Algebraic Number Theory

Home , Adele ring, Different ideal, Field norm, Product ring

Algebraic number theory

笔记整理

Jachin Chen

第 1 章 Valuation Theory 4 1.1 Valuation and valuation ring...... 4 1.2 Discret valuations...... 12 1.3 Complete the valuation field ...... 19 1.4 Absolute values and Berkovich spaces ...... 21 1.5 Extension of valuations ...... 27 1.6 Hensel’s Lemma...... 28 第 2 章 Dedekind domain 34 2.1 *Fractional ideals and ideal class groups...... 34 2.2 Etale´ algebras and lattices...... 36 2.3 Prime decomposition in an order ...... 46 2.4 Norm map of one-dimensional Schemes...... 52 2.5 Ramification in Dedekind domain ...... 59 2.6 *Different and discriminant...... 66 第 3 章 Ramification theory 73 3.1 Extensions of complete DVR ...... 73 3.2 Higher Ramification Groups ...... 80 3.3 *The Theory of Witt Vectors ...... 88 3.4 Structure of complete DVRs ...... 98 3.5 Lubin-Tate formal groups ...... 107 第 4 章 Local and Global 115 4.1 Topological ring and group ...... 115 4.2 Local fields and global fields...... 120 4.3 Adele ring and Idele group ...... 130 4.4 Ideal Class Groups and Unit Groups ...... 140 4.5 Some computations and applications ...... 144 目录 3

第 5 章 Number Fields 145 5.1 *Cyclotomic fields ...... 145 5.2 Dirichlet character and Gauss sum...... 150 5.3 The Minkowski bound ...... 157 5.4 Dirichlet Unit Theorem ...... 166 5.5 Moduli of a number field...... 173 5.6 Some examples and supplements...... 181 第 6 章 Zeta and L-functions 186 6.1 Prime number theorem ...... 186 6.2 The functional equation ...... 193 6.3 *Dirichlet L-functions...... 200 6.4 The analytic class number formula ...... 205 6.5 Dirichlet density and Polar density...... 210 第 7 章 Class field theory 217 7.1 *Ray class groups and ray class fields...... 217 7.2 *Class field theory: ideal-theoretic form...... 224 7.3 *Local class field theory ...... 227 7.4 *Global class field theory ...... 233 7.5 Tate Cohomology ...... 240 1 第 1 章 Valuation Theory

1.1 Valuation and valuation ring

Krull valuations

Let G be an Abelian group. A subset ∆ ⫅ G is said to be an ordering of the group G if:

(1) ∀s1, s2 ∈ ∆, s1 + s2 ∈ ∆; (2) ∀s ∈ ∆, s ∈ ∆ or −s ∈ ∆; (3) ∆ ∩ −∆ = {0}

For the given ordering ∆ of the group G we denote:

a ⩽ b :⇐⇒ b − a ∈ ∆

It is easy to check that the relation ⩽ is a linear ordering on G. We call G an ordered group.

注记. 1. ∆ = {a ∈ G | a ⩾ 0} is an ordering. 2. Ordered groups do not have elements of finite order.

For the ordered group G we define a projective group G ∪ {∞} consistent of the group G with its ordering and a symbol ∞ which satisfies the following conditions:

1. ∀a ∈ G, a < ∞; 2. ∀a ∈ G, a + ∞ = ∞ + a = ∞. 1.1. VALUATION AND VALUATION RING 5

♦ 定义 1.1.1. Krull valuation Let F be a field and G ∪ {∞} an ordered projective group. A surjective map v : F → G ∪ {∞} is said to be the Krull valuation when: 1. v(a) = ∞ ⇐⇒ a = 0; 2. v(ab) = v(a) + v(b); 3. v(a + b) ⩾ min{v(a), v(b)}, provided a + b ≠ 0.

注记. 1. v(1) = 0 and v(a−1) = −v(a); 2. If an = 1, then v(a) = 0; v(−1) = 0. 3. v(a) = v(−a).

引理 1.1.1. If v(a) ≠ v(b), v(a + b) = min{v(a), v(b)}.

Proof. We may assume that v(a) < v(b). Suppose that v(a + b) ≠ min{v(a), v(b)}. This implies v(a + b) > min{v(a), v(b)}, in particular v(a + b) > v(a). Thus:

v(a) = v((a + b)−b) ⩾ min{v(a + b), v(b)} > v(a) which is a contradiction.

Valuation rings

Let F be a field. A ring A ⊆ F is called the valuation ring if: ∀a ∈ F, a ∈ A or a−1 ∈ A. If the field F is not given we shall assume that F is the field of fractions of A. ♦ 定理 1.1.1 Let v : F → G ∪ {∞} be a valuation.

1. The set Av = {a ∈ F | v(a) ⩾ 0} is a valuation ring. We shall call it the valuation ring associated with v.

2. The set mv = {a ∈ F | v(a) > 0} is the only maximal ideal in the ring Av.

In particular, Av is a local ring and κ(v) := Av/mv is a field, which shall be called the residue field of v.

3. The set Uv = {a ∈ F | v(a) = 0} is a group consistent of all units of the

ring Av.

Proof. Easy check. 6 CHAPTER 1. VALUATION THEORY

注记. It is easy to see that R is a valuation ring iff the divisibility relation | is total.

a|b :⇐⇒ ba−1 ∈ R

♦ 定理 1.1.2 Let A be a valuation ring in F . There exists a Krull valuation v : F → G ∪ {∞} s.t. × Av = A, mv = A\A

Proof. Consider the quotient additive group G := F ×/A×. Define the relation by

a + A× ⩽ b + A× :⇐⇒ a|b such that the group G is an ordered abelian group. Define the mapping v : F → G ∪ {∞} by   × a + A if a ≠ 0 v(a) :=  ∞ if a = 0

We shall show that v is a valuation which will be called the canonical valuation. Fix a, b ∈ F . We may assume that v(a) ⩽ v(b). Note that v(1 + ba−1) ⩾ 0, therefore

− v(a + b) = v(a) + v(1 + ba 1) ⩾ v(a) = min{v(a), v(b)}

If v1 : F → G1 ∪ {∞} and v2 : F → G2 ∪ {∞} are two valuations, then we say that they are equivalent, written v1 ≃ v2, if there exists an order preserving group isomorphism g : G1 → G2 s.t. v2 = g ◦ v1 (we take g(∞) = ∞). Clearly such relation is an equivalence and we can state the following result:

The set of all equivalence classes of the relation ≃ is in a bijective correspondence with the family of all valuation rings in F .

Proof. Suppose that v1 : F → G1 ∪ {∞} and v2 : F → G2 ∪ {∞} are equivalent. ⩾ ⩾ Then v1(a) 0 iff v2(a) 0, so Av1 = Av2 .

Conversely, let A be a valuation ring and let A = Av for some valuation v : → ∪ {∞} F G . By the previous theorem A = AvA for the cannonical valuation 1.1. VALUATION AND VALUATION RING 7

× vA. We shall show that v ≃ vA. Observe that v|F × : F → G is a surjective × homomorphism and that ker v|F × = A . By the isomorphism theorem GA := × × F /A ≃ G. If g : GA → G is such isomorphism, then it is easy to verify that g preserves order and that v = g ◦ vA.

Rank of valuations

Let G be an ordered abelian group. A subgroup H of G is said to be the isolated subgroup if

∀ h ∈ H, {g ∈ G | 0 ⩽ g ⩽ h} ⊆ H

The set G(G) of all isolated subgroups of G is totally ordered by inclusion. The order type of the set G(G)\{G} is called the rank of G. If G is a value group of some valuation v, then the rank of valuation v is the rank of G. ⇐⇒ −1 ⇐⇒ −1 ∈ × ⇐⇒ Note that v(a) = v(b) v(ab ) = 0 ab Rv a, b associates −1 −1 in Rv, so v v(I) = I for ideal in Rv. On the other hand, vv (H) = H for any subset of G since v is surjective. Therefore we have established a bijection between

Spec(Rv) and the set of isolated subgroups of G by

p 7→ v(Rv\p) −1 H 7→ Rv\v (H+)

♦ 定义 1.1.2. Rank of valuation

The rank of valuation v is defined by rank of G or dimension of Rv.

注记. Since Spec(Rv) is ordered by inclusion, dim(Rv) = |Spec(Rv)| − 1.

Clearly v is a valuation of rank 0 iff G is the zero group. Moreover, we have

引理 1.1.2. v : F → G ∪ {∞} has rank less or equal that 1 iff G is Archimedean, that is: ∀ a, b ⩾ 0, ∃ n ∈ N s.t. na ⩾ b ∪ Proof. (Sketch) Suppose that G ≠ 0. Fix a ∈ G and let Ha = {b ∈ G | −na ⩽ n∈N b ⩽ na}. Note that Ha is an isolated subgroup of G. Next we shall show that if H an isolated subgroup of G and a ∈ H, then

Ha ⊆ H. 8 CHAPTER 1. VALUATION THEORY

Thus Ha is the smallest isolated subgroup containing a. This implies that the rank of G is equal to 1 iff:

∀ a > 0,Ha = G which is equivalent to ∀ a, b > 0, ∃ n ∈ N s.t. b ⩽ na.

♦ 命题 1.1.1. Holder Every ordered Archimedean commutative group is isomorphic to some subgroup of the (R, +).

Proof. (Sketch) Fix 0 ⩽ g ∈ G. Define the mapping Φ: G → R by: { } m Φ(h) := inf | mg > nh, m, n ∈ Z n

Since it is obvious that Φ is order-preserving, so we need to check that 1. Φ is well-defined; 2. Φ is a homomorphism; 3. Φ is injective.

The valuation whose value group is a subgroup of (R, +) is called the exponential valuation. For a given valuation v : F → G ∪ {∞} the following four conditions are equivalent: (a) v has rank less or equal than 1;

(b) the valuation ring Av has rank less or equal than 1; (c) G is Archimedean; (d) v is an exponential valuation.

Equivalent characterization of valuation ring

If (A, mA), (B, mB) are local rings contained in a field K, we say B dominates A if

A ⊆ B, and mB ∩ A = mA

引理 1.1.3. If b is any element of an A-algebra B, either b is integral over A, or A[b−1]/(b−1) is not zero. 1.1. VALUATION AND VALUATION RING 9

−1 × Proof. let c = b ∈ Bb, and suppose that the ring A[c]/(c) is zero. Then c ∈ A[c] , n n−1 and hence one can find elements ai of A s.t. c(anc + an−1c + ··· a0) = 1 in the n+1 ring Bb. Multiplying by b , we find that

n n+1 an + an−1b + ··· a0b = b

k k+1 n+k−1 n+k It follows that there exists an integer k ⩾ 1 s.t. anb +an−1b +··· a0b = b in B, proving that b is indeed integral over A.

Let R be an integral domain with fraction field F , the following are equivalent:

1. There is a valuation v of K for which R is the associated valuation ring. 2. For every element a of F , either a or a−1 belongs to R. 3. The set of principal ideals of R is totally ordered by inclusion. 4. The set of ideals of R is totally ordered by inclusion. 5. R is local and every finitely generated ideal of R is principal. 6. R is local, and is maximal among local subrings of K under the partial ordering of domination. 7. There exists an algebraically closed field L and a homomorphism θ : R → L (not necessarily injective) with respect to which R is maximal: if R ⊆ R′ ⊆ F and θ : R′ → L prolongs θ, then R = R′.

Proof. (3 ⇒ 4): Suppose that I and J are ideals of R and I is not contained in J. Choose some a ∈ I\J. Let b ∈ J. Since (a) ̸⊆ (b), (b) ⊆ (a) ⊆ I, it follows that J ⊆ I. (4 ⇒ 5): Since the set of ideals of R is totally ordered, R has a unique maximal ideal, and hence is local. To prove that every finitely generated ideal is principal, it will suffice to show that any ideal which is generated by two elements isinfact principle. But if I = (a, b), either (a) ⊆ (b) or (b) ⊆ (a), so certainly I = (a) or (g). (5 ⇒ 2): Suppose that a, b ∈ R, and let m be the maximal ideal of R, I := (a, b). Then since I is principal, I/mI is a one-dimensional vector space over the field k =: R/m, and hence the images of a and b are linearly dependent. Thus we can find u, v ∈ R s.t. ua + vb ∈ mI, with u, v not both in m. Furthermore, we can find x, y ∈ m s.t. ua + vb = xa + yb, i.e. a(u−x) = b(y−v). Now if for example u is a unit, so is u−x, and we see that a/b = (y−v)/(u−x) ∈ R. 10 CHAPTER 1. VALUATION THEORY

(2 ⇒ 6): Suppose that R ⊆ R′ is local. If x ∈ R′\R, then x−1 ∈ R. But then x−1 ∈ R′, so that x ∈ R′×\R×; this means that R′ does not dominate R. (6 ⇒ 7): Let k be the residue field of R, let k ,→ L be an algebraic closure of k, and let θ : R → L be the composite R → k → L. Suppose that R ⊆ R′ ⊆ F and θ′ : R′ → L extends θ. Let m′ : ker θ′. Then θ′ factors through the localization R” of R′ by θ′, so we may as well replace R′ by R” and assume that R′ is local, with maximal ideal m′. Since θ′ prolongs θ, m maps to m′, so R′ dominates R. Then R = R′, as claimed. (7 ⇒ 2) (non-trival): First we prove a general lemma about ring extensions. Suppose that B is an A-algebra and b ∈ B. Recall that b is integral over A iff the subalgebra A[b] is a finitely generated A-module. (Note that the map A → A[b] −1 might not be injective.) We can also consider the subalgebra A[b ] of Bb. First let us observe that R is a local ring, with maximal ideal the kernel of θ. Indeed, if θ(x) ≠ 0, then θ(x) ∈ L×, and hence θ prolongs to the localization of R by x. By the maximality property of θ, this localization is just R itself, so x ∈ R×. Now suppose that x is any element of K. If x is integral over R then R′ := R[x] is a finite extension of R and the maximal ideal m of R lifts to some maximal ideal m′ of of R′. Then R′/m′ is a finite extension of R/m and since L is algebraically closed, ϕ extends to R′/m′ and hence to R′. Then R′ = R and x ∈ R. On the other hand, if x is not integral over R, the lemma implies that y = x−1 is not a unit of R[y], so there is a maximal ideal m′ of R[y] containing y. Then R → R[y] → R[y]/m′ is surjective, and hence its kernel is a maximal ideal of R, which can only be the kernel of θ. This implies that θ prolongs to R[y], and hence that y ∈ R. In paricular it follows that K is the fraction field of R and that R is a valuation ring of K.

As results, we have the following important theorem.

♦ 定理 1.1.3. Existence If R is a local ring contained in a field K, there exists a valuation ring A of K which dominates R.

Proof. Consider the family of local subrings of K with the partial ordering of domination, it will suffice to show that every chain has an upper bound. If C is such a chain, let V be the union of the elements of C. It is clear that V is a subring of K. Furthermore, notice that if a ∈ S ∈ C, then a is a unit of S iff it is a unit of V . Now it is clear that if x and y elements of V which aren’t units, then x + y is also not a unit, so that V is a local ring, and also that V dominates R. 1.1. VALUATION AND VALUATION RING 11

♦ 定理 1.1.4 ∩ If A is an integral domain contained in a field K, then A = R where R A⊆R is valuation rings over A.

Proof. It is easy to see that a valuation ring is integrally closed in its field of fractions, and hence that A is contained in every valuation ring containing A. If x ∈ K is not integral over A, then let y = x−1; by the lemma we see that A[y]/(y) is not zero. Let n be a maximal ideal of A[y] containing y and let V be a valuation ring of K which dominates A[y]n. Then y belongs to the maximal ideal of V and hence x does not belong to V . 12 CHAPTER 1. VALUATION THEORY

1.2 Discret valuations

An exponential valuation whose value group is a discrete subspace of R (with respect to the usual topology in R) is called the discrete valuation. Let G be a subgroup of R. We shall prove that the following conditions are equivalent: (a) G is a discrete subspace of R; (b) G is not dense in R; (c) {g ∈ G | g > 0} has a minimal element; (d) G = ρ · Z for some ρ > 0.

Proof. (a) ⇒ (b): {{a} | a ∈ G} is a basis of the topology in G and {(b, c) | b, c ∈ Q, b < c} is a basis of the topology in R. On the other hand the topology in G is induced from R, so:

∀ a ∈ G, ∃ b, c ∈ Q, s.t. {a} = (b, c)

So if a ∈ G, then we may pick b ∈ Q s.t. (b, a) ∩ G = ∅. (b) ⇒ (c): suppose that {g ∈ G | g > 0} has no minimal element. We shall show that G is dense in R. Fix (a, b) ⊆ R. We may assume that a, b > 0. Since { ∈ | } ∈ b−a g G g > 0 has no minimal element, we may choose g G s.t. 0 < g < 2 . Since in the group R the Archimedean rule holds, there exists n ∈ Z s.t. ng < b ⩽ (n + 1)g. Observe that ng > a, otherwise, (n + 1)g < a + g < b, a contradiction.

By the Archimedean rule applied for R, there exists n ∈ Z s.t. ng0 ⩽ g < (n+1)g0, so 0 ⩽ g−ng0 < g0. By the choice of g0, g = ng0, so G = g0 · Z. (d) ⇒ (a): observe that for all a ∈ G(a−ρ, a + ρ) ∩ G = {a}. Therefore {{a} | a ∈ G} is a basis for the topology of G.

A discrete valuation is said to be the normalized discrete valuation when its value group is Z.

引理 1.2.1. Let v1, v2 be two discret valuations. Then v1, v2 are equivalent iff ∃ b ∈

R s.t. v1 = bv2.

Proof. Let G1 = ρ1Z,G2 = ρ2Z and ϕ : G1 → G2 be such group isomorphism that v1 = ϕ ◦ v2.

Suppose that ϕ(ρ1) = nρ2 for some n ∈ N\{1}. Since ϕ is an isomorphism, there exists m ∈ N s.t. ϕ(mρ1) = (n−1)ρ2. But ϕ(mρ1) = ϕ(ρ1) + ··· + ϕ(ρ1) = mnρ2. Hence mn = n − 1, that is n(1−m) = 1, so n = 1 and m = 0, a contradiction. 1.2. DISCRET VALUATIONS 13

ρ2 ρ2 Therefore ϕ(mρ1) = mρ2 = mρ1 and taking b = we obtain v1(a) = ρ1 ρ1 ρ2 ϕ(v2(a)) = v2(a). ρ1

推论. Every discrete valuation is equivalent to exactly one normalized discrete valuation.

Equivalent characterization of DVR ∩∞ 引理 1.2.2 (Krull). Let R be a noetherian ring and I an ideal in R. Then In = 0 n=1 iff no element of the set {1−a | a ∈ I} is a zero-divisor.

Proof. Suppose that 1−z, z ∈ I is a zero divisor. Then (1−z)y = 0 for some y ≠ 0. ∩∞ This implies that y = zy = z2y = ··· = zny, n ∈ N, that is y ∈ In = 0. n=1 For the reverse, one can take I = (a1, ··· , ak) for some a1, ··· , ak ∈ R.

♦ 定理 1.2.1 Let R be a local domain, K be its fractional field. Then the following are equivalent: 1. R is a discrete valuation ring in K; 2. R is a noetherian valuation ring in K; 3. R is a PID; 4. R is a noetherian ring and its only maximal ideal is a principal ideal; ∩∞ 5. the only maximal ideal I of R is principal and In = 0. n=1

Valuations of field of rational function

Let R be a UFD, let F be the field of fractions of R and P be the set of representatives of irreducible elements in R. For every P ∈ P the mapping vP : F → R ∪ {∞} given by   ∞ if a = 0 vP (a) := ∏  nQ × nP if a = u Q , u ∈ R Q∈P is a normalized discrete valuation in F . Moreover, 14 CHAPTER 1. VALUATION THEORY

∩ 1. AvP = R(P ), mvP R = (P );

2. ∀a ∈ F, {P ∈ P | vP (a) ≠ 0}. ⊆ 3. R/(P ) = κvP (R) κ(vP ). { } { } { ( ) } s s s s Proof. R = t ∈/ (P ) = P ̸ | t = v ⩾ 0 = A (P ) t t t P t vP ∩

推论. R = AvP P ∈P

♦ 命题 1.2.1 → R ∪ {∞} If R is a PID, then κvP (R) = κ(vP ). Moreover, if v : F is a

non-trivial exponential valuation s.t.R ⊆ Av, then v is equivalent to exactly

one valuation of the form vP for some P ∈ P.

∈ s ∈ Proof. Suppose that R is a PID, fix a + mvP κ(vP ). There exists an element t s ∈ AvP = R(P ), s.t. κvP ( t ) = a + mvP . Since R is a PID and t / (P ), we have that ∈ s s (P, t) = 1, so there exist c, d R s.t. cP + td = 1. Thus t −sd = t (1−dt) = s s ∈ s s t (cp + td−td) = t cp (P ) = mvP . Hence κvP ( t −sd) = κvP ( t )−κvP (sd) = mvP , that s ∈ is κvP ( t ) = κvP (sd) = a + mvP and sd R.

Let v : F → R ∪ {∞} be a non-trivial exponential valuation s.t.R ⊆ Av.

Then mv ∩ R is a prime ideal in the ring R, so mv ∩ R = (P ) for some P ∈ P. Since

R is a PID, mv ∩ R is also maximal and thus non-zero. Let v(P ) = ρ. Since v is ∈ × ∈ × non-trivial, we have that P / Av and hence P / R , which implies that ρ > 0.

Thus v(P ) = ρ = ρvP (P ). Moreover, for a ∈ R\(P ), v(a) = 0 = ρvP (a). Since any ̸ ∈ m a ∈ Z ∈ \ 0 = x F is of the form x = P b , m , a, b R (P ), we have:

m v(x) = v(P ) + v(a)−v(b) = mv(P ) = mρ = ρvP (x) which means that v is equivalent to vP .

As a corollary we shall state the following theorem describing normalized exponential valuations of a field of rational functions. Let F be a field, z a transcendental element over F and let us consider the field F (z). Let P be the set of all irreducible polynomials in the ring R = F [X].

For an arbitrary P ∈ P define the mapping vz,P : F (z) → R ∪ {∞} by:   ∞ if a = 0 vz,P (a) := ∏  nQ × nP if a = u Q(z) , u ∈ R Q∈P 1.2. DISCRET VALUATIONS 15

Define also the function vz,∞ : F (z) → R ∪ {∞} by (we set deg ∞ = 1):   ∞ if a = 0 vz,∞(a) :=  − f(z) ∈ \ deg g deg f if a = g(z) , f, g R 0

♦ 定理 1.2.2

The mapping P 7→ vz,P establishes a bijection between the set P∪{∞} and the set of all normalized exponential valuations of the field F (z) s.t. v(a) = 0 for a ∈ F . In particular, every non-trivial exponential valuation in F (z) is discrete.

Proof. We first observe the trivial fact that vz,∞ is discrete. By the previous results it suffices to show that if v : F (z) → R ∪ {∞} is such exponential valuation that

R ̸⊆ Av, then v is equivalent to the valuation vz,∞. Since R ̸⊆ Av and F ⊆ Av −1 −1 (because v(a) = 0 for a ∈ F ), we have that z ∈/ Av. Thus z ∈ Av. Let v(z ) = ρ. n Obviously ρ > 0 and v(z) = −ρ. Let f = anX + ··· + a0 ∈ R. Then, since k l v(akz ) = −ρk ≠ −ρl = v(alz ) for k ≠ l, we get:

n v(f(z)) = min{v(anz ), ··· , v(a0)} = min{−nρ, ··· , −ρ, 0} = −nρ

f(z) ∈ Thus, when x = g(z) F (z), we obtain: v(x) = v(f(z))−v(g(z)) = −ρ deg f +

ρ deg g = ρvz,∞(x). ∩ 推论. Az,P = F . P ∈P∪{∞}

推论. κ(vz,P ) is a simple extension of the field κvz,P (F ) with degree deg P .

∈ P ≃ Proof. Fix P and observe that F κvz,P (F ). Since R is a PID, we have that

κ(vz,P ) = κvz,P (R) = κvz,P (F )[κvz,P (z)] and κvz,P (P ) is a minimal polynomial for

κvz,P (z). When P = ∞, then ( ) ( ) n n anz + ··· + a0 anz + ··· + a0 ∞ − −1 vz, m = m n = vz ,X m bmz + ··· + b0 bmz + ··· + b0 and the result follows from the previous part of proof and the remark that deg ∞ = 1 = deg X. 16 CHAPTER 1. VALUATION THEORY

Valuations of algebraic function field

The algebraic function field over the field K is a finite extension F |K(t) there t is transcendental over K. The algebraic closure K in F shall be called the field of constans.

引理 1.2.3. Let F be an algebraic function field over K. Then z ∈ F is transcendental over K iff [F : K(t)] < ∞.

Proof. Suppose that F |K(t) is infinite, and z is transcendental over K. Consider the extensions F |K(z, t)|K(t). Then z is algebraic over K(t), therefore there exists f ∈ K(t)[X] s.t. f(z) = 0. Since K(t) = (K[t]), we may suppose that f ∈ K[t][X]. Let F (x, X) ∈ K[x, X] be such polynomial that F (t, X) = f(X). Since t is transcendental over K, we have that F is non-zero. Let g(x) = F (x, z) ∈ K(x)[x]. Since z is transcendental over K, g is non-zero. Moreover, g(t) = F (t, z) = f(z) = 0. So t is algebraic over K(z). Consider the extensions F |K(t, z)|K(z). Since K(z, t)|K(z) is algebraic and finitely generated, it is finite. Since F |K(z, x) is finite, F |K(z) is also finite.

引理 1.2.4. Let F be an algebraic function field over K, let R be such valuation ring in F that K ⊊ R ⊊ F . Then:

× K ⊆ R and K ∩ (R\R ) = {0}

Proof. Fix z ∈ K and suppose that z ∈/ R. Then z−1 ∈ R. Since z is algebraic over −1 K, we have that z is algebraic over K, so for some a1, ··· , ar ∈ K:

−1 r −1 ar(z ) + ··· + a1z + 1 = 0

−1 −1 r−1 −1 r−1 −1 hence −1 = z (ar(z ) +···+a1). Thus z = −(ar(z ) +···+a1) ∈ K(z ) ⊆ R, so z ∈ R, which is a contradiction. Suppose that there exists a ̸= 0 s.t. a ∈ K ∩(R\R×). We have that a−1 ∈ K ⊆ R. Since R\R× is an ideal, we have that aa−1 ∈ R\R×, which is a contradiction.

引理 1.2.5. Let F be an algebraic function field over K, let (R, m) be such valuation ring in F that K ⊊ R ⊊ F . Let 0 ≠ x ∈ m and x1, ··· , xn ∈ m s.t. x1 = x, xi ∈ xi+1m for i ∈ {1, ··· , n − 1}. Then n ⩽ [F : K(x)] < ∞.

Proof. By the previous lemmas x is not algebraic over K, so it is transcendental,

[F : K(x)] < ∞. It remains to show that x1, ··· , xn are linearly independent over 1.2. DISCRET VALUATIONS 17

K(x). Suppose that there exist f1, ··· , fn ∈ K(x) not all zero such that:

f1x1 + ··· + fnxn = 0

The elements fi are rational functions in one indeterminate x. Multiplying both sides of the above equality by the common denominator of f1, ··· , fn we may assume that f1, ··· , fn ∈ K[x]. Eventually dividing by the appropriate power of x we may also assume that not all fi are divisibe by x. Let ai = fi(0) and aj be the last non-zero element in the sequence a1, ··· , an. Then:

−fjxj = f1x1 + ··· + fj−1xj−1 + fj+1xj+1 + ··· + fnxn

Moreover, by the choice of j we get fi = xgi for some gi ∈ K[x], i ∈ {j +

1, ··· , n}. Dividing both sides of the above equality by xj yields:

x1 xj−1 xj+1 xn −fj = f1 + ··· + fj−1 + xgj+1 + ··· + xgn xj xj xj xj

Since x = x1 ∈ m and K ⊆ R, we get f1, ··· , fn ∈ K[x] ⊆ R. Since xi ∈ xi+1m, i ∈ {1, ··· , n−1}, we have in particular xi ∈ xjm for i ∈ {1, ··· , j−1}. Hence xi xi x x1 ∈ m and so fi ∈ m for i ∈ {1, ··· , j−1}. Similarly = ∈ m and since xj xj xj xj x xi, gi ∈ R for i ∈ {j + 1, ··· , n}, we have that gixi ∈ m for i ∈ {j + 1, ··· , n}. xj Therefore all summands on the right side belong to the ideal m, so fj ∈ m.

On the other hand fj = aj +xgj, where gj ∈ K[x] ⊆ R and x ∈ m (so xgj ∈ m).

Thus aj = fj−xgj ∈ m. But also aj ∈ K, aj ≠ 0, which is contradicts the result of the previous lemma.

♦ 定理 1.2.3 Let F be an algebraic function field over K, let (R, m) be such valuation ring in F that K ⊊ R ⊊ F . Then R is a DVR.

Proof. It suffices to show R is a PID.

m is a principal ideal: Suppose not and fix 0 ≠ x1 ∈ m. Then there exist x2 ∈ \ −1 ∈ −1 ∈ ∈ m s.t. m (x1), thus x2x1 / R, so x1x2 m, namely x1 x2m. By induction we may pick the infinite sequence x1, x2, ··· of the elements of m such that xi ∈ xi+1m, which is a contradiction with the previous lemma. Let m = (π), then every element z ∈ F • has a unique representation of the form z = πnu for some n ∈ Z and u ∈ R×: Fix z ∈ F , we may assume that z ∈ R and z ∈/ R×. We get z = πlx for x ∈ R. Observe that the sequence z, tl−1, tl−2, ··· , t satisfies the assumptions of the previous lemma, so its length is bounded. Let 18 CHAPTER 1. VALUATION THEORY k = max{l : z = πlx, x ∈ R}, we may assume that z = πkx. It remains to check whether x ∈ R× is true, otherwise x ∈ m, so x = πy for some y ∈ R, so z = πk+1y, which contradicts the definitoon of k. It is trivial to check that such presentation is unique. Let 0 ≠ I ◁R, then I = (πn): we may assume that I ⊆ m, note that if z ∈ I and z = πku, k > 0, u ∈ R×, then πk = zu−1 ∈ I, so the set {k : πk ∈ I} is non-empty. As a subset of the set of positive integers it has the smallest element, say n. We shall see that I = (πn). Fix z ∈ I and let z = πku. Thus k ⩾ n, that is πk−n ∈ R. Hence z = πku = πnπk−nu ∈ (πn).

Thus we have described almost all valuations in the field F (X). If v : F (X) →

G ∪ {∞} is a non-trivial valuation such that v(a) = 0 for a ∈ F , then F ⊊ Rv ⊊

F (X). So Rv is a DVR in its field of fractions. Since v is non-trivial, we may assume that X ∈ Rv, so F [X] ⊆ Rv and F (X) must be the field of fractions for R. 1.3. COMPLETE THE VALUATION FIELD 19

1.3 Complete the valuation field

Complete discret valuation rings

♦ 命题 1.3.1 Let K be a field with a discrete valuation v, A be the valuation ring of K, b b and π be a uniformizer. The valuation ring of the K is a complete DVR A with uniformizer π, and we have an isomorphism of topological rings A Ab ∼ = lim←− n n π A

b n Proof. We define a ring homomorphism ϕn : A → A/(π ) from which induce a ϕ : Ab → A/(πn) surjective ring homomorphism lim← . Note that ∩ n b ker ϕ = π A = {0} n

b 注记. 1. A is complete because it is closed and therefore contains all its limit b points in the complete field K, it is a DVR with uniformizer π because v b extends to a discrete valuation on A with v(π) = 1. 2. When R is a DVR with maximal ideal m, taking the completion of R with

respect to the absolute value | · |m is the same thing as taking the m-adic completion. This is not true in general. 3. In particular, the m-adic completion of a (not necessarily discrete) valuation ring R with respect to its maximal ideal m need not be complete. The key issue that arises is that the ker ϕ need not be trivial.

例 1.3.1. Let K = Q and v = vp be the p-adic valuation for some prime p. The valuation ring of Q corresponding to vp is the local ring Z(p). Taking π = p as our uniformizer, we get Z ∼ (p) ∼ Z Zd = = = Z (p) lim←− nZ lim←− nZ p n p (p) n p the ring of p-adic integers.

−vt(x) 例 1.3.2. Let K = Fq(t), v = vt be the t-adic valuation and let |x|t := q be the corresponding absolute value. 20 CHAPTER 1. VALUATION THEORY

The completion of Fq(t) with respect to |·|t is isomorphic to the field Fq((t)) of

Laurent series over Fq. The valuation ring of Fq(t) with respect to vt is the local ring

Fq[t](t) consisting of rational functions whose denominators have nonzero constant term. Taking π = t as our uniformizer, we get

F F \ ∼ q[t](t) ∼ q[t] ∼ F [t] = = = F [[t]] q (t) lim←− nF lim←− nF q n t q[t](t) n t q[t] where Fq[[t]] denotes the power series ring over Fq.

Z nZ Z The isomorphism lim←− /p = p gives us a canonical way to represent el- n n ements of Zp: we can write a ∈ Zp as a sequence (an) with an+1 ≡ an mod p , n n−1 where each an ∈ Z/p Z is uniquely represented by an integer in [0, p ].

♦ 定义 1.3.1. p-adic expansion

Let a = (an) be a p-adic integer with each an uniquely represented by n an integer in [0, p −1]. The sequence (b0, b1, b2, ··· ) with b0 = a1 and bn = n (an+1−an)/p is called the p-adic expansion of a.

注记. Every element of Zp has a unique p-adic expansion and every sequence

(b0, b1, b2, ··· ) of integers in [0, p−1] is the p-adic expansion of an element of Zp.

例 1.3.3 (p-adic expansion in Z7). 1. 2 = (2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ··· ) 2. 2002 = (0, 6, 5, 5, 0, 0, 0, 0, 0, 0, ··· ) 3. −2 = (5, 6, 6, 6, 6, 6, 6, 6, 6, 6, ··· ) 2−1 = (4, 3, 3, 3, 3, 3, 3, 3, 3, 3, ··· ) 4. √ 5. 5 2 = (4, 6, 1, 3, 6, 4, 3, 5, 4, 6, ··· ) 1.4. ABSOLUTE VALUES AND BERKOVICH SPACES 21

1.4 Absolute values and Berkovich spaces

Absolute values

♦ 定义 1.4.1

An absolute value on a field k is a map | · | : k → R⩾0 such that for all x, y ∈ k the following hold: 1. |x| = 0 iff x = 0; 2. |xy| = |x||y|; 3. |x + y| ⩽ |x| + |y|. If the stronger condition |x + y| ⩽ max(|x|, |y|) also holds, then the absolute value is nonarchimedean; otherwise it is archimedean.

例 1.4.1 (trivial absolute value). The map | · | : k → R⩾0 defined by   1 if x ≠ 0; | | x :=  0 if x = 0. is the trivial absolute value on k. It is nonarchimedean.

例 1.4.2 (p-adic absolute value on Q). The p-adic absolute value on Q is defined by

−vp(x) |x|p := p where vp is the p-adic valuation.

引理 1.4.1. The following are equivalent:

1. | · | is nonarchimedean; 2. {|n| : n ∈ Z} is bounded; 3. |n| ⩽ |1| = 1 for any n ⩾ 1; 4. | · | is induced by an exponential valuation, | · | = e−v(·).

例 1.4.3. The usual absolute value on Q (inherited from R) denoted by | · |∞ is an archimedean absolute value.

♦ 定理 1.4.1. Ostrowski’s Theorem

Every nontrivial absolute value on Q is equivalent to | · |p for some p ⩽ ∞. 22 CHAPTER 1. VALUATION THEORY

♦ 定理 1.4.2. Product Formula For every x ∈ Q× we have ∏ |x|p = 1 p⩽∞

Topological fields with an absolute value

If K is a field with an absolute value | · |, then it has a natural topology as a metric space under the distance metric d(x, y) := |x−y| induced by the absolute value. The field operations of addition, multiplication and inversion are both continuous, this means that K is a topological field.

Let | · |1 and | · |2 be absolute value over K, we say | · |1 and | · |2 is equivalent ∈ R | · | | · |c if there exists c >0 s.t. 1 = 2. 引理 1.4.2 (Weak approximation). Let K be a field with pairwise inequivalent nontrivial absolute values | · |1, ··· , | · |n. Let a1, ··· , an ∈ K and ϵ1, ··· , ϵn be positive real numbers. Then there exists an x ∈ K s.t.

|x−ai|i < ϵi for 1 ⩽ i ⩽ n

♦ 命题 1.4.1

Let K be a field with absolute values |·|1 and |·|2. The induced topologies

on K coincide ⇐⇒ | · |1 ≃ | · |2 ⇐⇒ |x|1 < 1 ⇔ |x|2 < 1 .

Proof. We assume | · |1 and | · |2 are nontrivial. The field K must then be infinite, and if | · |1 and | · |2 are inequivalent we can use weak approximation to construct a sequence that converges in one topology but not the other.

When the absolute value | | is nonarchimedean the topology it induces has some features:

1. Every point in an open ball is a center, that is, B(y, r) = B(x, r) for all y ∈ B(x, r). 2. Any pair of open balls are either disjoint or concentric. 3. Every open ball is closed and every closed ball is open, so the closure of B(x, r) is not necessarily B(x, r) unless these two sets are already equal. 4. The space is totally disconnected: every pair of distinct points lie in disjoint open neighborhoods whose union is the whole space. 1.4. ABSOLUTE VALUES AND BERKOVICH SPACES 23

Complete (k, | · |) b 对拓扑域 (k, | · |) 作完备化可得完备拓扑环 k. 以下性质成立:

b • k is a topological field;

b • k 上绝对值连续地延拓为 | · | : k → R⩾0.

b 因为度量空间总满足第一可数公理, k 既可以由极小 Cauchy 滤子来构造, 亦可仿照从 Q 造 R 的手法构造为 Cauchy 列的等价类; 在此仍采取滤子语言. 假定 xb 是 k 上的 Cauchy 滤子, 因此对每个 ϵ > 0 总存在 E ∈ xb 使得 |E − E| < ϵ; 由

||x|−|y||R ⩽ |x−y| 可知 {|E| : E ∈ xb} 构成 R⩾0 上的 Cauchy 滤子基, 于是有极限 −1 −1 −1 |xb| ∈ R⩾0; 这就说明了 | · | 的延拓方法. 由于 |x −y | = |xy| · |y−x|, 取逆运算可 b 连续地延拓到不以 0 为极限的极小 Cauchy 滤子. 从 (daibu) 导出 k 仍是拓扑域.

域对绝对值的完备化具有函子性: 设 ι : k1 ,→ k2 为域嵌入, 并且 k1, k2 上给定 b b 的绝对值满足 |ι(x)|2 = |x|1, 则 ι 连续地延拓为域嵌入 k1 ,→ k2.

注记. 若完备拓扑域 K 上的拓扑向量空间满足 Hausdorff 性质, 且 K 非离散, 则有限维 K-向量空间 E 具有唯一的拓扑结构, 满足完备性, 而且其间的线性映射自动连续.

♦ 定理 1.4.3. A.Ostrowski 若 K 对 Archimedes 绝对值 | · | 是完备的, 则存在从 K 到 R 或 C 的域同构

σ 以及 t ∈ R>0, 使得 t |σ(x)|∞ = |x| , x ∈ K

透过完备化, Archimedes 绝对值的分类也彻底明白了: 精确到等价, 所有 Archimedes 的 (K, | · |) 都来自于域嵌入 ι : K,→ C 和 | · | = |ι(·)|∞. 综上, 至少在技术层面, 探讨绝对值的性质时往往可以化约到非 Archimedes 情形, 继而过渡到秩 1 赋值的研究. 这并不是说 | · |∞ 的研究是简单或无关紧要的:

在数论的研究中, 根本的难点往往正在于寻觅一套自然的理论, 让 |·|2, |·|3, ··· , |·|∞ 能在其中各安生理.

Berkovich space

A Berkovich space, introduced by Berkovich (1990), is a version of an analytic space over a non-Archimedean field (e.g. p-adic field), refining Tate’s notion ofa rigid analytic space. 24 CHAPTER 1. VALUATION THEORY

In the complex case we start by defining the complex affine space tobe Cn. n For each U ⊆ C , we define OU , the ring of analytic functions on U to be the ring of holomorphic function. We then define a local model space for f1, ··· , fn ∈ OU to be X := {x ∈ U : f1(x) = ··· = fn(x) = 0} with OX = OU /(f1, ··· , fn).A complex analytic space is a locally ringed C-space (Y, OY ) which is locally isomorphic to a local model space. When k is a complete non-Archimedean field, we have that k is totally dis- conected. In such a case, if we continue with the same definition as in the complex case, we wouldn’t get a good analytic theory. Berkovich gave a definition which gives nice analytic spaces over such k, and also gives back the usual definition over C. In addition to defining analytic functions over non-Archimedean fields, Berkovich spaces also have a nice underlying topological space.

A seminorm on a ring A is a non-constant function A → R⩾0 : f 7→ |f| such that |0| = 0, |1| = 1, |f + g| ⩽ |f| + |g|, |fg| ⩽ |f||g|

It is called multiplicative if |fg| = |f||g| and is called a norm if |f| = 0 implies f = 0.

♦ 定义 1.4.2 If A is a normed ring then the Berkovich spectrum of A, M(A) is the set of multiplicative seminorms on A that are bounded by the norm of A. The Berkovich spectrum is topologized with the weakest topology such that for any

f ∈ A the map φf : M(A) → R, | · | 7→ |f| is continuous.

注记. The Berkovich spectrum of a normed ring A is non-empty if A is non-zero and is compact if A is complete.

If x ∈ M(A) then the elements f with |f|x = 0 form a prime ideal of A. The quotient field of the quotient by this prime ideal is a normed field, whose completion is a complete field with a multiplicative norm, this field is denoted by H(x) and the image of an element f ∈ A is denoted by f(x). The field H(x) is generated by the image of A. Conversely a bounded map from A to a complete normed field with a multiplicative norm that is generated by the image of A gives a point in the spectrum of A. n 1/n The spectral radius ρ(f) := lim ∥f ∥ of f is equal to sup |f|x. n→∞ x∈M(A) 1.4. ABSOLUTE VALUES AND BERKOVICH SPACES 25

例 1.4.4. The spectrum of a field complete with respect to a valuation is asingle point corresponding to its valuation.

例 1.4.5. If A is a commutative C∗-algebra then the Berkovich spectrum is the same as the Gelfand spectrum. A point of the Gelfand spectrum is essentially a homomorphism to C, and its absolute value is the corresponding seminorm in the Berkovich spectrum.

例 1.4.6. Ostrowski’s theorem shows that the Berkovich spectrum of the Z (with | · |ϵ the usual norm) consists of the powers p of the usual valuation, for p a prime or ∞. If p is a prime then 0 ⩽ ϵ ⩽ ∞, and if p = ∞ then 0 ⩽ ϵ ⩽ 1. When ϵ = 0 these all coincide with the trivial valuation that is 1 on all non-zero elements. For each p (prime or infinity) we get a branch which is homeomorphic to a real interval, the branches meet at the point corresponding to the trivial valuation. The open neighborhoods of the trivial valuations are such that they contains all but finitely many branches, and their intersection with each branch is open.

♦ 定义 1.4.3 If k is a field with a valuation, then the n-dimensional Berkovich affine n ··· space over k, Ak , is the set of multiplicative seminorms on k[x1, , xn] extending the norm on k. The Berkovich affine space is topologized with the weakest ∈ n → R | · | 7→ | | topology such that for any f k the map φf : Ak , f is continuous.

注记. This is not a Berkovich spectrum, but is an increasing union of the Berkovich spectrums of rings of power series that converge in some ball (so it is locally compact).

⊂ n We define an analytic function on an open subset U Ak is a map ∏ f : U → H(x), x 7→ H(x) x∈U which is a local limit of rational functions, i.e., such that every point x ∈ U has an open neighborhood U ′ ⊂ U with the following property: For every ε > 0 there exist g, h ∈ k[x1, ··· , xn] with

′ ′ ′ g(x ) ′ ′ h(x ) ≠ 0 and f(x ) − < ε for all x ∈ U h(x′)

Continuing with the same definitions as in the complex case one can define the ring of analytic function, local model space, and analytic spaces over any field 26 CHAPTER 1. VALUATION THEORY with a valuation (one can also define similar objects over normed rings). This gives reasonable objects for fields complete with respect to a nontrivial valuation and the ring of integers Z. In the case where k = C, this will give the same objects as described before. These analytic spaces are not all analytic spaces over non-Archimedean fields.

例 1.4.7 (Berkovich affine line). The 1-dimensional Berkovich affine space is called the Berkovich affine line.

1. When k algebraically closed non-Archimedean field, complete with respects to its valuation, one can describe all the points of the affine line.

→ 1 2. There is a canonical embedding k , Ak.

1 3. The space Ak is a locally compact, Hausdorff, and uniquely path-connected topological space which contains k as a dense subspace.

1 1 4. One can also define the Berkovich projective line Pk by adjoining to Ak in a suitable manner a point at infinity. The resulting space is a compact, Haus- dorff, and uniquely path-connected topological space which contains P1(k) as a dense subspace. 1.5. EXTENSION OF VALUATIONS 27

1.5 Extension of valuations 28 CHAPTER 1. VALUATION THEORY

1.6 Hensel’s Lemma

Complete fields are not countable and therefore are relatively huge; algebraic extensions of complete fields are not necessarily complete with respect to any natural extension of the valuation. One of the most important features of complete fields is the Hensel Lemma. Fields satisfying this lemma are called Henselian. They can be relatively small; and, as we shall see later, an algebraic extension of a Henselian field is a Henselian field.

Hensel’s Lemma ∑ d i ∈ ∈ 引理 1.6.1. Let R be a ring, f = i=0 fix R[x] be a polynomial, and let a R. Then f(x) = f(a) + f ′(a)(x−a) + g(x)(x−a)2 for a unique g ∈ R[x].

Proof. Write f(x) = f(a + (x − a)) and use Binomial theorem.

注记. The lemma can be viewed as giving the first two terms of a formal Taylor expansion of f(x) about a. And we obtain f(a) = f ′(a) = 0 iff a is (at least) a double root of f.

(Conventions in this section) Let A be a complete DVR with maximal ideal m and residue field k := A/m, and set K := Frac(A). If I is an ideal, by a lift of an element of A/I, we mean a preimage under the quotient map R ↠ R/I.

♦ 命题 1.6.1. Hensel’s Lemma I Suppose f ∈ A[x] is a monic polynomial whose reduction to k[x] has a simple root r ∈ k. Then r can be lifted to a root of f in A.

Proof. (Sketch) Let a0 be any lift of r to A; the element a0 is not necessarily a root · of f. In terms of the absolute value | · | := cvm( ) (for some 0 < c < 1) on K, we have

| ′ | f(a0) ⩽ f (a0) = 1, and ϵ := ′ 2 c < 1. f (a0) For each n ⩾ 0 we define

− f(an) an+1 := an ′ f (an) 1.6. HENSEL’S LEMMA 29

We can prove by induction on n that   |an| ⩽ 1 (so an ∈ A)   |an − a0| ⩽ ϵ < 1 (so an is a lift of r) | ′ | | ′ | ̸  f (an) = f (a0) = 0 (so an+1 is well defined)   2n ′ 2 |f(an)| ⩽ ϵ |f (a0)| (so f(an) converges rapidly to 1)

2n We have |an+1−an| ⩽ ϵ → 0 as n → ∞, and for a nonarchimedean absolute value this implies {an} is Cauchy. Thus a := lim an ∈ A, since A is complete. n→∞ Then f(a) = lim f(an) = 0, so a is a root of f, and |a−a0| = lim |an−a0| < 1, so a n→∞ n→∞ is a lift of r ≡ a0 mod m.

2n 注记. 1. an is a root of f modulo m . 2. The proof only requires ϵ < 1, so it actually works in many cases where r is not a simple root (such as next lemma); we also don’t need f to be monic. 3. The lemma shows that complete discret valuation field is Henselian. 4. The completeness is not strictly necessary. A local ring A in which the Lemma holds without the hypothesis that A is a complete DVR is called a henselian ring. 5. In general, any lemma that holds for a local ring if and only if it is a henselian ring may be called ”Hensel’s Lemma”, and there are at least a dozen candi- dates. 6. One can define the henselization of a noetherian local ring R as universal property. ; in many cases this turns out to be the subring of the completion b R consisting of elements that are algebraic over R.

2 2 例 1.6.1. Let A = Z5 and f(x) = x − 6 ∈ Z5[x]. Then f(x) = x − 1 ∈ F5[x] has 2 r = 1 as a simple root. By the lemma above, there is a unique a ∈ Z5 s.t. a −6 = 0 and a ≡ 1 mod 5. We could also have chosen r = −1, which would give another distinct root of f(x), which must be −a. Thus Z5 contains two distinct square roots of 6.

♦ 命题 1.6.2. Hensel’s Lemma II

Let a0 ∈ A, f ∈ A[x] satifies

′ 2 |f(a0)| < |f (a0)| 30 CHAPTER 1. VALUATION THEORY

and for n ⩾ 0 define {an} as above. The sequence {an} is well-defined and converges to the unique root a ∈ A of f for which | | | − | ⩽ f(a0) a a0 ϵ := ′ 2 |f (a0)|

2n ′ 2 Moreover, |f(an)| ⩽ ϵ |f (a0)| for all n ⩾ 0.

2 2 2 例 1.6.2. Let A = Z2 and f(x) = x − 17 ∈ Z2[x]. Then f(x) = x − 1 = (x − 1) ∈ F − | | 1 2[x] has no simple root. But if we let a0 = 1, then f(a0) = 16 and f(a0) = 16 , ′ | ′ | 1 | | | ′ |2 while f (a0) = 2 and f (a0) = 2 . We thus have f(a0) < f (a0) and can apply the above lemma to get a square root of 17 in Z2

♦ 命题 1.6.3. Hensel’s Lemma III

Suppose that f(x) ∈ A[x] s.t. f = gh for some coprime g, h ∈ k[x]. Then there exist polynomials g, h ∈ A[x] for which f = gh with g ≡ g, h ≡ h mod m s.t. deg g = deg g.

♦ 命题 1.6.4. Hensel’s Lemma IV Suppose that f(x) ∈ K[x] be an irreducible polynomial whose leading and constant coefficients lie in A, then f ∈ A[x].

∑n i Proof. Write f(x) = fix , then n > 0 since f is irreducible. Let m := min{vm(fi)}. i=0 Suppose for the sake of contradiction that m < 0, and let g := π−mf = ∑n i gix ∈ A[x]. Then Then g is an irreducible polynomial in A[x] with g0, gn ∈ m, i=0 and gi is a unit for some 0 < i < n. Note that the reduction g of g to k[x] is divisible by x. Let u := xd be the largest power of x dividing g. Then we can use the proposition 1.6.3 to find a contradiction.

推论. let L|K be a finite extension of degree n. Then α ∈ L is integral over A iff

NL|K (α) ∈ A.

∑d i Proof. Let f(x) = fix ∈ K[x] be the minimal polynomial of α. Note that i=0 NL|K (α) ∈ A ⇐⇒ f0 ∈ A.

引理 1.6.2. The absolute value | · | of K extends to algebraic closure K. Then

∀α ∈ K, σ ∈ AutK (K), we have |σ(α)| = |α|. 1.6. HENSEL’S LEMMA 31

Let α, β ∈ K. If |β − α| < |β − σ(α)| for all 1 ≠ σ ∈ AutK (K), then we say β belongs to α. This is equivalent to requiring that |β − α| < |α − σ(α)| for all

1 ≠ σ ∈ AutK (K) (non-Archimedean).

♦ 命题 1.6.5. Hensel’s Lemma V (Krasner’s lemma)

Let α, β ∈ K. If β belongs to α, then K(α) ⊆ K(β).

Proof. Suppose not. Then β belongs to α but α ∈/ K(β). The extension K(α, β)|K(β) is separable and non-trivial, so there is an automorphism σ ∈ AutK(β)(K|K(β)) for which σ(α) ≠ α (let σ send α to a different root of the minimal polynomial of

α over K(β)). For any σ ∈ AutK(β)(K|K(β)) we have

|β−α| = |σ(β−α)| = |σ(β)−σ(α)| = |β−σ(α)| since σ fixes β. But this is a contradiction. ∑ 1 We define a L -norm of f ∈ K[x] by ∥f∥1 := |fi|.

引理 1.6.3. Let f be a monic polynomial. Then |α| < ∥f∥1 for every root α ∈ L of f, where L|K is a field with an absolute value that extends | · |.

Proof. We may assume that ∥f∥1 > 1, |α| > 1, then

− − ∑n 1 ∑n 1 | | n i ⩾ | |n − | |n−1 | | ⩾ | | − ∥ ∥ − 0 = f(α) = α + fiα α α fi α ( f 1 1) i=0 i=0

♦ 定理 1.6.1. continuity of roots Let f ∈ K[x] be a monic irreducible separable polynomial. There exist

δ = δ(f) > 0 s.t. ∀ monic polynomial g ∈ K[x] with ∥f − g∥1 < δ the following holds: • Every root β of g belongs to a root α of f for which K(β) = K(α). • Every such g is separable, irreducible, and has the same splitting field as f.

