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The prime discriminant factorization of discriminants of algebraic number fields
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Davis, Danny Nevin, 1946-
Davis, Danny Nevin, 1946-
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ix, 58 leaves : ; 28 cm.
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Academic degrees ( jstor ) Algebra ( jstor ) Discriminants ( jstor ) Factorization ( jstor ) Index numbers ( jstor ) Integers ( jstor ) Logical givens ( jstor ) Numbers ( jstor ) Prime numbers ( jstor ) Uniqueness ( jstor ) Algebraic number theory ( lcsh ) ( lcsh ) Dissertations, Academic -- Mathematics -- UF Mathematics thesis Ph. D
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Thesis--University of Florida.
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Bibliography: leaf 57.
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Typescript.
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Vita.
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by Danny Nevin Davis.
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THE" PRSIME! DISCRlIMINANT FACTORIZA~TTON OF:

DISCRiMINANISS OF ALGEiRA-IC NUMBER FIELDS

BY

DANNY NEVIN7 DAVIS

A DISSERTATION PRESENT'ED TO TIE GRA2DUArTE COUNCIL OF

THIE rP:IVERSiTY CT FLORIDA

IN; AR9TIAL FU'LFILLMENTC~' I OF: THEl REQUIREMEN!TS~I FORi THEZ

DEGREE clF DOCTOR OF PHILIOSOPHIY

UNIVERSITY OF FLORIDA

Dedicated to

my parents,

Nevin B. and Frances L. Davis

ACKNOWLEDGMENiTS

The author wishes to express his appreciation to Robert Long for

his years of instruction and encouragement in the attainment of this

degree; he is owed a particular debt of graditude for his efforts

--too numerous to mention-- in connection with this dissertation.

Shorter College has generously provided xeroxing services, a

typewriter and various other encouragements. In this regard, the

author especially thanks Dr. Randall Minor, Tom Lagow and Bob Chisholm.

The author also acknowledges his indebtedness to his wife, Jenny;

she typed the dissertation; she advised, planned and edited; but,

mostly, she went the many miles between then and now with the constancy

that got us through.

TABLE OF CONT'ENTS

Acknowledgments ... .. .. ... .. .. .. . .. iii

List of Symbols ... .. .. .. .. .. .. .. . .. v

Abstract ... .. .. .. .. .. .. .. .. ... viii

Section I BASIC CONCEPTS AND TERMINOLOGY . ... .. 1

Section II UNIQUENESS IN THE FAICTORIZATION AND THE

CLASSICAL RESULT .. .. ... .. .. .. 7

Section III DIRECTIONS FOR GENiERALIZATIONS .. .. .. 10

Section IV THE FIRST GENERALIZATION .. .. .. . 11

Section V THE CASE OF QUADRATIC DISCRIMINANT IDEALS . 20

Section VI TRE. CASE OF QUADRATIC IDELE DISCRIMINANTS . 34

Section VII REMARKS ON A THEOREMI OF GOLDSTEIN . .. .. 44

Section VIII THE CASE OF CYCLIC e-DISCRIMINANT IDEALS, e >2 47

Bibliography ... .. . .... .. ... .. .. 57

Biographical Sketch ... .. .. .. . ... ... . 58

.. .. .. .. .. .. .algebraic number fields

.. .. .. .. .. .. rings of integers of K, L respectively

. . .. .. . unit groups of K, L. respectively

.~~~~~~~~ u U: u E 1 mad p'}

.. .. .. .. .. .. .totally positive units of K

.. .. .. .. .. .. .completion of K at the prime ideal p

.. .. .. .. .. .. -completion of K at prime v (used for

both finite and infinite primes)

.. .. .. .. .. .. .ring of integers of Kp

.. .. .. .. .. .. unit group of KI

.. .. .. .. .. .. unit group of Ky

.~ ~ ~ ~ ~ i . u U: u E 1 md (pA )m)

. . . . . . ideles of K

.. .. .. .. .. .. unit ideles of K

.. .. .. .. .. .. maximal abelian extension of K with

conductor dividing the cycle c

,... .. .. .. .. a divides b

.. .. .. .. .. .. .cyclotomic field of mt roots of 1

.. .. .. .. .. .. .numerical discriminant of L/K (in

case B is free as an A-module)

. . .. .. .idele discriminant of L/K

. . .. .. discriminantt ideal of L/K

. . .. .. .discriminnnt of any basis of

BeAAp as an A -module considered

mod O

LIST OF SYMBiOLS

K, L

A, B

U, V

K .

A .

U .

Up~

J

K

UK

E

a b

C

dL/K

DL/K

DL/K

d sr

dL/K .

. .. .. discriminantt of a y basis of L/K

considered mad K'

. . .. .. .ramification index of b over K

(b is a prime ideal in some extension

of K)

. . .. .. .ramification index of the prime in a

p-adic field L over the p-adic field K

. . inertial degree of 6 over K (b is a

prime ideal in some extension of K)

. .. .. .inertial degree of the prime in a

p-adic field i over the p-adic field K

. . conductor of an extension L/K

. . .. .. .order of a finite group G

. .. .. . -group component of an abelian group

e(b/K) ..

e(i/'l) ..

f(b/K) ..

f(i/K) ..

IGI . .

(G) . .

hK . . . . . . .

IK, I . . . . . .

K(1) . . . . . . .

N'p) NK/Qq ) . . . . .

NL/K . . . . . . .

TrL/K . . . .

v (a), v (a) . . . . .

