J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
F
RANZ
L
EMMERMEYER
On 2-class field towers of imaginary quadratic number fields
Journal de Théorie des Nombres de Bordeaux, tome
6, n
o2 (1994),
p. 261-272
<http://www.numdam.org/item?id=JTNB_1994__6_2_261_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
On 2-class field
towersof
imaginary quadratic
number fields
par FRANZ LEMMERMEYER
ABSTRACT. For a number field k, let k1 denote its Hilbert 2-class field, and
put k2 =
(k1)1.
We will determine all imaginary quadratic number fields k such that G =Gal(k2/k)
is abelian or metacyclic, and we will give G interms of generators and relations.
1. Introduction
Let k = be an
imaginary
quadratic
number field withdiscrimi-nant d 0. It is well known that the structure of the 2-class group
Cl2(k)
depends
on the factorization of d intoprime
discriminants: these aredis-criminants which are
prime
powers, i.e.-4, ~8,
-q,(q -
3 mod4),
andp
(p -
1 mod4).
We say that d has t factors if d is theproduct
ofexac-tly t prime
discriminants. Here we willstudy
thequestion
how far thesefactorizations determine the 2-class field tower of k.
To this end let
k1
denote the Hilbert 2-class field ofk,
i.e. the maximal unramified normal extension of k whose Galois group is an abelian2-group.
Moreover,
letk2
=(kl ) 1 _
We willclassify
the discriminants d ofimagi-nary
quadratic
number fieldsaccording
to the structure of G =Gal(k2/k),
and we will determine all d such that k = has abelian or
meta-cyclic
Gal(k 2 Ik).
Partial classifications have been obtained in[1]
and[10];
whereasBenjamin
andSnyder
used Koch’s Satz 1 of[9],
we willemploy
his Satz 2 instead. The formulation of this theorem contains some errors; its correction reads
(see
[10];
a G-extension of k is an extensionK/k
withGal(K/k) -
G,
the notation of the groups is the one used in[2]):
THEOREM 1. Let k be a
quadratic
nurrtberfield;
there exists a G-extensionK/k
which isunramified
at thefinite places
and such thatK/Q
is normalif
andonly if
there is afactorization
d = disc k =dld2d3
intorelatively
prise
discriminants such that the Kroneckersymbols
(dil Pj)
in(*)
belowequal
+1(here
p; runsthrough
allprimes
dividing
dj):
If
(di/pj)
= 1for
all i~
j,
there also exists anunramified
extensionL/k
such that 32.18 and 64.144.The
following
twopropositions
willhelp
us indeciding
if thep-class
field tower of aquadratic
number field terminates at somestage;
the basicobservation is due to Iwasawa
[5]:
PROPOSITION 1. Let k be a number
field
and suppose that itsp-class
field
tower terminates with K. Then
EkINKIkEK,
where9R(G)
denotes the Schurmultiplier of
a group G andEk
denotes the unit groupof
thering
of
integers
SJk of
k.If,
forexample,
K/k
is an unramified and normal 2-extension of animaginary
quadratic
number field k with4,
thenK’ 0
K. Forshowing
that a certain class field tower terminates we use thefollowing
result(cf.
[10]):
PROPOSITION 2. Let k be a number
field
and letKl
be a normalunramified
extensionscontaining
thep-class
field
kl
of
k;
if
9R(Gal(Klk))
=1,
thenthe
p-class
field
towerof
k terminates with K.Proof. Suppose
otherwise;
then there exists an unramified extensionL/K,
which is central with
respect
toK/k.
i.e. which satisfiesGal (L/K)
CZ(Gal(L/k)).
Moreover,
Gal(L/K)
CGal(Llk)’
because K containskl.
Recalling
the mostelementary
properties
of Schurmultipliers
(cf.
[13]
or[7]),
thisgives
the contradiction9X(Gal(Klk)) =,4
1.In
particular,
thep-class
field tower of a field withcyclic p-class
group terminates withkl,
becausecyclic
groups have trivialmultiplier.
2. Abelian groups as
Gal(k2/k)
The fields d
0,
such that G =Gal(k2/k)
isabelian,
have been determined in[10];
for fields withG/G’ -
(2, 2’"’~)
this result has been obtainedindependently by Benjamin
andSnyder
[1].
