Notes on class field theory

On one hand, suppose that $\alpha \in K^{\ast }$ is such that ${\alpha }^{1/d}\notin K$ for any proper divisor $d$ of $n\text{.}$ Then every automorphism of $K({\alpha }^{1/n})$ over $K$ must have the form $\alpha \mapsto \alpha {\zeta }_n^r$ for some $r\in \mathbb{Z}/n\mathbb{Z}\text{;}$ this defines a homomorphism $\mathrm{Gal}(K({\alpha }^{1/n})/K)\cong \mathbb{Z}/n\mathbb{Z}$ which is injective because any automorphism of $K({\alpha }^{1/n})$ over $K$ is uniquely determined by its effect on ${\alpha }^{1/n}\text{.}$ We claim that this map is also surjective. If $n$ is prime, we can see this from the fact that by hypothesis $K({\alpha }^{1/p})\ne K\text{,}$ so the map $\mathrm{Gal}(K({\alpha }^{1/n})/K)\cong \mathbb{Z}/n\mathbb{Z}$ is an injective map from a nonzero group into a prime cyclic group and hence must be surjective. In the general case, note that the definition of the map $\mathrm{Gal}(K({\alpha }^{1/n})/K)\cong \mathbb{Z}/n\mathbb{Z}$ is compatible with replacing $n$ with one of its prime factors $p\text{,}$ and this logic tells us that the image of $\mathrm{Gal}(K({\alpha }^{1/n})/K)$ in $\mathbb{Z}/n\mathbb{Z}$ cannot be contained in $p\mathbb{Z}/n\mathbb{Z}$ for any prime divisor $p$ of $n\text{.}$ So again we conclude that $\mathrm{Gal}(K({\alpha }^{1/n})/K)\cong \mathbb{Z}/n\mathbb{Z}\text{.}$ (As a corollary, we deduce that the polynomial $x^n-\alpha$ is irreducible over $K\text{;}$ see Exercise 4 and Exercise 5 for discussion of what happens when ${\zeta }_n\notin K\text{.}$)

On the other hand, let $L$ be an arbitrary $\mathbb{Z}/n\mathbb{Z}$-extension of $K\text{.}$ Choose a generator $g\in \mathrm{Gal}(L/K)\text{,}$ and let $f:\mathrm{Gal}(L/K)\to L^{\ast }$ be the map that sends $rg$ to ${\zeta }_n^r$ for $r\in \mathbb{Z}\text{.}$ Then $f\in H^1(\mathrm{Gal}(L/K),L^{\ast })\text{,}$ so there exists $t\in L$ such that $t^{rg}/t=f(rg)={\zeta }_n^r$ for $r\in \mathbb{Z}\text{.}$ In particular, $t^n$ is invariant under $\mathrm{Gal}(L/K)\text{,}$ so $t^n=\alpha$ for some $\alpha \in K$ and $L=K(t)=K({\alpha }^{1/n})\text{,}$ as desired.

The map comes from the pairing; we have to check that it is injective and surjective. If $y\in K^{\ast }\setminus (K^{\ast }{)}^n\text{,}$ then $K(y^{1/n})$ is a nontrivial Galois extension of $K\text{,}$ so there exists some element of $\mathrm{Gal}(K(y^{1/n})/K)$ that doesn’t preserve $y^{1/n}\text{.}$ Any lift of that element to $\mathrm{Gal}(\bar{K}/K)$ pairs with $y$ to give something other than 1; that is, $y$ induces a nonzero homomorphism of $\mathrm{Gal}(\bar{K}/K)$ to $\mathbb{Z}/n\mathbb{Z}\text{.}$ Thus injectivity follows.

On the other hand, suppose $f:\mathrm{Gal}(\bar{K}/K)\to \mathbb{Z}/n\mathbb{Z}$ is a homomorphism whose image is the cyclic subgroup of $\mathbb{Z}/n\mathbb{Z}$ of order $d\text{.}$ Let $H$ be the kernel of $f\text{;}$ then the fixed field $L$ of $H$ is a $\mathbb{Z}/d\mathbb{Z}$-extension of $K$ with Galois group $\mathrm{Gal}(\bar{K}/K)/H\text{.}$ By Kummer theory, $L=K(y^{1/d})$ for some $y\text{.}$ But now the homomorphisms induced by $y^{mn/d}\text{,}$ as $m$ runs over all integers coprime to $d\text{,}$ give all possible surjective homomorphisms of $\mathrm{Gal}(\bar{K}/K)/H$ to $\mathbb{Z}/d\mathbb{Z}\text{,}$ so one of them must equal $f\text{.}$ Thus surjectivity follows.

If $a^g/a^{\omega (g)}\in (L^{\ast }{)}^n$ for all $g\in \mathrm{Gal}(L/K)\text{,}$ then $a\text{,}$ $a^{\omega (g)}$ and $a^g$ all generate the same subgroup of $(L^{\ast })/(L^{\ast }{)}^n\text{.}$ Thus $L(a^{1/n})=L((a^g{)}^{1/n})$ for all $g\in \mathrm{Gal}(L/K)\text{,}$ so $M/K$ is Galois. Thus it suffices to assume $M/K$ is Galois, then prove that $M/K$ is abelian if and only if (1.2.1) holds. In this case, we must have $a^g/a^{\rho (g)}\in (M^{\ast }{)}^n$ for some map $\rho :\mathrm{Gal}(L/K)\to (\mathbb{Z}/n\mathbb{Z}{)}^{\ast }\text{,}$ whose codomain is cyclic by our assumption on $n\text{.}$

Note that $\mathrm{Gal}(M/K)$ admits a homomorphism $\omega$ to a cyclic group whose kernel $\mathrm{Gal}(M/L)\subseteq \mathbb{Z}/n\mathbb{Z}$ is also abelian. Thus $\mathrm{Gal}(M/K)$ is abelian if and only if $g$ and $h$ commute for any $g\in \mathrm{Gal}(M/K)$ and $h\in \mathrm{Gal}(M/L)\text{,}$ i.e., if $h=g^{-1}hg\text{.}$ (Since $g$ commutes with powers of itself, $g$ then commutes with everything.)

Thus $g$ and $h$ commute if and only if $⟨h,a^g⟩=⟨h,a^{\omega (g)}⟩\text{,}$ if and only if (by nondegeneracy) $a^g/a^{\omega (g)}\in (L^{\ast }{)}^n\text{,}$ as desired.