Proof. We may assume that deg g = deg f and fix an algebraic closure K, write ∏n ∑n i f(x) = (x − αi) = fix i=0 i=0

Let 0 < ϵ < min0⩽i,j⩽n{1, |αi − αj|} and define ( ) ϵ δ := ∈ (0, 1) 2(∥f∥1 + 1) 32 CHAPTER 1. VALUATION THEORY

Then ∥g∥1 ⩽ ∥f∥1 + ∥g − f∥1 = ∥f∥1 + δ and for any root β of g we have

∑n ∑n i i |f(β)| = |f(β) − g(β)| = | (fi − gi)β | ⩽ |fi − gi||β| i=0 i=0

| | ∥ − ∥ ∥ ∥n ∥ ∥ n ⩽ ϵ n Thus f(β) < f g 1 g 1 < δ( f 1 + 1) ( 2 ) and therefore

∏n ϵ |β − α | = |f(β)| < ( )n i 2 i=1

| − | ϵ It follows that β αi < 2 for at least one αi, and the triangle inequality implies that this αi must be unique since |αi−αj| ⩾ for i ≠ j. Therefore β belongs to α := αi. By Krasner’s lemma, K(α) ⊆ K(β), and we have n = [K(α): K] ⩽ [K(β): K] ⩽ n, so K(α) = K(β). It follows that g is the minimal polynomial of β thus g is irreducible. g is also separable, since β ∈ K(α) lies in a separable extension of K.

By this theorem ,we get a result:

♦ 命题 1.6.6 ′ Let Kv be a local field, if K |Kv is a finite extension, then there exists a ′ global filed L s.t.K ≃ Lw where w|v is an extension of v.

注记. 1. The non-Archimedean case is proved by theorem 1.6.1.

2. If Lw|Kv is Galois, not necessarily L|K does.

3 例 1.6.3. Let K = Q,Kv = Q7 and Lv = Q7[x]/(x − 2). The extension Lw|Kv is 2 Galois since Q7 contains ζ3 (we can lift the root 2 of x + x + 1 ∈ F7[x] to a root of 2 3 x +x+1 ∈ Q7[x] via Hensel’s lemma), and this implies that x −2 splits completely 3 in Lv. But L = K[x]/(x −2) is not a Galois extension of K because it contains only 3 one root of x −2. However, we can replace K with Q(ζ3) without changing Kv (take the completion of K with respect to the valuation induced by a prime above 3 7) or Lv, but now L = K[x]/(x −2) is a Galois extension of K.

Dedekind-Kummer theorem in a local setting

引理 1.6.4 (Nakayama’s lemma). Let A be a local ring with maximal ideal p, and let M be a finitely generated A-module. If the images of x1, ··· , xn ∈ M generate

M/pM as an (A/p)-vector space then x1, ··· , xn generate M as an A-module. 1.6. HENSEL’S LEMMA 33

推论. Let A be a local noetherian ring with maximal ideal p, let g ∈ A[x], and let B := A[x]/(g(x)). Every maximal ideal m of B contains the ideal pB.

Proof. Suppose not. Then m + pB = B for some maximal ideal m of B. The ring B is a noetherian A-module, so its A-submodules are all finitely generated.

Let z1, ··· , zn be A-module generators for m. Every coset of pB in B can be written as z + pB for some A-linear combination z of z1, ··· , zn, so the images of z1, ··· , zn generate B/pB as an (A/p)-vector space. By Nakayama’s lemma, z1, ··· , zn generate B, in which case m = B, a contradiction.

推论. Let A be a local noetherian ring with maximal ideal p, g ∈ A[x] be a polynomial with reduction g ∈ A/p[x], and let α be the image of x in B := A[x]/(g(x)) =

A[α]. The maximal ideal of B are (p, gi(α)), where gi, ··· , gm are lifts of the distinct ∈ irreducible polynomials gi A/p[x] that divide g.

Proof. B ≃ A[x] ≃ A/p[x] pB (p, g(x)) (g(x))

♦ 定理 1.6.2 Assume that B|A are DVRs (not necessarily complete). If l|k is separable then B = A[α] for some α ∈ B.

Proof. (Sketch) Write l = k(α) for some α ∈ l whose minimal polynomial g is separable of degree equal to f. We consider a lift α ∈ B, g ∈ A[x] which satisfy g(α) is a uniformizer for B by lemma 1.6.1. We now claim B = A[α], by Nakayama’s lemma, it suffices to show that the reductions of 1, α, ··· , αn−1 span B/pB as a vector space.

推论. If L|K is unramified this holds for any α ∈ B whose image generates the residue field extension l|k. 第 2 2 章 Dedekind domain

2.1 *Fractional ideals and ideal class groups

Definition of Dedekind domain

Let R denotes a domain, we consider the R-submodule of K := Frac(R).

♦ 定义 2.1.1. Fractional ideals Let a be a R-submodule of K. If there exista a non-zero element d ∈ R， s.t. da ⊆ R，then we call a the fractional ideal of R.

注记. 1. We have da ≃ a as R-module, and da is an R-ideal; 2. The fractional ideal of a PID is such as ⟨q|q ∈ K⟩; 3. A finitely generated R-submodule of K is a fractional ideal of R; 4. If R is a Noetherian domain, then the fractional ideal of R is finitely generated.

We define operations between fractional ideals as following: sum I + J := {a + b | a ∈ I, b ∈ J}, ∑ product IJ := { aibi | ai ∈ I, bi ∈ J}, colon (I : J) := {x ∈ K | xJ ⊆ I}.

We call Λ(I) := (I : I) the multiplier ring of I. If ab = R, we say a is invertible.

注记. 1. If a is invertible, then the inverse is unique; 2. a is invertible ⇐⇒ a(R : a) = R. In this case, we set a−1 := (R : a). 2.1. *FRACTIONAL IDEALS AND IDEAL CLASS GROUPS 35

3. Invertible is a local property since the three operations is. 4. If I is invertible, then Λ(I) = R.

Let F(R) denotes the set of non-zero fractional ideals, which is always a commutative monoid.

引理 2.1.1. If a ∈ F(R) is invertible, then a is finitely generated. ∑ −1 −1 Proof. If aa = R, then ∃ xi ∈ a, yi ∈ a s.t. xiyi = 1. Therefore ∀ x ∈ a, x = ∑ ∑ i (xyi)xi ∈ Rxi. i i

♦ 定义 2.1.2 If F(R) is a group, we call it Dedekind domain.

♦ 命题 2.1.1 A domain R is a Dedekind iff 1. R is Noetherian； 2. R has dimension 1； 3. R is integrally closed.

Proof. Let 0 ≠ p ∈ Spec(R), if ∃ a s.t. p ⊆ a ⊆ R, then a−1a ⊆ R. Note that p = a(a−1p), so p|a or a−1p, namely a = p or a = p. If α ∈ Frac(R) is integral in R, namely R[α] is a finitely generated module, thus a fractional ideal. We have R[α] = R since R[α]2 = R[α]. For the reverse, note that DVR is a PID, so each non-zero fractional ideal is invertible.

注记. 1. A localization of Dedekind domain is still Dedekind. 2. A nonzero fractional ideal I in a noetherian local domain R is invertible if and only if it is principal. 36 CHAPTER 2. DEDEKIND DOMAIN

2.2 E´tale algebras and lattices

E´tale algebras and base change

Let K be a field. An ´etale K-algebra is a K-algebra L that is isomorphic to a finite product of separable field extensions of K. It’s easy to see ´etale K-algebras are semisimple.

例 2.2.1. If K is a separably closed field then every ´etale K-algebra A is isomorphic to Kn = K × · · · × K.

♦ 命题 2.2.1 ∏ Let A = Ki be a K-algebra written that is a product of field extensions

Ki|K. Every surjective homomorphism φ : A → B of K-algebras corresponds to the projection of A on to a subproduct of its factors. ∏ Proof. The ideal ker φ is a subproduct of Ki, thus A ≃ ker φ×Imφ and B = Imφ is isomorphic to the complementary subproduct.

注记. The proposition can be viewed as a generalization of the fact that every surjective homomorphism of fields is an isomorphism.

推论. The decomposition of an ´etale algebra into field extensions is unique up to permutation and isomorphisms of factors.

Proof. Let A be an ´etale K-algebra and suppose A is isomorphic (as a K-algebra) to two products of field extensions of K, say

∏m ∏n Ki ≃ A ≃ Lj i=1 j=1

Composing with isomorphisms yields surjective K-algebra homomorphisms π : ∏ ∏ i Lj → Ki and πj : Ki → Lj. Then each Ki must be isomorphic to one of the

Lj and each Lj must be isomorphic to one of the Ki (and m = n).

Our main interest in ´etale algebras is that they naturally arise from and are stable under base change. 2.2. E´TALE ALGEBRAS AND LATTICES 37

♦ 定义 2.2.1 Let ψ : A → B be a homomorphism of rings (so B is an A-module), and let

M be any A-module. The tensor product of A-modules M ⊗A B is a B-module (with multiplication defined by b(m ⊗ b′) := m ⊗ bb′) called the base change (or extension of scalars) of M from A to B. If M is an A-algebra then its base change to B is a B-algebra.

注记. Each ψ : A → B determines a functor from the category of A-modules to the category of B-modules via base change. It has an adjoint functor called restriction of scalars that converts a B-module M into an A-module by the rule am = ψ(a)m (if ψ is inclusion this amounts to restricting the scalar multiplication by B to the subring A).

例 2.2.2 (localization of a module). If M is an A-module and p is a prime ideal of

A then Mp = M ⊗A Ap.

♦ 命题 2.2.2 Suppose L is a finite ´etale K-algebra and K′|K is any field extension. Then ′ ′ L ⊗K K is a finite ´etale K -algebra of the same dimension as L.

Proof. We may assume that L is a field, then L ≃ K[x]/(f) for some separable f ∈ ′ K[x], and if f = f1f2 ··· fm is the factorization of f in K [x], we have isomorphisms of K′-algebras ∏ ′ ′ ′ L ⊗K K ≃ K [x]/(f) ≃ K [x]/(fi) i

′ ′ 注记. 1. Each K [x]/(fi) is a finite separable field extension of K . 2. The proof shows that a seperable field extension under base change will become an ´etale algebra.

例 2.2.3. Any finite dimensional real vector space V is a finite ´etale R-algebra (with coordinate-wise multiplication with respect to some basis); the complex vector space V ⊗R C is then a finite ´etale C-algebra of the same dimension.

♦ 定理 2.2.1 Let L be a commutative K-algebra of finite dimension and assume that dim L < #K. The following are equivalent: 38 CHAPTER 2. DEDEKIND DOMAIN

1. L is a finite ´etale K-algebra; 2. Every element of L is separable over K; ′ ′ 3. L ⊗K K is reduced for every extension K |K; ′ ′ 4. L ⊗K K is semisimple for every extension K |K; 5. L = K[x]/(f)for some separable f ∈ K[x].

∏n Proof. To show (1) ⇒ (2), let L = Ki with each Ki|K separable, and consider i=1 α = (α1, ··· , αn) ∈ L. Each αi ∈ Ki is separable over K with separable minimal polynomial fi ∈ K[x], and α is a root of f := lcm{f1, ··· , fn}, which is separable, thus α is separable. To show (2) ⇒ (3), note that if α ∈ L is nonzero and separable over K it cannot be nilpotent, and separability is preserved under base change. To show (4) ⇒ (1), we can assume L is semisimple (take K′ = K), and it suffices to treat the case where L is a field. By base-changing to the separable closure of K in L, we can further reduce to the case that L|K is a purely inseparable field extension. If L = K we are done. Otherwise we may pick an inseparable n α ∈ L, and the minimal polynomial of α has the form f(x) = xp −a for some a ∈ K. Now consider

γ := α ⊗ 1−1 ⊗ α ∈ L ⊗K L

n n n We have γ ≠ 0, since γ ∈/ K, but γp = αp ⊗ 1−1 ⊗ αp = α ⊗ 1−1 ⊗ α = 0, so γ is a nonzero nilpotent and L ⊗K L is not reduced, contradicting (3) ⇔ (4). ∏n We have seen (5) ⇒ (1). For the converse, suppose L = Li with each Li|K i=1 a finite separable extension of K. Pick a monic irreducible separable polynomial f1(x) s.t.L1 ≃ K[x]/(f1(x)), and then do the same for i = 2, ··· , n ensuring that each polynomial fj we pick is not equal to fi for any i < j. This can be achieved × by replacing fj(x) with fj(x + a) for some a ∈ K if necessary. Here we use the fact that there are at least n distinct choices for a, under our assumption (note that if f(x) is irreducible then the polynomials f(x + a) are irreducible and pairwise coprime as a ranges over K). The polynomials f1, ··· , fn are then coprime and separable, so their product f is separable and L = K[x]/(f), as desired.

注记. 1. A commutative K-algebra of finite dimension is semisimple iff itis reduced (K is a field). 2. The implications (1) ⇔ (2) ⇔ (3) ⇔ (4) ⇐ (5) hold regardless of the dimension of L. 2.2. E´TALE ALGEBRAS AND LATTICES 39

♦ 命题 2.2.3 Suppose L is a finite ´etale K-algebra and Ω is a separably closed field extension of K. There is an isomorphism of finite ´etale Ω-algebras ∏ ∼ L ⊗K Ω −→ Ω

σ∈HomK (L,Ω)

that sends β ⊗ 1 to the vector (σ(β))σ for each β ∈ L.

Proof. We may reduce to the case that L = K[x]/(f) is a separable field extension, and we may then factor f(x) = (x−α1) ··· (x−αn) over Ω, with the αi are distinct. We have a bijection between HomK (K[x]/(f), Ω) and the set {αi}: each

σ ∈ HomK (K[x]/(f), Ω) is determined by σ(x) ∈ {αi}, and for each αi, the map x 7→ αi determines a K-algebra homomorphism σi ∈ HomK (K[x]/(f), Ω). We have Ω-algebra isomorphisms

∏n ∏n K[x] ∼ Ω ∼ Ω[x] ∼ ⊗ Ω −→ −→ −→ Ω (f) K (f) (f) i=1 i=1

x ⊗ 1 7→ x 7→ (α1, ··· , αn) 7→ (σ1(x), ··· , σn(x))

The element x ⊗ 1 generates L ⊗K Ω as an Ω-algebra, and it follows that β ⊗ 1 7→

(σ(β))σ for every β ∈ L.

注记. The proof does not require Ω to be separably closed, we could replace Ω with the compositum of the normal closure of the field extensions L |K in the ∏ i decomposition of L = Li into separable extensions of K (in the proof above we just needed f to split into linear factors).

例 2.2.4. Let L|K = Q(i)|Q and Ω = C. Then Q[x] C[x] C[x] C[x] Q(i) ⊗Q C ≃ ⊗Q C ≃ ≃ × ≃ C × C x2 + 1 (x2 + 1) (x − i) (x + i)

As C-algebra isomorphisms, the corresponding maps are determined by

i ⊗ 1 7→ x ⊗ 1 7→ x 7→ (x, x) ≡ (i, −i) 7→ (i, −i)

Taking the base change of Q(i) to C lets us see the two distinct embeddings of Q(i) in C. Note that Q(i) is canonically embedded in its base change Q(i) ⊗Q C to C via α 7→ α ⊗ 1. We have

−1 = i2 = (i ⊗ 1)2 = i2 ⊗ 12 = −1 ⊗ 1 = −(1 ⊗ 1) 40 CHAPTER 2. DEDEKIND DOMAIN

Thus as an isomorphism of C-algebras, the basis {1 ⊗ 1, i ⊗ 1} for Q(i) ⊗Q C is mapped to the basis {(1, 1), (i, −i)} for C × C. For any (α, β) ∈ C × C, the inverse image of α + β α − β (α, β) = (1, 1) + (i, −i) 2 2i in Q(i) ⊗ C under this isomorphism is α + β α − β α + β α − β (1 ⊗ 1) + (i ⊗ 1) = 1 ⊗ + i ⊗ 2 2i 2 2i Now R|Q is an extension of rings, so we can also consider the base change of the Q-algebra Q(i) to R. But note that R is not separably closed and in particular, it does not contain a subfield isomorphic to Q(i), thus the roposition above does 2 not apply. Indeed, as an R-module, we have Q(i) ⊗Q R ≃ R , but as an R-algebra, 2 Q(i) ⊗Q R ≃ C ̸≃ R .

Dual modules and perfect pairings

Let A be a commutative ring and M an A-module. We recall that the dual ∗ module of M is M := HomA(M,A).

♦ 命题 2.2.4 Let A be an integral domain with fraction field K and let M be a nonzero A-submodule of K. Then M ∗ ≃ (A : M) := {x ∈ K | xM ⊆ A}; in particular, if M is an invertible fractional ideal then M ∗ ≃ M −1 and M ∗∗ ≃ M.

Proof. For any x ∈ (A : M) the map m 7→ xm is an A-linear map from M to A, hence an element of M ∗, and this defines an A-module homomorphism ψ :(A : M) → M ∗, since the map x 7→ [m 7→ xm] is itself A-linear. Since M ⊆ K is a nonzero A-module, it contains some nonzero a ∈ A (if a/b ∈ M, so is ba/b = a). If f ∈ M ∗ and m = b/c ∈ M then ( ) ( ) ( ) ( ) ( ) ( ) b ac b b ac b f(a) f(m) = f = f = f = f(a) = m c ac c ac c ac a

It follows that f corresponds to multiplication by x = f(a)/a, which lies in (A : M) since xm = f(m) ∈ A for all m ∈ M. The map f 7→ f(a)/a defines an A-module homomorphism M ∗ → (A : M) inverse to ψ, so ψ is an isomorphism.

例 2.2.5. 1. As a Z-module, we have Q∗ = 0. 2. The dual of any finite Z-module (any finite abelian group) is 0, as is the double dual. 2.2. E´TALE ALGEBRAS AND LATTICES 41

3. If A is an integral domain every dual (and double dual) A-module must be torsion free.

♦ 命题 2.2.5 Let A be a commutative ring and let M be a free A-module of rank n. Then ∗ M is also a free A-module of rank n, and each basis {e1, ··· , en} of M uniquely { ∗ ··· ∗ } ∗ ∗ determines a dual basis e1, , en of M with the property ei (ej) = δij.

Proof. We may assume n ⩾ 1 and fix an A-basis e := (e1, ··· , en) for M. For each n ∗ a := (a1, ··· , an) ∈ A , define fa ∈ M by setting fa(ei) = ai and extending A- n ∗ linearly. The map a 7→ fa gives an A-module homomorphism A → M with inverse f 7→ (f(e1), ··· , f(en)) and is therefore an isomorphism. It follows that M ∗ ≃ An is a free A-module of rank n. ∗ ˆ ··· ··· ∈ n Now let ei := fˆi, where i := (0, , 0, 1, 0, , 0) A has a 1 in the i-th ∗ ∗ ··· ∗ ∗ ∗ ∗ position. Then e := (e1, , en) is a basis for M , and ei (ej) = δij. The basis e is uniquely determined by e: it must be the image of (1ˆ, ··· , nˆ) under the isomorphism a 7→ fa determined by e.

(Perfect pairing on module) Let A be a commutative ring and M an A-module. A A-bilinear pairing on M is ⟨·, ·⟩ : M × M → A. The pairing ⟨·, ·⟩ induces an A- module homomorphism

φ : M → M ∗ m 7→ [n 7→ ⟨m, n⟩]

If ker φ = 0 then ⟨·, ·⟩ is is nondegenerate, and if φ is an isomorphism then ⟨·, ·⟩ is perfect.

例 2.2.6 ( free modules of finite rank). Consider the pairing ⟨x, y⟩ := 2xy on Z, which is non-degenerate but not perfect.

If M is a free A-module with basis (e1, ··· , en) and ⟨·, ·⟩ is a perfect pairing, ∼ we can apply the inverse of the isomorphism φ : M −→ M ∗ induced by the pairing ∗ ··· ∗ ′ ··· ′ to the dual basis (e1, , en) to obtain a basis (e1, , en) for M that satisfies

⟨ ′ ⟩ ei, ej = δij 42 CHAPTER 2. DEDEKIND DOMAIN

♦ 命题 2.2.6 Let A be a commutative ring and let M be a free A-module of rank n with

a perfect pairing ⟨·, ·⟩. For each A-basis (e1, ··· , en) of M there is a unique basis ′ ··· ′ ⟨ ′ ⟩ (e1, , en) for M s.t. ei, ej = δij.

∗ ′ Proof. Existence follows from the discussion above, we then have ei = φ(ei) = 7→ ⟨ ′ ⟩ ⟨ ′ ⟩ ′ ∗ m ei, m and ei, ej = φ(ei)(ej) = ei (ej) = δij. ′ ··· ′ If (f1, , fn) is another basis for M with the same property then for each i ⟨ ′ − ′ ⟩ ′ ′ we have ei fi , ej = 0 for every ej, and therefore fi = ei for each i; uniqueness follows.

注记. Let A to denote a subring of a field K and M denote an A-submodule of K-linear space V . A perfect paring on the K-vector space V will typically not restrict to a perfect pairing on the A-module M.

例 2.2.7. The perfect pairing ⟨x, y⟩ = xy on Q does not restrict to a perfect pairing on the Z-module 2Z because the induced map φ : 2Z → 2Z∗ defined by φ(m) = ∗ [n 7→ mn] is not surjective: the map x 7→ x/2 lies in 2Z = HomZ(2Z, Z) but it is not in the image of φ.

Lattices

♦ 定义 2.2.2 Let A be an integral domain with fraction field K and let V be a K-vector space of finite dimension. A (full) A-lattice in V is a finitely generated A- submodule M of V that spans V as a K-vector space.

注记. When A is a PID, M is necessarily torsion-free, and therefore is a free module.

例 2.2.8 (Important). Let A be an integral domain with fraction field K and L|K be a finite extension, and let B be the integral closure of A in L.

K L

A B 2.2. E´TALE ALGEBRAS AND LATTICES 43

∈ ∃ ∈ ∈ b • L = Frac(B) and for all c L, a A, b B s.t. c = a . • There exists a K-basis of L lies in B, namely B is a A-lattice in L. • If A is normal, then K ∩ B = A and B is normal. • If A is normal, then a monic polynomial f ∈ A[x] is irreducible over A iff it is irreducible over K. • If A is normal, then the minimal polynomial of c ∈ B over K equals to its integral polynomial over A.

Let A be a noetherian domain with fraction field K, and let V be a K-vector space of finite dimension with a perfect pairing ⟨·, ·⟩. If M is an A-lattice in V , its dual lattice (with respect to the pairing ) is the A-module

M ∗ := {x ∈ V | ⟨x, M⟩ ⊆ A}

注记. Here we use the same symbol with dual module.

It is clear that dual lattice M ∗ is an A-submodule of V , but it is not clear that it is an A-lattice in V , nor is it obvious that it is isomorphic to the dual module M ∗. In order to justify the term dual lattice, let us now prove both facts. We will need to use the hypothesis that A is noetherian, since in general the dual of a finitely generated A-module need not be finitely generated. Notice that ⟨·, ·⟩ is a perfect pairing on the K-module V that need not restrict to a perfect pairing on the A-module M. ♦ 定理 2.2.2 Let A be a noetherian domain with fraction field K, V be a K-vector space with a perfect pairing ⟨·, ·⟩, and let M be an A-lattice in V . The dual lattice M ∗ is an A-lattice in V isomorphic to dual module of M.

Proof. Let e be a K-basis for V that lies in M, and e′ be the unique K-basis for V given by ⟨·, ·⟩. To show that M ∗ spans V we write a finite set S of generators for M in terms of the basis e with coefficients in K and let d be the product of all denominators that appear. We claim that de′ lies in M ∗: for each e′ and generator m ∈ S, if we ∑ i put m = mjej then j ⟨ ∑ ⟩ ∑ ⟨ ′ ⟩ ′ ⟨ ′ ⟩ ∈ dei, m = d ei, mjej = d mj ei, ej = dmi A j j

′ ∈ ∗ ∗ ′ by our choice of d, this implies dei M . Thus M contains a basis de for V . 44 CHAPTER 2. DEDEKIND DOMAIN

We now show M ∗ is finitely generated. Let

∑n n N := { aiei | ai ∈ A} ≃ A i=1 be the free A-submodule of M spanned by e. The A-module N is an A-lattice in V , and K-basis e′ for V lies in N ∗. We claim e′ is an A-basis for N ∗. ∑ ∈ ∗ ′ ⟨ ⟩ ⟨ ′ ⟩ ∈ Given x N , if we write x = xiei then x, ei = xi ei, ei = xi A. It i follows that N ∗ is a free A-module of rank n, and in particular, a finitely generated module over a noetherian ring and therefore a noetherian module. From the definition of the dual lattice we have N ⊆ M ⇒ M ∗ ⊆ N ∗, so M ∗ is a submodule of a noetherian module, hence finitely generated. Now we use M ∨ denotes dual module of M. Consider A-module homomorphism ϕ : M ∗ → M ∨, x 7→ [m 7→ ⟨x, m⟩] and its inverse ψ : M ∨ → M ∗, f 7→ ∑ ′ f(ei)ei. i

推论. If M1,M2 are A-lattices in K-vector spaces V1,V2 with perfect pairings ⟨·, ·⟩1, ∗ ⟨·, ·⟩2 (resp.), then ⟨·, ·⟩1 +⟨·, ·⟩2 defines a perfect pairing on V1 ⊕V2 and (M ⊕N) ≃ M ∗ ⊕ N ∗.

Proof. Note that (M ⊕ N)∨ ≃ M ∨ ⊕ N ∨.

引理 2.2.1. Let M be an A-lattice in V , and let S be a multiplicative subset of A. Then S−1M and S−1M ∗ are (S−1A)-lattices in V satisfying (S−1M)∗ = S−1M ∗.

Proof. It is clear that S−1M are S−1M ∗ are both S−1A-lattices. −1 Let m1, ··· mn be A-module generators for M (and therefore S A-module −1 −1 ∗ generators for S M). If x is an element of (S M) then for each mi we have

⟨x, mi⟩ = ai/si for some ai ∈ A and si ∈ S, and if we put s = s1 ··· sn then ∗ −1 ∗ ⟨sx, mi⟩ ∈ A for every mi, hence for all m ∈ M; thus sx ∈ M and x ∈ S M . −1 ∗ Conversely, if x = y/s ∈ S M , then ⟨y, mi⟩ ∈ A and ⟨x, mi⟩ = ⟨y, mi⟩/s ∈ −1 −1 −1 ∗ S A for every mi, hence for all m ∈ S M, and it follows that x ∈ (S M) .

♦ 命题 2.2.7 Let A be a Dedekind domain with fraction field K, V be a K-vector space of finite dimension with a symmetric perfect pairing ⟨·, ·⟩, and let M be an 2.2. E´TALE ALGEBRAS AND LATTICES 45

A-lattice in V . Then M ∗∗ = M.

∗∗ ∗∗ ∗∗ Proof. It suffices to show (M )p = Mp. By previous lemma, (M )p = Mp , so it is enough to show that the proposition holds when A is a DVR. Assume A is a DVR, then A is a PID and M and M ∗ are both torsion-free modules over a PID, hence free A-modules. So let us choose an A-basis (e1, ··· , en) for ∗ ··· ∗ ∗ ∗∗ ··· ∗∗ M, and let (e1, , en) be the unique dual A-basis for M . If we now let (e1 , , en ) ∗∗ ⟨ ∗∗ ∗⟩ ⟨ ∗⟩ be the unique A-basis for M that satisfies ei , ej = δij and note that ei, ej = δij ⟨· ·⟩ ∗∗ (since , is symmetric), by uniqueness, we must have ei = ei, and therefore M ∗∗ = M. 46 CHAPTER 2. DEDEKIND DOMAIN

2.3 Prime decomposition in an order

Convention: A is a Dedekind domain with fractional field K, and L|K is a separable extension of degree n. B is the integral closure of A in L.

Dedeking-Kummer theorem

If L|K is a finite separable extension then we can always write L = K(α) for some α ∈ B. Extensions L|K for which B = A[α] are said to be monogenic, it implies that B is a free A-module and has an integral power basis, one of the form {1, α, ··· , αn−1}. Let p is a prime of A, then we have

1:1 {primes above p} −→{primes of B/pB}

It follows that ♦ 命题 2.3.1 Assume L = K(α) and B = A[α], let f ∈ A[x] be the minimal polynomial

of α. Suppose g1, ··· , gr ∈ A[x] are monic polynomials for which

e1 ··· er f = g1 gr

is a complete factorization of f ∈ κ(p)[x], and let qi := (p, gi(α)) be the B-ideal. Then e1 ··· er pB = q1 qr

is the prime factorization of pB in B and the residue degree of qi is fi := deg gi.

Proof. (Sketch) By the following key point

A[α] ∼ A[x] ∼ κ(p)[x] = = (p, gi(α)) (p, gi(x)) (gi(x)) we obtain qi is a prime in B and [κ(qi): κ(p)] = deg gi. e1 ··· er e1 ··· er ≡ q1 qr lies in pB since f(α) = g1 (α) gr (α) 0 mod pB. We also have

e1 ··· er deg f n NB|A(q1 qr ) = p = p so in fact ei is the ramification index of qi. 2.3. PRIME DECOMPOSITION IN AN ORDER 47

例 2.3.1. Let A = Z,K = Q, and L = Q(ζ5), where α = ζ5 be a primitive 5-th root 4 3 2 of unity with minimal polynomial f(x) = x + x + x + x + 1. Then B = Z[ζ5]

• (2): f(x) is irreducible modulo 2, so 2Z[ζ5] is prime and (2) is inert in Q(ζ5).

4 4 • (5): f(x) ≡ (x − 1) mod 5, so 5Z[ζ5] = (5, ζ5 − 1) and (5) is totally ramified

in Q(ζ5).

• (11): f(x) ≡ (x − 4)(x − 9)(x − 5)(x − 3) mod 11, so 11Z[ζ5] = (11, ζ5 −

4)(11, ζ5 − 9) (11, ζ5 − 5)(11, ζ5 − 3), and (11) splits completely in Q(ζ5).

≡ 2 2 − Z 2 • (19): f(x) (x + 5x + 1)(x 4x + 1) mod 19, so 19 [ζ5] = (19, ζ5 + 5ζ5 + 2 − 11)(19, ζ5 4ζ5 + 1).

注记. 1. We can prejudge the number of irreducible divisors of f in Fp by proposition 5.1.3.

2. The four cases above actually cover every possible prime factorization pat-

tern in the cyclotomic extension Q(ζ5)|Q.

Orders in Dedekind domain

Let S|R be an ring extension. The conductor of R in S is the largest ideal of S contained in R. It can be written as

c := {α ∈ S | αS ⊆ R} = {α ∈ R | αS ⊆ R}

If R is an integral domain, the conductor of R is the conductor of R in its integral closure.

引理 2.3.1. Let R be a noetherian domain. The conductor of R in its integral closure S is nonzero iff S is finitely generated as an R-module.

注记. If noetherian domain R has nonzero conductor then its integral closure S 1 is a fractional ideal of R that is also a ring. This means we can write S as r I for some r ∈ R and R-ideal I, and the conductor c is precisely the set of denominators ∈ 1 1 r R for which S = r I for some R-ideal I (note that the representation r I is far from unique).

2.3.2. 1. The conductor of Z in Z[i] is 0. 例 √ √ 2. The conductor of Z[ −3] in Z[ζ3] is (2, 1 + −3) 48 CHAPTER 2. DEDEKIND DOMAIN

♦ 定义 2.3.1 An order O is a noetherian domain of dimension one whose conductor is nonzero, equivalently, whose integral closure is finitely generated as an O- module.

注记. 1. Dedekind domain that is not a field is also an order. 2. The integral closure of an order is always a Dedekind domain. 3. One can construct noetherian domains of dimension one with zero conductor. But in the case of interest to us the conductor is nonzero：2.3.

例 2.3.3. As convention 2.3, let A ⊆ O ⊆ B, then O is an order. In particular, if A[α] and B have the same fraction field (so L = K(α)), then A[α] is an order in B.

♦ 定义 2.3.2 Let A be a Noetherian domain with fraction field K, and let L be a (not necessarily commutative) K-algebra of finite dimension. We define an A-order in L which is both an A-lattice and a ring.

注记. An A-order in L is a maximal order if it is not properly contained in any other A-order in L. When A is a Dedekind domain one can show that every A- order in L lies in a maximal order. Maximal orders are not unique in general, but in the convention 2.3 B is the unique maximal order.

例 2.3.4. The endomorphism ring of an elliptic curve is isomorphic to a Z-order in a Q-algebra L of dimension 1, 2, or 4. This Z-order is necessarily commutative in dimensions 1 and 2, where L is either Q or an imaginary quadratic field, but it is non-commutative in dimension 4, where L is a quaternion algebra.

引理 2.3.2. As convention 2.3, let O be a subring of L. Then O is an A-order in L iff it is an order with integral closure B.

Proof. Suppose O is an A-order in L. Then O is an A-lattice, hence finitely generated as an A-module, and therefore integral over A. Thus O lies in the integral closure B of A in L. The fraction field of O is L, so O and B have the same fraction field and B is the integral closure of O. Thus O is a domain of dimension 1 (since B is), and it is noetherian because it is a finitely generated over the noetherian ring A. The integral closure B of O is initely generated over A, hence over O; therefore O is an order. 2.3. PRIME DECOMPOSITION IN AN ORDER 49

Now suppose O is an order with integral closure B. It is an A-submodule of the noetherian A-module B, hence finitely generated over A. It contains a K-basis for L because L is its fraction field (take any K-basis for L written as fractions over O and clear denominators). Thus O is an A-lattice in L that is also a ring, hence it is an A-order in L.

例 2.3.5. There may be subrings O of L that are orders but not A-orders in L, but these do not have B as their integral closure. Consider A = B = Z,K = L = Q, and O = Z(2). In this case O is a DVR, hence a Dedekind domain, hence an order, but it is not an A-order in L, because it is is not finitely generated over A. But its integral closure is not B (indeed, O ̸⊆ B).

引理 2.3.3. In any order O, only finitely many primes contain the conductor.

Proof. We can factor c into a product of powers of finitely many primes q in its integral closure B.

We now consider the surjection SpecB → SpecO, q 7→ q ∩ O. Note that c ⊆ q ⇐⇒ c ⊆ q ∩ O. It follows that the map is SpecB → SpecO is still well- defined if we restrict to primes that do not contain c.

♦ 命题 2.3.2 Let O be an order with integral closure B and conductor c and let p be a prime of O not containing c. Then pB is prime of B.

Proof. Let q be a prime of B lying above p, so that p = q ∩ O, and pick an element s ∈ c\p.

(Claim):Op = Bq.

If a/b ∈ Op, then b ∈ B\q, so a/b ∈ Bq. Conversely, if a/b ∈ Bq then sa ∈ O and sb ∈ O\p, so (sa)/(sb) = a/b ∈ Op; here we have used that sB ⊆ O and sb ∈/ q, so sb ∈/ p. ′ ′ We now note that q |p =⇒ Bq′ = Op = Bq =⇒ q = q, so there is only one prime q lying above p. It follows that pB = qe, and we claim that e = 1. Indeed, e we must have pOp = qBq , so q Bq = qBq and therefore e = 1.

推论. The restriction of the map SpecB → SpecO defined by q 7→ q ∩ O to prime ideals not containing c is a bijection with inverse p 7→ pB.

An invertible prime p of an order O is a regular point in X, a non-invertible prime is a singular point. 50 CHAPTER 2. DEDEKIND DOMAIN

♦ 定理 2.3.1 Let O be an order with integral closure B and conductor c and let p be a prime of O. The following are equivalent: (a) p does not contain c; (b) O = {x ∈ B | xp ⊆ p}; (c) p is invertible;

(d) Op is a DVR;

(e) pOp is principal. If any of these equivalent properties hold, then pB is a prime of B.

条件 2 是否是指的 p 的 multiplier ring 是 O？ Recall that two ideals I and J in a ring A are said to be coprime if I + J = A.

When A is a noetherian domain this is equivalent to requiring that Ip + Jp = Ap for every prime ideal p of A. For prime ideals p that do not contain J, we have

Jp = Ap, so we only need to consider the case where p contains J. In this case Jp is contained in pAp and Ip + Jp = Ap iff Ip ̸⊆ pAp, in which case Ip = Ap, equivalently,

IAp = Ap. This leads to the following definition: A fractional ideal I of A is prime to J if

IAp = Ap for all prime ideals p that contain J. The set of invertible fractional ideals F J prime to J is denoted A; it is a subgroup of the ideal group . ♦ 定理 2.3.2 Let O be an order with integral closure B. Let c be any ideal of B contained in the conductor of O. The map q 7→ q ∩ O induces a group isomorphism ∼ F c −→F c B O，and both groups are isomorphic to the free abelian group generated by their prime ideals.

Proof. The B-ideal c lies in the conductor of O and is therefore also an O-ideal, so F c F c 7→ ∩O the subgroups B and O are well defined and the map q q gives a bijection between the sets of prime ideals contained in these subgroups.

注记. Every fractional ideal of O prime to the conductor has a unique factorization ∏ ∏ ei ei into prime ideals pi which matches the factorization IB = qi with pi = qi ∩ O.

The assumption B = A[α] in the Dedekind-Kummer theorem can be replaced with the assumption that pB is prime to the conductor of A[α] in B. 2.3. PRIME DECOMPOSITION IN AN ORDER 51

♦ 定理 2.3.3. Dedekind-Kummer

(1) the prime ideals of A[α] that lie above p are the ideals qi; e1 ··· er ⊆ (2) we have an inclusion q1 qr pA[α]; e1 ··· er (3) the equality q1 qr = pA[α] holds if and only if every prime qi is invertible;

(4) writing ri ∈ A[x] for the remainder of f upon division by gi, one has 2 qi is singular ⇐⇒ ei > 1 and p divides ri

−1 推论. Suppose that f = qigi + ri, and qi is singular. Then p qi(α) is in the multiplier ring of qi but not in A[α] (normalization at a point). √ 例 2.3.6. Let K = Q(α) with α = 3 19, the minimal polynomial of α is f = x3 + 19. Now f and f ′ are coprime modulo all primes except 3 and 19. The factorization 19Z[α] = (α)3 shows that Z[α] is regular and totally ramified above 19. As for modulo 3, we find

f = (x + 1)3 mod 3 and the remainder of f upon division by x + 1 is f(−1) = 18 divided by 32. It follows that the unique prime q3 = (3, α + 1) above 3 is not invertible, and the 2 identity f = (x − x + 1)(x + 1) + 18 shows that the multiplier ring of q3 contains an element β = (α2 − α + 1)/3 outside Z[α]. Note that we have (α + 1)β = −6.

We claim that OK = Z[α, β] and [OK : Z[α]] = 3.

As we have Z(p)[α] = Z(p)[α, β] at all primes p ≠ 3, we know that Z[α, β] is regular at all primes p ≠ 3. We compute the minimal of β as

3 2 fβ = x − x − 6x − 12

2 The factorization fβ = x (x−1) mod 3 shows that the primes over 3 in Z[β] are

(3, β) and (3, β−1). The prime (3, β) is invertible as fβ has remainder −12 upon Z −6 2 division by x, so [β] is regular above 3. The relation α + 1 = β = (−β + β + 6)/2 shows that Z[β] is of index 2 in Z[α, β], so locally at 3 we find Z(3)[β] = Z(3)[α, β]. It follows that Z[α, β] is regular above 3 as well.

例 2.3.7 (prime decomposition in the above example). 2 • 2OK = p2p4 = (2, α − 1)(2, α + α + 1) O 2 2 − • 3 K = p3q3 = (2, β) (3, β 1) 2 • 5OK = p5p25 = (5, α − 1)(5, α + α + 1)

• 7OK = p343 = (7) 52 CHAPTER 2. DEDEKIND DOMAIN

2.4 Norm map of one-dimensional Schemes

We use module index define norm map, the basic truth is that B is a locally free module over A in convention 2.3.

Module index

Let A be a Dedekind domain with fraction field K, V be an n-dimensional

K-vector space, M,N be A-lattices in V . Then Ap is a PID and we must have ≃ n ≃ −→∼ Mp Ap Np. Choose an Ap-module isomorphism ϕp : Mp Np and there exists a unique K-linear isomorphism V → V extending ϕp. The module index

[Mp : Np]Ap is the principal fractional Ap-ideal generated by det ϕp:

[Mp : Np]Ap := (det ϕp)

♦ 定义 2.4.1

The module index [M : N]A is the A-module: ∩

[M : N]A := [Mp : Np]Ap p

注记. 1. The module index [Mp : Np]Ap does not depend on our choice of ϕp

because any other choice can be written as ϕ1ϕpϕ2 for some Ap-module au- ∼ ∼ tomorphisms ϕ1 : Mp −→ Mp and ϕ2 : Np −→ Np that necessarily have unit determinants.

2. Each module index [Mp : Np]Ap is an A-submodule of K (which need not be finitely generated), so their intersection is clearly an A-submodule of K.

♦ 命题 2.4.1. Property of module index

1. [M : N]A ∈ F(A)

2. ([M : N]A)p = [Mp : Np]Ap

3. [M : N]A [N : P ]A = [M : P ]A

Proof. When M and N are free A-modules: fix a global A-module isomorphism ∼ ϕ : M −→ N so that (det ϕ)p = (det ϕp) for all primes p (where ϕp is just the Ap- module isomorphism induced by ϕ). To prove the general case we apply a standard ”gluing” argument that will be familiar in algebraic geometry. 2.4. NORM MAP OF ONE-DIMENSIONAL SCHEMES 53

∼ We can write any isomorphism Mp −→ Pp as a composition of isomorphisms ∼ ∼ Mp −→ Np −→ Pp; we then note that the determinant map is multiplicative with respect to composition and multiplication of fractional ideals is compatible with localization.

注记. 1. [M : N]A [N : M]A = [M : M]A = A;

2. when N ⊆ M the module index [M : N]A ⊆ A is actually an ideal.

In the special case V = K, an A-lattice in V is simply a fractional ideal of A. Let M and N be nonzero fractional ideals of A then

1. [M : N]A = (N : M)

2. M[M : N]A = M(N : M) = N

3. If N ⊆ M, then M/N ≃ A/I ⇐⇒ I = [M : N]A.

m n Proof. let π be a uniformizer for Ap. Then Mp = (π ) and Np = (π ) for some n−m m, n ∈ Z, and we may take ϕp to be the multiplication by π map. We then have

n−m n m [Mp : Np]Ap = (det ϕp) = (π ) = (π /π ) = (Np : Mp)

So one can view module index as a generalization of colon operation. Now we generalize the third observation.

♦ 定理 2.4.1 Let A be a Dedekind domain with fraction field K, and let N ⊆ M be A- lattices in a K-vector space V of dimension r for which the quotient module M/N is a direct sum of cyclic A-modules:

M/N ≃ A/I1 ⊕ · · · ⊕ A/In

where I1, ··· ,In are ideals of A. Then

[M : N]A = I1 ··· In

Proof. Let p be a prime of A, π be a uniformizer for Ap, and let ej = vp(Ij).

Pick up a basis for Mp and an isomorphism ϕp : Mp → Np so that Mp/Np = cokerϕp.

The matrix of ϕp is an r × r matrix over the PID Ap with nonzero determinant. It therefore has Smith normal form UDV , with U, V ∈ GLr(Ap) and 54 CHAPTER 2. DEDEKIND DOMAIN

d1 dr D = diag(π , ··· , π ) for some uniquely determined nonnegative integers d1 ⩽

··· ⩽ dr. We then have

e1 en d1 dr Ap/(π ) ⊕ · · · ⊕ Ap/(π ) ≃ Mp/Np = cokerϕ ≃ Ap/(π ) ⊕ · · · ⊕ Ap/(π )

It follows from the structure theorem for modules over a PID that the non- trivial summands on each side are precisely the invariant factors of M /N , pos- ∑ ∑ p p sibly in different orders. We therefore have ej = di, and applying the definition of the module index yields

∑ ∑ di ej e1 ··· en ··· [Mp : Np]Ap = (det ϕp) = (det D) = (π ) = (π ) = (π ) (π ) = (I1 In)p

It follows that [M : N]A = I1 ··· In, since the localizations ([M : N]A)p = [Mp : ··· Np]Ap and (I1 In)p coincide for every prime p.

The ideal norm

In the convention 2.3, we have a homomorphism of ideal groups:

IB|A : F(A) → F(B) I 7→ IB

We wish define a homomorphism NB|A : F(B) → F(A) in the reverse direction. As every fractional B-ideal J is an A-lattice in L, so let us consider

NB|A : F(B) → F(A)

J 7→ [B : J]A

We extend NB|A to the zero ideal by defining NB|A((0)) = (0).

引理 2.4.1. Let α ∈ L,NB|A((α)) = (NL|K (α)).

Proof. We may assume α ≠ 0, then ∩ ( ) −−→×α NB|A((α)) = [B : αB]A = [Bp : αBp]Ap = det(L L) = (NL|K (α)) p

♦ 命题 2.4.2

The map NB|A : F(B) → F(A) is a group homomorphism. 2.4. NORM MAP OF ONE-DIMENSIONAL SCHEMES 55

Proof. Let p be a prime of A. Then Ap is a DVR and Bp is a semilocal Dedekind domain, hence a PID. Thus every element of IBp is a principal ideal (α) for some ∈ × α L , and the previous lemma implies that NBp|Ap is a group homomorphism. For any I,J ∈ F(B) we then have ∩ ∩

−1 推论. [I : J]A = NB|A(I J) = NB|A((J : I))

Proof. The second equality is immediate: (J : I) = I−1J since B is a Dedekind domain. The first follows from the previous proposition. Indeed, we have

−1 −1 −1 [I : J]A = [I : B]A[B : J]A = [B : I]A [B : J]A = NB|A(I )NB|A(J) = NB|A(I J)

推论. NB|A(I) = (NL|K (α) | α ∈ I).

Proof. Let J denote the RHS. For any nonzero prime p of A, the localization of the ideal NB|A(I) = (B : I)A at p is (Bp : Ip)Ap = NBp|Ap (Ip). The fractional ideal

NBp|Ap (Ip) of Ap is principal, so NBp|Ap (Ip) = Jp follows from the proposition, and ∩ ∩

NB|A(I) = NBp|Ap (Ip) = Jp = J p p

We have the following commutative diagrams, in which the downward ar- rows map nonzero field elements to the principal fractional ideals they generate.

K× L× K×

IB|A NB|A F(A) F(B) F(A)

We know that composing the maps K× → L× → K× along the top corresponds to exponentiation by n = [L : K]; we now show that this is also true for the composition of the bottom maps. 56 CHAPTER 2. DEDEKIND DOMAIN

♦ 定理 2.4.2

fq Let q be a prime lying above p. Then NB|A(q) = p .

Proof. The (A/p)-vector space B/q has dimension fq; we have B/q ≃ A/p ⊕ · · · ⊕

A/p, an fq-fold direct sum of cyclic A-modules A/p. Thus NB|A(q) = [B : q]A = pfq .

n 推论. For I ∈ F(A) we have NB|A ◦ IB|A(I) = I , where n = [L : K].

The ideal norm in algebraic geometry

Dedekind domains naturally arise in algebraic geometry as coordinate rings of smooth curves (which for the sake of this discussion one can take to mean geometrically irreducible algebraic varieties of dimension one with no singularities). In order to make this explicit, let us fix a perfect field k and a polynomial f ∈ k[x, y] that we will assume is irreducible in k[x, y]. The ring A = k[x, y]/(f) is a noetherian domain of dimension 1, and if we further assume that the algebraic variety X defined by f(x, y) = 0 has no singularities, then A is also integrally closed and therefore a Dedekind domain. We call A the coordinate ring of X, denoted by k[X], and its fraction field is the function field of X, denoted by k(X). Conversely, given a Dedekind domain A, we can regard X = SpecA as a smooth curve whose closed points are the maximal ideals of A (all of SpecA except the zero ideal, which is called the generic point). When the field of constants k is algebraically closed, Hilbert’s Nullstellensatz gives a one-to-one correspondence between maximal ideals (x−x0, y−y0) and points (x0, y0) in the affine plane, but in general closed points correspond to Gal(k|k)-orbits of k-points.

注记. Orders in Dedekind domains also have a geometric interpretation. If O is an order, the curve X = SpecO will have a singularity at each closed point P corresponding to a maximal ideal of O that contains the conductor. Taking the integral closure B of O yields a smooth curve Y = SpecB with the same function field as X and a morphism Y → X that looks like a bijection above non-singular points (a dominant morphism of degree 1). The curve Y is called the normalization of X.

The maps IB|A and NB|A are the pushforward and pullback maps on divisors associated to the morphism of curves SpecB → SpecA induced by the inclusion A,→ B. 2.4. NORM MAP OF ONE-DIMENSIONAL SCHEMES 57

Note that F(A) is isomorphic to the free abelian group generated by the nonzero prime ideals of A. The corresponding object in algebraic geometry is the divisor group Div(SpecA), the free abelian group generated by the closed points of SpecA. The group Div(SpecA) is written additively, so its elements have the ∑ form D = nP P with all but finitely many of the integers nP equal to 0. A finite extension of Dedekind domains B|A induces a surjective morphism

ϕ : SpecB → SpecA. Primes q of B above a prime p of A is the fiber. The map IB|A corresponds to the pullback map ϕ∗ : Div(SpecA) → Div(SpecB) induced by ϕ, which is defined by ∑ ∗ ϕ (P ) := eQQ ϕ(Q)=P

where eQ is the ramification index, geometrically we think of eQ as the“multiplicity” of Q in the fiber above P .

In the other direction, the norm map NB|A corresponds to pushforward map ϕ∗ : Div(SpecB) → Div(SpecA) induced by φ, which is defined by

ϕ∗(Q) := fQϕ(Q) = fQP

where fQ counts the number of k-points in the Gal(k|k)-orbit corresponding to the closed point Q, equivalently, the degree of the field extension of k needed to split Q into fQ distinct closed points after base extension (here we assume that k is perfect). This is precisely the residue field degree of Q as a prime in the Dedekind extension B|A. Note that when k = k we always have fQ = 1 (so over algebraically closed fields one typically omits fQ from the pushforward map and the degree formula below). If we compose the pushforward and pullback maps we obtain

∑ ∗ ϕ∗ϕ (P ) = eQfQP = deg(ϕ)P ϕ(Q)=P

Here deg(ϕ) is the degree of the morphism ϕ : SpecB → SpecA, which is typically defined as the degree of the function field extension [k(SpecB): k(SpecA)]. It is a weighted measure of the cardinality of the fibers of ϕ that reflects both the ramification and degree of each closed point in the fiber (and as a consequence, it is the same for every fiber and is an invariant of ϕ). 58 CHAPTER 2. DEDEKIND DOMAIN

The ideal norm in number fields

We now consider the special case A = Z,K = Q, where B = OL. In this situation we may simply write N in place of NB|A and call it the absolute norm.

N(q) = (pfq ) where p ∈ Z is the unique prime in q ∩ Z.

Note that Z is a PID, OL and OL-ideal a are both free Z-module, and have a Φ Z-module isomorphism OL −→ a.

(a) free module index [OL : a]Z = (det Φ)

(b) sub-lattice index [OL : a] = | det Φ|

Therefore one gets N(a) = ([OL : a]).

♦ 命题 2.4.3

If b ⊆ a are nonzero fractional ideals of OL, then

[a : b]Z = ([a : b])

× Proof. For any α ∈ L we have [a : b] = [αa : αb] and [a : b]Z = [αa : αb]Z, so we can assume without loss of generality that a and b are ideals in OL. We then have a tower of free Z-modules b ⊆ a ⊆ OL, and therefore

[OL : a][a : b] = [OL : b]

Replacing both sides with the Z-ideals they generate, we have

N(a)([a : b]) = [OL : b]

1 and therefore ([a : b]) = N(a− b) = [a : b]Z.

注记. We can view the absolute norm N : F(OL) → F(Z) as a homomorphism F O → Q× N : ( L) >0. 2.5. RAMIFICATION IN DEDEKIND DOMAIN 59

2.5 Ramification in Dedekind domain

Let p be a prime of A, then pB is a proper ideal of B. The prime decomposition of pB is e1 ··· er pB = q1 qr where qi is exactly the prime above p.

♦ 定义 2.5.1

1. ei is the ramification index of qi.

2. fi := [B/qi : A/p] is the residue degree (or inertia degree) of qi.

♦ 定理 2.5.1 ∑r If L|K is separable of degree n, then eifi = n. i=1

Proof. We may assume that A is a DVR with maximal ideal m. ⊕n Note that B is a free modle of rank n, we write B = Abi. Then {bi} is a i=1 basis of A/p-linear space B/pB. We have dimA/p B/pB = n. On the other hand, by CRT, we obtain ∏r ∼ ei B/pB = B/qi i=1

e e ei Let us consider dimension of A/p-linear space B/q (q := qi ). Note that pqj ⊆ qj+1, so qj/qj+1 could be thought as A/p-linear space. e−1 ∼ B/qe It’s easy to see that B/q = qe−1/qe , by induction, we get ⊕ ⊕ ⊕ e ∼ 2 e−1 e B/q = B/q q/q ··· q /q

We set q = (b) since B is PID, consider the map B/q → qj/qj+1, x 7→ bjx. It’s j j+1 a bijection. So dimA/p(q /q ) = dimA/p(B/q) = f.

♦ 定义 2.5.2

1. L|K is totally ramified at q if eq = [L : K] (equivalently, fq = 1 = gp = 1).

2. L|K is unramified at q if eq = 1 and B/q|A/p is a separable extension. 3. L|K is unramified above p if it is unramified at all q|p, equivalently, if B/pB is a finite ´etale algebra over A/p. When L|K is unramified above p we say that 60 CHAPTER 2. DEDEKIND DOMAIN

• p remains inert in L if pB is prime (equivalently, eq = gp = 1, and fq = [L : K]).

• p splits completely in L if gp = [L : K] (equivalently, eq = fq = 1 for all q|p).

Totally ramifid and Eisenstein Polynomial

n n−1 A (monic) polynomial f(x) = x +cn−1x +···+c1x+c0 ∈ A[x], is Eisenstein at a prime p when each coefficient ci ≡ 0 mod p and the constant term c0 ̸≡ 0 mod p2.

Let P be a prime ideal of OL which is above p and n = [L : K], e = e(P|p).

♦ 定理 2.5.2 A has a prime p which is totally ramified in L iff L = K(α) for some α which is the root of an Eisenstein polynomial at p.

