4A22 Local Fields

Shaun Stevens

Spring Semester 2003

Plan

Chapter 1 Foundations

§1.1 Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5§1.2 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7§1.3 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9§1.4 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Chapter 2 The rationals Q§2.1 Valuations on Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15§2.2 The p-adic numbers Qp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16

Chapter 3 Non-archimedean local fields

§3.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20§3.2 Hensel’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Chapter 4 Field extensions

§4.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26§4.2 Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Chapter 5 Algebraic extensions

§5.1 Normed vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30§5.2 Extension of valuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31

Chapter 6 Ramification

§6.1 Residue fields and unramified extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34§6.2 Totally ramified extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36§6.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Chapter 7 Algebraic closure

§7.1 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41§7.2 Incompleteness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44§7.3 Completion Cp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Chapter 8 Action of Galois

§8.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47§8.2 Kronecker-Weber Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1

Introduction

These are notes for the MTH4A22 Special Pure Mathematics Course in the spring semester of 2003.The subject for the course is Local Fields, in particular the p-adic fields and their extensions. Themain aim of the course is (hopefully) the Kronecker-Weber Theorem for abelian extensions of Qp.

We begin with some motivation, which led Hensel to the p-adic numbers in the early 20th century.If we try to solve the congruences

2x ≡ −1 (mod 3n), n > 0,

with x ∈ Z, we see that there is a solution for each n > 0, namely x = (3n − 1)/2, but there is nosimultaneous solution in Z. Similarly, look at the congruences

x2 ≡ 2 (mod 7n), n > 0,

with x ∈ Z; here again, there is a solution for each n, which we see by induction, by “lifting” asolution modulo 7n to one modulo 7n+1:

• for n = 1, put x1 = 3;

• as an example, we will do the n = 2 step. (Exercise: do the general inductive step.) Wehave our solution x1 = 3 modulo 7 so we try putting x2 = 3 + 7y2 into the equation modulo72 = 49. This gives

(3 + 7y2)2 = 9 + 42y2 + 49y22 ≡ 2 (mod 49).

The last term is zero and we rearrange to get

42y2 ≡ −7 ≡ 42 (mod 49) ⇒ 6y2 ≡ 6 (mod 7) ⇒ y2 ≡ 1 (mod 7),

so x2 = 3 + 7.1 = 10 is a solution modulo 49.

However, there is no simultaneous solution, since such an x would satisfy x2 = 2, which has nosolution even in Q.

The solutions to these two congruences can be written in a “base p” expansion: in the first case itis x = 1+3+32 + · · ·+3n−1, in the second it begins x = 3.1+1.7+2.72 +6.73 + · · · . In both cases,we can choose n to be arbitrarily large so, if we can let n → ∞, then we would get simultaneoussolutions 1 + 3 + 32 + · · · and 3.1 + 1.7 + 2.72 + 6.73 + · · · . Of course, this seems to be nonsensesince these sums don’t converge in R, but perhaps we can begin to make sense of it by replacingour prime p by an indeterminate X; then we are looking at some power series 1+X+X2 + · · · and3 +X + 2X2 + 6X3 + · · · , which are things which at least mean something. Of course, if a powerseries is actually finite (in other words, it is a polynomial), then we can reverse this by replacingX by p to get an integer in its “base p” expansion.

This thinking leads to an analogy between the ring of integers Z and the ring C[X] of polynomialswith coefficients in the complex numbers C, which is what Hensel was looking at. Attached to/BZ, we have its field of fractions Q, the rationals, which consists of quotients a/b, with a, b ∈ Zand b 6= 0; similarly, attached to C[X], we have the field of rational functions C(X), which consistsof quotients P (X)/Q(X) of polynomials P (X), Q(X), with Q(X) 6= 0. Both rings are also unique

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factorization domains: any integer can be uniquely expressed as ±1 times a product of primenumbers; any polynomial can be uniquely expressed (upto order) as

P (X) = a(X − α1)(X − α2) · · · (X − αn),

where a, α1, α2, . . . , αn ∈ C. This gives us the main point of Hensel’s analogy: the prime numbersare analogous to the linear polynomials X − α ∈ C[X]. Indeed, in the language of rings, bothgenerate prime ideals∗ in their respective rings.

The analogy goes even further than this: given m ∈ Z and a prime p, we can write m (uniquely)in “base p”

m = a0 + a1p+ a2p2 + · · ·+ anpn =n∑

i=0

aipi,

with ai ∈ Z and 0 ≤ ai ≤ p− 1. Similarly, given a polynomial P (X) and α ∈ C, we can write the(finite) Taylor expansion

P (X) = a0 + a1(X − α) + a2(X − α)2 + · · ·+ an(X − α)n =n∑

i=0

ai(X − α)i,

with ai ∈ C. Now, for polynomials and their quotients, we can push this much further: givenf(X) ∈ C(X), there is always an expression

f(X) =P (X)Q(X)

= an0(X − α)n0 + an0+1(X − α)n0+1 + · · · =∑i≥n0

ai(X − α)i.

This is just the Laurent expansion from complex analysis. Of course, in complex analysis we wouldalso worry about the convergence of this series, but here we are just interested in the series as aformal object . From an algebraic point of view, what this gives is an inclusion of fields

C(X) ↪→ C((X − α))

of the field of rational functions into the field of (finite-tailed) Laurent series in (X − α).Hensel’s idea was to try to extend the analogy to include such expansions for rationals Q. We dothis by doing arithmetic somehow regarding the prime p as a formal object (i.e. an indeterminate)but at the same time remembering to “carry”. As an example, take p = 3 and consider the rationalnumber 1/4; we have 4 = 1 + p so

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=1

1 + p= 1 + 2p+ 2p3 + 2p5 + · · · .

To check this, we multiply our expansion by 4 = 1 + p:

(1 + p)(1 + 2p+ 2p3 + 2p5 + · · · ) = 1 + p+ 2p︸ ︷︷ ︸+2p2 + 2p3 + 2p4 + 2p5 + p6 + · · ·= 1 + p2 + 2p2︸ ︷︷ ︸+2p3 + 2p4 + 2p5 + 2p6 + · · ·= · · · = 1

so that the higher powers of p disappear off “to the right”! (Exercise: check that our “3-adic”expansion for x = −12 satisfies 2x+ 1 = 0 and that our “7-adic” expansion for a square root y of 2satisfies y2 = 2.)

∗An ideal I / R is prime if ab ∈ I implies a ∈ I or b ∈ I.

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Treating the whole process formally, we can see that any positive rational x can be expanded inthis way (exercise?)

x =a

b=∑n≥n0

anpn, with an ∈ {0, 1, ..., p− 1}.

The value of n0 reflects the multiplicity with which p divides x

x = pn0a1b1, with (ab, p) = 1.

(cf. the degree of poles and zeros of Laurent series.)

For the negative rationals we use that

−1 = (p− 1) + (p− 1)p+ (p− 1)p2 + · · · .

(Exercise: check this.) So any rational number can be written as a “finite-tailed Laurent series inp” – which we call the p-adic expansion.

The set of all finite-tailed Laurent series in p is a field, which we call the field Qp of p-adic numbers;so we have an inclusion of fields

Q ↪→ Qp.

This definition of Qp is rather formal and unenlightening – and even proving that Qp is a field israther ugly – so we will introduce a more abstract and conceptual approach in the following. Wenote here that any definition which is going to make sense of these power series in p must havepn → 0 as n→∞!!

Recommended texts

The relevant section in the library is QA247.

[C] Cassels J., Local fields, LMS Student Texts 3, Cambridge University Press (1986).

[G] Gouvêa F.Q., p-adic numbers - an introduction, Springer-Verlag (1997).

[K] Koblitz N., p-adic numbers, p-adic analysis, and zeta-functions, GTM 58, Springer-Verlag(1977).

Less recommended texts

[CF] Cassels J. & Fröhlich A. eds.,Algebraic number theory, Associated Press (1967).

[S] Serre J.-P., Local fields, GTM 67, Springer-Verlag (1979).

[W] Weil A., Basic number theory, Springer-Verlag (1967).

I intend to assume a familiarity with the basic notions of abstract algebra – fields, groups, rings,ideals, etc. I may also assume some basic number theory – in particular, the structure of the group(Z/pnZ)×. It is also likely that I will leave some (important) results as exercises – this may wellmean you will sometimes need to look things up in the literature, but you should really be doingthis anyway, to get another point of view.

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Chapter 1 Foundations

In this chapter, we introduce the notion of a (multiplicative) valuation on a field, give some fun-damental examples, and develop some elementary properties. In particular, we investigate thetopology induced by a valuation, the independence of inequivalent valuations and the notion ofcompleteness with respect to a valuation.

F will denote an arbitrary field for now, though you may find it helpful to keep the example of Qin mind.

1.1 Valuations

Definition 1.1. A (multiplicative) valuation (or absolute value) on F is a map | · | : F → R+ suchthat, for all x, y ∈ F ,

(i) |x| = 0 ⇐⇒ x = 0;

(ii) |xy| = |x||y|;

(iii) |x+ y| ≤ |x|+ |y|. (The triangle inequality)

A valuation on F is called non-archimedean if also, for all x, y ∈ F ,

(iv) |x+ y| ≤ max{|x|, |y|}. (The ultrametric inequality)

Otherwise, we say that the valuation is archimedean.

Note that condition (iv) implies condition (iii).

Examples 1.2. (i) The most obvious example is the usual absolute value on Q – this is givenby the inclusion Q ↪→ R, with the usual absolute value on R. It is an archimedean valuationand is often called the valuation at infinity and denoted | · |∞.

(ii) Perhaps the least exciting example is given by the trivial valuation on any field F , given by|x| = 1, for all x 6= 0, and |0| = 0. It is non-archimedean but is also so unusual that it willhave to be excluded from most of the theory that we develop. Exercise: show that the onlyvaluation on a finite field is the trivial valuation.

(iii) Now we come to the crucial example, relating this to what we saw in the introduction. LetF = Q and let p be a prime. For x ∈ Q \ {0}, we write

x = pna

b, with (ab, p) = 1,

and put vp(x) = n; we also put vp(0) = +∞ (since 0 is divisible by an arbitrarily large powerof p). Then, for all x, y ∈ Q, we have (exercise)

vp(xy) = vp(x) + vp(y), and vp(x+ y) ≥ min{vp(x), vp(y)}. (1.3)

We now define the p-adic valuation of x ∈ Q to be

|x|p = p−vp(x),

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with the convention that |0|p = 0. That this is a non-archimedean valuation follows immedi-ately from (1.3). We also note that

|pn|p = p−n,

so that, with this p-adic notion of size, pn → 0 as n→∞.We remark that the map vp is often called an additive valuation and that the theory of non-archimedean local fields can be developed taking this notion as the primitive one and definingmultiplicative valuations from them – some books do this.

(iv) Let K be any field and put F = K(T ) = {P (T )Q(T ) : P (T ), Q(T ) ∈ K[T ], Q(T ) 6= 0}. ForP (T ) ∈ K[T ], we define v∞(P (T )) = −deg(P (T )) and we extend this to rational functionsby setting v∞(0) = +∞ and, for f(T ) = P (T )Q(T ) ,

v∞(f(T )) = v∞

(P (T )Q(T )

)= v∞(P (T ))− v∞(Q(T )) = deg(Q(T ))− deg(P (T )).

