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R´edei fields and dyadic extensions
David Brink
April 2010
Abstract. For an arbitrary non-square discriminant D, the R´edei field Γ0(D) is
introduced as an extension of K=Q(√D) analogous to the genus field and con-
nected with the R´edei-Reichardt Theorem. It is shown how to compute R´edei fields,
and this is used to find socles of dyadic extensions of Kfor negative D. Finally, a
theorem and two conjectures are presented relating the fields Q(√−p) and Q(√−2p)
for an odd prime p.
1. Introduction
Let Dbe a non-square discriminant, i.e. a non-square rational integer congruent to 0 or
1 modulo 4. The ring class field N=N(D) is a normal extension of the quadratic field
K=Q(√D) with Galois group Gal(N/K) canonically isomorphic to the form class group
C=C(D) via the Artin symbol. Moreover, N/Qis normal with Gal(N/Q)∼
=CoC2
generalized dihedral. The genus field Γ = Γ(D) is the compositum of all quadratic extensions
of Kinside N. Since Gal(Γ/K) is an elementary Abelian 2-group isomorphic to C/C2, its
rank equals C’s 2-rank, i.e. the number of cyclic factors of Cwhose order is divisible by 2.
The determination of this 2-rank goes back to Gauss. The famous R´edei-Reichardt Theorem
determines C’s 4-rank. In light of this, we propose the following terminology:
Definition 1. The R´edei field Γ0= Γ0(D) is the compositum of the quadratic subextensions
of all C4-extensions (i.e. cyclic extensions of degree four) of Kinside N.
Clearly, the R´edei field is contained in the genus field, and the rank of the elementary Abelian
2-group Gal(Γ0/K) equals C’s 4-rank. When Dis a fundamental discriminant, i.e. equal
to the field discriminant of K, the form class group Cis isomorphic to K’s (narrow) ideal
class group, and the ring class field Ncoincides with K’s (narrow) Hilbert class field. The
computation of Γ0is well known in this case, but has to the author’s best knowledge never
been carried out for non-fundamental D, cf. [9]. Theorem 1 in section 2 below fills this gap.
This investigation was motivated by [2] where the problem occurred of determining the R´edei
field over K=Q(√−7) for D=−1792.
1
Let Z2= lim
←− C2nbe the pro-cyclic group of dyadic (2-adic) integers, and let D2=
lim
←− D2n∼
=Z2oC2be the corresponding pro-dihedral group. If Dis negative, then there
exists a unique field MKsuch that MK/K is a Z2-extension and MK/Qis a D2-extension [5,
p. 360].
Definition 2. The socle Soc(K) of an imaginary quadratic number field Kis the unique
quadratic subextension of MK/K, cf. [5, Definition 6.1].
In section 3, we illustrate the theory of R´edei fields by determining Soc(K) in some new cases.
Finally, in section 4, we prove a theorem and present two conjectures relating the socles and
dyadic extensions of the fields K=Q(√−p) and K0=Q(√−2p) for an odd prime p.
2. R´edei fields
For an arbitrary non-square discriminant D, we write
|D|= 2e0pe1
1···per
r(1)
with distinct odd primes pi. Thus ris the number of odd primes dividing D. We start
by reviewing some basic concepts from the theory of binary quadratic forms. The reader is
referred to [4] for a thorough treatment of this subject.
All forms (a, b, c) := ax2+bxy +cy2considered in the following are tacitly assumed prim-
itive, of discriminant D, positive definite in case D < 0, and with rational integer coefficients
a,b,c. Two such forms are called properly equivalent if one can be transformed into the
other by a linear change of variables with determinant 1. The class of a form funder this
equivalence relation is denoted [f]. Dirichlet composition turns the set of all such classes into
a (finite) Abelian group, the form class group C=C(D). The class number h=h(D) is
the order of this group. The principal form is defined as (1,0,−D/4) or (1,1,(1 −D)/4)
according to whether Dis even or odd. The class containing the principal form is called the
principal class; it is the identity element in C.
