# In the ring $\mathbb{Z}[\sqrt{3}]=\{a+b\sqrt{3}\mid a,b \in \mathbb{Z}\}$, show the following.

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 can use \lbrace and \rbrace to write $\lbrace$ and $\rbrace$. Regarding your question; Hint: Norms – ÍgjøgnumMeg May 15 '18 at 19:22
• See my edits for proper MathJax usage, including but not limited to \mid. $\qquad$ – Michael Hardy May 15 '18 at 19:26

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!
• It does! It comes out to $2+\sqrt{3}$. Would you have something to say about the first part? – Matthijs Bjornlund May 15 '18 at 19:29
• Think about $1/(1-2\sqrt3)$. – Lord Shark the Unknown May 15 '18 at 19:29
• Rationalizing that leads to $\frac{1+2\sqrt{3}}{-11}$. Is that enough to conclude it is not a unit? Also, sorry to be a bother, but if you had any help with this, I'd be grateful: math.stackexchange.com/questions/2782591/… – Matthijs Bjornlund May 15 '18 at 19:33
• Is that an element of $R$? – Lord Shark the Unknown May 15 '18 at 19:34