Abstract Powerful Algebra

If $u = a + ib\sqrt{5} \in \mathbb{Z}[i\sqrt{5}]$ , then $|u|^2 = a^2 + 5b^2$ is an integer. If $2$ were reducible in $\mathbb{Z}[i\sqrt{5}]$ , we could find nonunits $u,v \in \mathbb{Z}[i\sqrt{5}]$ such that $2 = uv$ , so that $4 = |u|^2|v|^2$ . Since $u,v$ are not units, $|u|,|v| \neq 1$ and hence $|u|^2=|v|^2=2$ . But no pair of integers $a,b$ exist such that $a^2 + 5b^2 = 2$ . Thus we deduce that $2$ is irreducible in $\mathbb{Z}[i\sqrt{5}]$ .

On the other hand, $2$ divides $(1 +i\sqrt{5})(1-i\sqrt{5}) = 6$ , but $2$ divides neither $1+ i\sqrt{5}$ nor $1 - i\sqrt{5}$ in $\mathbb{Z}[i\sqrt{5}]$ . Thus $2$ is not prime in $\mathbb{Z}[i\sqrt{5}]$ .

Since $\mathbb{Z}[i\sqrt{5}]$ possesses elements which are irreducible, but not prime, we see that $\mathbb{Z}[i\sqrt{5}]$ is not a UFD.

That $\mathbb{Z}[X]$ is not a PID has already been shown.

Remark: * solution is taken from online resources *