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On the p-rank of tame kernel of number ﬁelds ✩ Chaochao Sun a , Kejian Xu b,∗ a b

School of Mathematics, Jilin University, Changchun, 130012, China College of Mathematics, Qingdao University, Qingdao, 266071, China

a r t i c l e

i n f o

Article history: Received 3 November 2014 Received in revised form 25 June 2015 Accepted 25 June 2015 Available online 5 August 2015 Communicated by David Goss Keywords: Number ﬁeld Cyclic quartic ﬁeld p-rank Tame kernel Ideal class group

a b s t r a c t In this paper, the relations between p-ranks of the tame kernel and the ideal class group for a general number ﬁeld are investigated. As a result, nearly all of Browkin’s results about quadratic ﬁelds are generalized to those for general number ﬁelds. In particular, a p-rank formula between the tame kernel and the ideal class group for a totally real number ﬁeld of odd degree is obtained when p is a Fermat prime. As an example, the case of cyclic quartic ﬁelds is considered in more details. More precisely, using the results on cyclic quartic ﬁelds, we give some results connecting the p-rank(K2 OF ) with the p-rank(Cl(OE )), where F is a cyclic quartic ﬁeld and E is an appropriate subﬁeld of F (ζp ). © 2015 Elsevier Inc. All rights reserved.

1. Introduction For a number ﬁeld F and an odd prime p, many authors investigated the p-rank of the tame kernel K2 OF . It has been shown that this value is related to the order of the ideal class group of some related number ﬁeld. So, some relations of p-ranks or the p-Sylow ✩

This research is supported by the National Natural Science Foundation of China (Grant No. 10871106).

* Corresponding author. E-mail address: [email protected] (K. Xu). http://dx.doi.org/10.1016/j.jnt.2015.06.009 0022-314X/© 2015 Elsevier Inc. All rights reserved.

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subgroups between the tame kernel K2 OF and the ideal class group of F (ζp ) have been established, see Tate [13], Keune [7], Browkin [1], Qin [11] and others [3,4,8–10,12], etc. In particular, using an appropriate reﬂection theorem, Browkin [1] proved several results connecting the p-rank(K2 OF ) with the p-rank(Cl(OE )), where E is an appropriate subﬁeld of F (ζp ), under the assumption that the ﬁelds F and Q(ζp ) are linearly disjoint, i.e. F ∩ Q(ζp ) = Q. In the case of quadratic ﬁelds F and p = 3, 5, the results are more explicit. In [11], Qin set up a new reﬂection theorem which generalizes the classical Scholz theorem (see [14]), and then he presented some formulas for p-rank(K2 OF ), which also connect the p-rank(K2 OF ) with the p-rank(Cl(OE )) for some appropriate subﬁeld E of F (ζp ) under similar assumption. Qin also considered the p-Sylow subgroup of K2 OF with F being some special extension of a quadratic ﬁeld. On the other hand, Zhou [15] once considered the p-rank(K2 OF ) when F is a multi-quadratic ﬁeld, and Browkin [2] studied the case of cyclic cubic ﬁelds. In the present paper, we consider the general number ﬁeld case and as results, nearly all of Browkin’s results about quadratic ﬁelds in [1] are generalized to those for general number ﬁelds. As an example, we consider the case of cyclic quartic ﬁelds in more details. More precisely, using the results on cyclic quartic ﬁelds [5,6] and also using Qin’s result we give some results connecting the p-rank(K2 OF ) with the p-rank(Cl(OE )), where F is a cyclic quartic ﬁeld and E is an appropriate subﬁeld of F (ζp ). In Section 2, we consider the case of general number ﬁelds and Browkin’s results are improved, in particular, a p-rank formula between the tame kernel and the ideal class group for totally real number ﬁeld of odd degree is obtained when p is a Fermat prime. In Section 3, we prove some properties about cyclic quartic ﬁelds which are needed in the sequel sections. In Section 4, the computation of the values |S | for cyclic quartic ﬁelds is given. In Section 5, we give the p-ranks of tame kernels for cyclic quartic ﬁelds. 2. General number ﬁelds Let p be an odd prime number, and ζp a p-th primitive root of unity. Then Gal(Q(ζp )/Q) = {σa : 1 ≤ a ≤ p − 1}, where σa (ζp ) = ζpa . For a ﬁxed primitive root g (mod p), let σ := σg . Then Gal(Q(ζp )/Q) is a cyclic group of order p − 1 generated by σ. In this section, we always assume that F/Q is a Galois extension of degree n such that ζp ∈ / F and F F ∩ Q(ζp ). Let K := F ∩ Q(ζp ). Then l := [K : Q] < n. Let ω be the Teichmüller character of the group Gal(Q(ζp )/Q) ∼ = (Z/p)∗ . We use the symbol ω ¯ to denote the restriction of ω on G1 := Gal(Q(ζp )/K) = σ l , i.e. ω ¯ := ω|G1 .

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So all the primitive characters on G1 are given by ω ¯ j , j = 0, · · · , p−1 l − 1. Consequently all the primitive idempotents in the group ring Zp [G1 ] are given by the formula p−1

p−1

l l −1 l l ε¯j = ω ¯ j (al )σa−1 = ω ¯ jk (g l )σ −lk , l p − 1 a=1 p−1

k=0

where j = 0, · · · , p−(l+1) . l Let E := F (ζp ) and denote G := Gal(E/F ). Then we have Gal(E/K) = Gal(F/K) × Gal(Q(ζp )/K), and the isomorphism G∼ = Gal(Q(ζp )/K). Hence all the primitive idempotents in the group ring Zp [G] are also given by p−1

l −1 l ε¯j = ω ¯ jk (g l )σ −lk , p−1

k=0

where j = 0, · · · , p−1 ¯p−1 = ε¯0 . l − 1. We identify ε l In the following, we will use the same notations as in [1]. In particular, for a subﬁeld K of E we denote Cl(K )p := Cl(OK )p (the Sylow p-subgroup of the class group of K ). Let λ : Cl(E) → Cl(OE [ p1 ]) be the homomorphism induced by the imbedding OE → OE [ p1 ]. Then λ : Cl(E)p → Cl(OE [ p1 ])p is a surjective homomorphism of Sylow p-subgroups. The symbol v ⊂ OF means that v is a nontrivial valuation of the ring OF . In [7], Keune has proved the following theorem. Theorem 2.1. Let F be number ﬁeld, p an odd prime and G = Gal(E/F ). Let S be the set of places of E dividing p. We have the following short exact sequences of G-modules (i) If E = F , then 1 1 → μp ⊗ Cl OE → K2 OF /p → p

μp → μp → 1.

v|p,v⊂OF

(ii) If E = F , then 1 G 1 → μp ⊗ Cl OE → K2 OF /p → p

μp (Fv ) → 1.