Proof. Taking ideal norms, N(P) = p, since e = n. We consider α ∈ P\P2. The characteristic polynomial of α over K, is f(x) = n n−1 n x + cn−1x + ··· + c1x + c0 ∈ OK [x]. The constant term is a0 = (−1) NL|K (α) ∈ p\p2, so f(x) is an Eisenstein polynomial at p. e For the reverse, it’s easy to see α ≡ 0 mod P. Since c0, ··· , cn−1 ≡ 0 mod P , i e+1 we get ciα ≡ 0 mod P for i = 1, ··· , n−1. Therefore

n e+1 α + c0 ≡ 0 mod P

e+1 n e+1 Note that c0 ̸≡ 0 mod P . So α ̸≡ 0 mod P , we get e + 1 > n.

Galois extensions

Now we consider Galois extension L|K. Let G = Gal(L|K). Note that B is G-invariant, so we can restrict σ ∈ G to B|A. Moreover, ∀σ ∈ G, σq ∩ A = σ(q ∩ A) = σp = p. Therefore G acts on {q|p}.

♦ 命题 2.5.1 1. The group G acts transitively on the set {q|p}.

2. fq and eq are the same for every q|p.

Proof. Let {q|p} = {q1, ··· , qn}. Assume that q1 and q2 lie in distinct G-orbits. By 2.5. RAMIFICATION IN DEDEKIND DOMAIN 61

CRT, B ≃ B × · · · × B q1 ··· qn q1 qn −1 and we may choose b ∈ B so that b ≡ 0 mod q2 and b ≡ 1 mod σ (q1) for all ∈ σ G. Then ∏ NL|K (b) = σ(b) ≡ 1 mod q1 σ∈G so NL|K (b)/∈ A ∩ q1 = p, a contradiction..

Now we fix a q|p and consider the stabilizer of it denoted by Dq. We call the subgroup Dq the decomposition group of q.

Reduction Let’s consider Dq acts on B/q, since q is Dq-invariant. We have a natural homomorphism

πq : Dq → Gal(κ(q)|κ(p)), σ 7→ [ b 7→ σ(b)] ⊔r Coset decomposion G = σiDq, where σiq = qi. Moreover, the decomposion i=1 −1 | | group of qi is exactly σiDqσi , thus Dq = ef.

♦ 命题 2.5.2 1. κ(q)|κ(p) is normal;

2. πq is surjective.

Proof. Let F be the separable closure of κ(p) in κ(q), so that restriction to F induces ∼ an isomorphism Autκ(p)(κ(q)) −→ Gal(F |κ(p)). Since F is a finite separable extension of κ(p), the primitive element theorem implies that F = κ(p)(α) for some α ∈ F ×. Let us now pick a ∈ B such that a ≡ α mod q and a ≡ 0 mod σ−1(q) for all σ ∈/ Dq; the CRT implies that such an a. Now define ∏ g(x) := (x − σ(a)) ∈ A[x] σ∈G

For each σ ∈/ Dq we have σ(a) = 0, so 0 is the root of g(x) with multiplicity at least m := #(G\Dq) and the remaining roots are σ(α) for σ ∈ Dq. It follows that g(x)/xm is equal to a power of the minimal polynomial of α, and this means that every Gal(F |κ(p))-conjugate of α has the form σ(α) for some σ ∈ Dq. It follows that πq is surjective. To show that κ(q)|κ(p) is a normal extension it suffices to show that each a ∈ κ(q) is the root of a monic polynomial in κ(p)[x] that splits completely in κ(q)[x]. 62 CHAPTER 2. DEDEKIND DOMAIN

For each a ∈ B, we can define g ∈ A[x] as above, showing that each a ∈ κ(q) is a root of the monic polynomial g, which splits completely in κ(q)[x] as desired.

♦ 定义 2.5.3

Iq := ker πq is the inertial group of q.

注记. We have an exact sequence

1 → Iq → Dq → Autκ(p)κ(q) → 1 and #Iq = ep[κ(q): κ(p)]i.

Iq = {σ ∈ Dq | σ = 1}

= {σ ∈ Dq | ∀ α ∈ κ(q), σ(α) = α}

= {σ ∈ Dq | ∀ α ∈ B, σ(α) ≡ α mod q} = {σ ∈ G | ∀ α ∈ B, σ(α) ≡ α mod q}

If κ(q)|κ(p) is separable, then κ(qI ) = κ(p). We have

Galois extension

r f e K LDq LIq L

1 f 1 κ(p) κ(p) κ(qI ) κ(q)

Galois extension

注记. If κ(p) is a perfect field, and in particular, whenever κ(p) is a finite field, then κ(q)|κ(p) is a Galois extension.

Actually, in the general case, κ(qI ) is the separable closure of κ(p) in κ(q). Let p denotes the characteristic of κ(p), we have

s r f0 ep K LDq LIq L

s 1 f0 p κ(p) κ(p) κ(qI ) κ(q) 2.5. RAMIFICATION IN DEDEKIND DOMAIN 63

Frobenius elements and Artin symbols

(Convention in this subsection): κ(p) are finite for all p. This holds, for example, whenever K is a global field.

When working with finite residue fields we may write Fq := κ(q) and

Fp := κ(p); these are finite fields of p-power order, where p is the characteristic

of Fq (and of Fp).

We always assume that p (equivalently, q) is unramified, then ep = eq = 1

and Iq is trivial.

In this case we have an isomorphism ∼ πq : Dq −→ Gal(Fq|Fp)

The Galois group Gal(Fq|Fp) is the cyclic group of order fp = [Fq : Fp] generated by the Frobenius automorphism

F x 7→ x# p

♦ 定义 2.5.4

The inverse image of the Frobenius automorphism of Gal(Fq|Fp) under πq

is the Frobenius element σq ∈ Dq.

注记. 1. The Frobenius element σq is the unique σ ∈ G such that for all x ∈ B F we have σ(x) ≡ x# p mod q.

2. For all {q|p}, the Frobenius elements σq are conjugate in G. The conjugacy

class is the Frobenius class of p, denoted Frobp.

There is another notation commonly used to denote Frobenius elements that includes the field extension in the notation. For each unramified prime q of L we define the Artin symbol ( ) L|K := σ q q ( ) L|K 引理 2.5.1. p splits completely iff q = 1. 注记. We will see that the extension L|K is completely determined by the set of primes p that split completely in L. Thus in some sense the Artin symbol captures the essential structure of L|K. 64 CHAPTER 2. DEDEKIND DOMAIN

♦ 命题 2.5.3

F F F F [ qE : p] Proof. For the first claim, note that # qE = (# p) . The second claim follows from the commutativity of the lower right square in the commutative diagram F 7→ # p F |F of Proposition 待补: the Frobenius automorphism x x of Gal( qE p) is the F 7→ # p F |F F restriction of the Frobenius automorphism x x of Gal( q p) to qE .

When L|K is abelian, the Artin symbol takes the same value for all q|p and we may instead write ( ) L|K := σ := σ p p q In this setting we now view the Artin symbol as a function mapping unramified primes p to Frobenius elements σp ∈ G. We wish to extend this map to a multiplicative homomorphism from the ideal group FA to the Galois group G = Gal(L|K), but ramified primes q|p cause problems: the homomorphism

πq : Dq → Gal(Fq|Fp) is not a bijection. F S F For any set S of primes of A, let A denote the subgroup of A generated by the primes of A that do not lie in S (a free abelian group).

♦ 定义 2.5.5 Assume that L|K is Abelian, and let S be the set of primes of A that ramify in L. The Artin map is the homomorphism ( ) L|K : F S → (L|K) · A Gal ( ) ∏ ∏ L|K ei ei 7→ pi pi

注记. One of the main results of class field theory is that the Artin map is surjective. √ 例 2.5.1 (Quadratic fields). Let K = Q and L = Q( d) for some square-free integer d ≠ 1. Then Gal(L|K) ≃ {1}. The only ramified primes p = (p) of A = Z are 2.5. RAMIFICATION IN DEDEKIND DOMAIN 65 those that divide the discriminant D. For p ∤ D, ( ) ( √ ) ( ) L|K Q( d)|Q D = = = 1 p (p) p

例 2.5.2 ((Cyclotomic fields)). Let K = Q and L = Q(ζn) for some n > 1. Let p a prime that does not divide n, and let [p] denote the residue class of p in (Z/nZ)× ≃

Gal(Q(ζn)|Q). Then ( ) Q(ζ )|Q n = [p] (p) 66 CHAPTER 2. DEDEKIND DOMAIN

2.6 *Different and discriminant

In this section, we follow the conventin 2.3.

Different ideal

The trace form Tr : L × L → K is a non-degenerate bilinear form. Every A-lattice M in L has a dual lattice

∗ M := {x ∈ L | Tr(xM) ⊆ A} which is an A-lattice in L isomorphic to the dual A-module HomA(M,A). Note that every element in F(B) is an A-lattice in L.

引理 2.6.1. F(B) is closed under the operation of taking dual lattice.

♦ 定义 2.6.1. Different The different ideal is the inverse of B∗. That is ∗ −1 ∗ ∗ DB|A := (B ) = (B : B ) = {x ∈ L | xB ⊆ B}

∗ 注记. DB|A is a B-ideal since B ⊆ B .

引理 2.6.2 (Property of different).

Transitivity Assume that M|L|K,C|B|A, then DC|A = DC|BDB|A. −1 Localization Let S be a multiplicative subset of A, then S DB|A = DS−1B|S−1A. b b b B |A B|A q|p D b b = D | B Completion Let q p be the completions of at then Bq|Ap B A q

例 2.6.1 (Important case). If B = A[α] for some α ∈ L with the minimal polynomial f, then ′ DB|A = (f (α))

Proof.

Now we define the different of a element α ∈ B as following   ′ fα(α) if L = K(α), δB|A(α) =  0 else.

Actually we can prove that DB|A is generated by all δB|A(α).

DB|A = (δB|A(α): α ∈ B) 2.6. *DIFFERENT AND DISCRIMINANT 67

Discriminant ideal

♦ 定义 2.6.2 Let S|R be a ring extension in which S is a free R-module of rank n. For

any x1, ··· , xn ∈ S we define the discriminant

disc(x1, ··· , xn) := det[TrS|R(xixj)]ij ∈ R

注记. If x1, ··· , xn ∈ S satisfy a non-trivial R-linear relation then the discriminant will be zero.

♦ 命题 2.6.1 As in convention 2.3, let Ω|K be a field extension for which there are dis-

tinct σ1, ··· , σn ∈ HomK (L, Ω). For any e1, ··· , en ∈ L we have ∏ 2 n−1 2 disc(1, x, x , ··· , x ) = (σi(x) − σj(x)) i

∑n Proof. For 1 ⩽ i, j ⩽ n we have TrL|K (eiej) = σk(eiej). Therefore k=1

disc(e1, ··· , en) = det[TrL|K (eiej)]ij

= det([σk(ei)]ik[σk(ej)]kj) 2 = det[σi(ej)]ij

i−1 Now let x ∈ L and put ei := x for 1 ⩽ i ⩽ n. Then ∏ 2 ··· n−1 j−1 2 − 2 disc(1, x, x , , x ) = det[σi(x) ]ij = (σi(x) σj(x)) i

注记. Such a field extension Ω|K always exists, since L|K is separable (Ω = Ksep works). ∏ 例 2.6.2. For a polynomial f(x) = i(x−αi), the discriminant of f is ∏ 2 disc(f) := (αi − αj) i

If A is a Dedekind domain, f ∈ A[x] is a monic separable polynomial, and α is the image of x in A[x]/(f(x)), then

− disc(f) = disc(1, α, ··· , αn 1) ∈ A 68 CHAPTER 2. DEDEKIND DOMAIN

Now let M be an A-lattice in L. Then M is a finitely generated A-module that contains a K-basis for L. We want to define the discriminant of M in a way that does not require us to choose a basis.

Let us first consider the case where M is a free A-lattice. If e1, ··· , en ∈ M ′ ··· ′ ∈ and e1, , en M are two A-bases for M, then

′ ··· ′ 2 ··· disc(e1, , en) = u disc(e1, , en) for some unit u ∈ A×. This actually gives us a basis independent definition when A = Z. In this case B is always a free Z-lattice, and u2 = 1.

♦ 定义 2.6.3 let M be an A-lattice in L, the discriminant D(M) of M is the A-module

generated by {disc(x1, ··· , xn): x1, ··· , xn ∈ M}.

引理 2.6.3. Let M ′ ⊆ M be free A-lattices in L. The discriminants D(M ′) ⊆ D(M) are nonzero principal fractional ideals. If D(M ′) = D(M) then M ′ = M.

Proof. Let e1, ··· , en be an A-basis for M, then D(M) = (disc(e1, ··· , en)).

♦ 命题 2.6.2 let M be an A-lattice in L, the discriminant D(M) ∈ F(A).

Proof. The A-module D(M) ⊆ K is nonzero because M contains a K-basis e for L. To show that D(M) is a finitely generated as an A-module (and thus a fractional ideal), we use the usual trick: make it a submodule of a noetherian module. Let N be the free A-lattice in L generated by e and then pick a nonzero a ∈ A such that M ⊆ a−1N. We then have D(M) ⊆ D(a−1N), and D(a−1N) is a principal fractional ideal of A, hence a noetherian A-module, so its submodule D(M) must be finitely generated.

♦ 定义 2.6.4 As in convention 2.3, the discriminant of B|A is the discriminant of B as 2.6. *DIFFERENT AND DISCRIMINANT 69

an A-module:

DB|A := D(B)

注记. DB|A is an A-ideal.

例 2.6.3. Consider the case A = Z,K = Q,L = Q(i),B = Z[i]. Then B is a free

A-lattice with basis (1, i) and we can compute DB|A in three ways:

• disc(1, i) = −4. (Dedinition)

• The non-trivial automorphism of L|K fixes 1 and sends i to −i, so we could instead compute disc(1, i) = −4.

• We have B = Z[i] = Z[x]/(x2 + 1) and can compute disc(x2 + 1) = −4.

引理 2.6.4 (Property of discriminant). [M:L] Transitivity Assume that M|L|K,C|B|A, then DC|A = (DB|A) NB|A(DC|B). −1 Localization Let S be a multiplicative subset of A, then S DB|A = DS−1B|S−1A.

♦ 定理 2.6.1

DB|A = NB|A(DB|A)

Proof. The different and discriminant are both compatible with localization, sowe may assume that A is a DVR and therefore B is a free A-lattice in L.

Ramification

Having defined the different and discriminant ideals we now want to understand how they relate to ramification. If p is a prime of A then we can factor the B-ideal pB as e1 ··· er pB = q1 qr The Chinese remainder theorem implies

e1 × · · · × er B/pB = B/q1 B/qr

This is a commutative A/p-algebra. It is a product of fields iff we have ei = 1 for all i, and it is a finite ´etale-algebra iff it is a product of fields that are separable extensions of A/p. The following lemma relates the discriminant to the property of being a finite ´etale algebra.

引理 2.6.5. Let k be a field and let R be a commutative k-algebra with k-basis r1, ··· , rn. Then R is a finite ´etale k-algebra iff disc(r1, ··· , rn) ≠ 0. 70 CHAPTER 2. DEDEKIND DOMAIN

♦ 定理 2.6.2 Let q be a prime of B lying above a prime p of A. The extension L|K is

unramified at q iff q does not divide DB|A, and it is unramified at p iff p does

not divide DB|A.

推论. Only finitely many primes of A (or B) ramify.

♦ 定理 2.6.3 Let q|p be a prime of B with B/q|A/p separable. Then

eq − 1 ⩽ vq(DB|A) ⩽ eq − 1 + vq(eq)

and the lower bound is an equality iff q is tamely ramified.

注记. Recall that unramified is a case of tamely ramified.

The discriminant of an order

Assume 2.3 and let O be an order with integral closure B. Note that such an order is precisely the A-lattice in L. The fact that O ⊆ B implies that D(O) ⊆ DB|A is an A-ideal.

例 2.6.4. If O is an order of the form A[α], where α ∈ B generates L = K(α) with minimal polynomial f ∈ A[x], then O is a free A-lattice with basis 1, α, ··· , αn−1, and we may compute its discriminant as

D(O) = (disc(f)) = (disc(1, α, ··· , αn−1)) which is a principal A-ideal contained in DB|A.

If O and B both are a free A-lattices, then we have

2 O 2 DO|A = (det P ) DB|A = [B : ]ADB|A where P is the matrix of the A-linear map ϕ : B → O that sends an A-basis for B to an A-basis for O and [B : O]A is the module index (a principal A-ideal). (Important case): Let A = Z and L be a number field, the integer (det P )2 is uniquely determined and it necessarily divides disc(f), the generator of the principal ideal D(O) = D(A[α]). It follows that if disc(f) is squarefree then we must have B = O = A[α]. More generally, any prime p for which vp(disc(f)) is odd 2.6. *DIFFERENT AND DISCRIMINANT 71 must be ramified, and any prime that does not divide disc(f) must be unramified. The module index [B : O]Z = ([B : O]) is the principal ideal generated by the index of O in B (as Z-lattices), and we have the relation

∆(O) = [B : O]2∆(B) between the absolute discriminant of the order O and its integral closure B.

例 2.6.5. Consider A = Z,K = Q with L = Q(α), where α3−α−1 = 0. We can compute the absolute discriminant of Z[α] as

disc(x3 − x − 1) = −23

The fact that −23 is squarefree immediately implies that 23 is the only prime of 2 A that ramifies, and we have ∆(Z[α]) = −23 = [OL : Z[]] ∆L, which forces [OL :

Z[α]] = 1, so ∆L = −23 and OL = Z[α].

♦ 定理 2.6.4 Assume 2.3 and let O be an order with integral closure B and conductor

c. Then D(O) = NB|A(c)DB|A

The geometric significance of the different

For an arbitrary extension B|A of commutative rings, consider the homomorphism

µ : B ⊗A B → B x ⊗ y 7→ xy whose kernel we denote by I. Then

1 2 ⊗ ΩB|A := I/I = I B⊗B B is a B ⊗ B-module, and hence in particular also a B-module, via the embedding B,→ B ⊗ B, b 7→ b ⊗ 1. It is called the module of differentials of B|A, and its elements are called Kahler differentials. If we put

dx = x ⊗ 1 − 1 ⊗ x mod I2 then we obtain a mapping → 1 d : B ΩB|A 72 CHAPTER 2. DEDEKIND DOMAIN satisfying d(xy) = xdy + ydx and da = 0 for a ∈ A. Such a map is called a derivation of B|A. One can show that d is universal among all derivations of B|A ∑ 1 with values in B-modules. ΩB|A consists of the linear combinations yidxi. The link with the different is now this. ♦ 命题 2.6.3 D 1 The different B|A is the annihilator of the B-module ΩB|A, i.e.

DB|A = {x ∈ B : xdy = 0 for all y ∈ B}

′ ′ ′ Proof. If A is any commutative A-algebra and B = B ⊗A A , then it is easy to 1 1 ⊗ ′ see that ΩB′|A′ = ΩB|A A A . Thus the module of differentials is preserved under localization and completion, and we may therefore assume that A is a complete discrete valuation ring. Then we find that B = A[x], and if f(x) ∈ A[x] is the minimal polynomial of x, then ΩB|A is generated by dx. The annihilator of dx is ′ ′ f (x). On the other hand, we have DB|A = (f (x)). This proves the claim. 第 3 章 Ramification theory 3

3.1 Extensions of complete DVR

We set L|K,B|A, note that an extension of DVR may be not a DVR.

♦ 定理 3.1.1 Let K be a complete discret valuation field with (A, p) as valaution ring. Assume that L|K is a finite separable extension. Then B is a DVR whose maximal ideal q is necessarily the unique prime above p.

Proof. It suffices to show that #{q|p} = 1. suppose for the sake of contradiction that q1, q2 ∈ {q|p} with q1 ≠ q2. Choose b ∈ q1\q2 and consider the ring A[b] ⊆ B.

The ideals q1 ∩ A[b] and q2 ∩ A[b] are distinct prime ideals of A[b] containing pA[b], and both are maximal, since they are nonzero and dim A[b] = dim A = 1 (note that A[b] ⊆ B is integral over A and therefore has the same dimension). The quotient ring A[b]/pA[b] thus has at least two maximal ideals. Let f ∈ A[x] be the minimal polynomial of b over K, and let f ∈ k[x] be its reduction to the residue field A/p. We have k[x] ≃ A[x] ≃ A[b] (f) (p, f) pA[b] thus the ring k[x]/(f) has at least two maximal ideals, which implies that f is divisible by two distinct irreducible polynomials. We can thus factor f = gh with g and h coprime. By Hensel’s Lemma, we can lift this to a non-trivial factorization f = gh, contradicting the irreducibility of f.

例 3.1.1. If A is the DVR Z(5) with fraction field K = Q and we take L = Q(i), then the integral B = Z(5)[i], which is a PID but not a DVR: the ideals (1+2i) and (1−2i) 74 CHAPTER 3. RAMIFICATION THEORY are both maximal (and not equal).

But if we take completions we get A = Z5 and K = Q5, and now L = Q5(i) = 2 Q5 = K, since x + 1 has a root in F5 ≃ Z5/5Z5 that we can lift to Z5 via Hensel’s lemma; thus if we complete A then B = A is a DVR as required.

引理 3.1.1. Let V be a vector space of finite dimension over a complete field K. Every norm on V induces the same topology, in which V is a complete metric space.

♦ 定理 3.1.2

Let (A, p) be a complete DVR with fraction field K, discrete valuation vp,

vp(x) and absolute value |x|p := c with 0 < c < 1. Let L|K be a finite extension of degree n. The following hold. 1 n 1. There is a unique absolute value | · | := |NL|K (·)|p on L that extends | · |p. 2. The field L is complete with respect to | · |, and its valuation ring {x ∈ L | |x| ⩽ 1} is equal to B. 3. If L|K is separable then B is a complete DVR whose maximal ideal q 1 vq(·) induces | · | = | · |q := c e where e is the ramification index .

Proof. Assuming for the moment that | · | is actually an absolute value, it’s easy to see that |·| extends |·|p and is therefore a norm on L. The fact that |·|p is nontrivial × a means that |x|p ≠ 1 for some x ∈ K , and |x| = |x|p = |x| only for a = 1, which implies that |·| is the unique absolute value in its equivalence class extending |·|p. Every norm on L induces the same topology, so | · | is the only absolute value on

L that extends | · |p. We now show | · | is an absolute value, only need to check the triangle inequality. It suffices to show

|x| ⩽ 1 =⇒ |x + 1| ⩽ |x| + 1 since we always have |y + z| = |z||y/z + 1| and |y| + |z| = |z|(|y/z| + 1), and without loss of generality we assume |y| ⩽ |z|. In fact the stronger implication |x| ⩽ 1 =⇒ |x + 1| ⩽ 1 holds:

|x| ⩽ 1 ⇐⇒ |NL|K (x)|p ⩽ 1 ⇐⇒ NL|K (x) ∈ A ⇐⇒ x ∈ B ⇐⇒ x+1 ∈ B ⇐⇒ |x+1| ⩽ 1

We now assume L|K is separable. Then B is a DVR and it is complete because it is the valuation ring of L. The valuation vq extends vp with index e, so vq(x) = 3.1. EXTENSIONS OF COMPLETE DVR 75

× 1/e 1/evq(x) evp(x) for x ∈ K . We have 0 < c < 1, so |x|q := c is an absolute value on

L induced by vq. To show it is equal to | · |, it suffices to show that it extends | · |p. For x ∈ K× we have 1 vq(x) vp(x) |x|q = c e = c = |x|p and the theorem follows.

1 ∈ 推论. vq(x) = f vp(NL|K (x)) for all x L.

注记. 1. The transitivity of NL|K in towers implies that we can uniquely extend the absolute value on K to an algebraic closure K. 2. A valuation ring A is Henselian iff the absolute value of its fraction field K can be uniquely extended to K.

Unramified extension

Let K be a complete discret valuation field. A finite extension L|K of degree n is unramified if l|k is separable and [l : k] = n.

注记. If L|K is unramified,then e(ω|v) = 1, f(ω|v) = n, Gω = Gv.

Cunr The finite unramified extensions L of K form a category K whose morphisms are K-algebra homomorphism; and the finite separable extensions l of k Csep form a category k whose morphisms are k-algebra homomorphisms. F Cunr → Csep Let : K k denotes the ”reduction” functor that sends L to its residue field l and K-homomorphism ϕ : L1 → L2 to its reduction ϕ : l1 → l2, α 7→ ϕ(α).

♦ 定理 3.1.3

Cunr Csep F The category K and k are equivalent via .

Proof. We need to Check the following:

• The reduction is well-defined; • F is essentially surjective: each separable l|k is isomorphic to the residue field of some unramified L|K;

• F is full and faithful: the induced map HomK (L1,L2) → Homk(l1, l2) is a bijection. 76 CHAPTER 3. RAMIFICATION THEORY

Given a finite separable extension l|k, we may write k[x] l ≃ k(α) = (g(x)) where g ∈ k[x] is necessarily monic, irreducible, separable, and of degree n := [l : k]. Let g ∈ A[x] be any monic lift of g, then g is also irreducible, separable, and of degree n. Now let K[x] L := = K(α) (g(x)) where α is the image of x in K[x]/(g(x)). Then L|K is a finite separable extension, (p, g(α)) is the unique maximal ideal of A[α] and B A[α] A[x] k[x] ≃ ≃ ≃ ≃ l q (p, g(α)) (p, g(x)) (g(x))

We now consider the induced map

∼ HomK (L1,L2) −→ HomA(B1,B2) → Homk(l1, l2)

As above, we write l1 = k(α) = k[x]/(g(x)) and have B1 = A[α]. Each A- module homomorphism in HomA(B1,B2) is uniquely determined by the image of α in B2, thus gives us a bijection between HomA(B1,B2) and the roots of g in

B2, similarly in the reduction.

Now g is separable, so each root of g in l2 lifts to a unique root of g in B2, by Hensel’s Lemma.

注记. In the proof above we actually only used the fact that L1|K is unramified.

The map HomK (L1,L2) → Homk(l1, l2) is a bijection even if L2|K is not unramified.

♦ 命题 3.1.1 L|K is unramified iff B = A[α] for some α ∈ L whose minimal polynomial g ∈ A[x] has separable image g in k[x]. Moreover, g is separable ⇐⇒ α is ′ ′ × ′ separable ⇐⇒ g (α) ≠ 0 ⇐⇒ g (α) ∈ B ⇐⇒ vp(g (α)) = 0.

Proof. For the reverse direction note that g must be irreducible, since otherwise we could use Hensel’s lemma to lift a non-trivial factorization of g to a non-trivial factorization of g, so the residue field extension is separable and has the same degree as L|K.

推论. 1. Unramified extension is ”good extension”; 3.1. EXTENSIONS OF COMPLETE DVR 77

2. Unramified extension is separable.

Proof. (Lifting) L = K(α), then EL = E(α).

推论. Suppose (n, char(k)) ≠ 1, then K(ζn)|K is unramified.

n Proof. K(ζn) is a splitting field of K of f(x) = x −1. We see that the reduction g(x) of minimal polynomial of ζn is irreducible by Hensel’s lemma. Note that f ∈ k[x] is separable, so is g(x).

推论. If the residue field k = Fq is finite, then an extension L|K is unramified iff

L ≃ K(ζqn−1), where n = [L : K].

When this holds, B ≃ A[ζqn−1] and L|K is a Galois extension with Gal(L|K) ≃ Z/nZ.

Proof. (Sketch) Use Hensel’s lemma.

推论. If the residue field k = Fpn and K does not contain a primitive p-th root of unity. Then extension K(ζm)|K is ramified iff p divides m.

Proof. If p|m then K(ζp) ⊆ K(ζm). By previous corollarry, K(ζp) is unramified n ⇐⇒ p|p − 1 where n = [K(ζp): K]. ¯ ¯ 例 3.1.2. Consider A = Zp,K = Qp, k = Fp, and fix Fp and Qp. Fp has a unique ¯ extension of degree n in Fp, namely, Fpn . Thus Qp has a unique unramified extension of degree n; it can be explicitly constructed by adjoining a primitive root of unity ζpn−1 to Qp. The element ζpn−1 will necessarily have minimal polynomial n of degree n dividing xp − 1.

If L|K is separable, we can define the maximal unramified subextension of L|K which is denoted by Lu|K. When L = Ksep is the separable closure of K, Lu is the maximal unramified extension of K, denoted by Kunr

1. u := κ(u) is the separable closure of l|k. 2. L|Lu is totally ramified.

f K −→ Lu −→1 L k −→1 u −→e l

Qunr Q 例 3.1.3. The field p is an infinite extension of p with Galois group ¯ b Gal(Fp|Fp) ≃ Z

Qunr Z F The field p has value group and residue field p. 78 CHAPTER 3. RAMIFICATION THEORY √ √ K = Q Q ( 2), Q ( 3) 例 3.1.4.√Let 3. There are just three quadratic extensions: 3 3 , Q ( 6) and 3 . √ The extension Q3( 2) is the unique unramified quadratic extension, and we note that it can be written as a cyclotomic extension Q3(ζ8). The other two are both ramified, and can be defined by the Eisenstein polynomials x2−3 and x2−6.

♦ 定理 3.1.4 Assume that A is a complete DVR with finite residue field. Then L|K is × × unramified iffL N |K (B ) = A .

Tamely ramified extension

If char(k) = p > 0, l|k is separable and (e, p) = 1, then L|K is called tamely ramified.

注记. e = [L : Lu].

♦ 命题 3.1.2 √ u m A finite extension L|K is tamely ramified ⇐⇒ L = L ( πA) for some

uniformizer πA ∈ A with (m, p) = 1.

Proof. We may assume L|K is totally tamely ramified, in which case (p, n) = 1. n Then vq extends vp with index e = n and vq(πB) = n = vq(πA). This implies that n ∈ × πB = uπA for some unit u B . We have f = 1, so B and A have the same residue field, and if we lift u ∈ l = k to a unit uA in A and replace πA with uAπA, we can assume that u ≡ 1 mod q. Now define g(x) := xn−u ∈ B[x] with reduction g = xn−1 ∈ l[x]. We have g′(1) = n ≠ 0, so by Hensel’s Lemma 1.6.1 we can lift the root 1 of g(x) ∈ l to a root r of g(x) in B. Let π := π /r. Then π is a uniformizer for B and B = A[π], so L = K(π), and B √ n n n n n π = πB/r = πB/u = πA, so L = K( πA) as desired.

注记. If K is a Henselian field with respect to a exponential valuation. Afinite √ √ u m m extension L|K is tamely ramified ⇐⇒ L = L ( 1 a1, ··· , r ar), where (mi, p) = 1. In this case we have [L : K] = ef .

推论. Tamely ramified extension satifies lifting and composition and sub-inherited. 3.1. EXTENSIONS OF COMPLETE DVR 79

Proof. (Lifting) Let E|K be a finite extension, by previous result, Lu ⊆ (EL)u, then we have a commutative diagram: E (EL)u EL

K Lu L √ u u u Since L ⊆ EL, (EL) ⊆ EL we obtain EL = (EL) L. Note that L = L ( m1 a , ··· , √ √ √ 1 m u m m r ar), so EL = (EL) ( 1 a1, ··· , r ar)

We can define the maximal tamely ramified subextension of some L|K which is denoted by Lt|K.

注记. 1. Let e = e′pa with (e′, p) = 1, then [Lt : Lu] = e′. 2. If Lt ≠ L, we call L|K wildly ramified (i.e. a ⩾ 1 or p|e).

Almost everything is contained in the following diagram:

unramified tamely ramified wildly ramified K Lu Lt L

separable κ(v) κ(u) = κ(t) κ(ω)

′ pa = e (p) Gv Gu Gω Gω

例 3.1.5. 1. If (n, p) = 1, then Q(ζn)|Q is unramified; s 2. If n = p , then Q(ζn)|Q is totally ramified; 3. Assume n = n′ps with (n′, p) = 1, then

unramified totally ramified Qp −−−−−−→ Qp(ζn′ ) −−−−−−−−→ Q(ζn)|Q 80 CHAPTER 3. RAMIFICATION THEORY

3.2 Higher Ramification Groups

Ramification Groups

We follow the previous section, let K be a complete descret valuation field, L|K be a finite Galois extension with Galois group G := Gal(L|K). For i ⩾ −1, the i-th ramification group is

Gi := {σ ∈ G : w(σ(α) − α) > i for all α ∈ B}

Note that for any σ ∈ G, we have that σ(α)−α ∈ B for all α ∈ L, and hence w(σ(α) − α) ⩾ 0 for all α ∈ L. Because w is discrete, we therefore have that G−1 =

G. As defined above, G0 is the inertia group of L|K. Also note that w(σ(α)−α) > i iff σ(α) − α ∈ qi+1 iff σ(α) ≡ α mod qi+1, so that

i+1 Gi = {σ ∈ G : σ(α) ≡ α mod q for all α ∈ B}

引理 3.2.1. The groups Gi are normal subgroups of G and give a filtration G ⊇

G0 ⊇ G1 ⊇ · · · .

Proof. Note that any σ ∈ G naturally induces an automorphism of the ring B/qi+1 i+1 i+1 i+1 (because σ(q ) = q for all i). Since Gi is the kernel of the map G → Aut(B/q ), we have that Gi is normal in G for all i ⩾ −1.

♦ 定义 3.2.1

For some β ∈ B such that B = A[β], define iL|K : G → Z ∪ {∞} by

iL|K (σ) := w(σ(β) − β)

引理 3.2.2. 1. σ ∈ Gi iff iL|K (σ) > i.

Proof. 1. Note that σ ∈ Gi iff w(σ(α) − α) > i for all α ∈ B, hence iL|K (σ) =

w(σ(β) − β) > i. Conversely, if iL|K (σ) > i, then w(σ(β) − β) > i, hence σ(β) ≡ β mod qi+1. But because σ ∈ G, we have that σ(α) ≡ α mod qi+1 for all α ∈ A, so σ(α) ≡ α mod qi+1 for all α ∈ B.

j + 1 and iL|K (τ) ⩾ j + 1, which by part (1) implies σ, τ ∈ Gj and, hence

στ ∈ Gj, hence iL|K (στ) ⩾ j + 1, and thus iL|K (στ) ⩾ inf(iL|K (σ), iL|K (τ)). 3.2. HIGHER RAMIFICATION GROUPS 81

−1 −1 3. σ ∈ Gi iff τστ ∈ Gi for all τ ∈ G, so iL|K (σ) ⩾ i + 1 iff iL|K (τστ ) ⩾ i + 1, −1 for all i. Thus iL|K (τστ ) = iL|K (σ).

注记. Part (1) of this proposition shows that, because the elements of each Gi are what they are, independent of our choice of generator a ∈ B in the definition of iL|K , we must have that for any choice, the function is the same. Furthermore, it shows that knowing the function iL|K is actually equivalent to knowing all the groups Gi.

引理 3.2.3. Gi = 1 for all sufficiently large i.

Proof. Note that iL|K (1) = w(0) = ∞, but if we set m = sup(iL|K (σ)), then no σ=1̸ σ ∈ G, σ ≠ 1 has iL|K (σ) ⩾ m + 1, so that Gm = 1.

For a subgroup H ⊆ G, let K′ be the fixed field of H, so that L|K′ is Galois with H = Gal(L|K′). Because the residue field k′ is an intermediate field of the extension l|k, which by assumption is separable, we also have that l|k′ is separable.

Thus we can apply our results above, so that we get a filtration Hi on H, and a function iL|K′ : H → Z ∪ {∞}. The following (obvious) result shows that the Hi are easy to compute:

♦ 命题 3.2.1

For any subgroup H ⊆ G, we have Hi = H ∩ Gi.

i+1 Proof. Because Gi is the kernel of the map G → Aut(B/q ) and Hi is the kernel i+1 of the map H → Aut(B/q ), we have that Hi = H ∩ Gi.

推论. L|K is unramified iff G0 = 1. In particular, the ramification degree of L|K is #G0 and the maximal unramified extension of K in L is the fixed field of G0 and has galois group G/G0.

推论. L|K is tamely ramified iff G1 = 1. In particular, if we write the ramification s s index as e = p m with (m, p) = 1 and p = chark, then #G1 = p and this divides

#l. Moreover, if K1 is the maximal tamely ramified extension of K inside L, then

K1 is the fixed field of G1, and is hence an extension of K0, and has Galois group

G/G1.

推论. We have a tower of fields L ⊇ K1 ⊇ K0 ⊇ K where K0|K is unramified of degree #G/G0, the extension K1|K0 is totally tamely ramified of degree #G0/G1 × (which divides #l ) and L|K1 is totally wildly ramified of degreeG # 1 (which divided #l). 82 CHAPTER 3. RAMIFICATION THEORY

The Structure of Gi/Gi+1

We define the following subgroups:

i × Ui := 1 + q ⊆ B

× Note that U0 = B . ♦ 命题 3.2.2 × The quotient map B → l induces an isomorphism U0/U1 ≃ l .

For each n ⩾ 1, the map u 7→ u − 1 induces an isomorphism Un/Un+1 ≃ qn/qn+1 ≃ l.

Proof. Because U0 = q, and B → l is surjective, and q is its kernel, we have that × the image of U0 is l . Forgetting about the additive structure, this is a surjective × homomorphism from the group U0 to the group l . The kernel consists of those units which get sent to 1 + q, which by definition is U1. n n+1 For each n ⩾ 1, we have that the map f : Un/Un+1 → q /q defined by n+1 f(u + Un+1) = (u − 1) + q is a homomorphism, because for any u, v ∈ Un,

(uv − 1) − (u − 1) − (v − 1) = uv − u − v + 1 = (u − 1)(v − 1) ∈ q2n ⊆ qn+1 and therefore

n+1 n+1 f(u + Un+1) + f(v + Un+1) = ((u − 1) + q ) + ((v − 1) + q ) = (u − 1) + (v − 1) + qn+1 = (uv − 1) + qn+1

= f(uv + Un+1)

Its inverse is similarly defined and seen to be a homomorphism.

⩾ p ⊆ 推论. If l is of characteristic p, then for n 1,Un Un+1.

n n+1 n n+1 Proof. Because Un/Un+1 ≃ q /q , and q /q is an B-module which is annihilated by q, we have that Un/Un+1 is a one-dimensional l-vector space. Therefore, p p if l is of characteristic p, then (u + Un+1) = u + Un+1 = 1 + Un+1 for all u ∈ Un, p ⊆ hence Un Un+1. 3.2. HIGHER RAMIFICATION GROUPS 83

♦ 定理 3.2.1 ⩾ ∈ σ(x) ∈ ∈ × Let i 0. Then σ Gi iff x Ui for all x L .

Proof. We may assume σ ∈ G0. By Proposition 3.2.1, we can WLOG replace K u by L , the fixed field of G0, because it will have the same ramification groups. The intermediate field Lu is the largest subfield such that Lu|K is unramified, and furthermore has the property that L|Lu is totally ramified. Thus, by Proposition 18 in Chapter 1, §6 of [Jean-Pierre Serre, Local Fields], we can now choose a generator u σ(x) ∈ ∈ × a of B over B that is an element of q. Suppose that x Ui for all x L . Then σ(a) ∈ σ(x) − ∈ i σ(a) − ⩾ in particular a Ui, so that x 1 q , hence w( a 1) i, and hence ( ) σ(a) i | (σ) = w(σ(a) − a) = w a( − 1) ⩾ i + 1 L K a

× Conversely, if σ ∈ Gi, then for any x ∈ L , we have that ( ) σ(x) w(σ(x) − x) w − 1 = = w(σ(x) − x) ⩾ i + 1 x w(x)

σ(x) − ∈ i+1 σ(x) ∈ ⊆ so that x 1 q , x Ui+1 Ui.

♦ 定理 3.2.2 Let i ⩾ 0. Then the function defined by

σ(πL) θi(σ) := mod Ui+1 πL

is a homomorphism θi : Gi → Ui/Ui+1 which is independent of the choice of

uniformizer πL, and whose kernel is Gi+1.

σ(πL) Proof. Let σ ∈ Gi, then ∈ Ui, so θi is a well-defined map from Gi to Ui/Ui+1. πL ′ ′ Note that any other uniformizer π for L differs from πL by a unit, say π = uπL, so that ′ σ(π ) σ(πL) σ(u) ′ = π πL u ≡ i+1 σ(u) ≡ We have σ(u) u mod q and thus u 1 mod Ui+1. Therefore, θi is independent of the choice of uniformizer. To see that θi is a homomorphism, note that for any σ, τ ∈ Gi, ( ) σ τ(πL) στ(πL) σ(πL) τ(πL) πL θi(στ) = mod Ui+1 = τ(π ) mod Ui+1 πL πL πL L πL 84 CHAPTER 3. RAMIFICATION THEORY ( ) τ(πL) × τ(πL) τ(πL) Since ∈ B , we again have that σ ≡ mod Ui+1, hence πL πL πL ( ) σ τ(πL) πL ≡ 1 mod Ui+1 τ(πL) πL

σ(πL) τ(πL) and thus θi(στ) = = θi(σ)θi(τ). Finally, note that σ is in the kernel of θi πL πL σ(πL) iff ∈ Ui+1, which is the case precisely when πL ( ) σ(πL) iL|K (σ) − 1 = w(σ(πL) − πL) − 1 = w − 1 ⩾ i + 1 πL i.e. iL|K (σ) ⩾ i + 2, which is equivalent to σ ∈ Gi+1.

引理 3.2.4. The group G0/G1 is cyclic, of order relatively prime to the characteristic of l.

× Proof. U0/U1 ≃ l . Because G0 is finite, we have that G0/G1 is isomorphic to a finite subgroup of l×. It is well-known that any finite subgroup of the multiplicative group of a field is cyclic, and it is clear that if the characateristic of l is p > 0, then the order of any finite subgroup of l× is relatively prime to p.

♦ 定理 3.2.3 If the characteristic of l is p > 0, then for every i ⩾ 1, the factor group

Gi/Gi+1 is a finite abelian p-group, and in fact a direct sum of cyclic groups of

order p; hence G1 is a p-group. If the characteristic of l is 0, then G1 is trivial,

and G0 is cyclic.

Proof. We have an injection from Gi/Gi+1 into Ui/Ui+1 for all i ⩾ 0. If the characteristic of k is p > 0, then for i ⩾ 1,Ui/Ui+1 is an abelian group annihilated by p, hence a direct sum of cyclic groups of order p. Because G = Gal(L|F ) is finite and

Gi ⊆ G for all i, we have that for all i ⩾ 1, Gi/Gi+1 is a finite abelian p-group, and indeed a finite direct sum of cyclic groups of order p. Thus #G1 is a power of p, and hence G1 is a p-group.

On the other hand, if the characteristic of l is 0, then Ui/Ui+1 will have no non-trivial finite subgroups for i ⩾ 1. But every Gi is finite, hence Gi/Gi+1 is finite, hence the image of Gi/Gi+1 in Ui/Ui+1 is finite, and therefore trivial. Thus, for i ⩾ 1, we have that Gi = Gi+1, and we know that eventually Gi is trivial, so that every Gi for i ⩾ 1 is trivial. Thus G1 is trivial. Finally, because G1 is trivial and G0/G1 is cyclic by the above Lemma, we have that G0 is cyclic. 3.2. HIGHER RAMIFICATION GROUPS 85

An Application to Puiseux Series

The field of Puiseux series over a field k is a generalization of k((T )), the field of Laurent series over k, which allows for rational powers of the indeterminate T instead of just integers. Specifically, the field of Puiseux series k{{T }} over ∪∞ 1/n 1/n k is defined to be Kn, where Kn = k((T )) and the T all live in some n=1 algebraic closure of k((T )). The following is Puiseux’s Theorem for an arbitrary field k (when k is complete with respect to a valuation, there is a separate theorem concerning the subfield of convergent series in k{{T }}).

♦ 定理 3.2.4. Puiseux’s Theorem Let k be an algebraically closed field of characteristic 0. Then the algebraic closure of k((T )) is k{{T }}.

Proof. Let K := k((T )), and let Kalg be an algebraic closure of K. Let L|K be a finite Galois subextension of Kalg|K, with Galois group G = Gal(L|K). Clearly, the residue field of K = k((T )) relative to the prime ideal (T ) in k[[T ]] is k, which is algebraically closed by hypothesis. Because L|K is finite, we must therefore have that l = k. Since Gal(l|k) ≃ G/G0, we have that G = G0. Because K is of characteristic 0, then we have that G is cyclic. Let L′|K be another finite Galois subextension of Kalg|K such that [L : K] divides [L′ : K]. Then LL′|K is also finite and Galois, and hence by the same argument Gal(LL′|K) is cyclic. Because the subgroups of a cyclic group are totally ordered by inclusion and

′ ′ ′ [LL : K] [LL : K] ′ #Gal(LL |L) = ⩽ = #Gal(LL |L) [L′ : K] [L : K] we must have that Gal(LL′|L′) ⊆ Gal(LL′|L) and hence L ⊆ L′. Thus, for any finite Galois subextension L|K of degree n = [L : K], we have that L ⊆ Kn alg because [Kn : K] = n, and hence L ⊆ k{{T }}. Because any element a ∈ K is an element of a finite Galois subextension, we have that every element of Kalg is in k{{T }}, and hence Kalg ⊆ k{{T }}. But

∪∞ 1/2 1/3 alg k{{T }} = Kn = K(T ,T , ··· ) ⊆ K n=1 by definition, so that k{{T }} = Kalg. 86 CHAPTER 3. RAMIFICATION THEORY

The Upper Numbering

Many results about higher ramification groups depend on giving them a special renumbering, called the upper numbering. This is calculated via the Her- brand function. For example, this is used to establish Herbrand’s Theorem, which describes the ramification groups of a Galois subextension F |K of L|K in terms of those of L|K itself. The proofs of the results in this section require many lemmas which are tech- nical and unnecessary for the rest of our discussion, so they will be omitted. First, we extend the usual (or lower ) numbering of the ramification groups to real numbers, by defining for any real u ⩾ −1

Gu := Gi, where i = [u]

♦ 定义 3.2.2

The Herbrand function ϕL|K :[−1, ∞) → [−1, ∞) is defined by ∫  u 1  dt if 0 ⩽ u ϕ | (u) = 0 [G0 : Gt] L K  u if − 1 ⩽ u ⩽ 0

From this definition, it’s clear that the function ϕL|K is continuous and strictly increasing, and therefore has an inverse ψL|K :[−1, ∞) → [−1, ∞). We then define the upper numbering of the ramification groups by, for any real s ⩾ −1,

s ϕL|K (u) G = GψL|K (s), or equivalently, Gu = G

Recall that subgroups H ⊆ G correspond to extensions L|F for K ⊆ F ⊆ L, and that to compute the higher ramification groups of L|F , we have the simple result:

Hi = H ∩Gi. Now let H ◁G be a normal subgroup with fixed field F , so that F |K is a Galois extension with Galois group isomorphic to G/H. The upper numbering is necessary to state the natural analog for the higher ramification groups of F |K:

♦ 定理 3.2.5 For a normal subgroup H◁G, we have (G/H)s = GsH/H for all real s ⩾ −1.

Indeed, as Serre states in [Local Fields], “the upper numbering is adapted to quotients, just as the lower numbering is adapted to subgroups.”We also have the following results. 3.2. HIGHER RAMIFICATION GROUPS 87

Herbrand’s Theorem If s = ϕL|K (u), then GuH/H = (G/H)s.

Hasse-Arf Theorem If G is abelian and s is a jump in the filtration in the upper s s+1 numbering, i.e. G ≠ G , then v is an integer. Equivalently, if Gu ≠ Gu+1,

then ϕL|K (u) is an integer. 88 CHAPTER 3. RAMIFICATION THEORY

3.3 *The Theory of Witt Vectors

Definition of the Witt Ring

For a ring A, we let AlgA denote the category of A-algebras. Let Ab denote the category of abelian groups. A subset P ⊆ N is a divisor-stable set provided that P ≠ ∅, and if n ∈ P , then all proper divisors of n are also in P . If P is a divisor-stable set, we let ℘(P ) denote the set of prime numbers contained in P .

注记. 1. 1 ∈ P , since P ≠ ∅. 2. The multiplicatively closed subset T generated by P is simply the set of products of primes in ℘(P ).

2 3 例 3.3.1. Let p be a prime number. The set Pp = {1, p, p , p , ···} is divisor-stable, 2 n as are the finite sets Pp(n) = {1, p, p , ··· , p }.

For any divisor-stable set P and any ring A, define the set ∏ P WP (A) := A = A n∈P and for x ∈ WP (A), we write xn for the n-th coordinate, so x = (xn)n∈P .

♦ 定义 3.3.1 Define the n-th Witt polynomial to be ∑ n/d ∈ Z { | } wn := dXd [ Xd : d n ] d|n

We consider the Witt polynomials wn as set-theoretic maps wn : WP (A) → A for n ∈ P , and we write

P w∗ = (wn)n∈P : WP (A) → A

For x ∈ WP (A), the values wn(x) for n ∈ P are called the ghost components of

x, and the coordinates xn are the Witt components.

注记. 1. If P = N we write W (A) for WP (A), and if P = Pp, we write Wp(A) for

WP (A).

2. The reason we do not write WP (A) for the codomain as well as the domain

of w∗ above is because they will soon have different ring structures. 3.3. *THE THEORY OF WITT VECTORS 89

3. If all elements of P have inverses in A, then we can solve for the Witt com-

ponents of x ∈ WP (A) in terms of its ghost components wn(x) for n ∈ P , P so w∗ : WP (A) → A is in fact a bijection. Similarly, if no element of P is a

zero-divisor in A, then w∗ is an injection.

Now we can state the main theorem of this section: ♦ 定理 3.3.1

There is a unique covariant functor WP : AlgZ → AlgZ, such that for any ring A,

P 1. WP (A) = A as sets, and for a ring homomorphism f : A → B,

WP (f)((an)n∈P ) = (f(an))n∈P

2. The maps wn : WP (A) → A are homomorphisms of rings for all n ∈ P .

3. The zero element of WP (A) is (0, 0, ··· ), and the unit element is (1, 0, 0, ··· ).

WP (f) WP (A) WP (B)

w∗ w∗

AP ∏ BP f

注记. 1. We call the ring WP (A) the ring of P -Witt vectors, or P -Witt ring, with

coefficients in A, W (A) the big Witt ring with coefficients in A, and Wp(A) the p-Witt ring with coefficients in A.

2. (Important) p-Witt ring Wp(A) is by far the most commonly used in practice (at least in number theory). In fact, in the presence of a fixed prime number

p, some authors will generally write W (A) for Wp(A), and refer to Wp(A) as ”the” ring of Witt vectors over A. In this case, people generally index

the Witt components and ghost components of a Witt vector x ∈ Wp(A)

by the exponent of p: in other words, people write x = (x0, x1, x2, ··· ) for

(xp0 , xp1 , xp2 , ··· ) and wn(x) for wpn (x).

注记. 1. If A is a K-algebra, it is not true in general that WP (A) is a K-algebra. 2 For example, if A = Fp and P = 1, p, p , ··· then WP (Fp) ≃ Zp by Theorem

3.4.3, which is not an Fp-algebra. We may still consider WP as a functor from

AlgK to AlgZ. 90 CHAPTER 3. RAMIFICATION THEORY

2. Let R = Z[{Xn : n ∈ P }]. Then for any ring A, Hom(R,A) is naturally iden-

tified with WP (A) as sets, and hence WP is representable. The ring structure

on WP (A) makes R into a ring object in AlgZ.

3. When all elements of P have an inverse in A, the condition that each wn be a ∼ P ring homomorphism implies that w∗ : WP (A) −→ A is a ring isomorphism, where AP has the product ring structure. Similarly, when the elements of P P are not zero-divisors in A, the map w∗ makes WP (A) into a subring of A ; P however, w∗(WP (A)) ≠ A in general.

4. Witt originally thought of the rings Wp(A) as inverse limits of the rings

WPp(n) (A). Since the elements of WPp(n) (A) have finitely many components, Witt thought of them as vectors. This is the only sense in which rings of Witt vectors are related to vectors; really they are rings, and in fact ring-valued functors.

Proof of the Existence of the Witt Rings

For a ring A, we let Λ(A) be the (multiplicative) abelian group

Λ(A) := 1 + tA[[t]]

∑∞ n 引理 3.3.1. Every element f = 1 + xnt ∈ Λ(A) can be written in the form n=1 ∏∞ n f = (1−ynt ), for unique elements yn ∈ A. Furthermore, there are polynomials n=1 ∈ Z ··· ′ ∈ Z ′ ··· ′ Yn [X1, ,Xn] and Xn [Y1 , ,Yn], independent of A, such that yn = ··· ′ ··· Yn(x1, , xn) and xn = Xn(y1, , yn).

Proof. (Sketch): First we will prove by induction on n that there are unique y1, ··· , yn such that f(t) ∏ = 1 + O(tn+1) n − i i=1(1 yit ) To show that Y ∈ Z[X , ··· ,X ], choose A = Z[X , ··· ,X ] and f = 1 + ∑ n 1 n 1 n n Xnt . In this case we necessarily have Yn ∈ A, and it is easy to see that Yn only depends on the first n + 1 terms of f, so in fact Yn ∈ Z[X1, ··· ,Xn]. For an arbitrary ring B, define a ring homomorphism A → B by Xn 7→ xn for some choice of x ∈ A, and extend to a map Λ(A) → Λ(B). Taking the image of the ∏n ∏ ∏ ∏ n n n n equality (1−Ynt ) = 1 + Xnt in Λ(B), we find that (1−ynt ) = 1 + xnt , where yn = Yn(x1, ··· , xn). 3.3. *THE THEORY OF WITT VECTORS 91

推论. The map x 7→ fx : W (A) → Λ(A) defined by ∏∞ n fx(t) := (1 − xnt ) n=1 is a bijection.

