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Let $\langle\sqrt{\mathbb{N}}\rangle=\mathbb{Z}[\sqrt2,\sqrt3,\sqrt5,\ldots]$ denote the ring generated by the square roots of all prime numbers. Is it it known whether $\langle\sqrt{\mathbb{N}}\rangle$ is a unique factorisation domain?

It seems hard to determine so because I can't think think of a 'norm-function' as we do have in 'finite' extensions of $\mathbb Z$, such as $\mathbb Z[\sqrt2]$.

user26857
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Bart Michels
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    Does it help: $(1+\sqrt{2})(-1+\sqrt{2})=1=(2+\sqrt{5})(-2+\sqrt{5})$ or $(1+\sqrt{3})(-1+\sqrt{3})=2=(2+\sqrt{6})(-2+\sqrt{6})$? – Janko Bracic Feb 02 '15 at 10:32
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    The first example you give are just units and their inverse so that doesn't exclude unique factorisation. I'm not sure about the second, I'll check. – Bart Michels Feb 02 '15 at 12:16
  • For any $a\in {\mathbb Z}$ one has, for instance, $(a+1-\sqrt{a^2+a+1})(a+1+\sqrt{a^2+a+1})=a$ and $a=(a+2-\sqrt{a^2+3a+4})(a+2+\sqrt{a^2+3a+4})$. Does it mean that any $0\ne a\in {\mathbb Z}$ is a unit? – Janko Bracic Feb 02 '15 at 12:47
  • What I wanted to say is that perhaps $a+1+\sqrt{a^2+a+1}$ and $a+2+\sqrt{a^2+3a+4}$ are unique up to a unit. But I doubt they are. – Bart Michels Feb 02 '15 at 12:54
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    Is this ring even Noetherian (i.e. are we even guaranteed factorizations into irreducibles, let alone unique such). – Tobias Kildetoft Feb 02 '15 at 12:54
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    @user26857 Sure, but if it is not noetherian, then it might be more natural to look for elements that do not factor into irreducibles before we look for ones with different factorizations. – Tobias Kildetoft Feb 02 '15 at 13:50
  • In the factorization $(1+\sqrt{3})(−1+\sqrt{3})=2=(2+\sqrt{6})(−2+\sqrt{6})$, if $1+\sqrt{3}$ and $2+\sqrt{6}$ equal up to a unit, the $\frac{2+\sqrt{6}}{1+\sqrt{3}}=-1+\sqrt{3}+\frac{3}{2}\sqrt{2}-\frac{1}{2}\sqrt{6}$ should be a unit. Is it? – Janko Bracic Feb 02 '15 at 15:00
  • @JankoBracic It isn't even contained in $\langle\sqrt{\mathbb N}\rangle$. Neither is $\frac{2+\sqrt{6}}{1-\sqrt{3}}$ so I think you have found us a counterexample. (Note that we need the fact that the representation of elements in $\langle\sqrt{\mathbb Q}\rangle$ is unique in some sense; Hagen von Eitzen gave a proof for that here.) – Bart Michels Feb 02 '15 at 16:15

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