vm * * * *

[K :Kl, the narrow class number of K

order of the (ideal) class group of K

group of fractional ideals of K

group of principal fractional ideals

of K

Gilbert class field of K

extended Hilbert class field of K

the order of A/p

absolute norm

norm from L, to K

rational field

cyclic group of order e

trace from L to K

power of the prime ideal p in the

ideal a or (a), respectively

formal product of all real primes

2 ... .. .. .. .. .. rational integers

5, . .... . ....... primitivemthroot ofl1

CC*]] .. ... .. .. .. greatest integer function

.. ... .. .. .. cyclic group generated by a
W(v) . . . . . an open neighborhood about 1
in K; (see p. 14)
Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the Requirements
for the Degree of Doctor of Philosoph~y
THIE PRIME DISCRIMINANT FACTORIZATION OF
DISCRIMINANTS OF ALGEBRAIC NUMBER FIELDS
By
Danny Nevin Davis
December 1978
Chairman: Mlark L. Teply
Major Department: Mathematics
Let Q(12) be a quadratic extension of the rationals with
discriminant d. In this case d is called a quadratic discriminant in
Q (or, an absolute quadratic discriminant); moreover, d is called a
prime quadratic discriminant in Q whenever d is divisible by exactly
one rational prime. It is a well known result in algebraic numbrler
theory that, in Q, every quadratic discriminan~t can, be f;tored uniquely
as a product of prime quadratic discriminants. In this dissertation
several generalizations of this classical result are proved.
The first step toward these generalizations is that of extending
to the discriminant ideal the concept of prime discriminant. Let K be
an algebraic number fiel~d. Call the discriminant ideal of any quadra-
tic extension of K a quadratic discriminant ideal in K. It is proved
thiat any quadratic discriminant ideal in K can be factored uniquely as
a product of prime quadratic discriminant ideals if and only if K is
totally real with add narrow class number.
viii
The concept of prime discriminanlt is then further extended to a
relatively new discriminant due to A. Frohlich which is here called
the idele discriminant. This discriminant is defined for any L/K to
be a certain class of the ideles of K miodulo the squares of the unit
ideles. The idele discriminant is paramount among discriminants
because while possessing the universal applicability of the discrimin-
ant ideal it does not lose the numerical specificality of the absolute
discriminant. In particular, when the base field is Q the idele
discriminate is effectively an absolute discriminant. Therefore, a
direct generalization of the classical result is achieved when it is
proved that the aforementioned result concerning the factorization of
discriminant ideals holds as well for idele discriminants.
Besides extending the concept of prime discriminant, one
additional direction for the generalization of the classical result is
pursued. In the factorization theorems mentioned thus far the
discriminants were generated in a fixed base field K by quadratic
extensions of K. An investigation is begun here into the possibility
of generating the discriminants from extensions over K with fixed
Galois group other than cyclic of order 2. The only case considered
is that in which the Galois group is cyclic of odd prime order. Let
e be an odd prime. It is proved, for example, that if K is an
imaginary quadratic field with class number prime to & and if K is not
the cyclotomic field of Cth roots of unity, then any discriminant ideal
of K that is generated by a cyclic b-extension of K has a unique
factorization as a product of prime discriminant ideals likewise
generated.
SECTION I
BASIC CONCEPTS AND TERMINOLOGY
Aln algebraic number field is any finite extension of the rational
number field Q. Of utmost significance to the study of algebraic num-
ber fields is the concept of integer in these fields. The ring of
integiers of an algebraic number field K is the integral closure of Z,
the rational integers, in K.
In general the ring of integers of an algebraic number field is
not a unique factorization domain. For example, the ring of integers
of Q(V-) is Z[/-] but (1 + / 5)(1 /E) = 2 3, where all four fac-
tors are irreducible elements in Z[/-E];. (For more details see
r19, ~pp 49-54].) Fortunately, however, the ring of integers of any
algebraic number field is a Dedekind domain. We shall now list the
important properties this fact insures for the ring of integers A
of an arbitrary number field K;.
1) Every prime ideal of A has an inverse in the monoid of frac-
tional ideals of A. (A fractional ideal of A is an A-submodule I of K
for which there exists an integer d EA {O} such that dI E A.)
2) Every (nonzero) fractional ideal of A has a unique factoriza-
tion as a product of powers of prime ideals of A.
3) The monoid of nonzero fractional ideals of A is a group.
Throughout this thesis the symbols K and L will stand for
algebraic number fields. The rings of integers of K and L are denoted
by A and B, respectively, and the groups of units of A and B are
dented by U and V, respectively. We refer to U (respectively V) as the
units of K (respectively L).
Ltx1,...,xn) be a basis for an extension L/K. The discriminant
of this basis is defined by the relation dL/K (x1,...,xn)
= det(trL/K(x x )), where trL/K stands for the trace from L, to K. The
discriminant ideal of L/K, denoted by DL/K, is the ideal of K (i.e.,A)
generated by the numbers dL/K(x1,...,xn aS {x1,...,xn) ranges over all
bases of L/X with x. E B for all i. Perhaps the most important
property of the discriminant ideal of L/K is that a prime ideal p of K
(called a prime K-ideal) divides DL/K if and only if p ramifies ~inL,
i.e., there is a prime L-ideal b such that the power of b dividing pB
is greater than 1. An interesting consequence of this property of DL/K
is that only a finite number of prime K-ideals ramily in L. Finding
the exact factorization of pB, for any prime K-ideal p, as a product
of prime L-ideals is an important problem in "ramification theory" and
the discriminant ideal plays a large role in this part of algebraic
number theory.
e.
Let pB = ii t. be the prime L-ideal factorization of the prime
i=1 i
K-ideal p. The positive integer ei is called the index of ramification
of b. over K and is denoted e(b./K). The index of finite fields
[B/bi: A/p] is called the degree of inertia of bi over K and is
denoted f'"./K).
The following identity is perhaps the central theorem of
ramiicaton teory: ETe(b./K)f(b./K) = [L:K]. (For a proof see
[9, p. 71].) This very useful result will be applied frequently
throughout the thesis often without specific reference to the general
identity.
We shall say that the primes bi are the prime L-ideals that lie
above p; it is, in fact, the case that b. n A = p for all i and these
are the only prime L-ideals with this property.
The symbol N(P), called the absolute norm of p, is used to denote
|A/p| and this definition extends to N(a) for any fractional K-ideal
by multiplicitivity using the properties of Dedekind domain listed
above. N(a) is called the absolute norm of the ideal a and is related
to the absolute norm map NK/Q: K'+f Q' by N(aA) = 'NK/Q(a) for all
a E K'. (We will always use R' to denote the group of units of a
commutative ring R which has a multiplicative identity.)
A frequently encountered class of extension L/K is that in which
B is free as an A-module. In this case, a basis for B as an A-module
is called a relative integral basis of L/K and the discriminant of a
relative integral basis of L/K is called a relative discriminant of L/K.
When we speak of the relative integral basis or relative discriminant
of an extension L/K we are assuming implicitly (if it's not stated
specifically) that B is a free A-module.
Ltx1,...,xn} and {yl"" n}Y, be two relative integral bases
for L over K[. There exists api E A such that yp = E =1 ap x ; hence
(letting tr denote the trace from L to K)
tr(y y ) = tr( ,~j=1 apia jxix )
= n aa txx).Thus we have the matrix equation
1 C,j=1 apl qj 1 3xj
(tr(y y )) = (En= ap tr(x x ))(a )~
= (a .) (tr(xx.j))(a .) .
Therefore dL/K Y 1,.' n) = det(aij) 2dL/K(xl ,..., xn). Since B is
a free A-module, (a..) is invertible in A; hence det(a..) 8 U. We
have thus shown that all relative discriminants of L over K are
congruent module U2. We take advantage of this relationship in making
the following definition (following Frohlich in [31):
In case B is a free A-module the numerical discriminant of L/K is
the caset of K/U2 generated by dL/K(xl...,xn) where {x1,...xn) is
any relative integral basis of L/K. The numerical discriminant of L/R
is denoted by dL/K'
Suppose now that {x1,...,,xn is a relative integral basis of L/K
and {yl"" n}Y, is just a basis of integers (i.e., yi e B) of L/K.
Keeping the notation of the above discussion we again have
dL/K Y 1.-. n) = (det(aij )2 dL/K(xl,...,xn
From this equation and the facts that 1) det(aij) C A and
2) DL/K is the g.c.d. of all discriminants of bases of L/K consisting
of integers, we deduce that DL/K = dL/K(Z ,...,xn) A. This actually
gives the relationship of DL/K to dL/K since dL/K = dL/K(X ,...,xn) U2
Further consideration of the equation dL/R Y 1' nl
2i n
= (det(aij))Z dL/K(x1... ,xn)) leads us to notice that {y i=1 s
relative integral basis if and only if the following equivalent
conditions hold (let v ~(x) be the power of pj dividing xA):
(a j) is invertible over A; det(aij) is a unit in A; v (dL/K Y 1,.. n)
= v(dL/K(xl,'...xn)) for all prime K-ideals p; for all p,
V (dL/K Y1.. '" )) is minimal in the set {v (dL/K(zl,...Zn))} as
{zl,..., zn ranges over all bases of L/K with zi CB for all i. This
criterion for determining relative integral bases will be most useful
later on.
Suppose now that {xl...,xn) and {y ,...',yn are both arbitrary
bases for L/K (here we no longer assume B is a free A-module). This
time aij c K ( not necessarily in A) but still we have as before
dL/K 1l"'r n) = (det(aij) 2dL/K(x ,...,xn). He draw from this
equation the following congruence
(1) dL/K Y1'.. n) dL/K(x ...,xn) mod K.2
which we will refer to from time to time.
When the base field K is replaced by Q, the rational numbers, some
changes in terminology occur. Relative integral basis becomes
absolute integral basis and numerical discriminant becomes absolute
discriminant. Since U2 = <1> the absolute discriminant is a well
defined interger; it is from this fact that the numerical discriminant
(which isn't usually a number) gets its name.
It is interesting to note that if A is a P.I.D. (principal ideal
domain), then the generators of DL/K; are a certain class of K'/U
whereas dL/K: is an element of K'/U 2. It would therefore appear that
dL/K is more sharply defined than is DL/K, at least in this instance.
This apparent shortcoming in DL/K has been resolved by A. Frohlich
in his creation of yet another discriminant. It is called here the
idele discriminant and is defined for all extensions L/K< as an element
of the ideles of K module the squares of the unit ideles (note the
analogy to dL/K)- The idele discriminant will be defined and dis-
cussed in a later section; it will be denoted DLK
The following terminology will be most useful in dealing with
discriminants of extensions L/K. For the sake of brevity only DL/K
is mentioned here but similar terminology will be applied in the
obvious way to dL/K; and OL/K. An extension L/K of degree & is called
an e-extension; discriminants of P-extensions will be called
P-discriminants. In particular, DL/K is an e-discriminant ideal of L/K.
In the special case tha~t(L:K] = 2, DL/K is called a quadratic
