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!

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.

• Hi Fretty, thanks for your help. Didn't notice that $\pi\cong -\overline{\delta}$. Are you able to prove the result in general? See edited question. Thanks once again! Jun 21 '13 at 9:43
• From what I see it boils down to noticing that there are $6$ possible generators since $|\mathfrak{O}_K^{\times}|=6$. Reducing each mod $3\mathfrak{O}_K$ gives a different unit and there are exactly $6$ of them in $\mathfrak{O}_K/3\mathfrak{O}_K$. In particular $2$ is a unit so a unique generator must produce this. Jun 21 '13 at 10:01
• So to put it into perspective I didn't just see that $\pi \equiv -\delta \bmod 3$ in my head...I knew beforehand that whatever it is mod $3$ it would be a unit and that the units in $\mathfrak{O}_K/3\mathfrak{O}_K$ would form a group including $2$...hence there would be some unit multiplying to give $2$. By basic group theory this multiplier would be unique. Jun 21 '13 at 10:58

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 $-1+3\delta\equiv 2 \bmod 3\mathcal{O}_K$.

• Thanks Alvaro! That helped as well. Like the example too. Jun 22 '13 at 8:37