Notes on class eld theory
Akhil Mathew
These are notes on algebraic number theory that I wrote for myself to help
understand the subject. If you nd them useful in any way at all, that’s great!
I have attempted to be as self-contained beyond elementary algebra. This
includes commutative algebra, and I frequently refer to books on the subject when I
treat results like Nakayama’s lemma as random bits of magic. Someday I will type up
notes on the subject, and perhaps these two PDFs will have a close relationship.
If you nd a mistake, I shall appreciate it if you email me at amathew@college.harvard.edu.
Notes: As of today (October 19, 2010), the notes are a mess. There are many
gaps in the part on elementary algebraic number theory. The notes were culled from mostly
disjointed and highly irregular blog posts on algebraic number theory, and many of them still
retain vestigial features from their evolutionary past. It will take several macro-mutations, and
possibly even a few extinctions of redundant material, before they can become properly
adapted to their surroundings.
Contents
Part 1. Algebraic numbers 5 1. Dedekind domains and discrete valuation rings 7
1. Absolute values and discrete valuation rings 7 2. A criterion for DVRs 9 3.
Dedekind domains 10 4. Extensions 12 5. Action of the Galois group 14
2. Complete elds 17 1. Completions 17 2. Hensel’s lemma 20 3. Approximation
theorems 21 4. A classi cation theorem for complete DVRs 24
3. Rami cation 27 1. Rami cation in general 27 2. Unrami ed extensions 30 3. Totally
rami ed extensions 33 4. Two homomorphisms 34 5. Completions revisited 36 6. The
di erent 37
4. Number elds 39 1. Introductory remarks 39 2. The Riemann-Roch problem in
number elds 41 3. The unit theorem 44 4. The adele ring 46 5. The idele group 47
5. Some applications 51 1. Fermat’s last theorem in a special case 51
Part 2. Class eld theory 53 6. The rst inequality 55
1. What’s class eld theory all about? 55 2. The Herbrand quotient 56 3. Shapiro’s
lemma 57 4. The local norm index for a cyclic extension 58 5. The cohomology of
the S-units in a global eld 60
3
4 CONTENTS
6. The rst inequality 61 7. The second inequality 63
1. A warm-up: the Riemann-zeta function and its cousins 63 2. The analytic proof
of the second inequality (sketch) 66 3. The idele-ideal correspondence 68 4. The
algebraic proof of the second inequality 70
8. Global class eld theory 77 1. The global Artin map 77 2. Strategy of the proof 78
3. The case of a cyclotomic extension 79 4. The basic reduction lemma 80 5. Proof of
the reciprocity law 84 6. Norm groups of Kummer extensions 86
Bibliography 89
Part 1
Algebraic numbers
1
2So
j1j
Dedekind domains and discrete valuation rings
for instance
=
12
= j1j, i.e. j1j= 1. The standard e
on R or C , buthere is another: Ex
absolute value on Q de ned as fo
highest powers of pdividing x;yres
1. Absolute values and discrete valuation rings 1.1. Absolute Values.
Actually, it is perhaps more logical to introduce discrete valuations as a special case of absolute values, which in turn generalize the
standard absolute value on R .
De nition 1.1. Let Fbe a eld. An absolute value on Fis a function j j: F! R ,
satisfying the following conditions:
(1) jxj>0 for x2Fwith x6= 0, and j0j= 0. (2) jxyj= jxjjyjfor all x;y2F. (3) jx+ yj jxj+
jyj:(Triangle inequality.)
= psr:
p7
(1.1) x y
( = 0.) It can be checked directly from the de nition that the p-adic absolute value
is
A
l
indeed an absolute value, though there are some strange properties: a number has
s
a small p-adic absolute value precisely when it is divisible by a high power of p.
o
Moreover, by elementary number theory, it satis es the nonarchimedean
property:
j
0
De nition 1.3. An absolute value j jon a eld Fis nonarchimedean if jx+
j
yj max(jxj;jyj) for all x;y2F. A nonarchimedean absolute value is sometimes called
p
a valuation.
This is a key property of the p-adic absolute value, and what distinguishes it
fundamentally from the regular absolute value restricted to Q. In general, there is
an easy way to check for this:
nonarchifleqone Proposition 1.4. The absolute value j jon F is non-archimedean if and only if there is
a Cwith jnj Cfor all n2N (by abuse of notation, we regard nas an element of Fas well, even when Fis
of nonzero characteristic and the map N !F is not injective). In this case, we can even take C= 1.
8 1. DEDEKIND DOMAINS AND DISCRETE VALUATION RINGS
n
Proof. One way is straightforward: if j jis non-archimedean, then j1j= 1,
j2j max(j1j;j1j) = 1, j3j max(j2j;j1j) = 1, and so on inductively.The other way is
slightly more subtle. Suppose jvj Cfor v2N . Then x x;y2F. We have:
njx+ yj =
nk k
n
nmax(jxj;jyj):
Xk=0
x
X
v
0.
k
y
n
=
n
0
k
k
Cnmax(jxj;jyj)
j R+
ow by the hypothesis, jx+ yj n
: Taking n-th roots and letting n!1gives the result. N
This also showsNthat the p-adic absolute value is nonarchimedean, since it it
automatically 1 on the integers.
Corollary 1.5. If F has nonzero characteristic, then any absolute value on F is
non-archimedean.
cal
led
a
val
uat
ion
Proof. Indeed, if Fis of characteristic p, take C:= max(j1j;j2j;:::;jp1j). N
).
1.2. Discrete Valuation Rings. The nonarchimedean absolute values we are Fu
rth
primarily interested in are
er
De nition 1.6. A discrete valuation is an absolute value j jon a eld F such that jFismo
a cyclic group. In other words, there is c2R such that, for each x6= 0, we can writere
we
jxj= c, where v= v(x) is an integer depending on x. We assume without loss of
generality that c<1, in which case v: F!Z is the order function (sometimes itself as
sume vsurjective by choosing cas na
generator of the cyclic group jFj.
Now, if j jis any nonarchimedean
absolute value on a eld F, de ne
the ringof integers Ras R:= fx2F:
jxj 1g:(This is a ring.) Note that
x2Ris a non-unit if and only if jxj<1,
so the sum of two non-units is a
non-unit and Ris a local ring with
maximal ideal
m := fx2F:
jxj<1g: When j jis a discrete
valuation, we call the ring of
integers so obtained adiscrete
valuation ring (DVR). In this case,
we choose such that j jgenerates
jFjand is <1; this is called a
uniformizing element and generates
m. The rst thing to notice is:
Proposition 1.7. A discrete valuation
ring is a principal ideal domain.
Conversely, a local domain whose
maximal ideal is principal is a
discrete valuation ring.
Proof. Indeed, if Ris a DVR and if
I Ris an ideal, let x 2I be an
element of minimal order v(x);
then I= Rx, since Rconsists of the
elements of F of nonnegative
order. It is easy to see that we
can take xto be a power of (if
necessary, by multiplying by a
unit). In particular, the non-zero
ideals form a monoid isomorphic
to Z
2. A CRITERION FOR DVRS 9
Conversely, if Ris a local PID, then let generate the maximal ideal m R; then since the Krull
intersection theorem implies
\ mn = \ ( n n+1m
as x= n
n
n
n
(p)
1
1
1
1
1
set of ideals I with I1
1
1
) = 0; we can write each nonzero x2R,
say x2mufor u=2m, i.e. u a unit. This is unique and we can de ne a discrete valuation
j jby jxj= 2for n as above. This extends to the quotient eld and makes Ra DVR. N
We de ne a function on the quotient eld Kof a DVR A, called the order. If x2Kcan
be written as ufor a uniformizer and ua unit, then the order is de ned to be n. The
previous statements can be recast using the order; for instance, to say that an
element is small is to say that its order is high.
Example 1.8. Let pbe a prime number. Then the ring Zof rational numbers x=ywhere
y6= 0 is not divisible by pis a discrete valuation ring, with uniformizing element p. The
associated discrete valuation is the p-adic valuation on Q.
Example 1.9. If k is a eld, the formal power series ring k[[X]] is a discrete valuation
ring, where the order of a power series is taken to be the degree of its rst non-zero
coe cient (and the absolute value is 2 raised to the negative of that). The quotient eld
is the set of all formal Laurent series with only nitely many negative terms.
2. A criterion for DVRs A much more
interesting (and nontrivial) result is the following:
niupmeansdvr Theorem 1.10. If the domain Ris Noetherian, integrally closed, and has a unique
nonzero prime ideal m, then Ris a DVR. Conversely, any DVR has those properties.
This is equivalent to the fact that Dedekind domains have unique factorization of
ideals, as we’ll see eventually. In particular, it will give a correspondence with the
notions of prime ideal and discrete valuation (for Dedekind domains).
Proof. First, let’s do the last part. We already know the part about DVRs being
PIDs, which are Noetherian; integrally closedness is true for any UFD (and thus for
any PID). The ideals of a DVR are generated by powers of the uniformizing element,
so the ideal m of non-units is the only prime ideal.
The converse is much harder. Let Rbe a ring satisfying the three conditions of the
theorem.
We will show that m is principal, by showing it is invertible (as will be seen below).
We divide the proof into steps:
2.1. Step one. For an ideal I R, let I:= fx2F: xI Rg, where Fis the quotient eld of R.
Then clearly I Rand Iis an R-module, but it isn’t clear that I
m
= R). There must be a
6= Ris
prime.
6= R:
The
The
set of
proof
such
runs
ideals
acros
is
sa
none
famili
mpty:
ar
it
line|s
contai
how
ns
that
any
any
(a) for
maxi
a2m
mal
(in
elem
which
ent in
case
the
(a)
6=
R.
Nev
erth
eles
s, I
clai
m
that
= Ra
10 1. DEDEKIND DOMAINS AND DISCRETE VALUATION RINGS
maximal element in this set of ideals by noetherianness, which is prime; thus, that
maximal element must be m, which proves our claim.
So to ll in the missing link, we must prove: Lemma 1.11. If S is a Noetherian
domain, any maximal element in the set of
S, then za;zb2J1
S. The J1
part follows since J1
ideals I Swith I1
1
claim that if z2J1
1
1
1
)1
R, write 1 =X
mini; mi
1
j
j
1=
1
1
j
i
1
is a S-module. By symmetry it’s enough to prove the other half for a, namely za=2S;
but thenif za2S, we’d have z((a) + J) S, which implies ((a) + J)
Then it follows that z(ab) = (za)b2JS, by applying the claim I just made twice. But
ab2J, so z(ab) 2S, contradiction. N
= R. Well, we know that mm 1
2.2. Step two. I now claim mm
too. So mm
langalgebra
claim. Indeed, since mm
Rby
de nition of inverses. and m mmis an ideal sandwiched between m and R. Thus we
only need to prove that mm= m is impossible. To do this, choose some a2mRwhich
must satisfy am m. Now by the next lemma, a is integral over R.
6=
S,
cont
radic
tion,
for
Jwa
s
maxi
mal.
6= Sis
prime.
Proof.
Let J
be a
maxim
al
elemen
t, and
suppos
e ab2J,
with a;b
=2J. I
Lemma 1.12. If N F is a nitely generated R-module satisfying bN N for some b2F,
then bis integral over R.
Proof. This follows by the Cayley-Hamilton theorem, and is a standard criterion for
integrality. (It can even be generalized to faithful nitely generated R-modules.)
Cf.[4]. N So returning back above, a=2Ris integral over R, a contradiction by
integrallyclosedness. 2.3. Step three. Now I claim m is principal, which by
yesterday proves the
2m; n 2m
1 2(mj): N
nj
j
, so x= mj(xnj)(mjnj
: At least one mis invertible,
since Ris local. Then I claim m = (m). Indeed, we need only prove m (m). For this, if
x2m, then xn2Rby de nition of m
So we’re done. Taking stock, we have an e ective way to say whether a ring is a
DVR. These three conditions are much easier to check in practice (noetherianness is
usually easy, integral closure is usually automatic, and the last one is not too hard
either for reasons that will follow) than the existence of an absolute value.
3. Dedekind domains Dedekind domains come up when
you take the ring of algebraic integers in any
nite extension of Q (i.e. number elds). In these, you don’t necessarily have unique
factorization. But you do have something close|namely, unique factorization of
ideals|which makes these crucial.
3. DEDEKIND DOMAINS 11
3.1. Basics. De nition 1.13. A Dedekind domain is a Noetherian integral domain
Athat
is integrally closed, and of Krull dimension one|that is, each nonzero prime ideal is
maximal.
niupmeansdvrniupmeansdvrA
DVR is a Dedekind domain, and the localization of a
Dedekind domain at a nonzero prime is a DVR by Theorem1.10. Another example
(Serre) is to take a nonsingular a ne variety V of dimension 1 and consider the ring of
globally regular functions k[V]; the localizations at closed points (or equivalently,
nonzero prime ideals) are DVRs, and it quickly follows that k[V] is Dedekind.
In particular, we can encapsulate what has already been proved as: Theorem
1.14. Let Abe a Dedekind domain with quotient eld K. Then there is
a bijection between the discrete valuations of Kthat assign nonnegative orders to
elements of Aand the nonzero prime ideals of A.
Proof. Indeed, every valuation gives a prime ideal of elements of positive order;
every prime ideal p gives a discrete valuation on A p, hence on K. N This result,
however trivial to prove, is the main reason we can work essentially
interchangeably with prime ideals in Dedekind domains and discrete valuations. 3.2.
Factorization of ideals. Now assume Ais Dedekind. A f.g. A-submodule
of the quotient eld Fis called a fractional ideal; by multiplying by some element of A,
we can always pull a fractional ideal into A, when it becomes an ordinary ideal. The
sum and product of two fractional ideals are fractional ideals.
Theorem 1.15 (Invertibility). If Iis a nonzero fractional ideal and I 11is a fractional ideal
and II1:= fx2 F: xI Ag, then I= A. Thus, the nonzero fractional ideals are an abelian
group under multiplication.
Proof. To see this, note that invertibility is preserved under localization: for a
multiplicative set S, we have S1(I1) = (S1I)1, where the second ideal inverse is with
respect to S1A; this follows from the fact that Iis nitely generated. Note also that
invertibility is true for discrete valuation rings: this is because the only ideals are
principal, and principal ideals (in any integral domain) are obviously invertible.
= Ap
So for all primes p, we have (II1)p
1
1
1
, which means the inclusion of
Amodules II!Ais an isomorphism at each localization. Therefore it is an isomorphism,
by general algebra. N
6=
Ican
be
writt
en
as a
prod
uct
of
prim
es,
by
the
indu
ctive
assu
mpti
on.
Whe
nce
so
can
I,
cont
radic
tion.
The next result says we have unique factorization of ideals: Theorem 1.16
(Factorization). Each ideal I Acan be written uniquely as aproduct of powers of prime
ideals. Proof. Let’s use the pseudo-inductive argument to obtain existence of a prime
factorization. Let Ibe the maximal ideal which can’t be written in such a manner, which
exists since Ais Noetherian. Then Iisn’t prime (evidently), so it’s contained in some
prime p. But I= (Ip)p, and Ip
Uniqueness of factorization follows by localizing at each prime. N
12 1. DEDEKIND DOMAINS AND DISCRETE VALUATION RINGS
De nition 1.17. Let P be the subgroup of nonzero principal ideals in the group Iof nonzero ideals. The
quotient I=Pis called the ideal class group.
The ideal class group of the integers, for instance (or any principal ideal domain) is clearly trivial. In
general, this is not the case, because Dedekind domains do not generally admit unique factorization.
Proposition 1.18. Let Abe a Dedekind domain. Then Ais a UFD if and only if its ideal class group is trivial.
Proof. If the ideal class group is trivial, then Ais a principal ideal domain, hence a UFD by elementary
algebra. Conversely, suppose Aadmits unique factorization. Then, by the following lemma, every prime
ideal is principal. Hence every ideal is principal, in view of the unique factorization of ideals. N
Lemma 1.19. Let Rbe a UFD, and let p be a prime ideal which contains no proper prime subideal except
for 0. Then p is principal.
eisenbudThe converse holds as well; a domain is a UFD if and only if every prime ideal of height one is
principal. Cf.
j
j).
(Note that ( j
1
4. Extensions We will show that
one can always valuations to bigger elds.
bourbaki
This can be generalized to the Krull-Azizuki theorem, cf.[?]. Proof. We need to
check that B is Noetherian, integrally, closed, and of
, where F obviously implies Noetherianness. The K-linear map (:;:) :
L L!K, a;b!Tr(ab) is nondegenerate since Lis separable over K. Let F B
be a free module spanned by a K-basis for L. Then since traces preserve
integrality and Ais integrally closed, we have B F := fx2 K: (x;F) Ag. Now
Fis A-free on the dual basis of Fthough, so Bis a submodule of a f.g.
Amodule, hence a f.g. A-module.
Integrally closed. It is an integral closure, and integrality is
transitive. Dimension 1. Indeed, if A Bis an integral extension of domains,
thenEisenbuddimA= dimB. This follows essentially from the theorems of \lying
over" and \going up." Cf.[?]. So, consequently the ring of algebraic integers
(integral over Z) in a number
Dedekind
intclosdedekind Theorem 1.20
). Since p is minimal La nite separable extension
among nonzero prime ideals, we have p = ( ) is prime by unique factorization.) N Dedekind.
4.1. Integral closure in a nite separable extension. One of the reasons
dimension
generated
eld ( nite extension of Q) is Dedekind. N
[?]. Proof. First, p contains an
irreducibles::: k. One of these
2
2
)
:
D
(
(and even A by integrality);it is nonzero by basic facts about vanderMonde
determinants. The next lemma clearly implies that Bis) contained in a nitely
generated A-module, hence is nitely generated (since Ais noetherian).
=
1Lemma 1.22. We have B D( )A[ ]. Proof. Indeed, suppose
x2B. We can write x=
c0(1)+c1( )+:::cn1( n1) where each ci2K. We will show that in fact, each ci12D( )A,
which will prove the lemma. Applying each ) + + cn1( i n1i, we have for each
i, ). Now by Cramer’s lemma, each ciix= c0(1) + c1( ican be written as a quotient
of determinants of matrices involving jxand the j. The denominator determinant
is in fact D( ). The numerator is in Kand must be integral, hence is in A. This
proves the claim and the lemma. N
Example 1.23. Let pi be a power of a prime pand consider the extension
Q( p
i1p(p1)
p
(p1) +X i1(p2)
jp
i a primitive ppi
pp
p
p
p
First of all,
p
p
i
p
p
p
p
1
4. EXTENSIONS 13
Note that the above proof actually implied (by the argument about traces) the
following useful fact:
Proposition 1.21. Let Abe a noetherian integrally domain with quotient eld K. Let
Lbe a nite separable extension and Bthe ring of integers. Then Bis a nitely
generated A-module.
1;:::;
W
e shall
give
anothe
r, more
n= [L:
K]
and th
e
distinct
embed
dings
of
Linto
the
algebr
aic
closure
of K.
De ne
the
discrim
inant
of to
be
i
)=
Q
for
i
)
is
pr
ec
is
el
y
Z[
-th
ro
ot
of
un
ity
.
Th
is
is
a
sp
ec
ial
ca
se
of
a
cy
cl
ot
o
mi
c
ex
te
ns
2 64 1 1 ( 1::: 1 2 ( 22 )::: .. .. . .This maps to the sa
each i. . .. . .3 71 C5, so is in KA
io
n,
an
im
po
rta
nt
ex
a
m
pl
e
in
th
e
su
bj
ec
t. I
cl
ai
m
th
at
th
e
rin
g
of
int
eg
er
s
(in
te
gr
al
ov
er
Z!
)
in
Q(
i ].
Th
is
is
tru
e
in
fa
ct
for
all
cy
cl
ot
o
mi
c
ex
te
ns
io
ns
,
bu
t
w
e
wil
l
no
t
be
ab
le
to
pr
ov
e
it
he
re.
pi jX
+X i satis es the equation X
i1
p
the form Q(1
+1
= 0.
This
is
beca
use
if is
a
p-th
root
of
unity
,
( i1(p2
)1 =
0. In
parti
cular
,Xi
divid
es
pin
the
ring
of
integ
ers
in
Q( +
+1
,
and
p
1)(1
p1 p+
+ )=
+
. Thus the norm to Q of 1
+
i
cons
equ
ently
(taki
ng
X=
1),
we
nd
that
1 ipi
)=Q.
This
is
true
for
any
primi
tive
p-th
root
of
unity
for
any
p
for
any
jis a
pow
er of
p. I
clai
m
that
this
impli
es
that
the
discr
imin
ant
D( i )
is a
pow
er of
p,
up
to
sign.
But
by
the
vand
erM
ond
e
form
ula,
this
discr
imin
ant
is a
prod
uct
of
term
s of
j pi)
up
to
ro
ot
s
of
un
ity
.
Th
e
no
rm
to
Q
of
ea
ch
fa
ct
or
is
th
us
a
po
w
er
of
p,
an
d
th
e
di
sc
ri
mi
na
nt
its
elf
pl
us
or
mi
nu
s
a
po
w
er
of
p.
By the lemma, it follows that the ring of integers is contained in Z[p
;
i
]
.
4
.
2
.
E
x
t
e
n
s
i
o
n
s
o
f
d
i
s
c
r
e
t
e
v
a
l
u
a
t
i
o
n
s
.
T
h
e
r
e
a
l
r
Th
eo
re
m
1.
24
.
Le
t
K
be
a
el
d,
La
nit
e
se
pa
ra
bl
e
ex
te
ns
io
n.
Th
en
a
di
sc
ret
e
va
lu
e
s
u
l
t
w
e
c
a
r
e
a
b
o
u
t
i
s
:
ati
on
on
Kc
an
be
ex
te
nd
ed
to
on
e
on
L.
14 1. DEDEKIND DOMAINS AND DISCRETE VALUATION RINGS
Proof. Indeed, let R Kbe the ring of integers. Then Ris a DVR, hence Dedekind, so
the integral closure S Lis Dedekind too (though in general it is not a DVR|it may have
several non-zero prime ideals) by Theorem intclosdedekindintclosdedekind1.20. Now as
above, Sis a nitely generated R-module, so if m Ris the maximal ideal, then
mS6= S by Nakayama’s lemma (cf. for
instanceEisenbud[?]). So mS is contained in a maximal ideal M of S with, therefore, M
\R= m. (This is indeed the basic argument behind lying over, which I could have just
invoked.) Now SM Rniupmeansdvrniupmeansdvris a DVR as it is the localization of a
Dedekind domain at a prime ideal, and one can appeal to Theorem1.10. So there is a
discrete valuation on SMm. Restricted to R, it will be a power of the given R-valuation,
because its value on a uniformizer is <1. However, a power of a discrete valuation is
a discrete valuation too. So we can adjust the discrete valuation on S Mif necessary to
make it an extension. This completes the proof. N
Note that there is a one-to-one correspondence between extensions of the
valuation on Kand primes of Slying above m. Indeed, the above proof indicated a way
of getting valuations on Lfrom primes of S. For an extension of the valuation on Kto L,
let M := fx2S: jxj<1g.
5. Action of the Galois group This section will basically be
commutative algebra. Suppose we have an integral
domain (we don’t even have to assume it Dedekind) Awith quotient eld K, a nite
Galois extension L=K, with Bthe integral closure in L. Then the Galois group G=
G(L=K) acts on B; it preserves Bbecause it preserves equations in A[X]. In particular,
if P Bis a prime ideal, so is P, and the set SpecBof prime ideals in Bbecomes a
G-set.
5.1. The orbits of the Galois group. It is of interest to determine the orbits; this
question has a very clean answer.
1Proposition
1.25. The orbits of Gon the prime ideals of Bare in bijection with the
primes of A, where a prime ideal p Acorresponds to the set of primes of B lying over
A.Alternatively, any two primes P;Q Blying over Aare conjugate by some element of
G.
In other words, under the natural map SpecB!SpecA= SpecBGG, the latter space is
the quotient under the action of G, while A= B2is the ring of invariants in B.
1Bis
the integral closure of S1A, and in S1A= AProof. We need only prove the second
statement. Let Sbe the multiplicative set Ap. Then S, the ideal p is maximal. Let Q;P
lie over p; then S1Q;S1P lie over S1pp and are maximal (to be added). If we prove that
S1Q;S1P are conjugate under the Galois group, then Q;P must also be conjugate by
the properties of localization. In particular, we can reduce to the case of p;Q;P all
maximal.
The rest of the proof is now an application of the Chinese remainder theorem.
Suppose that, for all 2G, we have P 6= Q. Then the ideals P;Q are distinct
1It is useful to note here that the lying over theorem works for arbitrary integral extensions.
2The reader who does not know about the Spec of a ring can disregard these remarks.
