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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.

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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.

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  • $\begingroup$ For the ring of the form $\mathbb{Z}[\sqrt{-a}]$, is 5 for $a$ the only case that is not UFD? $\endgroup$ – user170547 Aug 19 '14 at 12:39
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    $\begingroup$ @user170547 If $d>163$ is squarefree and $d\equiv1$ mod $4$ then $\Bbb Z[\sqrt{-d}]$ is not a UFD. See here $\endgroup$ – whacka Aug 19 '14 at 12:47
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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 $4620=42\times110=66\times70$.

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    $\begingroup$ 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 $2$. $\endgroup$ – Gro-Tsen Aug 19 '14 at 13:09
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    $\begingroup$ Since OP hasn't defined "number system", one could argue anything one pleases. I've made my guess. OP may choose to clarify. $\endgroup$ – Gerry Myerson Aug 19 '14 at 13:24
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Hint $ $ In the subring of $\,\Bbb Q[x]\,$ of integer-valued polynomials, $\ 2\mid x(x\!+\!1),\,$ but $\, 2\nmid x, x\!+\!1$.

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