Table Of Content4A22 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 Q ............................................................16
p
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 C ....................................................................45
p
Chapter 8 Action of Galois
§8.1 Definitions ........................................................................47
§8.2 Kronecker-Weber Theorem ........................................................47
1
Introduction
ThesearenotesfortheMTH4A22SpecialPureMathematicsCourseinthespringsemesterof2003.
The subject for the course is Local Fields, in particular the p-adic fields and their extensions. The
main aim of the course is (hopefully) the Kronecker-Weber Theorem for abelian extensions of Q .
p
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 no
simultaneous 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” a
solution modulo 7n to one modulo 7n+1:
• for n = 1, put x = 3;
1
• as an example, we will do the n = 2 step. (Exercise: do the general inductive step.) We
have our solution x = 3 modulo 7 so we try putting x = 3+7y into the equation modulo
1 2 2
72 = 49. This gives
(3+7y )2 = 9+42y +49y2 ≡ 2 (mod 49).
2 2 2
The last term is zero and we rearrange to get
42y ≡ −7 ≡ 42 (mod 49) ⇒ 6y ≡ 6 (mod 7) ⇒ y ≡ 1 (mod 7),
2 2 2
so x = 3+7.1 = 10 is a solution modulo 49.
2
However, there is no simultaneous solution, since such an x would satisfy x2 = 2, which has no
solution even in Q.
The solutions to these two congruences can be written in a “base p” expansion: in the first case it
is 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 simultaneous
solutions 1+3+32 +··· and 3.1+1.7+2.72 +6.73 +···. Of course, this seems to be nonsense
since these sums don’t converge in R, but perhaps we can begin to make sense of it by replacing
our prime p by an indeterminate X; then we are looking at some power series 1+X+X2+··· and
3+X +2X2+6X3+···, which are things which at least mean something. Of course, if a power
series is actually finite (in other words, it is a polynomial), then we can reverse this by replacing
X 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 polynomials
with 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 ∈ Z
and b 6= 0; similarly, attached to C[X], we have the field of rational functions C(X), which consists
of quotients P(X)/Q(X) of polynomials P(X),Q(X), with Q(X) 6= 0. Both rings are also unique
2
factorization domains: any integer can be uniquely expressed as ±1 times a product of prime
numbers; any polynomial can be uniquely expressed (upto order) as
P(X) = a(X −α )(X −α )···(X −α ),
1 2 n
where a,α ,α ,...,α ∈ C. This gives us the main point of Hensel’s analogy: the prime numbers
1 2 n
are analogous to the linear polynomials X − α ∈ C[X]. Indeed, in the language of rings, both
generate 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”
n
X
m = a +a p+a p2+···+a pn = a pi,
0 1 2 n i
i=0
with a ∈ Z and 0 ≤ a ≤ p−1. Similarly, given a polynomial P(X) and α ∈ C, we can write the
i i
(finite) Taylor expansion
n
X
P(X) = a +a (X −α)+a (X −α)2+···+a (X −α)n = a (X −α)i,
0 1 2 n i
i=0
with a ∈ C. Now, for polynomials and their quotients, we can push this much further: given
i
f(X) ∈ C(X), there is always an expression
P(X) X
f(X) = = a (X −α)n0 +a (X −α)n0+1+··· = a (X −α)i.
Q(X) n0 n0+1 i
i≥n0
This is just the Laurent expansion from complex analysis. Of course, in complex analysis we would
also worry about the convergence of this series, but here we are just interested in the series as a
formal 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 do
this 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 rational
number 1/4; we have 4 = 1+p so
1 1
= = 1+2p+2p3+2p5+··· .
4 1+p
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+···
| {z }
= 1+p2+2p2+2p3+2p4+2p5+2p6+···
| {z }
= ··· = 1
so that the higher powers of p disappear off “to the right”! (Exercise: check that our “3-adic”
expansion for x = −1 satisfies 2x+1 = 0 and that our “7-adic” expansion for a square root y of 2
2
satisfies y2 = 2.)
∗An ideal I/R is prime if ab∈I implies a∈I or b∈I.
