Exercise 9.36

Proof. Suppose that $q=a\wedge b>1$. Then $a=q{a}^{\prime },b=q{b}^{\prime },a^{\prime },b^{\prime }\in \mathbb{Z}$, so $\pi =q(a^{\prime }+i{b}^{\prime })$.

As $\pi$ is irreducible, and as $q$ is not an unit, $u=a^{\prime }+b^{\prime }i$ is an unit, and so $\pi =\mathit{uq}$ is associate to $q$ : the rational integer $q$ is then a prime in $D$, so a rational prime $q\equiv 3(\mathit{mod}4)$.

If $u=\pm i$, then $\pi =\pm q=a+\mathit{bi}$ is such that $b$ is odd, in contradiction with $\pi$ primary. Thus $u=\pm 1$, and $\pi =\mathit{𝜀q},𝜀=\pm 1$. As $\pi$ is primary, $𝜀=-1$, so $\pi =-q$.

Then ${\chi }_{\pi }(a)={\chi }_{-q}(-q)=0$, the result of Ex. 34 is false if $b=0$.

Conclusion : if $\pi =a+\mathit{bi}$ is a primary irreducible, and $b\ne 0$, then

(a)

if $\pi \equiv 1[4],{\chi }_{\pi }(a)=i^{(a-1)∕2}$,

(b)

if $\pi \equiv 3+2i[4]$, ${\chi }_{\pi }(a)=-i^{(-a-1)∕2}$.