Part III Algebraic number theory Lecturer: Prof. A.J. Scholl Lent term

Part III Algebraic number theory Lecturer: Prof. A.J. Scholl Lent term 2014 1 1 Transcribed by Stiofáin Fordham.

Contents Page Lecture 1 5 1.1. Algebraic preliminaries 6 Lecture 2 6 2.1. 1: Valuations and absolute values 8 Lecture 3 9 3.1. Valuation rings 9 Lecture 4 11 4.1. 2: Completions 13 Lecture 5 13 5.1. Inverse limits 15 Lecture 6 15 6.1. Hensel s lemma 15 Lecture 7 17 7.1. 3: Extensions of local fields 17 Lecture 8 18 8.1. Ramification degree and residue class degree 20 Lecture 9 20 9.1. Unramified extensions 20 Lecture 10 22 10.1. Totally ramified extensions 23 Lecture 11 24 11.1. 4: Ramification theory 24 Lecture 12 26 12.1. The ramification subgroups 26 Lecture 13 28 13.1. 5: Places 28 13.2. Extensions 29 13.3. Case where L/K is finite and Galois 30 Lecture 14 30 14.1. 6: Number fields 31 Lecture 15 32 15.1. Places and ideals 33 15.2. 7: Ideles and adeles 33 15.3. Idele norm 33 15.4. Content map 33 15.5. Topology on J K 34 Lecture 16 34 16.1. 8: Geometry of numbers 35 Lecture 17 36 17.1. Dirichlet s unit theorem and the finiteness of the ideal class group 37 Lecture 18 38 18.1. 9: The Dedekind zeta-function 39 3

Lecture 19 41 19.1. Fourier analysis 41 Lecture 20 43 20.1. Fourier inversion theorem 44 20.2. Local ζ-integrals 45 Lecture 21 45 21.1. The global theory 45 Lecture 22 47 22.1. Global zeta integrals 47 Lecture 23 49 23.1. Computation of the volume 50 Lecture 24 51 24.1. The fundamental domain for JK 1 /K 51 24.2. L-functions 52 4

Lecture 1 Lecture 1 16 th January 11.00 We are interested in algebraic number fields, finite field extensions of Q. Sometimes we are also interested in other, infinite, extensions of Q e.g. the cyclotomic field Q cyc = Q({ζ n = exp(2πi/n) for all n 1}) The classical approach consists of two parts the algebraic part: denote O K the ring of algebraic integers of K, we study the algebraic properties of this ring, we study the ideals in O K : there is a theorem that every non-zero ideal can be uniquely written as a product of prime ideals (O K is a Dedekind domain), define a fractional ideal to be a non-zero, finitely generated O K - submodule of K (these form a group under multiplication). The principal fractional ideals are the fractional ideals of the form xo K (x K, x 0) - these form a subgroup: the quotient group is the class group Cl(K) of K. This measures how far we are from having a principal ideal domain. We are also interested in OK, the group of units of K. Because we have xo K = O K iff x OK, there is some kind of relationship between these two groups (one such result is the so-called analytic class number formula - we will prove this later). Algebra alone says nothing about the relationship between Cl(K) and OK - if we do want to say something then we need some analytic input - then we need geometry of numbers. the other part: take K a field with [K, Q] = n, then there exists n embeddings σ 1,..., σ n K C which can be ordered so that where r 1 + 2r 2 = n and These fit together to give σ 1,..., σ r1 K R σ r1+1,..., σ r1+r 2 K C σ r1+r 2+1 = σ r1+1,..., σ n = σ r1+r 2 σ = (σ 1,..., σ r1+r 2 ) R R r1 C r2 R n called the Minkowski embedding. The key fact is that σ(o K ) is a lattice - i.e. a rank n discrete subgroup of R r1 C r2. This leads to the proof that Cl(K) is finite and (considering the action of K by multiplication) that OK is finitely generated of rank r 1 + r 2 1 (called the Dirichlet unit theorem). We will prove both of these in due course but as a result of a more general theorem. The modern approach is to take as a starting point the view that prime ideals and embeddings into R or C are two different cases of a general notion, that of absolute values (or places) of K. For example, for K = Q, we have the usual ( Archimedean ) absolute value x = x, the absolute value on R. But also, for every prime p, we have the p-adic absolute value 0 x = 0 x p = 1 if x = p r a for r, a, b, Z and (ab, p) = 1 p r b and it also satisfies the triangle inequality x + y p x p + y p and in fact it satisfies the strong triangle inequality max( x p, y p ). The real numbers are the completion 5

Lecture 2 of Q with respect to. For every prime p, we also have a completion of Q with respect to p - the field of p-adic numbers Q p. The replacement for Minkowski space is the embedding Q Q p R = Q p, p p where Q = R (actually this is not quite the right thing to consider; later we will define the correct object called the adele ring of Q (for any number field)). Regarding textbooks: the book of Cassels and Frohlich called Algebraic number theory see chapter 1 and (especially) chapter 2; the book of Neukirch 1 Algebraic number theory. 1.1. Algebraic preliminaries. First from Galois theory: we have the trace and norm. Let L/K be a finite extension of degree n (so L K n as a K-vector space). Then for all x L, the map u x L L given by y xy is a K-linear endomorphism of L. The norm (resp. trace) of x is N L/K (x) = det K u x K, Tr L/K (x) = tr K u x K, where the norm is multiplicative from L K and is a homomorphism from L K and the trace is an additive homomorphism from L K. If L/K is separable, then N L/K (x) = σ i (x), i Tr L/K (x) = σ i (x), i where {σ i } is the set of K-homomorphisms σ i L K an algebraic closure of K, and Tr L/K is not identically zero (in fact this is equivalent to L/K being separable). A consequence of this: for L/K separable 2, the map ψ L L K given by (x, y) Tr L/K (xy) is a non-degenerate symmetric K-bilinear form on L (i.e. ψ induces an isomorphism between L and its dual Hom K (L/K) - called the trace form). So in particular, if L/K is a finite Galois extension, then (if we take K to be an algebraic closure of L) each σ i is given by composing an element of Gal(L/K) with the embedding L K. So N L/K (x) = Tr L/K (x) = σ Gal(L/K) σ Gal(L/K) σ(x), σ(x). Lecture 2 18 th January 11:00 Now we will continue with the preliminaries - some commutative algebra. All rings are commutative with 1. Definition 2.1. Denote R = {invertible elements of R}. Definition 2.2. Let R, S be rings with R S then x S is integral over R if there exists f R[T ] monic with f(x) = 0. Remark. Equivalent conditions to the above are 1 Neukirch calls the thing that Scholl calls a valuation, an exponential valuation and calls something else a valuation but confusingly he calls (in his language) a valuation induced from (what he calls) an exponential valuation a multiplicative valuation or absolute value, see pg. 116 & 120. 6

