On Annihilators of the Class Group of an Imaginary Compositum of

13 Pages·2009·0.19 MB·English

On Annihilators of the Class Group of an
Imaginary Compositum of Quadratic Fields
Radan Kuˇcera
Kotla´ˇrska´ 2
Faculty of Science, Masaryk University
611 37 Brno, Czech Republic
E-mail: [email protected]
Abstract
Let k be a tamely ramified imaginary compositum of quadratic
fields. The aim of the paper is to construct new explicit annihilators
oftheclassgroupofk notbelongingtotheStickelbergerideal.These
new annihilators are obtained as quotients of elements of the Stick-
elberger ideal, the usual source of annihilators, by suitable powers
of 2.
1 Introduction
Let k be an imaginary compositum of quadratic fields and let −1 not be a
square in the genus field K of k in the narrow sense. This paper resumes the
study of the Stickelberger ideal of k that started in [1] in a similar way as
[2] does for circular units of a real compositum of quadratic fields. The aim
of the paper is to prove a divisibility relation for the relative class number
h− of k and, in the case that 2 does not ramify in k/Q, to construct new
explicitannihilatorsoftheclassgroupofk notbelongingtotheStickelberger
ideal. These new annihilators are obtained as quotients of elements of the
Stickelberger ideal, the usual source of annihilators, by suitable powers of 2.
The main result of this paper can be summarized as follows:
Theorem 1.1. Let k be an imaginary compositum of quadratic fields such
that 2 does not ramify in k/Q. Let X(cid:48) be the set of all odd Dirichlet char-
acters corresponding to k. Let S be the Stickelberger ideal of k defined by
k
Sinnott in [4] and let T ⊆ Z[Gal(k/Q)] be the subgroup defined below by
k
2010 Mathematics Subject Classification: Primary 11R20; Secondary 11R29.
Key words and phrases: Stickelberger ideal, imaginary compositum of quadratic
fields, annihilators of the class group.
1
2 R. Kuˇcera
means of explicit generators. Then S +2T annihilates the ideal class group
k k
Cl of k and the index
k
(cid:89) [K : (k ∩K )]
χ χ
[(S +2T ) : S ] = ,
k k k
2
χ∈X(cid:48)
Kχ(cid:54)=k∩Kχ
where K means the genus field of the quadratic field corresponding to χ.
χ
Hence this approach give explicit new annihilators of Cl if and only
k
if there is an odd Dirichlet character χ corresponding to k such that the
degree [K : (k ∩K )] ≥ 4.
χ χ
2 Definitions and basic results
Recall that k is a compositum of quadratic fields such that −1 is not a
square in the genus field K of k in the narrow sense. This condition can
be written equivalently as follows: either 2 does not ramify in k and k =
√ √
Q( d ,..., d ), where d , ..., d with s ≥ 1 are square-free integers all
1 s 1 s
congruent to 1 modulo 4, or 2 ramifies in k and there is uniquely determined
√ √
x ∈ {2,−2} such that k = Q( d ,..., d ), where d , ..., d with s ≥ 1
1 s 1 s
are square-free integers such that d ≡ 1 (mod4) or d ≡ x (mod8) for each
i i
i ∈ {1,...,s}. In the former case, let
J = {p ∈ Z; p ≡ 1 (mod4), |p| is a prime ramifying in k},
and, in the latter case, let
J = {x}∪{p ∈ Z; p ≡ 1 (mod4), |p| is a prime ramifying in k}.
We assume that k is imaginary, i.e. at least one of d ’s is negative. For any
i
p ∈ J, let
(cid:40)
|p| if p is odd,
n =
{p}
8 if p is even.
For any S ⊆ J let (by convention, an empty product is 1)
(cid:89) √
n = n , ζ = e2πi/nS, QS = Q(ζ ), K = Q( p; p ∈ S),
S {p} S S S
p∈S
and k = k ∩ K . It is easy to see that K = K and that n is the
S S J J
conductor of k. For any p ∈ J let σ be the non-trivial automorphism in
p
Gal(K /K ). Then G = Gal(K /Q) can be considered as a (multiplica-
J J\{p} J
tive) vector space over F with F -basis {σ ; p ∈ J}.
