# 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 $R$ is the ring of integers of the quadratic number field $\mathbb{Q}(\sqrt{-5})$). Notice that the only units (http://planetmath.org/UnitsOfQuadraticFields) of $R$ 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 $R$ 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 (http://planetmath.org/DivisibilityInRings) 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 $\mathbb{Q}$ (http://planetmath.org/PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ)). 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. (1) 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.

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