2

v|p,v⊂OF

It is easy to show that μp is included in Fv if and only if v splits completely in F (ζp ), where Fv is the completion of F with respect to the valuation v.

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By Theorem 2.1, we have the following short exact sequence: 1 G 1 → μp ⊗ Cl OE → K2 OF /p → μp → 1, p

(2.1)

S

where G = Gal(E/F ) acts on μp ⊗ Cl(OE [1/p]) by the formula (ζ ⊗ x)ρ = ζ ρ ⊗ xρ , for ζ ∈ μp , ρ ∈ G, x ∈ Cl(OE [1/p]), and S is the set of prime ideals of F which divide p and split in E. Obviously, in (2.1), we can replace Cl(OE [ p1 ]) by λ(Cl(E)p ). Since p |G| = (p − 1)/l, for every ﬁnite p-group M , we have M G = ε¯0 M . So Browkin’s result (see [1]) can be modiﬁed as follows. Lemma 2.2. rankp (K2 OF ) = rankp (¯ ε0 (μp ⊗ λ(Cl(E)p ))) + |S |.

2

Following Browkin [1], we also deﬁne a homomorphism γ as the composition of the following homomorphisms: δ

λ

ε

Cl(E)p − → λ(Cl(E)p ) − → μp ⊗ λ(Cl(E)p ) − → ε¯0 (μp ⊗ λ(Cl(E)p )), where δ(λ(a)) = ζp ⊗ λ(a) and ε(ζp ⊗ λ(a)) = ε0 (ζp ⊗ λ(a)) for a ∈ Cl(E)p , that is, for a ∈ Cl(E)p , we put γ(a) = εδλ(a) = ε¯0 (ζp ⊗ λ(a)). The homomorphisms λ, δ, ε, hence γ, are surjective. λ and ε are homomorphisms of G-modules, but in general δ is not a homomorphism of G-modules. In fact, according to Lemma 3.3 in [1], i.e. for z := ζpn , we have σ(z) = z ω(g) , so we get l

σ l (δx) = σ l (ζp ) ⊗ σ l x = ζpω(g ) ⊗ σ l x = ω(g l ) · δ(σ l (x)). Therefore p−1

l −1 l ζp ⊗ λ(¯ εj a) = ζp ⊗ λ ω jk (g l )σ −lk (a) p−1

k=0

p−(l+1) l

=

l ω (j+1)k (g l )σ −lk (ζp ⊗ λ(a)) p−1 k=0

= ε¯j+1 (ζp ⊗ λ(a)). Hence

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δ(λ(¯ εj a)) = ζp ⊗ λ(¯ εj a) = ε¯j+1 (ζp ⊗ λ(a)) = ε¯j+1 δ(λ(a)). Taking j =

p−(l+1) l

(2.2)

in (2.2) and noting that ε¯p−1 = ε¯0 , we get l

ε¯0 (ζp ⊗ λ(¯ ε p−(l+1) a)) = ζp ⊗ λ((¯ ε p−(l+1) )2 a) = ζp ⊗ ε¯p−(l+1) λ(a). l

l

(2.3)

l

Moreover γ(¯ εj a) = εδλ(¯ εj a) = ε¯ εj+1 δ(λ(a)) = ε¯j+1 εδ(λ(a)) = ε¯j+1 γ(a). Since p−(l+1) l

Cl(E)p =

ε¯j Cl(E)p ,

j=0

we have γ(¯ εj Cl(E)p ) = ε¯j+1 (γCl(E)p ) = ε¯j+1 ε¯0 (μp ⊗ λ(Cl(E)p )) = 0 for j =

p−(l+1) . l

(2.4)

Thus from (2.3) and (2.4) we conclude that

ε¯0 (μp ⊗ λ(Cl(E)p )) = γ(Cl(E)p ) = γ(¯ ε p−(l+1) Cl(E)p ) l

= ε¯0 (μp ⊗ λ(¯ ε p−(l+1) Cl(E)p )) = μp ⊗ ε¯p−(l+1) λ(Cl(E)p ). l

l

Since tensoring by μp preserves the p-rank, from Lemma 2.2, we get the more general form of Browkin’s result [1]: Theorem 2.3. rankp (K2 OF ) = rankp (¯ ε p−(l+1) λ(Cl(E)p )) + |S |. l

2

Assume that τ ∈ Gal(F/Q) is an automorphism of order 2 belonging to the center of Gal(F/Q). We also consider the idempotents η0 =

1 1 (1 + τ ), η1 = (1 − τ ) 2 2

of the group ring Zp [Gal(F/Q)]. Let e and f denote respectively the ramiﬁcation number and the residue class degree in E of a prime ideal of F over p. Obviously p ≥ ef l + 1. The following lemma is crucial in our discussions. Lemma 2.4. Let F/Q be a Galois extension and E = F (ζp ). Then the mapping λ : εj Cl(E)p → εj Cl(OE [ p1 ])p is an isomorphism for j = 0, ef, 2ef, . . . , ( p−1 ef l − 1)ef .

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Proof. Let [F : Q] = n > 1. From F ∩Q(ζp ) = K and l = [K : Q], we have [E : F ] = p−1 l . Since F/Q is a Galois extension, it can be assumed that we have the decomposition:

(p) = pe1 pe2 · · · pes , where pi , 1 ≤ i ≤ s ≤ n, are prime ideals of OF , and that

pi = Pei1 Pei2 · · · Peis , 1 ≤ i ≤ s, where Pii , 1 ≤ i ≤ s, 1 ≤ i ≤ s , are prime ideals of OE and 1 ≤ s ≤ [E : F ] = Clearly s := in OE :

[E:F ] ef

=

p−1 ef l .

p−1 l .

