On the other hand, in the non-split case, Gal (L/K) can be arbitrarily large. For example, in the case of the prime 2, the ray class number can be any arbitrary power of 2:
Proposition 2.4.9. Suppose that the class number of K = Q( √
d) is 1, where
d ≡ 5 ( mod 8) is a positive prime. Let u = a+b √
d
2 be the fundamental unit of K,
and assume that a, b are both odd. Then the ray class number of K mod 2k for any positive integer k is as follows:
|ClK2k|=
1 for k = 1, 2k for k ≥2.
Proof. Under the above assumption, we know 2 is inert in K with residue degree f = 2. Again we look at the following exact sequence:
1→ O+∗/O+2k →(OK/2k)∗ →Cl2Kk →ClK1 →1.
We know |(OK/2k)∗|= (pf −1)·pf(k−1) = 3·22(k−1) and O+∗ ={u2n}. For O2
k
+, let
un = an+bn√d
2 . Since a, b are odd, by Lemma 4 of page 173 of [Ri], an, bn are even
if and only if 3 divides n. So u3 = a3+b3
√ d
2 with a3, b3 even. Let a3 = 2a
0
3, b3 = 2b03.
So −1 = N(u3) = a02 3 −b
02
3d. Working modulo 4, we know a
02
3 ≡ 0 ( mod 4) and
b023d≡1 ( mod 4). So we can write a03 = 2a300, b03 = 2b003 + 1 and d= 8k+ 5. We get 4a0032−4b003(b003+ 1) = 8k+ 4, soa003 ∈/ 2OK. Now writeu3 = 2t+ 1 witht= a03−1+b03
√ d
2 .
Then N(t) = −a003 and N(t+ 1) =a003, which are both odd. So we have u6 = 4t(t+ 1) + 1≡1 ( mod 22) and u6 6= 1 ( mod 23). Using induction we can get
u3·2k−1 ≡1 ( mod 2k) and u3·2k−1 6= 1 ( mod 2k+1). So |O∗
+/O2
k
+|= 3·2k−2 and we have the following:
Case (k = 2). We know |O∗ +/O+2|= 3. So |Cl2K|= |(OK/2) ∗| · |Cl1 K| |O∗ +/O2+| = 1. Case (k ≥2). We know |O∗ +/O2 k +|= 3·2k−2, so |ClK2k|= |(OK/2 k)∗||Cl1 K| |O∗ +/O2 k +| = 3·2 2(k−1) 3·2k−2 = 2 k
Remark 2.4.10. In fact if we remove the condition the class number of K is 1, following the same proof above we can get
|Cl2Kk|=
cl1
K =|Cl1K| for k= 1, 2kcl1K = 2k|ClK1 | fork ≥2.
So the ray class field of K mod 2 is just the big Hilbert class field of K.
By Corollary 1.4.2, every group inπA(U2) is a quasi-2 group. We have a similar
statement below for Galois groups over quadratic fields ramified only at one prime. Define πA(UQ(√d),p) to be the set of finite quotients of π1(UQ(√d),p), where p is a prime in Q(√d), and U
Q(√d),p = SpecOQ(√d) − {p}. So πA(UQ(√d),p) consists of finite groups that can occur as a Galois group over Q(√d) with ramification only at p.
Proposition2.4.11. Letd≡1 ( mod 4), letu= a+b
√ d
2 be the fundamental unit
of K = Q(√d), and suppose a, b are both odd. Suppose p is the prime in Q(√d)
above p= 2. Then every group in πA(UQ(√
d),p) is a quasi-2 group.
Proof. Take any group G∈πA(UQ(√d),p); so G/p(G) is of odd order, thus solvable. If G/p(G) is nontrivial, then there exists a nontrivial abelian odd extension of
Q( √
d) ramified only at p. This is impossible, since we know by Proposition 2.4.2 and Proposition 2.4.9 that the ray class number of K mod pk for any integer k is a power of 2 , which cannot be odd. Thus G/p(G) is trivial, i.e. G is a quasi-2 group.
Theorem 2.4.12. Let K = Q(
√
d) be a real quadratic field, and suppose p re- mains prime in K. Let Rk be the ray class field of K mod pk.
a) Then for k >>0, Rk+1 =Rk(ζpk+1).
b) Suppose p = 2, and the fundamental unit of K is u = a+b √
d
2 with a, b both
odd. Let H be the big Hilbert class field of K. Then Rk = H(
√
2u, ζ2k) if k ≥ 2,
and Rk=H if k = 1.
Proof. First we look at the case p= 2. When k = 1, the ray class field is the big Hilbert class field by Remark 2.4.10. When k ≥ 2, the conductor of K(ζ2k)/K divides 2k−2, and the conductor of K(√2u) over K is 22, so the conductor of H(√2u, ζ2k)/Kdivides 2k. Also [H(√2u, ζ
2k) :K] = 2kwhich equals to the ray class
number, soH(√2u, ζ2k) is the ray class fieldRKofKmodpk, andRk+1 =Rk(ζpk+1).
Now we consider the casep >2. Again we look at the following exact sequence: 1→ O+∗/O+pk →(OK/pk)∗ →ClpKk →ClK1 →1.
We will computer the ratio |Cl pk+1 K | |ClpKk| , where |ClpKk|= |(OK/p k)∗||Cl1 K| |O∗ +/O pk +| = (p 2 −1)·p2(k−1)· |Cl1 K| |O∗ +/O pk +| . For |O∗ +/O pk
+|, we can always pick k big enough such that O+∗/O
pk + 6= O+∗/O pk+1 + . Take a generator ¯u0 in O+∗/O pk
+; and denote by us0 the smallest power of u0 such
that us
0 ≡ 1 mod pk. So we can write us0 = tpk+ 1, where t ∈ OK and p -t. Then usp0 = (tpk+ 1)p ≡1 mod pk+1, so |O ∗ +/O pk+1 + | |O∗ +/O pk +|
=pwhen k >>0. Now we have
|ClpKk+1| |ClpKk| = |(OK/pk+1)∗||Cl1 K| |(OK/pk)∗||Cl1K| |O∗ +/O pk+1 + | |O∗ +/O pk+1 + | = p 2 p =p,
so Rk+1/Rk is of degree p. While Q(ζpk+1)/Q(ζp) is of conductor pk, the conductor
of Rk(ζpk+1)/Rk divides pk and is of degreep, so Rk+1 =Rk(ζpk+1) fork >>0.