# Ireland-Rosen Hecke Character for $y^2=x^3-Dx$

I would like to refer you to page $310$ of Ireland-Rosen: A Classical Intro to Modern Number Theory.

Firstly, to construct the Hecke character, it is enough to specify $\chi(P)$ for prime ideals $P$ in $\mathbb{Z}[i]$. If $P$ divides $2D$, then define $\chi(P)=0$.
Suppose $P\nmid 2D$,

• if $NP=p$, then $p\equiv1\mod4$, and $P=(\pi)$, where $\pi\equiv1\mod2+2i$, and we define $$\chi(P)=\overline{\left(\frac{D}{\pi}\right)}_4\pi.$$
• if $NP=p^2$, then $p\equiv3\mod4$, and $P=(p)$ and we define $$\chi(P)=-p.$$

They then go now to prove a lemma which states that if $p\equiv3\mod4$, then $\left(\frac{D}{p}\right)_4=1$.
After which, they say that they can define $\chi(P)$ uniformly for prime ideals $P$ not dividing $2D$.

My understanding of the text is that for all primes $P$ not dividing $2D$, we can define $$\chi(P)=\overline{\left(\frac{D}{\pi}\right)}_4\pi.$$ But for $p\equiv3\mod4$, we have $\chi(P)=\overline{\left(\frac{D}{p}\right)}_4p=\overline{1}p=p\neq-p$.

Is my understanding of "define $\chi(P)$ uniformly" wrong?
Thanks for the help in advanced!

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For any prime ideal $P$ coprime with $(1+i)$, you can always pick a unique $\pi \equiv 1 \pmod {(2+2i)}$ such that $P= (\pi)$ and then define $\chi(P) = \overline{\left(\frac{D}{\pi}\right)}_4\pi$. (This is because every ideal is principal, $(2+2i) = (1+i)^3$, and the unit group $\{1,i,-1,-i\}$ is canonically isomorphic to$(\Bbb Z[i]/(1+i)^3)^*$. So given an $x \in \Bbb Z[i]$ coprime with $1+i$, there is exactly one way to multiply $x$ by a unit $u$ such that $ux \equiv 1 \pmod {2+2i}$)
$(2+2i)\cap \Bbb Z = 4\Bbb Z$, so when $NP = p^2$, we have $P = (p)$ with $p \equiv 3 \pmod 4$ a positive prime integer, and we must choose $\pi = -p$ instead of $p$. Finally $\chi(P) = \overline{\left(\frac{D}{\pi}\right)}_4\pi = \pi = -p$, as requested.