Algebraic Number Theory II: Valuations, Local Fields and Adeles [Lecture notes]

135 Pages·2017·0.966 MB·English

by  Pete L. Clark

Algebraic Number Theory II: Valuations, Local
Fields and Adeles
Pete L. Clark
Thanks to Tom Church, Rankeya Datta, John Doyle, David Krumm and Todd
Trimble for pointing out typos.
Contents
Chapter 1. Valuation Theory 5
1. Absolute values and valuations 5
2. Completions 19
3. Extending Norms 35
4. The Degree In/Equality 48
Chapter 2. Local Fields 53
1. Hensel’s Lemma 53
2. Locally Compact Fields 65
3. Structure Theory of CDVFs 74
Chapter 3. Adeles 83
1. Introducing the Adeles 83
2. The Adelic Approach to Class Groups and Unit Groups 91
3. Ray Class Groups and Ray Class Fields 97
Chapter 4. Complements and Applications 109
1. Mahler Series 109
2. Monsky’s Theorem 115
3. General Linear Groups Over Locally Compact Fields 120
4. The Cassels Embedding Theorem 126
5. Finite Matrix Groups 129
Bibliography 133
3

CHAPTER 1
Valuation Theory
1. Absolute values and valuations
1.1. Basic definitions.
All rings are commutative with unity unless explicit mention is made otherwise.
A norm, on a field k is a map |·|:k →R≥0 satisfying:
(V1) |x|=0 ⇐⇒ x=0.
(V2) ∀x,y ∈k, |xy|=|x||y|.
(V3) ∀x,y ∈k, |x+y|≤|x|+|y|.
Example 1.1.0: On any field k, define | | : k → R≥0 by 0 (cid:55)→ 0, x ∈ k\{0} (cid:55)→ 1.
0
Thisisimmediatelyseentobeanabsolutevalueonk (Exercise!),calledthetrivial
norm. In many respects it functions as an exception in the general theory.
√
Example 1.1.1: The standard norm on the complex numbers: |a+bi|= a2+b2.
The restriction of this to Q or to R will also be called “standard”.
Example 1.1.2: The p-adic norm on Q: write a = pnc with gcd(p,cd) = 1 and
b d
put |a| =p−n.
b p
We will see more examples later. In particular, to each prime ideal in a Dedekind
domain we may associate a norm. This remark serves to guarantee both the pleni-
tude of examples of norms and their link to classical algebraic number theory.
Exercise 1.1. Let |·| be a norm on the field k. Show that for all a,b ∈ k,
||a|−|b||≤|a−b|.
Exercise 1.2. Let R be a ring in which 1 (cid:54)= 0. Let |·| : R → R≥0 be a map
which satisfies (V1) and (V2) (with k replaced by R) above.
a) Show that |1|=1.
b) Show that R is an integral domain, hence has a field of fractions k.
c)Showthatthereisauniqueextensionof|·|tok, thefractionfieldofR, satisfying
(V1) and (V2).
d) Suppose that moreover R satisfies (V3). Show that the extension of part b) to k
satisfies (V3) and hence defines a norm on k.
e) Conversely, show that every integral domain admits a mapping |·| satisfying
(V1), (V2), (V3).
5
6 1. VALUATION THEORY
Exercise 1.3.
a) Let | | be a norm on k and x∈k a root of unity.1 Show |x|=1.
b) Show that for a field k, TFAE:
(i) Every nonzero element of k is a root of unity.
(ii) The characteristic of k is p>0, and k/F is algebraic.
p
c) If k/F is algebraic, show that the only norm on k is |·| .
p 0
Remark: In Chapter 2 we will see that the converse of Exercise 1.3c) is also true:
any field which is not algebraic over a finite field admits at least one (and in fact
infinitely many) nontrivial norm.
Exercise 1.4. Let (k,| · | be a normed field, and let σ : k → k be a field
automorphism. Define σ∗| |:k →R by x(cid:55)→|σ(x)|.
a) Show that σ∗|·| is a norm on k.
b) Show that this defines a left action of Aut(k) on the set of all norms on k which
preserves equivalence.
√
c) Let d be a squarefree integer, not equal to 0 or 1. Let k = Q( d) viewed as a
√
subfield of C (for specificity, when d < 0, we choose d to lie in the upper half
plane, as usual), and let |·| be the restriction of the standard valuation on C to k.
√ √
Let σ : d(cid:55)→− d be the nontrivial automorphism of k. Is σ∗| |=|·|? (Hint: the
answer depends on d.)