A form (a, b, c) is called ambic (or ambiguous) if adivides b, and a class of forms is called
ambic if it contains an ambic form. As shown by Gauss, the set of ambic classes coincides
with the subgroup C[2] = {a∈C|a2= 1}of self-inverse classes, and each ambic class
contains exactly two ambic forms (a, b, c) with bequalling 0 or a. We adopt the terminology
of [10] and call such forms ancipital. Table 1 defines, depending on D’s congruence class, a
number µand a list of ancipital forms f1, . . . , fµthat we call the assigned ancipital forms.
Each ancipital form is the product, by Dirichlet composition, of a unique subset of the fi(the
empty product is taken to be the principal form). Therefore, the number of ancipital forms
is 2µ, the number of ambic classes is 2µ−1, and C’s 2-rank is µ−1. It also follows that the
classes [f1],...,[fµ] generate C[2] and satisfy one non-trivial relation.
2
The group of characters χ:C→C×is denoted C∗. The elements of the subgroup
C∗[2] = {χ∈C∗|χ2= 1}are called quadratic characters. A prime discriminant is a number
of the form –4, 8, –8 or p∗= (−1)(p−1)/2pwhere pis any odd prime. Table 1 defines a list
of prime discriminants q1, . . . , qµthat we call the assigned prime discriminants. A (possibly
empty) product dof distinct assigned prime discriminants qj, where the product of –4 and
8 is taken to be –8, is called a fundamental discriminant divisor of D. Any such ddefines a
quadratic character χ∈C∗[2] by χ([f]) := (d/m) where the right-hand side is a Kronecker
symbol, and mis any integer prime to drepresented by f. Each quadratic character is defined
exactly twice in this way. The number of fundamental discriminant divisors is clearly 2µ, and
hence the number of quadratic characters is 2µ−1. The quadratic characters χjdefined by
the assigned prime discriminants qjare called the assigned (or generic)characters. It follows
that they generate C∗[2] and satisfy one non-trivial relation.
Table 1: Definition of the number µ, the assigned ancipital forms f1, . . . , fµand the assigned prime
discriminants q1, . . . , qµaccording to D’s congruence class. The indices iand jrun from 1 to r.
D µ f1, . . . , fµq1, . . . , qµ
1(4) r(pei
i, pei
i,(pei
i−D/pei
i)/4) p∗
j
4(16) r(pei
i,0,−D/4pei
i)p∗
j
12(16) r+ 1 (pei
i,0,−D/4pei
i),(2,2,(4 −D)/8) p∗
j,−4
8(32) r+ 1 (pei
i,0,−D/4pei
i),(2,0,−D/8) p∗
j,8
24(32) r+ 1 (pei
i,0,−D/4pei
i),(2,0,−D/8) p∗
j,−8
16(32) r+ 1 (pei
i,0,−D/4pei
i),(4,0,−D/16) p∗
j,−4
0(32) r+ 2 (pei
i,0,−D/4pei
i),(2e0−2,0,−D/2e0),(4,4,1−D/16) p∗
j,−4,8
The R´edei matrix is the µ-by-µmatrix R= (χj([fi]))ij .Thus the ijth entry is the Kro-
necker symbol (qj/mi) where qjis the jth assigned prime discriminant, and miis an integer
prime to qjrepresented by the ith assigned ancipital form fi. Note that (a, c) = 1 for all
fi= (a, b, c) such that at least one of a=fi(1,0) and c=fi(0,1) can serve as mi.
An ancipital form fis said to be of the second kind if the product of the rows of the
R´edei matrix Rcorresponding to the factors fiof fis the 1-row, i.e. if χj([f]) = 1 for all
j= 1, . . . , µ. It follows from the fact that the χjgenerate C∗[2] that
C2={a∈C|χj(a) = 1 for all assigned characters χj}.(2)
Hence an ancipital form fis of the second kind if and only if its class [f] is a square. Since
every ambic class contains exactly two ancipital forms, the number of ancipital forms of the
second kind is twice the order of C[2] ∩C2.
A fundamental discriminant divisor dis said to be of the second kind if the product of the
columns of Rcorresponding to the factors qjof dis the 1-column, i.e. if (d/mi) = 1 for all
3
i= 1, . . . , µ where miis an integer prime to drepresented by fi. It follows from the fact that
the [fi] generate C[2] that
C∗2={χ∈C∗|χ([fi]) = 1 for all assigned ancipital forms fi}.(3)
Hence a fundamental discriminant divisor dis of the second kind if and only if the quadratic
character χdefined by dis a square. Since every quadratic character χ∈C∗[2] is defined
by exactly two fundamental discriminant divisors, the number of fundamental discriminant
divisors of the second kind is twice the order of C∗[2] ∩C∗2.