5. ACTION OF THE GALOIS GROUP 15
L Kmaximal ideals, so by the remainder theorem, we can nd x 1 mod P for
all 2Gand x 0 mod Q. Now, consider the norm N(x); the rst condition implies that it
is congruent to 1 modulo p. But the second implies that the norm is in Q \K= p,
contradiction. N
1;:::;Pg5.2. The decomposition and inertia groups. Now, let’s zoom in on a given prime
p A. We know that Gacts transitively on the set Pof primes lying above p; in
particular, there are at most [L: K] of them.
De nition 1.26. If P is any one of the Pi, then the stabilizer in Gof this prime ideal is
called the decomposition group G. We have, clearly, (G: GPP) = g. Now if 2GP,
then acts on the residue eld B=P while xing the sub eld A=p. In this way, we get a
homomorphism ! from Ginto the automorphism group of B=P over A=p) (we don’t
call it a Galois group because we don’t yet know whether the extension is Galois).
The following result will be crucial in constructing the so-called \Frobenius
elements" of crucial use in class eld theory.
PProposition
1.27. Suppose A=p is perfect. Then B=P is Galois over A=p, and the
homomorphism ! is surjective from G!G(B=P=A=p). Proof. In this case, the
extension B=P=A=p is separable, and we can choose
x2B=P generating it by the primitive element theorem. We will show that x satis es a
polynomial equation P(X) 2A=p[X] all of whose roots lie in B=P,which will prove that
the residue eld extension is Galois. Moreover, we will show that all the nonzero roots
of Pin B=P are conjugates of xunder elements of G. This latter will imply surjectivity of
the homomorphism ! , because it shows that any conjugate of xunder G(B=P=A=p) is
a conjugate under GPP. We now construct the aforementioned polynomial. Let x2Blift
x. Choosey2Bsuch that y x mod P but y 0 mod Q for the other primes Q lying over p.
We take P(X) =Q 2G(X (y)) 2A[X]. Then the reduction Psatis es P(x) = P(y) = 0, and P
factors completely (viaQ P(X (t))) in B=P[X]. This implies that the residue eld
extension is Galois, as already stated. But it is also clear that the polynomial P(X) has
roots of zero and (y) = (x) for 2G. This completes the proof fo the other assertion,
and hence the proposition. N
De nition 1.28. The kernel of the map ! is called the inertia group TP. Its xed eld is
called the inertia eld
These groups will resurface signi cantly in the future. Remark. Although we shall
never need this in the future, it is of interest to
3see what happens when the extension L=K is purely inseparable.Suppose Ais
integrally closed in K, and Bis the integral closure in L. Let the characteristic
be p, and the degree [L: K] = pi
p. In this case, x2Bif and only if x i
ip
2A.
pif xis integral, then so is xi
3Cf. langalgebra [4], for instance.
2A.
Indeed, it is clear that the condition mentioned implies integrality. Conversely,
, which belongs to K (by basic facts about purely inseparable extensions). Since Ais
integrally closed, it follows that x
16 1. DEDEKIND DOMAINS AND DISCRETE VALUATION RINGS
Let now p Abe a prime ideal. I claim that there is precisely one prime ideal P of
Blying above A, and Pip= p. Namely, this ideal consists of x2Bwith xip2p! The proof
is straightforward; if P is any prime ideal lying over p, then x2P i x2L\P = p. In a
terminology to be explained later, p is totally rami ed.
ip
2
Complete elds
1. Completions The utility of completions is vast. Many properties
(e.g. rami cation) are
preserved by the completion, but, as we will see, in the complete case the criteria for such properties
is simpli ed by the uniqueness of extensions of valuations. Completions are also essential to
developing class eld theory via ideles.
1.1. Basic de nitions. Let Fbe a eld with the absolute value j j. De nition 2.1. The completion^ Fof
Fis de ned as the set of equivalence classes
xm
yn
n
ng;fyn
n
so we want to put on on ^ F. If fxng2 ^ F, de ne jfxn^
Fand the image of Fis dense.
p-adic absolute value is the called p-adic numbers Qp
nonarchifleqonenonarchifleqone
17
f Cauchy sequences: A Cauchy
sequence fxngis one that satis es
jxn
j!0 as n;m!1. Two Cauchy sequences fxgare e
First
o o , ^ Fis a eld, since we can add or multiply
division is also allowable if the denominator sequence
\long run" (i.e. is not in the equivalence class of (0;0;:::
justi cation to check here, but it is straighforward. Also
gj:= limjxj. Third, there is a natural map F!There are se
this, of which the most basic are: Example 2.2. The co
the usual absolute value is the
real numbers. Completion is an idempotent operator. A
embedding
in its completion is an isomorphism is called|surprise a
2.3. Let pbe a prime number. The completion of Q with
representative of what we care about: completions
with respect to nonarchimedean (especially discrete) v
criterion (Proposition1.4), completions
preserve nonarhimedeanness. Next, here
is a lemma about nonarchimedean elds:
nonarchdisks Lemma 2.4. Let F be a eld with a
nonarchimedean absolute value j:j. Then if x;y2F and
jxyj<jxj, then jxj= jyj.
(In English: Two elements very close together have the
same absolute value. Or, any disk in a
nonarchimedean eld has each interior point as a
center.)
18 2. COMPLETE FIELDS
Proof. Indeed, jyj= j(yx) + xj max(jxyj;jxj) = jxj:
Similarly jxj= j(xy) + yj max(jxyj;jyj);and since jxj>jxyjwe can write jxj jyj:^ F.N Corollary 2.5. If j:jis discrete on F,
it is discrete on
LemmaProof. Indeed, if V R+is the value group (absolute values of nonzero elments) of F, then it is the value
group ofnonarchdisksnonarchdisks^ F since F is dense in^ F, in view of12.4. N 1.2. (Aside) Completions of rings.
Take a eld Fwith a discrete valuation j:jand its completion ^ F. We can takethe ring of integers R Fand ^ R ^
F, and their maximal ideals m; ^ m.I claim that ^ Ris the completion of Rwith respect to the m-adic topolThis follows because ^ Rconsists of equivalence classes of sequences fxn
n
nj
1 by discreteness, so wlog all the xn
gof elements of F, the limit of whose absolute values jxjis 1. This meanso from
gy.2 some point on, the jx2R. This is just the de nition of an element of the completion of
R. I leave the remaining details to the reader.
1.3. Topologies determined by absolute values. Henceforth, all absolute values
are nontrivial|that is, we exclude the trivial absolute value that takes the value one
everywhere except at zero. We do not assume completeness.
It is clear that an absolute value on a eld Fmakes it into a topological eld,
which is to say a eld that is simutaneously a topological space, and where
the eld operations are continuous. The topology essentially determines the
absolute value, though:
1n, the sequence fx <1, and i jxj2
, j j2
1
1 1, jyj2
2
2
Now x
xwith
jxj1
2
logjyj1logj
xj1
1
j1
topdeterminesabsvalue Theorem 2.6. Let j j
aa (2.1) jxj
1
jyj
1
be absolute values on K inducing the same topology. Then j jis a power of j j.
Proof. So, rst of all, given x2Kgconverges to zero (in the common topology
induced by each absolute value) i jxj<1. Thus we see that the unit disks are the
same under both absolute values.
i jxj2 2 jyj; as it then follows that=logjyj2logjxj2, and then one gets that jyjdi er by the same <1 a
power as the x’s. And so j j;j j2aadi er by a constant power. First, I claim this is
y2F,
true for 2Z. Then (2.1) says jx=y<1 i jx=y j2<1
1This brief section is not necessary for understanding the rest of these notes.2We have not de ned this construction; cf. Eisenbud [?] for
instance.
2V. We can choose the norm k k1 civ1, then kvk= maxjcii1j. Then from the triangle inequality it is easy to check that for
all v, kvk Ckvkfor some large C(for instance, C= Pkv1ik). If we show the reverse,
that there exists a c>0 with kvk ckvkfor all v, then the assertion will be complete. By
homogeneity we may assume kvk 11= 1; the set of such v forms a compact subset
of V (under the kvk-topology!), on which by the previous inequality k kis a
continuous and nonzero function. It has a minimum, which is the creferred to. N
Fix a basis v1;:::;vn
We can apply this to the case of elds: Corollary 2.9. Given a complete eld
Kand a nite extension L K, the absolutevalue on Kcan be extended to Lin
precisely one way. Proof. Uniqueness is clear from
Theoremtopdeterminesabsvaluetopdeterminesabsvalue2.6 and Theoremtopuniquetopunique2.8
Existence needs to be shown. This has been done for discrete valuations, but the
general case is to be added. N
nx
+ c1 xn1
+
n
n1
+ cn1
= 0; 8ici
Consider the situation as in the previous corollary, and let R;Sbe the rings of
integers in K;L, respectively. The integral closure~ Sof Ris a Dedekind domain, as
is S; and both have the same quotient eld. It is no surprise that they are actually
equal (in the complete case):
Proposition 2.10. Sis the integral closure of Rin Lin the complete case. Proof.
Any element x2Lintegral over Rsatis es an equation
xn1
1
1. COMPLETIONS 19
on V: if v= P
aa
so (aa2.1) follows in this case by the rst paragraph. By raising each side to an
integer power, it follows that (2.1) is valid for 2Q+, and by continuity for 6= 0
arbitrary. N
1.4. Uniqueness of norms in extensions of complete elds. In the complete
case, there is a rigidity on absolute values: they are equivalent much more often
than in the non-complete case. We will approach this fact through a look at norms
on topological vector spaces.
So x a eld Kcomplete with respect to the absolute value j j. Let V be
a nite-dimensional vector space over K.
De nition 2.7. A norm on V is a map k k: V !R 0satisfying all the usual constraints,
i.e. scalar multiplicativity (with respect to j j) and the triangle inequality.(1) kvk 0 for
all v2V, with equality i v= 0 (2) kavk= jajkvkfor a2k;v2V (3) kv+ wk kvk+ kwk;for all
v;w2V
topunique Theorem 2.8. All norms on V induce the same topology. Proof. For the proof, I’ll make
the simplifying (but unnecessary) assumption
qthat Kis locally compact|this is the case for the real and complex numbers, nite extensions
of the p-adic elds, and nite extensions of power series elds F((X)) (all of which go under
the name \local elds").
2
R:
Ta
ki
ng
or
de
rs,
it
fol
lo
w
s
th
at
th
e
or
de
r
of
xi
s
th
e
sa
m
e
as
th
at
of
c+
+
c,
bu
t
thi
s
is
im
po
ss
ibl
e
un
le
ss
xh
as
no
nn
eg
ati
ve
or
de
r.
S
o
~
S
S.
20 2. COMPLETE FIELDS
By contrast, any element of y2Shas absolute value at most one, and ditto for all
its conjugates. Indeed, any K-automorphism of Linto the algebraic closure K(which
has a unique absolute value extending that of K) xes the absolute value by
uniqueness (the previous corollary), so yfor an isomorphism has absolute value at
most 1. Since the minimal polynomial of yhas coe cients which are the symmetric
functions of yand its conjugates, this polynomial is an integral one, and yis integral. N
So, if we use the correspondence between valuations and ideals, we see that the
unique maximal ideal p of Rprolongs uniquely to one of S.
31.5.
Further topological remarks. In the cases of interest, complete elds will always
be locally compact. In the archimedean case, this is because every complete eld is
isomorphic to R or C , by a theorem of Ostrowski.In the nonarchimedean case, it
follows from:
Proposition 2.11. Let Kbe a complete eld with respect to a discrete valuation. Let
Abe the ring of integers, m the maximal ideal, and suppose that the residue eld A=m
is nite. Then Kis locally compact.
Proof. Indeed, if is a uniformizer, then Ais topologically the inverse limit of
the nite groups lim A= iA, so A Kis compact. Now Ais just the closed unit disk at the
origin(!), so Kis locally compact. N
weilAs
a result, there is a Haar measure on K (with the additive structure). There is
an interesting interpretation of the absolute value in terms of it, which is the starting
point in[?]. Multiplication by a2Kis a homomorphism of Kinto itself, so there is a
constant kaksuch that for any measurable set E K,
(aE) = kak (E): This is because the maps
E! (E);E! (aE) are both invariant measures, andthe Haar measure is unique up to a
constant factor. Clearly, the map k k: K!Ris multiplicative. Proposition 2.12. k kis the
discrete valuation on Ksending to1 jA=mj 0. Proof. Let q= jA=mjbe the cardinality of the
residue eld. Then (A) =1q ( A) because Ais the unit of qcosets of A. Also, Ais open,
so (A) >0, and the fact that k k= qis clear. Similarly, if uis a unit, then uA= A, so kuk=
1. The result is now clear. N
2. Hensel’s lemma Hensel’s lemma allows us to lift
approximate solutions of equations to exact
psolutions. This is one of the reasons completions are so useful. It implies that one
can reduce questions about whether a polynomial has a root in, say, Qto whether it
has a root in Z=pZ.
0Theorem
2.13 (Hensel’s lemma). Let Rbe a complete DVR with quotient eld K.
Suppose f2R[X] and x2Rsatis es f(x0) = 0 2m while f00and f(x) = 0.(x0) 6= 0. Then
there is a unique x2Rwith x= x
3Citation needed.
3. APPROXIMATION THEOREMS 21
Here the bar denotes reduction. The result re nes the \approximate solution" x04to an actual
solution x. Proof.The idea is to use Newton’s method of successive approximation. Recall that
given an approximate root r, Newton’s method \re nes" it to
r f:= r f(r)0(r) :
0
So de ne xn 2Rinductively (x0
= (xn1 )0, the 0
) f(r) f0
is already de ned) as xn= f(r) f(r) + C(r0
0)j
c
0
2
:
0
as above. I claim that the xnapproach a limit x2Rwhich is as claimed. For r2Rby Taylor’s
formula we can write f(X) = f(r) + f2(r)(Xr) + C(X)(Xr), where C(X) 2R[X] depends on r. Then for any
r) = f fr f(r)0(r) (r)
2 and jf 0(r
Thus, if jf(r)j cand jf0(r)j= 1, we have jf(r0 rmod m.
0
We even have jrrnn2R (2) jf(x)j jf(x02)jnn. (3)
jxxn1j jf(x02)jn1. Now it follows that we may set x:=
limxn
0f(rnotation
0(x0
2
0
0
)j= 1, since rj c. This enables us
to claim inductively: (1) x
and we will have f(x) = 0. Uniqueness follows because xis
a simple root of f2k[X]. N There is a more general version of Hensel’s lemma that says if you have
jf(x)j jf)j, the conclusion holds. Also, there’s no need for discreteness of the absolute value|just
completeness is necessary.
Corollary 2.14. For n xed, any element of Rsu ciently close to 1 is a n-th power.
Proof. Use the polynomial Xn1. N Example 2.15. Using this, for instance, one can determine what
the squares in
Qp look like in terms of quadratic residues.
3. Approximation theorems 3.1. The Artin-Whaples approximation theorem.
The Artin-Whaples
approximation theorem is a nice extension of the Chinese remainder theorem to absolute values, to
which it reduces when the absolute values are discrete. It is extremely useful in the glboal case
(where there are di erent absolute values) as a type of density result.
So x pairwise nonequivalent absolute values j j1 ;:::;j jn
1;:::;an
topdeterminesabsvaluetopdeterminesabsvalue
2.6).
jaaiji
4See lang [3] for a more general version of
Hensel’s lemma.
on the eld K; this
means that they induce di erent topologies, so are not powers of each other
(Theorem
artinwhaples Theorem 2.16 (Artin-Whaples). Hypotheses as above, given a2Kand >0, there exists
a2Kwith
< ; 1 i n:
22 2. COMPLETE FIELDS
Consider the topological space Qn i=1iKwith the product topology, where the ith factor
has the topology induced by j j. Then the theorem states that the diagonal is dense in
this space.
Proof. I claim that rst of all, it is su cient to take the case where a= 1;ai1= 0 for i>1.
Indeed, if this case is proven, then (by symmetry) choose for each i,
biapproximating 1 at j jiand close to 0 at the other absolute values, and takea=
Xaibi: Before proving the approximation theorem, we prove:
>1 and jaji
01
topdeterminesabsvaluetopdetermin
esabsvalue
2.6. Suppose now n 3. By
induction on n, assume there is a0ji0<1 if 1 <i<n. Case 1. If
jajn<1, then we’re already done with a= a 00. Case 2. If
jajn>1, consider a0N
1 + a0N
which when Nis large is close to 1 at
;j jn
00j >1. Then take a= a00 a0N1+a0N
n
j j100002Kwith jaj1
i
0jn0N = 1, then choose a000as in Case 2 (i.e. with jaj1 0>1;jajn
00
00, jaj1
n
00
i
i
i
j
i
Y A=I
1
i
a1
N=(1 + a
<1 if i>1.
L
Proof of the lemma. The case of two absolute
nonequivalence; see the proof of Theorem
2Kwith ja
j
emma 2.17. Hypotheses as above, there exists a2Kwith jaj1
0
>1 and ja
and close to zero elsewhere. By the
case n= 2, there is a<1 and ja. Case 3. If ja<1) and let a= aa, where Nis large enough
to bring the absolute values of aat j j<1 because of a, 1 <i<n, down below 1. We
automatically have jaj>1 because of a, and jaj<1 (1 <i<n) because of N’s being taken
large. N Now nally we have to establish the theorem in the case we described,
where
= 1;aiN= 0 if i > 1. For this, simply choose aas in the lemma, and take a) for Nlarge.
This is close to 1 for j j, and close to zero at the other absolute values. N
3.2. The Chinese remainder theorem. The remainder theorem below was known
for the integers for thousands of years, but its modern form is elegant.
Theorem 2.18 (Chinese Remainder Theorem). Let Abe a ring and I;1 i n be ideals
with I+ I= Afor i6= j. Then the homomorphism A!
n
is the same thing as jaai
i
i
i
i
. The
i remainder theorem is related to the approximation theorem in the
s surjective with kernel following
I
reformulation. Surjectivity states that, given amod Ii. The congruence modulo
I2A;1 i n, there is a2Asuch that a aj being small if Iis a high power of a prime
ideal in a Dedekind domain and j jthe associated discrete valuation.
:::I
3.
AP
PR
OX
IM
ATI
ON
TH
EO
RE
MS
23
Proof. First we tackle surjectivity. If n= 1 the
assertion is trivial. If n= 2, then I1
1 + 2 where 1 2I2; 2
2I2
=
A.
W
e
th
en
ha
ve
1
=
.
S
o,
if
w
e
x
a;
b2
A
an
d
w
an
t
to
+ I2
2
i)
1
Q
A=I
1
1
+b2
i
j
1
+ I2 1 + Ij
:::I
2for
j 2, 1 2Y(Ij 211+ Ij) I1
+ I2 :::In;
1
2
=Q
n
1;I
:::I
2
2
(I1
i
2
1;I
ch
oo
se
2
A
wi
th
;
th
e
na
tur
al
ch
oi
ce
is
:=
a.
Fo
r
hi
gh
er
n,
it
wil
l
be
su
ci
en
t
to
pr
ov
e
th
at
ea
ch
ve
ct
or
(a
wi
th
all
ze
ro
s
ex
ce
i
.
e
.
I
a mod I
Q A=Iwhile a2I
2
; b mod I
i
Ii. For two ideals I
pt
for
on
e
1
in
th
e
pr
od
uc
t
oc
cu
rs
in
th
e
im
ag
e,
si
nc
e
th
e
im
ag
e
is
an
Asu
b
m
od
ul
e
of.
By
sy
m
m
etr
y,
w
e
ne
ed
on
ly
sh
o
w
th
at
th
er
e
is
a2
As
uc
h
th
at
a
1
m
od
Ifo
r
j2
.
Si
nc
e
1
2I
n = A. Now apply the n= 2 assertion to I
2
n
211
it follows because one
2I ; 2 2I
+ = 1, and notes that
chooses
\I ) = (I1 \I ) + (I \I ) 2 I2I1 + I1I
n :::I 2= I
2 :::I
I1 \ 1 I
= I :::I
2
nideals I1;:::;I
T Ias above, to nd a satisfying the conditions a 1 mod I. This handles surjectivity. Now for the n :::I
with
kernel assertion: we must show
:
Fo
r
na
rbi
tra
ry,
w
e
us
e
in
du
cti
on
.
As
su
m
e
th
e
re
su
lt
for
n1
id
ea
ls.
In
th
e
1
1;a2I2
ca
se
of,
w
e
ha
ve
I+I
=
A
as
m
en
tio
ne
d
pr
ev
io
us
ly
in
th
e
pr
oo
f.
C
on
se
qu
en
tly
,
n\
j
1
2
1
n
\I
1
1;:::;pk
1
1
p
I should cite Curtis-Reiner
pi
1
n
p
;I
n
, i.e. p1Ap1
x mod (p
j
(a) for each i, which we can do by
p
k.
Choose y2Awith ordpi
5
and the
b inductive hypothesis. N As a corollary, we get a criterion for when a :::I
Dedekind domain is principal:
y the n= 2 case applied to the pair
I
Theorem 2.19. A Dedekind domain Awith nitely many prime ideals is a principal
ideal domain.
Proof. To do this, we need only show that each prime is principal. Let the
(nonzero) primes be p; we show pis principal. Choose a uniformizer 2A at p= ( );
alternatively has order one in the associated valuation. Now choose x2Asuch
that
) (y) = ord ; x 1 mod (p
;
:
1
:
5
:
;
p
)
for
j 2:
The
n (x)
and
phav
e
equ
al
orde
rs at
all
prim
es,
or
equi
vale
ntly
beco
me
equ
21
al in
all
local
izati
ons,
henc
e
are
equ
al. N
I
n
gen
eral,
how
ever
,
Ded
ekin
d
dom
ains
are
not
princ
ipal
ideal
dom
ains,
or
even
uniq
ue
facto
rizati
on
dom
ains.
But
we
do
have
a
wea
ker
resul
t:
Th
eo
re
m
2.
20
.
In
a
D
ed
ek
in
d
do
m
ai
n,
an
y
id
ea
l
is
ge
ne
rat
ed
by
tw
o
el
e
m
en
ts.
Pr
oo
f.T
he
re
su
lt
wil
l
ne
ve
r
be
us
ed
ag
ai
n,
bu
t
th
e
pr
oo
f
is
a
m
us
in
g.
Le
ta
be
a
no
nz
er
o
id
ea
l
an
d
x2
a
f0
g.
Fo
r
ea
ch
p,
let
or
db
e
th
e
as
so
ci
at
ed
or
de
r
fu
nc
tio
n.
Th
en
w
e
ha
ve
or
d(
x)
=
or
d(
a)
=
0
for
all
bu
t
nit
el
y
m
an
y
pri
m
es
p
x=
si
2S:
Proof. Given x, we can nd by the de nitions s0 2Swith xs0
1;:::with x Pn i=0n i=0 0
1 with
;s
P1 i=0
2
i 1
x1
2( n+1
P1 i=0
i
s
i2f0;:::;p1g.
Proof of
Theorem
0
for x
xs 0
2Rand we can choose s
si i
0xs0
s
si i 2 R, or xs
n+1
s
2
i
ip
completedvrchar0completedvrchar0
24 2. COMPLETE FIELDS
i=0
X
ordthe Chinese remainder theorem (e.g. say that yis congruent modulo p p i(a)+1 iordp ito
something in p(a) iord(a)+1 ipp i). In addition, if q1;:::;qkare the ideals occurring in the
factorization of a, choose ysuch that in addition ordqi(y) = ordqi(a) for each i. It is
clear that these two requirements do not conict.
The rst requirement guarantees that (x;y) a since this is true at each prime.
The second requirement guarantees that y2a (and we already chose x2a). N
4. A classi cation theorem for complete DVRs We discuss the proof
of a classi cation theorem about complete DVRs, using
Hensel’s lemma. The proof will illustrate a useful technique about systems of
representatives.
completedvrchar0 Theorem 2.21. Let Rbe a complete DVR with maximal ideal m and residue eld k.
Suppose kis of characteristic zero. Then R’k[[X]], the power series ring in one variable, with respect to
the usual discrete valuation on k[[X]].
For a generalization of this theorem, see Serre’s Local Fields. To prove it,
we rst need to introduce another concept. A system of representatives is a set S Rsuch that the reduction map S!kis bijective. Then we have
the following useful fact:
sysrep Proposition 2.22. If Sis a system of representatives in a complete DVR Rand is a uniformizer,
we can write each x2Runiquely as 1
2
R.
Next
, we
have
R.
Rep
eatin
g
the
proc
ess
indu
ctive
ly to
obtai
n
s1),
we
have
x=si
nce
the
di er
ence
sx
P2
Rten
d to
zero
i;
where si
.
