Nonarchimedean Local Fields

Patrick Allen

Contents

Introduction 4 Local ﬁelds 4 The connection with number theory6 Global ﬁelds 8 This course 9

Part 1. Nonarchimedean fields 10 1. Valuations 11 2. Valued fields and the valuation subring 13 3. Completions 16 4. Hensel’s Lemma 19 5. Complete discretely valued fields 22 6. Extensions of complete valued fields 28 7. Ramification: first properties 33 8. Units 37 9. Norm groups 41

Part 2. Local class ﬁeld theory 44 10. The main theorems 45 11. Formal group laws 46 12. Lubin–Tate formal groups laws 53 13. Lubin–Tate extensions 57 14. The Artin map 61 15. Coleman’s norm operator 64 16. Base change 68 17. Ramiﬁcation groups 69 18. The Local Kronecker–Weber Theorem 74

Part 3. Further topics? 76 Appendix A. Inverse Limits 77 Appendix B. Integral extensions, norm and trace 82 Integral extensions 82 Norm and trace 83 Bibliography 87

3 Introduction

In this course we investigate the structure and Galois theory of nonarchimedean local fields. We assume the reader has had a course in graduate level algebra. Before beginning the course proper, we use this introduction to introduce general local fields, state their classification, and discuss the connections with number theory. Some of the things discussed in this introduction assume prerequisites beyond graduate level algebra. But these will not be necessary in understanding the material later, and are only meant to give the reader an initial idea of the bigger picture.

Local fields A topological field is a field K such that addition, negation, multiplication, and inversion K × K → KK → KK × K → KK× → K× (x, y) 7→ x + y x 7→ −x (x, y) 7→ xy x 7→ x−1 are all continuous (K × K is given the product topology, and K× = K r {0} the subspace topology). Note that both negation and inversion are homeomorphisms as they’re continuous involutions. Similarly, for any x ∈ K×, the map y 7→ xy is a homeomorphism with inverse y 7→ x−1y. Recall that a topological space X is called discrete if every subset of X is open, and is called locally compact if every element of X is contained in a compact neighbourhood. Definition 0.1. A local field is a nondiscrete, locally compact topological field.

Example 0.2. The most well known examples of local ﬁelds are R and C.

Example 0.3. Fix a prime p, and define a function vp : Z → Z ∪ {∞} by vp(n) = k, for n 6= 0, if k k+1 a p | n but p - n, and setting vp(0) = ∞. We can then extend vp to Q by setting vp( b ) = vp(a)−vp(b). This is called the p-adic valuation on Q, and we define the p-adic absolute value |·|p : Q → R≥0 by −vp(x) |x|p = p . One checks that this is an absolute value (i.e. a multiplicative norm) on Q, hence defines a metric on Q. The completion of Q with respect to this metric is a local field, called the field of p-adic numbers (or just the p-adics), and is denoted by Qp. P n Example 0.4. Let Fp be the finite field with p elements, and let Fp((T )) := { n−∞ anT | an ∈ } be the field of Laurent series over . Define |·| : ((T )) → by |f| = p−n0 if f = P a T n Fp Fp Fp R≥0 n≥n0 n with an0 =6 0. One checks that |·| is an absolute value on Fp((T )), and that the induced topology on Fp((T )) is nondiscrete and locally compact. Example 0.5. Let K be a local field, and let L be a finite field extension of K. Giving L the topology of a finite dimensional K-vector space, L is also a local field. Note that in each of Examples 0.2 to 0.4, the topology is induced from an absolute value. This is no accident, as we’ll see in the proof sketch of the following classification theorem, which shows the above examples exhaust all possibilities.

Theorem 0.6. Let K be a local field. If K has characteristic 0, then K is either isomorphic to R, to C, or to a finite extension of Qp for some prime p. If K has characteristic p, then K is isomorphic to a finite extension of Fp((T )).

4 LOCAL FIELDS 5

We will not use this result in the remainder of the course, so we only give an outline of the proof. The interested reader can consult [Wei67, Chapter 1] for the details.

Sketch. Since (K, +) is a locally compact abelian group, it admits a Haar measure: there is a measure µ on K such that µ(x + A) = µ(A) for any Borel subset A of K and any x ∈ K. Moreover, 0 it is unique up to scalar: if µ is another choice of Haar measure, then there is α ∈ R>0 such that µ0 = αµ. Fix a choice of Haar measure µ. Since multiplication by any x ∈ K× is an automorphism of (K, +), precomposing µ with multiplication by x yields another Haar measure µ0. By uniqueness, there 0 is a positive real α(x) such that µ = α(x)µ. Setting α(0) = 0, we have a function α : K → R≥0 that one proves is an absolute value 1 defining the topology on K, and moreover that K is complete with respect to α. Now assume that K has characteristic p. Since the topology on K is nondiscrete, we can find T ∈ K such that 0 < α(T ) < 1. The field Fp(T ) is contained in K, and since K is complete it contains the completion of Fp(T ) with respect to α|Fp(T ). One then proves that this is Fp((T )). If K has characteristic 0, then K contains Q. Since K is complete, it contains the completion of Q with respect to α|Q. Applying a theorem of Ostrowski, Theorem 0.14 below, one deduces that the restriction of α to Q is equivalent either to standard absolute value on Q, or to thep-adic absolute value for some prime p. Hence, K contains either R or Qp for some prime p. Finally, one concludes by showing that if L is a subfield of K such that the induced topology on L makes L a local field, then K is a finite extension of L.

Distinguishing fields according to their characteristic is certainly a natural thing to do. However with local fields, it turns out that Qp and Fp((T )) actually behave somewhat similarly and are drastically different from R and C. This distinction is formalized in the following definition. Definition 0.7. Let K be a field and let |·| be an absolute value on K. We say |·| is archimedean if the sequence (|n|)n≥1 is unbounded, and nonarchimedean if it is bounded. If K is complete with respect to |·|, then we say K is archimedean, resp. nonarchimedean, if |·| archimedean, resp. nonarchimedean.

Example 0.8. R and C are archimedean local ﬁelds, while Qp and Fp((T )) are nonarchimedean local ﬁelds.

In this course, we study the nonarchimedean local fields. We’ll see that their topology has a much more algebraic flavour than that of R of C. One dramatic difference is the following. If K is a nonarchimedean local field, complete for a nonarchimedean absolute value |·|, then |K×| is a discrete subgroup of R>0 (see Examples 0.3 and 0.4). This implies that open balls are also closed, so the topology on K has a base of clopen sets. In particular, this implies that K is totally disconnected. Another one of the most important differences is:

Proposition 0.9. Let |·| be a nonarchimedean absolute value on a ﬁeld K. For any x, y ∈ K (0.10) |x + y| ≤ max{|x|, |y|}.

The inequality (0.10) is known as the ultrametric inequality. Before proving Proposition 0.9, we prove an easy lemma.

Lemma 0.11. An absolute value |·| on a ﬁeld K is nonarchimedean if and only if |n| ≤ 1 for all n ∈ N. Proof. The “if” is obvious. Conversely, if there is a natural number m such that |m| > 1, then k (|m |)k≥1, and hence also (|n|)n≥1, is unbounded.

1 If K =∼ C, then α is the square of the usual absolute value |·| on C, hence isn’t technically an absolute value as |·|2 doesn’t satisfy the triangle inequality. However, it does satisfy |x + y|2 ≤ 2(|x|2 + |y|2), which is enough to do all of analysis (e.g. deﬁnes the same topology and the same notions of convergence and Cauchy sequences). THE CONNECTION WITH NUMBER THEORY 6

Proof of Proposition 0.9. Without loss of generality, we can assume |x| ≥ |y|. Then for any n−k k n n n ≥ 1 and any 0 ≤ k ≤ n, |x| |y| ≤ |x| and | k | ≤ 1 by Lemma 0.11. Thus, for any n ≥ 1, n n n X n n−k k n |x + y| = |(x + y) | ≤ |x| |y| ≤ (n + 1)|x| . k k=0 Taking nth roots, |x + y| ≤ (n + 1)1/n|x| for all n ≥ 1. The result follows from taking the limit n → ∞. One consequence of this is that one can “re-centre” open balls. Precisely, let |·| be a nonarchimedean absolute value on a ﬁeld K, and let B(x, r) ⊂ K be the open ball in K of radius r > 0 centred at x ∈ K. Then B(y, r) = B(x, r) for any y ∈ B(x, r).

The connection with number theory One of the motivations for studying local fields is their connection with number theory. For example, one might try to study questions about Q by studying these questions in the completions of Q with respect to various absolute values, where analytic tools are available. We discuss some examples of this below, but first one should understand all possible absolute values |·| on Q. This is answered by a theorem of Ostrowski, Theorem 0.14 below. Before stating it, we need a definition.

Definition 0.12. Let K be a ﬁeld. We say two absolute values |·|1 and |·|2 on K are equivalent if they generate the same topology on K. We say an absolute value is trivial if it deﬁnes the discrete topology on K.

Example 0.13. Let |·|p be the p-adic absolute value on Q from Example 0.3. Fix any real number 0 0 −vp(x) 0 r > 1, and deﬁne |·|p : Q → R≥0 by |x|p = r . Then |·|p and |·|p are equivalent. Indeed, log r log r 0 log p 0 log p |·|p = |·|p , so open balls of radius ε in the |·|p-topology are equal to open balls of radius ε in the |·|p-topology. On the other hand, if ` is a prime distinct from p, then |·|p and |·|` are not equivalent. n −n n n Indeed, |p |p = p and |p |` = 1 for all n ≥ 1, so the sequence (p )n≥1 converges to 0 with respect to |·|p but not with respect to |·|`. It is also easy to see that |·|p is not equivalent to the usual absolute value on Q given by ( x if x ≥ 0, |x|∞ := −x if x < 0.

Theorem 0.14 (Ostrowski). Any nontrivial absolute value on Q is equivalent to either |·|∞ or |·|p for some prime p.

Sketch. We content ourselves with showing that a nonarchimedean absolute value on Q is equivalent to |·|p for some prime p. For the remainder of the proof, i.e. showing that an archimedean absolute value on Q is equivalent to |·|∞, see [Kob84, Chapter I, §2] or [Neu99, Chapter II, §3]. Let |·| be a nonarchimedean absolute value on Q. By Lemma 0.11, we know that |n| ≤ 1 for all n ∈ N. It is not hard to show that if |n| = 1 for all nonzero natural numbers n, then |x| = 1 for all x ∈ Q×, hence is trivial. So there is a nonzero natural number n such that |n| < 1. By unique factorization and multiplicativity of |·|, there is a prime p such that |p| < 1. Let r = |p|−1. Using Proposition 0.9 and the multiplicativity of |·|, it is straightforward to verify that a := {x ∈ Z | |x| < 1} is an ideal in Z. Since p ∈ a and 1 ∈/ a, we must have a = pZ. Consequently, for any x ∈ Z, writing x = pkm with k ≥ 0 and m coprime to p, we have (0.15) |x| = |p|k|m| = r−k = r−vp(x).

Multiplicativity implies that (0.15) holds for all x ∈ Q, so |·| is equivalent to |·|p (see Example 0.13). THE CONNECTION WITH NUMBER THEORY 7

Philosophically, one might think of things as follows. The real numbers R remember the ordering on Q, but forgets all of its arithmetic (e.g. R has no way of knowing that 17 is prime). On the other hand, Qp does know that p is prime, but forgets all other primes ` 6= p as well as the order on Q. One might hope that using all the information of R, Q2, Q3, Q5,..., might give us some information about arithmetic questions over Q. A concrete example of this is the following.

Proposition 0.16. x ∈ Q is a square in Q if and only if it is a square in R and in Qp for all primes p. Proof. The only if part is obvious. To see the if direction, we may assume x 6= 0, and we can write Y x = pvp(x), p prime with ∈ {−1, 1}. Then x is a square if and only if = 1 and vp(x) is even for every prime p. If x is a square in R, then x > 0 and = 1. It remains to show that if x is a square in Qp, then vp(x) is even. × The absolute value |·|p on Q extends to Qp by continuity. But since |Q |p = pZ is a discrete × Z −vp(x) 2Z subgroup of R>0, we also have |Qp |p = p . Hence, if x is a square in Qp, p = |x|p ∈ p , and vp(x) is even. A more sophisticated example is the Hasse–Minkowski Theorem: Theorem 0.17 (Hasse–Minkowski). Let Q(x) be a rational quadratic form in n ≥ 1 variables (i.e. homogenous degree 2 polynomial in n variables with coefficients in Q). Then Q(x) = 0 has a nonzero n n n solution in Q if and only if Q(x) = 0 has a nonzero solution in R and in Qp for all primes p. We refer the reader to [Ser73, Part I, Chapter IV] for a proof. One can ask whether or not the conclusion of the above theorem, known as the Hasse principle, holds for more general systems of polynomial equations. This isn’t true in general. For example, one can show that the equation (x2 − 2)(x2 − 17)(x2 − 34) = 0 has solutions in R and in Qp for all primes p, but (visibly) does not have solutions in Q. Understanding for the failure of the Hasse principle in many situations is an active area of research. One such instance is for smooth projective curves over Q of genus one. In this context it is known that the Hasse principle does not hold, but it is conjectured that its failure is finite in a precise sense we briefly explain; the interested reader can consult [Sil86, Chapter X] for more details. We assume the reader has some knowledge of algebraic curves. Let C be a smooth projective curve of genus one over a field K. For simplicity, we’ll assume the characteristic of K is zero. By definition, C is an elliptic curve if C has a K-rational point. At the very least, we know C has points over an algebraic closure K of K, so C becomes an elliptic curve when base changed to K (i.e. viewing C as a curve over K). One can prove that the isomorphism class of this elliptic curve over K is actually defined over K. Precisely, there is an elliptic curve E over K, called the Jacobian of C, such that E and C are isomorphic over K. Thus, the isomorphism classes of smooth projective genus one curves are naturally partitioned by isomorphism classes of elliptic curves over K. For an elliptic curve E defined over K, the Weil–Châteletgroup of E, denoted WC(E/K), is the set of K-isomorphism classes of smooth projective genus one curves over K with Jacobian E. (It can be shown that WC(E/K) has a natural group structure, hence the name, but we don’t pursue that here.) Now let’s take an elliptic curve over E defined over Q. The Shafaravich–Tate group of E, denoted X(E/Q), is the subset (actually a subgroup) of the Weil–Châteletgroup consisting of the Q-isomorphism classes of curves C that have an R-point and a Qp-point for each prime p. Note that the isomorphism class of E is contained in X(E/Q), and X(E/Q) is a singleton if and only if the Hasse principle holds for every C ∈ WC(E/Q). So X(E/Q) measures the failure of the Hasse principle for smooth projective genus one curves with Jacobian E.

Conjecture 0.18 (Birch–Swinerton-Dyer [BSD65]). X(E/Q) is ﬁnite. GLOBAL FIELDS 8

This conjecture is notoriously difficult, and it took 20 years before it was known for even a single elliptic curve [Rub87]. There has been recent spectacular progress on this conjecture, and it is now known that it holds for > 66% of elliptic curves over Q (when ordered by height) [BSZ14]. Another situation in which it is useful and important to Q along with its completions Qp is in the study of Gal(Q/Q) and its representations. Fix an algebraic closure Q of Q. Then Gal(Q/Q) is naturally a topological group with the Krull topology. This is the topology where a neighbourhood base at the identity is given by the subgroups Gal(Q/F ) as F ranges over all finite extensions of Q. Equivalently, this is the coarsest topology that makes the restriction maps Gal(Q/Q) → Gal(F/Q) continuous for all F/Q finite and Galois, where we give the finite group Gal(F/Q) the discrete topology. For each prime p, we can also fix an algebraic closure Qp of Qp, and Gal(Qp/Q) is similarly a topological group. One can show that the natural embedding Q ,→ Qp extends to an embedding Q ,→ Qp, and that the image is dense. We’ll see later that Qp has a natural topology extending that of Qp and that the action of Gal(Qp/Qp) on Qp is continuous. Thus, an element of Gal(Qp/Qp) is uniquely determined by its restriction to Q under our fixed embedding Q ,→ Qp, and we get an embedding of groups Gal(Qp/Qp) ,→ Gal(Q/Q) that one can show has closed image. This image does depend on the choice of embedding Q ,→ Qp, but its conjugacy class does not. The following is a consequence of the Chebotarev density theorem.

Theorem 0.19. The union over all primes of the conjugacy class of Gal(Qp/Qp) is dense in Gal(Q/Q). For a more precise theorem and a proof, see [Neu99, Chapter VII, §13]. The local Galois groups Gal(Qp/Qp) are easier to understand than Gal(Q/Q), so Theorem 0.19 gives a useful way of understanding Gal(Q/Q). Further, the images of Gal(Qp/Qp) in Gal(Q/Q) carry arithmetic information. For example, whether or not a ﬁnite Galois extension F/Q is ramiﬁed at p is equivalent to knowing properties of the kernel of the composite

Gal(Qp/Qp) ,→ Gal(Q/Q) Gal(F/Q). These properties are crucial in deﬁnitions (sometimes modulo some conjectures) of higher rank L- functions associated to Galois representations and related objects studied in arithmetic geometry.

Global ﬁelds

We saw above that the local fields R and Qp, for any prime p, arise as completions of Q. It can be shown that any characteristic 0 local field arises as the completion of some (nonunique) number field with respect to some absolute value, and that there is a version of Ostrowski’s Theorem for arbitrary number fields. What about the characteristic p local fields? They arise in a similar way. First, a definition. Definition 0.20. A global field is a field F that is either – a finite extension of Q, or – a finite extension of Fp(T ) for some prime p. This definition may seem a bit ad hoc, and it is possible to give an axiomatic definition of a global field [AW45,AW46], but it is better to think in terms of Definition 0.20. Global fields of characteristic p are similar to global fields of characteristic 0 (i.e. number fields) in many ways. For instance, both are the field of fractions of Dedekind domains with finite residue fields (for example, compare Fp[T ] and Z). One can classify the absolute values on a characteristic p global field and show the completions are characteristic p local fields, and moreover show that any characteristic p local field is the completion of a (nonunique) global field with respect to some absolute value. As another example, the Hasse–Minkowski Theorem (Theorem 0.17) holds replacing Q with any global field F (and replacing R, Q2, Q3,... with all the completions of F ). However, the characteristic p global fields have some extra structure that the characteristic 0 local fields lack, which affords extra tools in their study. Namely, if F is a characteristic p global field, then THIS COURSE 9 there is a finite field F and a smooth projective geometrically irreducible curve C defined over F, unique up to isomorphism, such that F is the function field of C. This allows one to translate many arithmetic questions about F into geometric questions about C, allowing one to use tools in algebraic geometry. The analogy between number fields and function fields of curves over finite fields, largely expounded by Weil, has been incredibly useful in number theory. There have been many instances where one uses geometric techniques in the function field setting to prove theorems, and infer from this what might be true/provable in the number field case.

This course As mentioned in the beginning of this introduction, these notes cover the theory of nonarchimedean local fields, which we henceforth refer to as simply local fields. Part1 covers the basic structure of nonarchimedean local fields and their extensions. We first define complete nonarchimedean valued fields, and investigate their basic structure and arithmetic (e.g. their valuation subring and ideal, residue residue field, and group of units). We then discuss extensions of local fields, in particular the theory of ramification and how this is reflected in the structure of the Galois groups. Our principle references for this part are [Iwa86], [Ser79]. Part2 deals with local class field theory. Local class field theory completely describes the collection of all abelian Galois extensions of a local field (i.e. Galois extensions with abelian Galois group) purely in terms of the arithmetic of the local field, and in a canonical way. The main theorems are both incredibly elegant (in the author’s opinion) and incredibly useful. They do, however, take quite a bit of work to prove. There are two main approaches to proving the main theorems: via group cohomology as in [Ser79], or via formal groups as in [Iwa86]. We take the later approach. We first introduce formal groups and Lubin–Tate formal groups, proving the properties we need, and then state and prove the main theorems of local class field theory. Our principle references for this part are [Iwa86] and [Yos08]. Time permitting, we discuss some further topics in Part3. While I have some ideas of what will go here, I will wait to see how quickly we are moving through Parts1 and2 before deciding firmly. There will be several sections in the notes titled “Extra topic:. . . ” These are not necessary for any of the following material, but are inserted to give the reader a glimpse of some topics beyond the scope of this course related to the material in the previous sections. These sections will contain virtually no proofs, and will often assume more prerequisites than the rest of the notes; they are just meant to give the reader a glimpse down various rabbit holes. Part 1

Nonarchimedean ﬁelds 1. Valuations

Convention. R ∪ {∞} denotes the totally ordered set where a < ∞ for any a ∈ R. We extend addition to R ∪ {∞} by setting, for any a ∈ R ∪ {∞}, a + ∞ := ∞ and ∞ + a := ∞.

Definition 1.1. Let R be a commutative ring. A valuation on R is a functions v : R → R ∪ {∞} satisfying: (1) v(0) = ∞ and v(1) = 0, (2) v(xy) = v(x) + v(y) for all x, y ∈ R. (3) v(x + y) ≥ min{v(x), v(y)} for all x, y ∈ R.

Example 1.2. Let p be a prime ideal in a commutative ring R. Define a function v : R → R ∪ {∞} by ( ∞ if x ∈ p v(x) = 0 if x∈ / p. Let’s check that v is a valuation on R. Part1 of Definition 1.1 is immediate. Take x, y ∈ R. Since p is a prime ideal, v(xy) = ∞ ⇔ xy ∈ p ⇔ x ∈ p or y ∈ p ⇔ v(x) = ∞ or v(y) = ∞ ⇔ v(x) + v(y) = ∞, which shows part2 of Definition 1.1, and v(x + y) = 0 ⇔ x∈ / p and y∈ / p ⇔ v(x) = v(y) = 0, which shows part3. A special case of this example is that if R is an integral domain, we have a valuation v on R given by v(0) = ∞, and v(x) = 0 for all x 6= 0. This valuation is called the trivial valuation.

Example 1.3. Fix a prime number p. For nonzero x ∈ Z, we let vp(x) be the nonnegative integer vp(x) uniquely defined by x = p y with p - y. Setting vp(0) = ∞, we get a valuation vp : Z → R ∪ {∞}, called the p-adic valuation. Let’s check the details. Part1 of Definition 1.1 follows immediately from the definition of vp, as do parts2 and3 if either x or y are 0, so we assume x, y ∈ Z are nonzero. Writing x = pvp(x)m and y = pvp(y)n with m, n coprime to p, we have xy = pvp(x)+vp(y)mn and min{vp(x),vp(y)} vp(x) vp(y) p - mn, so vp(xy) = vp(x) + vp(y). Lastly, p divides x + y = p m + p n, so vp(x + y) ≥ min{vp(x), vp(y)}. Example 1.4. Let k be a field, and let f ∈ k[T ] be an irreducible polynomial. Any nonzero element g ∈ k[T ] can be written uniquely (up to order and units) as a product of irreducible polynomials, and we let vf (g) be the exponent of f in this decomposition. Setting vf (0) = ∞, we obtain a valuation vf on k[T ] called the f-adic valuation. The verification that is it a valuation is similar to that of Example 1.3. These two examples generalize to any unique factorization domain (see Exercise 1.1). It follows immediately form the definition that a valuation v : R → R ∪ {∞} on a commutative ring R satisfies – If xn = 1 for some n ≥ 1, then v(x) = 0. – v(−x) = v(x) for all x ∈ R. – v : R× → R is a group homomorphism. Another useful property that follows quickly from the definition is – if v(x) < v(y) then v(x + y) = v(x). Indeed, we know v(x + y) ≥ v(x) from the definition, and the converse follows from v(x) ≥ min{v(x + y), v(−y)} = min{v(x + y), v(y)} together with our assumption that v(x) < v(y). In this course, we will be interested in studying valuations on a field, and we’ll see shortly that the general case reduces to that of a field. Before showing this we first establish a lemma. 1. VALUATIONS 12

Lemma 1.5. Let R be an integral domain and let v be a valuation on R such that v(x) 6= ∞ for x 6= 0. Then v extends uniquely to the fraction field of R. Proof. Let K be the fraction field of R. The uniqueness is immediate, since any valuation w : K → R ∪ {∞} with w|R = v must satisfy a (1.6) w = v(a) − v(b) b for any a, b ∈ R with b 6= 0. So we show that (1.6) gives well-defined valuation on K. To see this is well defined, first note that v(a) − v(b) = v(a) + (−v(b)) is well defined according to our conventions a c on R ∪ {∞}, since we have assumed v(x) 6= ∞ if x 6= 0. Secondly, if b = d , then ad = bc so v(a) + v(d) = v(b) + v(c), and w is well-defined. We also see that w|R = v, and w(0) = ∞ and w(1) = 1. Fix a, b, c, d ∈ R with b, d nonzero. Then, a c a c w = v(ac) − v(bd) = v(a) − v(b) + v(c) − v(d) = w + w , b d b d and a c w + = v(ad + bc) − v(bd) ≥ min{v(a) + v(d), v(b) + v(c)} − v(b) − v(d) b d n a c o = min{v(a) − v(b), v(c) − v(d)} = min w , w . b d

Example 1.7. The p-adic valuation of Example 1.3 extends to Q. For nonzero x ∈ Q, vp(x) is the vp(x) a integer uniquely defined by x = p b with a, b integers coprime to p. Similarly, the f-adic valuation of Example 1.4 extends to k(T ). The uniqueness in 1.8 is often useful for checking two valuations on a field are the same, as one can check that they agree on any subring whose fraction field is the given field. Proposition 1.8. Let v be a valuation on a commutative ring R, and let p = {x ∈ R | v(x) = ∞}. Then p is a prime ideal and, letting k(p) denote the fraction field of R/p, there is a unique valuation w on k(p) such that v is the composite of the canonical homomorphism R → k(p) with w. Proof. By definition 0 ∈ p, and for any x, y ∈ p and z ∈ R, v(x ± y) ≥ min{v(x), v(y)} = ∞ and v(zx) = v(z) + v(x) = ∞, so p is an ideal. It is moreover prime since v(x) + v(y) = ∞ if and only if at least one of v(x), v(y) is ∞. For any x, y ∈ R with x∈ / p and y ∈ p, v(x + y) = v(x) since v(x) < v(y). It follows that v is the composite of the canonical map R → R/p with a unique valuation on R/p that we can uniquely extend to k(p) by Lemma 1.5 The prime ideal p in Proposition 1.8 is called the support of v. Definition 1.9. We say two valuations v and w on a commutative ring R are equivalent if for all x, y ∈ R, v(x) ≤ v(y) if and only if w(x) ≤ w(y). Example 1.10. If p and ` be distinct primes, the p-adic and `-adic valuations are not equivalent since vp(`) = 0 = vp(1), but v`(`) = 1 > 0 = v`(1). Similarly, if k is a field and f, g ∈ k[T ] are nonassociate irreducible elements, the f-adic and g-adic valuations on k[T ] are nonequivalent. One easy way to create two equivalent valuations is to multiply one by some fixed positive real number. In fact, this is the only way: Proposition 1.11. Two valuations v and w on a commutative ring R are equivalent if and only if there is a positive real r such that v = rw. 2. VALUED FIELDS AND THE VALUATION SUBRING 13

Proof. The equivalence of v and rv for any positive real number r is immediate. Let v and w be two equivalent valuations on R. Switching the roles of x and y in Definition 1.9, we see that v(x) = v(y) if and only if w(x) = w(y). In particular, {x ∈ R | v(x) = v(0)} = {x ∈ R | w(x) = w(0)}, i.e. they have equal supports. Applying Proposition 1.8, we reduce to the case that R is a field. Our assumption implies that v is trivial if and only if w is trivial, so we can assume they are both nontrivial. Then there is nonzero x ∈ R such that v(x) 6= 0, and replacing x by x−1 if necessary, we can v(x) assume v(x) > 0. Then we also have w(x) > 0, and r = w(x) is a positive real. We claim that v = rw. Replacing w with rw, we can assume that v(x) = w(x), and we want to show that v(y) = w(y) for all y ∈ R. Set α = v(x) = w(x), and assume for a contradiction that there is y ∈ R with v(y) 6= w(y). Note that y is necessarily nonzero, so v(y) and w(y) are both finite. Switching v and w if necessary, we can assume v(y) < w(y). We can find a positive integer k such that k(w(y) − v(y)) > α, and we can find an integer n such that kv(y) ≤ nα < kv(y) + α. But now v(yk) = kv(y) ≤ nα = v(xn), and w(yk) = kw(y) > kv(y) + α > nα = w(xn), a contradiction. Remark. Underpinning this proposition is the fact that the only automorphisms of the totally ordered abelian group R are multiplication by some fixed positive real.

Remark 1.12. Some references deﬁne a valuation as being a function |·|: K → R≥0 satisfying (1) |0| = 0 and |1| = 1, (2) |xy| = |x||y|, (3) |x + y| ≤ max{|x|, |y|}. The diﬀerence is only a matter of convention, as we can switch back and forth by |·| = e−v and v = − log|·|.

Remark 1.13. Strictly speaking, we have only defined valuations of rank at most one. For the general definition of a valuation, one replaces R ∪ {∞} in Definition 1.1 (resp. R≥0 in Remark 1.12) by Γ ∪ {∞} (resp. {0} ∪ Γ) where Γ is any totally ordered abelian group written additively (resp. multiplicatively). We don’t define the rank of a valuation here, but suffice to say that the rank of the valuation is ≤ 1 if and only if Γ can be embedded in R (resp. in R>0) as totally ordered abelian groups. We will not need the higher rank valuations in this course, so we have excluded them from the definition for simplicity.

Exercises. 1.1. Let D be a unique factorization domain, and let f ∈ D be an irreducible element. Show there is a unique valuation v : D → R ∪ {∞} satisfying v(f) = 1 and v(g) = 0 for any irreducible g not associate to f. 1.2. Let v be a valuation on a commutative ring R. Show that w : R[T ] → R ∪ {∞} deﬁned by n w(a0 + ··· + anT ) = min{v(a0), . . . , v(an)} is a valuation on R[T ] extending v. 1.3. Let k be a ﬁeld, and let v be the extension to k(T ) of the T −1-adic valuation on k[T −1]. Describe the restriction of v to k[T ]. 1.4. Construct two nonequivalent valuations v, w on Z[T ] satisfying v(T ) = w(T ) = 1. 1.5. Classify all equivalence classes of valuations on Z.

2. Valued fields and the valuation subring Definition 2.1. A valued field is a field equipped with a valuation. We say a valuation v on a field K is discrete if v(K×) is a discrete subgroup of R. 2. VALUED FIELDS AND THE VALUATION SUBRING 14

The only discrete subgroups of R are the cyclic groups Zα for some α ∈ R. (If you are not familiar with this fact, you should prove it for yourself.) We say a nontrivial discrete valuation v on a ﬁeld K is normalized if v(K×) = Z. If v is a nontrivial discrete valuation on a ﬁeld, there is a unique normalized discrete valuation in the equivalence class of v.

Example 2.2. The p-adic valuation on Q is discrete, nontrivial, and normalized. For k if a field and f ∈ k[T ] irreducible, the f-adic valuation on k(T ) is discrete, nontrivial, and normalized. Example 2.3. We construct an example of a nondiscrete valuation on a field. Let k be a field, and 1 let k(T ) be an algebraic closure of k(T ). Choose a compatible system of nth roots (T n )n≥1 of T , i.e. 1 m 1 1 1 (T m ) n = T n whenever n | m. For each n ≥ 1, k(T n ) is the fraction field of the polynomial ring k[T n ], 1 1 1 so we have the T n -adic valuation v 1 on k(T n ). Let vn be the valuation on k(T n ), equivalent to v 1 , T n T n 1 1 1 given by vn = v 1 . Note that for any n | m we have k(T n ) ⊆ k(T m ), and we claim vm| 1 = vn. n T n k(T n ) First, we have 1 m 1 m 1 1 v (T n ) = v (T m ) = = v (T n ). m n m n m n 1 1 1 1 Now let f ∈ k[T n ] be a nonzero element not divisible by T n . If f were divisible by T m in k[T m ], then 1 1 1 1 1 1 f would lie in the ideal T m k[T m ] ∩ k[T n ] of k[T n ], a proper ideal containing T n . But T n generates a 1 maximal ideal in k[T n ], so 1 1 1 1 1 T m k[T m ] ∩ k[T n ] = T n k[T n ], 1 contradicting our assumption that f is not divisible by T n . Thus vm(f) = 1 = vn(f). It follows that 1 1 1 vm agrees with vn on k[T n ], hence also on k(T n ). We now set K = ∪n≥1k(T n ) and we obtain a valuation v on K given by v| 1 = vn. This valuation is not discrete since k(T n )

1 × n 1 v(K ) ⊇ {0} ∪ {v(T ) | n ≥ 1} = {0} ∪ { n | n ≥ 1}.

Proposition 2.4. Let v be a valuation on a ﬁeld K. The subset Ov = {x ∈ K | v(x) ≥ 0} is a subring of K. Moreover Ov is local with maximal ideal mv = {x ∈ K | v(x) > 0}.

Proof. Clearly 0, 1 ∈ Ov and 0 ∈ mv. Both Ov and mv are subgroups of (K, +) since v(x ± y) ≥ min{v(x), v(y)}, and this is ≥ 0 if v(x), v(y) ≥ 0 and > 0 if v(x), v(y) > 0. For any x, y ∈ Ov, v(xy) = v(x) + v(y) ≥ 0 and is > 0 if either x or y is in mv. Thus Ov is a subring of K and mv is an ideal of Ov, proper since 1 ∈/ mv. Now Ov r mv = {x ∈ K | v(x) = 0}, and if v(x) = 0, then −1 v(x ) = −v(x) = 0. So any x ∈ Ov r mv is invertible. This shows Ov is local with maximal ideal mv.

Definition 2.5. For v a valuation on a ﬁeld K, the ring Ov = {x ∈ K | v(x) ≥ 0} is called the valuation ring of v, the maximal ideal mv = {x ∈ K | v(x) > 0} of Ov is called the maximal ideal of v, and the quotient Ov/mv is called the residue ﬁeld of v.

Example 2.6. The valuation ring of the p-adic valuation vp on Q is Z(p), the localization of Z at a the prime ideal (p) = pZ. Concretely, Z(p) = { b ∈ Q | p - b}. The maximal ideal of vp is the principal ideal generated by p, and the residue field is Fp. Similarly, if k is a field and f ∈ k[T ] is irreducible, the valuation ring of the f-adic valuation vf on g k[T ] is the localization of k[T ] at the prime ideal (f) = fk[T ], i.e. k[T ](f) = { h ∈ k(T ) | f - h}. The ∼ maximal ideal of vf is the principal ideal generated by f, and the residue field is k[T ](f)/fk[T ](f) = k[T ]/(f). Let v be a valuation on a field K. The following facts follow immediately from the definition and Proposition 2.4. – The valuation ring and maximal ideal of v depend only on the equivalence class of v. × × × ∼ × – Ov = {x ∈ K | v(x) = 0} and v induces an isomorphism K /Ov = v(K ). × y – For x, y ∈ K , v(x) ≤ v(y) if and only if x ∈ Ov. 2. VALUED FIELDS AND THE VALUATION SUBRING 15

This last fact implies that the equivalence class of v is completely determined by Ov. One of the most important examples for of this fact is the case when the valuation is nontrivial and discrete. Definition 2.7. A discrete valuation ring is a principal ideal domain with a unique nonzero prime ideal. An element that generates the unique nonzero prime ideal is called a uniformizer.

