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?
Proving that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain
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1You should write it out and check it yourself. – 2012-05-28
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0I 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? – 2012-05-28
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3@morphism, do you know that the mapping $a+b\sqrt2\mapsto a-b\sqrt2$ is a homomorphism of rings? – 2012-05-28
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0Erm not off the top of my head no.. – 2012-05-28
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2In that case prove that as an exercise first. The rest will follow. – 2012-05-28
3 Answers
Hint $\ $ Conjugation $\rm\:(a+b\sqrt{2})'\! = a - b\sqrt{2}\:$ is well-defined and multiplicative $\rm\:(xy)' = x'y',\:$ so
$$\rm v(xy) = |xy(xy)'| = |xx'yy'| = |xx'||yy'| = v(x) v(y)$$
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0i think this does not prove that $a, b\in$ the ring can be expressed as a = qb + r such that v(r) < v(b) right? does it? – 2015-04-17
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0@user136266 The answer only pertains to the OP's question about the multiplicativity of the norm. – 2015-04-17
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 $N(\delta) \leq N(\beta)$. 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.
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0Would it be true to say you could generalize this proof for $\mathbb{Z}[\sqrt{n}]$ for any $n \in \mathbb{Z}$ s.t $n^2 \not \in \mathbb{Z}$? – 2017-01-14
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0@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 $n=3$. – 2017-02-15
$$\nu (a + b\sqrt{2} )\nu (c+d\sqrt{2} ) = |[(a + b\sqrt{2})(c + d\sqrt{2})][(a - b\sqrt{2})(c - d\sqrt{2})]|\\ =|(ac+2bd)+(ad+bc)\sqrt 2||ac+2bd-(ad+bc)\sqrt{2}|\\ =\nu(ac+2bd+(ad+bc)\sqrt{2})\\ =\nu((a+b\sqrt2)(c+d\sqrt2)) $$
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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0But this seems like an awfully long calculation for something that is apparently 'clear'.. Is there a bit of intuition that I am missing? – 2012-05-28
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0(not to sound ungrateful for your answer) :) – 2012-05-28
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1It'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. – 2012-05-28
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0Ah ok, it seems clearer now, thanks! – 2012-05-28