similarly |−x| = |x|.

Example 17.1.2. We always have the trivial valuation, namely |x| = 1 for all x ∈ K×, and |0| = 0. In R and C, the usual absolute value is a valuation. We will usually denote it |·|∞ in order to avoid ambiguity. Let K be a number field and let σ : K,→ C be an embedding. Then |x|σ = |σ(x)|∞ is a valuation. (That this is the case follows from σ being a homomorphism, then using the fact that |·|∞ is a valuation on R.) Let K be a number field and let P ⊂ OK be a nonzero prime ideal. For × a x ∈ K , define (x) = P q1q2 ··· , with a ∈ Z, i.e. prime factor the ideal (x) and count the number of times P occurs (hence a = 0 if P isn’t a divisor of (x)). Define ordP (x) = a. Then

− ordP (x) |x|P = NK (P ) is a valuation. In particular, if K = Q, write x = pa · m/n with p - mn, a ∈ Z, then −a |x|p = p is a valuation, namely the so-called p-adic valuation. N

Definition 17.1.3. Two valuations |·|1 and |·|2 on K are equivalent if there c exists some c > 0 such that |x|1 = |x|2 for all x ∈ K. Remark 17.1.4. Let |·| be a valuation on K. We can then define the distance between two points x, y ∈ K as d(x, y) = |x − y|/ Then K is a metric space and hence a topological space. This makes K into a topological field, meaning that K × K → K by (x, y) 7→ x + y and (x, y) 7→ xy are continuous maps, and so are x 7→ −x and x 7→ x−1 (the last one of course on K×). In view of this, equivalent valuations induce the same topology.

It turns out that on Q, the only nontrivial valuations are |·|∞ and |·|p, for p prime. We will work toward proving this.

Proposition 17.1.5. Let |·|1 and |·|2 be two valuations on K with |·|1 nontrivial. Suppose |x|1 < 1 implies |x|2 < 1. Then |·|1 and |·|2 are equivalent.

Proof. First, we claim that if |x|2 < 1, then |x|1 < 1 as well, whereby |x|1 if and only if |x|2.

Date: March 6th, 2018. 17.1 Valuations 54

× Since |·|2 is nontrivial, there exists some y ∈ K such that |y|1 6= 1. We −1 may assume 0 < |y|1 < 1, since otherwise we can just work with y . Then by assumption |y|2 < 1. Now suppose there exists some x such that |x|2 < 1 but |x|1 ≥ 1. Then

y |y| = 1 ≤ |y| < 1 n n 1 x 1 |x|1

n for some suﬃciently big n. Now since |y/x |1 < 1, we must by assumption have n n |y/x |2 < 1, meaning that |y|2 < |x|2 , which goes to 0 as n → ∞. Hence |y|2 < 0, implying y = 0, which is a contradiction since 0 < |y|1 < 1. × B1 Now let z = 1/y. Then |z|1 > 1 and |z|2 > 1. For each x ∈ K , |x|1 = |z|1 B2 and |x|2 = |z|2 for some B1,B2 ∈ R, dependent on x. We claim that B1 = B2. To see this, consider any n/m ∈ Q with n/m < B1. B1 n/m This is equivalent with |x|1 = |z|1 > |z|1 since |z|1 > 1. This in turn is m n −m n equivalent with |x |1 > |z|1 , if and only if |x z |1 < 1. −m n n/m Bt the claim this is equivalent with |x z |2 < 1, if and only if |x|2 > |z|2 , B2 n/m if and only if |z|2 > |z|2 , if and only if B2 > n/m, since |z|2 > 1. Hence B1 = B2 = B. B B c So |x|1 = |z|1 and |x|2 = |z|2 with B depending on x. Now write |z|1 = |z|2 for some c > 0. Then

B c B B c c |x|1 = |z|1 = (|z|2) = (|z|2 ) = |x|2.

Notice how c is independent of x, so |·|1 and |·|2 are equivalent.

Corollary 17.1.6. Two valuations |·|1 and |·|1 are equivalent if and only if they deﬁne the same topology. Proof. The forward direction is trivial since if the valuations are equivalent c |x|1 = |x|2 for some c, so open sets in one will be open sets in the other. The reverse direction requires a little bit more work. Suppose |·|1 and |·|2 induce the same topology. Then

lim xn = 0 n→∞ with respect to |·|1 if and only if it does the same with respect to |·|2. Hence n n |x|1 → 0 in R, and |x|2 → 0 in R, whence |x|1 < 1 if and only if |x|2 < 1, so the valuations are equivalent. Deﬁnition 17.1.7 (Non-archimedean valuation). A valuation |·| is called nonarchimedean if |x + y| ≤ max{ |x|, |y| }, called the strong triangle inequality. Otherwise the valuation is called archimedean. Deﬁnition 17.1.8 (Discrete valuation). A valuation is discrete if there exists δ > 0 such that 1 − δ < |x| < 1 + δ implies |x| = 1. 17.1 Valuations 55

Suppose that |y| = a ∈ R. If a − δ0 < |x| < a + δ0, then δ0 |x| δ0 1 − < < 1 + . a |y| a

Hence if we choose δ0 small enough, so that δ0/a < δ, then this implies |x/y| = 1, i.e. |x| = |y|, so |·| takes discrete values not only around 1, but everywhere. There is a reasonably straight forward condition for a valuation being nonarchimedean: Proposition 17.1.9. The valuation |·| is non-archimedean if and only if |n| ≤ A for some A > 0 for all n ∈ N. Proof. The forward direction is simple. Consider

|2| = |1 + 1| ≤ max{ |1|, |1| } = 1.

By induction we can do the same thing and get |n| ≤ 1 for all n ∈ N. For the opposite direction, take x, y ∈ K with |x| ≥ |y|. Then we have

|x|m|y|n−m ≤ |x|m|x|n−m = |x|n for all 0 ≤ m ≤ n. Then

n n X n X n |x + y|n = |(x + y)n| = xmyn−m ≤ |x|m|y|n−m. m m m=0 m=0 Now the binomial coeﬃcient is an integer, so its valuation is by assumption bounded by A, and the two last factors are bounded by |x|n by the above, so

|x + y|n ≤ A(n + 1)|x|n.

Taking nth roots, |x + y| ≤ A1/n(n + 1)1/n|x| → |x| as n → ∞. Hence |x + y| ≤ |x| = max{ |x|, |y| }, so the valuation is non-archimedean.

Remark 17.1.10. The strong triangle inequality implies that if |x| 6= |y|, then |x + y| = max{ |x|, |y| }. To see this, suppose |x| > |y|. Then

|x| = |(x + y) + (−y)| ≤ max |x + y|, |y|, but |x| > |y|, so |x| ≤ |x+y|. The other inequality is true by the strong triangle inequality itself. OSTROWSKI’S THEOREM 56

Lecture 18 Ostrowski’s Theorem

18.1 Ostrowski’s Theorem

Theorem 18.1.1 (Ostrowski). Every nontrivial valuation on Q is equivalent to one of |·|∞, the usual absolute value, or |·|p, with p prime.

Remark 18.1.2. Note that |·|∞, |·|p are pairwise inequivalent. To see this, ﬁnd some x that violates |x|1 < 1 if and only if |x|2 < 1.

Proof. First, let |·| be a nontrivial non-archimedean valuation on Q. Then for n ∈ N, we have |n| ≤ 1, and there exists a prime p such that |p| < 1 (since otherwise |p| = 1 for all p, making the valuation trivial, since it means that |n| = 1 for all n ∈ Z, so |x| = 1 for all x ∈ Q \{ 0 }). Now consider I = { n ∈ Z | |n| < 1 }. This is an ideal of Z, clearly, since |k| ≤ 1 for all k ∈ Z, so kn ∈ I if n ∈ I, and by the strong triangle inequality n1 + n2 ∈ I if n1, n2 ∈ I. Moreover pZ ⊂ I (Z since |1| = 1, but pZ is prime, hence maximal (we’re in a Dedekind domain and it’s nonzero), so pZ = I. Now for any n ∈ Z, wriet n = mpa with p - m. Then |m| = 1 since a c a m 6∈ pZ = I, but |m| ≤ 1. Hence |n| = |p| , and also |n|p = |p| for some c, −a since |n|p = p . So log|p| c = − , log p c implying |x| = |x|p for every x ∈ Q, so |·| is equivalent to |·|p on Q. Next, let |·| be a nontrivial archimedean valuation on Q. We claim that for all integers n, m > 1, |m|1/ log m = |n|1/ log n. 2 r To see this, write m = a0 + a1n + a2n + ... + arn , with 0 ≤ ai ≤ n − 1 and ar 6= 0, i.e. expand m in base n. Hence nr ≤ m < nr+1 inplies log m r ≤ < r + 1, log n and |ai| = |1 + 1 + ... + 1| ≤ ai|1| = ai < n. By the triangle inequality,

r X log m |m| ≤ |a ||n|i ≤ (r + 1)nN r ≤ + 1 nN log m/ log n. i log n i=0 where N = max{ 1, |n| }. Replace m by mt, take t ∈ N, and then take tth roots, we get t log m 1/t |m| ≤ + 1 n1/tN log m/ log n log n for every t ∈ N. The ﬁrst two terms, the ones depending on t, go to 1 as t → ∞, whence |m| ≤ N log m/ log n.

Date: March 8th, 2018. 18.2 Completions 57

There are now two cases to consider. First, |n| > 1 for all n > 1, in which case N = |n|. Then |m| ≤ |n|log m/ log n whereby |m|1/ log m ≤ |n|1/ log n, where if we switch the roles of m and n and repeat we get equality. Secondly, suppose there exists some n > 1 such that |n| ≤ 1. Fixing this n, we have N = 1, and if we consider all integers m > 1 we get |m| ≤ 1, whence |·| is bounded on integers, which by our earlier proposition means that |·| is non-archimedean, a contradiction. Now let S = |n|1/ log n for some n > 1, which by our claim is independent of n. Then |n| = Slog n, whence we set S = ec, for c = log S. Then c log n c c |n| = e = n = |n|∞, c so |x| = |x|∞ for all x ∈ Q.

18.2 Completions In the discussion that follows we’ll have |·| be a valuation on K.

Deﬁnition 18.2.1 (Cauchy sequence). A sequence { an } is Cauchy if for every ε > 0 there exists some N ∈ N such that |an − am| < ε for every n, m ≥ N. Example 18.2.2. If lim an = a n→∞ in K, i.e. |an − a| → 0 as n → ∞, then { an } is Cauchy. N Deﬁnition 18.2.3 (Complete space). If every Cauchy sequence converges in K, then we say that K is complete with respect to |·|.

Deﬁnition 18.2.4 (Completion). A ﬁeld Kb ⊃ K with valuation |·|0 extending |·| on K is the completion of K with respect to |·| if

(i) Kb is complete with respect to |·|0, and

(ii) K is dense in Kb. The completion exists and is unique. We’ll prove this in two parts. Proof of existence. Let R be the set of all Cauchy sequences in K. Then R is a ring, since termwise operations { an } + { bn } = { an + bn } and { an }{ bn } = { anbn } maintain Cauchy by the triangle inequality. Moreover let M be the set of all null sequences in K, by which we mean sequences { an } such that lim an = 0. n→∞ Then M is a maximal ideal in R, and so we define Kb = R/M, which is a field since M is maximal. For a = { an } + M, define

0 |a| = lim |an| n→∞ OSTROWSKI’S THEOREM CONTINUED 58 on Kb. This limit exists since

|an| − |am| ∞ ≤ |an − am| → 0 as n, m → ∞, where the inequality is just the triangle inequality on R, and the convergence is by { an } being Cauchy in K. We can embed K into Kb by a 7→ { an } + M, where { an } is the constant sequence an = a. Then |·|0 is a valuation on Kb, Kb is complete with respect to |·|0, and K is dense in Kb. For the last one, given a = { an } + M ∈ Kb, take { an } in K and consider embedding each term in the sequence as above. Then this sequence converves to a in Kb.

Proof of uniqueness. Let (K,b |·|) and (Kb 0, |·|0) be two completions of K. Define 0 σ : Kb → Kb in the following way. Given x ∈ Kb, K is dense in Kb, so find { xn } in K such that xn → x in Kb. 0 0 Look at { xn } in K ⊂ Kb . Then xn → y for some y ∈ Kb , since the sequence is Cauchy and K is dense in Kb 0. Hence define σ(x) = y. Then σ is well-defined, i.e. it does not depend on the choice of sequence 0 { xn }, and moreover σ is a K-isomorphism, and |x| = |σ(x)| . Theorem 18.2.5 (Ostrowski). Let K be a number field and let |·| be some archimedean valuation on K. Let Kb be a completion of K with respect to |·|. s Then there exists an isomorphism σ from Kb to R or C satisfying |x| = |σ(x)|∞ for all x ∈ Kb for some fixed s ∈ R, S > 0.