Uniq
uen
ess
is
easy
to
see:
the
choi
ce of
s0wa
s
uniq
ue
sinc
e
the
redu
ction
map
S!kw
as
bijec
tive,
and
indu
ctive
ly it
is
easy
to
see
that
the
choi
ce of
s;:::;
was
uniq
ue
as
well.
N
Rem
ark.
In
the
p-ad
ic
num
bers
, we
can
take
f0;:::
;p1g
as a
syst
em
of
repr
esen
tativ
es,
so
we
nd
each
p-ad
ic
integ
er
has
a
uniq
ue
p-ad
ic
expa
nsio
n x=
.21.
The
point
of
the
proo
f is
that
ther
e is
a
syst
em
of
repr
esen
tativ
es in
Rtha
t is
a el
d
isom
orph
ic to
kund
er
redu
ction
.
This
eld
will
in
fact
be
the
maxi
mal
sub
eld
of
R(w
hich
exist
s in
view
of
Zorn
’s
lem
ma,
once
we
prov
e
2
that
ther
e is
a el
d
cont
aine
d in
R).
This
last
ques
tion
is
strai
ghtf
orwa
rd.
Note
that
Z
f0g
Rget
s
sent
to
nonz
ero
elem
ents
in
the
resid
ue
eld
k,
whic
h is
of
char
acte
ristic
zero
.
This
mea
ns
that
Z
f0g
Rco
nsist
s of
units
, so
Q
R.Th
us
the
maxi
mal
eld
inde
ed
exist
s;
call
it L. I
clai
m
that
L= k.
The
map
L!L=
kis
then
surje
ctive
,
and
injec
tive
(sinc
e it
is a
mor
phis
m
of e
lds),
and
this
will
prov
e
the
theo
rem
in
view
of
the
prop
ositi
on.
0S
o,
su
pp
os
e
t2
kL
,I
cl
ai
m
th
at
th
er
e
is
a
bi
gg
er
el
d
L.
Th
is
wil
l
co
ntr
ad
ict
m
ax
im
ali
ty.
R
co
nt
ai
ni
ng
L
wi
th
t2
L
4. A CLASSIFICATION THEOREM FOR COMPLETE DVRS 25
If t2k Lis transcendental, lift it to T2R; then Tis transcendental over L and is
invertible in R, so we can take L0:= L(T). Suppose now that tis algebraic. If the
minimal polynomial of tover Lis f(X) 2k[X], we have f(t) = 0.Moreover, f 0(t) 6= 0
because these elds are of characteristic zero and all extensions are separable. So
lift f(X) to f(X) 2R[X]; by Hensel lift tto u2Rwith f(u) = 0. 0Then fis irreducible in L[X]
(otherwise we could reduce a factoring to get one of f2L[X]), so L[u] = L[X]=(f(X)),
which is a eld Lwhose reduction contains t. In either case, we have obtained a
contradiction, which proves the theorem. N
3
Rami cation
1. Rami cation in general Fix a Dedekind domain Awith
quotient eld K; let Lbe a nite separable
intclosdedekindintclosdedekindextension of K, and B the integral closure of A in L. We know that
B is a Dedekind domain (Theorem1.20). Given a prime p A, there is a prime P Blying
above p. This prime is,however, not unique. We can do a prime factorization of pB B;say
pB= Pe11e:::Pg. The primes Pggcontain pBand consequently lie above p (since their
intersection with Ais prime and containing p). Conversely, any prime of Bcontaining
pBmust occur in the factorization of pB, since if Iis an ideal in a Dedekind domain
contained in a prime ideal P, then Poccurs in the prime factorization of I(to see this,
localize and work in a DVR).
Corollary 3.1. Only nitely many primes of Blie above a given prime of A. Proof.
Immediate from the preceding discussion. N
1.1. Basic de nitions. De nition 3.2. If P Blies above p A, we write eP =pfor the
number of times P occurs in the prime factorization of pB. We call this the rami cation
index.
We let fP =pbe the degree of the eld extension A=p !B=P. This is called the residue class
degree. The quotients A=p;B=P are called the residue elds.
1The
following remark is sometimes useful: given p;P as above, consider the multiplicative
set S = (Ap) and note that SA S1Bis an extension of Dedekind domains. Then the
rami cation index eS1P :S1p= eP =p. The proof is immediate from the de nitions in terms of
factorization. We could even localize Bwith respect to P (using the bigger multiplicative set
BP) and get the same result. Additionally, we will see below that rami cation behaves well
with respect to completions.
The rami cation index ehas an interpretation in terms of discrete valuations. Let j jbe
the absolute value on Kcorresponding to the prime p and, by abuse of notation, j jits
extension to Lcorresponding to P. Then I claim that:
1Proposition 3.3. Suppose A B is an extension of Dedekind domainswith quotient elds
K;Land associated absolute value j j
ee (3.1) eP =p
= (jLj: jK j):
We don’t assume that B is the integral closure of A. For then, we couldn’t localize enough to make B a
1
DVR|that
is, to zone in on one prime of B lying above p.
27
28 3. RAMIFICATION
Proof. By localization, it is su cient to prove this when Aand Bare discrete valuation rings.
Let K; Lbe uniformizers. Then ( K) = p;( ) = P. Therefore, by de nition ( K) = ( e L). Hence K e LLis an
L-unit. The value group of Kis generated by the valuation of K; that of Lis generated by the
valuation of L. The claim is now clear. N
1.2. Pef= n. A basic fact about eand f is that they are multiplicative in towers; that is, if L Mis
a nite separable extension, Cthe integral closure in L, Q Ca prime lying over P Bwhich lies
over p A, we have:
ee
= eQ=PeP =p; fQ=p
= fQ=PfP =p
P jpXeP
e
Q
=pfP =p
=
p
Aand Bwith S
A
i
1
P ef. B=pB=
e
g
By the Chinese remainder theorem, we have
g
^
A;
ei
i
B=P
i
1
: The assertion about
efollows from (3.1), and that about fby the multiplicativity of degrees of eld
extensions. The degree n:= [L: K] is also multiplicative in towers for the same reason.
There is a similarity.
efequalsn Proposition 3.4. Let A;Bbe an extension of Dedekind domains such that Bis the integral
closure of A. For p A, we have
1
= n:
Proof. Indeed, we may replace Awith S1
eIndeed, let the factorization be pB = P
:::P g
e
+ Pj
;P e i
i
1
i
Yi=1
P
i
ePii
B, where S:= Ap. Localization
preserves integral closure, and the localization of a Dedekind domain is one too (unless it is
a eld). Finally, eand fare stable under localization, as has been remarked.
In this case, Ais assumed to be a DVR, hence a PID. Thus Bis a torsion-free, hence free, nitely
generated A-module. Since L= KBis free over Kof rank n, the rank of Bover Ais ntoo. Thus
B=pBis a vector space over A=p of rank n. I claim that this rank is also
. Now we have for i6= j, P= B, so taking high
powers yields that Pare relatively prime as well.
ej
j
as rings (and A=p-algebras). We need to compute the dimension of each factor as an A=p-vector
space. Now there is a ltration:
P B and since Piis principal (see below) all the successive quotients are isomorphic to B=P i,
which has dimension fPi=p. So counting dimensions gives the proof. N Remark. This formula
simpli es to ef= nin the complete case.
It is to be noted that rami cation behaves rather well with respect to completion. Namely,
suppose Ais a Dedekind domain with quotient eld K, La separable extension, Bthe integral
closure, Pjp an extension of prime ideals, and^ Bthe associated completions. Let ebe the
rami cation index of Pjp; let ^ebethe rami cation index of the completion. Similarly for f; ^ f.
Proposition 3.5. We have e= ^e;f= ^ f.
1. RAMIFICATION IN GENERAL 29
nonarchdisksnonarchdisksProof.
Indeed, the value groups of Kand its completion are the
same since we are in the nonarchimedean case, and one can use Lemma ee2.4; the
same is clearly true for L. The assertion about eis thus clear in view of (3.1). I claim
next that completion does not change the residue eld, which implies
the corresponding assertion about f. In detail, if ^ p is the unique nonzero prime in ^A,
then the natural map A=p !^ A= ^ pis an isomorphism. This is because if ^x2 ^ A, then
we can approximate (modulo ^ p)it with an element xof A. This means that x;^xmap to
the same element in ^ A= ^ p.
But xis in the image of A=p, so the map is surjective. As a homomorphic map of elds,
it is injective as well. N
1.3. The splitting of polynomials. In certain cases, rami cation can be described in
terms of the splitting of polynomials. Let Abe a Dedekind domain with quotient eld K,
La nite extension, and Bthe integral closure in L. Suppose there is 2Bwith B= A[ ].
This is not always the case, but we shall see it holds at the localizations for all but
a nite number of primes. Fix a prime ideal p A, and write k= A=p.
Let P(X) 2A[X] be the irreducible monic polynomial of . I claim that the splitting of p
in Bcan be described in terms of how the reduction of P factors in k[X]. In detail:
Proposition 3.6. Suppose there exists a factoring P = P1 e1
:::Pg eg
ai 2k[X], each with degrees f1;:::;fg
j
1;:::;Pg
i
;:::;Pg
).
1
i
i ei
). Since Pi
e
i
21;:::;Pg
1
a
i
i
i i ei
1;:::;eg
g
e
i
) has only one maximal
ideal, M
i
i
iMk[X]=
(Pi ei):
into
irreducibles P. Then there are gprimes Pof Blying above p, with rami cation indices
eand residue class eld degrees f;:::;f. Proof. Indeed, the key step is to note that B=
A[X]=(P). We study the ring
B=pB; its maximal ideals correspond to the primes of Blying above p. This ring is
precisely k[X]=(P), though; this in turn is isomorphic by the Chinese remainder
theorem to
The maximal ideals of B=pBcorrespond precisely to the maximal ideals of all the
k[X]=( PBut each k[X]=( Pgenerated by P. It follows that there are gsuch maximal
ideals P, and their residue class eld degrees are the degrees of P= 0 in k[X]=(
Pgenerates Mis contained in pBif and only if M, it follows that this is true if and only
if aii. In particular, the eiare the rami cation indices. N
Note that the primes P can be obtained via pB+Pi( ), if Pi
2
2 + 1, = i.
j
I
f
B
1
=
L
B
i
is any polynomial
with A-coe cients lifting . This is a straightforward corollary of the proof.
This is very useful. For instance: Example 3.7. Consider Q(i). We will show (??)
that the ring of integers (that is,
the integral closure of Z) is precisely the ring of Gaussian integers Z[i]. We will apply
the preceding proposition with A= Z, P(X) = X
is a nite direct sum of rings, then a maximal ideal in B is of the form B B2
a maximal ideal of Bj.
Mj : : : Bn, where Mis
. For any sequence
30 3. RAMIFICATION
Let pbe a prime, p>2. Then pdoes not ramify in Q(i), because X 2+ 1 has no multiple
roots in Z=pZ. If 1 is a quadratic residue modulo p, then Psplits andthe prime psplits
in Q(i) into two factors, each with residue class degree 1. If not, then pdivides only
one prime of Q(i), whose residue class degree is 2.
We shall now show that in the complete case, two special cases can be classi ed
rather completely.
2. Unrami ed extensions We work, for convenience, in the
local case where all our DVRs are complete,
and all our residue elds are perfect (e.g. nite), which ensures separability of the
residue eld extensions. This is always true in the cases of interest.
An extension of prime ideals Mjm is said to be unrami ed if e= 1. Then, unrami ed
extensions can be described fairly explicitly.
unr Proposition 3.8. Notation as above, if L=Kis unrami ed and with rings of integers R;Sand
maximal ideals M;m, then we can write L= K(a) for some a2S with S= R[a]. The irreducible
monic polynomial P satis ed by remains irreducible upon reduction to the residue eld K. Proof. Fix DVRs R;Swith quotient elds
K;Land residue elds K;L. Let
ebe the rami cation index, f the residue class degree, and n= [L: K]. Recall that since
ef= n, unrami edness is equivalent to f= n, i.e.
[L: K] = [L: K]: Now by the primitive element theorem
(recall we assumed perfection of K), we
can write L= K( ) for some 2L. The goal is to lift to a generator of Sover R.
0There is a polynomial P(X) 2K[X] with P(a) = 0; we can choose Pirreducible and thus
of degree n. Lift P to P(X) 2R[X] and ato a2R; then of course P(a 0) 6= 0 in general,
but P(a) 0 mod M if M is the maximal ideal in S, say lying over m R. So, we use
Hensel’s lemma to nd areducing to awith P(a) = 0|indeed P0(a0) is a unit by
separability of [L: K]. It follows from the next lemma that S= R[a], since there is a
uniformizer for S
contained in Rbecause the extension is unrami ed. As a consequence, L= K(a), and
Pmust be irreducible (since it is of degree n). N
The next lemma is useful in nding generators for a ring:
generatedbytwoelements Lemma 3.9. Let A Bbe an extension of discrete valuation rings (not necessarily complete)
with maximal ideals p;P. Suppose 2Bhas image in B=P generating the eld extension B=P=A=p, and
suppose 2Bis a uniformizer. Finally, suppose Bis nitely generated as an A-module. Then B= A[ ; ].
3Proof.
Consider the A-submodule C = A[ ; ] B. We will prove that the map
C=pC!B=pBis surjective; this implies that B=C= p(B=C), and by Nakayama’s lemma,
we’ll have C= B. So far, we know that C= C!B= Bis surjective by assumption on . In
otherwords, we have B= C+ B. Repeating, we have B= C+ (C+ B) = C+ 2eB. And so
on, proceeding by induction, we get B= C+ Bfor ethe rami cation
3Cf. eisenbud [?].
2. UNRAMIFIED EXTENSIONS 31
index. But this says precisely that B= C+pB, and proves surjectivity of C=pC!
B=pBand the lemma. N
We briey digress to give an application of this lemma. Proposition 3.10. Suppose
A Bis an extension of (not necessary complete)DVRs such that Bis the integral
closure of Ain a nite extension eld. In addition, suppose the extension of
residue elds is separable. Then there is 2B with B= A[ ].
This shows that the situation in Proposition ?? is not uncommon at all, and in
particular is always true in the local case.
0Proof.
Let p;P be the prime ideals of A;B. Suppose y2Blifts a generator yof B=P=A=p. Then
there is a polynomial P2A[X] with P(y) = 0, but P2(y) 6= 0 by separability; we could take the
irreducible polynomial, for instance. This means that if P(y) 2P, and if it is not a uniformizer (i.e.
is in P), then P(y+ ) P02(y) mod Pis one for a uniformizer. So, if we take to be either y(if P(y)
is a uniformizer) or y+ (if P(y+ ) is a uniformizer), then generates the extension B=P=A=p, and
P( ) is a uniformizer.generatedbytwoelementsgeneratedbytwoelements
By Proposition ??, Bis a nitely generated A-module, so Lemma3.9 applies
(with ;P( )) and concludes the proof. N
There is a converse as well, about unrami ed extensions. We keep the same
notation: a nite extension of complete elds L=Kwith discrete valuatinons, rings of
integers R;S, and maximal ideals m;M.
unrconverse Proposition 3.11. If L= K( ) for 2S whose monic irreducible P remains irreducible upon
reduction to K, then L=K is unrami ed, and S = R[ ]. More
4generally, the conclusion holds if satis es a monic polynomial in R[X] whose
reduction has no multiple roots.
Proof. First suppose P irreducible. Consider T := R[X]=(P(X)); this is clearly an
integral domain and (by an obvious injection) a subring of Lcontained in S. I claim that
T’S. This will evidently prove the claims.First, we show T is a DVR. Now T is a nitely
generated R-module, so any maximal ideal Q of T must contain mT by the same
Nakayama-type argument. Indeed, if not, then Q + mT= T, which contradicts
Nakayama’s lemma.
In particular, a maximal ideal of T can be obtained as an inverse image of a
maximal ideal in
TRK= K[X]=(P(X)) by right-exactness of the tensor product. But this is a eld by the
assumptions, so
locprincipalmeansDVRlocprincipalmeansDVRmT is the only maximal ideal of T. This is
principal, so T is a DVR by Proposition??. It must thus be the integral closure S, since
the eld of fractions of T is L.
Now [L: K] = degP(X) = degP(X) = [L: K], so unrami edness follows. In the general
case (the last sentence), the reduction of the irreducible poly-nomial for has no
multiple roots. But this reduction is Q0(X ( )), where ranges over the embeddings
of in an algebraic closure of K. The hypothesis means that ( ) 6= ( ) for 6= 0 .
Since the map ! is surjective, it follows that has nconjugates over A=p, THIS
NEEDS WORK N
4No multiple roots in the algebraic closure of the residue eld.
32 3. RAMIFICATION
pExample
3.12. This example will be useful in class eld theory. Let kbe a local eld,
i.e. a nite extension of Q1=n. Suppose p- n, and in particular, nis a unit in the ring of
integers. Suppose 2kis a unit. Then the extension k( n)=kis unrami ed, because the
polynomial X has no multiple roots in the algebraic closure of the residue eld (since it
is prime to its derivative nX n1). Remark. This example, and all the previous results,
can apply to the global case
too. In particular, if Ais a Dedekind domain with quotient eld K, 2Kis a unit in Apfor p
a prime ideal, and nis a unit at p as well, then the extension K( 1=n)=K is unrami ed at
all primes lying above p. This follows from the preceding example applied to the
completions.
Proposition 3.13 (Le sorite5for unrami ed extensions). Let K Lbe an unrami ed
extension of local elds.
(1) If M=Lis unrami ed, then so is M=K(\composition") (2) If Eis an extension of
K, then LE=Eis unrami ed (\base extension") (3) If L0is another unrami ed
extension of K, then LL0=K is unrami ed (\ bered products")
unrunrProof.
(1) is immediate from multiplicativity in towers, and (3) follows from (1)
and (2). It remains only to prove (2). By Propositionunrconverseunrconverse3.8, we can
write L= K( ) where satis es a polynomial with no multiple roots in the algebraic
closure of the residue eld. Since LE= E( ), it follows by this property of and
Proposition3.11 that LE=Eis unrami ed too. N Remark. By (3) of the above
proposition, any extension L=K has a maximal
unrami ed subextension Lun=K. Moreover, the extension L=Lunhas no choice but to be
totally rami ed.
Digression 3.14. We briey digress, following Milne, to explain how these ideas are
useful in determining Galois groups of number elds. Let Kbe any eld and f2K[X].
Recall that the Galois group of f is the Galois group of a splitting eld for f.
Now suppose Kis a number eld with ring of integers OK and P2OK
K
P= P1
:::Pk
irreducible and distinct in K[X]. . Claim. The
Galois group of P(regarded as a subgroup
of the symmetric group Sn
[X] is a
monic separable polynomial of degree n. Let p Obe a prime ideal with residue eld
K= O=p. We shall connect the factorization of Pmodulo p with the Galois group of P.
Suppose modulo p (we use the bar to denote reduction mod p), we have
; where each Pihas degree di, for the Pi
::: k, where i is a di
of permutations of the roots of n) contains a
1
i
unrconverseunrconv
erse
5
-cycle and the ’s
are pairwise disjoint. Proof of the claim. First, this factorization implies that p does not
ramify
(or more precisely, any prime prolonging p does not ramify) in the splitting eld L of P,
in view of Proposition3.11 and the above remarks. Also, the roots of Pare distinct
modulo any prime dividing p.
I have no idea what this means, which is bad since I take French, but in the highly entertaining book
Introduction to Grothendieck Duality Theory by Altman and Kleiman, it is used for results that say
\Property X" is preserved under base extension, composition, etc.
3. TOTALLY RAMIFIED EXTENSIONS 33
Fix any prime P of Llying above p, and let Lbe the residue eld of Lmodulo P.
Let 2G(L=K) be the Frobenius element. Then generates G(L=K). Now acts on the
roots of P(all of which lie in L). So if decomposes into a product of r disjoint cycles of
degrees t1;:::;tr, it follows that Pmust factor into a product of r irreducible polynomials
of degrees t1;:::;tr6.This means in particular that r= k and up to rearrangement, ti= ri.
But is the reduction of some 2G(L=K) in view of Proposition ??. This means
that must be as described in the claim, since has such a description as an element
of Sn. N This result can be used in computing Galois groups. For instance, if P(X)
2Z[X] is irreducible of degree nand remains irreducible modulo a prime, then the
Galois group of P contains a n-cycle. If nis prime, and P has precisely one pair of
complex conjugate roots, then the Galois group is the full symmetric group Sn
because it also contains a transposition (namely, complex conjugation). 2.1. The
inertia eld. Recall that any extension of local eld L=K has a
maximal unrami ed subextension. In the global case, if L=Kis an extension and P a
prime of L,
3. Totally rami ed extensions We next consider the case of a
totally rami ed extension of complete elds
K L, with residue elds K;L|recall that this means e= [L: K] = n;f= 1. It turns out that
there is a similar characterization as for unrami ed extensions.
Proposition 3.15. Given a totally rami ed extension of complete elds K L with rings of
integers R;S, we can take a uniformizer 2Swith S= R[ ] and such that the irreducible
monic polynomial for is an Eisenstein polynomial.
sysrepsysrepProof. Choose a uniformizer of the DVR S. I claim that S= R[ ]. This
follows easily from Proposition2.22 or Propositiongenbytwoelementsgenbytwoelements??.
Your choice. Now, with that established, note that L= K( ) as a consequence.
Consider
+ an1Xn1
+ + a0
7is a uniformizer in R.
n1
n
0
i
0
0= n
i
ii
6
1;
: : : ; tr
7
minimal polynomial of , Xn
the:By de nition, we have + an1+ + a= 0: Because of the total rami cation
hypothesis, since each a2R, any two terms in the above sum must have di ere
orders in S|except potentially the rst and the last. Consequently the rst and
last must have the same orders in S(that is, n) if the sum is to equal zero, so a
unit, or a
Moreover, it follows that none of
the a;i6= 0, can be a
unit|otherwise the order of the
term awould be i<n, and the rst
such term would prevent the sum
from being zero. Hence this
polynomial is an Eisenstein
polynomial. N
In particular, what all this
means is that Pis an
Eisenstein polynomial: Now
we prove the converse:
Proposition 3.16. If K Lis an
extension with L= K( )
where 2Ssatis es an Eisenstein
polynomial, then S= R[ ] and L=Kis
totally rami ed.
This follows from a general fact,
easily proved: If the Galois group of a
polynomial P has r orbits of size t, then P
factors into r irreducible polynomials of
degrees t1; : : : ; tr.
We have repeatedly used the following
easy fact: given a DVR and elements of
pairwise distinct orders, the sum is
nonzero.
34 3. RAMIFICATION
Proof. Note rst of all that the Eisenstein polynomial P mentioned in the statement
is necessarily irreducible. As before, I claim that
T:= R[X]=(P(X)) is a DVR, which will establish one
claim, namely that S = R[ ]. By the same
Nakayama-type argument in the previous post, one can show that any maximal
ideal in T contains the image of the maximal ideal m R; in particular, it arises as
the inverse image of an ideal in
TR K= K[X]=(Xn
n a0
2mod
by (X;a0) for a0
n
n
); this ideal must be (X). In
particular, the unique maximal ideal n of Tis generated
2Aa generator of m, which without loss of generality we can take as
the last (constant) term in the polynomial P. But, since P(X) is Eisenstein and the
leading term is X, it follows that X. This also implies that X is nonnilpotent in T.
Now the following lemma implies that Tis a DVR: Lemma 3.17. Any commutative
ring Awith a unique principal maximal ideal m
serre
generated by a nonilpotent element is a DVR.
Proof of the lemma. This is a lemma in
n n+1m
n
eisenbud
0
A;Mod
1
8
M
k
i=Mi1So,
any A-module
M2Mod
0
M
i
0 i 00
for the pi
i
, we have (M) = (M0 !M0
A
A
!M!M00
8
[6], but we can take a slightly
quicker approach to prove this. We can always write a nonzero x2Aas x= u for ua
unit, because x2mfor some n|this is the Krull intersection theorem, cf.[?]. Thus from
this representation Ais a domain, and the result is then clear. N
We now resume the proof of the proposition. There is really only one more step,
viz. to show that L=Kis totally rami ed. But this is straightforward, because Pis
Eisenstein, and if there was anything less than total rami cation then one sees that
P( ) would be nonzero|indeed, it would have the same order as the last constant
coe cient a. N Proposition 3.18 (Le sorite).
