Nonarchimedean Local Fields. Patrick Allen

Transcription

1 Nonarchimedean Local Fields Patrick Allen

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3 Contents Introduction 4 Local fields 4 The connection with number theory 6 Global fields 8 This course 9 Part 1. Nonarchimedean fields Valuations Valued fields and the valuation subring Completions Hensel s Lemma Complete discretely valued fields Extensions of complete valued fields Ramification: first properties Units Norm groups 41 Part 2. Local class field theory The main theorems Formal group laws Lubin Tate formal groups laws Lubin Tate extensions The Artin map Coleman s norm operator Base change Ramification groups 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

4 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 K K K K K K K K (x, y) x + y x x (x, y) xy x x 1 are all continuous (K K is given the product topology, and K = K {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 xy is a homeomorphism with inverse y x 1 y. 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 v p : Z Z { } by v p (n) = k, for n 0, if p k n but p k+1 n, and setting v p (0) =. We can then extend v p to Q by setting v p ( a b ) = v p(a) v p (b). This is called the p-adic valuation on Q, and we define the p-adic absolute value p : Q R 0 by x p = p vp(x). 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 Q p. Example 0.4. Let F p be the finite field with p elements, and let F p ((T )) := { n a nt n a n F p } be the field of Laurent series over F p. Define : F p ((T )) R 0 by f = p n0 if f = n n 0 a n T n with a n0 0. One checks that is an absolute value on F p ((T )), and that the induced topology on F p ((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 Q p for some prime p. If K has characteristic p, then K is isomorphic to a finite extension of F p ((T )). 4

5 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, it is unique up to scalar: if µ is another choice of Haar measure, then there is α R >0 such that µ = 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 µ. By uniqueness, there 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 F p (T ) is contained in K, and since K is complete it contains the completion of F p (T ) with respect to α Fp(T ). One then proves that this is F p ((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 Q p 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 Q p and F p ((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 Q p and F p ((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 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 ( m k ) 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).

6 THE CONNECTION WITH NUMBER THEORY 6 Proof of Proposition 0.9. Without loss of generality, we can assume x y. Then for any n 1 and any 0 k n, x n k y k x n and ( n k) 1 by Lemma Thus, for any n 1, n ( ) x + y n = (x + y) n n x n k y k (n + 1) x n. k Taking nth roots, k=0 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 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 Let p be the p-adic absolute value on Q from Example 0.3. Fix any real number r > 1, and define p : Q R 0 by x p = r vp(x). Then p and p are equivalent. Indeed, log r log p p = p, so open balls of radius ε in the p-topology are equal to open balls of radius ε log r log p in the p -topology. On the other hand, if l is a prime distinct from p, then p and l are not equivalent. Indeed, p n p = p n and p n l = 1 for all n 1, so the sequence (p n ) n 1 converges to 0 with respect to p but not with respect to l. 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 = p k m 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).

7 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, Q p does know that p is prime, but forgets all other primes l p as well as the order on Q. One might hope that using all the information of R, Q 2, Q 3, Q 5,..., might give us some information about arithmetic questions over Q. A concrete example of this is the following. Proposition x Q is a square in Q if and only if it is a square in R and in Q p for all primes p. Proof. The only if part is obvious. To see the if direction, we may assume x 0, and we can write x = ɛ p vp(x), p prime with ɛ { 1, 1}. Then x is a square if and only if ɛ = 1 and v p (x) is even for every prime p. If x is a square in R, then x > 0 and ɛ = 1. It remains to show that if x is a square in Q p, then v p (x) is even. The absolute value p on Q extends to Q p by continuity. But since Q p = p Z is a discrete subgroup of R >0, we also have Q p p = p Z. Hence, if x is a square in Q p, p vp(x) = x p p 2Z, and v p (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 solution in Q n if and only if Q(x) = 0 has a nonzero solution in R n and in Q n p 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 (x 2 2)(x 2 17)(x 2 34) = 0 has solutions in R and in Q p 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 Châtelet group 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 Châtelet group consisting of the Q-isomorphism classes of curves C that have an R-point and a Q p -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.