3
Treating the whole process formally, we can see that any positive rational x can be expanded in
this way (exercise?)
a X
x = = a pn, with a ∈ {0,1,...,p−1}.
n n
b
n≥n0
The value of n reflects the multiplicity with which p divides x
0
a
x = pn0 1, with (ab,p) = 1.
b
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 in
p” – 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 Q of p-adic numbers;
p
so we have an inclusion of fields
Q ,→ Q .
p
This definition of Q is rather formal and unenlightening – and even proving that Q is a field is
p p
rather ugly – so we will introduce a more abstract and conceptual approach in the following. We
note here that any definition which is going to make sense of these power series in p must have
pn → 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] Gouveˆa 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. & Fro¨hlich 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 well
mean you will sometimes need to look things up in the literature, but you should really be doing
this anyway, to get another point of view.
4
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 the
topology induced by a valuation, the independence of inequivalent valuations and the notion of
completeness with respect to a valuation.
F will denote an arbitrary field for now, though you may find it helpful to keep the example of Q
in mind.
1.1 Valuations
Definition 1.1. A (multiplicative) valuation (or absolute value) on F is a map |·| : F → R such
+
that, 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 given
by the inclusion Q ,→ R, with the usual absolute value on R. It is an archimedean valuation
and 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 will
have to be excluded from most of the theory that we develop. Exercise: show that the only
valuation 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. Let
F = Q and let p be a prime. For x ∈ Q\{0}, we write
a
x = pn , with (ab,p) = 1,
b
and put v (x) = n; we also put v (0) = +∞ (since 0 is divisible by an arbitrarily large power
p p
of p). Then, for all x,y ∈ Q, we have (exercise)
v (xy) = v (x)+v (y), and v (x+y) ≥ min{v (x),v (y)}. (1.3)
p p p p p p
We now define the p-adic valuation of x ∈ Q to be
|x| = p−vp(x),
p
5
with the convention that |0| = 0. That this is a non-archimedean valuation follows immedi-
p
ately from (1.3). We also note that
|pn| = p−n,
p
so that, with this p-adic notion of size, pn → 0 as n → ∞.
We remark that the map v is often called an additive valuation and that the theory of non-
p
archimedean local fields can be developed taking this notion as the primitive one and defining
multiplicative valuations from them – some books do this.
P(T)
(iv) Let K be any field and put F = K(T) = { : P(T),Q(T) ∈ K[T],Q(T) 6= 0}. For
Q(T)
P(T) ∈ K[T], we define v (P(T)) = −deg(P(T)) and we extend this to rational functions
∞
P(T)
by setting v (0) = +∞ and, for f(T) = ,
∞ Q(T)
(cid:18) (cid:19)
P(T)
v (f(T)) = v = v (P(T))−v (Q(T)) = deg(Q(T))−deg(P(T)).
∞ ∞ ∞ ∞
Q(T)
Itiseasytocheckthatthisiswell-definedandthatv satisfiestheequations(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, a
nice 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 the
p(T)-adic valuation on F. [Hint: (iv) almost does this for the polynomial p(T) = T... watch
out for the signs]
(vi) Exercise: in (v), K is a subfield of F so the absolute value on F defines one on K. What is
it?
(vii) Exercise: find an archimedean valuation on Q(T); if K is a finite field, is it possible to find
an 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, which
is impossible). The rest follows easily (as −x = −1.x and every element of a finite field satisfies
xn = 1 for some n). (cid:4)
The following gives a criterion to determine whether or not a valuation is non-archimedean.
6
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
n (cid:18) (cid:19)
X n
|a+b|n = |(a+b)n| = ajbn−j by the binomial expansion
j
j=0
n (cid:12)(cid:18) (cid:19)(cid:12)
≤ X(cid:12)(cid:12) n (cid:12)(cid:12)|a|j|b|n−j by the ultrametric inequality
(cid:12) j (cid:12)
j=0
n
X
≤ |a|j|b|n−j
j=0
≤ (n+1)(max{|a|,|b|})n.
Taking nth roots of both sides, letting n → ∞ and observing that (n+1)1/n → 1 as n → ∞, we
get |a+b| ≤ max{|a|,|b|}, as required. (cid:4)
This helps to explain the word non-archimedean: a valuation is archimedean if it has the following
property:
Archimedean Property: given x,y ∈ F with x 6= 0, there exists a positive
integer 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. (cid:4)
Corollary 1.7. If K is a subfield of F and |·| is a valuation on F, then |·| is non-archimedean
on 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 us
to define open and closed sets, as, for example, in C. We will now investigate this topology, with
some 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 distance
d(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;
7
(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 ofdistance givenby non-archimedeanvaluations is a ratherstrange
one:
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. (cid:4)
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 open
if and only if, for any x ∈ U, there is an open ball around x contained inside U; a set S is closed if
and 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 of
+
radius 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.
