Math 254B, UC Berkeley, Spring 2002 (Kedlaya) The Local Kronecker-Weber Theorem

Reference. Washington, Introduction to Cyclotomic Fields, Chapter 14.

We now prove the local Kronecker-Weber theorem, modulo some steps which will be left as exercises. As shown previously, this will imply the original Kronecker-Weber theorem.

Theorem 1 (Local Kronecker-Weber). If K/ℚ_{p} is a finite abelian extension, then K ⊆ ℚ_{p}(ζ_{n}) for some positive integer n.

Since Gal(K/ℚ_{p}) decomposes into a product of cyclic groups of prime power order, by the structure theorem for finite abelian groups, we may write K as the compositum of extensions of ℚ_{p} whose Galois groups are cyclic of prime power order. In other words, it suffices to prove local Kronecker-Weber under the assumption that Gal(K/ℚ_{p}) ≅ ℤ/q^{r}ℤ for some prime q and some r ∈ ℕ.

First recall the following facts from last semester.

Lemma 2. Let L/K be an unramified extension of finite extensions of ℚ_{p}. Then L = K(ζ_{q−1}), where q is the cardinality of the residue field of L.

Lemma 3. Let L/K be a totally and tamely ramified extension of finite extensions of ℚ_{p} of degree e. (Recall that tamely ramified means that p does not divide e.) Then there exists a generator π of the maximal ideal of the valuation ring of K such that L = K(π^{1/e}).

Lemma 4. The fields ℚ_{p}((−p)^{1/(p−1)}) and ℚ_{p}(ζ_{p}) are equal.

We now proceed to the proof of the local Kronecker-Weber theorem. Case 1: q ≠ p.

Let L be the maximal unramified subextension of K. By Lemma 2, L = ℚ_{p}(ζ_{n}) for some n. Let e = [K : L]. Since e is a power of q, e is not divisible by p, so K is totally and tamely ramified over L. Thus by Lemma 3, there exists π ∈ L generating the maximal ideal of 𝔬_{L} such that K = L(π^{1/e}). Since L/ℚ_{p} is unramified, p also generates the maximal ideal of 𝔬_{L}, so we can write π = −pu for some unit u ∈ 𝔬_{L}^{∗}. Now L(u^{1/e})/L is unramified since e is prime to p and u is a unit. In particular, L(u^{1/e})/ℚ_{p} is unramified, hence abelian. Then K(u^{1/e})/ℚ_{p} is the compositum of the two abelian extensions K/ℚ_{p} and L(u^{1/e})/ℚ_{p}, so it’s also abelian. Hence any subextension is abelian, in particular ℚ_{p}((−p)^{1/e})/ℚ_{p}.

For ℚ_{p}((−p)^{1/e})/ℚ_{p} to be Galois, it must contain the e-th roots of unity (since it must contain all of the e-th roots of −p, and we can divide one by another to get an e-th root of unity). But ℚ_{p}((−p)^{1/e})/ℚ_{p} is totally ramified, whereas ℚ_{p}(ζ_{e})/ℚ_{p} is unramified. This is a contradiction unless ℚ_{p}(ζ_{e}) is actually equal to ℚ_{p}, which only happens if e|p − 1 (since the residue field 𝔽_{p} of ℚ_{p} contains only (p − 1)-st roots of unity).

Now K ⊆ L((−p)^{1/e}, u^{1/e}) as noted above. But on one hand, L(u^{1/e}) is unramified over L, so L(u^{1/e}) = L(ζ_{m}) for some m; on the other hand, because e|(p − 1), we have ℚ_{p}((−p)^{1/e}) ⊆ ℚ_{p}((−p)^{1/(p−1)}) = ℚ_{p}(ζ_{p}) by Lemma 4. Putting it all together,

K ⊆ L((−p)^{1/e}, u^{1/e}) ⊆ ℚ_{p}(ζ_{n}, ζ_{p}, ζ_{m}) ⊆ ℚ_{p}(ζ_{mnp}).

Case 2: q = p ≠ 2.

Before proceeding, recall a lemma from Kummer theory (previous handout).

Lemma 5. Let K be a field of characteristic coprime to n, let L = K(ζ_{n}), and let M = L(a^{1/n}) for some a ∈ L^{∗}. Define the homomorphism ω : Gal(L/K) → (ℤ/nℤ)^{∗} by the relation ζ_{n}^{ω(g)} = ζ_{n}^{g}. Then M/K is Galois and abelian if and only if

a^{g}/a^{ω(g)} ∈ (L^{∗})^{n}  ∀g ∈ Gal(M/K).  (1)

Suppose Gal(K/ℚ_{p}) ≅ ℤ/p^{r}ℤ. We can use roots of unity to construct two other extensions of ℚ_{p} with this Galois group. Namely, ℚ_{p}(ζ_{p^{p^{r}} −1})/ℚ_{p} is unramified of degree p^{r}, and automatically has cyclic Galois group; meanwhile, the index p − 1 subfield of ℚ_{p}(ζ_{p^{r+1}}) is ramified with Galois group ℤ/p^{r}ℤ. By assumption, K is not contained in the compositum of these two fields, so Gal(K(ζ_{p^{p^{r}} −1}, ζ_{p^{r+1}})/ℚ_{p}) ≅ (ℤ/p^{r}ℤ)^{2} ×ℤ/p^{s}ℤ for some s > 0. This group admits (ℤ/pℤ)^{3} as a quotient, so we have an extension of ℚ_{p} with Galois group (ℤ/pℤ)^{3}. It thus suffices to prove the following lemma.

