The prime discriminant factorization of discriminants of algebraic number fields

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The prime discriminant factorization of discriminants of algebraic number fields
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
bibliography ( marcgt )
non-fiction ( marcgt )


Thesis--University of Florida.
Bibliography: leaf 57.
General Note:
General Note:
Statement of Responsibility:
by Danny Nevin Davis.

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Dedicated to

my parents,

Nevin B. and Frances L. Davis


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.


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

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

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


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







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


K, L

A, B

U, V

K .

A .

U .





a b





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



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.


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



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.
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


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 .


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 )


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


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.


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

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,


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

Proof: The reader is referred to Artin-Tate [1], corollary

to Theorem 14 p. 135 and corollary to Theorem 18 p. 139.


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
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-


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

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-
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
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
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


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.


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)
vell known isomorphism: _for A, B subgroups of a given group.


(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


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
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'
Now notice that the natural map from KW onto KUhas kernel
v, K
Therefore Gal(K /R ) a
K'W K'W K'W U nK'Wm
v, V. Um v

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.

Proof: We will use the isomorphism -and the identity

AnBC = (AnB)C (where C < A) freely in this proof.
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

.The natural map of
K' W mSnUK (K' nUK ~p"m m UWp Om UTWrJmu
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
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).


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
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( -).


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.
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 ).
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
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
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

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

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

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

SO = {p: P 2, V (a) = 11
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
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).
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
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
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 ).
Let bEi SO. Then u0 u up mad 6 which gives
2 2b 2 2b
U0 = Y ubu mod b since H u E y mod b for some yEiA such that
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

2e6 2e6 2eb2
ar was chosen such that ab+6 -a+6 mad (A/6 b). (and here b k a).
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'
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


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

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.
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
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).
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
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~

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.


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


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.


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


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.


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 &.


1. Artin, A. and Tate, J., Class Field Theory, Benjamin, New York,

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,

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.


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
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