12

We're proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain, using the norm function $$\nu (a + b\sqrt{2} ) = |a^2 - 2b^2|$$ and the first part says that since $\nu (a + b\sqrt{2} ) = |(a + b\sqrt{2})(a - b\sqrt{2})|$ it's clear that $\nu (xy) = \nu(x) \nu(y)$? ... Can someone please explain to me how this is clear?

KCd
  • 46,062

3 Answers3

24

Let $\alpha = a_1 + a_2 \sqrt{2}$ and $\beta = b_1 + b_2 \sqrt{2}$ be elements of $\mathbb{Z}[\sqrt{2}]$ with $\beta \neq 0$. We wish to show that there exist $\gamma$ and $\delta$ in $\mathbb{Z}[\sqrt{2}]$ such that $\alpha = \gamma\beta + \delta$ and $N(\delta) < N(\beta)$. To that end, note that in $\mathbb{Q}(\sqrt{2})$ we have $\frac{\alpha}{\beta} = c_1 + c_2 \sqrt{2}$, where $c_1 = \dfrac{a_1b_1 - 2a_2b_2}{b_1^2 - 2b_2^2}$ and $c_2 = \dfrac{a_2b_1 - a_1b_2}{b_1^2 - 2b_2^2}$.

Let $q_1$ be an integer closest to $c_1$ and $q_2$ an integer closest to $c_2$; then $|c_1 - q_1| \leq 1/2$ and $|c_2 - q_2| \leq 1/2$. Now let $\gamma = q_1 + q_2 \sqrt{2}$; certainly $\gamma \in \mathbb{Z}[\sqrt{2}]$. Next, let $\theta = (c_1 - q_1) + (c_2 - q_2) \sqrt{2}$. We have $\theta = \frac{\alpha}{\beta} - \gamma$, so that $\theta\beta = \alpha - \gamma\beta$.

Letting $\delta = \theta\beta$, we have $\alpha = \gamma\beta + \delta$. It remains to be shown that $N(\delta) < N(\beta)$. To that end, note that $$N(\theta) = |(c_1 - q_1)^2 - 2(c_2 - q_2)^2| \leq |(c_1 - q_1)^2| + |-2(c_2 - q_2)^2|$$ by the triangle inequality. Thus we have $$N(\theta) \leq (c_1 - q_1)^2 + 2(c_2 - q_2)^2 \leq (1/2)^2 + 2(1/2)^2 = 3/4.$$ In particular, $N(\delta) \leq \frac{3}{4}N(\beta)$ as desired.

Alejandro Bergasa Alonso
  • 3,534
Kowser
  • 341
  • Would it be true to say you could generalize this proof for $\mathbb{Z}[\sqrt{n}]$ for any $n \in \mathbb{Z}$ s.t $n^2 \not \in \mathbb{Z}$? – gowrath Jan 14 '17 at 05:42
  • @gowrath no. If $n\le -3$, then $\Bbb{Z}[\sqrt{n}]$ is not a UFD, hence not a ED. On the other hand, this proof works for $n=-2, -1$ and $n=3$. – Xam Feb 15 '17 at 16:59
  • @Xam Could you choose $_1$ and $_2$ such that $|_1−_1|<1/$ and $|_2−_2∣<1/$ , $>1$ instead of $1/2$ ? – Maman Apr 21 '19 at 12:23
  • @Maman I don't know. It's been a long time since I did any abstract algebra. – Xam Apr 21 '19 at 16:36
  • @Xam Ok sorry mate ! – Maman Apr 28 '19 at 09:16
  • 1
    I might be missing something here, but couldn't you also just say you can always find an integer < 1 away from $\alpha / \beta$, since $\mathbb Z[\sqrt{2}] \subset \mathbb R$? – vacant Sep 23 '19 at 03:12
  • Why is it true that $N(\delta)=N(\theta\beta)=N(\theta)N(\beta)$? $\theta \notin \mathbb{Z}[\sqrt{2}]$, so shouldn't you only be working with $N(\delta)$? – Alex Mar 02 '24 at 22:40
10

$$\nu (a + b\sqrt{2} )\nu (c+d\sqrt{2} ) = |[(a + b\sqrt{2})(c + d\sqrt{2})][(a - b\sqrt{2})(c - d\sqrt{2})]|\\ =|(ac+2bd)+(ad+bc)\sqrt 2||ac+2bd-(ad+bc)\sqrt{2}|\\ =\nu(ac+2bd+(ad+bc)\sqrt{2})\\ =\nu((a+b\sqrt2)(c+d\sqrt2)) $$

Alternatively, using Jyrki's comment and noting that $\phi(a+b\sqrt 2)=a-b\sqrt 2$ is a ring homomorphism, we have $\nu(xy)=|xy\phi(xy)|=|x\phi(x)||y\phi(y)|=\nu(x)\nu(y)$.

SL2
  • 4,321
  • But this seems like an awfully long calculation for something that is apparently 'clear'.. Is there a bit of intuition that I am missing? –  May 28 '12 at 20:00
  • (not to sound ungrateful for your answer) :) –  May 28 '12 at 20:00
  • 3
    It's not all that long. We expand a product, apply the observation $\nu (a + b\sqrt{2} ) = |(a + b\sqrt{2})(a - b\sqrt{2})|$ in reverse and then factor... The only 'trick' is that we don't expand the product on the first line completely. If you do that, you'll get something very messy. – SL2 May 28 '12 at 20:03
  • Ah ok, it seems clearer now, thanks! –  May 28 '12 at 20:05
7

Hint $\ $ Conjugation $\rm\:(a+b\sqrt{2})'\! = a - b\sqrt{2}\:$ is well-defined and multiplicative $\rm\:(xy)' = x'y',\:$ so

$$\rm v(xy) = |xy(xy)'| = |xx'yy'| = |xx'||yy'| = v(x) v(y)$$

Bill Dubuque
  • 272,048