Exercise 9.24

Proof. (of Lemma.)

If $q$ is a rational prime, $q\equiv 2(\mathit{mod}3)$, and $q\wedge b=1$, then ${\chi }_q(b)=1$ (Prop. 9.3.4, Corollary).

If $p$ is a rational prime, $p\equiv 1(\mathit{mod}3)$ and $p\wedge b=1$, then $p=\pi \bar{\pi }$, with $\pi$ primary prime in $D$ (and also $\bar{\pi }$), and by definition of ${\chi }_p$, ${\chi }_p(b)={\chi }_{\pi }(b){\chi }_{\bar{\pi }}(b)$.

As ${\chi }_{\bar{\pi }}(b)={\chi }_{\bar{\pi }}(\bar{b})=\overline{{\chi }_{\pi }(b)}$ (Prop. 9.3.4(b)), then ${\chi }_p(b)={\chi }_{\pi }(b){\chi }_{\bar{\pi }}(b)={\chi }_{\pi }(b)\overline{{\chi }_{\pi }(b)}=1$.

$a$ has a decomposition in prime factors of the form :

where $q_i\equiv -1,p_j\equiv 1(\mathit{mod}3)$, and the ${\pi }_k$ are primary primes (since all these elements are primary, the symbol $\text{±}$ is ${(-1)}^{k-1}$). Thus, by definition of ${\chi }_a$,

The result remains true if $a=-1$ : then, by definition, ${\chi }_a(b)=1$. □

Proof. (of Ex 9.24.) By hypothesis, $\pi$ is a primary element, so $\pi =3m-1+3\mathit{n\pi },m,n\in \mathbb{Z}$. We don’t suppose in this proof that $\pi$ is a prime element, so $p=N(\pi )$ is not necessarily prime.

(a)

$p-1={(3m-1)}^2+{(3n)}^2-3n(3m-1)-1\equiv -6m+3n(\mathit{mod}9)$, thus $$\frac{p-1}{3}\equiv -2m+n(\mathit{mod}3).$$

(b)

$a^2-1={(3m-1)}^2-1\equiv -6m(\mathit{mod}9)$, thus $$\frac{a^2-1}{3}\equiv m(\mathit{mod}3).$$

(c)

As $\pi ,a$ are primary, by Exercise 9.20, ${\chi }_{\pi }(a)={\chi }_a(\pi )$.

Since $\pi \equiv \mathit{b\omega }(\mathit{mod}a)$, ${\chi }_a(\pi )={\chi }_a(b){\chi }_a(\omega )$.

By Exercise 9.18, as $a=3m-1$, ${\chi }_a(\omega )={\omega }^{M+N}$, where $M=m,N=0$, so

$${\chi }_a(\omega )={\omega }^m.$$

Here $a$ is relatively prime to $b$ in $\mathbb{Z}$ : if a rational prime $r$ divides $a,b$, then $r\mid \pi$ in $D$, thus $r\mid \bar{\pi }$, so $r^2\mid \pi \bar{\pi }=p$ in $D$, thus $r^2\mid p$ in $\mathbb{Z}$, which is absurd. The Lemma gives then ${\chi }_a(b)=1$.

We conclude that ${\chi }_a(b)=1,{\chi }_a(\omega )={\omega }^m$, so ${\chi }_{\pi }(a)={\chi }_a(\pi )={\chi }_a(b){\chi }_a(\omega )={\omega }^m$.

$${\chi }_{\pi }(a)={\omega }^m.$$

(d)

$$a+b=[(a+b)\omega ]{\omega }^{-1},$$

and

$$(a+b)\omega =(a+\mathit{b\omega })+\mathit{a\omega }-a\equiv a(\omega -1)(\mathit{mod}\pi ),$$

thus

$$a+b\equiv a(1-\omega ){\omega }^{-1}[\pi ],$$

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

${\chi }_{\pi }(a)={\omega }^m$ by (c), and ${\chi }_{\pi }(\omega )={\omega }^{m+n}$(as in Ex. 9.3), thus

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