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For homework, we were given a problem that asked to explain why $(\sqrt{-6})(-\sqrt{-6})$ implies $\mathbb{Q}(\sqrt{-6})$ does not have a unique factorization.

I understand the unique part in this case only means unique up to multiplication by a unit, but how does $(\sqrt{-6})(-\sqrt{-6})$ relate to that? Is it that each of the square roots can be factored in more than one way?

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  • $\begingroup$ Hi Mark! It means that for at least one $x \in \Bbb Q[\sqrt-6]$, this number $x$ can be written in two differents ways in terms of irreducible elements of the ring. Can you guess which number and which irreducibles work for this counterexample? $\endgroup$
    – hedphelym
    Commented Aug 10, 2020 at 5:09
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    $\begingroup$ Hello, Leo! Could it be $-2$ and $3$, for example? A prime factorization of $-6$? $\endgroup$
    – Mark
    Commented Aug 10, 2020 at 5:13
  • $\begingroup$ So that's two irreducibles (why?) and which are the other two? $\endgroup$
    – hedphelym
    Commented Aug 10, 2020 at 5:15
  • $\begingroup$ $2$ and $-3$ (i.e. the other possibility)? Are they irreducible because if they are prime in $\mathbb{Q}(\sqrt{-6})$? Since they divide $\sqrt{-6}$? $\endgroup$
    – Mark
    Commented Aug 10, 2020 at 5:18
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    $\begingroup$ Sorry, it's late here and I'm tired but I was talking as if it was $\Bbb Z[\sqrt{-6}]$ as that is a domain. Otherwise when we are dealing with $\Bbb Q[\sqrt{-6}]$ note that it is a field so checking if it is a UFD it's a rather vacuous affair. $\endgroup$
    – hedphelym
    Commented Aug 10, 2020 at 5:19

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We have $$ 10=2\cdot 5=(2-\sqrt{-6})(2+\sqrt{-6}). $$ Now show that these elements are irreducible in $\Bbb Z[\sqrt{-6}]$, so that this ring is not a UFD.

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