Question

In $\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{7}}\mathbf{\right]}$, factor 8 as a product of two irreducible elements and as a product of three irreducible elements.

Short answer

8 can be factorized as:

$\begin{array}{l}8=(1+\sqrt{-7})(1-\sqrt{-7})\\ 8=2\cdot 2\cdot 2\end{array}$

Step-by-step solution

Factor 8 as a product of two irreducible elements

Considering the hint:

$8=(1+\sqrt{-7})(1-\sqrt{-7})$

Here $(1\pm \sqrt{-7})$are irreducible in$\mathbb{Z}\left[\sqrt{-7}\right]$

If $(1\pm \sqrt{-7})$are further irreducible then there exists some a for whichrole="math" localid="1654702600651" $N\left(a\right)=2$ . But no such a exists, therefore$(1\pm \sqrt{-7})$ are irreducible.

Factor 8 as a product of three irreducible elements

Note that $8=2\cdot 2\cdot 2$

Here, 2 is irreducible in$\mathbb{Z}\left[\sqrt{-7}\right]$as $N\left(a\right)=2$is a prime in $\mathrm{\mathbb{Z}}$.

Conclusion

Here 8 is factorized as a product of two irreducible elements and as a product of three irreducible elements in$\mathbb{Z}\left[\sqrt{-7}\right]$ .