8 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 Q p 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 Q p of Q p, and Gal(Q p /Q) is similarly a topological group. One can show that the natural embedding Q Q p extends to an embedding Q Q p, and that the image is dense. We ll see later that Q p has a natural topology extending that of Q p and that the action of Gal(Q p /Q p ) on Q p is continuous. Thus, an element of Gal(Q p /Q p ) is uniquely determined by its restriction to Q under our fixed embedding Q Q p, and we get an embedding of groups Gal(Q p /Q p ) Gal(Q/Q) that one can show has closed image. This image does depend on the choice of embedding Q Q p, but its conjugacy class does not. The following is a consequence of the Chebotarev density theorem. Theorem The union over all primes of the conjugacy class of Gal(Q p /Q p ) is dense in Gal(Q/Q). For a more precise theorem and a proof, see [Neu99, Chapter VII, 13]. The local Galois groups Gal(Q p /Q p ) 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(Q p /Q p ) 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(Q p /Q p ) 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 Q p, 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 A global field is a field F that is either a finite extension of Q, or a finite extension of F p (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 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 F p [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, Q 2, Q 3,... 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

9 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. Part 1 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]. Part 2 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 Part 3. While I have some ideas of what will go here, I will wait to see how quickly we are moving through Parts 1 and 2 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.

10 Part 1 Nonarchimedean fields

11 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. by Example 1.2. Let p be a prime ideal in a commutative ring R. Define a function v : R R { } v(x) = { if x p 0 if x / p. Let s check that v is a valuation on R. Part 1 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 part 2 of Definition 1.1, and v(x + y) = 0 x / p and y / p v(x) = v(y) = 0, which shows part 3. 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 0. This valuation is called the trivial valuation. Example 1.3. Fix a prime number p. For nonzero x Z, we let v p (x) be the nonnegative integer uniquely defined by x = p vp(x) y with p y. Setting v p (0) =, we get a valuation v p : Z R { }, called the p-adic valuation. Let s check the details. Part 1 of Definition 1.1 follows immediately from the definition of v p, as do parts 2 and 3 if either x or y are 0, so we assume x, y Z are nonzero. Writing x = p vp(x) m and y = p vp(y) n with m, n coprime to p, we have xy = p vp(x)+vp(y) mn and p mn, so v p (xy) = v p (x) + v p (y). Lastly, p min{vp(x),vp(y)} divides x + y = p vp(x) m + p vp(y) n, so v p (x + y) min{v p (x), v p (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 v f (g) be the exponent of f in this decomposition. Setting v f (0) =, we obtain a valuation v f 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 x n = 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.

12 1. VALUATIONS 12 Lemma 1.5. Let R be an integral domain and let v be a valuation on R such that v(x) for x 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 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 on R { }, since we have assumed v(x) if x 0. Secondly, if a b = c 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 w b d) + c = v(ad + bc) v(bd) min{v(a) + v(d), v(b) + v(c)} v(b) v(d) { ( a ) ( c = 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, v p (x) is the integer uniquely defined by x = p vp(x) a 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 If p and l be distinct primes, the p-adic and l-adic valuations are not equivalent since v p (l) = 0 = v p (1), but v l (l) = 1 > 0 = v l (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 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.

13 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) 0, and replacing x by x 1 if necessary, we can assume v(x) > 0. Then we also have w(x) > 0, and r = v(x) 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) 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(y k ) = kv(y) nα = v(x n ), and a contradiction. w(y k ) = kw(y) > kv(y) + α > nα = w(x n ), 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 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 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 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 Let v be a valuation on a commutative ring R. Show that w : R[T ] R { } defined by w(a a n T n ) = min{v(a 0 ),..., v(a n )} is a valuation on R[T ] extending v 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 ] Construct two nonequivalent valuations v, w on Z[T ] satisfying v(t ) = w(t ) = 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.