Lecture 19 Ostrowski’s Theorem continued

19.1 Proof of Ostrowki’s Theorem Before we prove Ostrowski’s theorem we will prove the following:

Theorem 19.1.1 (Approximation theorem). Let |·|1, |·|2,..., |·|n be pairwise inequivalent nontrivial valuations on K. Let Ki be the completion of K with respect to |·|i. Consider K1 × K2 × ... × Kn with the product topology, i.e. the basis for the open sets are the open sets in the respective Ki. Deﬁne d: K → K1 × K2 × ... × Kn, the diagonal map deﬁned by x 7→ (x, x, . . . , x). Then d(K) is dense in K1 × K2 × ... × Kn. To prove this we require the following lemma.

Lemma 19.1.2. There exists a z ∈ K such that |z|1 > 1 and |z|i < 1 for all i = 2, 3, . . . , n.

Proof. We prove it by induction on n. For n = 1, since |·|1 and |·|2 are inequivalent, there exists some x ∈ K such that |x|1 < 1 and |x|2 ≥ 1, and similarly there is some y ∈ K such that |y|2 < 1 and |y|1 ≥ 1.

Date: March 20th, 2018. 19.1 Proof of Ostrowki’s Theorem 59

Hence if we take z = y/x we get

|y|1 |z|1 = > 1 |x|1 and |y|2 |z|2 = < 1. |x|2

For n > 2, suppose there exists x ∈ K such that |x|1 > 1 and |x|i < 1 for all i = 2, 3, . . . , n − 1. By the n = 2 case, pairing |·|1 and |·|n, there exists some y ∈ K such that |y|1 > 1 and |y|n < 1. N Now we have two options. If |x|n ≤ 1, then take z = x y, whence |z|1 = N N |x|1 |y|1 > 1 and |z|i = |x|i |y|i < 1 for i = 2, 3, . . . , n − 1, taking N to be N suﬃciently large. Moreover |z|n = |x|n |y|n < 1. N N If, on the other hand, |x|n > 1, we take z = x y/(1 + x ). Then if we consider the limit of xN y 1 + xN as N → ∞, we get the following: with respect to |·|1, it approaches y, and |y|1 > 1. With respect to |·|i, for i = 2, 3, . . . , n − 1, it goes to 0, since |x|i < 1. Finally with respect to |·|n it goes to y, and |y|n < 1.

Proof of Ostrowski’s theorem. By the lemma there exists zi ∈ K such that N N |zi|i > 1 and |zi|j < 1 for i 6= j, for i = 1, 2, . . . , n. Notice how zi /(1 + zi ) approaches 1 with respect to |·|i, but 0 with respect to |·|j, for j 6= i, as N → ∞. Given x1, x2, . . . , xn ∈ K, consider

N X zN y = x i . N i 1 + zN i=1 i

Then yN → xi with respect to |·|i. Hence given (x1, x2, . . . , xn) ∈ K ×K ×...×K ⊂ K1 ×K2 ×...×Kn, we have d(yN ) → (x1, x2, . . . , xn). But K × K × ... × K is dense in K1 × K2 × ... × Kn by deﬁnition of completion, whereby d(K) is dense in K1 × K2 × ... × Kn. The part of this result we really need is the following immediate corollary:

Corollary 19.1.3. Let |·|i, i = 1, 2, . . . , n be pairwise inequivalent valuations on K. Given ε > 0 and a1, a2, . . . , an ∈ K, there exists x ∈ K such that |x − ai|i < ε for all i = 1, 2, . . . , n.

Theorem 19.1.4 (Ostrowski). Let Kb be a field extension of Q that is complete with respect to an archimedian valuation |·|. Then there exists an isomorphism s σ from Kb to R or C satisfying |x| = |σ(x)|∞ for every x ∈ Kb for some fixed s > 0. Note that strictly speaking Ostrowski proved a slightly more general result, without the field being an extension of Q, and just required that it was of characteristic 0. 19.1 Proof of Ostrowki’s Theorem 60

Proof. We prove it in three parts. Step 1: Reduce to the case |·| = |·|∞ on Q. To see that this suffices, suppose |·| is an archimedean valuation on Q. c 1/c Hence |·| = |·|∞ on Q for some c. Now consider |·|1 = |·| on Kb. This is an archimedean valuation, and |·|1 = |·|∞ on Q. s Thus if the theorem is true for |·|1, i.e. |x| = |σ(x)|∞ for all x ∈ Kb, then c cs |x| = |x|1 = |σ(x)|∞ for all x ∈ Kb. Now let Qb be the completion of Q in Kb with respect to |·| = |·|∞ on Q. By the uniqueness of completion, and since completing Q with respect to |·|∞, we must have Qb =∼ R, so there exists some Q-isomorphism σ : Qb → R such that |x| = |σ(x)|∞ for every x ∈ Qb. Without loss of generality we can therefore assume R ⊂ Kb, and |·| = |·|∞ on R. Step 2: It suffices to show for each a ∈ Kb it√ satisfies a quadratic equation over R. This would mean that R ⊂ Kb ⊂ C = R[ −1]. But since [C : R] = 2, there can be no intermediate fields, so Kb = R or Kb = C. Consider the continuous map f : C → R≥0 defined by

2 f(z) = a − (z + z)a + zz , which is then the valuation on Kb. Notice how z + z = 2 Re(z) and zz = |z|2, so they are real. Now since this is a quadratic expression in z, we must have

lim f(z) = ∞. z→∞

Hence f(z) has a minimum, say m. Thus

S = { z ∈ C | f(z) = m }= 6 ∅, and S is bounded and closed in C, whence it is compact. Hence there exists some z0 ∈ S such that |z0|∞ ≥ |z|∞ for every z ∈ S (since continuous functions map compact sets to compact sets). We claim that m = 0. If this is the case we’re done, since it implies

2 a − (z0 + z0)a + z0z0 = 0, a quadratic over R, so a solves a quadratic polynomial over R and hence R ⊂ Kb ⊂ C. Suppose m > 0 and consider

2 g(x) = x − (z0 + z0)x + z0z0 + ε ∈ R[x] for 0 < ε < m. Then g(x) has two roots, z1, z1 ∈ C with z1z1 = z0z0 + ε, 2 2 meaning that |z1|∞ = |z0|∞ + ε. This implies |z1|∞ > |z0|∞, whence z1 6∈ S, wherefore f(z1) > m. On the other hand, ﬁx n ∈ N and consider

2n 2n n n Y Y G(x) = (g(x) − ε) − (−ε) = (x − αi) = (x − αi) ∈ R[x], i=1 i=1 19.1 Proof of Ostrowki’s Theorem 61

where α1, α2, . . . , αn ∈ C are the roots of G(x). Now n n G(z1) = (g(z1) − ε) − (−ε) = 0 so z1 is a root of G(x), say z1 = α1. Next,

2n 2n 2 Y Y 2 G(x) = (x − αi)(x − αi) = (x − (αi + αi)x + αiαi), i=1 i=1 which means that

2n 2 Y 2n−1 2n−1 |G(a)| = f(αi) ≥ f(α1)m = f(z1)m . i=1

From the deﬁnition of G(x) we also have

2 n n n n n n |G(a)| ≤ |a − (z0 + z0)a + z0z0| + |−ε| = f(z0) + ε = m + ε .

2n−1 n n 2 Therefore f(z1)m ≤ (m + ε ) so

f(z ) ε n2 1 ≤ 1 + → 1 m m as n → ∞, meaning that f(z1) ≤ m, which contradicts f(z1) > m. Hence m = 0, as claimed. Step 3: let us show that |·| = |·|∞ on Kb. If Kb = R, we are done. If Kb = C, then we claim that |·| and |·|∞ are equivalent on Kb = C. c Then we would have |·| = |·|∞ on Kb, but |·| = |·|∞ on R, so c = 1. Now for any z = a + bi ∈ C, with a, b ∈ R, we have

|z| = |a + bi| ≤ |a| + |i||b| = |a|∞ + |b|∞ ≤ |z|∞ + |z|∞ whereby |z| ≤ 2|z|∞. Suppose now that |·| and |·|∞ are inequivalent on C. By the approximation theorem, for any ε > 0 there exists some α ∈ C such that |α − 1| < ε and |α|∞ < ε. In other words, use the approximation theorem on C × C and ﬁnd α such that (1, 0) is close to (α, α). Hence

|α| − |1| ∞ ≤ |α − 1| < ε whereby 1 − ε < |α| < 1 + ε. By |z| ≤ 2|z|∞ above, then |α| ≤ 2|α|∞ < ε, but this is a contradiction since with ε small we can’t have |α| close to 1 and 0 simultaneously.

Example 19.1.5. Let K be a number ﬁeld and σ : K,→ C an embedding. Deﬁne a valuation on K by |x| = |σ(x)|∞ for x ∈ K. This is archimedean, so the completion of K with respect to |·| is Kb, which is isomorphic to R if σ(K) ⊂ R, and isomoprhic to C if σ(K) 6⊂ R. N LOCAL FIELDS 62

Lecture 20 Local Fields

20.1 Local Fields In what follows K is any ﬁeld and |·| is a nontrivial valuation on K.

Deﬁnition 20.1.1 (Valuation Ring). The set OK = { x ∈ K | |x| ≤ 1 } is called the valuation ring of K and MK = { x ∈ K | |x| < 1 } is the maximal ideal of OK .

× Lemma 20.1.2. (i) OK is an integrally closed integral domain, OK = { x ∈ K | |x| = 1 }, and K is the ﬁeld of fractions of OK .

(ii) MK is the unique maximal ideal of OK , i.e. OK is a local ring.

(iii) If |·| is discrete, then OK is a principal ideal domain. In fact, OK is a discrete valuation ring, and it is also a Dedekind domain.

Proof. (i) Let us first show that OK is a ring. Let x, y ∈ OK . Then |x| = |−x| ≤ 1, whereby −x ∈ OK . Similarly |xy| = |x||y| ≤ 1 · 1 ≤ 1, so xy ∈ OK , and finally |x + y| ≤ max{ |x|, |y| } ≤ 1, so x + y ∈ OK . Now since this is a subring of a field, it can’t contain any zero divisors, and so is an integral domain. To see that it is integrally closed, suppose x ∈ K is integral over OK . Then

n n−1 x + an−1x + ... + a1x + a0 = 0 with ai ∈ OK . Taking valuations and rearranging,

n n−1 n−2 i j |x| = |an−1x + an−2x + ... + a1x + a0| ≤ max |aix | ≤ |x| 0≤i≤n−1

n−j for some 0 ≤ j ≤ n − 1. Hence |x| ≤ 1, and taking roots |x| ≤ 1, so x ∈ OK . × −1 −1 Moreover x ∈ OK if and only if x, x ∈ OK , if and only if |x|, |x| ≤ 1, if and only if |x| = 1. Finally y ∈ K but y 6∈ OK if and only if |y| > 1, if and only if |1/y| < 1, which implies y ∈ OK , and hence K is the ﬁeld of fractions of OK . (ii) If x, y ∈ MK , then |x − y| ≤ max{ |x|, |y| } < 1, so x − y ∈ MK . Next if x ∈ MK and y ∈ OK , we have |xy| = |x||y| < 1, so xy ∈ MK . Hence MK is an ideal of OK . To see that it is maximal, suppose I ⊂ OK is an ideal but I 6⊂ MK . Then × there exists some x ∈ I \MK , meaning |x| = 1, so x ∈ OK is a unit, so I = OK . Hence MK is the unique maximal ideal of OK . (iii) To see that it is a principal ideal domain if |·| is discrete, let π ∈ MK such that |π| is maximal (which then exists speciﬁcally because |·| is discrete). Suppose α ∈ MK , then |α/π| ≤ 1 and so α/π ∈ OK , whereby α ∈ πOK ⊂ MK for every α ∈ MK . Now since it’s true for all α ∈ MK , we indeed have MK = πOK . For any ideal I ⊂ OK the same argument shows that I is generated by some α ∈ OK , so OK is a principal ideal domain.

Date: March 22nd, 2018. 20.2 p-adic Numbers 63

In fact we can do better: let n ∈ N such that |α/πn| ≤ 1 and |α/πn+1| > 1, i.e. α 1 α 1 < = , πn+1 |π| πn n n n × wherebyh |π| < |α/π | ≤ 1. Thus α/π ∈ OK \MK , so α/π ∈ OK , meaning n × that I = αOK = π OK = MK , so every ideal is some power of the maximal ideal. To see that OK is a Dedekind domain we need to check three things. First, that it is integrally closed. Second, that it is Noetherian. Thirdly, that it is of dimension 1. That it is integrally closed we’ve already established. To be Noetherian is equivalent with having ﬁnitely generated ideals, and since OK is a principal ideal domain all its ideals are generated by exactly one element. Finalyl dimension 1 means that all nonzero prime ideals are maximal, and in our case the only nonzero prime ideal is MK , which is maximal.

Definition 20.1.3. The residue field of K is defined as k = OK /MK . If |·| is discrete, then any π ∈ OK such that MK = πOK is called a uniformiser of OK . Definition 20.1.4 (Local field). A local field is a field which is complete with respect to a discrete non-archimedean valuation such that the residue field is finite. Remark 20.1.5. Some books instead base the above definition on the topology, i.e. they take K to be a complete locally compact field, in which case the definition inclues R and C, which the above does not.

