Factoring the ideal $(8)$ into a product of prime ideals in $\mathbb{Q}(\sqrt{-7})$

I am trying to factor the ideal $(8)$ into a product of prime ideals in $\mathbb{Q}(\sqrt{-7})$.

I am not exactly sure how to go about doing this, and I feel I am missing some theory in the matter. What I have noted is that we have $$8=(1+\sqrt{-7})(1-\sqrt{-7})=2^3$$ and so I surmise the factorization will involve these numbers somehow. I also noted that $N(8)=64$, so the prime ideal factors will have norms that multiply out to $64.$ Unfortunately I cannot proceed further than this. Any hints or references are appreciated.

Solution 1:

Let $K=\mathbb{Q}(\sqrt{-7})$, so that $\mathcal{O}_K=\mathbb{Z}[\frac{1+\sqrt{-7}}{2}]$ because $-7\equiv 1\bmod 4$. Let $(8)$ denote the ideal generated by $8$ in $\mathcal{O}_K$.

Because $(8)=(2)^3$, it will suffice to determine the factorization of $(2)$ in $\mathcal{O}_K$, and then the factorization of $(8)$ will be the same with the exponents multiplied by $3$.

Note that $$(\tfrac{1+\sqrt{-7}}{2})(\tfrac{1-\sqrt{-7}}{2})=(2).$$ The norm of $\frac{1+\sqrt{-7}}{2}$ is $$N\mathopen{\big(}\tfrac{1+\sqrt{-7}}{2}\mathclose{\big)}=\mathopen{\big(}\tfrac{1}{2}\mathclose{\big)}^2+7\mathopen{\big(}\tfrac{1}{2}\mathclose{\big)}^2=2,$$ and similarly with $\frac{1-\sqrt{-7}}{2}$, so that $\mathcal{O}_K/\mathopen{\big(}\frac{1+\sqrt{-7}}{2}\mathclose{\big)}$ and $\mathcal{O}_K/\mathopen{\big(}\frac{1-\sqrt{-7}}{2}\mathclose{\big)}$ have cardinality $2$, and they are therefore the field $\mathbb{F}_2$.

Thus the ideals $\mathopen{\big(}\frac{1+\sqrt{-7}}{2}\mathclose{\big)}$ and $\mathopen{\big(}\frac{1-\sqrt{-7}}{2}\mathclose{\big)}$ are prime, so that $$(8)=\mathopen{\big(}\tfrac{1+\sqrt{-7}}{2}\mathclose{\big)}^3\mathopen{\big(}\tfrac{1-\sqrt{-7}}{2}\mathclose{\big)}^3.$$

Solution 2:

As you noted, $N(8) = 64$. Also note that $N(1 \pm \sqrt{-7}) = 8$ and $N(2) = 4$. Since the ring of algebraic integers of $\textbf Q(\sqrt{-7})$ is said to be a unique factorization domain and, as you noticed, $(1 - \sqrt{-7})(1 + \sqrt{-7}) = 2^3 = 8$, this must mean that the two apparently distinct factorizations are in fact incomplete factorizations, just as $4^3$ would be an incomplete factorization of $64$ in $\textbf Z$.

As it turns out, $$\frac{1 + \sqrt{-7}}{2}$$ is an algebraic integer and it belongs in this domain, since its minimal polynomial is $x^2 - x + 2$. Therefore, as numbers, we have $$\left( \frac{1}{2} - \frac{\sqrt{-7}}{2} \right)^3 \left( \frac{1}{2} + \frac{\sqrt{-7}}{2} \right)^3 = 8.$$

However, for ideals, we need to verify that $$\left\langle \frac{1}{2} - \frac{\sqrt{-7}}{2} \right\rangle \not \subseteq \left\langle \frac{1}{2} + \frac{\sqrt{-7}}{2} \right\rangle$$ nor vice-versa. A couple of divisions will quickly confirm that $\langle 2 \rangle$ is a splitting, not ramifying ideal. Therefore, $$\langle 8 \rangle = \left\langle \frac{1}{2} - \frac{\sqrt{-7}}{2} \right\rangle^3 \left\langle \frac{1}{2} + \frac{\sqrt{-7}}{2} \right\rangle^3.$$