discriminant ideal.
WJhen the Z-exctension L./K is also cycl~c, DL/R is called a
cyclic e-discriminant ideal. The expression "X is a (cyclic)
L--discriminant ideal of K" means that there is a (cyclic) e-extension
L/K for which DL/ X.
A prime numerical discriminate is a numerical discriminant which
is represented in K/U2 by a power of a prime element of A (this
concept applies only when A is a P.I.D.). A prime discriminant ideal
of K is a discriminant ideal which is divisible by exactly one prime
K-ideal .
SECTION II
UNIQUENESS IN THE FACTORIZATION
AND THE CLASSICAL RESULT
The object of this thesis is to prove a generalization of a
classical result concerning the prime discriminant factorization
of absolute quadratic discriminants. In any generalization the
question of uniqueness will inevitably arise. There are two different
kinds of uniqueness that will be considered here. One involves the
question of how many extensions of K generate a given discriminant.
We will say that a discriminant X (be it ideal, numerical, or idele)
is uniquely generated if there is only one extension L/K (of the type
under consideration, e.g. cyclic -extension) such that DL/K (or dL/K'
or aL/K) = X. The other kind of uniqueness is the usual algebraic one,
i.e., the uniqueness of the prime discriminant factors in the factori-
zation of the given discriminant; this kind of uniqueness we call
simply uniqueness of the factorization.
Theorem 1. (The Classical Relsult) Every absolute quadratic
discriminant has a unique factorization as a product of prime absolute
quadratic discriminants that are uniquely generated.
Proof: Let L = Q(12) with d a square free integer. The following
is easily established by elementary methods (see for example Samuel
[9, p. 34]).
If d E 2,3 mad 4,an integral basis for L/Q is (1,V2); if
de 1 mod 4, {1,(1+/2)/2} will serve as an integral basis.
The definition of discriminant of a basis applied to the case
d 2,3 mod 4 yields dL/ 12 O d; applied to the case
L (0 2d
dE 1 miod 4 it yields dL/ 1 = d. (These discriminants can
1/ j (d+1)/2
also be calculated using the more general results of Proposition 8
that follows.)
The set of prime discriminants can easily be determined for these
two cases:
If d = 2,3 mod 4, dL/Q is a prime discriminant only if d = '2 or
d = -1. Therefore this case yields the prime discriminants '8 and -4.
If d 1 mod 4, dL/Q is a prime discriminate provided d is `p
where p is an odd prime. But for any- odd prime p, (-1)(p-1)/2
= 1 mod 4. Therefore the odd prime discriminants are of the form
(-1)(p1/ p where p is any add prime.
To show that d~/ can be written as a product of the prime dis-
criminants -4, 18, (-1)(p-1)/2 p (p prime), three cases are considered:
(The symbol 6 will be used to denote '1.)
1) d 1 mod 4. Here dL/Q = d = H m=1 Pi, where the pi are odd
primes. Therefore dL/Q = 6 H =1 (-1) Pi-1)/2 p ; so d 6 mod 4.
Thus 6 = 1.
2) dE 2 m~od 4. It follows that d~i = Ad = 86 Em l(-1(1) i-1)/
Either value of 6 is permissible.
3) d 3 mad 4. We see that dL/Q = Ad = 46 n@(1)(-1)/ i-)2i
from which d = 6 Em=1(-1) Pi1/ pi follows; so that d 6 mad 4.
Hence 6 = -1.
In each instance we have found a value of 6 that allows us to
write dL/Q as a product of prime discriminants.
To deal with the question of uniqueness of the factorization
suppose that dL/Q is an absolute quadratic discriminant with dL/
= H e. and dL/ = H e. where the a., e. are prime absolute quadratic
discriminants ordered so that a. 8 wihi osbesneec
ai 8 can itself be factored only trivially). We see that
H (a /B ) = 1. If for some i, ai = -B then there is an index j # i
such that a.j = -G. also. But this would imply cr. = '8 and aj = '8.
That aiar dL/Q is impossible follows since dL/Q = d or Ad where d
is a square free integer. Therefore a. = a. for all i and the
factorization is unique.
That these prime discriminants are uniquely generated follows
from the more general fact that, for .--ny quadratic extension L/Q,
L = Q(d )
SECTION III
DIRECTIONS FOR GENERALIZATION
In the generalizations of the classical result of Theorem 1 to be
pursued in this thesis the discriminants will be calculated over a
fixed field K from extensions L/K with specified Galois group G.
Therefore the variables to be considered in any generalization will
be the base field K, the Galois group C and the type of discriminant
involved.
Ideally we would like to find a satisfactory charaterization of
all fields K on which a given type of discriminant factorization holds.
In fact we have achieved this goal only in the instance of G = G(2)
(cyclic of order 2) for both the idele discriminant and the ideal
discriminant. (It will be shown later that the idele discriminant
encompasses conceptually both the numerical discriminant and the dis-
criminant ideal.) For the case G = o(1) with Ie an odd prime we will
find a large class of fields K on which discriminate ideal factor-
ization holds and we will find a large class on which it does not hold,
but a characterization of such base fields K is not found. Our first
generalization, which is taken up in the next section, deals with a
specific example in which G = o(e), e odd; K will be Q and the dis-
criminant will he the absolute discriminant.
SECTION IV
THE FIRST GENERALIZATION
We will state in this section some well known results from
algebraic number theory that will be used throughout the rest of the
thesis; however, a review of p-adic fields must precede.
Let p be a rational prime. Recall that for x E Q' v (x) is the
power of p dividing x. In addition we now define v (0) =
A (non-archemedian) absolute value is defined on Q by Ixl = (1/p)VP(x)
for X E Q. (In particular 0/ = 0). The field obtained by completing
Q with respect to this absolute value is called the completion of g at p
anld is denoted Q Any finite exten:ion of Qp (viewed as lying in
Q a fixed algebraic closure of Q if you will) is called a p-adic
field.
To see how p-adic fields arise from algebraic number fields
consider K/Q, a finite extension. For a prime K-ideal p we define
v (x) to be the power of p dividing xA~. Suppose that p nZ = pZ for
the rational prime p. We say that p lies above p and call
e = e~p/Q) = v (p) the absolut rmifiato inde of p. An absolute
value is defined on K, which extends I1-1 on Q, by
x = v (x for x E K'. The completion of K with respect to
*~l is called the completion of K at P and is denoted K Since
Q E Kand since Ix11 = Ixl for all x E Q, Kp contains (an isomorphic
copy of) Q If f =- [ A/p:Z/p ], the absolute inertial degree of p,
11
then [K : Q ] = ef. Therefore Kp is a p-adic field. Kp is called a
local field because it reveals the local behavior of K at the single
prime p. If Zp is the topological closure of Z in Qp with respect to
*I and if AQ is the topological closure of A in Kp with respect to
I~p, phe p h ig
he is the integral closure of Zp in K .Terng n
Ap are discete valuation rig (i.e., principal ideal domains with
exactly one nonzero prime ideal) and are called the integers of their
respective fields. The prime ideal of AI is PAI and there is an
isomorphism A /pAy -A/p (wre will have a closer look at this iso-
morphism in 55); thus [A/pA 3:Z /PpZ] = f. Also there is the relation
pAP = (pA )e
Now suppose L/K is a finite abelian extension of p-adic fields
with unit groups V/U. Let p be the prime ideal of Ki. Thle conductor of
L/K is the smallest power im of y for which NL/ (L ) 2 U(m), where
U;(m) = ix c U;: x 1 mad p }l (we define U;(0) = U;); the conductor of
L/K is denoted FL K. That there exists anl m for which N(L') 2 U(m)
follows from these facts: 1) the sets U(n) form a neighborhood
system of open sets about 1 and, hence, by translation these sets
form! a topology for K;'; 2) NiL/K (the norm map) is a continuous function
from L' +t K' with respect to this topology. Keeping the notation
of this discussion we have
Proposition 2. If L/K is cyclic of prime degree a then
P-1
DL/K L/K
Proof: The reader is referred to Artin-Tate [1], corollary
to Theorem 14 p. 135 and corollary to Theorem 18 p. 139.
I
It is usual to include among the local fields associated with an
algebraic number field those fields obtained by completing K at its
archimedian absolute values, i.e., for any imbedding o: K + C (the
complex numbers), we define, for each X E K, Ix 0 = lo(x) where *
denotes the usual absolute value on C. If a is a complex imbedding
the completion of K with respect to * 0, denoted K is C; otherwise
Ko = R (the real numben).The imbe~ddings a of K into C are called the
infinite primes of K while the prime ideals of K are called the
finite primes of K. (Since we will make statements about "primes"
without regard to which type (finite or infinite) it is convenient to
use v as a common symbol for both.) As with finite primes there is
a concept of ramification for the infinite primes of K. Given an
extension L/KC, when a real infinite prime a of K has an extension to
a complex infinite prime os' of L, a is said to ramify in L/K.
Let L/K be any abelian extension of algebraic number fields.
Let v be a prime of K (finite or infinite) for which v' is any prime
of L that lies over (extends) v. If v is a real prime that ramifies
in L/K: we set FR,/ v; if v is infinite and not ramified in L/K
V V
then we set FL =/ L. / has already been defined in case v
is finite. The conductors FL / are called the local conductors of
L/K. The conductor of L/K (called a global conductor) is defined to
be the formal product of all local conductors of K and is denoted FLK
Such a product is also called a cycle of K. (Although this is a
formal product we will speak of one conductor (cycle) dividing
another as thought the terms in the product had an algebraic signifi-
cance.)
We now wish to describe those aspects of class field theory
that are used in the remainder of this thesis. We follow the idele
approach of Lang in [7], as opposed to the ideal approach of Janusz
in [6].
Let v be a (finite or infinite) prime of K and let U~ be defined
by Uv = A; if v is finite,
= K; if v is infinite; Uv will be called the group of units of
K ~. The group of ideles of a number field K is the restricted direct
product of the K; with respect to the U i.e., it is the multipli-
cat-ive group JK = {x E H K;: x (the v-component of x) UV for all but
a finite number of v} where v ranges over all primes of K. It is
important to note that K' is a subgroup of JK via x +t (...x,x,x,...).
By a cycle c we mean the formal product of a finite number of
m (v)
primes of K to powers as in C = Hv where m (v) is a nonnegative
integer called the m~ul~tiplicity of v in c. Assume vlC. For v finite
m (v)
W (v) = {ac K : a 1 mod pv } where pv is the prime ideal of K .
For v real we set Wc(v) = R ; for v complex, We!v) = C. We now define
WC= CW (v) HCUV subgroup of JK.
Our use of the Artin Reciprocity Law, which is the central
theorem of class field theory, is limited to the following:
Theorem 3. (Global Class Field Theory) Let En be the maximal
abelian extension of K with conductor dividing a given cycle c of K.
Then Gal(E /K) = J. K'W .
(Ec is called the ray class field with cycle c and JK/K'WC is
called rhe class group corresponding to E ).
Theorem 4. (Local Class Field Theory)
a) Let L/K be an abelian extension of p-adic fields. Then
Gal(L/K) .
b) Let K be a given p-adic field. The map L +NiL/(L')
establishes an inclusion reversing bijection between the finite
abelian extensions of K and the open subgroups of K' of finite index.
If H = N /(L') we say that i is the class field belonging to H and that
H is the class group belonging to L,
