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.
Best Answer
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 $$(8)=\mathopen{\big(}\tfrac{1+\sqrt{-7}}{2}\mathclose{\big)}^3\mathopen{\big(}\tfrac{1-\sqrt{-7}}{2}\mathclose{\big)}^3.$$