Exercise 9.2.8

Proof.

(a)

By Exercise 1, $3^8\equiv 9^4=81^2\equiv {(-4)}^2\equiv -1\not\equiv 1(\mathit{mod}17)$, therefore the order of $[3]$ in ${(\mathbb{Z}∕17\mathbb{Z})}^{\text{∗}}$ is $16$, so 3 is a primitive root modulo $17$.

(b)

$$\begin{array}{llll}\hfill H_8=⟨3^2⟩& =\{1,9,9^2,9^3,-1,-9,-9^2,-9^3\}& \hfill & \\ \hfill & =\{1,9,-4,-2,-1,-9,4,2\}& \hfill & \\ \hfill & =\{1,9,13,15,16,8,4,2\},& \hfill & \end{array}$$

$H_4=⟨3^4⟩=\{1,13,16,4\}$, and $H_2=⟨3^8⟩=\{1,16\}$, so

$$\begin{array}{llll}\hfill H_8& =\{1,2,4,8,9,13,15,16\},& \hfill & \\ \hfill H_4& =\{1,4,13,16\},& \hfill & \\ \hfill H_2& =\{1,16\}.& \hfill & \end{array}$$

(c)

$\text{∙}$

Extension $\mathbb{Q}\subset L_8$.

The cosets of $H_8$ in ${(\mathbb{Z}∕17\mathbb{Z})}^{\text{∗}}$ are

$$\begin{array}{llll}\hfill H_8& =\{1,2,4,8,9,13,15,16\},& \hfill & \\ \hfill 3H_8& =\{3,6,12,7,10,5,11,14\}.& \hfill & \end{array}$$

(Verification Sage :

         a = Mod(3,17)
         lc = [sorted([a^j*a^(2*i) for i in range(8)]) for j in range(2)]; lc
         [[1, 2, 4, 8, 9, 13, 15, 16], [3, 5, 6, 7, 10, 11, 12, 14]] )

$L_8$ is generated over $\mathbb{Q}$ by the 8-periods $(8,1),(8,3)$, where $(8,1)+(8,3)=-1$, and

$$\begin{array}{llll}\hfill (8,1)(8,3)& =\underset{\lambda \in H_8}{\sum }(8,\lambda +3)& \hfill & \\ \hfill & =(8,4)+(8,5)+(8,7)+(8,11)+(8,12)+(8,16)+(8,1)+(8,2)& \hfill & \\ \hfill & =4(8,1)+4(8,3)& \hfill & \\ \hfill & =-4.& \hfill & \end{array}$$

The minimal polynomial over $\mathbb{Q}$ of the 8-periods $(8,1),(8,3)$ is so

$$(x-(8,1))(x-(8,3))=x^2+x-4.$$

$\text{∙}$

Extension $L_8\subset L_4$.

$$\begin{array}{llll}\hfill H_8& =\{1,4,13,16\}\cup \{2,8,9,15\}=H_4\cup 2H_4,& \hfill & \\ \hfill 3H_8& =\{3,5,12,14\}\cup \{6,7,10,11\}=3H_4\cup 6H_4.& \hfill & \end{array}$$ The 4-periods are so $(4,1),(4,2)$, and $(4,3),(4,6)$, where $$\begin{array}{llll}\hfill (4,1)+(4,2)& =(8,1),& \hfill & \\ \hfill (4,1)\times (4,2)& =\underset{\lambda \in H_4}{\sum }(4,\lambda +2)& \hfill & \\ \hfill & =(4,3)+(4,7)+(4,15)+(4,1)& \hfill & \\ \hfill & =-1.& \hfill & \end{array}$$

The minimal polynomial of $(4,1)$ and $(4,2)$ over $L_8$ is so

$$(x-(4,1))(x-(4,2))=x^2-(8,1)x-1.$$

Applying $\sigma :{\zeta }_{17}\mapsto {\zeta }_{17}^3$, we obtain the minimal polynomial of $(4,3)$ and $(4,6)$

$$(x-(4,3))(x-(4,6))=x^2-(8,3)x-1.$$

$\text{∙}$

Extension $L_4\subset L_2$. $$\begin{array}{llll}\hfill H_4& =\{1,16\}\cup \{4,13\}=H_2\cup 4H_2,& \hfill & \\ \hfill 3H_4& =\{3,14\}\cup \{5,12\}=3H_2\cup 5H_2,& \hfill & \\ \hfill & \cdots & \hfill & \end{array}$$

The $2$-periods $(2,1),(2,4)$ satisfy

$$\begin{array}{llll}\hfill (2,1)+(2,4)& =(4,1),& \hfill & \\ \hfill (2,1)\times (2,4)& =\underset{\lambda \in H_2}{\sum }(2,\lambda +4)& \hfill & \\ \hfill & =(2,5)+(2,3)& \hfill & \\ \hfill & =(4,3).& \hfill & \end{array}$$

The minimal polynomial of $(2,1)$ and $(2,4)$ over $L_4$ is so

$$(x-(2,1))(x-(2,4))=x^2-(4,1)x+(4,3).$$