THEOREM 2. Let
~(~)
be animaginary
quadratic
numberfield
with dis-crirrtinant d. ThenGal(k2/k)
is abelianif
andonly if
d has at most threefactors
andif
at most oneof
them ispositive.
Actually,
we have(p
and q,q’
etc. denoteprimes =-
1 3 mod4,
respectively):
The "only-If"-part
of the theorem can beproved
quite
simply:
supposethat G =
Gal(k2/k)
is abelian. Then the 2-class field tower of ktermi-nates with K =
kl.
Since k isimaginary
quadratic
and-3,
-4(recall
that we have assumed d to have threeprime
factors),
we findEk
={-1, +1},
so has order 2. Thisimplies
that the abelian2-groups possibly
occuring
asGal (k 2Ik)
must have Schurmultiplier
ofor-der 2. The
only
such2-groups
are thecyclic
groups and those oftype
(2, 2"’),
m > 1(see
[7]).
If
Gal (k2Ik)
iscyclic,
then so isCl2(k),
andby
genustheory
this isequivalent
to disc kbeing
theproduct
ofexactly
twoprime
discriminants. Thetheory
ofR6dei,
Reichardt and Scholz allows us to find all d such thatCl2 (k) - (2, 2"’~),
m >1;
we will outline the method used tocompute
in case d =
-4p ~ q
is thecorresponding
factorization of thesecond kind. Here we have
(-q/p)
= 1 andq =- 7
mod8;
the ideal classes of z =(2,1 +
and p =
(p,
-pq )
have order 2 inCl(k)
and differ from each other. Genustheory
shows that the ideal classof p
is no square inCl(k);
class fieldtheory implies
that thequadratic
unramified extensionK,
=&(B/~9)
belongs
to asubgroup
of index 2 inC12(k),
andsince p
splits
in thissubgroup
is notcyclic.
Therefore,
the fields K =k( A)
andbelong
to thecyclic subgroups
of index 2 inCl2(k),
i.e.NK/kCl2(K)
is acyclic
group of order 2m.Now we notice that the
quadratic
field has odd classnumber,
and that K =
k(A)
has unit groupEK
=>, where epq is
the fundamental unit of
(for
aproof,
see[8]).
The class numberformula now
gives
h2(K)
=2m,
and since the norm ofCl2 (K)
to k isterminates with
KI,
and theequality
(K1 :
k)
=2’‘~+1
=(kl :
k)
shows that in factK1
= k.Remark. The
knowledge
ofk2/k
allows us tocompute
the structure ofCl2(L)
for the subfields L ofkl/k;
we are also able tocompute
thesub-groups of those
Cl2 (L)
whichcapitulate
in anygiven
extensionL/K
con-tained in
k1/k.
For fieldswith kl
= k2,
however,
this ishardly interesting:
in this case,exactly
the ideal classes of order(L :
K)
capitulate
in any extensionL/K
such that k c K c LC kl
(this
fact wasalready
knownto Scholz and can be
proved immediately by
computing
the kernel of thecorresponding
transfermaps).
3.
Metacyclic
groups asGal(k2/k)
Our
primary
aim is to show:THEOREM 3. Let k = be an
imaginary
quadratic
numberfield
withdiscriminant d. Then
Gal(k 2Ik)
ismetacyclic if
andonly if
i)
G is abelian and k is oneof
thefields
described in Section2,
orii)
G/G’ - (2,2);
then G isdihedral,
semidihedral orquaterniontc,
and k is one
of
thefields
listed in[8]
and[10],
oriii)
d =-4pq,
where p and q areprimes -
5 mod 8.For groups G =
Gal(k2/k)
such thatGIG’ -- (2, 2’’n)
this has been obtainedby Benjamin
andSnyder
[1];
theproof
of theorem 3 that we willoffer
depends partly
on their results. Webegin
ourproof
with the wellknown observation that factor groups of
metacyclic
groups aremetacyclic.
If, therefore,
G =Gal (k 2 Ik)
ismetacyclic,
then so isGIG’ -
Cl2(k).
Genustheory
nowimplies
that d =dld2d3
forprime
discriminants IfCl2(k)
has asubgroup
oftype
(4,4),
then theorem 1 shows that G has afactor group - 32.18. Since 32.18 is not
metacyclic,
neither is G.So far we have seen: if G =
Gal (k 2Ik)
ismetacyclic,
then(2, 2"’~)
for some
dihedral,
semidihedral orquaternionic,
andthis case has
already
been settledby
Kisilevsky
[8]
(see
also(10)).