For proofs of Theorems 3 and 4 see Lang [7, Chapters 9,10, and 11].
Let i/K be an extension of p-adic fields with prime ideals b and p;
respectively. Since L has only one prime ideal it makes sense to denote
e( /K) by e(2/fRland f(&/K by f(il/K) -- which we will do. The extension
i/2 is said to be tamely ramified provided p / e(L/K); otherwise it is
widl ramified. When e(L/K) = [i:a], i/2 is said to be totygggy
ramified.
The following proposition will be most useful in determining the
conductor of an extension 1 R. Both tame and wild ramifications are
dealt with.
Proposition 5: Let 1 be a p-adic field with unit group O. Let i/2
be a cyclic e-extension where d is prime. Let N = Ni K'
a) iLK is unramified (i.e., e(L/K;) = 1) if and only if
b) i/K is tamely ramified if and only if N(L') U(1).
c) Let e = e(K/Q ). Let n be the smallest integer for which
U(n) c U(1) If (p-1) f e or if (p-1) e and 2 contains a primitive
pth root of 1, then n = [[ep/(p-1)J] + 1. ([[*]] means "greatest integer
in".) If (p-1) je but j! does not contain a primitive pth1 root of 1,
then n = ep/(p-1).
d) If L/K is wildly ramified and n is as in c), then
Proof: a) Let He~r be prime elements of i, K respectively.
Then I~' -. x O where is the cyclic group generated by rr.
N(H) = urr K/) for some U E U. Therefore if N(L') 2 U, then
[F': N(i')] = f(L/K) which by Theorem 4 implies f(i/R) = [fi:]. Thus
e(1K) 1.(Remember that the relation [L:K] = e(iL/)f(iLK) holds
for all p-adic fields L/K.) On the other hand if e(i/R) = 1 then
f(i^/R) = [K': N(L')]. But
K/N(L') E (/). There-
(N(L')nU)
fore N'(I/) 2 U.
b) If i/1 is unramified then the result holds by part a).
Assume that iL/ is ramified. Since [i:KZ] = e(ilK)f(L/K) = L, a prime,
it follows that e(L/K) = e and f(i/K) =1. Therefore c N(i') and
consequently [K': N'(i')] = [I: N(i') n 0]. This equality will be
applied after we have found values for [0: U(1)] and [U(1): U(2)].
Let A be the ring of integers of K and let p be the prime ideal
of 2. The natural epimorphism 6 + (1/Ji)' has kernel U(1) giving the
isomorphism ir/U(1) E (1/p)'. Moreover thle additive group of A maps
onto U(1)/U(2) homomorphically via a 1 + a mod U(2) with kernel p;
thus A/p U(1)/U(2). Now A/p is a finite extension of the field
Z /PpZ of degree f, say. But as we have remaerked before Z /PpZ Z/pZ
which results from the natural injection Z/pA + Z pZ~i and the fact
that Z is dense in Z_ with respect to the p-adic topology. Therefore
[e:o(1)]l = p -1 and [fl(1);U'(2)] = p .These indices together with the
fact that [i':N(i')] = [i:N(f') n 0] implies that N(L') 2 U(1) if and
onl ifp /[i'N~i)].By Theorem 4a) [R :N(i')] = [i:d]. The result
in b) now follows from a).
c) See Long in [8, Chnapter 4, Proposition 2.11].
d) Let C = Gal(iLK). N /(x) =0 Go(x) = xP for all x c- K' since
[L:K] = p. TherefaceNi K(x) U(1) The result in d) now follows
from c).
'j:Cggy g:'ya 6. Let L/K be an extension of algebraic number fields.
Let p be a prime K-ideal with prime ideal factorization in L given by
e.
pB = n 6.1. Let L /Kp be the extensions of the completions. Then
DL/K A = H1 DL /
Proof: See Long in [8, Chapter 5].
We now prove the first generalization of Theorem 1.
Theorem 7. Let e be an odd prime. Every absolute cyclic
e-discriminant has a unique factorization as a product of prime
absolute cyclic ~e-discriminants that are uniquely generated.
Proof: The theomem will be proved first for discriminant ideals.
Let DL/Q be a cyclic e-discriminant ideal. Suppose p is a
rational prime such that p Z e and p DLQ Let b lie above p in L
and further let L /Qp be the extension of completions obtained by
completing L at 6. Let N = NL /Q '
From1 Popo~sitionl Sa) and b) it follows that FL /Q Pp'
From proposition Sa) we see that -e = [Q':N(L,')] I [8 :O (1)] = p-1;
hence p E 1 mod -e. From Propositions 2 and 6 we have V (DL/Q) =e1
Supose1 D/Q.Keep the same notation as above with p = e.
In this case L /Qe is wildly ramified. Therefore from Proposition 5b)
N(L ) fi d(1) and from c) and d) N(L ) g i (2). Therefore FL Q
= (CZZ)2. From Propositions 2 and 6 we have v (DL/Q) = 2(e-1).
To summnarize, the only primes p that can divide DL/Q are p = e
and p 1 mod I. Also to any given prime of Q there corresponds at
most one prime cyclic e-discriminant: ideal. WJe need only exhibit
these prime discriminant ideals and show that the fields that generate
them are unique.
The ray class field to the cycle my (u_ = the product of all real
th
infinite primes) of Q is C the cyclotomic field of p roots of 1.
(For a proof see Lang [7, p. 209].) Jince the cyclic e-extensions of
Q in question that have absolute discriminants p -Z with p 1 mod e
have conductors dividing pum, it follows that these fields are
contained in C_ (if they exist). Likewise the cyclic e-extensions of
Q with discriminate e(C1) Z mut be contained in C 2 (if they exist).
But C ,, pE1mo and C 2 each contain exactly one cyclic
e-extension of Q, which, of course, must yield thle required prime
discriminant ideals in Q.
In what remains we will use the theorem of Stickleberger: every
absolute discriminant is congruent to either 0 or 1 mad 4.
Let dL/Q be a cyclic e-discriminant. First note that
2 / dL/ since 2 / 1 mod I and e f 2. Therefore dL/Q 1 mod 4. So,
in particular, the prime absolute cyclic e-discriminants are pf- for
19
p 1 mod e and 2(t-1). But dL/Q differs from a product of such
prime discriminants by at most a sign. Since every discriminant
involved is congruent to 1 mod 4 the result follows.
SECTION V
TRE CASE OF QUADRATIC DISCRIMINSANT IDEALS
In this section we will characterize the algebraic number fields
that have the property of prime discriminant factorization of quadratic
discriminant ideals. In the next section w~e will show that this same
characterization holds in the case of quadratic idele discriminants.
We shall prove some preliminary results, the first of which is
" well known but for which a detailed proof was not found by the
author in th~e literature.
Proposition 8. Let L = K(Jra) br a quadratic extension of K with
a a square free integer of K. Suppose that p is a prime K-ideal.
We assume (without loss of generality)* that either v (a) = 0 or 1.
Let v ~(2) = e.
1) If p 2 and p a, then v (DL/)=1
2) If p 2 and p a, then v (DL/) = 0.
3) If p 2 and p I then V (DL/K) 2e + 1.
4) If p 2 and p 4, then u (DL/K() = 2(e s), where s is the
largest integer r e for which a is a quadratic residue mod p2s.
(*) Note: We will show that generality is not lost: if we assume that-
V (a) = 0 o 1 for all primes p that divide a fixed integral ideal
bof K: Consider a particular p that divides b. LtA pr+2m~ hr
r = 0 or 1, m is some non-negative rational integer and p / e. Now
we can choose an integral n in K such that a and b are relatively
prime and a = (8)p for some B c K. (Every ideal class in the ideal
class group of K contains an infinite set of prime ideals. See
1~7, p. 167].) Therefore a8-2mA = pa2mc. 'Hence DB-2m is integral,
K(l$ ) =K(A), v (uii-2m) = r and up (00-2m)) = Pi (a) for all
pi 6,pi f p. If we proceed in like manner for each prime p b
the statement at the beginning of the note will be proved. The full
strength of this result, although excessive for the present application,
will be needed later.
Proof: As was mentioned in Section I, DL/K is the greatest
common divisor of the ideals {dL/K(x ,x2)A: {x1,x2) is a basis for
L/K and xl'x2 E B). In particular DL/K dL/K(1,VEU)A; hence DL/K 4uA.
Statement 2) of the proposition follows immediately from this relation.
Statement 1) follows from this relat on and the fact that all dis-
criminants of bases of L/K are congruent mod K.2. (See Equation (1)
of Section I).
We will next prove 4). In this proof we will find 6 = a-1(la -b)
with a,b c Kp so that (1,8} is a relative integral basis for
K (Ja)/K We will apply the following useful result: If [L:K] = n
and if L = K:(9)with f(X) = Irred(BK) thlen dL/K(1,8,82 n-"1)
'NL/K(f'(B)). (See Long [8,Corollary 1.4 of Chapter 5]). Here L and
K can be either algebraic number fields or local fields. Therefore
(from the criterion for determining relative integral bases given in
Section I, p. 4) it will suffice to find a,b c Kp so that
v (d(1,8))= v p(NK (J2)/X(f'(B))) is minimal subject to the condition
p p
that f be monic with~ integral coefficients. We see that f'(6) =
2a-1/a and (letting N stand for the norm from K (la) to K )
(1) N(f'(6)) = (2a-1)2a.
Therefore we seek a such that v (a) is maximal (but -e) subject to
the condition that f have integral coefficients. The constant term
of f(X) is (b2-a)/a2 which must be an integer of K So if
GE y2 mod p2s (y is a unit and s is as in the hypothesis) we can do
no better than set b = y and a = ns for some prime element ?r of K .
Checking the coefficient of X in f(X) we find v (2b/a) = e-s t 0.
Thus {l,0} is a relative integral basis for K (/a)/K where
6 = w-s(ra-Y). Therefore v (dK (la)/K ) = 2(e-s). Statement 4) of
the proposition now follows from Proposition 6.
A very similar argument to that just given for 4) also suffices
to prove 3): here note that v (N~f'(6))) < 2e + 1 only if pJ a.
(Consider Equation (1) above.) But then v ((b2-a)/a2) < 0. Since
the coefficients of f(X) must be integral in KP we conclude that
p / a. He may choose a = 1 and b = 0 since this choice will make
v (N(f' (8))) = 2e + 1, which must be minimal. Hence {l,Ja) is a
relative integral basis for K (ra)/K~ and v (DL/K) = 2e + 1, where
again we apply Tcoposition 6.
Corollary to Proposition 8. Let K, a, and p be as in the prop-
osition. Let d = dK (Ja)/K Then in cases 1) and 2) of the proposi-
tion d = aU In case 4), d = (2x-1)2a~ whr 2adu~)=s
In case 3) d = aU 2
Proof; Let d(xl,x2) = dK (Jg)/K(xl'x2). The criterion given in
p P
Section I (p. 4) for determining when an integral basis is a relative
integral basis applies equally well in this local case. Namely, if
v (d(01 82)) is minimal in the set {v (d(Yl Y2) ~1 2)~ is a basis
for K (Ja)/Kp and Yl'Y2 E B) then {61'82) is a relative integral basis
In cases 1) and 2) d(1,ia) E a mod Up It therefore follows
from Proposition 8 and the preceding criterion that fl,J-a) is a rela-
tive integral basis for K ~(,$)/Kp in this case.
In cases 3) and 4) integral bases were found in the proof of
Proposition 8 and their discriminants can be determined from Equation
(1): case 3) a = 1 so d = 4aU 2
case 4) a = us so d =(2n-s 2aU 2 where ?r is a prime
element in K .
A prime K-ideal p is totally ramified in L/K provided only one
prime L-ideal, say 6, lies over p and e(b/K) = [L:Kl. A fact that
is used in the next proposition is that the rational prime e is
totally ramified in C /Q where C~ = Q(5e) and ie is a primitive eth
root of 1. Since Gal(C /Q) = (Z/fZ)' we have for a the prime C -ideal
over , that e(a/Q) = e 1. We will also use the interesting fact
that a is the principal ideal of CQ generated by 1 5e. (The reader
is referred to [6, Chapter I, Section 9] for proofs of claims concern-
ing Cg made in this paragraph.) Now if K 2 Ce and if pJ is a prime Ki-
ideal lying above -e it follows that (C 1) e(p/Q) since the index
of ramification is multiplicative in towers of extensions. Therefore
the ideal (LA~)1f1 is defined in K.
Proposition 9. Let K; 2 C for some rational prime e. Let
G be the group of units in A module the ideal (-fA)/(-) Then
[G:G ] = Q[K:Q]
Proof: For a prime K-ideal a lying over C let Gn be the unit
group in the quotient ring A modolo the ideal ale(La/Q)/(e-1). Notice.
that G/ G /C Therefore we will have proved the proposition
when we. show that [G :CD"] = aef (for fixed but arbitrary bLe in K)
