Product of two non-principal ideals

Question

I have problems understanding why $$(6,2+\sqrt{-56})(6,-2+\sqrt{-56})=6(2,\sqrt{-56})$$ in $\mathbb{Z}[\sqrt{-14}]$.

By definition the product of two ideals $$IJ=\sum_{i,j}^{k}f_{i}g_{j}$$

we have

$(6)(6)=36$

$6(2+\sqrt{-56})=12+6\sqrt{-56}$

$6(2-\sqrt{-56})=12-6\sqrt{-56}$

$(2+\sqrt{-56})(-2+\sqrt{-56})=-60$

so $6$ divides all the products hence $IJ\subseteq (6)$

How to find other inclusion? Could someone explain why the product equals $6(2,\sqrt{-56})$

Thank you.

Answer 1

Divide by $6$, then in your ideal you have $$6, 2+\sqrt{-56}, 2-\sqrt{-56}, -10$$

so you also have $-10+ 2\cdot 6=2$. And then you get $\sqrt{-56}$

so you contain $(2,\sqrt{-56})$. The other direction is then clear.