Lemma 6. For p ≠ 2, there is no extension of ℚ_{p} with Galois group (ℤ/pℤ)^{3}.

Proof. For convenience, put π = ζ_{p} − 1. Then π is a uniformizer of ℚ_{p}(ζ_{p}).

If Gal(K/ℚ_{p}) ≅ (ℤ/pℤ)^{3}, then Gal(K(ζ_{p})/ℚ_{p}(ζ_{p})) ≅ (ℤ/pℤ)^{3} as well, and K(ζ_{p}) is abelian over ℚ_{p} with Galois group (ℤ/pℤ)^{∗} × (ℤ/pℤ)^{3}. We can apply Kummer theory to K(ζ_{p})/ℚ_{p}(ζ_{p}); it gives us a subgroup B ⊆ ℚ_{p}(ζ_{p})^{∗}/(ℚ_{p}(ζ_{p})^{∗})^{p} isomorphic to (ℤ/pℤ)^{3} such that K(ζ_{p}) = ℚ_{p}(ζ_{p}, B^{1/p}). Let ω : Gal(ℚ_{p}(ζ_{p})/ℚ_{p}) → (ℤ/pℤ)^{∗} be the canonical map; by the previous lemma,

b^{g}/b^{ω(g)} ∈ (ℚ_{p}(ζ_{p})^{∗})^{p} ∀b ∈ B, g ∈ Gal(ℚ_{p}(ζ_{p})/ℚ_{p}),

since ℚ_{p}(ζ_{p}, b^{1/p}) ⊆ K(ζ_{p}) is also abelian over ℚ_{p}.

Recall the structure of ℚ_{p}(ζ_{p})^{∗}. The maximal ideal of ℤ_{p}[ζ_{p}] is generated by π, while each unit of ℤ_{p}[ζ_{p}] is congruent to a p − 1-st root of unity modulo π. Thus

ℚ_{p}(ζ_{p})^{∗} = π^{ℤ} × (ζ_{p−1})^{ℤ} × U_{1},

where U_{1} denotes the set of units of ℤ_{p}[ζ_{p}] congruent to 1 modulo π, and correspondingly

(ℚ_{p}(ζ_{p})^{∗})^{p} = π^{pℤ} × (ζ_{p−1})^{pℤ} × U_{1}^{p}.

Now choose a representative a ∈ L^{∗} of some nonzero element of B; without loss of generality, we may assume a = π^{m}u for some m ∈ ℤ and u ∈ U_{1}. Then

\frac{a^{g}}{a^{ω(g)}} = \frac{(ζ_{p}^{ω(g)} − 1)^{m}}{π^{mω(g)}} \frac{u^{g}}{u^{ω(g)}} ;

but v_{π}(π) = v_{π}(ζ_{p}^{ω(g)} − 1) = 1. Thus the valuation of the right hand side is m(1 − ω(g)), which can only be a multiple of p for all g if m ≡ 0 (mod p). (Notice we just used that p is odd!) That is, we could have taken m = 0 and a = u ∈ U_{1}.

As for u^{g}/u^{ω(g)}, note that U_{1}^{p} is precisely the set of units congruent to 1 modulo π^{p+1} (see exercises). Since ζ_{p} = 1 + π + O(π^{2}), we can write u = ζ_{p}^{b}(1 + cπ^{d} + O(π^{d+1})), with c ∈ ℤ and d ≥ 2. Since π^{g}/π ≡ ω(g) (mod π), we get

u^{g} = ζ_{p}^{bω(g)}(1 + cω(g)^{d}π^{d} + O(π^{d+1})), u^{ω(g)} = ζ_{p}^{bω(g)}(1 + cω(g)π^{d} + O(π^{d+1})).

But these two have to be congruent modulo π^{p+1}. Thus either d ≥ p+1 or d ≡ 1 (mod p−1), the latter only occurring for d = p.

What this means is that the set of possible u is generated by ζ_{p} and by 1 + π^{p}. But these only generate a subgroup of U_{1}/U_{1}^{p} isomorphic to (ℤ/pℤ)^{2}, whereas B ≅ (ℤ/pℤ)^{3}. Contradiction.

Case 3: p = q = 2. This case is similar to the previous case, but a bit messier, because ℚ_{2} does admit an extension with Galois group (ℤ/2ℤ)^{3}. We defer this case to the exercises.

Exercises.

1. Prove that (in the notation of Lemma 6) U_{1}^{p} is the set of units congruent to 1 modulo π^{p+1}. (Hint: in one direction, write u ∈ U_{1} as a power of ζ_{p} times a unit congruent to 1 modulo π^{2}. In the other direction, use the binomial series for (1 + x)^{1/p}.)

2. Prove that for any r > 0, there is an extension of ℚ_{2} with Galois group ℤ/2ℤ×(ℤ/2^{r}ℤ)^{2} contained in ℚ_{2}(ζ_{n}) for some n > 0.

3. Suppose that K/ℚ_{2} is a ℤ/2^{r}ℤ-extension not contained in ℚ_{2}(ζ_{n}) for any n > 0. Prove that there exists an extension of ℚ_{2} with Galois group (ℤ/2ℤ)^{4} or (ℤ/4ℤ)^{3}.

4. Prove that there is no extension of ℚ_{2} with Galois group (ℤ/2ℤ)^{4}. (Hint: use Kummer theory.)

5. Prove that there is no extension of ℚ_{2} with Galois group (ℤ/4ℤ)^{3}. (Hint: reduce to showing that there exists no extension of ℚ_{2} containing ℚ_{2}(\sqrt{−1}) with Galois group ℤ/4ℤ.)