where e = e(b/Q) and f = f(b/Q).
Let m = e/(-1). From this point we will transfer to the
completion Kb via the ismorphnism Gb ED /0 ~(m). Let us prove this
important isomorphism in greater generality for our future reference:
Let r be any positive rational integer. Let b = ~bA. The map
a mod br + a mad br induces an injective homomorphism of A/br into
~b br. That this homomorphism is surjective follows from the fact that
Ab is the closure of A in KL under the p-adic topology defined by
*1 (hence A is dense in 1 ). Therefore we have the isomorphism
A/br a A /gr. The kernel of the natural epimorphism Ub + (Ab r ). is
Ub(r). Hence (A/br U / (r). Returning to the case at hand, we
have G -Gb Ubi
hv b v b/Ub(em). It follows that --- .(For the rest
GbUb(.em)Ur b
of this proof we will write U for U .)
Let A = A~ and b = bA Then the residue field A/b E A/b and has
order 4 A mCore general version of Hensel's L~emma than the one in
Lemma 12 (see, for example, [8, Chapter 3, Section 2]) would imply that
A contains a primitive (e -1)st root of 1, say 5. The cyclic group
<5> is a multiplicative set of representatives of U/U(1) (1/6) .
Thus we arrive at the decomposition U = <5> x U(1). It therefore
follows that
(1) Gb U(1)
Gb U(1) U(em)
because <5> -= <5>.
We will next establish the index [U(1):U(1) ]. Consider the
composite map U(1) -t U(1) -tU(1) /U(r) for some r 21 2, where the
first map is given by x +t x and the second is the natural map.
Letting 5e be a primitive eth root of I in Kb we have
U~(1) U~(1)
(Let Y be thle compposite map. That U(r) c ker Y
is obvious. Also B(u) = 1 implies u = ac for some G E U(r). There-
fore u = at some i.) That U(r)<5q> <5Q> follows from the
U~r) nU(r)
AB A
vell known isomorphism: _for A, B subgroups of a given group.
B AnB
Hence,
(2) (U(1):U(r)] = [U(1) :U(r)][ >< nU)]
Now we will show that if r is chosen sufficiently large then
U(r)~ = U(r+e). Choose r = Ae. (IfeZ 2, this argument works with
r = 2e.) Consider the following expansion:
(3)(1 xw1 =1 +Ex/ f~x 2 + ... + exf- (1) + x nl
where X E O, C iS 8 DOnnegative rational integer and + is any prime
element of K nefine v(c) = v ((1+xxje-1). Then l(c) r min(c+e, cP.)
with equality if e + c Z c, i.e., c Z m = e/(e-1). Let c = r.
Since 5e < 4ee it follows that U(r)e c U(r+e). To show that
U~(rte) c Ur(r) first notice that 4e > eel/(b-1) +1 and so byProposition 5c
U(r + e) c U(1) .Let y E U(rte).Threxsscadx Ufo
which y = (1 + xwe L~. If c < e/L-1 then v(c) = cet < et/e-1. < r
which is impossible. If a = e/e-1 then an examination of Equation (3)
shows that v(c) = 2c + e = 2e/(b-1) + e < 5e = r which again is
impossible. If c > e/(&-1), then v(c) = c + e r r + e (since
(1 + xwce I 6 (r+e)); thus 1 + x*e E O(r). We therefore have
1
UT(rte) c ir(r)e and equality of these groups follows from the reverse
containment proved above.
Let us choose r so that r > 4e and 5e C U(r). (Shortly we will
indicate that Se E U(m) \ U(m+1); hence r = Ae suffices here.)
Equation (2) now becomes [U(1):U(r)] = [U(1) :U(rte)]l.
= [U(1):U(rte)][U(1) :U(1) ]-1*. Therefore [U(1):U(1) ]
= [U~r):U(r+e)]*e = eef+1. (Much1 of the argument leading to this
index is from Artin via Lang (7, Chapter 2, Section 3].)
From Equation (1) and the last index we now have
[G :G ] = [ )( eef+1 [U(1~;) U~m:U(1) ]-1
[Ur(1) U(xm) :U(1) ]
Therefore to prove the proposition it will suffice to show that
[U(1) U(fm):U(1) ] = ,, or, equivalently, that
(4) [U(em) :U(1) ni~(.m)] = a
To this and let us first show that U(m). = U(1). nU(men).
Letx EA.That v(m) := vn((1+xam) -1) s min(im, e+m) = Em and hence
that UIm)e c U() follows easily. Let (1 + xrrcL E iT(Cm) for SOme
rational c and assume x E U. It follows that v(c) 2 km. But if c < m,
v(c) = e + c < em. Hence c Z m. We have proved that
UT(u)e = U~(1) ni(~em) which allows us to express Equation (4) as
[U(-m) :U(m) ] = e; this is the index we must establish to complete
the proof.
We have remarked that 1-Se generates a principal ideal in Cp
which is the prime ideal lying over d with index of ramification
equal to e-1. It follows that 1-rC is a prime element of Q ( e)
and, being totally ramified in Qp(Sq)/Q must satisy 1-SQ = une/-
where n is a prime element in Kb and U E iU. Henc Se C iU(m)\iU(m+1).
But this holds for arbitrary primitive th root of 1. So
< >e UU'm). Using this relation in Equation (2) with r = m we
have [fr(1) :i(m) ] = Zmf-f. From Proposition 5c it follows that
U(1)" 2 U(Em+1l). In fact [U(1) :Ui(- em+1)] = [UI(1):U(fa+1)] [U(1) :U(1) ]-1
= Emf-1, Putting the last two indices together we have [D~(m) :U~(em+1)]
= f-. Tht[(m)Um = LP follows immediately.
We will now give some special notation for certain important ray
class fields of K. Let K(1 be the ray class field to the unit cycle,
i.e., an empty product of primes. Let K+ be the ray class field to the
cycle um the formal product of all real infinite primes of K. K(1
is called the Hilbert Class Field of K; it is the maximal abelian
extension of K that is unramified over K at all primes. K~ is called
the Extended Hilbert Class Field of K; it is the maximal abelian
extension of K thiat is unramified ove~ K at all finite primes.
(1) K
From Theorem 3, Gal(K /K a-.Let I be the group of
K'
fractional ideals of K and let P be the subgroup of I of all principal
ideals of K. For an element of JK we can write (aq) where a~ E K .
If v is a finite prime we write Py in place of v to stand for the
prime ideal. Suppose aAt2 = pr We establish a h~omomorphism of JK
onto I by (qy) HI br Composing this map with the natural map
v finite
of I onto I/P, we arrive at a homomorphism of JK onto 1/P whose
kernel is K'UK. Thus Gal(K 1/K) I/P, the ideal class group of _K;
the order of I/P is finite and is denoted hK'
JK K
Now notice that the natural map from KW onto KUhas kernel
v, K
K'UK(1 K'UK< K'IJmUK UK(
Therefore Gal(K /R ) a
K'W K'W K'W U nK'Wm
v, V. Um v
AB A
where we use the isomorphism B ABfor A, B subgroups of some given
group G. Note also that if G is abelian and if C 5 A then
Ana BC = (AnB)C. Therefore Gal(K +/K ) (UKK
For a given cycle c of K let Uc = !u cU u 1 mod p for p
finite and o(u) > 0 for real a i c). In particular note that U is
the set of all totally positive units of K, i.e., {u cU o(u) > 0
for all real o:K +C}. Notice that Uv 5 W ; so [K :K(1)] = [U :UW~]
[UK: U~m vm] [UK evm3 r -
[Uq UM/O] [UUv) = 2 1[U.:Uv.] whee e rl is the number of
real infinite primes of K.
Proposition 10. Let p3 be a prime K-ideal and let m be a non-
negative rational integer. Then Gal(EPmum~v/K ) = (A/pm)./9(U m
where O: U -t (A/p i) is the natural map.
AB A
Proof: We will use the isomorphism -and the identity
AnBC = (AnB)C (where C < A) freely in this proof.
JK JK K~'UK
The natural map from onto has kernel
K' TJmum K UK :' W avel
(1) K'U
Therefore, by class field theory, Gal(E m~V/K ) K 9
UK UK UK UK
.The natural map of
K' W mSnUK (K' nUK ~p"m m UWp Om UTWrJmu
U UW UW,
ono hskene m .Therefore Gal(E /K )
UW UWJrlV m pm UW mv
= UW n UW W U o The last
pmv vmVmn: V (U~V)pmU UumT~m U Uz (m)
isomorphism is a consequence of the p component map applied to W f
Now consider the isomorphism (A/p m)' EU / 8 (m) where
a mod pm + a mod U (m). (See the proof of Proposition 9.) Composing
this last map withl the natural map S /8 (m~) + 6 /U 0 (m) we arrive
at the homomorphism that we shall call Y': (A/pm) + /UV p (m).
Suppose a mod p is in the kernel of Y. Then acU (m) which implies
v, p
m
that there is some u cUL such that a -u mod p .Sinice a and u cA
it follows that a ~u mod p m. It is now easy to see that the sequence
U e+(/m +0/U 0 (m) + 1 is exact where 0 is the homomorphism
in the hypothesis. The result of the proposition is a consequence of
this exact sequence and that Gal(E m /K ) /U ()
Several of the remaining theoremns will require the following
lemmas :
Lemma 11. Let G be a finite cyclic group and let -e be a prime.
If H is a subgroup of G such that e I G:H] then H I G the subgroup
of the et powers in G.
Proof: Let G = . Suppose H = .Tn GH=r=m.
Therefore or = (om 1 and H r Ge follows.
The next lemma is a special case of a famous result known as
Hensel's Lermma.
Lemma 12. Let d be a prime. Suppose K is a finite extension of
w~ith p f E. Let p7 be the prime ideal of K and suppose asE U, the
units of K. If a E Ye mod p for some y EUi then a E .~
Proof: We have ay- c U~(1), a Z -module. (Zp acts exponentially;
cf Long [8, p.61].) Since e is a unit in Z_ the result follows.
Lenmma13. Let e be any rational prime and let S be any finite
set of prime K-ideals. Suppose a sK and that 8 E K for all prime
K-ideals p L S. Then a sK.
Proof: See Artin-Tate [1, p. 82].
Theorem 14. Let K be an algebraic number field. The following
are equivalent:
1) Every quadratic discriminant ideal of K has a unique
factorization as a product of prime quadratic discriminant ideals of K.
2) The class number hK is odd and every totally positive unit is
a square, i.e., U = U .
3) K is totally real and [K :K] is odd.
4) The class number hK is odd and the natural map
4: U - (A/4A)'/(A/4A).2 is surjective with ker Q = U2
Proof: The proof of Theorem 14 will proceed as indicated:
1) 2) 3) 4) 1).
1) 2) Let p? be a prime K-ideal such that p k 2. Let
L = K(/a) with (a() = pb where b Z p is a prime K-ideal in the inverse
ideal class of p. From Proposition 8 we have v (DL/K 1
The class of any discriminant ideal is a square in the ideal
class group. (The proof of this statement is deferred until
Proposition 16.) So given that 1) holds it follows that P is in a
square ideal class of K; but this must be true for all p k 2. If we
consider the fact that each ideal class of K contains an infinite
number of prime ideals, it must be that every ideal class is a square.
Hence hK is odd.
Let EpV be the ray class field to the cycle py. of K. Then
Epym is the maximal abelian extension of K with conductor dividing pum.
If p k 2 is a prime quadratic discriminant ideal, say DLK it follows
from Proposition 2 that the conductor of L/K must divide py, and hence
thtL pvm By Proposition 10,Gal(E ~/K) ) (A/P)./6(Uz ".
Therefore 2 [(A/P) :0(U,,)]. But since (A/P)' is cyclic (A/p is a
finite field)- Up omUU follows from Lemrma 11.
p p
So each element of Um is a quadratic residue mod p. By
Lemma 12 (Hensel's Lemma) it follows that each element of Um is a
square in K j. Therefore for all p k 2, U_ c K2. So by Lemm~a 13 it
follows that U c U since only a finite number of prime K-i~deals
divide 2. But U 0 U2 is obvious. Therefore U =U
21 o 3) Let H be the group of all roots of unity in K. The
Dirichlet Unit Theorem states that Ui EH x Arlr- where rl (respec-
tively 2r2) is the number of real (complex) imbeddings K +t C. Since
[H:H2] 2 it follows that [U:U2] = 2rlr2. But we proved in the
discussion proceeding Proposition 10 that [K :K l] = 2rl/[:y ].
Since UV._ = U2, we have r2 = 0 and [K :K(1)]i = 1. As hK is odd by
hypothesis, it follows thiat [K :K 1] is odd.
3) 4) In case K is totally real it follows that [U:U2] = 2rl
Also by Proposition 9, [(A1/4A)':(A/4A).2] = 2rl We need show only th-at
the kernel of the homomorphism Q of the statement 4) is U2. Certainly
U2 c ker g. Suppose u s ker Q \ U2. Consider L = K(Ju). By
Proposition 8-4) no prime K-ideal above 2 ramifies in L/K; by
Proposition 8-2) no other prime K-ideals ramify in L/K. Therefore L/K
is unramified at all finite prime K-ideals. So L e K But
2 a [K :K]. This contradiction means that the relation u E ker 4 \ U2
is false so that U2 = ker 4.
4) + 1) Let L/R; be an arbitrary quadratic extension of K. For
each p DL/K we must produce a quadratic extension L(p)/K such that