Example 20.1.6. Let K = Fp((x)) be the ring of formal Laurent series, i.e. the field of fractions of Fp[[x]], the ring of formal power series over Fp, with Fp being the finite field of p elements. Define a valuation on K by |u| = c−r where c > 1 r and u = x (a0 + a1x + ...) and a0 6= 0, i.e. r is the order of the series. Then K is a local field with OK = Fp[[x]] and MK = (x), and k = OK /MK = Fp. N

Example 20.1.7. Let K = Qp, with p prime. This is a local ﬁeld with OK = Zp, the p-adic integers, and MK = pZp along with

OK Zp ∼ Z k = = = = Fp. N MK pZp pZ We won’t prove the following theorme because it requires delving deeper into the formal Laurent series, but for the record note that

Theorem 20.1.8. The local fields are preciselt the finite extensions of fields Qp or Fp((x)), with p a prime.

20.2 p-adic Numbers The p-adic numbers were introduced by Hensel (1861–1941). For each prime p ∈ Z, we deﬁne a valuation |·|p on Q in the following way. For a ∈ Q, write a = pm · b/c where m ∈ Z and p - bc. Then we deﬁne −m |a|p = p . 20.2 p-adic Numbers 64

This valuation |·|p is a nonarchimedean valuation on Q. Note that the valuation ring is

b { x ∈ | |x|p ≤ 1 } = ∈ p c = (p), Q c Q - Z i.e. the localisation of Z at P = (p). Let Qp be the completion of Q with respect to |·|p. Since this means that Qp is composed of equivalence classes of Cauchy sequences in Q, we ask ourselves what the Cauchy sequences in Q with respect to |·|p are. For this we consider the p-adic expansion. First, take f ∈ N. Then f can be written as 2 n f = a0 + a1p + a2p + ... + anp with 0 ≤ ai ≤ p − 1, i.e. we’ve written f in base p. If am 6= 0 and ai = 0 for all −m i < m, i.e. m is the index of the ﬁrst nonzero coeﬃcient, then |f|p = p . As m → ∞, this goes to 0. Hence for 0 ≤ ai ≤ p − 1,

N ∞ n X n o anp N=1 n=0 is a Cauchy sequence in Q. Hence this sequence converges in Qp. In order to extend this to negative integers, we need a p-adic expansion of −1. To accomplishthis, note that

−1 = (p − 1) − p = (p − 1) + (p − 1)p − p2 ∞ X = (p − 1) + (p − 1)p + (p − 1)p2 − p3 = ... = (p − 1)pn n=0 converges in Qp. In this way we can express all integers as p-adic expansions. For a = pm · b/c ∈ Q, m ∈ Z and p - bc with c > 0, we need to ﬁnd the p-adic expansion of 1/c, then multiply by the expansion of pmb. Note that since gcd(p, c) = 1, we have p ∈ (Z/cZ)×, so pk ≡ 1 (mod c) for some k ∈ N. Hence 1 − pk = cd for some d ∈ Z, and so 1 d d = = = d(1 + pk + p2k + ...) c cd 1 − pk by geometric series. Hence n o X n Qp = anp m ∈ Z, 0 ≤ an ≤ p − 1 n≥m and Zp = { x ∈ Qp | |x|p ≤ 1 } is the valuation ring of Qp, which means n o X n Zp = anp 0 ≤ an ≤ p − 1 . n≥0 HENSEL’S LEMMA 65

The maximal ideal n o X n pZp = { x ∈ Zp | |x|p < 1 } = anp 0 ≤ an ≤ p − 1 n≥1 and n o × X n Zp = anp 0 ≤ an ≤ p − 1, a0 6= 0 . n≥0 Finally the residue ﬁeld Zp ∼ Z k = = = Fp, pZp pZ whence (Qp, |·|p) is a local ﬁeld.

Lecture 21 Hensel’s Lemma

21.1 Generalisation of p-adic Numbers Let O be any Dedekind domain and P ⊂ O a nonzero prime ideal. Assume k = O/P is a finite field. Let K be the field of fractions of O, and define |·|P on K by

− ordP (x) − ordP (x) |x|P = (#k) = (N(P)) for x 6= 0, where ordP (x) is the power of P that appears in the prime factorisation of the ideal (x). This is a nonarchimedean discrete valuation on K, but it might not be complete.

Example 21.1.1. Take O = Z and K = Q, along with P = pZ. N 2 −1 Take π ∈ P \ P . Then ordP (π) = 1 so |π|P = (#k) . Let R be the set of representatives for O/P, a ﬁnite set. The valuation ring of K with respect to |·|P is

{ x ∈ K | |x|P ≤ 1 } = { x ∈ K | ordP (x) ≥ 0 }. The element π is a uniformiser of this ring, and M = (π) is the maximal ideal. Let KP be the completion of K with respect to |·|P . Then n X o O = r πn r ∈ R KP n n n≥0 and n o X n KP = rnπ rn ∈ R, m ∈ Z . n≥m

In addition MKP = πOKP is the maximal ideal of OKP . The residue ﬁeld is O O KP =∼ = k. MKP P Then KP is called a P-adic ﬁeld.

Remark 21.1.2. Let L be a local field. Set KP = L with P = ML, the unique maximal ideal. Thus every local field is a P-adic field. Date: March 27th, 2018. 21.2 Topology of Local Fields 66

21.2 Topology of Local Fields

Let K be a local ﬁeld and OK = { x | |x| ≤ 1 } its valuation ring.

Lemma 21.2.1. The valuation ring OK is compact.

Proof. First we note that OK is both open and closed. This is because ϕ = |·|: K → R is a continuous map, so

−1 OK = ϕ ([0, 1]) is closed since [0, 1] is closed in R. But on the other hand

−1 OK = ϕ ((−δ, 1 + δ)) for some δ > 0 since |·| is discrete, and this set is open. Secondly, OK is complete as a metric space with respect to |·|. If { xn } ⊂ OK is Cauchy, then xn → x in K. Since OK is closed, we moreover have x ∈ OK . A metrix space X is compact if and only if it is complete and totally bounded, i.e. for any ε > 0 the space X is covered by ﬁnitely many ε-balls. Note that OK has a unique maximal ideal MK = (π) with |π| < 1. Given any ε > 0, take n ∈ N such that |π|n < ε. Then O K n n = (#k) < ∞, π OK n and let x1, x2, . . . , xm ∈ OK be the representatives of OK /π OK . We claim that n [ OK = Bε(xi) i=1 with Bε(x) = { a ∈ OK | |a − x| < ε }. n For any a ∈ OK , in the quotient OK /π OK we have

n a = xi + π b for some i and some b ∈ OK . Then

n n |a − xi| = |π |b ≤ |π| < ε so a ∈ Bε(xi). Hence OK is totally bounded, which along with OK being complete hence means that it is compact. Corollary 21.2.2. Let K be a local ﬁeld and A ⊂ K a subset. Then A is compact if and only if A is closed and bounded. That is to say, local ﬁelds behave somewhat like Euclidean spaces. Proof. For the forward direction, A being compact implies A is complete, meaning that it is closed. Moreover |·|: K → R is continuous, so since |A| ⊂ R is compact, hence closed and bounded, |A| < C for some C ∈ R. For the reverse direction, assume A is closed and bounded, i.e. |A| < C for some C > 0. Take any z ∈ K such that |z| > C (this must exist because if 21.3 Hensel’s Lemma 67

|z| > 1 we raise it to some big power, and if |z| < 1 we raise its inverse to some −1 big power). Then |a| < |z| for all a ∈ A, hence |a/z| < 1, i.e. z A ⊂ OK , so z−1A is a closed subset of a compact set, meaning it’s compact, and therefore A is compact. Corollary 21.2.3. Let K be a local ﬁeld. Then K is locally compact (i.e. every element has a compact neighbourhood) and each fractional ideal of K is compact.

21.3 Hensel’s Lemma Recall how if |·| is a nonarchimedean valuations on a ﬁeld K that is complete with respect |·|, we have

|x + y| ≤ max{ |x|, |y| } and if |x| < |y|¡ then |x + y| = |y|. To see this, note that |x| < |y| implies |x/y| < 1, so x/y ∈ MK , whence × 1 + x/y ∈ OK . Hence |1 + x/y| = 1, and multiplying by |y| we get |x + y| = |y|. Theorem 21.3.1 (Hensel’s lemma, first form). Let K be a field complete with respect to a nonarchimedean valuation |·|. Ket f(x) ∈ OK [x]. Suppose there 0 2 exists α0 ∈ OK such that |f(α0)| < |f (α0)| . Then there exists a unique α ∈ OK such that (i) f(α) = 0 and f(α ) 0 (ii) |α − α0| ≤ 0 2 . f (α0) Proof. The proof essentially boils down to Newton’s method from first semester calculus. Consider the sequence

f(αi) αi+1 = αi − 0 f (αi) for i = 0, 1, 2,.... Let f(α ) 0 C = 0 2 < 1 f (α0) by assumption. We will show inductively that

(i) |αi| ≤ 1, i.e. αi ∈ OK ;

(ii) |αi − α0| ≤ C;

f(α ) i i 2 (iii) 0 2 ≤ C . f (αi) The case i = 0 is true by assumptions in the theorem. Assume then that it is true for i, and consider i + 1. We then have

f(α ) f(α ) i i i 2 |αi+1 − αi| = 0 ≤ 0 2 ≤ C < 1 f (αi) f (αi) 21.3 Hensel’s Lemma 68

0 0 since f(x) ∈ OK [x] implies f (x) ∈ OK [x], whence f (αi) ∈ OK and so 0 0 |f (αi)| ≤ 1 and hence 1/|f (αi)| ≥ 1. Therefore

|αi+1| ≤ max{ |αi+1 − αi|, |αi| } ≤ 1, ticking oﬀ the ﬁrst property. For the second property,

|αi+1 − α0| ≤ max{ |αi+1 − αi|, |αi − α0| } ≤ C

2i since the ﬁrst is bounded by C ≤ C by what we did above, and |αi − α0| ≤ C by our inductive hypothesis. Finally let fj(x) ∈ OK [x] be deﬁne dby

2 k f(x + y) = f(x) + f1(x)y + f2(x)y + ... + fk(x)y ;

0 then f1(x) = f (x)—think Taylor expansions. Hence

2 f(αi) 0 f(αi) f(αi) f(αi+1) = f αi − 0 = f(αi) + f (αi) − 0 + β 0 , f (αi) f (αi) f (αi) where β ∈ OK . The ﬁrst two terms make 0, so f(α ) 2 i |f(αi+1)| ≤ 0 . f (αi)

Now do the same with f1, i.e.

2 ` f1(x + y) = f1(x) + g1(x)y + g2(x)y + ... + g`(x)y , with gi(x) ∈ OK [x]. Then

f(αi) f(αi) f1(αi+1) = f1 αi − 0 = f1(α1) + γ 0 , f (αi) f (αi) with γ ∈ OK . Therefore 0 f (αi+1) f(αi) 0 = 1 + γ 0 2 . f (αi) f (αi) Note that f(α ) i i 2 γ 0 2 ≤ 1 · C < 1, f (αi) so f 0(α ) i+1 0 = 1, f (αi) 0 0 the maximum of the above sum, whereby |f (αi+1)| = |f (αi)|. Finally

f(α ) f(α ) 2 1 f(α ) 2 i i+1 i+1 i i 2 2 2 0 2 ≤ 0 0 2 = 0 2 ≤ (C ) = C . f (αi+1) f (αi) |f (αi) | f (αi) HENSEL’S LEMMA REVISITED 69

Using this we can ﬁnish the proof of the theorem, since from the third property

f(α ) f(α ) i i i 2 |αi+1 − αi| = 0 ≤ 0 2 ≤ C → 0 f (αi) f (αi) as i → ∞ since C < 1. Hence |αi| is Cauchy in OK since |·| is nonarchimedean (meaning that subsequent terms going to zero is enough for Cauchy). Ergo αi → α, with α ∈ OK , and by the second property |α − α0| < C, whence by the third property 0 2 2i |f(αi)| ≤ |f (αi)| C . Taking the limit as i → ∞, this goes to |f(α)| = 0, meaning that f(α) = 0. Corollary 21.3.2. Let K be a ﬁeld complete with respect to a nonarchimedean valuation |·|. Suppose f(x) ∈ OK [x] and α0 ∈ OK such that f(α0) ≡ 0 0 (mod MK ) and f (α0) 6≡ 0 (mod MK ). Then there exists α ∈ OK such that f(α) = 0 and α ≡ α0 (mod MK ).

0 Proof. The assumptions give |f(α0)| < 1 since f(α0) ∈ MK , and |f (α0)| = 1 since it’s in OK but not in MK . So we can apply Hensel’s lemma. This means that the existence of solutions to polynomial equations becomes easy in local ﬁelds—we need only check ﬁnitely many places.

