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.