Exercise 8.3.4

Proof. Recall the context: the extension $F_{i-1}\subset F_i$ is a Galois extension, where $F_i=F_{i-1}({\gamma }_i)=F(1,{\gamma }_i,{\zeta }^i{\gamma }_i,\cdots ,{\zeta }^{m_i-1}{\gamma }_i)$ is the splitting field of $x^{m_i}-a_i$ over $F_i$, $a_i={\gamma }_i^{m_i}\in F_{i-1}$ and ${\zeta }_i$ is a $m_i$th primitive root of unity.

(a)

Let $\sigma \in \mathrm{Gal}(F_i∕F_{i-1})$. As ${\gamma }_i$ is a root of $x^{m_i}-a_i\in F_{i-1}(x)$, $\sigma ({\gamma }_i)$ is another root, so $$\sigma ({\gamma }_i)={\zeta }_i^l{\gamma }_i,0\le l\le m_i-1.$$

Such an $l$ is unique, since ${\zeta }_i^l{\gamma }_i={\zeta }_i^{l^{\prime }}{\gamma }_i,1\le l,l^{\prime }\le m_i$ implies ${\zeta }_i^l={\zeta }_i^{l^{\prime }}$, thus $m_i\mid l^{\prime }-l$, so $\equiv l^{\prime }(\mathit{mod}{m}_i)$. As $|l^{\prime }-l|\le m_i-1$, then $l^{\prime }-l=0$.

(b)

Let $$\phi :\{\begin{array}{ccc}\hfill \mathrm{Gal}(F_i∕F_{i-1})\hfill & \hfill \to \hfill & \hfill \mathbb{Z}∕m_i\mathbb{Z}\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill [l]:\sigma ({\gamma }_i)={\zeta }_i^l{\gamma }_i\hfill \end{array}$$

This mapping is well defined, since $l$ is known modulo $m_i$.

$\text{∙}$ We verify that $\phi$ is a group homomorphism.

Let $\sigma ,\tau \in \mathrm{Gal}(F_i∕F_{i-1})$. Then $\sigma ({\gamma }_i)={\zeta }_i^l{\gamma }_i,\tau ({\gamma }_i)={\zeta }_i^k{\gamma }_i$.

Thus $(\mathit{\sigma \tau })({\gamma }_i)=\sigma ({\zeta }_i^k{\gamma }_i)=\sigma {({\zeta }_i)}^k\sigma ({\gamma }_i)={\zeta }_i^k\sigma ({\gamma }_i)$, since ${\zeta }_i\in F$, therefore ${\zeta }_i\in F_{i-1}$.

$(\mathit{\sigma \tau })({\gamma }_i)={\zeta }_i^k{\zeta }_i^l{\gamma }_i={\zeta }_i^{l+k}{\gamma }_i$, so $\phi (\mathit{\sigma \tau })=[l+k]=[l]+[k]=\phi (\sigma )+\phi (\tau )$.

$\text{∙}$ $\phi$ is an injective homomorphism:

if $\sigma \in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, $[l]=[0]$, so $\sigma ({\gamma }_i)={\zeta }_i^l{\gamma }_i={\gamma }_i$. As $\sigma$ fixes the elements of $F_{i-1}$ and also ${\gamma }_i$, this is the identity on $F_i=F_{i-1}({\gamma }_i)$. $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{e\}$, and $\phi$ is injective.

(c)

Therefore $\mathrm{Gal}(F_i∕F_{i-1})$ is isomorphic to a subgroup $H$ of $\mathbb{Z}∕m_i\mathbb{Z}$. As every subgroup of a cyclic group is cyclic, $H$ is cyclic, so

$\mathrm{Gal}(F_i∕F_{i-1})$ is a cyclic group.