Lecture 2 R[x] is a finite R-algebra (i.e. it is finitely generated as an R-module), there exists a subring S S containing R and x which is a finite R-algebra. It is left as an exercise to show that these are equivalent. Definition 2.3. The set R = {x S x is integral over R} is a ring called the integral closure of R in S. If R = R we say that R is integrally closed in S. If R is a domain, integrally closed in its field of fractions Frac(R), we say that R is normal (or integrally closed). Proposition 2.1. Let R be a normal domain which is Noetherian, K = Frac(R) and L/K a finite separable extension. If S is the integral closure of R in L, then S is a finite R-algebra. Remark. The standard example is the case R = Z, K = Q, L an algebraic number field, and S = O L is then the ring of integers of L. The proposition then says that S has a finite Z-basis. Proof. Let n = [L K]. Then Frac(S) = L, in fact for all x L, there exists a R/{0} such that ax S so x Frac(S) (take a = b n where b = any common denominator for coefficients of the minimal polynomial of x over K). If x S, then all of its conjugates {σ(x) σ L K} will be integral over R. So Tr L/K (x), N L/K (x) are integral over R, hence are in R, as R is normal. Now choose a basis {e i } for L/K with all e i S. The trace form (x, y) Tr L/K (xy) is non-degenerate: let {f i } be the dual basis, f i L. Let x S: write x = a i f i, a i K. So as e i x S and Tr L/K (e i f j ) = δ ij so R Tr L/K (xe i ) = a i. So S Rf i, so as R is Noetherian, S is a finitely generated R-module. Remark. The standard reference for commutative algebra is the book of Atiyah & Macdonald. Definition 2.4. R is a local ring if it has a unique maximal ideal. Example. Let R be a field, then the unique maximal ideal is (0). Let R = k[[x]] be the ring of power series with coefficients in a field k, then the unique maximal ideal is (X). Definition 2.5. Let R be an integral domain, P R a prime ideal. Then is a local ring with maximal ideal called the localisation of R at P. R P = {x/y x, y R, y P}, m P = {x/y x P, y R/P}, Example. If P = (0) is a prime ideal then R P = Frac(0). Example. We can define (not to be confused with Z p ). Z (p) = {x/y x, y Z, p y} Let s have some more notation. We write N = {0, 1, 2,... }, and Z, Q, R, C are defined as usual. I write and to mean the same thing. The notation a := b means that a is defined to be b. 7

Lecture 3 2.1. 1: Valuations and absolute values. Definition 2.6. A (rank 1) valuation of a field K is a non-trivial homomorphism v K R such that for all x, y K with x y, then v(x + y) min(v(x), v(y)). Remark. By convention, we extend v to K by setting v(0) = +. Some writers don t require v to be non-trivial. There are valuations of higher rank, where you replace R by some totally-ordered group for example R n under lexicographical ordering - these are useful in algebraic geometry and in rigid analysis, but we won t need them here. Example. The standard example is the p-adic valuation v p Q Z R given by v p (p n a ) = n if (ab, p) = 1. b Example. More generally, for any algebraic number field, P O K a non-zero prime ideal, let v P (x) = exponent of P occurring in the prime factorisation of xo K. Obviously this is a homomorphism. To check that it satisfies the condition of a valuation, we can replace x, y by xz, yz for any z K, so without loss of generality we can assume that x, y O K. Then v P (x) = n iff x P n /P n+1 thus the condition is satisfied trivially. Example. Let K be the field of meromorphic functions on C, then we can take which is a valuation. v(f) = ord z=0 f Remark. These three examples have in common that the value group is the integers. This is obvious in 1 and 3 and with a little thought also obvious for 2. Definition 2.7. We say that v is discrete if v(k) R (the value group of v) is a discrete subgroup of R, in which case v(k) = Z r for some r > 0. A discrete valuation is normalised if v(k) = Z. Example. Examples 1,2,3 above are normalised discrete valuations. Later we will see useful valuations with v(k ) = Q. Remark. If v is any valuation and α > 0, then αv is also a valuation. We say that v, αv are equivalent (so every discrete valuation is equivalent to a unique normalised one). Proposition 2.2 (1.1). Let v be a valuation on K. Then if v(x) v(y), v(x + y) = min(v(x), v(y)). Proof. Without loss of generality, v(x) < v(y) = v( y) (there are no torsion elements), so v(x) = v((x+y)+( y)) min(v(x+y), v(y)). Hence v(x) v(x+y) v(x) so we have equality. Definition 2.8. Let K be a field, R K a proper subring, then R is a valuation ring of K if for all x K/R, we have x 1 R. Remark. If x, y R/{0}, condition implies that at least one of x/y and y/x is in R. So in particular Frac(R) = K. Theorem 2.3 (1.2). Let R be a valuation ring of K. The following are true (1) R is local, (2) R is normal, (3) every finitely generated ideal of R is principal; in particular if R is Noetherian, then R is a principal ideal domain. Proof. (1) Let m = R/R. Trivially x m, y R then xy m. If x, y m/{0} then, without loss of generality, y/x R, hence x + y = x(1 + y ) m. So m is an x ideal, hence R is local. 8

Lecture 3 Lecture 3 21 st January 11:00 3.1. Valuation rings. Last time I introduced this deceptively innocuous idea of what a valuation ring is: a valuation ring R K satisfies x K/R then x 1 R. We continue with the proof from last time. We had proved part (1). Theorem 3.4 (1.2). Let R be a valuation ring of K. The following are true (1) R is local, (2) R is normal, (3) every finitely-generated ideal of R is principal; in particular if R is Noetherian then R is a principal ideal domain. Proof. (2) Let x K be integral over R so x n n 1 + i=0 a i x i = 0, with a i R. If x 1 R, then x R and we are done. Otherwise, x 1 R and 1 = x 1 n 1 ( i=0 a i (x 1 ) n i 1 ), but since x 1 R and a i R, this means that x 1 R so x R. (3) If x, y R then xr if y/x R xr + yr = yr if x/y R because if (y/x) R then (y/x) 1 = x/y R (and using xr + yr = x(r + y x R) = y( x y R + R)). Now a result that describes the relationship between valuations and valuation rings. Theorem 3.5 (1.3). (1) Let K be a field, v a valuation of K. Define R v = {x K v(x) 0} m v = {x K v(x) > 0}. Then R v is a valuation ring of K with maximal ideal m v and v induces an isomorphism between K /Rv v(k ) R. (2) R v is a maximal proper subring of K, depending only on the equivalence class of the valuation v. (3) If v, v are valuations of K, and R v R v then R v = R v and v, v are equivalent. In particular, for any valuation ring R of K, there exists at most one equivalent class of valuations v with R = R v. 3 Remark. N.b. the element 0 m v is defined to have v(0) = +. Remark. There are valuation rings which are not of this form R v - they are associated to valuations of rank > 1. Example. The example to bear in mind here is Z (p) which is R v with K = Q where v is the p-adic valuation, and more generally O K,P, the localisation of O K (where K is a number-field) at the prime ideal P, the valuation ring of v P. 3 Two valuations v1 and v 2 are called equivalent if v 1 = sv 2 for some real number s > 0. 9