Then we get the decomposition of the prime ideal (1 − ζp )

(1 − ζp ) =

s

Pei1 Pei2 · · · Peis ,

i=1

where e = e e /(p − 1), since E/Q(ζp ) is also a Galois extension. Obviously σ l ﬁxes F , so it ﬁxes every pi . While the Galois group Gal(E/F ) = σ l is transitive on the set of prime ideals {Pii }si =1 for each 1 ≤ i ≤ s, hence when s > 1, we should have σ l (Pii ) = Pii . Moreover, we have (σ l )s (Pii ) = Pii , 1 ≤ i ≤ s, 1 ≤ i ≤ s . Clearly, Ker(λ) is generated by the elements: aii = Cl(Pii ), 1 ≤ i ≤ s, 1 ≤ i ≤ s . So we have (σ l )s (aii ) = aii , 1 ≤ i ≤ s, 1 ≤ i ≤ s . Suppose that for some a ∈ Ker(λ), we have a ∈ ε¯j Cl(E)p . Then (σ l )s (a) = a. Noting that p − 1 = ef ls , we have p−1

l −1 l a = εj a = ω(g l )kj σ −kl (a) p−1

k=0

p−1 −1 s l

s −1 l ω(g l )(s k+t)j σ −(s k+t)l (a) = p − 1 t=0 k=0 p−1

−1

l s −1 l s ω(g l )s jk ω(g l )jt σ −tl (a) = p − 1 t=0

k=0

j(p−1) s −1 l (1 − ω(g l ) l ) · ω(g l )jt σ −tl (a) = 0 = p − 1 t=0 1 − ω(g l )s j

for j = 0, ef, 2ef, . . . , ( p−1 ef l − 1)ef . Hence in these cases we get ker λ = 0, that is, λ is an isomorphism. 2 Let FD denote the decomposition subﬁeld of p in F . Then Theorem 2.3 can be improved as follows.

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Theorem 2.5. Let F/Q be a Galois extension, E = F (ζp ) and F ∩ Q(ζp ) = K with l = [K : Q]. Assume that p = l + 1. (i) If ef > 1, then rankp (K2 OF ) = rankp (¯ ε p−1 −1 Cl(E)p ). Moreover, l

rankp (K2 OF ) = rankp (η0 ε¯p−1 −1 Cl(E)p ) + rankp (η1 ε¯p−1 −1 Cl(E)p ). l

l

(ii) If ef = 1, then rankp (K2 OF ) = rankp (¯ ε p−1 −1 Cl(E)p ) + [FD : Q], l

if (Kerλ) ∩ ε¯p−1 −1 Cl(E)p does not contain a nontrivial direct summand of l ε¯p−1 −1 Cl(E)p , and l

rankp (K2 OF ) ≤ rankp (¯ ε p−1 −1 Cl(E)p ) + [FD : Q] − 1 l

otherwise. / F at the beginning of this section, we could Proof. Note that from the assumption ζp ∈ have the cases ef > 1 and ef = 1. (i) Obviously, if ef > 1, then the prime ideals in F over p cannot split in E. So |S | = 0. Since p = l + 1, we have p−1 l − 1 = 0. From ef > 1, we have p−1 p−1 − 1 ef = − ef < − 1. ef l l l

p − 1 Hence we get

p − 1 p−1 − 1 = 0, ef, . . . , − 1 ef. l ef l Thus the result from Lemma 2.4 and Theorem 2.3. The second statement is clear. (ii) If ef = 1, then every prime ideal of F which divides p splits in E. Since the number of prime ideals over p is equal to [FD : Q]. So |S | = [FD : Q]. We have the exact sequence 0

(Kerλ) ∩ ε¯p−1 −1 Cl(E)p l

ε¯p−1 −1 Cl(E)p l

λ

λ(¯ ε p−1 −1 Cl(E)p ) l

0. (2.5)

If (Kerλ) ∩ ε¯p−1 −1 Cl(E)p does not contain a nontrivial direct summand of l ε¯p−1 −1 Cl(E)p , then the result follows from (2.5); otherwise from Lemma 2.2, and (2.5) l we get: rankp (K2 OF ) ≤ rankp (¯ ε p−1 −1 Cl(E)p ) + |S | − 1 l

ε p−1 −1 Cl(E)p ) + [FD : Q] − 1. = rankp (¯ l

2

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Corollary 2.6. Let F/Q be a Galois extension linearly disjoint from Q(ζp ) and E = F (ζp ). If ef > 1, then (i) rankp (K2 OF ) = rankp (εp−2 Cl(E)p ). (ii) rankp (K2 OF ) = rankp (η0 εp−2 Cl(E)p ) + rankp (η1 εp−2 Cl(E)p ). Proof. When F and Q(ζp ) are linearly disjoint, we have l = 1, so p > 2 = l + 1. 2 Corollary 2.7. Let F/Q be a totally real Galois extension and E = F (ζp ). If ([F : Q], p − 1) = 1, then rankp (K2 OF ) = rankp (εp−2 Cl(E)p ). Proof. When F/Q is a totally real Galois extension satisfying ([F : Q], p − 1) = 1, F and Q(ζp ) are linearly disjoint. So the prime ideals in F over p must be totally ramiﬁed in E. Thus ef > 1. Hence the result follows from Corollary 2.6 (i). 2 Corollary 2.8. Let F/Q be a totally real Galois extension of odd degree and E = F (ζp ). If p is a Fermat prime number, then rankp (K2 OF ) = rankp (εp−2 Cl(E)p ). In particular, we have p|#K2 OF =⇒ p|#Cl(E). n

Proof. A Fermat prime number is of the form p = 22 + 1. So the Galois group n Gal(Q(ζp )/Q) = (Z/p)∗ has order 22 , which is relatively prime to the odd number [F : Q]. 2 Corollary 2.9. Let F/Q be a Galois extension of degree n and E = F (ζp ). If p > n + 1, then p|#K2 OF =⇒ p|#Cl(E). Proof. When p > n+1, we must have ef > 1, otherwise if ef = 1, comparing ramiﬁcation numbers we have p − 1 ≤ n, which is a contradiction. Hence the result follows from Theorem 2.5 (i). 2 To give the following corollaries and the results in Section 5, we need the following lemma. Here, for a subﬁeld Ei of E, we use the symbols UEi , UE i to denote the groups of units, respectively, of singular primary units in Ei . See the deﬁnition in Section 3 in [1] for details.