2 2 p
For any positive integer n let
(cid:88) t
θ = τ−1
n n t
0<t≤n
(t,n)=1
On Annihilators of the Class Group 3
be the usual Stickelberger element in the rational group ring over the Galois
group of the n-th cyclotomic field; here τ means the automorphism sending
t
each root of unity to its t-th power. For any S ⊆ J we define
α = cor res θ ∈ Q[G],
S KJ/KS QS/KS nS
β = cor res θ ∈ Q[G ],
S k/kS QS/kS nS k
where G = Gal(k/Q). Here res and cor mean the usual restriction and
k
corestriction maps between group rings (see [4]). Let N = (cid:80) σ ∈ Z[G]
K σ∈G
and N = (cid:80) σ ∈ Z[G ]. Finally, let S(cid:48) be the G-module generated
k σ∈G k K
k
in Q[G] by {1N } ∪ {α ; T ⊆ J} and similarly let S(cid:48) be the G -module
2 K T k k
generated in Q[G ] by {1N }∪{β ; T ⊆ J}. We have proved in [1, Remark
k 2 k T
at page 159] that S(cid:48) and S(cid:48) are precisely the modules S(cid:48) for K and k
K k
used by Sinnott (see [4, page 189]) to define the Stickelberger ideal, so
S = S(cid:48) ∩Z[G] and S = S(cid:48) ∩Z[G ] are the Stickelberger ideals of K and
K K k k k
k, respectively.
Lemma 2.1. For any S ⊆ J and any σ ∈ G we have
(cid:88)
(1+σ)·α = a·N +2 a ·α
S K T T
T⊆S
for suitable a,a ∈ Z.
T
Proof. This is a direct consequence of [1, Lemma 18], because
(1+σ)α = 2α −(1−σ)α .
S S S
Proposition 2.2. For any S ⊆ J we have [K : k ]−1 ·cor β ∈ S(cid:48) .
S S K/k S K
Proof. It is easy to see that Gal(K/k) is a subspace of the (multiplicative)
vector space Gal(K/Q) over F . Let ρ ,...,ρ be a basis of Gal(K/k), then
2 1 r
(cid:80)
cor 1 = σ = (1 + ρ )···(1 + ρ ). Using [1, Lemma 17] we
K/k σ∈Gal(K/k) 1 r
obtain
[K : kK ]cor β = cor res α = (1+ρ )···(1+ρ )·α .
S K/k S K/k K/k S 1 r S
It is now easy to show by means of induction with respect to r using
Lemma 2.1 that
(cid:88)
[K : kK ]cor β = 2r−1a·N +2r a ·α
S K/k S K T T
T⊆S
for suitable a,a ∈ Z. The proposition follows from [K : k] = 2r and
T
[kK : k] = [K : k ].
S S S
4 R. Kuˇcera
Let τ mean the complex conjugation (both in G and G , but there is no
k
danger of confusion). Following Sinnott, we define
A = {δ ∈ Z[G]; (1+τ)δ ∈ N Z}
K K
and
A = {δ ∈ Z[G ]; (1+τ)δ ∈ N Z}.
k k k
We have S ⊆ A and S ⊆ A (see [4, Lemma 2.1]). Moreover, we shall
k k K K
need
(cid:40)
A + 1N Z if −3 ∈/ J,
A(cid:48) = K 2 K
K A + 1N Z+α Z if −3 ∈ J,
K 2 K {−3}
and
(cid:40)
A + 1N Z if −3 ∈/ J,
A(cid:48) = k 2 k
k A + 1N Z+β Z if −3 ∈ J.
k 2 k {−3}
Lemma 2.3. The indices [A(cid:48) : A ] and [A(cid:48) : A ] are equal to the number
K K k k
of roots of unity in K and k, respectively.
Proof. At first assume that −3 ∈/ J. Then ±1 are the only roots of unity
in both K and k and the lemma follows. Now, let −3 ∈ J. Then K has
exactly six roots of unity and α = cor √ (1 + 2σ ). On one hand,
√ {−3} K/Q( −3) 3√3 −3
if −3 ∈/ k then β = N . On the other hand, if −3 ∈ k then β =
{−3} k {−3}
cor √ (1 + 2σ ). The lemma follows in both cases.
k/Q( −3) 3 3 −3
Proposition 2.4. For any S ⊆ J we have γ = [K : k ]−1 ·β ∈ A(cid:48).