Lecture 3 Proof. (1) From the definition of the valuation, it is rather easy to see that R v is a subring of K, and is not equal to K because v / 0 4. If x R v, then v(x) < 0 so v(x 1 ) = v(x) > 0 thus x 1 R v. So R v is a valuation ring, whose non-units are {x v(x) > 0} = m v. Finally, ker(v) = R v: if x is a unit in R v, then v(x), v(x 1 ) 0 but v(x 1 ) = v(x) hence v(x) = 0 so R v ker(v); if x R v satisfies v(x) = 0 then v(x 1 ) = v(x) = 0 so actually x 1 R v (x 1 exists since K is a field!) so x is a unit in the ring R v, thus ker(v) R v. This gives the required isomorphism. (2) Let x K/R v. Then v(x) < 0, hence for all y K, there exists n Z such that v(y) > nv(x) so y/x n R v so y R v [x], therefore R v [x] = K, so R v is maximal. Trivially, if v, v are equivalent then R v = R v. (3) By maximality in (2), we get that R v = R v (hence m v = m v ). So for all x, y K, then v(x) v(y) iff x/y R v = R v iff v (x) v (y) (*). Let 0 π m v (so v(π) > 0). Then for p/q Q, q > 0 then v(x) v(π) p q v(xq ) v(π p ) v (x) v (π) p q by (*). So v(x)/v(π) = v (x)/v (π) for any x K, so v, v are equivalent. 5 Definition 3.9. R v is called the valuation ring of v. Definition 3.10. A discrete valuation ring is the valuation ring of a discrete valuation. Proposition 3.6 (1.4). A domain R is a discrete valuation ring iff it is a principal ideal domain with a unique non-zero prime ideal. Proof. Suppose that R is a principal ideal domain with unique prime ideal πr 0, K = Frac(R). If x R/{0}, xr = π n R for some n N 6, define v(x) = n and for x/y K then put v(x/y) = v(x) v(y). It is easy to see that v is a normalised discrete valuation. Conversely, let R be a discrete valuation ring. Choose π m R /{0} with v(π) minimal (it exists as v is discrete). Then m R = πr 7 ; if 0 I R is any ideal, then I contains π m for some minimal m 1, and it follows that I = π m R. Lemma 3.7 (1.5). Let R be a ring and π R not a zero-divisor. Then for all m, n 0, we have an R-module isomorphism Proof. Obvious. R/π n R π m π m R/π m+n R Theorem 3.8 (1.6). Any valuation of Q is equivalent to some v p, p prime. Any valuation of a number field K is equivalent to some v P for P O K a prime. Proof. Let O K be the ring of integers of K, v a valuation on K. R v is normal and contains Z, so R v O K. As Frac(O K ) = K 8, v is non-trivial on O K, so 4 Let x, y R with x y and v(x), v(y) > 0 then v(x + y) min(v(x), v(y)) > 0 and v(xy) = v(x) + v(y) > 0 because v K R is a homomorphism. Also 0 = v(1) = v(( 1) ( 1)) = v( 1) + v( 1) = 2v( 1) so v(x) = v( x). 5 Suppose that functions f, f satisfy f(x) > p iff f (x) > p. If there exists y such that f (y) f(y) (without loss of generality f(y) < f (y)), then define Q = (f (y) f(y))/2 and R = f (y). Then R Q = f (y)/2 + f(y)/2 > f(y) so f (y) < R Q which is false. 6 From Neukirch pg. 67: the prime ideal is of the form p = (π) = πo, for some prime element π. If π 1 is prime then (π 1 ) is a prime ideal, so (π) = (π 1 ) which is only possible if π 1 = uπ (u R ) i.e. they are associates (in a principal ideal domain, element is irreducible iff it is prime). If π 1 is not prime, then we may decompose it as a product of primes, and since π is the only prime (upto associates) then we can write π 1 = uπ n (u R, units form a group). 7 Let x mv then v(x) v(π) so v(xπ 1 ) 0 hence p = xπ 1 m v so x = pπ so x πr. 8 One direction is obvious, the other is described in Neukirch pg. 8. 10

Lecture 4 P := m v O K is a prime ideal (0) of O K. Then if x O K /P R v /m v = Rv, then v(x) = 0. So R v O K,P. Therefore by theorem 3.5 (3), as O K,P is the valuation ring of v P, we have that R v = O K,P and v is equivalent to v P. Definition 3.11. Let K be a field. A map K R 0 is an absolute value if for all x, y K we have (AV1) x = 0 iff x = 0, (AV2) xy = x y so K R >0 is a homomorphism, (AV3) x + y x + y, (AV4) there exists x with x 0, 1. If the stronger condition (AV3N) x + y max( x, y ) is satisfied, then we say that is non-archimedean; otherwise we say that is archimedean. Example. The usual absolute value on R, and modulus on C are (examples of) archimedean absolute values. Lecture 4 23 rd January 11:00 Theorem 4.9 (1.7). Fix ρ (0, 1), and let v be a valuation on K. Then 0 if x = 0 x = ρ v(x) if x 0 is a non-archimedean absolute value on K and the map v v is a bijection between the set of valuations and the set of non-archimedean absolute values on K. Proof. Immediate from the definitions - one recovers the valuation by v(x) = log x / log ρ. Remark. Note that people sometimes use valuation for non-archimedean absolute value. Example (v p, the p-adic valuation on Q). Usually we choose ρ = 1/p in the above, and we get the p-adic valuation p n u v p = 1 if u, v Z and (uv, p) = 1 pn Remark. If is a non-archimedean absolute value, then so is r for any r > 0 (not necessarily true for archimedean absolute values). We say that absolute values 1 and 2 are equivalent if 1 = r 2 for some r > 0. Proposition 4.10 (1.8). Let be an absolute value on K. Then the function d(x, y) = x y is a metric on K, translation-invariant, for which the field operations are continuous. Equivalent absolute values give equivalent metrics. Proof.(sketch). The proof is the same as showing that x y is a metric on R. Note that we use that xy = x y to show that multiplication and inversion are continuous. Remark. So (K, ) is a topological field. The following gives us an easier way of checking whether an absolute value is non-archimedean. Proposition 4.11 (1.9). An absolute value on K is non-archimedean iff a 1 K 1 for all a Z. If is archimedean, then is unbounded on Z 1 K K. 11

