characterization of abelian extensions of exponent n

Theorem 1.

Let $K$ be a field containing the $n^{\mathrm{th}}$ roots of unity, with characteristic not dividing $n$. Let $L$ be a finite extension of $K$. Then the following are equivalent:

1. $L/K$ is Galois with abelian Galois group of exponent dividing $n$

2. $L=K(\sqrt[n]{a_1},\mathrm{\dots },\sqrt[n]{a_k})$ for some $a_i\in K$.

Proof.

Let $\zeta \in K$ be a primitive $n^{\mathrm{th}}$ root of unity.

$(2\Rightarrow 1):$ Choose ${\alpha }_i\in L$ such that ${\alpha }_i^n=a_i\in K$. Then for each $i$, the elements ${\alpha }_i,\zeta {\alpha }_i,\mathrm{\dots },{\zeta }^{n-1}{\alpha }_i$ are distinct and are all the roots of $x^n-a_i$ in $L$. Thus $x^n-a_i$ is separable over $K$ and splits in $L$, so that $L$ is the splitting field of the set of polynomials $\{x^n-a_i\mid \mathrm{\hspace{0.25em}1}\le i\le k\}$. Thus $L/K$ is Galois. Given $\sigma \in \mathrm{Gal}(L/K)$, for each $i$ we have $\sigma ({\alpha }_i)={\zeta }^j{\alpha }_i$ for some $1\le j\le n$, so that ${\sigma }^k({\alpha }_i)={\zeta }^{kj}{\alpha }_i$. It follows that ${\sigma }^n$ is the identity for every $\sigma \in \mathrm{Gal}(L/K)$, so that the exponent of $\mathrm{Gal}(L/K)$ divides $n$. It remains to show that $\mathrm{Gal}(L/K)$ is abelian; this follows trivially from the simple definition of the Galois action as multiplication by some $n^{\mathrm{th}}$ root of unity: if $\sigma ,\tau \in \mathrm{Gal}(L/K)$ with $\sigma ({\alpha }_i)={\zeta }^r{\alpha }_i,\tau ({\alpha }_i)={\zeta }^s{\alpha }_i$, then

$(\sigma \tau )({\alpha }_i)$$=\sigma ({\zeta }^s{\alpha }_i)={\zeta }^s{\zeta }^r{\alpha }_i$

$(\tau \sigma )({\alpha }_i)$$=\tau ({\zeta }^r{\alpha }_i)={\zeta }^r{\zeta }^s{\alpha }_i$

Thus $\sigma \tau =\tau \sigma$ on each ${\alpha }_i$. But the ${\alpha }_i$ generate $L/K$, so $\sigma \tau =\tau \sigma$ on $L$ and $\mathrm{Gal}(L/K)$ is abelian.

$(1\Rightarrow 2):$ Let $G=\mathrm{Gal}(L/K)$, and write $G=C_1\times \mathrm{\dots }\times C_r$ where each $C_i$ is cyclic; $\left|C_i\right|=m_i\mid n$ for each $i$. For each $i$, define a subgroup $H_i\le G$ by

$$H_i=C_1\times \mathrm{\dots }\times C_{i-1}\times C_{i+1}\times \mathrm{\dots }\times C_r$$

Then $G/H_i\cong C_i$. Let $L_i$ be the fixed field of $H_i$. $L_i$ is normal over $K$ since $H_i$ is normal in $G$, and $\mathrm{Gal}(L_i/K)\cong G/H_i\cong C_i$ and thus $L_i/K$ is cyclic Galois of order $m_i$. $K$ contains the primitive $m_i^{\mathrm{th}}$ root of unity ${\zeta }^{n/m_i}$ and thus $L_i=K({\alpha }_i)$ for some ${\alpha }_i\in L$ with ${\alpha }_i^{m_i}\in K$ (by Kummer theory). But then also ${\alpha }_i^n\in K$. Then

$$\mathrm{Gal}(L:K({\alpha }_1,\mathrm{\dots },{\alpha }_r))=H_1\cap \mathrm{\dots }\cap H_r=\{1\}$$

since any element of the left-hand group fixes each ${\alpha }_i$ and thus fixes $L_i$ so is the identity in $G/H_i$. Thus $L=K({\alpha }_1,\mathrm{\dots },{\alpha }_r)$.∎

Corollary 2.

If $L\mathrm{/}K$ is the maximal abelian extension of $K$ of exponent $n$, where $n$ is prime to the characteristic of $K$, then $L\mathrm{=}K\mathit{}\mathrm{(}\mathrm{\{}\sqrt[n]{a}\mathrm{\}}\mathrm{)}$ for some set of $a\mathrm{\in }K$.

Proof.

Clearly $K(\{\sqrt[n]{a}\mid a\in K^{*})$ is an infinite abelian extension of exponent $n$. If $L$ is the maximal such extension, choose $b\in L$. Then $K(b)$ is a finite extension of exponent dividing $n$ and thus $K(b)$ is of the required form. Thus $L={\cup }_{b\in L}K(b)$ is also of the required form; for example, if $S\subset K^{*}$ is a set of coset representatives for $K^{*}/{(K^{*})}^n$, then $L=K(S)$.∎