Silverman Adv. Topics example I would like to refer you to Silverman's Advanced Topics in the Arithmetic of Elliptic Curves example 10.6:
Let $D$ be a nonzero integer, $E:y^2=x^3+D$ with complex multiplication by $\mathcal{O}_K$ where $K=\mathbb{Q}(\sqrt{-3})$. Let $\mathfrak{p}$ be a prime of $\mathcal{O}_K$ with $\mathfrak{p}\nmid 6D$. Since $\mathcal{O}_K$ is a PID, write $\mathfrak{p}=(\pi)$ and one can check that there is a unique $\pi$ generating $\mathfrak{p}$ which satisfies $\pi\equiv 2 (\text{mod }3)$.
My question is now this, if we let $\mathfrak{p}=(2+\delta)$, where $\delta=\frac{1+\sqrt{-3}}{2}$, then because it has norm 7, it is a prime. So, taking $\pi=2+\delta$ how to we get the result that $\pi\equiv 2 (\text{mod }3)$? I've tried to multiply $\pi$ by units in $\mathcal{O}_K$, but don't seem to get anything useful. Think may be missing something obvious here. 
Edit: Are you able to prove in general that there is a unique $\pi$ generating $\mathfrak{p}$ which satisfies $\pi\equiv 2 (\text{mod }3)$?
Thanks for the help!
 A: The short answer is...it isn't true for $\pi$.
It is the ideal that possesses such a generator. This doesn't mean all generators satisfy the congruence.
In fact $\pi \equiv -\bar{\delta} \bmod 3$ so you should find that $\pi\delta \equiv -\bar{\delta}\delta \equiv -1 \equiv 2 \bmod 3$.
Since $\delta$ is a unit in $\mathfrak{O}_K$ we find that $\mathfrak{p}=(\pi) = (\pi\delta)$ and so you have found the right generator.
A: In general, notice that the units of $\mathcal{O}_K$ are 
$$\{1,-1,\delta,-\delta,\delta^2,-\delta^2\}=\left\{1,-1,\frac{1+\sqrt{-3}}{2},-\frac{1+\sqrt{-3}}{2},\frac{-1+\sqrt{-3}}{2},\frac{1-\sqrt{-3}}{2}\right\}.$$
Therefore, the units $\mathcal{O}_K^\times$ modulo $3\mathcal{O}_K$ are the classes
$$\{1,-1,\delta,-\delta,\delta^2,-\delta^2\}\equiv \left\{1,2,2+2\sqrt{-3},1+\sqrt{-3},1+2\sqrt{-3},2+\sqrt{-3}\right\} \bmod 3\mathcal{O}_K,$$
which form a complete system of representatives for $(\mathcal{O}_K/3\mathcal{O}_K)^\times$, which has size $\varphi(9)=3\cdot 2=6$. Now let $\mathfrak{P}$ be an ideal generated by $\pi$, such that $\mathfrak{P}$ does not divide $3$. Then, $\pi \bmod 3\mathcal{O}_K$ is a unit in $(\mathcal{O}_K/3\mathcal{O}_K)^\times$. Then, there is a unit $u$ in $\mathcal{O}_K^\times$ which is the inverse of $\pi$ modulo $3\mathcal{O}_K$, so that $\pi\cdot u \equiv 1 \bmod 3\mathcal{O}_K$. Finally,
$$\pi\cdot(-u)\equiv - \pi\cdot u \equiv -1\equiv 2 \bmod 3\mathcal{O}_K,$$
as desired, because $-u$ is also a unit in $\mathcal{O}_K$.
Example: Take $\mathfrak{P}=(2+\delta)$. Then $$\pi = 2+\delta\equiv 2+(2+2\sqrt{-3})\equiv 4+2\sqrt{-3}\equiv 1+2\sqrt{-3}\equiv \delta^2 \bmod 3\mathcal{O}_K.$$
The multiplicative inverse $u$ of $\delta^2$ is $-\delta$. Hence, $\pi\cdot (-\delta)\equiv 1 \bmod 3\mathcal{O}_K$. And therefore $\pi\cdot \delta\equiv 2 \bmod 3\mathcal{O}_K$. Indeed: 
$$(2+\delta)\cdot(\delta)=-1+3\delta.$$ Clearly $\mathfrak{P}=(-1+3\delta)$ because $\delta$ is a unit. Moreover,
$$(-1+3\delta)-2 = -3+3\delta=3(-1+\delta).$$
Hence $-1+3\delta\equiv 2 \bmod 3\mathcal{O}_K$.
