CAPÍTULO III. ASPECTOS METODOLÓGICOS
3.3 Proceso de investigación
Proposition 3.5.3. Let K be a number field with [K : Q] = n. Let E/K be a
nilpotent extension. Suppose the primes in K that are ramified in E/K do not lie
over 2 or any integral prime that ramifies in K/Q. Let C be the class group of K
and a the number of primes that ramify in E/K. Then, the number of generators
of the Galois group of E/K is at most d(C) +n+a.
Proof. Since the number of generators required in a nilpotent group is the maximum
of the number of generators of one of its Sylow subgroups, it suffices to prove this for p-groups. By the Burnside basis theorem, it then suffices to prove this for abelian p-groups. Let m be the product of the primes ramifying in E/K. Any abelian
p-power extension ramified only at primes dividing m is in a ray class field for mk for some k. If some prime P overp divides m, then by Corollary 4.2.11 in [5],
OK/Pk
∗ ∼
= Z/(pf −1)Z×(Z/pqZ)(r+1)f × Z/pq−1Z(e−r−1)f
where k+e−2 =eq+r,0≤r < e. Since pis unramified by assumption,
OK/Pk
∗ ∼
= Z/(pf −1)Z× Z/pk−1Zf ∼= Z/(pk−1)(pf −1)Z× Z/pk−1Zf−1.
This contributes at most 1 +f−1 =f generators to OK/mk
∗
. Ifg is the number of primes p splits into in K, even if all of them divide m, this contributes at most gf =n generators to OK/mk
∗
. For the primes L overl that divide m but do not lie over p, we have
OK/Lk
∗ ∼
= Z/(lf −1)Z×(Z/lqZ)(r+1)f × Z/lq−1Z(e−r−1)f.
This contributes to the p-part of OK/mk
∗
if and only ifp divideslf−1, in which case it contributes at most one generator since Z/(lf −1)
Z is cyclic. Hence, the
number of generators of the p-part of OK/mk
∗
is at most n+a, where a is the number of primes dividing m. By Proposition 3.2.3 in [5], C is isomorphic to the ray class group modulo some homomorphic image of OK/mk
∗
, which implies that the p-part of the ray class group requires at mostd(C) +n+agenerators. Note that
in the case p= 2, we do not allow primes lying over 2 to ramify, so we can subtract n from this bound. However, the potentially n infinite places can each contribute a
Z/2Zfactor to the ray class group and so we need to add n back to the bound.
Remark 3.5.4. Proposition 3.5.3 in the case of K = Q says that for nilpotent
extensions, the number of generators required for the Galois groups is at most 1 plus the number of ramified primes. If we allow 2 to ramify and only count finite primes, this upper bound is realized as demonstrated by the example of Q(ζ8)/Q, in which 2 is the only ramified finite prime, with Galois group Z/2Z×Z/2Z.
Example 3.5.5. Let K be a number field with a cyclic class group C. Let p ≥ 3 be a prime that splits completely in K/Q with gcd (|C|, p·(p−1)) = 1. LetP be any prime in OK lying over p. Then, any nilpotent extension E/K ramified only at Pis cyclic. Note here that ifK is totally real, then we do not allow any infinite place to ramify in E/K.
First consider an abelian extension E/K ramified only at P. Then it is in the ray class group corresponding to the modulus Pk for some k ∈N. By Proposition 3.2.3 in [5], C is isomorphic to this ray class group modulo some homomorphic image of OK/Pk ∗ . By Corollary 4.2.11 in [5], OK/Pk ∗ ∼ = Z/(pf −1)Z×(Z/pqZ)(r+1)f × Z/pq−1Z(e−r−1)f.
completely inK/Q. So,r= 0 as well sincer < eandq =k−1. Hence, (e−r−1)f = 0, (r+ 1)f = 1, and
OK/Pk
∗ ∼
= (Z/(p−1)Z)× Z/pk−1Z∼=Z/ (p−1)·pk−1Z.
Any homomorphic image, say N, of OK/Pk
∗
is a quotient of OK/Pk
∗
and so is also cyclic of order dividing (p−1)·pk−1. Thus, the ray class group modulo a cyclic group of order dividing (p−1)·pk−1,N, is isomorphic toC, a cyclic group of order prime to (p−1)·p. By Schur-Zassenhaus, the ray class group is isomorphic to some semidirect product, NoC. Since the ray class group is abelian, the semidirect product must be a direct product,N×C. SinceN andCare both cyclic of coprime order, N ×C is also cyclic.
Now suppose that E/K is a nilpotent extension ramified only at P. By the Burnside basis theorem, if we take the quotient of the Galois group by the Frattini subgroup, we obtain an abelian extentsion of K ramified only at P. By the above argument, this extension is cyclic. Hence, the original Galois group must also be cyclic.
As an application of the above, consider a quadratic extension of Q with class number 1. Determining whether an unramified prime p ≥ 3 splits completely amounts to determining if it is a quadratic residue. If it is, then it splits and so the above applies to both primes above p in the quadratic field.
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