Question

Show that $\mathbf{\left(}\mathbf{6}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$ in$\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{\right]}$ . (The product of ideals is defined on page 349.)

Short answer

It is proved that $\left(6\right)=\left(2\right)\left(3\right)$.

Step-by-step solution

Product of Ideals

The product $\mathit{I}\mathit{J}$of ideals I and J is the set of all sums of elements in the form of $\mathit{a}\mathit{b}$, with$\mathit{a}\mathbf{\in }\mathit{I}$ and$\mathit{b}\mathbf{\in }\mathit{J}$ that is $\mathit{I}\mathit{J}\mathbf{=}\mathbf{\{}{\mathbf{a}}_{\mathbf{1}}{\mathbf{b}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}{\mathbf{b}}_{\mathbf{2}}\mathbf{+}\mathbf{\cdots }\mathbf{+}{\mathbf{a}}_{\mathbf{n}}{\mathbf{b}}_{\mathbf{n}}\mathbf{|}\mathbf{n}\mathbf{\ge }\mathbf{1}\mathbf{,}{\mathbf{a}}_{\mathbf{k}}\mathbf{\in }\mathbf{I}\mathbf{,}{\mathbf{b}}_{\mathbf{k}}\mathbf{\in }\mathbf{J}\mathbf{\}}$.

Show that (6)=(2)(3)  in ℤ[−5]

By the definition,$\left(2\right)\left(3\right)=\left\{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}2{\mathrm{b}}_{\mathrm{i}}3|\mathrm{n}\ge 1,{\mathrm{a}}_{\mathrm{i}},{\mathrm{b}}_{\mathrm{i}}\in \mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]\right\}$ . Since,

It is known that $\left(2\right)\left(3\right)\subseteq \left(6\right)$.

On the other hand $6=2\cdot 3\in \left(2\right)\left(3\right)$, so$\left(6\right)\subseteq \left(2\right)\left(3\right)$ . Therefore, it is proved that $\left(6\right)=\left(2\right)\left(3\right)$.