The case m > 1 has been studiedby
Benjamin
andSnyder
[1]:
they
have shown thatmetacyclic
groups G with(2, 2m)
occur as G =Gal(k2/k)
if andonly
if d =-4pq,
where p
and q areprimes -
5 mod 8.Using
theorem1,
we can prove this as follows: sinceCl(k)
is assumed to have acyclic subgroup
of order
4,
we must have d =dld2d3
forprime
discriminantsd~,
such thatd, - d2d3
is a factorization of the second kind(in
theterminology
of R6dei andReichardt):
this is to say that(d2d3/pi) = (di/P2) = (di /p3) =
+1,
where p; is theunique prime
dividing
6~.
Now we observe the
following:
ifdl
andd2
areprime
discriminants suchthat
(dl/p2) _
+1,
then we also have(d2/pi) =
+1except
when bothdi
and
d2
arenegative
or whendl
=-4, d2
= p - 5 mod 8(or
viceversa).
Returning
to the situation discussed above we see that there are three cases to consider:1. All
three dj
arenegative:
then G =Gal (k 2Ik)
isabelian,
as wehave seen in Section 2.
2.
Exactly
one dj
isnegative,
and d is not of the form-4pq,
wherep
and q
areprimes -
5 mod 8: then(d,/P2) = (di /p3) =
+1im-plies,
from what we have seen, that(d2/Pl) = (d3/pl)
= 1.By
theorem
1,
these relationsimply
the existence of an unramified16.09-extensions of k. Since 16.09 is not
metacyclic,
neither isG =
3. disc k =
-4pq,
p - q - 5 mod 8.
We will now use the
techniques
sketched in[10]
tocompute
the structureof
Gal(k2/k)
for the fields k in 3. To thisend,
suppose that p - q - 5 mod 8are
primes
such that(lro/q)
=1;
suppose moreover that the fundamentalunit epq of F =
Q(Vp-q)
has norm -1. Thetheory
ofR6dei,
Reichardt and Scholz nowgives
(p/q)4
=(q/P)4
andCl2(F) - Z/2"Z
for some n > 2. Theprime
ideal p
in F above p is notprincipal:
forif p =
(7r)
for some7r E
SJF,
then7r2 I P
would be a unit withpositive
norm; this wouldimply
that
±7r 2lp
and therefore~p
are squares in F. This contradiction showsthat the ideal class
[p]
has order2;
sinceCl2(F)
iscyclic,
it isgenerated by
an ideal class
[a]
such that p. In fact we may choose a as one ofthe two
prime
ideals above2Z,
because genustheory
shows that their ideal classes are not squares inCl(F).
Similarly,
theprime
ideals 2 =(2,
1 +-pq )
and p =(p,
in kare not
principal
(the
prime ideals p
in F and in k coincide inkF),
and from genustheory
we infer that the ideal class[p]
is a square inCl(k).
Kaplan
([6])
has shown that the condition(p/q)4
=(q/P)4
deduced aboveimplies
thatCL2(k) -
(2, 4).
Therefore,
2, b
>(we
will writez, b
> for the ratherclumsy
~2~,
[b]
>),
whereb2 I’V
p.Now let K =
pq );
thisquadratic
extension of k and F iscontained in the 2-class
field kl
of k. It is an easy exercise to show that the unit groupEK
= > and thath2 (K)
=2’~+2.
SinceNepq
=-1,
there existsexactly
one non-trivial ideal class whichcapitulates
inK/k;
infact,
= 2 >(here
xK/k denotes thesubgroup
of ideal classesin
Cl(k)
becoming principal
inK),
because2DK
=(1
+i).
From thiswe deduce that is not
principal,
and that the ideal class ofbDK
the class of
aOK generates
asubgroup
of order 2’~ inCl(K).
We claimthat the
subgroup
a, b
> ofCl(K)
has index 2 inCl2(K).
This indexis
certainly >
2 because~2.
Now suppose that a’ - b inK;
taking
the norm to kgives
1 Nb2
as a relation in which isclearly
acontradiction.