Lecture 4 Proof. Write a for a 1 K. Suppose it is non-archimedean. If a N, then a = a = 1 + + 1 1 = 1. Suppose conversely that a 1 for all a Z. Let x, y K, then if is non-archimedean, we have n x + y n = (x + y) n = ( n i )xi y n i (n + 1) max( x n, y n ) i=0 as we have ( n ) 1. The result follows by taking the nth root of both sides and i letting n. Last part: if is archimedean then a > 1 for some a K, then a n = a n as n to it cannot be bounded. It is convenient to weaken the definition of absolute value slightly by replacing (AV3) by (AV3 ) for some α (0, 1], x + y α x α + y α. With this definition, the square of the modulus on C is an absolute value (later on we ll see that this is a good idea). If satisfies (AV3 ) then α satisfies (AV3), so we haven t introduced anything new. Remark. If char(k) = p > 0, then any absolute value on K is non-archimedean - since Z 1 K is finite, so is bounded on it. Theorem 4.12 (1.10, Ostrowski). Every absolute value on Q is equivalent to (the usual archimedean value) or some p. Proof. 9 Consider a non-trivial absolute value on the rationals (Q, ). We consider two cases, (i) there exists n N such that n > 1 and (ii) for all n N, we have n 1. It suffices for us to consider the valuation of integers greater than one. For if we find some c R + for which n = n c for all naturals greater than one; then this relation trivially holds for 0 and 1, and for positive rationals since m/n = m / n = m c / n c = ( m / n ) c = m/n c and for negative rationals x = x = x c = x c. Case I: there exists n N such that n > 1. Consider the following calculation. Let a, b N > 1. Let n N > 0. Expressing b n in base a yields b n = i<m c i a i, where each c i {0, 1,..., a 1} and m n(log b/ log a) + 1. Then we see, by the properties of an absolute value that b n = b n am max{ a m, 1} a(n log a b + 1) max{ a n log a b, 1} b (a(n log a b + 1)) 1 n max{ a log a b, 1} 1 as n b max{ a log a b, 1}. Now choose b N > 1 such that b > 1. Using this in the above ensures that a > 1 regardless of the choice of a (otherwise a log a b 1 implying b 1). Thus for any log b/ log a choice of a, b > 1 above, we get b a, i.e. log b / log b log a / log a. By symmetry, this inequality is an equality. Since a and b were arbitrary, there is a constant, λ R + for which log n = λ log n, i.e. n = n λ = n λ for all naturals n > 1. As per the above remarks, we easily see that for all rationals, x = x λ, thus demonstrating equivalence to the real absolute value. Case II: for all n N we have n 1. As this valuation is non-trivial, there must be a natural number for which n < 1. Factorising this natural, n = i<r p ei i 9 The proof from Fesenko & Vostokov Local fields and their extensions. 12

Lecture 5 yields p must be less than 1, for at least one of the prime factors p = p j. We claim than in fact, that this is so for precisely one prime factor. Suppose per contra that p, q are distinct primes with absolute value less than 1. First, let e N + be such that p e, q e < 1/2. By the Euclidean algorithm, let m, n Z be integers for which mp e +nq e = 1. This yields 1 = 1 m p e + n q e < m + n 1, a contradiction. 2 So we must have p = α < 1 for some prime, and q = 1 for all other primes. Letting c = log α/ log p, we see that for general positive naturals n = i<r p ei n = i<r p i ei x = x c p one. = p j ej = (p ej ) c = n c p. As per the above remarks we see that all rationals, implying the absolute value is equivalent to the p-adic 4.1. 2: Completions. Let (X, d) be a metric space. The completion ( ˆX, ˆd) is defined ˆX = {Cauchy sequences in X}/ where (x n ) (y n ) iff d(x n, y n ) 0, and ˆd((x n ), (y n )) = lim n d(x n, y n ). We say that ˆX is complete if X ˆX is an isometry, and this is an isomorphism iff X is complete. Let K be a field, an absolute value. If is non-archimedean, let R be the valuation ring of the associated valuation. The metric d(x, y) = x y gives completions ˆK and ˆR (R = {x K x 1}). Theorem 4.13 (2.1). (1) ˆK is a field and ˆd(x, y) = x y ˆ, for an absolute value ˆ on ˆK, extending, (2) if is non-archimedean then so is ˆ, and ˆR is the valuation ring of ˆK. Proof. (1) The set {Cauchy sequences in K} = S K N is easily seen to be a ring (by the axioms for absolute values), and ˆK = S/I, where I = {(x n ) K N x n 0} (null sequences), an ideal. Extend to ˆ by (x n ) ˆ = lim x n - clearly this is an absolute value. It is sufficient to prove that I is maximal: let x = (x n ) S/I. As x I, x n is bounded below by some ɛ > 0, for n sufficiently large, say n > N. Let y n = 1/x n for n > N. Then y n y m = x n x m x n x m ɛ 2 x n x m so y = (y n ) S (we set y n = 1 for n N) then xy 1 I so everything not in I is a unit, hence I is maximal, hence S/I is a field (the fact that xy = x y is crucial here). (2) The first part follows from proposition 4.11. The second part: we have that x ˆ 1 for all x ˆR, so ˆR valuation ring of ˆK. Suppose x ˆK, represented by (x n ) S, and x ˆ = lim x n 1. By (AV3N) some x n ˆ 0, x n = x ˆ, for n sufficiently large (if x 0) implies that x n R for n 0 which implies that x ˆR. Example. For on Q then the completion is ˆK = R. Next time we will look at p-adic completions. Lecture 5 25 th January 11:00 Example. If we consider Q with then the completion is R. If we take Q with p, the p-adic absolute value then its completion is Q p the field of p-adic numbers 13 i ;

Lecture 5 and Z p = {x Q p x p 1} is the valuation ring of Q p, called the ring of p-adic integers. Proposition 5.14 (2.2). Every element of Z p has a unique representation as a series x = a 0 + a 1 p + a n p n, a i {0, 1,..., p 1} n 2 Every element of Q p has a unique representation (5.1) x = a N p N + + a 1 p + a n p n n 0 In either case, v p (x) = min{n Z a n 0}. Proof. 5.1 is a Cauchy sequence, so converges to an element of Q p, and by the strong triangle inequality applied to its partial sums, v p (x) = min{n Z a n 0}. Given two representations n N a n p n = n N b n p n of x Q p, multiplying through by powers of p, we may assume that N = 0. Thus a 0 b 0 = (b n a n )p n n 1 so v p (a 0 b 0 ) 1 thus a 0 = b 0 so a n = b n for all n. It remains to show that every element of Q p has a representation of the form 5.1. The set of partial sums of the series 5.1 is precisely Z[1/p] 0 = {x Z[1/p] x 0}. As Q is dense in Q p, it is sufficient to prove that Z[1/p] 0 is dense in Q (for the topology inducsed by p ). Let x = p n a Q, with (b, p) = 1 and b > 0. Let m > 0, b and choose c N with bc a (mod p m+n ). Then x p n c p = a bc p n b p m p and p n c Z[1/p] 0, and therefore Z[1/p] 0 is dense in Q. In elementary terms, the p-adic numbers are backward decimals, and it is easy to see what the arithmetical operations are. For the remainder of the section, we will only consider non-archimedean absolute values, associated to some valuation v. Let K be a field, and R the valuation ring. Observe if we take π R with 0 < π < 1 (so it belongs to the maximal ideal m v = m R ) then π n R = {x K x π n } is open for the p-adic topology induced by - because if x π n R and 0 < ɛ < π n, then for all y with y < ɛ, then x + y π n (AV3N) so x + y π n R, hence it is open. Remark. Analysis is rather easier (in some ways) for non-archimedean values. For example, the series x n converges (in a complete field with respect to nonarchimedean absolute values) iff x n 0 (by the strong triangle inequality). Proposition 5.15 (2.3). Let R be the valuation ring of (K, ) and ˆR its completion. Let π R with 0 < π < 1. Then for every n 1, the canonical map R/π n R ˆR/π n ˆR is an isomorphism. Proof. We have that R ˆR is dense, and π n ˆR ˆR is open. So R+π n ˆR = ˆR, so it is surjective. Also R π n ˆR = {x R x π n } = π n R, so it is an isomorphism. Remark. Since π n R R is open, hence so is every translate x + π n R, so π n R is also closed (or see directly by the continuity of ) so it follows that the induced topology on R/π n R is discrete. 14

