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III 2-class groups of CM fields 125

8.4 Proof of the Structure Theorem

haveλ= 1. Thus,λ=λ= 1 and the first claim follows

For the second claim recall from Lemma 8.2.2 that λ(Ad) = λ(A). Note that the groups Acn ∩A+n are of exponent 2. So if we know that the 2-class group of Ln,d is cyclic and that λ(Ad) = 1, then A+n contains at most 2 elements. As the capitulation kernelAn(L+n,d)→ An(Ln,d) contains at most 2 elements due to [Wash, Theorem 10.3], we see that the 2-class group ofL+n,dis uniformly bounded.

In order to prove our main result (Theorem 8.1.1) we will also need [C-A-Z, Theorem 5] and [C, Theorem 1] which are summarized in the following Theorem.

Theorem 8.3.6. Let n≥1 and assume thatd takes one of the following forms:

a) d=pq, for two distinct primes p and q congruent to 3 (mod 8).

b) d, is a prime congruent to 9 (mod 16).

Then the rank of the 2-class group of Ln,d is2.

8.4 Proof of the Structure Theorem

We will first compute the cardinality of h2(L1,d) and then use this result to prove Theorem 8.1.1.

Lemma 8.4.1. Let dbe an odd positive square-free integer. We have:

a) h2(L1,d) = 2·h2(−d), if d = pq, for two distinct primes p ≡ 5 (mod 8) and q≡3 (mod 8).

b) h2(L1,d) = h2(−2d), if d= pq, for two distinct primes p ≡ q ≡3 (mod 8) or d=p for a primep such that p≡9 (mod 16) and (2p)4= 1.

Proof. The proof is basically a computation in units and class number formulas.

a) Denote by ε2pq the fundamental unit of the quadratic fieldQ(√

2pq). We have ε2pq =x+y√

2pqfor some integers x andy. Sinceε2pq has a positive norm we obtain x2−2pqy2 = 1. Thusx2−1 = 2pqy2. Set y=y1y2 for yi∈Z. Assume which is impossible. Sox±1 is not square inN. From the third and the fourth item of [A-Z-T, Proposition 3.3], we deduce thatq(L1,d) = 4. By Kuroda’s class

number formula (cf. [Wa, p. 201]), we have h2(L1,d) = 1

25q(L1,d)h2(pq)h2(−pq)h2(2pq)h2(−2qp)h2(2)h2(−2)h2(−1)

= 1

25q(L1,d)h2(pq)h2(−pq)h2(2pq)h2(−2qp)

= 1

25 ·4·2·h2(−pq)·2·4

= 2·h2(−pq), which proves the first claim.

b) Suppose now that dis the product of two primes p and q that congruent to 3 mod 8 then we obtain the desired result by [A-C-Z2, Corollary 2]. Ifd=p is a prime congruent to 9 (mod 16) the result follows from the proof of Theorem 1 of [A-C-Z2, p. 7].

Now we have all ingredients to provide a partial proof of our first main Theorem:

Theorem 8.4.2. Let d be of one of the following forms:

ˆ d=p be a prime congruent to9 (mod 16) and assume that(2p)4= 1.

ˆ d=pq for two primes congruent to3 (mod 8).

Let2r=h2(−2d).Then for n≥1 the2-class group ofLn,d is isomorphic to the group Z/2Z×Z/2n+r−2Z. In the projective limit we obtainZ2×Z/2Z.

Note that Theorem 8.4.2 is point a) of Theorem 8.1.1

Proof. By Theorem 8.3.6 we know that the 2-rank of the 2-class group ofLn,dequals 2 forn≥1. Furthermore, we have λ = 1 due to Theorem 8.3.2, andh2(L1,d) = 2r by Lemma 8.4.1. By Theorem 8.3.3 the class number ofL+n,dis odd for alln. As there is no capitulation inAn according to Theorem 7.3.1 andλ= 1, we see thatAn has rank one for n large enough (see also Lemma 8.2.1). This implies that the second generator of the 2-class group ofLn,d is a class of a ramified prime in Ln,d/L+n,d. As the class number of L+n,d is odd, these ramified classes have order 2 and we obtain that the 2-class group ofLn,d is isomorphic toZ/2Z×Z/2lnZ.

