Exercise 9.40

Proof. Here $p=6m+1,m\in \mathbb{Z}$, and $p=\pi \bar{\pi }$, where $\pi =a+\mathit{b\omega }$ is a primary prime.

We have proved in Exercise 39 that

Write $J({\chi }_{\pi },\rho )=c+\mathit{d\omega }$. The Exercise 8.27(c) shows that

(a)

If ${\chi }_{\pi }(2)=\omega$, then (1) gives $$\begin{array}{llll}\hfill a+\mathit{b\omega }& =\omega (c+\mathit{d\omega })=-d+\omega (c-d),& \hfill & \end{array}$$

so $a=-d,b=c-d$, therefore the equality (2) gives

$$2b-a=2(c-d)+d=2c-d\equiv -\left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)(\mathit{mod}p).$$

(b)

If ${\chi }_{\pi }(2)={\omega }^2$, then $$a+\mathit{b\omega }={\omega }^2(c+\mathit{d\omega })=d-c-\mathit{c\omega },$$

so $a=d-c,b=-c$, and

$$a+b=d-2c\equiv \left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)(\mathit{mod}p).$$

(c)

Suppose that ${\chi }_{\pi }(2)=\omega$, and put $A=2a-b,B=b∕3$, so $$4p=A^2+27B^2,A\equiv 1[3],$$

which shows that $A,B$ have same parities. Then, by part (a),

$$\begin{array}{llll}\hfill \frac{A-9B}{2}& =\frac{2a-b-3b}{2}& \hfill & \\ \hfill & =a-2b& \hfill & \\ \hfill & \equiv \left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)(\mathit{mod}p)& \hfill & \end{array}$$

(d)

Suppose that ${\chi }_{\pi }(2)={\omega }^2$, and put $A=2a-b,B=-b∕3$, so we have again $$4p=A^2+27B^2,A\equiv 1[3].$$

In this case, by part (b)

$$\begin{array}{llll}\hfill \frac{A-9B}{2}& =\frac{2a-b+3b}{2}& \hfill & \\ \hfill & =a+b& \hfill & \\ \hfill & \equiv \left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)(\mathit{mod}p)& \hfill & \end{array}$$

(e)

The conditions $4p=A^2+27B^2,A\equiv 1[3],$ determine $A,B$, except the sign of $B$. So $4p=A^2+27B^2={(2a-b)}^2+3b^2$, implies $A=2a-b$ and $B=\pm \frac{b}{3}$.

By Exercise 39, since $A,B$ have same parity, the condition $A,B$ odd is equivalent to ${\chi }_{\pi }(2)\in \{\omega ,{\omega }^2\}$. We choose this sign of $B$ so that

$$\frac{A-9B}{2}\equiv \left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)(\mathit{mod}p).$$

By parts (d) and (e), where $A,B$ are odd, this choice is given by $B=b∕3$ if ${\chi }_{\pi }(2)=\omega$, and $B=-b∕3$ if ${\chi }_{\pi }(2)={\omega }^2$. We show that these conditions are equivalent to $A\equiv B(\mathit{mod}4)$.

$\text{∙}$ If ${\chi }_{\pi }(2)=\omega$, then $A=2a-b,B=b∕3$.

By cubic reciprocity, ${\chi }_{\pi }(2)\equiv \pi (\mathit{mod}2)$ (see section 6). Here ${\chi }_{\pi }(2)=\omega$, so $\omega \equiv a+\mathit{b\omega }(\mathit{mod}2)$, therefore $a\equiv 0(\mathit{mod}2),b\equiv 1(mod2)$,

$$A=2a-b\equiv -b\equiv \frac{b}{3}=B(\mathit{mod}4),$$

so $A\equiv B(\mathit{mod}4)$.

$\text{∙}$ If ${\chi }_{\pi }(2)={\omega }^2$, then $A=2a-b,B=-b∕3$. In this case,

$${\omega }^2=-1-\omega \equiv a+\mathit{b\omega }(\mathit{mod}2),$$

therefore $a\equiv 1\equiv b(\mathit{mod}2)$, and

$$A=2a-b\equiv 2-b\equiv b\equiv -\frac{b}{3}=B(\mathit{mod}4).$$

In both cases, the choice of the sign of $B$ implies that $A\equiv B(\mathit{mod}4)$.

Conversely, suppose that $A\equiv B(\mathit{mod}4)$. Write $B=𝜀\frac{b}{3}$, where $𝜀=\pm 1$. Then $A\equiv B(\mathit{mod}4)$ gives

$$2a-b\equiv 𝜀\frac{b}{3}\equiv -\mathit{𝜀b}(\mathit{mod}4),$$

thus $a\equiv \frac{1-𝜀}{2}b(\mathit{mod}2)$. Then

$$\begin{array}{llll}\hfill {\chi }_{\pi }(2)& \equiv \pi =a+\mathit{b\omega }& \hfill & \\ \hfill & \equiv b\left(\frac{1-𝜀}{2}+\omega \right)(\mathit{mod}2)& \hfill & \end{array}$$

If ${\chi }_{\pi }(2)=\omega$, since $b=3B$ is odd, $\frac{1-𝜀}{2}\equiv 0(\mathit{mod}2)$, therefore $𝜀=1$, and $B=\frac{b}{3}$.

If ${\chi }_{\pi }(2)={\omega }^2=-1-\omega$, $\frac{1-𝜀}{2}\equiv 1(\mathit{mod}2)$, therefore $𝜀=-1$, and $B=-\frac{b}{3}$.

The normalisation given in parts (c) and (d) for the choice of the sign of $B$ is equivalent to $A\equiv B(\mathit{mod}4)$ (where $A,B$ are odd).