Lecture 6 5.1. Inverse limits. Let (X n ) n N be a collection of sets (say) and π n X n+1 X n a collection of maps. Then (X n, π n ) is called an inverse system. Its inverse limit is defined to be lim (X n, π n ) = lim X n = {(x n ) n N π n (x n+1 ) = x n for all n} X n. Typically we will consider such systems where all π n are surjective. If X n are groups (or rings) and π n are homomorphisms, then lim X n is a group (or ring), in fact a subgroup (or ring) of X n. Example. Let X n = Z/p n Z and π n Z/p n+1 Z Z/p n Z is reduction mod p n. Our claim is that lim Z/p n Z = Z p : using the standard bijection Z/p n Z {0, 1,..., p n 1} and writing in base p, then it follows from proposition 5.14. Example (I-adic completion). Let R be a ring, I an ideal. Consider R/I n with maps R/I n+1 R/I n the canonical maps and IJ = { x α y α x α I, y α J} and I n = I(I n 1 ). The I-adic completion of R is RÎ = lim R/In (e.g. Zˆ (p) = Z p). The canonical map R RÎ is x ((x mod I n ) R/I n ) n and the kernel is I m. We have a topology on RÎ, given as follows: more generally, suppose that X = lim(x n, π n ) an inverse limit of sets. Let p m X X m be the map ((x n ) lim X n ) x m. The inverse limit topology on X is the smallest topology (smallest meaning least number of sets) for which the maps p m are continuous (for the discrete topology on X m ) i.e. open sets in lim X n are arbitrary unions of sets of the form U m,a = p 1 m ({a}). 6.1. Hensel s lemma. Lecture 6 28 th January 11:00 Proposition 6.16 (2.4). (1) lim X n is totally disconnected (so the only connected subsets are the singleton sets). (2) Suppose that X n is finite. Then lim X n is compact. Proof. (1) Take elements x = (x n ) y = (y n ) in lim X n. Choose m such that x m y m. Then lim X n is the disjoint union of U m,xm and Um,x c m = U m,a y a X m a x m So X is totally disconnected. (2) X n is finite implies that X n is compact (for the discrete topology). Thus X n is compact with the product topology (by Tychonoff s theorem). Then lim X n X n is closed (by definition - that π n (x n+1 ) = x n holds, is a closed condition) so is compact. Example. Z p = lim Z/p n Z is compact and totally disconnected. Remark. One can define the inverse limit more generally (we could replace the index set N by some other partially ordered set, possibly satisfying some other properties depending on what you want to do with it e.g. take G to be some infinite group and consider the family of normal subgroups of finite index, partially ordered by reverse inclusion. If N N are normal subgroups of G then we have maps G/N G/N, and we can form lim N G/N, called the profinite completion of G. 15 n

Lecture 6 Remark. Sometimes (but not here) it is useful to consider some other topology on {X n }; we can then form an induced topology on X n (product) and on lim X n (assuming that the π n are continuous). An example: take the topological circle X n = R/Z with the usual topology for all n and π n R/Z 2 R/Z then we have that lim (R/Z, 2) is a topological group, called the 2-adic solenoid. Theorem 6.17 (2.5). Let K be a field, v a valuation associated to an absolute value, and a valuation ring R. Let ˆK be the completion of K with respect to, ˆR its valuation ring. Let π 0 with π R and π < 1. Then (1) K is complete (i.e. K = ˆK) iff R is π-adically complete i.e. R = lim R/π n R, (2) in general, ˆR equals the π-adic completion of R. Proof. First assume that K = ˆK. Obviously n π n R = n {x R x π n } = {0}. This means that R lim R/π n R is injective. Let (x n ) lim R/π n R. Pick y n R with y n ( mod π n ) x n. Then y n+1 y n ( mod π n ) thus (y n ) is a Cauchy sequence, with a limit y, say. It is sufficient to prove that y ( mod π n ) x n for all n. If not, say y / y m ( mod π m ) thus for all n m we have y / y n ( mod π m ) so y y n / 0. Hence R is π-adically complete. (2) What we have just proved shows that ˆR is π-adically complete i.e. ˆR lim ˆR/π n ˆR lim R/π n R by proposition 5.15, i.e. ˆR is the the π-adic completion of R. In particular, if R is π-adically complete, then R = ˆR, so K = ˆK is complete (hence the other direction of (1) follows). Remark. The origin of the idea of completion of the p-adic numbers goes back to Hensel and congruences (see the textbook of Borevich & Shafarevich). Consider the following problem: suppose that you have f Z[T ], and have a Z with f(a) 0 ( mod p n ). When does there exist b Z with f(b) 0 ( mod p n+1 )? (and b a ( mod p n )). Or we could take a variant of the problem with f(t 1,..., T r ) and a Q r, e.g. take p = 2 and f(t ) = T 2 + 1, then f(1) 0 ( mod 2) but there is no solution mod 4 since -1 is a quadratic non-residue mod 4. Suppose that you have a n Z with f(a n ) 0 ( mod p n ) and a n+1 a n ( mod p n ) for all n. Then x = lim(a n ) Z p satisfies f(x) = 0 (conversely given x with f(x) = 0 then we have elements a n as above). So this is reminiscent of successive approximation (Newton s method). Theorem 6.18 (2.6, Hensel s lemma). Let R be a complete, discrete, valuation ring and πr its maximal ideal (we say that π is a uniformiser of R). Suppose that f, g 1, h 1 R[T ] with g 1 monic, f g 1 h 1 (mod π) and (ḡ 1, h 1 ) = 1 where an overline denotes the image in (R/πR)[T ] = k[t ]. Then there exists a unique g, h R[T ] with g monic, f = gh and g g 1 (mod π) and h h 1 ( mod π). Remark. N.b. we do not assume that f is monic here. Proof. Let N = deg(f) and d = deg(g 1 ) and without loss of generality we assume that deg(h 1 ) N d (there might be some terms at the beginning that depend on π but we can forget about these). We are inductively going to construct elements g n, h n R[T ] such that g n is monic of degree d with deg(h n ) N d and f g n h n ( mod π n ) and g n+1 g n (mod π n ) and h n+1 h n (mod π n ) and such that (g n, h n ) is unique mod π n (this doesn t use the completion). 16