Lecture 22 Hensel’s Lemma revisited

22.1 Second Form of Hensel’s Lemma Theorem 22.1.1 (Hensen’s lemma, second form). Let K be a ﬁeld complete with respect to a nonarchimedean discrete1 valuation |·|. Let f(x) ∈ OK [x] be a monic polynomial and let k = OK /MK be the residue ﬁeld of K. ¯ ¯ Suppose h, y¯ ∈ k[x] such that f(x) ≡ h(x)¯g(x) (mod MK ) in k[x] and ¯ gcd(h, g¯) = 1. Then there exists h(x), g(x) ∈ OK [x] such that f(x) = h(x)g(x) ¯ and h(x) ≡ h(x) (mod MK ) and g(x) ≡ g¯(x) (mod MK ). Moreover g(x) and h(x) are unique up to multiplication by scalar.

Proof. Let MK = (π) (note that OK is a discrete valuation ring and hence a principal ideal domain since |·| is discrete; otherwise we would take π ∈ 2 MK \MK ). We start by proving existence. For n ≥ 0 we will construct, inductively, gn(x), hn(x) ∈ OK [x] with leading coeﬃcients being units such that

n+1 f(x) − gn(x)hn(x) ≡ 0 (mod (π )), ¯ along with gn(x) ≡ g¯(x) (mod π), hn(x) ≡ h(x) (mod π) and gn(x) ≡ gn+1(x) n n (mod π ) and hn(x) ≡ hn−1(x) (mod π ). ¯ For n = 0, let g0(x) and h(x) be any lifts ofg ¯(x) and h(x) in OK [x].

Date: March 29th, 2018. 1The assumption of |·| being discrete is not essential, but simpliﬁes the proof. 22.1 Second Form of Hensel’s Lemma 70

For n > 0, suppose we have gn−1(x) and hn−1(x) constructed. Let f(x) − g (x)h (x) p(x) = n−1 n−1 ∈ O [x] πn K ¯ by the inductive hypothesis. Since gcd(¯g, h) = 1 in k[x] there exists some un(x) and vn(x) in OK [x] such that

un(x)gn−1(x) + vn(x)hn−1(x) ≡ p(x) (mod π). Put n gn(x) = gn−1(x) + π vn(x) ∈ OK [x] and n hn(x) = hn−1(x) + π un(x) ∈ OK [x]. n n Then hn(x) ≡ hn−1(x) (mod π ) and gn(x) ≡ gn−1(x) (mod π ), and gn(x) ≡ ¯ g¯(x) (mod π) and hn(x) ≡ h(x) (mod π) as well. Then

n n f(x) − gn(x)hn(x) = f(x) − gn−1(x) + π vn(x) hn−1(x) + π un(x) n = f(x) − gn−1(x)hn−1(x) −π un(x)gn−1(x) + vn(x)hn−1(x) | {z } = p(x)πn ≡ p(x)πn − πnp(x) ≡ 0 (mod πn+1). since the second part of the middle line is p(x) + πw(x) for some w(x) ∈ OK [x]. Hence k |gn(x) − gn−1(x)| < |π| → 0 and k |hn(x) − hn−1(x)| < |π| → 0 as n → ∞, meaning that { gn(x) } and { hn(x) } are Cauchy in OK [x], so they ap- proach some g(x), h(x) ∈ OK [x] such that f(x) = g(x)h(x) and g ≡ g¯ (mod π) and h ≡ h¯ (mod π). Next let us prove uniqueness. Since gcd(¯g, h¯) = 1 in k[x], we have that the ideal (¯g, h¯) = k[x], meaning that

(g, h) + MK OK [x] = OK [x] ¯ for any lifts ofg ¯ and h. Let M = OK [x]/(g, h), which is an OK -module. Note that the Jacobson radical (i.e. the intersection of all maximal ideals) of OK is MK since there is only one maximal ideal. Now MK M = M by the above, whence by Nakayama’s lemma we must have M = 0. Hence (g, h) = OK [x], so any lifts are coprime. Now suppose f(x) = g(x)h(x) = g1(x)h1(x) in OK [x], so that g ≡ g1 ≡ g¯ ¯ (mod π) and h ≡ h1 ≡ h (mod π). Then since any two lifts are coprimne, (g1, h) = OK [x], and so there exists r(x), s(x) ∈ OK [x] such that r(x)g1(x) + s(x)h(x) = 1. Thus h1(x) = h1(x)(r(x)g1(x) + s(x)h(x)) which we can distribute and switch h1(x)g1(x) for h(x)g(x) so

h1(x) = r(x)g(x)h(x) + h1(x)s(x)h(x) = h(x)(r(x)g(x) + h1(x)s(x)) 22.2 Extension of Valuations 71

meaning that h(x) | h1(x) in OK [x]. Repeating the same argument with the roles of h(x) and h1(x) switched we get h1(x) | h(x), so h(x) = uh1(x) for some × unit u ∈ OK . The exact same argument applies to g(x).

22.2 Extension of Valuations Theorem 22.2.1. Suppose K is a field complete with respect to a discrete nonarchimedean valuation |·|. Suppose L is a finite separable extension of K (note that it is unnecessary to assume separability if we are in characteristic 0 since every algebraic extension of a field of characteristic 0 is separable). Then we have

(i) There exists a unique valuation |·|L on L extending |·|, and |·|L is discrete and nonarchimedean.

(ii) The valuation ring OL of |·|L is the integral closure of OK in L. 1/n (iii) For β ∈ L, |β|L = |NL/K (β)| , where n = [L : K].

(iv) L is complete with respect to |·|L. (v) If K is local, then L is local.

Proof. Let π be a uniformiser of K, i.e. MK = (π). Let OL be the integral closure of OK in L. First we note that OL is a Dedekind domain and a ﬁnite rank free OK - module. The proof of this is identical to the proof of OK being the same when K is a number ﬁeld, and bouls down to OK (respectively OL) being a principal ideal domain. Secondly, OL has a unique maximal ideal. To see this, let ML be any maximal ideal in OL. Then since OL is integral over OK , ML ∩ OK is a maximal ideal in OK , so ML ∩ OK = MK . In particular all maximal ideals in OL appear in the factorisation of MK OL, whence

s1 s2 sr MK OL = M1 M2 ···Mr , where Mi are all maximal ideals in OL and si > 0 for all i. Hence r OL M OL = M O Msi K L i=1 i by the Chinese remainder theorem. Suppose r ≥ 2 and let e1 = (1, 0,..., 0) and e2 = (0, 1, 0,..., 0) above. Then choose α ∈ OL such that α 7→ e1 + e2 in the direct sum, and let f(x) ∈ OK [x] be the monic minimal polynomial of α over K. Then we have

∼ 2 OK = OL L OL OK [x] OL si (f(x)) MK OL M i=1 i

x α α α¯ (1, 1)

OK [x] =∼ k[x] (MK ,f(x)) (f¯(x)) x¯ α¯ KRASNER’S LEMMA 72

This implies f¯(x) has nontrivial coprime factors in k[x] since things split at ¯ ¯ the end of the diagram. In other words, f(x) ≡ g¯(x)h(x) (mod MK ) with gcd(¯g, h¯) = 1 in k[x]. By Hensel’s lemma therefore f(x) = g(x)h(x) in OK [x], which is a contradiction since f is irreducible in OK [x], being the minimal polynomial of α over K. Hence r = 1, i.e. there is only one maximal ideal in OL. Thirdly, |·| extends to a discrete valuation on L. To see this, let ML be the unique maximal ideal on OL. By the Chinese remainder theorem any Dedekind domain with a finite number of primes is a principal ideal domain, so ML = (ω). e e ordM (x)/e Suppose MK OL = ML = (ω) . For x ∈ L, define |x|L = |π| L , ordML (x) where MK = πOK . So xOL = ML . This gives a discrete nonarchimedean valuation on L. Moreover |π|L = e/e |π| = |π|, meaning that |·|L agrees with |·| on K. 1/n Fourthly, for β ∈ L, |β|L = |NL/K (β)| . To show this it suffices to show that this is true for β = ω. f Note that NL/K (ω) = π · u, where f = f(ML/MK ) is the inertial degree, × e e and u ∈ OK since (π) = (ω) , with π = ω · u and fe = n. Then πn = N(π) = N(ω)e N(u), × n/e with the norm of u being in OK , so N(ω) = π . Hence

1/n f 1/(fe) 1/e ordM (ω)/e |NL/K (ω)| = |π u| = |π| = |π| L = |ω|L.

Lecture 23 Krasner’s Lemma

23.1 Proof continued We start by ﬁnishing up the proof from last time.

Proof continued. Step ﬁve: Show that OL is the valuation ring of |·|L. r Suppose x ∈ L with |x|L ≤ 1. If x 6∈ OL, then x = ω · u with r < 0 and × u ∈ OL . Then r r/e |x|L = |ω|L = |π| > 1 since |π| < 1 and r < 0. This is a contradiction; |x| can’t simultaneously be less than or equal to 1 and greater than 1. Hence x ∈ OL if and only if |x|L ≤ 1, meaning that x is in the valuation ring of L with respect to |·|L. It remains to show that |·|L is the unique extension of |·| and that L is complete with respect to |·|L. Step six: Show that L is complete with respect to |·|L. Let { αN } be a Cauchy sequence in L, and let e1, e2, . . . , en be an OK -basis of OL (since it is a ﬁnite rank free OK -module), and hence also a K-basis for L. Date: April 3rd, 2018. 23.1 Proof continued 73

Write αN = βN,1e1 + βN,2e2 + ... + βN,nen with βN,i ∈ K. Then for each r > 0 there exists some M ∈ N such that for every N,N 0 ≥ M we have

r |αN − αN 0 | < |π| < 1

r since the sequence if Cauchy. Therefore αN − αN 0 ∈ π OL, meaning that r βN,i − βN 0,i ∈ π OK since the ei are linearly independent. r Hence |βN,i − βN 0,i| < |π| , so { βN,i } is Cauchy in K, and K is complete with respect to |·|, so βN,i → βi in K. Hence let α = β1e1 + β2e2 + ... + βnen and αN → α in L. Step seven: Show that if k·k is an extension of |·| to L, then if x ∈ OL, we × must have kxk ≤ 1. In particular if x ∈ OL , then kxk = 1. To see this, write

OL = OK e1 ⊕ OK e2 ⊕ ... ⊕ OK en.

Take r > 0 such that

r (ke1k + ke2k + ... + kenk) ≤ 2.

Then for x ∈ OL, write x = β1e1 + β2e2 + ... + βnen with βi ∈ OK . Then by the triangle inequality

r r r kxk ≤ (kβ1kke1k+kβ2kke2k+...+kβnkkenk) ≤ (ke1k+ke2k+...+kenk) ≤ 2, since kβ1k = |β1| due to k·k being an extension of |·|, and hence kβ1k ≤ 1, and likewise for β2 and so forth. 1/r Thus kxk ≤ 2 is uniformly bounded, meaning that if there exists x ∈ OL such that kxk > 1, then kxnk → ∞, which would contradict the uniform bound. Hence kxk ≤ 1 for all x ∈ OL, and the unit property follows. Finally, step eight: Show that |·|L is unique. To do this, suppose k·k is an extension of |·| to L alongside |·|L. It suﬃces to show that they agree on ω, a uniformiser for OL, i.e. kωk = |ω|L. e × Write π = ω · u, with u ∈ OL and e = e(ML/MK ). Then

e e kωk = kπk = absπ = |ω|L, meaning that kωk = |ω|L. Corollary 23.1.1. Suppose K is a field complete with respect to a discrete nonarchimedean valuation |·|. If L is any separable algebraic extension of K (not necessarily finite), then |·| extends uniquely to a non-archimedean valuation on L. Proof. Since L is the union of finite extensions of K, we can apply the previous theorem step by step. 23.2 Krasner’s Lemma 74

Remark 23.1.2. The extension of valuations might not be discrete in the inﬁnite extension case, and L might not be complete with respect to the extended valuation in this situation either.

Example 23.1.3. Consider (Qp, |·|p), and let Qp be its algebraic closure. Then (Qp, |·|p) is neither discrete nor complete. N Corollary 23.1.4. Let K be a ﬁeld complete with respect to a discrete nonarchimedean valuation. Let L be a ﬁnite separable extension of K, and let σ : L → Lsep be a K-embedding, where by Lsep = Ksep we mean the separable closure of L. Then for α ∈ L we have

|α|Lsep = |σ(α)|Lsep.

Hence if α ∈ Ksep and σ ∈ Gal(Ksep/K), then

|α|Ksep = |σ(α)|Ksep , so all Galois conjugates have the same valuation.