8
(ii) Two open (resp. closed) balls have non-empty intersection if and only if one is contained in
the 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) so
neither ball is contained in the other and hence they are disjoint. Hence B(b,r)⊆F \B(a,r) and
we see that the complement of B(a,r) is open. (cid:4)
The fact that there are so many sets which are both open and closed is what makes the topology
so 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 sets
that contain x; this is the largest connected set containing x, since the union of two non-disjoint
connected sets is connected (Exercise). For example, if F = R with the usual valuation then the
connected component of x is R (since it is connected). In the non-archimedean case, things are
very different:
Proposition 1.15. Let F be a field with a non-archimedean valuation | ·| and let x ∈ F. The
connected component of x is {x}.
Proof Let S be a set containing x and some other point y and put r = |y −x|. Then, putting
U = B(x,r) and V = F \U, we see that S is disconnected. (cid:4)
In the language of topology, this says that F is a totally disconnected topological space. Provided
the valuation is not trivial (in which case the topology is discrete – all sets are open), this means
there 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 as
being the same – that is the notion of equivalence of valuations. We will also show that, when two
valuations are not equivalent, they are really very different.
9
Definition 1.16. Two valuations |·| and |·| on a field F are equivalent if they define the same
1 2
topology on F – that is, every set that is open with respect to one is also open with respect to the
other.
Note that, given a valuation |·| on a field F, a sequence {a } in F converges to a in the induced
n
topology if and only if, for all ε > 0, there exists N such that, for n > N, |a −a| < ε; equivalently,
n
if and only if, for all open sets U containing a, there exists N such that, for n > N, a ∈ U. In
n
particular, the notion of convergence of a sequence depends only on the topology induced by the
valuation.
Proposition 1.17. Let |·| and |·| be valuations on a field F, with |·| non-trivial. The following
1 2 1
are equivalent:
(i) |·| and |·| are equivalent;
1 2
(ii) for any x ∈ F, we have |x| < 1 ⇒ |x| < 1;
1 2
(iii) there exists a positive real number α such that, for all x ∈ F,
|x| = |x|α.
1 2
Proof (iii)⇒(i) We have |x−a| < r ⇐⇒ |x−a| < rα so that any open ball with respect to
2 1
|·| is also a open ball with respect to |·| (albeit of a different radius).
2 1
(i)⇒(ii) We have |x| < 1 if and only if xn → 0 with the |·| -topology, if and only if xn → 0 with
1 1
the |·| -topology, if and only if |x| < 1.
2 2
(ii)⇒(iii) If |a| > 1 then |a−1| < 1 so we have |a−1| < 1 and hence |a| > 1. We also claim that
1 1 2 2
|a| = 1 ⇒ |a| = 1. (∗)
1 2
For suppose |a| = 1 but |a| 6= 1. By considering a−1 if necessary, we may (and do) assume
1 2
|a| > 1. Now let b ∈ F be such that |b| < 1, which is possible since |·| is non-trivial. Then,
2 1 1
putting c = ban, we have |c| < 1 but, for n large enough, |c| > 1, contradicting the assumption
1 2
of (ii).
Now pick (and fix) a ∈ F \{0} such that |a| < 1, so that |a| < 1 also, and put α = log|a|1 > 0,
1 2 log|a|2
so that |a| = |a|α. For b ∈ F \{0} we have three possibilities:
1 2
(1) |b| = 1; then |b| = 1 also, by (∗), so |b| = |b|α.
1 2 1 2
(2) |b| < 1; then |b| < 1 also and we put β = log|a|1 and β = log|a|2. We will show that β = β ,
1 2 1 log|b|1 2 log|b|2 1 2
for then log|b|1 = log|a|1 = α and |b| = |b|α.
log|b|2 log|a|2 1 2
Suppose β > β (the case β < β is very similar); then there exists a rational number m such
1 2 1 2 n
that β < m < β . Now consider x = anb−m with respect to the two valuations:
2 n 1
log|x| = nlog|a| −mlog|b| = nlog|b| (cid:0)β − m(cid:1) < 0
1 1 1 1 1 n
so that |x| < 1; on the other hand, the same calculation for the second valuation gives |x| > 1,
1 2
contradicting our assumption.
(3) |b| > 1; then |b| > 1 also and the proof is essentially the same as case (2). (cid:4)
1 2
10
