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$$\alpha,\beta\in\mathbb{Z}[\sqrt{2}]=\{\textbf{a}+\textbf{b}\sqrt{2}:\textbf{a, b}\in\mathbb{Z} \}$$ $$\text{Proof}\quad \rm{Norm(\alpha\beta)=Norm(\alpha)Norm(\beta)}$$ $$\text{let }\alpha=\textbf{a+b}\sqrt{2}\quad\beta=\textbf{c+d}\sqrt{2}$$ $$\alpha\beta=(\textbf{a+b}\sqrt{2})(\textbf{c+d}\sqrt{2})$$ $$\alpha\beta=\textbf{ac+ad}\sqrt{2}+\textbf{bc}\sqrt{2}+2\textbf{bd}$$ $$\alpha\beta=(\textbf{ac+2bd})+(\textbf{ad+bc})\sqrt{2}$$ $$\rm{Norm(\alpha\beta)}=(\textbf{ac+2bd})^2-2(\textbf{ad+bc})^2$$ $$\rm{Norm(\alpha\beta)}=((\textbf{ac})^2 + 4(\textbf{bd})^2 + 4\textbf{abcd})-2((\textbf{ad})^2+(\textbf{bc})^2+2\textbf{abcd})$$ $$\rm{Norm(\alpha\beta)}=(\textbf{ac})^2+4(\textbf{bd})^2-2(\textbf{ad})^2-2(\textbf{bc})^2$$ $$\rm{Norm(\alpha)Norm(\beta)}=(\textbf{a}^2-2\textbf{b}^2)(\textbf{c}^2-2\textbf{d}^2)$$ $$\rm{Norm(\alpha)Norm(\beta)}=(\textbf{ac})^2+4(\textbf{bd})^2-2(\textbf{ad})^2-2(\textbf{bc})^2$$ $$\Rightarrow\quad \rm{Norm(\alpha\beta)=Norm(\alpha)Norm(\beta)}$$

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  • $\begingroup$ Looks good to me. $\endgroup$ – Prahlad Vaidyanathan Dec 2 '13 at 10:00
  • $\begingroup$ this looks fine! $\endgroup$ – user87543 Dec 2 '13 at 10:00
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I'd do it rather this way for convenience, but your proof is OK:

$\mathbb{Z}[\sqrt{2}]$ as a ring is isomorphic to the set of matrices $\{\begin{bmatrix} a & b \\ 2b &a\end{bmatrix}: a,b \in \mathbb{Z}\}$ where isomorphism is given by $a+\sqrt{2}b \mapsto $$\begin{bmatrix} a & b \\ 2b &a\end{bmatrix}$.

Notice that $\operatorname{norm}(a+b\sqrt{2})=\det\begin{bmatrix} a & b \\ 2b &a\end{bmatrix}$.

Now the fact that $\operatorname{norm(\alpha)} \cdot \operatorname{norm(\beta)}=\operatorname{norm(\alpha \cdot \beta)}$ becomes obvious by the property of determinant.

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    $\begingroup$ This isomorphism comes from the representation of $\alpha=a+b\sqrt 2$ as the linear map $z \mapsto \alpha z$. $\endgroup$ – lhf Dec 2 '13 at 10:56
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If you use the fact that the norm is by definition the product of the element by its conjugate, the multiplicativity of norm is immediate from the commutativity of $Q(\sqrt{2})$.

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    $\begingroup$ +1, but note that this does not avoid all computations, because you still need to prove that $(\alpha \beta)'=\alpha' \beta'$, where $'$ is conjugation. $\endgroup$ – lhf Dec 2 '13 at 10:32
  • $\begingroup$ Good point. But showing that conjugation is a field automorphism seems more basic. You are going to have trouble doing Galois theory without it :-) $\endgroup$ – Mikhail Katz Dec 2 '13 at 10:35
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Your proof is fine. You could be a bit less detailed about the algebra, if you write it down. But it is good, that you know the small steps.

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