∼ Let A be a Q-algebra. The Mercator series defines a bijection log : Λ(A) −→ ∼ tA[[t]], whose inverse is given by the standard exponential series exp : tA[[t]] −→ Λ(A). Of course the log map takes products to sums, as this is a formal property of the Mercator series, so log is an isomorphism of abelian groups. It is clear that ∼ f 7→ t df : tA[[t]] −→ tA[[t]] ∫ − dt is also an isomorphism of abelian groups, with inverse −t−1(·)dt. Set

df ∼ D := −t log : Λ(A) −→ tA[[t]] dt 引理 3.3.2. Let A be a Q-algebra, and let x ∈ W (A). Then ∑∞ n D(fx(t)) = wn(x)t n=1

引理 3.3.3. Let A be a Q-algebra, and let x, y ∈ W (A). Let ∏ − m/d m/e m de/m f(t) = (1 xd ye t ) , where m = lcm(d, e) d,e∈N

Then ∑∞ n D(f(t)) = wn(x)wn(y)t n=1

Now we give a proof of 3.3.1.

Proof. We will show that the big Witt functor W exists. Let A be a Q-algebra. ∼ N There is a unique ring structure on W (A) making w∗ : W (A) −→ A into a ring homomorphism, where the codomain has the product ring structure. Since w∗(0) =

(0, 0, ··· ) and w∗(1, 0, 0, ··· ) = (1, 1, ··· ), the zero element of W (A) is (0, 0, ··· ) and the unit element is (1, 0, 0, ··· ). As this construction is obviously functorial in

A, we have proved that W exists and is unique on the category AlgQ ⊆ AlgZ. We must show that the ring laws are in fact defined over the integers.

Let R = Q[X1,Y1,X2,Y2, ··· ], where the Xi and Yi are indeterminates. Let

X = (X1,X2, ··· ),Y + (Y1,Y2, ··· ) ∈ W (R) 92 CHAPTER 3. RAMIFICATION THEORY

Let S = (S1,S2, ··· ) ∈ W (R) be such that fX (t)fY (t) = fS(t), i.e., ∏∞ ∏∞ ∏∞ n n n (1 − Xnt ) (1 − Ynt ) = (1 − Snt ) n=1 n=1 n=1

By lemma 3.3.1 we have Sn ∈ Z[X1,Y1,X2,Y2, ··· ] and by lemma 3.3.3, ∑∞ ∑∞ n n wn(S)t = D(fS(t)) = D(fX (t)) + D(fY (t)) = (wn(X) + wn(Y ))t n=1 n=1

Hence w∗(S) = w∗(X)+w∗(Y ), so S = X +Y in W (R). Now let Z = (Z1,Z2, ··· ) ∈ W (R) be such that ∏ − m/d m/e m de/m fZ (t) = (1 Xd Ye t ) , where m = lcm(d, e) d,e∈N

Then by lemma 3.3.1 we have Zn ∈ Z[X1,Y1,X2,Y2, ··· ], and by lemma 3.3.3, ∑∞ ∑∞ n n wn(Z)t = D(fZ (t)) = wn(X)wn(Y )t n=1 n=1 so w∗(Z) = w∗(X)w∗(Y ), and hence Z = X · Y in W (R).

Let A be an arbitrary ring, and let x, y ∈ W (A). Define s = x + y by sn =

Sn(x, y), and z = x · y by zn = Zn(x, y). It is clear that when A is a Q-algebra, these recover the ring laws on W (A). In any case, we have constructed well-defined addition and multiplication maps on W (A), which are functorial in A. We have not yet shown that W (A) is a ring when equipped with these addition and multiplication laws. Suppose that A embeds into a Q-algebra A′, which is to say, that A is torsionfree as a Z-module. Then the inclusion W (A) → W (A′) respects addition and multiplication, i.e., W (A) is a subring of W (A′). Hence W (A) is a ring for such

A. Now let B be an arbitrary ring, and choose a set {xi}i∈I of generators of B as a Z-algebra. Set A = Z[{xi}i∈I ], and let φ : A ↠ B be the surjective ring homomorphism such that φ(Xi) = xi. Then W (φ): W (A) ↠ W (B) also respects the addition and multiplication laws, which is to say that W (B) is a quotient ring of W (A). As A is a torsionfree Z-module, W (A) is a ring, so W (B) is a ring. This completes the construction of a functor W satisfying the properties of Theorem 3.3.1. The unicity of the ring structure on W (A) is proved in the same way as the previous paragraph: namely, we know that W (A) has only one ring structure such that w∗ is a ring homomorphism when A is a Q-algebra; hence it is determined when A embeds into a Q-algebra, and therefore, when A is the quotient of a ring embedding into a Q-algebra. 3.3. *THE THEORY OF WITT VECTORS 93

Construct the functor WP for an arbitrary divisor-stable set P , using the same addition and multiplication polynomials Sn,Zn above.

Standard Topology on the Witt Rings

There is a natural inverse limit topology on the Witt rings, which is an important piece of structure, as almost all maps between Witt rings that we will see are continuous. This topology allows one to make sense of infinite sums of Witt vectors, which will be very useful in the sequel. Let P be a divisor-stable set. For n ∈ N write

P (n) := {m ∈ P : m ⩽ n}

It is clear that P (n) is a divisor-stable set. Let πn := πP,n : WP → WP (n) be the projection. It is obvious from the definitions that for any P ,

WP (A) = ←−−lim WP (n)(A) n∈N as rings, under the maps πn. Let A be a ring, equipped with the discrete topology. The standard topology W (A) W (A) on P is by definition the inverse limit topology←− on lim P (n) , which is the P same as the product topology on WP (A) = A .

• The standard topology makes WP (A) into a topological ring.

• The filtered set of ideals {ker(πn): n ∈ N} forms a neighborhood base of the

identity in WP (A).

• WP (A) is complete and Hausdorff with respect to the standard topology.

(n) (n) • The sequence x ∈ WP (A) is Cauchy iff, for all m ∈ P, πm(x ) is constant (n) for n ≫ 0; the sequence converges to y ∈ WP (A) iff, for all m ∈ P, πm(x ) =

πm(y) for n ≫ 0.

• The standard topology on WP (A) is discrete iff P is finite.

The following maps are continuous:

1. wn : WP (A) → A for n ∈ P , where A has the discrete topology.

Proof. Since ker(πn) ⊆ ker(wn) we have that ker(wn) is open. 94 CHAPTER 3. RAMIFICATION THEORY

P P 2. w∗ : WP (A) → A , where A has the product topology induced by the discrete topology on A.

Proof. The product topology is defined to be the finest topology such thata product of continuous maps is continuous.

′ 3. The projection WP (A) → WP ′ (A) for P ⊆ P .

−1 ′ → ′ ′ ⊇ Proof. Let πP,P : WP (A) WP (A) be the projection. Then πP,P ′ (ker(πP ,n))

ker(πP,n).

4. The homomorphism WP (f): WP (A) → WP (B) for a ring B and a ring homomorphism f : A → B.

Proof. W (f) = W (f) P lim←− P (n) .

The Artin-Hasse Exponential

For any prime p and any ring A there is a natural quotient map W (A) →

Wp(A). It is natural to ask if that map has a section. It is too much to expect that such a section would be a ring homomorphism, but it turns out to be true that for

Z(p)-algebras A, there is a natural homomorphism of abelian groups ιp : Wp(A) →

W (A) splitting W (A) ↠ Wp(A).

We will define ιp, and show how it is a kind of p-adic analogue of an exponential map, which is interesting since is difficult to make sense of the ordinary exponential series exp(x) over a ring in which there exist nonzero integers that are zero divisors. Cartier uses ιp in his theory of modules classifying p-divisible groups, in which certain modules over the rings Wp(A) are a kind of ”lineariza- tion at the origin” of a p-divisible group (like a tangent space, or a Jet space); in this philosophy, ιp is in a sense the analogue of the exponential map Lie(G) → G, where G is a Lie group. Cartier theory, as well as a more thorough treatment of the Artin-Hasse exponential, can be found in [Michiel Hazewinkel, Formal groups and applications]. Cartier’s construction rests on the following amazing power series: The Artin-Hasse exponential power series is defined by ( ) xp xp2 xp3 hexp(x) = exp x + + + + ··· ∈ Q[[x]] p p2 p3 3.3. *THE THEORY OF WITT VECTORS 95

♦ 命题 3.3.1 The Artin-Hasse exponential has p-integral coefficients, i.e.,

hexp(x) ∈ Z(p)[[x]]

The proposition can be proved in many ways, including:

• Dwork’s criterion states that a power series f ∈ 1 + xQ[[x]] has p-integral p p p p coefficients iff f(x )/f(x) ∈ Z(p)[[x]] and f(x )/f(x) ≡ 1 mod p.

• The coefficient of xn in n!hexp(x) is the number of elements of the symmetric group on n letters whose order is a power of p; the result then follows from a general (rather difficult) group theory fact.

See [Alain M. Robert, A course in p-adic analysis] for details. We will prove this in a different way.

引理 3.3.4. We have the following identity in Q[[x]]: ∏ n − µ(n) hexp(x) = (1 − x ) n

p∤n∈N+ where µ is the Mobius¨ function.

引理 3.3.5. Let f(x) ∈ 1 + xZ(p)[[x]] have p-integral coefficients, and let g(x) ∈ 1 + xQ[[x]] be such that g(x)n = f(x) for an integer n not divisible by p. Then g has p-integral coefficients as well, i.e., g(x) ∈ 1 + xZ(p)[[x]].

Using Lemma 3.3.4 and applying Lemma 3.3.5 to f(x) = 1 − xn, we obtain the proof. + → + Our goal is to find a natural section ιp : Wp (A) W (A) of the quotient + → + Z W (A) Wp (A) for (p)-algebras A, and to show that ιp resembles an exponential map. The construction follows the general strategy for Witt vectors: namely, we will make a universal construction for Q-algebras A (where the logarithm and exponential do make sense), then show that the polynomials in our construction have p-integral coefficients, so that it makes sense for Z(p)-algebras as well.

The first thing one might try is to define ιp(x) to be the Witt vector y such that ypr = xpr and yn = 0 when n is not a p-th power. Whereas ιp is certainly a section of W (A) → Wp(A), it is unfortunately not a homomorphism of additive groups.

What one really wants is wn(ιp(x)) = 0 for n not a p-th power, and wpr (ιp(x)) = wpr (x). 96 CHAPTER 3. RAMIFICATION THEORY

∼ Let A be a Q-algebra. Consider the isomorphism D : Λ(A) −→ tA[[t]]. Define → εp : tA[[t]] tA[[t]] by ( ) ∑ ∑ n pr εp ant = apr t n⩾1 r⩾0

Then εp is an endomorphism of abelian groups, so there is a unique endomorphism of Λ(A), which we also denote by εp, such that D ◦ εp = εp ◦ D. Note that

εp ◦ εp = εp. ∏ n 引理 3.3.6. Let A be a Q-algebra, and let f = (1 − xnt ) ∈ Λ(A). Then n⩾1 ∏ pr εp(f) = hexp(xpr t ) r⩾1

Using the canonical identification W (A) ≃ Λ(A), we may think of εp as an additive endomorphism of W (A).

♦ 定理 3.3.2

Let A be a Z(p)-algebra, and let π : W (A) → Wp(A) be the projection. + + Define εp : W (A) → Λ(A) ≃ W (A) by ∏ pr εp(x1, x2, x3, ··· ) = hexp(xpr t ) r⩾

Then εp is an endomorphism of abelian groups, functorial in A, satisfying:

1. εp ◦ εp = εp.

2. If εp(x) = y, then xpr = ypr for all r ⩾ 0.

3. π(εp(x)) = π(x) for all x ∈ W (A).

4. wpr (εp(x)) = wpr (x).

Furthermore, the restriction of the canonical projection W (A) → Wp(A) in- + −→∼ + duces an isomorphism εpW (A) Wp (A) of abelian groups. Therefore, the group homomorphism

+ ≃ + → + ιp : Wp (A) εpW (A) , W (A)

+ → + is a section of the projection W (A) Wp (A), satisfying wpr (ιp(x)) = wpr (x).

What Theorem 3.3.2 says is that, given (xpr )r⩾0 ∈ Wp(A), there is a canonical choice of the coordinates xn where n is not a power of p, which respects the addition law. Explicitly, if we set ypr = xpr and yn = 0 when n is not a power of p, then

ιp(x) = εp(y). 3.3. *THE THEORY OF WITT VECTORS 97

Now we will try to indicate in what sense ιp is a p-adic analogue of an exponential map. Let N be a ring without unit, such that every element of N is nilpotent. Suppose that N has the structure of Q-algebra. Then we can think of ∼ the exponential map as a group isomorphism exp : N −→ (1 + N ), where 1 + N is the abelian group with the law (1+a)(1+b) := 1+(a+b+ab). Now suppose that N is a Z -algebra. The statement analogous to Lemma 3.3.1 says that every element (p) ∏ b n of Λ(N ) := 1 + N [t] can be written uniquely as a finite product (1 − ant ). This c gives an identification of the additive group W +(N ) of finite-length Witt vectors b with entries in N , with the multiplicative group Λ(N ), analogous to the identification W +(A) ≃ Λ(A) for a ring A. The finite version of Theorem 3.3.2 says that ∼ c+ N −→ c+ N εpW ( ) Wp ( ). Consider the composition

c+ N ≃ c+ N → c+ N N −−→t7→1 N Ep : Wp ( ) εpW ( ) , W ( ) = 1 + t [t] (1 + ) given explicitly by

∏m Ep(xp0 , xp1 , ··· , xpm , 0, 0, ··· ) = hexp(xpn ) n=0

It is clear that Ep is a homomorphism of abelian groups, and one can show that its − c+ N kernel is equal to (Id Vp)Wp ( ). This ”exponential” map is now an isomorphism

c+ N − −→∼ N Ep : Wp ( )/(Id Vp) (1 + ) defined for any nilpotent Z(p)-algebra N . The preceding discussion is a special case of an isomorphism from Zink’s theory of displays. 98 CHAPTER 3. RAMIFICATION THEORY

3.4 Structure of complete DVRs p-rings and strict p-rings

The theory of Witt vectors arises from the study of the properties of certain p-adic rings, and indeed, provides a construction of the unramified extensions of the p-adic integers, for example. These rings are defined as follows:

♦ 定义 3.4.1 1. A p-ring: a ring R, with topology induced by a system of ideals

a1 ⊇ a2 ⊇ · · · satisfying an · am ⊆ anm, is complete and Hausdorff, i.e. ∩ ≃ R lim←− R/an, an = 0 n n

and the residue ring R/a1 is a perfect ring of characteristic p. n 2. A strict p-ring: a ring R with an = p R, and p is not a zero-divisor.

R (0, p) p 例 3.4.1. A complete DVR with characteristic is√ a -ring. If the maximal m pR R = Z Z [ 1] p ≡ 3 4 ideal √is , it is also strict, for instance p, or p − if mod . The ring Z5[ 5] is a p-ring, but not a strict one.

例 3.4.2. Fp[[X]] is not a p-ring.

例 3.4.3. Zp[[X]] with topology induced by the ideal (p, X) is a p-ring that is not strict. Its residue ring Fp is a field, but it’s not a DVR. Its maximal ideal isnot principal.

It is easy to show that

• There exists one and only one system of representatives τ : k → A which commutes with p-th powers: τ(λp) = τ(λ)p. • a ∈ A belong to S = τ(k) iff a is a pn-th power for all n ⩾ 0. • This system of representatives is multiplicative, i.e., τ(xy) = τ(x)τ(y) for all x, y ∈ k. • If A has characteristic p, this system of representatives is additive, i.e., τ(x + y) = τ(x) + τ(y) for all x, y ∈ k.

This implies that when A is a p-ring, it always has the system of multiplicative representatives τ : k → A, and for every sequence α0, ··· , αn, ··· , of elements 3.4. STRUCTURE OF COMPLETE DVRS 99 of k, the series ∑∞ i τ(αi)p (∗) i=0 converges to an element a ∈ A. If furthermore A is a strict p-ring, every element ∈ ∗ pi a A can be uniquely expressed in the form of a series of type ( ). Let βi = αi , ∑∞ p−i i { } then a = τ(βi )p . We call βi the coordinates of a. i=0 The following are the important properties of strict p-rings:

♦ 定理 3.4.1 There’s an equivalence of the categories of strict p-rings and perfect rings of characteristic p.

p − Strict p − Perfict

R k := R/a1

The formation of R and τ is functorial in k, in that if f : k → k′ is a homomorphism of perfect rings of characteristic p, and R′ is the strict p-ring with residue ring k′ and section τ ′, then there is a unique homomorphism F : R → R′ making the following squares commute:

F F R R′ R R′

τ τ ′

f f k k′ k k′

The map F is given by ( ) ∑∞ ∑∞ n ′ n F τ(xn)p = τ (f(xn))p n=0 n=0

例 3.4.4. Let R be an unramified extension of Zp, with residue field k =≃ Fq. Then

R is a strict p-ring, and is hence the unique strict p-ring with residue field Fq. The F× ≃ Z Z Teichmüller representatives are constructed as follows: we have q /(q−1) , q−1 so that the nonzero elements of Fq are the roots of the polynomial X −1. By F× ∈ Hensel’s Lemma, each element x of q has a unique lift τ(x) R also satisfying τ(x)q−1−1 = 0. Setting τ(0) = 0 completes the definition of the map τ. In other words, the Teichmüller representatives are exactly the (q−1)st roots of unity in R, union {0}. 100 CHAPTER 3. RAMIFICATION THEORY

Motivation for Witt vectors

The result above is an abstract fact, which may be proved without yielding a useful construction of the strict p-ring R with given residue ring k. But the strong unicity and functoriality of R indicates that one should be able to construct it algebraically in terms of k. We can certainly reconstruct the set underlying R as all sums of the form ∑∞ n τ(xn)p . But in order to reconstruct the ring structure on R, we need to under- n=0 stand the addition and multiplication laws in terms of the arithmetic of k. ∑ ∑ ∑ n n n If τ(xn)p + τ(yn)p = τ(sn)p , we need to write the sn in terms of the xn and yn, and similarly for multiplication. By unicity, the answer should not depend on R, and by functoriality, the answer should not even depend on k, in that the same addition and multiplication laws will have to work for every k. This suggests that the sn will be given by polynomials in the xn and yn with p- integral rational coefficients; that is, by polynomials whose coefficients are rational numbers with nonnegative p-adic valuation. In fact the sn will be given by integer polynomials in the xn and yn. The following lemma will fundamental in proving the p-integrality of polynomials:

引理 3.4.1. Let A be a ring, and let x, y ∈ A be such that x ≡ y mod pA. Then for i i all i ⩾ 0 we have xp ≡ yp mod pi+1A.

Proof. We proceed by induction on i; the case i = 0 is clear. Let i ⩾ 1, and write i+1 i 1 xp = yp − + piz for z ∈ A. Raising both sides side to the pth power, we obtain

∑p−1 ( ) i i p i xp = yp + yp (p−n)pinzn + pipzp n n=1 The lemma follows because p divides all of the binomial coefficients, and ip ⩾ i + 1. ∑ Let R be a strict p-ring with residue ring k, and suppose that τ(x )pn + ∑ ∑ n n n τ(yn)p = τ(sn)p . To calculate the sn, we proceed inductively. Looking mod p, we have

τ(x0) + τ(y0) ≡ τ(s0) mod p since τ(x) = x mod p, we have x0 + y0 = s0. The second step is to write

2 τ(x0) + pτ(x1) + τ(y0) + pτ(y1) ≡ τ(s0) + pτ(s1) ≡ τ(x0 + y0) + pτ(s1) mod p 3.4. STRUCTURE OF COMPLETE DVRS 101 then rewrite to find

2 pτ(s1) ≡ τ(x0) + τ(y0) − τ(x0 + y0) + p(τ(x1) + τ(y1)) mod p

But whereas we know that τ(x0) + τ(y0)−τ(x0 + y0) ≡ 0 mod p, we have no idea 2 what its residue mod p is. The trick to calculating s1 is as follows. Since k is 1/p perfect, every x ∈ k has a unique p-th root, written x . Since x0 + y0 = s0, we 1/p 1/p 1/p must have x0 + y0 = s0 . By the lemma above and the fact that τ commutes with multiplication, we can write ( ) ( ) p p 1/p p 1/p 1/p ≡ 1/p 1/p 2 τ(s0) = τ(s0 ) = τ x0 + y0 τ(x0 ) + τ(y0 ) mod p

Therefore, ( ) p ≡ 1/p p 1/p p 1/p 1/p 2 pτ(s1) τ(x0 ) + τ(y0 ) − τ(x0 ) + τ(y0 ) + p(τ(x1) + τ(y1)) mod p

Expanding out the above equation and dividing by p, we obtain

p−1 ( ) ∑ 1 p τ(s ) ≡ τ(x ) + τ(y )− τ(xn/p)τ(y(p−n)/p) mod p 1 1 1 p n 0 0 n=1 and therefore, p−1 ( ) ∑ 1 p s = x + y − xn/py(p−n)/p 1 1 1 p n 0 0 n=1 The above bit of formal manipulation has nothing to do with k. Indeed, let X0,X1,Y0,Y1 be indeterminates, and define polynomials w1(X0) = X0 and p wp(X0,X1) = X0 + pX1; then solve the polynomial equations

S0 = w1(S0) = w1(X0) + w1(Y0) = X0 + Y0 p p p S0 + pS1 = wp(S0,S1) = wp(X0,X1) + wp(Y0,Y1) = X0 + pX1 + Y0 + pY1 for S0 and S1. This is the same bit of algebra as above, except with X0 replacing 1/p τ(x0 ), etc., so we have

S0 = X0 + Y0 ∑p−1 ( ) 1 p − S = X + Y − XnY p n 1 1 1 p n 0 0 n=1

∈ Z ∈ Z 1/p In particular, S0 [X0,Y0] and S1 [X0,X1,Y0,Y1]. Substituting τ(x0 ) 1/p 1/p back in for X0, etc., we see that s0 = S0(x0, y0) and s1 = S1(x0 , y0 , x1, y1). 102 CHAPTER 3. RAMIFICATION THEORY

Witt realized this, and also discovered the pattern. Define

n ∑ − − ··· i pn i pn pn 1 ··· n−1 p n wpn (X0,X1, ,Xn) = p Xi = X0 + pX1 + + p Xn−1 + p Xn i=0

Letting X1,Y1,X2,Y2, ··· be indeterminates, inductively find Sn that solve the polynomial equations

wpn (S0,S1, ··· ,Sn) = wpn (X0,X1, ··· ,Xn) + wpn (Y0,Y1, ··· ,Yn)

n As the only term of wpn (S0,S1, ··· ,Sn) involving Sn is p Sn, it is clear that there are unique polynomials Sn with rational coefficients satisfying the above identity.

What Witt showed is that in fact, Sn ∈ Z[X0,Y0,X1,Y1, ··· ,Xn,Yn]. Assuming this, we have: ♦ 定理 3.4.2 Let R be a strict p-ring, let k = R/p be its residue ring, and let τ : k → R the system of Teichmüller representatives. Suppose that ∑∞ ∑∞ ∑∞ n n n τ(xn)p + τ(yn)p = τ(sn)p n=0 n=0 n=0

Then with the Sn as above, we have ( ) 1/pn 1/pn 1/pn−1 1/pn−1 ··· 1/p 1/p sn = Sn x0 , y0 , x1 , y1 , , xn−1, yn−1, xn, yn

Witt also showed the analogous result for multiplication: if we solve the polynomial equations

wpn (Z0,Z1, ··· ,Zn) = wpn (X0,X1, ··· ,Xn)wpn (Y0,Y1, ··· ,Yn) then Zn ∈ Z[X0,Y0,X1,Y1, ··· ,Xn,Yn]. Setting

1/pn 1/pn 1/pn−1 1/pn−1 ··· 1/p 1/p zn := Zn(x0 , y0 , x1 , y1 , , xn−1, yn−1, xn, yn) one can verify (following the proof of theorem above) that (∑ )(∑ ) ∑ n n n τ(xn)p τ(yn)p = τ(zn)p

Hence the ring laws on R are described entirely by the polynomials Sn and

Zn, which were obtained algebraically. This motivates the definition of the Witt vectors: given any ring A, we will construct a ring Wp(A), whose addition and multiplication laws are somehow given by the polynomials Sn and Zn. Taking

A = k, we will show that Wp(k) is the strict p-ring with residue ring k. 3.4. STRUCTURE OF COMPLETE DVRS 103

Structure of complete DVRs

We want to classify complete DVRs.

• This is done by first splitting according to characteristics:(0, 0), (p, p), (0, p).

• Equal case (0, 0) and (p, p): we saw these are just k[[T ]], k residue field.

• Case (0, p): based on absolute ramification index e = v(p).

(Equal characteristic: (p, p) or (0, 0)): it turns out in the equal characteristic case one can always pick a field in A isomorphic to k. The case (p, p) is immediately followed. We need to show the case (0, 0).

引理 3.4.2. Let A be a ring, complete and Hausdorff with respect to a system {an}, such that k := A/a1 is a field of characteristic 0. Then A ↠ k admits a section.

Proof. If A has characteristic zero, it contains Z. Since k has characteristic zero, every nonzero integer n must be a unit in A, otherwise n = 0 in k. Then A contains a copy of Q. Now any transcendental element a ∈ k can be lifted to an element a ∈ A such that Q(a) maps isomorphic to its image in k. That’s because a necessarily satisfies no algebraic relations. Hence the maximal transcendental subextension L of k|Q lifts to a subfield in A. Now any element in k algebraic over L has a minimal polynomial which is separable, since char(k) = 0. Then by Hensel’s lemma one can lift all the finite algebraic extensions also. Passing to the limit, the entire algebraic extension k|L also lifts to A.

推论. A complete DVR of characteristic (0, 0) is isomorphic to k[[T ]] where k is the residue field.

As an application of Witt vectors, we discuss the structure of DVRs in the unequal characteristic case.

♦ 定义 3.4.2 Let (A, k) be a complete DVR with characteristic (0, p). The integer e = v(p) is called the absolute ramification index of A. A is called absolutely unramified if e = 1, i.e., if p is a local uniformizer of A. 104 CHAPTER 3. RAMIFICATION THEORY

(Convention): absolutely unramified complete DVR is short for aucDVR, aucDVR always means its characteristic is (0, p).

注记. (A, k) is a strict p-ring with residue ring k is a field iff A is an aucDVR of characteristic (0, p).

We want to show that there’s an equivalence of categories

aucDVR of characteristic (0, p) perfect fields of with perfect residue field characteristic p

R k := R/p with the inverse functor given by some k 7→ W (k). This will clarify the total picture: Any complete DVRs of char. (0, p) with residue field k, is a totally ramified extension of W (k), of degree e = v(p).

注记. The equivalence of categories is immediately derived from the general case.

Any totally ramified extension of Qp can be defined by an Eisenstein polynomial. The non-leading coefficients of such a polynomial must must liein pZp, a compact open subset of Qp. But recall that two irreducible polynomials whose coefficients are close enough will define the same extension. This shows thereare only finitely many ramified extensions of Qp of a given degree. In fact any aucDVR of characteristic (0, p) will have only finitely many totally ramified extensions of a given degree n. The argument is the same.

Universal construction by Witt vectors

Let R = Z[{Xn,Yn : n ∈ P }], and let X = (Xn)n∈P ,Y = (Yn)n∈P ∈ WP (R),S := X + Y . By definition,

wn(S) = wn(X) + wn(Y ) for n ∈ P , so the solutions Sn to the polynomial equations

wn((Sn)n∈P ) = wn((Xn)n∈P ) + wn((Yn)n∈P ) are contained in Z[{Xi,Yi : i ∈ P }]. A similar result holds for multiplication.

These polynomials in fact give the ring laws for WP (A) for any ring A (and any P , for that matter), as the following corollary shows: 3.4. STRUCTURE OF COMPLETE DVRS 105

推论. Let R,P,X,Y,Z, and S be as above. The polynomials Sn and Zn do not depend on the choice of P , and in addition,

Sn,Zn ∈ Z[{Xi,Yi : i|n}]

Let A be an arbitrary ring, and let x, y ∈ WP (A). Let sn = Sn evaluated at the xi and yi, and similarly for zn. Then

s = (sn)n∈P = x + y, z = (zn)n∈P = x · y

where the addition and multiplication takes place in WP (A).

Proof. Solving the equation wn(S) = wn(X) + wn(Y ) explicitly for Sn shows that

Sn only depends on {Xi,Yi : i|n}. As the equation wn(S) = wn(X) + wn(Y ) does not depend on P , neither does S. The same statements hold with Z replacing S.

Define a ring homomorphism f : R → A by f(Xi) = xi and f(Yi) = yi. Since

WP (f) is a ring homomorphism, we have

s = WP (f)(S) = WP (f)(X) + WP (f)(Y ) = x + y

注记. The corollary essentially shows that the ring laws of WP (A) for an arbitrary ring A can be calculated in WP (R) where R = Z[{Xn,Yn : n ∈ P }]. Since R is a ring which is torsion-free as a Z-module, one can often prove statements about WP (R) P using the injection w∗ : WP (R) ,→ R , and then derive facts about WP (A). This often-used trick is called ”reduction to the universal case”, and it is extremely powerful —one often cares about Witt rings over rings A of characteristic p, but in order to prove theorems about these rings, one reduces to the characteristic-0 case.

♦ 定理 3.4.3 Let R be the strict p-ring with perfect residue ring k, with Teichmüller

reprsentatives τ : k → R. Then the map f : Wp(k) → R given by ∑∞ ··· 1/pn n f(x1, xp, xp2 , ) = τ(xpn )p n=0

is a ring isomorphism. 106 CHAPTER 3. RAMIFICATION THEORY

Proof. It is clear that f is a well-defined bijection, and that f(1) = τ(1) = 1. Let x, y ∈ Wp(K), and let s = x+y. We will show that f(s) = f(x)+f(y). By Corollary 3.4,

spi = Spi (x1, y1, xp, yp, ··· , xpi , ypi ) and hence, since Spi is a polynomial with integer coefficients,

1/pi 1/pi 1/pi 1/pi 1/pi ··· 1/pi 1/pi spi = Spi (x1 , y1 , xp , yp , , xpi , ypi )

1/pj Let x˜j := xpj , and similarly for y˜j and s˜j. Substituting into the above equation, by Theorem 3.4.2,

∑n ( ) ∑n ( ) ∑n ( ) i 1/pi i 1/pi i 1/pi n+1 p τ xpi + p τ ypi = p τ spi mod p i=0 i=0 i=0

Thus f(x + y) ≡ f(x) + f(y) mod pn+1 for any n, completing the proof.

注记. One amazing aspect of the universal construction of WP (A) is that, not only can we recover the standard generalization of the ring Zp = Wp(Fp) to any perfect residue field K of characteristic p (namely, the unique unramified extension of Zp with residue field K), but we can in fact define the ring Wp(A) for any ring A, of arbitrary characteristic. In other words, the same ring laws used to construct

Zp can be used to define its analogue for an arbitrary residue ring, although now

A = Wp(A)/ ker(w1) instead of Wp(A)/p. Perhaps even more amazingly, these Witt rings will come equipped with Frobenius and Verschiebung maps associated to numbers n ∈ P which are not necessarily prime.

推论. Let P and P ′ be divisor-stable sets with P ′ ⊆ P . The quotient map

WP (A) → WP ′ (A):(xn)n∈P 7→ (xn)n∈P ′ is a ring homomorphism for any ring A, and hence defines a natural transformation WP → WP ′ of ring-valued functors. 3.5. LUBIN-TATE FORMAL GROUPS 107

3.5 Lubin-Tate formal groups

Formal groups

Let A be a commutative ring and A[[T ]] the ring of formal power series f(T ) = ∑ n anT with coefficients in A. we could equivalently view elements of A[[T ]] n⩾0 as sequences indexed by Z⩾ that we add component wise and multiply using ∑ 0 convolutions: (fg)k := figj. i+j=k In addition to the ring operations, we also have a composition operation f ◦ g defined whenever the constant term of g is zero (without this restriction the constant term of f ◦ g would be undefined; we cannot formally sum infinitely many elements of A). This still make sense when one of f or g is a power series in several variables. For any f ∈ A[[T ]] and g ∈ A[[X1, ··· ,Xr]] both with constant term zero we define

(f◦g)(X1, ··· ,Xr) := f(g(X1, ··· ,Xr)), (g◦f)(X1, ··· ,Xr) := g(f(X1), ··· , f(Xr))

We note that the ideal TA[[T ]] generated by T is precisely the set of univariate power series with constant term 0.

引理 3.5.1. 1. For all f ∈ A[[T ]] and g, h ∈ TA[[T ]] we have f ◦(g◦h) = (f ◦g)◦h. 2. For each f ∈ TA[[T ]] there exists g ∈ TA[[T ]] such that f ◦ g = T iff the coefficient of T in f is a unit in A. 3. The elements of TA[[T ]] for which the coefficient of T is a unit form a group under composition, with identity T . In particular, f ◦ g = T iff g ◦ f = T .

♦ 定义 3.5.1 A (one parameter) formal group law over a commutative ring A is a power series F ∈ A[[X,Y ]] in two variables such that ∑ i j 1. F (X,Y ) = X + Y + aijX Y ; i+j>1 2. F (X,F (Y,Z)) = F (F (X,Y ),Z). A homomorphism of formal group laws ϕ : F → G is a power series ϕ ∈ TA[[T ]] such that ϕ ◦ F = G ◦ ϕ. When G = F we call ϕ an endomorphism of formal group laws.

If there exist homomorphisms of formal group laws ϕ : F → G and ψ : G → F such that ϕ ◦ ψ = ψ ◦ ϕ = T , then we call ϕ (and ψ) isomorphisms of formal 108 CHAPTER 3. RAMIFICATION THEORY group laws and write F ≃ G, and in the case G = F we call ϕ an automorphism of formal group laws.

引理 3.5.2. A homomorphism ϕ : F → G of formal group laws is an isomorphism iff the coefficient of T in ϕ is a unit in A.

Proof. In order for ϕ to be an isomorphism the coefficient of T in ϕ must be a unit, which we now assume. Let ψ ∈ TA[[T ]] be the inverse of ϕ under composition. Then ϕ ◦ ψ = T = ψ ◦ ϕ, we just need to check that ψ ∈ Hom(G, F ). We have G = G ◦ ϕ ◦ ψ = ϕ ◦ F ◦ ψ, so ψ ◦ G = ψ ◦ ϕ ◦ F ◦ ψ = F ◦ ψ as desired.

If ϕ : F → G and ψ : G → H are homomorphisms of formal group laws, then so is their composition ψ ◦ ϕ : F → H, and ϕ(T ) = T is an automorphism of formal group laws that acts as the identity with respect to composition. If φ : A → B is a homomorphism of commutative rings and F (X,Y ) = ∑ i j X +Y + aijX Y is a formal group law over A, then the power series φ∗(F ) := i+j>∑1 ∑ i j i X + Y + φ(aij)X Y is a formal group law over B, and if ϕ(T ) = aiT i+j>1 i⩾1 is a homomorphism of formal group laws ϕ : F → G, then the power series ∑ i φ∗(ϕ) := φ(ai)T is a homomorphism of formal group laws φ∗(ϕ): φ∗(F ) → i⩾1 φ∗(G).

♦ 命题 3.5.1 Let F ∈ A[[X,Y ]] be a formal group law. The following hold: 1. F (X, 0) = X and F (0,Y ) = Y ;

2. The is a unique iF ∈ TA[[T ]] such that F (T, iF (T )) = 0; 3. If A contains no nonzero torsion elements that are also nilpotent then we also have F (X,Y ) = F (Y,X).

注记. Formal groups laws that satisfy property (3) of the Proposition are commutative; the Proposition implies if A is a reduced ring (an integral domain, for example), then all formal group laws over A are commutative. This applies in all the rings of interest to us.

例 3.5.1. The additive formal group law Ga defined by Ga(X,Y ) = X + Y and the multiplicative formal group law Gm defined by

Gm(X,Y ) = X + Y + XY = (1 + X)(1 + Y ) − 1 are examples of formal group laws over any commutative ring A. 3.5. LUBIN-TATE FORMAL GROUPS 109

例 3.5.2. Let F be a commutative formal group law over a commutative ring A. For each integer n inductively define the power series [n]F ∈ TA[[T ]] by putting [0]F :=

0, inductively defining [n]F (T ) := F ([n − 1]F (T ),T ) and [−n]F (T ) := iF ([n]F (T )) for n ⩾ 1. One can show that [n]F (T ) is an endomorphism of the formal group law F , and that it is an automorphism iff n is a unit in A.

If F (X,Y ) is any formal group law on a commutative ring A, the binary operation

ϕ +F ψ := F (ϕ(T ), ψ(T )) makes the set TA[[T ]] into a group: closure and associativity follow from the definition of a formal group law, the identity element is 0 (by part (1) of Proposition

3.5.1), and inverses are given by −F ϕ := iF ◦ ϕ, by part (2) of Proposition 3.5.1.

♦ 命题 3.5.2 Let A be a commutative ring with no nonzero torsion nilpotents and let F,G be formal group laws over A. The set of all homomorphisms ϕ : F → G

is an abelian group Hom(F,G) under the operation +G, and the set of all en-

domorphisms ϕ : F → F with the addition operation +F and multiplication given by composition is a (not necessarily commutative) ring End(F ) with multiplicative identity T and unit group Aut(F ) consisting of all automorphisms of the formal group law F .

Proof. (Sketch) The hypothesis on A implies G(X,Y ) = G(Y,X), by part (3) of

Proposition 3.5.1, so +G is commutative. To prove the first statement we only need to show that Hom(F,G) is closed under +G, since TA[[T ]] is an abelian group under +G. For the second statement, the associativity of composition of power series in TA[[T ]] is given by Lemma 3.5.1, and it is clear that T is the identity with respect to composition, so we just need to check the distributive law.

Formal group laws over complete DVRs

(Convention in this subsection): Let (A, m) be a complete DVR, F (X,Y ) be a formal group law over A.

引理 3.5.3. For any x, y ∈ m the series F (x, y) converges to an element of m. 110 CHAPTER 3. RAMIFICATION THEORY

n n Proof. If we define Fn(X,Y ) := F (X,Y ) mod (X ,Y ), the sequence (Fn(x, y))n is Cauchy, since v(Fm(x, y)−Fn(x, y)) ⩾ N for all m, n ⩾ N, and therefore converges in our complete ring A, and it converges to an element with positive valuation (hence an element of m), since the constant term of F (X,Y ) is zero.

The binary operation

x +F y := F (x, y) makes the set m into an abelian group with identity element 0 and inverse −F x := iF (x) via parts (1) and (2) of Proposition 3.5.1; note that associativity is implied by the definition of a formal group law and commutativity is given by part(3) of Proposition 3.5.1, since A is an integral domain. The group F (m) := (m, +F ) is the group associated to F /A. Note that if x, y lie in an ideal mn, then so does n n F (x, y), thus we have a filtration of F (m) by subgroups F (m ) := (m , +F ). The group F (m) is also a topological group (in the subspace topology from A), since the group operation is defined by the power series F , which is continuous as a map m × m → m, as is the map m → m defined by the power series iF . If φ : A → B is a homomorphism of complete DVRs (as topological rings), then we have an induced homomorphism F (φ): F (mA) → φ∗(F )(mB) of topological groups. This applies in particular when φ is an inclusion map, so we can view a formal group law over A as a functor from the category of complete DVRs extending A to the category of topological abelian groups. If ϕ : F → G is a homomorphism of formal group laws over A, then ϕ(x) converges to an element of m for all x ∈ m (since ϕ has constant term zero), and we have an induced group homomorphism

ϕ : F (m) → G(m) a 7→ ϕ(a)

We have a commutative diagram

F (φ) F (mA) F (mB)

ϕ φ∗(ϕ)

G(φ) G(mA) G(mB)

We can thus view ϕ as a morphism of functors (a natural transformation). 3.5. LUBIN-TATE FORMAL GROUPS 111

例 3.5.3. Let k be residue field of A. Then Ga(m) = m ⊆ A and we have an exact sequence of topological groups

0 → Ga(m) −→ A −→ k → 0

× We have Gm(m) = 1 + m ⊆ A and an exact sequence of topological groups

a7→1+a × × 1 → Gm(m) −−−−→ A −→ k → 1

n − The endomorphisms [n]Ga (T ) = nT and [n]Gm (T ) = (1 + T ) 1 corresponds to the multiplication-by-n and n-power maps on m and 1 + m, respectively.

Lubin-Tate group laws

(Convention in this subsection): Let (A, m) be a complete DVR with finite residue field k of cardinality q, this is equivalent to assuming that A is the valuation ring of a nonarchimedean local field; and let π be a uniformizer of A.

Define the set

Φ(π) := {ϕ ∈ TA[[T ]] : ϕ(T ) ≡ πT mod T 2, ϕ(T ) ≡ T q mod π}

One should think of elements of Φ(π) as ”Frobenius endomorphisms”; we will show that for each ϕ ∈ Φ(π) there is a unique formal group law Fϕ(X,Y ) such that ϕ ∈ End(Fϕ). If φ : A → k is the natural map, then φ∗(ϕ) is the q-power Frobenius map x 7→ xq. For a power series ring R over a commutative ring A (in any number of variables), let R denote the A-submodule of consisting of homogeneous polynomials n ∏ of degree n; we have an obvious A-module isomorphism R ≃ R given by col- n n ∏ lecting terms of the same degree. We define the A-submodules R⩽n := Ri and ∏ i⩽n R>n := Ri; the latter is simply the R-ideal generated by Rn+1. i>n

♦ 命题 3.5.3

Let ϕ, ψ ∈ Φ(π), let r be a positive integer, and let R := A[[X1, ··· ,Xr]].

For every F1 ∈ R1 there is a unique F ∈ R such that F ≡ F1 mod R>1 and ϕ ◦ F = F ◦ ψ. 112 CHAPTER 3. RAMIFICATION THEORY

Proof. We will show by induction that there is a unique Fn ∈ R⩽n for which we have (i) Fn ≡ F1 mod R>1, (ii) ϕ ◦ Fn ≡ Fn ◦ ψ mod R>n, and (iii) Fn ≡ Fn−1 mod R>n−1 if n > 1. We may then take F := lim Fn and the proposition follows. n→∞ For n = 1 we have ϕ ◦ F1 ≡ πF1 ≡ F1 ◦ ψ mod R>1, so (ii) holds, (i) is given, and (iii) is vacuous; it is clear that F1 is the unique solution for n = 1. For n > 1 the inductive hypothesis implies that there is a unique homogeneous polynomial Pn+1 ∈ Rn+1 such that

ϕ ◦ Fn − Fn ◦ ψ ≡ Pn+1 mod R>n+1

Since ϕ, ψ ∈ Φ(π) we have

◦ − ◦ ≡ ··· q − q ··· q ≡ ϕ Fn Fn ψ Fn(X1, ,Xr) Fn(X1 , ,Xr ) 0 mod π

q since x 7→ x is an automorphism modulo π, so π divides ϕ ◦ Fn − Fn ◦ ψ and n therefore Pn+1. We also note that π − 1 has valuation 0 and is thus invertible in A, so we may define Pn+1 F := F + ∈ R⩽ n+1 n πn+1 − π n+1

Now Pn+1 ≡ 0 mod R>n, so Fn+1 ≡ Fn mod R>n, thus (iii) and (i) hold for n + 1. We have

ϕ ◦ Pn+1 − ψ ◦ Pn+1 ≡ πPn+1(X1, ··· ,Xr) − Pn+1(πX1, ··· , πXr) mod R>n+1 n+1 ≡ (π − π )Pn+1 mod R>n+1 since Pn+1 is homogeneous of degree n + 1, and therefore

ϕ ◦ Pn+1 Pn+1 ◦ ψ ϕ ◦ F − F ◦ ψ ≡ ϕ ◦ F + − F ◦ ψ − mod R n+1 n+1 n πn+1 − π n πn+1 − π >n+1 ϕ ◦ Pn+1 − Pn+1 ◦ ψ ≡ P + mod R n+1 πn+1 − π >n+1

≡ 0 mod Rn+1 so (ii) holds as well, and the uniqueness of Pn+1 implies the uniqueness of Fn+1.

♦ 命题 3.5.4

For every ϕ ∈ Φ(π) there is a unique formal group law Fϕ over A such that

ϕ ∈ End(Fϕ).

Proof. By Proposition above, there is a unique Fϕ ∈ R := A[[X,Y ]] satisfying the constraints Fϕ ≡ X +Y mod R>1 and ϕ(Fϕ(X,Y )) = Fϕ(ϕ(X), ϕ(Y )) which must 3.5. LUBIN-TATE FORMAL GROUPS 113 hold for any formal group law F over A for which ϕ ∈ End(F ). We only need to check that Fϕ is actually a formal group law: we also require F (X,F (Y,Z)) = F (F (X,Y ),Z). ′ The power series G(X,Y,Z) := Fϕ(X,Fϕ(Y,Z) and G (X,Y,Z) := Fϕ(Fϕ(X,Y ),Z) ′ satisfy G ≡ X + Y + Z ≡ G mod R>1 and

ϕ ◦ G = ϕ(Fϕ(X,Fϕ(Y,Z)) = Fϕ(ϕ(X),Fϕ(ϕ(Y ), ϕ(Z))) = G ◦ ϕ ′ ′ ϕ ◦ G = ϕ(Fϕ(Fϕ(X,Y ),Z)) = Fϕ(Fϕ(ϕ(X), ϕ(Y )), ϕ(Z)) = G ◦ ϕ Proposition above implies that there is a unique G ∈ A[[X,Y,Z]] congruent to

X + Y + Z modulo R>1 that satisfies ϕ(G(X,Y,Z)) = G(ϕ(X), ϕ(Y ), ϕ(Z)), so we must have G′ = G and therefore F (X,F (Y,Z)) = F (F (X,Y ),Z) as desired.

Formal group laws of the form Fϕ given by Proposition above, where ϕ ∈ Φ(π) for some uniformizer π of a complete DVR A with finite residue field) are known as Lubin-Tate formal group laws (for the uniformizer π).

♦ 定义 3.5.2

For ϕ, ψ ∈ Φ(π) and a ∈ A, let [a]ϕ,ψ be the unique element of TA[[T ]] that 2 satisfies [a]ϕ,ψ ≡ aT mod T and ϕ ◦ [a]ϕ,ψ = [a]ϕ,ψ ◦ ψ given by Proposition

3.5.3. Let [a]ϕ := [a]ϕ,ϕ.

♦ 命题 3.5.5 For all ϕ, ψ ∈ Φ(π) the following hold:

1. [a]ϕ,ψ ∈ Hom(Fϕ,Fψ) for all a ∈ A; ∼ 2. [1]ϕ,ψ gives a canonical isomorphism Fψ −→ Fϕ.

Here Fϕ and Fψ are the Lubin-Tate formal group laws for the uniformizer π corresponding to ϕ and ψ, respectively.

2 Proof. (1). Let φ := [a]ϕ,ψ and R := A[[X,Y ]]. We have φ ≡ aT mod T , so φ ∈ TA[[T ]], and

φ ◦ Fψ ≡ aX + aY ≡ Fϕ ◦ φ mod R>1

We have ϕ ◦ φ = φ ◦ ϕ and ϕ ∈ End(Fϕ) and ψ ∈ End(Fψ), so

ϕ ◦ (φ ◦ Fψ) = (ϕ ◦ φ) ◦ Fψ = (ϕ ◦ ψ) ◦ Fψ = φ ◦ (ψ ◦ Fψ) = φ ◦ (Fψ ◦ ψ) = (φ ◦ Fψ) ◦ ψ

ϕ ◦ (Fϕ ◦ φ) = (ϕ ◦ Fϕ) ◦ φ = (Fϕ ◦ ϕ) ◦ φ = Fϕ ◦ (ϕ ◦ φ) = Fϕ ◦ (φ ◦ ψ) = (Fϕ ◦ φ) ◦ ψ

Proposition 3.5.3 now implies φ ◦ Fψ = Fϕ ◦ φ, so [a]ϕ,ψ = φ ∈ Hom(Fψ,Fϕ).

(2). [1]ϕ,ψ is an isomorphism Fψ → Fϕ, and it is clearly canonical (since 1 is). 114 CHAPTER 3. RAMIFICATION THEORY

♦ 命题 3.5.6

For every ϕ ∈ Φ(π) the map a 7→ [a]ϕ is an injective ring homomorphism

A,→ End(Fϕ) that sends π to ϕ and A into the centralizer of ϕ.

Proof. It follows from Proposition 3.5.5 that [a]ϕ ∈ End(Fϕ) for all a ∈ A; the map 2 a 7→ [a]ϕ is clearly injective, since [a]ϕ ≡ aT mod T . It follows from Proposition

3.5.3 that every φ ∈ End(Fϕ) for which ϕ ◦ φ = φ ◦ ϕ is uniquely determined by its 2 reduction modulo T . This applies in particular to every φ of the form [a]ϕ, since the condition ϕ ◦ [a]ϕ = [a]ϕ ◦ ϕ was used to define [a]ϕ. For all a, b ∈ A we have

≡ ≡ 2 [a]ϕ +Fϕ [b]ϕ aT + bT [a + b]ϕ mod T 2 [a]ϕ ◦ [b]ϕ ≡ abT ≡ [ab]ϕ mod T

◦ ≡ and therefore [a]ϕ +Fϕ [b]ϕ = [a + b]ϕ and [a]ϕ [b]ϕ = [ab]ϕ. We also have [1]ϕ T 2 mod T , and ϕ ◦ T = ϕ ◦ T , so we must have [1]f = T . It follows that the map 2 a 7→ [a]ϕ is a ring homomorphism. Finally, [π]ϕ ≡ πT ≡ ϕ mod T , and [π]ϕ = ϕ, and [a]ϕ commutes with ϕ (both by construction and because A is commutative) for all a ∈ A.

注记. 1. For any choice of uniformizer π and any ϕ ∈ Φ(π), the group Fϕ(m) = ∈ (m, +Fϕ ) has an A-module structure defined by am := [a]ϕ(m), for any a A and m ∈ m, in which π corresponds to the endomorphism ϕ whose reduction modulo π is the Frobenius map x 7→ xq. 2. Proposition 3.5.5 implies that up to a canonical isomorphism, the A-module

Fϕ(m) depends only on π, not on the choice of ϕ. 3. If B|A is a finite extension of complete DVRs with finite residue fields, then

Fϕ(mB) is also an A-module, via the embedding A,→ End(Fϕ). 第 4 章 Local and Global 4 4.1 Topological ring and group

Restricted topology

Let {Xi | i ∈ I} be a family of topological spaces and Ui is open in Xi, set

∏′ ∏ (Xi,Ui) := {(xi) ∈ Xi | for almost all i ∈ I, xi ∈ Ui} i∈I i=I ∏ with the open base B := { Vi | Vi ⊆ Xi is open and for almost all i, Vi = Ui}, we get a topological space.

∏′ 注记. 1. projection map πi : (Xi,Ui) → Xi is continuous. ∏ 2. res-topo is generally not the same as the sub-topo which is inherited. U ∏ i is open in res-topo, but not necessarily in Xi. ∏′ ∏ 3. If Ui = Xi for almost all i ∈ I, then (Xi,Ui) = Xi. (The smaller Ui is, the finer the res-topo is). ∏ ∏ ′ ′ ′ ′ ∈ 4. (Xi,Ui) = (Xi,Ui ) whenever Ui = Ui for almost all i I.

∏′ Let X = (Xi,Ui), ∀x ∈ X, x correspond to a finite set S(x) ⊆ I such as:

x 7→ S(x) := {i ∈ I | xi ∈/ Ui} ∏ ∏ ∪ Let XS := Xi × Ui , note that XS ∈ B is open and X = XS. It is i∈S i∈I\S S finite easy to see that XS has a base as a subspace of X: ∏ BS := { Vi | Vi ⊆ πi(XS) is open and Vi = Ui = πi(XS) for almost all i ∈ I} 116 CHAPTER 4. LOCAL AND GLOBAL which is standard base of product topology. We consider the category I in which the object is the finite set of I and the homorphism is the inclusion map. we have a direct system

1. For S, T ∈ ob(I), S ⊆ T ⇐⇒ XS ⊆ XT

2. U ⊆ X is open ⇐⇒ ∀S ∈ ob(I),U ∩ XS is open in XS.

♦ 命题 4.1.1

As above, supposed that XS equip with product topology, then we have a canonical homeomorphism of topological spaces:

−→∼ ϕ : X lim−→ XS S ⨿ that sends x ∈ X to the equivalence class of x ∈ XS(x) ⊆ XS.

By this proposition, we can think restricted topology as a direct limit of product topology (universal property).

♦ 命题 4.1.2

If Xi is a locally compact group, and Ui is compact and open in Xi for almost all i ∈ I. then X is locally compact.

Proof. Note that XS is locally compact by Tychonoff’s theorem (key point is that in a finite product, products of open sets are open). It follows that X is locally compact, since each x ∈ XS has a compact neighborhood x ∈ U ⊆ C ⊆ XS that is also a compact neighborhood in X.

注记. We can assume that almost all Ui are compact.

Profinite groups

♦ 定义 4.1.1 A profinite group is a topological group that is an inverse limit of finite groups with the discrete topology.

Given any (topological) group G, we can construct a profinite group by taking the profinite completion ∏ b ⊆ G := lim←− G/N G/N N N 4.1. TOPOLOGICAL RING AND GROUP 117 where N ranges over (open) normal subgroups of finite index, ordered by con- tainment.

例 4.1.1 (Profinite integer). Zb Z Z := lim←− /n for which ew have: n b 1. Z = {(a1, a2, a3, ··· ) | ai ∈ Z/iZ, ai mod j = aj if j|i}, then we have a in- b jection Z ,→ Z, a 7→ (a mod 1, a mod 2, a mod 3, ··· ). This map cannot be b surjective, as Z is uncountable while Z is countable. b ∼ ∏ 2. By Chinese remainder theorem , we obtain Z = Zp. With this in mind p prime we can define the adele of Q. ∪ ¯ ¯ 3. Fix any prime p and consider Gal(Fp|Fp), where Fp = Fpn . Note that n⩾ ∼ ¯ ∼ b Gal(Fpn |Fp) = Z/nZ, we see Gal(Fp|Fp) = Z. b 4. There is a canonical pairing Q/Z × Z → U(1), (q, a) 7→ χ(qa) where χ is the fin 2πiα character of AQ induced by Q/Z → U(1), α 7→ e . The pairing identifies b Z with the Pontrjagin dual of Q/Z.