Example 2.8. The localization Z(p) of Z at the prime ideal pZ is a discrete valuation ring and p is a uniformizer. For k a field and f ∈ k[T ] irreducible, the localization k[T ](f) of k[T ] at the prime ideal fk[T ] is a discrete valuation ring and f is uniformizer. Generalizing both these examples, if R is a unique factorization domain and $ ∈ R is irreducible, then $R is a prime ideal, the localization R($) of R at $R is a discrete valuation ring, and $ is a uniformizer. Note that a discrete valuation ring is local, and its unique nonzero prime ideal is its unique maximal ideal. Further, if $ is a uniformizer, then any nonzero element x ∈ R can be written uniquely as x = u$n with n ∈ N and u ∈ R×. Indeed, any principal ideal domain is a unique factorization domain, so we can write x as a product of irreducibles, unique up to order and units. But every irreducible in a principal ideal domain is a prime element, so up to units, $ is the only irreducible in R. Then v(u$n) = n defines a valuation v on R (this is a special case of Exercise 1.1), which we can then extend to the fraction field K of R. Since $R is the unique nonzero prime ideal in R, the zero ideal is the 1 1 × n unique prime ideal of R[ $ ], and R[ $ ] = K. Hence, any x ∈ K can be written uniquely as x = u$ with n ∈ Z and u ∈ R×, and v(u$n) = n. We say that v is the valuation on K associated to R. It does not depend on the choice of $, and we have Ov = R and mv = $R.

Proposition 2.9. Let v be a nontrivial discrete valuation on a field K. The valuation ring Ov of v is a discrete valuation ring, and the valuation on K associated to Ov is equivalent to v. This defines a bijection between normalized nontrivial discrete valuations on K and discrete valuation subrings of K. Moreover, an element $ ∈ Ov is a uniformizer if and only if v($) = min{v(x) | x ∈ mv}. Proof. Let α = inf{v(x) | v(x) > 0}. Since v is nontrivial, α exists. Since v is discrete, α > 0, × there is $ ∈ K such that v($) = α, and v(K ) = Zα. For any nonzero x ∈ Ov, v(x) = nα for some x n n ∈ N, and v( $n ) = 0. Hence, any nonzero x ∈ Ov can be written in the form x = u$ with n ∈ N × and u ∈ Ov . It follows that mv = $Ov is the unique nonzero prime ideal. So Ov is a discrete valuation 0 0 0 ring, and $ is a uniformizer. Further, the valuation v associated to Ov satisfies v = αv , so v and v are equivalent. If v is normalized, then v0 = v. On the other hand, if R is a discrete valuation subring 0 of K with associated valuation v , then Ov0 = R. Remark. One can define an intrinsic notion of a valuation ring, generalizing discrete valuation rings, and prove a generalization of Proposition 2.9 that includes all valuations on K provided one uses the more general definition of valuation in Remark 1.13. We don’t pursue this here, and refer the interested reader to [Mat89, Chapter 4]. We finish by introducing some natural notation. Let R be a discrete valuation ring with fraction field K, and let v be the (normalized) valuation on K associate to R. Then for any integer n ≥ 1 and n n any choice of uniformizer $, mv = $ R = {x ∈ K | v(x) ≥ n}. We extend this to all n ∈ Z, i.e. for n n an integer n ≤ 0, we let mv denote {x ∈ K | v(x) ≥ n} = $ R. This is an R-submodule of K (see Exercise 2.2 for a more general statement), and does not depend on the choice of $. Exercises. 2.1. Let k be a field, and let α ∈ R be positive and irrational. Show that v : k[X,Y ] → R ∪ {∞} P n m given by v( an,mX Y ) = min{n + mα | an,m 6= 0} defines a valuation on k(X,Y ) and that v is nondiscrete.

2.2. Let v be a valuation on a ﬁeld K, and let Ov and mv be the valuation ring and maximal ideal, respectively, of v. For r ∈ R, let Br = {x ∈ K | v(x) > r} and Cr = {x ∈ K | v(x) ≥ r}. 3. COMPLETIONS 16

(a) Prove that Br and Cr are Ov-submodules of K. n (b) Assume that v is nontrivial discrete and normalized; so Cn = mv for any n ∈ Z. Show m m+n ∼ n mv /mv = Ov/mv as Ov-modules for any m, n ∈ Z with n ≥ 1.

2.3. Let v be a valuation on a ﬁeld K, and let Ov and mv be the valuation ring and maximal ideal, (r) × (r) × respectively, of v. For r ∈ R≥0, let V = {x ∈ O | v(x − 1) > r} and U = {x ∈ Ov | v(x − 1) ≥ r}. (r) (r) × (a) Prove that V and U are subgroups of Ov . (n) n (b) Assume that v is nontrivial discrete and normalized; so U = 1 + mv for any integer × (n) ∼ n × (n) (n+1) ∼ n ≥ 1. Show Ov /U = (Ov/mv ) and U /U = O/mv for any integer n ≥ 1.

2.4. Let v be a valuation on a ﬁeld K, and let Ov be the valuation ring of v. (a) Show that the ideals of Ov are totally ordered by inclusion. (b) Show that any ﬁnitely generated ideal of Ov is principal.

2.5. Let v be a nondiscrete valuation on a field K. Show that the maximal ideal mv of v is not finitely generated, hence the valuation ring Ov of v is not Noetherian. 2.6. Let k be a finite field. (a) Show that any valuation on k(T ) is discrete. 1 (b) Viewing k(T ) as the function field of one-dimensional projective space Pk over k, show that the maximal ideal of a valuation determines a bijection between equivalence classes 1 of valuations on k(T ) and points of Pk, and that it restricts to a bijection between 1 equivalence classes of nontrivial valuation on k(T ) and the closed points of Pk. 2.7. Let K be a valued field with valuation ring O, and let f, g ∈ O[X]. Write f = gh + r in K[X] with deg(r) < deg(g). Show that if the leading coefficient of g is a unit in O, then h, r ∈ O[X].

3. Completions

Definition 3.1. An absolute value on a field K is a function |·|: K → R≥0 satisfying (1) |x| = 0 if and only if x = 0, (2) |xy| = |x||y| for all x, y ∈ K, (3) |x + y| ≤ |x| + |y| for all x, y ∈ K. We say an absolute value |·| is ultrametric if it further satisfies (3’) |x + y| ≤ max{|x|, |y|} for all x, y ∈ K. Note that parts1 and2 imply |1| = 1. Conversely, if |·| is a multiplicative seminorm on a field K satisfying |1| = 1, then |·| is an absolute value, since 1 = |x||x−1| for x ∈ K×. Example 3.2. The trivial absolute value on a field K is the absolute value given by |0| = 0 and |x| = 1 for all x ∈ K×. An absolute value |·| on a field K defines a metric d, hence a topology, by d(x, y) = |x − y|. Proposition 3.3. The topology defined by an absolute value |·| on a field K is discrete if and only if |·| is the trivial absolute value. If |·| is not trivial, then 0 is a limit point of |K|. Proof. If |·| is trivial, then for any x ∈ K, the singleton {x} equals the open ball of radius 1 centred at x. So |·| defines the discrete topology. If |·| is nontrivial, there is x ∈ K× such that |x|= 6 1. −1 n n Replacing x by x , if necessary, we can assume |x| < 1. Then |x | = |x| → 0 as n → ∞. Given a valuation v : K → R ∪ {∞} on a field K we can associate to it an ultrametric absolute −v value |·|: R → R≥0 by fixing any real number r > 1 and setting |·| = r . The absolute value |·| depends on the choice of r, but only in a mild way: different choices of r yield the same topology, Cauchy sequences, and limits. To see this, first note that if |·| is an ultrametric absolute value on K and α > 0, then |·|α is another ultrametric absolute value on K that defines the same topology, Cauchy 3. COMPLETIONS 17 sequences, and limits as |·|, since an open ball of radius ε with respect to |·| is an open ball of radius εα log r2 α log r2 −v −v with respect to |·| . If r1, r2 are real numbers > 1, then > 0 and r = (r ) log r1 . Because of log r1 1 2 this, we will sometimes refer to |·| as the absolute value associated to v, even if we don’t specify the constant r. Note that to specify r it is equivalent to specify the absolute value of some nonzero element in the maximal ideal of the valuation.

Example 3.4. Let p be a prime and let vp be the p-adic valuation on Q. The p-adic absolute value −vp on Q is the ultrametric absolute value given by |·|p = p . On the other hand, if |·| is an ultrametric absolute value on a field K, then for any real number r > 1, v = − logr|·| is a valuation on K. Different choices of r yield equivalent valuations. So valuations and ultrametric absolute values on a field define equivalent theories, and can be thought of as different optics for viewing the same phenomena. For example, we showed in §1 that if v is a valuation on a field K and x, y ∈ K satisfy v(x) < v(y), then v(x + y) = v(x). In terms of ultrametric absolute values, this translates into the property that if |x| > |y|, then |x + y| = |x|. Under this correspondence, the trivial valuation corresponds to the trivial absolute value. The additive theory of valuations appears naturally from some algebraic perspectives. For example, the valuation associated to a discrete valuation ring has a canonical normalization, the one valued in Z ∪ {∞}, whereas its associated absolute value does not in general have a canonical normalization. On the other hand, the ultrametric absolute values are useful in analytic arguments as we can rely on results and intuition from the general theory of metric spaces.

Proposition 3.5. Let v1, v2 be two valuations on a ﬁeld K. Fix a positive real number r, and let −vi |·|i = r , for i = 1, 2. Then |·|1 and |·|2 generate the same topology on K if and only if v1 and v2 are equivalent.

Proof. If v1 and v2 are not equivalent, there are x, y ∈ K such that v1(x) ≤ v1(y) but v2(x) > v2(y). × y n y n y n Necessarily, x, y ∈ K . Then for all n ≥ 1, |( x ) |1 = | x |1 ≥ 1, so the sequence (( x ) ) is bounded y y n y n away from 0 in the |·|1-topology. On the other hand | x |2 < 1, so |( x ) |2 → 0 as n → ∞, and (( x ) ) converges to 0 in the |·|2-topology. Now assume that v1 and v2 are equivalent. One can conclude quickly that |·|1 and |·|2 define the same topology on K by appealing to Proposition 1.11. We give a different argument using only Definition 1.9. We need to show that open balls with respect to |·|1 contain open balls with respect to |·|2, and vice versa. As the argument is symmetric, we only show open balls with respect to |·|1 contain open balls with respect to |·|2. By Proposition 3.5, we can assume v1 and v2 are nontrivial. Fix x ∈ K and a positive real ε. By Proposition 3.5, there is z ∈ K such that 0 < |z|1 ≤ ε. Then using the equivalence of v1 and v2, we have

{y ∈ K | |x − y|1 < ε} ⊇ {y ∈ K | |x − y|1 < |z|1}

= {y ∈ K | v1(x − y) > v1(z)}

= {y ∈ K | v2(x − y) > v2(z)} = {y ∈ K | |x − y|2 < |z|2} Given a valuation v on a field K, one can define the associated topology on K without recourse to the associated absolute value by declaring the sets {y ∈ K | v(x − y) > r}, for x ∈ K and r ∈ R, to be a basis for the topology. In particular, if v is nontrivial and discrete with valuation ring Ov and n maximal ideal mv, then the cosets x + mv , for x ∈ K and n ∈ Z, form a basis for the topology. On the other hand, given an ultrametric absolute value |·| on a field K, we can define the valuation ring O and maximal ideal m, respectively, without recourse to the associated valuation by O = {x ∈ K | |x| ≤ 1} and m = {x ∈ K | |x| < 1}. We will refer to O and m as the valuation ring and maximal ideal, respectively, of |·|. Further, O is a discrete valuation ring if and only if |K×| is a nontrivial discrete subgroup of R>0, and $ ∈ O is a uniformizer if and only if |$| = max{|x| | x ∈ m}. Any metric space X admits a completion Xb, i.e. a complete metric space containing X as a dense subspace, which is unique up to unique isomorphism. We briefly recall the construction. One 3. COMPLETIONS 18 extends the metric d on X to a pseudometric d˜ on the set X˜ of all Cauchy sequences in X, by ˜ d((xn), (yn)) = limn d(xn, yy). Then d((xn), (yn)) = 0 defines an equivalence relation on X˜, and we define Xb to be the set of equivalence classes with respect to this equivalence relation. The pseudometric d˜ defines a metric db on Xb, and identifying x ∈ X with the class of the constant sequence (x, x, . . .), identifies X with a dense subspace of Xb. Proposition 3.6. Let |·| be an ultrametric absolute value on a field K. The completion Kb of K is a field, |·| extends to Kb, and |Kb| = |K|. Proof. The proof that Kb is a field is identical to the proof that R is a field. One shows that Q that the set of all Cauchy sequences in K is a subring of n≥1 K, and that the set of all Cauchy sequences (xn) such that limn xn = 0 is a maximal ideal in this ring. The details are left to the reader (Exercise 3.1). Let K˜ be the ring all Cauchy sequences in K. We can extend |·| to a multiplicative seminorm on K˜ by |(xn)| = limn|xn|. To see that it is ultrametric, take (xn) and (yn) Cauchy sequences in K. If limn|xn| = limn|yn|, then max{|xn|, |yn|} also converges to this common value. If limn|xn|= 6 limn|yn|, then either |xn| > |yn| for all n sufficiently large, or |yn| > |xn| for all n sufficiently large. In all of the above cases lim|xn + yn| ≤ lim max{|xn|, |yn|} = max{lim|xn|, lim|yn|}. n n n n If (xn) and (yn) are Cauchy sequences in K and limn yn = 0, then

lim|xn + yn| ≤ max{lim|xn|, lim|yn|} = lim xn. n n n n So |·| descends to Kb. On the field Kb, it is a multiplicative ultrametric seminorm satisfying |1| = 1, hence is an ultrametric absolute value. It remains to prove |Kb| = |K|, equivalently |Kb ×| = |K×|. Take any x ∈ Kb ×, so |x| > 0. Since K is dense in Kb, there is y ∈ K such that |y − x| < |x|. Then |y| = |x + y − x| = |x| since |·| is ultrametric and |y − x| < |x|. Corollary 3.7. Let v be a valuation on a field K. Choose a real number r > 1 and set |·| = r−v. The completion Kb as a topological field does not depend on r. The valuation v extends to Kb and satisfies v(Kb) = v(K). Proof. This follows immediately from Proposition 3.6 and the fact that different choices of r yield the same topology, limits, and Cauchy sequences. Because of this corollary, we also refer to Kb as the completion with respect to v. Corollary 3.8. Let R be a discrete valuation ring, let v be the associated valuation on its fraction field K, and let Kb denote the completion of K with respect to v. The valuation ring Rb of v in Kb is again a discrete valuation ring, and a uniformizer for R is a uniformizer for Rb. Proof. This follows from Propositions 2.9 and 3.6.

Example 3.9. The ﬁeld of p-adic numbers, denoted Qp, is the completion of Q with respect to the p-adic absolute value. Its valuation ring is called the ring of p-adic integers, and is denoted by Zp.

Example 3.10. Let k be a field, and let vT be the T -adic valuation on k(T ). We’ll show the completions of k(T ) with respect to vT is the field of Laurent series k((T )) in T with coefficients in k. First note that k(T ) ⊆ k((T )) and that v extends to k((T )) by v (P a T n) = n if a 6= 0. It T T n≥n0 n n0 is easy to see that k(T ) is dense in k((T )), since for any f = P a T n ∈ k((T )) and N ≥ n , the n≥n0 n 0 element g = PN a T n ∈ k(T ) satisfies v (f − g) > N. It remains to show that k((T )) is complete. n=n0 n T P n Let (fk) be a Cauchy sequence in k((T )), and write fk = n−∞ ak,nT . Since (fk) is Cauchy, for any N ≥ 0, there is M ≥ 1 such that v(fk − fm) > N for all k, m ≥ M. This implies that ak,n = ak,m for all n ≤ N if k, m ≥ M. Hence, for each n ∈ Z, the sequence (ak,n)k is eventually constant, so we can P n define an = limk ak,n and the element f = n−∞ anT ∈ k((T )) is the limit of (fn). 4. HENSEL’S LEMMA 19

Exercises. 3.1. Prove that the completion of a field equipped with a metric is again a field. 3.2. Let |·| be an ultrametric absolute value on a field K. (a) Show that a sequence (xn) in K is Cauchy if limn(xn+1 − xn) = 0. P (b) Show that if K is complete, then a series n−∞ xn converges in K if limn xn = 0. 3.3. Let |·| be an ultrametric absolute value on a field K. (a) Show that for any x ∈ K and real ε > 0, the open ball of radius ε centred at x is closed. Hence, K has a basis of clopen sets. (b) Deduce that K is totally disconnected. 3.4. Let |·| be an ultrametric absolute value on a field K, and denote by O and m the valuation ring and maximal ideal, respectively, of |·|. Let Kb be the completion of K with respect to |·|, and denote by Ob and mb the valuation ring and maximal ideal, respectively, of |·| in Kb. (a) Show that Ob and mb are the closures of O and m, respectively, in Kb. (b) Show that the natural map O/m → Ob/mb on residue fields is an isomorphism. 3.5. Let |·| be an ultrametric absolute value on a field K, and let O be the valuation ring. Show that K is complete with respect to |·| if and only if O is complete with respect to |·|.

3.6. Without using the results of the next section, show that Zp is compact.

4. Hensel’s Lemma We start investigating properties that are special to complete valued fields. Throughout this section K is a field complete with respect to a nontrivial ultrametric absolute value |·|, O is the valuation ring, m the maximal ideal, and k the residue field. One of the most important tools is Hensel’s Lemma, which gives a very useful criterion to determine when a polynomial in a complete valued field factors. Theorem 4.1 (Hensel’s Lemma). Let f ∈ O[X], and let f be its image in k[X]. Assume that f 6= 0 and factors as f = gh with g, h ∈ k[X] coprime. Then f factors as f = gh with g, h ∈ O[X] such that g mod m = g, h mod m = h, and deg(g) = deg(g).

Proof. Let d = deg(f) and e = deg(g). So deg(h) ≤ d − e. Choose any g0, h0 ∈ O[X] with g0 mod m = g, h0 mod m = h, deg(g0) = e, and deg(h0) ≤ d − e. Note that f − g0h0 has coefficients in m. Since g and h are coprime in k[X], we can find polynomials a, b ∈ O[X] such that ag0 + bh0 − 1 has coefficients in m. Let $ ∈ m be a coefficient of one of the polynomials f − g0h0 and ag0 + bh0 − 1 that has maximal absolute value. So f ≡ g0h0 mod $ and ag0 + bh0 ≡ 1 mod $. We inductively construct polynomials gn, hn ∈ O[X] such that for all n ≥ 0, n n (i) gn ≡ gn−1 mod $ and hn ≡ hn−1 mod $ , (ii) deg(hn) ≤ d − e, deg(gn) = e, and deg(gn − g0) < e. n+1 (iii) f ≡ gnhn mod $ .

Granting for the moment the construction of the gn and hn with the desired properties, we see that the congruences of property (i) and the boundedness of the degrees in property (ii), imply that we have well defined polynomials g = limn gn, whose jth coefficient is the limit of the jth coefficients of the gn, for each 0 ≤ j ≤ e. Since the eth coefficient in constant in this sequence, g has degree e. Similarly, there is a well defined polynomial h = limn hn of degree at most d − e. Then by property (iii), we have f = gh. It remains to inductively construct the polynomials gn and hn. We have chosen g0 and h0 above, n so we assume the result holds for some n − 1 ≥ 0. Then f − gn−1hn−1 ∈ $ O[X], and we let −n fn = $ (f − gn−1hn−1) ∈ O[X]. With a, b ∈ O[X] as above, we have

fn ≡ ag0fn + bh0fn mod $. 4. HENSEL’S LEMMA 20

Write bfn = qg0 + r in K[X] with deg(r) < deg(g0) = e. Since g0 mod m = g and deg(g0) = deg(g), the eth coeﬃcient of g0 is a unit in O. By Exercise 2.7, q, r ∈ O[X]. We now have

g0(afn + h0q) + h0r = g0fna + h0fnb ≡ fn mod $.

Let p be the polynomial obtained from afn + h0q by deleting all terms with coefficients that are 0 modulo $. Then we still have g0p+h0r ≡ fn mod $, and deg(p) ≤ d−e since deg(fn) ≤ d, deg(g0) = e, n n and deg(h0r) < d. We now set gn = gn−1 + $ r and hn = hn−1 + $ p. Property (i) above clearly holds. Property (ii) holds by inductive hypothesis and since deg(r) < e and deg(p) ≤ d − e. Finally, we check property (iii). Working mod $n+1, we have n gnhn ≡ gn−1hn−1 + $ (gn−1p + hn−1r) n n ≡ f − $ fn + $ (g0p + h0r) ≡ f. The following corollary is often called Hensel’s Lemma. Corollary 4.2. Let f ∈ O[X], and let f be its image in k[X]. If f 6= 0 and a ∈ k is a simple root of f, then f has a root a ∈ O with a mod m = a. Moreover a is the unique root of f in O satisfying a mod m = a. Proof. Applying Theorem 4.1 with g = X − a, we have g, h ∈ O[X] with deg(g) = 1 and f = gh. The leading coefficient of g is a unit in O, so we can assume g is monic and g = X −a with a mod m = a. If there were another root b ∈ O of f with b mod m = a, then f would be divisible by (X − a)(X − b) 2 and f would be divisible by (X − a) , contradicting the simplicity of a. Example 4.3. If the residue field k has finite of cardinality q, then K contains all (q − 1)st roots of unity. Indeed, the polynomial Xq−1 − 1 modulo m splits completely in k[X] and has distinct roots, so we can apply Corollary 4.2. In particular, Zp contains all (p − 1)st roots of 1. It is occasionally useful to have a version of Hensel’s Lemma that generalizes Corollary 4.2 in a direction different than that of Theorem 4.1.

0 2 Proposition 4.4. Let f ∈ O[X]. If there is a0 ∈ O such that |f(a0)| < |f (a0)| , then the sequence (an) inductively constructed by f(an−1) an = an−1 − 0 f (an−1) converges to a root a of f in O. Moreover

f(a0) (1) |a − a0| = | 0 | < 1, f (a0) 0 0 (2) |f (a)| = |f (a0)|, 0 (3) a is the unique root of f in O satisfying |a − a0| < |f (a0)|.

f(a0) Proof. Let α = | 0 2 |, which is < 1 by assumption. We show by induction on n ≥ 0 that f (a0)

f(a0) (i) an ∈ O and |an − a0| = | 0 | < 1 if n ≥ 1; f (a0) 0 0 (ii) |f (an)| = |f (a0)|; 0 2 2n (iii) |f(an)| ≤ |f (a0)| α . The n = 0 case follows from our assumptions, so we assume true for some n ≥ 0. f(a0) We ﬁrst show property (i) for n + 1. First note that |an+1 − a0| = | 0 | < 1 and a0 ∈ O implies f (a0) f(a0) an+1 ∈ O. If n = 0, |an − a0| = | 0 | by deﬁnition, so assume n ≥ 1. Property (iii) of the inductive f (a0) 2n−1 hypothesis implies |f(an)| ≤ |f(a0)|α < |f(a0)| since n ≥ 1. This together with properties (i) and (ii) of the inductive hypothesis imply

|f(a0)| |f(an)| |an − a0| = 0 > 0 = |an+1 − an|, |f (a0)| |f (an)| 4. HENSEL’S LEMMA 21 so |f(a0)| |an+1 − a0| = |an+1 − an + an − a0| = |an − a0| = 0 . |f (a0)| This shows the inductive step for property (i). m To show the inductive step for properties (ii) and (iii), we ﬁrst note that for any g = b0+···+bmX ∈ O[X] and x, h ∈ O, we have a Taylor expansion m g(x + h) = b0 + b1(an + h) + ··· bm(an + h) m m−1 2 = (b0 + ··· + bmx ) + (b1x + ··· + bmmx )h + bh = g(x) + g0(x)h + bh2,

f(an) 0 0 for some b ∈ O. We apply this with x = an and h = − 0 , to both f and f . Applying it ﬁrst to f , f (an) we have 0 0 f (an+1) = f (an) + ch for some c ∈ O. By our inductive hypothesis, 0 2n 0 |ch| ≤ |h| ≤ |f (an)|α < |f (an)|. 0 0 Thus, |f (an+1)| = |f (an)|. Now applying the Taylor expansion to f, 0 2 f(an+1) = f(an) + f (an)h + bh 0 2 for some b ∈ O. But f(an) + f (an)h = 0, so f(an+1) = bh and 2 0 4 2n+1 2 |f(an)| |f (a0)| α 0 2 2n+1 |f(an+1)| ≤ |h | = 0 2 ≤ 0 2 = |f (a0)| α . |f (an)| |f (a0)| This completes the induction. We now use properties (i) to (iii) to prove the proposition. First

|f(an)| 0 2n |an+1 − an| = 0 ≤ |f (a0)|α |f (an)| for all n ≥ 0, which tends to 0 as n → ∞ since α < 1. Thus (an) is Cauchy, so we can let a = limn an. Since an ∈ O for all n ≥ 0, so is a. Taking limits in property (iii), we see |f(a)| = 0, so a is root of f. Parts1 and2 of the proposition follow from taking limits in properties (i) and (ii). 0 It remains to show part3 of the proposition. Let b ∈ O be another root of f with |b − a0| < |f (a0)|. 0 Then |b − a| < |f (a0)|. Letting y = b − a, and applying the Taylor expansion to f(b) = f(a + y), we have f(b) = f(a) + f 0(a)y + cy2, for some c ∈ O. But f(a) = f(b) = 0, so |f 0(a)y| = |cy2|. If y 6= 0, then this implies |f 0(a)| ≤ |y|. But 0 0 |y| < |f (a0)| = |f (a)|, so we must have y = 0. Remark. Both Theorem 4.1 and Proposition 4.4 hold true more generally than we have discussed here (see [Bou62, Chapter III, §4]). We finish this section by giving a pleasant application of Hensel’s Lemma. Namely, that the only field automorphism of Qp is the identity. The reader may be familiar with the same result for R, i.e. that the only field automorphism of R is the identity. In both cases, the main point is to prove that a field automorphism is necessarily continuous. One then concludes using the density of Q and the fact that the only field automorphism of Q is the identity. In the case of R, one proves the continuity by proving a field automorphism must preserve the ordering on R, and this is because the ordering can be described algebraically: a ≤ b if and only if b − a is a square. Hensel’s Lemma allows us to prove something reminiscent for Qp: × Lemma 4.5. (1) Let p be an odd prime. An element x ∈ Qp is a (p − 1)m-th power in Qp for all integers m ≥ 1 coprime with p, if and only if x ∈ 1 + pZp. 5. COMPLETE DISCRETELY VALUED FIELDS 22

× (2) An element x ∈ Q2 is a 2m-th power in Qp for all odd integers m ≥ 1, if and only if x ∈ 1 + 8Z2.

Proof. First note that if x ∈ Qp is an rth power, for some integer r ≥ 1, then vp(x) ∈ rZ. So if x r is nonzero and an rth power for inﬁnitely many r, it must be a unit in Zp. Note also that if x = a × × × p−1 × and x ∈ Zp , then a ∈ Zp . Say p is odd and x ∈ Zp is a (p − 1)st power, so x = a for some a ∈ Zp . p−1 × 2 Then x = a ≡ 1 mod p, so x ∈ 1 + pZp. Now say p = 2 and x ∈ Zp is a square, so x = a for some × 2 a ∈ Zp . Then x = a ≡ 1 mod 8, so x ∈ 1 + 8Z2. (p−1)m Assume p is odd. Take any x ∈ 1+pZp and any integer m ≥ 1 coprime with p. Let f = X −a. Then f ≡ X(p−1)m − 1 mod p, so f(1) ≡ 0 mod p. Because m is coprime with p, we have f 0(1) ≡ (p − 1)m 6≡ 0 mod p. So 1 is a simple root of f mod p. By Hensel’s Lemma 4.2, the polynomial (p−1)m X − a has a root in Zp. 2m We turn to p = 2. Take any x ∈ 1 + 8Z2 and any odd integer m ≥ 1. Let f = X − a. Then 1 0 1 |f(1)|2 ≤ |8|p = 8 . Since m is odd, we have |f (1)|2 = |2m|2 = 2 . By Hensel’s Lemma 4.4, the 2m polynomial X − a has a root in Zp.

Proposition 4.6. The only ﬁeld automorphism of Qp is the identity.

Proof. Let σ : Qp → Qp be a field automorphism. Note that σ|Q is the identity. Since Q is dense in Qp, we will be finished if we show that σ is continuous. n The topology on Qp is generated by sets of the form a + p Zp with a ∈ Q and n ∈ Z. Thus σ is n n continuous if and only if σ(a + p Zp) = a + p Zp for all a ∈ Q and n ∈ Z. But since σ|Q is the identity, n n for any a ∈ Q and n ∈ Z, we have σ(a + p Zp) = a + p σ(Zp). Thus, σ is continuous if and only if k k σ(Zp) = Zp, and this happens if and only if σ(a + p Zp) = a + p Zp for some a ∈ Q and k ∈ Z. For any nonzero x ∈ Qp and integer r ≥ 1, x is an rth power if and only if σ(x) is an rth power. Lemma 4.5 then implies that σ(1 + pZp) = 1 + pZp for p odd, and σ(1 + 8Z2) = 1 + 8Z2. Exercises. 4.1. Given an example of a field K that is complete with respect to a nontrivial valuation with algebraically closed residue field, but such that K is not algebraically closed. Explain why this does not violate Hensel’s Lemma. What does this say about irreducible polynomials in K[X]? 4.2. Let K be a field complete with respect to a nontrivial discrete valuation, and let O be its valuation ring. Assume the residue field of K is finite. Let P be the set of elements x ∈ K× such that x is an mth power for infinitely many integers m ≥ 1. Prove P = O×. × × 2 × 4.3. (a) Let p be an odd prime. Prove that Qp /(Qp ) is a group of order 4, and that if c ∈ Zp × × is any element whose image is not a square in Fp , then {1, c, p, pc} ∈ Qp is a complete set of coset representatives. × × × 2 (b) Prove that u ∈ Z2 is a square if and only if u ∈ 1 + 8Z2. Deduce that Q2 /(Q2 ) has order 8, and that {1, −1, 5, −5, 2, −2, 10, −10} is a complete set of representatives for × × 2 Q2 /(Q2 ) . 4.4. (a) Let p be an odd prime. Show that the (p − 1)st roots of unity are the only roots of unity in Qp. (b) Show that ±1 are the only roots of unity in Q2. 4.5. (This exercises assumes the reader is familiar with quadratic reciprocity.) Prove that for every 2 2 2 prime p, the polynomial (X − 2)(X − 17)(X − 34) has solutions in Qp. (Note that it also has a solution in R, but not in Q. So this polynomial does not satisfy the Hasse principal from the introduction.)

5. Complete discretely valued fields In this section we focus principally on the discretely valued case. Throughout this section K is a field complete with respect to a nontrivial valuation v, O is the valuation ring, m is the maximal ideal, 5. COMPLETE DISCRETELY VALUED FIELDS 23 and k is the residue field. We say K has equal characteristic, or is equicharacteristic, if k has the same characteristic as K. Otherwise, we say K is mixed characteristic. Note that if K has characteristic p > 0, then p = 0 in O, hence also in k. Thus, if K has mixed characteristic, it has characteristic 0 with positive characteristic residue field. P Recall from Exercise 3.2 that a series n−∞ xn converges in a field complete with respect to a valuation if and only if limn xn = 0. A particularly useful case of this is when the elements xn are of n the form xn = any , where the an are any elements in the valuation ring and y is in the maximal ideal. Proposition 5.1. Assume v is discrete. Let $ ∈ O be a uniformizer and set α = v($), so v(K×) = Zα. Let S ⊂ O be a system of representatives for k with 0 ∈ S. (1) Every element x of K can be uniquely expressed as

X n (5.2) x = an$ with an ∈ S. n−∞

Moreover, if an = 0 for n < n0 and an0 6= 0, then v(x) = nα. P n P n (2) Let x = n−∞ an$ and y = n−∞ bn$ with an, bn ∈ S. Then v(x − y) ≥ Nα if and only if an = bn for all n < N. Proof. Without loss of generality, we can assume that v is normalized, so α = 1. We ﬁrst show existence of the expression (5.2). If x = 0, then we set an = 0 for all n, so we assume x 6= 0. Let n0 n0 = v(x), so x ∈ m . Since S is a system of representatives for k in O, we have O = S + m. Then for any integer N ≥ n0, mn0 = $n0 O = $n0 S + mn0+1 = ··· = $n0 S + $n0+1S + ··· + $N S + mN+1. We can thus inductively ﬁnd elements a ∈ S, for n ≥ n , such that x − PN a $n ∈ mN+1 for all n 0 n=n0 n N ≥ n0. Then N X n X n x = lim an$ = an$ . N n=n0 n≥n0 The uniqueness, as well as the claim on the valuation in part1, follow from part2 by taking y = 0, so we now focus on part2. Let n1 = min{n ∈ Z | an 6= 0}, n2 = min{n ∈ Z | bn 6= 0}, and m = min{n ∈ Z | an =6 bn}. Part2 follows if we show v(x − y) = m. We have

N N N X n X n X n (5.3) x − y = lim an$ − lim bn$ = lim (an − bn)$ . N N N n=n1 n=n1 n=m

Since S ⊂ O is a system of representatives of k and am 6= bm, we have am − bm ∈/ m. So v(am − bm) = 0, m n and v((am − bm)$ ) = m. For every n > m, we have v((an − bn)$ ) = v(an − bn) + n ≥ n > m. PN n Hence, v( n=m(an − bn)$ ) = m for all N ≥ m. This together with (5.3) implies v(x − y) = m.

P n Example 5.4. Any element in Qp can be expressed uniquely as n−∞ anp with an ∈ {0, . . . , p− P n 1}, and any element in Zp can be expressed uniquely as n≥0 anp with an ∈ {0, . . . , p − 1}. Alternatively, since Zp contains all (p − 1)st roots of unity, and these are a complete set of × P n representatives for Fp , we can write any element in Qp as n−∞ anp , with an either 0 or a (p − 1)st root of unity.

