Notes on class field theory

Choose \(u \in \gotho_L\) generating the residue field of \(L\) over the residue field of \(K\text{,}\) and let \(P(x)\) be the minimal polynomial of \(u\) over \(K\text{.}\) Then over the residue field of \(K\text{,}\) \(P(x)\) divides the \((q-1)\)-st cyclotomic polynomial, so by Hensel's lemma it splits over \(K(\zeta_{q-1})\text{.}\) Hence \(L \subseteq K(\zeta_{q-1})\text{,}\) and equality follows by comparing degrees.

Let \(\pi_L\) be a uniformizer of \(L\text{.}\) Then \(\pi_L^e\) can be written as a product of a uniformizer \(\pi\) of \(K\) times an element \(u\) of \(\gotho_L\) congruent to 1 modulo \(\pi_L\text{.}\) By Hensel's lemma, \(u\) has an \(e\)-th root in \(L\text{,}\) as then does \(\pi\text{.}\)

Since \(\Gal(K/\QQ_p)\) 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 \(\QQ_p\) whose Galois groups are cyclic of prime-power order. In other words, it suffices to prove local Kronecker-Weber under the assumption that \(\Gal(K/\QQ_p) \cong \ZZ/q^r \ZZ\) for some prime \(q\) and some positive integer \(r\text{.}\) We split this discussion into three cases; see Lemma 1.3.5, Lemma 1.3.6, and Lemma 1.3.7.

To begin, note that \(\QQ_p(\zeta_p)/\QQ_p\) is totally tamely ramified of degree \(p-1\text{,}\) so by Lemma 1.3.3 it has the form \(\QQ_p(c^{1/(p-1)})\) for some \(c \in p \ZZ_p^*\text{.}\) (The value of \(c\) won't be critical here, but see Lemma 1.3.8 for later reference.)

Let \(L\) be the maximal unramified subextension of \(K\text{.}\) By Lemma 1.3.2, \(L = \QQ_p(\zeta_n)\) for some \(n\text{.}\) Let \(e = [K:L]\text{.}\) Since \(e\) is a power of \(q\text{,}\) \(e\) is not divisible by \(p\text{,}\) so \(K\) is totally and tamely ramified over \(L\text{.}\) Thus by Lemma 1.3.3, there exists \(\pi \in L\) generating the maximal ideal of \(\gotho_L\) such that \(K = L(\pi^{1/e})\text{.}\) Since \(L/\QQ_p\) is unramified, \(p\) also generates the maximal ideal of \(\gotho_L\text{,}\) so we can write \(\pi = cu\) for some unit \(u \in \gotho_L^*\text{.}\) Now \(L(u^{1/e})/L\) is unramified since \(e\) is prime to \(p\) and \(u\) is a unit. In particular, \(L(u^{1/e})/\QQ_p\) is unramified, hence abelian. Then \(K(u^{1/e})/\QQ_p\) is the compositum of the two abelian extensions \(K/\QQ_p\) and \(L(u^{1/e})/\QQ_p\text{,}\) so it's also abelian. Hence any subextension is abelian, in particular \(\QQ_p(c^{1/e})/\QQ_p\text{.}\)

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

Now \(K \subseteq L(c^{1/e}, u^{1/e})\) as noted above. But on one hand, \(L(u^{1/e})\) is unramified over \(L\text{,}\) so \(L(u^{1/e}) = L(\zeta_m)\) for some \(m\text{;}\) on the other hand, because \(e|(p-1)\text{,}\) we have \(\QQ_p(c^{1/e}) \subseteq \QQ_p(c^{1/(p-1)}) = \QQ_p(\zeta_p)\text{.}\) Putting it all together,

Suppose \(\Gal(K/\QQ_p) \cong \ZZ/p^r\ZZ\text{.}\) We can use roots of unity to construct two other extensions of \(\QQ_p\) with this Galois group. Namely, \(\QQ_{p}(\zeta_{p^{p^r}-1})/\QQ_p\) is unramified of degree \(p^r\text{,}\) and automatically has cyclic Galois group; meanwhile, the index \(p-1\) subfield of \(\QQ_p(\zeta_{p^{r+1}})\) is totally ramified with Galois group \(\ZZ/p^r\ZZ\text{.}\) By assumption, \(K\) is not contained in the compositum of these two fields, so for some \(s>0\text{,}\)

This group admits \((\ZZ/p\ZZ)^3\) as a quotient, so we have an extension of \(\QQ_p\) with Galois group \((\ZZ/p\ZZ)^3\text{.}\) We rule this out using Lemma 1.3.9.