DL(p)/K is a prime discriminant and v (DL(p)/K) =p(DL/K '
By Proposition 8 if p DL/K but p / we must exhibit L(p) such
that DL(p)/K = p. To this end let a~A = p .(Recall that hk is the
class number of K.) Since the natural map 9: U + (A/4A\)'/(A/4A).2 is
surjective, we may assume without loss of generality that a_ is a unit
quadratic residue (i.e., congruent to a square) mod AA. Therefore by
Proposition 8 we may choose L(p) = K(Vap).
We now take up the case of prime K-ideals that lie over 2. Here
it will be convenient to set L = K(V5). In considering a particular
prime p3 | 2 we will assume that either v (a) = 0 or v (a) = 1. (See
the note following the statement of Proposition 8.)
Suppose first that P is a prime K-ideal lying above 2 and that
p a a. Suppose a y2 mod p2s for some y E A., but a is not. a
quadratic residue mod p2(+) (Thus we define a nonnegative
integer s.) Let e_ = e(p/Q). From Proposition 8 we have:
1) If s < e then v (DL/K) = 2(e -s).
2) If s 2 e then v (DL/K) = 0.
He need consider only case 1). Since Q is surjective we may
choose a E U such that
2 2e 2e
up + + p mad (A/p )2
2e 2e
S1 + p )"mod (A/b b).2 for all other primes b 2 and
b f p. Thus referring to Proposition 8 again, we may set L(p3) = K(lap).
33
Suppose now p is a prime K-ideal above 2 and p a.
Let G = (A/4A) and let GI = (A/ II 2e ) ', where b are prime
b|2
bip
K-ideals. Let \': C +t G be the natural map. Yi induces
'Y: G/G2 + G /Gr2 since ?(G2) c Gr2. But 4 is surjective and F is
surjective; therefore Yo@1 is surjective. If aAn = p we deduce
exacly s ws dne n te cse / tht a may be chosen such that
cc E G12. It follows from Proposition 8 that we may set L(p) = K( -).
SECTION VlI
THE CASE OF QUADRATIC IDELE DISCRIMIINANTS
Let L be a finite extension of the algebraic number field K with
ringof itegrs Bandlet b } be the set of prime L-ideals that
lie above p~ a prime of K. Let Lb be the completion of L at bi viewed
as lying in the algebraic closure of K Let ~B denote the ring of
integers of Lb.
1i
Let ai be the imbedding of L into Lb~. The map given by
ge +. (al(k)a, 02(C)a,...,o (t)a) provides an isomorphism of K
algebras L mK K H P=1 L .. (For a proof see [8, Chapter 3, Proposi-
tion 3.12].) This same mapping leads to B erA A3 n- =1 Bb. as
Ap modules. (See [8, Chapter 3, Corollary 3.13].) Thus Bl OA Ap is
the maximal order of L OK Kp (siince iBi is the maximal order of L ).
The discriminant of any basis for B $AA as an A -module determines a
uniquely defined element of Al /U~ that is denoted dL@ Kp (or dLeK '.
SiceB A H=1b. it follows that
i 1
(1) dLeK = "=1 dL /K'$
But from Proposition 6, DL/K p =1 dL / A so
(2) dLeK p = DL/K Ap.
Frohlich defines the idele discriminant of L/K, DLK to be the
2 th'
element of JK/UK whose p~t component is given by (1)L/K = dLeK E K.
That this actually defines an element of JK/UK; follows from the fact
that (DL )p EUp / 2 for all but a finite number of p as is seen from
Equation (2) and the fact that only a finite number of primes ramify
in L/K.
The natural map 0: JK/UK i K/UK takes DL/K into DL/K where this
ideal is viewed as an element of JK/UK: let y: JK + I be the map
given in the discussion proceeding Proposition 10; ker y = UK, thus
JK/UK a I. An idele discriminant of K is said to be a prime idele
discriminant of K provided its image under ai is a prime discriminant
ideal of K.
For any extension L/K let dL/K denote the unique element of
K./K.2 with representative dL/K(xl.,..,x) where {x1..,xn) is any
basis of L/K. (See Equation (1) of Section I.)
Propsitin 15 LetL, (.}m be given finite extensions of K.
Then L/ =1 L./ if and only if DL/K m=1 DL /Kan
dL/K i= dL./K
Proof: Let 6 and the 6. be fixed butarbitrary representatives
of DL/K and the nL/,rsetvli K Let w and the oi be fixed
but arbitrary representatives of dL/K and the dtL~./ Trepectively, in
K' .
Assume DL/ =1 L./K That DL/ =1DKfllw from
properties of the homnomorphism o: JK/UK; 'K/UK discussed above.
Let p be a fixed but arbitrary finite prime of K. Let {xi n=1 be
a basis for L/K. From the theory of tensor products we know that
fx 01) is a basis for L 0K Kp over K It is easy to see that
(3) dL@K /K (x 01) = dL/K(x ).
P P
Using an exactly similar argument to that used in the derivation of
Equation (1) of Section I we can conclude that d /K(x el)
L0 K i
B 6 md Kp2 which together witih Eqluaion (3) yields 6 E dL/K(x )
Em mod JK2. Thus we have (since this congruence also must hold for
the pairs 6., Wi) o Em e. 6 Em 6.mdJ2. 6 Em cJ2
1 1 i=11 11 1 K' i=1 K'
wec m 2 m K2
whnc a 1 e E J By Lemma 13 we conclude that al nm 8 '
Therefore dL/K Ti=1 dtL./
Nowassme L/K= i~=1 DL/K and dL/K ni=1 dL./K. Since
DL/ =1 DL/ it follow that DL/ =1L/ hr EU
Therefore 6 H.l~~ 6. r mad JK and as in the first part of this proof
6 i=/i es Hi=1 i modn JK But w li=1Ui ; l mod JK2 by hypothesis.
Hence y c U2 and it follows that OL/ 1L/
K L/K ~l L.1
We now prove some results of Frohlich (3J as
Proposition 16.
1) I]L/K E JK2K'/UK2 for all L/K.
2) The ideal class of DL/K (i.e., the coset DL/Ki;
group IK PK) is a square.
3) If B c JK2 K'/UK2 there exists at most one quadratic extension
L/K such that OL/K =B
Proof: 1) Let {x ) =1 be a basis for L/K. In the proof of
Proposition 15 we showed that dL/xi) 6mod J2 where 6UK L/ D
This yields DL/K K2K'ZK/UK2
2) Consider the isomorphism JK/UK + I composed with the natural
map I +~ I/P. The kernel of this composite map is K'UK/UK; thus
JK/K'UK K I/P Asbfr et KU2 K/UK It follows from
1) that o(DL/K() E JK2K./UK but (identifying JK UK and I) o(VL/K)
= DL/K .Therefore the ideal class of DL/K is in KK JK_. The
K'UK K'UK
result in 2) follows.
3) Suppose L and E are both quadratic extensions of K with
DL/K = E/K To show that L = E we will employ the following
theorem: Let L1, L2 be two normal extensions of K and let S1,S2 be
the sets of primes of K that split completely in L1,L2, respectively.
If S1 C S2 (except possibly for a finite set) then L2 c Ll. (A proof
of this may be found in Janusz [6, Chapter IV, 5.5].) We will now
show that a prime K-ideal p splits in L ( i.e., pB = b 62 with bl b2
prime L-ideals) if and only if deK=U.
Let L = K(la). As in Proposition 8 we may assume that
v (U) = 0 or 1. Assume pJ splits and bl b2 are the prime L-ideals
above p. We have seen that [Lb.:K ] = e(bi/K)f(b /K) and
:=2 e(b./K)f(ib./K) = [L:K] = 2. Therefore e(bi/R)f(b./K) = 1and
[L :Kp ] = Hence dLoK = lUp follows from Equation (1).
To prove the converse suppose that p remains prime or ramifies
in L. Then K (la) is a quadratic extension and therefore o i K .
(Here (VE) is K p-isomorphic to the completion of L at the prime
above p). From the corollary to Proposition 8 and Equation (1) it
follows that d = aUT or Rcri~ where a is not a unit in K~ In
LQK P P
i2 2
any case it is clear that dLeK J- lUp therefore if dLDK = le then
P P"
p must split in L.
Since DL/K = E/K it follows from the preceding discussion that
the prime Kt-ideals that split in L/K are exactly those that split in
E/K. From the theorem mentioned at the beginning of the proof, we
have L = E.
Theorem 17. Every quadratic idole discriminant of K has a unique
factorization as a product of uniquely generated prime quadratic
idele discriminants if and only if hK =[I:KisodndKsttly
real.
Proof: Necessity follows from properties of o: JK/UK2 K JK/
and Theorem 14. Now let L/K be an arbitrary quadratic extension of K.
L = K((a) for a EA where we assume that v (a) = 0 or 1 for all
p DL/K (See the note following the statement of Proposition 8.)
By Theorem 14 there exist fields L(p)/K such that L(P) = K(Vap),
with ax~ as defined in the proof of Theorem 14 (4+1l), for which
DL/K =pl H KDL(p)/K and DL(p)/K is a prime quadratic discriminant ideal
of K for all prime K-ideals p that divide DL/K
We will show that dL/K =H dL(p)/K .Theorem 17 (except for
piDL/K
the uniqueness properties) will then follow from Proposition 15.
If we choose {1,/2p} and {l,/a) as bases for L(p)/K and L/K,
respectively, we see that dL(p)/R(1,/a) = 4ao and dL/K(1,/a) = 4cL,
therefore ( H a ) aK.2 = (n dL(p)/K) 'dL/K Thus we must show
p|DL/K p DL/K
that ( H a ) a E K .2
plL/K
From Equation (1) of Section I we know that dL/K;(1,72) is
congruent to the discriminant of any other basis of L/K module K.2
Therefore dL/K(1,72)A E DL/K mod 12 This holds also with L and a
replaced by L(p) and a respectively, therefore
&L/K(1,a) EdL(p)/K AE DL/K H DL/~(p)/KI mod I2 Thus
(arPDa) 2 2
(aH )A l .Since hK is add, (l a )A = rA for some
p L/K pDLn/K
r E A'. Therefore there exists a unit u cU such that "TeH p a=ur2
It will be convenient to distinguish four classes of prime
K-ideals:
SO = {p: P 2, V (a) = 11
2e
S1 = {p: p 2, pke, a is not a quadratic residue mad p }~
where e_ = e(p/Q).
S2 = p: p 2}
S3 = {P: P DL/K but pf2}.
We now decompose the left side of the last equation to obtain
2 2eb
aa lIa la =n ur. If u is not a quadratic residue mod b
pcs ps p s p
with b c SO we must modify ab in the manner that is now described.
2eb leb~o (/2eb)~ .2 dl
Let ub E U be such that uq+b -u+6 o A6 n e
2e
ub be a quadratic residue mod p P orllp S2 b.Such a
choice of ug is possible since by Theorem 14 4) the natural map
4: U +t (A/4A)'/ (A/4A).2 is surjective. We can now redefine
L(b) to be K(/uba).
2eb
If on the other hand u is a quadratic residue mad b
set ub 5 1. In this way we choose a u for each pEs S.
We have
(4) a u a n a H a =u0r
PcSOP p ES1P pS3P1
where u0 = u up He now show u0 r U We will do this by
pes
application of theorem 14 3) -t 4) which asserts that the natural map
Q: U +t (A/4A)'/(A/4A).2 has kernel C12. Thus the proof is complete if
we show that u0 is a quadratic residue mod 4A. Clearly it will suffice
(in fact, is equivalent) to show that ug is a quadratic residue
2e
mod p for all p 2. This we will do by considering separately the
cases p c SO, p e S1, and p e S2\(SO U S ).
2eb
Let bEi SO. Then u0 u up mad 6 which gives
PES0
2 2b 2 2b
U0 = Y ubu mod b since H u E y mod b for some yEiA such that
pcSO
p~b
bkv2 2 2 2b
b .Because of the choice of u :10 Y Bua mod 6 for some
3 e A.
Suppose that b e Sl. Let us recall that for all p DL/K such
2eb 2eb 2eb2
that p # b we chose a +b -1+6 mod (A/b b).2 Therefore there is
P 2e
some y C A such that b J v and n a 2 md bb.As elltt
pD/KP
2e6 2e6 2eb2
ar was chosen such that ab+6 -a+6 mad (A/6 b). (and here b k a).
22e
Threo re Euthere4 yexiss 6 Asuhthtb adaa 2 2 mo
NowEqutin () yels a 2 Eu0rl2 mod 62 But b J aBu, so b k rl'
2eb
hence uO is a quadratic residue mod b
Finally suppose b e SO \ (SO U Sl). Here b does not divide the
left side of Equation (4) and each term is a quadratic residue
mod 62e (b does not ramify in L or L(p)). Therefore
2e
quadratic residue mod b
We have thus shown that u0 is a quadratic residue mod 4A. By
Theorem 14 3) 4) u0 E U2. Thus dL/K = H dL(p-)/K and, by
pDLn/K
Proposition 15, DL/K = L,(p)/K .Uniqueness of generation of all
p DL/K
quadratic idele discriminants follows from Proposition 16. The
uniqueness of the factorization is the content of the next therorm.
Theorem 183. Let K be an algebraic number field and for a
prime K-ideal p let L(p) and E(p) denote quadratic extensions of K
that are ramified at p but are unramified at all other finite
primes of K. Let DL/ P L/(p)/K and DL/ ~ )Kb w
factorizations of the idele discriminant of a quadratic extension
L/K;. If K is totally real and if h = [K :K] is odd then
OL(p)/K = E(p)/K; for all p? | DL/K and (equivalently) L(p) = E(p)
for all p DL/K'
Proof: Consider for the moment that p is a fixed prime K-ideal
for which an extension L(p)/K exists. Let L(p) = K(Jla ) where ap e A.