14 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 let k(t ) be an algebraic closure of k(t ). Choose a compatible system of nth roots (T 1 n ) n 1 of T, i.e. (T 1 m ) m n = T 1 n whenever n m. For each n 1, k(t 1 n ) is the fraction field of the polynomial ring k[t 1 n ], so we have the T 1 n -adic valuation v T 1 n on k(t 1 n ). Let v n be the valuation on k(t 1 n ), equivalent to v T 1 n, given by v n = 1 n v T 1 n. Note that for any n m we have k(t 1 n ) k(t 1 m ), and we claim v m k(t 1 n ) = v n. First, we have v m (T 1 m n ) = n v m(t 1 m 1 m ) = n m = v n(t 1 n ). Now let f k[t 1 n ] be a nonzero element not divisible by T 1 n. If f were divisible by T 1 m in k[t 1 m ], then f would lie in the ideal T 1 m k[t 1 m ] k[t 1 n ] of k[t 1 n ], a proper ideal containing T 1 n. But T 1 n generates a maximal ideal in k[t 1 n ], so T m k[t m ] k[t n ] = T n k[t n ], contradicting our assumption that f is not divisible by T 1 n. Thus v m (f) = 1 = v n (f). It follows that v m agrees with v n on k[t 1 n ], hence also on k(t 1 n ). We now set K = n 1 k(t 1 n ) and we obtain a valuation v on K given by v k(t 1 n ) = v n. This valuation is not discrete since v(k ) {0} {v(t 1 n ) n 1} = {0} { 1 n n 1}. Proposition 2.4. Let v be a valuation on a field K. The subset O v = {x K v(x) 0} is a subring of K. Moreover O v is local with maximal ideal m v = {x K v(x) > 0}. Proof. Clearly 0, 1 O v and 0 m v. Both O v and m v 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 O v, v(xy) = v(x) + v(y) 0 and is > 0 if either x or y is in m v. Thus O v is a subring of K and m v is an ideal of O v, proper since 1 / m v. Now O v m v = {x K v(x) = 0}, and if v(x) = 0, then v(x 1 ) = v(x) = 0. So any x O v m v is invertible. This shows O v is local with maximal ideal m v. Definition 2.5. For v a valuation on a field K, the ring O v = {x K v(x) 0} is called the valuation ring of v, the maximal ideal m v = {x K v(x) > 0} of O v is called the maximal ideal of v, and the quotient O v /m v is called the residue field of v. Example 2.6. The valuation ring of the p-adic valuation v p on Q is Z (p), the localization of Z at the prime ideal (p) = pz. Concretely, Z (p) = { a b Q p b}. The maximal ideal of v p is the principal ideal generated by p, and the residue field is F p. Similarly, if k is a field and f k[t ] is irreducible, the valuation ring of the f-adic valuation v f on k[t ] is the localization of k[t ] at the prime ideal (f) = fk[t ], i.e. k[t ] (f) = { g h k(t ) f h}. The maximal ideal of v f 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. O v = {x K v(x) = 0} and v induces an isomorphism K /O v = v(k ). For x, y K, v(x) v(y) if and only if y x O v.

15 2. VALUED FIELDS AND THE VALUATION SUBRING 15 This last fact implies that the equivalence class of v is completely determined by O v. 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 ϖ R is irreducible, then ϖr is a prime ideal, the localization R (ϖ) of R at ϖr is a discrete valuation ring, and ϖ is a uniformizer. Note that a discrete valuation ring is local, and its unique nonzero prime ideal is its unique maximal ideal. Further, if ϖ is a uniformizer, then any nonzero element x R can be written uniquely as x = uϖ 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, ϖ is the only irreducible in R. Then v(uϖ 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 ϖr is the unique nonzero prime ideal in R, the zero ideal is the unique prime ideal of R[ 1 ϖ ], and R[ 1 ϖ ] = K. Hence, any x K can be written uniquely as x = uϖ n with n Z and u R, and v(uϖ n ) = n. We say that v is the valuation on K associated to R. It does not depend on the choice of ϖ, and we have O v = R and m v = ϖr. Proposition 2.9. Let v be a nontrivial discrete valuation on a field K. The valuation ring O v of v is a discrete valuation ring, and the valuation on K associated to O v is equivalent to v. This defines a bijection between normalized nontrivial discrete valuations on K and discrete valuation subrings of K. Moreover, an element ϖ O v is a uniformizer if and only if v(ϖ) = min{v(x) x m v }. Proof. Let α = inf{v(x) v(x) > 0}. Since v is nontrivial, α exists. Since v is discrete, α > 0, there is ϖ K such that v(ϖ) = α, and v(k ) = Zα. For any nonzero x O v, v(x) = nα for some n N, and v( x ϖ ) = 0. Hence, any nonzero x O n v can be written in the form x = uϖ n with n N and u O v. It follows that m v = ϖo v is the unique nonzero prime ideal. So O v is a discrete valuation ring, and ϖ is a uniformizer. Further, the valuation v associated to O v satisfies v = αv, so v and v are equivalent. If v is normalized, then v = v. On the other hand, if R is a discrete valuation subring of K with associated valuation v, then O v = 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 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 any choice of uniformizer ϖ, m n v = ϖ n R = {x K v(x) n}. We extend this to all n Z, i.e. for an integer n 0, we let m n v denote {x K v(x) n} = ϖ 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 ϖ. Exercises Let k be a field, and let α R be positive and irrational. Show that v : k[x, Y ] R { } given by v( a n,m X n Y m ) = min{n + mα a n,m 0} defines a valuation on k(x, Y ) and that v is nondiscrete Let v be a valuation on a field K, and let O v and m v be the valuation ring and maximal ideal, respectively, of v. For r R, let B r = {x K v(x) > r} and C r = {x K v(x) r}.