Lecture 7 Suppose that we have constructed g n, h n, so f g n h n = π n q with q R[T ] for some n. Well the degree of q is N and we have assumed that (g n, h n ) is unique mod π n. Let g n+1 = g n + π n u, h n+1 = h n + π n v with deg(u) d 1 and deg(v) N d. Then f g n+1 h n+1 (mod π n+1 ) iff g n v+h n u q (mod π). So it is sufficient to prove that there exists a unique ū, v k[t ] such that (6.2) ḡ 1 v + h 1 ū = q with deg ū d 1 and deg v N d. But (ḡ 1, h 1 ) = 1 so there exists ū, v satisfying (6.2), unique up to transformations (ū, v) (ū + rḡ 1, v r h 1 ) with r k[t ], so unique with the condition on the degree. To complete the proof we take g = lim g n and h = lim h n and this is unique by the uniqueness at every stage. Lecture 7 1 st February 11:00 Let R be a complete discrete valuation ring and π a uniformiser. Let f, g 1, h 1 R[T ] with g 1 monic and (ḡ 1, h 1 ) = 1 and f g 1 h 1 (mod π). Then Hensel s lemma says that there exists a unique g monic and h R[T ] with f = gh and g 1 g (mod π) and h 1 h (mod π). A special case of this is where we just take the linear case. Corollary 7.19 (2.7). Let f R[T ], a R with f(a) 0 (mod π) and f (a) / 0 (mod π). Then there exists a unique b R with f(b) = 0. Proof. Let g 1 = T a and let h 1 be any polynomial with f(t ) = (T a)h 1 (T )+ f(a) then we have f = (T ā) h 1. As f (a) 0 10 we have (T ā, h 1 ) = 1. Then apply Hensel s lemma. Example. Let f = T p 1 1 (T 1)(T 2)... (T (p 1)) (mod p) 11 so each a F p has a unique lifting to a (p 1) th root of 1, [a] Z p (so Z p contains all (p 1) th roots of 1). Similarly, let R be a complete discrete valuation ring and suppose that the residue field k contains a finite field F q. The same idea applied to T q 1 1 shows that R has all (q 1) th roots of 1. Remark. There is a wider class of (discrete valuation, or more generally local) rings for which Hensel s lemma holds (mod π implies mod m R ). They are called Henselian rings (and the definition is what was just given, i.e. a local ring in which Hensel s lemma holds). E.g. Let R be the set of all elements of Z p that are algebraic over Q. 7.1. 3: Extensions of local fields. For now, our local field is going to be a field complete with respect to an absolute value (usually people are more restrictive, e.g. they might be referring to finite extensions of Q p or the reals or the complex numbers). Theorem 7.20 (3.1). Let K be complete with respect to an absolute value, and L/K an algebraic extension. (1) There exists a unique absolute value L on L extending. (2) If [L K] = n < and x L, then x L = N L/K (x) 1/n 10 Recall that f (a) / 0 implies that f does not have a double root at a, thus h 1 does not have a root at a, thus gcd(t ā, h) = 1. 11 Ireland and Rosen, pg. 40, proposition 4.1.1 17

Lecture 8 (3) Suppose that K is non-archimedean with valuation ring R. Then L is non-archimedean, and its valuation ring is the integral closure of R in L (in particular, the integral closure of R in L is local). Remark. If K = R or C then this is easy - left as an exercise (i.e. to show that any extension of the usual absolute value to the complex numbers must be the usual complex absolute value). We will prove this for discretely valued non-archimedean absolute values (the general case requires an appropriate version of Hensel s lemma - see Cassel s book Local fields ). Lemma 7.21 (3.2). Let R be a discrete valuation ring, K = Frac(R), π a uniformiser and k = R/(π), and assume that K is complete. (1) Let f K[T ] be monic and irreducible. Suppose that f(0) R, then f R[T ]. (2) If L/K is finite and z L with N L/K (z) R, then z is integral over R. Remark. If we take R = Z and L = Q(i) and z = (1 + 2i)/(1 2i) then N L/K (z) = 1 Z but z is not integral over Z (Q is not complete - this shows that the condition that K be complete is necessary here). Proof. (1) Let d = deg(f), and let m be minimal with π m f = f (T ) = d i=0 a i T i R[T ]. If m 0 then we are finished (because then f R[T ]). Otherwise m > 0 so let j be the largest integer with a j R. So in k[t ], we have f (T ) = ā j T j + + ā 0 = ḡ h 12 (0 < j < d by hypothesis) where ḡ = ā j T j + + ā 0 and h = 0.T d j + + 1 (you may argue that this is stupid, but it does satisfy the conditions of Hensel s lemma) or better, we could take ḡ = (T j + a j 1 T j 1 + + ā0 ), h = (0.T d j + + 0.T + ā j ) ā j ā j To this we can apply Hensel s lemma (as a j R ), so we get a non-trivial factorisation of f. (2) Apply part (1) to the minimal polynomial of z over K. Proof: (of thm. 7.20). (For the (discrete) non-archimedean absolute value case). If is non-archimedean then it is bounded on Z.1 K K, so L is bounded on Z.1 L if it exists. First assume that [L K] = n <. First we prove existence. Define x L = N L/K (x) 1/n. We need to check the triangle inequality. Let x, y L with x L y L. It is sufficient to prove that x + y L y L. Equivalently, it is sufficient to prove that if z L 1 then z + 1 L 1 (set z = x/y). Let f be the minimum polynomial of z over K and m its degree. Then f(0) 1/m = z 13 L, so if z L 1 then f(0) 1 i.e. f(0) R which implies that f R[T ] by the previous thing, and f(t 1) is the minimum polynomial of z + 1, so 1 + z L = f( 1) 1/m 1. This proves existence. We were proving a theorem. Lecture 8 4 th February 11:00 12 Because since ak R (for k > j) we have v(a k ) > 0 so a k 0 (mod π) and thus they are 0 in the residue field (if a k R then v(a k ) = 0 so they can t be 0(mod π)). 13 The trace and norm of an element x L are defined to be the trace and determinant, respectively, of the endomorphism T x L L given by T x(α) = xα, of the K-vector space L: tr L/K (x) = tr(t x), N L/K (x) = det(t x). In the characteristic polynomial f x(t) = det(t1 T x) = t n a 1 t n 1 + + ( 1) n a n K[t] of T x, n = [L K], we recognise the trace and the norm as a 1 = tr L/K (x) and a n = N L/K (x). When L/K is separable, the characteristic polynomial is a power f x(t) = p x(t) d where d = [L K(x)], of the minimal polynomial of x. See Neukirch pp. 8-9. 18

