Exercise 9.25

Proof.

(a)

$\pi =a+\mathit{b\omega }$ and $a\equiv -b(\mathit{mod}a+b)$, thus $\pi \equiv -b(1-\omega )(\mathit{mod}a+b)$. So $${\chi }_{a+b}(\pi )={\chi }_{a+b}(b){\chi }_{a+b}(1-\omega ).$$

Since $a\wedge b=1$, $(a+b)\wedge b=1$ : as in Ex. 9.24, ${\chi }_{a+b}(b)=1$. So

$${\chi }_{a+b}(\pi )={\chi }_{a+b}(1-\omega ).$$

(b)

Since the character ${\chi }_{a+b}$ has order 3, $$\begin{array}{llll}\hfill {\chi }_{a+b}(1-\omega )& ={({\chi }_{a+b}({(1-\omega )}^2))}^2& \hfill & \\ \hfill & ={({\chi }_{a+b}(-3\omega ))}^2& \hfill & \\ \hfill & ={[{\chi }_{a+b}(3){\chi }_{a+b}(\omega )]}^2& \hfill & \end{array}$$

${\chi }_{a+b}(3)=1$ because $(a+b)\wedge 3=(3(m+n)-1)\wedge 3=1$.

${\chi }_{a+b}(\omega )={\omega }^{m+n}$ (Ex. 9.19).

Conclusion :

$${\chi }_{a+b}(1-\omega )={\omega }^{2(m+n)}.$$