Corollary 5.5. With the notation and assumptions as in Proposition 5.1, the map ∞ Y X n ϕ: S → O given by ϕ(a0, a1,...) = an$ n≥0 n=0 Q is a homeomorphism, where we give S the discrete topology and n≥0 S the product topology. 5. COMPLETE DISCRETELY VALUED FIELDS 24

Proof. Without loss of generality, we can assume v is normalized. The fact that ϕ it is bijective follows immediately form part1 of Proposition 5.1. The topology on O is generated by sets of the form P n 0 B(x, N) = {y ∈ O | v(x−y) > N}, for x ∈ O and N ≥ 0. If x = n≥0 an$ , then B(x, N) = B(x ,N) 0 PN n 0 with x = an$ by part2 of Proposition 5.1. And B(x ,N) is the image under ϕ of the set n=0 Q BN;a1,...,aN = {(bn) ∈ n≥0 S | bn = an for n ≤ N}. The sets BN;a1,...,aN generate the topology on Q n≥0 S, so ϕ is a homeomorphism. Better, using inverse limits2 we can describe the ring of integers in a complete discretely valued field in an algebraic way. Proposition 5.6. Assume v is discrete. The canonical map ϕ: O → lim O/mn given by ϕ(x) = (x mod m, x mod m2,...) ←− n is an isomorphism of topological rings, where we give each O/mn the discrete topology. Conversely, if R is a discrete valuation ring with maximal ideal p, and the natural map R → lim R/pn is an isomorphism of rings, then the fraction field of R is complete with respect to the ←−n valuation associated to R. Proof. The map ϕ is a ring morphism with kernel {x ∈ O | x ∈ mn for all n ≥ 1} = {0}. Fix (a ) ∈ lim O/mn. For each n ≥ 1, take x ∈ O with x mod mn = a . Then for any N ≥ 1 n ←−n n n n N and m, n ≥ N, we have xn − xm ∈ m since the sequence (an) is compatible. It follows that (xn) is Cauchy, hence has a limit x since K is complete. Further, x ∈ O since O is closed in K. For N any N ≥ 1, x − xn ∈ m for n sufficiently large. In particular, taking n ≥ N sufficiently large, N N x mod m = xn mod m = aN . So ϕ(x) = (an). To see that it’s a homeomorphism, we can assume that v is normalized, in which case the topology on O is generated by sets of the form B(x, N) = {y ∈ O | v(x − y) > N}, for x ∈ O and N ≥ 0. Letting a = x mod mn for all n ≤ N, the set B(x, N) is taken to the set B = {(b ) ∈ lim O/mn | n N;a1,...,aN n ←−n n bn = an for all n ≤ N}. The sets BN;a1,...,aN for N ≥ 0 and an ∈ O/m , 1 ≤ n ≤ N, generate the topology on lim O/mn. So ϕ is a homeomorphism. ←−n Now assume that R is a discrete valuation ring with maximal ideal p such that the natural map R → lim R/pn is an isomorphism of rings. By Exercise 3.5, it suffices to show R is complete. Let ←−n N (xn) be a Cauchy sequence in R. Then for all N ≥ 1, xn − xm ∈ p for all m, n sufficiently large. N So for all N ≥ 1, xn and xm have the same image in R/p for all m, n sufficiently large, and we denote this common image by a . Then the sequence (a ) lies in lim R/pN , and we let x denote N N ←−N N the corresponding element in R. Then for all N ≥ 1, x and xn have the same image in R/p for all n N sufficiently large, i.e. x − xn ∈ p for all n sufficiently large. This shows x = limn xn. Definition 5.7. A complete discrete valuation ring is a discrete valuation ring R such that the canonical map R → lim R/pn is an isomorphism of rings, where p denotes the maximal ideal of R, ←−n equipped with the inverse limit topology.

Example 5.8. Zp is a complete discrete valuation ring, and we have canonical isomorphisms =∼ lim /pn =∼ lim /pn. So can be (canonically) thought of as the set of sequences (a ) with Zp ←−n Zp ←−n Z Zp n n n an ∈ Z/p and an+1 mod p = an. Using what we have proved so far, we can completely characterize when the topology on a ﬁeld complete with respect to a nontrivial valuation is locally compact. Proposition 5.9. The following are equivalent: (1) K is locally compact; (2) O is compact; (3) v is discrete and k is ﬁnite.

2See appendixA for a primer on inverse limits. 5. COMPLETE DISCRETELY VALUED FIELDS 25

Proof. Part3 implies part2 by Corollary 5.5 and Tychonoff’s Theorem. If O is compact, then x + O is a compact neighbourhood of x for any x ∈ K, so part2 implies part1. We now show part1 implies part3. We first show that the valuation v is discrete. Let V ⊆ K be a compact neighbourhood of 0 ∈ K. Then there is some real r0 > 0 such that {x ∈ K | v(x) > r0} ⊆ V . Choosing a real r > r0, {x ∈ K | v(x) ≥ r} is a closed ball contained in the compact set V , hence is compact, and has nonempty interior. We can thus assume V = {x ∈ K | v(x) ≥ r}. Choose a real s > r such that (r, s)∩v(K×) 6= ∅, which is possible since v is nontrivial. Letting U = {x ∈ K | v(x) > s}, U is an open subgroup of V . Choose a system of representatives S ⊂ V for V/U. Then tx∈S(s + U) is a disjoint open cover of V . By compactness of V , S is finite. On the other hand, any element y ∈ K with v(y) ∈ [r, s] can be written as y = x + u with x ∈ S and u ∈ U, and v(y) = v(x) since v(u) > s ≥ v(u). Thus, v(K×) ∩ (r, s) is finite, and it is nonempty by choice of s. It follows that we can find α ∈ v(K×) ∩ (r, s) and an open interval (a, b) ⊆ (r, s) such that (5.10) v(K×) ∩ (a, b) = {α}. Now for any β ∈ v(K×), we can translate (5.10) by β − α, and we have v(K×) ∩ (β − α + a, β − α + b) = {β}. This shows v is discrete. It only remains to show k is finite. Let V = {x ∈ K | v(x) ≥ r} = $rO be the compact set above. multiplication by $−r is a homeomorphism on K, so O = $−rV is also compact. Letting R ⊂ O be a set of representatives for k, O = tx∈R(x + m) is a disjoint open cover of O. Since O is compact, R, hence also k, is finite.

Example 5.11. The field of p-adic numbers is locally compact and Zp is compact. Let k be a field, let f ∈ k[T ] be irreducible, and let K be the f-adic completion of k(T ). The residue field of K is k[T ]/(f), which is a finite extension of k. Thus, K is locally compact if and only if k is finite. Definition 5.12. A nonarchimedean local field is a field complete with respect to a nontrivial valuation that is locally compact. We will sometimes (maybe often) refer to a nonarchimedean local field simply as a local field. Definition 5.13. If K is a nonarchimedean local field, and q is the cardinality of its residue field k, the normalized absolute value on K is the absolute value |·| = q−v.

−vp Example 5.14. The p-adic absolute value |·|p = p is the normalized absolute value on Qp. Remark. If K is a nonarchimedean local field, then (K, +) admits Haar measure. Fix a choice of Haar measure µ, and fix a Borel set A ⊂ K with nonzero finite measure. One can prove that the µ(xA) function |·|: K → R≥0 given by |x| = µ(A) is a nonarchimedean absolute value defining the topology on K, and that it does not depend on the normalization of µ nor on the choice of A. Thus, |·| is a canonically normalized absolute value associated to the valuation on K. We claim that it is the normalized absolute value on K. Taking for granted that it is a nonarchimedean absolute value defining the topology on K, it only remains to show that |$| = q−1 for a uniformizer $. We are free to choose A, and we take A = O. Then if S is a set of representatives for k in O, we have a disjoint union P O = tx∈S(x + $O). Then µ(O) = µ(tx∈S(x + $O)) = x∈S µ(x + $O) = qµ($O), since µ is Haar µ($O) −1 measure. Thus |x| = µ(O) = q . Although Proposition 5.1 shows that every element in a field complete with respect to a discrete valuation looks like a field of Laurent series, and allows us to characterize the topology in the same way, the operations of addition and multiplication on the field may not behave at all like those of Laurent series. For example, if K has characteristic 0, but its residue field k has characteristic p > 0 (for example, Qp), then it is impossible to find a set of representatives S for k that is closed under addition. Indeed, for any nonzero x ∈ O, the elements nx, with n ∈ N, will all be distinct since K has 5. COMPLETE DISCRETELY VALUED FIELDS 26 characteristic 0. But their images in k will not be distinct since k has positive characteristic. We’ll see that this is essentially the only obstruction. Proposition 5.15. Assume v is discrete. K =∼ k((T )) as valued fields if and only if O contains a subfield k0 mapping isomorphically onto k via the surjection O → k. Proof. The only if direction is clear. Assume that O contains a subfield k0 mapping isomorphically onto k via the surjection ϕ: O → k, and let $ ∈ O be a uniformizer. Then there is an injection of fields −1 ψ : k(T ) → K such that ψ|k = ϕ and ψ(T ) = $. The topology induced on k(T ) from the valuation on K is equivalent to the T -adic topology on k(T ). Then ψ extends to an injection of fields on the completion ψ : k((T )) → K. This is also surjective by Proposition 5.1 since ψ(k) is a complete system of representatives containing 0 for the residue field k, and ψ(T ) = $. Theorem 5.16. If v is discrete and K has equal characteristic, then K =∼ k((T )) as valued fields.

Proof when k has characteristic 0. Since k has characteristic 0, the canonical map Z → O → k extends to Q, so Q ⊂ O. Let k0 be a maximal subfield of O. It exists by Zorn’s Lemma. The induced map k0 → k is injective, as it is a map of fields. If we prove surjectivity, we are done by Proposition 5.15. Let k00 be the image of k0 in k, and take arbitrary nonzero a ∈ k. Assume first a is transcendental over k0. Choose any lift x ∈ O of a. Then x ∈ O×, so k0(x) ⊆ O. But x∈ / k0 since a∈ / k00. This contradicts the maximality of k0, so a must be algebraic over k00. Let f ∈ O[X] be such that its reduction f ∈ k[X] modulo m is the minimal polynomial of a. Then a is a simple root of f since k has characteristic 0, so we can apply Hensel’s Lemma (Corollary 4.2) and there is x ∈ O such that f(x) = 0 and x mod m = a. Then x ∈ O×, and k0(x) is a subfield of O. By maximality of k0, we have 0 00 x ∈ k and a ∈ k . We don’t prove the general case of equal positive characteristic, and refer the reader to [Mat89, Chapter 11] for this and more general statements. However, we do prove it when K has equal positive characteristic and the residue field is perfect; it will follow from Proposition 5.15 and Theorem 5.17 below. p A commutative Fp-algebra A is called perfect if the map Φ: x 7→ x is surjective. Note that Φ is a ring p homomorphism in characteristic p, since the binomial coefficients j vanish for all 1 ≤ j ≤ p − 1. Note also that Φ is injective on characteristic p fields, since if Xp − a has a root α, then Xp − a = (X − α)p, and α is the unique root. As we noted above, nothing like Theorem 5.16 can hold in mixed characteristic because the additive structure of K and k are different. It’s remarkable then, that one can relate their multiplicative structures. As a concrete example, recall that Zp contains all (p − 1)st roots of unity by Hensel’s Lemma. Letting S be the group of (p − 1)st roots of unity in Zp together with 0, we have a system of representatives of Fp closed under multiplication. This holds more generally. Theorem 5.17. Assume that v is discrete, and that k has characteristic p > 0 and is perfect. Then (1) there is a unique section ω : k → O of the surjection O → k that commutes with pth powers, i.e. ω(ap) = ap for all a ∈ k. Moreover, ω satisfies the following: (2) ω is multiplicative, ω(0) = 0, and ω(1) = 1. Hence, ω restricts to a homomorphism ω : k× → R×. (3) If K also has characteristic p, then ω is additive. Hence, ω defines an isomorphism of k with a subfield of K. Definition 5.18. With the notation and assumptions as in Theorem 5.17, the element ω(a) ∈ O is called the Teichmüller lift of a ∈ k, and is often written [a]. The homomorphism ω : k× → R× is called the Teichmüller character. Before proving Theorem 5.17, we establish a lemma. 5. COMPLETE DISCRETELY VALUED FIELDS 27

Lemma 5.19. Assume that v is discrete and that k has characteristic p, so v(p) > 0. Let a, b ∈ K n n be such that v(a − b) > 0, and set α = min{v(a − b), v(p)}. Then for all n ∈ N, we have v(ap − bp ) ≥ (n + 1)α. Proof. We induct on n ≥ 0, the n = 0 case being by assumption. Assuming true for some n−1 n−1 n − 1 ≥ 0, there is x ∈ K with ap = bp + x and v(x) ≥ nα. Then p n n X p ap = bp + xj, j j=1 p and since p | j for each 1 ≤ j ≤ p − 1, we have pn pn v(a − b ) ≥ min{v(p) + v(x), . . . , v(p) + (p − 1)v(x), pv(x)} ≥ (n + 1)α. Proof of Theorem 5.17. Without loss of generality, we can assume v is normalized. Since k is pn perfect, for any a ∈ k and n ≥ 0, there is unique an ∈ k such that an = a. For each n ≥ 0, choose pm añ ∈ O witha ñ mod m = an. For any m, n ≥ 0, we have v(ãn+m − añ) ≥ 1. Then by Lemma 5.19,

pn+m pn v(ãn+m − añ ) ≥ (n + 1), pn for all m, n ≥ 0. Thus, the sequence (ãn ) is Cauchy, and we let pn ω(a) = lim añ . n ˜ This limit does not depend on the choice of añ. Indeed, if bn is another choice of element in O with ˜ ˜ bn mod m = an, then v(ãn − bn) ≥ 1, and we can again apply Lemma 5.19 to conclude pn ˜pn lim(ãn − bn ) = 0. n pn pn So ω : k → O is well-defined. It is a section of O → k since añ mod m = an = a for all n ≥ 0, so the pn p same holds for ω(a) = limn añ . It also commutes with pth powers, since an ∈ k is the unique element p pn p p p satisfying (an) = a , anda ñ lifts an for all n ≥ 1, so p p p pn pn p ω(a ) = lim(ãn) = lim añ = ω(a) . n n 0 Now assume that ω : k → O is another section of O → k. As above, for each n ≥ 0, let an ∈ k be n 0 0 the unique p th root of a. Note that v(ω (an) − ω(an)) ≥ 1 for all n ≥ 0. If ω commutes with pth powers, then for all n ≥ 0, 0 0 pn pn v(ω (a) − ω(a)) = v(ω (an) − ω(an) ) ≥ (n + 1), again by Lemma 5.19. So ω0(a) = ω(a). We now show parts2 and3. For all n ≥ 0, the unique pnth root of 0 ∈ k is 0. Since we are free n to choose any lift añ ∈ O of these p th roots, we can choose añ = 0 ∈ O for all n ≥ 0 in the above construction. Thus ω(0) = limn 0 = 0. Similarly for 1 ∈ k. For any a, b ∈ k, if an and bn are the unique n n ˜ p th roots of a and b, respectively, then anbn is the unique p th root of ab. Thus, if añ and bn are any ˜ lifts in O of an and bn, respectively,a ñbn is a lift of anbn. Then ˜ pn pn ˜pn ω(ab) = lim(ãnbn) = lim añ lim bn = ω(a)ω(b). n n n Any multiplicative function k× → O× sending 1 to 1 is automatically a group homomorphism. For n ˜ each n ≥ 0, an + bn is the unique p th root of a + b, since k has characteristic p. Then añ + bn is a lift of an + bn, and if K also has characteristic p, we have ˜ pn pn ˜pn ω(a + b) = lim(ãn + bn) = lim(ã + b ) = ω(a) + ω(b). n n Finally, any additive and multiplicative map k → K sending 0 to 0 and 1 to 1 is necessarily a field homomorphism. 6. EXTENSIONS OF COMPLETE VALUED FIELDS 28

Exercises. P∞ n 1 5.1. Show that n=0 p converges to 1−p in Qp. 5.2. Let |·| be an absolute value on a field K, and give K the topology induced by |·|. Prove that if K is locally compact, then K is complete. (This holds more generally for any metric group.) 5.3. Let k be a field of characteristic p. Show that any purely inseparable extension of k((T )) is 1 isomorphic to k((T q )) with q a power of p. 5.4. Let K be a complete discretely valued field of equal characteristic, and let O and k denote its valuation ring and residue field, respectively. By Proposition 5.15, there is a subfield k0 ⊆ O that maps isomorphically onto k via the canonical surjection O → k. Is k0 necessarily unique? Prove or give a counterexample. 5.5. Show that the construction in the proof of Theorem 5.17 does not work for nondiscrete valuations. More specifically, give an example of a field K, complete with respect to a nontrivial valuation and with perfect characteristic p > 0 residue field k, an element a ∈ k, n and elements añ in the valuation ring that lift the p th roots of a in k, such that the sequence pn (ãn ) is not Cauchy. 6. Extensions of complete valued fields If L/K is an extension of fields and v is a valuation on K, an extension of v to L is a valuation w on L such that w|K = v. By an extension of valued fields, we mean a field extension L of K equipped with a valuation that extends the valuation on v. The fundamental result, Proposition 6.4 below, is that if K is complete with respect to v and L/K is algebraic, then v can be uniquely extended to L. Throughout this section, we let K denote a field complete with respect to a nontrivial valuation v. We let O denote the valuation ring of K, m the maximal ideal, and k the residue field. n Lemma 6.1. If f = anX + ··· + a0 ∈ K[X] is irreducible, then v(ai) ≥ min{v(a0), v(an)} for each 0 ≤ i ≤ n. Proof. After multiplying f by a suitable nonzero element of K, we can assume that f ∈ O[X] and that there is some 1 ≤ i ≤ n with v(ai) = 0. We then want to show that min{v(a0), v(an)} = 0. Assume otherwise, and let r = min{i | v(ai) = 0}. Then our assumption implies 1 ≤ r ≤ n − 1, and we can write f mod m = Xrh with h ∈ k[X] not divisible by X. But then Hensel’s Lemma (Theorem 4.1) implies there is a factorization f = gh in O[X] with deg(g) = r, which contradicts the irreducibility of f since 1 ≤ r ≤ n − 1. Lemma 6.2. Let L/K be a finite extension of fields. An element x ∈ L is integral over O if and 3 only if NL/K (x) ∈ O.

Proof. The fact that NL/K (x) ∈ O if x is integral over O is proved more generally in Corollary B.15. k We now prove the converse. Take x ∈ L such that NL/K (x) ∈ O, and let f = X + ··· + c0 ∈ K[X] [L:K(x)] be the minimal polynomial of x. Since c0 = ± NL/K (x) ∈ O, we have c0 ∈ O by considering valuations (or using the normality of O). Then Lemma 6.1 implies f ∈ O[X], so x is integral over O. Remark. This is not true without the assumption that K is complete. For example, let v be √ √ the 2-adic valuation on , and consider the extension ( −7)/ . The element 1+3 −7 has minimal Q Q Q √ 4 2 1 1+3 −7 polynomial X − 2 X + 4 over Q, which does not lie in Z(2)[X]. So 4 is not integral over Z(2), but has norm = 4. Proposition 6.3. Let F be a ﬁeld complete with respect to an ultrametric absolute value |·|. Let V be a ﬁnite dimensional normed vector space over F . Then for any basis B = {e1, . . . , en}, the norm

ka1e1 + ··· + anenkB := max{|a1|,..., |an|}

3See appendixB for a primer on integral elements, and on the norm of a ﬁnite ﬁeld extension. 6. EXTENSIONS OF COMPLETE VALUED FIELDS 29 is equivalent to the given norm on V . In particular, V is complete and the map F n → V given by (a1, . . . , an) 7→ a1e1 + ··· + anen is a homeomorphism. Proof. Let k·k denote the given norm on V . We want to show there are real numbers s ≥ r > 0 such that

rkxkB ≤ kxk ≤ skxkB for all x ∈ V. Pn We can take s = i=1keik, since for any a1, . . . , an ∈ F , n n X X ka1e1 + ··· + anenk ≤ |ai|keik ≤ max{|a1|,..., |an|} keik. i=1 i=1 To prove the existence of r, we induct on n = dim V . 0 When n = 1, we can take r = ke1k . Assume n > 1 and that the result holds for (n−1)-dimensional spaces. For each 1 ≤ i ≤ n, let Vi be the K-span of {e1, . . . , ei−1, ei+1, . . . , en}. By induction, each Vi is complete with respect to the restriction of k·k to Vi. In particular, each Vi is closed in V , hence so is n ei + Vi for each 1 ≤ i ≤ n. Then ∪i=1(ei + Vi) is closed in V , and does not contain 0, so we can ﬁnd a real s > 0 such that the open ball of radius s centred at 0 is disjoint from each ei + Vi. Take nonzero Pn −1 v = i=1 aiei, and let 1 ≤ j ≤ n be such that |aj| = max{|a1|,..., |an|}. Then aj v ∈ ej + Vj, so −1 kaj vk ≥ r and kvk ≥ rkvkB.

Proposition 6.4. If L/K be an algebraic extension, then (1) v extends uniquely to a valuation w on L. Moreover, (2) The valuation ring of w is the integral closure of O in L. 1 (3) If L/K is finite, then w = [L:K] v ◦ NL/K , and L is complete with respect to w. Proof. Since an infinite algebraic extension is the union of its finite subextensions, the claim in the infinite degree case follows from the finite degree case. So we assume n = [L : K] < ∞. We first show the uniqueness of w. Let w1 and w2 be two valuations on L extending v, and let OL,1 and OL,2 denote their respective valuation rings. Then for each i = 1, 2, OL,i is the subset × −n of elements x ∈ L such that x does not converge to 0 in the wi-topology, together with 0. By Proposition 6.3, the w1 and w2 topologies coincide, so OL,1 = OL,2. This implies w1 and w2 are equivalent, and w2 = rw1 for some real r > 0 by Proposition 1.11. Since they both restrict to v on K and v is nontrivial, r = 1. 1 Now define w : K → R ∪ {∞} by w = [L:K] v ◦ NL/K . We show that w is a valuation extending v. Since the norm is multiplicative, sending 0 to 0 and 1 to 1, we immediately have that w(0) = ∞, w(1) = 1, and w(xy) = w(x)+w(y) for all x, y ∈ K. It remains to show that w(x+y) ≥ min{w(x), w(y)} for any x, y ∈ K. This is clearly true if either x or y is zero, so we can assume they are both nonzero, x and that w(x) ≥ w(y). Then w(x + y) = w(y) + w( y + 1), and min{w(x), w(y)} = w(y). So we are x x x reduced to showing that if w( y ) ≥ 0, then w( y + 1) ≥ 0. This follows from Lemma 6.2, since y + 1 is x integral over O if y is integral over O. So w is a valuation. It is clear form the definition of w that the valuation ring of w is the set of x ∈ L such that NL/K (x) lies in O. This is the integral closure of O in L by Lemma 6.2. Finally, the fact that L is complete with respect to w follows from Proposition 6.3.

Because of the uniqueness, we will often refer to the valuation ring of w as the valuation ring of L, and similarly for the maximal ideal and residue field. Note that if we let OL, mL, and kL denote the valuation ring, maximal ideal, and residue field, respectively, of L, then we have inclusions O ⊆ OL and m ⊆ mL. We therefore get an inclusion k ⊆ kL, i.e. an induced extension of residue fields. Note × × also that since w|K = v, the value group v(K ) of v is a subgroup of the value group w(K ) of w. We will return to these observations shortly. 6. EXTENSIONS OF COMPLETE VALUED FIELDS 30

Remark. The uniqueness is not necessarily true if K is not complete. For example, in the Gaussian integers Z[i] we have a factorization 5 = (2 + i)(2 − i), with 2 + i and 2 − i nonassociate irreducibles. The (2 + i)-adic and (2 − i)-adic valuations on Q(i) both extend the 5-adic valuation on Q, and are nonequivalent.

Corollary 6.5. Let L/K be an algebraic extension, and let w be a valuation on L such that w|K is equivalent to v. For any field automorphism σ of L inducing the identity on K, we have w ◦ σ = w. In particular, σ is continuous, and if OL, mL, and kL denote the valuation ring, maximal ideal, and residue field, respectively, of w, then σ(OL) = OL and σ(mL) = mL, and σ induces a field automorphism of kL restricting to the identity on k. Proof. We are free to replace w by a valuation in its equivalence class. Hence we can assume that w|K = v, and the result follows from the uniqueness in Proposition 6.4. Corollary 6.6. Let L/K be a finite field extension. If O is a discrete valuation ring, then so is its integral closure OL in L. Proof. Let w denote the unique extension of v to L. Part2 of Proposition 6.4 implies that the integral closure OL of O in L equals the valuation ring of w. Since O is a discrete valuation ring, v is × 1 discrete. part3 of Proposition 6.4 implies that w(L ) = [L:K] Z, so w is also discrete. Proposition 2.9 implies OL is a discrete valuation ring. Remark. It is not true in general that an infinite algebraic extension L/K is complete (you are asked to give an example in Exercise 6.3). However, one can always take the completion with respect to the unique valuation w on L extending v and obtain a field Lb that contains L densely and is complete extending the original valuation v on K. In the case that L is an algebraic closure of K, the completion of L is often denoted by CK . One can show (Exercise 6.5) that CK is algebraically closed. It is thus the smallest (up to isomorphism) complete algebraically closed field containing K, which is the reason for the notation CK . In the case that K = Qp, we denote this field by Cp, and think of it as a p-adic version of the complex numbers.

Proposition 6.7. Assume that v is discrete, and let L/K be an algebraic extension. Let OL, mL, and kL denote the valuation ring, valuation ideal, and residue field, respectively, of L. Let w be the × × unique valuation on L with w|K = v, let e denote the index of the subgroup v(K ) in w(L ), and let f denote the degree of the induced extension of residue fields kL/k. If e and f are both finite, then OL is a finite free O-module of rank ef. Moverover, if {b1, . . . , bf } and {y0, . . . , ye−1} are subsets of OL such that {bi mod mL}1≤i≤f is a j j+1 basis for kL over k and yj ∈ mL rmL for each 0 ≤ j ≤ e−1, then B = {biyj | 1 ≤ i ≤ f, 0 ≤ j ≤ e−1} is a basis for OL over O.

Proof. If e is finite, then OL is also a discrete valuation ring, so there exists {y0, . . . , ye−1} ⊆ OL j j+1 such that yj ∈ mL r mL for each 0 ≤ j ≤ e − 1. Also, if f is finite, then clearly there exists {b1, . . . , bf } ⊆ OL such that {bi mod mL}1≤i≤f is a basis for kL over k. So the first claim follow from the second, and it remains to show B = {biyj | 1 ≤ i ≤ f, 0 ≤ j ≤ e − 1} is a basis for OL over O. e Let $ be a uniformizer for K, and note that $OL = mL. For each 0 ≤ j ≤ e − 1, the one- j j+1 j+1 j+1 dimensional kL-vector space mL/mL is spanned by {yj mod mL }, so {biyj mod mL }1≤i≤f is a j j+1 basis for mL/mL as a k-vector space. This implies that e e B mod mL = {biyj mod mL | 1 ≤ i ≤ f, 0 ≤ j ≤ e − 1}

e (0) is a basis for the k-vector space OL/mL = OL/$OL. So given any x ∈ OL, we can elements aij ∈ O and x1 ∈ OL such that f e−1 X X (0) x = aij biyj + $x1. i=1 j=0 6. EXTENSIONS OF COMPLETE VALUED FIELDS 31

(k) Repeating this process for x1 ∈ OL, and so on, we construct elements aij ∈ O for all k ≥ 0, 1 ≤ i ≤ f, P (k) k and 0 ≤ j ≤ e − 1, such that if we denote by aij ∈ O the limit of the convergent series k≥0 aij $ , we have f e−1 X X x = aijbiyj. i=1 j=0

This shows that B generates OL over O. Now say we have a nontrivial relation e−1 f X X aijbiyj = 0. j=0 i=1 × Then by dividing the aij by a suitable power of $, we can assume some aij ∈ O . Then taking this e e relation mod $OL = mL yields a nontrivial relation among the elements of B mod mL, contradicting e e that B mod mL is a basis for OL/mL. Corollary 6.8. Let the notation be as in Proposition 6.7. Then L/K is finite if and only if both e and f are finite. If this is the case, then [L : K] = ef. Proof. Assume that L/K is finite. Then by part3 of Proposition 6.4, e ≤ [L : K]. Now consider a finite subset {b1 . . . , br} ⊂ OL that is linearly dependent over K. Take a1, . . . , ar ∈ K, not all zero, such that

(6.9) a1b1 + ··· + arbr = 0. × × Scaling by an element of K , we can assume each ai ∈ O and that some ai ∈ O . Then taking (6.9) modulo mL, we see that {bi mod mL}1≤i≤r ⊆ kL is linearly dependent over k. Hence f ≤ [L : K]. The converse and the equality [L : K] = ef follow from Proposition 6.7 since any basis B for OL over O is also a basis for L over K. To see this, first note that for any x ∈ L, we can find a ∈ K× with sufficiently large valuation such that ax ∈ OL. Then ax can be written as an O-linear combination of elements in B. Multiplying this by a−1, we see x can be written as a K-linear combination of elements of B. This expression is also unique, since given two such expressions we an multiply it by an a ∈ K× of large enough valuation to obtain two O-linear expressions, which then must be equal. Multiplying −1 by a , the original two expressions are then equal. 3 2 Example 6.10. Let f = X + 2X − X − 3 ∈ Q5[X]. If f were reducible in Q5[X], it would be reducible in Z5[X] and f = f mod 5 would be reducible in F5[X]. But it is easy to check that f has not roots in F5, so f is irreducible over F5. So L = Q5[X]/f is a degree 3 field extension of Q5. Letting α ∈ L be a root of f, α is integral over Z5, hence belongs to the valuation ring of L. Its image in kL is a root of f, which is irreducible over F5, so [kL : F5] ≥ 3. Corollary 6.8 then implies that × × [kL : F5] = 3. Further, if w denotes the extension of the valuation v on Q5, then w(L ) = v(Q5 ) and 5 is a uniformizer for L. We can use Proposition 6.7 to prove existence of so-called power bases in most cases. We first isolate some special cases that come up often. Corollary 6.11. Let the notation be as in Proposition 6.7, and assume that L/K is finite.

(1) If f = 1, then for any uniformizer $L of L, we have OL = O[$L]. (2) If e = 1 and x ∈ OL is such that x mod mL generates kL over k, then OL = O[x].

Proof. For the ﬁrst part, applying Proposition 6.7 with {b1, . . . , bf } = {1} and {y0, . . . , ye−1} = e−1 e−1 {1, $L, . . . , $L } shows that {1, $L, . . . , $L } is a basis for OL as an O-module. For the seond f−1 par, applying Proposition 6.7 with {b1, . . . , bf } = {1, x, . . . , x } and {y0, . . . , ye−1} = {1} shows that f−1 {1, x, . . . , x } is a basis for OL as an O-module. To handle more general cases, we ﬁrst prove a Lemma. 6. EXTENSIONS OF COMPLETE VALUED FIELDS 32

Lemma 6.12. Assume that v is discrete. Let L/K be a finite extension with valuation ring OL and residue field kL. Let x ∈ OL be such that x mod mL generates kL over k. If there is a monic polynomial g ∈ O[X] of degree f = [kL : k] such that g(x) is a unifmorizer for L, then OL = O[x]. Proof. Let w denote the unique valuation of L extending v, and let e be the index of v(K×) is × i w(L ). Let mL be the valuation ideal of L. Our assumptions imply that {x mod mL | 0 ≤ i ≤ f − 1} j j j+1 is a basis for kL over k, and that g(x) ∈ mL r mL for each 0 ≤ j ≤ e − 1. Proposition 6.7 then i j implies that the products x g(x) with 0 ≤ i ≤ f − 1 and 0 ≤ j ≤ e − 1, are a basis for OL as an O-module. It follows that OL = O[x]. Proposition 6.13. Assume that v is discrete. Let L/K be a finite extension with valuation ring OL and residue field kL. If kL/k is separable, then there is x ∈ OL such that OL = O[x].

Proof. Let mL denote the valuation ideal of L. Since kL/k is a finite separable extension, there is x ∈ OL, such that x mod mL generates kL over k. We claim that we can find such a x so that there is a monic polynomial g ∈ O[X] of degree f = [kL : k], such that g(x) is a uniformizer for L. Then Lemma 6.12 implies OL = O[x]. First choose any y ∈ OL, such that y mod mL generates kL over k, and let g ∈ O[X] be any monic polynomial of degree f such that g mod mL is the minimal polynomial of y mod mL. Then g(y) ∈ mL. 2 If g(y) ∈/ mL, we take x = y. Otherwise, let $ ∈ OL be any uniforizer for L. They we have a Taylor expansion, g(y + $) = g(y) + $g0(y) + $2z, 0 2 for some z ∈ OL. Since kL/k is separable, we have g (y) 6≡ 0 mod mL, so g(y + $) ∈ mL r mL. We then take x = y + $. The quantities e and f in Proposition 6.7 and Corollary 6.8, and some notions that we derive from them, will be quite important in what follows. Before pursuing this further, we complete the classification of nonarchimedean local fields. Theorem 6.14. If K is a nonarchimedean local field of characteristic 0, then K is a finite extension of Qp, where p is the characteristic of the residue field k. Proof. By Proposition 5.9, the valuation v on K is discrete and the residue field k is finite. Without loss of generality, we can assume that v is normalized. Since K has characteristic 0, we know that Q ⊆ K. The valuation on K restricts to a nontrivial valuation on Q, since otherwise we would have Q× ⊆ O× and the induced map Q → k would be injective. Then v must restrict to the p-adic valuation on Q for some prime p. Since K is complete, the inclusion Q ⊂ K extends to Qp ⊆ K. Now the induced extension of residue fields k/Fp is finite, since k is a finite field, and the index of v(Q×) in v(K×) is v(p) < ∞, since v is discrete and normalized on K. The corollary then follows from Corollary 6.8. Recall that by Proposition 5.15 and Theorem 5.17, we know that if K is a nonarchimedean local field of characteristic p > 0, then K =∼ k((T )) as valued fields with k a finite field. Combining this with Theorem 6.14, we have the following pleasant classification theorem. Theorem 6.15. Any nonarchimedean local field is isomorphic, as valued fields, to either a finite extension of Qp or a finite extension of Fp((T )) for some prime p. Exercises. 6.1. Let K be a field complete with respect to a nontrivial absolute value |·|. Let L/K be a finite field extension equipped with a norm k·k restricting to |·| on K. Show that the trace and norm, TrL/K : L → K and NL/K : L → K, are continuous. (Apologies for the two different uses of “norm”!) 6.2. Let K be the fraction field of a discrete valuation ring. Show that an algebraic closure K of K has infinite degree over K. 7. RAMIFICATION: FIRST PROPERTIES 33

6.3. Let K be a nonarchimedean local field. Give an example of an infinite algebraic extension L of K such that L is not complete with respect to the unique valuation extending the one on K. 6.4. Let v be a valuation on an algebraically closed field K. Define a valuation w on K[X] by n w(a0 + ··· + anX ) = min{v(a0), . . . , v(an)} (this is a valuation by Exercise 1.2). (a) For monic f ∈ K[X] and root α ∈ K of f, show that v(α) ≥ w(f). (b) Show that if f, g ∈ K[X] are monic polynomials of the same degree n, and α ∈ K is a root of f, then v(g(α)) ≥ w(f − g) + (n − 1)w(f). (c) Let f, g ∈ K[X] be monic polynomials of the same degree n. Show that for any root α ∈ K of f, there is a root β ∈ K of g such that 1 v(α − β) ≥ w(f − g) + w(f). n (This property is known as continuity of roots.) 6.5. Let v be a nontrivial valuation on an algebraically closed field K. Show that the completion of K with respect to v is algebraically closed.