The
prime
ideal 2splits
in Let =2l2l’;
thenNK/k
a, ~
> =p >, because
NK~ka
and~2 N
p, and therefore thesubgroup
ofCl2(K)
generated by
the classes of a and b does not contain the ideal class of2~
because not contained in p >. Sincea, b
>has index 2 in
Cl2 (K),
we must haveCl2 (I~) _
>.Let p denote the
automorphism
ofK/Q
which fixesF;
then is aprime
ideal in F above2Z,
and without loss ofgenerality
we may assumethat it
equals
a.Similarly,
let u denote theautomorphism
fixing
I;
then2(1+u = 2.
We thereforeget
thefollowing
relations between ideal classes inK,
keeping
in mind that 1-f- ~ acts onCl(K),
whereasNK~k
mapsCl(K)
to
Cl(k):
the least relation holds because pQ fixes
Q(I)
which has class number one.Now we see that
which shows that the ideal class of 2l has order
2n+1
inCl(K),
and thatCl2 (K) _
2t, b
>. The relation = p reveals that(2, 2n+I).
.The fact that we know the structure of
C12(K)
as aGal(K/k)-module
allows us to
compute
Gal(Kl/k).
To thisend,
let(L/k, a)
denote the Artinsymbol
of a normal extensionL/K (for
properties
of Artinsymbols
see Hasse’s Zahlbericht[3]),
and let Q =(Kllk,
b)
be an extension of theautomorphism
(k1/k,
b);
thenU2
=b)
has order4,
andQ2
andT =
(KIIK,
2t)
generate
Gal(KIIK).
Therefore,
Gal(Kl/k)
= Q,T >,and we have the relations
Now it is easy to see that = 1
(there
isactually
a formula for the order offlR(G)
formetacyclic
groupsG;
cf.[7]);
thisimplies
thatK’
= k2
the 2-class field tower of k terminates with
k2.
We note that the group G =Gal(k 2Ik)
in theorem 4 is the group oftype
2 in[1,
prop.2],
witha = n - 1. We have
proved:
THEOREM 4. Let p - q - 5 mod 8 be
primes
such that(plq)
=1,
andsuppose that the
fundamental
unit êpqof
has norm -1. ThenExamples:
Now we will examine the case where
Nepq
= +1. In F =Q(,rp-q-),
theprime
ideals 21 and 22 above 2Z are notprincipal
because theequation x2
-pqy2
= ±8 has no solutions mod p.Moreover,
genustheory
shows that[21]
is no square inCl(F);
sinceCl2(F)
iscyclic,
the ideal class[zi]
generates
Cl2(F).
We will need the fact that there is no non-trivialcapitulation
inK/F,
where K = This will follow from theslightly
moregeneral
PROPOSITION 3. Let F be a real
quadratic
numberfield
and K =F(i).
There is non-trivialcapitulation
inKIF
if
andonly if
25JF
=Z2,
i. e. 2 isramified
inF/Q; in
this case, the ideal classof
2capitulates.
Proof.
Let a be anon-principal
ideal ini7F
whichcapitulates
in K. Thenthere is an a E
JK
such thataD K
=(a).
Let a denote the non-trivialautomorphism
ofKIF;
then = c is a unit in and sinceK/Q
isabelian,
we have16 1
= 1. NowE2
is a root ofunity
times a unit inD F
(this
comes from theidentity
62 =
keeping
in mind that the units of the twocomplex quadratic
fields are roots ofunity);
theonly
units in
5JF
with absolute valueequal
to 1 are tl.Therefore,
mustbe a root of
unity.
There are thefollowing possibilities:
a be
non-principal
inDF;
2.
a’-’
= 20131: then ia ~F,
andagain
a =(ia)
would beprincipal
inD F ;
i3.
a~"~
= +i: then(1 - i)a
~F;
since a and are ideals inDF,
it follows that Z =( 1-
.0 F
must be an ideal butthis
implies
z2
=2JDF;
4.
a~-1= -i:
then EF,
andagain
we must have22
=2.oF.
Now assume that22
=2DF
and that 2 is anon-principal
idealin DF;
then2DK =
(1
+i)
isprincipal
inK,
i.e. 2capitulates.
Let 2’~ be the order of
[21]
inCl2(F);
since there is nocapitulation
inK/F,
theprime
ideal 2Hn K above 21generates
an ideal class of order2n+1,
because =
2IDK.
In k = we haveC12(k)
_2, b
>, whereb2m-l rv 1’.
