Example 1   We define a ring $R=\Ints[\sqrt{-5}]=\{ n+m\sqrt{-5} : n,m\in\Ints\}$ with addition and multiplication inherited from $\Complex$ (notice that $R$ is the ring of integers of the quadratic number field $\Rats(\sqrt{-5})$ ). Notice that the only units of $R$ are $R^\times=\{ \pm 1 \}$ . Then: \begin{eqnarray} \label{eq1} 6=2\cdot 3 = (1+\sqrt{-5})\cdot (1-\sqrt{-5}).\end{eqnarray}Moreover, $2,\ 3,\ 1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible elements of $R$ and they are not associates (to see this, one can compare the norm of every element). Therefore, $R$ is not a UFD.

However, the ideals of $R$ factor uniquely into prime ideals. For example: $$(6)=(2,1+\sqrt{-5})^2\cdot (3,1+\sqrt{-5})\cdot (3,1-\sqrt{-5})$$ where $\mathfrak{P}=(2,1+\sqrt{-5})$ , $\mathfrak{Q}=(3,1+\sqrt{-5})$ , and $\overline{\mathfrak{Q}}=(3,1-\sqrt{-5})$ are all prime ideals (see prime ideal decomposition of quadratic extensions of $\mathbb{Q}$ ). Notice that: $$\mathfrak{P}^2=(2),\quad \mathfrak{Q}\cdot\overline{\mathfrak{Q}}=(3),\quad \mathfrak{P}\cdot\mathfrak{Q}=(1+\sqrt{-5}),\quad \mathfrak{P}\cdot \overline{\mathfrak{Q}}=(1-\sqrt{-5}).$$ Thus, Eq. () above is the outcome of different rearrangements of the product of prime ideals: $$(6)=\mathfrak{P}^2\cdot(\mathfrak{Q}\cdot \overline{\mathfrak{Q}})=(\mathfrak{P}\cdot \mathfrak{Q})\cdot (\mathfrak{P}\cdot \overline{\mathfrak{Q}}).$$ Notice also that if $\mathfrak{P}$ was a principal ideal then there would be an element $\alpha \in R$ with $(\alpha)=\mathfrak{P}$ and $(\alpha)^2 = (2)$ . Thus such a number $\alpha$ would have norm $2$ , but the norm of $n+m\sqrt{-5}$ is $n^2+5m^2$ so it is clear that there are no algebraic integers of norm $2$ . Therefore $\mathfrak{P}$ is not principal. Thus $R$ is not a PID.