7. Ramification: first properties Throughout this section, we let K denote a field complete with respect to a nontrivial discrete valuation v. We let O denote the valuation ring of K, m the maximal ideal, and k the residue field. Definition 7.1. Let L/K be an algebraic extension, and denote again by v the unique extension of v to L. The residue degree of L/K, denoted f(L/K), is the degree of the induced extension of residue fields. The ramification index of L/K, denoted e(L/K), is the index of the subgroup v(K×) in v(L×). We say L/K is unramified if e(L/K) = 1 and the induced extension of residue fields in separable. Otherwise we say L/K is ramified. We say L/K is totally ramified if f(L/K) = 1. For an algebraic extension L/K, Corollary 6.8 implies that [L : K] = e(L/K)f(L/K). In particular, an algebraic extension L/K is both unramified and totally ramified if and only if it is the trivial extension. If the residue field is either characteristic 0 or is positive characteristic and perfect, then any algebraic extension is automatically separable, and the condition of being unramified reduces to e(L/K) = 1. We are most interested in the case of local fields, which satisfy this property as their residue fields are finite.

2 Example 7.2. The polynomial X − 2 ∈ Q3[X] is irreducible. Indeed, otherwise we would have 2 × a ∈ Q3 with a = 2. Taking valuations, we see a ∈ Z3 . But then a mod 3 would be a root of 2 2 X − 2 ∈ F3[X] in F3, which is a contradiction as 2 is not a square in F3. Thus, L = Q3[X]/(X − 2) × is a degree 2 extension of Q3. By the same argument, a root a ∈ L lies in OL and reduces to a root 2 of X − 2 ∈ F3[X] in kL. So [kL : F3] ≥ 2, and we must have f(L/Q3) = 2 = [L : Q3] and L/Q3 is unramified. Example 7.3. Let k be a field. For any integer n ≥ 1, consider the field extension k((T ))/k((T n)). Letting v denote the normalized valuation on k((T )), we have v(k((T n))×) = nZ, so the ramification index is n. On the other hand, the residue extension is the trivial extension k/k. So k((T ))/k((T n)) is a totally ramified extension of degree n. For a finitely ramified algebraic extension L/K, the ramification index can also be interpreted as follows. Since L/K is finitely ramified, the valuation on L extending K is also discrete, so the valuation ring OL of L is also a discrete valuation ring. Hence, letting mL denote the maximal ideal of e L, the ramification index e = e(L/K) is the positive integer such that mOL = mL. Indeed, if $ is a × uniformizer for K and $L is a uniformizer for L, then v(K ) is the cyclic subgroup of R generated by × × × v($) and v(L ) is the cyclic subgroup of R generated by v($L). So the index of v(K ) in v(L ) is 7. RAMIFICATION: FIRST PROPERTIES 34 the positive integer e such that v($) = w($) = ew($L). In particular, if L/K is unramified then a uniformizer for K is also a uniformizer for L. Recall that for an algebraic extension L/K, an element a ∈ K lies in the valuation ring of L if and only if it is integral over O (Proposition 6.4), which happens if and only if its minimal polynomial lies in O[X] (Proposition B.12).

Lemma 7.4. Let L/K be an algebraic extension, and denote by L, mL, and kL the valuation ring, maximal ideal, and residue field, respectively, of L. Let a ∈ OL and let g ∈ O[X] be the minimal polynomial of a. Let a = a mod mL ∈ kL and g = g mod mL ∈ k[X]. If g is irreducible over k and separable, then K(a)/K is unramified and its corresponding extension of residue fields is k(a)/k. Proof. The residue field of K(a) certainly contains k(a). Since g ∈ O[X] is monic, deg(g) = deg(g), so if g is irreducible, [k(a): k] = deg(g) = deg(g) = [K(a): K]. Corollary 6.8 then implies e(K(a)/K) = 1, and k(a)/k is separable by assumption. Proposition 7.5. (1) Let k0/k be a finite separable extension. There is an unramified extension L/K whose corresponding extension of residue fields is isomorphic to k0/k. (2) Let L/K be a finite unramified extension and let M/K be an algebraic extension Let kL/k and kM /k denote the respective extensions of residue fields. For any injection of fields σ : kL → kM restricting to the identity on k, there is a unique injection of fields σ : L → M restricting to the identity on K and inducing σ on the residue fields. Proof. We first prove part1. Since k0/k is separable, there is a ∈ k0 such that k0 = k(a). Let g ∈ O[X] be a monic polynomial such that g = g mod m ∈ k[X] is the minimal polynomial of a over k. Since g is irreducible over k, g is irreducible over K by Gauss’s Lemma, and we let L = K[X]/(g). Note [L : K] = [k0 : k]. Also, if we let b denote the image of X in L, then b is integral over O and its image b in the residue field of L is a root of g. By Lemma 7.4, L/K is unramified with residue extension k(b)/k, which is isomorphic to k0/k since a and b have the same minimal polynomial. We now show part2. Note first that since L/K is finite, any injection of fields L,→ M, resp. kL ,→ kM , restricting to the identity on K, resp. on k, takes image in a subfield of M, resp. of kM , that is finite over K. Morover, given two injections of field σ, σ0 : L,→ M that restrict to the identiy on K, they both take values in a subfield of M that is finite over K, namely the composite of σ(L) and 0 σ (L). Hence, in proving part2 of the proposition, we can assume M/K is finite. Take a ∈ kL such that kL = k(a), and let b = σ(a). Then k(b) ⊆ kM , and a and b have the same minimal polynomial g ∈ k[X]. Choose any monic g ∈ O[X] lifting g. Since L and M are both complete with repsect to their corresponding valuations (Proposition 6.4), Hensel’s Lemma (Corollary 4.2) implies there are unique roots a ∈ L and b ∈ M of g lifting a and b, respectively. By Lemma 7.4, K(a)/K is an unramified subextension of L/K with the same residue extension, so [K(a): K] = [L : K] and L = K(a). Then there is a unique injection of fields σ : L → M with σ(a) = b and σ|K the identity. This is the unique injection of fields restricting to the identity on K and inducing σ on the residue fields, since a ∈ L and b ∈ M are the unique roots of g lifting a and b = σ(a), respectively. We easily get many important corollaries.

Corollary 7.6. Let M/K be a algebraic extension with corresponding residue extension kM /k. The finite unramified subextensions L/K of M/K are in bijection with the finite separable subextensions of kM /k 0 Proof. Given a finite separable extension subextension k /k of kM /k, Part1 of Proposition 7.5 ∼ 0 implies there is a finite unramified extension L/K and an isomorphism σ : kL −→ k restricting to the identity on k, where kL is the residue field of L. Then part2 of Proposition 7.5 gives an isomorphism of L with a subfield K0 of M that is unramified over K and with residue field k0. Part2 of Proposition 7.5 also implies K0 is unique, since two different such extensions would yield two different embeddings of L into M lifting the same embedding of residue fields. 7. RAMIFICATION: FIRST PROPERTIES 35

Corollary 7.7. Let M/K be an algebraic extension with corresponding residue extension kM /k. Then there is a maximal unramiﬁed subextension L/K of M/K, whose residue extension is the maximal separable subextension of kM /k. We have e(M/K) = e(M/L), and if kM /k is separable, then f(M/L) = 1 and M/L is totally ramiﬁed.

Proof. The maximal separable subextension of kM /k is the union of all finite separable subextensions of kM /k, each of which corresponds to a unique finite unramified subsection of M/K by Corollary 7.6. It follows that the union of all finite unramified subextensions of M/K is an unramified extension of K with residue field equal to the maximal separable subextension of kM /k, and is the maximal unramified of M/K. The last claims follow from this and Exercise 7.1.

Corollary 7.8. Let L/K be an unramiﬁed extension with residue extension kL/k. If kL/k is Galois, then so is L/K and reduction modulo the maximal ideal of the valuation ring of L induces an ∼ isomorphism Gal(L/K) −→ Gal(kL/k).

Proof. First assume that L/K is finite. Then |AutK (L)| ≤ [L : K] with equality if and only if L/K is Galois. Since kL/k is Galois and L/K is unramified, |Autk(kL)| = [kL : k] = [L : K]. By part2 of Proposition 7.5, every element of Autk(kL) lifts uniquely to an element of AutK (L). So |AutK (L)| = [L : K], and L/K is Galois. Further, reduction modulo the maximal ideal of the valuation ring of L induces an isomorphism Gal(L/K) → Gal(kL/k), which is bijective by part2 of Proposition 7.5. Now assume that [L : K] = [kL : k] = ∞. Since kL/k is Galois, kL is the union of its subfields that are finite degree and Galois over k. By Corollary 7.6 and what we proved in the finite degree case, L is the union of its subfields that are finite degree and Galois over K. Hence, L/K is Galois. The fact that reduction modulo the maximal ideal of the valuation ring of L induces an isomorphism ∼ Gal(L/K) −→ Gal(kL/k) follows from what we proved in the finite degree case upon taking inverse limits (see appendixA).

Corollary 7.9. Assume that K is a local field. Then for any positive integer n ≥ 1 there is an unramified extension L/K of degree n; it is unique up to isomorphism. Moreover, L = K(ζ) with ζ a primitive (qn − 1)st root of unity, L/K is Galois, and Gal(L/K) =∼ Z/n canonically. Proof. Let q = |k|. Since k is finite, for any positive integer n ≥ 1, there is a finite extension k0/k of degree n, which is unique up to isomorphism and is separable over k. By Proposition 7.5, there is an 0 unramified extension L/K of degree n with residue field kL isomorphic to k , and it is unique up to × n isomorphism. Choose a generator ζ of the cyclic group kL , which has order q − 1. The Teichmüller × × n character ω : kL → OL is an injective homomorphism, so letting ζ = ω(ζ), ζ is a primitive (q −1)st root of unity in L lifting ζ. Since L/K is unramified, K(ζ) ⊆ L, and the residue field of K(ζ) is k(ζ) = kL, ∼ we have K(ζ) = L. Finally, by Corollary 7.8, L/K is Galois and Gal(L/K) = Gal(kL/k) canonically. q But Gal(kL/k) is cyclic with a canonical generator: the Frobenius automorphism x 7→ x .

Corollary 7.10. Assume that K is a local field and let K be an algebraic closure of K. For each ur n ≥ 1, let µn denote the group of nth roots of unity in K. The maximal unramified subextension K /K ur ur ur ∼ of K/K is given by K = ∪p-nK(µn). Moreover, K /K is Galois and Gal(K /K) = Zb canonically. Proof. By Exercise 7.2, the residue field k of K is an algebraic closure of k. Then by Corollary 7.6 there is a unique unramified subextension Kf /K of K/K of degree f for any f ≥ 1, and by Corollary 7.9 ur Kf = K(µqf −1), where q = |k|. So K = ∪f≥1K(µqf −1). This is clearly contained in ∪p-nK(µn). On the other hand, for every n coprime with p there is f sufficiently large such that n | (qf − 1). So ∪f≥1K(µqf −1) = ∪p nK(µn). - ∼ Further, for each f ≥ 1, the extension Kf /K is Galois Gal(Kf /K) = Z/f canonically by Corol- lary 7.9. And if d | f we have an inclusion Kf ⊆ Kd, and the natural surjection Gal(Kf /K) → Gal(Kd/K) given by restriction corresponds to the natural surjection Z/f → Z/d, since this is the case 7. RAMIFICATION: FIRST PROPERTIES 36 for the Galois groups of the corresponding residue extensions. From this we have canonical isomorphisms Gal(Kur/K) ∼ lim Gal(K /K) ∼ lim /f = . = ←− f = ←− Z Zb f f Definition 7.11. Assume that K is a local field, and let L/K be an unramified extension. The Frobenius element (or just Frobenius) in Gal(L/K) is the element that maps to the Frobenius q ∼ automorphism x 7→ x , where q = |k|, under the isomorphism Gal(L/K) −→ Gal(kL/k) given by reduction modulo the maximal ideal of the valuation ring of L. For an unramified extension L/K, the Frobenius element σ ∈ Gal(L/K) is the unique element such q that σ(x) ≡ x mod mL for every element x in the valuation ring OL of L, where mL is the maximal ideal of OL and q = |k|. Under the canonical isomorphisms of Corollaries 7.9 and 7.10, Frobenius is mapped to 1. Note also that in the case of Corollary 7.10, the cyclic subgroup of Gal(Kur/K) generated by Frobenius is dense (with Gal(Kur/K) given the profinite topology; see appendixA). The above corollaries give us a good handle on unramified extensions (at least in the case of local fields), so we now wish to understand the ramified extensions. By Corollary 7.7, it is essentially enough to understand totally ramified extensions. These can be described in a uniform way via Eisenstein polynomials. Definition 7.12. Let A be a discrete valuation ring, and let p denote its maximal ideal. A monic n n polynomial X + an−1X + ··· + a0 ∈ A[X] is called an Eisenstein polynomial if ai ∈ p for each 2 0 ≤ i ≤ n − 1, and a0 ∈/ p . Proposition 7.13. Let A be a discrete valuation ring with field of fractions F . Any Eisenstein polynomial is irreducible over F .

n n−1 Proof. Let g = X + an−1X + ··· + a0 ∈ A[X] be an Eisenstein polynomial. Since A is a unique factorization domain, it suffices to show that g is irreducible in A[X], by Gauss’s Lemma. Assume g = fh with f, h ∈ A[X]. Since g is monic, we can assume f and h are monic, and write i i−1 j j−1 2 f = X +bi−1X +···+b0 and h = X +cj−1X +···+c0, with i+j = n. Then a0 = b0c0 ∈ prp × implies precisely one of b0 and c0 is a unit in A. Without loss of generality, assume b0 ∈ A . Letting f = f mod p and h = h mod p, we have Xn = fh in A/p[X]. Since the constant term of f is nonzero, this is only possible if f ∈ A/p. Since f is monic, this is only possible if f ∈ A, and moreover f = 1. Proposition 7.14. Let g ∈ O[X] be an Eisenstein polynomial and let L = K[X]/(g). Then L/K is a totally ramified field extension and the image of X in L is a uniformizer for L. Conversely, if L/K is a finite totally ramified extension and $L is a uniformizer for L, then L = K($L) and the minimal polynomial for $L is an Eisenstein polynomial in O[X]. Proof. Without loss of generality, we can assume that the valuation v on K is normalized. n n−1 Let g = X + an−1X + ··· + a0 ∈ O[X] be an Eisenstein polynomial. It is irreducible by Proposition 7.13, so L/K is a field extension of degree n. Denote again by v the unique extension 2 of v to L, and let $L denote the image of X in K. Note that c0 ∈ m r m , so v(c0) = 1. Since 1 NL/K ($L) = ±a0, Proposition 6.4 implies v($L) = n . It follows that e(L/K) ≥ n, and then by Corollary 6.8, that e(L/K) = n and L/K is totally ramified. Now assume that L/K is totally ramified, and denote again by v the unique extension of v to L. Let n n−1 $L be a uniformizer for L, and let g = X + an−1X + ··· + a0 ∈ O[X] be its minimal polynomial. 1 Since L/K is totally ramified, v($L) = [L:K] . By Proposition 6.4, v(NL/K ($L)) = [L : K]v($L) = 1 m and v(NL/K ($L)) is a uniformizer for K. On the other hand, NL/K ($L) = ±a0 with m = [L : K($L)]. 2 Since this is a uniformizer, we must have m = 1, L = K($L), and a0 ∈ m r m . It remains to show that ai ∈ m for each 1 ≤ i ≤ n − 1. Let g = g mod m ∈ k[X]. Since a0 ∈ m, we see that X | g. Write g = Xjh with h ∈ k[X] and X - h. If we had j < n, then Hensel’s Lemma (Theorem 4.1) would imply n that g is reducible. Hence, g = X , and ai ∈ m for all 0 ≤ i ≤ n − 1. 8. UNITS 37

We finish with a definition, which gives us a Galois theoretic way of talking about ramification.

Definition 7.15. Let L/K be a Galois extension, and let K0/K be the maximal unramified subextension. The inertia subgroup, often denoted I(L/K), of Gal(L/K) is the subgroup Gal(L/K0) of Gal(L/K). If L is a separable closure of K, then the inertia subgroup of Gal(L/K) is often denoted by IK . Note that a Galois extension L/K is unramified if and only if the inertia subgroup of Gal(L/K) is trivial. Further, if L/K is a Galois extension with separable residue extension, then the cardinality of the inertia group subgroup I(L/K) equals the ramification index of L/K, and the index of I(L/K) in Gal(L/K) equals the residue degree of L/K.

Exercises. 7.1. Let K be a field complete with respect to a nontrivial valuation v. Let L/K and M/L be finite extensions. Show e(M/K) = e(M/L)e(L/K) and f(M/K) = f(M/L)f(L/K). 7.2. Let K be a field complete with respect to a nontrivial valuation v. Let L/K be an algebraic extension. (a) Show that the induced extension of residue fields is algebraic. (b) Show that if L is an algebraic closure of K, then the residue field of L is an algebraic closure of the residue field of K. 7.3. Let K be a nonarchimedean local field and let q denote the cardinality of its residue field. n (a) Show that for any integer n ≥ 1, a splitting field for Xq −1 − 1 over K is an unramified extension of degree n. (b) Let K be an algebraic closure of K. Show that the maximal unramified subextension Kur/K of K/K is given by adjoining to K all root of unity of order prime to p. n 7.4. Consider an extension Qp(ζ)/Qp with ζ a p th root of unity. Show the following. (a) Q(ζ)/Q is a totally ramified extension of degree (p − 1)pn−1. (b) Q(ζ)/Q is Galois, and Q(ζ)/Q =∼ (Z/pn)×. (c) Zp[ζ] is the valuation ring of Qp(ζ), and ζ − 1 is a uniformizer. 7.5. Give an example of a nonarchimedean local field K and two finite totally ramified extensions L/K and M/K, such that the composite extension LM/K is not totally ramified.

8. Units Before proceeding with our study of ramification, we first need to understand the units in a nonarchimedean local field. Throughout this section K will denote a nonarchimedean local field. We let O, m, and k denote the valuation ring, maximal ideal, and residue field, respectively, of K. Let p be the characteristic of k, and let q = |k|. Finally, let v denote the normalized valuation on K. × × n We introduce a filtration {Un}n≥0 on O . Set U0 = O , and for any n ≥ 1 set Un = 1 + m . Note × n × that Un is the kernel of the homomorphism O → (O/m ) . Proposition 8.1. The topology on K× as a subspace of K makes K× a nondiscrete, totally × disconnected, locally compact abelian group. The subgroups {Un}n≥0 are compact, U0 = O is the × × unique maximal compact subgroup of K , and the collection {xUn | x ∈ K , n ≥ 0} is a basis for the topology.

Proof. The fact that K× is a nondiscrete, totally disconnected topological group follows from the fact that K is a nondiscrete, totally disconnected topological ﬁeld. For any n ≥ 1, since mn is compact n × × and open in K, Un = 1 + m is open and compact in K . Since U0 = O is a disjoint union of q − 1 × many sets, each homeomorphic to U1, we deduce that U0 is also compact and open. For any x ∈ K and n ≥ 0, × × xUn = {xy ∈ K | v(y − 1) ≥ n} = {z ∈ K | v(z − x) ≥ n + v(x)}, 8. UNITS 38

× × which generates the topology on K . It remains to show that U0 = O is the unique maximal compact × × n subgroup of K . For any compact subgroup U of K and any x ∈ U, the sequence (x )n≥1 has a × convergent subsequence. This is only possible if v(x) ≥ 0, so U ⊆ O . Let $ be a uniformizer for K. The choice of $ defines a splitting of the homomorphism v : K× → Z. × n Moreover, K is the disjoint union of the open sets {$ U0}n∈Z, so we have a decomposition of topological groups × K = $Z × U0. × × ∼ × By Exercise 2.3, the map U0 = O → k defines an isomorphism U0/U1 = k , and the Teichmüller character (Theorem 5.17; or by Hensel’s Lemma) gives a splitting of the surjection O× → k×. Let µq−1 ⊂ U0 be the image of the Teichmüllercharacter; it consists of all (q − 1)st roots of unity. Since U0 is the disjoint union of the open sets {ζU1}ζ∈µq−1 , we get a further decomposition of topological groups

U0 = µq−1 × U1. × So to determine the structure of K , it only remains to determine the structure of U1. ∼ p By Exercise 2.3, Un/Un+1 = k for any n ≥ 1. In particular, for any n ≥ 1 and x ∈ Un, x ∈ Un+1. pm Repeating this, we see that for any m, n ≥ 1 and x ∈ Un, we have x ∈ Un+m. So the natural n Z-module structure on the abelian group U1/Un+1 factors through Z/p . Composing with the canonical n projection Zp → Z/p , we obtain a canonical Zp-module structure on U1/Un+1 for any n ≥ 1, and for any n ≥ m ≥ 1, the surjections U1/Un+1 → U1/Um+1 are surjections of Zp-modules. Then we can define a Zp-module structure on lim U /U = {(a ) | a ∈ U /U and a mod U = a for all n ≥ 1}. ←− 1 n+1 n n≥1 n 1 n+1 n+1 n+1 n n≥1 But an argument similar to Proposition 5.6 shows that the canonical map U → lim U /U 1 ←− 1 n+1 n≥1 given by a 7→ (a mod Un+1)n≥1 is an isomorphism of topological groups. We thus get a canonical a Zp-module structure on U1. That is, for any u ∈ U1 and a ∈ Zp, u makes sense and is an element of U1. Moreover, that this defines the structure of a Zp-module is equivalent to saying that the usual exponential rules hold: (uv)a = uava, ua+b = uaub, (ua)b = uab, u0 = 1, etc. a Another way to interpret this element u is as follows. Let (am)m≥1 be a sequence of integers with am a a = limm am. Then one can show that (u )m≥1 is Cauchy, hence has a limit. This limit is u . To determine the structure of U1, we will describe it as a Zp-module. When K has characteristic 0, this can be done via the p-adic logarithm and exponential. By Exercise 8.1, if the characteristic of K is 0, we have a continuous function log: U1 → K P m−1 xm n e called the logarithm, given by log(1 + x) = m≥0(−1) m , and log(Un) ⊆ m if n > p−1 , where e = e(K/Qp) is the ramification index of K/Qp.

Lemma 8.2. Assume K has characteristic 0. The function log: U1 → K is a Zp-module homomorphism. Proof. There is an equality of formal power series log((1 + X)(1 + Y )) = log(1 + X) + log(1 + Y ) in Q[[X,Y ]]. This implies an equality of formal power series in K[[X,Y ]], hence also an equality at any points at which both sides converge. By deﬁnition log(1) = 0. This shows log is a group homomorphism. Now let a ∈ Zp, and let u ∈ U1. Choose a sequence of integers (an) with an → a. Since log is a an group homomorphism, log(u ) = an log(u) for any n ≥ 1. Since log is continuous

a an log(u ) = lim log(u ) = lim an log(u) = a log(u). n n 8. UNITS 39

By Exercise 8.1 again, if the characteristic of K is 0, we a continuous function n exp: m → Un e called the exponential, for any n > p−1 with e = e(K/Qp) the ramification index of K over Qp (which is finite by Theorem 6.14). e Lemma 8.3. Assume that the characteristic of K is zero, and let n > p−1 with e = e(K/Qp) the n n ramification index of K over Qp. Then log: Un → m and exp: m → Un are inverse isomorphisms of topological groups. Proof. There are equalities of formal power series exp(log(1 + X)) = 1 + X and log(exp(X)) = X in Q[[X]]. This implies an equality of formal power series in K[[X]], hence also an equality at any points at which both sides converge. This together with Exercise 8.1 shows that log and exp determine n e continuous inverse bijections between Un and m when n > p−1 . Since log is a homomorphism by n Lemma 8.2, so is exp (this also follows from an equality of formal power series), and log: Un → m and n exp: m → Un are inverse isomorphisms of topological groups. Proposition 8.4. Assume that K has characteristic 0, and let P denote the group of all p-power roots of unity in K. Then P is a finite cyclic subgroup of U1 and U1/P is a free Zp-module of rank [K : Qp]. Proof. Taking n sufficiently large, Lemma 8.3 implies that the logarithm defines an isomorphism n of topological groups log: Un → m . This is further an isomorphism of Zp-modules by Lemma 8.2. Letting $ denote a uniformizer for K, multiplication by $n defines an isomorphism of topological ∼ n O-modules O = m . Applying Corollary 6.8, we deduce that Un is a free Zp-module of rank [K : Qp]. Since Un is finite index in U1, we see that U1 is a finitely generated Zp-module. Since Zp is a principal ideal domain, we can apply the structure theorem for finitely generated modules over principal ideal domains. So letting W be the torsion submodule of U1, it is finitely generated, and the quotient U1/W is a free Zp-module of some finite rank d. Since Un is a finite index submodule of U1, and is free of m rank [K : Q], we must have d = [K : Q]. The only nonzero proper ideals in Zp are of the form p Zp, so the torsion submodule of U1 is P , the set of p-power roots of unity. Since every element in P is finite order and P is finitely generated, P is finite. It is further cyclic since any finite subgroup of the group of roots of unity in a field is cyclic. × ∼ × × Plugging this into our decompositions K = $Z ×O and O = µq−1 ×U1, we obtain the following corollary.

Corollary 8.5. Assume that K has characteristic 0. Let d = [K : Qp], and let r = max{s | K contains a primitive psth root of unity}. Then we have isomorphisms of topological groups × ∼ r d × ∼ r d O = Z/(q − 1) × Z/p × Zp and K = Z × Z/(q − 1) × Z/p × Zp.

Proposition 8.6. Assume that K has characteristic p. Then U1 is isomorphic, as topological groups, to the direct product of countably many copies of Zp. ∼ Proof. By Proposition 5.15 and Theorem 5.17, K = k((T )). Fix a basis β1, . . . , βf for k over Fp. Let n be a natural number coprime to p, and deﬁne a continuous homomorphism f f Y n ai gn : Zp → Un by gn(a1, . . . , af ) = (1 + βiT ) . i=1 Let s ≥ 0 be an integer, and set m = nps. We claim the following: s f (1) Um = gn(p Zp )Um+1; f s s (2)( a1, . . . , af ) ∈/ pZp if and only if gn(p a1, . . . , p af ) ∈/ Um+1. 8. UNITS 40

f Say x = (a1, . . . , af ) ∈ Zp , and bi ∈ Z with bi ≡ ai mod p. Then setting β = b1β1 + ··· + bf βf and working modulo mn+1, we have f Y n bi n gn(x) ≡ (1 + βiT ) ≡ 1 + βT . i=1 Further, since K has characteristic p, s ps ps m m+1 gn(p x) = gn(x) ≡ 1 + β t mod m . ps The map β 7→ β is an isomorphism on k, and for any u ∈ Um there is α ∈ k such that u ≡ m m+1 f s m+1 1 + αT mod m . So by above, we can find x = (a1, . . . , af ) ∈ Zp such that u = g(p x) mod m , which shows part1. Moreover u ∈ Um+1 if and only if α = 0, if and only if β = 0, if and only if f bi ≡ 0 mod p for each 1 ≤ i ≤ f, if and only if x ∈ pZp , which shows part2. f Q For each n ≥ 1 coprime with p, set An = Zp , and let A = (n,p)=1 An, the product taken over all positive integers n coprime with p. For any (xn) ∈ A, we have gn(xn) ∈ Un and the product Q (n,p)=1 gn(xn) converges in U1. So we get a continuous homomorphism of Zp-modules Y g = gn : A → U1. (n,p)=1 We will show that g is an isomorphism of topological groups. By property1 above, for any integer m ≥ 1, every coset Um/Um+1 can be represented by an element of g(A), so g(A) is dense in U1. On the other hand, A is compact and g is continuous, so g(A) is a compact dense subset of U1. This implies that g is surjective. s s Take nonzero ξ = (xn) ∈ A. If xn 6= 0, write xn = p yn with yn ∈ An r pAn, and set m(xn) = np . 0 Then property2 above implies gn(xn) ∈/ Um(xn). Because the n are coprime to p, if n and n are two distinct indices with xn 6= 0 and xn0 6= 0, we have m(xn) 6= m(xn0 ). Thus, there is a unique n 0 coprime with p with xn 6= 0 and m = m(xn) minimal. Then for any n 6= n coprime with p, we have gn0 (xn0 ) ∈ Um+1 and g(ξ) ≡ gn(xn) 6≡ 1 mod Um+1. This shows g is injective. Finally g is a homeomorphism because it’s a continuous bijection between compact Hausdorff topological spaces. Corollary 8.7. Assume that K has characteristic p. Then we have isomorphisms of topological groups × ∼ N × ∼ N O = Z/(q − 1) × Zp and K = Z × Z/(q − 1) × Zp . 8.1. Exercises. 8.1. Let K be a characteristic 0 field, complete with respect to a nontrivial valuation, and with characteristic p > 0 residue field. Let m denote the maximal ideal of the the valuation ring of K. (a) Show that for any x ∈ m, the series X xm log(1 + x) := (−1)m−1 m m≥1 is convergent, and defines a continuous function log: 1 + m → K. (b) Assume that v is discrete and that K/Qp is finitely ramified. Let e = e(K/Qp) be the e n ramification index. Show that if n > p−1 and x ∈ Un, then log(u) ∈ m . (c) Assume that v is discrete and that K/Qp is finitely ramified. Let e = e(K/Qp) be the n e ramification index. Show that for any x ∈ m with n > p−1 , the series X xm exp(x) := m! m≥0 9. NORM GROUPS 41

is convergent, and deﬁnes a continuous function exp: mn → 1 + mn.

9. Norm groups We again let K denote a nonarchimedean local field with valuation ring O, valuation ideal m, and residue field k. We let q = |k|. Recall that we have defined a filtration of the units O× of O by × n U0 = O and Un = 1 + m for every n ≥ 1. × Lemma 9.1. Let L/K be a finite extension, and let OL be the valuation ring of L. Then NL/K (OL ) × × × is a compact subgroup of O , and NL/K (L ) is a closed subgroup of K .

× × × Proof. Proposition 8.1 implies that OL is a compact subgroup of K . Then since NL/K : L → × × K is a homomorphism with NL/K (OL) ⊆ NL/K (O) and is continuous by Exercise 6.1, NL/K (OL ) is a compact subgroup of K× and is contained in O×. By Lemma 6.2, × × × NL/K (L ) ∩ O = NL/K (OL ), × × × × so NL/K (OL ) is an open subgroup of NL/K (L ) and NL/K (L )/ NL/K (OL ) is a discrete subgroup of × × × × K / NL/K (OL ). Hence NL/K (L ) is closed in K . Given any algebraic extension M/K, we set × × N(M/K) = ∩L NL/K (L ) and NU(M/K) = ∩L NL/K (OL ) where the intersections are taken over all subextensions L/K of M/K with [L : K] < ∞. We call N(M/K) and NU(M/K) the norm group and unit norm group, respectively, of the extension. It follows from Lemma 9.1 that N(M/K) is a closed subgroup of K×, that NU(M/K) is compact subgroup of × OL , and that NU(M/K) = N(M/K) ∩ O×.

Recall that if M/K is ﬁnite, then NM/K = NL/K ◦ NM/L for any subextension L/K. It follows × × that if M/K is ﬁnite, then N(M/K) = NM/K (M ) and NU(M/K) = NL/K (OL ). It also follows that for any algebraic extensions L/K and M/L, that N(M/K) ⊆ N(L/K) and NU(M/K) ⊆ NU(L/K).

0 0 × × 0 Lemma 9.2. Let k /k be a finite extension. Both Nk0/k : (k ) → k and Trk0/k : k → k are surjective. Proof. For any finite separable extension of fields, the trace is a nonzero linear functional (see Proposition B.20), hence is surjective. Since k is a finite field, k0/k is separable. Choose a generator x of the cyclic group (k0)×, and let f = [k0 : k]. Then x has order qf − 1, so qf −1 x q−1 is a generator of the subgroup k×. Since Gal(k0/k) is cyclic of order f, generated by y 7→ yq, we have f−1 f Y qi Pf−1 qi q −1 Nk0/k(x) x = x i=0 = x q−1 . i=0 0 0 Lemma 9.3. Let K /K be a finite unramified extension. Then NU(K /K) = U0. Proof. Denote by O0, m0, and k0 the ring of integers, maximal ideal, and residue field, respectively, 0 0 0 0 × 0 of K . Let {Un}n≥0 be the unit filtration of U0 = (O ) . Choose a uniformizer $ for K. Since K /K is unramified, $ is also a uniformizer for L, and we have commutative diagrams

0 0 ∼ 0 × 0 0 ∼ 0 U0/U1 (k ) Un/Un+1 k and for each n ≥ 1,

∼ × ∼ U0/U1 k Un/Un+1 k with the vertical arrows induced from inclusion. 9. NORM GROUPS 42

For any u ∈ U0, we have Y NK0/K (u) = σ(u) σ∈Gal(K0/K) n 0 0 and if u = 1 + x$ ∈ Un with n ≥ 1 and x ∈ O , Y n X n n+1 NK0/K (u) = (1 + σ(x)$ ) ≡ 1 + σ(x)$ mod $ σ∈Gal(K0/K) σ Since K0/K is unramiﬁed, reduction modulo $ deﬁnes an isomorphism Gal(K0/K) =∼ Gal(k0/k), and we deduce the existence of commutative diagrams

0 0 ∼ 0 × 0 0 ∼ 0 U0/U1 (k ) Un/Un+1 k

Tr 0 NK0/K Nk0/k and NK0/K k /k for each n ≥ 1,

∼ × ∼ U0/U1 k Un/Un+1 k 0 By Lemma 9.2, both Nk0/k and Trk0/k are surjective. We deduce that U0 = NU(K /K)Un for every 0 0 n ≥ 1, i.e. NU(K /K) is dense in U0. On the other hand NU(K /K) is a compact, hence closed, 0 subgroup of U0, so U0 = NU(K /K). Proposition 9.4. Let L/K be an unramiﬁed extension. Then the following hold.