In analogy with µ, we define νas the number satisfying that the 4-rank of Cis ν−1. It
follows directly from this definition and the isomorphism C∼
=C∗that
2ν−1=|C[2] ∩C2|=|C∗[2] ∩C∗2|.(4)
We have thus proved the R´edei-Reichardt Theorema[11, 12, 13] for an arbitrary non-square
discriminant Dusing only the elementary theory of quadratic forms:
There are 2νancipital forms of the second kind. There are 2νfundamental discriminant
divisors of the second kind. The matrix obtained from the R´edei matrix by replacing each
entry (−1)nby nhas rank µ−νover F2.
We continue considering a non-square discriminant Dand write D=dKf2where dKis the
field discriminant of K=Q(√D). The reader is referred to [3, 4] regarding definition and
basic properties of ring class fields. The ring class field N=N(D) is the class field over K
corresponding by class field theory to the ring modulo f(in the narrow sense if D > 0). It is
the unique Galois extension of Qwith the property
psplits in N⇔pis represented by the principal form of discriminant D(5)
for every prime p-D. Being canonically isomorphic to the (narrow) ideal class group C+(O)
of the order O⊆OKof conductor f, the form class group C=C(D) is canonically isomorphic
to Gal(N/K) via the Artin symbol. Moreover, N/Qis normal with
Gal(N/Q)∼
=CoC2(6)
generalized dihedral (i.e. the non-trivial element of C2acts on Cby inversion). For an
intermediate field Lin N/K corresponding, via Galois theory and Artin symbol, to a subgroup
Aof C, one has the “reciprocity law”
psplits in L⇔pis represented by some form (class) in A(7)
for every prime p-D. For example, since Kcorresponds to C, a prime p-Dsplits in Kif
and only if pis represented by some primitive form of discriminant D.
aNamed after L´aszl´o R´edei (1900–1980) and Hans Reichardt (1908–1991).
4
The genus field Γ = Γ(D) is the maximal elementary Abelian 2-extension of K(or Q)
inside N. It corresponds, via Galois theory and Artin symbol, to the principal genus, i.e. the
subgroup C2of squares in C. Hence
Gal(Γ/K)∼
=C/C2∼
=
µ−1 times
z }| {
C2× · ·· × C2.(8)
Being a Galois extension of Q, Γ is determined by the set of primes that split in it. Consider
a prime p-Dthat splits in K. Then pis represented by a unique form class [f]∈C. Now
psplits in Γ ⇔[f]∈C2
⇔χj([f]) = 1 for all assigned characters χj
⇔(qj/p) = 1 for all assigned prime discriminants qj
by (7), (2) and the definition of the assigned characters χj. The well-known fact follows that
Γ is obtained by adjoining to Qthe square roots of all assigned prime discriminants qj.
As already noted, the R´edei field Γ0is contained in the genus field Γ. Let A0be the
subgroup of Ccorresponding, via Galois theory and Artin symbol, to Γ0. Thus A0contains
C2, and the R´edei-Reichardt Theorem gives
Gal(Γ0/K)∼
=C/A0∼
=
ν−1 times
z }| {
C2× · ·· × C2.(9)
It follows from Definition 1 that A0is the intersection of all those subgroups H⊂Cof
index two which themselves contain a subgroup H0⊂Hwith C/H0∼
=C4. The condition
[C:H] = 2 means that His the kernel of a quadratic character χ∈C∗[2], and the
conditions H0⊂Hand C/H0∼
=C4mean that χis a square, i.e. χ∈C∗2. Therefore, we
have the characterization
A0={a∈C|χ(a) = 1 for all χ∈C∗[2] ∩C∗2}.(10)
Incidentally, it follows that A0is the subgroup of Cgenerated by C2and C[2]. For a prime
p-Dthat splits in Kand is represented by the form class [f]∈C, we thus get
psplits in Γ0⇔[f]∈A0
⇔χ([f]) = 1 for all χ∈C∗[2] ∩C∗2
⇔(d/p) = 1 for all dof the second kind
by (7), (10) and the fact that a quadratic character χ∈C∗[2] defined by a fundamental
discriminant divisor dis a square if and only if dis of the second kind, cf. (3). We have
proved:
Theorem 1. For an arbitrary non-square discriminant D, the R´edei field Γ0= Γ0(D)is
5
obtained by adjoining to Qthe square roots of all fundamental discriminant divisors dof D
of the second kind.