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Lemma 2.10. (See [1].) Denote NK := NE/K . Then (i) For j even, we have 1 ω(g)kj σ −k NE0 p−1 t−1

εj =

k=0

where E0 = E σ , t = p−1 2 . (ii) For k + j even, we have t

ηk εj =

t−1 1 ηk ω(g)rj σ −r NE1 p − 1 r=0

η1 ε2 =

1 η1 ω(g)2k σ −k NE2 p−1

where E1 = E τ σ . (iii) For p = 4s + 1, we have t

s−1

k=0

where E2 = E τ σ . (iv) If F is a CM-ﬁeld, then s

rankp (η1 εp−2 Cl(E)p ) ≤ rankp (η1 ε2 Cl(E)p ) ≤ rankp (η1 εp−2 Cl(E)p ) + rankp (η1 εp−2 (UE 1 /UEp 1 )). (v) If F is totally real ﬁeld, then rankp (ηk ε2 Cl(E)p ) ≤ rankp (ηk εp−2 Cl(E)p ) ≤ rankp (ηk ε2 Cl(E)p ) + rankp (ηk ε2 (UE 0 /UEp 0 )), where E0 is the subﬁeld of E ﬁxed by σ t . (vi) If K ⊂ L are subﬁelds of E such that p [L : K] and jL/K : Cl(K)p → Cl(L)p , NL/K : Cl(L)p → Cl(K)p are canonical mappings, then jL/K is injective and NL/K is surjective. The same holds if we replace Cl(K)p and Cl(L)p by UK /p and UL /p. Proof. See Lemma 3.4, Theorem 3.8 and Lemma 2.1 in [1].

2

Corollary 2.11. Let F/Q be a Galois extension linearly disjoint from Q(ζp ), F a CM ﬁeld and E = F (ζp ). Assume that ef > 1. Then we have

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(i) If p > 3, then rankp (K2 OF ) ≤ rankp (Cl(E1 )) + rankp (Cl(E τ )), where E1 = E τ σ . (ii) If p = 4s + 1, then we have t

rankp (K2 OF ) ≤ rankp (Cl(E2 )) + rankp (Cl(E τ )), where E2 = E τ σ . s

Proof. By Corollary 2.6, we have rankp (K2 OF ) = rankp (εp−2 Cl(E)p ) = rankp (η0 εp−2 Cl(E)p ) + rankp (η1 εp−2 Cl(E)p ).

(2.6)

(i) Let E1 denote the maximal real subﬁeld of E, i.e., the subﬁeld of E ﬁxed by τ σ t , where t = p−1 2 . From Lemma 2.10 (ii), we have η1 εp−2 =

t−1 1 η1 ω(g)(p−2)r σ −r (1 + τ σ t ), p − 1 r=0

so we get η1 εp−2 Cl(E)p ⊆ Cl(E1 )p , η0 εp−2 Cl(E)p ⊆ η0 Cl(E)p = Cl(E τ )p .

(2.7)

Hence by Lemma 2.10 (iv), and (2.6) and (2.7), we get rankp (K2 OF ) ≤ rankp (Cl(E1 )) + rankp (Cl(E τ )). (ii) From Lemma 2.10 (iv), we have rankp (η1 εp−2 Cl(E)p ) ≤ rankp (η1 ε2 Cl(E)p ), From Lemma 2.10 (iii), for p = 4s + 1 we have η1 ε2 =

s−1 1 η1 ω(g)2r σ −r (1 + τ σ s ). p − 1 r=0

So from (2.6), (2.7) and (2.8), we have rankp (K2 OF ) ≤ rankp (Cl(E2 )) + rankp (Cl(E τ )), where E2 is the subﬁeld of E ﬁxed by τ σ s .

2

(2.8)

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Corollary 2.12. Let F/Q be a totally real Galois extension linearly disjoint from Q(ζp ), and E = F (ζp ). Assume that ef > 1. Then we have (i) If p > 3, then rankp (K2 OF ) ≤ rankp (Cl(E0 )) + rankp (UE 0 /UEp 0 ), where E0 is the maximal real subﬁeld of E. (ii) If p = 4s + 1, then we have rankp (K2 OF ) ≤ rankp (Cl(E τ )) + rankp (Cl(E2 )) + rankp (UE 2 /UEp 2 ), where E2 is the subﬁeld of E ﬁxed by τ σ s . Proof. (i) From Lemma 2.10 (v), we have rankp (εp−2 Cl(E)p ) ≤ rankp (ε2 Cl(E)p ) + rankp (ε2 (UE 0 /UEp 0 )). From Lemma 2.10 (i) and Theorem 2.5 (i), we get rankp (K2 OF ) ≤ rankp (Cl(E0 )) + rankp (UE 0 /UEp 0 ), where E0 is the maximal real subﬁeld of E. (ii) From Lemma 2.10 (v), the proof is analogous to Corollary 2.11.

2

Corollary 2.13. Let F/Q be a Galois extension linearly disjoint from Q(ζ3 ), and E = F (ζ3 ). (i) If ef > 1, we have rank3 (K2 OF ) = rank3 (Cl(E τ )) + rank3 (Cl(E τ σ )) − 2 · rank3 (Cl(E τ,σ )). (ii) If ef = 1, we have rank3 (K2 OF ) = rank3 (Cl(E τ ))+rank3 (Cl(E τ σ ))−2·rank3 (Cl(E τ,σ ))+[FD : Q], if (Kerλ) ∩ Cl(E τ σ )3 does not contain a nontrivial direct summand of Cl(E τ σ )3 , and rank3 (K2 OF ) ≤ rank3 (Cl(E τ )) + rank3 (Cl(E τ σ )) − 2 · rank3 (Cl(E τ,σ )) + [FD : Q] − 1, otherwise.

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Proof. By Lemma 2.10 (vi), we have 1−τ 1−σ 1 − τ 1 + τσ 1−τ · · Cl(E)3 = · · Cl(E)3 = · Cl(E τ σ )3 , 2 2 2 2 2 1+τ 1−σ 1−σ · · Cl(E)3 = · Cl(E τ )3 , η0 ε1 Cl(E)3 = 2 2 2

η1 ε1 Cl(E)3 =

1+τ · Cl(E τ σ )3 = Cl(E τ,σ )3 , 2 1+σ · Cl(E τ )3 = Cl(E τ,σ )3 , 2

Cl(E τ,σ )3 ⊕ η1 ε1 Cl(E)3 = Cl(E τ σ )3 , Cl(E τ,σ )3 ⊕ η0 ε1 Cl(E)3 = Cl(E τ )3 .