(1) NU(L/F ) = U0. n (2) If L/K has finite degree n and $ is a uniformizer for K, then N(L/K) = $ Z × U0 as a × subgroup of K = $Z × U0. (3) If L/K is infinite degree, then N(L/K) = U0. Proof. The first part follows from Lemma 9.3 and the definition of NU(L/F ). For the second part, since L/K is unramified, $ is also a uniformizer for L, and

× Z × L = $ × OL . n Since $ ∈ K, we have NL/K ($) = $ . Then Z × nZ N(L/K) = NL/K ($) × NL/K (OL ) = $ × U0, again by Lemma 9.3. Finally, for the last part note that if L/F is infinite degree and unramified, the 0 residue field kL of L is an infinite degree extension of k. Thus there are finite subextensions k /k of kL/k of arbitrarily large degree. By Corollary 7.6, there are finite degree unramified subextensions K0/K of L/K with n = [K0 : K] arbitrarily large. Parts1 and2 then imply 0 n U0 ⊆ N(L/K) ⊆ N(K /K) = $ Z × U0, for arbitrarily large n. Part2 follows. Proposition 9.5. An algebraic extension L/K is totally ramified if and only if N(L/K) contains a uniformizer of K. Proof. First assume that L/K is not totally ramified. Then L/K contains a finite unramified subextension K0/K of some degree n ≥ 2. Then, letting $ be a uniformizer for K, Proposition 9.4 implies 0 n N(L/K) ⊆ N(K /K) = $ Z × U0, so N(L/K) does not contain a uniformizer since n ≥ 2. Now assume that L/K is totally ramified, and let $ be a uniformizer for K. For every finite subextension K0/K of L/K, let P (K0) denote the set of uniformizers of K that are contained in 0 N(K /K), and note that it is contained in the compact subset $U0 of K. Since a finite subextension K0/K of L/K is totally ramified, the formula for the extension of the valuation on K to K0 implies 0 0 0 that if $ is a uniformizer for K , then NK0/K is a uniformizer for K, and letting U0 denote the group of units in the valuation ring of K0, 0 0 0 0 0 P (K ) = NK0/K ($ U0) = NK0/K ($ ) NU(K /K). 9. NORM GROUPS 43

0 0 0 So P (K ) is a compact nonempty subset of the compact set $U0. If K1/K and K2/K are two finite 0 0 0 subextension of L/K with K ⊆ K , since N 0 0 applied to a uniformizer of K is a uniformizer for 1 2 K2/K1 2 0 0 0 0 K1, we see that P (K2) ⊆ P (K1). Then for any pair of finite subextensions K1/K of L/K, we have 0 0 0 0 P (K1K2) ⊆ P (K1) ∩ P (K2), from which it follows that 0 P (L) = ∩K0 P (K ), 0 with the intersection ranging over all finite subextensions K /K of L/K. By compactness of $U0, this intersection is nonempty. Part 2

Local class field theory 10. The main theorems Let K be a nonarchimedean local field, and let O denote its valuation ring. Let Ks be a separable closure of K, and let Kab be the maximal abelian subextension, i.e. Kab is the maximal Galois subextension of Ks/K such that Gal(Kab/K) is an abelian group. The Galois group Gal(Kab/K) is a topological group with the profinite topology (see appendixA). Recall that a basis for this topology is {g Gal(Kur/L)} with g ∈ Gal(Kab/K) and L/K a finite subextension of Kab/K. Recall also (Corollary 7.10) that there exists a maximal unramified subextension Kur/K in Ks. It is Galois ur ∼ and Gal(K /K) = Zb with a canonical topological generator FrobK , the Frobenius (Definition 7.11). Since Gal(Kur/K) is abelian, Kur is a subextension of Kab/K. We now have the setup to state the main theorems of Local Class Field Theory. We will prove them in pieces, but combine them all into one theorem with multiple parts. Theorem 10.1 (Local Class Field Theory). There is a canonical map × ab ArtK : K → Gal(K /K) called the Artin map (or Artin reciprocity map or just reciprocity map) satisfying the following.

(1) ArtK is a continuous injective homomorphism with dense image. ur n (2) The image of ArtK consists of all elements σ ∈ Gal(K /K) such that σ|Kur = FrobK for some n ∈ Z. × ∼ ab ur (3) ArtK induces an isomorphism of topological groups O −→ Gal(K /K ). (4) Let L/K be any ﬁnite extension in Ks, and let Lab be the maximal abelian extension of L inside of Ks. Note that Kab ⊆ Lab, so we have a restriction homomorphism res: Gal(Lab/L) → ab Gal(K /K) given by σ 7→ σ|Kab . The following diagram commutes: L× ArtL Gal(Lab/L)

NL/K res

K× ArtK Gal(Kab/K).

(5) Let L/K be any ﬁnite extension in Ks, and let Lab be the maximal abelian extension of L inside of Ks. Let transfer: Gal(Kab/K) → Gal(Lab/L) be the transfer map (to be deﬁned later). The following diagram commutes:

L× ArtL Gal(Lab/L)

inclusion transfer K× ArtK Gal(Kab/K).

(6) Let σ : K −→∼ L be an isomorphism of local ﬁelds. Choose a separable closure Ls of L and let Lab be the maximal abelian subextension. Choose an extension of σ to an isomorphism σ : Kab −→∼ Lab and deﬁne σ∗ : Gal(Kab/K) −→∼ Gal(Lab/L) by σ∗(τ) = στσ−1 (σ∗ does not depend on the choice of extension). The following diagram commutes:

L× ArtL Gal(Lab/L)

σ σ∗ K× ArtK Gal(Kab/K).

× ab (7) ArtK is the unique homomorphism K → Gal(K /K) satisfying the following: (i) For each uniformizer $ of K, ArtK ($)|Kur = FrobK . ab (ii) For each finite extension L/K in K , ArtK (N(L/K))|L = 1. The map is named after Emil Artin, who defined it. There are two main routes to proving local class field theory (besides deducing it from global class field theory): 11. FORMAL GROUP LAWS 46

– Via Galois cohomology as in [Ser79]. – Via Lubin–Tate formal groups as in [Iwa86]. Both have their merits, but we will take the later approach (with some updates due to Yoshida [Yos08]). The formal group approach has the advantage that it is constructive. Let’s ﬁrst consider the example of Qp (without proof) that will motivate some of what we do later. ur Example 10.2. Take K = Qp. Recall that Qp = ∪p-nQp(µn), where we have written µn for the group of nth roots of unity in Qp. By Exercise 7.4, for each m ≥ 1, the extension Qp(µpm )/Qp is totally ramiﬁed and Galois, with ∼ m × Gal(Qp(µpm )/Qp) = (Z/p ) . This isomorphism is canonical, being given as follows. Choose any primitive pmth root of unity ζ. m Then for any σ ∈ Gal(Qp(µpm )/Qp), σ(ζ) is another primitive p th root of unity. There is unique j ∈ (Z/pm)× such that σ(ζ) = ζj, and this j does not depend on the choice of ζ. The map σ 7→ j gives the desired isomorphism. Further, for m ≥ n ≥ 1, the surjection

Gal(Qp(µpm )/Qp) → Gal(Qp(µpn )/Qp) given by restriction corresponds to the canonical surjection (Z/pm)× → (Z/pn)× under the above isomorphism. Thus, setting Qp(µp∞ ) = ∪m≥1Qp(µpm ), we have that Qp(µp∞ ) is Galois over Qp, and we have canonical isomorphisms m × × (10.3) Gal( (µ ∞ )/ ) ∼ lim Gal( (µ m )/ ) ∼ lim( /p ) ∼ . Qp p Qp = ←− Qp p Qp = ←− Z = Zp m m We will see that ab ur Qp = ∪n≥1Qp(µn) = Qp(µp∞ )Qp . ur And since Qp(µp∞ ) ∩ Qp = Qp (as one is totally ramified and the other unramified), we have a splitting ab ∼ ur ∼ × Gal(Qp /Qp) = Gal(Qp /Qp) × Gal(Qp(µp∞ )/Qp) = Zb × Zp . × Z × In this case the Artin reciprocity map ArtQp is the map on Qp = p × Zp that sends p to FrobQp , and × is the inverse of the isomorphism (10.3) on Zp . ab ur In the above, we constructed a totally ramified extension Qp(µp∞ ) such that Qp = Qp(µp∞ )Qp , and defined the Artin map using this. This is not the unique totally ramified extension with this property, so it seems like a choice was involved in the defining the Artin map. But when defining the × Artin map in this case, we also made a choice in the domain, namely we split Qp using the uniformizer p. These two choice are intimately related. This totally ramified extension Qp(µp∞ ) is given by adjoining × to Qp the p-power torsion elements inside the group Qp . Lubin and Tate had the remarkable idea that this generalizes if the following (currently vague) way. We will show that for any choice of uniformizer $ of K, using a certain type of group operation we can construct a canonical totally ramified extension K$ of K, built out of “$-power torsion” such that we have a canonical isomorphism × ∼ O −→ Gal(K$/K), ab ur and such that K = K$K .

11. Formal group laws Fix a nonzero commutative ring R. We ﬁrst introduce some notation that we will use when dealing with power series. Given power series F,G ∈ R[[X1,...,Xn]], we write F ≡ G mod deg d if F −G has no term of total degree < d. Say we have a sequence (Fm)m≥1 of elements in R[[X1,...,Xn]] such that for each d ≥ 1, there is some M ≥ 1 such that

Fr ≡ Fs mod deg d 11. FORMAL GROUP LAWS 47

i1 in for all r, s ≥ M. Then for any i1, . . . , in ≥ 0, the coefficient of X1 ··· Xn in Fm is eventually constant as m gets sufficiently large. We thus have a well-defined power series in R[[X1,...,Xn]] whose i1 in X1 ··· Xn -coefficient is this eventual constant value, and we denote this power series by

lim Fm. m

Let f ∈ R[[X]] and F ∈ R[[X1,...,Xn]]. Assume that both f and F have no constant term. Then we have well-deﬁned power series in R[[X1,...,Xn]], denoted by f ◦ F and F ◦ f, given by

f ◦ F := f(F (X1,...,Xn)) and F ◦ f := F (f(X1), . . . , f(Xn)). An important special case of this is when f, g ∈ XR[[X]], then both f ◦ g and g ◦ f are also in XR[[X]]. Further, if h ∈ XR[[X]] as well, then (f ◦ g) ◦ h = f ◦ (g ◦ h). So XR[[X]] becomes a monoid under composition with identity element X. Proposition 11.1. An element f ∈ XR[[X]] is invertible with respect to composition if and only if the coefficient of X in f is in R×. Proof. Let a ∈ R be the coefficient of X in f. Assume f has an inverse f −1 ∈ XR[[X]] with respect to composition, and let b be the coefficient of X in f −1. Then X = f ◦ f −1 ≡ abX mod deg 2, so ab = 1 in R, and a ∈ R×. × Now assume that a ∈ R . We inductively construct polynomials gn ∈ R[X] such that for all n ≥ 1,

f ◦ gn ≡ X mod deg(n + 1) and gn+1 ≡ gn mod deg(n + 1). −1 Setting g1 = a X, we have f ◦ g1 = X mod deg 2. Now assume that gn−1 has been constructed, for n some n ≥ 2. We want to find α ∈ R such that gn = gn−1 + αX satisfies the inductive step. We have n n f ◦ (gn−1 + αX ) ≡ f ◦ gn−1 + aαX mod deg(n + 1) ≡ X + bXn + aαXn mod deg(n + 1), for some b ∈ R. Taking α = −a−1b establishes the inductive step. Since gn+1 ≡ gn mod deg(n + 1) for all n ≥ 1, we can define g = limn gn ∈ R[[X]]. Then since −1 f ◦ gn ≡ X mod deg(n + 1) for all n ≥ 1, we have f ◦ g = X. The coefficient of X in g is a by construction, which is invertible in R, so we can apply what we’ve shown so far to g and obtain h ∈ XR[[X]] such that g ◦ h = X. Then f = f ◦ X = f ◦ (g ◦ h) = (f ◦ g) ◦ h = X ◦ h = h. So f is invertible. Example 11.2. Assume that R is a Q-algebra, so every integer is invertible in R. The power series X Xn f = (−1)n−1 ∈ XR[[X]], n n≥1 has linear coefficient −1, which is invertible in R. So f has an inverse with respect to composition. Of course, this power series is usually written log(1 + X), and we know that exp(log(1 + X)) = log(exp(X)) = 1 + X as formal power series, with X Xn exp = 1 + . n! n≥1 −1 P Xn So f = exp −1 = n≥1 n! . We now come to the main definition of this section. Definition 11.3. A formal group law over R is a power series F ∈ R[[X,Y ]] satisfying the following axioms. 11. FORMAL GROUP LAWS 48

(1) F = X + Y mod deg 2. (2) F (F (X,Y ),Z) = F (X,F (Y,Z)) (associativity). (3) F (X, 0) = X and F (0,Y ) = Y (identity element). (4) There is a unique power series iF ∈ R[[T ]] such that F (T, iF (T )) = 0 and F (iF (T ),T ) = 0 (inverse). (5) F (X,Y ) = F (Y,X) (commutativity). Remark 11.4. We have actually only defined a commutative formal group laws of dimension one. For the general definition of a formal group law of dimension one, one omits axiom5. To define formal group laws of arbitrary dimensions, one uses a power series F in a 2n variable power series ring R[[X1,...,Xn,Y1,...,Yn]]. In this course we will only encounter commutative formal groups laws of dimension one, so we refer to them simply as formal groups laws. Formal group laws are often referred to simply as formal groups. Remark 11.5. In fact we have given an overdetermined list of axioms: axioms3 and4 follow from the other axioms (this is Exercise 11.1 below).

Example 11.6. The additive formal group law over R is the formal group law Ga ∈ R[[X,Y ]] given by Ga = X + Y. In many references this formal group law is denoted by Gb a.

Example 11.7. The multiplicative formal group law over R is the formal group law Gm ∈ R[[X,Y ]] given by Gm = (1 + X)(1 + Y ) − 1 = X + Y + XY. One should think of Gm as the multiplication of 1 + X and 1 + Y using X and Y as parameters. We have −1 X n n iGm (T ) = (1 + T ) − 1 = (−1) T . n≥1

In many references this formal group law is denoted by Gb m. Example 11.8. Let E be an elliptic curve defined over a field K (or more generally over any commutative ring). It is a commutative group variety. Let A be the local ring of E at the identity element of E. Since E is smooth, A is a discrete valuation ring a choice of local parameter T (i.e. a uniformizer) determines an isomorphism between the completion Ab of A and K[[T ]]. The group structure E × E → E yield a K-algebra map A → A ⊗K A that induces a map K[[T ]] → K[[X,Y ]] on completions (taking an appropriate completion on the right hand side). The image of T under this map is a formal group law FE ∈ K[[X,Y ]] called the formal group law of E. One should think of FE as giving the group structure on an infinitesimal neighbourhood of the identity in E. For more details, see [Sil86, Chapter IV]. Let A be an commutative R-algebra and let I be an ideal in A. We say that A is I-adically complete if the canonical map of R-algebras A → lim A/In given by a 7→ (a mod In) ←− n≥1 n is an isomorphism. For us, the two most important examples are the following: – R[[X]] is (X)-adically complete, where (X) = XR[[X]]; – a complete discrete valuation ring is m-adically complete, where m is its maximal ideal. 11. FORMAL GROUP LAWS 49

Let A be an I-adically complete commutative R-algebra, with I an ideal of A. If F ∈ R[[X1,...,Xm]] and a1, . . . , am ∈ I, evaluation of F at the tuple (a1, . . . , am) is a well deﬁned element F (a1, . . . , am) ∈ A, and if F has no constant term, then F (a1, . . . , am) ∈ I. To see this, let Fn be the sum of all monomials of F of total degree < n; it is an element in R[X1,...,Xm] so we can evaluate it on (a1, . . . , am) using the R-algebra structure of A. Then the sequence (F (a , . . . , a )) is an element of lim A/In, n 1 m n≥1 ←−n so corresponds to a unique element F (a , . . . , a ) ∈ A under the isomorphism A −→∼ lim A/In. 1 m ←−n Moreover, this element belongs to I if and only if its image in A/I is zero, which happens if and only if F1(a1, . . . , am) = 0, i.e. if and only if F has no constant term. So given an I-adically complete commutative R-algebra A, and a formal group law F ∈ R[[X,Y ]] over R, we have a bilinear pairing

+F : I × I → I given by a +F b = F (a, b).

The axioms of a formal group law imply that +F gives I the structure of an abelian group with identity 0 ∈ I, and the inverse of an element a ∈ I given by iF (a). We denote this abelian group (I, +F ) by IF . Further, if B is another commutative R-algebra that is J-adically complete, with J an ideal of B, then an R-algebra homomorphism φ: A → B induces a homomorphism IF → JF of abelian groups.

Example 11.9. For any I-adically complete commutative R-algebra A, the operation +Ga is the usual addition on I, and IGa is just I with its usual group structure as a subgroup of A with respect to addition. Example 11.10. Let K be a nonarchimedean local ﬁeld with valuation ring O and valuation ideal × m. Recall that U1 = 1 + m is a subgroup of O . The map

φ: mGm → U1 given by φ(x) = 1 + x is an isomorphism of groups. Indeed, it’s clearly bijective, takes 0 to 1, and for any x, y ∈ m,

φ(x +Gm y) = φ((1 + x)(1 + y) − 1) = (1 + x)(1 + y) = φ(x)φ(y). Definition 11.11. Let F,G ∈ R[[X,Y ]] be two formal groups over R.A homomorphism from F to G is a power series f ∈ XR[[X]] such that f ◦ F = G ◦ f, i.e. f(F (X,Y )) = G(f(X), f(Y )). We write f : F → G. If f has an inverse f −1 ∈ XR[[X]] with ∼ respect to composition, we say f is an isomorphism and write f : F −→ G. We denote by HomR(F,G) the set of homomorphisms from F to G, and set EndR(F ) = HomR(F,F ).

Example 11.12. For any formal group F ∈ R[[X,Y ]] and any integer n ∈ Z, we deﬁne the multiplication by n homomorphism [n]: F → F inductively as follows: – [0](X) = 0, –[ n + 1](X) = [n](X) +F X = F ([n](X),X), –[ n − 1](X) = [n](X) +F iF (X) = F ([n](X), iF (X)). Inducting on n ∈ Z, it is straightforward to show [n] ≡ nX mod deg 2. So by Proposition 11.1,[n]: F → F is an isomorphism if and only if n ∈ R×. Example 11.13. As a speciﬁc case of Example 11.12, it is straightforward to show by induction that the multiplication by n homomorphism on Gm is [n] = (1 + X)n − 1. 11. FORMAL GROUP LAWS 50

Example 11.14. Assume that R is a Q-algebra, so every integer is invertible in R. Deﬁne X Xn log := (−1)n−1 ∈ XR[[X]]. Gm n n≥1 This power series is invertible with respect to composition, with inverse X Xn exp := Gm n! n≥1 (see Example 11.2). The power series log (X) is usually written log(1 + X), and we know that there Gm is an equality of formal power series in two variables log((1 + X)(1 + Y )) = log(1 + X) + log(1 + Y ), which can be rewritten as log ( (X,Y )) = log (X) + log (Y ). Gm Gm Gm Gm We conclude that log is an isomorphism of formal group laws −→∼ . Gm Gm Ga Proposition 11.15. Let F and G be formal group laws over R.

(1) HomR(F,G) is a subgroup of (X)G. (2) EndR(F ) is a ring with addition given by +F and multiplication given by composition.

Proof. We ﬁrst show part1. Take f, g ∈ HomR(F,G); we know f +G g ∈ (X)G. We have

(f +G g) ◦ F = G(f ◦ F, g ◦ G) = G(G ◦ f, G ◦ g), where the second equality uses that f, g ∈ HomR(F,G). We have G(G ◦ f, G ◦ g) = G(G(f(X), f(Y )),G(g(X), g(Y ))) = G(G(f(X), g(X)),G(f(Y ), g(Y ))), where the second equality follows from repeated use of the associativity and commutativity axioms for G, and

G(G(f(X), g(X)),G(f(Y ), g(Y ))) = G ◦ (f +G g) by deﬁnition. So (f +G g)◦F = G◦(f +G g), i.e. f +G g ∈ HomR(F,G). We clearly have 0 ∈ HomR(F,G). Now take some f ∈ HomR(F,G); it has an inverse iG(f) in (X)G. Again using the commutativity and associativity axioms for G, we have

G(G(X,Y ),G(iG(X), iG(Y ))) = G(G(X, iG(X)),G(Y, iG(Y ))) = G(0, 0) = 0.

Since R[[X,Y ]] is I-adically complete for I the ideal generated by X and Y , the above shows G +G (G ◦ iG) = 0 in IG. On the other hand, we know that the inverse of G in IG is iG ◦ G, so we deduce that G ◦ iG = iG ◦ G. Then

iG(f) ◦ F = iG ◦ f ◦ F = iG ◦ G ◦ f = G ◦ iG ◦ f = G ◦ iG(f), so iG(f) ∈ HomR(F,G). This ﬁnishes the proof of part1. For part2, we saw above that EndR(F ) is a group with respect to +F . It is immediate that EndR(F ) is closed under composition, that composition is associative, and that X is the multiplicative identity. It also follows immediately from the deﬁnition of composition that for any f, g, h ∈ EndR(F ), we have (g +F h) ◦ f = g ◦ f +F h ◦ f. It only remains to check that f ◦ (g +F h) = f ◦ g +F f ◦ h. Since f ∈ EndR(F ),

f ◦ (g +F h) = f(F (g(X), h(X))) = F (f(g(X)), f(h(X))) = f ◦ g +F f ◦ h.

11. FORMAL GROUP LAWS 51

We will use formal group laws to construct totally ramified abelian extensions of local fields. Formal group laws are a very useful tool in many other areas of math as well (e.g. the study of abelian schemes, homotopy theory), and they are interesting objects in their own right. We finish this section with a smattering of results about formal group laws that are useful and important in their study, but that we will not need in this course (so the reader interested only in local class field theory can safely skip the remainder of this section). We saw that in Example 11.14, that Gm is isomorphic to Ga if our base ring R is a Q-algebra. In fact, this is a much more general phenomena: Proposition 11.16. Let F ∈ R[[X,Y ]] be a formal group law over R. If R is a Q-algebra, then there is a unique isomorphism of formal group laws ∼ logF : F −→ Ga such that logF ≡ X mod deg 2. The interested reader can see [Iwa86, Lemma 4.2] or [Sil86, Chapter IV, Proposition 5.2] for a proof. This shows that the theory of formal group laws is more complicated and interesting over base rings that are not Q-algebras. Naturally, the easiest place to start is over fields of positive characteristic. Lemma 11.17. Let k be a field of characteristic p > 0, let F and G be formal group laws over k, and let f : F → G be a nonzero homomorphism. There is an integer h ≥ 1 such that h f ≡ aXp mod deg(ph + 1) with a 6= 0. Proof. Let n ≥ 1 and a ∈ k× be such that f ≡ aXn mod deg(n + 1). Write n = phm with p - m. We want to show m = 1. Taking the equality f ◦ F = G ◦ f modulo degrees n + 1, we have a(X + Y )n ≡ aXn + aY n mod deg(n + 1), which implies n−1 X n XjY n−j = 0, j j=1 n mph which in turn implies j = 0 in k for all 1 ≤ j ≤ n − 1. If m > 1, the binomial coefficient ph is not divisible by p, hence is nonzero in k. Definition 11.18. With the notation as in Theorem 11.22, we call h the height of f. We define the height of the zero homomorphism to be ∞. The height of the formal group law F is the height of the multiplication by p homomorphism [p]: F → F .

Example 11.19. The additive group law Ga over a ﬁeld of characteristic p has height ∞, since [p] = 0.

Example 11.20. The multiplicative group Gm over a ﬁeld of characteristic p has height 1, since the multiplication by p map is given by [p] = (1 + X)p − 1 = Xp

Example 11.21. Let E be an elliptic curve defined over a field of characteristic p, and let FE be the formal group law attached to E (see Example 11.8). Then the height of FE is either 1 or 2 (see [Sil86, Chapter IV, Corollary 7.5]). Moreover, the dichotomy between height 1 and height 2 elliptic curves has many equivalent and important formulations (see [Sil86, Chapter 5, Theorem 3.1]). Here is one: since E is a group variety, we can ask if it has nontrivial torsion points of a given order (defined over an algebraic closure k of k). Then the subgroup of p-torsion points of E is trivial if and only if the height of FE is 2. We call these elliptic curves supersingular. The elliptic curves whose formal groups have height 1 (equivalently that have a nontrivial p-torsion point), are called ordinary. 11. FORMAL GROUP LAWS 52

The above example of elliptic curves indicates that the height of a formal group law is an important invariant. If our base ﬁeld is algebraically closed, it turns out that the height uniquely determines the formal group law:

Theorem 11.22 (Lazzard, [Laz55]). Let k be an algebraically closed ﬁeld of characteristic p > 0. (1) For any h ≥ 1 (or h = ∞), there is a formal group law over k of height h. (2) Two formal group laws over k are isomorphic if and only if they have the same height.

Now say we are more ambitious, and want to understand formal group laws defined over more general rings in which p is not invertible. One way to attempt this is to start with a formal group law defined over a field of characteristic p, and then try to bootstrap up. Let’s make this more precise. Let R be a commutative ring, and let F ∈ R[[X,Y ]] be a formal group law. Then given any ideal I in R, we can reduce the coefficients of F modulo I, to formal group law F mod I over R/I. Generalizing this, given any homomorphism ψ : R → S, with S another commutative ring, applying ψ to the coefficients of F gives a formal group law ψ(F ) ∈ S[[X,Y ]]. For example, say k is a field of characteristic p > 0, and F is a formal group law over k. Can one understand formal group laws F over (say) the Artinian ring k[X]/(Xn) such that F mod X = F ? Or, say k = Fp, then can one understand formal group laws F over Zp with F mod p = F ? One can put both these questions into the following more general framework. Let R be a complete Noetherian local ring. Recall that saying R is a local ring means that it has a unique maximal ideal mR. Saying it is a complete local ring means that it is mR-adically complete. Let k = R/mR, it is called the residue field of R. We say a commutative ring S is a complete Noetherian local R-algebra if it is a complete Noetherian local ring equipped with a homomorphism of rings φ: R → S such that φ(mR) ⊆ mS. We say that S has residue field k, if the induced map k = R/mR → S/mS is an isomorphism. For example, the power series ring R[[X1,...,Xn]] is a complete Noetherian local R-algebra with residue field k (its maximal ideal is (mR,X1,...,Xn)). It can be shown that any complete local Noetherian R-algebra with residue field k is a quotient of a power series ring over R is some number of variables.

Theorem 11.23 (Lubin–Tate, [LT66, Theorem 3.1]). Let R be a complete Noetherian local ring with residue field k of characteristic p > 0. Let F be a formal group law over k of finite height h. Then there is a formal group law Funiv over R[[T1,...,Th−1]] such that for any complete local Noetherian R-algebra S with residue field k, and any formal group law F over S with FS mod mS = F , there is a unique local R-algebra homomorphism ψ : R[[T1,...,Th−1]] → S such that ψ(Funiv) is isomorphic to FS. Moreover, this isomorphism is unique subject to the condition that it reduces to the identity modulo mS.

So the formal group law Funiv is thought of as the universal formal group law lifting F . Note that the choice of a local R-algebra homomorphism R[[T1,...,Th−1]] → S is equivalent to the choice of elements s1, . . . , sh−1 ∈ mS, by Ti 7→ si. So the theorem also says that there are h − 1 “free parameters” in the universal formal group law Funiv. In particular, if F has height 1, then for any complete Noetherian local R-algebra, there is a unique (up to isomorphism) formal group law F over S with F mod mS = F . Trying to understand this universal formal group law when the height is ≥ 2 is very interesting, and has many important consequences that we won’t go into here.

11.1. Exercises. 11.1. Show that axioms3 and4 in Deﬁnition 11.3 follow from axioms1 and2. 11.2. Let R be an nonzero commutative ring, and let F be a formal group law over R. Let A be an I-adically complete commutative R-algebra, with I an ideal of A. Show that the maps

EndR(F ) × I → I given by (g, a) 7→ g(a)

gives the abelian group IF the structure of an EndR(F )-module. 12. LUBIN–TATE FORMAL GROUPS LAWS 53

12. Lubin–Tate formal groups laws Throughout this section we fix a nonarchimedean local field K. We denote by O the valuation ring of K, by m the maximal ideal of O, by k the residue field of K, and we let q = |k|. We also fix a complete unramified extension L/K, by which we mean one of two things: (1) L/K is a finite (hence complete) unramified extension, or (2) there is an infinite algebraic extension L0/K such that L is the completion of L0 with respect to the unique discrete valuation on L0 extending the one on K. Note that in either case the value group of L equals the value group of K, and a uniformizer for K is also a uniformizer for L. In particular, letting OL and mL denote the valuation ring and valuation ideal, respectively, of L, we have mOL = mL. Further, in either case the residue field kL of L is an algebraic extension of k (in the second case it is the same as the residue field of L0). Fixing a choice of uniformizer $ for L, every element a ∈ L can be written uniquely as a convergent series X n a = xen$ , n−∞ where xen ∈ OL denotes the Teichmüllerlift of xn ∈ kL. ur Example 12.1. Let L = Qdp be the completion of the maximal unramified extension of Qp (inside of some algebraic closure Qp of Qp). Then every element a ∈ L can be written uniquely as X n a = ynp n−∞ where each yn is either 0 or a root of unity of order prime to p. There is a unique continuous automorphism ϕ: L −→∼ L, called the Frobenius, that induces the q isomorphism x 7→ x on the residue field kL. Indeed, we saw this in section7 in the case that L/K is finite. If L is the completion of an infinite unramified extension L0/K, then we also saw the existence of the Frobenius automorphism ϕ on L0 in section7. Since ϕ is continuous on L0 it extends uniquely to L.

i Notation 12.2. For any a ∈ L and i ∈ Z, we set aϕ := ϕi(a). Given two integers i, j ∈ Z, we ϕi+ϕj ϕi ϕj i j ϕ set a = a a = ϕ (a)ϕ (a). For any F ∈ L[[X1,...,Xn]], we let F denote the power series in L[[X1,...,Xn]] obtained by applying ϕ to all the coeﬃcients of F . The following is the main lemma of this section.

Lemma 12.3. Let $1 and $2 be uniformizers for L, and let f1, f2 ∈ OL[[X]] satisfy

(a) fi ≡ $iX mod deg 2, q (b) fi ≡ X mod mL, ϕ for each i = 1, 2. Let θ1, . . . , θn ∈ OL be such that $1θ = $2θ for each 1 ≤ i ≤ n. Then there is a unique F ∈ OL[[X1,...,Xn]] such that ϕ f1 ◦ F = F ◦ f2 and F ≡ θ1X1 + ··· + θnXn mod deg 2.

Proof. We show by induction that there are unique polynomials Fm ∈ OL[X1,...,Xn] of total degree ≤ m such that ϕ f1 ◦ Fm ≡ Fm ◦ f2 mod deg(m + 1) and Fm ≡ θ1X1 + ··· + θnXn mod deg 2.

Given this, F = limm Fm ∈ OL[[X1,...,Xn]] is the unique power series we want. For m = 1, we must take F1 = θ1X1 + ··· + θnXn. By our assumption on the θi, we have

f1 ◦ F1 ≡ $1θ1X1 + ··· + $1θnXn mod deg 2 ϕ ϕ ≡ $2θ1 X1 + ··· + $2θn Xn mod deg 2 ϕ ≡ F1 ◦ f2 mod deg 2. 12. LUBIN–TATE FORMAL GROUPS LAWS 54

ϕ Now assume we have constructed Fm for some m ≥ 1. Set Gm+1 := f1 ◦Fm −Fm ◦f2. By assumption (b), modulo mL we have q ϕ q q Gm+1 ≡ Fm − Fm(X1 ,...,Xn) ≡ 0 mod mL. So every coeﬃcient of Gm+1 lies in mL. We want to show there is a unique homogeneous polynomial Hm+1 such that ϕ (12.4) f1 ◦ (Fm + Hm+1) = (Fm + Hm+1) ◦ f2 ≡ 0 mod deg(m + 2), since then setting Fm+1 = Fm + Hm+1 establishes the inductive step. Equation (12.4) is equivalent to ϕ Gm+1 + f1 ◦ Hm+1 − Hm+1 ◦ f2 ≡ 0 mod deg(m + 2).

To show this has a unique solution in Hm+1, we can work monomial by monomial. Fix J = (j1, . . . , jn) J j1 jn with j1, . . . , jn ≥ 0 and j1 +···+jn = m+1, and write X := X1 ··· Xn . We know that the coeﬃcient J J of X lies in mL, so there is βJ ∈ OL such that $1βJ is the coeﬃcient of X in Gm+1. We then want to show that there is unique αJ ∈ OL such that J J ϕ J $1βJ X + f1(αJ X ) − αJ X (f2) ≡ 0 mod deg(m + 2), and by assumption (a), this is equivalent to m+1 ϕ $1βJ + $1αJ − $2 αJ = 0. m+1 $2 ϕ This holds if and only if αJ = −βJ + α . We can repeatedly substitute this equation into itself, $1 J and we obtain the unique solution

i−1 m+1 1+ϕ+···+ϕ X ϕi $2 αJ = −βJ + βJ , $1 i≥0 which is convergent since m ≥ 1.

Proposition 12.5. Let $ be a uniformizer for L, and let f ∈ OL[[X]] satisfy q f ≡ $X mod deg 2 and f ≡ X mod mL. ϕ There exists a unique formal group law Ff over OL such that f ∈ HomOL (Ff ,Ff ).

Proof. Applying Lemma 12.3 with $1 = $2 = $, f1 = f2 = f, n = 2, and θ1 = θ2 = 1, we have a unique Ff ∈ OL[[X,Y ]] such that ϕ f ◦ Ff = Ff ◦ f and Ff ≡ X + Y mod deg 2.

Note that Ff (Y,X) also satisﬁes the same properties, so by uniqueness Ff (X,Y ) = Ff (Y,X). By Exercise 11.1, it only remains to show the associativity axiom for for Ff . We have

Ff (X,Ff (Y,Z)) ≡ X + Y + Z mod deg 2, and ϕ ϕ ϕ ϕ Ff (X,Ff (Y,Z)) ◦ f = Ff (f(X),Ff (Y,Z) ◦ f) = Ff (f(X), f ◦ Ff (Y,Z)) = f ◦ Ff (X,Ff (Y,Z)). Similarly, Ff (Ff (X,Y ),Z) ≡ X + Y + Z mod deg 2 and ϕ Ff (Ff (X,Y ),Z) ◦ f = f ◦ Ff (Ff (X,Y ),Z).

Applying the uniqueness in Lemma 12.3 with n = 3 and θ1 = θ2 = θ3, yields

Ff (X,Ff (Y,Z)) = Ff (Ff (X,Y ),Z).

Definition 12.6. For f and Ff as in Proposition 12.5, we call Ff the Lubin–Tate formal group law associated to f. 12. LUBIN–TATE FORMAL GROUPS LAWS 55

p Pp p j Example 12.7. If K = Qp, $ = p, and f = (1 + X) − 1 = j=1 j X , then the Lubin–Tate ϕ formal group law associated to f is Gm. By the uniqueness it suﬃces to show f ◦ Gm = Gm ◦ f. We ϕ have Gm = Gm, p p p f ◦ Gm = (1 + (1 + X)(1 + Y ) − 1) − 1 = (1 + X) (1 + Y ) − 1, and p p p p Gm ◦ f = (1 + (1 + X) − 1)(1 + (1 + Y ) − 1) − 1 = (1 + X) (1 + Y ) − 1.

Proposition 12.8. Let $1 and $2 be uniformizers for L, and let f1, f2 ∈ OL[[X]] satisfy q (12.9) fi ≡ $iX mod deg 2 and fi ≡ X mod mL, ϕ for each i = 1, 2. Let Θ($1, $2) = {θ ∈ OL | $1θ = $2θ }. There is a unique map

[·]f1,f2 : Θ($1, $2) → XOL[[X]] satisfying f ◦ [θ] = [θ]ϕ ◦ f and [θ] ≡ θX mod deg 2, 1 f1,f2 f1,f2 2 f1,f2 for all θ ∈ Θ($1, $2). It moreover satisﬁes the following properties. (1) [θ + θ0] = [θ] + [θ0] for all θ, θ0 ∈ Θ($ , $ ). f1,f2 f1,f2 Ff1 f1,f2 1 2

(2) For $3 another uniformizer for L, and f3 ∈ OL[[X]] satisfying (12.9), we have [θ]f1,f2 ◦ 0 0 0 [θ ]f2,f3 = [θθ ]f1,f3 for all θ ∈ Θ($1, $2) and θ ∈ Θ($2, $3).

