I am studying quadratic number rings and I have a problem with multiplying and reducing ideals, for example:
Let $w=\sqrt{-14}$. Let $a=(5+w,2+w)$, $b=(4+w,2-w)$ be ideals in $\mathbb Z[w]$. Now, allegedly, the product of ideals $a$ and $b$ in $\mathbb Z[w]$ is $(6,3w)$. Please explain clearly, how to get to $(6,3w)$.
In general, if you multiply the ideals $\mathbf{a} = (a_1, a_2, \dots, a_n)$ and $\mathbf{b} = (b_1, b_2, \dots, b_m)$ then the result is $$\mathbf{a} \cdot \mathbf{b} = (a_i b_j)_{1 \le i \le n, 1 \le j \le m}$$ Thus, in this special case we have \begin{align*} \mathbf{a} \cdot \mathbf{b} & = ((5 + w) (4 + w), (5 + w) (2 - w), (2 + w) (4 + w), (2 + w) (2 - w))\\ & = (6 + 9w, 24 - 3w, -6 + 6w, 18). \end{align*} It is not hard to reduce this further. First, we see that the difference of the second and fourth generator is $6 - 3w$. Adding this to the third generator gives $3w$. Now if we subtract $3 \cdot 3w$ from the first generator we get $6$. Hence $$(6, 3w) \subset \mathbf{a} \cdot \mathbf{b}.$$ It is also easy to see that the four generators above are all generated by $6$ and $3w$, which gives equality.