The profinite completion of G is (by construction) a profinite group, and it comes equipped with a natural homomorphism

ϕ : G → Gb ∏ that sends each element of G to G/N. We have the following universal property: For every continuous homomorphism ψ : G → H of topological groups there is a unique continuous homomorphism that makes the following diagram commute:

ϕ G Gb

ψ ∃1

♦ 命题 4.1.3 b The image of G in G is a dense subgroup.

b Proof. Consider the structure of open set in G.

引理 4.1.1. Let G be a compact group, and H a subgroup. Then H is open if and only if H is closed and of finite index. 118 CHAPTER 4. LOCAL AND GLOBAL

引理 4.1.2. Suppose that G is a Hausdorff, compact, totally disconnected topological group. Then if {Hs} is the set of open normal subgroups of G, we have ∩ Hs = {1} s

♦ 定理 4.1.1 A topological group is profinite if and only if it is Hausdorff, compact, and totally disconnected.

Some useful properties and facts:

• Every product of (arbitrarily many) profinite groups is profinite; the topology arising from the profiniteness agrees with the product topology. The inverse limit of an inverse system of profinite groups with continuous tran- sition maps is profinite and the inverse limit functor is exact on the category of profinite groups. Further, being profinite is an extension property.

• Every closed subgroup of a profinite group is itself profinite; the topology arising from the profiniteness agrees with the subspace topology. If N is a closed normal subgroup of a profinite group G, then the factor group G/N is profinite; the topology arising from the profiniteness agrees with thequo- tient topology.

• Since every profinite group G is compact Hausdorff, we have a Haar measure on G, which allows us to measure the ”size” of subsets of G, compute certain probabilities, and integrate functions on G.

• A subgroup of a profinite group is open iff it is closed and has finite index.

• According to a theorem of Nikolay Nikolov and Dan Segal, in any topologically finitely-generated profinite group (that is, a profinite group thathasa dense finitely-generated subgroup) the subgroups of finite index are open. This generalizes an earlier analogous result of Jean-Pierre Serre for topologically finitely-generated pro-p groups. The proof uses the classification of finite simple groups.

• As an easy corollary of the Nikolov-Segal result above, any surjective discrete group homomorphism ϕ : G → H between profinite groups G and H is continuous as long as G is topologically finitely-generated. Indeed, any 4.1. TOPOLOGICAL RING AND GROUP 119

open subgroup of H is of finite index, so its preimage in G is also of finite index, hence it must be open.

• Suppose G and H are topologically finitely-generated profinite groups which are isomorphic as discrete groups by an isomorphism ι. Then ι is bijective and continuous by the above result. Furthermore, ι−1 is also continuous, so ι is a homeomorphism. Therefore the topology on a topologically finitely- generated profinite group is uniquely determined by its algebraic structure.

注记. 1. The homomorphism ϕ is injective iff G is residually finite (intersection of all finite-index open normal subgroups of G is the trivial) 2. The profinite completion of G as a topological group is not necessarily the same thing as the profinite completion of G as a group (forgetting its topology); this depends on whether the original topology on G contains the profinite topology or not. 3. A profinite group need not equal to its profinite completion , for example,the group Gal(Q|Q) endowed with the Krull topology. 4. Profinite groups that are isomorphic to their profinite completion as groups are said to be strongly complete; this is equivalent to requiring every finite index subgroup to be open. It was recently proved that if G is finitely generated as a topological group (meaning it contains a finitely generated dense subgroup), then G is strongly complete [http://annals.math.princeton.edu/wp- content/uploads/annals-v165-n1-p05.pdf]. unit group of topological ring

Given a topological ring R, its unit group R× is not necessarily to be a topological group in the subspace topology unless R happens to be a subring of a topological field (the definition of which requires inversion to be continuous). There is a standard solution to this problem, which is to give the group R× the weakest topology that makes it a topological group. This is done by embedding R× in R × R via the injective group homomorphism ϕ : R× ,→ R × R r 7→ (r, r−1) The topology on R× is determined by the subspace topology on ϕ(R×) ⊆ R × R the inversion map r 7→ r−1 is then continuous: R× → ϕ(R×) → R×, r 7→ (r, r−1) 7→ r−1 120 CHAPTER 4. LOCAL AND GLOBAL

4.2 Local fields and global fields

Claasification of local field

♦ 定义 4.2.1 A local field is a field K with a nontrivial absolute value that is locally compact under the induced topology.

引理 4.2.1. Let K be a field with a nontrivial absolute value | · |. Then K is a local field iff every closed ball in K is compact.

Proof. Suppose K is a local field. The point 0 ∈ K lies in a compact neighborhood that contains some compact closed ball B(0, r). Pick c ∈ K× with |c| > 1 (this is possible because | · | is nontrivial). The map x 7→ cx is continuous and | · | is multiplicative, so B(0, |c|nr) is compact for every n ∈ N. We thus have compact balls about 0 of arbitrarily large radii, implying that every closed ball B(0, r) is a closed subset of a compact set, hence compact. The translation map x 7→ x + z is continuous, so every closed ball B(z, r) is compact. This proves the forward implication, and the reverse implication is immediate. For the parenthetical, note that the argument above applies to any compact closed ball.

推论. Let K be a local field with absolute value | · |, then K is complete.

Proof. Suppose not. Then there is a Cauchy sequence (xn) in K that converges to ∈ b\ ∈ N | − | 1 ⩾ a limit x K K. Pick N s.t. xn x < 2 for all n N, and consider the set S := B(xN , 1), which is compact by previous Lemma. The Cauchy sequence

(xn)n⩾N in S must converge to a limit in S, since S is compact, and this limit must be x ∈/ K, a contradiction.

♦ 命题 4.2.1 Let K be a field with discrete valuation v, valuation ring A, and uniformizer π. Then K is a local field iff K is complete and the residue field κ(A) := A/πA is finite.

v(x) Proof. We fix an absolute value |x|v := c induced by v (with 0 < c < 1). Suppose K is a local field. Then the valuation ring A = B(0, 1) is a closed ball, hence compact.

Each coset x + πA of πA is an open ball B(x, 1) since y ∈ x + πA ⇐⇒ |x−y|v ⩽

|π|v = c < 1. Two cosets x + πA and y + πA are either equal or disjoint, so we can 4.2. LOCAL FIELDS AND GLOBAL FIELDS 121 cover A with disjoint open balls B(x, 1) = x + πA. This disjoint open cover must be finite, since A is compact, so A/πA =: k is finite. b Now suppose that K is complete and that k = A/πA is finite. Then A = A is complete. By the isomorphism of topological rings

A Ab ∼ = lim←− n n π A Each quotient A/πnA is finite and therefore compact; it follows that A is compact. Thus K contains a compact closed ball B(0, 1) = A and is therefore locally compact.

推论. Let L be a global field with a nontrivial absolute value | · |v. The completion

Lv of L with respect to | · |v is a local field.

引理 4.2.2. A locally compact topological vector space over a nondiscrete locally compact field has finite dimension.

♦ 定理 4.2.1 Let L be a local field. If L is archimedean then it is isomorphic to R or C,

and otherwise L is isomorphic to a finite extension of Qp or Fp((t)) for some prime p.

Proof. If char(L) = 0, then the prime subfield of L is Q, and L contains the completion of Q by the universal property of completions. By Ostrowski’s theorem,

L contains a subfield K isomorphic to Qp for some prime p, or to R.

If char(L) = p, then the prime subfield of L is Fp, and L must contain a transcendental element t, since no algebraic extension of Fp has a nontrivial absolute value. Thus L contains the completion of F = Fp(t), and every completion of Fp(t) is isomorphic to Fq((t)) for some q a power of p, each of which is a finite extension of Fp((t)).

注记. The only finite extension of R is C. archimedean, characteristic 0 : R or C. nonarchimedean, characteristic 0 : finite extension of Qp. nonarchimedean, characteristic p : finite extension of Fq((t)).

Places of a field 122 CHAPTER 4. LOCAL AND GLOBAL

♦ 定义 4.2.2 A place of a field K is an equivalence class of nontrivial absolute values on K.

注记. Recall that the completion of K at an absolute value depends only on its equivalence class, so there is a one-to-one correspondence between places of K and completions of K.

We use MK to denote the set of places of K, and for each place v ∈ MK we use Kv to denote the corresponding completion of K and | · |v to denote the corresponding absolute value.

Now let K be a global field. For each place v ∈ Mk the completion Kv is a local field. From our classification of local fields, if Kv is archimedean then Kv ≃

R or Kv ≃ C, and otherwise the absolute value of Kv is induced by a discrete −v(x) valuation that we also denote v; note that while the absolute value |x|v := c depends on a choice of c ∈ [0, 1], the discrete valuation v : Kv → Z is uniquely determined. We now introduce the following terminology:

• if Kv ≃ R then v is a real place;

• if Kv ≃ C then v is a complex place;

• if | · |v is induced a discrete valuation vp corresponding to a prime p of K then v is a finite place; otherwise v is an infinite place.

注记. Every finite place is nonarchimedean, but infinite places may be archimedean or nonarchimedean (the latter is possible only for global function fields).

例 4.2.1. MQ consists of finite places p corresponding to p-adic absolute values

| · |p, and a single archimedean infinite place ∞ corresponding to the Euclidean absolute value | · |∞.

例 4.2.2. MFq(t) consist of finite places corresponding to irreducible polynomials in

Fq[t] and a single nonarchimedean infinite place ∞ corresponding to the absolute (·) value | · |∞ := qdeg .

注记. If we put z := 1/t and consider Fq(z) ≃ Fq(t), the absolute value | · |∞ on

Fq(t) is the same as the absolute value | · · · |z on Fq(z) corresponding to the irreducible polynomial z ∈ Fq[z]. 4.2. LOCAL FIELDS AND GLOBAL FIELDS 123

♦ 定义 4.2.3 If L|K is an extension of global fields, for every place w of L, any absolute

value |·|w restricts to an absolute value on K that represents a place v of K. We write w|v to indicate this relationship and say that w extends v or that w lies above v.

注记. v independent of the choice of | · |w

♦ 定理 4.2.2 Let L|K be a finite separable extension of global fields andlet v be a place

of K. We have an isomorphism of finite etale Kv-algebras ∏ ∼ L ⊗K Kv −→ Lw w|v

defined by l ⊗ x 7→ (lx, ··· , lx).

Proof. (Sketch) The separable extension L|K is a finite etale K-algebra, so the base change L ⊗ K is a finite etale K -algebra, and is therefore isomorphic to a finite ∏ K v v product i∈I Li of finite separable extensions Li of Kv, each of which is a local field. We just need to show that there is a one-to-one correspondence between the sets of local fields {Li | i ∈ I} and {Lw : w|v}. Each L is a local field extending K , and therefore has a unique absolute i v ∏ | · | | · | → ⊗ ≃ ↠ value w that restricts to v. The map L, L K Kv i Li Li allows us to view L as a subfield of each Li, so the absolute value | · |w on Li restricts to an absolute value on L that uniquely determines a place w|v. This defines a map ϕ : {Li | i ∈ I} → {Lw : w|v} that we will show is a bijection satisfying

ϕ(Li) ≃ Li.

推论. Let f ∈ K[x] be an irreducible polynomial such that L ≃ K[x]/(f(x)). There is a one-to-one correspondence between the irreducible factors of f in Kv[x] and the places of L lying above v. If f = f1 ··· fr is the factorization of f in Kv[x], { | } { ··· } ≃ then we can order the set w v = w1, , wr so that Lwi Kv[x]/(fi(x)) for 1 ⩽ i ⩽ r.

The Galois group Gal(Kv|Kv) acts on the set HomK (L, Kv) via composition: given σ ∈ Gal(Kv|Kv) and τ ∈ HomK (L, Kv), we have σ ◦ τ ∈ HomK (L, Kv), and this clearly defines a group action. 124 CHAPTER 4. LOCAL AND GLOBAL

♦ 命题 4.2.2 Let L|K be a finite separable extension of global fields andlet v be a place of K. We have a bijection

HomK (L, Kv)/Gal(Kv|Kv) ↔ {w|v}

between Gal(Kv|Kv)-orbits of K-embeddings of L into Kv and the places of L above v. Proof. We may assume L ≃ K(α) = K[x]/(f) by the primitive element theorem.

We then have a bijection between HomK (L, Kv) and the roots αi of f in Kv that is compatible with the action of Gal(Kv|Kv) on both sets. If f = f1 ··· fr is the factorization of f in Kv[x], each fi corresponds to an orbit of the action of Gal(Kv|Kv) on the roots of f, and by the previous corollary, these are in one-to-one correspondence with the places of L above v.

例 4.2.3. For K = Q and v = ∞, HomQ(L, C)/Gal(C|R) is in bijection with the set {w|∞} of infinite places of the number field L; note that Gal(C|R) is the cyclic group of order 2, so the orbits of HomQ(L, C) all have size 1 or 2, depending on whether the embedding of L into C is fixed by complex conjugation or not. Each real place w corresponds to a Gal(C|R)-orbit of size 1; this occurs for the elements of HomQ(L, C) whose image lies in R and may also be viewed as elements of HomQ(L, R). Each complex place corresponds to a Gal(C|R)-orbit of size 2 in HomQ(L, C); these are conjugate pairs whose images do not lie in R.

Let K be a number field. Elements of HomQ(K, R) are real embeddings, and elements of HomQ(K, C) whose image does not lie in R are complex embeddings. There is a one-to-one correspondence between real embeddings and real places, but complex embeddings come in conjugate pairs; each pair of complex embeddings corresponds to a single complex place.

♦ 命题 4.2.3 Let K be a number field with r real places and s complex places. Then [K : Q] = r + 2s.

Proof. We may write K ≃ Q[x]/(f) for some irreducible separable f ∈ Q[x], and we then have [K : Q] = deg f = #HomQ(K, C), since there is a one-to-one correspondence between HomQ(K, C) and the roots of f. The action of Gal(C|R) on HomQ(K, C) has r orbits of size 1, and s orbits of size 2. 4.2. LOCAL FIELDS AND GLOBAL FIELDS 125

♦ 命题 4.2.4

Let K be a number field with s complex places. Then the sign of ∆K is (−1)s.

Proof. Linear algebra.

Haar measures

A locally compact group is a topological group that is Hausdorff and locally compact; a compact group is a locally compact group that is compact.

♦ 定义 4.2.4 A (left) Haar measure µ on a locally compact group X is a nonzero Radon measure that is translation invariant, meaning that

µ(E) = µ(x + E)

for all x ∈ X and measurable E ⊆ X.

注记. 1. One defines a right Haar measure analogously, but in most cases they coincide and in our situation we are working with an abelian group (the additive group of a field), in which case they necessarily do. 2. Every measurable set in compact groups has finite measure. 3. Local fields are σ-compact .

♦ 定理 4.2.3. Weil Every locally compact group G has a Haar measure. If µ and µ′ are two Haar measure on G, then there is a positive real number λ such that µ′ = λµ.

例 4.2.4. The standard Lebesgue measure µ on Rn is the unique Haar measure on Rn for which the unit cube has measure 1.

The additive group of a local field K is a locally compact group (it is a metric space, hence Hausdorff). For compact groups G, it is standard to normalize the Haar measure so that µ(G) = 1, but local fields are never compact, and we will always have µ(K) = ∞. 126 CHAPTER 4. LOCAL AND GLOBAL

For nonarchimedean local fields the valuation ring A = B(0, 1) is a compact group, and it is then natural to normalize the Haar measure on K so that µ(A) = 1. The key point is that there is a unique absolute value on K that is compatible with every Haar measure µ on K, no matter how it is normalized.

♦ 命题 4.2.5 Let K be a local field with discrete valuation v, residue field k, and absolute −v(·) value | · |v := (#k) , and let µ be a Haar measure on K. For every x ∈ K and measurable set S ⊆ K we have

µ(xS) = |x|vµ(S)

Moreover, the absolute value | · |v is the unique absolute value compatible with the topology on K for which this is true.

Proof. Let (A, p) be the valuation ring of K and x ≠ 0. The map ϕx : y 7→ xy is an automorphism of (K, +), and it follows that the composition µx := µ◦ϕx is a Haar × measure on K, hence µx = λxµ for some λx > 0. Define the function χ : K → R⩾0 × by χ(x) := λx. Then µx = χ(x)µ, and for all x, y ∈ K we have

µ (A) χ(x)µ (A) χ(x)χ(y)µ(A) χ(xy) = xy = y = = χ(x)χ(y) µ(A) µ(A) µ(A)

× Thus χ is multiplicative, and we claim that in fact χ(x) = |x|v for all x ∈ K , it suffices to consider x ∈ A•. For x ∈ A•, A/xA is a k-vector space of dimension v(x) and has cardinality [A : xA] = (#k)v(x). Writing A as a finite disjoint union of cosets of xA, we have

µ(A) = [A : xA]µ(xA) = (#k)v(x)χ(x)µ(A) and therefore χ(x) = |x|v as claimed. It follows that

µ(xS) = µx(S) = χ(x)µ(S) = |x|vµ(S)

| · | ∼ | · | | | | |c To prove uniqueness, if v then for some 0 < c < 1 we have x = x v × × for all x ∈ K . Let us fix x ∈ K with |x|v ≠ 1 (take any x with v(x) ≠ 0). If | · | also satisfies µ(xS) = |x|µ(S) then ( ) µ(xA) µ(xA) c = |x| = |x|c = µ(A) v µ(A) which implies c = 1. 4.2. LOCAL FIELDS AND GLOBAL FIELDS 127

The product formula for global fields

♦ 定义 4.2.5 Let K be a global field. For each place v of K the normalized absolute

value ∥ · ∥v : Kv → R⩾0 is defined by

µ(xS) ∥x∥ := v µ(S)

where µ is a Haar measure on Kv and S is any measurable set with µ(S) ≠ 0.

注记. 1. This definition is independent of the choice of µ and S. One can al-

ways take S = Av.

2. If v is nonarchimedean then the normalized absolute value ∥ · ∥v is precisely

the absolute value | · |v defined in Proposition 4.2.5. 3. If v is a real or complex place then the normalized absolute value is just the usual Euclidean absolute value on R or square of the Euclidean absolute value C.

4. When v is a complex place the normalized absolute value ∥ · ∥v is not an absolute value, because it does not satisfy the triangle inequality.

例 4.2.5. Let K = Q(i) and v a complex place, then ∥1∥v = 1 but

∥1 + 1∥v = 4 > 2 = ∥1∥v + ∥1∥v

Nevertheless, the normalized absolute value is always multiplicative and compatible with the topology on Kv in the sense that the open balls B(x, r) are a basis for the topology on Kv; these are the properties that we care about for the product formula (and for the topology on the ring of adeles AK ).

引理 4.2.3. Let L|K be a finite separable extension of global fields, let v be a place of K and let w|v be a place of L. Then

∥ ∥ ∥ ∥ x w = NLw|Kv (x) v

Proof. We may assume [Lw : Kv] > 1. If v is archimedean then Lw ≃ C and

Kv ≃ R, in which case for any x ∈ Lw we have

µ(xS) 2 ∥x∥ = = |x|C = |xx|R = |NC|R(x)|R = ∥N | (x)∥ w µ(S) Lw Kv v

Now assume v is nonarchimedean. Let πw and πv be uniformizers for the local fields Kw and Lv, respectively, and let f := [Kw : kv]. Without loss of generality, 128 CHAPTER 4. LOCAL AND GLOBAL

w(x) we may assume x = πw , since ∥x∥v = |x|v depends only on w(x). Then

∥ ∥ ∥ f ∥ −f NLw|Kv (πw) v = πv v = (#kv)

∥ ∥ −fw(x) so NLw|Kv (x) v = (#kv) . Proposition 4.2.5 then implies

∥ ∥ −w(x) −fw(x) ∥ ∥ x w = (#kw) = (#kv) = NLw|Kv (x) v

注记. Note that if v is a nonarchimedean place of K extended by a place w|v of

L|K, the absolute value ∥ · ∥w is not the unique absolute value on Lw that extends the absolute value on | · |v on Kv, it differs by a power of n = [Lw : Kv], but it is equivalent to it.

♦ 定理 4.2.4. product fomula Let L be a global field. For all x ∈ L× we have ∏ ∥x∥v = 1

v∈ML

Proof. The global field L is a finite separable extension of K = Q or K = Fq(t). Let p be a place of K. By Theorem 4.2.2, any basis for L as a K-vector space is also a basis for ∏ L ⊗K Kp ≃ Lv v|p as a Kv-vector space. Thus ∏

We now take the product of both sides over all places p ∈ MK to obtain ∏ ∏ ∏ ∏ ∥NL|K (x)∥p = ∥x∥v = ∥x∥v

p∈MK p∈MK v|p p∈ML The LHS is equal to 1, by the product formula for K.

With the product formula in hand, we can now give an axiomatic definition of a global field, which up to now we have simply defined as a finite extension of

Q or Fq(t), due to Emil Artin and George Whaples. 4.2. LOCAL FIELDS AND GLOBAL FIELDS 129

♦ 定义 4.2.6

A global field is a field K whose completion at each of its places v ∈ MK is a local field, and which has a product formula of the form ∏ ∥x∥v = 1

v∈MK

∥ · ∥ → R ∥ · ∥ | · |mv where each normalized absolute value v : Kv ⩾0 satisfies v = v

for some absolute value | · |v representing v and some fixed mv ∈ R>0.

♦ 定理 4.2.5. Artin-Whaples

Every global field is a finite extension of Q or Fq(t). 130 CHAPTER 4. LOCAL AND GLOBAL

4.3 Adele ring and Idele group

Convention：We assume that K always denotes a number field unless declaim. • The set of all finite places (non-Archemedean aabsolute, prime, or discret valuation) writed by P . • The set of all infinite places (Archemedean aabsolute) writed by Σ.

• Let S denotes the finite subset of MK and assume always Σ ⊆ S.

• Kv denotes the completion of K at v. If v ∈ P , let Ov be its valuation ring.

We define the ring OS of S-integers of K as following

OS := {x ∈ K | ∀v ∈/ S, x ∈ Ov}

注记. 1. If S ≠ ∅, then OS is a Dedekind domain that is not a field.

2. Understanding OS is the most important thing, ”adele and idele” is just a tool or a method.

例 4.3.1. 1. OΣ =: OK is a especially important case. Q {∞ } O Z 1 2. K = ,S = , p , then S = [ p ]. −1 3. K = Fq(t),S = {(t )}, then OS = Fq[t].

4. If K is a function field and S = ∅, the OS is the maximal finite subfield of

K (or, in other words, the “algebraic closure of Fp in K”).

O O× Since S is a Dedekind domain, we can consider its unit group S and ideal class group Cl(OS). To summarize, we have exact sequences

→ O× → × → P O → F O → O → 1 S K ( S) ( S) Cl( S) 1

Adele ring

We consider the product ring equipped with the product topology ∏ ∏ AS O × K := v Kv v∈/S v∈S 4.3. ADELE RING AND IDELE GROUP 131

O AS Note that v is compact open subring of Kv, and by Tychonoff theorem, K is locally compact ring. We set { } ∪ ∏ A AS ∈ | ∈ O K := K = (av) Kv for almost all finite place v, av v

S finite v∈MK A ⊆ A ⇐⇒ ∀ ∩ AS • K is equipped with topology as follow: A K is open S, A K is AS open in K .

• a ∈ K belongs to AK by identify a = (a, a, ··· ), since the v(a) = 0 for almost

all places. The elements of K ⊂ AK are the so-called principal adeles of AK . ∏′ • AK = (Kv, Ov).

AS 注记. The ring of S-adeles K may be viewed as an algebra over ring of S-integers

OS.

(p) ··· 1 ··· ∈ A 例 4.3.2. The sequence x = (1, , 1, p , 1, ) Q converges in the product topology with limit (1, 1, ··· ); however, it doesn’t converge in the restricted product topology. ∏ For each adele a = (ap)p ∈ AQ and for each basic open set U = Up × ∏ p∈S Z 1 − ∈ Z ∈ Z 1 − ∈ Z ∈ p we have p ap / p for ap p and therefore p ap / p for all p / S. p∈MQ\S As a result, it stands, that (x(p) − a)/∈ U for almost all p.

Norm and measure

We extend the normalized absolute value ∥ · ∥v of Kv to AK via ∥a∥v := ∥av∥v, and define the adelic absolute value (adelic norm): ∏ ∥a∥ := ∥a∥v

v∈MK

注记. 1. For the finite place v, Ov = {x ∈ Kv : ∥x∥v ⩽ 1} is a closed ball and hence compact;

2. for almost all v ∈ MK , ∥a∥v ⩽ 1, so the adelic norm is well-defined; 3. For the principal adele a, we have ∥a∥ = 1 by the product formula.

♦ 命题 4.3.1

The adele ring AK is locally compact and Hausdorff (LCH).

Proof. If x, y ∈ AK are distinct then xv ≠ yv for some v ∈ MK , and since Kv is

Hausdorff we can separate xv and yv by open sets whose inverse images under the projection map πv : AK → Kv are open sets separating x and y. 132 CHAPTER 4. LOCAL AND GLOBAL

By the propsition above, (AK , +) is a a locally compact group, and thus has a Haar measure that is unique up to scaling.

Each of the completions Kv is a local field with a Haar measure µv, which we normalize as follows:

• µv(Ov) = 1 for all nonarchimedean v;

• µv = µR for Kv ≃ R, where µR is the Lebesgue measure on R;

• µv = 2µC for Kv ≃ C, where µC is the Lebesgue measure on C ≃ R × R.

Note that the normalization of µv at the archimedean places is consistent with r s n the measure µ on KR ≃ R × C ≃ R induced by the canonical inner product on KR ⊆ KC. We now define a measure µ on A as follows. We take as a basis for the K ∏ σ-algebra of measurable sets all sets of the form v Bv, where each Bv is a measurable set in Kv with µv(Bv) < ∞ such that Bv = Ov for almost all v (the σ-algebra is then generated by taking countable intersections, unions, and complements in A K ). We then define ( ) ∏ ∏ µ Bv := µv(Bv) v v It follows from uniqueness of the Haar measure (up to scaling) that µ is a Haar measure on AK which we henceforth adopt as our normalized Haar measure on

AK .

Base change

Note that the canonical embedding K,→ AK makes AK a K-vector space, and if L|K is a finite separable extension of K (also a K-vector space), we may consider the tensor product

AK ⊗K L which is also an L-vector space. As a topological K-vector space, the topology on

AK ⊗ L is just the product topology on [L : K] copies of of AK .

♦ 命题 4.3.2 There is a canonical isomorphism of topological rings: ∼ AK ⊗ L −→ AL 4.3. ADELE RING AND IDELE GROUP 133

Proof. We have a isomorphism of topological rings ∏′ A ⊗ −→∼ ⊗ O ⊗ O ϕ : K L (Kv K L, v OK L)

(av) ⊗ x 7→ (av ⊗ x) ∏ ∏ ⊗ ≃ O ⊗ O ≃ O Note that Kv L Lw and v OK L w, so w|v w|v

∏ ′ AK ⊗ L ≃ (Lw, Ow)

w∈ML

推论. The canonical embeddings of L ≃ K ⊗K L into AK ⊗K L and L into AL agree, see diagram:

∼ L K ⊗K L

∼ AL AK ⊗K L

Another definition b Define the ring AZ := Z × R. We have the properties:

∼ ∅ • AZ = AQ • AZ ∩ Q = Z • AZ + Q = AQ

Proof. For the forth one: ⊕ ⊕ 1 ∼ Q/Z = Z[ ]/Z −→ Q /Z = AQ/AZ p p p p p

∼ 引理 4.3.1. AQ = AZ ⊗Z Q ∼ b ∼ b Proof. AZ ⊗Z Q = (Z ⊗Z Q) × (R ⊗Z Q) = (Z ⊗Z Q) × R. As a result, it suffices to b ∼ fin show that Z ⊗Z Q = AQ . b fin Define a Z-bilinear function Φ: Z × Q → AQ , ((ap)p, q) 7→ (apq)p. This function is obviously well-defined, because only a finite number of prime numbers divide the denominator of q ∈ Q. b Let M be another Z-module together with a Z-bilinear function Ψ: Z × Q → fin M. We have to show that there exists one and only one Z-linear function θ : AQ → M s.t. θ ◦ Φ = Ψ. 134 CHAPTER 4. LOCAL AND GLOBAL

b Φ fin Z × Q AQ

Ψ ∃1θ

We define the function θ in the following way: For a given ((up)p) there exists ∈ N ∈ Zb 1 · 1 a u and a (vp)p s.t. up = u vp for all p. Define θ((up)p) := Ψ((vp)p, u ). It can be shown that θ is well-defined, Z-linear and satisfies θ ◦Φ = Ψ. Furthermore, θ is unique with these properties.

For an algebraic number field K with degree n, we use base change to define:

AK := AQ ⊗Q K

n We give the right hand side the product topology of (AQ) and transport this topology via the isomorphism onto AQ ⊗Q K. (AZ 赋予 prfinite group 拓扑，如何在 AZ ⊗Z Q 上定义拓扑？)

Core conclusion

♦ 定理 4.3.1

K,→ AK is a discret subgroup, and AK /K is compact.

Proof. We may assume that K = Q. Zb × − 1 1 Zb × − 1 1 Let U := ( 2 , 2 ),W := [ 2 , 2 ], then

U ∩ Q = {0},W + Q = AQ

Therefore W → AQ/Q is surjective and hence AQ/Q is compact.

Strong approximation

引理 4.3.2 (Adelic Blichfeldt-Minkowski lemma). There is a positive constant BK such that for any a ∈ AK with ∥a∥ > BK there exists a nonzero principal adele x for which ∥x∥v ⩽ ∥a∥v for all v ∈ MK .

♦ 定理 4.3.2. Strong Approximation ⊔ ⊔ Let MK = S T {w} be a partition with S finite. Given any av ∈ K and 4.3. ADELE RING AND IDELE GROUP 135

ϵv ∈ R>0 with v ∈ S, there exists an x ∈ K for which

∥x − av∥v ⩽ ϵv for all v ∈ S

∥x∥v ⩽ 1 for all v ∈ T

Idele group

(p) 例 4.3.3. We consider the sequence x = (1, ··· , 1, p, 1, ··· ) ∈ AQ. ∏ ∏ (p) The basic open set U = Up × Zp contains x for all sufficiently p∈S p∈MQ\S (p) large p. It follows that the sequence {x } converges to 1 in the topology of AQ. But notice that the inverse of x(p) is not convergent by Example 4.3.2. Thus the function x 7→ x−1 is not continuous in the subspace topology for A× K .

Idele group and norm 1 idele

A× { ∈ A | ∀ ∈ ∈ × ∈ O× ∈ } K = (av) K v MK , av Kv , av v for almost all v MK A× → A × A A× By K , K K we install K a topology which has a open base of the form: ∏ ∏ ′ × O× U = Uv v v∈S v∈/S ⊆ × A× → A × A where Uv Kv . To see this, note that in terms of the embedding K , K K , −1 each ϕ(a) = (a, a ) lies in a product U × V of basic open sets U, V ⊆ AK , and this forces both a and a−1 to lie in O×, for almost all v. The open sets U ′ are precisely v ∏ ′ × O× the open sets in the restricted product (Kv , v ). This leads to the following definition. ♦ 定义 4.3.1 Let K be a global field. The idele group of K is the topological group

∏ ′ I × O× K = (Kv , v )

v∈MK

with multiplication defined component-wise, which we view as the subgroup A× A K of K endowed with the restricted product topology rather than the subspace topology. ∏ ∏ IS O× × × I IS 注记. Let K := v Kv , then K = lim K . ∈ −→ v∈/S v S S 136 CHAPTER 4. LOCAL AND GLOBAL

O× × Note that when K is a number field the unit group K is not cocompact in K O× Rr+s because log K is not a (full) lattice in ; it lies inside the trace zero hyperplane Rr+s I 0 . In order to get a cocompact subgroup we need to restrict K to a subgroup corresponding to the trace zero hyperplane.

♦ 定义 4.3.2 The group of 1-ideles is the topological group

I1 | · | { ∈ I ∥ ∥ } K := ker = x K : x = 1

which we note contains K×, by the product formula.

∥ ∥ ∈ I ∈ O× 注记. 1. a > 0 for all a K , because av v for almost all v, and this means

that |a|v = 1 for almost all v;

2. ∥ · ∥ : IK → (R>0, +) is a continuous homomorphism of topological groups.

Idele class group

× The canonical embedding K,→ AK restricts to a canonical embedding of K into IK , and we have a surjective homomorphism

i : I ↠ F(O ) S K ∏ S vp(a) (av) 7→ p

where vp(a) := vp(av), the prime p of OS corresponds to the place v ∈/ S. The composition

× K ,→ IK ↠ F(OS)

has a image P(OS), the subgroup of principal fractional ideals, and thus induces a × homomorphism of the idele class group CK := IK /K onto the ideal class group

Cl(OS), we have the following commutative diagram of exact sequences: 4.3. ADELE RING AND IDELE GROUP 137

1 1 1

O× IS IS O× 1 S K K / S 1

× 1 K IK CK 1

1 P(OS) F(OS) Cl(OS) 1

1 1 1

注记. 1. Consider (a) 7→ (v(a))v∈/S, we see that iS is surjective. IS IS K× IS /O× = K = K 2. K S IS ∩ × × . K ∏K K∏ IS × × O× I 3. ker iS = K = Kv v is open in K .(同时闭？) v∈S v∈/S 4. iS does not depend on archimedean places.

Topological result about idele group

♦ 命题 4.3.3

Let K be a global field. The idele group IK is a locally compact group (LCH).

I × Proof. We need to show that K is compact Hausdorff. Each Kv is Hausdorff, so ∏ ∏′ K× is Hausdorff in the product topology, and this implies that (K×, O×) ⊆ ∏ v v v × Kv is Hausdorff, since its topology is finer. For each nonarchimedean place v, O× { ∈ × | | | } the set v = x Kv x v = 1 is closed, hence compact (because the local field K is Hausdorff); this applies to almost all v, and the K× are all locally compact, v ∏ v ′ × O× so the restricted product (Kv , v ) is locally compact, by Proposition 4.1.2.

♦ 命题 4.3.4 × Let K be a global field. Then K is a discrete subgroup of IK .

× Proof. We have K ,→ K × K,→ AK × AK . K is a discrete subset of AK , and it × follows that K × K is a discrete subset of AK × AK . The image of K in AK × AK A× → A × A A× I lies in K , K K , hence it is discrete in K = K . 138 CHAPTER 4. LOCAL AND GLOBAL

注记. Discrete sets need not be closed, in general, but discrete subgroups of a topological Hausdorff group are.

I1 A I 引理 4.3.3. The group of 1-ideles K is a closed subset of K and K , and the two I1 subspace topologies on K coincide.

I1 A I Proof. We first show that K is closed in K , and therefore also in K , since it has a finer topology. ∈ A \I1 Consider any x K K . We will construct an open neighborhood Ux of x I1 that is disjoint from K . The union of the Ux is then the open complement of the I1 closed set K .

For any finite S ⊆ MK and x ∈ AK we define

Uϵ(x, S) := {u ∈ AK : ∥u − x∥ < ϵv for v ∈ S and ∥u∥v ⩽ 1 for v ∈/ S}

The case ∥x∥ < 1:Let Σ ⊆ S be a finite set containing all v for which ∥x∥v > 1, ∏ such that ∥x∥v < 1: such an S exists since ∥x∥ < 1 and ∥x∥v ⩽ 1 for almost all v∈S v. For all sufficiently small ϵ > 0 the set Ux := Uϵ(x, S) is an open neighborhood I1 ∈ ∥ ∥ of x disjoint from K because every y Ux must satisfy y < 1.

The case ∥x∥ > 1: Let B be twice the product of all the ∥x∥v greater than 1, Let Σ ⊆ S be the finite set containing all nonarchimedean v with residue field cardinality less than 2B, and all v for which ∥x∥v > 1. For all sufficiently small I1 ϵ > 0 the set Ux := Uϵ(x, S) is an open neighborhood of x disjoint from K . To prove that the subspace topologies coincide, it suffices to show that for every x ∈ IK and open U ⊆ IK containing x there exists open sets V ⊆ IK and ⊆ A ∈ ⊆ ∩ I1 ∩ I1 W K such that x V U and V K = W K ; this implies that every I1 ⊆ I neighborhood basis in the subspace topology of K K is a neighborhood basis I1 ⊆ A in the subspace topology of K K (the latter is a priori coarser than the former). ∈ I1 ⊆ I So consider any x K and open U K of u. Then U contains a basic open set

V = {u ∈ AK : ∥u−x∥v < ϵv for v ∈ S and ∥u∥v = 1 for v ∈/ S} for some ϵ > 0 and S ⊆ MK (take S = {v ∈ MK : ∥x∥ ̸= 1} and ϵ > 0 small ∈ ⊆ ∩ I1 ∩ I1 enough). If we now put W := Uϵ(x, S) then x V U and V K = W K as desired. 4.3. ADELE RING AND IDELE GROUP 139

♦ 定理 4.3.3. Fujisaki’s lemma × I1 I1 × K is a discrete closed subgroup of K , and K /K is compact.

× I I1 Proof. K is discrete in K , and therefore discrete in the subspace K . × I1 ⊆ To prove that K is cocompact in K it suffices to exhibit a compact set W A ∩ I1 I1 × K for which W K surjects onto K /K under the quotient map (here we are I1 ∩ I1 using Lemma above: K is closed so W K is compact).

To construct W we first choose a ∈ AK such that ∥x∥ > BK , where BK is the Blichfeldt-Minkowski constant in Lemma 4.3.2, and let

W := L(a) = {x ∈ AK ∀ v ∈ MK , ∥x∥v ⩽ ∥a∥v}

u ∈ I1 ∥u∥ = 1 a = ∥a∥ > B Now consider any K . We have , so u K , and by Lemma ∈ × ∥ ∥ ⩽ a ∈ ∈ 4.3.2 there is a z K for which z v u v for all v MK . Therefore zu W . ∈ I1 −1 ∩ I1 I1 × Thus every u K can be written as u = z zu. Thus W K surjects onto K /K I1 → I1 × under the quotient map K K /K , which is continuous, and it follows that I1 × K /K is compact.

♦ 定义 4.3.3 1 I1 × For a global field K the compact group CK := K /K , is the norm-1 idele class group.

1 注记. When K is a function field the norm-1 idele class group CK is totally disconnected, in addition to being compact Hausdorff; this makes it a profinite group. 140 CHAPTER 4. LOCAL AND GLOBAL

4.4 Ideal Class Groups and Unit Groups

We consider the Dedekind domain OS with its ideal class group Cl(OS) and O× unit group S .

Ideal class groups

A celebrated theorem of L. Claborn asserts that for any abelian group A , ∼ there exists a Dedekind domain R s.t. Cl(R) = A.

引理 4.4.1. Let H be a subgroup of topological group G, then

• H is open ⇐⇒ G/H is discret; • H is closed ⇐⇒ G/H is Hausdorff.

♦ 定理 4.4.1

Cl(OS) is a finite abelian group.

I1 ↠ F O Proof. (Number field case) We see that iS : K ( S) is also surjective, since we can modify an idele with its archimedean component. We get a diagram: × I1 1 1 K K CK 1

1 P(OS) F(OS) Cl(OS) 1

注记. A conjecture of Kummer-Vandiner, p does not divide the class number of Q −1 (ζp + ζp ).

Unit groups

引理 4.4.2. Let 0 < c ⩽ C. Then the set S = S(S, c, C) of S-units x with c ⩽ |x|v ⩽ C for all v ∈ S is finite.

Proof. The set W of ideles x = (xv) with |x|v = 1 for all v ∈/ S and c ⩽ |x|v ⩽ C for all v ∈ S is visibly compact. We have S = W ∩ K×, so S is compact and discrete, thus finite. 4.4. IDEAL CLASS GROUPS AND UNIT GROUPS 141

♦ 命题 4.4.1

The set ωK := {x ∈ K | ∀v ∈ MK , |x|v = 1} is precisely the group of roots of unity of K, which is a finite abelian group.

Proof. Applying Lemma 4.4.2 with c = C = 1 shows that ωK is finite. A unit root n x satisfies |x| = 1 for some n, so |x| = 1. Each element of ωK has finite order thus is a unit root.

注记. Any locally compact field except C, the group of roots of unity is finite. This is obvious for R. For a non-Archimedean field of residue characteristic p... For p-adic fields, that the group of roots of unity of p-power is finite follows from the fact that for all n, the Eisenstein criterion can be used to show that the cyclotomic polynomial Φpn (t) is irreducible over Qp and generates a totally ramified extension. For local fields of positive characteristic, there is nothing toworry about because there are no p-power roots of unity even over the algebraic closure.

引理 4.4.3. Let r, s ∈ Z with s ⩾ r ⩾ 0, and let G = Rr × Zs+1−r. Let λ : G → (R, +) be a nontrivial homomorphism of topological groups. Moreover, ∼ • When r = 0, we assume that λ(Zs+1−r) = (Z, +) ∼ • When r > 0, we assume that λ|Rr = (R, +). ∼ Let K := ker(λ) and Γ be any discrete, cocompact subgroup of K. Then Γ = Zs.

Proof. Write λ = λ1 + λ2, where λ1 = λ|Rr and λ2 = λ|Zs+1−r . s+1 ∼ ∼ s Case 1: r = 0. Then λ1 = 0, so Z /K = Z and thus K = Z . In this case every subgroup of K is discrete and is cocompact iff it has maximal rank. s+1−r ∼ r−1 s+1−r Case 2: λ2 = 0. Then K = ker(λ1)⊕Z = R ⊕Z . A discrete cocompact subgroup of this is obtained by choosing a rank r−1 lattice of Rr−1 together with Zs+1−r, hence is isomorphic to Zs.

Case 3: we assume that λ1 and λ2 are both nontrivial. Then the image of λ2 is isomorphic to Zt for some 1 ⩽ t ⩽ s + 1−r. Then

K0 := {(x, y) ∈ G | λ1(x) = λ2(x)}

∼ t ∼ r−1 s+1−r−t is a subgroup of K and K/K0 = Z . Explicitly, K0 = R × Z , and thus ∼ − K = Rr−1 × Zs r+1, so again a discrete cocompact subgroup must have rank s.

For v ∈ M , let C := {x ∈ K | |x| = 1}. This is a compact subgroup of K×. K v ∏ v v v IS Therefore, by Tychonoff, C := Cv is a compact subgroup of K , the adelic circle 142 CHAPTER 4. LOCAL AND GLOBAL group. We have a short exact sequence of topological groups ∏ → → IS → × → 1 C K Kv /Cv 1 v∈S Note that   (R , ×) ≃ (R, +), v is Archimedean × ≃ + Kv /Cv  Z, v is non-Archimedean

If we put r = |Σ|, s = |S|−r, then we may rewrite the exact sequence as

→ → IS → Rr ⊕ Zs → 1 C K 0

♦ 定理 4.4.2 O× S is a finitely generated abelian group. More precisely, its torsion subgroup is the finite group of roots of unity in K, and its rankis |S|−1.

O× ∼ × Z|S|−1 S = WK

× ∩ × ∩ I1 O× Proof. Since K C = ωK ,K K = S ，we get a exact sequence

→ → O× → → 1 ωK S Γ 1

O× Rr ⊕ Zs where Γ denotes the image of S in . ∼ O× × I1 Note that Γ = S /ωK . Moreover, since K is discrete and cocompact in K IS I O× × ∩IS S 1 I1 ∩IS and K is closed in K , S = K K is discrete and cocompact in (IK ) = K K . Now we consider the sequence (which is the norm map factors through Rr ⊕ log Zs −→ (R, +)) IS 1 → Rr ⊕ Zs −→log R ( K ) ( , +) By Lemma 4.4.3, we get the desired result.

Q {∞ } O× { n | ∈ Z} ∼ Z Z ⊕ Z 例 4.4.1. K = ,S = , p , then S = p n = /2 .

Regulator

Rr Zr Let S = Σ. Now s = 0, so Γ is a lattice in 0 isomorphic to 0.

♦ 定义 4.4.1 ∼ −→ Zr The determinant of the isomorphism ϕ :Γ 0 is the regulator of K,

denoted by RK . 4.4. IDEAL CLASS GROUPS AND UNIT GROUPS 143

Rr 注记. Let µ be the invariant measure on 0, then µ(Rr/Γ) R = 0 K Rr Zr µ( 0/ 0) Now we imploy the fundermental system of units of K to compute the regulator.

If the images of ϵ1, ··· , ϵr−1 ∈ OK in Γ forms a basis of Γ, we call ϵ1, ··· ,

ϵr−1 fundermental units. Let vi, ··· , vr−1 be any r − 1 places in Σ, then ( ) | | | | RK = det log ϵi vj

| | | | Note that log ϵi vj = cj log σj(ϵi) , where cj = 1, 2 which is depends on σj. The ∈ O× choice of embedding to omit does not affect the result, because for any α F we have ∑r ∏r cj cj log |σj(α)| = log |σj(α)| = 0 j=1 j so the omitted row with entries cr log |σr(αi)| is (minus) the sum of the rows of the matrix. The regulators of different sets of units have the following relation.

引理 4.4.4. Suppose B = ⟨β1, ··· , βr−1⟩ ⊆ A = ⟨α1, ··· , αr−1⟩ for βi and αi Z- independent sets of units. Then

Reg(β ) i = [A : B] Reg(αi) 144 CHAPTER 4. LOCAL AND GLOBAL

4.5 Some computations and applications

real places r1, imaginary places r2.

unit roots ωK . class number h(K).

Units in quadratic field

例 4.5.1 (fundamental√ unit of real quadratic field). K = Q( d)(d > 0) r = 2, r = 0, ω = {1} O× = {ϵn | Let , then 1 √ 2 k , therefore K n ∈ Z} ϵ Q( d) , is the fundamental unit√ of .√ The fundamental unit of Q( 2) is 1 + 2. − Q O× 例 4.5.2. If r1 + r2 1 = 0, then K = or imaginary quadratic field. So K is finite ⇐⇒ K = Q or imaginary quadratic field.

例 4.5.3 (Pell equation). 2 2 Let N > 0 be a square-free integer, PN := {(x, y) ∈ Z × Z | x − Ny = 1} + { ∈ | ⩾ ⩾ } and PN := (x, y)√ PN x 1, y 1 . Then θ : P → Z( N)×, (x, y) 7→ x + yN 1. N is bijiective; √ ∈ + ∀ ∈ + ⩽ ⩽ Z × 2. There exists (x0, y0) PN , s.t. (x, y) PN , x x0, y y0 and ( N) = { n | ∈ Z} + { n | ∈ N} (x0 + y0N) n , θ(PN ) = (x0 + y0N) n

Class number fomula of imaginary quadratic field √ Let d < 0 be a square-free integer,the discriminant of K = Q( d) is denoted by D. Then we have a fomula of class number of K:

♦ 命题 4.5.1 √ ω ∑D ω ω D h = − K aχ (a) = K L(0, χ ) = K L(1, χ ) K 2D d 2 d 2π d a=1

引理 4.5.1. ωK

Now computate several examples.

例 4.5.4 (d = −1).

χ−1(1) = 1, χ−1(3) = −1, hK = 1

Fermat’s Last Theorem 第 5 章 Number Fields

5.1 *Cyclotomic fields

Q Let ζn(where n > 2) denotes a primitive n-th root of unity, then the field (ζn) 5 is called n-th cyclotomic field. Always let p denotes a prime number in Z.

Prime decomposion

♦ 命题 5.1.1 ( ) Q |Q ∤ (ζn) 7→ If p n, then p is unramified. Moreover, (p) =: σp p ( mod n) is well-defined, and we have a bijiection:

Q |Q −→∼ Z× 7→ Gal( (ζn) ) n , σp p mod n

′ n−1 ∈ Proof. Let q denotes a lying-over of (p), then f (ζn) = nζn / q, otherwise n = nζn−1ζ ∈ q ∩ Q = (p). So q is non-ramified, therefore p is non-ramified. n n ∏ n−1 − i ∈ − i ∈ ⩽ ⩽ − Notice that i=1 (1 ζn) = n / q, then (1 ζn)/ q for 1 i n 1, therefore i ≡ j ≡ ζn ζn ( mod q) imply i j ( mod n). ∏ 推论. Φn(x) = (x − σp(ζn)) is an irreducible polynomial since it is minimal (p,n)=1 polynomial of ζn.

例 5.1.1. The regular 17-gon(seventeen-sided polygon) can be constructed by ruler- and-compass.

Z× ⊇ { } ⊇ { } ⊇ { } ⊇ { } 17 1, 2, 4, 8 1, 4 1 1 146 CHAPTER 5. NUMBER FIELDS

By Galois theory, Q(ζ17) has a series of quadratic field extensions by Q.

♦ 定理 5.1.1 Every finite abelian group is the Galois group of some Galois extension of Q.

例 5.1.2. Construct a Galois extension of Q with Galois group G = Z/7 × Z/13 × Z/13. The idea is to construct Galois extensions which realize each of the cyclic factors of G. If we can do this in such a way so that these extensions are linearly disjoint, then we can just take the compositum of these extensions. So let us first construct an extension with Galois group Z/7. The trick is to choose a prime p such that p ≡ 1( mod 7). The first prime that works is p = 29.

Now consider the extension Q(ζ29)|Q, it is an abelian extension with Galois group

Z/28. Clearly, since 7|28, it has a sub-field K1 whose Galois group is Z/7. We can similarly construct another extension K2|Q with Galois group Z/13.

This time we note that 53 ≡ 1( mod 13), that Gal(Q(ζ53)|Q) = Z/52, and that

13|52. So the field K2 with Gal(K2|Q) = Z/13 exists. Now we have only one more factor to worry about, the ’second’ factor of Z/13 in G. This time we choose a different prime congruent to 1( mod 13). In fact 79 seems to work. As above, in Q(ζ79) there is a sub-field K3 with Gal(K3|Q) = Z/13.

Now note that K1,K2 and K3 are linearly disjoint, that is, the intersection of any two of these fields is Q. This is because they each lie in cyclotomic fields Q(ζp), for different p, which themselves are linearly disjoint (use ramification theory to prove this-note that only p ramifies in Q(ζp)). We now choose K = K1K2K3, Then Galois theory shows Gal(K|Q) = G.

注记. Inverse Galois Problem: Is every finite group the Galois group of a finite Galois extension of Q?

♦ 命题 5.1.2 Q Z× Suppose that subfield L of (ζn) corresponds to the subgroup H of n and p ∤ n, then we have 1. p is unramified in L. 2. p is totally split⇐⇒ p mod n ∈ H. 5.1. *CYCLOTOMIC FIELDS 147

f ∈ 1 Q 3. Let f denotes the smallest integer s.t. p mod n H, then p has f [L : ] prime divisors.

Proof.

♦ 命题 5.1.3. prime decomposition in cyclotomic field

1. If p ∤ n, then (p) = q1 ··· qg in Q(ζn), where fg = ϕ(n), f = f(qi|p) =

ord(σp); moreover, p is totally split⇐⇒ p ≡ 1 mod n. s ϕ(ps) 2. If n = p , then p is totally ramified in Q(ζn), (p) = (1 − ζn) . s ϕ(ps) 3. If n = p m, (m, p) = 1, s ⩾ 1, then (p) = (q1 ··· qg) in Q(ζn), where f fg = ϕ(m), f = f(qi|p) is smallest integer s.t. p ≡ 1 mod m.

Proof. 1. By proposition 5.1.2. 2. We set x = 1 in ps − x 1 (p−1)ps−1 ps−1 Φ s = = x + ··· + x + 1 p xps−1 − 1

and obtain ∏ − i p = (1 ζn) (i,ps)=1 − i − j Z Note that 1 ζn = u(1 ζn) where u is a unit in [ζn], therefore (p) = ϕ(ps) (1 − ζn) . ∀ ∃ i j lij i, j, lij s.t. ζn = (ζn) , so

1 − (ζj )lij n ∈ Z[ζ ] − i n 1 ζn

3. Q(ζn) = Q(ζps )Q(ζm).

注记. The decomposion field of p in Q(ζn) is exactly Q(ζm).

例 5.1.3 (Inertial field). Let K = Q(ζ20). We invoke 5.2.1. √ (1) p = 7 we find g = 2, the inertial field of 7 is a quadratic field Q( −5), since 7 √ ( ) − ⇐⇒ −5 is split in 5 7 = 1. √ 7 : Q → Q( −5) → Q(ζ20)

(2) p = 11 Now g = 4, the inertial field of 7 is a quartic field. By 2.3.1, we see 11 is

split in Q(ζ5).

11 : Q → Q(ζ5) → Q(ζ20) 148 CHAPTER 5. NUMBER FIELDS

More on cyclotomic field

Let K = Q(ζn).

ϕ(n)−1 1. OK = Z[ζn], namely {1, ζn, ··· , ζn } is an integral base of K/Q. s a s−1 2. If n = p , then ∆K = p where a = p (ps − s − 1); especially when s = 1,

∆K = p − 2.

Kummer theory

Let K be a field of characteristic prime to n that contains ζ . √ n For any a ∈ K, the field L = K( n a) is the splitting field of f(x) = xn−a √ over K; the notation n a denotes a particular n-th root of a, but it does not matter which root we pick because all the n-th roots of a lie in L (if f(α) = f(β) = 0 then i ∈ ⩽ α/β = ζn K for some 0 i < n and K(α) = K(β)). The polynomial f(x) is separable, since n is prime to the characteristic of K, so L is a Galois extension of K, and Gal(L|K) is cyclic, since we have an injective homomorphism

Gal(L|K) ,→ ⟨ζ ⟩ n√ σ( n a) σ 7→ √ n a

This homomorphism is an isomorphism if and only if xn−a is irreducible. Kummer’s key observation is that the converse holds. In order to prove this we first recall a basic lemma from Galois theory, originally due to Dedekind.