It is easy to check that this is well-defined and that v∞ satisfies the equations (1.3) (Exercise).If we choose c > 1 arbitrarily, this gives us a non-archimedean valuation on F by

|f(T )|∞ = c−v∞(f(T )).

[We will see that, for most purposes, the choice of c > 1 is irrelevant. If K is a finite field, anice choice is the number of elements of K.]

(v) Exercise: let F = K(T ) as in (iv) and let p(T ) be an irreducible polynomial; define thep(T )-adic valuation on F . [Hint: (iv) almost does this for the polynomial p(T ) = T ... watchout for the signs]

(vi) Exercise: in (v), K is a subfield of F so the absolute value on F defines one on K. What isit?

(vii) Exercise: find an archimedean valuation on Q(T ); if K is a finite field, is it possible to findan archimedean valuation on K(T )?

Now we will develop some basic properties:

Lemma 1.4. Let | · | be a valuation on a field F . Then

(i) |1| = 1;

(ii) if x ∈ F and xn = 1 then |x| = 1;

(iii) if x ∈ F then | − x| = |x|;

(iv) if F is a finite field then the valuation is trivial.

Proof Remember that if x 6= 0 then |x| > 0. We have |1| = |12| = |1|2 so |1| = 1 (or 0, whichis impossible). The rest follows easily (as −x = −1.x and every element of a finite field satisfiesxn = 1 for some n). �

The following gives a criterion to determine whether or not a valuation is non-archimedean.

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Proposition 1.5. A necessary and sufficient condition for | · | on F to be non-archimedean is that|e| ≤ 1, for all e in the ring generated by 1.

Proof (⇒) This is easy: e = ±(1 + 1 + · · · 1) so |e| ≤ max{|1|, |1|, ..., |1|} = 1.(⇐) For a, b ∈ F and n ∈ N, we have

|a+ b|n = |(a+ b)n| =n∑

j=0

(nj

)ajbn−j by the binomial expansion

≤n∑

j=0

∣∣∣∣(nj)∣∣∣∣ |a|j |b|n−j by the ultrametric inequality

≤n∑

j=0

|a|j |b|n−j

≤ (n+ 1) (max{|a|, |b|})n .

Taking nth roots of both sides, letting n → ∞ and observing that (n + 1)1/n → 1 as n → ∞, weget |a+ b| ≤ max{|a|, |b|}, as required. �

This helps to explain the word non-archimedean: a valuation is archimedean if it has the followingproperty:

Archimedean Property: given x, y ∈ F with x 6= 0, there exists a positiveinteger n such that |nx| > |y|.

Corollary 1.6. If charF 6= 0 then all valuations on F are non-archimedean.

Proof The ring in the lemma is a finite field so the valuation is trivial on it. �

Corollary 1.7. If K is a subfield of F and | · | is a valuation on F , then | · | is non-archimedeanon F if and only if it is non-archimedean on K.

1.2 Topology

Given a field F with a valuation | · |, we get a metric on F (a notion of distance) and this allows usto define open and closed sets, as, for example, in C. We will now investigate this topology , withsome surprising (perhaps counter-intuitive) results when the valuation is non-archimedean.

Definition 1.8. Let F be a field an | · | a valuation on F . Given x, y ∈ F , we define the distanced(x, y) between them to be

d(x, y) = |x− y|.

The function d : F × F → R+ is called the metric induced by the valuation.

We leave the following as a simple exercise; it says that the function d really is a metric:

Lemma 1.9. For any x, y, z ∈ F ,

(i) d(x, y) ≥ 0 and d(x, y) = 0 if and only if x = y;

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(ii) d(x, y) = d(y, x);

(iii) d(x, z) ≤ d(x, y) + d(y, z). (The triangle inequality.)

That a valuation is non-archimedean can also be expressed in terms of the metric:

Lemma 1.10. The valuation | · | is non-archimedean if and only if, for any x, y, z ∈ F ,

d(x, y) ≤ max{d(x, z), d(z, y)}.

Now we will see that the notion of distance given by non-archimedean valuations is a rather strangeone:

Lemma 1.11. Let F be a field and | · | a non-archimedean valuation on F . If x, y ∈ F and |x| 6= |y|then

|x+ y| = max{|x|, |y|}.

Proof By exchanging x and y if necessary, we assume |x| > |y|. We certainly have

|x+ y| ≤ |x| = max{|x|, |y|}.

On the other hand, x = (x+ y)− y and | − y| = |y|, so that

|x| ≤ max{|x+ y|, |y|}.

But |x| > |y|, so this implies |x| ≤ |x+ y| also. �

Corollary 1.12. In the situation of the lemma, all “triangles” are isosceles.

From complex analysis, you should remember the notions of open and closed sets: a set U is openif and only if, for any x ∈ U , there is an open ball around x contained inside U ; a set S is closed ifand only if its complement is open. Here we have the same notions of open and closed (a topology)once we have defined open balls:

Definition 1.13. Let F be a field with valuation | · |, let a ∈ F and let r ∈ R+. The open ball ofradius r and centre a is the set

B(a, r) = {x ∈ F : |x− a| < r}.

The closed ball of radius r and centre a is the set

B(a, r) = {x ∈ F : |x− a| ≤ r}.

Exercise: Show that open balls are open and closed balls are closed.

Now we will examine the topology in the case that | · | is non-archimedean, with surprising results(or perhaps not so surprising, since Corollary 1.12)

Proposition 1.14. Let F be a field with non-archimedean valuation | · |.

(i) Every point of an open (resp. closed) ball is a centre of that ball.

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(ii) Two open (resp. closed) balls have non-empty intersection if and only if one is contained inthe other.

(iii) Every open ball (resp. closed ball with r > 0) is both open and closed.

Proof We will only give the proofs for open balls, leaving those for closed balls as an exercise.

(i) Suppose b ∈ B(a, r) and let c ∈ B(b, r); then

|c− a| ≤ max{|c− b|, |b− a|} < r.

Hence B(b, r)⊆B(a, r), and the reverse inclusion follows by symmetry since a ∈ B(b, r).(ii) Any point in the intersection is the centre of both balls!

(iii) If b ∈ B(a, r) then B(b, r)⊆B(a, r) so any open ball is open. If b 6∈ B(a, r) then a 6∈ B(b, r) soneither ball is contained in the other and hence they are disjoint. Hence B(b, r)⊆F \ B(a, r) andwe see that the complement of B(a, r) is open. �

The fact that there are so many sets which are both open and closed is what makes the topologyso peculiar. Recall that a set S is called disconnected if there exist open sets U, V such that

• U ∩ V = ∅;

• S⊆U ∪ V ;

• neither S ∩ U nor S ∩ V is empty.

The idea is that such an S is made up of two pieces. Otherwise S is called connected .

If x ∈ F then the connected component of x is defined to be the union of all the connected setsthat contain x; this is the largest connected set containing x, since the union of two non-disjointconnected sets is connected (Exercise). For example, if F = R with the usual valuation then theconnected component of x is R (since it is connected). In the non-archimedean case, things arevery different:

Proposition 1.15. Let F be a field with a non-archimedean valuation | · | and let x ∈ F . Theconnected component of x is {x}.

Proof Let S be a set containing x and some other point y and put r = |y − x|. Then, puttingU = B(x, r) and V = F \ U , we see that S is disconnected. �

In the language of topology, this says that F is a totally disconnected topological space. Providedthe valuation is not trivial (in which case the topology is discrete – all sets are open), this meansthere are no open connected sets.

1.3 Independence

In this section we examine when two valuations on a field are sufficiently similar to be regarded asbeing the same – that is the notion of equivalence of valuations. We will also show that, when twovaluations are not equivalent, they are really very different.

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Definition 1.16. Two valuations | · |1 and | · |2 on a field F are equivalent if they define the sametopology on F – that is, every set that is open with respect to one is also open with respect to theother.

Note that, given a valuation | · | on a field F , a sequence {an} in F converges to a in the inducedtopology if and only if, for all ε > 0, there exists N such that, for n > N , |an−a| < ε; equivalently,if and only if, for all open sets U containing a, there exists N such that, for n > N , an ∈ U . Inparticular, the notion of convergence of a sequence depends only on the topology induced by thevaluation.

Proposition 1.17. Let | · |1 and | · |2 be valuations on a field F , with | · |1 non-trivial. The followingare equivalent:

(i) | · |1 and | · |2 are equivalent;

(ii) for any x ∈ F , we have |x|1 < 1 ⇒ |x|2 < 1;

(iii) there exists a positive real number α such that, for all x ∈ F ,

|x|1 = |x|α2 .

Proof (iii)⇒(i) We have |x − a|2 < r ⇐⇒ |x − a|1 < rα so that any open ball with respect to| · |2 is also a open ball with respect to | · |1 (albeit of a different radius).(i)⇒(ii) We have |x|1 < 1 if and only if xn → 0 with the | · |1-topology, if and only if xn → 0 withthe | · |2-topology, if and only if |x|2 < 1.(ii)⇒(iii) If |a|1 > 1 then |a−1|1 < 1 so we have |a−1|2 < 1 and hence |a|2 > 1. We also claim that

|a|1 = 1 ⇒ |a|2 = 1. (∗)

For suppose |a|1 = 1 but |a|2 6= 1. By considering a−1 if necessary, we may (and do) assume|a|2 > 1. Now let b ∈ F be such that |b|1 < 1, which is possible since | · |1 is non-trivial. Then,putting c = ban, we have |c|1 < 1 but, for n large enough, |c|2 > 1, contradicting the assumptionof (ii).

Now pick (and fix) a ∈ F \ {0} such that |a|1 < 1, so that |a|2 < 1 also, and put α = log |a|1log |a|2 > 0,so that |a|1 = |a|α2 . For b ∈ F \ {0} we have three possibilities:(1) |b|1 = 1; then |b|2 = 1 also, by (∗), so |b|1 = |b|α2 .

(2) |b|1 < 1; then |b|2 < 1 also and we put β1 = log |a|1log |b|1 and β2 =log |a|2log |b|2 . We will show that β1 = β2,

for then log |b|1log |b|2 =log |a|1log |a|2 = α and |b|1 = |b|

α2 .

Suppose β1 > β2 (the case β1 < β2 is very similar); then there exists a rational number mn suchthat β2 < mn < β1. Now consider x = a

nb−m with respect to the two valuations:

log |x|1 = n log |a|1 −m log |b|1 = n log |b|1(β1 − mn

)< 0

so that |x|1 < 1; on the other hand, the same calculation for the second valuation gives |x|2 > 1,contradicting our assumption.

(3) |b|1 > 1; then |b|2 > 1 also and the proof is essentially the same as case (2). �

10

Now we look at just how independent inequivalent valuations are.

Lemma 1.18. Let | · |1, ..., | · |J be non-trivial inequivalent valuations on a field F . Then thereexists x ∈ F with

|x|1 > 1, |x|j < 1 for 2 ≤ j ≤ J.

Proof By induction on J , starting with J = 2. Since | · |1 is non-trivial, there exists y ∈ F with|y|1 < 1; since | · |2 is not equivalent, we can choose y such that |y|2 ≥ 1. Similarly, there existsz ∈ F such that |z|2 < 1 and |z|1 ≥ 1. We put x = zy−1.Now let J > 2. By the induction hypothesis, there exist y, z ∈ F such that

|y|1 > 1, |y|j < 1 for 2 ≤ j < J ;|z|1 > 1, |z|J < 1.

