On the p-rank of tame kernel of number fields

On the p-rank of tame kernel of number fields

Journal of Number Theory 158 (2016) 244–267 Contents lists available at ScienceDirect Journal of Number Theory www.elsevier.com/locate/jnt On the p...

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Journal of Number Theory 158 (2016) 244–267

Contents lists available at ScienceDirect

Journal of Number Theory www.elsevier.com/locate/jnt

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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262

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

Acknowledgment We are grateful to Professor Jerzy Browkin for his many helpful suggestions. References [1] J. Browkin, On the p-rank of the tame kernel of algebraic number fields, J. Reine Angew. Math. 432 (1992) 135–149. [2] J. Browkin, Tame kernels of cubic cyclic fields, Math. Comp. 74 (2005) 967–999. [3] J. Browkin, H. Gangl, Table of tame and wild kernels of quadratic imaginary number fields, Math. Comp. 68 (1999) 291–305. [4] J. Coates, On K2 and some classical conjectures in algebraic number theory, Ann. of Math. 95 (1972) 99–116. [5] K. Hardy, R. Hudson, D. Richman, K. Williams, N. Holtz, Calculation of the class numbers of imaginary cyclic quartic fields, Math. Comp. 49 (1987) 615–620. [6] R. Hudson, K. Williams, The integers of a cyclic quartic field, Rocky Mountain J. Math. 20 (1990) 145–150. [7] F. Keune, On the structure of K2 of the ring of the integers in a number field, K-Theory 2 (1989) 625–645. [8] M. Kolster, Odd torsion in the tame kernel of totally real number fields, in: Algebraic K-Theory: Connect. with Geom. Topol., 1989, pp. 177–188. [9] M. Kolster, K2 of rings of algebraic integers, J. Number Theory 42 (1992) 103–122. [10] H. Qin, Tame kernels and Tate kernels of quadratic number fields, J. Reine Angew. Math. 530 (2001) 105–144. [11] H. Qin, Reflection theorems and the p-Sylow subgroup of K2 OF for a number field F , J. Pure Appl. Algebra 214 (2010) 1181–1192. [12] H. Qin, H. Zhou, The 3-Sylow subgroup of the tame kernel of real number fields, J. Pure Appl. Algebra 209 (2007) 245–253. [13] J. Tate, Relation between K2 and Galois cohomology, Invent. Math. 36 (1976) 257–274. [14] L. Washington, Introduction to Cyclotomic Fields, Graduate Texts in Mathematics, vol. 83, Springer, 1982. [15] H. Zhou, The tame kernel of multiquadratic number fields, Comm. Algebra 37 (2009) 630–638.