# Proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain

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?

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You should write it out and check it yourself. –  Brandon Carter May 28 '12 at 19:51
I tried to computer $\nu((a+b\sqrt{2})(x+y\sqrt{2}))$ but it got really complicated, and apparently it's 'clear'? Am I missing something intuitive? –  user26069 May 28 '12 at 19:56
@morphism, do you know that the mapping $a+b\sqrt2\mapsto a-b\sqrt2$ is a homomorphism of rings? –  Jyrki Lahtonen May 28 '12 at 20:00
Erm not off the top of my head no.. –  user26069 May 28 '12 at 20:05
In that case prove that as an exercise first. The rest will follow. –  Jyrki Lahtonen May 28 '12 at 20:12

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

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

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But this seems like an awfully long calculation for something that is apparently 'clear'.. Is there a bit of intuition that I am missing? –  user26069 May 28 '12 at 20:00
(not to sound ungrateful for your answer) :) –  user26069 May 28 '12 at 20:00
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! –  user26069 May 28 '12 at 20:05
Hint $\$ Conjugation $\rm\:(a+b\sqrt{2})'\! = a - b\sqrt{2}\:$ is well-defined and multiplicative $\rm\:(xy)' = x'y',\:$ so
$$\rm v(xy) = |xy(xy)'| = |xx'yy'| = |xx'||yy'| = v(x) v(y)$$