23.2 Krasner’s Lemma Lemma 23.2.1 (Krasner). Let K be a ﬁeld complete with respect to a discrete non-archimedean valuation |·|. Suppose α, β ∈ Ksep such that

|α − β|Ksep < |α − σ(α)|Ksep for every σ ∈ Gal(Ksep/K) with σ(α) 6= α, i.e. β is closer to α than all of its Galois conjugates. Then K(α) ⊂ K(β). Proof. It suﬃces to show K(α, β) = K(β). Note that we of course know K(α, β) ⊃ K(β) out of the box. Suppose therefore that K(α, β) ) K(β). Then there exists some σ ∈ Gal(Ksep/K) such that σ(β) = β but σ(α) 6= α. Then since Galois conjugates have the same valuation, as shown above, we have

Thus by the strong triangle inequality

|α − σ(α)|Ksep = |(α − β) + (β − σ(α))|Ksep

≤ max{ |α − β|Ksep , |β − σ(α)|Ksep } = |α − β|Ksep , which contradicts the assumption of β being closer to α than all of its Galois conjugates. Hence K(α, β) = K(β). Deﬁnition 23.2.2. Let K be a ﬁeld complete with respect to a discrete nonarchimedean valuation |·|. Let α, β ∈ Ksep. We say that β belongs to α if |α − β|Ksep < |σ(α) − α|Ksep for all σ ∈ Gal(Ksep/K) such that σ(α) 6= α. 23.2 Krasner’s Lemma 75

Proposition 23.2.3. Let K be a ﬁeld complete with respect to a discrete, nonarchimedean valuation |·|. Let

n n−1 f(x) = x + a1x + ... + an−1x + an ∈ K[x] be separable and irreducible. Let α be a root of f(x). Then there exists a constant C(f) > 0 such that if g(x) ∈ K[x] is a monic polynomial wuch that kg(x) − f(x)k ≤ C(f) m m−1 where for h(x) = cmx + cm−1x + ... + c1x + c0 we deﬁne

kh(x)k = max |ci|, 0≤i≤m i.e. the largest coeﬃcient. Then g(x) has a root β that belongs to α. Moreover g(x) is irreducible and deg g = deg f, so K(α) = K(β). Proof. Take C(f) > 0 such that

(i) C(f) < min{ 1, kfk }, where kfk= 6 0 since at least one coeﬃcient, the leading one, is 1. (ii) C(f) < min { kfk−n|σ(α) − α|n }. σ(α)6=α Since C(f) < 1 and f and g are both monic, we must have deg f = deg g since otherwise the biggest coeﬃcients in f −g would be 1, making kf(x)−g(x)k = 1. Note that kgk ≤ max{ kfk, kf − gk } ≤ kfk. Write n n−1 g(x) = x + b1x + ... + bn.

If β0 is any root of g(x), then

n n X n−i n−i n−j |β0| = biβ0 ≤ max|bi|β0 = |bj||β0| i i=1 if we call j the index that maximises this. Then |βj| ≤ kgk ≤ kfk, whence the n−j above is less than or equal to kfk|β0| . Rearranging, we have j |β0| ≤ kfk.

Note moreover that since g(β0) = 0,

m m/j |f(β0)| = |(f − g)(β0)| ≤ C(f) max |β0| ≤ C(f) max kfk 1≤m≤n 1≤m≤n < kfk−n|σ(α) − α|n max kfkm/j ≤ |σ(α) − α|n 1≤m≤n since m/j − n < 0 and kfk ≥ 1. EISENSTEIN EXTENSIONS 76

Lecture 24 Eisenstein Extensions

24.1 Proof continued We ﬁnish oﬀ the proof started last time. Proof continued. Now write

n Y f(x) = (x − αi) i=1 where by αi we mean the Galois conjugates of α = α1. Thus

n Y n |f(β0)| = |β0 − αi| < |σ(α) − α| i=1 for all σ(α) 6= α. Take j such that

|β0 − αj| = min|β0 − αi|, i whence |β0 − αj| < |σ(α) − α| for all σ(α) 6= α. Since f is irreducible, there exists σj such that σj(αj) = α. Set β = σj(β0). Then |β − α| = |σj(β − α)| = |β0 − αj| < |σ(α) − α| for all σ(α) 6= α, meaning that β belongs to α. Hence by Krasner’s lemma K(α) ⊂ K(β), and since

K(β)

K(α)

K with [K(α): K] = deg f = n, we have moreover [K(β): K] ≤ deg g = n, so K(α) = K(β) and hence g(x) is irreducible.

Corollary 24.1.1. Take the same assumptions as in the previous proposition. Then every root of g(x) belongs to exactly one root of f(x), so in particular the roots of g(x) yield the same extensions as the roots of f(x). Proof. Let α be a root of f(x). By Proposition 23.2.3, there exists a root of g(x), say β, that belongs to α. Let β1 = β, β2, . . . , βn be the roots of g(x). sep Since g(x) is irreducible, there exists σi ∈ Gal(K /K) such that σi(β) = βi. Then βi belongs to σi(α), which is a root of f(x).

Date: April 5th, 2018. 24.2 Eisenstein Extension 77

Suppose β is a root of g(x) belonging to two roots of f(x), say α and σ(α) with σ(α) 6= α. Then |α − β| < |σ(α) − α| by deﬁnition, and since Galois conjugates don’t aﬀect valuations, we can apply σ−1 to the right-hand side, so

|σ(α) − β| < |σ−1(σ(α)) − σ(α)| = |σ(α) − α| since σ−1(σ(α)) = α 6= σ(α). Hence |σ(α) − α| = |(σ(α) − β) − (β − α)| < |σ(α) − α|, a contradiciton. So roots of f(x) and g(x) are in one-to-one correspondence.

Corollary 24.1.2. Let K be a ﬁnite extension of Qp. Then there exists a ﬁnite extension F of Q contained in K such that K = F · Qp. Proof. We are considering the situation

K = F · Qp

∃ Qp

Suppose K = Qp(α) (this is possible since the ﬁeld has characteristic 0 and is separable, meaning that it must be a simple extension). Let f(x) ∈ Qp[x] be the monic minimal polynomial α over Qp. Now take g(x) ∈ Q[x] such that kf − gk < C(f), so g(x) is irreducible and has a root β such that Qp(β) = Qp(α) = K, since β belongs to α. Set F = Q(β), then F Qp = Qp(β) = K.

24.2 Eisenstein Extension The basis of this discussion is a generalisation of Eisenstein’s criterion for irreducible polynomials. Proposition 24.2.1 (Eisenstein’s criterion). Suppose A is a commutative ring n n−1 with unity. Let f(x) = a0x + a1x + ... + an−1x + an ∈ A[x]. Suppose there exists a prime ideal P ⊂ A such that a0 6∈ P, Ai ∈ P for all i = 1, 2, . . . , n, and 2 an 6∈ P . Then f(x) is irreducible. Moreover, if f(x) is monic and A is integrally closed, then (f(x)) is a prime ideal in A[x].

Compare this to the classical Eisenstein criterion over Q, where instead of a prime ideal P we have a prime number p ∈ Z. In the discussion that follows, let K be a ﬁeld complete with respect to a non-archimedean valuation |·|. Let OK be the valuation ring and MK be the maximal ideal. 24.2 Eisenstein Extension 78

n n−1 Definition 24.2.2 (Eisenstein polynomial). Let f(x) = x + a1x + ... + an−1x + an ∈ OK [x]. If |ai| < 1 (i.e. ai ∈ MK ) for all i = 1, 2, . . . , n, and 2 an 6∈ MK , then f(x) is called an Eisenstein polynomial. Remark 24.2.3. Note that Eisenstein polynomials are necessarily irreducible since they satisfy Eiseinstein’s criterion. Moreover, if |·| is not discrete (meaning that we aren’t a discrete valuation ring and hence might not have a unique maximal ideal), then there are no 2 Eisenstein polynomials since MK = MK . For α ∈ MK , |α| < 1, and since the valuation isn’t discrete there exists { αn } ⊂ MK such that |αn| → 1. Hence |αn| > |α| for n > M large enough. Thus |α/αn| < 1, but α = αn · α/αn is a product of two elements in MK , 2 so α ∈ MK as well. Definition 24.2.4. An extension L of K is Eisenstein if L = K(α) when α is a root of an Eisenstein polynomial. Lemma 24.2.5. Let K be a local field. Then there are only finitely many Eisenstein extensions of K of fixed degree prime to char(K). Proof. Let 2 X = MK × MK × ... × MK × (MK \MK ) be the space of coefficients of Eisenstein polynomials of degree n, where we have n sets multiplied together above. If a = (a1, a2, . . . , an) ∈ X, then

n n−1 fa(x) = x + a1x + ... + an−1x + an ∈ OK [x] is an Eisenstein polynomial and hence irreducible. Crucially, every Eisenstein extension of K of degree n arises from adjoining a root of some fa(x) for a ∈ X, since they give us all Eisenstein polynomials. 0 Since gcd(char(K), n) = 1, fa(x) 6= 0, since the leading coeﬃcient of the formal derivative is n. Hence fa(x) is separable, and thus Proposition 23.2.3 applies since we are deaing with a ﬁnite separable extension. Now for each a ∈ X there exists a neighbourhood Ua ⊂ X of a such that if b ∈ Ua, then the rotos of fa(x) and fb(x) yield the same extensions of K since their roots belong to each other. Now MK is compact (since K is local) and hence

2 2 2 MK = (π ) = { x | |x| ≤ |π| } = { x | |x| < |π| }

2 is both closed and open. Hence MK \MK is closed in MK and therefore compact. Thus X is the product of compact sets and hence is compact in the product topology, so there exists a1, a2,..., am ∈ X such that

m [ X = Uai . i=1 Therefore every Eisenstein extension of K of degree n is of the form K(α) with α a root of fai (x) for i = 1, 2, . . . , m. Hence there are only ﬁnitely many Eisenstein extensions of K of degree n. TOTALLY RAMIFIED EXTENSIONS 79

Lemma 24.2.6. Let K be a ﬁeld complete with respect to a discrete nonarchimedean valuation |·|. Suppose L is a separable Eisenstein extension of K. Then we have

(i) L is totally ramiﬁed over K, i.e. MK = (πK ) is totally ramiﬁed on OL. (ii) If L = K(α) with α a root of an Eisenstein polynomial over K, then α is × a uniformiser of L, i.e. ML = (πL) = (α), so α = πL · u with u ∈ OL .

(iii) If πL is any uniformiser of L, then OL = OK [πL].

Lecture 25 Totally Ramiﬁed Extensions

25.1 Proof of Lemma Proof. Suppose

n n−1 f(x) = x + a1x + ... + an−1x + an ∈ OK [x] is Eisenstein and separable. Let α = α1, α2, . . . αn be the roots of f(x). Then L = K(α). By Corollary 23.1.4, |αi| = |αj| for every i and j, with |·| being the sep extension of the valuation on K to K . Let MK = (πK ). Since n Y f(x) = (x − αi), i=1 we must have n n Y an = (−1) αi, i=1 2 and since f(x) is Eisenstein an ∈ Mk \MK , implying |an| = |πK |. Hence n 1/n |αi| = |πK | for all i, and so |α| = |πK | ¿ e Let πL be a uniformiser of OL, i.e. ML = (πL), and write πK = πL · u with × r × u ∈ OL and e = e(ML/MK ) and α = πL · v with v ∈ OL . Then

1/n r 1/e r r/e |πK | = |α| = |πL| = (|πK | ) = |πK | , implying 1/n = r/e, i.e. e = n · r. But we also have n = e · f, so r = 1 and e = n, so L is a totally ramified extension of K, and α is a uniformiser of OL since r = 1. For the third and final part we want to prove OL = OK [πL]. Since L is a totally ramified extension of K, the inertial degree f(ML/MK ) = 1, so O O K =∼ L MK ML implying OL = OK + πLO=OK [πL] + πLOL. For α ∈ OL, write α = α1 + β1 with α1 ∈ OK [πL] and β1 ∈ πLOL, then write β1 = πL(α2 + β2) with α2 ∈ OK [πL] and β2 ∈ πLOL, and so forth. Hence in the end

2 r−1 r α = α1 + πLα2 + πLα3 + ... + πL αr + πLβr, Date: April 10th, 2018. 25.1 Proof of Lemma 80

k so OL = OK [πL] + πLOL for all k ∈ N. 2 n−1 Now note that { 1, πL, πL, . . . , πL } is an integral basis for L over K, with n = [L : K]. Hence

n−1 m mn OK 3 ∆ = disc(1, πL, . . . , πL ) = πK · u = πL · v

× × with u ∈ OK and v ∈ OL . Note that ∆OL ⊂ OK [πL] (the proof of this is the same as that for the ring of integers being a finite rank Z-module). Take πk with k ≥ mn. We then have k k πLOL ⊂ ∆OL ⊂ OK [πL], so for k large enough πL ⊂ OK [πL], so the last βr term above goes away. Lemma 25.1.1. Let K be a field complete with respect to a discrete, nonarchimedean valuation |·|. Then a finite separable extension L of K is totally ramified if and only of L is an Eisenstein extension of K.