So the corollary follows from Theorem 2.5. 2 √ Remark 2.14. In this remark, we assume that F = Q( d) is a quadratic ﬁeld. To deduce Browkin’s results established for quadratic ﬁelds in [1], we need the following two claims: Claim 1. If p > 3, then ef > 1. This has been proved in the proof of Corollary 2.9. Claim 2. If p = 3, then ef = 2 ⇐⇒ d ≡ 6(mod 9). √ ⇒: Assume that ef = 2. Then 3 is ramiﬁed in E = F ( −3). If e = 1, f = 2, then 3 is ramiﬁed in F and there is only one prime in E over 3, hence 3 is inert in Q( −d/3), so we have d ≡ 3(mod 9); if e = 2, f = 1, then 3 splits or is inert in F , i.e. d ≡ 1, 4 or 7(mod 9), or d ≡ 2, 5 or 8(mod 9). Summarily we have d ≡ 6(mod 9). ⇐: Assume that p = 3 and d ≡ 6(mod 9). If ef = 1, since 3 is ramiﬁed in E, we know that 3 ramiﬁes in F , thus 3|d and 3 splits in Q( −d/3). Hence we have d ≡ 6(mod 9), a contradiction. So we get ef = 2. Claim 2 is proved. Now we can deduce Browkin’s results as follows. (i) From Theorem 2.5, and Claims 1 and 2, we get Theorem 5.3 in [1]. (ii) If F is an imaginary quadratic ﬁeld, then we have E τ = Q(ζp ). If p is regular prime, we have Cl(E τ )p = 0. Since p > 3, by Claim 1 we have ef > 1, which satisﬁes the assumption in Corollary 2.11. Hence, from Corollary 2.11, we get Theorem 5.4 (i), (ii) in [1]. (iii) If F is a real quadratic ﬁeld, from Corollary 2.12 and Claim 1, similar to (i), we get Theorem 5.5 (i), (ii) in [1]. (iv) For the quadratic ﬁeld F , we have √ E τ,σ = Q, E τ = Q(ζ3 ), E τ σ = E1 = Q( −3d), so rank3 (Cl(E τ )) = rank3 (Cl(E τ,σ )) = 0. Thus from Corollary 2.13 (i) and the above claims we get Theorem 5.6 (i) in [1].

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For the quadratic ﬁeld F , we have 1 ≤ ef ≤ 2. From Claim 2 we have that p = 3 and ef = 1 if and only if d ≡ 6(mod 9). On the other hand, p = 3 and ef = 1 implies that p ramiﬁes in F , so [FD : Q] = 1. In virtue of the well-known result rank3 (K2 OF ) ≥ 1, from Corollary 2.13 (ii) we get Theorem 5.6 (ii) in [1]. 3. Cyclic quartic ﬁelds From now on, we assume that F is a cyclic quartic extension of Q. Then it follows from [5] that there exist integers A, B, C, D, satisfying the conditions B > 0, C > 0, D > 0, (A, D) = 1, and D = B 2 + C 2 , so that F can be written in the form of F =Q

√ A(D + B D) ,

where A is odd and A, D are both squarefree. Conversely, any ﬁeld satisfying the above conditions is a cyclic quartic extension of Q. Thus, the representation of F is uniquely expressed by the above form. It is shown in [6] that the discriminant d(F ) of F is given by ⎧ ⎪ 28 A 2 D 3 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 26 A2 D3 , ⎪ ⎪ ⎪ ⎨24 A2 D3 , d(F ) = ⎪ ⎪ ⎪ ⎪ ⎪ 2 3 ⎪ ⎪ A D , ⎪ ⎪ ⎪ ⎩

if D ≡ 0 (mod 2), if D ≡ 1 (mod 2), B ≡ 1 (mod 2), if D ≡ 1 (mod 2), B ≡ 0 (mod 2), A + B ≡ 3 (mod 4),

(3.1)

if D ≡ 1 (mod 2), B ≡ 0 (mod 2), A + B ≡ 1 (mod 4).

Now, we study some properties of the ﬁeld F , which will be used in the following sections.

√ Proposition 3.1. Let F = Q A(D + B D) be the cyclic quartic extension over Q √ with p A. Then the prime p is inert in Q( D) if and only if it is inert in F . Proof. “⇐”: This direction is easy. √ “⇒”: Assume that the prime p is inert in Q( D). Then the decomposition subﬁeld FD of F with respect to the prime p is not equal to F . But F is a cyclic quartic ﬁeld, √ √ so Q( D) is the only quadratic subﬁeld of F . Hence FD ⊆ Q( D), therefore FD = Q √ since p is inert in Q( D). But p A, so p does not ramify in F , which implies that p is inert in F/Q. 2

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Proposition 3.2. Let F = Q

257

√ A(D + B D) be a cyclic quartic ﬁeld. Let p be an odd

prime and p AD. Then p splits in F if and only if there exists an integer E such that D ≡ E 2 (mod p) and

A(D + BE) p

= 1.

Proof. From p AD, by (3.1), it follows that p does not ramify in F/Q. Since the extension F/Q is Galois, it is well known that the prime p splits in F if and only if there is an injection

√ λ:F =Q A(D + B D) −→ Qp . From D = B 2 + C 2 it follows that √ √ A(D + B D) · A(D − B D) = A2 C 2 D. Therefore F =Q

√ √ A(D + B D) = Q A(D − B D) .

If p splits in F , we can assume that the injection λ maps √

D −→ u ∈ Qp ,

√ A(D + εB D) −→ A(D + εBu) =: ωε ∈ Qp

for ε ∈ {−1, 1}. From p AD it follows that A(D + εBu) ≡ 0 (mod p) for ε = −1 or ε = 1. For this ε we have εu ≡ E(mod p) where E ∈ Z, p E. Then A(D + εBu) ≡ A(D + BE)(mod p). Thus we have D = u2 = (εu)2 ≡ E 2 (mod p), and A(D + BE) ≡ A(D + εBu) ≡ 2 ωε (mod p). Hence A(D + BE) is a nonzero square modulo p. This proves

A(D + BE) p

= 1.

Conversely, assume that there exists an integer E such that D ≡ E 2 (mod p) and

A(D + BE) p

= 1.

By Hensel’s lemma, x2 − D has a root u ∈ Qp , such that u ≡ E(mod p). Hence the congruence x2 ≡ A(D + Bu) ≡ A(D + BE)(mod p) has a solution modulo p. So, by Hensel’s lemma, x2 − A(D + Bu) has a root in Qp . This gives an injection F → Qp , which implies that p splits in F . 2

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Corollary 3.3. Let F = Q

√ A(D + B D) be a cyclic quartic extension of Q. Then 3

splits in F if and only if ( D 3 ) = 1 and A 3

=

1,

if 3|B,

(3.2)

−1, if 3 B.

Proof. Put p = 3 in Proposition 3.2. From ( D 3 ) = 1 it follows that D ≡ 1(mod 3). So we can take E = ±1. It is suﬃcient to prove that (3.2) is equivalent to A(D ± B) 3

=1

(3.3)

with an appropriate sign. If 3|B, then A(D ± B) 3

=

AD 3

=

A 3

.

