Notes on class field theory

Choose $u\in {\mathfrak{o}}_L$ 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 ${\mathfrak{o}}_L$ 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 $\mathrm{Gal}(K/{\mathbb{Q}}_p)$ 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 ${\mathbb{Q}}_p$ whose Galois groups are cyclic of prime-power order. In other words, it suffices to prove local Kronecker-Weber under the assumption that $\mathrm{Gal}(K/{\mathbb{Q}}_p)\cong \mathbb{Z}/q^r\mathbb{Z}$ 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 ${\mathbb{Q}}_p({\zeta }_p)/{\mathbb{Q}}_p$ is totally tamely ramified of degree $p-1\text{,}$ so by Lemma 1.3.3 it has the form ${\mathbb{Q}}_p(c^{1/(p-1)})$ for some $c\in p{\mathbb{Z}}_p^{\ast }\text{.}$ (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, $L={\mathbb{Q}}_p({\zeta }_n)$ 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 ${\mathfrak{o}}_L$ such that $K=L({\pi }^{1/e})\text{.}$ Since $L/{\mathbb{Q}}_p$ is unramified, $p$ also generates the maximal ideal of ${\mathfrak{o}}_L\text{,}$ so we can write $\pi =cu$ for some unit $u\in {\mathfrak{o}}_L^{\ast }\text{.}$ Now $L(u^{1/e})/L$ is unramified since $e$ is prime to $p$ and $u$ is a unit. In particular, $L(u^{1/e})/{\mathbb{Q}}_p$ is unramified, hence abelian. Then $K(u^{1/e})/{\mathbb{Q}}_p$ is the compositum of the two abelian extensions $K/{\mathbb{Q}}_p$ and $L(u^{1/e})/{\mathbb{Q}}_p\text{,}$ so it’s also abelian. Hence any subextension is abelian, in particular ${\mathbb{Q}}_p(c^{1/e})/{\mathbb{Q}}_p\text{.}$

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

This is similar to Lemma 1.3.6, but a bit messier because ${\mathbb{Q}}_2$ does admit an extension with Galois group $(\mathbb{Z}/2\mathbb{Z}{)}^3\text{.}$ 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 ${\mathbb{Q}}_p({\zeta }_p)\text{.}$

But these two have to be congruent modulo ${\pi }^{p+1}\text{.}$ Thus either $d\ge p+1$ or $d\equiv 1(\mathrm{mod}p-1)\text{,}$ 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 $(\mathbb{Z}/p\mathbb{Z}{)}^2\text{,}$ whereas $B\cong (\mathbb{Z}/p\mathbb{Z}{)}^3\text{.}$ Contradiction.