Proof. The reverse direction is the last Lemma. For the forward direction, suppose L is a totally ramiﬁed extension of K, and let n = [L : K]. Let πL be a uniformiser of L. n We claim L = K(πL). To see this, let E = K(πL). Then πK OL = πLOL since L is a totally ramiﬁed extension of K, and therefore

1/n 1/[E:K] |πK | = |πL| ≥ |πK | since [E : K] ≤ n and |πK | < 1. Also

e(ME/MK ) = e = e(ML/MK ) = n since ordM (πL)/e(ME /MK ) |πL| = |πK | E , with ordME (πL) = 1 since πLOE = ME. Now e(ME/MK ) ≤ [E : K], so [E : K] = n, implying L = K(πL). n n−1 Now let f(x) = x + a1x + ... + an−1x + an ∈ OK [x] be the monic minimal polynomial of πL over K. Let πL = π1, π2, . . . , πn be the roots of f(x). Then |πi| = |πj| < 1 for all i and j, an |ai| < 1 by the strong triangle inequality since ai are polynomials in πj. Moreover since

n n Y an = (−1) πi i=1

n × we have |an| = |πL| = |πK |, implying an = πK · u for some u ∈ OK , and 2 an ∈ MK \MK , so f(x) is an Eisenstein polynomial, meaning that L = K(πL) is an Eistenstein extension of K. Corollary 25.1.2. Let K be a local field. Then there exists only finitely many extensions of K of a fixed degree prime to char(K) that are totally ramified.

Proof. By Lemma 25.1.1 totally ramified extensions are the same as Eisenstein extensions, and by Lemma 24.2.5 there are only finitely many Eisenstein extensions of a fixed degree prime to char(K). 25.2 Unramified Extensions 81

25.2 Unramified Extensions Let K be a local field with L a finite separable extension of K. Then L is an unramified extension of K if and only if e(L/K) = e(ML/MK ) = 1 if and only if MK OL = ML if and only if πK is a uniformiser of K implies πK is a uniformiser of L¿

Proposition 25.2.1. Let K be a local field with residue field k = OK /MK . Then there exists a bijection between the set of finite separable unramified extensions L of K and finite extensions ` of k, sending L to `, the residue field of L, satisfying (i) if K0 ↔ k0 and K00 ↔ k00, then K0 ⊂ K00 if and only if k0 ⊂ k00; ∼ (ii) Gal(L/K) = Gal(`/k) with the isomorphism given by σ 7→ σ|OL = σ which sends ML 7→ ML (in general it must send maximal ideal to maximal ideal, but there is only one here). So it induces σ on OL/ML which fixes OK /MK . Before we prove this, note that L being a finite, separable, unramified extension of K implies that L is a Galois extension of K, whence Gal(L/K) in the theorem is well-defined. To see this, take α ∈ OL and let f(x) be the monic minimal polynomial of α over K, so f(x) ∈ OK [x] is irreducible, reducing to f(x) ∈ OK /MK [x] = k[x]. Now deg f ≤ deg f, and in particular since f is monic they are equal, and α being the reduction of α modulo MK is a root of f(x). Then f(x) splits in `[x] since its a finte extension of a finite field, so it is Galois, and hence f(x) splits in L[x] by Hensel’s lemma, so L is a Galois extension of K.

Lecture 26 Unramiﬁed Extensions

26.1 Classifying Unramiﬁed Extensions Proof of Proposition 25.2.1. Let p = char(k). Step one: Suppose gcd(p, m) = 1. The irreducible factors of xm − 1 in k[x] m are the irreducible factors of x − 1 in OK [x] modulo MK . That is,

m e1 e2 er x − 1 = f1(x) f2(x) · ... · fr(x) ∈ k[x] with fi irreducible in k[x] and

m n1 n2 nr x − 1 = g1(x) g2(x) · ... · gr(x) ∈ OK [x] with gi irreducible in OK [x], then for each fi there exists a gi such that fi ≡ gi (mod MK ) and deg fi = deg gi. To prove this it suﬃces to show the reduction of an irreducible factor g(x) m of x − 1 in OK [x] remains irreducible in k[x]. Let g(x) ≡ g(x) (mod MK ) in k[x]. We want to show that g(x) is irreducible in k[x] and deg g = deg g.

Date: April 12th, 2018. 26.1 Classifying Unramiﬁed Extensions 82

m m Note that g(x) | x − 1 in OK [x] implies g(x) | x − 1 in k[x], and the latter is separable in k[x] since gcd(p, m) = 1 implies its derivative mxm−1 6= 0. Hence g(x) is separable. Let f0(x) ∈ k[x] be a monic irreducible factor of g(x). Let f(x) ∈ OK [x] be a monic lift of f0(x), i.e. f(x) ≡ f0(x) (mod MK ). Since the lift is monic we have deg f = deg f0. Let α be any root of f(x) and let L = K(α). Then

[L : K] = deg f(x) = deg f0(x) ≤ deg g(x) ≤ deg g(x).

Moreover g(x) ≡ g(x) (mod MK ), and f(α) = 0 implying f(α) = f0(α) = 0, hence g(α) = 0. Hence g has a root α ≡ α (mod MK ). 0 0 Now g (α) ≡ g (α) 6≡ 0 (mod MK ) since g is separable, and so by Hensel’s lemma g(x) has a zero, say β, in OK , and L ⊃ K(β). Moreover

deg g(x) ≥ [L : K] ≥ [K(β): K] = deg g(x) since g is irreducible, and hence the above degrees are all equal, so deg f0 = deg g = deg g, so g is also irreducible, as desired. Step two: Suppose [` : k] = n. Then there exists a unique unramified extension Kn of K that is finte, separable, and Kn has residue field `. × To prove this, write ` = k(α0) for some α0. Since ` is a cyclic group of m order |`|−1, α0 is a root of x −1 in k[x], where m = |`|−1, and gcd(p, m) = 1. m Let f0(x) be the monic minimal polynomial of x − 1. By step one, there exists m an irreducible factor f(x) ∈ OK [x] of x −1 such that f(x) ≡ f0(x) (mod MK ) and deg f = deg f0. Let α be any root of f(x) and put Kn = K(α). Then [Kn : K] = deg f = deg f0 = [` : k] = n, and remark that

OKn ef = n ≥ f(Kn/K) = dimk . MKn

Let α ≡ α (mod MK ). Then for f0(α) ≡ f(α) (mod MK ) and f(α) = 0. Now [k(α): k] = deg f0 = n, and

OKn dimk = n MKn ∼ implying OKn /MKn = `, so Kn is an unramified extension of K since n = ef = f. Now on to uniqueness. Suppose L is a finite, separable, unramified extension of K such that the residue field of L is `. Since f(x) has a root α in ` = OL/ML, we have f(α) ≡ 0 (mod ML). Hensel’s lemma thus implies f(x) ∈ OK [x] has a root in OL, and since Kn is a Galois extension of K we have Kn ⊂ L, because it means Kn = K(α) must be equal to K(αi) for the Galois conjugates of α. Now L is an unramified extension of K, meaning e = 1, so f = [L : K] whence [L : K] = [` : k] = [Kn : K], 0 00 0 00 implying that Kn = L and moreover k ⊂ k implies K ⊂ K (the opposite inclusion is trivial). ∼ Step three: Let us show that Gal(Kn/K) = Gal(`/k) sending σ 7→ σ. 26.1 Classifying Unramified Extensions 83

Note that |Gal(Kn/K)| = |Gal(`/k)| since their degrees are equal, so it suﬃces to show that the map between these two is injective. Let f(x) and f0(x) be as above. By Hensel’s lemma, any root of f0(x) in ` lifts to a root of f(x) in

OKn . Let α1, α2, . . . , αn be the roots of f(x) and α1, α2,..., αn be the roots of f0(x) such that αi ≡ αi (mod MK ). Suppose σ ∈ Gal(Kn/K) such that σ 6= Id. Then there exists i and j such that σ(αi) = αj with i 6= j, implying σ(αi) = αj 6= αi, so α 6= Id in Gal(`/k). So ker(Gal(Kn/K) 7→ Gal(`/k)) = { Id }, so it is injective, and hence ∼ Gal(Kn/K) = Gal(`/k). Corollary 26.1.1. Let K be a local field. (i) The composition of two finite, separable, unramified extensions of K is unramified. (ii) If L is any separable, algebraic extension of K (possibly of infinite degree), then there exists a unique maximal unramified subextension K ⊂ Lur ⊂ L.

(iii) Given any integer n, there exists a unique unramiﬁed extension Kn of K ∼ of degree n. Moreover Gal(Kn/K) = Z/nZ = Gal(`/k). Proof. We have the following situation

L `1`2

L1 L1L2 L2 `1 `2

K k

Some of this needs to be proven. First, `1`2 is a finite extension of k, so by the previous proposition `1`2/k is in one to one correspondence with some L/K. Now `1 ⊂ `1`2 implies L1 ⊂ L, and similarly `2 ⊂ `1`2 implies L2 ⊂ L¡ so L1L2 ⊂ L. Also L is an unramified extension of K, so L1L2 is as well, because ramification index is multiplicative, so

1 = e(L/K) = e(L/L1L2)e(L1L2/K).

For the second property, take Y Lur = Ei i where Ei are finite, separable, unramified extensions of K. The third property we proved in the course of the last proof. Corollary 26.1.2. Let K be a local field. Then there exists finitely many extensions of K of fixed degree with ramification index that is prime to char(K). UNRAMIFIED EXTENSIONS 84

Proof. Suppose L is such an extension of K of degree n. Then we have [Lur : K] | n and by the previous corollary, for each d | n there exists a unique unramified extension Kd of K. So Lur is in the set of Kd for d | n, and that’s a finite set. Moreover [L : Lur] = e(L/K) which is coprime to char(K), and by Corollary 25.1.2 there are only finitely many totally ramified extensions L of Lur with gcd([L : Lur], p) = 1 for every Lur. Hence there are finitely many L of the sort desired.

Example 26.1.3. There are only finitely many quadratic extensions of Qp (note that char(Qp) = 0, so the coprimality condition holds automatically). Moreiver we have either e = 1 and f = 2, for unramified extensions, or e = 2 and f = 1, for totally ramified ones. √ Compare this to quadratic extensions Q( D) of Q; here there are infinitely many. N

Lecture 27 Unramiﬁed Extensions

27.1 Tamely and Wildly Ramiﬁed Extensions

Let K be a local field with residue field k = OK /MK and let p = char(k). Let L be a finite, separable extension of K, and let πK be a uniformiser of K. Definition 27.1.1. (i) L is an unramified extension of K if e(L/K) = 1, meaning that prime ideals in K remain prime in L.

(ii) L is a tamely ramiﬁed extension of K if e(L/K) 6= 1 and p - e(L/K). (iii) L is a wildly ramiﬁed extension of K if p | e(L/K). What we want to get at with this is the following:

L pr

tamely ramiﬁed

Lur

unramiﬁed K

Lemma 27.1.2. Suppose L is a totally ramiﬁed extension of K. Suppose there × e exists β0 ∈ L such that |β0| = |πK | for some e, p - e. Then there exists β ∈ L and a uniformiser α of K such that βe = α.

e × Proof. Write β0 = πK · u with u ∈ OK . Since L is a totally ramiﬁed extension of K, f(L/K) = 1, i.e. O O L =∼ K MK MK Date: April 17th, 2018. 27.1 Tamely and Wildly Ramiﬁed Extensions 85

so there exists some u0 ∈ OK such that u (mod ML) = u0 (mod MK ). Put α = πK · u0 ∈ OK . Note that

e |β0 − α| = |πK · u − πK · u0| = |πK (u − u0)| < |πK | = |α| since u − u0 ∈ OK , making its valuation bounded by 1. Let β1, β2, . . . , βe be the roots of e e Y f(x) = x − α = (x − βi). i=1 Hence e e Y |α| > |β0 − α| = |f(β0)| = |β0 − βi|, i=1 1/e so there exists some i0 such that |β0 − βi0 | < |α| = |βi0 | since f(βi0 ) = 0 if and only if βe − α = 0. i0

We now claim that K(βi0 ) ⊂ K(β0) ⊂ L (letting β = βi0 ). To prove this, note that p - e means that e ∈ MK , so |e| = 1. Hence

e e−1 e−1 0 Y |βi0 | = |e||βi0 | = |f (βi0 )| = |βi0 − βj|. j=1 j6=i0

Since

|βi0 − βj| ≤ max{ |βi0 |, |βj| } = |βi0 | for every j we have

e e−1 Y e−1 |βi0 | = |βi0 − βj| ≤ |βi0 | , j=1 j6=i0

so |βi0 − βj| = |βi0 | for all j. This means that |β0 − βi0 | < |βi0 − βj| for all j 6= i0, so by Krasner’s lemma K(βi0 ) ⊂ K(β0) ⊂ L whence βi0 ∈ L. Corollary 27.1.3. (i) L is a totally tamely ramiﬁed extension of K if and n only if there exist uniformisers πK of K and πL of L such that πL = πK where n = [L : K] and char(k) = p - n.