1.2. Absolute values and the Artin constant.
For technical reasons soon to be seen, it is convenient to also entertain the fol-
lowing slightly version of (V3): for C ∈R>0, let (V3C) be the statement
(V3C) ∀x∈k, |x|≤1 =⇒ |x+1|≤C.
A mapping |·| : k → R≥0 satisfying (V1), (V2) and (V3C) for some C will be
called an absolute value.
For an absolute value | | on a field k, we define the Artin constant C to be
k
the infimum of all C ∈R>0 such that |·| satisfies (V3C).
Exercise 1.5. Let |·| be an absolute value on k.
a) Show that |·| satisfies (V3C) for some C, then C ≥1.
b) Let C be the Artin constant. Show that |·| satisfies (V3C ).
k k
c) Compute C for the standard norm on C and the p-adic norms on Q.
k
Lemma 1.1. Let k be a field and |·| an absolute value, and C ∈[1,∞). Then
the following are equivalent:
(i) ∀x∈k, |x|≤1 =⇒ |x+1|≤C.
(ii) ∀x,y ∈k, |x+y|≤Cmax(|x|,|y|).
Proof. Assume(i)andletx,y ∈k. Withoutlossofgeneralitywemayassume
that 0<|x|≤|y|. Then |x|≤1, so |x +1|≤C. Multiplying through by |y| gives
y y
|x+y|≤C|y|=Cmax(|x|,|y|).
1I.e.,thereexistsk∈Z+ suchthatxk=1.
1. ABSOLUTE VALUES AND VALUATIONS 7
Now assume (ii) and let x∈k be such that|x|≤1. Then
|x+1|≤Cmax(|x|,|1|)=Cmax(|x|,1)=C.
(cid:3)
Lemma 1.2. Let k be a field and |·| an absolute value with Artin constant C.
Then |·| is a norm iff C ≤2.
Proof. (=⇒) Let |·| be a norm on k, and let x ∈ k be such that |x| ≤ 1.
Then
|x+1|≤|x|+|1|=|x|+1≤1+1=2.
(⇐=) Suppose C ≤ 2. Let x,y ∈ k. Without loss of generality, we may assume
that 0 < |x| ≤ |y|. Then |x| ≤ 1, so |1+ x| ≤ C ≤ 2. Multiplying through by y,
y y
we get |x+y|≤2|y|=2max(|x|,|y|). Applying this reasoning inductively, we get
that for any x ,...,x ∈k with 0<|x |≤...≤|x |, we have
1 2n 1 2n
|x +...+x |≤2nmax|x |.
1 2n i
i
Let r be an integer such that n≤2r <2n. Then
(1) |x +...+x |=|x +...+x +0+...+0|≤2rmax|x |≤2nmax|x |.
1 n 1 n i i
i i
Applying this with x = ... = x = 1 gives that |n| ≤ 2n. Moreover, by replacing
1 n
the max by a sum, we get the following weakened version of (9):
n
(cid:88)
|x +...+x |≤2n |x |.
1 n i
i=1
Finally, let x,y ∈k be such that 0<|x|≤|y|. Then for all n∈Z+,
(cid:18) (cid:19) n (cid:18) (cid:19)
(cid:88) n (cid:88) n
|x+y|n =| xiyn−i|≤2(n+1) | ||x|i|y|n−i
i i
i=0n i=0
n (cid:18) (cid:19)
(cid:88) n
≤4(n+1) |x|i|y|n−i =4(n+1)(|x|+|y|)n.
i
i=0
Taking nth roots and the limit as n→∞ gives |x+y|≤|x|+|y|. (cid:3)
Why absolute values and not just norms?
Lemma 1.3. Let |·| : k → R≥0 be an absolute value with Artin constant C.
Put
|·|α :k →R≥0, x(cid:55)→|x|α.
a) The map |·|α is an absolute value with Artin constant Cα.
b) If |·| is a norm, |·|α need not be a norm.
Exercise 1.6. Prove Lemma 1.3.
This is the point of absolute values: the set of such things is closed under the op-
eration of raising to a power, whereas the set of norms need not be.
Moreover, Lemma 1.3 suggests a dichotomy for absolute values. We say an ab-
solute value is non-Archimedean if the Artin constant is equal to 1 (the smallest
possible value). Conversely, if the Artin constant is greater than one, we say that
the Artin constant is Archimedean.
