1

Question:

Let $\alpha = \sqrt{-2}$ and let $R = \{ x + y \alpha | x, y \in \mathbb{Z}\}$

Show that $R$ is a Euclidean domain.

My attempt I'm not entirely sure where to start so I have tried to mimic the proof that Gaussian integers are Euclidean.

I've tried to use $\nu(z) = |z|$ so for $z = x + y\alpha$, $\nu(z) = x^2 + 2y^2$

I have said let $a, b \in R$

I've been fiddling around the with the algebra but I can't seem to prove that $\nu(ab) = \nu(a)\nu(b)$ or that $\nu(ab) \geq \nu(a)$ I keep getting a linear term $a_1b_1a_2b_2$ which I can't prove is positive.

For the second part, which is to calculate the remainder I have assumed that $a = qb+r$, and that $r = b(\frac{a}{b} - q)$ and thus $\nu(r) = \nu(b)\nu(\frac{a}{q} - r)$. I have set $p$ to be the closest Gaussian integer near $\frac{a}{b}$ but have calculated that the distance between $p$ and $\frac{a}{b}$ could potentially be $1$ so this doesn't seem to work either.

user26857
  • 52,094
Kevin
  • 311
  • Related:http://math.stackexchange.com/questions/166449 – Watson Jun 05 '16 at 19:25
  • I used the norm suggested but could not derive the inequality $\nu(ab) \geq \nu(a)$ I am left with a linear term that I can not prove is positive. – Kevin Jun 05 '16 at 19:42
  • you should get $|ab| = |a||b|$, this is true in $\Bbb C$, and should be true in restriction to your case. The terms $x_1x_2y_1y_2$ cancel out in my expansion of $|ab| = (x_1x_2 - 2y_1y_2)^2 + 2(x_1y_2+x_2y_1)^2$. – Rolf Hoyer Jun 05 '16 at 19:48
  • Oh I see, I made a silly mistake and got $(x_1x_2 + 2y_1y_2)$ instead of the negative. Thanks.

    So does this norm function hold for any any $\sqrt{-a}, a \in \mathbb{N}?$

     – Kevin Jun 05 '16 at 19:54
  • It works for any subring of $\Bbb C$, although in general this norm does not satisfy the Euclidean property. – Rolf Hoyer Jun 05 '16 at 19:56
  • Does it satisfy the euclidean property as long as $a < 3, a \in \mathbb{R}$ – Kevin Jun 05 '16 at 19:58

0 Answers0