S S S S k
Proof. One can see immediately from definitions that S + 1N Z ⊆ S(cid:48)
K 2 K K
and that in the case −3 ∈ J we have S + 1N Z+α Z ⊆ S(cid:48) . Sinnott
K 2 K {−3} K
proved in [4, Proposition 2.1] that [S(cid:48) : S ] is equal to the number of roots
K K
of unity in K and a similar discussion as in the proof of Lemma 2.3 gives
that
(cid:40)
S + 1N Z if −3 ∈/ J,
S(cid:48) = K 2 K
K S + 1N Z+α Z if −3 ∈ J.
K 2 K {−3}
It implies that S(cid:48) ⊆ A(cid:48) . We can prove similarly S(cid:48) ⊆ A(cid:48). We have β ∈
K K k k S
S(cid:48) ⊆ A(cid:48), i.e. [K : k ]·γ ∈ A(cid:48). To avoid distinguishing two cases, in the
k k S S S k
case −3 ∈/ J we put α = 0. Proposition 2.2 gives that there are c,d ∈ Z
{−3}
such that
[K : k ]−1 ·cor β + c ·N +d·α ∈ A .
S S K/k S 2 K {−3} K
In both cases we have 3α ∈ A and so
{−3} K
3[K : k ]−1 ·cor β + 3c ·N ∈ A .
S S K/k S 2 K K
Hence
cor (3γ + 3c ·N ) ∈ A ,
K/k S 2 k K
which means 3γ + 3c ·N ∈ A and so 3γ ∈ A(cid:48). The proposition follows
S 2 k k S k
as [K : k ] and 3 are relatively prime.
S S
On Annihilators of the Class Group 5
Lemma 2.5. For any S ⊆ J and any σ ∈ G we have
k
(cid:88)
(1−σ)·γ = a·N +2 a ·γ
S k T T
T⊆S
for suitable a,a ∈ Z.
T
Proof. In [1, Proof of Lemma 19] we have derived the following identity
(cid:88)
(1−σ)·β = a[kK : k]·N +2 a [kK : kK ]·β
S S k T S T T
T⊆S
with a,a ∈ Z. Dividing by [kK : k] = [K : k ] gives the lemma.
T S S S
Lemma 2.6. Let p ∈ S ⊆ J. Then
(1+res σ )γ = (1−Frob(|p|,k))γ +[QS : K ]N ,
KJ/k p S S\{p} S k
where Frob(|p|,k) is any extension to k of the Frobenius automorphism of
|p| in k /Q.
J\{p}
Proof. [1, Lemma 20] states
(1+res σ )β = (1−Frob(|p|,k))[kK : kK ]β
KJ/k p S S S\{p} S\{p}
+[QS : K ][kK : k]N .
S S k
Using [kK : kK ] = [kK : k][kK : k]−1 the lemma follows.
S S\{p} S S\{p}
Lemma 2.7. For any non-empty S ⊆ J we have
(1+τ)γ = [QS : K ]N .
S S k
Proof. The lemma follows from the identity
(1+τ)β = [QS : k ]N
S S k
given by [1, Lemma 21].
3 Divisibility of the relative class number h−
of k by a power of 2
Let
X = {ξ ∈ G(cid:98); ξ(τ) = 1, ξ(σ) = 1 for all σ ∈ Gal(K /k)},
J
X(cid:48) = {ξ ∈ G(cid:98); ξ(τ) = −1, ξ(σ) = 1 for all σ ∈ Gal(K /k)},
J
where G(cid:98) is the character group of G. Then X and X(cid:48) can be viewed also as
the sets of all even and of all odd Dirichlet characters corresponding to k,
respectively. For any χ ∈ X ∪X(cid:48) let
S = {p ∈ J; χ(σ ) = −1},
χ p
hence n is the conductor of χ.
Sχ
Let T (cid:48) be the G -module generated in Q[G ] by {1N }∪{γ ; S ⊆ J}.
k k k 2 k S
Proposition 2.4 states that T (cid:48) ⊆ A(cid:48).
k k
6 R. Kuˇcera
Theorem 3.1. The set B = {γ ; χ ∈ X(cid:48)}∪{1N } is a Z-basis of T (cid:48) and
Sχ 2 k k
h− (cid:18) 2 (cid:19)[k:Q]/4
[A(cid:48) : T (cid:48)] = · ,
k k Q [K : K(cid:48)]·[K : k]
where K(cid:48) is the genus field in narrow sense of k+ = k ∩ R, h− = h /h
k k+
is the relative class number of k and Q is the Hasse unit index of k, to wit
Q = [E : (E ∩ R)W]; here E and W are the group of units of k and the
group of roots of unity in k, respectively.