4. Two homomorphisms Let Abe a Dedekind domain with
B
quotient eld K. 4.1. Some abstract nonsense. Consider the
categories Mod
= M such that the quotients Mare isomorphic to R=m i for the
maximal
mi
ideals.
of modules of nite length over A;B, respectively. It is a well-known fact that for any
noetherian ring R, a module Mof nite length has a nite ltration
has a ltration whose quotients are isomorphic to A=pnonzero. Let (M) = Qp. It is
clear, by the Jordan-Holder theorem, that for an exact sequence
!0 in Mod) (M). Consequently, there is a homomorphism : K(Mod) !Fr(A) for
Fr(A) the group of fractional ideals. Here Kdenotes the Grothendieck group. This
A map sends A=I!Ifor Ia nonzero integral ideal.
With only a nite generation hypothesis, one would only have the m
prime.
4. TWO HOMOMORPHISMS 35
Proposition 3.19. is an isomorphism. Proof. Mod Ais an artinian category, so the
Grothendieck group is free on the simple objects, which are of the form A=p for p
prime. Similarly, Fr(A) is free on the prime ideals p, and the appropriate bases of the
two groups get mapped to each other. N
4.2. De nitions of N;I. Let Lbe a nite separable extension and B the integral closure
in L. There is an exact functor
Mod A!Mod
B
L Kgiven
by restriction. Indeed, it is easy to see that a module over a Dedekind
domain is of nite length if and only if it is nitely generated and annihilated by a
nonzero element, and if x2Bannihilates M, N(x) 2Aannihilates M. Moreover, B is
torsion-free over A, so hence is at (cf. ??); there is thus
B!Mod;M!B BA
);K(Mod A A) !K(Mod
A
B
:
B
A
S1A
S1BS 1A
S
n exact functor Mod A
A
1
A
S1B
S
M. These induce homomorphisms
on the Grothendieck groups K(Mod) !K(Mod),
a
and corresponding homomorphisms Fr(B) !Fr(A);Fr(A) !Fr(B). These will be called,
respectively, the norm and inclusion homomorphisms.
It is easy to see that these commute with localization. Indeed, if S Ais a
multiplicative set, we have a commutative diagram
Mod
B
//
Mod
A
Mod
1B
//
M
o
d
The next lemma could have been taken as the de nition of N and i, since both
are multiplicative and consequently determined by their values on the prime ideals.
fProposition
3.20. Let p;P be prime ideals of A;B, with P jp. (1) We have
NP = p, for fthe residue class degree [B=P : A=p]. (2) We have i(p) = pB.
Proof. The proof of (2) is easy: p corresponds to the image of A=p in K(Mod A). Under
the tensor product functor, this is mapped to B=pB.
The proof of (1) is also straightforward. The A=p-module B=P is a f-dimensional
vector space. So as an A-module, B=P has a ltration of length fwith quotients
isomorphic to A=p. This implies (1). N
We now study these homomorphisms further; the proofs of their properties will
rely on elementary facts about rami cation.
nProposition 3.21. We have N i(a) = afor any ideal a A. Proof. It is enough to
prove this for prime ideals p. But i(p) = pB =
1 e:::Pgg for primes P1;:::;Pg
P eifi
n= p
g
e
of B. Let the residue class eld degrees
be f;:::;f. Then the norm of this product is
P11 equal to pefequalsnefequalsnby Proposition3.4. N We next show that the ideal norm is
a generalization of the usual norm. to be
xed!!
36 3. RAMIFICATION
Proposition 3.22. If b Bis an ideal, we have N(b ) = Q b Proof. The norm commutes
with localization, as is easily seen (to be added).
Therefore, by localizing at each prime, it su ces to prove the proposition when A and
Bare principal ideal domains N
5. Completions revisited Let kbe a eld with a valuation
v(possibly archimedean) and La nite separable extension. Proposition 3.23. We have an isomorphism of L-algebras and
topological rings,
Lk kv wjv’ M Lw;
where kv;Lw kkvdenote completions and wranges over all extensions of vto L. First, Lis a
direct product of elds. Indeed, we can write L= k(Alpha) = k[X]=(P(X)) for
suitable Alpha2L;P(X) 2k[X] by the primitive element theorem. Let P split in k v[X]
as P = P1:::Pr; there are no repeated factors by separability. Then
Lk kv ’kvi[X]=(P(X)) ’ M
kv[X]=(Pi):
So Lk kv is isomorphic to a direct product (i.e. direct sum) of elds kv[X]=(P). I claim now that
the elds Li= kv[X]=(Pi) are precisely the Lwifor wprolonging v. Indeed, rst note that
kv[X]=(P), as a nite extension of the complete eld kvikkv, has a unique extension
valuation with respect to which it is complete. The composite map L!L!k v[X]=(Pi)
induces a map of L!Land a valuation on Lextending v. Lis dense in Lii, since
Alphagets mapped to the image of Xin k v[X]=(Pi). So Liis the completion of Lwith
respect to the valuation of L induced on it.
What we now need to show is that the valuations induced on Lare distinct and
that every L-valuation prolonging vcan be realized in this way. Start with the rst
task.
i
i
kv is a nite-dimensional space over kv
Lk
;L!Lj
r
k
k
v
i
i
, so it has a canonical topology induced by a (non-unique) norm, and the topologies
on the Lare the subspace topologies. Also, Lis dense in it. If the morphisms
L!Linduced the same valuation on Lfor di erent i;j, then we would not be able to
approximate a vector inL Lwhose i-th coordinate was 1 and whose j-th coordinate
was zero arbitrarily closely by elements of L.Suppose we have a valuation w
prolonging v on L. Then w extends to a multiplicative function on Lby continuity.
This must be the norm on one of the L, since it is not identically zero. So we can
get any wby pulling back from some L. Corollary 3.24. For a nite separable
extension L=k,
: kv] = [L: k]
w
ef= nequality.
j
v
X
[
L
w
In the case of va discrete valuation, this is just the P
6. THE DIFFERENT 37
6. The di erent We begin with generalities on duals. 6.1. Dual
modules. In reality, we shall only care about the speci c case of
the di erent, but it it useful to start things more generally. Let Abe a Dedekind domain with
quotient eld K, V a nite-dimensional K-vector space. Suppose M is a torsion-free, nitely generated
submodule of V that spans V; we call such a module a lattice. For instance, A n K nsatis es this
condition. In general, however, Mneed not be free.
Suppose given an K-linear, symmetric bilinear form B: V V !K, which we always assume
nondegenerate. De ne the dual module M0of Mto be v2V : B(v;M) A:
We start by handling the easiest case. Proposition 3.25. Suppose M is a free A-module with
basis e1;:::;e. Let f1;:::;fnbe the dual basis to e1;:::;en, i.e. such that B(ei;fj) = ijn. Then M0is A-free on
f1;:::;fn. Proof. It is clear that each fi, and hence their span, is in M0. Conversely, any x2V can be
written as x=P aififor ai2K. If in fact x2M0, then taking B(x;ei) 2Afor each ishows that a2Aby the dual
bases relations. N Corollary 3.26. M0iis a lattice if Mis one. Proof. First of all, there are This is
because M. N
We shall often have to deal with localization to reduce problems to the case of a PID, so let Sbe a
multiplicative subset. Then S1Ais a Dedekind domain, S1Msatis es the conditions with respect to
S1A. It is clear that
locgoodduallocgooddual (3.2) S1(M0) = (S1
M)0
in view of the nite generation of M. are lattices in
with bilinear forms B1;B2
Proposition 3.27. Suppose M1;M2 V1;V2
1
1
V2 0 =
M0 2
M2
M
0
i
1
0
B=A
T
r
. Then M M2is a
lattice in V1, and with the obvious bilinear form Bon V V2, we have (M1): Proof. Clear. N
L=KDe
nition-Example 3.28. Let Abe a Dedekind domain with quotient eld K, L=Kbe a nite
separable extension and Bthe integral closure in L. Consider the form (a;b) !Tr(ab). The dual module
of B Litself, that is B, is a lattice of L. It is in fact a B-submodule, since
x2B0L=K(xy) 2A8y2B: It is thus a fractional ideal, and its inverse is called the di erent D. In
particular,
= DS1B=S1A
S1
locgoodduallocgooddualB. Proof. Clear from (3.2). N 6.2. The di erent and
Proposition 3.29. S1DB=A
rami cation. We now want to prove that the discriminant is divisible by all the rami ed primes.
4
Number elds
In this chapter, we drop the generality of the previous ones, and restrict ourselves
to the case of primary importance in the rest of these notes.
1. Introductory remarks De nition 4.1.
A number eld is a nite extension of Q.
It is of interest to nd the absolute values on a number eld. These necessarily
extend ones of Q. And we have:
1Theorem
4.2 (Ostrowski). The only nontrivialabsolute values on Q (up to
equivalence) the p-adic valuations and the usual archimedean one.
Proof. Omitted. N In particular, the nontrivial absolute values on a number eld
Kfall into two
categories. 1.1. Valuations. There are some absolute values j jthat are
nonarchimedean,in which case they extend j jpon Q for some prime p. Let OKbe the
integral closure of Z in K, i.e. the ring of integers; this is a Dedekind domain
(Proposition ??).
Then the fact that j jis 1 on Z and <1 on (p) Z means that jOKKj [0;1] and that
OKdoes not consist solely of units with respect to j j. In particular, j jdetermines a
prime ideal of OKprolonging (p), and is the discrete valuation associated to this prime
ideal. Conversely, every prime ideal of Odetermines a distinct nonarchimedean
valuation of K.
21.2.
Archimedean absolute values. There are some absolute values j jthat are
archimedean, and consequently prolong the usual absolute value on Q (up to a
power). I claim that there is a distinct one corresponding to each real embedding K!R
and each pair of complex conjugate embeddings K!C (where a complex embedding is
required not to fall entirely into R !).
Indeed, it is clear that every such embedding determines an absolute value on Kby
pulling back. Two conjugate embeddings K!C obviously determine the same
absolute value. Conversely, given an archimedean absolute value on K, we
have an embedding into the completion, K! ^ K. But
^ Kis a nite extension of
In particular, if there are r1 real embeddings and
r2
+
r
2
archimedean
absolute
values.
1
39
1
2
^ Q = R . Thus we get an embedding into either R or C preserving the absolute
value.
complex embeddings, there
are r
There is a trivial absolute value taking the value 1 at all nonzero elements.
Note that the reals have no nontrivial absolute-value-preserving automorphisms (or any others, for that
matter, but it is irrelevant), while the complexes have one: complex conjugation. This explains the
somewhat strange asymmetry.
40 4. NUMBER FIELDS
1.3. The product formula. There is a striking global relation between the absolute
values on a number eld. The setup is as follows. Kwill be a number eld, and van
absolute value (which, by abuse of terminology, we will use interchangeably with
\place") extending one of the absolute values on Q (which are always normalized in
the standard way), that is to say jpjfor the output at x2K. Suppose v0is a valuation
of Q; we write vjv0p=if vextends v01 p; we will write jxjv. Recall the following important
formula from the theory of absolute values an extension elds:
K QjN(x)jv0 v
[Kjxjv:Qv 0] v:
j
P
1
v
r
3
=
o
4
o
Y
f
0
.
v
: Qv0
Pvjv0 Nv
v
Write N
In the nonarchimedean case, it follows from ??. In the archimedean case, it follows
because the norm is the product
The ’s each correspond to one extension of
j jQ xfor ranging over the eld
embeddings of K!C .
:= [Kv
for x2K. Theorem 4.3 (Product formula). If
x6= 0,
on Q to K, and the only way two di erent embeddings yield the same absolute value
is if they are complex conjugates. N
]; these are the local degrees. From ?? (or the same reasoning in the archimedean
case), we have= N := [K : Q]:This is essentially a version of theP ef= Nformula.The
above formalism allows us to deduce an important global relation between the
absolute values jxj
v = 1:
Njxjvv = 1:
vY
p
l
1
Nv
Nv
KQ
v
v v
v0
j
vjv0 0
x
j
Proof. The proof of this theorem starts with the case K= Q, in which case it is an
immediate consequence of unique factorization. For instance, one can argue as
follows. Let x= p, a prime. Then jpj= 1=pby the standard normalization, jpj= 1 if l6=
p, and jpj1= pwhere j jis the standard archimedean absolute value on Q. The
formula is thus clear for x= p2Q, and it follows in general by multiplicativity.
By the above reasoning, it follows that if x2K, Y=
YYjxj= YjNxj
v
0
K
K
0
f
3Cf. langalgebra [4] for a
generalization.
K Q(a)
Z.
4This includes both real and complex embeddings.
This completes the proof. N 1.4. The function N. For future reference, we pause to
de ne a Z-valued function N on the integral ideals of O. Speci cally, we let Na =
Card(O=a): In particular, if a is a prime p dividing p2Z with residue degree f, then Np
= p. The next proposition shows that we are already familiar with N.
Proposition 4.4. The function N is multiplicative: N(ab) = NaNb . In fact, if a is an
ideal, then Na is the positive integer generating the ideal N
2. THE RIEMANN-ROCH PROBLEM IN NUMBER FIELDS 41
ZProof. Clearly it is su cient to prove the second statement, since the ideal norm is
multiplicative. Let Mbe a Z-module of nite length, i.e. M2Mod. Then I claim that the
positive integral generator of the ideal Z Z(M) is CardM(which is nite!). This is clear
for M= Z=pZ for a prime Z, and since every M2Modhas a ltration with subquotients of
this form, the claim is clear.
K=a)) = ZNow, if a is an ideal, we have the following equality of Z-ideals: (Na) =
(Card(O(OK=a) = NK Q(a) by a look at the de nitions of the norm and inclusion
homomorphisms in terms of
the Grothendieck group. This completes the proof. N Incidentally, it would not be hard
to prove this directly. For instance, the
2proposition is a direct corollary of the Chinese remainder theorem when a;b are
relatively prime. Moreover, by localization it is easy to check that p=p 22’A=p, so that
Np= (CardA=p)(Cardp=p2) = (Np)K: Corollary 4.5. Let a2O. Then the order of O=(a) is
the absolute value of NK Q(a). NK
2. The Riemann-Roch problem in number elds We shall be interested in
studying elements x2Kwith certain distributionsof its absolute values jxj v. In particular,
let us pose the following problem. Problem. Suppose we have constants c for each
place v. When does there exist x2Kwith jxjv cvvfor all v? There is an analogy here to
algebraic geometry. Suppose C is a projectivenonsingular curve. Let Dbe a divisor,
i.e. a formal nite sum PP2CnP. When does there exist a rational function fon Cwith
ordP(f) nPPfor all P? Information on this problem is provided by the Riemann-Roch
theorem. Inparticular, if jDj:= PnPis very large, then we can always nd a lot of such
fs. In particular, when jDjis very large, the dimension of the space L(D) of such fis
approximately proportional to jDj.Note the strength of the analogy: the valuation rings
of the function eld of C are precisely the local rings of the points of C. Moreover, the
product formula is analogous to the fact that the principal divisor associated to a
nonzero meromorphic function has degree zero.
In our case, we will de ne a K-divisor as follows: De nition 4.6. A K-divisor c to be
an assignment of constants cfor the valuations vof K. We will assume that c vv= 1
almost everywhere (i.e. for a co nite set); this is analogous to a divisor in algebraic
geometry being nite. Also, when v is nonarchimedean, we assume cvvin the value
group of v. De ne L(c) to be the set of x2Kwith jxj cvfor all v. We will show that
cardL(c) is approximately proportional to
kck:= Y Ncv
v
: Theorem 4.7. There are positive
constants a;bdepending on Konly such that
akck cardL(c) b(1 + kck):
42 4. NUMBER FIELDS
Note that we need the 1 on the right side, because 0 2L(c) always! The proof of
this theorem will not use any of the fancy machinery as in the
Riemann-Roch theorem (in particular, no Serre duality). It only gives bounded
proportionality; one can prove a stronger asymptotic result, but we shall not do so
here.
Proof. We rst do the right inequality. The idea is simple. If there were a lot of
points in L(c), then by the pigeonhole principle we could nd ones that very close
together. But the di erence of any two points in L(c) has bounded absolute values
everywhere, so if the di erence is very small at even one valuation, then the product
formula will be violated.
5In detail, it is as follows. Suppose wis a real archimedean absolute value of
K.w;cwLet n= cardL(c), and suppose n>1. The place wis obtained by pulling back
the usual absolute value on R by a real embedding : K!R that sends L(c) into [c].
Then, by the pigeonhole principle, there are distinct points x;y2L(c) with
2cw
2c
j
v
x
y
j
w
v nonarch
2c
N
[K:Q] Y
N . kck=n:
=
v)
jxyj
v
v6=w
v6=w
w
v1
=
Yjxyjv
w
v
v
jxyj
2
v
m
2
v
p 2c 2
=n: We will now apply the product formula to xy. Note that jxyj
w
N
w
Y
c v
c
archY(2
n
c
2
v
v
2c
Nv
v6=w
v
at the archimedean absolute values, but we do not need the factor of 2 at the
nonarchimedean ones. Then, by splitting the product into wand non-wvaluations,
and those into the archimedean and non-archimedean ones, we nd:
Here the symbol . means up to a constant factor depending only on K. This proves
one side of the equality when there is a real archimedean absolute value of K.
We still need to prove that side of the inequailty when there are no real absolute
values. In this case, Kmust have a complex absolute value warising from a complex
embedding : K!C . The embedding carries L(c) into the square region Sof side
2ccentered at the origin. If n= L(c), and m<nis the largest square less than n, we
can slice the square region Sinto msubsquares of side length. One such subsquare
must contain two distinct elements x;y2L(c) by the pigeonhole principle. Then by
elementary geometry,
2c
m:
At the other archimedean places v, we have jxyj
v cv
v N nonarch N
p 2c [K:Q] Y N . kck=n:
w
2
= jxyj
v1 =
Yjxyjv
Yv6=w
v6=w
v
. At nonarchimedean places v, we have jxyj. The same product formula can now be
repeated, since m>n=2:
2 arch
2
) v Y
c v
c v
Nv
w
v6=w
2 wn=
2
2
(2c
vcv.
ThenNow, we prove the left hand inequality, which bounds below cardL(c). This
will be the necessary part for the unit theorem. First o , note that we can de ne the
product divisor xc if c is a K-divisor and x2c as the family v!jxj
5That is, one arising from a real embedding. The case where K admits only complex embeddings will be
handled below.
2. THE RIEMANN-ROCH PROBLEM IN NUMBER FIELDS 43
cardL(c) and kckare una ected by scalar multiplication by elements of K, in view of the product formula. It will
be necessary to do a bit of normalization, only after which we can use a Chinese remainder theorem-type
argument, and hence this observation will be useful.
artinwhaplesartinwhaplesBy multiplying c by an element of K, we may assume 1 cv 2 at the archimedean places,
by the approximation theorem2.16. Next, we choose d 2 Z divisible by a lot of prime numbers, so that dc has
absolute value 1 at all nonarchimedean primes. Replace c with dc.
So, here is the situation. There is a c satisfying cvv 1 for vnonarchimedean and d c 2d for v archimedean.
We need to show that there are & kck elements in L(c). But, L(c) has a nice interpretation in this case,
since c 1 for nonarchimedean v. There is an ideal a =Q pordpcpvvsuch that the elements of a are precisely the
x2Kwith jxj cv6for vnonarchimedean. The points of L(c) are just elements of a whose archimedean absolute
values satisfy certain bounds. In particular, we are counting points of a lattice (namely, a) in a certain
space region.
Nd
while kc
K
Indeed, let cp
f
nonarchvYConsequently,
kck d
v
1
w
p
)
=
(p1=
K
i
w
x= Xaixi;8ai
e =
. It follows that Np =
f
pf kc
p
:
e
N
N
N
a
:
1=
e
N
p
)
v
c
Na =
If we show that
atinowneedtoprove (4.1) CardL(c) cK
=Na for an absolute constant c>0 (depending only on the eld K) w
the normalization of the absolute values is carried out such that kc
as we now show.
whatinowneedtoprovewh
be the divisor with 1’s everywhere except the a
associated to the prime ideal p, but with order 1 at p. Suppose p j(
rami cation index e, residue class eld degree f. Then Np = pk= (p
k: More generally, we see by the same reasoning that
;:::;xN
ij
We shall now bound below CardL(c). Let x
2Z;ja
(with N= [K: Q]) be a basis of the free Z-module O. Then if Wdenotes the the maximum of Njxfor
warchimedean and 1 i N, then Wdepends only on K. Moreover, any sum
K
N
Nd
WN
N
N
a
t
l
e
a
s
t
. Also, since there are Na di erent classes of Omodulo a, it follows that th
least
Na belonging to the same class modulo a. Subtracting these from each o
j
get
d=W satis es jxj dfor warchimedean. The number of such xas
in the above expression with 0 a
d Na distinct elements of a whose norms at the archimedean places
W
are
6This remark will be elucidated later.
44 4. NUMBER FIELDS
whatinowneedtoprovewhatinowneedtoproveat
most d. All these belong to L(c) since they
belong to a and satisfy the condition on the archimedean places. This proves that
(4.1) with cK= W N, and completes the proof of the theorem. N
3. The unit theorem Following the philosophy of
examples rst, let us motivate things with an example. Example 4.8 (Units in quadratic elds). Consider the ring Z[i] of Gaussian integers. Claim. Z[i] is the ring of integers in the quadratic eld Q(i).
2+ b2Proof. To see this, suppose a+ib;a;b2Q is integral; then so is its conjugate aib,
and thus it follows that 2a;2bare algebraic integers, hence rational integers. Since (a+
ib)(aib) = amust be an integer, neither a;bcan be of the form k=2 for kodd. N
2+b2What are the units in Z[i]? If xis a unit, so is x, so the norm N(x) must be a unit
in Z[i] (and hence in Z). So if x= a+ib, then a= 1 and x= 1 or i. So, the units are
just the roots of unity.In general, however, the situation is more complicated.
Consider Z[ p 2], which
7is again integrally
Then x= a+ b p
closed.
p
2 2b2
K
, O=p\OpK
places S1
7
2 is a unit if and only if its norm to Q, i.e.
ais equal to 1. Indeed, the norm N(x) of a unit xis still a unit, and since Z is integrally
closed, we nd that N(x) is a Z-unit. In particular, the units correspond to the solutions
to the Pell equation. There are in nitely many of them.
But the situation is not hopeless. We will show that in any number eld, the unit
group is a direct sum of copies of Z and the roots of unity. We will also determine the
rank.
3.1. S-units. We start with an easy proposition. Let K be a number eld with ring
of integers O. Then: Proposition 4.9. x2Kis a unit if and only if its order at any
nonarchimedeanvaluation is zero. Proof. A unit must obviously have order zero at
each prime. Conversely,
suppose xhas order zero at each prime. Then xbelongs to the localizations Ofor each
nonzero prime ideal p. But, if we abbreviate O= O
1from
a basic result in commutative algebra true of any integral domain. Hence if x
satis es the order condition, then x2O, and by symmetry x2O, proving the result. N
Motivated by this result, we de ne a generalization of the notion of unit: De nition
4.10. Let Sbe a nite set of places of Kcontaining the archimedean
. We say that x2Kis a S-unit if xhas order zero at the places outside S;
denote the set of S-units by U(S).
The reader may either prove this directly or skip ahead to ?? to see the general case.
3. THE UNIT THEOREM 45
This is clearly a generalization of the usual notion of unit, and reduces to it when
S= S1. Alternatively, if we consider the ring OSSof elements of Kintegral outside S,
then U(S) is the unit group of O. 3.2. The unit theorem. The principal goal in this post
is to prove the easy
half of: Theorem 4.11 (Dirichlet unit theorem). The group U(S) is isomorphic to a
direct
v L G, where Gis the group of roots of unity in K.
sum ZjSj1
h(x) = (Nv
v
Proof. De ne a mapping h: U(S) !RS
by (where N
S
logjxj )v2S
=xv
n
de ned by (yv)v2S
the image of his a lattice8
v
1
m
S
v.
Thus h(U(S)) is a lattice in RS
8
], as usual)
so that by the product formula, h(U(S)) is contained in the subspace W R2Wi Pyv=
0. I claim that: rst, the kernel of hconsists of the roots of unity; and second,in W.
Together, these will prove the unit theorem after we have computed the rank (which
we will do later).
First, it is clear that if x2ker h, then jxj= 1 for all places v, because x2U(S). We now
v
= [Kv
:Q
need a lemma.
Lemma 4.12. If x2Kfor Ka number eld and jxjv= 1 for all absolute values v, then xis a
root of unity.
Proof. Note that every power of xsatis es the same condition. We now need a
sublemma.