Lecture 8 Proof: (thm. 7.20). We had L/K finite of degree n, and K complete with respect to (a non-archimedean absolute value). Let R L be the valuation ring of L : {x L N L/K (x) R} 14. Recall lemma 7.21 (2) which was that N L/K (x) R implies that x is integral over R. So every element of R L is integral over R. Since R L is normal (as a valuation ring) thus R L is the integral closure of R in L. If is any other absolute value on L extending, let R be its valuation ring. As extends, we have R R. R is normal, so it contains R L. So by theorem 3.5 (3), we have R = R L and, are equivalent. Now let L/K be an arbitrary algebraic extension. Then L = subfields finite over K = L α say. Define x L, for x L, to be x Lα for any L α containing x. This doesn t depend on α: because if K(x) L α then K(x), Lα are two extensions of to K(x) which are therefore the same. So this is well defined and is an absolute value. Remark. Some consequences of this: let K be the algebraic closure of K, a field complete with respect to some absolute value. Then there exists a unique extension of to K e.g. K = Q p, = p, there is a unique absolute value on Q p extending p. We write it as p. It is not discrete: p 1/n p = p 1/n p = 1/p 1/n (in fact, it is easy to see that Q p = p Z Q p = p Q, where p Q means rational powers of p and so on). Proposition 8.22 (3.3). Let K be complete with respect to a discrete valuation, and L/K a finite, separable extension. Then L is complete (with respect to the unique extension of the valuation) and discretely valued. Moreover R L R [L K] as an R-module. Proof. Let n = [L K]. Then L K 1/n, so L is discretely valued. As R L is the integral closure of R in a finite separable extension, R L is a finite separable extension, therefore R L is a finite R-module. R is a discrete valuation ring, so it is a principal ideal domain 15, so R L is free (being torsion-free) 16 and rank R (R L ) = dim K L = n. Let π K R and π L R L be uniformisers. Then π K R L = π e L R L for some e 1 as R L is a discrete valuation ring. So we have lim R L /π m R L = lim R L /π m L em R L = lim R L /π m KR m L lim ( R n m m πk mr ) as R-modules. Then if we look at the left-most and the right-most objects in the diagram above, if follows that R L R n as R is π-adically complete 17 i.e. R L is complete. Remark. If L/K is finite, then L is complete (without the assumption that the valuation is discrete 18 ). If L/K is infinite: L/K is typically not complete e.g. Q p is not complete. If L/K is finite, and the valuation is not discrete, then R L need not be a finitely-generated (or free) R-module. 14 The fact that an exponential valuation v on K associated with extends uniquely to L is a trivial consequence of a previous theorem. The extension w is given by the formula w(α) = 1 n v(n L/K(α)) if n = [L K] <. 15 See chapter III, proposition 3.9 in Neukirch. 16 Recall proposition 12.10 from Wilkin s notes: let M be a finitely-generated torsion-free module over a principal ideal domain R. Then M is a free module of finite rank over R. 17 Because K is complete, see theorem 6.17. 18 In the proof we did not actually use completeness of K except for the last part. 19

Lecture 9 8.1. Ramification degree and residue class degree. Now we are going to examine (more algebraically) finite (separable) extensions of complete discretely valued fields. Let K be such a field. Notation: we denote O K the valuation ring of K (also called the ring of integers of K ), π K a uniformiser, v K the normalised valuation on K (v K (π K ) = 1), k K = O K /π K O K the residue field. Let L/K be a finite extension. We use the same notation for the respective constructions in L, and O L O K. As π K π L O L, the inclusion O K O L induces an extension k K k L. Definition 8.12. The residue class degree is defined f = f(l/k) = [k L k K ] The ramification degree of the field extension is e = e(l/k) = v L (π K ) i.e. π e L O L = π K O L (n.b. as they re normalised, v L K is not necessarily v K ). Proposition 8.23 (3.4). Let L/K be a finite separable extension. Then (1) e(l/k)f(l/k) = [L K], (2) L K [L K] as topological K-vector spaces. Proof. (1) We have π K O L = πlo e L πl e 1 O L π L O L O L therefore 0 πe 1 L O L πl e O π LO L L πl e O O L L πl e O L as k K -vector spaces. So we have 19 e 1 dim kk O L /π K O L = i=0 dim kk πl i O L πl i+1o L = e dim kk O L π L O L = ef using lemma 3.7, and recalling that O L /π K O L O n K /π KO n K so dim k K O L /π K O L = n. (2) Follows from the proof of proposition 8.22. Lecture 9 6 th February 11:00 9.1. Unramified extensions. For the moment we are assuming (until the end of this section) that all valuations are discrete. Let K be complete with respect to v K (normalised), ring of integers O K, uniformiser π K and residue field k K. If L/K is finite then there is an associated valuation v L from before and we define e(l/k) = v L (π K ) and f(l/k) = [k L k K ] - we saw that [L K] = f(l/k)e(l/k). Definition 9.13. We say that L/K is unramified if e(l/k) = 1 (i.e. π K is also a uniformiser of L) and k L /k K is separable. Remark. In most cases of interest here, k K will be finite so the second condition is automatic. These types of extensions are easy to classify. 19 One has e 1 i=0 dim k K (πl i O L/πL i+1o L) = e 1 i=0 [dim k K (πl i O L) dim kk (πl i+1o L)] which is a telescoping sum and is equal to dim kk (O L ) dim kk (πl e O L) = dim kk (O L /π K O L ). 20

Lecture 9 Proposition 9.24 (3.5). Let L/K be finite and separable. The following are equivalent. (1) L/K is unramified. (2) L = K(x) for some x O L whose minimal polynomial over K is separable mod π K. If so, then O L = O K [x] for any x as in (2). Proof. Let L/K be unramified of degree n. So k L /k K is separable of degree n. So k L = k K ( x), x is separable over k K. Let x O L be any lifting of x to O L, g O K [T ] its minimal polynomial. Then ḡ( x) = 0 implies that ḡ is the minimal polynomial of x and so it is separable and g has degree n 20 so L = K(x). Conversely, suppose we have x as in (2) with minimal polynomial f O K [T ]. As f k K [T ] is separable, it must be irreducible (if not, then its factorisation would lift to a factorisation of f, by using Hensel s lemma). So if x k L is the image of x and f( x) = 0 with f irreducible and separable of degree n then k L = k K (x) is separable of degree n, i.e. L/K is unramified. If O L O K [x], then there exists y O L with π K y O K [x] but y O K [x] (viewing O K [x] as an O K -submodule of O L ). We write n 1 π K y = i=0 for a i O K. We have that 1, x,..., x n 1 is a basis for k L /k K. As y O L, thus π K y π L O L, so π K y = 0 in k L thus all ā i = 0 i.e. all a i are divisible by π K thus y O K [x]. Let L/K, M/K be finite and separable, with K as above. Any K-homomorphism L M maps O L to O M, so it induces a map k L k M (a k K -homomorphism). So L k L is a functor of categories: a i x i finite separable ( extensions of K ) ( finite ) extensions of k K ( unramified finite separable ) ( ) extensions of K extensions of k K Theorem 9.25 (3.6). (1) Let L/K be unramified, M/K any finite extension (separable). Then the map Hom K-alg (L, M) Hom kk -alg(k L, k M ) is bijective. (2) Let k /k K be any finite separable extension. Then there exists L/K unramified with k L k (as k K -algebras), and it is unique up to isomorphism. Proof. (1) Write L = K(x) as in proposition 9.24 (2), and f the minimal polynomial of x over K. We have Hom K-alg (L, M) = {y M f(y) = 0} = {y O M f(y) = 0} {ȳ k M f(ȳ) = 0} = Hom kk -alg(k L = k K ( x), k M ) where we have used Hensel s lemma. (2) Let k = k K ( x) with x separable over k K with minimal polynomial ḡ k K [T ]. Lift ḡ to some monic g O K [T ]. Let L = K(x), with g(x) = 0. Then by 20 Because it is unramified of degree n so [kl k K ] = n and k L = k K ( x) so g must have degree n since it is the minimum polynomial. 21