Proof. We can prove that B is a system of generators of T (cid:48) just in the same
k
way as [1, Lemma 22] was proved – it is enough to use Lemmas 2.5, 2.6, 2.7
instead of Lemmas 19, 20, 21 of loc. cit. In [1, Theorem 3] we have proved
that the set
B(cid:48) = {β ; χ ∈ X(cid:48)}∪{1N }
Sχ 2 k
is a basis of S(cid:48). As S(cid:48) ⊆ T (cid:48) and the sets B and B(cid:48) have the same number
k k k
of elements, we see that B is in fact a basis of T (cid:48). The transition matrix
k
from B(cid:48) to B is the diagonal matrix whose diagonal consists of [K : k ],
Sχ Sχ
for all χ ∈ X(cid:48), and 1. Thus
(cid:89)
[T (cid:48) : S(cid:48)] = [K : k ].
k k Sχ Sχ
χ∈X(cid:48)
Lemma 2.3 and [4, Proposition 2.1] give that both [A(cid:48) : A ] and [S(cid:48) : S ]
k k k k
are equal to the number of roots of unity in k and so [A(cid:48) : S(cid:48)] = [A : S ].
k k k k
This equality and [1, Theorem 3] imply
h− (cid:89)
[A(cid:48) : S(cid:48)] = ·(#X(cid:48))−1(#X(cid:48)) [k : k ],
k k Q 2 Sχ
χ∈X(cid:48)
therefore
[A(cid:48) : T (cid:48)] = h− ·(#X(cid:48))−1(#X(cid:48)) (cid:89) [k : kSχ]
2
k k Q [K : k ]
χ∈X(cid:48) Sχ Sχ
h− (cid:89) [k : Q]
= ·[k+ : Q]−[k+:Q]/2 .
Q [K : Q]
χ∈X(cid:48) Sχ
In [2, Lemma 8] we have proved
(cid:89)
[K : Q] = [K(cid:48) : Q][k+:Q]/2
Sχ
χ∈X
and we can prove
(cid:89)
[K : Q] = [K : Q][k:Q]/2
Sχ
χ∈X∪X(cid:48)
On Annihilators of the Class Group 7
just in the same way. Therefore
(cid:89)
(3.1) [K : Q] = [K : K(cid:48)][k+:Q]/2 ·[K : Q][k+:Q]/2
Sχ
χ∈X(cid:48)
and the theorem follows.
Corollary 3.2. The relative class number h− of k is divisible by the follow-
ing power of 2:
(cid:18)[K : K(cid:48)][K : k](cid:19)[k:Q]/4 (cid:12)
Q· (cid:12) h−.
2 (cid:12)
Proof. This follows from T (cid:48) ⊆ A(cid:48).
k k
Remark. Let us mention that the strength of Corollary 3.2 consists mainly
in the algebraic interpretation of the divisibility result because if [k : Q] ≥ 8
then one can get even stronger divisibility result (3.2) using analytical class
number formula and genus theory as follows. For any χ ∈ X(cid:48) let B be
1,χ
the first generalized Bernoulli number and k be the imaginary quadratic
χ
field corresponding to χ. Let h and w be the class number of k and the
χ χ χ
number of roots of unity in k , respectively. The analytical class number
χ
formula (for example, see [5, Theorem 4.17]) gives
(cid:89)
h− = Qw (−1B ) and h = w (−1B ),
2 1,χ χ χ 2 1,χ
χ∈X(cid:48)
where w = #W ∈ {2,6} is the number of roots of unity in k. It is easy to
see that each w = 2 up to at most one exception. This exceptional case
χ
w = 6 appears if and only if w = 6, and in any case
χ
(cid:89) h
h− = 2Q χ.
2
χ∈X(cid:48)
The genus field of k is K , so genus theory gives
χ Sχ
1[K : Q] = [K : k ] | h ,
2 Sχ Sχ χ χ
and using (3.1) we obtain that
(3.2) 2Q (cid:89) [KSχ : Q] = 2Q·(cid:18)[K : K(cid:48)][K : Q](cid:19)[k:Q]/4 (cid:12)(cid:12) h−.