Remarks. (a) For a fundamental discriminant D, R´edei and Reichardt [13] defines a fac-
torization D=d1·d2with discriminants d1and d2to be of the second kind (D-Zerf¨allung
zweiter Art) if (d1/p) = 1 for all primes p|d2, and (d2/p) = 1 for all primes p|d1. It is easily
seen that this is the case exactly when d1and d2are fundamental discriminant divisors of the
second kind in the sense defined here. Also, the above definition of R´edei matrices coincides
(essentially) with R´edei’s original definition [12] for fundamental D.
(b) If Dis fundamental with prime divisors p1, . . . , pµ, there is a simple, alternative de-
scription of the ijth entry of the R´edei matrix as the Hilbert symbol pi,D
pj[1, problem 25,
p. 250].
(c) The following application is due to R´edei and Reichardt [13]: Let nbe a non-square
positive integer not divisible by 4 or by any prime p≡3 (mod 4). Assume that the class group
C(4n) has 4-rank zero. Then f= (−1,0, n) is the second ancipital form in the principal class
other than the principal form (1,0,−n) itself, since [f] is both self-inverse and, as follows
from (2) since (qj/−1) = 1 for all j, a square. Hence the principal form represents −1, i.e.
the non-Pellian equation x2−ny2=−1is solvable. See also [9] for a thorough discussion of
this interesting problem.
(d) Returning to the case D=−1792 = −28·7 mentioned in the introduction, we have
K=Q(√−7) and find ourselves in the last case of Table 1 with assigned ancipital forms
(7,0,64), (64,0,7), (4,4,113) and assigned prime discriminants –7, –4, 8. The genus field
is thus Γ = K(√−4,√8). Since the class number is h= 8, it follows that the class group
is C∼
=C4×C2. Hence the R´edei field Γ0is the unique field among K(√−4), K(√8) and
K(√−8) which is embeddable in a C4-extension of Kinside the ring class field N, but a priori
it is not clear which. The R´edei matrix is
(−7/64) (−4/7) (8/7)
(−7/64) (−4/7) (8/7)
(−7/4) (−4/113) (8/113)
=
1−1 1
1−1 1
1 1 1
.
Therefore, the fundamental discriminant divisors of Dof the second kind are 1, –7, 8 and
−7·8, and we conclude Γ0=K(√8).
3. Dyadic extensions
Let Kbe an imaginary quadratic number field, and recall the definition of MKand Soc(K)
from the introduction. It is easy to show that the fields Q(√−1) and Q(√−2) both have
socle Q(√−1,√2), but even in the simplest cases when Kis of type Q(√−p) or Q(√−2p)
with an odd prime p, the determination of Soc(K) leads to substantial problems. We apply
Theorem 1 to show two results of which the first extends [5, Propositions 1 and 3].
6
Theorem 2. Let K=Q(√−pq)with distinct odd primes pand q. Then Soc(K) = K(√a)
with
a=
pif p≡3 (mod 8), q ≡7 (mod 8) and (p/q) = +1;
pif p≡5 (mod 8), q ≡7 (mod 8) and (p/q) = +1;
−1if p≡1 (mod 8), q ≡5 (mod 8) and (p/q) = −1;
−pif p≡3 (mod 8), q ≡7 (mod 8) and (p/q) = −1.
Proof. Since MK/Qis pro-dihedral, Soc(K) is of the form K(√a) with a rational integer a.
Only primes above 2 can be ramified in a Z2-extension by a result of Iwasawa [8]. Hence a
can be taken to be a divisor of 2p. It is a theorem of Bruckner [3, Satz 8] that any Abelian
extension of Kwhich is (generalized) dihedral over Qis contained in some ring class field.