We distinguish three cases:

(i) |y|J < 1. Then we take x = y.(ii) |y|J = 1. Then x = ynz will do, for large enough n.(iii) |y|J > 1. Then we have

yn

1 + yn=

11 + y−n

→j

{1 for j = 1, J,0 otherwise,

so x =(

yn

1+yn

)z will do, for large enough n. �

Theorem 1.19. With notation as in the previous lemma, let bj ∈ F , for j = 1, ..., J and let ε > 0.Then there exists x ∈ F such that

|x− bj |j < ε.

Proof By the lemma, there exist xj ∈ F such that |xj |j > 1 but |xj |i < 1 for i 6= j. Then

xnj1 + xnj

→i

{1 for j = i,0 forj 6= i.

We put wn =∑J

j=1 bjxnj

1+xnjso that wn →j bj as n→∞. Then, for large enough n, x = wn will do.

�

We remark that this is closely related to the Chinese Remainder Theorem: given distinct primesp1, ..., pJ , positive integer m(j) and bj ∈ Z, there exists x ∈ Z such that

x ≡ bj (mod pm(j)j ), for j = 1, ..., J.

We can write this as|x− bj |j < p−m(j)j ,

where | · |j is the pj-adic valuation. The theorem says that there is an x ∈ Q satisfying this but,in general, there is no analogue of the additional information provided by the Chinese RemainderTheorem, that x can be chosen to be an integer.

11

1.4 Completion

When we consider Q with the usual absolute value, a big problem with regard to doing sensibleanalysis is that there are “holes” where numbers are “missing” – that is, there are sequences {xn}of rational numbers which “ought to converge” but which don’t converge to any rational number.To get round this, we “complete” the rationals by filling in these “holes”, to get the real numbersR. Of course, the notion of convergence is determined by the absolute value | · |∞ so if we did thesame with a different valuation we might (and do) get a different completion. In this section wewill describe this completion for a general valuation | · | on a field F .

Definition 1.20. A sequence {xn} in F is called Cauchy if, for all ε > 0, there exists N such that,for all m,n > N , |xm − xn| < ε.A sequence {xn} converges to x ∈ F if, for all ε > 0, there exists N such that, for all n > N ,|xn − x| < ε.

We note that every convergent sequence is certainly Cauchy. The converse is true, for example, ifF = R and | · | = | · |∞ but is not true in general.

Definition 1.21. F, | · | is complete if every Cauchy sequence is convergent.

For example, Q is not complete with respect to | · |v for v = p prime or v = ∞. (See exercise sheet1, question 2.)

Definition 1.22. A subset S of F is called dense in F if, for all x ∈ F and ε > 0, B(x, ε)∩S 6= ∅.

Equivalently, S is dense in F if, for all x ∈ F , there exists a sequence {xn} in S such thatlimn→∞ xn = x.

For example, Q is dense in R with the usual absolute value since, given any irrational number,there are rational numbers arbitrarily close to it.

Definition 1.23. We say that a field F̄ with valuation ‖ · ‖ is the completion of F, | · | if

(i) there is an inclusion i : F ↪→ F̄ respecting the valuations;

(ii) the image i(F ) of F in F̄ is dense;

(iii) F̄ , ‖ · ‖ is complete.

For example, R, | · |∞ is the completion of Q, | · |∞.Before proving that there is a completion, we will derive some properties of it (if it exists):

Proposition 1.24. Let F, | · | be a field with completion F̄ , ‖·‖. Suppose K, ‖| ·‖| is a complete fieldand I : F → K is an injection. Then I factors through i, that is, there exists a map λ : F̄ → Ksuch that I = λ ◦ i:

FI−→ K

F̄

commutes.

In particular, the completion (if it exists) is unique upto unique isomorphism.

12

Proof Let x ∈ F̄ and let {xn} be a sequence in F such that {i(xn)} converges to x. Then we putλ(x) = limn→∞ I(xn). Exercise: check that this is well-defined.

In particular, if F̄1, ‖ · ‖1 and F̄2, ‖ · ‖2 are completions of F then we have a commuting diagram

F̄1F λ↑ ↓µ

F̄2

�

Theorem 1.25. Let F be a field with valuation | · |. Then there exists a completion F̄ of F and itis unique upto unique isomorphism.

Proof We will give a sketch of the proof, leaving various details to be filled in.

Let F be the set of all Cauchy sequences in F ; this is a ring, with elementwise multiplication andaddition:

{an}+ {bn} = {an + bn}, {an}{bn} = {anbn}.

We define a function ‖ · ‖ : F → R+ by

‖{an}‖ = limn→∞

|an|,

where the limit on the right is the limit in R with the usual topology.Now let N be the subset of F of null sequences, that is, sequences which converge to 0 (equivalently,‖{an}‖ = 0). Then N is an ideal in F , and indeed a maximal ideal, so that the quotient ringF̄ = F/N is a field. Then ‖ · ‖ depends only on {an} ∈ F modulo N , so it induces a well-definedfunction (also denoted ‖ · ‖) on F̄ ; moreover, it is a valuation on F̄ .Finally, we define an embedding i : F → F̄ by i(a) = {a, a, a, ...} (mod N ), the constant sequence.Then i(F ) is dense in F and F is complete with respect to ‖ · ‖. �

Given a (the) completion F̄ of F with respect to | · |, we will identify F with its image i(F ) andwill denote the valuation on F̄ by | · | also. We also remark that, since the notion of a sequencebeing Cauchy depends only on the equivalence class of the valuation, the completion also dependsonly on the valuation upto equivalence.

We also remark that, if |·| is non-archimedean on F , then its extension to F̄ is also non-archimedean(by Corollary 1.7) and the value group ΓF = {|x| : x ∈ F \{0}} of F is the same as the value groupΓF̄ of F̄ : we certainly have ΓF⊆ΓF̄ so suppose r ∈ ΓF̄ , so |x| = r > 0, for some x ∈ F̄ ; by densityof F in F̄ , there exists y ∈ F such that |x− y| < r = |x|; but then, by Lemma 1.11, |y| = r also.

Corollary 1.26. Let | · |1, ..., | · |J be non-trivial inequivalent valuations on a field F , let F̄j , | · |j bethe completion of F with respect to | · |j, for j = 1, ..., J , and let cj ∈ F̄j. Then, for all ε > 0, thereexists a ∈ F such that

|a− cj |j < ε, for j = 1, ..., J.

Proof Since F is dense in F̄j , there exist bj ∈ F such that |cj − bj |j < ε/2. Then, by theapproximation Theorem 1.19, there exists a ∈ F such that |a− cj |j < ε/2, for j = 1, ..., J and, bythe triangle inequality, this a will do. �

13

More picturesquely, the corollary says that the diagonal map

∆ : F →J∏

j=1

F̄j

is dense (where the product is given the product topology).

Example Let K be a field and F = K(T ), the field of rational functions in one variable, and let| · | be the non-archimedean valuation on F given by p(T ) = T in Example 1.2(v): for

14

Chapter 2 The rationals Q

In this chapter we will look at valuations on the field of rationals Q. In particular, we will classify(upto equivalence) all valuations on Q. We will then make a preliminary investigation of thecompletions of Q with respect to these valuations.

2.1 Valuations on Q

Theorem 2.1 (Ostrowski). Every non-trivial valuation on Q is equivalent to | · |v, where v = p,a prime number, or v = ∞.

Proof Let | · | be a valuation on Q and let a > 1 and b > 0 be integers. We put A =max{|0|, |1|, ..., |a− 1|}. We can write

b = bmam + bm−1am−1 + · · ·+ b0,

with bj ∈ {0, ..., a− 1}, bm 6= 0 and m ≤ log blog a . Then, by the triangle inequality, we have

|b| ≤m∑

i=0

|biai| ≤ (m+ 1)Amax{1, |a|m}.

Hence we have|b| ≤

(log blog a + 1

)A (max{1, |a|})log b/ log a .

Now if we replace b by bn and take nth roots, we get

|b| ≤(n log blog a + 1

)1/nA1/n (max{1, |a|})log b/ log a ,

so, letting n→∞, we get

|b| ≤ (max{1, |a|})log b/ log a . (†)

Now we split into two cases:

(i) | · | is archimedean. Then |b| > 1 for some integer b so, by (†), |a| > 1 for all integers a > 1.Then, for a, b > 1, (†) gives |b| ≤ |a|log b/ log a. Reversing the roles of a and b gives |a| ≤ |b|log a/ log bso we get |a|1/ log a = |b|1/ log b and hence

log |a|log a

=log |b|log b

= α > 0,

independent of a and b. Hence |a| = aα = |a|α∞, for all a ∈ Z, with a > 1. But we also have| ± 1| = 1 = | ± 1|α∞ and, since any rational is ±1 times the quotient of two positive integers, weget |q| = |q|α∞, for any q ∈ Q.(ii) | · | is non-archimedean. Since the valuation is non-trivial, |a| < 1 for some positive integer a; letb be the least such. If b is not prime then b = uv, with 1 < u, v < b so, by minimality, |u| = |v| = 1;but then |b| = |u||v| = 1, which is absurd. Hence b = p is prime.Now, if a = up is a multiple of p then |a| = |u||p| < 1. On the other hand, the converse is also true:if a is not a multiple of p then write a = up+ r, with 0 < r < p; then |r| = 1, by minimality of p,while |up| < 1 so |a| = |up+ r| = max{|up|, |r|} = 1, by Lemma 1.11.

15

Now put α = − log |p|log p , so that |p| = |p|αp . Then, for a ∈ Z, we have a = pra′, with (a′, p) = 1, so

that |a′| = |a′|p = 1 and|a| = |pra′| = |p|r|a′| = |p|rαp |a′|αp = |a|αp .

That |q| = |q|αp for all rationals q follows as in case (i). �

2.2 The p-adic numbers Qp

For each prime number p, we define the field Qp of p-adic numbers to be the completion of Q withrespect to the p-adic valuation | · |p. We also denote by | · |p the valuation on Qp which restricts tothe usual p-adic valuation on Q; then Q is dense in Qp and Qp is complete with respect to | · |p.Also, the value group

ΓQp = {r ∈ R+ : r = |x|p for some x ∈ Qp \ {0}} = {pn : n ∈ Z}.

So, for each x ∈ Qp \ {0}, there is an integer n = vp(x) such that |x|p = p−n – that is, the additivevaluation vp extends to a function on Qp.Now we define the ring of p-adic integers

Zp = {x ∈ Qp : |x|p ≤ 1}.

This really is a ring: for x, y ∈ Zp, we have |xy|p = |x|p|y|p ≤ 1 and |x+ y|p ≤ max{|x|p, |y|p} ≤ 1,so xy, x + y ∈ Zp. It is also the closed unit ball with centre 0 so we already know it is both openand closed.

Zp is a local ring , that is, it has a unique maximal ideal, I = {x ∈ Qp : |x|p < 1}: for suppose J isan ideal and J 6⊆ I; then there is y ∈ J such that |y|p = 1 so that |y−1|p = 1 and y−1 ∈ Zp; butthen 1 = y−1y ∈ J so J = Zp.In fact, the maximal ideal is the principal ideal pZp: if x ∈ I then |x|p ≤ p−1 so |p−1x|p ≤ 1 andx = p(p−1x) ∈ pZp.The units of the ring Zp are precisely

U(Zp) = Z×p = {x ∈ Zp : |x|p = 1}.

