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. 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 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 $\{x^n-a_i\ \mid\ 1\leq i\leq k\}$ . Thus $L/K$ is Galois. Given $\sigma\in \Gal(L/K)$ , for each $i$ we have $\sigma(\alpha_i) = \zeta^j\alpha_i$ for some $1\leq j\leq n$ , so that $\sigma^k(\alpha_i) = \zeta^{kj}\alpha_i$ . It follows that $\sigma^n$ is the identity for every $\sigma\in\Gal(L/K)$ , so that the exponent of $\Gal(L/K)$ divides $n$ . It remains to show that $\Gal(L/K)$ is abelian; this follows trivially from the simple definition of the Galois action as multiplication by some $n^{\mathrm{th}}$ root of unity: if $\sigma,\tau\in\Gal(L/K)$ with $\sigma(\alpha_i) = \zeta^r\alpha_i,\quad\tau(\alpha_i) = \zeta^s\alpha_i$ , then

Thus $\sigma\tau=\tau\sigma$ on each $\alpha_i$ . But the $\alpha_i$ generate $L/K$ , so $\sigma\tau=\tau\sigma$ on $L$ and $\Gal(L/K)$ is abelian.

$(1 \Rightarrow 2):$ Let $G=\Gal(L/K)$ , and write $G=C_1\times \dots \times C_r$ where each $C_i$ is cyclic; $\Order{C_i}=m_i\mid n$ for each $i$ . For each $i$ , define a subgroup $H_i\leq G$ by$$ H_i = C_1 \times \dots \times C_{i-1} \times C_{i+1} \times \dots \times C_r$$ 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 $\Gal(L_i/K) \cong G/H_i \cong C_i$ 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

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 .

Corollary 2   If $L/K$ is the maximal abelian extension of $K$ of exponent $n$ , where $n$ is prime to the characteristic of $K$ , then $L = K(\{\sqrt[n]{a}\})$ for some set of $a\in K$ .

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 $L=K(S)$ .