# A number system that is not unique factorization domain

Can anyone present a number system that is not unique factorization domain and is a commutative ring?

So I want the case that does not involve polynomials/monomials or some trivial cases.

• – lhf Aug 19 '14 at 12:54

The ring $\mathbb{Z}[\sqrt{-5}]$ of complex numbers of the form $a+b\sqrt{-5}$ with $a,b\in \mathbb{Z}$ is not a UFD because $6=2\cdot 3$ and $6=(1+\sqrt{-5})(1-\sqrt{5})$; none of $2,3,1+\sqrt{-5},1-\sqrt{-5}$ are associates, so even when we make these irreducible factorizations (they actually are already), they won't be the same.
• For the ring of the form $\mathbb{Z}[\sqrt{-a}]$, is 5 for $a$ the only case that is not UFD? – user170547 Aug 19 '14 at 12:39
• @user170547 If $d>163$ is squarefree and $d\equiv1$ mod $4$ then $\Bbb Z[\sqrt{-d}]$ is not a UFD. See here – whacka Aug 19 '14 at 12:47
Consider the set of all multiples of 2. This is a commutative ring. 42, 66, 70, and 110 are all irreducible, and not associates of each other, and $4620=42\times110=66\times70$.
• Since $2\mathbb{Z}$ is not a ring-with-unit, one could argue that it does not form a "number system". On the other hand, the same idea works for a non-maximal order in a number field, say, $\mathbb{Z}[2\sqrt{-1}]$, where $-4$ can be written as $-1 \times 2 \times 2$ or $(2\sqrt{-1}) \times (2\sqrt{-1})$ with factors being irreducible or units, and $2\sqrt{-1}$ not associate to $2$. – Gro-Tsen Aug 19 '14 at 13:09
Hint  In the subring of $\,\Bbb Q[x]\,$ of integer-valued polynomials, $\ 2\mid x(x\!+\!1),\,$ but $\, 2\nmid x, x\!+\!1$.