3.4 Examples

Example 3.21

We analyse the extension $\mathbb{Q}[\zeta]/\mathbb{Q}$ , where $\zeta$ is a primitive $7$ th root of $1$ , say $\zeta=e^{2\pi i/7}$ .

Note that $\mathbb{Q}[\zeta]$ is the splitting field of the polynomial $X^{7}-1$ , and that $\zeta$ has minimal polynomial

X^{6}+X^{5}+X^{4}+X^{3}+X^{2}+X+1

(see 1.41). Therefore, $\mathbb{Q}{}[\zeta]$ is Galois of degree $6$ over $\mathbb{Q}$ . For any $\sigma\in\operatorname{Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ , $\sigma\zeta=\zeta^{i}$ , some $i$ , $1\leq i\leq 6$ , and the map $\sigma\mapsto i$ defines an isomorphism $\operatorname{Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})\rightarrow(\mathbb{Z}/7% \mathbb{Z})^{\times}$ . Let $\sigma$ be the element of $\operatorname{Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ such that $\sigma\zeta=\zeta^{3}$ . Then $\sigma$ generates $\operatorname{Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ because the class of $3$ in $(\mathbb{Z}/7\mathbb{Z})^{\times}$ generates it (the powers of $3$ mod $7$ are $3,2,6,4,5,1$ ). We investigate the subfields of $\mathbb{Q}[\zeta]$ corresponding to the subgroups $\langle{}\sigma^{3}\rangle$ and $\langle{}\sigma^{2}\rangle$ .

Note that $\sigma^{3}\zeta=\zeta^{6}=\bar{\zeta}$ (complex conjugate of $\zeta)$ , and so $\zeta+\bar{\zeta}=2\cos\frac{2\pi}{7}$ is fixed by $\sigma^{3}$ . Now $\mathbb{Q}{}[\zeta]\supset\mathbb{Q}{}[\zeta]^{\langle\sigma^{3}\rangle}% \supset\mathbb{Q}{}[\zeta+\bar{\zeta}]\neq\mathbb{Q}{}$ , and so $\mathbb{Q}{}[\zeta]^{\langle\sigma^{3}\rangle}=\mathbb{Q}{}[\zeta+\bar{\zeta}]$ (look at degrees). As $\langle{}\sigma^{3}\rangle$ is a normal subgroup of $\langle{}\sigma\rangle$ , $\mathbb{Q}[\zeta+\bar{\zeta}]$ is Galois over $\mathbb{Q}$ , with Galois group $\langle{}\sigma\rangle/\langle{}\sigma^{3}\rangle.$ The conjugates of $\alpha_{1}\overset{\smash{\lower 1.31395pt\hbox{{def}}}}{=}\zeta+\bar{\zeta}$ are $\alpha_{3}=\zeta^{3}+\zeta^{-3}$ , $\alpha_{2}=\zeta^{2}+\zeta^{-2}$ . Direct calculation shows that

\displaystyle\alpha_{1}+\alpha_{2}+\alpha_{3}

\displaystyle=\sum\nolimits_{i=1}^{6}\zeta^{i}=-1,

\displaystyle\alpha_{1}\alpha_{2}+\alpha_{1}\alpha_{3}+\alpha_{2}\alpha_{3}

\displaystyle=-2,

\displaystyle\alpha_{1}\alpha_{2}\alpha_{3}

\displaystyle=(\zeta+\zeta^{6})(\zeta^{2}+\zeta^{5})(\zeta^{3}+\zeta^{4})

\displaystyle=(\zeta+\zeta^{3}+\zeta^{4}+\zeta^{6})(\zeta^{3}+\zeta^{4})

\displaystyle=(\zeta^{4}+\zeta^{6}+1+\zeta^{2}+\zeta^{5}+1+\zeta+\zeta^{3})

\displaystyle=1.

Hence the minimal polynomial²² 2 More directly, on setting $X=\zeta+\bar{\zeta}$ in $(X^{3}-3X)+(X^{2}-2)+X+1$ one obtains $1+\zeta+\zeta^{2}+\cdots+\zeta^{6}=0$ . of $\zeta+\bar{\zeta}$ is

g(X)=X^{3}+X^{2}-2X-1.

