PlanetMath: example of ring which is not a UFD

Example 1 We define a ring $R=\mathbb{Z}[\sqrt{-5}]=\{ n+m\sqrt{-5} : n,m\in\mathbb{Z}\}$ with addition and multiplication inherited from $\mathbb{C}$ (notice that

is the ring of integers of the quadratic number field $\mathbb{Q}(\sqrt{-5})$ ). Notice that the only units of

are $R^\times=\{ \pm 1 \}$ . Then:

$\displaystyle 6=2\cdot 3 = (1+\sqrt{-5})\cdot (1-\sqrt{-5}).$

(1)

Moreover, $2,\ 3,\ 1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible elements of

and they are not associates (to see this, one can compare the norm of every element). Therefore,

is not a UFD.

However, the ideals of factor uniquely into prime ideals. For example:

$\displaystyle (6)=(2,1+\sqrt{-5})^2\cdot (3,1+\sqrt{-5})\cdot (3,1-\sqrt{-5})$

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:

$\displaystyle \mathfrak{P}^2=(2),\quad \mathfrak{Q}\cdot\overline{\mathfrak{Q}}... ...Q}=(1+\sqrt{-5}),\quad \mathfrak{P}\cdot \overline{\mathfrak{Q}}=(1-\sqrt{-5}).$

Thus, Eq. (1) above is the outcome of different rearrangements of the product of prime ideals:

$\displaystyle (6)=\mathfrak{P}^2\cdot(\mathfrak{Q}\cdot \overline{\mathfrak{Q}}... ...athfrak{P}\cdot \mathfrak{Q})\cdot (\mathfrak{P}\cdot \overline{\mathfrak{Q}}).$

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

, but the norm of $n+m\sqrt{-5}$ is

so it is clear that there are no algebraic integers of norm

. Therefore $\mathfrak{P}$ is not principal. Thus

is not a PID.