LetE be the elementary Λ-module associated toA. Then according to [Wash, page 282-283]νn,0E= 2νn−1,0E for all n≥2. Indeed, νn,0n,n−1νn−1,0. AsE has Z2-rank 1, we know that T acts as 2v on E for some v ∈ Z2. For n ≥ 2 we have νn,n−1 = (T + 1)2n−1−1 + 2. Forn≥2 the term (T + 1)2n−1 −1 = T2n−1 +O(2T) acts as 4v0 on E for some v0 ∈ Z2. Hence, νn,n−1 acts as 2(1 + 2v0) = 2u on E for some unitu∈Z2 and all n≥2.

Consequently,

|E/νn,0E|=|E/2n−1E||E/ν1,0E|= 2n−1+c0

8.4. PROOF OF THE STRUCTURE THEOREM 145 for n ≥ 1 and some constant c0 ≥ 1 independent of n. Note that we can rewrite this as |E/νn,0E| = 2n+c. Since E has only one Z2-generator we can assume that the pseudo-isomorphism φ : A → E is surjective. The maximal finite submodule of A is generated by the classes (cn)n∈N of the ramified primes above 2. Let τ be a generator of Gal(Ld,∞/L0,d). Then τ(cn) = cn as the primes above 2 are totally ramified in L∞,d/L0,d. It follows that T cn = 0. Using that νn,0 is coprime to the characteristic ideal of A for all n, we obtain for every n ≥ 1 that the kernel of φ : An,0A → E/νn,0E is isomorphic to the maximal finite submodule in A

and contains 2 elements. Let Y be such that A/Y ∼= A0. Then An ∼= An,0Y (see [Wash, page 281]). We obtain

|An|=|An,0Y|=|An,0A||νn,0An,0Y|= 2n+c+1n,0An,0Y|forn≥1.

As the maximal finite submodule inAis annihilated byνn,0, we see that the size of νn,0An,0Y is constant independent ofn. Hence, we obtain that the 2-class group of Ln,d is of size 2n+ν for all n≥1. Using that h2(L1,d) = 2r, we obtain ν =r−1.

This yields 2·2ln = 2n+r−1 and we obtain ln = n+r−2. Noting that Ln,d is the n-th step of the cyclotomic Z2-extension of L0,d finishes the proof of the first claim.

As the direct term Z/2Z is norm coherent the second claim is immediate.

Corollary 8.4.3. Let d be of one of the following two forms:

a) d=p a prime congruent to9 (mod 16) and assume that(2p)4= 1, b) d=pq for two distinct primes congruent to 3 (mod 8).

Ifdtakes the first form, setp=u2−2v2 whereu andv are two positive integers such thatu≡1 (mod 8).

Ifdtakes the second form set(pq) = 1and let the integersX, Y, k, l andm be such that 2q=k2X2+ 2lXY + 2mY2 and p=l2−2k2m (see [Ka, p. 356] for their existence).

For all n≥1, we have:

a) Ifdis of the first form, then the 2-class group of Ln,d is isomorphic toZ/2Z× Z/2n+1Z, if and only if (up)4=−1.

Outside of this particular case it is isomorphic to the groupZ/2Z×Z/2n+r−2Z, where r≥4 was defined in Theorem 8.4.2.

b) If d is of the second form then the 2-class group of Ln,d is isomorphic to the groupZ/2Z×Z/2n+1Z if and only if (|X+lY−2 |) =−1.

Outside of this case it is isomorphic to Z/2Z×Z/2n+r−2Z, where r ≥ 4 was defined in Theorem 8.4.2.

Proof. By Lemma 8.4.1 we know that h2(L1,d) = h2(−2d). Since the 2-rank of Cl(L1,d) equals 2 and |Cl2(Ln,d)| 6= 4 (see [A-C-Z1, Theorem 5.7]) it follows that h2(−2d) is divisible by 8. Thus if dis as in point a) [L-W 1, Theorem 2] shows that h2(−2d) is not divisible by 16 and we obtain r= 3.

If d is in the second case then [Ka, pp. 356-357]) implies that h2(−2d) is not divisible by 16 and againr= 3.

Let us give the following example for Corollary 1.

Example 8.4.4. ˆ Set p = 89, u = 17 and v = 10. We have p =u2−2v2 and 2

p

4 = −

u p

4 = 1. So the 2-class group of Ln,p is isomorphic to Z/2Z× Z/2n+1Z for all n≥1.

ˆ Let p = 11, q = 19, k = 1, l = 3, m = −1, X = 4 and Y = 1. We have:

p=l2−2k2m and2q=k2X2+ 2lXY + 2mY2. Since −2

|X+lY|

= −27

=−1.

By Corollary 8.4.32-class group of Ln,p is isomorphic to Z/2Z×Z/2n+1Z for alln≥1.

Now we finish the proof of Theorem 8.1.1.

