# Factorize $(9+11\sqrt{-5})$ as a product of prime ideals in $\mathcal{O}_K$ where $K=\mathbb{Q}(\sqrt{-5})$

Factorize $(9+11\sqrt{-5})$ as a product of prime ideals in $\mathcal{O}_K$, where $K=\mathbb{Q}(\sqrt{-5})$.

The ring of integers in this case is $\mathbb{Z}[\sqrt{-5}]$.

I have calculated that the norm of $(9+11\sqrt{-5})$ is $686=2\times 7^3$ and therefore its prime factorization must contain a prime ideal of norm 2, and $(2,1+\sqrt{-5})$ has norm 2 and contains (hence divides) $(9+11\sqrt{-5})$. I don't know how to explicitly divide $(9+11\sqrt{-5})$ by $(2,1+\sqrt{-5})$. I don't know how to go further than this.

• You could try to find an ideal $J$ of norm $7$, and see whether $(2,1+\sqrt{-5})J^3=(9+11\sqrt{-5})$. – Gerry Myerson Mar 4 '13 at 12:04
• But be aware there are two prime ideals of norm $7$ ! – user18119 Mar 4 '13 at 15:13

We have $\mathbb Z[\sqrt{-5}]=\mathbb Z[X]/(X^2+5)$. Mod $7$, we have $X^2+5=(X+3)(X-3)$. So there are two primes in $\mathbb Z[\sqrt{-5}]$ above $7$: $\mathfrak q_1=(7, -3+\sqrt{-5})$ and $\mathfrak q_2=(7, 3+\sqrt{-5})$.
Note that $3(3-\sqrt{-5})=(9+11\sqrt{-5})-14\sqrt{-5}$. So $9+11\sqrt{-5}\in \mathfrak q_1$.
We have $9+11\sqrt{-5}\notin \mathfrak q_2$ because otherwise $8\sqrt{-5}=(9+11\sqrt{-5})-3(3+\sqrt{-5})\in \mathfrak q_2$. Hence, taking the norm, we find $64\times 5\in \mathfrak q_2\cap \mathbb Z=7\mathbb Z$. Contradiction.
Therefore the decomposition is $$(9+11\sqrt{-5})=(2, 1+\sqrt{-5})(7, -3+\sqrt{-5})^3.$$