4 16 (cid:12)
χ∈X(cid:48)
4 The case of tame ramification
Let us assume that k/Q is not wildly ramified, i.e. 2 does not ramify in k,
which means that the conductor n = n of k is odd. Thus the parity of a
J
character χ ∈ X ∪X(cid:48) is determined by its conductor n . Moreover n for
Sχ S
8 R. Kuˇcera
all S ⊆ J runs over all positive divisors of n without repetition. So we shall
simplify our notation and if d = n we shall write K , k , Q , α , β , γ
S d d d d d d
instead of K , k , QS, α , β , γ etc.
S S S S S
We want to construct annihilators of the class group Cl of k outside of
k
the Stickelberger ideal S . Let T = T (cid:48) ∩Z[G ]. The aim of this section is
k k k k
to show that elements of T annihilate the principal genus PG of k, i.e.
k k
the subgroup of Cl of all classes containing the prime ideals of k whose
k
Frobenius on K/k is trivial. (Note that PG is also sometimes called “the
k
non-genus part” of Cl .)
k
Lemma 4.1. Each ideal class in the principal genus PG contains infinitely
k
many prime ideals above primes p ≡ 1 (modn).
Proof. As K is the maximal absolutely abelian subfield of the Hilbert class
field H of k and K is a subfield of the nth cyclotomic field Q , we have
k n
H ∩ Q = K. Therefore for any class C ∈ PG there is an element in
k n k
Gal(H Q /k) whose restriction to Q is trivial and restriction to H is the
k n n k
ˇ
ArtinsymbolofC.ThelemmafollowsfromCebotarevdensitytheorem.
LetusfixaclassC ∈ PG andaprimeidealP ∈ C abovep ≡ 1(modn).
k
Ofcourse,toshowthatelementsofT annihilateC weshalluseStickelberger
k
factorization of Gauss sums. Let χ:(Z/pZ)∗ → Q∗ be the nth power residue
n
symbol modulo a prime ideal P of Q above P, i.e. for any t ∈ Z relatively
n
prime to p we have χ(t) ≡ t(p−1)/n (modP). For any a ∈ Z, n (cid:45) a, we
consider the following Gauss sum
p−1
(cid:88)
x = − χ(t)−aζt,
a p
t=1
where ζ is a fixed primitive pth root of unity. If n | a ∈ Z we put x = 1.
p a
Lemma 4.2. For any positive integer r|n and any a ∈ Z we have
r−1
(cid:89)
x = χ(r)ar ·p(r−1)/2 ·x .
a+in/r ar
i=0
Proof. Davenport-Hasserelation(forexample,see[3,Theorem10.2ofChap-
ter 2]) gives
r−1 r−1
(cid:89) (cid:89)
x = χ(r)ar ·x · x .
a+in/r ar in/r
i=0 i=1
Moreover if n (cid:45) b then x ·x = χ(−1)p (for example, see [3, GS 2 in §1 of
b −b
Chapter 1]). But χ(−1) = 1 as n is odd and the lemma follows.
Well-known properties of Gauss sums show that xd is a nonzero ele-
n/d
ment of the dth cyclotomic field Q for any positive d|n. We define y =
d d
N (xd ). Let p be the prime ideal of K below P. Recall that P is the
Qd/Kd n/d
prime ideal of k below P.
On Annihilators of the Class Group 9
Lemma 4.3. We have
(y ) = pdαd as ideals of K
d
and
(cid:0) (cid:1)
N (y ) = Pdβd as ideals of k.
K /k d
d d
Proof. It has been proved by Sinnott, see [4, formulae (2.2), (3.2),(3.6), and
(3.7)], in the former case for Sinnott’s k being our K, so Sinnott’s g(cid:48)(−1,p)
d
andθ(cid:48)(−1)correspondtooury andα ,andinthelattercaseforSinnott’sk
d d d
being our k, so Sinnott’s g(cid:48)(−1,P) and θ(cid:48)(−1) correspond to our N (y )
d d Kd/kd d
and β .
d
Lemma 4.4. For any d|n and any prime q|d we have
N (y ) = pd·[Qd:Kd] ·yq·(1−Frob(q,Kd/q)).
Kd/Kd/q d d/q
Proof. It is easy to see that
(cid:89)
N (xd ) = xd .