SinceNêpq
=+1,
the ideal classes[2]
and[1’]
bothcapitulate
inK/k,
and so bgenerates
asubgroup
of order2’~-1
inCl(K).
Thecomputa-tion of the 2-class number of K
yields
h2(K)
= =2m+n;
we claim that the ideal classes of 6 and 2t
generate
C12 (K) -
For sup-pose that bS rv thenb2,
=N K/kbs rv
=2t
shows that t is even and that s - 0 mod2’~-1
(recall
that 2 and bgenerate
differentsubgroups
ofCl2(l~)).
But the order of[b]
inC12(K)
equals
2m-1,
and therefore the relation bs N2(t
isnecessarily
trivial. This proves thatin-1 > is an abelian
subgroup
of index 2 in G =Gal(Kllk).
Let a E(K1lk, b)
be an extension of(kl/k,
b)
EGal(kl/k);
then we have G = p, a, T >,and we find the relations
a 2
=(KIIK,
b)
=r,a-Ipa
=(KIIK,
2to) =
p ;
to prove the last relation we use the fact that2{0"+1 =
z -(1
~-i) ~
1lies in the kernel of the Artin
symbol
ofKI I K.
Wefinally
remark that it follows from[6]
that we either have n = 1 or m =2,
and thatZ/2Z.
This is inagreement
with prop.1,
because -1 EEk B
The fact that
k2
=K’
follows from[1]:
they proved
thatGal(k2/k)
hasan abelian
subgroup
of order2,
whichnecessarily
must beGal (k 2 IK) -
We also observe that G is theirmetacyclic
group oftype
1. We have shown:THEOREM 5. Let p - q - 5 mod 8 be
primes
such that(plq)
=1,
andThe
only
case left to consider is d= -4pq, p -
q - 5 mod8,
(p/~) = 20131.
Then we haveNepq
= -1 andh2(F)
= 2 for F = we leave it as an exercise to the reader to prove thefollowing
theorem, using
the methodsdescribed in this paper. Theorem 6 has
already
been announced in[10],
and a
proof
using
Koch’s Satz 1 can be found in[1].
Examples:
THEOREM 6. Let p - q - 5 mod 8 be
primes
such thatObviously,
theorems4,
5 and 6 prove thepart
of theorem3. (iii~
that wasstill open.
4. Hasse’s rank formula
On p. 52 of his book
"Uber
die Klassenzahl abelscherZahlk6rper"
[4],
Hasse gave a formula forcomputing
the 2-rank of class groups in CM-fieldsKIKO
which reduces to the well knownambiguous
class number formulain case
.Ko
has odd class number. IfKo
has even classnumber,
however,
Hasse’s formula does not
always
give
the correct rank: thecyclic
quartic
field~
-10
+3 10
hascyclic
2-class group of order4,
and it iseasily
seen that this contradicts Hasse’s formula. Infact,
the fields listed informula stated in
[4]
readswhere
~y where t is the number of
(finite)
prime
idealsramifying
inK/Ko
r*
denotes the 2-rank ofCl(Ko)j,
j :
is the transfer of idealclasses,
so
= rankCl(Ko)j ICl(Ko)j n Cl(K)2
denotes the rank of the groupof ideal classes
becoming
squares inCl(K).
It follows from ourproof
below,
that we alsohave s*
= rankH/H
nCl(K)2,
where H =x =
I ker j is
the order of thecapitulation
kernel.To see that the formula is
incorrect,
we letKo
= as in theorem5,
i.e. we assume that p - q = 5 mod 8 areprimes
such that(p/q)
= 1 andNepq
= -1;
if we assume moreover that m =2,
then we find1,
toverify
thisclaim,
it is sufficient to show that Epq is norm from K. This can be done as follows: since epqhas norm
+1,
theprime ideal p
above p isprincipal,
i.e. p =
(7r).
Obviously, 77
=1r0"-1
is a unit inDK; if
were a square, so were p =~r°+1.
Now it iseasily
seen that we can choose 7r in sucha way that
1r0"-1 =
Epq. Thisimplies
and since p= a2
+b2
= NK~k (a
+bi),
we find that epq =NK~k ( a+bi )
isindeed a norm.
q
= t -1 + 1= 2,
becauseonly
the twoprime
ideals above 2ramify;
r*
=1
aH = = 1: we have 1 and
2l1-0" rv
1. Inparticular,
wehave s*
=0;
x =
0,
because there isonly
trivialcapitulation
inK/F.