8 1. VALUATION THEORY
Forexample,onk =Q,p-adicnorm|·| isnon-Archimedean,whereasthestandard
p
absolute value |·| is Archimedean with Artin constant 2.
∞
Exercise 1.7. Let | | be an absolute value on l, and let k be a subfield of l.
a) Show that the restriction of |·| to k is an absolute value on k.
b) If |·| is a norm on l, then the restriction to k is a norm on k.
1.3. Equivalence of absolute values.
Two absolute values |·| , |·| on a field k are equivalent if there exists α∈R>0
1 2
such that |·| =|·|α. When convenient, we write this as |·| ∼|·| .
2 1 1 2
By a place on a field k, we mean an equivalence class of absolute values.2 It
is easy to check that this is indeed an equivalence relation on the set of absolute
values on a field k. Moreover, immediately from Lemma 1.3 we get:
Corollary 1.4. Each absolute value on a field is equivalent to a norm.
Theorem 1.5. Let |·| , | | be two nontrivial absolute values on a field k.
1 2
TFAE:
(i) There exists α∈R>0 such that |·|α =|·| .
1 2
(ii) ∀x∈k, |x| <1 =⇒ |x| <1.
1 2
(iii) ∀x∈k, |x| ≤1 =⇒ |x| ≤1.
1 2
(iv) ∀x∈k, all of the following hold:
|x| <1 ⇐⇒ |x| <1,
1 2
|x| >1 ⇐⇒ |x| >1,
1 2
|x| =1 ⇐⇒ |x| =1.
1 2
Remark: Thismayseemlikeastrangewaytoorganizetheequivalences, butitwill
be seen to be helpful in the proof, which we give following [Wei, Thm. 1-1-4].
Proof. We shall show (i) =⇒ (ii) =⇒ (iii) =⇒ (iv) =⇒ (i). That (i)
=⇒ (ii) (and, in fact, all the other properties) is clear.
(ii) =⇒ (iii): let x∈k be such that |x| =1. We must show that |x| =1. Since
1 2
|·| is nontrivial, there exists a ∈ k with 0 < |a| < 1, and then by (ii) we have
1 1
0 < |a| < 1. Then, for all n ∈ Z+, |xna| < 1, so |xna| < 1, so |x| < |a|−n1.
2 1 2 2 2
Taking n to infinity gives |x| ≤ 1. We may apply the same argument to x−1,
2
getting |x| ≥1.
2
(iii) =⇒ (iv): Choose c∈k such that 0<|c| <1. Then for sufficiently large n,
2
xn
|x| <1 =⇒ |x|n ≤|c| =⇒ | | ≤1 =⇒ |x|n ≤|c| <1 =⇒ |x| <1.
1 1 1 c 1 2 2 2
So far we have shown (iii) =⇒ (ii). As in the proof of (ii) =⇒ (iii) we have
|x| =1 =⇒ |x| =1. Moreover
1 2
1 1
|x| >1 =⇒ | | <1 =⇒ | | <1 =⇒ |x| >1.
1 x 1 x 2 2
2Warning: inmoreadvancedvaluationtheory,onehasthenotionofaK-placeofafieldk,a
related but distinct concept. However, in these notes we shall always use place in the sense just
defined.
1. ABSOLUTE VALUES AND VALUATIONS 9
This establishes (iv).
(iv) =⇒ (i): Fix a∈k such that |a| <1. Then |a| <1, so
1 2
log|a|
α= 2 >0.
log|a|
1
We will show that |·| =|·|α. For this, let x∈k, and put, for i=1,2,
2 1
log|x|
γ = i.
i log|a|
i
It suffices to show γ =γ . Let r = p be a rational number (with q >0). Then
1 2 q
p
r = ≥γ ⇐⇒ plog|a| ≤qlog|x|
q 1 1 1
xq xq
⇐⇒ |ap| ≥|xq| ⇐⇒ | | ≤1 ⇐⇒ | | ≤1
1 1 ap 1 ap 2
p
⇐⇒ plog|a| ≥qlog|x| ⇐⇒ ≥γ . (cid:3)
2 2 q 2
Exercise 1.8. Let | · | be an absolute value on a field k. Show that | · |
is Archimedean (resp. non-Archimedean) iff every equivalent absolute value is
Archimedean (resp. non-Archimedean).
