# Why don't these different factorizations of $7$ contradict number ring is UFD?

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?

• The factors differ by units and order. Try $\frac{5+\sqrt{-3}}{4-2\sqrt{-3}}$
– lhf
Nov 23, 2021 at 16:16
• The unit might be $$\frac{5+\sqrt{-3}}{4-2\sqrt{-3}}$$ Nov 23, 2021 at 16:18
• @ThomasAndrews well actually $\frac{5+\sqrt{-3}}{4+2\sqrt{-3}}$ is also a unit, right? It has norm $1$.
– kubo
Nov 23, 2021 at 16:24
• @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. Nov 23, 2021 at 16:42
• 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. Nov 23, 2021 at 16:42

$$\frac{5+\sqrt{-3}}{4+2\sqrt3}$$ 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.
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.