What about the quotient field Zp/pZp? Well, we will see more: for each x ∈ Zp and n > 0, weknow that there is a q = a/b ∈ Q such that |x − a/b|p ≤ p−n, by density. But then |a/b|p ≤ 1,so p 6 |b and, by the elementary theory of congruences (the Euclidean algorithm), there is a uniqueb′ ∈ Z such that bb′ ≡ 1 (mod pn). But then∣∣∣a

b− ab′

∣∣∣p

=∣∣∣ab(1− bb′)

∣∣∣p≤ p−n,

so that |x− ab′|p ≤ max{|x− a/b|p, |a/b− ab′|p} ≤ p−n. Now let α ∈ {0, ..., pn − 1} be the uniqueinteger such that ab′ ≡ α (mod p), and we have |x− α|p < p−n.So we have seen that,

for any x ∈ Zp, there is an α ∈ {0, ..., pn − 1} such that x ≡ α (mod pnZp). (‡)

16

Proposition 2.2. For any n ≥ 1, we have an exact sequence of topological rings

0 −→ Zppn−→ Zp

ϕn−→ Z/pnZ −→ 0,

where the first map is multiplication by pn and, for x ∈ Zp, ϕn(x) = α (mod pn), where α is theinteger given by (‡). In particular

Zp/pnZp ' Z/pnZ.

In particular, Zp/pZp ' Fp, the field with p elements.

Proof We have ker pn = {z ∈ Zp : pnz = 0} = {0} and, since {0, ..., pn − 1}⊆Zp, the map ϕn issurjective.

Suppose now x ∈ im pn so x ∈ Zp and x = pny, for some y ∈ Zp; then |pny−0| ≤ p−n so ϕn(x) = 0.Conversely, suppose x ∈ kerϕn; then |x| = |x− 0| ≤ p−n so |p−nx| ≤ 1 and x = pn(p−nx) ∈ im pn.Finally, the continuity of pn is because F is a topological field (see Sheet 1 q.14). For the continuityof ϕn, we need to show that ϕ−1n ({α}) is open, for α ∈ Z/pnZ (note that we put the discretetopology on Z/pnZ – all sets are open); but, if ϕn(x) = α then ϕ−1n ({α}) = B̄(x, p−n), which isopen. �

So what do elements of Qp look like? We start by looking at x ∈ Zp.From (‡), for each n ≥ 0, there exists a unique αn ∈ {0, ..., pn−1} such that |x−αn|p < p−n. Thenαn → x as n→∞. Notice also that, for m > n, |αm − αn|p ≤ max{|αm − x|p, |x− αn|p} = p−n so

αm ≡ αn (mod pn). (∗∗)

We call a sequence of integers satisfying (∗∗) coherent so that any x ∈ Zp gives rise to a uniquecoherent sequence.

Conversely, given a coherent sequence {αn}, the property (∗∗) implies that it is Cauchy so itconverges to some x ∈ Qp. Exercise: show that, in fact x ∈ Zp.

So we can identify Zp with coherent sequences: for n ≥ 1, put An = Z/pnZ and let ϕn : Zp → Anbe the map defined above, using (‡). For m ≥ n, there is an obvious map

τn,m : Am → An;a (mod pm) 7→ a (mod pn).

Then the An, together with the maps τm,n, form a projective (or inverse) system

· · · −→ An −→ · · · −→ A2τ1,2−→ A1

with all triangles commuting (e.g. τ1,2 ◦ τ2,n = τ1,n).Then Zp has the following properties (left as slightly harder exercises):

• The projection maps ϕn induce an inclusion

ϕ : Zp ↪→∏n≥1

An,

17

which identifies Zp as a topological ring with the (closed) subring of∏

n≥1An consisting ofcoherent sequences, i.e.{αn} ∈ ∏

n≥1An : τn,m(αm) = αn, for m ≥ n ≥ 1

.(Note that the topology on An is the discrete topology and the topology on the product∏

n≥1An is then the product topology.)

• Given any ring R with homomorphisms πn : R→ An such that all the triangles

AmR ↓τn,m

Ancommute, there exists a unique homomorphism π : R→ Zp such that, for all n ≥ 1, πn factorsthrough π, that is πn = ϕn ◦ π.

R

↓π

Zp

· · · −→ An −→ · · · −→ A2 −→ A1

These say that Zp is the projective limit of the system (An, τn,m). In fact, the theory can bedeveloped from this point of view.

Now an x ∈ Zp corresponds uniquely to a coherent sequence {αn}. We can write αn in base p

αn = b0 = b1p+ b2p2 + · · ·+ bn−1pn−1, 0 ≤ bi ≤ p− 1,

and, by the coherence property (∗∗), the bi do not depend on n > i. Then the series∑∞

i=0 bipi

converges to x, since the partial sums are the αn and αn → x. In particular, each element of Zpcan be uniquely written as such a power series.

Now, for y ∈ Qp \ {0}, put n0 = vp(y); then x = p−n0y ∈ Zp so y = pn0x and we get

y =∞∑

i=n0

bipi.

Every element of Qp can be written as a Laurent series expansion in p.

Finally, we look at a topological property. Let x ∈ Qp and put n = vp(x). Then |p−nx|p = pnp−n =1 so we see that x = pnu, with u ∈ Z×p . In particular, we see that⋃

n∈ZpnZp = Qp

and we also have ⋂n∈Z

pnZp = {0}.

Together, these (almost) say that the open sets pnZp form a fundamental system of neighbourhoodsof 0 in Qp which covers Qp. Recall:

18

• a neighbourhood of x ∈ F is a set U which contains some open ball B(x, r), r > 0;

• a fundamental system of neighbourhoods of x is a collection of neighbourhoods {Uλ : λ ∈ Λ}such that any neighbourhood of x contains some Uλ;

• a collection of sets covers X⊆F if its union contains (or equals) X.

Note that pnZp = B(0, p−n) = B(0, p−n+1/2).Exercise: Find a fundamental system of neighbourhoods of 0 in R which covers R.

19

Chapter 3 Non-archimedean local fields

In this chapter we will examine more closely the general theory of fields which are complete withrespect to a non-archimedean valuation, in particular those which are locally compact, like Qp. Wewill look at the algebraic structure of such fields and see that, for example, the elements of the fieldhave power series type expansions, like in Qp. Then we will look at the problem of finding rootsof equations, and we will prove Hensel’s Lemma, which essentially says that Newton’s method ofapproximation of roots works very well in our situation.

Before we begin, just a few words about the archimedean case: why are we only interesting our-selves in the non-archimedean valuations? The answer is that there are not very many completearchimedean fields:

Theorem 3.1 (Ostrowski). Let F be a field complete with respect to an archimedean valuation| · |. Then F is isomorphic to either R or C and the valuation is equivalent to the usual absolutevalue | · |∞.

We will not prove this but a proof can be found, for example, in [C] Chapter 3.

3.1 Basics

Let F be a field with non-trivial non-archimedean valuation | · |. We put

oF = {x ∈ F : |x| ≤ 1}, the ring of integers of F ;

pF = {x ∈ F : |x| < 1}.

Then oF is an integral domain and pF is a maximal ideal: for suppose J ) pF is an ideal of oF ;then, for some x ∈ J , |x| = 1 so |x−1| = 1 and x−1 ∈ oF ; but then 1 = xx−1 ∈ J so J = oF .

We also put

UF = o×F = {x ∈ F : |x| = 1} = oF \ pF , the group of units of oF ;

kF = oF /pF , the residue class field ; the characteristic p = char kF is called the residualcharacteristic – note that we need not have char kF = charF ;

ΓF = {|x| : x ∈ F×}, the value group of | · | on F (a multiplicative subgroup of R×+).

Definition 3.2. A non-archimedean valuation | · | is discrete if ΓF is a discrete subgroup of R×+.

Lemma 3.3. A non-archimedean valuation | · | is discrete if and only if the maximal ideal pF isprincipal.

Proof By Exercise Sheet 1 q.6, ΓF is discrete if and only if it is cyclic.

Suppose first pF = (πF ) and put γ = |πF | < 1; hence, for all x ∈ pF , |x| ≤ γ. Then, for x ∈ F ,there exists n ∈ Z such that γn ≤ |x| < γn−1, so that γ ≤ |xπF 1−n| < 1; but then xπF 1−n ∈ pF , sowe must have |xπF 1−n| = γ and |x| = γn.Conversely, suppose | · | is discrete so ΓF is cyclic, with generator γ < 1, say, and let πF ∈ F besuch that |πF | = γ. Clearly, we have (πF )⊆pF . Conversely, for x ∈ pF , |x| = γn for some n ≥ 1 so|xπF−1| = γn−1 ≤ 1, i.e. xπF−1 ∈ oF so x ∈ (πF ). �

20

From now on, let | · | be a discrete non-archimedean valuation on a field F . If pF = (πF ) thenwe call πF a uniformizer or prime element for the valuation. Then, for x ∈ F×, x = πF nε, forsome n ∈ Z, ε ∈ o×F . We then write vF (x) = n (the order of x) and obtain an additive valuationvF : F → Z ∪ {∞} by putting vF (0) = ∞.

Lemma 3.4. Let 0 6= J⊆oF be an ideal of oF . Then J = pF n, for some n ∈ N, where pF n is theideal generated by products x1...xn, for x1, ..., xn ∈ pF .

Proof The subset {|x| : x ∈ J} of ΓF is bounded above so attains its upper bound at x0 = πF nε,say; then J = (x0) = pF n. �

In particular, pF is the unique non-zero prime ideal and oF is a principal ideal domain and a localring (a ring with a unique maximal ideal).

Now we will move on to the completion:

Let F̄ be the completion of F with respect to | · | and let oF̄ , pF̄ denote its ring of integers andmaximal ideal respectively. We clearly have oF = oF̄ ∩ F and pF = pF̄ ∩ F .There is then an inclusion oF ↪→ oF̄ and hence a map

oF → kF̄ = oF̄ /pF̄ ,x 7→ x+ pF̄ .

But pF is in the kernel of this map so it induces a natural map

kF → kF̄ (3.5)

of residue class fields.

Lemma 3.6. The map (3.5) is an isomorphism.

Proof We need only show that it is surjective. Let x∗ ∈ oF̄ so, by density of F in F̄ , thereexists x ∈ F such that |x − x∗| < 1. Then x − x∗ ∈ pF̄ and |x| ≤ max{|x − x∗|, |x∗|} ≤ 1 sox ∈ oF̄ ∩ F = oF . �

Definition 3.7. By a non-archimedean local field , we mean a field F that is complete with respectto a non-trivial discrete non-archimedean valuation | · | such that the residue class field kF is finite.

For example, Qp and Fq((T )) are non-archimedean local fields (and, it turns out, that all such fieldsare finite extensions of these).

From now on F will denote a non-archimedean local field.

We say that the infinite sum∑∞

n=0 xn, xn ∈ F , converges to the sum s if s = limN→∞∑N

n=0 xn.Clearly, if the sum converges then we have |

∑∞n=0 xn| ≤ maxn |xn|, since this is true for all the

partial sums.

Lemma 3.8. The sum∑∞

n=0 xn converges if and only if xn → 0 as n→∞.

Note that this is certainly not the case for an archimedean valuation!