The minimal polynomial of $\cos\frac{2\pi}{7}=\frac{\alpha_{1}}{2}$ is therefore

\frac{g(2X)}{8}=X^{3}+X^{2}/2-X/2-1/8.

The subfield of $\mathbb{Q}[\zeta]$ corresponding to $\langle{}\sigma^{2}\rangle$ is generated by $\beta=\zeta+\zeta^{2}+\zeta^{4}$ . Let $\beta^{\prime}=\sigma\beta$ . Then $(\beta-\beta^{\prime})^{2}=-7$ . Hence the field fixed by $\langle{}\sigma^{2}\rangle$ is $\mathbb{Q}[\sqrt{-7}].$

Example 3.22

We compute the Galois group of a splitting field $E$ of $X^{5}-2\in\mathbb{Q}[X]$ .

Recall from Exercise 2-3 that $E=\mathbb{Q}[\zeta,\alpha]$ where $\zeta$ is a primitive $5$ th root of $1$ , and $\alpha$ is a root of $X^{5}-2$ . For example, we could take $E$ to be the splitting field of $X^{5}-2$ in $\mathbb{C}$ , with $\zeta=e^{2\pi i/5}$ and $\alpha$ equal to the real $5$ th root of $2$ . We have the picture at right, and

[\mathbb{Q}[\zeta]:\mathbb{Q}]=4,\quad[\mathbb{Q}[\alpha]:\mathbb{Q}]=5.

Because $4$ and $5$ are relatively prime,

[\mathbb{Q}[\zeta,\alpha]:\mathbb{Q}]=20.

Hence $G=\operatorname{Gal}(\mathbb{Q}[\zeta,\alpha]/\mathbb{Q})$ has order $20$ , and the subgroups $N$ and $H$ fixing $\mathbb{Q}[\zeta]$ and $\mathbb{Q}[\alpha]$ have orders $5$ and $4$ respectively. Because $\mathbb{Q}[\zeta]$ is normal over $\mathbb{Q}$ (it is the splitting field of $X^{5}-1$ ), $N$ is normal in $G$ . Because $\mathbb{Q}[\zeta]\cdot\mathbb{Q}[\alpha]=\mathbb{Q}[\zeta,\alpha]$ , we have $H\cap N=1$ , and so $G=N\rtimes_{\theta}H$ . Moreover, $H\simeq G/N\simeq(\mathbb{Z}/5\mathbb{Z})^{\times}$ , which is cyclic, being generated by the class of $2$ . Let $\tau$ be the generator of $H$ corresponding to $2$ under this isomorphism, and let $\sigma$ be a generator of $N$ . Thus $\sigma(\alpha)$ is another root of $X^{5}-2$ , which we can take to be $\zeta\alpha$ (after possibly replacing $\sigma$ by a power). Hence:

\left\{\begin{array}[c]{rcl}\tau\zeta&=&\zeta^{2}\\ \tau\alpha&=&\alpha\end{array}\right.\left\{\begin{array}[c]{rcl}\sigma\zeta&=% &\zeta\\ \sigma\alpha&=&\zeta\alpha.\end{array}\right.

Note that $\tau\sigma\tau^{-1}(\alpha)=\tau\sigma\alpha=\tau(\zeta\alpha)=\zeta^{2}\alpha$ and it fixes $\zeta$ ; therefore $\tau\sigma\tau^{-1}=\sigma^{2}$ . Thus $G$ has generators $\sigma$ and $\tau$ and defining relations

\sigma^{5}=1,\quad\tau^{4}=1,\quad\tau\sigma\tau^{-1}=\sigma^{2}.

The subgroup $H$ has five conjugates, which correspond to the five fields $\mathbb{Q}[\zeta^{i}\alpha]$ ,

\sigma^{i}H\sigma^{-i}\leftrightarrow\sigma^{i}\mathbb{Q}[\alpha]=\mathbb{Q}[% \zeta^{i}\alpha],\qquad 1\leq i\leq 5.