(3) [θ]f1,f2 ∈ HomOL (Ff2 ,Ff1 ) for all θ ∈ Θ($1, $2).

Proof. The existence and uniqueness of [·]f1,f2 follows from Lemma 12.3 with n = 1. To simplify notation, for the remainder of the proof we write [θ]ij for [θ]fi,fj , and Fi for Ffi . We now show parts1 and2. First note that 0 0 [θ]12 +F1 [θ ]12 ≡ (θ + θ )X mod deg 2, and

0 0 ϕ 0 f1 ◦ ([θ]12 +F1 [θ ]12) = f1 ◦ F1([θ]12, [θ ]12) = F1 (f1 ◦ [θ]12, f1 ◦ [θ ]12) ϕ ϕ 0 ϕ 0 ϕ 0 ϕ = F1 ([θ]12 ◦ f2, [θ ]12 ◦ f2) = F1([θ]12, [θ ]12) ◦ f2 = ([θ]12 +F1 [θ ]12) ◦ f2. 0 0 By the uniqueness in Lemma 12.3, we have [θ]12 +F1 [θ ]12 = [θ + θ ]12. The proof of part2 is similar. For part2, we have 0 0 [θ]12 ◦ [θ ]23 ≡ θθ X mod deg 2, and 0 ϕ 0 ϕ 0 ϕ 0 ϕ f1 ◦ ([θ]12 ◦ [θ ]23) = [θ]12 ◦ f2 ◦ [θ ]23 = [θ]12 ◦ [θ ]23 ◦ f3 = ([θ]12 ◦ [θ ]23) ◦ f3. 0 0 By Lemma 12.3 again, we have [θ]12 ◦ [θ ]23 = [θθ ]13. We now show part3. We want to show [ θ]12 ◦ F2 = F1 ◦ [θ]12. First note that

[θ]12 ◦ F2 ≡ θX + θY ≡ F1 ◦ [θ]12 mod deg 2. Further, ϕ ϕ ϕ ϕ f1 ◦ ([θ]12 ◦ F2) = [θ]12 ◦ f2 ◦ F2 = [θ]12 ◦ F2 ◦ f2 = ([θ]12 ◦ F2) ◦ f2, and ϕ ϕ ϕ ϕ f1 ◦ (F1 ◦ [θ]12) = F1 ◦ f1 ◦ [θ]12 = F1 ◦ [θ]12 ◦ f2 = (F1 ◦ [θ]12) ◦ f2. By the uniqueness in Lemma 12.3 with n = 2 and θ1 = θ2 = θ, we have [θ]12 ◦ F2 = F1 ◦ [θ]12.

Corollary 12.10. Let the notation be as in Proposition 12.8. If θ ∈ Θ($1, $2) a unit in OL, −1 then [θ]f1,f2 is invertible with inverse [θ ]f2,f1 . −1 Proof. By part2 of Proposition 12.8, we know [ θ]f1,f2 ◦ [θ ]f2,f1 = [1]f1,f1 . We have f ◦ [1] = [1]ϕ ◦ f and [1] ≡ X mod deg 2. 1 f1,f1 f1,f1 1 f1,f2

But X also satisﬁes these properties, so by uniqueness [1]f1,f1 = X. A similar argument shows −1 [θ ]f2,f1 ◦ [θ]f1,f2 = X. 12. LUBIN–TATE FORMAL GROUPS LAWS 56

Corollary 12.11. Let $ be a uniformizer for L, and let f ∈ OL[[X]] satisfy q f ≡ $X mod deg 2 and f ≡ X mod mL.

The map [·]f,f of Proposition 12.8 induces an injective ring homomorphism O → EndOL (Ff ). Proof. Since aϕ = a for any a ∈ O, we see that O is a subset of the set Θ($, $) of Proposition 12.8. Then Proposition 12.8 implies that

[·]f,f : O → EndOL (Ff ,Ff ) is a ring homomorphism, and it is injective since

[a]f,f ≡ aX mod deg 2 for any a ∈ O.

Example 12.12. Let $ be a uniformizer for L, and let f ∈ OL[[X]] satisfy q f ≡ $X mod deg 2 and f ≡ X mod mL. ϕ Then $ ∈ Θ($, $ ), and [$]f ϕ,f = f. This follows from uniqueness, since f ≡ $X mod deg 2 and (trivially) f ϕ ◦ f = f ϕ ◦ f.

In light of Corollary 12.11, we say that the Lubin–Tate formal group law Ff is a formal O-module law. This will allow us to construct modules of $-power torsion as alluded to at the end of section 10. Corollary 12.10 will allow us to compare these results for different choices of initial data, in order to show the resulting Artin map is canonical. To do this, one needs to know that the set Θ($1, $2) is nonempty, which isn’t always the case. However, the following proposition shows that it does hold in one particular case. Proposition 12.13. Assume that L is the completion of a maximal unramified extension of K. × × θϕ Then the map ψ : OL → OL given by ψ(θ) = θ is surjective. × × Proof. Fix u ∈ OL . By induction, we construct elements θm ∈ OL such that ψ(θm) ≡ u m m mod mL and θm+1 ≡ θm mod mL , for all m ≥ 1. If we do so, then we can set θ = limm θm, since L is complete, and we have ψ(θ) = limm ψ(θm) = u, since ψ is continuous. q−1 The map ψ induces x 7→ x on the residue field kL of L. The residue field of L is the same as the residue field of the maximal unramified extension, hence is an algebraic closure of k. Since kL is × algebraically closed, any element has a (q − 1)st root, and we can find θ0 ∈ OL such that ψ(θ0) ≡ u mod mL. Now assume we have constructed θm, for some m ≥ 0. Fix a uniformizer $ ∈ K. Then m m −1 m mL = $ OL, and we can find α ∈ OL such that u ψ(θm) = 1 + α$ . Using the algebraic closure of q kL, given any y ∈ kL, we can find x ∈ kL such that x − x = y. We can thus find β ∈ OL such that ϕ m β − β ≡ α mod mL, and we set θm+1 = θm(1 + β$ ). We then compute m ϕ m m −1 ψ(θm+1) = u(1 + α$ )(1 + β $ )(1 + β$ ) ϕ m m+1 ≡ u(1 + (α + β − β)$ ) mod mL m+1 ≡ u mod mL . Remark 12.14. This proposition is false for finite extensions L/K, as well as for a maximal unramified extension Kur/K, hence it was crucial that we used the completion of a maximal unramified θϕ extension. To see this, first consider the case when L/K is finite. Then an element of the form u = θ , × with θ ∈ L satisfies NL/K (u) = 1. Indeed

[L:K]−1 i+1 Y Y θϕ N (u) = σ(u) = = 1. L/K θϕi σ∈Gal(L/K) i=0 × × × But by Lemma 9.3, we know that NL/K (OL ) = O , so there are many u ∈ OL such that NL/K (u) 6= 1. In particular, for such u and any choice of uniformizer $, we have Θ($, u$) = {0}. Further, say we 13. LUBIN–TATE EXTENSIONS 57

× θϕ ur × choose u ∈ O such that u is not a root of unity. Then u cannot be of the form θ for any θ ∈ (K ) , because otherwise we would have [K(θ):K] u = NK(θ)/K (u) = 1. θϕ In particular, for this u, the element θ ∈ L with θ = u in Proposition 12.13 is necessarily transcendental over K (we will show later that an element in the completion of Kur that is algebraic over K lies in Kur).

The fact that the sets Θ($1, $2) may be empty for finite L/K, but are never empty when L is the completion of the maximal unramified extension is related to the fact (discussed at the end of §10) that we will construct a totally ramified extension K$/K that depends on the choice of a uniformizer $, ur ur but such that the extension K$K /K will be independent of the choice of $. 13. Lubin–Tate extensions We continue to let K be a nonarchimedean local field with valuation ring O and valuation ideal m, and we let q denote the cardinality of its residue field. We again fix a complete unramified extension L/K, and let OL denote its valuation ring. Fix an algebraic closure L of L. Throughout this section, we fix a uniformizer $ of L, and f ∈ OL[[X]] satisfying q f ≡ $X mod deg 2 and f ≡ X mod mL.

We let Ff be the Lubin–Tate formal group law associated to f. We will write +f for +Ff . 0 0 Let L /L be a ﬁnite extension in L. Then L is complete, so letting mL0 be the valuation ideal of 0 L , we have an abelian group (mL0 , +f ), whose underlying set is mL0 and whose binary operation is

x +f y = Ff (x, y).

Recall also (see Proposition 11.15) that EndOL (F ) ⊆ XOL[[X]] is a ring with addition given by +f and multiplication given by composition, and the map

0 0 EndOL (F ) × mL → mL given by (g, x) 7→ g(x)

0 0 gives (mL , +f ) the structure of an EndOL (F )-module by Exercise 11.2. By Corollary 12.11,(mL , +f ) is a O-module, and we denote the action of O on mL0 by ·f , i.e.

a ·f α = [a]f,f (α). 00 0 If L /L is a further finite extension in L, then the inclusion (mL0 , +F ) ⊆ (mL00 , +F ) is a homomorphism of EndOL (F )-modules, hence of O-modules. Now L is the union of its finite separable 0 subextensions L /L, and its valuation ideal mL it the union of the valuation ideals mL0 . Thus, even though L is not complete, the operation a +f b = Ff (a, b) on mL, and the action ·f of O on mL, are both well defined and give mL the structure of an O-module. n Notation 13.1. For each integer n ≥ 0, we let µf,n denote the m -torsion in mL under ·f , i.e. n µf,n = {α ∈ mL | a ·f α = 0 for all a ∈ m }. n n−1 n For n ≥ 0 and any a ∈ m r m , we have m = aO, so µf,n is equal to the a-torsion in mL,

µf,n = {α ∈ mL | a ·f α = 0}.

For example, we can choose a to be the nth power of a uniformizer for K. Note also that µf,n ⊆ µf,n+1 for all n ≥ 0. p Example 13.2. Say K = Qp, $ = p, and f = (1 + X) − 1. We saw (Example 12.7) that Ff = Gm. n For each n ≥ 1, [p ]f,f is the unique power series satisfying n n n n ϕ [p ]f,f ≡ p X mod deg 2 and f ◦ [p ]f,f = [p ]f,f ◦ f. pn n pn It is straightforward to check that (1 + X) − 1 satisﬁes these properties, so [p ]f,f = (1 + X) − 1, and n pn µf,n = {α ∈ mL | p ·f α = 0} = {ζ − 1 ∈ L | ζ = 1}. 13. LUBIN–TATE EXTENSIONS 58

Note that the automorphism group of the O-module O/mn is canonically isomorphic to (O/mn)×. Our goal is to produce totally ramiﬁed Galois extensions of L that have Galois group canonically n × isomorphic to (O/m ) . We can (and will) use the O-modules µf,n to generate these extensions provided we can show that the elements of µf,n generate totally ramiﬁed Galois extensions of the correct degree. In order to do this, we need to get a better handle on these modules µf,n. For example, it is not even a priori obvious that they are nontrivial.

ϕn−1 ϕ Lemma 13.3. Set f0 = X, and for each n ≥ 1, set fn = f ◦ · · · ◦ f ◦ f. Then µf,n = {α ∈ mL | fn(α) = 0}. n n−1 Proof. Since µf,0 = 0, the result is clear for n = 0, and we assume n ≥ 1. Fix some a ∈ m rm . Then we want to show that for any α ∈ mL, we have a ·f α = [a]f,f (α) = 0 if and only if fn(α) = 0. We will use the map [·]f ϕi ,f ϕj of Proposition 12.8 for varying i, j, so to simplify notation we write ϕi+1 ϕi it as [·]i,j. Note in particular that [a]0,0 = [a]f,f . Let 0 ≤ i ≤ n − 1. Taking $1 = $ and $2 = $ ϕi in the statement of Proposition 12.8, we see there is a unique power series [$ ]i+1,i satisfying ϕi+1 ϕi ϕi ϕ ϕi ϕi ϕi f ◦ [$ ]i+1,i = [$ ]i+1,i ◦ f and [$ ]i+1,i ≡ $ X mod deg 2. On the other hand, i+1 i i i i i f ϕ ◦ f ϕ = (f ϕ )ϕ ◦ f ϕ and f ϕ ≡ $ϕ X mod deg 2. ϕi ϕi So [$ ]i+1,i = f . Then part2 of Proposition 12.8 further implies that ϕn−1 ϕ fn = [$ ]n,n−1 ◦ · · · ◦ [$ ]2,1 ◦ [$]1,0 = [$n]n,0, Qn−1 ϕi ϕi n where $n = i=0 $ . Note that each $ is a uniformizer for L, so $nOL = mL. Since L is a n $n × complete unramiﬁed extension of K, we have aOL = mL = $nOL, so a ∈ OL . Then Corollary 12.10 $n implies that the power series [ a ]n,0 has an inverse under composition. It follows that for any α ∈ mL, $n [a]0,0(α) = 0 ⇐⇒ [ a ]n,0 ◦ [a]0,0(α) = 0 ⇐⇒ [$n]n,0(α) = 0. If f ∈ O[[X]], in particular $ ∈ K, then f ϕ = f, and the proof of Lemma 13.3 shows that n fn = [$ ]f,f . So in this case it is immediate that µf,n is the set of zeros of fn. p Example 13.4. Returning to the example of K = Qp and f = (1 + X) − 1, so Ff = Gm, we have n pn fn = [p ]f,f = (1 + X) − 1 (see Example 13.2). Recall that pn pn Y X − 1 = Φpi (X), i=0 i with Φpi (X) the p th cyclotomic polynomial. We have Φ1(X) = X − 1, each Φpi (X) is of degree (p − 1)pi−1 if i ≥ 1, and a pnth root of unity is a primitive pith root of unity if and only if it is a root of Φpi (X). Thus, we have a factorization pn pn Y fn = (1 + X) − 1 = X Φpi (1 + X), i=1 n−1 such that α ∈ µf,n r µf,n−1 if and only if it is a root of the degree (p − 1)p polynomial Φpn (1 + X). This example generalizes:

ϕn−1 ϕ Lemma 13.5. Assume that f is a monic polynomial, and let fn = f ◦ · · · ◦ f ◦ f, for each ϕn−1 n ≥ 1. Write f = h1X, and let hn = h1 ◦ fn−1, for each n ≥ 2. Then hn is a separable Eisenstein polynomial of degree (q − 1)qn−1, and

fn = hnhn−1 ··· h1X.

Moreover α ∈ µf,n r µf,n−1 if and only if hn(α) = 0. 13. LUBIN–TATE EXTENSIONS 59

q q−1 2 Proof. Our assumptions imply f = X + aq−1X + ··· + a2X + $X, with ai ∈ mL for each 2 ≤ i ≤ q − 1. So h1 ∈ OL[X] is monic and Eisenstein of degree q − 1. By induction, we have

ϕn−1 ϕn−1 fn = f ◦ fn−1 = (h1 X) ◦ fn−1 = hnfn−1 = hnhn−1 ··· h1X. n−1 Fix some n ≥ 1. The polynomial fn−1 is monic of degree q , has trivial constant term, and qn−1 ϕn−1 ϕn−1 q−1 fn−1 ≡ X mod mL. Since h is monic of degree q − 1, and h ≡ X mod mL, we have ϕn−1 n−1 (q−1)qn−1 that hn = h ◦ fn−1 is monic of degree (q − 1)q , and hn ≡ X mod mL. Further, since ϕn−1 ϕn−1 fn−1 has no constant term, the constant term of hn equals the constant term of h , which is $ . This is a uniformizer for L, so we have shown that hn is Eisenstein. Proposition 7.14 implies that hn is irreducible. It is thus separable if the characteristic of K is 0, or if the characteristic of K is positive dhn and dX 6= 0. Assume the characteristic of K is p > 0, and note that q is a power of p. We have n−1 ! dh dhϕ df n = (f ) n−1 . dX dX n−1 dX Firstly, n−1 dhϕ (f ) = ((q − 1)Xq−2 + terms of lower degree) ◦ f dX n−1 n−1 n−1 has leading term −X(q−2)q (since q is a power of the characteristic), so is nonzero. Secondly, dfn−1 dX = 1 if n = 1, and if n > 1, df n−1 = (terms of degree ≥ 1) + $ dX n−1 i Qn−1 ϕ dfn−1 dhn with $n−1 = i=0 $ . In either case, dX has nontrivial constant term, so is nonzero. Then dX is the product of two nonzero polynomials, so is nonzero. It remains to show that α ∈ µf,n r µf,n−1 if and only if hn(α) = 0. Set h0 = f0 = X. Then for any n ≥ 1, Lemma 13.3 and what we showed above imply that µf,n r µf,n−1 is the set of zeros of fn = hnhn−1 ··· h0 that are not zeros of fn−1 = hn−1 ··· h0. Proposition 13.6. Assume that f is a monic polynomial, and let n ≥ 1. The following hold. n n−1 (1) |µf,n| = q and |µf,n r µf,n−1| = (q − 1)q . For any α ∈ µf,n r µf,n−1, the map O → µf,n n ∼ given by a 7→ a ·f α induces an O-module isomorphism O/m = µf,n. (2) If α ∈ µf,n r µf,n−1, then L(µf,n) = L(α). This is a totally ramiﬁed extension of L of degree n−1 ϕn−1 (q − 1)q , NL(µf,n)/L(−α) = $ , and α is a uniformizer for L(µf,n).

Proof. We first prove part1. Let fn and hn be as in Lemma 13.5. Then by Lemma 13.5, n−1 µf,n r µf,n−1 is the set of zeros of hn, which has cardinality (q − 1)q since hn is separable of n−1 degree (q − 1)q . Then fn is the product of irreducible polynomials of different degrees, each of n n which is separable, so fn has distinct roots. Since deg fn = q , Lemma 13.3 implies |µf,n| = q . Now choose some α ∈ µf,n r µf,n−1. The O-module map ψ : O → µf,n given by ψ(a) = a ·f α contains n n−1 m in its kernel since α ∈ µf,n. On the other hand, if a ∈ m , then a ·f α 6= 0 since α∈ / µf,n−1. n n So ker ψ = m , and ψ induces an O-module injection O/m → µf,n, which must be surjective since n n |O/m | = q = |µf,n|. n−1 We now prove part2. Since α is a root of the degree (q − 1)q Eisenstein polynomial hn, the extension L(α)/L is totally ramified of degree (q − 1)qn−1, and α is a uniformizer for L(α) (see ϕn−1 Proposition 7.14). Also, NL(α)/L(−α) is the constant term of hn, which is $ . Since L(α) is finite over L, it is complete. So for any a ∈ O, the power series [a]f,f ∈ OL[[X]] can be evaluated on α, and this converges in L(α). By part1, every element of µf,n is of the form a ·f α = [a]f,f (α). So µf,n ⊆ L(α), and L(µf,n) = L(α). We now remove the assumption that f is a polynomial, and, more importantly, show that the resulting field extensions do not depend on the choice of f, but only on the choice of $. 13. LUBIN–TATE EXTENSIONS 60

0 0 0 q × Lemma 13.7. Let $ be another uniformizer of L, and let f = $ X + X . Assume there is θ ∈ OL θϕ $ such that θ = $0 , and let [θ]f,f 0 ∈ OL[[X]] be the power series of Proposition 12.8. Then for all n ≥ 1, the following hold. ∼ (1) α 7→ [θ]f,f 0 (α) deﬁnes an isomorphism of O-modules µf 0,n −→ µf,n. (2) L(µf,n) = L(µf 0,n).

0 0 0 Proof. By part3 of Proposition 12.8,[ θ]f,f ∈ HomOL (Ff ,Ff ). This implies that α 7→ [θ]f,f (α) is a homomorphism (mL, +f 0 ) → (mL, +f ) of abelian groups. By part2 of Proposition 12.8, for any a ∈ O we have

[θ]f,f 0 ◦ [a]f 0,f 0 = [aθ]f,f 0 = [a]f,f ◦ [θ]f,f 0 . × This implies that [θ]f,f 0 is a homomorphism of O-modules (mL, +f 0 , ·f 0 ) → (mL, +f , ·f ). Since θ ∈ OL , −1 Corollary 12.10 imlpies that [θ]f,f 0 is invertible with inverse [θ ]f 0,f . Thus, [θ]f,f 0 is an isomorphism of ∼ −1 O-modules, which restricts to an isomorphism of O-modules [θ]f,f 0 : µf 0,n −→ µf,n with inverse [θ ]f 0,f . Lemma 13.5 implies that µf 0,n is finite, so L(µf 0,n) is a finite extension of L, hence is complete. Then since [θ]f,f 0 ∈ OL[[X]], we see that [θ]f,f 0 (α) converges in L(µf 0,n) for any α ∈ µf 0,n. Every element of µf,n is of this form, so µf,n ⊆ L(µf 0,n) and L(µf,n) ⊆ L(µf 0,n). The same argument using −1 [θ ]f 0,f shows L(µf 0,n) ⊆ L(µf,n). Note that the assumption of this lemma holds if $ = $0, since we can take θ = 1. In particular, this shows that the field extensions L(µf,n)/L depend only on $ and not on f.

n n Theorem 13.8. Let n ≥ 1, and set L$ = L(µf,n). The extension L$/L is a ﬁnite totally ramiﬁed Galois extension of degree (q − 1)qn−1. There is a canonical isomorphism of abelian groups

n n ∼ n × ρ$ : Gal(L$/L) −→ (O/m ) n uniquely characterized by σ(α) = ρ$(σ) ·f α for all α ∈ µf,n. n Setting L$ = ∪n≥1L$, L$ is a totally ramiﬁed Galois extension of L, and there is a canonical isomorphism of topological abelian groups

∼ × ρ$ : Gal(L$/L) −→O , uniquely characterized by σ(α) = ρ$(σ) ·f α for all α ∈ µf,n and all n ≥ 1. n n Moreover, the ﬁelds L$ and L$, and the isomorphisms ρ$ and ρ$, depend only on the choice of $, and not on the choice of f.

0 q n Proof. Let f = $X + X . Then we can apply Lemma 13.7 with θ = 1. In particular L$ = 0 n L(µf 0,n). Applying Lemma 13.3 and Proposition 13.6 to f , we see that L$/L is a totally ramified extension of degree (q − 1)qn−1, and is the the splitting field of a degree qn polynomial with the qn n n distinct roots µf 0,n, hence is also Galois. Since any σ ∈ Gal(L$/L) acts continuously on L$, and the power series for +f and ·f have coefficients in L, we have

σ(α +f β) = σ(α) +f σ(β) and σ(a ·f α) = a ·f σ(α) n for all α, β ∈ µf,n and a ∈ O. In particular, Gal(L$/L) acts on µf,n by O-module automorphisms. We n thus have a homomorphism Gal(L$/L) → AutO(µf,n), which is necessarily injective since µf,n generates n n L$ over L. Since µf,n is isomorphic to O/m as O-modules, by part1 of Lemma 13.7 and part1, the n × map O → EndO(µf,n) given by a 7→ (α 7→ a ·f α) induces an isomorphism (O/m ) → AutO(µf,n). We thus obtain an injective homomorphism n n n × ρ$ : Gal(L$/L) → (O/m ) , n n−1 n × characterized by σ(α) = ρf,n(σ) ·f α for all α ∈ µf,n. Since [L$ : L] = (q − 1)q = |(O/m ) |, ρf,n is also surjective. n We have already seen that L$ depends only on the choice of $, and not on f. We now show the n n n same for ρ$. Since [θ]f,f 0 ∈ OL[[X]], and Gal(L$/L) acts continuously on L$, we have [θ]f,f 0 (σ(α)) = 14. THE ARTIN MAP 61

n σ([θ]f,f 0 (α)) for any σ ∈ Gal(L$/L) and α ∈ µf 0,n. We also know that [θ]f,f 0 is an isomorphism of O-modules µf 0,n → µf,n. We deduce that there is a commutative diagram

n n × Gal(L$/L) AutO(µf 0,n) (O/m )

AutO(µf,n)

n with all arrow isomorphisms. This shows that ρ$ is the same regardless of whether we use µf,n or µf 0,n. n n n+1 The extension L$/L is totally ramified since for all n ≥ 1, L$/L is totally ramified and L$ ⊆ L$ . n n+1 n It follows from the definition of ρ$ that the restriction map Gal(L$ /L) → Gal(L$/L) corresponds to n+1 × n × n the canonical surjection (O/m ) → (O/m ) under ρn+1 and ρ$. We obtain canonical isomorphisms n lim ρ$ Gal(L /L) ∼ lim Gal(Ln /L) −−−−→←− lim(O/mn)× ∼ O×, $ = ←− $ ←− = n n n n and we let ρ denote the composite. Since ρ(σ) mod m = ρ$(σ) for all n ≥ 1, the fact that ρ is uniquely characterized by σ(α) = ρ(σ) ·f α for all α ∈ µf,n and all n ≥ 1, follows from the similar claim for n ρ$. n Definition 13.9. The extensions L$/L and L$/L are called the Lubin–Tate extensions associated to $. In the case when L is the completion of a maximal unramified extension, the Lubin–Tate extensions and the isomorphisms of Theorem 13.8 do not even depend on $. Theorem 13.10. Assume that L is the completion of a maximal unramified extension of K. The n n extensions L$/L and L$/L, and the isomorphisms ρ$ and ρ$, of Theorem 13.8 do not depend on the choice of $. Proof. Since L = ∪ Ln , and ρ = lim ρn , it suffices to prove that Ln and ρn do not $ n≥1 $ $ ←−n $ $ $ depend on $. n The fact that L$ does not depend on $ follows from Proposition 12.13 and part2 of Lemma 13.7. n The fact that ρ$ does not depend on the choice of $ follows fromProposition 12.13 and part1 of Lemma 13.7 by the same argument that showed the independence of f in the proof of Theorem 13.8.

14. The Artin map In this section, we fix a nonarchimedean local field K. We let O and m denote its ring of integers and residue field, respectively. We let k denote its residue field, and q = |k|. Let K be an algebraic closure of K, and let Kur be the maximal unramified subextension. Many of the results of the previous section that hold for the completion of Kur can be transferred to Kur via the following lemma.

Lemma 14.1. Let F and E be algebraic extensions of K with F ⊆ E, and let Fb and Eb denote their respective completions. Then the following hold. (1) If E/F is finite, then EFb = Eb. (2) If E/F is separable, then E ∩ Fb = F . (3) If E/F finite and Galois, then so is E/b Fb and the restriction map σ 7→ σ|E defines an ∼ isomorphism Gal(E/b Fb) −→ Gal(E/F ). Proof. The extension EF/b Fb is finite, so the valuation on Fb extends uniquely to EFb, and EFb is complete with respect to this valuation. This is the valuation induced on EFb as a subfield of Eb, so EFb is closed in Eb. On the other hand, since E ⊆ EFb ⊆ Eb, we see that EFb is dense in Eb, hence EFb = Eb. To prove part2, note that E ∩ Fb = F if and only if E0 ∩ Fb = F for every finite subextension E0/F of E/F . So we may assume E/F is finite. Moreover, if we let E00 be the Galois closure of E/F , we 14. THE ARTIN MAP 62 have F ⊆ E ∩ Fb ⊆ E00 ∩ Fb. So it suffices to show E00 ∩ Fb = F . We are thus redued to proving part2 in the case when E/F is finite and Galois, and we prove this at the same time as proving part3. By part1, Eb = EFb, hence is a finite Galois extension of Fb, and (14.2) [Eb : Fb] = [EFb : Fb] = [E : E ∩ Fb] ≤ [E : F ]. Any element σ ∈ Gal(E/b Fb) induces an isomorhpism from E to a subfield of Eb that is the identiy on F . Since E/F is Galois, it maps E to itself and we have a well defined restriction Gal(E/b Fb) → Gal(E/F ). This a homomorphism, and is injective since any element of Gal(E/b Fb) acts continuously on Eb, hence is determined by its restriction to the dense subfield E. It is surjective since any element of Gal(E/F ) extends by continuity to an element of Gal(E/b Fb). We deduce that [E/b Fb] = [E : F ], so E ∩ Fb = F by ∼ (14.2), and that the restriction map Gal(E/b Fb) −→ Gal(E/F ) is an isomorphism. n n−1 Proposition 14.3. Let $ be a uniformizer for K, let n ≥ 1, and let K$ be the degree (q − 1)q n ur ur Lubin–Tate extensions of K associated to $. The extension K$K of K is a totally ramified Galois n−1 n extension of degree (q − 1)q that does not depend on the choice of $. Restriction to K$ defines an n ur ur ∼ n isomorphism Gal(K$K /K ) −→ Gal(K$/K), and composing this with the canonical isomorphism n n ∼ n × ρ$ : Gal(K$/K) −→ (O/m ) we obtain a canonical isomorphism n n ur ∼ n × ρ : Gal(K$K /K) −→ (O/m ) that does not depend on the choice of $.

ur n n−1 Proof. Let L be the completion of K , and let L$ be the degree (q−1)q Lubin–Tate extension n n n n associated to $. By Theorem 13.10, L$ does not depend on the choice of $. Note L$ = K$L, so L$ n ur s is the completion of K$K by part1 of Lemma 14.1. Then letting K denote the separable closure s n ur of K inside of K, we apply part2 of Lemma 14.1 with E = K and F = K$K , and deduce that n ur s n K$K = K ∩ L$, so does not depend on the choice of $. n n ur n ur Since K$/K is totally ramified, K$ ∩K = K, and K$K /K is a totally ramified Galois extension n−1 n n ur ur ∼ of degree (q − 1)q . Moreover, the restriction to K$ defines an isomorphism Gal(K$K /K ) −→ n n Gal(K$/K), so we obtain the canonical isomorphism ρ as in the statement of the Proposition. It only remains to see that ρn is independent of the choice of $. Applying part3 of Lemma 14.1, n ur n ∼ n ur n restriction to K$K defines an isomorphism Gal(L$/L) −→ Gal(K$K /K). Since L$ = L(µf,n) n n and K$ = K(µf,n) for appropriately chosen f, the formula for ρ$ of Theorem 13.8 imlies that the composite n n ∼ n ur ur ∼ n ρ$ n × Gal(L$/L) −→ Gal(K$K /K ) −→ Gal(K$/K) −−→ (O/m ) n n n is canonical isomorphism of Theorem 13.8 for the extension L$/L, call it ρL. It follows that ρ equals n n ur ∼ n ρL precomposed with the isomorphism Gal(K$K /K) = Gal(L$/L), so doesn’t depend on $ by Theorem 13.10. We are now ready to define the Artin map. First some notation.

LT ur LT Notation 14.4. Let $ be a uniformizer for K, and set K = K$K . Note that K = n ur ∪n≥1K$K , so does not depend on the choice of $ by Proposition 14.3. ur LT Let $ be a uniformizer for K. Since both K$/K and K /K are Galois, K /K is also Galois. ur Since K$ ∩ K = K, we have a splitting of the Galois group LT ∼ ur Gal(K /K) = Gal(K /K) × Gal(K$/K)

ur given by σ 7→ (σ|K , σ|K$ ). This is an isomorphism of topological groups. Note that this splitting depends on the choice of the field the field K$, hence on the choice of $, but otherwise is canonical. The choice of $ also determines a splitting K× =∼ $Z × O×, 14. THE ARTIN MAP 63 which is again an isomorphism of topological groups. Using these two splittings, we we can define a continuous homomorphism × LT ψ$ : K → Gal(K /K) by × ur $Z × O Gal(K /K) × Gal(K$/K) n n −1 −1 ($ , u) (FrobK , ρ$ (u )). The reader should compare this with Example 10.2.

Theorem 14.5. The homomorphism ψ$ does not depend on the choice of $. Before proving the theorem, we make a deﬁnition and a couple of remarks.

× LT Definition 14.6. The Artin map ArtK : K → Gal(K /K) (or Artin reciprocity map or just reciprocity map) is ArtK = ψ$, with ψ$ as above for any choice of uniformizer $ of K. Remark 14.7. The Artin map is completely canonical, and independent of all choices. There is, however, another normalization for it. Namely, we could have deﬁned it by

× ur $Z × O Gal(K /K) × Gal(K$/K) n −n −1 ($ , u) (FrobK , ρ$ (u)). This is sometimes referred to as the geometric normalization, whereas we are using what is sometimes called the arithmetic normalization. Recall that we have deﬁned FrobK to be the unique automorphism of Kur that induces x 7→ xq on the residue ﬁeld of Kur. This automorphism of Kur is called the arithmetic Frobenius, and its inverse is called the geometric Frobenius. So the normalization of the Artin map is determined by specifying whether a uniformizer is sent to the arithmetic Frobenius, or the geometric Frobenius.

Remark 14.8. Since ρ$ is an isomorphism, it follows immediately from the deﬁnition that the LT n image of ArtK consists of all elements σ ∈ Gal(K /K) such that σ|Kur = FrobK for some n ∈ Z. In particular, ArtK has dense image. Remark 14.9. We will prove later that KLT is the maximal abelian extension of K.

To show that ψ$ is independent of $, it will actually suffice to see how ψ$ behaves when applied to other choices of uniformizers. For this, we use a special case of the following lemma, that will also be important when we discuss base change later. First note that if L/K is any finite unramified extension in K, we have LLT = KLT, since Lur = Kur and any uniformizer for K is a uniformizer for L. In LT ur particular, since L = LπL does not depend on the choice of uniformizer π of L, the Lubin–Tate LT extension Lπ is contained in K even if π∈ / K. × LT Lemma 14.10. Let $ be a uniformizer for K, and let ψ$ : K → Gal(K /K) be as above. Let x ∈ K× have normalized valuation m > 0, and let L/K be the degree m unramified subextension of K/K. Choose a uniformizer let π for L such that NL/K (π) = x (this is possible by Proposition 9.4), and let Lπ be the Lubin–Tate extension of L associated to π. Then ψ$(x) is the identity on Lπ. −m × × Proof. Let u = x$ ∈ O . Since $ is a uniformizer for L, there is v ∈ OL such that π = v$. q q Since x = NL/K (π), we then have u = NL/K (v). Let f = $X + X , let g = πX + X , and let µf,n n and µg,n denote the m -torsion O-modules for the Lubin–Tate formal group laws associated to f and m g, respectively. Since x has normalized valuation m, we have ψ$(x)|Kur = FrobK . This acts as the identity on L, since L/K is the unramified extension of degree m. Since Lπ = ∪n≥1L(µg,n), it then suffices to show that ψ$(x) acts as the identity on µg,n for all n ≥ 1. Let Kdur be the completion of Kur, and let ϕ denote the Frobenius automorphism of Kdur. Fix ur n ≥ 1, and let σ denote the unique extension of ψ$(x) to a continuous automorphism of Kd (µg,n). It suffices to show that σ(α) = α for any α ∈ µ . By Proposition 12.13, there is θ ∈ O [[X]] such g,n Kdur 15. COLEMAN’S NORM OPERATOR 64

ϕ that θ = π = v. Then the power series [θ] ∈ O [[X]] of Proposition 12.8 deﬁnes an isomorphism θ $ g,f Kdur ∼ ur ur [θ]g,f : µf,n −→ µg,n, and Kd (µg,f ) = Kd (µf,n) by Theorem 13.10. It thus suﬃces to show that σ([θ]g,f (β)) = [θ]g,f (β) for all β ∈ µf,n. m ur −1 ur Since ψ$(x) acts as FrobK on K , and as [u ]f,f on µf,n, the extension σ to Kd (µf,n) acts as m ur −1 ϕ on Kd , and again as [u ]f,f on µf,n. So for any β ∈ µf,n, we have ϕm −1 σ([θ]g,f (β)) = [θ]g,f ([u ]f,f (β)). The power series [θ] ∈ O [[X]] is the unique one satisfying g,f Kdur ϕ g ◦ [θ]g,f = [θ]g,f ◦ f and [θ]g,f ≡ θX mod deg 2. m m Since f, g ∈ OL[X], they are both stable under ϕ , so applying ϕ to the above, we have ϕm ϕm+1 ϕm ϕm g ◦ [θ]g,f = [θ]g,f ◦ f and [θ]g,f ≡ θ X mod deg 2. m ϕ ϕm The uniqueness in Proposition 12.8 implies [θ]g,f = [θ ]g,f . Since ϕm ϕm−1 ϕm−1 ϕ θ = (vθ) = ··· = (v ··· v v)θ = NL/K (v)θ = uθ, we conclude that ϕm −1 −1 −1 [θ]g,f ([u ]f,f (β)) = [uθ]g,f ◦ [u ]f,f (β) = [θ]g,f ◦ [u]f,f ◦ [u]f,f (β) = [θ]g,f (β), for all β ∈ µf,n.