21

Proof Suppose first∑∞

n=0 xn converges to s; then

limN→∞

xN = limN→∞

(N∑

n=0

xn −N−1∑n=0

xn

)= s− s = 0.

Conversely, suppose xn → 0. Let ε > 0 and let R be such that |xn| < ε for n > R. Then, forM > N > R, ∣∣∣∣∣

M∑n=0

xn −N∑

n=0

xn

∣∣∣∣∣ ≤ maxN 0.

Proposition 3.11. Let F be a non-archimedean local field. Then oF is compact and hence F islocally compact.

Proof Let Uλ (λ ∈ Λ) be a family of open sets covering oF and suppose there is no finite subcover-ing. Now oF can be written as the finite union

⋃x∈A(x+πF oF ), where A is a set of representatives

for oF /pF , so, for at least one x0 ∈ A, the set x0 + πF oF is not covered by finitely many Uλ.

22

Similarly, there is an x1 ∈ A such that x0 + x1πF + πF 2oF is not finitely covered, etc. . . Letx∗ = x0 + x1πF + · · · ; then x∗ ∈ Uλ0 , for some λ0 ∈ Λ, and, since Uλ0 is open, x∗ + πF NoF⊆Uλ0 ,for some N , which is absurd.

That F is locally compact now follows immediately: for x ∈ F , put Vx = B(x, 1) = x+ oF . �

We remark that the converse is also true: if F is a field which is locally compact with respect to anon-trivial non-archimedean valuation then F is a non-archimedean local field.

3.2 Hensel’s Lemma

This is a term given to a number of results, which are essentially Newton’s method of approximation.It turns out that this works much better than in the real case.

Theorem 3.12. Let F be a non-archimedean local field and let f(X) ∈ oF [X]. Suppose x0 ∈ oFsatisfies |f(x0)| < |f ′(x0)|2, where f ′(X) is the formal derivative of f(X). Then there exists aunique x ∈ oF such that

f(x) = 0,|x− x0| ≤ |f(x0)|/|f ′(x0)|.

Proof Define fj(X) ∈ oF [X] by

f(X + Y ) = f(X) + f1(X)Y + f2(X)Y 2 + · · · (3.13)

where X,Y are independent indeterminates, so f1(X) = f ′(X). Define y0 ∈ oF by f(x0) +y0f1(x0) = 0 so, by (3.13),

|f(x0 + y0)| ≤ maxj≥2

|fj(x0)yj0|

≤ maxj≥2

|yj0|

≤ |y20| = |f(x0)|2/|f ′(x0)|2 < |f(x0)|.

Similarly,|f1(x0 + y0)− f1(x0)| ≤ |y0| < |f1(x0)|

so |f1(x0 + y0)| = |f1(x0)|.We put x1 = x0 + y0; then

|f(x1)| ≤ |f(x0)|2/|f1(x0)|2;|f1(x1)| = |f1(x0)|;|x1 − x0| ≤ |f(x0)|/|f ′(x0)|.

We repeat the process and obtain a sequence of xn+1 = xn + yn such that |f1(xn)| = |f1(x0)| and

|f(xn+1)| ≤ |f(xn)|2/|f1(xn)|2 = |f(xn)|2/|f1(x1)|2

so f(xn) → 0. Further, |xn+1 − xn| = |yn| = |f(xn)|/|f1(x0)| → 0 so {xn} is Cauchy and has alimit as required.

23

Now suppose we have another solution x + z, with z 6= 0 and |z| ≤ |f(x0)|/|f ′(x0)| < |f1(x0)| =|f1(x)|. Then, putting X = x, Y = z in (3.13), we get

0 = f(x+ z)− f(x) = zf1(x) + z2f2(x) + · · · . (3.14)

But then the first term of (3.14) has valuation strictly greater than the other terms, as |fj(x)| ≤ 1,a contradiction. �

Examples 3.15. (i) Squares in Qp:We recall some things about finite fields: suppose F = Fq is a (the) finite field with q = pf elements.Then F× is cyclic, of order q − 1, with generator g, say. Then the squares in F× are the elements{1 = g0, g2, ..., g2(q−2)} = {g2n : n ∈ Z}. Note

• If p 6= 2 then q − 1 is even and, since gq−1 = 1, the set of squares is {1, g2, ..., gq−3} – that is,exactly half the elements are squares. If p = 2 then all elements of F are squares.

• If a, b ∈ F then at least one of a, b, ab is a square – for if a, b are not squares then a = gr,b = gs, with r, s odd, so r + s is even and ab =

(g(r+s)/2

)2.

Now let p 6= 2 and y ∈ Z×p . Suppose there exists x0 ∈ Zp such that |x20 − y| < 1. Then there existsx ∈ Zp such that x2 = y. [Take f(X) = X2 − y so |f(x0)| < 1 but f ′(X) = 2X and |f ′(x0)| = 1.]In particular, a rational integer z ∈ Z which is coprime to p is a square in Zp if and only if andonly if there is a solution to the congruence X2 ≡ z (mod p). (Which is possible for precisely halfof {1, ..., p− 1} – if there is a solutions, z is called a quadratic residue modulo p.)For p = 2, y ∈ Z×p is a square in Zp if and only if y ≡ 1 mod 8Zp. (Exercise)This means we have

• For p 6= 2, Q×p /(Q×p )2 has four elements, represented by 1, c, p, cp, where c ∈ {1, ..., p − 1} issuch that there is no solution to X2 ≡ c (mod p). [Suppose x ∈ Q×p . Multiplying x by asuitable power of p2 ∈ (Q×p )2, we may assume x = u or pu, with u ∈ Zp. Let α ∈ {1, ..., p−1}be such that u ≡ α (mod pZp); then u = α(1 + α−1u) and 1 + α−1u ≡ 1 (mod pZp) so it isa square and we may assume u = α. If now u is a quadratic residue then it is a square, sou ∈ (Q×p )2; otherwise, uc is a quadratic residue and u ∈ c(Q×p )2.

• For p = 2, Q×2 /(Q×2 )

2 has eight elements, represented by ±1,±2,±5,±10. (Exercise)

(ii) Since the residue field kF is finite, kF× is cyclic, of order q − 1, q = pr for some prime p andr ∈ N. Note then that p.1 ∈ pF so q − 1 ∈ o×F . For each α ∈ kF

×, let x0 ∈ o×F be such that x0mod pF = α and consider f(X) = Xq−1 − 1. Then |f(x0)| < 1 and |f ′(α)| = |q − 1||x0|q−2 = 1 sothere exists (a unique) α̂ ∈ o×F such that f(α̂) = 1 and α̂ ≡ x0 mod pF , i.e. α̂ is a (q− 1)th root ofunity in F and α̂ (mod pF ) = α. In particular, F has a full set of (q − 1)th roots of unity.Then one can take {0} ∪ {α̂ : α ∈ k×F } to be a set of representatives for oF /pF , and sometimes it ismore convenient to take these than 0, 1, ..., q − 1.Note that α̂ is called the Teichmüller representative for α.

Now we will look at the multiplicative group F× of F (with the subspace topology). In the subgroupUF = o×F , we have the principal congruence subgroups

UnF = {u ∈ o×F : u− 1 ∈ pFn} = 1 + pF n.

24

Then UF and the UnF are open-closed and compact in F×. We have isomorphisms of topological

groups

F×/o×F → Zxo×F 7→ vF (x),

(where Z is endowed with the discrete topology), and

o×F /U1F → kF×

ξvU1F 7→ gv,

where ξ is a primitive (q− 1)-th root of unity in F , kF× is endowed with the discrete topology andg is a generator for kF×. (So ξ is the Teichmüller lift of g.) Any element of F× can be uniquelywritten as πF uξvε, where ε ∈ U1F . i.e. F× is isomorphic to the direct product Z×Z/(q− 1)Z×U1F .

3.3 Local-global

Suppose we are looking for rational solutions to a Diophantine equation, e.g. xn + yn = zn, for n afixed positive integer. If there are rational solutions then certainly there are solutions in R = Q∞and Qp, for each prime p. So, if there are no solutions in Qv, for some v, then there are none in Q.For example

• x2 + y2 + z2 = 0 has no non-trivial solutions in Q as it has none in Q∞.

• 3x2 + 2y2 + z2 = 0 has no non-trivial solutions in Q as it has none in Q3. (Exercise)

What about the converse? If we have solutions in Qv for all v then is there necessarily a solutionin Q? This is called the local-global (or Hasse) principle.The answer is: sometimes; it depends on the type of equation.

For example, suppose u ∈ Q× and x2−u has a solution in Qv for every v; then x2−u has a solutionin Q:

Proof Write u = ±∏

p pvp(u). Then, since u is a square in Q∞, u > 0; since u is a square in Qp.

vp(u) is even. �

However, it is not always true. For example, (X2 − 2)(X2 − 17)(X2 − 34) = 0 has solutions in Qv,for every v but none in Q.

Proof For p 6= 2, 17, either 2, 17 or 34 = 2.17 is a quadratic residue modulo p, so is a square. Forp = 2, 17 ≡ 1 (mod 8) so is a square. For p = 17, 2 is a quadratic residue so is a square. �

See the problem sheet for a more interesting example.

The biggest (reasonably accessible) result in this area is:

Theorem 3.16 (Hasse-Minkowski). Let F (X1, ..., Xn) ∈ Q[X1, ..., Xn] be a homogeneous poly-nomial of degree 2 in n variables – i.e.

F (X1, ..., Xn) =∑

1≤i≤j≤naijXiXj , with aij ∈ Q.

Then F (X1, ..., Xn) = 0 has a non-trivial solution in Q if and only if it has non-trivial solutions inQv, for each v.

25

Chapter 4 Field extensions

4.1 Basics

Recall:

• Let F be a field. An extension K of F is a field K containing F . An extension is calledalgebraic if every element α ∈ K satisfies some polynomial equation f(α) = 0, with coefficientsin F . For example, C is algebraic over R (any z ∈ C satisfies f(X) = X2 − (z + z) + zz);Q(√

2) := {a+ b√

2 : a, b ∈ Q} is algebraic over Q (any a+ b√

2 satisfies f(X) = X2− 2aX +(a2 − 2b2)).

• If K is an extension of F then it can be thought of as a vector space over F . It is called a finiteextension if it is finite-dimensional and, in that case, we write [K : F ] for that dimension, andcall it the degree of K over F . For example, [C : R] = [Q(

√2) : Q] = 2. Any finite extension

is algebraic, but the converse is not true.

• If K/F is a field extension and L/K is another field extension then [L : F ] = [L : K][K : F ](the Tower Law).

• F [X] has a Euclidean algorithm, based on the degree of polynomials. If f(X) ∈ F [X] isan irreducible polynomial (cannot be factorized into polynomials of strictly smaller degree)then the principal ideal (f(X)) is maximal so the quotient K = F [X]/(f(X)) is a field whichcontains a root α = X + (f(X)) of f(X). In fact K = F (α) is of degree n = deg f over F(one possible basis is {1, α, ..., αn−1}). We call this process “adjoining a root of f(X)”.

• Conversely, if L/F is any algebraic extension and α ∈ L, then α satisfies a unique monicirreducible polynomial f(X) (called the minimal polynomial of α); n = deg f is called thedegree of α and the subfield F (α) := {rational functions in α} has degree n over F . In fact,F (α) is isomorphic to F [X]/(f(X)), the isomorphism given by α 7→ X + (f(X)).