16 3. COMPLETIONS 16 (a) Prove that B r and C r are O v -submodules of K. (b) Assume that v is nontrivial discrete and normalized; so C n = m n v for any n Z. Show m m v /m m+n v = O v /m n v as O v -modules for any m, n Z with n Let v be a valuation on a field K, and let O v and m v be the valuation ring and maximal ideal, respectively, of v. For r R 0, let V (r) = {x O v(x 1) > r} and U (r) = {x O v v(x 1) r}. (a) Prove that V (r) and U (r) are subgroups of O v. (b) Assume that v is nontrivial discrete and normalized; so U (n) = 1 + m n v for any integer n 1. Show O v /U (n) = (O v /m n v ) and U (n) /U (n+1) = O/m v for any integer n Let v be a valuation on a field K, and let O v be the valuation ring of v. (a) Show that the ideals of O v are totally ordered by inclusion. (b) Show that any finitely generated ideal of O v is principal Let v be a nondiscrete valuation on a field K. Show that the maximal ideal m v of v is not finitely generated, hence the valuation ring O v of v is not Noetherian Let k be a finite field. (a) Show that any valuation on k(t ) is discrete. (b) Viewing k(t ) as the function field of one-dimensional projective space P 1 k over k, show that the maximal ideal of a valuation determines a bijection between equivalence classes of valuations on k(t ) and points of P 1 k, and that it restricts to a bijection between equivalence classes of nontrivial valuation on k(t ) and the closed points of P 1 k 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 parts 1 and 2 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 1. Replacing x by x 1, if necessary, we can assume x < 1. Then x n = x n 0 as n. Given a valuation v : K R { } on a field K we can associate to it an ultrametric absolute value : R R 0 by fixing any real number r > 1 and setting = r v. 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

17 3. COMPLETIONS 17 sequences, and limits as, since an open ball of radius ε with respect to is an open ball of radius ε α with respect to α log r2. If r 1, r 2 are real numbers > 1, then log r 1 > 0 and r1 v = (r2 v ) log r 2 log r 1. Because of 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 v p be the p-adic valuation on Q. The p-adic absolute value on Q is the ultrametric absolute value given by p = p vp. On the other hand, if is an ultrametric absolute value on a field K, then for any real number r > 1, v = log r 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 v 1, v 2 be two valuations on a field K. Fix a positive real number r, and let i = r vi, for i = 1, 2. Then 1 and 2 generate the same topology on K if and only if v 1 and v 2 are equivalent. Proof. If v 1 and v 2 are not equivalent, there are x, y K such that v 1 (x) v 1 (y) but v 2 (x) > v 2 (y). Necessarily, x, y K. Then for all n 1, ( y x )n 1 = y x n 1 1, so the sequence (( y x )n ) is bounded away from 0 in the 1 -topology. On the other hand y x 2 < 1, so ( y x )n 2 0 as n, and (( y x )n ) converges to 0 in the 2 -topology. Now assume that v 1 and v 2 are equivalent. One can conclude quickly that 1 and 2 define the same topology on K by appealing to Proposition 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 v 1 and v 2 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 v 1 and v 2, we have {y K x y 1 < ε} {y K x y 1 < z 1 } = {y K v 1 (x y) > v 1 (z)} = {y K v 2 (x y) > v 2 (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 O v and maximal ideal m v, then the cosets x + m n v, 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 ϖ O is a uniformizer if and only if ϖ = max{ x x m}. Any metric space X admits a completion X, 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