It then follows from Iwasawa’s ramification restraint that MKis contained in the union of
the ring class fields N(−22npq), n≥1. Consequently, Soc(K) is contained in the R´edei field
Γ0(−22npq) for nsufficiently large. Now note that for all n≥4 we are in the last case of
Table 1, and the R´edei matrix and the R´edei field remain constant. Hence n= 4 is “sufficiently
large”, i.e. Soc(K) is contained in Γ0(−256pq). We therefore put D=−256pq and find the
assigned ancipital forms (p, 0,64q), (q, 0,64p), (64,0, pq), (4,4,1 + 16pq) and assigned prime
discriminants p∗,q∗,−4, 8. The R´edei matrix thus becomes
(p∗/64q) (q∗/p) (−4/p) (8/p)
(p∗/q) (q∗/64p) (−4/q) (8/q)
(p∗/64) (q∗/64) (−4/pq) (8/pq)
(p∗/4) (q∗/4) (−4/1 + 16pq) (8/1 + 16pq)
=
−1 1 −1−1
−1 1 −1 1
1 1 1 −1
1 1 1 1
,
1 1 1 −1
1 1 −1 1
1 1 −1−1
1 1 1 1
,
−1−1 1 1
−1−1 1 −1
1 1 1 −1
1 1 1 1
,
1−1−1−1
1−1−1 1
111−1
1111
,
respectively. In all four cases, the 4-rank is ν−1 = 1, and consequently Soc(K) = Γ0. The
determination of Γ0follows from Theorem 1. In the first case, for example, the fundamental
discriminant divisors of the second kind are 1, p∗·(−4) = 4p,q∗=−qand p∗·q∗·(−4) = −4pq,
and therefore Γ0=K(√p).
The proof of the following theorem is completely analogous.
Theorem 3. Let K=Q(√−2pq)with distinct odd primes pand q. Then Soc(K) = K(√a)
7
with
a=
−2if p≡3 (mod 8), q ≡3 (mod 8) and (p/q) = +1;
pif p≡3 (mod 8), q ≡5 (mod 8) and (p/q) = +1;
2pif p≡3 (mod 8), q ≡7 (mod 8) and (p/q) = +1;
−1if p≡5 (mod 8), q ≡5 (mod 8) and (p/q) = +1;
2pif p≡5 (mod 8), q ≡7 (mod 8) and (p/q) = +1;
−2if p≡1 (mod 8), q ≡3 (mod 8) and (p/q) = −1;
−1if p≡1 (mod 8), q ≡5 (mod 8) and (p/q) = −1;
2if p≡1 (mod 8), q ≡7 (mod 8) and (p/q) = −1;
2pif p≡3 (mod 8), q ≡5 (mod 8) and (p/q) = −1;
−2pif p≡3 (mod 8), q ≡7 (mod 8) and (p/q) = −1;
−1if p≡5 (mod 8), q ≡5 (mod 8) and (p/q) = −1;
−pif p≡5 (mod 8), q ≡7 (mod 8) and (p/q) = −1.
4. Two conjectures
Let pbe an odd prime, and let hand h0be the class numbers of K=Q(√−p) and
K0=Q(√−2p), respectively. The socles of Kand K0were determined in [5, Propositions
2, 4, 5] in all cases except when p≡1 (mod 16) and 8|h, h0. Note that the two conditions
8|hand 8|h0are equivalent when p≡1 (mod 16) by an old, seemingly forgotten result of
Glaisher [6]. Write Soc(K) = K(√a) and Soc(K0) = K0(√a0) with a, a0∈ {−1,2,−2}(cf.
the proof of Theorem 2). This gives nine a priori possibilities for the ordered pair (a, a0).
Conjecture 1. Let the notation be as above with a prime p≡1 (mod 16), and denote by v2
the dyadic valuation.
(i) If v2(h)=3and v2(h0) = 3, then (a, a0) = (2,2) or (−2,−1).
(ii) If v2(h)=3and v2(h0) = 4, then (a, a0) = (2,−1),(2,2) or (−2,−2).
(iii) If v2(h) = 3 and v2(h0)≥5, then (a, a0) = (2,−1),(2,2),(2,−2) or (−2,−2).
(iv) If v2(h)=4and v2(h0) = 3, then (a, a0)=(−1,−1),(2,2) or (−2,2).
