Question

Find two different factorizations of 9 as a product of irreducible in $\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{\right]}$.

Short answer

9 can be factorized as:

$\begin{array}{l}9=(2+\sqrt{-5})(2-\sqrt{-5})\\ 9=3\cdot 3\end{array}$

Step-by-step solution

First factorization of 9

Let $9=pq$

$N\left(p\right)N\left(q\right)=N\left(pq\right)=81$

Now $N\left(p\right)|81$ implies $N\left(p\right)=9$

Let $p=x+iy$

This means $N\left(p\right)=x^2+y^2=9$

On solving it is clear that

$9=(2+\sqrt{-5})(2-\sqrt{-5})$

Here $(2\pm \sqrt{-5})$ are irreducible in $\mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]$

If $(2\pm \sqrt{-5})$are further irreducible then there exists some a for which $N\left(a\right)=3$. But no such a exists, therefore$(2\pm \sqrt{-5})$ are irreducible.

Second factorization of 9

Note that $9=3\cdot 3$

Here, 3 is irreducible in $\mathrm{\mathbb{Z}}\left[\sqrt{-5}\right]$ as norm is 3 which is a prime in $\mathrm{\mathbb{Z}}$.

Conclusion

Here, 9 can be factorized as:

$\begin{array}{l}9=(2+\sqrt{-5})(2-\sqrt{-5})\\ 9=3\cdot 3\end{array}$