Sublemma 4.12.1. Suppose Cis a constant. Then there are only nitely many points
xof Ksuch that jxj Cfor all places v. Proof of the sublemma. The characteristic
polynomial of x(as an endoNmorphism of K) has degree N = [K: Q] and rational coe cients which are the
symmetric functions of the conjugates of x, and which thus are bounded at all rational
places by D= D(C); for instance, we could take D= N!Cp. However, there are
only nitely many rational numbers rsatisfying jrj Dfor all pand jrj D, as is easily
checked, so there are only nitely many possible characteristic polynomials. Now
xsatis es its characteristic polynomial, so there are only nitely many possibilities for
it. N
To nish the proof of the lemma, note that all powers of xhave norm 1 at all
places, so by the sublemma we must get xfor some m6= n, so xis a root of unity. N
We have shown that the kernel of hconsists of the roots of unity. In addition, we
can use the lemma again to show that h(U(S)) must be a discrete set, since any
bounded region in Rcan contain only nitely many points of h(U(S))|-indeed, if x2U(S)
is mapped by hinto a bounded region, then that yields a bound on all of the jxj,
consequently free abelian. The exact sequence
U(S) !h(U(S)) !0
A lattice is a discrete abelian subgroup of a euclidean space. It is a theorem that a lattice is nitely
generated, hence free. Cf. ..uh, I dunno.
46
4.
NU
MB
ER
FIE
LD
S
must
ther
efor
e
split,
and
U(S)
is
the
direc
t
sum
of
a ni
te
grou
p
(the
root
s of
unity
)
and
a
lattic
e.
I
t
rem
ains
to
dete
rmin
e
the
rank
of
the
lattic
e,
whic
h we
will
do
after
idele
s
have
bee
n
intro
duce
d. N
4
.
T
h
e
a
d
e
l
e
r
i
n
g
L
e
t
K
b
e
a
n
u
m
b
e
r
e
l
d
.
I
n
t
h
e
f
o
l
l
o
w
i
n
g
of
K,
a
se
t
de
no
te
d
by
V.
Th
e
co
m
pl
eti
on
of
K
at
v2
V
wil
l
,
v
w
i
l
l
r
a
n
g
e
o
v
e
r
t
h
e
a
b
s
o
l
u
t
e
v
a
l
u
e
s
be
de
no
te
d
Kv
;
th
e
rin
g
of
int
eg
er
s
in
thi
s
co
m
pl
eti
on
wil
l
be
wr
itt
en
O.
D
e
nit
io
n
4.
13
.
Th
e
ad
el
e
rin
g
is
th
e
re
str
ict
ed
dir
ec
t
pr
od
uc
tv
AK :=
0v
Kv:
Y
vThe word \restricted" means that any vector (x v)v2Vjv 2AK
K
v
v
T Yv=2SOv:
AK
is
re
qu
ire
d
to
sa
tis
fy
jx
1
for
al
m
os
t
all
v.
Th
is
ca
n
be
m
ad
e
int
o
a
to
po
lo
gi
v2S
K
K
v
1
K
1
v
SK
SK
v)
K
v
= 1 for v=2S1
v
K
Q
Q
First, recall that AQ
0
p
, and AQ
Qp:
ca
l
rin
g
in
th
e
fol
lo
wi
ng
m
an
ne
r.
Fo
r
a
ni
te
e
t
S
c
o
n
t
a
i
n
i
n
g
S
1
s
A
a
=
Y for v2S, consider the set Y
Q n
Such sets
d clearly form an
R admissible basis for A. It is clear that addition and
multiplication
are continuous, and that Ais locally compact, since Ois compact
o
p
v=2S. Moreover,
unlike a regular direct product, there is a convenient ltration
e
.For S nnite and containing the archimedean absolute values S, there is a sub
A= Qv2S
K
v Qv=2SOv, and Ais the union of these subrings. Since any x2K is
s in Ofor almost all v(this is analogous to a rational function on a curv
contained
havinge
only nitely many poles), there is an injective homomorphism K!A. Nex
may dets ne a Haar measure on Aby taking the product of the Haar measures
Kv, normalized such that (O. Thus one gets a (i.e., the) Haar measure on Aits
4.1. Investigation
of A. We want to prove compactness results about the adele
and it isT
v easiest to start with the case K= Q.
Theorem 4.14. Q is a discrete subset of A=Q, with the quotient topology, is
compact.
is a restricted direct product
We will nd a closed neighborhood of zero containing no other element of Q. For N= [1=2;1=2]
pY Zp:
this, consider
0
5. THE IDELE GROUP 47
= 1 (where v ranges over all normalized absolute values of Q), we cannot have x2N
or elseQvjxjv 1 2v. This proves discreteness. The proof generalizes, since the
product formula, and shows that Kis discrete in Afor any number eld K. Now, we
show that any (xv) 2AQKcan be represented as r+ cwhere r2Q and c2Nwhere Nis
the compact set as above. Indeed, let Sbe the nite set of nonarchimedean
absolute values where xv=2Ov. Then set r1= P2Q. Then clearly
Ak: Here ALis as a topological group Ak
is discrete and AK
0vYkv k
L= 0v wjv Lw = AL:
Y Y
Since any nonzero x2Q satis es the product formula Qv
jxj
pord (x )
p2S
0
1
x
r
1p
p
2
R
p
Y
1
Zp
2
2
since 1=p2Oq = Zq for q6= p. Also, by subtracting an integer r, we can arrange it so that
xrr22N(since r22Zp;8p). Then take r= r+ r. Since Nis compact (by Tychono ), it
follows that AQ=Q is compact as well. 4.2. Changing elds. We now want to
prove an extension of the previous
=Kis compact.
The fact about discreteness
follows from the produ
r
esult, namely: Theorem 4.15. If Kis a number eld, thenCompactness is tricker. We will need a discussion abou
completions rst with which we can reduce to the case
K AK
Corollary 4.16. Let L=kbe an extension of number elds
L-algebras,
k
L’AL
k
Oplus:::OplusAk
. This is the restricted direct product
Now, we shall prove that AK=K is compact for any number eld K. As a group, this is
topologically isomorphic toL[K:Q]=Q under the identi cation K=L Q though, and this last
group is compact.AQ5. The idele group As usual, let Kbe a global eld. Now we do the
same thing that we did last
time, but for the ideles. First of all, we have to de ne the ideles. These are only a
group, and are de ned
as the restricted direct product
0v = ; this has the product topology and is an
= Qv2S
JK
K v
Y
SK
v
U
)v
relative to the unit subgroups Uv of v-units (which are de ned to be Kif vis archimedean). In other words, an id
(xvis required to satisfy jxvj= 1 for almost all v.
KIf S is a nite set of places containing the archimedean ones, we can de ne
subset Jv Qv=2Sv
48 4. NUMBER FIELDS
open subgroup of JK. These are called the S-ideles. As we will see, they form an
extremely useful ltration on the whole idele group.
Dangerous bend: Note incidentally that while the ideles are a subset of the
adeles, the induced topology on JKis not the J(n)K(n)of ideles where xis pnat vpn(where
p-topology. For instance, take K= Q. Consider the sequence x (n)is the n-th prime) and
1 everywhere else. Then x!0 2AQbut not in J. However, we still do have a canonical
\diagonal" embedding K Q!JKnK., since any nonzero element of Kis a unit almost
everywhere. This is analogous to the embedding K!A
5.1. Idele classes. In analogy to the whole business of adele classes modulo the
ground eld, we will consider the idele class group CK= JK=K . This will be
super-important in class eld theory. In fact, class eld theory exhibits a bijection
between the open nite-index subgroups of CKand the abelian extensions of K.
v
First, though, let’s prove that K is discrete as a subset of JK
1=2(1)
YvnonarchUv
N= YvarchD
v jx N
and if x2K
v
k(xv
. Cassels-Frohlich
suggest that, rather than an actual proof, a metamathematical argument is more
convincing: it is impossible to imagine any natural topology on the global eld K other
than the discrete one!
But, let’s give a legitimate proof. Consider the unit idele 1 = (1). We can nd a small
neighborhood of it of the form
f1gbelonged to N, then x1 would not satisfy the product formula,
contradiction. This completes the proof of the claim.
5.2. Compactness. In the adele case, we had AK=Kcompact. This doesn’t work in
the idele case. Indeed, rst de ne the idele norm
v
)k= Y
vj
so that by the product formula, kxk= 1 for x2K . This is evidently continuous since it is
continuous on each JS K, and it factors to become a continuous map CK!R+whose
image is easily seen to be unbounded. However, we do have:
Theorem 4.17. The group C0 Kof idele classes of norm 1 is compact in the induced
topology. More generally, JS;0 K=U(S) of S-ideles of norm 1 modulo S-units is compact
whenever Scontains the archimedean places.
We have an embedding JS;0 K=U(S) !Cin an obvious manner, so it su ces to prove
the rst claim. However, it even su ces to prove that the topological group Ct Kof ideles
of norm t(where t2RK+) is compact for some t. We can take tvery large if we want.
This is because the groups C0 Kand Ct Kare topologically isomorphic, the isomorphism
being given by multiplication by an idele which is t at one archimedean place and 1
everywhere else.
Lemma 4.18. There is tso large such that if i2JK has norm t, then there is x2Kwith
1 jxivjv tfor all v.
5. THE IDELE GROUP 49
This will be an application of the business of counting lattice points in
parallelotopes, i.e. the \Riemann-Roch problem" in number elds.
1 wj
1,
Note that the family Cv = jivjv
vwith jyj jivjv
1ivjv
w
jy1i vjv QQw v jy1i1i j j
t
v
K
kik: i
fx2K
m
p
v
l
tg i
e
s
v
K
tK
j
x
j
K
v
t
w
K
vj
S;0 K
K
K
v)
0K
for any idele iis a divisor c in the sense of that post, and kikis just the norm of c. So
if kikis large enough (depending only on K), say at least t, then there exists y2K
; 8v which means that jy 1 for each v. In addition, since each jy
i
w w
w6=
v
jy w w
So if kik= t, we get the lemma with x= y1. We can nish the proof of the compactness
theorem. Now consider the groupG= Y: 1 jxjvwith the product topology. It is evidently
compact, by Tychono ’s theorem. (Incidentally, does this mean algebraic number
theory depends on the axiom of choice?) I claim that it is actually a topological
subgroup of the idele group J. But this is because, for almost all v, the inequality
1 jxj, so C= 1 by the way these absolute values are normalized. Then Gsurjects on
Cis compact, proving the theorem.
5.3. Application I: The unit theorem. We can now nish the proof of the unit
theorem. Consider the continuous map h: JS;0 K!RSsending i!fNlogjiv2Sg: The image V
is contained in the subspace W R of vectors such that the sumof the coordinates is
zero. Also, it is easy to see that V contains a lattice Lof maximal rank in W(and much
more, actually). The image VKof the S-units U(S) is a lattice in this subspace, as we
discussed already. Now V=Vis compact (as the continuous image of the compact set
J=VK. This means that (L+ VK)=VK
is compact, but it is also discrete (Lbeing a lattice), which means that VKmust be of
the same rank as L. In particular, V= h(U(S)) has maximal rank in W. Since all that
was left in the prooof of the unit theorem was determining the rank of h(U(S)), we
have completed the proof.
5.4. Application II: Finiteness of the class number. This is a big theorem:
Theorem 4.19. If Kis a number eld, then the ideal class group of Kis nite. Indeed,
we have a map gfrom J(subgroup of ideles of norm 1) to the group F of fractional
ideals of K. This sends (ito QvnonarchordpvviIf we endow Fwith the discrete topology,
then gis easily seen to be continuous. Moreover, gbecomes a continuous map
J=K v!F=P, where Pis the subgroup of principal ideals. But gis surjective. In fact,
since the values at archimedean places don’t a ect
g, it is even surjective as a map C 0 K!F=P. Since C
is compact, its image (i.e. the ideal class group) is both compact and discrete,
hence nite.
5
Some applications
1. Fermat’s last theorem in a special case Theorem 5.1. Suppose pdoes not divide
the class number of Q( p). Then the equation
px + yp = zp
;p- xyz has
no solutions in the integers.
Proof. The idea is actually simple. First o , assume x;y;z are relatively prime,
without loss of generality, and p 7. If we had such an equation, we’d necessarily
have
(x+ i y) = zp:
p
1
We will
next show
that the
ideals (x+ i
p
jp
Y
i
=
0
p
py=
urp
paa
y);(x+ y) are relatively prime for i6= j (Lemma2.1). This means in particular that
each, in particular (x+ y), is a p-th power as an ideal, by the hypothesis on the class
number. As a result, x+
for ua unit in Z[ p]. We will show that it is possible to write u= j pvfor va real a
a
unit (Lemma
a
p
p
p y=
a
But in fact any p-th power in Z[ p
p] to something, which we call t, in Z (Lemma
p
j purp.
jp
aaaaaa
??). The same is then true for r
j p )y, which up to associates is (1 p)y. Similarly, it must divide
p). But (1 p) jp, and p- z; since necessarily (x+ i p
jij py)
= ( ji p
ip
ip
??). In
this case, it follows that x ] such as ris congruent modulo pZ[
. Whence, we can write x+ pj py vt mod p; x y vt mod p: This will
lead to a contradiction as follows. N
It now only remains to prove the three lemmas below. Lemma 5.2. The ideals (x+ i
py);(x+ j py) for i6= jare relatively prime. Proof. Any common divisor of the two
numbers x+ i py;x+ j pymust divide ( (x+ y)(x+ 1)x. Since x;yare relatively prime, the
common divisor must in fact divide (1 y) jz, this is a contradiction. N
] is a unit, then we can write u= j p
p
51
p)=Q),
and
emma 5.3. If u2Z[
vfor va real unit and jan integer.
L
Proof. We know that w= u=uis a unit in Z[ ]. Moreover, I claim that it an all its
conjugates have norm 1; this is because for any , we have w= u= u (since
complex conjugation commutes with the automorphisms of Q(
p
52 5. SOME APPLICATIONS
bbthis
last has norm 1. In particular, wis a root of unity. This means that, by Sublemma j
below, w= for some j. I claim that only the minus sign can occur. This is
because u= wu. Now,
p
modulo (1 p), conjugation is trivial: this is because p
1p
pj
p jis congruent to 1 modulo this ideal, while each
b Sublemma 5.3.1. The only roots of unity.
belongs to that ideal. So reducing the aforementioned ideal gives that w 1 mod (1 ).
But each is not; this means only the plus sign can occur. N
p5.3.1
Part 2
Class eld theory
6
The rst inequality
1. What’s class eld theory all about? Class eld theory is about
the abelian extensions of a number eld Kand its
completions. Speci cally, in the global case, it provides striking relations between
generalized ideal class groups and Galois groups. Stated another way, it gives a
correspondence between certain subgroups of the ideles and the abelian extensions.
Let us, however, start by discussing the situation for local elds|which we will later
investigate more|as follows. Suppose kis a local eld and Lan unrami ed extension.
Then the Galois group L=kis isomorphic to the Galois group of the residue eld
extension, i.e. is cyclic of order fand generated by the Frobenius. But I claim that
the group k =NL is the same. Indeed, NUL= Uby a basic theorem about local elds
that we will prove using abstract nonsense later (but can also be easily proved
using successive approximation and facts about nite elds). So k =NL Kis cyclic,
generated by a uniformizer of k, which has order fin this group. Thus we get an
isomorphism
w
=NL
k
=NL
k=k
NJ
Jk=k
w
w
Q 2G(L
=k)
k=k
NJL
55
’G(L=k)
sending a uniformizer to the Frobenius.
According to local class eld theory, this isomorphism holds for L=kabelian more
gneerally, but it is harder to describe.
Now suppose Land kare number elds. The group k
L
will generally be
in nite, so there is no hope for such an isomorphism as before. The remedy is to use
the ideles. The fundamental theorem of class eld theory is a bijection between the
open subgroups of Jof nite index and the abelian extensions of k. For a nite abelian
extension L=k, class eld theory gives an isomorphism
’G(L=k): What exactly is the norm on ideles though? I never de ned this, so I may
as
well now. Note that the Galois group G(L=k) maps each
completion L
to L
in some
manner (when is in the decomposition group, then this is just the usual action on L).
In this way, G(L=k) acts on the idele group, and even on the idele classes because
the action reduces to the usual action on L. So the norm is de ned by N(i) = i. This
result, called the Artin reciprocity law, is somewhat complicated to state
and takes a large amount of e ort to prove. In outline, here will be the strategy.
First, we will compute the order J. We will prove that it is equal to [L: k] by proving
the and inequalities. The proof uses some cohomological machinery (the
Herbrand quotient) and a detailed look at the cohomology of local elds and units.
Speci cally, one shows that the Herbrand quotient (which divides
56 6. THE FIRST INEQUALITY
the order of Jk=k NJL) is equal to [L: k] (at least for a cyclic extension). The proof
uses a nifty trick manipulation in the land of analysis, speci cally the basic properties
of L-functions.
We then describe a map, called the Artin symbol, from the ideles to the Galois
group and prove it surjective. The Artin symbol sends an idele which is a uniformizer
at an unrami ed valuation and su ciently 1 everywhere else to a Frobenius element
with respect to said valuation, though more generally it is harder to describe. By the
dimension count, we have proved the isomorphism. We will then show that
any nite-index subgroup corresponds to some eld by reducing to a special case
(where enough roots of unity are contained) and then give explicit constructions of
abelian extensions using Kummer theory.
This is a very loose outline of the proof that utterly ignores many details and
subtleties, but these will be discussed in due time.
2. The Herbrand quotient In class eld theory, it will be important to compute and
keep track of the orders
G
: NL
iT
00.
By de nition, H(G;A) = AG
g
=NL
1
Q(A) = jH0 T1 T(G;A)j
jH(G;A)j :
) (where L
f groups such as (K
), where
o L=Kis a Galois extension of local elds. A convenient piece of machinery
for doing this is the Herbrand quotient, which we discuss today. I only sketch the
proofs though, a little familiarity with the Tate cohomology groups (but is not strictly
necessary if one accepts the essentially results without proof or proves them
directly).
2.1. De nition. Let Gbe a cyclic group generated by and Aa G-module. It is
well-known that the Tate cohomology groups H(G;A) are periodic with period two
and thus determined by Hand H1
=NA; where Aconsists of the
elements of A xed by G, and N: A!Ais the norm map, a!Pga. Moreover, H(G;A) =
kerN=( 1)A: (Normally, for Gonly assumed nite, we would quotient by the sum of
( 1)Afor 2Garbitrary, but here it is enough to do it for a generator|easy exercise.) If
both cohomology groups are nite, de ne the Herbrand quotient Q(A) as
Why is this so useful? Well, local class eld theory gives an isomorphism
K=NL!G(L=K) for L=Ka nite abelian extension of local elds. To prove this, one
needs rst to
K. I claim that when G= G(L=K) is cyclic, this is in fact Q(L is regarded as
compute the order of G-module). Indeed,
Q(L ) = (K
1 T(G;L ) jH)j = (K
: NL
: NL )
since H1 T (G;L
) = 0 by Hilbert’s Theorem 90! (I never proved HT 90 or really covered any of this
cohomology business in much detail, but it basically states that if you have a cyclic
extension of elds L=Kand a2Lhas norm 1 to K, then it can be written as b=bfor a
generator of the Galois group and b2L. This precisely means that the cohomology
group at -1 vanishes.)
3. SHAPIRO’S LEMMA 57
So instead of computing actual orders of cohomology groups, one just has to
compute the Herbrand quotient. And the Herbrand quotient has many nice properties
that make its computation a lot easier, as we will see below.
2.2. Proporties of the Herbrand quotient. 1. Qis an Euler-Poincar e function. In
other words, if 0 !A !B !C !0 is an exact sequence of G-modules, and if Qis de ned on
two of A;B;C, then it is de ned on all three and
Q(B) = Q(A)Q(C): This follows because there is always
a long exact sequence of Tate cohomology
groups associated to the short exact sequence, which becomes an exact hexagon
since Gis cyclic and we have periodicity. When one has an exact hexagon of abelian
groups, the orders satisfy a well-known multiplicative relation (essentially an extension
of jAj= (A: B)jBjfor B Aa subgroup) This is how one proves multiplicativity.
2. Q(A) = 1 if Ais nite. This is somewhat counterintuitive, but is is incredibly useful.
To prove it, we shall write some exact sequences. De ne = A=( 1)A, AG
AG
G
G 0 !AG
!A!1
!AG A!AG
!AG
j= jAG
G
G
has order jGj.
i
0T
i
0
i
G G0
iT
the xed points of (i.e. those of G). Then there is an exact sequence of abelian
groups:
!0; which proves that jAj, if
Ais nite. Then 0 !ker(N)=( 1)A!A=NA!0 is exact, for A!Athe norm map N. It now
follows that jH1 T(G;A)j= jH(G;A)j, since these are precisely the extreme terms of the
last exact sequence, and the two middle terms have the same order.
3. If A= Z with trivial action, then Q(A) = jGj. Indeed, in this case the norm is just
multiplication by G, so that H1 T= Z and NA= jGjZ, so H0 Tis trivial. AG
3. Shapiro’s lemma The following situation|namely, the
cohomology of induced objects|occurs
very frequently, and we will devote a post to its analysis. Let Gbe a cyclic group
acting on an abelian group A. Suppose we have a decomposition A=Li2IAsuch that
any two Aare isomorphic and Gpermutes the Aiiwith each other. It turns out that the
computation of the cohomology of Acan often be simpli ed.
Then let G be the stabilizer of Ai0 for some xed i0 2I, i.e. G0 = fg: gAi0
i
(G;A) ’H
0
0
A
(G0;A 0i0
H
iT
=
Ag:Then, we have A= Ind. This is what I meant about Abeing induced. I claim that
); i= 1;0: In particular, we get an equality
of the Herbrand quotients Q(A);Q(A). (This is actually true for all i, and is a
general fact about the Tate cohomology
groups. It can be proved using abstract nonsense.) We shall give a direct
elementary (partial) proof.
58 6. THE FIRST INEQUALITY
3.1. i= 0. We compute AG; I claim it is isomorphic to AG. Indeed, we shall de ne
maps A0!A, A!A0i00which are inverses when restricted to the xed points.
i0 2AG. Then gai0
i
= ai0 for g2G0, i.e. ai0
2AGi00
iSuppose
)
0
2AG
(ai)i2I)i2I) = ai0
G0
i
2
I
i0
iAi0
i
=
0
0
:
for each i2I. Then we
set aiG0G
i
= giai0
i0
is xed by G0
A
i
A
i
0
. So, set f((a. Conversely, given a2A, we can de ne an extension (aby selecting
coset representatives (note that Gis abelian!) for G, and noting that these coset
representatives map bijectively onto I. In detail, we choose g2Gsuch that g, and in
this way get a map f!A. This map sends Ainto A. Indeed, because ain this case, this
is G-invariant.
G
I
induces an
!AGi00
G
0
isomorphism AG
G
So, the map (ai) !ai0
. Now the norm map NgiNG0:: A!Acan be computed as follows: it
is X
0
In particular, the norm maps N ;NG0
i0 =NAi0 are compatible with the maps f;f
’AG
G
;IG0
k=k
norm index of a cyclic extension.
interested in groups of the form J
cation e. Then Q(UL
=x
of the additive group OL
isomorphic to OK
a
such that x
a
iT
. As a result, we get the claimed isomorphism A=NA. 3.2. i= 1. To be honest, I’m
not terribly inclined to go through the detailshere, especially as I’m short on time; if
anyone has a burning urge to see the proof, I refer you to Lang’s Algebraic Number
Theory. The idea is to prove an isomorphism between
i)i2I A’kerNG0=IG0Ai0
=I
to Pai
G
kerNG ; where Iare the augmentation ideals (generated by terms of the form g1). This
isomorphism is de ned by sending (a. It is not too bad to check that everything
works out...
Class eld theory is more fun, and I want to get to it ASAP. Next time, we’re 2Gfor all ; 2G. It is known (th
multiplying by a high power o
going to use just these three properties to compute the local
V[G], i.e. is induced. We see
cohomology and Herbrand qu
induced from the subgroup 1
4. The local norm index for a cyclic extension We shall now take
the rst steps in class eld theory. Speci cally, since we are
NJL, we will need their orders. And the rst place to begin is with a local analog.
4.1. The cohomology of the units. Theorem 6.1. Let L=Kbe a cyclic extension
of local elds of degree nand rami) = 1. Let Gbe the Galois group. We will start by showing that Q(UL) = 1.
Indeed, rst of all let us choose a normal basis of L=K, i.e. a basis (x )
4. THE LOCAL NORM INDEX FOR A CYCLIC EXTENSION 59
But if Vais taken su ciently close to zero, then there is a G-equivariant map exp :
Va!UL, de ned via exp(x) =kXkxk! which converges appropriately at su ciently small x.
(Proof omitted, but standard.