If 3 B, then ±B ≡ 1(mod 3) if we choose an appropriate sign. Therefore A(D ± B) 3

=

A 2 A · =− . 3 3 3

Thus (3.2) is equivalent to (3.3). 2

√ Proposition 3.4. Let F = Q A(D + B D) be a cyclic quartic extension of Q. If q is an odd prime divisor of D, then q is totally ramiﬁed in F/Q.

Proof. The minimal polynomial for

√ A(D + B D) over Q is

x4 − 2ADx2 + A2 D(D − B 2 ). It is an Eisenstein polynomial with respect to q, since (A, D) = 1 and D = B 2 + C 2 is squarefree. Therefore q is totally ramiﬁed in F . 2 Proposition 3.5. If F is a cyclic quartic ﬁeld F contained in Q(ζp ), then F =Q

√ p−1 (−1) 4 (p + B p) ,

where B ≡ 0 (mod 2).

Proof. We know that F is of the form F = Q

√ A(D + B D) , where D = B 2 + C 2 ,

B, C ∈ N, and A is squarefree. By (3.1), we get AD = ±p since AD is squarefree and only p ramiﬁes, so D = p since D = B 2 + C 2 ≥ 2, and therefore A = ±1.

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259

Moreover, by (3.1), B is even and A ≡ 1 − B(mod 4). Then p = B 2 + C 2 implies that C is odd. Hence p ≡ B 2 + 1(mod 8). Consequently, 1 (mod 8), if B ≡ 0 (mod 4), i.e. A ≡ 1 (mod 4), p≡ 5 (mod 8), if B ≡ 2 (mod 4), i.e. A ≡ −1 (mod 4). This is equivalent to A = (−1)

p−1 4

, as required. 2

4. The computations of |S | In this section, for a cyclic quartic ﬁeld F , we give the computations of the number |S |, where S denotes the set of prime ideals of F which divide p and split in E := F (ζp ).

√ Lemma 4.1. Let F = Q A(D + B D) be a cyclic quartic ﬁeld, and E = F (ζp ), where p is an odd prime. (i) If p D, then

|S | =

⎧ ⎪ 2, ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 0,

if p = 3, D ≡ 1(mod 3), 3|A and −1 (mod 3), if 3|B, A 3 ≡ 1 (mod 3), if 3 B,

(4.1)

otherwise.

(ii) If p|D but p = D, then |S | = 0; (iii) If p = D, then 1, if p = 5, A = −1, ( A5 ) = 1, |S | = 0, otherwise. Proof. (i) Firstly we assume that |S | > 0. For arbitrary subﬁelds E1 , E2 of E satisfying E1 ⊆ E2 let e(E2 /E1 ) and f (E2 /E1 ) be the ramiﬁcation index and the residue ﬁelds degree, respectively, of a prime ideal over p in the extension E2 /E1 . Claim 1. p = 3, 3|A and D ≡ 1(mod 3). In fact, by assumption, e(E/F ) = f (E/F ) = 1. On the other hand, the extension √ Q(ζp )/Q is totally ramiﬁed at p, and Q( D)/Q is not ramiﬁed at p, since p D. Hence √ e(Q( D)/Q) = 1. Consequently, √ √ p − 1 | e(E/Q) = e(F/Q( D)) ≤ (F : Q( D)) = 2.

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260

√ √ It follows that p = 3, and e(E/Q) = e(F/Q( D)) = 2, so f (F/Q( D)) = 1. Therefore the extension F/Q is ramiﬁed at p = 3, hence 3|A, by (3.1). Clearly, the decomposition subﬁeld ED of the extension E/Q at 3 is nontrivial, since primes over 3 in F split in E. It follows that ED contains a quadratic subﬁeld of E. There are three quadratic subﬁelds: √ √ √ Q( D), Q(ζ3 ) = Q( −3), Q( −3D). In the last two subﬁelds the prime 3 ramiﬁes, so they are not contained in the decom√ √ position ﬁeld ED . Hence Q( D) ⊆ ED , consequently, 3 splits in Q( D), which implies D 3

= 1, so D ≡ 1(mod 3).

Claim 2. A − ≡ 3 Denote A = −A/3. Since √

−3 ·

Consequently, F := Q

1 (mod 3),

if 3|B,

−1(mod 3), if 3 B.

√ −3 ∈ E, the following product belongs to E:

√ √ A(D + B D) = −3 A (D + B D).

√ √ A (D + B D) is a subﬁeld of E containing Q( D). More-

over, 3 does not ramify in F , since 3 A . Hence F = F . The prime 3 is not inert in F /Q, as well, because from the proof of Claim 1 we have: √ √ f (F /Q) | f (E/Q) = f (E/F ) · f (F/Q( D)) · f (Q( D)/Q) = 1. Consequently, 3 splits in F /Q, and applying Corollary 3.3 we get the claim. Secondly, to ﬁnish the proof of the lemma it is suﬃcient to prove that the necessary condition for |S | > 0 given in Claims 1 and 2 is also suﬃcient for |S | = 2. Applying Corollary 3.3 to (3.1) we get that 3 splits in F /Q in the product of four √ prime ideals. Since 3 splits in Q( D)/Q in the product of two prime ideals, which ramify √ in F/Q( D) since 3|A, it follows that the two prime divisors of 3 in F split in E/F . Thus |S | = 2. This completes the proof of (i). (ii) Assume that p|D. Then by Proposition 3.4, p totally ramiﬁes in F , so (p) = p4 for some prime p of OF . Moreover, if p = D, then F and Q(ζp ) are linearly disjoint. Hence [F (ζp ) : F ] = p − 1. If p > 5, then the ramiﬁcation degree of p in F (ζp ) is greater than 1, since (p) = (1 − ζp )p−1 , so p does not split in F (ζp ). Therefore |S | = 0 if p > 5. If p = 5, then [F (ζp ) : F ] = 4. Hence if p splits in F (ζ5 ), the decomposition ﬁeld of p in F (ζ5 ) must be a quartic ﬁeld. But from Proposition 3.5, in F (ζ5 ) there are only three

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261

quartic ﬁelds: F, Q

√

√ √ D, 5 , Q −(5 + B 5) ,

where B is some positive integer, and 5 splits in none of these ﬁelds. This is a contradiction. So p cannot split in F (ζp ) in the case p = 5. Thus |S | = 0 if p = 5. This completes the proof of (ii). p−1 (iii) Assume that p = D. First we consider the case A = (−1) 4 . From Proposition 3.5, the cyclic quartic ﬁeld F is not contained in Q(ζp ). It is easy to see that

p−1 F (ζp ) = Q ζp , (−1) 4 A ,

p−1 Q(ζp ) ∩ Q (−1) 4 A = Q, and [F (ζp ) : F ] = p−1 2 . If p > 5, then F (ζp )/Q(ζp ) is a quadratic extension. Hence there are at most two diﬀerent prime ideals in F (ζp ) over p. But [F (ζp ) : F ] = p−1 2 > 2, so p cannot split in F (ζp ). Thus |S | = 0. √ If p = 5, then F (ζ5 )/F is a quadratic extension. If ( A splits in Q( −A), 5 ) = 1, then 5 √ so p splits in F (ζ5 ), hence |S | = 1; if ( A5 ) = −1, then 5 is inert in Q( −A), so, p is inert in F (ζ5 ), hence |S | = 0. Summarily, we have proved |S | =

1, if p = 5, A = −1, ( A5 ) = 1, 0, if p > 5 and A = (−1)

Now we turn to the case A = (−1) F =Q

p−1 4

p−1 4

(−1)

, or p = 5, A = −1 and ( A5 ) = −1.