Theorem 8.4.5. Assume that d = pq is the product of two primes p ≡ −q ≡ 5 (mod 8)and2r= 2·h2(−pq). Then forn≥1the 2-class group of Ln,dis isomorphic toZ/2n+r−1Z.

Note that this is point 2 of Theorem 8.1.1.

Proof. We know already from Theorem 8.3.4 that the 2-class group of Ln,d is cyclic, and by Theorem 8.3.2 we know thatλ(L0,d) = 1. In particular, the module A does not contain a finite submodule and is hence isomorphic to its elementary moduleE.

LetY be defined as in the proof of Theorem 8.4.2, then there is no νn,0-torsion and we obtain that the size ofνn,0An,0Y is constant independent of n. As before we obtain |An|=|An,0A||νn,0An,0Y|= 2n+d.In particular, Iwasawa’s formula holds for alln≥1. Hence, h2(L1,d) = 2r= 21+ν and ν =r−1. From this the claim follows.

Corollary 8.4.6. Let d = pq be the product of two primes p and q such that p ≡

−q ≡5 (mod 8). Then for all n≥1, we have

a) If (pq) =−1, the 2-class group of Ln,d is isomorphic to Z/2n+1Z.

b) If (pq) = 1 and (qp)4= 1, the 2-class group of Ln,d is isomorphic to Z/2n+2Z. Outside of these two cases, the 2-class group of Ln,d is isomorphic to Z/2n+r−1Z, where r≥4 was defined in Theorem 8.4.5.

Proof. Assume that d is of the first form. By [Co-Hu, 19.6 Corollary] we have h2(−pq) ≡ 2 (mod 4). Hence, r = 2. If d is of the second form we know that Cl2(Q(√

−d)) is cyclic and divisible by 4 [Qi 1, page 1427]. Asp≡5 mod 8 we see that (2p) = −1 and therefore (4p)4 = −1. Then (4qp) = −1 and [Qi 1, Theorem 3.9]

implies thath2(−pq) is not divisible by 8. Hence, h2(−pq) = 4 andr = 3.

For the above corollary we provide the following Example 8.4.7. ˆ Let d= 13·19. We have 1319

=−1. So the 2-class group of Ln,p is isomorphic to Z/2n+1Z for all n≥1.

8.5. APPLICATIONS 147

ˆ Let d= 5·11. We have 115

= 1 and 115

4 = 1. So the 2-class group of Ln,p

is isomorphic to Z/2n+2Z for all n≥1.

Let nowX0,Y0 andZ be three positive integers satisfying the Legendre equation

pX02+qY02−Z2= 0 (8.1)

such that

(X0, Y0) = (Y0, Z) = (Z0, X0) = (p, Y0Z) = (q, X0Z) = 1, (8.2) and

X0 odd, Y0 even andZ ≡1 (mod 4). (8.3) (see [L-W2] for more details)

Corollary 8.4.8. Let d = pq be the product of two primes p and q satisfying p ≡

−q ≡ 5 (mod 8), (pq) = 1 and (−qp )4 = 1. Let X0, Y0 and Z be three positive integers satisfying equation(8.1)and the conditions(8.2)and (8.3). If (Zp)4 6= (2XZ0), the 2-class group of Ln,d is isomorphic to Z/2n+3Z. Otherwise, it is isomorphic to Z/2n+r−1Z, for r ≥5 defined as in Theorem 8.4.5.

Proof. It is immediate that the assumptions of Corollary 8.4.6 are not satisfied.

Hence, we have r ≥ 4. By [L-W2, Theorem 2] h2(−pq) is divisible by 16 if and only if

Z p

4 =

2X0 Z

. Hence if (Zp)46= (2XZ0) we get 4≤r <5. Otherwise we obtain that 2·16|2r andr ≥5.

Now we close this section with some numerical examples illustrating the above corollary

Example 8.4.9. ˆ Let p = 5, q = 19 and d =−pq. Then X0 = 1, Y0 = 2 and Z = 9 are solutions of equation (8.1) satisfying the condition (8.2) and (8.3).

Furthermore,(95)4=−(29) =−1. Thus, the 2-class group of Ln,d is isomorphic toZ/2n+3Z.

ˆ Let p = 37, q = 11 and d = −pq = −407. Then X0 = 1, Y0 = 56518 and Z = 187449 are solutions of equation (8.1) satisfying the condition (8.2) and (8.3). Furthermore, (18744937 )4 = (1874492 ) = 1. Thus, the 2-class group of Ln,d is isomorphic toZ/2n+r−1Z for somer ≥5. Indeed with these settingsr = 5 (see [L-W2, p. 230]).