引理 5.1.1. Let L|K be a finite extension of fields. The set AutK (L) is a linearly independent subset of the L-vector space of functions L → L.

推论. Let L|K be a cyclic field extension of degree n with Galois group ⟨σ⟩ and suppose L contains ζn. Then σ(α) = ζnα for some α ∈ L.

注记. This is a special case of Hilbert’s Theorem 90, which replaces ζn with any element u of norm NL|K (u) = 1.

♦ 命题 5.1.4 Assume that n be prime to the characteristic of K, and ζ ∈ K. If L|K is a √ n cyclic extension of degree n then L = K( n a) for some a ∈ K. 5.1. *CYCLOTOMIC FIELDS 149

Proof. By corollary 5.1, there exists an element α ∈ L for which σ(α) = ζnα. We have σ(αn) = σ(α)n = αn thus a = αn is invariant under the action of Gal(L|K) and therefore lies in K. √ { ··· n−1 } n Moreover, the orbit α, ζnα, , ζn α has order n, so L = K(α) = K( a) as desired.

Let K be a field with algebraic closure K, let n be prime to the characteristic of K, and assume ζn ∈ K. The Kummer pairing is the map

× ⟨·, ·⟩ : Gal(K|K) × K ,→ ⟨ζ ⟩ n√ σ( n a) ⟨σ, a⟩ 7→ √ n a

√ × where n a is any n-th root of a ∈ K . If α and β are two n-th roots of a, then ( α )n = 1 , so α ∈ ⟨ζ ⟩ ⊆ K is fixed by σ and σ(β) = σ(α) , so the value of ⟨σ, a⟩ does β β n √ β α not depend on the choice of n a. If a (K×)n, then ⟨σ, a⟩ = 1 for all σ ∈ Gal(K,K), so the Kummer pairing depends only on the image of a in K×/(K×)n; thus we may also view it as a pairing on Gal(K,K) × K×/(K×)n.

♦ 定理 5.1.2

Let K be a field with ζn ∈ K, n is prime to the characteristic of K. Then the Kummer pairing induces an isomorphism

× × n Φ: K × (K ) → Hom(Gal(K|K), ⟨ζn⟩) a 7→ (σ 7→ ⟨σ, a⟩) 150 CHAPTER 5. NUMBER FIELDS

5.2 Dirichlet character and Gauss sum

Dirichlet character Z× → C× A Dirichlet character of modulus n means a group character χ : n . | Z× −→ϕ Z× Z For d n, there is natural surjection n d ( nis finite). If we have a commutative diagram:

Z× ϕ Z× n d

χ χ′

C× we say χ is induced by χ′.

♦ 定义 5.2.1 1. A Dirichlet character is primitive if it is not induced by any other Dirichlet character. 2. A Dirichlet character induced by I is called principal character.

注记. 1. I is the trival Dirichlet character, which is the Dirichlet character of

modulus 1. We use Im denote the principal character of modulus m. 2. Every surjective homomorphism of finite rings induces a surjective homomorphism of unit groups, but this does not hold in general (Z → Z/5Z). ∑ 3. Since χ is a finite Abelian group character, we have χ = In ⇐⇒ χ(a) ≠ 0. a∈Zn We call periodic totally multiplicative function χ : Z → C Dirichlet function. Suppose that χ has a period n > 0, then χ(a) = 0 iff (a, n) ≠ 1. So we can think f as a Dirichlet character of modulus n.

引理 5.2.1. Let χ be a Dirichlet function of period m. Then χ is a Dirichlet character of modulus m′ iff m|m′|mk for some k.

Proof. If m|m′|mk, we have

∈ Z× ⇐⇒ ∈ Z× ⇐⇒ ̸ n m′ n m χ(n) = 0

Conversely, if χ is a Dirichlet character of modulus m′, then χ is m′-periodic, and therefore m|m′. And since χ is a Dirichlet character of modulus m and of 5.2. DIRICHLET CHARACTER AND GAUSS SUM 151 modulus m′, for each prime p we have

∈ Z× ⇐⇒ ⇐⇒ ∈ Z× p / m χ(p) = 0 p / m′ thus the prime divisors of m and m′ coincide.

引理 5.2.2. A Dirichlet character χ of modulus n is induced by χ′ of modulus d ⇐⇒ ′ ∈ Z× χ(a) = χ (a) for a n ; ⇐⇒ Z× χ is constant on residue classes in n that are congruent modulo d; ⇐⇒ ker ϕ ⊆ ker χ.

′ where ϕ is induced by Zn ,→ Zd, and χ is uniquely determined.

Proof. We check that ϕ is well-defined surjection. ∈ Z× n Suppose u1 d . Let a be the product of all primes dividing d , but not u1.

Then u2 = u1 + da is not divisible by any prime p|d, nor is it divisible by any prime | n p d : by construction, such a p divides exactly one of u1 and da. Thus u2 is a unit modulo n that reduces to u1 modulo d and ϕ is surjective.

♦ 定理 5.2.1 Every Dirichlet character χ is induced by a primitive Dirichlet character χ′ that is uniquely determined by χ.

Proof. We define a partial ordering by χ1 ≼ χ2 :⇔ χ2 is induced by χ2. It’s easy to see that the relation is reflexive and transitive. Let χ be a Dirichlet character of period m and consider the set X := {χ′ : ′ χ ≼ χ}. Suppose χ1, χ2 ∈ X have periods d1 and d2, and d := gcd(d1, d2). We have a commutative square of surjective unit group homomorphisms induced by reduction maps:

Z× Z× m d1

Z× Z× d2 d

It follows that there is a unique Dirichlet character χ′ of modulus d that induces

χ, χ1, χ2. 152 CHAPTER 5. NUMBER FIELDS

♦ 定义 5.2.2 The conductor of a Dirichlet character χ is the period of the unique prim- ′ itive Dirichlet character χ that induces χ, denoted by fχ. ∑ 推论. χ(a) ≠ 0 ⇐⇒ χ has conductor 1. a∈Zm

推论. Let D(m) denote the set of Dirichlet characters of modulus m, X(m) denote b the set of primitive Dirichlet characters of conductor dividing m, and G(m) denote Z× the character group of m. We have

∼ ∼ D(m) −→ X(m) −→ Gb(m) χ 7→ χ′ 7→ (n 7→ χ′(n))

Gauss sum Z× → C× ∈ Z× For Dirichlet character χ : n , letχ(a) = 0 for a / n , then the Gauss sum of χ and ζn is defined by

∑n i G(χ, ζn) := χ(i)ζn i=1

a 引理 5.2.3. For primitive Dirichlet character χ, χ(a)G(χ, ζn) = G(χ, ζn).

Proof. Notice that χ(a) = χ−1(a). ∑n ∑n a ai ai If (a, n) = 1, then G(χ, ζn) = χ(i)ζn = χ(a) χ(ai)ζn = χ(a)G(χ, ζn). i=1 i=1 If (a, n) ≠ 1, then ζais a primitive d-th root(d < n). Let H denote the kernel n ∑ Z× → Z× | ̸ I of n d . By definition,χ H = ,then indicate χ(i) = 0. Consider the coset i∈H Z× a decompsosion of n , we get G(χ, ζn) = 0 = χ(a)G(χ, ζn).

♦ 定理 5.2.2 For primitive Dirichlet character χ,we have √ |G(χ, ζn)| = n

Proof. By the previous lemma, ∑ | |2 a −a (i−j)a χ(a)G(χ, ζn) = G(χ, ζn)G(χ, ζn ) = χ(i)χ(j)ζn i,j 5.2. DIRICHLET CHARACTER AND GAUSS SUM 153

We sum it up for a = 1, ··· , n, and notice that the item which i ≠ j will vanish,we get ∑n Z×| |2 | |2 |Z×| n G(χ, ζn) = χ(a) n = n n. a=1

Hecke character

♦ 定义 5.2.3

A Hecke character of K is a continuous character of IK that is trivial on K×, × × χ : IK → C , χ(K ) = 1

注记. We can view χ as a character of the group CK .

Dirichlet character of quadratic field √ Let d be a square-free integer,the discriminant of Q( d) is denoted by D, more explicitly,   d d ≡ 1 (mod 4) D =  4d d ≡ 2, 3 (mod 4)

♦ 定义 5.2.4 ∏ Z× For (a, D) = 1, the Dirichlet character of D defined by χd(a) := (a, d)p √ p|D is called the character of Q( d).

注记. 1. D may have a multi-divisor 2, but the multipicity is ignored. Z× 2. χd is a primitive Dirichlet character of D. ( ) | a 3. (a, d)p is Hilbert symbol. If p is a odd prime, then p d,we have (a, d)p = p . a−1 d−1 4. For a, d is both odd, we have (a, d)2 = (−1) 2 2 ; for d is even,   1 a ≡ 1, 1 − d ( mod 8) (a, d)2 =  −1 else

5. When d ≡ 2, 3( mod 4), a is coprime to D, so a is not even, (a, d)2 is well- defined. 154 CHAPTER 5. NUMBER FIELDS

Z× → C× 6. χ−1 : 4   1 a ≡ 1 ( mod 4) χ−1(a) =  −1 a ≡ 3 ( mod 4) Z× −1 notice that 4 has two element, so χ−1(a) = χ−1(a ).

− d 引理 5.2.4. χd( 1) = χ−1( |d| ) = sgn(d).   χ−1(d)(d, 2) = 1 Proof. (1) (−1, d)2 =  d ̸ χ−1( 2 )(d, 2) = 1 −1 p−1 − − 2 (2) For odd prime p, ( 1, d)p = ( p ) = ( 1) = χ−1(p) Therefore we have  ∏ ∏  χ−1(|d|)(d, 2) = 1 (−1, d)p = χ−1(p) =  |d| ̸ p odd prime p odd prime χ−1( 2 )(d, 2) = 1 .   d d ≡ 1 ( 4) 2 mod 引理 5.2.5. G(χd, ζD) =  . 4d d ≡ 2, 3 (mod 4)

− −1 Proof. Notice that χd = χd , by the lemma 5.2.3 we have χd( 1)G(χd, ζD) = G(χd, ζD ) 2 = G(χd, ζD), and by the theorem 5.2.2 we get D = χd(−1)G(χd, ζD) .

♦ 定理 5.2.3 √ Q( d) ⊆ Q(ζ ) D We√ have a field inclusion D , and is the smallest integer s.t. Q( d) ⊆ Q(ζD).

Q Z× 例 5.2.1 (subfield of (ζ20)). The group 20 has three subgroups of order 4 are

• H1 = {1, 3, 9, 7} (cyclic)

• H2 = {1, 13, 9, 17} (cyclic) • H = {1, 9, 11, 19} (Klein four-group) 3 √ √ √ √ √ Note that 5, −5, −1 ∈ Q(ζ20) since 5 ∈ Q(ζ5), −1 ∈ Q(ζ4). Now we see that σ is exactly the complex conjugation, so the fixed field √ 19 √ of H must be Q( 5). Q( −1) is a fixed point for all of the automorphisms in 3 √ H since −1 = ζ5 . By a process of elimination , the fixed field of H must be 2√ 20 1 Q( −5). × Z has three elements of order 2, yielding three quartic subfields: 20 √ √ • K1 = {1, 9} = H1 ∩ H2 ∩ H3, so its fixed field must be Q( 5, −1). 5.2. DIRICHLET CHARACTER AND GAUSS SUM 155

{ } Q −1 • K2 = 1, 19 , its fixed field is the maximal real subfield (ζ20 + ζ20 ). { } 4 • K3 = 1, 11 , σ11 maps ζ5 = ζ20 to itself, so the fixed field of K3 is the quartic

field Q(ζ5).

−1 4− 2 注记. The minimal polynomial of ζ20+ζ20 is x 5x +5 (irreducible by Eisenstein).

σ ∈ (Q(ζ )|Q) By the previous√ theorem, we may consider Gal D restricted to Q r the subfield ( d). Let σ(ζD) = ζD, we calculate √ σ( d) σ(G(χ , ζ )) √ = d D d G(χd, ζD) G(χ , ζr ) = d D G(χd, ζD) (5.1)

= χd(r)

So we get a diagram as follow:

Q |Q ∼ Z× Gal( (ζD) ) D

restrict χd

√ ∼ Gal(Q( d)|Q) 1

√ prime decomposition in Q( d)

p Let denote a prime√ number p Q( d) ⇐⇒ p|D (1) is ramified √in . ( ) d (2) p is split in Q( d) ⇐⇒ χd(p) = 1 ⇐⇒ = 1. √ p ( ) Q ⇐⇒ − ⇐⇒ d − (2) p is inertial in ( d) χd(p) = 1 p = 1. √ 例 5.2.2. Consider Q( −6),D = 24. • ramified: 2, 3 • split: 1, 5, 7, 11 • inertial: 13, 17, 19, 23

♦ 定理 5.2.4 156 CHAPTER 5. NUMBER FIELDS

Let p, q be two diffrrent odd prime, then ( ) ( ) q p p−1 q−1 = (−1) 2 2 p q ( ) ( ) Proof. We have d = χ (p) for odd prime since d = 1 ⇐⇒ χ (p) = 1. There- ( ) p d ( ) ( ) p d d q p p−1 q−1 − 2 2 fore χd(p) = p . Especially, p = χq(p) = q ( 1) .

Local and global Kronecker-Weber theorems

Kronecker-Weber theorem can be derived from class field theoy. In this subsetction, we give a statement of this theorem, and a sketch of another proof.

♦ 定理 5.2.5. Kronecker-Weber

1. Every finite abelian extension of Q lies in a cyclotomic field Q(ζm).

2. Every finite abelian extension of Qp lies in a cyclotomic field Qp(ζm).

The local Kronecker-Weber theorem implies the global Kronecker-Weber theorem. To prove the local Kronecker-Weber theorem we first reduce to the case of cyclic extensions of prime-power degree. It follows from the structure theorem for finite abelian groups that we may decompose any finite abelian extension L|K into a compositum L = L1 ··· Lr of linearly disjoint cyclic extensions Li|K of prime-power degree. If each Li lies in a Q ⊆ Q ··· Q Q cyclotomic field (ζmi ), then so does L. Indeed, L (ζm1 ) (ζmr ) = (ζm), where m := m1 ··· mr. To prove the local Kronecker-Weber theorem it thus suffices to consider cyclic r extensions K|Qp of prime power degree l . There two distinct cases:l ≠ p and l = p. 5.3. THE MINKOWSKI BOUND 157

5.3 The Minkowski bound

Lattices in real vector spaces

Recall the definition of A-lattice in a K-vector space V . We now want to spe- cialize to the case A = Z, but rather than working with the fraction field K = Q we will instead work with its archimedean completion R. Now V is an R-vector space of some finite dimension n, and has a canonical structure as a topological metric space isomorphic to Rn. This topology makes V a LCH. ♦ 定义 5.3.1 Let V be an R-vector space of finite dimension. A (full) lattice in V is a Z- submodule generated by an R-basis for V ; equivalently, a discrete cocompact subgroup.

注记. 1. A discrete subgroup of a Hausdorff topological group is always closed. This implies that the quotient of a Hausdorff topological group by a normal discrete subgroup is Hausdorff (which is false for topological spaces in general). It follows that the quotient of a Hausdorff topological group by a discrete cocompact subgroup is a compact group. 2. The reason why we are using the archimedean completion R rather than

some other completion Qp is that Z is not a discrete subset of Qp (elements of Z can be arbitrarily close to 0 under the p-adic metric).

Any basis v1, ··· , vn for V determines a parallelepiped

F (v1, ··· , vn) := {t1v1 + ··· + tnvn | t1, ··· , tn ∈ [0, 1)}

∼ that we may view as the unit cube by fixing an isomorphism ϕ : V −→ Rn that n maps (v1, ··· , vn) to the standard basis of unit vectors for R . It then makes sense to normalize the Haar measure µ so that µ(F (v1, ··· , vn)) = 1, and we then have

µ(S) = µRn (ϕ(S)) for every measurable set S ⊆ V .

For any other basis e1, ··· , en of V , if we let E = [eij] be the transformation matrix between e and v, then √ t µ(F (v1, ··· , vn)) = | det E| = det(E E) 158 CHAPTER 5. NUMBER FIELDS

♦ 命题 5.3.1 Let T : V → V be a linear transformation. For any Haar measure µ on V and every measurable set S ⊆ V we have

µ(T (S)) = | det T |µ(S)

(Convention):Let Λ be a lattice in V ≃ Rn and fix a Haar measure µ on V .

If Λ = e1Z + ··· + enZ in V , the quotient V /Λ is a compact group that we may identify with the parallelepiped F (e1, ··· , en) ⊆ V , which forms a set of unique coset representatives. More generally, we make the following definition

♦ 定义 5.3.2 1. A fundamental domain for Λ is a measurable set F ⊆ V such that ⊔ V = (F + λ) λ∈Λ

2. The covolume covol(Λ) ∈ R>0 of Λ is the measure µ(F ) of any fundamental domain F for Λ.

引理 5.3.1. Every fundamental domain for Λ has the same measure, and this measure is finite and nonzero.

Proof. Let F and G be two fundamental domains for Λ. Using the translation invariance and countable additivity of µ (note that Λ ≃ Zn is a countable set) along with the fact that Λ is closed under negation, we obtain ( ) ⊔ µ(F ) = µ(F ∩ V ) = µ F ∩ (G + λ) ( λ)∈Λ ⊔ = µ (F ∩ (G + λ)) ∑ λ∈Λ ∑ = µ(F ∩ (G + λ)) = µ((F − λ) ∩ G) ∑λ∈Λ λ∈Λ = µ(G ∩ (F + λ)) = µ(G) λ∈Λ

If we fix a Z-basis e1, ··· , en for Λ, the parallelepiped F (e1, ··· , en) is a fundamental domain for Λ, and its closure is compact, so µ(F (e1, ··· , en)) is finite, and it is nonzero because there is an isomorphism V ≃ Rn that maps the closure of n F (e1, ··· , en) to the unit cube in R whose Lebesgue measure is nonzero. 5.3. THE MINKOWSKI BOUND 159

注记. Whether a set has zero measure or not does not depend on the normalization of the Haar measure and is therefore preserved by isomorphisms of locally compact groups.

♦ 命题 5.3.2 If Λ′ ⊆ Λ are lattices in V , then covol(Λ′) = [Λ : Λ′]covol(Λ).

Proof. Fix a fundamental domain F for Λ and a set of coset representatives S for Λ/Λ′ . Then ⊔ F ′ : (F + λ) λ∈S is a fundamental domain for Λ′, and #S = [Λ : Λ′] = µ(F ′)/µ(F ) is finite. We then have ′ ′ ′ covol(Λ ) = µ(F ) = (#S)µ(F ) = [Λ : Λ ]covol(Λ)

The set S ⊆ V is symmetric if it is closed under negation.

♦ 定理 5.3.1. Minkowski’s Lattice Point Theorem If S ⊆ V is a symmetric convex measurable set that satisfies

µ(S) > 2ncovol(Λ)

then S contains a nonzero element of Λ.

注记. Note that the inequality bounds the ratio of the measures of two sets (S and a fundamental domain), and is thus independent of the choice of µ.

The canonical inner product

(Convention):Let K|Q be a number field of degree n with r real places and s complex places.

Consider the base change of K to R and C:

r s KR := K ⊗Q R ≃ R × C

n KC := K ⊗Q C ≃ C 160 CHAPTER 5. NUMBER FIELDS

s s 注记. 1. The isomorphism KR ≃ R × C follows from Theorem 4.2.2 and the n isomorphism KC ≃ C follows from the fact that C is separably closed. n 2. Note that KR is an R-vector space of dimension n, thus KR ≃ R , but this is an isomorphism of R-vector spaces and is not an R-algebra isomorphism unless s = 0.

We have a sequence of injective homomorphisms of topological rings

OK ,→ K,→ KR ,→ KC (∗)

r s r 2s r the map KR ≃ R × C ,→ C × C ≃ KC embeds each factor of R in a corresponding factor of Cr via inclusion and each C in Cs is mapped to C × C in C2s via z 7→ (z, z). To better understand the last map, note that each C in Cs arises as R[α] = R[x]/(f) ≃ C for some monic irreducible f ∈ R[x] of degree 2, but when we base- change to C the field R[α] splits into the etale algebra C[x]/(x−α) × C[x]/(x−α) ≃ C × C. The composition K,→ KR ,→ KC is given by the map

x 7→ (σ1(x), ··· , σn(x)) where HomQ(K, C) = {σ1, ··· , σn}. If we put K = Q(α) := K[x]/(f) and let

α1, ··· , αn ∈ C be the roots of f in C, each σi is the Q-algebra homomorphism

K → C defined by α 7→ αi.

If we fix a Z-basis for OK , its image under the maps in (∗) is a Q-basis for K, a R-basis for KR, and a C-basis for KC, all of which are vector spaces of dimension n = [K : Q]. We may thus view the injections in (∗) as inclusions of topological groups (but not rings!) Zn ,→ Qn ,→ Rn ,→ Cn

n The ring of integers OK is a lattice in the real vector space KR ≃ R , which n inherits an inner product from the canonical Hermitian inner product on KC ≃ C defined by ∑n ⟨ ′⟩ ′ ∈ C z, z := zizi i=1

For elements x, y ∈ K,→ KR ,→ KC the Hermitian inner product can be computed as ∑ ⟨x, y⟩ := σ(x)σ(y) ∈ R

σ∈HomQ(K,C) 5.3. THE MINKOWSKI BOUND 161 which is a real number because the non-real embeddings in HomQ(K, C) come in complex conjugate pairs. The inner product defined above agrees with the restriction of the Hermitian inner product on KR ,→ KC. The topology it induces r s on KR is the same as the Euclidean topology on the R-vector space R × C , but the corresponding norm ∥x∥ := ⟨x, x⟩ has a different normalization, as we now explain. n If we write elements z ∈ KC ≃ C as vectors (zσ) indexed by the set σ ∈ HomQ(K, C) in some fixed order, we may identify KR with its image in KC as the set

KR = {z ∈ KC : zσ = zσ, ∀σ ∈ HomQ(K, C)}

For real embeddings σ = σ we have zσ ∈ R ⊆ C, and for pairs of conjugate complex embeddings (σ, σ) we get the embedding z 7→ (zσ, zσ) = (zσ, zσ) of C into C × C used to defined the map KR ,→ KC above. Each z ∈ KR can be uniquely written in the form

(w1, ··· , wr, x1 + iy1, x1−iy1, ··· , xs + iys, xs−iys) with wi, xj, yj ∈ R. Each wi corresponds to a zσ with σ = σ, and each (xj + iyj, xj−iyj) corresponds to a complex conjugate pair (zσ, zσ) with σ ≠ σ. The canonical inner product on KR can then be written as

∑r ∑ ⟨ ′⟩ ′ ′ ′ z, z := wiwi + 2 (xjxj + yjyj) i=1 j=1

n Thus if we take w1, ··· , wr, x1, y1, ··· , xs, ys as coordinates for KR ≃ R (as R-vector spaces), in order to normalize the Haar measure µ on KR so that it is consistent with the Lebesgue measure Rn we define

s µ(S) := 2 µRn (S)

n for any measurable set S ⊆ KR that we may view as a subset of R by expressing it in wi, xj, yj coordinates as above.

Covolumes of fractional ideals

Fix a normalized Haar measure for KR as above.

Let I be a fractional ideal of OK , and suppose aOK ⊆ I with a ≠ 0. If e1, ··· , en is a Z-basis for OK then ae1, ··· , aen is a Z-basis for aOK that is an R- basis for KR that lies in I. 162 CHAPTER 5. NUMBER FIELDS

♦ 命题 5.3.3 √ (O ) = |∆ | 1. covol K √K 2. covol(I) = N(I) |∆K |

Proof. Linear algebra.

The Minkowski bound

引理 5.3.2. Let K be a number field of degree n with s complex places. For each t ∈ R>0, the measure of the convex symmetric set { ∑ } St := (zσ) ∈ KR : |zσ| ⩽ t ⊆ KR

n r s t with respect to the normalized Haar measure µ on KR is µ(S ) = 2 π . t n! Proof. (Sketch): we may uniquely write each z = (z ) in the form (w , ··· , w , x + ∑ σ 1 r 1 iy1, x1−iy1, ··· , xs + iys, xs−iys). Then |zσ| ⩽ t iff ∑r ∑s √ 2 2 |wi| + 2 |xj| + |yj| ⩽ t i=1 j=1 We now need to compute the volume of this region in Rn...

♦ 定理 5.3.2. Minkowski bound Let K be a number field of degree n with s complex places. Define the

Minkowski constant mK for K as the positive real number ( ) n! 4 s √ m := |∆ | K nn π K

For every nonzero fractional ideal I of OK there is a nonzero a ∈ I for which

N(a) ⩽ mK N(I)

n Proof. If we choose t s.t. µ(St) > 2 covol(I), then St will contain a nonzero a ∈ I. So it suffices( ) to choose t so that ( ) n n s √ t n!µ(St) n!2 n! 4 = > covol(I) = |∆ |N(I) = m N(I) n nn2rπs nn2rπs nn π K K ( ) n Let us now pick t so that t > m N(I). Then S contains a ∈ I with ∑ n K t |σ(a)| ⩽ t. Recalling that the geometric mean is bounded above by the arithmetic mean, we then have ( ) ( ) n n ( ) ∏ 1 ∑ t n N(a) = |σ(a)|1/n ⩽ |σ(a)| ⩽ n n σ σ 5.3. THE MINKOWSKI BOUND 163 ( ) t n → ⩽ Taking the limit as n mK N(I) from above yields N(a) mK N(I).

Finiteness of the ideal class group

♦ 命题 5.3.4

Let K be a number field. Every ideal class contains an ideal I ⊆ OK of

absolute norm N(I) ⩽ mK .

Proof. Let [J] be an ideal class of OK represented by the nonzero fractional ideal J. Then J −1 contains a nonzero element a for which

−1 −1 N(a) ⩽ mK N(J ) = mK N(J)

−1 −1 and therefore N(aJ) = N(a)N(J) ⩽ mK . We have a ∈ J , thus aJ ⊆ J J = OK , so I = aJ is an OK -ideal in the ideal class [J] with N(I) ⩽ mK as desired.

引理 5.3.3. Let C be a real number. The set of ideals I ⊆ OK with N(I) ⩽ C is finite.

n n Proof. As a lattice in KR ≃ R , the additive group OK ≃ Z has only finitely many subgroups Iof index m for each positive integer m ⩽ C, since [Zn : I] = m implies

(mZ)n ⊆ I ⊆ Zn and (mZ)n n has finite index mn = [Zn : mZn] = [Z : mZ]n in Zn.

As a corollary of Proposition 5.3.4 and Lemma 5.3.3, we get

♦ 定理 5.3.3

Let K be a number field. The ideal class group of OK is finite.

The discriminant of K

If I is an ideal and a ∈ I is nonzero, then N(a) ⩾ N(I), then Theorem 5.3.2 implies m ⩾ 1, namely K ( ) ( ) nn 2 π 2s |∆ | ⩾ K n! 4 By Stirling’s approximation ( ) √ n n n! ⩽ e n e we finally get 164 CHAPTER 5. NUMBER FIELDS

( ) ( ) ( ) n nn 2 π 2s 1 πe2 |∆K | ⩾ > n! 4 e2n 4

πe2 ≈ | | Note that 4 5.8 > 1, so the minimum value of ∆K increases exponen- tially with n = [K : Q].

♦ 命题 5.3.5

If K is a number field other than Q then |∆K | > 1; equivalently, there are no nontrivial unramified extensions of Q.

♦ 定理 5.3.4

For every real number C the set of number fields K with |∆K | < C is finite.

Proof. (Sketch): It suffices to prove this for fixed n := [K : Q], since for all sufficiently large n we will have |∆K | > C. First we may assume that K is totally real, namely r = n, s = 0. Consider the convex symmetric set √ n S := {(x1, ··· , xn) ∈ KR ≃ R : |x1| ⩽ C, |xi| < 1 for i > 1} √ n n with measure µ(S) = 2 C > 2 covol(OK ).

By Minkowski’s Lattice Point Theorem, S contains a nonzero a ∈ OK with N(a) ⩾ 1. We claim that K = Q(a)... f ∈ Z[x] a f √ Let be the minimal polynomial . The roots of are bounded by C. Note that each coefficient fi of f is an elementary symmetric functions of its roots, hence also bounded. The fi are integers, so there are only finitely many possibilities for f, hence only finitely many totally real number fields K of degree n. r s If s > 0, then KR ≃ R × C . Let √ 2 S := {(w1, ··· , wr, z1, ··· , zs) ∈ KR : |z1| ⩽ c C, |wi|, |zj| < 1 for j > 1}

n with c chosen so that µ(S) > 2 covol(OK ).

引理 5.3.4. For each prime number p we have

⩽ − vp(∆K ) n logp n + n 1 5.3. THE MINKOWSKI BOUND 165

Proof. We have ∏ | | | D | |D | ∆K p = NK|Q( K|Q) p = Kv|Qp v v|p

e 注记. The bound is tight; it is achieved by K = Q[x]/(xp −p), for example.

♦ 定理 5.3.5. Hermite Let S be a finite set of places of Q. The number of extensions K|Q of degree n unramified outside of S is finite.

Proof. Since n is fixed, the valuation vp(∆K ) is bounded for each p ∈ S and must be zero for p ∈/ S. Thus |∆K | is bounded, and the theorem then follows from Theorem 5.3.4.

(Remark on history): These theorems are of fundamental importance for number theory. Their significance is seen especially clearly in the light of higher dimensional analogues. For instance, let us replace the finite field extensions L|K of a number field K by all smooth complete (i.e., proper) algebraic curves defined over K of a fixed genus g. If p is a prime ideal of K, then for any such curve X, one may define the ”reduction mod p”. This is a curve defined over the residue class field of p. One says that X has good reduction at the prime p if its reduction mod p is again a smooth curve. This corresponds to an extension L|K being unramified. In analogy to Theorem 5.3.5, the Russian mathematician Shafarevich formulated the conjecture that there xist only finitely many smooth complete curves of genus g over K with good reduction outside a fixed finite set of primes S. This conjecture was proved in 1983 by the mathematician Faltings. The impact of this result can be gauged from the fact that it was the basis for Faltings’s proof of the famous Mordell Conjecture: Every algebraic equation f(x, y) = 0 of genus g > 1 with coefficients in K admits only finitely many solutions in K. A 1-dimensional analogue of Minkowski’s theorem 5.3.5 was proved in 1985 by the French mathematician Fontaine: over the field Q, there are no smooth proper curves with good reduction mod p for all prime numbers p. 166 CHAPTER 5. NUMBER FIELDS

5.4 Dirichlet Unit Theorem

The group of multiplicative divisors of a global field

Let K be a global field.

♦ 定义 5.4.1

An MK -divisor (or Arakelov divisor) is a sequence of positive real num-

bers c = (cv) indexed by v ∈ MK with all but finitely many cv = 1 and cv ∈ ∥ ×∥ {∥ ∥ ∈ ×} Kv := x v : x Kv . The set Div(K) of all MK -divisors is an abelian

group under pointwise multiplication (cv)(dv) := (cvdv).

注记. 1. ∥ · ∥v is the normalized absolute value defined in 4.2.5. × 2. K is canonically embedded in Div(K) via the map x 7→ (∥x∥v). 3. Many authors define Div(K) as an additive group by taking logarithms (for −v(c) nonarchimedean places v, one replaces cv = (#kv) with the integer v(c)); the multiplicative convention we use here is due to Weil and better suited to O× our application to the multiplicative group K .

× The MK -divisors that lie in the image of the embedding K → Div(K) are said to be principal, and they form a subgroup. The quotient of Div(K) by its subgroup of principal MK -divisors is denoted Pic(K).

♦ 定义 5.4.2

The size of an MK -divisor c is the real number ∏ ∥c∥ := cv ∈ R>0

v∈MK

→ R× 7→ ∥ ∥ The map Div(K) >0, c c is a group homomorphism that contains the subgroup of principal MK -divisors in its kernel by the product formula. Cor- responding to each MK -divisor c is a subset L(c) of K defined by

L(c) := {x ∈ K : ∥x∥v ⩽ cv for all v ∈ MK } and a nonzero fractional ideal of OK defined by ∏ v(c) Ic := qv v∤∞ v(c) := − (c ) ∈ Z v(x) = v(c) ∥x∥ = c L(c) ⊆ where log#kv v (so iff v v). We have

Ic ⊆ K, and the map c 7→ Ic defines a group homomorphism Div(K) → FK . The 5.4. DIRICHLET UNIT THEOREM 167 homomorphism sends principal MK -divisors to principal fractional ideals, and it follows that the ideal class group ClK is a quotient of Pic(K), and we have a commutative diagram

Div(K) FK

Pic(K) ClK

If we now restrict our attention to MK -divisors of size 1, these form a subgroup of Div(K) denoted Div0(K) that contains the subgroup of principal divisors and surjects onto FK via the map Div(K) → FK (we are free to choose any

Ic ∈ FK because we can always choose the cv at infinite places to ensure ∥c∥ = 1). 0 The quotient of Div (K) by the subgroup of principal MK -divisors is the Arakelov class group Pic0(K), and the ideal class group is also a quotient of the Arakelov class group.

注记. The set L(c) associated to an MK -divisor c is directly analogous to the Riemann- Roch space

L(D) := {f ∈ k(X): vP (f) ⩾ −nP for all closed points P ∈ X} associated to a divisor D ∈ Div(X) of a smooth projective curve X/k, which is a k- ∑ vector space of finite dimension. Recall that a divisor is a formal sum D = nP P over the closed points (Gal(k|k)-orbits) of the curve X with nP ∈ Z and all but finitely many nP zero. If k is a finite field then K = k(X) is a global field and there is a one-to-one correspondence between closed points of X and places of K, and a normalized absolute value ∥ · ∥P for each closed point P (indeed, one can take this as a defi-

nP nition). The constraint vP (f) ⩾ −nP is equivalent to ∥f∥P ⩽ (#kP ) , where kP is

nP the residue field corresponding to P . If we put cP := (#kP ) then c = (cP ) is an

MK -divisor with L(c) = L(D). The Riemann-Roch space L(D) is finite (since k is finite), and we will prove below that L(c) is also finite (when K is a number field the finite set L(c) is not a vector space). In subsection 2.4 we described the divisor group Div(X) as the additive analog of the ideal group F(A) of the ring of integers A = OK (equivalently, the coordinate ring A = k[X]) of the global function field K = k(X). This is correct when X is an affine curve, but here X is a smooth projective curve and has ”points 168 CHAPTER 5. NUMBER FIELDS at infinity” that correspond to infinite places. Taking the projective closure ofan affine curve corresponds to including all the factors in the product formulaand is precisely what is needed to ensure that principal divisors have degree 0 (every function f ∈ k(X) has the same number of zeros and poles, when counted correctly).

The case of number field

Now let K be a number field. Note that ∏ ∏ ∏ v(c) v(c) −1 N(Ic) = N(qv) = (#kv) = cv v∤∞ v∤∞ v∤∞ and therefore ∏ −1 ∥c∥ = N(Ic) cv v|∞ We also define

Rc := {x ∈ KR : |x|v ⩽ cv for all v | ∞} which we note is a compact, convex, symmetric subset of the real vector space KR.

If we view Ic and L(c) as subsets of KR via the canonical embedding K,→ KR, then

L(c) = Ic ∩ Rc

引理 5.4.1. Let c be an MK -divisor of a global field K. The set L(c) is finite.

Proof. (number field): The fractional ideal Ic is a lattice in KR, and is thus a closed discrete subset of KR. In KR we may view L(c) = Ic ∩ Rc as the intersection of a discrete closed set with a compact set, which is a compact discrete set and therefore finite.

× 推论. Let K be a global field, and let wK denote the torsion subgroup of K (equivalently, the roots of unity in K). The group wK is finite and equal to the kernel of × → O× the map K Div(K); it is also the torsion subgroup of K .

n n Proof. Each ζ ∈ wK satisfies ζ = 1. For every place v ∈ MK we have ∥ζ ∥v = ∥ ∥n ∥ ∥ ⊆ × → ζ v = 1, and therefore ζ v = 1. It follows that wK ker(K Div(K)). Let c × be the MK -divisor with cv = 1 for all v ∈ MK . Then ker(K → Div(K)) ⊆ L(c) × is a finite subgroup of K and is therefore contained in its torsion subgroup wK .

Every element of wK is an algebraic integer, so wK is also the torsion subgroup of O× K . 5.4. DIRICHLET UNIT THEOREM 169

It follows that for any global field K we have the following exact sequence of abelian groups

× 1 → wK −→ K −→ Div(K) −→ Pic(K) → 1

♦ 命题 5.4.1 Let K be a number field with s complex places, define ( ) 2 s √ B := |∆ | K π K

and let c be any MK -divisor for which ∥c∥ > BK . Then L(c) contains an element of K×.

Proof. We apply Minkowski’s lattice point theorem to the convex symmetric set

Rc and the lattice Ic ⊆ K ⊆ KR.

注记. The bound in the Proposition can be turned into an asymptotic, that is, for c ∈ Div(K), as ∥c∥ → ∞ we have ( ) 2r(2π)s #L(c) = √ + o(1) ∥c∥ (∗) |∆K | This can be viewed as a multiplicative analog of the Riemann-Roch theorem for ∑ ∑ function fields, which states that for divisors D = nP P , as deg D := nP → ∞ we have dim L(D) = 1 − g + deg D (∗∗) The nonnegative integer g is the genus, an important invariant of a function field that is often defined by the (∗∗); one could similarly use (∗) to define the nonnegative integer |∆K |. For all sufficiently large ∥c∥ the o(1) error term will be small enough so that (∗) uniquely determines |∆K |. Conversely, with a bit more work one can adapt the proofs of Lemma 5.4.1 and Proposition above to give a proof of the Riemann-Roch theorem for global function field.

The unit group of a number field

As a ring, the finite ´etale R-algebra KR = K ⊗Q R also has a unit group, and we have an isomorphism of topological group ∏ × × ≃ R× r × C× s KR = Kv ( ) ( ) v|∞ 170 CHAPTER 5. NUMBER FIELDS

× Writing elements of KR as vectors x = (xv) indexed by the infinite places v of K, we now define a surjective homomorphism of locally compact groups

× r+s Log : KR → R

(xv) 7→ (log ∥xv∥v)

It is surjective and continuous because each of the maps xv 7→ log ∥xv∥v is, and it is a group homomorphism. Recall that there is a one-to-one correspondence between the infinite places of K and the Gal(C|R)-orbits of HomQ(K, C). For each v|∞ let us now pick a representative σv of its corresponding Gal(C|R)-orbit; for real places v there is a unique choice for σv, while for complex places there are two choices, σv and its complex conjugate. Regardless of our choices, we then have   |σ (x)|R if v|∞ is real ∥ ∥ v x v =  |σv(x)σv(x)|R if v|∞ is complex × → Q× The absolute norm N : K >0 extends naturally to a continuous homomorphism of locally compact groups

× × N : K → R R ∏ (xv) 7→ ∥xv∥v v|∞

× × which is compatible with the canonical embedding K → KR . Indeed, we have

∏ ∏ | | ∥ ∥ N(x) = NK|Q(x) = σ(x) = x v σ R v|∞

We thus have a commutative diagram

Log × × r+s K KR R

N N T

Q× R× log >0 >0 R

∑ Rr+s → R where T : is defined by T (x) = i xi. We may view Log as a map from × r+s × × K to R via the embedding K ,→ KR , and similarly view N as a map from 5.4. DIRICHLET UNIT THEOREM 171

× R× K to >0. We can succinctly summarize the commutativity of the above diagram by the identity T(Log x) = log N(x)

The norm of a unit in OK must be a unit in Z, hence have absolute value 1. O× 7→ Thus K lies in the kernel of the map x log N(x) and therefore also in the 7→ O× kernel of the map x T(Log x). It follows that Log( K ) is a subgroup of the trace zero hyperplane

Rs+t { ∈ Rs+t } 0 := x : T(x) = 0 which we note is both a subgroup of Rs+t, and an R-vector subspace of dimension r + s−1. The proof of Dirichlet’s unit theorem amounts to showing that ΛK := O× Rs+t Log( K ) is a lattice in 0 . ♦ 命题 5.4.2 1. We have a split exact sequence of finitely generated abelian groups

→ −→ O× −−→Log → 1 wK K ΛK 0

Rs+t 2. ΛK is a lattice in the trace zero hyperplane 0 .

Dirichlet’s unit theorem follows immediately

♦ 定理 5.4.1. Dirichlet’s unit theorem Let K be a number field with r real and s complex places. Then

O× ≃ × Zr+s−1 K wK

is a finitely generated abelian group.

O× Proof. The image of the torsion-free part of the unit group K under the Log map Rr+s − is the lattice ΛK in the trace-zero hyperplane 0 , which has dimension r+s 1.

The regulator of a number field

Choose any coordinate projection π : Rr+s → Rr+s−1, and use the induced Rr+s −→∼ Rr+s−1 Rr+s isomorphism 0 to endow 0 with a Euclidean measure. Since ΛK Rr+s is a lattice in 0 , we can measure its covolume using the Euclidean measure on Rr+s 0 . 172 CHAPTER 5. NUMBER FIELDS

♦ 定义 5.4.3 The regulator of a number field K is

RK := covol (π(ΛK )) ∈ R>0

where π : Rr+s → Rr+s−1 is any coordinate projection.

注记. RK does not depend on the choice of π, since we use π to normalize the Rr+s ≃ Rr+s−1 Haar measure on 0 .

If ϵ1, ··· , ϵr+s−1 is a fundamental system of units (a Z-basis for the free part O× of K ), then RK can be computed as the absolute value of the determinant of any (r + s−1)×(r + s−1) minor of the (r + s)×(r + s−1) matrix whose columns are the r+s vectors Log(ϵi) ∈ R . √ √ Q x+y d 例 5.4.1. Let K = ( d) with fundamental unit ϵ = 2 , then r + s = 2 and the product of the two real embeddings σ1(ϵ), σ2(ϵ) ∈ R is N(ϵ) = 1. Thus log |σ2(ϵ)| = − log |σ1(ϵ)| and

Log(ϵ) = (log |σ1(ϵ)|, log |σ2(ϵ)|) = (log |σ1(ϵ)|, − log |σ1(ϵ)|)

The 1 × 1 minors of the 2 × 1 transpose of Log(ϵ) have determinant log |σ1(ϵ)|; the absolute value of the determinant is the same in both cases, and since we have require the fundamental unit to satisfy ϵ > 1 (which forces a choice of embedding), the regulator of K is simply RK = log ϵ. 5.5. MODULI OF A NUMBER FIELD 173

5.5 Moduli of a number field

Moduli and ray class groups

Let K be a number field. A modulus (or cycle) m for K is a function MK → N with finite support such that for v|∞ we have m(v) ⩽ 1 with m(v) = 0 unless v is a ∏ m(v) real place. We view m as a formal product v over MK , which we may factor as ∏ ∏ m(p) m(v) m = m0m∞, where m0 := p , m∞ := v p∤∞ v|∞

As we see, m0 is an OK -ideal, and m∞ represents a subset of the real places of K. If m and n are two moduli for K we say that m divides n if m(v) ⩽ n(v) for all v ∈ MK and define gcd(m, n) and lcm(m, n) in the obvious way. ∈ F | • a (K) is prime to m if vp(a) = 0 for all p m0. F m ⊆ F • K K is the subgroup of fractional ideals coprime to m. m ⊆ × ∈ × ∈ F m • K K is the subgroup of elements α K for which (α) K . m,1 m m • K ⊆ K is the subgroup of elements α ∈ K for which vp(α−1) ⩾ vp(m0) | | → ≃ for all p m0 and αv > 0 for v m∞ (here αv is the image of α under K, Kv R). Pm ⊆ F m ∈ F m ∈ • K K is the subgroup of principal fractional ideals (α) K with α Km,1. Pm The groups K are called rays or ray groups. ♦ 定义 5.5.1 The ray class group for the modulus m is the quotient

m F m Pm ClK := K / K

A finite abelian extension L|K that is unramified at all places not in the support of m for which the kernel of the Artin map is equal to the ray group is a ray class field for the modulus m.

注记. 1. When m is the trivial modulus, the ray class group is the same as the usual class group 2. The definitions above make sense for any global field, but in ideal-theoretic treatment of class field theory we will mostly restrict our attention to number fields. 174 CHAPTER 5. NUMBER FIELDS

m 3. If m(v) = 1 for every real place v of K then ClK is a narrow ray class group.

The narrow ray class group with m0 = (1) is the narrow class group; for number fields with no real places (imaginary quadratic fields, in particular) there is no distinction to the usual class group.

Q m { a | ̸≡ 例 5.5.1. For K = with the modulus m = (5) we have K = b a, b 0 } m,1 { a | ≡ ̸≡ } mod 5 and K = b a b 0 mod 5 . Thus { ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) } 1 1 2 3 1 3 4 1 F m = (1), , (2), , , , (3), , , , (4), , (6), ··· K 2 3 3 2 4 4 3 6 { ( ) ( ) ( ) ( ) ( ) ( ) } 2 3 1 1 2 7 Pm = (1), , , , (4), , (6), , , ··· K 3 2 4 6 7 2 2 ∈ Pm − 2 ∈ m,1 Note that ( 3 ) K because of 3 K . The ray class group is

m { } ≃ Z Z × ClK = [(1)], [(2)] ( /5 ) / 1

+ which is isomorphic to the Galois group of the totally real subfield Q(ζ5) of Q(ζ5), which is the ray class field for this modulus. If we change the modulus to m = (5)∞ Pm { ···} m ≃ Z Z × we instead get K = (1), (6), (1/6), (2/7), (7/2), , ClK ( /5 ) , and the ray class field is Q(ζ5).

引理 5.5.1. Let A be a Dedekind domain and let a be an A-ideal. Every ideal class in Cl(A) contains an A-ideal coprime to a.

♦ 定理 5.5.1 Let m be a modulus for a number field K. We have an exact sequence

→ O× ∩ m,1 → O× → m m,1 → m → → 1 K K K K /K ClK ClK 1

and a canonical isomorphism

m m,1 #m × K /K ≃ {1}∞ × (OK /m0)

Proof. Let us consider the composition of the maps Km ⊆ Km,1 and α 7→ (α):

m,1 −→f m −→Fg m K K K

◦ O× ∩ m,1 O× ◦ m We see ker(g f) = K K , ker(g) = K , coker(g f) = ClK and coker(g) = ClK . Applying the snake lemma to the following commutative diagram with exact rows 5.5. MODULI OF A NUMBER FIELD 175

f 1 Km,1 Km Km/Km,1 1

g ◦ f g π

∼ F m F m 1 K K 1 yields the exact sequence

→ O× ∩ m,1 → O× → m m,1 → m → → 1 K K K K /K ClK ClK 1

m We can write each α ∈ K as α = a/b with a, b ∈ OK s.t. (a) and (b) are coprime to m0 and to each other. The ideals (a) and (b) are uniquely determined by α. We now define the homomorphism   ∏ m   × φ : K → {1} × (OK /m0) | v m∞  ∏   α 7→ sgn(αv) × (α) v|m∞ ∏ × m(p) × The ring (OK /m0) is isomorphic to (OK /p ) , by the Chinese remainder | p m0 theorem, and weak approximation implies that φ is surjective. Since ker(φ) = m,1 m m,1 #m × K , φ induces an isomorphism K /K ≃ {1}∞ × (OK /m0) . This isomorphism is canonical, because α depends only on the uniquely determined ideals (a) and (b).

m 推论. Let m be a modulus for a number field K. The ray class group ClK is a finite m m abelian group whose cardinality hK := #ClK is given by ϕ(m)h hm = K K O× O× ∩ m,1 [ K : K K ] m m,1 where ϕ(m) := #(K /K ) = ϕ(m∞)ϕ(m0) with ∏ #m∞ × −1 ϕ(m∞) = 2 , ϕ(m0) = #(OK /m0) = N(m0) (1 − N(p) ) | p m0

m m In particular, hK divides hK and hK divides hK ϕ(m).

O× O× ∩ m,1 m Proof. The exact sequence implies ϕ(m)/[ K : K K ] = hK /hK

m 注记. Computing the ray class number hK is not a trivial problem. See Henri Cohen, Advanced topics in computational number theory. 176 CHAPTER 5. NUMBER FIELDS

Artin map

Let L|K be a finite Galois extension of global fields, andlet p be a prime of K. Recall that the Galois group Gal(L|K) acts on the set {q|p} and the stabilizer of q|p is the decomposition group Dq ⊆ Gal(L|K). We have an exact sequence

πq 1 → Iq → Dq −→ Gal(Fq|Fp) → 1

If q is unramified then Iq is trivial and πq is an isomorphism. The Artin symbol is defined by ( ) ( ) L|K − F :=:= σ = π 1 x 7→ x# p q q q ( ) #Fp where x 7→ x ∈ Gal(Fq|Fp) is the Frobenius automorphism, a canonical generator for the cyclic group Gal(Fq|Fp). Equivalently, σq is the unique element of Gal(L|K) for which

#Fp σq(x) ≡ x mod q for all x ∈ OL. For q|p the Frobenius elements σq are all conjugate (they form the

Frobenius class Frobp), and when L|K is abelian they coincide, in which case we may write σ instead of σ , and we may write the Artin symbol as p q ( ) L|K := σ p p

Now assume L|K is abelian, let m be an OK -ideal divisible by every ramified F m ∈ F prime of K, and let K denote the subgroup of fractional ideals I (K) for which vp(I) = 0 for all p|m. The Artin map is the homomorphism

m m ψ | : F → Gal(L|K) L K K ( ) ∏ ∏ L|K np pnp 7→ p p∤m p∤m

♦ 命题 5.5.1 Let K ⊆ L ⊆ M be a tower of finite abelian extension of global fields and

let m be an OK -ideal divisible by all primes p of K that ramify in M. We have a commutative diagram

(M|K) m Gal ψM|K

F m K res

m ψL|K Gal(L|K) 5.5. MODULI OF A NUMBER FIELD 177

Proof. It suffices to check commutativity at primes p ∤ m, which are necessarily unramified.

Congruence subgroups

Let K be a number field and let m be a modulus for K. ♦ 定义 5.5.2 C F m A congruence subgroup for the modulus m is a subgroup of K that Pm C C m contains K . We use to denote the image of in ClK under the quotient map.

例 5.5.2. Let K = Q, and consider the cyclic cubic field L := Q[x]/(x3−3x−1), m which is ramified only at 3. The Artin map ψL|K is well-defined for any modulus m divisible by (3). The ray class field for m = (3) is Q(ζ )+ = Q, and the ray class √ 3 ∞ Q Q − m field for m = (3) is (ζ3) = ( 3), neither of which contains L, so ker ψL|K Pm does not contain K for either of these moduli and is not a congruence subgroup. Q + m On the other hand, L is equal to (ζ9) , the ray class field for m = (9), so ker ψL|K Pm contains (and is equal to) K , and is thus a congruence subgroup for this modulus.

m If ker ψL|K is a congruence subgroup for the modulus m, then it is also a con- | ⇒ Pn ⊆ Pm gruence subgroup for every modulus n divisible by m, since m n K K and n m F n Pn ψL|K is just the restriction of ψL|K to K , which contains K . If m and n are sup- F m F n m n ported on the same primes, then K = K and ψL|K = ψL|K , but the ray groups Pm Pn K and K may differ. To deal with these complications, we are going to define an equivalence relation on congruence subgroups and show that each equivalence class has a canonical representative whose modulus divides the modulus of every equivalent congruence subgroup.

♦ 定义 5.5.3

If C1 is a congruence subgroup for m1 and C2 is a congruence subgroup for

m2, then we say that (C1, m1) and (C2, m2) are equivalent and write (C1, m1) ∼

(C2, m2) whenever F m1 ∩ C F m2 ∩ C K 1 = K 2

引理 5.5.2. The relation (C1, m1) ∼ (C2, m2) is an equivalence relation.

Proof. The relation ∼ is clearly symmetric, and reflexive To show that it is transitive, let Ci be congruence subgroups for moduli mi for 1 ⩽ i ⩽ 3 and suppose 178 CHAPTER 5. NUMBER FIELDS

C ∼ C C ∼ C ∈ F m3 ∩ C ∈ m1m3,1 ( 1, m1) ( 2, m2) and ( 2, m2) ( 3, m3). Let I K 1 and pick α K ∈ F m1m2m3 ∈ Pm1m3 ⊆ Pm1 ⊆ C ⊆ C ∈ C so that αI K . Then (α) K K 1 and I 1, so αI 1, and ∈ F m1m2m3 ⊆ F m2 we also have αI K K , so

∈ F m2 ∩ C F m1 ∩ C ⊆ C αI K 1 = K 2 2

C ∼ C ∈ F m1m2m3 ⊆ F m3 since 1 2, and αI K K , so

∈ F m3 ∩ C F m2 ∩ C ⊆ C αI K 2 = K 3 3

C ∼ C ∈ Pm1m3 ⊆ Pm3 ∈ C −1 ∈ C since 2 3. We have (α) K K , so (α) 3 and therefore (α) 3, C −1 ∈ C ∈ C ⊆ F m1 since 3 is a group. Thus α αI = I 3, and we also have I 1 K , so ∈ F m1 ∩ C ∈ F m3 ∩ C I K 3. Since I K 1 was chosen arbitrarily, this proves that

F m3 ∩ C ⊆ F m1 ∩ C K 1 K 3

The reverse inclusion follows by symmetry.

♦ 命题 5.5.2

C ∼ C F m1 C ≃ F m2 C If ( 1, m1) ( 2, m2) then K / 1 K / 2are related by a canonical iso-

morphism that preserves cosets of fractional ideals prime to both m1 and m2.