(4.2)

. From Proposition 3.5, we have

p−1 4

√ (p + B p) ⊆ Q(ζp ).

It is easy to see that |S | = 0, since p is totally ramiﬁed in F (ζp ) = Q(ζp ). In other words, we have |S | = 0, if A = (−1)

p−1 4

.

(4.3)

Combining (4.2) and (4.3), we get the proof. 2 5. The p-rank of tame kernel of cyclic quartic ﬁelds We will use the results in the previous sections to study the p-rank of tame kernel of cyclic quartic ﬁelds.

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5.1. The case of p = D

√ Theorem 5.1. Let F = Q A(D + B D) be a cyclic quartic ﬁeld and E = F (ζp ). (i) If p D and p > 3, then rankp (K2 OF ) = rankp (εp−2 Cl(E)p ). (ii) If p|D but p = D, then rankp (K2 OF ) = rankp (εp−2 Cl(E)p ). Proof. In these cases we have F and Q(ζp ) are linearly disjoint, so the results follow from Corollary 2.6 (i) and Lemma 4.1. 2 To give more precise results for the cases of p = 3 and 5, we need the following result. Theorem 5.2. (See [11].) Assume that K is a totally real number ﬁeld with [K : Q] = d. √ Let p be an odd prime and n a positive integer. Let F = K( δ), with δ ∈ K ∗ \ K ∗2 be a relative quadratic extension such that F is a totally real or CM ﬁeld. Assume further that F ∩ Q(ζpn ) = Q and p h(K(ζpn + ζp−1 n )). Let r be the p-rank of the ideal class group √ √ −1 of K δ, ζpn + ζpn and s be the p-rank of the ideal class group of K δ(ζpn − ζp−1 n ) . If δ > 0, then 1 r ≤ s ≤ r + dpn−1 (p − 1). 2

2

For p = 3 and 5, we have:

√ Theorem 5.3. Let F = Q A(D + B D) be a quartic ﬁeld. √ (i) If 3 h(Q( −3D)), then ⎧ ⎪ ≤ rank3 (Cl(E1 )) + 2, if D ≡ 1(mod 3), 3|A and ⎪ ⎪ ⎪ ⎨ −1 (mod 3), if 3|B, A rank3 (K2 OF ) 3 ≡ ⎪ 1 (mod 3), if 3 B, ⎪ ⎪ ⎪ ⎩ otherwise. = rank3 (Cl(E1 )),

√ −3A(D + B D) .

√ (ii) If A < 0 and 5 h Q −D(5 + 2 5 , then where E1 = Q

rank5 (K2 OF ) ≤ rank5 (Cl(E )) where E = Q

√

5A(D + B D) .

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263

√ A(D + B D), ζ3 . By Corollary 3.3, if

Proof. (i) For p = 3, let E = Q

A D ≡ 1(mod 3), 3|A and − ≡ 3

1 (mod 3), if 3|B, −1 (mod 3), if 3 B,

we have ED F , ef = 1 and [FD : Q] = 2, hence by Theorem 2.5 (ii) we get rank3 (K2 OF ) ≤ rank3 (ε1 Cl(E)3 ) + 2. Let

√ √ A(D + B D) −→ A(D + B D);

√ √ τ1 : ζ3 − → ζ3 , A(D + B D) −→ − A(D + B D). σ : ζ3 −→ ζ32 ,

Then we have E1 = E τ1 σ = Q Clearly we have

√ −3A(D + B D) .

ε1 Cl(E)3 = η0 ε1 Cl(E)3 ⊕ η1 ε1 Cl(E)3 , √ √ η0 ε1 Cl(E)3 ⊆ η0 Cl(E)3 = Cl(Q( D, −3))3 . From the class number product formula for a biquadratic ﬁeld, we have √ √ √ √ √ Q h(Q( D, −3)) = h(Q( D))h(Q( −3D))h(Q( −3)) 2 where Q = 1 or 2. And from the reﬂection theorem of Scholz [14], we have √ √ rank3 (Cl(Q( D))) ≤ rank3 (Cl(Q( −3D))). √ √ √ √ Since 3 h(Q( −3D)) and h(Q( −3)) = 1, we have Cl(Q( D, −3))3 = 0, so η0 ε1 Cl(E)3 = 0. Hence we get ε1 Cl(E)3 = η1 ε1 Cl(E)3 . By Lemma 2.10 (ii), we get that η1 ε1 Cl(E)3 =

1 1 η1 NE1 Cl(E)3 = η1 Cl(E1 )3 = η1 Cl(E1 )3 . 2 2

√ While η0 Cl(E1 )3 = Cl(Q( D))3 = 0, hence η1 Cl(E1 )3 = η0 Cl(E1 )3 ⊕ η1 Cl(E1 )p = Cl(E1 )3 . Therefore, we get ε1

Cl(E)3 = Cl(E1 )3 . √ (ii) For p = 5, let E = Q A(D + B D), ζ5 . Let

Then we have E = E τ2 σ = Q

√

√ A(D + B D);

√ √ τ2 : ζ5 − → ζ5 , A(D + B D) −→ − A(D + B D). σ : ζ5 −→

ζ52 ,

A(D + B D) −→

√ 5A(D + B D) .

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By Theorem 5.1, we have rank5 (K2 OF ) = rank5 (ε3 Cl(E)5 ).

(5.1)

√ Note that η0 Cl(E)5 = Cl(Q( D, ζ5 ))5 . √ √ √ For the ﬁeld Q( D), we have Q( D) ∩ Q(ζ5 ) = Q and 5 h(Q( 5)). Hence, applying Theorem 5.2, we get

√ √ √ rank5 (Cl(Q( D, 5))) ≤ rank5 Cl Q −D(5 + 2 5) .