1.4. Artin-Whaples Approximation Theorem.
Theorem1.6. (Artin-Whaples) Letkbeafieldand|| ,...,|| beinequivalent
1 n
nontrivial norms on k. Then for any x ,...,x ∈ k and any (cid:15) > 0, there exists
1 n
x∈k such that
∀ i, 1≤i≤n, |x−x | <(cid:15).
i i
Proof. Our proof closely follows [A, §1.4].
Step1: Weestablishthefollowingspecialcase: thereexistsa∈ksuchthat|a| >1,
1
|a| <1 for 1<i≤n.
i
Proof: Wegobyinductiononn. Firstsupposen=2. Then, since|·| and|·| are
1 2
inequivalent and nontrivial, by Theorem 1.5 there exist b,c∈k such that |b| <1,
1
|b| ≥1, |c| ≥1, |c |<1. Put a= c.
2 1 2 b
Now suppose the result holds for any n−1 norms, so that there exists b ∈ k
with |b| > 1 and |b| < 1 for 1 < i ≤ n−1. Using the n = 2 case, there is c ∈ k
1 i
such that |c| >1 and |c| <1.
1 n
Case 1: |b| ≤ 1. Consider the sequence a = cbr. Then for all r ∈ Z+ we have
n r
|a | >1 while |a | <1. For sufficiently large r, |a | <1 for all 2≤i≤n, so we
r 1 r n r i
may take a=a .
r
Case 2: |b| >1. This time, for r ∈Z+, we put
n
cbr
a = .
r 1+br
Then for i=1 and i=n,
|br−(1+br)| |c|
lim |a −c| = lim |c| i = lim i =0,
r→∞ r i r→∞ i |1+br|i r→∞|1+br|i
so for sufficiently large r we have
|a | =|c| >1 and |a | =|c| <1.
r 1 1 r n n
10 1. VALUATION THEORY
On the other hand, for 1<i<n,
|c| |b|r
|a | = i i ≤|c| |b |r <1.
r i |1+br| i i
i
Therefore we may take a=a for sufficiently large r.
r
Step 2: We claim that for any δ >0, there exists a∈k such that ||a| −1|<δ and
1
|a| <δ for 1<i≤n.
i
Proof: If b is such that |b| > 1 and |b| < 1 for 1 < i ≤ n, then the computations
1 i
of Step 1 show that we may take a = br for sufficiently large r.
r 1+br
Step 3: Fix δ > 0. By Step 2, for each 1 ≤ i ≤ n, there exists a ∈ k such that
i
||a| −1|<δ and for all j (cid:54)=i, |a | <δ. Put A=max |x | . Take
i i j i,j i j
x=a x +...+a x .
1 1 n n
Then
(cid:88)
|x−x | ≤|a x −x | + |a x | ≤Aδ+(n−1)Aδ =nAδ.
i i i i i i j j i
j(cid:54)=i
Thus taking δ < (cid:15) does the job. (cid:3)
nA
Remark: Theorem1.6alsogoesbythenameweakapproximation. Byanyname,
itisthemostimportantelementaryresultinvaluationtheory,playingarolehighly
analogous to that of the Chinese Remainder Theorem in commutative algebra. On
other hand, when both apply the Chinese Remainder Theorem is subtly stronger,
in a way that we will attempt to clarify at little later on.
1.5. Archimedean absolute values.
In the land of Archimedean absolute values, there is one theorem to rule them
all. It is as follows.
Theorem1.7. (BigOstrowskiTheorem) Letkbeafieldand|·| anArchimedean
1
absolute value on k. Then there exists a constant α ∈ R>0 and an embedding
ι:k (cid:44)→C such that for all x∈k, |x| =|ι(x)|α . In other words, up to equivalence,
1 ∞
every Archimedean absolute value arises by embedding k into the complex numbers
and restricting the standard norm.
Theorem 1.7 is a rather deep result: every known proof from first principles takes
several pages. It immediately implies all of the other results in this section, and
conversely these results – and more! – are used in its proof. Indeed, to prove Big
Ostrowski it is convenient to use aspects of the theory of completions, so the proof
is deferred until Chapter 2. (In fact, the proof of Big Ostrowski was not presented
in the 2010 course at all and was only added to these notes in July of 2010. The
implicit message here – that this is an important result whose proof may neverthe-
less be safely skipped on a first reading – seems valid.)
For a field k, let Z·1 be the additive subgroup generated by 1. Recall that if
k has characteristic 0, then Z·1 is isomorphic to the integers, whereas if k has
characteristic p>0, Z·1∼=F .3
p
3ThuseitherwayZ·1isasubringofk,oftencalledtheprime subring.