∈ F m1 ∈ m1,1 ∈ F m2 Proof. For any fractional ideal I K we can pick α K so that αI K . F m2 C The image of α in K / 2 does not depend on the choice of α. This defines a F m1 → F m2 C ∈ C ∈ F m2 ∩ C group homomorphism φ : K K / 2. For I 1, we have αI K 1 = F m1 ∩ C ⊆ C ∈ F m1 \C ∈ F m2 \C ∈ C K 2 2, but for I K 1 we have αI K 1 and therefore αI / 2, C F m1 C → so ker φ = 1. It follows that φ induces an injective homomorphism K / 1 F m2 C K / 2, and by symmetry we have an injective homomorphism in the opposite F m1 C ≃ F m2 C direction, so K / 1 K / 2 as claimed. This isomorphism is independent of the choice of α used to define it (hence canonical), and for fractional ideals I coprime to both m1 and m2 we can choose F m1 C α = 1, in which case the coset of I in K / 1 will be identified with the coset of I F m2 C in K / 2.

We now observe that if C is a congruence subgroup for two moduli m1 and m2, then (C, m1) ∼ (C, m2). In particular, each subgroup of FK lies in at most one equivalence class of congruence subgroups. We can thus view the equivalence relation (C1, m1) ∼ (C2, m2) as an equivalence relation on the congruence subgroups of FK and write C1 ∼ C2 without ambiguity. It follows from Proposition above 5.5. MODULI OF A NUMBER FIELD 179 that each equivalence class of congruence subgroups uniquely determines a finite abelian group that is the quotient of a ray class group. Within an equivalence class of congruence subgroups there can be at most one congruence subgroup for each modulus (since C1 ∼ C2 ⇔ C1 = C2 whenever

C1 and C2 are congruence subgroups for the same modulus). The following lemma gives a criterion for determining when there exists a congruence subgroup of a given modulus within a given equivalence class.

引理 5.5.3. Let C1 be a congruence subgroup of modulus m1 for a number field K.

There exists a congruence subgroup C2 of modulus m2|m1 equivalent to C1 iff

F m1 ∩ F m2 ⊆ C K K 1

C C Pm2 in which case 2 = 1 K

Proof.

♦ 命题 5.5.3

Let C1 ∼ C2 be congruence subgroups of modulus m1 and m2, respec-

tively. There exists a congruence subgroup C ∼ C1 ∼ C2 with modulus n :=

gcd(m1, m2).

D F m2 ∩ C F m1 ∩ C Proof. Put m := lcm(m1, m2) and := K 1 = K 2; then

Pm Pm1 ∩ Pm2 ⊆ D ⊆ F m K = K K K so D is a congruence subgroup of modulus m, and we have

F m ∩ Pm1 ⊆ D F m ∩ Pm2 ⊆ D K K and K K so D ∼ C1 ∼ C2. To prove the existence of an equivalent congruence subgroup C F m ∩ Pn ⊆ D of modulus n it suffices to show K K . ∈ F m ∩ Pn ∈ m ∩ m2,1 ∈ m1,1 So let a = (α) K K , and choose β K K so that αβ K . ∈ D ∈ F m ∩ Pm1 ⊆ D −1 ∈ D F m ∩ Pn ⊆ D Then (β) and βa K K , so β βa = a . Thus K K C DPn and therefore = K is a congruence subgroup of modulus n equivalent to

D ∼ C1 ∼ C2.

推论. Let C be a congruence subgroup of modulus m for a number field K. There is a unique congruence subgroup in the equivalence class of C whose modulus c divides the modulus of every congruence subgroup equivalent to C. 180 CHAPTER 5. NUMBER FIELDS

♦ 定义 5.5.4 Let C be a congruence subgroup for a number field K. The unique modulus c := c(C) given by Corollary 5.5 is the conductor of C, and we say that C C CPc is primitive if = K (this is the unique primitive congruence subgroup equivalent to C).

♦ 命题 5.5.4 Let C be a primitive congruence subgroup of modulus m for a number field K. Then m is the conductor of every congruence subgroup of modulus m C Pm contained in ; in particular, m is the conductor of K .

Proof. Let C0 ⊆ C be a congruence subgroup of modulus m and let c be its conduc- | F m ∩ Pc ⊆ C ⊆ C tor. Then c m and K K 0 , and this implies that there is a congruence subgroup of modulus c equivalent to C, and therefore m|c, so c = m.

Pm The proposition implies that a modulus m occurs as a conductor iff K is primitive.

(2) 例 5.5.3. Consider K = Q and m = (2). The conductor of PQ = FQ is (1), so (2) is not a conductor. 5.6. SOME EXAMPLES AND SUPPLEMENTS 181

5.6 Some examples and supplements

Absolute discriminant and integral basis

For number field K|Q, K has an integral since Z is a PID. Let B = {b1, ··· , n} ⊆

OK be a basis of K over Q.

2 ∆(B) = [OK : Z[B]] ∆(K) By the indentity, we have an algorithm for finding an integral basis.

1. Start with any basis B ⊆ OK . 2. Calculate ∆(B) and let N be the largest natural number whose square divides ∆(B). 3. For each element of the form 1 ∑ θ = n b , n ∈ {0, ··· ,N − 1} N i i i determine whether θ is an algebraic integer. 4. If it is then replace one of the basis vectors by θ to get a new basis with discriminant of smaller absolute value, and go back to step 2. 5. If none of the θ are algebraic integers or N = 1, then B is an integral basis.

Now we sum up several tricks for calculating integral bases.

⃝1 In order to prove θ ∈/ O , it is therefore sufficient to check for each prime p ∑K 2| B 1 ∈ O with p ∆( ) that p nibi / K . n−1 ⃝2 ∀r ∈ Q we have N(r − α) = f(r), and ∆(1, α, ··· , α ) = ∆(1, r + α, ··· , (r + α)n−1). ⃝3 Employ the conclusions in above subsection, such as 5.6.1.

例 5.6.1 (Cubic fields). Let K = Q(α) for some α with minimal polynomial f(x) = x3 + ax + b. Note that ∆(1, α, α2) = −4a3 − 27b2. Let a = b = −1, then {1, α, α2} is an integral basis.

例 5.6.2 (Pure cubic fields ). √ K = K = Q( 3 m) K = Q(θ ) = Q(θ ) Let√ √ , we may assume that 1 2 where 3 2 3 2 θ1 = a b, θ2 = ab , a, b ∈ Z, (a, b) = 1, ab ≠ 1. Let   1 if a ̸≡ b mod 9 θ0 :=  1+θ1+θ2 2 ≡ 2 ≡ 2 if a b ab 1 mod 9 182 CHAPTER 5. NUMBER FIELDS then {θ0, θ1, θ2} is an integral basis of K. And   −27a2b2 if a ̸≡ b mod 9 ∆(K) =  −3a2b2 if a2b ≡ ab2 ≡ 1 mod 9 √ If m = 19, we obtain ∆(Q( 3 19)) = −3 · 192. √ Now we give another method to compute ∆(Q( 3 19)), which is more general.

例 5.6.3 (Important case). Let f = x3 − x2 − 6x − 12 be the irreducible polynomial of the element β 1 − ′ − 38 occurring in 2.3.6. By computation we have f = 9 (x 1)f 9 (x + 3), then employ resultant to obtain ( ) ( ) 2 ′ ′ 38 38 ′ ∆(f) = −R(f, f ) = 32R f , − (x + 3) = −32 · · f (−3) = −22 · 3 · 192 9 9

As [OK : Z[β]] = 2 (we can use the conlusion of 2.3.6, or prove it as 2.3.6 did), so ∆(K) = −3 · 192. By the same argument of 2.3.6, we find the prime (2, β) is singular. γ := β2−β−6 Z { } 2 is outside [β]. 1, β, γ is exactly an integral basis of K since ∆(1, β, γ) = 1 2 4 ∆(1, β, β ) = ∆(K). p-divisibility of coefficients

(Convention in this subsection) Let K|Q be a number field with degree

n. Assume K = Q(α), where α ∈ OK and its minimal polynomial over Q is Eisenstein at p.

引理 5.6.1. For a0, a1, ··· , an−1 ∈ Z, if

n−1 a0 + a1α + ··· + an−1α ≡ 0 mod pOK (∗) then ai ≡ 0 mod pZ for all i.

Proof. We will argue by induction from a0 up to an−1. n n−1 Since α ≡ 0 mod pOK , we get a0α ≡ 0 mod pOK . n n−1 ≡ nZ Take norms, a0 NK/Q(α) 0 mod p . Notice that NK/Q(α) is divisible n | n | by p exactly once, so the above congruence modulo p implies p a0 , so p a0. 5.6. SOME EXAMPLES AND SUPPLEMENTS 183

n−1 Now the congruence (∗) becomes a1α + ··· + an−1α ≡ 0 mod pOK . n−2 n−1 Multiply this by α to get a1α ≡ 0 mod pOK and take norms again. The conclusion now will be p|a1.

We can iterate this all the way to the last term, so each ai is divisible by p.

♦ 命题 5.6.1

n−1 If r0 + r1α + ··· + rn−1α ∈ OK with ri ∈ Q, then each ri has no p in its denominator.

Proof. Assume some ri has a p in its denominator. Let d be the least common denominator, so dri ∈ Z for all i, and some dri ̸≡ 0 mod p. This is a contradiction by the previous lemma.

♦ 命题 5.6.2

p ∤ [OK : Z[α]].

Proof. Suppose p | [OK : Z[α]]. Then OK /Z[α], viewed as a finite abelian group, has an element of order p: there is some γ ∈ OK s.t. γ ∈/ Z[α] but p ∈ Z[α]. Write

n−1 γ = r0 + r1γ + ··· + rn−1γ hence some ri has a p as its denominator.

By previous result, we obtain a useful method to determine the integer domain in several cases. ♦ 命题 5.6.3 If p ∤ n, then pn−1|∆(K); If p | n, then pn|∆(K).

2 Proof. Since ∆(Z[α]) = [OK : Z[α]] ∆(K), the highest power of p in ∆(K) and ∆(Z[α]) is the same. Note that

′ ∆(Z[α]) = ∆(f(x)) = NK|Q(f (α))

Write (p) = qn and (α) = qa with q ∤ a. Let the minimal polynomial of α over Q be ∑ n i n ′ f(x) = i=0 cix . Since each ci is divisible by p and thus by q , all terms in f (α) except the leading term are divisible by qn. Thus

′ − f (α) ≡ nαn 1 mod qn 184 CHAPTER 5. NUMBER FIELDS

We know αn is divisible by qn−1 and not by qn. Therefore if n ≠ 0 mod q, which is the same as n ≠ 0 mod p, then (f ′(α)) is divisibile by q exactly n − 1 times, while if n ≡ 0 mod p then (f ′(α)) is divisible by q at least n times. In the first case, (f ′(α)) = qn−1b where q ∤ b, so taking norms gives us ′ n−1 n ′ n NK|Q(f (α)) = p b where p ∤ b. In the second case, q | (f (α)), so pn p | ′ NK|Q(f (α)). √ √ √ K = Q( 3 2) ∆(Z[ 3 2]) = [O : Z[ 3 2]]2∆(K) 例 5.6.4. Let√ , note that K . By calcu- ∆(Z[ 3 2]) = −108 = −2233 2 3 lation, √ , so and are the only primes which could [O : Z[ 3 2]] divide K√ . 3 2 x3 − 2 2 2 Since√ is the√ root of , which is Eisenstein at , does not divide [O : Z[ 3 2]] 3 2 + 1 (x − 1)3 − 2 = x3 − 3x2 + 3x − 3 K . And is the root of √ √ , which 3 3 [O : Z[ 3 2 + 1]] = [O : Z[ 3 2]] is Eisenstein√ at ,so does not divide√ K K . Hence 3 3 [OK : Z[ 2]] = 1, namely OK = Z[ 2].

Compute class group and unit group √ 例 5.6.5. K = Q( 3 19). O = Z[α, β], ∆(K) = −3 · 192 As we showed, K √ , as 2.3.6. The Minkowski 4 · 6 · 2 ≈ constant M(K) = π 27 3 19 9.3, so Cl(K) is generated by the primes of norm ⩽ 7.

From our set of generators {p2, p4, p3, q3, p5} we can discard p4 and q3, since the factorizations of 2 and 3 yield the relations [p4] = −[p2] and [q3] = −2[p3] in Cl(K). 2 We have N(α−1) = −fα(1) = −20 = −2 ·5, and it is clear that α−1 is contained 2 in p2 and p5. We deduce that there is a factorization (α−1) = p2p5, and this allows 2 us to discard the generator [p5]. The element β = (α −α + 1)/3 is contained in p3 and p4 and of norm N(β) = −fβ(0) = 12, so we have (β) = p3p4 and a resulting relation [p3] = −[p4] = [p2]. It follows that Cl(K) is cyclic and generated by [p2]. 3 3 The factorization of (α + 3) = p2 or (β + 1) = p2 show that Cl(K) has order 1 or 2 3. Factoring additional elements yield no new relations: we have (α−2) = p3q3 in 2 the class of −3[p3] and (β−3) = p2p3 in the class of 3[p3]. This suggests that Cl(K) has order 3. Showing that an ideal is not principal involves the knowledge of the unit group. In this case, finding a unit is easy as the 8 elements β + j with j ∈ Z between −4 and 3 and the 5 elements α + k with k ∈ {1, −2, 3, 4} all factor into primes lying over 2, 3 and 5. 5.6. SOME EXAMPLES AND SUPPLEMENTS 185

3 As α + 3 and β + 1 both generate p2, the element η = (α + 3)/(β + 1) = 1−α−β O× is a unit in K . Before showing that η is fundamental, let us see how we can apply this to prove that the order of [p2] in the class group is 3. 3 Suppose that p2 = (x) is a principal ideal. Then x and α+3 generate the same O 3 O× { } × ⟨ ⟩ ideal in K , so we have α + 3 = εx for some unit ε. We have K = 1 η , so ε = ηk for some k ∈ Z. In order to derive a contradiction, we reduce the equation modulo a suitable prime of norm p ≡ 1 mod 3. Modulo p19 = (α), we find that

β maps to 1/3 = −6 ∈ OK /p19 = F19 and η = 1−α−β to 7 ∈ F19. As 7 is a cube in 3 F19, we find that εx maps to a cube in OK /p19 for all k. However, α + 3 maps to

3, which is not a cube in F19. This shows that p2 is not principal.

It remains to show that η is a fundamental unit in OK . In fact, the argument O× O× 3 above only uses that η generates the cyclic group K /( K ) , and this can again be proved by exhibiting a prime p in OK for which η is not a cube in OK /p. ⩾ 1 3·192 ≈ O → R We have RK 3 log( 4 ) 1.86. Under the unique embedding K , we have log(η) ≈ −2.63, so the inequality Reg(η)/Reg(K) < 2 shows that η is fundamental. 第 6 章 Zeta and L-functions

6.1 Prime number theorem

As we shall see, every global field has a zeta function that is intimately related to the distribution of its primes. We begin with the zeta function of the rational Q field , ∑ 6 ζ(s) := n−s n⩾1 which we will use to prove the prime number theorem.

注记. This series converges absolutely and locally uniformly on ℜ(s) > 1.

The Riemann zeta function

♦ 定理 6.1.1. Euler product For ℜ(s) > 1 we have ∑ ∏ ζ(s) = n−s = (1 − p−s)−1 n⩾1 p

where the product converges absolutely. In particular, ζ(s) ≠ 0 for ℜ(s) > 1.

Proof. We have ∑ ∑ ∏ ∏ ∑ ∏ − − − − − n s = p vp(n)s = p es = (1 − p s) 1 n⩾1 n⩾1 p p v⩾0 p

To justify the second equality, consider the partial zeta function ζm(s), which

restricts the summation in ζ(s) to the set Sm of m-smooth integers (those with 6.1. PRIME NUMBER THEOREM 187 no prime factors p > m). If p1, ··· , pk are the primes up to m, then absolutely convergence implies ∑ ∑ ∏ ∑ ∏ − − − − − s e1 ··· ek s s ei − s 1 ζm(s) := n = (p1 pk ) = (pi ) = (1 p ) n∈Sm e1,··· ,ek⩾0 1⩽i⩽k ei⩾0 p⩽m

For any δ > 0 the sequence of functions ζm(s) converges uniformly on ℜ(s) > 1 + δ to ζ(s); indeed, for any ϵ > 0 and any such s we have ∫ ∑ ∑ ∑ ∞ − − −ℜ − − 1 − |ζ (s) − ζ(s)| ⩽ n s ⩽ |n s| = n (s) ⩽ x 1 δdx ⩽ m δ < ϵ m δ n⩾m n⩾m n⩾m m for all sufficiently large m. It follows that the sequence ζ (s) converges locally m ∏ −s −1 uniformly to ζ(s) on ℜ(s) > 1. The sequence of functions Pm(s) := (1 − p ) ∏ p⩽m clearly converges locally uniformly to (1−p−2)−1 on any region in which the latter function is absolutely convergent (or even just convergent). For any s in ℜ(s) > 1 we have ∑ ∑ ∑ ∑ ∑ ∑ − − 1 − − − | log(1 − p s) 1| = p es ⩽ |p s|e = (|ps| − 1) 1 < ∞ e p p e⩾1 p e⩾1 p ∑ − 1 n | | where we have used the identity log(1 z) = − n z , valid for z < 1. It ∏ n⩾1 follows that (1−p−s)−1 is absolutely convergent (and in particular, nonzero) on p ℜ(s) > 1.

♦ 定理 6.1.2. Analytic continuation I For ℜ(s) > 1 we have 1 ζ(s) = + ϕ(s) s − 1 where ϕ(s) is a holomorphic function on ℜ(s) > 0. Thus ζ(s) extends to a meromorphic function on ℜ(s) > 0 that has a simple pole at s = 1 with residue 1 and no other poles.

Proof. For ℜ(s) > 1 we have ∫ ( ∫ ) ∫ ∑ ∞ ∑ n+1 ∑ n+1 1 − − − − − − ζ(s)− = n s− x sdx = n s − x sdx = (n s−x s)dx s − 1 n⩾1 1 n⩾1 n n⩾1 n ∫ n+1 −s −s For each n ⩾ 1 the function ϕn(s) := (n − x )dx is holomorphic on n ℜ(s) > 0. For each fixed s in ℜ(s) > 0 and x ∈ [n, n + 1] we have ∫ ∫ ∫ x x |s| x |s| |s| |n−s − x−s| = st−s−1 t ⩽ t = t ⩽ d | 1+s|d 1+ℜ(s) d 1+ℜ(s) n n t n t n 188 CHAPTER 6. ZETA AND L-FUNCTIONS and therefore ∫ n+1 |s| | | ⩽ | −s − −s| ⩽ ϕn(s) n x dx 1+ℜ(s) n n ℜ ℜ (s0) For any s0 with (s0) > 0, if we put ϵ := 2 and U := B(s0, ϵ), then for each n ⩾ 1, | | | | ⩽ s0 + ϵ sup ϕn(s) 1+ϵ =: Mn s∈U n ∑ ∑ and Mn = (|s0| + ϵ)ζ(1 + ϵ) converges. The series ϕn thus converges locally normally on ℜ(s) > 0. By the Weierstrass M-test, ℜ(s) > 0 converges to a function ϕ(s) = ζ(s) 1 ℜ(s) > 0 − s−1 that is holomorphic on .

We now show that ζ(s) has no zeros on ℜ(s) = 1; this fact is crucial to the prime number theorem. For this we use the following ingenious lemma, at- tributed to Mertens.

引理 6.1.1 ((Merten). For x, y ∈ R with x > 1 we have |ζ(x)3ζ(x+iy)4ζ(x+2iy)| ⩾ 1.

Proof. (Sketch): For ℜ(s) > 1 we have

Note that the trigonometric identity cos(2θ) = 2 cos2(θ) − 1 implies

3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ⩾ 0

Taking θ = ny log p yields the lemma.

推论. ζ(s) has no zeros on ℜ(s) ⩾ 1.

Proof. Suppose ζ(1 + iy) = 0 for some y ∈ R. Then y ≠ 0, and we know that ζ(s) does not have a pole at 1 + 2iy ≠ 1, by Theorem 6.1.2. We therefore must have

lim |ζ(x)3ζ(x + iy)4ζ(x + 2iy)| = 0 x→1 since ζ(s) has a simple pole at s = 1, a zero at 1 + iy, and no pole at 1 + 2iy. But this contradicts Lemma above. 6.1. PRIME NUMBER THEOREM 189

The Prime Number Theorem

The prime counting function π : R → N is defined by ∑ π(x) := 1 p⩽x it counts the number of primes up to x. The prime number theorem (PNT) states that x π(x) ∼ , as x → ∞ log x This conjectured growth rate for π(x) dates back to Gauss and Legendre in the late 18th century. In fact Gauss believed the asymptotically equivalent but more accurate statement ∫ ∞ dx π(x) ∼ Li(x) := , as x → ∞ 2 log x

However it was not until a century later that the prime number theorem was independently proved by Hadamard and de la Vall´ee Poussin in 1896. Their proofs are both based on the work of Riemann, who in 1860 showed that there is a precise connection between the zeros of ζ(s) and the distribution of primes (we shall say more about this later), but was unable to prove the prime number theorem. The proof we will give is more recent and due to Newman, but it relies on the same properties of the Riemann zeta function that were exploited by both Hadamard and de la Vall´ee, the most essential of which is the fact that ζ(s) has no zeros on ℜ(s) ⩽ 1. We will follow Zagier’s outline but be slightly more expansive in our presentation. We should note that there are also ”elementary” proofs of the prime number theorem independently obtained by Erdos¨ and Selberg in the 1940s that do not use the Riemann zeta function, but they are elementary only in the sense that they do not use complex analysis; the details of these proofs are considerably more complicated than the one we will give. Rather than work directly with π(x), it is more convenient to work with the log-weighted prime- counting function defined by Chebyshev: ∑ ϑ(x) := log p p⩽x whose growth rate differs from that of π(x) by a logarithmic factor. 190 CHAPTER 6. ZETA AND L-FUNCTIONS

♦ 定理 6.1.3. Chebyshev π(x) ∼ x ϑ(x) ∼ x log x iff .

In view of Chebyshev’s result, the prime number theorem is equivalent to ϑ(x) ∼ x. Let us first show that lim ϑ(x) bounded, which is indicated by the x→∞ x asymptotic notation ϑ(x) = O(x).

引理 6.1.2 (Chebyshev). For x ⩾ 1 we have ϑ(x) ⩽ (4 log 2)x, thus ϑ(x) = O(x).

In order to prove ϑ(x) ∼ x, we will use a general analytic criterion applicable to any non-decreasing real function f(x). ∫ ∞ f(t) − t R⩾ → R 引理 6.1.3. Let f : 1 be a nondecreasing function. If the integra 2 dt 1 t converges then f(x) ∼ x.

In order to show that the hypothesis of Lemma above is satisfied for f = ϑ, we will work with the function H(t) := ϑ(et)e−t − 1; the change of variables t = eu shows that ∫ ∫ ∞ − ∞ ϑ(t) t ⇐⇒ 2 dt converges H(u)du converges 1 t 0 We now recall the Laplace transform.

♦ 定义 6.1.1

Let h : R>0 → R be a piecewise continuous function. The Laplace transform Lh of h is the complex function defined by

∫ ∞ − Lh(s) := e sth(t)dt 0

which is holomorphic function ℜ(s) > c for any c ∈ R for which h(t) = O(ect).

The following properties of the Laplace transform are easily verified.

• L is linear transformation over R.

∈ R L a • If h(t) = a is constant then h(s) = s .

• L(eath(t))(s) = L(h)(s − a) for all a ∈ R.

Now define the auxiliary function ∑ − Φ(s) := p s log p p which is related to ϑ(x) by the following lemma. 6.1. PRIME NUMBER THEOREM 191

L t Φ(s) ℜ 引理 6.1.4. (ϑ(e ))(s) = s is holomorphic on (s) > 1.

Let us now consider the function H(t) := ϑ(et)e−t − 1. It follows from the lemma and standard properties of the Laplace transform that on ℜ(s) > 0 we have Φ(s + 1) 1 LH(s) = − s + 1 s Φ(s) 1 ℜ(s) > 1 引理 6.1.5. The function − s−1 extends to a meromorphic function on 2 that is holomorphic on ℜ(s) ⩾ 1.

1 L Φ(s+1) − 1 推论. The functions Φ(s + 1)− s and H(s) = s+1 s both extend to meromor- ℜ 1 ℜ ⩾ phic functions on (s) > − 2 that are holomorphic on (s) 0.

The final step of the proof relies on the following analytic result duetoNew- man。 ♦ 定理 6.1.4

Let f : R⩾0 → R be a bounded piecewise continuous function, and suppose its Laplace transform extends to a holomorphic function g(s) on ℜ(s) ⩾ 0. ∫ ∞ Then the integral f(t)dt converges and is equal to g(0). 0

注记. The theorem is an example of what is known as a Tauberian theorem. For a piecewise continuous function f : R⩾0 → R, The Laplace transform Lf is typically not defined on ℜ(s) ⩽ c, where c ∈ R is the least c for which f(t) = O(ect). Now it may happen that the function Lf has an analytic continuation to a larger domain; f(t) = t (Lf)(s) = 1 for example, if e then s−1 extends to a holomorphic function on C\{1}. But plugging values of s with ℜ(s) ⩽ c into the integral usually does not work; in our f(t) = et example, the integral diverges on ℜ(s) ⩽ 1. The theorem says that when Lf extends to a holomorphic function on the entire half-plane ℜ(s) ⩾ 0, its value at s = 0 is exactly what we would get by simply plugging 0 into the integral defining Lf. More generally, Tauberian theorems refer to results related to transforms f → T (f) that allow us to deduce properties of f (such as the convergence of ∫ ∞ f(t)dt) from properties of T (f) (such as analytic continuation to ℜ(s) ⩾ 0). 0 The term ”Tauberian” was coined by Hardy and Littlewood and refers to Alfred Tauber, who proved a theorem of this type as a partial converse to a theorem of Abel. 192 CHAPTER 6. ZETA AND L-FUNCTIONS

♦ 定理 6.1.5. Prime Number Theorem x π(x) ∼ log x

Proof. H(t) = ϑ(et)e−t − 1 is piecewise continuous and bounded, by Lemma 6.1.2, and its Laplace transform extends to a holomorphic function on ℜ(s) ⩾ 0, by Corollary 6.1. Theorem 6.1.4 then implies that the integral

∫ ∞ ∫ ∞ H(t)dt = (ϑ(et)e−t − 1)dt 0 0 converges. Replacing t with log x, we see that ∫ ∞ ϑ(x) − x 2 dx 1 x ϑ(x) ∼ x π(x) ∼ x converges. Lemma 6.1.3 implies , equivalently, log x .

One disadvantage of our proof is that it does not give us an error term. Using more sophisticated methods, Korobov and Vinogradov independently obtained the bound ( ) x π(x) = Li(x) + O exp((log x)3/5+o(1)) in which we note that the error term is bounded by O(x/(log x)n) for all n but not by O(x1−ϵ) for any ϵ > 0. Assuming the Riemann Hypothesis, which states that ℜ ℜ 1 the zeros of ζ(s) in the critical strip 0 < (s) < 1 all lie on the line (s) = 2 , one can prove π(x) = Li(x) + O(x1/2+o(1))

More generally, if we knew that ζ(s) has no zeros in the critical strip with ⩾ 1 real part greater than c, for some c 2 strictly less than 1, we could prove π(x) = Li(x) + O(xc + o(1)). There thus remains a large gap between what we can prove about the distribution of prime numbers and what we believe to be true. Remarkably, other than refinements to the o(1) term appearing in the Korobov-Vinogradov bound, essentially no progress has been made on this problem in the last 60 year. 6.2. THE FUNCTIONAL EQUATION 193

6.2 The functional equation

Recall that the Riemann zeta function ζ(s) has an Euler product and an analytic continuation to the right half-plane Re(s) > 0. We now want complete the picture by deriving a functional equation that relates the values of ζ(s) to those of ζ(1−s). This will then also allow us to extend ζ(s) to a meromorphic function on C that is holomorphic except for a simple pole at s = 1.

Fourier transforms and Poisson summation

A key tool we will use to derive the functional equation is the Poisson summation formula. ♦ 定义 6.2.1 A Schwartz function f : R → C is a C∞-function that decays rapidly to zero: for all m, n ∈ N we have

m (n) sup x f (x) < ∞ x∈R

The Schwartz space S(R) of all Schwartz functions is a C-vector space.

p 注记. For any p ∈ R⩾1, the Schwartz space S(R) is contained in the space L (R). The Swchartz space S(R) is not complete under the Lp norm, but it is dense in Lp(R) under the subspace topology. One can equip the Schwartz space with a translation-invariant metric of its own under which it is a complete metric space.

It follows immediately from the definition and standard properties of the derivative that the Schwartz space S(R) is closed under differentiation, multiplication by polynomials, sums and products, and linear change of variable. It is also closed under convolution.

例 6.2.1. All compactly supported functions C∞ functions are Schwartz functions, 2 as is the Gaussian g(x) := e−πx .

♦ 定义 6.2.2 The Fourier transform of a Schwartz function f ∈ S(R) is the function ∫ b −2πixy f(y) := f(x)e dx R 194 CHAPTER 6. ZETA AND L-FUNCTIONS

b which is also a Schwartz function. We can recover f(x) from f(y) via the inverse transform ∫ b 2πixy f(x) := f(y)e dy R

b b 注记. 1. The maps f 7→ f and f 7→ f are thus inverse linear operators on S(R) (they are also continuous in the metric topology of S(R) and thus homeo- morphisms). 2. The invertibility of the Fourier transform on the Schwartz space S(R) a key motivation for its definition. For functions in L1(R) (the largest space of functions for which our definition of the Fourier transform makes sense), the Fourier transform of a smooth function decays rapidly to zero, and the Fourier transform of a function that decays rapidly to zero is smooth; this leads one to consider the subspace S(R) of smooth functions that decay rapidly to zero. 3. One can show that S(R) is the largest subspace of L1(R) closed under multiplication by polynomials on which the Fourier transform is invertible.

The Fourier transform changes convolutions into products, and vice versa. We have [ b c b f ∗ g = f · gb and fg = f ∗ gb for all f, g ∈ S(R). One can thus view the Fourier transform as an isomorphism of (non-unital) C-algebras that sends (S(R), +, ×) to (S(R), +, ∗).

∈ R ∈ S R \ 1 b y 引理 6.2.1. For all a >0 and f ( ), we have f(ax)(y) = a f( a ). \ f ∈ S(R) d fb(y) = −2πixf\(x)(y) d f(x)(y) = 2πiyfb(y) 引理 6.2.2. For we have dy and dx .

2 例 6.2.2. Let g(x) = e−πx , then gb = g.

Proof. The function g(x) satisfies the first order ordinary differential equation

g′ + 2πxg = 0 with initial value g(0) = 1. Multiplying both sides by i and taking Fourier transforms yields i(gb′ + 2πxgc) = i(2πixgb − igb′) = gb′ + 2πxgb = 0 by the Lemma above. So gb satisfies the differential equation and gb(0) = 1. 6.2. THE FUNCTIONAL EQUATION 195 ∫ ⟨ ⟩ The Fourier transform is compatible with the inner product f, g := R f(x)g(x)dx. Indeed, we can easily derive Parseval’s identity: ∫ ∫ ∫ ∫ b 2πixy b b ⟨f, g⟩ = f(x)g(x)dx = f(y)g(x)e dxdy = f(y)gb(y)dy = ⟨f, gb⟩ R R R R which when applied to g = f yields Plancherel’s identity: b b b ∥f∥2 = ⟨f, f⟩ = ⟨f, f⟩ = ∥f∥2

2 where ∥ · ∥2 is the L -norm. For number-theoretic applications there is an analogous result due to Poisson.

♦ 定理 6.2.1. Poisson Summation Formula For all f ∈ S(R) we have the identity ∑ ∑ f(n) = fb(n) n∈Z n∈Z

Proof. We first note that both sums are well defined; the rapid decay property of ∑ Schwartz functions guarantees absolute convergence. Let F (x) := f(x + n). n∈Z Then F is a periodic C∞-function, so it has a Fourier series expansion ∑ 2πinx F (x) = cne n∈Z with Fourier coefficients ∫ ∫ ∫ 1 1 ∑ −2πint −2πiny −2πiny b cn = F (x)e dt = f(x + m)e dy = f(x)e dy = f(n) R 0 0 m∈Z

We then note that ∑ ∑ f(n) = F (0) = fb(n) n∈Z n∈Z

Jacobi’s theta function

We now define the theta function ∑ 2 Θ(τ) := eπin τ n∈Z The sum is absolutely convergent for Imτ > 0 and thus defines a holomorphic function on the upper half plane. It is easy to see that Θ(τ) is periodic modulo 2, that is, Θ(τ + 2) = Θ(τ) 196 CHAPTER 6. ZETA AND L-FUNCTIONS

∈ R Θ(√i/a) 引理 6.2.3. For all a >0 we have Θ(ia) = a .

2 √ 2 Proof. Put g(x) := e−πx and h(x) := g( ax) = e−πx a. We have √ √ \√ gb(y/ a) g (y/ a) bh(y) = g( ax)(y) = √ = √ a a

Plugging τ = ia into Θ(τ) and applying Poisson summation yields ∑ ∑ ∑ ∑ √ πn2a b g(n/ a) Θ(i/a) Θ(ia) = e− = h(n) = h(n) = √ = √ a a n∈Z n∈Z n∈Z n∈Z

Euler’s gamma function

♦ 定义 6.2.3

The Mellin transform of a function f : R>0 → C is the complex function defined by ∫ ∞ − M(f)(s) := f(t)ts 1dt 0 (s) ∈ (a, b) whenever this integral converges. It∫ is holomorphic on Re for any ∞ | | σ−1 ∈ interval (a, b) in which the integral 0 f(t) t dt converges for all σ (a, b).

∫ ∞ −t | −t| σ−1 注记. The Gamma function Γ(s) is the Mellin transform of e . Since 0 e t dt converges for all σ > 0, the integral defines a holomorphic function on Re(s) > 0.

Integration by parts yields Γ(s + 1) Γ(s) = s for Re(s) > 0, thus Γ(s) has a simple pole at s = 0 with residue 1. The equation Γ(s + 1) = sΓ(s) allows us to extend Γ(s) to a meromorphic function on C with simple poles at s = 0, −1, −2, ··· , and no other poles.

♦ 命题 6.2.1. Euler’s Reflection Formula We have π Γ(s)Γ(1 − s) = sin(πs) as meromorphic functions on C with simple poles at each integer s ∈ Z. √ 1 1 2 1 推论. Putting s = 2 in the reflection formula yields Γ( 2 ) = π, so Γ( 2 ) = π.

推论. The function Γ(s) has no zeros on C. 6.2. THE FUNCTIONAL EQUATION 197

Proof. Suppose Γ(s0) = 0. The RHS of the reflection formula is never zero, since sin(πs) has no poles, so Γ(1−s) must have a pole at s0. Therefore 1−s0 ∈ Z⩽0, equivalently, s0 ∈ Z⩾1, but then Γ(s0) = (s0 − 1)! ≠ 0.

Completing the zeta function

Let us now consider the function

F (s) := π−sΓ(s)ζ(2s)

C 1 which is a meromorphic on and holomorphic on Re(s) > 2 . In the region 1 Re(s) > 2 we have an absolutely convergent sum ∫ ∑ ∑ ∞ − − − − F (s) = Γ(s) (πn2) s = (πn2) sts 1e tdt n⩾1 n⩾1 0 and the substitution t = πn2y with dt = πn2dy yields ∫ ∑ ∞ − − 2 F (s) = ys 1e πn ydy n⩾1 0

By the Fubini-Tonelli theorem, we can swap the sum and the integral to obtain ∫ ∞ ∑ − − 2 F (s) = ys 1 e πn ydy 0 n⩾1

∑ 2 We have Θ(iy) = 1 + 2 e−πn y, thus n⩾1 (∫ ∫ ) 1 ∞ 1 − 1 − F (s) = ys 1Θ(iy)dy − + ys 1(Θ(iy) − 1)dy 2 0 s 1

1 We now focus on the first integral on the RHS. The change of variable t = y yields ∫ ∫ ( ) 1 ∞ − − − i ys 1Θ(iy)dy = t s 1Θ dt 0 1 t By Lemma 6.2.3 we yields ∫ ∫ 1 ∞ s−1 −s− 1 1 y Θ(iy) y = t 2 (Θ(it) − 1) t − d d 1 − 0 1 2 s

Plugging this back into our equation for F (s) we obtain the identity

∫ ∞ 1 s−1 −s− 1 1 1 2 − − − F (s) = (y + y )(Θ(iy) 1)dy − 2 1 2s 1 2s 198 CHAPTER 6. ZETA AND L-FUNCTIONS

1 1 ̸ 1 valid on Re(s) > 2 . We now observe that F (s) = F ( 2 −s) for s = 0, 2 , which allows us to analytically extend F (s) to a meromorphic function on C with poles only at 1 s s = 0, 2 . Replacing s with 2 leads us to define the completed zeta function ( ) − s s Z(s) := π 2 Γ ζ(s) 2 which is meromorphic on C and satisfies the functional equation

Z(s) = Z(1 − s)

Zeros: It has simple poles at 0 and 1 (and no other poles). The only zeros of Z(s) on Re(s) > 0 are the zeros of ζ(s). Thus the zeros of Z(s) on C all lie in the critical strip 0 < Re(s) < 1. The functional equation also allows us to analytically extend ζ(s) to a meromorphic function on C whose only pole is a simple pole at s s = 1; the pole of Z(s) at s = 0 comes from the pole of Γ( 2 ) at s = 0. The function s ··· Γ( 2 ) also has poles at −2, −4, where Z(s) does not, so our extended ζ(s) must have zeros at −2, −4, ··· . These are trivial zeros; all the interesting zeros of ζ(s) lie 1 in the critical strip and are conjectured to lie only on the critical line Re(s) = 2 (this is the Riemann hypothesis). Compute ζ(0): We have ( ) ( ) − s−1 1−s s− 1 1−s Z(s) Z(1 − s) π 2 Γ π 2 Γ ( ) ( ) ( 2) − ( ) 2 − ζ(s) = − s s = − s s = − s s ζ(1 s) = s ζ(1 s) 2 2 2 π Γ 2 π Γ 2 π Γ 2 Γ 2

We know that ζ(s) has a simple pole with residue 1 at s = 1, so ( ) s− 1 1−s (s − 1)π 2 Γ lim (s − 1)ζ(s) = lim ( ) 2 ζ(1 − s) = 1 s→1+ s→1+ s Γ 2 √ 1 1 When s = 1, note that Γ( 2 ) = π and Γ(z) = z Γ(z + 1), we get ( ) 2 3 − s 1 = lim (s − 1) Γ ζ(1 − s) = −2Γ(1)ζ(0) = −2ζ(0) s→1+ 1 − s 2

− 1 Thus ζ(0) = 2 .

注记. By the reflection formul and the duplication formula we obtain an another he functional equation for ζ(s): ( ) − πs ζ(s) = 2sπs 1 sin Γ(1 − s)ζ(1 − s) 2 6.2. THE FUNCTIONAL EQUATION 199

Gamma factors and a holomorphic zeta function

If we write out the Euler product for the completed zeta function, we have ( ) ∏ − s s −s −1 Z(s) = π 2 Γ (1 − p ) 2 p

One should think of this as a product over the places of the field Q; the leading factor ( ) − s s ΓR(s) := π 2 Γ 2 that distinguishes the completed zeta function Z(s) from ζ(s) corresponds to the real archimedean place of Q. When we discuss Dedekind zeta functions we will see that there are Gamma factors ΓR and ΓC associated to each of the real and complex places of a number field. ( ) s s(s−1) If we insert an additional factor of 2 := 2 in Z(s) we can remove the poles at 0 and 1, yielding a function ξ(s) holomorphic on C. This yields Riemann’s seminal result ♦ 定理 6.2.2. Analytic Continuation II The function ( ) s ξ(s) := ΓR(s)ζ(s) 2 is holomorphic on C and satisfies the functional equation

ξ(s) = ξ(1 − s)

The zeros of ξ(s) all lie in the critical strip 0 < Re(s) < 1.

注记. We will usually work with Z(s) and deal with the poles rather than making it holomorphic by introducing additional factors; some authors use ξ(s) to denote our Z(s). 200 CHAPTER 6. ZETA AND L-FUNCTIONS

6.3 *Dirichlet L-functions

Mertens’ estimate

It is known that ζ(s) has a pole at s = 1, now we want to clarify how quickly it diverges as s → 1+. Note that ∑ ∑ − − log ζ(s) = − log(1 − p s) = p s + O(1), as s → 1+ p p ∑ 1 We can estimate p via Mertens’ second theorem. p⩽x

♦ 定理 6.3.1. Mertens 1874 As x → ∞ we have ∑ log p | | 1. p = log x + R(x), where R(x) < 2. p⩽x ∑ ( ) 1 = x + B + O 1 B = 0.2614 ··· 2. p log log log x , where is Mertens’ con- p⩽x stant. ∑ ( ) ( ) 1 − 1 = − x − γ + O 1 γ = 0.5772 ··· 3. log p log log log x , where is Eu- p⩽x ler’s constant.

注记. The Prime Number Theorem is equivalent to the statement ( ) ∑ 1 1 = log log x + B + o p x p⩽x log which is (ever so slightly) sharper than Mertens’ estimate.

推论. There are infinitely many primes.

Dirichlet L-functions

♦ 定义 6.3.1 The Dirichlet L-function associated to a Dirichlet character χ is

( )− ∏ χ(p) 1 ∑ χ(n) L(s, χ) := 1 − = ps ns p n⩾1

注记. 1. The sum and product converge absolutely for Re s > 1, since |χ(n)| ⩽ 1, thus L(s, χ) is holomorphic on Re s > 1. 6.3. *DIRICHLET L-FUNCTIONS 201

2. For the trivial Dirichlet character I have L(s, I) = ζ(s). ∏ −s −1 3. For the principal character Im we have ζ(s) = L(s, Im) (1 − p ) . p|m

♦ 定理 6.3.2 The L-series L(χ, s) has a meromorphic continuation to the whole complex plane. If χ is not the trivial character then L(χ, s) is in fact holomorphic.

Special Values

We are interested in these L-functions for their special values (at integers). In order to compute these special values we introduce Bernoulli numbers, along with a slight generalization.

Define the Bernoulli numbers Bn for n ⩾ 0 by t ∑ tn = B et − 1 n n! n⩾0

The inverse of this power series is

et − 1 ∑ tn = t (n + 1)! n⩾0 and we can use this fact to inductively compute the Bn.

n 0 1 2 3 4 5 6 7 8 9 10 − 1 1 − 1 1 − 1 5 Bn 1 2 6 0 30 0 42 0 30 0 66

Note in particular that Bn = 0 for all odd n > 1 (which can be seen by showing t + 1 t that et−1 2 is an even function). For χ a primitive Dirichlet character, define generalized Bernoulli numbers

Bn,χ for n ⩾ 0 by f ∑χ teat ∑ tn χ(a) = Bn,χ efχt − 1 n! a=1 n⩾0

In fact in this definition we can replace fχ by any multiple of it, using the identity

− ∑r 1 xk 1 = xr − 1 x − 1 k=0

Now we assume that χ is primitive. Note that χ is odd and Bn,χ = 0 for even n. 202 CHAPTER 6. ZETA AND L-FUNCTIONS

n 0 1 2 3 4 5 √ √ √ − 4 − 2 3 445 − 565 3 Bn,χ 0 7 7 i 0 3 + 3 3i 0 7 7 i n 6 7 8 9 10 √ √ 22249 − 30049 3 − − Bn,χ 0 7 7 i 0 281223 385551 3i 0 − − Bn,χ By complex analysis, we have L(χ, 1 n) = n for primitive Dirichlet character. ♦ 定理 6.3.3 Let χ be a non-trivial primitive Dirichlet character. Then  πiτ(χ)  B1,χ if χ is odd, fχ L(χ, 1) = f  τ(χ) ∑χ 2πia − χ(a) log 1 − e fχ if χ is even. fχ a=1

Cyclotomic zeta function

♦ 定义 6.3.2 The Dedekind ζ-function associated to a number field K is ∏ ( ) ∑ −s −1 −s ζK (s) := 1 − N(p) = N(a)

p⊆OK a⊆OK

where the product is over non-zero prime ideals, the sum is over non-zero ide-

als of OK .

注记. ζF (s) converge absolutely on Re(s) > 1, and has an analytic continuation to all of C and satisfies a functional equation.

Now let F be an Ablian number field. Suppose that Q ⊆ F ⊂ Q(ζn), and (Z/nZ)× Gal(F |Q) ≃ Gal(Q(ζn)|F ) realizes Gal(F |Q) as a quotient of (Z/nZ)×. Given a character of Gal(F |Q), we can lift it to a character of (Z/nZ)×, which has an associated primitive Dirichlet character. Define X(F ) to be the set of Dirichlet characters produced in this way, i.e. the set of primitive Dirichlet characters associated to characters of (Z/nZ)× that factor through Gal(F |Q). We have the following relationship between the ζ-function of F and the L- functions of these Dirichlet characters. 6.3. *DIRICHLET L-FUNCTIONS 203

♦ 命题 6.3.1 If F is an Abelian number field, we have ∏ ζF (s) = L(χ, s) χ∈X(F )

e Proof. (Sketch) Let p be a prime which decomposes in F as p = (p1 ··· pr) , with N(p ) = f. Then p contributes (1 − p−fs)−r to the Euler product of ζ (s), and i ∏ F contributes (1 − χ(p)p−s)−1 to the product of L-functions. χ∈X(F ) We want to show that these are the same.

注记. Note by examining both sides’ poles at s = 1, we can see that L(χ, 1) ≠ 0 for a non-trivial character χ, and this can be used to prove Dirichlet’s theorem on primes in arithmetic progressions.

例 6.3.1. Let F = Q(i). Then Gal(Q(i)|Q) ≃ (Z/4Z)×, which has two associated primitive Dirichlet characters: the trivial character I, and the character χ defined by   1 if n ≡ 1 mod 4 χ(n) =  −1 if n ≡ 3 mod 4 In this field 2 ramifies, primes p ≡ 1 mod 4 split completely, and primes p ≡ 3 mod 4 are inert. Thus 1 ∏ 1 ∏ 1 ζQ (s) = (i) 1 − 2−s (1 − p−s)2 1 − p−2s p≡1 mod 4 p≡3 mod 4 1 ∏ 1 ∏ 1 1 = 1 − 2−s (1 − p−s)2 1 − p−s 1 + p−s p≡1 mod 4 p≡3 mod 4 ∏ 1 ∏ 1 = L(1, s) (1 − p−s) 1 + p−s p≡1 mod 4 p≡3 mod 4 = L(1, s)L(χ, s).

♦ 定理 6.3.4 Let χ be any non-principal Dirichlet character. Then L(1, χ) ≠ 0.

Proof. Let ψ be a non-principal Dirichlet character, say of modulus m. Then ψ is induced by a non-trivial primitive Dirichlet character ψ˜ of conductor m˜ dividing m. The L-functions of ψ and ψ˜ differ at only finitely many Euler factors (1−χ(p)p−s)−1 (corresponding to primes p dividing m/m ˜ ), and these factors are clearly nonzero 204 CHAPTER 6. ZETA AND L-FUNCTIONS at s = 1, since p > 1. We thus assume without loss of generality that ψ = ψ˜ is primitive.

Let K be the m-th cyclotomic field Q(ζm). Then ∏ ζK (s) = L(s, χ) χ where χ ranges over the primitive Dirichlet characters of conductor dividing m, including ψ. By the analytic class number formula, the LHS has a simple pole at s = 1, and the same must be true of the RHS. Thus ∏ ords=1ζK (s) = ords=1 L(s, χ) χ ∏ −1 = ords=1L(s, I) L(s, χ) ̸ I ∏χ= −1 = ords=1ζ(s) L(s, χ) ̸ I ∑ χ= −1 = −1 + ords=1L(s, χ) χ≠ I

Each χ ≠ 1 in the sum is necessarily non-principal (since it is primitive). Note that for non-principal χ the Dirichlet L-series L(s, χ) is holomorphic on Re(s) > 0, thus ords=1L(s, χ) ⩾ 0 for all χ appearing in the sum, which can therefore be zero iff every term ords=1L(s, χ) is zero. So L(1, χ) ≠ 0 for every non-trivial primitive Dirichlet character χ of conductor dividing m, including ψ.

注记. This is a key claim needed to proof Dirichlet’s theorem on primes in arithmetic progression. 6.4. THE ANALYTIC CLASS NUMBER FORMULA 205

6.4 The analytic class number formula

Lipschitz parametrizability

♦ 定义 6.4.1 A set B in a metric space X is d-Lipschitz parametrizable if it is the union of d the images of a finite number of Lipschitz continuous functions fi : [0, 1] → B.

引理 6.4.1. Let S ⊆ Rn be a set whose boundary ∂S is (n−1)-Lipschitz parametrizable. Then − #(tS ∩ Zn) = µ(S)tn + O(tn 1) as t → ∞

推论. Let Λ be a lattice in an R-vector space V ≃ Rn and let S ⊆ V be a set whose boundary is (n − 1)-Lipschitz parametrizable. Then

µ(S) − #(tS ∩ Λ) = tn + O(tn 1) as t → ∞ covol(Λ)

Proof. Note that the normalization of the Haar measure µ is irrelevant, since we are taking a ratio of volumes which is necessarily preserved under the isomorphism of topological vector spaces V ≃ Rn. We now note that if the corollary holds for sΛ, for some s > 0, then it also holds for Λ, since tS∩sΛ = (t/s)S∩Λ. For any lattice Λ, we can choose s > 0 so that sΛ is arbitrarily close to an integer lattice (for example, take s to be the LCM of all denominators appearing in rational approximations of the coordinates of a basis for Λ). The corollary follows.

Counting algebraic integers of bounded norm

Recall that we have a natural embedding

× × K ,→ KR

x 7→ (xv) where v ranges over the r + s archimedean places of K. We want to estimate the quantity

#{a : N(a) ⩽ t}

where a ranges over the nonzero ideals of OK , as t → ∞. 206 CHAPTER 6. ZETA AND L-FUNCTIONS

As a first step, let us restrict our attention to nonzero principal ideals (α) ⊆

OK . We then want to estimate the cardinality of #{(α): N(α) ⩽ t}. We have ′ ′ ∈ O× (α) = (α ) iff α/α K , so this is equivalent to

{ ∈ × ∩ O× ⩽ } O× α K K : N(α) t / K

If we now define

× { ∈ × ⩽ } ⊆ × KR,⩽t := x KR : N(x) t KR ( ) × ∩ O O× then we want to estimate the cardinality of the finite set KR,⩽t K / K . To O× U ⊆ O× simplify matters, let us replace( K with) the free group K , It suffices to × ∩ O estimate the cardinality of KR,⩽t K /U and divide the result by wK . We now define a surjective homomorphism

× ↠ × v : KR KR,1 − x 7→ xN(x) 1/n

× Rr+s The image of KR,1 under the Log map is precisely the trace zero hyperplane 0 Rr+s O× in in which Log(U) = Log( K ) = ΛK is a lattice. Let us fix a fundamen- − tal domain F for the lattice so that S := v−1(Log 1(F )) is a set of unique coset × representatives for the quotient KR /U. If we now define

S⩽t := {x ∈ S : N(x) ⩽ t} ⊆ KR we want to estimate the cardinality of the finite set

S⩽t ∩ OK

(Sketch): OK is a lattice in the R-vector space KR of dimension n. We have 1/n tS⩽1 = S⩽tn , so we can estimate the cardinality of S⩽t = t S⩽1 via Corollary 6.4 1/n with S = S⩽1 and Λ = OK by replacing t with t , provided that the boundary of

S⩽1 is (n − 1)-Lipschitz parametrizable, which can be proved. This yield ( ) µ(S⩽1) 1/n n 1/n n−1 µ(S⩽1) 1− 1 ⩽ ∩ O n ∗ #(S t K ) = (t ) + O((t ) ) = 1/2 t + O(t )( ) covol(OK ) |∆K |

Our major task is compute µ(S⩽1), we must use the normalized Haar measure r s µ on KR. It turned out to be µ(S⩽1) = 2 (2π) RK . Plugging this into (∗) yields ( ) r s 2 (2π) RK 1− 1 ⩽ ∩ O n ∗∗ #(S t K ) = 1/2 t + O(t )( ) |∆K | 6.4. THE ANALYTIC CLASS NUMBER FORMULA 207

Proof of the analytic class number formula

We are now ready to prove the analytic class number formula. Our main tool is the following theorem, which uses our analysis in the previous section to give a precise asymptotic estimate on the number of ideals of bounded norm.

♦ 定理 6.4.1 Let K be a number field of degree n. As t → ∞, the number of nonzero

OK -ideals a of absolute norm N(a) ⩽ t is ( ) r s 2 (2π) hK RK − 1 1 n 1/2 t + O(t ) wK |∆K |

Proof. (Sketch): Dividing (∗∗) by wK we get ( ) r s 2 (2π) RK 1− 1 { ⊆ O ⩽ } n # (α) K : N(α) t = 1/2 t + O(t ) wK |∆K | To complete the proof we now show that we get the same answer for every ideal class; the nonzero ideals a of norm N(a) ⩽ t are asymptotically equidis- tributed among ideal classes. Fix an ideal class [a], with a ⊆ OK nonzero. Multi- plication by a gives a bijection between

• {ideals b ∈ [a−1]: N(b) ⩽ t}

• {nonzero principal ideals (α) ⊆ a : N(α) ⩽ tN(a)}

{ ∈ ⩽ } O× • nonzero α a : N(α) tN(a) / K .