Proof of Theorem 14.5. Let $ and π be two uniformizers for K. We will show ψ$(π) = ψπ(π). If we show this, then since $ and π were arbitrary, we obtain 0 0 0 ψ$($ ) = ψ$0 ($ ) = ψπ($ ) 0 × for any third choice of uniformizer $ . Since uniformizers generate K as a group, this shows ψ$ = ψπ. LT ur Since K = KπK , to show ψ$(π) = ψπ(π), it suffices to show they induce the same isomorphism ur ur on each of Kπ and K . They both induce FrobK on K , and ψπ(π) is the identity on Kπ, so it only remains to show that ψ$(π) is the identity on Kπ. This is a special case of Lemma 14.10 (taking x = π in the statement). 15. Coleman’s norm operator In this section, we fix a nonarchimedean local field K, let O and m denote it’s valuation ring and ideal, respectively, and let q denote the cardinality of its residue field. We fix a finite unramified extension L/K, a uniformizer $ for L, and letting OL denote the valuation ring of L, a monic polynomial f ∈ OL[X] satisfying f ≡ $X mod deg 2 and f ≡ X mod mL. n n−1 So for each n ≥ 1, we have the m -torsion O-modules µf,n, and the degree (q − 1)q Lubin–Tate n extension L$ = L(µf,n). To simplify the notation slightly, we will write µn for µf,n throughout this section.

Lemma 15.1. Let g ∈ OL[[X]]. If g(α) = 0 for all α ∈ µ1, then g = fh for some h ∈ OL[[X]]. P i P j Proof. Write g = i≥0 aiX . Fix α ∈ µ1, and let bi = j≥0 ai+j+1α , for each i ≥ 0. This n series is convergent, since α lies in the valuation ideal for L and b lies in the valuation ring O n of $ i L$ n L$. If g(α) = 0, then X i g(X) = (X − α) biX . i≥0

Since the elements of µ1 are the q distinct roots of f, if g(α) = 0 for all α ∈ µ1, we can repeat the above process, and g = fh for some h ∈ O n [[X]]. Since the coeﬃcients of f and g both belong to L, L$ so do the coeﬃcients of h, which then lie in O n ∩ L = O . L$ L r r Lemma 15.2. Let h ∈ OL[[X]], and let r ≥ 1. If h ◦ f ≡ 0 mod mL, then h ≡ 0 mod mL. 15. COLEMAN’S NORM OPERATOR 65

q Proof. We induct on r. Since h ◦ f ≡ h(X ) mod mL, the claim holds for r = 1. Assume it holds r+1 r 0 0 for some r ≥ 1, and that h ◦ f ≡ 0 mod mL . By the inductive hypothesis, h = $ h with h ∈ OL[[X]]. 0 0 r+1 Then h ◦ f ≡ 0 mod mL. By the r = 1 case, h ≡ 0 mod mL, hence h ≡ 0 mod mL .

Let Ff be the Lubin–Tate formal group law associated to f. For a power series g ∈ OL[[X]] and α ∈ µn, we write g(X +f α) = g ◦ Ff (X, α).

For example, if α ∈ µ1, then ϕ ϕ f(X +f α) = f ◦ Ff (X, α) = Ff (f(X), f(α)) = Ff (f(X), 0) = f(X). 1 This makes intuitive sense, since over L$, Y f(X) = (X − α),

α∈µ1 and µ1 is permuted under +f .

Lemma 15.3. Let g ∈ OL[[X]] satisfy g(X +f α) = g(X) for all α ∈ µ1. Then g = h ◦ f for a unique power series h ∈ OL[[X]]. 0 0 Proof. If g(X +f α) = g(X) for all α ∈ µ1, then g (X) = g(X) − g(0) satisﬁes g (α) = 0 for all α ∈ µ1. By Lemma 15.1, there is g1 ∈ OL[[X]] such that

g(X) = g(0) + g1(X)f(X).

For all α ∈ µ1, we have g(X +f α) = g(X) and f(X +f α) = f(X), so we must also have g1(X +f α) = g1(X). By what we showed above, we can ﬁnd g2(X) ∈ OL[[X]] such that

g1(X) = g1(0) + g2(X)f(X).

Repeating this procedure, we can inductively construct gi ∈ OL[[X]] for i ≥ 0, with g0 = g, such that X i g(X) = gi(0)f(X) . i≥0 P i 0 0 Setting h(X) = i≥0 gi(0)X ∈ OL[[X]], we have h ◦ f = g. If h ∈ OL[[X]] also satisﬁes h ◦ f = g, 0 0 then (h − h ) ◦ f = 0, and Lemma 15.2 implies h − h = 0.

Proposition 15.4. For any g ∈ OL[[X]], there is a unique power series Nf (g) ∈ OL[[X]] such that Y Nf (g) ◦ f = g(X +f α).

α∈µ1

Moreover, Nf (gh) = Nf (g)Nf (h) for any g, h ∈ OL[[X]]. The coeﬃcients of Q g(X + α) are O -polynomials in α ∈ µ . They are stable Proof. α∈µ1 f L 1 1 under permutations of µ1, so these coeﬃcients are stable under the action of Gal(L$/L), hence lie in O 1 ∩ L = O . Lemma 15.3 then implies the existence and uniqueness of N (g). L$ L f Then for g, h ∈ OL[[X]], Y (Nf (g)Nf (h)) ◦ f = (Nf (g) ◦ f)(Nf (h) ◦ f) = g(X +f α)h(X +f α).

α∈µ1

Uniqueness then implies Nf (gh) = Nf (g)Nf (h).

Definition 15.5. Nf is called Coleman’s norm operator.

ϕn−1 ϕ Lemma 15.6. Let g ∈ OL[[X]]. For n ≥ 1, let fn = f ◦ · · · ◦ f ◦ f, and inductively deﬁne n n−1 ϕ−1 ϕ 0 N (g) = N Nf (g) , where we take N (g) = g. Then we have

n Y N (g) ◦ fn = g(X +f α).

α∈µn 15. COLEMAN’S NORM OPERATOR 66

1 Proof. We prove this by induction on n ≥ 1. Since N (g) = Nf (g), the n = 1 case holds by deﬁnition. We assume true for some n − 1 ≥ 1, and let C be a set of coset representatives for µn/µ1. Then Y Y Y Y g(X +f α) = g(X +f β +f α) = Nf (g) ◦ f(X +f β).

α∈µn β∈C α∈µ1 β∈C n −1 Choose any liftϕ ˜ ∈ Gal(L$/K) of ϕ = FrobK ∈ Gal(L/K). Applyingϕ ˜ to ϕ f(X +f β) = f ◦ Ff (X, β) = Ff (f(X), f(β)), we have ϕ˜−1 ϕ−1 ϕ˜−1 ϕ−1 ϕ˜−1 f(X +f β) = Ff (f (X), f(β) ) = f (X) +f f(β) . ϕ By Lemma 13.3, µn is the set of roots of fn = fn−1 ◦ f, and µn−1 is the set of roots of fn−1. This ϕ˜−1 ∼ implies that the map β 7→ f(β) is a bijection C −→ µn−1. So we have !ϕ˜ Y Y ϕ−1 ϕ˜−1 Nf (g) ◦ f(X +f β) = Nf (g) ◦ f(X +f β) β∈C β∈C !ϕ˜ Y ϕ−1 ϕ−1 ϕ˜−1 = Nf (g) ◦ f (X) +f f(β) β∈C !ϕ˜ Y ϕ−1 ϕ−1 = Nf (g) ◦ f (X) +f α

α∈µn−1 ϕ n−1 ϕ−1 ϕ−1 = N Nf (g) ◦ fn−1 ◦ f n = N ◦ fn, with the second to last equality by the inductive hypothesis. n Lemma 15.7. Let g ∈ OL[[X]], let n ≥ 1, and let N (g) be as in Lemma 15.6. ϕ × × (1) Nf (g) ≡ g mod mL. If particular, if g ∈ OL[[X]] , then Nf (g) ∈ OL[[X]] . n n+1 (2) If g ≡ 1 mod mL, then Nf (g) ≡ 1 mod mL . × n n−1 ϕ n (3) If g ∈ OL[[X]] , then N (g)/N (g) ≡ 1 mod mL. 1 We ﬁrst prove part1. Since µ lies in the valuation ideal m 1 of L , we have Proof. 1 L$ $ Y q g(X + α) ≡ g(X) mod m 1 . f L$ α∈µ1

Since the coeﬃcients of this product lie in O , this equality further hold modulo m 1 ∩ O = m . On L L$ L L q the other hand, Nf (g) ◦ f(X) ≡ Nf (g)(X ) mod mL, so we have q q Nf (g)(X ) ≡ g(X) mod mL, ϕ which implies Nf (g) ≡ g mod mL. n n n For part2, write g = 1 + $ h, with h ∈ O [[X]]. Note that $ α ∈ m m 1 for any α ∈ µ , and L L L$ 1 jnq n that $ ∈ m m 1 for any 1 ≤ j ≤ q − 1. Hence, j L L$

Y n Nf (g) ◦ f = (1 + $ h(X +f α))

α∈µ1 n q n ≡ (1 + $ h(X)) mod m m 1 L L$ n ≡ 1 mod m m 1 . L L$ n So we have (N (g) − 1) ◦ f ≡ 0 mod m m 1 . Since (N (g) − 1) ◦ f ∈ O [[X]], this congruence holds f L L$ f L n n+1 n+1 modulo m m 1 ∩ O = m . Then Lemma 15.2 implies N (g) − 1 ≡ 0 mod m . L L$ L L f L 15. COLEMAN’S NORM OPERATOR 67

We now prove part3. Since both ϕ and Nf are multiplicative (see Proposition 15.4), it follows that N n is multiplicative for any n ≥ 0. So

−1 ϕ −1 !ϕ N n(g) N n−1N (g)ϕ N (g)ϕ = f = N n−1 f . N n−1(g)ϕ N n−1(g)ϕ g

ϕ ϕ−1 By part1, Nf (g) ≡ g mod mL, so Nf (g) /g ≡ 1 mod mL. In particular, this shows the result for n = 1, since N 0(g) = g. If n ≥ 2, then unwinding the inductive deﬁnition of N n−1, we have

−1 −1 n−2 n−1 ϕ−1 ϕ ϕ ϕ N (·) = Nf Nf ··· Nf (·) ··· . | {z } n−1 times

ϕ−1 Then since Nf (g) /g ≡ 1 mod mL, we can apply part2 to see that each of the n − 1 applications Nf increases the exponent of the congruence by 1, hence

−1 !ϕ N (g)ϕ N n−1 f ≡ 1 mod mn . g L

× Recall from section9 that given a finite separable extension F/K, we let N(F/K) = NF/K (F ) ⊆ K×, and that if F/K is any separable extension, we define \ N(F/K) = N(F 0/K), F 0/K with the intersection taken over all finite subextensions F 0/K of F/K. Also recall the notation n × Un = 1 + m ⊆ O , for any n ≥ 1.

n Proposition 15.8. Let x = NL/K ($). Then for any n ≥ 1, we have N(L$/K) ⊆ hxi × Un. n Proof. Fix some α ∈ µn r µn−1. By part2 of Proposition 13.6, −α is a uniformizer for L$ and ϕn−1 N n (−α) = N ($ ) = x. L$ /K L/K × So it remains to show that NLn /K (O n ) ⊆ Un. $ L$ × n Fix some u ∈ O n . Since α is a uniformizer for L , given a set S of representatives containing 0 L$ $ n P i n for the residue ﬁeld of L$, we can write u = i≥0 aiα , with ai ∈ S and a0 =6 0. Since L$/L is totally × j ramiﬁed, we can assume S ⊆ OL. Hence u = g(α) for some g ∈ OL[[X]] . For each j ≥ 0, let N be as in Lemma 15.6 and set u = N j(g)(0). By Lemma 15.6, we have u = Q g(β). Then j j β∈µj

Y Y un NLn /L(u) = σ(g(α)) = g(β) = . $ u n n−1 σ∈Gal(L$ /L) β∈µnrµn−1 ϕ n By part3 of Lemma 15.7, un/un−1 ∈ 1 + mL, so ϕ n n N n (u) = N (u /u ) = N (u /u )) ∈ N (1 + m ) ⊆ 1 + m . L$ /K L/K n n−1 L/K n n−1 L/K L

Corollary 15.9. Let x = NL/K ($), and let F/L be a totally ramiﬁed extension containing L$. Then N(F/K) = hxi. Proof. Proposition 9.5 implies that N(F/L) contains a uniformizer π, so N(F/K) contains × NL/K (π). Then since π and $ diﬀer by a unit in OL, NL/K (π) = u$ for some u ∈ O . On the other hand, Proposition 15.8 implies that N(F/K) ⊆ hxi × Un for all n ≥ 1, hence N(F/K) ⊆ hxi. It follows that u = 1 and N(F/K) = hxi. 16. BASE CHANGE 68

16. Base change As usual, we fix a nonarchimedean local field K, let O and m denote it’s valuation ring and ideal, respectively. We let Ks denote a separable closure of K, and Kur the maximal unramfied extension of K in Ks. Recall we have the Lubin–Tate extension KLT/K in Ks, and the Artin map of §14, × LT ArtK : K → Gal(K /K). s n s Proposition 16.1. Let σ ∈ Gal(K /K) be such that σ|Kur = FrobK with n ∈ Z>0, and let Fσ ⊂ K × −1 be the fixed field of σ. Then N(Fσ/K) is the cyclic subgroup of K generated by ArtK (σ|KLT ). −1 s Proof. Set x = ArtK (σ|KLT ), and let L be the degree n unramified extension of K in K . Since n ur ur σ acts as FrobK on K , L is the maximal subfield of K on which σ acts as the identity. Hence ur Fσ ∩K = L, and F/L is totally ramified. Since the normalized valuation of x is n, there is a uniformizer $ of L such that NL/K ($) = x by Proposition 9.4. Lemma 14.10 implies that σ|KLT = ArtK (x) acts as the identity on the Lubin–Tate extension L$, so L$ ⊆ Fσ. We can now apply Corollary 15.9, and N(Fσ/K) = hxi. We now have all the necessary preparation to prove the Base Change Theorem: Theorem 16.2 (Base Change). Let L/K be an finite extension in Ks. Then KLT ⊆ LLT and the following diagram commutes L× ArtL Gal(LLT/L)

N L/K σ7→σ|KLT K× ArtK Gal(KLT/K).

Proof. Let y be any nonzero element in the valuation ideal mL of L. Choose any lift σ ∈ s LT LT LT Gal(K /L) of ArtL(y) ∈ Gal(L /L). To show K ⊆ L , it suffices to show that σ|KLT depends only on σ|LLT = ArtL(y). This will follow from showing σ|KLT = ArtK (NL/K (y)), which also implies × commutativity of the above diagram since nonzero elements in mL generate L . −1 s Let x = ArtK (σ|KLT ), and let Fσ be subfield of K fixed by σ. We want to show NL/K (y) = x. × × By Proposition 16.1, N(Fσ/L) = hyi ⊂ L and N(Fσ/K) = hxi ⊆ K , so

hNL/K (y)i = NL/K (hyi) = NL/K (N(Fσ/L)) = hN(Fσ/K)i = hxi.

Then to see that NL/K (y) = x, it suﬃces to show they have the same valuation. Let vK and vL denote the normalized valuations on K and L, respectively, and let d denote the residue degree of the extension 1 d L/K. Then vL = d vK ◦ NL/K , and FrobL|Kur = FrobK . We obtain

vK (x) vL(y) dvL(y) FrobK = ArtK (x)|Kur = ArtL(y)|Kur = FrobL |Kur = FrobK .

So vK (x) = dvL(y) = vK (NL/K (y)). Corollary 16.3 (Existence Theorem). For any ﬁnite index open subgroup V ⊆ K×, there is a × ﬁnite abelian extension L/K such that NL/K (L ) = V .

Proof. Choose a uniformizer $ for K, and let K$ be the maximal totally ramified Lubin–Tate LT ur × ∼ × extension associated to $; so K = K$K . The splitting K = h$i × O is an isomorphism of topological groups, so under this isomorphism V = h$ni × U for some positive integer n, and some × finite index open subgroup U of O . Letting L be the fixed field of ArtK (V ), it is then the composite ur n of the subfield L1 of K fixed by FrobK and the subfield L2 of K$ fixed by U. Since ArtK (U) is ∼ ur open in Gal(K$/K), L2/K is finite and Galois with Gal(K$/L2) = U. The subfield L1 of K fixed n ur ur by FrobK is the unique degree n extension of K in K , and Gal(K /L1) is the closure of the cyclic n subgroup generated by FrobK . Then setting L = L1L2, the subgroup LT LT ∼ ur Gal(K /L) ⊆ Gal(K /K) = Gal(K /K) × Gal(K$/K) 17. RAMIFICATION GROUPS 69

n is hFrobK i × U, and the composite σ7→σ| L× −−−→ArtL Gal(LLT/L) −−−−−−→KLT Gal(KLT/K) surjects onto V . The corollary now follows from Theorem 16.2. We can now also prove the uniqueness theorem. Theorem 16.4 (Uniqueness of the Artin map). The Artin map is the unique homomorphism × LT ArtK : K → Gal(K /K) satisfying the following.

(i) For each uniformizer $ of K, ArtK ($)|Kur = FrobK . LT (ii) For each finite extension L/K in K , ArtK (N(L/K))|L = 1. Proof. The Artin map satisfies property (i) by definition, and property (ii) by Theorem 16.2. 0 Now let ArtK be another homomorphism satisfying properties (i) and (ii). Since uniformizers generate × 0 K as a group, it suffices to show ArtK ($) = ArtK ($) for any uniformizer $. n n−1 Fix a uniformizer $ of K. For each n ≥ 1, let K$ be the totally ramified degree (q − 1)q n Lubin–Tate extension of K associated to $, and let K$ = ∪n≥1K$. By property (i), ArtK ($)|Kur = 0 LT ur ArtK ($)|Kur . Since K = K$K , and ArtK ($) acts as the identity on K$, it only remains to show 0 that ArtK ($) acts as the identity on K$. n By part2 of Proposition 13.6, for each n ≥ 1, there is α ∈ K such that N n (−α) = $. So $ K$ /K n 0 n $ ∈ N(K$/K) for each n ≥ 1. Then by property (ii), ArtK ($) acts as the identity on K$ for all n ≥ 1. 17. Ramification groups In order to prove the Local Kronecker–Weber Theorem, i.e. the assertion that KLT is the maximal abelian extension of K, we need to understand subgroups of the inertia group, called the ramification subgroups. Throughout this section, K will denote a nonarchimedean local field. We let O, m, and k denote the valuation ring, valuation ideal, and residue field, respectively, of K. We let U 0 = O×, and U i = 1 + mi for any i ≥ 1. We also fix a finite Galois extension L/K, and set G = Gal(L/K). We denote by OL, mL, and 0 × kL, the valuation ring, valuation ideal, and residue field, respectively, of L. We let UL = OL , and i i i i UL = 1 + mL for any i ≥ 1. Recall that any σ ∈ G satisfies σ(mL) = mL for all i ∈ Z. Hence, σ induces i an O-algebra automorphism on OL/mL for any i ≥ 0.

Lemma 17.1. Let vL be the normalized valuation on L, and let y ∈ OL be such that OL = O[y] (this is possible by Proposition 6.13). Fix an integer i ≥ 0. For any σ ∈ G, the following are equivalent. i+1 (1) σ induces the identity on OL/mL . (2) vL(σ(x) − x) ≥ i + 1 for all x ∈ OL. (3) vL(σ(y) − y) ≥ i + 1. Proof. Parts1 and2 are easily seen to be equivalent, and part2 trivially implies part3. Part3 i+1 i+1 implies part1 since y mod mL generates OL/mL as an OL-algebra.

Proposition 17.2. For each integer i ≥ 0, let Gi be the set set of σ ∈ G satisfying the equivalent conditions of Lemma 17.1. Then (Gi)i≥0 form a decreasing filtration of normal subgroups of G; G0 is the inertia subgroup of G, and Gi = {1} for i sufficiently large. i+1 i+1 Proof. Reduction modulo mL induces a homomorphism G → AutO(OL/mL ), and Gi is its kernel by part1 of Lemma 17.1. So each Gi is a normal subgroup of G. The fact that Gi are decreasing is obvious. An element belongs to the inertia subgroup if and only if it induces the trivial automorphism on kL, which is the same as the characterization of G0. To see that Gi = {1} for i sufficiently large, choose y ∈ OL such that OL = O[y], and let vL denote the normalized valuation on L. Since L = K(y), σ(y) 6= y for any nonidentity σ ∈ G. Then choosing i ∈ Z such that i + 1 > vL(σ(y) − y) for all nonidentity σ ∈ G, we have Gi = {1}. 17. RAMIFICATION GROUPS 70

Definition 17.3. For i ≥ 0, the subgroup Gi of Proposition 17.2 is called the ith ramiﬁcation group of L/K (or the ith ramiﬁcation subgroup of G). n Example 17.4. Assume that K = Qp, and that L = Qp(ζ), where ζ is a primitive p th root of n × a unity. Then G = Gal(Qp(ζ)/Qp) is canonically isomorphic to (Z/p ) via σ 7→ a where σ(ζ) = ζ . For each 1 ≤ m < n, let (1 + pmZ)/pn denote the image of 1 + pmZ in (Z/pn)×. Then under the isomorphism G =∼ (Z/pn)×, we claim  ( /pn)× if i = 0,  Z ∼ m n m−1 m Gi = (1 + p Z)/p if p ≤ i ≤ p − 1 and 1 ≤ m ≤ n − 1, {1} if i ≥ pn−1.

Since Qp(ζ)/Qp is totally ramified, G0 = G. By Exercise 7.4, OL = Zp[ζ] and ζ − 1 is a uniformizer for Qp(ζ). Let v denote the normalized valuation on Qp(ζ). We have σ ∈ Gi if and only if v(σ(ζ)−ζ) ≥ i+1. Let a ∈ (Z/pn)× corresponds to σ under the canonical isomorphism. Then σ(ζ) = ζa, and v(σ(ζ) − ζ) = v(ζa − ζ) = v(ζa−1 − 1). Letting 0 ≤ m ≤ n be maximum such that a − 1 ∈ pmZ/pn ⊆ Z/pn, we have that ζa−1 is a primitive n−m a−1 m a−1 p th root of unity. So Qp(ζ)/Qp(ζ ) is totally ramified extension of degree p , and ζ − 1 is a a−1 a−1 m m uniformizer for Qp(ζ ). It follows that v(ζ − 1) = p , and σ ∈ Gi if and only if p ≥ i + 1. Important to the understanding of the ramification is the following relation between the filtration i (Gi)i≥0 and the filtration (UL)i≥0. σ($) i+1 Proposition 17.5. Let $ be a uniformizer for L. For each i ≥ 0, the map σ 7→ $ mod UL i i+1 induces an injective homomorphism Gi/Gi+1 ,→ UL/UL . This homomorphism is independent of the choice of $.

σ($) × σ($) i i+1 Proof. For any σ ∈ G, $ ∈ OL and $ − 1 ∈ mL if and only if σ($) − $ ∈ mL . It follows σ($) i+1 i i+1 that g(σ) = $ mod UL is a well deﬁned map g : Gi → UL/UL . It also follows that if g is a i i+1 homomorphism, then it induces an injection Gi/Gi+1 ,→ UL/UL . × i+1 σ(u) i+1 × For any u ∈ OL , if σ(u) − u ∈ mL , then u − 1 ∈ mL . So for any u ∈ OL and σ ∈ Gi, we have σ(u$) σ($) i+1 u$ ≡ $ mod UL . This shows that g is independent of the choice of $. Then for any σ, τ ∈ Gi, since τ($) is also a uniformizer for L, we have στ($) σ(τ($)) τ($) g(στ) = mod U i+1 = mod U i+1 = g(σ)τ(σ). $ L τ($) $ L

Corollary 17.6. G0/G1 has order prime to p, and Gi is a p-group for any i ≥ 1. 0 1 ∼ × Proof. This follows immediately from Proposition 17.5 and the fact that UL/UL = kL , which i i+1 ∼ has order prime to p, and the fact that for any i ≥ 1, UL/UL = kL, which is a p-group. Corollary 17.7. The group Gal(L/K) is solvable.

Proof. The quotient G/G0 is isomorphic to Gal(kL/k), which is abelian. Proposition 17.5 implies that Gi/Gi+1 is abelian for each i ≥ 0, and by Proposition 17.2 we know Gi = {1} for i sufficiently large. The ramification filtration behaves well upon passing to subgroups. More specifically, if F/K is a subextension of L/K and H = Gal(L/F ), then Hi = H ∩ Gi for any i ≥ 0. Passing to quotients, however, is more subtle. To that end, we introduce some notation. Define the function iG : G0 → Z ∪ {∞} by iG(σ) = i + 1 if σ ∈ Gi r Gi+1 and iG(1) = ∞. So if x is a generator for OL as an O-algebra and vL is the normalized valuation on L, then iG(σ) = vL(σ(x) − x). For any real number r ≥ 0, set Gr = Gi where i is the smallest integer ≥ r. Thus, σ ∈ Gu if and only if iG(σ) ≥ u + 1. 17. RAMIFICATION GROUPS 71

Proposition 17.8. Let H be a normal subgroup of G with ﬁxed ﬁeld F . Then for every σ ∈ G,

1 X i (σH) = i (στ), G/H e G L/F τ∈H with eL/F the ramiﬁcation index of L/F .

Proof. When σ ∈ H, this equality is to be interpreted as ∞ = ∞, so we assume σ 6∈ H. Letting vF and vL denote the normalized valuations on F and L, respectively, we have vF = eL/F vL. Letting x and y be generators for OF and OL, respectively, as O-algebras, we have X X eL/F iG/H (σH) = vL(σ(x) − x) and iG(στ) = vL(στ(y) − y). τ∈H τ∈H

So it suffices to show that the elements Y a = σ(x) − x and b = (στ(y) − y) τ∈H generate the same ideal in OL. Q Let f ∈ OF [X] be the minimal polynomial of y over F . Then f(X) = τ∈H (X − τ(y)). Let σ(f) ∈ OL[X] be the polynomial obtained by applying σ to all coefficients of f. Then σ(f)(X) = Q τ∈H (X − στ(y)). Since OF = O[x], for any z ∈ OF the element a = σ(x) − x divides σ(z) − z in OL. In particular, a divides all coefficients of σ(f) − f. Thus, a divides

σ(f)(y) − f(y) = σ(f)(y) = ±b in OL. To see that b | a in OL, ﬁrst take g ∈ O[X] such that x = g(y). Then g(X) − x ∈ OF [X] and has x as a root, so g(X) − x = f(X)h(X) for some h ∈ OF [X]. Applying σ, and then evaluating at y, we have x − σ(x) = σ(f)(y)σ(h)(y), which shows b = ±σ(f)(y) divides a = σ(x) − x.

Deﬁne a real valued function φ = φL/K by

Z r 1 φ(r) = dt. 0 (G0 : Gt) The function φ is piecewise linear, and for m ≤ r ≤ m + 1,

1 (17.9) φ(r) = (|G1| + ··· + |Gm| + (r − m)|Gm+1|). |G0|

In particular, for any integer m ≥ 0,

m 1 X φ(m) = −1 + |G |. |G | i 0 i=0

A useful formula for φ is given by the following lemma.

1 P Lemma 17.10. φ(r) = −1 + min{iG(σ), r + 1} for any r ∈ ≥0. |G0| σ∈G R 17. RAMIFICATION GROUPS 72

Proof. Let m be the nonnegative integer satisfying m ≤ r < m + 1. Note that if σ 6∈ G0, then iG(σ) = 0, so X X min{iG(σ), r + 1} = min{iG(σ), r + 1}

σ∈G σ∈G0 m X X X = (i + 1) + (r + 1)

i=0 σ∈GirGi+1 σ∈Gm+1 m X = |G0| + (i|Gi| − i|Gi+1|) + r|Gm+1| i=1 = |G0| + |G1| + ··· + |Gm| + (r − m)|Gm+1|

= |G0|(φ(r) + 1), with the last equality by (17.9). The reason for introducing the function φ is that it describes how the ramification filtration behaves with respect to quotients. Proposition 17.11 (Herbrand’s Theorem). Let H be a normal subgroup of G with fixed field F .

Then for any r ∈ R≥0, we have GrH/H = (G/H)φL/F (r).

Proof. For any σ ∈ G, let σ = σH. Deﬁne a function j : G/H → Z∪{∞} by j(σ) = max{iG(στ) | τ ∈ H}. Then σ ∈ GrH/H if and only if j(σ) ≥ r + 1. On the other hand, σ ∈ (G/H)s if and only if iG/H (σ) ≥ s + 1. So it suﬃces to show

iG/H (σ) − 1 = φL/F (j(σ) − 1). Both sides are inﬁnite if σ = 1, so we can assume σ 6= 1. Take a representative σ ∈ G for σ such that iG(σ) = j(σ). Putting m = iG(σ), we have σ ∈ Gm−1 and στ∈ / Gm for any τ ∈ H. If τ ∈ Hm−1 = H ∩ Gm−1, then στ ∈ Gm−1 and iG(στ) = m. On the other hand, if τ ∈ H but τ∈ / Hm−1, then στ∈ / Gm−1 and iG(στ) = iG(τ). In either case, we have iG(στ) = min{iG(τ), m}. By Proposition 17.8, we have 1 X i (σ) = min{i (τ), m} G/H e G L/F τ∈H

But iG(τ) = iH (τ), and eL/F = |H0|. So we can apply Lemma 17.10 to the group H, and we obtain

iG/H (σ) = 1 + φL/F (m − 1) = 1 + φL/F (j(σ) − 1). Since the function φ is piecewise linear and strictly increasing, it has an inverse that we denote by ψ = ψL/K . s Definition 17.12. The ramification groups in the upper numbering are G = Gψ(s), for any φ(r) s ∈ R≥0. Equivalently, G = Gr for any r ∈ R≥0. 0 s It follows from the definition that G = G0, and since φ is strictly increasing, G = {1} for s sufficiently large. One can check that Z s ψ(s) = (G0 : Gt)dt. 0 The upper number filtration allows a version of Herbrand’s Theorem that is cleaner to state (Proposi- tion 17.14 below). Lemma 17.13. Let L/K be a finite Galois extension, let H be a normal subgroup of G, and let F be the field field of H. Then

φL/K = φF/K ◦ φL/F and ψL/K = ψL/F ◦ ψF/K . 17. RAMIFICATION GROUPS 73

Proof. The result for ψ follows from the result for φ. Both φL/K and φF/K ◦ φL/F are piecewise linear functions that are diﬀerentiable at non-integers r ∈ R≥0 and that take 0 to 0. It thus suﬃces to show that the derivatives of φL/K and φF/K ◦ φL/F are the same at any non-integer r ∈ R≥0. Setting s = φL/F (r), the derivative of φF/K ◦ φL/F at r is

0 0 |(G/H)s| |Hr| |(G/H)s| |Hr| φF/K (s)φL/F (r) = = . |(G/H)0| |H0| eF/K eL/F

By Proposition 17.11, |(G/H)s||Hr| = |Gr|. Then since eL/F eF/K = eL/K = |G0|,

0 0 |Gr| 0 φF/K (s)φL/F (r) = = φL/K (r). |G0| Proposition 17.14. Let H be a normal subgroup of G. Then (G/H)s = GsH/H for any s ≥ 0. s Proof. Letting F be the fixed field of H, we have (G/H) = (G/H)t with t = ψF/K (s) by definition. By Proposition 17.11,(G/H)t = GrH/H with r = ψL/F (t). By Lemma 17.13, r = ψL/K (s), s so Gr = G .

Definition 17.15. Given an group Γ and a ﬁltration (Γj)i∈I of Γ by subgroups Γi indexed by some totally ordered set I, a jump in the ﬁltration is an index i ∈ I such that Γi 6= Γj for any j > i in I.

Note that by definition, the jumps in the filtration (Gr)r≥0 are all integers. This is not true for the s filtration (G )s≥0 (see [Ser79, Chapter IV, §4, Exercise 2]), but the Theorem of Hasse–Arf states that it is true when G is abelian. s Theorem 17.16 (Hasse–Arf). If G is abelian, then the jumps in the filtration (G )s≥0 are integers. We refer the reader to [Ser79, Chapter V, §7] or [Yos08, §6] for a proof. The most involved part of the proof is to first show it when G is cyclic. The most efficient way of doing this is via a lemma of Sen as in [Yos08, §6]. One then proceeds by induction, showing that if Gn 6= Gn+1, then we can find a subgroup H of G such that G/H is cyclic and such that GnH/H 6= Gn+1H/H. This is where the abelian hypothesis is used. In general, there is no subgroup H satisfying both these properties, but if G is abelian, we can split it into a product of cyclic groups and it is then not hard to see that such an H exists. The following corollary will be important in our proof the the Local Kronecker–Weber Theorem in the next section.

Corollary 17.17. Assume that L/K is totally ramiﬁed and abelian. If n ∈ Z≥1 is such that Gn = {1}, then [L : K] divides (q − 1)qn−1. n Proof. Let m ∈ Z≥0 be such that m − 1 < ψ(n) ≤ m. Then Gm = G = {1}, and we consider the usual ﬁltration G = G0 ⊇ G1 ⊇ · · · ⊇ Gm−1 ⊇ Gm = {1}.

By Proposition 17.5, G0/G1 has order dividing q − 1, and Gi/Gi+1 has order dividing q for each 0 < i ≤ m − 1. Note that if 0 < i ≤ m − 1, we have 0 < φ(i) ≤ φ(m − 1) < n. By Theorem 17.16, if i satisfies Gi 6= Gi+1, then φ(i) is an integer, so there are at most n − 1 indicies 0 < i ≤ m − 1 such that n−1 Gi 6= Gi+1. It follows that the order of G divides (q − 1)q . We finish this section by revisiting the example Example 17.4, this time using the upper numbering filtration. n Example 17.18. Consider the extension Qp(ζ)/Qp with ζ a primitive p th root of unity as in 2 n−1 Example 17.4. Then we saw that the jumps in the filtration (Gr)r≥0 occur at {0, p−1, p −1, . . . , p −1}. For any integer 1 ≤ m ≤ n − 1, we compute pm−1 m 1 X 1 X φ(pm − 1) = |G | = (pj − pj−1)pn−j = m. |G | i (p − 1)pn−1 0 i=1 j=1 18. THE LOCAL KRONECKER–WEBER THEOREM 74

We thus have  ( /pn)× if s = 0  Z s ∼ m n G = (1 + p Z)/p if m − 1 < s ≤ m and 1 ≤ m ≤ n − 1, {1} if s > n − 1.

