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Does all non-zero, not invertible elements in $\mathbb{Z[\sqrt{-3}]}$ have an irreducible factorization?

I know $\mathbb{Z[\sqrt{-3}]}$ is not UFD, for example: $4 = (1 - \sqrt{-3})(1 + \sqrt{-3}) = 2.2$.

But how one does prove that every non-zero, not invertible element has at least one irreducible factorization?

My idea is using some proof by contradiction where we have such an element but not sure how to proceed with it. Hints are welcome.

Thanks.

meerkat
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  • As you suggested, consider an element which does not factor into irreducibles. In particular, it is not irreducible, so you can decompose it. Try to argue that you can keep splitting this expression further and further. The contradiction will come from the fact that the ring $\Bbb Z [\sqrt{-3}]$ is Northerian. More generally, the same argument shows that any element of a Noetherian ring has a decomposition into irreducibles. – François Gatine Jul 27 '24 at 07:13
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    It's the same as the linked classical proof in $\Bbb Z$ except here we uses the norm as a size measure in the inductive step, i.e. if $,a, $is not irreducible the $,a = bc,$ for nonunits $,b,c,$ so $,b,c,$ have norm $> 1$ so $Na = (Nb)(Nc) \Rightarrow ,b,c,$ have smaller norm than $a$ so by induction they have irr. factorizations, which appended yield one for $a.\ \ $ – Bill Dubuque Jul 27 '24 at 08:00
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    Or note divsibility is well-founded since any infinite proper chain $,\ldots a_3\mid a_2 \mid a_1,$ would yield the same in $,\Bbb N,$ upon taking norms. – Bill Dubuque Jul 27 '24 at 08:07
  • @BillDubuque Thanks guys, I am not sure I understand Northerian argument, but with the idea that if $a_1 = a_2*b$ (where norm of $a_1$ is some natural number) does not have irreducible factorization, then for example also $a_2$ does not have irreducible factorization with smaller norm than $a_1$, but $a_i$ goes to infinity, but norm cannot be infinitely smaller, thus contradiction. – meerkat Jul 27 '24 at 08:18
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    In ideal language the above infinite proper divisor chain becomes an ascending chain of principal ideals $,\cdots (a_3)\supsetneq (a_2)\supsetneq (a_1),$ contra Noetherian (or ACCP = Ascending Chain Condition for Principal ideals) $\ \ $ – Bill Dubuque Jul 27 '24 at 15:49

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