引理 6.4.2. Let a1, a2, ··· be a sequence of complex numbers and let σ be a real number. Suppose that

σ a1 + ··· + at = O(t ) as t → ∞ ∑ −s Then the Dirichlet series ann defines a holomorphic function on ℜ(s) > σ.

引理 6.4.3. Let a1, a2, ··· be a sequence of complex numbers that satisfies

σ a1 + ··· + at = ρt + O(t ) as t → ∞ ∑ × −s for some σ ∈ [0, 1) and ρ ∈ C . Then the Dirichlet series ann converges on ℜ(s) > 1 and has a meromorphic continuation to ℜ(s) > σ that is holomorphic except for a simple pole at s = 1 with residue ρ. 208 CHAPTER 6. ZETA AND L-FUNCTIONS

We are now ready to prove the analytic class number formula.

♦ 定理 6.4.2. Analytic Class Number Formula

Let K be a number field of degree n. The Dedekind zeta function ζK (z) ℜ 1 extends to a meromorphic function on (z) > 1− n that is holomorphic except for a simple pole at z = 1 with residue

r s 2 (2π) hK RK Resz=1ζK (z) = √ wK |∆(K)|

Proof. We have ∑ ∑ −z −z ζK (z) = N(a) = att a t⩾1 where at := #{a : N(a) = t} with t ∈ Z⩾1. If we now define

r s 2 (2π) hK RK ρK := √ wK |∆(K)|

then we have

1− 1 a1 + ··· + at = #{a : N(a) = t} = ρK t + O(t n ) as t → ∞

1 Applying Lemma above with σ = 1− n , get what we desired.

注记. Since the Riemann ζ-function has a simple pole at s = 1 with residue 1, we obtain the analytic class number formula

∏ 2r(2π)sh R L(χ, 1) = √ F F ω |∆(F )| 1≠ χ∈X(F ) F

Hecke proved that ζK (z) extends to a meromorphic function on C with no poles other than the simple pole at z = 1, and it satisfies a functional equation. If we define the gamma factors ( ) −z/2 z −z ΓR(z) := π Γ , ΓC(z) := (2π) Γ(z) 2 and the completed zeta function

z/2 r s ξK (z) := |∆K | ΓR(z) ΓC(z) ζK (z) then ξK (z) is holomorphic except for simple poles at z = 0, 1 and satisfies the functional equation

ξK (z) = ξK (1 − z) 6.4. THE ANALYTIC CLASS NUMBER FORMULA 209

In the case K = Q, we have r = 1 and s = 0, so

z/2 ξQ(z) = ΓR(z)ζ(z) = π ΓQ(z) which is precisely the completed zeta function Z(z) we defined for the Riemann zeta function ζ(z) = ζQ(z) (without any extra factors to remove the zeros at z = 0, 1). 210 CHAPTER 6. ZETA AND L-FUNCTIONS

6.5 Dirichlet density and Polar density

(Convention in this section): For functions f(s) and g(s), we define an equivalence relation by writing f(s) ∼ g(s) when

f(s) lim = 1 s→1+ g(s)

Let K be a global field and let S be a set of non-archimedean places of K.

The Dirichlet density

例 6.5.1 (Important). Replacing the (Dedekind) zeta function with its Euler product, and applying log to the Euler expansion, we see that ∑ − 1 p s ∼ log 1 − s p

Similarly, let K be a number field, ∑ − 1 N(p) s ∼ log 1 − s p

♦ 定义 6.5.1 A set S of rational primes has Dirichlet density d(S) if ∑ − 1 p s ∼ d(S) log 1 − s p∈S

注记. We can generalize the Definition, and say a set S of prime ideals of OK has Dirichlet density d(S) if ∑ − 1 N(p) s ∼ d(S) log 1 − s p∈S Now we have ∑ ∑ (p)−s (p)−s ∑p∈S N p∈S N • Dirichlet density: d(S) := lim − = lim 1 s→1+ s s→1+ p N(p) log s−1 #{p ∈ S : N(p) ⩽ x} • natural density: δ(S) := lim x→∞ #{p : N(p) ⩽ x} If S has a natural density then it has a Dirichlet density and the two coincide. 6.5. DIRICHLET DENSITY AND POLAR DENSITY 211

Dirichlet’s Theorem on Arithmetic Progressions

Let us fix a positive number m. We are interested in the set Pa of prime numbers in {a + nm : n ∈ Z>0} for a relatively prime to m.

♦ 定理 6.5.1. Dirichlet’s Theorem on Arithmetic Progressions

1 The density d(Pa) exists and is equal to ϕ(m) , i.e. the densities of the sets

Pa and Pb are the same for any a and b relatively prime to m.

Proof. For each Dirichlet character χ modulo m, define

∑ χ(p) f (s) := χ ps p∤m

For χ ≠ 1, Theorem 6.3.4 implies that fχ(s) is bounded near s = 1. Let D(m) denote the set of Dirichlet characters of modulus m. We have ∑ ∑ ∑ χ(a−1p) χ(a)−1f (s) = χ ps χ∈D(m) p∤m χ∈D(m) By the theory of group character, one gets

∑ χ(a−1p) = ϕ(m) ⇐⇒ p ≡ a mod m ps χ∈D(m) and equals 0 otherwise, so the above sum becomes

∑ ∑ ∑ χ(a−1p) ∑ χ(a)−1f (s) = = ϕ(m) p−s χ ps χ∈D(m) p≡a mod m χ∈D(m) p≡a mod m

But for any χ ≠ 1, the term indexed by χ in the above sum is bounded near s = 1, so ∑ 1 p−s ∼ f (s) ϕ(m) 1 p≡a mod m 1 ∑ = p−s ϕ(m) p∤m 1 ∑ ∼ p−s ϕ(m) p 1 1 ∼ log ϕ(m) 1 − s 212 CHAPTER 6. ZETA AND L-FUNCTIONS

The Polar density

The polar density can be used to prove the surjectivity of the Artin map for finite abelian extensions L|K of number fields.

♦ 定义 6.5.2 Let K be a number field and S be a set of primes of K. The partial Dedekind zeta function associated to S is the complex function ∏ −s −1 ζK,S(s) := (1 − N(p) ) p∈S

which converges to a holomorphic function on Re(s) > 1.

注记. 1. If S is finite then ζK,S(s) is certainly holomorphic (and nonzero) on a neighborhood of 1.

2. If S contains all but finitely many primes of K then it differs from ζK (s) by a holomorphic factor and therefore extends to a meromorphic function with a simple pole at s = 1.

Generally, ζK,S(s) may not extend to a function that is meromorphic on a neighborhood of 1, but if it does, then we can use the order of the pole at 1 (or the absence of a pole) to measure the density of S.

♦ 定义 6.5.3 ⩾ n If for some integer n 1 the function ζK,S(s) extends to a meromorphic function on a neighborhood of 1, the polar density of S is defined by m ρ(S) := , m = −ord ζn (s) n s=1 K,S

n1 n2 注记. If ζK,S(s) and ζK,S(s) both extend to a meromorphic function on a neighborhood of 1 then we necessarily have

n1 n1n2 n2 n2ords=1ζK,S(s) = ords=1ζK,S (s) = n1ords=1ζK,S(s) which implies that ρ(S) does not depend on the choice of n.

♦ 命题 6.5.1 If S has a polar density then it has a Dirichlet density and the two are 6.5. DIRICHLET DENSITY AND POLAR DENSITY 213

equal. In particular, ρ(S) ∈ [0, 1] whenever it is defined.

m Proof. Suppose S has polar density ρ(S) = n . By taking the Laurent series ex- n pansion of ζK,S(s) at s = 1 and factoring out the leading nonzero term we can write ( ) a ∑ ζn (s) = 1 + a (s − 1)n K,S (s − 1)m n n>1 × for some a ∈ C . We must have a ∈ R>0, since ζK,S(s) ∈ R>0 for s ∈ R>1 and m n therefore lim (s−1) ζK,S(s) is a positive real number. Taking logs of both sides s→1+ yields ∑ − 1 n N(p) s ∼ m log (as s → 1+) s − 1 p∈S m which implies that S has Dirichlet density d(S) = n .

推论. If S has both a polar density and a natural density then the two coincide.

注记. Not every set of primes with a natural density has a polar density, since the later is always a rational number while the former need not be.

Recall that a degree-1 prime in a number field K is a prime with residue field degree 1 over Q, equivalently, a prime p whose absolute norm N(p) = [OK : p] =

#Fp is prime.

引理 6.5.1. Let S and T denote sets of primes in a number field K, let P be the set of all primes, and let P1 be the set of degree-1 primes of K. The following hold: 1. If S is finite then ρ(S) = 0; if P−S is finite then ρ(S) = 1. 2. If S ⊆ T both have polar densities, then ρ(S) ⩽ ρ(T ). 3. If two sets S and T have finite intersection, and any two of the sets S, T and S ∪T have polar densities then so does the third and ρ(S ∪T ) = ρ(S)+ρ(T ).

4. ρ(P1) = 1, and ρ(S ∩ P1) = ρ(S) whenever S has a polar density.

Proof. We first note that for any finite set S, the function ζK,S(s) is a finite product of nonvanishing entire functions and therefore holomorphic and nonzero every- where (including at s = 1). If the symmetric difference of S and T is finite, then

ζK,S(s)f(s) = ζK,T (s)g(s) for some nonvanishing functions f(s) and g(s) holomorphic on C. Thus if S and T differ by a finite set, then ρ(S) = ρ(T ) whenever either set has a polar density.

Part 1 follows, since ρ(∅) = 0 and ρ(P) = 1 (note that ζK,P (s) = ζK (s), and ords=1ζK (s) = −1). 214 CHAPTER 6. ZETA AND L-FUNCTIONS

Part 2 follows from the analogous statement for Dirichlet density. For 3 we may assume S and T are disjoint (by the argument above), in which n n n ⩾ case ζK,S∪T (s) = ζK,S(s) = ζK,T (s) for all n 1, and the claim follows.

For 4, let P2 := P\P1. For each rational prime p there are at most n := [K : Q] 2 (in fact n/2) primes p|p in P2, each of which has absolute norm N(p) ⩾ p . It n follows by comparison with ζ(2s) that the product defining ζK,P2 (s) converges absolutely to a holomorphic function on Re(s) > 1/2 and is therefore holomorphic (and nonvanishing, since it is an Euler product) on a neighborhood of 1; thus

ρ(P2) = 0. We therefore have ρ(S ∩ P2) = 0, so ρ(S) = ρ(S ∩ P1) whenever ρ(S) exists, by 3.

For a finite Galois extension of number fields L|K, let Spl(L|K) denote the set of primes of K that split completely in L. When K is clear from context we may just write Spl(L).

♦ 定理 6.5.2 Let L|K be a Galois extension of number fields of degree n. Then 1 ρ(Spl(L)) = n

Proof. Let S be the set of degree-1 primes of K that split completely in L; it suffices to show ρ(S) = 1/n by the lemma above.

Recall that p splits completely in L iff both the ramification index ep and residue field degree fp are equal to 1. Let T be the set of primes q of L that lie above some p ∈ S. For each q ∈ T lying above p ∈ S we have NL|K (q) = p, so

N(q) = N(NL|K (q)) = N(p), thus q is a degree-1 prime, since p is. On the other hand, if q is any unramified degree-1 prime of L and p = q ∩

fp OK , then N(q) = N(NL|K (q)) = N(p ) is prime, so we must have fp = 1, and ep = 1 since q is unramified, which implies that p is a degree-1 prime that splits completely in L and is thus an element of S. Only finitely many primes ramify, so all but finitely many of the degree-1 primes in L lie in T , thus ρ(T ) = 1. Each p ∈ S has exactly n primes q ∈ T lying above it (since p splits completely), and we have ∏ ∏ ∏ −s −1 −s −1 −s −n n ζL,T (s) = (1−N(q) ) = (1−N(NL|K (q)) ) = (1−N(p) ) = ζK,S(s) q∈T q∈T p∈S

1 1 It follows that ρ(S) = n ρ(T ) = n as desired. 6.5. DIRICHLET DENSITY AND POLAR DENSITY 215

推论. If L|K is a finite extension of number fields with Galois closure M|K of degree n, then ρ(Spl(L)) = ρ(Spl(M)) = 1/n.

Proof. A prime p of K splits completely in L iff it splits completely in all the conjugates of L in M; the Galois closure M is the compositum of the conjugates of L, so p splits completely in L iff it splits completely in M.

推论. Let L|K be a finite Galois extension of number fields andlet H be a normal subgroup of Gal(L|K). The set S of primes for which Frobp ⊆ H has polar density ρ(S) = #H #G .

Proof. Let F = LH ; then F |K is Galois and Gal(F |K) ≃ G/H. For each unramified prime p of K, the Frobenius class Frobp lies in H iff every σq ∈ Frobp acts trivially on LH = F , which occurs iff p splits completely in F . By the theorem above, the 1 = #H density of this set of primes is [F :K] #G .

If S and T are sets of primes whose symmetric difference is finite, then either ρ(S) = ρ(T ) or neither set has a polar density. Let us write S ∼ T to indicate that two sets of primes have finite symmetric difference (this is clearly an equivalence relation), and partially order sets of primes by defining S ≼ T ⇔ S ∼ S ∩ T (in other words, S\T is finite). If S and T have polar densities, then S ≼ T implies ρ(S) ⩽ ρ(T ).

♦ 定理 6.5.3 If L|K and M|K are two finite Galois extensions of number fields then

L ⊆ M ⇐⇒ Spl(M) ≼ Spl(L) ⇐⇒ Spl(M) ⊆ Spl(L) L = M ⇐⇒ Spl(M) ∼ Spl(L) ⇐⇒ Spl(M) = Spl(L)

and the map L 7→ Spl(L) is an injection from the set of finite Galois extensions of K (inside some fixed algebraic closure) to sets of primes of K that have a positive polar density.

Proof. The implications L ⊆ M =⇒ Spl(M) ≼ Spl(L) =⇒ Spl(M) ⊆ Spl(L) are clear, so it suffices to show that( Spl M) ⊆ Spl(L) =⇒ L ⊆ M. A prime p of K splits completely in the compositum LM iff it splits completely in both L and M: the forward implication is clear and for the reverse, note that if p splits completely in both L and M then it certainly splits completely in L ∩ M, so we may assume K = L ∩ M; we then have Gal(LM|K)s ≃ Gal(L|K) × 216 CHAPTER 6. ZETA AND L-FUNCTIONS

Gal(M|K), and if the decomposition subgroups of all primes above p are trivial in both Gal(L|K) and Gal(M|K) then the same applies in Gal(LM|K). Thus Spl(LM) = Spl(L) ∩ Spl(M). It follows that Spl(M) ≼ Spl(L) =⇒ Spl(LM) ∼ Spl(M). By Theorem 6.5.2, we have ρ(Spl(M)) = 1/[M : K] and ρ(Spl(LM) = 1/[LM : K], thus Spl(LM) ∼ Spl(M) implies

[LM : K] = ρ(Spl(LM)) = ρ(Spl(M)) = [M : K] in which case LM = M and L ⊆ M. This proves Spl(M) ∼ Spl(L) =⇒ L ⊆ M, so the three conditions in the first line of biconditionals are all equivalent, and this immediately implies the second line of biconditionals. The last statement of the theorem is clear, since Spl(L) has positive polar density, by Theorem 6.5.2. 第 7 章 Class field theory

7.1 *Ray class groups and ray class fields

Ray class fields and Artin reciprocity

As a special case of Corollary 6.5, if F |K is a finite extension of number fields in which all but finitely many primes split completely, then [F : K] = 1 and therefore F = K. We will use this fact to prove that the Artin map is surjective.

♦ 定理 7.1.1 Let L|K be a finite abelian extension of number fields andlet m be a mod- 7 ulus for K that is divisible by all primes of K that ramify in L. Then the Artin m F m → | map ψL|K : K Gal(L K) is surjective.

⊆ | m H Proof. Let H Gal(L K) be the image of ψL|K and let F = L be its fixed field, ∈ F m which we note is a Galois extension of K. For each prime p K the automor- m ∈ ∈ m phism ψL|K (p) H acts trivially on F ; therefore p ker ψF |K (p) and p splits F m completely in F . The group K contains all but finitely many primes p of K, so the polar density of the set of primes of K that split completely in F is 1. Thus [F : K] = 1 and H = Gal(L|K).

m We now show that the kernel of the Artin map ψL|K uniquely determines the field L. ♦ 定理 7.1.2 Let m be a modulus for a number field K and let L and M be finite abelian 218 CHAPTER 7. CLASS FIELD THEORY

m extensions of K unramified at all primes not in the support of m. If ker ψL|K = m ker ψM|K then L = M.

Proof. Let S be the set of primes of K that do not divide m. Each prime p in S is unramified in both L and M, and p splits completely in L (resp. M) iff it lies in the m m m m kernel of ψL|K (resp. ψM|K ). If ker ψL|K = ker ψM|K then

∼ ∩ m ∩ m ∼ Spl(L) (S ker ψL|K ) = (S ker ψM|K ) Spl(M) and therefore L = M, by Theorem 6.5.3.

注记. Ray class fields are unique whenever they exist, which denoted by K(m).

Thus we have an exact sequence

→ m → F m → | → 1 ker ψL|K K Gal(L K) 1

One of the key results of class field theory is that for a suitable choice ofthe Pm ⊆ m modulus m, we have K ker ψL|K . This implies that the Artin map induces | m an isomorphism between Gal(L K) and a quotient of the ray class group ClK = F m Pm K / K . When L is the ray class field for the modulus m, the Artin map allows m ≃ | us to relate subfields of L to quotients of the ray class group ClK Gal(L K) in a way that we will make more precise later; this is known as Artin reciprocity.

The conductor of an abelian extension

We now introduce another notion of conductor, one attached to an abelian extension of number fields, which is defined as a product of local conductors attached to corresponding abelian extensions of the local field Kv for each place v ∈ MK . Let L|K be a finite abelian extension of local fields. The conductor c(L|K) is defined as follows.

• If K is archimedean then c(L|K) = 1 when K ≃ R and L ≃ C and c(L|K) = 0 otherwise.

• If K is nonarchimedean and p is the maximal ideal of its valuation ring OK , n × then c(L|K) := min{n : 1 + p ⊆ NL|K (L )}

n O× 0 O× 注记. Here 1 + p is a subgroup of K , with 1 + p := K . 7.1. *RAY CLASS GROUPS AND RAY CLASS FIELDS 219

♦ 定义 7.1.1 If L|K is a finite abelian extension of global fields then its conductor isthe modulus

c(L|K): MK → Z

v 7→ c(Lw|Kv)

where w|v is an extension of place.

注记. 1. The fact that L|K is Galois ensures that c(Lw|Kv) is the same for every w|v.

2. As with any modulus, we may view the finite part of c(L|K) as an OK -ideal and the infinite part as a subset of ramified infinite places.

♦ 命题 7.1.1 Let L|K be a finite abelian extension. For each prime p of K we have   0 iff p is unramified | vp(c(L K)) = 1 iff p is tamely ramified   ⩾ 2 iff p is tamely ramified

The finite part of the conductor of an abelian extension divides the discriminant ideal and is divisible by the same set of primes, but the valuation of the conductor at these primes is typically smaller than that of the discriminant. For p−2 example, the discriminant of the extension Q(ζp)|Q is (p) , but its conductor is (p)∞.

引理 7.1.1. Let L1|K and L2|K be two finite abelian extensions of a local or global field K. If L1 ⊆ L2 then c(L1|K) divides c(L2|K).

Proof. If K ≃ R, C the result is clear, and for nonarchimedean local K we may × × ⊆ × apply NL2|K (L2 ) = NL1|K (NL2|K (L2 )) NL1|K (L1 ). The global case follows.

Ray class characters

We will now prove a generalization of Dirichlet’s theorem on primes in arithmetic progressions.

Let K be a number field and let χ : FK → C be a totally multiplicative function with finite image; so χ(OK ) = 1, χ(IJ) = χ(I)χ(J) for all I,J ∈ FK , and χ 220 CHAPTER 7. CLASS FIELD THEORY restricts to a homomorphism from a subgroup of FK to a finite subgroup of U(1) whose kernel we denote ker χ.

♦ 定义 7.1.2 −1 F m Pm ⊆ If m is a modulus for K such that χ (U(1)) = K and K ker χ, then χ is a ray class character of modulus m = m(χ) and its kernel is a congruence subgroup of modulus m. Equivalently, χ is the extension by zero of a character m ∈ F m of the finite abelian group ClK defined by setting χ(I) = 0 for I / K .

例 7.1.1. For K = Q there is a one-to-one correspondence between Dirichlet char- ′ ′ acters χ : Z → C and ray class characters χ : FQ → C with χ(a) = χ ((a)) for all a ∈ Z⩾1. Each Dirichlet character χ of modulus m corresponds to a ray class character of modulus m = (m)∞ whose conductor divides (m) iff χ is an even Dirichlet character, meaning that χ(−1) = 1.

Let χ1, χ2 be ray class characters of moduli m1, m2 of a number field K, with | ∈ F m2 m1 m2. If χ2(I) = χ1(I) for all I K , then χ2 is induced by χ1. ♦ 定义 7.1.3 1. A ray class character is primitive if it is not induced by any ray class character other than itself. 2. The conductor of a ray class character χ is the conductor c = c(χ) of its kernel (as a congruence subgroup).

♦ 定理 7.1.3 A ray class character is primitive iff its kernel is primitive, equivalently, iff its modulus is equal to its conductor. Every ray class character χ is induced by a unique primitive ray class character χ˜.

F m → Proof. Let χ be a ray class character of modulus m, let κ : K /(ker χ) U(1) be the group character induced by χ, and let C be the primitive congruence subgroup equivalent to ker χ with modulus c = c(χ) dividing m given by Corollary 5.5. We F c C −−→Fsim m have a canonical isomorphism φ : K / K /(ker χ) that we can use to define a ray class character χ˜ of modulus c as the extension by zero of the character κ ◦ φ F c C F m ⊆ F c of K / . The isomorphism φ preserves cosets of fractional ideals in K K , so ∈ F m χ˜(I) = χ(I) for all I K and χ is induced by χ˜.

If χ2 is a ray class character of conductor m2 induced by a ray class character ∩ F m2 ∩ F m1 ∼ χ1 of conductor m1, then ker χ1 K = ker χ2 = ker χ2 K and ker χ1 ker χ2, 7.1. *RAY CLASS GROUPS AND RAY CLASS FIELDS 221

m1 m2 and we also note that if ker χ1 ≠ ker χ2 then F ≠ F and m1 ≠ m2. It follows that χ˜ is primitive, it is the unique primitive ray class character that induces χ. Thus χ is primitive iff it is equal to χ˜, which holds iff ker χ = ker χ˜ is primitive, equivalently, iff m(χ) = m(˜χ) = c(˜χ) = c(χ).

For a modulus m of K we use X(m) to denote the set of primitive ray class characters of conductor dividing m, which we note is in bijection with the charac- m ter group of ClK , and thus has a group structure given by χ˜1χ˜2 =χ1 ˜χ2. Indeed, for m each character of ClK , its extension by zero is a ray class character χ of modulus m induced by a primitive ray class character χ˜ whose conductor divides m, and each primitive ray class character χ˜ of conductor dividing m induces a ray class m character χ of modulus m that determines a character of ClK ; these two maps are inverses, hence bijections.

♦ 定义 7.1.4 If ker χ = χ−1(U(1)) then χ is principal, and we use I to denote the unique primitive principal ray class character.

For Dirichlet characters, I is also the unique Dirichlet character with conductor 1, but for ray class characters this holds only when the class group ClK is trivial.

In general, the extension by zero of any character of ClK is a ray class character of conductor (1) which need not be principal but is necessarily primitive. Like Dirichlet characters, each ray class character has an associated L-function.

♦ 定义 7.1.5 The Weber L-function L(s, χ) of a ray class character χ for a number field K is the function ∏ ( ) ∑ − −1 − L(s, χ) := 1 − χ(p)N(p) s = χ(a)N(a) s p a

where the the product is over prime ideals of OK and the sum is over

nonzero OK -ideals.

注记. 1. The product and sum both converge to a non-vanishing holomorphic function on Re(s) > 1, this follows from comparison with the Dedekind zeta

function ζK (s), since |χ(a)| ⩽ 1. 2. For K = Q a Weber L-function is the same thing as a Dirichlet L-function.

For any number field K, we have L(s, I) = ζK (s). 222 CHAPTER 7. CLASS FIELD THEORY

If C is a congruence subgroup of modulus m, then the Dirichlet characters of modulus m whose kernels contains C form a group under pointwise multiplica- F m C tion corresponding to the character group of K / , and there is a corresponding subgroup of X(m) consisting of primitive Dirichlet characters whose kernels contain C. ♦ 命题 7.1.2 Let χ be a ray class character of modulus m for a number field K of degree 1 n. Then L(s, χ) extends to a meromorphic function on Re(s) > 1− n that has at most a simple pole at s = 1 and is holomorphic if χ is non-principal.

∈ m Proof. Associated to each ray class γ ClK we have a partial Dedekind zeta function ∏ −s −1 ζK,γ(s) := (1 − N(p) ) p∈γ that is holomorphic on ℜ(s) > 1. For the trivial modulus m, our proof of analytic class number formula immediately implies that ζK,γ(s) has a meromorphic con- 1 tinuation to 1− n with a simple pole at s = 1 that has the same residue ρ as the

Dedekind zeta function ζK (s).....

For a congruence subgroup C, let X(C) denote the set of primitive Dirichlet characters whose kernels contain C. If C is a congruence subgroup of modulus m C F m C then X( ) is a subgroup of X(m) isomorphic to the character group of K / and C F m C we may view X( ) as the the character group of K / .

(Convention in this subsection): Let C be a congruence subgroup of F m C modulus m for a number field K and let n := [ K : ].

♦ 定理 7.1.4 The set of primes S := {p ∈ C} has Dirichlet densit   1 ̸ ∀ ̸ I C n if L(1, χ) = 0 χ = in X( ) d(S) =  0 otherwise.

∈ F m { ∈ C} 推论. For every I K the set S := p I has Dirichlet density   1 ̸ ∀ ̸ I C n if L(1, χ) = 0 χ = in X( ) d(S) =  0 otherwise. 7.1. *RAY CLASS GROUPS AND RAY CLASS FIELDS 223

推论. Let L|K be an abelian extension of number fields. If Spl(L) ≼ {p ∈ C} then

F m C ⩽ [ K : ] [L : K] and if Spl(L) ∼ {p ∈ C} then equality holds. 224 CHAPTER 7. CLASS FIELD THEORY

7.2 *Class field theory: ideal-theoretic form

Let m be a modulus for a number field K. Assuming the ray class field m F m → K(m) exists, it follows from the surjectivity of the Artin map ψK(m)|K : K Gal(K(m)|K) proved in Theorem 7.1.1 that we have a canonical isomorphism

m F m Pm ≃ | ClK = K / K Gal(K(m) K) between the ray class group and the Galois group of the ray class field. More generally, if L is any intermediate field between K and K(m), the kernel of the m C ⊆ F m Pm ⊆ C ⊆ Artin map ker ψL|K is a subgroup K that contains the ray group: K F m K (Definition 5.5.2), and we have an isomorphism

F m C ≃ m C ≃ | K / ClK / Gal(L K)

To prove that a given abelian extension L|K lies in a ray class field, it is enough m Pm to show that there exists a modulus m for K such that ker ψL|K contains K , since we then have Spl(K(m)) ≼ Spl(L) and therefore L ⊆ K(m). This is the other half of Artin reciprocity (along with the surjectivity). We want to better understand the structure of congruence subgroups, and to specify a minimal modulus m for which we should expect a given finite abelian extension L|K to lie in a subfield of the ray class field K(m); this minimal modulus is know as the conductor of the extension. So far we have not addressed this question even for K = Q; our proof of the Kronecker-Weber theorem showed that every abelian extension lies in some cyclotomic field Q(ζm), but we made no attempt to determine such an integer m (or more precisely, a modulus m of the form m = (m)∞ or m = (m)).

The main theorems of class field theory

Artin reciprocity gives us a commutative diagram of canonical bijections: 7.2. *CLASS FIELD THEORY: IDEAL-THEORETIC FORM 225

abelian L|K with congruence C ⊆ F m c(L|K)|m subgroup K

quotient of m quotient of ClK Gal(K(m)|K)

Norm groups

(Convention in this subsection): Let L|K be a finite abelian extension of number fields and let m be a modulus for K divisible by the conductor of L|K.

♦ 定义 7.2.1 The norm group (or Takagi group) associated to m is the congruence subgroup m Pm F m TL|K := K NL|K ( L ) F m F where L denotes the subgroup of fractional ideals in L that are coprime

to mOL.

m ⊆ m 注记. ker ψL|K TL|K .

To prove Artin reciprocity we need to establish the reverse inclusion, which requires a different approach. But we can record the following theorem, histori- cally known as the second fundamental inequality of class field theory (in modern terminology it is typically known as the second, even though it was proved first, by Weber).

♦ 定理 7.2.1. second fundamental inequality Let L|K be a Galois extension of number fields and let m be a modulus for K divisible by the conductor of L|K. Then

F m m ⩽ [ K : TL|K ] [L : K]

The Hilbert class field 226 CHAPTER 7. CLASS FIELD THEORY

♦ 定义 7.2.2 Let K be global field. The Hilbert class field of K is the maximal unramified abelian extension of K (the compositum of all finite unramified abelian extensions of K inside a fixed separable closure of K).

While it is not obvious from the definition, it follows from the completeness theorem of class field theory that the Hilbert class field must be the ray class field for the trivial modulus, and in particular, that it is a finite extension of K. This is a remarkable result, since infinite unramified extensions of number fields doexist (they are necessarily nonabelian). Indeed, one way to construct such an extension is by considering a tower of

Hilbert class fields. Starting with a number field K0 := K, for each integer n ⩾ 0 define Kn+1 to be the be the Hilbert class field of Kn. This yields an infinite tower of finite abelian extensions

K0 ⊆ K1 ⊆ K2 ⊆ · · · ∪ and we may then consider the field L := n Kn. There are two possibilities: either we eventually reach a field Kn with class number 1, in which case Km = Kn for all m ⩾ n and L|K is a finite unramified extension of K, or this does not happen and L|K is an infinite unramified extension of K (which is necessarily nonabelian). It was a longstanding open question as to whether the latter could occur, but in 1964 Golod and Shafarevich proved that indeed it can; in particular, the field √ √ K0 = Q( −30030) = Q( −2 · 3 · 5 · 7 · 11 · 13) is the base of an infinite tower of Hilbert class field extensions. 7.3. *LOCAL CLASS FIELD THEORY 227

7.3 *Local class field theory

The goal of local class field theory is to classify all finite abelian extensions of a given local field K. Note that we thus have a tower of field extensions

K ⊆ Kunr ⊆ Kab ⊆ Ksep

Proof. This is obvious in the archimedean case, where we have K = Kunr and Ksep = C since C|R is ramified. In the nonarchimedean case, Kunr is isomorphic to the algebraic closure of the residue field of K, which is an abelian extension because it is pro-cyclic (every finite extension of the residue field is cyclic because the residue field isfinite).

Local Artin reciprocity

Local class field theory is based on the existence of a continuous homomorphism × ab θK : K → Gal(K |K) known as the local Artin homomorphism (or local reciprocity map), which is described by the following theorem.

♦ 定理 7.3.1. Local Artin reciprocity Let K be a local field. There is a unique continuous homomorphism

× ab θK : K → Gal(K |K)

with the property that for each finite extension L|K in Kab, the homomorphism

× θL|K : K → Gal(L|K)

ab given by composing θK with the natural map Gal(K |K) ↠ Gal(L|K) satisfies:

• if K is nonarchimedean and L|K is unramified then θL|K (π) = FrobL|K

for every uniformizer π of OK ;

× × × • θL|K is surjective with kernel NL|K (L ), inducing K /NL|K (L ) ≃ Gal(L|K).

It is first worth contrasting local Artin reciprocity with the more complicated global version of Artin reciprocity:

• There is no modulus m; working in Kab addresses all abelian extensions of K at once.

m × • The ray class groups ClK are replaced by quotients of K .

F m Pm ⊆ F m × ⊆ × • The Takagi group NL|K ( L ) K K is replaced by NL|K (L ) K .

Norm groups

♦ 定义 7.3.1 A norm group of a local field K is a subgroup of the form

× × × N(L ) := NL|K (L ) ⊆ K

for some finite abelian extension L|K.

注记. Removing the word abelian does not change the definition above. If L|K is any finite extension (not necessarily Galois), thenN(L×) = N(F ×), where F is the maximal abelian extension of K in L; this result is known as the Norm Limitation Theorem. So we could have defined norm groups more generally. This is not relevant to classifying the abelian extension of K, but it demonstrates a key limitation of local class field theory (which extends to global class field theory): norm groups tell us nothing about nonabelian extensions of K.

Local Artin reciprocity implies that the Galois group of any finite abelian ex- × × tension L|K of a local fields is canonically isomorphic to the quotient K /NL|K (L ). In order to understand the finite abelian extensions of a local field K, we just need to understand its norm groups.

推论. The map L 7→ N(L×) defines an inclusion reversing bijection between the finite abelian extensions L|K in Kab and the norm groups in K× which satisfies

× × ∩ × N((L1L2) ) = N(L1 ) N(L2 ) ∩ × × × N((L1 L2) ) = N(L1 )N(L2 ) 7.3. *LOCAL CLASS FIELD THEORY 229

In particular, every norm group of K has finite index in K×, and every subgroup of K× that contains a norm group is a norm group.

引理 7.3.1. Let L|K be any extension of local fields. If N(L×) has finite index in K× then it is open.

注记. If K is a local field of characteristic zero then one can show that in fact every finite index subgroup of K× is open (whether it is a norm group or not), but this is not true in positive characteristic.

The main theorems of local class field theory

Corollary 7.3 implies that all norm groups of K have finite index in K×, and Lemma 7.3.1 then implies that all norm groups are finite index open subgroups of K×. The existence theorem of local class field theory states that the converse also holds. ♦ 定理 7.3.2. Local Existence Theorem Let K be a local field and let H be a finite index open subgroup of K×. ab × There is a unique extension L|K in K with NL|K (L ) = H.

× ab The local Artin homomorphism θK : K → Gal(K |K) is not an isomorphism; indeed, it cannot be, because Gal(Kab|K) is compact and K× is not. How- ever, the local existence theorem implies that after taking profinite completions the local Artin homomorphism becomes an isomorphism.

♦ 定理 7.3.3. Main Theorem of Local Class Field Theory Let K be a local field. The local Artin homomorphism induces a canonical isomorphism ∼ c d× ab θK : K −→ Gal(K |K)

of profinite groups.

In view of this Theorem, we would like to better understand the profinite d× d× group K . If K is archimedean then K is either trivial or the cyclic group of order 2, so let us assume that K is nonarchimedean. If we pick a uniformizer π for × the maximal ideal p of OK , then we can uniquely write each x ∈ K in the form v(x) ∈ O× ∈ Z uπ , with u K and v(x) . This defines an isomorphism ∼ K× −→O× × Z ( K ) x x 7→ , v(x) πv(x) 230 CHAPTER 7. CLASS FIELD THEORY

Taking profinite completions (which commutes with products), we obtain an isomorphism d× ≃ O× × Zb K K since the unit group

O× ≃ F× × O ≃ F× × O n K p K p lim←− K /p n d× ≃ O× ×Zb is already profinite. Note that the isomorphism K K is far from canonical; it depends on our choice of π, and there are uncountably many π to choose from. We have a commutative diagram of exact sequences of topological groups

O× × v 1 K K Z 0

≃ θK ϕ

1 Gal(Kab|Kunr) Gal(Kab|K) Gal(Kunr|K) 1 in which the bottom row is the profinite completion of the top row. The map ϕ on the right is given by

注记. . The Frobenius element ϕ(1) ∈ Gal(Kunr|K) corresponds to the Frobenius

#Fp automorphism x 7→ x of Gal(Fp|Fp); both are canonical topological generators of the Galois groups in which they reside, and both are sometimes referred to as the arithmetic Frobenius. There is another obvious generator for Gal(Kunr|K) ≃

Gal(Fp|Fp), namely ϕ(−1), which is called the geometric Frobenius (for reasons we won’t explain here).

ab| unr ≃ O× ab| The group Gal(K K ) K corresponds to the inertia subgroup of Gal(K K). The top sequence splits (but not canonically), hence so does the bottom, and we have ab| ≃ ab| unr × unr| ≃ O× × Zb Gal(K K) Gal(K K ) Gal(K K) K 7.3. *LOCAL CLASS FIELD THEORY 231

ab unr For each choice of a uniformizer π ∈ OK we get a decomposition K = KπK × O× Z ab ∈ corresponding to K = K π . The field Kπ is the subfield of K fixed by θK (π) ab Gal(K |K). Equivalently, Kπ is the compositum of all the totally ramified finite extensions L|K in Kab for which π ∈ N(L×).

ab unr 例 7.3.1. Let K = Qp and pick π = p. The decomposition K = KπK is ∪ ∪ Qab Q · Q p = p(ζpn ) p(ζm) n m|p where the first union on the RHS is fixed by θK (p) and the second is fixed by O× θK ( K ).

Constructing the local Artin homomorphism is the difficult part of local class field theory. However, assuming the local existence theorem, it is easy toshow that the local Artin homomorphism is unique if it exists.

♦ 命题 7.3.1 Let K be a local field and assume every finite index open subgroup of K× is a norm group. There is at most one homomorphism θ : K× → Gal(Kab|K) of topological groups that has the properties given in Local Artin Reciprocity.

注记. One approach to proving local class field theory uses the theory of for- ∪ mal groups due to Lubin and Tate to explicitly construct the fields Kπ = n Kπ,n used in the proof of Proposition above, along with a continuous homomorphism O× → | θπ : K Gal(Kπ K) that extends uniquely to a continuous homomorphism × unr ab unr : K → Gal(KπK |K). One then shows that K = KπK (using the Hasse- Arf Theorem), and that θ does not depend on the choice of π.

Finite abelian extensions

Local class field theory gives us canonical bijections between the following sets:

(1) finite-index open subgroups of K× (which are necessarily normal);

(2) open subgroups of Gal(Kab|K) (which are necessarily normal and of finite index);

(3) finite extensions of K in Kab (which are necessarily normal). 232 CHAPTER 7. CLASS FIELD THEORY

d× The bijection from (1) to (2) is induced by the isomorphism K ≃ Gal(Kab|K) given by Theorem 7.3.3 and is inclusion preserving. The bijection from (2) to (3) follows from Galois theory (for infinite extensions), and is inclusion reversing, while the bijection from (3) to (1) is via the map L 7→ N(L×), which is also inclusion reversing. 7.4. *GLOBAL CLASS FIELD THEORY 233

7.4 *Global class field theory

(Convention in this section): Let K denote a global field. For each place sep ab v of K, we fix a separable closure Kv and a maximal abelian extension Kv

of the local field Kv.

The idele norm

♦ 定义 7.4.1 Let L|K is a finite separable extension of global fields. The idele norm

NL|K : IL → IK is defined by NL|K (bw) = (av), where each ∏

av := NLw|Kv (bw) w|v

is a product over places w of L that extend the place v of K.

× × The idele norm NL|K : IL → IK agrees with the field norm NL|K : L → K × on the subgroup of principal ideles L ⊆ IL. The field norm is also compatible with the ideal norm NL|K : FL → FK , and we have the following commutative diagram:

× L IL FL

NL|K NL|K NL|K

× K IK FK

× The image of L in IL under the composition of the maps on the top row is precisely the group PL of principal ideals. Taking quotients yields induced norm maps on the idele and ideal class groups, both of which we also denote NL|K , and we have a commutative square

CL ClL

NL|K NL|K

CK ClK 234 CHAPTER 7. CLASS FIELD THEORY

The Artin homomorphism

We now construct the global Artin homomorphism using the local Artin homomorphisms we defined in the previous section.

By the Local Artin Reciprocity, each local field Kv is equipped with a local Artin homomorphism × → ab| θKv : Kv Gal(Kv Kv) | | For each finite abelian extension L K and each place w v of L, composing θKv with ab| → | the natural map Gal(Kv Kv) Gal(Lw Kv) yields a surjective homomorphism

× → | θLw|Kv : Kv Gal(Lw Kv)

× | with kernel NLw|Kv (Lw). When Kv is nonarchimedean and Lw Kv is unramified we have θLw|Kv (πv) = FrobLw|Kv for all uniformizers πv of Kv. Note that by Proposition

1.6.6 every finite separable extension of Kv is of the form Lw for some place w|v. We now define an embedding of Galois groups

ιw : Gal(Lw|Kv) ,→ Gal(L|K)

注记. The map ιw is well defined and injective because every element of Lw can be written as lx for some l ∈ L and x ∈ Kv (any K-basis for L spans Lw as a Kv vector space), so each σ ∈ Gal(Lw|Kv) is uniquely determined by its action on L, which fixes K ⊆ Kv.

If v is archimedean then ιw(Gal(Lw|Kv)) is either trivial or generated by the involution corresponding to complex conjugation in Lw ≃ C.

If v is a finite place and q is the prime of L corresponding to w|v, then ιw(Gal(Lw|Kv)) is the decomposition group Dq ⊆ Gal(L|K).

注记. More generally, for any place v of K, the Galois group Gal(L|K) acts on the set {w|v}, and ιw(Gal(Lw|Kv)) is the stabilizer of w under this action. It thus makes sense to call ιw(Gal(Lw|Kv)) the decomposition group of the place w.

◦ × → | Moreover, the composition ιw θLw|Kv defines a map Kv Gal(L K) that is independent of the choice of w|v: this is easy to see when v is an unramified ◦ nonarchimedean place, since then ιw θLw|Kv (πv) = Frobv for every uniformizer ◦ × πv of Kv, and this determines ιw θLw|Kv since the πv generate Kv . × Now embed Kv into the idele group IK whose image intersects K ⊆ IK trivially. This embedding is compatible with the idele norm in the following sense: 7.4. *GLOBAL CLASS FIELD THEORY 235

N | × Lw Kv × Lw Kv

NL|K IL IK

Now let L|K be a finite abelian extension. For each place v of K, let us pick a place w of L extending v and define

◦ v(av) ιw θLw|Kv (av) = Frobv = 1

It is clear that θ | is a homomorphism, since each θ | is, and θ | is con- L K ∏ L∏w Kv L K × O× I tinuous because its kernel is a basic open set Uv v of K . v|DL|K v∤DL|K ⊆ | If L1 L2 are two finite abelian extensions of K, then θL1|K (x) = θL2|K (x) L1 for all x ∈ IK . The θL|K form a compatible system of homomorphisms from I | ≃ ab| K to the inverse limit lim←− Gal(L K) Gal(K K), where L ranges over finite L abelian extensions of K in Kab ordered by inclusion. By the universal property of the profinite completion, they uniquely determine a continuous homomorphism.

♦ 定义 7.4.2 Let K be a global field. The global Artin homomorphism is the continuous homomorphism

I → | ≃ ab| θK : K lim←− Gal(L K) Gal(K K) L

defined by the compatible system of homomorphisms θL|K : IK → Gal(L|K), where L ranges over finite abelian extensions of K in Kab. 236 CHAPTER 7. CLASS FIELD THEORY

♦ 命题 7.4.1

The global Artin homomorphism θK is the unique continuous homomor- ab phism IK → Gal(K |K) with the property that for every finite abelian extension L|K in Kab and every place w of L lying over a place v of K the diagram

θ × Lw|Kv | Kv Gal(Lw Kv)

ιw

θL|K IK Gal(L|K)

commutes, where the homomorphism θL|K is defined by θL|K (x) := θK (x)|L.

Proof. That 4θK has this property follows directly from its construction. Now sup- ′ I → ab| pose θK : K Gal(K K) has the same property. I × The idele group K is generated by the images of the embeddings Kv , so if θK ′ ··· ··· and θK are not identical, then they disagree at a point a := (1, 1, , 1, av, 1, 1, ) × → I in the image of one the embeddings Kv , K . We must have θL|K (a) = θLw|Kv (av) = ′ | ab θL|K (a) for every finite abelian extension L K in K and every place w of L ex- ′ tending v. This implies that θK (a) = θK (a), since the image of a under any ho- (Kab|K) ≃ (L|K) momorphism to Gal lim← Gal is determined by its images in the Gal(L|K).

The main theorems of global class field theory

In the global version of Artin reciprocity, the idele class group CK plays the × role that the multiplicative group Kv plays in local Artin reciprocity.

♦ 定理 7.4.1. Global Artin Reciprocity × The kernel of the global Artin homomorphism θK contains K , and we thus have a continuous homomorphism

ab θK : CK → Gal(K |K)

with the property that for every finite abelian extension L|K in Kab the homo-

isomorphism

CL/NL|K (CL) ≃ Gal(L|K)

注记. When K is a number field, θK is surjective but not injective; its kernel is the connected component of the identity in CK . When K is a global function field, θK is injective but not surjective; its image consists of automorphisms σ ∈ Gal(Kab|K) corresponding to integer powers of the Frobenius automorphism of Gal(ksep|k), b where k is the constant field of K (this is precisely the dense image of Z in Z ≃ Gal(ksep|k).

♦ 定理 7.4.2. Global Existence Theorem

For every finite index open subgroup H of CK there is a unique finite ab abelian extension L|K in K for which NL|K (CL) = H.

♦ 定理 7.4.3. Main theorem of global class field theory

The global Artin homomorphism θK induces a canonical isomorphism

∼ c c ab θK : CK −→ Gal(K |K)

of profinite groups.

We then have an inclusion reversing bijection

ab {finite index open subgroups H of CK } ↔ {finite abelian extensions L|K in K } H 7→ (Kab)θK (H)

NL|K (CL) ←[ L and corresponding isomorphisms CK /H ≃ Gal(L|K), where H = NL|K (CL). We also note that the global Artin homomorphism is functorial in the following sense.

♦ 定理 7.4.4. Functoriality Let L|K be any finite separable extension (not necessarily abelian). Then the following diagram commutes 238 CHAPTER 7. CLASS FIELD THEORY

θ L ab CL Gal(L |L)

NL|K res

θ K ab CK Gal(K |K)

Relation to ideal-theoretic version of global class field theory ∏ m(v) Let m = v v be a modulus for K. For each place v we define the open subgroup

 O× ∤ O× ×  v if v m, where v := Kv when v is infinite, U m(v) := R | R ⊆ R× ≃ O× × K  >0 if v m is real, where >0 v := Kv ,   m(v) 1 + p if v|m is finite, where p = {x ∈ Ov : |x|v < 1}. ∏ m m ⊆ I I and let UK := v UK (v) K denote the corresponding open subgroup of K . m m The image U K of UK in the idele class group CK is a finite index open subgroup. The idelic version of a ray class group is the quotient

m I m × m CK : K /(UK K ) = CK /U K and we have isomorphisms

m ≃ m ≃ | CK ClK Gal(K(m) K)

K(m) is the corresponding ray class field, which we can now define as the finite | m abelian extension L K for which NL|K (CL) = U K , whose existence is guaranteed by Theorem 7.4.2. | m If L K is any finite abelian extension, then NL|K (CL) contains U L for some m modulus m; this follows from the fact that the groups U L form a fundamental system of open neighborhoods of the identity. Indeed, the conductor of the extension L|K is precisely the minimal modulus m for which this is true. It follows that every finite abelian extension L|K lies in a ray class field K(m), with Gal(L|K) m isomorphic to a quotient of a ray class group CK .

The Chebotarev density theorem

If C be a union of conjugacy classes of a group G, we say C is stable under conjugation. 7.4. *GLOBAL CLASS FIELD THEORY 239

♦ 定理 7.4.5. Chebotarev density theorem Let L|K be a finite Galois extension of number fields. Let C ⊆ Gal(L|K) be stable under conjugation, and let S be the set of primes p of K unramified

in L with Frobp ⊆ C. Then d(S) = #C/#G. 240 CHAPTER 7. CLASS FIELD THEORY

7.5 Tate Cohomology

Group cohomology

Let G be a group. A G-module is a module over the group ring Z[G]. Note that we can always form a Z[G] module from any abelian group A by giving it the trivial action under Z[G], that is, ga = a for all g ∈ G and a ∈ A. These are known as trivial Z[G] modules.

注记. For our purposes we will usually consider Z as a trivial Z[G]-module.

♦ 定义 7.5.1 Let G be a group and let A be a G-module. We have the invariant subgroup AG := {a ∈ A : ga = a for all g ∈ G}

Let M be the G-module generated by elements of the from ga−a for g ∈ G and a ∈ A. Then we also have the coinvariant quotient group

AG := A/M

G 注记. − and −G are functors form the category of G-modules to the category of abelian groups.

G 引理 7.5.1. A ≃ HomZ[G](Z,A)

Proof. Let ϕa ∈ HomZ(Z,A) be the group homomorphism such that ϕa(1) = a.

Note that this completely determines ϕa. Consider the map Φ: a 7→ ϕa. Note then that Φ: A → HomZ(Z,A) is an isomorphism. ′ G Now let Φ be the restriction of Φ to A . Let ϕa ∈ HomZ[G](Z,A). Then since ′ Z[G] has trivial action on Z, for g ∈ G we have Φ (a) = ϕa(1) = ϕa(g ·1) = gϕa(1) = G ′ g · a. Thus a ∈ A so Φ surjects onto HomZ[G](Z,A).

At this point we are able to define group cohomology. Our definition arises from the theory of derived functors. It should be noted that there is an equivalent definition of group cohomology which makes no use of derived functors andin- stead relies on the familiar coboundary and cocycle constructions. We will soon use some of these ideas to give explicit descriptions of the first cohomology group. 7.5. TATE COHOMOLOGY 241

♦ 定义 7.5.2 Let A be a G-module. The homology groups of G with coefficients in A

are defined to be the left derived functors Li(−G)(A) and are written Hi(G, A). Likewise, the cohomology groups of G with coefficients in A are defined to be the right derived functors Ri(−G)(A) and are written Hi(G, A).

0 G 注记. By definition H0(G, A) = AG and H (G, A) = A .

i ≃ i Z 推论. H (G, A) ExtZ[G]( ,A).

G Proof. From Lemma above we know that the functor − is equivalent to HomZ[G](Z, −).

Let A be a (left) G-module. A crossed homomorphism or 1-cocyle is a map d : G → A which satisfies d(gh) = gd(h) + d(g). The collection of 1-cocyles will be denoted Z(G, A). It’s easy to show Z(G, A) is a group with the operation (d + d′)(g) := d(g) + d′(g). For each a ∈ A we define the principle crossed homomorphism or 0-coboundary da : G → A such that da(g) = ga−a. The set of 0-coboundaries will be denoted

B(G, A). It should be noted that da is a 1-cocycle for each a. Furthermore B(G, A) is a subgroup of Z(G, A).

♦ 定义 7.5.3 In the group ring Z[G] we have the augmentation map ϵ : Z[G] → Z given by ( ) ∑ ∑ ϵ nigi := ni i i This gives the exact sequence

0 → IG → Z[G] → Z → 0

where the kernel IG is the augmentation ideal of Z[G].

注记. 1. IG is clearly generated by elements of the form g−gi with g ≠ gi. Multi- −1 plying by gi shows that it’s equivalently generated by elements of the form g−1 with g ≠ 1.

2. The group of G-coinvariants of A is the G-module AG := A/IGA, it is the largest trivial G-module that is a quotient of A. 242 CHAPTER 7. CLASS FIELD THEORY

3. The augmentation ideal IG is precisely the annihilator of the Z[G]-module Z; therefore

Z ⊗Z[G] A ≃ A/IGA

For each ϕ ∈ HomZG(IG,A) we have an associated map dϕ : G → A where dϕ(g) := ϕ(g−1). Note that dϕ ∈ Z(G, A). We now give an important lemma which demonstrates the importance of the augmentation ideal.

引理 7.5.2. Let Φ: HomZG(IG,A) → Z(G, A) be given by Φ(ϕ) := dϕ. Then Φ is an isomorphism.

The next result gives an explicit description of H1(G, A). Incidentally, this theorem shows that the group cohomology can be described in terms of cocycles and coboundaries in a similar fashion to the singular cohomology of topological spaces. While this result only gives a description of H1(G, A), it’s impossible to give similar descriptions of Hi(G, A) so that this theorem is only a special case.

♦ 定理 7.5.1

H1(G, A) ≃ Z(G, A)/B(G, A)

1 推论. If G acts trivially on A then H (G, A) ≃ Z(G, A) = HomGrp(G, A).

Tate cohomology

Tate cohomology groups are a slightly modified form of the usual cohomology groups of a finite group that combine homology and cohomology groups into one sequence. They were introduced by John Tate (1952), and are used in class field theory.

♦ 定义 7.5.4 ∑ Z The norm element of [G] is NG := g∈G g.

Let A be a G-module and let NG : A → A be the G-module endomorphism a 7→ NGa.

G 引理 7.5.3. We have IGA ⊆ ker NG and imNG ⊆ A , thus NG induces a morphism b G NG : AG → A of trivial G-modules. 7.5. TATE COHOMOLOGY 243

G Proof. We have gNG = NG for all g ∈ G, so imNG ⊆ A , and NG(g − 1) = 0 for all g ∈ G, so NG annihilates the augmentation ideal IG and IGA ⊆ ker NG. The lemma follows.

♦ 定义 7.5.5 For n ⩾ 0 the Tate cohomology and homology groups are defined by b • Hn(G, A) := Hn(G, A) for n ⩾ 1, b 0 • H (G, A) := coker NG, b −1 • H (G, A) := ker NG, b n • H (G, A) := H−(n+1)(G, A) for n ⩽ −2.

For every short exact sequence of G-modules

β 0 → A −−→α B −−→ C → 0 induces a long exact sequence of Tate cohomology groups

bn βbn bn · · · → Hb n(G, A) −−→α Hb n(G, B) −−→ Hb n(G, C) −−→δ Hb n+1(G, A) → · · ·