8.5 Applications

Note that Theorem 7.3.1 and Theorem 7.2.5 only hold for CM fields containing the 4-th root of unityi. Therefore, we cannot compute the 2-class groups of layers of the cyclotomic Z2-extension of imaginary quadratic fields using the same techniques as in the proof of Theorem 8.1.1. But we can still use Theorem 8.4.2 to deduce results on the class field tower of imaginary quadratic fields.

Ln,d

Kn L+n,d Kn,d

K+n

Figure 8.1: Subfields ofLn,d/K+n.

Theorem 8.5.1. Letdbe a positive square-free integer andr such that2r=h2(−2d).

Let K0,d=Q(√

−d) and denote by Kn,d the n-th step of the cyclotomic Z2-extension of K0,d. Suppose that dtakes one of the following forms:

a) d=pq, for two distinct primes p and q congruent to 3 (mod 8).

b) d=p, for a prime p≡9 (mod 16) such that that (2p)4 = 1.

Then for alln≥1 the 2-class group of Kn,d is isomorphic to Z/2Z×Z/2n+r−1Z. In the projective limit we obtainZ2×Z/2Z.

Note that this is point 1 of Theorem 8.1.2.

Proof. LetKn=Q(ζ2n+2) andKn,d=Q(ζ2n+22−1n+2,√

−d) =K+n(√

−d) (see Figure 8.1). By the class number formula (cf. [Le 1, Proposition 3 and equation (1)]) we have:

h2(Ln,d) = QLn,d

QKnQKn,d · µLn,d

µKnµKn,d ·h2(Kn)h2(Kn,d)h2(L+n,d) h2(K+n)2 .

It is known thath2(Kn) = 1 and by Theorems 8.3.3 and 8.4.2 we have h2(L+n,d) = 1 andh2(Ln,d) = 2n+r−1, respectively. Therefore

2·2n+r−1= QLn,d

QKnQKn,d ·h2(Kn,d). (8.4) It is also known thatQKn = 1. Let k=Q(i,√

d). As the natural norm NL1,d/k :WL1,d/WL2

1,d →Wk/Wk2

is surjective, we obtain that QL1,d divides Qk (cf. [Le 1, Proposition 1]). Since Qk = 1 (cf. [Az, p. 19] and the proof of [A-C-Z2, Lemma 4]), we have QL1 = 1.

SinceNLn,d/Ln−1,d :WLn,d/W2

Ln →WLn−1/W2

Ln−1 is onto, it follows that QLn,d divides QLn−1,d. Thus, by inductionQLn,d = 1.

Note that the extensions Kn,d are essentially ramified (cf. [Le 1, p. 349]) for all n ≥ 1. Since µKn,d = 2 we obtain QKn,d = 1 by [Le 1, Theorem 1] . Hence, h2(Kn,d) = 2n+rfor alln≥1. Since the rank of the 2-class group ofK1,dequals 2 (cf.

[M-C-R, Proposition 4]) and the 2-class group of Kn,d is of type (2,2) for n large enough (cf [Miz, p. 119]), we achieve the result.

8.5. APPLICATIONS 149 Now using Theorem 8.4.5 we can finish the proof of Theorem 8.1.2.

Theorem 8.5.2. Assume that d = pq is the product of two primes p ≡ −q ≡ 5 (mod 8)and 2r= 2·h2(−pq). LetK0,d =Q(√

−d) and denote for n≥3 by Kn,d the n-th step of the cyclotomic Z2-extension of K0,d. Then forn≥1the 2-class group of Kn,d is isomorphic to Z/2n+r−1Z.

Note that this is point 2 of Theorem 8.1.2

Proof. We keep similar notations and proceed as in the proof of Theorem 8.5.1. Note that by [A-C-Z2, Proposition 3], we haveh2(L+n,d) = 2. So as above we obtain

h2(Ln,d) = QLn,d

QKnQKn,d · µLn,d

µKnµKn,d ·h2(Kn)h2(Kn,d)h2(L+n,d) h2(K+n)2 · Thus,

2n+r−1 = 1

1·1· 2n

2n·2 ·1·h2(Kn,d)·2

1 ·

It follows that h2(Kn,d) = 2n+r−1 for all n. Since Ln,d/Kn,d is ramified, it is obvi-ous that 2-rank(Cl(Kn,d)) ≤ 2-rank(Cl(Ln,d)) = 1 (Theorem 8.4.5). In particular, Cl(Kn,d) is cyclic.

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