(v) If v2(h)≥5and v2(h0)=3, then (a, a0)=(−1,−1),(−1,2),(2,2) or (−2,2).
(vi) If v2(h)=4and v2(h0) = 4, then (a, a0)=(−1,−2),(2,−1),(2,2),(−2,−1) or (−2,2).
(vii) If v2(h)=4and v2(h0)≥5, then (a, a0)=(−1,−2),(2,−1),(2,2),(2,−2),(−2,−1),
(−2,2) or (−2,−2).
(viii) If v2(h)≥5and v2(h0)=4, then (a, a0)=(−1,−1),(−1,2),(−1,−2),(2,−1),(2,2),
(−2,−1) or (−2,2)
The author has verified this conjecture for all pup to 14,000,000. These computations also
show that all of the pairs (a, a0) stated above actually occur (with certain conjectural densi-
ties), and that all nine pairs occur when hand h0are both divisible by 32.
8
As in the case of Q(√−1) and Q(√−2), it may happen that Kand K0have common socle
Soc(K) = Soc(K0), i.e. that (a, a0) = (2,2). We then have the following:
Theorem 4. Let pbe an odd prime. Suppose K=Q(√−p)and K0=Q(√−2p)have
common socle. Then the compositum of the Z2-extensions MKand MK0contains a square
root of either √2+2 or √2−2.
Proof. As remarked above, the common socle is K(√2) = K0(√2). Let Land L0be the
unique subextensions of degree four of MK/K and MK0/K0, respectively. Both Land L0are
D4-extensions of Q. The group G= Gal(LL0/Q), therefore, is a fibre product of two copies of
D4(the dihedral group of order eight). It is the ninth group of order 16 in Hall and Senior’s
classification [7], and the third in the Magma library. One has G/G0∼
=C4×C2such that
the maximal Abelian subfield Aof LL0has Gal(A/Q)∼
=C4×C2. Consequently, LL0has
two subfields A0and A00 that are C4-extensions of Q. Both A0and A00 contain Q(√2) since
neither Knor K0(being imaginary quadratic) is C4-embeddable. The situation is summed
up in Figure 1.
9
Figure 1: Various subfields of the compositum of MKand MK0.
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r
r r r
r
r r r
r
r r
r r
Q
KQ(√2) K0
K(√2)
L0A L
A0A00
LL0
MK0MK
The inertia field T=Tp(A/Q) is the maximal subfield of Ain which pis unramified. Since p
is odd, it does not divide [A:Q] = 8 and is hence tamely ramified in A. It follows that the
inertia group Ip(A/Q) = Gal(A/T ) is cyclic. Since pramifies in both Kand K0, this leaves
only two possibilities: T=A0or T=A00. In either case, T/Qis a C4-extension unramified
outside {2,∞}. It follows from the Kronecker-Weber Theorem that there are only two such
extensions, namely the two subfields Qp√2+2and Qp√2−2of the 16th cyclotomic
field. This finishes the proof.
The determination of the right square root in Theorem 4 seems to be non-trivial. The as-
sumption of the theorem is satisfied in three cases: When p≡7 (mod 8) [5, Proposition 2],
when p≡1 (mod 16) and 8 -h, h0[5, Proposition 5], and sometimes when p≡1 (mod 16)
and 8|h, h0(cf. Conjecture 1; in fact, (2,2) is the only ubiquitous pair). We have the following
conjectural result regarding the first two cases:
Conjecture 2. Let K=Q(√−p)and K0=Q(√−2p)with an odd prime p.
(i) If p≡7 (mod 16), then MKMK0contains a square root of √2−2.
(ii) If p≡15 (mod 16), then MKMK0contains a square root of √2+2.
(iii) If p≡1 (mod 16) and 8-h, h0, then MKMK0contains a square root √2+2.
This conjecture has been verified by the author for all pup to 8,000,000. Both square roots
can occur when p≡1 (mod 16) and 8|h, h0(the smallest examples being p= 337 for the
10
square root of √2−2, and p= 593 for the square root of √2 + 2), and there seems to be no
simple criterion deciding which in this case.
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[13] L. R´edei and H. Reichardt, Die Anzahl der durch 4 teilbaren Invarianten der Klassen-
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