• If f(X) ∈ F [X] is any polynomial then, by successively adjoining the roots of f(X) we obtaina field K in which f(X) splits into linear factors. The smallest such field K is called thesplitting field of f(X) over F . It has degree at most n!, where n = deg f .

• If F = Fq is a finite field with q = pf elements then F is the splitting field of Xq−X over Fp.In particular, there is, upto isomorphism, a unique field with q elements. The multiplicativegroup F×q is cyclic of order q − 1.

4.2 Factorization

Let F be a non-archimedean local field, with valuation |·|. We begin by introducing some valuationson F (X), which extend the valuation on F .

Lemma 4.1. Let c > 0 be arbitrary. For f(X) = fnXn + · · ·+ f1X + f0 ∈ F [X], put

‖f‖ = ‖f‖c = maxici|fi|.

For h(X) = f(X)/g(X) ∈ F (X), with f(X), g(X) ∈ F [X], we put

‖h‖ = ‖f‖/‖g‖.

Then ‖ · ‖ is a valuation on F (X) which coincides with | · | on F .

26

Proof Let f(X), g(X) ∈ F [X]. Clearly,

‖f + g‖ = maxici|fi + gi| ≤ max

ici(|fi|+ |gi|) ≤ max

ici|fi|+ max

jcj |gj | = ‖f‖+ ‖g‖.

Similarly,

‖fg‖ = maxk

ck

∣∣∣∣∣∣∑

i+j=k

figj

∣∣∣∣∣∣ ≤ maxi,j ci|fi|cj |gj | = maxi ci|fi|maxj cj |gj | = ‖f‖‖g‖.To show we have a valuation, we need to show we have equality here and that ‖h‖ = ‖f/g‖ iswell-defined, for then the rest is immediate. There are integers I, J such that

‖fIXI‖ = ‖f‖; ‖fiXi‖ < ‖f‖, for i < I;‖gJXJ‖ = ‖g‖; ‖gjXj‖ < ‖g‖, for j < J.

The coefficient of XI+J in fg is∑

i+j=I+J figj . Then

• if i < I then |fi| < c−i‖f‖; since also ‖gjXj‖ ≤ ‖g‖, we have |gj | ≤ c−j‖g‖ so

|figj | < c−I−J‖f‖‖g‖;

• if j < J then we get the same inequality.

• otherwise, i = I and j = J and we have

|fIgJ | = c−I−J‖f‖‖g‖.

Hence fIgJ is the larger than all other terms in the sum and we get cI+J∣∣∣∑i+j=I+J figj∣∣∣ ≥ ‖f‖‖g‖.

Hence‖fg‖ ≥ ‖f‖‖g‖,

and we have equality, as required.

Now let h(X) ∈ F (X) and suppose h(X) = f(X)/g(X) = F (X)/G(X), with f, g, F,G ∈ F [X].Then f(X)G(X) = F (X)g(X) so ‖f‖‖G‖ = ‖F‖‖g‖ and the definition of ‖h‖ is independent ofthe choice of f, g. �

Remark: This can be extended, by induction, to rational functions in several variables.

We can now use this to prove “Gauss’s Lemma” for polynomials over oF ; this too can be generalisedto several variables.

Lemma 4.2. Suppose that f(X) ∈ oF [X] is the product of two non-constant polynomials in F [X].Then it is the product of two non-constant polynomials in oF [X].

Proof We use the valuation ‖ · ‖ = ‖ · ‖1 given by the previous lemma. Then oF [X] is just the setof elements of F [X] which are valuation integers (have valuation at most 1). Further, | · | and ‖ · ‖have the same value group.

Suppose that f = gh, with g, h ∈ F [X]. There is an a ∈ F such that |a| = ‖g‖; by replacing g, hby a−1g, ah, we may assume ‖g‖ = 1 (so that g ∈ oF [X]). Then

‖h‖ = ‖g‖‖h‖ = ‖f‖ ≤ 1,

so h ∈ oF [X] also. �

27

Remark: Gauss’s Lemma for polynomials over Z can be deduced from this. (See Cassels, p.96.)Now we look at another irreducibility criterion. For this we recall that we have the reduction map

oF → oF /pF = kF ,

which we denote by a 7→ a, for a ∈ oF . For a polynomial f(X) = fnXn + · · ·+ f1X + f0 ∈ oF [X],we denote by f(X) ∈ kF (X) the polynomial obtained by reducing the coefficients modulo pF :f(X) = fnXn + · · ·+ f1X + f0.

Theorem 4.3 (Eisenstein irreducibility criterion). Suppose πF is a uniformizer of F andf(X) = fnXn + · · ·+ f1X + f0 ∈ oF [X] has

|fn| = 1; |fj | < 1 for j < n; |f0| = |πF |.

Then f(X) is irreducible in F [X].

Proof By Gauss’s Lemma, if f(X) is reducible in F [X] then it is reducible in oF [X], say f(X) =g(X)h(X), where

g(X) = grXr + · · ·+ g0,h(X) = hsXs + · · ·+ h0,

and r + s = n. Then, since f(X) = fnXn, we must have g(X) = grXr and h(X) = hsXs. Inparticular, |g0| < 1 and |h0| < 1 so the are both at most |πF |. But then |f0| = |g0h0| ≤ |πF |2, acontradiction. �

A polynomial which satisfies the hypotheses of Eisenstein’s criterion is called an Eisenstein poly-nomial .

Corollary 4.4. The polynomial

φp(X) = Xp−1 +Xp−2 + · · ·+ 1 = (Xp − 1)/(X − 1)

is irreducible in Qp.

Proof The polynomial

φp(Y + 1) = Y p−1 +(

pp− 1

)Y p−2 + · · ·+

(p2

)Y +

(p1

)is Eisenstein. �

Corollary 4.5. The polynomial

φpn(X) =(Xp

n − 1)/(Xp

n−1 − 1)

= φp(Xp

n−1)

is irreducible in Qp.

28

Proof We again put X = Y + 1, say θ(Y ) = ψpn(Y + 1). We have θ(0) = φp(1) = p. Further,((Y + 1)p

n−1 − 1)θ(Y ) =

((Y + 1)p

n − 1).

If we reduce this modulo p, we get]Y p

n−1θ(Y ) = Y p

n,

so θ(Y ) = Y pn−pn−1 and θ is Eisenstein. �

To end this section, we give another way of telling whether a polynomial is irreducible or not. Solet f(X) = f0 + f1X + · · ·+ fnXn be a polynomial in F [X], with f0 6= 0 and fn 6= 0. The Newtonpolygon of f is the convex cover in R2 of the points

P (j) = (j, log |fj |), for fj 6= 0.

So it consists of line segments σs, for 1 ≤ s ≤ r, say, where σs joins P (ms−1) to P (ms) and

0 = m0 < m1 < · · · < mr = n.

The slope of σs is

γs =log |fms | − log |fms−1 |

ms −ms−1,

and, by convexity,γ1 > γ2 > · · · > γr.

We shall say f is of type(l1, γ1; l2, γ2; ...; lr, γr),

wherels = ms −ms−1, for 1 ≤ s ≤ r.

If r = 1, we say that f is pure.

The main theorem here is the following (see e.g. Cassels, p100, for a proof):

Theorem 4.6. Suppose that f(X) ∈ F [X] is complete of type (l1, γ1; l2, γ2; ...; lr, γr). Then

f(X) = g1(X) · · · gr(X),

where gs(X) is pure of type (ls, γs), for 1 ≤ s ≤ r.

Note that the gs(X) are not necessarily irreducible.

29

Chapter 5 Algebraic extensions

In this chapter, we begin to look at field extensions of a non-archimedean local field more closely.In particular, we would like to see that any finite extension is a non-archimedean local field. Thisrequires us to find a valuation on such an extension, and we would like to find such a valuationwhich extends the valuation on our original field. It turns out that there is such a valuation and,moreover, it is unique.

5.1 Normed vector spaces

We let F be a non-archimedean local field, as usual.

Definition 5.1. Let V be a vector space over F . A function ‖ · ‖ : V → R+ is called a norm if

(i) ‖v‖ = 0 if and only of v = 0;

(ii) ‖v + w‖ ≤ ‖v‖+ ‖w‖, for all v,w ∈ V ;

(iii) ‖λv‖ = |λ|‖v‖, for all λ ∈ F , v ∈ V .

A vector space V with a norm ‖ · ‖ will be called a normed vector space.

Note that a norm on a vector space V induces a metric d(v,w) = ‖v −w‖ on V and so we havenotions of open and closed balls – i.e. a topology.

Definition 5.2. Two norms ‖ · ‖1 and ‖ · ‖2 on a vector space V are said to be equivalent if thereexist positive real numbers C1, C2 such that

‖v‖1 ≤ C2‖v‖2, ‖v‖2 ≤ C1‖v‖1, for all v ∈ V ;

this is if and only if the two norms induce the same topology on V (Exercise).

The main result of this section is the following, which we will need to prove uniqueness of theextension of a valuation.

Lemma 5.3. Let V be a finite-dimensional vector space over F . Then any two norms on V areequivalent. Moreover, V is complete with respect to the induced metric.

Proof First we define the sup-norm on V and then we will show that any norm is equivalent toit. We will do this by induction on the dimension n = dim FV , noting that the case n = 1 is trivial.So let v1, ...,vn be a basis for V . For v = λ1v1 + · · ·+ λnvn ∈ V , with λi ∈ F , we put

‖v‖0 = maxi|λi|.

It is easy to check that this is a norm and that V is complete with respect to it (Exercise).

Now let ‖ · ‖ be any norm on V and put C0 =∑

i ‖vi‖. Then, for v = λ1v1 + · · ·+ λnvn ∈ V , wehave

‖v‖ =

∥∥∥∥∥∑i

λivi

∥∥∥∥∥ ≤∑i

|λi|‖vi‖ ≤ C0‖v‖0.

30

It remains to show that there is some C such that

‖v‖0 ≤ C‖v‖, for all v ∈ V. (†)

Suppose not, so for every m ≥ there is a wm ∈ V such that

‖wm‖ < 1m‖wm‖0.

Now, since ‖wm‖0 is the sup norm, it is just the valuation of the ith coefficient, for some i. Inparticular, since there are only finitely many coefficients, there is an index i0 such that ‖wm‖0 isthe the valuation of the ith0 coefficient for infinitely many m. After permuting the basis vectors, wemay assume i0 = n and let {wmk} be the subsequence consisting of those vectors whose sup normis the valuation of the nth coefficient, which we denote by µk.

We consider now the vectors µ−1k wmk , (which have nth coefficient 1) and let W be the (n − 1)-

dimensional subspace of V spanned by v1, ...,vn−1. Then we have

(i) µ−1k wmk = uk + vn, for some uk ∈W ;

(ii) ‖uk + vn‖ = |µk|−1‖wmk‖ < 1mk .

Now the sequence {uk} in W is Cauchy (from (ii) and the triangle inequality). Since W is n− 1-dimensional it is complete (by the inductive hypothesis) so {uk} converges to some u ∈ W . Butthen

‖u + vn‖ = limk→∞

‖uk + vn‖ = limk→∞

1mk

= 0,

while u + vn 6= 0, as vn 6∈W , contradicting the definition of a norm.Hence (†) holds and the norms ‖ · ‖ and ‖ · ‖0 are equivalent. �

Corollary 5.4. Let V be a finite-dimensional normed vector space over F . Then V is locallycompact (i.e. every v ∈ V has a compact neighbourhood).