18 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((x n ), (y n )) = lim n d(x n, y y ). Then d((x n ), (y n )) = 0 defines an equivalence relation on X, and we define X to be the set of equivalence classes with respect to this equivalence relation. The pseudometric d defines a metric d on X, and identifying x X with the class of the constant sequence (x, x,...), identifies X with a dense subspace of X. Proposition 3.6. Let be an ultrametric absolute value on a field K. The completion K of K is a field, extends to K, and K = K. Proof. The proof that K is a field is identical to the proof that R is a field. One shows that that the set of all Cauchy sequences in K is a subring of n 1 K, and that the set of all Cauchy sequences (x n ) such that lim n x n = 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 (x n ) = lim n x n. To see that it is ultrametric, take (x n ) and (y n ) Cauchy sequences in K. If lim n x n = lim n y n, then max{ x n, y n } also converges to this common value. If lim n x n lim n y n, then either x n > y n for all n sufficiently large, or y n > x n for all n sufficiently large. In all of the above cases lim x n + y n lim max{ x n, y n } = max{lim x n, lim y n }. n n n n If (x n ) and (y n ) are Cauchy sequences in K and lim n y n = 0, then lim n x n + y n max{lim n x n, lim n y n } = lim n x n. So descends to K. On the field K, it is a multiplicative ultrametric seminorm satisfying 1 = 1, hence is an ultrametric absolute value. It remains to prove K = K, equivalently K = K. Take any x K, so x > 0. Since K is dense in K, 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 K as a topological field does not depend on r. The valuation v extends to K and satisfies v( K) = 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 K 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 K denote the completion of K with respect to v. The valuation ring R of v in K is again a discrete valuation ring, and a uniformizer for R is a uniformizer for R. Proof. This follows from Propositions 2.9 and 3.6. Example 3.9. The field of p-adic numbers, denoted Q p, 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 Z p. Example Let k be a field, and let v T be the T -adic valuation on k(t ). We ll show the completions of k(t ) with respect to v T is the field of Laurent series k((t )) in T with coefficients in k. First note that k(t ) k((t )) and that v T extends to k((t )) by v T ( n n 0 a n T n ) = n if a n0 0. It is easy to see that k(t ) is dense in k((t )), since for any f = n n 0 a n T n k((t )) and N n 0, the element g = N n=n 0 a n T n k(t ) satisfies v T (f g) > N. It remains to show that k((t )) is complete. Let (f k ) be a Cauchy sequence in k((t )), and write f k = n a k,nt n. Since (f k ) is Cauchy, for any N 0, there is M 1 such that v(f k f m ) > N for all k, m M. This implies that a k,n = a k,m for all n N if k, m M. Hence, for each n Z, the sequence (a k,n ) k is eventually constant, so we can define a n = lim k a k,n and the element f = n a nt n k((t )) is the limit of (f n ).

19 4. HENSEL S LEMMA 19 Exercises Prove that the completion of a field equipped with a metric is again a field Let be an ultrametric absolute value on a field K. (a) Show that a sequence (x n ) in K is Cauchy if lim n (x n+1 x n ) = 0. (b) Show that if K is complete, then a series n x n converges in K if lim n x n = 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 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 K be the completion of K with respect to, and denote by Ô and m the valuation ring and maximal ideal, respectively, of in K. (a) Show that Ô and m are the closures of O and m, respectively, in K. (b) Show that the natural map O/m Ô/ m on residue fields is an isomorphism 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 Without using the results of the next section, show that Z p 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 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 g 0, h 0 O[X] with g 0 mod m = g, h 0 mod m = h, deg(g 0 ) = e, and deg(h 0 ) d e. Note that f g 0 h 0 has coefficients in m. Since g and h are coprime in k[x], we can find polynomials a, b O[X] such that ag 0 + bh 0 1 has coefficients in m. Let ϖ m be a coefficient of one of the polynomials f g 0 h 0 and ag 0 + bh 0 1 that has maximal absolute value. So f g 0 h 0 mod ϖ and ag 0 + bh 0 1 mod ϖ. We inductively construct polynomials g n, h n O[X] such that for all n 0, (i) g n g n 1 mod ϖ n and h n h n 1 mod ϖ n, (ii) deg(h n ) d e, deg(g n ) = e, and deg(g n g 0 ) < e. (iii) f g n h n mod ϖ n+1. Granting for the moment the construction of the g n and h n 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 = lim n g n, whose jth coefficient is the limit of the jth coefficients of the g n, 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 = lim n h n of degree at most d e. Then by property (iii), we have f = gh. It remains to inductively construct the polynomials g n and h n. We have chosen g 0 and h 0 above, so we assume the result holds for some n 1 0. Then f g n 1 h n 1 ϖ n O[X], and we let With a, b O[X] as above, we have f n = ϖ n (f g n 1 h n 1 ) O[X]. f n ag 0 f n + bh 0 f n mod ϖ.