# 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. 1.

$L/K$$n$

2. 2.

$L=K(\sqrt[n]{a_{1}},\ \ldots,\ \sqrt[n]{a_{k}})$ 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 $\alpha_{i},\ \zeta\alpha_{i},\ \ldots,\ \zeta^{n-1}\alpha_{i}$ 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    $\{x^{n}-a_{i}\ \mid\ 1\leq i\leq k\}$. Thus $L/K$ is Galois. Given $\sigma\in\operatorname{Gal}(L/K)$, 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 $\sigma\in\operatorname{Gal}(L/K)$, so that the exponent of $\operatorname{Gal}(L/K)$ divides $n$. It remains to show that $\operatorname{Gal}(L/K)$ is abelian; this follows trivially from the simple definition of the Galois action as multiplication  by some $n^{\mathrm{th}}$ root of unity: if $\sigma,\tau\in\operatorname{Gal}(L/K)$ with $\sigma(\alpha_{i})=\zeta^{r}\alpha_{i},\quad\tau(\alpha_{i})=\zeta^{s}\alpha_{i}$, then

 $\displaystyle(\sigma\tau)(\alpha_{i})$ $\displaystyle=\sigma(\zeta^{s}\alpha_{i})=\zeta^{s}\zeta^{r}\alpha_{i}$ $\displaystyle(\tau\sigma)(\alpha_{i})$ $\displaystyle=\tau(\zeta^{r}\alpha_{i})=\zeta^{r}\zeta^{s}\alpha_{i}$

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

$(1\Rightarrow 2):$ Let $G=\operatorname{Gal}(L/K)$, and write $G=C_{1}\times\dots\times C_{r}$ where each $C_{i}$ is cyclic; $\left\lvert C_{i}\right\rvert=m_{i}\mid n$ 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 $\operatorname{Gal}(L_{i}/K)\cong G/H_{i}\cong C_{i}$ 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

 $\operatorname{Gal}(L:K(\alpha_{1},\ldots,\alpha_{r}))=H_{1}\cap\dots\cap H_{r}% =\{1\}$

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 $L=K(\alpha_{1},\ldots,\alpha_{r})$. ∎

###### 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 $L=K(S)$. ∎

## References

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Title characterization of abelian extensions of exponent n CharacterizationOfAbelianExtensionsOfExponentN 2013-03-22 18:42:13 2013-03-22 18:42:13 rm50 (10146) rm50 (10146) 4 rm50 (10146) Theorem msc 12F10