Proof Exercise: by Lemma 5.3, we may as well assume ‖ · ‖ is the sup-norm with respect to somefixed basis v1, ...,vn of V . Now imitate the proof of 3.11 to show that B̄ = {v ∈ V : ‖v‖ ≤ 1} iscompact. �

5.2 Extension of valuations

F is a non-archimedean local field, as usual. If E is an extension of F then we say that a valuation‖ · ‖ on E is an extension of | · | to E if ‖λ‖ = |λ| for all λ ∈ F .The main theorem here will be the following:

Theorem 5.5. Let E be a finite extension of F . Then there is precisely one extension ‖ · ‖ of | · |to F . Moreover, E, ‖ · ‖ is a non-archimedean local field.

Proof We will actually construct the extension ‖ · ‖ explicitly, in terms of | · | and properties ofthe extension E/F . Uniqueness is already quite easy:

Let ‖ · ‖1, ‖ · ‖2 be two extensions of | · | to E. Then, regarding E as a finite-dimensional vectorspace over F , they are both norms on E so, by Lemma 5.3, are equivalent as norms, i.e. they define

31

the same topology on E. But then they are equivalent as valuations so, by Proposition 1.17, thereis a positive real number α such that

‖a‖1 = ‖a‖α2 , for all a ∈ E.

But the two valuations coincide on F , so taking any a ∈ F with |a| 6= 1, we get α = 1, as required.The last statement of the Theorem also follows immediately, from 5.4 and the converse of 3.11.

To show the existence, we must first look at some properties of finite field extensions E/F – thesewill be true for any field F . The main thing we need is the notion of the Norm map NE/F : E → F ,of which we can give a first definition now: (N.B. Don’t confuse norms as in the previous sectionwith the Norm map as we are about to define.)

Let E/F be an extension of degree n and let α ∈ E. Thinking of E as a vector space over F ,multiplication by α is a linear map mα : E → E, with matrix Aα ∈ M(n, F ) with respect to somebasis. Then we put

NE/F (α) = detAα,

so we get a map NE/F : E → F , the norm map. Note that, although the matrix Aα depends onthe choice of a basis for E, it’s determinant does not; indeed, if the characteristic polynomial ofthe linear map mα is

det (XIn −Aα) = Xn + an−1Xn−1 + · · ·+ a1X + a0 ∈ F [X],

where In is the n× n identity matrix, then

NE/F (α) = detAα = (−1)na0.

Since determinants are multiplicative, this means we have

NE/F (αβ) = NE/F (α)NE/F (β), for all α, β ∈ E.

We also have that, if in fact α ∈ F , then

NE/F (α) = αn,

where n = [E : F ].

Now, for any α ∈ E, we have a tower of fields F ⊆ F (α) ⊆ E and α ∈ F (α). If the minimalpolynomial of α is f(X) = Xr + fr−1Xr−1 + · · ·+ f0 ∈ F [X], then [F (α) : F ] = r and 1, α, ..., αr−1is a basis for F (α)/F . With respect to this basis, the matrix of the multiplication by α on F (α) is

Bα =

0 0 · · · 0 −f01 0 · · · 0 −f10

. . . . . ....

......

. . . . . . 0 −fr−20 · · · 0 1 −fr−1

,

the characteristic polynomial of Bα is f(X) and NF (α)/F (α) = (−1)rf0. Now let e1, ..., es be a basisfor E/F (α), so that n = [E : F ] = rs. Then e1, e1α, ..., e1αr−1, e2, e2α, ..., esαr−1 is a basis for E/F

32

and, with respect to this basis, multiplication mα by α on E has matrix

Aα =

Bα 0 · · · 0

0 Bα. . .

......

. . . . . . 00 · · · 0 Bα

.In particular, the characteristic polynomial of Aα is f(X)s and

NE/F (α) =(NF (α)/F (α)

)s = (−1)nfs0 .Note that this means that, in order to find NE/F (α), we actually only need to find NF (α)/F (α).

Now we are ready to define the extension of the valuation to E. (However, there is another way todefine the Norm map, which we may see later, which helps to explain how we arrive at this formulaand why this is the only possibility.)

Claim In the situation of Theorem 5.5, the extension ‖ · ‖ of | · | to E is given by

‖α‖ = |NE/F (α)|1/n, for α ∈ E,

where n = [E : F ].

Proof We need to show that this defines a valuation on E which extends | · |.First note that, for α ∈ F ,

‖α‖ = |NE/F (α)|1/n = |αn|1/n = |α|,

so it does indeed extend | · |.If α ∈ E×, then α is invertible so the linear map mα : E → E (corresponding to multiplicationby α) is invertible, with inverse mα−1 . In particular, the corresponding matrix Aα is invertible soNE/F (α) = detAα 6= 0 and we have ‖α‖ 6= 0.That ‖ · ‖ is multiplicative follows immediately from multiplicativity of the Norm map so we onlyneed to prove

‖α‖ ≤ 1 ⇒ ‖1 + α‖ ≤ 1.[Exercise: why does this imply the ultrametric inequality?] So suppose ‖α‖ ≤ 1, let χ(X) be thecharacteristic polynomial of multiplication by α on E and let f(X) = Xr + fr−1Xr−1 + · · ·+ f0 bethe minimal polynomial of α, so that χ(X) = f(X)s, for s = n/r. Then

|fs0 | = | ±NE/F (α)| ≤ 1,

so |f0| ≤ 1. Now f(X) is irreducible so it is pure (i.e. the Newton polygon consists of a single linesegment). Since |1| = 1 and |f0| ≤ 1, this means |fi| ≤ 1, for all i, so that f(X) ∈ oF [X]. Henceχ(X) ∈ oF [X] also. Now

NE/F (1 + α) = det (In +Aα) = (−1)nχ(−1),

where In is the n× n identity matrix, so

‖1 + α‖ = |χ(−1)|1/n ≤ 1,

as required. �

This now completes the proof of Theorem 5.5. Since the valuation on E is unique, we will write itas | · | also, instead of ‖ · ‖. �

33

Chapter 6 Ramification

In this chapter, we investigate extensions of a non-archimedean local field F more closely. Sincesuch an extension E is itself a non-archimedean local field, we can compare the residue fields kEand kF and the value groups ΓE and ΓF . It turns out that kE is a finite extension of kF and ΓFhas finite index in ΓE .

6.1 Residue fields and unramified extensions

Let E be a finite extension of the non-archimedean local field F , of degree n = [E : F ]. Let ‖·‖ be avaluation on E, let oE , oF be the ring of integers in E,F respectively, let pE , pF be their respectivemaximal ideals and let kE , kF be their respective residue fields.

Lemma 6.1. There is a natural injection

kF ↪→ kE ,

so kE is an extension of kF . Further, the degree f = [kE : kF ] ≤ n.

Note that the notation f = f(E/F ) is standard; it is called the residue class degree of E/F .

Proof There is certainly an inclusion oF ↪→ oE ; but pF = oF ∩ pE so this induces the injectionkF ↪→ kE .Now let α1, ..., αn+1 be in kE ; we will show that they are linearly dependent over kF , so thatf = dim kF kE ≤ n. So let α̂1, ..., α̂n+1 be any elements of oE such that αi = α̂i + pE . Sincedim FE = n, they are linearly dependent so there are λi ∈ F , not all zero, such that

n+1∑i=1

λiα̂i = 0.

Without loss of generality, we may assume λn+1 6= 0 so, multiplying by λ−1n+1, we get

n∑i=1

µiα̂i + α̂n+1 = 0,

where µi = λiλ−1n+1. Now, if ¯ denotes the reduction modulo pE map, we get

n∑i=1

µ̄iαi + αn+1 = 0,

a non-trivial linear dependence. �

Definition 6.2. With notation as above:

(i) if f = f(E/F ) = n, we say that E/F is unramified ;

(ii) if f = f(E/F ) = 1, we say that E/F is totally ramified .

34

Lemma 6.3. Let F ⊆E ⊆K be a tower of finite extensions. Then the residue class degrees satisfy

f(K/F ) = f(K/E)f(E/F ).

Because we have restricted ourselves to looking at non-archimedean local fields, where the residuefields are finite (more pertinently, perfect), we avoid pathological behaviour and unramified exten-sions are easy to characterize. The important point is that, if α ∈ kE and q = #kE , then αq−1 = 1so the minimal polynomial φ(X) ∈ kF [X] of α is a factor of Xq−1 − 1; but this has no repeatedroots (as all q−1 elements of kE are roots) so φ(X) also has no repeated roots and hence f ′(α) 6= 0.

Theorem 6.4. Let α ∈ kE. Then there is an α̂ ∈ oE such that α = α̂+ pE and

[F (α̂) : F ] = [kF (α) : kF ].

Further, the field F (α̂) depends only on α.

Proof Let φ(X) ∈ kF [X] be the minimal polynomial of α and let Φ(X) ∈ oF [X] be any fixed liftof φ(X), that is:

• φ(X) and Φ(X) have the same degree;

• φ(X) = Φ̄(X), where ¯ represents the map reducing the coefficients modulo pE .

Note first that any root of Φ(X) reduces modulo pE to a root of φ(X), so the roots are in distinctresidue classes modulo pE . In particular, there is at most one root of Φ(X) which reduces to α.

Now let α̂0 ∈ oE be any element of the residue class α. Then Φ(α̂0) = φ(α) = 0, while Φ′(α̂0) =φ′(α) 6= 0 so

|Φ(α̂0)| < 1, |Φ′(α̂0)|.Now, applying Hensel’s Lemma to Φ(X), with F (α̂0) as groundfield, we find that there existsα̂ ∈ F (α̂0) ⊆ E such that

Φ(α̂) = 0; |α̂− α̂0| < 1.Hence α̂ is the unique root of Φ(X) in the residue class α so, since Φ(X) is irreducible, this α̂ is asrequired.

Now suppose α̂0 in the residue class of α also satisfies [F (α̂0) : F ] = [kF (α) : kF ]. By the above,α̂ ∈ F (α̂0) so F (α̂) ⊆ F (α̂0) and we have equality since the degrees are the same. �

Corollary 6.5. There is a bijection between the intermediate fields K (F ⊆ K ⊆ E) which areunramified over F and the fields k with kF ⊆ k ⊆ kE. The field corresponding to K is

k = kK = K ∩ oE mod pE .

Proof Let k be an intermediate field kF ⊆ k ⊆kE and let q = #k. Then k = kF (α), for someprimitive (q − 1)th root of unity α, so we can apply the previous Theorem. �

Corollary 6.6. There is an intermediate field L with the property that L/F is unramified andevery K⊆E which is unramified over F is contained in L. Further, E/L is totally ramified.

Proof L corresponds to kE in the previous Corollary. �

35

Corollary 6.7. Let F be a non-archimedean local field. For each n ≥ 1, there is a unique (uptoisomorphism) unramified extension E of F of degree n. It is the splitting field over F of Xq −X,where q = qnF and qF = #kF .