We will show that ap can be chosen in the following way:
1) If V (DL(p)/K) is odd then we can choose ap such that
hK 2eb
arA = p and ap is a (unit) quadratic residue mod 6 for all b 2
and b # p (eb = e(b/Q)).
2) If V (D ) is even then we canechoose a such that
ay E U and ap is a quadratic residue mad b for all b 2 and b / p.
From Proposition 8 and the note that follows the statement of
the proposition we see that, for any prime b of K, Vb(DL(p)/K) is odd
if and only if Vb(Up) is odd.
Thus in case 1) v (a ) is odd. Let opA = PC2 for some integral
ideal C. Then ahA = ph Ch 2. (Here and for the rest of this proof
h h -2\ ph
we let h = h .) But C = cA for some c e A. Therefore ac A=
and of course, K(h -) L(p). From Proposition 8 it follows that
b -2 2b
a c is a (unit) quadratic residue mad b for all primes b 2 and
b f p (since these b do not ramify in L(P)/K by assumption).
In case 2) qb(a ) must be even for all primes b of K. Therefore
a=2. -2
aA= But hK is odd so C = cA for some C E A. Thus ace A = A.
-2
Clearly we can replace ap with ae E U and again from Proposition 8
-2 2e
we have that acr must be a quadratic residue mod b for all
b 2, 6 # p.
Now let L(p) = K(V~) and E(p) = K(B p) for each p DL/K,
where we assume that the a Bp have been chosen in the manner just
described, that is, we will assume that ap E U and Bp E U, Or elSe
a A = ph and B A = P In either case we have that ap and ap are
unit quadratic residues mod b for all b 2 and b p. In
-1
particular we have apB E o alp
From Proposition 15 it follows that dL/K(14) H dL(p)/K(1, /ap)
p DL/K
c K. and likewise dL/K(1, la) n dE(p)/K(1, /Ep) K. 2.Tu
H aSp c U2 since2 dL(p)/K(1, /a) = 4np (and similarly for the
other discriminants).
-1
Let w = a 6 .Since cp and Bp are unit quadratic residues
mod b for all b 2 with b / p it follows that w_ is also a unit
2eg
quadratic residue mod b for all b 2 with b Z p. But if we let
2 -1 2
Hl w = v for some v eU then w = (H w )v and so
p DL/KP 1 pIDL/K~
p/pl
2e
w r2 mod pi Pl for some r E K' We see then that w is a
pl p
quadratic residue mod 4A for all P DL,/K and hence wp E ker Q.
(See Theroem 14 3) +t 4)). Since ker + = U2 it follows that L(p) = E(p)
for all p.
We close this section with a look at the relationship of dL/K to
DL/K. Suppose that B is free as an A-module so that dL/K is defined.
Let {x } =1 be a relative. integral basis for L/K. Since the map
o: JK/UK Rg/UK has the property D(DL/K) = DL/K and since
DL/K = dL/K(x )A (5 I), we have v (6 ) = v (dL/K(x )) for all p3 where
we let 6 Ul = (3L/K p = d,,K But we also showed in Proposition 15
tha 6 dL/K(x ) mod Kp2; fo~r all p3. Therefore dL/K(x )UK~ =L/K'
Since dL/K E A/U2 it may be considered as an element- of JK/UK via
th~e imbedding of K' + JK We have
Proposition 19. If B is free as an A-module then dL/K = L/K'
It follows from Proposition 19 that the classical result of
Theorem 1 is a special case of Theorem 17.
SECTION VII
REMARKS ON A THEOREM OF GOLDSTEIN
The purpose of this section is to show that the results of
Goldstein in [4] are a special case of Theorem 17.
Goldstein has shown
Theorem Gl. If K is totally real of narrow class number 1
(i~., K =1)then every numerical quadratic discriminant in K is a
product of prime numerical quadratic discriminants that are uniquely
generated.
Let L/K be an arbitrary abelian extension. Denote by L" the
maximal abelian extension of K which is unratified over L at all
finite prime L-ideals.
Theorem G2. Let K be totally real of narrow class number 1.
Let dL/K be a numerical quadratic discriminant of K which is divisible
by exactly t prime K-ideals. Then
]) [L :L] > 2t-1 and
2) The factorization of dL/K given in Theorem G1 is unique if
and only if [L*:L] = 2t-
Now Theorem G1 is an immediate consequence of Theorem 17 and
Proposition 19. But Theorem 17 and Proposition 19 also imply a
uniqueness of the factorization given in Theorem G1 which is independent
of [L :L]. Thus our result in Theorem 17 is seemingly at odds with
Theorem G2. We will resolve this conflict by showing that the case
[L*:L] > 2t-1 never occurs and hence that Theorem G2 is consistent with
our results.
An additional result of Goldstein will be required,but first
we present some facts about Galois extensions. Suppose that L/K is
Galois and C = Gal(L/K). Then G acts transitively on the prime L-ideals
lying above a given prime K-ideal p. 18, Chapter 1, Section 4.1]
It follows immediately from this that if b1 b2 are prime L-ideals
over p then e(bl/K() = e(b2/K). In case L/K is Galois we will denote
the common ramification index of all prime L-ideals over p by e(p,L/K).
Proposition G3. Let S be the set of all finite primes of K
and let rl be the number of real primes of K. Then
,~~ h2 e(p,L/K;)
[L^:L] =where UL/K is the set of units of K that
[L:K] [U:UL/K
are local norms at all finite primes of K and are totally positive.
A unit u of K is a local norm alt p provided u cN b/K(~L ')
where Lb is the completion of L at any prime b that lies above p.
(Since L/K is abelian NL /K: (L ') = NL ,/K(Lb,) if b' is any other
prime of L lying over p.) Hasse's Norm Theorem states that if L/K
is cyclic then X E n NL /KL ") implieS X E NL/K(L'). Thus in the
p finite Lb p
cyclic case n NL /K(L ') n K =NL/K(L'). Furthermore,
p finite b p
UL/K = L/K(L') nUg
Let us now assume the hypothesis of Theorem G2; that is, that
.+
[L:K) = 2 with K totally real and hK = 1. W~e have shown previously
(5 5) thant hilK K = [K:K(1)]i = 2rl[U:U~ -1. But here hK = 1, so
[U:UY ] = 2rl. Also since K is totally real r2 = 0 and therefore,
m r1
by Dirichlet's unit theorem, (U:U'] = 2 Bu lalyU E
Theefoe U = 2 and UL/K = NL/K(L.) n U2. It follows from~
Proposition G3 now that [L :L] = 2r 2t 2t
Thus Goldstein's Theorem G2 is consistent with Theorem 17 and
uniqueness of the factorization in Theorem G1 always holds.
Remark: Goldstein [4] also proved that the prime discriminant
factors in the factorizations achieved in his results were distinct.
That this kind of distinctness holds in the prime discriminant
factorizations produced in this thesis is a trivial consequence of the
method of our proofs. Herein we prove that given an extension L/K
and a prime p that divides DL/K then there exists subject to certain
conditions--another extension L(p3)/K (of the same type as L.), ramified
only at p3, such that U (DL/K) Vp L(p)/R;). Distinctness of the
DL(p-)/K for p DL/K follows immediately. This illustration carries
over easily to the idele discriminants.
SECTION VIII
THE CA.SE OF CYCLIC e-DISCRIMINANUT IDEALS, g > 2
We begin with a characterization of those prime K-ideals that
ramify in some cyclic e-extension of K with I > 2.
Proposition 20. Let K be an algebraic number field and let
k be an odd rational prime. Given a K-prime p~, p it, there exists a
cyclic e-extension of K in which p~ ramifies if and only if
NK~/Q(p) E 1 mod .
Proof: Suppose L/K is a cyclic &-extension and p DL/K is a
prime K-ideal. Let b be the prime L-ideal that lies over p. Then
Lb/Kp is a tamely ramified cyclic -extension. It follows from
Theorem 4 that [Kp' : N(L ')] = &. (Here we are letting N stand for
NLs/Kg .) Fromc Prcposition 5b we have e [ :U (1)]. But, as we
saw in the proof of Proposition 9, (A/p)' U /U (1). Since
I(A/p)' I = NK/Qg(p-) 1, it follows that NK/Q(pJ) ? mod b.
Now assume that NK;/Q(p) = 1 mod 1. This implies that
[ U :U ~(1)J. Therefore there exists a subgroup H of K of
index e which contains U (1) but does not contain U Biy Theorem 4,
there exists an extension L/Kp such that Gal(L/K ) K lj/H and
NC/K (L') =H. Therefore~ Fi/K= p,~ and, by Proposition 2, DL,/Kp
= (pA )e-1. Therefore L/Kp i s atamely ramified cyclic C-extension.
A special case of Grunwald's Theorem is needed: Let F be an
algebraic number field; q, a rational prime; and 6, a prime F-ideal.
Let E be a cyclic q-extension of Fb. Then there exists a cyclic
q-extension E of F' such that for any prime E-ideal C that lies over 6,
Ec is KP~-isomorphic to E. (Shianghaw Wang found an error in the
original Grunwald's Theorem which he pointed out in a paper of 1948.
For the corrected version see [101.) Note: If E/F is ramified it
follows, of course, that b DE/F, but DE/F is not necessarily a
prime discriminant ideal.
This case of Grunwald's Theorem fits our problem exactly (K = F,
e = q, p = 6, L = 2) and the result of the proposition follows.
In the next proposition we will refer to the natural map
6: U + (A/pm)' which was introduced in Proposition 10.
Proposition 21. Let K be an algebraic number field and let e
be an add rational prime. Suppose e k hK and let p be a prime K-
ideal such that p / -LA. There exists a cyclic ~e-extension L/K
that is ramified at p and unramified at all other finite primes if
andony f N/Qp)E 1mo Iand B(U m) c (A/p)' Whenever such
an extension L/K: exists it is unique.
Proof: Suppose that L exists. Let b be a prime L-ideal lying
over p. We have from Proposition 5b (letting N stand for the norm
from Lg to K ) that N(Lb') Uy(1), but N(Lb') Up since L /Kp is
ramified. Therefore the conductor FL /K = 1pA Since p is the
only finite prime of K that ramifies in L we must have FL/K PV"
(vm is the product of all real primes of K).
Epv, is the ray class field to the cycle pum and is the maximal
abelian extension of K with conductor dividing Pvm. Since e a hK
= [K(1):K] and since [~K :K(1)] is a power of 2 it follows from simple
Galois considerations that L exists if and only if i [E :K ].l
(Note: e b hK is not needed for =>.) Let O: U + (A/p)' be the homo-
morphnism given by 0(u) = u mad p. Then,as was shown in Proposition 10,
L CEpmI:K ] if and only if I (/):0U.] But
e [(A/p). 9(U, )] is equivalent to L 1(A/p).I and 0(U ) c (A/p).
by Lemma 11 since (A/p)' is cyclic. Also because (A/J)' is cyclic
it follows from Proposition 10 that Glal(Epym/K ) is cyclic. There-
fore the -group component of Gal(E ~/K) is cyclic because
1 [ K(1):K] (= hK) and [K :K(1)-] is a power of 2. Hence if L/K
exists then it is unique.
Theorem 22. L~et K be either the rational field or an imaginary
quadratic field. Let e be an odd rational prime. If e / hK and if
K does not contain a primitive eth root of 1, then for each p with
N(p) = 1 mad e there is a unique cyclic L-extension L(p3)/K ramified
at p and unramified at all other finite primes.
Proof: Since :4 d K, it follou~s from Dirichlet's unit theorem
(see Theorem 14, proof of 2) => 3)) that [U:U ] .elrZl r2- here
rl is the number of real primes of K and 2r2 is the number of complex
primes. For K as given in the hypothesis either rl =1, r2 = 0, when
K =Q, r lserl= 0, r2 = 1, when K = Q(12) for some d < 0.
Therefore for any such K; we have Ui = U In particular it follows
that 0(Uem_) c (A/p) &(0: U -t (A/p) the natural map) for all primes
p of K. Application of Proposition 21 completes the proof.
We will now take up the case that was excluded from Theorem 22
by the hypothesis that 5e C K, i.e., the case of K = Q(53) and C = 3.
The class number is hK = 1. (Refer to [2,Tables]) Let 3A = B2A be
the prime ideal factorization of 3 in Ki (R = 1 5 ). It is known
that K( E)/K is unramified at eA if and if only if a y3 mod B3A for
some y E A'. (For a proof see [5, Section 39].)
Lemm 23 Letas \ .3.K( a)/K is ramified at BA if and
only if as 1 1 mod B A.
Proof: We will show that (A/B3A).' (2) x o(3) x o(3). The
lemma then follows by counting the number of cubes in the abstract
group on the right-hand side. (Recall o(m) = cyclic of order m.)
We have shown before in this thesis (e.g., see the proof of
Proposition 9) that: (A/B3A)' 6/U(3) where we let U = U g;