Note that k! !0 in the nonarchimedean case though!) In other words, the additive and
multiplicative groups are locally isomorphic. This map (for Vasu ciently small) is an
injection, the inverse being given by the logarithm power series. Its image is an open
subgroup V of the units, and since the units are compact, of nit index. So we have
1 = Q(Va) = Q(V) = Q(UL): This proves the theorem.
4.2. The cohomology of L . Theorem 6.2. Hypotheses as above (cyclic extension of
local elds), we have
: NL
= Z UL
Q(L ) = (K
Q(L
L
L
) = (K
) = n: The equality Q(L) = nfollows
because as G-modules, we have L, where Z has the trivial action. Since Q(Z) = n,
we nd that
) = nQ(U): We then use the previous computation of Q(U). Then, recall that (by HT
90) Q(L: NL ). This proves the theorem.
4.3. The norm index of the units. The following will also be a useful bit of the story:
: NUL
Indeed, we have to show that jH1 T(G;UL
1 are precisely of the form c=c, where c2Lf
)j= ((L )1
: U1 L ) = ((L )1
1 T(G;UL
L
jH1 T
: (K UL )1
: Bf )(Af
f
f
f
heorem 6.3. Hypotheses as above, we have (U
) = e, for ethe
T
rami cation index. In particular, every unit is a
case (since unrami edness implies cyclicity).
)j= esince the H
But, this is equal to
(units of norm 1 : quotients b=bfor ba unit in
the Galois group. By Hilbert’s Theorem 90, the units of norm
(not necessarily of norm 1). Let us use the notation Ato deno
Aunder a map f. Then we have jH) Let A, similarly, denote th
easy to show that (A: B) = (A: B) for B Aa subgroup. It follow
: K UL
: K ) = e:
(G;U
This completes the
computation of H
)j
=
(
L
1T
)K
and thus the proof.
60 6. THE FIRST INEQUALITY
5. The cohomology of the S-units in a global eld Let Sbe a nite set of
places of a number eld L, containing the archimedean
ones. Suppose L=kis a cyclic extension with Galois group G. Then, if Gkeeps S
invariant, Gkeeps the group LSof S-units invariant. We will need to compute its
Herbrand quotient, and that is the purpose of this post.
Up to a nite group, LS is isomorphic to a lattice in RjSj
jSj
Q[G]
0 Q= L0
R [G]Q,
W RjSj
Proposition 6.4. Let L;L0
0
;L
RQ
Commutative Algebra, for instance.
Now L;L0 R
Q[G](LQ;LQ
to L
0
Q
, though|this is the unit
theorem. This isomorphism is by the log map, and it is even a G-isomorphism if Gis
given an action on Rcoming from the permutation of S(i.e. the permutation
representation). This lattice is of maximal rank in the G-invariant hyperplane
. Motivated by this, we study the Herbrand quotient on lattices
next.
5.1. The cohomology of a lattice. Fix a cyclic group G. We will suppose given a
G-lattice L, that is to say a Z-free module of nite rank on which Gacts. One way to
get a G-lattice is to consider a representation of Gon some real vector space V, and
choose a lattice in V that is invariant under the action of G.
be two lattices in V of maximal rank. Then Q(L) = Q(L).
The way Lang (following Artin-Tate’s Class Field Theory) approaches this result seems a little unwieldy to me. I will follow Cassels-Frohlich (actually, AtiyahWall
in their article on group cohomology in that excellent conference volume). We have
Q[G]-modules L= LQ;Lde ned similarly. Moreover, we have
hom
and R -modules LR
R
(LQ;LQ) ’hom
(LR
0R
;
L
0Q
0
R [G](L
R
S
0
0
Let the image h(LS
0
S
0
1 nQ(M0
w)
): This follows by basic
properties of at base extension. Cf. Chapter 1 of Bourbaki’s
are isomorphic to each other (and to V!). So x bases of L;L. We can construct a
polynomial function on hom R), namely the determinant, which does not always
vanish. In particular, it cannot always vanish on the dense subspace hom). So, there
is a G-isomorphism f of V that sends L
, where both are regarded as subspaces of V. By multiplying by a highly
divisible integer, we may assume that the image of Lis contained in L. Then Q(L) =
Q(f(L)) = Q(L) because f(L) is of nite index in L. 5.2. The main formula. Now there is
a bit of confusion since Lis a eld,
yet I have used Lin the last section for a lattice. I will no longer do that. Now,
Ldenotes just the eld, and a lattice will be denoted by M. We are now ready to
resume our problem as in the beginning of this post. Let n= [L: k].
!W RS
Now recall that we
have the log map h: L
S
0
w2S
, where the last space has
the permutation representation of Gon it. W is the G-invariant space where the sum of
the coordinates is zero.
) be the lattice M; the map hhas nite kernel, so by the properties of the Herbrand
quotient we need only compute Q(h(L)). Then the lattice M, in view of the unit
theorem. Also, G xes v, so Q(M) == M Zv, for vthe vector of coordinate 1
everywhere, is of full rank in R) But Mis of full rank, so we can replace the
computation of Q(M) with Qof another lattice of full rank. For this, we choose the
standard basis vectors (e
6.
TH
E
FIR
ST
INE
QU
ALI
TY
61
for RS, i.e. the basis for the permutation representation. The lattice M00 = L Zew
w
Lv2T L
Zew
j
w
0theory: L
v
j
wjv
@
v
w
0
0
0
M
Z
e
v
w
1
A
w
wjv
G Gw =
0
[Lw0 (Z) for w ]
= nv
:k
:
has full rank, so Q(M0) = Q(M00 ). However, we can compute the Herbrand quotient directly.
L
Zew
is isomorphic to Ind2Sprolonging v xed and Gthe inertia
Shapiro’s lemma,Q 0
If we put all this together, we nd: Theorem 6.5. For a nit
L=kand a set Sof places as above,
Let T be the set of k-valuations that S prolongs. As a G-module, Mis isomorphic to.
The Herbrand quotient is multiplicative, so we are reduced to computing Q
Q(LS ) = 1n
nv:
Yv2T
Ze
6
.
T
h
e
r
s
t
i
n
e
q
u
a
l
i
t
y
W
e
a
r
e
n
o
w
(
n
a
l
l
y
)
r
e
a
d
y
t
o
s
t
a
r
t
h
a
n
d
l
i
n
g
t
h
e
c
o
h
o
m
o
l
o
g
y
o
f
t
h
e
Le
t
L=
kb
e
a
ni
te
cy
cli
c
ex
te
ns
io
n
of
gl
ob
al
el
ds
of
de
gr
ee
n.
In
th
e
fol
lo
wi
ng
,
th
e
H
er
br
an
d
qu
oti
en
i
d
e
l
e
c
l
a
s
s
e
s
.
t
wil
l
al
w
ay
s
be
re
sp
ec
t
to
th
e
G
al
oi
s
gr
ou
p
G
=
G(
L=
k).
Theorem 6.6. We have Q(JL
L
L
, and also on L
. There is a map Ck
=L ) = n. In particular,
(Jk: k NJL
= Jk=k !CL
L
L
)
n
:
6
.
1
.
S
o
m
e
r
e
m
a
r
k
s
.
T
h
k
!JL
!CL
e
p
o
i
n
t
o
f
c
l
a
s
s
e
l
d
t
h
e
o
r
y
,
o
f
c
o
u
r
s
e
,
i
s
t
h
a
t
t
h
e
r
e
is
exac
tly
an
equ
ality
in
the
abov
e
state
men
t,
whic
h is
indu
ced
by
an
isom
orph
ism
betw
een
the
two
grou
ps,
and
whic
h
hold
s for
an
arbit
rary
abeli
an
exte
nsio
n of
num
ber
elds.
B
ef
or
e
w
e
pr
ov
e
thi
s
th
eo
re
m,
let
’s
re
vi
e
w
a
litt
le.
W
e
kn
o
w
th
at
G
ac
ts
on
th
e
id
el
es
J(
cl
ea
rly
).
As
a
re
su
lt,
w
e
ge
t
an
ac
tio
n
on
th
e
id
el
e
cl
as
se
s
C
=
J=
L;
I
cl
ai
m
th
at
it
is
an
inj
ec
tio
n,
an
d
th
e
x
ed
po
int
s
of
Gi
n
C
ar
e
pr
ec
is
el
y
th
e
po
int
s
of
C.
Th
is
ca
n
be
pr
ov
ed
us
in
g
gr
ou
p
co
ho
m
ol
og
y.
W
e
ha
ve
an
ex
ac
t
se
qu
en
ce
0
!L!
0,
an
d
co
ns
eq
ue
ntl
y
on
e
ha
s
a
lo
ng
ex
ac
t
se
qu
en
ce
0 !H0
)=
0
by
Hil
be
rt’
s
Th
eo
re
m
90
,
an
d
w
he
re
Hi
s
th
e
or
di
na
ry
(n
on
-T
at
e)
co
ho
m
ol
og
(G;L ) !H0(G;JL) !H0(G;CL) !H1
L)
k
8(G;L
00
k
(G;L ) = k
y
gr
ou
ps
,
th
at
is
to
sa
y
ju
st
th
e
Gst
ab
le
po
int
s.
Si
nc
e
w
e
kn
o
w
th
at
H
) = J , we nd that (C
Ck=NC
and H0(G;JL
w
e
ge
t
th
e
ot
he
r
cl
ai
m
of
th
e
th
eo
re
m.
W
e
G
=C
L
0 T(G;C
L
k=k
NCL
L
= J , q.e.d. So, anyhow, this Big Theorem today computes the Herbrand quotient
Q(C). It in particular implies that H) n, and since this group is none other than
ar
e
re
du
ce
d
to
co
m
pu
tin
g
thi
s
m
es
sy
H
er
br
an
d
qu
oti
en
t,
an
d
it
wil
l
us
e
all
th
e
to
ol
s
th
at
w
e
ha
ve
de
ve
lo
pe
d
up
to
no
w.
62 6. THE FIRST INEQUALITY
6.2. A lemma. First of all, the whole idele group JLis really, really big, and we want to
cut it down. Hence, we need:
Lemma 6.7. Let Lbe a number eld. There is a nite set Sof places of Lsuch that
JL= L JS, where JSdenotes the S-ideles (units outside of S). Indeed, rst consider
the map JL!Cl, where Cl is the ideal class group, and the map proceeds by taking
an idele (xvL)v, nding the orders at all the nonarchimedean places, and taking the
product of the associated prime ideals of the ring of integers raised to those orders.
The kernel is clearly L JS1, so JL=L Jis nite by the niteness of the ideal class
group. Since JL= SSJSS1, the lemma follows.
= L JS
6.3. Reduction. Now choose S so large such that JL
S)=Q(LS
S
). We have Q(CL) = Q(L LJS=L ) = Q(JS=(L\JS))
= Q(JS=LS) = Q(J
S). We need to compute Q(J
6.4. The cohomology of the ideles. We have JS
= Qw2S L
Qv=2S Uw
w
v2SQ(
Yk
wjv
L wv2S) =
Y Yk
nv
Uw) = Hi T (Gw0;Uw0
iT
outside S, we have Hi T(G; Qwjv
iT
v2S
Q
(
C
L
S
)
nv
S) =
1
n
Q
k
nv
=
Q
v
2
S
k
, as in the
previous section, and enlarge it if necessary to contain all rami ed primes and so that
Sis stable under G(so that Jis a G-module). We use this to simply the computation of
Q(C
): We have already computed Q(LS). Once we do so (and show that it is de ned!),
we will have completed our goal of computing the cohomology of the idele classes
and thereby ensured that the fate of humanity is secure.
= A B.
Now we rst compute Q(A). Let Tbe the set of places of kthat Sextends. Then Q(A) is
in view of Shapiro’s lemma and also the computation of the local Herbrand quotient.
Next we compute Q(B). But for each place vof kwhose prolongations lie
) = 0 for i= 0;1 because the
rami ed primes are contained in S. It follows that Q(B) = 1 since Hcommutes with
arbitrary direct products for i= 0;1: indeed, H(G;B) = 0 for i= 0;1. So Q(J. 6.5. The nal
computation. We have Q(L, so if we put everything together, we nd
) = n: Bingo.
7
The second inequality
1. A warm-up: the Riemann-zeta function and its cousins 1.1. Ramblings on the
Riemann-zeta function. Recall that the Riemann-zeta function is de ned by (s) = Pns,
and that it is intimately connected with the distribution of the prime numbers because
of the product formulap (s) = Y(1 ps)1
valid for Re(s) >1, and which is a simple example of unique factorization. In particular,
we haveplog (s) = Xps+ O(1); s!1+:It is known that (s) has an analytic continuation to
the whole plane with a simple pole with residue one at 1. The easiest way to see this
is to construct the analytic continuation for Re(s) > 0. For instance, (s) 1 s1can be
represented as a certain integral for Re(s) >1 that actually converges for Re(s) >0
though. (The
functional equation is then used for the rest of the analytic continuation.) The
details are here LINK for instance.
+:
ppsAs a corollary, it follows that X= s1 + O(1); s!1
log 1
This fact can be used in deducing properties about the prime numbers. (Maybe
sometime I’ll discuss the proof of the prime number theorem on this blog.) Much
simpler than that, however, is the proof of Dirichlet’s theorem on the in nitude of
primes in arithmetic progressions. I will briey outline the proof of this theorem, since it
will motivate the idea of L-functions.
Theorem 7.1 (Dirichlet). Let fan+ bgn2Zbe an arithmetic progression with a;b relatively
prime. Then it contains in nitely many primes.
The idea of this proof is to note that the elements of the arithmetic progression
fan+ bgcan be characterized by so-called \Dirichlet characters." This is actually a
general and very useful (though technically trivial) fact about abelian groups, which I
will describe now.
; the characters themselves form a group, called the dual G _1.2. Characters. Let Gbe
an abelian group. A character of Gis a homomorphism : G!C. It is a general fact
thatXg2G (g) = 0
63
64 7. THE SECOND INEQUALITY
unless 1 (in which case the sum is obviously jGj). We can rephrase this in another
form. Consider the vector space Fun(G;C ) of
complex-valued functions from G!C . This space has a Hermitian inner product (_;_ )
de ned via (f;f0) =1 jGjPg2Gf(g)f0(g). So, in particular, we see that the characters of
Gform an orthonormal set with respect to this inner product. ByFourier theory, they
are actually an orthonormal basis! But, we don’t need Fourier theory for this. We can
directly show that there
_are precisely jGjcharacters. Indeed, Gis a direct sum of cyclic groups, and it is easy
to check that there is a noncanonical isomorphism G’Gfor Gcyclic (and hence for
G nite abelian).
Abstract nonsensical aside: Although the isomorphism G!G _is noncanonical, the
isomorphism between Gand (G_)_is actually canonical. This is similar to the situation
in Eilenberg and Maclane’s paper where category theory began.
The key fact we aim to prove is that: Proposition
7.2. If a2G, then the function g! X (a)1 (g)
is equal to the characteristic function of fag. This is now simply the fact that any
f2Fun(G;C ) has a Fourier expansionf= P (f; ) applied to the characteristic function of
a! Although this is true only for Gabelian, there is still something very interesting _, but
group-homomorphisms into linear groups GLnthat holds for Gnonabelian. One has to
look not simply at group-homorphisms into C(C ), i.e. group representations. A
character is obtained by taking the trace of such a group-homomorphism. It turns out
that the irreducible characters form an orthonormal basis for the subspace of Fun(G;C
) of functions constant on conjugacy classes. All this is covered in any basic text on
representation theory of nite groups, e.g. Serre’s.
1.3. Dirichlet’s theorem and L-functions. The way to prove Dirichlet’s theorem is
to consider sums of the formXp b mod aps
and to prove they are unbounded as s!1+. But, this is badly behaved because of the
additivity in the notion of congruence. Here is what one does to accomodate the
intrinsically multiplicative nature of the primes.
Fix a;bnow as in the statement of Dirichlet’s theorem. By the previous section, we
now know the identity
Xp b mod aps
= 1 (a) X (b)1 pX (p)ps
where ranges over the characters of (Z=aZ) , extended to N as functions taking the
value zero one numbers not prime to a. The beauty of this is that the sum on the left,
which before involved the ugly additive notion of congruence, has been replaced by
sums of the form
p ( ;s) = X (p)ps;
1. A WARM-UP: THE RIEMANN-ZETA FUNCTION AND ITS COUSINS 65
and the function is multiplicative. This in fact looks a lot like our expression for log ,
but with terms introduced.
In fact, this leads us to introduce the L-function L(s; ) =X (n)ns: It follows similarly
using the multiplicativity of and unique factorization that we
have a product formula L(s; ) =
pY(1 (p)ps
)1
s1and the last term is Cn
@X A
k
w
+
in particular, ( ;s) logL(s; ). It thus becomes crucial to study the behavior of
L-functions as s!1pja(1 ps) which is basically the zeta function itself, so logL(s; ) ’log1
s1. When is the unit character, then L(s; ) is Q. The rst thing to notice is that
analytic continuation is much easier for L-nsfunctions when is nontrivial. We have
by summation by parts: X (n) = X(nss(n1)) 0k n (k) 1
L(s; )
LINK The trick is to consider the
product
N(s) = (s) Y
which by analytic continuation, is at least a meromorphic function in Re(s) >0 with at
most a pole of order 1 at s= 1. If any one of L(1; ) = 0, then Nis actually analytic in the
entire half-plane.
p But, let us look at what this product looks like: it is
YY(1 (p)ps)1
where = 1 is allowed. Suppose phas order fmodulo a. Then (p) ranges over the
f-th roots of unity, taking each one (a)=ftimes, as ranges over the characters
6=1
6=1
+and
th
c
T
p
w
th
66 7. THE SECOND INEQUALITY
of a. By the identity Q k=1(1 X ) = Xk1, we nd Np(s) = Y(1
pfs) (a)=f:
p(1 p (a)s)This holds for s>1. This product, however, is bigger
than Y
1 (a)for s>1. This last product can be expanded as a Dirichlet series, however, when
s!it blows up to 1since Qp(1 p1) diverges to zero. Using facts about Dirichlet series, it
follows that N(s) must have a pole on the positive real axis, which is impossible if it is
analytic in the entire half-plane. Bingo.
2. The analytic proof of the second inequality (sketch) 2.1. A Big Theorem.
We shall use one key fact from the theory of L-series.
Namely, it is that: Theorem 7.4. If kis a number eld, we have
pX sNp
log 1s1 ( )
+. Here p ranges over the primes of k. The notation means that the two di er by a
+as s!1
bounded quantity as s!1. This gives a qualititative expression for what the
distribution of primes must
kinda look like|with the aid of some Tauberian theorems, one can deduce that the
number of primes of norm at most Nis asymptotically N=logNfor N!1, i.e. an analog
of the standard prime number theorem. In number elds.
We actually need a slight re nement thereof. Theorem 7.5. More generally,
if is a character of the group I(c)=Pp6jcs (p)Np log 1, we have Xs1 if 1, and
otherwise it tends either to a nite limit or .c
Instead of just stating this as a random, isolated fact, I’d like to give some sort
of context for people who know less about analytic number theory than I (who
probably don’t number very many among readers of this blog, but oh well). So
recall that the Riemann-zeta function was de ned as (s) =Pnns. There is a
generalization of this to number elds, called the Dedekind zeta function. The
Dedekind-zeta function is not de ned by summing overP jN( )jfor in the ringof
integers (minus 0). Why not? Because the ring of integers is not a unique
factorization domain in general, and therefore we don’t get a nice product formula.
k 2.2. Dedekind zeta and L-functions. We do, however, have unique factorization
of ideals. And we have a nice way to measure the size of an ideal: the norm N
(to Q). So, bearing this in mind, de ne the Dedekind zeta function(s) = Xideals
asNa:
2. THE ANALYTIC PROOF OF THE SECOND INEQUALITY (SKETCH) 67
I claim that this extends analytically to Re(s) >0 with a simple pole at s= 1. The
reasons are too detailed for me to sketch here without really leaving the topic that I’m
trying to give an exposition of, namely class eld theory, but nonetheless, here goes.
So the key fact is that the number of ideals a (integral ideals, that is) with Na N is
asymptotically proportional to N when N is large. This fact, utterly trivial when k= Q,
takes much more work for general k. The idea is to look at each ideal class T
separately (and there are nitely many of them!) and show that the number of a with
Na N and a in said class is asymptotically proportional to N.
kWhence, if we express But such a are basically in one-to-one correspondence
with principal ideals ( ) with N cN by multiplying each ( ) by a representative of the
class T. Then the question becomes of counting integers in the domain of having
norms N. That domain is a homogeneously expanding smooth domain and the
integers are a lattice, so the number of lattice points contained in it is approximately
proportional to the volume of the domain. (I am ignoring a subtletly because one
has to look not just at integers, but integers up to the action of the units.) So in this
way we get an estimate for the number of such lattice points, which when plugged
back in yields the approximate linear growth in the number of ideals a.(s) as a
Dirichlet series Pan=ns, we nd that Pk nakis basically proportional to n. And from
this it is possible to show using some analysis that kminus a suitable multiple
of extends to the full half-plane Re(s) > 0. And thus khas a pole of order 1 at s= 1.
By using the product formula (a corollary of unique ideal factorization)
s )1
kp(s) = Y (1 Np
we nd that log k
p(s)
XNp s
; there is a similar product expansion, and we nd
p6jalogL(s; )
X s (p)Np:
and this implies the claim made in the rst theorem. Similarly,
P(a;c)=1 (a)Nas
for such characters of I(c), we can de ne L(s; ) =
It now turns out that L(s; ) is actually analytic in a neighborhood of Re(s) 1 if is not
the unit character. This follows by further lemmas on Dirichlet series, and summation
by parts; basically the idea is that the partial sumsP (a) are uniformly bounded
because a falls into each ideal class approximately the same number of times when
Nis large. As a result, we get the second theorem (whose variability depends on
whether L(1; ) = 0 or not; we will show below that this is not the case when is a
character of I(c)=PcNL kN a N(c): 2.3. The proof of the second inequality. I’m going to
stop vaguely babbling now and get down to a detailed proof. To prove the second inequality, it su ces
to prove the idealic version, namely:
Theorem 7.6. Let c be an admissible cycle for an abelian extension L=kof degree n.
Then I(c)=PcNL k(c) has order at most n.
68 7. THE SECOND INEQUALITY
To prove this theorem, we consider characters of I(c)=PcNL k(c) and use a few
facts about them. If ris the order of this group, then there are precisely w
characters. Summing all the associated L-functions, we have
Xp; L(s; ) = X
s (p)Np:
But by basic character theory for nite abelian groups (note that I(c)=Pis nite), this
becomesp2PcNw XL k(c)sNp:c
1 s1 +as s!1
+
I claim now that this last sum is wlog
cN
s
s:
P 2L;f(L=Q)=1
L kP.
In particular, we have
w
p2PcNw XL k(c)
; the notation here is
analogous to , and means \greater than up to a constant added when s!1." Indeed, to
see this we will rst describe what primes in PL k(c) can look like; we will show that
there are a lot of them.
Suppose we know that P is a prime of Ldividing p of kand the residue class eld
degree f over Q of P is 1, then the same is true over k, and that implies p = N
Np
nX
NP
This last sum on the right, however, is a subsum of the PPs NPs
log 1 s1
p [L: Q] X
p2s = O(1)
sNP
f>1
+as
s!1
It follows that X
owing to the properties of the Dedekind zeta function for L. However, the terms NPfor
f>1 are bounded because the norm is large; in particular, we have X
. This is an easily proved but often-used fact: the terms with f>1 don’t contribute
signi cantly.
L(s; ) wn log 1 s1 :
1 s1 +as s!1
Since L(1; ) is nite for 6= 1 and L(s;1) log
, it follows that w n, and also that the L(1; ) 6= 0 for 6= 1.
N
(c). It turns out that we can harv
C
orollary 7.7. L(1; ) 6= 0 whenever is a nontrivial character of I(c)=P theorem on arithmetic progressions,
for that.
3. The idele-ide
thing called the Artin map on the
set of primes. But we really care abo
relate ideals and ideles. In this post,
idealic and ideleic framework
3. THE IDELE-IDEAL CORRESPONDENCE 69
3.1. Some subgroups of the ideles. Fix a number eld k. Let’s rst look at the open
subgroups of Jk. For this, we determine a basis of open subgroups in k v
when vis a place. When vis real, k+ will do. When vis complex,
v
k v
1
;
:
:
c, then
:
(xv)v
;
c
l
v
(a)
v
l
)
k
Jc=kcU(c) = I(c)=P
cU(c).
Conversely, if an
idele (x
o
r
d
(the full thing) is the smallest it gets. When vis p-adic, we can use the subgroups U.