Proof Let E/F be an unramified extension of degree n. Then kE has q = qnF elements so Econtains a full set of (q − 1)th roots of unity, by Hensel’s Lemma (see Example 3.15(ii)). Inparticular, Xq −X splits in E so E contains its splitting field L. On the other hand qE = qL so,by Corollary 6.5, E = L. �

Corollary 6.8. Let f(X) ∈ oF [X] be a monic polynomial of degree n whose reduction modulo pFis irreducible. Then:

(i) if E = F (α) and α has minimal polynomial f(X) then E/F is unramified of degree n;

(ii) the splitting field of f(X) over F is unramified of degree n.

Proof First note that f(X) is irreducible (by Sheet 2, q.10).

(i) Note that kE ⊃ kF (α), which has degree n over kF so f(E/F ) ≥ n = [E : F ] and hence we haveequality.

(ii) Let L be the splitting field of f(X) over F and let α, β be roots of f(X) in L. Then, by (i),F (α) and F (β) are both unramified subextensions of degree n so, by Corollary 6.5, they are equal.This is true for any pair of roots so all the roots lie in F (α) and we have L = F (α). �

So we see that the every unramified extension of F is obtained by adjoining a root of unity of ordercoprime to p, and every such extension is unramified.

6.2 Totally ramified extensions

Now, for E/F a finite extension, we look at the relation between the value groups. Note first that,since E is a non-archimedean local field, its value group ΓE = {|x| : x ∈ E×} is a discrete (cyclic)subgroup of R×+. Hence we can define

Definition 6.9. The indexe = e(E/F ) = [ΓE : ΓF ]

is called the ramification index of E/F . The notation e is standard here.

So, if πF , πE are uniformizers for F,E respectively (that is, |πF | < 1 is a generator for the valuegroup ΓF ) then we have

|πF | = |πE |e.

We immediately have

Lemma 6.10. Let F ⊆E ⊆K be a tower of finite extensions. Then the ramification indices satisfy

e(K/F ) = e(K/E)e(E/F ).

The main result linking ramification indices and residue class degrees is the following:

36

Theorem 6.11. Let E/F be a finite extension of non-archimedean local fields of degree n. Then

n = ef,

where e = e(E/F ) and f = f(E/F ).

Proof Let πE be any uniformizer for E and let α̂1, ..., α̂f be any lift to oE of a basis for kE/kF(as in the proof of Theorem 6.4). Then we will prove that

B = {α̂iπjE : 1 ≤ i ≤ f, 0 ≤ j ≤ e− 1}

is a basis for E/F . [In fact, we will prove that it is an oF -basis for oE .]

First suppose that B is not linearly independent over F . Then we have aij ∈ F , not all zero, suchthat ∑

i,j

aijα̂iπjE = 0. (†)

Dividing by the largest non-zero coefficient aij , we may assume

maxi,j

|aij | = 1,

so there are integers I, J such that

|aij | ≤ |πF |, for 1 ≤ i ≤ f, j < J|aIJ | = 1.

Then, if we reduce∑

i aiJ α̂i modulo pE , we have a non-zero coefficient āIJ so, since the α̂i (mod pE)are (in particular) linearly independent over kF , this reduction is non-zero. Hence

|∑

i

aiJ α̂i| = 1

and we get

|∑

i

aijα̂iπjE |

≤ |πF | = |πE |e for j < J ;= |πE |J for j = J ;≤ |πE |J+1 for j > J.

But then one term in the sum (†) is bigger than all the rest, a contradiction.Now let x ∈ E and we will show that x is in the linear span of B. Multiplying by a suitable power ofπF , we reduce to the case x ∈ oE . [If πnFx =

∑i,j aijα̂iπ

jE , with aij ∈ F then, putting bij = π

−nF aij ,

we get x =∑

i,j bijα̂iπjE .] Since the αi = α̂i (mod pE) form a basis for kE/kF , there are ci0 ∈ kF

such thatx =

∑i

ci0αi,

where x denotes the reduction modulo pE , as usual. Choosing any lifts ĉi0 ∈ oF , we have

x−∑

i

ĉi0α̂i = πEx1 ∈ pE = πEoE ,

37

for some x1 ∈ oE . Now we repeat the process with x1 and so on, until we have obtained aij ∈ oFsuch that

x−e−1∑j=0

∑i

ĉijα̂iπjE = π

eExe,

for some xE ∈ oE . Now |πE |e = |πF | so we have

πeExe = πFx(1),

for some x(1) ∈ oE . Now we start again with x(1) in place of x; we find in succession linearcombinations Cr =

∑e−1j=0

∑i ĉ

(r)ij α̂iπ

jE of B with coefficients in oF such that

x− C0 − πFC1 − ...− πsFCs ∈ πs+1F oE ,

for every s ≥ 0. Now we let s→∞ and, using completeness, we put aij =∑∞

r=0 ĉ(r)ij π

rF ; then

x =∑i,j

aijα̂iπjE ,

as required. �

Corollary 6.12. A finite extension E/F of degree n is unramified if and only e(E/F ) = 1; it istotally ramified if and only if e(E/F ) = n.

We end this section by seeing that totally ramified extensions are, like unramified ones, fairly easyto describe (though not quite as explicitly as in the unramified case). Since every finite extensionis built up of an unramified one and a totally ramified one, this is quite useful.

Recall that f(X) = fnXn + · · ·+ f1X + f0 ∈ oF [X] is called an Eisenstein polynomial if

|fn| = 1; |fj | < 1 for j < n; |f0| = |πF |. (‡)

Theorem 6.13. Let E/F be a finite extension of non-archimedean local fields. Then E/F is totallyramified if and only if E = F (β), where β is a root of an Eisenstein polynomial.

Proof Suppose first that E = F (β), where β is the root of an Eisenstein polynomial, say

fnβn + · · ·+ f1β + f0 = 0,

where the coefficients satisfy (‡). Then |βn| < 1 so |β| < 1. Hence the last term in the sum isbigger than all other terms, except possibly the first term. Since the sum is zero, the first and lastterms must then have the same size so |β|n = |πF |. Hence e(E/F ) ≥ n = [E : F ] and we must haveequality, with E/F totally ramified.

Conversely, suppose E/F is totally ramified of degree n and let β = πE be a uniformizer for E.Then 1, πE , ..., πn−1E are linearly independent over F (as in the proof of Theorem 6.11); hence theremust be an equation

βn + fn−1βn−1 + · · ·+ f0 = 0,with fj ∈ F . Two of the summands must have the same valuation and this must be the first and thelast (since the others all lie in different cosets of ΓE/ΓF ). Hence |f0| = |πE |n = |πF | and |fj | < 1,for all j. �

Note that this does NOT give an element β ∈ E such that βn = πF . If E/F is tamely ramified(that is e(E/F ) is coprime to p = char kF ), then it is possible to get some way towards this (seeproblem sheet).

38

6.3 Examples

We will look at the splitting fields E of polynomials over Qp, for various p. The questions we willbe asking are:

• What are the degree [E : Qp], the ramification index e(E/Qp) and the residue class degreef(E/Qp)?

• What is the maximal unramified subextension L? (That is, the field F ⊆ L ⊆ E such thatL/F is unramified and E/L is totally ramified.)

f1(X) = X4 − 2X3 − 17X2 + 22X + 66

We begin with p = 2. The first thing to see is whether the given polynomial is irreducible. For adegree 4 polynomial, this is not so easy but we can use the Newton polygon to help us. We seethat f1(X) is of type (2, 12 log 2; 2, 0) so it factorizes as the product of two quadratics. Given this(and assuming the question is designed not to be too horrible) we can look for such a factorizationwith integer coefficients and we see that

f1(X) = (X2 − 2X − 6)(X2 − 11) = g(X)h(X).

These two quadratics are irreducible in Q2, g(X) by Eisenstein’s criterion and h(X) because 11 6≡ 1(mod 8) so 11 is not a square in Q2.Let α be a root of g(X) and β a root of h(X) in E. Then Q2(α)/Q2 is totally ramified of degree2, by Theorem 6.13.

What about Q2(β)/Q2? We notice that β − 1 satisfies the polynomial

h(X + 1) = (X + 1)2 − 11 = X2 + 2X − 10,

and Eisenstein polynomial. Hence Q2(β) = Q2(β − 1) is also totally ramified of degree 2 over Q2.So what does this tell us about the extension E/Q2? Well, certainly e(E/Q2) is a multiple of 2and the degree is at most 4 but that’s all we know for now.

Consider next γ = α − 1; this satisfies g(X + 1) = X2 − 7 so γ2 = 7. Now put δ = βγ, so thatδ2 = 77 and Q2(δ) is the “other” intermediate extension of degree 2. We claim that Q2(δ)/Q2 isunramified of degree 2. In that case, we must have

[E : Q2] = 4, e(E/Q2) = 2, f(E/Q2) = 2, L = Q2(δ).

To prove that Q2(δ)/Q2 is unramified we need to show that Q2(δ) is the field obtained from Q2 byadjoining a primitive (22 − 1)th root of unity, i.e. we need to show that Q2(δ) contains a root of

φ3(X) = X2 +X + 1.

We can do this by applying Hensel’s Lemma to x0 = δ−12 . (Exercise)

For odd p, we first use that E = Qp(γ, β) (notation as in the case p = 2), with γ2 = 7 and β2 = 11.(So E is the splitting field of (X2 − 7)(X2 − 11).)p=3 7 is a quadratic residue mod 3, so γ ∈ Q3, while 11 is not, so X2 − 11 is irreducible in F3[X]and E = Q3(β) is unramified of degree 2 over Q3, by Corollary 6.8.

39

p=5 Here 11 is a quadratic residue but 7 is not so, as above, E = Q5(γ) is unramified of degree 2over Q5.p=7 Here 11 is a quadratic residue. On the other hand, X2 − 7 is an Eisenstein polynomial soE = Q7(γ) is totally ramified of degree 2 over Q7.p=11 Here 7 is a quadratic non-residue so Q11(γ) is unramified of degree 2 over Q11. On the otherhand, X2 − 11 is an Eisenstein polynomial so E = Q11(β) is totally ramified of degree 2 over Q11.Hence

[E : Q11] = 4, e(E/Q11) = 2, f(E/Q11) = 2, L = Q11(γ).

p=13 Both 7 and 11 are quadratic non-residues so Q13(γ) and Q13(β) are both unramified ofdegree 2 over Q13. But there is a unique unramified extension of degree 2 over Q13 so we haveE = Q13(γ) = Q13(β) is unramified of degree 2 over Q13.p=19 Both 7 and 11 are quadratic residues so γ, β ∈ Q19 and E = Q19 is a degree 1 extension!Every other odd prime p will behave like 3, 5, 13 or 19.

f2(X) = X3 − 2

We note first that E = Qp(α, ωα, ω2α) = Qp(α, ω), where α3 = 2 and ω is a primitive cube root of1 (that is, ω2 + ω + 1 = 0). In particular, this gives that the degree of the extension is at most 6.(In fact, divides 6 = 3!, from Galois Theory.)

p=2 f2(X) is an Eisenstein polynomial so Q2(α) is totally ramified of degree 3 over Q2. On theother hand, Q2(ω) is unramified of degree 2 over Q2. Hence

[E : Q2] = 6, e(E/Q2) = 3, f(E/Q2) = 2, L = Q2(ω).

p=3 There is no root of X3 − 2 in Q3 so it is irreducible and Q3(α)/Q3 is of degree 3. In fact,γ = α+ 1 satisfies

f2(X − 1) = X3 − 3X2 + 3X − 3,

an Ei