(ii) If L1 and L2 are both tamely ramiﬁed extensions of K, then so is L1L2.

n Proof. (i) For the reverse direction, consider f(x) = x −πK . This is irreducible by Eisenstein’s criterion and has a root πL with L = K(πL). This is totally ramified since it’s Eisenstein, and e(L/K) = n, and p - n by assumption. For the forward direction, suppose L is totally tamely ramified over K. Then 0 0e πK OL = πL OK with e = [L : K] = n and p - e. Then by Lemma 27.1.2 there e n e exists a πK in K and πL in L such that πL = πL = πK . Since |πL| = |πK |, πL is a uniformiser of L. 27.1 Tamely and Wildly Ramified Extensions 86

(ii) First we reduce to the case where L1 and L2 are not only tamely ramiﬁed, but totally tamely ramiﬁed. To see this, we have

L1 L2

L1,ur L2,ur

unramiﬁed unramiﬁed K K and consider F = L1L2

Fur

K Then F = L1L2

L1Fur L2Fur

totally ramiﬁed totally ramiﬁed Fur

e=1 unramified K 0 0 0 0 0 Call L1 = L1Fur, L2 = L2Fur, so F = L1L2 = L1L2, and treat Fur as K . Then F is tamely ramified over K0, so likewise over K since ramification indices are multiplicative. Hence assume L1 and L2 are totally ramified over K. By (i), there exist uniformisers π ∈ L and π ∈ L and π , π ∈ K such that π[L1:K] = π and L1 1 L2 2 1 2 L1 1 π[L2:K] = π . Then L = K(π ) = K(π1/ei ) ⊂ K(ξ , π1/ei ), and e = [L : K] L2 2 i Li i ei i i i since the extensions are totally ramified.

Now p - ei means that K(ξei )/K is unramiﬁed (cf. the proof of Q(ξei )/Q of 1/ei ei same). Hence K(ξei , πi )/K(ξei ) is totally tamely ramiﬁed since x − πi is Eisenstein and p - ei. The situation then is

1/e1 1/e2 K(ξei , ξe2 , π1 , π2 )

tamely ramiﬁed

L1L2 K(ξe1 , ξe2 )

unramified K where L1L2 is a tamely ramified extension of K since its ramification index divides that of the labelled tamely ramified extension above. 27.2 Decomposition Group and Intertia Group 87

Proposition 27.1.4. Let L be a finte, separable extension of a local field K. Then there exist unique subfields K ⊂ Lur ⊂ Lt ⊂ L such that

(i) Lur is an unramiﬁed extension of K,

(ii) Lt is a totally tamely ramiﬁed extension of K, and

r (iii) L is a degree p extension of Lt for some r, where p = char(k) and L is a totally ramified extension of Lt. Proof. We’ve done (i) before. For (ii), Y Lt = Ei i for every Lur ⊂ Ei ⊂ L being tamely ramified. r 0 0 For (iii), write [L : Lt] = p e with p - e . Let πL be a uniformiser in L, pr e0 and πt a uniformiser in Lt. Since L is totally ramified over L, πtOL = πL OL, r p e0 0 0 so |πL | = |πt|, p - e . By Lemma 27.1.2 there exists β ∈ L and π ∈ Lt, 0 uniformisers, such that βe = π0. (Note that β might not be a uniformiser since 0 e 6= [Lt : L].) e1 0 Now β is a root of f(x) = x − π ∈ OLt [x], which is Eisenstein, so

L pr

Lt(β)

Lur where Lt(β) is tamely ramiﬁed over Lur. But Lt is the maximal tamely ramiﬁed 0 extension of Lur, so Lt = Lt(β), i.e. e = 1.

27.2 Decomposition Group and Intertia Group Let L and K be number fields with [L : K] = n, and assume L is a Galois e extension of K. Let P ⊂ OK be a prime ideal and P OL = (Q1Q2 ··· Qr) (the extension being Galois implies that they all have the same ramification index), where e = e(Qi/P ) and f = f(Qi/P ), and ref = n. The Galois group Gal(L/K) acts transitively on Qj. Fix Q = Q1, say. Then for each Q lying above P we define the decomposition group

D = DP = D(Q/P ) = { σ ∈ Gal(L/K) | σ(Q) = Q } and the intertia group

I = IP = I(Q/P ) = { σ ∈ Gal(L/K) | σ(α) ≡ α (mod Q) for all α ∈ OL }. DECOMPOSITION AND INERTIA GROUP 88

Remark 27.2.1. We clearly have I ≤ D ≤ Gal(L/K) are subgroups, and for σ ∈ D we have σ OL OL

σ¯ OL/Q OL/Q so σ inducesσ ¯ whereσ ¯ ∈ Gal(`/k), with ` = OL/Q and k = OK /P .

Lecture 28 Decomposition and Inertia Group

28.1 Decomposition and Inertia From last time we have the diagram

σ OL OL

σ¯ OL/Q OL/Q = `

OK /P OK /P = k This gives us the group homomorphism ψ : D → Gal(`/k) by σ 7→ σ¯, which has ker ψ = I since I by deﬁnition are those elements of Gal(L/K) that act as the identity modulo Q on OL. This in fact gives rise to the short exact sequence

ψ 1 I D Gal(`/k) 1.

The last step in the sequence, the surjectivity of ψ, is nontrivial, and we shall endeavour to prove it in a moment. In the discussion that follows we will, for a subgrou H ≤ G = Gal(L/K), refer to LH = { x ∈ L | σ(x) = x for all σ ∈ H } as the ﬁxed ﬁeld of H.

Definition 28.1.1. The field LD is the decomposition field and LI is the inertia field. Set QD = Q ∩ LD and QI = Q ∩ LI . Theorem 28.1.2. With the notation and conditions as above, we have

{ 1 } L Q ramiﬁcation index inertial degree

≤ e e 1

I LI QI ≤ f 1 f

D LD QD ≤ r 1 1 G K P Date: April 19th, 2018. 28.1 Decomposition and Inertia 89

Proof. We’ll prove these indices and degrees one at a time. Step one: [LD : K] = r. By Galois theory, [LD : K] = [G : D]. Let G/D denote the set of left cosets of D (D might not be a normal subgroup, so this might not be a proper quotient group) and define ϕ: G/D → { Q1 = Q, Q2,...,Qr }, the image being the set of primes lying above P , by ϕ(σD) = σ(Q). This is well-defined because if σ1D = σ2D, then σ1(Q) = σ2(ψ(Q)), but ψ fixes Q, and moreover it’s surjective since G acts transitively on primes above P . We claim that ϕ is also injective: if ϕ(σD) = ϕ(τD), then σ(Q) = τ(Q), so τ −1σ ∈ D, so σD = τD. Hence ϕ is bijective, and so |G/D| = r. Step two: e(QD/P ) = 1 and f(QD/P ) = 1. By step one, [LD : K] = r, and this implies [L : LD] = ef since [L : K] = erf. Now D = Gal(L/LD) acts transitively on primes in L lying above QD, but D fixes Q, so Q is the only prime lying above QD, so r(Q/QD) = 1. Hence ef = [L : LD] = e(Q/QD)f(Q/QD) where the first factor is bounded by e and the second bounded by f since ramification indices and intertial degrees are multiplicative. Hence we must have e(Q/QD) = e and f(Q/QD) = f, so e(QD/P ) = f(QD/P ) = 1. Step three: f(Q/QI ) = 1 and therefore f(QI /QD) = f (since their product is f). ∼ To see this, note that f(Q/QI ) = 1 if and only if ` = OL/Q = OLI /QI = kI , so it suffices to show Gal(`/kI ) = { Id }. For any θ ∈ `, let α ∈ OL correspond to θ, so α (mod Q) = θ. Then Y g(x) = (x − σ(α)) ∈ OLI [x].

σ∈Gal(L/LI )

m Modulo Q this isg ¯(x) ∈ kI [x]. Since σ(α) ≡ α (mod Q),g ¯(x) = (x − θ) , with m = |Gal(L/LI )|. Hence the only Galois conjugate of θ is θ itself, so for τ ∈ Gal(`/kI ), τ(θ) = θ for all θ ∈ `, so θ = Id. Step four: [LI : LD] = f and hence e(QI /QD) = 1, e(Q/QI ) = e, and [L : LI ] = e. Now ψ : D → Gal(`/k) by σ 7→ σ¯ has ker ψ = I. We don’t know they if this is surjective, so all we know so far is

|D/I| ≤ |Gal(`/k)| = [` : k] = f, the last equality by deﬁnition. On the other hand,

|D/I| = [LI : LD] ≥ e(QI /QD)f(QI /QD), but f(QI /QD) = f from above, so [LI : LD] = f, whence e(QI /QD) = 1. By the sizes of the sets in the very last step we then ﬁnally conﬁrm Corollary 28.1.3. We have the following short exact sequence

ψ 1 I D Gal(`/k) 1 28.1 Decomposition and Inertia 90

since ψ is surjective. Corollary 28.1.4. Suppose D is a normal subgroup of G. Then P splits into r distinct primes in LD. 0 If I is also normal in G, then each prime QP ⊂ OLD lying above P remains 0 e prime in LI and QDOL = QL in OL. Proof. For the ﬁrst claim, simply note

0 e0 P OLD = (Q1Q2 ··· Qr) , where by the above results e0 = e and r0 = r. For the second part, LI is a Galois extension of L since I is normal in G, and hence LI is a Galois extension of LD as well. Hence

0 0 0 0 0 0 f = r(QI /QD)e(QI /QD)f(QI /QD)

0 0 0 where r(QI /QD) is the number of primes lying above QD in OLI . But we know 0 0 0 0 0 that e(QI /QD) = 1 and f(QI /QD) = f, so the number of primes above QD 0 0 0 0 e must be exactly 1, whence QDOLI = QI in OLI and QDOL = QI OL = QL in OL. These decomposition and inertia ﬁelds are special, as we will discuss in what 0 0 follows. Let K be an intermediate ﬁeld, i.e. K = LH for some H ≤ G, and

{ 1 } L Q ≤

H K0 P 0 ≤

G K P

With this setup,

0 0 Theorem 28.1.5. (i) LD is the largest intermediate ﬁeld K , K ⊂ K ⊂ L, such that e(P 0/P ) = f(P 0/P ) = 1.

0 0 (ii) LD is the smallest K such that Q is the only prime in OL lying above P .

0 0 (iii) LI is the largest K such that e(P /P ) = 1.

0 0 (iv) LI is the smallest K such that Q is totally ramified over P , in other words e(Q/P 0) = [L : K0]. Definition 28.1.6. Let F and K be number fields, with F an extension of K, and let P ⊂ OK be a prime ideal. We say that P splits completely in F is P splits into [F : K] distinct primes, i.e. e(Q/P ) = f(Q/P ) = 1 for all Q ⊂ OF lying above P . Corollary 28.1.7. Assume D = D(Q/P ) is normal in Gal(L/K). Then P 0 0 0 splits completely in K , K ⊂ K ⊂ L, if and only if K ⊂ LD. MORE ON RAMIFICATION OF EXTENSIONS 91

Proof. The forward direction is immediate by the maximality of LD in the theorem above. For the reverse direction, D is normal in G means that LD is a Galois ex- 0 tension over K, and so P splits completely in LD and K ⊂ K ⊂ LD. So e(QD/P ) = f(QD/P ) = 1, and we know that each of those can be written 0 0 0 0 as e(QD/P )e(P /P ) and f(QD/P )f(P /P ) respectively, since they are multiplicative, and they too must then be 1.