√ Since 5 h Q −D(5 + 2 5) , we get

√ √ √ rank5 (Cl(Q( D, 5))) = rank5 Cl Q −D(5 + 2 5) = 0.

√ √ √ √ Let E1 = Q( 5), E2 = Q( D, 5), E3 = Q −D(5 + 2 5) and E4 = Q(ζ5 ). From the proof of Theorem 5.2 [11], we have √ Cl(Q( D, ζ5 ))5 = Cl(E2 )5 ⊕ Cl(E3 )5 ⊕ Cl(E4 )5 . √ Hence we have Cl(Q( D, ζ5 ))5 = 0 and rank5 (ε3 Cl(E)5 ) = rank5 (η1 ε3 Cl(E)5 ). By Lemma 2.10 (iv), we know rank5 (ε3 Cl(E)5 ) = rank5 (η1 ε3 Cl(E)5 ) ≤ rank5 (η1 ε2 Cl(E)5 ). By Lemma 2.10 (iii), from (5.1) and (5.2) we have rank5 (K2 OF ) ≤ rank5 Cl(E ).

2

√ Corollary 5.4. Let F = Q A(D + B D) be as above. (i) If the integers A, B, D do not satisfy the conditions: A D ≡ 1(mod 3), 3|A, − ≡ 3

1 (mod 3), if 3|B,

−1 (mod 3), if 3 B,

then we have 3|#K2 OF ⇐⇒ 3|#Cl(E1 ), where E1 = Q

√ −3A(D + B D) .

(5.2)

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265

√ (ii) If A < 0 and 5 h Q −D(5 + 2 5 , then we have 5| #K2 OF =⇒ 5| #Cl(E2 ), where E2 = Q

√

5A(D + B D) .

2

5.2. The case of p = D

In this section we consider the cyclic quartic ﬁeld of the form F = Q Clearly we have p ≥ 5, since p ≡ 1(mod 4), and we have

√ A(p + B p) .

√ E = F (ζp ) = Q ζp , A∗ , where A∗ = (−1)

p−1 4

A.

√ Theorem 5.5. Let F = Q A(p + B p) be a cyclic quartic ﬁeld. Then ⎧ ⎨= rank (¯ if p > 5, or p = 5, ( A5 ) = −1, p ε p−3 Cl(E)p ), 2 rankp (K2 OF ) ⎩≤ rankp (¯ ε p−3 Cl(E)p ) + 1, if p = 5, A = −1, ( A5 ) = 1. 2

Proof. If p > 5 or p = 5, ( A5 ) = −1, then we have |S | = 0, i.e. ef > 1, so it follows from Theorem 2.5 (i). √ If p = 5, A = −1, ( A5 ) = 1, then we have ED = Q( A∗ ) F , so [FD : Q] = 1 and 5 totally ramiﬁes in F , hence ef = 1. Then by Theorem 2.5 (ii), we have rankp (K2 OF ) ≤ ε p−3 Cl(E)p ) + 1. 2 rankp (¯ 2

When p = 5, A = −1, we can give more precise computation. In this case, we have

√ A(5 + 2 5) ,

√ √ A(5 + 2 5), −A , E = F (ζ5 ) = Q √ √ Gal(E/F ) = {1, ς | ς( −A) = − −A, ς|F = id}. F =Q

It is easy to see that ς is the complex conjugation of Q(ζ5 ). Let

√ √ √ √ τ3 : A(5 + 2 5) −→ − A(5 + 2 5), −A −→ −A. Then we have

√ √ √ E τ3 = Q( 5, −A), E ς = Q A(5 + 2 5) , E ςτ3 = Q(ζ5 ).

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266

From Theorem 5.5, we have rank5 (K2 OF ) = rank5

1 − ς

A Cl(E)5 , for = −1. 2 5

We can see that the restriction of ς on Q(ζ5 ) is the σ 2 of the Galois group Gal(Q(ζ5 )/Q). The next result shows that ε¯1 Cl(E)5 can be expressed by the class group of the biquadratic subﬁeld of E using the automorphism of order 2 of Gal(F/Q) i.e. τ3 . Proposition 5.6.

1−ς 2 Cl(E)5

= Cl(E τ3 )5 .

Proof. Firstly, we have 1−ς 1 − ς 1 + τ3 1 − τ3 1 − ς Cl(E)5 = · Cl(E)5 ⊕ · Cl(E)5 . 2 2 2 2 2 It is easy to see that 1 + τ3 1 − ς 1 + τ3 1 − ς 1 + τ3 · Cl(E)5 = · Cl(E)5 ⊆ Cl(E)5 2 2 2 2 2 1 = NE/E τ3 Cl(E)5 = Cl(E τ3 )5 . 2 On the other hand, if x ∈ Cl(E τ3 )5 , then τ3 (x) = x. Since 1 + ς acts on Cl(E τ3 ) as 1+τ3 the norm NE τ3 /Q(√5) , so (1 + ς)x = 0, i.e. ςx = −x. Hence we have x = 1−ς 2 · 2 x. 1−ς 1+τ3 τ3 Consequently, 2 · 2 Cl(E)5 = Cl(E )5 . Clearly we have (1 − ς)(1 − τ3 ) = (1 − ς)(1 + ςτ3 ). So we get 1 − τ3 1 + ςτ3 1 − τ3 1 − ς · Cl(E)5 = · Cl(E)5 ⊆ Cl(E ςτ3 )5 = Cl(Q(ζ5 ))5 = 0. 2 2 2 2 Hence

1−ς 2 Cl(E)5

= Cl(E τ3 )5 .

2

√ √ From Proposition 5.6 and in virtue of Cl(E τ3 )5 = Cl(Q( −A))5 ⊕ Cl(Q( −5A))5 , we have

√ Theorem 5.7. Let F = Q A(5 + 2 5) . Then (i) If A = −1, ( A5 ) = 1, then √ √ rank5 (K2 OF ) ≤ rank5 Cl(Q( −A)) + rank5 Cl(Q( −5A)) + 1. (ii) If ( A5 ) = −1, then √ √ rank5 (K2 OF ) = rank5 Cl(Q( −A)) + rank5 Cl(Q( −5A)).

2

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Corollary 5.8. Let F = Q

267

√ A(5 + 2 5) . If ( A5 ) = −1, then

√ √ 5| #K2 OF ⇐⇒ 5| max{#Cl(Q( −A))5 , #Cl(Q( −5A))5 }.

2

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