Qd/Qd/q n/d bn/d
b≡1 (modn/q)
1≤b≤n, q(cid:45)b
Frob(q,Q )
If an integer b satisfies b ≡ 1 (modn/q) and q|b then x n/q = x .
bn/d qn/d
Hence
q−1
N (xd ) = x−d·Frob(q,Qn/q)−1 ·(cid:89)xd
Qd/Qd/q n/d qn/d (n/d)+in/q
i=0
and Lemma 4.2 gives
N (xd ) = x−d·Frob(q,Qn/q)−1 ·(cid:0)χ(q)qn/d ·p(q−1)/2 ·x (cid:1)d
Qd/Qd/q n/d qn/d qn/d
= xd·(1−Frob(q,Qn/q)−1) ·pd(q−1)/2.
qn/d
Therefore
N (y ) = N (xd ) = N (xd·(1−Frob(q,Qn/q)−1) ·pd(q−1)/2)
Kd/Kd/q d Qd/Kd/q n/d Qd/q/Kd/q qn/d
and the lemma follows.
Let Y be the subgroup of the multiplicative group K∗ generated by −1,
by p, and by all conjugates of y for d|n. The following lemma shows that
d
the action of the augmentation ideal of Z[G] on elements of Y gives the
square of an element of Y multiplied by a power of p:
Lemma 4.5. For any d|n and σ ∈ G we have
(cid:89)
y1−σ = pa · y2at
d t
t|d
for suitable integers a, a .
t
10 R. Kuˇcera
Proof. If d = 1 then y = 1 and the statement is trivial. Let us suppose
d
that d > 1 and that the lemma has been proved for all divisors of n smaller
(cid:81)
than d. Let R ⊆ J be determined by σ = σ . If (n ,d) = 1 then
σ r∈Rσ r Rσ
y1−σ = 1. Let us suppose that (n ,d) > 1 and that the lemma has also
d Rσ
been proved for this d with all ρ ∈ G having (n ,d) < (n ,d). Let us
Rρ Rσ
fix any q ∈ R , q|d. Then ρ = σ σ satisfy n = n /|q|. On one hand, if
σ q Rρ Rσ
n = 1 then
Rρ
y1−σ = y1−σq = y2 ·(cid:0)N (y )(cid:1)−1
d d d Kd/Kd/|q| d
and Lemma 4.4 together with the induction hypothesis gives what we need.
On the other hand, if n > 1 then the lemma has already been proved for
Rρ
d with both ρ and σ . Hence
q
y1−σ = y1−σq ·(y1−ρ)σq = (cid:0)pa ·(cid:89)y2at(cid:1)·(cid:0)pb ·(cid:89)y2bt(cid:1)σq
d d d t t
t|d t|d
= (cid:0)pa+b ·(cid:89)y2(at+bt)(cid:1)·(cid:89)(cid:0)y1−σq(cid:1)−2bt
t t
t|d t|d
and the lemma follows from the induction hypothesis.
Theorem 4.6. The set
B = {p}∪{y ; d|n, d > 0, d ≡ 3 (mod4)}
d
is a Z-basis of Y, more precisely B ∪{−1} is a system of generators of Y
and B is multiplicatively independent over Z.
Proof. Lemma 4.5 gives that Y is generated by {−1,p} ∪ {y ; d|n}. Let
d
(cid:81)
d|n, d ≡ 1 (mod4) and put S = {q ∈ J ; q < 0, q|d} and ρ = σ .
q∈S q
Then the number of elements of S is even and ρ acts on y as the complex
d
conjugation τ. Hence
ydρ = ydτ = NQd/Kd(xd−n/d) = NQd/Kd(pd ·x−n/dd) = pd·[Qd:Kd] ·yd−1.
Therefore
y2 = pd·[Qd:Kd] ·y1−ρ = pd·[Qd:Kd] ·(cid:89)(y1+σq)Qt∈S,t<q(−σt)
d d d
q∈S
= pd·[Qd:Kd] ·(cid:89)NK /K (yd)Qt∈S,t<q(−σt).
d d/|q|
q∈S
Lemma 4.4 and Lemma 4.5 give
(cid:89)
y2 = pa · y2at
d t
t|d,t<d
for suitable integers a, a . But p is not a square in K, so a is even and
t
(cid:89)
y = ±pa/2 · yat.
d t
t|d,t<d