18. The Local Kronecker–Weber Theorem Our goal in this section will be to prove the Local Kronecker–Weber Theorem. This is the statement that the maximal abelian extension of a nonarchimedean local field is the Lubin–Tate extension we constructed previously. Throughout this section, K will denote a nonarchimedean local field. We let O, m, and k denote the valuation ring, valuation ideal, and residue field, respectively, of K. We let U 0 = O×, and U i = 1 + mi for any i ≥ 1. We fix a separable closure Ks of K. We let Kur be the maximal unramified extension of K in Ks, let KLT be the maximal Lubin–Tate extension (Notation 14.4) of K in Ks, and let Kab be the maximal abelian extension of K in Ks. We have inclusions Kur ⊂ KLT ⊆ Kab. Our goal is to prove KLT = Kab. The two main inputs in the proof of the Kronecker–Weber Theorem are the Hasse–Arf Theorem via Corollary 17.17, and the following computation of the ramification groups for ramified Lubin–Tate extensions that generalizes Example 17.18. Proposition 18.1. Let L/K be a finite unramified extension. Let $ be a uniformizer for L, let n n−1 L$/L be the degree (q − 1)q Lubin–Tate extension of L associated to $, for some n ≥ 1, and let n n ∼ n × 0 n G = Gal(L$/L). Then the isomorphism ρ$ : G −→ (O/m ) = U /U induces isomorphisms  U 0/U n if s = 0  Gs =∼ U m/U n if m − 1 < s ≤ m and 1 ≤ m ≤ n − 1, {1} if s > n − 1.

n n Proof. Let v denote the normalized valuation on L$. Note that L$/L is totally ramified, so 0 G0 = G = G. Let iG : G → Z ∪ {∞} be the function iG(σ) = i + 1 if σ ∈ Gi r Gi+1 and ıG(1) = ∞. q n Let f = $X + X , and let α ∈ µf,n r µf,n−1, with µf,n the Lubin–Tate modules of §13. Since L$/L n is totally ramified and α is a uniformizer for L , we have O n = O[α] and i (σ) = v(σ(α) − α). $ L$ G n n n Recall that ρ$ is given by ρ$(σ) = u mod m if σ(α) = u ·f α. Fix σ 6= 1 in G, and take u ∈ O n n m m+1 such that ρ$(σ) = u mod m . Note there is 0 ≤ m < n such that u − 1 ∈ m r m . Then letting n−m m β = (u − 1) ·f α, we have β ∈ µf,n−m r µf,n−m+1. So β is a uniformizer for L$ and v(β) = q since n n−m m L$/L$ is totally ramified of degree q . We then have

σ(α) = u ·f α = α +f β ≡ α + β mod αβ. and m iG(σ) = v(σ(α) − α) = v(β) = q . n We deduce that ρ$ induces isomorphisms  U 0/U n if i = 0,  ∼ m n m−1 m Gi = U /U if q ≤ i ≤ q − 1 and 1 ≤ m ≤ n − 1, {1} if i ≥ qn−1. For any integer 1 ≤ m ≤ n − 1, we compute qm−1 m 1 X 1 X φ(qm − 1) = |G | = (qj − qj−1)qn−j = m, |G| i (q − 1)qn−1 i=1 j=1 from with the proposition follows. 18. THE LOCAL KRONECKER–WEBER THEOREM 75

Lemma 18.2. Let L and F be two ﬁnite Galois extension of K inside our ﬁxed separable closure. If Gal(L/K)s = {1} and Gal(F/K)s = {1} for some s ≥ 0, then Gal(LF/K)s = {1}.

Proof. Let G = Gal(LF/K), let H1 = Gal(LF/F ) and H2 = Gal(LF/L). Then by Proposi- tion 17.14, s s s s G H1/H1 = Gal(F/K) = {1} and G H2/H2 = Gal(L/K) = {1}, s s s so G ⊆ H1 and G ⊆ H2. Then G ⊆ H1 ∩ H2 = {1}. Theorem 18.3 (Local Kronecker–Weber Theorem). Kab = KLT. ab ab σ Proof. Choose some σ ∈ Gal(K /K) such that σ|Kur = FrobK , and let Lσ = (K ) be the ab ur fixed field of σ. Then Lσ/K is Galois, since K /K is abelian. Further, Lσ ∩ K = K, so Lσ/K is ur ab ab totally ramified. We claim that LσK = K . To see this, let F/Lσ be any finite extension in K ; we ur ab want to show F ⊆ LσK . Let d = [F : Lσ]. Since Gal(K /Lσ) is procyclic with topological generator σ, the quotient Gal(F/Lσ) is a cyclic group of order d generated by σ|F , and F/Lσ is the unique degree ab ur d subextension of K /Lσ. Let Kd be the unique degree d extension of K in K and consider the restriction homomorphisms ab Gal(K /Lσ) → Gal(KdLσ/Lσ) → Gal(Kd/K).

The first map is surjective, and the second is an isomorphism since Kd ∩ Lσ = K. Since Gal(Kd/K) is ur cyclic of order d, so is Gal(KdLσ/Lσ). This implies F = LσKd ⊂ LσK . Since σ|Kur = FrobK , there is a uniformizer $ of K such that ArtK ($) = σ|KLT . Let K$ be the s LT ur infinite totally ramified Lubin–Tate extension of K in K associated to $. Then K = K$K , and ab ur since we saw above that K = LσK , it only remains to show that Lσ = K$. By the definition of the Artin map, ArtK ($) = σ|KLT is the identity on K$, so K$ ⊆ Lσ. Let L be an arbitrary finite Galois extension of K contained in Lσ. To finish the proof, it suffices to show L ⊆ K$. Take n ≥ 1 n n n−1 sufficiently large so that Gal(L/K) = {1}. Let K$ be the degree (q − 1)q Lubin–Tate extension LT n n n of K in K . By Proposition 18.1 and Lemma 18.2, Gal(LK$/K) = {1}. Since the composite LK$ is contained in Lσ, it is totally ramified over K. Then Corollary 17.17 implies that n n−1 n [LK$ : K] ≤ (q − 1)q = [K$ : K]. n Hence, L ⊆ K$ ⊂ K$, which finishes the proof. Part 3

Further topics? APPENDIX A

Inverse Limits

Recall that a directed set is a partially ordered set I such that for any i, j ∈ I there is some k ∈ I with i ≤ k and j ≤ k. The most important examples for us will be the following.

Example A.1. The set of positive integers Z≥1 with its natural ordering is a directed set.

Example A.2. The set of positive integers Z≥1 with partial order given by divisibility is a directed set. Example A.3. Let G be any group. The set of finite index normal subgroups of G with order given by reverse containment (i.e. N1 ≤ N2 if N1 ⊇ N2) is a directed set. Example A.4. Let F be a field with separable closure F s. The set of all finite Galois extensions of F in F s ordered by inclusion is a directed set. We define inverse (or projective) systems and limits for arbitrary categories. However, in these notes we will only need them for the category of groups, the category of topological groups, the category of rings, and the category of topological rings. Definition A.5. Let C be a category. A inverse system (or projective system) in C consists of: (1) a directed set I, (2) an object Gi of C for every i ∈ I, (3) a morphism ϕij : Gj → Gi in C for every i ≤ j in I, such that the following hold:

(a) ϕii is the identity for all i ∈ I, (b) ϕij ◦ ϕjk = ϕik for every i ≤ j ≤ k in I.

The maps (ϕij)i≤j∈I are often omitted from the notation if they are understood. We ﬁrst dispense with two trivial examples.

Example A.6. Let G be an object in a category C, and let I be any directed set. Setting Gi = G for all i ∈ I, and ϕij = idG for all i ≤ j ∈ I, we obtain an inverse system ((Gi)i∈I , (ϕij)i≤j∈I ).

Example A.7. Let I be any directed set, and for each i ∈ I, let Gi be a group. Setting ϕij : Gj → Gi to be the trivial map ϕij(g) = 1 for all i ≤ j in I, we obtain obtain an inverse system ((Gi)i∈I , (ϕij)i≤j∈I ). If each Gi is a topological group, then this is an inverse system of topological groups. Now some less trivial examples. j i Example A.8. Fix a prime p. For each pair of integers i, j with j ≥ i ≥ 1, let ϕij : Z/p → Z/p i be the canonical projection. Then ((Z/p )i≥1, (ϕij)j≥i≥1) is an inverse system of rings.

Example A.9. The set of positive integers Z>0 with the order deﬁned by divisibility is a directed set. For each m | n in Z>0, we let ϕmn : Z/n → Z/m be the canonical projection. Then ((Z/n)n≥1, (ϕmn)m|n) is an inverse system of rings. Example A.10. Let R be any commutative ring, I be any directed set, and for each i ∈ I, let ai be an ideal in R such that aj ⊆ ai for any i ≤ j in I. Then letting ϕij : R/aj → R/ai be the canonical projection for any i ≤ j in I, we obtain an inverse system ((R/ai)i∈I , (ϕij)i≤j∈I ) of rings. A n particularly common example of this is when I = Z>0, and an = a for each n ≥ 1.

77 A. INVERSE LIMITS 78

Example A.11. Let G be any group and let I be the directed set of all finite index normal subgroups of G, ordered by reverse inclusion. For each H,N ∈ I with H ⊇ N, we let ϕHN : G/N → G/H be the canonical projection. Then ((G/N)N∈I , (ϕHN )N⊆H ) is an inverse system of groups. Example A.12. Let F be a field and let F s be a separable closure of F . Let I be the directed set s of all finite Galois extensions of F inside F . For each K,L ∈ I with K ⊆ L, let ϕKL : : Gal(L/F ) → Gal(K/F ) be the canonical projection. Then (Gal(K/F )K∈I , (ϕKL)K⊆L) is an inverse system of groups. Definition A.13. Let C be a category. An inverse limit (or projective limit, or just limit) of an inverse system ((Gi)i∈I , (ϕij)i≤j∈I ) in C is a pair (G, (ϕi)i∈I ), where (a) G is an object in C, and (b) for each i ∈ I, ϕi : G → Gi is morphism in C such that ϕi = ϕij ◦ ϕj for any i ≤ j in I, such that (G, (ϕi)i∈I ) is the universal object satisfying parts (a) and (b). More specifically, if (G, (ϕi)i∈I ) is an inverse limit, then for any other pair (H, (ψi)i∈I ) satisfying parts (a) and (b), there exists a unique morphism ψ : H → G in C such that ψi = ϕi ◦ ψ for each i ∈ I. We often omit the (ϕ ) from the notation if they are understood, and write lim G for the i i∈I ←− i object G is the definition of an inverse limit. An often useful observation is that inverse limits only depend on “sufficiently large” elements in the index set I. More precisely, let ((Gi)i∈I , (ϕik)i≤k∈I ) be an inverse system, and let J ⊆ I be a sub-directed set with the property that for every i ∈ I there is some j ∈ J with i ≤ j. If (G, (ϕi)i∈I ) is an inverse limit for ((Gi)i∈I , (ϕik)i≤k∈I ), then (G, (ϕj)j∈J ) is an inverse limit for ((Gj)j∈J , (ϕjm)j≤m∈J ). Conversely, say (G, (ϕj)j∈J ) is an inverse limit for ((Gj)j∈J , (ϕjm)j≤m∈J ). Then for each i ∈ I, choose some j ∈ J with i ≤ j, and let ϕi : G → Gi be defined by ϕi = ϕijϕj. Then ϕi is independent of the choice of j, and (G, (ϕi)i∈I ) is an inverse limit of ((Gi)i∈I , (ϕik)i≤k∈I ). We leave it to the reader to check the details. As is usual with things defined via a universal property, we need to prove existence, and if it exists, then it is unique up to unique isomorphism. As mentioned above, in these notes we are only concerned with the categories of groups, topological groups, rings, and topological rings. In these cases we can prove existence. Proposition A.14. The inverse limit of an inverse system of groups (resp. of rings) exists. More specifically, let ((Gi)i∈I , (ϕij)i≤j∈I ) be an inverse system of groups (resp. of rings). Then Y (A.15) G = {(gi) ∈ Gi | gi = ϕij(gj) for all i ≤ j ∈ I} i∈I Q together with the maps ϕi : G → Gi induced from the canonical projection j∈I Gj → Gi, is an inverse limit. If ((Gi)i∈I , (ϕij)i≤j∈I ) is an inverse system of topological groups (resp. of topological rings), then giving G the coarsest topology that makes each ϕi continuous makes G an inverse limit of topological Q groups (resp. of topological rings). This topology coincides with the subspace topology of Gi, where Q Q i∈I i∈I Gi is given the product topology, and it is closed in i∈I Gi.

Proof. We first consider the nontopological case. Since the ϕij are group (resp. ring) homomor- Q phisms, G is a subgroup (resp. subring) of i∈I Gi, and the maps ϕi : G → Gi are group (resp. ring) homomorphisms. If (H, (ψi)i∈I ) is a group (resp. ring) H with group (resp. ring) homomorphisms Q ψi : H → Gi for all i ∈ I, then there is a unique group (resp. ring) homomorphism ψ : H → Gi Q i∈I such that ψi is composite of ψ with the natural projection j∈I Gj → Gi. If (H, (ψi)i∈I ) further satisfies ψi = ψj ◦ ϕij for each i ≤ j in I, then the image of ψ is contained in G. This shows (G, (ϕi)i∈I ) satisfies the universal property. We now turn to the topological case. Let H be a topological group (resp. ring) and let ψi : H → Gi be continuous homomorphisms such that ψi = ψj ◦ ϕij for each i ≤ j in I. Forgetting the topologies and using what we’ve already proved, there is a unique group (resp. ring) homomorphism ψ : H → G such that ψi = ϕi ◦ ψ for all i ∈ I. Since each ψi is continuous, and we have given G the coarsest A. INVERSE LIMITS 79 topology making each ϕi continuous, ψ is also continuous. This shows (G, (ϕi)i∈I ) equipped with this topology is an inverse limit in the category of topological groups (resp. topological rings). By definition of the product topology, the coarsest topology on G that makes each ϕi continuous is Q Q the topology induced on G as a subspace of i∈I Gi. Finally, we check that G is closed in i∈I Gi. For any j ≤ k ∈ I, the subset Xjk ⊆ Gj ×Gk given by X = {(ϕjk(g), g) | g ∈ Gk} is closed in Gj ×Gk, since Q ϕjk is continuous. Then the inverse image Yjk of Xjk under the canonical map i∈I Gi → Gj × Gk is also closed. Thus so is G = ∩j≤k∈I Yjk. Example A.16. In the case of Example A.6, Y lim G = {(g ) ∈ G | g = g for all i ≤ j ∈ I} ∼ G. ←− i i i∈I i j = i∈I Example A.17. In the case of Example A.17, Y lim G = {(g ) ∈ G | g = 1 for all i ∈ I} ∼ {1}. ←− i i i∈I i i = i∈I Example A.18. In the case of Example A.10, if a is an ideal in a commutative ring R, then Y lim R/an = {(r ) ∈ R/an | a = a mod an for all n ≥ 1} ←− n n≥1 n n+1 n≥1 is called the a-adic completion of R. Note that in each of Examples A.8, A.9, A.11 and A.12, the groups and rings in the inverse system are all finite. So their inverse limits inherit a canonical topology, by giving each of the objects in the inverse system the discrete topology and viewing the invese limit in the category of topological groups (resp. rings). Definition A.19. Let G = lim G be an inverse limit of finite groups (resp. rings). The profinite ←−i∈I i topology (also known as the Krull topology) on G is the coarsest topology on G making all the maps G → Gi continuous, where each Gi is given the discrete topology. This make G the inverse limit of (Gi)i∈I in the category of topological groups (resp. topological rings). A profinite group (resp. profinite ring) is a topological group (resp. topological ring) that is isomorphic as topological groups (resp. topological rings) to an inverse limit lim G of finite groups ←−i∈I i (resp. rings), where each Gi is given the discrete topology. Example A.20. In the case of Example A.8, Y lim /pn = {(a ) ∈ /pn | a mod pn = a for all n ≥ 1} ←− Z n Z n+1 n n≥1 is a profinite ring. This is sometimes given as the definition of ring of p-adic integers Zp. We showed in Proposition 5.6 that this is canonically isomorphic to the definition given in terms of the completion of Z with respect to the p-adic absolute value. Example A.21. In the case of Example A.9, Y lim /n = {(a ) ∈ /n | a mod m = a for all m | n}, ←− Z n Z n m n≥1 is a profinite ring denoted by Zb. Note that there is a canonical map Z → Zb, which is injective with dense image. There is also a canonical isomorphism ∼ Y Zb = Zp p a prime of topological rings, which should be thought of as a “pro–Chinese Remainder Theorem.” By the Chinese ∼ e1 ek Remainder Theorem, for any n ≥ 2 we have have a canonical isomorphism Z/n = Z/p × · · · Z/pk , e1 ek where n = p1 ··· pk and the pi are distinct primes. This decomposition is compatible with divisibility, A. INVERSE LIMITS 80

∼ Qk ei so we have a canonical isomorphism of inverse systems ( /n)n≥2 = ( /p ) e1 ek . We Z i=1 Z n=p1 ···pk Q Qk ei have canonical maps ϕ e1 ek : p → /p given by the product of the canonical maps p1 ···pk p Z i=1 Z i ei Zpi → Z/pi for each 1 ≤ i ≤ k, and the trivial map for p∈ / {p1, . . . , pk}. It suffices now to show Q Qk ei p with the maps ϕ e1 ek is the inverse limit of the system ( /p ) e1 ek . Let H be a p Z p1 ···pk i=1 Z p1 ···pk Qn ei topological ring, and let ψ e1 ek : H → /p be continuous ring homomorphisms for each p1 ···pk i=1 Z i e1 ek p1 ··· pk ≥ 2, compatible with divisibility. In particular, for each prime p, we have continuous ring e homomorphisms ψpe : H → Z/p , compatible with varying e. By the universal property of Zp, we get a unique continuous ring map Ψp : H → Zp such that ψpe = Ψp ◦ ϕpe . We have this for each prime p, so Q we get a unique continuous ring homomorphism Ψ: H → p Zp such that Ψp is the composite of Ψ Q with the projection onto Zp. Then Ψ is the unique continuous ring homomorphism H → p Zp such that ψ e1 ek = ϕ e1 ek ◦ Ψ. p1 ···pk p1 ···pk Example A.22. Let G be any group and let I be the directed set consisting of all finite index normal subgroups of G, ordered by reverse inclusion. Then Y lim G/N = {(g ) ∈ G/N | g = g mod H for any finite index normal subgroups N ⊆ H}. ←− N H N N∈I N∈I

This inverse limit is called the profinite completion of G, and is denoted Gb. As with the case of Zb, there is a canonical map G → Gb with dense image. It may not be injective, however. For example, if G is an infinite simple group then Gb = {1}. s Example A.23. Let F be a field with separable closure F , and let (Gal(K/F ))K/F be the inverse system of Example A.12, indexed by all finite Galois extensions K/F in F s. Then Gal(F s/F ) ∼ lim Gal(K/F ). = ←− K/F finite Galois canonically. In particular Gal(F s/F ) is canonically a profinite group. s To show this isomorphism, first note any σ ∈ Gal(F /F ) restricts to an element σ|K ∈ Gal(K/F ) for every finite Galois extension K/F in F s, compatibly with subextensions. So we have a canonical map Gal(F s/F ) → lim Gal(K/F ), which is clearly a group homomorphism. It is injective since σ ←−K/F is identity on F s if and only if it is identity on every element of F s if and only if it is identity on every finite Galois extension in F s. To see that it is surjective, let (σ ) ∈ lim Gal(K/F ), and define K ←−K/F s s s σ : F → F by σ(a) = σ|K (a) if K/F is any finite Galois extension in F containing a. This is well s defined by the compatibility of (σK ), and an automorphism fixing F , so σ ∈ Gal(F /F ) and satisfies σ|K = σK . Proposition A.24. A profinite group is Hausdorff, compact, and totally disconnected. Proof. Let G = lim G with each G finite. If (g ) and (h ) are distinct elements in G, then ←−i∈I i i i i there is some index j ∈ I such that gj =6 hj. The inverse image in G of the open sets {gi} and {hi} in Gi under the map G → Gi from the definition of G are disjoint opens sets containing (gi) and (hi), respectively. So G is Hausdorff. The fact that G is compact follows from Tychonoff’s Theorem and the last claim in Proposition A.14. To prove the total disconnected statement, first note that any finite index open subgroup N of a topological group H is closed, since if H/N = {h1N, . . . , hrN}, then each hiN is open and

N = H r ∪hiN6=N hiN. For each i ∈ I, let Ni = ker(G → Gi). Then Ni is a finite index open, hence also closed, normal subgroup. Further, {gNi}g∈G,i∈I is a basis for the topology on G. So G has a basis of clopen sets, which implies it is totally disconnected. What may be surprising at first, is that the converse is also true: Theorem A.25. A topological group is profinite if and only if it is Hausdorff, compact, and totally disconnected. A. INVERSE LIMITS 81

The reader is encouraged to try to prove Theorem A.25 for themselves. Otherwise, see [NSW00, Chapter I, §1] for a proof. Q∞ Remark. A finite index subgroup of a profinite group need not be open. For example, n=1 Z/2 with the product topology is profinite, as ∞ k Y Y /2 ∼ lim /2 Z = ←− Z n=1 k≥1 n=1 Qm Qk where for m ≥ k, the map n=1 Z/2 → n=1 Z/2 is the projection onto the first k factors. The set of Q∞ open index two subgroups in n=1 Z/2 is countable, since they are given by choosing some k ≥ 1 and Q∞ Qk taking the inverse image of an index two subgroup under the canonical map n=1 Z/2 → n=1 Z/2. On the other hand, the set of all index two subgroups in uncountable, as they are in bijection with Q∞ the set surjections λ: n=1 Z/2 → Z/2. The set of these surjections is uncountable because it is in bijection with sequences of 0s and 1s with at least one 1, via λ 7→ (λ(en))n≥1, where en is the nth basis Q∞ element of n=1 Z/2. APPENDIX B

Integral extensions, norm and trace

All rings in this appendix are assumed to be commutative.

Integral extensions We review the basics of integral ring extensions. Definition B.1. Let A ⊆ B be a ring extension. An element b ∈ B is integral over A, if it satisﬁes a monic equation n n−1 b + an−1b + ··· + a0 = 0 with n ≥ 1 and ai ∈ A. We say that B is integral over A if every element of B is integral over A. √ √ 1+ 5 Example B.2. 2 ∈ Q( 5) is integral over Z. Let A ⊆ B be a ring extension. It is clear that any element of A is integral over A. Also easy to see is that if b ∈ B is integral over A, then so is −b, since if n n−1 b + an−1b + ··· + a0 = 0, then n n−1 n (−b) − an−1(−b) + ··· + (−1) a0 = 0. What’s less obvious is that being integral is closed under addition and multiplication. To show this, one uses the following proposition.

Proposition B.3. Let A ⊆ B be an extension of rings. If b1, . . . , bk ∈ B are integral over A, then the subring A[b1, . . . , bk] of B is a finitely generated A-module. Conversely, if C ⊆ B is a subring containing A that is finitely generated as an A-module, then C is integral over A. Sketch. We give a sketch. For details, the reader can see [Neu99, Chapter 1, §2] or [Mat89, §9] (or any other gradate level text on Algebra, Commutative Algebra, or Algebraic Number Theory). If b is integral over A, then there is an n ≥ 1 such that bn is an A-linear combination of 1, . . . , bn−1, and A[b] is generated as an A-module by 1, . . . , bn−1. This proves the first claim when k = 1, and one proceeds by induction for general k ≥ 1. For the second claim, if C ⊆ B is a subring containing A that is finitely generated as an A-module, and c ∈ C, we can view multiplication by c as an A-linear endomorphism of C. One then uses the assumption that C is a finitely generated A-module to apply the Cayley–Hamilton Theorem, which produces a monic equation with coefficients in A for which c is a root.

Corollary B.4. Let A ⊆ B be a ring extension. The set of all elements in B that are integral over A is a subring of B containing A.

Proof. It suﬃces to show that for any b1, b2 that are integral over A, the subring A[b1, b2] ⊆ B is integral over A. Applying both directions of Proposition B.3, we see that A[b1, b2] is a ﬁnitely generated A-module, hence is integral over A. Corollary B.5. Let A ⊆ B ⊆ C be two ring extensions. If B is integral over A and C is integral over B, then C is integral over A.

82 NORM AND TRACE 83

n n−1 Proof. Let c ∈ C. Then there is n ≥ 1 and b0, . . . , bn−1 ∈ B such that c +bn−1c +···+b0 = 0. Then c is integral over A[b0, . . . , bn−1]. By Proposition B.3, A[b0, . . . , bn−1, c] is a finite A[b0, . . . , bn−1]- module and A[b0, . . . , bn−1] is a finite A-module. It follows that A[b0, . . . , bn−1, c] is a finite A-module, and that c is itegral over A by again applying Proposition B.3. Definition B.6. Let A ⊆ B be a ring extension. The set of all elements in B that are integral over A is called the integral closure of A in B. If A equals its integral closure in B, we say A is integrally closed in B. If B is the fraction field of A, we call the integral closure of A in B the normalization of A, and if A equals its normalization, we say A is normal (or integrally closed).

Example B.7. Z is normal. Example B.8. If k is a field, k[T ] is normal. The ring k[T 2,T 3] is not normal, since it’s field of fractions is k(T ) and T is integral over but does not belong to k[T 2,T 3]. The normalization of k[T 2,T 3] is k[T ]. √ √ Example B.9. Let d ∈ be square free. The integral closure of in ( d) is [ d] if d 6≡ 1 mod 4. Z √ √ Z Q Z 1+ d If d ≡ 1 mod 4, then the integral closure of Z in Q( d) is Z[ 2 ]. It follows from Corollary B.5 that if A is a subring of a field L, the integral closure of A in L is normal. A particular situation where this arises frequently is if L is an extension of the field of fractions of A. √ Example B.10. For d ∈ Z square-free, Z[ d] is normal if and only if d 6≡ 1 mod 4. Example B.11. Any unique factorization domain is normal. Indeed, let A be a unique factorization domain, and let K denote its fraction field. Let x =∈ K be integral over A; so there are a0, . . . , an−1 ∈ A n n−1 c such that x + an−1x + ··· + a0 = 0. Write x = d , with c, d ∈ A having no common irreducible divisors. Then n n−1 n−2 n−1 c = d(−an−1c − an−2dc − · · · − a0d ). So d | cn in A. Since A is a UFD and c and d are assumed to have no common divisors, d must be a unit in A, and x ∈ A. In particular, the above example shows that any principal ideal domain, and hence any discrete valuation ring, is normal. A useful property of normal domains is that one can characterized integral elements in extensions of their fraction field via minimal polynomials.

Proposition B.12. Let A be a normal domain with ﬁeld of fractions K, let L/K be a ﬁeld extension, and let x ∈ L be algebraic over K. Then x is integral over A if and only if the minimal polynomial of x over K lies in A[X].

Proof. Let g ∈ K[X] be the minimal polynomial of x over K. If g ∈ A[X], then x is integral over A as it is a root of g. Assume that x is integral over A. Enlarging L if necessary, we can assume g splits over L. Then there is f ∈ A[X] of degree at least 1 for which x is a root. Then g | f in K[X]. In particular, every root of g is a root of f, hence integral over A. Since the coefficients of g lie in the subring of L generated over A by the roots of g, the coefficients of g are integral over A. Since A is normal, the coefficients of g lie in A.

It can be shown that if A is a normal domain, then its localization Ap at any prime ideal p is also a normal domain. One then deﬁnes a normal ring to be a (commutative) ring A such that its localization at any prime ideal is a normal domain. We will not need this in this course.

Norm and trace We now briefly review the norm and trace of a finite field extension. NORM AND TRACE 84

Definition B.13. Let L/K be a ﬁnite extension of ﬁelds. For x ∈ L, the trace and norm of x, denoted TrL/K (x) and NL/K (x), respectively, are the trace and determinant, respectively, of the K-linear endomorphism Tx : L → L given by Tx(y) = xy, i.e.

TrL/K (x) = TrL/K (Tx) and NL/K (x) = det(Tx).

Note that the endomorphism Tx in the definition satisfies Tx+y = Tx + Ty and Txy = Tx ◦ Ty. From this it follows that TrL/K is an additive map L → K, and NL/K is a multiplicative map L → K. Moreover × × TrL/K : L → K and NL/K : L → K are homomorphisms. It is also easy to see that for any x ∈ K, if d = [L : K], we have TrL/K (x) = dx d ∼ and NL/K (x) = x . It also follows from the definition that if σ : L −→ σ(L) is an isomoprhism of fields, then Trσ(L)/σ(K) = σ ◦ TrL/K and Nσ(L)/σ(K) = σ ◦ NL/K . To study norms and traces of elements, it is useful to establish a relationship between the minimal polynomial of x ∈ L over K, and the characteristic polynomial of the endomorphism Tx. Proposition B.14. Let L/K be a finite field extension, let x ∈ L, and let g be the minimal polynomial of x over K. Let Tx : L → L be the K-linear endomoprhism given by Tx(y) = xy, and m let f be the characteristic polynomial of Tx. Then f = g , with m = [L : K(x)]. In particular, if k k−1 k+m m g = X + ck−1X + ··· + c0 ∈ K[X], then TrL/K (x) = −mck−1 and NL/K (x) = (−1) c0 . k k−1 Proof. Let k = [K(x): K], and let g = X + ck−1X + ··· + c0 ∈ K[X] be the minimal i polynomial for x over K. Choosing any basis e1, . . . , em for L over K(x), the elements (x ej), with 0 ≤ i ≤ k − 1 and 1 ≤ j ≤ m, form a basis for L over K. For each 1 ≤ j ≤ m, the subspace with basis k−1 ej, xej, . . . , x ej is stable under multiplication by x, and the corresponding matrix is   0 0 0 · · · −c0 1 0 0 · · · −c1    0 1 0 · · · −c2    , ......  . . . . .  0 0 0 · · · −cn−1 m which has characteristic polynomial g. It follows that f = g . Corollary B.15. Let A be a normal domain with field of fractions K, let L/K be a finite extensions of fields, and let B be the integral closure of A in L. Then TrL/K (B) ⊆ A and NL/K (B) ⊆ A. Proof. This follows from Propositions B.12 and B.14. In the case that L/K is separable, a very useful characterization of the norm and trace is given by the following. Proposition B.16. Let L/K be a finite separable extension. Letting K be an algebraic closure of K, X Y TrL/K (x) = σ(x) and NL/K (x) = σ(x) σ σ where the sum and product run over all K-linear field embedding σ : L,→ K.

Proof. Let x ∈ L, let Tx be the K-linear endomorphism of L given by multiplication by x, and let f ∈ K[X] be its characteristic polynomial. We will show Y f = (X − σ(x)), σ from which the proposition follows. Letting k = [K(x): K], m = [L : K(x)], and letting g ∈ K[X] be the minimal polynomial of x over K, we have f = gm by Proposition B.14. On the other hand, since Q L/K is separable, we know that g = τ (X − τ(x)), with the product running over all K-linear field embeddings K(x) ,→ K. Also, the set of K-linear field embeddings L,→ K that restrict to a fixed NORM AND TRACE 85

τ : K(x) ,→ K has size m, and this gives a partition of the set of all K-linear ﬁeld embeddings of L in K. Hence, Y Y Y (X − σ(x)) = (X − σ(x))

σ : L,→K τ : K(x),→K σ|K(x)=τ Y = (X − τ(x))m τ : K(x),→K = gm.

Corollary B.17. If L/K is a ﬁnite Galois extension, then for any x ∈ L, X Y TrL/K (x) = σ(x) and NL/K (x) = σ(x). σ∈Gal(L/K) σ∈Gal(L/K)

Proof. If we choose K in Proposition B.16 to be an algebraic closure of L, then the set of K-linear field embeddings L,→ K equals Gal(L/K). √ √ Example B.18. Let d ∈ Z be square-free. Then for any a + b d ∈ Q( d), √ √ Tr √ (a + b d) = 2a and N √ (a + b d) = a2 + b2d. Q( d)/Q Q( d)/Q Proposition B.19. For finite extensions of fields L/K and M/L,

TrM/K = TrL/K ◦ TrM/L and NM/K = NL/K ◦ NM/L .

Proof when L/F is separable. Fix an algebraic closure K of K. The set of K-linear ﬁeld embeddings M,→ K is partitioned into k = [L : K] equivalence classes by the equivalence relation σ ∼ τ if σ|L = τ|L. Let σ1, . . . , σm be a set of representatives for the m equivalence classes. Then the set of K-linear ﬁeld embedding L,→ K is σ1|L, . . . , σm|L, and for each 1 ≤ i ≤ m, we have X σ(x) = Trσi(M)/σi(L)(σi(x)) = σi(TrM/L(x)). σ∼σi So, m m X X X TrM/K (x) = σ(x) = σi(TrM/L(x)) = TrL/K (TrM/L(x)).

i=1 σ∼σi i=1 The proof for the norm is identical, replacing sums with products. To prove Proposition B.19 in the general case, one needs an adjustment to the statement and proof of Proposition B.16. In general, one can show that for any ﬁnite extension of ﬁelds L/K, letting [L : K]i be the inseparability degree (so = 1 if L/K is separable, and otherwise is a positive power of the characteristic p > 0), then

X Y [L:K]i TrL/K (x) = [L : K]i σ(x) and NL/K (x) = σ(x) σ σ where the sum and product run over all K-linear ﬁeld embedding σ : L,→ K. In particular, if L/K is not separable, then TrL/K is identically zero. See [Gri07, Chapter 5, §7] for details.

Proposition B.20. Let L/K be a ﬁnite separable extension. Then TrL/K : L → K is nonzero. Further, the pairing

L × L → K given by (x, y) 7→ TrL/K (xy) is bilinear and perfect. NORM AND TRACE 86

Proof. By Proposition B.19, to prove TrL/K is nonzero, we can replace L with its normal closure P and assume that L/K is Galois. Bu Corollary B.17, TrL/K is the function x 7→ σ∈Gal(L/K) σ(x). Then linear independence of characters [Lan02, Chapter IV, §4] implies that the functions x 7→ σ(x), for σ ∈ Gal(L/K) are linearly independent over K, so TrL/K is not the zero function. Since TrL/K is linear, it follows that (x, y) 7→ TrL/K (xy) is bilinear. Fix nonzero x ∈ L, and let Trx : L → K be given by Trx(y) = TrL/K (xy). Since x 6= 0, we have xL = L, so Trx is nonzero by the ﬁrst part. Then x 7→ Trx is an injective map from L to its K-linear dual. Counting dimensions, it is an isomorphism. Bibliography

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