U/0(3) U/U(1) x 0(1)/8(3); U/U(1) = (A/BA)'; and U(i)/ir(i+1)
" A/BA for i > 1.
Since A/SA Z /3Z we will know the structure of A/BJA as
soon as we know the structure of U(1)/U(3). That U(1)/U(3)
S0(3) x o(3) will be the content of Lemma 27 which follows. There-
fore (A/R34) a o(2) x o(3) x 0(3).
It follows from elementary ramification theory and the quadratic
reciprocity law that a rational prime P splits in R/Q (i.e., it has
prime factorization pA. = blb2) if and only if p = 1 mod 3. (Samuel
E~9, Section 5.43.) For such a p w~ith prime factorization pA = b ,
let b1 = alA. Now NK/Q(al) = alo(al) = 'p where = Gal(K/Q). That
NK/Q(al) = p follows from the next lemma.
Lemrma 24. N'K/Q(x) is positive for all x c A'.
Proof: Here A = Z[S3]. Let N = N .q Let x = a + bll with
a, b c Q. N(a + b/-) = a2 + 3b2 > 0.
Proposition 25. Let p be a rational prime and let aA be a
prime K-ideal lying over p. If p 1 mod 3 but p / 1 mod 9, then BA
ramifies in K(3/a)/K.
Proof: From the fact that p splits in K and from Lermma 24, we
have p = aco(a) where = Gal(K/Q). Assume that BA does not ramify in
K(3Ja)/K (note that a cannot be a cube in K). Lemma 23 implies
aE 1 1 mod G3A. But then o(a) = '1 mod B3A. Hence p = ac(a)
E 1 mod 8 A which implies that 3 (~p-1) E A and 3 (~p-1) 0 mod BA.
It follows that 3 (~p-1) E Z and 3 (9,-1) = 0 mod 3. Hence p r 1 mod 9.
In the case at hand every cyclic 3-extension of K is of the
form K(3J4) for some a t A'\A 3. Fix one such a. and set L = K(3/E).
Suppose that L/K is ramified at some prime K-ideal p but is
unramified at all other finite primes. k'e wish to show that L = K(3J;)
where uA = p. It is well known that a prime K-ideal p ramifies in L/K
whenever v (a) > 0 and 3 / v (a). (For a proof see Hecke
[5, Section 39].) It follows that aA = pr 3 where r = '1. Since
hK = 1, b = bA for some b E A'. Hence ab-A=p n (b)A=p
Therefore we may set 9 = (ab-3 q.
It is now clear that Proposition 25 asserts that whenever p
is a prime K-ideal lying over a rational prime p such that p 1 mod 3
but p / 1 mod 9 then there does not exist a cyclic 3-extension over K
in which p is the only ramified (finite) prime. Therefore
Proposition 25 provides us with ample justification for having excluded
the. case K = Q(7-5), e = 3 from Theorem 22.
The case of K = Q(V-3), e = 3 which we have just finished by an
explicit construction is also covered, in part, by the following
general theorem; however, the proof of the tneorem is by an indirect
argument.
Theorem 26. L~et e be an odd prime. The property of prime cyclic
e-discriminant factorization of cyclic e-discriminant ideals does not
hold in algebraic number fields that contain a primitive kth root of 1.
Proof: We need the following result: If b is a prime C e-ideal
such that b J e then N(b) l mod e. (N is the absolute norm.) To
see this consider the polynomial X -1 mod b. Let B be the ring of
integers of C Since 5e E B, X -1 splits in B/b into L linear
factors. Since b k e each of these linear factors must be distinct.
Therefore e |(B/b)'|, i.e., N(b) 1 mod e.
Let us assume that the factorization property in the hypothesis
does hold for some K 2 Cg We will look for a contradiction.
Let p be a prime K-ideal such that p / LA. Since
N(pn@C~) 1 mod I it follows that NT(P) = 1 mod 8. Hence by
Proposition 20 some power of p must be a prime cyclic e-discriminant.
It is pointed out in the proof of Proposition 21 that the "necessary"
part of that result does not require the hypothesis that e / hK. We
use this fact here to conclude that 0(U m) (A/P)' But this means
thateachelemnt o U s cogruet toan th pwrmdP
tht ah lmetofU iscnguettoa I pwe odp
It follows from Lemma 12 that Um c U But this must be true for
all primes p3 / eA. Therefore by Lemma 13 Um C U Now [U:U _]
is a power of 2, whereas [U:U ] is a power of I and yet
(U:U ] [UL:Uv_ ]. Therefore either U = U or e = 2. If U = U we
contradict the assumption that C~ c K. If I = 2 we contradict the
hypothesis that I is odd.
Lemma 27. Let e be an odd rational prime and let K~ be a ramified
quadratic extension of Q e. Let 6 be the units of 2. Then U(1)/i(3)
o(L) x o(1). (o(1) denotes the cyclic group of order L.)
Proof: Let e be the ramification index of K~/Q = 2. If I f 3
then the greatest integer in ee/(C-1) = [[et/(e-1)]] = 2. If 53 / K
and e = 3, r[[8/(e-1)]] = 3. We have from Proposition 5c that
0(3) c 0(1)* in all of the cases mentioned. If 53 E R and e = 3, we
conclude from Proposition 5c that U(3) d U(1)d but U(4) c U(1).
Considering an arbitrary case again we let a be a prime element
of K. Let x = 1 + an with as A Thus x = 1 + an + ..+ a n
S1 mod a". So in all the cases of the first paragraph we have
Ur(3) 2 Ur(1) InI the cases e P 3 and B = 3 with 53 K w~ e conclude
tha U() =U(1 .If e = 3 and 13 c K we have 0(4) = U(1) .Thus in
all cases I(U()/U(3):(U(1)/U(3)) ] = 2. The result follows readily.
(We have used implicitly the fact that U(i)/U(i+1) A/p which holds
for all i > 1. The proof for i = 1 given in the proof of Proposition 5b
suffices with little change to prove the general case.)
Proposition 28. Let I be an odd rational prime. Let K be either
the rational field or an imaginary quadratic field. Suppose that
e k hK. Let p be a prime K-ideal such that P I A4. If p ramifies in
a cyclic e-extension L/K thenI there exists a cyclic e-extension L(p)
ramified only at p3 among finite primes andv (DL/K) =p(DL(p)/K '
Proof: The case of K = Q is dealt with in Theorem 7 where a
stronger result is obtained. So let K = Q(12) where d is a square free
negative integer. Let b be a prime L-ideal above p. We seek the
conductor of L /K .
Since L /K;I is wildly ramified it follows from Proposition 5b
thatN(L') (1) (N= N /K.) Suppose first that k is unramified
in K/Q. Then [[ee/(e-1)]] = 1 where e = e(K /Q ) = 1. From
Propostion 50 we have that U p(2) 5 U(1) Therefore N(Lb. ") 2U (2);
so the conductor of L /Ki is (pa )2. In this case by Propositions 2
and 6,v (DL/K) = 2( 1).
Now suppose that e is ramified in K. Let us handle as a
separate case that of K = Q(7-) and e = 3. In the proof of Lemma 27
we saw tha3t (N(L ') 2 ) U~(1)3 = U(4). Hence FL /K = (P~p) so that
v (DL/K) = 4(e 1). Wh~at is important here is that there is only one
p-power cyclic 3-discriminant in K. Thus if we produce just one
cyclic 3-extension L(p)/K in which p is the only ramified finite prime
the case of K = Q(rClj) and e = 3 is complete. In fact though we will
list all such extensions since there are only 3. From the discussion
of the case K = Q(1V- ), -e = 3 that follows Theorem 22, we know that
L.(p) = K(3Ja) where a is a unit of K or aAl = p = (1-r; )A. If G E \ U ,
it follows from Theorem 3 that (1-53)A ramifies in K(3/a)/K since hK =i
Therefore the fiedfomhchL) may be chosen are: K(3J- = 'g
K( ), and K(~t ).c Lp
Suppose now that e is ramified in K/Q and 5q / K. From the
proof of Lemma 27 we havce U(3) = U (1)e so that N(L ') 7 (3).
Therefoe FL / (ppl 2 r (pi )3. In the latter case
V (DL/K) = 3(e 1).
In order to conclude the proof it will suffice to exhibit cyclic
e-extensions L2 and Lj of K. with conductors p2 and p3. In fact since
b / hK it will suffice to find L2 c EPPYm with [L2:K ] = Le and
L3 EE 3v_ such that Lj e Ep29m and [L :K t] = e. (We are using
implicitly the fact that [k :K] = 2 hK[U:U ]]- proved just before
Proposition 10.)
Let G = Gal(E mu/K ). By Proposition 10,Gm (A/pJm)l( V
Since we are now assuming 5e 1 K, U = <-1>; it follows that
e / [U:U m]; hence B 4 10(Uym) I. Let (G); denote the C-group
component of the abelian group G. He have (Gm) ((A/pm .)
U i(1)/U (m). If mn = 2, (G2) o(L) and L2 exists as required
(uniquely so, in fact). If m = 3, (G3 IE 0() x o(e) by Lemma 27.
This group has e + 1 subgroups of index e. Therefore 1 of these
generate fields of type L'3. (The other one is L2.)
Theorem 29. Let K be either the rational field or an
imaginary quadratic field. Let E be an odd rational prime. If I k hK
and if K does not contain a primitive eth root of 1 than every cyclic
e-discriminant ideal of KC has a factorization as a product of prime
cyclic e-discriminant ideals.
56
Proof; This theorem is just the composite of Proposition 20,
Theorem 22, and Proposition 28.
Remark: As the proof of Proposition 28 shows there is much
duplication in the fields that generate prime cyclic e-discriminants
that divide a power of &.
BIBLIOGRAPHY
1. Artin, A. and Tate, J., Class Field Theory, Benjamin, New York,
1967.
2. Borevich, Z. I. and Shafarevich, I. R., Number Theory, Academic
Press, New York, 1966.
3. Fr~ihlich, Albrecht, "DiscriminarntS Of Algebraic Number Fields,"
Math. Zeitschnr., 74(1960), pp. 18-28.
4. Goldstein, Larry, "On Prime Discriminants," Nagycas Math.. J.,
45(1971), pp. 119-127.
5. Hecke, Erich, Vodlesungen uber die theories der algebraischen
Zahlen, Chelsea, New York, 1970.
6. Janusz, Gerald, Algebraic Numbecr Feld, Academic Press,
New York, 1973.
7. Lang, Serge, Algebraic Number Theory, Addison-!Jesley, Reading,
1970.
8. Long, Robert L., Algebraic Number Theory, M~arcel Dekker,
New York, 1977.
9. Samuel, Pierre, Algebraic Theory of Numbers, Hermann, Houghton
Mifflin, Boston, 1970.
10. Wang, Shianghaw, "On Grunwald's Theorem," Ann. of Mlath. (2),
51(1950), pp. 471-484.
BIOGR~APHICAL SKETCH
Danny Nevin Davis
Born August 11, 1946, in Rome, Georgia, to Nevin B. and Frances L,
Davis. Elementary and secondary education received in Rome and Floyd
County public schools; graduated from Pepperall H~igh School, 1964.
Attended Georgia Institute of Technology, 1964-68; graduated,
June 1968, with the B.S. degree in Physics. Employed at Lockheed
Aircraft Corporation in the Nuclear Analysis Department, 1968-70.
Attended graduate school in mathematics and worked as a teaching
assistant at the University of Florida, 1.971-76; obtained the H~.S.
degree in Mlathematics, 1973. Employed as instructor in mathematics at
Shorter College, Rome, Georgia, 1976 to the present.
I certify that I have read this study and that in my opinion
it conforms to acceptable standards of scholarly presentation and
is fully adequate, in scope and quality, as a dissertation for
the degree of Doctor of Philosophy.
Mark L. Teply
Chairman
Associate Professor of Mathematics
I certify that I have read this study and that in my opinion
it conforms to acceptable standards of scholarly presentation and
is fully adequate, in scope and quality, as a dissertation for the
degree of Doctor of Philosophy.
Robert L. Long
Associate Professor of M;Khematics
I certify that I have read this study and that in my opinion
it conforms to acceptable standards f scholarly presentation and
is fully adequate, in scope and quality, as a dissertation for the
degree of Doctor of Philosophy.
Beverly L. Brechner
Associate Professor of Mathematics
I certify that I have read this study and that in my opinion
it conforms to acceptable standards of scholarly presentation and
is fully adequate, in scope and quality, as a dissertation for the
degree of Doctor of Philosophy.
Jorge Martinez
Associate Professor of Mathematics
I certify that I have read this study and that in my opinion~
it conforms to acceptable standards of scholarly presentation and
is fully adequate, in scope and quality, as a, dissertation for
the degree of Doctor of Philosophy. '1
Elroy Boldue
Associate Pro essor of Education
This dissertation was submitted to the Graduate Faculty of the
Department of Mathematics in the College of Liberal Arts and Sciences
and to the Graduate Council, and was accepted as partial fulfillment
of the requirements for the degree of Doctor of Philosophy.
December 1978
Dean, Graduate School