This is similar to Lemma 1.3.6, but a bit messier because \(\QQ_2\) does admit an extension with Galois group \((\ZZ/2\ZZ)^3\text{.}\) We defer this case to the exercises; see Exercise 4, Exercise 5, and Exercise 6.

See Exercise 1.

For convenience, put \(\pi = \zeta_p - 1\text{.}\) Then \(\pi\) is a uniformizer of \(\QQ_p(\zeta_p)\text{.}\)

If \(\Gal(K/\QQ_p) \cong (\ZZ/p\ZZ)^3\text{,}\) then \(\Gal(K(\zeta_p)/\QQ_p(\zeta_p)) \cong (\ZZ/p\ZZ)^3\) as well, and \(K(\zeta_p)\) is abelian over \(\QQ_p\) with Galois group \((\ZZ/p\ZZ)^* \times (\ZZ/p\ZZ)^3\text{.}\) Applying Kummer theory to \(K(\zeta_p)/\QQ_p(\zeta_p)\) produces a subgroup \(B \subseteq \QQ_p(\zeta_p)^*/(\QQ_p(\zeta_p)^*)^p\) isomorphic to \((\ZZ/p\ZZ)^3\) such that \(K(\zeta_p) = \QQ_p(\zeta_p, B^{1/p})\text{.}\) Let \(\omega: \Gal(\QQ_p(\zeta_p)/\QQ_p) \to (\ZZ/p\ZZ)^*\) be the canonical map; since \(\QQ_p(\zeta_p, b^{1/p}) \subseteq K(\zeta_p)\) is also abelian over \(\QQ_p\text{,}\) by Lemma 1.2.11,

Recall the structure of \(\QQ_p(\zeta_p)^*\text{:}\) the maximal ideal of \(\ZZ_p[\zeta_p]\) is generated by \(\pi\text{,}\) while each unit of \(\ZZ_p[\zeta_p]\) is congruent to a \((p-1)\)-st root of unity modulo \(\pi\text{,}\) and so

where \(U_1\) denotes the set of units of \(\ZZ_p[\zeta_p]\) congruent to 1 modulo \(\pi\text{.}\) Correspondingly,

Now choose a representative \(a \in L^*\) of some nonzero element of \(B\text{;}\) without loss of generality, we may assume \(a = \pi^m u\) for some \(m \in \ZZ\) and \(u \in U_1\text{.}\) Then

but \(v_\pi(\pi) = v_\pi(\zeta_p^{\omega(g)}-1) = 1\text{.}\) Thus the valuation of the right hand side is \(m(1-\omega(g))\text{,}\) which can only be a multiple of \(p\) for all \(g\) if \(m \equiv 0 \pmod{p}\text{.}\) (Notice we just used that \(p\) is odd!) That is, we could have taken \(m=0\) and \(a = u \in U_1\text{.}\)

As for \(u^g/u^{\omega(g)}\text{,}\) note that \(U_1^p\) is precisely the set of units congruent to 1 modulo \(\pi^{p+1}\) (see Exercise 2). Since \(\zeta_p = 1 + \pi + O(\pi^2)\text{,}\) we can write \(u = \zeta_p^b(1 + c\pi^d + O(\pi^{d+1}))\text{,}\) with \(c \in \ZZ\) and \(d \geq 2\text{.}\) Since \(\pi^g/\pi \equiv \omega(g) \pmod{\pi}\text{,}\) we get

But these two have to be congruent modulo \(\pi^{p+1}\text{.}\) Thus either \(d \geq p+1\) or \(d \equiv 1 \pmod{p-1}\text{,}\) the latter only occurring for \(d=p\text{.}\)

What this means is that the set of possible \(u\) is generated by \(\zeta_p\) and by \(1 + \pi^p\text{.}\) But these only generate a subgroup of \(U_1/U_1^p\) isomorphic to \((\ZZ/p\ZZ)^2\text{,}\) whereas \(B \cong (\ZZ/p\ZZ)^3\text{.}\) Contradiction.