Lecture 10 proposition 9.24 (2), L/K is unramified of degree equal to the degree of g which is equal to the degree of [k k K ]. If L is any other unramified extension with k L k, applying (1) shows that any k K -isomorphism k L k L lifts to a (unique) isomorphism L L. Remark. A consequence of this is that the functor L k L gives an equivalence of categories finite separable ( extensions of K ) finite ( ) extensions of k K and it preserves degrees. Remark. Let L/K be an arbitrary algebraic extension (L need not be complete). We can extend the normalised valuation v K to a v K of L, which is of course not necessarily discrete or normalised. Then we say that L/K is unramified if its value group is the integers, that is v K (L ) = Z, and k L /k K is separable (and if you think about it, this is the same as requiring that every finite subextension L L K (with L finite) is unramified). The same equivalence holds (i.e. we can delete finite in the equivalence). Corollary 9.26 (3.7). Suppose that k K = F q is finite. Then K has a unique unramified extension of degree n 1, for every n 1, namely the splitting field of X q 1 1. Proof. These are the finite extensions of F q. Corollary 9.27 (3.8). (1) Let L/K be unramified. Then L/K is Galois iff k L /k K is, and if so then Gal(L/K) Gal(k L /k K ) canonically. (2) Let k K = F q finite. Then every finite unramified extension L/K is Galois, and there exists a unique σ L/K Gal(L/K) (called the arithmetic Frobenius) such that for all x O L, we have and it generates Gal(L/K). σ L/K (x) x q (mod π L ) Proof. This is an immediate consequence of the equivalence of categories. Lecture 10 8 th February 11:00 Last time we discussed unramified extensions which you recall: L/K is unramified if e(l/k) = 1 and k L /k K is separable i.e. π K O L = π L O L. We established an equivalence of categories finite unramified ( extensions L/K ) finite separable ( ) extensions of k K Remark. Let K/Q p be a finite extension (so complete with respect to the p-adic absolute value), k K = F q say. We have (where q = p n ) F q = F q n = (m,p)=1 F q (µ m ) (where µ m is the group of m th roots of unity). Thus if K nr = (m,p)=1 K(µ m ), then we see that K nr is the union of all finite unramified extensions of K - it is called the maximal unramified extension of K (it is not complete). It is rather easy to see what its Galois group is Gal(K nr /K) Gal(F q /F q ) = lim Gal(F q n/f q ) lim Z/nZ Ẑ 22

Lecture 10 One has the map (x x q ) Gal(F q /F q ) which corresponds with φ K (the arithmetic Frobenius of K nr /K) in Gal(K nr /K). The map φ K generates a dense subgroup of Gal(K nr /K). Often it is more convenient to consider F K = φ 1 K instead - called the geometric Frobenius. We will assume (for considerable simplification) that all L/K have k L /k K separable (e.g. k K perfect, or finite). Theorem 10.28 (3.9). Let L/K be a finite separable field extension with k L /k K separable. Then there exists a unique intermediate field K L 0 L such that L 0 /K is unramified and L/L 0 is totally ramified (i.e. f(l/l 0 ) = 1). If K F L, then L 0 F iff F /K is unramified (call L 0 the maximal unramified subfield of L/K). Proof. By theorem 9.25 (2), there exists K /K unramified with k K k L, and this induces K L - let L 0 be the image. Then L 0 /K is unramified of degree equal to [k L k K ] = f(l/k), so f(l/l 0 ) = [k L k L0 ] = 1 21. If F L 0, then F /K is unramified (trivial). Conversely, if F /K is unramified then we just by the same procedure look at k F k L0 = k L and then theorem 9.25 (1) gives F L 0. Remark. So the idea is that we can break or fields up into a totally ramified part and an unramified part. So to understand our extensions, we just need to know all about totally ramified extensions and unramified extensions. 10.1. Totally ramified extensions. Definition 10.14. A monic polynomial g T n + n 1 i=0 a i T i O K [T ] is said to be Eisenstein if v K (a i ) 1 for all i and v K (a 0 ) = 1 (so g is irreducible by the Eisenstein criterion). Theorem 10.29 (3.10). (1) If g O K [T ] is an Eisenstein polynomial and g(x) = 0 then L/K is totally ramified where L = K(x), and O L = O K [x] and v L (x) = 1. (2) Conversely, if L/K is totally ramified, and π L is any uniformiser of L, then its minimal polynomial is Eisenstein, and O L = O K [π L ]. Example. Let K = Q p and L = Q p (µ q ) = Q p (ζ q ) where ζ q q = 1, and q = p r. Let Φ q (T ) = T q 1 T p 1 q 1 and Φ q (T + 1) is an Eisenstein polynomial so O L = Z p [ζ q ] and π L = ζ q 1 is a uniformiser of L. 22 Proof. (1) Let g = T n + a n 1 T n 1 + + a 1 T + a 0. Let v K be a normalised valuation on K, extended to a valuation on L (not necessarily Z-valued), L = K(x). Then x n = n 1 i=0 a i x i implies that v K (x) > 0 (as v K (a i ) 1 for all i) thus for all i 0, we have v K (a i x i ) > 1 and v K (a 0 ) = 1. So v K ( n 1 i=0 a i x i ) = 1 by the triangle inequality, thus v K (x) = 1/n and recall that n = [L K]. Thus, since ef = n, this implies that e = n (i.e. that x is a uniformiser) 23 and f = 1 so L/K is totally ramified, π L = x say. Consider y = n 1 i=0 b i πl i L with b i K. Then v L (b i π i L) = i + nv K (b i ) i (mod n) 21 By the Tower Law - since kk k L one has [k K k K ] = [k L k K ] = f(l/k) and [k L k K ] = [k L k L0 ][k L0 k K ]? 22 A different proof of this is given in Neukirch II. 7, Proposition 13, pg. 159. 23 We set vl = nv K so then v L (x) = nv K (x) = 1, so it is a uniformiser in L and if π K is a uniformiser in K then v L (π K ) = nv K (π K ) = n so e = e(l/k) = n so f = 1. 23