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I have the following factorizations of $7$ in the number ring of $\mathbb{Q}(\sqrt{-3})$, which is $\mathbb{Z}\left[ \frac{1+\sqrt{-3}}{2} \right]=\mathcal{O}_{\mathbb{Q}(\sqrt{-3})}$ $$7=(2+\sqrt{-3})(2-\sqrt{-3})=\frac{5+\sqrt{-3}}{2}\cdot\frac{5-\sqrt{-3}}{2}$$ I have already checked that all the factors are irreducible elements of $\mathcal{O}_{\mathbb{Q}(\sqrt{-3})}$. However, $\mathcal{O}_{\mathbb{Q}(\sqrt{-3})}$ is a UFD, so I have to find a reason for these two different factorizations not being a contradiction with the UFD status. If the factors were different by a unit, the problem would be solved. Let $$(2+\sqrt{-3})u_1=\frac{5+\sqrt{-3}}{2} \quad , \quad (2-\sqrt{-3})u_1=\frac{5-\sqrt{-3}}{2}$$ where $u_1, u_2$ are units of $\mathcal{O}_{\mathbb{Q}(\sqrt{-3})}$. I get that $$u_1=\frac{5+\sqrt{-3}}{4+2\sqrt{-3}}$$ However this isn't a unit. What am I doing wrong?

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    $\begingroup$ The factors differ by units and order. Try $\frac{5+\sqrt{-3}}{4-2\sqrt{-3}}$ $\endgroup$
    – lhf
    Nov 23, 2021 at 16:16
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    $\begingroup$ The unit might be $$\frac{5+\sqrt{-3}}{4-2\sqrt{-3}}$$ $\endgroup$ Nov 23, 2021 at 16:18
  • $\begingroup$ @ThomasAndrews well actually $\frac{5+\sqrt{-3}}{4+2\sqrt{-3}}$ is also a unit, right? It has norm $1$. $\endgroup$
    – kubo
    Nov 23, 2021 at 16:24
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    $\begingroup$ @MichaelCohen here, apparently, $\mathbb Z[\sqrt{-3}]$ is being used for the algebraic integers in $\mathbb Q[\sqrt{-3}],$ and it is a unique factorization domain. The notation is infusing, but not uncommon, and we can tell it is in use here by the values in the factorization. $\endgroup$ Nov 23, 2021 at 16:42
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    $\begingroup$ Maybe the OP means the algebraic integers in $\mathbb Q[\sqrt{-3}]$. These include quantities where the rational part and the coefficient of $\sqrt{-3}$ are each an integer plus one-half, and that inclusion enables UF for Stark-Heegner number ($>2$) radicands. $\endgroup$ Nov 23, 2021 at 16:42

2 Answers 2

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$$\frac{5+\sqrt{-3}}{4+2\sqrt3}$$ has norm $1,$ but it isn’t in $\mathbb Z[\omega],$ where $\omega=\frac{1+\sqrt{-3}}2.$ You can check this by rationalizing the denominator:

$$ \begin{align} \frac{5+\sqrt{-3}}{4+2\sqrt3}&= \frac{(5+\sqrt{-3})(4-2\sqrt{-3})}{(4+2\sqrt{-3})(4-2\sqrt{-3})}\\&=\frac{26-6\sqrt{-3}}{28}\\&=\frac{13}{14}+\frac{3}{14}\sqrt{-3}\notin\mathbb Z[\omega]. \end{align} $$

Instead, try: $$\frac{5+\sqrt{-3}}{4-2\sqrt{-3}}=\omega$$


This is like $$5=(2+i)(2-i )=(1+2i)(1-2i)$$ in $\mathbb Z[i].$ $$\frac{2+i}{1+2i}=\frac45-\frac35i$$ is not in $\mathbb Z[i],$ so even though the norm is $1,$ it is not a unit in $\mathbb Z[i].$

But $\mathbb Z[i]$ is a UFD, we just need a different quotient. $$\frac{2+i}{1-2i}=i$$ is a unit.

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Try writing your factors in terms of $\omega=\frac{-1+\sqrt{-3}}2$, where $\omega^2=-\omega-1=\frac1\omega$: $$7=(3+2\omega)(1-2\omega)=(3+\omega)(2-\omega)$$ Now multiply $3+2\omega$ by the unit $-\omega$ to get $-3\omega-2(-\omega-1)=2-\omega$, and $1-2\omega$ by its inverse $1+\omega$ to get $1-\omega-2(-\omega-1)=3+\omega$. This shows that the factorisations are the same up to units.

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