Hasse’s formula
gives
r =3,
although
rankCl2(K)
= 2.The correct rank formula can be deduced as follows: let a denote
complex
conjugation
(hence
is the non-trivialautomorphism
ofK/Ko),
andput
C =Cl2 (K),
r = rankC,
Co
=Cl2 (Ko),
and H =Cl2 (K) 1-°.
Then the formula 2 = 1 + 7 + 1- u shows thatC2C1+0"
= =C03H
(the
fact that alsoC2CI-0"
= proves ourprevious
claimthat Sô
= rank, and now we find
Now,
(C : H)
equals
the number of ideal classes in C fixedby a,
becauseis a short exact sequence
(CG
is thesubgroup
of C fixedby
G = Q> ) .
Moreover,
where
2 G
denotes thesubgroup
of elements of order 2 of an abelian groupG. Let c E
Co
n H;
then a fixes c, andapplying
u to theequation
c =yields
c =c-1,
i.e. c E2~.
Hasse claimed that in factC-01
f1 H = but an ideal class of order 2 is notnecessarily
of the formdl-O"
for somed E C.
Actually,
in thecounterexamples
presented
above,
1- ~ annihilates the whole 2-class group of K.Using
theambiguous
class number formula(C :
H)
=21’h2(Ko),
whereq was defined
above,
we find thatwith
2X
=(2Coj : Co
rlH).
The main
problem
with this rank formula is the fact that Ausually
isquite
hard tocompute.
The trivial bounds 0 Arõ
yield
It should be noted that Hasse
applied
his formulaonly
to fieldsKo
with odd classnumber;
we havealready
observed that in this case it coincides with theambiguous
class numberformula,
because thenobviously
r*
=ACKNOWLEDGMENTS
I would like to thank E.
Benjamin
and C.Snyder
forsending
me theirpreprint,
and Prof. Kida forproviding UBasic;
theexamples
in thispa-per
(and
manymore)
have beencomputed
with version 8.65. Prof. M.Olivier has
kindly
verified that the 2-class group of thequartic
cyclic
fieldQfB/-10+3VTo)
iscyclic
of order 4using
PARI,
and Prof. J. Martinet hassuggested
thepossibility
ofbicyclic counterexamples
to Hasse’s rank formula.REFERENCES
[1]
E. Benjamin, C. Snyder, Number fields with 2-class groups isomorphic to (2, 2m),Austr. J. Math.
[2]
M. Hall, J. K. Senior, The groups of order 2n(n ~ 6);, Macmillan, New York (1964).[3]
H. Hasse, Zahlbericht, Physica Verlag, Würzburg, 1965.[4]
H. Hasse, Über die Klassenzahl abelscher Zahlkörper, Springer Verlag, Heidelberg.[5]
K. Iwasawa, A note on the group of units of an algebraic number field, . Math. puresappl. 35 (1956), 189-192.
[6]
P. Kaplan, Sur le 2-groupe des classes d’idéaux des corps quadratiques, J. reine angew. Math. 283/284 (1974), 313-363.[7]
G. Karpilovsky, The Schur multiplier, London Math. Soc. monographs (1987), Ox-ford.[8]
H. Kisilevsky, Number fields with class number congruent to 4 mod 8 and Hilbert’s theorem 94, J. Number Theory 8 (1976), 271-279.[9]
H. Koch, Über den 2-Klassenkörperturm eines quadratischen Zahlkörpers, J. reine angew. Math. 214/215 (1963), 201-206.[10]
F. Lemmermeyer, Die Konstruktion von Klassenkörpern, Diss. Univ. Heidelberg(1994).
[11]
L. Rédei, H. Reichardt, Die Anzahl der durch 4 teilbaren Invarianten derKlas-sengruppe eines beliebigen quadratischen Zahlkörpers, J. reine angew. Math. 170
(1933), 69-74.
[12]
A. Scholz, Über die Lösbarkeit der Gleichung t2 - du2 = -4, Math. Z. 39 (1934), 95-111.[13]
A. Scholz, Abelsche Durchkreuzung, Monatsh. Math. Phys. 48 (1939), 340-352.Franz Lemmermeyer Institut fur Mathematik
Universitat Heidelberg
Im Neuenheimer Feld 288 69120 HEIDELBERG ALLEMAGNE