In the ring $\mathbb{Z}[\sqrt{3}]=\{a+b\sqrt{3}\mid a,b \in \mathbb{Z}\}$, show the following:
a) $1 - 2\sqrt{3}$ is not a unit,
b) $1 - 2\sqrt{3}$ and $8-5\sqrt{3}$ are associate elements.
Firstly, how would you show it is not a unit? It seems to fit the form $a+b\sqrt{3}$ quite well. For the second part, as associate elements, $1 - 2\sqrt{3}\mid 8-5\sqrt{3}$ and vice versa should hold, but neither do. Could someone help explain this?
You say that $(1-2\sqrt3)\nmid(8-5\sqrt3)$. Let's test this out. Consider $$\frac{8-5\sqrt3}{1-2\sqrt3}=\frac{(8-5\sqrt3)(1+2\sqrt3)}{(1-2\sqrt3)(1+2\sqrt3)}=\frac{-22+11\sqrt3}{-11}=2-\sqrt3.$$ I reckon that actually $(1-2\sqrt3)\mid(8-5\sqrt3)$. But does $(8-5\sqrt3)\mid(1-2\sqrt3)$? Over to you!