Motivated by this, we de ne the notion of a cycle c: by this we mean a formal product of an ideal a and real places v ;:::;v
: k!R . Say that an idele (xpfor all
primes p ja and xv
We de ne the subgroup U(c) Jc
Jc is congruent to 1 modulo c if x
= Jk
induced by real embeddings 1
mod pi>0 for 1 i l. We have subgroups Jconsisting of ideles congruent to 1 modulo c.
Note that kin view of the approximation theorem.
consisting of ideles that are congruent to 1
modulo c and units everywhere. Fix a nite Galois extension M=k. If c is large enough
(e.g. contains the rami ed primes and to a high enough power), then U(c) consists of
norms|this is because any unit is a local norm, and any idele in Y(c) is very close to 1
(or positive) at the rami ed primes. These in fact form a basis of open subgroups of J.
3.2. Ideal class groups. Similarly, we can say that x2k c= Jis congruent to 1 modulo c,
written x 1 mod c. We will denote this group by k\k. We shall denote the principal
ideals generated by elements of k cby Pc. First, we de ne the generalized ideal class
groups. For a cycle c, we de ne
I(c) as the group of ideals prime to c|or more precisely, to the associated ideal (the
archimedean places mean nothing here). Then we can de ne the generalized ideal
class group
c1
)
Jk
I(c)=Pc: Just as we expressed the regular ideal class group as a quotient J k=k JS1, we
can do the same for the generalized ideal class groups. These results are essentially
exercises. I claim that
c: The proof of this is straightforward. There is an obvious map Jcto the generalized
ideal class group, and it is evidently surjective (by unique factorization). It also factors
through kv2Jmaps to the idele (c) for c2kis clearly a unit everywhere and congruent to
1 mod c, hence in U(c).
3.3. Norm class groups. The previous ideas were fairly straightforward, but now
things start to get a little more subtle. Recall that, for global class eld theory, we’re
not so much interested in the idele class group Jk=k but rather the quotient group
= NJL
k
for L=ka nite abelian extension. There is a way to represent this as quotient group of
the ideals, which we shall explore next. This will be very useful in class eld theory,
because itis not at all obvious how to de ne the Artin map on the ideles, while it is
much easier to de ne it on these generalized ideal class groups.
So, let’s say that a cycle c is admissible for an extension L=kif it is divisible by the
rami ed primes, and also if whenever x 1 mod c, then xis a local norm at the primes
dividing c. As a result, it follows that U(c) consists of norms only| indeed, this is true
because every unit at an unrami ed prime is a local norm. (Cf.
70 7. THE SECOND INEQUALITY
the computation of the local norm index, and note that an unrami ed extension of
local elds is cyclic.)
Let N(c) denote the group of ideals in kwhich are norms of ideals in Lprime to c (i.e.
to the primes dividing c). I claim that
’I(c)=Pc
Jk=k NJL
instead of x.
Then xy1
k=k NJL
and (xy1) 2Pc
Suppose there are two ideles i1;i2
1)(yi)
i
1
;i
xy
is in U(c), which implies it is a norm, and i
k : k NJL
E
and prove that NJ
k
N(c)
when c is admissible.
In the next post, we will use facts about the (idealic, inherently) Dedekind zeta
and yi2
function to bound the index of the latter group. This is another reason to look at these
norm class groups.
How do we de ne this map? Well, rst pick an idele i. Choose x2ksuch that xiis very
close to 1 at the primes dividing c, namely xi 1 mod c. Of course, xineed not be a unit
at the other primes. Then map xito the associated ideal (xi), which is prime to c. This
de nes a map J!I(c)=PN(c). But need to check that this map is well-de ned. Ok, so
suppose we used a di erent y2k 1 mod c, so (xi) = (xyc. So it is well-de ned as a map
into the quotient. We do something similar if we replace an idele iwith something that
di ers from it by a norm (in view of the approximation theorem).
We need to show, of course, that this map is both injective and surjective. First,
let’s do surjectivity. Let a be an ideal, prime to c; then we can choose an idele with
component 1 at the primes dividing c, and this idele will map to the class of a:
appropriately, we have xithat both map to the class of a. Then, choosing
x;y2kmapping to the same ideal, which means that they must have the same
valuation at the primes outside c. (At the primes in c, by de nition they are both
units|in fact close to 1.) So
1 1i2 2
L
(J
1
map to the same place in the norm
group.
4. The algebraic proof of the second inequality So, it turns out there’s
another way to prove the second inequality, due to
Chevalley in 1940. It’s purely arithmetical, where "arithmetic" is allowed to include
cohomology and ideles. But the point is that no analysis is used, which was
apparently seen as good for presumably the same reasons that the standard proof of
the prime number theorem is occasionally shunned. I’m not going into the proof so
much for the sake of number-theory triumphalism but rather because I can do it more
completely, and because the ideas will resurface when we prove the existence
theorem.
Anyhow, the proof is somewhat involved, and I am going to split it into steps. The
goal, remember, is to prove that if L=kis a nite abelian extension of degree n, then
) n: Here is an outline of the proof: 1.
Technical abstract nonsense: Reduce to the
case of L=kcyclic of a prime degree pand kcontaining the p-th roots of unity 2.
Explicitly construct a group E J
4. THE ALGEBRAIC PROOF OF THE SECOND INEQUALITY 71
:k
3. Compute the index (Jk
index (k: k n
k
k
k)
(k : k n) n
= Q(k ) = nQ(Uk
: k n) is computed as n2Q(Ok
=
k
(k
k
(
U
E). The whole proof is too long for one blog post, so
I will do step 1 (as well as
some preliminary index computations|yes, these are quite fun|today). 4.1. Some index
computations. We are going to need to know what the
) looks like for a local eld k; this is nite, because (by Hensel’s lemma) anything
close enough to 1 is an n-th power. Suppose the n-th roots of unity are in k. To do
this, we shall use the magical device called the Herbrand quotient. We have an
action of the cyclic group G = Z=nZ on k, where each element acts as the identity.
Then k= Z Uin a noncanonical manner (depending on a uniformizing parameter),
so we have (the denominator in the rst term is nbecause kcontains the n-th roots
of unity)
): Now Uhas a subgroup
of nite index which is isomorphic to the ring of integers Oby the exponential map.
(Cf. the computations of the local norm index.) So we have that the index in question,
(k). But this last Herbrand quotient is easy to compute directly. The denominator in
the Herbrand quotient is 1 because we are in characteristic zero. The numerator is
(O: nO1 jnjwith respect to the absolute value on k(assuming it is suitably normalized, of
course). So we nd
: k n) = n2jnj :
Next, we need to do the same thing for the units Uk. In this case we nd
: Un k) = njnj
k
:
The reason there is no nis that when computing Q(k ), we had a decomposition k = Z Uk, while here thereis
is no additional Z. 4.2. Reduction to the cyclic case. The case of a cyclic extensionmu
L=k(of
ch
nic
number elds) of prime degree nwhere the n-th roots of unity are in the ground eld
2
er, because it means that
the extension is obtained
by adjoining the n-th root
of something in k. In
general, it isn’t known
how to generate the
abelian extensions of a
number eld k(this is a
Hilbert problem, number
12), though it can be done
in certain cases (e.g. k=
Q, where every abelian
extension is contained in
a cyclotomic
extension|this is the
famous Kronecker-Weber
theorem that will be a
corollary of class eld
theory).
So, let’s rst formally
state the theorem:
Theorem 7.8 (The
second inequality). Let
L=kbe a nite abelian
extension of
: k NJL) = jH0 T (G;JL
)j n. The second inequality
d
is also true for non-Galois extensions. W
egree nwith Galois group G. Then (Jk see
what the norm index actually means for non-abelian extensions. This m
general result can also be reduced to the cyclic prime order case, but w
not do so (since it requires a few facts about group cohomology that I h
covered). For a cyclic extension, this together with the rst inequality im
the global
72 7. THE SECOND INEQUALITY
norm index is precisely the degree of the eld extension|which hints closely at the
reciprocity law (for which there is still much more work, however).
4.2.1. D evissage. We will prove this weaker result by d evissage: i.e., an
unscrewing technique.
So, here is the key lemma. Lemma 7.9 (D evissage). Suppose the result is true
for M=Land L=k, where M=k
is abelian. Then it is true for M=k. This is a straightforward multiplicativity
computation:
(Jk : k NM JL) = (Jk : k NL kJL)(k NL kJL : k NM JM
NL k
k
L
k
M
NM k(L JM
0
0
L
: LN
JM
0. Because k0
0
n
): In the product, the rst term is bounded by [L: k] by assumption. I claim that
abelian group A. This last term is bounded by J : N N (L
(JM L
00
No
w
the
ext
en
sio
n
L0
=k
= L( n);k
0
=
k
0
i
s
c
y
c
l
i
c
.
= =
k( k
n
: k0
)) by a general (and easy) inequality (AC: BC) (A: B) for B;Csub
t
) [M: L] since (f(A) :
he second is bounded by [M: L]. Indeed, it is
assumption the inequality is true for M=k.Well, now any nite ab
(kJ: k
a ltration whose quotients are cyclic of degree p. By the Galois
between subgroups and sub elds, it now follows that if we prove
inequality for extensions L=kof prime degree n, then it is true in g
4.2.2. Roots of unity. We need to make one more extension
of unity are in k. This we do next.
So consider a cyclic extension L=kof degree n. Let be an n-th r
consider L), and the extensions L=k, L=k must have order prime
most n1), it follows that [L] = [L: k]. In particular L
is precisely in the form we want: cyclic of prime order n, with the
unity in the ground eld.
k=
=k0
k=k
k
k
NJ
L
L
most that of
L0=k0
k
Ck=NCL
L0 !C
L =NC
to NC
k
0
because G(L =k
0
L
So: Lemma 7.10. If the second inequality is true for it is true for L=k. First o , it is easy to see that Jis an n-torsi
L
any n-th power is a norm. It is also nite; this is true becaus
quotient Q(J) is de ned (and has been computed!). So the
We will now prove that the order of the norm idele class gro
by exhibiting it as a factor group. To abbreviate, I will
for J
change notation slightly, and write C
k
L
0
!C 0 =NC
L
k0 =NCL
=NC
. There are maps:
: Here the rst one is the inclusion, and it sends NC) ’ G(L=k) in the natural way
(restriction to L), and the second is the norm map on the ideles. The composition is
raising to the power [k0: k], so it is surjective. As a result, the norm map C0 !Cis
surjective; since the former group has order at most n(since the second inequality
is assumed true for it), then so does the latter.
4. THE ALGEBRAIC PROOF OF THE SECOND INEQUALITY 73
The lemma is now proved. And we are now in the following situation. We need to
prove the second inequality of a cyclic extensio of prime degree, where the roots of
unity are contained in the ground eld. Then, we will have proved (by the above
reasoning) the second inequality in general. We will nish this in the next post.
4.3. Construction of E. So, here’s the situation. We have a cyclic extension L=k of
degree n, a prime number, and k contains the n-th roots of unity. In
particular, we can write (by Kummer theory) L= k(D1=n) for Da subgroup of k
such that (k nD: k n) = n, in particular Dcan be taken to be generated by one element
a.
We are going to prove that the norm index of the ideles is at most n. Then, by the
reductions made earlier, we will have proved the second inequality.
4.3.1. Setting the stage. Now take a huge but nite set Sof primes such that: 1.
Jk = k JS
2. ais an S-unit 3. Scontains all the primes dividing n 4. Scontains the
rami ed primes We will now nd a bigger extension of Lwhose degree is a
prime power.
1=n SWe consider the tower k L M= k(U) for USthe S-units. We have the extension
[M: k] whose degree we can easily compute; it is
L : k n) = (US Y : Un S) = njSj
(k nUS
v
because US
jSj1
k
v
n
Now, I claim that the group
E=
S
is up to roots of unity a free abelian group of rank jSj1, and the units are a
cyclic group of order divisible by n(since kcontains the n-th roots of
unity). In particular, [M: L] = n. The extension M is only of an auxiliary nature in
constructing the group E.
4.3.2. First attempt. We will now describe a lower bound on the norm subgroup of
L. Then, we will modify this to get a lower bound on the norm subgroup of L. It is
easy to see that
S
Y
nv
NJ
n
U :
v=2S
Yk
1
s1
1;:::;ws1
v
k
nv
T
Indeed, L=kis unrami ed outside Sbecause we are adjoining n-th roots of a S-unit a:
the equation xa= 0 has no multiple root when reduced modulo a place not in S. Also,
[L: k] = nso n-th powers are norms.
4.3.3. E, at last. This is too small a group for our E, though. For it, we will need
more than the U’s outside S. This we tackle next; we shall use the extension Mto
obtain places where L=ksplits completely. So, choose places w=2 Sof kwhose
Frobenius elements ;:::; form a basis for the Galois group G(M=L) G(M=k). We
can do this by surjectivity of the Artin map into G(M=k). Note that G(M=k) is a
Z-vector space, and this is what I mean by a basis. These places form a set T. This
will be important in constructing the set E.
Y
Y(S[T)c U
v
contains the norms NJL . Indeed, this is clear for the factors k(indeed, we are working with an extension ofvn
degree n).
I claim now that the places T split completely in L, which would mean that the
k v;v2T are local norms from Ltoo. Let w;w0be places of M;Lextending
74
7.
TH
E
SE
CO
ND
INE
QU
ALI
TY
w
v
: Lw0
Lw0 kv
w
v
v2T. Then Mw
(Jk , and this is a cyclic extension, since these are local elds and the extension
M=kis unrami ed. The extension [M] has order n because the Frobenius element
is nonzero, whence Lw0 = kvor otherwise we could not have a cyclic group
G(M=k) since nis prime. Finally, it is clear that U;v=2S[Tis contained in the norm
group. 4.4. The index of E. We will show that
E) n which will prove the result, because we know that Econtains the norm
:k
subgroup.
S[T
(k v= k J, so by the general formula (ADD THIS) (Jk:
Now we have Jk
kS[T E) = (JS[T: E)=(US[T: k
: k n v) = n2jSj
\E): Indeed,
: E) = Y
there is a general formula, proven by a bit of diagram chasing:
(AC: BC)(A\C: B\C) = (A: B) for
B;Csubgroups of the abelian group A.
4.4.1. Computation of the rst index. This is the easiest. We just have (J
n
jnjv
Y v2S
\E) = n
(US[T
k
k
: k = 1 by the product formula. The second index, by contrast, will be considerably
more subtle; we don’t have an obvious expression for k\E. 4.4.2. The second index.
The point is that we will develop a nice expression for
n\E; namely we will show that it is equal to U(S[T). By the unit theorem, this clearly
implies that
b
ecause Qv2S
N
Jk(
2jSj1
p
b
)
k
T
v
Un
v
Uv:
: This will, in particular, yieldS Y
the global bound for the norm index and prove the
second inequality. This is a bit tricky, and will use the rst inequality. First, let’s
suppose b2\E; the key is to prove that k(np b) = kor that bis. actually an n-th
power.nWe see this supposed equality of elds is true locally at the primes in
Sbecause b is an n-th power there. Also, k(p b) is unrami ed outside S[T, so it
follows that
C
all
th
e
gr
ou
p
on
th
e
rig
ht
W
.I
cl
ai
m
th
at
k
W
=
Jk .
Th
is
wil
l
fol
lo
w
fro
m
th
e
ne
xt
Y U =Un v
Lemma 7.11. The map U(S) !
v
v
2
T
W = Jk
Thus k W= Jk
S[T
n\E= U(S[T)|but for the lemma. So it all boils down to proving this lemma.
is
surje
ctive
.
Inde
ed,
once
we
have
prov
ed
Y(S[T)c
this
lem
ma,
the
clai
m
kfoll
ows
easil
y,
beca
use
any
idele
clas
s,
we
can
nd
a
repr
esen
tativ
e
idele
in J.
This
woul
d be
in
Wex
cept
for
the
fact
that
it
may
not
be
an
n-th
pow
er at
T.
But
by
multi
plyin
g by
som
e
S-un
it,
we
can
arra
nge
this.
,
im
pl
yi
ng
bi
s
an
n-t
h
po
w
er,
an
d
co
m
pl
eti
ng
th
e
cl
ai
m
th
at
k
4. THE ALGEBRAIC PROOF OF THE SECOND INEQUALITY 75
4.4.3. The last technical step. We must now prove this lemma. But it
is|how tting|another index computation! Consider the kernel Hof the map out of U(S).
We will obtain a clean description of it. Then we will show that the image of U(S) has
the same cardinality as the productQv2TUv=Un v. Namely, suppose bis a S-unit and a
power at the primes of T; I claim then thatbis an n-th power in L. For b1=n2Mis
then xed by 1;:::; s1(because b1=n nis in the appropriate local elds corresponding to
T), and these ’s generate G(M=L). Conversely, suppose b2U(S) \L; then bis an n-th
power at the primes of T (because they split completely in L=k), and an S-unit, so it
belongs to the kernel. We have obtained our description of H.
nIn particular, the kernel Hconsists of U(S) \L, and we have (U(S) :
H) = (L nU(S)=L n) = [M: L] = njSj1
by Kummer theory. We have computed the index of H. Meanwhile the cardinality
ofQv2TUv=Un vis njTj= njSj1in view of the power index computations for the units. So we
must have surjectivity, which proves the lemma.
The proof of the second inequality is thus complete.
8
Global class eld theory
1. The global Artin map We shall now consider a
number eld kand an abelian extension L. Let S
be a nite set of primes (nonarchimedean valuations) of kcontaining the rami ed
primes, and consider the group I(S) of fractional ideals prime to the elements of S.
This is a free abelian group on the primes not in S. We shall de ne a map, called the
Artin map from I(S) !G(L=k).
1.1. How does this work? Speci cally, let p =2Sbe a prime in k. There is a prime P
of Llying above it. If A;Bare the rings of integers in k;L, respectively, then we have
a eld extension A=p !B=P. As is well-known, there is a surjective homomoprhism
of the decomposition group GPof P onto G(B=P=A=p) whose kernel, called the
inertia group, is of degree e(Pjp).jA=pj. We can lift this to an element pof GPBut, we
know that the extension B=P=A=p is cyclic, because these are nite elds. The
Galois group is generated by a canonically determined Frobenius element which
sends a!a, still called the Frobenius element.
First of all, p does not depend on the choice of lifting to GP
L = (NE k P
77
p
L
P
|indeed, the
rami cation index is one, so G!G(B=P=A=p) is even an isomorphism. Moreover, Gis
independent of P, because any two decomposition groups are conjugate (since any
two primes of Llying over the same prime of kare conjugate), and we are working with
an abelian extension. It follows similarly that is independent of the extension P.
By multiplicativity, we get a homomorphism I(S) !G(L=k). This is called the Artin
map. We write (a;L=k) to denote the image of the fractional ideal a (prime to S).
Eventually, we will de ne this as a map on the ideles, and this is how we will get the
isomorphism of class eld theory.
1.2. Basic properties. There are a few easy properties of it that we may note.
First of all, suppose k L Mis a tower with M=kabelian. If a is an ideal prime to the
primes rami ed in M=k, then we have (a;M=k)= (a;L=k). To see this, we may assume
a = p, a prime ideal. Then this is because both induce the Frobenius automorphism
on the extension of residue elds and lie in appropriate decomposition groups
(formally, take P of Lextending p and Q extending P, and look at the actions of these
on the residue class elds of the rings of integers in L;Mquotiented by P;Q|it is the
same).
b ;L=k). So the ArtinNext, suppose we have an abelian extension L=kand a nite
extension E=k. Then LE=E is an abelian extension too and the Galois group is a
subgroup of G(L=k). Moreover, I claim that we have (b ;LE=E)
78 8. GLOBAL CLASS FIELD THEORY
map behaves specially with respect to the norm. To see this, we need only check on
prime ideals P of E (whose restriction p to k is unrami ed in L=k). Then (P;LE=E)LN
Pinduces the map a!aon the residue elds. By contrast, NE kfNpP = pfor fthe residue
class eld degree and (P;L=k) induces the map a!afon the residue class elds. But
Np= NP, so we are done. In particular, the following important fact follows: (NL kb
;L=k) = 1 when b is an ideal of L(not divisible by the rami ed primes).
1.3. The Artin map is surjective. This is the primary thing we shall prove today,
and it is far from trivial. In fact, it will use the rst inequality.
So, let Sbe a nite set of primes containing the ones rami ed in a nite abelian
extension L=k. Consider the subgroup Hof G(L=k) generated by the Frobenius
elements p;p =2S. Then the xed eld Zof this satis es the following: Z=kis Galois,
with group G(L=k)=H.
I claim now that the primes p;p =2Sall split completely in Z. Indeed, they are
unrami ed by assumption. In addition, the residue class eld extension must be trivial.
When p =2S, we have by consistency (p;Z=k) = 1, because xes Z. But (p;Z=k)
induces on the residue class eld extension the Frobenius, which must be the identity;
thus Z=ksatis es fpp= 2 for p =2S. Thus all these primes split completely.
So, we now prove: Theorem 8.1. Let M=kbe an abelian extension. Then if all
but nitely many
primes of ksplit completely in M, we have M= k. First, since splitting completely is
de ned by e= f = 1, it is preserved by
subextensions. So it is enough to prove the result when M=kis cyclic, in view of Galois
theory.
I claim that if all but nitely many primes split completely, then we have Jk= k NJM.
To see this, rst note that complete splitting implies that M w= kv
kfor wjv;voutside a nite set Sof bad v’s. Now pick some idele x2J. We can multiply it
by c2k so that xchas component near 1 at all v2S, because of the approximation
theorem. Locally, any element of k Mvclose enough to 1 is a norm, because the norms
form an open subgroup of nite index (which we have computed!). So if xis chosen
appropriately, xcwill be a local norm at all v2S, and it is so at all v=2Sby complete
splitting. So xc2NJ. We nd in particular that (Jk: k NJM) = 1, which means that M= kby
the rst inequality.
Corollary 8.2. The Artin map I(S) !G(L=k) is surjective for L=kan arbitrary abelian
extension.
2. Strategy of the proof It is now time to begin the nal
descent towards the Artin reciprocity law,
which states that for an abelian extension L=k, there is an isomorphism
Jk=k NJL’G(L=k): We will actually prove the Artin reciprocity law in the idealic form,
because we
have only de ned the Artin map on idelas. In particular, we will show that if c is
3. THE CASE OF A CYCLOTOMIC EXTENSION 79
a suitable cycle in k, then the Artin map induces an isomorphism I(c)=PcN(c) !G(L=k):
The proof is a bit strange; as some have said, the theorems of class eld theory are true because they could not be otherwise. In fact, the approach I will take
(which follows Lang’s Algebraic Number Theory, in turn following Emil Artin himself).
So, rst of all, we know that there is a map I(c) !G(L=k) via the Artin symbol, and
we know that it vanishes on N(c). It is also necessarily surjective (a consequence of
the rst inequality). We don’t know that it factors through Pc, however.
Once we prove that Pc(for a suitable c) is in the kernel, then we see that the Artin map
actually factors through this norm class group. By the second inequality, the norm
class group has order at most that of G(L=k), which implies that the map must be an
isomorphism, since it is surjective.
In particular, we will prove that there is a conductor for the Artin symbol. If xis
su cieintly close to 1 at a large set of primes, then the ideal (x) has trivial Artin symbol.
This is what we need to prove.
Our strategy will be as follows. We will rst analyze the situation for
cyclotomic elds, which is much simpler. Then we will use some number theory to
reduce the general abelian case to the cyclotomic case (in a kind of similar manner as
we reduced the second inequality to the Kummer case). Putting all this together will
lead to the reciprocity law.
3. The case of a cyclotomic extension Today, we will begin by
analyzing what goes on in the cyclotomic case. By the
\functoriality" of the Artin map, we can start by looking at Q( n)=Q and analyzing the
Artin map (for which, as we shall see, there is an explicit expression)
, where the residue class of rinduces the automorphism nIt is known that the only
primes rami ed in thi sextension are the primes dividing n. I shall not give the proof
here; the interested reader can refer for instance to Washington’s Introduction to
Cyclotomic Fields. The proof is not very di cult. In addition, I shall use the fact that the
Galois group is the multiplicative group (Z=nZ)r n! p= . So let p6jn. I claim that the
Frobenius element is just the automorphism np n! (p). Indeed, it is the \raise to the
p-th power" automorphism on the residue eld, so p( n) = p n. In addition, p( n) is an
n-th root of unity. But I claim that the primitive n-th roots of unity are distinct modulo
any prime of the ring of integers in Q( kk n(1 ) prolonging (p); indeed, Yn) = n6 0 mod
p;
so no n-th root of unity (other than 1 itself) is congruent to 1 modulo a prime
extending (p). It follows that if mis a rational integer prime to n, then ((m);Q( ))
corresponds to nm n! n. In particular, if m>0 is a rational number with no prime dividing
nin the
numerator or denominator, and m 1 mod n, then ((m);Q( nn)) = 1. So there exists a
conductor in the case of Q( )=Q, and the reciprocity law is true in this case.
