1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You can use \lbrace and \rbrace to write $\lbrace$ and $\rbrace$. Regarding your question; Hint: Norms $\endgroup$ – ÍgjøgnumMeg May 15 '18 at 19:22
  • $\begingroup$ See my edits for proper MathJax usage, including but not limited to \mid. $\qquad$ $\endgroup$ – Michael Hardy May 15 '18 at 19:26
2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ It does! It comes out to $2+\sqrt{3}$. Would you have something to say about the first part? $\endgroup$ – Matthijs Bjornlund May 15 '18 at 19:29
  • $\begingroup$ Think about $1/(1-2\sqrt3)$. $\endgroup$ – Lord Shark the Unknown May 15 '18 at 19:29
  • $\begingroup$ 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/… $\endgroup$ – Matthijs Bjornlund May 15 '18 at 19:33
  • $\begingroup$ Is that an element of $R$? $\endgroup$ – Lord Shark the Unknown May 15 '18 at 19:34
  • $\begingroup$ Oops, brainfart. Understood! $\endgroup$ – Matthijs Bjornlund May 15 '18 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.