Lecture 29 More on Ramiﬁcation of Extensions

29.1 Splitting of Primes Theorem 29.1.1. Let K be a number field. Let F and M be two finite extensions of K. Fix a prime P in K. (i) If P is unramified in both F and M, then P is unramified in FM. (ii) If P splits completely in both F and M, then P splits completely in FM. Proof. (i) Let L be anuy normal extension of K such that FM ⊂ L. In other words we have the situation L

F M

0 Let P be any prime in FM lying above P , and let QF and QM be primes above P in F and M, respectively. By assumption e(QF /P ) = 1. Now let I = I(Q/P ) for some Q in L lying over P . Since P is unramified in F , F ∈ LI , the largest fiend with P unramified. Similarl M ∈ LI , so FM ⊂ LI , so P is unramified in FM. (ii) Having e(QF /P ) = f(QF /P ) = q implies F ∈ LD, and likewise M ⊂ 0 0 LD, so FM ⊂ LD, whence e(P /P ) = f(P /P ) = q, meaning that P splits completely in FM. Corollary 29.1.2. Let L/K be number fields. Let P be a prime in K. Then P is unramified (respectively splits completely) in L if and only if P is unramified (respectively splits completely) in the normal closure M of L. Suppose we have a Galois extension L of K, with two different primes Q and Q0 lying above a prime P in K. In other words,

L Q Q0

Galois K P

Date: April 24th, 2018. 29.1 Splitting of Primes 92

0 Then in other words P OL = QQ ··· , and since the Galois group acts transitively on these, there exists some σ ∈ Gal(L/K) such that σ(Q) = Q0. The decomposition and intertia groups are also related; namely D(Q0/P ) = σD(Q/P )σ−1 and I(Q0/P ) = σI(Q/P )σ−1. In particular, if Gal(L/K) is abelian, then D(Q/P ) and I(Q/P ) depend only on P , not Q. Recall that

1 I(Q/P ) D(Q/P ) Gal(`/k) 1 is a short exact sequence. Since the Galois group above is the Galois group of finite fields it is cyclic, and it is generated by the Frobenius automorphism. f Let Fpf be a finite field of order p and let F be a finite field extension of Fpf of degree n. Then Gal(F/Fpf ) = hτi

f where τ(x) = xp is the Frobenius automorphism. Let L/K be number ﬁelds. Assume L is a Galois extension of K, and let P be a prime in K and Q a prime in L lying above P . Assume P is unramiﬁed in L. Then LI = L, so I = Gal(L/LI ) = { 1 } is trivial. So we have an isomorphism ψ : D(Q/P ) → Gal(`/k), σ 7→ σ¯, and in partic- ¯ ¯ N(P ) ular φ 7→ φ, where φ = x , N(P ) = [OK : P ]. Thus D(Q/P ) = hφi, and N(P ) φ(α) ≡ α (mod Q) for all α ∈ OL. Remark 29.1.3. The above φ is the only element in D(Q/P ) with this property. Correspondingly, we call φ = φ(Q | P ) the Frobenius automorphism of Q over P . This has the special property that

φ(σ(Q | P ) = σφ(Q | P )σ−1 for all σ ∈ Gal(L/K). Remark 29.1.4. The conjugacy class of φ(Q | P ) in Gal(L/K) is uniquely de- termined by the unramified prime in P . In particular, if G = Gal(L/K) is abelian, then φ(Q | P ) is uniquely de- termined by P . We have φ = φ(Q | P ) for all primes Q lying over P , and φ(α) ≡ αN(P ) (mod Q) for all Q lying above P . N(P ) Thus φ(α) ≡ α (mod P OL), with P OL = Q1Q2 ··· being the product of all primes above P . Theorem 29.1.5. Let L and K be number fields with L a Galois extension of K. Let P be a prime in K that is unramified in L. Then for each prime Q in L lying above P , there is a unique φ ∈ Gal(L/K) such that φ(α) ≡ αN(P ) (mod Q) for every α ∈ OL. Moreover if Gal(L/K) is abelian, then φ depends only on P and φ(α) ≡ N(P ) α (mod P OL) for all α ∈ OL. Exercise 29.1.6. Let L and K be number fields, with L being a Galois extension of K. Let P be a prime in K unramified in L. Then P splits completely in L if and only if φ(Q | P ) = Id for all Q in L above P . 29.1 Splitting of Primes 93

Theorem 29.1.7. Let K be a number ﬁeld. Let p ∈ Z be a prime. Then p is ramiﬁed in K if and only if p | disc(K).

Proof. We’ve done the forward direction a long time ago. For the opposite direction, assume p | disc(K). Fir an integral basis α1, α2, . . . , αn for OK . Then disc(K) = det(Tr(αiαj)), and consider all integers modulo p. We have det(Tr(αiαj)) = 0 in Fp since p | disc(K), so the row vectors

vi = (Tr(αiα1), Tr(αiα2),..., Tr(αiαn)) for i = 1, 2, . . . , n are linearly dependent over Fp. Hence there exists some mi ∈ Z, not all mi ≡ 0 (mod p), such that

m1v1 + m2v2 + ... + mnvn ≡ 0 (mod p).

The jth coordinate of this is

n n X X mi Tr(αiαn) = Tr miαi αj ≡ 0 (mod p) i=1 i=1 for every j = 1, 2, . . . , n, since trace is linear. Set

n X α = miαi, i=1 then Tr(ααj) ≡ 0 (mod p) for all j, and since αj form an integral basis for OK this is true for all of OK , whence Tr(αOK ) ⊂ pZ. Note that α 6∈ P OK since not all mi ≡ 0 (mod p). Now assume p is unram- iﬁed in K. Then pOK = Q1Q2 ··· Qr, with Qi all distinct primes in K. Thus α 6∈ Qi for some i, say, α 6∈ Q1 = Q, and let L be the normal closure of K over Q. Hence P is unramiﬁed in L by the previous corollary. Let QL be any prime in L lying above Q. Since α 6∈ Q, we also have α 6∈ QL (else α ∈ QL ∩ K = Q).

We claim TrL/Q(αOL) ⊂ pZ. To prove this, note that

TrL/Q(αOL) = TrK/Q(TrL/K (αOL)) ⊂ TrK/Q(αOK ) ⊂ pZ, since TrL/K (αOL) = α TrL/K (OL) ⊂ K. Now ﬁx any β ∈ Qi, i = 2, 3, . . . , r, with β 6∈ Q = Q1. (Such a β exists by the Chinese remainder theorem since Qi are coprime.) We make two claims: For each γ ∈ OL, we have

(i) TrL/Q(αβγ) ∈ Q, (ii) σ(αβγ) ∈ Q for each σ ∈ Gal(L/Q) = G, σ 6∈ D = D(Q/P ).

(i) Since TrL/Q(αOL) ⊂ pZ ⊂ Q, we also have TrL/Q(αβγ) ∈ Q since βγ ∈ OL. (ii) For σ ∈ G \ D, σ−1(Q) 6= Q. Thus β ∈ σ−1(Q), whence σ(β) ∈ Q. Hence σ(αβγ) ∈ Q since σ(αγ)σ(β) ∈ Q since Q is prime. THE DIFFERENT IDEAL 94

From the claim, therefore, X X σ(αβγ) = TrL/Q(αβγ) − σ(αβγ). σ∈D σ∈G\D

If we modulo this by Q, X σ¯(¯αβ¯γ¯) ≡ 0 σ∈D in OL/Q for everyγ ¯ ∈ OL/Q. ¯ Nowα ¯β 6= 0 in OL/Q since α, β 6∈ Q, so X σ¯(x) = 0 σ∈D for every x ∈ OL/Q, whence X σ¯ ≡ 0 σ∈D as a function. Recall that D =∼ Gal(`/k) since p is unramiﬁed. This is a contradiction since distinct Galois objects are linearly independent, so p is ramiﬁed.

Lecture 30 The Diﬀerent Ideal

30.1 Duals

3 Example 30.1.1. Let K = Q(α) with α − α − 1 = 0. Then OK = Z[α]. We have disc(K) = −23, so p = 23 is the only prime ramified in K. 2 Now 23OK = PQ , where P = (23, α − 3) and Q = (23, α − 10). Hence e(Q/p) = 2 > 1, and e(P/p) = 1. We want to detect that Q is the prime with e > 1, in general. N Definition 30.1.2. Let K be a number field of degree n. Then L is called a lattice in K if L is of the form

L = Zα1 + Zα2 + ... + Zαn where { α1, α2, . . . , αn } is a basis for K over Q. Consider the bilinear form K × K → C defined by (x, y) 7→ TrK/Q(xy). This is called the trace product, and is an analogue of the inner product in Rn. Definition 30.1.3. Let L be a lattice in K. The dual lattice of L is defined as ∨ L = { α ∈ K | TrK/Q(αL) ⊂ Z }. Theorem 30.1.4. Let K be a number field. Let L ⊂ K be a lattice with Z-basis α1, α2, . . . , αn. Then

∨ ∨ ∨ ∨ L = Zα1 + Zα2 + ... + Zαn , Date: April 26th, 2018. 30.1 Duals 95

∨ where { αi } is the dual basis of { αi } with respect to the trace product (and is indeed also a basis for K over Q). In particular, L∨ is a lattice. ∨ Here αi is deﬁned by ( ∨ 1, if i = j Tr (α αj) = δij = K/Q i 0, if i 6= j.

Theorem 30.1.5. Let K = Q(α) and let f(x) ∈ Q[x] be the minimal polynomial of α over Q. Write

n−1 f(x) = (x − α)(cn−1(α)x + ... + c1(α)x + c0(α))

2 n−1 with ci(α) ∈ K. Then the dual basis to { 1, α, α , . . . , α } with respect to the trace product is n c (α) c (α) c (α) o 0 , 1 ,..., n−1 . f 0(α) f 0(α) f 0(α)

In particular, if K = Q(α) with α ∈ OK , then

∨ n−1 ∨ Z[α] = (Z · 1 + Zα + ... + Zα ) 1 1 = ( · 1 + α + ... + αn−1) = [α]. f 0(α) Z Z Z f 0(α)Z

Note that this does not say ci(α) = 1, but there exists a transformation of basis that makes it so. Theorem 30.1.6. For any fractional ideal I in K, I∨ is also a fractional ideal ∨ −1 ∨ and I = I OK . ∨ Theorem 30.1.7. The dual lattice OK is the largest fractional ideal in K whose elements all have trace in Z. ∨ Remark 30.1.8. The theorem does not say OK is the set of all elements in K with integral trace. In fact,

{ α ∈ K | TrK/Q(α) ∈ Z } is an additive group, but it is not a fractional ideal. √ √ Example 30.1.9. Take K = Q( −1). Then OK = Z[ −1] and 1 √ O∨ = [ −1], K 2Z but 1 √ { α ∈ K | Tr (α) ∈ } = + −1 K/Q Z 2Z Q is not a fractional ideal. N

Deﬁnition 30.1.10. The diﬀerent ideal DK of K is

∨ −1 ∨ DK = (OK ) = { x ∈ K | xOK ⊂ OK },

−1 ∨ so DK = OK . 30.1 Duals 96

∨ ∨ −1 −1 Remark 30.1.11. Since OK ⊂ OK ,(OK ) ⊂ OK = OK , i.e. DK ⊂ OK . In other words DK is an integral ideal.

Theorem 30.1.12. Let K be a number ﬁeld. Then N(DK ) = [OK : DK ] = |disc(K)|.

Theorem 30.1.13. The prime factors of DK are the primes in K that ramify over Q. More precisely, for any prime ideal P ⊂ Q lying above p ∈ Z with e(P/pZ) = e, if s DK = P Q with gcd(P,Q) = 1, then ( s = e − 1 if p - e s ≥ e if p | e.

So if e(P/pZ) = 1, i.e. P is unramified, then P is not in DK . √ Example 30.1.14. Let K = Q( D) with D square free. Then ( √ [ D], if D 6≡ 1 (mod 4) Z √ OK = 1+ D Z[ 2 ] if D ≡ 1 (mod 4), and ( 4D, if D 6≡ 1 (mod 4) disc(K) = D if D ≡ 1 (mod 4). √ Two cases: if D 6≡ 1√ (mod 4),√ the minimal polynomial of D over Q is f(x) = x2 − D. We have f 0( D) = 2 D, so √ √ ∨ 0 OK = frac1f ( D)Z[ D], √ ∨ −1 so DK = (OK ) = (2 D)OK . √ In the second case, D ≡ 1 (mod√ 4). Let α =√ 1 + D/2. We have f(x) = 2 0 x − x + (1 − D)/4, and f (α) = D, so DK = DOK . Hence ( √ 2 D if D 6≡ 1 (mod 4) DK = √ D if D ≡ 1 (mod 4). and ( 4|D|, if D 6≡ 1 (mod 4) N(DK ) = = |disc(K)|. |D| if D ≡ 1 (mod 4). N Index algebraic inertia field, 88 element, 1 inertia group, 87 integer, 2 inertial degree, 27 integral basis, 16 belongs to, 74 integrally closed, 21 class group, 25 lattice, 38, 94 class number, 25 lies over, 26 complete, 57 lies under, 26 completion, 57 local field, 63 composition field, 18 convex set, 42 measurable set, 42 cyclotomic field, 2 Noetherian ring, 21 decomposition field, 88 norm, 8 decomposition group, 87 relative, 10 Dedekind domain, 21 number field, 1 diagonal map, 58 different ideal, 95 P-adic field, 65 dimension 1, 21 p-adic valuation, 53 discriminant, 11, 18 Pell’s equation, 49 dual lattice, 94 prime intert, 34 Eisenstein extension, 78 split, 34 Eisenstein polynomial, 78 embedding quadratic field complex, 8 imaginary, 2 real, 8 real, 2 extension tamely ramified, 84 ramification index, 26 unramified, 84 ramified, 26 wildly ramified, 84 residue field, 27, 63 ring of integers, 3 field extension root of unity, 2 degree, 1 finite, 1 sequence fixed field, 88 Cauchy, 57 Frobenius automorphism, 92 splits completely, 90 fundamental parallelotope, 39 strong triangle inequality, 54 fundamental system, 49 symmetric set, 42

Galois conjugate, 6 topological ﬁeld, 53 totally bounded, 66 ideal trace, 8 divisibility, 22 relative, 10 fractional, 24 trace product, 94 integral, 24 norm, 28 uniformiser, 63 principal fractional, 25

97 INDEX 98 valuation, 53 archimedean, 54 discrete, 54 equivalence, 53 non-archimedean, 54 valuation ring, 62