More generally,
80 8. GLOBAL CLASS FIELD THEORY
Theorem 8.3. The reciprocity law is true for an extension L=kif Lis contained in a
cyclotomic extension of k.
By the consistency property, if there exists a conductor for the Artin symbol for
a superextension of L, then that conductor works for Ltoo. So it is enough to show
that there is a conductor for the Artin symbol in the case of L= k( n). Now if x2kis
close to 1 at all primes of kdividing nand totally positive at the real places, then Nk
Q(x) is positive and close to 1 at the primes dividing n. So ((x);k( n)=k)) = ((Nk
Q(x);Q( n)=Q) = 1: 4. The basic reduction lemma
So, the key to proving the Artin reciprocity law in general is to reduce it to the
cyclotomic case (which has already been handled, cf. this). To carry out this
reduction, start with a number eld kand a cyclic extension L. We will prove that
there is an extension k0of kand L0of Lsuch that we have a lattice of elds L0
@@@@
(where lattice means L\k0
0=k0
@
@
@
k0
= k, Lk0 = L0 ) such that L0=k0
0
0
0
kL
~~~~
~~~
0d
0
d
q
d
0
0
1
f
o
r
s
o
m
e
d
q
d
0
‘‘
>>~~
@ ~~~~
@ ~
@
__
@
@
??
@
~
@
@
is cyclotomic and a given prime
p of ksplits completely in k. This means that, in a sense, the Artin law for L=kbecomes
reduced to that of L=k, at least for the prime p, in that (p;L) = (p;L=k). From this, we shall be
able to deduce the Artin law for cyclic extensions, whence the general case will be a \mere"
corollary.
4.1. A funny lemma. The thing is, though, nding the appropriate root of unity to use is not
at all trivial. In fact, it requires some tricky number-theoretic reasoning, which we shall carry
out in this post. The rst step is a lemma in elementary number theory. Basically, we will need
two large cyclic subgroups of multiplicative groups of residue classes modulo an integer, one
of which is generated by a xed integer known in advance. We describe now how to do this.
Proposition 8.4. Let a;n2N f1g. Then there exists m2N , prime to a, such that a2(Z=mZ)has
order divisible by n. In addition, there exists b2(Z=mZ)of order divisible by nsuch that the
cyclic subgroups generated by a;bhave trivial intersection.
Finally if N2N , we can assume that mis divisible only by primes >N. The proof is not
terribly di cult, but we rst need a lemma that will allow us
to construct mso that ahas a large order. Lemma 8.5. Let a;d2N f1gand let qbe a prime.
Then there exists a prime psuch that the order of amodulo pis q d. The order of ais qmodulo
pif pja1; p6ja1:
4. THE BASIC REDUCTION LEMMA 81
qwhich
we write as (x 1)=(x1) for x= aq
We will nd such a pwith a little trick. Consider the ratio
0T(dq) = ad 0qd 0 11 a1 ;
q1i=0iX
x = X ci i(x1)
+q
T(d0) = i
q
0
0
when d0
0
:::rsk
0
q
for
suita
ble
integ
ral
coe
cient
s c.
This
mea
ns
that
the
grea
test
com
mon
divis
or of
x1
and
x1
must
be a
pow
er of
q.
So if
s
1
2 0000where
the r
. I claim
have
d01
we
sho
w
that
T(d)
is
not
a
pow
er of
q,
then
we
get
a
prim
e
p6=
qwit
h
pjT(
d),
impl
ying
(by
the
abov
e
reas
onin
g)
that
p6jx
1
and
pis
our
prim
e.
But
if
T(d)
were
a
pow
er of
qfor
all
d d,
we’d
then
have
x1
divis
ible
by
q d+
1.
So a
look
at
the
expa
nsio
n of
T(d)
in
pow
ers
of x1
impli
es
that
for
dver
y
larg
e,
we
have
T(d)
divis
ible
only
to
orde
r1
by q.
In
parti
cular
, x1
=
q(x1
).
Sinc
e
x!1a
s
d!1,
this
is
evid
ently
impo
ssibl
e.
This
reas
onin
g
can
be
impr
oved
to
yield
a
bett
er
resul
t,
but
we
don’t
reall
y
nee
d it.
Cf.
Lan
g’s
book
. I’m
bein
g
rath
er
lazy
here
and
takin
g
the
path
of
least
proo
f
lengt
h.
Next
, we
will
be
looki
ng
at
cycli
c
subg
roup
s of
multi
plica
tive
grou
ps
(Z=
mZ),
and
the
follo
wing
lem
ma
will
be
indis
pens
able.
Le
m
m
a
8.
6.
Le
t
a;
n2
N
.
Th
en
th
er
e
ex
ist
s
m
2
N
,
w
hi
ch
ca
n
be
ch
os
en
di
vi
si
bl
e
on
ly
by
ar
bit
ra
ril
y
lar
ge
pri
m
es
,
su
ch
th
at
ah
as
or
de
r
di
vi
si
bl
e
by
ni
n
(Z
=
m
Z)
.
M
or
eo
ve
r,
th
er
e
ex
ist
s
b2
(Z
=
m
Z)
of
or
de
r
di
vi
si
bl
e
by
ns
uc
h
th
at
th
e
cy
cli
c
su
bg
ro
up
s
ge
ne
rat
ed
by
a;
bh
av
e
tri
vi
al
int
er
se
cti
on
.
1
k
i
such that the order of ain (Z=m iZ)
We rst factor n= r
i
0 i>si
i . If we take s0 i Q m iZ)
i
>s0 i
ar
e
pri
m
e.
By
th
e
pr
ev
io
us
le
m
m
a,
w
e
ca
n
ch
oo
se
for
ea
ch
i,
a
pri
m
e
m,
w
he
re
sl
ar
ge
(w
hi
is rsi0
i
i
iZ)
i
is rs00
i
ch
ca
n
be
do
ne
by
th
e
pr
ev
io
us
le
m
m
a).
th
en
m
wil
l
be
a
lar
ge
pri
m
e.
W
e
m
ay
as
su
m
e
th
at
no
tw
o
of
th
e
m
ar
e
th
e
sa
m
e.
N
o
w
th
e
or
de
r
of
ai
n
(Z
=i
s
di
vi
si
bl
e
by
n,
by
th
e
C
hi
ne
se
re
m
ai
nd
er
th
eo
re
m.
H
o
w
ev
er,
w
e
do
n’t
ha
ve
by
et.
Fo
r
thi
s
w
e
wil
l
ad
d
in
m
or
e
pri
m
es
.
So choose even bigger
such that the order of ain (Z=l
primes l
li;b 1 mod Q = (Z=li
(Z= Q liZ)
0
. We let m= Q m iQ
c
0i
1 mod Q
Q liQ rsi00 i
s00 i
or s
.
Th
en
by
th
e
C
hi
ne
se
re
m
ai
nd
er
th
eo
re
m,
w
e
ca
n
n
d
bs
uc
h
th
at
b
a
m
od
Q
m.
I
cl
ai
m
th
at
m;
bs
ati
sf
y
th
ri
iZ)
i
li
c
d
f
e
hy
po
th
es
es
of
th
e
le
m
m
a.
Fir
st,
w
e
ha
ve
(Z
=
m
Z)
m,
so
it
fol
lo
w
s
th
at
a;
b
bo
th
ha
ve
or
de
r
di
vi
si
bl
e
by
n.
Suppose now that we had
a. Thus b. Thus Q
ince
b1
modj
c,
and
in
parti
cular
a1
bd
cmod
m. Then a
s
mod
mbe
caus
e
the
orde
r of
ais 1
mod
mto
o,
and
the
cycli
c
grou
ps
gen
erat
ed
by
a;bh
ave
trivia
l
inter
secti
on.
4.
2.
Ar
tin
’s
le
m
m
a.
Th
is
is
th
e
le
m
m
a
w
e
sh
all
us
e
in
th
e
pr
oo
f
of
th
e
re
ci
pr
oc
ity
la
w:
82 8. GLOBAL CLASS FIELD THEORY
Lemma 8.7 (Artin). Let L=k be a cyclic extension of degree nand p a prime of
kunrami ed in L. Then we can nd a eld E, a subextension of L( m), with E\L= ksuch
that in the lattice of elds
L( m)
LE
FF E
L
F
FF
x
x
x
x
x
x
x
x
FF kxxxx
FF F x x x x
F
m
primes.
The proof of this will use the previous number-theory lemmas and the basic
tools of Galois theory.
So, rst of all, we know that Eis a subextension of some L( m). We don’t know what
mis, but pretend we do, and will start carrying out the proof. As we do so, we will
learn more and more about what mhas to be like, and eventually choose it.
4.2.1. The lattice condition. The rst thing we will want is for k( m) = k, so there is
a lattice) and Lto satisfy L\k(
L( m) ), so that LE=Eis cyclotomic 3. p is unrami ed in LE=k Moreover, we c
choose msuch that it is divisible only by arbitrarily large
:
we have: 1. p
splits completely in E=k 2. E( m) =
L( m
HHH
zz
HH
zz
m)
zz
zz
L
E
EE
E
E
E
u
k
(
u
k
u
OO
u
<
<x
cc
FF
FF
u
u
;;
x
u
bb
FF
F
u
u
::
u
bb
EE
E
=
=z
dd
HH
HH
The reason is that we then have a nice description of G(L( mm)=k) (as a product of
Galois groups G(L=k) G(k( )=k)), and we will de ne Eas a xed eld of a certain
subgroup.
I claim that this condition will happen once mis divisible only by su ciently large
primes. For in this case, L\Q( mm) = Q. Now in the lattice of elds) is unrami ed
everywhere over Q (because khas xed, nite rami cation and the cyclotomic part is
rami ed precisely at the primes dividing m). It is a theorem, depending on the
Minkowski bound for the discriminant, that any number eld di erent from Q is
rami ed somewhere, so it follows that L\Q(
4. THE BASIC REDUCTION LEMMA 83
L(
9
9
s
ssss
ssss
s
:
ee
KK
KK
;;
v
KKK
KK K
k( m)
cc
H
H
H
m)
vv
L
vv
vv
vv
v
H H L\k( m)
HH H
HH
m)
I I I II
L
(
=
=z
dd
III
I
::
u
DD Quuuu
DD
uuuu
DD
u
L z z z z )z: L\k( m)] = [L( m) : L] = (m), the last equality in view of the lattice of elds
z z z Q( m)
bb
D
D
D
we have that [k( m
However, we also know (by the same reasoning) that [k( mm) : k] = (m). It follows that k= L\k( ), as claimed, when
mis not divisible by the primes of Q rami ed in L.
4.2.2. Unrami cation in LE. Since we are going to de ne Eas a sub eld of L( m), this one is easy: just choose
mnot divisible by the rational prime p prolongs. 4.2.3. Complete splitting. We are now going to choose a subgroup
H G(L=k) G(k( m)=k), whose xed eld will be E. Now since p splits completely in E=k, we know that we are going
to require H
to contain the decomposition group of some extension of it, i.e. the group generated by the Frobenius elements
(p;L=k) (p;k( m)=k): This, however, is still not enough. We need L\E= kand L( m) = E( ). 4.2.4. L\E. So L\Eis
the xed eld of the subgroup of G= G(L=k) G(k( m)=k) generated by Hand by G(L( m)=L) = 1 G(k( mm)=k). If we
want L\E= k, we will needH(1 G(k( m)=k)) = G: We can arrange this if we choose a generator of G(L=k) and
some 2G(k( m)=k) and add to H, so that His generated by two elements. So do this. We will choose and
mlater.
4.2.5. LE is cyclotomic; choice of m; . This is the most subtle part of the proof, and where the number theory
developed earlier will come in handy. We will prove that L( m) = E( m), so that LE=Eis a subextension of a
cyclotomic extension.
The eld E( m) = Ek( ) is the xed eld of the intersection of H and G(L( m)=k( mm)) = G(L=k) 1, and we therefore
want this intersection to be the trivial subgroup. We have de ned Hto be generated by
(p;L=k) (p;k( m)=k);
84 8. GLOBAL CLASS FIELD THEORY
where 2G(k( m)=k) = (Z=mZ) is yet to be chosen. Suppose this subgroup
intersected with G(L=k) 1; we’d then be able to write for some i;j,
j
(p;k( m i)=k) =
m; m
ahas order in (Z=mZ)
(p;k( m
j
=
i)=k)
j
=
mi)=k)
i
; which is to say that the cyclic
groups in (Z=mZ)generated by Np (since this is how p acts on Np m! ) and intersect.
OK. Now it’s starting to become clear how this intersects (groan) with our
previous number-theoretical shenanigans. So, let a= Np. Choose mdivisible only by
super large primes and such that
dividing n, and such that there also exists b2(Z=mZ)of order dividing nsuch that the
cyclic groups generated by a;brespectively are disjoint.
We have now chosen m. For ;take it to correspond to b. I claim now that
H\G(L=k) 1 = f1g, which will prove the last remaining claim in Artin’s lemma.Indeed, if
we had some element in the intersection, then rst of all we’d be able to write
; as mentioned earlier; this is
because we’d have an expression ((p) (p))= ( )j ( 1) for some in the intersection
of the two groups. This implies that (p;k( = 1 by independence and i;jare consequently
divisible by n. But then, because G(L=k) has order n, it follows that ((p) (p))= ( )= 1,
and = 1 too. Thus the intersection of the two subgroups is the trivial subgroup as
claimed. This proves Artin’s lemma.
5. Proof of the reciprocity law It is now
time to prove the reciprocity law.
5.1. The cyclic reciprocity law. Theorem 8.8 (Reciprocity law, cyclic case). Let L=k
be a cyclic extension of
number elds of degree n. Then the reciprocity law holds for L=k: there is an
admissible cycle c such that the kernel of the map I(c) !G(L=k) is PN(c), and the
Artin map consequently induces an isomorphism
Jk=k NJL
i
c
c
’I(c)=Pc
c
We will prove that a 2Pc
N(c) ’G(L=k): The
proof of this theorem is a little sly and devious.
Recall that, for any admissible cycle c, we have (I(c) : PN(c)) = n by the conjunction of
the rst and second inequalities, and the Artin map I(c) !G(L=k) is surjective. If we
prove that the kernel of the Artin map is contained in PN(c), then we’ll be done by the
obvious count. This is what we shall do. Let a be a fractional ideal prime to c which is
in the kernel of the Artin map.
N(c) in a slow, careful way. The rst step is to factor
5. PROOF OF THE RECIPROCITY LAW 85
a
We will apply Artin’s lemma for each prime p, leading to a lattice
L( mi)i =
npii.
Q
i
LE x x x x
xxxx
i
G G Ei
G
GG
kw w w w i =Ei
w w w w successively so that m i
w
L F F FF
FFF
splits
completely in
Ei
, and LE
i( m i
j
K
k
c
i;L=k)
i
=
qi
i
)
=
q
i
is cyclotomic, contained in E) for some suitable
w
m i. We can choose the mis not
here each pi divisible by any of the primes dividing m;j<i. Lemma 8.9. The eld E, the
compositum of all the Ei, satis es E\L= k. OK, cool. So we still need to do the
reduction though.
5.1.1. The ideal d. We can choose d, an ideal of E, such that (d;LE=E)
= for 2G(L=k) a generator. We can do this by the surjectivity of the Artin map,
and moreover so that d is prime to the primes above c. Then dkE kd satis es
(d;L=k) = too. This ideal d:= Nis going to be a generic bookkeeping device with
which we adjust a to get something in PN(c). 5.1.2. Bookkeeping. Suppose (p.
Then if Piis a prime of Eprolonging p, we have (Pi;LEi=Eias well by complete
splitting. In particular, we have
i=Ei
(P NE i
qdi;LEi=Ei
Ei
) = 1 but since LEis cyclotomic, we have the reciprocity law, and
this means we can write
l
Ei
So
e
,
i
i
t’s take the norm down to k; then piq=
dki(NEkiyi)(NLEki
yi
i
i
i
i
k
LEi(l
OO
that a is dk PniqiP niqk Pniqii
i
cc
FF
F
) for yvery
close to 1 at all the primes rami ed in L=k, and lnot divisible by
those primes.
)): But now Nis close to 1 at the primes of c, i.e. in c, if k was
chosen properly (i.e. REALLY close to 1 at those primes, which
is kosher in view of the reciprocity
<
<x
cc
GG
GG
(l
i
qis
dki
;;
w
Pi = (NE Ei qd)i(yi)N
E
i
c
i
c
) is
(l LE
l a norm from L. The proof is now almost over. We have represented pas a
k
power
of
a
xed
ideal
times
something
in
PN(c).
5.1.3.
Putting
everything
together.
aw for cyclotomic extensions). Moreover N
In particular, we have ptimes some norm times something close to 1 at relevant
primes. Taking the product, we see
times some norm times something close to 1 at relevant primes. But now = 1
because this is equal to (a;L=k), so that sum is divisible by n, and dis a norm. We
have proved that a is in the PN(c) subgroup, and this completes the proof. Bam.
86 8. GLOBAL CLASS FIELD THEORY
5.2. The general reciprocity law. Theorem 8.10 (Artin reciprocity). For a nite
abelian extension L=kof number
elds, the Artin map has a conductor c and so factors into an isomorphism
Jk=k NJL’I(c)=PcN(c) !G(L=k): The proof is now surprisingly easy. We can represent
L as a compositum
1such that G(L=k)=Hi
such that each Pci is cyclic and taking the xed elds Li
L :::Lj
1
i
j
of Hi
i
i
i
i
of cyclic extensions over k, by writing 1 G(L=k) as an intersection of
subgroups H. The compositum L:::Lcorresponds by Galois theory to the intersection 1
of H, and G(L=k) ’G(L=k)=Hiis cyclic. Now, there exist cycles c1;:::;cjmaps to 1 in
G(L=k) by the reciprocity law (and existence of conductor) for cyclic extensions.
Taking the product c =Q ci, we nd a conductor c such that the Artin map on Pcacts
trivially on each L, hence on L. Thus c is a conductor for the Artin symbol on L=k, so
the Artin map factors
through I(c)=PcN(c) !G(L=k) and since, rst of all, it is surjective; and second of all, the
former group is of order
at most that of the latter (by the rst and second inequalities), we have that the Artin
map is an isomorphism.
The proof of the reciprocity law, the prime result of class eld theory, is now
complete.
WOO HOO. Next time, we will take stock of how far we have come, discuss
corollaries, and
plan for the future.
6. Norm groups of Kummer extensions We shall now compute the
norm class subgroups for certain Kummer extensions.
This will enable us to prove the existence theorem (every open subgroup of nite
index in the idele class group corresponds to an abelian extension) as well as an
important lemma in developing local class eld theory.
k= k JConsider a number eld kcontaining the n-th roots of unity; we will consider
extensions of kobtained by adjoining n-th roots of other elements. So, x nite sets
S;Tof places such that S[Tis very large in the following sense: all the primes
dividing nare contained in S[T, and J. Consider the two subgroups of J, D 1S=
Yk vkT Yk n vS[T Y(S[T)cUv
and D2
S= Y k n T Y k
Y(S[T)c Uv:
v
v
We want to adjoin n-th roots of elements of kin these. So take = Di\k
Bi
Theorem 8.11. k D2
is the norm subgroup of
and k D1 of L2.
L1
;i= 1;2, and make the Kummer extensi
(T
he
re
is
a
re
ve
rs
al
of
in
di
ce
s.)
Th
e
re
as
on
for
thi
s
is,
ho
w
ev
er,
ve
6
.
N
O
R
M
G
R
O
U
P
S
O
F
K
U
M
M
E
R
E
X
T
E
N
S
I
O
N
S
8
7
ry
ni
ce
an
d
int
uit
iv
e.
W
he
n
w
e
m
ak
e
an
ex
te
ns
io
n
by
ad
joi
ni
ng
n-t
h
ro
ot
s
of
thi
ng
s
in
B,
w
e
do
n’t
ex
te
nd
an
yt
hi
ng
at
T;
i.e
.,
th
e
pri
m
es
of
T
sp
lit
co
m
pl
et
el
y
in
L1
1.
In
de
ed
,
th
e
el
e
m
en
ts
of
B1
ar
e
n-t
h
po
w
er
s
at
th
e
po
w
er
s
of
T,
so
thi
s
is
ev
id
en
t.
Si
mi
lar
ly,
th
e
pri
m
es
of
Ss
pli
t
co
m
pl
et
el
y
in
L2
.
In
ad
dit
io
n,
th
e
el
e
m
en
ts
of
kn
vfo
r
an
y
va
re
in
th
e
gr
ou
p
k
N
Jf
or
i=
1;
2.
Th
is
is
be
ca
us
e
th
e
G
al
oi
s
gr
ou
ps
ar
e
of
ex
po
ne
nt
n,
an
d
so
kn
vLi
m
us
t
lie
in
th
e
ke
rn
el
of
th
e
Ar
tin
m
ap
.
(
W
e
ju
st
us
ed
th
e
re
ci
pr
oc
ity
la
w.
)
Finally, L1;L2
NJLi
1
: k D ) = (k
S[T
1are
unramifeid outside S[T so U
S[T 1) (U(S[T) : k
k D1 k D2; k NJL2
1
k NJL1 J
S[T
k
;
: kD
1
Q jnj ) = (J
:D
\D
v
=2(S
[T)
belo
ngs
to
k;i=
1;2.
By
this
reas
onin
g, it
follo
ws
thatv
:
W
e
wil
l
no
w
pr
ov
e
th
at
th
er
e
ar
e
ac
tu
all
y
eq
ua
liti
es
,
an
d
thi
s
w
e
do
by
co
m
pu
tin
g
indices (surprise surprise). First, we do the familiar technique:
(J
Th
e
to
)
:
p
in
de
x
is
th
e
ea
si
es
t
to
co
m
pu
te
si
nc
e
w
e
kn
o
w
th
e
lo
ca
l
po
w
er
in
di
ce
s
by
no
w
in
ou
r
bl
oo
d.
S
o,
w
e
n
d
Th
e
de
no
mi
na
:)v = n2jTj
:
(
J
T
D
tor
ap
pe
ar
s
le
ss
tra
ct
ab
le
be
ca
us
e
of
th
e
int
er
se
cti
on
,
bu
t
(y
ay
!)
it
ca
n
be
co
m
pu
te
d
in
ter
m
s
of
K
u
m
m
er
th
eo
ry:
(U(S[T) : k
\D1n) = U(S[T) : U(S[T)) k \D1n:
U(S[T)) :
The top index can be computed via the unit theorem: it is n
: k n) = [L1
(B1 : B1 \k n) = (k nB1
:
: k] :
(Jk
k
:
k]:
S
jSj+jTj.
The denominator is
o,
if
w
e
pu
t
ev
er
yt
hi
ng
to
ge
th
er,
w
e
n
d
in
tot
al
th
e
in
de
x
T jnj [L
D
1
Q ) = n2jTjnjSj+jTj
We can also do the same for
D2
k
v
1
in a parallel manner:
jnjv[L2
jnjv
: k][L
1
2
S[T
2
(Jk
L
(Jk : k D2) = n2jSjnjSj+jTj
: k D1)(Jk
;k]
in
vi
e
w
of
th
e
re
cii
pr
oc
ity
la
w
: k D ) (Jk
;D2
QS : k] :When we multiply these and use the product formual Q= 1, we nd (J:
k D1)(Jk: k D2) = [L1: k][L: k]: But we also have a bound
1: k NJ
2
: k NJL2)(Jk
1) = [L
an
d
th
e
fa
ct
th
at
D
ar
e
co
nt
ai
ne
di
n
th
e
no
rm
cl
as
s
su
bg
ro
up
s,
so
it
fol
lo
w
s
th
at
eq
ua
lit
y
m
us
t
ho
ld
ev
er
y
w
he
re,
an
d
in
pa
rti
cu
alr
,
th
e
ap
pr
op
ria
te
gr
ou
ps
in
th
e
st
at
e
m
en
t
of
th
e
th
eo
re
m
ar
e
eq
ua
l.
Bibliography
cf 1. J. W. S. Cassels and A. Frohlich, Algebraic number theory, Thompson, 1967.
cr 2. Charles Curtis and Irving Reiner, Representation theory of nite groups and associative algebras, to be
added.
lang 3. Serge Lang, Algebraic number theory, Springer-Verlag, add.
langalgebra 4. , Algebra, to be added.
milne 5. James Milne, Class eld theory, Available at www.jmilne.org .
serre 6. Jean-Pierre Serre, Local elds, 1979.
89